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.

• "...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). – Suvrit Oct 29 '13 at 2:51
• 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). – Bill Johnson Oct 29 '13 at 3:01
• 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 $n\to \infty$? – Bill Johnson Oct 29 '13 at 3:05
• @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 $M={\mathcal B}(\ell_2)$. – Narutaka OZAWA 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?