Disclaimer : I found here https://mathoverflow.net/editing-help in the spoilers paragraph that putting >! would hide following things, which was a way for me to alleviate my question's presentation by hiding paragraphs that could be read if and only if details are needed, but it did not work... So that my question is very long... Sorry for that.

My question is about the assertion starting with "Note that" from the last paragraph of 1.1 of "Spectral theory and analytic geometry over non-Archimedean fields" of Vladimir Berkovich. The assertion is the following :

Let $A$ is a normed ring, $\varphi : M'\rightarrow M$ and $\psi : N'\rightarrow N$ admissible morphisms of semi-normed $A$-modules. Then the induced morphism $\varphi\otimes\psi : M' \otimes_{A} N' \rightarrow M \otimes_{A} N$ defined by $(\varphi\otimes\psi) (m'\otimes n') = \varphi(m')\otimes\psi(n')$ for all $(m',n')\in M'\times N'$ is admissible.

Let me recall as briefly as I can what the previous notions in italic mean in Berkovich's context : >! a normed ring $A$ is commutative ring with unit endowed with a (non necessarily non-archimedean) norm $\|\cdot\|$ such that $\|1\| = 1$ and $\|fg\| \leq \|f\|\|g\|$ for all $f,g\in A$. Then a semi-normed $A$-module is an $A$-module $M$ endowed with a (non necessarily non-archimedean) seminorm $\|\cdot\|$ having the following property : there exists $c\in\mathbf{R}$ such that $\|f m\| \leq c \|f\| \|m\|$ for all $f\in A$ and $m\in M$. Let $M,N$ be seminormed $A$-modules. One can define a seminorm on $M \otimes_{A} N$ by defining the semi-norm $\|x\|$ of $x$ as the quantity $$\inf\left\{ \left.\sum_{i=0}^d \|m_i\|\|n_i\|\;\right|\; \exists d\in\mathbf{N}, \exists m_1,\ldots,m_d\in M,\exists n_1,\ldots,n_d\in \mathbf{N}, x = \sum_{i=0}^d m_i \otimes n_i\right\}$$ Let $\varphi : M \rightarrow N$ be a morphism of semi-normed $A$-modules. It is said to be bounded if there exists a $c\in\mathbf{R}_{+}^{\times}$ such that $\|\varphi(f)\|\leq c \|f\|$ for all $f\in M$. It is said to be admissible if the residue seminorm on $M/\textrm{Ker}(\varphi)$ is equivalent (through the canonical isomorphism $M/\textrm{Ker}(\varphi) \rightarrow \textrm{Im}(\varphi)$) to the restriction to $\textrm{Im}(\varphi)$ of the seminorm of $N$. This is equivalent to the following fact : there exist $c,c'\in\mathbf{R}_{+}^{\times}$ such that for all $x\in M$ one has $$c \| \varphi(x) \| \leq \inf\left\{ \| x+h \|\;|\;h\in\textrm{Ker}(\varphi)\right\} \leq c' \| \varphi(x) \| \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(E) $$ This is also equivalent to say that $M/\textrm{Ker}(\varphi) \rightarrow \textrm{Im}(\varphi)$, isomorphism in the category of $A$-modules, is an isomorphism in the category whose objects are semi-normed $A$-modules and arrows are the bounded morphisms. Finally, If $N$ is a sub-$A$-module of a semi-normed $A$-module $M$ one can define the residue seminorm on the quotient $A$-module $M/N$ as follows : $\|f\| = \inf\{ \|g\|\;|\; g\in {\pi}^{-1}\left( \{f\} \right) \}$ where $\pi : M \rightarrow M/N$ is the canonical epimorphism.

To prove Berkovich assertion, we have to prove both inequalities from the reformulation (E) in the above reminder. This first one is rather easy and goes like this >! We know that there exist $c,c',d,d'\in \mathbf{R}_{+}^{\times}$ such that $$\forall m'\in M', c \| \varphi(m') \| \leq \inf\left\{ \| m'+h \|\;|\;h\in\textrm{Ker}(\varphi)\right\} \leq c' \| \varphi(m') \|$$ and $$\forall n'\in N', d \| \psi(n') \| \leq \inf\left\{ \| n'+k \|\;|\;k\in\textrm{Ker}(\psi)\right\} \leq d' \| \psi(n') \|$$ Let now $\xi'\in M' \otimes_{A} N'$ and $\varepsilon\in\mathbf{R}_{+}^{\times}$. By definition of $\|\xi'\|$ we know that there exist $d\in\mathbf{N}$, $m'_1,\ldots,m'_d\in M'$ and $n'_1,\ldots,n'_d\in N'$ such that $\xi' = \sum_{i=0}^d m'_i \otimes n'_i$ and $\sum_{i=0}^d \|m'_i\|\|n'_i\| \leq \|\xi'\|+\varepsilon$. Now $$\| \left( \varphi\otimes\psi \right) (\xi') \| = \left\| \left( \varphi\otimes\psi \right) \left( \sum_{i=0}^d m'_i \otimes n'_i \right) \right\| \\ = \left\| \sum_{i=0}^d \varphi(m'_i) \otimes \psi(n'_i) \right\| \\ \leq \sum_{i=0}^d \left\| \varphi(m'_i) \right\| \left\| \psi(n'_i) \right\|\textrm{ by definition of the seminorm on $M \otimes_{A} N$}\\ \leq \frac{1}{c c'} \sum_{i=0}^d \|m'_i\|\|n'_i\| \\ \leq \frac{1}{c c'} \left(\|\xi'\|+\varepsilon\right)$$ for all $\varepsilon\in\mathbf{R}_{+}^{\times}$. Letting $\varepsilon\rightarrow 0$ shows that $$\| \left( \varphi\otimes\psi \right) (\xi') \| \leq \frac{1}{c c'} \|\xi'\|$$ for all $\xi'\in M' \otimes_{A} N'$ : the morphism $\varphi\otimes\psi$ is bounded. Now, replacing $\xi'$ by any $\xi'+h$ with $h\in \textrm{Ker}(\varphi\otimes\psi)$ in the previous inequality shows that $$\| \left( \varphi\otimes\psi \right) (\xi') \| \leq \frac{1}{c c'} \|\xi'+h\|$$ for all $\xi'\in M' \otimes_{A} N'$ and $h\in \textrm{Ker}(\varphi\otimes\psi)$, from which we deduce that $$\| \left( \varphi\otimes\psi \right) (\xi') \| \leq \frac{1}{c c'} \inf\left\{\|\xi'+h\|\;|\;h\in \textrm{Ker}(\varphi\otimes\psi)\right\}$$ for all $\xi'\in M' \otimes_{A} N'$, showing thereby "half" of the admissibility of $\varphi\otimes\psi$.

I have tried to prove the second inequality of the reformulation (E) without success. I wrote to Berkovich who answered that this assertion was used in his book only for $k$-affinoid algebras $A$ and finite Banach $A$-modules, with $\varphi$ and $\psi$ admissible epimorphisms, in which case it is really easy to prove, not telling anything about the general case. By the way, without more hypothesis on $A$ and the modules than the initial ones, if the two arrows are epimorphisms, then the assertion it's true. (Easy to prove also.)

But still, the general case remains, so if anyone has an idea, or a counter-example... Thanks !


1 Answer 1


I do not know how to prove the result you want in full generality, but here is a way to do it when $A = K$ is a non-archimedean valued field and $M,M',N,N'$ are Banach spaces over $K$ (and with completed tensor products).

First, remark that the composition of two admissible maps is still admissible whenever the first map is surjective or the second injective. Using this and writing the map you want as the composition $$M'\hat{\otimes}_K N' \to \mathrm{Im}(\varphi)\hat{\otimes}_K N' \to \mathrm{Im}(\varphi)\hat{\otimes}_K N \to M \hat{\otimes}_K N,$$ you see that it is enough to prove that an admissible map remains admissible after tensor product by a single Banach space. Then, you can find this result for instance in [Poineau - Pulita, The convergence Newton polygon of a $p$-adic differential equation IV : local and global index theorems, arXiv:1309.3940, Proposition A.3.8 i]. (The language is different since we try to do more general things. What you call admissible is there called strict in $\mathrm{Ban}^b_{M,K}$ and in our case $M$ is a singleton.)

The strategy is the following. It comes from [Gruson, Théorie de Fredholm $p$-adique, Bulletin de la Société Mathématique de France 94 (1966)] and you can also find a short and nice account of it, especially in you do not read French, in [Rémy - Thuillier - Werner, Bruhat-Tits theory from Berkovich's point of view. I: realizations and compactifications of buildings. Ann. Sci. École Norm. Sup. 43 (2010), proof of Lemma A.5].

Now to the point. Say we start with an admissible morphism $M'\to M$ and want to tensorize with $N$. Remark that the result is clear if $N$ is finite-dimensional since it is then isomorphic to some $k^n$ with the sup-norm by a well-known result. Then you write $N$ as an direct limit of finite-dimensional subspaces and check that direct limits preserve admissibility (direct computation) and commute with tensor products (right-exactness).

Playing a little with completion, you should get the same result for normed spaces and usual tensor products. Maybe it also works with semi-normed spaces and separated completions but I do not have the time to do the computations right now.

On the other hand, at the very basis of this strategy lies the fact that norms of finite-dimensional vector spaces are all equivalent. So it is not going to work this way when $A$ is not a valued field.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.