Question

The Baumslag-Solitar group $BS(m,n)$ is given by the group presentation $BS(m,n)=(a,b|ba^{m}b^{-1}=a^{n})$. When does it embed into a linear group? Thanks!

Answer 1

Presumably you've consulted the Wikipedia page on the Baumslag-Solitar group, which states that $BS(m,n)$ is not residually finite (and therefore not linear) if $|m|\neq |n|$ and $|m|>1, |n|>1$. This leaves the case that $|m|=|n|$. In this case, one can show that the group is the fundamental group of a compact Seifert-fibered 3-manifold, which is known to be linear.

Consider $BS(m,\pm m)$. Take a solid torus, and two parallel annuli in the boundary whose cores run $|m|$ times around the core of the solid torus. Attach an annulus $\times I$ to these two annuli, with opposite orientation for $BS(m,-m)$, to get a Seifert-fibered 3-manifold with fundamental group $BS(m,\pm m)$. For example, $BS(1,-1)$ is the fundamental group of the Klein bottle, which when thickened up is an interval bundle over a Klein bottle.

A Seifert 3-manifold with boundary has a finite-sheeted cover which is a product $S^1\times \Sigma^2$. The fundamental group is $\mathbb{Z}\times F$, where $F$ is a free group, and thus this group is linear. Take the induced representation to get a linear representation of the original 3-manifold group.

Answer 2

The metabelian groups $BS(n,1)\simeq BS(1,n)=\langle a,b\mid aba^{-1}=b^n \rangle$ are also linear (this seems not mentionened in the Wikipedia article). A faithful, linear representation $BS(1,n)\hookrightarrow GL_2(\mathbb{R})$ is given by $$ a\mapsto \begin{pmatrix} n^{\frac{1}{2}} & 0 \cr 0 & n^{-\frac{1}{2}} \end{pmatrix}, \quad b\mapsto \begin{pmatrix} 1 & 1 \cr 0 & 1 \end{pmatrix}. $$ This representation is not discrete, and the groups are not polycyclic (except for $n=\pm 1$).