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Suppose that $\overline G$ is a Lie group such that the connected component of $1$ is $\mathbb C^*$. Assume that $\mathbb C^*$ is central in $\overline{G}$, and set $G := \overline G/\mathbb C^*$. There are two ways of constructing a class in $\mathrm H^3(G, \mathbb Z)$.

The first is to consider the central extension $$ 1 \longrightarrow \mathbb C^{*} \longrightarrow \overline G \longrightarrow G \longrightarrow 1 $$ as an extension of discrete groups; this gives a class in $\mathrm H^{2}(G, \mathbb C^{*})$, and we can take its image in $\mathrm H^{3}(G, \mathbb Z)$ via the connecting homomorphism $\mathrm H^{2}(G, \mathbb C^{*}) \to \mathrm H^{3}(G, \mathbb Z)$.

The other involves the fibration $B \overline{G} \to B G$, with fiber $B \mathbb C^{*}$. Since $B \mathbb C^{*}$ is a $K(\mathbb Z, 2)$ and the action of $G$ on $B \mathbb C^{*}$ is trivial, we obtain an obstruction to the existence of a section $BG \to B\overline{G}$, which lives in $\mathrm H^{3}(G, \mathbb Z)$. This is the second way.

An alternate construction of the second class (up to sign) is via the Leray-Serre spectral sequence $$ E^{ij}_{2} = \mathrm H^{i}(G,\mathbb Z)\otimes \mathrm H^{j}(B\mathbb C^{*}, \mathbb Z) \Longrightarrow \mathrm H^{i+j}(B \overline{G}, \mathbb Z) $$ in which the first possibly non-zero differential is $$ d_{3}\colon \mathrm H^{2}(B\mathbb C^{*}, \mathbb Z) \longrightarrow \mathrm H^{3}(G,\mathbb Z)\,. $$ Then we take the image of the canonical generator of $\mathrm H^{2}(B\mathbb C^{*}, \mathbb Z)$ under $d_{3}$.

Do these two classes coincide, up to sign? I am fairly sure that they do, and it's only my deep ignorance in algebraic topology that prevents me from seeing clearly why this is so. A reference would be better than an argument, but I'll be grateful for any hint.

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There's a difference though, in your first construction you obtain a class in $H^3(B'G,\mathbb Z)$, where $B'G$ denotes the classifying space of $G$ as a discrete group, while in your second construction you get a class in $H^3(BG,\mathbb Z)$ where $BG$ is the classifying space of the topological group $G$. Of course, you have a canonical map $B'G\rightarrow BG$ and your question may be whether the first class coincides with the pull-back along this map of the second class. – Fernando Muro Apr 3 '13 at 15:48
To Fernando: you are right; but I am assuming that $\overline G$ is locally connected, or, equivalently, that is a Lie group, so $G$ is discrete. I edited the question to reflect this. – Angelo Apr 3 '13 at 18:51
Just in case you haven't already looked in this direction (I got stuck): $H^3(A,B)$ corresponds to crossed module extensions. So your extension $\alpha\in H^2(G,\mathbb{C}^*)$ maps under the connecting homomorphism to a particular $\mathbb{Z}\to N\to E\to G$, which ideally you can read off from the definitions (with the help of MacLane's paper on this notion). – Chris Gerig Apr 4 '13 at 0:06
Also, your extension $\alpha$ corresponds to the $B\mathbb{C}^*$-fibration that you write, and so we want an explicit description, in terms of the underlying groups, of the failure of a set-theoretic section to be the desired map. Ideally this will have either a cocyle-description (and then check it with $\delta\alpha$) or a crossed module extension description (and then compare extensions). On afterthought, these comments are probably recasting your question into a harder one. – Chris Gerig Apr 4 '13 at 0:11
Addendum: This fibration I think is a principal bundle, and so existence of a section would mean the bundle is trivial, which should mean that the cohomology class in $H^2$ is zero, so that its image in $H^3$ is also zero, giving agreement here. – Chris Gerig Apr 4 '13 at 0:14

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