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This is a followup to my previous question, reproduced here.

Let$$0 \to R \to F \to G \to 0$$be a short exact sequence of groups. Is it possible to construct an associated fibration of spaces$$K(R, 1) \to K(F, 1) \to K(G, 1)?$$

Qiaochu Yuan gave the following answer.

Yes. In fact a fiber sequence of Eilenberg-MacLane spaces is the same thing as a short exact sequence of groups.

Take a surjective group homomorphism $f: G \to H$. This induces a map $Bf : BG \to BH$ on classifying spaces/Eilenberg-MacLane spaces (how to see this at the point-set level depends on what your point-set construction of Eilenberg-MacLane spaces is). Now take the homotopy fiber of $Bf$. A computation with the long exact sequence in homotopy shows that it is $B\text{ker}(f)$.

My question is, how do we calculate the $E_{p, 0}^2$ and the $E_{0, 1}^2$ terms in this sequence, in terms of the homology of $K(G, 1)$, the homology of $K(R, 1)$, and the action of $G$ on $R$ by conjugation?

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  • $\begingroup$ In what sequence? The Serre spectral sequence of the fibration? Look up the Lyndon-Hochschild-Serre spectral sequence. $\endgroup$ – Qiaochu Yuan Feb 16 '16 at 20:30
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The easiest way to see it from a direct construction is through milnor's constructon of $E_G$ and $BG$ - it has nice functoriality properties.

But the easiest way to get the maps is to note that we know that the deloopings of all of these groups exist. Thus we can continue the fibration $ R \to F \to G$ to get the fibration sequence $R \to F \to G \to BR \to BF \to BG$. Therefore we have the fibration $BF$ \to $BG$ with fiber $BR$.

Remark: In general the action of $G$ on $BR$ is not just given 'by conjugation' when $R$ is not abelian. This is because there is no action of $\pi_1(BG)$ on $\pi_*(BR)$ - basepoints matter.

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