Let us have a look on the proof of Theorem 2 in [P. G. Spain, Boolean algebras of projections, Proceedings of the Edinburgh Mathematical Society (Series 2) 19, 03, March 1975, 287-289]

The author claims in the proof of Theorem 2 that if $B$ is a (Bade) complete Boolean algebra of projections on a Banach space which is relatively weakly compact subset of the algebra of operators, then the weak (WOT?) closure of the algebra generated by $B$ is a W*-algebra.

I don't think the proof is OK. Take $p\neq 2$, $p\in (1,\infty)$ and consider $X=\ell_p$ with its canonical basis, which is 1-unconditional. The family $B$ of basis projections forms a complete Boolean algebra (actually isomorphic to $\wp(\omega)$) and is relatively weakly compact because all the projections have norm one and the unit ball of the space of operators on a reflexive space is WOT-compact. Though, $\overline{\mbox{alg }B}^{WOT}=B(\ell_p)$ which is not a von Neumann algebra (it is not even Banach-space isomorphic to a von Neumann algebra).

My question is (before it's closed by the math police). Is Theorem 2 correct?

  • $\begingroup$ Not sure this is a question- voting to close. $\endgroup$ – Daniel Moskovich Feb 18 '13 at 13:49
  • 2
    $\begingroup$ If this turns into a protracted tug-of-war then of course the discussion should be moved to meta. But, for the record, it seems to me that this is a real question. If anything, the OP suffers from having thought too productively about it, by producing a (putative) counterexample. Without that, Spain's claim could just be rephrased as a question. $\endgroup$ – HJRW Feb 18 '13 at 15:19
  • 2
    $\begingroup$ Why not email Philip Spain first? His email is available on his homepage: maths.gla.ac.uk/~pgs $\endgroup$ – Dmitri Pavlov Feb 18 '13 at 15:42
  • 2
    $\begingroup$ I find mention of the "math police" tiresome and unhelpful. $\endgroup$ – Yemon Choi Feb 18 '13 at 19:45
  • 1
    $\begingroup$ Also, having had a quick look at Theorem 2, I can't see where there might be a mistake. So I would respectfully venture that the answer to your question is YES. $\endgroup$ – Yemon Choi Feb 18 '13 at 19:51

I haven't looked at the paper but your counterexample is mistaken. The basis projections generate not $B(l^p)$ but the algebra of multiplication operators, which is isometrically isomorphic to $l^\infty$ and hence is a von Neumann algebra.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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