\input{"preamble.tex"} \addbibresource{ModuliSpaces.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{ Moduli Spaces } \\ {\normalsize University of Georgia, Spring 2020} \\ \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 course in Moduli Spaces taught by Ben Bakker at the University of Georgia in Spring 2020. Any errors or inaccuracies are almost certainly my own. } } \\ \end{flushleft} \end{minipage} \hfill \begin{minipage}{.65\linewidth} \end{minipage} } \begin{document} \date{} \author{D. Zack Garza} \maketitle \begin{flushleft} \textit{D. Zack Garza} \\ \textit{University of Georgia} \\ \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\ {\tiny \textit{Last updated:} 2021-01-05 } \end{flushleft} \newpage % Note: addsec only in KomaScript \addsec{Table of Contents} \tableofcontents \newpage \hypertarget{references}{% \section{References}\label{references}} \begin{itemize} \item Course notes \autocite{bakker_8330} \item General reference \autocite{hartshorne_2010} \item Hilbert schemes/functors of points: \autocite{stromme}, \autocite{hartshorne_def}. \begin{itemize} \tightlist \item Slightly more detailed: \autocite{fantechi_2005} \end{itemize} \item Curves on surfaces: \autocite{mumford_1985} \item Moduli of Curves: \autocite{harris_morrison_1998} (chatty and less rigorous) \end{itemize} \hypertarget{schemes-vs-representable-functors-thursday-january-9th}{% \section{Schemes vs Representable Functors (Thursday January 9th)}\label{schemes-vs-representable-functors-thursday-january-9th}} Last time: fix an \(S{\hbox{-}}\)scheme, i.e.~a scheme over \(S\). Then there is a map \begin{align*} {\operatorname{Sch}}_{_{/S}} &\to {\operatorname{Fun}}( {\operatorname{Sch}}_{_{/S}}^{\operatorname{op}}, {\operatorname{Set}}) \\ x &\mapsto h_x(T) = \hom_{{\operatorname{Sch}}_{_{/S}} }(T, x) .\end{align*} where \(T' \xrightarrow{f} T\) is given by \begin{align*} h_x(f): h_x(T) &\to h_x(T') \\ \qty{ T \mapsto x } &\mapsto \text{triangles of the form} .\end{align*} \begin{center} \begin{tikzcd} T' \arrow[rr] \arrow[rdd] & & X \\ & & \\ & T \arrow[ruu] & \end{tikzcd} \end{center} \hypertarget{representability}{% \subsection{Representability}\label{representability}} \begin{theorem}[?] \begin{align*}\hom_{{\operatorname{Fun}}}(h_x, F) = F(x).\end{align*} \end{theorem} \begin{corollary}[?] \begin{align*}\hom_{{\operatorname{Sch}}_{/S}}(x, y) \cong \hom_{{\operatorname{Fun}}}(h_x, h_y).\end{align*} \end{corollary} \begin{definition}[Moduli Functor] A \textbf{moduli functor} is a map \begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ F(x) &= \text{ "Families of something over $x$" } \\ F(f) &= \text{"Pullback"} .\end{align*} \end{definition} \begin{definition}[Moduli Space] A \textbf{moduli space} for that ``something'' appearing above is an \(M \in \mathrm{Obj}({\operatorname{Sch}}_{/S})\) such that \(F \cong h_M\). \end{definition} \begin{remark} Now fix \(S = \operatorname{Spec}(k)\), and write \(h_m\) for the functor of points over \(M\). Then \begin{align*} h_m(\operatorname{Spec}(k)) = M(\operatorname{Spec}(k)) \cong \text{families over } \operatorname{Spec}k = F(\operatorname{Spec}k) .\end{align*} \end{remark} \begin{remark} \(h_M(M) \cong F(M)\) are families over \(M\), and \(\operatorname{id}_M \in \mathrm{Mor}_{{\operatorname{Sch}}_{/S}}(M, M) = \xi_{Univ}\) is the universal family. Every family is uniquely the pullback of \(\xi_{\text{Univ}}\). This makes it much like a classifying space. For \(T\in {\operatorname{Sch}}_{/S}\), \begin{align*} h_M &\xrightarrow{\cong} F \\ f\in h_M(T) &\xrightarrow{\cong} F(T) \ni \xi = F(f)(\xi_{\text{Univ}}) .\end{align*} where \(T\xrightarrow{f} M\) and \(f = h_M(f)(\operatorname{id}_M)\). \end{remark} \begin{remark} If \(M\) and \(M'\) both represent \(F\) then \(M \cong M'\) up to unique isomorphism. \begin{center} \begin{tikzcd} \xi_M & & \xi_{M'} \\ M \arrow[rr, "f"] & & M' \\ & & \\ M' \arrow[rr, "g"] & & M \\ \xi_{M'} & & \xi_M \end{tikzcd} \end{center} which shows that \(f, g\) must be mutually inverse by using universal properties. \end{remark} \begin{example}[?] A length 2 subscheme of \({\mathbb{A}}^1_k\) (??) then \begin{align*} F(S) = \left\{{ V(x^2 + bx + c)}\right\} \subset {\mathbb{A}}^5 \end{align*} where \(b, c \in {\mathcal{O}}_s(s)\), which is functorially bijective with \(\left\{{b, c \in {\mathcal{O}}_s(s)}\right\}\) and \(F(f)\) is pullback. Then \(F\) is representable by \({\mathbb{A}}_k^2(b, c)\) and the universal object is given by \begin{align*} V(x^2 + bx + c) \subset {\mathbb{A}}^1(?) \times{\mathbb{A}}^2(b, c) \end{align*} where \(b, c \in k[b, c]\). Moreover, \(F'(S)\) is the set of effective Cartier divisors in \({\mathbb{A}}_5'\) which are length 2 for every geometric fiber. \(F''(S)\) is the set of subschemes of \({\mathbb{A}}_5'\) which are length 2 on all geometric fibers. In both cases, \(F(f)\) is always given by pullback. \end{example} Problem: \(F''\) is not a good moduli functor, as it is not representable. Consider \(\operatorname{Spec}k[\varepsilon]\), for which we have the following situation: \begin{center} \begin{tikzpicture}[scale=2.0] \begin{axis}[ hide axis, xmin=-12, xmax=18, ymin=-4, ymax=10, xtick = {0}, ytick = {0}, disabledatascaling] \draw[-][black][opacity=1] (axis cs:-10.0, 9) -- (axis cs:-10, -0); \draw[-][black][opacity=1] (axis cs:-14.0, 0) -- (axis cs:-6, -0); \draw[-][black][opacity=1] (axis cs:0.0, 9) -- (axis cs:-0, -0); \draw[-][black][opacity=1] (axis cs:-4.0, 0) -- (axis cs:4, -0); \draw[-][black][opacity=1] (axis cs:10.0, 9) -- (axis cs: 10, -0); \draw[-][black][opacity=1] (axis cs:6.0, 0) -- (axis cs:14, -0); \node[draw, circle, blue, scale=0.4, fill=blue](left1) at (axis cs:-10, 6) [anchor=center] {}; \node[right=1mm of left1,font=\tiny] {$(\varepsilon+ x - 1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](center1) at (axis cs:0, 6) [anchor=center] {}; \node[right=1mm of center1,font=\tiny] {$(x)(\varepsilon, x-1)$}; \node[draw, circle, blue, scale=0.4, fill=blue](right1) at (axis cs:10, 6) [anchor=center] {}; \node[right=1mm of right1,font=\tiny] {$(x(x-1), \varepsilon)$}; \node[draw, circle, blue, scale=0.4, fill=blue](left2) at (axis cs:-10, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](center2) at (axis cs:0, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](right2) at (axis cs:10, 3) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](left3) at (axis cs:-10, 0) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](center3) at (axis cs:0, 0) [anchor=center] {}; \node[draw, circle, blue, scale=0.4, fill=blue](right3) at (axis cs:10, 0) [anchor=center] {}; \draw[-][blue, very thick][opacity=0.9] (axis cs:-10.0, 6) -- (axis cs:-7, 9); \draw[-][blue, very thick][opacity=0.9] (axis cs:-10.0, 3) -- (axis cs:-7, 3); \draw[-][blue, very thick][opacity=0.9] (axis cs:0, 3) -- (axis cs:3, 3); \draw[-][blue, very thick][opacity=0.9] (axis cs:0, 0) -- (axis cs:3, 0); \draw[-][blue, very thick][opacity=0.9] (axis cs:10, 0) -- (axis cs:13, 0); \end{axis} \end{tikzpicture} \end{center} \begin{table}[H] \centering \begin{tabular}{l|lll} \hline \\ $F$ & $\checkmark$ & x & x \\ $F'$ & $\checkmark$ & x & x \\ $F''$ & $\checkmark$ & $\checkmark$ & $\checkmark$ \end{tabular} \end{table} \begin{center} \begin{tikzcd} \operatorname{Spec}k \arrow[rrr, "i", hook] & & & {\operatorname{Spec}k[\varepsilon]} & & =F'(\operatorname{Spec}k) \arrow[rd] & \\ {F(\operatorname{Spec}k[\varepsilon])} \arrow[rrr, "F(i)"] & & & F(\operatorname{Spec}k) \arrow[rru] & & & =F''(\operatorname{Spec}k) \\ & & & & & & \\ {T_p F^{', ''}}\arrow[uu, "\subset", hook]& & & P = V(x(x-1)) \arrow[uu, "\in", hook] & && \end{tikzcd} \end{center} We think of \(T_p F^{', ''}\) as the tangent space at \(p\). If \(F\) is representable, then it is actually the Zariski tangent space. \begin{center} \begin{tikzcd} {M(\operatorname{Spec}k[\varepsilon])} \arrow[rr] & & M(\operatorname{Spec}k) \\ & & \\ T_p M \arrow[rr] \arrow[uu, "\subset", hook]& & p \arrow[uu, "\subset", hook] \end{tikzcd} \end{center} \begin{center} \begin{tikzcd} & & \operatorname{Spec}k \arrow[rdd, "?"] \arrow[lldd, hook] & \\ & & & \\ {\operatorname{Spec}k[\varepsilon]} \arrow[rrr] & & & {\operatorname{Spec}{\mathcal{O}}_{M, p} \subset M} \\ & & & k \\ & {{\mathcal{O}}_{M, p}} \arrow[rru] \arrow[rr] & & {k[\varepsilon]} \arrow[u] \\ & {\mathfrak{m}}_p \arrow[u, hook] & & (\varepsilon) \arrow[u, hook] \\ & {\mathfrak{m}}_p^2 \arrow[u, hook] & & 0 \arrow[u, hook] \end{tikzcd} \end{center} Moreover, \(T_p M = ({\mathfrak{m}}_p / {\mathfrak{m}}_p^2)^\vee\), and in particular this is a \(k{\hbox{-}}\)vector space. To see the scaling structure, take \(\lambda \in k\). \begin{align*} \lambda: k[\varepsilon] & \to k[\varepsilon] \\ \varepsilon &\mapsto \lambda \varepsilon \\ \\ \lambda^*: \operatorname{Spec}(k[\varepsilon]) &\to \operatorname{Spec}(k[\varepsilon]) \\ \\ \lambda: M(\operatorname{Spec}(k[\varepsilon])) &\to M(\operatorname{Spec}(k[\varepsilon])) .\end{align*} \begin{center} \begin{tikzcd} M(\operatorname{Spec}(k[\varepsilon])) \ar[r, "\lambda"] & M(\operatorname{Spec}(k[\varepsilon])) \\ T_pM \ar[r] \ar[u, "\subseteq"] & T_pM \ar[u, "\subseteq"] \end{tikzcd} \end{center} \textbf{Conclusion}: If \(F\) is representable, for each \(p\in F(\operatorname{Spec}k)\) there exists a unique point of \(T_p F\) that are invariant under scaling. \begin{remark} If \(F, F', G \in {\operatorname{Fun}}( ({\operatorname{Sch}}_{/S})^{\operatorname{op}}, {\operatorname{Set}})\), there exists a fiber product \begin{center} \begin{tikzcd} F \times_G F' \arrow[rr, dotted] \arrow[dd, dotted] & & F' \arrow[dd] \\ & & \\ F \arrow[rr] & & G \end{tikzcd} \end{center} where \begin{align*} (F \times_G F')(T) = F(T) \times_{G(T)} F'(T) .\end{align*} \end{remark} \begin{remark} This works with the functor of points over a fiber product of schemes \(X \times_T Y\) for \(X, Y \to T\), where \begin{align*} h_{X \times_T Y}= h_X \times_{h_t} h_Y .\end{align*} \end{remark} \begin{remark} If \(F, F', G\) are representable, then so is the fiber product \(F \times_G F'\). \end{remark} \begin{remark} For any functor \begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}} ,\end{align*} for any \(T \xrightarrow{f} S\) there is an induced functor \begin{align*} F_T: ({\operatorname{Sch}}_{/T}) &\to {\operatorname{Set}}\\ x &\mapsto F(x) .\end{align*} \end{remark} \begin{remark} \(F\) is representable by \(M_{/S}\) implies that \(F_T\) is representable by \(M_T = M \times_S T / T\). \end{remark} \hypertarget{projective-space}{% \subsection{Projective Space}\label{projective-space}} Consider \({\mathbb{P}}^n_{\mathbb{Z}}\), i.e.~``rank 1 quotient of an \(n+1\) dimensional free module''. \begin{proposition}[?] \({\mathbb{P}}^n_{/{\mathbb{Z}}}\) represents the following functor \begin{align*} F: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto \left\{{ {\mathcal{O}}_S^{n+1} \to L \to 0 }\right\} / \sim .\end{align*} where \(\sim\) identifies diagrams of the following form: \begin{center} \begin{tikzcd} {\mathcal{O}}_s^{n+1} \arrow[dd, equal] \arrow[rr] & & L \arrow[dd, "\cong"] \arrow[rr] & & 0 \\ & & & & \\ {\mathcal{O}}_s^{n+1} \arrow[rr] & & M \arrow[rr] & & 0 \end{tikzcd} \end{center} and \(F(f)\) is given by pullbacks. \end{proposition} \begin{remark} \({\mathbb{P}}^n_{/S}\) represents the following functor: \begin{align*} F_S: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ T &\mapsto F_S(T) = \left\{{ {\mathcal{O}}_T^{n+1} \to L \to 0}\right\} / \sim .\end{align*} This gives us a cleaner way of gluing affine data into a scheme. \end{remark} \hypertarget{proof-of-proposition}{% \subsubsection{Proof of Proposition}\label{proof-of-proposition}} \begin{remark} Note that \({\mathcal{O}}^{n+1} \to L \to 0\) is the same as giving \(n+1\) sections \(s_1, \cdots s_n\) of \(L\), where surjectivity ensures that they are not the zero section. So \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\}/\sim ,\end{align*} with the additional condition that \(s_i \neq 0\) at any point. There is a natural transformation \(F_i \to F\) by forgetting the latter condition, and is in fact a subfunctor. \footnote{\(F\leq G\) is a subfunctor iff \(F(s) \hookrightarrow G(s)\).} \end{remark} \begin{claim} It is enough to show that each \(F_i\) and each \(F_{ij}\) are representable, since we have natural transformations: \begin{center} \begin{tikzcd} F_i \arrow[rr] & & F \\ & & \\ F_{ij} \arrow[rr] \arrow[uu] & & F_j \arrow[uu] \end{tikzcd} \end{center} and each \(F_{ij} \to F_i\) is an open embedding on the level of their representing schemes. \end{claim} \begin{example}[?] For \(n=1\), we can glue along open subschemes \begin{center} \begin{tikzcd} & & F_0 \\ F_{01} \arrow[rru] \arrow[rrd] & & \\ & & F_1 \end{tikzcd} \end{center} For \(n=2\), we get overlaps of the following form: \begin{center} \begin{tikzcd} & & & & F_0 \arrow[rrdd] & & \\ & & & F_{01} \arrow[rd] \arrow[ru] & & & \\ F_{012} \arrow[rr] \arrow[rrru] \arrow[rrrd] & & F_{02} \arrow[ru] \arrow[rd] \arrow[rruu, dotted, bend left=49] \arrow[rrdd, bend right=49] & & F_1 \arrow[rr] & & F \\ & & & F_{12} \arrow[ru] \arrow[rd] & & & \\ & & & & F_2 \arrow[rruu] & & \end{tikzcd} \end{center} This claim implies that we can glue together \(F_i\) to get a scheme \(M\). We want to show that \(M\) represents \(F\). \(F(s)\) (LHS) is equivalent to an open cover \(U_i\) of \(S\) and sections of \(F_i(U_i)\) satisfying the gluing (RHS). Going from LHS to RHS isn't difficult, since for \({\mathcal{O}}_s^{n+1} \to L \to 0\), \(U_i\) is the locus where \(s_i \neq 0\) and by surjectivity, this gives a cover of \(S\). The RHS to LHS comes from gluing. \end{example} \begin{proof}[of claim] We have \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_S^{n+1} \to L \cong {\mathcal{O}}_s \to 0, s_i \neq 0}\right\} ,\end{align*} but there are no conditions on the sections other than \(s_i\).\\ So specifying \(F_i(S)\) is equivalent to specifying \(n-1\) functions \(f_1 \cdots \widehat{f}_i \cdots f_n \in {\mathcal{O}}_S(s)\) with \(f_k \neq 0\). We know this is representable by \({\mathbb{A}}^n\). We also know \(F_{ij}\) is obviously the same set of sequences, where now \(s_j \neq 0\) as well, so we need to specify \(f_0 \cdots \widehat{f}_i \cdots f_j \cdots f_n\) with \(f_j \neq 0\). This is representable by \({\mathbb{A}}^{n-1} \times{\mathbb{G}}_m\), i.e.~\(\operatorname{Spec}k[x_1, \cdots, \widehat{x}_i, \cdots, x_n, x_j^{-1}]\). Moreover, \(F_{ij} \hookrightarrow F_i\) is open. What is the compatibility we are using to glue? For any subset \(I \subset \left\{{0, \cdots, n}\right\}\), we can define \begin{align*} F_I = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0, s_i\neq 0 \text{ for } i\in I}\right\} = {\prod{i\in I}}_F F_i ,\end{align*} and \(F_I \to F_J\) when \(I \supset J\). \end{proof} \hypertarget{functors-as-spaces-tuesday-january-14th}{% \section{Functors as Spaces (Tuesday January 14th)}\label{functors-as-spaces-tuesday-january-14th}} Last time: representability of functors, and specifically projective space \({\mathbb{P}}_{/{\mathbb{Z}}}^n\) constructed via a functor of points, i.e. \begin{align*} h_{{\mathbb{P}}^n_{/{\mathbb{Z}}} }: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ s &\mapsto {\mathbb{P}}^n_{/{\mathbb{Z}}}(s) = \left\{{ {\mathcal{O}}_s^{n+1} \to L \to 0}\right\} .\end{align*} for \(L\) a line bundle, up to isomorphisms of diagrams: \begin{center} \begin{tikzcd} {\mathcal{O}}_{s}^{n+1} \arrow[dd, no head, Rightarrow] \arrow[rr] & & L \arrow[rr] \arrow[dd, "\cong"] & & 0 \\ & & & & \\ {\mathcal{O}}_{s}^{n+1} \arrow[rr] & & M \arrow[rr] & & 0 \end{tikzcd} \end{center} That is, line bundles with \(n+1\) sections that globally generate it, up to isomorphism. The point was that for \(F_i \subset {\mathbb{P}}_{/{\mathbb{Z}}}^n\) where \begin{align*} F_i(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0 {~\mathrel{\Big|}~}s_i \text{ is invertible}}\right\} \end{align*} are representable and can be glued together, and projective space represents this functor. \begin{remark} Because projective space represents this functor, there is a universal object: \begin{center} \begin{tikzcd} {\mathcal{O}}_{{\mathbb{P}}_{{\mathbb{Z}}}^n}^{n+1} \arrow[rr] & & L \arrow[dd, equal] \arrow[rr] & & 0 \\ & & & & \\ & & {\mathcal{O}}_{{\mathbb{P}}_{{\mathbb{Z}}}^n}(1) & & \end{tikzcd} \end{center} and other functors are pullbacks of the universal one. (Moduli Space) \end{remark} \begin{exercise}[?] Show that \({\mathbb{P}}_{/{\mathbb{Z}}}^n\) is proper over \(\operatorname{Spec}{\mathbb{Z}}\). Use the evaluative criterion, i.e.~there is a unique lift \begin{center} \begin{tikzcd} \operatorname{Spec}k \arrow[dd] \arrow[rrr] & & & {\mathbb{P}}^n_{{\mathbb{Z}}} \arrow[dd] \\ & & & \\ \operatorname{Spec}R \arrow[rrr] \arrow[rrruu, dashed] & & & \operatorname{Spec}{\mathbb{Z}} \end{tikzcd} \end{center} \end{exercise} \hypertarget{generalizing-open-covers}{% \subsection{Generalizing Open Covers}\label{generalizing-open-covers}} \begin{definition}[Equalizer] For a category \(C\), we say a diagram \(X \to Y \rightrightarrows Z\) is an \emph{equalizer} iff it is universal with respect to the following property: \begin{center} \begin{tikzcd} X \arrow[rr] & & Y \arrow[rr, shift left=.5ex] \ar[rr, shift right=.5ex] & & Z \\ & & & & \\ & & S \arrow[lluu, dashed, "\exists!"] \arrow[uu] \arrow[rruu] & & \end{tikzcd} \end{center} where \(X\) is the universal object. \end{definition} \begin{example}[?] For sets, \(X = \left\{{y {~\mathrel{\Big|}~}f(y) = g(y)}\right\}\) for \(Y \xrightarrow{f, g} Z\). \end{example} \begin{definition}[?] A \textbf{coequalizer} is the dual notion, \begin{center} \begin{tikzcd} & & S & & \\ & & & & \\ Z \arrow[rruu] \arrow[rr, shift left=.5ex] \ar[rr, shift right=.5ex] & & Y \arrow[uu] \arrow[rr] & & X \arrow[lluu, "\exists!"', dashed] \end{tikzcd} \end{center} \end{definition} \begin{example}[?] Take \(C = {\operatorname{Sch}}_{/S}\), \(X_{/S}\) a scheme, and \(X_\alpha \subset X\) an open cover. We can take two fiber products, \(X_{\alpha \beta}, X_{\beta, \alpha}\): \begin{center} \begin{tikzcd} X_\alpha \arrow[rr] & & X & & & X_\beta \arrow[rr] & & X \\ & & & & & & & \\ X_{\alpha\beta} \arrow[uu] \arrow[rr] & & X_\beta \arrow[uu] & & & X_{\beta\alpha} \arrow[uu] \arrow[rr] & & X_\alpha \arrow[uu] \end{tikzcd} \end{center} These are canonically isomorphic. \end{example} In \({\operatorname{Sch}}_{/S}\), we have \begin{center} \begin{tikzcd} {\coprod}_{\alpha\beta} X_{\alpha\beta} \arrow[rr, shift left=.5ex, "f_{\alpha\beta}"] \arrow[rr, shift right=.5ex,"g_{\alpha\beta}", swap] & & {\coprod}_{\alpha} X_\alpha \arrow[rr] & & X \end{tikzcd} \end{center} where \begin{align*} f_{\alpha\beta}: X_{\alpha\beta} &\to X_\alpha \\ g_{\alpha\beta}: X_{\alpha\beta} &\to X_\beta ;\end{align*} form a coequalizer. Conversely, we can glue schemes. Given \(X_\alpha \to X_{\alpha\beta}\) (schemes over open subschemes), we need to check triple intersections: \begin{center} \begin{tikzpicture}[scale=0.25] \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[xshift=-1.25cm, yshift=-1.0cm] $X_\alpha$ }] at (-3,3) {}; \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[xshift=1.25cm, yshift=-1.0cm] $X_\beta$ }] at (3,3) {}; \node[draw,circle,minimum size=5cm,inner sep=0pt, label={[yshift=-4.25cm] $X_\gamma$ }] at (0,-3) {}; \end{tikzpicture} \end{center} Then \(\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}\) must satisfy the \textbf{cocycle condition}: \begin{definition}[Cocycle Condition] Maps \(\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}\) satisfy the \textbf{cocycle condition} iff \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item \begin{align*}\varphi_{\alpha\beta}^{-1}\qty{ X_{\beta\alpha} \cap X_{\beta\gamma} } = X_{\alpha\beta} \cap X_{\alpha \gamma},\end{align*} noting that the intersection is exactly the fiber product \(X_{\beta\alpha} \times_{X_\beta} X_{\beta \gamma}\). \item The following diagram commutes: \end{enumerate} \begin{center} \begin{tikzcd} X_{\alpha\beta} \cap X_{\alpha\gamma} \arrow[rdd, "\varphi_{\alpha\beta}"'] \arrow[rr, "\varphi_{\alpha\gamma}"] && X_{\gamma\alpha} \cap X_{\gamma\beta} \\ && \\ & X_{\beta\alpha}\cap X_{\beta\gamma} \arrow[ruu, "\varphi_{\beta\gamma}"'] & \end{tikzcd} \end{center} \end{definition} Then there exists a scheme \(X_{/S}\) such that \({\coprod}_{\alpha\beta} X_{\alpha\beta} \rightrightarrows {\coprod}X_\alpha \to X\) is a coequalizer; this is the gluing. Subfunctors satisfy a patching property because morphisms to schemes are locally determined. Thus representable functors (e.g.~functors of points) have to be (Zariski) sheaves. \begin{definition}[Zariski Sheaf] A functor \(F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}\) is a \textbf{Zariski sheaf} iff for any scheme \(T_{/S}\) and any open cover \(T_\alpha\), the following is an equalizer: \begin{align*} F(T) \to \prod F(T_\alpha) \rightrightarrows \prod_{\alpha\beta} F(T_{\alpha\beta}) \end{align*} where the maps are given by restrictions. \end{definition} \begin{example}[?] Any representable functor is a Zariski sheaf precisely because the gluing is a coequalizer. Thus if you take the cover \begin{align*} {\coprod}_{\alpha\beta} T_{\alpha\beta} \to {\coprod}_{\alpha}T_\alpha \to T ,\end{align*} since giving a local map to \(X\) that agrees on intersections if enough to specify a map from \(T\to X\). Thus any functor represented by a scheme automatically satisfies the sheaf axioms. \end{example} \begin{definition}[Subfunctors and Open/Closed Functors] Suppose we have a morphism \(F' \to F\) in the category \({\operatorname{Fun}}({\operatorname{Sch}}_{/S}, {\operatorname{Set}})\). \begin{itemize} \item This is a \textbf{subfunctor} if \(\iota(T)\) is injective for all \(T_{/S}\). \item \(\iota\) is \textbf{open/closed/locally closed} iff for any scheme \(T_{/S}\) and any section \(\xi \in F(T)\) over \(T\), then there is an open/closed/locally closed set \(U\subset T\) such that for all maps of schemes \(T' \xrightarrow{f} T\), we can take the pullback \(f^* \xi\) and \(f^*\xi \in F'(T')\) iff \(f\) factors through \(U\). \end{itemize} \end{definition} \begin{remark} This says that we can test if pullbacks are contained in a subfunctors by checking factorization. This is the same as asking if the subfunctor \(F'\), which maps to \(F\) (noting a section is the same as a map to the functor of points), and since \(T\to F\) and \(F' \to F\), we can form the fiber product \(F' \times_F T\): \begin{center} \begin{tikzcd} F' \ar[r] & F \\ & \\ F' \times_F T \ar[r, "g"] \ar[uu] & T \ar[uu, "\xi" swap] \end{tikzcd} \end{center} and \(F' \times_F T \cong U\). Note: this is almost tautological! Thus \(F' \to F\) is open/closed/locally closed iff \(F' \times_F T\) is representable and \(g\) is open/closed/locally closed. I.e. base change is representable. \end{remark} \begin{exercise}[?] \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item If \(F' \to F\) is open/closed/locally closed and \(F\) is representable, then \(F'\) is representable as an open/closed/locally closed subscheme \item If \(F\) is representable, then open/etc subschemes yield open/etc subfunctors \end{enumerate} \end{exercise} \begin{slogan} Treat functors as spaces. \end{slogan} We have a definition of open, so now we'll define coverings. \begin{definition}[Open Covers] A collection of open subfunctors \(F_\alpha \subset F\) is an \textbf{open cover} iff for any \(T_{/S}\) and any section \(\xi \in F(T)\), i.e.~\(\xi: T\to F\), the \(T_\alpha\) in the following diagram are an open cover of \(T\): \begin{center} \begin{tikzcd} F_\alpha \ar[r] & F \\ & \\ T_\alpha \ar[uu] \ar[r] & T \ar[uu, "\xi" swap] \end{tikzcd} \end{center} \end{definition} \begin{example}[?] Given \begin{align*} F(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\} \end{align*} and \(F_i(s)\) given by those where \(s_i \neq 0\) everywhere, the \(F_i \to F\) are an open cover. Because the sections generate everything, taking the \(T_i\) yields an open cover. \end{example} \hypertarget{results-about-zariski-sheaves}{% \subsection{Results About Zariski Sheaves}\label{results-about-zariski-sheaves}} \begin{proposition}[?] A Zariski sheaf \(F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}\) with a representable open cover is representable. \end{proposition} \begin{proof}[?] Let \(F_\alpha \subset F\) be an open cover, say each \(F_\alpha\) is representable by \(x_\alpha\). Form the fiber product \(F_{\alpha\beta} = F_\alpha \times_F F_\beta\). Then \(x_\beta\) yields a section (plus some openness condition?), so \(F_{\alpha\beta} = x_{\alpha\beta}\) representable. Because \(F_\alpha \subset F\), the \(F_{\alpha\beta} \to F_\alpha\) have the correct gluing maps. This follows from Yoneda (schemes embed into functors), and we get maps \(x_{\alpha\beta} \to x_\alpha\) satisfying the gluing conditions. Call the gluing scheme \(x\); we'll show that \(x\) represents \(F\). First produce a map \(x\to F\) from the sheaf axioms. We have a map \(\xi \in \prod_\alpha F(x_\alpha)\), and because we can pullback, we get a unique element \(\xi \in F(X)\) coming from the diagram \begin{align*} F(x) \to \prod F(x_\alpha) \rightrightarrows \prod_{\alpha\beta} F(x_{\alpha\beta}) .\end{align*} \end{proof} \begin{lemma}[?] If \(E \to F\) is a map of functors and \(E, F\) are Zariski sheaves, where there are open covers \(E_\alpha \to E, F_\alpha \to F\) with commutative diagrams \begin{center} \begin{tikzcd} E \ar[r] & F \\ & \\ E_\alpha \ar[uu] \ar[r, "\cong"] & F_\alpha \ar[uu] \end{tikzcd} \end{center} (i.e.~these are isomorphisms locally), then the map is an isomorphism. \end{lemma} With the following diagram, we're done by the lemma: \begin{center} \begin{tikzcd} X \ar[r] & F \\ & \\ X_\alpha \ar[uu] \ar[r, "\cong"] & F_\alpha \ar[uu] \end{tikzcd} \end{center} \begin{example}[?] For \(S\) and \(E\) a locally free coherent \({\mathcal{O}}_s\) module, \begin{align*} {\mathbb{P}}E(T) = \left\{{f^* E \to L \to 0}\right\} / \sim \end{align*} is a generalization of projectivization, then \(S\) admits a cover \(U_i\) trivializing \(E\). Then the restriction \(F_i \to {\mathbb{P}}E\) were \(F_i(T)\) is the above set if \(f\) factors through \(U_i\) and empty otherwise. On \(U_i\), \(E \cong {\mathcal{O}}_{U_i}^{n_i}\), so \(F_i\) is representable by \({\mathbb{P}}_{U_i}^{n_i - 1}\) by the proposition. Note that this is clearly a sheaf. \end{example} \begin{example}[?] For \(E\) locally free over \(S\) of rank \(n\), take \(r 1\), we have \(H^i(I(m_i - i)) = 0\). We now need to find \(m_0 \geq m_1\) such that \(H^1(I(m_0 - 1)) = 0\) (trickiest part of the proof). \begin{lemma}[?] The sequence \(\qty{\dim H^1(I(k))}_{k\geq m_i - 1}\) is \emph{strictly} decreasing.\footnote{Note: \(h^1 = \dim H^1\).} \end{lemma} \begin{remark} Given the lemma, it's enough to take \(m_0 \geq m_1 + h^1(I(m_1 - 1))\). Consider the LES we have a surjection \begin{align*} H^0({\mathcal{O}}_Z(m_1 - 1)) \to H^1(I(m_1 - 1)) \to 0 .\end{align*} So the dimension of the LHS is equal to \(P_Z(m_1 - 1)\), using the fact that terms vanish and make the Euler characteristic equal to \(P_Z\). Thus we can take \(m_0 = m_1 + P(m_1 - 1)\). \end{remark} \begin{proof}[of Lemma] Considering the LES \begin{align*} H^0(I(k+1)) \xrightarrow{\alpha_{k+1}} H^0(I_H(k+1)) \to H^1(I(k)) \to H^1(I(k+1)) \to 0 ,\end{align*} where the last term is zero because \(I_H\) is \(m_1{\hbox{-}}\)regular. So the sequence \(h^1(I(k))\) is non-increasing. \begin{observation} If it does \emph{not} strictly decrease for some \(k\), then there is an equality on the RHS, which makes \(\alpha_{k+1}\) surjective. This means that \(\alpha_{k+2}\) is surjective, since \begin{align*} H^0({\mathcal{O}}(1)) \otimes H^0(I_H(k+1)) \twoheadrightarrow H^0(I_H(k+2)) .\end{align*} \end{observation} So if one is surjective, everything above it is surjective, but by Serre vanishing we eventually get zeros. So \(\alpha_{k+i}\) is surjective for all \(i\geq 1\), contradicting Serre vanishing, since the RHS are isomorphisms for all \(k\). \end{proof} Thus for any \(Z\subset {\mathbb{P}}^n_k\) with \(P_Z = P\), we uniformly know that \(I_Z\) is \(m_0{\hbox{-}}\)regular for some \(m_0\) depending only on \(P\). \begin{claim} \(Z\) is determined by the degree \(m_0\) polynomials vanishing on \(Z\), i.e.~\(H^0(I_z(m_0))\) as a subspace of all degree \(m_0\) polynomials \(H^0({\mathcal{O}}(m_0))\) and has fixed dimension. We have \(H^i(I_Z(m_0)) = 0\) for all \(i> 0\), and in particular \(h^0(I_Z(m_0)) = P(m_0)\) is constant. \end{claim} It is determined by these polynomials because we have a sequence \begin{align*} 0 \to I_Z(m_0) \to {\mathcal{O}}(m_0) \to {\mathcal{O}}_Z(m_0) \to 0 .\end{align*} We can get a commuting diagram over it \begin{align*} 0 \to H^0(I_Z(m_0)) \otimes{\mathcal{O}}_{{\mathbb{P}}^n} \to H^0({\mathcal{O}}(m_0)) \otimes{\mathcal{O}}_{{\mathbb{P}}^n} \to \cdots \end{align*} where the middle map down is just evaluation and.the first map down is a surjection. Hence \(I_Z(m_0)\), hence \({\mathcal{O}}_Z\), hence \(Z\) is determined by \(H^0(I_Z(m_0))\). \begin{quote} Next time: we'll show that this is a subfunctor that is locally closed. \end{quote} \hypertarget{thursday-february-6th}{% \section{Thursday February 6th}\label{thursday-february-6th}} \begin{quote} Review base-change! \end{quote} For \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\), and \(C_{/k}\) a smooth projective curve, then \(\operatorname{Hilb}_{C_{/k}}^n = \operatorname{Sym}^n C\). \begin{definition}[The Hilbert-Chow Map] For \(X_{_{/k}}\) a smooth projective \emph{surface}, \(\operatorname{Hilb}_{X_{_{/k}}}^n \neq \operatorname{Sym}^n X\), there is a map (the Hilbert-Chow map) \begin{align*} \operatorname{Hilb}_{X_{_{/k}}}^n &\to \operatorname{Sym}^n X \\ Z &\mapsto {\operatorname{supp}}(Z) \\ U = \text{reduced subschemes} &\mapsto U' = \text{ reduced multisets } \\ {\mathbb{P}}^1 &\mapsto (x, x) .\end{align*} \end{definition} \begin{example}[?] Consider \({\mathbb{A}}^2 \times{\mathbb{A}}^2\) under the \({\mathbb{Z}}/2{\mathbb{Z}}\) action \begin{align*} ( (x_1, y_1), (x_2, y_2)) \mapsto ((x_2, y_2), (x_1, y_1)) .\end{align*} Then \begin{align*} ({\mathbb{A}}^2)^2 / {\mathbb{Z}}/2{\mathbb{Z}} &= \operatorname{Spec}k[x_1, y_1, x_2, y_2]^{{\mathbb{Z}}/2{\mathbb{Z}}} \\ &= \operatorname{Spec}k[x_1 x_2, y_1 y_2, x_1 + x_2, y_1 + y_2, x_1 y_2 + x_2 y_1, \cdots] \end{align*} with a bunch of symmetric polynomials adjoined. \end{example} \begin{example}[?] Take \({\mathbb{A}}^2\) and consider \(\operatorname{Hilb}_{{\mathbb{P}}^2}^3\). If \(I\) is a monomial ideal in \({\mathbb{A}}^2\), there is a nice picture. We can identify the tangent space \begin{align*} T_Z \operatorname{Hilb}_{{\mathbb{P}}^2}^n = \hom_{{\mathcal{O}}_{{\mathbb{P}}^2}} ( I_2, {\mathcal{O}}_Z) = \bigoplus \hom(I_{Z_i}, {\mathcal{O}}_{Z_i}) .\end{align*} if \(Z = {\coprod}Z_i\). If \(I\) is supported at 0, then we can identify the ideal with the generators it leaves out. \end{example} \begin{example}[?] \(I = (x^2, xy, y^2)\): \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-12:48.png} \caption{Image} \end{figure} \end{example} \begin{example}[?] \(I = (x^6, x^2y^2, xy^4, y^5)\): \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-12:49.png} \caption{Image} \end{figure} \end{example} \begin{example}[?] \(I = (x^2, y)\). Let \(e=x^2, f = y\). \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-12:54.png} \caption{Image} \end{figure} By comparing rows to columns, we obtain a relation \(ye = x^2 f\). Write \({\mathcal{O}}= \left\{{1, x}\right\}\), then note that this relation is trivial in \({\mathcal{O}}\) since \(y=x^2=0\). Thus \(\hom(I, {\mathcal{O}}) = \hom(k^2, k^2)\) is 4-dimensional. \end{example} \begin{remark} Note that \(C_{_{/k}}\) for curves is an important case to know. Take \(Z \subset C \times C^n\), then quotient by the symmetric group \(S^n\) (need to show this can be done), then \(Z/S^n \subset C \times\operatorname{Sym}^n C\) and composing with the functor \(\operatorname{Hilb}\) represents yields a map \(\operatorname{Sym}^n C \to \operatorname{Hilb}_{C_{/k}}^n\). This is bijective on points, and a tangent space computation shows it's an isomorphism. \end{remark} \begin{example}[?] Consider the nodal cubic in \({\mathbb{P}}^2\): \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-13:01.png} \caption{Nodal cubic} \end{figure} \begin{quote} The nodal cubic \(zy^2 = x^2(x+z)\). \end{quote} Consider the open subscheme \(V \subset \operatorname{Hilb}_{C_{/k}}^2\) of points \(z \subset U\) for \(U \subset C\) open. We can normalize: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-13:03.png} \caption{Normalized cubic} \end{figure} This yields a map fro \({\mathbb{P}}^1 \setminus\text{2 points}\). This gives us a stratification, i.e.~a locally closed embedding \begin{align*} (\text{z supported on U}) {\coprod}(\text{1 point at p}) {\coprod}(\text{both points at p}) \to \operatorname{Hilb}_{C_{/k}}^2 .\end{align*} The first locus is given by the complement of two lines: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-13:08.png} \caption{Locus 1} \end{figure} The third locus is given by arrows at \(p\) pointing in any direction, which gives a copy of \({\mathbb{P}}^1\). The second is \({\mathbb{P}}^1\) minus two points. Above each point is a nodal cubic with two marked points, and moving the base point towards a line correspond to moving one of the points toward the node: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/2020-02-06-13:11.png} \caption{Moving base toward the point} \end{figure} More precisely, we're considering the cover \({\mathbb{P}}^1 \setminus\text{2 points} \to C\) and thinking about ways in which two points and approach the missing points. These give specific tangent directions at the node on the cubic, depending on how this approach happens -- either both points approach missing point \#1, both approach missing point \#2, or each approach a separate missing point. \end{example} \begin{remark} Useful example to think about. Not normal, reduced, but glued in a weird way. Possibly easier to think about: cuspidal cubic. \end{remark} \hypertarget{representability-1}{% \subsection{Representability}\label{representability-1}} Recall the following definition: \begin{definition}[$m\dash$Regularity] A coherent sheaf \(F\) on \({\mathbb{P}}_k^n\) for \(k\) a field is \(m{\hbox{-}}\)regular iff \(H^i(F(m-i)) = 0\) for all \(i> 0\). \end{definition} \begin{proposition}[?] For every Hilbert polynomial \(P\), there exists some \(m_0\) depending on \(P\) such that any \(Z \subset {\mathbb{P}}^n_k\) with \(P_Z = P\) satisfies \(I_Z\) is \(m{\hbox{-}}\)regular. \end{proposition} \begin{remark}[1] \(F\) is \(m{\hbox{-}}\) regular iff \(\mkern 1.5mu\overline{\mkern-1.5muF\mkern-1.5mu}\mkern 1.5mu = F \times_{\operatorname{Spec}k} \operatorname{Spec}\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) is \(m{\hbox{-}}\)regular. \end{remark} \begin{remark}[2] The \(m_0\) produced does not depend on \(k\). \end{remark} \begin{lemma}[?] For \(m_0 = m_0(P)\) and \(N = N(P)\), we have an embedding as a subfunctor \begin{align*} \operatorname{Hilb}_{{\mathbb{P}}^m_{\mathbb{Z}}}^P \to {\operatorname{Gr}}(N, H^0( {\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0) )^\vee) .\end{align*} \end{lemma} For any \(Z \subset {\mathbb{P}}^n_S\) flat over \(S\) with \(P_{Z_s} = P\) for all \(s\in S\) points, we want to send this to \begin{align*} 0\to R^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0))^\vee\to Q \to 0 \end{align*} or equivalently \begin{align*} 0 \to Q^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n_{\mathbb{Z}}, {\mathcal{O}}(m_0)) \to R \to 0 \end{align*} with \(R\) locally free. So instead of the quotient \(Q\) being locally free, we can ask for the sub \(Q^\vee\) to be locally free instead, which is a weaker condition. We thus send \(Z\) to \begin{align*} 0 \to \pi_{s*} I_Z(m_0) \to \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) = {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0)) \end{align*} which we obtain by taking the pushforward from this square: \begin{center} \begin{tikzcd} {\mathbb{P}}^n_s \arrow[dd, "\pi_s"] \arrow[rr] & & {\mathbb{P}}^n_Z \arrow[dd] \\ & & \\ S \arrow[rr] & & \operatorname{Spec}{\mathbb{Z}} \end{tikzcd} \end{center} We have a sequence \(0 \to I_Z(m_0) \to {\mathcal{O}}(m_0) \to {\mathcal{O}}_Z(m_0) \to 0\). Thus we get a sequence \begin{align*} 0 \to \pi_{s*}I_Z(m_0) \to \pi_{s*}{\mathcal{O}}(m_o) \to \pi_{s*} {\mathcal{O}}_Z(m_0) \to R^1 \pi_{s*}I_Z(m_0) \to \cdots .\end{align*} \hypertarget{step-1}{% \subsubsection{Step 1}\label{step-1}} \begin{align*} R^1\pi_* I_Z(m_0) = 0 .\end{align*} By base change, it's enough to show that \(H^1(Z_s, I_{Z_s}(m_0)) = 0\). This follows by \(m_0{\hbox{-}}\)regularity. \hypertarget{step-2}{% \subsubsection{Step 2}\label{step-2}} \(\pi_{s*}I_Z(m_0)\) and \(\pi_{s*} {\mathcal{O}}_Z(m_0)\) are locally free. For all \(i>0\), we have \begin{itemize} \tightlist \item \(R^i \pi_{s*} I_Z(m_0) = 0\) by \(m_0{\hbox{-}}\)regularity, \item \(R^i \pi_{s*} {\mathcal{O}}(m_0) = 0\) by base change, \item and thus \(R^i \pi_{s*} {\mathcal{O}}_Z(m_0) = 0\). \end{itemize} \hypertarget{step-3}{% \subsubsection{Step 3}\label{step-3}} \(\pi_{s*}I_Z(m_0)\) has rank \(N = N(P)\). Again by base change, there is a map \(\pi_* I_Z(m_0) \otimes k(s) \to H^0(Z_S, I_{Z_s}(m_0))\) which we know is an isomorphism. Because \(h^i ( I_{Z_S}(m_0) ) = 0\) for \(i>0\) by \(m{\hbox{-}}\)regularity and \begin{align*} h^0(I_{Z_S}(m_0)) = P_{\mathcal{O}}(m_0) - P_{{\mathcal{O}}_{Z_s}}(m_0) = P_{\mathcal{O}}(m_0) - P(m_0) .\end{align*} This yields a well-defined functor \begin{align*} \operatorname{Hilb}_{{\mathbb{P}}^n_{\mathbb{Z}}}^P \to {\operatorname{Gr}}(N, H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0))^\vee) .\end{align*} \begin{remark} Note that we've just said what happens to objects; strictly speaking we should define what happens for morphisms, but they're always give by pullback. \end{remark} We want to show injectivity, i.e.~that we can recover \(Z\) from the data of a number f polynomials vanishing on it, which is the data \(0 \to \pi_{s*} I_Z(m_0) \to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0))\). Given \begin{align*} 0 \to Q^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(m_0)) = \pi_{s*} {\mathcal{O}}_{{\mathbb{P}}^n_S}(m_0) \end{align*} we get a diagram \begin{center} \begin{tikzcd} \pi_{s}^* Q^\vee\arrow[rrdd] \arrow[rrr] & & & {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd} \end{center} where \(Q^\vee= \pi_{s*} I_Z(m_0)\), so we're looking at \begin{center} \begin{tikzcd} Q^\vee= \pi_{s*}^* \pi_{s*} I_Z(m_0) \arrow[rrdd, twoheadrightarrow] \arrow[rrr] & & & {\mathcal{O}}_{{\mathbb{P}}^n_s}(m_0) \\ & & & \\ & & I(m_0) \arrow[ruu, hook] & \end{tikzcd} \end{center} The surjectivity here follows from \({\mathcal{O}}_{Z_s} \otimes H^0(I_{Z_s}(m_0)) \to I_{Z_s}(m_0)\) (?). Given a universal family \(G = {\operatorname{Gr}}( N, H^0({\mathcal{O}}(m_0))^\vee)\) and \(Q^\vee\subset {\mathcal{O}}_G \otimes H^0({\mathcal{O}}(m_0))^\vee\), we obtain \(I_W \subset {\mathcal{O}}_G\) and \(W \subset {\mathbb{P}}^n_G\). \hypertarget{tuesday-february-18th}{% \section{Tuesday February 18th}\label{tuesday-february-18th}} \begin{theorem}[?] Let \(X/S\) be a projective subscheme (i.e.~\(X\subset {\mathbb{P}}^n\) for some \(n\)). The Hilbert functor of flat families \(\operatorname{Hilb}_{X/S}^p\) is representable by a projective \(S{\hbox{-}}\)scheme. \end{theorem} \begin{remark} Note that without a fixed \(P\), this is \emph{locally} of finite type but not finite type. After fixing \(P\), it becomes finite type. \end{remark} \begin{example}[?] For a curve of genus \(g\), there is a smooth family \({\mathcal{C}}\xrightarrow{\pi} S\) with \(S\) finite-type over \({\mathbb{Z}}\) where every genus \(g\) curve appears as a fiber. I.e., genus \(g\) curves form a \emph{bounded family} (here there are only finitely many algebraic parameters to specify a curve). How did we construct? Take the third power of the canonical bundle and show it's very ample, so it embeds into some projective space and has a Hilbert polynomial. \end{example} In fact, there is a finite type \emph{moduli stack} \({\mathcal{M}}_g / {\mathbb{Z}}\) of genus \(g\) curves. There will be a map \(S \twoheadrightarrow{\mathcal{M}}_g\), noting that \({\mathcal{C}}\) is not a moduli space since it may have redundancy. We'll use the fact that a finite-type scheme surjects onto \({\mathcal{M}}_g\) to show it is finite type. \begin{remark} If \(X/S\) is proper, we can't talk about the Hilbert polynomial, but the functor \(\operatorname{Hilb}_{X/S}\) is still representable by a locally finite-type scheme with connected components which are proper over \(S\). \end{remark} \begin{remark} If \(X/S\) is \emph{quasiprojective} (so locally closed, i.e.~\(X\hookrightarrow{\mathbb{P}}^n_S\)), then \(\operatorname{Hilb}_{X/S}^P(T) \coloneqq\left\{{z\in X_T \text{ projective, flat over S with fiberwise Hilbert polynomial P }}\right\}\) is still representable, but now by a quasiprojective scheme. \end{remark} \begin{example}[?] Length \(Z\) subschemes of \({\mathbb{A}}^1\): representable by \({\mathbb{A}}^2\). \includegraphics{figures/2020-02-18-12:46.png}\\ Upstairs: parametrizing length 1 subschemes, i.e.~points. \end{example} \begin{remark} If \(X\subset {\mathbb{P}}_S^n\) and \(E\) is a coherent sheaf on \(X\), then \begin{align*} \operatorname{Quot}_{E, X/S}^{P}(T) = \left\{{ j^*E \to F \to 0, \text{ over } X_T \to T,~F \text{ flat with fiberwise Hilbert polynomial } P }\right\} \end{align*} where \(T \xrightarrow{g} S\) is representable by an \(S{\hbox{-}}\)projective scheme. \end{remark} \begin{example}[?] Take \(E = {\mathcal{O}}_x\), \(X\) and \(S\) a point, and \(E\) is a vector space, then \(\operatorname{Quot}_{E/S}^P = {\operatorname{Gr}}(\operatorname{rank}, E)\). \end{example} \begin{warnings} The Hilbert scheme of 2 points on a surface is more complicated than just the symmetric product. \end{warnings} \begin{example}[?] \begin{align*} \qty{{\mathbb{A}}^2}^3 &\to \qty{{\mathbb{A}}^2}^2 \\ \supseteq \Delta\coloneqq\Delta_{01} \times\Delta_{02} &\to \qty{{\mathbb{A}}^2}^2 \end{align*} where \(\Delta_{ij}\) denote the diagonals on the \(i, j\) factors. Here all associate points of \(\Delta\) dominate the image, but it is not flat. Note that if we take the complement of the diagonal in the image, then the restriction \(\Delta' \to \qty{{\mathbb{A}}^2}^2\setminus D\) is in fact flat. \end{example} \begin{example}[Mumford] The Hilbert scheme may have nontrivial scheme structure, i.e.~this will be a ``nice'' Hilbert scheme with is generally not reduced. We will find a component \(H\) of a \(\operatorname{Hilb}_{{\mathbb{P}}^3_C}^P\) whose generic point corresponds to a smooth irreducible \(C\subset {\mathbb{P}}^3\) which is generically non-reduced. \end{example} \hypertarget{cubic-surfaces}{% \subsection{Cubic Surfaces}\label{cubic-surfaces}} \begin{quote} See Hartshorne Chapter 5. \end{quote} Let \(X\subset {\mathbb{P}}^3\) be a smooth cubic surface, then \({\mathcal{O}}(1)\) on \({\mathbb{P}}^3\) restricts to a divisor class \(H\) of a hyperplane section, i.e.~the associated line bundle \({\mathcal{O}}_x(H) = {\mathcal{O}}_x(1)\). \begin{fact}[Important fact 1] \(X\) is the blowup of \({\mathbb{P}}^2\) minus 6 points (replace each point with a curve). There is thus a blowdown map \(X \xrightarrow{\pi} {\mathbb{P}}^2\). \begin{figure} \centering \includegraphics{figures/2020-02-18-13:07.png} \caption{Image} \end{figure} Let \(\ell = \pi^*(\text{line})\), then a fact is that \(3\ell - E_1 -\cdots - E_6\) (where \(E_i\) are the curves about the \(p_i\)) is very ample and embeds \(X\) into \({\mathbb{P}}^3\) as a cubic. \end{fact} \begin{fact}[Important fact 2] Every smooth cubic surface \(X\) has \emph{precisely} 27 lines. Any 6 pairwise skew lines arise as \(E_1, \cdots, E_6\) as in the previous construction. \end{fact} Take an \(X\) and a line \(L\subset X\). Consider any \(C\) in the linear system \({\left\lvert {4H + 2L} \right\rvert}\). Fact: \({\mathcal{O}}(4H + 2L)\) is very ample, so embeds into a big projective space, and thus \(C\) is smooth and irreducible by Bertini. Then the Hilbert polynomial of \(C\) is of the form \(at + b\) where \(b = \chi({\mathcal{O}}_c)\), the Euler characteristic of the structure sheaf of \(C\), and \(a = \deg C\). So we'll compute these. We have \(\deg C = H \cdot C\) (intersection) \(= H \cdot(4H + 2L) = 4H^2 + 2H\cdot L = 4\cdot 3 + 2 = 14\). The intersections here correspond to taking hyperplane sections, intersecting with \(X\) to get a curve, and counting intersection points: \includegraphics{figures/2020-02-18-13:14.png}\\ In general, for \(X\) a surface and \(C\subset X\) a smooth curve, then \(\omega_C = \omega_X(C)\mathrel{\Big|}_C\). Since \(X\subset {\mathbb{P}}^3\), we have \begin{align*} \omega_X &= \omega_{{\mathbb{P}}^3}(X) \mathrel{\Big|}_X \\ &= {\mathcal{O}}(-4) \oplus {\mathcal{O}}(3)\mathrel{\Big|}_X \\ &= {\mathcal{O}}_X(-1) \\ &= {\mathcal{O}}_X(-H) .\end{align*} We also have \begin{align*} \omega_C &= \omega_X(C)\mathrel{\Big|}_X \\ &= { \left.{{ \qty{ {\mathcal{O}}_X(-H) \oplus {\mathcal{O}}_X(4H + 2L)}}} \right|_{{C}} } \\ \\ &\Downarrow \qquad \text{taking degrees} \\ \\ \deg \omega_C &= C\cdot(3H + 2L) \\ &= (4H+2L)(3H+2L) \\ &= 12H^2 + 14HL + 4L^2 \\ &= 36 + 14 + (-4) \\ &= 46 .\end{align*} Since this equals \(2g(C) - 2\), we can conclude that the genus is given by \(g(C) = 24\). Thus \(P\) is given by \(14t + (1-g) = 14t - 23\). \begin{remark} Good to know: moving a cubic surface moves the lines, you get a monodromy action, and the Weyl group of \(E_6\) acts transitively so lines look the same. \end{remark} \begin{claim}[1] There is a flat family \(Z\subset {\mathbb{P}}^3_S\) with fiberwise Hilbert polynomial \(P\) of cures of this form such that the image of the map \(S \to \operatorname{Hilb}_{{\mathbb{P}}^3}^P\) has dimension 56. \end{claim} \begin{proof}[of claim] We can compute the dimension of the space of smooth cubic surfaces, since these live in \({\mathbb{P}}H^0({\mathbb{P}}^3, {\mathcal{O}}(3))\), which has dimension \({3+3\choose 3} -1 = 19\). Since there are 27 lines, the dimension of the space of such cubics with a choices of a line is also 19. Choose a general \(C\) in the linear system \({\left\lvert {4H + 2L} \right\rvert}\) will add \(\dim {\left\lvert {4H + 2L} \right\rvert} = \dim {\mathbb{P}}H^0(x, {\mathcal{O}}_x(C))\). We have an exact sequence \begin{align*} 0 \to {\mathcal{O}}_X \to {\mathcal{O}}_X(C) \to {\mathcal{O}}_C(C) \to 0 \\ H^0\qty{ 0 \to {\mathcal{O}}_X \to {\mathcal{O}}_X(C) \to {\mathcal{O}}_C(C) \to 0 } \\ .\end{align*} Since the first \(H^0\) vanishes (?) we get an isomorphism. By Riemann-Roch, we have \begin{align*} \deg {\mathcal{O}}_C(C) = C^2 = (4H+2L)^2 = 16H^2 + 16 HL + 4L^2 = 64 - 4 = 60 .\end{align*} We can also compute \(\chi({\mathcal{O}}_C(C)) = 60 - 23 = 37\). We have \begin{align*} h^0({\mathcal{O}}_C(C)) - h^1({\mathcal{O}}_C(C)) = h^0({\mathcal{O}}_C(C)) - h^0(\omega_C(-C))) = 2(23) - 60 < 0 ,\end{align*} so there are no sections. So \(\dim {\left\lvert {4H + 2L} \right\rvert} = 37\). Thus letting \(S\) be the space of cubic surfaces \(X\), a line \(L\), and a general \(C \in {\left\lvert {4H + 2L} \right\rvert}\), \(\dim S = 56\). We get a map \(S \to \operatorname{Hilb}_{{\mathbb{P}}^3}^P\), and we need to check that the fibers are 0-dimensional (so there are no redundancies). We then just need that every such \(C\) lies on a unique cubic. Why does this have to be the case? If \(C \subset X, X'\) then \(C \subset X\cap X'\) is degree 14 curve sitting inside a degree 6 curve, which can't happen. Thus if \(H\) is a component of \(\operatorname{Hilb}_{{\mathbb{P}}^3}^P\) containing the image of \(S\), the \(\dim H \geq 56\). \end{proof} \begin{claim}[2] For any \(C\) above, we have \(\dim T_C H = 57\). \end{claim} When the subscheme is smooth, we have an identification with sections of the normal bundle \(T_C H = H^0(C, N_{C/{\mathbb{P}}^3})\). There's an exact sequence \begin{align*} 0 \to N_{C/X} = {\mathcal{O}}_C(C) \to N_{C/{\mathbb{P}}^3} \to N_{X/{\mathbb{P}}^3}\mathrel{\Big|}_C = {\mathcal{O}}_C(x)\mathrel{\Big|}_C = {\mathcal{O}}_C(3H)\mathrel{\Big|}_C \to 0 .\end{align*} \begin{quote} Note \(\omega_C = {\mathcal{O}}_C(3H + 2L)\). \end{quote} As we computed, \begin{align*} H^0({\mathcal{O}}_C(C)) &= 37 \\ H^1({\mathcal{O}}_C(C)) &= 0 .\end{align*} So we need to understand the right-hand term \(H^0({\mathcal{O}}_C(3H))\). By Serre duality, this equals \(h^1(\omega_C(-3H)) = h^1({\mathcal{O}}_C(3L))\). We get an exact sequence \begin{align*} 0 \to {\mathcal{O}}_X(2L-C) \to {\mathcal{O}}_X(2L) \to {\mathcal{O}}_C(2L) \to 0 .\end{align*} Taking homology, we have \(0\to 0 \to 1 \to 1 \to 0\) since \(2L-C = -4H\). Computing degrees yields \(h^0 ({\mathcal{O}}_C(3H)) = 20\). Thus the original exact sequence yields \begin{align*} 0 \to 37 \to ? \to 20 \to 0 ,\end{align*} so \(? = 57\) and thus \(\dim N_{C/{\mathbb{P}}^3} = 57\). \begin{claim}[3] \begin{align*} \dim H = 56 .\end{align*} \end{claim} \hypertarget{proof-that-the-dimension-is-56}{% \subsubsection{Proof That the Dimension is 56}\label{proof-that-the-dimension-is-56}} Suppose otherwise. Then we have a family over \(H^\mathrm{red}\) of \emph{smooth} curves, where \(f(S) \subset H^\mathrm{red}\), where the generic element is not on a cubic or any lower degree surface. Let \(C'\) be a generic fiber. Then \(C'\) lies on a pencil of quartics, i.e.~2 linearly independent quartics. Let \(I = I_{C'}\) be the ideal of this curve in \({\mathbb{P}}^3\), there is a SES \begin{align*} 0\to I(4) \to {\mathcal{O}}(4) \to {\mathcal{O}}_C(4) \to 0 .\end{align*} It can be shown that \(\dim H^0(I(4)) \geq 2\). \begin{fact} A generic quartic in this pencil is \emph{smooth} (can be argued because of low degree and smoothness). \end{fact} We can compute the dimension of quartics, which is \({4+3 \choose 3} - 1 = 35 - 1 = 34\). The dimension of \(C'\)s lying on a fixed quartic is \(24\). But then the dimension of the image in the Hilbert scheme is at most \(24 + 34 - 1 = 57\). It can be shown that the picard rank of such a quartic is 1, generated by \({\mathcal{O}}(1)\), so this is a \emph{strict} inequality, which is a contradiction since \(\dim \operatorname{Hilb}= 56\). This proves the theorem. \begin{remark} Use the fact that these curves are \(K3\) surfaces? Get the fact about the generator of the Picard group from Hodge theory. So we can deform curves a bit, but not construct an algebraic family that escapes a particular cubic. \end{remark} \hypertarget{obstruction-and-deformation-tuesday-february-25th}{% \section{Obstruction and Deformation (Tuesday February 25th)}\label{obstruction-and-deformation-tuesday-february-25th}} Let \(k\) be a field, \(X_{_{/k}}\) projective, then the \(k{\hbox{-}}\)points \(\operatorname{Hilb}_{X_{_{/k}}}^P(k)\) corresponds to closed subschemes \(Z\subset X\) with hilbert polynomial \(P_z = P\). Given a \(P\), we want to understand the local structure of \(\operatorname{Hilb}_{X_{_{/k}}}^p\), i.e.~diagrams of the form \begin{center} \begin{tikzcd} & & & & \operatorname{Hilb}_{X_{_{/k}}}^P \arrow[dd] \\ & & & & \\ \operatorname{Spec}(k) \arrow[rrrruu, "p"] \arrow[rr] & & \operatorname{Spec}(A) \arrow[rruu, "?", dashed] \arrow[rr] & & \operatorname{Spec}(k) \\ & & & & \\ & & A_{/k} \text{ Artinian local} \arrow[uu] & & \end{tikzcd} \end{center} \begin{example}[?] For \(A = k[\varepsilon]\), the set of extensions is the Zariski tangent space. \end{example} \begin{definition}[Category of Artinian Algebras] Let \((\operatorname{Art}_{/k})\) be the category of local Artinian \(k{\hbox{-}}\)algebras with local residue field \(k\). \end{definition} Note that these will be the types of algebras appearing in the above diagrams. \begin{remark} This category has fiber coproducts, i.e.~there are pushouts: \begin{center} \begin{tikzcd} C \arrow[dd] \arrow[rr] & & A \arrow[dd, dashed] \\ & & \\ B \arrow[rr, dashed] & & A \otimes_C B \end{tikzcd} \end{center} There are also fibered products, \begin{center} \begin{tikzcd} A \times_C B \arrow[rr, dashed] \arrow[dd, dashed] & & B \arrow[dd] \\ & & \\ A \arrow[rr] & & C \end{tikzcd} \end{center} Here, \(A \times_C B \coloneqq\left\{{(a, b) {~\mathrel{\Big|}~}f(a) = g(b)}\right\} \subset A\times B\). \end{remark} \begin{example}[?] If \(A = B = k[\varepsilon]/(\varepsilon^2)\) and \(C = k\), then \(A\times_C B = k[\varepsilon_1, \varepsilon_2]/(\varepsilon_1, \varepsilon_2)^2\) Note that on the \(\operatorname{Spec}\) side, these should be viewed as \begin{align*} \operatorname{Spec}(A) {\coprod}_{\operatorname{Spec}(C)} \operatorname{Spec}(B) = \operatorname{Spec}(A\times_C B) .\end{align*} \end{example} \begin{definition}[Deformation Functor (Preliminary Definition)] A \emph{deformation functor} is a functor \(F: (\operatorname{Art}_{/k}) \to {\operatorname{Set}}\) such that \(F(k) = {\{\operatorname{pt}\}}\) is a singleton. \end{definition} \begin{example}[?] Let \(X_{_{/k}}\) be any scheme and let \(x\in X(k)\) be a \(k{\hbox{-}}\)point. We can consider the deformation functor \(F\) such that \(F(A)\) is the set of extensions \(f\) of the following form: \begin{center} \begin{tikzcd} & & & & X \arrow[dd] \\ & & & & \\ \operatorname{Spec}(k) \arrow[rrrruu, "x"] \arrow[rr, hook] & & \operatorname{Spec}(A) \arrow[rruu, "f", dashed] \arrow[rr] & & \operatorname{Spec}(k) \end{tikzcd} \end{center} If \(A' \to A\) is a morphism, then we define \(F(A') \to F(A)\) is defined because we can precompose to fill in the following diagram \begin{center} \begin{tikzcd} & & & & & & & & X \arrow[ddd] \\ & & & & & & & & \\ & & & & & & & & \\ \operatorname{Spec}(k) \arrow[rrd] \arrow[rrrrrrrruuu] & & & & & & & & \operatorname{Spec}(k) \\ & & \operatorname{Spec}(A) \arrow[rr] \arrow[rrrrrruuuu, "\exists \tilde f"] & & \operatorname{Spec}(A') \arrow[rrrru] \arrow[rrrruuuu, "f", dashed] & & & & \end{tikzcd} \end{center} So this is indeed a deformation functor. \end{example} \begin{example}[a motivating example] The Zariski tangent space on the nodal cubic doesn't ``see'' the two branches, so we allow ``second order'' tangent vectors. \end{example} We can consider parametrizing the functors above as \(F_{X, x}(A)\), which is isomorphic to \(F_{\operatorname{Spec}({\mathcal{O}}_x)_{X, x}}\) and further isomorphic to \(F_{\operatorname{Spec}\widehat{{\mathcal{O}}_x}_{x, X} }\). This is because for Artinian algebras, we have maps \begin{align*} \operatorname{Spec}({\mathcal{O}}_{x, X})/{\mathfrak{m}}^N \to \operatorname{Spec}{\mathcal{O}}_{X, x} \to X .\end{align*} \begin{remark} \(\widehat{ {\mathcal{O}}}_{X, x}\) will be determined by \(F_{X, x}\). \end{remark} \begin{example}[?] Consider \(y^2 = x^2(x+1)\), and think about solving this over \(k[t]/t^n\) with solutions equivalent to \((0, 0) \pmod t\). \includegraphics{figures/2020-02-25-13:20.png}\\ Note that the `second order' tangent vector comes from \(\operatorname{Spec}k[t]/t^3\). We can write \(F_{X, x}(A) = \pi^{-1}(x)\) where \begin{align*} \hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}k, X) \xrightarrow{\pi} \hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}k, x) \ni x .\end{align*} Thus \begin{align*} F_{X, x}(A) = \hom_{{\operatorname{Sch}}_{/k}}(\operatorname{Spec}A, \operatorname{Spec}{\mathcal{O}}_{x, X}) = \hom_{k{\hbox{-}}\mathrm{Alg}}(\widehat{{\mathcal{O}}}_{X, x}, A) .\end{align*} \end{example} \begin{example}[?] Given any local \(k{\hbox{-}}\)algebra \(R\), we can consider \begin{align*} h_R: (\operatorname{Art}_{/k}) &\to {\operatorname{Set}}\\ A &\mapsto \hom(R, A) .\end{align*} and \begin{align*} h_{\operatorname{Spec}R}: (\operatorname{Art}{\operatorname{Sch}}_{/k})^{\operatorname{op}}\to {\operatorname{Set}}\\ \operatorname{Spec}(A) &\mapsto \hom(\operatorname{Spec}A, \operatorname{Spec}R) .\end{align*} \end{example} \begin{definition}[Representable Deformation] A deformation \(F\) is \textbf{representable} if it is of the form \(h_R\) as above for some \(R \in \operatorname{Art}_{/k}\). \end{definition} \begin{remark} There is a Yoneda Lemma for \(A\in \operatorname{Art}_{/k}\), \begin{align*} \hom_{\mathrm{Fun}}(h_A, F) = F(A) .\end{align*} We are thus looking for things that are representable in a larger category, which restrict. \end{remark} \begin{definition}[Pro-Representability] A deformation functor is \emph{pro-representable} if it is of the form \(h_R\) for \(R\) a complete local \(k{\hbox{-}}\)algebra (i.e.~a limit of Artinian local \(k{\hbox{-}}\)algebras). \end{definition} \begin{remark} We will see that there are simple criteria for a deformation functor to be pro-representable. This will eventually give us the complete local ring, which will give us the scheme representing the functor we want. \end{remark} \begin{remark} It is difficult to understand even \(F_{X, x}(A)\) directly, but it's easier to understand small extensions. \end{remark} \begin{definition}[Small Extensions] A \emph{small extension} is a SES of Artinian \(k{\hbox{-}}\)algebras of the form \begin{align*} 0 \to J \to A' \to A \to 0 .\end{align*} such that \(J\) is annihilated by the maximal ideal fo \(A'\). \end{definition} \begin{lemma}[?] Given any quotient \(B\to A \to 0\) of Artinian \(k{\hbox{-}}\)algebras, there is a sequence of small extensions (quotients): \begin{center} \begin{tikzcd} 0 & & & & & & \\ & & & & & & \\ B_0 \arrow[uu] & & B_1 \arrow[lluu] & & \cdots & & B_n = A \arrow[lllllluu] \\ & & & & & & \\ B \arrow[uu] \arrow[rruu] \arrow[rrrrrruu] & & & & & & \end{tikzcd} \end{center} This yields \begin{center} \begin{tikzcd} \operatorname{Spec}A \arrow[rrrr, hook] \arrow[rrrrdddddd, Rightarrow] & & & & \operatorname{Spec}B \\ & & & & \\ & & & & \operatorname{Spec}B_0 \arrow[uu, hook] \\ & & & & \\ & & & & \vdots \arrow[uu, hook] \\ & & & & \\ & & & & \operatorname{Spec}B_n \arrow[uu, hook] \end{tikzcd} \end{center} where the \(\operatorname{Spec}B_i\) are all small. \end{lemma} \begin{remark} In most cases, extending deformations over small extensions is easy. \end{remark} \hypertarget{first-example-of-deformation-and-obstruction-spaces}{% \subsection{First Example of Deformation and Obstruction Spaces}\label{first-example-of-deformation-and-obstruction-spaces}} Suppose \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) and let \(X_{_{/k}}\) be connected. We have a picard functor \begin{align*} {\operatorname{Pic}}_{X_{_{/k}}}: ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto {\operatorname{Pic}}(X_S) / {\operatorname{Pic}}(S) .\end{align*} If we take a point \(x\in {\operatorname{Pic}}_{X_{_{/k}}}(k)\), which is equivalent to line bundles on \(X\) up to equivalence, we obtain a deformation functor \begin{align*} F \coloneqq F_{{\operatorname{Pic}}_{ X_{_{/k}}, x }} &\to {\operatorname{Set}}\\ A \mapsto \pi^{-1}(x) \end{align*} where \begin{align*} \pi: {\operatorname{Pic}}_{X_{_{/k}}}(\operatorname{Spec}A) &\to {\operatorname{Pic}}_{X_{_{/k}}} (\operatorname{Spec}k) \\ \pi^{-1}(x) &\mapsto x .\end{align*} This is given by taking a line bundle on the thickening and restricting to a closed point. Thus the functor is given by sending \(A\) to the set of line bundles on \(X_A\) which restrict to \(X_x\). That is, \(F(A) \subset {\operatorname{Pic}}_{X_{_{/k}}}(\operatorname{Spec}A)\) which restrict to \(x\). So just pick the subspace \({\operatorname{Pic}}(X_A)\) (base changing to \(A\)) which restrict. There is a natural identification of \({\operatorname{Pic}}(X_A) = H^1(X_A, {\mathcal{O}}_{X_A}^*)\). If \begin{align*} 0\to J \to A' \to A \to 0 .\end{align*} is a thickening of Artinian \(k{\hbox{-}}\)algebras, there is a restriction map of invertible functions \begin{align*} {\mathcal{O}}_{X_A}^* \to {\mathcal{O}}_{X_A'}^* \to 0 .\end{align*} which is surjective since the map on structure sheaves is surjective and its a nilpotent extension. The kernel is then just \({\mathcal{O}}_{X_{A'}} \otimes J\). If this is a small extension, we get a SES \begin{align*} 0 \to {\mathcal{O}}_X \otimes J \to {\mathcal{O}}_{X_{A'}}^* \to {\mathcal{O}}_{x_A}^* \to 0 .\end{align*} Taking the LES in cohomology, we obtain \begin{align*} H^1 {\mathcal{O}}_X \otimes J \to H^1 {\mathcal{O}}_{X_{A'}}^* \to H^1{\mathcal{O}}_{x_A}^* \to H^0 {\mathcal{O}}_X \otimes J .\end{align*} Thus there is an obstruction class in \(H^2\), and the ambiguity is detected by \(H^1\). Thus \(H^1\) is referred to as the \textbf{deformation space}, since it counts the extensions, and \(H^2\) is the \textbf{obstruction space}. \hypertarget{deformation-theory-thursday-february-27th}{% \section{Deformation Theory (Thursday February 27th)}\label{deformation-theory-thursday-february-27th}} Big picture idea: We have moduli functors, such as \begin{align*} F_{S'}: ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ \operatorname{Hilb}: S &\to \text{flat subschemes of } X_S \\ {\operatorname{Pic}}: S &\to {\operatorname{Pic}}(X_S)/{\operatorname{Pic}}(S) \\ \mathrm{Def}: S &\to \text{flat families } / S,~ \text{smooth, finite, of genus } g .\end{align*} \begin{definition}[Deformation Theory] Choose a point \(f\) the scheme representing \(F_{S'}\) with \(\xi_0 \in F_{gl}(\operatorname{Spec}K)\). Define \begin{align*} F_{\text{loc}}: (\text{Artinian local schemes} / K)^{\operatorname{op}}\to {\operatorname{Set}} .\end{align*} \begin{center} \begin{tikzcd} \operatorname{Spec}(K) \arrow[rr, "i", hook] & & \operatorname{Spec}(A) \arrow[rr] & & F(i)^{-1}(\xi_0) \arrow[rr] & & F_{gr}(\operatorname{Spec}K) \arrow[dd, "F(i)"] \\ & & & & & & \\ & & & & & & F_{gl}(\operatorname{Spec}K) \end{tikzcd} \end{center} \end{definition} \begin{definition}[Deformation Functors] Let \(F: (\operatorname{Art}_{/k}) \to {\operatorname{Set}}\) where \(F(k)\) is a point. Denote \(\widehat{\operatorname{Art}}_{/k}\) the set of complete local \(k{\hbox{-}}\)algebras. Since \(\operatorname{Art}_{/k} \subset \widehat{\operatorname{Art}} / k\), we can make extensions \(\widehat{F}\) by just taking limits: \begin{center} \begin{tikzcd} & \operatorname{Art}_{/k} \arrow[rrr, "F"] & & & {\operatorname{Set}}\\ & & & & \\ \lim_{\leftarrow} R/{\mathfrak{m}}_R^n = R \in & \widehat{\operatorname{Art}}_{/k} \arrow[uu] \arrow[rrruu, "\widehat{F}"] & & & \end{tikzcd} \end{center} where we define \begin{align*} \widehat{F}(R) \coloneqq\varprojlim F(R/{\mathfrak{m}}_R^n) .\end{align*} \end{definition} \begin{question} When is \(F\) pro-representable, which happens iff \(\widehat{F}\) is representable? In particular, we want \(h_R \xrightarrow{\cong} \widehat{F}\) for \(R\in \widehat{\operatorname{Art}}_{/k}\), so \begin{align*} h_R = \hom_{\widehat{\operatorname{Art}}_{/k}}(R, {\,\cdot\,}) = \hom_{?}({\,\cdot\,}, \operatorname{Spec}k) .\end{align*} \end{question} \begin{example}[?] Let \(F_{\text{gl}} = \operatorname{Hilb}_{X_{_{/k}}}^p\), which is represented by \(H_{/k}\). Then . \begin{align*} \xi_0 = F_{\text{gl}}(k) = H(k) = \left\{{Z\subset X {~\mathrel{\Big|}~}P_z = f}\right\} .\end{align*} Then \(F_{\text{loc} }\) is representable by \(\widehat{{\mathcal{O}}}_{H/\xi_0}\). \end{example} \begin{definition}[Thickening] Given an Artinian \(k{\hbox{-}}\)algebra \(A \in \operatorname{Art}_{/k}\), a \emph{thickening} is an \(A' \in \operatorname{Art}_{/k}\) such that \(0 \to J \to A' \to A \to 0\), so \(\operatorname{Spec}A \hookrightarrow\operatorname{Spec}A'\). \end{definition} \begin{definition}[Small Thickening] A \textbf{small thickening} is a thickening such that \(0 = {\mathfrak{m}}_{A'} J\), so \(J\) becomes a module for the residue field, and \(\dim_k J = 1\). \end{definition} \begin{lemma}[?] Any thickening of \(A\), say \(B\to A\), fits into a diagram: \begin{center} \begin{tikzcd} & & & & 0 & & & & \\ & & & & & & & & \\ & & J \arrow[rr] & & A' \arrow[uu] \arrow[rr] & & A \arrow[dd, Rightarrow] \arrow[rr] & & 0 \\ & & & & & & & & \\ 0 \arrow[rr] & & I \arrow[rr] \arrow[uu] & & B \arrow[uu] \arrow[rr] & & A \arrow[rr] & & 0 \\ & & & & & & & & \\ & & I' \arrow[rr, Rightarrow] \arrow[uu] & & I' \arrow[uu] & & & & \\ & & & & & & & & \\ & & 0 \arrow[uu] & & 0 \arrow[uu] & & & & \end{tikzcd} \end{center} \end{lemma} \begin{proof}[of lemma] We just need \(I' \subset I\) with \({\mathfrak{m}}_S I \subset J' \subset I \iff J {\mathfrak{m}}_B = 0\). Choose \(J'\) to be a preimage of a codimension 1 vector space in \(I/{\mathfrak{m}}_B I\). Thus \(J = I/I'\) is 1-dimensional. \end{proof} Thus any thickening \(A\) can be obtained by a sequence of small thickenings. By the lemma, in principle \(F\) and thus \(\widehat{F}\) are determined by their behavior under small extensions. \hypertarget{example}{% \subsubsection{Example}\label{example}} Consider \({\operatorname{Pic}}\), fix \(X_{_{/k}}\), start with a line bundle \(L_0 \in {\operatorname{Pic}}(x) /{\operatorname{Pic}}(k) = {\operatorname{Pic}}(x)\) and the deformation functor \(F(A)\) being the set of line bundles \(L\) on \(X_A\)with \({\left.{{L}} \right|_{{x}} } \cong L_0\), modulo isomorphism. Note that this yields a diagram \begin{center} \begin{tikzcd} x \arrow[rr] \arrow[dd, hook] & & k \arrow[dd, "\text{unique closed point}"] \\ & & \\ X_A \arrow[rr] & & \operatorname{Spec}A \end{tikzcd} \end{center} This is equal to \((I_x)^{-1}(L_0)\), where \({\operatorname{Pic}}(X_a) \xrightarrow{I_x} {\operatorname{Pic}}(x)\). If \begin{align*} 0 \to J \to A' \to A \to 0 .\end{align*} is a small thickening, we can identify \begin{center} \begin{tikzcd} 0 \arrow[rr] & & J \otimes_x {\mathcal{O}}_{x} \cong {\mathcal{O}}_x \arrow[rr] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] & & {\mathcal{O}}_{X_{A}} \arrow[rr] & & 0 & & \\ & & & & & & & & & &\in\text{AbSheaves} \\ 0 \arrow[rr] & & {\mathcal{O}}_x \arrow[rr, "f\mapsto 1+f"] & & {\mathcal{O}}_{X_{A'}}^* \arrow[rr] \arrow[uu, hook] & & {\mathcal{O}}_{X_{A}}^* \arrow[rr] & & 0 & & \end{tikzcd} \end{center} This yields a LES \begin{center} \begin{tikzcd}[column sep=tiny] 0 \arrow[rr] & & {H^0(X, {\mathcal{O}}_x) = k} \arrow[rr] & & {H^0(X_{A'}, {\mathcal{O}}_{x_{A'}}^*) = {A'}^*} \arrow[rr] & & {H^0(X_{A'}, {\mathcal{O}}_{x_{A}}^*) = A^*} \arrow[lllldd] \arrow[rr] & & \therefore 0 \\ & & & & & & & & \\ \therefore 0 \arrow[rr] & & {H^1(X, {\mathcal{O}}_{x})} \arrow[rr] & & {H^1(X_{A'}, {\mathcal{O}}_{x_{A'}}^*) = {\operatorname{Pic}}(X_{A'})} \arrow[rr, "\scriptsize\text{restriction to } X_A", outer sep=1em] & & {H^1(X_{A}, {\mathcal{O}}_{x_{A}}^*) = {\operatorname{Pic}}(X_A)} \arrow[lllldd, "\text{obs}"] & & \\ & & & & & & & & \\ & & {H^2(X, {\mathcal{O}}_x)} \ar[rr] & &\cdots & & & & \end{tikzcd} \end{center} \begin{remark} To understand \(F\) on small extensions, we're interested in \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item Given \(L \in F_{\text{loc}}(A)\), i.e.~\(L\) on \(X_A\) restricting to \(L_0\), when does it extend to \(L' \in F_{\text{loc}}(A')\)? I.e., does there exist an \(L'\) on \(X_{A'}\) restricting to \(L\)? \item Provided such an extension \(L'\) exists, how many are there, and what is the structure of the space of extensions? \end{enumerate} \end{remark} \begin{question} We have an \(L\in {\operatorname{Pic}}(X_A)\), when does it extend? \end{question} By exactness, \(L'\) exists iff \(\text{obs}(L) = 0\in H^2(X, {\mathcal{O}}_x)\), which answers 1. To answer 2, \((I_x)^{-1}(L)\) is the set of extensions of \(L\), which is a torsor under \(H^1(x, {\mathcal{O}}_x)\). Note that these are fixed \(k{\hbox{-}}\)vector spaces. \begin{remark} \(H^1(X, {\mathcal{O}}_x)\) is interpreted as the \textbf{tangent space} of the functor \(F\), i.e.~\(F_{\text{loc}}(K[\varepsilon])\). Note that if \(X\) is projective, line bundles can be unobstructed without the group itself being zero. \end{remark} For (3), just play with \(A = k[\varepsilon]\), which yields \(0 \to k \xrightarrow{\varepsilon} k[\varepsilon] \to k \to 0\), then \begin{center} \begin{tikzcd} 0 \arrow[rr] & & {H^1(X, {\mathcal{O}}_x)} \arrow[rr] & & {H^1(X_{k[\varepsilon]}, {\mathcal{O}}_{k[\varepsilon]}^*)} \arrow[rr, "I_x"] & & {H^1(X, {\mathcal{O}}_x^*)} \arrow[ll, bend right=49] \\ & & & & {(I_x)^{-1}(L_0) \in {\operatorname{Pic}}(X_{k[\varepsilon]})} & & L_0 \in {\operatorname{Pic}}(x) \end{tikzcd} \end{center} i.e., there is a canonical trivial extension \(L_0[\varepsilon]\). \begin{example}[?] Let \(X \supset Z_0 \in \operatorname{Hilb}_{X_{_{/k}}}(k)\), we computed \begin{align*} T_{Z_0} \operatorname{Hilb}_{X_{_{/k}}} = \hom_{{\mathcal{O}}_x}(I_{Z_0}, {\mathcal{O}}_z) .\end{align*} We took \(Z_0 \subset X\) and extended to \(Z' \subset X_{k[\varepsilon]}\) by base change. In this case, \(F_{\text{loc}}(A)\) was the set of \(Z'\subset X_A\) which are flat over \(A\), such that base-changing \(Z' \times_{\operatorname{Spec}A} \operatorname{Spec}k \cong Z\). This was the same as looking at the preimage restricted to the closed point, \begin{align*} \operatorname{Hilb}_{X_{_{/k}}}(A) \xrightarrow{i^*} \operatorname{Hilb}_{X_{_{/k}}}(k) \\ (i^*)^{-1}(z_0) \mapsfrom z_0 .\end{align*} Recall how we did the thickening: we had \(0 \to J \to A' \to A \to 0\) with \(J^2 = 0\), along with \(F\) on \(X_A\) which is flat over \(A\) with \(X_{_{/k}}\) projective, and finally an \(F'\) on \(X_{A'}\) restricting to \(F\). The criterion we had was \(F'\) was flat over \(A'\) iff \(0 \to J\otimes_{A'} F' \to F'\), i.e.~this is injective. Suppose \(z\in F_{\text{loc}}(A)\) and an extension \(z' \in F_{\text{loc}}(A')\). By tensoring the two exact sequences here, we get an exact grid: \begin{center} \begin{tikzcd} 0 \arrow[rr] \arrow[dd] & & I_{Z'} \arrow[rr] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] & & {\mathcal{O}}_{Z'} \arrow[rr] & & 0 \\ & & 0 \arrow[d] & & 0 \arrow[d] & & 0 \arrow[d] & & \\ J \arrow[dd] & 0 \arrow[r] & I_{Z_0} \arrow[dd] \arrow[rr] & & {\mathcal{O}}_X \arrow[dd] \arrow[rr] & & {\mathcal{O}}_{Z_0} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A' \arrow[dd] & 0 \arrow[r] & I_{Z'} \arrow[rr] \arrow[dd] & & {\mathcal{O}}_{X_{A'}} \arrow[rr] \arrow[dd] & & {\mathcal{O}}_{Z'} \arrow[dd] \arrow[r] & 0 & \\ & & & & & & & & \\ A \arrow[dd] & 0 \arrow[r] & I_Z \arrow[d] \arrow[rr] & & {\mathcal{O}}_{X_A} \arrow[rr] \arrow[d] & & {\mathcal{O}}_Z \arrow[d] \arrow[r] & 0 & \\ & & 0 & & 0 & & 0 & & \\ 0 & & & & & & & & \end{tikzcd} \end{center} The space of extension should be a torsor under \(\hom_{{\mathcal{O}}_X}(I_{Z_0}, {\mathcal{O}}_{Z_0})\), which we want to think of as \(\hom_{{\mathcal{O}}_X}(I_{Z_0}, {\mathcal{O}}_{Z_0})\). Picking a \(\phi\) in this hom space, we want to take an extension \(I_{Z'} \xrightarrow{\phi} I_{Z''}\). \end{example} \begin{quote} We'll cover how to make this extension next time. \end{quote} \hypertarget{tuesday-march-31st}{% \section{Tuesday March 31st}\label{tuesday-march-31st}} See notes on Ben's website. We'll review where we were. \hypertarget{deformation-theory}{% \subsection{Deformation Theory}\label{deformation-theory}} We want to represent certain moduli functors by schemes. If we know a functor is representable, it's easier to understand the deformation theory of it and still retain a lot of geometric information. The representability of deformation is much easier to show. We're considering functors \(F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\). \begin{example}[?] The Hilbert functor \begin{align*} \operatorname{Hilb}_{X_{_{/k}}} ({\operatorname{Sch}}_{/k})^{\operatorname{op}}\to {\operatorname{Set}}\\ S \mapsto \left\{{ Z \subset X \times S \text{ flat over } S}\right\} .\end{align*} This yields \begin{align*} F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\\ ??? .\end{align*} \end{example} \begin{figure} \centering \includegraphics{figures/2020-03-31-12:44.png} \caption{Image} \end{figure} Recall that we're interested in pro-representability, where \(\widehat{F}(R) = \varprojlim F(R\mu_R^n)\) is given by a lift of the form \begin{center} \begin{tikzcd} \operatorname{Art}_{/k} \ar[r, "F"] & {\operatorname{Set}} \\ \widehat{\operatorname{Art}_{/k}} \ar[u, hook] \ar[ur, "\widehat{F}"'] & \end{tikzcd} \end{center} \begin{question} Is \(\widehat{F}\) representable, i.e.~is \(F\) pro-representable? \end{question} \begin{example}[?] The \(F\) in the previous example is pro-representable by \(\widehat{F} = \hom({\mathcal{O}}_{\operatorname{Hilb}, z_0}, {\,\cdot\,})\). \end{example} \begin{definition}[Pro-Representable Hull] \(F\) has a \emph{pro-representable hull} iff there is a formally smooth map \(h_R \to F\). \end{definition} \begin{question} Does \(F\) have a pro-representable hull? \end{question} Recall that a map of functors on artinian \(k{\hbox{-}}\)algebras is \textbf{formally smooth} if it can be lifted through nilpotent thickenings. That is, for \(F, G: \operatorname{Art}_{/k} \to {\operatorname{Set}}\), \(F \to G\) is \emph{formally smooth} if for any thickening \(A' \twoheadrightarrow A\), we have \begin{center} \begin{tikzcd} & & F \ar[d] \\ h_{A} \ar[rru] \ar[r] & h_{A'} \ar[ru, dotted] \ar[r] & G \\ \operatorname{Spec}A \ar[u, equal] \ar[r] & \operatorname{Spec}A' \ar[u, equal] \ar[r] & G \ar[u, equal] \end{tikzcd} \end{center} We proved for \(R, A\) finite type over \(k\), \(\operatorname{Spec}R \to \operatorname{Spec}A\) smooth is formally smooth. Given a complete local \(k{\hbox{-}}\)algebra \(R\) and a section \(\xi \in \widehat{F}(R)\), we make the following definitions: \begin{definition}[Versal, Miniversal, Universal] The pair \((R, \xi)\) is \begin{itemize} \item \emph{Versal} for \(F\) iff \(h_R \xrightarrow{\xi} F\) is formally smooth.\footnote{Not a unique map, but still a pullback} \item \emph{Miniversal} for \(F\) iff versal and an isomorphism on Zariski tangent spaces. \item \emph{Universal} for \(F\) if \(h_R \xrightarrow{\cong} F\) is an isomorphism, i.e.~\(h_R\) pro-represents \(F\). \begin{itemize} \tightlist \item Pullback by a unique map \end{itemize} \end{itemize} \end{definition} \begin{remark} Note that \textbf{versal} means that any formal section \((s, \eta)\) where \(\eta \in \widehat{F}(s)\) comes from pullback, i.e there exists a map \begin{align*} R &\to S \\ \widehat{F}(R) &\to \widehat{F}(s) \\ \xi &\mapsto \eta .\end{align*} \textbf{Miniversal} means adds that the derivative is uniquely determined, and universal means that \(R\to S\) is unique. \end{remark} \begin{definition}[Obstruction Theory] An \textbf{obstruction theory} for \(F\) is the data of \(\mathrm{def}(F), \mathrm{obs}(F)\) which are finite-dimensional \(k{\hbox{-}}\)vector spaces, along with a functorial assignment of the following form: \begin{align*} (A' \twoheadrightarrow A) \quad \text{a small thickening } \mapsto \\ \mathrm{def}(F) {\circlearrowleft}F(A') \to F(A) \xrightarrow{\mathrm{obs}} \mathrm{obs}(F) \end{align*} that is exact\footnote{Recall that right-exactness was a transitive action.} and if \(A=k\), it is exact on the left (so the action was faithful on nonempty fibers). \end{definition} \begin{example}[?] We have \begin{align*} {\operatorname{Pic}}_{X_{/k}} : ({\operatorname{Sch}}_{/k})^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto {\operatorname{Pic}}(X\times X) / {\operatorname{Pic}}(S) .\end{align*} This yields \begin{align*} F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\\ A \mapsto L\in {\operatorname{Pic}}(X_A),~ L\otimes k \cong L_0 \end{align*} where \(X_{_{/k}}\) is proper and irreducible. Then \(F\) has an obstruction theory with \(\mathrm{def}(F) = H^1({\mathcal{O}}_x)\) and \(\mathrm{obs}(F) = H^2({\mathcal{O}}_x)\). The key was to look at the LES of \begin{align*} 0 \to {\mathcal{O}}_x \to {\mathcal{O}}_{X_{A'}}^* \to {\mathcal{O}}_{X_A}^* \to 0 .\end{align*} for \(0 \to k \to A' \to A \to 0\) small. \end{example} \begin{remark}[Summary] In both cases, the obstruction theory is exact on the left for any small thickening. We will prove the following: \begin{itemize} \item \(F\) has an obstruction \(\iff\) it has a pro-representable hull, i.e.~a versal family \item \(F\) has an obstruction theory which is always exact at the left \(\iff\) it has a universal family. \end{itemize} \end{remark} \hypertarget{schlessingers-criterion}{% \subsection{Schlessinger's Criterion}\label{schlessingers-criterion}} Let \(F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\) be a deformation functor (and it only makes sense to talk about deformation functors when \(F(k) = {\{\operatorname{pt}\}}\)). This theorem will tell us when a miniversal and a universal family exists. \begin{theorem}[Schlessinger] \(F\) has a miniversal family iff \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item Gluing along common subspaces: ror any small \(A' \to A\) and \(A'' \to A\) any other thickening, the map \begin{align*} F(A' \times_A A'') \to F(A') \times_{F(A)} F(A'') \end{align*} is surjective. \item Unique gluing: if \((A' \to A) = (k[\varepsilon] \to k)\), then the above map is bijective. \item \(t_F = F(k[\varepsilon])\) is a finite dimensional \(k{\hbox{-}}\)vector space, i.e. \begin{align*} F(k[\varepsilon] \times_k k[\varepsilon]) \xrightarrow{\cong} F(k[\varepsilon]) \times F(k[\varepsilon]) .\end{align*} \item For \(A' \to A\) small, \end{enumerate} \begin{center} \begin{tikzcd} F(A') \ar[r, "f"] & F(A) \\ t_f\, {\circlearrowleft}f^{-1}(\eta) \ar[u, hook, "\subseteq"] & \eta \ar[u, "\in"] \end{tikzcd} \end{center} where the action is simply transitive. \(F\) has a miniversal family iff (1)-(3) hold, and universal iff all 4 hold. \end{theorem} \begin{exercise}[?] Show that the existence of an obstruction theory which is exact on the left implies (1)-(4). \end{exercise} The following diagram commutes: \begin{center} \begin{tikzcd} \mathrm{def} {\circlearrowleft}F(A' \times_A A'') \ni \eta \ar[r] \ar[d] & F(A'') \ni \xi'' \ar[r, "\mathrm{obs}"] \ar[d] & \mathrm{obs} \\ \mathrm{def} {\circlearrowleft}F(A')\ni \eta'm \xi' \ar[r] & F(A')\ni \xi \ar[r, "\mathrm{obs}"] & \mathrm{obs} \\ \end{tikzcd} \end{center} So we have a map \(F(A' \times_A A'') \to F(A') \times_{F(A)} F(A'') \ni (\xi',\xi'')\). Using transitivity of the \(\mathrm{def}\) action, we can get \(\xi' = \eta' + \theta\) and thus \(\eta + \theta\) is the lift. \hypertarget{abstract-deformation-theory}{% \subsection{Abstract Deformation Theory}\label{abstract-deformation-theory}} \begin{example}[?] We start with \(\qty{X_0}_{/k}\) and define the functor \(F\) sending \(A\) to \(X/A\) flat families over \(A\) with \(X_0 \hookrightarrow^i X\) such that \(i \otimes k\) is an isomorphism. The punchline is that \(F\) has an obstruction theory if \(X_0\) is smooth with \begin{itemize} \tightlist \item \(\mathrm{def}(F) = H^1(T_{X_0})\) \item \(\mathrm{obs}(F) = H^2(T_{X_0})\) \end{itemize} \end{example} \begin{remark} \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item If \(X\) is a deformation of \(X_0\) over \(A\) and we have a small extension \(k \to A'\to A\) with \(X'\) over \(A'\) a lift of \(X\). Then there is an exact sequence \begin{align*} 0 \to \text{Der}_R({\mathcal{O}}_{X_0}) \to\operatorname{Aut}_{A'}(X') \to \operatorname{Aut}_A(X) .\end{align*} \item If \(\qty{X_0}_{/k}\) is smooth and \emph{affine}, then any deformation \(X\) over \(A\) (a flat family restricting to \(X_0\)) is trivial, i.e.~\(X \cong X_0 \times_k \operatorname{Spec}(A)\). \end{enumerate} \begin{center} \begin{tikzcd} & & X_0 \times\operatorname{Spec}(A) \ar[d] \\ X_0 \ar[r, hook] & X \ar[r] \ar[ru, "f", dotted] & \operatorname{Spec}(A) \end{tikzcd} \end{center} Thus \(X_0 \hookrightarrow X\) has a section \(X\to X_0\), and the claim is that this forces \(X\) to be trivial. \end{remark} We have \begin{center} \begin{tikzcd} 0 \ar[r] & J \otimes{\mathcal{O}}_X \ar[r] & {\mathcal{O}}_x \ar[r] & {\mathcal{O}}_{X_0} \ar[r] \ar[l, bend right] & 0 \end{tikzcd} \end{center} yielding \begin{align*} 0 \to K \to {\mathcal{O}}_{X_0} \otimes A \to {\mathcal{O}}_X \to 0 \\ ({\,\cdot\,}\otimes k) \\ 1 \to k\otimes k = 0 \to {\mathcal{O}}_{X_0} \xrightarrow{\cong} {\mathcal{O}}_{X_0} \to 0 .\end{align*} \begin{remark} Why does this involve cohomology of the tangent bundle? For \(X_0\) smooth, \(\operatorname{Der}_k({\mathcal{O}}_{X_0}) = \mathcal{H}(T_{X_0})\), but the LHS is equal to \(\hom( \Omega_{ \qty{X_0}_{/k}}, {\mathcal{O}}_{X_0}) = H^0 (T_{X_0})\). \end{remark} \begin{quote} Upcoming: proof of Schlessinger so we can use it! \end{quote} \hypertarget{thursday-april-2nd}{% \section{Thursday April 2nd}\label{thursday-april-2nd}} \hypertarget{abstract-deformations}{% \subsection{Abstract Deformations}\label{abstract-deformations}} Let \(X_0\) be smooth and consider the deformation functor \begin{align*} F : \operatorname{Art}_{_{/k}} &\to {\operatorname{Set}}\\ A &\mapsto (X_{/A} , \iota) \end{align*} where \(X\) is flat (and thus smooth) and \(i\) is a closed embedding \(i: X_0 \hookrightarrow X\) with \(i\otimes k\) an isomorphism. Then \(F\) has an obstruction theory with \begin{itemize} \tightlist \item \(\mathrm{def}(F) = H^1(X_0, T_0)\) of the tangent bundle \item \(\mathrm{obs}(F) = H^2(X_0, T_0)\). \end{itemize} Additionally assume \(X_0\) is smooth and projective, which will force the above cohomology groups to be finite-dimensional over \(k\). \begin{remark}[Key points] \envlist \begin{itemize} \tightlist \item All deformations of smooth affine schemes are trivial \item Automorphisms of a deformation \(X/A\) which are the identity on \(X_0\) are \(\operatorname{id}+ \delta\) for \(\delta\) a derivation in \(\operatorname{Der}_k({\mathcal{O}}_{X_0}) = \hom_{{\mathcal{O}}_{X_0}}(\Omega_{\qty{X_0}_{_{/k}}}, {\mathcal{O}}_{X_0})\). \end{itemize} \begin{quote} See screenshot. \end{quote} \end{remark} Suppose we have a small thickening \(k \to {\mathbb{A}}^1 \to {\mathbb{A}}\) and \(X/{\mathbb{A}}\) with an affine cover \(X_\alpha\) of \(X\). This comes with gluing information \(\phi_{\alpha\beta}: X_{\alpha\beta} \to X_{\beta\alpha} = X_\alpha \cap X_\beta\). These maps satisfy a cocycle condition: \begin{center} \begin{tikzcd} X_{\alpha\beta} \cap X_{\alpha\gamma} \ar[rr] \ar[rd] & & X_{\gamma\alpha} \cap X_{\gamma\beta} \ar[ld] \\ & X_{\beta\alpha} \cap X_{\beta\gamma} & \end{tikzcd} \end{center} \begin{question} Can we extend this to \(X'/{\mathbb{A}}\)? \end{question} We have \(X_\alpha \cong X_\alpha^\mathrm{red} \times{\mathbb{A}}\)? Choose \(\phi'_{\alpha\beta}\) such that \begin{center} \begin{tikzcd} X'_{\alpha\beta} \ar[r, "\phi'_{\alpha\beta}"] & X'_{\beta\alpha} = X_{\beta\alpha}^\mathrm{red} \times{\mathbb{A}} \\ X_{\alpha\beta}\ar[u, hook] \ar[r, "\phi_{\alpha\beta}"] & X_{\beta\alpha} \ar[u, hook] \end{tikzcd} \end{center} We need \(\phi'_{\alpha\beta}\) to satisfy the cocycle condition in order to glue. We want the following map to be the identity: \((\phi'_{\alpha\gamma})^{-1}\phi'_{\beta\gamma} \phi'_{\alpha\beta}\). This is an automorphism of \(X'_{\alpha\beta} \cap X'_{\alpha\beta}\) and is thus the identity in \(\operatorname{Aut}(X_{\alpha\beta} \cap X_{\alpha\gamma})\). So it makes sense to talk about \begin{align*} \delta_{\alpha\beta\gamma} \coloneqq (\phi'_{\alpha\gamma})^{-1} \phi'_{\beta\gamma} \phi'_{\alpha\beta} - \operatorname{id}\in M^0(T_{X^\mathrm{red}_{\alpha\beta\gamma}}) .\end{align*} \begin{exercise}[?] In parts, \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item \(\delta_{\alpha\beta\gamma}\) is a \(2{\hbox{-}}\)cocycle for \(T_{X_0}\), so it has trivial boundary in terms of Cech cocycles. Thus \([\delta_{\alpha\beta\gamma}] \in H^2(T_{X_0})\). \item The class \([\delta_{\alpha\beta\gamma}]\) is independent of choice of \(\phi'_{\alpha\beta}\), i.e.~\(\phi'_{\alpha\beta} - \phi_{\alpha\beta}'' \in H^0((T_X)_{\alpha\beta})\) gives a coboundary \(\eta\) and thus \(\delta = \delta' + \eta\). This yields \(\mathrm{obs}(X) \in H^2(T_{X_0})\). \item \(\mathrm{obs}(X) = 0 \iff X\) lifts to some \(X'\) (i.e.~a lift exists) \end{enumerate} \end{exercise} \begin{remark} For the sufficiency, we have \(\delta_{\alpha\beta\gamma} = {{\partial}}\eta_{\alpha\beta} \in H^0(T_{X_{\alpha\beta}})\). Let \(\phi_{\alpha\beta}'' = \phi_{\alpha\beta}' - \eta_{\alpha\beta}\), the claim is that \(\phi_{\alpha\beta}''\) satisfies the gluing condition. This covers the obstruction, so now we need to show that the set of lifts is a torsor for the action of the deformation space \(\mathrm{def}(F) = H^1(T_{X_0})\). From an \(X'\), we obtain \(X_{\alpha\beta}' \xrightarrow{\phi_{\alpha\beta}'} X_{\beta\alpha}'\) where the LHS is isomorphic to \((X_{\alpha\beta}')^\mathrm{red} \times{\mathbb{A}}^r\)? Given \(\eta_{\alpha\beta} \in H^0(T_{X_{\alpha\beta}})\), then \(\phi'_{\alpha\beta} + \eta_{\alpha\beta} = \phi_{\alpha\beta}''\) is another such identification. \end{remark} \begin{exercise}[?] In parts \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \({{\partial}}\eta_{\alpha\beta} = 0\). \item Given an \(X'\) and 1-coboundary \(\eta\), we get a new lift \(X'' = X' + \eta\). If \([\eta] = [\eta'] \in H^1(T_{X_0})\), then \(X' + \eta \cong X' + \eta'\). \end{enumerate} By construction, \((X' + \eta)_\alpha \cong (X' + \eta')_\alpha\), but these may not patch together. However, if \([\eta] = [\eta']\) then this isomorphism can be modified by by \(\varepsilon\) defined by \(\eta-\eta' = {{\partial}}\varepsilon\), and it patches. \end{exercise} \begin{remark} This kind of patching is ubiquitous -- essentially patching together local obstructions to get a global one. In general, there is a local-to-global spectral sequence that computes the obstruction space \end{remark} \hypertarget{proving-schlessinger}{% \subsection{Proving Schlessinger}\label{proving-schlessinger}} \hypertarget{the-schlessinger-axioms}{% \subsubsection{The Schlessinger Axioms}\label{the-schlessinger-axioms}} \hypertarget{h1}{% \paragraph{H1}\label{h1}} For any two small thickenings \begin{align*} A' &\to A \\ A'' &\to A \end{align*} we have a natural map \begin{align*} F(A' \times_A A'') \to F(A') \times_{F(A)} F(A'') \end{align*} and we require that this map is surjective. So deformations agreeing on the sub glue together. \hypertarget{h2}{% \paragraph{H2}\label{h2}} When \((A' \to A) = (k[\varepsilon] \to k)\) is the trivial extension, the map in H1 is an isomorphism. \begin{quote} Doing things to first order is especially simple. \end{quote} \hypertarget{h3}{% \paragraph{H3}\label{h3}} The tangent space of \(F\) is given by \(t_F = F(k[\varepsilon])\), and we require that \(\dim_k t_F < \infty\), which makes sense due to H2. \hypertarget{h4}{% \paragraph{H4}\label{h4}} If we have two equal small thickenings \((A' \to A) = (A'' \to A)\), then the map in H1 is an isomorphism. \hypertarget{h4-1}{% \paragraph{H4'}\label{h4-1}} For \(A' \to A\) small, \begin{align*} t_F {\circlearrowleft}F(A') \to F(A) \end{align*} is exact in the middle and left. \begin{remark} Note that the existence of this action uses H2. \end{remark} \begin{warnings} For \((R, \xi)\) a complete local ring and \(\xi \in \widehat{F}(R)\) a formal family, this is a hull \(\iff\) miniversal, i.e.~for \(h_R \xrightarrow{\xi} F\), this is smooth an isomorphism on tangent spaces. \end{warnings} \begin{theorem}[1, Schlessinger] \envlist \begin{enumerate} \def\labelenumi{\alph{enumi}.} \tightlist \item \(F\) has a miniversal family \((R, \xi)\) with \(\dim t_R < \infty\), noting that \(t_R = {\mathfrak{m}}_R / {\mathfrak{m}}_R^2\), iff H1-H3 hold. \item \(F\) has a universal family \((R, \xi)\) with \(\dim t_R < \infty\) iff h1-H4 hold. \end{enumerate} \end{theorem} \begin{theorem}[2] \envlist \begin{enumerate} \def\labelenumi{\alph{enumi}.} \tightlist \item \(F\) having an obstruction theory implies H1-H3. \item \(F\) having a strong obstruction theory (exact on the left) is equivalent to H1-H4. \end{enumerate} \end{theorem} Some preliminary observations: \begin{exercise}[Easy, fun, diagram chase] If \(F\) has an obstruction theory, then H1-H3 hold. \end{exercise} \begin{exercise}[?] An obstruction theory being exact on the left implies H4. \end{exercise} \hypertarget{example-1}{% \subsubsection{Example}\label{example-1}} \begin{exercise}[?] For \(R\) a complete local \(k{\hbox{-}}\)algebra with \(t_R\) finite dimensional has a strong obstruction theory. \end{exercise} Can always find a surjection from a power series ring: \begin{align*} S \coloneqq k[[t_R^\vee]] \twoheadrightarrow R \end{align*} which yields an obstruction theory \begin{itemize} \tightlist \item \(\mathrm{def} = t_R\) \item \(\mathrm{obs} = I/{\mathfrak{m}}_S I\) \end{itemize} i.e., if \(F\) is pro-representable, then it has a strong obstruction theory. Suppose that \((R, \xi)\) is versal for \(F\), this implies H1. We get \(F(A' \times_A A'') \twoheadrightarrow F(A') \times_{F(A)} F(A'')\) For versal, if we have \(h_R \xrightarrow{\xi} F\) smooth, we have \begin{center} \begin{tikzcd} & & & h_r \ar[d] \\ h_k \ar[r] \ar[rrru] & h_A \ar[r] \ar[rru, dotted] \ar[rr, bend right, "\eta"] & h_{A'} \ar[ur, dotted] \ar[r] & F \end{tikzcd} \end{center} and we can find a lift from \(h_{A''}\) as well, so we get a diagram \begin{center} \begin{tikzcd} & & F \\ h_{A''} \ar[r] & h_R \ar[ru] \\ h_A \ar[u] \ar[r] & h_{A'} \ar[u] \end{tikzcd} \end{center} and thus \begin{center} \begin{tikzcd} {A''} \ar[r] & R \\ A \ar[u] \ar[r] & {A'} \ar[u] \end{tikzcd} \end{center} So we get the left \(\tilde \eta\) of \((\eta', \eta'')\) we want from \begin{center} \begin{tikzcd} h_{A' \times_A A''} \ar[r, "f"] \ar[rr, "\tilde\eta", bend right] & h_R \ar[r] & F \end{tikzcd} \end{center} If \((R, \xi)\) is miniversal, then H2 holds. We want to show that the map \begin{align*} F(A'' \times_K k[\varepsilon]) \xrightarrow{\sim} ?? \end{align*} is a bijection. Suppose we have two maps \begin{center} \begin{tikzcd} & & h_R \\ h_{A''} \ar[rru, bend left] \ar[r] & h_{A'' \times k[\varepsilon]} \ar[ru, shift left=0.75ex] \ar[ru, shift right=0.75ex] \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & F \\ & h_{k[\varepsilon]} \ar[ur, bend right] & \end{tikzcd} \end{center} Then the two lifts are in fact equal, and \begin{center} \begin{tikzcd} R \ar[r, shift left=0.75ex] \ar[r, shift right=0.75ex] & A'' \times k[\varepsilon] \ar[r] \ar[d] & k[\varepsilon] \\ & A'' & \end{tikzcd} \end{center} If \((R, \xi)\) is miniversal with \(t_R\) finite dimensional, then H3 holds immediately. If \((R, \xi)\) is universal, then H4 holds. \begin{question} Why are H4 and H4' connected? \end{question} \begin{answer} Let \(A' \to A\) be small, then \begin{align*} A' \times_A A' &= A' \times_k k[\varepsilon] \\ (x, y) &\mapsto ?? .\end{align*} Using H2, we can identify \(F(A; \times_A A') \cong t_F \times F(A')\). We can thus define an action \begin{align*} (\theta, \xi) &\mapsto (\theta + \xi, \xi) .\end{align*} If this is an isomorphism, then this action is simply transitive. The map \(\theta \mapsto \theta + \xi\) gives an isomorphism on the fiber of \(F(A') \to F(A)\). \end{answer} \begin{quote} Next time we'll show the interesting part of the sufficiency proof. \end{quote} \hypertarget{tuesday-april-7th}{% \section{Tuesday April 7th}\label{tuesday-april-7th}} \begin{quote} (Missing first few minutes.) \end{quote} Take \(I_{q+1}\) to be the minimal \(I\) such that \({\mathfrak{m}}_q I_q \subset I \subset I_1\) and \(\xi_q\) lifts to \(S/I\). \begin{claim} Such a minimal \(I\) exists, i.e.~if \(I, I'\) satisfy the two conditions then \(I \cap I'\) does as well. So \(I, I'\) are determined by their images \(v, v'\) in the vector space \(I_q \otimes k\). \end{claim} So enlarge either \(v\) or \(v'\) such that \(v + v' = I_q \otimes k\) but \(v \cap v'\) is the same. We can thus assume that \(I + I' = I_q\), and so \begin{align*} S / I \cap I' = S/I \times_{S/I_q} S/I' \end{align*} which by H1 yields a map \begin{align*} F(S/I\cap I') &\to F(S/I) \times_{F(S/I_q)} F(S/I') \end{align*} So \(I\cap I'\) satisfies both conditions and thus a minimal \(I_{q+1}\) exists. Let \(\xi_{q+1}\) be a lift of \(\xi_q\) over \(S/I_{q+1}\) (noting that there may be many lifts). \hypertarget{showing-miniversality}{% \subsection{Showing Miniversality}\label{showing-miniversality}} \begin{claim} Define \(R = \varinjlim R_q\) and \(\xi = \varinjlim\xi_q\), the claim is that \((R, \xi)\) is miniversal. \end{claim} We already have \(h_R \xrightarrow{\xi} F\) and thus \(t_R \xrightarrow{\cong}t_F\) is fulfilled. We need to show formal smoothness, i.e.~for \(A' \to A\) a small thickening, suppose we have a lift \begin{center} \begin{tikzcd} & & h_R \ar[d, "\xi"] \\ h_a \ar[rru, "n"] \ar[rr, bend right] \ar[r] & h_{A'} \ar[r] & F \end{tikzcd} \end{center} If we have a \(u'\) such that commutativity in square 1 holds (?) then we can form a lift \(u'\) satisfying commutativity in both squares 1 and 2. We can restrict sections to get a map \(F(A') \to F(A)\) and using representability obtain \(h_R(A') \to h_R(A)\). Combining H1 and H2, we know \(t_F\) acts transitively on fibers, yielding \begin{center} \begin{tikzcd} t_R {\circlearrowleft}\ar[d, "\cong"] & u'\in h_R(A') \ar[r] \ar[d] \ar[r] & u\in h_R(A) \ar[d]\\ t_F {\circlearrowleft}& \eta' \in F(A') \ar[r] \ar[r] & \eta \in F(A) \\ \end{tikzcd} \end{center} Then \(u' \mapsto u\) is equivalent to (1), and \(u' \mapsto \eta'\) is equivalent to (2). Let \(\eta_0\) be the image of \(u'\) and define \(\eta' = \eta_0 + \theta, \theta \in t_F\) then \(u' = u' + \theta, \theta \in t_R\). So we can modify the lift to make these agree. Thus it suffices to show \begin{center} \begin{tikzcd} A' \ar[r] & A & R_q \ar[l] \\ S \ar[u, "v"] \ar[r] & \ar[lu, dotted, "\exists_? u'"] \ar[u, "u"] \ar[ur] & \end{tikzcd} \end{center} We get a diagram of the form \begin{center} \begin{tikzcd} S \ar[d] \ar[r, "w"] & A' \times_A R_1 \ar[r] \ar[d, "{ \pi_2, \text{small} }"] & A' \ar[d, "{ \text{small} }"] \\ R \ar[r] & R_q \ar[r] & A \end{tikzcd} \end{center} \begin{observation} \envlist \begin{itemize} \item \(S \to R_q\) is surjective. \item \(\operatorname{im}(w) \subset A' \times_A R_1\) is a subring, so either \begin{itemize} \item \(\operatorname{im}(w) \xrightarrow{\cong} R_q\) if it doesn't meet the kernel, or \item \(\operatorname{im}(w) = A' \times_A R_q\) \end{itemize} \end{itemize} In case (a), this yields a section of the middle map and we'd get a map \(R_q \to A'\) and thus the original map we were after \(R \to A\). \end{observation} So assume \(w\) is surjective and consider \begin{center} \begin{tikzcd} 0 \ar[r] & I \ar[r] & S \ar[r] & A' \times_A R_q \ar[r]\ar[d, "\text{small}"] & 0 \\ & & & R_q & \end{tikzcd} \end{center} and we have \({\mathfrak{m}}_S I_1 \subset I \subset I_q\) where the second containment is because \(I\) a quotient of \(R_q\) factors through \(S/I\) and the first is because \(S/I\) is a small thickening of \(R_q\). But \(\xi_q\) lifts of \(S/I\), and we have \begin{align*} \xi \in F(S/I) \twoheadrightarrow\xi = \xi' \times\xi_q ? .\end{align*} Therefore \(I_{q+1} \subset I\) and we have a factorization \begin{center} \begin{tikzcd} S \ar[rr] \ar[dr, dotted] & & S/I \\ & R_{q+1}\ar[ur, dotted] & \end{tikzcd} \end{center} Recall that we had \includegraphics{figures/image_2020-04-07-13-17-11.png} \todo[inline]{Image to diagram} where the diagonal map \(u'\) gives us the desired lift, and thus \begin{center} \begin{tikzcd} R \ar[r] \ar[rr, bend left] & R_{q+1} \ar[r] & A' \end{tikzcd} \end{center} exists. This concludes showing miniversality. \hypertarget{part-of-proof}{% \subsection{Part of Proof}\label{part-of-proof}} To finish, we want to show that H4 implies that the map on sections \(h_R \xrightarrow{\xi} F\) is bijective. \begin{center} \begin{tikzcd} & & & h_R \ar[d, "\xi"] \\ & h_A \ar[rr, bend right, "\eta"] \ar[rru, bend left, "u"] & h_{A'} \ar[r, "\eta'"] \ar[ru, "\exists ! u'"] & F \end{tikzcd} \end{center} where the map \(\xi\) is ``formal etale'', which will necessarily imply that it's a bijection over all artinian rings. So we just need to show formal étaleness. We have a diagram \begin{center} \begin{tikzcd} t_R {\circlearrowleft}u'\in h_R(A') \ar[r] \ar[d] & u\in h_R(A) \ar[d] \\ t_F {\circlearrowleft}\eta' \in h_R(A') \ar[r] & \eta \in h_R(A) \end{tikzcd} \end{center} where \(u'\) exists by smoothness. Assume that are two \(u', u''\), then \(u' = u'' + \theta\) and \(\operatorname{im}(u') = \operatorname{im}(u'') + \theta \implies \theta = 0\) and thus \(u' = u''\). \hypertarget{revisiting-goals}{% \subsection{Revisiting Goals}\label{revisiting-goals}} We originally had two goals: \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item Given a representable moduli functor (such as the Hilbert functor), we wanted to understand the local structure by analyzing the deformation functor at a given point. \item We want to use representability of the deformation functors to get global representability of the original functor. \end{enumerate} \begin{question} What can we now deduce about the local structure of functors using their deformation theory? \end{question} \begin{fact}[1] Any two hulls \(h_R \to F\) are isomorphic but not canonically. We can lift maps at every finite level and induct up, which is an isomorphism on tangent spaces and thus an isomorphism. The sketch: use smoothness to get the map, and the tangent space condition will imply the full isomorphism. \end{fact} \begin{fact}[3] Suppose that \(F\) has an obstruction theory (not necessarily strong). This implies there exists a hull \(h_R \xrightarrow{\xi }F\). The obstruction theory of \(F\) \emph{gives} an obstruction theory of \(h_R\): given \(A' \to A\) a small thickening, we need a functorial assignment \begin{align*} t_R = \mathrm{def} {\circlearrowleft}h_R(A') \to h_R(A) \xrightarrow{\mathrm{obs}} \mathrm{obs} \\ \mathrm{def} {\circlearrowleft}F(A') \to F(A) \xrightarrow{\mathrm{obs}} \mathrm{obs} \end{align*} where there are vertical maps with equality on the edges. \begin{figure} \centering \includegraphics{figures/image_2020-04-07-13-35-00.png} \caption{Vertical maps} \end{figure} By formal smoothness, \(\eta'\) lifts to some \(\xi'\), but using the transitivity of the action of the tangent space can fix this. We already had an obstruction theory of \(R\), since we can always find a quotient \begin{align*} I \to S = k[[t_R^\vee]] \twoheadrightarrow R \end{align*} and \(h_K\) has an obstruction theory \begin{itemize} \tightlist \item \(\mathrm{def} = t_R = \qty{{\mathfrak{m}}_R/{\mathfrak{m}}_R^2}^\vee\) \item \(\mathrm{obs} = \qty{I/{\mathfrak{m}}_S I}^\vee\) \end{itemize} \end{fact} \begin{fact}[proof can be found in FGA] Any other obstruction theory \((\mathrm{def}', \mathrm{obs}')\) of \(h_R\) admits an injection \(\qty{I/{\mathfrak{m}}_S I}^\vee\hookrightarrow\mathrm{obs}'\). \end{fact} Combining these three facts, we conclude the following: If \(F\) has an obstruction theory \(\mathrm{def}(F), \mathrm{obs}(F)\), then \(F\) has a miniversal family \(h_R \xrightarrow{\xi }F\) with \(R = S/ I\) a quotient of the formal power series ring over some ideal, where \(S = k[[t_F^\vee]]\). It follows that \(\dim(I/{\mathfrak{m}}_S I) \leq \dim \mathrm{obs}(F)\), and thus the minimal number of generators of \(I\) (equal to the LHS by Nakayama) is bounded by the RHS. Thus \begin{align*} \dim_k \mathrm{def}(F) \geq \dim R \geq \dim \mathrm{def}(F) - \dim \mathrm{obs}(F) .\end{align*} In particular, if \(\dim(R) = \dim \mathrm{def}(F) - \dim \mathrm{obs}(F)\), then \(R\) is a complete intersection. If \(\dim(R) = \dim \mathrm{def}(R)\), the ideal doesn't have any generators, and \(R \cong S\). In particular, if \(\mathrm{obs}(F) = 0\), then \(R \cong S\) is isomorphic to this power series ring. Finally, if \(F\) is the deformation functor for a global representable functor, then \(R = \widehat{{\mathcal{O}}}_{{\mathfrak{m}}, p}\) is the completion of this local ring and the same things hold for this completion. Thus regularity can be checked on the completion. So if you have a representable functor with an obstruction theory (e.g.~the Hilbert Scheme) with zero obstruction, then we have smoothness at that point. If we know something about the dimension at a point relative to the obstruction, we can deduce information about being a local intersection. So the deformation tells you the dimension of a minimal smooth embedding, and the obstruction is the maximal number of equations needed to cut it out locally. \begin{remark} The content here: see Hartshorne's \emph{Deformation Theory}. The section in FGA is in less generality but has many good examples. See ``Fundamental Algebraic Geometry''. See also representability of the Picard scheme. \end{remark} \hypertarget{thursday-april-9th}{% \section{Thursday April 9th}\label{thursday-april-9th}} Let \(F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\) be a deformation functor with an obstruction theory. Then H1-H3 imply the existence of a miniversal family, and gives us some control on the hull \(h_{R} \to F\), namely \begin{align*} \dim \mathrm{def}(F) \geq \dim R \geq \dim \mathrm{def}(F) - \dim \mathrm{obs}(F) .\end{align*} In particular, if \(\mathrm{obs}(F) = 0\), then \(R \cong k[[\mathrm{def}(F)^\vee]] = k[[ t_{F}^\vee]]\). \begin{example}[?] Let \(M = \operatorname{Hilb}_{{\mathbb{P}}^n_{/k}}^{dt + (1-g)}\) where \(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\), and suppose \([Z] \in M\) is a smooth point. Then \begin{align*} \mathrm{def} = \hom_{ {{ {\mathcal{O}}_{x} }{\hbox{-}}\operatorname{mod}} }(I_{Z}, {\mathcal{O}}_{Z}) = \hom_{Z}(I_{Z}/I_{Z}^2, {\mathcal{O}}_{Z}) = H^0(N_{Z/X}) .\end{align*} the normal bundle \(N_{Z/X} = (I/I^2)^\vee\) of the regular embedding, and \(\mathrm{obs} = H^1(N_{Z/X})\). \begin{claim} If \(H^1({\mathcal{O}}_{Z}(1)) = 0\) (e.g.~if \(d > 2g-2)\) then \(M\) is smooth. \end{claim} \begin{proof}[of claim] The tangent bundle of \({\mathbb{P}}^n\) sits in the Euler sequence \begin{align*} 0 \to {\mathcal{O}}\to {\mathcal{O}}(1)^{n+1} \to T_{{\mathbb{P}}^n} \to 0 .\end{align*} And the normal bundles satisfies \begin{align*} 0 \to T_{Z} &\to T_{{\mathbb{P}}^n}\mathrel{\Big|}_{Z} \to N_{Z/{\mathbb{P}}^n} \to 0 \\ \\ &\Downarrow \text{ is the dual of }\\ \\ 0 \to I/I^2 &\to \Omega \mathrel{\Big|}_{Z} \to \Omega \to 0 .\end{align*} There is another SES: \begin{align*} ????? .\end{align*} Taking the LES in cohomology yields \begin{align*} H^1({\mathcal{O}}_{Z}(1)^{n+1})=0 \to H^1(N_{Z/{\mathbb{P}}^n}) =0 \to 0 \end{align*} and thus \(M\) is smooth at \([Z]\). We can compute the dimension using Riemann-Roch: \begin{align*} \dim_{[Z]} M &= \dim H^0(N_{Z/{\mathbb{P}}^n}) \\ &= \chi(N_{Z/{\mathbb{P}}^n}) \\ &= \deg N + {\operatorname{rank}}N(1-g) \\ &= \deg T_{{\mathbb{P}}^n} \mathrel{\Big|}_Z - \deg T_{Z} + (n-1)(1-g) \\ &= d(n+1) + (2-2g) + (n-1)(1-g) .\end{align*} \end{proof} \end{example} \begin{remark} This is one of the key outputs of obstruction theory: being able to compute these dimensions. \end{remark} \begin{example}[?] Let \(X \subset {\mathbb{P}}^5\) be a smooth cubic hypersurface and let \(H = \operatorname{Hilb}_{X_{/k}}^{\text{lines} = t+1} \subset \operatorname{Hilb}_{{\mathbb{P}}^5/k}^{t+1} = {\operatorname{Gr}}(1, {\mathbb{P}}^5)\), the usual Grassmannian. \begin{claim} Let \([\ell] \in H\), then the claim is that \(H\) is smooth at \([\ell]\) of dimension 4. \end{claim} \begin{proof}[of claim] We have \begin{itemize} \tightlist \item \(\mathrm{def} = H^0(N_{\ell/X})\) \item \(\mathrm{obs} = H^1(N_{\ell/X})\) \end{itemize} We have an exact sequence \begin{align*} 0 \to N_{\ell/X} \to N_{\ell/{\mathbb{P}}} \to N_{X/{\mathbb{P}}}\mathrel{\Big|}_\ell \to 0 \\ .\end{align*} There are surjections from \({\mathcal{O}}_\ell(1)^6\) onto the last two terms. \begin{claim}[Subclaim] For \(N = N_{\ell/{\mathbb{P}}}\) or \(N_{X/{\mathbb{P}}}\mathrel{\Big|}_\ell\), we have \(H^1(N) = 0\) and \({\mathcal{O}}(1)^6 \twoheadrightarrow N\) is surjective on global sections. \end{claim} \begin{proof}[of subclaim] Because \(\ell\) is a line, \({\mathcal{O}}_\ell(1) = {\mathcal{O}}(1)\) and \(H^1({\mathcal{O}}_\ell(1)) = 0\) and the previous proof applies, so \(H^1(N) = 0\). \end{proof} We thus have a diagram: \begin{figure} \centering \includegraphics{figures/image_2020-04-09-12-51-51.png} \caption{Image} \end{figure} In particular, \(T_\ell = {\mathcal{O}}(2)\), and the LES for \(0 \to {\mathcal{O}}\to K \to T_\ell\) shows \(H^1(K) = 0\). Looking at the horizontal SES \(0 \to K \to {\mathcal{O}}_\ell(1)^6 \twoheadrightarrow N_{\ell/{\mathbb{P}}}\) yields the surjection claim. We have \begin{figure} \centering \includegraphics{figures/a.png} \caption{Diagram} \end{figure} and taking the LES in cohomology yields \includegraphics{figures/image_2020-04-09-12-55-01.png}\\ Therefore \(H\) is smooth at \(\ell\) and \begin{align*} \dim_\ell H &= \chi(N_{\ell/X}) \\ &= \deg T_{X} - \deg T_\ell + 3 \\ &= \deg T_{\mathbb{P}}- \deg N_{X/{\mathbb{P}}} - \deg T_\ell + 3 \\ &= 6 - 3 - 2 + 3 = 4 .\end{align*} \end{proof} \end{example} \begin{remark} It turns out that the Hilbert scheme of lines on a cubic has some geometry: the Hilbert scheme of two points on a K3 surface. \end{remark} \hypertarget{abstract-deformations-revisited}{% \subsection{Abstract Deformations Revisited}\label{abstract-deformations-revisited}} Take \(X_{0} / k\) some scheme and consider the deformation functor \(F(A)\) taking \(A\) to \(X/A\) flat with an embedding \(\iota: X_{0} \hookrightarrow X\) with \(\iota \otimes k\) an isomorphism. Start with H1, the gluing axiom (regarding small thickenings \(A' \to A\) and a thickening \(A'' \to A\)). Suppose \begin{align*} X_{0} \hookrightarrow X' \in F(A') \to F(A) .\end{align*} which restricts to \(X_{0} \hookrightarrow X\). Then in \(F(A)\), we have \(X_{0} \hookrightarrow X' \otimes_{A'} A\), and we obtain a commutative diagram where \(X' \otimes A \hookrightarrow X'\) is a closed immersion: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/abcdefg.png} \caption{???} \end{figure} The restriction \(X' \to X\) means that there exists a diagram \begin{center} \begin{tikzcd} X' & & X \ar[ll, dotted, "\exists"] \\ & X \ar[ur, hook] \ar[ul, hook] \end{tikzcd} \end{center} Note that this is not necessarily unique. We have \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-06-40.png} \caption{Diagram?} \end{figure} This means that we can find embeddings such that \begin{center} \begin{tikzcd} X'' & \ar[l, "\exists", hook] X \ar[r, "\exists", hook] & X' \\ & X_{0} \ar[ul, hook] \ar[u, hook] \ar[ur, hook] \end{tikzcd} \end{center} \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-08-19.png} \caption{Diagram} \end{figure} And thus if we have \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-08-42.png} \caption{Diagram} \end{figure} then \(X_{0} \hookrightarrow Z\) is \textbf{a} required lift (again not unique). \begin{question} When is such a lift unique? \end{question} Suppose \(X_{0} \hookrightarrow W\) is another lift, then it restricts to both \(X, X'\) and we can fill in the following diagrams: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-10-44.png} \caption{Diagram} \end{figure} Using the universal property of \(Z\), which is the coproduct of this diagram: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-11-13.png} \caption{Diagram} \end{figure} However, there may be no such way to fill in the following diagram: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-11-58.png} \caption{Diagram} \end{figure} But if there exists a map making this diagram commute: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-12-25.png} \caption{Diagram} \end{figure} Then there is a map \(Z\to W\) which is flat after tensoring with \(k\), which is thus an isomorphism.\footnote{Recall that by Nakayama, a nonzero module tensor \(k\) can not be zero.} \begin{remark} Thus the lift is unique if \begin{itemize} \item \(X = X_{0}\), then the following diagrams commute by taking the identity and the embedding you have. Note that in particular, this implies H2. \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-15-09.png} \caption{Diagram} \end{figure} \item Generally, these diagrams can be completed (and thus the gluing maps are bijective) if the map \begin{align*} \operatorname{Aut}(X_{0}\hookrightarrow X') \to \operatorname{Aut}(X_{0} \hookrightarrow X) .\end{align*} of automorphisms of \(X'\) commuting with \(X_{0} \hookrightarrow X\) is surjective. \end{itemize} \end{remark} So in this situation, there is only \emph{one} way to fill in this diagram up to isomorphism: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-18-59.png} \caption{Diagram} \end{figure} If we had two ways of filling it in, we obtain bridging maps: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-09-13-20-07.png} \caption{Diagram} \end{figure} \begin{lemma}[?] If \(H^0(X_{0}, T_{X_{0}}) = 0\) (where the tangent bundle always makes sense as the dual of the sheaf of Kahler differentials) which we can identify as derivations \(D_{{\mathcal{O}}_{k}}({\mathcal{O}}_{X_{0}}, {\mathcal{O}}_{X_{0}})\), then the gluing map is bijective. \end{lemma} \begin{proof}[?] The claim is that \(\operatorname{Aut}(X_{0} \hookrightarrow X) = 1\) are always trivial. This would imply that all random choices lead to triangles that commute. Proceeding by induction, for the base case \(\operatorname{Aut}(X_{0} \hookrightarrow X_{0}) = 1\) trivially. Assume \(X_{0} \hookrightarrow X_{i}\) lifts \(X_{0} \hookrightarrow X\), then there's an exact sequence \begin{align*} 0 \to \operatorname{Der}_{k}({\mathcal{O}}_{X_{0}}, {\mathcal{O}}_{X_{0}}) \to {\operatorname{Aut}}(X_{0} \hookrightarrow X_0') \to {\operatorname{Aut}}(X_{0} \hookrightarrow X) .\end{align*} \end{proof} Thus \(F\) always satisfies H1 and H2, and \(H^0(T_{X_{0}}) = 0\) (so no ``infinitesimal automorphism'') implies H4. Recall that the dimension of deformations of \(F\) over \(k[\varepsilon]\) is finite, i.e.~\(\dim t_{F} < \infty\) This is where some assumptions are needed. If \(X_{/K}\) is either \begin{itemize} \tightlist \item Projective, or \item Affine with isolated singularities, \end{itemize} this is enough to imply H3. Thus by Schlessinger, under these conditions \(F\) has a miniversal family. Moreover, if \(H^0(T_{X_{0}}) = 0\) then \(F\) is pro-representable. \begin{example}[?] If \(X_{0}\) is a smooth projective genus \(g\geq 2\) curve, then \begin{itemize} \tightlist \item Obstruction theory gives the existence of a miniversal family \item We have \(\mathrm{obs} = H^2(T_{X_{0}}) = 0\), and thus the base of the miniversal family is smooth of dimension \(\mathrm{def}(F) \dim H^1(T_{X_{0}})\), \item \(H^0(T_{X_{0}}) = 0\) and \(\deg T_{X_{0}} = 2-2g < 0\), which implies that the miniversal family is universal. \end{itemize} We can conclude \begin{align*} \dim H^1(T_{X_{0}}) = -\chi(T_{X_{0}}) = -\deg T_{X_{0}} + g-1 = 3(g-1) .\end{align*} \end{example} \begin{remark} Note that the global deformation functor is not representable by a scheme, and instead requires a stack. However, the same fact shows smoothness in that setting. \end{remark} \hypertarget{hypersurface-singularities}{% \subsection{Hypersurface Singularities}\label{hypersurface-singularities}} Consider \(X(f) \subset {\mathbb{A}}^n\), and for simplicity, \((f=0) \subset {\mathbb{A}}^2\), and let \begin{itemize} \item \(S = {\mathbb{C}}[x, y]\). \item \(B = {\mathbb{C}}[x, y] / (f)\) \end{itemize} \begin{question} What are the deformations over \(A \coloneqq k[\varepsilon]\)? \end{question} This means we have a ring \(B'\) flat over \(k\) and tensors to an isomorphism, so tensoring \(k\to A\to k\) yields the following: \begin{center} \begin{tikzcd} 0 \ar[r] & B \ar[r] & B' \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & S \ar[u] \ar[r] & S[\varepsilon] \ar[u, "\exists", twoheadrightarrow] \ar[r] & S \ar[r] \ar[u] & 0 \\ 0 \ar[r] & S \cong I \ar[u] \ar[r] & I'= \left\langle{f'}\right\rangle \ar[u] \ar[r] & I = \left\langle{f}\right\rangle \ar[u, "\cong"] \ar[r] & S \end{tikzcd} \end{center} Thus any such \(B'\) is the quotient of \(S[\varepsilon]\) by an ideal, and we have \(f' = f + \varepsilon g\). \begin{question} When do two \(f'\)s give the same \(B'\)? \end{question} We have \(\varepsilon f' = \varepsilon f\), so \(\varepsilon f \in (f')\) and we can modify \(g\) by any \(cf\) where \(c\in S\), where only the equivalence class \(g\in S/(f)\) matters. Now consider \(\operatorname{Aut}(B \hookrightarrow B')\), i.e.~maps of the form \begin{align*} x &\mapsto x + \varepsilon a \\ y &\mapsto y + cb \end{align*} for \(a, b\in S\). Under this map, \begin{align*} f_0' = f + \varepsilon g \mapsto & f(x + \varepsilon a, y + \varepsilon b) + \varepsilon g(x ,y) \\ \\ &\Downarrow \quad\text{implies} \\ \\ f(x, y) &= \varepsilon a {\frac{\partial }{\partial x}\,} f + \varepsilon b {\frac{\partial }{\partial y}\,} f + \varepsilon g(x ,y) ,\end{align*} so in fact only the class of \(g\in S/(f, {\partial}_{x} f, {\partial}_{y} f)\). This is the ideal of the singular locus, and will be Artinian (and thus finite-dimensional) if the singularities are isolated, which implies H3. We can in fact exhibit the miniversal family explicitly by taking \(g_{i} \in S\), yielding a basis of the above quotient. The hull will be given by setting \(R = {\mathbb{C}}[[t_{1}, \cdots, t_{m} ]]\) and taking the locus \(V(f + \sum t_{i} g_{i}) \subset {\mathbb{A}}_{R}^2\). \begin{example}[simple] For \(f = xy\), then the ideal is \(I = (xy, y, x) = (x, y)\) and \(C/I\) is 1-dimensional, so the miniversal family is given by \(V(xy + t) \subset {\mathbb{C}}[[t_{1}]][x, y]\). The greater generality is needed because there are deformation functors with a hull but no universal families. \end{example} \hypertarget{tuesday-april-14th}{% \section{Tuesday April 14th}\label{tuesday-april-14th}} Recall that we are looking at \((X_{0})_{/k}\) and \(F: \operatorname{Art}_{/k} \to {\operatorname{Set}}\) where \(A\) is sent to \(X_{/A}\) flat with \(i: X_{0} \hookrightarrow X\) where \(i\otimes k\) is an isomorphism. The second condition is equivalent to a cartesian diagram \begin{center} \begin{tikzcd} X_{0} \ar[r, hook] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , very near start, color=black] & X \ar[d] \\ \operatorname{Spec}k \ar[r, hook] & \operatorname{Spec}A \end{tikzcd} \end{center} We showed we always have H1 and H2, and H3 if \(X_{0}/k\) is projective or \(X_{0}\) is affine with isolated singularities. In this situation we have a miniversal family. This occurs iff for \(A' \to A\) a small thickening and \((X_{0} \hookrightarrow X) \in F(A)\), we have a surjection \begin{align*} {\operatorname{Aut}}_{A'}(X_{0} \hookrightarrow X') \twoheadrightarrow{\operatorname{Aut}}_{A}(X_{0} \hookrightarrow X) .\end{align*} where the RHS are automorphisms of \(X_{/A}\), i.e.~those which commute with the identity on \(A\) and \(X_{0}\). We had a naive functor \(F_{n}\) where we don't include the inclusion \(X_{0} \hookrightarrow X\). When \(F\) has a hull then the naive functor has a versal family, since there is a forgetful map that is formally smooth. If it's the case that for all \(A' \to A\) small and \(F_{\text{n}} \to F_{n}(A)\) we have \({\operatorname{Aut}}_{A'}(X') \twoheadrightarrow{\operatorname{Aut}}_{A} (X)\), then \(F = F_{n}\) and both are pro-representable. The forgetful map is smooth because given \(X_{/A}\) in \(F_{n}(A)\), we have some inclusion \(X_{0} \hookrightarrow X\), so one gives surjectivity. Using the surjectivity on automorphisms, we get \begin{center} \begin{tikzcd} X_{0}\ar[rd, hook] \ar[rr, hook] & & X\ar[ld, dotted] \\ & X & \end{tikzcd} \end{center} Deformation theory is better at answering when the following diagrams exist: \begin{center} \begin{tikzcd} X \ar[r, dotted, hook, "\exists?"] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{$\ulcorner$}" , shift right=0.4em, very near start, color=black] & X' \ar[d, dotted, "\exists?"] \\ \operatorname{Spec}A \ar[r] & \operatorname{Spec}A' \end{tikzcd} \end{center} i.e., the existence of an extension of \(X\) to \(A'\). This is different than understanding diagrams of the following type, where we're considering isomorphism classes of the squares, and deformation theory helps understand the blue one: \begin{center} \begin{tikzpicture} [ greenbox/.style={ draw=green, fill=green!3, thick, rounded corners, rectangle }, redbox/.style={ draw=red, fill=red!3, thick, rounded corners, rectangle }, ] \node[ greenbox, minimum height=0.9cm, minimum width=1.2cm ] at (-0.1, 1.3) {}; \node[ redbox, minimum height=0.9cm, minimum width=1.2cm ] at (2.35, 1.3) {}; \node[ greenbox, minimum height=2.4cm, minimum width=8.2cm ] at (0, -0.5) {}; \node[ draw=red, thick, rectangle, minimum height=0.8cm, minimum width=1.2cm ] at (-1.2, -0.6) {}; \node[ draw=blue, thick, rectangle, dotted, minimum height=0.8cm, minimum width=1.2cm ] at (1.2, -0.6) {}; \node at (0, 0) {% \begin{tikzcd} & F(A') \ar[r] & F(A) \\ X_0 \ar[r, hook] \ar[d] & X \ar[r, hook] \ar[d] & X' \ar[d] \\ \operatorname{Spec}k \ar[r] & \operatorname{Spec}A \ar[r] & \operatorname{Spec}A' \end{tikzcd} }; \end{tikzpicture} \end{center} \begin{example}[Hypersurface Singularities] Take \(S = k[x, y]\) and \(B = S/(f)\), then deformations of \(\operatorname{Spec}B\) to ? Given \(k \to k[\varepsilon] \to k\) we can tensor\footnote{For flat maps, tensoring up to an isomorphism implies isomorphism.} to obtain \begin{center} \begin{tikzcd} {0} & {B} & {B'} & {B} & {0} \\ {0} & {S} & {S[\varepsilon]} & {S} & {0} \\ {0} & {I} & {I'} & {I} & {0} \\ && {\tiny \left\langle{f'}\right\rangle} & {\tiny \left\langle{f}\right\rangle} \arrow[from=3-1, to=3-2] \arrow[from=3-2, to=3-3] \arrow[from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-2, to=2-2] \arrow["{\pi}", from=2-2, to=1-2] \arrow[from=3-3, to=2-3] \arrow["{\pi'}", from=2-3, to=1-3] \arrow[from=3-4, to=2-4] \arrow["{\pi}", from=2-4, to=1-4] \arrow["{\subseteq}" description, from=4-3, to=3-3, no head] \arrow["{\subseteq}" description, from=4-4, to=3-4, no head] \end{tikzcd} \end{center} \begin{quote} \href{https://q.uiver.app/?q=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}{Link to diagram.} \end{quote} \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-12-55-58.png} \caption{Diagram} \end{figure} We want to understand \(F(k[\varepsilon])\). We know \(f' = f + \varepsilon g\) for some \(g\in S\). \begin{observation} \envlist \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item \(g\in B\) and \(f'' = f + \varepsilon(g + cf)\) generates the same ideal. \item We're free to reparameterize, i.e.~\(x \mapsto x + \varepsilon a\) and \(y \mapsto y + \varepsilon b\) and thus\\ \begin{align*} g \mapsto g + a f_{x} + b f_{y} \end{align*} , i.e.~the partial derivatives. \end{enumerate} \end{observation} Thus isomorphism classes of \(B'\) in deformations \(B' \to B\) only depend on the isomorphism classes \(g\in B/(f_{x}, f_{y}) B\). When the singularities are isolated, this quotient is finite-dimensional as a \(k{\hbox{-}}\)vector space. \end{example} \begin{example}[?] \(F(k[\varepsilon]) = B/(f_{x}, f_{y})B\). Thus H3 holds and there is a miniversal family \(h_{R} \to F\). We can describe it explicitly: take \(g_{i} \in S\), yielding a \(k{\hbox{-}}\)basis in \(S/(f, f_{x}, f_{y})\). Then \begin{align*} V(f + \sum t_{i} g_{i}) \subset \operatorname{Spec}k[[t_{1}, \cdots, t_{n}]][x, y] .\end{align*} Set \(R = k[[t_{1}, \cdots, t_{n}]]\), then this lands in \({\mathbb{A}}_{R}^2\). \end{example} \begin{example}[?] The nodal curve \(y^2 = x^3\), take . \begin{align*} S/(y^2-x^3, 2y, -3x^2) = S/(y, x^2) .\end{align*} So take \(g_{1} = 1, g_{2} = x\), then the miniversal family is . \begin{align*} V(y^2 - x^3 + t + t_{2} x) \subset {\mathbb{A}}^2_{k[[t_{1}, t_{2}]]} .\end{align*} This gives all ways of smoothing the node. \end{example} \begin{remark} Note that none of these are pro-representable. \end{remark} Given \(X\) and \(A\), we obtain a miniversal family over the formal spectrum \(\mathrm{Spf}(R) = (R, \xi)\) and a unique map: \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-13-10-21.png} \caption{Diagram} \end{figure} We can take two deformations over \(A = k[\xi]/ S^n\): \begin{itemize} \tightlist \item \(X_{1} = V(x + y)\)?? \item \(X_{2} = V(x + uy)\)?? \end{itemize} As deformations over \(A\), \(X_{1} \cong X_{2}\) where we send , \begin{align*} s&\mapsto s, \\ y&\mapsto y, \\ x&\mapsto ux .\end{align*} since \begin{align*} (xy + us) = (uxy + us) = (u(xy + s)) = (xy + s) .\end{align*} But we have two different classifying maps, which do commute up to an automorphism of \(A\), but are not equal. Since they pullback to different elements (?), \(F\) can not be pro-representable. \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-13-20-05.png} \caption{Diagram} \end{figure} So reparameterization in \(A\) yield different objects in \(F(A)\). In other words, \({\mathcal{X}}\to \mathrm{Spf}(R)\) has automorphisms inducing reparameterizations of \(R\). This indicates why we need maps restricting to the identity. \hypertarget{the-cotangent-complex}{% \subsection{The Cotangent Complex}\label{the-cotangent-complex}} For \(X \xrightarrow{f} Y\), we have \(L_{X/Y} \in D {\mathrm{QCoh}}(X)\), the derived category of quasicoherent sheaves on \(X\). This answers the extension question: \begin{answer} For any square-zero thickening \(Y \hookrightarrow Y'\) (a closed immersion) with ideal \(I\) yields an \({\mathcal{O}}_{Y}{\hbox{-}}\)module. \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \tightlist \item An extension exists iff \(0 = \mathrm{obs} \in \operatorname{Ext}^2(L_{X/Y}, f^* I)\) \item If so, the set of ways to do so is a torsor over this ext group. \item The automorphisms of the completion are given by \(\hom(L_{X/Y}, f^* I)\). \end{enumerate} \end{answer} \begin{remark} Some special cases: \(X \to Y\) smooth yields \(L_{X/Y} = \Omega_{X/Y}[0]\) concentrated in degree zero. \end{remark} \begin{example}[?] \(Y = \operatorname{Spec}k\) and \(Y' = \operatorname{Spec}k[\varepsilon]\) yields \begin{align*} \mathrm{obs} \in \operatorname{Ext}_{x}^2(\Omega_{X/Y}, {\mathcal{O}}_{x})= H^2(T_{X_{/k}}) .\end{align*} For \(X\hookrightarrow Y\) is a regular embedding (closed immersion and locally a regular sequence) \(L_{X/Y} = \qty{I/I^2}[1]\), the conormal bundle. \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-13-32-13.png} \caption{Diagram} \end{figure} \end{example} \begin{example}[?] For \(Y\) smooth, \(X \hookrightarrow Y\) a regular embedding, \(L_{X_{/k}} = \Omega_{X_{/k}}\) with \(\mathrm{obs}/\mathrm{def} = \operatorname{Ext}^{2/1}(\Omega_{x}, {\mathcal{O}})\) and the infinitesimal automorphisms are the homs. \end{example} \begin{example}[?] For \(Y = \operatorname{Spec}k[x, y] = {\mathbb{A}}^2\) and \(X = \operatorname{Spec}B = V(f) \subset {\mathbb{A}}^2\) we get \begin{align*} 0 \to I/I^2 \to \Omega_{X_{/k}} \otimes B &\to \Omega_?{X_{/k}} \to 0 \\ \\ & \Downarrow \quad \text{equals} \\ \\ 0 \to B \xrightarrow{1 \mapsto (f_{x}, f_{y})} &B^2 \to \Omega_{B_{/k}} = L_{X_{/k}} \to 0 .\end{align*} Taking \(\hom({\,\cdot\,}, B)\) yields \begin{center} \begin{tikzcd} 0 \ar[r] & \hom(\Omega, B) \ar[r] & B^2 \ar[lld, "{(f_{x}, f_{y})^t}"] \\ \operatorname{Ext}^1(\Omega, B) \ar[r] & 0 \ar[r] & 0 \ar[lld] \\ \operatorname{Ext}^2(\Omega, B) \ar[r] & 0 \ar[r] & 0 \end{tikzcd} \end{center} So , \begin{align*} \mathrm{obs} &= 0 \\ \mathrm{def} &= B/(f_{x}, f_{y})B \\ \operatorname{Aut}&\neq 0 .\end{align*} and \end{example} \begin{remark} We have the following obstruction theories: \begin{itemize} \item For abstract deformations, we have \begin{align*} X_{0} {}_{/k} \text{ smooth } \implies \operatorname{Aut}/\mathrm{def}/\mathrm{obs} = H^{0/1/2}(T_{X_{0}}) .\end{align*} \item For embedded deformations, \(Y_{0}/k\) smooth, \(X_{0} \hookrightarrow Y_{0}\) regular, we have \begin{align*} \operatorname{Aut}/\mathrm{def}/\mathrm{obs} = 0, H^{0/1}(N_{X_{0}/Y_{0}}) .\end{align*} \begin{quote} As an exercise, interpret this in terms of \(L_{X_{0}/Y_{0}}\). \end{quote} \item For maps \(X_{0} \xrightarrow{f_{0}} Y_{0}\), i.e.~maps \begin{align*} X_{0} \times k[\varepsilon] \xrightarrow{f} Y_{0} \times k[\varepsilon] .\end{align*} we consider the graph \(\Gamma(f_{0}) \subset X_{0} \times Y_{0}\). \end{itemize} \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-13-43-40.png} \caption{Diagram} \end{figure} Since all of these structures are special cases of the cotangent complex, they place nicely together in the following sense: Given \(X \hookrightarrow_{i} Y\) we have \begin{align*} 0 \to T_{X} \to i^* T_{Y} \to N_{X/Y} \to 0 .\end{align*} Yielding a LES \begin{align*} 0 &\to H^0(T_{X}) \to H^0(i^* T_{Y}) \to H^0(N_{X/Y}) \\ &\to H^1(T_{X}) \to H^1(i^* T_{Y}) \to H^1(N_{X/Y}) \\ &\to H^2(T_{X}) .\end{align*} \begin{figure} \centering \includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-04-14-13-47-05.png} \caption{Diagram} \end{figure} \end{remark} \begin{exercise}[?] Consider \(X \subset {\mathbb{P}}^3\) a smooth quartic, and show that \(\mathrm{def}(X) \cong k^{20}\) but \(\mathrm{def}_{\text{embedded}} \cong k^{19}\). This is a quartic K3 surface for which deformations don't lift (non-algebraic, don't sit inside any \({\mathbb{P}}^n\)). \end{exercise} \begin{quote} Next time: Obstruction theory of sheaves, T1 lifting as a way to show unobstructedness. \end{quote} \hypertarget{characterization-of-smoothness-thursday-april-16th}{% \section{Characterization of Smoothness (Thursday April 16th)}\label{characterization-of-smoothness-thursday-april-16th}} \begin{quote} Recap from last time: the cotangent complex answers an extension problem. \end{quote} Given \(X \xrightarrow{f} Y\) and \(Y \hookrightarrow Y'\) a square zero thickening. When can the pullback diagram be filled in? \begin{center} \begin{tikzcd} X \ar[r, dotted] \ar[d] \arrow[dr, phantom, "\scalebox{1.5}{\color{black}$\lrcorner$}" , very near start, color=black] & X' \ar[d, dotted] \\ Y \ar[r] & Y' \end{tikzcd} \end{center} \begin{itemize} \tightlist \item The existence is governed by \(\mathrm{obs} \in \operatorname{Ext}^2( L_{X/Y}, f^* I)\) \item The number of extensions by \(\operatorname{Ext}^1( L_{X/Y}, f^* I)\) \item The automorphisms by \(\operatorname{Ext}^0( L_{X/Y}, f^* I)\) \end{itemize} Suppose we're considering \(k[\varepsilon] \to k\), where \(L_{X_{/k}} = \Omega_{X_{/k}}\), and \(H^*(T_{X_{/k}})\) houses the obstruction theory. For an embedded deformation \(X \hookrightarrow Y\), we have \begin{center} \begin{tikzcd} X \ar[r, dotted] & X' \ar[d, dotted] \\ Y \ar[r] & Y \times_{\operatorname{Spec}k} \operatorname{Spec}k[\varepsilon] \end{tikzcd} \end{center} then \(L_{X/Y} = I/I^2 [1] = N_{X/Y}^\vee[1]\) and \begin{align*} \mathrm{obs} \in \operatorname{Ext}^2(N^\vee[1], {\mathcal{O}}) = \operatorname{Ext}^1(N^\vee, {\mathcal{O}}) = H^1(N) .\end{align*} and similarly \(\mathrm{def} = H^0(N)\) and \(\operatorname{Aut}= 0\). For \(X \xrightarrow{f} Y\), we can think of this as an embedded deformation of \(\Gamma \subset X \times Y\), in which case \(N^\vee= F^* \Omega_{Y_{/k}}\). Then \(\mathrm{obs}, \mathrm{def} \in H^{1, 0}(f^* T_{X_{/k}})\) respectively and \(\operatorname{Aut}= 0\). There is an exact triangle \begin{align*} f^* L_{Y_{/k}} \to L_{X_{/k}} \to L_{X/Y} \to f^* L_{Y_{/k}}[1] .\end{align*} \hypertarget{t1-lifting}{% \subsection{T1 Lifting}\label{t1-lifting}} This will give a criterion for a pro-representable functor to be smooth. We've seen a condition on \(F\) with obstruction theory for the hull to be smooth, namely \(\mathrm{obs}(F) = 0\). However, often \(F = h_{R}\) will have \(R\) smooth with a natural obstruction theory for which \(\mathrm{obs}(F) \neq 0\). \begin{example}[?] For \(X_{/k}\) smooth projective, the picard functor \({\operatorname{Pic}}_{X_{/k}}\) is smooth because we know it's an abelian variety. We also know that the natural obstruction space is \(\mathrm{obs} = H^2({\mathcal{O}}_{X})\), which may be nonzero. We could also have abstract deformations given by \(H^2(T_{X})\) Given \(A \in \operatorname{Art}_{/k}\) and \(M\) a finite length \(A{\hbox{-}}\)module, we can form the ring \(A \oplus M\) where \(M\) is square zero and \(A\curvearrowright M\) by the module structure. This yields \begin{align*} 0 \to M \to A \oplus M \to A \to 0 \end{align*} The explicit ring structure is given by \((x, y) \cdot (x, y') = (xx', x'y + xy')\). \end{example} \begin{proposition}[Characterization of Smoothness] Assume \(\operatorname{ch}k =0\) and \(F\) is a pro-representable deformation functor, so \(F = \hom(R, \cdot)\) where \(R\) is a complete local \(k{\hbox{-}}\)algebra with \(\dim t_{R} < \infty\). Then \(R\) is smooth\footnote{I.e. \(R \cong k[[t_{R}^\vee]]\).} over \(k\) \(\iff\) for all \(A\in \operatorname{Art}_{/k}\) and all \(M, M' \in A{\hbox{-}}\text{mod}\) finite dimensional with \(M \twoheadrightarrow M'\), we have \begin{align*} F(A\oplus M) \twoheadrightarrow F(A\oplus M') .\end{align*} \end{proposition} \hypertarget{proof-of-proposition-2}{% \subsubsection{Proof of Proposition}\label{proof-of-proposition-2}} \begin{observation} First observe that \(\ker(F(A\oplus M) \to F(A)) = \ker(\hom(R, A\oplus M) \to \hom(R, A))\), note that if we have two morphisms \begin{center} \begin{tikzcd} R \ar[r] & R \ar[r, shift left=0.75ex, "g \oplus g"] \ar[r, shift right=0.75ex, "f \oplus g'"'] & A \oplus M \end{tikzcd} \end{center} denoting these maps \(h, h'\) we have \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item \(g-g' \in \operatorname{Der}_{k}(R, M)\), since \begin{align*} (h-h')(x, y) &= h(x)h(y) - h'(x) h'(y) \\ &= (f(x)f(y), f(x)g(y) + f(y)g(x) ) - (f(x)f(y), f(x) g'(y) + f(y) g'(x)) \\ &= f(x)(g-g')(y) + f(y)(g-g')(x) .\end{align*} \item Given \(g: R\to A\oplus M\) and \(\theta \in \operatorname{Der}_{k}(R, M)\), then \(g + \theta: R \to A\oplus M\). \end{enumerate} \end{observation} We conclude that the fibers are naturally torsors for \(\operatorname{Der}_{k}(R, M)\) if nonempty. It is in fact a canonically trivial torsor, since there is a distinguished element in each fiber. Thus to show the following, it is enough to show surjection on fibers and trivial extensions go to trivial ones, then \(\operatorname{Der}_{k}(R, M) \to \operatorname{Der}_{k}(R, M')\) with \(0\mapsto 0\). \begin{center} \begin{tikzcd} F(A\oplus M) \ar[rr] \ar[rd] & & F(A\oplus M') \ar[ld] \\ & F(A) & \end{tikzcd} \end{center} The criterion for \(F\) being surjective is equivalent to \begin{align*} \operatorname{Der}_{k}(R, M) &\twoheadrightarrow\operatorname{Der}_{k}(R, M') \\ \\ &\Downarrow \qquad \text{identified as }\\ \\ \hom_{R}(\Omega_{R_{/k}}, M) &\twoheadrightarrow\hom(\Omega_{R_{/k}'}, M') .\end{align*} \begin{warnings} \(\Omega_{R_{/k}}\) is complicated. An example is \begin{align*} \Omega_{k[[x]]/k} \otimes k((x)) = \Omega_{k((x))/k} .\end{align*} which is an infinite dimensional \(k((x))\) vector space. \end{warnings} Here we only need to consider the completions \(\hom_{R}(\widehat{\Omega}_{R_{/k}}, M) \twoheadrightarrow\hom(\widehat{\Omega}_{R_{/k}'}, M') = k[[x]]~dx\). \begin{fact} In characteristic zero, \(R?k\) is smooth iff \(\widehat{\Omega}_{R_{/k}}\) is free. \end{fact} Thus the surjectivity condition is equivalent to checking that \(\hom(\widehat{\Omega}_{R_{/k}}, {\,\cdot\,})\) is right-exact on finite length modules. This happens iff \(\widehat{\Omega}\) are projective iff they are free. \begin{fact}[from algebra] Uses an algebra fact: for a complete finitely-generated module \(M\) over a complete ring, then \(M\) is free if \(M\) projective with respect to sequences of finite-length modules. Over a local ring, finitely-generated and projective implies free. \end{fact} \begin{remark} This is powerful -- allows showing deformations of Calabi-Yaus are unobstructed! \end{remark} \begin{definition}[Calabi-Yau] A smooth projective \(X_{/k}\) is \textbf{Calabi-Yau} iff \begin{align*} \omega_{x} \cong {\mathcal{O}}_{x} ,\end{align*} i.e.~the canonical bundle is trivial. \end{definition} \begin{proposition}[?] \(X_{/k}\) CY with \(H^0(T_{X}) = 0\) (implying that the deformation functor \(F\) of \(X\) is pro-representable, say by \(R\), and has no infinitesimal automorphisms) has unobstructed deformations, i.e.~\(R\) is smooth of dimension \(H^1(T_{X})\). \end{proposition} Note that \(H^2(T_{X}) \neq 0\) in general, so this is a finer criterion. \begin{example}[?] Take \(X \subset {\mathbb{P}}^4\) a smooth quintic threefold. \begin{itemize} \item By adjunction, this is Calabi-Yau since \begin{align*} \omega_{x} = \omega_{{\mathbb{P}}^4}(5) \mathrel{\Big|}_{X} = {\mathcal{O}}_{x} .\end{align*} \item By Lefschetz, \begin{align*} H^i_\mathrm{sing} ({\mathbb{P}}^4, {\mathbb{C}}) &\xrightarrow{\cong} H^i_{\mathrm{sing}}(X, {\mathbb{C}}) && \text{except in middle dimension} \\ \\ &\Downarrow \quad \text{ implies} \\ \\ H^{3, 1} &= H^{1, 3} = 0 .\end{align*} \item By Serre duality, \begin{align*} H^0(T_{x}) &= 0 \cong H^4(\Omega_{x} \otimes\omega_{x}) \\ \\ &\Downarrow \quad\text{implies} \\ \\ H^3(\Omega_{x}) &= H^{3, 1} = 0 .\end{align*} \end{itemize} \end{example} \begin{exercise}[?] There are nontrivial embedded deformations that yield the same abstract deformations, write them down for the quintic threefold. \end{exercise} \begin{claim} The abstract moduli space here is given by \(\operatorname{PGL}(5) \setminus\operatorname{Hilb}\) where \(\operatorname{Hilb}\) is smooth. \end{claim} \hypertarget{proof-that-obstructions-to-deformations-of-calabi-yaus-are-unobstructed}{% \subsubsection{Proof that obstructions to deformations of Calabi-Yaus are unobstructed}\label{proof-that-obstructions-to-deformations-of-calabi-yaus-are-unobstructed}} We need to show that for any \(M \twoheadrightarrow M'\) that \begin{align*} F(A\oplus M) \twoheadrightarrow F(A\oplus M') .\end{align*} The fibers of the LHS are extensions from \(A\) to \(A\oplus M\), and the RHS are extensions of \(X/A\)? By dualizing, we need to show \(H^1(T_{X/A}\otimes M ) \twoheadrightarrow H^1(T_{X/A} \otimes M')\) since the LHS is \(\operatorname{Ext}^1(\Omega_{X/A}, M)\). We want the bottom map here to be surjective: \begin{center} \begin{tikzcd} X \ar[d] & X' \ar[d] \\ \operatorname{Spec}A \ar[r, hook] & \operatorname{Spec}A \oplus M \end{tikzcd} \end{center} \begin{fact}[Important] For \(X/A\) a deformation of a CY, \(H^*(T_{X/A})\) is free. This will finish the proof, since the map is given by \(H^1(T_{X/A}) \otimes M \twoheadrightarrow H^1(T_{X/A}) \otimes M'\) by exactness. This uses the fact that there's a spectral sequence \begin{align*} \operatorname{Tor}_{q}(H^p(T_{X/A}), M) \implies H^{p+q} (T_{X/A} \otimes M) \end{align*} which follows from base change and uses the fact that \(T_{X/A}\) is flat. \end{fact} We'll be looking at \(\operatorname{Tor}_{1}(H^0(T_{X/A}), M)\) which is zero by freeness. Hodge theory is now used: by Deligne-Illusie, for \(X\xrightarrow{f} S\) smooth projective, taking pushforwards \(R^p f_* \Omega^q_{X_{/S}}\) are free (coming from degeneration of Hodge to de Rham) and commutes with base change. \begin{remark} This implies that \(\omega_{X/A} = {\mathcal{O}}_{X}\) is trivial. Using Deligne-Illusie, since \(\omega\) is trivial on the special fiber, \(H^0(\omega_{X/A}) = A\) is free of rank 1. We thus have a section \({\mathcal{O}}_{X} \to \omega_{X/A}\) which is an isomorphism by flatness, since it's an isomorphism on the special fiber. \end{remark} \begin{remark} By Serre duality, \(H^1(T_{X/A}) = H^{n-1}(\Omega_{X/A} \otimes\omega_{X/A}) ^\vee= H^{n-1}(\Omega_{X/A})^\vee\), which is free by Deligne-Illusie. This also holds for \(H^0(T_{X/A}) = H^n(\Omega_{X/A})^\vee\) is free. \end{remark} Thus deformations of Calabi-Yaus are unobstructed. \hypertarget{remarks}{% \subsubsection{Remarks}\label{remarks}} \begin{remark} In fact we need much less. Take \(A_{n} = k[t] / t^n\), then consider \begin{center} \begin{tikzcd} 0 \ar[r] & A_n \ar[r] & A_n[\varepsilon] \ar[r] & A_n \\ 0 \ar[r] \ar[u, equal] & A_n \ar[r] \ar[u, equal] & A_n \oplus \varepsilon A_n \ar[r] \ar[u, equal] & A_n \ar[u, equal] \end{tikzcd} \end{center} For a deformation \(X/A_{n}\), let \(T^1(X/A_{n}) = \ker(F(A_{n}[\varepsilon]) \to F(A_{n}) )\), the fiber above \(X/A_{n}\). Then Kuramata shows that one only needs to show surjectivity for these kinds of extensions, which is quite a bit less. \end{remark} In the T1 lifting theorem, the condition is equivalent to the following: For any deformation \(X/A_{n+1}\), there is a map \begin{align*} T^1(X/A_{n+1}) \to T^1(X\otimes A_{n} / A_{n}) .\end{align*} and surjectivity is equivalent to the lifting condition. In the CY situation, the extension group \(T^1(X/A_{n+1}) = H^1(T_{X/A_{n+1}})\) and the RHS is \(H^1(T_{X\otimes A_{n} / A_{n}})\). So the slogan for the T1 lifting property is the following: \begin{slogan} If the deformation space is free and commutes with base change, then deformations are unobstructed. \end{slogan} Commuting with base change means the RHS is \(H^1(T_{X/A_{n}}) \otimes A_{n}\), so we just need to show it's free? \hypertarget{monday-april-27th}{% \section{Monday April 27th}\label{monday-april-27th}} \hypertarget{principle-of-galois-cohomology}{% \subsection{Principle of Galois Cohomology}\label{principle-of-galois-cohomology}} Let \(\ell_{/k}\) a galois extension and \(X_{/k}\) some ``object'' for which it makes sense to associate another object over \(\ell\). We'll prove that there's a correspondence \begin{align*} \left\{{\substack{ \ell_{/k}, \text{ twisted forms} \\ Y \text{ of } X_{/k} }}\right\} &\rightleftharpoons H^1(\ell_{/k}, \operatorname{Aut}(X_{/\ell})) .\end{align*} Recall that \(\operatorname{PGL}(n ,\ell) \coloneqq\operatorname{GL}(n ,\ell) / \ell^{\times}\). \begin{example}[?] Let \(X = {\mathbb{P}}^{n-1}/k\), then \(H^1(\ell_{/k}, \operatorname{PGL}(n, \ell)\) parameterizes twisted forms of \({\mathbb{P}}^{n-1}\), e.g.~for \(n=2\) twisted forms of \({\mathbb{P}}^1\) and plane curves. \end{example} \begin{example}[?] Take \(X = M_{n}(k)\) the algebra of \(n\times n\) matrices. Then by a theorem (Skolern-Noether) \(\operatorname{Aut}(M_{n}(k)) = \operatorname{PGL}(n, k)\). Thus \(H^1(\ell_{/k}, \operatorname{PGL}(n, k))\) also parameterizes twisted forms of \(M_{n}(k)\) in the category of unital (not necessarily commutative) \(k{\hbox{-}}\)algebras. These are exactly central simple algebras \(A_{/k}\) where \(\dim_{k} A = n^2\) with center \(Z(A) = k\) with no nontrivial two-sided ideals. By taking \(\ell = k^{s}\), we get a correspondence \begin{align*} \left\{{\substack{\text{CSAs} A_{/k} \text{ of degree } n}}\right\} &\rightleftharpoons \left\{{\substack{\text{ Severi-Brauer varieties of dimension n-1} }}\right\} .\end{align*} Taking \(n=2\) we obtain \begin{align*} \left\{{\substack{\text{Quaternion algebras } A_{/k}}}\right\} &\rightleftharpoons \left\{{\substack{\text{Genus 0 curves } \ell_{/k}}}\right\} .\end{align*} \end{example} \hypertarget{the-weil-descent-criterion}{% \subsection{The Weil Descent Criterion}\label{the-weil-descent-criterion}} Fix \(\ell_{/k}\) finite Galois with \(g \coloneqq\operatorname{Aut}(\ell_{/k})\). \begin{enumerate} \def\labelenumi{\arabic{enumi}.} \item \(X_{/k} \to X_{/\ell}\) with a \(g{\hbox{-}}\)action. \item What additional data on an \(\ell{\hbox{-}}\)variety \(Y_{/\ell}\) do we need in order to ``descend the base'' from \(\ell\) to \(k\)? \end{enumerate} For \(\sigma \in g\), write \(\ell^\sigma\) to denote \(\ell\) given the structure of an \(\ell{\hbox{-}}\)algebra via \(\sigma: \ell \to \ell^\sigma\). If \(X_{/\ell}\) is a variety, so is \(X^\sigma_{/\ell}\)? \begin{center} \begin{tikzcd} X^\sigma\ar[dr, dotted] \ar[r]\ar[d] & X \ar[d] \\ \operatorname{Spec}\ell^\sigma \ar[r, "f"] & \operatorname{Spec}\ell \end{tikzcd} \end{center} where \(f\) is the map induced on \(\operatorname{Spec}\) by \(\sigma\). We can also think of these on defining equations: \begin{align*} X &= \operatorname{Spec}\ell[t_{1}, \cdots, t_{n}] / \left\langle{p_{1}, \cdots, p^n}\right\rangle \\ X^\sigma &= \operatorname{Spec}\ell[t_{1}, \cdots, t_{n}] / \left\langle{\sigma_{p_{1}}, \cdots, \sigma{p^n}}\right\rangle \\ .\end{align*} For \(X_{/k}, X_{/\ell}\), we canonically identify \(X\) with \(X^\sigma\) by the map \(f_\sigma: X \xrightarrow{\cong} X^\sigma\), a canonical isomorphism of \(\ell{\hbox{-}}\)varieties. We thus have \begin{center} \begin{tikzcd} X \ar[r, "f_\sigma"] \ar[rr, bend left, "f_{\sigma \tau}"] & X^\sigma \ar[r, "f_\sigma"] & X^{\sigma \tau} \end{tikzcd} \end{center} under a ``cocycle condition'' \(f_{\sigma \tau} = {}^\sigma f_\tau \circ f_\sigma\). \begin{theorem}[Weil] Given \(Y_{/\ell}\) quasi-projective and \(\forall \sigma \in g\) we have descent datum \(f_\sigma: Y\xrightarrow{\cong} Y^\sigma\) satisfying the above cocycle condition, and there exists a unique \(X_{/k}\) such that \(X_{/\ell} \xrightarrow{\cong} Y_{/\ell}\) and the descent data coincide. \end{theorem} \hypertarget{an-application}{% \subsubsection{An Application}\label{an-application}} Let \(X_{/k}\) be a quasiprojective variety and \(Y_{/k}\) and \(\ell_{/k}\) twisted forms. Then \(a_{0} \in Z' (\ell_{/k}, \operatorname{Aut}X)\). Conversely, we have the following: \begin{definition}[Twisted Descent Data] Let \(a_{0}\) be such a cocycle and \(\left\{{s_\sigma: X\to X^\sigma}\right\}\) be descent datum attached to \(X\). Define twisted descent datum \(g_\sigma \coloneqq f_\sigma \circ a_\sigma\) from \begin{align*} X /\ell\xrightarrow{a_\sigma} X_{/\ell} \xrightarrow{f_\sigma} X^\sigma / \ell .\end{align*} \end{definition} \begin{exercise}[?] Check that \(g_\sigma\) satisfies the cocycle condition, so by Weil uniquely determines a (\(k{\hbox{-}}\)model) \(Y_{/k}\) of \(X_{/\ell}\). \end{exercise} \begin{example}[?] Let \(G_{/k}\) be a smooth algebraic group and \(X_{/k}\) a torsor under \(G\). Then \({\operatorname{Aut}}(G) \supset \operatorname{Aut}_{G{\hbox{-}}\text{torsor}} (G) = G\), since in general the translations will only be a subgroup of the full group of automorphisms. Then \begin{align*} H^1(\ell_{/k}, G) \to H^1(\ell_{/k}, \operatorname{Aut}G) \end{align*} defines a twisted form \(X\) of \(G\). How do you descend the torsor structure? This is possible, but not covered in Bjoern's book! This requires expressing the descent data more functorially -- see the book on Neron models. \end{example} \hypertarget{the-cohomology-theory}{% \subsection{The Cohomology Theory}\label{the-cohomology-theory}} \hypertarget{motivation}{% \subsubsection{Motivation}\label{motivation}} Let \(G_{/k}\) be a smooth connected commutative algebraic group where \(\operatorname{ch}k\) does not divide \(n\), so the map \([n]: G \to G\) is an isogeny. Then \begin{align*} 0 \to G[n] (k^{s} ) \to G(k^{s} ) \xrightarrow{[n]} G(k^{s} ) \to 0 \end{align*} is a SES of \(g = \operatorname{Aut}(k^{s}_{/k}){\hbox{-}}\)modules. \begin{claim} Taking the associated cohomology sequence yields the Kummer sequence: \begin{align*} 0 \to G(k) / nG(k) \to H^1(k, G[n]) \to H^1(k, G)[n] \to 0 \end{align*} where the RHS is the \textbf{Weil--Châtelet} group and the LHS is the \textbf{Mordell-Weil} group. \end{claim} For \(g\) a profinite group, a commutative discrete \(g{\hbox{-}}\)group is by definition a \(g{\hbox{-}}\)module. These form an abelian category with enough injectives, so we can take right-derived functors of left-exact functors. We will consider the functor \begin{align*} A \mapsto A^g \coloneqq\left\{{x\in A {~\mathrel{\Big|}~}\sigma x = x ~\forall \sigma\in g}\right\} ,\end{align*} then define \(H^i(g, A)\) to be the \(i\)th right-derived functor of \(A \mapsto A^\sigma\). This is abstractly defined by taking an injective resolution, applying the functor, then taking cohomology. A concrete description is given by \(C^n(g, A) = {\operatorname{Map}}(g^n, A)\) with \begin{align*} d: C^n(g, A) &\to C^{n+1}(g, A) \\ (df)(\sigma_{1}, \cdots, \sigma_{n+1} &\coloneqq \sigma_{1} f(\sigma_{2}, \cdots, \sigma_{n+1}) \\ &\qquad + \sum_{i=1}^n (-1) f(\sigma _1, \cdots, \sigma_{i-1}, \sigma_{i}, \sigma_{i+1}, \cdots, \sigma_{n+1}) \\ &\qquad + (-1)^{n+1} f(\sigma_{1}, \cdots, \sigma_{n}) .\end{align*} Then \(d^2 = 0\), \(H^n\) is kernels mod images, and this agrees with \(H^1\) as defined before with \(H^0 = A^g\). We'll see that that \begin{align*} H^i(g, A) = \varinjlim_{U} G^i(g/U, A^U) .\end{align*} If \(g\) is finite, \(A\) is a \(g{\hbox{-}}\)module \(\iff\) \(A\) is a \({\mathbb{Z}}[g]{\hbox{-}}\)module, and thus \begin{align*} A^g = \hom_{{\mathbb{Z}}[g]{\hbox{-}}\text{mod}}({\mathbb{Z}}, A) .\end{align*} where \({\mathbb{Z}}\) is equipped with a trivial \(g{\hbox{-}}\)action. We can thus think of \begin{align*} H^i(g, A) = \operatorname{Ext}^i_{{\mathbb{Z}}[g]}({\mathbb{Z}}, A) .\end{align*} \begin{quote} The end! \end{quote} \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}