9
$\begingroup$

In the beginning of the 7th chapter of the book "Spectral theory and analytic geometry over non-Archimedean fields" by Vladimir Berkovich one can find the phrase "...tensor product functor is exact on the category of Banach spaces...". He gave no clue how to prove it, but it is known that the same fact is not true for Archimedean Banach spaces. Is the statement correct, and how can it be proved?

UPD I know that tensor product of complex Banach spaces is not left exact. I'm interested in the proof (or counterexample) for non-Archimedean Banach spaces.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

To make things precise, let me add the end of the quoted sentence: "with admissible linear operators as morphisms". Moreover, I believe that Berkovich refers here to tensor products over a fixed base field.

In this case, the exactness result you are looking for may be found in Gruson's paper "Théorie de Fredholm $p$-adique" (Bulletin de la Société Mathématique de France, Tome 94 (1966), p. 67-95, http://www.numdam.org/item/BSMF_1966__94__67_0/). It is part of Theorem 1 on page 79. To be sure that it means what you want, you will have to unravel a few definitions and you should in particular have a look at the first section of the paper for the definition of the category (enc) and strict morphisms (which are Berkovich's admissible morphisms).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .