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). Commented Oct 29, 2013 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). Commented Oct 29, 2013 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 $n\to \infty$? Commented Oct 29, 2013 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 $M={\mathcal B}(\ell_2)$. Commented Oct 29, 2013 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?