\input{"preamble.tex"} \addbibresource{FourManifolds.bib} \let\Begin\begin \let\End\end \newcommand\wrapenv[1]{#1} \makeatletter \def\ScaleWidthIfNeeded{% \ifdim\Gin@nat@width>\linewidth \linewidth \else \Gin@nat@width \fi } \def\ScaleHeightIfNeeded{% \ifdim\Gin@nat@height>0.9\textheight 0.9\textheight \else \Gin@nat@width \fi } \makeatother \setkeys{Gin}{width=\ScaleWidthIfNeeded,height=\ScaleHeightIfNeeded,keepaspectratio}% \title{ \rule{\linewidth}{1pt} \\ \textbf{ 4-Manifolds } \\ {\normalsize Lectures by Philip Engel. University of Georgia, Spring 2021} \\ \rule{\linewidth}{2pt} } \titlehead{ \begin{center} \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png} \end{center} \begin{minipage}{.35\linewidth} \begin{flushleft} \vspace{2em} {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd from a graduate course in 4-Manifolds taught by Philip Engel at the University of Georgia in Spring 2021. As such, any errors or inaccuracies are almost certainly my own. } } \\ \end{flushleft} \end{minipage} \hfill \begin{minipage}{.65\linewidth} \end{minipage} } \begin{document} \date{} \maketitle \begin{flushleft} \textit{D. Zack Garza} \\ \textit{University of Georgia} \\ \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\ {\tiny \textit{Last updated:} 2021-08-02 } \end{flushleft} \newpage % Note: addsec only in KomaScript \addsec{Table of Contents} \tableofcontents \newpage \hypertarget{tuesday-january-12}{% \section{Tuesday, January 12}\label{tuesday-january-12}} \hypertarget{background}{% \subsection{Background}\label{background}} From Phil's email: Personally, I found the following online references particularly useful: \begin{itemize} \item Dietmar Salamon: Spin Geometry and Seiberg-Witten Invariants \autocite{Dietmar99} \item Richard Mandelbaum: Four-dimensional Topology: An Introduction \autocite{Mandelbaum1980} \begin{itemize} \tightlist \item This book has a nice introduction to surgery aspects of four-manifolds, but as a warning: It was published right before Freedman's famous theorem. For instance, the existence of an exotic R\^{}4 was not known. This actually makes it quite useful, as a summary of what was known before, and provides the historical context in which Freedman's theorem was proven. \end{itemize} \item Danny Calegari: Notes on 4-Manifolds \autocite{Calegari} \item Yuli Rudyak: Piecewise Linear Structures on Topological Manifolds \autocite{Rudyak} \item Akhil Mathew: The Dirac Operator \autocite{Matthew} \item Tom Weston: An Introduction to Cobordism Theory \autocite{Weston} A wide variety of lecture notes on the Atiyah-Singer index theorem, which are available online. \end{itemize} \hypertarget{introduction}{% \subsection{Introduction}\label{introduction}} \begin{definition}[Topological Manifold] Recall that a \textbf{topological manifold} (or $$C^0$$ manifold) $$X$$ is a Hausdorff topological space \emph{locally homeomorphic} to $${\mathbb{R}}^n$$ with a countable topological base, so we have charts $$\phi_u: U\to {\mathbb{R}}^n$$ which are homeomorphisms from open sets covering $$X$$. \end{definition} \begin{example}[The circle] $$S^1$$ is covered by two charts homeomorphic to intervals: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{45pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures/}{2021-01-16_21-54.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \begin{remark} Maps that are merely continuous are poorly behaved, so we may want to impose extra structure. This can be done by imposing restrictions on the transition functions, defined as \begin{align*} t_{uv} \coloneqq\varphi_V \to \varphi_U ^{-1} : \varphi_U(U \cap V) \to \varphi_V(U \cap V) .\end{align*} \end{remark} \begin{definition}[Restricted Structures on Manifolds] \envlist \begin{itemize} \item We say $$X$$ is a \textbf{PL manifold} if and only if $$t_{UV}$$ are piecewise-linear. Note that an invertible PL map has a PL inverse. \item We say $$X$$ is a \textbf{$$C^k$$ manifold} if they are $$k$$ times continuously differentiable, and \textbf{smooth} if infinitely differentiable. \item We say $$X$$ is \textbf{real-analytic} if they are locally given by convergent power series. \item We say $$X$$ is \textbf{complex-analytic} if under the identification $${\mathbb{R}}^n \cong {\mathbb{C}}^{n/2}$$ if they are holomorphic, i.e.~the differential of $$t_{UV}$$ is complex linear. \item We say $$X$$ is a \textbf{projective variety} if it is the vanishing locus of homogeneous polynomials on $${\mathbb{CP}}^N$$. \end{itemize} \end{definition} \begin{remark} Is this a strictly increasing hierarchy? It's not clear e.g.~that every $$C^k$$ manifold is PL. \end{remark} \begin{question} Consider $${\mathbb{R}}^n$$ as a topological manifold: are any two smooth structures on $${\mathbb{R}}^n$$ diffeomorphic? \end{question} \begin{remark} Fix a copy of $${\mathbb{R}}$$ and form a single chart $${\mathbb{R}}\xrightarrow{\operatorname{id}} {\mathbb{R}}$$. There is only a single transition function, the identity, which is smooth. But consider \begin{align*} X &\to {\mathbb{R}}\\ t &\mapsto t^3 .\end{align*} This is also a smooth structure on $$X$$, since the transition function is the identity. This yields a different smooth structure, since these two charts don't like in the same maximal atlas. Otherwise there would be a transition function of the form $$t_{VU}: t\mapsto t^{1/3}$$, which is not smooth at zero. However, the map \begin{align*} X &\to X \\ t &\mapsto t^3 .\end{align*} defines a diffeomorphism between the two smooth structures. \end{remark} \begin{claim} $${\mathbb{R}}$$ admits a unique smooth structure. \end{claim} \begin{proof}[sketch] Let $$\tilde {\mathbb{R}}$$ be some exotic $${\mathbb{R}}$$, i.e.~a smooth manifold homeomorphic to $${\mathbb{R}}$$. Cover this by coordinate charts to the standard $${\mathbb{R}}$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-16_22-31.pdf_tex} }; \end{tikzpicture} } \end{figure} \begin{fact} There exists a cover which is \emph{locally finite} and supports a \emph{partition of unity}: a collection of smooth functions $$f_i: U_i \to {\mathbb{R}}$$ with $$f_i \geq 0$$ and $${\operatorname{supp}}f \subseteq U_i$$ such that $$\sum f_i = 1$$ (\emph{i.e., bump functions}). It is also a purely topological fact that $$\tilde {\mathbb{R}}$$ is orientable. \end{fact} So we have bump functions: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-16_22-37.pdf_tex} }; \end{tikzpicture} } \end{figure} Take a smooth vector field $$V_i$$ on $$U_i$$ everywhere aligning with the orientation. Then $$\sum f_i V_i$$ is a smooth nowhere vector field on $$X$$ that is nowhere zero in the direction of the orientation. Taking the associated flow \begin{align*} {\mathbb{R}}&\to \tilde {\mathbb{R}}\\ t &\mapsto \varphi(t) .\end{align*} such that $$\varphi'(t) = V(\varphi(t))$$. Then $$\varphi$$ is a smooth map that defines a diffeomorphism. This follows from the fact that the vector field is everywhere positive. \end{proof} \begin{slogan} To understand smooth structures on $$X$$, we should try to solve differential equations on $$X$$. \end{slogan} \begin{remark} Note that here we used the existence of a global frame, i.e.~a trivialization of the tangent bundle, so this doesn't quite work for e.g.~$$S^2$$. \end{remark} \begin{question} What is the difference between all of the above structures? Are there obstructions to admitting any particular one? \end{question} \begin{answer} \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item (Munkres) Every $$C^1$$ structure gives a unique $$C^k$$ and $$C^ \infty$$ structure.\footnote{Note that this doesn't start at $$C^0$$, so topological manifolds are genuinely different! There exist topological manifolds with no smooth structure.} \item (Grauert) Every $$C^ \infty$$ structure gives a unique real-analytic structure. \item Every PL manifold admits a smooth structure in $$\dim X \leq 7$$, and it's unique in $$\dim X\leq 6$$, and above these dimensions there exists PL manifolds with no smooth structure. \item (Kirby--Siebenmann) Let $$X$$ be a topological manifold of $$\dim X\geq 5$$, then there exists a cohomology class $$\operatorname{ks}(X) \in H^4(X; {\mathbb{Z}}/2{\mathbb{Z}})$$ which is 0 if and only if $$X$$ admits a PL structure. Moreover, if $$\operatorname{ks}(X) = 0$$, then (up to concordance) the set of PL structures is given by $$H^3(X; {\mathbb{Z}}/2{\mathbb{Z}})$$. \item (Moise) Every topological manifold in $$\dim X\leq 3$$ admits a unique smooth structure. \item (Smale et al.): In $$\dim X\geq 5$$, the number of smooth structures on a topological manifold $$X$$ is finite. In particular, $${\mathbb{R}}^n$$ for $$n \neq 4$$ has a unique smooth structure. So dimension 4 is interesting! \item (Taubes) $${\mathbb{R}}^4$$ admits uncountably many non-diffeomorphic smooth structures. \item A compact oriented smooth surface $$\Sigma$$, the space of complex-analytic structures is a complex orbifold \footnote{Locally admits a chart to $${\mathbb{C}}^n/ \Gamma$$ for $$\Gamma$$ a finite group.} of dimension $$3g-2$$ where $$g$$ is the genus of $$\Sigma$$, up to biholomorphism (i.e.~\emph{moduli}). \end{enumerate} \end{answer} \begin{remark} Kervaire-Milnor: $$S^7$$ admits 28 smooth structures, which form a group. \end{remark} \hypertarget{friday-january-15}{% \section{Friday, January 15}\label{friday-january-15}} \begin{remark} Let \begin{align*} V &\coloneqq\left\{{a^2 + b^2 + c^2 + d^3 + e^{6k-1} = 0}\right\} \subseteq {\mathbb{C}}^5 \\ S_{\varepsilon}&\coloneqq\left\{{ {\left\lvert {a} \right\rvert}^2 + {\left\lvert {b} \right\rvert}^2 + {\left\lvert {c} \right\rvert}^2 + {\left\lvert {d} \right\rvert}^2 + {\left\lvert {e} \right\rvert}^2 = 1}\right\} .\end{align*} Then $$V_k \cap S_{\varepsilon}\cong S^7$$ is a homeomorphism, and taking $$k=1,2,\cdots, 28$$ yields the 28 smooth structures on $$S^7$$. Note that $$V_k$$ is the cone over $$V_k \cap S_{\varepsilon}$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{25pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures/}{2021-01-15_13-54.pdf_tex} }; \end{tikzpicture} } \end{figure} \begin{quote} ? Admits a smooth structure, and $$\mkern 1.5mu\overline{\mkern-1.5muV\mkern-1.5mu}\mkern 1.5mu_k \subseteq {\mathbb{CP}}^5$$ admits no smooth structure. \end{quote} \end{remark} \begin{question} Is every triangulable manifold PL, i.e.~homeomorphic to a simplicial complex? \end{question} \begin{answer} No! Given a simplicial complex, there is a notion of the \textbf{combinatorial link} $$L_V$$ of a vertex $$V$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures/}{2021-01-15_13-57.pdf_tex} }; \end{tikzpicture} } \end{figure} It turns out that there exist simplicial manifolds such that the link is not homeomorphic to a sphere, whereas every PL manifold admits a PL triangulation'' where the links are spheres. \end{answer} \begin{remark} What's special in dimension 4? Recall the \textbf{Kirby-Siebenmann} invariant $$\operatorname{ks}(x) \in H^4(X; {\mathbb{Z}}_2)$$ for $$X$$ a topological manifold where $$\operatorname{ks}(X) = 0 \iff X$$ admits a PL structure, with the caveat that $$\dim X \geq 5$$. We can use this to cook up an invariant of 4-manifolds. \end{remark} \begin{definition}[Kirby-Siebenmann Invariant of a 4-manifold] Let $$X$$ be a topological 4-manifold, then \begin{align*} \operatorname{ks}(X) \coloneqq\operatorname{ks}(X \times{\mathbb{R}}) .\end{align*} \end{definition} \begin{remark} Recall that in $$\dim X\geq 7$$, every PL manifold admits a smooth structure, and we can note that \begin{align*} H^4(X; {\mathbb{Z}}_2) = H^4(X \times{\mathbb{R}}; {\mathbb{Z}}_2) = {\mathbb{Z}}_2, .\end{align*} since every oriented 4-manifold admits a fundamental class. Thus \begin{align*} \operatorname{ks}(X) = \begin{cases} 0 & X \times{\mathbb{R}}\text{ admits a PL and smooth structure} \\ 1 & X \times{\mathbb{R}}\text{ admits no PL or smooth structures }. \end{cases} \end{align*} \end{remark} \begin{remark} $$\operatorname{ks}(X) \neq 0$$ implies that $$X$$ has no smooth structure, since $$X \times{\mathbb{R}}$$ doesn't. Note that it was not known if this invariant was ever nonzero for a while! \end{remark} \begin{remark} Note that $$H^2(X; {\mathbb{Z}})$$ admits a symmetric bilinear form $$Q_X$$ defined by \begin{align*} Q_X: H^2(X; {\mathbb{Z}})^{\otimes 2} &\to {\mathbb{Z}}\\ \alpha \otimes\beta &\mapsto \int_X \alpha\wedge \beta \coloneqq(\alpha \smile\beta)([X]) .\end{align*} where $$[X]$$ is the fundamental class. \end{remark} \hypertarget{main-theorems-for-the-course}{% \section{Main Theorems for the Course}\label{main-theorems-for-the-course}} \begin{remark} Proving the following theorems is the main goal of this course: \end{remark} \begin{theorem}[Freedman] If $$X, Y$$ are compact oriented topological 4-manifolds, then $$X\cong Y$$ are homeomorphic if and only if $$\operatorname{ks}(X) = \operatorname{ks}(Y)$$ and $$Q_X \cong Q_Y$$ are isometric, i.e.~there exists an isometry \begin{align*} \varphi: H^2(X; {\mathbb{Z}}) \to H^2(Y; {\mathbb{Z}}) .\end{align*} that preserves the two bilinear forms in the sense that $${\left\langle {\varphi \alpha},~{ \varphi \beta} \right\rangle} = {\left\langle { \alpha},~{ \beta} \right\rangle}$$. Conversely, every \textbf{unimodular} bilinear form appears as $$H^2(X; {\mathbb{Z}})$$ for some $$X$$, i.e.~the pairing induces a map \begin{align*} H^2(X; {\mathbb{Z}}) &\to H^2(X; {\mathbb{Z}}) {}^{ \vee }\\ \alpha &\mapsto {\left\langle { \alpha },~{ {-}} \right\rangle} .\end{align*} which is an isomorphism. This is essentially a classification of simply-connected 4-manifolds. \end{theorem} \begin{remark} Note that preservation of a bilinear form is a stand-in for being an element of the orthogonal group'', where we only have a lattice instead of a full vector space. \end{remark} \begin{remark} There is a map $$H^2(X; {\mathbb{Z}}) \xrightarrow{PD} H_2(X; {\mathbb{Z}})$$ from Poincaré , where we can think of elements in the latter as closed surfaces $$[\Sigma]$$, and \begin{align*} {\left\langle { \Sigma_1 },~{ \Sigma_2 } \right\rangle} = \text{signed number of intersections points of } \Sigma_1 \pitchfork\Sigma_2 .\end{align*} Note that Freedman's theorem is only about homeomorphism, and is not true smoothly. This gives a way to show that two 4-manifolds are homeomorphic, but this is hard to prove! So we'll black-box this, and focus on ways to show that two \emph{smooth} 4-manifolds are \emph{not} diffeomorphic, since we want homeomorphic but non-diffeomorphic manifolds. \end{remark} \begin{definition}[Signature] The \textbf{signature} of a topological 4- manifold is the signature of $$Q_X$$, where we note that $$Q_X$$ is a symmetric nondegenerate bilinear form on $$H^2(X; {\mathbb{R}})$$ and for some $$a, b$$ \begin{align*} (H^2(X; {\mathbb{R}}), Q_x) \xrightarrow{\text{isometric}} {\mathbb{R}}^{a, b} .\end{align*} where $$a$$ is the number of $$+1$$s appearing in the matrix and $$b$$ is the number of $$-1$$s. This is $${\mathbb{R}}^{ab}$$ where $$e_i^2 = 1, i=1\cdots a$$ and $$e_i^2 = -1, i=a+1, \cdots b$$, and is thus equipped with a specific bilinear form corresponding to the Gram matrix of this basis. \begin{align*} \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1 \end{bmatrix} = I_{a\times a} \oplus -I_{b \times b} .\end{align*} Then the signature is $$a-b$$, the dimension of the positive-definite space minus the dimension of the negative-definite space. \end{definition} \begin{theorem}[Rokhlin's Theorem] Suppose $${\left\langle { \alpha},~{\alpha} \right\rangle} \in 2{\mathbb{Z}}$$ and $$\alpha\in H^2(X; {\mathbb{Z}})$$ and $$X$$ a simply connected \textbf{smooth} 4-manifold. Then 16 divides $$\operatorname{sig}(X)$$. \end{theorem} \begin{remark} Note that Freedman's theorem implies that there exists topological 4-manifolds with no smooth structure. \end{remark} \begin{theorem}[Donaldson] Let $$X$$ be a smooth simply-connected 4-manifold. If $$a=0$$ or $$b=0$$, then $$Q_X$$ is diagonalizable and there exists an orthonormal basis of $$H^2(X; {\mathbb{Z}})$$. \end{theorem} \begin{remark} This comes from Gram-Schmidt, and restricts what types of intersection forms can occur. \end{remark} \hypertarget{warm-up-mathbbr2-has-a-unique-smooth-structure}{% \subsection{\texorpdfstring{Warm Up: $${\mathbb{R}}^2$$ Has a Unique Smooth Structure}{Warm Up: \{\textbackslash mathbb\{R\}\}\^{}2 Has a Unique Smooth Structure}}\label{warm-up-mathbbr2-has-a-unique-smooth-structure}} \begin{remark} Last time we showed $${\mathbb{R}}^1$$ had a unique smooth structure, so now we'll do this for $${\mathbb{R}}^2$$. The strategy of solving a differential equation, we'll now sketch the proof. \end{remark} \begin{definition}[Riemannian Metrics] A \textbf{Riemannian metric} $$g\in \Gamma( \operatorname{Sym}^2 T {}^{ \vee }X)$$ for $$X$$ a smooth manifold is a metric on every $$T_p X$$, so $$g_p \in (T_p X^{\otimes 2}) {}^{ \vee }$$, such that \begin{align*} g_p: T_pX \otimes T_p X &\to {\mathbb{R}}&& g(v, v) \geq 0, \quad g(v,v) = 0 \iff v=0 .\end{align*} \end{definition} \begin{definition}[Almost complex structure] An \textbf{almost complex structure} is a morphism $$J\in \mathop{\mathrm{End}}_{{ \mathsf{Vect} }(X)}(TX)$$ of vector bundles over $$X$$ such that $$J^2 = -\operatorname{id}_{TX}$$. \end{definition} \begin{definition}[Integrable] An almost-complex structure is \textbf{integrable} $$J$$ if it comes from a complex structure in the following sense: for a complex manifold $$M\in {\mathsf{Mfd}}({\mathbb{C}})$$, take holomorphic coordinates $$z = x+iy$$ and set $$J {\frac{\partial }{\partial x}\,} \coloneqq{\frac{\partial }{\partial y}\,}$$ and $$J{\frac{\partial }{\partial y}\,} \coloneqq-{\frac{\partial }{\partial x}\,}$$. \end{definition} \begin{remark} A manifold $$M\in {\mathsf{sm}}{\mathsf{Mfd}}({\mathbb{R}})$$ admits an almost-complex structure iff $$TM$$ admits a reduction of structure group $$\operatorname{GL}_{2n}({\mathbb{R}}) \to \operatorname{GL}_n({\mathbb{C}})$$. \end{remark} \begin{remark} Let $$e\in T_p X$$ and $$e\neq 0$$, then if $$X$$ is a surface then $$\left\{{e, J_p e}\right\}$$ is a basis of $$T_p X$$, where $$J_p$$ is the restriction of $$J$$ to $$T_p X$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{25pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures/}{2021-01-15_14-33.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{exercise}[?] Show that $$\left\{{ e, J_p e }\right\}$$ are linearly independent in $$T_p X$$. In particular, $$J_p$$ is determined by a point in $${\mathbb{R}}^2\setminus\left\{{\text{the }x{\hbox{-}}\text{axis}}\right\}$$ \end{exercise} \begin{proof}[That R2 admits a unique smooth structure (sketch)] Let $$\tilde {\mathbb{R}}^2$$ be an exotic $${\mathbb{R}}^2$$. \hypertarget{step-1}{% \subsubsection{Step 1}\label{step-1}} Choose a metric on $$\tilde {\mathbb{R}}^2$$, say $$g \coloneqq\sum f_I g_i$$ with $$g_i$$ metrics on coordinate charts $$U_i$$ and $$f_i$$ a partition of unity. \hypertarget{step-2}{% \subsubsection{Step 2}\label{step-2}} Find an almost complex structure on $$\tilde {\mathbb{R}}^2$$. Choosing an orientation of $$\tilde {\mathbb{R}}^2$$, the metric $$g$$ defines a unique almost complex structure $$J_p e \coloneqq f\in T_p \tilde {\mathbb{R}}^2$$ such that \begin{itemize} \tightlist \item $$g(e, e) = g(f, f)$$ \item $$g(e, f) = 0$$. \item $$\left\{{e, f}\right\}$$ is an oriented basis of $$T_p \tilde {\mathbb{R}}^2$$ \end{itemize} This is because after choosing $$e$$, there are two orthogonal vectors, but only one choice yields an \emph{oriented} basis. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{25pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures/}{2021-01-15_14-39.pdf_tex} }; \end{tikzpicture} } \end{figure} \hypertarget{step-3}{% \subsubsection{Step 3}\label{step-3}} We then apply a theorem: \begin{theorem}[Almost-complex structures on surfaces come from complex structures] Any almost complex structure on a surface comes from a complex structure, in the sense that there exist charts $$\varphi_i: U_i \to {\mathbb{C}}$$ such that $$J$$ is multiplication by $$i$$. \end{theorem} So \begin{align*} d \varphi(J \cdot e) = i \cdot d \varphi_i (e) ,\end{align*} and $$(\tilde {\mathbb{R}}^2, J)$$ is a complex manifold. Since it's simply connected, the Riemann Mapping Theorem shows that it's biholomorphic to $${\mathbb{D}}$$ or $${\mathbb{C}}$$, both of which are diffeomorphic to $${\mathbb{R}}^2$$. \end{proof} \begin{remark} See the Newlander-Nirenberg theorem, a result in complex geometry. \end{remark} \hypertarget{sheaves-bundles-connections-lecture-3-wednesday-january-20}{% \section{Sheaves, Bundles, Connections (Lecture 3, Wednesday, January 20)}\label{sheaves-bundles-connections-lecture-3-wednesday-january-20}} \hypertarget{sheaves}{% \subsection{Sheaves}\label{sheaves}} \begin{definition}[Presheaves and Sheaves] Recall that if $$X$$ is a topological space, a \textbf{presheaf} of abelian groups $$\mathcal{F}$$ is an assignment $$U\to \mathcal{F}(U)$$ of an abelian group to every open set $$U \subseteq X$$ together with a restriction map $$\rho_{UV}: \mathcal{F}(U) \to \mathcal{F}(V)$$ for any inclusion $$V \subseteq U$$ of open sets. This data has to satisfying certain conditions: \begin{enumerate} \def\labelenumi{\alph{enumi}.} \item $$\mathcal{F}(\emptyset) = 0$$, the trivial abelian group. \item $$\rho_{UU}: \mathcal{F}(U) \to \mathcal{F}(U) = \operatorname{id}_{\mathcal{F}(U) }$$ \item Compatibility if restriction is taken in steps: $$U \subseteq V \subseteq W \implies \rho_{VW} \circ \rho_{UV} = \rho_{UW}$$. \end{enumerate} We say $$\mathcal{F}$$ is a \textbf{sheaf} if additionally: \begin{enumerate} \def\labelenumi{\alph{enumi}.} \setcounter{enumi}{3} \tightlist \item Given $$s_i \in \mathcal{F}(U_i)$$ such that $$\rho_{U_i \cap U_j} (s_i) = \rho_{U_i \cap U_j} (s_j)$$ implies that there exists a unique $$s\in \mathcal{F}\qty{\displaystyle\bigcup_i U_i}$$ such that $$\rho_{U_i}(s) = s_i$$. \end{enumerate} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{45pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-20_13-59.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{definition} \begin{example}[?] Let $$X$$ be a topological manifold, then $$\mathcal{F}\coloneqq C^0({-}, {\mathbb{R}})$$ the set of continuous functionals form a sheaf. We have a diagram \begin{center} \begin{tikzcd} U && {C^0(U; {\mathbb{R}})} \\ \\ V && {C^0(V; {\mathbb{R}})} \arrow[hook, from=3-1, to=1-1] \arrow["{\text{restrict cts. functions}}", dashed, hook, from=1-3, to=3-3] \arrow["{\mathcal{F}}", from=1-1, to=1-3] \arrow["{\mathcal{F}}"', from=3-1, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJVIl0sWzAsMiwiViJdLFsyLDAsIkNeMChVOyBcXFJSKSJdLFsyLDIsIkNeMChWOyBcXFJSKSJdLFsxLDAsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMywiXFx0ZXh0e3Jlc3RyaWN0IGN0cy4gZnVuY3Rpb25zfSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDIsIlxcbWF0aGNhbHtGfSJdLFsxLDMsIlxcbWF0aGNhbHtGfSIsMl1d}{Link to diagram} \end{quote} Property (d) holds because given sections $$s_i \in C^0(U_i; {\mathbb{R}})$$ agreeing on overlaps, so $${ \left.{{s_i}} \right|_{{U_i \cap U_j}} } = { \left.{{s_j}} \right|_{{U_i \cap U_j}} }$$, there exists a unique $$s\in C^0\qty{\displaystyle\bigcup_i U_i; {\mathbb{R}}}$$ such that $${ \left.{{s}} \right|_{{U_i}} } = s_i$$ for all $$i$$ -- i.e.~continuous functions glue. \end{example} \begin{remark} Recall that we discussed various structures on manifolds: PL, continuous, smooth, complex-analytic, etc. We can characterize these by their sheaves of functions, which we'll denote $${\mathcal{O}}$$. For example, $${\mathcal{O}}\coloneqq C^0({-}; {\mathbb{R}})$$ for topological manifolds, and $${\mathcal{O}}\coloneqq C^ \infty ({-}; {\mathbb{R}})$$ is the sheaf for smooth manifolds. Note that this also works for PL functions, since pullbacks of PL functions are again PL. For complex manifolds, we set $${\mathcal{O}}$$ to be the sheaf of holomorphic functions. \end{remark} \begin{example}[Locally Constant Sheaves] Let $$A\in {\mathsf{Ab}}$$ be an abelian group, then $$\underline{A}$$ is the sheaf defined by setting $$\underline{A}(U)$$ to be the locally constant functions $$U\to A$$. E.g. let $$X \in {\mathsf{Mfd}}_{{\mathsf{Top}}}$$ be a topological manifold, then $$\underline{{\mathbb{R}}}(U) = {\mathbb{R}}$$ if $$U$$ is connected since locally constant $$\implies$$ globally constant in this case. \end{example} \begin{warnings} Note that the presheaf of constant functions doesn't satisfy (d)! Take $${\mathbb{R}}$$ and a function with two different values on disjoint intervals: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{41pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-20_14-11.pdf_tex} }; \end{tikzpicture} } \end{figure} Note that $${ \left.{{s_1}} \right|_{{U_1 \cap U_2}} } = { \left.{{s_2}} \right|_{{U_1 \cap U_2}} }$$ since the intersection is empty, but there is no constant function that restricts to the two different values. \end{warnings} \hypertarget{bundles}{% \subsection{Bundles}\label{bundles}} \begin{remark} Let $$\pi: \mathcal{E}\to X$$ be a \textbf{vector bundle}, so we have local trivializations $$\pi ^{-1} (U) \xrightarrow{h_u} Y^d \times U$$ where we take either $$Y={\mathbb{R}}, {\mathbb{C}}$$, such that $$h_v \circ h_u ^{-1}$$ preserves the fibers of $$\pi$$ and acts linearly on each fiber of $$Y\times(U \cap V)$$. Define \begin{align*} t_{UV}: U \cap V \to \operatorname{GL}_d(Y) \end{align*} where we require that $$t_{UV}$$ is continuous, smooth, complex-analytic, etc depending on the context. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{47pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-20_14-17.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{example}[Bundles over $S^1$] There are two $${\mathbb{R}}^1$$ bundles over $$S^1$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{32pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-20_14-20.pdf_tex} }; \end{tikzpicture} } \end{figure} Note that the Mobius bundle is not trivial, but can be locally trivialized. \end{example} \begin{remark} We abuse notation: $$\mathcal{E}$$ is also a sheaf, and we write $$\mathcal{E}(U)$$ to be the set of sections $$s: U\to \mathcal{E}$$ where $$s$$ is continuous, smooth, holomorphic, etc where $$\pi \circ s = \operatorname{id}_U$$. I.e. a bundle is a sheaf in the sense that its sections \emph{form} a sheaf. \end{remark} \begin{example}[?] The trivial line bundle gives the sheaf $${\mathcal{O}}$$ : maps $$U \xrightarrow{s} U\times Y$$ for $$Y={\mathbb{R}}, {\mathbb{C}}$$ such that $$\pi \circ s = \operatorname{id}$$ are the same as maps $$U\to Y$$. \end{example} \begin{definition}[$\OO\dash$modules] An \textbf{$${\mathcal{O}}{\hbox{-}}$$module} is a sheaf $$\mathcal{F}$$ such that $$\mathcal{F}(U)$$ has an action of $$\mathcal{O}(U)$$ compatible with restriction. \end{definition} \begin{example}[?] If $$\mathcal{E}$$ is a vector bundle, then $$\mathcal{E}(U)$$ has a natural action of $${\mathcal{O}}(U)$$ given by $$f\curvearrowright s \coloneqq fs$$, i.e.~just multiplying functions. \end{example} \begin{example}[Non-example] The locally constant sheaf $$\underline{{\mathbb{R}}}$$ is not an $${\mathcal{O}}{\hbox{-}}$$module: there isn't natural action since the sections of $${\mathcal{O}}$$ are generally non-constant functions, and multiplying a constant function by a non-constant function doesn't generally give back a constant function. \end{example} \begin{remark} We'd like a notion of maps between sheaves: \end{remark} \begin{definition}[Morphisms of Sheaves] A \textbf{morphism} of sheaves $$\mathcal{F} \to \mathcal{G}$$ is a group morphism $$\varphi(U): \mathcal{F}(U) \to \mathcal{G}(U)$$ for all opens $$U \subseteq X$$ such that the diagram involving restrictions commutes: \begin{center} \begin{tikzcd} \mathcal{F}(U) \ar[r, "\phi(U)"] \ar[d, "\rho_{UV}"] & \mathcal{G}(U) \ar[d, "\rho_{UV}"] \\ \mathcal{F}(V) \ar[r, "\phi(V)"] & \mathcal{F}(V) \end{tikzcd} \end{center} \end{definition} \begin{example}[An $\OO\dash$module that is not a vector bundle.] Let $$X = {\mathbb{R}}$$ and define the \textbf{skyscraper sheaf} at $$p \in {\mathbb{R}}$$ as \begin{align*} {\mathbb{R}}_p(U) \coloneqq \begin{cases} {\mathbb{R}}& p\in U \\ 0 & p\not\in U. \end{cases} .\end{align*} The $${\mathcal{O}}(U){\hbox{-}}$$module structure is given by \begin{align*} {\mathcal{O}}(U) \times{\mathcal{O}}(U) &\to {\mathbb{R}}_p(U) \\ (f, s) &\mapsto f(p) s .\end{align*} This is not a vector bundle since $${\mathbb{R}}_p(U)$$ is not an infinite dimensional vector space, whereas the space of sections of a vector bundle is generally infinite dimensional (?). Alternatively, there are arbitrarily small punctured open neighborhoods of $$p$$ for which the sheaf makes trivial assignments. \end{example} \begin{example}[of morphisms] Let $$X = {\mathbb{R}}\in {\mathsf{sm}}{\mathsf{Mfd}}$$ viewed as a smooth manifold, then multiplication by $$x$$ induces a morphism of structure sheaves: \begin{align*} (x \cdot): {\mathcal{O}}&\to {\mathcal{O}}\\ s & \mapsto x\cdot s \end{align*} for any $$x\in {\mathcal{O}}(U)$$, noting that $$x\cdot s\in {\mathcal{O}}(U)$$ again. \begin{exercise}[The kernel of a sheaf morphism is a sheaf] Check that $$\ker \varphi$$ is naturally a sheaf and $$\ker(\varphi)(U) = \ker (\varphi(U)): \mathcal{F}(U) \to \mathcal{G}(U)$$ \end{exercise} Here the kernel is trivial, i.e.~on any open $$U$$ we have $$(x\cdot):{\mathcal{O}}(U) \hookrightarrow{\mathcal{O}}(U)$$ is injective. Taking the cokernel $$\operatorname{coker}(x\cdot)$$ as a presheaf, this assigns to $$U$$ the quotient presheaf $${\mathcal{O}}(U) / x{\mathcal{O}}(U)$$, which turns out to be equal to $${\mathbb{R}}_0$$. So $${\mathcal{O}}\to {\mathbb{R}}_0$$ by restricting to the value at $$0$$, and there is an exact sequence \begin{align*} 0 \to {\mathcal{O}}\xrightarrow{(x\cdot)} {\mathcal{O}}\to {\mathbb{R}}_0 \to 0 .\end{align*} This is one reason sheaves are better than vector bundles: the category is closed under taking quotients, whereas quotients of vector bundles may not be vector bundles. \end{example} \hypertarget{lecture-4-friday-january-22}{% \section{Lecture 4 (Friday, January 22)}\label{lecture-4-friday-january-22}} \hypertarget{the-exponential-exact-sequence}{% \subsection{The Exponential Exact Sequence}\label{the-exponential-exact-sequence}} \begin{remark} Let $$X = {\mathbb{C}}$$ and consider $${\mathcal{O}}$$ the sheaf of holomorphic functions and $${\mathcal{O}}^{\times}$$ the sheaf of \emph{nonvanishing} holomorphic functions. The former is a vector bundle and the latter is a sheaf of abelian groups. There is a map $$\exp: {\mathcal{O}}\to {\mathcal{O}}^{\times}$$, the \textbf{exponential map}, which is the data $$\exp(U): {\mathcal{O}}(U) \to {\mathcal{O}}^{\times}(U)$$ on every open $$U$$ given by $$f\mapsto e^f$$. There is a kernel sheaf $$2\pi i \underline{{\mathbb{Z}}}$$, and we get an exact sequence \begin{align*} 0 \to 2\pi i \underline{{\mathbb{Z}}} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times}\to \operatorname{coker}(\exp) \to 0 .\end{align*} \end{remark} \begin{question} What is the cokernel sheaf here? \end{question} \begin{remark} Let $$U$$ be a contractible open set, then we can identify $${\mathcal{O}}^{\times}(U) / \exp({\mathcal{O}}^{\times}(U)) = 1$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{44pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-22_13-58.pdf_tex} }; \end{tikzpicture} } \end{figure} Any $$f\in {\mathcal{O}}^{\times}(U)$$ has a logarithm, say by taking a branch cut, since $$\pi_1(U) =0 \implies \log f$$ has an analytic continuation. Consider the annulus $$U$$ and the function $$z\in {\mathcal{O}}^{\times}(U)$$, then $$z\not\in \exp({\mathcal{O}}(U))$$ -- if $$z=e^f$$ then $$f=\log(z)$$, but $$\log(z)$$ has monodromy on $$U$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{44pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-22_14-02.pdf_tex} }; \end{tikzpicture} } \end{figure} Thus on any sufficiently small open set, $$\operatorname{coker}(\exp) = 1$$. This is only a presheaf: there exists an open cover of the annulus for which $${ \left.{{z}} \right|_{{U_i}} }$$, and so the naive cokernel doesn't define a sheaf. This is because we have a locally trivial section which glues to $$z$$, which is nontrivial. \end{remark} \begin{exercise}[Fixing the sheaf cokernel] Redefine the cokernel so that it is a sheaf. Hint: look at sheafification, which has the defining property \begin{align*} \mathop{\mathrm{Hom}}_{ \underset{ \mathsf{pre} } {\mathsf{Sh} }}(\mathcal{G}, \mathcal{F}^{ \underset{ \mathsf{pre} } {\mathsf{Sh} }} ) =\mathop{\mathrm{Hom}}_{{\mathsf{Sh}}}( \mathcal{G}, \mathcal{F}^{{\mathsf{Sh}}}) \end{align*} for any sheaf $$\mathcal{G}$$. \end{exercise} \hypertarget{global-sections}{% \subsection{Global Sections}\label{global-sections}} \begin{definition}[Global Sections Sheaf] The \textbf{global sections} sheaf of $$\mathcal{F}$$ on $$X$$ is given by $$H^0( X; \mathcal{F}) = \mathcal{F}(X)$$. \end{definition} \begin{example}[?] \envlist \begin{itemize} \tightlist \item $$C^ \infty (X) = H^0(X, C^ \infty )$$ are the smooth functions on $$X$$ \item $$VF(X) = H^0(X; T)$$ are the smooth vector fields on $$X$$ for $$T$$ the tangent bundle \item If $$X$$ is a complex manifold then $${\mathcal{O}}(X) = H^0(X; {\mathcal{O}})$$ are the globally holomorphic functions on $$X$$. \item $$H^0(X; {\mathbb{Z}}) = \underline{{\mathbb{Z}}}(X)$$ are ?? \end{itemize} \end{example} \begin{remark} Given vector bundles $$V, W$$, we have constructions $$V \oplus W, V \otimes W, V {}^{ \vee }, \mathop{\mathrm{Hom}}(V, W) = V {}^{ \vee }\otimes W, \operatorname{Sym}^n V, \bigwedge^p V$$, and so on. Some of these work directly for sheaves: \begin{itemize} \tightlist \item $$\mathcal{F} \oplus \mathcal{G}(U) \coloneqq\mathcal{F}(U) \oplus \mathcal{G}(U)$$ \item For tensors, duals, and homs $$\mathscr{H}\kern-2pt\operatorname{om}(V, W)$$ we only get presheaves, so we need to sheafify. \end{itemize} \end{remark} \begin{warnings} $$\mathop{\mathrm{Hom}}(V, W)$$ will denote the \emph{global} homomorphisms $$\mathscr{H}\kern-2pt\operatorname{om}(V, W)(X)$$, which is a sheaf. \end{warnings} \begin{example}[?] Let $$X^n \in {\mathsf{Mfd}}_{{\mathsf{sm}}}$$ and let $$\Omega^p$$ be the sheaf of smooth $$p{\hbox{-}}$$forms, i.e $$\bigwedge^p T {}^{ \vee }$$, i.e.~$$\Omega^p(U)$$ are the smooth $$p$$ forms on $$U$$, which are locally of the form $$\sum f_{i_1, \cdots, i_p} (x_1, \cdots, x_n) dx_{i_1} \wedge dx_{i_2} \wedge \cdots dx_{i_p}$$ where the $$f_{i_1, \cdots, i_p}$$ are smooth functions. \begin{example}[Sub-example] Take $$X= S^1$$, writing this as $${\mathbb{R}}/{\mathbb{Z}}$$, we have $$\Omega^1(X) \ni dx$$. There are two coordinate charts which differ by a translation on their overlaps, and $$dx(x + c) =dx$$ for $$c$$ a constant: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{44pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-22_14-22.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \begin{exercise}[?] Check that on a torus, $$dx_i$$ is a well-defined 1-form. \end{exercise} \end{example} \begin{remark} Note that there is a map $$d: \Omega^p \to \Omega^{p+1}$$ where $$\omega\mapsto d \omega$$. \end{remark} \begin{warnings} $$d$$ is \textbf{not} a map of $${\mathcal{O}}{\hbox{-}}$$modules: $$d(f\cdot \omega) = f\cdot \omega + {\color{red} df \wedge \omega}$$, where the latter is a correction term. In particular, it is not a map of vector bundles, but is a map of sheaves of abelian groups since $$d ( \omega_1 + \omega_2) = d( \omega_1 ) + d ( \omega_2)$$, making $$d$$ a sheaf morphism. \end{warnings} \begin{remark} Let $$X \in {\mathsf{Mfd}}_{\mathbb{C}}$$, we'll use the fact that $$TX$$ is complex-linear and thus a $${\mathbb{C}}{\hbox{-}}$$vector bundle. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \node (node_one) at (0,0) { \fontsize{44pt}{1em} \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-22_14-27.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{remark}[Subtlety 1] Note that $$\Omega^p$$ for complex manifolds is $$\bigwedge^p T {}^{ \vee }$$, and so if we want to view $$X \in {\mathsf{Mfd}}_{\mathbb{R}}$$ we'll write $$X_{{\mathbb{R}}}$$. $$TX_{\mathbb{R}}$$ is then a real vector bundle of rank $$2n$$. \end{remark} \begin{remark}[Subtlety 2] $$\Omega^p$$ will denote \emph{holomorphic} $$p{\hbox{-}}$$forms, i.e.~local expressions of the form \begin{align*} \sum f_I(z_1, \cdots, z_n) \bigwedge dz_I .\end{align*} For example, $$e^zdz\in \Omega^1({\mathbb{C}})$$ but $$z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu dz$$ is not, where $$dz = dx + idy$$. We'll use a different notation when we allow the $$f_I$$ to just be smooth: $$A^{p, 0}$$, the sheaf of $$(p, 0){\hbox{-}}$$forms. Then $$z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu dz\in A^{1, 0}$$. \end{remark} \begin{remark} Note that $$T {}^{ \vee }X_{\mathbb{R}}\otimes _{\mathbb{C}}= A^{1, 0} \oplus A^{0, 1}$$ since there is a unique decomposition $$\omega = fdz + gd\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$ where $$f,g$$ are smooth. Then $$\Omega^d X_{\mathbb{R}}\otimes_{\mathbb{R}}{\mathbb{C}}= \bigoplus _{p+q=d} A^{p, q}$$. Note that $$\Omega_{{\mathsf{sm}}}^p \neq A^{p, q}$$ and these are really quite different: the former are more like holomorphic bundles, and the latter smooth. Moreover $$\dim \Omega^p(X) < \infty$$, whereas $$\Omega_{{\mathsf{sm}}}^1$$ is infinite-dimensional. \end{remark} \hypertarget{principal-ghbox-bundles-and-connections-monday-january-25}{% \section{\texorpdfstring{Principal $$G{\hbox{-}}$$Bundles and Connections (Monday, January 25)}{Principal G\{\textbackslash hbox\{-\}\}Bundles and Connections (Monday, January 25)}}\label{principal-ghbox-bundles-and-connections-monday-january-25}} \begin{definition}[Principal Bundles] Let $$G$$ be a (possibly disconnected) Lie group. Then a \textbf{principal $$G{\hbox{-}}$$bundle} $$\pi:P\to X$$ is a space admitting local trivializations $$h_u: \pi ^{-1} (U) \to G \times U$$ such that the transition functions are given by left multiplication by a continuous function $$t_{UV}: U \cap V \to G$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{40pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-25_13-55.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{definition} \begin{remark} Setup: we'll consider $$TX$$ for $$X\in {\mathsf{Mfd}}_{\operatorname{Sm}}$$, and let $$g$$ be a metric on the tangent bundle given by \begin{align*} g_p: T_pX^{\otimes 2} \to {\mathbb{R}} ,\end{align*} a symmetric bilinear form with $$g_p(u, v) \geq 0$$ with equality if and only if $$v=0$$. \end{remark} \begin{definition}[The Frame Bundle] Define $$\mathop{\mathrm{Frame}}_p(X) \coloneqq\left\{{\text{bases of } T_p X}\right\}$$, and $$\mathop{\mathrm{Frame}}(X) \coloneqq\displaystyle\bigcup_{p\in X} \mathop{\mathrm{Frame}}_p(X)$$. \end{definition} \begin{remark} More generally, $$\mathop{\mathrm{Frame}}(\mathcal{E})$$ can be defined for any vector bundle $$\mathcal{E}$$, so $$\mathop{\mathrm{Frame}}(X) \coloneqq\mathop{\mathrm{Frame}}(TX)$$. Note that $$\mathop{\mathrm{Frame}}(X)$$ is a principal $$\operatorname{GL}_n({\mathbb{R}}){\hbox{-}}$$bundle where $$n\coloneqq\operatorname{rank}(\mathcal{E})$$. This follows from the fact that the transition functions are fiberwise in $$\operatorname{GL}_n({\mathbb{R}})$$, so the transition functions are given by left-multiplication by matrices. \end{remark} \begin{remark}[Important] A principal $$G{\hbox{-}}$$bundle admits a $$G{\hbox{-}}$$action where $$G$$ acts by \emph{right} multiplication: \begin{align*} P \times G \to P \\ ( (g, x), h) \mapsto (gh, x) .\end{align*} This is necessary for compatibility on overlaps. \textbf{Key point}: the actions of left and right multiplication commute. \end{remark} \begin{definition}[Orthogonal Frame Bundle] The \textbf{orthogonal frame bundle} of a vector bundle $$\mathcal{E}$$ equipped with a metric $$g$$ is defined as $$\mathop{\mathrm{OFrame}}_p(\mathcal{E}) \coloneqq\left\{{\text{orthonormal bases of } \mathcal{E}_p}\right\}$$, also written $$O_r({\mathbb{R}})$$ where $$r \coloneqq\operatorname{rank}( \mathcal{E})$$. \end{definition} \begin{remark} The fibers $$P_x \to \left\{{x}\right\}$$ of a principal $$G{\hbox{-}}$$bundle are naturally \textbf{torsors} over $$G$$, i.e.~a set with a free transitive $$G{\hbox{-}}$$action. \end{remark} \begin{definition}[Hermitian metric] Let $$\mathcal{E}\to X$$ be a complex vector bundle. Then a \textbf{Hermitian metric} is a hermitian form on every fiber, i.e.~ \begin{align*} h_p: \mathcal{E}_p \times\overline{\mathcal{E}_p } \to {\mathbb{C}} .\end{align*} where $$h_p(v, \mkern 1.5mu\overline{\mkern-1.5muv\mkern-1.5mu}\mkern 1.5mu ) \geq 0$$ with equality if and only if $$v=0$$. Here we define $$\overline{\mathcal{E}_p}$$ as the fiber of the complex vector bundle $$\overline{\mathcal{E}}$$ whose transition functions are given by the complex conjugates of those from $$\mathcal{E}$$. \end{definition} \begin{remark} Note that $$\mathcal{E}, \overline{\mathcal{E}}$$ are genuinely different as complex bundles. There is a \emph{conjugate-linear} map given by conjugation, i.e.~$$L(cv) = \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu L(v)$$, where the canonical example is \begin{align*} {\mathbb{C}}^n &\to {\mathbb{C}}^n \\ (z_1, \cdots, z_n) &\mapsto (\mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu, \cdots, \mkern 1.5mu\overline{\mkern-1.5muz_n\mkern-1.5mu}\mkern 1.5mu) .\end{align*} \end{remark} \begin{definition}[Unitary Frame Bundle] We define the \textbf{unitary frame bundle} $$\mathop{\mathrm{UFrame}}(\mathcal{E}) \coloneqq\displaystyle\bigcup_p \mathop{\mathrm{UFrame}}(\mathcal{E})_p$$, where at each point this is given by the set of orthogonal frames of $$\mathcal{E}_p$$ given by $$(e_1, \cdots, e_n)$$ where $$h(e_i , \mkern 1.5mu\overline{\mkern-1.5mue_j\mkern-1.5mu}\mkern 1.5mu ) = \delta_{ij}$$. \end{definition} \begin{remark} This is a principal $$G{\hbox{-}}$$bundle for $$G = U_r({\mathbb{C}})$$, the invertible matrices $$A_{/{\mathbb{C}}}$$ satisfy $$A \overline{A}^t = \operatorname{id}$$. \end{remark} \begin{example}[of more principal bundles] For $$G={\mathbb{Z}}/2{\mathbb{Z}}$$ and $$X= S^1$$, the Möbius band is a principal $$G{\hbox{-}}$$bundle: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-25_14-25.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \begin{example}[more principal bundles] For $$G={\mathbb{Z}}/2{\mathbb{Z}}$$, for any (possibly non-oriented) manifold $$X$$ there is an \textbf{orientation principal bundle} $$P$$ which is locally a set of orientations on $$U$$, i.e.~ \begin{align*} P\coloneqq\left\{{(x, O) {~\mathrel{\Big|}~}x\in X,\, O \text{ is an orientation of }T_p X}\right\} .\end{align*} Note that $$P$$ is an oriented manifold, $$P\to X$$ is a local isomorphism, and has a canonical orientation. (?) This can also be written as $$P = \mathop{\mathrm{Frame}}(X) / \operatorname{GL}_n^+({\mathbb{R}})$$, since an orientation can be specified by a choice of $$n$$ linearly independent vectors where we identify any two sets that differ by a matrix of positive determinant. \end{example} \begin{definition}[Associated Bundles] Let $$P\to X$$ be a principal $$G{\hbox{-}}$$bundle and let $$G\to \operatorname{GL}(V)$$ be a continuous representation. The \textbf{associated bundle} is defined as \begin{align*} P\times_G V = \left\{{(p, v){~\mathrel{\Big|}~}p\in P,\, v\in V}\right\} / \sim && \text{where } (p, v) \sim (pg, g ^{-1} v) ,\end{align*} which is well-defined since there is a right action on the first component and a left action on the second. \end{definition} \begin{example}[?] Note that $$\mathop{\mathrm{Frame}}(\mathcal{E})$$ is a $$\operatorname{GL}_r({\mathbb{R}}){\hbox{-}}$$bundle and the map $$\operatorname{GL}_r({\mathbb{R}}) \xrightarrow{\operatorname{id}} \operatorname{GL}({\mathbb{R}}^r)$$ is a representation. At every fiber, we have $$G \times_G V = (p, v)/\sim$$ where there is a unique representative of this equivalence class given by $$(e, pv)$$. So $$P\times_G V_p \to \left\{{p}\right\} \cong V_x$$. \begin{exercise}[?] Show that $$\mathop{\mathrm{Frame}}(\mathcal{E}) \times_{\operatorname{GL}_r({\mathbb{R}})} {\mathbb{R}}^r \cong \mathcal{E}$$. This follows from the fact that the transition functions of $$P \times_G V$$ are given by left multiplication of $$t_{UV}: U \cap V \to G$$, and so by the equivalence relation, $$\operatorname{im}t_{UV} \in \operatorname{GL}(V)$$. \end{exercise} \end{example} \begin{remark} Suppose that $$M^3$$ is an oriented Riemannian 3-manifold. Them $$TM\to \mathop{\mathrm{Frame}}(M)$$ which is a principal $${\operatorname{SO}}(3){\hbox{-}}$$bundle. The universal cover is the double cover $${\operatorname{SU}}(2) \to {\operatorname{SO}}(3)$$, so can the transition functions be lifted? This shows up for spin structures, and we can get a $${\mathbb{C}}^2$$ bundle out of this. \end{remark} \hypertarget{wednesday-january-27}{% \section{Wednesday, January 27}\label{wednesday-january-27}} \hypertarget{bundles-and-connections}{% \subsection{Bundles and Connections}\label{bundles-and-connections}} \begin{definition}[Connections] Let $$\mathcal{E}\to X$$ be a vector bundle, then a \textbf{connection} on $$\mathcal{E}$$ is a map of sheaves of abelian groups \begin{align*} \nabla: \mathcal{E}\to \mathcal{E}\otimes\Omega^1_X \end{align*} satisfying the \emph{Leibniz rule}: \begin{align*} \nabla (fs) = f \nabla s + s\otimes ds \end{align*} for all opens $$U$$ with $$f\in {\mathcal{O}}(U)$$ and $$s\in \mathcal{E}(U)$$. Note that this works in the category of complex manifolds, in which case $$\nabla$$ is referred to as a \textbf{holomorphic connection}. \end{definition} \begin{remark} A connection $$\nabla$$ induces a map \begin{align*} \tilde{\nabla}: \mathcal{E}\otimes\Omega^p &\to \mathcal{E}\otimes\Omega^{p+1} \\ s \otimes \omega &\mapsto \nabla s \wedge w + s\otimes d \omega .\end{align*} where $$\wedge: \Omega^p \otimes\Omega^1 \to \Omega^{p+1}$$. The standard example is \begin{align*} d: {\mathcal{O}}&\to \Omega^1 \\ f &\mapsto df .\end{align*} where the induced map is the usual de Rham differential. \end{remark} \begin{exercise}[?] Prove that the \emph{curvature} of $$\nabla$$, i.e.~the map \begin{align*} F_{\nabla} \coloneqq\nabla \circ \nabla: \mathcal{E}\to \mathcal{E}\otimes\Omega^2 \end{align*} is $${\mathcal{O}}{\hbox{-}}$$linear, so $$F_{\nabla}(fs) = f\nabla \circ \nabla(s)$$. Use the fact that $$\nabla s \in \mathcal{E}\otimes\Omega^1$$ and $$\omega \in \Omega^p$$ and so $$\nabla s \otimes \omega \in \mathcal{E} \Omega^1 \otimes \Omega^p$$ and thus reassociating the tensor product yields $$\nabla s \wedge \omega \in \mathcal{E}\otimes\Omega^{p+1}$$. \end{exercise} \begin{remark} Why is this called a connection? \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{25pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-27_14-05.pdf_tex} }; \end{tikzpicture} } \end{figure} This gives us a way to transport $$v\in \mathcal{E}_p$$ over a path $$\gamma$$ in the base, and $$\nabla$$ provides a differential equation (a flow equation) to solve that lifts this path. Solving this is referred to as \textbf{parallel transport}. This works by pairing $$\gamma'(t) \in T_{ \gamma(t) } X$$ with $$\Omega^1$$, yielding $$\nabla s = ( \gamma'(t)) = s( \gamma(t))$$ which are sections of $$\gamma$$. Note that taking a different path yields an endpoint in the same fiber but potentially at a different point, and $$F_\nabla = 0$$ if and only if the parallel transport from $$p$$ to $$q$$ depends only on the homotopy class of $$\gamma$$. \begin{quote} Note: this works for any bundle, so can become confusing in Riemannian geometry when all of the bundles taken are tangent bundles! \end{quote} \end{remark} \begin{example}[A classic example] The Levi-Cevita connection $$\nabla^{LC}$$ on $$TX$$, which depends on a metric $$g$$. Taking $$X=S^2$$ and $$g$$ is the round metric, there is nonzero curvature: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-27_14-15.pdf_tex} }; \end{tikzpicture} } \end{figure} In general, every such transport will be rotation by some vector, and the angle is given by the area of the enclosed region. \end{example} \begin{definition}[Flat Connection and Flat Sections] A connection is \textbf{flat} if $$F_\nabla = 0$$. A section $$s \in \mathcal{E}(U)$$ is \textbf{flat} if it is given by \begin{align*} L(U) \coloneqq\left\{{ s\in \mathcal{E}(U) {~\mathrel{\Big|}~}\nabla s = 0}\right\} .\end{align*} \end{definition} \begin{exercise}[?] Show that if $$\nabla$$ is flat then $$L$$ is a \emph{local system}: a sheaf that assigns to any sufficiently small open set a vector space of fixed dimension. An example is the constant sheaf $$\underline{{\mathbb{C}}^d}$$. Furthermore $${\operatorname{rank}}(L) = {\operatorname{rank}}(\mathcal{E})$$. \end{exercise} \begin{remark} Given a local system, we can construct a vector bundle whose transition functions are the same as those of the local system, e.g.~for vector bundles this is a fixed matrix, and in general these will be constant transition functions. Equivalently, we can take $$L\otimes_{\mathbb{R}}{\mathcal{O}}$$, and $$L\otimes 1$$ form flat sections of a connection. \end{remark} \hypertarget{sheaf-cohomology}{% \subsection{Sheaf Cohomology}\label{sheaf-cohomology}} \begin{definition}[Čech complex] Let $$\mathcal{F}$$ be a sheaf of abelian groups on a topological space $$X$$, and let $$\mathfrak{U} \coloneqq\left\{{U_i}\right\} \rightrightarrows X$$ be an open cover of $$X$$. Let $$U_{i_1, \cdots, i_p} \coloneqq U_{i_1} \cap U_{i_2} \cap\cdots \cap U_{i_p}$$. Then the \textbf{Čech Complex} is defined as \begin{align*} C_{\mathfrak{U}}^p(X, \mathcal{F}) \coloneqq\prod_{i_1 < \cdots < i_p} \mathcal{F}(U_{i_1, \cdots, i_p}) \end{align*} with a differential \begin{align*} {{\partial}}^p: C_{\mathfrak{U}}^p(X, \mathcal{F}) &\to C_{\mathfrak{U}}^{p+1}(X \mathcal{F}) \\ \sigma &\mapsto ({{\partial}}\sigma)_{i_0, \cdots, i_p} \coloneqq\prod_j (-1)^j { \left.{{\sigma_{i_0, \cdots, \widehat{i_j}, \cdots, i_p}}} \right|_{{U_{i_0, \cdots, i_p}}} } \end{align*} where we've defined this just on one given term in the product, i.e.~a $$p{\hbox{-}}$$fold intersection. \end{definition} \begin{exercise}[?] Check that $${{\partial}}^2 = 0$$. \end{exercise} \begin{remark} The Čech cohomology $$H^p_{\mathfrak{U}}(X, \mathcal{F})$$ with respect to the cover $$\mathfrak{U}$$ is defined as $$\ker {{\partial}}^p/\operatorname{im}{{\partial}}^{p-1}$$. It is a difficult theorem, but we write $$H^p(X, \mathcal{F})$$ for the Čech cohomology for any sufficiently refined open cover when $$X$$ is assumed paracompact. \end{remark} \begin{example}[?] Consider $$S^1$$ and the constant sheaf $$\underline{{\mathbb{Z}}}$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-27_14-40.pdf_tex} }; \end{tikzpicture} } \end{figure} ere we have \begin{align*} C^0(S^1, \underline{{\mathbb{Z}}}) = \underline{{\mathbb{Z}}}(U_1) \oplus \underline{{\mathbb{Z}}}(U_2) = \underline{{\mathbb{Z}}} \oplus \underline{{\mathbb{Z}}} ,\end{align*} and \begin{align*} C^1(S^1, {\mathbb{Z}}) = \bigoplus_{\substack{ \text{double} \\ \text{intersections}} } \underline{{\mathbb{Z}}}(U_{ij}) \underline{{\mathbb{Z}}}(U_{12}) = \underline{{\mathbb{Z}}}(U_1 \cap U_{2}) = \underline{{\mathbb{Z}}} \oplus \underline{{\mathbb{Z}}} .\end{align*} We then get \begin{align*} C^0(S^1, \underline{{\mathbb{Z}}}) &\xrightarrow{{{\partial}}} C^1(S^1, \underline{{\mathbb{Z}}}) \\ {\mathbb{Z}}\oplus {\mathbb{Z}}&\to {\mathbb{Z}}\oplus {\mathbb{Z}}\\ (a, b) &\mapsto (a-b, a-b) ,\end{align*} Which yields $$H^*(S^1, \underline{{\mathbb{Z}}}) = [{\mathbb{Z}}, {\mathbb{Z}}, 0, \cdots]$$. \end{example} \hypertarget{sheaf-cohomology-friday-january-29}{% \section{Sheaf Cohomology (Friday, January 29)}\label{sheaf-cohomology-friday-january-29}} Last time: we defined the Čech complex $$C_{\mathfrak{U} }^p(X, \mathcal{F} ) \coloneqq\prod_{i_1, \cdots, i_p} \mathcal{F} (U_{i_1} \cap\cdots \cap U_{i_p})$$ for $$\mathfrak{U}\coloneqq\left\{{U_i}\right\}$$ is an open cover of $$X$$ and $$F$$ is a sheaf of abelian groups. \begin{fact} If $$\mathfrak{U}$$ is a sufficiently fine cover then $$H^p_{\mathfrak{U}}(X, \mathcal{F})$$ is independent of $$\mathfrak{U}$$, and we call this $$H^p(X; \mathcal{F})$$. \end{fact} \begin{remark} Recall that we computed $$H^p(S^1, \underline{{\mathbb{Z}}} = [{\mathbb{Z}}, {\mathbb{Z}}, 0, \cdots]$$. \end{remark} \begin{theorem}[When sheaf cohomology is isomorphic to singular cohomology] Let $$X$$ be a paracompact and locally contractible topological space. Then $$H^p(X, \underline{{\mathbb{Z}}}) \cong H^p_{{\operatorname{Sing}}}(X, \underline{{\mathbb{Z}}})$$. This will also hold more generally with $$\underline{{\mathbb{Z}}}$$ replaced by $$\underline{A}$$ for any $$A\in {\mathsf{Ab}}$$. \end{theorem} \begin{definition}[Acyclic Sheaves] We say $$\mathcal{F}$$ is \emph{acyclic} on $$X$$ if $$H^{> 0 }(X; \mathcal{F}) = 0$$. \end{definition} \begin{remark} How to visualize when $$H^1(X; \mathcal{F}) = 0$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-01-29_14-01.pdf_tex} }; \end{tikzpicture} } \end{figure} On the intersections, we have $$\operatorname{im}{{\partial}}^0 = \left\{{ (s_{i} - s_{j})_{ij} {~\mathrel{\Big|}~}s_i \in \mathcal{F}(U_i)}\right\}$$, which are \emph{cocycles}. We have $$C^1(X; \mathcal{F})$$ are collections of sections of $$\mathcal{F}$$ on every double overlap. We can check that $$\ker {{\partial}}^1 = \left\{{ (s_{ij}) {~\mathrel{\Big|}~}s_{ij} - s_{ik} + s_{jk} = 0}\right\}$$, which is the cocycle condition. From the exercise from last class, $${{\partial}}^2 = 0$$. \end{remark} \begin{theorem}[(Important!)] Let $$X$$ be a paracompact Hausdorff space and let \begin{align*} 0 \to \mathcal{F}_1 \xrightarrow{\varphi} \mathcal{F}_2 \to \mathcal{F}_3 \to 0 \end{align*} be a SES of sheaves of abelian groups, i.e.~$$\mathcal{F}_3 = \operatorname{coker}(\varphi)$$ and $$\varphi$$ is injective. Then there is a LES in cohomology: \begin{center} \begin{tikzcd} 0 && {H^0(X; \mathcal{F}_1)} && {H^0(X; \mathcal{F}_2)} && {H^0(X; \mathcal{F}_3)} \\ \\ && {H^1(X; \mathcal{F}_1)} && {H^1(X; \mathcal{F}_2)} && {H^1(X; \mathcal{F}_3)} \\ \\ && \cdots \arrow[from=1-7, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3] \end{tikzcd} \end{center} \end{theorem} \begin{example}[?] For $$X$$ a manifold, we can define a map and its cokernel sheaf: \begin{align*} 0 \to \underline{{\mathbb{Z}}} \xrightarrow{\cdot 2} \underline{{\mathbb{Z}}} \to \underline{{\mathbb{Z}}/2{\mathbb{Z}}} \to 0 .\end{align*} Using that cohomology of constant sheaves reduces to singular cohomology, we obtain a LES in homology: \begin{center} \begin{tikzcd} 0 && {H^0(X; {\mathbb{Z}})} && {H^0(X; {\mathbb{Z}})} && {H^0(X; {\mathbb{Z}}/2{\mathbb{Z}})} \\ \\ && {H^1(X; {\mathbb{Z}})} && {H^1(X; {\mathbb{Z}})} && {H^1(X; {\mathbb{Z}}/2{\mathbb{Z}})} \\ \\ && \cdots \arrow[from=1-7, to=3-3] \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=1-7] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=3-7] \arrow[from=3-7, to=5-3] \end{tikzcd} \end{center} \end{example} \begin{corollary}[of theorem] Suppose $$0 \to \mathcal{F}\to I_0 \xrightarrow{d_0} I_1 \xrightarrow{d_1} I_2 \xrightarrow{d_2} \cdots$$ is an exact sequence of sheaves, so on any sufficiently small set kernels equal images., and suppose $$I_n$$ is acyclic for all $$n\geq 0$$. This is referred to as an \textbf{acyclic resolution}. Then the homology can be computed at $$H^p(X; \mathcal{F}) = \ker (I_p(X) \to I_{p+1}(X)) / \operatorname{im}(I_{p-1}(X) \to I_p(X) )$$. \begin{quote} Note that locally having kernels equal images is different than satisfying this globally! \end{quote} \end{corollary} \begin{proof}[of corollary] This is a formal consequence of the existence of the LES. We can split the LES into a collection of SESs of sheaves: \begin{align*} 0 \to \mathcal{F}\to I_0 \xrightarrow{d_0} \operatorname{im}(d_0) \to 0 && \operatorname{im}(d_0) = \ker(d_1) \\ 0 \to \ker(d_1) \hookrightarrow I_1 \to I_1/\ker (d_1) = \operatorname{im}(d_1) && \operatorname{im}(d_1) = \ker(d_2) \\ .\end{align*} Note that these are all exact sheaves, and thus only true on small sets. So take the associated LESs. For the SES involving $$I_0$$, we obtain: \begin{center} \begin{tikzcd} {} \\ \\ {} &&&& \cdots \\ \\ {H^{p-1}(\mathcal{F})} && {H^{p-1}(\mathcal{I_0}) = 0} && {H^{p-1}(\mathcal{\operatorname{im}(d_0)})} \\ \\ {H^p(\mathcal{F})} && {\cdots = 0} \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow["\cong", from=5-5, to=7-1] \arrow[from=7-1, to=7-3] \arrow[from=3-5, to=5-1] \end{tikzcd} \end{center} The middle entries vanish since $$I_*$$ was assumed acyclic, and so we obtain $$H^p(\mathcal{F}) \cong H^{p-1}(\operatorname{im}d_0) \cong H^{p-1}(\ker d_1)$$. Now taking the LES associated to $$I_1$$, we get $$H^{p-1}(\ker d_1) \cong H^{p-2}(\operatorname{im}d_1)$$. Continuing this inductively, these are all isomorphic to $$H^p(\mathcal{F}) \cong H^0(\ker d_p)/ d_{p-1}(H^0(I_{p-1}))$$ after the $$p$$th step. \end{proof} \begin{corollary}[of the previous corollary] Suppose $$\mathfrak{U}\rightrightarrows X$$, then if $$\mathcal{F}$$ is acyclic on each $$U_{i_1, \cdots, i_p}$$, then $$\mathfrak{U}$$ is sufficiently fine to compute Čech cohomology, and $$H^p_{\mathfrak{U}}(X; \mathcal{F}) \cong H^p(X; \mathcal{F})$$. \end{corollary} \begin{proof}[?] See notes. \end{proof} \begin{corollary}[of corollary] Let $$X \in {\mathsf{Mfd}}_{\mathsf{sm}}$$, then $$H^p(X, \underline{{\mathbb{R}}}) = H^p_{\mathrm{dR}}(X;\ RR)$$. \end{corollary} \begin{proof}[?] Idea: construct an acyclic resolution of the sheaf $$\underline{{\mathbb{R}}}$$ on $$M$$. The following exact sequence works: \begin{align*} 0 \to \underline{{\mathbb{R}}} \to {\mathcal{O}}\xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \to \cdots .\end{align*} So we start with locally constant functions, then smooth functions, then smooth 1-forms, and so on. This is an exact sequence of sheaves, but importantly, not exact on the total space. To check this, it suffices to show that $$\ker d^p = \operatorname{im}d^{p-1}$$ on any contractible coordinate chart. In other words, we want to show that if $$d \omega=0$$ for $$\omega\in \Omega^p({\mathbb{R}}^n)$$ then $$\omega= d \alpha$$ for some $$\alpha\in \Omega^{p-1}({\mathbb{R}}^n)$$. This is true by integration! Using the previous corollary, $$H^p(X; \underline{{\mathbb{R}}}) = \ker(\Omega^p(X) \xrightarrow{d} \Omega^{p+1}(X) ) / \operatorname{im}(\Omega^{p-1}(X) \xrightarrow{d} \Omega^p(X))$$. \end{proof} \begin{quote} Check Hartshorne to see how injective resolutions line up with derived functors! \end{quote} \hypertarget{monday-february-01}{% \section{Monday, February 01}\label{monday-february-01}} \begin{remark} Last time $$\underline{{\mathbb{R}}}$$ on a manifold $$M$$ has a resolution by vector bundles: \begin{align*} 0 \to \underline{{\mathbb{R}}} \hookrightarrow\Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \cdots .\end{align*} This is an exact sequence of sheaves of any smooth manifold, since locally $$d \omega = 0 \implies \omega = d \alpha$$ (by the \emph{Poincaré $$d {\hbox{-}}$$lemma}). We also want to know that $$\Omega^k$$ is an acyclic sheaf on a smooth manifold. \end{remark} \begin{exercise}[?] Let $$X\in Top$$ and $$\mathcal{F}\in {\mathsf{Sh}}({\mathsf{Ab}})_{/X}$$. We say $$\mathcal{F}$$ is \textbf{flasque} if and only if for all $$U \supseteq V$$ the map $$\mathcal{F}(U) \xrightarrow{\rho_{UV}} \mathcal{F}(V)$$ is surjective. Show that $$\mathcal{F}$$ is acyclic, i.e.~$$H^i(X; \mathcal{F}) = 0$$. This can also be generalized with a POU. \end{exercise} \begin{example}[?] The function $$1/x\in {\mathcal{O}}({\mathbb{R}}\setminus\left\{{0}\right\})$$, but doesn't extend to a continuous map on $${\mathbb{R}}$$. So the restriction map is not surjective. \end{example} \begin{remark} Any vector bundle on a smooth manifold is acyclic. Using the fact that $$\Omega^k$$ is acyclic and the above resolution of $$\underline{{\mathbb{R}}}$$, we can write $$H^k(X; {\mathbb{R}}) = \ker(d_k) / \operatorname{im}d_{k-1} \coloneqq H^k_{dR}(X; {\mathbb{R}})$$. \end{remark} \begin{remark} Now letting $$X \in {\mathsf{Mfd}}_{\mathbb{C}}$$, recalling that $$\Omega^p$$ was the sheaf of holomorphic $$p {\hbox{-}}$$forms. Locally these are of the form $$\sum_{{\left\lvert {I} \right\rvert} = p} f_I(\mathbf{z}) dz^I$$ where $$f_I(\mathbf{z})$$ is holomorphic. There is a resolution \begin{align*} 0 \xrightarrow{} \Omega^p \xrightarrow{} A^{p, 0} ,\end{align*} where in $$A^{p, 0}$$ we allowed also $$f_I$$ are \emph{smooth}. These are the same as bundles, but we view sections differently. The first allows only holomorphic sections, whereas the latter allows smooth sections. What can you apply to a smooth $$(p, 0)$$ form to check if it's holomorphic? \end{remark} \begin{example}[?] For $$p=0$$, we have \begin{align*} 0 \to {\mathcal{O}}\to A^{0, 0} .\end{align*} where we have the sheaf of holomorphic functions mapping to the sheaf of smooth functions. We essentially want a version of checking the Cauchy-Riemann equations. \end{example} \begin{definition}[$\del$ and $\delbar$ operators] Let $$\omega\in A^{p, q}(X)$$ where \begin{align*} d \omega = \sum {\frac{\partial f_I}{\partial z_j}\,} dz^j \wedge dz^I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^J + \sum_j {\frac{\partial f_I}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_j}\,} d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^j \wedge dz^I d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^J\coloneqq{\partial}+ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu \end{align*} with $${\left\lvert {I} \right\rvert} = p, {\left\lvert {J} \right\rvert} = q$$. \end{definition} \begin{example}[?] The function $$f(z) = z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \in A^{0, 0}({\mathbb{C}})$$ is smooth, and $$df = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu dz + z d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$. This can be checked by writing $$z^j = x^j + iy^j$$ and $$\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu^j = x^j - iy_j$$, and $${\frac{\partial }{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}\,} g = 0$$ if and only if $$g$$ is holomorphic. Here we get $${\partial}\omega \in A^{p+1, q}(X)$$ and $$\mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu \in A^{p, q+1}(X)$$, and we can write $$d(z \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) = {\partial}(z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)$$. \end{example} \begin{definition}[Cauchy-Riemann Equations] Recall the Cauchy-Riemann equations: $$\omega$$ is a holomorphic $$(p, 0) {\hbox{-}}$$form on $${\mathbb{C}}^n$$ if and only if $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\omega = 0$$. \end{definition} \begin{remark} Thus to extend the previous resolution, we should take \begin{align*} 0 \to \Omega^p \hookrightarrow A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 2} \to \cdots .\end{align*} The fact that this is exact is called the \emph{Poincaré $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{\hbox{-}}$$lemma}. \end{remark} \begin{remark} There are no bump functions in the holomorphic world, and since $$\Omega^p$$ is a holomorphic bundle, it may not be acyclic. However, the $$A^{p, q}$$ \emph{are} acyclic (since they are smooth vector bundles and thus admit POUs), and we obtain \begin{align*} H^q(X; \Omega^p) = \ker( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_q) / \operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{q-1}) .\end{align*} Note the similarity to $$H_{\mathrm{dR}}$$, using $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$ instead of $$d$$. This is called \textbf{Dolbeault cohomology}, and yields invariants of complex manifolds: the \textbf{Hodge numbers} $$h^{p, q}(X) \coloneqq\dim_{\mathbb{C}}H^q(X; \Omega^p)$$. These are analogies: \begin{longtable}[]{@{}ll@{}} \toprule Smooth & Complex \\ \midrule \endhead $$\underline{{\mathbb{R}}}$$ & $$\Omega^p$$ \\ $$\Omega^k$$ & $$A^{p, q}$$ \\ Betti numbers $$\beta_k$$ & Hodge numbers $$h^{p, q}$$ \\ \bottomrule \end{longtable} Note the slight overloading of terminology here! \end{remark} \begin{theorem}[Properties of Singular Cohomology] Let $$X\in {\mathsf{Top}}$$, then $$H_{{\operatorname{Sing}}}^i(X; {\mathbb{Z}})$$ satisfies the following properties: \begin{itemize} \item Functoriality: given $$f\in \mathop{\mathrm{Hom}}_{\mathsf{Top}}(X, Y)$$, there is a pullback $$f^*: H^i(Y; {\mathbb{Z}}) \to H^i(X; {\mathbb{Z}})$$. \item The cap product: a pairing \begin{align*} H^i(X; {\mathbb{Z}}) \otimes_{\mathbb{Z}}H_j(X; {\mathbb{Z}}) &\to H_{j-i}(X; {\mathbb{Z}}) \\ \varphi\otimes\sigma &\mapsto \varphi\qty{{ \left.{{\sigma}} \right|_{{\Delta_{0, \cdots, j}}} }} { \left.{{ \sigma}} \right|_{{\Delta_{i, \cdots, j}}} } .\end{align*} This makes $$H_*$$ a module over $$H^*$$. \item There is a ring structure induced by the cup product: \begin{align*} H^i(X; {\mathbb{R}}) \times H^j(X; {\mathbb{R}})\to H^{i+j}(X; {\mathbb{R}}) && \alpha\cup \beta &= (-1)^{ij} \beta \cup \alpha .\end{align*} \item Poincaré Duality: If $$X$$ is an oriented manifold, there exists a fundamental class $$[X] \in H_{n}(X; {\mathbb{Z}}) \cong {\mathbb{Z}}$$ and $$({-})\cap X: H^i \to H_{n-i}$$ is an isomorphism. \end{itemize} \end{theorem} \begin{remark} Let $$M \subset X$$ be a submanifold where $$X$$ is a smooth oriented $$n{\hbox{-}}$$manifold. Then $$M \hookrightarrow X$$ induces a pushforward $$H_n(M; {\mathbb{Z}}) \xrightarrow{\iota_*} H_n(X; {\mathbb{Z}})$$ where $$\sigma \mapsto \iota \circ \sigma$$. Using Poincaré duality, we'll identify $$H_{\dim M}(X; {\mathbb{Z}}) \to H^{\operatorname{codim}M}(X; {\mathbb{Z}})$$ and identify $$[M] = PD( \iota_*( [M]))$$. In this case, if $$M\pitchfork N$$ then $$[M] \cap [N] = [M \cap N]$$, i.e.~the cap product is given by intersecting submanifolds. \end{remark} \begin{warnings} This can't always be done! There are counterexamples where homology classes can't be represented by submanifolds. \end{warnings} \hypertarget{wednesday-february-03}{% \section{Wednesday, February 03}\label{wednesday-february-03}} Consider an oriented surface, and take two oriented submanifolds \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{39pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-03_13-54.pdf_tex} }; \end{tikzpicture} } \end{figure} We can then take the fundamental classes of the submanifolds, say $$[\alpha], [\beta] \in H^1(X; {\mathbb{Z}}) \xrightarrow{PD} H^1(X, {\mathbb{Z}})$$. Here $$T_p \alpha \oplus T_p \beta = T_p X$$, since the intersections are transverse. Since $$\alpha, \beta$$ are oriented, let $$\left\{{ e }\right\}$$ be a basis of $$T_p \alpha$$ (up to $${\mathbb{R}}^+$$) and similarly $$\left\{{ f }\right\}$$ a basis of $$T_p \beta$$. We can then ask if $$\left\{{ e, f }\right\}$$ constitutes an \emph{oriented} basis of $$T_pX$$. If so, we write $$\alpha \cdot_p \beta \coloneqq+1$$ and otherwise $$\alpha \cdot_p \beta = - 1$$. We thus have \begin{align*} [ \alpha] \smile[ \beta] \in H^2(X; {\mathbb{Z}}) \xrightarrow{PD} H_0(X; {\mathbb{Z}}) = {\mathbb{Z}} \end{align*} since $$X$$ is connected. We can thus define the \textbf{intersection form} $$\alpha\cdot \beta\coloneqq[ \alpha] \smile[ \beta]$$. In general if $$A, B$$ are oriented transverse submanifolds of $$M$$ which are themselves oriented, we'll have $$[A] \smile[B] = [A \cap B]$$. We need to be careful: how do we orient the intersection? This is given by comparing the orientations on $$A$$ and $$B$$ as before. \begin{example}[?] If $$\dim M = \dim A + \dim B$$, then any $$p\in A \cap B$$ is oriented by comparing $$\left\{{ \mathrm{or}_A, \mathrm{or}_B}\right\}$$ to $$\mathrm{or}_M$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-03_14-03.pdf_tex} }; \end{tikzpicture} } \end{figure} Here it suffices to check that $$\left\{{ e, f_1, f_2 }\right\}$$ is an oriented basis of $$T_p M$$. \end{example} \begin{example}[?] In this case, $$[\alpha] \smile[\beta] = 0$$ and so $$\alpha\cdot \beta = 0$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-03_14-06.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \begin{remark} Note that cohomology with $${\mathbb{Z}}$$ coefficients can be defined for any topological space, and Poincaré duality still holds. \end{remark} \begin{remark} We'll be considering $$M = M^4$$, smooth 4-manifolds. How to visualize: take a 3-manifold and cross it with time! \begin{figure} \centering \includegraphics{figures/time_manifold_glitch_workaround.png} \caption{Picking one basis element in the time direction} \end{figure} Here $$?$$ is oriented in the forward time'' direction, and this is a surface at time $$t=0$$. Where $$A\cdot B = +1$$, since $$\left\{{ e_1, e_2, f_1, f_2 }\right\} = \left\{{ e_x, e_y, e_z, e_t }\right\}$$ is a oriented basis for $${\mathbb{R}}^4$$. For $$?^2$$, switching the order of $$\alpha, \beta$$ no longer yields an oriented basis, but in this case it is $$?$$ and $$A\cdot B = B \cdot A$$. This is because \begin{align*} A \coloneqq \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \implies \operatorname{det}(A) =-1 && \operatorname{det} \begin{bmatrix} A & \\ & A \end{bmatrix} = 1 .\end{align*} \end{remark} \begin{remark} Let $$M^{2n}$$ be an oriented manifold, then the cup product yields a bilinear map $$H^n(M; {\mathbb{Z}}) \otimes H^n(M; {\mathbb{Z}}) \to {\mathbb{Z}}$$ which is symmetric when $$n$$ is odd and antisymmetric (or symplectic) when $$n$$ is even. This is a \textbf{perfect} (or \textbf{unimodular}) pairing (potentially after modding out by torsion) which realizes an isomorphism: \begin{align*} \qty{ H^n(M; {\mathbb{Z}})/{\operatorname{tors}}} {}^{ \vee }&\xrightarrow{\sim} H^n(M; {\mathbb{Z}})/{\operatorname{tors}}\\ \alpha \smile{-}&\mapsfrom \alpha ,\end{align*} where the LHS are linear functionals on cohomology. \end{remark} \begin{remark} Recall the universal coefficients theorem: \begin{align*} H^i(X; {\mathbb{Z}})/{\operatorname{tors}}\cong \qty{ H_i(X; {\mathbb{Z}})/{\operatorname{tors}}} {}^{ \vee } .\end{align*} The general theorem shows that $$H^i(X; {\mathbb{Z}})_{\operatorname{tors}}= H_{i-1}(X; {\mathbb{Z}})_{\operatorname{tors}}$$. \end{remark} \begin{remark} Note that if $$M$$ is an oriented 4-manifold, then \begin{center} \begin{tikzcd} && {\operatorname{tors}}& {\text{torsionfree}} &&&&& {\operatorname{tors}}& {\text{torsionfree}} \\ {H^0} && 0 & {\mathbb{Z}}&&& {H_0} && 0 & {\mathbb{Z}}\\ {H^1} && 0 & \textcolor{rgb,255:red,214;green,92;blue,92}{{\mathbb{Z}}^{\beta_1}} &&& {H_1} && A & \textcolor{rgb,255:red,214;green,92;blue,214}{{\mathbb{Z}}^{\beta_1}} \\ {H^2} && A & \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{Z}}^{\beta_2}} & {} & {{}} & {H_2} && A & \textcolor{rgb,255:red,92;green,92;blue,214}{{\mathbb{Z}}^{\beta_2}} \\ {H^3} && A & \textcolor{rgb,255:red,214;green,92;blue,214}{{\mathbb{Z}}^{\beta_1}} &&& {H_3} && 0 & \textcolor{rgb,255:red,214;green,92;blue,92}{{\mathbb{Z}}^{\beta_1}} \\ {H^4} && 0 & {\mathbb{Z}}&&& {H_4} && 0 & {\mathbb{Z}} \arrow["PD", from=4-5, to=4-6] \end{tikzcd} \end{center} In particular, if $$M$$ is simply connected, then $$H_1(M) = {\mathsf{Ab}}(\pi_1(M)) = 0$$, which forces $$A = 0$$ and $$\beta_1 = 0$$. \end{remark} \begin{definition}[Lattice] A \textbf{lattice} is a finite-dimensional free $${\mathbb{Z}}{\hbox{-}}$$module $$L$$ together with a symmetric bilinear form \begin{align*} \cdot: L^{\otimes 2} &\to {\mathbb{Z}}\\ \ell \otimes m &\mapsto \ell \cdot m .\end{align*} The lattice $$(L, \cdot)$$ is \textbf{unimodular} if and only if the following map is an isomorphism: \begin{align*} L &\to L {}^{ \vee }\\ \ell &\mapsto \ell \cdot ({-}) .\end{align*} \end{definition} \begin{remark} How to determine if a lattice is unimodular: take a basis $$\left\{{ e_1, \cdots, e_n }\right\}$$ of $$L$$ and form the \emph{Gram matrix} $$M_{ij} \coloneqq( e_i \cdot e_j) \in \operatorname{Mat}(n\times n, {\mathbb{Z}})^{\operatorname{Sym}}$$. Then $$(L, \cdot)$$ is unimodular if and only if $$\operatorname{det}(M) = \pm 1$$ if and only if $$M ^{-1}$$ is integral. In this case, the rows of $$M ^{-1}$$ will form a basis of the dual basis. \end{remark} \begin{definition}[Index of a lattice] The \textbf{index} of a lattice is $${\left\lvert { \operatorname{det}M} \right\rvert}$$. \end{definition} \begin{exercise}[?] Prove that $${\left\lvert {\operatorname{det}M} \right\rvert} = {\left\lvert { L {}^{ \vee }/ L } \right\rvert}$$. \end{exercise} \begin{remark} In general, for $$M^{4k}$$, the $$H^{2k}/{\operatorname{tors}}$$ is unimodular. For $$M^{4k+2}$$, the $$H^{2k+1}/{\operatorname{tors}}$$ is a unimodular \emph{symplectic} lattice, which is obtained by replacing the word symmetric'' with antisymmetric'' everywhere above. \end{remark} \begin{example}[?] For the torus, since the dimension is $$2 \pmod 4$$, you get the skew-symmetric matrix \begin{align*} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} .\end{align*} \todo[inline]{Check!} \end{example} \begin{definition}[Nondegenerate lattices] A lattice is \textbf{nondegenerate} if $$\operatorname{det}M \neq 0$$. \end{definition} \begin{definition}[Base change of lattices] The tensor product $$L \otimes_{\mathbb{Z}}{\mathbb{R}}$$ is a vector space with an $${\mathbb{R}}{\hbox{-}}$$valued symmetric bilinear form. This allows extending the lattice from $${\mathbb{Z}}^n$$ to $${\mathbb{R}}^n$$. \end{definition} \begin{remark} If $$(L, \cdot)$$ is nondegenerate, then Gram-Schmidt will yield an orthonormal basis $$\left\{{ v_i }\right\}$$. The number of positive norm vectors is an invariant, so we obtain $${\mathbb{R}}^{p, q}$$ where $$p$$ is the number of $$+1$$s in the Gram matrix and $$q$$ is the number of $$-1$$s. The \textbf{signature} of $$(L, {-})$$ is $$(p, q)$$, or by abuse of notation $$p-q$$. This is an invariant of the 4-manifold, as is the lattice itself $$H^2(X; {\mathbb{Z}})/{\operatorname{tors}}$$ equipped with the intersection form. \end{remark} \begin{remark} There is a perfect pairing called the \textbf{linking pairing}: \begin{align*} H^i(X; {\mathbb{Q}}/{\mathbb{Z}}) \otimes H^{n-i-1}(X; {\mathbb{Q}}/{\mathbb{Z}}) \to {\mathbb{Q}}/{\mathbb{Z}} .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-03_14-43.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{remark} $$A \cdot B \coloneqq\sum_{p\in A \cap B} \operatorname{sgn}_p(A, B)$$, where $$A \pitchfork B$$ and this turns out to be equal to the cup product. This works for topological manifolds -- but there are no tangent spaces there, so taking oriented bases doesn't work so well! You can also view \begin{align*} [A] \smile[\omega] = \int_A \omega .\end{align*} \end{remark} \hypertarget{friday-february-05}{% \section{Friday, February 05}\label{friday-february-05}} \begin{remark} Recall that a lattice is \textbf{unimodular} if the map $$L\to L {}^{ \vee }\coloneqq\mathop{\mathrm{Hom}}(L, {\mathbb{Z}})$$ is an isomorphism, where $$\ell \mapsto \ell \cdot ({-})$$. To check this, it suffices to check if the Gram matrix $$M$$ of a basis $$\left\{{e_i}\right\}$$ satisfies $${\left\lvert { \operatorname{det}M } \right\rvert} = 1$$. \end{remark} \begin{example}[Determinant 1 Integer Matrices] The matrices $$[1]$$ and $$[-1]$$ correspond to the lattice $${\mathbb{Z}}e$$ where either $$e^2 \coloneqq e\cdot e = 1$$ or $$e^2 = -1$$. If $$M_1, M_2$$ both have absolute determinant $$1$$, then so does \begin{align*} \begin{bmatrix} M_1 & 0 \\ 0 & M_2 \end{bmatrix} .\end{align*} So if $$L_1, L_2$$ are unimodular, then taking an orthogonal sum $$L_1 \oplus L_2$$ also yields a unimodular lattice. So this yields diagonal matrices with $$p$$ copies of $$+1$$ and $$q$$ copies of $$-1$$. This is referred to as $$rm{1}_{p, q}$$, and is an \emph{odd} unimodular lattice of signature $$(p, q)$$ (after passing to $${\mathbb{R}}$$). Here \emph{odd} means that there exists a $$v\in L$$ such that $$v^2$$ is odd. \end{example} \begin{example}[Even unimodular lattices] An even lattice must have no vectors of odd norm, so all of the diagonal elements are in $$2{\mathbb{Z}}$$. This is because $$(\sum n_i e_i)^2 = \sum_i n_i^2 e_i^2 + \sum_{i N} = 0$$. For any $$L \in {\operatorname{Gr}}_d({\mathbb{R}}^{\infty })$$, since $${\mathbb{R}}^d$$ has a standard basis, there is a natural $$\operatorname{GL}_d$$ torsor: the set of ordered bases of linear subspaces. So define \begin{align*} EG \coloneqq\left\{{ \text{bases of linear subspaces } L \in {\operatorname{Gr}}_d({\mathbb{R}}^{\infty }) }\right\} ,\end{align*} then any $$A\in \operatorname{GL}_d({\mathbb{R}})$$ acts on $$EG$$ by sending $$(L, \left\{{e_i}\right\}) \mapsto (L, \left\{{ Le_i}\right\} )$$. We can identify $$EG$$ as $$d{\hbox{-}}$$tuples of linearly independent elements of $${\mathbb{R}}^{\infty }$$, and there is a map \begin{align*} EG &\to BG \\ \left\{{e_i}\right\} &\mapsto {\operatorname{span}}_{\mathbb{R}}\left\{{e_i}\right\} .\end{align*} Thus there is a universal vector bundle over $$BGL_d$$: \begin{center} \begin{tikzcd} \mathcal{E}_L \coloneqq L \ar[r] & \mathcal{E} \ar[d] \\ & BGL_d \end{tikzcd} \end{center} So $$\mathcal{E} \subseteq BGL_d \times{\mathbb{R}}^{\infty }$$, where we can define $$\mathcal{E} \coloneqq\left\{{(L, p) {~\mathrel{\Big|}~}p\in L}\right\}$$. In this case, $$EG = \mathop{\mathrm{Frame}}( \mathcal{E})$$ is the frame bundle of this universal bundle. The same setup applies for $$G \coloneqq\operatorname{GL}_d({\mathbb{C}})$$, except we take $${\operatorname{Gr}}_d({\mathbb{C}}^{\infty })$$. \end{example} \begin{example}[?] Consider $$G = O_d$$, the set of orthogonal transformations of $${\mathbb{R}}^d$$ with the standard bilinear form, and $$U_d$$ the set of unitary such transformations. To be explicit: \begin{align*} U_d \coloneqq\left\{{ A \in \operatorname{Mat}(d \times d, {\mathbb{C}}) {~\mathrel{\Big|}~}{\left\langle {Av},~{Av} \right\rangle} = {\left\langle {v},~{v} \right\rangle} }\right\} ,\end{align*} where \begin{align*} {\left\langle { {\left[ {v_1, \cdots, v_n} \right]}},~{{\left[ {v_1, \cdots, v_n } \right]} } \right\rangle} = \sum {\left\lvert {v_i} \right\rvert}^2 .\end{align*} Alternatively, $$A^t A = I$$ for $$O_d$$ and $${\overline{{A^t}}} A = I$$ for $$U_d$$. In this case, $$BO_d = {\operatorname{Gr}}_d( {\mathbb{R}}^{\infty } )$$ and $$BU_d = {\operatorname{Gr}}_d( {\mathbb{C}}^{ \infty })$$, but we'll make the fibers smaller: set the fiber over $$L$$ to be \begin{align*} (EO_d)_L \coloneqq\left\{{ \text{orthogonal frames of } L }\right\} \end{align*} and similarly $$(EU_d)_L$$ the unitary frames of $$L$$. That there are related comes from the fact that $$\operatorname{GL}_d$$ retracts onto $$O_d$$ using the Gram-Schmidt procedure. \end{example} \begin{remark} Recall that there is a bijective correspondence \begin{align*} \left\{{\substack{ \text{Principal $G{\hbox{-}}$ bundles} \\ \text{on } X }}\right\} &\rightleftharpoons [X, BG] \end{align*} and there is also a correspondence \begin{align*} \left\{{\substack{ \text{Principal $\operatorname{GL}_d{\hbox{-}}$bundles }\\ \text{on } X }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Principal ${\mathcal{O}}_d{\hbox{-}}$bundles } \\ \text{on } X }}\right\} \end{align*} Using the associated bundle construction, on the LHS we obtain vector bundles $$\mathcal{E}\to X$$ of rank $$d$$, and on the RHS we have bundles with a metric. In local trivializations $$U \times{\mathbb{R}}^d \to {\mathbb{R}}^d$$, the metric is the standard one on $${\mathbb{R}}^d$$. This is referred to as a \textbf{reduction of structure group}, i.e.~a principal $$\operatorname{GL}_d$$ bundle admits possibly different trivializations for which the transition functions lie in the subgroup $$O_d$$. \end{remark} \begin{example}[?] Given any trivial principal $$G{\hbox{-}}$$bundle, it has a reduction of structure group to the trivial group. But the fact that the bundle is trivial may not be obvious. \includegraphics{figures/forbidden_donut.png} \end{example} \begin{remark} We want to compute $$H^*(BU_d; {\mathbb{Z}})$$. Why is this important? Given any complex vector bundle $$\mathcal{E}\to X$$ there is an associated principal $$U_d$$ bundle by choosing a metric, so we get a homotopy class $$[X, BU_d]$$. Given any $$f\in [X, BU_d]$$ and any $$\alpha\in H^k(BU_d; {\mathbb{Z}})$$, we can take the pullback $$f^* \alpha \in H^k(X; {\mathbb{Z}})$$, which are \textbf{Chern classes}. \end{remark} \begin{exercise}[?] Show that $$H^*(BU_d; {\mathbb{Z}})$$ stabilizes as $$d\to \infty$$ to an infinitely generated polynomial ring $${\mathbb{Z}}[c_1, c_2, \cdots]$$ with each $$c_i$$ in cohomological degree $$2i$$, so $$c_i \in H^{2i}(BU_d, {\mathbb{Z}})$$. \end{exercise} \begin{definition}[Chern class] There is a map $$BU_{d-1} \to BU_d$$, which we can identify as \begin{align*} {\operatorname{Gr}}_{d-1}(C^{\infty }) &\to {\operatorname{Gr}}_d({\mathbb{C}}^{\infty }) \\ \left\{{v_1, \cdots, v_{d-1}}\right\} &\mapsto {\operatorname{span}}\left\{{ (1, 0, 0, \cdots), sv_1, \cdots, sv_{d-1} }\right\} .\end{align*} This is defined by sending a basis where $$s: {\mathbb{C}}^{\infty } \to {\mathbb{C}}^{\infty}$$ is the map that shifts every coordinate to the right by one. \todo[inline]{ Question: does ${\operatorname{Gr}}_d({\mathbb{C}}^{\infty})$ deformation retract onto the image of this map? } This will yield a fiber sequence \begin{align*} S^{2d-1} \to BU_{d-1} \to BU_d \end{align*} and using connectedness of the sphere and the LES in homotopy this will identify \begin{align*} H^*(BU_d) = H^*(BU_{d-1})[c_d] && \text{where } c_d \in H^{2d}(BU_d) .\end{align*} The \textbf{Chern class} of a vector bundle $$\mathcal{E}$$ , denoted $$c_k( \mathcal{E} )$$, will be defined as the pullback $$f^* c_k$$. \end{definition} \hypertarget{wednesday-february-10}{% \section{Wednesday, February 10}\label{wednesday-february-10}} \begin{theorem}[Stable cohomology of BOn] As $$n\to \infty$$, we have \begin{align*} H^*(BO_n, {\mathbb{Z}}/2{\mathbb{Z}}) = {\mathbb{Z}}/2{\mathbb{Z}}[w_1, w_2, \cdots] && w_i \in H^i .\end{align*} \end{theorem} \begin{definition}[Stiefel-Whitney class] Given any principal $$O_n{\hbox{-}}$$bundle $$P\to X$$, there is an induced map $$X \xrightarrow{f} BO_n$$, so we can pull back the above generators to define the \textbf{Stiefel-Whitney classes} $$f^* w_i$$. \end{definition} \begin{remark} If $$P \coloneqq\mathop{\mathrm{OFrame}}TX$$, then $$f^* w_1$$ measures whether $$X$$ has an orientation, i.e.~$$f^* w_1 = 0 \iff X$$ can be oriented. We also have $$f^* w_i(P) = w_i( \mathcal{E} )$$ where $$P = \mathop{\mathrm{OFrame}}( \mathcal{E} )$$. In general, we'll just write $$w_i$$ for Stiefel-Whitney classes and $$c_i$$ for Chern classes. \end{remark} \begin{definition}[Pontryagin Classes] The \textbf{Pontryagin classes} of a real vector bundle $$\mathcal{E}$$ are defined as \begin{align*} p_i( \mathcal{E} ) = (-1)^i c_{2i}( \mathcal{E} \otimes_{\mathbb{R}}{\mathbb{C}}) .\end{align*} Note that the complexified bundle above is a complex vector bundle with the same transition functions as $$\mathcal{E}$$, but has a reduction of structure group from $$\operatorname{GL}_n({\mathbb{C}})$$ to $$\operatorname{GL}_n({\mathbb{R}})$$. \end{definition} \begin{observation} $${\mathbb{RP}}^{\infty }$$ and $${\mathbb{CP}}^{\infty }$$ are examples of $$K(\pi, n)$$ spaces, which are the unique-up-to-homotopy spaces defined by \begin{align*} \pi_k K (\pi, n) = \begin{cases} \pi & k=n \\ 0 & \text{else}. \end{cases} \end{align*} \end{observation} \begin{theorem}[Brown Representability] \begin{align*} H^n(X; \pi) \cong [X, K( \pi, n) ] .\end{align*} \end{theorem} \begin{example}[?] \begin{align*} [X, {\mathbb{RP}}^{\infty } ] &\cong H^1(X; {\mathbb{Z}}/2{\mathbb{Z}}) \\ [X, {\mathbb{CP}}^{\infty } ] &\cong H^2(X; {\mathbb{Z}}) .\end{align*} \end{example} \begin{proposition}[Classification of complex line bundles] There is a correspondence \begin{align*} \left\{{\substack{ \text{Complex line bundles} }}\right\} \rightleftharpoons [X, {\mathbb{CP}}^{\infty }] = [X, BC^{\times}] \rightleftharpoons H^2(X; {\mathbb{Z}}) \end{align*} Importantly, note that for $$X \in {\mathsf{Mfd}}_{\mathbb{C}}$$, $$H^2(X; {\mathbb{Z}})$$ measures \emph{smooth} complex line bundles and not holomorphic bundles. \end{proposition} \begin{proof}[of proposition] We'll take an alternate direct proof. Consider the exponential exact sequence on $$X$$: \begin{align*} 0 \to \underline{Z} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times} .\end{align*} Note that $$\underline{{\mathbb{Z}}}$$ consists of locally constant $${\mathbb{Z}}{\hbox{-}}$$valued functions, $${\mathcal{O}}$$ consists of smooth functions, and $${\mathcal{O}}^{\times}$$ are ???. \todo[inline]{Can't read screenshot! :(} This yields a LES in homology: \begin{center} \begin{tikzcd} {H^0(X; \underline{{\mathbb{Z}}})} && {H^0(X; {\mathcal{O}})} && {H^0(X; {\mathcal{O}}^{\times})} \\ \\ {H^1(X; \underline{{\mathbb{Z}}})} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(X; {\mathcal{O}})} && {H^1(X; {\mathcal{O}}^{\times})} \\ \\ {H^2(X; \underline{{\mathbb{Z}}})} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(X; {\mathcal{O}})} && {H^2(X; {\mathcal{O}}^{\times})} \arrow[from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=1-5, to=3-1, out=0, in=180] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=5-1, out=0, in=180] \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMCwwLCJIXjAoWDsgXFxjb25zdGFudHtcXFpafSkiXSxbMCwyLCJIXjEoWDsgXFxjb25zdGFudHtcXFpafSkiXSxbMCw0LCJIXjIoWDsgXFxjb25zdGFudHtcXFpafSkiXSxbMiwwLCJIXjAoWDsgXFxPTykiXSxbMiwyLCJIXjEoWDsgXFxPTykiLFswLDYwLDYwLDFdXSxbMiw0LCJIXjIoWDsgXFxPTykiLFswLDYwLDYwLDFdXSxbNCwwLCJIXjAoWDsgXFxPT1xcdW5pdHMpIl0sWzQsMiwiSF4xKFg7IFxcT09cXHVuaXRzKSJdLFs0LDQsIkheMihYOyBcXE9PXFx1bml0cykiXSxbMCwzXSxbMyw2XSxbNiwxXSxbMSw0XSxbNCw3XSxbNywyXSxbMiw1XSxbNSw4XV0=}{Link to Diagram} \end{quote} Since $${\mathcal{O}}$$ admits a partition of unity, $$H^{>0}(X; {\mathcal{O}}) = 0$$ and all of the red terms vanish. For complex line bundles $$L$$, $$H^1(X, {\mathcal{O}}^{\times}) \cong H^2(X; {\mathbb{Z}})$$. Taking a local trivialization $${ \left.{{L}} \right|_{{U}} } \cong U \times{\mathbb{C}}$$, we obtain transition functions \begin{align*} t_{UV} \in C^{\infty }(U \cap V, \operatorname{GL}_1({\mathbb{C}}) ) \end{align*} where we can identify $$\operatorname{GL}_1({\mathbb{C}}) \cong {\mathbb{C}}^{\times}$$. We then have \begin{align*} (t_{U_{ij}}) \in \prod_{i < j} {\mathcal{O}}^{\times}(U_i \cap U_j) = C^1(X; {\mathcal{O}}^{\times}) .\end{align*} Moreover, \begin{align*} \qty{ t_{U_{ij}} t_{U_{ik}} ^{-1} t_{U_{jk}} }_{i,j,k} = {{\partial}}( t_{U_{ij} } ) _{i, j} = 0 ,\end{align*} since transitions functions satisfy the cocycle condition. So in fact $$(t_{U_{ij}}) \in Z^1(X; {\mathcal{O}}^{\times}) = \ker {{\partial}}^1$$, and we can take its equivalence class $$[ ( t_{U_{ij} } ) ] \in H^1(X; {\mathcal{O}}^{\times}) = \ker {{\partial}}^1 / \operatorname{im}{{\partial}}^0$$. Changing trivializations by some $$s_i \in \prod_i {\mathcal{O}}^{\times}(U_i)$$ yields a composition which is a different trivialization of the same bundle: \begin{center} \begin{tikzcd} {{ \left.{{L}} \right|_{{U_i}} }} && {U_i \times{\mathbb{C}}} &&& {U_i \times{\mathbb{C}}} \arrow["{h_i}", from=1-1, to=1-3] \arrow["{\cdot s_i}", from=1-3, to=1-6] \arrow[curve={height=30pt}, from=1-1, to=1-6] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXHJve0x9e1VfaX0iXSxbMiwwLCJVX2kgXFxjcm9zcyBcXENDIl0sWzUsMCwiVV9pIFxcY3Jvc3MgXFxDQyJdLFswLDEsImhfaSJdLFsxLDIsIlxcY2RvdCBzX2kiXSxbMCwyLCIiLDIseyJjdXJ2ZSI6NX1dXQ==}{Link to Diagram} \end{quote} So the $$(t_{ U_{ij}}$$ change \emph{exactly} by an $${{\partial}}^0( s_i)$$. Thus the following map is well-defined: \begin{align*} L \mapsto [ (t_{U_{ij}} ) ] \in H^1(X; {\mathcal{O}}^{\times}) .\end{align*} There is another construction of the map \begin{align*} \left\{{L}\right\} &\to H^2(X; {\mathbb{Z}}) \\ L &\mapsto c_1(L) .\end{align*} Take a smooth section of $$L$$ and $$s\in H^0(X; L)$$ that intersects an $${\mathcal{O}}{\hbox{-}}$$section of $$L$$ transversely. Then \begin{align*} V(s) \coloneqq\left\{{ x\in X {~\mathrel{\Big|}~}s(x) = 0 }\right\} \end{align*} is a submanifold of real codimension 2 in $$X$$, and $$c_1(L) = [ V(s) ] \in H^2(X; {\mathbb{Z}})$$. \end{proof} \begin{theorem}[Splitting Principle for Complex Vector Bundles] \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item Suppose that $$\mathcal{E} = \bigoplus_{i=1}^r L_i$$ and let $$c(\mathcal{E}) \coloneqq\sum c_i(\mathcal{E}$$. Then \begin{align*} c(\mathcal{E}) = \prod_{i=1}^r \qty{ 1 + c_i (L_i) } .\end{align*} \item Given any vector bundle $$\mathcal{E} \to X$$, there exists some $$Y$$ and a map $$Y\to X$$ such that $$f^*: H^k(X; {\mathbb{Z}}) \hookrightarrow H^k(Y; {\mathbb{Z}})$$ is injective and $$f^* \mathcal{E} = \bigoplus_{i=1}^r L_i$$. \end{enumerate} \end{theorem} \begin{slogan} To verify any identities on characteristic classes, it suffices to prove them in the case where $$\mathcal{E}$$ splits into a direct sum of line bundles. \end{slogan} \begin{example}[?] \begin{align*} c( \mathcal{E} \oplus \mathcal{F}) = c( \mathcal{E} ) c( \mathcal{F} ) .\end{align*} To prove this, apply the splitting principle. Choose $$Y, Y'$$ splitting $$\mathcal{E}, \mathcal{E}'$$ respectively, this produces a space $$Z$$ and a map $$f:Z\to X$$ where both split. We can write \begin{align*} f^* \mathcal{E} &= \bigoplus L_i && c(f^* \mathcal{E} ) = \prod \qty{ 1 + c_1(L_i) } \\ f^* \mathcal{F} &= \bigoplus M_j && c(f^* \mathcal{E} ) = \prod \qty{ 1 + c_1(M_j) } .\end{align*} We thus have \begin{align*} c( f^* \mathcal{E} \oplus f^* \mathcal{F} ) &= \prod \qty{1 + c_1(L_i) } \qty{1 + c_1(M_j)} \\ &= c(f^* \mathcal{E} ) c(f^* \mathcal{F} ) ,\end{align*} and $$f^* (c( \mathcal{E} \oplus \mathcal{F} ) = f^* (c (\mathcal{E}) c( \mathcal{F}))$$. Since $$f^*$$ is injective, this yields the desired identity. \end{example} \begin{example}[?] We can compute $$c(\operatorname{Sym}^2 \mathcal{E})$$, and really any tensorial combination involving $$\mathcal{E}$$, and it will always yield some formula in the $$c_i( \mathcal{E} )$$. \end{example} \hypertarget{friday-february-12}{% \section{Friday, February 12}\label{friday-february-12}} \begin{remark} Last time: the splitting principle. Suppose we have $$\mathcal{E} = L_1 \oplus \cdots \oplus L_r$$ and let $$x_i \coloneqq c_i(L_i)$$. Then $$c_k(\mathcal{E})$$ is the degree $$2k$$ part of $$\prod_{i=1}^r (1 + x_i )$$ where each $$x_i$$ is in degree $$2$$. This is equal to $$e_k(x_1, \cdots, x_r)$$ where $$e_k$$ is the $$k$$th elementary symmetric polynomial. \end{remark} \begin{example}[?] For example, \begin{itemize} \item $$e_1 = x_1 + \cdots x_r$$. \item $$e_2 = x_1 x_2 + x_1 x_3 + \cdots = \sum_{i < j} x_i x_j$$ \item $$e_3 = \sum_{i 0}(X; {\mathcal{O}}) = 0$$, but $$H^0(X; {\mathcal{O}})$$ is the space of all holomorphic functions on $${\mathbb{C}}$$, making $$\dim_{\mathbb{C}}h^0(X; {\mathcal{O}})$$ infinite. \end{example} \begin{example}[?] Take $$X = {\mathbb{P}}^1$$ with $${\mathcal{O}}$$ as above, $$h^0({\mathbb{P}}^1; {\mathcal{O}}) = 1$$ since $${\mathbb{P}}^1$$ is compact and the maximum modulus principle applies, so the only global holomorphic functions are constant. We can write $${\mathbb{P}}^1 = {\mathbb{C}}_1 \cup{\mathbb{C}}_2$$ as a cover and $$h^i({\mathbb{C}}, {\mathcal{O}}) = 0$$, so this is an acyclic cover and we can use it to compute $$h^1({\mathbb{P}}^1; {\mathcal{O}})$$ using Čech cohomology. We have \begin{itemize} \item $$C^0({\mathbb{P}}^1; {\mathcal{O}}) = {\mathcal{O}}({\mathbb{C}}_1) \oplus {\mathcal{O}}({\mathbb{C}}_2)$$ \item $$C^1({\mathbb{P}}^1; {\mathcal{O}}) = {\mathcal{O}}({\mathbb{C}}_1 \cap{\mathbb{C}}_2) = {\mathcal{O}}({\mathbb{C}}^{\times})$$. \item The boundary map is given by \begin{align*} {\partial}_0: C^0 &\to C^1 \\ ( f(z), g(z) ) &\mapsto g(1/z) - f(z) \end{align*} and there are no triple intersections. \end{itemize} Is every holomorphic function on $${\mathbb{C}}^{\times}$$ of the form $$g(1/z) - f(z)$$ with $$f,g$$ holomorphic on $${\mathbb{C}}$$. The answer is yes, by Laurent expansion, and thus $$h^1 = 0$$. We can thus compute $$\chi({\mathbb{P}}^1; {\mathcal{O}}) = 1-0 = 1$$. \end{example} \hypertarget{monday-february-15}{% \section{Monday, February 15}\label{monday-february-15}} \begin{remark} Last time: we saw that $$\chi({\mathbb{P}}^1, {\mathcal{O}}) = 1$$, and we'd like to generalize to holomorphic line bundles on a Riemann surface. This will be the main ingredient for Riemann-Roch. \end{remark} \begin{theorem}[Euler characteristic and homological vanishing for holomorphic vector bundles] Let $$X \in {\mathsf{Mfd}}_{\mathbb{C}}$$ be compact and let $$\mathcal{F}$$ be a holomorphic vector bundle on $$X$$ \footnote{Or more generally a finitely-generated $${\mathcal{O}}{\hbox{-}}$$module, i.e.~a coherent sheaf.} Then $$\chi$$ is well-defined and \begin{align*} h^{> \dim_{\mathbb{C}}X}(X; \mathcal{F} ) = 0.\end{align*} \end{theorem} \begin{remark} The locally constant sheaf $$\underline{{\mathbb{C}}}$$ is not an $${\mathcal{O}}{\hbox{-}}$$module, i.e.~$$\underline{{\mathbb{C}}}(U) \not\in {\mathsf{{\mathcal{O}}(U)}{\hbox{-}}\mathsf{Mod}}$$. In fact, $$h^{2i}(X, \underline{{\mathbb{C}}}) = {\mathbb{C}}$$ for all $$i$$. \end{remark} \begin{proof}[of theorem] We'll can resolve $$\mathcal{F}$$ as a sheaf by first mapping to its smooth sections and continuing in the following way: \begin{align*} 0 \to \mathcal{F} \to C^{\infty } \mathcal{F} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} F \otimes A^{0, 1} \to \cdots ,\end{align*} where $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}f = \sum_i {\frac{\partial f}{\partial {\overline{{z}}}_i}\,} \, d{\overline{{z}}}_i$$. Suppose we have a holomorphic trivialization of $${ \left.{{\mathcal{F} }} \right|_{{U}} } \cong {\mathcal{O}}_{U}^{\oplus r}$$ and we have sections $$(s_1, \cdots, s_r) \in C^{\infty } \mathcal{F}(U)$$, which are smooth functions on $$U$$. In local coordinates we have \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s \coloneqq({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s_1, \cdots, { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s_r) ,\end{align*} but is this well-defined globally? Given a different trivialization over $$V \subseteq X$$, the $$s_i$$ are related by transition functions, so the new sections are $$t_{UV}(s_1, \cdots, s_r)$$ where $$t_{UV}: U \cap V \to \operatorname{GL}_r({\mathbb{C}})$$. Since $$t_{UV}$$ are holomorphic, we have \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}( t_{UV} (s_1, \cdots, s_r)) = t_{UV} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(s_1, \cdots, s_r) .\end{align*} This makes $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}: C^{\infty } \mathcal{F} \to F\otimes A^{0, 1}$$ a well-defined (but not $${\mathcal{O}}{\hbox{-}}$$linear) map. We can thus continue this resolution using the Leibniz rule: \begin{align*} 0 \to \mathcal{F} \to C^{\infty } \mathcal{F} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} F \otimes A^{0, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \cdots F \otimes A^{0, 2} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \cdots ,\end{align*} which is an exact sequence of sheaves since $$(A^{0, {-}}, { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$ is exact. \todo[inline]{Why? Split into line bundles?} We can identify $$C^{\infty }\operatorname{\mathcal{F}} = \operatorname{\mathcal{F}} \otimes A^{0, 0}$$, and $$\operatorname{\mathcal{F}} \otimes A^{0, q}$$ is a smooth vector bundle on $$X$$. Using partitions of unity, we have that $$\operatorname{\mathcal{F}} \otimes A^{0, q}$$ is acyclic, so its higher cohomology vanishes, and \begin{align*} H^i(X; \operatorname{\mathcal{F}} ) \cong \frac { \ker ( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}: \operatorname{\mathcal{F}}\otimes A^{0, i} \to \operatorname{\mathcal{F}} \otimes A^{0, i+1} } { \operatorname{im}( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}: \operatorname{\mathcal{F}}\otimes A^{0, i-1} \to \operatorname{\mathcal{F}} \otimes A^{0, i} } .\end{align*} However, we know that $$A^{0, p} = 0$$ for all $$p> n \coloneqq\dim_{\mathbb{C}}X$$, since any wedge of $$p>n$$ forms necessarily vanishes since there are only $$n$$ complex coordinates. \end{proof} \begin{warnings} This only applies to holomorphic vector bundles or $${\mathcal{O}}{\hbox{-}}$$modules! \end{warnings} \hypertarget{riemann-roch}{% \subsection{Riemann-Roch}\label{riemann-roch}} \begin{theorem}[Riemann-Roch] Let $$C$$ be a compact connected Riemann surface, i.e.~$$C\in {\mathsf{Mfd}}_{\mathbb{C}}$$ with $$\dim_{\mathbb{C}}(C) = 1$$, and let $$\mathcal{L}\to C$$ be a holomorphic line bundle. Then \begin{align*} \chi(C, \mathcal{L}) = \deg(L) + (1-g) && \text{where } \deg(L) \coloneqq\int_C c_1(\mathcal{L}) \end{align*} and $$g$$ is the genus of $$C$$. \end{theorem} \begin{proof}[of Riemann-Roch] We'll introduce the notion of a point bundle'', which are particularly nice line bundles, denoted $${\mathcal{O}}(p)$$ for $$p\in {\mathbb{C}}$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{34pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-15_14-16.pdf_tex} }; \end{tikzpicture} } \end{figure} Taking $${\mathbb{D}}$$ to be a disc of radius $$1/2$$ and $$V$$ to be its complement, we have $$t_{uv}(z) = z^{-1}\in {\mathcal{O}}^*(U \cap V)$$. We can take a holomorphic section $$s_p \in H^0( C, {\mathcal{O}}(p) )$$, where $${ \left.{{s_p}} \right|_{{U}} } = z$$ and $${ \left.{{s_p}} \right|_{{V}} } = 1$$. Then $$t_{uv}( { \left.{{s_p}} \right|_{{U}} } ) = { \left.{{s_p}} \right|_{{V}} }$$ on the overlaps. We have a function which precisely vanishes to first order at $$p$$. Recall that $$c_1( {\mathcal{O}}(p) )$$ is represented by $$[ V(s) ] = [p]$$, and moreover $$\int_C c_1 ( {\mathcal{O}}(p) ) = 1$$. We now want to generalize this to a \textbf{divisor}: a formal $${\mathbb{Z}}{\hbox{-}}$$linear combination of points. \begin{example}[?] Take $$p, q,r\in C$$, then a divisor can be defined as something like $$D \coloneqq 2[p] - [q] + 3[r]$$. \end{example} Define $${\mathcal{O}}(D) \coloneqq\bigotimes_{i} {\mathcal{O}}(p_i)^{\otimes n_i}$$ for any $$D = \sum n_i [p_i]$$. Here tensoring by negatives means taking duals, i.e.~$${\mathcal{O}}( -[p] ) \coloneqq{\mathcal{O}}^{\otimes_{-1}} \coloneqq{\mathcal{O}}(p) {}^{ \vee }$$, the line bundle with inverted transition functions. $${\mathcal{O}}(D)$$ has a meromorphic section given by \begin{align*} s_D \coloneqq\prod s_{p_i}^{n_i} \in \operatorname{Mero}(C, {\mathcal{O}}(D) ) \end{align*} where we take the sections coming from point bundles. We can compute \begin{align*} \int_C c_1 ( {\mathcal{O}}(D) ) = \sum n_i \coloneqq\deg(D) .\end{align*} . \begin{example}[?] \begin{align*} \deg( 2[p] -[q] + 3[r]) = 4 .\end{align*} \end{example} \begin{remark} Assume our line bundle $$L$$ is $${\mathcal{O}}(D)$$, we'll prove Riemann-Roch in this case by induction on $$\sum {\left\lvert {n_i} \right\rvert}$$. The base case is $${\mathcal{O}}$$, which corresponds to taking an empty divisor. Then either \begin{itemize} \tightlist \item Take $$D = D_0 + [p]$$ with $$\deg(D_0) < \sum {\left\lvert {n_i} \right\rvert}$$ (for which we need some positive coefficient), or \item Take $$D_0 = D + [p]$$. \end{itemize} \end{remark} \begin{claim} There is an exact sequence \begin{align*} 0 \to {\mathcal{O}}(D_0) &\to {\mathcal{O}}(D) \to {\mathbb{C}}_p \to 0\\ s\in {\mathcal{O}}(D_0)(U) &\mapsto s \cdot s_p \in {\mathcal{O}}( D_0 + [p] ) (U) ,\end{align*} where the last term is the skyscraper sheaf at $$p$$. \end{claim} \begin{proof}[of claim] The given map is $${\mathcal{O}}{\hbox{-}}$$linear and injective, since $$s_p\neq 0$$ and $$s s_p=0$$ forces $$s=0$$. Recall that we looked at $${\mathcal{O}}\xrightarrow{\cdot z} {\mathcal{O}}$$ on $${\mathbb{C}}$$, and this section only vanishes at $$p$$ (and to first order). The same situation is happening here. \end{proof} Thus there is a LES \begin{center} \begin{tikzcd} &&&& 0 \\ \\ {H^0( {\mathcal{O}}(D_0) )} && {H^0( {\mathcal{O}}(D) )} && {H^0( {\mathcal{O}}({\mathbb{C}}_p) )} \\ \\ {H^1( {\mathcal{O}}(D_0) )} && {H^1( {\mathcal{O}}(D) )} && {H^1( {\mathcal{O}}({\mathbb{C}}_p) ) = 0} \\ \\ 0 \arrow[from=3-5, to=5-1, out=0, in=180] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=5-1, to=5-3] \arrow[from=5-3, to=5-5] \arrow[from=1-5, to=3-1, out=0, in=180] \arrow[from=5-5, to=7-1, out=0, in=180] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOCxbMCwyLCJIXjAoIFxcT08oRF8wKSApIl0sWzIsMiwiSF4wKCBcXE9PKEQpICkiXSxbNCwyLCJIXjAoIFxcT08oXFxDQ19wKSApIl0sWzAsNCwiSF4xKCBcXE9PKERfMCkgKSJdLFsyLDQsIkheMSggXFxPTyhEKSApIl0sWzQsNCwiSF4xKCBcXE9PKFxcQ0NfcCkgKSA9IDAiXSxbNCwwLCIwIl0sWzAsNiwiMCJdLFsyLDNdLFswLDFdLFsxLDJdLFszLDRdLFs0LDVdLFs2LDBdLFs1LDddXQ==}{Link to Diagram} \end{quote} We also have $$h^1({\mathbb{C}}_p) = 0$$ by taking a sufficiently fine open cover where $$p$$ is only in one open set. So just checking Čech cocycles yields $$C_U^1(C, {\mathbb{C}}_p) \coloneqq\prod_{i1} = 0$$. We also have the sheaves $$A^{1, 0}, A^{0, 1}, A^{1, 1},$$ the sheaves of smooth $$(p, q){\hbox{-}}$$forms. Here the only nonzero combinations are $$(0, 0), (0, 1), (1, 0), (1, 1)$$ by dimensional considerations. Let $$L$$ be a holomorphic line bundle on $$C$$, then \begin{align*} \chi(C, L) = h^0(L) - h^1(L) = \deg(L) + 1 - g .\end{align*} \end{example} \begin{remark} In general it can be hard to compute $$h^1(L)$$, since this is sheaf cohomology (sections over double overlaps, cocycle conditions, etc). On the other hand, $$h^0$$ is easy to understand, since $$h^0( \Omega^1_C)$$ is the dimension of the global holomorphic sections $$H^0(C, L) = L(C)$$. A key tool here is the following: \end{remark} \hypertarget{serre-duality}{% \subsubsection{Serre Duality}\label{serre-duality}} \begin{proposition}[Serre Duality] \begin{align*} H^1(C, L) \cong H^0(C, L ^{-1} \otimes\Omega_C^1) {}^{ \vee } ,\end{align*} noting that these are both global sections of a line bundle. \end{proposition} \begin{proof}[of Serre Duality] Recall that we had a resolution of the sheaf $$L$$ given by by smooth vector bundles: \begin{align*} 0 \to L \hookrightarrow L\otimes A^{0, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} L \otimes A^{0, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} 0 .\end{align*} So we know that \begin{align*} H^1(C, L) = H^0(L\otimes A^{0, 1}) / { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}H^0(L\otimes A^{0, 0}) .\end{align*} Choose a Hermitian metric $$h$$ on $$L$$, i.e.~a map $$h: L\otimes{\overline{{L}}} \to {\mathcal{O}}$$. On fibers, we have $$h_p: L_p \otimes\mkern 1.5mu\overline{\mkern-1.5mu L_p \mkern-1.5mu}\mkern 1.5mu \to {\mathbb{C}}$$. We'll also choose a metric on $$C$$, say $$g$$. Since $$C$$ is a Riemann surface, we have an associated volume form $$\nu$$ on $$C$$ (essentially the determinant), so we can define a pairing between sections of $$L\otimes A^{0, 0}$$: \begin{align*} {\left\langle {s},~{t} \right\rangle} \coloneqq\int_C h(s, {\overline{{t}}} ) \,d\nu .\end{align*} Note that \begin{align*} {\left\langle {s},~{s} \right\rangle} = \int_C h(s, {\overline{{s}}}) \,d\nu \geq 0 && \text{since } h(s, {\overline{{s}}})(p) = 0 \iff s_p = 0 ,\end{align*} and moreover this integral is zero if and only if $$s=0$$. So we have an inner product on $$H^0(L\otimes A^{0, 0})$$. We can also define a pairing on sections of $$L\otimes A^{0, 1}$$, say \begin{align*} {\left\langle { s \otimes \alpha},~{ t \otimes \beta} \right\rangle} = \int_C h(s, {\overline{{t}}}) \alpha\wedge {\overline{{\beta}}} .\end{align*} Note that $$h$$ is a smooth function and $$\alpha\wedge {\overline{{\beta}}}$$ is a $$(1, 1){\hbox{-}}$$form. Moreover, this is positive and nondegenerate. We want to understand the cokernel of the linear map \begin{align*} H^0(L \otimes A^{0, 0}) \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} H^0( L \otimes A^{0, 1}) .\end{align*} To compute $$\operatorname{coker}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$, we can look at the kernel of the adjoint, and it suffices to find the orthogonal complement of $$\operatorname{im}( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$, i.e.~ \begin{align*} \operatorname{coker}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) = \left\{{ t\in H^0(L\otimes A^{0, 1}) {~\mathrel{\Big|}~}{\left\langle {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s},~{t} \right\rangle} = 0 \, \forall s}\right\} .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-19_14-18.pdf_tex} }; \end{tikzpicture} } \end{figure} So we want to understand sections $$t\in H^0(L\otimes A^{0, 1})$$ such that \begin{align*} \int_C ({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s){\overline{{t}}} = 0 && \forall s\in H^0(L\otimes A^{0, 0}) ,\end{align*} where $${{\partial}}C = \emptyset$$. We'll basically want to do integration by parts on this. Note that $$h(s, t) = hst$$ here where we view $$h$$ as a certain section. Note that $${\overline{{t}}} \in H^0({\overline{{L}}} \otimes A^{1, 0})$$, so we can replace $${\partial}$$ with $$d = { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}+ {\partial}$$ and apply Stokes' theorem: \begin{align*} \int_C s d(h {\overline{{t}}}) &= 0 && \forall s\in H^0(L\otimes A^{0, 0}) \\ 0 &= \int_C s{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(h {\overline{{t}}}) \\ &= \int_C s {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(h {\overline{{t}}}) \over d\nu}d\nu\\ &= {\left\langle {s},~{{\overline{{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(h {\overline{{t}}}) \over d\nu }}}} \right\rangle} \end{align*} where $$h \in C^{\infty }(L ^{-1} \otimes{\overline{{L}}}^{-1})$$ and $$h{\overline{{t}}} \in C^{\infty }(L^{-1}\otimes A^{1, 0})$$. But the right-hand side is in $$H^0(L \otimes A^{0, 0} )$$ and by nondegeneracy we can conclude \begin{align*} {\overline{{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(h {\overline{{t}}}) \over d\nu }}} = 0 \iff { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(h{\overline{{t}}}) = 0 .\end{align*} We thus have $$h{\overline{{t}}} \in H^0( L ^{-1}\otimes A^{1, 0}$$ which is a holomorphic line bundle tensored with $$A^{0, 0}$$. Thus $$\operatorname{coker}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \cong_h H^0( L ^{-1} \otimes\Omega^1)$$. \end{proof} \begin{remark} We showed $${\left\langle {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s},~{t} \right\rangle} = {\left\langle {s},~{Y (t)} \right\rangle}$$ where $$Y$$ is the adjoint given above. Then the kernel of $$Y$$ wound up being where $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$ vanishes, i.e.~holomorphic sections of a separate bundle. Here we had \begin{itemize} \tightlist \item $$t \in H^0(L\otimes A^{0, 1})$$ \item $${\overline{{t}}} \in H^0({\overline{{L}}}\otimes A^{1,0})$$ \item $$h\in H^0( L ^{-1} \otimes{\overline{{ L ^{-1} }}})$$ \end{itemize} \end{remark} \hypertarget{monday-february-22}{% \section{Monday, February 22}\label{monday-february-22}} \begin{remark} Last time: Serre duality, and we'll review Riemann-Roch. Recall that this depended on the statement that every holomorphic line bundle $$L\to C$$ for $$C$$ a complex curve is of the form $$L = {\mathcal{O}}(D)$$ for some divisor $$D$$. Then \begin{align*} \chi(C, L) = h^0(L) - h^1(L) = \deg L + 1 - g, && \deg L = \int_C c_1(L) ,\end{align*} Serre duality said that the space of sections $$H^1(C; L)$$ is naturally isomorphic to $$H^0(C, L ^{-1} \otimes\Omega_C^1) {}^{ \vee }$$. Notation: given $$X \in {\mathsf{Mfd}}_{\mathbb{C}}^n$$ of complex, dimension $$n$$, the \textbf{canonical bundle} is written $$K_X \coloneqq\Omega_X^n$$ and is the sheaf of holomorphic $$n{\hbox{-}}$$forms. Serre duality will generalize: if $$\mathcal{E}\to X$$ is a holomorphic vector bundle, then $$H^i(X; \mathcal{E}) \cong H^{n-i}(X; \mathcal{E} {}^{ \vee }\otimes K_X) {}^{ \vee }$$. Note that only $$H^0, H^1$$ are the only nontrivial degrees for a curve. For 4-manifolds, we'll have an $$H^2$$ as well. \end{remark} \hypertarget{applications-of-riemann-roch-1}{% \subsection{Applications of Riemann-Roch}\label{applications-of-riemann-roch-1}} \begin{proposition}[The 2-sphere has a unique complex structure] There is a unique complex $$X\in {\mathsf{Mfd}}_{\mathbb{C}}$$ diffeomorphic to $$S^2$$. \end{proposition} \begin{proof}[of proposition] Note existence is clear, since we can take $${\mathbb{CP}}^1 \coloneqq({\mathbb{C}}^2 \setminus\left\{{0}\right\}) / \mathbf{x} \sim \lambda\mathbf{x}$$ for $$\lambda\in {\mathbb{C}}^{\times}$$, which is identified as the set of complex lines through $$0$$ in $${\mathbb{C}}^2$$. This decomposes as $${\mathbb{C}}\cup{\mathbb{C}}= \left\{{ [1, *] }\right\} \cup\left\{{ [*, 1] }\right\}$$. We now want to show that any two such complex manifolds are biholomorphic. Let $$X \in {\mathsf{Mfd}}_{\mathbb{C}}^1$$ with $$X\cong_{C^{\infty }} S^2$$, and consider for $$p\in X$$ the point bundle $${\mathcal{O}}(p) \to X$$. The defining property was that there exists a section $$s_p \in H^0(X; {\mathcal{O}}(p) )$$ which vanishes at first order at $$p$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-22_14-03.pdf_tex} }; \end{tikzpicture} } \end{figure} We have \begin{align*} \chi(X; {\mathcal{O}}(p)) = \deg {\mathcal{O}}(p) + 1 - g(x) = 1 + 1 - 0 = 2 .\end{align*} \begin{exercise}[?] Check that $$\deg {\mathcal{O}}(p) = 1$$. \end{exercise} On the other hand we have \begin{align*} \chi(X; {\mathcal{O}}(p)) = h^0({\mathcal{O}}(p)) - h^1( {\mathcal{O}}(p) ) .\end{align*} We have $$h^1( {\mathcal{O}}(p) ) = H60( K \otimes{\mathcal{O}}(-p)$$, and $$K_X = \Omega_X^1 = T {}^{ \vee }X$$, so the question is: what is the degree of $$TX$$ for $$X\cong S^2$$? We need to compute $$\int_X c_1(TX)$$. How many zeros does a vector field on the sphere have? You can take the gradient vector field for a height function to get $$2$$, noting that the two zeros come in with a positive orientation \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-22_14-10.pdf_tex} }; \end{tikzpicture} } \end{figure} In coordinates on $${\mathbb{CP}}^1$$, the coordinate is given by $$z$$ and $$z {\frac{\partial }{\partial z}\,} \mapsto -2 {\frac{\partial }{\partial w}\,}$$ for the coordinate $$w = 1/z$$. We get $$\int_X c_1(TX) = 2$$ and thus $$\deg K_X = -2$$ by dualizing. \begin{fact} $$\deg K_X = 2g-2$$. Use the existence of a smooth vector field on $$X$$. \end{fact} \begin{lemma}[When h0 of a line bundle on a curve vanishes] If $$\deg L < 0$$ on $$C$$, thne $$h^0(C, L) = 0$$. \end{lemma} \begin{proof}[of lemma] If $$s\in H^0(C, L)$$ is nonzero, then since $$s$$ is a holomorphic section, \begin{align*} 0 \leq \sum_{p\in C} {\operatorname{Ord}}_P (s) = \deg L .\end{align*} \end{proof} By this lemma, $$h^1({\mathcal{O}}(p)) = 0$$. We have $$H^0(X; {\mathcal{O}}(p)) = {\mathbb{C}}s_p \oplus {\mathbb{C}}s$$ for our specific section $$s_p$$ and some other section $$s \neq \lambda s_p$$. Note that $$s/s_p$$ is a meromorphic section of $${\mathcal{O}}(p) \times{\mathcal{O}}(-p) = {\mathcal{O}}$$, so we have a map \begin{align*} \varphi: {s \over s_p} : X\to {\mathbb{P}}^1 .\end{align*} Note that $$P\mapsto \infty \in {\mathbb{P}}^1$$ under this $$\varphi$$, and it's only the ratio that is well-defined. We have $$\varphi ^{-1} (u) = \left\{{ s/s_p = u }\right\} = \left\{{ s - us_p =0 }\right\}$$ which is a single point. So $$\varphi$$ is a degree 1 map, and $$X$$ is biholomorphic to $${\mathbb{P}}^1$$ via $$\varphi$$. \end{proof} \begin{remark} So there is only one genus 0 Riemann surface. What about genus 1? \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-22_14-23.pdf_tex} }; \end{tikzpicture} } \end{figure} By Riemann-Roch we know \begin{align*} \chi(C; {\mathcal{O}}) = \deg {\mathcal{O}}+ l - 1 = 0 = h^0({\mathcal{O}}) - h^1({\mathcal{O}}) .\end{align*} We know $$h^0({\mathcal{O}}) = 1$$ by the maximum modulus principle and $$h^1(C; {\mathcal{O}}) = 1$$. By Serre duality, $$h^0(C, K) = 1$$, and since $$\deg K = 2g-2 = 0$$. So let $$s\in H^0(C, K)$$ by a nonzero section, which we know exists. We then get $${\operatorname{Ord}}_p s = 0$$ for all $$p$$, so $$s$$ vanishes nowhere. But then we get an isomorphism of sheaves, since $$s$$ everywhere nonvanishing implies trivial cokernel: \begin{align*} {\mathcal{O}}\xrightarrow{\cdot s} K .\end{align*} So $$K_C = {\mathcal{O}}_C$$ if $$g(C) = 1$$, and such a Riemann surface is an \textbf{elliptic curve}. \end{remark} \begin{example}[?] Let $$C \coloneqq{\mathbb{C}}/ \Lambda$$ for $$\Lambda$$ some lattice. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{34pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-22_14-28.pdf_tex} }; \end{tikzpicture} } \end{figure} All transition functions are of the form $$z \mapsto z + \lambda$$ for some $$\lambda\in \Lambda$$. What is a nonvanishing section of $$K_C$$, i.e.~a holomorphic one form $$\omega \coloneqq f(z) dz$$ on $${\mathbb{C}}$$ that descends to $${\mathbb{C}}/\Lambda$$. We would need $$f(z)dz = f(z + \lambda)d(z+ \lambda)$$ for all $$\lambda$$. Something like $$f=1$$ works, so $$\omega= dz$$ descends. In fact, $$f$$ must be constant, since $$H^0( {\mathbb{C}}/ \Lambda, {\mathcal{O}}) = {\mathbb{C}}\,dz$$ by the maximum modulus principle. Now let $$p, q\in C$$ and apply Riemann-Roch to the line bundle $${\mathcal{O}}(p+q)$$ yields \begin{align*} \chi( {\mathcal{O}}(p+q) ) &= h^0( {\mathcal{O}}(p+q) )- h^1( {\mathcal{O}}(-p-q) ) \\ &= h^0( {\mathcal{O}}(p+q) )-0\\ &= \deg {\mathcal{O}}(p+q)+1-1 \\ &=2 .\end{align*} Thus there is a section $$s_{p+q} \in H^0( {\mathcal{O}}(p+q)) \ni s$$ that vanishes at $$p+q$$, and similarly a map \begin{align*} {s\over s_{p+q}}: C \xrightarrow{\varphi} {\mathbb{P}}^1 .\end{align*} We can check $$\varphi ^{-1} ( \infty ) = p+q$$ and $$\deg \varphi = 2$$. Thus genus 1 surfaces have a generically 2-to-1 map to $${\mathbb{P}}^1$$. \begin{figure} \centering \includegraphics{figures/image_2021-02-25-20-41-53.png} \caption{image\_2021-02-25-20-41-53} \end{figure} Note that homothetic lattices define an isomorphism between the elliptic curves, and lattices mod homothety are in correspondence of elliptic curves. By acting $$\operatorname{PGL}_2(C) \curvearrowright{\mathbb{P}}^1$$ since $$\operatorname{GL}_2$$ acts on lines since scaling an element fixes a line. This is dimension 3. So elliptic curves are also in correspondence with $$\left\{{ 4 \text{ points on } {\mathbb{P}}^1}\right\} / \operatorname{PGL}_2({\mathbb{C}})$$ since this is now dimension 1. Note that by applying homothety, the two basis vectors for a lattice can be rescaled so one is length 1 and the other is a complex number $$\tau$$, and we can identify this space with $${\mathbb{H}}/ {\operatorname{SL}}_2({\mathbb{Z}})$$. \end{example} \begin{exercise}[?] Show that any $$g(C) = 2$$ curve has a degree 2 map to $${\mathbb{P}}^1$$. \end{exercise} \begin{remark} Similarly $$g(C) = 3$$ are usually a curve of degree $$4$$ in $${\mathbb{CP}}^2$$. Severi proof in the 50s: false! issues with building moduli space for $$g\geq 23$$. Need to use orbifold structure to take into account automorphisms. \end{remark} \hypertarget{wednesday-february-24}{% \section{Wednesday, February 24}\label{wednesday-february-24}} Last time: \begin{align*} \chi(C, L) &= h^0(C, L) - h^1(C, L) \\ &= h^0(C, L) - h^0(C, L ^{-1} \otimes K_C) \\ &= \deg L + 1 -g ,\end{align*} which is determined by purely topological information. We can generalize this to arbitrary ranks of the bundle and arbitrary dimensions of manifold: \begin{theorem}[Hirzebruch-Riemann-Roch (HRR) Formula] Let $$X$$ be a compact complex manifold and let $$\mathcal{E} \to X$$ be a holomorphic vector bundle. Then \begin{align*} \chi( \mathcal{E} ) = \int_C \operatorname{ch}( \mathcal{E} ) \mathrm{td}(X) .\end{align*} The constituents here: \begin{itemize} \item The \textbf{Chern character}, summed over $$R$$ the \emph{Chern roots}, which is in mixed cohomological degree. \begin{align*} \operatorname{ch}( \mathcal{E} ) \coloneqq\sum_{x_i \in R} e^{x_i} = \operatorname{ch}_0( \mathcal{E} ) + \operatorname{ch}_1( \mathcal{E} ) + \cdots + \operatorname{ch}_i( \mathcal{E} ) \in H^{2i}(X; {\mathbb{Q}}) .\end{align*} \item The \textbf{Todd class}, defined as \begin{align*} \mathrm{td}( F) \coloneqq\prod_{x_i \in R} {x_i \over 1 - e^{-x_i} } \end{align*} where $$\mathrm{td}(X) \coloneqq\mathrm{td}(TX)$$ is viewed as a complex vector bundle, which is again in mixed cohomological degree. \end{itemize} \end{theorem} \begin{remark} Note that integrating over cohomology classes in mixed degree is just equal to the integral over the top degree terms. Applying this to $$X = C$$ a curve and $$\mathcal{E} \coloneqq{\mathcal{O}}$$, we obtain \begin{align*} \chi(C, {\mathcal{O}}) = \int_C \operatorname{ch}( {\mathcal{O}}) \mathrm{td}(C) .\end{align*} We have \begin{itemize} \item $$\operatorname{ch}({\mathcal{O}}) = e^{c_1({\mathcal{O}})} = e^0 = 1$$ \item $$\mathrm{td}(C) \coloneqq\mathrm{td}(TC) = c_1(TC) / (1- e^{ - c_1(TC) } )$$, whose Taylor coefficients are the Bernoulli numbers. We can expand $$x/(1 -e^{-x}) = 1 + (x/2) + (x^2/12) - x^4(720) + \cdots$$, and since terms above degree 2 vanish, we have \begin{align*} \cdots &= \int_C 1 + \qty{ 1 + {c_1(TC) \over 2} } \\ &= \int_C \qty{c_1(TC) \over 2 }\\ &= {1\over 2} \chi_{\mathsf{Top}}(C) && \text{Chern-Gauss-Bonnet} \\ &= {2-2g \over 2} \\ &= 1-g .\end{align*} \end{itemize} We thus obtain \begin{align*} \chi(C, L) &= \int_C \operatorname{ch}(L) \mathrm{td}(C) \\ &= \int_C (1 + c_1(L) ) \qty {1 + {c_1(L) \over 2} }\\ &= \int_C c_1(L) + {c_1(TC) \over 2} \\ &= \deg L + 1-g .\end{align*} \end{remark} \begin{remark} Note that this is a better definition of genus than the previous one, which was just the correction term in Riemann-Roch. Here we can define it as $$g \coloneqq h^1/2$$. \end{remark} \begin{exercise}[?] Try to state and prove a Riemann-Roch formula for vector bundles on curves. \end{exercise} \begin{proposition}[Formula for Euler characteristic of a line bundle on a complex surface] Let $$S$$ be a compact complex surface, i.e.~$$S\in {\mathsf{Mfd}}_{\mathbb{C}}^2$$. An example might be $$C\times D$$ for $$C,D$$ two complex curves, or $${\mathbb{CP}}^2$$. Let $$L\to S$$ be a holomorphic vector bundle. Then \begin{align*} \chi(L) = \chi({\mathcal{O}}_S) + {1\over 2} \qty{ L^2 - L \cdot K} .\end{align*} Note that $$L^2 \coloneqq\int_S c_1(L) c_1(L)$$ is just shorthand for taking the intersection of $$L$$ with itself. Recall that $$K \coloneqq\Omega_S^2$$ is the space of holomorphic top forms. \end{proposition} \begin{proof}[?] Let $$x_1, x_2$$ be the Chern roots of $$TS$$. By HRR, we have \begin{align*} \chi(L) &= \int_S \operatorname{ch}(L) \mathrm{td}(S) \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{ {x_1 \over 1 - e^{-x_1} } {x_2 \over 1-e^{-x_2}} }\\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {x_1 \over 2} + {x_1^2 \over 12} }\qty{ 1 + {x_2 \over 2} + {x_2^2\over 12}} \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {x_1 + x_2 \over 2} + {x_1^2 + x_2^2 + 3x_1 x_2 \over 12} } \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + {c_1(x_1, x_2) \over 2} + {c_1(x_1, x_2)^2 + c_2(x_1, x_2) \over 12 } } \\ &= \int_S \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2!} } \qty{1 + { c_1(T) \over 2} + {c_1(T)^2 + c_2(T) \over 2 } } \\ &= \int_S {c_1(L)^2 \over 2} + {c_1(L) c_1(T) \over 2} + {c_1(T)^2 \over 2} + {c_2(T) \over 12} \quad \text{Take deg 4} \\ &= \int_S \qty{ c_1(L)^2 + c_1(L) c_1(T) \over 2} + \chi({\mathcal{O}}_S) \quad \text{HRR on last two terms} .\end{align*} where we've applied HRR to $${\mathcal{O}}_S$$. It remains to show that $$c_1(T) = -c_1(K)$$. We have \begin{align*} K = \Omega_S^2 = \bigwedge\nolimits^2 T {}^{ \vee } .\end{align*} Note that $$\bigwedge\nolimits^{\text{top}} \mathcal{E} \coloneqq\operatorname{det}( \mathcal{E} )$$ for any bundle $$\mathcal{E}$$ since this is a 1-dimensional bundle. We have $$c_1(T) = -c_1(T {}^{ \vee })$$ since the Chern roots of $$T {}^{ \vee }$$ are $$-x_1, -x_2$$. So it suffices to show $$c_1(T {}^{ \vee }) = c_1(K)$$, but there is a general result that $$c_1(\mathcal{E}) = c_1( \operatorname{det}\mathcal{E} )$$. This uses the splitting principle $$\mathcal{E} = \bigoplus_{i=1}^r L_i$$ with $$x_i = c_1(L_i)$$. We have $$c_1(\mathcal{E}) = \sum x_i$$ and $$\operatorname{det}\mathcal{E} = \bigotimes_{i=1}^r L_i$$, so $$\sum x_i = c_1(L_1\otimes\cdots \otimes L_r)$$. \end{proof} \begin{remark} We want to use the following formula: \begin{align*} \chi(S, L) = \chi({\mathcal{O}}_S) = {1\over 2}(L^2 - L\cdot K) .\end{align*} This requires knowing $$\chi({\mathcal{O}}_S)$$. Applying HRR yields \begin{align*} \chi({\mathcal{O}}_S) &= \int_S {c_1(T)^2 + c_2(T) \over 12}\\ &= \int_S { (-c_1(K))^2 + c_2(T) \over 12}\\ &= {K^2 + \displaystyle\int_S c_2(T) \over 12} ,\end{align*} so we just need to understand $$\int_S c_2(T)$$. But for $$n=\operatorname{rank}\mathcal{E}$$, $$c_n( \mathcal{E} )$$ (the top Chern class) is the fundamental class of a zero locus of a section of $$\mathcal{E}$$. Note that $$S \in {\mathsf{Mfd}}_{\mathbb{R}}^4$$ is oriented, so $$\int_S c_2(T)$$ is the signed number of zeros of a smooth vector field. \begin{figure} \centering \includegraphics{figures/image_2021-02-25-20-42-49.png} \caption{image\_2021-02-25-20-42-49} \end{figure} \todo[inline]{Check.} Looking at the tangent bundle of the surface, the local sign of an intersection will be the number of incoming directions $$\pmod 2$$, i.e.~the index of the critical point. Then the signed number of zeros here yields $$1-6+1 = -4 = \chi_{{\mathsf{Top}}}(C)$$. More generally, we have \begin{align*} \chi_{{\mathsf{Top}}}(M^n) = \int_C c_{n}(TM) ,\end{align*} the \textbf{Chern-Gauss-Bonnet} formula. We can thus write \begin{align*} \chi({\mathcal{O}}_S) = {K^2 + \chi_{\mathsf{Top}}(S) \over 12 } .\end{align*} \end{remark} \hypertarget{friday-february-26}{% \section{Friday, February 26}\label{friday-february-26}} \begin{remark} Last time: Riemann-Roch for surfaces, today we'll discuss some examples. Recall that if $$S \in {\mathsf{Mfd}}_{\mathbb{C}}^2$$ is closed and compact (noting that $$S\in {\mathsf{Mfd}}_{\mathbb{R}}^4$$) and $$L\to S$$ is a holomorphic line bundle then \begin{align*} \chi(S, L) = \chi({\mathcal{O}}_S) + {1\over 2}(L^2 - L \cdot K) \end{align*} where $$K = c_1(K_S)$$ for $$K_S \coloneqq\Omega_S^2$$ the canonical bundle and $$L = c_1(L)$$. We also saw \begin{align*} \chi({\mathcal{O}}_S) = {1\over 12}(K^2 + \chi_{{\mathsf{Top}}}(S)) ,\end{align*} where $$\chi_{\mathsf{Top}}$$ is the Euler characteristic and is given by \begin{align*} \chi_{\mathsf{Top}}(S) = 2 h^0(S; {\mathbb{C}}) - 2 h^1(S, {\mathbb{C}}) + h^2(S; {\mathbb{C}}) .\end{align*} \end{remark} \begin{example}[?] Let $$S = {\mathbb{CP}}^2$$, which can be given in local coordinates by \begin{align*} \left\{{ [x_0: x_1: x_2 ] {~\mathrel{\Big|}~}(x_0, x_1, x_2) \in {\mathbb{C}}^3\setminus\left\{{0}\right\}}\right\} \end{align*} where we only take equivalence classes of ratios $$[x,y,z] = [\lambda x, \lambda y, \lambda z]$$ for any $$\lambda\in {\mathbb{C}}^{\times}$$. This decomposes as \begin{align*} {\mathbb{CP}}^2 \cup{\mathbb{C}}\cup\left\{{ {\operatorname{pt}}}\right\} = \left\{{ [1: x_1: x_2] }\right\} \cup\left\{{ [0 : x_1: x_2] }\right\} \cup\left\{{ [0:0:1] }\right\} ,\end{align*} i.e.~we take $$x_0 \neq 0$$, then $$x_0 = 0, x_1\neq 0$$, then $$x_0 = x_1 = 0$$. Note that \begin{align*} h^i({\mathbb{CP}}^n; {\mathbb{Z}}) = \begin{cases} {\mathbb{Z}}& 0 \leq i \leq 2n \text{ even} \\ 0 & \text{else}. \end{cases} \end{align*} We can use this to conclude that $$\chi_{\mathsf{Top}}({\mathbb{CP}}^n) = n+1$$ and $$\chi_{\mathsf{Top}}({\mathbb{CP}}^2) = 3$$. Over $${\mathbb{CP}}^n$$ we have a \textbf{tautological line bundle} $${\mathcal{O}}(-1)$$ given by sending each point to the corresponding line in $${\mathbb{C}}^{n+1}$$, i.e.~$${\mathcal{O}}(-1) \to {\mathbb{CP}}^n$$ given by \begin{align*} \lambda (x_0, \cdots, x_n) \mapsto [x_0: \cdots: x_n] .\end{align*} Note that the total space is $$\mathop{\mathrm{Bl}}_0({\mathbb{C}}^{n+1})$$ is the \textbf{blowup} at zero, which separates the tangents at 0. \end{example} \begin{remark} Let $$X$$ be an algebraic variety, i.e.~spaces cut out by polynomial equations, for example $$\left\{{ xy = 0 }\right\} \subseteq {\mathbb{C}}^2$$ which has a singularity at the origin. A \textbf{divisor} is a $${\mathbb{Z}}{\hbox{-}}$$linear combination of subvarieties of codimension 1. Note that for a curve $$X$$, this recovers the definition involving points. For $$D$$ a divisor on $$X$$, we associated a bundle $${\mathcal{O}}_X(D)$$ which had a meromorphic section with a zero/pole locus whose divisor was precisely $$D$$. Recall the construction: we chose a point, then a trivializing neighborhood where the transition functions where $$V$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{41pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-26_14-12.pdf_tex} }; \end{tikzpicture} } \end{figure} For a higher dimensional algebraic variety or complex manifold, for $$D$$ a complex submanifold, pick a chart around a point that the nearby portion of $$D$$ to a coordinate axis in $${\mathbb{C}}^n$$, which e.g.~can be given by $$\left\{{ z_1 = 0 }\right\}$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-02-26_15-58.pdf_tex} }; \end{tikzpicture} } \end{figure} As before there's a distinguished section $$s_D \in H^0(X; {\mathcal{O}}_X(D) )$$ vanishing along $$D$$. Note that a line bundle is a free rank 1 $${\mathcal{O}}{\hbox{-}}$$module, and analogously here the functions vanishing along $$D$$ are $${\mathcal{O}}{\hbox{-}}$$modules generated by (here) $$z_1$$. \end{remark} \begin{definition}[Hyperplane] A \textbf{hyperplane} in $${\mathbb{CP}}^n$$ is any set of the form \begin{align*} H = \left\{{ [x_0: \cdots : x_1 ] {~\mathrel{\Big|}~}\sum a_i x_i = 0 }\right\} \cong {\mathbb{CP}}^{n-1} .\end{align*} \end{definition} \begin{example}[?] Take $${\mathbb{CP}}^{n-1} \subseteq {\mathbb{CP}}^n$$, e.g.~$$\left\{{ x_0 = 0 }\right\}$$. This is an example of a \textbf{divisor} on $${\mathbb{CP}}^n$$, i.e.~a complex codimension 1 submanifold''. We can take the line bundle constructed above to get $${\mathcal{O}}_{{\mathbb{CP}}^n}({\mathbb{CP}}^{n-1})$$ which vanishes along $${\mathbb{CP}}^{n-1}$$. More generally, for any hyperplane $$H$$ we can take $${\mathcal{O}}_{{\mathbb{CP}}^n}(H)$$, and these are all isomorphic, so we'll denote them all by $${\mathcal{O}}_{{\mathbb{CP}}^n}(1)$$. The implicit claim is that is the inverse line bundle of the tautological bundle, so $${\mathcal{O}}(1) \otimes{\mathcal{O}}(-1)$$ is the trivial bundle since the transition functions are given by reciprocals and multiplying them yields 1. We can classify complex line bundles on $${\mathbb{CP}}^n$$ using the SES \begin{align*} 0 \to \underline{{\mathbb{Z}}} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times}\to 1 .\end{align*} We know that $$H^1(X; {\mathcal{O}}^{\times})$$ were precisely holomorphic line bundles, since they were functions agreeing on double overlaps with a cocycle condition. We have a LES coming from sheaf cohomology: \begin{center} \begin{tikzcd} &&&& \cdots \\ \\ {H^1(X; {\mathcal{O}})} && {H^1(X; {\mathcal{O}})} && {H^1(X; {\mathcal{O}}^{\times})} \\ \\ {H^2(X; {\mathcal{O}})} && \cdots \arrow["{c_1}", from=3-5, to=5-1, out=0, in=180] \arrow[from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=5-1, to=5-3] \arrow[from=1-5, to=3-1, out=0, in=180] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMiwyLCJIXjEoWDsgXFxPTykiXSxbNCwyLCJIXjEoWDsgXFxPT1xcdW5pdHMpIl0sWzAsNCwiSF4yKFg7IFxcT08pIl0sWzAsMiwiSF4xKFg7IFxcT08pIl0sWzQsMCwiXFxjZG90cyJdLFsyLDQsIlxcY2RvdHMiXSxbMSwyLCJjXzEiXSxbMywwXSxbMCwxXSxbMiw1XSxbNCwzXV0=}{Link to Diagram} \end{quote} Applying this to $$X\coloneqq{\mathbb{CP}}^n$$, we have $$H^1({\mathcal{O}}) = H^2({\mathcal{O}}) = 0$$. This can be computed directly using that $${\mathbb{CP}}^n = \cup_{n\geq 1} {\mathbb{C}}^n$$ by taking charts $$x_i\neq 0$$, and this yields an acyclic cover. Thus $$c_1$$ is an isomorphism above, and $${\operatorname{Pic}}({\mathbb{CP}}^n) \cong {\mathbb{Z}}$$, where $${\operatorname{Pic}}$$ denotes isomorphism classes of line bundles. We can identify $${\operatorname{Pic}}({\mathbb{CP}}^n) = \left\{{ {\mathcal{O}}_{{\mathbb{CP}}^n}(k) {~\mathrel{\Big|}~}k\in {\mathbb{Z}}}\right\}$$. \end{example} \hypertarget{monday-march-01}{% \section{Monday, March 01}\label{monday-march-01}} \begin{remark} Last time: we defined $${\operatorname{Pic}}({\mathbb{CP}}^n)$$ as the set of line bundles on $${\mathbb{CP}}^n$$. \end{remark} \begin{definition}[Picard Group of a Manifold] Given any $$X\in {\mathsf{Mfd}}_{\mathbb{C}}$$, define $${\operatorname{Pic}}(X)$$ as the set of isomorphism classes of holomorphic line bundles on $$X$$. This is an abelian group given by $$L \otimes L'$$ and inversion $$L\to L^{-1}$$. \end{definition} \begin{remark} We saw that $${\operatorname{Pic}}(X) \cong H^1(X; {\mathcal{O}}^{\times})$$ as groups, noting that $$H^1$$ has a natural group structure here. We defined a \textbf{tautological bundle} on $${\mathbb{CP}}^n$$ and saw it was isomorphic to $${\mathcal{O}}(-1)$$, and moreover $${\mathcal{O}}(H) \cong {\mathcal{O}}(1)$$ for $$H$$ a hyperplane. The fiber was given by \begin{align*} \mathrm{Taut} &\to {\mathbb{CP}}^n \\ \left\{{ \lambda (x_0, \cdots, x_n) {~\mathrel{\Big|}~}\lambda\in {\mathbb{C}}}\right\} &\mapsto [x_0: \cdots : x_n] ,\end{align*} i.e.~the entire line corresponding to the given projective point. We also have $${\mathcal{O}}(H)(U)$$ is the sect of rational homogeneous functions $$\phi$$ on $$U$$ of degree 1 such that $$\operatorname{Div}\phi + H \geq 0$$ where $$H \coloneqq\left\{{x_0 = 0}\right\}$$. We want $$\phi/x_0$$ to be a well-defined function, so $$\phi$$ should scale like $$x_0$$ in the sense that \begin{align*} \phi( \lambda x_0, \cdots, \lambda x_n) = \lambda\phi( x_0, \cdots, x_n) .\end{align*} Note that there is a natural map \begin{align*} \mathop{\mathrm{Taut}}\otimes{\mathcal{O}}(H) \xrightarrow{} {\mathcal{O}} ,\end{align*} given by taking the line over a point and evaluating the homogeneous function on that line. Thus $$\mathop{\mathrm{Taut}}$$ is the inverse of $${\mathcal{O}}(H)$$. \end{remark} \begin{remark} We want to understand what Noether's formula says for $${\mathbb{CP}}^2$$, which requires understanding the canonical bundle $$K_{{\mathbb{CP}}^n}$$. We'll do this by writing down a meromorphic section $$\omega$$ (since it's a meromorphic volume form) which will yield $$K_{{\mathbb{CP}}^n} = {\mathcal{O}}(\operatorname{Div}\omega)$$. So take \begin{align*} \omega \coloneqq x_1^{-1}dx_1 \wedge \cdots \wedge x_n^{-1}dx_n ,\end{align*} noting that we leave out the first coordinate $$x_0$$ and divide by coordinates to make this scale-invariant. Here we work in a $${\mathbb{C}}^n$$ chart of points of the form $$[1: x_1 : \cdots : x_n]$$. Where does $$\omega$$ have poles? Along $$x_i = 0$$ for any $$1\leq i \leq n$$, and similarly in any other coordinate chart. We also have a 1st order pole along $$x_0 = 0$$. We then get \begin{align*} K_{{\mathbb{CP}}^n} = {\mathcal{O}}(\operatorname{Div}\omega) = {\mathcal{O}}( -H_0 -H_1 - \cdots - H_n) = {\mathcal{O}}(-n-1) ,\end{align*} where $$H_i = \left\{{x_i = 0}\right\}$$. Note that $${\mathbb{CP}}^n$$ is like a simplex: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-01_14-12.pdf_tex} }; \end{tikzpicture} } \end{figure} Applying this to $${\mathbb{CP}}^2$$, we obtain \begin{align*} K_{{\mathbb{CP}}^2} = {\mathcal{O}}(-3) .\end{align*} What is the intersection form? We know $$H^2({\mathbb{CP}}^2; {\mathbb{Z}}) \cong {\mathbb{Z}}$$ and the intersection form is unimodular. So write $${\mathbb{Z}}\coloneqq{\mathbb{Z}}\alpha$$ for $$\alpha$$ some generator. Then $$\alpha \cdot \alpha = \pm 1$$ since $$\operatorname{det}G = \pm 1$$ for the Gram matrix for this to be unimodular. Note that $$(- \alpha) \cdot (- \alpha) = \pm 1$$ with the same sign. \begin{claim} $${\mathcal{O}}(1) = {\mathcal{O}}(H)$$ generates $${\operatorname{Pic}}({\mathbb{CP}}^2) = H^2({\mathbb{CP}}^2; {\mathbb{Z}})$$. \end{claim} This is because $$c_1 {\mathcal{O}}(H) \cdot c_1 {\mathcal{O}}(H) = H\cdot H = \left\{{ x_0 = 0 }\right\} \pitchfork\left\{{ x_1 = 0 }\right\} = \left\{{ [0:0:1] }\right\}$$ here we note that the two hyperplanes can be oriented transversely and intersected. This is an oriented intersection. Recall Noether's formula, which was HRR applied to $${\mathcal{O}}$$ and the Chern-Gauss-Bonet theorem: \begin{align*} \chi({\mathcal{O}}) &= {1\over 12}(K^2 + \chi_{\mathsf{Top}})\\ &= h^0({\mathcal{O}}) - h^1({\mathcal{O}}) + h^2({\mathcal{O}})\\ &= 1 -1 + 1\\ &= 1 .\end{align*} The right-hand side can be written as \begin{align*} {1\over 12} \qty{ (-3H) \cdot (-3H) + 3} = {1\over 12}(9+3) = 1 .\end{align*} \end{remark} \begin{proposition}[The 4-sphere has no complex structure] $$S^4$$ has no complex structure. \end{proposition} \begin{proof}[?] We know that $$\chi_{\mathsf{Top}}(S^4) = 2$$. If $$S^4$$ had a complex structure, then $$c_1(K_{S^4}) \in H^2(S^4; {\mathbb{Z}}) = 0$$. Thus would make $$K_{S^4}^2 = 0$$, and so \begin{align*} \chi( {\mathcal{O}}_{S^4} ) = {1\over 12}( 0 + 2) = {1\over 6} \not\in {\mathbb{Z}} ,\end{align*} which is a contradiction. $$\contradiction$$ \end{proof} \begin{example}[?] Consider $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2$$, a 4-manifold diffeomorphic to $${\mathbb{CP}}^2$$ with the opposite orientation. What is the intersection form? Taking $$H\cdot H = -1$$ since the orientations aren't compatible, and more generally the Gram matrix is negated when the orientation is reversed. \end{example} \begin{proposition}[Barred projective 2-space is not orientably diffeomorphic to a complex surface] $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2$$ is not diffeomorphic to a complex surface by an orientation-preserving diffeomorphism (or any homeomorphism). \end{proposition} \begin{proof}[?] We have $$\chi_{\mathsf{Top}}= 3$$, and $$K_{\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2} = -c_1(T \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2) = \pm 3H$$. Then \begin{align*} \chi({\mathcal{O}}) = {1\over 12}\qty{ K_{\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2}^2 + \chi_{\mathsf{Top}}} = {1\over 12}(-9+3) \not\in {\mathbb{Z}} .\end{align*} \end{proof} \begin{remark} Consider $${\mathcal{O}}_{{\mathbb{CP}}^n}(d)$$, what are its global sections $$H^0({\mathbb{CP}}^n, {\mathcal{O}}_{{\mathbb{CP}}^n}(d))$$. Locally we have $${\mathcal{O}}_{{\mathbb{CP}}^n}(d)(U)$$ given by holomorphic functions in $$(x_0, \cdots, x_n) \in \pi^{-1}(U)$$ where $$\pi: {\mathbb{C}}^{n+1} \to {\mathbb{CP}}^n$$ and the functions satisfy $$f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})$$. The global sections will be the homogeneous degree $$d$$ polynomials in the coordinates of $$\mathbf{x}$$. \end{remark} \begin{remark} Why does a holomorphic function $$f: {\mathbb{C}}^{n+1} \to {\mathbb{C}}$$ such that $$f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})$$ necessarily a polynomial? Use the result that any such function with at most polynomial growth is itself a polynomial. If $${ \left.{{f}} \right|_{{S^{2d+1}}} }$$ is bounded by $$C$$, we have $${\left\lVert {f} \right\rVert}_{L^2} \leq C {\left\lvert {x} \right\rvert}^{2d}$$. Since $$({{\partial}}_{x_1} \cdots {{\partial}}_{x_k})^d f$$ is globally bounded $$k\geq 2d$$, applying Liouville's theorem makes it constant, and so a finite number of derivatives kill $$f$$ and this forces it to be polynomial. \end{remark} \begin{remark} So how many homogeneous degree $$d$$ functions are there? Here $$h^0({\mathbb{CP}}^n, {\mathcal{O}}(d)) =$$ will be the number of linearly independent degree $$d$$ polynomials in the variables $$x_0, \cdots, x_n$$, which is $${\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n+1}{d}\right)\kern-.3em\right)} = {n + d\choose n}$$, using the fact that monomials span this space. \end{remark} \begin{exercise}[?] Using that $$h^0({\mathbb{CP}}^2; {\mathcal{O}}(k))= h^2({\mathbb{CP}}^2; {\mathcal{O}}(-3-k) )$$ by Serre duality and Riemann-Roch, compute $$h^i({\mathbb{CP}}^2; {\mathcal{O}}(k))$$ for all $$i, k$$. \end{exercise} \begin{fact} $$h^i({\mathbb{CP}}^n; {\mathcal{O}}(k)) = 0$$ unless $$i=0, n$$. \end{fact} \hypertarget{wednesday-march-03}{% \section{Wednesday, March 03}\label{wednesday-march-03}} Find first 5m. \begin{remark} When we considered $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2$$, we implicitly assumed $$T\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2$$ was a complex rank 2 vector bundle with some purported complex structure. \end{remark} \begin{claim} \begin{align*} c_1( T\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2) = \pm 3H ,\end{align*} although it's not clear that $$c_1(K) \in H^2( \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2; {\mathbb{Z}}) \cong ({\mathbb{Z}}, [-1] )$$. \end{claim} \begin{remark} We had $$\chi({\mathcal{O}}) = {1\over 12} \qty{ K^2 + \chi_{\mathsf{Top}}} = {1\over 12}(3-n^2)$$, and since $$3-n^2 \in 12{\mathbb{Z}}$$, we have $$n^2 \in 3 + 12{\mathbb{Z}}\subset 3 + 4{\mathbb{Z}}$$ and this forces $$n^2 \equiv 3 \pmod 4$$. \end{remark} \begin{definition}[Differential Complex] Let \begin{align*} 0 \to \mathcal{E}^0 \xrightarrow{d_0} \mathcal{E}^1 \xrightarrow{d_1} \cdots \to \mathcal{E}^n \to 0 \end{align*} be a complex (so $$d^2 = 0$$) of smooth vector bundles on a smooth manifold $$X\operatorname{im}{\mathsf{Mfd}}_{\mathbb{R}}^{C^\infty}$$. Suppose that the $$d_i$$ are \textbf{differential operators}, i.e.~in local trivializing charts over $$U$$ we have \begin{align*} \mathcal{E}^i \cong {\mathcal{O}}^{\oplus r_i} {\mathcal{O}}^{\oplus r_{i+1}} \cong \mathcal{E}^{i+1} \end{align*} where in every matrix coordinate, $$d_i$$ is of the form $$\sum_{{\left\lvert {I} \right\rvert} < N} g_I {{\partial}}_I$$ where $${{\partial}}_I \coloneqq{{\partial}}_{i_1} \cdots {{\partial}}_{i_N}$$ is a partial derived and the $$g_I$$ are smooth functions. \end{definition} \begin{example}[?] For $$X\in {\mathsf{Mfd}}_{\mathbb{R}}^{C^ \infty }$$, we can take \begin{align*} 0 \to {\mathcal{O}}\xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \cdots .\end{align*} In local coordinates, \begin{itemize} \tightlist \item $$\Omega^1$$ is spanned over $${\mathcal{O}}$$ by $$dx_1, \cdots, dx_n$$ where $$n = \dim_{\mathbb{R}}(X)$$ \item $$\Omega^2$$ is spanned over $${\mathcal{O}}$$ by $$dx_i \wedge dx_j$$ for $$1\leq i, j \leq n$$. \end{itemize} Then the component of $$d$$ sending $$dx_i \to dx_i \wedge dx_j$$ is of the form \begin{align*} fdx_i &\mapsto -{\frac{\partial f}{\partial x_j}\,} dx_i \wedge dx_j .\end{align*} \end{example} \begin{example}[?] For $$X\in {\mathsf{Mfd}}_{\mathbb{C}}$$ and $$\mathcal{E} \to X$$ a holomorphic vector bundle, take \begin{align*} \mathcal{E} \otimes A^{0,0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \mathcal{E} \otimes A^{0, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \mathcal{E} \otimes A^{0, 2} \to \cdots .\end{align*} This is because for $$s_i$$ local holomorphic sections and $$\omega$$ a smooth form we have \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\qty{ (s_1, \cdots, s_r) \otimes\omega } = \qty{s_1, \cdots, s_r} \otimes{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\omega .\end{align*} \end{example} \begin{definition}[Order of an operator] The maximal $$N$$ that appears in $$\sum_{ {\left\lvert {I} \right\rvert} \leq N} g_I {{\partial}}_I$$ is the \textbf{order}. \end{definition} \begin{definition}[Symbol Complex] The \textbf{symbol complex} is a sequence of vector bundles on $$T {}^{ \vee }X$$. Noting that we have $$\pi: T {}^{ \vee }X\to X$$, and using pullbacks we can obtain bundles over the cotangent bundle: \begin{align*} 0 \to \pi^* \mathcal{E}_0 \xrightarrow{\sigma(d_0)} \pi^* \mathcal{E}_1 \xrightarrow{\sigma(d_1)} \cdots \to \pi^* \mathcal{E}_n \to 0 .\end{align*} The \textbf{symbol} of the differential operator $$d_i$$ is $$\sigma(d_i)$$. It is defined by replacing $${\partial}_i$$ in $$\sum_{{\left\lvert {I} \right\rvert} {\color{red} =} N } g_I {\partial}_I$$ with $$y_i$$ where \begin{align*} y_i: T {}^{ \vee }U \to {\mathbb{R}} \end{align*} is the coordinate function on the second factor of $$T {}^{ \vee }U = U \times{\mathbb{R}}^n$$ associated to the local coordinate $$i$$. Using that $$TU = (T {}^{ \vee }) {}^{ \vee }U$$, we can view $${\partial}_i$$ as functions on the cotangent bundle, $$\sigma(d_i)$$ is given in local trivializations by multiplication by a smooth function $$\sum_{{\left\lvert {I} \right\rvert} = N} g_I y^I$$. \end{definition} \begin{example}[?] Consider $${\mathcal{O}}\xrightarrow{d} \Omega^1$$. In local coordinates, this is given by $$d = \qty{{\partial}_1, \cdots, {\partial}_n}$$, i.e.~coordinate-wise differentiation, since we can write a local trivialization $$\Omega^1 = {\mathcal{O}}dz_1 \oplus \cdots \oplus {\mathcal{O}}dz_n$$. Then the symbol of $$d$$ is given by \begin{align*} \sigma(d): \pi^* {\mathcal{O}}&\to \pi^* \Omega^1 \\ 1 &\mapsto (y_1, \cdots, y_n) ,\end{align*} thought of as vector bundles over $$T {}^{ \vee }X$$, and this is projection onto to cotangent factor. Locally, the image of 1 is given by $$y_1 dx_1 + \cdots y_n dx_n$$, which is a point in $$T_p {}^{ \vee }X$$ for all $$(p, \alpha) \in T {}^{ \vee }X$$ which is an assignment to every point $$(p, \alpha) \in T_p {}^{ \vee }X$$ a point in $$(\pi^* \Omega^1)_{p, \alpha} \cong T_p {}^{ \vee }X$$. There is a tautological section $$(p, \alpha) \to \alpha\in T_p {}^{ \vee }X\in (\pi^* \Omega^1)_{p, \alpha}$$, or really $$(p, \alpha) \mapsto ( (p, \alpha), \alpha)$$. \end{example} \begin{remark} See similarly to the canonical symplectic structure of the cotangent bundle. \end{remark} \begin{remark} More generally, for $$d: \Omega^p \to \Omega^{p+1}$$, $$\sigma(d)$$ acts on the frame $$dx_{i_1} \wedge \cdots dx_{i_p}$$ in the following way: \begin{align*} \sigma(d)(dx_{i_1} \wedge \cdots \wedge dx_{i_p}) = \sum_y y_y dx_j \wedge dx_{i_1} \wedge \cdots dx_{i_p} \end{align*} where \begin{align*} d: fdx_{i_1} \wedge \cdots \wedge dx_{i_p} \mapsto \sum_j {\frac{\partial f}{\partial x_j}\,} dx_j \wedge \qty{dx_{i_1} \wedge \cdots \wedge dx_{i_p}} .\end{align*} The symbol complex is \begin{align*} \pi^* {\mathcal{O}}\xrightarrow{\sigma(d)} \pi^* \Omega^1 \xrightarrow{\sigma(d)} \pi^* \Omega^2 \to \cdots \to \pi^* \Omega^n \to 0 \end{align*} for $$n$$ the dimension. In this case, $$\sigma(d)$$ has the same formula everywhere, since it's $$C^ \infty {\hbox{-}}$$linear: \begin{align*} \sigma(d) = \sum_j y_j dx_j \wedge \qty{\cdots} .\end{align*} \end{remark} \begin{definition}[Elliptic Complex] A differential complex $$({\mathcal{E}}_{*}, d)$$ is \textbf{elliptic} if the symbol complex $$(\pi^* {\mathcal{E}}_{*}, \sigma(d))$$ is an exact sequence of sheaves (importantly) on $$T {}^{ \vee }X \setminus\left\{{s_z}\right\}$$ for $$s_z$$ the zero section. \end{definition} \begin{claim} $$({\Omega}_{*}, d)$$ is elliptic. To check exactness of a sequence of vector bundles, it suffices to check exactness on every fiber. Fix $$(p, \alpha) \in T {}^{ \vee }X \setminus\left\{{ s_z }\right\}$$, then \begin{align*} 0 \to {\mathbb{C}}\xrightarrow{\wedge \alpha} T {}^{ \vee }_p X \xrightarrow{\wedge \alpha} \bigwedge^2 T_p {}^{ \vee }X \xrightarrow{\wedge \alpha} \bigwedge^3 T_p {}^{ \vee }X \to \cdots .\end{align*} Moreover, if $$\alpha\wedge \beta = 0$$ implies that $$\beta = \alpha\wedge \gamma$$ for some $$\gamma$$, which implies that this sequence is exact. \end{claim} \hypertarget{friday-march-05}{% \section{Friday, March 05}\label{friday-march-05}} \begin{remark} Recall that we set up a differential complex, whose objects were vector bundles and differentials were differential operators (i.e.~linear combinations of partial derivatives) in local trivializations. We pulled back to tangent bundles (?) and defined the \emph{symbol} of an operator, and saw that when taking the symbol complex of the deRham complex. the sequence of maps was given by wedging against a tautological one-form. This was an \emph{elliptic complex} because the maps became wedging with a covector. \end{remark} \begin{example}[of an elliptic complex] Let $$X\in {\mathsf{Mfd}}_{\mathbb{C}}$$ and $$\mathcal{E}\to X \in { {\mathsf{Bun}}_{\operatorname{GL}_r} }_{\mathbb{C}}$$ be holomorphic. There is a resolution \begin{align*} 0 \to \mathcal{E} \xrightarrow{i} \mathcal{E} \otimes A^{0, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \mathcal{E} \otimes A^{0, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \cdots .\end{align*} What is the symbol complex? Consider the projection $$\pi: T {}^{ \vee }X\to X$$, and use pullbacks to get a sequence \begin{align*} 0 \to \pi^* \mathcal{E} \otimes A^{0, 0} \xrightarrow{\sigma( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})} \pi^* \mathcal{E} \otimes A^{0, 1} \xrightarrow{\sigma( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})} \cdots .\end{align*} Here the symbol $$\sigma({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$ replace $${\frac{\partial }{\partial t {\overline{{z}}}_i}\,}$$ with the corresponding function on $$T {}^{ \vee }X$$, say $${\overline{{y}}}_i$$. Then $$\sigma( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) = \sum_i {\overline{{y}}}_i \, d{\overline{{z}}}_i \wedge ({-}) = {\overline{{ \alpha }}} \wedge ({-})$$. As before, at a point $$(p, \alpha)$$ where $$\alpha\neq 0$$ in $$T {}^{ \vee }X$$, we get \begin{align*} 0 \to \mathcal{E}_p \xrightarrow{{\overline{{ \alpha}}} \wedge ({-})} \mathcal{E}_p \otimes\bigwedge^{0, 1}_p X \xrightarrow{{\overline{{ \alpha }}} \wedge ({-})} \mathcal{E}_p \otimes\bigwedge^{0, 2} X \to \cdots ,\end{align*} which is an exact sequence of vector spaces. So $$( \mathcal{E} \otimes A^{0, p}, { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$ is an elliptic complex. \end{example} \begin{slogan} The symbol being exact is approximately the top-order part being nowhere-vanishing. \end{slogan} \begin{remark} The next theorem computes the cohomology of an elliptic complex using Chern and Todd classes. \end{remark} \begin{theorem}[Atiyah-Singer Index Theorem] If $$( {\mathcal{E}}_{*}, d)$$ is an elliptic complex of smooth vector bundles on a compact oriented $$X\in {\mathsf{Mfd}}^n_{\mathbb{R}}$$, then \begin{align*} \chi({ \mathcal{E} }_{*}, d) = \sum (-1)^i \dim \qty{\ker d^i \over \operatorname{im}d^{i-1} } = (-1)^{\dim(X) \choose 2} \int_X {\operatorname{ch}\over {\operatorname{eul}}}( {\mathcal{E}}_{*} ) \mathrm{td}(TX \otimes_{\mathbb{R}}{\mathbb{C}}) .\end{align*} \end{theorem} \begin{remark} Here we define $$\operatorname{ch}( {\mathcal{E}}_{*} ) \coloneqq\sum_i (-1)^i \operatorname{ch}( \mathcal{E}^i )$$. What does it mean to divide by the Euler class? Let $$\left\{{ x_i, -x_i }\right\}$$ be the Chern roots of the complexified tangent bundle $$TX\otimes{\mathbb{C}}$$, then $${\operatorname{eul}}(X) \coloneqq\prod x_i$$ is the product where we pick one of each of the Chern roots from each of the pairs. The preferred sign to choose is the one for which $$\int_X \prod x_i = \chi_{\mathsf{Top}}(X)$$. Dividing just means to take the Chern character, then if it's divisible by $$\prod x_i$$, we do so. We have \begin{align*} \mathrm{td}(TX\otimes{\mathbb{C}}) = \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} .\end{align*} Thus \begin{align*} {\mathrm{td}(TX\otimes{\mathbb{C}}) \over {\operatorname{eul}}(X) } = \prod_i {1\over x_i} \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} ,\end{align*} but note that this doesn't necessarily make sense. However, all all computations we'll see, there will be enough cancellation to make this well-defined. \end{remark} \begin{exercise}[Chern character of the de Rham complex] $$\operatorname{ch}( {\Omega}_{*}X \otimes{\mathbb{C}}) = \prod_i (1-e^{x_i}) (1 - e^{-x_i})$$ for $$X\in {\mathsf{Mfd}}_{\mathbb{R}}^{2n}$$ even dimensional. \end{exercise} \begin{example}[?] Supposing $$X\in {\mathsf{Mfd}}_{\mathbb{R}}^2$$ is a genus $$g$$ surface, we have \begin{align*} {\mathcal{O}}\to \Omega^1\otimes{\mathbb{C}}\to \Omega^2 \otimes{\mathbb{C}} ,\end{align*} and $$\operatorname{ch}({ \Omega }_{*}) = \operatorname{ch}( {\mathcal{O}}) - \operatorname{ch}( \Omega^1 \otimes{\mathbb{C}}) + \operatorname{ch}(\Omega^2\otimes{\mathbb{C}})$$. The Chern roots of $$TX \otimes{\mathbb{C}}$$ are $$\left\{{ x_i, -x_i }\right\}$$, which come in pairs. So \begin{align*} \operatorname{ch}( {\Omega}_{*} ) = 1 - e^{x_i} - e^{x_i} + e^{-x_i + x_i} = (1 - e^{-x_i})( 1 - e^{x_i} ) .\end{align*} From the theorem, we're supposed to have \begin{align*} \chi( {\Omega}_{*}, d) &= (-1)^{n(n-1) \over 2} \int_X {\prod_i (1 - e^{-x_i})( 1 - e^{x_i} ) \over \prod_{i=1}^n x_i } \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} \\ &= (-1)^{n(n-1) \over 2} \int_X \prod_{i=1}^n (-x_i)\\ & = \int_X \prod_i x_i \\ &= \chi_{\mathsf{Top}}(X) && \text{C-G-B} .\end{align*} Letting $$d=\dim X = 2n$$, we have \begin{align*} (-1)^n (-1)^{d(d-1) \over 2} = (-1)^n (-1)^{n(2n-1)} = (-1)^2n = 1 . \end{align*} \end{example} \begin{example}[?] We can prove HRR using this theorem: we have \begin{align*} \chi(X, \mathcal{E} ) = \chi( \mathcal{E} \otimes A^{0, {-}}, { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \overset{\text{ASIT}}{=} \int_X { \operatorname{ch}(\mathcal{E} \otimes A^{0, {-}} ) \over {\operatorname{eul}}(X) } \mathrm{td}(TX \otimes_R {\mathbb{C}}) .\end{align*} We have $$\operatorname{ch}( \mathcal{E} \otimes A^{0, {-}} ) = \operatorname{ch}(\mathcal{E}) \operatorname{ch}( A^{0, {-}} )$$ where $$\operatorname{ch}(A^{0, 1}) = \sum_I (-1)^i \operatorname{ch}(\bigwedge^i A^{0, 1} )$$. The Chern roots of \begin{itemize} \tightlist \item $$TX$$ are $$\left\{{ x_i }\right\}$$\\ \item $$A^{1, 0} = T {}^{ \vee }X$$ are $$\left\{{ -x_i }\right\}$$\\ \item $$A^{0, 1}$$ are $$\left\{{ -x_i }\right\}$$ \end{itemize} So we obtain \begin{align*} \chi( \mathcal{E} ) &= (-1)^n \int_X {\prod ( 1- e^{x_i}) \over \prod x_i} \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} \\ &= \int_X \operatorname{ch}( \mathcal{E} ) \prod_i {x_i \over 1-e^{-x_i} } \\ &= \int_X \operatorname{ch}( \mathcal{E} ) \mathrm{td}(TX) ,\end{align*} which is HRR. \end{example} \hypertarget{monday-march-08}{% \section{Monday, March 08}\label{monday-march-08}} \begin{remark} Recall that given a differential complex $$({ \mathcal{E} }_{*}, d)$$ we had a symbol complex $$( \pi^* {\mathcal{E}}_{*}, \sigma(d) )$$ where $$\pi: T {}^{ \vee }X\to X$$ and \begin{align*} \sigma\qty{ \sum_{{\left\lvert {I} \right\rvert} \leq N} f_I {{\partial}}_I } \coloneqq\sum_{{\left\lvert {I} \right\rvert} = N} f_I y^I ,\end{align*} where we take the top-order differentials, $${\frac{\partial }{\partial x_j}\,} \mapsto y_j$$ and \begin{align*} T {}^{ \vee }X &\to {\mathbb{R}}\\ \alpha &\mapsto \alpha\qty{{\frac{\partial }{\partial x_j}\,} } .\end{align*} We say that $$( {\mathcal{E} }_{*}, d )$$ is \textbf{elliptic} if the symbol complex is exact on $$T {}^{ \vee }X \setminus\left\{{0}\right\}$$ where we delete the zero section. The Atiyah-Singer index theorem stated \begin{align*} \chi( {\mathcal{E}}_{*}, d) = \int_X { \operatorname{ch}( { \mathcal{E} }_{*}) \over {\operatorname{eul}}(X) } \mathrm{td}( TX\otimes_{\mathbb{R}}{\mathbb{C}}) .\end{align*} What's the connection to elliptic operators? Given a 2-term complex \begin{align*} 0 \to \mathcal{E}^0 \xrightarrow{D} \mathcal{E}^1 \to 0 ,\end{align*} then $$D$$ is an \textbf{elliptic operator} if this is an elliptic complex. This means the symbol complex is an isomorphism, i.e.~ \begin{align*} 0 \to \pi^* \mathcal{E}^0 \xrightarrow{\sigma(D)} \pi^* \mathcal{E}^1 \to 0 \end{align*} where $$\sigma(D)$$ is an isomorphism away from the zero section. \end{remark} \begin{remark} Every elliptic complex can be converted into a 2-term complex using a hermitian metric. Given \begin{align*} \mathcal{E}^0 \xrightarrow{d^0} \mathcal{E}^1 \xrightarrow{d^1} \mathcal{E}^2 \to \cdots ,\end{align*} we map this to \begin{align*} 0 \to \mathcal{E}^{\text{even}} \coloneqq\bigoplus_{i \text{ even} } \mathcal{E}^i \mathrel{\operatorname*{\rightleftharpoons}_{D^{\text{odd} }}^{D^\text{even}}} \mathcal{E}^{\text{odd}} \coloneqq\bigoplus_{i \text{ odd}} \to 0 \end{align*} where \begin{align*} D \coloneqq((d^{2i-1})^{\dagger} , d^{2i} ) : \mathcal{E}^{2i} \to \mathcal{E}^{2i-1} \oplus \mathcal{E}^{2i+2} \\ \end{align*} and $$(d^{2i-1})^{\dagger}$$ is defined by the following property: for $$\alpha\in \mathcal{E}^{2i-1}$$ and $$\beta \in \mathcal{E}^{2i}(X)$$, \begin{align*} {\left\langle { d^{2i-1} \alpha},~{\beta} \right\rangle}_h = {\left\langle { \alpha },~{ ( (d^{2i-1})^{\dagger} \beta} \right\rangle}_h .\end{align*} Here this pairing depends on a hermitian metric $$h$$, which is a hermitian form on each fiber: \begin{align*} h_i: \mathcal{E}^i \otimes{\overline{{ \mathcal{E}^i}}} \to {\mathbb{C}} .\end{align*} Using this, we can fix a volume form $$dV$$ on $$X$$ and define \begin{align*} {\left\langle {u},~{v} \right\rangle}_h \coloneqq\int_X h_i(u, {\overline{{v}}}) \, dV && u, v\in \mathcal{E}^i(X) .\end{align*} This yields the desired two-term complex, and $$( {\mathcal{E}}_{*}, d)$$ is elliptic if and only if $$D^e \circ D^o: \mathcal{E}^o {\circlearrowleft}$$ and $$D^o \circ D^e: \mathcal{E}^e {\circlearrowleft}$$ are elliptic operators. \end{remark} \begin{example}[?] Taking the de Rham complex \begin{align*} 0 \to {\mathcal{O}}\xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \to \cdots ,\end{align*} one can define \begin{align*} \Omega^{\text{even}} \mathrel{\operatorname*{\rightleftharpoons}_{ d + d^{\dagger}}^{d + d^\dagger}} \Omega^{\text{odd}} .\end{align*} Then using adjoint properties, we have \begin{align*} {\left\langle {\alpha},~{ d^\dagger d^\dagger \beta} \right\rangle} = {\left\langle { d \alpha},~{ d^\dagger \beta} \right\rangle} = {\left\langle { d^2 \alpha},~{ \beta } \right\rangle} = 0 ,\end{align*} using that $$d^2 = 0$$, and since this is true for all $$\alpha, \beta$$ we have $$(d^\dagger)^2 \beta = 0$$ for all $$\beta$$. Noting that $$d d^\dagger + d^\dagger d: \Omega^i(X) {\circlearrowleft}$$, and this operator is \textbf{the Laplacian}. Moreover $$\ker (d d^\dagger + d^\dagger d )$$ is the space of \textbf{harmonic $$i{\hbox{-}}$$forms}. \end{example} \begin{remark} Note that this space of harmonic forms depended on the Hermitian metrics on $$\mathcal{E}^i$$ and the volume form $$dV$$. In the case $$\mathcal{E}^i \coloneqq\Omega^i$$, there is a natural metric determined by any Riemannian metric on $$X$$. Recall that this is given by a metric \begin{align*} g: TX \otimes TX \to {\mathbb{R}} .\end{align*} This determines an isomorphism \begin{align*} T_p X &\xrightarrow{\sim} T_p {}^{ \vee }X\\ v &\mapsto g(v, {-}) ,\end{align*} which we can invert to get a metric on the cotangent bundle $$T {}^{ \vee }X$$. This induces a metric on $$i{\hbox{-}}$$forms using the identification $$\Omega^i \coloneqq\bigwedge^i T {}^{ \vee }X$$ and induces a volume form \begin{align*} dV \coloneqq\sqrt{ \operatorname{det}g}: \bigwedge^{\text{top}} TX \to {\mathbb{R}} .\end{align*} In this case, $$d d^\dagger + d^\dagger d$$ on $$\Omega^i(X)$$ is called the \textbf{metric Laplacian}. \end{remark} \begin{remark} Let $$(X, g)$$ be a Riemannian manifold. We thus have a symmetric bilinear form on $$\Omega^p(X)$$ given by pairing sections: \begin{align*} {\left\langle { \alpha},~{ \beta} \right\rangle} \coloneqq\int_X g( \alpha, \beta) .\end{align*} Note that we have orthonormal frames on $$\Omega^p(X)$$ of the form $$e_{i_1} \wedge \cdots \wedge e_{i_p}$$ where the $$\left\{{ e_i }\right\}$$ are orthonormal frames on $$T {}^{ \vee }X$$. \end{remark} \begin{definition}[Hodge Star Operator] Let $$n\coloneqq\dim(X)$$. The \textbf{Hodge star} operator is a map \begin{align*} \star: \Omega^p \to \Omega^{n-p} .\end{align*} defined by the property \begin{align*} \alpha\wedge \star\beta= g( \alpha, \beta) dV .\end{align*} Concretely, we have \begin{align*} \star\qty{ \sum f_I dx_{i_1} \wedge \cdots \wedge dx_{i_p} } &= \star\qty{ \sum f_I e_{i_1} \wedge \cdots \wedge e_{i_p} } \\ &= (-1)^\ell \sum_{j_k \in \left\{{ 1, \cdots, n }\right\} \setminus I} f_I e_{j_1} \wedge \cdots \wedge e_{j_{n-p}} \end{align*} for some sign $$\ell$$. \end{definition} \begin{example}[?] Let $$X\coloneqq{\mathbb{R}}^4$$ and $$g$$ the standard metric, i.e.~$$d = dx_1^2 + \cdots + dx_4^2$$. Take an orthonormal basis of $$T {}^{ \vee }{\mathbb{R}}^4$$, say $$\left\{{ e_1, e_2, e_3, e_4 }\right\}$$ where $$e_i \coloneqq dx_i$$. Then the induced volume form is $$dV \coloneqq e_1 \wedge e_2 \wedge e_3 \wedge e_4$$. We can then compute $$\star(e_1 \wedge e_2)$$ which is defined by the property \begin{align*} \alpha\wedge \star( e_1 \wedge e_2) = g( \alpha, e_1 \wedge e_2) dV .\end{align*} On the right-hand side, $$g( \alpha, e_1 \wedge e_2) = c_{12}(\alpha) e_1 \wedge e_2 \wedge e_3 \wedge e_4$$ where $$c_{12}$$ is the coefficient of $$e_1 \wedge e_2$$. To extract that coefficient, we can take $$\alpha( e_3 \wedge e_4$$, writing $$\alpha = \sum c_{ij} e_i \wedge e_j$$. Similarly, $$\star)e_1 \wedge e_3) = -e_2 \wedge e_4$$. This follows from writing \begin{align*} \alpha \wedge \star(e_1 \wedge e_3) = c_{13}(\alpha) {\color{blue} e_1} \wedge e_2 \wedge {\color{blue} e_3 } \wedge e_4 = (-1) c_{13}(\alpha) e_1 \wedge {\color{blue} e_3 \wedge e_2} \wedge e_4 .\end{align*} From this, $$\star: \Omega^p \to \Omega^{n-p}$$ is defined fiber-wise as \begin{align*} {\left\langle { \alpha},~{ \beta} \right\rangle} = \int_X \alpha\wedge \star\beta .\end{align*} \end{example} \begin{exercise}[?] Show that $$\star^2 = (-1)^{p(n-p)}$$. \end{exercise} \begin{proposition}[Formula for the adjoint of the Hodge star] Let $$d^\dagger \coloneqq(-1)^{n(p-1) +1} \star d \star$$. Then \begin{align*} {\left\langle {\alpha},~{ d \beta} \right\rangle} = {\left\langle {d^\dagger \alpha},~{ \beta} \right\rangle} && \alpha\in \Omega^p(X), \beta\in \Omega^{p-1}(X) .\end{align*} \end{proposition} \begin{proof}[?] A slick application of Stokes' theorem! Using that $$\star$$ is an isometry, we have \begin{align*} {\left\langle { \alpha},~{ d \beta} \right\rangle} &= \int_X \alpha\wedge \star d \beta \\ &= \int_X \star\alpha \wedge d \beta (-1)^{p(n-p)} && \text{applying $\star$ to both} \\ &= -\int_X d( \star\alpha) \wedge \beta (-1)^{p(n-p)} && \text{Stokes/IBP} \\ &= (-1)^{p(n-p)+1} \int_X \star d \star\alpha \wedge \star\beta && \text{isometry}\\ &= (-1)^{p(n-p)+1} {\left\langle {\star d \star\alpha},~{ \beta} \right\rangle} ,\end{align*} which shows that the term in the left-hand side of the inner product above is the adjoint of $$d^\dagger$$. \end{proof} \hypertarget{wednesday-march-10}{% \section{Wednesday, March 10}\label{wednesday-march-10}} \begin{warnings} Missing some stuff from the first few minutes here! \end{warnings} \begin{remark} Can we always get a Hermitian metric? Let $$X \in {\mathsf{Mfd}}_{C^{\infty }({\mathbb{R}})}$$ and $$\mathcal{E} \to X \in { {\mathsf{Bun}}_{\operatorname{GL}_r} }_{{\mathbb{C}}}$$ a smooth complex vector bundle. Then any section $$h\in \mathcal{E} {}^{ \vee }\otimes{\overline{{\mathcal{E}}}} {}^{ \vee }(X)$$, we have \begin{align*} h: \mathcal{E} \otimes{\overline{{\mathcal{E}}}} \to {\mathcal{O}}\\ h( e\otimes f) .\end{align*} for $$e, f \in \mathcal{E}_p$$ is a Hermitian form for all $$p$$. In local trivializations, $${ \left.{{\mathcal{E}}} \right|_{{U}} } \cong {\mathcal{O}}_U^{\oplus r}$$, and one can take the standard Hermitian form here. Then for $$(f_1, \cdots, f_r) \in {\mathcal{O}}^{\oplus r}(U)$$, we have $$\sum f_i \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_i\in {\mathcal{O}}(U)$$. This can be extended to all of $$X$$ using a partition of unity subordinate to the coordinate charts. The thing to check here is that on $${\mathbb{C}}^r$$, for any collection $$h_1, \cdots, h_n$$, any positive linear combination $$\sum a_i h_i$$ is again a Hermitian metric for any $$a_i \in {\mathbb{R}}^+$$. One can regard these as skew-symmetric matrices, which are closed under addition, and the positive-definite property ensures it's still a metric since $$h(v, v) = \sum a_i h_i(v, v) > 0$$ for $$v\neq 0$$. \end{remark} \begin{remark} Recall that we start with a Riemannian manifold $$(X, g)$$ where $$g: TX^{\otimes 2} \to {\mathcal{O}}$$ is a metric on the tangent bundle. Locally choose $$f_1,\cdots, f_n$$ an orthogonal frame of $$TX$$, then setting $$e_i \coloneqq f_i {}^{ \vee }$$ yields an orthogonal frame of $$T {}^{ \vee }X$$ and thus an orthogonal frame $$e_{i_1} \wedge \cdots e_{i_p}$$ of $$\bigwedge^p T {}^{ \vee }X \coloneqq\Omega^p X$$. So we get a metric on the smooth $$p{\hbox{-}}$$forms $$\Omega^p X$$. We defined the Hodge star operator \begin{align*} \star: \Omega^p &\to \Omega^{n-p} \\ e_{i_1} \wedge \cdots e_{i_p} &\mapsto \pm e_{j_1} \wedge \cdots \wedge e_{j_{n-p}} .\end{align*} where $$\left\{{ i_1, \cdots, i_p, j_1, \cdots, j_{n-p} }\right\} = \left\{{ e_1, \cdots, e_n }\right\}$$. We saw that \begin{align*} e_{i_1} \wedge \cdots \wedge e_{i_p} \star\qty{ e_1 \wedge \cdots e_{i_p}} &= e_1 \wedge \cdots \wedge e_n \\ \star\qty{ \sum_{{\left\lvert {I} \right\rvert} = p } f_I e_I} &= \sum_{ {\left\lvert {I} \right\rvert} = p} e_{I^c} (-1)^{\operatorname{sign}(I)} .\end{align*} Moreover, \begin{align*} {\left\langle { \alpha},~{ \beta} \right\rangle} = \int_X g( \alpha, \beta) dV = \int_X \alpha\wedge \qty{\star\beta} ,\end{align*} and we showed that \begin{align*} {\left\langle { \alpha},~{ d \beta} \right\rangle} = \pm {\left\langle { d^\dagger \alpha},~{ \beta} \right\rangle} && d^\dagger \coloneqq\star d \star, \beta\in \Omega^{p-1}(X), \alpha\in \Omega^p(X) ,\end{align*} yielding an adjoint operator \begin{align*} d^\dagger: \Omega^p(X) \to \Omega^{p-1}(X) .\end{align*} \end{remark} \begin{definition}[Laplacian] The \textbf{Laplacian} is the differential operator \begin{align*} \Delta \coloneqq dd^\dagger + d^\dagger d: \Omega^p(X) \to \Omega^p(X) .\end{align*} \end{definition} \begin{definition}[Harmonic Forms] A $$p{\hbox{-}}$$form $$\omega$$ is \textbf{harmonic} if and only if $$\Delta \omega = 0$$. We define $$\mathcal{H}^p(X)$$ as the space of harmonic $$p{\hbox{-}}$$forms. \end{definition} \begin{remark} This operator is $${\mathbb{R}}{\hbox{-}}$$linear, so $$\mathcal{H}^p(X) \in { \mathsf{Vect} }_{\mathbb{R}}$$. Note that this whole construction can be made to work over $${\mathbb{C}}$$ by adding conjugates in appropriate places. \end{remark} \begin{proposition}[Characterization of when a smooth p-form is harmonic] A smooth $$p{\hbox{-}}$$form $$\omega$$ is harmonic if and only if $$d \omega = d^\dagger \omega = 0$$. \end{proposition} \begin{proof}[?] $$\impliedby$$: This direct is easy, since $$\Delta \omega \coloneqq(dd^\dagger + d^\dagger d) \omega = d(0) + d^\dagger 0 = 0$$. $$\implies$$: A nice trick! Using the adjunction $$d, d^\dagger$$ we have \begin{align*} {\left\langle { \Delta \omega},~{ \omega} \right\rangle} &= {\left\langle { d d^\dagger \omega},~{ \omega} \right\rangle} + {\left\langle {d^\dagger \omega},~{ \omega} \right\rangle} \\ &= {\left\langle { d^\dagger \omega},~{ d^\dagger \omega} \right\rangle} + {\left\langle {d \omega},~{ d \omega} \right\rangle} .\end{align*} We now use that since $$g$$ is positive definite, it is a non-negative smooth function, and \begin{align*} {\left\langle { \alpha},~{ \alpha} \right\rangle} \coloneqq\int_X g( \alpha, \alpha) \, dV \geq 0 \text{ with equality } \iff \alpha \equiv 0 \text{ on } X .\end{align*} So we can conclude that $$d^\dagger \omega = d \omega = 0$$. \end{proof} \begin{warnings} Note that we've used that the inner product is symmetric over $${\mathbb{R}}$$. Over $${\mathbb{C}}$$, there are bars introduced from conjugation when swapping the variables. \end{warnings} \begin{proposition}[Orthogonal decomposition of p-forms] The following three subspaces of $$\Omega^p(X)$$ are mutually orthogonal: \begin{align*} d \Omega^{p-1}(X), \mathcal{H}^p(X), d^\dagger \Omega^{p+1}(X) .\end{align*} \end{proposition} \begin{proof}[?] We can write \begin{align*} {\left\langle { d \alpha},~{ d^\dagger } \right\rangle} = {\left\langle { d^2 \alpha},~{ \beta} \right\rangle} = {\left\langle {0},~{ \beta} \right\rangle} ,\end{align*} showing that the 1st and 3rd spaces are orthogonal. If $$\alpha\in \mathcal{H}^p(X)$$ then by the above proposition, $$d \alpha = d^\dagger \alpha = 0$$, and so \begin{align*} {\left\langle { \alpha },~{ d \beta} \right\rangle} = {\left\langle {d^\dagger \alpha},~{ \beta} \right\rangle} = 0 \\ {\left\langle { \alpha },~{ d^\dagger \beta} \right\rangle} = {\left\langle {d \alpha},~{ \beta} \right\rangle} = 0 .\end{align*} Thus the 2nd space is orthogonal to the 1st and 3rd. \end{proof} \begin{observation} Suppose something false ($$\danger$$): that $$\Omega^p(X)$$ is a \emph{complete} vector space with respect to the inner product. Remember that it is \textbf{not}! But if it were, there would be a decomposition \begin{align*} \Omega^p(X) = d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \oplus d^\dagger \Omega^{p+1}(X) .\end{align*} Let $$\alpha\in \qty{ d \Omega^{p-1}(X) \oplus d^\dagger \Omega^{p+1}(X)}^\perp$$ where we take the orthogonal complement with respect to the inner product. Then \begin{align*} {\left\langle { \alpha },~{ d \beta } \right\rangle} &= 0 \forall \beta \\ {\left\langle { \alpha },~{ d^\dagger \gamma } \right\rangle} &= 0 \forall \gamma\\ \\ \implies {\left\langle { d^\dagger \alpha},~{ \beta} \right\rangle} = 0 \forall \beta \\ \implies d^\dagger \alpha \equiv 0 && \text{setting} \beta\coloneqq d^\dagger \alpha .\end{align*} Similarly, $$d \alpha = 0$$ and so $$\alpha\in \mathcal{H}^p(X)$$. The conclusion (which is true \emph{without} the false assumption) is that \begin{align*} \qty{ d \Omega^{p-1}(X) \oplus d^\dagger \Omega^{p+1}(X)}^\perp = \mathcal{H}^p .\end{align*} However, this doesn't yield the full direct sum decomposition: if $$W \subseteq V$$, then it's not necessarily true that $$V \cong W \oplus W^\perp$$, which only holds if \begin{itemize} \item $$V$$ is complete, \item $$W$$ is closed. \end{itemize} \end{observation} \begin{fact} For smooth $$p{\hbox{-}}$$forms, this decomposition \textbf{does} hold despite the false assumption: \begin{align*} \Omega^p(X) = d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \oplus d^\dagger \Omega^{p+1}(X) .\end{align*} \end{fact} \begin{corollary}[p-forms have harmonic representatives] Thus $$\mathcal{H}^p(X)$$ represents $$H^p(X; {\mathbb{R}})$$. \end{corollary} \begin{remark} We have \begin{align*} H^p(X; {\mathbb{R}}) &= {\ker d \over \operatorname{im}d} \\ &= { d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \over d \Omega^{p-1}(X) } \\ &= \mathcal{H}^p(X) .\end{align*} Note that there is a map \begin{align*} \mathcal{H}^p(X) \to H^p(X; {\mathbb{R}}) \end{align*} since $$\alpha\in \mathcal{H}^p(X)$$ satisfies $$d \alpha = 0$$ in addition to $$d^\dagger \alpha = 0$$. \end{remark} \begin{remark} Note that one can complete these spaces using Sobolev spaces, but there are issues. Take $$S^1$$, then \begin{align*} L_2(S^1) \coloneqq\left\{{ \sum a_n e^{2\pi i n z} {~\mathrel{\Big|}~}\sum {\left\lvert {a} \right\rvert}_i < \infty }\right\} ,\end{align*} but for $$f\in L_2(S^1)$$ we have $$df = \sum 2\pi i n a_n e^{2\pi i n z}$$ which may not converge. \end{remark} \hypertarget{review-monday-march-15}{% \section{Review (Monday, March 15)}\label{review-monday-march-15}} \begin{remark} Recall that a \emph{sheaf of rings} $$\operatorname{\mathcal{F}}$$ on $$X\in {\mathsf{Top}}$$ is an assignment of a ring $$\operatorname{\mathcal{F}}(U)$$ to each open set $$U\subseteq X$$ and restriction maps $$\operatorname{\mathcal{F}}(U) \xrightarrow{\rho_{UV}} \operatorname{\mathcal{F}}(V)$$ for $$V \subseteq U$$ that is a presheaf, so \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item This diagram commutes: \end{enumerate} \begin{center} \begin{tikzcd} U && V && W \arrow["{\rho_{UV}}", from=1-1, to=1-3] \arrow["{\rho_{VW}}", from=1-3, to=1-5] \arrow["{\rho_{UW}}"', curve={height=30pt}, from=1-1, to=1-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJVIl0sWzIsMCwiViJdLFs0LDAsIlciXSxbMCwxLCJcXHJob197VVZ9Il0sWzEsMiwiXFxyaG9fe1ZXfSJdLFswLDIsIlxccmhvX3tVV30iLDIseyJjdXJ2ZSI6NX1dXQ==}{Link to Diagram} \end{quote} \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \setcounter{enumi}{1} \tightlist \item $$\phi_{UU} = \one_{\operatorname{\mathcal{F}}(U)}$$ and $$\operatorname{\mathcal{F}}(\emptyset) = 0$$. \end{enumerate} That additionally satisfies unique gluing on double overlaps. \end{remark} \begin{example}[?] Any reasonable class of functions whose behavior is only locally restricted. Examples are being smooth or continuous, but e.g.~being constant is a global condition. Other examples include $$X\in {\mathsf{Mfd}}^n(C^\infty({-}, {\mathbb{R}}))$$, denoting $${\mathcal{O}}$$ the sheaf of smooth functions. This also carries a sheaf of \emph{abelian groups} $$\Omega^p$$. In the special case where $$U$$ is a coordinate chart, we have functions $$\varphi_U: U\to {\mathbb{R}}^n$$. Writing $$S \coloneqq\varphi_U(U)$$, we can define \begin{align*} \Omega^p(U) \cong \Omega^p(S) \coloneqq\left\{{ \sum f_I(\mathbf{x}) dx_I {~\mathrel{\Big|}~}f_I \in C^\infty({\mathbb{R}}^n, {\mathbb{R}})}\right\} .\end{align*} \end{example} :::\{.remark\} More generally, for an arbitrary open $$U$$, cover it by coordinate charts $$\left\{{ U_i }\right\} \rightrightarrows U$$. Then we want $$\omega_i \in \Omega^p(U_i)$$ which are compatible on double overlaps, so such a collection defines a section $$\left\{{ \omega_i {~\mathrel{\Big|}~}i\in I }\right\} \in \Gamma( \Omega^p(U) )$$. The compatibility is given by taking coordinate charts $$\varphi_i: U_i \to {\mathbb{R}}^n$$ with $$\omega_i \in \Omega^p(U_i)$$, we consider \begin{align*} t_{ij}: \varphi_i \circ \varphi_2 ^{-1} : \varphi_j(U_i \cap U_j) \to \varphi_i( U_i \cap U_j) ,\end{align*} and we require that the pullback satisfies $$t_{ij}^*(\omega_1) = \omega_2$$ This pullback can be thought of as a coordinate change for the forms. Writing $$x_I$$ as coordinates on $$U_i$$ and $$y_J$$ on $$U_j$$, we can write \begin{align*} x_1 &= h_1(y_J) \\ x_2 &= h_2(y_J) \\ \vdots& \\ x_n &= h_n(y_J) \\ \end{align*} which expresses $$t_{ij}$$ in coordinates. This allows us to give meaning to the formal symbols $$dx_I$$: \begin{align*} dx_1 &\coloneqq\sum_{i=1}^n {\frac{\partial h_1}{\partial y_i}\,} dy_i \\ dx_2 &\coloneqq\sum_{i=1}^n {\frac{\partial h_2}{\partial y_i}\,} dy_i \\ \vdots& \\ dx_k &\coloneqq\sum_{i=1}^n {\frac{\partial h_k}{\partial y_i}\,} dy_i \\ ,\end{align*} and under these substitutions in the original expression we obtain \begin{align*} \omega_1 = \sum_{{\left\lvert {I} \right\rvert} = p} f_I(\mathbf{x}) dx_I \mapsto \omega_2 .\end{align*} \begin{remark} For $$X \in {\mathsf{Mfd}}(\mathop{\mathrm{Hol}}({-}, {\mathbb{C}}))$$ such that $$\varphi_V \circ \varphi_U ^{-1} : \varphi_U( U \cap V) \to \varphi_V(U \cap V)$$ is holomorphic, so $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}z_i = 0$$. Then $$\Omega^p(U) = \left\{{ \sum_{{\left\lvert {I} \right\rvert} = p} f_I( \mathbf{z}) dz_I }\right\}$$, and the \emph{key difference} is that the $$f_I$$ be holomorphic. This matters since POUs exist in the smooth setting but not the complex setting. Note that $${\mathcal{O}}, \Omega^p$$ denote smooth/holomorphic functions and smooth/holomorphic $$p{\hbox{-}}$$forms in the smooth/complex settings. So we need a new notation for \emph{smooth holomorphic} $$p{\hbox{-}}$$forms in the complex setting. We defined $$A^{p, 0}$$ to be the smooth $$p{\hbox{-}}$$forms, and $$A^{p, q}$$ the smooth $$(p, q){\hbox{-}}$$forms. In local coordinates, these look like \begin{align*} A^{p, q}(U) = \left\{{ \sum_{{\left\lvert {I} \right\rvert} = p, {\left\lvert {J} \right\rvert} = q} f_{I, J} (\mathbf{z}) dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J }\right\} .\end{align*} \end{remark} \begin{example}[?] \envlist \begin{itemize} \tightlist \item $$\Re(z) \,dz\in A^{1, 0}({\mathbb{C}})$$ is a smooth $$(1, 0){\hbox{-}}$$form. \item $$z\,dw- w\,dz\in \Omega^1({\mathbb{C}}^2)$$ is a holomorphic 1-form. \item On $${\mathbb{C}}^3$$, $$z_1 dz_2 \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_3 - \Re(z_3) dz_1 d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_1 \in A^{1, 1}({\mathbb{C}}^3)$$. \end{itemize} \end{example} \begin{remark} Why are these $$A^{p, q}$$ useful? They give a resolution of $$\Omega^p$$ on a complex manifold. There are maps of sheaves \begin{align*} 0 \to \Omega^p \xrightarrow{i} A^{p, 0} ,\end{align*} where being a map of sheaves means there are maps $$\Omega^p(U) \to A^{p, 0}(U)$$ for all opens $$U$$ which are compatible with restriction: \begin{center} \begin{tikzcd} {\Omega^p(U)} && {A^{p, 0}(U)} \\ \\ {\Omega^p(V)} && {A^{p, 0}(U)} \arrow["{i_U}", from=1-1, to=1-3] \arrow["{i_V}", from=3-1, to=3-3] \arrow["{\rho_{UV}^*}"{description}, no head, from=1-1, to=3-1] \arrow["{\rho_{UV}^*}", no head, from=1-3, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXE9tZWdhXnAoVSkiXSxbMCwyLCJcXE9tZWdhXnAoVikiXSxbMiwwLCJBXntwLCAwfShVKSJdLFsyLDIsIkFee3AsIDB9KFUpIl0sWzAsMiwiaV9VIl0sWzEsMywiaV9WIl0sWzAsMSwiXFxyaG9fe1VWfV4qIiwxLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDMsIlxccmhvX3tVVn1eKiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link to Diagram} \end{quote} It's clear that this works for $$i$$, since any holomorphic function simply \emph{is} smooth. We could continue this resolution: \begin{align*} 0 \to \Omega^p \xrightarrow{i} A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \end{align*} where \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\qty{ \sum_{I, J} f_{I, J} dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J } \coloneqq \sum_{I, J, K} {\frac{\partial f_{I, J}}{\partial z_k}\,} d \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_k \wedge dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J .\end{align*} We then defined Dolbeaut cohomology, $$H^q(X, \Omega^p) = \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{p, q} / \operatorname{im}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{p, q-1}$$. \end{remark} \hypertarget{wednesday-march-17}{% \section{Wednesday, March 17}\label{wednesday-march-17}} \hypertarget{inverting-bundles}{% \subsection{Inverting Bundles}\label{inverting-bundles}} \begin{remark} Continuing review: let $$\mathcal{E} \to X \in {\mathsf{Bun}}({\mathbb{R}}^n)$$. A \textbf{metric} on $$\mathcal{E}$$ is a smoothly varying positive definite inner product on the fibers. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-17_13-55.pdf_tex} }; \end{tikzpicture} } \end{figure} \todo[inline]{Fix this diagram! Need to remember what it was demonstrating.} For $$v, w\in \mathcal{E}_p$$, we want a pairing $$g_p(v, w): \mathcal{E}_p^{\otimes 2} \to {\mathbb{R}}$$. To think about this globally, this should be a map \begin{align*} g: \mathcal{E}^{\otimes 2} \to {\mathcal{O}} .\end{align*} where $$g_p: \mathcal{E}_p^{\otimes 2} \to {\mathbb{R}}$$. Note that this map is $${\mathcal{O}}{\hbox{-}}$$linear, which follows from the fact that it's $${\mathbb{R}}{\hbox{-}}$$linear on each fiber, or equivalently it is a map of vector bundles. We should also have that $$g(s\otimes s) \in {\mathcal{O}}(X)$$ is a smooth function, and we require $$g(s\otimes s) \geq 0$$. We also require $$g(s\otimes s)(p) = 0 \iff s_0 = 0$$ and $$g(s\otimes t) = g(t\otimes s)$$. This implies that $$g\in (\mathcal{E}^{\otimes 2}) {}^{ \vee }\otimes{\mathcal{O}}= (\mathcal{E} {}^{ \vee })^{\otimes 2}(X)$$. The symmetric condition means that $$g\in \operatorname{Sym}^2 \mathcal{E} {}^{ \vee }(X)$$. \end{remark} \begin{remark} For Hermitian forms, we take \begin{align*} h: ({\mathbb{C}}^n)^{\otimes 2}\to {\mathbb{C}} \end{align*} where $$h$$ is conjugate linear, so $$h(cv, c'w) = \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5muc' h(v, w)$$. Note that we can write $$h(v, w) = {\overline{{v}}}^t H w$$ where $$H$$ is Hermitian, so $${\overline{{H}}}^t = H$$. This implies that $$h(v,v) \in {\mathbb{R}}^{\geq 0}$$ and $$h(v,v) = 0 \iff v=0$$ with $$h(v, w) = {\overline{{h(v, w)}}}$$ The great thing about metrics: we can identify zero sections by self-pairing, multiplying by a volume form, and integrating. For $$\mathcal{E}\to X \in {\mathsf{Bun}}({\mathbb{C}})$$, there is another bundle $${\overline{{\mathcal{E}}}} \to X \in {\mathsf{Bun}}({\mathbb{C}})$$. Supposing that $${ \left.{{ \mathcal{E}}} \right|_{{U}} } \xrightarrow{\varphi_U} {\mathcal{O}}_U^{\oplus n}$$ in a local trivialization, conjugating all of the transition functions gives the transition functions $${ \left.{{ {\overline{{ \mathcal{E}}}} }} \right|_{{U}} } \xrightarrow{\mathrm{conj} \circ \varphi_U} {\mathcal{O}}_U^{\oplus n}$$. This yields a map \begin{align*} h: {\overline{{ \mathcal{E} }}} \otimes_{\mathbb{C}}\mathcal{E} \to {\mathcal{O}}\in ( {\overline{{\mathcal{E}}}} \otimes\mathcal{E} ) {}^{ \vee } .\end{align*} In local trivializations we have $${ \left.{{ \mathcal{E} }} \right|_{{U}} } = {\mathcal{O}}_U^{\oplus n} = {\mathbb{C}}^n \times U$$, and $$h$$ is described by $$h_U \in ({\overline{{ {\mathcal{O}}}}}^{\oplus n} \otimes{\mathcal{O}}^{\oplus n})(U)$$. \end{remark} \begin{remark} When $$\operatorname{rank}\mathcal{E} = 1$$ we abuse notation! For $$h\in ({\overline{{\mathcal{E}}}} {}^{ \vee }\otimes\mathcal{E} {}^{ \vee })(X)$$, this is locally a $$1\times 1$$ Hermitian matrix, thus of the form $$[a]$$ for $$a\in {\mathbb{R}}^{\geq 0}$$. So we write \begin{align*} h(s, t) = hs{\overline{{t}}} \coloneqq h\otimes s \otimes{\overline{{t}}} \in ({\overline{{\mathcal{E}}}} {}^{ \vee }\otimes\mathcal{E} {}^{ \vee }) \otimes\mathcal{E} \otimes{\overline{{\mathcal{E}}}} = {\mathcal{O}} \end{align*} if $$\mathcal{E}$$ is a line bundle. Why is $$V\otimes V {}^{ \vee }= {\mathcal{O}}$$ in this case? There is a pairing $$v\otimes\lambda \mapsto \lambda(v)$$, or more generally a trace pairing. \end{remark} \hypertarget{serre-duality-revisited}{% \subsection{Serre Duality Revisited}\label{serre-duality-revisited}} \begin{remark} Let $$X$$ be a Riemann surface, so $$X\in {\mathsf{Mfd}}^1({\mathbb{C}})$$. Let $$L\to X \in {\mathsf{Bun}}^1(\mathop{\mathrm{Hol}})$$, then we have a resolution \begin{align*} 0 \to L \hookrightarrow L\otimes A^{0, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} L \otimes A^{0, 1} \to 0 ,\end{align*} where the first map is inclusion of smooth holomorphic sections into smooth sections. What is this cut out by? We had $$s\mapsto { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s$$ and thus $$f \mapsto {\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\,} \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$. Note that $$H_1(L) = \operatorname{coker}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$. \end{remark} \begin{remark} Serre duality said that \begin{align*} h^1(L) = \dim H^1(L) = h^0( L {}^{ \vee }\otimes K) && K = \Omega^1 ,\end{align*} where $$\Omega^1$$ is the sheaf of holomorphic 1-forms. Choose a metric to identify $$H^1(L)$$ and $$H^0(L {}^{ \vee }\otimes K)$$. Choose a hermitian metric on $$L$$ and take $$s, t\in H^0(L\otimes A^{0, 0}) = C^\infty(L; {\mathbb{C}})$$, then we get $$h(s, t) \in C^{\infty }(X; {\mathbb{C}})$$ a smooth complex function. We abuse notation by writing this as $$h(s, t) = hs{\overline{{t}}}$$, viewing $$h\in C^{\infty }(L {}^{ \vee }\otimes{\overline{{L}}} {}^{ \vee })$$ locally. Note that we can't integrate a function on a manifold without a form, so choosing a volume for $$dV$$ we can define a pairing on sections \begin{align*} {\left\langle {s},~{t} \right\rangle} \coloneqq\int_X hs{\overline{{t}}} dV .\end{align*} Now for two sections $$\alpha, \beta\in H^0(L\otimes A^{0, 1})$$ we can write \begin{align*} \int_X h \alpha {\overline{{ \beta}}} = \int_X \omega ,\end{align*} where $$\omega$$ is a smooth $$(1, 1){\hbox{-}}$$form since $$h\in {\overline{{L}}} {}^{ \vee }\otimes L {}^{ \vee }$$, $$\alpha\in L\otimes A^{0, 1}$$, and $${\overline{{ \beta}}} \in \mkern 1.5mu\overline{\mkern-1.5muL\mkern-1.5mu}\mkern 1.5mu \otimes A^{1, 0}$$. We now have metric on both the source and target spaces here: \begin{align*} H^0( L\otimes A^{0, 0}) \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} H^0(L\otimes A^{0, 1}) ,\end{align*} where on the left-hand side we take $$(s, t) \mapsto \int_X hs{\overline{{t}}}dV$$ and on the right-hand side we have $$(\alpha, \beta) \mapsto \int_X h \alpha{\overline{{\beta}}}$$. \end{remark} \begin{remark} Given a map of metric vector spaces $$V \xrightarrow{\varphi} W$$, the \emph{adjoint} $$\varphi^\dagger$$ satisfies \begin{align*} {\left\langle { \varphi(v) },~{w} \right\rangle} = {\left\langle {v},~{ \varphi^\dagger(w)} \right\rangle} .\end{align*} and $$\operatorname{coker}( \varphi) = \ker( \varphi^\dagger)$$. So $$H^1(L) = \operatorname{coker}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger$$, and after integrating by parts we have \begin{align*} {\left\langle { \alpha},~{ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s} \right\rangle} &\coloneqq\int_X \alpha{\overline{{ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s }}} h \\ &= \int_X \alpha{\partial}({\overline{{s}}}) h \\ &= -\int_X {\overline{{s}}} {\partial}( \alpha h) && \text{IBP} \\ &= -\int_X {\overline{{s}}} {{\partial}(\alpha h) \over dV} dV \\ &= {\left\langle { - { {\partial}( \alpha h ) \over dV}},~{s} \right\rangle} .\end{align*} So we could define \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger \alpha = {\overline{{- {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}({\overline{{\alpha}}} h )} \over dV }}} .\end{align*} Note that $$\alpha\mapsto {\overline{{ \alpha}}} h$$, so $$\alpha\in \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger \iff {\overline{{ \alpha}}}h\in \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$. Then $$\ker ({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger) = H^0(L {}^{ \vee }\otimes K)$$. \end{remark} \hypertarget{friday-march-19}{% \section{Friday, March 19}\label{friday-march-19}} \begin{remark} Recall Serre duality: let $$C\in {\mathsf{Mfd}}_{\mathbb{C}}({ \operatorname{compact} } ,{ \operatorname{oriented} } )$$ and $$L\to C \in {\mathsf{Bun}}(\mathop{\mathrm{Hol}})$$. Then \begin{align*} h^1(L) = h^0(L {}^{ \vee }\otimes K_C) .\end{align*} We also have Riemann-Roch, a very important tool: \begin{align*} h^0(L) - h^1(L) = \deg L + 1 - g(C) ,\end{align*} where $$\deg L = \int_C c_1(L)$$, which is also equal to $$\deg [ \left\{{ s = 0 }\right\}] = \deg(\operatorname{Div}s)$$. Note that $$c_1$$ is the most important Chern class to know, thanks to the splitting principle. How was it defined? There are several definitions: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$L$$ defines an element of \begin{align*} H^1(C, {\mathcal{O}}^{\times}) &= \left\{{ t_{UV}: U \cap V \to {\mathbb{C}}^{\times}{~\mathrel{\Big|}~}t_{UV} t_{UW}^{-1}t_{VW} = 1 }\right\} / {{\partial}}\left\{{ h_u: U\to {\mathbb{C}}^{\times}}\right\} \\ &= \ker {{\partial}}^1 / \operatorname{im}{{\partial}}^0 \end{align*} in Čech cohomology. By definition $${{\partial}}\left\{{ h_U {~\mathrel{\Big|}~}U\in \mathcal{U} }\right\} = \left\{{ h_u h_v^{-1}{~\mathrel{\Big|}~}U, V \in \mathcal{U} }\right\}$$, where $${{\partial}}^2 = 1$$ since \begin{align*} (h_U h_V)^{-1}\qty{h_U h_W ^{-1}}^{-1}(h_V h_W ^{-1}) = 1 && \text{on } U \cap V \cap W .\end{align*} By assigning $$L$$ to its transition functions, we get a map $$L\to H^1$$. We have the exponential exact sequence: \begin{align*} 0 \to \underline{{\mathbb{Z}}} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times}\to 1 ,\end{align*} which induces a map \begin{align*} H^1(C, {\mathcal{O}}^{\times}) &\to H^2(C, {\mathbb{Z}}) \\ L &\mapsto c_1(L) .\end{align*} \item $$L$$ defines an element $$\mathop{\mathrm{Fr}}L \in \mathrm{Bun}^{\mathrm{prin}}({\mathbb{C}}^{\times})$$ (which only works for line bundles), which is defined by $$\mathop{\mathrm{Fr}}L = L \setminus s_0$$ where $$s_0$$ is the zero section of $$L$$. By topology, we get a classifying map \begin{align*} C \xrightarrow{\phi_L} B{\mathbb{C}}^{\times}= {\mathbb{CP}}^\infty = ({\mathbb{C}}^{\infty} \setminus\left\{{0}\right\}) / {\mathbb{C}}^{\times} .\end{align*} There is a universal $$c_1\in H^2({\mathbb{CP}}^{\infty}; {\mathbb{Z}})$$, so we take the pullback to define $$c_1(L) \coloneqq\phi_L^*(c_1)$$. We can use that there is a cell decomposition $${\mathbb{CP}}^{\infty } = {\mathbb{C}}^0 \cup{\mathbb{C}}^1 \cup{\mathbb{C}}^2 \cup\cdots$$, and so there is a unique generator in its $$H^2$$. \item Consider a smooth section $$s\in C^{\infty }(L)$$, then we can define $$c_1(L) \coloneqq[ \left\{{ s = 0 }\right\} ]$$ by taking the fundamental class, assuming that $$s$$ is transverse to the zero section $$s_z$$ of $$L$$. Here we view the zero set as an oriented submanifold. See picture: in this case $$[\left\{{ s = 0 }\right\} ] = [p] - [q] + [r]$$. \end{enumerate} \todo[inline]{Add picture.} \end{remark} \begin{remark} Applying Serre duality to the left-hand side in Riemann-Roch yields the dimension of the space of holomorphic sections of some \emph{other} bundle, $$L {}^{ \vee }\otimes K$$. \end{remark} \begin{example}[The structure sheaf] Applying Riemann-Roch to $$L \coloneqq{\mathcal{O}}$$, we get \begin{align*} \chi({\mathcal{O}}) = h^0({\mathcal{O}}) - h^1({\mathcal{O}}) = 0 + 1 - g ,\end{align*} which is equal to $$h^0({\mathcal{O}}) - h^0(K)$$. But the only holomorphic functions on $${\mathbb{C}}$$ are constant, so $$h^0({\mathcal{O}}) = 1$$. In particular, $$h^0(K) = g$$, so any Riemann surface of genus $$g$$ has a $$g{\hbox{-}}$$dimensional space of holomorphic 1-forms. \end{example} \begin{example}[The Canonical Bundle] Applying Riemann-Roch to $$L\coloneqq K$$, we get \begin{align*} \chi(K) = h^0(K) - h^0(K {}^{ \vee }\otimes K) = \deg(K) + 1 - g .\end{align*} Since $$K {}^{ \vee }\otimes K = {\mathcal{O}}$$, we obtain $$g-1 = \deg(K) + 1 - g$$, so $$\deg(K) = 2g-2$$. We also proved this using that $$K$$ was the dual of holomorphic vector fields, i.e.~$$\int_C c_1(K) = -\int_C c_1(T)$$, which by Gauss-Bonnet equals $$-\chi_{\mathsf{Top}}(C) = -(2-2g) = 2g-2$$. \end{example} \begin{example}[Genus 2 Riemann Surfaces] Taking $$C$$ of genus 2, we have $$h^0(K_C) = g= 2$$, so $$\deg K_C = 2(2) - 2 = 2$$. Thus there exist linearly independent sections $$s, t \in H^0(K_C)$$, i.e.~two linearly independent holomorphic 1-forms. We can take the ratio $$s/t$$, which defines a map \begin{align*} {s\over t}: C\to {\mathbb{P}}^1 .\end{align*} Locally we have $$s = f(z) \,dz$$ for $$z$$ a local holomorphic coordinate on $$C$$ and $$f\in \mathop{\mathrm{Hol}}(C, {\mathbb{C}})$$, and similarly $$t = g(z) \,dz$$. So $$s/t = f(z) / g(z)$$ is meromorphic in this chart. Choosing a new coordinate chart $$w$$, this yields a transition function $$z(w)$$ -- not of $$L$$, but from the atlas on $$C$$. We can write $$s =f(z(w)) \, d(z(w)) = f(z(w)) z'(w) \, dw$$ by the chain rule. Thus \begin{align*} {s\over t}(z) = {f(z(w)) z'(w) \,dw\over g(z(w)) z'(w) \,dw} = {s \over t}(w) .\end{align*} So although $$s/t$$ was only defined in a coordinate chart, it winds up being independent of coordinates. This works in general for any holomorphic line bundle: for $$s, t\in H^0(L)$$, there is a map $${s\over t}: C\to {\mathbb{P}}^1$$ since writing $$s_V = \varphi_{UV} s_U, t_V = \varphi_{UV} t_U$$ where $$\varphi_{UV}$$ is the transition function for $$L$$. \begin{fact} Important fact: we can take these ratios to get maps to $${\mathbb{P}}^1$$. \end{fact} \begin{slogan} The canonical bundle is the line bundle whose transition functions are the Jacobians of the change of variables for the atlas. \end{slogan} \begin{question} What is the degree of this map generically? I.e. given $$[x_0: x_1] \in {\mathbb{P}}^1$$ fixed, what is the size of the inverse image $$\qty{s\over t}^{-1}\qty{ [x_0: x_1] }$$? \end{question} \begin{answer} Writing $$s/t = x_1/x_0$$, we have $$x_0 s - x_1 t=0$$. This is in $$H^0(K_C)$$, and we computed $$\deg K_C = 2$$, meaning there are two zeros of this function. Thus is $$g(C) = 2$$, there is a generically 2-to-1 map $$C \to {\mathbb{P}}^1$$, a degree 2 meromorphic function. Note that this section could have a double zero. \end{answer} \end{example} \begin{example}[?] Consider the curve $$y^2 = (z-1)(z-2)\cdots (z-5)$$, where we think of $$z, y\in {\mathbb{P}}^1$$. This has roots $$z=1,\cdots, 5$$, and is equal to $$\infty$$ if $$z=\infty$$. These are the only points of $${\mathbb{P}}^1$$ with just one square root, all other points have two square roots. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-19_14-37.pdf_tex} }; \end{tikzpicture} } \end{figure} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-19_14-40.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \hypertarget{monday-march-22}{% \section{Monday, March 22}\label{monday-march-22}} \begin{remark} Last time: we reviewed Riemann-Roch, Serre duality, sheaves of $$p{\hbox{-}}$$forms. Recall a theorem from a few weeks ago: \end{remark} \begin{theorem}[The Hodge Theorem] If $$(X,g)$$ is a compact oriented Riemannian manifold, then there is a decomposition of the smooth $$p{\hbox{-}}$$forms on $$X$$: \begin{align*} \Omega^p(X) = d \Omega^{p-1}(X) \oplus {\mathcal{H}}^p(X) + d^\dagger \Omega^{p+1}(X) .\end{align*} \end{theorem} \begin{remark} Note that $$\mathcal{H}$$ was the space of harmonic $$p{\hbox{-}}$$forms, and $$d^\dagger: \coloneqq(-1)^? \star d\star$$ where \begin{align*} \star: \Omega^{p}(X) &\to \Omega^{n-p}(X) \\ e_{i_1} \wedge \cdots \wedge e_{i_p} &\mapsto \pm e_{j_1} \wedge \cdots e_{j_{n-p}} \end{align*} where $$\left\{{ e_i}\right\}$$ is an orthonormal basis of basis of $$T {}^{ \vee }X$$. Note that this formula is replacing the $$e_i$$ that do appear with the $$e_i$$ that don't appear, up to a sign. The harmonic forms were defined as $${\mathcal{H}}^p(X) = \ker (dd^\dagger + d^\dagger d ) = \ker (d) \cap\ker(d^\dagger)$$. We proved that assuming this decomposition, there is an isomorphism \begin{align*} {\mathcal{H}}^p(X) \cong H^p_{\mathrm{dR}}(X; {\mathbb{R}}) .\end{align*} \end{remark} \begin{example}[The circle $S^1$] There's a standard flat metric $$g_\text{std}$$ on $$S^1$$ where $$g_\text{std}= \,dx^2$$ with $$x$$ the coordinate on $${\mathbb{R}}$$ which is the universal cover of $$S^1$$. We can write \begin{align*} \Omega^1(S^1) = \left\{{ f(x)\,dx{~\mathrel{\Big|}~}f \in C^{\infty }(S^1, {\mathbb{R}}) }\right\} ,\end{align*} since every 1-form $$\omega$$ looks like this. Then $$d \omega = 0$$ since this is a 2-form on $$S^1$$. On the other hand, what is $$d^\dagger$$? We know that $$\star\omega$$ is a 0-form, so a function. The volume form is given by $$\sqrt{ \operatorname{det}g_\text{std}} = \sqrt{ [\,dx^2 ] }$$, and you can wedge $$1\wedge dx = dx$$, so $$\star\omega = f(x)$$. Then $$d \star\omega = f'(x) \,dx$$ and $$d^\dagger x \omega = f'(x)$$. If this is zero, $$f'(x) = 0$$ and $$f$$ is a constant function. So in this metric, $${\mathcal{H}}^1(S^1) = {\mathbb{R}}\left\langle{ \,dx}\right\rangle \cong H^1(S^1; {\mathbb{R}})$$. \end{example} \begin{remark}[Important] The harmonic forms $${\mathcal{H}}^p(X)$$ depend on the metric $$g$$, despite mapping isomorphically to de Rham cohomology. \end{remark} \begin{remark} This was just in the case of a real smooth Riemannian manifold. What extra structure to we have for $$X \in {\mathsf{Mfd}}(\mathop{\mathrm{Hol}}({-}, {\mathbb{C}}) )$$? \end{remark} \begin{definition}[Kähler Forms (Important!)] Let $$X\in {\mathsf{Mfd}}( \mathop{\mathrm{Hol}}({-}, {\mathbb{C}}) )$$ be a complex manifold. A \textbf{Kähler form} $$\omega\in \Omega^2(X_{\mathbb{R}})$$ is a closed real (possibly needed: $$J{\hbox{-}}$$invariant) 2-form on the underlying real manifold of $$X$$ for which $$\omega(v, Jw) \coloneqq g(v, w)$$ is a metric on $$TX_{\mathbb{R}}$$ where $$J$$ is an almost complex structure. The associated \textbf{hermitian metric} is $$h\coloneqq g + i \omega$$, which defines a hermitian form on $$TX \in { \mathsf{Vect} }_{\mathbb{C}}$$. \end{definition} \begin{example}[?] Take $$X \coloneqq{\mathbb{C}}^n$$ and $$J(v) \coloneqq i\cdot v$$. Note that $$X_{\mathbb{R}}= {\mathbb{R}}^{2n}$$, so write its coordinates as $$x_k, y_k$$ for $$k = 1, \cdots, n$$ where $$z_k = x_k + iy_k$$ are the complex coordinates. Consider $$g = g_\text{std}$$ on $${\mathbb{R}}^{2n}$$ -- does this come from a closed 2-form $$g_\text{std}= \sum (\,dx_k)^2 + (dy_k)^2$$? Using $$\omega(v, Jw) = g(v, w)$$, we have $$\omega(v, J^2 w) = g(v, Jw)$$. The left-hand side is equal to $$- \omega(v, w)$$ and the right-hand side is $$\omega(v, w) = -g(v, Jw)$$. What 2-form does this give? We have \begin{align*} \omega\qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial x_\ell}\,} } &= -g \qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial y_\ell}\,} } = 0 \\ \omega\qty{ {\frac{\partial }{\partial y_k}\,}, {\frac{\partial }{\partial x_\ell}\,} } &= -g \qty{ {\frac{\partial }{\partial y_k}\,}, {\frac{\partial }{\partial y_\ell}\,} } = 0 \\ \omega\qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial y_\ell}\,} } &= -g \qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial y_\ell}\,} } = 0 && \forall k\neq \ell \\ \omega\qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial y_k}\,} } &= -g \qty{ {\frac{\partial }{\partial x_k}\,}, {\frac{\partial }{\partial y_k}\,} } \\ &= (-1)^2 g \qty{ {\frac{\partial }{\partial x_k}\,} , {\frac{\partial }{\partial x_k}\,} } \\ &= 1 \\ \omega\qty{ {\frac{\partial }{\partial y_k}\,}, {\frac{\partial }{\partial x_k}\,} } &= -1 .\end{align*} So we can write this in block form using blocks \begin{align*} M = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} && \omega = \begin{bmatrix} M & & \\ & M & \\ & & M \end{bmatrix} ,\end{align*} which is a closed ($$d\omega = 0$$) antisymmetric 2-form, i.e.~a symplectic form, and \begin{align*} \omega_\text{std}= dx_1 \wedge dy_1 + dx_2 \wedge dy_2 + \cdots + dx_n \wedge dy_n ,\end{align*} \end{example} \begin{remark} So the Kähler geometry is determined by the data $$({\mathbb{C}}^n, g_\text{std}, J, \omega_\text{std})$$, i.e.~a metric, an almost complex structure, and a symplectic form. Note that the relation $$\omega(x, y) = g(x, Jy)$$ can be used to determine the 3rd piece of data from any 2. This is the fiberwise/local model, i.e.~every tangent space at a point looks like this. \end{remark} \begin{warnings} But note that a form being closed is not a tensorial property! So this local data (looking at a single fiber) is not quite enough to determine the global geometry. \end{warnings} \begin{remark} Given $$g$$ and $$J$$, $$\omega$$ is automatically a 2-form. That it's antisymmetric follows from \begin{align*} -\omega(w, v) &= -g(w, Jv) \\ &= -g(Jv, w) \\ &= -g(J^2 v, Jw)\\ &= g(v, Jw)\\ &= \omega(v, w) .\end{align*} Conversely, we can always define $$g(v, w) \coloneqq- \omega(v, Jw)$$, but a priori this may not be a metric. This will be symmetric, but potentially not positive-definite. \end{remark} \begin{definition}[$\omega\dash$tame almost complex structures] An almost complex structure $$J$$ is \textbf{$$\omega{\hbox{-}}$$tame} if $$g(v, w) = - \omega(v, Jw)$$ is positive definite. \end{definition} \begin{remark} Next time: we'll see that if $$X$$ is Kähler, then \begin{align*} {\mathcal{H}}^k(X) = \bigoplus_{p+q=k} {\mathcal{H}}^{p, q}(X), \end{align*} so this is compatible with the Hodge decomposition. This is what people usually call the Hodge decomposition theorem, and gives some invariants of complex manifolds. By a miracle, this decomposition only depends on $$g$$ and the complex structure. \end{remark} \begin{remark} Note that there is a notion of \emph{hyperkähler} manifolds, which have 3 complex structures $$I, J, K$$ such that $$I^2=J^2=K^2 = IJK = -\one$$, yielding 3 parallel'' 2-forms $$\omega_I, \omega_J, \omega_K$$ such that the covariant derivative vanishes, i.e.~$$\nabla_g \left\{{ \omega_I, \omega_J, \omega_K }\right\} = 0$$. With respect to the complex structure $$I$$, $$\omega_J + \omega_K$$ is a holomorphic 2-form. There is a sphere's worth of almost complex structures, and there is an action $${\operatorname{SO}}(4, b_2 - 4) \curvearrowright H^*(X)$$. There's no known example where the hyperkähler metric has been explicitly written down. \end{remark} \hypertarget{wednesday-march-24}{% \section{Wednesday, March 24}\label{wednesday-march-24}} \begin{remark} Last time: we defined a \textbf{Kähler manifold}: $$X\in {\mathsf{Mfd}}({\mathbb{C}})_{ \operatorname{compact} }$$ and $$\omega \in \Omega^2(X_{\mathbb{R}})$$ a closed real 2-form such that $$g(x, y) \coloneqq\omega(x, Jy)$$ is a metric. By the Hodge theorem, we have a space $${\mathcal{H}}^k(X)$$ of harmonic $$k{\hbox{-}}$$forms for $$(X, g)$$ which represents $$H^k_{\mathrm{dR}}(X; {\mathbb{R}})$$. We can consider the $${\mathbb{C}}{\hbox{-}}$$valued harmonic forms $${\mathcal{H}}^k_{\mathbb{C}}\coloneqq{\mathcal{H}}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}$$, which represents $$H^k_\mathrm{dR}(X; {\mathbb{C}})$$ \end{remark} \begin{question} How does this interact with the decomposition of the smooth $$k{\hbox{-}}$$forms \begin{align*} \Omega^k(X_{\mathbb{R}})\otimes_{\mathbb{R}}{\mathbb{C}}= \bigoplus_{p+q=k}^K A^{p, q}(X) ,\end{align*} where $${\mathcal{H}}^k_{\mathbb{C}}(X)$$ is contained in this. Note that this is a small finite dimensional space in an infinite dimensional space! The following miracle occurs: \end{question} \begin{theorem}[Kähler manifolds admit a Hodge decomposition?] If $$X \in {\mathsf{Mfd}}(\operatorname{Kähler})$$, \begin{align*} {\mathcal{H}}^k_{\mathbb{C}}= \bigoplus_{p+q = k} {\mathcal{H}}^{p, q}(X) ,\end{align*} where \begin{align*} {\mathcal{H}}^{p, q}(X) \coloneqq \qty{ {\mathcal{H}}^K(X) \otimes_{\mathbb{R}}{\mathbb{C}}} \cap A^{p, q}(X) \subseteq \Omega^k(X_{\mathbb{R}}) .\end{align*} \end{theorem} \begin{example}[?] Let $$X = {\mathbb{C}}/ \Lambda$$ be an elliptic curve where $$\Lambda$$ is a lattice. The standard metric $$dx^2 + dy^2$$ on $${\mathbb{C}}$$ descends to a metric on $$X$$ since translation is an isometry on the metric space $$({\mathbb{C}}, dx^2 + dy^2)$$. Let $$z=x+iy$$ be a complex coordinate on $${\mathbb{C}}$$ so $$dz = dx + idy$$ and $$d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = dx - idy$$, then $$dx^2 + dy^2 = dz d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \in \operatorname{Sym}^2({\mathsf{T}}{\mathbb{C}})$$. The symplectic form is given by \begin{align*} \omega(v, w) = \pm g(v, Jw) = i \,dz\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu (v, w) \end{align*} since $$J$$ is given by $$i$$ on $${\mathbb{C}}$$. Then $$\omega(v, w) = i \,dz(v) \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu (w)$$, i.e.~$$\omega = i\,dz\wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$. So \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu \omega\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mui\mkern-1.5mu}\mkern 1.5mu \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \wedge \,dz= -i \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \wedge \,dz= i\,dz\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = \omega , \end{align*} and this determines the Kähler geometry on $$X$$. What are the harmonic 1-forms on $$X$$, $${\mathcal{H}}^1(X) \otimes_{\mathbb{R}}{\mathbb{C}}$$? Note that $$\omega= \,dV$$ is the volume form. The smooth 1-forms are given by \begin{align*} \Omega^1(X_{\mathbb{R}}) \otimes_{\mathbb{R}}{\mathbb{C}}= A^{1, 0}(X) \oplus A^{0, 1}(X) = \left\{{ f(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu )\,dz}\right\} \oplus \left\{{ g(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\right\} ,\end{align*} where $$f,g$$ are smooth and $$\Lambda{\hbox{-}}$$periodic on $${\mathbb{C}}$$ to make them well-defined. We can find the Hodge star: \begin{align*} \star: ? &\to ? \\ \,dz& \mapsto i\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \\ \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu &\mapsto -i\,dz .\end{align*} Writing $$\alpha\coloneqq f(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\,dz+ g(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$, this is harmonic if $$d \alpha = 0$$ and $$\star{d} \mkern-5mu \star\alpha = 0$$. The first implies $$\partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} f - \partial_{z} g = 0$$. What does the second imply? We can compute \begin{align*} \star\alpha &= if(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu ) \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu - ig(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) \,dz\\ \implies \partial_z f + \partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} g &= 0 ,\end{align*} and so $$\partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} f = \partial_z g$$ and $$\partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}^2 f = \partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} \partial_z g = - \partial_z^2 f$$, so \begin{align*} \qty{ \partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}^2 + \partial_z^2 }f &= 0 \\ \qty{ \partial_{\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}^2 + \partial_z^2 }g &= 0 .\end{align*} Note that this recovers the usual notion of harmonic functions on $${\mathbb{C}}$$, i.e.~being in the kernel of the Laplacian. The only biperiodic functions that satisfy these equations are constants, since there is a maximum modulus principle for harmonic functions. Thus \begin{align*} {\mathcal{H}}^1(X) \otimes_{\mathbb{R}}{\mathbb{C}}= \left\{{ c_1 \,dz+ c_2 \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\right\} = {\mathbb{C}}\,dz\oplus {\mathbb{C}}\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = H^{1, 0}(X) \oplus H^{0, 1}(X) .\end{align*} \end{example} \begin{remark} There is a generalization to higher genus curves. Recall the following theorem: \end{remark} \begin{theorem}[Uniformization] Let $$C \in {\mathsf{Mfd}}^1({\mathbb{C}})_{ \operatorname{compact} }$$ of genus $$g\geq 2$$. Then the universal cover admits a biholomorphism \begin{align*} \tilde C \cong {\mathbb{H}}\coloneqq\left\{{ z\in {\mathbb{C}}{~\mathrel{\Big|}~}\Im(z) > 0 }\right\} .\end{align*} \end{theorem} \begin{remark} This essentially follows from the Riemann mapping principle. \end{remark} \begin{corollary}[Every curve of genus g>1 is the plane mod a subgroup of biholomorphisms] Any curve $$C$$ of genus $$g\geq 2$$ is of the form $$C = {\mathbb{H}}/ \Gamma$$ where $$\Gamma \leq \mathop{\mathrm{BiHol}}({\mathbb{H}})$$ is a subgroup that acts freely. By covering space theory, $$\Gamma = \pi_1(C)$$, and it's known that $$\mathop{\mathrm{BiHol}}({\mathbb{H}}) \cong {\operatorname{PSL}}_2({\mathbb{R}})$$ by the map \begin{align*} \begin{bmatrix} a & b \\ c & d \end{bmatrix} z \mapsto {az + b \over cz + d} .\end{align*} \end{corollary} \begin{proposition}[The upper half-plane admits a PSL-invariant hyperbolic metric] The upper half plane $${\mathbb{H}}$$ admits a \textbf{hyperbolic metric} which is invariant under $${\operatorname{PSL}}_2({\mathbb{R}})$$ given by \begin{align*} g_{{\operatorname{hyp}}} = {dx^2 + dy^2 \over y^2 } = {\,dz\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over \Im(z)^2 } .\end{align*} \end{proposition} \begin{proof}[?] This follows from a computation: \begin{align*} d\qty{ az + b \over cz + d} &= {a\,dz\over cz + d} - {c (az+b)\,dz\over (dz+d)^2 } \\ &= {a (cz+d) - c(az+b) \,dz\over (cz+d)^2} \\ &= { (ad-bc)\,dz\over (cz + d)^2 } \\ &= {\,dz\over (cz + d)^2 } \\ &= { d\qty{ az+b \over cz+d} d \qty{\mkern 1.5mu\overline{\mkern-1.5mu az + b \over cz + d\mkern-1.5mu}\mkern 1.5mu} \over \Im\qty{az+b \over cz+d}^2 } \\ &= { \,dz\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over (cz+d)^2(c \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu + d)^2 \Im\qty{az+b \over cz+d} } \\ &= { \,dz\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \over \Im(z)^2 } .\end{align*} \end{proof} \begin{remark} It's miraculous! The biholomorphisms of $${\mathbb{H}}$$ preserve a metric. So $$C$$ has a canonical metric, $$g_{\operatorname{hyp}}$$, which descends along the quotient map $${\mathbb{H}}\to {\mathbb{H}}/\Gamma \cong {\mathbb{C}}$$. \end{remark} \begin{question} What are the harmonic 1-forms on $$(C, g_{\operatorname{hyp}})$$? \end{question} \begin{remark} By lifting we can write \begin{align*} \Omega^1(C_{\mathbb{R}}) \otimes_{\mathbb{R}}{\mathbb{C}}= A^{1, 0}(C) \oplus A^{0, 1}(C) = \left\{{ f(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\,dz+ g(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu {~\mathrel{\Big|}~}z\in {\mathbb{H}}, \, f,g\in C^{\infty }({\mathbb{C}}, {\mathbb{R}}) }\right\} \end{align*} But $$\,dz$$ is \emph{not} invariant under the map $$z\mapsto {az+b \over cz+d}$$, since $$\,dz\mapsto {\,dz\over (cz+d)^2 }$$. In order to descend $$f(z)$$ to $$C$$, we need \begin{align*} f\qty{az +b \over cz + d} = (cz+d)^2 f(z) && \text{ for all } \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \Gamma \end{align*} This says that $$f$$ is a \textbf{modular form of weight 2}. \end{remark} \begin{exercise}[?] Check that this implies that $$f$$ must be holomorphic and $$g$$ must be antiholomorphic. \end{exercise} \begin{fact} There is a decomposition \begin{align*} {\mathcal{H}}^1(C_{\mathbb{R}}) \otimes_{\mathbb{R}}{\mathbb{C}}= {\mathcal{H}}^{1, 0}(C) \oplus {\mathcal{H}}^{0, 1}(C) ,\end{align*} and the first space will be the space of holomorphic 1-forms $$H^0(K_C)$$, and the second term will be $$\mkern 1.5mu\overline{\mkern-1.5muH^0(K_C)\mkern-1.5mu}\mkern 1.5mu$$. This shows the power of the Hodge decomposition theorem! \end{fact} \hypertarget{friday-march-26th}{% \section{Friday, March 26th}\label{friday-march-26th}} \begin{remark} Recall the Hodge decomposition theorem. Let $$(M, g) \in {\mathsf{Mfd}}_{\mathbb{R}}^n(\mathsf{Riem}, { \operatorname{compact} } )$$, then choosing an orthonormal basis $$\left\{{ v_j }\right\}$$ for $$T_p M$$ yields a corresponding orthonormal basis in $$T_p {}^{ \vee }M \coloneqq\mathop{\mathrm{Hom}}_{\mathbb{R}}(T_p M, {\mathbb{R}})$$ given by taking $$\left\{{ e_i {~\mathrel{\Big|}~}e_i(v_j) = \delta_{ij} }\right\}$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-15_23-21.pdf_tex} }; \end{tikzpicture} } \end{figure} There is a map \begin{align*} \star: \bigwedge^k T_p {}^{ \vee }M &\to \bigwedge^{n-k} T_p {}^{ \vee }M \\ \bigwedge_{j=1}^k e_{i_j} &\mapsto \pm \bigwedge_{\ell = 1}^{n-k} e_{j_\ell} \end{align*} where the $$e_{j}$$ are defined such that $$\bigwedge_{j=1}^k e_{i_j} \wedge \bigwedge_{\ell = 1}^{n-k} e_{j_\ell} \coloneqq\,dV$$, where $$\,dV$$ is the volume form on $$M$$ at $$p$$. Thus we have a map \begin{align*} \star: \Omega^k &\to \Omega^{n-k} \\ 1 &\mapsto \,dV .\end{align*} We defined $$d^\dagger \coloneqq\star{d} \mkern-5mu \star$$, and said a form $$\omega$$ was \emph{harmonic} iff $$\Delta \omega=0$$, where $$\Delta \coloneqq dd^\dagger + d^\dagger d$$. The space of such forms was denoted $$\mathcal{H}^k(M) \subseteq \Omega^k(M)$$. \end{remark} \begin{theorem}[Hodge Theorem] \begin{align*} \mathcal{H}^k(M) \cong H^k_{\mathrm{dR}}(M; {\mathbb{R}}) .\end{align*} \end{theorem} \begin{question} What kinds of extra structure can we put on a complex manifold? \end{question} \begin{definition}[Kähler Form] A \textbf{Kähler form} is a closed 2-form $$\omega\in \Omega^2_{\mathbb{R}}$$ such that the following equation defines a metric on $$T_p M$$: \begin{align*} g(u, v) \coloneqq\omega(u, iv) .\end{align*} I.e., this is a closed symplectic form that defines a metric. \end{definition} \begin{example}[?] Consider $$M = {\mathbb{C}}^n$$ with holomorphic coordinates $${ {z}_1, {z}_2, \cdots, {z}_{n}}$$, where $$z_{j} \coloneqq x_j + iy_j$$. Then take \begin{align*} \omega \coloneqq\sum_{j=1}^n \,dx_j \wedge \,dy_j .\end{align*} Note that multiplication by $$i$$ induces a map \begin{align*} \cdot i: T_p {\mathbb{C}}^n &{\circlearrowleft}\\ {\frac{\partial }{\partial x_j}\,} &\mapsto {\frac{\partial }{\partial y_j}\,} \\ {\frac{\partial }{\partial y_j}\,} &\mapsto - {\frac{\partial }{\partial x_j}\,} \\ .\end{align*} Moreover, $$\omega(u, iv)$$ recovers the standard metric on $${\mathbb{C}}^n$$ given by \begin{align*} g_{\text{std}} = \sum (\,dx_j)^2 + (\,dy_j)^2 \in \operatorname{Sym}^2 T {}^{ \vee }{\mathbb{C}}^n ,\end{align*} which is incidentally positive-definite, where $$(\,dx)^2(u, v) \coloneqq({\frac{\partial }{\partial x_j}\,})u \cdot *({\frac{\partial }{\partial y_j}\,}) v$$. Is this closed? We need to check to see if $$d\omega = 0$$, but this is true: applying $$d$$ to all of the coefficients yields the constant 1. \end{example} \begin{remark} So for $$M\in {\mathsf{Mfd}}({\mathbb{C}})$$ a complex manifold, we have a decomposition \begin{align*} \Omega^k(M) &= \bigoplus_{p+q=k} A^{p, q}(M) \\ \\ A^{p, q} &\coloneqq\left\{{ \sum_{\substack{ {\left\lvert {I} \right\rvert} = p \\ {\left\lvert {J} \right\rvert} = q}} \qty{\,dz_{i_1} \wedge \cdots \,dz_{i_p} } \wedge \qty{ \,dz_{j_1} \wedge \cdots \,dz_{j_q} } }\right\} .\end{align*} For $$M$$ a Kähler manifold, we have \begin{align*} \mathcal{H}^k(M) = \bigoplus _{p+q = k} \mathcal{H}^{p, q}(M) \\ \\ \mathcal{H}^{p, q}(M) = \mathcal{H}^k(M) \cap A^{p, q}(M) .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-17_19-13.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{remark} Why is this true? We have a map \begin{align*} d: A^{p, q}(M) \to A^{p+1, q}(M) \oplus A^{p, q+1}(M) ,\end{align*} where for example if $$f(z) \coloneqq z \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \in A^{0, 0}({\mathbb{C}})$$, we have $$df = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \,dz+ z\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu$$ where the first is a $$(1, 0)$$ form and the latter is a $$(0, 1)$$ form. Write $$d = {\partial}+ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$ where $${\partial}\coloneqq\sum \,dz_j$$ and $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= \sum \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _j$$, as well as \begin{align*} d^\dagger: A^{p, q}(M) \to A^{p-1, q}(M) \oplus A^{p, q-1}(M) .\end{align*} Now $$\star$$ of a $$(p, q)$$ form is an $$(n-p, n-q)$$ form, and so \begin{align*} \star\qty{ \,dz_{i_1} \wedge \cdots \wedge \,dz_{i_r} \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _{j_1} \wedge \cdots \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _q } \coloneqq\star(\,dz_I \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _J) = \pm \,dz_{I^c} \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _{J^c} ,\end{align*} and we have $$d^\dagger = {\partial}^\dagger + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger$$. We can thus move around the bigraded group in several ways: \begin{center} \begin{tikzcd} {A^{2, 0}} && {A^{2, 1}} && {A^{2, 2}} \\ \\ {A^{1, 0}} && {A^{1, 1}} && {A^{1, 2}} \\ \\ {A^{0, 0}} && {A^{0, 1}} && {A^{0, 2}} \arrow["{\partial}", dashed, from=5-1, to=3-1] \arrow["{\partial}", dashed, from=3-1, to=1-1] \arrow["{\partial}", dashed, from=3-3, to=1-3] \arrow["{\partial}", dashed, from=5-5, to=3-5] \arrow["{\partial}", dashed, from=3-5, to=1-5] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=1-1, to=1-3] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=1-3, to=1-5] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=3-1, to=3-3] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=3-3, to=3-5] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=5-1, to=5-3] \arrow["{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}", dashed, from=5-3, to=5-5] \arrow["{\partial}", dashed, from=5-3, to=3-3] \arrow["{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-18pt}, from=5-3, to=5-1] \arrow["{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger}", color={rgb,255:red,214;green,92;blue,92}, curve={height=-18pt}, from=5-5, to=5-3] \arrow["{{\partial}^\dagger}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, from=1-1, to=3-1] \arrow["{{\partial}^\dagger}"', color={rgb,255:red,92;green,92;blue,214}, curve={height=24pt}, from=3-1, to=5-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to Diagram} \end{quote} \end{remark} \begin{theorem}[Kähler Identities] Let \begin{align*} \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} &\coloneqq{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\\ \Delta_{\partial}&\coloneqq{\partial}{\partial}^\dagger + {\partial}^\dagger {\partial}\\ \Delta_d &\coloneqq dd^\dagger + d^\dagger d .\end{align*} Then \begin{align*} {1\over 2} \Delta_d = \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} = \Delta_{\partial} .\end{align*} \end{theorem} \begin{remark} See Griffiths-Harris for details. Note that this is a local statement, i.e.~it can be checked in coordinate charts. \end{remark} \begin{remark} The upshot: \begin{align*} \mathcal{H}^k(M) = \ker \Delta_d = \ker \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} ,\end{align*} and moreover \begin{align*} \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}: A^{p, q}(M) {\circlearrowleft} \end{align*} which implies that on $$\Omega^k(M)$$, \begin{align*} \ker \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \circ \bigoplus _{p+q=k} \ker \qty{ A^{p, q}(M) \xrightarrow{\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}} A^{p, q}(M) } &= \bigoplus _{p+q = k} \ker \Delta_d ,\end{align*} which yields the Hodge decomposition theorem \begin{align*} \mathcal{H}^k(M) = \bigoplus _{p+q=k} \mathcal{H}^{p, q}(M) .\end{align*} \end{remark} \begin{remark} This is a strong restriction on what manifolds can admit a Kähler structure. Moreover, since $$\Delta_d$$ is a real operator, we obtain $$\mkern 1.5mu\overline{\mkern-1.5mu \mathcal{H}^{p, q}(M) \mkern-1.5mu}\mkern 1.5mu \cong \mathcal{H}^{p, q}(M)$$. \end{remark} \begin{remark} Some consequences: For $$M$$ a Kähler manifold, the odd Betti numbers $$\beta_{2i+1}(M) \coloneqq\dim H_{\mathrm{dR}}^{2i+1}(M; {\mathbb{C}})$$ are even. This is because \begin{align*} \bigoplus _{p+q=k} \mathcal{H}^{p, q} \cong \mathcal{H}^{2i+1}(M) \cong H_{\mathrm{dR}}^{2i+1}(M) .\end{align*} If we define $$h^{p, q}(M) \coloneqq\dim_{\mathbb{C}}\mathcal{H}^{p, q}(M)$$, we clearly have \begin{align*} \beta_{2i+1} = \sum_{p+q = 2i+1} h^{p, q}(M) .\end{align*} Now using that $$\mkern 1.5mu\overline{\mkern-1.5mu\mathcal{H} \mkern-1.5mu}\mkern 1.5mu \cong \mathcal{H}$$, we can rewrite this as \begin{align*} \beta_{2i+1} &= \sum_{p+q = 2i+1} h^{p, q}(M) \\ &= 2 \sum_{\substack{ p+q = 2i+1 \\ p < q} } h^{p, q}(M) .\end{align*} \end{remark} \begin{remark} Is this just some fact about arbitrary complex manifolds, with no extra structure? The answer is no, and the counterexample is the \emph{Hopf surface} \begin{align*} X \coloneqq\qty{ {\mathbb{C}}^2 \setminus\left\{{\mathbf{0}}\right\}} / (x,y)\sim (2x, 2y) ,\end{align*} which we can roughly identify as $${\mathbb{R}}^4$$ modulo doubling''. We can take a fundamental domain $$1\leq {\left\lvert {r} \right\rvert} \leq 3$$, this yields an annulus-like sphere with the inner shell glued to the outer: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-17_19-49.pdf_tex} }; \end{tikzpicture} } \end{figure} This is homeomorphic to $$S^1 \times S^3$$, but $$\beta_1(M) = 1$$, so this won't yield a Kähler structure. \end{remark} \hypertarget{monday-march-29}{% \section{Monday, March 29}\label{monday-march-29}} \begin{remark} Last time: the Hodge decomposition theorem. Let $$(X, g) \in {\mathsf{Mfd}}_{\mathbb{C}}^{ \operatorname{compact} } ({ \operatorname{Kähler} } )$$, then the space of harmonic $$k{\hbox{-}}$$forms $$\mathcal{H}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}$$ decomposes as $$\bigoplus_{p+q = k} \mathcal{H}^{p, q}(X)$$. There is also a symmetry $$\mkern 1.5mu\overline{\mkern-1.5mu\mathcal{H}^{p, q}(X) \mkern-1.5mu}\mkern 1.5mu = \mathcal{H}^{q, p}(X)$$. We have an isomorphism to the de Rham cohomology $$\mathcal{H}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}\cong H^k_\mathrm{dR}(X; {\mathbb{C}})$$. We know the constituent pieces as well, as well as several relationships: \begin{align*} \mathcal{H}^{p, q}(X) &= \ker (\Delta_d: A^{p, q}(X) {\circlearrowleft}) \\ \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} &= { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\\ \Delta_d &= 2 \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} .\end{align*} There was a proposition that $$\ker(\Delta_d) = \ker(d) \cap\ker(d^\dagger)$$, and the same proposition holds for $$\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}$$. In this case we have $$\ker(\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}) = \ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \cap\ker( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger)$$ on $$A^{p, q}(X)$$, and this is isomorphic to $$\ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) / \operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})$$. Recall that we resolved the sheaf $$\Omega^p$$ of holomorphic $$p{\hbox{-}}$$forms by taking the Dolbeault resolution \begin{align*} 0 \to \Omega^p \to A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 2} \to \cdots .\end{align*} Thus we can identify $$\ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})/\operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \cong { \mathcal{H} }( X; \Omega^p)$$ as sheaf cohomology. We defined $$h^{p, q}(X) \coloneqq\dim_{\mathbb{C}}H^{p, q}(X)$$. \end{remark} \begin{corollary}[Homology is independent of the choice of Kähler form] $$h^{p,q }(X)$$ is independent of the Kähler form, noting that the isomorphism to sheaf cohomology doesn't involve taking adjoints, and $$\dim_{\mathbb{C}}{ \mathcal{H} }^q(X; \Omega^p)$$ doesn't depend on the complex structure. \end{corollary} \begin{remark} A priori, one could vary the Kähler form and have some $$h^{p, q}$$ jump or drop dimension. It also turns out that varying the complex structure will also not change these dimensions. \end{remark} \begin{remark} Whenever the Hodge-de Rham spectral sequence degenerates, one generally gets $$\sum_{p+q} h^{p,q } = h^k$$. Note that there is a resolution: \begin{align*} 0 \to \underline{{\mathbb{C}}} \to {\mathcal{O}}\xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \cdots ,\end{align*} which is not acyclic and thus has homology. In general, the spectral sequence is \begin{align*} E^1_{p,q} = { \mathcal{H} }^q(X; \Omega^p) \Rightarrow{ \mathcal{H} }^{p+q}(X; \underline{{\mathbb{C}}}) .\end{align*} \end{remark} \begin{fact} A fact about the cohomology of vector bundles: given a family of Kähler manifolds $$X_t$$, one can consider $$H^q(X_t; \mathcal{E}_t$$ where $$\mathcal{E}_t$$ is a family of holomorphic vector bundles. This can only jump upward in dimension, i.e.~$$\dim_{\mathbb{C}}H^q(X_t; \mathcal{E}_t)$$ is \textbf{lower semicontinuous}. \end{fact} \begin{example}[?] Consider \begin{align*} X_t \coloneqq\left\{{ x^3 + y^3 + z^3 + txyz = 0 }\right\} \subseteq {\mathbb{CP}}^2 ,\end{align*} where $$t$$ varies in $${\mathbb{C}}$$. These all admit a line bundle $$\mathcal{L}_t \coloneqq{ \left.{{ {\mathcal{O}}(1) }} \right|_{{X_t}} }$$, the anti-tautological line bundle on $${\mathbb{P}}^2$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-29_14-14.pdf_tex} }; \end{tikzpicture} } \end{figure} The real points of this vanishing locus form an elliptic curve, and each $$X_t$$ is a Riemann surface of genus 1. Note that $$h^{0, 1}$$ can jump on closed sets, but $$H^1$$ is constant since Riemann-Roch involves genus and degree. What is $$\deg { \left.{{{\mathcal{O}}(1)}} \right|_{{X_t}} }$$? Take a section $$s \in H^0({\mathbb{P}}^2; {\mathcal{O}}(1))$$ which vanishes on a line in $${\mathbb{P}}^2$$. How many points lie in a line intersected with $$X_t$$? Looking at fundamental classes, we have $$[X_t] = 3\ell$$, and by Bezout $$3\ell \cdot \ell = 3$$. The point is that $$H^q(X_t; \Omega^p)$$ can only possibly increase at special values of $$t$$. Assuming the $$X_t$$ are all diffeomorphic, then $$h^k(X_t)$$ is constant and $$h^{p, q}(X_t)$$ can't jump. So the $$h^{p, q}$$ are invariants of families. \end{example} \begin{definition}[Hodge Diamond] The \textbf{Hodge Diamond} of $$X \in {\mathsf{Mfd}}(\operatorname{Kähler})$$ (which won't depend on the choice of Kähler form) is given by \begin{center} \begin{tikzcd} &&& {h^{n, n}} \\ && {h^{n-1, n}} && {h^{n, n-1}} \\ & \ddots &&&& \ddots \\ \ddots &&& \vdots &&& \ddots \\ & {h^{2, 0}} && {h^{1, 1}} && {h^{0, 2}} \\ && {h^{1, 0}} && {h^{0, 1}} \\ &&& {h^{0, 0}} \arrow["\star"{pos=0}, dotted, tail reversed, from=6-3, to=2-5] \arrow["{z\mapsto \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}"', dotted, tail reversed, from=6-5, to=2-5] \arrow["{z\mapsto \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu}", dotted, tail reversed, from=6-3, to=2-3] \arrow["\star"{description, pos=0.1}, dotted, tail reversed, from=6-5, to=2-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTQsWzMsNiwiaF57MCwgMH0iXSxbMiw1LCJoXnsxLCAwfSJdLFs0LDUsImheezAsIDF9Il0sWzEsNCwiaF57MiwgMH0iXSxbMyw0LCJoXnsxLCAxfSJdLFs1LDQsImheezAsIDJ9Il0sWzAsMywiXFxkZG90cyJdLFszLDAsImhee24sIG59Il0sWzIsMSwiaF57bi0xLCBufSJdLFs0LDEsImhee24sIG4tMX0iXSxbNiwzLCJcXGRkb3RzIl0sWzEsMiwiXFxkZG90cyJdLFs1LDIsIlxcZGRvdHMiXSxbMywzLCJcXHZkb3RzIl0sWzEsOSwiXFxob2RnZXN0YXIiLDAseyJsYWJlbF9wb3NpdGlvbiI6MCwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiYXJyb3doZWFkIn0sImJvZHkiOnsibmFtZSI6ImRvdHRlZCJ9fX1dLFsyLDksInpcXG1hcHN0byBcXGJhcnt6fSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6ImFycm93aGVhZCJ9LCJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XSxbMSw4LCJ6XFxtYXBzdG8gXFxiYXJ7en0iLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifSwiYm9keSI6eyJuYW1lIjoiZG90dGVkIn19fV0sWzIsOCwiXFxob2RnZXN0YXIiLDEseyJsYWJlbF9wb3NpdGlvbiI6MTAsInN0eWxlIjp7InRhaWwiOnsibmFtZSI6ImFycm93aGVhZCJ9LCJib2R5Ijp7Im5hbWUiOiJkb3R0ZWQifX19XV0=}{Link to Diagram} \end{quote} Note that there are symmetries, e.g.~$$\star$$ takes $$h^{1, 0} = h^{n-1, n}$$ and $$\mkern 1.5mu\overline{\mkern-1.5muh^{p, q}\mkern-1.5mu}\mkern 1.5mu = h^{q, p}$$. \end{definition} \begin{proposition}[CYs have extra Hodge diamond symmetry] If $$X$$ is \textbf{Calabi-Yau}, so $$K_X = {\mathcal{O}}_X$$ (i.e the canonical bundle is trivial), then the Hodge diamond has an orientation preserving $$({\mathbb{Z}}/2)^2$$ symmetry, i.e.~there is a rotation by $$\pi/2$$. \begin{quote} Note: this isn't extra symmetry! Just a proof of the symmetry in this case. \end{quote} \end{proposition} \begin{proof}[?] Let $$\Omega^k_X$$ be the sheaf of holomorphic $$k{\hbox{-}}$$forms, then there is a map \begin{align*} \Omega_X^k \otimes\Omega_X^{n-k} &\to \Omega_X^n \coloneqq K_X \\ \alpha \otimes\beta &\mapsto \alpha \wedge \beta .\end{align*} Fiberwise, this is a perfect pairing. If one takes $$\alpha \coloneqq e_{i_1} \wedge \cdots e_{i_k} \in \bigwedge^k T_x {}^{ \vee }X$$, there is a unique basis wedge $$\beta \coloneqq e_{j_1} \wedge \cdots \wedge e_{j_n - k}$$ then $$\alpha\wedge \beta$$ is a basis wedge $$e_1 \wedge \cdots \wedge e_n$$. So $$\Omega_X^k \cong ( \Omega_X^{n-k} ) {}^{ \vee }$$ if $$X$$ is Calabi-Yau. By Serre duality, \begin{align*} { \mathcal{H} }^p(X; \Omega_X^q) {}^{ \vee }\cong { \mathcal{H} }^{n-p}(X; (\Omega_X^q) {}^{ \vee }\otimes K_X ) .\end{align*} \end{proof} \begin{example}[?] In dimension 3, take \begin{align*} X \coloneqq\left\{{ x_0^5 + \cdots + x_4^5 = 0 }\right\} \subseteq {\mathbb{P}}^4 \in {\mathsf{Mfd}}^3({\mathbb{C}}) .\end{align*} See Hodge diamond. \end{example} \begin{remark} Note that $$K3$$s are special CYs. An example is $${\mathbb{C}}^2 / \Lambda$$ for $$\Lambda$$ a rank 4 lattice. This is diffeomorphic to $$(S^1)^4$$, for example $$E\times E$$. \end{remark} \hypertarget{wednesday-march-31}{% \section{Wednesday, March 31}\label{wednesday-march-31}} \hypertarget{polyvector-fields}{% \subsection{Polyvector Fields}\label{polyvector-fields}} \begin{remark} We have a perfect pairing \begin{align*} \Omega^k \otimes\Omega^{n-k} \to K ,\end{align*} and thus $$\Omega^{n-k} \cong K \otimes(\Omega^{k}) {}^{ \vee }$$. So we have \begin{align*} H^p( \Omega^k ) {}^{ \vee }\cong H^{n-p}( (\Omega^{k}) {}^{ \vee }\otimes K ) = H^{n-p}( \Omega^{n-k}) ,\end{align*} and thus $$h^{p, k} = h^{n-p, n-k}$$, which recovers what we knew about $$\star: \mathcal{H}^{p, q} \to \mathcal{H} ^{n-p, n-q}$$. So we don't get anything new from the Serre duality argument. What is special when $$X\in { \text{CY} }$$ is that \begin{align*} \Omega^{n-k} \cong ( \Omega^k ) {}^{ \vee }= \bigwedge^k TX \end{align*} for $$TX$$ the tangent bundle. Note that taking the cotangent bundle gives forms, and instead this gives a bundle of \emph{polyvector fields}. For $$k=1$$, we get a holomorphic vector field, which one might think of as an infinitesimal biholomorphism. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-31_13-58.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{example}[?] $${\mathbb{P}}^1$$ has a holomorphic vector field in coordinate charts $${\mathbb{C}}\cong \left\{{ [z: 1] \in {\mathbb{P}}^1 }\right\}$$ which we'll write as $$z{\frac{\partial }{\partial z}\,}$$. The coordinate chart is $${\mathbb{P}}^1 \setminus\infty$$, so we obtain \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-31_14-00.pdf_tex} }; \end{tikzpicture} } \end{figure} Does this vector field $$V$$ extend over $$\infty$$? The local coordinate at $$\infty$$ is $$w = 1/z$$, so $$z=1/w$$ and we can compute \begin{align*} {1\over w} {\frac{\partial }{\partial {1\over w} }\,} = {1\over w} {\partial \over {-1\over w^2} \partial w} = -w {\frac{\partial }{\partial w}\,} .\end{align*} We have $${\operatorname{Ord}}_0V = 1$$ and $${\operatorname{Ord}}_{\infty } V = 1$$, and so $$\deg T{\mathbb{P}}^1 = 2$$. \end{example} \begin{example}[?] For $$\bigwedge^2 T$$, the local sections are of the form $$\sum f_I {\frac{\partial }{\partial x_I}\,} \wedge {\frac{\partial }{\partial x_J}\,}$$ instead of e.g.~$${d\over d x_I}$$. This yields a \textbf{Poisson structure} $$H^0(X, \bigwedge^2 T)$$, which is a generalization of symplectic structure, which would be a section $$\omega \in H^0( X, \bigwedge^2 T {}^{ \vee })$$ which is nondegenerate. This would yield an isomorphism $$\omega: T\xrightarrow{\sim} T {}^{ \vee }$$ which is alternating, in which case $$\omega^{-1}: T {}^{ \vee }\xrightarrow{\sim} T$$ which is also alternating, so $$\omega ^{-1}\in H^0(X, \bigwedge^2 T)$$. However the Poisson structure need not be nondegenerate. \end{example} \begin{remark} Polyvector fields show up in Hochschild homology! \end{remark} \hypertarget{algebraic-surfaces}{% \subsection{Algebraic Surfaces}\label{algebraic-surfaces}} \begin{definition}[Algebraic Surface] An \textbf{algebraic surface} is a compact complex 2-fold (so of complex dimension and real dimension 4, admitting local charts to $${\mathbb{C}}^2$$) which admits a holomorphic embedding into $${\mathbb{CP}}^N$$ for some $$N$$. \end{definition} \begin{remark} This implies that $$S$$ is a \textbf{projective variety} cut out by homogeneous polynomials in $$N+1$$ variables in $${\mathbb{CP}}^N$$. \end{remark} \begin{example}[?] A non-example would be $${\mathbb{C}}^2 \setminus\left\{{ (0, 0) }\right\} / (x, y) \sim (2x, 2y)$$, The \emph{Hopf surface}. This is a complex manifold of complex dimension 2. It is compact, but has no projective embedding! \end{example} \begin{example}[?] Another non-example is $${\mathbb{C}}^2 \setminus\left\{{0}\right\}/ (x, y) \sim (2x, 2 e^{i\theta} y)$$, a \emph{twisted Hopf surface}. This admits no nontrivial holomorphic line bundles. \end{example} \begin{remark} What makes having a projective embedding special? If $$S \hookrightarrow{\mathbb{CP}}^N$$, it admits a line bundle: $${\mathcal{O}}_S(1) \coloneqq{ \left.{{ {\mathcal{O}}_{{\mathbb{CP}}^N}(1) }} \right|_{{S}} }$$. \end{remark} \begin{proposition}[Existence of the Fubini-Study form/metric] $${\mathbb{CP}}^N$$ is a Kähler manifold, and admits a distinguished 2-form $$\omega \coloneqq\omega_{\text{FS}}$$ the \textbf{Fubini-Study form} which induces the Fubini-Study metric $$g_{\text{FS}}$$. \end{proposition} \begin{remark} This can be written down as $${i\over 2} {\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\log( \sum_{i=1}^N z_i \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_i )$$, which is well-defined since scaling comes out as a constant. Being closed follows from $${\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= d{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}$$ since $${ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^2 = 0$$, which implies $$d({\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\cdots) = d^2 { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(\cdots) = 0$$. This defines a metric: this follows from checking in local coordinate charts, say $$z_0 = 1$$, and checking that $$g(x ,y) \coloneqq\omega(x, Jy)$$ yields a metric. This involves taking a fussy derivative! \end{remark} \begin{remark} Thus given $$S\xhookrightarrow{\phi} {\mathbb{CP}}^N$$, we can restrict or take the pullback of $$\omega_{{ \text{FS} }}$$ to $$S$$. Then $$\omega \coloneqq\phi^* \omega_{ \text{FS} }$$ is still Kähler: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item $$\omega$$ is closed: this is true for any smooth map at the level of smooth manifolds because of the chain rule. \item $$\omega$$ defines a metric: this is true because $$S$$ is a complex submanifold. Suppose $$v,w \in T_p S$$, and we want to check if $$g(v, w) \coloneqq\omega(v, Jw)$$. This equals $$\omega_{{ \text{FS} }}(v, JW)$$, viewing $$T_p S \subseteq T_p {\mathbb{CP}}^N$$, so this is equal to $$g_{ \text{FS} }(v, w)$$. \end{enumerate} \end{remark} \begin{remark} Note that a submanifold of a \emph{symplectic} manifold is not necessarily a symplectic submanifold, since there are Lagrangian submanifolds for which the symplectic form restricts to 0 and isn't nondegenerate. However, Kähler forms do restrict. \end{remark} \begin{remark} So we get a Hodge diamond: \begin{center} \begin{tikzcd} && {h^{2, 2}} \\ & {h^{2, 1}} && {h^{1, 2}} \\ {h^{2, 0}} && {h^{1, 1}} && {h^{0, 2}} \\ & {h^{1, 0}} && {h^{0, 1}} \\ && {h^{0, 0}} \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiw0LCJoXnswLCAwfSJdLFsxLDMsImheezEsIDB9Il0sWzMsMywiaF57MCwgMX0iXSxbMCwyLCJoXnsyLCAwfSJdLFsyLDIsImheezAsIDB9Il0sWzQsMiwiaF57MCwgMn0iXSxbMSwxLCJoXnsyLCAxfSJdLFszLDEsImheezEsIDJ9Il0sWzIsMCwiaF57MiwgMn0iXV0=}{Link to Diagram} \end{quote} Here $$h^{2, 0} = h^0( \Omega^2) = h^0(K) = g$$ is called the \emph{genus} in analogy with curves. Similarly, $$h^{1, 0} = h^0( \Omega^1)$$ is the space of holomorphic 1-forms, sometimes referred to as the \emph{irregularity}. There is some symmetry: \begin{center} \begin{tikzcd} && 1 \\ & q && q \\ g && {h^{0, 0}} && g \\ & q && q \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiw0LCIxIl0sWzEsMywicSJdLFszLDMsInEiXSxbMCwyLCJnIl0sWzIsMiwiaF57MCwgMH0iXSxbNCwyLCJnIl0sWzEsMSwicSJdLFszLDEsInEiXSxbMiwwLCIxIl1d}{Link to Diagram} \end{quote} \end{remark} \begin{exercise}[?] Solve for $$h^{1, 1}$$ in terms of $$q$$ and $$g$$. \end{exercise} \hypertarget{friday-april-02}{% \section{Friday, April 02}\label{friday-april-02}} \hypertarget{when-line-bundles-are-mathcalo-of-a-divisor}{% \subsection{\texorpdfstring{When Line Bundles are $${\mathcal{O}}$$ of a Divisor}{When Line Bundles are \{\textbackslash mathcal\{O\}\} of a Divisor}}\label{when-line-bundles-are-mathcalo-of-a-divisor}} \begin{remark} Last time: if we have such a Hodge diamond, can we solve for $$h^{1, 1}$$? \begin{center} \begin{tikzcd} && 1 \\ & q && q \\ p && {h^{1, 1}} && p \\ & q && q \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwyLCJoXnsxLCAxfSJdLFswLDIsInAiXSxbNCwyLCJwIl0sWzEsMywicSJdLFszLDMsInEiXSxbMiw0LCIxIl0sWzEsMSwicSJdLFszLDEsInEiXSxbMiwwLCIxIl1d}{Link to Diagram} \end{quote} Recall Noether's formula \begin{align*} \chi(S, {\mathcal{O}}_S) &= \int \operatorname{ch}({\mathcal{O}}_S) \mathrm{td}(S) \\ &= \int_S {x_1 \over 1-e^{-x_1} } {x_2 \over 1- e^{-x_2} } \\ &= {K^2 + \chi_{\mathsf{Top}}(S) \over 12} ,\end{align*} where $$c_1(TS) = - K$$ and $$\chi_{\mathsf{Top}}$$ is due to the Chern-Gauss-Bonet formula. We have \begin{align*} \chi({\mathcal{O}}_S) = h^0({\mathcal{O}}_S) - h_1({\mathcal{O}}_S) + h^2({\mathcal{O}}_S) = 1-q+p .\end{align*} On the other hand, \begin{align*} \chi_{\mathsf{Top}}(S) = 1 -2q + (2p + h^{1, 1}) -4q = 1-4q + 2^p + h^{1, 1} ,\end{align*} so \begin{align*} 12(1-q+p) = K^2 + 2-4g + 2p + h^{1, 1} \implies h^{1, 1} = 110 - 8q + 10p-K^2 .\end{align*} \end{remark} \begin{remark} Recall the extraordinarily important exact sequence \begin{align*} 0 \to {\mathcal{O}}(-p) \to {\mathcal{O}}\to {\mathcal{O}}_p \to 0 ,\end{align*} where the right-hand side is the sheaf of holomorphic functions vanishing at $$p$$ and this is an inclusion into the sheaf of holomorphic functions, and the right-hand term is the skyscraper sheaf. There is a similar exact sequence for an embedded curve $$C\hookrightarrow S$$ in a surface: \begin{align*} 0 \to {\mathcal{O}}_S(-C) \to {\mathcal{O}}_S \to {\mathcal{O}}_C \to 0 ,\end{align*} where the left term is the sheaf of holomorphic functions vanishing on $$C$$. Note that this has no global sections! Any function vanishing along a compact subset (?) are constant (?). Locally on an open set $$U$$, one can write $$C \cap U = V(f_u)$$, since algebraically this ring is locally a PID. So this is a line bundle, where we can map into the trivial bundle by $$\phi \mapsto \phi/f_u$$. Thus \begin{align*} {\mathcal{O}}_S(U) / {\mathcal{O}}_S(-C)(U) \cong {\mathcal{O}}_C(C \cap U ) .\end{align*} We then get surjectivity since every holomorphic function on $$C$$ extends to a holomorphic function on $$S$$. Now letting $$\mathcal{E} \in{ \mathsf{Vect} }(\mathop{\mathrm{Hol}})$$, we can tensor this exact sequence to get \begin{align*} 0 \to \mathcal{E}(-C) \to \mathcal{E} \to { \left.{{\mathcal{E}}} \right|_{{C}} } \to 0 ,\end{align*} which is also exact since locally we have the splitting principle. \end{remark} \begin{proposition}[Every line bundle over a smooth projective complex manifold is O of a divisor] Let $$X$$ be a smooth projective \footnote{So $$X$$ admits an embedding into some $${\mathbb{CP}}^N$$.} complex manifold. Then every line bundle over $$X$$ is of the form $$L = {\mathcal{O}}_X(D)$$ for some divisor $$D = \sum n_i D_i \in {\mathbb{Z}}[\mathop{\mathrm{SubMfds}}(\operatorname{codim}_1)]$$. \end{proposition} \hypertarget{proof}{% \subsection{Proof}\label{proof}} \begin{proof}[?] Let $$H$$ be a \textbf{hyperplane section}, i.e.~an intersection of $$X$$ with a generic hyperplane in $${\mathbb{CP}}^N$$. \begin{lemma}[Serre Vanishing Theorem] For any vector bundle $$\mathcal{E}$$ and all $$i>0$$, for $$k\gg 0$$ we have \begin{align*} h^i(X, \mathcal{E} \otimes{\mathcal{O}}(kH) ) = 0 .\end{align*} \end{lemma} \begin{remark} We'll not prove this! It requires some heavy analysis and the Kähler identities, see Huybrechts complex geometry Prop 5.27. \end{remark} We can write \begin{align*} \chi(L\otimes{\mathcal{O}}(kH)) &= \int_X \operatorname{ch}(L\otimes{\mathcal{O}}(kH)) \mathrm{td}(X) \\ &= \int_X \operatorname{ch}(L) \operatorname{ch}(H)^k \mathrm{td}(X) \\ &= \int_X \qty{ 1 + c_1(L) + {c_1(L)^2 \over 2} + \cdots } \cdot \qty{1+ kh + {(kh)^2 \over 2 } + \cdots + {(kh)^{\dim X} \over (\dim X)!} } \cdot \qty{1 + \mathrm{td}_1(X) + \mathrm{td}_2(X) + \cdots } .\end{align*} where $$h$$ is the restriction of the generator of $$H^2({\mathbb{CP}}^N; {\mathbb{Z}})$$ to $$X$$. Note that for $$k$$ large, the dominating term grows like $$(kh)^{\dim X}$$, so asymptotically we have \begin{align*} \cdots \sim \int_X { k^{\dim X} h^{\dim X} \over (\dim X)! } .\end{align*} What is this $$\dim(X){\hbox{-}}$$fold intersection? \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-02_14-25.pdf_tex} }; \end{tikzpicture} } \end{figure} We can slice $$X$$ by multiple hyperplanes, each homologically perturbed, and so $$\int_X h^{\dim X}$$ is the number of points where $$\dim X$$ generic hyperplanes intersect $$X$$, which is called the \textbf{degree} $$\deg X$$. This roughly follows from $$\int_X \omega_{ \text{FS} }^{\dim X} > 0$$. Alternatively, suppose $$X \cap H = \emptyset$$, then $$X \hookrightarrow H^c = {\mathbb{A}}^N$$. Then each holomorphic coordinate restricts to a constant on $$X$$ by the maximal principle. Back to what we were proving: we have \begin{align*} \chi(L \otimes{\mathcal{O}}(kH) ) \sim ck^{\dim X} ,\end{align*} for $$c$$ some constant. By Serre Vanishing, $$h^i(L\otimes{\mathcal{O}}(kH)) = 0$$ for $$k\gg 0$$, and so we obtain \begin{align*} h^0(L\otimes{\mathcal{O}}(kH)) \sim ck^{\dim X} \implies \exists k \text{ s.t. } h^0(L\otimes{\mathcal{O}}(kH)) > 0 .\end{align*} We conclude that there is some nonzero section $$s\in { \mathcal{H} }^0(X; L\otimes{\mathcal{O}}(kH))$$ for which $${\mathcal{O}}(\operatorname{Div}s) \cong L\otimes{\mathcal{O}}(kH)$$. Thus $$L \cong {\mathcal{O}}(\operatorname{Div}s - kH)$$, where $$\operatorname{Div}s- kH$$ is some divisor. \end{proof} \begin{remark} With some more work, one can show $$L\cong {\mathcal{O}}(C-D)$$ for $$C,D$$ \emph{smooth} divisors. \end{remark} \hypertarget{aside}{% \subsection{Aside}\label{aside}} \begin{remark} Felix Klein has a proof'' of the existence of a meromorphic function on a Riemann surface. The argument roughly goes as follows: take your Riemann surface and make it out of metal. Attach it to a battery: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-02_14-39.pdf_tex} }; \end{tikzpicture} } \end{figure} This induces an electric potential function $$V: C\to {\mathbb{R}}$$, where $$V$$ is the real part of the meromorphic function. Here $$V$$ is a harmonic function away from $$p$$ and $$q$$. \end{remark} \hypertarget{monday-april-05}{% \section{Monday, April 05}\label{monday-april-05}} \begin{remark} Last time: line bundles are of the form $${\mathcal{O}}(D)$$ for $$D$$ a divisor, and the extremely important SES \begin{align*} 0 \to {\mathcal{O}}_S(-D) \to {\mathcal{O}}_S \to {\mathcal{O}}_D \to 0 .\end{align*} We now want to discuss an alternative characterization of the intersection form on an algebraic surface. The next result comes from Beauville's Complex Algebraic Surfaces'': \end{remark} \begin{proposition}[Formula for computing intersection numbers between complex curves] Let $$S \in {\mathsf{Mfd}}^2({\mathbb{C}})^{ \operatorname{compact} }$$, then the intersection number between complex curves $$C, D$$ can be computed in the following ways: \begin{align*} C\cdot D = \deg {\mathcal{O}}_S(C) { \left.{{}} \right|_{{D}} } = \sum_{p\in C \cap D } \mathop{\mathrm{len}}_p(C \cap D ) ,\end{align*} where we'll define $$\mathop{\mathrm{len}}_p$$ soon. \end{proposition} \begin{remark} This will count intersection points after a small perturbation. Note that not every two curves will intersect transversely: consider $${\mathbb{P}}_2$$ with a line $$C$$ and a tangent conic $$D$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{41pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_13-57.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{proof}[?] We have the first equality because \begin{align*} C\cdot D = \int_S [C] \frown[D] = \int_C i^* [D] ,\end{align*} where $$i: C\hookrightarrow S$$ is the inclusion. This equality holds because if $$\alpha\in \Omega^2$$ is a 2-form, \begin{align*} \int_S [C] \cdot \alpha = \int { \left.{{ \alpha}} \right|_{{C}} } .\end{align*} Using the pullback commutes with taking Chern classes, we can write the \begin{align*} \int_C i^* [D] = \int_C i^* (c_1 ({\mathcal{O}}(D))) = \int_C c_1( i^* {\mathcal{O}}(D) ) = \int_C {\mathcal{O}}(D){ \left.{{}} \right|_{{C}} } = \deg {\mathcal{O}}(D) { \left.{{}} \right|_{{C}} } .\end{align*} Note that this formula was symmetric, so we could have done this the other way to obtain $$\deg {\mathcal{O}}_S(C){ \left.{{}} \right|_{{D}} } = \deg {\mathcal{O}}_S(D) { \left.{{}} \right|_{{C}} }$$. For the second equality, consider the following 4-term exact sequence: \begin{align*} 0 \to {\mathcal{O}}_S(-C -D) \underset{p_1}{ \xhookrightarrow{{\left[ { s_D, s_C} \right]}} } {\mathcal{O}}_S(-C) \oplus {\mathcal{O}}_S(-D) \underset{p_2}{ \xrightarrow{{\left[ {s_D, -s_C} \right]}^t} } {\mathcal{O}}_S \underset{p_3}{\to} {\mathcal{O}}_{C \cap D} \to 0 .\end{align*} For the first map, we have \begin{align*} \left\{{ \text{Functions vanishing on } C+D}\right\} \hookrightarrow \left\{{ \text{Functions vanishing on } C}\right\} \oplus \left\{{ \text{Functions vanishing on } D}\right\} .\end{align*} Locally we can write $$C = V(f)$$ and $$D = V(g)$$ for some holomorphic functions $$f,g\in \mathop{\mathrm{Hol}}(U, {\mathbb{C}})$$. We have the following picture: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_14-06.pdf_tex} }; \end{tikzpicture} } \end{figure} We have $$s_C \in H^0(S; {\mathcal{O}}_S(C))$$ and $$s_D \in H^0(S; {\mathcal{O}}_S(D))$$ as global sections where $$V(s_c) = C, V(s_D) = D$$. In a local trivialization, we can assume $${ \left.{{s_C}} \right|_{{U}} } = f$$ and $${ \left.{{s_D}} \right|_{{U}} } = g$$. So the first map is $$(s_D, s_C)$$. The next map is $${\left[ {s_C, -s_D} \right]}^t$$ as a column vector, i.e.~given a section we map it in the following way: \begin{align*} (\varphi_1, \varphi_2) \in H^0(U, {\mathcal{O}}_S(-C) \oplus {\mathcal{O}}_S(-D) ) \mapsto \phi_1 \cdot s_D - \phi_2 \cdot s_C .\end{align*} Why is this exact? Considering the composition, we have \begin{align*} \phi \xrightarrow{p_1} (\phi s_D, \phi s_C) \xrightarrow{p_2} (\phi s_D) s_C - (\phi s_C) s_D = 0 .\end{align*} So we get $$\operatorname{im}p_1 \subseteq \ker p_2$$. Why do we have the reverse containment for exactness? Looking locally, given a pair $$\phi_1, \phi_2 \in \mathop{\mathrm{Hol}}(U; {\mathbb{C}})$$ such that $$\phi_1 \varphi- \phi_2 g = 0$$ and locally $$(\phi_1, \phi_2) \in \ker p_2$$, we want to show that $$\phi_1 = g \phi, \phi_2 = f\phi$$ for some $$f,g \in \mathop{\mathrm{Hol}}(U; {\mathbb{C}})$$. Equivalently, we want to show that \begin{align*} \phi_1 f = \phi_2 g \implies g\divides \phi_1 .\end{align*} If this is true, then we can set $$\phi \coloneqq{\phi_1 \over g}$$, since this would yield $$g\phi = \phi_1$$ and $$f\phi = {f\phi_1 \over g} = \phi_2$$. Note that we can divide here because the ring $$\mathop{\mathrm{Hol}}(U;{\mathbb{C}})$$ is a domain (i.e.~it has no zero divisors) on small sets. \begin{question} Is $$\mathop{\mathrm{Hol}}(U, {\mathbb{C}})$$ a PID in general? \end{question} \begin{answer} No! Take $$U \subseteq {\mathbb{C}}^2$$ a ball around $$z=0$$, then $$\left\langle{x, y}\right\rangle$$ is not principal. \end{answer} However, this will form a UFD, which is weaker but still enough here. This is not obvious, but can be proved using the Weierstrass preparation theorem. This should be believable since $$R$$ a UFD implies $$R[x]$$ is a UFD, and $${\mathbb{C}}[x, y] \subsetneq \mathop{\mathrm{Hol}}(U; {\mathbb{C}}) \subsetneq {\mathbb{C}}[[x, y]]$$, and the latter is a UFD. So we do get exactness at this position. For exactness at the next position $${\mathcal{O}}_S(-C) \oplus {\mathcal{O}}_S(-D) \to {\mathcal{O}}_S$$, locally we have $$(\varphi_1, \varphi_2 ) \mapsto \varphi_1 f- \varphi_2 g$$ where $$V(f) = C \cap U$$ and $$V(g) = D \cap U$$. We can write $$\phi_1 f- \phi_2 g = \left\langle{f, g}\right\rangle$$ locally, so the cokernel sheaf of $$p_2$$ is given by \begin{align*} \operatorname{coker}p_2(U) \coloneqq{{\mathcal{O}}_S(U) \over \operatorname{im}p_2} = { {\mathcal{O}}_S(U) \over \left\langle{f, g}\right\rangle} .\end{align*} By definition, this is equal to $${\mathcal{O}}_{V(f, g)} = {\mathcal{O}}_{C \cap D}$$, and if $$C \cap D \cap U = \emptyset$$ then $${\mathcal{O}}_{C \cap D}(U) = 0$$. So let $$p \in {\mathcal{O}}_{C \cap D}$$ and let $$U_p \ni p$$ which contains no other points $$q\in C \cap D$$, since the set of intersection points is isolated (and thus finite). Note that compactness here prevents accumulation of intersection points. In this case, $${\mathcal{O}}_{C \cap D}(U_p)$$ will be a finite-dimensional vector space $${\mathbb{C}}^d$$, and we'll define $$\mathop{\mathrm{len}}_p(C \cap D) \coloneqq d$$. \end{proof} \begin{example}[?] Let $$U={\mathbb{C}}^2$$ and take $$f=y$$ so $$C\coloneqq V(f)$$ is the x-axis, and set $$g = y-x^2$$ so $$D\coloneqq V(g)$$ is a parabola. We're then considering \begin{align*} { \mathop{\mathrm{Hol}}({\mathbb{C}}^2) \over y \mathop{\mathrm{Hol}}({\mathbb{C}}^2) + (y-x^2) \mathop{\mathrm{Hol}}({\mathbb{C}}^2)} = {\mathop{\mathrm{Hol}}({\mathbb{C}}^2) \over \left\langle{ y, x^2 }\right\rangle } .\end{align*} Elements in the ideal can be expanded as power series of the form $$a_{0,1}y + a_{2, 0}x^2 + a_{1, 1} xy + a_{2,2} y^2$$, where there is no $$a_{1, 0} \sim x^1 y_0$$ coefficient, nor any $$a_{0, 0} \sim x^0 y^0$$ coefficient. So this quotient is isomorphic to $${\mathbb{C}}1 \oplus {\mathbb{C}}x$$, which is 2-dimensional, so $$\mathop{\mathrm{len}}_{(0, 0)} V(y) \cap V(x) = 2$$. Geometrically we have the following, where this is picking up the multiplicity 2 intersection: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em}3 \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-05_14-35.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \begin{remark} What's the payoff of this algebraic work? We can compute the Euler characteristic as \begin{align*} \chi( {\mathcal{O}}_{C \cap D}) = h^0({\mathcal{O}}_{C \cap D}) = \sum_{p\in C \cap D} \mathop{\mathrm{len}}_p(C \cap D ) .\end{align*} But by additivity of $$\chi$$ over exact sequences, we also have \begin{align*} \chi({\mathcal{O}}_{C \cap D}) &= \chi({\mathcal{O}}_S) - \chi({\mathcal{O}}_S(-C)) - \chi({\mathcal{O}}_S(-D)) + \chi( {\mathcal{O}}_S( -C -D))\\ &\overset{HRR}{=} \int_S \qty{ \operatorname{ch}({\mathcal{O}}_S) - \operatorname{ch}({\mathcal{O}}_S(-C)) - \operatorname{ch}( {\mathcal{O}}_S(-D)) + \operatorname{ch}( {\mathcal{O}}_S(-C-D)) } \mathrm{td}(S) \\ &= c_1( {\mathcal{O}}_S(-C)) \cdot c_1({\mathcal{O}}_S(-D)) \\ &= \qty{ -[C] } \cdot \qty{ -[D] } \\ &= C\cdot D .\end{align*} \end{remark} \begin{remark} Next time: adjunction formula that allows computing genus for surfaces. \end{remark} \hypertarget{wednesday-april-07}{% \section{Wednesday, April 07}\label{wednesday-april-07}} \begin{remark} Last time: let $$C, D \subset S$$ be distinct curves, then the intersection number is given by \begin{align*} C\cdot D = \deg {\mathcal{O}}_S(C) { \left.{{}} \right|_{{D}} } = \sum_{p \in C \cap D } \mathop{\mathrm{len}}_p( C \cap D) \end{align*} where $$\mathop{\mathrm{len}}_p(C \cap D) \coloneqq\dim_{\mathbb{C}}{\mathcal{O}}(U) / \left\langle{ f, g }\right\rangle$$ where $$V(f) = C \cap U$$ and $$V(g) = D \cap U$$ with $$f, g \in {\mathcal{O}}(U) = \mathop{\mathrm{Hol}}(U)$$. Here we're also assuming that $$C \cap D \cap U = \left\{{ p }\right\}$$. \end{remark} \hypertarget{adjunction-formula}{% \subsection{Adjunction Formula}\label{adjunction-formula}} \begin{remark} We'll now discuss a way to compute the genus of a curve in a surface. \end{remark} \begin{proposition}[Adjunction Formula] Let $$C \subset S$$ be a smooth curve, then $$K_C = (K_S \otimes{\mathcal{O}}_S(C)) { \left.{{}} \right|_{{C}} }$$, which is restriction of a line bundle. Note that $$K_C = \Omega^1_C$$ is the sheaf of holomorphic 1-forms, but $$K_S = \Omega^2_S$$ since we take the sheaf of top forms. \end{proposition} \begin{proof}[?] Let $$s\in \Omega^2_S \otimes{\mathcal{O}}(C)(U)$$ be a section, then $$s_C$$ is a section of $${\mathcal{O}}_C$$ vanishing along $$c$$ and have $$s/s_C$$ a meromorphic section of $$\Omega^2_S(U)$$. Here dividing by $$s_C$$ is like tensoring with $${\mathcal{O}}(-C)$$. This can have poles along $$\left\{{ s_C = 0 }\right\} = C$$ up to first order. There is a residue map: let $$p\in C$$ be a point and $$\gamma_p(r)$$ be an oriented loop in $$S\setminus C$$ around $$p\in C$$ of radius $$r$$ (a meridian): \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-07_14-01.pdf_tex} }; \end{tikzpicture} } \end{figure} We can assemble a 1-form from the following contour integral: \begin{align*} \mathop{\mathrm{Res}}_C {s\over s_C} \coloneqq\lim_{r\to 0} {1\over 2\pi i} \oint_{\gamma_p(r)} {s\over s_C} \in \Omega^1(U \cap C) .\end{align*} Locally $$C = V(x)$$ in a coordinate chart of $${\mathbb{C}}^2$$ where $$s_C = x$$, so this is roughly of the form $$\oint_{{\left\lvert {x} \right\rvert} = r} {f(x, y) \over x} \,dx\wedge \,dy$$, which is a one form in the variable $$y$$. Note that if $$f$$ were analytic, writing $$f = a_{0,0} + a_{0, 1} y + a_{0, 2} y^2 + \cdots + a_{1, 0}x + \cdots$$, we would have \begin{align*} \mathop{\mathrm{Res}}_C {s\over s_C} = (a_{0, 0} + a_{0, 1}y + a_{0, 2}y^2 + \cdots)\,dy= f(0, y)\,dy\text{locally} ,\end{align*} which picks out all components not involving $$x$$. This defines an $${\mathcal{O}}{\hbox{-}}$$linear map \begin{align*} \Omega^2_S \otimes{\mathcal{O}}_C &\to \Omega^1_C \\ s & \mapsto \mathop{\mathrm{Res}}_C {s\over s_C} ,\end{align*} since it doesn't involve any derivatives of $$f$$. Note that this only depends on the restriction of $$s$$ to $$C$$. What is the kernel of $$\mathop{\mathrm{Res}}_C$$? We claim it is $$\Omega2_S$$, which follows from the fact that the contour integral of any holomorphic form $$\omega$$ will integrate to zero. We thus get a SES of sheaves \begin{align*} 0 \to \Omega^2_S \xrightarrow{\cdot s_C} \Omega_S^2 \otimes{\mathcal{O}}(C) \to \Omega^1(C) \to 0 .\end{align*} where we send holomorphic forms to meromorphic forms with at most order 1 poles along $$C$$ to holomorphic 1-forms on $$C$$. The residue map is surjective since we can take \begin{align*} \mathop{\mathrm{Res}}_{x=0} {g(y) \over x} \,dx\wedge \,dy= g(y) \,dy ,\end{align*} so locally an arbitrary 1-form is a residue of some 2-form with simple poles along $$C$$. We have a SES \begin{align*} 0 \to {\mathcal{O}}(-C) \xrightarrow{\cdot s_C} {\mathcal{O}}\to {\mathcal{O}}_C \to 0 ,\end{align*} and tensoring with the line bundle $$\Omega^2 \otimes{\mathcal{O}}(C)$$ we obtain \begin{align*} 0 \to \Omega_S^2 \to \Omega^2_S \otimes{\mathcal{O}}(C) \to \Omega^2_S \otimes{\mathcal{O}}(C) { \left.{{}} \right|_{{C}} } \to 0 .\end{align*} Since cokernels are unique, we have $$\Omega^1_C \cong \Omega^2_S \otimes{\mathcal{O}}(C) { \left.{{}} \right|_{{C}} }$$, which yields the adjunction formula. \end{proof} \begin{corollary}[The Genus Formula] We have \begin{align*} \deg \Omega_S^2 \otimes{\mathcal{O}}(C) { \left.{{}} \right|_{{C}} } = \deg \Omega_C^1 = 2g-2 \end{align*} where $$g = g(C)$$ is the genus of $$C$$. On the other hand, the left-hand side is equal to \begin{align*} (K_S + C) \cdot C = 2g(C) - 2 .\end{align*} \end{corollary} \begin{example}[?] We showed $$K_{{\mathbb{P}}^n} = {\mathcal{O}}(-n-1)$$ where $${\mathcal{O}}(-1)$$ was the tautological line bundle over $${\mathbb{P}}^n$$. So for example $$K_{{\mathbb{P}}^2} = {\mathcal{O}}(-3) = -3 L$$ where $$L \in H^2({\mathbb{P}}^2, {\mathbb{Z}})$$ is a hyperplane (here a line) in $${\mathbb{P}}^2$$. \end{example} \begin{corollary}[Formula for genus of a curve in terms of degree] Let $$f$$ be a degree $$d$$ homogeneous polynomial in $$x,y,z$$, then $$V(f) \subseteq {\mathbb{P}}^2 = \left\{{ [x:y:z] }\right\}$$. If $$C\coloneqq V(f)$$ is a smooth complex curve, then applying the genus formula yields \begin{align*} 2g(C) - 2 = (-3L + dL) \cdot dL .\end{align*} Using that $$L^2 = 1$$, this equals $$d(d-3)$$ and thus \begin{align*} g(C) = {d^2 - 3d + 2 \over 2} = {d-1 \choose 2} .\end{align*} \end{corollary} \begin{example}[?] If $$d=3$$ and say $$f(x,y,z) = x^3 + y^3 + z^3$$, then $$V(f) \subseteq {\mathbb{P}}^2$$ has genus $${3-1 \choose 2} = 1$$. So this is diffeomorphic to a torus. \end{example} \begin{example}[?] If $$d=2$$ then $$g(C) = 0$$, so conics in $${\mathbb{P}}^2$$ have genus zero, and we proved that every genus zero curve is isomorphic to $${\mathbb{P}}^1$$. So conics in $${\mathbb{P}}^2$$ are isomorphic to $${\mathbb{P}}^1$$ (as are lines of course!). \end{example} \begin{example}[?] If $$d=4$$ then $$g(C) = 3$$ \end{example} \begin{theorem}[Harnack Curve Theorem] Noting that $${\mathbb{RP}}^2 \subset {\mathbb{CP}}^2 = {\mathbb{P}}^2$$, the number $$n_C$$ of connected components of a curve $$C \cap{\mathbb{RP}}^2$$ satisfies \begin{align*} n_C \leq 1 + g(C) .\end{align*} \end{theorem} \begin{remark} See the Trott curve: \begin{align*} 144(x^4 + y^4) - 225(x^2 + y^2) + 350x^2 y^2 + 81 = 0 ,\end{align*} whose plot looks like the following: \begin{verbatim} f(x,y) = 12^2*(x^4 + y^4) - 15^2*(x^2 + y^2) + 350*x^2*y^2 + 81 implicit_plot(f, (x,-1,1), (y,-1,1)) \end{verbatim} \begin{figure} \centering \includegraphics{figures/image_2021-04-09-16-40-49.png} \caption{image\_2021-04-09-16-40-49} \end{figure} \end{remark} \begin{example}[?] Consider $$S \coloneqq{\mathbb{P}}^1 \times{\mathbb{P}}^1$$, which is homeomorphic to $$S^2 \times S^2$$. The homology is given by $${\mathbb{Z}}$$ in degrees 0 and 4, $${\mathbb{Z}}^{\oplus 2}$$ in degree 3, and 0 elsewhere. What is the intersection form on $${\mathbb{Z}}^{\oplus 2} = H^2({\mathbb{P}}^1 \times{\mathbb{P}}^1; {\mathbb{Z}})$$? The two generators are $$f_1 = [S^2 \times{\operatorname{pt}}], f_2 = [ {\operatorname{pt}}\times S^2]$$. We can compute \begin{itemize} \tightlist \item $$f_1 \cdot f_1 = 0$$ \item $$f_1 \cdot f_2 = 1$$ \item $$f_2 \cdot f_2 = 0$$ \end{itemize} This is because we can perturb these to be transverse: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-07_14-39.pdf_tex} }; \end{tikzpicture} } \end{figure} Since $$f_2 \cap f_2 ' = \emptyset$$, we have $$f_2 \cdot f_2' = f_2 \cdot f_2 = 0$$, and similarly with 1. So the Gram matrix is \begin{align*} G = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} .\end{align*} So setting $$C = {\mathbb{P}}^1 \times{\mathbb{P}}^1 = V(f_{2, 3})$$, a function of bidegree $$(2, 3)$$, writing the coordinates as $$[x: y], [z: w]$$, we can write this as $$x^2 z^3 + y^2 z^2 w + xy w^3 = 0$$. We get \begin{align*} 2g(C) - 2 = (K_{{\mathbb{P}}^1 \times{\mathbb{P}}^1} + 2f_1 + 3f_2) \cdot (2f_1 + 3f_2) = f_2 \cdot (2f_1 + 3f_2) = 2 ,\end{align*} since $$K_{{\mathbb{P}}^1} \circ = -2f_1 - 2f_2$$ and so $$g(C) = 2$$. \end{example} \hypertarget{friday-april-09}{% \section{Friday, April 09}\label{friday-april-09}} \begin{remark} Recall the adjunction formula: for $$D \subset X \in {\mathsf{Mfd}}_{\mathbb{C}}$$ a codimension 1 complex submanifold, we have \begin{align*} K_D = (K_x + {\mathcal{O}}_x(0)) { \left.{{}} \right|_{{D}} } .\end{align*} We'll apply this to curves $$C$$ in a surface $$S$$. Recall the genus formula, which was given by $$2g(C) - 2= (C+ K_S)\cdot C$$. For example, a degree 4 equation in $${\mathbb{P}}^2$$ carves out a genus $$g(C) = 3$$ complex curve. \end{remark} \begin{remark} Recall that line bundles on $${\mathbb{CP}}^n$$ were in bijection with $${\mathbb{Z}}$$, where send $$d$$ to a bundle $${\mathcal{O}}(d) \coloneqq{\mathcal{O}}_{{\mathbb{CP}}^N}(d)$$. We produced the tautological line bundle $${\mathcal{O}}(-1)$$ whose fiber over $$\mathbf{x} \subseteq {\mathbb{CP}}^n$$ is the line in $${\mathbb{C}}^n$$ spanned by its coordinates. We have $${\mathcal{O}}(-1) {}^{ \vee }\coloneqq{\mathcal{O}}(1)$$, and $${\mathcal{O}}(n)\coloneqq{\mathcal{O}}(1)^{\otimes n}$$. Alternatively, it was characterized in terms of homogeneous functions, where the fiber $${\mathcal{O}}(n)_{\mathbf{x}}$$ are the linear functions $$L$$ on lines $$\left\{{\lambda \mathbf{x}}\right\} \to {\mathbb{C}}$$ such that $$L(\lambda p) = \lambda^n L(p)$$. Noting that these are linear functions, such $$L$$ form a 1-dimensional $${\mathbb{C}}{\hbox{-}}$$vector space. \end{remark} \begin{example}[K3 Surfaces] The classic example is $$x_0 \in {\mathcal{O}}(1)_{\mathbf{x}}$$ since $$x_0( \lambda p) = \lambda x_0 (p)$$. Similarly, $$x_0^2 + x_1 x_2 \in {\mathcal{O}}(2)_{\mathbf{x}}$$ since \begin{align*} x_0^2 + x_1 x_2 (\lambda p) = \lambda^2 (x_0^2 + x_1 x_2(p)) .\end{align*} \end{example} \begin{remark} Note that the global sections were given by $$\Gamma^0({\mathbb{P}}^n, {\mathcal{O}}(d)) = H^0({\mathbb{P}}^n; {\mathcal{O}}(d))$$ was the span of degree $$d$$ monomials in $$x_0, \cdots, x_n$$. For example $$x_0^2 + x_1 x_2$$ is a well-defined element of $${\mathcal{O}}(2)_p$$ which varies holomorphically with $$p$$, yielding a section: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-09_14-11.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{example}[?] For a K3 surface, consider $$S = \left\{{ \sum_{i=0}^4 x_i^4 = 0 }\right\} \subset {\mathbb{CP}}^3$$. By the adjunction formula, \begin{align*} K_S = { \left.{{\qty{ K_{{\mathbb{CP}}^3} \otimes{\mathcal{O}}_{{\mathbb{CP}}^3}(S) }}} \right|_{{S}} } .\end{align*} Note that if $$s\in H^0(\mathcal{L})$$, we can recover $${\mathcal{O}}(\operatorname{Div}S) = \mathcal{L}$$. Moreover, $$K_{{\mathbb{CP}}^3} = {\mathcal{O}}(-4)$$ and $${\mathcal{O}}_{{\mathbb{CP}}^3}(S) = {\mathcal{O}}(4)$$ since we can view the formula as a function on the tautological line, which yields a section. So we get $$K_S = {\mathcal{O}}(-4) \otimes{\mathcal{O}}(4) = {\mathcal{O}}(0) = {\mathcal{O}}$$, i.e.~these yield actual functions on $${\mathbb{CP}}^n$$ since they're products of functions that scale by $$\lambda^{-4}$$ and functions that scale by $$\lambda^4$$. We're using the fact that $${\mathcal{O}}_{\mathbf{p} = [x_0: \cdots : x_n]}$$ are functions $$L$$ such that $$L(\lambda p) = \lambda^0 L(p) = L(p)$$, which yields a well-defined function on $${\mathbb{CP}}^n$$. So quartics in $${\mathbb{P}}^3$$ have trivial canonical bundle, i.e.~$$K_S = {\mathcal{O}}_S$$ for $$S = V(x_0^4 + x_1^4 + x_2^4 + x_3^4)$$. \end{example} \begin{remark} We know that $$H^0(S, K_S)$$ are the globally holomorphic 2-forms on $$S$$, and here this is isomorphic to $$H^0(S, {\mathcal{O}}_S) = {\mathbb{C}}\Omega_S$$ for some single holomorphic 2-form. Moreover $$\operatorname{Div}(\Omega_S) = 0$$ since $${\mathcal{O}}( \operatorname{Div}( \Omega_S)) = K_S = {\mathcal{O}}_S$$. So these are the analogs of elliptic curves in dimension 2, since for example $$E \coloneqq{\mathbb{C}}/ \Lambda$$ has a nonvanishing section $$\,dz\in H^0(E, K_E)$$, and we can write $$E = V(f)$$ for $$f$$ a cubic in $${\mathbb{P}}^3$$, and we computed the genus of cubics. Moreover, every genus 1 curve is $${\mathbb{C}}$$ mod a lattice. \end{remark} \begin{remark} Recall an exercise from the notes: computing the Hodge diamond of a genus 5 curve. We'll compute the diamond for a K3 surface: \begin{center} \begin{tikzcd} && {h^{2, 2}} \\ & {h^{3, 1}} && {h^{1, 3}} \\ {h^{2, 0}} && {h^{1, 1}} && {h^{0, 2}} \\ & {h^{1, 0}} && {h^{0, 1}} \\ && {h^{0, 0}} \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwwLCJoXnsyLCAyfSJdLFsxLDEsImheezMsIDF9Il0sWzMsMSwiaF57MSwgM30iXSxbMCwyLCJoXnsyLCAwfSJdLFsyLDIsImheezEsIDF9Il0sWzQsMiwiaF57MCwgMn0iXSxbMSwzLCJoXnsxLCAwfSJdLFszLDMsImheezAsIDF9Il0sWzIsNCwiaF57MCwgMH0iXV0=}{Link to Diagram} \end{quote} We know $$h^{2, 0} = H^0( S, \Omega_S^2)$$, which yields 1s: \begin{center} \begin{tikzcd} && 1 \\ & {h^{3, 1}} && {h^{1, 3}} \\ 1 && {h^{1, 1}} && 1 \\ & {h^{1, 0}} && {h^{0, 1}} \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiaF57MywgMX0iXSxbMywxLCJoXnsxLCAzfSJdLFswLDIsIjEiXSxbMiwyLCJoXnsxLCAxfSJdLFs0LDIsIjEiXSxbMSwzLCJoXnsxLCAwfSJdLFszLDMsImheezAsIDF9Il0sWzIsNCwiMSJdXQ==}{Link to Diagram} \end{quote} We'll use the following theorem: \begin{theorem}[Lefschetz Hyperplane Theorem] Let $$X \subset {\mathbb{P}}^n$$ with $$\dim X > 3$$. Then $$\pi_1(X) \cong \pi_1(X \cap H)$$ for $$H$$ a hypersurface intersection $$X$$ at a smooth codimension 1 complex manifold. \end{theorem} \begin{remark} Applying this to $$X = {\mathbb{P}}^3$$, we have $$V(x_0^4 + \cdots + x_3^4) = S$$, we have $$\pi_1({\mathbb{P}}^3) = \pi_1(S)$$. We can write $${\mathbb{P}}^3 = {\mathbb{C}}\cup{\mathbb{C}}^2 \cup{\mathbb{C}}^4$$, which is a cell decomposition with cells only in degrees 0,2,4, and so in fact $$\pi_1({\mathbb{P}}^n) = 0$$. \end{remark} \begin{corollary}[h1 of K3 surfaces] K3 surfaces are simply connected, and $$h^1(S; {\mathbb{C}}) = 0$$. \end{corollary} Note that anything embedded in projective space as a complex submanifold is Kähler by restricting the Fubini-Study form. Using simple connectedness and Serre duality, we have \begin{center} \begin{tikzcd} && 1 \\ & 0 && 0 \\ 1 && {h^{1, 1}} && 1 \\ & 0 && 0 \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiMCJdLFszLDEsIjAiXSxbMCwyLCIxIl0sWzIsMiwiaF57MSwgMX0iXSxbNCwyLCIxIl0sWzEsMywiMCJdLFszLDMsIjAiXSxbMiw0LCIxIl1d}{Link to Diagram} \end{quote} We know $$\chi({\mathcal{O}}_S) = (1/12)(K^2+\chi_{\mathsf{Top}})$$, and since $$K_S = {\mathcal{O}}_S$$ is trivial, we have $$c_1({\mathcal{O}}_S) = 0$$. Noting that $$h^{p, q} = H^( \Omega^p)$$, so we can sum the lower-right part of the diamond to get $$\chi({\mathcal{O}}_S) = 1 - 0 + 1 = 2$$, since we take $$p=0$$ to get $$\Omega^p = {\mathcal{O}}$$. Computing $$\chi_{\mathsf{Top}}$$, we get $$h_{1, 1} = 20$$. \end{remark} \hypertarget{monday-april-12}{% \section{Monday, April 12}\label{monday-april-12}} \begin{remark} Last time: the Lefschetz hyperplane theorem. Intersecting a projective variety of dimension $$d\geq 3$$ with a hypersurface $$S$$, the map $$\pi_1({\mathbb{P}}^3) \to \pi_1(S)$$ is an isomorphism. We saw that K3 surfaces were thus simply connected, and $$h^1(S; {\mathbb{C}}) = 0$$, so we could compute the Hodge diamond. \end{remark} \begin{example}[?] What is the Hodge diamond for a cubic surface $$S \subseteq {\mathbb{P}}^3$$, such as $$\sum x_i^3 = 0$$? We first need to compute the canonical bundle $$K$$, for which we have a useful tool: the adjunction formula. This say $$K_S = \qty{K_{{\mathbb{P}}^3} \otimes{\mathbb{P}}_{{\mathbb{P}}^3}(S) } { \left.{{}} \right|_{{S}} } = \qty{ {\mathcal{O}}(-4) \otimes{\mathcal{O}}(3) }{ \left.{{}} \right|_{{S}} } = { \left.{{ {\mathcal{O}}(-1)}} \right|_{{S}} }$$. \begin{proposition}[If a holomorphic line bundle has a section, its inverse doesn't] Let $$\mathcal{L} \to X$$ be a holomorphic line bundle. If $$h^0( \mathcal{L} ^{-1}) > 0$$, then either $$\mathcal{L} = {\mathcal{O}}$$ or $$h^0(\mathcal{L}) = 0$$. \end{proposition} \begin{slogan} If a line bundle has a section, its inverse does not. \end{slogan} \begin{proof}[?] Suppose that both $$\mathcal{L}, \mathcal{L}^{-1}$$ have a section, so $$h^0(\mathcal{L}), h^0( \mathcal{L} ) > 0$$. Let $$s, t$$ be sections of each, then $$st\in H^0( \mathcal{L} \otimes\mathcal{L}^{-1}) = H^0({\mathcal{O}}) = {\mathbb{C}}$$. So taking zero loci yields $$\operatorname{Div}(s) + \operatorname{Div}(t) = 0$$ Writing these as $$\operatorname{Div}(s) \coloneqq\sum n_D D, \operatorname{Div}(t) \coloneqq\sum n_C C$$, we have $$n_D, n_C \geq 0$$, which implies that $$\operatorname{Div}(s) = \operatorname{Div}(t) = 0$$. So $$s, t$$ are nowhere vanishing, making $${\mathcal{O}}\xrightarrow{\cdot s} \mathcal{L}$$ is an isomorphism. \end{proof} \begin{corollary}[H0 of cubic surfaces] For $$S$$ a cubic surface, $$H^0(S; K_S) = 0$$. \end{corollary} \begin{proof}[?] This follows because $$K_S = {\mathcal{O}}_S(-1)$$, so $$K_S^{-1}= {\mathcal{O}}_S(1)$$ which has a nontrivial section: namely $${\mathcal{O}}_{{\mathbb{CP}}^1}(1)$$ which has sections vanishing along hyperplanes. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-12_14-06.pdf_tex} }; \end{tikzpicture} } \end{figure} Letting $$H$$ be a hyperplane containing $$S$$, there exists an $$f\in H^0({\mathbb{P}}^3; {\mathcal{O}}_{{\mathbb{CP}}^3}(1))$$. Since $$\operatorname{Div}(f) = H$$, the restriction $${ \left.{{f}} \right|_{{S}} }$$ is a section of $${\mathcal{O}}_S(-1) = K_S^{-1}$$ which is not identically zero and vanishes along $$H \cap S$$. \end{proof} We now know $$h^0(S; K_S) = 0$$, and this equals $$h^0(S, \Omega^2) = h^{2, 0}(S)$$, so we have the following Hodge diamond: \begin{center} \begin{tikzcd} && 1 \\ & 0 && 0 \\ 0 && {h^{1, 1}} && 0 \\ & 0 && 0 \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiMCJdLFszLDEsIjAiXSxbMCwyLCIwIl0sWzIsMiwiaF57MSwgMX0iXSxbNCwyLCIwIl0sWzEsMywiMCJdLFszLDMsIjAiXSxbMiw0LCIxIl1d}{Link to Diagram} \end{quote} We have $$h^{0, 1} + h^{1, 0} = h^1 = 0$$ since $$S$$ is simply connected. We can now apply Noether's formula as before: $$\chi({\mathcal{O}}_S) = {1\over 12} (K_S^2 + \chi_{\mathsf{Top}}(S))$$. We have $$K_S = {\mathcal{O}}_S(-1)$$, so $$K_S^2 = c_1( {\mathcal{O}}(-1))^2$$, and $$\chi({\mathcal{O}}_S) = 1-0+1 = 1$$. We now want to compute $$\int_S \qty{- c_1({\mathcal{O}}_S(1)) }^2$$. We know $$c_1(\mathcal{L}) = [ \operatorname{Div}s]$$ where $$s\in H^0( \mathcal{L} )$$ is a section of a line bundle. This equals $$[H \cap S]$$. On the other hand, $$\int_S c_1 \qty{ {\mathcal{O}}_S(1)}^2$$ is the self-intersection number of $$H \cap S$$. \hfill\break Take $$H_1 \coloneqq\left\{{ x_0 = 0 }\right\}$$ and $$H_2 \coloneqq\left\{{ x_1 = 0 }\right\}$$. Points in this intersection are of the form $$[0:0:1:\zeta_6^a]$$ where $$a=1,3,5$$ since this is in the triple intersection $$H_1 \cap H_2 \cap S$$. So there are exactly 3 points here, and in fact $$\deg S =3$$. This is the same as integrating $$\int_{{\mathbb{P}}^3} c_1(S) c_1( {\mathcal{O}}(1)) c_1( {\mathcal{O}}(2))$$, which contains 3 elements in $$H^2$$ and lands in $$H^6$$, so this yields a number. \hfill\break We thus have $$K_S = {\mathcal{O}}_S(-1) \coloneqq{\mathcal{O}}_{{\mathbb{CP}}^3}(-1) { \left.{{}} \right|_{{S}} }$$. Thus $$\chi_{\mathsf{Top}}(S) = 9$$ and $$h^{1, 1} = 7$$. \end{example} \begin{example}[Hypersurfaces] Note that a degree 5 surface (a quintic) such as $$x_0^5 + x_3^5 = 0$$ would be harder, since $$h^{2, 0} \neq 0$$. We would get $$K_S = {\mathcal{O}}(-4) \otimes{\mathcal{O}}(5) { \left.{{}} \right|_{{S}} } = {\mathcal{O}}_S(1)$$, and there are nontrivial sections so $$h^0(K_S) = {\operatorname{span}}{x_0, x_1, x_2, x_3}$$. This follows because there is a map given by restriction which turns out to be an isomorphism \begin{align*} 0 \to H^0({\mathbb{P}}^3; {\mathcal{O}}(1)) & \xrightarrow{\mathop{\mathrm{res}}_S} H^0(S; {\mathcal{O}}(1)) \to 0 \\ f &\mapsto { \left.{{f}} \right|_{{S}} } .\end{align*} Injectivity isn't difficult, surjectivity is harder. We have a SES \begin{align*} 0 \to {\mathcal{O}}_{{\mathbb{CP}}^3}(-S) \to {\mathcal{O}}_{{\mathbb{CP}}^3} \to {\mathcal{O}}_S \to 0 .\end{align*} Tensor all of these with $${\mathcal{O}}(1)$$ to obtain \begin{align*} 0 \to {\mathcal{O}}_{\mathbb{CP}}^3(-4) \to {\mathcal{O}}_{{\mathbb{CP}}^3}(1) \to {\mathcal{O}}_S(1) \to 0 .\end{align*} Taking the associated LES yields \begin{center} \begin{tikzcd} {H^1({\mathcal{O}}_{{\mathbb{CP}}^3}(-4)) =_? 0} \\ \\ {H^0({\mathcal{O}}_{{\mathbb{CP}}^3}(-4)) = 0} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0({\mathcal{O}}_{{\mathbb{CP}}^3}(-1)) = {\mathbb{C}}\left\langle{x_0, x_1, x_2, x_3}\right\rangle} && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0({\mathcal{O}}_{S}(1))} \\ 0 \arrow[from=4-1, to=3-1] \arrow[from=3-1, to=3-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, from=3-3, to=3-5] \arrow[from=3-5, to=1-1, out=0, in=180] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwyLCJIXjAoXFxPT197XFxDUF4zfSgtNCkpID0gMCJdLFsyLDIsIkheMChcXE9PX3tcXENQXjN9KC0xKSkgPSBcXENDIFxcZ2Vuc3t4XzAsIHhfMSwgeF8yLCB4XzN9IixbMjQwLDYwLDYwLDFdXSxbNCwyLCJIXjAoXFxPT197U30oMSkpIixbMjQwLDYwLDYwLDFdXSxbMCwwLCJIXjEoXFxPT197XFxDUF4zfSgtNCkpID1fPyAwIl0sWzAsMywiMCJdLFs0LDBdLFswLDFdLFsxLDIsIiIsMCx7ImNvbG91ciI6WzI0MCw2MCw2MF19XSxbMiwzXV0=}{Link to Diagram} \end{quote} This gives us a way to relate things back to the cohomology of $${\mathbb{CP}}^3$$. Showing that the indicated term is zero involves computing Čech cohomology. It turns out that $$h^0(K_S) = 4$$ here, and it turns out that the Hodge diamond is the following: \begin{center} \begin{tikzcd} && 1 \\ & 0 && 0 \\ 4 && {h^{1, 1} = 45} && 4 \\ & 0 && 0 \\ && 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsOSxbMiwwLCIxIl0sWzEsMSwiMCJdLFszLDEsIjAiXSxbMCwyLCI0Il0sWzQsMiwiNCJdLFsxLDMsIjAiXSxbMywzLCIwIl0sWzIsNCwiMSJdLFsyLDIsImheezEsIDF9ID0gNDUiXV0=}{Link to Diagram} \end{quote} Here $$K_S^2 = c_1({\mathcal{O}}_S(1))^2 = 5$$ and $$\chi_{\mathsf{Top}}= 55$$. \end{example} \begin{example}[Products] Consider now a product of curves $$C \times D$$ of genera $$g, h$$ respectively. Computing the Hodge diamond is easy here due to the Kunneth formula: \begin{align*} H^k(S; {\mathbb{C}}) = \bigoplus_{i+j = k} H^i(C; {\mathbb{C}}) \otimes H^j(D; {\mathbb{C}}) .\end{align*} What is the actual map? Take cohomology classes $$[\alpha], [\beta]$$, closed $$i$$ and $$j$$ forms respectively. The surface has two maps: \begin{center} \begin{tikzcd} S && C \\ \\ D \arrow["{\pi_C}", from=1-1, to=1-3] \arrow["{\pi_D}"', from=1-1, to=3-1] \end{tikzcd} \end{center} Here we send $$[\alpha] \otimes[ \beta] \mapsto [ \pi_C^* \alpha\wedge \pi_D^* \beta]$$ where we take pullbacks. Note that $$\pi_D, \pi_C$$ are holomorphic maps, and pullbacks of $$(p, q)$$ forms are still $$(p, q)$$ forms. Thus the Kunneth formula gives a decomposition \begin{align*} H^{p, q}(S; {\mathbb{C}}) = \sum_{\substack{ i_1 + j_1 = p \\ i_2 + j_2 = q }} H^{i_1, j_1}(C) \oplus H^{i_2, j_2}(D) .\end{align*} So we can tensor'' the Hodge diamonds: \begin{center} \begin{tikzcd} & 1 &&&& 1 \\ g && g && h && h \\ & 1 &&&& 1 \\ \\ &&& 1 \\ && {g+h} && {g+h} \\ & gh && {2 + 2gh} && gh \\ && {g+h} && {g+h} \\ &&& 1 \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTcsWzEsMCwiMSJdLFswLDEsImciXSxbMiwxLCJnIl0sWzEsMiwiMSJdLFs1LDAsIjEiXSxbNCwxLCJoIl0sWzYsMSwiaCJdLFs1LDIsIjEiXSxbMyw0LCIxIl0sWzIsNSwiZytoIl0sWzQsNSwiZytoIl0sWzIsNywiZytoIl0sWzQsNywiZytoIl0sWzEsNiwiZ2giXSxbNSw2LCJnaCJdLFszLDgsIjEiXSxbMyw2LCIyICsgMmdoIl1d}{Link to Diagram} \end{quote} \end{example} \begin{remark} Check out complete intersections. \end{remark} \hypertarget{blowups-and-blowdowns-wednesday-april-14}{% \section{Blowups and Blowdowns (Wednesday, April 14)}\label{blowups-and-blowdowns-wednesday-april-14}} \begin{definition}[Blowup] Let $$S \in {\mathsf{Mfd}}_{\mathbb{C}}^2$$ be a complex surface and $$p\in S$$ a point, and let $$(x, y)$$ be local holomorphic coordinates on a neighborhood of $$U$$ containing $$p$$. Without loss of generality, $$p = (0, 0)$$ in these coordinates. Set $$U^* \coloneqq U \setminus\left\{{p}\right\}$$, and consider the holomorphic map \begin{align*} \phi: U^* &\to U \times{\mathbb{CP}}^2 \\ (x, y) & \mapsto ( (x, y), [x: y] ) .\end{align*} We'll define the \textbf{blowup at $$p$$} to be $$\mathop{\mathrm{Bl}}_p(U) { \operatorname{cl}} (\phi(U^*))$$ to be the closure of the image of $$U^*$$. \end{definition} \begin{observation} There is a map $$\mathop{\mathrm{Bl}}_p(U) \to U$$ given by projection onto the first coordinate which is the identity on $$U^*$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-14_13-56.pdf_tex} }; \end{tikzpicture} } \end{figure} Here $$q$$ maps to the pair $$(q, s)$$ where $$s$$ is the slope of a line through $$q$$, and this will be continuous. \todo[inline]{? Missed part} We claim that $$\pi_U^{-1}(0, 0) \subset \mathop{\mathrm{Bl}}_p(U) = \left\{{ p }\right\} \times{\mathbb{CP}}^1$$, and for a fixed $$9x_0, y_0) \in U^*$$, considering $$\phi(x_0 t, y_0 t)$$ as $$t\to 0$$, we can write \begin{align*} ( (x_0 t, y_0 t), [x_0: y_0] ) \in U \times{\mathbb{CP}}^1 \\ \overset{t\to 0} ( (0, 0) [x_0: y_0] ) \subset { \operatorname{cl}} (\phi(U^*)) .\end{align*} So approaching $$(0, 0)$$ along any slope $$s$$ just yields the point $$(0, s)$$ in the blowup. \end{observation} \begin{remark} We can thus write \begin{align*} \mathop{\mathrm{Bl}}_pS S \setminus\left\{{p}\right\} {\textstyle\coprod}_{U^*} \mathop{\mathrm{Bl}}_p U .\end{align*} Writing $$\pi: \mathop{\mathrm{Bl}}_pS\to S$$, we have $$\pi^{-1}(p) \cong {\mathbb{CP}}^1$$ and $$\pi^{-1}(q)$$ is a point for all $$q\neq p$$. Then all limits approaching $$p$$ in $$S$$ turn into distinct limit points in $$\mathop{\mathrm{Bl}}_p(S)$$ \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-14_14-06.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{remark} \begin{slogan} The blowup separates all tangent directions at $$p$$. \end{slogan} \begin{example}[?] Consider \begin{align*} \left\{{ y^2 = x^3 - x^2 }\right\} \subseteq {\mathbb{C}}^2 .\end{align*} This yields a nodal curve with a double-point: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-14_14-11.pdf_tex} }; \end{tikzpicture} } \end{figure} Here we'll consider $$\mathop{\mathrm{Bl}}_{(0, 0)} {\mathbb{C}}^2$$. \begin{definition}[Strict Transform] Letting $$C \subset S$$ be a curve, define the \textbf{strict transform} \begin{align*} \widehat{C} \coloneqq{ \operatorname{cl}} ( \pi^{-1}(C \setminus\left\{{p}\right\} ) ) .\end{align*} \end{definition} Note that approaching by different sequences yields different limiting slopes \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-14_14-15.pdf_tex} }; \end{tikzpicture} } \end{figure} The curve in the blowup is called the \textbf{exceptional divisor}. \end{example} \begin{example}[?] Consider all lines in $${\mathbb{CP}}^2$$ through $$[0:0:1]$$, which we can model in the following way: \begin{figure} \centering \includegraphics{figures/image_2021-04-14-14-18-15.png} \caption{image\_2021-04-14-14-18-15} \end{figure} These are in bijection with $${\mathbb{CP}}^1$$ since there is always a unique line through $$[0:0:1]$$ and $$[s:t:0]$$, where the latter is a copy of $${\mathbb{CP}}^1$$ as $$s,t$$ are allowed to vary. So consider $$\mathop{\mathrm{Bl}}_{p}{\mathbb{CP}}^2$$ for $$p=[0:0:1]$$, and consider the strict transforms of the lines $$L$$ to obtain $$\widehat{L} \subset \mathop{\mathrm{Bl}}_p {\mathbb{CP}}^2$$. Any two are disjoint since they pass through different slopes of the exceptional divisor. Thus the red lines in the blowup go through distinct slopes, yielding a fibration of $${\mathbb{CP}}^1$$s: \begin{figure} \centering \includegraphics{figures/image_2021-04-14-14-24-31.png} \caption{image\_2021-04-14-14-24-31} \end{figure} So consider the map \begin{align*} \sigma: \mathop{\mathrm{Bl}}_p {\mathbb{CP}}^2 &\to {\mathbb{CP}}^2 \\ p \in \widehat{L} &\mapsto [0:s:t] .\end{align*} which projects points to the boundary copy of $${\mathbb{CP}}^1$$: \includegraphics{figures/image_2021-04-14-14-25-44.png} We can't necessarily project from the blue point itself, but if we add in the data of a tangent vector at that point, the map becomes well-defined. Thus the blowup makes projecting from a point in $${\mathbb{CP}}^2$$ to a line in $${\mathbb{CP}}^2$$ a well-defined map on $$\mathop{\mathrm{Bl}}{\mathbb{CP}}^2$$. \end{example} \begin{remark} This is referred to as $${\mathbb{F}}_1$$, the \textbf{first Hirzebruch surface}. \end{remark} \begin{proposition}[Blowup for smooth manifolds is connect-sum with CP2] For $$S\in {\mathsf{Mfd}}_{\mathbb{R}}(C^\infty)$$ a smooth manifold, we can identify \begin{align*} \mathop{\mathrm{Bl}}_p S = S \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu .\end{align*} \end{proposition} \begin{proof}[?] It suffices to work in coordinate charts and prove this for $$p=0$$. \begin{claim} \begin{align*} \mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2 = { \operatorname{Tot} }( {\mathcal{O}}_{{\mathbb{CP}}^1}(-1) ) .\end{align*} \end{claim} Recall that this was the tautological line bundle that whose fibers at a point $$p\in {\mathbb{CP}}^1$$ was the line in $${\mathbb{C}}^2$$ spanned by $$p$$. We can write this as $$\left\{{ [x:y] {~\mathrel{\Big|}~}(x, y) \in L_{[x:y]} }\right\}$$: \begin{figure} \centering \includegraphics{figures/image_2021-04-14-14-32-58.png} \caption{image\_2021-04-14-14-32-58} \end{figure} We have $${\mathcal{O}}(-1) \xrightarrow{\sim} \mkern 1.5mu\overline{\mkern-1.5mu{\mathcal{O}}(1)\mkern-1.5mu}\mkern 1.5mu$$, where this map is a diffeomorphism that can be constructed using a Hermitian metric. However we can identify $${\mathcal{O}}(1)$$ with the set of lines in $${\mathbb{CP}}^2$$ through $$[0:0:1]$$, leaving out the point $$[0:0:1]$$ itself. This follows by checking that there exists a section that vanishes at only one point. In fact $${ \operatorname{Tot} }{\mathcal{O}}(1)$$ is diffeomorphic to the complement of a ball in $${\mathbb{CP}}^2$$, which ends up precisely being taking a connect-sum. So we obtain $$\mathop{\mathrm{Bl}}_{0} {\mathbb{C}}^2 \cong {\mathbb{C}}^2 \mathop{ \Large\scalebox{0.8}{\raisebox{0.4ex}{\#}}}\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu$$. \end{proof} \begin{proof}[Alternative] Cut out a ball $$B^4 \subseteq {\mathbb{C}}^2$$, so $${{\partial}}B^4 = S^3 = \left\{{ {\left\lvert {x} \right\rvert}^2 + {\left\lvert {y} \right\rvert}^2 = {\varepsilon}}\right\}$$. Then $$\mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2$$ is the result of collapsing $$S^3$$ along an $$S^1{\hbox{-}}$$foliation $$(e^{i\theta} x, e^{i\theta}y)$$. This has an $$S^2$$ quotient, yielding the Hopf fibration \begin{align*} S^1 \hookrightarrow S^3 \to S^2 .\end{align*} \end{proof} \begin{exercise}[?] Show that the blowup over $${\mathbb{R}}$$ is gluing in a mobius strip. \end{exercise} See the Tate curve! \hypertarget{friday-april-16}{% \section{Friday, April 16}\label{friday-april-16}} \begin{remark} Last time: we defined the blowup $$\mathop{\mathrm{Bl}}_0{\mathbb{C}}^2$$ as the closure of \begin{align*} \mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2 \coloneqq{ \operatorname{cl}} \left\{{ (x, y), [x:y] {~\mathrel{\Big|}~}(x, y) \neq 0 }\right\} \subseteq {\mathbb{C}}^2 \times{\mathbb{CP}}^2 .\end{align*} This had the effect of adding in all limits of slopes as points approach $$(0, 0) \in {\mathbb{C}}^2$$. We defined this using local holomorphic coordinate charts to $${\mathbb{C}}^2$$. Why is this a complex manifold? We can cover it with charts: given a point $$(x, \mu)$$ where $$\mu = {y \over x}\in {\mathbb{P}}^1$$ is a slope, we can form a first chart by sending \begin{align*} (x, \mu) \mapsto \left\{{ (x, x\mu), [1: \mu] }\right\} .\end{align*} This yields the first chart, as long as the slope is not infinite, so this applies to all finite slopes. The second chart will work for all nonzero slopes, where we take \begin{align*} (v, y)\in {\mathbb{C}}^2 \mapsto \left\{{ (yv, y), [v: 1] }\right\} .\end{align*} Note that restricting to $$(x, y) = (0, 0)$$, these give the standard $${\mathbb{C}}{\hbox{-}}$$charts on $${\mathbb{CP}}^2$$. How do these two charts glue? When $$\mu, \nu \neq 0$$, we have well-defined transition functions $$\mu = \nu ^{-1}$$ and $$x=y\nu$$. \end{remark} \begin{remark} Recall that for a complex curve $$C \in {\mathsf{Mfd}}_{\mathbb{C}}^2$$, we have the blowup morphism $$\pi: \mathop{\mathrm{Bl}}_pS\to S$$ and we defined the \textbf{strict transform} $$\widehat{C} \coloneqq{ \operatorname{cl}} \pi^{-1}(C\setminus\left\{{{\operatorname{pt}}}\right\})$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-02.pdf_tex} }; \end{tikzpicture} } \end{figure} Here $$E={\mathbb{CP}}^1$$ is the exceptional curve of the blowup, and intersects the curve twice. This has the effect of changing $$D$$ into an embedded curve. \begin{quote} Note that here $$\pi^* D = \widehat{D} + 2E$$, where we'll define this next. \end{quote} \end{remark} \begin{definition}[Pullback of a Curve] The \textbf{pullback} of $$C$$, denoted $$\pi^* C$$, is constructed by writing $$C = V(f)$$ locally. We then set $$\pi^* C \coloneqq V( \pi^* f)$$. \end{definition} \begin{example}[?] Take $$C \coloneqq\left\{{ y=x }\right\} \subset {\mathbb{C}}^2$$ and consider $$\mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2$$. Then \begin{align*} \widehat{C} \coloneqq{ \operatorname{cl}} \left\{{ \qty{ (x, x), [x:x]} {~\mathrel{\Big|}~}x\neq 0 }\right\} = { \operatorname{cl}} \left\{{ \qty{ (x, x), [1:1] } {~\mathrel{\Big|}~}x\neq 0 }\right\} \subset \mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2 .\end{align*} By projecting onto the first component, $$\pi:\widehat{C} \xrightarrow{\sim} C$$ is an isomorphism. We can compute the pullback: we first have $$\pi^* C = \pi^* V(y-x) = V( \pi^*(y-x))$$, so consider $$\pi^*(y-x)$$ in the coordinate chart $$(x, \mu)$$. In this chart, $$y=x\mu$$, and so $$\pi^*(y-x) = x\mu - x = x(\mu - 1)$$, and so \begin{align*} V(\pi^*(y-x)) = V(x) + V(\mu -1) \implies \pi^* C = E + \widehat{C} \text{ as a divisor} .\end{align*} \end{example} \begin{example}[A nodal curve] Take the nodal curve $$C = \left\{{ y^2 - x^3 + x^2 }\right\}$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-15.pdf_tex} }; \end{tikzpicture} } \end{figure} The pullback is then given by \begin{align*} \pi^* C &= V( \pi^* (y^2 - x^3 + x^2) ) \\ &= V(\mu^2x^2 -x^3 + x^2) \\ &= V(x^2) + V(\mu^2 - x + 1) \\ &= 2V(x) + V(\mu^2 -x + 1) .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-17.pdf_tex} }; \end{tikzpicture} } \end{figure} In the second coordinate chart, we have \begin{align*} \pi^* C = V(y^2 - y^4 \nu^3 +y^2 \nu^2) = 2V(y) + V(1-y\nu^3 + \nu^2 .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-21.pdf_tex} }; \end{tikzpicture} } \end{figure} Gluing along $$\mu, \nu \neq 0$$ we get the following picture for $$\pi^* C$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-22.pdf_tex} }; \end{tikzpicture} } \end{figure} Writing $$C = \left\{{ x=0 }\right\}$$, note that $$\widehat{C}$$ doesn't intersect the first coordinate chart. In the $$\mu, x$$ coordinate chart, for example, we can't get an infinite slop: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-25.pdf_tex} }; \end{tikzpicture} } \end{figure} \end{example} \hypertarget{change-in-canonical-bundle-formula}{% \subsection{Change in Canonical Bundle Formula}\label{change-in-canonical-bundle-formula}} \begin{question} Given $$\Omega^2_S = K_S \to S$$ the canonical line bundle, can we relate $$K_{\mathop{\mathrm{Bl}}_p S}$$ to $$K_S$$? \end{question} \begin{proposition}[Canonical of a blowup] \begin{align*} K_{\mathop{\mathrm{Bl}}_p S} = \pi^* K_S \otimes{\mathcal{O}}_S(E) .\end{align*} \end{proposition} \begin{proof}[?] We'll abbreviate $$\widehat{S} \coloneqq\mathop{\mathrm{Bl}}_p(S)$$. Let $$\omega$$ be a local section of $$K_S$$ near $$p$$, and in coordinate charts $$(x, y)$$, write $$\omega = dx \wedge dy$$. In the first coordinate chart on the blowup, we can write \begin{align*} \pi^* \omega = dx \wedge d(x\mu) = dx \wedge (\mu dx + x d\mu) = x\,dx\wedge d\mu .\end{align*} Note that $$V(x) = E$$, and that pulling back the canonical bundle yields something vanishing to order 1 (?). So $$\pi^* K_S$$ is isomorphic to the subsheaf of $$K_{\widehat{S}}$$ whose sections vanish along $$E$$, which is isomorphic to $$K_{\widehat{S}} \otimes{\mathcal{O}}(-E)$$, since the latter are the functions which vanish along $$E$$. Tensoring both sides with $${\mathcal{O}}(E)$$ yields \begin{align*} K_{\widehat{S}} = \pi^* K_S \otimes{\mathcal{O}}_{\widehat{S}}(E) \end{align*} as a line bundle, or in divisor notation $$K_{\widehat{S}} = \pi^* K_S + E$$ where we take the divisor representing the line bundle instead. \end{proof} \begin{remark} Using $$\pi: \widehat{S}\to S$$, we get pullback maps \begin{align*} \pi^*: H^2(S; {\mathbb{Z}}) &\to H^2(\widehat{S}; {\mathbb{Z}}) \\ \pi^*: \operatorname{Div}(S) &\to \operatorname{Div}(\widehat{S}) .\end{align*} These are compatible in the sense that \begin{align*} [\pi^* C] = \pi^* [C] .\end{align*} . This can be seen by expressing $${\mathcal{O}}_S(C) \cong {\mathcal{O}}_S(A-b)$$ for $$A, B$$ hyperplane section. We can assume $$A, B$$ avoid $$p$$ in their projective embeddings, making $$[C] = [A] - [B]$$ since $$c_1({\mathcal{O}}_S(c)) = [C]$$ is the fundamental class of $$C$$. So it suffices to prove the formula for curves \emph{not} passing through $$p$$, but this is obvious! It follows from the fact that $$\pi: \widehat{S} \setminus E \xrightarrow{\sim} S\setminus\left\{{p}\right\}$$ is an isomorphism. \end{remark} \begin{remark} In fact, \begin{align*} H^2(\widehat{S}; {\mathbb{Z}}) \cong \pi^* H^2(S, {\mathbb{Z}}) \oplus {\mathbb{Z}}[E] .\end{align*} , which follows from Mayer-Vietoris. So this adds one to the rank. \end{remark} \hypertarget{monday-april-19}{% \section{Monday, April 19}\label{monday-april-19}} \begin{remark} Recall that we have the following: \begin{align*} H^2(\widehat{S}; {\mathbb{Z}}) = \pi^* H^2(S; {\mathbb{Z}}) \oplus {\mathbb{Z}}[E] \end{align*} where $$E$$ is the exceptional curve, which follows from Mayer-Vietoris. We can write $$\widehat{S} = S \# \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu$$, and by excision $$H^2(S\setminus{\mathbb{B}}^4) = H^2(S)$$. So we get a LES \begin{center} \begin{tikzcd} {H^3(S, S\setminus B)} \\ \\ {H^2(S\setminus{\mathbb{B}}^4)} && {H^2(S)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(S, S\setminus B)=0} \\ \\ && {H^1(S)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(S, S\setminus B)=0} \arrow[from=5-3, to=5-5] \arrow[from=5-5, to=3-1] \arrow["\sim", from=3-1, to=3-3] \arrow[from=3-3, to=3-5] \arrow[from=3-5, to=1-1] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNixbMiw0LCJIXjEoUykiXSxbNCw0LCJIXjEoUywgU1xcc20gQik9MCIsWzAsNjAsNjAsMV1dLFswLDIsIkheMihTXFxzbSBcXEJCXjQpIl0sWzIsMiwiSF4yKFMpIl0sWzQsMiwiSF4yKFMsIFNcXHNtIEIpPTAiLFswLDYwLDYwLDFdXSxbMCwwLCJIXjMoUywgU1xcc20gQikiXSxbMCwxXSxbMSwyXSxbMiwzLCJcXHNpbSJdLFszLDRdLFs0LDVdXQ==}{Link to Diagram} \end{quote} We have $$H^i(S, S\setminus{\mathbb{B}}^4) = H^i(T, T\setminus{\mathbb{B}}^4) = H^i( {\mathbb{B}}^4, {{\partial}})$$, and by Poincaré-Lefschetz duality, this is isomorphic to $$H_{4-i}({\mathbb{B}}^4)$$. This is equal to 0 if $$i\neq 0$$ or $$4$$. Writing $$\widehat{S} = (S\setminus{\mathbb{B}}^4) {\textstyle\coprod}_{S^3} (\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu\setminus{\mathbb{B}}^4)$$ and applying Mayer-Vietoris yields \begin{center} \begin{tikzcd} {H^2(\widehat{S})} && {H^2(S\setminus{\mathbb{B}}^4) \oplus H^2(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu \setminus{\mathbb{B}}^4)} && \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(S^3) =0} \\ \\ && \cdots && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(S^3) =0 } \arrow[from=3-5, to=1-1] \arrow["\sim", from=1-1, to=1-3] \arrow[from=1-3, to=1-5] \arrow[from=3-3, to=3-5] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbNCwyLCJIXjEoU14zKSA9MCAiLFswLDYwLDYwLDFdXSxbMCwwLCJIXjIoXFxoYXR7U30pIl0sWzIsMCwiSF4yKFNcXHNtIFxcQkJeNCkgXFxvcGx1cyBIXjIoXFxiYXJ7XFxDUF4yfSBcXHNtIFxcQkJeNCkiXSxbNCwwLCJIXjIoU14zKSA9MCIsWzAsNjAsNjAsMV1dLFsyLDIsIlxcY2RvdHMiXSxbMCwxXSxbMSwyLCJcXHNpbSJdLFsyLDNdLFs0LDBdXQ==}{Link to Diagram} \end{quote} Combining this with the isomorphisms from earlier, we can write the direct sum as $$H^2(S) \oplus H^2( \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu)$$ where the latter is equal to $${\mathbb{Z}}\ell = [E]$$ for $$\ell$$ a line class. \end{remark} \begin{question} What is the intersection form on $$H^2(\widehat{S}; {\mathbb{Z}})$$? \end{question} \begin{remark} Using the proposition, along with the fact that \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item its an orthogonal decomposition, \item $$\pi^*$$ is an isometry, and \item $$[E]^2 = -1$$, \end{enumerate} we know that the Gram matrix for $$H^2(\widehat{S})$$ is the same as that for $$H^1(S) \oplus [-1]$$, i.e.~it is of the form \begin{align*} \begin{bmatrix} A & 0 \\ 0 & -1 \end{bmatrix} .\end{align*} \end{remark} \begin{proof}[of 2] Consider $$[\Sigma_1], [\Sigma_2]\in H^2(S; {\mathbb{Z}})$$ where the $$\Sigma_i$$ are real surfaces, and suppose $$\Sigma_1 \pitchfork\Sigma_2$$ and $$p\not\in \Sigma_1, \Sigma_2$$. We then have \begin{align*} [ \pi^{-1}( \Sigma_i ) ] = \pi^* [ \Sigma_i ] .\end{align*} \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{35pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-19_14-07.pdf_tex} }; \end{tikzpicture} } \end{figure} The intersection number is preserved because $$\pi$$ is generically injective. \end{proof} \begin{proof}[of 1] It also follows that if $$p\not\in \Sigma$$, $$\pi^*[\Sigma] = [\pi^{-1}\Sigma]$$ where the latter is disjoint from $$E$$. So $$\pi^*[\Sigma] \cdot E = 0$$. \end{proof} \begin{proof}[of 3] Since $$[E] \sim [\text{line}] \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu\setminus{\mathbb{B}}^4$$, and $$E^2 = [E] \cdot [E] = -1$$ since the orientations disagree in $$\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu$$. \end{proof} \begin{proposition}[Computing the pullback of a curve] Let $$C \subset S$$ be a curve on a surface and suppose $$C$$ is locally cut out by \begin{align*} f(x, y) = a_{m, 0} x^m + a_{n-1, 1} x^{m-1} y + \cdots + a_{0, m} y^m + O(x^{m+1}, y^{m+1}) ,\end{align*} near $$p\in S$$, so the lowest order terms in the Taylor expansion are degree $$m$$. Then \begin{align*} \pi^* C = \widehat{C} + mE .\end{align*} \end{proposition} \begin{proof}[?] On the blowup, take local coordinates $$(x, \mu)$$ where $$y = x\mu$$ and write \begin{align*} V(\pi^* f) &= V( x^m \qty{a_{m, 0} + a_{m-1, 1} \mu + \cdots + a_{0, m} \mu^m + O(x^{m+1}, \mu^{m+1} ) } ) \\ &= m V(x) + V( a_{m, 0} + \cdots ) \\ &= E + \widehat{C} .\end{align*} \end{proof} \begin{example}[?] Take \begin{align*} C = \left\{{ y^2 = x^3-x^2 }\right\} \subseteq {\mathbb{C}}^2 ,\end{align*} where $$\mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2 \to C$$. Then $$\pi^* C = \widehat{C} + 2E$$, so \begin{align*} C = V(x^2 + y^2 + O(\deg(3)) .\end{align*} . \end{example} \begin{corollary}[Computing the square of the strict transform] $$\widehat{C}^2 = C^2 - m^2$$. \end{corollary} \begin{proof}[?] Write $$\pi^* C = \widehat{C} + mE$$, then $$\widehat{C} = \pi^* C - mE$$ implies that $$\widehat{C}^2 = (\pi^* C - mE)^2$$. This equals \begin{align*} (\pi^* C)^2 - 2m \pi^* C\cdot E + m^2 E^2 &= C^2 - 0 - m^2 \\ &= C^2 - m^2 ,\end{align*} where we've used (2), (1), and (3) respectively to identity these terms. \end{proof} \begin{example}[?] Let \begin{align*} C\coloneqq\left\{{ zy^2 = x^3 - x^2 z }\right\} \subset {\mathbb{CP}}^2 ,\end{align*} then $$C^2 = (3\ell)^2 = 9$$. The multiplicity of $$C$$ at the point $$[0:0:1]$$ is 2. Taking the coordinate chart $$\left\{{ z=1 }\right\} \cong {\mathbb{C}}^2$$, we recover the curve $$y^2 = x^3 - x^2$$ which has multiplicity 2 at $$(0, 0)$$. We can conclude $$\widehat{C} = \mathop{\mathrm{Bl}}_{[0:0:1]} {\mathbb{CP}}^2$$ has self-intersection number $$\widehat{C}^2 = 9-2^2 = 5$$. \end{example} \begin{theorem}[Castelnuovo Contractibility Criterion] Let $$S$$ be a complex surface and let $$E \subset S$$ be a holomorphically embedded $${\mathbb{CP}}^2$$ such that $$E^2 = -1$$ Then there exists a smooth surface $$\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu$$ and $$p\in \mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu$$ such that $$S = \mathop{\mathrm{Bl}}_p \mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu$$ with $$E$$ as the exceptional curve. \end{theorem} \begin{definition}[Blowdown] This $$\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu$$ is called the \textbf{blowdown} of $$S$$ along $$E$$. \end{definition} \begin{remark} Note that this is the exact situation when we blow things up. This is a converse: if we have something that looks like a blowup, we can find something that blows up to it. \end{remark} \begin{exercise}[?] Show that the category $${\mathsf{Mfd}}_{\mathbb{C}}$$ is not closed under blowdowns, i.e.~there is no blowdown of a holomorphically embedded $${\mathbb{CP}}^1$$, say $$E$$, with $$E^2 = 1$$. \begin{quote} Hint: think about $${\mathbb{CP}}^2$$. \end{quote} \end{exercise} \begin{remark} This is interesting because there does exist a blowdown in the smooth category $${\mathsf{Mfd}}(C^\infty({\mathbb{R}}))$$. This is because $$S \to S\# \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu$$ and $$S\to S\# {\mathbb{CP}}^2$$ are indistinguishable here. One can just reverse orientations. \end{remark} \begin{example}[?] A complex surface with a holomorphically embedded $${\mathbb{CP}}^1$$ of self intersection $$-1$$. Let $$p, q\in {\mathbb{CP}}^2$$ be distinct points, and let $$\mathop{\mathrm{Bl}}_{p, q} {\mathbb{CP}}^2 \coloneqq\mathop{\mathrm{Bl}}_p \mathop{\mathrm{Bl}}_q {\mathbb{CP}}^2$$. Note that these two operations commute since these are distinct points and blowing up is a purely local operation. Let $$\ell \subset {\mathbb{CP}}^2$$ be the unique line through $$p$$ and $$q$$. Viewing $$p, q$$ as lines in $${\mathbb{C}}^3$$, they span a unique plane, which is a line in projective space, so this makes sense and we can write $$\ell \approx {\operatorname{span}}\left\{{ p, q }\right\}$$. Since $$\ell$$ is defined by a linear equation in local coordinates near $$p, q$$, we have $${\operatorname{mult}}_p \ell = {\operatorname{mult}}_q \ell = 1$$. We hve \begin{align*} \widehat{\ell} = \pi^* \ell - E_p - E_q \\ \widehat{\ell}^2 = \ell^2 - 1^2 - 1^2 = 1-1-1 = - 1 .\end{align*} Under $$\pi: \mathop{\mathrm{Bl}}_{p, q} {\mathbb{CP}}^2 \to {\mathbb{CP}}^2$$, we have $$\widehat{\ell} \xrightarrow{\sim} \ell$$. \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{35pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-19_14-40.pdf_tex} }; \end{tikzpicture} } \end{figure} Here since all of the lower order terms have degree 1, there is a well-defined tangent line. Since $$\ell \cong {\mathbb{CP}}^2$$, we have $$\widehat{\ell }\cong {\mathbb{CP}}^2$$. Letting $$\sigma$$ be the blowdown of $$\widehat{\ell}$$, we have \begin{center} \begin{tikzcd} & {\mathop{\mathrm{Bl}}_{p, q}{\mathbb{CP}}^2} \\ \\ {{\mathbb{CP}}^1 \times{\mathbb{CP}}^2} && {{\mathbb{CP}}^2} \arrow["\sigma"', from=1-2, to=3-1] \arrow["\pi", from=1-2, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMSwwLCJcXEJsX3twLCBxfVxcQ1BeMiJdLFsyLDIsIlxcQ1BeMiJdLFswLDIsIlxcQ1BeMSBcXGNyb3NzIFxcQ1BeMSJdLFswLDIsIlxcc2lnbWEiLDJdLFswLDEsIlxccGkiXV0=}{Link to Diagram} \end{quote} \end{example} \begin{remark} There's a way to do this with Kirby Calculus. \end{remark} \hypertarget{wednesday-april-21}{% \section{Wednesday, April 21}\label{wednesday-april-21}} \begin{remark} Why can't one blow down a curve $$E\cong {\mathbb{CP}}^1$$ with $$E^2 = 1$$ in a complex surface? Disproof: consider $$S\coloneqq{\mathbb{CP}}^2$$ and $$E$$ a line, where $$E^2 = 1$$. If there were a blowdown in the complex analytic category \begin{align*} S &\to \mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \\ E &\mapsto {\operatorname{pt}} .\end{align*} But $$\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \cong_{\mathsf{Top}}S^4$$, since $$S^4 \# {\mathbb{CP}}^2 \cong {\mathbb{CP}}^2$$, and this would yield a complex structure on $$S^4$$ -- a contradiction. This also follows because $$\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \in \mathbb{Z}\operatorname{HS}^4$$, and Noether's formula implies that every $$\mathbb{Z}\operatorname{HS}^4$$ has no complex structure. \end{remark} \begin{remark} Recall that we were considering the following: \begin{center} \begin{tikzcd} & {\mathop{\mathrm{Bl}}_{p, q}{\mathbb{CP}}^2} \\ \\ {{\mathbb{CP}}^1 \times{\mathbb{CP}}^2} && {{\mathbb{CP}}^2} \arrow["\sigma"', from=1-2, to=3-1] \arrow["\pi", from=1-2, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMyxbMSwwLCJcXEJsX3twLCBxfVxcQ1BeMiJdLFsyLDIsIlxcQ1BeMiJdLFswLDIsIlxcQ1BeMSBcXGNyb3NzIFxcQ1BeMSJdLFswLDIsIlxcc2lnbWEiLDJdLFswLDEsIlxccGkiXV0=}{Link to Diagram} \end{quote} Let $$\mkern 1.5mu\overline{\mkern-1.5mu\ell\mkern-1.5mu}\mkern 1.5mu \subset \mathop{\mathrm{Bl}}_{p, q}({\mathbb{CP}}^2)$$ the strict transform of a line through $$p, q$$ with $$\widehat{\ell}^2 = -1$$. Goal: we want to construct the map $$\sigma$$ sending $$\widehat{\ell}$$ to a single point. Let $$r\in \mathop{\mathrm{Bl}}_{p, q}{\mathbb{CP}}^2$$, then there are three possibilities: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$r\in {\mathbb{CP}}^2 \setminus\left\{{ p, q }\right\}$$ \item $$r\in E_p$$ \item $$r\in E_q$$ \end{enumerate} If a point $$r \neq p, q$$, we can take lines $$\ell_{pr}. \ell_{qr}$$. We can take slopes of these lines to get points in $${\mathbb{CP}}^1$$, and in fact it's the exceptional divisor (since these are sets of slopes through a point). So we can map \begin{align*} r \mapsto \begin{cases} ( {\mathrm{slope}}_p \ell_{pr}, {\mathrm{slope}}_q \ell_{qr} ) \in {\mathbb{CP}}^2 \times{\mathbb{CP}}^2 & \text{Case 1} \\ (r, {\mathrm{slope}}_q \ell_{qp} ) & \text{Case 2} \\ ({\mathrm{slope}}_p \ell_{pq}, r) &\text{Case 3} . \end{cases} \end{align*} This is clearly continuous, is this injective? The outputs will be the same for any point on the line between $$p$$ and $$q$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-21_14-16.pdf_tex} }; \end{tikzpicture} } \end{figure} So this realizes the blowdown map, since $$\Phi \widehat{\ell}_{pq} ) = {\operatorname{pt}}$$ and restricting it to the complement of the line is injective. \end{remark} \hypertarget{spin-and-spinc-groups}{% \subsection{Spin and Spinc Groups}\label{spin-and-spinc-groups}} \begin{remark} Goal: show that $$3[\ell]$$ can't be realized by a sphere, we'll need Rohklin's theorem for this. Let $$(V, {\left\langle {{-}},~{{-}} \right\rangle})$$ be an inner product space, and assume the inner product is positive-definite. Recall that the tensor algebra is defined as $$T(V) \coloneqq\bigoplus _{n\geq 0} V^{\otimes n}$$. \end{remark} \begin{definition}[Clifford Algebra] Define the \textbf{Clifford Algebra} of $$V$$ as \begin{align*} { \operatorname{Cl}} (V) \coloneqq T(V) / \left\langle{ v\otimes v + {\left\lVert {v} \right\rVert}^2 1 }\right\rangle .\end{align*} \end{definition} \begin{example}[The reals] Take $${\mathbb{R}}$$ with the standard inner product, so $${\left\langle {x},~{y} \right\rangle} \coloneqq xy$$. Then $$T({\mathbb{R}}) = \bigoplus _{n\geq 0} {\mathbb{R}}$$. Letting $$\left\{{ e }\right\}$$ be a basis of $${\mathbb{R}}$$, we have $$T({\mathbb{R}}) = {\mathbb{R}}\oplus {\mathbb{R}}e \oplus {\mathbb{R}}(e^2) \oplus \cdots \cong {\mathbb{R}}[x]$$ by sending $$e^n\mapsto x^n$$. Since $${\left\lVert {e} \right\rVert} = 1$$, and we mod out by $$e^2 + {\left\lVert {e} \right\rVert}^2 1$$ where $$e^2 = -1$$ and thus \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}, {\left\langle {{-}},~{{-}} \right\rangle}_\text{std}) \cong {\mathbb{R}}[x] / \left\langle{ x^2 = -1 }\right\rangle\cong {\mathbb{C}} .\end{align*} The denominator is referred to as the \textbf{Clifford relation}. \end{example} \begin{example}[More reals] Take $${\mathbb{R}}^2$$ with the standard inner product and an orthonormal basis $$\left\{{ e_1, e_2 }\right\}$$. Then \begin{align*} T({\mathbb{R}}) = {\mathbb{R}}\oplus {\mathbb{R}}\left\langle{ e_1, e_2 }\right\rangle \oplus {\mathbb{R}}\left\langle{ e_1^2, e_1e_2, e_2 e_1, e_2^2 }\right\rangle \oplus \cdots .\end{align*} Note that there are $$2^k$$ terms in the $$k$$th graded piece. It suffices to mod out only by the relations on the orthonormal basis. This is of the form $$(v+w)^2 = - {\left\lVert {v+w} \right\rVert}^2 = -{\left\lVert {v} \right\rVert}^2 - 2{\left\langle {v},~{w} \right\rangle} - {\left\lVert {w} \right\rVert}^2$$. On the other hand, this equals $$v^2 + vw + wv + w^2$$. So we obtain \begin{align*} vw + wv = 2{\left\langle {v},~{w} \right\rangle} ,\end{align*} and setting $$v=w$$ and dividing by 2 yields the original Clifford relation. For $${\mathbb{R}}^2$$, we can explicitly check \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item $$e_1^2 = -1$$, \item $$e_2^2 = -1$$, \item $$e_1e_2 + e_2 e_1 = -2e_1 e_2 = 0$$, \item $$e_1 e_2 = -e_2 e_1$$. \end{enumerate} Here (1), (2), and (4) generate all of the relations, so \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^2) - {\mathbb{R}}\left\langle{ e_1, e_2 }\right\rangle / \left\langle{ e_1^2 = -1, e_2^2 =-1, e_1e_2 = -e_2 e_1 }\right\rangle \cong HH .\end{align*} We can form this map by \begin{align*} 1 &\mapsto 1 \\ e_1 &\mapsto i \\ e_2 &\mapsto j \\ e_1 e_2 &\mapsto k ,\end{align*} and then checking that the appropriate relations hold. These hold since $$i^2 = j^2 = -1$$ and $$ij=-ji = k$$. These suffice, but you can check the rest: for example, does $$jk=i$$ hold? We can write this as \begin{align*} e_2(e_1e_2) = -e_2 (e_2 e_1) = -e_2^2 e_1 = -(-1)e_1 = e_1 .\end{align*} \end{example} \begin{exercise}[?] Check that $$\dim_{\mathbb{R}}{ \operatorname{Cl}} (V) = 2^{\dim V} < \infty$$. \end{exercise} \hypertarget{friday-april-23}{% \section{Friday, April 23}\label{friday-april-23}} \begin{remark} Given $$(V, \cdot)$$ an inner product space, we defined \begin{align*} { \operatorname{Cl}} (V) \coloneqq{ \bigoplus _{n\geq 0} V^{\otimes n} \over \left\langle{ v\otimes w + w\otimes v = 2v\cdot w }\right\rangle } .\end{align*} \end{remark} \begin{example}[?] We saw that\\ \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}, \cdot) &\cong {\mathbb{R}}[e] / e^2 =-1 \cong {\mathbb{C}}\\ { \operatorname{Cl}} ({\mathbb{R}}^2, \cdot) &= {\mathbb{R}}\left\langle{ e_1, e_2 }\right\rangle / \left\langle{ e_1^2 = e_2^2 = -1, e_1e_2 = -e_2 e_1 -}\right\rangle \cong {\mathbb{H}} \end{align*} where $$e_1\mapsto i, e_2\mapsto j, e_3 = e_1 e_2 \mapsto k$$. Can we describe $${ \operatorname{Cl}} ({\mathbb{R}}^n, \cdot)$$ in general? Choose an orthonormal basis $$\left\{{ e_i }\right\}$$, then \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^n, \cdot) = { {\mathbb{R}}\left\langle{ e_1, \cdots, e_n }\right\rangle \over \left\langle{ e_i^2 = -1, e_i e_j = -e_j e_i {~\mathrel{\Big|}~}i\neq j }\right\rangle } .\end{align*} We saw that replacing $$2$$ with $$\epsilon$$ in the defining relation recovers $$\bigwedge\nolimits^* V$$. \end{example} \begin{definition}[Degree Filtration] Define the \textbf{degree filtration} on $${ \operatorname{Cl}} (V, \cdot)$$ as the filtration induced by the degree filtration on $$T(V) \coloneqq\bigoplus _{n\geq 0} V^{\otimes n}$$. \end{definition} \begin{example}[?] Consider $${ \operatorname{Cl}} ({\mathbb{R}}^2, \cdot)$$. Then \begin{itemize} \tightlist \item Degree 0: $${\mathbb{R}}$$. \item Degree 1: $${\mathbb{R}}\oplus {\mathbb{R}}e_1 \oplus {\mathbb{R}}e_2$$ \item Degree 2: $${\mathbb{R}}\oplus {\mathbb{R}}e_1 \oplus {\mathbb{R}}e_2 \oplus {\mathbb{R}}e_1 e_2$$ \end{itemize} \end{example} \begin{definition}[Grading and Filtration] Recall that there's a distinction between gradings and filtration: \begin{itemize} \tightlist \item Gradings: $$R^i R^j \subset R^{i+j}$$ and $$R = \bigoplus_i R^i$$. \item Filtrations: $$F^1 \subset F^2 \subset \cdots$$ with $$F^i F^j \subseteq F^{i+j}$$ \end{itemize} An algebra equipped with a grading is a \textbf{graded algebra}, and similarly an algebra equipped with a filtration is a \textbf{filtered algebra}. \end{definition} \begin{remark} Note that \begin{itemize} \tightlist \item $$k[x_1, \cdots, x_n]$$ is graded (by monomials of uniform degree) and filtered (by polynomials of a bounded degree) \item $$T(V)$$ is graded and filtered, since multiplying a pure $$p$$ tensor with a pure $$q$$ tensor yields a pure $$p+q$$ tensor \item $${ \operatorname{Cl}} (V)$$ is a quotient of $$T(V)$$, but one can't simply define $${ \operatorname{Cl}} (V, \cdot)^i = \operatorname{im}T(V)^i$$ since the relations have mixed degree: for example $$e_1^2 = -1$$ So $${ \operatorname{Cl}} (V)$$ isn't graded, but is still filtered: take the filtration $$F$$ on $$T(V)$$ defined by $$F^i \coloneqq\bigoplus _{j\leq i} V^{\otimes j}$$ and descend it through the quotient map. The relations can only decrease degree, so this is well defined. \end{itemize} \end{remark} \begin{definition}[Filtration on the Clifford Algebra] Define a filtration $$F^{-}$$ on $${ \operatorname{Cl}} (V)$$ by the following: \begin{align*} F^i { \operatorname{Cl}} (V) \coloneqq{\operatorname{span}}\left\{{ { {e_j}_1, {e_j}_2, \cdots, {e_j}_{i}} }\right\} .\end{align*} \end{definition} \begin{definition}[The associated graded] The \textbf{associated graded} ring $${\mathsf{gr}\,}_{F^{-}} R$$ is the graded ring defined by \begin{align*} ({\mathsf{gr}\,}_{F^{-}})^i \coloneqq F^i R / F^{i-1} R .\end{align*} This induces a decomposition \begin{align*} {\mathsf{gr}\,}_{F^{-}} \cong \bigoplus _{i\geq 0} F^i R/ F^{i-1} R = \bigoplus _{i\geq 0} ({\mathsf{gr}\,}_{F^{-}})^i ,\end{align*} which has a multiplicative structure \begin{align*} F^i/F_{i-1} \cdot F^j/F_{j-1} \to F^{i+j} / F^{i+j-1} .\end{align*} \end{definition} \begin{remark} Note that if $$R \in {\mathsf{gr}\,}\mathsf{Ring}$$, then $${\mathsf{gr}\,}(R) = R$$, so taking the associated graded recovers the ring itself. What's happening: taking the smallest homogeneous ideal. \end{remark} \begin{fact} If one has relations of mixed degree, the associated graded also has the top degree part of each relation. \end{fact} \begin{remark} In our case, the Clifford relation relates degree $$k$$ pieces to degree $$k-2$$ pieces, so we obtain \begin{align*} {\mathsf{gr}\,}_{F^{-}} { \operatorname{Cl}} (V) \cong T(V) / \left\langle{ v\otimes w + w\otimes v = 0 }\right\rangle \coloneqq\bigwedge\nolimits^* V .\end{align*} There is an isomorphism of $$k{\hbox{-}}$$vector spaces \begin{align*} { \operatorname{Cl}} (V) & \xrightarrow{\sim} {\mathsf{gr}\,}{ \operatorname{Cl}} (V) \\ x\in F^i &\mapsto \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu \in F^i / F^{i-1} .\end{align*} This is because $$F^0 \subseteq \cdots \subseteq \cdots$$ with $$\cup_i F^i = { \operatorname{Cl}} (V)$$. We can conclude $$\dim_{\mathbb{R}}{ \operatorname{Cl}} (V) = \dim_{\mathbb{R}}\bigwedge\nolimits^* V = 2^{\dim_k V}$$ and use this to construct a basis for $${ \operatorname{Cl}} (V)$$. The relevant map is \begin{align*} { {e_j}_1, {e_j}_2, \cdots, {e_j}_{i}} &\mapsto e_{j_1} \wedge \cdots \wedge e_{j_i} .\end{align*} \end{remark} \begin{corollary}[of the fact] The following set forms an $${\mathbb{R}}{\hbox{-}}$$basis for $${ \operatorname{Cl}} ({\mathbb{R}}^n, \cdot)$$: \begin{align*} \left\{{ { {e_j}_1, {e_j}_2, \cdots, {e_j}_{i}} {~\mathrel{\Big|}~}j_1 < j_2 < \cdots < j_i,\, i\leq n }\right\} .\end{align*} \end{corollary} \begin{example}[?] Consider \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^3, \cdot) \cong {\operatorname{span}}_{\mathbb{R}}\left\{{ 1, e_1, e_2, e_3, e_1e_2, e_1 e_3, e_1 e_2 e_3 }\right\} .\end{align*} Then \begin{align*} e_1 e_2 \cdot e_1 e_3 &= -e_1 e_1 e_3 e_3 && e_2 e_1 = -e_1 e_2 \\ &= e_2 e_3 && e_1^2 = - 1 .\end{align*} \end{example} \begin{exercise}[?] Show that $${ \operatorname{Cl}} ({\mathbb{R}}^3) \cong {\mathbb{H}}\oplus {\mathbb{H}}$$. \end{exercise} \begin{definition}[Even and odd parts of the Clifford algebra] $${ \operatorname{Cl}} (V)$$ has a $${\mathbb{Z}}/2$$ (super'') grading, so \begin{align*} { \operatorname{Cl}} (V) \circ { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V) && { \operatorname{Cl}} _i(V) \cdot { \operatorname{Cl}} _j(V) \subset { \operatorname{Cl}} _{i+j\pmod 2}(V) .\end{align*} The \textbf{even} subalgebra is given by \begin{align*} { \operatorname{Cl}} _0(V) = {\operatorname{span}}_k \left\{{ { {e_i}_1, {e_i}_2, \cdots, {e_i}_{2k}} {~\mathrel{\Big|}~}2k\leq n }\right\} ,\end{align*} where we take an even number of basis elements, which makes sense because the Clifford relation $$vw + 2v = -2v\cdot w$$ preserves degree mod 2. This is still an algebra. The \textbf{odd} sub-vector space (not an algebra) is given by \begin{align*} { \operatorname{Cl}} _1(V) = {\operatorname{span}}_k \left\{{ { {e_i}_1, {e_i}_2, \cdots, {e_i}_{2k+1}} {~\mathrel{\Big|}~}2k+1\leq n }\right\} .\end{align*} \end{definition} \begin{example}[?] \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^3) = {\operatorname{span}}_{\mathbb{R}}\left\{{ 1, e_1 e_2, e_1 e_3, e_2 e_3 }\right\} ,\end{align*} and we saw $$e_1 e_2 = e_1 e_3 = e_2 e_3$$. This product has degree 4, and when we applied the relation $$e_1^2=1$$ we dropped the degree by 2. For the odd part, $$e_3 \in Cl_1({\mathbb{R}}^3)$$ and $$e_1 e_2 \in { \operatorname{Cl}} _0({\mathbb{R}}^3)$$, and we have \begin{align*} e_3 \cdot (e_1 e_2) = -e_1 e_3 e_2 = e_1 e_2 e_3 \in { \operatorname{Cl}} _1({\mathbb{R}}^3) .\end{align*} \end{example} \begin{proposition}[Decomposing the Clifford algebra of V] \begin{align*} { \operatorname{Cl}} (V) \cong { \operatorname{Cl}} _0(V \oplus {\mathbb{R}}) .\end{align*} \end{proposition} \begin{proof}[?] Let $$e\in {\mathbb{R}}$$ be a unit vector. Given $$x\in { \operatorname{Cl}} (V)$$, decompose $$x = x_0 + x_1 \in { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V)$$. Define an isomorphism \begin{align*} \phi: { \operatorname{Cl}} (V) &\to { \operatorname{Cl}} _0(V \oplus {\mathbb{R}}) \\ x &\mapsto x_0 + x_1 e ,\end{align*} which is well-defined since $$x_0$$ was odd degree, and both $$x_1, e$$ were odd degree and thus $$x_1 e$$ is even. One checks that this preserves multiplication: \begin{align*} x\cdot y = (x_0 + x_1) \cdot (y_0 + y_1) = (x_0 y_0 + x_1 y_1) + (x_0 y_1 + x_1 y_0) \in { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V) ,\end{align*} and so \begin{align*} \phi(x) \cdot \phi(y) &= (x_0 +x_1 e)(y_0 + y_1 e) \\ &= x_0 y_0 + x_0 y_1 e + x_1 e y_0 + x_1 e y_1 e_1 .\end{align*} The question is if this equals \begin{align*} \phi(xy) \coloneqq(x_0 y_0 + x_1 y_1) + (x_0 y_1 + x_1 y_0)e .\end{align*} But for example, $$x_1 e y_0 = (-1)^{|y_0|} x_1 y_0 e$$, and $$y_0$$ is even. Similarly, $$x_1 e y_1 e = -x_1 y_1 e^2 = x_1 y_1$$. \end{proof} \hypertarget{wednesday-april-28}{% \section{Wednesday, April 28}\label{wednesday-april-28}} \begin{remark} Last time: we defined $${\operatorname{Pin}}(n) \subseteq { \operatorname{Cl}} ({\mathbb{R}}^n)$$ which was generated by $$S^1({\mathbb{R}}^n)$$. These were units because $$v^2 = -{\left\lVert {v} \right\rVert}^2 = -1$$, so $$v^{-1}= -v$$, and formed a group contained in $${ \operatorname{Cl}} ({\mathbb{R}}^n)^{\times}$$. There is a decomposition $${ \operatorname{Cl}} (V) = { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V)$$ with a $${\mathbb{Z}}/2{\hbox{-}}$$grading, and we defined \begin{align*} {\operatorname{Spin}}(V) \coloneqq{\operatorname{Pin}}(V) \cap{ \operatorname{Cl}} _0(V) = \left\langle{ vw {~\mathrel{\Big|}~}v,w\in S^1({\mathbb{R}}^n) }\right\rangle .\end{align*} There is a map \begin{align*} {\operatorname{Pin}}(n) &\twoheadrightarrow O(n) \\ v & \mapsto (u\mapsto vuv^{-1}) = -R_{v^\perp} ,\end{align*} which preserves $$V^{\otimes 1} \subset { \operatorname{Cl}} (V)$$, and was reflection about the hyperplane $$v^\perp$$. There is also a SES \begin{align*} 0 \to {\mathbb{Z}}/2 \to {\operatorname{Spin}}(n) \xrightarrow{\pi} {\operatorname{SO}}(n) \to 0 ,\end{align*} where we used the fact that $$\ker \pi \subset Z{ \operatorname{Cl}} ({\mathbb{R}}^n)$$. It turns out that $${\operatorname{Spin}}(n) = \overline{ {\operatorname{SO}}(n) }$$, using that $$\pi_1( {\operatorname{SO}}(n), {\operatorname{pt}}) = {\mathbb{Z}}/2$$ and checking that $$\pm 1\in {\operatorname{Spin}}(n)$$, yielding a nontrivial kernel. \end{remark} \begin{remark} This is local, at a single vector space, so we'll now try to globalise this to the tangent space of a manifold. \end{remark} \begin{definition}[Clifford Bundle] Let $$(V, g)$$ be an oriented smooth Riemannian manifold where $$g$$ is a metric on $$TX$$. Define the \textbf{Clifford bundle} of $$X$$ by \begin{align*} { \operatorname{Cl}} (X) \coloneqq{ \operatorname{Cl}} (T {}^{ \vee }X, g {}^{ \vee }) ,\end{align*} where we've used the dual metric $$g {}^{ \vee }$$ on the cotangent bundle. \end{definition} \begin{remark} We showed that $${\mathsf{gr}\,}{ \operatorname{Cl}} ({\mathbb{R}}^n) = \bigwedge\nolimits{\mathbb{R}}^n$$, and so there is a bundle isomorphism \begin{align*} { \operatorname{Cl}} (X) \xrightarrow{\sim} \bigwedge\nolimits^* T {}^{ \vee }X ,\end{align*} but the ring structure is different. On the right, we have a way of multiplying sections, namely $$\omega_1 \wedge \omega_2$$, but on the left we have the Clifford multiplication $$\alpha_1 \cdot \alpha_2$$. Note that $$\omega^{\wedge 2} = 0$$, but $$\alpha^{\cdot 2} \in {\mathbb{R}}$$ is some scalar. We define $$\omega\cdot \omega= g^*( \omega, \omega)$$, so we use the metric fiberwise to define a Clifford multiplication. \end{remark} \begin{definition}[The principal oriented frame bundle] Given an oriented bundle with a metric, there is a principal $${\operatorname{SO}}(n)$$ bundle $$P \coloneqq\mathop{\mathrm{OFrame}}$$, the space of orthogonal oriented frames. \end{definition} \begin{remark} This is principal since any two elements are related by a unique element of $${\operatorname{SO}}(n)$$. Recall that we had an \emph{associated bundle} construction, so taking the standard representation $$\rho: {\operatorname{SO}}(n) \to ({\mathbb{R}}^n, g)$$ where elements act by their transformations (?), there is an oriented bundle $$P { \underset{\scriptscriptstyle {\rho} }{\times} } {\mathbb{R}}^n$$. If the bundle is $$TX$$ with a metric $$g$$, this yields a distinguished $${\operatorname{SO}}(n)$$ bundle $$P\to X$$. \end{remark} \begin{definition}[Spin Structures] A \textbf{spin structure} is a lift $$\tilde P$$ of $$P$$ to a principal $${\operatorname{Spin}}(n)$$ bundle. \end{definition} \begin{proposition}[Spin iff nontrivial $w_2$] $$X$$ admits a spin structure iff the second Stiefel--Whitney class $$w_2(X) = 0$$ in $$H^2(X; {\mathbb{Z}}/2)$$. If $$w_2(X) = 0$$, then the spin structures are torsors over $$H^1(X; {\mathbb{Z}}/2)$$. \end{proposition} \begin{remark} Recall that a $$G{\hbox{-}}$$torsor is a set with a free transitive $$G{\hbox{-}}$$action. For example, the fibers of a principal bundle are torsors. Given any two torsors, we can compare them using elements of $$G$$, but there is no distinguished element. For example, $${\mathbb{A}}_n$$ is a torsor over the vector space $$k^n$$. \end{remark} \begin{proof}[?] Consider transitions for $$P\to X$$: \begin{align*} t_{ij}: U_i \cap U_j \to {\operatorname{SO}}(n) \\ ,\end{align*} where $$t_{ij} = t^{-1}_{ji}$$ and the cocycle condition $$t_{ij} t_{jk} t_{ki} = 1$$ is satisfied. We want a lift: \begin{center} \begin{tikzcd} && {{\mathbb{Z}}/2 \in Z{\operatorname{Spin}}(n)} \\ \\ {U_i \cap U_j} && {{\operatorname{Spin}}(n)} \\ \\ {U_i \cap U_j} && {{\operatorname{SO}}(n)} \arrow["\one", from=3-1, to=5-1] \arrow["{t_{ij}}", from=5-1, to=5-3] \arrow["{\tilde t_{ij}}", from=3-1, to=3-3] \arrow["{2:1}", from=3-3, to=5-3] \arrow[hook, from=1-3, to=3-3] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNSxbMCwyLCJVX2kgXFxpbnRlcnNlY3QgVV9qIl0sWzIsMiwiXFxTcGluKG4pIl0sWzIsNCwiXFxTTyhuKSJdLFsyLDAsIlxcWlovMiBcXGluIFpcXFNwaW4obikiXSxbMCw0LCJVX2kgXFxpbnRlcnNlY3QgVV9qIl0sWzAsNCwiXFxvbmUiXSxbNCwyLCJ0X3tpan0iXSxbMCwxLCJcXHRpbGRlIHRfe2lqfSJdLFsxLDIsIjI6MSJdLFszLDEsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link to Diagram} \end{quote} We can always lift to \emph{some} $$\tilde t_{ij}$$ using the path-lifting property of covers if $$U_i \cap U_j$$ is contractible, using that $${\mathbb{Z}}/2$$ is discrete. We can arrange $$\tilde t_{ij} = \tilde t_{ji}^{-1}$$ since $$U_i \cap U_j = U_j \cap U_i$$, but we may not have the cocycle condition on the lift. We have $$t_{ij} t_{jk} t_{ki} = 1$$, so \begin{align*} \tilde t_{ij} \tilde t_{jk} \tilde t_{ki} \in \ker ({\operatorname{Spin}}(n) \to {\operatorname{SO}}(n)) = \left\{{ \pm 1 }\right\} ,\end{align*} using that everything in sight needs to be a group morphism. So define \begin{align*} \tilde t_{ijk} \coloneqq( \tilde t_{ij} \tilde t_{jk} \tilde t_{ki} )_{i,j,k} \in {\check{C}}^2_{\mathcal{U}}(X, \underline{{\mathbb{Z}}/2}) .\end{align*} The claim is that $${{\partial}}^2( \tilde t_{ijk}) = 0$$, but it turns out that regardless of choice of lift we obtain \begin{align*} {{\partial}}^2(\tilde t_{ijk}) = \tilde t_{ijk} \tilde t_{ikl}^{-1}\tilde t_{ijl} \tilde t_{ijk}^{-1}= 0 \implies [ \tilde t_{ijk} ] \in {\check{H}}^2(X, \underline{{\mathbb{Z}}/2}) .\end{align*} Is this class well-defined? Consider replacing $$\tilde t_{ij}$$ with $$-\tilde t_{ij}$$. In general, we have \begin{align*} i, j\in \left\{{ a,b,c }\right\} \implies \tilde t_{abc} \mapsto -\tilde t_{abc} ,\end{align*} and so this is a Čech coboundary in $${{\partial}}^1(1, \cdots, 1, -1, 1, \cdots, 1)$$ where the $$-1$$ occurs in the $$t_{ij}$$ coordinate. Thus $$\tilde t_{ijk}$$ is well-defined moduli $${{\partial}}^1 C_{\mathcal{U}}^1(X, \underline{{\mathbb{Z}}/2})$$.\\ Note that $$w_2(X)$$ was produced from the pair $$(X, g)$$, but the space of metrics is connected and thus $$w_2(X)$$ depends only on $$X$$. Suppose $$w_2(X) = 0$$, then $$[ \tilde t_{ijk} ] = 0$$ which implies that there is some $$(s_{ij})$$ with $${{\partial}}^1 (s_{ij}) = (\tilde t_{ijk})$$. So replace each $$\tilde t_{ij}$$ with $$\tilde{ \tilde t}_{ij} \coloneqq s_{ij} \tilde t_{ij}$$ is a new lift which satisfies the cocycle condition. Thus they define the transition functions of a principal $${\operatorname{Spin}}(n)$$ bundle lifting $$P \to X$$. \hfill\break To see the claim about torsors, given any $$\ell_{ij} \in \ker {{\partial}}^1$$, note that any $$\tilde{ \tilde t}_{ij} \ell_{ij}$$ also satisfies the cocycle condition. There is a map \begin{align*} \left\{{ \text{Spin structures} }\right\} &\leftarrow\ker {{\partial}}^1 \\ \tilde{\tilde t}_{ij} \ell_{ij} &\mapsfrom \ell_{ij} ,\end{align*} which is a torsor because we needed to start with a given lift $$\tilde{\tilde t}_{ij}$$. Then $$\tilde P_1 \cong \tilde P_2$$ iff there exists an $$(m_i) \in {\check{C}}^0_{\mathcal{U}}(X, \underline{{\mathbb{Z}}/2})$$ such that $$(\ell_{ij})_1 = (\ell_{ij})_2 + {{\partial}}^0 (m_i)$$, which are different trivializations of the same bundle. \end{proof} \begin{remark} This is a nice example to get a hang of the use and importance of Čech cohomology. We then use the isomorphism $${\check{H}}\to H_{{\operatorname{Sing}}}$$. \end{remark} \begin{theorem}[Existence of spin representation of Clifford algebras in even dimension] Assume $$n \coloneqq\dim V$$ is even, then $${ \operatorname{Cl}} (V)$$ has a unique nontrivial irreducible finite dimensional complex representation $$S$$ of dimension $$\dim S = 2^{n/2}$$, the \textbf{spin representation}. \end{theorem} \begin{remark} It turns out that $${ \operatorname{Cl}} (V) \otimes_{\mathbb{R}}{\mathbb{C}}\cong \mathop{\mathrm{End}}(S)$$. The left-hand side contains $${\operatorname{Spin}}(n)$$, so given $$\rho: { \operatorname{Cl}} (V) \to \mathop{\mathrm{End}}(S)$$ a representation (i.e.~a ring homomorphism) in matrices, we can restrict $$\rho$$ to $${\operatorname{Spin}}(n)$$ to get $$\rho { \left.{{}} \right|_{{{\operatorname{Spin}}(n)}} }: {\operatorname{Spin}}(n) \to \operatorname{GL}(S)$$. Next time: spin representations. Spinor bundle will be sections of associated bundle of the Clifford bundle. \end{remark} \hypertarget{friday-april-30}{% \section{Friday, April 30}\label{friday-april-30}} \begin{remark} Last time: we defined \begin{align*} { \operatorname{Cl}} (V, \cdot) &\coloneqq\bigoplus_n V^{\otimes n} / \left\langle{ v\otimes v = -{\left\lVert {v} \right\rVert}^2 1 }\right\rangle \\ {\operatorname{Pin}}(V) &\coloneqq\left\langle{ v {~\mathrel{\Big|}~}{\left\lVert {v} \right\rVert} = 1 }\right\rangle \subseteq { \operatorname{Cl}} (V) .\end{align*} There is a $${\mathbb{Z}}/2$$ grading $${ \operatorname{Cl}} (V) = { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V)$$ where $${ \operatorname{Cl}} _0(V)$$ is the image of even tensors and $${ \operatorname{Cl}} _1(V)$$ is the image of odd tensors. We also had \begin{align*} {\operatorname{Spin}}(V) \coloneqq{\operatorname{Pin}}(V) \cap{ \operatorname{Cl}} _0(V) = \left\langle{ v\cdot w {~\mathrel{\Big|}~}v,w \in V, \, {\left\lVert {v} \right\rVert} = {\left\lVert {w} \right\rVert} = 1 }\right\rangle .\end{align*} There was a map \begin{align*} {\operatorname{Pin}}(V) &\to O(V) \\ v &\mapsto -R_v ,\end{align*} where $$R_v$$ was reflection about $$v^\perp$$, where we identified this as an action on $$V^{\otimes 1} \subset { \operatorname{Cl}} (V)$$ where $$u\to vuv^{-1}$$. For any Riemannian manifold $$(X, g)$$, we could define the Clifford bundle $${ \operatorname{Cl}} (X) = { \operatorname{Cl}} (T {}^{ \vee }X, g {}^{ \vee })$$ to globalise this from vector spaces to bundles with metrics. We defined a spin structure on $$X$$ as any lift of the principal $${\operatorname{SO}}(n)$$ bundle over $$(T {}^{ \vee }X, g)$$ (namely $$\mathop{\mathrm{Frame}}(X)$$) to a $${\operatorname{Spin}}(n)$$ bundle. \end{remark} \begin{warnings} Each fiber is a metric space, so what happens if you just try to define\\ \begin{align*} Y \coloneqq\displaystyle\coprod_{x\in X} \left\langle{ v {~\mathrel{\Big|}~}{\left\lVert {v} \right\rVert}^2 = 1,\, v\in T_x {}^{ \vee }X }\right\rangle \,\,?\end{align*} This seems to be isomorphic to a spin structure, but we do not have a distinguished action of any \emph{fixed} group $${\operatorname{Spin}}(n)$$. We would have to choose isomorphisms to the standard spin group at each fiber, but the isomorphisms are not unique -- there is ambiguity up to the entire spin group. So this does not define a spin structure. \end{warnings} \begin{remark} We showed that there exists a spin structure iff some cohomology class $$w_2(K) \in H^2(X; {\mathbb{Z}}/2)$$ vanishes. \end{remark} \begin{theorem}[Classification of complex representations of Clifford algebras] If $$\dim_k V$$ is even, there is a unique finite-dimensional complex irreducible $${ \operatorname{Cl}} (V)$$ representation of dimension $$2^{n/2}$$. If $$\dim_k V$$ is odd, there are two complex conjugate representations of dimension $$2^{{\left\lfloor n/2 \right\rfloor}}$$. \end{theorem} \begin{example}[?] Consider $${ \operatorname{Cl}} ({\mathbb{R}}^2) \cong {\mathbb{H}}$$. There is an irreducible complex representation of dimension 2: \begin{align*} 1 &\mapsto { \begin{bmatrix} {1} & {0} \\ {0} & {1} \end{bmatrix} } \\ i &\mapsto \sigma_1 \coloneqq{ \begin{bmatrix} {0} & {i} \\ {i} & {0} \end{bmatrix} } \\ j &\mapsto \sigma_2 \coloneqq{ \begin{bmatrix} {i} & {0} \\ {0} & {-i} \end{bmatrix} } \\ k &\mapsto \sigma_3 \coloneqq{ \begin{bmatrix} {0} & {1} \\ {-1} & {0} \end{bmatrix} } .\end{align*} \end{example} \begin{definition}[Pauli matrices] The $$\sigma_i$$ defined above are referred to as the \textbf{Pauli matrices}. \end{definition} \begin{example}[?] Consider $${ \operatorname{Cl}} ({\mathbb{R}}^4)$$. By the theorem, there is a unique complex representation of $$2^{4/2} = 2^2 = 4$$, although the 4 here matching the dimension of $${\mathbb{R}}^4$$ is coincidental. We'd like to find an isomorphism \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^4) \xrightarrow{ \sim} \mathop{\mathrm{End}}( ({\mathbb{C}}^2)^{\otimes 2} ) \cong \mathop{\mathrm{End}}({\mathbb{C}}^4) = \operatorname{Mat}(4\times 4; {\mathbb{C}}) .\end{align*} Note that $${ \operatorname{Cl}} ({\mathbb{R}}^4) \xrightarrow{\sim} \mathop{\mathrm{End}}( ({\mathbb{C}}^2)^{\otimes 3})$$, which is why the dimensions multiply. We can write \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^4) = { {\mathbb{R}}\left\langle{ e_1, e_2, e_3 ,e_4 }\right\rangle \over e_i e_j + e_j e_i = 2\delta_{ij} } .\end{align*} So define a map \begin{align*} e_1 &\mapsto \gamma_1 \coloneqq 1\otimes\sigma_1 \\ e_2 &\mapsto \gamma_2 \coloneqq 1\otimes\sigma_2 \\ e_3 &\mapsto \gamma_3 \coloneqq\sigma_1 \otimes i \sigma_3 \\ e_4 &\mapsto \gamma_4 \coloneqq\sigma_2 \otimes i \sigma_3 .\end{align*} \begin{definition}[Dirac matrices] The matrices appearing above are called the \textbf{Dirac matrices}. \end{definition} \begin{exercise}[?] Determine a similar map for $${ \operatorname{Cl}} ({\mathbb{R}}^6)$$ continuing this pattern. \end{exercise} We can check that this is a representation. Note that we can tensor matrices in a simple way: \begin{center} \begin{tikzcd} & {e_1} & {e_2} && {} & {f_1} & {f_2} \\ {e_1} & a & b && {f_1} & e & f \\ {e_2} & c & d && {f_2} & g & h \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsMTcsWzAsMSwiZV8xIl0sWzAsMiwiZV8yIl0sWzEsMCwiZV8xIl0sWzIsMCwiZV8yIl0sWzEsMSwiYSJdLFsyLDEsImIiXSxbMSwyLCJjIl0sWzIsMiwiZCJdLFs0LDBdLFs1LDAsImZfMSJdLFs2LDAsImZfMiJdLFs0LDEsImZfMSJdLFs0LDIsImZfMiJdLFs1LDEsImUiXSxbNiwxLCJmIl0sWzUsMiwiZyJdLFs2LDIsImgiXV0=}{Link to Diagram} \end{quote} Checking $$e_2 \cdot e_2 = -1$$, we have \begin{align*} (1 \otimes\sigma_2) \cdot (1 \otimes\sigma_2) &= ? \\ 1_2 \otimes\sigma_2^2 &= -I_2 \oplus I_2 \\ \gamma_2 \gamma_3 &= - \gamma_3 \gamma_2 .\end{align*} \begin{quote} Todo: messed up! \end{quote} One can similarly check \begin{align*} (1 \otimes\sigma_2 ) \cdot (\sigma_1 \otimes i \sigma_3) = - ( \sigma_1 \otimes i \sigma_2) (1\otimes\sigma_2) .\end{align*} \end{example} \begin{remark} We thus have $${ \operatorname{Cl}} ({\mathbb{R}}^4) \curvearrowright{\mathbb{C}}^4$$ by sending $$e_i \mapsto \delta_i$$, the Dirac matrices. Using that $${\operatorname{Pin}}(4) \cap{ \operatorname{Cl}} ({\mathbb{R}}^4) = {\operatorname{Spin}}(4) \subseteq { \operatorname{Cl}} ({\mathbb{R}}^4)$$, we can a spin representation, but this may no longer be irreducible. In fact, as a $${\operatorname{Spin}}(4)$$ representation this splits into two irreducible representations. We know that $${\operatorname{Spin}}(4) \subseteq { \operatorname{Cl}} _0({\mathbb{R}}^4) = { \operatorname{Cl}} ({\mathbb{R}}^3)$$ which has two complex conjugate irreducible representations. The key is to define an element $$\omega_{\mathbb{C}}\in { \operatorname{Cl}} (V) \otimes_{\mathbb{R}}{\mathbb{C}}$$ with $$\omega_{\mathbb{C}}^2 = 1$$, which yields a decomposition of $${\mathbb{S}}= {\mathbb{S}}^+ \oplus {\mathbb{S}}^-$$ as the $$\pm 1$$ eigenspaces of the action. Here $$\omega_C \coloneqq-e_1 e_2 e_3 e_4 \mapsto \gamma_5$$. One can define \begin{align*} \gamma_5 \coloneqq\operatorname{im}(\omega_{\mathbb{C}}) = - \gamma_1 \gamma_2 \gamma_3 \gamma_4 = - \sigma_3 \otimes\sigma_3 \end{align*} and one obtains the matrix \begin{align*} -\sigma_3 \otimes\sigma_3 = \begin{bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0& 1 & 0 \\ 0 & 1& 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} .\end{align*} One can check that $$\gamma_5$$ anticommutes with the $$\delta_i$$ for $$1\leq i \leq 4$$, and thus commutes with $${ \operatorname{Cl}} _0({\mathbb{R}}^4)$$. We can write $${\mathbb{S}}^+$$, the positive 1 eigenspace of $$\gamma_5$$, as $${\mathbb{C}}(s_1 - s_4 ) \oplus {\mathbb{C}}(s_1 + s_2)$$. So we have $${\operatorname{Spin}}(4) = { \operatorname{Cl}} ({\mathbb{R}}^4) \curvearrowright{\mathbb{C}}^2 \oplus {\mathbb{C}}^2 = {\mathbb{S}}$$, which splits into $$\gamma_5{\hbox{-}}$$eigenspaces $${\mathbb{S}}^+ \oplus {\mathbb{S}}^-$$, the \textbf{positive and negative spinors}. This means that $$\gamma_5$$ commutes with the image of $${\operatorname{Spin}}(4) \hookrightarrow\operatorname{GL}( {\mathbb{C}}^2 \oplus {\mathbb{C}}^2)$$. \end{remark} \begin{fact} If the action commutes with everything in the representation, the representation splits. (??? missed) \end{fact} \begin{remark} Let $$g\in {\operatorname{Spin}}(4)$$, and $$v^+ \in {\mathbb{S}}^+ \subseteq {\mathbb{S}}$$. Question: is it true that $$g\cdot v^+ \in {\mathbb{S}}^+$$? If so, this yields a subrepresentation. If so, $$\gamma_5 v^+ = v^+$$ since we're in the $$+1$$ eigenspace, and on the other hand, $$g\cdot v^+ = g \cdot \gamma_5 v^+ = g \omega_{\mathbb{C}}\cdot v^+$$ where the last identification comes from the map $$\gamma_5 \mapsto \omega_{\mathbb{C}}$$, and this is equal to $$\omega_{\mathbb{C}}g \cdot v^+$$ using commutativity. So $$g\cdot v^+$$ is in the $$+1$$ eigenspace of $$\gamma_5$$. \end{remark} \begin{remark} Now take $$\gamma_i$$. This actually switches spinors: by anticommutativity of the $$\gamma_i$$ with $$\gamma_5$$, we have \begin{align*} \gamma_i \cdot v^+ = \gamma_i \gamma_5 v^+ = - \gamma_5 \gamma_i v^+ ,\end{align*} which means $$\gamma_i v^+$$ is in the $$-1$$ eigenspace for $$\gamma_5$$, i.e.~$$\gamma_i v^+ \in {\mathbb{S}}^-$$. \end{remark} \begin{remark} Suppose one has a spin structure and $$\tilde P \to X$$ is a principal $${\operatorname{Spin}}(n)$$ bundle. There are bundles over this of the form $$\rho: {\operatorname{Spin}}(n) \to \operatorname{GL}({\mathbb{S}}^\pm)$$, yielding the \textbf{spinor bundle} \begin{align*} \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}= {\mathbb{S}}_x^+ \oplus {\mathbb{S}}_x^- .\end{align*} \end{remark} \begin{remark} Let $$G \xrightarrow{\rho} \operatorname{GL}(V)$$ be any representation. If $$\phi \in \mathop{\mathrm{End}}(V)$$ commutes with $$\rho(G)$$, then the eigenspaces of $$\phi$$ are subrepresentations. In other words, $$G\curvearrowright V = \bigoplus _{i=1}^n V_{\lambda_i}$$, then $$G\curvearrowright V_{\lambda_i}$$ is a subrepresentation, using that \begin{align*} \phi(v) = \lambda v \implies gv = g\phi(\lambda ^{-1}v) = \phi \rho(g) \lambda^{-1}v ,\end{align*} which says $$\phi( \rho(g) \cdot v) = \lambda (\rho(g) \cdot v) \implies \rho(g) \cdot v \in V_{\lambda}$$. We used it here by This rephrases Schur's lemma! \end{remark} \hypertarget{spin-bundles-and-dirac-operators-monday-may-03}{% \section{Spin Bundles and Dirac Operators (Monday, May 03)}\label{spin-bundles-and-dirac-operators-monday-may-03}} \begin{remark} Last time: we defined a Spin structure on an oriented manifold $$M$$ as a lift of the principal $${\operatorname{SO}}(n)$$ bundle $$P\to M$$ (\emph{unassociated} to $$TM$$) to a $${\operatorname{Spin}}(n)$$ bundle $$\tilde P$$. There was a \textbf{spin representation} $${\operatorname{Spin}}(n) \curvearrowright{\mathbb{S}}$$, which is irreducible for $${ \operatorname{Cl}} ({\mathbb{R}}^n)$$ and splits as $${\mathbb{S}}= {\mathbb{S}}^+ \oplus {\mathbb{S}}^-$$, which are $${\operatorname{Spin}}(n)$$ subrepresentations. We defined \textbf{spinor bundles} \begin{align*} \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}= {\mathbb{S}}_M = {\mathbb{S}}_M^+ \oplus {\mathbb{S}}_M^- .\end{align*} \end{remark} \begin{example}[Dimension 4] If $$\dim_{\mathbb{R}}M = 4$$, then $${\mathbb{S}}_M^\pm \in { \mathsf{Vect} }_{\mathbb{C}}^{{\operatorname{rank}}= 2}$$, i.e.~they are complex vector bundles of rank 2. Consider the eigenspaces $$-e_1e_2 e_3 e_4 \curvearrowright{\mathbb{S}}$$, then $$e_i\cdot({-}): {\mathbb{S}}^{\pm} \to {\mathbb{S}}^{\mp}$$. \end{example} \begin{remark} Principal bundle: fibers are left $$G{\hbox{-}}$$torsors. In the fiber product, the group sits in the middle and acts on each factor. So $$\tilde P$$ eats the right $$G{\hbox{-}}$$action, and $${\mathbb{S}}$$ eats the left action. Remarkably, for Spin bundles, there is an action leftover. \end{remark} \begin{proposition}[The spin bundle is a Clifford module] The spin bundle $${\mathbb{S}}_M$$ naturally has the structure of a $${ \operatorname{Cl}} (M){\hbox{-}}$$module. \end{proposition} \begin{proof}[?] We have a Clifford action \begin{align*} { \operatorname{Cl}} ({\mathbb{R}}^n) \otimes{\mathbb{S}}&\to {\mathbb{S}}\\ x\otimes s &\mapsto x\cdot s .\end{align*} Recall that we have a natural conjugation action $${\operatorname{Spin}}(n) \curvearrowright{ \operatorname{Cl}} ({\mathbb{R}}^n)$$ where $$g \mapsto g({-})g^{-1}$$, and similarly $${\operatorname{Spin}}(n) \curvearrowright{\mathbb{S}}$$ by $$g\mapsto g\cdot({-})$$. Given any $$V\to W$$ of $$G{\hbox{-}}$$modules, any $$P\in \mathrm{Bun}^{\mathrm{prin}}(G)$$ yields an induced module \begin{align*} P { \underset{\scriptscriptstyle {G} }{\times} } V \to P { \underset{\scriptscriptstyle {G} }{\times} } W ,\end{align*} and moreover $$\tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } { \operatorname{Cl}} ({\mathbb{R}}^n) = { \operatorname{Cl}} (M)$$. We then conclude that there is an action $${ \operatorname{Cl}} (M) \otimes{\mathbb{S}}_M \to {\mathbb{S}}_M$$, the \textbf{Clifford multiplication}. \end{proof} \begin{remark} We have an isomorphism of bundles (not of algebras) $${ \operatorname{Cl}} (M) \cong \bigwedge\nolimits T {}^{ \vee }M$$, and any one form $$\omega$$ is an analogue of an element of $$V^{\otimes 1}$$, and $$\omega \cdot ({\mathbb{S}}^+, {\mathbb{S}}^-) \in {\mathbb{S}}_M^- \oplus {\mathbb{S}}_M^+$$. \end{remark} \begin{definition}[Clifford connection] A connection $$\nabla$$ on $${\mathbb{S}}$$ is a \textbf{Clifford connection} if \begin{align*} \nabla(x\cdot s) = x \cdot \nabla (s) + d(x) \cdot s && x\in H^0 { \operatorname{Cl}} (M) = H^0\qty{ \bigwedge\nolimits^* T {}^{ \vee }M} ,\,\, s\in H^0 ({\mathbb{S}}_M) ,\end{align*} where $$d$$ is the de Rham differential. \end{definition} \begin{remark} It is not obvious that a Clifford connection exists! We have $${\mathbb{S}}_M = \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}$$, so it suffices to give a connection on $$\tilde P$$ which is $${\operatorname{Spin}}(n)$$ invariant, since any associated bundle will inherit the connection. Idea: we need a notion of parallel transport. This is a principal $${\operatorname{Spin}}(n)$$ bundle, so the fibers look like $${\operatorname{Spin}}(n)$$, and we want to lift paths in $$M$$ to paths in $$\tilde P$$: \begin{figure} \centering \resizebox{\columnwidth}{!}{% \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-05-03_14-16.pdf_tex} }; \end{tikzpicture} } \end{figure} It suffices to give a connection on $$P$$, and using that $$\tilde P\to P$$ is a 2 to 1 covering map, we can take a connecting on $$P$$ coming from $$\mathop{\mathrm{OFrame}}(T {}^{ \vee }M, g {}^{ \vee })$$. So it further suffices to produce a connection on $$T {}^{ \vee }M$$ preserving orthogonality of frames under parallel transport, which is essentially the definition of the Levi-Cevita connection $$\nabla^{\mathrm{LC}}$$. Then the $$\nabla$$ associated to $$\nabla^{\mathrm{LC}}$$ on $$P$$ is a Clifford connection, yielding existence. \end{remark} \begin{remark} The set of Clifford connections is a torsor over $$\Omega^1(M)$$. The association is $$\nabla \mapsto \nabla - \nabla^{\mathrm{LC}}$$, and one can compute \begin{align*} (\nabla - \nabla^{\mathrm{LC}})(x\cdot s) = x \cdot( \nabla - \nabla^{\mathrm{LC}})(s) ,\end{align*} which exactly says that this is a $${ \operatorname{Cl}} (M){\hbox{-}}$$linear map $${\mathbb{S}}_M \to {\mathbb{S}}_M \otimes\Omega^1$$. We can write $${ \operatorname{Cl}} (M) \cong \mathop{\mathrm{End}}({\mathbb{S}}_M)$$, and one can check that $$[\mathop{\mathrm{End}}{\mathbb{S}}_M, \mathop{\mathrm{End}}{\mathbb{S}}_M]$$ consists only of scalars. \end{remark} \begin{definition}[Dirac Operator] Let $$\nabla$$ be a Clifford connection on $${\mathbb{S}}_M$$ and $$s\in H^0({\mathbb{S}}_M)$$, so $$\nabla(s) \in {\mathbb{S}}_M \otimes\Omega^1(M)$$. Then the \textbf{Dirac operator} is defined as \begin{align*} \mkern-3mu \not{ \partial} : H^0({\mathbb{S}}) &\to H^0({\mathbb{S}}) \\ s &\mapsto \sum _{e_i \in \mathop{\mathrm{Fr}}(T {}^{ \vee }M) } e_i \cdot \nabla_{e_i {}^{ \vee }}(s) \end{align*} where \begin{itemize} \tightlist \item $$\nabla(s) = H^0({\mathbb{S}}_M \otimes\Omega^1)$$ \item $$\nabla_{e_i {}^{ \vee }}(s) = \nabla(s)(e_i {}^{ \vee }) \in H^0({\mathbb{S}}_M)$$ \end{itemize} \end{definition} \begin{remark} This makes sense locally, and is well-defined independent of choice of frame. Henceforth, we'll take $$\nabla = \nabla^{\mathrm{LC}}$$ -- in this case, if $$s^+ \in H^0({\mathbb{S}}^\pm)$$ then $$\nabla_v^{{\mathrm{LC}}}(s^\pm) \in H^0({\mathbb{S}}^\pm)$$. This is an order 1 differential operator: \begin{align*} \mkern-3mu \not{ \partial} _{ \nabla^{\mathrm{LC}}} = \mkern-3mu \not{ \partial} : H^0({\mathbb{S}}^\pm) \to H^p({\mathbb{S}}^{\mp}) .\end{align*} \end{remark} \begin{proposition}[Relation between Dirac operator and Laplacian] \begin{align*} \mkern-3mu \not{ \partial} ^2 = - \Delta .\end{align*} \end{proposition} \begin{proof}[?] Given $$\psi \in H^0({\mathbb{S}})$$, write $$\psi = \sum_{i=1}^4 \psi_i s_i$$ with the $$s_i$$ forming a local frame of $${\mathbb{S}}= {\mathbb{S}}^+ \oplus {\mathbb{S}}^-$$. We can write \begin{align*} \mkern-3mu \not{ \partial} \psi = \sum e_i \partial_{x_i} \psi = \sum_{i=1}^4 \gamma_i \psi_{x_i} .\end{align*} where $$\psi_{x_i} = {\left[ { (\psi_1)_{x_i}, (\psi_2)_{x_i}, \cdots } \right]}$$. We then have \begin{align*} \mkern-3mu \not{ \partial} ^2 \psi &= \sum_{i, j} \gamma_i \gamma_j \psi_{x_i x_j} \\ &= -\sum_{ij} 2 (e_i \cdot_g e_j) \psi_{x_i x_j} \\ &= -2 \sum_{ij} \delta_{ij} \psi_{x_i x_j} \\ &= -2 \sum_i \psi_{x_i x_i} \\ &= -2 \qty{ \sum_{i=1}^4 \partial^2_{x_i} }\psi \\ &= -2 \Delta .\end{align*} where we sum over \emph{all} $$i, j$$ and can pair terms, and we use that $$\gamma_i \gamma_j + \gamma_j \gamma_i = -2 e_1 \cdot e_j$$\\ Upshot: $$\mkern-3mu \not{ \partial} \in \sqrt{ \Delta}$$, which is why the Dirac is an invariant in quantum mechanics. This reduces the 2nd order Schrödinger operator a 1st order operator. Note that $$\mkern-3mu \not{ \partial} \psi = 0$$ is the equation for a massless particle. \end{proof} \begin{quote} See maybe Lawson's spin geometry? Or Salamon. \end{quote} \hypertarget{wednesday-may-05}{% \section{Wednesday, May 05}\label{wednesday-may-05}} \hypertarget{fun-physics-aside}{% \subsection{Fun Physics Aside}\label{fun-physics-aside}} \begin{remark} Last time: we showed $${ \operatorname{Cl}} (X) \coloneqq{ \operatorname{Cl}} (\dualof{T} X, \dualof{g})$$ acts on the spinor bundle $${\mathbb{S}}_X \coloneqq\tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}$$ by Clifford multiplication. For $$\dim_{\mathbb{R}}X = 4$$, we have a splitting $${\mathbb{S}}^+ \oplus {\mathbb{S}}^-$$ as complex rank 2 vector bundles. If $$\omega\in H^0 { \operatorname{Cl}} (X)$$ is a one form, then $$\omega{\mathbb{S}}_X^\pm \subset {\mathbb{S}}^{\mp}$$ . \end{remark} \begin{definition}[Clifford Connection] A \textbf{Clifford connection} is a map \begin{align*} \nabla: {\mathbb{S}}_X \to {\mathbb{S}}_X \otimes\Omega^1 \\ .\end{align*} where $$\alpha \cdot s \mapsto \alpha\cdot \nabla s + dx \cdot s$$. \end{definition} \begin{remark} There is a distinguished Clifford connection associated to $$\nabla^{\mathrm{LC}}$$. Also recall that we defined a Dirac operator $$\mkern-3mu \not{ \partial}$$ and showed $$\mkern-3mu \not{ \partial} ^2 = -2 \Delta$$. \end{remark} \begin{definition}[The Dirac Equation] The \textbf{Dirac equation} is defined on $$\psi\in H^0(X, {\mathbb{S}})$$ as \begin{align*} (i \mkern-3mu \not{ \partial} + m \omega)\psi = 0 .\end{align*} Here $$m$$ denotes a mass, $$\omega = \omega_{\mathbb{C}}= \prod_{i=1}^4 \gamma_i$$. \end{definition} \begin{remark} This describes fermions in a vacuum, e.g.~an electron where $$\psi$$ is its wave function. Applying this to $${\mathbb{R}}^4$$ with $$g = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2$$, then this equation in $$\psi$$ is invariant under the Lorentz group $$O({\mathbb{R}}^4, g)$$. \end{remark} \hypertarget{rohklins-theorem}{% \subsection{Rohklin's Theorem}\label{rohklins-theorem}} \begin{theorem}[Rohklin's Theorem] Let $$X$$ be a smooth closed oriented spin 4-manifold. Then the signature $$\sigma(X) \coloneqq b_2^+(X) - b_2^-(X)$$ (the dimensions of positive/negative definite subspaces of $$H^2(X; {\mathbb{R}})$$ is divisible by 16. \end{theorem} \begin{remark} This restricts what topological manifolds can admit smooth structures. Freedman constructed a topological manifold of dimension 4 with signature 8, which thus can not admit a smooth structure. Recall that having a spin structure was having a lift of a principal $${\operatorname{SO}}(n)$$ bundle over $$(\dualof{T} X, g)$$ (namely $$\mathop{\mathrm{Frame}}(X)$$) to a $${\operatorname{Spin}}(n)$$ bundle. \begin{center} \begin{tikzcd} {\tilde P_{{\operatorname{SO}}(V)} \coloneqq P_{{\operatorname{Spin}}(V)}} \\ \\ {P_{{\operatorname{SO}}(V)}} & {} \\ \\ & X \arrow["\pi"', from=3-1, to=5-2] \arrow["{\tilde p}"', from=1-1, to=3-1] \arrow["\exists", dashed, from=1-1, to=5-2] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=WzAsNCxbMSw0LCJYIl0sWzEsMl0sWzAsMiwiUF97XFxTTyhWKX0iXSxbMCwwLCJcXHRpbGRlIFBfe1xcU08oVil9IFxcZGEgUF97XFxTcGluKFYpfSJdLFsyLDAsIlxccGkiLDJdLFszLDIsIlxcdGlsZGUgcCIsMl0sWzMsMCwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link to Diagram} \end{quote} \todo[inline]{Diagram doesn't match definition, check Phil's notes!} \end{remark} \hypertarget{proof-1}{% \subsubsection{Proof}\label{proof-1}} Consider $${\mathbb{S}}_X \coloneqq\tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}$$, then define \begin{align*} \mkern-3mu \not{ \partial} ^\pm: H^0({\mathbb{S}}_X^\pm) \to H^0({\mathbb{S}}^{\mp}) .\end{align*} Note that we can write $$\mkern-3mu \not{ \partial} = \mkern-3mu \not{ \partial} ^+ + \mkern-3mu \not{ \partial} ^-$$; \begin{itemize} \tightlist \item Step 1: Show $$\mathop{\mathrm{ind}}\mkern-3mu \not{ \partial} ^+ = -\sigma(X) / 8$$, \item Step 2: Show $$\mathop{\mathrm{ind}}\mkern-3mu \not{ \partial} ^+$$ is even. \end{itemize} \hypertarget{step-1-1}{% \subsubsection{Step 1}\label{step-1-1}} What is the symbol $$\mathop{\mathrm{Symb}}(\mkern-3mu \not{ \partial} )$$? By definition \begin{align*} \mathop{\mathrm{Symb}}\mkern-3mu \not{ \partial} : \pi^* {\mathbb{S}}\to \pi^* {\mathbb{S}} .\end{align*} where $$\pi:\dualof{T} X\to X$$, and the symbol was defined by replacing $${\frac{\partial }{\partial x_i}\,}$$ with a function $$y_i: \dualof{T} X\to {\mathbb{R}}$$. We can write \begin{align*} \mkern-3mu \not{ \partial} \phi = \sum_{e_i \in \mathop{\mathrm{Fr}}} e_i \cdot \nabla_{e_i {}^{ \vee }} \psi ,\end{align*} and so \begin{align*} \mathop{\mathrm{Symb}}\mkern-3mu \not{ \partial} (\psi) = \sum_i y_i e_i = \psi .\end{align*} We have a tautological form $$\alpha\in H^0(\dualof{T} X, \pi^* \Omega^1)$$ where $$(p, \alpha) \mapsto \alpha$$, and so $$\mathop{\mathrm{Symb}}(\mkern-3mu \not{ \partial} )({-}) = \alpha\cdot({-})$$. \begin{claim} \begin{align*} \mkern-3mu \not{ \partial} : H^0({\mathbb{S}}) {\circlearrowleft}&& \text{is an elliptic operator} .\end{align*} \end{claim} We need to check that the map $$\alpha\cdot({-})$$ is exact if $$\alpha\neq 0$$. We have $$\alpha\cdot({-}): {\mathbb{S}}\to {\mathbb{S}}$$ and \begin{align*} (- \alpha)({-}) \, \alpha({-}) = (- \alpha \cdot \alpha) = {\left\lVert { \alpha} \right\rVert}^2 \neq 0 ,\end{align*} which makes the operator invertible away from zero. Thus we can apply Atiyah-Singer. \begin{lemma}[Formula for Chern characters] There is a nice formula for Chern characters: \begin{align*} \operatorname{ch}{\mathbb{S}}^+ - \operatorname{ch}{\mathbb{S}}^- = \prod_{i=1}^n( e ^{x_i/2} - e^{-x_i/2}) .\end{align*} where $$\left\{{ \pm x_i }\right\}$$ are the Chern roots of $$\dualof{T} X$$. \end{lemma} \begin{proof}[?] Use the splitting principle to write \begin{align*} \dualof{T} X \otimes_{\mathbb{R}}{\mathbb{C}}= \bigoplus _{i=1}^n L_i \otimes L_i^{-1} .\end{align*} Then $${\mathbb{S}}^+$$ is a sum of all tensor products of $$L_i \otimes L_i^{-1}$$ where the number of $$-1$$s appearing is even. \end{proof} \begin{remark} Note there is ambiguity up to 2-torsion in the formula, but this gets moved into the choice of spin structure, which amounts to choice of a square root of each of these line bundles. \end{remark} Setting $$2n\coloneqq\dim X$$, we have \begin{align*} \mathop{\mathrm{ind}}\mkern-3mu \not{ \partial} ^+ &= (-1)^n \int_X { \operatorname{ch}{\mathbb{S}}^+ - \operatorname{ch}{\mathbb{S}}^- \over {\operatorname{eul}}X} \mathrm{td}(TX\otimes{\mathbb{C}}) \\ \\ &= \int_X { \prod e^{x_i/2} - e^{-x_i /2} \over (-1)^n \prod x_i} \prod {x_i \over 1 - e^{x_i} } \prod {x_i \over 1 - e^{-x_i}} \\ \\ &= \int_X \prod { (e^{x_i/2} - e^{-x_i / 2 )} x_i \over (1-e^{x_i} ) (1 - e^{-x_i}) } \\ \\ &= (-1)^n \int_X \prod_I {x_i \over e^{x_i/2} - e^{-x_i/2}} \\ \\ &= \int_X \qty{ 1 - {x_1^2 \over 24} } \qty{ 1 - {x_2^2 \over 24}} \\ \\ &= -{1\over 24} \int_X x_1^2 + x_2^2 + (x_1 + x_2)^2 - 2x_1 x_2 \\ \\ &= -{1\over 24} \qty{c_1^2 - 2c_2} .\end{align*} \begin{remark} See the $$\widehat{A}$$ genus. \end{remark} \begin{claim} \begin{align*} c_1^2 - 2c^2 = 3\cdot \sigma(X) .\end{align*} \end{claim} This is another application of Atiyah-Singer, applied to a slightly different operator. Recall the Hodge star operator, \begin{align*} \star: \Omega^k(X) \to \Omega^{4-k}(X) .\end{align*} Defining $$\tau \coloneqq i^{k(k-1) + 4 \over 2}$$, we get $$\tau^2 = 1$$, so define an operator $$\tau \star$$. This yields a splitting into $$\pm 1$$ eigenspaces: \begin{align*} \Omega(X) = \Omega^+(X) \oplus \Omega^-(X) .\end{align*} Recalling that $$d^\dagger$$ was the adjoint of $$d$$, one can check that $$d+d^\dagger: \Omega^{\pm}(X) \to \Omega^{\mp}(X)$$ interchanges these. It turns out that $$\mathop{\mathrm{ind}}(d + d^\dagger) = \sigma(X)$$, which by Atiyah-Singer and Hermite forms will equal $$c_1^2 -2 c_2 \over 3$$. This yields the desired formula for step 1. \hypertarget{step-2-1}{% \subsection{Step 2}\label{step-2-1}} We now want to show $$\mathop{\mathrm{ind}}\mkern-3mu \not{ \partial} ^+$$ is divisible by 2. The key point is that $$\ker \mkern-3mu \not{ \partial} ^+$$ and $$\operatorname{coker}\mkern-3mu \not{ \partial} ^+ = \ker \mkern-3mu \not{ \partial} ^-$$ admit a quaternionic vector space structure. This comes from the fact that \begin{align*} {\operatorname{Spin}}(4) \cong {\operatorname{SU}}(2) \times{\operatorname{SU}}(2) \cong S^1({\mathbb{H}}) \oplus S^1({\mathbb{H}}) \coloneqq{\mathbb{S}}^+ \oplus {\mathbb{S}}^- ,\end{align*} so we have a splitting into subspaces of unit quaternions. It turns out that $$\mkern-3mu \not{ \partial}$$ is $${\mathbb{H}}{\hbox{-}}$$linear. So we get an equality \begin{align*} -\sigma(X) / 8 = \mathop{\mathrm{ind}}\mkern-3mu \not{ \partial} ^+ = 2\lambda \end{align*} for some $$\lambda$$, yielding $$8\divides \sigma(X)$$. \hypertarget{remarks}{% \subsection{Remarks}\label{remarks}} \begin{remark} If $$H_1(X; {\mathbb{Z}})$$ has no 2-torsion, e.g.~if $$\pi_1X = 0$$, then $$w_2(X) = 0$$ iff the intersection form on $$H^2$$ is even, where $$w_2$$ is the obstruction to existence of spin structures. Note that this makes sense for topological manifolds and not just smooth manifolds, and in this case $$\sigma(X)$$ is divisible by 8. This restriction comes from number theory: since we have a unimodular lattice, it breaks into sums of $$E_8, -E_8$$, and $$H$$ if indefinite, and any even unimodular lattice has signature divisible by 8. So this can work as an obstruction to the existence of smooth structures. \end{remark} \begin{remark} Note that $${\mathbb{CP}}^2$$ has no spin structure, and $$\sigma({\mathbb{CP}}^2) = 1$$. There's a way to modify the invariant to set $$\sigma(X)/8 = ? \pmod 2$$. \end{remark} \addsec{ToDos} \listoftodos[List of Todos] \cleardoublepage % Hook into amsthm environments to list them. \addsec{Definitions} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={definition}, numwidth=3.5em] \cleardoublepage \addsec{Theorems} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em] \cleardoublepage \addsec{Exercises} \renewcommand{\listtheoremname}{} \listoftheorems[ignoreall,show={exercise}, numwidth=3.5em] \cleardoublepage \addsec{Figures} \listoffigures \cleardoublepage \printbibliography[title=Bibliography] \end{document}