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This is inspired partly by this question, especially Tom Leinster's answer.

Let me start with some background. I apologize that this will be rather long, since I'm hoping for input from people who probably don't much about the following.

Classically, there are two obvious categories whose objects are Banach spaces (let's say all Banach spaces are real for simplicity): in the first, morphisms are bounded linear maps and isomorphisms are exactly what are usually called isomorphisms of Banach spaces; in the second, morphisms are linear contractions (bounded linear maps with norm at most 1) and isomorphisms are linear isometries. These categories distinguish between the "isomorphic" and "isometric" theories of Banach spaces.

Now if I'm interested in finite-dimensional spaces, then the "isomorphic" category is not rigid enough, because any two n-dimensional Banach spaces are isomorphic. But the isometric category is too rigid for most purposes. So we get more quantitative about our isomorphisms. One way to do this is with the Banach-Mazur distance. If X and Y are both n-dimensional Banach spaces, $$d(X,Y) = \inf_T (\lVert T \rVert \lVert T^{-1} \rVert),$$ where the infimum is over all linear isomorphisms $T:X\to Y$. Then $\log d$ is a metric on the class of isometry classes of n-dimensional Banach spaces.

Theorems about spaces of arbitrary dimension which include some constants independent of the dimension are characterized as "isomorphic results". One example is Kashin's theorem: There exists a constant c>0 such that for every n, $\ell_1^n$ has a subspace X with $\dim X = m= \lfloor n/2 \rfloor$ such that $d(X,\ell_2^m) < c$. (Here $\ell_p^n$ denotes R^n with the $\ell_p$ norm $\lVert x \rVert_p = (\sum |x_i|^p)^{1/p}$.) Thus $\ell_1^n$ contains an n/2-dimensional subspace isomorphic to Hilbert space in a dimension-independent way.

On the other hand there are "almost isometric" results, typified by Dvoretzky's theorem: There exists a function $f$ such that for every n-dimensional Banach space X and every $\varepsilon > 0$, X has a subspace Y with $\dim Y = m \ge f(\varepsilon) \log (n+1)$ such that $d(Y,\ell_2^m) < 1+\varepsilon$. Thus any space contains subspaces, of not too small dimension, which are arbitrarily close to being isometrically Hilbert spaces.

So my question, finally, is: are there natural categories in which to interpret such results? I suppose that the objects should not be individual spaces, but sequences of spaces with increasing dimensions. In particular, as the two results quoted above highlight, the sequence of n-dimensional Hilbert spaces $\ell_2^n$ should play a distinguished role. But I have no idea what the morphisms should be to accomodate quantitative control over norms in these ways.

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3 Answers 3

This is a rewording of Matthew's answer, but one can use nonstandard analysis to obtain a category that achieves what Mark wants. After taking ultrapowers, one obtains non-standard integers (ultralimits of standard integers), nonstandard finite dimensional Banach spaces (ultralimits of standard finite dimensional Banach spaces), nonstandard linear transformations (ultralimits of standard linear transformations), etc. One can then separate these non-standard objects into various classes, e.g. bounded non-standard reals (ultralimits of uniformly bounded standard reals, or equivalently non-standard reals that are bounded in magnitude by a standard real), bounded nonstandard linear transformations, poly(n)-bounded nonstandard linear transformations, polylog(n)-bounded nonstandard linear transformations, etc., where n is an unbounded nonstandard natural number (which, in practice, would be used to bound dimensions of things). Each of these form a category.

Thus, for instance Kashin's theorem becomes the statement that every nonstandard Banach space of some nonstandard finite dimension N has an M-dimensional subspace which is isomorphic (in the category of bounded nonstandard linear transformations) to $\ell^2(M)$ whenever $M \leq N/2$ (or more generally when $M \leq (1-\epsilon)N$ for some standard $\epsilon > 0$.

Dvoretsky's theorem is trickier. Here, I guess one needs to work with the category of almost contractions: operators whose operator norm is at most $1+o(1)$ (i.e. bounded by $1+\epsilon$ for every standard $\epsilon > 0$. Then I think the theorem says that any nonstandard finite dimensional Banach space with some nonstandard dimension N has a M-dimensional subspace that is almost isometric to $\ell^2(M)$, whenever $M = o(\log N)$. (I may have messed up the quantifiers slightly, but this is pretty close to what the nonstandard translation of things should be.)

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The thing that I don't see in your formulation of Dvoretzky is where the dependence of M on $\varepsilon$ is. In the classical formulation, I may need to take a much smaller subspace of X to be 1.01-isomorphic to Hilbert space than I need to be 1.1-isomorphic. But I haven't taken the time to absorb your and Matthew's answers, so maybe it's there and I just need to think some more. –  Mark Meckes Nov 19 '09 at 19:27
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A (reasonably) standard trick to convert finite-dimensional limiting type problems into single, infinite dimensional problems is to use Ultraproducts (or non-standard hulls, if you know more model theory than I do).

Let $(E_i)$ be a family of Banach spaces on an index set $I$, and let $\ell^\infty(I,E_i)$ be the Banach space of all bounded families $(x_i)_{i\in I}$ where $x_i\in E_i$. Let $U$ be a non-principal ultrafilter on $I$, and consider $N_U=\{ (x_i)\in\ell^\infty(I,E_i) : \lim_{i\rightarrow U} \|x_i\|=0 \}$. This is a closed subspace, so we can form the quotient Banach space $\ell^\infty(I,E_i) / N_U$. This is the ultraproduct on the $(E_i)$. If $E_i=E$ for all $i$, we get an ultrapower of $E$. (If you know about ultraproducts from logic, there are the obvious "normed" definitions).

Heinrich has a nice paper in Crelles on this. For example, Dvoretzky's theorem now becomes: for any infinite dimensional Banach space $X$, all ultrapowers of $X$ contain an isometric copy of $\ell^2$. (Okay, so I've lost some constants).

I'm not sure quite how to get this into a category theory setting, however.

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Well, you can take ultralimits of morphisms also. A bunch of uniformly bounded linear transformations $T_i: E_i \to F_i$ induces a bounded linear transformation $T: \ell^\infty(I,E_i)/N_U \to \ell^\infty(I,F_i)/N_U$. This seems to give the right category if one wants dimension-independent bounds. If instead one wants (say) bounds that are polynomial on the dimension, one replaces "bounded" by "growing polynomial in the dimension" throughout, etc. –  Terry Tao Nov 19 '09 at 16:41
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The best solution that I can think of is one that analysts have already embraced: Use the category of linear maps, but enrich the category over itself. In other words, make $\mathrm{Hom}(X,Y)$ into a Banach space as well using operator norm. Then it isn't so bad that all $n$-dimensional Banach spaces are isomorphic. Since the isomorphisms have norms themselves, you can ask how well they are isomorphic.

Strictly speaking, you can't enrich a category over itself, because that's circular. However, there is a way out of that in category theory; self-enrichment is called "internal Hom" and there are axioms for it. It is closely related to making the category into a tensor category, and finite-dimensional Banach spaces are that too.

Another way of saying is this: A category in which everything is isomorphic is a groupoid, plus some non-invertible maps. That may seem unsatisfying. The $n$-dimensional Banach spaces are indeed a groupoid, but they are a normed groupoid, which is not so bad.


Okay, that last paragraph is something that Mark already was saying, so let me make a slightly different point. Organizing objects of study into a category is a kind of justification. If it is an enriched category, then it is a relative justification: The modules of an algebra are a category over the category of vector spaces, so modules are justified relative to vector spaces. (And not just relative to sets in that case.)

Sometimes the best justification is a self-justification, and that's what internal homs do for you. The categories Set and Vect are already like this. Since in the category of Banach spaces, $\mathrm{Hom}(X,Y)$ is also a Banach space, that's a self-justification in much the same spirit. I suppose that to fully capture the idea, you should introduce Banach algebras and mixed multiplication between Banach spaces. The bilinear composition of $\mathrm{Hom}(X,Y)$ and $\mathrm{Hom}(Y,X)$ is what tells you how close $X$ and $Y$ are to isometric.

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