## H_2(Gamma_N(p),Z)

In this post (which is a follow-up to the last post), I wanted to compute the group $H_2(\Gamma_N(p),\mathbf{Z})$, where $\Gamma_N(p)$ is the congruence subgroup of $\mathrm{SL}_N(\mathbf{Z})$ for large enough $N$ and $p$ is prime. In fact, to make my life easier, I will also assume that $p > 3,$ and in addition, ignore $2$-torsion. The first problem is to compute the prime to $p$ torsion. By Charney’s theorem, this will come from the cohomology of the homotopy fibre $X$ of the map

$\displaystyle{ SK(\mathbf{Z}) \rightarrow SK(\mathbf{F}_p).}$

The relevant part of the Serre long exact sequence is, using classical computations of the first few K-groups of the integers together with Quillen’s computation of $K_*(\mathbf{F}_p),$

$\displaystyle{ 0 \rightarrow \pi_3(X) \rightarrow \mathbf{Z}/48 \mathbf{Z} \rightarrow \mathbf{Z}/(p^2 - 1) \mathbf{Z} \rightarrow \pi_2(X) \rightarrow \mathbf{Z}/2 \mathbf{Z} \rightarrow 0.}$

Here is where it is convenient to invert primes dividing $6;$ from Hurewicz theorem and Charney’s theorem we may deduce that, where $\sim$ denotes an equality up to a finite group of order dividing $48,$

$\displaystyle{ H_2(\Gamma_N(p),\mathbf{Z}[1/p]) \sim \mathbf{Z}/(p^2 - 1) \mathbf{Z}.}$

In order to deal with $3$-torsion, then we also have to show that the map $K_3(\mathbf{Z}) \otimes \mathbf{Z}/3 \mathbf{Z} \rightarrow K_3(\mathbf{F}_p)$ is injective for $p \ne 3.$ I have a sketch of this which I will omit from this discussion but it is not too hard (assuming Quillen-Lichtenbaum). It remains to compute the homology with coefficients in $\mathbf{Z}_p$. I previously computed that there was an isomorphism

$\displaystyle{ H_2(\Gamma_N(p),\mathbf{F}_p) = \wedge^2 \mathfrak{g} \oplus \mathbf{F}_p = H_2(G_N(p),\mathbf{F}_p) \oplus \mathbf{F}_p,}$

where $G_N = \mathrm{SL}_N(\mathbf{Z}_p)$ and $\mathfrak{g} = H_1(\Gamma_N(p),\mathbf{F}_p)$ is the adjoint representation.

Some facts concerning the cohomology of $G_N(p)$:

There are short exact sequences:

$\displaystyle{ 0 \rightarrow H_2(G_N(p),\mathbf{Z}_p)/p \rightarrow H_2(G_N(p),\mathbf{Z}/ p \mathbf{Z}) \rightarrow H_1(G_N(p),\mathbf{Z}_p)[p] \rightarrow 0,}$

$\displaystyle{ 0 \rightarrow H_2(G_N(p),\mathbf{Z}_p)/p^2 \rightarrow H_2(G_N(p),\mathbf{Z}/ p^2 \mathbf{Z}) \rightarrow H_1(G_N(p),\mathbf{Z}_p)[p^2] \rightarrow 0.}$

Since $H_1(G_N(p),\mathbf{Z}_p) = \mathfrak{g}$ is annihilated by $p,$ we may deduce that

$\displaystyle{ H_2(G_N(p),\mathbf{Z}_p)/p = H_2(G_N(p),\mathbf{Z}_p)/p^2}$

as long as

$\displaystyle{ |H_2(G_N(p),\mathbf{Z}/p^2 \mathbf{Z}) | = | H_2(G_N(p),\mathbf{Z}/ p \mathbf{Z})|.}$

Such an equality (for any group) is a claim about the Bockstein maps having a big an image as possible. Indeed, for any group $\Phi,$ there is an exact sequence:

$\displaystyle{ H_3(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p^2 {\mathbf{Z}}) \rightarrow H_2(\Phi,{\mathbf{Z}}/p {\mathbf{Z}}) \rightarrow H_1(\Phi,{\mathbf{Z}}/p {\mathbf{Z}})}$

The first and last maps here are the Bockstein maps $\beta_2$ and $\beta_1$. Since $p$ is odd, $\beta_1 \circ \beta_2 = 0$. On the other hand, we see that the orders of the cohomology groups with coefficients in $\mathbf{Z}/p \mathbf{Z}$ and $\mathbf{Z}/p^2 \mathbf{Z}$ will have the same order if and only if

$\displaystyle{ \ker(\beta_1) = \mathrm{im}(\beta_2).}$

Hence we have reduced to the following claim. Take the complex

$\displaystyle{ H_*(G_N(p),\mathbf{F}_p) = \wedge^* \mathfrak{g} }$

where the differentials are given by the Bockstein maps. Then we have to show that the cohomology of this complex vanishes in degree two. But what are the Bockstein map is in this case? Note that since $H_1(G_N(p),\mathbf{Z}_p) = \mathfrak{g}$ is annihilated by $p$, the Bockstein map $\beta_1$ will be a surjective map:

$\displaystyle{ \beta_1: \wedge^2 \mathfrak{g} \rightarrow \mathfrak{g}.}$

To compute this explicitly, recall that the isomorphism $H_1(G_N(p),{\mathbf{Z}}_p) = \mathfrak{g}$ comes from the identification of $\mathfrak{g}$ with $G_N(p)/G_N(p^2).$ Then, computation omitted due to laziness, we find that the Bockstein is precisely the Lie bracket. Moreover, since the (co-)homology is generated in degree one, the higher Bockstein maps can be computed from the first using the cup product formula. So the Bockstein complex above is, and I haven’t checked this because it must be true, the complex computing the mod-$p$ Lie algebra cohomology of $\mathfrak{g}$. And this cohomology vanishes in degrees one and two, so we are done. One consequence of this computation is that

$\displaystyle{ H_2(G_N(p),\mathbf{Z}_p) = H_2(G_N(p),\mathbf{Z}_p)/p}$

is annihilated by $p.$ Moreover, the last term can be identified with the kernel of the Lie bracket (Bockstein) on $H_2(G_N(p),\mathbf{F}_p) = \wedge^2 \mathfrak{g}.$

Returning to the main computation:

From the Hochschild–Serre spectral sequence and the computation of stable completed cohomology, one has an exact sequence:

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}_p) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}_p) \leftarrow \mathbf{Z}_p \leftarrow H_3(G_N(p),\mathbf{Z}_p).}$

From known results in characteristic zero, we immediately deduce that there is some $\alpha$ such that there is an exact sequence

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}_p) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}_p) \leftarrow \mathbf{Z}/p^{\alpha} \mathbf{Z} \leftarrow 0.}$

we also deduce that there is an exact sequence:

$\displaystyle{ 0 \leftarrow H_2(G_N(p),\mathbf{Z}/p^n \mathbf{Z}) \leftarrow H_2(\Gamma_N(p),\mathbf{Z}/p^n \mathbf{Z}) \leftarrow \mathbf{Z}/p^{\mathrm{min}(\alpha,n)} \mathbf{Z} \leftarrow 0,}$

There are spectral sequences:

$\displaystyle{ H_i(\mathrm{SL}_N(\mathbf{F}_p),H_j(\Gamma_N(p),A)) \Rightarrow H_{i+j}(\Gamma_N,A)}$

for $A = \mathbf{Z}/p^n \mathbf{Z}$ and $A = \mathbf{Z}_p.$
For both of these rings, we have

$\displaystyle{ H_1(\Gamma_N(p),A) = \mathfrak{g}, \qquad H_0(\Gamma_N(p),A) = A.}$

Moreover, for sufficiently large $N,$ we have

$\displaystyle{ H_i(\mathrm{SL}_N(\mathbf{F}_p),A) = 0,}$

this follows from and is equivalent to Quillen’s computation which implies that the $K$-groups of finite fields have order prime to $p.$ Since $H_2(\mathrm{SL}_N(\mathbf{Z}),\mathbf{Z}_p)$ is trivial for $p > 2,$ we deduce that

$\displaystyle{ H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),A)) = H_2(\mathrm{SL}_2(\mathbf{F}_p),\mathfrak{g}) = \mathbf{F}_p,}$

where the last equality was already used in my paper. The compatibility of the spectral sequence above for different $A$ implies that we also get an isomorphism

$\displaystyle{ H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),\mathbf{Z}_p)) = H_0(\mathrm{SL}_N(\mathbf{F}_p),H_2(\Gamma_N(p),\mathbf{Z}/p \mathbf{Z})).}$

On the other hand, the invariant class must be an element of order $p$ in

$\displaystyle{ \mathbf{Z}/p^{\alpha} \mathbf{Z};}$

and hence the reduction map

$\displaystyle{ \mathbf{Z}/p^{\alpha} \mathbf{Z} \rightarrow \mathbf{Z}/p \mathbf{Z}}$

sends an element of order $p$ to an element of order $p,$ and so $\alpha = 1.$

Putting things back together:

Assembling all the pieces, we see that we have proven the following:

Theorem: Let $p > 3,$ and let $N$ be sufficiently large. Let $\mathfrak{g}$ be the Lie algebra $\mathfrak{sl}_N$ over $\mathbf{F}_p.$ Then, up to a finite group of order dividing $48,$ we have

$\displaystyle{ \displaystyle{ H_2(\Gamma_N(p),\mathbf{Z}) \sim \mathbf{Z}/(p^2 - 1) \mathbf{Z} \oplus \mathbf{Z}/p \mathbf{Z} \oplus \ker \left( [\ , \ ] \wedge^2 \mathfrak{g} \rightarrow \mathfrak{g} \right)}.}$

Moreover, still with $p > 3,$ then up to a group of order dividing $16,$ we should have the same equality with $p^2 - 1$ replaced by $(p^2 - 1)/3.$