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I am interested in the classifying space $BG$ of a finite group $G$.

A real representation $V$ of $G$ of dimension $r$ defines a real vector bundle over $BG$ of rank $r$. If the determinant of this representation is trivial, then this bundle is orientable and a choice of orientation determines an Euler class $e(V) \in H^r(BG,\mathbb{Z})$.

Do these classes generate $H^*(BG,\mathbb{Z})$ as a ring?

This is easy to show when $G$ is abelian. In that case we need only consider $G = \mathbb{Z}_n$ and the cohomology ring is generated in degree 2 by the Euler class of the $2\pi/n$-rotation representation.

More generally, for any $G$, the map $H^1(BG,SO(2)) \to H^2(BG,\mathbb{Z})$ is an isomorphism since the latter is torsion and this is a Bockstein-type operation. We can consider the element of $H^1(BG,SO(2))$ as a homomorphism $G \to SO(2)$ giving us a 2d representation and it is easy to see from the definition of the Bockstein that the above map is the Euler class of this representation.

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    $\begingroup$ This is a great question. $\endgroup$ – Bombyx mori Feb 8 at 17:13
  • $\begingroup$ @PraphullaKoushik: I am not an expert on this as well, I was reading Atiyah's paper earlier (before I heard the news of his passing). So this came naturally within the framework. $\endgroup$ – Bombyx mori Feb 9 at 21:10
  • $\begingroup$ @Bombyxmori oh... ok ok.. :) $\endgroup$ – Praphulla Koushik Feb 10 at 3:11
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Let $G = \mathbb{Z}/p \times \mathbb{Z}/p$ for $p$ odd. Then $H^3(BG; \mathbb{Z}) \cong \mathbb{Z}/p$ is not in the subring generated by Euler classes, since the non-trivial irreducible representations are of rank $2$.

EDIT: The problem of determining the subring of $H^*(BG; \mathbb{Z})$ generated by all Chern classes has been much studied. You might start with Atiyah's 1961 paper.

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  • $\begingroup$ Thanks! I forgot the Tor term in the Kunneth sequence. This is correct. $\endgroup$ – user404153 Feb 8 at 18:52
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Here is another short argument for why this cannot hold in general:

The cohomology of the alternating groups $H^*(A_n)$ stabilizes, and I think the first degree in which it stays nontrivial is 3 (But the precise degree doesn't matter for the argument). So there is a nontrivial element in $H^3(A_n)$ for large enough $n$.

However, since the $A_n$ are simple, the dimension of their smallest irreducible representation goes to infinity with $n$. So for large enough $n$, there simply is no $3$-dimensional representation!

If the $3$ above is indeed the correct degree, you can just get this statement from the knowledge of the discrete subgroups of $SO(3)$, since any nontrivial homomorphism $A_n\to SO(3)$ would have to be injective.

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  • $\begingroup$ Thanks! Do you know a reference for the stable $H^*(A_n)$? I could only find one for $\mathbb{Z}/p$ coefficients. $\endgroup$ – user404153 Feb 9 at 13:56
  • $\begingroup$ Sadly, I don't know one, but I'm not really an expert in this, so there might very well exist a good one. $\endgroup$ – Achim Krause Feb 9 at 15:48
  • $\begingroup$ @user404153 This is a very special case of the result of this article of Palmer. He cites a 1978 article of Hausmann as a much earlier proof of this case. $\endgroup$ – Mike Miller Feb 10 at 0:02
  • $\begingroup$ @MikeMiller Thanks, although the Palmer paper seems to prove homological stability but doesn't compute any of the groups (although does give the slope so in principle I suppose any given degree is computable?). I looked at the reference in Hausmann as well and I don't see how to tease the calculation out of what he did, even though Palmer refers to it as an "explicit calculation", lol. Cheers. $\endgroup$ – user404153 Feb 10 at 8:43

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