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Suppose $G$ is a Lie group, with $\pi_0(G)$ not necessarily finite, but might as well assume $G_0$, the connected component of the identity, is compact.

In the case that $\pi_0(G)$ is finite, then we know that there is an injection $H^*(BG,\mathbb{Q})\to H^*(BG_0,\mathbb{Q})$, and this can apparently be seen via a spectral sequence argument, using the fact that the rational cohomology of $B\pi_0(G)$ is concentrated in degree zero. So this is some kind of Leray–Serre spectral sequence argument on either $\pi_0(G)\to BG_0\to BG$ or $BG_0\to BG\to B\pi_0(G)$ (and I suspect the latter), probably using the degeneration and some kind of "edge homomorphism is injective" argument.

I suspect that in the case that we know something strong about the rational cohomology of $B\pi_0(G)$, then we might be able to say something in the case where $\pi_0(G)$ is not finite.

Unfortunately my spectral sequence knowledge is limited, and I can't find a treatment of spectral sequences that seems general enough to deal with this setup in general (namely non-simply-connected base, and possibly non-connected fibre, plus non-finiteness issues, depending on which fibration is used).

Is my intuition correct, that $H^*(B\pi_0(G),\mathbb{Q}) = H^0(B\pi_0(G),\mathbb{Q})$ can let us conclude something about how the cohomology of $BG$ relates to that of $BG_0$?

Also, what would be a good reference that covers a general-enough version of the relevant spectral sequence?

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    $\begingroup$ Hello, down-voter! Care to explain? If it's super obvious a reference would be appreciated. $\endgroup$ – David Roberts May 4 at 8:50
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Think about the case where $\pi_0(G)=\mathbb{Z}$, so $B(\pi_0(G))=S^1$, so we have a fibre bundle $BG_0\to BG\to S^1$. In this case $G$ is always a semidirect product formed using an automorphism $\alpha$ of $G_0$. By considering the preimages of the the complements of two points in $S^1$, we can express $BG$ as $U\cup V$, where $U$ and $V$ are each open and homotopy equivalent to $BG_0$, and $U\cap V$ is homotopy equivalent to $BG_0\amalg BG_0$. This gives a Mayer-Vietoris sequence. If we set up the identifications carefully we find that $\alpha^*$ appears in one of the Mayer-Vietoris maps, and we deduce that there is a short exact sequence $C^{*-1}\to H^*(BG)\to K^*$, where $K^*$ and $C^*$ are respectively the kernel and cokernel of $\alpha^*-1\colon H^*(BG_0)\to H^*(BG_0)$. All this is valid integrally as well as rationally.

For more general $\pi_0(G)$, we have a spectral sequence $$ H^p(\pi_0(G);H^q(BG_0)) \Longrightarrow H^{p+q}(BG). $$ Note that the $E_2$ term involves group cohomology of the group $\pi_0(G)$ with coefficients in the module $H^*(BG_0)$, which typically has nontrivial action of $\pi_0(G)$. This is the same as the cohomology of the space $B\pi_0(G)$ with coefficients in a local system that typically has nontrivial twisting.

In the case $\pi_0(G)=\mathbb{Z}$ the only group cohomology groups are $E_2^{0q}=K^q$ and $E_2^{1q}=C^q$. The differentials are like $d_r\colon E_r^{pq}\to E_r^{p+r,q-r+1}$ so there is no room for them to be nonzero. Thus $E_\infty=E_2$ and we recover the picture in the first paragraph above.

On the other hand, if $\pi_0(G)$ is finite and we use rational coefficients then $H^p(\pi_0(G);H^q(BG_0))=0$ for $p>0$ and the spectral sequence collapses to an isomorphism $H^n(BG)=H^n(BG_0)^{\pi_0(G)}$.

Note also that $G$ could just be $G_0\times\Gamma$ for an arbitrary discrete group $\Gamma=\pi_0(G)$. In this case $BG=BG_0\times B\Gamma$, and the Kan-Thurston Theorem tells us that $H^*(B\Gamma)$ can be essentially anything.

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  • $\begingroup$ Ok, excellent. I was concerned about issues of convergence of the spectral sequence you gave, since checking eg Whitehead's book for the cohomological Serre SS it isn't clear the hypotheses are satisfied. But I'm happy to assume that the cohomology of $\pi_0(G)$, with values in a representation on a rational vector space, is trivial. This seems to me to give the collapse just the same as the finite case, no? $\endgroup$ – David Roberts May 4 at 10:33
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    $\begingroup$ Why do you think that $H^{>0}(\pi_0(G);V)$ should be zero for rational $V$? That is an unusual property unless $\pi_0(G)$ is a torsion group. But you are right that if it holds, then the spectral sequence will collapse. There will never be any problems with convergence of this kind of spectral sequence. $\endgroup$ – Neil Strickland May 4 at 10:53
  • $\begingroup$ Because I intend to state a theorem where that is the hypothesis, not that I have an a priori given group and I somehow have to check that. I know $\pi_0(G)=\mathbb{Z}$ gives me an explicit counterexample already to what I'm showing. $\endgroup$ – David Roberts May 4 at 11:07
  • $\begingroup$ And, anyway, the result I'm after doesn't need all the $H^n(\pi_0(G),V)$ to vanish, just a couple of low-dimensional ones... $\endgroup$ – David Roberts May 7 at 5:39

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