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The invariant subspace problem (ISP) for Hilbert spaces asks whether every bounded linear operator $A$ on $l^2$ (with complex scalars) must have a closed invariant subspace other than $\{0\}$ and $l^2$. A subspace $E$ is invariant for $A$ if $A(E) \subseteq E$.

Some time ago I noticed a reformulation that has a set-theoretic flavor. Let $P$ be the set of all linear operators $A$ from a finite-dimensional subspace of $l^2$ into $l^2$ such that (1) $\|A\| < 1$ and (2) if $E$ is a nonzero subspace of the domain of $A$ then $A(E) \not\subseteq E$. Order $P$ by reverse inclusion and for any unit vectors $v,w \in l^2$ define $$D_{v,w} = \{A \in P: \langle A^nv,w\rangle \neq 0\mbox{ for some $n$ such that $A^nv$ is defined}\}.$$ It is easy to see that each $D_{v,w}$ is dense in $P$. Having a counterexample to the ISP is the same as having a filter of $P$ that meets each $D_{v,w}$.

My question for the set theory experts on MO is simply whether this version of the problem suggests any possible approaches?

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19  
Two obvious observations are that the problem is absolute (Shoenfield) and that one can essentially recast your $P$ as a version of Cohen forcing (by using separability of $l^2$ carefully). Both observations seem to be folklore, but I haven't seen any reasonable way of taking advantage of them. I once attended a talk at the analysis seminar at Berkeley where the speaker began by basically describing a version of $P$, and forcing (and $\mathsf{MA}$), and then quickly hitting a wall and not getting anywhere. This was about 12 years ago, and I haven't seen any scenarios building on it since. – Andrés Caicedo Jul 9 '13 at 6:08
21  
I suspect I was the speaker at that talk. – Nik Weaver Jul 9 '13 at 6:26
4  
I don't think so, but based on the timing there's little doubt you're remembering a talk I gave. It was in a little room on the east side of the building, Don Sarason was sitting in the front row, and you were in the back row. – Nik Weaver Jul 9 '13 at 6:32
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I have nothing intelligible to say about the question, but the comment thread is awesome. – Asaf Karagila Jul 9 '13 at 7:48
6  
@Sébastien: your comments in this thread have not been helpful. – Nik Weaver Jul 17 '13 at 18:19

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