Hodge theory: Lecture 14

We saw the last time that the following lets us show that the Dolbeault complex is exact everywhere, which lets us construct a resolution for the sheaf of $p$-forms (on a complex manifold).

Lemma ($\bar{\partial}$-lemma). Let $U \subset X$ be a nonempty open subset. Fix $p \geq 0.$ Given any $q \geq 1$ and any smooth $(p, q)$-form $\omega$ such that $\bar{\partial}\omega = 0$ on $U.$ Then we may find a local section $\beta \in \Gamma(V, \mathscr{A}^{p, q-1})$ such that $\bar{\partial} \beta = \omega$ on a nonempty open subset $V \subset U.$

Proof. We may work locally, so denote by $z_{1}, \dots, z_{n}$ a complex chart.

Step 1. Reduction to the case $p = 0.$

We may write $$\omega = \sum_{|I|=p, |J|=q}f_{I,J} dz_{I} \wedge d\bar{z_{J}}.$$ We have $$0 = \bar{\partial}\omega = \sum_{|I|=p, |J|=q}\frac{\partial f_{I,J}}{\partial \bar{z_{j}}} d\bar{z_{j}} \wedge dz_{I} \wedge d\bar{z_{J}}.$$ This implies that $\bar{\partial} \omega_{I} = 0$ for any $I,$ where $\omega_{I} := \sum_{|J|=q} f_{I,J} d\bar{z_{J}},$ which is a $(0, q)$-form.

Hence, if we have proven the case $p = 0,$ we can apply it for $\omega_{I}$ to choose a $(0, q-1)$-form $\beta_{I}$ such that $\bar{\partial} \beta_{I} = \omega_{I}.$ Thus, if we write $$\beta := \sum_{|I| = p} (-1)^{p} dz_{I} \wedge \beta_{I},$$ then $$\bar{\partial}\beta = \omega,$$ as desired.

Step 2. The case $p = 0.$

Since $p = 0,$ we have $$\omega = \sum_{|J|=q}f_{J} d\bar{z_{J}},$$ which is a $(0,q)$-form. We may assume that $$\omega = f d\bar{z_{j_{1}}} \wedge \cdots \wedge d\bar{z_{j_{q}}},$$ with $j_{1} < \cdots < j_{q}.$ Note that $\bar{\partial}\omega = 0$ precisely means that $\partial f / \partial \bar{z_{j}} = 0$ for all $j \in \{1, 2, \dots, n\} \setminus \{j_{1}, \dots, j_{q}\}.$ In other words, we are given that $f$ is holomorphic with respect to $z_{j}$ for all such $j.$

By single variable $\bar{\partial}$-lemma (in this previous posting--see "Observation"), by possible shrinking $U$ (but still nonempty open in $X$), we may find a smooth $g$ that is holomorphic with respect to $z_{j}$ for all such $j \in \{1, 2, \dots, n\} \setminus \{j_{1}, \dots, j_{q}\}$ such that $\partial g / \partial \bar{z_{j_{1}}} = f.$ This implies that $$\bar{\partial} (g d\bar{z_{j_{2}}} \wedge \cdots \wedge d\bar{z_{j_{q}}}) = \omega.$$ This finishes the proof. $\Box$

Remark. We may repeat the whole argument above when $M$ is a smooth (real) manifold to show that if $\omega$ is a closed $p$-form with $p \geq 1,$ then $\omega$ is locally exact. This is called Poincaré's lemma. This lets us have the following resolution called the de Rham complex: $$0 \rightarrow \underline{\mathbb{R}} \rightarrow \mathscr{A}^{0}_{M} \overset{d}{\longrightarrow} \mathscr{A}^{1}_{M} \overset{d}{\longrightarrow} \cdots \overset{d}{\longrightarrow} \mathscr{A}^{n}_{M} \rightarrow 0,$$ where $n = \dim_{\mathbb{R}}(M).$

Next time, we shall see that for any $p \geq 0,$ we have $$H^{i}(M, \mathscr{A}^{p}_{M}) = 0,$$ for $i \geq 1.$ In particular, this implies that $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})).$$ This is particularly interesting because for any $i \geq 0,$ we have $$H^{i}(M, \underline{\mathbb{R}}) \simeq H^{i}_{\mathrm{sing}}(M, \mathbb{R}),$$ the singular cohomology of $M$ with $\mathbb{R}$-coefficients. (We will not prove the last isomorphism, but the details are written here.) The isomorphism $$H^{i}(\Gamma(M, \mathscr{A}^{\bullet}_{M})) \simeq H^{i}_{\mathrm{sing}}(M, \mathbb{R})$$ is often referred as de Rham's theorem, and the left-hand side of this isomorphism is often denoted as $H^{i}_{\mathrm{dR}}(M),$ called $i$-th de Rham cohomology.

Hodge theory: Lecture 13

Dolbeault complex. Let $M$ be a complex manifold of dimension $n.$ Then the canonical almost complex structure on $TM$ gives us $$(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M,$$ and $T^{1,0}M$ turned out to be a holomorphic vector bundle, the holomorphic tangent bundle of $M$ to be more specific. The decomposition is functorial in the sense that for any holomorphic map $\phi : M \rightarrow M'$ between two complex manifolds, its push forward map $(TM)_{\mathbb{C}} \rightarrow \phi^{*}(TM')_{\mathbb{C}}$ restricts to $T^{1,0}M \rightarrow \phi^{*}T^{1,0}M'.$

Dually, we have $$(T^{\vee}M)_{\mathbb{C}} = A^{1,0}_{M} \oplus A^{0,1}_{M},$$ but we also note that $$(T^{\vee}M)_{\mathbb{C}} \simeq \mathrm{Hom}_{\mathbb{C}}((TM)_{\mathbb{C}}, \mathbb{C}),$$ where at each fiber, we have $(v \otimes 1)^{\vee} \mapsto v^{\vee} \otimes 1.$ Under this isomorphism we have $$A^{1,0}_{M} \simeq \mathrm{Hom}_{\mathbb{C}}(T^{1,0}M, \mathbb{C})$$ and $$A^{0,1}_{M} \simeq \mathrm{Hom}_{\mathbb{C}}(T^{0,1}M, \mathbb{C}).$$ We also have the complexified $k$-th exterior power $$\begin{align*}\left(\bigwedge^{k} T^{\vee}M\right)_{\mathbb{C}} &\simeq \bigwedge^{k} (T^{\vee}M)_{\mathbb{C}} \\ &= \bigwedge^{k} (A^{1,0}_{M} \oplus A^{0,1}_{M}) \\ &= \bigoplus_{p+q = k}\left(\bigwedge^{p} A^{1,0}_{M} \otimes_{\mathbb{R}} \bigwedge^{q} A^{0,1}_{M}\right).\end{align*}$$ We shall write $$A^{p,q}_{M} := \bigwedge^{p} A^{1,0}_{M} \otimes_{\mathbb{R}} \bigwedge^{q} A^{0,1}_{M},$$ and call it $(p,q)$-th exterior power.

Sections of the exterior power $\bigwedge^{k} T^{\vee}M,$ which are called (real) differential $k$-forms on $M$, form a sheaf $\mathscr{A}_{M}^{k},$ and we complexify this sheaf, meaning $$\mathscr{A}_{M, \mathbb{C}}^{k} := \mathscr{A}_{M}^{k} \otimes_{\underline{\mathbb{R}}} \underline{\mathbb{C}},$$ and this matches the sheaf of the complexified $k$-exterior of $M,$ so we have $$\mathscr{A}_{M, \mathbb{C}}^{k} = \bigoplus_{p+q = k}\left(\bigwedge^{p} \mathscr{A}^{1,0}_{M} \otimes_{\mathscr{C}^{\infty}_{M}} \bigwedge^{q} \mathscr{A}^{0,1}_{M}\right),$$ where $\mathscr{A}^{1,0}_{M}$ is the sheaf of sections of $A^{1,0}_{M},$ while $\mathscr{A}^{0,1}_{M}$ is the sheaf of sections of $A^{0,1}_{M}.$ We write $$\mathscr{A}^{p,q}_{M} := \bigwedge^{p} \mathscr{A}^{1,0}_{M} \otimes_{\otimes_{\mathscr{C}^{\infty}_{M}}} \bigwedge^{q} \mathscr{A}^{0,1}_{M}$$ and say the sections of them are $(p,q)$-sections of $M.$ Note that $$\overline{\mathscr{A}_{M}^{p,q}} = \mathscr{A}_{M}^{q,p},$$ where the conjugation happen in the target, namely the (complexified) $k$-th exterior power. Denote by $\Omega^{p}_{M}$ the sheaf holomorphic sections of the holomorphic vector bundle $A^{p,0}.$ We call the sections of $\Omega^{p}_{M}$ holomorphic $p$-forms on $M$ Note that we have $$\Omega^{p}_{M} \hookrightarrow \mathscr{A}^{p,0}_{M},$$ where the right-hand side is the sheaf of smooth (not necessarily holomorphic) sections of the holomorphic vector bundle $A^{p,0}.$ Note that $A^{1,0}$ is identified to be the (complex) dual of $T^{1,0}M,$ which is a holomorphic vector bundle whose local frame is given by $dz_{1}, \dots, dz_{n}$ for each local chart $z_{1}, \dots, z_{n}$ for $M.$ With such a local chart, say on an open subset $U \subset M,$ we note that $\mathscr{A}^{p,q}_{U}$ is a free $\mathscr{C}^{\infty}_{U, \mathbb{C}}$-module with basis $dz_{I} \wedge d\bar{z_{J}}$ with $I = (i_{1} < \cdots < i_{p})$ and $J = (j_{1} < \cdots < j_{q}).$

Consider the de Rham differential $$d : \mathscr{A}^{k}_{M} \rightarrow \mathscr{A}^{k+1}_{M}.$$ That is, on each open subset $U \subset M,$ the map $$d : \mathscr{A}^{k}_{M}(U) \rightarrow \mathscr{A}^{k+1}_{M}(U)$$ is given by the exterior derivative. Considering the complexified version $$\mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}}.$$ For any $p, q \geq 0$ such that $p + q = k,$ this gives $$\mathscr{A}^{p,q} \hookrightarrow \mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}} \twoheadrightarrow \mathscr{A}^{p+1,q},$$ which we call $\partial^{p,q}$ and $$\mathscr{A}^{p,q} \hookrightarrow \mathscr{A}^{k}_{M, \mathbb{C}} \overset{d}{\longrightarrow} \mathscr{A}^{k+1}_{M, \mathbb{C}} \twoheadrightarrow \mathscr{A}^{p,q+1},$$ which we call $\bar{\partial}^{p,q}.$

Proposition. When $M$ is complex manifold, for $p + q = k,$ we have $d = \partial + \bar{\partial}$ as a map $$\mathscr{A}^{p,q}_{M} \overset{d}{\longrightarrow} \mathscr{A}^{p+1,q}_{M} \oplus \mathscr{A}^{p,q+1}_{M}.$$
Personal remark. We don't really need this because the identity can be checked at the stalks, but one can check that $$(\mathscr{A}^{k} \otimes_{\underline{\mathbb{R}}} \underline{\mathbb{C}})(U) = \mathscr{A}^{k}(U) \otimes_{\mathbb{R}} \mathbb{C},$$ even without sheafification after tensoring.

Proof. We suppress $M$ in the notations for sheaves for simplicity. We can check the identity at an open subset $U \subset M$ with the coordinate functions $$x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n},$$ which induces the coordinate functions $$z_{1}, \dots, z_{n}, \bar{z_{1}}, \dots, \bar{z_{n}}.$$ Fix any $\omega \in \mathscr{A}^{p,q}(U).$ Our goal is to show that $$d\omega = \partial\omega + \bar{\partial}\omega \in \mathscr{A}^{k+1}(U) \otimes_{\mathbb{R}} \mathbb{C}.$$ We may assume that $\omega = f dz_{I} \wedge d\bar{z_{J}}$ with $|I| = p$ and $|J| = 1$ because $\mathscr{A}^{p,q}(U)$ is $\mathbb{C}$-linearly generated by such $k$-forms. We have $$df = \sum_{j=1}^{n} \frac{\partial f}{\partial z_{j}} dz_{j} + \sum_{j=1}^{n} \frac{\partial f}{\bar{\partial z_{j}}} d\bar{z_{j}}$$ because $f$ is holomorphic. By definition, we have $$\partial f = \sum_{j=1}^{n} \frac{\partial f}{\partial z_{j}} dz_{j} \in \mathscr{A}^{1,0}(U)$$ and $$\bar{\partial} f = \sum_{j=1}^{n} \frac{\partial f}{\partial \bar{z_{j}}} d\bar{z_{j}} \in \mathscr{A}^{0,1}(U).$$ Thus, we have $$\begin{align*}d \omega &= df \wedge dz_{I} \wedge d\bar{z_{J}} \\ &= \partial f \wedge dz_{I} \wedge d\bar{z_{J}} + \bar{\partial} f \wedge dz_{I} \wedge d\bar{z_{J}} \\ &= \partial \omega + \bar{\partial} \omega \end{align*}.$$ This finishes the proof. $\Box$

As in the above proof, we shall continue drop $M$ in the notations for sheaves, unless it causes too much confusion.

Corollary. Assuming the previous notations, at every $(p, q)$ level, we have

• $\partial^{2} = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p+2,q};$
• $\bar{\partial}^{2} = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p,q+2};$
• $\partial\bar{\partial} + \bar{\partial}\partial = 0 : \mathscr{A}^{p,q} \rightarrow \mathscr{A}^{p+1,q+1}.$

Proof. We have $$0 = d^{2} = (\partial + \bar{\partial})^{2} = \partial^{2} + (\partial\bar{\partial} + \bar{\partial}\partial) + \bar{\partial}^{2}.$$ Looking at the different degrees in the grading, the result follows. $\Box$

Derivation property for $\partial$ and $\bar{\partial}$. Recall that for $\omega \in \mathscr{A}^{k}(U)$ and $\eta \in \mathscr{A}^{l}(U),$ we have $$d(\omega \wedge \eta) = d\omega \wedge \eta + (-1)^{k} \omega \wedge d \eta.$$ Thus, the same formula holds for the complexification. This implies that $$\partial(\omega \wedge \eta) = \partial\omega \wedge \eta + (-1)^{k} \omega \wedge \partial \eta$$ and $$\bar{\partial}(\omega \wedge \eta) = \bar{\partial}\omega \wedge \eta + (-1)^{k} \omega \wedge \bar{\partial} \eta,$$ for any $\omega \in \mathscr{A}^{p+q}(U)$ and $\eta \in \mathscr{A}^{p'+q'}(U)$ such that $p + q = k.$

Dolbeault complex. Given a complex manifold $M$ of dimension $n,$ for every $p \in \mathbb{Z}_{\geq 0},$ we have the following complex of sheaves: $$0 \rightarrow \mathscr{A}^{p,0} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,1} \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,n-1} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,n} \rightarrow 0,$$ taking the global sections, we have the following complex of $\mathbb{C}$-vector spaces: $$0 \rightarrow \Gamma(M, \mathscr{A}^{p,0}) \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,1}) \overset{\bar{\partial}}{\longrightarrow} \cdots \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,n-1}) \overset{\bar{\partial}}{\longrightarrow} \Gamma(M, \mathscr{A}^{p,n}) \rightarrow 0,$$ which is called the $p$-th Dolbeault complex of $M.$ The $q$-th cohomology of $p$-th Dolbeault complex is called the Dolbeault cohomology group of degree $(p, q).$ For notation, we write $$H^{p,q}(M) := H^{q}(\Gamma(M, \mathscr{A}^{p, \bullet}).$$ We now argue that we can compute the cohomology of the sheaf $\Omega^{p}_{M}$ of holomorphic $p$-forms on $M$ by the $p$-th Dolbeault complex.

Lemma 1. We have $$\ker(\mathscr{A}^{p,0}_{M} \overset{\bar{\partial}}{\longrightarrow} \mathscr{A}^{p,1}_{M}) = \Omega^{p}_{M},$$ for any $p \geq 0.$

Proof. We may prove this locally on open subsets $U \subset M$ with chart maps. An element of $\mathscr{A}^{p,0}(U)$ looks like $$\omega = \sum_{|I| = p} f_{I} dz_{I}$$ where $f_{I} \in \mathscr{C}^{\infty}(U).$ We have $$\bar{\partial}\omega = \sum_{|I| = p} \sum_{j=1}^{n} \frac{\partial f_{I}}{\partial \bar{z_{j}}} d\bar{z_{j}} \wedge dz_{I}.$$ On $U,$ saying that $\omega$ is holomorphic is equivalent to $f_{I}$ are holomorphic for all $I$ which is precisely when all $\partial f_{I} / (\partial \bar{z_{j}}) = 0$ for all $I$ and $j.$ This is precisely the same as saying that $\bar{\partial} \omega = 0,$ and this finishes the proof. $\Box$

Lemma 2 ($\bar{\partial}$-lemma). Let $U \subset X$ be a nonempty open subset. Fix $p \geq 0.$ Given any $q \geq 1$ and any smooth $(p, q)$-form $\omega$ such that $\bar{\partial}\omega = 0$ on $U.$ Then we may find a local section $\beta \in \Gamma(V, \mathscr{A}^{p, q-1})$ such that $\bar{\partial} \beta = \omega$ on a nonempty open subset $V \subset U.$

The point of Lemma 1 and Lemma 2 is that we now have a resolution (i.e., everywhere exact): $$0 \rightarrow \Omega^{p}_{M} \rightarrow \mathscr{A}^{p,0}_{M} \rightarrow \cdots \rightarrow \mathscr{A}^{p,n}_{M} \rightarrow 0.$$ What is this good for? We need one more lemma to see this.

Lemma 3. Every $\mathscr{A}^{p,q}_{M}$ is acyclic. Namely, we have $$H^{i}(X, \mathscr{A}^{p,q}) = 0$$ for all $i \geq 1.$

Hence, by a previous posing that says any acyclic resolution of a sheaf computes its cohomology, we have:

Corollary. We have $$H^{p,q}(M) = H^{q}(M, \Omega^{p}_{M}) \simeq H^{q}(\Gamma(M, \mathscr{A}^{p, \bullet})),$$ where the first equality is just a definition.

We will discuss the proof of Lemma 2 next time, and Lemma 3 later.

Any acyclic resolution computes sheaf cohomology

Let $X$ be any topological space and $\mathscr{F}$ a sheaf on $X$ valued in $\textbf{Ab}$, the category of abelian groups. For now, any sheaf we will discuss will be valued in $\textbf{Ab}.$

A resolution of $\mathscr{F}$ is an exact sequence of the form $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{R}^{0} \rightarrow \mathscr{R}^{1} \rightarrow \cdots.$$ Let's state facts we will use without proofs.

Fact 1. The category of sheaves on $X$ (valued in $\textbf{Ab}$) has enough injectives. For our purpose, this just means the following: given any sheaf $\mathscr{F}$on $X,$ we may construct an injective resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{I}^{0} \rightarrow \mathscr{I}^{1} \rightarrow \cdots$$ of sheaves on $X,$ meaning that this is a resolution where $\mathscr{I}^{j}$ are injective objects in the category of sheaves.

Fact 2. Given any two injective resolutions $0 \rightarrow \mathscr{F} \rightarrow \mathscr{I}^{\bullet}$ and $0 \rightarrow \mathscr{F} \rightarrow \mathscr{J}^{\bullet},$ we have $$H^{i}(\Gamma(X, \mathscr{I}^{\bullet})) \simeq H^{i}(\Gamma(X, \mathscr{J}^{\bullet}))$$ for all $i \geq 0.$ Hence, we will just define $$H^{i}(X, \mathscr{F}) := H^{i}(\Gamma(X, \mathscr{I}^{\bullet})),$$ which only make sense up to isomorphisms.

Remark. It is immediate that $$H^{0}(X, \mathscr{F}) = \Gamma(X, \mathscr{F})$$ because the global section functor is left-exact.

Fact 3. Given any short exact sequence $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0$$ of sheaves on $X,$ we have a long exact sequence $$\cdots \rightarrow H^{i}(X, \mathscr{A}) \rightarrow H^{i}(X, \mathscr{B}) \rightarrow H^{i}(X, \mathscr{C}) \rightarrow H^{i+1}(X, \mathscr{A}) \rightarrow \cdots.$$

Remark. We will sketch the proof of Fact 3 at the end of this posting. The definition using the Godement resolution of a sheaf makes Fact 3 almost automatically true, but then we need to switch the order of the exposition.

Lemma. Let $\mathscr{I}$ be any injective sheaf on $X.$ Then $$H^{i}(\Gamma(X, \mathscr{I})) = 0$$ for $i \geq 1.$

Proof. We may use the following injective resolution: $$0 \rightarrow \mathscr{I} \rightarrow \mathscr{I} \rightarrow 0 \rightarrow 0 \rightarrow \cdots,$$ which computes what we want. $\Box$

We say that a sheaf $\mathscr{A}$ is acyclic if $$H^{i}(X, \mathscr{A}) = 0$$ for all $i \geq 1.$ Note that we have just shown that injective sheaves are acyclic.

Theorem. Let $\mathscr{F}$ be a sheaf on a topological space $X.$ For any resolution $$0 \rightarrow \mathscr{F} \rightarrow \mathscr{A}^{\bullet},$$ we have a natural map $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) \rightarrow H^{i}(X, \mathscr{F}),$$ meaning it is a map in $\textbf{Ab}$ functorial in sheaves along with choices of resolutions.

Moreover, if each $\mathscr{A}^{j}$ is acyclic, then this map is an isomorphism.

Proof. We follow Theorem 3.13 in Wells. Take the exact sequence $$0 \rightarrow \mathscr{K}^{j} \rightarrow \mathscr{A}^{i} \rightarrow \mathscr{A}^{i+1},$$ which gives the exact sequence $$0 \rightarrow \Gamma(X, \mathscr{K}^{i}) \rightarrow \Gamma(X, \mathscr{A}^{i}) \rightarrow \Gamma(X, \mathscr{A}^{i+1}).$$ (Note that $\mathscr{K}^{0} = \mathscr{F}.$) Consider the induced short exact sequences $$0 \rightarrow \mathscr{K}^{i-1} \rightarrow \mathscr{A}^{i-1} \rightarrow \mathscr{K}^{i} \rightarrow 0.$$ Each such sequence induces the long exact sequence $$0 \rightarrow \Gamma(X, \mathscr{K}^{i-1}) \rightarrow \Gamma(X, \mathscr{A}^{i-1}) \rightarrow \Gamma(X, \mathscr{K}^{i}) \rightarrow H^{1}(X, \mathscr{K}^{i-1}) \rightarrow H^{1}(X, \mathscr{A}^{i}) \cdots.$$ Thus, we have $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) = \frac{\Gamma(X, \mathscr{K}^{i})}{\mathrm{im}(\Gamma(X, \mathscr{A}^{i-1}) \rightarrow \Gamma(X, \mathscr{K}^{i}))} \rightarrow H^{1}(X, \mathscr{K}^{i-1}),$$ which we call $\alpha_{i,1},$ which is injective due to the suitable part of the long exact sequence above. Note that $\alpha_{i,1}$ is also surjective (and hence an isomorphism) if $H^{1}(X, \mathscr{A}^{i}) = 0.$

Fix $2 \leq j \leq i$ and consider the induced short exact sequences $$0 \rightarrow \mathscr{K}^{i-j} \rightarrow \mathscr{A}^{i-j} \rightarrow \mathscr{K}^{i-j+1} \rightarrow 0.$$ We get a long exact sequence $$\cdots \rightarrow H^{j-1}(X, \mathscr{A}^{i-j}) \rightarrow H^{j-1}(X, \mathscr{K}^{i-j+1}) \overset{\alpha_{i,j}}{\longrightarrow} H^{j}(X, \mathscr{K}^{i-j}) \rightarrow H^{j}(X, \mathscr{A}^{i-j}) \cdots.$$ Note that if $$H^{j-1}(X, \mathscr{A}^{i-j}) = 0 = H^{j}(X, \mathscr{A}^{i-j}),$$ then $\alpha_{i,j}$ is an isomorphism. Thus, we have constructed the composition $$H^{i}(\Gamma(X, \mathscr{A}^{\bullet})) \xrightarrow{\alpha_{i,1}} H^{1}(X, \mathscr{K}^{i-1}) \xrightarrow{\alpha_{i,2}} \cdots \xrightarrow{\alpha_{i,i-1}} H^{i-1}(X, \mathscr{K}^{1}) \xrightarrow{\alpha_{i,i}} H^{i}(X, \mathscr{F})$$ of the maps, which of which is an isomorphism if all $\mathscr{A}^{0}, \mathscr{A}^{1}, \dots$ are acyclic. Naturality should follow from Snake Lemma. This finishes the proof. $\Box$

Sketch of proof of Fact 3. Let $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{C} \rightarrow 0$$ be a short exact sequences of sheaves on $X.$ Choose injective resolutions $$0 \rightarrow \mathscr{A} \rightarrow \mathscr{A}^{0} \rightarrow \mathscr{A}^{1} \rightarrow \cdots$$ and $$0 \rightarrow \mathscr{C} \rightarrow \mathscr{C}^{0} \rightarrow \mathscr{C}^{1} \rightarrow \cdots$$ using Fact 1. Considering the exact sequence $0 \rightarrow \mathscr{A} \rightarrow \mathscr{B},$ the injectivity of $\mathscr{A}^{0}$ induces a map $\mathscr{B} \rightarrow \mathscr{A}^{0}$ such that the map $\mathscr{A} \rightarrow \mathscr{A}^{0}$ factors as $$\mathscr{A} \rightarrow \mathscr{B} \rightarrow \mathscr{A}^{0}.$$ Using $\mathscr{B} \rightarrow \mathscr{A}^{0}$ and $\mathscr{B} \rightarrow \mathscr{C} \rightarrow \mathscr{C}^{0},$ we may induce a unique map $$\mathscr{B} \rightarrow \mathscr{A}^{0} \times \mathscr{C}^{0}$$ using the universal property of the product. If we pretend as if we can chase elements, we can show that $$0 \rightarrow \mathscr{B} \rightarrow \mathscr{A}^{0} \times \mathscr{C}^{0}$$ is exact. I remember Chapter 8 of MacLane's book allows us to do this in any general abelian category setting. I think in our case, it is easier because exactness can be checked at the level of stalks (which are abelian groups). Once we show this, we can construct a short exact sequence of injective resolutions, where taking the global sections at each level (except for the given sheaves) gives exact sequences. Then by applying Snake Lemma, we are done. $\Box$

Remark. The category $\textbf{Ab}$ can probably be replaced by any abelian category. The argument used in the (sketch of) proof of Fact 3 should be called "injective Horseshoe lemma".

Symplectic forms and symplectic groups: Part 1

We follow Chapter 2 of a paper by Garton (and perhaps some other sources, for example these notes written by Elkies), but I will also write bunch more gibberish so that I can learn better. Please correct me if I have made mistakes!

Let $R$ be a commutative ring and fix $n \in \mathbb{Z}_{\geq 1}.$ A bilinear from $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R$$ is

• symmetric if $\langle v, w \rangle = \langle w, v \rangle$ for all $v, w \in R^{n};$
• alternating if $\langle v, v \rangle = 0$ for all $v \in R^{n};$ 
• nondegenerate if two maps $R^{n} \rightarrow (R^{n})^{\vee}$ given by $v \mapsto \langle v, - \rangle$ and $w \mapsto \langle -, w \rangle$ are injective;
• symplectic if it is alternating and nondegenerate. 

Remark. If the bilinear form is alternating, then $$0 = \langle v + w, v + w \rangle = \langle v, w \rangle + \langle w, v\rangle,$$ so $$\langle w, v \rangle = -\langle v, w\rangle.$$ This condition is generally weaker than the alternating condition, unless (the image of) $2$ is invertible in $R.$ Then we have $$\langle v, v \rangle = \frac{1}{2}(\langle v, v \rangle + \langle v, v \rangle),$$ so the above condition with $w = v$ gives the alternating condition back.

If the given bilinear form is symmetric or alternating, we only need to check one of the two maps $R^{n} \rightarrow (R^{n})^{\vee}$ is injective when it comes to checking nondegeneracy.

Matrix representation. Abstract definition of bilinear forms may sound sleek, but to show certain forms exist, it is helpful to use matrix representation of the forms. In particular, we want to see whether we can construct a symplectic bilinear form on $R^{n}.$ Denote by $\mathrm{Bi}(R^{n})$ the set of all bilinear forms $R^{n} \times R^{n} \rightarrow R$ and $\mathrm{Mat}_{n}(R)$ the set of all $n \times n$ matrices over $R.$ Then we have a bijection $$\mathrm{Bi}(R^{n}) \leftrightarrow \mathrm{Mat}_{n}(R)$$ given by $$\langle \cdot, \cdot \rangle \mapsto [\langle e_{i}, e_{j} \rangle]_{1 \leq i, j \leq n},$$ and whose inverse is given by $$A \mapsto ((v,w) \mapsto v^{T}Aw).$$ This bijection is not additive but it preserves the multiplication by $R.$

Remark. Here, the correspondence depends the standard basis $e_{1}, \dots, e_{n}$ for $R^{n}.$ If $B : R^{n} \rightarrow R^{n}$ induces an $R$-linearly free elemenets $Be_{1}, \dots, Be_{n},$ then we cannot use the same matrix for bilinear forms on $R^{n}.$ If the orignal matrix is $A,$ the new matrix is $B^{T}AB.$ If it happens that $\det(B) \in R^{\times},$ which is always true when $R$ is a field, then this observation shows that we get a self-bijection $\mathrm{Bi}(R^{n}) \leftrightarrow \mathrm{Bi}(R^{n})$ given by $$\langle \cdot, \cdot \rangle \mapsto \langle B(\cdot), B(\cdot) \rangle.$$ Since $(B^{T}AB)^{T} = B^{T}A^{T}B$ and $\det(B^{T}AB) = \det(B)^{2} \det(A),$ we see that

• $\langle \cdot, \cdot \rangle$ is symmetric if and only if $\langle B(\cdot), B(\cdot) \rangle$ is symmetric;
• $\langle \cdot, \cdot \rangle$ is alternating if and only if $\langle B(\cdot), B(\cdot) \rangle$ is alternating;
• $\langle \cdot, \cdot \rangle$ is nondegenerate if and only if $\langle B(\cdot), B(\cdot) \rangle$ is nondegenerate,

when $\det(B) \in R^{\times},$ or equivalently $B \in \mathrm{GL}_{n}(R).$

Lemma. Let $A$ be the matrix corresponding to a bilnear form $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R.$$ The bilinear form is

• symmetric if and only if $A^{T} = A;$
• alternating if and only if $A^{T} = -A.$

Proof. We assume $A^{T} = \epsilon A$ for some $\epsilon \in R.$ We have $$\begin{align*} \langle w, v \rangle &= w^{T}Av \\ &= (v^{T}A^{T}w)^{T} \\ &= v^{T}A^{T}w \\ &= \epsilon v^{T}Aw \\ &= \epsilon \langle v, w \rangle. \end{align*}$$ This proves the two statements. $\Box$


Lemma. Let $A$ be the matrix corresponding to a bilnear form $$\langle \cdot, \cdot \rangle : R^{n} \times R^{n} \rightarrow R.$$ The following are equivalent:

• $v \mapsto \langle v, -\rangle$ is injective;
• $\ker(A^{T}) = 0.$

Moreover, the following are equivalent as well:

• $w \mapsto \langle -, w\rangle$ is injective;
• $\ker(A) = 0.$

In particular, if $\det(A) \in R^{\times}$ (i.e., $A$ is invertible in $\mathrm{Mat}_{n}(R)$ by Cramer's rule), then the form is nondegnerate.

Proof. Saying that $v \mapsto \langle v, -\rangle$ is injective means that if $v^{T}Ae_{1} = \cdots = v^{T}Ae_{n} = 0,$ then $v = 0.$ Equivalently, this means that if $v^{T}A = 0,$ then $v = 0,$ or better yet, if $A^{T}v = 0,$ then $v = 0.$ This is equivalent to saying that $\ker(A^{T}) = 0.$

Saying that $w \mapsto \langle -, w\rangle$ is injective means that if $e_{1}^{T}Aw = \cdots = e_{n}^{T}Aw = 0,$ then $w = 0.$ Equivalently, this means that if $Aw = 0,$ then $w = 0.$ This is equivalent to saying that $\ker(A) = 0,$ and we are done $\Box$

Symplectic forms force parity in $n$ over a field. Let $k$ be a field with characteristic $\neq 2.$ For any $n \in \mathbb{Z}_{\geq 1},$ the following are equivalent:

• there is a symplectic bilinear form on $k^{n};$
• $n$ is even. 

Proof. We prove the converse first. Assume that $n$ is even so that we can write $n = 2g$ for some $g \geq 1.$ Then consider the matrix $$A = \begin{bmatrix} 0 & I_{g} \\ -I_{g} & 0 \end{bmatrix},$$ where $I_{g} \in \mathrm{Mat}_{g}(R)$ is the identity matrix. We have $A^{T} = -A$ and $$\det(A) = (-1)^{g(g+1)/2} \in R^{\times},$$ so the bilinear from on $k^{n}$ corresponding to $A$ is symplectic. Note that this direction works for any commutative ring in place of $k.$

Now, assume that there is a symplectic bilinear from $\langle \cdot, \cdot \rangle : k^{n} \times k^{n} \rightarrow k.$ Denote by $A = (a_{i,j})_{1 \leq i, j \leq n}$ the corresponding matrix. Since $A^{T} = -A,$ we have $a_{i,i} = -a_{i,i},$ which implies that $a_{i,i} = 0$ because $\mathrm{char}(k) \neq 2.$ Since $A$ must be invertible, observation already says $n \geq 2.$ We have so far only used that $2$ is invertible in $k,$ and we did not use the assumption that $k$ is a field.

Fix any $v_{1} \neq 0$ in $k^{n}.$ Since the form is nondegenerte, there must be some $w_{1} \neq 0$ in $k^{n}$ such that $\langle v_{1}, w_{1} \rangle \neq 0.$ We are over a field, so we may renormalize so that $\langle v_{1}, w_{1} \rangle = 1.$ Being over a field also implies that $v_{1}, w_{1}$ are linearly independent. Hence, we can construct an even dimensional subspace $H_{1} := kv_{1} + kw_{1} \subset k^{n}.$ Consider $$H_{1}^{\perp} := \{x \in k^{n} : \langle x, H_{1} \rangle = 0\}.$$ We claim that $k^{n} = H_{1} \oplus H_{1}^{\perp}.$

To see this, note that the bilinear form on $k^{n}$ restricts to $H_{1}$ with the matrix $$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},$$ with respect to the basis $v_{1}, w_{1}.$ We know this restirction gives a symplectic form on $H_{1}.$ In particular, it is non-degenerate, and this is enough to show that $H_{1} \cap H_{1}^{\perp} = 0,$ as follows: fix any $v \in H_{1} \cap H_{1}^{\perp}.$ Then $\langle v, v_{1} \rangle = \langle v, w_{1} \rangle = 0.$ Since the form restricted to $H_{1}$ is already nondegenerte, this ensures that $v = 0,$ as desired.

Since the form on $k^{n}$ is nondegenerate, we are given with the injection $k^{n} \hookrightarrow (k^{n})^{\vee}$ given by $v \mapsto \langle v, - \rangle.$ Since we have $\dim_{k}((k^{n})^{\vee}) = n = \dim_{k}(k^{n}),$ this injection must be an isomorphism, which uses the fact that $k$ is a field. Under this isomorphism, we have $$H_{1}^{\perp} \simeq \{f : V \rightarrow k : f(H_{1}) = 0\},$$ and the right-hand side has dimension $n - \dim_{k}(H_{1}) = n - 2.$ Since $H_{1} \cap H_{1}^{\perp} = 0,$ we have $$H_{1} + H_{1}^{\perp} = H_{1} \oplus H_{1}^{\perp} = k^{n},$$ due to the matching dimensions. 

Note that the form from $k^{n}$ restricts to $H_{1}^{\perp}$ as a symplectic form, so we may repeat our argument to $H_{1}^{\perp}$ and then to their subspaces to deduce that $n$ is even. $\Box$

Existence of symplectic forms for even $n$. If we are over a general commutative ring $R,$ the proof above at least ensures that there exists a specific symplectic form given by the matrix $$\begin{bmatrix} 0 & I_{g} \\ -I_{g} & 0 \end{bmatrix},$$ when $n = 2g$ is even.

Question. For general $R,$ does the existence of symplectic form on $R^{n}$ forces $n$ to be even?

Remark. Note that if $2 = 0$ in $R$ (e.g., $R = \mathbb{F}_{2}$), then the $n \times n$ identity matrix represents a symplectic form for every $n \geq 1,$ not just even ones. Thus, we can at least tell that the characteristic assumption on $R$ is crucial.

Equivalence among forms. We say two bilinear forms $\langle \cdot, \cdot, \rangle, \langle \cdot, \cdot, \rangle' \in \mathrm{Bi}(R^{n})$ are equivalent if there is $g \in \mathrm{GL}_{n}(R)$ such that $\langle v, w, \rangle' = \langle gv, gw \rangle$ for all $v, w \in R^{n}.$

It is conceptually easier to think about the action of $\mathrm{GL}_{n}(R)$ on $\mathrm{Bi}(R^{n})$ given by $$(gF)(v, w) := F(gv, gw),$$ where $g \in \mathrm{GL}_{n}(R)$ and $F \in \mathrm{Bi}(R^{n}).$ With this action, two bilinear forms are equivalent if and only if they are in the same orbit. The last proof also shows that when $R = k$ is a field with characteristic $\neq 2,$ all symplectic forms are equivalent, meaning they all lie in the same orbit. We might try to classify symplectic forms over other commutative rings next time.

In any case, we are able to define the notion we would like to think about. Fix the symplectic form $F \in \mathrm{Bi}(R^{2g})$ given by the matrix above. Then the symplectic group of genus $g$ is defined to be $$\mathrm{Sp}_{2g}(R) := \{g \in \mathrm{GL}_{n}(R) : g F = F\}.$$ The general symplectic group of genus $g$ is defined to be $$\mathrm{GSp}_{2g}(R) := \{g \in \mathrm{GL}_{n}(R) : (g F)/F \in R^{\times}\},$$ where the condition means that there is an element $m(g) \in R^{\times}$ such that $(gF)(v, w) = m(g) F(v, w)$ for all $v, w \in R^{2g}.$ We immediately note that $$\mathrm{Sp}_{2g}(R) \subset \mathrm{GSp}_{2g}(R).$$ The map $m : \mathrm{GSp}_{2g}(R) \rightarrow R^{\times}$ defines a group homomorphism.

Hodge theory: Lecture 12

We also follow some parts of a book by Wells.

Almost complex structure at fiber level. A module structure on $\mathbb{R}^{m}$ over $\mathbb{R}[t]/(t^{2} + 1) = \mathbb{C}$ is precisely a matrix $J \in \mathrm{Mat}_{m}(\mathbb{R})$ such that $J^{2} + \mathrm{id} = 0.$ We call such $J$ an almost complex structure on $\mathbb{R}^{m}.$

Remark. Existence of such $J$ implies that $m$ is even by considering the $\mathbb{R}[t]$-module structure given on $\mathbb{R}^{m}$ due to $J.$ Hence, we may write $m = 2n.$

We may consider $$J_{\mathbb{C}} := J \otimes_{\mathbb{R}} \mathrm{id} : \mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} \rightarrow \mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C},$$ given by $v \otimes c \mapsto J(v) \otimes c.$ Identifying $\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C}$ with $\mathbb{C}^{2n},$ we can consider $J_{\mathbb{C}} \in \mathrm{Mat}_{2n}(\mathbb{C}),$ and as a matrix it can be thought of just $J$ but as the image of the embedding $$\mathrm{Mat}_{2n}(\mathbb{R}) \hookrightarrow \mathrm{Mat}_{2n}(\mathbb{C})$$ given by $$\mathbb{R} \hookrightarrow \mathbb{C}.$$ Hence, we will just write $J$ without $\mathbb{C}$ in the notation. We note that $J$ gives $\mathbb{C}^{2n}$ a module structure over $$\mathbb{C}[t]/(t^{2} + 1) \simeq \mathbb{C}[t]/(t - i) \times \mathbb{C}[t]/(t + i),$$ and this implies that we have a decomposition $$\mathbb{C}^{2n} = V^{1,0} \oplus V^{0,1},$$ where on $V^{1,0},$ the linear map $J$ acts as the multiplication by $i,$ while on $V^{0,1},$ the linear map $J$ acts as the multiplication by $-i.$ We write $Q : \mathbb{C}^{2n} \rightarrow \mathbb{C}^{2n}$ to denote the map given by complex conjugating each entry. (Not that important: $\mathbb{R}^{2n}$ is the set of points in $\mathbb{C}^{2n}$ fixed by $Q.$)

Lemma. We continue to assume that there is $J \in \mathrm{Mat}_{2n}(\mathbb{R})$ with $J^{2} + \mathrm{id} = 0.$ The map $\mathbb{R}^{2n} \hookrightarrow \mathbb{C}^{2n} \twoheadrightarrow V^{1,0}$ is an isomorphism of $\mathbb{C}$-vector spaces, while the map $\mathbb{R}^{2n} \hookrightarrow \mathbb{C}^{2n} \twoheadrightarrow V_{0,1}$ is a conjugate isomorphism.

Proof. Since $J^{2} + \mathrm{id} = 0,$ the $\mathbb{C}[t]$-module structure on $\mathbb{C}^{2n}$ given by $J$ is necessarily of the form $$\mathbb{C}^{2n} \simeq (\mathbb{C}[t]/(t^{2} + 1))^{\oplus n} \simeq (\mathbb{C}[t]/(t - i))^{\oplus n} \oplus (\mathbb{C}[t]/(t + i))^{\oplus n}.$$ Hence, we have $n = \dim_{\mathbb{C}}(V^{1,0}) = \dim_{\mathbb{C}}(V^{0,1}).$ This implies that $$2n = \dim_{\mathbb{R}}(V^{1,0}) = \dim_{\mathbb{R}}(V^{0,1}).$$ Fix any $v \in \mathbb{R}^{2n}.$ Then we have $$v = v^{1,0} + v^{0,1} \in \mathbb{C}^{n}$$ with $v^{1,0} \in V^{1,0}$ and $v^{0,1} \in V^{0,1}.$ Note that $J(v) = i v^{1,0} - i v^{0,1}$ and thus $iJ(v) = -v^{1,0} + v^{0,1}.$ This implies that $$v - iJ(v) = 2v^{1,0}$$ and $$v + iJ(v) = 2v^{0,1}.$$ Since $v \in \mathbb{R}^{2n},$ we have $J(v) \in \mathbb{R}^{2n}.$ Thus, if $v^{1,0},$ then $v = 0.$ Likewise, if $v^{0,1} = 0,$ then $v = 0.$ This implies that the maps in question are both injective. Since they are $\mathbb{R}$-linear maps, by dimension counting they are $\mathbb{R}$-isomoprhisms. The first map is given by $f^{1,0} : v \mapsto v^{1,0},$ so $$f^{1,0}(iv) = f^{1,0}(J(v)) = f^{1,0}(iv^{1,0} - iv^{0,1}) = iv^{1,0} = if^{1,0}(v).$$ Hence $f^{1,0}$ is $\mathbb{C}$-linear. Likewise, the map $f^{0,1} : v \mapsto v^{0,1}$ is conjugate linear. This finishes the proof. $\Box$

Remark. In the proof above, given any $v \in \mathbb{R}^{2n},$ we realized the following explicit formulas: $$v^{1,0} = \frac{v - iJ(v)}{2}$$ and $$v^{0,1} = \frac{v + iJ(v)}{2}.$$ Apparently this does not work in characteristic $2,$ but isn't it okay because we did not divide by $2$ in the proof? Characteristic $2$ does not work even before that, because we have $t^{2} + 1 = (t + 1)^{2}$ over any characteristic $2$ field. This explicit description lets us see the following.

Corollary. Keeping the above notations, we have $\overline{V^{1,0}} = V^{0,1},$ where $\overline{W} := Q(W).$

Dual. Given any $A \in \mathrm{Mat}_{m}(\mathbb{R}),$ we can let $A$ act on $(\mathbb{R}^{m})^{\vee} = \mathrm{Hom}_{\mathbb{R}}(\mathbb{R}^{m}, \mathbb{R})$ by defining $(A \varphi)(v) := \varphi(Av).$ This imbues $(\mathbb{R}^{m})^{\vee}$ an $\mathbb{R}[t]$-module structure.

 Any $J \in \mathrm{Mat}_{2n}(\mathbb{R})$ with $J^{2} + \mathrm{id} = 0,$ gives $(\mathbb{R}^{2n})^{\vee}$ a module structure over $\mathbb{R}[t]/(t^{2} + 1) = \mathbb{C}.$ Thus, we get a decomposition $$(\mathbb{R}^{2n})^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}$$ like before. 

Exterior power. We also have $$\begin{align*}\bigwedge^{r} ((\mathbb{R}^{2n})^{\vee} \otimes_{\mathbb{R}} \mathbb{C}) &= \bigwedge^{r} (U^{1,0} \oplus U^{0,1}) \\ &= \bigoplus_{\substack{p, q \geq 1 \\ p+q = r } } \left( \bigwedge^{p} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{q} U^{0,1} \right),\end{align*}$$ and by looking element-wise, the complex conjugation map $Q$ induces a conjugation map on this $r$-th wedge product that maps $\bigwedge^{p} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{q} U^{0,1}$ to $\bigwedge^{q} U^{1,0} \otimes_{\mathbb{R}} \bigwedge^{p} U^{0,1}.$

Description of basis at fiber level. Let $V$ be a finite dimensional real vector space with an almost complex structure $J,$ so that in particular $\dim_{\mathbb{R}}(V) = 2n.$ This gives $$V_{\mathbb{C}} := V \otimes_{\mathbb{R}} \mathbb{C} = V^{1,0} \oplus V^{0,1},$$ which is $2n$-dimensional complex vector space. Let $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}$ be a basis for $V.$ We already know from previous computations that writing $$z_{j} = \frac{x_{j} - iy_{j}}{2}$$ and $$\bar{z_{j}} = \frac{x_{j} + iy_{j}}{2},$$ we have $$V^{1,0} = \mathbb{C}z_{1} \oplus \cdots \oplus \mathbb{C}z_{n}$$ and $$V^{0,1} = \mathbb{C}\bar{z_{1}} \oplus \cdots \oplus \mathbb{C}\bar{z_{n}}.$$ Let $x_{1}^{\vee}, \dots, x_{n}^{\vee}, y_{1}^{\vee}, \dots, y_{n}^{\vee}$ the basis of $$V^{\vee} = \mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R})$$ dual to $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}.$ We have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} = U^{1,0} \oplus U^{0,1}.$$ We will make use of the following isomorphism: $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq  \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee}$ for any $v \in V,$ where $0^{\vee} = 0$ and $v^{\vee}$ is the map sending $v \mapsto 1$ and everything else to $0.$ This isomorphism is useful in getting our hands on $U^{1,0}$ and $U^{0,1}.$ This is because the right-hand side gives us $$\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}) = \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \oplus V^{0,1}, \mathbb{C}) \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \times \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$$ We have:

Lemma. Under the isomorphism $$\mathrm{Hom}_{\mathbb{R}}(V, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}),$$ and the first takes the action of $J$ as the multiplication by $i$ and the second the multiplication by $-i.$

Proof. Denote by $\phi$ the given isomorphism, so $$\phi(v^{\vee} \otimes 1) = (v \otimes 1)^{\vee}$$ for any basis element $v$ of $V.$ We have $$\begin{align*}\phi(J(v^{\vee} \otimes 1)) &= \phi((Jv)^{\vee} \otimes 1) \\ &= (Jv \otimes 1)^{\vee} \\ &= (J(v \otimes 1))^{\vee} \\ &= J(v \otimes 1)^{\vee}.\end{align*}$$ Hence, to show the claim it is enough to show that the $i$-eigenspace of the action of $J$ on $\mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ is $\mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}),$ while the $(-i)$-eigenspace is $\mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}).$ Given $$(f^{1,0}, 0) \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ we have $$\begin{align*}(J \cdot (f^{1,0}, 0)) (v \otimes 1) &= (f^{1,0}, 0) (Jv \otimes 1) \\ &= (f^{1,0}, 0) (iv^{1,0} - iv^{1,0}) \\ &= if^{1,0}(v^{1,0}) \\ &= (i \cdot (f^{1,0}, 0))(v \otimes 1), \end{align*}$$ so we have $J (f^{1,0}, 0) = i (f^{1,0}, 0).$ Likewise, we can deduce $J (0, f^{0,1}) = -i (0, f^{0,1})$ for any $$(0, f^{0,1}) \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1}, \mathbb{C}) \hookrightarrow \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ This finishes the proof. $\Box$

Note that $x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee}, x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ form a dual basis for $z_{1}, \dots, z_{n}, \bar{z_{1}}, \dots, \bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$

Lemma. We have $$x_{1}^{\vee} + iy_{1}^{\vee}, \dots, x_{n} + iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$x_{1}^{\vee} - iy_{1}^{\vee}, \dots, x_{n} - iy_{n}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$$ In particular, these elements form $\mathbb{C}$-bases for $\mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$ and $\mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}).$

Proof. Fix a general element $$w = a_{1}z_{1} + \cdots + a_{n}z_{n} + b_{1}\bar{z_{1}} + \cdots + b_{n}\bar{z_{n}} \in V \otimes_{\mathbb{R}} \mathbb{C}.$$ We check $$\begin{align*}(J (x_{j}^{\vee} + iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(ia_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(a_{j}z_{j}) \\ &= (i(x_{j}^{\vee} + iy_{j}^{\vee}))(w)\end{align*}$$ and $$\begin{align*}(J (x_{j}^{\vee} - iy_{j}^{\vee}))(w) &= (x_{j}^{\vee} + iy_{j}^{\vee})(-ib_{j}\bar{z}_{j}) \\ &= (-i(x_{j}^{\vee} - iy_{j}^{\vee}))(b_{j}\bar{z}_{j}) \\ &= (i(x_{j}^{\vee} - iy_{j}^{\vee}))(w).\end{align*}$$ This implies that $J (x_{j}^{\vee} + iy_{j}^{\vee}) = i (x_{j}^{\vee} + iy_{j}^{\vee})$ so that $$x_{j}^{\vee} + iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ due to the previous lemma. Similarly, we get $$x_{j}^{\vee} - iy_{j}^{\vee} \in \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C}),$$ as desired. $\Box$

What's the upshot? Again, we have $$V^{\vee} \otimes_{\mathbb{R}} \mathbb{C} \simeq \mathrm{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ determined by $v^{\vee} \otimes 1 \mapsto (v \otimes 1)^{\vee},$ which gives $$U^{1,0} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{1,0} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ and $$U^{0,1} \simeq \mathrm{Hom}_{\mathbb{C}}(V^{0,1} \otimes_{\mathbb{R}} \mathbb{C}, \mathbb{C})$$ the $\mathbb{C}$-bases we found for the right-hand sides produce those of the left-hand sides. Moreover, for any pure tensors $f \otimes a \in V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}$ and $v \otimes b \in V \otimes_{\mathbb{R}} \mathbb{C},$ we can make sense of the evaluation $$(f \otimes a)(v \otimes b) := f(v) \otimes ab \in \mathbb{C},$$ which gives a definition of evaluation of any element of $V \otimes_{\mathbb{R}} \mathbb{C}$ by any element of $V^{\vee} \otimes_{\mathbb{R}} \mathbb{C}.$

Globalization. Let $M$ be a smooth real manifold and $E$ be a smooth real vector bundle on $M.$ An almost complex structure on $E$ is a map $J : E \rightarrow E$ of vector bundles such that $J^{2} = -\mathrm{id}.$ (This makes sense fiber-wise, but it seem okay globally as well.) Note that $J$ will force $E$ have an even rank, say $2r.$ The above discussion globalizes, so we have $$E_{\mathbb{C}} := E \otimes_{\mathbb{R}} \mathbb{C} = E^{1,0} \oplus E^{0,1},$$ coming from fiber-wise description. Everything we discussed in at fiber level works.

What do we mean by tensor product? Write $n = \mathrm{rk}(E)$ for now. We can describe $E$ by various smooth transition functions of the bundle $$(U \cap V) \times \mathbb{R}^{n} \simeq (U \cap V) \times \mathbb{R}^{n},$$ where $U, V$ are from a trivializing chart, which can be presented by smooth maps $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}).$ We can then consider the smooth maps given by $U \cap V \rightarrow \mathrm{GL}_{n}(\mathbb{R}) \hookrightarrow \mathrm{GL}_{n}(\mathbb{C}),$ which will give us smooth transition functions $$(U \cap V) \times \mathbb{C}^{n} \simeq (U \cap V) \times \mathbb{C}^{n},$$ and these glue back to give $E_{\mathbb{C}}.$

Almost complex structure on tangent bundles. Let $M$ be a smooth real manifold. An almost complex structure on $M$ is an almost complex structure on its tangent bundle $TM$ (i.e., it is a map $J : TM \rightarrow TM$ of bundles with $J^{2} = -\mathrm{id}.$) We write $(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M$ to mean that $T^{1,0}M$ is where $J$ acts as the multiplication by $i,$ while $T^{0,1}M$ by $-i.$ (We will use this notation for other vector bundles and vector spaces from now on.) Note that if $M$ is a complex manifold, then $TM,$ as a real bundle, is already equipped with the complex structure given by multiplication of $i.$

Proposition. Let $M$ be a complex manifold of dimension $n.$ Then there is an almost complex structure on $M$ (or $TM$).

Proof. It is enough to treat the case of (nonempty) open subsets of $\mathbb{C}^{n}$ and then show that biholomorphic maps preserve this structure. Fix any nonempty open $U \subset \mathbb{C}^{n},$ with complex coordinates $z_{j} = x_{j} + i y_{j}$ with $1 \leq j \leq n.$ Then $TU$ has frame (i.e., global sections $U \rightarrow TU$ giving a real basis at each fiber) given by $$\frac{\partial}{\partial x_{1}}, \dots, \frac{\partial}{\partial x_{n}}, \frac{\partial}{\partial y_{1}}, \dots, \frac{\partial}{\partial y_{n}},$$ which gives a trivialization. Define $J : TU \rightarrow TU$ by $$\frac{\partial}{\partial x_{j}} \mapsto \frac{\partial}{\partial y_{j}}$$ and $$\frac{\partial}{\partial y_{j}} \mapsto -\frac{\partial}{\partial x_{j}}.$$ Since $J^{2} = -\mathrm{id},$ this $J$ is an almost complex structure on $TU.$ Hence, we get a decomposition $$TU = T^{1,0}U \oplus T^{0,1}U,$$ where the first component is trivialized by the frame $$\frac{\partial}{\partial z_{1}}, \dots, \frac{\partial}{\partial z_{n}},$$ while the second component is trivialized by the frame $$\frac{\partial}{\partial \bar{z_{1}}}, \dots, \frac{\partial}{\partial \bar{z_{n}}}.$$ Let $V \subset \mathbb{C}^{n}$ be another nonempty open subset. In an earlier lecture, we showed that given any holomorphic map $\phi : U \rightarrow V,$ the map $(TU)_{\mathbb{C}} \rightarrow (\phi^{*}TV)_{\mathbb{C}}$ (which fiber-wise looks like $T_{x}U \otimes_{\mathbb{R}} {\mathbb{C}} \rightarrow T_{\phi(x)}V \otimes_{\mathbb{R}} {\mathbb{C}}$) restricts to $$T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$$ and $$T^{0,1}U \rightarrow \phi^{*}T^{0,1}V.$$ Recall that this actually uses the fact that $\phi$ is holomorphic. In particular, if $\phi$ is biholomorphic, then all the maps above will be isomorphisms of vector bundles preserving $\mathbb{C}$-linearity fiber-wise. $\Box$

Remark. The almost complex structure given on a complex manifold in the proof above is said to be canonical.

Corollary. Let $M$ be a complex manifold of dimension $n.$ Give $M$ the canonical almost complex structure so that we can have the decomposition $$(TM)_{\mathbb{C}} = T^{1,0}M \oplus T^{0,1}M.$$ Then $T^{1,0}M$ is a holomorphic vector bundle.

Proof. It is enough to show that given two nonempty subsets $U, V \subset \mathbb{C}^{n}$ and a holomorphic map $\phi : U \rightarrow V,$ the induced map $T^{1,0}U \rightarrow \phi^{*}T^{1,0}V$ is holomorphic. The induce map $T^{1,0}U \rightarrow T^{1,0}V$ is given by the matrix $$\left[ \frac{\partial \phi_{j}}{\partial z_{k}} \right]_{1 \leq j,k \leq n},$$ and thus each entry is holomorphic. $\Box$

Symmetric and alternating tensors

This is a topic in undergraduate algebra that comes up a lot whenever I deal with some sort of differential forms and vector bundles. I am just summarizing it here to prevent my own confusion. We follow Chapter 11 of Dummit and Foote.

Fix a commutative ring $R.$ Given any $n \geq 0,$ we have an endomorphism functor on $\mathrm{Mod}_{R}$ given by $M \mapsto M^{\otimes n}.$ Note that $M^{\otimes 0} = R$ and given an $R$-module map $\varphi : M \rightarrow N,$ the map $M^{\otimes n} \rightarrow N^{\otimes n}$ is determined by $$x_{1} \otimes \cdots \otimes x_{n} \mapsto \varphi(x_{1}) \otimes \cdots \otimes \varphi(x_{n}).$$ We write $$\mathrm{Sym}^{n}(M) := \frac{M^{\otimes n}}{(x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}},$$ and note that we also get an endomorphism functor on $\mathrm{Mod}_{R}$ given by $M \mapsto \mathrm{Sym}^{n}(M).$ We write $$\bigwedge^{n}M := \frac{M^{\otimes n}}{(x_{1} \otimes \cdots \otimes x_{n} : x_{i} = x_{j} \text{ for some } i \neq j)},$$ which also gives a functor. If (the image of) $2$ is invertible in $R,$ then the ideal in the definition the $n$-the exterior power can be replaced by $$(x_{1} \otimes \cdots \otimes x_{n} - \mathrm{sgn}(\sigma)x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}.$$ Note that there are evident universal properties associated to constructions of $M^{\otimes n}, \mathrm{Sym}^{n}(M),$ and $\bigwedge^{n}M.$

Symmetric and alternating tensors. Given a module $M$ over $R,$ the $n$-tensor is an element in $M^{\otimes n}.$ Note that $S_{n}$ acts on $M^{\otimes n}$ by $$\sigma \cdot (x_{1} \otimes \cdots \otimes x_{n}) := x_{\sigma^{-1}(1)} \otimes \cdots \otimes x_{\sigma^{-1}(n)}$$ for each $\sigma \in S_{n}.$ An $n$-tensor $t \in M^{\otimes n}$ is

• symmetric if $\sigma t = t$ for all $\sigma \in S_{n}$;
• alternating if $\sigma t = \mathrm{sgn}(\sigma)t$ for all $\sigma \in S_{n}.$

Given an $n$-tensor $t,$ we define $$S(t) := \sum_{\sigma \in S_{n}}\sigma t,$$ and $$A(t) := \sum_{\sigma \in S_{n}}\mathrm{sgn}(\sigma)\sigma t.$$ It is not difficult to check that $S(t) = n!t$ if $t$ is an $n$-symmetric tensor and $A(t) = n!t$ if $t$ is an $n$-alternating tensor. Denote by $S(M^{\otimes n})$ the submodule of symmetric $n$-tensors and $A(M^{\otimes n})$ the submodule of alternating $n$-tensors in $M^{\otimes n}.$

Symmetric powers are symmetric tensors. Let $n!$ be invertible in $R.$ Then the image of the map $(n!)^{-1}S : M^{\otimes n} \rightarrow M^{\otimes n}$ is $S(M^{\otimes n})$, and its kernel is $(x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})_{\sigma \in S_{n}}.$ Thus, this map induces an isomorphism $$\mathrm{Sym}^{n}(M) \simeq S(M^{\otimes n})$$ such that $x_{1} \otimes \cdots \otimes x_{n} \mapsto x_{1} \otimes \cdots \otimes x_{n}.$

Proof. Since $n!$ is a unit in $R,$ we have $(n!)^{-1}S(t) = t$ for all $t \in S(M^{\otimes n}),$ so the image of $(n!)^{-1}S$ is $S(M^{\otimes n}).$ It is clear that $x_{1} \otimes \cdots \otimes x_{n} - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)}$ is in the kernel for any $\sigma \in S_{n},$ so we may define the desired map $$\mathrm{Sym}^{n}(M) \rightarrow S(M^{\otimes n}).$$ To check it is an isomorphism, let $t$ be any $n$-tensor such that $(n!)^{-1}S(t) = (n!)^{-1}\sum_{\sigma \in S_{n}}\sigma t = 0.$ We have $$t = t - \frac{1}{n!}\sum_{\sigma \in S_{n}}\sigma t = \frac{1}{n!}\sum_{\sigma \in S_{n}}\left(t - \sigma t \right),$$ but the image of $t - \sigma t$ is $0$ in $\mathrm{Sym}^{n}(M),$ so the image of $t$ must be $0$ as well. $\Box$

Exterior powers are alternating tensors. Let $n!$ be invertible in $R.$ Then the image of the map $(n!)^{-1}A : M^{\otimes n} \rightarrow M^{\otimes n}$ is $A(M^{\otimes n})$, and its kernel is $(x_{1} \otimes \cdots \otimes x_{n} : x_{i} = x_{j} \text{ for some } i \neq j).$ Thus, this map induces an isomorphism $$\bigwedge^{n}M \simeq A(M^{\otimes n})$$ such that $x_{1} \wedge \cdots \wedge x_{n} \mapsto x_{1} \otimes \cdots \otimes x_{n}.$

Proof. Since $n!$ is a unit in $R,$ we have $(n!)^{-1}A(t) = t$ for all $t \in A(M^{\otimes n}),$ so the image of $(n!)^{-1}A$ is $A(M^{\otimes n}).$ It is clear that $x_{1} \otimes \cdots \otimes x_{n}$ is in the kernel whenever we have some $i \neq j$ with $x_{i} = x_{j},$ by using the transpose switching $i$ and $j.$ Hence, we may define the desired map $$\bigwedge^{n}M \rightarrow A(M^{\otimes n}).$$ To check it is an isomorphism, let $t$ be any $n$-tensor such that $(n!)^{-1}A(t) = (n!)^{-1}\sum_{\sigma \in S_{n}}\mathrm{sgn}(\sigma)\sigma t = 0.$ We have $$t = t - \frac{1}{n!}\sum_{\sigma \in S_{n}} \mathrm{sgn}(\sigma) \sigma t = \frac{1}{n!}\sum_{\sigma \in S_{n}}\left(t - \mathrm{sgn}(\sigma)\sigma t \right),$$ but the image of $t - \mathrm{sgn}(\sigma)\sigma t$ is $0$ in $\bigwedge^{n}M,$ because $$0 = (x + y) \wedge (x + y) = x \otimes y + y \otimes x$$ so that $x \otimes y = - y \otimes x.$ This implies that the image of $t$ is $0$ in $\bigwedge^{n}M.$ $\Box$

Geometric tensors. Now, let $M$ be a free $R$-module with finite rank $m.$ Fix a basis $e_{1}, \dots, e_{m} \in M.$ Then we have an $R$-module isomorphism $$M \overset{\sim}{\longrightarrow} M^{\vee} = \mathrm{Hom}_{R}(M, R)$$ given by $e_{j} \mapsto e_{j}^{\vee}.$ This gives us an $R$-module isomorphism $$M^{\otimes n} \overset{\sim}{\longrightarrow} (M^{\otimes n})^{\vee}$$ given by $e_{j_{1}} \otimes \cdots \otimes e_{j_{n}} \mapsto (e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee}.$ We have $$(M^{\otimes n})^{\vee} = \mathrm{Hom}_{R}(M^{\otimes n}, R) \simeq \mathrm{Mult}_{R}(M^{n}, R)$$ given by $(e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee} \mapsto (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}).$ We call the elements in $\mathrm{Mult}_{R}(M^{n}, R)$ geometric $n$-tensors of $M.$

Remark. The terminology is non-standard.

We have $M^{\otimes n} \simeq \mathrm{Mult}_{R}(M^{n}, R)$ given by $$(e_{j_{1}} \otimes \cdots \otimes e_{j_{n}})^{\vee} \mapsto (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}).$$ We may define the product $$\mathrm{Mult}_{R}(M^{l}, R) \times \mathrm{Mult}_{R}(M^{n}, R) \rightarrow \mathrm{Mult}_{R}(M^{l+n}, R)$$ by concatenation $$(e_{i_{1}}^{\vee}, \dots, e_{i_{l}}^{\vee}) \cdot (e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}) = (e_{i_{1}}^{\vee}, \dots, e_{i_{l}}^{\vee}, e_{j_{1}}^{\vee}, \dots, e_{j_{n}}^{\vee}),$$ which is compatible with the product $M^{\otimes l} \times M^{\otimes n} \rightarrow M^{\otimes (l + n)}$ given by the tensor product. Note that given geometric $l$-tensor $f$ and $n$-tensor $g,$ we have $$(f \cdot g)(x_{1}, \dots, x_{l+n}) = f(x_{1}, \dots, x_{l}) g(x_{l+1}, \dots, x_{l+n}).$$ Considering the correspondence $e_{j_{1}} \otimes \cdots \otimes e_{j_{n}} \leftrightarrow (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}),$ we may consider the action of $S_{n}$ on $\mathrm{Mult}_{R}(M^{n}, R)$ given by $$\sigma \cdot (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}) := (e_{\sigma^{-1}(j_{1})}^{\vee}, \cdots, e_{\sigma^{-1}(j_{n})}^{\vee}).$$ Note that $$\begin{align*}(\sigma \cdot (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee}))(x_{1}, \dots, x_{n}) &= (e_{\sigma^{-1}(j_{1})}^{\vee}, \cdots, e_{\sigma^{-1}(j_{n})}^{\vee})(x_{1}, \dots, x_{n}) \\ &=  (e_{j_{1}}^{\vee}, \cdots, e_{j_{n}}^{\vee})(x_{\sigma(1)}, \dots, x_{\sigma(n)}),\end{align*}$$ because if $x_{i} = e_{\sigma^{-1}(i)},$ then $x_{\sigma(i)} = e_{i}.$ Thus, for $f \in \mathrm{Mult}_{R}(M^{n}, R),$ we have $$(\sigma f)(x_{1}, \dots, x_{n}) = f(x_{\sigma(1)}, \dots, x_{\sigma(n)}).$$ We note that under the isomorphism $M^{\otimes n} \simeq \mathrm{Mult}_{R}(M, R),$ the geometric $n$-tensor $f$ correponds to

• a symmetric $n$-tensor if $\sigma f = f$ for all $\sigma \in S_{n}$;
• an alternating $n$-tensor if $\sigma f = \mathrm{sgn}(\sigma)f$ for all $\sigma \in S_{n}.$

If $f$ corresponds to a symmetric (resp. alternating) $n$-tensor, we say $f$ is symmetric (resp. alternating). The set of symmetric geometric $n$-tensors is denoted as $S_{R}(M^{n}, R),$ and the set of alternating geometric $n$-tensors is denoted as $A_{R}(M^{n}, R),$ both of which are $R$-submodules of $\mathrm{Mult}_{R}(M^{n}, R).$ The discussions above imply the following:

Theorem. If $n!$ is invertible in $R,$ then we have $$\mathrm{Sym}^{n}(M) \simeq S_{R}(M^{n}, R)$$ given by $$e_{i_{1}} \otimes \cdots \otimes e_{i_{n}} \mapsto e_{i_{1}}^{\vee} \otimes \cdots \otimes e_{i_{n}}^{\vee} = (e_{i_{1}}^{\vee}, \dots, e_{i_{n}}^{\vee}),$$ where for $(f, g) \in \mathrm{Mult}_{R}(M^{r}, R) \times \mathrm{Mult}_{R}(M^{s}, R),$ we used the notation $f \otimes g := f \cdot g,$ namely $$(f \otimes g)(x_{1}, \dots, x_{r+s}) := f(x_{1}, \dots, x_{r})g(x_{r+1}, \dots, x_{r+s}).$$ Moreover, we have $$\bigwedge^{n}M \simeq A_{R}(M^{n}, R),$$ given by $e_{i_{1}} \wedge \cdots \wedge e_{i_{n}} \mapsto e_{i_{1}}^{\vee} \wedge \cdots \wedge e_{i_{n}}^{\vee} = (e_{i_{1}}^{\vee}, \dots, e_{i_{n}}^{\vee}),$ where for $(f, g) \in \mathrm{Mult}_{R}(M^{r}, R) \times \mathrm{Mult}_{R}(M^{s}, R)$ (assuming $r!, s! \in R^{\times}$) we used the notation $$f \wedge g := \frac{A(f \otimes g)}{r!s!} := \frac{1}{r!s!} \sum_{\sigma \in S_{r + s}}\sigma (f \otimes g).$$ That is, we have $$(f \wedge g)(x_{1}, \dots, x_{r+s}) := \frac{1}{r!s!}\sum_{\sigma \in S_{r+s}}f(x_{\sigma(1)}, \dots, x_{\sigma(r)})g(x_{\sigma(r+1)}, \dots, x_{\sigma(r+s)}).$$ The whole posting was due to extract the following purpose of mine:

Corollary. Algebraic definitions of symmetric power and exterior power over characteristic $0$ match with those of definitions in Chapter 3 of Tu's book.

Hodge theory: Lecture 11

We are following the notes taken by Alex Horawa, which are closer to lectures. The notes here might be slightly different from the actual lectures as it is mainly for my personal learning. We also substantially use this link (an English translation of Serre's GAGA) for the parts where I could not understood in the original lecture notes on my own.

Now that we know local rings of the structure sheaves are Noetherian, we now want to establish that the local map given by the (un)analytification map is faithfully flat. Just like last time, we will only discuss the smooth case.

Proposition. Let $X$ be a smooth complex variety. Then the local map $\mathscr{O}_{X, x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ induced by the analytification is faithfully flat for all $x \in X(\mathbb{C}).$

Proof. We consider the affine case first as the first step.

Step 1. Let $X = \mathbb{A}^{n}.$

Then consider the $\mathbb{C}$-algebra $R := \mathbb{C}\{z_{1}, \dots, z_{n}\}.$ It is a local ring with its unique maximal ideal being $$\mathfrak{m} := \{f \in R : f(0) = 0\} = (z_{1}, \dots, z_{n}).$$ Moreover, we have $$R/\mathfrak{m}^{N} \simeq \mathbb{C}[z_{1}, \dots, z_{n}]/(z_{1}, \dots, z_{n})^{N}$$ given by $z_{j} \mapsto z_{j},$ and this gives us $$\begin{align*}\hat{R} &= \lim_{N \geq 0} R/\mathfrak{m}^{N} \\ &\simeq \lim_{N \geq 0}\mathbb{C}[z_{1}, \dots, z_{n}]/(z_{1}, \dots, z_{n})^{N} \\ &\simeq \mathbb{C}[[z_{1}, \dots, z_{n}]]. \end{align*}$$ Recall that for any Noetherian local ring $(A, \mathfrak{m}),$ its completion $A \rightarrow \hat{A}$ is faitufully flat (e.g., Theorem on p.161 on Hochster's notes; he says "local" to mean Noetherian local). Thus, we deduce that $R \rightarrow \hat{R}$ is faithfully flat. We want to show that $$S := \mathbb{C}[z_{1}, \dots, z_{n}]_{(z_{1}, \dots, z_{n})} \rightarrow \mathbb{C}\{z_{1}, \dots, z_{n}\} = R$$ is faithfully flat. Fix any map $$M \rightarrow M'$$ of modules over $S.$ Then the following are equivalent:

1. $M \rightarrow M'$ is injective;
2. $M \otimes_{S} \hat{S} \rightarrow M' \otimes_{S} \hat{S}$ is injective;
3. $M \otimes_{S} R \otimes_{R} \hat{R} \rightarrow M' \otimes_{S} R \otimes_{R} \hat{R}$ is injective;
4. $M \otimes_{S} R \rightarrow M' \otimes_{S} R$ is injective,

because $S \rightarrow \hat{S}$ is faithfully flat and $\hat{S} \simeq \hat{R}.$ It follows that $S \rightarrow R$ is faithfully flat.

Remark. The fact that $R = \mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Noetherian local ring whose completion is $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ also implies that $R$ is a regular local ring of dimension $n$ (see the paragraph following 29.2.D in Vakil).

To tackle the general case, we need the following statement.

Lemma 1. Let $X \hookrightarrow Y$ be a smooth algebraic varieties, where $X$ is a closed subvariety of $Y$ with this inclusion, and let $\mathscr{I} \subset \mathscr{O}_{Y}$ be the ideal sheaf associated to $X.$ Then the ideal sheaf of $X^{\mathrm{an}}$ is given by $\mathscr{I}^{\mathrm{an}} \subset \mathscr{O}_{Y^{\mathrm{an}}}.$

Step 2. Let $X$ be a (smooth) closed subvariety of $\mathbb{A}^{n}.$

Let $I$ be the (radical) ideal of $\mathbb{C}[z_{1}, \dots, z_{n}]$ defining $X.$ Fix $x \in X(\mathbb{C}).$ We have $$\begin{align*}\mathscr{O}_{X,x} &= (\mathscr{O}_{\mathbb{A}^{n}}/\tilde{I})_{x} \\ &= \mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x}.\end{align*}$$ Using Lemma 1, we have $$\tilde{I}^{\mathrm{an}}_{x} = \tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ and $$\begin{align*}\mathscr{O}_{X^{\mathrm{an}},x} &= (\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}}/\tilde{I}^{\mathrm{an}})_{x} \\ &= \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}^{\mathrm{an}}_{x}.\end{align*}$$ Lemma also ensures that the map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is given by taking the quotient by $\tilde{I}_{x}$ of the map $$\mathscr{O}_{\mathbb{A}^{n}, x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x},$$ which we already showed to be faithfully flat. We want to show that $$\mathscr{O}_{\mathbb{A}^{n}, x}/\tilde{I}_{x} \rightarrow \mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}/\tilde{I}_{x}\mathscr{O}_{(\mathbb{A}^{n})^{\mathrm{an}}, x}$$ is faithfully flat. Note that taking the completion of this map would give us an isomorphism. Hence, it is enough to show the following:

Lemma 2. Let $A \rightarrow B$ be a local map of Notherian local rings. If their completion is flat, then the map is flat.

Before proving this lemma, note that this is enough to ensure faithful flatness because any flat local ring map is faithfully flat: fix any flat local map $(A, \mathfrak{m}) \rightarrow (B, \mathfrak{n})$. Say $M \rightarrow M'$ is an $A$-module map, denote by $K$ its kernel. We assume that $K \otimes_{A} B = 0$ and want to show $K = 0.$ Fix any finitely generated $A$-submodule $Q \subset K,$ and it is enough to show that $Q = 0.$ Since $A \rightarrow B$ is flat, we have $Q \otimes_{A} B \hookrightarrow K \otimes_{A} B = 0.$ Hence $Q \otimes_{A} B = 0.$ Because $A \rightarrow B$ is a local map, this implies that $$Q \otimes_{A} (A/\mathfrak{m}) = 0.$$ Since $Q$ is finitely generated, we may apply Nakayama's lemma to conclude that $Q = 0,$ as desired.

Proof of Lemma 2. Denote by $C = \hat{A} = \hat{B}.$ Fix any injective $A$-module map $M \hookrightarrow M'.$ Denote by $Q$ the kernel of $M \otimes_{A} B \rightarrow M' \otimes_{A} B.$ We want to show $Q = 0.$ Since $B \rightarrow C$ is flat, we have the following exact sequence $$0 \rightarrow Q \otimes_{B} C \rightarrow M \otimes_{A} C \rightarrow M' \otimes_{A} C.$$ Since $A \rightarrow \hat{A} = C$ is flat, we have $Q \otimes_{B} C = 0.$ Since $B \rightarrow \hat{B} = C$ is a flat local map, the same trick we have used above implies that $Q = 0,$ as desired. $\Box$

Step 3. Let $X$ be any smooth complex variety.

Since the statement is only about the stalks, Step 2 takes care of this general case. This finishes the proof (given Lemma 1) of Proposition. $\Box$

A reference for proof of Lemma 1. This seems a bit more convoluted than I initially thought. I decided to skip this for now, but a proof is given for Proposition 4 in this link, which seems to be an English translation of Serre's GAGA. The hardest part seems to be an analytic version of Nullstellensatz, but I did not look into the details.

Let $\iota : X \hookrightarrow Y$ be a closed embedding of smooth complex varieties and $\mathscr{F}$ an $\mathscr{O}_{X}$-module. We have

• the analytification $\iota^{\mathrm{an}} : X^{\mathrm{an}} \hookrightarrow Y^{\mathrm{an}}$ of the closed embedding,
• the unanalytification $u_{X} : X^{\mathrm{an}} \rightarrow X$ of $X,$ and
• the unanalytification $u_{Y} : Y^{\mathrm{an}} \rightarrow Y$ of $Y.$

Let $V \subset Y$ be any affine open. We have $$(\iota_{*}\mathscr{F})(V) = \mathscr{F}(V \cap X) \rightarrow (u_{X}^{*}\mathscr{F})(V^{\mathrm{an}} \cap X^{\mathrm{an}}) = ((u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F})(V),$$ which gives a map $$\iota_{*}\mathscr{F} \rightarrow (u_{Y})_{*}\iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y}$-modules. Hence, we get a map $$u_{Y}^{*}\iota_{*}\mathscr{F} \rightarrow \iota^{\mathrm{an}}_{*}u_{X}^{*}\mathscr{F}$$ of $\mathscr{O}_{Y^{\mathrm{an}}}$-modules.

Fact. The map above is an isomoprhism (at least when $\mathscr{F}$ is coherent).

Personal remark. The above fact will be used soon. However, I must have missed how it is proven. This is significantly stronger than Lemma 1, which is the statement for the case $\mathscr{F} = \mathscr{O}_{X}.$ Given Lemma 1 being convoluted, this fact should not be so easy, or perhaps one can deduce this from Lemma 1, although I cannot see this right away.

We now show the 1st GAGA for smooth projective varieties over $\mathbb{C}.$

Theorem. Let $X$ be a smooth projective variety over $\mathbb{C}.$ Then for any coherent $\mathscr{O}_{X}$-modules $\mathscr{F}, \mathscr{G},$ we have $$\mathrm{Ext}_{\mathscr{O}_{X}}^{i}(\mathscr{F}, \mathscr{G}) \simeq \mathrm{Ext}_{\mathscr{O}_{X^{\mathrm{an}}}}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ Before we prove this, we will organize some facts we will use without proofs. Fix any closed embedding $\iota : X \hookrightarrow \mathbb{P}^{n}.$

Fact 1. We repeat the fact we have said above: $$(\iota_{*}\mathscr{G})^{\mathrm{an}} \simeq \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}.$$ The next fact should be true, but I cannot find any reference just yet.

Fact 2. We have $$H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}),  \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}}).$$ Personal remark. We can probably track down what the map actually is, but let's not think about it now. The proof will probably require using Čech complex, but again, let's skip this detail for now.

Fact 3. We have $h^{0}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 1$ and $h^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}) = 0$ for all $i > 0.$

Remark. We were told that we will later deduce Fact 3 from the singular Betti numbers of $\mathbb{P}^{n}(\mathbb{C})$ and Hodge theory.

Proof of Theorem. We have several steps.

Step 1. Consider the case $\mathscr{F} = \mathscr{O}_{X}.$ Then our statement to show reduces to $$H^{i}(X, \mathscr{G}) \simeq H^{i}(X^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ By Fact 2, we may show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \iota^{\mathrm{an}}_{*}\mathscr{G}^{\mathrm{an}})$$ instead. By Fact 1, this means that we only need to show $$H^{i}(\mathbb{P}^{n}, \iota_{*}\mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), (\iota_{*}\mathscr{G})^{\mathrm{an}}),$$ so all in all, we reduce to the case where $X = \mathbb{P}^{n}.$

First, assume that $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$ More specifically, we assume $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ for some $m \in \mathbb{Z}.$ We then argue by induction on $n \geq 0.$ When $n = 0,$ we have $\mathbb{P}^{0} = \{*\},$ so either side has $1$-dimensional $H^{0},$ as the space of functions from a point consist of constant functions, and all the higher degree cohomology groups vanish. Hence, we get the base case $n = 0$ when $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m).$

Remark. The vanishing uses Čech cohomology on either side. A statement (not a proof) we need here can be found as (b) and (c) on p.64 in Wells.

Assume that the statement works for any complex projective spaces of dimension $0, 1, \dots, n-1.$ Assuming $n \geq 1,$ consider the exact sequence $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}} \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \rightarrow 0,$$ where $i : \mathbb{P}^{n-1} \hookrightarrow \mathbb{P}^{n}$ is the inclusion map for the closed subscheme (e.g., 14.3.B in Vakil). Tensoring with $\mathscr{O}_{\mathbb{P}^{n}}(m),$ we get $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow i_{*}\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow 0.$$ But then note that $$\begin{align*} (i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}) \otimes_{\mathscr{O}_{\mathbb{P}^{n}}} \mathscr{O}_{\mathbb{P}^{n}}(m) &\simeq i_{*}(\mathscr{O}_{\mathbb{P}^{n-1}} \otimes_{\mathscr{O}_{\mathbb{P}^{n-1}}} i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m)) \\ &\simeq i_{*}i^{*}\mathscr{O}_{\mathbb{P}^{n}}(m) \\ &= i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m),\end{align*}$$ where the first isomorphism is given by 16.3.H (b) in Vakil. Thus, we have the following exact sequence:  $$0 \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m-1) \rightarrow \mathscr{O}_{\mathbb{P}^{n}}(m) \rightarrow  i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m) \rightarrow 0.$$ Applying the analytification functor with Fact 1, we also have an exact sequence $$0 \rightarrow (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}} \rightarrow \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m) \rightarrow  i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}} \rightarrow 0.$$ The long exact sequences of cohomology groups give the following commuting diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) @>{}>>  H^{i}(\mathbb{P}^{n}, i_{*}\mathscr{O}_{\mathbb{P}^{n-1}}(m)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m-1)) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}}(m)) \\ @V{}VV @V{}VV @V{\simeq}VV @V{}VV @V{}VV \\ H^{i}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)) @>{}>>  H^{i}(\mathbb{P}^{n}, i^{\mathrm{an}}_{*}(\mathscr{O}_{\mathbb{P}^{n-1}}(m))^{\mathrm{an}}) @>{}>>  H^{i+1}(\mathbb{P}^{n}, (\mathscr{O}_{\mathbb{P}^{n}}(m-1))^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{O}_{\mathbb{P}^{n}(\mathbb{C})}(m)),\end{CD}$$ where the isomorphism in the middle is given by the induction hypothesis. In the lower row, the notations $\mathbb{P}^{n}$ should be replaced with $\mathbb{P}^{n}(\mathbb{C}),$ but we kept them for the sake of margin. Applying five lemma, we see that showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m)$ is equivalent to showing the statement for the case $\mathscr{G} = \mathscr{O}_{\mathbb{P}^{n}}(m+1).$ Thus, working with induction on $|m|,$ we may reduce to the case $m = 0,$ which holds by Fact 3. This finishes the case where $\mathscr{G}$ is a line bundle on $\mathbb{P}^{n}.$

For general coherent $\mathscr{G},$ we want to get $$H^{i}(\mathbb{P}^{n}, \mathscr{G}) \simeq H^{i}(\mathbb{P}^{n}(\mathbb{C}), \mathscr{G}^{\mathrm{an}})$$ for any $i \geq 0.$ Using Čech complex, both sides vanishes for $i \gg 0,$ so we may proceed with downward induction on $i$ (with all general coherent sheaves).

We may find an exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{G} \rightarrow 0,$$ where $\mathscr{E}$ is a finite direct sum of sheaves of the form $\mathscr{O}_{\mathbb{P}^{n}}(m)$ with finitely many various $m \in \mathbb{Z}.$ (For example, see Theorem 15.3.1 Vakil.) Taking direct sum of sheaves commutes with taking cohomology, so the statement must be true for $\mathscr{E}$ as well. Now, we have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} H^{i}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}) @>{}>>  H^{i}(\mathbb{P}^{n}, \mathscr{G}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}) \\ @V{\alpha}VV @V{\simeq}VV @V{\beta}VV @V{\simeq}VV @V{\simeq}VV \\ H^{i}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}) @>{}>>  H^{i}(\mathbb{P}^{n}, \mathscr{G}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{K}^{\mathrm{an}}) @>{}>> H^{i+1}(\mathbb{P}^{n}, \mathscr{E}^{\mathrm{an}}),\end{CD}$$ where the fourth vertical isomoprhism is given by induction hypothesis. By five lemma, the map $\beta$ is surjective. Again by five lemma, this implies that $\alpha$ is injective. Since $\mathscr{G}$ is an arbitrary coherent sheaf, this implies that $\alpha$ is also surjective. By five lemma, this implies that $\beta$ is bijective, as desired. This finishes the case where $\mathscr{F} = \mathscr{O}_{X}.$

Step 2. Let $\mathscr{F}$ be locally free.

Then $\mathscr{F}^{\mathrm{an}}$ is locally free. We have $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \simeq H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G})$$ functorially in $\mathscr{F}$ (e.g., Vakil 30.2.I). Hence, by Step 1, we may resolve this situation.

Personal remark. Presumably, a similar fact is available for manifolds, and the map $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{i}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}})$$ is given by $$H^{i}(X, \mathscr{F}^{\vee} \otimes \mathscr{G}) \rightarrow H^{i}(X^{\mathrm{an}}, (\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}}).$$ Moreover, we need to check $$(\mathscr{F}^{\mathrm{an}})^{\vee} \otimes \mathscr{G}^{\mathrm{an}} \simeq (\mathscr{F}^{\vee} \otimes \mathscr{G})^{\mathrm{an}},$$ which is probably true due to the fact that the analytification functor gives a flat base change. Still, I will leave these untouched for now.

Step 3. Let $\mathscr{F}$ be any coherent $\mathscr{O}_{X}$-module.

Choose any exact sequence $$0 \rightarrow \mathscr{K} \rightarrow \mathscr{E} \rightarrow \mathscr{F} \rightarrow 0$$ of coherent sheaves, where $\mathscr{E}$ is locally free. We have the following commutative diagram with exact rows $$\require{AMScd} \minCDarrowwidth6pt \begin{CD} \mathrm{Ext}^{i-1}(\mathscr{F}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}, \mathscr{G}) @>{}>>  \mathrm{Ext}^{i-1}(\mathscr{K}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}, \mathscr{G}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}, \mathscr{G}) \\ @V{\simeq}VV @V{\simeq}VV @V{\simeq}VV @V{\gamma}VV @V{\simeq}VV \\ \mathrm{Ext}^{i-1}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i-1}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>>  \mathrm{Ext}^{i-1}(\mathscr{K}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}) @>{}>> \mathrm{Ext}^{i}(\mathscr{E}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}),\end{CD}$$ where the isomorphisms on degree $i-1$ are given by induction hypothesis. Note that the base case is free because the negative degrees all vanish. By five lemma, we see $\gamma$ is injective (although the first vertical map is not necessary). We may apply five lemma again by introducing the next vertical map, which can now be assumed to be injective to conclude $\gamma$ is surjective. This finishes the proof (modulo a bunch of stuff I did not check). $\Box$

Hodge theory: Lecture 10

Some photos I have taken are too blurry, so I am also following some nicely organized notes by Aleksander Horawa. Of course, this blog is all about my personal learning notes, so they are not identical to the actual lectures.

We have used the following two facts.

1. Given a complex analytic space, the local ring at any point is Noetherian.
2. Let $X$ be a complex variety. Given any $\mathbb{C}$-point $x$ of $X,$ the local map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is faithfully flat.

Our goal is to prove the first for the case of complex manifolds and the second for the case of smooth complex varieties. (In this posting, we only deal with the first.)

Ring of convergent power series. We write $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ to mean the $\mathbb{C}$-subalgebra of $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ that consists of convergent power series, meaning $$f(z_{1}, \dots, z_{n}) = \sum_{d_{1} \geq 0} \cdots \sum_{d_{n} \geq 0} a_{d_{1}, \dots, d_{n}}z_{1}^{d_{1}} \cdots z_{n}^{d_{n}}$$ such that there exits some $\epsilon > 0$ such that $f(z_{1}, \dots, z_{n})$ converges uniformly and absolutely for $|z_{j}| < \epsilon$ for all $1 \leq j \leq n.$

Remark. Note that $f(z_{1}, \dots, z_{n})$ given above is convergent if and only if there is $\epsilon > 0$ such that the sequence $(|a_{d_{1}, \dots, d_{n}}|\epsilon^{d_{1} + \cdots + d_{n}})_{d_{1}, \dots, d_{n} \geq 0}$ is bounded. By Cauchy-Hadamard, this is also equivalent to $$\limsup_{d_{1} + \cdots + d_{n} \rightarrow \infty}|a_{d_{1}, \dots, d_{n}}|^{1/(d_{1} + \cdots + d_{n})} < \infty,$$ although I cannot quite comprehend this just yet.

The map $\mathscr{O}_{\mathbb{C}^{n}, \boldsymbol{0}} \rightarrow \mathbb{C}[[z_{1}, \dots, z_{n}]]$ given by $$f(z_{1}, \dots, z_{n}) \mapsto \sum_{\boldsymbol{d} \in (\mathbb{Z}_{\geq 0})^{n}}\frac{1}{\boldsymbol{d}!} \frac{\partial^{|\boldsymbol{d}|}f}{\partial \boldsymbol{z}^{\boldsymbol{d}}}(\boldsymbol{0}) \boldsymbol{z}^{\boldsymbol{d}}$$ is injective $\mathbb{C}$-algebra map, and its image is $\mathbb{C}\{z_{1}, \dots, z_{n}\}.$ Hence, for any complex manifold $X$ and a point $x \in X,$ we have $$\mathscr{O}_{X,x} \simeq \mathscr{O}_{\mathbb{C}^{n}, \boldsymbol{0}} \simeq \mathbb{C}\{z_{1}, \dots, z_{n}\}.$$ Thus, to show local rings are Noetherian, it is enough to show:

Theorem. The $\mathbb{C}$-algebra $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is Noetherian.

This will be a direct consequence of another theorem, called "Weierstrass Preparation". For this, we define that an element in $\mathbb{C}\{z_{1}, \dots, z_{n}\}$ is a Weierstrass polynomial in $z_{n}$ if it can be written as the following form: $$z_{n}^{d} + a_{1}(z_{1}, \dots, z_{n-1})z_{n}^{d-1} + \cdots + a_{d}(z_{1}, \dots, z_{n-1}) \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ such that $a_{j}(0, \dots, 0) = 0$ for $1 \leq j \leq d.$

Weierstrass Preparation Theorem. For $f(z_{1}, \dots, z_{n}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that $f(0, \dots, 0, z_{n}) \neq 0 \in \mathbb{C}\{z_{n}\},$ there are unique $g(\boldsymbol{z}), h(\boldsymbol{z}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}$ such that

1. $h(0, \dots, 0) \neq 0,$
2. $g(\boldsymbol{z})$ is a Weierstrass polynomial in $z_{n},$ and
3. $f(\boldsymbol{z}) = g(\boldsymbol{z})h(\boldsymbol{z}) \in \mathbb{C}\{z_{1}, \dots, z_{n}\}.$

Remark. Note that Weierstrass Preparation basically says that if a convergent power series has a monomial consisting of a single variable, then it is divisible by a Weierstrass polynomial (in somewhat unique fashion). For the case $n = 1,$ this is already interesting: any nonzero convergent power series $f(z)$ over $\mathbb{C}$ can be written as $$f(z) = z^{d} h(z)$$ with $h(0) \neq 0.$ This shows that zeros of $f \in \Gamma(X, \mathscr{O}_{X})$ do not accumulate in $X$ for any complex manifold $X$ of dimension $1.$

We first apply Weierstrass Preparation to prove Theorem.

Proof of Theorem. We proceed by induction on $n \geq 0.$ For $n = 0,$ we have $\mathbb{C},$ a field, which is Noetherian. Now, let $n \geq 1,$ and suppose that the result is true for any convergent power series $\mathbb{C}$-algebras with the number of variables $< n.$ Fix any nonzero ideal $I \subset \mathbb{C}\{z_{1}, \dots, z_{n}\}$ and a nonzero element $f(z_{1}, \dots, z_{n}) \in I.$ We may apply a $\mathbb{C}$-linear map to the variables $z_{1}, \dots, z_{n}$ to ensure that $f(z_{1}, \dots, z_{n})$ has a monomial that looks like $z_{n}^{d}$ for some $d \in \mathbb{Z}_{\geq 0}.$ Indeed, if there is any other $z_{j}^{d},$ we may permute the indices. If there is no monomial consisting of a single variable, consider any term, which must look like $c z_{i_{1}}^{d_{1}} \cdots z_{i_{r}}^{d_{r}}$ for some $c \neq 0.$ Permuting indices and dividing by $c,$ we may assume that the term we have is $z_{n}^{d_{1}} \cdots z_{n-r}^{d_{r}},$ and we just need to consider the invertible $\mathbb{C}$-linear map given by $z_{j} \mapsto z_{n}$ for $1 \leq j < n$ and $z_{n} \mapsto z_{n}.$ Thus, we are in the situation where $f(0, \dots, 0, z_{n}) \neq 0.$ Now, take any generating set $S$ of $I$ containing $f(\boldsymbol{z}),$ which is our modified version. For any $g(\boldsymbol{z}) \in S,$ if $g(0, \dots, 0, z_{n}) = 0,$ then replace $g$ with $g + f$ in $S$ so that we may assume $g(0, \dots, 0, z_{n}) \neq 0.$ This constructs another generating set for $I$ because any element in $I$ can be written as a linear sum of finitely many generators over the base ring. Hence, after this modification, we may assume that $S$ is a generating set for $I$ such that for any $f \in S,$ we have $$f(0, \dots, 0, z_{n}) \neq 0.$$ But then now note that Weierstrass Preparation says that we have $f \in I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$ This shows that $$S \subset I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$$ By the induction hypothesis $\mathbb{C}\{z_{1}, \dots, z_{n-1}\}$ is Noetherian, so by Hilbert basis, $\mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}].$ is Noetherian as well. This implies that $I \cap \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$ is finitely generated, so $(S) = I$ must be too. $\Box.$

Proof of Weierstrass Preparation. We will use the following residue theorem: consider any nonempty open $U \subset \mathbb{C}$ and fix distinct $p_{1}, \dots, p_{r} \in D,$ an open disc such that $\bar{D} \subset U.$ Given any $f \in \Gamma(U \smallsetminus \{p_{1}, \dots, p_{r}\}, \mathscr{O}_{\mathbb{C}}),$ we have $$\frac{1}{2\pi i}\int_{\partial D}f(z)dz = \sum_{j=1}^{r}\mathrm{Res}_{p_{j}}(f).$$ The proof of this fact is in any usual complex analysis text (e.g., Ch. 3 Corollary 2.2 of Stein and Shakarchi). We will apply this in the following specific situation.

Lemma. Let $U \subset \mathbb{C}$ be nonempty and open and say $f \in \Gamma(U, \mathscr{O}_{\mathbb{C}}).$ Say we have an open disk $D$ in $\mathbb{C}$ such that $\bar{D} \subset U$ and $f$ has no roots on $.$ For any $j \in \mathbb{Z}_{\geq 0},$ we have $$\int_{\partial D} z^{j} \frac{f'(z)}{f(z)} dz = \lambda_{1}^{j} + \cdots + \lambda_{m}^{j},$$ where $\lambda_{1}, \dots, \lambda_{m}$ are the roots of $f(z)$ in $D$ counted with multiplicity.

Proof. If there are no roots of $f$ in $D,$ then it has no roots in $\bar{D}$ due to the hypothesis. Since $\bar{D}$ is closed in $\mathbb{C},$ this means that there is an open subset $W$ of $\mathbb{C}$ containing $\bar{D}$ such that $f$ takes no roots in $W.$ On $W,$ the expression $z^{j} f'(z)/f(z)$ defines a holomorphic function in $z \in W,$ so by Cauchy's theorem the integral in question is evaluated to be $0.$

Now, suppose that we have some roots of $f$ in $D.$ Let $\lambda$ be any one of them. We may write $$f(z) = (z - \lambda)^{m}h(z),$$ where $m \in \mathbb{Z}_{\geq 1}$ and $h(z)$ is a holomorphic function defined in a small enough open disc centered at $\lambda$ such that $h(\lambda) \neq 0.$ Then $$\frac{f'(z)}{f(z)} = \frac{m(z - \lambda)^{m-1}h(z) + (z - \lambda)^{m}h'(z)}{(z - \lambda)^{m}h(z)} = \frac{m}{z - \lambda} + \frac{h'(z)}{h(z)},$$ so $$\begin{align*}\mathrm{Res}_{\lambda}\left(z^{j}\frac{f'(z)}{f(z)}\right) &= \mathrm{Res}_{\lambda}\left(\frac{mz^{j}}{z - \lambda}\right) \\ &= \lim_{z \rightarrow \lambda} (z - \lambda) \left(\frac{mz^{j}}{z - \lambda}\right) \\ &= m\lambda^{j}.\end{align*}$$ Since $m$ is the multiplicity of the root $\lambda.$ Applying the residue theorem, this finishes the proof. $\Box$

We are now ready to prove Weierstrass Preparation.

Proof of Weierstrass Preparation. Fix an open neighborhood $U$ of $(0, \dots, 0) \in \mathbb{C}.$ Since $f(0, \dots, 0, z_{n}) \neq 0,$ we know that zeros of $f(0, \dots, 0, z_{n})$ in $z_{n}$ stay outside a small disc centered at the origin. Hence, we may take $\epsilon_{n} > 0$ such that $f(0, \dots, 0, z_{n})$ has no zeros in $z_{n}$ for $0 < |z_{n}| \leq \epsilon_{n}.$ Since $z_{n} \in \mathbb{C}$ with $|z_{n}| = \epsilon_{n}$ defines a compact set in $\mathbb{C},$ we may find $\epsilon > 0$ such that whenever $|z_{1}|, \dots, |z_{n-1}| < \epsilon$ and $|z_{n}| = \epsilon_{n},$ we have $f(z_{1}, \dots, z_{n-1}, z_{n}) \neq 0.$ We may also assume that $\epsilon > 0$ is small enough so that $$D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0) \subset U.$$ Fix any $\boldsymbol{z} = (z_{1}, \dots, z_{n-1}) \in D_{\epsilon}(0)^{n-1},$ and let $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m_{\boldsymbol{z}}}(\boldsymbol{z})$ be the roots of $f(\boldsymbol{z}, z_{n})$ in $z_{n} \in D_{\epsilon_{n}}(0),$ counted with multiplicities. (The can only be finitely many zeros because otherwise, they will accrue in $\bar{D},$ which will contradict that $f(\boldsymbol{z}, z_{n}) \neq 0.$) Applying Lemma, for any $j \in \mathbb{Z}_{\geq 0},$ we get $$\sum_{l=1}^{m_{\boldsymbol{z}}}\lambda_{l}(\boldsymbol{z})^{j} = \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} w^{j}\frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw,$$ The upshot is that the right-hand side is a holomorphic function in $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1}.$ Taking $j = 0,$ we have $$ \frac{1}{2\pi i} \int_{|w| = \epsilon_{n}} \frac{\frac{\partial f}{\partial w}(\boldsymbol{z}, w)}{f(\boldsymbol{z}, w)} dw = m_{\boldsymbol{z}},$$ but since the right-hand side is always integer no matter where we vary $\boldsymbol{z}$ in the connected set $D_{\epsilon}(0)^{n-1},$ it must be the same integer regardless of the choice of $\boldsymbol{z}.$ Hence, we may write $m = m_{\boldsymbol{z}}.$

Denote by $\sigma_{1}(\boldsymbol{z}), \dots, \sigma_{n}(\boldsymbol{z})$ the elementary symmetric polynomials in $\lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$ meaning, we have $$\begin{align*}g(z_{1}, \dots, z_{n}) &:= z_{n}^{m} - \sigma_{1}(\boldsymbol{z})z_{n}^{m-1} + \cdots + (-1)^{m}\sigma_{m}(\boldsymbol{z}) \\ &= (z_{n} - \lambda_{1}(\boldsymbol{z})) \cdots (z_{n} - \lambda_{m}(\boldsymbol{z})).\end{align*}$$ By definition this is a Weierstrass polynomial in $z_{n}.$ Note that $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ defines a holomorphic function on $$(D_{\epsilon}(0)^{n-1} \times D_{\epsilon_{n}}(0)) \smallsetminus \{(\boldsymbol{z}, z_{n}) : z_{n} \neq \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z})\}.$$ For every fixed $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ the fraction $f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n})$ extends to a unique holomorphic function in $z_{n} \in D_{\epsilon_{n}}(0),$ since the numerator can cancel out all the roots of the denominator. Calling this function $h(\boldsymbol{z}, z_{n}),$ we check that $h(0, \dots, 0) \neq 0$ (since the roots of the numerator is precisely given by those of the denominator) and since $h(\boldsymbol{z}, z_{n}) = f(\boldsymbol{z}, z_{n})/g(\boldsymbol{z}, z_{n}),$ fixing $z_{n},$ we see that $h(\boldsymbol{z}, z_{n})$ is holomorphic in $\boldsymbol{z}$ as well. This gives us the existence of the desired factorization: $$f(z_{1}, \dots, z_{n}) = g(z_{1}, \dots, z_{n})h(z_{1}, \dots, z_{n}).$$ For the uniqueness, suppose that we have another factorization $$f(z_{1}, \dots, z_{n}) = g'(z_{1}, \dots, z_{n})h'(z_{1}, \dots, z_{n}),$$ where $$g'(z_{1}, \dots, z_{n}) = z_{n}^{m'} + \cdots \in \mathbb{C}\{z_{1}, \dots, z_{n-1}\}[z_{n}]$$ with $h'$ holomorphic such that $h'(0, \dots, 0) \neq 0.$ Then $$z_{n}^{m}h(0, \dots, 0, z_{n}) = f(0, \cdots, 0, z_{n}) = z_{n}^{m'}h'(0, \dots, 0, z_{n}),$$ and both sides must have the same number of roots in $z_{n}$ counting with multiplicities, so $m' = m.$ Fix $\boldsymbol{z} \in D_{\epsilon}(0)^{n-1},$ where we may have to make $\epsilon > 0$ smaller. Then $g'(\boldsymbol{z}, z_{n})$ vanishes at $$z_{n} = \lambda_{1}(\boldsymbol{z}), \dots, \lambda_{m}(\boldsymbol{z}),$$ counting with multiplicities. Hence, we must have $g(\boldsymbol{z}, z_{n}) = g'(\boldsymbol{z}, z_{n}),$ and $h = h'$ follows from here. $\Box$

Hodge theory: Lecture 9 - Part 2

Major warning. From this posting, we deal with some homological algebra, which I have not used it so long. Thus, mistakes are bound to happen, so please let me know if you find them!

Given a complex variety $X,$ we have a functor $\mathrm{Mod}_{\mathscr{O}_{X}} \rightarrow \mathrm{Mod}_{\mathscr{O}_{X^{\mathrm{an}}}}$ given by $\mathcal{F} \mapsto \mathcal{F}^{\mathrm{an}}.$ Recall that $\mathscr{F}^{\mathrm{an}} = u^{*}\mathscr{F}$ by definition, where $u : X^{\mathrm{an}} \rightarrow X$ is the unanalytification map. Using the adjoint bijection $$\mathrm{Hom}_{\mathscr{O}_{X^{\mathrm{an}}}}(u^{*}\mathscr{F}, \mathscr{G}) \simeq \mathrm{Hom}_{\mathscr{O}_{X}}(\mathscr{F}, u_{*}\mathscr{G}),$$ taking $\mathscr{G} = u^{*}\mathscr{F} = \mathscr{F}^{\mathrm{an}},$ we may obtain an $\mathscr{O}_{X}$-linear map $\mathscr{F} \rightarrow u_{*}\mathscr{F}^{\mathrm{an}}$ corresponding to the identity map of $\mathscr{F}^{\mathrm{an}}.$

What does the map look like? It is easier to use the more concrete definition $$u^{*}\mathscr{F} = u^{-1}\mathscr{F} \otimes_{u^{-1}\mathscr{O}_{X}} \mathscr{O}_{X^{\mathrm{an}}}.$$ Suppose that we are given a map $u^{*}\mathscr{F} \rightarrow \mathscr{G}.$ For any open $V \subset X^{\mathrm{an}},$ we have $$(u^{-1}\mathscr{F})(V) \rightarrow (u^{-1}\mathscr{F})(V) \otimes_{(u^{-1}\mathscr{O}_{X})(V)} \mathscr{O}_{X^{\mathrm{an}}}(V) \rightarrow \mathscr{G}(V),$$ so for any open $U \subset X,$ taking $V = U^{\mathrm{an}} = u^{-1}(U),$ we have $$(u^{-1}\mathscr{F})(u^{-1}(U)) \rightarrow \mathscr{G}(u^{-1}(U)).$$ Recall that $$(u^{-1}\mathscr{F})(V) = \mathrm{colim}_{W : u^{-1}(W) \subset V}\mathscr{F}(W),$$ where $W \subset X$ runs over all open subsets (with the specified condition), which gives us a map $\mathscr{F}(U) \rightarrow (u^{-1}\mathscr{F})(u^{-1}(U)).$ Hence, we get a map $$\mathscr{F}(U) \rightarrow (u^{-1}\mathscr{F})(u^{-1}(U)) \rightarrow \mathscr{G}(u^{-1}(U)) = (u_{*}\mathscr{G})(U).$$ This describes how to get $\mathscr{F} \rightarrow u_{*}\mathscr{G}.$

Explicit localization. Now, if $\mathscr{G} = u^{*}\mathscr{F}$ and we started with the identity map $u^{*}\mathscr{F} \rightarrow u^{*}\mathscr{F},$ then from the above description, the map $\mathscr{F} \rightarrow u_{*}u^{*}\mathscr{F}$ is given by $$\mathscr{F}(U) \rightarrow (u^{-1}\mathscr{F})(u^{-1}(U)),$$ followed by the tensor product and the sheafifications. Hence, if we compute the loacalization of $\mathscr{F} \rightarrow u_{*}u^{*}\mathscr{F}$ at $x \in X(\mathbb{C}),$ we get $$\mathscr{F}_{x} \rightarrow \mathscr{F}_{x} \otimes_{\mathscr{O}_{X,x}} \mathscr{O}_{X^{\mathrm{an}},x},$$ which is the same map as what we get by applying $\mathscr{F}_{x} \otimes_{\mathscr{O}_{X,x}} (-)$ to the local map $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}.$

Next goal. We will show that $\mathscr{O}_{X^{\mathrm{an}},x}$ is Noetherian and the local map $$\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$$ is flat. Due to the description above about the localization of $\mathscr{F} \rightarrow u_{*}\mathscr{F}^{\mathrm{an}},$ it would follow that this analytification map for $\mathscr{F}$ is exact.

Before we go into proving Noetherianity and flatness of the localization of the unanalytification map, we consider some statements from GAGA. For any ringed space $(X, \mathscr{O}_{X}),$ the category $\mathrm{Mod}_{\mathscr{O}_{X}}$ has enough injectives (Vakil Theorem 23.4.1), so the functor $\mathscr{F} \rightarrow u_{*}\mathscr{F}^{\mathrm{an}},$ induces $$H^{i}(X, \mathscr{F}) \rightarrow H^{i}(X, u_{*}\mathscr{F}^{\mathrm{an}}) = H^{i}(X^{\mathrm{an}}, \mathscr{F}^{\mathrm{an}}),$$ where the equality follows from the definition of the $i$-th right derived functor of global section functors of $u_{*}\mathscr{F}^{\mathrm{an}}$ and $\mathscr{F}^{\mathrm{an}}.$

1st GAGA. Let $X$ be a complete variety over $\mathbb{C}.$ Then the functor $$\mathrm{Mod}_{\mathscr{O}_{X}} \rightarrow \mathrm{Mod}_{\mathscr{O}_{X^{\mathrm{an}}}}$$ given by $\mathscr{F} \mapsto \mathscr{F}^{\mathrm{an}}$ restricts to a fully faithful functor $$\mathrm{Coh}(X) \rightarrow \mathrm{Coh}(X^{\mathrm{an}})$$ on coherent sheaves. Furthermore, for any coherent $\mathscr{F}$ and $\mathscr{G}$ on $X,$ the map $$\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{i}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}})$$ is an isomorphism.

Remark. We first need to explain what it means to be "coherent" means in $\mathrm{Mod}_{\mathscr{O}_{X}}$ for a general ringed space $(X, \mathscr{O}_{X}).$ We say an $\mathscr{O}_{X}$-module $\mathscr{F}$ is locally finitely generated if there is an open cover $X = \bigcup_{i \in I}U_{i}$ with some exact sequences $$\mathscr{O}_{U_{i}}^{n_{i}} \rightarrow \mathscr{F}|_{U_{i}} \rightarrow 0$$ for some $n_{i} \in \mathbb{Z}_{\geq 0}.$ A locally finitely generated $\mathscr{F}$ is coherent if

• $\mathscr{F}$ is locally finitely generated and 
• for any open $U \subset X,$ the kernel of any morphism $\mathscr{O}_{U}^{\oplus n} \rightarrow \mathscr{F}|_{U}$ with any $n \in \mathbb{Z}_{\geq 0}$ is locally finitely generated.

When $X$ is a locally Noetherian scheme, the notion of coherence coincides with the notion of coherence in scheme theory (e.g., Chapter 5 Theorem 1.7 in Liu). Here $X$ being locally Noetherian makes the second condition above moot.

Why do we bother to think about more general definition of cohernece? There are two reasons I can think about.

1. We do need the notion for the category of analytic spaces over $\mathbb{C}$.
2. With this definition, coherent sheaves form an abelain category, so one can work with any scheme without any Noetherian conditions.

Personal remark. The above remarks are just cited without much of my soul involved.

We also need to understand how the maps between the Ext's are given. We first know the functor $\mathscr{F} \mapsto \mathscr{F}^{\mathrm{an}}$ gives $$\mathrm{Hom}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Hom}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ This is the $0$-th map $$\mathrm{Ext}^{0}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{0}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$$ The category $\mathrm{Mod}_{\mathscr{O}_{X}}$ has enough injectives (Vakil Theorem 23.4.1), so the $i$-th right derived functor $\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, -)$ of the left exact functor $$\mathrm{Hom}_{\mathscr{O}_{X}}(\mathscr{F}, -) : \mathrm{Mod}_{\mathscr{O}_{X}} \rightarrow \textbf{Ab}$$ is a universal $\delta$-functor (Vakil Corollary 23.2.10). Hence, the morphism $$\mathrm{Hom}_{\mathscr{O}_{X}}(\mathscr{F}, -) \rightarrow \mathrm{Hom}_{\mathscr{O}_{X^{\mathrm{an}}}}(u_{*}\mathscr{F}^{\mathrm{an}}, u_{*}(-))$$ of functors giving the degree $0$ map must extend to higher degrees.

Still to do. I have not completed the description of the maps between Ext groups, so the 1st GAGA only makes sense to me for the case $i = 0$ yet. This may or may not work.

Personal remark. Quite a bit of the remarks in the preceding paragraphs are obtained by skimming various sources and conversations with some folks I know without detailed justifications, so I should perhaps come back to this at some point to check the details. Thanks to Nancy Wang, I realized that we cannot use free resolutions in this scenario. The reason is that the structure sheaf is rarely projective (e.g. the paragraph after 30.2.D in Vakil).

What I am told. I have consulted an expert, who suggested me to think about the description of $\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G})$ as an equivalence class of exact sequences of the form: $$0 \rightarrow \mathscr{G} \rightarrow \mathscr{E}_{1} \rightarrow \cdots \rightarrow \mathscr{E}_{n} \rightarrow \mathscr{F} \rightarrow 0.$$ This is great because since the analytification functor on sheaves is exact, so we get $$0 \rightarrow \mathscr{G}^{\mathrm{an}} \rightarrow \mathscr{E}_{1}^{\mathrm{an}} \rightarrow \cdots \rightarrow \mathscr{E}_{n}^{\mathrm{an}} \rightarrow \mathscr{F}^{\mathrm{an}} \rightarrow 0.$$ This does give a map $\mathrm{Ext}^{i}_{\mathscr{O}_{X}}(\mathscr{F}, \mathscr{G}) \rightarrow \mathrm{Ext}^{i}_{\mathscr{O}_{X^{\mathrm{an}}}}(\mathscr{F}^{\mathrm{an}}, \mathscr{G}^{\mathrm{an}}).$ This description of elements of Ext seems quite general. (See Construction of Ext in abelian categories in this Wikipedia page.)

Remark. The future lectures will prove the 1st GAGA for the case when $X$ is projective over $\mathbb{C},$ and this case is due to Serre. The version for complete varieties over $\mathbb{C}$ is due to Grothendieck. There are (relative) versions for proper morphisms.

Given an analytic space $X,$ checking that $\mathscr{O}_{X}$ is coherent seems quite difficult. That is, how an earth would we check every map $\mathscr{O}_{X}^{n} \rightarrow \mathscr{O}_{X}$ with any $n \in \mathbb{Z}_{\geq 0}$ has a kernel that is locally finitely generated? Thankfully, this seems like a famous theorem:

Theorem (Oka). Given any analytic space $X,$ its structure sheaf $\mathscr{O}_{X}$ is coherent. In particular, vector bundles on $X$ give coherent $\mathscr{O}_{X}$-modules.

Personal remark. Where is a reference for Oka's theorem for analytic spaces? Maybe this "easily" follows from the Wikipedia version when $X$ is complex manifold?

Analytification of a coherent module is coherent. Let $X$ be a complex variety and $\mathscr{F}$ a coherent $\mathscr{O}_{X}$-module. Then $\mathscr{F}^{\mathrm{an}}$ is a coherent $\mathscr{O}_{X^{\mathrm{an}}}$-module.

Proof. Note that for any complex variety $X$ and a coherent $\mathscr{O}_{X}$-module $\mathscr{F},$ we can choose an exact sequence $$\mathscr{O}_{X}^{\oplus m} \rightarrow \mathscr{O}_{X}^{\oplus n} \rightarrow \mathscr{F} \rightarrow 0.$$ This induces $$\mathscr{O}_{X^{\mathrm{an}}}^{\oplus m} \rightarrow \mathscr{O}_{X^{\mathrm{an}}}^{\oplus n} \rightarrow \mathscr{F}^{\mathrm{an}} \rightarrow 0,$$ where the exactness can be checked at the stalks. Thus, by Oka's theorem, we see that $\mathscr{F}^{\mathrm{an}}$ is the cokernel of a morphism between coherent $\mathscr{O}_{X^{\mathrm{an}}}$-modules. Using the fact that coherent $\mathscr{O}_{X^{\mathrm{an}}}$-modules form an abelian category (which does not seem obvious), this proves that $\mathscr{F}^{\mathrm{an}}$ is coherent. $\Box$

Remark. The above proves a tiny bit of the 1st GAGA.

2nd GAGA. Let $X$ be a complete complex variety. The functor $\mathscr{F} \mapsto \mathscr{F}^{\mathrm{an}}$ gives an equivalence of the categories $$\mathrm{Coh}(X) \simeq \mathrm{Coh}(X^{\mathrm{an}}).$$ Here is an interesting consequence. We will prove that $\mathscr{O}_{X,x} \rightarrow \mathscr{O}_{X^{\mathrm{an}},x}$ is faithfully flat for every $x \in X(\mathbb{C}),$ so this tells us that the functor given for the 2nd GAGA restricts to the following equivalence of categories: $$\{\text{locally free coherent } \mathscr{O}_{X}\text{-modules}\} \simeq \{\text{locally free coherent } \mathscr{O}_{X^{\mathrm{an}}}\text{-modules}\}.$$ This equivalence evidently preserves the rank, so it gives an equivalence between vector bundles of fixed rank.