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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!

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2–Solitar_group has your answer. – Benjamin Steinberg Jan 24 '13 at 2:54
Hmm the link didn't copy properly. – Benjamin Steinberg Jan 24 '13 at 2:56
Looks like you got an en-dash in there somehow. – HJRW Jan 25 '13 at 14:06

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.

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Algebraically we can view linearity of $BS(n,en)$ ($e\in\{1,-1\}$) as follows: consider its diagonal embedding into $(\mathbf{Z}*(\mathbf{Z}/n\mathbf{Z})\times \mathbf{Z}\rtimes\mathbf{Z}$, where the action is trivial or by flip according to whether $e=1$ or $-1$. Here the left homomorphism is given by modding out by the common $n\mathbf{Z}$ and the second is the obvious homomorphism (which for $BS(p,q)$ goes into $\mathbf{Z}[1/pq]\rtimes_{p/q}\mathbf{Z}$. Then both factors on the right are linear (over $\mathbf{Z}$, or alternatively over $\mathbf{C}$ in dimension 2), so $BS(n,en)$ is linear. – YCor Jan 24 '13 at 10:01
Parentheses are missing, I mean $(Z*(Z/nZ))\times (Z\rtimes Z)$. By By By "both factors on the right" I mean both factor in this direct decomposition. – YCor Jan 24 '13 at 10:03
It also leaves the case $|m|=1$ (or $|n|=1$), when the representation described in the article is faithful. – HJRW Jan 25 '13 at 14:05
The representation describes in this article is not faithful then, see [mathoverflow]… – Dietrich Burde Apr 3 '13 at 12:21
@D Burde: It looks like the questioner in that link you provided never corrected his question to cover the case $|m|=|n|$, as he had intended. I edited that question just now to make the correction. – Lee Mosher Apr 3 '13 at 13:27

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 $n=\pm 1$).

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Although technically this is different from the Wikipedia representation, it is closely related since it has the same image in $PGL_2(\mathbb{R})$, so they differ by twisting by the square root of the determinant. Also, the Wikipedia representation is also clearly faithful (as Lee Mosher points out in a comment above, his answer should have excluded $|m|=1$ or $|n|=1$). – Ian Agol Apr 3 '13 at 17:48
I agree, thank you for the clarification. The above representation unlikely leads to the (false) belief, that $BS(1,n)$ is a subgroup of $GL_2(\mathbb{ℤ})$. For the Wikipedia representation, the appearance reminds one of integer matrices at first sight. – Dietrich Burde Apr 3 '13 at 19:01

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