Hodge theory: Lecture 8

Now, we prove how connectedness of a complex variety implies connectedness of its analytification. Note that we are not assuming smoothness of the variety, and the statement is just a statement about two different topologies. However, we will somehow reduce our proof to the case of a smooth projective curve, where we will use that its analytification is a smooth real manifold of dimension $\geq 2.$

Theorem. Let $X$ be any complex variety. If $X$ is connected, then $X^{\mathrm{an}}$ is connected.

Remark. Before we proceed with the proof, note how astonishing this theorem is. For instance, it proves that for any irreducible polynomial $f(x, y) \in \mathbb{C}[x, y],$ the zero set $\{(z, w) \in \mathbb{C} : f(z, w) = 0\} \subset \mathbb{R}^{4}$ is connected in the Euclidean topology. I am not entirely sure how to prove this directly! (Maybe some compactification with Mayer-Vietoris argument will do? I am not sure.)

To assist our proof, we give a separate lemma:

Lemma. Let $X$ be any connected algebraic variety over an algebraically closed field $k$. For any $p, q \in X(k),$ there is an irreducible curve $C \subset X$ as a closed subset such that $p, q \in C.$

We will use this lemma to show Theorem first and then go back to proving the lemma.

Proof that Lemma implies Theorem. There will be several steps.

Step 1. We reduce to the case where $X$ is irreducible. 

Suppose that the statement is proven for irreducible components. To show $X^{\mathrm{an}} = X(\mathbb{C})$ is connected, write $X(\mathbb{C}) = U \sqcup V,$ where $U, V$ are clopen sets in $X(\mathbb{C}).$ Then consider the unique irreducible decomposition $$X = X_{1} \cup \cdots \cup X_{r},$$ where $X_{i}$'s are irreducible components. Our assumption says $X_{i}(\mathbb{C})$ are connected, so it is contained in exactly of $U$ and $V,$ so we can eventually write $$X(\mathbb{C}) = (X_{1}(\mathbb{C}) \cup \cdots \cup X_{s}(\mathbb{C})) \sqcup (X_{s+1}(\mathbb{C}) \cup \cdots \cup X_{r}(\mathbb{C}))$$ after some rearrangements, where unionands are equal to $U$ and $V.$ Since each $X_{i}$ is quasi-compact, this necessarily implies that $$X = (X_{1} \cup \cdots \cup X_{s}) \sqcup (X_{s+1} \cup \cdots \cup X_{r}).$$ But then $X$ is connected, so either $X_{1} \cup \cdots \cup X_{s}$ or $X_{s+1} \cup \cdots \cup X_{r}$ must be empty. This implies that either $U$ or $V$ is empty, showing $X(\mathbb{C})$ is connected. Hence, this reduces the problem to the case where $X$ is irreducible. This finishes this step.

Step 2. We reduce to the case where $X$ is an irreducible curve.

Suppose that the statement is proven for irreducible curves. For any $p, q \in X(\mathbb{C}),$ we use Lemma to choose an irreducible curve $C \subset X$ that contains both $p$ and $q.$ Analytifiying the curve, we see that $C(\mathbb{C})$ is connected, so it follows that $X(\mathbb{C})$ is connected (e.g., Munkres Theorem 23.3). This finishes this step (except that we need to prove Lemma later).

Step 3. We reduce to the case where $X$ is a smooth irreducible curve.

Take the normalization $\tilde{X} \twoheadrightarrow X$ (Vakil 9.7.C for surjectivity). Since $X$ is irreducible, we also have $\tilde{X}$ as irreducible (Vakil, beginning of Section 9.7). Normal (Noetherian) curves are regular, so the fact that we are over $\mathbb{C}$ tells us that $\tilde{X}$ is smooth.

Step 4. We reduce to the case where $X$ is a smooth irreducible projective curve.

For this, we will use that there is an open embedding $X \hookrightarrow \bar{X},$ where $\bar{X}$ is a smooth irreducible projective curve (e.g., Hartshorne I.6 Corollary 6.10). Hence, if we assume the statement for the case of smooth irreducible curves, then $\bar{X}^{\mathrm{an}}$ is connected. Now, note that $\bar{X} \setminus X$ is a $0$-dimensional Noetherian scheme, so it must have finitely many points. Thus, we see $\bar{X}(\mathbb{C}) \setminus X(\mathbb{C})$ is also a finite set. As $\bar{X}(\mathbb{C})$ is a connected real manifold of dimension $2,$ taking finitely many points away will keep its connectivity.

Personal remark. Shubhodip Mondal kindly explained how Hartshorne's exposition constructs a desired projective curve, using discrete valuations as points. I have not gone through this in detail, so I cannot explicitly tell in which generality this fact is true, although for the case that we deal with does not require us to understand the proof. As Shubhodip also mentioned, this is can be seen as a special case of Nagata's compactification theorem because proper curves are projective (Vakil 18.7.2).

Remark. We have used that given a real manifold $M$ of dimension $\geq 2,$ if $M$ is connected, for any $p \in M,$ the punctured manifold $M \smallsetminus \{p\}$ is connected. To see this, we may argue by contradiction, so say $M \smallsetminus \{p\} = U \sqcup V$ with nonempty open $U, V \subset M \smallsetminus \{p\} .$ Since $M$ is connected and $M = \bar{U} \cup \bar{V},$ the intersection $\bar{U} \cap \bar{V}$ must be nonempty. For any $q \in \bar{U} \cap \bar{V},$ as any open neighborhood of $q$ in $M$ should intersect both of $U$ and $V,$ which are open in $M,$ we cannot have $q \in U \cup V.$ This implies that $q = p,$ so $\bar{U} \cap \bar{V} = \{p\}.$ Now, take any open neighborhood $W \ni p$ in $M$ that is homeomorphic to an open disk in $\mathbb{R}^{2}.$ We now have $W \smallsetminus \{p\} = (W \cap U) \sqcup (W \cap V),$ and each unionand is nonempty while the left-hand side is connected. This is a contradiction, so $M \smallsetminus \{p\}$ must have been connected.

Step 5. We prove the statement for the case where $X$ is an irreducible smooth projective curve.

For contradiction, let $X^{\mathrm{an}} = U \sqcup V,$ where $U, V$ are nonempty open subsets in the classical topology. Let $g(X)$ be the genus of the curve $X.$ Fix any $p \in U \subset X(\mathbb{C}).$ If $\deg(\mathscr{O}_{X}(np)) = n \geq 2g(X),$ then there is a global section $$s \in \Gamma(X, \mathscr{O}_{X}(np)) = \{t \in K(X)^{\times} : \mathrm{div}(t) + np \geq 0\}$$ that does not vanish at $p$ (Vakil 19.2.11). Thus, we have $$np = \mathrm{div}(s) = q_{1} + \cdots + q_{m} \in \mathrm{Cl}(X),$$ for some finite sequence of points $q_{1}, \dots, q_{m} \in X(\mathbb{C})$ that are not equal to $p.$ (We have $\mathrm{div}(s) \neq np$ as Weil divisors; it is only their divisor classes that match.) Thus, we have some $f \in K(X)^{\times}$ such that $\mathrm{div}(f) = q_{1} + \cdots + q_{m} - np.$ Since the only pole of $f$ is at $p,$ it defines a regular function $f : X \smallsetminus \{p\} \rightarrow \mathbb{A}^{1}.$ This implies that $f : X(\mathbb{C}) \setminus \{p\} \rightarrow \mathbb{C}$ defines a holomorphic function, and since $X$ is projective, its analytification $X^{\mathrm{an}} = X(\mathbb{C})$ is compact. As $p \notin V,$ we may take any connected component $T$ of $V$ (which is a closed subset in $X(\mathbb{C})$ so that it is compact) and realize that $f|_{T}$ is constant. Since $T$ is connected and Hausdorff, it must be an infinite set, so $f$ is constant on infinitely many points. Since $T$ is compact, this implies that $f$ is constant on the whole $X(\mathbb{C}) \smallsetminus \{p\}.$ This is enough to deduce that $f \in \mathbb{C} \subset K(X)^{\times},$ so $\mathrm{div}(f) = 0,$ which is a contradiction as the earlier one had a pole at $p.$ This finishes the proof. $\Box$

Proof of Lemma. We now discuss a proof of Lemma. When $\dim(X) = 1,$ the statement is tautological, so assume $n := \dim(X) \geq 2.$

Step 1. Reduction to the case where $X$ is an irreducible projective $k$-variety.

By Nagata's compactification theorem (e.g., Lütkebohmert's note), our variety $X$ can be regarded as a dense open subset of some complete variety $Y$ over $k.$ Since $X$ is irreducible, this implies that $Y$ must be irreducible as well. Applying Chow's lemma (Vakil 18.9.2 and 18.9.6) to $Y,$ we get a birational surjective $k$-scheme map $\pi : \tilde{X} \twoheadrightarrow X,$ where $\tilde{X}$ is a nonempty open subset of an irreducible projective variety over $k.$ Birationality means that there is a dense open subset $U \subset X$ such that $$\pi : \pi^{-1}(U) \overset{\sim}{\longrightarrow} U.$$ Since the image of any irreducible set under a continuous map is irreducible, this implies that if we take an irreducible curve closed in $\tilde{X},$ the closure of its image is an irreducible curve closed in $X.$ This lets us reduce to the case where $X$ is an open subset of some irreducible projective $k$-variety $\bar{X}$. Then note that proving the statement for $\bar{X}$ is enough, so we have reduced the problem to the case where $X$ is an irreducible projective $k$-variety.

Step 2. Proof for the case where $X$ is an irreducible projective $k$-variety.

Now, fix any two distinct points $p, q \in X(k),$ and consider the blow-up $$\pi : \mathrm{Bl}_{\{p,q\}}(X) \twoheadrightarrow X.$$ Since $X$ is projective over $k,$ so is $\tilde{X} = \mathrm{Bl}_{\{p,q\}}(X)$ (e.g., Hartshorne II Proposition 7.16 or Vakil Theorem 22.3.2 + 17.3.B, both of which use $\pi$ is a projective morphism). Using the universal property of the blow-ups (Section 22.2 of Vakil), we may see that $\pi^{-1}(p)$ and $\pi^{-1}(q)$ are effective Cartier divisors of $\tilde{X},$ so they must have dimension $n - 1.$ (Recall that we are in the case $n \geq 2.$) Since $\tilde{X}$ is projective (of dimension $n$), we have a closed embedding $\pi : \tilde{X} \hookrightarrow \mathbb{P}^{N}$ for some $N \geq n = \dim(X).$ Then we apply Bertini's theorem to choose $n-1$ distinct hyperplanes $H_{1}, \dots, H_{n-1} \subset \mathbb{P}^{N}$ such that

• $C = \tilde{X} \cap H_{1} \cap \cdots \cap H_{n-1}$ is an irreducible curve, and 
• none of $\pi^{-1}(p)$ and $\pi^{-1}(q)$ are contained in any of $H_{1}, \dots, H_{n-1}.$

Since $\dim(\pi^{-1}(p)) = \dim(\pi^{-1}(q)) = n - 1,$ both fibers intersect $H_{1} \cap \cdots \cap H_{n-1}$ Since $\pi$ is projective, it is closed, so its image $\pi(C) \subset X$ is a closed subset and an irreducible curve that contain $p$ and $q$. $\Box$

Personal remark about Bertini. When $X$ is smooth, Theorem 8.18 in Hartshorne (note: this theorem seems to use the hypothesis that $k$ is algebraically closed) does the job, but the general version we use is harder. (I need to ask the lecturer again where this is written.) I was told that one can use Hironaka's theorem on the resolution of singularities for the case when the characteristic is $0.$ This will be okay as we only care about the case $k = \mathbb{C}$ in this class, but I am now unsure whether the statement is true for any algebraically closed $k,$ not just characteristic $0$ ones.

Hodge theory: Lecture 7

When we study a space, it is often important to study line bundles on the space, which reads off crucial information about the space. More generally, we may consider vector bundles.

Vector bundles. Given a real smooth manifold $M,$ a real (or respectively complex) vector bundle on $M$ of rank $r \geq 0$ is a smooth map $\pi : E \rightarrow M$ with an open cover $E = \bigcup_{i \in I}U_{i}$ such that for each $i \in I,$ we have a diffeomorphism $\pi^{-1}(U_{i}) \simeq U_{i} \times k^{r}$ over $U_{i}$ (where $k$ is $\mathbb{R}$ or $\mathbb{C}$) and at each $x \in U_{i},$ the diffeomorphism restricts to $k$-linear isomorphism $\pi^{-1}(x) \simeq \{x\} \times k^{r}.$ Note that $E$ gives us a sheaf $\mathcal{S}_{E}$ defined on $M$ given by $$\mathcal{S}_{E}(U) = \{s : U \rightarrow E \text{ smooth } | \ \pi \circ s = \mathrm{id}_{U}\}.$$ This gives an equivalence between the category of $k$-vector bundles on $M$ of rank $r$ and the category of locally free sheaves of rank $r$ of $\mathscr{C}^{r}_{M,k}$-modules.

Similarly, for a complex manifold $X$, we have an equivalence between the category of holomorphic vector bundles on $X$ of rank $r$ and the category of locally free sheaves of rank $r$ of $\mathscr{O}_{X}$-modules.

Personal remark. I have checked this years ago, and it was really about understanding formal gluing well enough.

Remark. For a holomorphic vector bundle, we will study both of its smooth sections and holomorphic sections.

Submanifolds. Let $X$ be a complex manifold of dimension $n.$ A closed submanifold of $X$ of codimension $0 \leq r \leq n$ is a closed subset $Y \subset X$ with an open covering $X = \bigcup_{i \in I}U_{i}$ and chart maps $\varphi_{i} : U_{i} \overset{\sim}{\longrightarrow} V_{i} \overset{\text{open}}{\hookrightarrow} \mathbb{C}^{n}$ such that $$\varphi_{i}(U_{i} \cap Y) = \{z = (z_{1}, \dots, z_{n}) \in V_{i} : z_{1} = \cdots = z_{r} = 0\}.$$ Note that by restricting such charts to $Y,$ we can get a holomorphic atlas on $Y,$ making it a complex manifold of dimension $n - r.$

Universal property. Let $\iota : Y \hookrightarrow X$ be a complex closed submanifold. Given any holomorphic $g : Z \rightarrow X$ such that $g(Z) \subset Y,$ there is a unique holomorphic map $g' : Z \rightarrow Y$ such that $\iota \circ g' = g.$

Personal remark. The above remark just seems to say that we can restrict the target of any holomorphic map to a closed submanifold containing the image. Perhaps there are better philosophical interpretations of this, but I am not able to come up with any at the moment.

Maximum rank Jacobian. We now study a criterion that tells us whether given holomorphic functions cut out closed submanifold. This should remind us the definition of smoothness of a variety over a field (Vakil 12.2.6).

Proposition. Let $0 \leq r \leq n.$ If $U \subset \mathbb{C}^{n}$ is an open subset and $f_{1}, \dots, f_{r} \in \mathscr{O}_{\mathbb{C}^{n}}(U),$ where the rank of the Jacobian matrix $$\left[\frac{\partial f_{i}}{\partial z_{j}} \right]_{\substack{1 \leq i \leq r \\ 1 \leq j \leq n}}$$ has rank $r$ everywhere in $U.$ Then $$Z = \{z \in U : f_{1}(z) = \cdots = f_{r}(z) = 0\} \hookrightarrow U$$ defines a closed submanifold of codimension $r.$

Proof. Given $p \in Z,$ rearranging the orders of $f_{1}, \dots, f_{r}$ if necessary, we may assume that $$\det\left[\frac{\partial f_{i}}{\partial z_{j}}(p) \right]_{1 \leq i, j \leq r} \neq 0.$$ Consider $\varphi : U \rightarrow \mathbb{C}^{n}$ defined by $$\varphi(z) = (f_{1}(z), \dots, f_{r}(z), z_{r+1}, \dots, z_{n}).$$ Then $$\det\left[\frac{\partial \varphi_{i}}{\partial z_{j}}(p) \right]_{1 \leq i, j \leq n} = \det\left[\frac{\partial f_{i}}{\partial z_{j}}(p) \right]_{1 \leq i, j \leq r} \neq 0,$$ so by the Inverse Function Theorem, we see $\varphi$ is biholomorphic in some neighborhood of $p,$ which gives a desired holomorphic chart map near $p.$ $\Box$

Complex manifolds from smooth complex varieties. Let $X$ be a smooth variety over $\mathbb{C}$ of pure dimension $n.$ Let $U \subset X$ be an affine open subset and $U \hookrightarrow \mathbb{A}^{N}$ a closed immersion. Denote $r = N - n,$ and consider an affine open cover $\mathbb{A}^{N} = \bigcup_{\alpha \in I}V_{\alpha}.$ If $V_{\alpha} \cap U$ is nonempty, then $$V_{\alpha} \cap U \hookrightarrow V_{\alpha}$$ is a closed subscheme cut out by $r$ sections $f_{1}, \dots, f_{r} \in \mathscr{O}_{V_{\alpha}}(U \cap V_{\alpha})$ such that $$\left[\frac{\partial f_{i}}{\partial z_{j}} \right]_{\substack{1 \leq i \leq r \\ 1 \leq j \leq n}}$$ has rank $r$ everywhere in $U,$ by the definition of smoothness. Applying the previous proposition, we see that the each $$V_{\alpha}(\mathbb{C}) \cap U(\mathbb{C}) \hookrightarrow V_{\alpha}(\mathbb{C})$$ defines a complex submanifold of codimension $r,$ or equivalently, dimension $n.$ The resulting transition maps are holomorphic, because sections are given as rational maps without singularities in their domain. This defines a complex manifold structure on $X^{\mathrm{an}} = X(\mathbb{C}).$

Remark. Note that if we had any number of sections in $\mathscr{O}_{V_{\alpha}}(V_{\alpha} \cap U)$ cutting out $V_{\alpha} \cap U \hookrightarrow V_{\alpha}$ bounds the codimension $r$ from above by Krull's Height Theorem. The fact that we can have precisely $r$ sections is the part of the definition of smoothness. I think that the last part of the previous paragraph can be seen by formal gluing as well. Moreover, also note that at least for any smooth variety $X$ over $\mathbb{C},$ we may say write $$\mathscr{O}_{X}^{\mathrm{an}} := \mathscr{O}_{X^{\mathrm{an}}},$$ and think of it as if the structure sheaf $\mathscr{O}_{X}$ is "analytified".

Analytification functor on smooth varieties. Now we note that any morphism for $f : X \rightarrow Y$ of varieties smooth over $\mathbb{C},$ the induced map $f^{\mathrm{an}} : X^{\mathrm{an}} \rightarrow Y^{\mathrm{an}}$ on complex manifolds is holomorphic. If I am not mistaken, checking this is surprisingly easy: for any affine open $V \subset Y,$ a section $\mathscr{O}_{Y^{\mathrm{an}}}(V)$ is given by analytifying a $\mathbb{C}$-scheme map $V \rightarrow \mathbb{A}^{1}.$ The pullback of such a section in $\mathscr{O}_{X^{\mathrm{an}}}(f^{-1}(V))$ is given by analytifying $$f^{-1}(V) \rightarrow V \rightarrow \mathbb{A}^{1},$$ which is holomorphic because (locally) rational maps are holomorphic. This phenomenon can be seen as the analytification functor $\textbf{Var}_{\mathbb{C}} \rightarrow \textbf{Top}$ taking smooth varieties into the category of complex manifolds.

Maximum modulus principle. If $f : U \rightarrow \mathbb{C}$ is a holomorphic function on an open and connected subset $U \subset \mathbb{C}^{n}$ such that $|f|$ has a local maximum, then $f$ is constant on the whole $U.$

Proof. Let $p \in U$ be a point where $|f|$ attains a local maximum. Since $U$ is connected, we can make the statement local by a property of holomorphic functions on $U$ discussed in a previous posting. Namely, it is enough to find an open neighborhood $U_{\epsilon} \ni p$ in $U$ such that $f$ is constant on $U_{0}.$ We do this in two steps.

Step 1: Reduction to the case $n = 1$. Take $\epsilon > 0$ small enough so that $$U_{\epsilon} := D_{\epsilon}(p_{1}) \times \cdots \times D_{\epsilon}(p_{n}) \subset U.$$ Of course, we note that $p = (p_{1}, \dots, p_{n}) \in U_{\epsilon}.$ We assume that the statement is proven for the case of $n = 1$ and use this to show that $f : U_{\epsilon} \rightarrow \mathbb{C}$ is constant. Take any $z \in U_{\epsilon},$ and we would like to prove that $f(z) = f(p).$ Take $\delta > 0$ small enough so that $p + \delta (z - p) \in U_{\epsilon}.$ Consider the map $L_{p,z} : D_{1+\delta}(0) \rightarrow U_{\epsilon}$ given by $w \mapsto (1-w)p + wz,$ which gives us $L_{p,z}(0) = p$ and $L_{p,z}(1) = z.$ By the special case $n = 1,$ we see that the composition $f \circ L_{p,z} : D_{1+\delta}(0) \rightarrow \mathbb{C}$ is constant, so $$f(p) = f(L_{p,z}(0)) = f(L_{p,z}(1)) = f(z).$$ Hence, we now focus on the spacial case $n = 1.$

Step 2: The case $n = 1$. Let $\epsilon > 0$ be small enough so that $\overline{D_{\epsilon}(p)} \subset U$ and $|f(p)|$ is maximum in $\overline{D_{\epsilon}(p)}.$ By Cauchy's formula (and taking $w = p + \epsilon e^{2\pi i \theta}$), we have $$\begin{align*} f(p) &= \frac{1}{2\pi i} \int_{\partial D_{\epsilon}(p)} \frac{f(w)}{w - p}dw \\ &= \frac{1}{2\pi i}\int_{0}^{1} \frac{f(p + \epsilon e^{2\pi i \theta})}{\epsilon e^{2\pi i \theta}} \epsilon 2\pi e^{2\pi i \theta} d\theta \\ &= \int_{0}^{1} f(p + \epsilon e^{2\pi i \theta})  d\theta \end{align*}.$$ Thus, we have $$|f(p)| \leq \int_{0}^{1} |f(p + \epsilon e^{2\pi i \theta})| d\theta \leq \int_{0}^{1} |f(p)| d\theta = |f(p)|.$$ Therefore, we have $$|f(p)| = \int_{0}^{1} |f(p + \epsilon e^{2\pi i \theta})| d\theta.$$ This implies that we have $$\int_{0}^{1} |f(p) - f(p + \epsilon e^{2\pi i \theta})| d\theta = \int_{0}^{1} |f(p)| - |f(p + \epsilon e^{2\pi i \theta})| d\theta = 0,$$ so continuity of $f$ ensures that we must have $f(p) = f(p + \epsilon e^{2\pi \theta})$ for all $0 \leq \theta \leq 1.$ This shows that $f$ is constant on $D_{\epsilon}(0),$ which is enough to conclude that $f$ is constant on $U,$ because $U$ is connected. $\Box$

Remark. As a corollary, given a connected complex manifold $X,$ if $f$ is a global section of $\mathscr{O}_{X}$ such that $|f|$ attains a local maximum, the section $f$ must be constant on whole $X.$ In particular, if $X$ is a compact connected complex manifold, we have $\Gamma(X, \mathscr{O}_{X}) = \mathbb{C}.$

Of course, we are using that given a connected complex manifold $X,$ if $f : X \rightarrow \mathbb{C}$ is a global section that vanishes on a nonempty open subset of $X,$ then $f = 0.$ This follows from a similar argument in the last proposition of this previous posting.

Hodge theory: Lecture 6

Our main interest lies on the geometry of varieties $X$ over $\mathbb{C}$ (i.e., reduced and separated schemes finite type over $\mathbb{C}$), and we wanted to understand them by analytic geometry of $X^{\mathrm{an}} = X(\mathbb{C}).$ We realized that this gives the analytification functor $\textbf{Var}_{\mathbb{C}} \rightarrow \textbf{Top},$ but we can vividly see from its definition that $X(\mathbb{C})$ has more structure than just its topology. In general, it seems that we can give the structure of analytic spaces to $X^{\mathrm{an}},$ but for smooth varieties (which will probably be our main focus), such structures will be turn out to be the same notion as complex manifolds.

Complex manifolds. Given a nonempty open subset $U \subset \mathbb{C}^{n},$ let $$\mathscr{O}_{U}(V) := \{f : V \rightarrow \mathbb{C} : f\text{ holomorphic}\},$$ for any nonempty open $V \subset U.$ Note that this is a $\mathbb{C}$-algebra. We define $\mathscr{O}_{U}(\emptyset) = 0,$ the zero $\mathbb{C}$-algebra. Note that these form a sheaf $\mathscr{O}_{U},$ where we could use the actual restrictions to define restriction maps. (The key here is that holomorphicity can be checked open-locally.)

Remark. I think we have $\mathscr{O}_{\emptyset} = 0,$ in any reasonable definition. In any case, let's declare this.

A complex manifold is a pair $(X, \mathscr{O}_{X})$ such that

1. $X$ is a topological space that is Hausdorff and second countable, and
2. $\mathscr{O}_{X} \hookrightarrow \mathscr{C}_{X, \mathbb{C}},$ the sheaf of $\mathbb{C}$-algebras given by $\mathbb{C}$-valued continuous functions, is a subsheaf over $X$ with is an open cover $X = \bigcup_{i \in I}U_{i}$ such that for each $i \in I,$ we have some open subset $V_{i} \subset \mathbb{C}^{n}$ with an isomorphism of ringed spaces $(U_{i} \mathscr{O}_{U_{i}}) \simeq (V_{i} \mathscr{O}_{V_{i}}).$

Remark. The last condition is more concrete than the way it may look. It is saying that we have a homeomorphism $\varphi : U_{i} \overset{\sim}{\longrightarrow} V_{i} \subset \mathbb{C}^{n}$ with a sheaf isomoprhism $\varphi_{*}\mathscr{O}_{U_{i}} \overset{\sim}{\longrightarrow} \mathscr{O}_{V_{i}}$ over $V_{i}.$ Thus, for any open $W \subset V_{i},$ we have a $\mathbb{C}$-algebra isomorphism $$\mathscr{O}_{V_{i}}(W) \overset{\sim}{\longrightarrow} \mathscr{O}_{U_{i}}(\phi^{-1}(W)),$$ so each element in the right-hand side corresponds to a holomorphic function $W \rightarrow \mathbb{C},$ and we can treat it as if it is the composition $\varphi^{-1}(W) \simeq W \rightarrow \mathbb{C},$ which we declare to be holomorphic on $\varphi^{-1}(W).$ This way, we obtain some concrete description of $\mathscr{O}_{X}.$ By definition, global sections of $\mathscr{O}_{X}$ are called holomorphic functions on $X,$ but we also have a concrete meaning namely, each $f \in \Gamma(X, \mathscr{O}_{X})$ can be realized as a continuous map $f : X \rightarrow \mathbb{C}$ such that for every $p \in X,$ there is an open neighborhood $U \ni p$ in $X,$ with open embedding $U \hookrightarrow \mathbb{C}^{n}$ such that $f : U \rightarrow \mathbb{C}$ is holomormophic. (The gluing procedure in $\textbf{Top}$ will do the job, and the same holds for any open subset of $X$ in place of the whole $X.$)

A holomorphic map $(X, \mathscr{O}_{X}) \rightarrow (Y, \mathscr{O}_{Y})$ is a continuous map $\phi : X \rightarrow Y$ together with a sheaf map $\mathscr{O}_{Y} \rightarrow \phi_{*}\mathscr{O}_{X}$ given as follows: for any open $V \subset Y,$ we have a $\mathbb{C}$-algebra map $$\Gamma(V, \mathscr{O}_{Y}) \rightarrow \Gamma (\phi^{-1}(V), \mathscr{O}_{X})$$ given by $g \mapsto g \circ \phi.$

Remark. If $X \subset \mathbb{C}^{n}$ and $Y \subset \mathbb{C}^{m}$ are open subsets, the above coincides with our previous definition of holomorphic maps.

Remark. Note that a complex manifold $(X, \mathscr{O}_{X})$ is a locally ringed space. To see this fix $p \in X$ and consider the stalk $\mathscr{O}_{X, p}$ at $p.$ Since any germ not vanishing at $p$ does not vanish nearby $p,$ we see that $$\mathfrak{m}_{X,p} = \{f \in \mathscr{O}_{X,p} : f(p) = 0\}$$ is the unique maximal ideal of $\mathscr{O}_{X, p},$ so $\mathscr{O}_{X,p}$ is a local ring.

Given any holomorphic map $\phi : X \rightarrow Y,$ for any $p \in X,$ we get an induced $\mathbb{C}$-algebra map on stalks $\mathscr{O}_{Y,\phi(p)} \rightarrow \mathscr{O}_{X,p}$ given by $$g \mapsto g \circ \phi.$$ Hence, we have $\mathfrak{m}_{p} \rightarrow \mathfrak{m}_{\phi(p)}$ under this map, so the map on the stalks is a local map. Thus, a holomorphic map of complex manifolds is a morphism of locally ringed spaces.

Personal question. In the case of varieties, any morphism $\phi : X \rightarrow Y$ of locally ringed spaces gives the map $\mathscr{O}_{Y} \rightarrow \phi_{*}\mathscr{O}_{X}$ by the post composition $g \mapsto g \circ \phi.$ I have a feeling that this is a very special feature of varieties, but I cannot tell whether there is a counterexample for real/complex manifolds. Are there?

Remark. For any $n$-dimensional manifold, it is not difficult to see that any of its stalk is isomorphic to the subalgebra of the formal power series ring $\mathbb{C}[[z_{1}, \dots, z_{n}]]$ consisting of convergent power series at $(0, \dots, 0).$

Atlas definition. One can define complex manifolds using atlases. That is, an $n$-dimensional complex manifold $X$ is a Hausdorff and 2nd countable topological space with an open cover $X = \bigcup_{i \in I}U_{i}$ equipped with homemomorphisms $$\varphi_{i} : U_{i} \overset{\sim}{\longrightarrow} V_{i} \overset{\text{open}}{\hookrightarrow} \mathbb{C}^{n}$$ such that for all $i, j \in I,$ the map $$\varphi_{j} \circ \varphi_{i}^{-1} : \varphi_{i}(U_{i} \cap U_{j}) \rightarrow \varphi_{j}(U_{i} \cap U_{j})$$ is a biholomorphic map. (Just perform gluing of sheaves $\mathscr{O}_{U_{i}}$ according to $\varphi_{i}.$)

Based on the atlas definition, since holomorphic maps between the open subsets of Euclidean spaces are smooth, any complex manifold $X$ of dimension $n$ has an underlying real smooth manifold structure $X_{\mathbb{R}}$ of dimension $2n.$ We also have the inclusion of two different sheaves $\mathscr{O}_{X} \hookrightarrow \mathscr{C}^{\infty}_{X, \mathbb{C}}$ for $X,$ and we will use both in a tidy way to understand $X$ better.

Personal remark. We did discuss the Maximum Modulus Theorem, but I decided not to include it in these notes because we are not using the theorem here.

Hodge theory: Lecture 5

Our imminent goal is to construct a meaningful structure sheaf on the analytification of a complex variety. We start with the affine space. Given an open subset $U \subset \mathbb{C}^{n},$ we denote by $\mathscr{O}_{\mathbb{C}^{n}}(U)$ the set of holomorphic functions $U \rightarrow \mathbb{C}.$ The algebraic structure of $\mathbb{C}$ gives a $\mathbb{C}$-algebra structure on the set of all smooth functions $U \rightarrow \mathbb{C},$ and to note that $\mathscr{O}_{\mathbb{C}^{n}}(U)$ is a $\mathbb{C}$-subalgebra, we just need to note that $\partial / \partial \bar{z_{j}}$ is a $\mathbb{C}$-linear derivation.

Given $f \in \mathscr{O}_{\mathbb{C}^{n}}(U)$ not vanishing anywhere in $U,$ we have $$\frac{\partial}{\partial \bar{z_{j}}}\frac{1}{f} = \frac{-(f_{x_{j}} + if_{y_{j}})}{f^{2}} = \frac{-f_{\bar{z_{j}}}}{f^{2}} = 0,$$ for all $1 \leq j \leq n,$ so $1/f \in \mathscr{O}_{\mathbb{C}^{n}}(U).$

More generally, we say $f = (f_{1}, \dots, f_{m}) : U \rightarrow \mathbb{C}^{m}$ is holomorphic if all $f_{1}, \dots, f_{m}$ are holomorphic. 

Notations. We give 

• $\mathbb{C}^{n}$ the complex coordinates $z_{1}, \dots, z_{n}$ (with $z_{j} = x_{j} + iy_{j}$ for the real coordinates);
• $\mathbb{C}^{m}$ the complex coordinates $z'_{1}, \dots, z'_{m}$ (with $z'_{j} = x'_{j} + iy'_{j}$ for the real coordinates).

Fix $p \in U.$ Then writing each $f_{j} = u_{j} + iv_{j},$ we have $df_{p} : T_{p}\mathbb{R}^{2n} \rightarrow T_{f(p)}\mathbb{R}^{2m}$ given by (e.g., Tu Proposition 8.11)

• $\left.\frac{\partial}{\partial x_{j}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial u_{k}}{\partial x_{j}}(p) \left.\frac{\partial }{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial x_{j}}(p) \left.\frac{\partial }{\partial y'_{k}}\right|_{f(p)} \right)$ and
• $\left.\frac{\partial}{\partial y_{j}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial u_{k}}{\partial y_{j}}(p) \left.\frac{\partial}{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial y_{j}}(p) \left.\frac{\partial}{\partial y'_{k}}\right|_{f(p)}\right).$

Consider the "complexified" differential, given by applying $(-) \otimes_{\mathbb{R}} \mathbb{C}$: $$T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} \rightarrow T_{f(p)}\mathbb{R}^{2m} \otimes_{\mathbb{R}} \mathbb{C}.$$ For $T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C},$ denoting $\left.\frac{\partial}{\partial x_{j}}\right|_{p} = \left.\frac{\partial}{\partial x_{j}}\right|_{p} \otimes 1$ and $\left.\frac{\partial}{\partial y_{j}}\right|_{p} = \left.\frac{\partial}{\partial y_{j}}\right|_{p} \otimes 1,$ the vectors $\left.\frac{\partial}{\partial x_{1}}\right|_{p}, \dots, \left.\frac{\partial}{\partial x_{n}}\right|_{p}, \left.\frac{\partial}{\partial y_{1}}\right|_{p}, \dots, \left.\frac{\partial}{\partial y_{n}}\right|_{p}$ form a basis for $T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C}.$ Again, we write $$\left.\frac{\partial}{\partial z_{j}}\right|_{p} = \frac{1}{2}\left(\left.\frac{\partial}{\partial x_{j}}\right|_{p} + i \left.\frac{\partial}{\partial y_{j}}\right|_{p}\right)$$ and $$\left.\frac{\partial}{\partial \bar{z_{j}}}\right|_{p} = \frac{1}{2}\left(\left.\frac{\partial}{\partial x_{j}}\right|_{p} - i \left.\frac{\partial}{\partial y_{j}}\right|_{p}\right).$$ By computing a nonzero determinant, we see that $\left.\frac{\partial}{\partial z_{1}}\right|_{p}, \dots, \left.\frac{\partial}{\partial z_{n}}\right|_{p}, \left.\frac{\partial}{\partial \bar{z_{1}}}\right|_{p}, \dots, \left.\frac{\partial}{\partial \bar{z_{n}}}\right|_{p}$ form a $\mathbb{C}$-basis for $T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C}.$

Remark. It is interesting to note that the differentiation operators were just symbols before, and now they are treated as actual mathematical objects (i.e., tangent vectors or complexified tangent vectors).

Note that the complexified differential at $p$ gives 

• $\left.\frac{\partial}{\partial z_{j}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial u_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial y'_{k}}\right|_{f(p)} \right)$ and
• $\left.\frac{\partial}{\partial \bar{z_{j}}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial u_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial}{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial}{\partial y'_{k}}\right|_{f(p)}\right).$

Since $$\frac{\partial}{\partial x'_{j}} = \frac{\partial}{\partial z'_{j}} + \frac{\partial}{\partial \bar{z'_{j}}}$$ and $$\frac{\partial}{\partial y'_{j}} = -i\left(\frac{\partial}{\partial z'_{j}} - \frac{\partial}{\partial \bar{z'_{j}}}\right),$$ we have $$\frac{\partial u_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial y'_{k}}\right|_{f(p)} = \frac{\partial f_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial z'_{k}}\right|_{f(p)} + \frac{\partial \bar{f_{k}}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial \bar{z'_{k}}}\right|_{f(p)}$$ and $$\frac{\partial u_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial }{\partial x'_{k}}\right|_{f(p)} + \frac{\partial v_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial }{\partial y'_{k}}\right|_{f(p)} = \frac{\partial f_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial }{\partial z'_{k}}\right|_{f(p)} + \frac{\partial \bar{f_{k}}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial }{\partial \bar{z'_{k}}}\right|_{f(p)}$$ merely due to definitions, so we conclude that the complexified differential at $p$ gives 

• $\left.\frac{\partial}{\partial z_{j}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial f_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial z'_{k}}\right|_{f(p)} + \frac{\partial \bar{f_{k}}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial \bar{z'_{k}}}\right|_{f(p)} \right)$ and
• $\left.\frac{\partial}{\partial \bar{z_{j}}}\right|_{p} \mapsto \sum_{k=1}^{m} \left( \frac{\partial f_{k}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial}{\partial z'_{k}}\right|_{f(p)} + \frac{\partial \bar{f_{k}}}{\partial \bar{z_{j}}}(p) \left.\frac{\partial}{\partial \bar{z'_{k}}}\right|_{f(p)}\right).$

Induced maps on complexified tangent spaces. Note that if $f = (f_{1}, \dots, f_{m})$ is holomorphic, then $$\frac{\partial f_{k}}{\partial \bar{z_{j}}} = 0 = \frac{\partial \bar{f_{k}}}{\partial z_{j}}.$$ Thus, we see that $$df_{p}\left( \left.\frac{\partial}{\partial z_{j}}\right|_{p} \right) = \sum_{k=1}^{m} \frac{\partial f_{k}}{\partial z_{j}}(p) \left.\frac{\partial }{\partial z'_{k}}\right|_{f(p)}$$ and $$df_{p}\left( \left.\frac{\partial}{\partial \bar{z_{j}}}\right|_{p} \right) = \sum_{k=1}^{m} \overline{\frac{\partial f_{k}}{\partial z_{j}}}(p) \left.\frac{\partial }{\partial \bar{z'_{k}}}\right|_{f(p)}.$$ Now, if $V \subset \mathbb{C}^{m}$ be an open subset such that $f(U) \subset V$ and $g : V \rightarrow \mathbb{C}$ any smooth function, then feeding the germ of $g$ at $f(p)$ to the last identity gives $$\left.\frac{\partial}{\partial \bar{z_{j}}}\right|_{p}(g \circ f) = \sum_{k=1}^{m} \overline{\frac{\partial f_{k}}{\partial z_{j}}}(p) \frac{\partial g}{\partial \bar{z'_{k}}}(f(p)).$$ This shows that if $f$ and $g$ are holomorphic, then $g \circ f$ is holomorphic. Even when $g = (g_{1}, \dots, g_{r}) : V \rightarrow \mathbb{C}^{r},$ the same argument would show that if $f$ and $g$ are holomorphic, then $g_{j} \circ f$ is holomorphic for all $1 \leq j \leq r,$ which implies that $g \circ f$ is holomorphic (by definition).

Holomorphicity in reverse. A similar argument also proves that if

• $f$ and $g \circ f$ are holomorphic at $p,$ 
• $n = m,$ and
• the $n \times n$ matrix $[\partial f_{i}/\partial z_{j}]_{i,j=1}^{n}$ at $p$ is invertible, 

then $g$ is holomorphic.

Remark. We have $$T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} = \mathbb{C} \frac{\partial}{\partial z_{1}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial z_{n}} \oplus \mathbb{C} \frac{\partial}{\partial \bar{z_{1}}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial \bar{z_{n}}},$$ and if $f : U \rightarrow \mathbb{C}^{m}$ is holomorphic, then the induced map $$df_{p} : T_{p}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} \rightarrow T_{f(p)}\mathbb{R}^{2m} \otimes_{\mathbb{R}} \mathbb{C}$$ can be decomposed as $$\mathbb{C} \frac{\partial}{\partial z_{1}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial z_{n}} \rightarrow \mathbb{C} \frac{\partial}{\partial z'_{1}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial z'_{m}}$$ and $$\mathbb{C} \frac{\partial}{\partial \bar{z_{1}}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial \bar{z_{n}}} \rightarrow \mathbb{C} \frac{\partial}{\partial \bar{z'_{1}}} \oplus \cdots \oplus \mathbb{C} \frac{\partial}{\partial \bar{z'_{m}}}.$$

Comparison between real Jacobian and complex Jacobian. We have a very interesting relationship between the real and the complex Jacobians of a holomorphic map $f : U \rightarrow \mathbb{C}^{n},$ where $U \subset \mathbb{C}^{n}$ is an open subset:

Theorem. If $f$ is holomorphic, the determinant of the real Jacobian matrix with respect to the coordinates $x_{1}, \dots, x_{n}, y_{1}, \dots, y_{n}$ is precisely $$\left|\det\left[ \frac{\partial f_{i}}{\partial z_{j}} \right]_{1 \leq i,j \leq n}\right|^{2}.$$

Before proving this, let's discuss how we can use this fact. First, note that the determinant of the $n \times n$ complex Jacobian matrix of a holomorphic map is always a nonnegative real number. Second, we see that it is $0$ precisely when the real Jacobian is $0$ (even true pointwise).

Corollary (Holomorphic Inverse Function Theorem). Let $U \subset \mathbb{C}^{n}$ be a nonempty open subset and $f : U \rightarrow \mathbb{C}^{n}$ be a holomorphic map. If there is a point $p \in U$ such that $$\det\left[ \frac{\partial f_{i}}{\partial z_{j}}(p) \right]_{1 \leq i,j \leq n} \neq 0,$$ then there is an open subsets $U' \ni p$ in $U$ and $V' \supset f(U')$ in $\mathbb{C}^{n}$ such that $f : U' \rightarrow V'$ is biholomorphic.

Proof. Based on the preceding paragraph, we know that the real Jacobian does not vanish, so we may apply the Inverse Function Theorem for real smooth maps (e.g., Tu Theorem 6.25) to conclude that we can obtain such $f : U' \rightarrow V'$ except the biholomorphic condition. So far we can only tell that the inverse of this local function is smooth. Since $f$ and $f^{-1} \circ f = \mathrm{id}_{U}$ are both smooth with invertible complex Jacobian matrix, by the discussion about "holomorphicity in reverse", it follows that $f^{-1}$ is holomorphic as well. $\Box$

Now, let's prove the theorem about Jacobian matrices.

Proof of Theorem. For now, let $f : U \rightarrow \mathbb{C}^{n}$ be smooth, not necessarily holomorphic. We will do everything at the given point $p \in U,$ which we will suppress in our writing. Then (e.g., by Proposition 18.3, 18.11 in Tu) $$\begin{align*}f^{*}(dx_{1} \wedge dy_{1} \wedge \cdots \wedge dx_{n} \wedge dy_{n}) &= df^{*}x_{1} \wedge df^{*}y_{1} \wedge \cdots \wedge df^{*}x_{n} \wedge df^{*}y_{n} \\ &= du_{1} \wedge dv_{1} \wedge \cdots \wedge du_{n} \wedge dv_{n} \\ &= \det (J_{\mathbb{R}}(f)) dx_{1} \wedge dy_{1} \wedge \cdots \wedge dx_{n} \wedge dy_{n},\end{align*}$$ where $J_{\mathbb{R}}(f)$ denotes the real Jacobian matrix of $f.$ We may consider the complexified version $$f^{*} : \bigwedge^{2n}T_{f(p)}^{\vee}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C} \rightarrow \bigwedge^{2n}T_{p}^{\vee}\mathbb{R}^{2n} \otimes_{\mathbb{R}} \mathbb{C},$$ and knowing $$dz_{j} \wedge d\bar{z_{j}} = -2i \cdot dx_{j} \wedge dy_{j},$$ we have $$\begin{align*}f^{*}(dz_{1} \wedge d\bar{z}_{1} \wedge \cdots \wedge dz_{n} \wedge d\bar{z}_{n}) &= -2if^{*}(dx_{1} \wedge dy_{1} \wedge \cdots \wedge dx_{n} \wedge dy_{n})\\ &= -2i\det (J_{\mathbb{R}}(f)) dx_{1} \wedge dy_{1} \wedge \cdots \wedge dx_{n} \wedge dy_{n} \\ &= \det (J_{\mathbb{R}}(f)) dz_{1} \wedge d\bar{z}_{1} \wedge \cdots \wedge dz_{n} \wedge d\bar{z}_{n}.\end{align*}$$ But then since $$df \left( \frac{\partial}{\partial z_{j}} \right) = \sum_{k=1}^{m} \frac{\partial f_{k}}{\partial z_{j}} \frac{\partial }{\partial z'_{k}}$$ and $$df\left( \frac{\partial}{\partial \bar{z_{j}}} \right) = \sum_{k=1}^{m} \overline{\frac{\partial f_{k}}{\partial z_{j}}} \frac{\partial }{\partial \bar{z'_{k}}},$$ we have $$\begin{align*}f^{*}(dz_{1} \wedge d\bar{z}_{1} \wedge \cdots \wedge dz_{n} \wedge d\bar{z}_{n}) &= -2i du_{1} \wedge dv_{1} \wedge \cdots \wedge du_{n} \wedge dv_{n} \\ &= df_{1} \wedge d\bar{f_{1}} \wedge \cdots \wedge df_{n} \wedge d\bar{f_{n}} \\ &= \det(J_{\mathbb{C}}(f))\overline{\det(J_{\mathbb{C}}(f))} dz_{1} \wedge d\bar{z_{1}} \wedge \cdots \wedge dz_{n} \wedge d\bar{z_{n}} \\ &= |\det(J_{\mathbb{C}}(f))|^{2} dz_{1} \wedge d\bar{z_{1}} \wedge \cdots \wedge dz_{n} \wedge d\bar{z_{n}}, \end{align*}$$ where $J_{\mathbb{C}}(f)$ denotes the $n \times n$ complex Jacobian matrix of $f$ (in $z_{1}, \dots, z_{n}$). Comparing the two computations above, we have $$\det (J_{\mathbb{R}}(f)) = |\det(J_{\mathbb{C}}(f))|^{2},$$ as desired. $\Box$

More properties about holomorphic functions. If $f : U \rightarrow \mathbb{C}$ is a holomorphic function, then any $$\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}$$ is a holomorphic function because partials commute so that we have $$\frac{\partial}{\partial \bar{z_{j}}}\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}} = \frac{\partial^{r}}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}\frac{\partial f}{\partial \bar{z_{j}}} = 0.$$ The following is another interesting property, that is not true for many smooth functions:

Lemma. Given $p \in U,$ if $$\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}(p) = 0$$ for all choices of $j_{1}, \dots, j_{r} \in \mathbb{Z}_{\geq 1},$ then there is a neighborhood of $p$ where $f$ vanishes.

Proof. Choose $\epsilon > 0$ small enough so that $\overline{D_{\epsilon}(p)} \subset U.$ By Cauchy's formula, we have $$f(z_{1}, \dots, z_{n}) = \left(\frac{1}{2 \pi i}\right)^{n} \int_{\partial D_{\epsilon}(p)} \frac{f(w)}{(w_{1} - z_{1}) \cdots (w_{n} - z_{n})} dw_{1} \wedge \cdots \wedge dw_{n}.$$ Using this, applying the argument in the single variable case $n$ times, we get $$f(z_{1}, \dots, z_{n}) = \sum_{j_{1}, \dots, j_{n} \geq 0} c_{j_{1}, \dots, j_{n}}(z_{1} - p_{1})^{j_{1}} \cdots (z - p_{n})^{j_{n}},$$ for $(z_{1}, \dots, z_{n}) \in D_{\epsilon}(p),$ where $$c_{j_{1}, \dots, j_{n}} = \frac{1}{(2\pi i)^{n}}\int_{\partial D_{\epsilon}(p)} \frac{f(w) dw_{1} \wedge \cdots \wedge dw_{n}}{(w_{1} - z_{1})^{j_{1} + 1} \cdots (w_{n} - z_{n})^{j_{n}+1}}.$$ Since all $$c_{j_{1}, \dots, j_{n}} = \frac{1}{j_{1}! \cdots j_{n}!}\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}(p) = 0,$$ the result follows. $\Box$

Proposition. Let $f : U \rightarrow \mathbb{C}$ be a holomorphic function and $U$ is connected. Then if $f$ vanishes on some nonempty open $V \subset U,$ then $f = 0.$

Proof. Let $U' := \{z \in U : f \text{ vanishes on some open neighborhood of }z\}.$ The existences of $V$ tells us that $U'$ is nonempty, and a little thought would reveal that $U'$ is open in $U.$ Hence, to show $U' = U,$ it is enough to show that $U' \subset U$ is closed. Fix any sequence $(w_{n})$ in $U'$ convergent to some $w \in U.$ We have $$\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}(w_{n}) = 0$$ for all $n$ by definition of $U',$ so $$0 = \lim_{n \rightarrow \infty}\frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}(w_{n}) = \frac{\partial^{r} f}{\partial z_{j_{1}} \cdots \partial z_{j_{r}}}(w),$$ for any choices of $j_{1}, \dots, j_{n} \geq 0.$ This implies that $f$ vanishes at a neighborhood $w.$ This means that $w \in U',$ so $U'$ is indeed closed in $U,$ as desired. $\Box$

Hodge theory: Lecture 4

Single-variable holomorphic functions. We continue our discussion from the previous notes.

Theorem 1. Let $f : U \rightarrow \mathbb{C}$ be a smooth function. The following are equivalent:

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

Proof. First, assume that $f$ is holomorphic. Fix any $a \in U$ and take an open disk $D$ containing $a$ such that $\overline{D} \subset U.$ For any $z \in D,$ Cauchy's integral formula tells us that $$f(z) = \frac{1}{2\pi i}\int_{\partial D}\frac{f(w)}{w - z} dw.$$ Let's look at the integrand: $$\begin{align*}\frac{f(w)}{w - z} &= \frac{f(w)}{(w - a) - (z - a)} \\ &= \left(\frac{f(w)}{w - a}\right) \frac{1}{1 - (z-a)/(w-a)} \\ &= \left(\frac{f(w)}{w - a}\right) \sum_{n=0}^{\infty} \frac{(z-a)^{n}}{(w-a)^{n}} \\ &= \sum_{n=0}^{\infty}\frac{f(w)}{(w-a)^{n+1}}(z-a)^{n},\end{align*}$$ where $w \in \partial D,$ which ensures that $|z - a| < |w - a|.$ Note that this series is absolutely and uniformly convergent as it is given by the geometric series. Thus, (even a weaker version) Dominate Convergence implies that $$\begin{align*} \frac{1}{2\pi i} \int_{\partial D}\frac{f(w)}{w - z} dw &= \frac{1}{2\pi i} \int_{\partial D} \left(\sum_{n=0}^{\infty}\frac{f(w)}{(w-a)^{n+1}}(z-a)^{n}\right)dw \\ &= \sum_{n=0}^{\infty} \left( \frac{1}{2\pi i} \int_{\partial D}\frac{f(w)}{(w-a)^{n+1}}dw \right) (z-a)^{n} \end{align*}$$ for any $z \in D.$ The uniform convergence and the absolute convergence of the last expression follow from the second expression, but we also may check explicitly as follows. Denote by $r > 0$ the radius of $D.$ If we take a disc $D'$ with radius $r - \epsilon > 0,$ for any $z \in D',$ we have $$\begin{align*}\left|\left( \int_{\partial D}\frac{f(w)}{(w-a)^{n+1}}dw \right) (z - a)^{n}\right| &\leq |z - a|^{n}\int_{\partial D}\frac{|f(w)|}{|w-a|^{n+1}}dw \\ &< r^{-1} (1 - \epsilon/r)^{n}\int_{\partial D}|f(w)|dw \\ &\leq r^{-1}(1 - \epsilon/r)^{n}C,\end{align*}$$ for some constant $C.$

Remark. A more general version for the first direction is available as Theorem 10.7 in Rudin. It is funny to note that $\partial D$ actually means topological boundary. It is equal to the boundary of the $\overline{D}$ in the manifold theory sense, which is really what we are using when we think about Stokes' theorem.

Conversely, let $f$ be analytic. Given $a \in U,$ take a small disc $D$ in $U$ centered at $a$ such that $\overline{D} \subset U$ and $$f(z) = \sum_{n=0}^{\infty}c_{n} (z - a)^{n}$$ for $z \in D,$ where the sum is uniformly and absolute convergent. The partial sum $$P_{n}(z) = \sum_{d=0}^{n}c_{d} (z - a)^{d}$$ is a polynomial, so it is holomorphic. Hence, by Cauchy's integral formula, we have $$P_{n}(z) = \frac{1}{2\pi i}\int_{\partial D} \frac{P_{n}(w)}{w - z}dw.$$ This implies that $$\begin{align*} f(z) &= \lim_{n \rightarrow \infty}P_{n}(z) \\ &= \frac{1}{2\pi i} \lim_{n \rightarrow \infty}\int_{\partial D} \frac{P_{n}(w)}{w - z}dw \\ &= \frac{1}{2\pi i} \int_{\partial D} \frac{f(w)}{w - z}dw\end{align*}$$ by Dominate Convergence. Now an application of Dominate Convergence (e.g., Theorem 2.27 in Folland) lets us have $$\frac{\partial f}{\partial \bar{z}} = \frac{1}{2\pi i} \int_{\partial D} \frac{\partial}{\partial \bar{z}} \frac{f(w)}{w - z}dw = 0,$$ since $1/(w - z)$ is holomorphic in $z \in D,$ for any $w \in \partial D.$ Thus $f$ is holomoprhic, as desired. $\Box$

Theorem 2 ($\bar{\partial}$-lemma in single variable). Let $D$ be a nonempty open disk in an open subset $U \subset \mathbb{C}$ such that $\overline{D} \subset U.$ Then for any smooth function $g : U \rightarrow \mathbb{C},$ there is a function $f : D \rightarrow \mathbb{C}$ such that $$g = \frac{\partial f}{\partial \bar{z}}.$$ More specifically, we can construct such an $f$ as follows: $$f(z) = \frac{1}{2\pi i} \int_{\overline{D}}g(w) \frac{dw \wedge d\bar{w}}{w - z}$$ for $z \in D.$

Remark. From a change of variable argument into the polar coordinate from the previous notes, we note that the integral above is well-defined.

Proof. Fix any $z_{0} \in D.$ Take any open discs $D_{1}, D_{2}$ centered at $z_{0}$ in $D$ such that $$\overline{D_{1}} \subsetneq \overline{D_{2}} \subsetneq D.$$ Then $\mathbb{C} \smallsetminus D_{1}$ and $D_{2}$ cover $\mathbb{C} = \mathbb{R}^{2},$ so (e.g., by Proposition 13.6 in Tu's book) we may find smooth (partition of unity) $\rho_{1}, \rho_{2} : \mathbb{C} \rightarrow \mathbb{R}$ such that

• $\rho_{1} + \rho_{2} = 1$ on $\mathbb{C},$
• $\rho_{1} = 0$ on $D_{1},$ and
• $\rho_{2} = 0$ on $\mathbb{C} \setminus D_{2}.$

Thus, we have $g = g_{1} + g_{2}$ with $g_{i} := \rho_{i}g$ smooth functions $U \rightarrow \mathbb{C}$ such that 

• $g_{1} = 0$ on $\overline{D_{1}},$ and
• $g_{2} = 0$ on $\mathbb{C} \setminus D_{2}.$

Hence, we may write $f = f_{1} + f_{2}$ with $$f_{j}(z) := \frac{1}{2 \pi i}\int_{\overline{D}} g_{j}(w) \frac{dw \wedge d\bar{w}}{w - z}.$$ We have $$f_{1}(z) = \frac{1}{2 \pi i}\int_{D \setminus \overline{D_{1}}} g_{1}(w) \frac{dw \wedge d\bar{w}}{w - z},$$ so for $z \in D_{1},$ we have $$\frac{\partial f_{1}}{\partial \bar{z}} = \frac{1}{2\pi i}\int_{D \setminus \overline{D_{1}}}\frac{\partial}{\partial \bar{z}} \left(\frac{g_{1}(w)}{w - z}\right) dw \wedge d\bar{w} = 0$$ because $$\frac{\partial}{\partial \bar{z}} \left(\frac{g_{1}(w)}{w - z}\right) = 0$$ when $w \in D \smallsetminus \overline{D_{1}}.$ (Again, we are using a consequence of Dominated Convergence, say Theorem 2.27 in Folland.) Note that $$f_{2}(z) = \frac{1}{2 \pi i}\int_{\mathbb{C}} g_{2}(w) \frac{dw \wedge d\bar{w}}{w - z} = -\frac{1}{\pi} \int_{0}^{2\pi}e^{-i\theta}\int_{0}^{\infty}g_{2}(z + re^{i\theta})drd\theta,$$ which can be computed with $w = z + re^{i \theta}$ where $r \in (0, \infty)$ and $\theta \in (0, 2\pi).$ Since $g_{2} = 0$ outside $D_{2},$ its derivative is also $0$ outside $\overline{D_{2}}.$ Thus, we have $$\begin{align*}\frac{\partial f_{2}}{\partial \bar{z}} &= -\frac{1}{\pi} \int_{0}^{2\pi}e^{-i\theta}\int_{0}^{\infty}\frac{\partial g_{2}(z + re^{i\theta})}{\partial \bar{z}}drd\theta \\ &= \frac{1}{2 \pi i}\int_{\mathbb{C}} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z} \\ &= \frac{1}{2 \pi i}\int_{D} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z}, \end{align*}$$ because $g_{2} = 0$ outside $D.$ Now, Cauchy's formula gives us $$g_{2}(z) = \frac{1}{2\pi i} \int_{\partial D} \frac{g_{2}(w)}{w - z}dw + \frac{1}{2\pi i} \int_{D} \frac{\partial g_{2}}{\partial \bar{w}} \frac{dw \wedge d\bar{w}}{w - z}dw.$$ But the first integral vanishes because $g_{2} = 0$ on $\partial D,$ as it is outside $D_{2}.$ Therefore, it follows that $$\frac{\partial f}{\partial \bar{z}} = \frac{\partial f_{2}}{\partial \bar{z}} = g$$ on $D_{1}.$ Since $D_{1} \subset D$ is an arbitrary open disk, this shows that the above equality is true on $D.$ $\Box$

Observation (for later). The proof can be modified so that if $$g = g(z_{1}, \dots, z_{n}) : U_{1} \times \cdots \times U_{n} \rightarrow \mathbb{C}$$ is a smooth function where each $U_{j}$ is open and holomorphic in variables $z_{2}, \dots, z_{n}$ (separably), then so is (modified) $f.$

Holomorphic functions of several variables. Let $U \subset \mathbb{C}^{n}$ be a nonempty open subset. The coordinate functions $(z_{1}, \dots, z_{n}),$ where we write $z_{j} = x_{j} + iy_{j} = (x_{j}, y_{j}) \in \mathbb{R}^{2}$ as usual.

A smooth function $f : U \rightarrow \mathbb{C}$ is said to be holomorphic if it is holomorphic in each $z_{j},$ which in other words, we have $$\frac{\partial f}{\partial \bar{z_{j}}} = 0$$ on $U.$ We say $f$ is analytic if for every $a \in U,$ we have a polydisk $$D(a) = D_{r_{1}}(a_{1}) \times \cdots \times D_{r_{n}}(a_{n})$$ such that $$f(z) = \sum_{k_{1}=0}^{\infty} \cdots \sum_{k_{n}=0}^{\infty}c_{k_{1},\dots,k_{n}}(z_{1} - a_{1})^{k_{1}} \cdots (z_{n} - a_{n})^{k_{n}}$$ for $z \in D,$ absolutely and uniformly.

Theorem. Given a smooth function $f : U \rightarrow \mathbb{C},$ the following are equivalent:

(1) $f$ is holomorphic;
(2) $f$ is analytic;
(3) For every polydisc $D_{1} \times \cdots \times D_{n} \subset U,$ we have $$f(z) = \left(\frac{1}{2 \pi i}\right)^{n} \int_{\prod_{j=1}^{n}\partial D_{j}} \frac{f(w)}{(w_{1} - z_{1}) \cdots (w_{n} - z_{n})} dw_{1} \wedge \cdots \wedge dw_{n},$$ where the integral is over the product of circles with product orientation.

Proof. Assume (2). Then $f$ is analytic in each variable, so it must be holomorphic in each variable. This implies (1).

To see that (1) implies (3), we use Cauchy's integral formula for single variable $n$ times, and apply Fubini (e.g., 2.37 b in Folland) as $f$ is continuous.

We can apply the same argument as in the single variable case to ensure that (3) implies (2). This finishes the proof. $\Box$

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 $g = \frac{\partial f}{\partial \bar{z}}.$

Hodge theory: Lecture 2

We continue from the last time.

Theorem. Let $X$ be an irreducible $\mathbb{C}$-variety and $U \subset X$ a nonempty open subset. Then $U^{\mathrm{an}}$ is dense in $X^{\mathrm{an}}.$

We also recall from the last time that the theorem had an immediate corollary, which we will use later:

Corollary. Let $X$ be any $\mathbb{C}$-variety. If $U \subset X$ is Zariski-open and Zariski-dense, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.

Reference. The lecturer kindly notified us that we are following Mumford's Red Book (Theorem 1 on p.58).

Proof. It is a convoluted proof (in my opinion), so we have several steps.

Step 1. Reduction to the case where $X$ is affine.

Recall that our definition of variety includes finite type over $\mathbb{C},$ which includes quasi-compactness. Thus, we can choose a finite open cover $X = \bigcup_{i=1}^{r} U_{i}$ with every $U_{i} \neq \emptyset.$ Since $X$ is irreducible, we have $U \cap U_{i} \neq \emptyset$ for all $1 \leq i \leq r.$ Thus, if the theorem is true for affine case, then $(U \cap U_{i})^{\mathrm{an}} = U(\mathbb{C}) \cap U_{i}(\mathbb{C})$ would be dense in $U_{i}^{\mathrm{an}} = U_{i}(\mathbb{C}).$ Since $U = \bigcup_{i=1}^{r} (U \cap U_{i}),$ this implies that $$\overline{U(\mathbb{C})} = \bigcup_{i=1}^{r} (\overline{U(\mathbb{C})} \cap U_{i}(\mathbb{C})) = \bigcup_{i=1}^{r} U_{i}(\mathbb{C}) = X(\mathbb{C}),$$ with respect to the classical topology. Hence, for the rest of the proof, we may assume that $X$ is affine.

Step 2. Apply Nother normalization.

Recall (e.g., Vakil 11.2.4) the following lemma:

Lemma (Nother normalization). Given any integral domain finitely generated over a field $k,$ if the transcendental degree of $A$ over $k$ is $n$ (which is necessarily finite), then there exist $x_{1}, \dots, x_{n} \in A,$ algebraically independent over $k,$ such that $k[x_{1}, \dots, x_{n}] \hookrightarrow A$ is a finite extension of rings (which is necessarily a $k$-algebra map). In particular, by Lying Over, we have a surjective $k$-scheme map $\mathrm{Spec}(A) \twoheadrightarrow \mathbb{A}^{n}_{k}.$

Remark. We also recall that for any such domain $A,$ the transcendental degree of $A$ over $k$ is precisely the dimension of $\mathrm{Spec}(A),$ which can be proved using Noether normalization (e.g., Vakil 11.2.7). Without this approach, it does not seem clear (at least to me) how to show that $\dim(A)$ is finite (cf. Vakil 11.1.K).

Now, we have $X = \mathrm{Spec}(A)$ and writing $\dim(X) = n,$ we have a finite $\mathbb{C}$-algebra extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A.$ Denote by $\pi : X \twoheadrightarrow \mathbb{A}^{n}$ the surjective $k$-scheme map induced by this finite extension. Given any $\mathbb{C}$-point $q \in \mathbb{A}^{n}(\mathbb{C}),$ the fiber $\pi^{-1}(q) \subset X$ is a (nonempty) closed subset of dimension $0$ (e.g., Vakil 11.1.D), so all of its elements are closed points of $X,$ which are $\mathbb{C}$-points by Nullstellensatz. Thus, the induced map $\pi^{\mathrm{an}} : X(\mathbb{C}) \rightarrow \mathbb{A}^{n}(\mathbb{C}) = \mathbb{C}^{n}$ is surjective as well.

Let $Z := X \setminus U.$ We know from the last time that $Z(\mathbb{C}) = X(\mathbb{C}) \setminus U(\mathbb{C})$ not only as sets but as topological spaces with the classical topologies. Our goal is to show $$\overline{U(\mathbb{C})} = X(\mathbb{C}),$$ so it is enough to show that given $p \in Z(\mathbb{C}),$ we can find a sequence $(y_{m})$ in $U(\mathbb{C})$ such that $$\lim_{m \rightarrow \infty}y_{m} = p,$$ with respect to the Euclidean topology of $X(\mathbb{C}).$ (Note that $\pi(p) \in \mathbb{A}^{n}(\mathbb{C}) = \mathbb{C}^{n}$.)

Since $\pi$ is a finite map, it sends closed subsets to closed subsets (e.g., Vakil 7.3.M), so $\pi(Z) \subset \mathbb{A}^{n}$ is Zariski-closed. Note that $\pi(Z) \subsetneq \mathbb{A}^{n}$ because if it were that $\pi(Z) = \mathbb{A}^{n},$ then the generic fiber must be contained $Z,$ but then in this case (because the ring map is injective) the generic fiber contains the generic point of $X,$ which would imply that $Z = X.$ This does not happen because $U = X \setminus Z$ is nonempty, so $\pi(Z)$ is a proper (closed) subset of $\mathbb{A}^{n}.$ Hence, we can write $$\pi(Z) = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/I) \simeq V(I) \subset \mathbb{A}^{n}$$ for some nonzero (radical) ideal $I \subset \mathbb{C}[x_{1}, \dots, x_{n}].$ Take any nonzero polynomial $g \in I$ so that we have $$\pi(Z) = V(I) \subset V(g) \subsetneq \mathbb{A}^{n}.$$ In other words, we have $Z \subset \pi^{-1}(V(g)).$ Next, we go analytic.

Consider $\varphi : \mathbb{R} \rightarrow \mathbb{C} = \mathbb{R}^{2}$ given by $$\varphi(t) := g(t\pi(p) + (1-t)w),$$ where we fix $w \in \mathbb{C}^{n}$ such that $g(w) \neq 0.$ (Note that $g(\pi(p)) = 0$ ) We have $\varphi(0) = g(w) \neq 0,$ and $\varphi$ is a polynomial in $t$ with complex coefficients, so it only vanishes at finitely many elements in $\mathbb{R},$ and in particular in the segment $[0, 1].$

Take any sequence $(t_{m})$ in $[0, 1]$ such that $\varphi(t_{m}) \neq 0$ for all $m,$ but $t_{m} \rightarrow 0,$ as $m \rightarrow \infty.$ Writing $u_{m} = t_{m}\pi(p) + (1 - t_{m})w \in \mathbb{C}^{n},$ we see that $g(u_{m}) \neq 0$ for all $m$ and $$\lim_{m \rightarrow \infty}u_{m} = \pi(p).$$ What do we need to do now?

Goal. We will replace $(u_{m})$ with a subsequence of it, and then find $y_{m} \in \pi^{-1}(u_{m})$ for each $m$ such that $y_{m} \rightarrow p$ as $m \rightarrow \infty.$ Once we are done with this, we have $\pi(y_{m}) = u_{m} \in D(g)$ so that $y_{m} \in \pi^{-1}(D(g)).$ Since $Z \subset \pi^{-1}(V(g)),$ which is disjoint to $\pi^{-1}(D(g)),$ we must have $y_{m} \in U = X \setminus Z$ so that $y_{m} \in U(\mathbb{C}).$

Let's take a break and think about what's been going on. After formally reducing the statement to the case where $X$ is affine. Our goal is to show that any $p \in Z(\mathbb{C}) = X(\mathbb{C}) \setminus U(\mathbb{C})$ can be analytically approached by points in $U(\mathbb{C}).$ To do so, we constructed a nice finite surjection $X \twoheadrightarrow \mathbb{A}^{n}$ via Noether normalization, and this gave us an analytic surjection $X(\mathbb{C}) \twoheadrightarrow \mathbb{C}^{n}.$ We took a nonzero polynomial $g \in \mathbb{C}[x_{1}, \dots, x_{n}]$ such that $\pi(Z) \subset V(g) \subsetneq \mathbb{A}^{n}$ and constructed a sequence $u_{m} \in D(g)(\mathbb{C}) \subset \mathbb{C}^{m}$ such that $u_{m} \rightarrow \pi(p)$ as $m \rightarrow \infty.$ We are now hoping to lift this convergence in $$\pi^{-1}(D(g))(\mathbb{C}) \subset U(\mathbb{C}) \subset X(\mathbb{C})$$ to achieve a convergence to $p,$ by possibly throwing away many $m$'s.

OK. Let's get back to the proof.

Step 3. Construction of such a subsequence.

Since $\pi$ is a finite map, any of its fiber has finite size (e.g., Vakil 7.3.K), so we may write $$\pi^{-1}(\pi(p)) = \{\mathfrak{p}_{1}, \mathfrak{p}_{2}, \dots, \mathfrak{p}_{r}\}.$$ Without loss of generality, we may assume $\mathfrak{p}_{1} = p.$ Recall that we have $X = \mathrm{Spec}(A)$ with a finite $\mathbb{C}$-algebra extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A.$ Since $\mathfrak{p}_{i}$ are maximal ideals of $A,$ by the Chinese Remainder, we get a surjection $$A \twoheadrightarrow (A/\mathfrak{p}_{1}) \times \cdots \times (A/\mathfrak{p}_{r})$$ given by the ring projections. Thus, we may pick $f(\boldsymbol{x}) = f \in \mathfrak{p}_{1} = p$ such that $f \notin \mathfrak{p}_{i}$ for all $i \geq 2.$ Since the extension is finite, it is also integral. Hence, we may find $$F \in \mathbb{C}[x_{1}, \dots, x_{n}][t] = \mathbb{C}[x_{1}, \dots, x_{n}, t]$$ such that $$F = F(x_{1}, \dots, x_{n}, t)  = t^{d} + a_{1}(\boldsymbol{x})t^{d-1} + \cdots + a_{d-1}(\boldsymbol{x})t + a_{d}(\boldsymbol{x})$$ and $$F(x_{1}, \dots, x_{n}, f) = 0 \in A.$$ Since $\mathbb{C}[x_{1}, \dots, x_{n}, t]$ is a UFD, we may write $$F(\boldsymbol{x}, t) = F_{1}(\boldsymbol{x}, t) \cdots F_{s}(\boldsymbol{x}, t)$$ in it, where each $F_{i}(\boldsymbol{x}, t) \in \mathbb{C}[\boldsymbol{x}, t]$ is a monic irreducible polynomial. We have $$F(\boldsymbol{x}, f) = F_{1}(\boldsymbol{x}, f) \cdots F_{s}(\boldsymbol{x}, f) = 0 \in A$$ and since $A$ is a domain, there must be at least one $i$ such that $F_{i}(\boldsymbol{x}, f) = 0 \in A$ by replacing $F(\boldsymbol{x}, t)$ with $F_{i}(\boldsymbol{x}, t)$ if necessary, we may assume that $F(\boldsymbol{x}, t)$ is an irreducible polynomial.

This means that $(F)$ is a prime ideal of $\mathbb{C}[x_{1}, \dots, x_{n}, t],$ so $V(F) \subset \mathbb{A}^{n+1}$ is an irreducible Zariski-closed subset. Since $F(x_{1}, \dots, x_{n}, f) = 0$ in $A,$ we can define a $\mathbb{C}$-algebra map $$\mathbb{C}[x_{1}, \dots, x_{n}, t]/(F(x_{1}, \dots, x_{n}, t)) \rightarrow A$$ given by $t \mapsto f.$ Since the extension $\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A$ factors as $$\mathbb{C}[x_{1}, \dots, x_{n}] \rightarrow \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \rightarrow A,$$ where the second one is the map given by $t \mapsto f,$ the second map must be a finite map. Now, consider the induced maps of $\mathbb{C}$-schemes $$X = \mathrm{Spec}(A) \overset{\pi_{2}}{\longrightarrow} \mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \right) \overset{\pi_{1}}{\longrightarrow} \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]) = \mathbb{A}^{n}$$ such that $\pi = \pi_{1} \circ \pi_{2}.$ We claim that $\pi_{2}$ is surjective.

Proof for surjectivity of $\pi_{2}$. First note that $\pi_{2}$ is given by a finite $\mathbb{C}$-algebra extension to its image, which is a closed subset of the scheme in the middle. By Krull's principal ideal theorem that $$\dim\left(\mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n}, t]}{(F(x_{1}, \dots, x_{n}, t))} \right)\right) = n.$$ Hence, if $\pi_{2}$ is not surjective, then $\pi_{2}(X)$ is a proper closed subset of this $n$-dimensional irreducible scheme (which is where we use our smart choice of $F$ being irreducible), so $\dim(\pi_{2}(X)) < n.$ However, denoting $B := \mathbb{C}[\boldsymbol{x}, t]/(F(\boldsymbol{x}, t)),$ the finite $\mathbb{C}$-algebra map $B \rightarrow A$ inducing $\pi_{2}$ factors as $$B \twoheadrightarrow B/J \hookrightarrow A,$$ where the latter one is a finite extension. This gives $\pi_{2},$ which breaks into $$X = \mathrm{Spec}(A) \twoheadrightarrow \mathrm{Spec}(B/J) \hookrightarrow \mathrm{Spec}(B),$$ so $\mathrm{Spec}(B/J) = \pi_{2}(X).$ Since $B/J \hookrightarrow A$ is a finite extension, we have $\dim(\pi_{2}(X)) = \dim(B/J) = \dim(A) = n$ (e.g., Vakil 11.1.E), so it would contradict what we had earlier: $\dim(\pi_{2}(X)) < n.$ This means that $\pi_{2}$ must be surjective.

There is more to the argument above. That is, we see that $$V(J) = V(0) = \mathrm{Spec}(B),$$ so $\sqrt{J} = \sqrt{(0)}$ in $B.$ Since $B$ is a domain, this implies that $\sqrt{J} = (0),$ so $J = (0).$ Thus, after all $\pi_{2}$ is induced by a finite extension $B \hookrightarrow A$ of $\mathbb{C}$-algebras.

Now, note that for any closed point $\mathfrak{m} \in X = \mathrm{Spec}(A),$ we have $$\pi_{2}(\mathfrak{m}) = (\pi(\mathfrak{m}), f(\mathfrak{m})) \in V(F)(\mathbb{C}) \subset \mathbb{C}^{n+1},$$ where $f(\mathfrak{m})$ is the image of $f$ in $A/\mathfrak{m} \simeq \mathbb{C}.$ This gives us $F(\pi(\mathfrak{m}), f(\mathfrak{m})) = 0.$ Since $f(p) = 0,$ we have $F(\pi(p), 0) = 0.$ Recall that $$F(\boldsymbol{x}, t) = t^{d} + a_{1}(\boldsymbol{x})t^{d-1} + \cdots + a_{d-1}(\boldsymbol{x})t + a_{d}(\boldsymbol{x}),$$ which lets us have $a_{d}(\pi(p)) = 0.$ For any $q \in \mathbb{C}^{n},$ denote by $w_{1}(q), \dots, w_{d}(q) \in \mathbb{C}$ the roots of $F(q, t)$ in $t$ allowing multiplicity. We have $$|a_{d}(q)| = |w_{1}(q)| \cdots |w_{d}(q)|.$$ Since $u_{m} \rightarrow \pi(p),$ we have $$a_{d}(u_{m}) \rightarrow a_{d}(\pi(p)) = 0,$$ as $m \rightarrow \infty.$ In other words, we have $$\lim_{m \rightarrow \infty} \left( |w_{1}(u_{m})| \cdots |w_{d}(u_{m})| \right) = 0.$$ For each $m,$ pick $t_{m} \in \{w_{1}(u_{m}), \dots, w_{d}(u_{m})\}$ so that $$|t_{i}| = \min\{|w_{1}(u_{m})|, \dots, |w_{d}(u_{m})|\}.$$ This means that $$|t_{m}|^{d} \leq |w_{1}(u_{m})| \cdots |w_{d}(u_{m})|,$$ so $t_{m} \rightarrow 0$ as $m \rightarrow \infty.$ Since $t_{m} = w_{i}(u_{m})$ for some $i,$ we have $F(u_{m}, t_{m}) = 0$ as well. Let's rewrite these two properties:

• $\lim_{m \rightarrow 0}t_{m} = 0$ in $\mathbb{C}$ and
• $F(u_{m}, t_{m}) = 0$ for all $m \geq 0.$

The second property tells us that $(u_{m}, t_{m}) \in V(F)(\mathbb{C}).$ Since $\pi_{2}$ induced a surjective map $X(\mathbb{C}) \twoheadrightarrow V(F)(\mathbb{C}),$ there exists some $y_{m} \in \pi_{2}^{-1}((u_{m}, t_{m})) \subset X(\mathbb{C}).$ Note that this means two things:

• $\pi(y_{m}) = u_{m}$ and
• $f(y_{m}) = t_{m}.$

In particular, the second property gives us $\lim_{m \rightarrow \infty}f(y_{m}) = 0.$

Claim. The sequence $(y_{m})_{m_{\geq 0}}$ is sits inside a compact subset of $X(\mathbb{C}).$ In particular, it has a convergent subsequence (e.g., Munkres Theorem 28.2, because the topology on $X(\mathbb{C})$ is given by the Euclidean metric).

For now, suppose that we have proven the claim. Then we exchange $(y_{m})$ with its convergent subsequence provided by the claim to assume that $y_{m} \rightarrow y$ for some $y \in X(\mathbb{C}),$ as $m \rightarrow \infty.$ Since $\pi$ is continuous with respect to the classical topology on the sets of $\mathbb{C}$-points, we have $$\pi(p) = \lim_{m \rightarrow \infty}u_{m} = \lim_{m \rightarrow \infty}\pi(y_{m}) = \pi(y).$$ That is, we have $$y \in \pi^{-1}(\pi(p)) = \{p = \mathfrak{p}_{1}, \dots, \mathfrak{p}_{r}\}.$$ What's surprising is that $y$ is necessarily equal to $p.$

To see this, recall that we have chosen $f \in A$ such that $f \in \mathfrak{p}_{1} = p$ and $f \notin \mathfrak{p}_{2}, \dots, \mathfrak{p}_{r}.$ The corresponding regular function $$f : X = \mathrm{Spec}(A) \rightarrow \mathbb{A}^{1}$$ induces a continuous function $f : X(\mathbb{C}) \rightarrow \mathbb{C}$ with respect to the classical topologies, and $f(p) = 0$ while $f(\mathfrak{p}_{i}) \neq 0$ for $i \geq 2.$ Therefore, we have $$f(y) = \lim_{m \rightarrow \infty}f(y_{m}) = 0,$$ but as $y \in \pi^{-1}(\pi(p))$ and the only point at which $f$ vanishes is $p,$ so we must have $y = p.$ Hence, it only remains to show the claim.

Step 4. Proof of the claim.

Now, since $A$ is finitely generated $\mathbb{C}$-algebra, we may choose finitely many generators $h_{1}, \dots, h_{N} \in A.$ Then we have the $\mathbb{C}$-algebra surjection $$\mathbb{C}[y_{1}, \dots, y_{N}] \twoheadrightarrow \mathbb{C}[h_{1}, \dots, h_{N}] = A,$$ where each indeterminate $y_{i}$ mapsto $h_{i},$ which gives us a closed embedding $$X = \mathrm{Spec}(A) \hookrightarrow \mathbb{A}^{N}.$$ More concretely, on the sets of $\mathbb{C}$-points, each $x \in X(\mathbb{C})$ maps to $(h_{1}(x), \dots, h_{N}(x)) \in \mathbb{C}^{n}$ under this map. This map is a closed homeomorphic embedding $X(\mathbb{C}) \hookrightarrow \mathbb{C}^{n},$ with respect to the classical topology. To show the claim, it is enough to show that the image $(h_{1}(y_{m}), \dots, h_{N}(y_{m}))_{m \geq 0}$ of the sequence $(y_{m})_{m \geq 0}$ under this embedding is bounded in $\mathbb{C}^{n}.$ (This will finish the proof because any closed subset of a compact set is compact.) For this, we are in $\mathbb{C}^{n},$ so it is enough to bound the sequence $(h_{i}(y_{m}))_{m \geq 0}$ in $\mathbb{C}$ for each $1 \leq i \leq N.$

Recall that we have a finite (and hence integral) extension $$\mathbb{C}[x_{1}, \dots, x_{n}] \hookrightarrow A$$ given by Nother normalization. Thus, for each $h_{i},$ we may find a monic polynomial $$H_{i}(\boldsymbol{x}, t) = t^{d_{i}} + a_{i,1}(\boldsymbol{x})t^{d_{i}-1} + \cdots + a_{i,d_{i}-1}(\boldsymbol{x})t + a_{i,d_{i}}(\boldsymbol{x}) \in \mathbb{C}[\boldsymbol{x}, t]$$ such that $H_{i}(\boldsymbol{x}, h_{i}) = 0$ in $A.$ This factors $\mathbb{C}[\boldsymbol{x}] \hookrightarrow A$ as $$\mathbb{C}[\boldsymbol{x}] \rightarrow \mathbb{C}[\boldsymbol{x}, t]/(H_{i}(\boldsymbol{x}, t)) \rightarrow A,$$ where the second map is given by $t \mapsto h_{i}.$ This induces $\mathbb{C}$-scheme maps $$X = \mathrm{Spec}(A) \rightarrow \mathbb{C}[\boldsymbol{x}, t]/(H_{i}(\boldsymbol{x}, t)) \rightarrow \mathbb{A}^{n},$$ which compose to $\pi,$ and the first map gives $$x \mapsto (\pi(x), h_{i}(x)) \in V(H_{i})(\mathbb{C}) \subset \mathbb{C}^{n+1},$$ for any $x \in X(\mathbb{C}).$ Therefore, taking $x = y_{m},$ we see that $h_{i}(y_{m}) \in \mathbb{C}$ is a root of $$H_{i}(\pi(y_{m}), t) = t^{d_{i}} + a_{i,1}(\pi(y_{m}))t^{d_{i}-1} + \cdots + a_{i,d_{i}-1}(\pi(y_{m}))t + a_{i,d_{i}}(\pi(y_{m})) \in \mathbb{C}[t].$$ Since each $a_{i,j}(\boldsymbol{x})$ is a polynomial in $\mathbb{C}[\boldsymbol{x}]$ we have continuity to help us conclude that $$\lim_{m \rightarrow \infty}a_{i,j}(\pi(y_{m})) = \lim_{m \rightarrow \infty}a_{i,j}(u_{m}) = \lim_{m \rightarrow \infty}a_{i,j}(\pi(p)) \in \mathbb{C}.$$ In particular, each $(a_{i,j}(\pi(y_{m})))_{m \geq 0}$ is a bounded sequence in $\mathbb{C}.$ Thus, the following lemma finishes the proof.

Lemma. Let $(a_{1,m})_{m \geq 0}, \dots, (a_{d,m})_{m \geq 0}$ be bounded sequences in $\mathbb{C}.$ Then the roots of $f(z) = z^{d} + a_{1,m}z^{d-1} + \cdots + a_{d-1,m}z + a_{d,m} \in \mathbb{C}[z]$ are bounded in $\mathbb{C}.$

Proof of Lemma. We will use Rouché's theorem. Let $R \gg 1$ such that $|a_{1,m}| + \dots + |a_{d,m}| < R/d$ for all $m \geq 0.$ Then we have $$\begin{align*}|a_{1,m}R^{d-1} + \cdots + a_{d-1,m}z + a_{d,m}| &\leq |a_{1,m}|R^{d-1} + \cdots + |a_{d-1,m}|R + |a_{d,m}| \\ & \leq |a_{1,m}|R^{d-1} + \cdots + |a_{d-1,m}|R^{d-1} + |a_{d,m}|R^{d-1} \\ &< R^{d}.\end{align*}$$ Thus, by Rouché's theorem, the number of roots of the polynomial $z^{d}$ is equal to the number of the roots of $f(z)$ in the open disk $D(0; R) = \{a \in \mathbb{C} : |a| < R\},$ counting with multiplicity. The first one has $d$ roots in $D(0; R),$ so all the roots of the polynomial $f(z)$ must be in $D(0; R).$ This finishes the proof. $\Box$

We are done! $\Box$

Constructable sets. Recall that constructible subsets of a Noetherian topological space $X$ are generated by open subsets of $X$ by taking finite intersections and complements. A constructible subset is precisely a finite disjoint union of locally closed subsets, each of which is the intersection of one open subset and one closed subset (Vakil 7.4.A).

Remark. A locally closed subscheme of a scheme $X$ seems to be a more subtle notion: $Z \overset{\text{closed}}{\hookrightarrow} U \overset{\text{open}}{\hookrightarrow} X$. Vakil 8.1.M discusses why it is subtle. He references Stacks Project for an example that prohibits us from changing the order of open and closed embeddings, but I have never taken a look at this.

What's important about constructible subsets is that their structures are preserved when we push them forward with reasonable morphisms.

Chevalley's theorem (Vakil 7.4.2). Let $\pi : X \rightarrow Y$ be finite type morphism of Noetherian schemes. For any constructible subset $W \subset X,$ the image $\pi(X) \subset Y$ is constructible.

Now, let's come back to our discussion.

Lemma. Let $X$ be any $\mathbb{C}$-variety and $W \subset X$ a constructible subset. Then taking closure of $W(\mathbb{C})$ in the classical topology yields the same set as taking closure of it in the Zariski topology.

Proof. Denote by $\overline{W(\mathbb{C})}^{\mathrm{an}}$ the closure of $W(\mathbb{C})$ with respect to the classical topology and $\overline{W(\mathbb{C})}^{\mathrm{Zar}}$ using the Zariski topology. In any topology, the closure of a subset $S$ of a topological space $T$ means that the smallest closed subset of $T$ containing $S.$ Since the classical topology is finer (i.e., having more closed subsets) taking the closure in it can get only smaller, so we have $$\overline{W(\mathbb{C})}^{\mathrm{an}} \subset \overline{W(\mathbb{C})}^{\mathrm{Zar}}.$$ Note that so far $W$ could have been any subset of $X.$

For the opposite inclusion, we use that $W$ is constructible. This lets us write $$W = (U_{1} \cap Z_{1}) \sqcup \cdots \sqcup (U_{r} \cap Z_{r}),$$ where $U_{i}$ are open and $Z_{i}$ are closed in $X.$ Without loss of generality, we may assume each $U_{i} \cap Z_{i} \neq \emptyset.$ We have $$\overline{W} = \overline{(U_{1} \cap Z_{1})} \cup \cdots \cup \overline{(U_{r} \cap Z_{r})}$$ in $X$ with respect to the Zariski topology. Each $Z_{i}$ is a finite union of irreducible components say $Z_{i} = Z_{i,1} \cup \cdots \cup Z_{i, s_{i}},$ so we may collect all nonempty $U_{i} \cap Z_{i,j}$ and take the union and call it $D.$ Note that $D$ is a dense open subset of $\overline{W},$ so Corollary on top of this posting implies that $D(\mathbb{C})$ is dense in $\overline{W}(\mathbb{C}) = \overline{W(\mathbb{C})}^{\mathrm{Zar}}$ in the classical topology. Since we have $$D(\mathbb{C}) \subset W(\mathbb{C}) \subset \overline{W(\mathbb{C})}^{\mathrm{Zar}},$$ this implies that we have $$\overline{W(\mathbb{C})}^{\mathrm{an}} = \overline{\overline{W(\mathbb{C})}^{\mathrm{Zar}}}^{\mathrm{an}} \supset \overline{W(\mathbb{C})}^{\mathrm{Zar}},$$ as desired. $\Box$

Theorem. Let $X$ be any $\mathbb{C}$-variety.

(1) $X$ is separated if and only if $X^{\mathrm{an}}$ is Hausdorff.

(2) $X$ is complete if and only if $X^{\mathrm{an}}$ is compact and Hausdorff.

(3) Given a morphism $f : X \rightarrow Y$ of separated varieties, the following are equivalent:

• $f$ is proper (algebraically)
• $f^{\mathrm{an}}$ is proper (topologically)

Remark. We say a continuous map $\phi : Y \rightarrow Z$ is proper if any compact Hausdorff subset of $Z$ pulls back to a compact Hausdorf subset of $Y.$

Proof of (1). In general, the diagonal map $\Delta : X \rightarrow X \times X$ is a locally closed embedding (i.e., a closed embedding and then an open embedding; see Vakil 10.1.3). Hence $\Delta$ is a closed embedding if and only if $\Delta(X) \subset X \times X$ is Zariski-closed. (This is generally true; see Vakil 10.1.4). Since $\Delta(X)$ is a constructible subset of $X \times X,$ we have the following chain of logical equivalences:

• $X$ is separated;
• $\Delta$ is a closed embedding;
• $\Delta(X) \subset X \times X$ is Zariski-closed;
• $\Delta(X)^{\mathrm{an}} \subset (X \times X)^{\mathrm{an}} = X^{\mathrm{an}} \times X^{\mathrm{an}}$ is closed;
• $X$ is Hausdorff.

This finishes the proof (for (1)). $\Box$

Remark. A relevant result to the first theorem in this posting is as follows: if $X$ is an irreducible $\mathbb{C}$-variety, then $X^{\mathrm{an}}$ is connected.

My thought. The last remark will probably be proven by comparision of cohomologies.

Hodge theory: Lecture 1

Before starting. These are notes for a class I sit in. I decided to spend some time retyping it so that I can distinguish what I know and what I blackbox. I would like to thank Devlin Mallory who was willing to his live-TeXed notes for the class. I won't reveal the name of the lecturer in case they did not want to be responsible for mistakes. Of course, there is a chance that I can stop going to the class, but I will try my best to keep up with it.

Chapter I. Classical topology on varieties over $\mathbb{C}$

By a variety over $\mathbb{C}$, we mean a reduced scheme of finite type over $\mathbb{C}$.

Classical topology on an affine variety. If $X$ is an affine complex variety, then $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})) \hookrightarrow \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]) = \mathbb{A}^{n}.$$ Note that $$X(\mathbb{C}) = \{\boldsymbol{a} = (a_{1}, \dots, a_{n}) \in \mathbb{C}^{n} : f_{1}(\boldsymbol{a}) = \cdots = f_{r}(\boldsymbol{a}) = 0\} \subset \mathbb{C}^{n} = \mathbb{A}^{n}(\mathbb{C}).$$ We define the classical topology on $X(\mathbb{C})$ to be the subspace topology of the Euclidean topology on $\mathbb{C}^{n}$. So far, this topology depends on the closed embedding $X \hookrightarrow \mathbb{A}^{n}$ of $\mathbb{C}$-schemes.

Classical topology on affine complex variety does not depend on the choice of a closed embedding. To see this, say $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})) \simeq \mathrm{Spec}(\mathbb{C}[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{s}))$$ as $\mathbb{C}$-schemes. This is given by an isomorphism $$\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}) \simeq \mathbb{C}[y_{1}, \dots, y_{m}]/(g_{1}, \dots, g_{s})$$ of $\mathbb{C}$-algebras. Say $x_i \mapsto G_i(\boldsymbol{y})$ and $y_j \mapsto F_j(\boldsymbol{x})$ under this isomorphism. Then we have two maps $G : \mathbb{C}^{n} \rightarrow \mathbb{C}^{m}$ and $F : \mathbb{C}^{m} \rightarrow \mathbb{C}^{n}$ that when restricted to $V(f_{1}, \dots, f_{r})(\mathbb{C})$ and $V(g_{1}, \dots, g_{s})(\mathbb{C})$, they are set-theoretically inverses to each other. Since these maps are given by polynomials, they are continuous with respect to the Eucleadian topologies coming from $\mathbb{C}^{n}$ and $\mathbb{C}^{m}.$ Thus, we have constructed a homeomorphism $V(f_{1}, \dots, f_{r})(\mathbb{C}) \simeq V(g_{1}, \dots, g_{s})(\mathbb{C}),$ as desired.

Properties. So far $X$ is an affine variety over $\mathbb{C}.$

(1) The classical topology on $X(\mathbb{C})$ is finer than the Zariski topology on $X(\mathbb{C})$ (i.e., the subspace topology $X(\mathbb{C}) \subset X$).

This is quite evident. Every zero set of a set of finitely many polynomials in $n$ variables in $\mathbb{C}^{n}$ is closed in the Euclidean topology of $\mathbb{C}^{n}$, and intersecting such locus with $X(\mathbb{C})$ is another zero set of finitely many polynomials.

(2) If $Z \hookrightarrow X$ is a closed subvariety, then the classical topology on $Z(\mathbb{C})$ is the subspace topology given by $Z(\mathbb{C}) \subset X(\mathbb{C}).$

First, for any $\mathbb{C}$-scheme map $\mathrm{Spec}(\mathbb{C}) \rightarrow Z,$ we get the $\mathbb{C}$-scheme map $$\mathrm{Spec}(\mathbb{C}) \rightarrow Z \hookrightarrow X.$$ This gives us $$Z(\mathbb{C}) \hookrightarrow X(\mathbb{C}).$$ We may choose a closed embedding $$X \hookrightarrow \mathbb{A}^{n}$$ of $\mathbb{C}$-schemes, which gives a closed embedding $Z \hookrightarrow X \hookrightarrow \mathbb{A}^{n}.$ We can choose a presentation of $Z$ by adding more polynomials in the presentation of $X$. Say $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})$$ and $$Z = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}, f_{r+1}, \dots, f_{s}).$$ Then $$Z(\mathbb{C}) = V(f_{1}, \dots, f_{s})(\mathbb{C}) \subset V(f_{1}, \dots, f_{r})(\mathbb{C}) = X(\mathbb{C}) \subset \mathbb{C}^{n}.$$ The classical topology of $Z(\mathbb{C})$ is given by $\mathbb{C}^{n}$ and it is closed in this sense. Since $X(\mathbb{C})$ is also closed in $\mathbb{C}^{n},$ this implies that the classical topology of $Z(\mathbb{C})$ can be regarded as the subspace $X(\mathbb{C}).$

(3) If $U \hookrightarrow X$ is an affine open subvariety, then the classical topology on $U(\mathbb{C})$ is given by the subspace topology of $X(\mathbb{C}).$

My first thought. We can take $Z := X \setminus U.$ We give the unique reduced closed subscheme structure on $Z.$ Since $Z(\mathbb{C})$ gets a subspace topology $X(\mathbb{C}),$ I thought $U(\mathbb{C}) = X(\mathbb{C}) \setminus Z(\mathbb{C})$ would get one too.

No! But the "classical topology" by definition mean that we are choosing a closed embedding from $U$ to an affine space, and get a topology from there. How do we know that such a topology would match with the topology on $U(\mathbb{C})$ given by that of $X(\mathbb{C})$ and $Z(\mathbb{C})$? This requires a proof.

Oh no, but $U$ is open, how can we find a closed embedding? Well, the key here is that $U$ is an affine scheme! Thus, it is quasi-compact, so $$U = D(g_{1}) \cup \cdots \cup D(g_{s}) = D(g) \subset X,$$ where $g = g_{1} \cdots g_{s}$ with $g_{i} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})$ if we choose our presentation of $X$ as follows: $$X = \mathrm{Spec}(\mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r})).$$ Now, note that $$D(g) \simeq \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{g}.$$ For computation, let us write $\bar{g}$ instead of $g$ and say $g \in \mathbb{C}[x_{1}, \dots, x_{n}]$ is a lift of $\bar{g} \in \mathbb{C}[x_{1}, \dots, x_{n}]/(f_{1}, \dots, f_{r}).$ Denote $\boldsymbol{x} = (x_{1}, \dots, x_{n}).$ Then we have a $\mathbb{C}$-algebra isomorphism $$\left(\frac{\mathbb{C}[x_{1}, \dots, x_{n}]}{(f_{1}, \dots, f_{r})}\right)_{\bar{g}} \simeq \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)},$$ given by $x_{i} \mapsto x_{i}.$ The inverse is given by $x_{n+1} \mapsto 1/\bar{g}.$ Thus, we can construct a closed embedding $$U = D(\bar{g}) \simeq \mathrm{Spec}\left( \frac{\mathbb{C}[x_{1}, \dots, x_{n},x_{n+1}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)} \right) \hookrightarrow \mathbb{A}^{n+1}.$$ This implies that the classical topology on $U(\mathbb{C})$ is given by the subspace topology $$U(\mathbb{C}) \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \subset \mathbb{C}^{n+1}.$$ The isomorphism written here is a bit funny precisely because the affine scheme structure of $D(\bar{g})$ is funny. It is more like this isomorphism is declaring the structure on $U(\mathbb{C})$ with the classical topology.

What do I mean? A priori, when we first learn schemes, we don't know whether $D(g)$ is an affine scheme for a given element $g$ in a commutative ring $R$. We understand that $D(g) \simeq \mathrm{Spec}(R_{g})$ is a homeomorphism and declare the affine scheme structure on $D(g)$ using this homeomorphism.

As Zariski topological spaces and also $\mathbb{C}$-schemes (because the reduced structure for a closed subscheme is unique), we have $$U = D(\bar{g}) \simeq D(g) \cap V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n},$$ so we may assume $$U = D(g) \cap V(f_{1}, \dots, f_{r}).$$ Similarly, we may assume $$X = V(f_{1}, \dots, f_{r}) \subset \mathbb{A}^{n}.$$ If we follow the $\mathbb{C}$-scheme isomorphism $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1) \simeq D(g(\boldsymbol{x})) \cap V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))$$ we get $$(x_{1} - a_{1}, \dots, x_{n} - a_{n}, x_{n+1} - a_{n+1}) \mapsto (x_{1} - a_{1}, \dots, x_{n} - a_{n}),$$ on the sets of $\mathbb{C}$-points. This coincides with the map given by the restriction of $\mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n}$ forgetting the last coordinate. This map is continuous, so the following bijection from left to right is continuous: $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), x_{n+1}g(\boldsymbol{x})-1)(\mathbb{C}) \simeq D(g)(\mathbb{C}) \cap X(\mathbb{C}) = U(\mathbb{C}),$$ where the right-hand side gets the subspace topology of the classical topology on $X(\mathbb{C}).$ When $g = 0 \in \mathbb{C}[\boldsymbol{x}],$ both sides are empty, so the bijection is a homeomorphism. Otherwise, we can take $(a_{1}, \dots, a_{n}) \mapsto (a_{1}, \dots, a_{n}, 1/g(a_{1}, \dots, a_{n}))$ as the continuous inverse. Thus, the bijection is a homeomorphism, so the classical topology on $U(\mathbb{C})$ coincides with the subspace topology in $X(\mathbb{C}).$

Classical topology on global variety (cf. 5.3.G of Vakil's book). Now let $X$ be any variety over $\mathbb{C}$. For instance, this includes reduced projective schemes over $\mathbb{C},$ which are rarely affine. To define the classical topology on $X(\mathbb{C}),$ let $X = \bigcup_{i \in I}U_{i}$ be an affine open cover. For any distinct $i, j \in I,$ we may have two different subspace topologies on $U_{i}(\mathbb{C}) \cap U_{j}(\mathbb{C}),$ one from $U_{i}(\mathbb{C})$ and another from $U_{j}(\mathbb{C}).$ Both induce the same topology because for any affine open $V \subset U_{i} \cap U_{j},$ we know that the classical topology on $V(\mathbb{C})$ is the same as the subspace topology either from $U_{i}(\mathbb{C})$ or $U_{j}(\mathbb{C}).$ We may use the same argument for the triple intersections, so we can glue $U_{i}(\mathbb{C})$ with the classical topology we studied above to give a unique topology on $X(\mathbb{C}).$ This topology is called the classical topology on $X(\mathbb{C}).$ More specifically, if we fix any affine open cover $X = \bigcup_{i \in I}U_{i},$ any subset $S \subset X(\mathbb{C})$ is open in the classical topology if and only if $S \cap U_{i}(\mathbb{C})$ is open for all $i \in I.$ The classical topology on $X(\mathbb{C})$ does not depend on the choice of an affine open cover of $X$ because if there are two, we may take the common refinement. In particular, for any affine open subset $U \subset X,$ we see that $U(\mathbb{C})$ is open in $X(\mathbb{C}).$

Note that (1) the classical topology on $X(\mathbb{C})$ is finer than the Zariski topology, (2) for any closed subvariety $Z \hookrightarrow X,$ the classical topology on $Z(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}),$ and (3) for any open subvariety $U \hookrightarrow X,$ the classical topology on $U(\mathbb{C})$ coincides with the subspace topology of $X(\mathbb{C}).$ (Proof: take an affine open cover of $X$ and intersect!)

Functoriality (cf. 6.3.N of Vakil's book). Any morphism $X \rightarrow Y$ of $\mathbb{C}$-varieties induces $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ by $$\mathrm{Spec}(\mathbb{C}) \rightarrow X \rightarrow Y.$$ Due to the description above, it is evident that this is a functor. The source of this functor is the category $\mathbf{Var}_{\mathbb{C}}$ of $\mathbb{C}$-varieties. What is the target category of this functor? Surely, we can start with the category $\textbf{Set}$ of sets. One can check that we can narrow down to the category $\textbf{Top}$ of topological spaces (i.e., the induced morphism $X(\mathbb{C}) \rightarrow Y(\mathbb{C})$ is continuous). Whichever category it sits in we denote this functor by $X \mapsto X^{\mathrm{an}}$ or maybe $$(\pi : X \rightarrow Y) \mapsto (\pi^{\mathrm{an}} : X^{\mathrm{an}} \rightarrow Y^{\mathrm{an}}).$$ Let's call this the analytification functor.

Guess. It seems like the class will soon talk about narrowing the target category down to the category of $\mathbb{C}$-analytic spaces with holomorphic mappings, whatever those mean. Then it will probably discuss GAGA theorems, which seems to be saying that this fuctor becomes something close to a fully faithful functor. (Probably not exactly, but something more interesting!)

In particular, a regular function on $X$ can also get analytified (and we know that it is at least continuous with the classical topology. Indeed, given any $f \in \Gamma(X, \mathscr{O}_{X}),$ the $\mathbb{C}$-algebra map $\mathbb{C}[t] \rightarrow \Gamma(X, \mathscr{O}_{X})$ given by $t \mapsto f$ gives rise to a $\mathbb{C}$-scheme map $X \rightarrow \mathbb{A}^{1},$ which sends all points where $f$ vanishes (not just closed points) are sent to $(t),$ the origin. We then analytify this.

Thought. It would be interesting if this turn out to be (something close to) the way we use to analytify the structure sheaf $\mathscr{O}_{X}.$

Remark. Every point in $X(\mathbb{C}) = X^{\mathrm{an}}$ has a countable basis of open neighborhoods in the classical topology. (I don't think I understand what this means. Perhaps this just means that I can just take an affine Zariski open subset containing that point and take its analytification.)

Analytification commutes with product. This asserts that $$(X \times Y)^{\mathrm{an}} \simeq X^{\mathrm{an}} \times Y^{\mathrm{an}}.$$ Of course, this would only make sense as a homeomorphism so far, unless we think about morphisms in a narrower category than $\textbf{Top}.$

First, note that we get $$(X \times Y)(\mathbb{C}) \simeq X(\mathbb{C}) \times Y(\mathbb{C})$$ in $\textbf{Set}.$ Why? It is because getting a map from $\mathrm{Spec}(\mathbb{C})$ to $X \times Y$ in $\textbf{Var}_{\mathbb{C}}$ is precisely the same as getting a pair of maps to $\mathrm{Spec}(\mathbb{C}) \rightarrow X$ and $\mathrm{Spec}(\mathbb{C}) \rightarrow Y.$ Next, we show that the classical topologies on both side matches. Given any affine opens $U \subset X$ and $V \subset Y,$ we choose closed embeddings $U \hookrightarrow \mathbb{A}^{m}$ and $V \hookrightarrow \mathbb{A}^{n}.$ Then we get the closed embedding $U \times V \hookrightarrow \mathbb{A}^{m + n}.$ The last map is induced categorically, but we check that this is a closed embedding using affine presentations. That is, write $$U = \mathrm{Spec}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right)$$ and $$V = \mathrm{Spec}\left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right).$$ We have $$\begin{align*}\left(\frac{\mathbb{C}[x_{1}, \dots, x_{m}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))}\right) & \otimes_{\mathbb{C}} \left(\frac{\mathbb{C}[y_{1}, \dots, y_{n}]}{(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}\right) \\ &\simeq \frac{\mathbb{C}[x_{1}, \dots, x_{m}, y_{1}, \dots, y_{n}]}{(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))}, \end{align*}$$ so the closed embedding is given by $$U \times V \simeq V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})) \subset \mathbb{A}^{m+n}.$$ Hence, the classical topology on $(U \times V)(\mathbb{C})$ is given by the subspace topology of $$V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C}) \subset \mathbb{C}^{m+n}.$$ Since $$V\left(\begin{array}{c}f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}), \\ g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y})\end{array}\right)(\mathbb{C}) = V(f_{1}(\boldsymbol{x}), \dots, f_{r}(\boldsymbol{x}))(\mathbb{C}) \times V(g_{1}(\boldsymbol{y}), \dots, g_{s}(\boldsymbol{y}))(\mathbb{C})$$ in $\mathbb{C}^{m+n},$ we see that the classical topology on $(U \times V)(\mathbb{C})$ is given by the product topology on $U(\mathbb{C}) \times V(\mathbb{C}).$ Since $X \times Y$ can be covered by $U \times V,$ the topology on $(X \times Y)(\mathbb{C})$ can be given by gluing those of $(U \times V)(\mathbb{C}) = U(\mathbb{C}) \times V(\mathbb{C}),$ and since this identity can be seen as the restriction of the bijection $$(X \times Y)(\mathbb{C}) \leftrightarrow X(\mathbb{C}) \times Y(\mathbb{C}),$$ we see that this bijection is a homemomorphism. Thus, the classical topology on $(X \times Y)(\mathbb{C})$ can be given by the product topology on $X(\mathbb{C}) \times Y(\mathbb{C}),$ where $X(\mathbb{C})$ and $Y(\mathbb{C})$ are given the classical topologies.

More thought. I am sure there is more structure to the product identity we have seen than merely as a homeomorphism, but so far at least we know the following. If we consider the analytification functor $\mathrm{an} : \textbf{Var}_{\mathbb{C}} \rightarrow \textbf{Top},$ with the target being just the category of topological spaces with continuous maps, then the following diagram of functors commute:

$\require{AMScd} \begin{CD}  \textbf{Var}_{\mathbb{C}}^{2} @>{\times}>> \textbf{Var}_{\mathbb{C}} \\ @V\mathrm{an}^{2}VV @V\mathrm{an}VV  \\ \textbf{Top}^{2} @>{\times}>> \textbf{Top}. \end{CD}$

Of course, one needs to check that the functors commute for the suitable morphisms as well, but this seems to go through without any pain.

Next time. We preview what we will discuss next time.

Theorem. Let $X$ be an irreducible $\mathbb{C}$-variety. If $U \subset X$ is nonempty and Zariski-open, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.

Corollary. Let $X$ be a $\mathbb{C}$-variety. If $U \subset X$ is Zariski-open and Zariski-dense, then $U(\mathbb{C}) \subset X(\mathbb{C})$ is dense in the classical topology.

Proof. Since $X$ is Noetherian, we may write $X = X_{1} \cup \cdots \cup X_{r},$ where $X_{i}$ are distinct irreducible components of $X.$ Write $$U = (U \cap X_{1}) \cup \cdots \cup (U \cap X_{r}).$$ Since $U \subset X$ is dense, taking the closure in $X$ gives $$X = \overline{(U \cap X_{1})} \cup \cdots \cup \overline{(U \cap X_{r})}.$$ This implies that $U \cap X_{i} \neq \emptyset$ for all $i,$ so  $U(\mathbb{C}) \cap X_{i}(\mathbb{C})$ is dense in $X_{i}(\mathbb{C})$ for all $i.$ Therefore, we have $U(\mathbb{C})$ dense in $X(\mathbb{C}),$ as we have the finite unions. $\Box$