Non-torsion part of the abelianisation of congruence subgroups











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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?










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    MathSE original post: math.stackexchange.com/questions/3006331
    – YCor
    Nov 22 at 11:59















up vote
4
down vote

favorite












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?










share|cite|improve this question


















  • 2




    MathSE original post: math.stackexchange.com/questions/3006331
    – YCor
    Nov 22 at 11:59













up vote
4
down vote

favorite









up vote
4
down vote

favorite











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?










share|cite|improve this question













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?







finite-fields abelian-groups congruences linear-groups






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asked Nov 22 at 10:38









Liam Baker

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  • 2




    MathSE original post: math.stackexchange.com/questions/3006331
    – YCor
    Nov 22 at 11:59














  • 2




    MathSE original post: math.stackexchange.com/questions/3006331
    – YCor
    Nov 22 at 11:59








2




2




MathSE original post: math.stackexchange.com/questions/3006331
– YCor
Nov 22 at 11:59




MathSE original post: math.stackexchange.com/questions/3006331
– YCor
Nov 22 at 11:59










2 Answers
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up vote
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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 $mathbb Q$).






share|cite|improve this answer























  • The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
    – YCor
    Nov 23 at 17:40










  • @YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
    – Venkataramana
    Nov 24 at 1:48




















up vote
1
down vote













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).






share|cite|improve this answer





















  • I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
    – YCor
    Nov 22 at 18:13










  • By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
    – YCor
    Nov 22 at 18:27











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2 Answers
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2 Answers
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up vote
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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 $mathbb Q$).






share|cite|improve this answer























  • The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
    – YCor
    Nov 23 at 17:40










  • @YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
    – Venkataramana
    Nov 24 at 1:48

















up vote
9
down vote













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 $mathbb Q$).






share|cite|improve this answer























  • The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
    – YCor
    Nov 23 at 17:40










  • @YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
    – Venkataramana
    Nov 24 at 1:48















up vote
9
down vote










up vote
9
down vote









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 $mathbb Q$).






share|cite|improve this answer














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 $mathbb Q$).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 at 17:39









YCor

27k380132




27k380132










answered Nov 22 at 11:27









Venkataramana

8,92412950




8,92412950












  • The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
    – YCor
    Nov 23 at 17:40










  • @YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
    – Venkataramana
    Nov 24 at 1:48




















  • The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
    – YCor
    Nov 23 at 17:40










  • @YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
    – Venkataramana
    Nov 24 at 1:48


















The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
– YCor
Nov 23 at 17:40




The result in Serre's book is actually for arbitrary finite index subgroups. For congruence subgroups, it remains to see whether the rational homology in degree 1 always vanishes.
– YCor
Nov 23 at 17:40












@YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
– Venkataramana
Nov 24 at 1:48






@YCor: I do not know about that. All I said was that the rational homology is finite dimensional. Whether it vanishes for congruence subgroups I do not know.
– Venkataramana
Nov 24 at 1:48












up vote
1
down vote













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).






share|cite|improve this answer





















  • I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
    – YCor
    Nov 22 at 18:13










  • By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
    – YCor
    Nov 22 at 18:27















up vote
1
down vote













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).






share|cite|improve this answer





















  • I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
    – YCor
    Nov 22 at 18:13










  • By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
    – YCor
    Nov 22 at 18:27













up vote
1
down vote










up vote
1
down vote









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).






share|cite|improve this answer












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).







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answered Nov 22 at 18:09









Andrei Smolensky

1,1311022




1,1311022












  • I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
    – YCor
    Nov 22 at 18:13










  • By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
    – YCor
    Nov 22 at 18:27


















  • I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
    – YCor
    Nov 22 at 18:13










  • By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
    – YCor
    Nov 22 at 18:27
















I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
– YCor
Nov 22 at 18:13




I wouldn't call it "more general setting" since in the setting of the question precisely the most delicate case is $n=2$.
– YCor
Nov 22 at 18:13












By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
– YCor
Nov 22 at 18:27




By the way it's been proved by Shalom-Vaserstein that $E(n,A)$ has Property T for all $nge 3$ and every finitely generated commutative (unital associative) ring $A$, so all its finite index subgroups have Property T. It's been extended to $A$ arbitrary finitely generated (associative unital) ring $A$ by Ershov-Jaikin.
– YCor
Nov 22 at 18:27


















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