Is it true that the universal cover of SL2(ℝ) has no non-trivial central extensions… as an abstract group?
(that's certainly true as a Lie group)
Motivation:
I have a projective action of SL2(ℝ) on some Hilbert space H
and I'd like to know that it induces an honest action of its universal cover.
But it's a hassle to show that the action is continuous.
So I'm wondering if there is an alternative argument that uses solely group theory.
Best Answer
The answer should be negative, because the $K_2$ of the reals is humongous. That is, there are nontrivial central extensions. (Please do not ask a question and then explain its negation.) Algebraic $K$ theory detects transcendentals. There is a Chern class map from $K_2(\Bbb R)$ towards $\Omega^2_{\Bbb R}$, where the Kähler differentials are taken over the integers. It maps the Steinberg symbol {t,u} to $dlog\ t\wedge dlog\ u$, with $dlog\ t=dt/t$. It hits much more than a cyclic group. The map from the Schur multiplier of $SL_2(\Bbb R)$ to the stable $K$ group is surjective in this case, by Steinberg. So the universal cover as a Lie group realizes only a very small part of the Schur multiplier.