# Kernel of linear representation of Baumslag-Solitar group

Let $BS(m,n)$ be the Baumslag-Solitar group defined by $B(m,n) = < a,b ~|~ b a^m b^{-1} = a^n >$, $mn \neq 0$. There is a linear representation of $BS(m,n)$ by mapping $a$ to the matrix $\left(\begin{matrix} 1&1 \cr 0&1\end{matrix}\right)$ and $b$ to the matrix $\left(\begin{matrix} \frac{n}{m}&0 \cr 0&1\end{matrix}\right)$. Denote this representation homomorphism as $f$, assume $|m| \neq |n|$, my main question is:

What is the kernel of $f$ ?

Some observations

(1) Commutator of the form $[a, a^b], [a,a^{b^2} ], [a,a^{b^3}] \ldots$, are in the kernel. Do these elements generate the Kernel of $f$? Do they form an infinite generated free group? .

(2) If $|m| \ne |n|$ and either $|m| = 1$ or $|n| = 1$ then $f$ is known to be injective.

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The answer is No, if you take $|n|=1=|m|$. – Anton Klyachko Nov 13 '12 at 23:35
Hi, Anton Klyachko. Your answer is contained in my second observation. – Xiaolei Wu Nov 13 '12 at 23:52
If $|m|=|n|=1$, then $f$ is not injective and the kernel is not generated by the commutators (1). – Anton Klyachko Nov 13 '12 at 23:57
Anton, you are right. I will edit the question to rule out those cases. – Xiaolei Wu Nov 14 '12 at 0:35
Actually, if $|m|=|n|$, then $f$ is not njective and its kernel is not generated by commutators (1). – Anton Klyachko Nov 14 '12 at 0:42

The kernel $K$ is a free group of infinite rank.

To see that it is a free group, take the action of $BS(m,n)$ on the Bass-Serre tree $T$ of the usual graph of groups presentation for the usual HNN decomposition $Z*_Z$, where one of the $Z \mapsto Z$ edge-to-vertex homomorphisms is multiplication by $m$ and the other is multiplication by $n$. Each vertex stabilizer of $BS(m,n)$ is conjugate to the $\langle a \rangle$ subgroup, whose intersection with $K$ is trivial. So, all vertex stabilizers of the action of $K$ on $T$ are trivial, implying that all edge stabilizers are trivial. $K$ therefore acts properly discontinuously on the tree $T$.

One can show that the rank is infinite by exhibiting arbitrarily long simple closed edge paths in the quotient graph $T/K$, although maybe there is a slicker way.

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Yes, but how does it actually sit in the mother group? – Igor Rivin Nov 14 '12 at 2:44
@Igor: Would a good answer to your followup question be a listing of a free basis in terms of a,b? If so, that could be worked out by going back and forth between the action of BS(m,n) on $\mathbb{H}^2$ and the action on T. – Lee Mosher Nov 14 '12 at 13:16
@ Lee, I will try to work it out by myself using your method. If I could answer all the question I asked, I will accept your answer. – Xiaolei Wu Nov 18 '12 at 5:18

This isn't a complete answer, but here is an observation that I hope is helpful.

Since linear groups are residually finite, the kernel clearly contains the finite residuum, ie the intersection of all the finite-index subgroups. By Theorem 4.1 of this paper by Jack Button and m and n are coprime, the finite residuum is equal to the second derived subgroup.

(Thanks to Yves Cornulier for pointing out the missing hypothesis.)

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if $m,n$ are not coprime, I think there are non-metabelian finite quotients. (blast, why don't we have a word to mean the "residuallyfinitization"...) – YCor Nov 14 '12 at 7:40
Sorry, yes of course you're right. If m and n are not coprime then in fact B(m,n) is large, ie has a finite-index subgroup that surjects a non-abelian free group. My answer holds in the coprime case. – HJRW Nov 14 '12 at 10:18
(Yves, re: the word, we do; it's the profinite completion! It's just in the wrong category.) – HJRW Nov 14 '12 at 10:20
No, I mean the quotient by the intersection of the finite index subgroup. It's distinct from profinite completion. – YCor Nov 14 '12 at 10:40
Yves - I know! But it's morally the same thing as the profinite completion - just in a different category. Just my bad joke. – HJRW Nov 14 '12 at 13:03