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. $\Box$