3.2. Quotient metric [026R]

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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, $\mathscr{H}=(H,\|\raisebox{1.72218pt}{.}\|)_{\leq 1}$ .

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

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