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Consider the category of Banach spaces with contractions as morphisms (weak, so $\|T\| \le 1$). Is this an algebraic theory?

I suspect that this is true. The "operations" will be weighted sums, where the sum of the weights is at most $1$. The "free Banach space" on a set $X$ should be $\ell^1(X)$. (Note that the "underlying set" functor sends a Banach space to its unit ball.) So, part one:

  • Is this correct? If so, can anyone supply a reference (unfortunately, searching for "algebraic theory" and "Banach" doesn't turn up anything obvious).
  • Has anything useful/unusual come out of this point of view?
  • If this is correct, then the algebraic theory seems to be commutative, in which case it's a symmetric closed category. Has this angle been used?
  • The norm can't be encoded as an operation, can it be categorically recovered?

Part two says: can we do this for Hilbert spaces?

Edit: Anyone even vaguely intrigued by this question should read the paper linked in Yemon's answer. In particular, it also answers a question that I was going to ask as a follow-up: what's the nearest algebraic theory to Banach spaces (the answer being totally convex spaces).

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I suspect that we will lose the topology, so this algebraic theory will not contain all of the information of the Banach space. – Harry Gindi Dec 11 '09 at 9:18
You don't lose the topology since the "underlying set" is the unit ball and that contains all the information about the topology that you need to know. – Loop Space Dec 11 '09 at 9:53
Well, if you've got the metrizable topology, and you've got the module structure, doesn't that give you the norm up to equivalence? – Harry Gindi Dec 11 '09 at 10:04
The norm can be recovered exactly via a simple supremum computation, but this doesn't feel very "categorical". – Loop Space Dec 11 '09 at 10:57
up vote 6 down vote accepted

I think this is some kind of infinitary algebraic theory, but that it is not a monadic adjunction. That is, if you take the "closed unit ball functor" $B$ from ${\bf Ban}\_1$ to ${\bf Set}$ and the "free Banach space functor" $L: {\bf Set} \to {\bf Ban}_1$, then $L$ is left adjoint to $B$ but this adjunction is not monadic (IIRC, and I often don't).

See, for instance, the first few pages of this paper by Pelletier and Rosicky.

You say something about a closed structure on ${\bf Ban}_1$, if I understand this right then this is symmetric monoidal with the tensor being the projective tensor product of Banach spaces. That seems to be well known but little-used, although IMHO having this kind of perspective takes some of the tedium/clutter out of certain computations/constructions in my corner of functional analysis.

I think the ball functor from (Hilbert spaces & contractions) to Set doesn't have a left adjoint, but that's more of a guess than an intuition. Certainly the `natural' attempt to build a left adjoint falls over.

As for putting a closed structure on Hilb .... well, the fact that the natural norm on B(H) is not Hilbertian suggests to me that this won't work. (Put another way, the natural Hilbertian tensor product would dualise to only considering Hilbert-Schmidt class maps between your Hilbert spaces, which in infinite dimensions rules out the identity morphism.)

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I know a closed structure won't work on Hilbert spaces, I meant the question to be about whether or not there's an algebraic theory there. – Loop Space Dec 12 '09 at 21:39
OK, misunderstood that bit of your question. Well, on a quick glance doesn't Mike Shulman's answer tell us "no"? – Yemon Choi Dec 12 '09 at 21:46
Part of the reason that I'm accepting this answer is the reference to Pelletier and Rosicky. That paper answers all my questions (about Banach spaces), is very readable, and is packed with nuggets of information and links to further reading. – Loop Space Dec 17 '09 at 8:48

One of the major-league experts on this topic is Michael Barr. In a message to the categories list (dated October 22, 2003), he writes:

"For Banach spaces, if you take as underlying functor the closed unit ball, it has an adjoint. It is not tripleable, however, but C^*-algebras are (with the unit ball underlying functor)."

("Tripleable" means the same as monadic. See the discussion in the book by Barr-Wells, Toposes, Theories, and Triples, pages 105-106 [pp. 118-119 of 303.) I think that I saw somewhere though that the forgetful functor from Banach spaces and linear contractions to metric spaces with basepoint and contractions is monadic.

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Thanks! 2003 is before I joined the mailing list so I missed that one. – Loop Space Jan 5 '10 at 18:53

You may also be interested in the discussion on p59-61 of Makkai-Pare's "Accessible categories," which shows that Banach spaces and Hilbert spaces are both accessible categories by giving an explicit axiomatization in terms of relations and operations. They claim that Banach spaces are actually locally presentable, although their proof doesn't show this; it could also be deduced from (co)completeness since any complete or cocomplete accessible category is locally presentable. Local presentability is not the same as being algebraic, of course. Hilbert spaces, on the other hand, are not locally presentable (and hence not complete or cocomplete, and thus certainly not algebraic), since their category is self-dual and no locally presentable category (other than a poset) has a locally presentable dual.

In "Basic concepts of enriched category theory" Kelly mentions the category of Banach spaces as one of his examples of a symmetric monoidal closed category one can enrich over. I don't know whether there are interesting applications.

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The reference (to part one) is Nikolai Durov's dissertation. So what has come out of this point of view is a new approach to Arakelov's geometry, Durov's compactification of $Spec\:\mathbb Z$, a point of view on $F_1$, etc. The commutativity of algebraic theories (or generalized rings, in his terminology), including this particular generalized ring, has been used by Durov extensively, yes.

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A quick look through the dissertation isn't very enlightening. Can you give a more specific reference (it is 568 pages long!). I searched for "Banach" and it was almost always near the words "finite dimensional". – Loop Space Dec 11 '09 at 14:42
That's right, he does it "algebraically", i.e., considers finite weighted sums only. It does indeed appear that finite-dimensional Banach spaces are more important to him than infinite-dimensional ones. As to the more specific references, see subsection 0.2 and section 2. Banach spaces are related to what he calls "generalized ring $\mathbb Z_\infty$" and "lattices" or "modules" over it. – Leonid Positselski Dec 11 '09 at 15:36

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