# Is the cohomology ring $H^*(BG,\mathbb{Z})$ generated by Euler classes?

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.

• This is a great question. – Bombyx mori Feb 8 at 17:13
• @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. – Bombyx mori Feb 9 at 21:10
• @Bombyxmori oh... ok ok.. :) – Praphulla Koushik Feb 10 at 3:11

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.

• Thanks! I forgot the Tor term in the Kunneth sequence. This is correct. – user404153 Feb 8 at 18:52

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.

• Thanks! Do you know a reference for the stable $H^*(A_n)$? I could only find one for $\mathbb{Z}/p$ coefficients. – user404153 Feb 9 at 13:56
• Sadly, I don't know one, but I'm not really an expert in this, so there might very well exist a good one. – Achim Krause Feb 9 at 15:48
• @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. – Mike Miller Feb 10 at 0:02
• @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. – user404153 Feb 10 at 8:43