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The Kadison-Singer problem is considered in relation to the separable Hilbert space:

KS: Does every pure state on the diagonal (atomic) masa of $B(\ell_2)$ has a unique extension to $B(\ell_2)$?

What is the status of this problem for non-separable Hilbert spaces? Most of operator-algebraits are uninterested in (starting a day with) non-separable Hilbert spaces, so probably they don't care. But if the (classical) KS problem has a negative solution, this solution must depend on the underlying ultrafilter $p\in \beta \mathbb{N}\setminus \mathbb{N}$ (it cannot be, for example, rare).

Sometimes weird ultrafilter over uncountable sets are (consistently) easier to grasp. This indicates that perhaps one should look at $\ell_2(\kappa)$ for some $\kappa$ big enough.

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Maybe the following links are helpful on this problem: aimath.org/WWN/kadisonsinger math.missouri.edu/~pete/index.html zentralblatt-math.org/portal/en/zmath/search/… –  Jiang Jan 10 '13 at 22:09
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It just got proven by Adam Marcus, Dan Spielman & Nikhil Srivastava arxiv.org/abs/1306.3969 - here's come commentaries: noncommutativeanalysis.wordpress.com/2013/06/20/… and gilkalai.wordpress.com/2013/06/19/… –  pageman Jun 26 '13 at 14:06
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1 Answer

up vote 8 down vote accepted

I think it's the same problem. If a pure state on $l^\infty$ had distinct extensions to states on $B(l^2)$, then you could embed $B(l^2)$ into $B(l^2(\kappa))$ for $\kappa > \aleph_0$ and extend those extensions to $B(l^2(\kappa))$. Conversely, if every pure state on $l^\infty$ has a unique extension then every operator in $B(l^2)$ can be paved. This implies that we can pave the compression of any operator in $B(l^2(\kappa))$ to $l^2(X)$ for any countable subset $X$ of $\kappa$, and by an easy compactness argument you can pave the entire operator. So pure states would have to extend uniquely to $B(l^2(\kappa))$.

(If that's too cryptic, you can think of a $k$-paving of the compression of $A$ to $l^2(X)$ as a $k$-coloring of the set $X$. Order the countable subsets $X \subset \kappa$ by inclusion, for each $X$ let $C_X$ be the set of $k$-pavings of $P_{l^2(X)}A|_{l^2(X)}$, and get $\bigcap C_X \neq \emptyset$ by compactness.)

However, to editorialize: as you say, if KS has a negative solution (some pure states have nonunique extensions) then there may well be set-theoretic issues involved in saying which ultrafilters extend uniquely. But KS can be reformulated as an arithmetical question, so if it has a positive solution (which I believe is the answer that expert opinion now favors) then it is unlikely to have anything to do with set theory. For instance, see my paper The Kadison-Singer problem in discrepancy theory for a combinatorial reformulation in terms of finite sets of vectors in ${\bf C}^n$.

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Actually, I guess you can just compress $A$ to the finite subsets of $\kappa$ ... –  Nik Weaver Jan 11 '13 at 0:25
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