MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Maybe the following links are helpful on this problem:… – Jiang Jan 10 '13 at 22:09
It just got proven by Adam Marcus, Dan Spielman & Nikhil Srivastava - here's come commentaries:… and… – pageman Jun 26 '13 at 14:06
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 ${\bf C}^n$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.