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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.

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    $\begingroup$ 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
    Commented Jul 16, 2012 at 20:58
  • $\begingroup$ @Bugs Bunny: The quotient you describe is not $3$-connected. $\endgroup$
    – Mark Grant
    Commented 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
    Commented Jul 17, 2012 at 21:26
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    $\begingroup$ @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
    Commented Jul 18, 2012 at 7:16
  • $\begingroup$ I guess it is even an answer then.... $\endgroup$
    – Bugs Bunny
    Commented Jul 18, 2012 at 7:16

2 Answers 2

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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.

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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).

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