Does anyone have a reference for a fairly direct proof that the second Borel cohomology group for $(\mathbb R, +)$ (with the trivial action on the circle group) is trivial? The motivation is to show that every Borel representation of $(\mathbb R, +)$ by Hilbert space projective automorphisms is induced by a Borel representation of $(\mathbb R, +)$ by Hilbert space automorphisms per se, i.e., unitary operators. Barry Simon, in the paper "Quantum dynamics: From automorphism to hamiltonian" sketches a proof using what appear to be standard tools, but I can't make it work.
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ It might be good to include a clarification about what Borel cohomology is; I think this is unfamiliar to most mathematicians and I've only seen physicists ask questions about it so far. $\endgroup$– Qiaochu YuanCommented Feb 5, 2015 at 6:43
-
$\begingroup$ It's group cohomology where the the cocycles $G^n\to A$ are Borel measurable. $\endgroup$– Chris GerigCommented Feb 5, 2015 at 9:39
-
$\begingroup$ Yes, Chris, that is my meaning. If we assume continuity, Simon's sketch can be (fairly) easily completed, but up to that point in his presentation he has gone to some trouble to manage with the assumption of Borel measurability, not continuity. $\endgroup$– Robert VanWesepCommented Feb 5, 2015 at 12:06
-
$\begingroup$ OK I don't know what you want, now that I looked at the paper of Simon. That is the direct proof (Theorem 8.1), and I don't see how you can get any more direct. If you're confused on the proof, Simon says (pun not intended) that he took the proof from a paper of Wigner (provided in the references), so check that out. $\endgroup$– Chris GerigCommented Feb 6, 2015 at 0:22
-
$\begingroup$ It seems that one must show that there is a vector $\delta$ in the Hilbert space completion of $L^1$ such that for all $a \in \mathbb R$, $(\delta, U(a) \delta) = \phi(a)$. This should be the limit of an approximate identity sequence, but I don't see how to complete the argument. It would seem that the purpose of completing to a Hilbert space is solely so that such a vector can exist in the space; there is no such vector in $L^1$ typically. The Wigner paper deals with continuous multipliers and proceeds by a very different method not applicable to the Borel case. $\endgroup$– Robert VanWesepCommented Feb 6, 2015 at 3:08
Add a comment
|