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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

By Venkov's proof of the Evens-Venkov theorem, if you take a faithful representation $G\rightarrow SO(m)$, then the cohomology $H^*(BG)$ is a finitely generated module for the image of $H^*(BSO(m))$. But the situation is quite different if you only allow yourself Euler classes, and allowing Euler classes of all representations doesn't help much.

There are insufficiently many such classes in general to come close to generating the whole cohomology ring.

Consider the group $G$ defined as the affine group $G=AGL(1,q)$ for $q$ a power of a prime $p$, say $q=p^n$. This is the group of all self-maps of the field $F_q$ of the form $x\mapsto ax+b$ with $a\neq 0$. This $G$ has a normal subgroup isomorphic to $(C_p)^n$ and quotient a cyclic group of order $p^n-1$. The irreducible complex representations of this group are easily computed: there are $p^n$ 1-dimensional representations and one other one of dimension $p^n-1$. The single large representation is already realized over the rationals. So the irreducible real representations are one of dimension $p^n-1$ and lots of others of dimension at most 2. The low-dimensional representations factor through the projection from $G$ onto $C_{q-1}$. So by taking Euler classes of irreducible real representations, you get a subring of $H^*(BG)$ in which only one of your generators has order divisible by $p$. The other generators have image 0 under the restriction map $H^*(BG)\rightarrow H^*(B(C_p)^n)$.

By the Evens-Venkov theorem, the cohomology of $(C_p)^n$ is a finitely-generated module for $H^*(BG)$ under this map. When $n>1$, $H^*(B(C_p)^n)$ cannot be a finitely-generated module for a subring generated by 1 element, so in this sense Euler classes do not come close to generating $H^*(BG)$.

One more remark: the smallest group to which my argument applies is the affine group $AGL(1,4)$, which is isomorphic to the alternating group $A_4$ mentioned in some of the other solutions.