1.2. Extension obstruction index [0255]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

1.2. Extension obstruction index

In this subsection, we introduce an invariant to describe the obstruction to the extension property. Let $X$ be a projective scheme over $\operatorname{Spec}k$ , $L$ be an invertible sheaf on $X$ equipped with a continuous metric $h$ , and $Y$ be a closed subscheme of $X$ . For any non-zero element $l$ of $H^{0}(Y,L|_{Y})$ , we denote by $\lambda_{h}(l)$ the following number (if there does not exist any section $s\in H^{0}(X,L^{\otimes n})$ extending $l^{\otimes n}$ , then the infimum in the formula is defined to be $+\infty$ by convention)

(3)

\lambda_{h}(l)=\limsup_{n\rightarrow+\infty}\inf_{\begin{subarray}{c}s\in H^{0}(X,L^{\otimes n})\\ {\left.{s}\right|_{{Y}}}=l^{\otimes n}\end{subarray}}{\bigg(\frac{\log\|s\|_{h^{n}}}{n}-\log\|l\|_{Y,h}\bigg)}\in[0,+\infty].

This invariant allows to describe in a numerically way the obstruction to the metric extendability of the section $l$ . In fact, the following assertions are equivalent:

(a)

$\lambda_{h}(l)=0$ ,
(b)

for any $\epsilon>0$ , there exists $n_{0}\in\mathbb{N}_{\geq 1}$ such that, for any integer $n\geq n_{0}$ , the element $l^{\otimes n}$ extends to a section $s\in H^{0}(X,L^{\otimes n})$ such that $\|s\|_{h}\leq e^{\epsilon n}\|l\|_{Y,h}^{n}$ .

The following proposition shows that, if $l^{\otimes n}$ extends to a global section of $L^{\otimes n}$ for sufficiently positive $n$ (it is the case notably when the line bundle $L$ is ample), then the limsup defining $\lambda_{h}(l)$ is actually a limit.

Proposition 1.1.

For any integer $n\geqslant 1$ , let

a_{n}=\inf_{\begin{subarray}{c}s\in H^{0}(X,L^{\otimes n})\\ {\left.{s}\right|_{{Y}}}=l^{\otimes n}\end{subarray}}{\Big(}\log\|s\|_{h^{n}}-n\log\|l\|_{Y,h}{\Big)}.

Then the sequence $(a_{n})_{n\geq 1}$ is sub-additive, namely one has $a_{m+n}\leq a_{m}+a_{n}$ for any $(m,n)\in\mathbb{N}_{\geq 1}$ . In particular, if for sufficiently positive integer $n$ , the section $l^{n}$ lies in the image of the restriction map $H^{0}(X,L^{\otimes n})\rightarrow H^{0}(Y,L|_{Y}^{\otimes n})$ , then “ $\limsup$ ” in (3) is actually “ $\lim$ ”.

Proof.

By (1), one has $a_{n}\geq 0$ for any integer $n\geq 1$ . Moreover, $a_{n}<+\infty$ if and only if $l^{n}$ lies in the image of the restriction map $H^{0}(X,L^{\otimes n})\rightarrow H^{0}(Y,L|_{Y}^{\otimes n})$ . To verify the inequality $a_{m+n}\leq a_{m}+a_{n}$ , it suffices to consider the case where both $a_{m}$ and $a_{n}$ are finite. Let $s_{m}$ and $s_{n}$ be respectively sections in $H^{0}(X,L^{\otimes m})$ and $H^{0}(X,L^{\otimes n})$ such that ${\left.{s_{m}}\right|_{{Y}}}=l^{\otimes m}$ and ${\left.{s_{n}}\right|_{{Y}}}=l^{\otimes n}$ , then the section $s=s_{m}\otimes s_{n}\in H^{0}(X,L^{\otimes(m+n)})$ verifies the relation ${\left.{s}\right|_{{Y}}}=l^{\otimes(n+m)}$ . Moreover, one has

\|s\|_{h}=\sup_{x\in X^{\mathrm{an}}}|s|_{h}(x)=\sup_{x\in X^{\mathrm{an}}}|s_{m}|_{h}(x)\cdot|s_{n}|_{h}(x)\leqslant\|s_{m}\|_{h}\cdot\|s_{n}\|_{h}.

Since $s_{m}$ and $s_{n}$ are arbitrary, one has $a_{m+n}\leq a_{m}+a_{n}$ . Finally, by Fekete’s lemma, if $a_{n}<+\infty$ for sufficiently positive integer $n$ , then the sequence $(a_{n}/n)_{n\geq 1}$ actually converges in $\mathbb{R}_{+}$ . The proposition is thus proved. ∎

Corollary 1.2.

Assume that the invertible sheaf $L$ is ample, then the following conditions are equivalent.

(a)

$\lambda_{h}(l)=0$ ,
(b)

for any $\epsilon>0$ , there exists $n\in\mathbb{N}_{\geq 1}$ and a section $s\in H^{0}(X,L^{\otimes n})$ such that ${\left.{s}\right|_{{Y}}}=l^{n}$ and that $\|s\|_{h}\leq e^{\epsilon n}\|l\|_{Y,h}$ .

Proof.

We keep the notation of the previous proposition. By definition the second condition is equivalent to

(4)

\liminf_{n\rightarrow+\infty}\frac{a_{n}}{n}=0.

Since $L$ is ample, Proposition 1.1 leads to the convergence of the sequence $(a_{n}/n)_{n\geq 1}$ in $\mathbb{R}_{+}$ . Hence the condition (4) is equivalent to $\lambda_{h}(l)=0$ . ∎