3.3. Semipositive metric [0277]

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

3.3. Semipositive metric

We assume that $L$ is semiample, namely certain tensor power of $L$ is generated by global sections. We say that a continuous metric $h=\{|\raisebox{1.72218pt}{.}|_{h}(x)\}_{x\in X^{\mathrm{an}}}$ is semipositive if there are a sequence $\{e_{n}\}$ of positive integers and a sequence $\{(V_{n},\|\raisebox{1.72218pt}{.}\|_{n})\}$ of normed finite-dimensional vector spaces over $k$ such that there is a surjective homomorphism $V_{n}\otimes_{k}\mathscr{O}_{X}\to L^{\otimes e_{n}}$ for every $n$ , and that the sequence

\left\{\frac{1}{e_{n}}\log\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(V_{n},\|\raisebox{1.20552pt}{.}\|_{n})}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}\right\}_{n=1}^{\infty}

converges to $0$ uniformly on $X^{\mathrm{an}}$ .

Proposition 3.10.

If $X$ is projective, $L$ is generated by global sections, and $h$ is semipositive, then the sequence

\left\{\frac{1}{m}\log\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{m}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{m}}(x)}\right\}_{m=1}^{\infty}

converges to $0$ uniformly on $X^{\mathrm{an}}$ .

Proof.

We set

a_{m}=\max_{x\in X^{\mathrm{an}}}\left\{\log\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{m}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{m}}(x)}\right\}.

Then $a_{m+m^{\prime}}\leq a_{m}+a_{m^{\prime}}$ by (3) in Lemma 3.5, and hence $\lim_{m\to\infty}a_{m}/m=\inf\{a_{m}/m\}$ by Fekete’s lemma. For $\epsilon>0$ , there is $e_{n}$ such that

e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)

for all $x\in X^{\mathrm{an}}$ , where $h_{n}=\big\{|\raisebox{1.72218pt}{.}|^{\operatorname{quot}}_{(V_{n},\|\raisebox{1.20552pt}{.}\|_{n})}(x)\big\}_{x\in X^{\operatorname{an}}}$ . Thus

e^{-e_{n}\epsilon}\|\raisebox{1.72218pt}{.}\|_{h^{e_{n}}}\leq\|\raisebox{1.72218pt}{.}\|_{h_{n}}\leq e^{e_{n}\epsilon}\|\raisebox{1.72218pt}{.}\|_{h^{e_{n}}},

so that $e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)$ . Thus, by Proposition 3.6,

e^{-e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)\leq|\raisebox{1.72218pt}{.}|_{h_{n}}(x)\leq e^{e_{n}\epsilon}|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x).

Therefore,

1\leq\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}=\frac{|\raisebox{1.72218pt}{.}|_{h_{n}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{e_{n}}}(x)}{|\raisebox{1.72218pt}{.}|_{h_{n}}(x)}\leq e^{2e_{n}\epsilon},

that is, $0\leq a_{e_{n}}/e_{n}\leq 2\epsilon$ , and hence $0\leq\lim_{m\to\infty}a_{m}/m\leq 2\epsilon$ , as required. ∎

Corollary 3.11.

A continuous metric $h$ is semipositive if and only if, for any $\epsilon>0$ , there is a positive integer $n$ such that, for all $x\in X^{\mathrm{an}}$ , we can find $s\in H^{0}(X,L^{\otimes n})_{\hat{\kappa}(x)}\setminus\{0\}$ with $\|s\|_{h^{n},\hat{\kappa}(x)}\leq e^{n\epsilon}|s|_{h^{n}}(x)$ .

Proof.

First we assume that $h$ is semipositive. By using Proposition 3.10, we can find a positive integer $n$ such that $L^{\otimes n}$ is generated by global sections and

|\raisebox{1.72218pt}{.}|_{h^{n}}(x)\leq|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h^{n}}(x)\leq e^{n\epsilon/2}|\raisebox{1.72218pt}{.}|_{h^{n}}(x)

for all $x\in X^{\mathrm{an}}$ . On the other hand, there is $s\in H^{0}(X,L^{\otimes n})_{\hat{\kappa}(x)}\setminus\{0\}$ such that $\|s\|_{h^{n},\hat{\kappa}(x)}\leq e^{n\epsilon/2}|s|^{\mathrm{quot}}_{h^{n}}(x)$ . Thus,

\|s\|_{h^{n},\hat{\kappa}(x)}\leq e^{n\epsilon/2}|s|^{\mathrm{quot}}_{h^{n}}(x)\leq e^{n\epsilon}|{s}|_{h^{n}}(x).

Next we consider the converse. For a positive integer $m$ , there is a positive integer $e_{m}$ such that, for any $x\in X^{\mathrm{an}}$ , we can find $s\in H^{0}(X,L^{\otimes e_{m}})_{\hat{\kappa}(x)}\setminus\{0\}$ with $\|s\|_{h^{e_{m}},\hat{\kappa}(x)}\leq e^{e_{m}/m}|s|_{h^{e_{m}}}(x)$ . Clearly $L^{\otimes e_{m}}$ is generated by global sections. Moreover,

|s|_{h^{e_{m}}}(x)\leq|s|^{\mathrm{quot}}_{(H^{0}(X,L^{\otimes e_{m}}),\|\raisebox{1.20552pt}{.}\|_{h^{e_{m}}})}(x)\leq e^{e_{m}/m}|s|_{h^{e_{m}}}(x),

that is,

0\leq\frac{1}{e_{m}}\log\left(\frac{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H^{0}(X,L^{\otimes e_{m}}),\|\raisebox{1.20552pt}{.}\|_{h^{e_{m}}})}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{m}}}(x)}\right)\leq\frac{1}{m}.

Thus $h$ is semipositive. ∎

Corollary 3.12.

Let $h$ be a continuous metric of $L^{\operatorname{an}}$ . If there are a sequence $\{e_{n}\}$ of positive integers and a sequence $\{h_{n}\}$ of metrics such that $h_{n}$ is a semipositive metric of $(L^{\otimes e_{n}})^{\operatorname{an}}$ for each $n$ and

\frac{1}{e_{n}}\log\frac{|\raisebox{1.72218pt}{.}|_{h_{n}}(x)}{|\raisebox{1.72218pt}{.}|_{h^{e_{n}}}(x)}

converges to $0$ uniformly as $n\to\infty$ , then $h$ is semipositive.

Proof.

For a positive number $\epsilon>0$ , choose a positive integer $n$ such that

e^{-\epsilon e_{n}/3}h^{e_{n}}\leq h_{n}\leq e^{\epsilon e_{n}/3}h^{e_{n}}.

As $h_{n}$ is semipositive, by Corollary 3.11, there is a positive integer $m$ such that, for all $x\in X^{\mathrm{an}}$ , we can find $s\in H^{0}(X,L^{\otimes me_{n}})_{\hat{\kappa}(x)}\setminus\{0\}$ with $\|s\|_{h_{n}^{m},\hat{\kappa}(x)}\leq e^{me_{n}\epsilon/3}|s|_{h_{n}^{m}}(x)$ , so that

\|s\|_{h^{me_{n}},\hat{\kappa}(x)}\leq e^{\epsilon me_{n}/3}\|s\|_{h_{n}^{m},\hat{\kappa}(x)}\leq e^{2me_{n}\epsilon/3}|s|_{h_{n}^{m}}(x)\leq e^{me_{n}\epsilon}|s|_{h^{me_{n}}}(x).

Therefore, the assertion follows from Corollary 3.11. ∎