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I am sure the answer to this question is well-known, but

It is well known that the group cohomology $H^2(G,\mathbb Z)$ classifies group extensions $0\to \mathbb Z\to E\to G\to 1$ and that for a topological space $X$ elements of $H^2(X,\mathbb Z)$ are in natural bijection with complex line bundles on $X$.

My question is thus:

What is the direct correspondence between extensions of $G$ and line bundles on $BG$?

That is, given an explicit line bundle $L$ how does one construct an explicit group extension $E$ such that the two give the same cohomology class and vice versa?

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  • $\begingroup$ I guess you meant line bundles on $G$ and not $BG$. And assuming that the action of $G$ on $\mathbb{Z}$ is trivial, $H^2(G;\mathbb{Z})$ classifies central extensions of $G$ by $\mathbb{Z}$. And finally, if your $G$ is a simply connected Lie group then $H^2(G;\mathbb{Z})=0$. $\endgroup$
    – Somnath Basu
    Commented Dec 6, 2010 at 21:46
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    $\begingroup$ I think the statements in the question are correct given that $H^2(G,\mathbb{Z})$ is supposed to denoted group cohomology, while $H^2(X,\mathbb{Z})$ denotes ordinary cohomology. I'd better replace $H^2(G,\mathbb{Z})$ right away by $H^2(BG,\mathbb{Z})$. $\endgroup$
    – Konrad Waldorf
    Commented Dec 6, 2010 at 22:01
  • $\begingroup$ ... and "line bundle" should probably be "hermitian line bundle". $\endgroup$
    – Konrad Waldorf
    Commented Dec 6, 2010 at 22:05

3 Answers 3

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If $L \to BG$ is the complex line bundle, take the unit sphere bundle $S^1 \to S(L) \to BG$ and take $\pi_1$.

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    $\begingroup$ And for the reverse direction, given an extension take classifying spaces. This gives you a $K(\mathbb{Z},1)$ bundle over BG, which you can replace (up to homotopy) with a line bundle. $\endgroup$
    – Chris Schommer-Pries
    Commented Dec 6, 2010 at 22:03
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    $\begingroup$ Chris, the $K\mathbb Z,1)$ bundle you can replace with an $S^1$-bundle up to homotopy, and then use that as principal bundle to obtain the line bundle, rather. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 6, 2010 at 22:05
  • $\begingroup$ @ Mariano, yes, there is a natural bijection between (iso. classes of) $S^1$-bundles and complex line bundles so I tend to equate the two. $\endgroup$
    – Chris Schommer-Pries
    Commented Dec 7, 2010 at 3:18
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Suppose $G$ is finite. From the short exact sequence of trivial $G$-modules $$0\to\mathbb Z\to\mathbb C\to\mathbb C^\times\to0$$ where the map $\mathbb C\to\mathbb C^\times$ is the exponential map, using the fact that $H^p(G,\mathbb C)=0$ for $p>0$ because $\mathbb C$ is divisible, you get that $$H^2(G,\mathbb Z)\cong H^1(G,\mathbb C^\times)\cong\hom_{\mathrm{groups}}(G,\mathbb C^\times),$$ and this last group is precisely the set of one-dimensional complex $G$-modules.

Therefore an extension of $G$ by $\mathbb Z$ determines a one dimensional complex $G$-module $V$. You can now get a line bundle on $BG$ from $V$ by constructing $EG\times_G V\to BG$.

This works for other groups $G$, like Lie groups by the same reasoning but with subtler justifications.

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Your line bundle $L$ over $BG$ can be seen as a $G$-equivariant line bundle over a point. That is, up to isomorphism, just a continuous group homomorphism $f:G \to \mathbb{C}^{\times}$. Try to lift $f$ along $\exp: \mathbb{C} \to \mathbb{C}^{\times}$ on an open cover of $G$. The error is a $\mathbb{Z}$-valued Cech-1-cocycle on $X$, and $E$ is the total space of the associated $\mathbb{Z}$-bundle over $X$.

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