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:



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> What is the kernel of $f$ ?
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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.



Kernel of linear representation of Baumslag-Solitar group
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.


