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The Kadison-Singer problem is the following statement: for any $\epsilon >0$, there exists $r\in \mathbb N$ such that for any bounded operator $A$ on $\ell^2(\mathbb Z)$, there exists a partition $(\mathcal P_s)_{1\le s\le r}$ of $\mathbb Z$, with $$ \max_{1\le s\le r}\Vert P_s(A-\text{diag}A)P_s\Vert_{\mathcal B(\ell^2(\mathbb Z))}\le \epsilon \Vert A-\text{diag}A\Vert_{\mathcal B(\ell^2(\mathbb Z))}\quad\tag {PC} $$ where $P_s=\sum_{j\in \mathcal P_s} p_j$ and $p_j$ is the orthogonal projection onto $e_j=(\delta_{j,k})_{k\in \mathbb Z}$; the point is that $r$ depends only on $\epsilon$. This problem is sometimes quoted as the Kadison-Singer conjecture, a rather inaccurate denomination since these authors were inclined to think that the answer should be negative. We have given here the formulation of the Paving Conjecture, known to be equivalent to KS. Since (PC) seems to be now solved, I will read the paper and I withdraw my question.

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Any reference for that recent result? –  François G. Dorais Jul 27 '13 at 19:49
    
@FrançoisG.Dorais I think Bazin is referring to pnas.org/content/105/14/5313.full although one has to read that paper with a little care, and not just the MR whose 1st version had problems –  Yemon Choi Jul 27 '13 at 19:53
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BTW, it seems a bit misleading to say KS is open without referring to the recent paper claiming a solution –  Yemon Choi Jul 27 '13 at 19:53
    
If you read french, please consider having a look at: bourbaki.ens.fr/TEXTES/1088.pdf –  Alain Valette Sep 7 at 9:02
    
@AlainValette Merci bien, ca me donne la chance de m'exercer –  Yemon Choi Sep 7 at 17:22

1 Answer 1

Virtually everything stated in the question is wrong. The Kadison-Singer problem has been solved positively by Marcus, Spielman, and Srivastava, see:

http://arxiv.org/pdf/1306.3969v3.pdf

The question apparently also refers to an earlier paper of Akemann and me, in which assuming CH we falsified a different conjecture of Kadison and Singer. This problem also has to do with pure states on $B(H)$ so there is some connection with the well-known Kadison-Singer problem. But it's a different question. See:

http://www.pnas.org/content/105/14/5313.full

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I thought it might be that paper, though I guess we need the OP to confirm this. Confusion may have arisen because the original MathReview of this PNAS paper had an error/mis-statement, which I pointed out to the reviewer, I don't know if the MR has since been revised. –  Yemon Choi Jul 27 '13 at 20:07
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More precisely, Marcus, Spielman, and Srivastava proved a statement that Nik had previously shown to imply Kadison-Singer. This is not the end of the story as it remains to get a good dependence of $r$ on $\epsilon$. AFAIK, Nik's argument does not give anything explicit. –  Bill Johnson Jul 27 '13 at 20:07
    
@Yemon: yes, the original MathReview was a bit embarassing! It was corrected a little while ago. –  Nik Weaver Jul 28 '13 at 0:40
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Yes the paving conjecture is solved. –  Nik Weaver Jul 28 '13 at 9:26
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Bill, the proof of Kadison-Singer works by showing how to pave projection operators. Going from projections to general operators was originally accomplished by Akemann-Anderson using completely nonconstructive infinite-dimensional techniques. Recently Tristan Bice found a more elementary route to this reduction (see his paper "Filters in C*-algebras"), so maybe more information can be extracted there. –  Nik Weaver Jul 28 '13 at 16:12

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