1.10 Complex Variety

? Some Properties

1. Zariski open/closed $\subseteq$ Analytic open/closed

2. If $f:X\to Y$ is a morphism, then $f:X^{\mathrm{an}}\to Y^{\mathrm{an}}$ is a morphism.
3. locally closed yields induces topology
4. $X^{\mathrm{an}}\times Y^{\mathrm{an}}$ product topology is homeomorphic to $(X\times Y{)}^{\mathrm{an}}$.
5. $X$ variety yields $X^{\mathrm{an}}$ Hausdorff, because ${\Delta }_X$ is Zariski closed.

Theorem

Let $X$ be an variety, and let $U$ be a nonempty open subvariety. Then $U^{\mathrm{an}}\subseteq X^{\mathrm{an}}$ is dense.

In fact, if $U=X\setminus ({\cup }_{\text{countable}}V(I))$ instead of open, then $\overline{U^{\mathrm{an}}}=X^{\mathrm{an}}$. 贝尔纲定理？

\begin{proof} For an affine variety $X$ and $U=X\setminus V(I)$, we have $U\supseteq X\setminus V(f)=X_f$. $\overline{(X_f{)}^{\mathrm{an}}}=X$ means for any $f(x_0)=0$, there exists $z_n\in X_f$ such that $z_n\to z_0$.

next time: complete iff compact.

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补了上节课卡住的：X variety embedding ${\mathbb{C}}^n$ and ${\mathbb{C}}^m$ are equivalent.

Proposition

Let $X$ be a prevariety over $\mathbb{C}$. Then $X$ is variety iff $X^{\mathrm{an}}$ is Hausdorff.

\begin{proof} "→" Assume that $X$ is a variety, then $\Delta \subseteq X\times X$ is closed in Zariski topology and ${\Delta }^{\mathrm{an}}\subseteq (X\times X{)}^{\mathrm{an}}$ is closed in analytic topology. Note that $(X\times X{)}^{\mathrm{an}}=X^{\mathrm{an}}\times X^{\mathrm{an}}$, then $X^{\mathrm{an}}$ is Hausdorff.

"←" $\Delta \subseteq X\times X$ is a locally closed subvariety. Since affine variety is a variety, then for each affine $X_i$, ${\Delta }_i\subseteq X_i\times X_i$ is closed. Write $X={\cup }_{i=1}^{\ell }{X}_i$, then $\Delta ={\cup }_{i=1}^{\ell }{\Delta }_i$ where ${\Delta }_i$ is closed in $X_i\times X_i$.

Theorem

Let $X$ be a variety over $\mathbb{C}$. Then $X$ is complete iff $X^{\mathrm{an}}$ is compact.

\begin{proof} "←" Assume that $X^{\mathrm{an}}$ is compact. It suffices to show for any variety $Y$, closed $Z\subseteq X\times Y$ and $p:X\times Y\to Y$, we have $p(Z)$ is closed.

Claim that $p(Z{)}^{\mathrm{an}}$ is closed. It is enough to show for $\{y_n\}\subseteq p(Z)$ and $y_n\to y$ in $Y$, there is $y\in p(Z)$. Hence there exist $(x_n,y_n)\in Z$ such that $p(x_n,y_n)=y_n$. Since $X$ is compact, there exists a convergent subsequence $\{x_{n_k}\}$ and we assume that $x_{n_k}\to x$. Hence $(x_{n_k},y_{n_k})\to (x,y)$. Since $Z$ is closed, $(x,y)\in Z$ and so $y\in p(Z)$. Therefore, $p(Z{)}^{\mathrm{an}}$ is closed.

Then $p(Z)$ is a constructible set. There exist open $U_i$ and irreducible closed $Y_i$ such that $p(Z)={\cup }_{i=1}^{\ell }(U_i\cap Y_i)$. (这里好像不是因为constructible，但是Delta_i确实可以写成开集交闭集什么的？) Since ${\cup }_{i=1}^{\ell }{Y}_i\supseteq p(Z)=\overline{p(Z{)}^{\mathrm{an}}}\supseteq \cup {\overline{U_i\cap Y_i}}^{\mathrm{an}}=\cup Y_i^{\mathrm{an}}=\cup Y_i$. Hence $p(Z)=\cup Y_i$ is closed.

"→" We first prove if $X$ is projective, then $X^{\mathrm{an}}$ is compact. Define

$$\theta :S=\{(x_0,\cdots ,x_n):|x_0{|}^2+\cdots +|x_n{|}^2=1\}\to ({\mathbb{P}}^n)\mathrm{an},(x_0,\cdots ,x_n)\mapsto [x_0:\cdots :x_n]$$

then $\theta$ is surjective continuous in analytic topology, Since $S$ is compact, then $\theta (S)$ is compact.

If $X$ is complete, by Chow’s lemma, there exists a projective variety $Y$ such that $Y\stackrel{\pi }{\to }X$ is a birational morphism. Since $Y^{\mathrm{an}}$ is compact and $\pi$ is a continuous map, ${\pi }^{\mathrm{an}}(Y^{\mathrm{an}})=X^{\mathrm{an}}$ is compact. \end{proof}

15:20 下黑板：

• $X,Y$ birational, $U_x\subseteq X$ and$U_y\subseteq Y$ Zariski open such that $U_x\simeq U_y$（15:31 好像是证明
• $X$ variety yields $Y$ complete variety such that $X\subseteq Y$ is an open set

$X$ is a closed analytic variety (locally zeros of analytic functions), by Chow’s theorem, $Z$ is a Zariski subset.

15:48 从15:20到现在什么也没听懂。