MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
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 $n=\pm 1$).

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.