Subsection 3.6. (00N8) · ScalingStacks

Verified tagged author-source HTML · 1904.03696v1 · cited publication edition alignment unverified.

3.6. Comparison of algebra norms

We compare the quotient algebra norm and the supremum algebra norm on the restricted section algebra, and get directly a (non-uniform) extension theorem.

00LA

Proposition 3.24. Let $\phi$ be an upper-semicontinuous metric on $L$ . Then

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi})|_{Y}=\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y}).

00LB

Proof. By definition, for any $y\in Y^{\mathrm{an}}$ , one has

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi})(y)=\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi})(y),

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y})(y)=\lim_{\begin{subarray}{c}n\to\infty\end{subarray}}\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi,X|Y})(y).

Since the $k$ -linear map $V_{n}(L)\rightarrow V_{n}(L_{X|Y})$ is surjective for all large $n\in\mathbb{N}$ , and $\lVert\mathord{\cdot}\rVert_{n\phi,X|Y}$ is the quotient norm of $\lVert\mathord{\cdot}\rVert_{n\phi}$ , one has

\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi})(y)=\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n\phi,X|Y})(y).

Hence the two envelop metrics are equal. ∎

00LC

Lemma 3.25. Let $\phi$ be an asymptotic Fubini-Study metric on $L$ . Then $\phi|_{Y}$ is an asymptotic Fubini-Study metric on $L|_{Y}$ .

00LD

Proof. Suppose that $\phi$ is the pointwise limit on $X^{\mathrm{an}}$ of $\{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n})\}$ , where $\{\lVert\mathord{\cdot}\rVert_{n}\}$ are norms on $V_{n}(L)$ . Then $L|_{Y}$ is the pointwise limit of $\{\frac{1}{n}\mathrm{FS}(\lVert\mathord{\cdot}\rVert_{n,X|Y})\}$ on $Y^{\mathrm{an}}$ . ∎

00LE

Proposition 3.26. Let $\phi$ be a asymptotic Fubini-Study metric on $L$ . Consider two algebra norms $\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}}$ and $\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y}$ on $V_{{\scriptscriptstyle\bullet}}(L_{X|Y})$ . Then the three metrics are equal

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y})=\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}})=\phi|_{Y}.

00LF

Proof. By Proposition 3.24,

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y})=\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi})|_{Y}=\phi|_{Y}.

It suffices to show the second equality. By Lemma 3.25, $\phi|_{Y}$ is an asymptotic Fubini-Study metric on $L|_{Y}$ . By Proposition 3.12,

\mathcal{P}(\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}})=\phi|_{Y}.

∎

00LG

Corollary 3.27. Let $\phi$ be a asymptotic Fubini-Study metric on $L$ . Then on $V_{{\scriptscriptstyle\bullet}}(L_{X|Y})$ , the spectral algebra seminorm of $\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y}$ is equal to $\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}}$ . There exists a canonical homeomorphism

\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y}))\simeq\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y})).

00LH

Proof. By Theorem 2.33, one has a homeomorphism

\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y;\mathrm{sp}}))\simeq\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y}))=\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi_{X|Y})),

by Proposition 3.26, one has

\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y;\mathrm{sp}}))\simeq\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}}))=\mathfrak{M}(\widehat{V}_{{\scriptscriptstyle\bullet}}(L_{X|Y},\phi|_{Y})).

By Proposition 2.30, the two power-multiplicative algebra seminorms $\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}}$ and $\vvvert\mathord{\cdot}\vvvert_{\phi,X|Y;\mathrm{sp}}$ on $V_{{\scriptscriptstyle\bullet}}(L_{X|Y})$ are equal since they are both supremum norms on the same spectrum. ∎

00LI

Theorem 3.28. Let $\phi$ be an asymptotic Fubini-Study metric on $L$ , then for any $\epsilon>0$ , and any $t_{1}\in V_{1}(L|_{Y})$ , there exists $n_{Y}\in\mathbb{N}$ such that for any $n\geq n_{Y}$ , there exists $s_{n}\in V_{n}(L)$ with $s_{n}|_{Y}=t_{1}^{\otimes n}$ and

\lVert s_{n}\rVert_{n\phi}\leq\mathrm{e}^{n\epsilon}\cdot(\lVert t_{1}\rVert_{\phi|_{Y}})^{n}.

00LJ

Proof. For any $M\leq m\leq 2M-1$ , we have $t_{1}^{\otimes m}\in V_{m}(L_{X|Y})$ . By Corollary 3.27, for any $\epsilon>0$ , there exists $N_{m}\in\mathbb{N}$ such that for $l\geq N_{m}$ ,

\vvvert t_{1}^{\otimes ml}\vvvert_{\phi,X|Y}^{\frac{1}{l}}/\vvvert t_{1}^{\otimes m}\vvvert_{\phi|_{Y}}\leq\epsilon.

It is easy to see that there exists $n_{Y}\in\mathbb{N}$ such that the set of integers

\{ml:l\geq N_{m},M\leq m\leq 2M-1\}

contains a subset of form $\mathbb{N}-\{0,\dots,n_{Y}-1\}$ : the case $M=1$ is clear; if $M\geq 2$ , the fact that $M$ and $M+1$ are coprime guarantees the existence of $n_{Y}$ . Note that $\vvvert\mathord{\cdot}\vvvert_{\phi|_{Y}}$ is power-multiplicative, so for any $n\geq n_{Y}$ , there exists $s_{n}\in V_{n}(L)$ with $s_{n}|_{Y}=t_{1}^{\otimes n}$ such that

\lVert s_{n}\rVert_{n\phi}\leq\mathrm{e}^{n\epsilon}\cdot\lVert t_{1}^{\otimes n}\rVert_{n\phi|_{Y}}=\mathrm{e}^{n\epsilon}\cdot(\lVert t_{1}\rVert_{\phi|_{Y}})^{n}.

∎

00LK

Remark 3.29. This result is first obtained in [CMor18], by using approximation of $\phi$ by model metrics. Here we give another proof. Note that a slight unsatisfactory point of this version of metric extension theorem is that the degree $n_{Y}$ depends a priori on the choice of initial data, the restricted section $t_{1}$ . We will remove this dependence in the following sections.