4
$\begingroup$

Suppose I have a central extension $1 \to U(1) \to \hat{G} \to G \to 1$ of a topological group $G$ by the circle group $U(1)$ in such a way that $\hat{G} \to G$ is a principal $U(1)$-bundle. Moreover, suppose that $G$ is $3$-connected. In particular, $\hat{G} \cong G \times U(1)$ as topological spaces.

Is the $3$-connectedness of $G$ enough to deduce that $\hat{G} \cong G \times U(1)$ as topological groups?

Such a central extension is described by a class in continuous group cohomology, but I can't see whether the connectedness of $G$ is any help in proving that this class is actually trivial.

In my case, $\hat{G}$ and $G$ are Banach Lie groups. So, feel free to assume that as well.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Sounds dodgy! Why cannot you just do a quotient of the Heisenberg group? I mean $R\times R \times S^1$ with the product $(u,v,a)\cdot (x,y,b)=(u+x,v+y,ab(uy-vx+Z))$... $\endgroup$
    – Bugs Bunny
    Jul 16, 2012 at 20:58
  • $\begingroup$ @Bugs Bunny: The quotient you describe is not $3$-connected. $\endgroup$
    – Mark Grant
    Jul 17, 2012 at 14:23
  • $\begingroup$ You can induce from the exponential sequence a long exact sequence in continuous group cohomology. If $G$ is compact, it is known that $H^n(BG,\mathbb{R})=0$, and you get $H^2(BG,U(1))=H^3(BG,\mathbb{Z})$. The latter vanishes by your connectedness assumptions. But I guess your $G$ is not compact... $\endgroup$
    – Konrad Waldorf
    Jul 17, 2012 at 21:26
  • 1
    $\begingroup$ @Mark Grant, so? The quotient is $\widehat{G}$ and it cannot be 3-connected at all. My $G$ is 3-connected: it was even contractible this morning... $\endgroup$
    – Bugs Bunny
    Jul 18, 2012 at 7:16
  • $\begingroup$ I guess it is even an answer then.... $\endgroup$
    – Bugs Bunny
    Jul 18, 2012 at 7:16

2 Answers 2

Reset to default
7
$\begingroup$

No, $G=({\mathbb R}^2 ,+)$ and a certain quotient $\widehat{G}$ of the Heiseberg group is an example. Namely, $\widehat{G}={\mathbb R}^2\times S^1$ with the product $(x,a)⋅(y,b)=(x+y,ab(\omega(x,y)+{\mathbb Z}))$ where $\omega$ is a nonzero bilinear form.

Improve this answer
$\endgroup$
5
$\begingroup$

Bugs is right. If $G$ is locally compact abelian and multiplication by $2$ (or squaring) is an automorphism of $G$, then $H^2(G,U(1))$ is isomorphic to the group of alternating bi-characters $G\times G\to U(1)$. This is discussed, for example, in Mumford's Tata Lectures on Theta III, and in my paper Locally Compact Abelian Groups with Symplectic Self-duality.

This points to a nice criterion for the vanishing of $H^2$ in this case, namely, $G$ should not admit non-trivial alternating bicharacters. $\mathbf R$ satisfies this condition. So do all groups which have dense locally cyclic subgroups (see Section 5 of my article).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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