Hodge theory: Lecture 3

Finite type assumption on morphisms. Listening to the last two lectures, I was thinking that morphisms are always $\mathbb{C}$-scheme maps so that I can consider $\textbf{Var}_{\mathbb{C}} \hookrightarrow \textbf{Sch}_{\mathbb{C}}$ as a full subcategory. However, I realized that in class, morphisms in $\textbf{Var}_{\mathbb{C}}$ are often assumed to be of finite type. Perhaps, one can work with only finite type maps, but this will require quite a bit of tedious checks whether various constructions (e.g., fiber product) work even if we only allow finite maps. I will just stick with $\mathbb{C}$-scheme maps, as they are easier to work with and try my best to catch where the lecture is using the finite type assumption. However, in this lecture we will change the notion of $\mathbb{C}$-varieties by adding the assumption that it is separated. In this case, all $\mathbb{C}$-scheme maps between varieties are necessarily of finite type (Vakil 10.1.19 (ii)), so there will be no problems!

Surjective morphism admits surjective analytification. Another thing I have not thought about earlier is that the analytification of any morphism (i.e., $\mathbb{C}$-scheme map) $\pi : X \rightarrow Y$ preserves surjectivity. Indeed, if $\pi$ is surjective, then for any $y \in Y,$ the fiber $\pi^{-1}(y)$ is a nonempty closed subset of a quasi-compact scheme $X,$ so there must be a closed point (i.e., a $\mathbb{C}$-point) in $\pi^{-1}(y)$ (Vakil 5.1.E).

Back to the lecture now.

Last time. A $\mathbb{C}$-variety $X$ is separated if and only if $X^{\mathrm{an}}$ is Hausdorff.

What about completeness? A $\mathbb{C}$-variety $X$ is complete (i.e., proper over $\mathrm{Spec}(\mathbb{C})$) if and only if $X^{\mathrm{an}}$ is compact Hausdorff.

Proof. Note first that we have $$\left\{(z_{0}, \dots, z_{n}) \in \mathbb{C}^{n+1} : \sum_{i=0}^{n}|z_{i}|^{2} = 1\right\} \twoheadrightarrow \mathbb{P}^{n}(\mathbb{C})$$ given by $$(z_{0}, \dots, z_{n}) \mapsto [z_{0} : \cdots : z_{n}],$$ which is a continuous map, where the right-hand side gets the classical topology. Since the left-hand side is compact (closed and bounded in $\mathbb{C}^{n+1}$), we notice that $\mathbb{P}^{n}(\mathbb{C})$ is compact with the classical topology.

If $X$ is complete, then it is separated (just by definition), so $X^{\mathrm{an}}$ is Hausdorff. Now, Chow's lemma (Vakil 18.9.2) implies that we have a $\mathbb{C}$-scheme map $\pi : \tilde{X} \rightarrow X$ such that

• $\tilde{X}$ is a projective $\mathbb{C}$-scheme,
• $\pi$ is surjective,
• $\pi$ is projective,
• $\pi$ is birational (i.e., it is an isomorphism on a dense open subset of $X$).

We won't use the last two properties. Note that we may take the reduction of $\tilde{X}$ (e.g., Vakil 8.3.10) if necessary to assume that $\tilde{X}$ is reduced. Hence, we see $\tilde{X}$ is a projective $\mathbb{C}$-variety. Thus, we have a closed embedding $\tilde{X} \hookrightarrow \mathbb{P}^{n},$ so $\tilde{X}(\mathbb{C}) \hookrightarrow \mathbb{P}^{n}(\mathbb{C})$ is closed in the classical topology. We know that $\mathbb{P}^{n}(\mathbb{C})$ is compact, so its closed subset $\tilde{X}(\mathbb{C})$ is compact. Since we have a continuous surjection $$\tilde{X}(\mathbb{C}) \twoheadrightarrow X(\mathbb{C}),$$ we see that $X(\mathbb{C})$ is compact in the classical topology.

Remark for myself. I have not ever read any proof of Chow's lemma yet, but the proof given in 18.9.4 of Vakil does not seem impossible to be read.

Conversely, let $X^{\mathrm{an}}$ be compact Hausdorff. We already know that $X$ is separated due to the Hausdorffness. Since $X$ is finite type over $\mathbb{C},$ to show that $X$ is complete, we only need to show that its structure morphism $X \rightarrow \mathrm{Spec}(\mathbb{C})$ is universally closed (e.g., Vakil 10.3.1). Thus, now given any $\mathbb{C}$-scheme $Y,$ we want to show that the projection map $\pi : X \times Y \rightarrow Y$ is closed in the Zariski topology. Fix any closed subset $Z \subset X \times Y.$ We want to show that $\pi(Z) \subset Y$ is closed. It is enough to show that $\pi(Z)(\mathbb{C}) \subset Y(\mathbb{C})$ is Zariski-closed.

Why? First, note that given closed subsets $Z_{1}, Z_{2}$ in $Y,$ having $Z_{1}(\mathbb{C}) \subset Z_{2}(\mathbb{C})$ implies that $Z_{1} = Z_{2}.$ We may reduce to the case where $Z_{1}$ and $Z_{2}$ are contained in the same affine space $\mathbb{A}^{n}$ so that we may write $Z_{1} = V(f_{1}, \dots, f_{r})$ and $Z_{2} = V(g_{1}, \dots, g_{s}),$ where $(f_{1}, \dots, f_{r})$ and $(g_{1}, \dots, g_{s})$ are radical ideals in $\mathbb{C}[x_{1}, \dots, x_{n}].$ By a formulation of Nullstellensatz, our hypothesis $Z_{1}(\mathbb{C}) \subset Z_{2}(\mathbb{C})$ implies that $(f_{1}, \dots, f_{r}) = (g_{1}, \dots, g_{s}).$ This implies that $Z_{1} = Z_{2}.$

Remark. I think the above detail might be an ingredient at the end the proof of Proposition 2.6 of Hartshorne's book.

Maybe we cannot quite do this right away, but Chevalley's theorem at least tells us that $\pi(Z)$ is a constructible subset of $Y.$ From the previous lecture, we see that it is enough to show that $\pi(Z)(\mathbb{C}) \subset Y(\mathbb{C})$ is closed in the classical topology. Given a sequence $(y_{n})$ in $\pi(Z)(\mathbb{C})$ such that $y_{n}$ converges to some $y \in Y(\mathbb{C}),$ we need to show that $y \in \pi(Z).$ Since $\pi$ is the projection map, we may find $x_{n} \in X(\mathbb{C})$ such that $$(x_{n}, y_{n}) \in Z(\mathbb{C}) \subset X(\mathbb{C}) \times Y(\mathbb{C}).$$ Since $X(\mathbb{C})$ is compact Hausdorff, hence it is sequentially compact (e.g., we may apply Theorem 34.1 in Munkres to consider $X(\mathbb{C})$ as a metric space and then apply Theorem 28.2 in the same book), so we may replace $(x_{n})$ with a subsequence to assume that it converges to some $x \in X(\mathbb{C}).$ Thus, we have $(x_{n}, y_{n}) \rightarrow (x, y) \in X \times Y$ as $n \rightarrow \infty.$ But then we have $$(x_{n}, y_{n}) \in Z(\mathbb{C}),$$ which is closed in $X(\mathbb{C}) \times Y(\mathbb{C}),$ so $(x, y) \in Z.$ This implies that $y \in \pi(Z),$ as desired. $\Box$

Exercise. Let $f : X \rightarrow Y$ be a morphism of separate varieties over $\mathbb{C}.$ Then the following are equivalent:

• $f$ is proper;
• $f^{\mathrm{an}}$ is proper (i.e., compact subsets pull back to compact subsets).

Change of terminology. We assume that all varieties over $\mathbb{C}$ are separated. (This is great, because all $\mathbb{C}$-scheme morphisms between them are going to be of finite type.)

Now, we are done with Chapter I.

Chapter II. Holomorphic functions (Reference: Griffiths-Harris)

We first start with the case of single variable.

Setup. Let $U \subset \mathbb{C} = \mathbb{R}^{2}$ be an open subset. By saying that a function $f : U \rightarrow \mathbb{R}^{2} = \mathbb{C}$ is smooth, when mean the function is $\mathscr{C}^{\infty}.$ (See this Wikipedia page for reminder of the definition.)

Coordinate functions on $U$. We have

• $z := x + yi$ and
• $\bar{z} := x - yi.$

Differential forms on $U$. We have

• holomorphic 1-forms: $dz := dx + idy$ and $d\bar{z} := dx - idy.$
• holomorphic 2-form: $dz \wedge d\bar{z} = (-2i) dx \wedge dy.$

Dual basis (cf. p.231 of Rudin). We have

$$\frac{\partial}{\partial z} = \frac{1}{2}\left( \frac{\partial}{\partial x} - i\frac{\partial}{\partial y} \right)$$ and 

$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left( \frac{\partial}{\partial x} + i\frac{\partial}{\partial y} \right).$$

Remark. We check $$dz\left(\frac{\partial}{\partial z}\right) = 1 = d\bar{z}\left(\frac{\partial}{\partial \bar{z}}\right)$$ and $$dz\left(\frac{\partial}{\partial \bar{z}}\right) = 0 = d\bar{z}\left(\frac{\partial}{\partial z}\right).$$

Quick check. For $m \in \mathbb{Z}_{\geq 1},$ we have $$\frac{\partial z^{m}}{\partial z} = mz^{m-1}$$ and $$\frac{\partial z^{m}}{\partial \bar{z}} = 0.$$ (An easy way to check this is to use the product rule to reduce it to the case $m = 1.$)

Given any smooth $f : U \rightarrow \mathbb{C},$ we have $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = \frac{\partial f}{\partial z} dz + \frac{\partial f}{\partial \bar{z}} d\bar{z}.$$ The first equality is based on the fact that $dx$ and $dy$ form a basis for the $\mathbb{R}$-vector space of differential $1$-forms on $U$ (i.e., smooth sections of the cotangent bundle of $U$). To see the second equality, we have $$\frac{\partial f}{\partial z} dz = \frac{1}{2}\left( \frac{\partial f}{\partial x} - i\frac{\partial f}{\partial y} \right)(dx + idy)$$ and $$\frac{\partial f}{\partial \bar{z}} d\bar{z} = \frac{1}{2}\left( \frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y} \right)(dx - idy),$$ and we expand the right-hand sides and add them up to see what we want.

We also note that $$\frac{\partial \bar{f}}{\partial \bar{z}} = \overline{\frac{\partial f}{\partial z}}.$$

Proposition (Cauchy's integral formula). Let $D \subset \mathbb{C}$ be a nonempty open disk. If $U$ is an open subset of $\mathbb{C}$ containing $\overline{D},$ then for any smooth function $f : U \rightarrow \mathbb{C}$ and $z \in D$ we have $$f(z) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(w)}{w - z} dw + \frac{1}{2\pi i} \int_{\overline{D}} \frac{\partial f}{\partial \bar{w}}\frac{dw \wedge d\bar{w}}{w - z},$$ where $\partial D$ is oriented counterclockwise.

The second integral makes sense. We first see why the integral $$\frac{1}{2\pi i} \int_{\overline{D}} \frac{\partial f}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z}$$ makes sense. Take $w = z + re^{i \theta}$ with $r \in (0, \text{radius of }D)$ and $\theta \in (0, 2\pi).$ Then we have 

• $dw = e^{i\theta} dr + ir e^{i\theta} d\theta,$ 
• $d\bar{w} = e^{-i\theta} dr - ir e^{-i\theta} d\theta,$
• $dw \wedge d\bar{w} = - 2ir dr \wedge d\theta,$

so $$\frac{dw \wedge d\bar{w}}{w - z} = -2i e^{-i\theta} dr \wedge d\theta.$$ Since we only removed a measure zero set away from $\overline{D},$ this shows that the integral is well-defined and gives us a complex number as a result, regardless of $f$.

Proof of Proposition. Let $D_{\epsilon}(z)$ be an open disk of tiny radius $\epsilon > 0$ centered at $z$ so that $z \in D_{\epsilon}(z) \subset D.$ We will apply Stokes' theorem for the (real and complex parts of) $1$-form $$\eta := \frac{f(w)}{w - z} dw$$ on $\overline{D} \smallsetminus D_{\epsilon}(z).$ We compute $$\begin{align*}d\eta &= d\left(\frac{f(w)}{w - z} dw\right) \\ &= d\left( \frac{f(w)}{w - z} \right) \wedge dw \\ &= \left( \frac{\partial}{\partial w}\frac{f(w)}{w - z} dw + \frac{\partial}{\partial \bar{w}}\frac{f(w)}{w - z} d\bar{w} \right) \wedge dw \\ &= -\frac{\partial}{\partial \bar{w}}\frac{f(w)}{w - z} dw \wedge d\bar{w}. \end{align*}$$ Stokes' theorem says that we have $$\int_{\partial D} \eta- \int_{\partial D_{\epsilon}(z)} \eta = \int_{\overline{D} \smallsetminus D_{\epsilon}(z)} d\eta.$$ Using the computations above, we have $$\int_{\partial D} \frac{f(w)}{w - z}dw - \int_{\partial D_{\epsilon}(z)} \frac{f(w)}{w - z}dw = - \int_{\overline{D} \smallsetminus D_{\epsilon}(z)} \frac{\partial}{\partial \bar{w}}\frac{f(w)}{w - z} dw \wedge d\bar{w}.$$ Writing $w = z + \epsilon e^{i \theta}$ for $\theta \in (0, 2\pi),$ we have $$dw = i\epsilon e^{i\theta} d\theta$$ so that $$\int_{\partial D_{\epsilon}(z)} \frac{f(w)}{w - z}dw = \int_{0}^{2\pi} \frac{f(z + \epsilon e^{i\theta})}{\epsilon e^{i\theta}} \cdot i\epsilon e^{i\theta} d\theta = \int_{0}^{2\pi} if(z + \epsilon e^{i\theta}) d\theta.$$ Therefore, by Dominated Convergence, we have $$\begin{align*} \lim_{\epsilon \rightarrow 0} \int_{\partial D_{\epsilon}(z)} \frac{f(w)}{w - z}dw &= \lim_{\epsilon \rightarrow 0} \int_{0}^{2\pi} if(z + \epsilon e^{i\theta}) d\theta \\ &= \int_{0}^{2\pi} \lim_{\epsilon \rightarrow 0} if(z + \epsilon e^{i\theta}) d\theta \\ &= 2\pi i f(z). \end{align*}$$ Again by Dominated Convergence, we have $$\lim_{\epsilon \rightarrow 0}\int_{\overline{D} \smallsetminus D_{\epsilon}(z)} \frac{\partial}{\partial \bar{w}}\frac{f(w)}{w - z} dw \wedge d\bar{w} = \int_{\overline{D}} \frac{\partial}{\partial \bar{w}}\frac{f(w)}{w - z} dw \wedge d\bar{w},$$ so this finishes the proof. $\Box$

Given a smooth function $f : U \rightarrow \mathbb{C},$ we say $f$ is holomorphic if $$\frac{\partial f}{\partial \bar{z}} = 0.$$ Writing $f = u + iv,$ we note that $f$ is holomorphic if and only if $u_{x} = v_{y}$ and $u_{y} = - v_{x}.$

Remark. Note that Cauchy's integral formula has the usual form when $f$ is holomorphic in the statement.

We say $f$ is analytic if for any $a \in U$ there is and open disk $D_{\epsilon}(a)$ with $\epsilon > 0$ such that we can write $f(z) = \sum_{n=0}^{\infty}c_{n} (z - a)^{n}$ with $c_{n} \in \mathbb{C}$ so that the power series converges absolutely and uniformly in $z \in D_{\epsilon}(a).$

Next time (?):

Theorem 1. Using the notations above, the following are equivalent:

• $f$ is holomorphic;
• $f$ is analytic.

Theorem 2 ($\bar{\partial}$-lemma in single variable). Given any disk $D \subset U,$ where $U \subset \mathbb{C}$ is a nonempty open subset, for any $\mathbb{C}$-valued smooth function $g$ on $U,$ there is a smooth function $f$ on $D$ such that $g = \frac{\partial f}{\partial \bar{z}}.$