Let $(X,L)$ be a compact polarized complex manifold of dimension $n$. Let $\varphi$ be a smooth positive metric on $L$. Define $\omega=dd^c\varphi$. We shall use $MA(\varphi)=\omega^n$ as the measure on $X$. Then there is a natural $L^2$-inner product on $H^0(X,L)$. Now fix $s\in H^0(X,L)$ of norm $1$. Consider the following set $$ A_\epsilon=\{s'\in H^0(X,L): \|s'\|=1, |(s,s')|<\epsilon\}, $$ where $\epsilon>0$. My question is: how can we get an upper bound of $$ MA(\varphi)\left(\bigcup_{s'\in A_{\epsilon}} Z_{s'}\right), $$ where $Z$ denotes the zero locus.


1 Answer 1


If I am not mistaken, it seems the set you are asking about is almost always $X$ itself (and the argument does not use anything about complex geometry).

For any $p$ in $X$, the vector subspace

$$A_p := \{ \sigma \in H^0(X,L) : \sigma(p) = 0 \}$$

has codimension at most 1 in $H^0(X,L)$.

On the other hand, for any $\epsilon$ your set $A_\epsilon$ contains the subset

$$A_0=\{s'\in H^0(X,L): \|s'\|=1, \ (s,s')=0\}.$$

Consider now the larger set

$$B_0=(s)^\perp = \{s'\in H^0(X,L): \ (s,s')=0\}.$$

Every element of $B_0\setminus \{0\}$ is a nonzero multiple of an element of $A_0$, so we have

$$\bigcup_{s'\in B_0 \setminus \{0\}} Z_{s'} = \bigcup_{s'\in A_0} Z_{s'} \subseteq \bigcup_{s'\in A_{\epsilon}} Z_{s'}.$$

But now $B_0$ is a vector subspace of $H^0(X,L)$ of codimension 1. So as long as $H^0(X,L)$ has dimension at least 3, for any $p \in X$ the intersection $A_p \cap (B_0 \setminus \{0\})$ will be nonempty, and so $p$ will be contained in $\bigcup_{s'\in B_0 \setminus \{0\}} Z_{s'}$ and hence in the set you are asking about.

  • $\begingroup$ Thanks. You are right. I was thinking about the case of (P^1,O(1)) and thought the general case would be similar. $\endgroup$ Apr 9, 2018 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.