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](http://arxiv.org/abs/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. 



Operator Valued Kadison--Singer Problem
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.


