I am trying to compute the group $H_1(\mathrm{SL}_2(\mathbb{Z}_2),M)$, where $\mathbb{Z}_2$ are $2$-adic integers and M is a module $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. I suppose that the group acts on $M$ by matrix multiplication.

I found a similar-looking computation in the paper of Dupont and Sah "Homology of Euclidean groups of motion made discrete and Euclidean scissors congruences". It was shown there that $H_1(\mathrm{SO}_3(\mathbb{R}),\mathbb{R^3}) = \Omega^1_\mathbb{R}.$

I would be very grateful for any help with computing the group or for any interpretation of its elements.


First, some general remarks on the situation for $R$ an arbitrary commutative ring. Since ${\rm diag}(-1,-1)$ acts by multiplication by $-1$ on $R^{\oplus 2}$, the homology groups ${\rm H}_i({\rm SL}_2(R),R^{\oplus 2})$ are $2$-torsion for $i\geq 1$. (This is called the center-kills-argument.) This doesn't happen in the case ${\rm SO}(3)$ because the center of ${\rm SO}(3)$ is trivial. This is a significant difference between the case of ${\rm SL}_2(R)\looparrowright R^{\oplus 2}$ in the question and the case ${\rm SO}(3)\looparrowright \mathbb{R}^3$ in the paper of Dupont and Sah.

As a particular case, if $R=K$ is a field, then the homology groups will be vector spaces over the field and therefore the homology group ${\rm H}_1({\rm SL}_2(K),K^{\oplus 2})$ can only be nontrivial in characteristic $2$. (In particular, these are trivial for $R=\mathbb{Q}_2$; the title said something about local fields.)

In the specific situation of $R=\mathbb{Z}_2$, we can say something more about the homology group in the question, by iterated application of the Hochschild-Serre spectral sequence (well, the five-term exact sequence suffices).

First, we compute the homology ${\rm H}_1({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})$. We have ${\rm SL}_2(\mathbb{F}_2)=S_3$, so we can use the Hochschild-Serre spectral sequence for the extension $0\to\mathbb{Z}/3\to S_3\to\mathbb{Z}/2\to 0$. To compute ${\rm H}_i(\mathbb{Z}/3,\mathbb{F}_2^{\oplus 2})$ we can use the standard resolution $$ \cdots\to M\xrightarrow{N} M\xrightarrow{t-1}M $$ where $M=\mathbb{F}_2^{\oplus 2}$, $t$ is a generator and $N=1+t+t^2$ is the norm element. In this case, $N$ is the zero map so the even homology groups are the cokernel of $t-1$ and the odd homology groups are the kernel of $t-1$. But $t-1$ is an isomorphism, so ${\rm H}_i(\mathbb{Z}/3,\mathbb{F}_2^{\oplus 2})=0$ for all $i$. The Hochschild-Serre spectral sequence then also implies that ${\rm H}_i({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})=0$ for all $i$.

Now we can use the Hochschild-Serre spectral sequence associated to the extension $1\to\mu_2\to{\rm SL}_2(\mathbb{Z}_2)\to {\rm PSL}_2(\mathbb{Z}_2)\to 1$. The homology groups ${\rm H}_i(\mathbb{Z}/2,\mathbb{Z}_2^{\oplus 2})$ can also be computed using the standard resolution as above. The norm map is again the zero map, and $t-1$ is multiplication by $-2$. In particular, odd homology is trivial and even homology is given by $\mathbb{F}_2^{\oplus 2}$. It remains to compute the homology ${\rm H}_i({\rm PSL}_2(\mathbb{Z}_2),\mathbb{F}_2^{\oplus 2})$ with the action by left multiplication of matrices. This can be done using the extension $1\to \Gamma\to{\rm PSL}_2(\mathbb{Z}_2)\to {\rm SL}_2(\mathbb{F}_2)\to 1$. The congruence subgroup $\Gamma$ acts trivially on $\mathbb{F}_2^{\oplus 2}$. By the above computation for ${\rm SL}_2(\mathbb{F}_2)$, the contribution ${\rm H}_1({\rm SL}_2(\mathbb{F}_2),{\rm H}_0(\Gamma,\mathbb{F}_2^{\oplus 2}))\cong {\rm H}_1({\rm SL}_2(\mathbb{F}_2),\mathbb{F}_2^{\oplus 2})$ is trivial. So we have identified
$$ {\rm H}_1({\rm SL}_2(\mathbb{Z}_2),\mathbb{Z}_2^{\oplus 2})\cong {\rm H}_0({\rm SL}_2(\mathbb{F}_2),{\rm H}_1(\Gamma,\mathbb{F}_2^{\oplus 2})). $$

At this point it's not quite clear to me what the abelianization of the congruence subgroup would be (this is the relevant thing to complete the calculation). In the higher-rank cases, the abelianization is the corresponding Lie algebra over the finite field. For ${\rm SL}_2$, the abelianization would still surject onto $\mathfrak{sl}_2(\mathbb{F}_2)$ but could be bigger, the higher-rank arguments don't work because they depend on the identification of the commutator subgroups and elementary subgroups. (Note however that a direct computation shows that if the abelianization of the congruence subgroup is $\mathfrak{sl}_2(\mathbb{F}_2)$ then the coinvariants and hence the homology group in the question would be trivial.)


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