Take 2: a public repost of something already shared with a limited circle, to whom I apologize for the repetition.Amenable operator algebras#spnetwork
arXiv:1309.2145 #amenability #Banachalgebras #operatoralgebras
(Not sure if I've correctly understood this Selected Papers Network tagging malarkey -- a concept which surely demands a compound noun in German. Experts, and fans of Spaced
, may wish to "skip to the end".)
Amenable Banach algebras can be thought about in many different ways (or indeed written about without thinking, sad to say). My own preference is the homological one: they are those which have a bounded approximate identity, and are flat as bimodules over themselves in a suitable sense.
This seems recondite but it has the following consequence: if you represent an amenable Banach algebra A on a Hilbert space H, and H has a closed A-invariant subspace V, then one can find a closed A-invariant subspace W with H=V\oplus W (a direct sum, but not necessarily an orthogonal
one). So in finite-dimensional settings, we are just talking about the complex associative algebras that are completely reducible
, which (Wedderburn theorem) means such examples are isomorphic to sums of full matrix algebras.
Which closed subalgebras of B(H) are amenable? If one restricts attention to subalgebras that are closed under the involution of B(H), the so-called C^\ast-algebras, then there is a deep result (using hard results of Connes/Popa together with details of various other people) that an amenable C^\ast-algebra is nuclear. Nuclear C^\ast-algebras, in this context, should be thought of as those which can be approximated linearly by finite-dimensional matrix algebras in a particularly tractable way. Conversely, Haagerup showed that every nuclear C^\ast-algebra is amenable as a Banach algebra.
If one want amenable subalgebras of B(H) that are not closed under the involution of B(H), one way to do it is to take a nuclear C^\ast-subalgebra A in B(H), and conjugate it with some suitable invertible operator, which will interact badly with the involution of B(H), but result in an algebra that is isomorphic to A and hence is also amenable. The question naturally arises: do all amenable subalgebras of B(H) appear this way?
The paper http://arxiv.org/abs/1309.2145
finally answers this question, with a negative answer. The counterexample is as an extension by the compact operators of a carefully chosen, commutative, non-separable subalgebra of the Calkin algebra. (In fact, this subalgebra of the Calkin is itself isomorphic to a commutative C^\ast-algebra.) Remarkably, the proof that one can obtain something not isomorphic to a C^\ast-algebra boils down, after a clever use of standard ideas in amenability, to the fact that 2^c > c where c is the cardinality of the continuum!
The construction has other features which run somewhat counter to one's naive intuition/guesses, but I think it will take some time for people in this area to absorb what this all means in the bigger picture. One thing to note is that the original question remains open for separable amenable operator algebras, and for commutative amenable operator algebras...