Question

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.

Answer 1

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.

Answer 2

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).