3.4. Fubini-Study metrics [00N6]

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3.4. Fubini-Study metrics

We study distances between Fubini-Study metrics, and gives explicit expressions for Fubini-Study metrics admitting non-Archimedean orthogonal basis. Many of the results here are also obtained in [CMor18] or in [BE18, Section 6].

Lemma 3.7.

Assume that $L$ is globally generated. Let $\lVert\mathord{\cdot}\rVert_{1}$ be a norm on $V_{1}(L)$ and let $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{1})$ be the associated Fubini-Study metric. Then for any $x\in X^{\mathrm{an}}$ and $e(x)\in L(x)\setminus 0$ ,

\lvert e(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{1})}=\inf_{\begin{subarray}{c}\lambda\in\widehat{\kappa}(x),\ s_{1}\in V_{1}(L)\\ s_{1}(x)=\lambda\cdot e(x)\end{subarray}}\lvert\lambda\rvert^{-1}\cdot\lVert s_{1}\rVert.

(with the convention that $0^{-1}=+\infty$ )

Proof.

This follows from Lemma 2.11. ∎

Lemma 3.8.

Let $\lVert\mathord{\cdot}\rVert_{n}$ be a norm on $V_{n}(L)$ , then $\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})$ is a continuous metric on $L$ . ([CMor18, Proposition 3.1])

Lemma 3.9.

Let $\phi_{1}$ and $\phi_{2}$ be two upper-semicontinuous metrics on $L$ , then

\dist(\lVert\mathord{\cdot}\rVert_{\phi_{1}},\lVert\mathord{\cdot}\rVert_{\phi_{2}})\leq\dist(\phi_{1},\phi_{2}).

(see also [BE18, Lemma 6.10])

Proof.

For any $s\in V_{1}(L)$ , if $\lVert s\rVert_{\phi_{1}}\geq\lVert s\rVert_{\phi_{2}}$ , let $x_{1}\in X^{\mathrm{an}}$ be a point where $\lVert s\rVert_{\phi_{1}}$ is attained. One has

\Big|\log\frac{\lVert s\rVert_{\phi_{1}}}{\lVert s\rVert_{\phi_{2}}}\Big|\leq\Big|\log\Big|\frac{s(x_{1})}{s(x_{1})}\Big|_{\widehat{\kappa}(x_{1})}\Big|\leq\dist(\phi_{1},\phi_{2}).

Otherwise, let $x_{2}\in X^{\mathrm{an}}$ be a point where $\lVert s\rVert_{\phi_{2}}$ is attained. Then

\Big|\log\frac{\lVert s\rVert_{\phi_{2}}}{\lVert s\rVert_{\phi_{1}}}\Big|\leq\Big|\log\Big|\frac{s(x_{2})}{s(x_{2})}\Big|_{\widehat{\kappa}(x_{2})}\Big|\leq\dist(\phi_{1},\phi_{2}).

Hence the desired inequality holds. ∎

Lemma 3.10.

Let $\lVert\mathord{\cdot}\rVert$ and $\lVert\mathord{\cdot}\rVert^{\prime}$ be two norms on $V_{1}(L)$ , then

\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert),\mathrm{FS}(\lVert\mathord{\cdot}\rVert^{\prime}))\leq\dist(\lVert\mathord{\cdot}\rVert,\lVert\mathord{\cdot}\rVert^{\prime}).

(see also [BE18, Equation (6.2)])

Proof.

For any $x\in X^{\mathrm{an}}$ , let $e(x)\in L^{\mathrm{an}}(x)\setminus\{0\}$ . Let $s,s^{\prime}\in V_{1}(L)$ and $\lambda,\lambda^{\prime}\in\hat{\kappa}(x)$ be elements such that

s(x)=\lambda\cdot e(x),\ \lVert e(x)\rVert_{X|x}=\lvert\lambda\rvert^{-1}\lVert s\rVert,

s^{\prime}(x)=\lambda^{\prime}\cdot e(x),\ \lVert e(x)\rVert_{X|x}=\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}.

If $\lVert e(x)\rVert_{X|x}>\lVert e(x)\rVert^{\prime}_{X|x}$ , one has

\begin{split}\dist(\lVert e(x)\rVert_{X|x},\lVert e(x)\rVert^{\prime}_{X|x})&=\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\\ &\leq\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda\rvert^{-1}\lVert s\rVert^{\prime}}\Big|\leq\dist(\lVert\mathord{\cdot}\rVert,\lVert\mathord{\cdot}\rVert^{\prime}).\end{split}

Otherwise, one has

\begin{split}\dist(\lVert e(x)\rVert_{X|x},\lVert e(x)\rVert^{\prime}_{X|x})&=\Big|\log\frac{\lvert\lambda\rvert^{-1}\lVert s\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\\ &\leq\Big|\log\frac{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert}{\lvert\lambda^{\prime}\rvert^{-1}\lVert s^{\prime}\rVert^{\prime}}\Big|\leq\dist(\lVert\mathord{\cdot}\rVert,\lVert\mathord{\cdot}\rVert^{\prime}).\end{split}

Varying $x$ and taking the supremum, one gets the desired inequality. ∎

Proposition 3.11.

Assume that there exist a norm $\lVert\mathord{\cdot}\rVert_{1}$ on $V_{1}(L)$ such that $\phi$ is equal to $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{1})$ . Then for any $n\in\mathbb{N}$ , $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi})$ is equal to $n\phi$ . ([CMor18, Proposition 3.3])

Proposition 3.12.

Assume that $\phi$ is an asymptotic Fubini-Study metric on $L$ , then the envelop metric $\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi})$ is equal to $\phi$ . (see also [BE18, Theorem 6.15 (iii)])

Proof.

By assumption, there exists a familly of norms $\{\lVert\mathord{\cdot}\rVert_{n}\}_{n\in\mathbb{N}}$ such that uniformly for $x\in X^{\mathrm{an}}$ ,

\quad\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})(\ast)(x)=\lvert\ast\rvert_{\phi}(x),

so for any $\epsilon>0$ , there exists $N_{0}\in\mathbb{N}$ such that for any $n\geq N_{0}$

\dist(n\phi,\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,

hence by Lemma 3.9 and 3.10,

\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))\leq n\epsilon.

By Proposition 3.11, one has $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}))=\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})$ , so

\dist(\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq n\epsilon,

and

\dist(\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi}),\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n}))\leq\epsilon.

Taking limit for $n\to\infty$ and then for $\epsilon\to 0$ , one has

\dist(\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi},\phi)=0,

so the two metrics are equal. ∎

If the ultrametric norm $\lVert\mathord{\cdot}\rVert_{n}$ admits an orthogonal basis, one can calculate explicitly the associated Fubini-Study metric $\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})$ .

Lemma 3.13.

Let $(K,\lvert\mathord{\cdot}\rvert_{K})$ be a complete ultrametric valued field extension of $(k,\lvert\mathord{\cdot}\rvert)$ . Then for any $\{r_{j}\}_{j\in\{0,\dots,d\}}$ elements in $\mathbb{R}_{+}$ , one has

\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{K}\cdot r_{j}\Big\}=\min_{j\in\{0,\dots,d\}}\{r_{j}\},

where $\{\kappa_{j}\}_{j\in\{0,\dots,d\}}$ are elements in $K$ .

Proof.

On the one hand, let $j_{0}\in\{0,\dots,d\}$ be an index such that $r_{j_{0}}$ is minimal. By taking $\kappa_{j}=0$ for $j\neq j_{0}$ and $\kappa_{j_{0}}=1$ , one sees that

\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{K}\cdot r_{j}\Big\}\leq r_{j_{0}}=\min_{j\in\{0,\dots,d\}}\{r_{j}\}.

On the other hand, by the ultrametricity of $\lvert\mathord{\cdot}\rvert_{K}$ , if $\sum_{j}\kappa_{j}=1$ , then there exist at least one $j_{1}\in\{0,\dots,d\}$ such that $\lvert\kappa_{j_{1}}\rvert\geq 1$ , so

\begin{split}\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{K}\cdot r_{j}\Big\}&\geq\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}\lvert\kappa_{j_{1}}\rvert_{K}\cdot r_{j_{1}}\\ &\geq\inf_{\sum_{j\in\{0,\dots,d\}}\kappa_{j}=1}r_{j_{1}}=r_{j_{1}}\geq\min_{j\in\{0,\dots,d\}}\{r_{j}\}.\end{split}

Hence the two sides are equal. ∎

Proposition 3.14.

Assume that $L^{\otimes n}$ is globally generated. Let $\{s_{n,j}\}_{j\in\{0,\dots,d_{n}\}}$ be a basis of $V_{n}(L)$ . Let $\lVert\mathord{\cdot}\rVert_{n}$ be a ultrametric norm on $V_{n}(L)$ with respect to which $\{s_{n,j}\}_{j\in\{0,\dots,d_{n}\}}$ is orthogonal. Then for any $x\in X^{\mathrm{an}}$ and $e_{n}(x)\in L^{\otimes n}(x)\setminus 0$ ,

\lvert e_{n}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}=\min_{j\in\{0,\dots,d_{n}\}}\Big\{\Big|\frac{e_{n}(x)}{s_{n,j}(x)}\Big|_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}\Big\},

with the convention that $0^{-1}=+\infty$ . (see also [CMor18, Lemma 3.3])

Proof.

For $j\in\{0,\dots,d_{n}\}$ , let $\lambda_{j}\in\widehat{\kappa}(x)$ be such that $s_{n,j}(x)=\lambda_{j}\cdot e_{n}(x)$ . By construction,

\begin{split}\lvert e_{n}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})}=&\inf_{\mu_{j}\in\widehat{\kappa}(x),\ \sum\mu_{j}\cdot s_{n,j}(x)=e_{n}(x)}\Big\|\sum_{j\in\{0,\dots,d_{n}\}}\mu_{j}\cdot s_{n,j}\Big\|\\ =&\inf_{\mu_{j}\in\widehat{\kappa}(x),\ \sum\mu_{j}\cdot s_{n,j}(x)=e_{n}(x)}\max_{j}\Big\{\lvert\mu_{j}\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\inf_{\kappa_{j}\in\widehat{\kappa}(x),\ \sum\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\cdot\lambda_{j}^{-1}\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\inf_{\kappa_{j}\in\widehat{\kappa}(x),\ \sum\kappa_{j}=1}\max_{j}\Big\{\lvert\kappa_{j}\rvert_{\widehat{\kappa}(x)}\cdot\lvert\lambda_{j}\rvert_{\widehat{\kappa}(x)}^{-1}\cdot\lVert s_{n,j}\rVert\Big\}\\ =&\min_{j\in\{0,\dots,d_{n}\}}\Big\{\lvert\lambda_{j}\rvert_{\widehat{\kappa}(x)}^{-1}\cdot\lVert s_{n,j}\rVert_{n}\Big\}.\end{split}

The last equality is obtained with Lemma 3.13. ∎

Corollary 3.15.

With the same hypothesis as above, for $e_{n}^{\vee}(x)\in(L^{\otimes n})^{\vee}(x)\setminus 0$ , one has

\lvert e_{n}^{\vee}(x)\rvert_{\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})^{\vee}}=\max\Big\{\lvert e_{n}^{\vee}(x)(s_{n,j})\rvert_{\widehat{\kappa}(x)}\cdot\lVert s_{n,j}\rVert_{n}^{-1}\Big\}.

Proof.

It suffices to note that $e_{n}^{\vee}(x)(s_{n,j})=\lambda_{j}$ . ∎