# bornological vector spaces over a non-archimedean field

Let $k$ be a complete non-archimedean field. In definitions I have seen of bornological vector spaces over $k$ there are usually some extra assumptions on the non-archimedean field. For instance in 'Espaces analytiques relatives et theorem de finitude' by Houzel it is assumed that the valuation is non-discrete and that $k$ is maximally complete. On page 43 of 'Seminaire Banach' published as Springer Lecture notes in Mathematics volume 277, it is assumed that the valuation is non-discrete. What is the main reason behind these restrictions? I am interested in bornological vector spaces over a field with trivial valuation. Banach spaces over such a field make sense and therefore one gets a metric space and therefore a bornological set where the bornology is compatible with the linear structure. This should still be a (complete) bornological vector space hopefully. For what parts of the theory are these extra restrictions needed or useful? Are there some pathologies about the category of bornological vector spaces over a general complete non-archimedean field that are not present when you add these extra assumptions?

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I think that Francesco Baldassarri was interested by this kind of things recently. Maybe you should ask him. – Jérôme Poineau Feb 25 '13 at 20:01
In P. Schneider's Nonarchimedean Functional Analysis (§6) the notion of a bornological locally convex vector space is introduced over any nonarchimedean field with non-trivial valuation. – Torsten Schoeneberg Feb 26 '13 at 14:04
Thanks, I was not sure what goes wrong in Schneider's theory if you try to apply it when the valuation is trivial. – Oren Ben-Bassat Feb 27 '13 at 5:22
I am mainly interested in tensor products in the category of complete bornological sapces and comparing them to tensor products in the category of Banach spaces. Is there a fully faithful functor Ban---> Born that does the job in general? In the Archimedean setting over the complex numbers I think there is such an assignment, discussed by Meyer. If you drop conditions of completeness I think there is also such a functor, due to Houzel. – Oren Ben-Bassat Feb 28 '13 at 12:44

1) In the first part of "Espaces analytiques relatives et theorem de finitude" Houzel says that the field under consideration is supposed to be maximally compact. I don't see any point where he use this hypothesis in his article neither in the result that he recall from the "Seminarie Banach". Moreover, in the second part, where he expose the sheaf-theoretic (global) version of the result obtained in the first part he recall the notation and remove this hypothesis (see on page 29 the first two lines of the paragraph "Faisceaux bornologiques"). Hence my idea with respect to this hypothesis is that it is simply a misprint.

2) For sure you can develop a theory of bornological or topological vector spaces over any valued field (also over any field with more exotic structures). The problem is that already in the case of trivially valued fields you find pathologies. First, the notion of convexity is quite strange: since $k = k^\circ$ then you find that the natural generalization of the absolute convex hull of a subset $X \subset E$ (where $E$ is a $k$-vector space) is given by $\Gamma(X) = \left \{ \sum \lambda_i x_i | \lambda_i \in k^\circ = k, x_i \in X \right \}$. Hence $\Gamma(X)$ is the linear span of $X$.

The main problem in this situation is that there is no analog of the duality between bornological and topological vector spaces, which is the main topic of the seminaire Banach. Houzel construct two adjoint functors $t: Born \to Top$ and $b: Top \to Born$ where $Born$ is the category of bornological vector spaces of convex type and $Top$ is the category of locally convex topological vector spaces. For the $t$ functor you consider on a bornological vector space of convex type the vector space topology given by bornivorous sets (i.e. sets which absorbs all bounded sets). And for the $b$ functor you consider on locally convex topological vector space the Von Neumann bornology.

So if you perform this construction on a seminormed vector space $E$ over a non-trivially valued field you get that $E \cong b(t(E))$ and $E \cong t(b(E))$ (but this is also true for more general $k$-vector space topologies and bornologies). This is false for the trivial valued field even thinking of $k$ as a $1$-dimensional vector space over itself. Cause if $k$ is trivially valued, then all subsets of $k$ are bounded. In particular $k$ itself is bounded, so the only bornivorus set for this bornology is $k$. Therefore, in this case, $t(k)$ has the indiscrete topology but the trivial valuation gives to $k$ the discrete topology. Moreover, if you try to describe the Von Neumann bornology for the discrete topology, you don't find a bornology cause $0$ is a neighborhood of itself and $0$ doesn't absorb nothing else than itself. This mean the $0$ would be the only bounded set for the Von Neaumann bornology but this doesn't make sense.

This lack of duality is (i think) the main issue for which Houzel exclude the trivial valuation in his work.

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My name is not displayed in the answer... so I am Federico Bambozzi :) – Federico Bambozzi Feb 26 '13 at 20:11
Dear Federico, Thanks for the great answer. I will probably accept it unless someone comes up with something in the next few days which could improve on that. Also, thanks for using the word bornivorous, what a great word! It seems that with the standard definition of completeness that every bornological vector space over a trivially valued field is complete. This seems a little strong but maybe it makes sense. – Oren Ben-Bassat Feb 27 '13 at 5:19
Yes, because if you try to develop the theory of Mackey convergence in this settings you find that a sequence is convergent if and only if become stationary. The word bornivorous is stolen from "Bornologies and functional analysis" by Henri Hogbe Nlend, North-Holland Mathematics Studies 26; which is another good reference for the theory of bornological vector spaces and the duality with topological vector spaces. – Federico Bambozzi Feb 27 '13 at 8:46
I can see now that the assumption of being non-trivially valued is used. However, I really wonder how much of it works if we assume the field is non-trivially valued but do not assume it is not discrete. – Oren Ben-Bassat Mar 1 '13 at 18:02
Everything works over non-trivially valued fields of any kind. When Houzel says "non-discritely valued field" he really mean "non-trivially valued field". He is using an old-fashioned way of call a trivially valued field a "discrete valued field" since the trivial valuation equip the field with the discrete topology. I think that functional analysts uses this terminology sometimes, for example you can find it also in the GTM number 3 on topological vector spaces. – Federico Bambozzi Mar 3 '13 at 13:46