3. Continuous metrics of invertible sheaves [026I]

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. Continuous metrics of invertible sheaves

In this section, we consider several properties of continuous metrics of invertible sheaves. Let $h=\{|\raisebox{1.72218pt}{.}|_{h}(x)\}_{x\in X^{\mathrm{an}}}$ and $h^{\prime}=\{|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)\}_{x\in X^{\mathrm{an}}}$ be continuous metrics of $L^{\mathrm{an}}$ (cf. §1.1.4). As ${L(x)}:=L\otimes_{\mathscr{O}_{X}}\hat{\kappa}(x)$ is a $1$ -dimensional vector space over $\hat{\kappa}(x)$ , $h+h^{\prime}:=\{|\raisebox{1.72218pt}{.}|_{h}(x)+|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)\}_{x\in X^{\mathrm{an}}}$ forms a continuous metric of $L^{\mathrm{an}}$ . Indeed, we can find a continuous positive function $\varphi$ on $X^{\mathrm{an}}$ such that $|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)=\varphi(x)|\raisebox{1.72218pt}{.}|_{h}(x)$ for any $x\in X^{\mathrm{an}}$ . Thus

h+h^{\prime}=\{(1+\varphi(x))|\raisebox{1.72218pt}{.}|_{h}(x)\}_{x\in X^{\mathrm{an}}}

is a continuous metric of $L^{\mathrm{an}}$ .

Lemma 3.1.

There is a continuous metric of $L^{\mathrm{an}}$ .

Proof.

Let us choose an affine open covering $X=\bigcup_{i=1}^{N}U_{i}$ together with a local basis $\omega_{i}$ of $L$ on each $U_{i}$ . Let $h_{i}$ be a metric of $L^{\mathrm{an}}$ over $U_{i}^{\mathrm{an}}$ given by $|\omega_{i}|_{h_{i}}(x)=1$ for $x\in U_{i}^{\mathrm{an}}$ . As $X^{\mathrm{an}}$ is paracompact (locally compact and $\sigma$ -compact), we can find a partition of unity $\{\rho_{i}\}_{i=1,\ldots,N}$ of continuous functions on $X^{\mathrm{an}}$ such that $\mathrm{supp}(\rho_{i})\subseteq U_{i}^{\mathrm{an}}$ for all $i$ . If we set $|\raisebox{1.72218pt}{.}|_{h}(x)=\sum_{i=1}^{N}\rho_{i}(x)|\raisebox{1.72218pt}{.}|_{h_{i}}(x)$ , then $h=\{|\raisebox{1.72218pt}{.}|_{h}(x)\}_{x\in X^{\mathrm{an}}}$ yields a continuous metric of $L^{\mathrm{an}}$ . ∎

3.1. Extension theorem for a metric arising from a model

We assume that $X$ is projective. Let $\mathscr{X}\rightarrow\operatorname{Spec}\mathfrak{o}_{k}$ be a model of $X$ . We let $\mathscr{L}$ be an invertible sheaf on $\mathscr{X}$ such that $\left.{\mathscr{L}}\right|_{{X}}=L$ . We have seen in §1.1.6 that $\mathscr{L}$ induces a continuous metric $h=\{|\raisebox{1.72218pt}{.}|_{\mathscr{L}}(x)\}_{x\in X^{\mathrm{an}}}$ of $L^{\mathrm{an}}$ .

Theorem 3.2.

We assume that $|\raisebox{1.72218pt}{.}|$ is non-trivial and $\mathscr{L}$ is an ample invertible sheaf. Fix a closed subscheme $Y$ of $X$ , $l\in H^{0}(Y,\left.{L}\right|_{{Y}})$ and a positive number $\epsilon$ . Then there are a positive integer $n$ and $s\in H^{0}(X,L^{\otimes n})$ such that $\left.{s}\right|_{{Y}}=l^{\otimes n}$ and

\|s\|_{h^{n}}\leq e^{n\epsilon}\left(\|l\|_{Y,h}\right)^{n}.

Proof.

Clearly, we may assume that $l\not=0$ . Let $\mathscr{Y}$ be the Zariski closure of $Y$ in $\mathscr{X}$ (cf. §1.1.7).

Claim 3.2.1.

There are a positive integer $a$ and $\alpha\in k^{\times}$ such that

e^{-a\epsilon/2}\leq\|\alpha l^{\otimes a}\|_{Y,h^{a}}\leq 1.

Proof.

First we assume that $|\raisebox{1.72218pt}{.}|$ is discrete. We take a positive integer $a$ such that $e^{-\epsilon a/2}\leq|\varpi|$ . We also choose $\alpha\in k^{\times}$ such that

|\alpha^{-1}|=\min\{|\gamma|\mid\text{$\gamma\in k^{\times}$ and $\|l^{\otimes a}\|_{Y,h^{a}}\leq|\gamma|$}\}.

Then, as $\|l^{\otimes a}\|_{Y,h^{a}}\leq|\alpha^{-1}|\leq|\varpi|^{-1}\|l^{\otimes a}\|_{Y,h^{a}}$ , we have

e^{-a\epsilon/2}\leq|\varpi|\leq\|\alpha l^{\otimes a}\|_{Y,h^{a}}\leq 1.

Next we assume that $|\raisebox{1.72218pt}{.}|$ is not discrete. In this case, $|k^{\times}|$ is dense in $\mathbb{R}_{>0}$ by Lemma 1.15, so that we can choose $\beta\in k^{\times}$ such that

e^{-\epsilon/2}\leq\|l\|_{Y,h}/|\beta|\leq 1.

Thus if we set $\alpha=\beta^{-1}$ and $a=1$ , we have the assertion. ∎

By Corollary 2.2, there is $\beta\in\mathfrak{o}_{K}\setminus\{0\}$ such that

\beta(\alpha l^{\otimes a})^{\otimes m}\in H^{0}(\mathscr{Y},\left.{\mathscr{L}^{\otimes am}}\right|_{{\mathscr{Y}}})

for all $m\geq 0$ . We choose a positive integer $m$ such that $|\beta|^{-1}\leq e^{am\epsilon/2}$ and

H^{0}(\mathscr{X},\mathscr{L}^{\otimes am})\to H^{0}(\mathscr{Y},\left.{\mathscr{L}^{\otimes am}}\right|_{{\mathscr{Y}}})

is surjective, so that we can find $l_{m}\in H^{0}(\mathscr{X},\mathscr{L}^{\otimes am})$ such that $\left.{l_{m}}\right|_{{\mathscr{Y}}}=\beta(\alpha l^{\otimes a})^{\otimes m}$ . Note that $\|l_{m}\|_{h^{am}}\leq 1$ . Thus, if we set $s=\beta^{-1}\alpha^{-m}l_{m}$ , then $\left.{s}\right|_{{\mathscr{Y}}}=l^{\otimes am}$ and

	$\displaystyle\\|s\\|_{h^{am}}$	$\displaystyle=\|\beta\|^{-1}\|\alpha\|^{-m}\\|l_{m}\\|_{h^{am}}\leq e^{am\epsilon/2}\|\alpha\|^{-m}$
		$\displaystyle\leq e^{am\epsilon/2}\|\alpha\|^{-m}\left(e^{a\epsilon/2}\\|\alpha l^{\otimes a}\\|_{Y,h^{a}}\right)^{m}=e^{am\epsilon}\left(\\|l\\|_{Y,h}\right)^{am},$

as required. ∎

3.2. Quotient metric

Let $V$ be a finite-dimensional vector space over $k$ . We assume that there is a surjective homomorphism

\pi:V\otimes_{k}\mathscr{O}_{X}\to L.

For each $e\in V$ , $\pi(e\otimes 1)$ yields a global section of $L$ , that is, $\pi(e\otimes 1)\in H^{0}(X,L)$ . We denote it by $\tilde{e}$ . Let $\|\raisebox{1.72218pt}{.}\|$ be a norm of $V$ and $\overline{V}:=(V,\|\raisebox{1.72218pt}{.}\|)$ . Let $\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x)}$ be a norm of $V\otimes_{k}\hat{\kappa}(x)$ obtained by the scalar extension of $\|\raisebox{1.72218pt}{.}\|$ (cf. Definition 1.8). Let $|\raisebox{1.72218pt}{.}|_{\overline{V}}^{\mathrm{quot}}(x)$ be the quotient norm of $L(x):=L\otimes\hat{\kappa}(x)$ induced by $\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x)}$ and the surjective homomorphism $V\otimes_{k}\hat{\kappa}(x)\to{L(x)}$ .

Lemma 3.3.

Let $h$ be a continuous metric of $L^{\mathrm{an}}$ (cf. Lemma 3.1). Let $(e_{0},\ldots,e_{r})$ be an orthogonal basis of $V$ with respect to $\|\raisebox{1.72218pt}{.}\|$ . Then, for $s\in H^{0}(X,L)$ ,

|s|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{|s|_{h}(x)}{{\displaystyle\max_{i=0,\ldots,r}\left\{\frac{|\tilde{e}_{i}|_{h}(x)}{\|e_{i}\|}\right\}}}

on $X^{\mathrm{an}}$ .

Proof.

We set $I:=\{i\mid\text{$\tilde{e}_{i}\not=0$ in $H^{0}(X,L)$}\}$ and $U_{i}:=\{p\in X\mid\text{$\tilde{e}_{i}\not=0$ at $p$}\}$ for $i\in I$ .

Claim 3.3.1.

For a fixed $j\in I$ , if we set $\tilde{e}_{i}=a_{ij}\tilde{e}_{j}$ on $U_{j}$ ( $a_{ij}\in\mathscr{O}_{U_{j}}$ ), then

|\tilde{e}_{j}|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{1}{\displaystyle{\max_{i=0,\ldots,r}\left\{\frac{|a_{ij}|_{x}}{\|e_{i}\|}\right\}}}

on $U_{j}^{\mathrm{an}}$ .

Proof.

We set $c_{i}=\|e_{i}\|$ for $i=0,\ldots,r$ . Without loss of generality, we may assume that $j=0$ , that is, we need to show that

|\tilde{e}_{0}|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{1}{\max\{1/c_{0},|a_{10}|_{x}/c_{1},\ldots,|a_{r0}|_{x}/c_{r}\}}.

Since

\ker(\pi_{x}:V\otimes_{k}\hat{\kappa}(x)\to L\otimes_{\mathscr{O}_{X}}\hat{\kappa}(x))=\langle e_{1}-a_{10}(x)e_{0},\ldots,e_{r}-a_{r0}(x)e_{0}\rangle

for $x\in U^{\mathrm{an}}_{0}$ , we have

|\tilde{e}_{0}|_{\overline{V}}^{\mathrm{quot}}(x)=\inf\left.\left\{f(\lambda_{1},\ldots,\lambda_{r})\ \right|\ (\lambda_{1},\ldots,\lambda_{r})\in\hat{\kappa}(x)^{r}\right\},

where $f(\lambda_{1},\ldots,\lambda_{r}):=\big\|e_{0}+\sum_{i=1}^{r}\lambda_{i}(e_{i}-a_{i0}(x)e_{0})\big\|_{\hat{\kappa}(x)}$ . Note that

f(\lambda_{1},\ldots,\lambda_{r})=\max\left\{c_{0}\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x},c_{1}|\lambda_{1}|_{x},\ldots,c_{r}|\lambda_{r}|_{x}\right\}.

\max\{\alpha_{0},\ldots,\alpha_{r}\}\max\{\beta_{0},\ldots,\beta_{r}\}\geq\max\{\alpha_{0}\beta_{0},\ldots,\alpha_{r}\beta_{r}\}

for $\alpha_{0},\ldots,\alpha_{r},\beta_{0},\ldots,\beta_{r}\in\mathbb{R}_{\geq 0}$ , we have

f(\lambda_{1},\ldots,\lambda_{r})\cdot\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}\\ \geq\max\left\{\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x},|\lambda_{1}a_{10}(x)|_{x},\ldots,|\lambda_{r}a_{r0}(x)|_{x}\right\}\\ \geq\left|1-\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)+\sum\nolimits_{i=1}^{r}\lambda_{i}a_{i0}(x)\right|_{x}=1.

Therefore, we obtain

\inf\left.\left\{f(\lambda_{1},\ldots,\lambda_{r})\ \right|\ (\lambda_{1},\ldots,\lambda_{r})\in\hat{\kappa}(x)^{n}\right\}\\ \geq\frac{1}{\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}}.

We need to see that

f(\eta_{1},\ldots,\eta_{r})=\frac{1}{\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}}.

for some $\eta_{1},\ldots,\eta_{r}\in\hat{\kappa}(x)$ . As $f(0,\ldots,0)=c_{0}$ , the assertion holds if

\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}=1/c_{0}.

Next we assume that

\max\{1/c_{0},|a_{10}(x)|_{x}/c_{1},\ldots,|a_{r0}(x)|_{x}/c_{r}\}=|a_{i0}(x)|_{x}/c_{i}

for some $i$ . Clearly $a_{i0}(x)\not=0$ . If we set

\eta_{j}=\begin{cases}0&\text{if $j\not=i$},\\ 1/a_{i0}(x)&\text{if $j=i$},\end{cases}

then $f(\eta_{1},\ldots,\eta_{n})=c_{i}/|a_{i0}(x)|_{x}$ , as required. ∎

If we set $s=f\tilde{e}_{j}$ on $U_{j}$ ( $f\in\mathscr{O}_{U_{j}}$ ), then $|s|_{\overline{V}}^{\mathrm{quot}}(x)=|f|_{x}|\tilde{e}_{j}|_{\overline{V}}^{\mathrm{quot}}(x)$ on $U_{j}^{\mathrm{an}}$ , so that, by Claim 3.3.1,

|s|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{|f|_{x}}{\displaystyle{\max_{i=0,\ldots,r}\left\{\frac{|a_{ij}|_{x}}{\|e_{i}\|}\right\}}}.

On the other hand, $|s|_{h}(x)=|f|_{x}|\tilde{e}_{j}|_{h}(x)$ and $|\tilde{e}_{i}|_{h}(x)=|a_{ij}|_{x}|\tilde{e}_{j}|_{h}(x)$ for $i=0,\ldots,r$ . Thus

|s|_{\overline{V}}^{\mathrm{quot}}(x)=\frac{|s|_{h}(x)}{{\displaystyle\max_{i=0,\ldots,n}\left\{\frac{|\tilde{e}_{i}|_{h}(x)}{\|e_{i}\|}\right\}}}

on $U_{j}^{\mathrm{an}}$ . Therefore, the assertion follows because $X=\bigcup_{j\in I}U_{j}$ . ∎

Corollary 3.4.

$\left\{|\raisebox{1.72218pt}{.}|_{\overline{V}}^{\mathrm{quot}}(x)\right\}_{x\in X^{\mathrm{an}}}$ yields a continuous metric of $L^{\mathrm{an}}$ .

Proof.

If $V$ has an orthogonal basis with respect to $\|\raisebox{1.72218pt}{.}\|$ , then the assertion follows from Lemma 3.3.

In general, by Proposition 1.3, for each $n\in\mathbb{Z}_{>0}$ , we choose a basis

(e_{n,0},e_{n,1},\ldots,e_{n,r})

of $V$ such that

(1-1/n)\max\{|c_{0}|\|e_{n,0}\|,\ldots,|c_{r}|\|e_{n,r}\|\}\leq\|c_{0}e_{n,0}+\cdots+c_{r}e_{n,r}\|

for all $c_{0},\ldots,c_{r}\in k$ . If we set

\|c_{0}e_{n,0}+\cdots+c_{r}e_{n,r}\|_{n}:=\max\{|c_{0}|\|e_{n,0}\|,\ldots,|c_{r}|\|e_{n,r}\|\}

for $c_{0},\ldots,c_{r}\in k$ . Then $(1-1/n)\|\raisebox{1.72218pt}{.}\|_{n}\leq\|\raisebox{1.72218pt}{.}\|\leq\|\raisebox{1.72218pt}{.}\|_{n}$ , so that

(1-1/n)|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\leq|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\leq|\raisebox{1.72218pt}{.}|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)

for all $x\in X^{\mathrm{an}}$ . Let $\omega$ be a local basis of $L$ over an open set $U$ . Then the above inequalities imply that

\log(1-1/n)\leq\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)-\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\right)\leq 0

for all $x\in U^{\mathrm{an}}$ , which shows that the sequence $\left\{\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|_{n})}^{\mathrm{quot}}(x)\right)\right\}_{n=1}^{\infty}$ converges to $\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)$ uniformly on $U^{\mathrm{an}}$ . Thus, by the previous observation, $\log\left(|\omega|_{(V,\|\raisebox{1.20552pt}{.}\|)}^{\mathrm{quot}}(x)\right)$ is continuous on $U^{\mathrm{an}}$ . ∎

From now on and until the end of the subsection, we assume that $X$ is projective and $L$ is generated by global sections. Let $h=\{|\raisebox{1.72218pt}{.}|_{h}(x)\}_{x\in X^{\mathrm{an}}}$ be a continuous metric of $L^{\mathrm{an}}$ . As $H^{0}(X,L)\otimes_{k}\mathscr{O}_{X}\to L$ is surjective, by Corollary 3.4,

h^{\mathrm{quot}}=\left\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H^{0}(X,L),\|\raisebox{1.20552pt}{.}\|_{h})}(x)\right\}_{x\in X^{\mathrm{an}}}

yields a continuous metric of $L^{\mathrm{an}}$ . For simplicity, we denote $|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H^{0}(X,L),\|\raisebox{1.20552pt}{.}\|_{h})}(x)$ by $|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h}(x)$ . Moreover, the supreme norm of $H^{0}(X,L)$ arising from $h^{\mathrm{quot}}$ is denoted by $\|\raisebox{1.72218pt}{.}\|_{h}^{\mathrm{quot}}$ , that is, $\|\raisebox{1.72218pt}{.}\|_{h}^{\mathrm{quot}}:=\|\raisebox{1.72218pt}{.}\|_{h^{\mathrm{quot}}}$ .

Lemma 3.5.

(1)

$|\raisebox{1.72218pt}{.}|_{h}(x)\leq|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{h}(x)$ for all $x\in X^{\mathrm{an}}$ .
(2)

$\|\raisebox{1.72218pt}{.}\|_{h}=\|\raisebox{1.72218pt}{.}\|_{h}^{\mathrm{quot}}$ .

(3)

Let $(L^{\prime},h^{\prime})$ be a pair of an invertible sheaf $L^{\prime}$ on $X$ and a continuous metric $h^{\prime}=\{|\raisebox{1.72218pt}{.}|_{h^{\prime}}(x)\}_{x\in X^{\operatorname{an}}}$ of ${L^{\prime}}^{\operatorname{an}}$ such that $L^{\prime}$ is generated by global sections. Then

|l\cdot l^{\prime}|_{h\otimes h^{\prime}}^{\mathrm{quot}}(x)\leq|l|_{h}^{\mathrm{quot}}(x)|l^{\prime}|_{h^{\prime}}^{\mathrm{quot}}(x)

for $l\in{L(x)}$ and $l^{\prime}\in{L^{\prime}(x)}$ .

Proof.

(1) Fix $l\in{L(x)}\setminus\{0\}$ . For $\epsilon>0$ , let $(e_{1},\ldots,e_{n})$ be an $e^{-\epsilon}$ -orthogonal basis of $H^{0}(X,L)$ with respect to $\|\raisebox{1.72218pt}{.}\|_{h}$ . There is $s\in H^{0}(X,L)\otimes_{k}\hat{\kappa}(x)$ such that $s(x)=l$ and $\|s\|_{h,\hat{\kappa}(x)}\leq e^{\epsilon}|l|^{\mathrm{quot}}_{h}(x)$ . We set $s=a_{1}e_{1}+\cdots+a_{n}e_{n}$ ( $a_{1},\ldots,a_{n}\in\hat{\kappa}(x)$ ). Then, by Proposition 1.9,

	$\displaystyle\\|s\\|_{h,\hat{\kappa}(x)}$	$\displaystyle\geq e^{-\epsilon}\max\{\|a_{1}\|_{x}\\|e_{1}\\|_{h},\ldots,\|a_{n}\|_{x}\\|e_{n}\\|_{h}\}$
		$\displaystyle\geq e^{-\epsilon}\max\{\|a_{1}\|_{x}\|e_{1}\|_{h}(x),\ldots,\|a_{n}\|_{x}\|e_{n}\|_{h}(x)\}\geq e^{-\epsilon}\|l\|_{h}(x),$

so that $|l|_{h}(x)\leq e^{2\epsilon}|l|^{\mathrm{quot}}_{h}(x)$ , and hence the assertion follows because $\epsilon$ is an arbitrary positive number.

(2) By (1), we have $\|\raisebox{1.72218pt}{.}\|_{h}\leq\|\raisebox{1.72218pt}{.}\|_{h}^{\mathrm{quot}}$ . On the other hand, as $|s|^{\mathrm{quot}}_{h}(x)\leq\|s\|_{h}$ for $s\in H^{0}(X,L)$ , we have $\|s\|^{\mathrm{quot}}_{h}\leq\|s\|_{h}$ .

(3) For $\epsilon>0$ , there are $s\in H^{0}(X,L)\otimes_{k}\hat{\kappa}(x)$ and $s^{\prime}\in H^{0}(X,L^{\prime})\otimes_{k}\hat{\kappa}(x)$ such that

s(x)=l,\ s^{\prime}(x)=l^{\prime},\ \|s\|_{h,\hat{\kappa}(x)}\leq e^{\epsilon}|l|_{h}^{\mathrm{quot}}(x)\ \text{and}\ \|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}\leq e^{\epsilon}|l^{\prime}|_{h^{\prime}}^{\mathrm{quot}}(x).

Here let us see that $\|s\cdot s^{\prime}\|_{h\otimes h^{\prime},\hat{\kappa}(x)}\leq e^{2\epsilon}\|s\|_{h,\hat{\kappa}(x)}\|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}$ . Let $(s_{1},\ldots,s_{m})$ and $(s^{\prime}_{1},\ldots,s^{\prime}_{m^{\prime}})$ be $e^{-\epsilon}$ -orthogonal bases of $H^{0}(X,L)$ and $H^{0}(X,L^{\prime})$ , respectively. If we set $s=t_{1}s_{1}+\cdots+t_{m}s_{m}$ and $s^{\prime}=t^{\prime}_{1}s^{\prime}_{1}+\cdots+t^{\prime}_{m^{\prime}}s^{\prime}_{m^{\prime}}$ ( $t_{1},\ldots,t_{m},t^{\prime}_{1},\ldots,t^{\prime}_{m^{\prime}}\in\hat{\kappa}(x)$ ), then

s\cdot s^{\prime}=\sum_{i,j}t_{i}t^{\prime}_{j}s_{i}\cdot s^{\prime}_{j}.

Thus,

	$\displaystyle\\|s\cdot s^{\prime}\\|_{h\otimes h^{\prime},\hat{\kappa}(x)}$	$\displaystyle\leq\max_{i,j}\left\{\|t_{i}\|_{x}\|t^{\prime}_{j}\|_{x}\\|s_{i}\cdot s^{\prime}_{j}\\|_{h\otimes h^{\prime}}\right\}\leq\max_{i,j}\left\{\|t_{i}\|_{x}\|t^{\prime}_{j}\|_{x}\\|s_{i}\\|_{h}\\|s^{\prime}_{j}\\|_{h^{\prime}}\right\}$
		$\displaystyle\leq\max_{i}\left\{\|t_{i}\|_{x}\\|s_{i}\\|_{h}\right\}\max_{j}\left\{\|t^{\prime}_{j}\|_{x}\\|s^{\prime}_{j}\\|_{h^{\prime}}\right\}$
		$\displaystyle\leq e^{2\epsilon}\\|s\\|_{h,\hat{\kappa}(x)}\\|s^{\prime}\\|_{h^{\prime},\hat{\kappa}(x)}.$

Therefore, we have $(s\cdot s^{\prime})(x)=l\cdot l^{\prime}$ and

|l\cdot l^{\prime}|_{h\otimes h^{\prime}}^{\mathrm{quot}}(x)\leq\|s\cdot s^{\prime}\|_{h\otimes h^{\prime},\hat{\kappa}(x)}\leq e^{2\epsilon}\|s\|_{h,\hat{\kappa}(x)}\|s^{\prime}\|_{h^{\prime},\hat{\kappa}(x)}\leq e^{4\epsilon}|l|_{h}^{\mathrm{quot}}(x)|l^{\prime}|_{h^{\prime}}^{\mathrm{quot}}(x),

as required. ∎

Proposition 3.6.

If there are a normed finite-dimensional vector space $(V,\|\raisebox{1.72218pt}{.}\|)$ and a surjective homomorphism $V\otimes_{k}\mathscr{O}_{X}\to L$ such that $h$ is given by $\left\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(V,\|\raisebox{1.20552pt}{.}\|)}(x)\right\}_{x\in X^{\mathrm{an}}}$ , then $|\raisebox{1.72218pt}{.}|_{h^{n}}(x)=|\raisebox{1.72218pt}{.}|_{h^{n}}^{\mathrm{quot}}(x)$ for all $n\geq 1$ .

Proof.

First we consider the case $n=1$ . Fix $l\in{L(x)}\setminus\{0\}$ . For $\epsilon>0$ , there is $s\in V\otimes_{k}\hat{\kappa}(x)$ such that $\tilde{s}(x)=l$ and $\|s\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|l|_{h}(x)$ .

Note that $\|\tilde{e}\|_{h}\leq\|e\|$ for all $e\in V$ . Let $(e_{1},\ldots,e_{r})$ be an $e^{-\epsilon}$ -orthogonal basis of $V$ with respect to $\|\raisebox{1.72218pt}{.}\|$ . If we set $s=a_{1}e_{1}+\cdots+a_{r}e_{r}$ ( $a_{1},\ldots,a_{r}\in\hat{\kappa}(x)$ ), then, by Proposition 1.9,

	$\displaystyle\\|\tilde{s}\\|_{h,\hat{\kappa}(x)}$	$\displaystyle\leq\max\{\|a_{1}\|_{x}\\|\tilde{e}_{1}\\|_{h},\ldots,\|a_{r}\|_{x}\\|\tilde{e}_{r}\\|_{h}\}$
		$\displaystyle\leq\max\{\|a_{1}\|_{x}\\|e_{1}\\|,\ldots,\|a_{r}\|_{x}\\|e_{r}\\|\}$
		$\displaystyle\leq e^{\epsilon}\\|s\\|_{\hat{\kappa}(x)},$

so that

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

and hence $|l|_{h}^{\mathrm{quot}}(x)\leq|l|_{h}(x)$ by taking $\epsilon\to 0$ . Thus the assertion for $n=1$ follows from (1) in Lemma 3.5.

In general, by using (3) in Lemma 3.5,

|l^{n}|_{h^{n}}(x)=\left(|l|_{h}(x)\right)^{n}=\left(|l|_{h}^{\mathrm{quot}}(x)\right)^{n}\geq|l^{n}|_{h^{n}}^{\mathrm{quot}}(x),

and hence we have the assertion by (1) in Lemma 3.5. ∎

Lemma 3.7.

We assume that there are a normed finite-dimensional vector space $(V,\|\raisebox{1.72218pt}{.}\|)$ and a surjective homomorphism $V\otimes_{k}\mathscr{O}_{X}\to L$ such that $h$ is given by $\left\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(V,\|\raisebox{1.20552pt}{.}\|)}(x)\right\}_{x\in X^{\mathrm{an}}}$ . Let $k^{\prime}$ be an extension field of $k$ , and let $|\raisebox{1.72218pt}{.}|^{\prime}$ be a complete absolute value of $k^{\prime}$ as an extension of $|\raisebox{1.72218pt}{.}|$ . We set

X^{\prime}:=X\times_{\operatorname{Spec}(k)}\operatorname{Spec}(k^{\prime}),\quad L=L\otimes_{k}k^{\prime}\quad\text{and}\quad V^{\prime}:=V\otimes_{k}k^{\prime}.

Let $\|\raisebox{1.72218pt}{.}\|^{\prime}$ be a norm of $V^{\prime}$ obtained by the scalar extension of $\|\raisebox{1.72218pt}{.}\|$ . Moreover, let $h^{\prime}$ be a continuous metric of ${L^{\prime}}^{\mathrm{an}}$ given by the scalar extension of $h$ . Then $h^{\prime}$ coincides with $\left\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(V^{\prime},\|\raisebox{1.20552pt}{.}\|^{\prime})}(x^{\prime})\right\}_{x^{\prime}\in{X^{\prime}}^{\mathrm{an}}}$ .

Proof.

Let $f:X^{\prime}\to X$ be the projection. For $x^{\prime}\in{X^{\prime}}^{\mathrm{an}}$ , we set $x=f^{\mathrm{an}}(x^{\prime})$ . Then $\hat{\kappa}(x)\subseteq\hat{\kappa}(x^{\prime})$ and $(L\otimes_{k}\hat{\kappa}(x))\otimes_{\hat{\kappa}(x)}\hat{\kappa}(x^{\prime})=L^{\prime}\otimes_{k^{\prime}}\hat{\kappa}(x^{\prime})$ , that is, $L(x)\otimes_{\hat{\kappa}(x)}\hat{\kappa}(x^{\prime})=L^{\prime}(x^{\prime})$ . Moreover, $V^{\prime}\otimes_{k^{\prime}}\hat{\kappa}(x^{\prime})=(V\otimes_{k}\hat{\kappa}(x))\otimes_{\hat{\kappa}(x)}\hat{\kappa}(x^{\prime})$ , and by Lemma 1.10, $\|\raisebox{1.72218pt}{.}\|^{\prime}_{\hat{\kappa}(x^{\prime})}=\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x^{\prime})}=\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x),\hat{\kappa}(x^{\prime})}$ . Thus the assertion follows from Lemma 1.11. ∎

Proposition 3.8.

We assume that there is a subspace $H$ of $H^{0}(X,L)$ such that $H\otimes_{k}\mathscr{O}_{X}\to L$ is surjective and the morphism $\phi_{H}:X\to\mathbb{P}(H)$ induced by $H$ is a closed embedding. We identify $X$ with $\phi_{H}(X)$ , so that $L=\left.{\mathscr{O}_{\mathbb{P}(H)}(1)}\right|_{{X}}$ . Let $\|\raisebox{1.72218pt}{.}\|$ be a norm of $H$ such that $H$ has an orthonormal basis $(e_{1},\ldots,e_{r})$ with respect to $\|\raisebox{1.72218pt}{.}\|$ . We set

h:=\left\{|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H,\|\raisebox{1.20552pt}{.}\|)}(x)\right\}_{x\in X^{\mathrm{an}}}\quad\text{and}\quad\mathscr{H}:=\mathfrak{o}_{k}e_{1}+\cdots+\mathfrak{o}_{k}e_{r}=(H,\|\raisebox{1.72218pt}{.}\|)_{\leq 1}.

Let $\mathscr{X}$ be the Zariski closure of $X$ in $\mathbb{P}(\mathscr{H})$ (cf. §1.1.7) and $\mathscr{L}:=\left.{\mathscr{O}_{\mathbb{P}(\mathscr{H})}(1)}\right|_{{\mathscr{X}}}$ . Then $|\raisebox{1.72218pt}{.}|_{h}(x)=|\raisebox{1.72218pt}{.}|_{\mathscr{L}}(x)$ for all $x\in X^{\mathrm{an}}$ .

Proof.

First let us see that $|s|_{h}(x)\leq|s|_{\mathscr{L}}(x)$ for $s\in H$ . Let $\omega_{\xi}$ be a local basis of $\mathscr{L}$ at $\xi=r_{\mathscr{X}}(x)$ . If we set $s=s_{\xi}\omega_{\xi}$ , then

|s|_{\mathscr{L}}(x)=|s_{\xi}|_{x}.

As $s_{\xi}^{-1}s\in\mathscr{L}_{\xi}$ and $\mathscr{H}\otimes_{\mathfrak{o}_{k}}\mathscr{O}_{\mathscr{X},\xi}\to\mathscr{L}_{\xi}$ is surjective, there are $l_{1},\ldots,l_{r}\in\mathscr{H}$ and $a_{1},\ldots,a_{r}\in\mathscr{O}_{\mathscr{X},\xi}$ such that $s_{\xi}^{-1}s=a_{1}l_{1}+\cdots+a_{r}l_{r}$ . Therefore,

	$\displaystyle\left\|s_{\xi}^{-1}s\right\|_{h}(x)$	$\displaystyle\leq\max\left\{\|a_{1}l_{1}\|_{h}(x),\ldots,\|a_{r}l_{r}\|_{h}(x)\right\}$
		$\displaystyle=\max\left\{\|a_{1}\|_{x}\|l_{1}\|_{h}(x),\ldots,\|a_{r}\|_{x}\|l_{r}\|_{h}(x)\right\}\leq 1,$

so that $|s|_{h}(x)\leq|s_{\xi}|_{x}=|s|_{\mathscr{L}}(x)$ , as required.

Next let us see that $|l|_{\mathscr{L}}(x)\leq\|l\|_{\hat{\kappa}(x)}$ for all $l\in H\otimes\hat{\kappa}(x)$ . By Proposition 1.9, $(e_{1},\ldots,e_{r})$ is an orthonormal basis of $H\otimes\hat{\kappa}(x)$ with respect to $\|\raisebox{1.72218pt}{.}\|_{\hat{\kappa}(x)}$ . Thus, if we set $l=a_{1}e_{1}+\cdots+a_{r}e_{r}$ ( $a_{1},\ldots,a_{r}\in\hat{\kappa}(x)$ ), then

	$\displaystyle\|l\|_{\mathscr{L}}(x)$	$\displaystyle\leq\max\{\|a_{1}\|_{x}\|e_{1}\|_{\mathscr{L}}(x),\ldots,\|a_{r}\|_{x}\|e_{r}\|_{\mathscr{L}}(x)\}$
		$\displaystyle\leq\max\{\|a_{1}\|_{x},\ldots,\|a_{r}\|_{x}\}=\\|l\\|_{\hat{\kappa}(x)}.$

Finally let us see that $|s|_{\mathscr{L}}(x)\leq|s|_{h}(x)$ for $s\in H$ . For $\epsilon>0$ , we choose $l\in H\otimes\hat{\kappa}(x)$ such that $l(x)=s(x)$ and $\|l\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|s|_{h}(x)$ . Then, by the previous observation,

|s|_{\mathscr{L}}(x)=|l|_{\mathscr{L}}(x)\leq\|l\|_{\hat{\kappa}(x)}\leq e^{\epsilon}|s|_{h}(x).

Thus the assertion follows. ∎

Remark 3.9.

We assume that $|\raisebox{1.72218pt}{.}|$ is non-trivial and $\|\raisebox{1.72218pt}{.}\|=\|\raisebox{1.72218pt}{.}\|_{\mathscr{H}}$ for some finitely generated lattice $\mathscr{H}$ of $H$ . Then a free basis $(e_{1},\ldots,e_{r})$ of $\mathscr{H}$ yields an orthonormal basis of $H$ with respect to $\|\raisebox{1.72218pt}{.}\|$ (cf. Proposition 1.14). Moreover, $\mathscr{H}=(H,\|\raisebox{1.72218pt}{.}\|)_{\leq 1}$ .

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. ∎

3.4. The functions $\sigma$ and $\mu$ on $X^{\operatorname{an}}$

Throughout this subsection, we assume that $X$ is projective. Let $\widehat{\operatorname{Pic}}_{C^{0}}(X)$ denote the group of isomorphism classes of pairs $(L,h)$ consisting of an invertible sheaf $L$ on $X$ and a continuous metric $h$ of $L^{\operatorname{an}}$ . Fix $\overline{L}=(L,h)\in\widehat{\operatorname{Pic}}_{C^{0}}(X)$ . We assume that $L$ is generated by global sections. We define $\sigma_{\overline{L}}(x)$ to be

\sigma_{\overline{L}}(x):=\log\left(\frac{|\raisebox{1.72218pt}{.}|_{h}^{\operatorname{quot}}(x)}{|\raisebox{1.72218pt}{.}|_{h}(x)}\right).

Lemma 3.13.

For $\overline{L}$ and $\overline{L}^{\prime}\in\widehat{\operatorname{Pic}}_{C^{0}}(X)$ such that both $L$ and $L^{\prime}$ are generated by global sections, we have the following:

(1)

$\sigma_{\overline{L}}\geq 0$ on $X^{\operatorname{an}}$ .
(2)

$\sigma_{\overline{L}\otimes\overline{L}^{\prime}}(x)\leq\sigma_{\overline{L}}(x)+\sigma_{\overline{L}^{\prime}}(x)$ for $x\in X^{\operatorname{an}}$ .
(3)

If $\overline{L}\simeq\overline{L}^{\prime}$ , then $\sigma_{\overline{L}}=\sigma_{\overline{L}^{\prime}}$ on $X^{\operatorname{an}}$ .

Proof.

(1) and (3) are obvious. (2) follows from (3) in Lemma 3.5. ∎

We assume that $L$ is semiample. We set

{\mathbb{N}}(L):=\left\{n\in{\mathbb{Z}}_{\geq 1}\mid\text{$L^{\otimes n}$ is generated by global sections}\right\}.

Note that ${\mathbb{N}}(L)\not=\emptyset$ and ${\mathbb{N}}(L)$ forms a subsemigroup of ${\mathbb{Z}}_{\geq 1}$ with respect to the addition of ${\mathbb{Z}}_{\geq 1}$ . For $x\in X^{\operatorname{an}}$ , we define $\mu_{\overline{L}}(x)$ to be

\mu_{\overline{L}}(x):=\inf\left\{\left.\frac{\sigma_{\overline{L}^{\otimes n}}(x)}{n}\ \right|\ n\in{\mathbb{N}}(L)\right\}.

Note that $\mu_{\overline{L}}$ is upper-semicontinuous on $X^{\operatorname{an}}$ because $\sigma_{\overline{L}^{\otimes n}}$ is continuous for all $n\in{\mathbb{N}}(L)$ . We set

\widehat{\operatorname{Pic}}_{C^{0}}^{+}(X):=\{(L,h)\in\widehat{\operatorname{Pic}}_{C^{0}}(X)\mid\text{$L$ is semiample}\}.

Note that $\widehat{\operatorname{Pic}}_{C^{0}}^{+}(X)$ forms a semigroup with respect to $\otimes$ .

Lemma 3.14.

Let $\overline{L}=(L,h)$ and $\overline{L}^{\prime}=(L^{\prime},h^{\prime})$ be elements of $\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)$ . Then we have the following:

(1)

$\mu_{\overline{L}}\geq 0$ on $X^{\operatorname{an}}$ .
(2)

${\displaystyle\mu_{\overline{L}}(x)=\lim\limits_{\begin{subarray}{c}n\to\infty\\ n\in{\mathbb{N}}(L)\end{subarray}}\frac{\sigma_{\overline{L}^{\otimes n}}(x)}{n}}$ for $x\in X^{\operatorname{an}}$ .
(3)

$\mu_{\overline{L}\otimes\overline{L}^{\prime}}(x)\leq\mu_{\overline{L}}(x)+\mu_{\overline{L}^{\prime}}(x)$ for $x\in X^{\operatorname{an}}$ .
(4)

If $\overline{L}\simeq\overline{L}^{\prime}$ , then $\mu_{\overline{L}}=\mu_{\overline{L}^{\prime}}$ on $X^{\operatorname{an}}$ .
(5)

For $n\geq 0$ , $\mu_{\overline{L}^{\otimes n}}=n\mu_{\overline{L}}$ on $X^{\operatorname{an}}$ .

Proof.

(1) follows from (1) in Lemma 3.13.

(2) Since $\sigma_{\overline{L}^{\otimes(n+n^{\prime})}}(x)\leq\sigma_{\overline{L}^{\otimes n}}(x)+\sigma_{\overline{L}^{\otimes n^{\prime}}}(x)$ for $n,n^{\prime}\in{\mathbb{N}}(L)$ by (2) in Lemma 3.13, the assertion follows from Fekete’s lemma.

(3) and (4) follow from (2) and (3) in Lemma 3.13 together with (2), respectively.

(5) If $n=0$ , then the assertion is obvious, so that we may assume that $n\geq 1$ . We fix $n_{0}\in{\mathbb{N}}(L)$ . Then $n_{0}\in{\mathbb{N}}(L^{\otimes n})$ . Thus, by (2),

\mu_{\overline{L}^{\otimes n}}(x)=\lim_{m\to\infty}\frac{\sigma_{L^{\otimes mn_{0}n}}(x)}{mn_{0}}=n\lim_{m\to\infty}\frac{\sigma_{L^{\otimes mn_{0}n}}(x)}{mn_{0}n}=n\mu_{\overline{L}}(x).

∎

We set $\widehat{\operatorname{Pic}}_{C^{0}}(X)_{{\mathbb{Q}}}:=\widehat{\operatorname{Pic}}_{C^{0}}(X)\otimes_{{\mathbb{Z}}}{\mathbb{Q}}$ and

\widehat{\operatorname{Pic}}_{C^{0}}^{+}(X)_{{\mathbb{Q}}}:=\{(L,h)\in\widehat{\operatorname{Pic}}_{C^{0}}(X)_{{\mathbb{Q}}}\mid\text{$L$ is semiample}\}.

Let $\iota:\widehat{\operatorname{Pic}}_{C^{0}}(X)\to\widehat{\operatorname{Pic}}_{C^{0}}(X)_{{\mathbb{Q}}}$ be the canonical homomorphism. For $\overline{L}\in\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)_{{\mathbb{Q}}}$ , we choose a positive integer $n$ and $\overline{L}_{n}\in\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)$ with $\iota(\overline{L}_{n})=\overline{L}^{\otimes n}$ . Then $\mu_{\overline{L}_{n}}(x)/n$ does not depend on the choice of $n$ and $\overline{L}_{n}$ . Indeed, let us choose another $n^{\prime}\in{\mathbb{Z}}_{\geq 1}$ and $\overline{L}_{n^{\prime}}\in\widehat{\operatorname{Pic}}_{C^{0}}^{+}(X)$ with $\iota(\overline{L}_{n^{\prime}})=\overline{L}^{\otimes n^{\prime}}$ . As $\iota(\overline{L}_{n}^{\otimes n^{\prime}})=\iota(\overline{L}_{n^{\prime}}^{\otimes n})=\overline{L}^{\otimes nn^{\prime}}$ , there is a positive integer $m$ such that $\overline{L}_{n}^{\otimes mn^{\prime}}=\overline{L}_{n^{\prime}}^{\otimes mn}$ . By (5) in Lemma 3.14,

mn^{\prime}\mu_{\overline{L}_{n}}(x)=\mu_{\overline{L}_{n}^{\otimes mn^{\prime}}}(x)=\mu_{\overline{L}_{n^{\prime}}^{\otimes mn}}(x)=mn\mu_{\overline{L}_{n^{\prime}}}(x),

that is, $\mu_{\overline{L}_{n}}(x)/n=\mu_{\overline{L}_{n^{\prime}}}(x)/n^{\prime}$ , as required. By abuse of notation, it is also denoted by $\mu_{\overline{L}}(x)$ .

Lemma 3.15.

For $\overline{L},\overline{L}^{\prime}\in\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)_{{\mathbb{Q}}}$ , we have the following:

(1)

$\mu_{\overline{L}\otimes\overline{L}^{\prime}}(x)\leq\mu_{\overline{L}}(x)+\mu_{\overline{L}^{\prime}}(x)$ for $x\in X^{\operatorname{an}}$ .
(2)

For $a\in{\mathbb{Q}}_{\geq 0}$ , $\mu_{\overline{L}^{\otimes a}}=a\mu_{\overline{L}}$ on $X^{\operatorname{an}}$ .

(3)

Let $\overline{L}_{1},\ldots,\overline{L}_{r}$ be elements of $\widehat{\operatorname{Pic}}_{C_{0}}(X)_{{\mathbb{Q}}}$ . We assume that there are open intervals $I_{1},\ldots,I_{r}$ of ${\mathbb{R}}$ such that

\overline{L}\otimes\overline{L}_{1}^{\otimes t_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t_{r}}\in\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)_{{\mathbb{Q}}}

for all $(t_{1},\ldots,t_{r})\in(I_{1}\times\cdots\times I_{r})\cap{\mathbb{Q}}^{r}$ . Then, for a fixed $x\in X^{\operatorname{an}}$ , there is a continuous function $f:I_{1}\times\cdots\times I_{r}\to{\mathbb{R}}$ such that

f(t_{1},\ldots,t_{r})=\mu_{\overline{L}\otimes\overline{L}_{1}^{\otimes t_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t_{r}}}(x)

for all $(t_{1},\ldots,t_{r})\in(I_{1}\times\cdots\times I_{r})\cap{\mathbb{Q}}^{r}$ .

Proof.

(1) and (2) are consequences of (3) and (5) in Lemma 3.14, respectively.

(3) We set

f_{0}(t_{1},\ldots,t_{r}):=\mu_{\overline{L}\otimes\overline{L}_{1}^{\otimes t_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t_{r}}}(x)

for $(t_{1},\ldots,t_{r})\in(I_{1}\times\cdots\times I_{r})\cap{\mathbb{Q}}^{r}$ . By (1) and (2), for $\lambda\in[0,1]\cap{\mathbb{Q}}$ and $(t_{1},\ldots,t_{r}),(t^{\prime}_{1},\ldots,t^{\prime}_{r})\in(I_{1}\times\cdots\times I_{r})\cap{\mathbb{Q}}^{r}$ , we have

f_{0}(\lambda(t_{1},\ldots,t_{r})+(1-\lambda)(t^{\prime}_{1},\ldots,t^{\prime}_{r}))\\ \hskip-50.00008pt=\mu^{\operatorname{a}}_{(\overline{L}\otimes\overline{L}_{1}^{\otimes t_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t_{r}})^{\otimes\lambda}\otimes(\overline{L}\otimes\overline{L}_{1}^{\otimes t^{\prime}_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t^{\prime}_{r}})^{\otimes(1-\lambda)}}(x)\\ \leq\lambda\mu_{\overline{L}\otimes\overline{L}_{1}^{\otimes t_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t_{r}}}(x)+(1-\lambda)\mu_{\overline{L}\otimes\overline{L}_{1}^{\otimes t^{\prime}_{1}}\otimes\cdots\otimes\overline{L}_{r}^{\otimes t^{\prime}_{r}}}(x)\\ =\lambda f_{0}(t_{1},\ldots,t_{r})+(1-\lambda)f_{0}(t^{\prime}_{1},\ldots,t^{\prime}_{r}),

that is, $f_{0}$ is concave on $(I_{1}\times\cdots\times I_{r})\cap{\mathbb{Q}}^{r}$ . Therefore, the assertion (3) follows from [7, Corollary 1.3.2]. ∎

Let $(L,h)$ be an element of $\widehat{\operatorname{Pic}}_{C^{0}}^{+}(X)_{{\mathbb{Q}}}$ . We say that $h$ is semipositive if there is a positive integer $n$ such that $L^{\otimes n}\in\operatorname{Pic}(X)$ and $h^{n}$ is semipositive. The following characterization of the semipositivity of $h$ is a consequence of Proposition 3.10.

Proposition 3.16.

For $\overline{L}=(L,h)\in\widehat{\operatorname{Pic}}^{+}_{C^{0}}(X)_{{\mathbb{Q}}}$ , $h$ is semipositive if and only if $\mu_{\overline{L}}=0$ on $X^{\operatorname{an}}$ .

We assume that $|\raisebox{1.72218pt}{.}|$ is non-trivial. Let ${\mathscr{X}}$ be a model of $X$ over $\operatorname{Spec}(\mathfrak{o}_{k})$ . Let $L\in\operatorname{Pic}(X)\otimes{\mathbb{Q}}$ and ${\mathscr{L}}\in\operatorname{Pic}({\mathscr{X}})\otimes{\mathbb{Q}}$ with $\left.{{\mathscr{L}}}\right|_{{X}}=L$ . Let $m$ be a positive integer such that $L^{\otimes m}\in\operatorname{Pic}(X)$ . Then we define $\overline{L}=(L,h)$ to be

(L,h):=\left(L^{\otimes m},\left\{|\raisebox{1.72218pt}{.}|_{{\mathscr{L}}^{\otimes m}}(x)\right\}_{x\in X^{\operatorname{an}}}\right)^{\otimes 1/m}.

Proposition 3.17.

If $L$ is ample and ${\mathscr{L}}$ is nef, then $h$ is semipositive.

Proof.

First we assume that ${\mathscr{L}}$ is ample. We choose a positive integer $n$ such that ${\mathscr{L}}^{\otimes n}\in\operatorname{Pic}({\mathscr{X}})$ and ${\mathscr{L}}^{\otimes n}$ is very ample. Then we have an embedding $\iota:{\mathscr{X}}\to{\mathbb{P}}(H^{0}({\mathscr{X}},{\mathscr{L}}^{\otimes n}))$ and ${\mathscr{L}}^{\otimes n}=\iota^{*}({\mathscr{O}}_{{\mathbb{P}}(H^{0}({\mathscr{X}},{\mathscr{L}}^{\otimes n}))}(1))$ . Let $(e_{1},\ldots,e_{r})$ be a free basis of $H^{0}({\mathscr{X}},{\mathscr{L}}^{\otimes n})$ . We define a norm $\|\raisebox{1.72218pt}{.}\|$ of $H^{0}(X,L^{\otimes n})$ to be

\|a_{1}e_{1}+\cdots+a_{r}e_{r}\|:=\max\{|a_{1}|,\ldots,|a_{r}|\}.

Note that $(H^{0}(X,L^{\otimes n}),\|\raisebox{1.72218pt}{.}\|)_{\leq 1}=H^{0}({\mathscr{X}},{\mathscr{L}}^{\otimes n})$ , so that, by Proposition 3.8, we have $|\raisebox{1.72218pt}{.}|^{\mathrm{quot}}_{(H,\|\raisebox{1.20552pt}{.}\|)}(x)=|\raisebox{1.72218pt}{.}|_{\mathscr{L}^{\otimes n}}(x)$ for $x\in X^{\operatorname{an}}$ . Thus $h$ is semipositive.

In general, let ${\mathscr{A}}$ be an ample invertible sheaf on ${\mathscr{X}}$ and $A:=\left.{{\mathscr{A}}}\right|_{{X}}$ . We choose $\delta\in{\mathbb{Q}}_{>0}$ such that $L\otimes A^{\otimes a}$ is ample for all $a\in(-\delta,\delta)\cap{\mathbb{Q}}$ . Note that $\overline{L}\otimes\left(A,|\raisebox{1.72218pt}{.}|_{{\mathscr{A}}}\right)^{\otimes\epsilon}=\left(L\otimes A^{\otimes\epsilon},|\raisebox{1.72218pt}{.}|_{{\mathscr{L}}\otimes{\mathscr{A}}^{\otimes\epsilon}}\right)$ , so that $\mu_{\overline{L}\otimes\left(A,|\raisebox{1.20552pt}{.}|_{{\mathscr{A}}}\right)^{\otimes\epsilon}}=0$ for $\epsilon\in(0,\delta)\cap{\mathbb{Q}}$ by the previous observation together with Proposition 3.16. On the other hand, by (3) in Lemma 3.15,

\mu_{\overline{L}}(x)=\lim_{\begin{subarray}{c}\epsilon\downarrow 0\\ \epsilon\in{\mathbb{Q}}\end{subarray}}\mu_{\overline{L}\otimes(A,|\raisebox{1.20552pt}{.}|_{{\mathscr{A}}})^{\otimes\epsilon}}(x).

Therefore, $\mu_{\overline{L}}=0$ , and hence $h$ is semipositive by Proposition 3.16. ∎

Remark 3.18.

Assume that the absolute value $|\raisebox{1.72218pt}{.}|$ is non-trivial. Let $L$ be an ample invertible sheaf on $X$ , equipped with a semipositive continuous metric $h$ . Then there exists a sequence $\{(\mathscr{X}_{n},\mathscr{L}_{n})\}_{n\geqslant 1}$ , where $\mathscr{X}_{n}$ is a model of $X$ and $\mathscr{L}_{n}$ is a nef invertible sheaf on $\mathscr{X}_{n}$ such that $\mathscr{L}_{n}|_{X}=L^{\otimes n}$ and that $h_{n}=(|\raisebox{1.72218pt}{.}|_{\mathscr{L}_{n}}(x)^{1/n})_{x\in X^{\mathrm{an}}}$ converges uniformly to $h$ . This follows from Proposition 3.10 and the comparison between quotient metrics and model metrics (via the embedding into the projective spaces of lattices). Combining with Proposition 3.17 and Corollary 3.11, we obtain that, in the non-trivial valuation case, our semipositivity coincides with that of Zhang [12] and Moriwaki [8]. We refer the readers to [6, §6] and to [2, §6.8] for the descriptions of the semipositivity in terms of plurisubharmonic currents. Note that their semipositivity is also equivalent to our semipositivity.

	$\displaystyle\\|s\cdot s^{\prime}\\|_{h\otimes h^{\prime},\hat{\kappa}(x)}$	$\displaystyle\leq\max_{i,j}\left\{\|t_{i}\|_{x}\|t^{\prime}_{j}\|_{x}\\|s_{i}\cdot s^{\prime}_{j}\\|_{h\otimes h^{\prime}}\right\}\leq\max_{i,j}\left\{\|t_{i}\|_{x}\|t^{\prime}_{j}\|_{x}\\|s_{i}\\|_{h}\\|s^{\prime}_{j}\\|_{h^{\prime}}\right\}$
		$\displaystyle\leq\max_{i}\left\{\|t_{i}\|_{x}\\|s_{i}\\|_{h}\right\}\max_{j}\left\{\|t^{\prime}_{j}\|_{x}\\|s^{\prime}_{j}\\|_{h^{\prime}}\right\}$
		$\displaystyle\leq e^{2\epsilon}\\|s\\|_{h,\hat{\kappa}(x)}\\|s^{\prime}\\|_{h^{\prime},\hat{\kappa}(x)}.$

	$\displaystyle\\|\tilde{s}\\|_{h,\hat{\kappa}(x)}$	$\displaystyle\leq\max\{\|a_{1}\|_{x}\\|\tilde{e}_{1}\\|_{h},\ldots,\|a_{r}\|_{x}\\|\tilde{e}_{r}\\|_{h}\}$
		$\displaystyle\leq\max\{\|a_{1}\|_{x}\\|e_{1}\\|,\ldots,\|a_{r}\|_{x}\\|e_{r}\\|\}$
		$\displaystyle\leq e^{\epsilon}\\|s\\|_{\hat{\kappa}(x)},$

3. Continuous metrics of invertible sheaves [026I]

3. Continuous metrics of invertible sheaves

Lemma 3.1.

Proof.

3.1. Extension theorem for a metric arising from a model

Theorem 3.2.

Proof.

Claim 3.2.1.

Proof.

3.2. Quotient metric

Lemma 3.3.

Proof.

Claim 3.3.1.

Proof.

Corollary 3.4.

Proof.

Lemma 3.5.

Proof.

Proposition 3.6.

Proof.

Lemma 3.7.

Proof.

Proposition 3.8.

Proof.

Remark 3.9.

3.3. Semipositive metric

Proposition 3.10.

Proof.

Corollary 3.11.

Proof.

Corollary 3.12.

Proof.

3.4. The functions σ\sigma and μ\mu on XanX^{\operatorname{an}}

Lemma 3.13.

Proof.

Lemma 3.14.

Proof.

Lemma 3.15.

Proof.

Proposition 3.16.

Proposition 3.17.

Proof.

Remark 3.18.

3.4. The functions $\sigma$ and $\mu$ on $X^{\operatorname{an}}$