Long story short; I posted in MSE


It is well known that if $G$ is a lie group and $H$ is a closed subgroup of $G$, the inclusion $H \hookrightarrow G$ induces a fiber bundle on the classifying spaces

$$ G/H \rightarrow BH \rightarrow BG$$

I am interested in the cohomological Serre spectral sequence (with coefficients in $\mathbb{Z}/2\mathbb{Z}$ for simplicity) associated to this fibration; namely, the $E_2$ term is given by

$$E_2^{p,q} = H^p(BG;\mathcal{H}^q(G/H))$$

If $G$ is path-connected, $BG$ is simply connected and therefore $\mathcal{H}^q(G/H)$ is thus the usual cohomology $H^q(G/H)$; However, what can I say in the case of $G$ being not path-connected? Is there any example where the twisted coefficients are not trivial? or they are trivial in my setting.

but I hope to get better ideas here; Moreover, I want to add the following:

If I work in the ground now, let's say $G = S^1 \times S^1 \times \mathbb{Z}/2\mathbb{Z}$ and $H = \mathbb{Z}/2\mathbb{Z}$ the right factor of $G$, we have a fibre bundle $$ S^1 \times S^1 \rightarrow\mathbb{R}P^\infty \rightarrow \mathbb{C}P^\infty \times\mathbb{C}P^\infty \times \mathbb{R}P^\infty$$

and an action of $\mathbb{Z}/2\mathbb{Z} = \pi_1(BG)$ on $H^1(S^1\times S^1) $. I know the abstract definition of such action (Following Lecture notes in Algebraic Topology; J. Davis, P. Kirk for instance), but I don't know how to make it work in an explicit case .

As a bouns inquiry; is there any good reference to learn how to compute these actions in the case of non-simply connected spaces. Most references that I have reached rather assume that the action is trivial and barely mention the local coefficient system.

  • $\begingroup$ The element of $\pi_1(BG)=\pi_0(G)$ represented by $g\in G$ acts on the homology of $G/H$ by the map $xH\mapsto gxH$ from $G/H$ to $G/H$.This can be nontrivial. $\endgroup$ Commented Nov 2, 2017 at 21:56
  • $\begingroup$ Would you mind to illustrate a little bit why the action becomes what you say? it isn't that clear for me $\endgroup$
    – C. Zhihao
    Commented Nov 3, 2017 at 0:46
  • $\begingroup$ You can think of $G/H\to BH\to BG$ as a special case of the associated bundle for an action of $G$ on a space $F$, $F\to EG\times_G F\to BG$. Let $F$ be $G/H$ with the action I mentioned. Then the bundle is $G/H\to EG\times_G(G/H)\to BG$, where $G\times_G(G/H)=(EG)/H\sim BH$. $\endgroup$ Commented Nov 3, 2017 at 1:06
  • $\begingroup$ In general in such an "associated bundle with fiber $F$" the action of $\pi_1 BG=\pi_0G$ on homology of the fiber $F$ is given by the given action of $G$ on $F$. $\endgroup$ Commented Nov 3, 2017 at 1:07

1 Answer 1


Lie groups include finite groups. Here is likely the simplest example.

Let $H = C_3$ be the cyclic subgroup of order 3 in the symmetric group $G = S_3$, so $S_3/C_3 = C_2$. So your fiber bundle is the double cover $C_2 \rightarrow BC_3 \rightarrow BS_3$. The resulting Serre spectral sequence converging to $H^*(BC_3;\mathbb Z/2)$ has $E_2^{p,q} = 0$ if $q>0$ and $E_2^{p,0} = H^p(BS_3;\mathbb Z/2[C_2])$, where $S_3$ acts on $C_2$, and thus on $\mathbb Z/2[C_2]$, in the interesting way. It is so interesting, in fact, that the twisted coefficients form a projective $\mathbb Z/2[S_3]$-module, and thus $E_2^{p,0} = 0$ unless $p=0$.

Conclusion: $H^p(BC_3;\mathbb Z/2) = \mathbb Z/2$ if $p=0$ and 0 otherwise. And happily this is the correct answer!

  • $\begingroup$ Even more simple, and familiar would be $G = \mathbb Z$ and $H = 2\mathbb Z$, which I leave as an exercise! $\endgroup$ Commented Nov 2, 2017 at 22:12

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