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
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