The Paving Conjecture, which is equivalent to the famous Kadison--Singer Problem, was spectacularly settled in the affirmative by Marcus--Spielman--Srivastava (arxiv:1306.3969). Let $E$ denote the canonical projection from ${\mathcal B}(\ell_2{\mathbb N})$ onto the subalgebra $\mathcal D$ of the diagonal operators (which is isomorphic to $\ell_\infty\mathbb N$). The Paving Conjecture asserts that for every $\epsilon>0$ there is $K=K(\epsilon)\in{\mathbb N}$ which satisfies the following property: For any $x\in{\mathcal B}(\ell_2{\mathbb N})$ such that $E(x)=0$, there is a partition ${\mathbb N}=S_1\sqcup\cdots\sqcup S_K$ such that $\|P_{S_i} x P_{S_i}\|<\epsilon\|x\|$ for every $i$. Here $P_S$ is the orthogonal projection from $\ell_2{\mathbb N}$ onto $\ell_2S$, which belongs to $\mathcal D$.

Given that the Paving Conjecture is settled, I think it is natural to wonder whether it can be generalized to the operator valued setting. Let $M$ be a von Neumann algebra. Is the following true? For every $\epsilon>0$ there is $K=K(M,\epsilon)$ which satisfies the following property: For any $x\in{\mathcal B}(\ell_2{\mathbb N})\bar\otimes M$ such that $(E\otimes\mathrm{id})(x)=0$, there is a prtition $1\otimes 1 = \sum_{i=1}^K P_i$ of the identity by orthogonal projections $P_i$ in ${\mathcal D}\bar\otimes M$ such that $\|P_i x P_i\|<\epsilon\|x\|$ for every $i$. Sorin Popa pointed out to me that even the case where $M=L^\infty[0,1]$ is not clear. Also, what is the growth of $K(M_n({\mathbb C}),\epsilon)$ as $n\to\infty$?

As an operator algebraist, I am frustrated by the fact that the important problem in operator algebra theory was solved by outsiders by linear algebra. My motivation is that, although I don't have an application, the above generalization, if it's true, would require a new proof of the Paving Conjecture (now the Marcus--Spielman--Srivastava theorem) that involves operator algebra theory.

  • 5
    $\begingroup$ "...frustrated by the fact that the important problem...solved by outsiders..." --- I guess that's part of the reason why it got solved, that too using tools that non-ultra-specialists can understand --- so one reason may be that the jargon or prevalent techniques were obstructing a clean look at the problem! (though this clean look was also made possible by the nice clear formulation of N. Weaver, that even non oa people can easily understand). $\endgroup$
    – Suvrit
    Oct 29 '13 at 2:51
  • 1
    $\begingroup$ Since the paving conjecture is equivalent to doing the paving for $B(\ell_\infty^m)$, with $K(\epsilon)$ independent of $m$, the paving conjecture really is a problem in linear algebra (which happens to be equivalent to the extension of pure states problem of Kadison and Singer). $\endgroup$ Oct 29 '13 at 3:01
  • $\begingroup$ Taka, an important question is how $K(\epsilon)$ grows as $\epsilon \to 0$. What is the interest in knowing the growth of $K(M_n,\epsilon)$ as $n\to \infty$? $\endgroup$ Oct 29 '13 at 3:05
  • $\begingroup$ @Bill: I don't know the estimate of $K(\epsilon)$, because the MSS theorem directly implies the Paving Conjecture only for projections with small diagonals. I'm interested in the limit $n\to\infty$, because it has something to do with the case $M={\mathcal B}(\ell_2)$. $\endgroup$ Oct 29 '13 at 5:27

The case $M = L^\infty[0,1]$ seems like a simple measurability question. If every $x \in B(l^2)$ can be $K$-paved to $\|\sum P_ixP_i\| < \epsilon \|x\|$, then every $x \in B(l^2)\otimes L^\infty[0,1] \cong L^\infty([0,1],B(l^2))$ has a pointwise a.e. $K$-paving that does the same thing. So we just need a measurable selection. Surely one of the standard measurable selection theorems suffices here, have you checked this?

I really disagree with your characterization of the problem as having been solved without operator algebra theory. You didn't mention the brilliant reduction of Akemann and Anderson that gets us down to this combinatorial question. That part of the solution is completely operator algebraic and it is quite nontrivial. In his paper "Filters in C*-algebras" Tristan Bice develops some very beautiful machinery for achieving this reduction, among other operator algebraic applications. It sounds as if you are not even aware of this part of the story.

  • $\begingroup$ Thank you. Probably you are right about the selection theorem. You are also right about Kadison--Singer problem. I didn't mean to ignore contributions by Akemann, Anderson and you. Just I wanted to stress the surprise solution. $\endgroup$ Oct 30 '13 at 2:35
  • $\begingroup$ Fair enough. (Anyway my minor contribution was already overemphasized.) $\endgroup$
    – Nik Weaver
    Oct 30 '13 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.