8
votes
0answers
235 views

Are there only finitely many maximal subfactors of a fixed finite index ?

A subfactor $N \subset M $ is maximal if it admits no non-trivial intermediate subfactors $N \subset P \subset M $. Question: are there only finitely many maximal subfactors of a fixed finite ...
45
votes
2answers
2k views

vector balancing problem

I believe the solution posted to the arXiv on June 17 by Marcus, Spielman, and Srivastava is correct. This problem may be hard, so I don't expect an off-the-cuff solution. But can anyone suggest ...
13
votes
1answer
598 views

Amenability of groups in terms of a perturbation condition

Let $G$ be a countable group and $\lambda \colon G \to U(\ell^2 G)$ its left-regular representation. Suppose that there exists a constant $C>0$ such that for all $T \in B(\ell^2 G)$ $$\inf ...
18
votes
2answers
4k views

Is the Invariant Subspace Problem interesting?

There's an amusing comment in Peter Lax's Functional Analysis book. After a brief description of the Invariant Subspace Problem, he says (paraphrasing) "...this question is still open. It is also an ...
20
votes
0answers
1k views

Finite-dimensional subalgebras of $C^\star$-algebras

Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. ...
8
votes
3answers
1k views

Conjugacy classes and reduced group $C^*$-algebra of an amenable group

The reduced $C^*$-algebra of a non-abelian free group $G$ has a unique trace. Hence, there is no chance to separate conjugacy classes of group elements using traces on $C^\star_{red} G$. On the other ...
7
votes
0answers
368 views

Saito-Wright definition of Rickart C*-algebras

A C*-algebra is Rickart if for each $x\in A$ there is a projection $p\in A$ so that $R(x)=pA$. Here the right-annihilator $R(S)$ of $S\subset A$ is defined as $R(S)=${$a\in A\mid xa=0\, \forall ...
6
votes
2answers
551 views

Do torsion-free groups give projectionless group ($C^\ast$) algebras?

One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the ...
7
votes
0answers
279 views

Is there a finite-index finite-depth II$_1$ subfactor which is more than $7$-super-transitive?

Background: See Noah and Emily's posts on subfactors and planar algebras on the Secret Blogging Seminar. There are plenty of examples of $3$-super-transitive (3-ST) subfactors; Haagerup, $S_4 < ...