Cohomology ring of $H^*(A_5;\mathbb Z_2)$?

Solution 1:

I do not know a systematic way to calculate the cohomology of alternating groups. Maybe there is some method involving the LHSSS for $A_n \to S_n \to \Bbb Z/2$, as well as the combinatorially-known facts about the cohomology of $S_n$, or maybe some modification of all the Young tableaux stuff to work for alternating groups as well. I don't know it.

In any case, you are still in a small enough range that you have something exceptional about your group (which is why your computation for $n = 4$ could work at all). Today, the exceptional features are the following collection of facts.

1) $A_5$ embeds as a finite subgroup of $SO(3)$, given by the symmetries of the icosahedron.

2) The quotient $SO(3)/A_5$ is a manifold called the Poincare homology sphere, whose integral homology groups have $H_*(P) = H_*(S^3)$. The easiest way to argue this is to show that $P = SO(3)/A_5$ is the same as $SU(2)/2I$, where $p: SU(2) \to SO(3)$ is the double-cover and the group $2I = p^{-1}(A_5)$; it is called the binary icosahedral group, and one may check that it is perfect. Because $S^3$ is simply connected and $2I$ acts freely (by orientation-preserving transformations: all are isotopic to the identity!), we have $\pi_1(P) = 2I$, and therefore $H_1(P) = \pi_1(P)^{\text{ab}} = 0$. Because this is an oriented manifold Poincare duality implies that $H_2(P) = 0$ as well (and of course $H_3(P) = \Bbb Z$).

Given a short exact sequence of groups $N \to H \to G$, the Lyndon-Hochschild-Serre spectral sequence is the Leray-Serre spectral sequence for the fibration $BN \to BH \to BG$. What we are going to exploit is a slightly more general version of this: suppose $G$ is a Lie group with closed subgroup $H$, not necessarily normal. Thinking of $BH = EG/H$ (after all, $BH$ is the quotient of a contractible space by a free and proper $H$-action; the particular choice of space doesn't matter), we obtain a fibration $$G/H \to BH \to BG.$$ We will run the Leray-Serre spectral sequence for $P \to BA_5 \to BSO(3).$

Because of the calculation in (2) above, the spectral sequence takes the form $E^{*,0} = E^{*,3} = H^*(BSO(3);\Bbb Z)$, where all other $E^{p,q} = 0$. This is a multiplicative spectral sequence, which means every page is a module over $H^*(BSO(3);\Bbb Z)$. In particular, the $E_2$ page is a free module over this ring (with two generators, one in degree $E^{0,0}$ and one in degree $E^{0,3}$). Clearly the only possible nonzero differential is $d_4$, for degree reasons; I claim it suffices to determine the map $d_4: E^{0,3} \to E^{4,0}$. This is because if we write the generator of $E^{0,3}$ as $1_{0,3}$, then any element in $E^{p,3}$ may be written as $c \cdot 1_{0,3}$ for some $c \in H^p(BSO(3))$. Therefore, because the differentials in the spectral sequence are derivations for the action of the cohomology ring, we have $$d_4(c \cdot 1_{0,3}) = d_4(c) \cdot 1_{0,3} + c \cdot d_4(1_{0,3}) = c \cdot d_4(1_{0,3}),$$ because $c$ is contained in the bottom row and hence $d_4(c) = 0$. Therefore the $E_5$ page is given on the $0$-line as $H^{*}(BSO(3))/d_4(1_{0,3})$ and on the $3$-line as $\text{ker}(d_4)$.

As for actually calculating this, I could in principle try to understand the transgression in the Serre spectral sequence, but it's easier to cheat and just use the computer calculation of the first four homology groups here. We see using the universal coefficient theorem that $H^3(A_5; \Bbb Z) = \Bbb Z/2$ and $H^4(A_5; \Bbb Z) = \Bbb Z/30$. One concludes that $d_4(1_{0,3})$ must be $30 p_1$, writing $p_1 \in H^4(BSO(3))$ for the generator in degree 4.

This map $d_4$ has no kernel, because multiplication by $p_1$ has no kernel in the cohomology ring $H^*(BSO(3))$. So the $E_5$ and hence $E_\infty$ page is contained in the $q = 0$ line; there are thus no filtration problems, and we find that $E_\infty = H^*(A_5;\Bbb Z)$. In particular, we find $$H^*(A_5;\Bbb Z) = H^*(BSO(3);\Bbb Z)/(30p_1) = \Bbb Z[e, p_1]/(2e, 30p_1).$$

The following has been edited from the original post, where I claimed a calculation of the cohomology ring.

The case $p = 2$ above requires slightly more care with regards to filtration problems because the $q = 3$ line survives (in fact, there are no differentials in the spectral sequence), but you're working over a field: this means that as groups, $H^k(BA_5; \Bbb F_2) \cong \oplus_i E_\infty^{k, i-k}$, and the product in $H^*(BA_5)$ agrees up to lower-filtration terms with that on $E_\infty$. That means that if $c_1 \in E_\infty^{i, k-i}$ and $c_2 \in E_\infty^{j,\ell-j}$, then the product $c_1 \cdot c_2$ in $H^{k+\ell}(BA_5)$ agrees with $c_1 \cdot c_2 + c_3$, where $c_1 \cdot c_2 \in E_\infty^{i+j,\ell+k-i-j}$ is the product on the $E_\infty$ page, and $$c_3\in \oplus_{s < i+j} E_\infty^{s, k+\ell - s}.$$ This means the product can be written in terms of the product on the $E_\infty$ page and some "terms which are further to the left".

What the spectral sequence shows is that $H^*(BA_5; \Bbb F_2)$ has a subalgebra $H^*(BSO(3); \Bbb F_2)$, and the quotient $H^*(BA_5; \Bbb F_2)/H^*(BSO(3);\Bbb F_2)$ as modules is a free rank 1 $H^*(BSO(3);\Bbb F_2)$-module. If $k$ corresponds to the element of $E^{0,3}$, it projects to a generator of the latter. But it is not immediately clear what the relations are on these products. (I believe a theorem of Symonds implies that there are no further module relations, but there may be further algebra relations; in particular, it is not clear what $k^2$ is.)

Similarly, let me remark that the above argument just as well shows that $H^*(A_5; \Bbb F_p)$ has $H^*(A_5;\Bbb F_p)/H^*(BSO(3);\Bbb F_p)$ a 1-dimensional free module over $H^*(BSO(3);\Bbb F_p).$ Of course, this is completely sufficient for determining the individual cohomology groups.

Finally,

$$H^*(A_5; F) = F \;\;\; F \text{ of characteristic} \geq 7 \text{ or } 0,$$ where this means there is no cohomology except for in degree $0$.

Solution 2:

There is a much easier answer that I missed on a first pass, but uses some technical theory as opposed to just doing a spectral sequence.

Let $G$ be a group and $P$ be the $p$-Sylow subgroup. If $P$ is abelian, then the map $$H^*(G;\Bbb F_p) \to H^*(P;\Bbb F_p)^{N_G(P)/P}$$ given by restriction is an isomorphism; here the exponent means we take fixed points under the natural action of $N_G(P)/P$ by conjugation (we may quotient by $P$ because it acts trivially: $P$ is abelian). The proof uses the double coset formula for the transfer operator; you may find such a proof in Theorem 6.8 of Adem and Milgram's book on group cohomology.

In particular, we have $$H^*(A_5; \Bbb F_2) = H^*(V_4; \Bbb F_2)^{\Bbb Z/3}$$ $$H^*(A_5; \Bbb F_3) = H^*(\Bbb Z/3; \Bbb F_3)^{\Bbb Z/2}$$ $$H^*(A_5; \Bbb F_5) = H^*(\Bbb Z/5; \Bbb F_5)^{\Bbb Z/2}$$

I'll let you calculate the actions of $N_G(P)/P$ on $P$ in each case.

The most interesting is the top. I will cite for you the full computation from Adem-Milgram, Theorem 1.3: $$H^*(A_4; \Bbb F_2) = \Bbb F_2[w_2, w_3, k]/(k^2 + kw_3 + w_3^2 + w_2^3).$$

The other two are quick calculations, and give $$H^*(A_5; \Bbb F_3) = \Bbb F_3[k, p_1]/(k^2)$$ and $$H^*(A_5; \Bbb F_5) = \Bbb F_5[k, p_1]/(k^2)$$ as expected.