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Let $T= \mathbb R^n/\Gamma$ a torus, where $\Gamma$ is the standard lattice in $\mathbb R^n$. A matrix $A\in SL(n,\mathbb Z)$ induces an automorphism $A_T$ on $T$. I want to calculate the order of $A_T$.

Since the identity induces the identity, it is clear that $\operatorname{ord} (A_T)\le \operatorname{ord}(A)$. But is there something more one can say? In particular, are there sufficient criteria on $A$ for $A_T$ being of infinite order?

I have trouble extracting useful information from the equality $$A^m x = x + \gamma, ~~~~~\gamma \in \Gamma.$$

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    $\begingroup$ I think that by looking at homology, one can prove that $(A_T)_\ast:H_1(T^n)\rightarrow H_1(T^n)$ is given by the matrix $A$ when using the standard basis. In particular, it would then follow that $\operatorname{ord}(A)\leq \operatorname{ord}(A_T)$. $\endgroup$ – Jason DeVito Aug 19 at 14:55
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There's a bunch more things to say.

First of all, the opposite inequality $\text{ord}(A) \le \text{ord}(A_T)$ is also true, for the simple reason that the function $A \mapsto A_T$ is an injective homomorphism from the group $SL(n,\mathbb Z)$ to the group of automorphisms of $T$.

It follows that you can convert your problem entirely into algebra, by working in the group $SL(n,\mathbb Z)$, in order to investigate this question.

Next, given a matrix $M \in SL(n,\mathbb Z)$, you can bring tools of linear algebra to bear on this problem, for example this theorem:

If $\lambda_1,...,\lambda_k \in \mathbb C$ are the eigenvalues of $M$ then $\lambda^n_1,\ldots,\lambda^n_k$ are the eigenvalues of $M^k$.

This gives a powerful sufficient condition for infinite order, because if $M$ has finite order then each of $\lambda_1,...,\lambda_k$ is a root of unity.

Here's an example. The characteristic polynomial of $M = \begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ is $\lambda^2 - \lambda - 1$, the roots of which are $\lambda = \frac{1}{2} \pm \frac{\sqrt{5}}{2}$, and neither of these has absolute value $1$ so they are not roots of unity, hence $M$ has infinite order.

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  • $\begingroup$ Very nice. Indeed, I was hoping to obtain the result on the eigenvalues. $\endgroup$ – klirk Aug 19 at 15:39
  • $\begingroup$ Actually, I just realized that my example matrix is in $GL(n,\mathbb Z)$ but not $SL(n,\mathbb Z)$, but I think I'll just leave it in place anyway, because my entire answer holds word-for-word with $GL(n,\mathbb Z)$ in place of $SL(n,\mathbb Z)$, and because I like that matrix. $\endgroup$ – Lee Mosher Aug 19 at 15:44

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