# Isomorphism of fundamental groups

This is some kind of follow-up to this computation of the fundamental group of a mapping torus.

For $A \in SL_2 (\mathbb{Z})$, let $G_A := \mathbb{Z}^2 \rtimes_A \mathbb{Z}$ be the fundamental group of the mapping torus corresponding to $A$; that is, the group law is defined as:

$$(p,n) * (q,m) = (p+A^n q,n+m).$$

Let $A$, $B \in SL_2 (\mathbb{Z})$. If $B$ is conjugated to $A$ or $A^{-1}$, then $G_A$ and $G_B$ are isomorphic.

Question : is the converse true (at least for hyperbolic matrices) ?

I've heard that (some version of) this proposition was proved by Poincaré, but I don't know where to look, and my meagre algebra skills are not sufficient to see an easy solution.

• If you replace $A$ by $-A$ do you get isomorphic groups? – Charlie Frohman Dec 20 '16 at 21:40
• @Charlie Frohman : quick sanity check, $G_I = \mathbb{Z}^3$ while $G_{-I}$ is not Abelian. – D. Thomine Dec 20 '16 at 22:01
• Thoughts in passing : in general, the center is trivial. The abelianisation looks more interesting (it is $\mathbb{Z}^2_{/(A-I)\mathbb{Z}^2} \times \mathbb{Z}$), and one can recover $|Tr(A)-2|$. Still far from being enough. Note that, by the way, $G_A$ is not isomorphic to $G_{-A}$ whenever $Tr(A) \neq 0$. – D. Thomine Dec 20 '16 at 22:26

Yes, this is true provided that "conjugated" means "conjugate in $GL(2, {\mathbb Z})$". I will leave it to you to construct examples where "conjugate in $SL(2, {\mathbb Z})$" is not enough.

Here is a proof, which is a heavy edit of the original argument that was missing several special cases.

I will use the notation $K={\mathbb Z}^2$ for the normal subgroup in the semidirect product $G_A={\mathbb Z}^2\rtimes _A {\mathbb Z}$ and the notation $Q$ for the quotient $G_A/K$. Note that

1. $G_A$ is abelian if and only if $A=I$.

2. $G_A$ is nonabelian but contains an abelian subgroup of index $2$ iff $A=-I$. (Recall that $A\in SL(2, {\mathbb Z})$.)

Therefore, from now on, I will assume that $A\ne \pm I, B\ne \pm I$.

Under the assumption $A\ne \pm I$, the following dichotomy holds:

(i) Either $A$ fixes a primitive integer vector $v\in {\mathbb Z}^2$; equivalently, $G_A$ contains a unique maximal normal infinite cyclic subgroup, generated by $v$, or

(ii) $A$ has no eigenvalues $\pm 1$, equivalently, $G_A$ contains a unique maximal normal rank 2 abelian subgroup, namely, the subgroup $K$.

Equivalence in (i) is clear; to prove equivalence in (ii), note that
any abelian subgroup $H$ strictly containing $K$ will project to an infinite cyclic subgroup of the quotient group $Q:=G_A/{\mathbb Z}^2\cong {\mathbb Z}$, which will imply that $H$ is abelian of rank 3. To show uniqueness, suppose that $H< G_A$ is an abelian subgroup of rank 2. Project $H$ to the quotient group $Q$. If the projection is trivial then $H$ is contained in our subgroup $K$. If the projection is nontrivial, then it is infinite cyclic. Since $H$ has rank 2, $H\cap K$ is also infinite cyclic. Since $H$ is normal in $G_A$, $H\cap K$ is generated by a primitive integer vector, an eigenvector of $A$.

Consider Case (i). Then either $Av=v$ or $Av=-v$. Which of these two cases occurs depend only on the isomorphism class of the group $G_A$, since $Av=v$ is equivalent to the assumption that $G_A$ has nontrivial center.

(ia) If $Av= v$, then $A$ is conjugate in $GL(2, {\mathbb Z})$ to the matrix $$\left[\begin{array}{cc} 1 & n\\ 0 & 1\end{array}\right]$$ In this case, the center $Z(G_A)$ of $G_A$ is infinite cyclic generated by $v\in {\mathbb Z}^2$ and $G_A/Z(G_A)\cong {\mathbb Z}^2$. For any choice of a basis $\{v, w\}$ of ${\mathbb Z}^2$, a generator $q$ of $Q$ and its lift $u$ to $G_A$, we have $$[u, w]= v^{\pm n}.$$ Therefore, $|n|$ is uniquely determined by the group $G_A$ and, so is the $GL(2, {\mathbb Z})$-conjugacy class of the matrix $A$.

(ib) If $Av=-v$ then $A$ is conjugate in $GL(2, {\mathbb Z})$ to the matrix $$\left[\begin{array}{cc} -1 & n\\ 0 & -1\end{array}\right]$$ Consider then the subgroup $G_{A^2} < G_A$: It is a characteristic subgroup of $G_A$ since it equals the kernel of the action of $G_A$ on $<v>$ (given by conjugation). The subgroup $G_{A^2}$ has nontrivial center and we recover $|2n|$ from $G_{A^2}$ as in (ia). Thus, again, the group $G_A$ determines the number $|n|$ and, hence, the conjugacy class of $A$ in $GL(2, {\mathbb Z})$.

Case (ii). Thus, $A$ (and $B$) does not have primitive integer eigenvectors in ${\mathbb Z}^2$. Then the subgroup $K$ is the unique maximal normal abelian subgroup of rank $2$ in $G_A$. In order to prove maximality, note that any strictly larger abelian subgroup $H$ will project to an infinite cyclic subgroup of the quotient group $Q:=G_A/{\mathbb Z}^2\cong {\mathbb Z}$, which will imply that $H$ is abelian of rank 3. To show uniqueness, suppose that $H< G_A$ is a normal abelian subgroup of rank 2. Project $H$ to the quotient group $Q$. If the projection is trivial then $H$ is contained in our subgroup $K$. If the projection is nontrivial, then it is infinite cyclic. Since $H$ has rank 2, $H\cap K$ is also infinite cyclic. Since $H$ is normal in $G_A$, $H\cap K$ is generated by a primitive integer vector, an eigenvector of $A$, a contradiction.

Thus, $K$ is a characteristic subgroup of $G_A$. Hence, any isomorphism $\phi: G_A\to G_B$ will preserve the semidirect product decomposition. The isomorphism $\phi$ also projects to an automorphism of quotient cyclic groups $\bar{\phi}: {\mathbb Z}\to {\mathbb Z}$. There will be two cases to consider: (i) $\overline\phi: 1\mapsto 1$ and (ii) $\overline\phi: 1\mapsto -1$.

I will identify matrices $A$ and $B$ in your question with corresponding automorphisms of $K$. Each isomorphism $\phi: G_A\to G_B$ will restrict to an equivariant (with respect to the action of $Q$ on $K$) isomorphism $\psi: {\mathbb Z}^2\to {\mathbb Z}^2$ such that (due to equivariance) either

(i) $\psi \circ A= B\circ \phi$, or

(ii) $\psi \circ A= B^{-1} \circ \phi$

(depending on whether we are in Case (i) or Case (ii) above).

Now, use the fact that $Aut({\mathbb Z}^2)= GL(2, {\mathbb Z})$ which allows us to identify the automorphism $\psi$ with a matrix $C\in GL(2, {\mathbb Z})$. Then the above equations translate into:

(i) $CA C^{-1}= B$ or (ii) $C A C^{-1}= B^{-1}$.

qed

• If $A$ has a nonzero fixed point in $K$, then $K$ is not the unique maximal normal abelian subgroup of rank two. For example, if $K$ has a basis $\{k_1,k_2\}$ and $A$ acts by $k_1\mapsto k_1$ and $k_2\mapsto k_1+k_2$, then $k_1$ and any element that maps to a generator of $Q$ will generate another such subgroup. – Jeremy Rickard Dec 25 '16 at 9:27
• Thank you. There are a few fuzzy points (e.g. Jeremy Rickard's comment), but that arguments seems to work in most cases, and I'll check the remaining ones (rotations, transvections...). – D. Thomine Dec 25 '16 at 17:08
• @D.Thomine Jeremy is wrong. – Moishe Kohan Dec 25 '16 at 19:18
• @Moishe Cohen: I was rather thinking about $A = Diag(1,-1)$, for which $\mathbb{Z}^2 \times 2\mathbb{Z}$ is abelian and normal. – D. Thomine Dec 25 '16 at 20:05
• @D.Thomine This example is not in SL(2,Z). It is also not rank 2. – Moishe Kohan Dec 25 '16 at 20:11