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Notations and definitions.

  • For all $A\in\textrm{GL}_n(\mathbb{Z})$, $A$ is hyperbolic if and only if none of its complex eigenvalue has module $1$.

  • For all $(A,B)\in\textrm{GL}_n(\mathbb{Z})^2$, one says that: $$A\sim B\iff\exists P\in\textrm{GL}_n(\mathbb{Z}),\exists\varepsilon\in\{\pm 1\}\textrm{ s.t. }A=PB^{\varepsilon}P^{-1}.$$

  • For all $A\in\textrm{GL}_n(\mathbb{Z})$, one defines the following group: $$G_A:=\mathbb{Z}^n\rtimes_{\alpha_A}\mathbb{Z},$$ where $\alpha_A\colon\mathbb{Z}\rightarrow\textrm{Aut}(\mathbb{Z}^n)$ is the group homomorphism given by: $$\alpha_A(n)(x):=A^nx.$$

Theorem. Let $(A,B)\in\textrm{GL}_n(\mathbb{Z})^2$ be two hyperbolic matrices such that $G_A\cong G_B$, then $A\sim B$.

Proof. Let $\phi\colon G_A\rightarrow G_B$ a group isomorphism, I have already proved that $\mathbb{Z}^n\times\{0\}$ is the only maximal abelian subgroup of rank $n$ of $G_A$. From there, I am asked to establish that $\phi$ gives rise through quotient to an isomorphism $\overline{\phi}\colon\mathbb{Z}\rightarrow\mathbb{Z}$ and conclude. $\Box$

I get that $\overline{\phi}$ will be either $\textrm{id}$ or $-\textrm{id}$, but I don't understand how to build $\overline{\phi}$ and how to conclude. Any help will be greatly appreciated.

C. Falcon
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1 Answers1

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The fact that $\bar\phi:\mathbb{Z}\rightarrow\mathbb{Z}$ is an isomorphism implies that $\phi(0,1)=(u,1)$ or $\phi(0,1)=(u,-1)$.

Let $x\in\mathbb{Z}^n$, you have $(0,1)(x,0)(0,-1)=(A(x),0)$. Suppose $\phi(0,1)=(u,1)$

$\phi$ preserves the maximal ideals implies that $\phi$ induces a linear map $P:\mathbb{Z}^n\times\{0\}\rightarrow \mathbb{Z}^n\times\{0\}$. You have: \begin{align}\phi((0,1)(x,0)(0,-1))&=\phi(A(x),0),\\&=(PA(x),0),\\&=\phi(0,1)\phi(x,0)\phi(0,-1),\\&=\phi(0,1)\phi(x,0)\phi(0,1)^{-1},\\&= (u,1)(P(x),0)(-B^{-1}(u),-1),\\&=(BP(x)+u,1)(-B^{-1}(u),-1),\\&=(BP(x),0). \end{align} We deduce that $BP(x)=PA(x)$.

If $\phi(0,1)=(u,-1)$. The same method shows that $B^{-1}P=PA$.

C. Falcon
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Tsemo Aristide
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  • Thank you kindly for your help! I took the liberty to fix the minor typos of your answer and add some more details. :) – C. Falcon Jul 30 '17 at 19:19
  • Do you have any idea if this result remains true without the hypothesis of $A$ and $B$ being hyperbolic or if there is a counterexample? In my case, $A$ and $B$ are in $SL(n,\mathbb{Z})$ with all their eigenvalues being roots of unity. – Alejandro Tolcachier Jul 27 '19 at 17:21