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I was looking carefully at all the definitions, trying to understand exactly what was going on in this question on categorical duals in Banach spaces. It seems that in the category of Banach spaces with contractions, only finite dimensional Banach spaces have duals in the sense of category theory. This is because one needs morphisms

$$ I \to A \otimes A^* $$

and

$$ A \otimes A^* \to I $$

where the first is like the inclusion of the identity and the second like the trace. This means that the identity map on $A$ has to have a trace, and that only works for finite dimensional Banach spaces.

So, my question: am I right? Or have I misunderstood something here?

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2 Answers 2

My internet access at the moment is limited & sluggish, so I haven't been able to look up all the details; but I think your reasoning is correct. Certainly my impression is that duals in the sense beloved by (S)MC people only work for finite-dimensional Banach spaces.

By the way, for arbitrary Banach spaces the first map you describe wants to land in the injective tensor product, while the second eants to come out of the projective tensor product. Thus the failure to get categorical duals for inf-dim Banach spaces is surely related to, though perhaps neither implying nor implied by, the following old result which I think is due to Grothendieck: if X is a Banach space and, for each Banach space E, the usual tensor product of X with E (in the category Vect) has a unique Banach completion, then X is finite-dimensional. More pithily, the only nuclear Banach spaces are the finite-dimensional ones.

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From the recent discussion in the question you linked to, it seems that even (some? most?) finite-dimensional Banach spaces do not have categorical duals in the precise sense you describe. In fact it's clear what goes wrong here with the construction used in the algebraic setting of finite-dimensional vector spaces: the composition of the two maps you describe would be multiplication on $I$ by the dimension of $A$, but that is not a contraction unless $\dim A \le 1$!

You might have more luck with a category of Banach spaces and all continuous linear maps, although I do not know whether that also has a closed symmetric monoidal structure.

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1  
Yes, that's a good point: the compact objects in Ban with all bounded linear maps as morphisms are indeed the finite-dimensional ones. This category is smc for the same reason as Ban_1 is (on general grounds, the only "extra" issue would be whether the structural constraints used for Ban_1 are natural with respect to all bounded linear maps, not just wrt the contractions, but that's obvious from linearity -- actually I think of it as the other way around: the smc structure on Ban is easy, and that the structural constraints belong to Ban_1 is a small extra bonus). –  Todd Trimble Jan 6 '10 at 1:32

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