Argthtjtr

I've posted this question on math.stackexchange, but haven't gotten any responses so I'm trying here instead.

Let $A = F_q[T]$ be the ring of polynomials in one variable with coefficients in a finite field, and let $r>1$ be an integer.
I'm currently looking for the abelianisation of the congruence subgroup $Γ(N)$ of the special linear group $SL(r,A)$, i.e. the kernel of the 'modulo $N$' map $SL(r,A) to SL(r,A/N)$, where $N in A$ is a nonconstant polynomial; more particularly, I'm looking for the torsion-free part of the abelianisation, but I wouldn't complain about knowing the abelianisation itself if that's possible.

So, here are my questions, in reverse order of importance:

1. What are the commutator subgroups of $SL(r,A)$ and $Γ(N)$?


2. What are the abelianisations of these groups?


3. What are the torsion-free abelianisations of these groups?

Suppose $r geq 3$. Then one can show that $Gamma=mathrm{SL}(r,A)$ is a lattice in the group $G=mathrm{SL}big(r, {mathbb F}_q(!(1/t)!)big)$. The group $G$ has Kazhdan's property T and hence $Gamma$ itself has property T. In particular, it is finitely generated and its abelianization is finite. The same conclusion holds for finite index subgroups of $Gamma$ and hence for the congruence subgroups $Gamma (N)$.
(You do not need property T for the whole group $Gamma$ since $Gamma$ is generated by the elementary matrices which have finite order; but for subgroups of finite index, you can use property T.)

The group $SL(2,A)$ again is generated by elementary matrices, but for finite index subgroups (the non-torsion part of ) the abelianization might be infinite; Serre's book on trees has results on this. (edited: it is proved Corollary 4 in section II.2.8 in Serre's book on trees that the first homology with rational coefficients of a congruence subgroup of $SL(2,A)$ is a finite dimensional vector space over $mathbb Q$).

In a more general setting, let $A$ be a commutative ring, $I, J$ its ideals, $ngeqslant3$, then
$$[E(n,A,I),E(n,A,J)]geqslant E(n,R,IJ),$$
where $E(n,A,I)$ is the normal closure in $E(n,A)$ of the subgroup $E(n,I)$ generated by the elementary generators $x_{ij}(xi)=1+xi e_{ij}$ of level $I$, that is, with $xiin I$. In particular, $SL(n,A)$ is perfect once it coincides with $E(n,R)$ (which is the case for polynomials over a field).

This is well-known since Bass' 1964 paper. Since $mathbb{F}_q[T]$ is Euclidean (and thus of stable rank $2$), one has $E(n,R,I)=SL(n,R,I)$ by the $K_1$-stability result of Vaserstein (Theorem 3.2).

It is also known that $E(n,R,I)=langle z_{ij}(xi,eta)colon xiin I, etain A rangle$, where $z_{ij}(xi,eta)=x_{ji}(eta)x_{ij}(xi)x_{ji}(-eta)$.

Now the projection $SL(n,A,I)to SL(n,A,I)_{mathrm{ab}}$ factors through $S=SL(n,R,I)/SL(n,r,I^2)$, and it is not hard to show for $A=mathbb{F}_q[T]$ and $I=fA$ that the image of $z_{ij}(xi,eta)$ is $S$ has finite order, hence this abelianization is also finite.

Alternatively, one can do it with an even more straightforward computation. A reasonably good set of generators is known for $[E(n,A,I),E(n,A,J)]$ (see this paper). Namely, it is generated (as a normal subgroup) by the elements of the following three types:

1. 
   $[x_{ij}(xi),z_{ij}(zeta,eta)]$,


2. 
   $[x_{ij}(xi),x_{ji}(zeta)]$,


3. 
   $x_{ij}(xizeta)$,

where $xi,zetain I$, $etain A$.

Now $z_{ij}(xi)^k=x_{ji}(eta)x_{ij}(kxi)x_{ji}(-eta) equiv x_{ij}(kxi)$, so taking $k=operatorname{char}(mathbb{F}_q)$ gives you the finiteness of orders of the generators, while the third relation shows that it suffices to take only finitely many of them and allows to calculate the abelianization explicitly (similar to this answer).