8
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.