Question

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.

Answer 1

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.

Answer 2

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.)