Maps that are merely continuous are poorly behaved, so we may want to impose extra structure. This can be done by imposing restrictions on the transition functions, defined as \begin{align*} t_{uv} \mathrel{\vcenter{:}}=\varphi_V \to \varphi_U ^{-1} : \varphi_U(U \cap V) \to \varphi_V(U \cap V) .\end{align*}

Is this a strictly increasing hierarchy? It’s not clear e.g. that every \(C^k\) manifold is PL.

Fix a copy of \({\mathbb{R}}\) and form a single chart \({\mathbb{R}}\xrightarrow{\operatorname{id}} {\mathbb{R}}\). There is only a single transition function, the identity, which is smooth. But consider \begin{align*} X &\to {\mathbb{R}}\\ t &\mapsto t^3 .\end{align*} This is also a smooth structure on \(X\), since the transition function is the identity. This yields a different smooth structure, since these two charts don’t like in the same maximal atlas. Otherwise there would be a transition function of the form \(t_{VU}: t\mapsto t^{1/3}\), which is not smooth at zero. However, the map \begin{align*} X &\to X \\ t &\mapsto t^3 .\end{align*} defines a diffeomorphism between the two smooth structures.

Note that here we used the existence of a global frame, i.e. a trivialization of the tangent bundle, so this doesn’t quite work for e.g. \(S^2\).

Kervaire-Milnor: \(S^7\) admits 28 smooth structures, which form a group.

Let \begin{align*} V &\mathrel{\vcenter{:}}=\left\{{a^2 + b^2 + c^2 + d^3 + e^{6k-1} = 0}\right\} \subseteq {\mathbb{C}}^5 \\ S_{\varepsilon}&\mathrel{\vcenter{:}}=\left\{{ {\left\lvert {a} \right\rvert}^2 + {\left\lvert {b} \right\rvert}^2 + {\left\lvert {c} \right\rvert}^2 + {\left\lvert {d} \right\rvert}^2 + {\left\lvert {e} \right\rvert}^2 = 1}\right\} .\end{align*} Then \(V_k \cap S_{\varepsilon}\cong S^7\) is a homeomorphism, and taking \(k=1,2,\cdots, 28\) yields the 28 smooth structures on \(S^7\). Note that \(V_k\) is the cone over \(V_k \cap S_{\varepsilon}\).

? Admits a smooth structure, and \(\mkern 1.5mu\overline{\mkern-1.5muV\mkern-1.5mu}\mkern 1.5mu_k \subseteq {\mathbb{CP}}^5\) admits no smooth structure.

What’s special in dimension 4? Recall the Kirby-Siebenmann invariant \(\operatorname{ks}(x) \in H^4(X; {\mathbb{Z}}_2)\) for \(X\) a topological manifold where \(\operatorname{ks}(X) = 0 \iff X\) admits a PL structure, with the caveat that \(\dim X \geq 5\). We can use this to cook up an invariant of 4-manifolds.

Recall that in \(\dim X\geq 7\), every PL manifold admits a smooth structure, and we can note that \begin{align*} H^4(X; {\mathbb{Z}}_2) = H^4(X \times{\mathbb{R}}; {\mathbb{Z}}_2) = {\mathbb{Z}}_2, .\end{align*} since every oriented 4-manifold admits a fundamental class. Thus \begin{align*} \operatorname{ks}(X) = \begin{cases} 0 & X \times{\mathbb{R}}\text{ admits a PL and smooth structure} \\ 1 & X \times{\mathbb{R}}\text{ admits no PL or smooth structures }. \end{cases} \end{align*}

\(\operatorname{ks}(X) \neq 0\) implies that \(X\) has no smooth structure, since \(X \times{\mathbb{R}}\) doesn’t. Note that it was not known if this invariant was ever nonzero for a while!

Note that \(H^2(X; {\mathbb{Z}})\) admits a symmetric bilinear form \(Q_X\) defined by \begin{align*} Q_X: H^2(X; {\mathbb{Z}})^{\otimes 2} &\to {\mathbb{Z}}\\ \alpha \otimes\beta &\mapsto \int_X \alpha\wedge \beta \mathrel{\vcenter{:}}=(\alpha \smile\beta)([X]) .\end{align*} where \([X]\) is the fundamental class.

Proving the following theorems is the main goal of this course:

Note that preservation of a bilinear form is a stand-in for “being an element of the orthogonal group,” where we only have a lattice instead of a full vector space.

There is a map \(H^2(X; {\mathbb{Z}}) \xrightarrow{PD} H_2(X; {\mathbb{Z}})\) from Poincaré , where we can think of elements in the latter as closed surfaces \([\Sigma]\), and \begin{align*} {\left\langle { \Sigma_1 },~{ \Sigma_2 } \right\rangle} = \text{signed number of intersections points of } \Sigma_1 \pitchfork\Sigma_2 .\end{align*} Note that Freedman’s theorem is only about homeomorphism, and is not true smoothly. This gives a way to show that two 4-manifolds are homeomorphic, but this is hard to prove! So we’ll black-box this, and focus on ways to show that two smooth 4-manifolds are not diffeomorphic, since we want homeomorphic but non-diffeomorphic manifolds.

Note that Freedman’s theorem implies that there exists topological 4-manifolds with no smooth structure.

This comes from Gram-Schmidt, and restricts what types of intersection forms can occur.

Last time we showed \({\mathbb{R}}^1\) had a unique smooth structure, so now we’ll do this for \({\mathbb{R}}^2\). The strategy of solving a differential equation, we’ll now sketch the proof.

A manifold \(M\in {\mathsf{sm}}{\mathsf{Mfd}}({\mathbb{R}})\) admits an almost-complex structure iff \(TM\) admits a reduction of structure group \(\operatorname{GL}_{2n}({\mathbb{R}}) \to \operatorname{GL}_n({\mathbb{C}})\).

Let \(e\in T_p X\) and \(e\neq 0\), then if \(X\) is a surface then \(\left\{{e, J_p e}\right\}\) is a basis of \(T_p X\), where \(J_p\) is the restriction of \(J\) to \(T_p X\):

See the Newlander-Nirenberg theorem, a result in complex geometry.

Recall that we discussed various structures on manifolds: PL, continuous, smooth, complex-analytic, etc. We can characterize these by their sheaves of functions, which we’ll denote \({\mathcal{O}}\). For example, \({\mathcal{O}}\mathrel{\vcenter{:}}= C^0({-}; {\mathbb{R}})\) for topological manifolds, and \({\mathcal{O}}\mathrel{\vcenter{:}}= C^ \infty ({-}; {\mathbb{R}})\) is the sheaf for smooth manifolds. Note that this also works for PL functions, since pullbacks of PL functions are again PL. For complex manifolds, we set \({\mathcal{O}}\) to be the sheaf of holomorphic functions.

Let \(\pi: \mathcal{E}\to X\) be a vector bundle, so we have local trivializations \(\pi ^{-1} (U) \xrightarrow{h_u} Y^d \times U\) where we take either \(Y={\mathbb{R}}, {\mathbb{C}}\), such that \(h_v \circ h_u ^{-1}\) preserves the fibers of \(\pi\) and acts linearly on each fiber of \(Y\times(U \cap V)\). Define \begin{align*} t_{UV}: U \cap V \to \operatorname{GL}_d(Y) \end{align*} where we require that \(t_{UV}\) is continuous, smooth, complex-analytic, etc depending on the context.

We abuse notation: \(\mathcal{E}\) is also a sheaf, and we write \(\mathcal{E}(U)\) to be the set of sections \(s: U\to \mathcal{E}\) where \(s\) is continuous, smooth, holomorphic, etc where \(\pi \circ s = \operatorname{id}_U\). I.e. a bundle is a sheaf in the sense that its sections form a sheaf.

We’d like a notion of maps between sheaves:

Let \(X = {\mathbb{C}}\) and consider \({\mathcal{O}}\) the sheaf of holomorphic functions and \({\mathcal{O}}^{\times}\) the sheaf of nonvanishing holomorphic functions. The former is a vector bundle and the latter is a sheaf of abelian groups. There is a map \(\exp: {\mathcal{O}}\to {\mathcal{O}}^{\times}\), the exponential map, which is the data \(\exp(U): {\mathcal{O}}(U) \to {\mathcal{O}}^{\times}(U)\) on every open \(U\) given by \(f\mapsto e^f\). There is a kernel sheaf \(2\pi i \underline{{\mathbb{Z}}}\), and we get an exact sequence \begin{align*} 0 \to 2\pi i \underline{{\mathbb{Z}}} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times}\to \operatorname{coker}(\exp) \to 0 .\end{align*}

Let \(U\) be a contractible open set, then we can identify \({\mathcal{O}}^{\times}(U) / \exp({\mathcal{O}}^{\times}(U)) = 1\).

Any \(f\in {\mathcal{O}}^{\times}(U)\) has a logarithm, say by taking a branch cut, since \(\pi_1(U) =0 \implies \log f\) has an analytic continuation. Consider the annulus \(U\) and the function \(z\in {\mathcal{O}}^{\times}(U)\), then \(z\not\in \exp({\mathcal{O}}(U))\) – if \(z=e^f\) then \(f=\log(z)\), but \(\log(z)\) has monodromy on \(U\):

Thus on any sufficiently small open set, \(\operatorname{coker}(\exp) = 1\). This is only a presheaf: there exists an open cover of the annulus for which \({ \left.{{z}} \right|_{{U_i}} }\), and so the naive cokernel doesn’t define a sheaf. This is because we have a locally trivial section which glues to \(z\), which is nontrivial.

Given vector bundles \(V, W\), we have constructions \(V \oplus W, V \otimes W, V {}^{ \vee }, \mathop{\mathrm{Hom}}(V, W) = V {}^{ \vee }\otimes W, \operatorname{Sym}^n V, \bigwedge^p V\), and so on. Some of these work directly for sheaves:

• \(\mathcal{F} \oplus \mathcal{G}(U) \mathrel{\vcenter{:}}=\mathcal{F}(U) \oplus \mathcal{G}(U)\)
• For tensors, duals, and homs \(\mathscr{H}\kern-2pt\operatorname{om}(V, W)\) we only get presheaves, so we need to sheafify.

Note that there is a map \(d: \Omega^p \to \Omega^{p+1}\) where \(\omega\mapsto d \omega\).

Let \(X \in {\mathsf{Mfd}}_{\mathbb{C}}\), we’ll use the fact that \(TX\) is complex-linear and thus a \({\mathbb{C}}{\hbox{-}}\)vector bundle.

Note that \(\Omega^p\) for complex manifolds is \(\bigwedge^p T {}^{ \vee }\), and so if we want to view \(X \in {\mathsf{Mfd}}_{\mathbb{R}}\) we’ll write \(X_{{\mathbb{R}}}\). \(TX_{\mathbb{R}}\) is then a real vector bundle of rank \(2n\).

\(\Omega^p\) will denote holomorphic \(p{\hbox{-}}\)forms, i.e. local expressions of the form \begin{align*} \sum f_I(z_1, \cdots, z_n) \bigwedge dz_I .\end{align*} For example, \(e^zdz\in \Omega^1({\mathbb{C}})\) but \(z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu dz\) is not, where \(dz = dx + idy\). We’ll use a different notation when we allow the \(f_I\) to just be smooth: \(A^{p, 0}\), the sheaf of \((p, 0){\hbox{-}}\)forms. Then \(z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu dz\in A^{1, 0}\).

Note that \(T {}^{ \vee }X_{\mathbb{R}}\otimes _{\mathbb{C}}= A^{1, 0} \oplus A^{0, 1}\) since there is a unique decomposition \(\omega = fdz + gd\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\) where \(f,g\) are smooth. Then \(\Omega^d X_{\mathbb{R}}\otimes_{\mathbb{R}}{\mathbb{C}}= \bigoplus _{p+q=d} A^{p, q}\). Note that \(\Omega_{{\mathsf{sm}}}^p \neq A^{p, q}\) and these are really quite different: the former are more like holomorphic bundles, and the latter smooth. Moreover \(\dim \Omega^p(X) < \infty\), whereas \(\Omega_{{\mathsf{sm}}}^1\) is infinite-dimensional.

Setup: we’ll consider \(TX\) for \(X\in {\mathsf{Mfd}}_{\operatorname{Sm}}\), and let \(g\) be a metric on the tangent bundle given by \begin{align*} g_p: T_pX^{\otimes 2} \to {\mathbb{R}} ,\end{align*} a symmetric bilinear form with \(g_p(u, v) \geq 0\) with equality if and only if \(v=0\).

More generally, \(\mathop{\mathrm{Frame}}(\mathcal{E})\) can be defined for any vector bundle \(\mathcal{E}\), so \(\mathop{\mathrm{Frame}}(X) \mathrel{\vcenter{:}}=\mathop{\mathrm{Frame}}(TX)\). Note that \(\mathop{\mathrm{Frame}}(X)\) is a principal \(\operatorname{GL}_n({\mathbb{R}}){\hbox{-}}\)bundle where \(n\mathrel{\vcenter{:}}=\operatorname{rank}(\mathcal{E})\). This follows from the fact that the transition functions are fiberwise in \(\operatorname{GL}_n({\mathbb{R}})\), so the transition functions are given by left-multiplication by matrices.

A principal \(G{\hbox{-}}\)bundle admits a \(G{\hbox{-}}\)action where \(G\) acts by right multiplication: \begin{align*} P \times G \to P \\ ( (g, x), h) \mapsto (gh, x) .\end{align*} This is necessary for compatibility on overlaps. Key point: the actions of left and right multiplication commute.

The fibers \(P_x \to \left\{{x}\right\}\) of a principal \(G{\hbox{-}}\)bundle are naturally torsors over \(G\), i.e. a set with a free transitive \(G{\hbox{-}}\)action.

Note that \(\mathcal{E}, \overline{\mathcal{E}}\) are genuinely different as complex bundles. There is a conjugate-linear map given by conjugation, i.e. \(L(cv) = \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu L(v)\), where the canonical example is \begin{align*} {\mathbb{C}}^n &\to {\mathbb{C}}^n \\ (z_1, \cdots, z_n) &\mapsto (\mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu, \cdots, \mkern 1.5mu\overline{\mkern-1.5muz_n\mkern-1.5mu}\mkern 1.5mu) .\end{align*}

This is a principal \(G{\hbox{-}}\)bundle for \(G = U_r({\mathbb{C}})\), the invertible matrices \(A_{/{\mathbb{C}}}\) satisfy \(A \overline{A}^t = \operatorname{id}\).

Suppose that \(M^3\) is an oriented Riemannian 3-manifold. Them \(TM\to \mathop{\mathrm{Frame}}(M)\) which is a principal \({\operatorname{SO}}(3){\hbox{-}}\)bundle. The universal cover is the double cover \({\operatorname{SU}}(2) \to {\operatorname{SO}}(3)\), so can the transition functions be lifted? This shows up for spin structures, and we can get a \({\mathbb{C}}^2\) bundle out of this.

A connection \(\nabla\) induces a map \begin{align*} \tilde{\nabla}: \mathcal{E}\otimes\Omega^p &\to \mathcal{E}\otimes\Omega^{p+1} \\ s \otimes \omega &\mapsto \nabla s \wedge w + s\otimes d \omega .\end{align*} where \(\wedge: \Omega^p \otimes\Omega^1 \to \Omega^{p+1}\). The standard example is \begin{align*} d: {\mathcal{O}}&\to \Omega^1 \\ f &\mapsto df .\end{align*} where the induced map is the usual de Rham differential.

Why is this called a connection?

This gives us a way to transport \(v\in \mathcal{E}_p\) over a path \(\gamma\) in the base, and \(\nabla\) provides a differential equation (a flow equation) to solve that lifts this path. Solving this is referred to as parallel transport. This works by pairing \(\gamma'(t) \in T_{ \gamma(t) } X\) with \(\Omega^1\), yielding \(\nabla s = ( \gamma'(t)) = s( \gamma(t))\) which are sections of \(\gamma\).

Note that taking a different path yields an endpoint in the same fiber but potentially at a different point, and \(F_\nabla = 0\) if and only if the parallel transport from \(p\) to \(q\) depends only on the homotopy class of \(\gamma\).

Note: this works for any bundle, so can become confusing in Riemannian geometry when all of the bundles taken are tangent bundles!

Given a local system, we can construct a vector bundle whose transition functions are the same as those of the local system, e.g. for vector bundles this is a fixed matrix, and in general these will be constant transition functions. Equivalently, we can take \(L\otimes_{\mathbb{R}}{\mathcal{O}}\), and \(L\otimes 1\) form flat sections of a connection.

The Čech cohomology \(H^p_{\mathfrak{U}}(X, \mathcal{F})\) with respect to the cover \(\mathfrak{U}\) is defined as \(\ker {{\partial}}^p/\operatorname{im}{{\partial}}^{p-1}\). It is a difficult theorem, but we write \(H^p(X, \mathcal{F})\) for the Čech cohomology for any sufficiently refined open cover when \(X\) is assumed paracompact.

Recall that we computed \(H^p(S^1, \underline{{\mathbb{Z}}} = [{\mathbb{Z}}, {\mathbb{Z}}, 0, \cdots]\).

How to visualize when \(H^1(X; \mathcal{F}) = 0\):

On the intersections, we have \(\operatorname{im}{{\partial}}^0 = \left\{{ (s_{i} - s_{j})_{ij} {~\mathrel{\Big|}~}s_i \in \mathcal{F}(U_i)}\right\}\), which are cocycles. We have \(C^1(X; \mathcal{F})\) are collections of sections of \(\mathcal{F}\) on every double overlap. We can check that \(\ker {{\partial}}^1 = \left\{{ (s_{ij}) {~\mathrel{\Big|}~}s_{ij} - s_{ik} + s_{jk} = 0}\right\}\), which is the cocycle condition. From the exercise from last class, \({{\partial}}^2 = 0\).

Last time \(\underline{{\mathbb{R}}}\) on a manifold \(M\) has a resolution by vector bundles: \begin{align*} 0 \to \underline{{\mathbb{R}}} \hookrightarrow\Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \cdots .\end{align*} This is an exact sequence of sheaves of any smooth manifold, since locally \(d \omega = 0 \implies \omega = d \alpha\) (by the Poincaré \(d {\hbox{-}}\)lemma). We also want to know that \(\Omega^k\) is an acyclic sheaf on a smooth manifold.

Any vector bundle on a smooth manifold is acyclic. Using the fact that \(\Omega^k\) is acyclic and the above resolution of \(\underline{{\mathbb{R}}}\), we can write \(H^k(X; {\mathbb{R}}) = \ker(d_k) / \operatorname{im}d_{k-1} \mathrel{\vcenter{:}}= H^k_{dR}(X; {\mathbb{R}})\).

Now letting \(X \in {\mathsf{Mfd}}_{\mathbb{C}}\), recalling that \(\Omega^p\) was the sheaf of holomorphic \(p {\hbox{-}}\)forms. Locally these are of the form \(\sum_{{\left\lvert {I} \right\rvert} = p} f_I(\mathbf{z}) dz^I\) where \(f_I(\mathbf{z})\) is holomorphic. There is a resolution \begin{align*} 0 \xrightarrow{} \Omega^p \xrightarrow{} A^{p, 0} ,\end{align*} where in \(A^{p, 0}\) we allowed also \(f_I\) are smooth. These are the same as bundles, but we view sections differently. The first allows only holomorphic sections, whereas the latter allows smooth sections. What can you apply to a smooth \((p, 0)\) form to check if it’s holomorphic?

Thus to extend the previous resolution, we should take \begin{align*} 0 \to \Omega^p \hookrightarrow A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 2} \to \cdots .\end{align*} The fact that this is exact is called the Poincaré \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{\hbox{-}}\)lemma.

There are no bump functions in the holomorphic world, and since \(\Omega^p\) is a holomorphic bundle, it may not be acyclic. However, the \(A^{p, q}\) are acyclic (since they are smooth vector bundles and thus admit POUs), and we obtain \begin{align*} H^q(X; \Omega^p) = \ker( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_q) / \operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{q-1}) .\end{align*} Note the similarity to \(H_{\mathrm{dR}}\), using \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\) instead of \(d\). This is called Dolbeault cohomology, and yields invariants of complex manifolds: the Hodge numbers \(h^{p, q}(X) \mathrel{\vcenter{:}}=\dim_{\mathbb{C}}H^q(X; \Omega^p)\). These are analogies:

Note the slight overloading of terminology here!

Let \(M \subset X\) be a submanifold where \(X\) is a smooth oriented \(n{\hbox{-}}\)manifold. Then \(M \hookrightarrow X\) induces a pushforward \(H_n(M; {\mathbb{Z}}) \xrightarrow{\iota_*} H_n(X; {\mathbb{Z}})\) where \(\sigma \mapsto \iota \circ \sigma\). Using Poincaré duality, we’ll identify \(H_{\dim M}(X; {\mathbb{Z}}) \to H^{\operatorname{codim}M}(X; {\mathbb{Z}})\) and identify \([M] = PD( \iota_*( [M]))\). In this case, if \(M\pitchfork N\) then \([M] \cap [N] = [M \cap N]\), i.e. the cap product is given by intersecting submanifolds.

Note that cohomology with \({\mathbb{Z}}\) coefficients can be defined for any topological space, and Poincaré duality still holds.

We’ll be considering \(M = M^4\), smooth 4-manifolds. How to visualize: take a 3-manifold and cross it with time!

Here \(?\) is oriented in the “forward time” direction, and this is a surface at time \(t=0\). Where \(A\cdot B = +1\), since \(\left\{{ e_1, e_2, f_1, f_2 }\right\} = \left\{{ e_x, e_y, e_z, e_t }\right\}\) is a oriented basis for \({\mathbb{R}}^4\). For \(?^2\), switching the order of \(\alpha, \beta\) no longer yields an oriented basis, but in this case it is \(?\) and \(A\cdot B = B \cdot A\). This is because \begin{align*} A \mathrel{\vcenter{:}}= \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \implies \operatorname{det}(A) =-1 && \operatorname{det} \begin{bmatrix} A & \\ & A \end{bmatrix} = 1 .\end{align*}

Let \(M^{2n}\) be an oriented manifold, then the cup product yields a bilinear map \(H^n(M; {\mathbb{Z}}) \otimes H^n(M; {\mathbb{Z}}) \to {\mathbb{Z}}\) which is symmetric when \(n\) is odd and antisymmetric (or symplectic) when \(n\) is even. This is a perfect (or unimodular) pairing (potentially after modding out by torsion) which realizes an isomorphism: \begin{align*} \qty{ H^n(M; {\mathbb{Z}})/{\operatorname{tors}}} {}^{ \vee }&\xrightarrow{\sim} H^n(M; {\mathbb{Z}})/{\operatorname{tors}}\\ \alpha \smile{-}&\mapsfrom \alpha ,\end{align*} where the LHS are linear functionals on cohomology.

Recall the universal coefficients theorem: \begin{align*} H^i(X; {\mathbb{Z}})/{\operatorname{tors}}\cong \qty{ H_i(X; {\mathbb{Z}})/{\operatorname{tors}}} {}^{ \vee } .\end{align*} The general theorem shows that \(H^i(X; {\mathbb{Z}})_{\operatorname{tors}}= H_{i-1}(X; {\mathbb{Z}})_{\operatorname{tors}}\).

Note that if \(M\) is an oriented 4-manifold, then

In particular, if \(M\) is simply connected, then \(H_1(M) = {\mathsf{Ab}}(\pi_1(M)) = 0\), which forces \(A = 0\) and \(\beta_1 = 0\).

How to determine if a lattice is unimodular: take a basis \(\left\{{ e_1, \cdots, e_n }\right\}\) of \(L\) and form the Gram matrix \(M_{ij} \mathrel{\vcenter{:}}=( e_i \cdot e_j) \in \operatorname{Mat}(n\times n, {\mathbb{Z}})^{\operatorname{Sym}}\). Then \((L, \cdot)\) is unimodular if and only if \(\operatorname{det}(M) = \pm 1\) if and only if \(M ^{-1}\) is integral. In this case, the rows of \(M ^{-1}\) will form a basis of the dual basis.

In general, for \(M^{4k}\), the \(H^{2k}/{\operatorname{tors}}\) is unimodular. For \(M^{4k+2}\), the \(H^{2k+1}/{\operatorname{tors}}\) is a unimodular symplectic lattice, which is obtained by replacing the word “symmetric” with “antisymmetric” everywhere above.

If \((L, \cdot)\) is nondegenerate, then Gram-Schmidt will yield an orthonormal basis \(\left\{{ v_i }\right\}\). The number of positive norm vectors is an invariant, so we obtain \({\mathbb{R}}^{p, q}\) where \(p\) is the number of \(+1\)s in the Gram matrix and \(q\) is the number of \(-1\)s. The signature of \((L, {-})\) is \((p, q)\), or by abuse of notation \(p-q\). This is an invariant of the 4-manifold, as is the lattice itself \(H^2(X; {\mathbb{Z}})/{\operatorname{tors}}\) equipped with the intersection form.

There is a perfect pairing called the linking pairing:

\begin{align*} H^i(X; {\mathbb{Q}}/{\mathbb{Z}}) \otimes H^{n-i-1}(X; {\mathbb{Q}}/{\mathbb{Z}}) \to {\mathbb{Q}}/{\mathbb{Z}} .\end{align*}

\(A \cdot B \mathrel{\vcenter{:}}=\sum_{p\in A \cap B} \operatorname{sgn}_p(A, B)\), where \(A \pitchfork B\) and this turns out to be equal to the cup product. This works for topological manifolds – but there are no tangent spaces there, so taking oriented bases doesn’t work so well! You can also view \begin{align*} [A] \smile[\omega] = \int_A \omega .\end{align*}

Recall that a lattice is unimodular if the map \(L\to L {}^{ \vee }\mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}(L, {\mathbb{Z}})\) is an isomorphism, where \(\ell \mapsto \ell \cdot ({-})\). To check this, it suffices to check if the Gram matrix \(M\) of a basis \(\left\{{e_i}\right\}\) satisfies \({\left\lvert { \operatorname{det}M } \right\rvert} = 1\).

Thus

• \(\mathbf{I}_{p, q}\) is odd, unimodular, of signature \((p, q)\).
• \(\mathbf{II}_{p, q}\) is even, unimodular, of signature \((p, q)\) only for \(p \equiv q \pmod 8\).

So there are obstructions to the existence of even unimodular lattices. Other than that, the number of (say) positive definite even unimodular lattices is

Note that the signature of a definite lattice must be divisible by 8.

There is an isometry: \(f: E_8 \to E_8\) where \(f\in O(E_8)\), the linear maps preserving the intersection form (i.e. the Weyl group \(W(E_8)\), given by \(v\mapsto v + (v, e_i) e_i\). The Leech lattice also shows up in the sphere packing problems for dimensions \(2,4,8,24\). See Hale’s theorem / Kepler conjecture for dimension 3! This uses an identification of \(L\) as a subset of \({\mathbb{R}}^n\), namely \(L \otimes_{\mathbb{Z}}{\mathbb{R}}\cong {\mathbb{R}}^{24}\) for example, and the map \(L \hookrightarrow({\mathbb{R}}^{24}, \cdot)\) is an isometric embedding into \({\mathbb{R}}^n\) with the standard form. Connection to classification of Lie groups: root lattices.

If \(M^4\) is a compact oriented 4-manifold and if the intersection form on \(H^2(M; {\mathbb{Z}})\) is indefinite, then the only invariants we can extract from that associated lattice are

• Whether it’s even or odd, and
• Its signature

If the lattice is even, then the signature satisfies \(8\divides p-q\). So Poincaré duality forces unimodularity, and then there are further number-theoretic restrictions. E.g. this prohibits \(\beta_2 =7\), since then the signature couldn’t possibly be 8 if the intersection form is even.

Thus there is a map \(EG \to BG \mathrel{\vcenter{:}}= EG/G\) which has the structure of a principal \(G{\hbox{-}}\)bundle.

Here we use a point \(p\) depending on \(U\) in an orbit to identify orbits \(g\cdot p\) with \(g\), and we want to take transverse slices to get local trivializations of \(U\in BG\). It suffices to know where \(\pi ^{-1} (U) \cong U \times G\), and it suffices to consider \(U \times\left\{{e}\right\}\). Moreover, \(EG\to BG\) is a universal principal \(G{\hbox{-}}\)bundle in the sense that if \(P\to X\) is a universal \(G{\hbox{-}}\)bundle, there is an \(f:X\to BG\).

Link to Diagram

Here bundles will be classified by homotopy classes of \(f\), so \begin{align*} \left\{{\text{Principal $G{\hbox{-}}$bundles} {}_{/ X} }\right\} \rightleftharpoons[X, BG] .\end{align*}

Recall that there is a bijective correspondence \begin{align*} \left\{{\substack{ \text{Principal $G{\hbox{-}}$ bundles} \\ \text{on } X }}\right\} &\rightleftharpoons [X, BG] \end{align*} and there is also a correspondence \begin{align*} \left\{{\substack{ \text{Principal $\operatorname{GL}_d{\hbox{-}}$bundles }\\ \text{on } X }}\right\} &\rightleftharpoons \left\{{\substack{ \text{Principal ${\mathcal{O}}_d{\hbox{-}}$bundles } \\ \text{on } X }}\right\} \end{align*} Using the associated bundle construction, on the LHS we obtain vector bundles \(\mathcal{E}\to X\) of rank \(d\), and on the RHS we have bundles with a metric. In local trivializations \(U \times{\mathbb{R}}^d \to {\mathbb{R}}^d\), the metric is the standard one on \({\mathbb{R}}^d\). This is referred to as a reduction of structure group, i.e. a principal \(\operatorname{GL}_d\) bundle admits possibly different trivializations for which the transition functions lie in the subgroup \(O_d\).

We want to compute \(H^*(BU_d; {\mathbb{Z}})\). Why is this important? Given any complex vector bundle \(\mathcal{E}\to X\) there is an associated principal \(U_d\) bundle by choosing a metric, so we get a homotopy class \([X, BU_d]\). Given any \(f\in [X, BU_d]\) and any \(\alpha\in H^k(BU_d; {\mathbb{Z}})\), we can take the pullback \(f^* \alpha \in H^k(X; {\mathbb{Z}})\), which are Chern classes.

If \(P \mathrel{\vcenter{:}}=\mathop{\mathrm{OFrame}}TX\), then \(f^* w_1\) measures whether \(X\) has an orientation, i.e. \(f^* w_1 = 0 \iff X\) can be oriented. We also have \(f^* w_i(P) = w_i( \mathcal{E} )\) where \(P = \mathop{\mathrm{OFrame}}( \mathcal{E} )\). In general, we’ll just write \(w_i\) for Stiefel-Whitney classes and \(c_i\) for Chern classes.

Last time: the splitting principle. Suppose we have \(\mathcal{E} = L_1 \oplus \cdots \oplus L_r\) and let \(x_i \mathrel{\vcenter{:}}= c_i(L_i)\). Then \(c_k(\mathcal{E})\) is the degree \(2k\) part of \(\prod_{i=1}^r (1 + x_i )\) where each \(x_i\) is in degree \(2\). This is equal to \(e_k(x_1, \cdots, x_r)\) where \(e_k\) is the \(k\)th elementary symmetric polynomial.

The theorem is that any symmetric polynomial is a polynomial in the \(e_i\). For example, \(p_2 = \sum x_i^2\) can be written as \(e_1^2 - 2e_2\). Similarly, \(p_3 = \sum x_i^3 = e_1^3 - 3e_1 e_2 -3e_3\) Note that the coefficients of these polynomials are important for representations of \(S_n\), see Schur polynomials.

Due to the splitting principle, we can pretend that \(x_i = c_i(L_i)\) exists even when \(\mathcal{E}\) doesn’t split. If \(\mathcal{E} \to X\), the individual symbols \(x_i\) don’t exist, but we can write ’ \begin{align*} x_1^3 + \cdots + x_r^3 = e_1^3 - 3e_1 e_2 - 3e_3 \mathrel{\vcenter{:}}= c_1(\mathcal{E})^3 + 3c_1(\mathcal{E}) c_2(\mathcal{E}) + \cdots ,\end{align*} which is a well-defined element of \(H^6(X; {\mathbb{Z}})\). So this polynomial defines a characteristic class of \(\mathcal{E}\), and this can be done for any symmetric polynomial. We can change basis in the space of symmetric polynomials to now define different characteristic classes.

\(\operatorname{ch}(\mathcal{E} \oplus \mathcal{F}) = \operatorname{ch}(\mathcal{E}) + \operatorname{ch}(\mathcal{F})\) and \(\operatorname{ch}( \mathcal{E} \otimes\mathcal{F} ) = \sum_{i,j} e^{x_i + y_j} = \operatorname{ch}( \mathcal{E} ) \operatorname{ch}(\mathcal{F} )\) using the fact that \(c_1(L_1 \otimes L_2) = c_1(L_1) c_1(L_2)\). So \(\operatorname{ch}\) is a “ring morphism” in the sense that it preserves multiplication \(\otimes\) and addition \(\oplus\), making the Chern character even better than the total Chern class.

Let \(X\in {\mathsf{Top}}\) and let \(\operatorname{\mathcal{F}}\) be a sheaf of vector spaces. Suppose \(h^i(X; \operatorname{\mathcal{F}}) \mathrel{\vcenter{:}}=\dim H^i(X; \operatorname{\mathcal{F}}) < \infty\) for all \(i\) and is equal to 0 for \(i \gg 0\).

Last time: we saw that \(\chi({\mathbb{P}}^1, {\mathcal{O}}) = 1\), and we’d like to generalize to holomorphic line bundles on a Riemann surface. This will be the main ingredient for Riemann-Roch.

The locally constant sheaf \(\underline{{\mathbb{C}}}\) is not an \({\mathcal{O}}{\hbox{-}}\)module, i.e. \(\underline{{\mathbb{C}}}(U) \not\in {\mathsf{{\mathcal{O}}(U)}{\hbox{-}}\mathsf{Mod}}\). In fact, \(h^{2i}(X, \underline{{\mathbb{C}}}) = {\mathbb{C}}\) for all \(i\).

Why is every \(L \cong {\mathcal{O}}(D)\) for some \(D\)? Easy to see if \(L\) has meromorphic sections: if \(s\) is a meromorphic section of \(L\), then the following works: \begin{align*} D = \operatorname{Div}(s) = \sum_p {\operatorname{Ord}}_p(s) [p] .\end{align*} Then \({\mathcal{O}}\cong L\otimes{\mathcal{O}}(-D)\) has a meromorphic section \(s s_{-D}\), a global nonvanishing section with \(\operatorname{Div}(s s_{-D} ) = \emptyset\). Proving that every holomorphic line bundle has a meromorphic section is hard!

In general it can be hard to compute \(h^1(L)\), since this is sheaf cohomology (sections over double overlaps, cocycle conditions, etc). On the other hand, \(h^0\) is easy to understand, since \(h^0( \Omega^1_C)\) is the dimension of the global holomorphic sections \(H^0(C, L) = L(C)\). A key tool here is the following:

We showed \({\left\langle {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s},~{t} \right\rangle} = {\left\langle {s},~{Y (t)} \right\rangle}\) where \(Y\) is the adjoint given above. Then the kernel of \(Y\) wound up being where \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\) vanishes, i.e. holomorphic sections of a separate bundle. Here we had

• \(t \in H^0(L\otimes A^{0, 1})\)
• \({\overline{{t}}} \in H^0({\overline{{L}}}\otimes A^{1,0})\)
• \(h\in H^0( L ^{-1} \otimes{\overline{{ L ^{-1} }}})\)

Last time: Serre duality, and we’ll review Riemann-Roch. Recall that this depended on the statement that every holomorphic line bundle \(L\to C\) for \(C\) a complex curve is of the form \(L = {\mathcal{O}}(D)\) for some divisor \(D\). Then \begin{align*} \chi(C, L) = h^0(L) - h^1(L) = \deg L + 1 - g, && \deg L = \int_C c_1(L) ,\end{align*} Serre duality said that the space of sections \(H^1(C; L)\) is naturally isomorphic to \(H^0(C, L ^{-1} \otimes\Omega_C^1) {}^{ \vee }\). Notation: given \(X \in {\mathsf{Mfd}}_{\mathbb{C}}^n\) of complex, dimension \(n\), the canonical bundle is written \(K_X \mathrel{\vcenter{:}}=\Omega_X^n\) and is the sheaf of holomorphic \(n{\hbox{-}}\)forms. Serre duality will generalize: if \(\mathcal{E}\to X\) is a holomorphic vector bundle, then \(H^i(X; \mathcal{E}) \cong H^{n-i}(X; \mathcal{E} {}^{ \vee }\otimes K_X) {}^{ \vee }\). Note that only \(H^0, H^1\) are the only nontrivial degrees for a curve. For 4-manifolds, we’ll have an \(H^2\) as well.

So there is only one genus 0 Riemann surface. What about genus 1?

By Riemann-Roch we know \begin{align*} \chi(C; {\mathcal{O}}) = \deg {\mathcal{O}}+ l - 1 = 0 = h^0({\mathcal{O}}) - h^1({\mathcal{O}}) .\end{align*} We know \(h^0({\mathcal{O}}) = 1\) by the maximum modulus principle and \(h^1(C; {\mathcal{O}}) = 1\). By Serre duality, \(h^0(C, K) = 1\), and since \(\deg K = 2g-2 = 0\). So let \(s\in H^0(C, K)\) by a nonzero section, which we know exists. We then get \({\operatorname{Ord}}_p s = 0\) for all \(p\), so \(s\) vanishes nowhere. But then we get an isomorphism of sheaves, since \(s\) everywhere nonvanishing implies trivial cokernel: \begin{align*} {\mathcal{O}}\xrightarrow{\cdot s} K .\end{align*} So \(K_C = {\mathcal{O}}_C\) if \(g(C) = 1\), and such a Riemann surface is an elliptic curve.

Similarly \(g(C) = 3\) are usually a curve of degree \(4\) in \({\mathbb{CP}}^2\). Severi proof in the 50s: false! issues with building moduli space for \(g\geq 23\). Need to use orbifold structure to take into account automorphisms.

Note that integrating over cohomology classes in mixed degree is just equal to the integral over the top degree terms. Applying this to \(X = C\) a curve and \(\mathcal{E} \mathrel{\vcenter{:}}={\mathcal{O}}\), we obtain \begin{align*} \chi(C, {\mathcal{O}}) = \int_C \operatorname{ch}( {\mathcal{O}}) \mathrm{td}(C) .\end{align*}

We have

• \(\operatorname{ch}({\mathcal{O}}) = e^{c_1({\mathcal{O}})} = e^0 = 1\)

• \(\mathrm{td}(C) \mathrel{\vcenter{:}}=\mathrm{td}(TC) = c_1(TC) / (1- e^{ - c_1(TC) } )\), whose Taylor coefficients are the Bernoulli numbers. We can expand \(x/(1 -e^{-x}) = 1 + (x/2) + (x^2/12) - x^4(720) + \cdots\), and since terms above degree 2 vanish, we have \begin{align*} \cdots &= \int_C 1 + \qty{ 1 + {c_1(TC) \over 2} } \\ &= \int_C \qty{c_1(TC) \over 2 }\\ &= {1\over 2} \chi_{\mathsf{Top}}(C) && \text{Chern-Gauss-Bonnet} \\ &= {2-2g \over 2} \\ &= 1-g .\end{align*}

We thus obtain \begin{align*} \chi(C, L) &= \int_C \operatorname{ch}(L) \mathrm{td}(C) \\ &= \int_C (1 + c_1(L) ) \qty {1 + {c_1(L) \over 2} }\\ &= \int_C c_1(L) + {c_1(TC) \over 2} \\ &= \deg L + 1-g .\end{align*}

Note that this is a better definition of genus than the previous one, which was just the correction term in Riemann-Roch. Here we can define it as \(g \mathrel{\vcenter{:}}= h^1/2\).

We want to use the following formula: \begin{align*} \chi(S, L) = \chi({\mathcal{O}}_S) = {1\over 2}(L^2 - L\cdot K) .\end{align*} This requires knowing \(\chi({\mathcal{O}}_S)\). Applying HRR yields \begin{align*} \chi({\mathcal{O}}_S) &= \int_S {c_1(T)^2 + c_2(T) \over 12}\\ &= \int_S { (-c_1(K))^2 + c_2(T) \over 12}\\ &= {K^2 + \displaystyle\int_S c_2(T) \over 12} ,\end{align*} so we just need to understand \(\int_S c_2(T)\). But for \(n=\operatorname{rank}\mathcal{E}\), \(c_n( \mathcal{E} )\) (the top Chern class) is the fundamental class of a zero locus of a section of \(\mathcal{E}\). Note that \(S \in {\mathsf{Mfd}}_{\mathbb{R}}^4\) is oriented, so \(\int_S c_2(T)\) is the signed number of zeros of a smooth vector field.

Looking at the tangent bundle of the surface, the local sign of an intersection will be the number of incoming directions \(\pmod 2\), i.e. the index of the critical point. Then the signed number of zeros here yields \(1-6+1 = -4 = \chi_{{\mathsf{Top}}}(C)\). More generally, we have \begin{align*} \chi_{{\mathsf{Top}}}(M^n) = \int_C c_{n}(TM) ,\end{align*} the Chern-Gauss-Bonnet formula. We can thus write \begin{align*} \chi({\mathcal{O}}_S) = {K^2 + \chi_{\mathsf{Top}}(S) \over 12 } .\end{align*}

Last time: Riemann-Roch for surfaces, today we’ll discuss some examples. Recall that if \(S \in {\mathsf{Mfd}}_{\mathbb{C}}^2\) is closed and compact (noting that \(S\in {\mathsf{Mfd}}_{\mathbb{R}}^4\)) and \(L\to S\) is a holomorphic line bundle then \begin{align*} \chi(S, L) = \chi({\mathcal{O}}_S) + {1\over 2}(L^2 - L \cdot K) \end{align*} where \(K = c_1(K_S)\) for \(K_S \mathrel{\vcenter{:}}=\Omega_S^2\) the canonical bundle and \(L = c_1(L)\). We also saw \begin{align*} \chi({\mathcal{O}}_S) = {1\over 12}(K^2 + \chi_{{\mathsf{Top}}}(S)) ,\end{align*} where \(\chi_{\mathsf{Top}}\) is the Euler characteristic and is given by \begin{align*} \chi_{\mathsf{Top}}(S) = 2 h^0(S; {\mathbb{C}}) - 2 h^1(S, {\mathbb{C}}) + h^2(S; {\mathbb{C}}) .\end{align*}

Let \(X\) be an algebraic variety, i.e. spaces cut out by polynomial equations, for example \(\left\{{ xy = 0 }\right\} \subseteq {\mathbb{C}}^2\) which has a singularity at the origin. A divisor is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of subvarieties of codimension 1. Note that for a curve \(X\), this recovers the definition involving points. For \(D\) a divisor on \(X\), we associated a bundle \({\mathcal{O}}_X(D)\) which had a meromorphic section with a zero/pole locus whose divisor was precisely \(D\).

Recall the construction: we chose a point, then a trivializing neighborhood where the transition functions where \(V\).

For a higher dimensional algebraic variety or complex manifold, for \(D\) a complex submanifold, pick a chart around a point that the nearby portion of \(D\) to a coordinate axis in \({\mathbb{C}}^n\), which e.g. can be given by \(\left\{{ z_1 = 0 }\right\}\).

As before there’s a distinguished section \(s_D \in H^0(X; {\mathcal{O}}_X(D) )\) vanishing along \(D\). Note that a line bundle is a free rank 1 \({\mathcal{O}}{\hbox{-}}\)module, and analogously here the functions vanishing along \(D\) are \({\mathcal{O}}{\hbox{-}}\)modules generated by (here) \(z_1\).

Last time: we defined \({\operatorname{Pic}}({\mathbb{CP}}^n)\) as the set of line bundles on \({\mathbb{CP}}^n\).

We saw that \({\operatorname{Pic}}(X) \cong H^1(X; {\mathcal{O}}^{\times})\) as groups, noting that \(H^1\) has a natural group structure here. We defined a tautological bundle on \({\mathbb{CP}}^n\) and saw it was isomorphic to \({\mathcal{O}}(-1)\), and moreover \({\mathcal{O}}(H) \cong {\mathcal{O}}(1)\) for \(H\) a hyperplane. The fiber was given by \begin{align*} \mathrm{Taut} &\to {\mathbb{CP}}^n \\ \left\{{ \lambda (x_0, \cdots, x_n) {~\mathrel{\Big|}~}\lambda\in {\mathbb{C}}}\right\} &\mapsto [x_0: \cdots : x_n] ,\end{align*} i.e. the entire line corresponding to the given projective point. We also have \({\mathcal{O}}(H)(U)\) is the sect of rational homogeneous functions \(\phi\) on \(U\) of degree 1 such that \(\operatorname{Div}\phi + H \geq 0\) where \(H \mathrel{\vcenter{:}}=\left\{{x_0 = 0}\right\}\). We want \(\phi/x_0\) to be a well-defined function, so \(\phi\) should scale like \(x_0\) in the sense that \begin{align*} \phi( \lambda x_0, \cdots, \lambda x_n) = \lambda\phi( x_0, \cdots, x_n) .\end{align*} Note that there is a natural map \begin{align*} \mathop{\mathrm{Taut}}\otimes{\mathcal{O}}(H) \xrightarrow{} {\mathcal{O}} ,\end{align*} given by taking the line over a point and evaluating the homogeneous function on that line. Thus \(\mathop{\mathrm{Taut}}\) is the inverse of \({\mathcal{O}}(H)\).

We want to understand what Noether’s formula says for \({\mathbb{CP}}^2\), which requires understanding the canonical bundle \(K_{{\mathbb{CP}}^n}\). We’ll do this by writing down a meromorphic section \(\omega\) (since it’s a meromorphic volume form) which will yield \(K_{{\mathbb{CP}}^n} = {\mathcal{O}}(\operatorname{Div}\omega)\). So take \begin{align*} \omega \mathrel{\vcenter{:}}= x_1^{-1}dx_1 \wedge \cdots \wedge x_n^{-1}dx_n ,\end{align*} noting that we leave out the first coordinate \(x_0\) and divide by coordinates to make this scale-invariant. Here we work in a \({\mathbb{C}}^n\) chart of points of the form \([1: x_1 : \cdots : x_n]\). Where does \(\omega\) have poles? Along \(x_i = 0\) for any \(1\leq i \leq n\), and similarly in any other coordinate chart. We also have a 1st order pole along \(x_0 = 0\). We then get \begin{align*} K_{{\mathbb{CP}}^n} = {\mathcal{O}}(\operatorname{Div}\omega) = {\mathcal{O}}( -H_0 -H_1 - \cdots - H_n) = {\mathcal{O}}(-n-1) ,\end{align*} where \(H_i = \left\{{x_i = 0}\right\}\).

Note that \({\mathbb{CP}}^n\) is like a simplex:

Applying this to \({\mathbb{CP}}^2\), we obtain \begin{align*} K_{{\mathbb{CP}}^2} = {\mathcal{O}}(-3) .\end{align*} What is the intersection form? We know \(H^2({\mathbb{CP}}^2; {\mathbb{Z}}) \cong {\mathbb{Z}}\) and the intersection form is unimodular. So write \({\mathbb{Z}}\mathrel{\vcenter{:}}={\mathbb{Z}}\alpha\) for \(\alpha\) some generator. Then \(\alpha \cdot \alpha = \pm 1\) since \(\operatorname{det}G = \pm 1\) for the Gram matrix for this to be unimodular. Note that \((- \alpha) \cdot (- \alpha) = \pm 1\) with the same sign.

This is because \(c_1 {\mathcal{O}}(H) \cdot c_1 {\mathcal{O}}(H) = H\cdot H = \left\{{ x_0 = 0 }\right\} \pitchfork\left\{{ x_1 = 0 }\right\} = \left\{{ [0:0:1] }\right\}\) here we note that the two hyperplanes can be oriented transversely and intersected. This is an oriented intersection.

Recall Noether’s formula, which was HRR applied to \({\mathcal{O}}\) and the Chern-Gauss-Bonet theorem: \begin{align*} \chi({\mathcal{O}}) &= {1\over 12}(K^2 + \chi_{\mathsf{Top}})\\ &= h^0({\mathcal{O}}) - h^1({\mathcal{O}}) + h^2({\mathcal{O}})\\ &= 1 -1 + 1\\ &= 1 .\end{align*} The right-hand side can be written as \begin{align*} {1\over 12} \qty{ (-3H) \cdot (-3H) + 3} = {1\over 12}(9+3) = 1 .\end{align*}

Consider \({\mathcal{O}}_{{\mathbb{CP}}^n}(d)\), what are its global sections \(H^0({\mathbb{CP}}^n, {\mathcal{O}}_{{\mathbb{CP}}^n}(d))\). Locally we have \({\mathcal{O}}_{{\mathbb{CP}}^n}(d)(U)\) given by holomorphic functions in \((x_0, \cdots, x_n) \in \pi^{-1}(U)\) where \(\pi: {\mathbb{C}}^{n+1} \to {\mathbb{CP}}^n\) and the functions satisfy \(f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})\). The global sections will be the homogeneous degree \(d\) polynomials in the coordinates of \(\mathbf{x}\).

Why does a holomorphic function \(f: {\mathbb{C}}^{n+1} \to {\mathbb{C}}\) such that \(f(\lambda \mathbf{x}) = \lambda^d f(\mathbf{x})\) necessarily a polynomial? Use the result that any such function with at most polynomial growth is itself a polynomial. If \({ \left.{{f}} \right|_{{S^{2d+1}}} }\) is bounded by \(C\), we have \({\left\lVert {f} \right\rVert}_{L^2} \leq C {\left\lvert {x} \right\rvert}^{2d}\). Since \(({{\partial}}_{x_1} \cdots {{\partial}}_{x_k})^d f\) is globally bounded \(k\geq 2d\), applying Liouville’s theorem makes it constant, and so a finite number of derivatives kill \(f\) and this forces it to be polynomial.

So how many homogeneous degree \(d\) functions are there? Here \(h^0({\mathbb{CP}}^n, {\mathcal{O}}(d)) =\) will be the number of linearly independent degree \(d\) polynomials in the variables \(x_0, \cdots, x_n\), which is \({\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{n+1}{d}\right)\kern-.3em\right)} = {n + d\choose n}\), using the fact that monomials span this space.

When we considered \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2\), we implicitly assumed \(T\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}\mkern-1.5mu}\mkern 1.5mu^2\) was a complex rank 2 vector bundle with some purported complex structure.

We had \(\chi({\mathcal{O}}) = {1\over 12} \qty{ K^2 + \chi_{\mathsf{Top}}} = {1\over 12}(3-n^2)\), and since \(3-n^2 \in 12{\mathbb{Z}}\), we have \(n^2 \in 3 + 12{\mathbb{Z}}\subset 3 + 4{\mathbb{Z}}\) and this forces \(n^2 \equiv 3 \pmod 4\).

See similarly to the canonical symplectic structure of the cotangent bundle.

More generally, for \(d: \Omega^p \to \Omega^{p+1}\), \(\sigma(d)\) acts on the frame \(dx_{i_1} \wedge \cdots dx_{i_p}\) in the following way: \begin{align*} \sigma(d)(dx_{i_1} \wedge \cdots \wedge dx_{i_p}) = \sum_y y_y dx_j \wedge dx_{i_1} \wedge \cdots dx_{i_p} \end{align*} where \begin{align*} d: fdx_{i_1} \wedge \cdots \wedge dx_{i_p} \mapsto \sum_j {\frac{\partial f}{\partial x_j}\,} dx_j \wedge \qty{dx_{i_1} \wedge \cdots \wedge dx_{i_p}} .\end{align*} The symbol complex is \begin{align*} \pi^* {\mathcal{O}}\xrightarrow{\sigma(d)} \pi^* \Omega^1 \xrightarrow{\sigma(d)} \pi^* \Omega^2 \to \cdots \to \pi^* \Omega^n \to 0 \end{align*} for \(n\) the dimension. In this case, \(\sigma(d)\) has the same formula everywhere, since it’s \(C^ \infty {\hbox{-}}\)linear: \begin{align*} \sigma(d) = \sum_j y_j dx_j \wedge \qty{\cdots} .\end{align*}

Recall that we set up a differential complex, whose objects were vector bundles and differentials were differential operators (i.e. linear combinations of partial derivatives) in local trivializations. We pulled back to tangent bundles (?) and defined the symbol of an operator, and saw that when taking the symbol complex of the deRham complex. the sequence of maps was given by wedging against a tautological one-form. This was an elliptic complex because the maps became wedging with a covector.

The next theorem computes the cohomology of an elliptic complex using Chern and Todd classes.

Here we define \(\operatorname{ch}( {\mathcal{E}}_{*} ) \mathrel{\vcenter{:}}=\sum_i (-1)^i \operatorname{ch}( \mathcal{E}^i )\). What does it mean to divide by the Euler class? Let \(\left\{{ x_i, -x_i }\right\}\) be the Chern roots of the complexified tangent bundle \(TX\otimes{\mathbb{C}}\), then \({\operatorname{eul}}(X) \mathrel{\vcenter{:}}=\prod x_i\) is the product where we pick one of each of the Chern roots from each of the pairs. The preferred sign to choose is the one for which \(\int_X \prod x_i = \chi_{\mathsf{Top}}(X)\). Dividing just means to take the Chern character, then if it’s divisible by \(\prod x_i\), we do so. We have \begin{align*} \mathrm{td}(TX\otimes{\mathbb{C}}) = \prod_i \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} .\end{align*}

Thus \begin{align*} {\mathrm{td}(TX\otimes{\mathbb{C}}) \over {\operatorname{eul}}(X) } = \prod_i {1\over x_i} \qty{x_i \over 1 - e^{-x_i}} \qty{-x_i \over 1 - e^{-x_i}} ,\end{align*} but note that this doesn’t necessarily make sense. However, all all computations we’ll see, there will be enough cancellation to make this well-defined.

Recall that given a differential complex \(({ \mathcal{E} }_{*}, d)\) we had a symbol complex \(( \pi^* {\mathcal{E}}_{*}, \sigma(d) )\) where \(\pi: T {}^{ \vee }X\to X\) and \begin{align*} \sigma\qty{ \sum_{{\left\lvert {I} \right\rvert} \leq N} f_I {{\partial}}_I } \mathrel{\vcenter{:}}=\sum_{{\left\lvert {I} \right\rvert} = N} f_I y^I ,\end{align*} where we take the top-order differentials, \({\frac{\partial }{\partial x_j}\,} \mapsto y_j\) and \begin{align*} T {}^{ \vee }X &\to {\mathbb{R}}\\ \alpha &\mapsto \alpha\qty{{\frac{\partial }{\partial x_j}\,} } .\end{align*} We say that \(( {\mathcal{E} }_{*}, d )\) is elliptic if the symbol complex is exact on \(T {}^{ \vee }X \setminus\left\{{0}\right\}\) where we delete the zero section. The Atiyah-Singer index theorem stated \begin{align*} \chi( {\mathcal{E}}_{*}, d) = \int_X { \operatorname{ch}( { \mathcal{E} }_{*}) \over {\operatorname{eul}}(X) } \mathrm{td}( TX\otimes_{\mathbb{R}}{\mathbb{C}}) .\end{align*} What’s the connection to elliptic operators? Given a 2-term complex \begin{align*} 0 \to \mathcal{E}^0 \xrightarrow{D} \mathcal{E}^1 \to 0 ,\end{align*} then \(D\) is an elliptic operator if this is an elliptic complex. This means the symbol complex is an isomorphism, i.e.  \begin{align*} 0 \to \pi^* \mathcal{E}^0 \xrightarrow{\sigma(D)} \pi^* \mathcal{E}^1 \to 0 \end{align*} where \(\sigma(D)\) is an isomorphism away from the zero section.

Every elliptic complex can be converted into a 2-term complex using a hermitian metric. Given \begin{align*} \mathcal{E}^0 \xrightarrow{d^0} \mathcal{E}^1 \xrightarrow{d^1} \mathcal{E}^2 \to \cdots ,\end{align*} we map this to \begin{align*} 0 \to \mathcal{E}^{\text{even}} \mathrel{\vcenter{:}}=\bigoplus_{i \text{ even} } \mathcal{E}^i \mathrel{\operatorname*{\rightleftharpoons}_{D^{\text{odd} }}^{D^\text{even}}} \mathcal{E}^{\text{odd}} \mathrel{\vcenter{:}}=\bigoplus_{i \text{ odd}} \to 0 \end{align*} where \begin{align*} D \mathrel{\vcenter{:}}=((d^{2i-1})^{\dagger} , d^{2i} ) : \mathcal{E}^{2i} \to \mathcal{E}^{2i-1} \oplus \mathcal{E}^{2i+2} \\ \end{align*} and \((d^{2i-1})^{\dagger}\) is defined by the following property: for \(\alpha\in \mathcal{E}^{2i-1}\) and \(\beta \in \mathcal{E}^{2i}(X)\), \begin{align*} {\left\langle { d^{2i-1} \alpha},~{\beta} \right\rangle}_h = {\left\langle { \alpha },~{ ( (d^{2i-1})^{\dagger} \beta} \right\rangle}_h .\end{align*} Here this pairing depends on a hermitian metric \(h\), which is a hermitian form on each fiber: \begin{align*} h_i: \mathcal{E}^i \otimes{\overline{{ \mathcal{E}^i}}} \to {\mathbb{C}} .\end{align*} Using this, we can fix a volume form \(dV\) on \(X\) and define \begin{align*} {\left\langle {u},~{v} \right\rangle}_h \mathrel{\vcenter{:}}=\int_X h_i(u, {\overline{{v}}}) \, dV && u, v\in \mathcal{E}^i(X) .\end{align*} This yields the desired two-term complex, and \(( {\mathcal{E}}_{*}, d)\) is elliptic if and only if \(D^e \circ D^o: \mathcal{E}^o {\circlearrowleft}\) and \(D^o \circ D^e: \mathcal{E}^e {\circlearrowleft}\) are elliptic operators.

Note that this space of harmonic forms depended on the Hermitian metrics on \(\mathcal{E}^i\) and the volume form \(dV\). In the case \(\mathcal{E}^i \mathrel{\vcenter{:}}=\Omega^i\), there is a natural metric determined by any Riemannian metric on \(X\). Recall that this is given by a metric \begin{align*} g: TX \otimes TX \to {\mathbb{R}} .\end{align*} This determines an isomorphism \begin{align*} T_p X &\xrightarrow{\sim} T_p {}^{ \vee }X\\ v &\mapsto g(v, {-}) ,\end{align*} which we can invert to get a metric on the cotangent bundle \(T {}^{ \vee }X\). This induces a metric on \(i{\hbox{-}}\)forms using the identification \(\Omega^i \mathrel{\vcenter{:}}=\bigwedge^i T {}^{ \vee }X\) and induces a volume form \begin{align*} dV \mathrel{\vcenter{:}}=\sqrt{ \operatorname{det}g}: \bigwedge^{\text{top}} TX \to {\mathbb{R}} .\end{align*}

In this case, \(d d^\dagger + d^\dagger d\) on \(\Omega^i(X)\) is called the metric Laplacian.

Let \((X, g)\) be a Riemannian manifold. We thus have a symmetric bilinear form on \(\Omega^p(X)\) given by pairing sections: \begin{align*} {\left\langle { \alpha},~{ \beta} \right\rangle} \mathrel{\vcenter{:}}=\int_X g( \alpha, \beta) .\end{align*} Note that we have orthonormal frames on \(\Omega^p(X)\) of the form \(e_{i_1} \wedge \cdots \wedge e_{i_p}\) where the \(\left\{{ e_i }\right\}\) are orthonormal frames on \(T {}^{ \vee }X\).

Can we always get a Hermitian metric? Let \(X \in {\mathsf{Mfd}}_{C^{\infty }({\mathbb{R}})}\) and \(\mathcal{E} \to X \in { {\mathsf{Bun}}_{\operatorname{GL}_r} }_{{\mathbb{C}}}\) a smooth complex vector bundle. Then any section \(h\in \mathcal{E} {}^{ \vee }\otimes{\overline{{\mathcal{E}}}} {}^{ \vee }(X)\), we have \begin{align*} h: \mathcal{E} \otimes{\overline{{\mathcal{E}}}} \to {\mathcal{O}}\\ h( e\otimes f) .\end{align*} for \(e, f \in \mathcal{E}_p\) is a Hermitian form for all \(p\). In local trivializations, \({ \left.{{\mathcal{E}}} \right|_{{U}} } \cong {\mathcal{O}}_U^{\oplus r}\), and one can take the standard Hermitian form here. Then for \((f_1, \cdots, f_r) \in {\mathcal{O}}^{\oplus r}(U)\), we have \(\sum f_i \mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu_i\in {\mathcal{O}}(U)\). This can be extended to all of \(X\) using a partition of unity subordinate to the coordinate charts.

The thing to check here is that on \({\mathbb{C}}^r\), for any collection \(h_1, \cdots, h_n\), any positive linear combination \(\sum a_i h_i\) is again a Hermitian metric for any \(a_i \in {\mathbb{R}}^+\). One can regard these as skew-symmetric matrices, which are closed under addition, and the positive-definite property ensures it’s still a metric since \(h(v, v) = \sum a_i h_i(v, v) > 0\) for \(v\neq 0\).

Recall that we start with a Riemannian manifold \((X, g)\) where \(g: TX^{\otimes 2} \to {\mathcal{O}}\) is a metric on the tangent bundle. Locally choose \(f_1,\cdots, f_n\) an orthogonal frame of \(TX\), then setting \(e_i \mathrel{\vcenter{:}}= f_i {}^{ \vee }\) yields an orthogonal frame of \(T {}^{ \vee }X\) and thus an orthogonal frame \(e_{i_1} \wedge \cdots e_{i_p}\) of \(\bigwedge^p T {}^{ \vee }X \mathrel{\vcenter{:}}=\Omega^p X\). So we get a metric on the smooth \(p{\hbox{-}}\)forms \(\Omega^p X\). We defined the Hodge star operator \begin{align*} \star: \Omega^p &\to \Omega^{n-p} \\ e_{i_1} \wedge \cdots e_{i_p} &\mapsto \pm e_{j_1} \wedge \cdots \wedge e_{j_{n-p}} .\end{align*} where \(\left\{{ i_1, \cdots, i_p, j_1, \cdots, j_{n-p} }\right\} = \left\{{ e_1, \cdots, e_n }\right\}\). We saw that \begin{align*} e_{i_1} \wedge \cdots \wedge e_{i_p} \star\qty{ e_1 \wedge \cdots e_{i_p}} &= e_1 \wedge \cdots \wedge e_n \\ \star\qty{ \sum_{{\left\lvert {I} \right\rvert} = p } f_I e_I} &= \sum_{ {\left\lvert {I} \right\rvert} = p} e_{I^c} (-1)^{\operatorname{sign}(I)} .\end{align*}

Moreover, \begin{align*} {\left\langle { \alpha},~{ \beta} \right\rangle} = \int_X g( \alpha, \beta) dV = \int_X \alpha\wedge \qty{\star\beta} ,\end{align*} and we showed that \begin{align*} {\left\langle { \alpha},~{ d \beta} \right\rangle} = \pm {\left\langle { d^\dagger \alpha},~{ \beta} \right\rangle} && d^\dagger \mathrel{\vcenter{:}}=\star d \star, \beta\in \Omega^{p-1}(X), \alpha\in \Omega^p(X) ,\end{align*} yielding an adjoint operator \begin{align*} d^\dagger: \Omega^p(X) \to \Omega^{p-1}(X) .\end{align*}

This operator is \({\mathbb{R}}{\hbox{-}}\)linear, so \(\mathcal{H}^p(X) \in { \mathsf{Vect} }_{\mathbb{R}}\). Note that this whole construction can be made to work over \({\mathbb{C}}\) by adding conjugates in appropriate places.

We have \begin{align*} H^p(X; {\mathbb{R}}) &= {\ker d \over \operatorname{im}d} \\ &= { d \Omega^{p-1}(X) \oplus \mathcal{H}^p(X) \over d \Omega^{p-1}(X) } \\ &= \mathcal{H}^p(X) .\end{align*} Note that there is a map \begin{align*} \mathcal{H}^p(X) \to H^p(X; {\mathbb{R}}) \end{align*} since \(\alpha\in \mathcal{H}^p(X)\) satisfies \(d \alpha = 0\) in addition to \(d^\dagger \alpha = 0\).

Note that one can complete these spaces using Sobolev spaces, but there are issues. Take \(S^1\), then \begin{align*} L_2(S^1) \mathrel{\vcenter{:}}=\left\{{ \sum a_n e^{2\pi i n z} {~\mathrel{\Big|}~}\sum {\left\lvert {a} \right\rvert}_i < \infty }\right\} ,\end{align*} but for \(f\in L_2(S^1)\) we have \(df = \sum 2\pi i n a_n e^{2\pi i n z}\) which may not converge.

Recall that a sheaf of rings \(\operatorname{\mathcal{F}}\) on \(X\in {\mathsf{Top}}\) is an assignment of a ring \(\operatorname{\mathcal{F}}(U)\) to each open set \(U\subseteq X\) and restriction maps \(\operatorname{\mathcal{F}}(U) \xrightarrow{\rho_{UV}} \operatorname{\mathcal{F}}(V)\) for \(V \subseteq U\) that is a presheaf, so

1. This diagram commutes:

Link to Diagram

2. \(\phi_{UU} = \one_{\operatorname{\mathcal{F}}(U)}\) and \(\operatorname{\mathcal{F}}(\emptyset) = 0\).

That additionally satisfies unique gluing on double overlaps.

For \(X \in {\mathsf{Mfd}}(\mathop{\mathrm{Hol}}({-}, {\mathbb{C}}))\) such that \(\varphi_V \circ \varphi_U ^{-1} : \varphi_U( U \cap V) \to \varphi_V(U \cap V)\) is holomorphic, so \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}z_i = 0\). Then \(\Omega^p(U) = \left\{{ \sum_{{\left\lvert {I} \right\rvert} = p} f_I( \mathbf{z}) dz_I }\right\}\), and the key difference is that the \(f_I\) be holomorphic. This matters since POUs exist in the smooth setting but not the complex setting. Note that \({\mathcal{O}}, \Omega^p\) denote smooth/holomorphic functions and smooth/holomorphic \(p{\hbox{-}}\)forms in the smooth/complex settings. So we need a new notation for smooth holomorphic \(p{\hbox{-}}\)forms in the complex setting. We defined \(A^{p, 0}\) to be the smooth \(p{\hbox{-}}\)forms, and \(A^{p, q}\) the smooth \((p, q){\hbox{-}}\)forms. In local coordinates, these look like \begin{align*} A^{p, q}(U) = \left\{{ \sum_{{\left\lvert {I} \right\rvert} = p, {\left\lvert {J} \right\rvert} = q} f_{I, J} (\mathbf{z}) dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J }\right\} .\end{align*}

Why are these \(A^{p, q}\) useful? They give a resolution of \(\Omega^p\) on a complex manifold. There are maps of sheaves \begin{align*} 0 \to \Omega^p \xrightarrow{i} A^{p, 0} ,\end{align*} where being a map of sheaves means there are maps \(\Omega^p(U) \to A^{p, 0}(U)\) for all opens \(U\) which are compatible with restriction:

Link to Diagram

It’s clear that this works for \(i\), since any holomorphic function simply is smooth. We could continue this resolution:

\begin{align*} 0 \to \Omega^p \xrightarrow{i} A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \end{align*} where \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\qty{ \sum_{I, J} f_{I, J} dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J } \mathrel{\vcenter{:}}= \sum_{I, J, K} {\frac{\partial f_{I, J}}{\partial z_k}\,} d \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_k \wedge dz_I \wedge d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_J .\end{align*} We then defined Dolbeaut cohomology, \(H^q(X, \Omega^p) = \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{p, q} / \operatorname{im}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}_{p, q-1}\).

Continuing review: let \(\mathcal{E} \to X \in {\mathsf{Bun}}({\mathbb{R}}^n)\). A metric on \(\mathcal{E}\) is a smoothly varying positive definite inner product on the fibers.

For \(v, w\in \mathcal{E}_p\), we want a pairing \(g_p(v, w): \mathcal{E}_p^{\otimes 2} \to {\mathbb{R}}\). To think about this globally, this should be a map \begin{align*} g: \mathcal{E}^{\otimes 2} \to {\mathcal{O}} .\end{align*} where \(g_p: \mathcal{E}_p^{\otimes 2} \to {\mathbb{R}}\). Note that this map is \({\mathcal{O}}{\hbox{-}}\)linear, which follows from the fact that it’s \({\mathbb{R}}{\hbox{-}}\)linear on each fiber, or equivalently it is a map of vector bundles. We should also have that \(g(s\otimes s) \in {\mathcal{O}}(X)\) is a smooth function, and we require \(g(s\otimes s) \geq 0\). We also require \(g(s\otimes s)(p) = 0 \iff s_0 = 0\) and \(g(s\otimes t) = g(t\otimes s)\). This implies that \(g\in (\mathcal{E}^{\otimes 2}) {}^{ \vee }\otimes{\mathcal{O}}= (\mathcal{E} {}^{ \vee })^{\otimes 2}(X)\). The symmetric condition means that \(g\in \operatorname{Sym}^2 \mathcal{E} {}^{ \vee }(X)\).

For Hermitian forms, we take \begin{align*} h: ({\mathbb{C}}^n)^{\otimes 2}\to {\mathbb{C}} \end{align*} where \(h\) is conjugate linear, so \(h(cv, c'w) = \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5muc' h(v, w)\). Note that we can write \(h(v, w) = {\overline{{v}}}^t H w\) where \(H\) is Hermitian, so \({\overline{{H}}}^t = H\). This implies that \(h(v,v) \in {\mathbb{R}}^{\geq 0}\) and \(h(v,v) = 0 \iff v=0\) with \(h(v, w) = {\overline{{h(v, w)}}}\) The great thing about metrics: we can identify zero sections by self-pairing, multiplying by a volume form, and integrating. For \(\mathcal{E}\to X \in {\mathsf{Bun}}({\mathbb{C}})\), there is another bundle \({\overline{{\mathcal{E}}}} \to X \in {\mathsf{Bun}}({\mathbb{C}})\). Supposing that \({ \left.{{ \mathcal{E}}} \right|_{{U}} } \xrightarrow{\varphi_U} {\mathcal{O}}_U^{\oplus n}\) in a local trivialization, conjugating all of the transition functions gives the transition functions \({ \left.{{ {\overline{{ \mathcal{E}}}} }} \right|_{{U}} } \xrightarrow{\mathrm{conj} \circ \varphi_U} {\mathcal{O}}_U^{\oplus n}\). This yields a map \begin{align*} h: {\overline{{ \mathcal{E} }}} \otimes_{\mathbb{C}}\mathcal{E} \to {\mathcal{O}}\in ( {\overline{{\mathcal{E}}}} \otimes\mathcal{E} ) {}^{ \vee } .\end{align*} In local trivializations we have \({ \left.{{ \mathcal{E} }} \right|_{{U}} } = {\mathcal{O}}_U^{\oplus n} = {\mathbb{C}}^n \times U\), and \(h\) is described by \(h_U \in ({\overline{{ {\mathcal{O}}}}}^{\oplus n} \otimes{\mathcal{O}}^{\oplus n})(U)\).

When \(\operatorname{rank}\mathcal{E} = 1\) we abuse notation! For \(h\in ({\overline{{\mathcal{E}}}} {}^{ \vee }\otimes\mathcal{E} {}^{ \vee })(X)\), this is locally a \(1\times 1\) Hermitian matrix, thus of the form \([a]\) for \(a\in {\mathbb{R}}^{\geq 0}\). So we write \begin{align*} h(s, t) = hs{\overline{{t}}} \mathrel{\vcenter{:}}= h\otimes s \otimes{\overline{{t}}} \in ({\overline{{\mathcal{E}}}} {}^{ \vee }\otimes\mathcal{E} {}^{ \vee }) \otimes\mathcal{E} \otimes{\overline{{\mathcal{E}}}} = {\mathcal{O}} \end{align*} if \(\mathcal{E}\) is a line bundle. Why is \(V\otimes V {}^{ \vee }= {\mathcal{O}}\) in this case? There is a pairing \(v\otimes\lambda \mapsto \lambda(v)\), or more generally a trace pairing.

Let \(X\) be a Riemann surface, so \(X\in {\mathsf{Mfd}}^1({\mathbb{C}})\). Let \(L\to X \in {\mathsf{Bun}}^1(\mathop{\mathrm{Hol}})\), then we have a resolution \begin{align*} 0 \to L \hookrightarrow L\otimes A^{0, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} L \otimes A^{0, 1} \to 0 ,\end{align*} where the first map is inclusion of smooth holomorphic sections into smooth sections. What is this cut out by? We had \(s\mapsto { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s\) and thus \(f \mapsto {\frac{\partial f}{\partial \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu }\,} \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\). Note that \(H_1(L) = \operatorname{coker}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\).

Serre duality said that \begin{align*} h^1(L) = \dim H^1(L) = h^0( L {}^{ \vee }\otimes K) && K = \Omega^1 ,\end{align*} where \(\Omega^1\) is the sheaf of holomorphic 1-forms. Choose a metric to identify \(H^1(L)\) and \(H^0(L {}^{ \vee }\otimes K)\). Choose a hermitian metric on \(L\) and take \(s, t\in H^0(L\otimes A^{0, 0}) = C^\infty(L; {\mathbb{C}})\), then we get \(h(s, t) \in C^{\infty }(X; {\mathbb{C}})\) a smooth complex function. We abuse notation by writing this as \(h(s, t) = hs{\overline{{t}}}\), viewing \(h\in C^{\infty }(L {}^{ \vee }\otimes{\overline{{L}}} {}^{ \vee })\) locally. Note that we can’t integrate a function on a manifold without a form, so choosing a volume for \(dV\) we can define a pairing on sections \begin{align*} {\left\langle {s},~{t} \right\rangle} \mathrel{\vcenter{:}}=\int_X hs{\overline{{t}}} dV .\end{align*}

Now for two sections \(\alpha, \beta\in H^0(L\otimes A^{0, 1})\) we can write \begin{align*} \int_X h \alpha {\overline{{ \beta}}} = \int_X \omega ,\end{align*} where \(\omega\) is a smooth \((1, 1){\hbox{-}}\)form since \(h\in {\overline{{L}}} {}^{ \vee }\otimes L {}^{ \vee }\), \(\alpha\in L\otimes A^{0, 1}\), and \({\overline{{ \beta}}} \in \mkern 1.5mu\overline{\mkern-1.5muL\mkern-1.5mu}\mkern 1.5mu \otimes A^{1, 0}\). We now have metric on both the source and target spaces here: \begin{align*} H^0( L\otimes A^{0, 0}) \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} H^0(L\otimes A^{0, 1}) ,\end{align*} where on the left-hand side we take \((s, t) \mapsto \int_X hs{\overline{{t}}}dV\) and on the right-hand side we have \((\alpha, \beta) \mapsto \int_X h \alpha{\overline{{\beta}}}\).

Given a map of metric vector spaces \(V \xrightarrow{\varphi} W\), the adjoint \(\varphi^\dagger\) satisfies \begin{align*} {\left\langle { \varphi(v) },~{w} \right\rangle} = {\left\langle {v},~{ \varphi^\dagger(w)} \right\rangle} .\end{align*} and \(\operatorname{coker}( \varphi) = \ker( \varphi^\dagger)\). So \(H^1(L) = \operatorname{coker}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger\), and after integrating by parts we have \begin{align*} {\left\langle { \alpha},~{ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s} \right\rangle} &\mathrel{\vcenter{:}}=\int_X \alpha{\overline{{ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}s }}} h \\ &= \int_X \alpha{\partial}({\overline{{s}}}) h \\ &= -\int_X {\overline{{s}}} {\partial}( \alpha h) && \text{IBP} \\ &= -\int_X {\overline{{s}}} {{\partial}(\alpha h) \over dV} dV \\ &= {\left\langle { - { {\partial}( \alpha h ) \over dV}},~{s} \right\rangle} .\end{align*} So we could define \begin{align*} { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger \alpha = {\overline{{- {{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}({\overline{{\alpha}}} h )} \over dV }}} .\end{align*} Note that \(\alpha\mapsto {\overline{{ \alpha}}} h\), so \(\alpha\in \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger \iff {\overline{{ \alpha}}}h\in \ker { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\). Then \(\ker ({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger) = H^0(L {}^{ \vee }\otimes K)\).

Recall Serre duality: let \(C\in {\mathsf{Mfd}}_{\mathbb{C}}({ \operatorname{compact} } ,{ \operatorname{oriented} } )\) and \(L\to C \in {\mathsf{Bun}}(\mathop{\mathrm{Hol}})\). Then \begin{align*} h^1(L) = h^0(L {}^{ \vee }\otimes K_C) .\end{align*} We also have Riemann-Roch, a very important tool: \begin{align*} h^0(L) - h^1(L) = \deg L + 1 - g(C) ,\end{align*} where \(\deg L = \int_C c_1(L)\), which is also equal to \(\deg [ \left\{{ s = 0 }\right\}] = \deg(\operatorname{Div}s)\). Note that \(c_1\) is the most important Chern class to know, thanks to the splitting principle. How was it defined? There are several definitions:

1. \(L\) defines an element of \begin{align*} H^1(C, {\mathcal{O}}^{\times}) &= \left\{{ t_{UV}: U \cap V \to {\mathbb{C}}^{\times}{~\mathrel{\Big|}~}t_{UV} t_{UW}^{-1}t_{VW} = 1 }\right\} / {{\partial}}\left\{{ h_u: U\to {\mathbb{C}}^{\times}}\right\} \\ &= \ker {{\partial}}^1 / \operatorname{im}{{\partial}}^0 \end{align*} in Čech cohomology. By definition \({{\partial}}\left\{{ h_U {~\mathrel{\Big|}~}U\in \mathcal{U} }\right\} = \left\{{ h_u h_v^{-1}{~\mathrel{\Big|}~}U, V \in \mathcal{U} }\right\}\), where \({{\partial}}^2 = 1\) since \begin{align*} (h_U h_V)^{-1}\qty{h_U h_W ^{-1}}^{-1}(h_V h_W ^{-1}) = 1 && \text{on } U \cap V \cap W .\end{align*} By assigning \(L\) to its transition functions, we get a map \(L\to H^1\). We have the exponential exact sequence: \begin{align*} 0 \to \underline{{\mathbb{Z}}} \to {\mathcal{O}}\xrightarrow{\exp} {\mathcal{O}}^{\times}\to 1 ,\end{align*} which induces a map \begin{align*} H^1(C, {\mathcal{O}}^{\times}) &\to H^2(C, {\mathbb{Z}}) \\ L &\mapsto c_1(L) .\end{align*}

2. \(L\) defines an element \(\mathop{\mathrm{Fr}}L \in \mathrm{Bun}^{\mathrm{prin}}({\mathbb{C}}^{\times})\) (which only works for line bundles), which is defined by \(\mathop{\mathrm{Fr}}L = L \setminus s_0\) where \(s_0\) is the zero section of \(L\). By topology, we get a classifying map \begin{align*} C \xrightarrow{\phi_L} B{\mathbb{C}}^{\times}= {\mathbb{CP}}^\infty = ({\mathbb{C}}^{\infty} \setminus\left\{{0}\right\}) / {\mathbb{C}}^{\times} .\end{align*} There is a universal \(c_1\in H^2({\mathbb{CP}}^{\infty}; {\mathbb{Z}})\), so we take the pullback to define \(c_1(L) \mathrel{\vcenter{:}}=\phi_L^*(c_1)\). We can use that there is a cell decomposition \({\mathbb{CP}}^{\infty } = {\mathbb{C}}^0 \cup{\mathbb{C}}^1 \cup{\mathbb{C}}^2 \cup\cdots\), and so there is a unique generator in its \(H^2\).

3. Consider a smooth section \(s\in C^{\infty }(L)\), then we can define \(c_1(L) \mathrel{\vcenter{:}}=[ \left\{{ s = 0 }\right\} ]\) by taking the fundamental class, assuming that \(s\) is transverse to the zero section \(s_z\) of \(L\). Here we view the zero set as an oriented submanifold. See picture: in this case \([\left\{{ s = 0 }\right\} ] = [p] - [q] + [r]\).

Applying Serre duality to the left-hand side in Riemann-Roch yields the dimension of the space of holomorphic sections of some other bundle, \(L {}^{ \vee }\otimes K\).

Last time: we reviewed Riemann-Roch, Serre duality, sheaves of \(p{\hbox{-}}\)forms. Recall a theorem from a few weeks ago:

Note that \(\mathcal{H}\) was the space of harmonic \(p{\hbox{-}}\)forms, and \(d^\dagger: \mathrel{\vcenter{:}}=(-1)^? \star d\star\) where \begin{align*} \star: \Omega^{p}(X) &\to \Omega^{n-p}(X) \\ e_{i_1} \wedge \cdots \wedge e_{i_p} &\mapsto \pm e_{j_1} \wedge \cdots e_{j_{n-p}} \end{align*} where \(\left\{{ e_i}\right\}\) is an orthonormal basis of basis of \(T {}^{ \vee }X\). Note that this formula is replacing the \(e_i\) that do appear with the \(e_i\) that don’t appear, up to a sign. The harmonic forms were defined as \({\mathcal{H}}^p(X) = \ker (dd^\dagger + d^\dagger d ) = \ker (d) \cap\ker(d^\dagger)\). We proved that assuming this decomposition, there is an isomorphism \begin{align*} {\mathcal{H}}^p(X) \cong H^p_{\mathrm{dR}}(X; {\mathbb{R}}) .\end{align*}

The harmonic forms \({\mathcal{H}}^p(X)\) depend on the metric \(g\), despite mapping isomorphically to de Rham cohomology.

This was just in the case of a real smooth Riemannian manifold. What extra structure to we have for \(X \in {\mathsf{Mfd}}(\mathop{\mathrm{Hol}}({-}, {\mathbb{C}}) )\)?

So the Kähler geometry is determined by the data \(({\mathbb{C}}^n, g_\text{std}, J, \omega_\text{std})\), i.e. a metric, an almost complex structure, and a symplectic form. Note that the relation \(\omega(x, y) = g(x, Jy)\) can be used to determine the 3rd piece of data from any 2. This is the fiberwise/local model, i.e. every tangent space at a point looks like this.

Given \(g\) and \(J\), \(\omega\) is automatically a 2-form. That it’s antisymmetric follows from \begin{align*} -\omega(w, v) &= -g(w, Jv) \\ &= -g(Jv, w) \\ &= -g(J^2 v, Jw)\\ &= g(v, Jw)\\ &= \omega(v, w) .\end{align*}

Conversely, we can always define \(g(v, w) \mathrel{\vcenter{:}}=- \omega(v, Jw)\), but a priori this may not be a metric. This will be symmetric, but potentially not positive-definite.

Next time: we’ll see that if \(X\) is Kähler, then \begin{align*} {\mathcal{H}}^k(X) = \bigoplus_{p+q=k} {\mathcal{H}}^{p, q}(X), \end{align*} so this is compatible with the Hodge decomposition. This is what people usually call the Hodge decomposition theorem, and gives some invariants of complex manifolds. By a miracle, this decomposition only depends on \(g\) and the complex structure.

Note that there is a notion of hyperkähler manifolds, which have 3 complex structures \(I, J, K\) such that \(I^2=J^2=K^2 = IJK = -\one\), yielding 3 “parallel” 2-forms \(\omega_I, \omega_J, \omega_K\) such that the covariant derivative vanishes, i.e. \(\nabla_g \left\{{ \omega_I, \omega_J, \omega_K }\right\} = 0\). With respect to the complex structure \(I\), \(\omega_J + \omega_K\) is a holomorphic 2-form. There is a sphere’s worth of almost complex structures, and there is an action \({\operatorname{SO}}(4, b_2 - 4) \curvearrowright H^*(X)\). There’s no known example where the hyperkähler metric has been explicitly written down.

Last time: we defined a Kähler manifold: \(X\in {\mathsf{Mfd}}({\mathbb{C}})_{ \operatorname{compact} }\) and \(\omega \in \Omega^2(X_{\mathbb{R}})\) a closed real 2-form such that \(g(x, y) \mathrel{\vcenter{:}}=\omega(x, Jy)\) is a metric. By the Hodge theorem, we have a space \({\mathcal{H}}^k(X)\) of harmonic \(k{\hbox{-}}\)forms for \((X, g)\) which represents \(H^k_{\mathrm{dR}}(X; {\mathbb{R}})\). We can consider the \({\mathbb{C}}{\hbox{-}}\)valued harmonic forms \({\mathcal{H}}^k_{\mathbb{C}}\mathrel{\vcenter{:}}={\mathcal{H}}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}\), which represents \(H^k_\mathrm{dR}(X; {\mathbb{C}})\)

There is a generalization to higher genus curves. Recall the following theorem:

This essentially follows from the Riemann mapping principle.

It’s miraculous! The biholomorphisms of \({\mathbb{H}}\) preserve a metric. So \(C\) has a canonical metric, \(g_{\operatorname{hyp}}\), which descends along the quotient map \({\mathbb{H}}\to {\mathbb{H}}/\Gamma \cong {\mathbb{C}}\).

By lifting we can write \begin{align*} \Omega^1(C_{\mathbb{R}}) \otimes_{\mathbb{R}}{\mathbb{C}}= A^{1, 0}(C) \oplus A^{0, 1}(C) = \left\{{ f(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu)\,dz+ g(z, \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu) \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu {~\mathrel{\Big|}~}z\in {\mathbb{H}}, \, f,g\in C^{\infty }({\mathbb{C}}, {\mathbb{R}}) }\right\} \end{align*} But \(\,dz\) is not invariant under the map \(z\mapsto {az+b \over cz+d}\), since \(\,dz\mapsto {\,dz\over (cz+d)^2 }\). In order to descend \(f(z)\) to \(C\), we need \begin{align*} f\qty{az +b \over cz + d} = (cz+d)^2 f(z) && \text{ for all } \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \Gamma \end{align*} This says that \(f\) is a modular form of weight 2.

Recall the Hodge decomposition theorem. Let \((M, g) \in {\mathsf{Mfd}}_{\mathbb{R}}^n(\mathsf{Riem}, { \operatorname{compact} } )\), then choosing an orthonormal basis \(\left\{{ v_j }\right\}\) for \(T_p M\) yields a corresponding orthonormal basis in \(T_p {}^{ \vee }M \mathrel{\vcenter{:}}=\mathop{\mathrm{Hom}}_{\mathbb{R}}(T_p M, {\mathbb{R}})\) given by taking \(\left\{{ e_i {~\mathrel{\Big|}~}e_i(v_j) = \delta_{ij} }\right\}\).

There is a map \begin{align*} \star: \bigwedge^k T_p {}^{ \vee }M &\to \bigwedge^{n-k} T_p {}^{ \vee }M \\ \bigwedge_{j=1}^k e_{i_j} &\mapsto \pm \bigwedge_{\ell = 1}^{n-k} e_{j_\ell} \end{align*} where the \(e_{j}\) are defined such that \(\bigwedge_{j=1}^k e_{i_j} \wedge \bigwedge_{\ell = 1}^{n-k} e_{j_\ell} \mathrel{\vcenter{:}}=\,dV\), where \(\,dV\) is the volume form on \(M\) at \(p\). Thus we have a map \begin{align*} \star: \Omega^k &\to \Omega^{n-k} \\ 1 &\mapsto \,dV .\end{align*} We defined \(d^\dagger \mathrel{\vcenter{:}}=\star{d} \mkern-5mu \star\), and said a form \(\omega\) was harmonic iff \(\Delta \omega=0\), where \(\Delta \mathrel{\vcenter{:}}= dd^\dagger + d^\dagger d\). The space of such forms was denoted \(\mathcal{H}^k(M) \subseteq \Omega^k(M)\).

So for \(M\in {\mathsf{Mfd}}({\mathbb{C}})\) a complex manifold, we have a decomposition \begin{align*} \Omega^k(M) &= \bigoplus_{p+q=k} A^{p, q}(M) \\ \\ A^{p, q} &\mathrel{\vcenter{:}}=\left\{{ \sum_{\substack{ {\left\lvert {I} \right\rvert} = p \\ {\left\lvert {J} \right\rvert} = q}} \qty{\,dz_{i_1} \wedge \cdots \,dz_{i_p} } \wedge \qty{ \,dz_{j_1} \wedge \cdots \,dz_{j_q} } }\right\} .\end{align*}

For \(M\) a Kähler manifold, we have \begin{align*} \mathcal{H}^k(M) = \bigoplus _{p+q = k} \mathcal{H}^{p, q}(M) \\ \\ \mathcal{H}^{p, q}(M) = \mathcal{H}^k(M) \cap A^{p, q}(M) .\end{align*}

Why is this true? We have a map \begin{align*} d: A^{p, q}(M) \to A^{p+1, q}(M) \oplus A^{p, q+1}(M) ,\end{align*} where for example if \(f(z) \mathrel{\vcenter{:}}= z \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \in A^{0, 0}({\mathbb{C}})\), we have \(df = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu \,dz+ z\,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\) where the first is a \((1, 0)\) form and the latter is a \((0, 1)\) form. Write \(d = {\partial}+ { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\) where \({\partial}\mathrel{\vcenter{:}}=\sum \,dz_j\) and \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= \sum \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _j\), as well as \begin{align*} d^\dagger: A^{p, q}(M) \to A^{p-1, q}(M) \oplus A^{p, q-1}(M) .\end{align*} Now \(\star\) of a \((p, q)\) form is an \((n-p, n-q)\) form, and so \begin{align*} \star\qty{ \,dz_{i_1} \wedge \cdots \wedge \,dz_{i_r} \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _{j_1} \wedge \cdots \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _q } \mathrel{\vcenter{:}}=\star(\,dz_I \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _J) = \pm \,dz_{I^c} \wedge \,d\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu _{J^c} ,\end{align*} and we have \(d^\dagger = {\partial}^\dagger + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger\). We can thus move around the bigraded group in several ways:

Link to Diagram

See Griffiths-Harris for details. Note that this is a local statement, i.e. it can be checked in coordinate charts.

The upshot: \begin{align*} \mathcal{H}^k(M) = \ker \Delta_d = \ker \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} ,\end{align*} and moreover \begin{align*} \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}: A^{p, q}(M) {\circlearrowleft} \end{align*} which implies that on \(\Omega^k(M)\), \begin{align*} \ker \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} \circ \bigoplus _{p+q=k} \ker \qty{ A^{p, q}(M) \xrightarrow{\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}} A^{p, q}(M) } &= \bigoplus _{p+q = k} \ker \Delta_d ,\end{align*} which yields the Hodge decomposition theorem \begin{align*} \mathcal{H}^k(M) = \bigoplus _{p+q=k} \mathcal{H}^{p, q}(M) .\end{align*}

This is a strong restriction on what manifolds can admit a Kähler structure. Moreover, since \(\Delta_d\) is a real operator, we obtain \(\mkern 1.5mu\overline{\mkern-1.5mu \mathcal{H}^{p, q}(M) \mkern-1.5mu}\mkern 1.5mu \cong \mathcal{H}^{p, q}(M)\).

Some consequences:

For \(M\) a Kähler manifold, the odd Betti numbers \(\beta_{2i+1}(M) \mathrel{\vcenter{:}}=\dim H_{\mathrm{dR}}^{2i+1}(M; {\mathbb{C}})\) are even. This is because \begin{align*} \bigoplus _{p+q=k} \mathcal{H}^{p, q} \cong \mathcal{H}^{2i+1}(M) \cong H_{\mathrm{dR}}^{2i+1}(M) .\end{align*} If we define \(h^{p, q}(M) \mathrel{\vcenter{:}}=\dim_{\mathbb{C}}\mathcal{H}^{p, q}(M)\), we clearly have \begin{align*} \beta_{2i+1} = \sum_{p+q = 2i+1} h^{p, q}(M) .\end{align*} Now using that \(\mkern 1.5mu\overline{\mkern-1.5mu\mathcal{H} \mkern-1.5mu}\mkern 1.5mu \cong \mathcal{H}\), we can rewrite this as \begin{align*} \beta_{2i+1} &= \sum_{p+q = 2i+1} h^{p, q}(M) \\ &= 2 \sum_{\substack{ p+q = 2i+1 \\ p < q} } h^{p, q}(M) .\end{align*}

Is this just some fact about arbitrary complex manifolds, with no extra structure? The answer is no, and the counterexample is the Hopf surface \begin{align*} X \mathrel{\vcenter{:}}=\qty{ {\mathbb{C}}^2 \setminus\left\{{\mathbf{0}}\right\}} / (x,y)\sim (2x, 2y) ,\end{align*} which we can roughly identify as \({\mathbb{R}}^4\) “modulo doubling.” We can take a fundamental domain \(1\leq {\left\lvert {r} \right\rvert} \leq 3\), this yields an annulus-like sphere with the inner shell glued to the outer:

This is homeomorphic to \(S^1 \times S^3\), but \(\beta_1(M) = 1\), so this won’t yield a Kähler structure.

Last time: the Hodge decomposition theorem. Let \((X, g) \in {\mathsf{Mfd}}_{\mathbb{C}}^{ \operatorname{compact} } ({ \operatorname{Kähler} } )\), then the space of harmonic \(k{\hbox{-}}\)forms \(\mathcal{H}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}\) decomposes as \(\bigoplus_{p+q = k} \mathcal{H}^{p, q}(X)\). There is also a symmetry \(\mkern 1.5mu\overline{\mkern-1.5mu\mathcal{H}^{p, q}(X) \mkern-1.5mu}\mkern 1.5mu = \mathcal{H}^{q, p}(X)\). We have an isomorphism to the de Rham cohomology \(\mathcal{H}^k(X) \otimes_{\mathbb{R}}{\mathbb{C}}\cong H^k_\mathrm{dR}(X; {\mathbb{C}})\). We know the constituent pieces as well, as well as several relationships: \begin{align*} \mathcal{H}^{p, q}(X) &= \ker (\Delta_d: A^{p, q}(X) {\circlearrowleft}) \\ \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} &= { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger + { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\\ \Delta_d &= 2 \Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} .\end{align*} There was a proposition that \(\ker(\Delta_d) = \ker(d) \cap\ker(d^\dagger)\), and the same proposition holds for \(\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}\). In this case we have \(\ker(\Delta_{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}}) = \ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \cap\ker( { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^\dagger)\) on \(A^{p, q}(X)\), and this is isomorphic to \(\ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) / \operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})\). Recall that we resolved the sheaf \(\Omega^p\) of holomorphic \(p{\hbox{-}}\)forms by taking the Dolbeault resolution \begin{align*} 0 \to \Omega^p \to A^{p, 0} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 1} \xrightarrow{{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}} A^{p, 2} \to \cdots .\end{align*} Thus we can identify \(\ker({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu})/\operatorname{im}({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}) \cong { \mathcal{H} }( X; \Omega^p)\) as sheaf cohomology. We defined \(h^{p, q}(X) \mathrel{\vcenter{:}}=\dim_{\mathbb{C}}H^{p, q}(X)\).

A priori, one could vary the Kähler form and have some \(h^{p, q}\) jump or drop dimension. It also turns out that varying the complex structure will also not change these dimensions.

Whenever the Hodge-de Rham spectral sequence degenerates, one generally gets \(\sum_{p+q} h^{p,q } = h^k\). Note that there is a resolution: \begin{align*} 0 \to \underline{{\mathbb{C}}} \to {\mathcal{O}}\xrightarrow{d} \Omega^1 \xrightarrow{d} \Omega^2 \xrightarrow{d} \cdots ,\end{align*} which is not acyclic and thus has homology. In general, the spectral sequence is \begin{align*} E^1_{p,q} = { \mathcal{H} }^q(X; \Omega^p) \Rightarrow{ \mathcal{H} }^{p+q}(X; \underline{{\mathbb{C}}}) .\end{align*}

Note that \(K3\)s are special CYs. An example is \({\mathbb{C}}^2 / \Lambda\) for \(\Lambda\) a rank 4 lattice. This is diffeomorphic to \((S^1)^4\), for example \(E\times E\).

We have a perfect pairing \begin{align*} \Omega^k \otimes\Omega^{n-k} \to K ,\end{align*} and thus \(\Omega^{n-k} \cong K \otimes(\Omega^{k}) {}^{ \vee }\). So we have \begin{align*} H^p( \Omega^k ) {}^{ \vee }\cong H^{n-p}( (\Omega^{k}) {}^{ \vee }\otimes K ) = H^{n-p}( \Omega^{n-k}) ,\end{align*} and thus \(h^{p, k} = h^{n-p, n-k}\), which recovers what we knew about \(\star: \mathcal{H}^{p, q} \to \mathcal{H} ^{n-p, n-q}\).

So we don’t get anything new from the Serre duality argument.

What is special when \(X\in { \text{CY} }\) is that \begin{align*} \Omega^{n-k} \cong ( \Omega^k ) {}^{ \vee }= \bigwedge^k TX \end{align*} for \(TX\) the tangent bundle. Note that taking the cotangent bundle gives forms, and instead this gives a bundle of polyvector fields. For \(k=1\), we get a holomorphic vector field, which one might think of as an infinitesimal biholomorphism.

Polyvector fields show up in Hochschild homology!

This implies that \(S\) is a projective variety cut out by homogeneous polynomials in \(N+1\) variables in \({\mathbb{CP}}^N\).

What makes having a projective embedding special? If \(S \hookrightarrow{\mathbb{CP}}^N\), it admits a line bundle: \({\mathcal{O}}_S(1) \mathrel{\vcenter{:}}={ \left.{{ {\mathcal{O}}_{{\mathbb{CP}}^N}(1) }} \right|_{{S}} }\).

This can be written down as \({i\over 2} {\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\log( \sum_{i=1}^N z_i \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu_i )\), which is well-defined since scaling comes out as a constant. Being closed follows from \({\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}= d{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\) since \({ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}^2 = 0\), which implies \(d({\partial}{ \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}\cdots) = d^2 { \mkern 1.5mu\overline{\mkern-1.5mu{\partial}\mkern-1.5mu}\mkern 1.5mu}(\cdots) = 0\). This defines a metric: this follows from checking in local coordinate charts, say \(z_0 = 1\), and checking that \(g(x ,y) \mathrel{\vcenter{:}}=\omega(x, Jy)\) yields a metric. This involves taking a fussy derivative!

Thus given \(S\xhookrightarrow{\phi} {\mathbb{CP}}^N\), we can restrict or take the pullback of \(\omega_{{ \text{FS} }}\) to \(S\). Then \(\omega \mathrel{\vcenter{:}}=\phi^* \omega_{ \text{FS} }\) is still Kähler:

1. \(\omega\) is closed: this is true for any smooth map at the level of smooth manifolds because of the chain rule.

2. \(\omega\) defines a metric: this is true because \(S\) is a complex submanifold. Suppose \(v,w \in T_p S\), and we want to check if \(g(v, w) \mathrel{\vcenter{:}}=\omega(v, Jw)\). This equals \(\omega_{{ \text{FS} }}(v, JW)\), viewing \(T_p S \subseteq T_p {\mathbb{CP}}^N\), so this is equal to \(g_{ \text{FS} }(v, w)\).

Note that a submanifold of a symplectic manifold is not necessarily a symplectic submanifold, since there are Lagrangian submanifolds for which the symplectic form restricts to 0 and isn’t nondegenerate. However, Kähler forms do restrict.

So we get a Hodge diamond:

Link to Diagram

Here \(h^{2, 0} = h^0( \Omega^2) = h^0(K) = g\) is called the genus in analogy with curves. Similarly, \(h^{1, 0} = h^0( \Omega^1)\) is the space of holomorphic 1-forms, sometimes referred to as the irregularity. There is some symmetry:

Link to Diagram

Last time: if we have such a Hodge diamond, can we solve for \(h^{1, 1}\)?

Link to Diagram

Recall Noether’s formula \begin{align*} \chi(S, {\mathcal{O}}_S) &= \int \operatorname{ch}({\mathcal{O}}_S) \mathrm{td}(S) \\ &= \int_S {x_1 \over 1-e^{-x_1} } {x_2 \over 1- e^{-x_2} } \\ &= {K^2 + \chi_{\mathsf{Top}}(S) \over 12} ,\end{align*} where \(c_1(TS) = - K\) and \(\chi_{\mathsf{Top}}\) is due to the Chern-Gauss-Bonet formula. We have \begin{align*} \chi({\mathcal{O}}_S) = h^0({\mathcal{O}}_S) - h_1({\mathcal{O}}_S) + h^2({\mathcal{O}}_S) = 1-q+p .\end{align*}

On the other hand, \begin{align*} \chi_{\mathsf{Top}}(S) = 1 -2q + (2p + h^{1, 1}) -4q = 1-4q + 2^p + h^{1, 1} ,\end{align*} so \begin{align*} 12(1-q+p) = K^2 + 2-4g + 2p + h^{1, 1} \implies h^{1, 1} = 110 - 8q + 10p-K^2 .\end{align*}

Recall the extraordinarily important exact sequence \begin{align*} 0 \to {\mathcal{O}}(-p) \to {\mathcal{O}}\to {\mathcal{O}}_p \to 0 ,\end{align*} where the right-hand side is the sheaf of holomorphic functions vanishing at \(p\) and this is an inclusion into the sheaf of holomorphic functions, and the right-hand term is the skyscraper sheaf. There is a similar exact sequence for an embedded curve \(C\hookrightarrow S\) in a surface: \begin{align*} 0 \to {\mathcal{O}}_S(-C) \to {\mathcal{O}}_S \to {\mathcal{O}}_C \to 0 ,\end{align*} where the left term is the sheaf of holomorphic functions vanishing on \(C\). Note that this has no global sections! Any function vanishing along a compact subset (?) are constant (?). Locally on an open set \(U\), one can write \(C \cap U = V(f_u)\), since algebraically this ring is locally a PID. So this is a line bundle, where we can map into the trivial bundle by \(\phi \mapsto \phi/f_u\). Thus \begin{align*} {\mathcal{O}}_S(U) / {\mathcal{O}}_S(-C)(U) \cong {\mathcal{O}}_C(C \cap U ) .\end{align*} We then get surjectivity since every holomorphic function on \(C\) extends to a holomorphic function on \(S\).

Now letting \(\mathcal{E} \in{ \mathsf{Vect} }(\mathop{\mathrm{Hol}})\), we can tensor this exact sequence to get \begin{align*} 0 \to \mathcal{E}(-C) \to \mathcal{E} \to { \left.{{\mathcal{E}}} \right|_{{C}} } \to 0 ,\end{align*} which is also exact since locally we have the splitting principle.

With some more work, one can show \(L\cong {\mathcal{O}}(C-D)\) for \(C,D\) smooth divisors.

Felix Klein has a “proof” of the existence of a meromorphic function on a Riemann surface. The argument roughly goes as follows: take your Riemann surface and make it out of metal. Attach it to a battery:

This induces an electric potential function \(V: C\to {\mathbb{R}}\), where \(V\) is the real part of the meromorphic function. Here \(V\) is a harmonic function away from \(p\) and \(q\).

Last time: line bundles are of the form \({\mathcal{O}}(D)\) for \(D\) a divisor, and the extremely important SES \begin{align*} 0 \to {\mathcal{O}}_S(-D) \to {\mathcal{O}}_S \to {\mathcal{O}}_D \to 0 .\end{align*} We now want to discuss an alternative characterization of the intersection form on an algebraic surface. The next result comes from Beauville’s “Complex Algebraic Surfaces”:

This will count intersection points after a small perturbation. Note that not every two curves will intersect transversely: consider \({\mathbb{P}}_2\) with a line \(C\) and a tangent conic \(D\):

What’s the payoff of this algebraic work? We can compute the Euler characteristic as \begin{align*} \chi( {\mathcal{O}}_{C \cap D}) = h^0({\mathcal{O}}_{C \cap D}) = \sum_{p\in C \cap D} \mathop{\mathrm{len}}_p(C \cap D ) .\end{align*} But by additivity of \(\chi\) over exact sequences, we also have \begin{align*} \chi({\mathcal{O}}_{C \cap D}) &= \chi({\mathcal{O}}_S) - \chi({\mathcal{O}}_S(-C)) - \chi({\mathcal{O}}_S(-D)) + \chi( {\mathcal{O}}_S( -C -D))\\ &\overset{HRR}{=} \int_S \qty{ \operatorname{ch}({\mathcal{O}}_S) - \operatorname{ch}({\mathcal{O}}_S(-C)) - \operatorname{ch}( {\mathcal{O}}_S(-D)) + \operatorname{ch}( {\mathcal{O}}_S(-C-D)) } \mathrm{td}(S) \\ &= c_1( {\mathcal{O}}_S(-C)) \cdot c_1({\mathcal{O}}_S(-D)) \\ &= \qty{ -[C] } \cdot \qty{ -[D] } \\ &= C\cdot D .\end{align*}

Next time: adjunction formula that allows computing genus for surfaces.

Last time: let \(C, D \subset S\) be distinct curves, then the intersection number is given by \begin{align*} C\cdot D = \deg {\mathcal{O}}_S(C) { \left.{{}} \right|_{{D}} } = \sum_{p \in C \cap D } \mathop{\mathrm{len}}_p( C \cap D) \end{align*} where \(\mathop{\mathrm{len}}_p(C \cap D) \mathrel{\vcenter{:}}=\dim_{\mathbb{C}}{\mathcal{O}}(U) / \left\langle{ f, g }\right\rangle\) where \(V(f) = C \cap U\) and \(V(g) = D \cap U\) with \(f, g \in {\mathcal{O}}(U) = \mathop{\mathrm{Hol}}(U)\). Here we’re also assuming that \(C \cap D \cap U = \left\{{ p }\right\}\).

We’ll now discuss a way to compute the genus of a curve in a surface.

See the Trott curve: \begin{align*} 144(x^4 + y^4) - 225(x^2 + y^2) + 350x^2 y^2 + 81 = 0 ,\end{align*} whose plot looks like the following:

f(x,y) = 12^2*(x^4 + y^4) - 15^2*(x^2 + y^2) + 350*x^2*y^2 + 81
implicit_plot(f, (x,-1,1), (y,-1,1))

Recall the adjunction formula: for \(D \subset X \in {\mathsf{Mfd}}_{\mathbb{C}}\) a codimension 1 complex submanifold, we have \begin{align*} K_D = (K_x + {\mathcal{O}}_x(0)) { \left.{{}} \right|_{{D}} } .\end{align*} We’ll apply this to curves \(C\) in a surface \(S\). Recall the genus formula, which was given by \(2g(C) - 2= (C+ K_S)\cdot C\). For example, a degree 4 equation in \({\mathbb{P}}^2\) carves out a genus \(g(C) = 3\) complex curve.

Recall that line bundles on \({\mathbb{CP}}^n\) were in bijection with \({\mathbb{Z}}\), where send \(d\) to a bundle \({\mathcal{O}}(d) \mathrel{\vcenter{:}}={\mathcal{O}}_{{\mathbb{CP}}^N}(d)\). We produced the tautological line bundle \({\mathcal{O}}(-1)\) whose fiber over \(\mathbf{x} \subseteq {\mathbb{CP}}^n\) is the line in \({\mathbb{C}}^n\) spanned by its coordinates. We have \({\mathcal{O}}(-1) {}^{ \vee }\mathrel{\vcenter{:}}={\mathcal{O}}(1)\), and \({\mathcal{O}}(n)\mathrel{\vcenter{:}}={\mathcal{O}}(1)^{\otimes n}\). Alternatively, it was characterized in terms of homogeneous functions, where the fiber \({\mathcal{O}}(n)_{\mathbf{x}}\) are the linear functions \(L\) on lines \(\left\{{\lambda \mathbf{x}}\right\} \to {\mathbb{C}}\) such that \(L(\lambda p) = \lambda^n L(p)\). Noting that these are linear functions, such \(L\) form a 1-dimensional \({\mathbb{C}}{\hbox{-}}\)vector space.

Note that the global sections were given by \(\Gamma^0({\mathbb{P}}^n, {\mathcal{O}}(d)) = H^0({\mathbb{P}}^n; {\mathcal{O}}(d))\) was the span of degree \(d\) monomials in \(x_0, \cdots, x_n\). For example \(x_0^2 + x_1 x_2\) is a well-defined element of \({\mathcal{O}}(2)_p\) which varies holomorphically with \(p\), yielding a section:

We know that \(H^0(S, K_S)\) are the globally holomorphic 2-forms on \(S\), and here this is isomorphic to \(H^0(S, {\mathcal{O}}_S) = {\mathbb{C}}\Omega_S\) for some single holomorphic 2-form. Moreover \(\operatorname{Div}(\Omega_S) = 0\) since \({\mathcal{O}}( \operatorname{Div}( \Omega_S)) = K_S = {\mathcal{O}}_S\). So these are the analogs of elliptic curves in dimension 2, since for example \(E \mathrel{\vcenter{:}}={\mathbb{C}}/ \Lambda\) has a nonvanishing section \(\,dz\in H^0(E, K_E)\), and we can write \(E = V(f)\) for \(f\) a cubic in \({\mathbb{P}}^3\), and we computed the genus of cubics. Moreover, every genus 1 curve is \({\mathbb{C}}\) mod a lattice.

Recall an exercise from the notes: computing the Hodge diamond of a genus 5 curve. We’ll compute the diamond for a K3 surface:

Link to Diagram

We know \(h^{2, 0} = H^0( S, \Omega_S^2)\), which yields 1s:

Link to Diagram

We’ll use the following theorem:

Note that anything embedded in projective space as a complex submanifold is Kähler by restricting the Fubini-Study form. Using simple connectedness and Serre duality, we have

Link to Diagram

We know \(\chi({\mathcal{O}}_S) = (1/12)(K^2+\chi_{\mathsf{Top}})\), and since \(K_S = {\mathcal{O}}_S\) is trivial, we have \(c_1({\mathcal{O}}_S) = 0\). Noting that \(h^{p, q} = H^( \Omega^p)\), so we can sum the lower-right part of the diamond to get \(\chi({\mathcal{O}}_S) = 1 - 0 + 1 = 2\), since we take \(p=0\) to get \(\Omega^p = {\mathcal{O}}\). Computing \(\chi_{\mathsf{Top}}\), we get \(h_{1, 1} = 20\).

Last time: the Lefschetz hyperplane theorem. Intersecting a projective variety of dimension \(d\geq 3\) with a hypersurface \(S\), the map \(\pi_1({\mathbb{P}}^3) \to \pi_1(S)\) is an isomorphism. We saw that K3 surfaces were thus simply connected, and \(h^1(S; {\mathbb{C}}) = 0\), so we could compute the Hodge diamond.

Check out complete intersections.

We can thus write \begin{align*} \mathop{\mathrm{Bl}}_pS S \setminus\left\{{p}\right\} {\textstyle\coprod}_{U^*} \mathop{\mathrm{Bl}}_p U .\end{align*}

Writing \(\pi: \mathop{\mathrm{Bl}}_pS\to S\), we have \(\pi^{-1}(p) \cong {\mathbb{CP}}^1\) and \(\pi^{-1}(q)\) is a point for all \(q\neq p\). Then all limits approaching \(p\) in \(S\) turn into distinct limit points in \(\mathop{\mathrm{Bl}}_p(S)\)

This is referred to as \({\mathbb{F}}_1\), the first Hirzebruch surface.

Last time: we defined the blowup \(\mathop{\mathrm{Bl}}_0{\mathbb{C}}^2\) as the closure of \begin{align*} \mathop{\mathrm{Bl}}_0 {\mathbb{C}}^2 \mathrel{\vcenter{:}}={ \operatorname{cl}} \left\{{ (x, y), [x:y] {~\mathrel{\Big|}~}(x, y) \neq 0 }\right\} \subseteq {\mathbb{C}}^2 \times{\mathbb{CP}}^2 .\end{align*} This had the effect of adding in all limits of slopes as points approach \((0, 0) \in {\mathbb{C}}^2\). We defined this using local holomorphic coordinate charts to \({\mathbb{C}}^2\). Why is this a complex manifold? We can cover it with charts: given a point \((x, \mu)\) where \(\mu = {y \over x}\in {\mathbb{P}}^1\) is a slope, we can form a first chart by sending \begin{align*} (x, \mu) \mapsto \left\{{ (x, x\mu), [1: \mu] }\right\} .\end{align*} This yields the first chart, as long as the slope is not infinite, so this applies to all finite slopes. The second chart will work for all nonzero slopes, where we take \begin{align*} (v, y)\in {\mathbb{C}}^2 \mapsto \left\{{ (yv, y), [v: 1] }\right\} .\end{align*} Note that restricting to \((x, y) = (0, 0)\), these give the standard \({\mathbb{C}}{\hbox{-}}\)charts on \({\mathbb{CP}}^2\). How do these two charts glue? When \(\mu, \nu \neq 0\), we have well-defined transition functions \(\mu = \nu ^{-1}\) and \(x=y\nu\).

Recall that for a complex curve \(C \in {\mathsf{Mfd}}_{\mathbb{C}}^2\), we have the blowup morphism \(\pi: \mathop{\mathrm{Bl}}_pS\to S\) and we defined the strict transform \(\widehat{C} \mathrel{\vcenter{:}}={ \operatorname{cl}} \pi^{-1}(C\setminus\left\{{{\operatorname{pt}}}\right\})\).

Here \(E={\mathbb{CP}}^1\) is the exceptional curve of the blowup, and intersects the curve twice. This has the effect of changing \(D\) into an embedded curve.

Note that here \(\pi^* D = \widehat{D} + 2E\), where we’ll define this next.

Using \(\pi: \widehat{S}\to S\), we get pullback maps \begin{align*} \pi^*: H^2(S; {\mathbb{Z}}) &\to H^2(\widehat{S}; {\mathbb{Z}}) \\ \pi^*: \operatorname{Div}(S) &\to \operatorname{Div}(\widehat{S}) .\end{align*} These are compatible in the sense that \begin{align*} [\pi^* C] = \pi^* [C] .\end{align*} . This can be seen by expressing \({\mathcal{O}}_S(C) \cong {\mathcal{O}}_S(A-b)\) for \(A, B\) hyperplane section. We can assume \(A, B\) avoid \(p\) in their projective embeddings, making \([C] = [A] - [B]\) since \(c_1({\mathcal{O}}_S(c)) = [C]\) is the fundamental class of \(C\). So it suffices to prove the formula for curves not passing through \(p\), but this is obvious! It follows from the fact that \(\pi: \widehat{S} \setminus E \xrightarrow{\sim} S\setminus\left\{{p}\right\}\) is an isomorphism.

In fact, \begin{align*} H^2(\widehat{S}; {\mathbb{Z}}) \cong \pi^* H^2(S, {\mathbb{Z}}) \oplus {\mathbb{Z}}[E] .\end{align*} , which follows from Mayer-Vietoris. So this adds one to the rank.

Recall that we have the following: \begin{align*} H^2(\widehat{S}; {\mathbb{Z}}) = \pi^* H^2(S; {\mathbb{Z}}) \oplus {\mathbb{Z}}[E] \end{align*} where \(E\) is the exceptional curve, which follows from Mayer-Vietoris. We can write \(\widehat{S} = S \# \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu\), and by excision \(H^2(S\setminus{\mathbb{B}}^4) = H^2(S)\). So we get a LES

Link to Diagram

We have \(H^i(S, S\setminus{\mathbb{B}}^4) = H^i(T, T\setminus{\mathbb{B}}^4) = H^i( {\mathbb{B}}^4, {{\partial}})\), and by Poincaré-Lefschetz duality, this is isomorphic to \(H_{4-i}({\mathbb{B}}^4)\). This is equal to 0 if \(i\neq 0\) or \(4\). Writing \(\widehat{S} = (S\setminus{\mathbb{B}}^4) {\textstyle\coprod}_{S^3} (\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu\setminus{\mathbb{B}}^4)\) and applying Mayer-Vietoris yields

Link to Diagram

Combining this with the isomorphisms from earlier, we can write the direct sum as \(H^2(S) \oplus H^2( \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu)\) where the latter is equal to \({\mathbb{Z}}\ell = [E]\) for \(\ell\) a line class.

Using the proposition, along with the fact that

1. its an orthogonal decomposition,
2. \(\pi^*\) is an isometry, and
3. \([E]^2 = -1\),

we know that the Gram matrix for \(H^2(\widehat{S})\) is the same as that for \(H^1(S) \oplus [-1]\), i.e. it is of the form \begin{align*} \begin{bmatrix} A & 0 \\ 0 & -1 \end{bmatrix} .\end{align*}

Note that this is the exact situation when we blow things up. This is a converse: if we have something that looks like a blowup, we can find something that blows up to it.

This is interesting because there does exist a blowdown in the smooth category \({\mathsf{Mfd}}(C^\infty({\mathbb{R}}))\). This is because \(S \to S\# \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{CP}}^2\mkern-1.5mu}\mkern 1.5mu\) and \(S\to S\# {\mathbb{CP}}^2\) are indistinguishable here. One can just reverse orientations.

There’s a way to do this with Kirby Calculus.

Why can’t one blow down a curve \(E\cong {\mathbb{CP}}^1\) with \(E^2 = 1\) in a complex surface? Disproof: consider \(S\mathrel{\vcenter{:}}={\mathbb{CP}}^2\) and \(E\) a line, where \(E^2 = 1\). If there were a blowdown in the complex analytic category \begin{align*} S &\to \mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \\ E &\mapsto {\operatorname{pt}} .\end{align*} But \(\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \cong_{\mathsf{Top}}S^4\), since \(S^4 \# {\mathbb{CP}}^2 \cong {\mathbb{CP}}^2\), and this would yield a complex structure on \(S^4\) – a contradiction. This also follows because \(\mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu \in \mathbb{Z}\operatorname{HS}^4\), and Noether’s formula implies that every \(\mathbb{Z}\operatorname{HS}^4\) has no complex structure.

Recall that we were considering the following:

Link to Diagram

Let \(\mkern 1.5mu\overline{\mkern-1.5mu\ell\mkern-1.5mu}\mkern 1.5mu \subset \mathop{\mathrm{Bl}}_{p, q}({\mathbb{CP}}^2)\) the strict transform of a line through \(p, q\) with \(\widehat{\ell}^2 = -1\). Goal: we want to construct the map \(\sigma\) sending \(\widehat{\ell}\) to a single point. Let \(r\in \mathop{\mathrm{Bl}}_{p, q}{\mathbb{CP}}^2\), then there are three possibilities:

1. \(r\in {\mathbb{CP}}^2 \setminus\left\{{ p, q }\right\}\)
2. \(r\in E_p\)
3. \(r\in E_q\)

If a point \(r \neq p, q\), we can take lines \(\ell_{pr}. \ell_{qr}\). We can take slopes of these lines to get points in \({\mathbb{CP}}^1\), and in fact it’s the exceptional divisor (since these are sets of slopes through a point).

So we can map \begin{align*} r \mapsto \begin{cases} ( {\mathrm{slope}}_p \ell_{pr}, {\mathrm{slope}}_q \ell_{qr} ) \in {\mathbb{CP}}^2 \times{\mathbb{CP}}^2 & \text{Case 1} \\ (r, {\mathrm{slope}}_q \ell_{qp} ) & \text{Case 2} \\ ({\mathrm{slope}}_p \ell_{pq}, r) &\text{Case 3} . \end{cases} \end{align*} This is clearly continuous, is this injective? The outputs will be the same for any point on the line between \(p\) and \(q\):

So this realizes the blowdown map, since \(\Phi \widehat{\ell}_{pq} ) = {\operatorname{pt}}\) and restricting it to the complement of the line is injective.

Goal: show that \(3[\ell]\) can’t be realized by a sphere, we’ll need Rohklin’s theorem for this. Let \((V, {\left\langle {{-}},~{{-}} \right\rangle})\) be an inner product space, and assume the inner product is positive-definite. Recall that the tensor algebra is defined as \(T(V) \mathrel{\vcenter{:}}=\bigoplus _{n\geq 0} V^{\otimes n}\).

Given \((V, \cdot)\) an inner product space, we defined \begin{align*} { \operatorname{Cl}} (V) \mathrel{\vcenter{:}}={ \bigoplus _{n\geq 0} V^{\otimes n} \over \left\langle{ v\otimes w + w\otimes v = 2v\cdot w }\right\rangle } .\end{align*}

Note that

• \(k[x_1, \cdots, x_n]\) is graded (by monomials of uniform degree) and filtered (by polynomials of a bounded degree)
• \(T(V)\) is graded and filtered, since multiplying a pure \(p\) tensor with a pure \(q\) tensor yields a pure \(p+q\) tensor
• \({ \operatorname{Cl}} (V)\) is a quotient of \(T(V)\), but one can’t simply define \({ \operatorname{Cl}} (V, \cdot)^i = \operatorname{im}T(V)^i\) since the relations have mixed degree: for example \(e_1^2 = -1\) So \({ \operatorname{Cl}} (V)\) isn’t graded, but is still filtered: take the filtration \(F\) on \(T(V)\) defined by \(F^i \mathrel{\vcenter{:}}=\bigoplus _{j\leq i} V^{\otimes j}\) and descend it through the quotient map. The relations can only decrease degree, so this is well defined.

Note that if \(R \in {\mathsf{gr}\,}\mathsf{Ring}\), then \({\mathsf{gr}\,}(R) = R\), so taking the associated graded recovers the ring itself. What’s happening: taking the smallest homogeneous ideal.

In our case, the Clifford relation relates degree \(k\) pieces to degree \(k-2\) pieces, so we obtain \begin{align*} {\mathsf{gr}\,}_{F^{-}} { \operatorname{Cl}} (V) \cong T(V) / \left\langle{ v\otimes w + w\otimes v = 0 }\right\rangle \mathrel{\vcenter{:}}=\bigwedge\nolimits^* V .\end{align*} There is an isomorphism of \(k{\hbox{-}}\)vector spaces \begin{align*} { \operatorname{Cl}} (V) & \xrightarrow{\sim} {\mathsf{gr}\,}{ \operatorname{Cl}} (V) \\ x\in F^i &\mapsto \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu \in F^i / F^{i-1} .\end{align*} This is because \(F^0 \subseteq \cdots \subseteq \cdots\) with \(\cup_i F^i = { \operatorname{Cl}} (V)\). We can conclude \(\dim_{\mathbb{R}}{ \operatorname{Cl}} (V) = \dim_{\mathbb{R}}\bigwedge\nolimits^* V = 2^{\dim_k V}\) and use this to construct a basis for \({ \operatorname{Cl}} (V)\). The relevant map is \begin{align*} { {e_j}_1, {e_j}_2, \cdots, {e_j}_{i}} &\mapsto e_{j_1} \wedge \cdots \wedge e_{j_i} .\end{align*}

Last time: we defined \({\operatorname{Pin}}(n) \subseteq { \operatorname{Cl}} ({\mathbb{R}}^n)\) which was generated by \(S^1({\mathbb{R}}^n)\). These were units because \(v^2 = -{\left\lVert {v} \right\rVert}^2 = -1\), so \(v^{-1}= -v\), and formed a group contained in \({ \operatorname{Cl}} ({\mathbb{R}}^n)^{\times}\). There is a decomposition \({ \operatorname{Cl}} (V) = { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V)\) with a \({\mathbb{Z}}/2{\hbox{-}}\)grading, and we defined \begin{align*} {\operatorname{Spin}}(V) \mathrel{\vcenter{:}}={\operatorname{Pin}}(V) \cap{ \operatorname{Cl}} _0(V) = \left\langle{ vw {~\mathrel{\Big|}~}v,w\in S^1({\mathbb{R}}^n) }\right\rangle .\end{align*} There is a map \begin{align*} {\operatorname{Pin}}(n) &\twoheadrightarrow O(n) \\ v & \mapsto (u\mapsto vuv^{-1}) = -R_{v^\perp} ,\end{align*} which preserves \(V^{\otimes 1} \subset { \operatorname{Cl}} (V)\), and was reflection about the hyperplane \(v^\perp\). There is also a SES \begin{align*} 0 \to {\mathbb{Z}}/2 \to {\operatorname{Spin}}(n) \xrightarrow{\pi} {\operatorname{SO}}(n) \to 0 ,\end{align*} where we used the fact that \(\ker \pi \subset Z{ \operatorname{Cl}} ({\mathbb{R}}^n)\). It turns out that \({\operatorname{Spin}}(n) = \overline{ {\operatorname{SO}}(n) }\), using that \(\pi_1( {\operatorname{SO}}(n), {\operatorname{pt}}) = {\mathbb{Z}}/2\) and checking that \(\pm 1\in {\operatorname{Spin}}(n)\), yielding a nontrivial kernel.

This is local, at a single vector space, so we’ll now try to globalise this to the tangent space of a manifold.

We showed that \({\mathsf{gr}\,}{ \operatorname{Cl}} ({\mathbb{R}}^n) = \bigwedge\nolimits{\mathbb{R}}^n\), and so there is a bundle isomorphism \begin{align*} { \operatorname{Cl}} (X) \xrightarrow{\sim} \bigwedge\nolimits^* T {}^{ \vee }X ,\end{align*} but the ring structure is different. On the right, we have a way of multiplying sections, namely \(\omega_1 \wedge \omega_2\), but on the left we have the Clifford multiplication \(\alpha_1 \cdot \alpha_2\). Note that \(\omega^{\wedge 2} = 0\), but \(\alpha^{\cdot 2} \in {\mathbb{R}}\) is some scalar. We define \(\omega\cdot \omega= g^*( \omega, \omega)\), so we use the metric fiberwise to define a Clifford multiplication.

This is principal since any two elements are related by a unique element of \({\operatorname{SO}}(n)\). Recall that we had an associated bundle construction, so taking the standard representation \(\rho: {\operatorname{SO}}(n) \to ({\mathbb{R}}^n, g)\) where elements act by their transformations (?), there is an oriented bundle \(P { \underset{\scriptscriptstyle {\rho} }{\times} } {\mathbb{R}}^n\). If the bundle is \(TX\) with a metric \(g\), this yields a distinguished \({\operatorname{SO}}(n)\) bundle \(P\to X\).

Recall that a \(G{\hbox{-}}\)torsor is a set with a free transitive \(G{\hbox{-}}\)action. For example, the fibers of a principal bundle are torsors. Given any two torsors, we can compare them using elements of \(G\), but there is no distinguished element. For example, \({\mathbb{A}}_n\) is a torsor over the vector space \(k^n\).

This is a nice example to get a hang of the use and importance of Čech cohomology. We then use the isomorphism \({\check{H}}\to H_{{\operatorname{Sing}}}\).

It turns out that \({ \operatorname{Cl}} (V) \otimes_{\mathbb{R}}{\mathbb{C}}\cong \mathop{\mathrm{End}}(S)\). The left-hand side contains \({\operatorname{Spin}}(n)\), so given \(\rho: { \operatorname{Cl}} (V) \to \mathop{\mathrm{End}}(S)\) a representation (i.e. a ring homomorphism) in matrices, we can restrict \(\rho\) to \({\operatorname{Spin}}(n)\) to get \(\rho { \left.{{}} \right|_{{{\operatorname{Spin}}(n)}} }: {\operatorname{Spin}}(n) \to \operatorname{GL}(S)\). Next time: spin representations. Spinor bundle will be sections of associated bundle of the Clifford bundle.

Last time: we defined \begin{align*} { \operatorname{Cl}} (V, \cdot) &\mathrel{\vcenter{:}}=\bigoplus_n V^{\otimes n} / \left\langle{ v\otimes v = -{\left\lVert {v} \right\rVert}^2 1 }\right\rangle \\ {\operatorname{Pin}}(V) &\mathrel{\vcenter{:}}=\left\langle{ v {~\mathrel{\Big|}~}{\left\lVert {v} \right\rVert} = 1 }\right\rangle \subseteq { \operatorname{Cl}} (V) .\end{align*} There is a \({\mathbb{Z}}/2\) grading \({ \operatorname{Cl}} (V) = { \operatorname{Cl}} _0(V) \oplus { \operatorname{Cl}} _1(V)\) where \({ \operatorname{Cl}} _0(V)\) is the image of even tensors and \({ \operatorname{Cl}} _1(V)\) is the image of odd tensors. We also had \begin{align*} {\operatorname{Spin}}(V) \mathrel{\vcenter{:}}={\operatorname{Pin}}(V) \cap{ \operatorname{Cl}} _0(V) = \left\langle{ v\cdot w {~\mathrel{\Big|}~}v,w \in V, \, {\left\lVert {v} \right\rVert} = {\left\lVert {w} \right\rVert} = 1 }\right\rangle .\end{align*} There was a map \begin{align*} {\operatorname{Pin}}(V) &\to O(V) \\ v &\mapsto -R_v ,\end{align*} where \(R_v\) was reflection about \(v^\perp\), where we identified this as an action on \(V^{\otimes 1} \subset { \operatorname{Cl}} (V)\) where \(u\to vuv^{-1}\). For any Riemannian manifold \((X, g)\), we could define the Clifford bundle \({ \operatorname{Cl}} (X) = { \operatorname{Cl}} (T {}^{ \vee }X, g {}^{ \vee })\) to globalise this from vector spaces to bundles with metrics. We defined a spin structure on \(X\) as any lift of the principal \({\operatorname{SO}}(n)\) bundle over \((T {}^{ \vee }X, g)\) (namely \(\mathop{\mathrm{Frame}}(X)\)) to a \({\operatorname{Spin}}(n)\) bundle.

We showed that there exists a spin structure iff some cohomology class \(w_2(K) \in H^2(X; {\mathbb{Z}}/2)\) vanishes.

We thus have \({ \operatorname{Cl}} ({\mathbb{R}}^4) \curvearrowright{\mathbb{C}}^4\) by sending \(e_i \mapsto \delta_i\), the Dirac matrices. Using that \({\operatorname{Pin}}(4) \cap{ \operatorname{Cl}} ({\mathbb{R}}^4) = {\operatorname{Spin}}(4) \subseteq { \operatorname{Cl}} ({\mathbb{R}}^4)\), we can a spin representation, but this may no longer be irreducible. In fact, as a \({\operatorname{Spin}}(4)\) representation this splits into two irreducible representations. We know that \({\operatorname{Spin}}(4) \subseteq { \operatorname{Cl}} _0({\mathbb{R}}^4) = { \operatorname{Cl}} ({\mathbb{R}}^3)\) which has two complex conjugate irreducible representations. The key is to define an element \(\omega_{\mathbb{C}}\in { \operatorname{Cl}} (V) \otimes_{\mathbb{R}}{\mathbb{C}}\) with \(\omega_{\mathbb{C}}^2 = 1\), which yields a decomposition of \({\mathbb{S}}= {\mathbb{S}}^+ \oplus {\mathbb{S}}^-\) as the \(\pm 1\) eigenspaces of the action. Here \(\omega_C \mathrel{\vcenter{:}}=-e_1 e_2 e_3 e_4 \mapsto \gamma_5\). One can define \begin{align*} \gamma_5 \mathrel{\vcenter{:}}=\operatorname{im}(\omega_{\mathbb{C}}) = - \gamma_1 \gamma_2 \gamma_3 \gamma_4 = - \sigma_3 \otimes\sigma_3 \end{align*} and one obtains the matrix \begin{align*} -\sigma_3 \otimes\sigma_3 = \begin{bmatrix} 0 & 0 & 0 & -1 \\ 0 & 0& 1 & 0 \\ 0 & 1& 0 & 0 \\ -1 & 0 & 0 & 0 \end{bmatrix} .\end{align*} One can check that \(\gamma_5\) anticommutes with the \(\delta_i\) for \(1\leq i \leq 4\), and thus commutes with \({ \operatorname{Cl}} _0({\mathbb{R}}^4)\). We can write \({\mathbb{S}}^+\), the positive 1 eigenspace of \(\gamma_5\), as \({\mathbb{C}}(s_1 - s_4 ) \oplus {\mathbb{C}}(s_1 + s_2)\). So we have \({\operatorname{Spin}}(4) = { \operatorname{Cl}} ({\mathbb{R}}^4) \curvearrowright{\mathbb{C}}^2 \oplus {\mathbb{C}}^2 = {\mathbb{S}}\), which splits into \(\gamma_5{\hbox{-}}\)eigenspaces \({\mathbb{S}}^+ \oplus {\mathbb{S}}^-\), the positive and negative spinors. This means that \(\gamma_5\) commutes with the image of \({\operatorname{Spin}}(4) \hookrightarrow\operatorname{GL}( {\mathbb{C}}^2 \oplus {\mathbb{C}}^2)\).

Let \(g\in {\operatorname{Spin}}(4)\), and \(v^+ \in {\mathbb{S}}^+ \subseteq {\mathbb{S}}\). Question: is it true that \(g\cdot v^+ \in {\mathbb{S}}^+\)? If so, this yields a subrepresentation. If so, \(\gamma_5 v^+ = v^+\) since we’re in the \(+1\) eigenspace, and on the other hand, \(g\cdot v^+ = g \cdot \gamma_5 v^+ = g \omega_{\mathbb{C}}\cdot v^+\) where the last identification comes from the map \(\gamma_5 \mapsto \omega_{\mathbb{C}}\), and this is equal to \(\omega_{\mathbb{C}}g \cdot v^+\) using commutativity. So \(g\cdot v^+\) is in the \(+1\) eigenspace of \(\gamma_5\).

Now take \(\gamma_i\). This actually switches spinors: by anticommutativity of the \(\gamma_i\) with \(\gamma_5\), we have \begin{align*} \gamma_i \cdot v^+ = \gamma_i \gamma_5 v^+ = - \gamma_5 \gamma_i v^+ ,\end{align*} which means \(\gamma_i v^+\) is in the \(-1\) eigenspace for \(\gamma_5\), i.e. \(\gamma_i v^+ \in {\mathbb{S}}^-\).

Suppose one has a spin structure and \(\tilde P \to X\) is a principal \({\operatorname{Spin}}(n)\) bundle. There are bundles over this of the form \(\rho: {\operatorname{Spin}}(n) \to \operatorname{GL}({\mathbb{S}}^\pm)\), yielding the spinor bundle \begin{align*} \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}= {\mathbb{S}}_x^+ \oplus {\mathbb{S}}_x^- .\end{align*}

Let \(G \xrightarrow{\rho} \operatorname{GL}(V)\) be any representation. If \(\phi \in \mathop{\mathrm{End}}(V)\) commutes with \(\rho(G)\), then the eigenspaces of \(\phi\) are subrepresentations. In other words, \(G\curvearrowright V = \bigoplus _{i=1}^n V_{\lambda_i}\), then \(G\curvearrowright V_{\lambda_i}\) is a subrepresentation, using that \begin{align*} \phi(v) = \lambda v \implies gv = g\phi(\lambda ^{-1}v) = \phi \rho(g) \lambda^{-1}v ,\end{align*} which says \(\phi( \rho(g) \cdot v) = \lambda (\rho(g) \cdot v) \implies \rho(g) \cdot v \in V_{\lambda}\). We used it here by This rephrases Schur’s lemma!

Last time: we defined a Spin structure on an oriented manifold \(M\) as a lift of the principal \({\operatorname{SO}}(n)\) bundle \(P\to M\) (unassociated to \(TM\)) to a \({\operatorname{Spin}}(n)\) bundle \(\tilde P\). There was a spin representation \({\operatorname{Spin}}(n) \curvearrowright{\mathbb{S}}\), which is irreducible for \({ \operatorname{Cl}} ({\mathbb{R}}^n)\) and splits as \({\mathbb{S}}= {\mathbb{S}}^+ \oplus {\mathbb{S}}^-\), which are \({\operatorname{Spin}}(n)\) subrepresentations. We defined spinor bundles \begin{align*} \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}= {\mathbb{S}}_M = {\mathbb{S}}_M^+ \oplus {\mathbb{S}}_M^- .\end{align*}

Principal bundle: fibers are left \(G{\hbox{-}}\)torsors. In the fiber product, the group sits in the middle and acts on each factor. So \(\tilde P\) eats the right \(G{\hbox{-}}\)action, and \({\mathbb{S}}\) eats the left action. Remarkably, for Spin bundles, there is an action leftover.

We have an isomorphism of bundles (not of algebras) \({ \operatorname{Cl}} (M) \cong \bigwedge\nolimits T {}^{ \vee }M\), and any one form \(\omega\) is an analogue of an element of \(V^{\otimes 1}\), and \(\omega \cdot ({\mathbb{S}}^+, {\mathbb{S}}^-) \in {\mathbb{S}}_M^- \oplus {\mathbb{S}}_M^+\).

It is not obvious that a Clifford connection exists! We have \({\mathbb{S}}_M = \tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}\), so it suffices to give a connection on \(\tilde P\) which is \({\operatorname{Spin}}(n)\) invariant, since any associated bundle will inherit the connection. Idea: we need a notion of parallel transport. This is a principal \({\operatorname{Spin}}(n)\) bundle, so the fibers look like \({\operatorname{Spin}}(n)\), and we want to lift paths in \(M\) to paths in \(\tilde P\):

It suffices to give a connection on \(P\), and using that \(\tilde P\to P\) is a 2 to 1 covering map, we can take a connecting on \(P\) coming from \(\mathop{\mathrm{OFrame}}(T {}^{ \vee }M, g {}^{ \vee })\). So it further suffices to produce a connection on \(T {}^{ \vee }M\) preserving orthogonality of frames under parallel transport, which is essentially the definition of the Levi-Cevita connection \(\nabla^{\mathrm{LC}}\). Then the \(\nabla\) associated to \(\nabla^{\mathrm{LC}}\) on \(P\) is a Clifford connection, yielding existence.

The set of Clifford connections is a torsor over \(\Omega^1(M)\). The association is \(\nabla \mapsto \nabla - \nabla^{\mathrm{LC}}\), and one can compute \begin{align*} (\nabla - \nabla^{\mathrm{LC}})(x\cdot s) = x \cdot( \nabla - \nabla^{\mathrm{LC}})(s) ,\end{align*} which exactly says that this is a \({ \operatorname{Cl}} (M){\hbox{-}}\)linear map \({\mathbb{S}}_M \to {\mathbb{S}}_M \otimes\Omega^1\). We can write \({ \operatorname{Cl}} (M) \cong \mathop{\mathrm{End}}({\mathbb{S}}_M)\), and one can check that \([\mathop{\mathrm{End}}{\mathbb{S}}_M, \mathop{\mathrm{End}}{\mathbb{S}}_M]\) consists only of scalars.

This makes sense locally, and is well-defined independent of choice of frame. Henceforth, we’ll take \(\nabla = \nabla^{\mathrm{LC}}\) – in this case, if \(s^+ \in H^0({\mathbb{S}}^\pm)\) then \(\nabla_v^{{\mathrm{LC}}}(s^\pm) \in H^0({\mathbb{S}}^\pm)\). This is an order 1 differential operator: \begin{align*} \mkern-3mu \not{ \partial} _{ \nabla^{\mathrm{LC}}} = \mkern-3mu \not{ \partial} : H^0({\mathbb{S}}^\pm) \to H^p({\mathbb{S}}^{\mp}) .\end{align*}

Last time: we showed \({ \operatorname{Cl}} (X) \mathrel{\vcenter{:}}={ \operatorname{Cl}} (\dualof{T} X, \dualof{g})\) acts on the spinor bundle \({\mathbb{S}}_X \mathrel{\vcenter{:}}=\tilde P { \underset{\scriptscriptstyle {{\operatorname{Spin}}(n)} }{\times} } {\mathbb{S}}\) by Clifford multiplication. For \(\dim_{\mathbb{R}}X = 4\), we have a splitting \({\mathbb{S}}^+ \oplus {\mathbb{S}}^-\) as complex rank 2 vector bundles. If \(\omega\in H^0 { \operatorname{Cl}} (X)\) is a one form, then \(\omega{\mathbb{S}}_X^\pm \subset {\mathbb{S}}^{\mp}\) .

There is a distinguished Clifford connection associated to \(\nabla^{\mathrm{LC}}\). Also recall that we defined a Dirac operator \(\mkern-3mu \not{ \partial}\) and showed \(\mkern-3mu \not{ \partial} ^2 = -2 \Delta\).

This describes fermions in a vacuum, e.g. an electron where \(\psi\) is its wave function. Applying this to \({\mathbb{R}}^4\) with \(g = (dt)^2 - (dx)^2 - (dy)^2 - (dz)^2\), then this equation in \(\psi\) is invariant under the Lorentz group \(O({\mathbb{R}}^4, g)\).

This restricts what topological manifolds can admit smooth structures. Freedman constructed a topological manifold of dimension 4 with signature 8, which thus can not admit a smooth structure. Recall that having a spin structure was having a lift of a principal \({\operatorname{SO}}(n)\) bundle over \((\dualof{T} X, g)\) (namely \(\mathop{\mathrm{Frame}}(X)\)) to a \({\operatorname{Spin}}(n)\) bundle.

Link to Diagram

Note there is ambiguity up to 2-torsion in the formula, but this gets moved into the choice of spin structure, which amounts to choice of a square root of each of these line bundles.

See the \(\widehat{A}\) genus.

If \(H_1(X; {\mathbb{Z}})\) has no 2-torsion, e.g. if \(\pi_1X = 0\), then \(w_2(X) = 0\) iff the intersection form on \(H^2\) is even, where \(w_2\) is the obstruction to existence of spin structures. Note that this makes sense for topological manifolds and not just smooth manifolds, and in this case \(\sigma(X)\) is divisible by 8. This restriction comes from number theory: since we have a unimodular lattice, it breaks into sums of \(E_8, -E_8\), and \(H\) if indefinite, and any even unimodular lattice has signature divisible by 8. So this can work as an obstruction to the existence of smooth structures.

Note that \({\mathbb{CP}}^2\) has no spin structure, and \(\sigma({\mathbb{CP}}^2) = 1\). There’s a way to modify the invariant to set \(\sigma(X)/8 = ? \pmod 2\).