This post is about a result that has recently been proved in my old stamping ground of the theory of Banach spaces. When I set up this blog, I wasn’t expecting to write a post about Banach spaces, but the result I want to talk about is one of those rare and delightful events when a problem that you thought you might well never live to see solved is solved. And since Banach space theory is one of the less fashionable areas of mathematics, the result may well not get the publicity it deserves: this is an attempt to counteract that to a small extent.
To explain the significance of the result, it will help to talk about a Banach space that Bernard Maurey and I constructed almost 20 years ago. At the time, this was a good candidate for the “nastiest known Banach space”, in a sense that it had almost no non-trivial structure.
What does this mean? Well, one obvious sort of structure that one likes in a Banach space is symmetry, which we detect by finding operators from the space to itself. A Hilbert space is an extreme example of a space with lots of structure: one has a huge group of unitary matrices, as well as an abundance of projections. Moreover, all the subspaces of a Hilbert space look like the space itself. Thus, one way of saying that a Banach space has structure is to say that the space of continuous linear operators from to is large.
The space that Maurey and I constructed has the property that every operator in is of the form , where is a scalar, is the identity, and is a strictly singular operator. The formal definition of a strictly singular operator is that you cannot restrict to an infinite-dimensional subspace on which it is an isomorphic embedding. To put that another way, for every infinite-dimensional subspace of and every you can find such that and . (The idea is that you can’t “make it non-singular” by restricting to a subspace.)
Now there are all sorts of theorems about strictly singular operators that show that we can regard them as in some sense “small”. For example, it is easy to see that a non-trivial projection (by which I mean that both and are infinite-dimensional) is not of the form . Also, if is Fredholm with index and is strictly singular, then is Fredholm with index . These two facts implied that could not be decomposed as a direct sum in such a way that the projections on to and were continuous (and in fact every subspace of had that property), and also that was not isomorphic to any proper subspace of itself (since any isomorphism from to a proper subspace would have to be Fredholm with non-zero index, whereas is Fredholm with zero index).
Now most people, if you ask them what they would understand by an operator being “small”, would not say “strictly singular”. In fact, most people haven’t even heard of strictly singular operators. The usual notion of smallness is compactness: an operator is considered to be small if the closure of the image of the unit ball is compact. In particular, if an operator has finite rank, then it is compact, and one can think of compact operators as “almost” having finite rank (though this raises subtle questions to do with the approximation property, which I won’t go into here). It turns out that the strictly singular operators and the compact operators on a Hilbert space are the same, which perhaps explains why strict singularity is not such a well-known concept. In general, a compact operator must be strictly singular but there are strictly singular operators that are not compact.
An obvious question this raised was whether the space Maurey and I had constructed admitted an operator that was strictly singular but not compact. If not, then we would have a space with the even stronger property that every operator was of the form with compact. However, as Androulakis and Schlumprecht eventually showed in 2003, that space does have non-compact strictly singular operators.
Since the construction of the original space, many variants have been constructed, which managed to combine its lack-of-structure properties with “nice” properties such as uniform convexity. (That particular result was proved by Valentin Ferenczi.) In particular, one person, Spiros Argyros, made this area his own, and with various collaborators proved a variety of remarkable results, both positive and negative, about spaces of this kind.
All this brings me back to the main subject of this post. When I thought about how one might try to build a space on which every operator is of the form , I came to the conclusion that this was such a strong property for the space to have that there might even be a proof that no such space existed. But I was unable to get anywhere at all with the problem — in either direction.
Fast forward to the end of 2008, when I received an email from Spiros Argyros to say that he and Richard Haydon had constructed a space on which every operator is a multiple of the identity plus a compact operator. Apparently they did this some months earlier, but the news hadn’t filtered through to me.
Now you might ask, if the strictly singular operators are the same as the compact operators on a Hilbert space, and if strictly singular operators do much of what compact operators do, does the new space of Argyros and Haydon have properties (besides the basic property they proved) that have not yet been shown for any other Banach space? The answer to this question is a resounding yes, but to explain this I need to give a bit more background.
One of the biggest problems in functional analysis is the invariant subspace problem , which asks whether for every operator on a Hilbert space there is a proper closed subspace such that . Even the corresponding question for Banach spaces is very hard, but operators without invariant subspaces have been constructed for various Banach spaces in amazing work of Per Enflo and Charles Read. Meanwhile, the problem for Hilbert spaces remains stubbornly open.
Now one might speculate that the result is hard to prove because it is in fact false. And one might even speculate that it is false for every Banach space. However, the example of Argyros and Haydon shows that the situation is more complicated. A famous result of Lomonosov shows that every operator that commutes with a non-zero compact operator must have an invariant subspace. And obviously commutes with if is non-zero, or trivially has an invariant subspace if . From this we conclude that every continuous linear operator on the space of Argyros and Haydon has a non-trivial invariant subspace, the first space for which such a result is known.
What this shows is that you can’t hope to find a counterexample for a general Banach space, because in a sense a general Banach space doesn’t have to have enough operators for there to be any chance at all of a counterexample. (This reminds me slightly of a problem that a different space Maurey and I “solved”. A Banach space is called prime if for every continuous projection with infinite rank, is isomorphic to . It was known that spaces and were prime, and the question was whether there were any other prime spaces. Maurey and I constructed a variant of our original space that was isomorphic to all its finite-codimensional spaces but did not admit any non-trivial projections. So it was prime by virtue of there not being enough projections for it to have a decent chance of not being prime.)
Thus, the Argyros-Haydon space has very definitely taken over as the new “nastiest known Banach space”.
If you want to join this race, there is still one unsolved problem, which arises when one tries to imagine what the strongest possible nonexistence-of-operators result could be. If you’re given a Banach space , then you automatically have all multiples of the identity. Also, the Hahn-Banach theorem allows you to define a wide range of finite-rank operators, and they are all continuous. (To define a rank-one operator, pick a vector and a linear functional and map to . Then add these rank-one operators together to get arbitrary finite-rank operators.) Thirdly, since is a Banach space, if you take a bunch of rank-one operators such that , then the sum is continuous. The question that’s still open is whether there must be any operators besides these trivial examples. An absolutely convergent sum of rank-one operators is called a nuclear operator, so we can ask this question in a different way: does there exist a Banach space such that every operator is of the form , where is nuclear?
It would be interesting to know what Argyros and Haydon think the answer to this question is likely to be: I have no idea myself, but would secretly quite like it if someone managed to construct just a very slightly nontrivial operator on an arbitrary (separable) Banach space. On the other hand, if someone gave an example where that couldn’t be done, then it would be the ultimately structureless Banach space, and that would be pretty good as well. (It’s tempting to say that the Argyros-Haydon space is already the ultimate space, and that this one would be beyond ultimate.)
I think the preprint of Argyros and Haydon isn’t yet up on the web. As soon as it is, I’ll add a link to it.
Update 24/3/09: the paper is now available on the arXiv.