infinite order automorphism on torus

Question

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 $$A^m x = x + \gamma, ~~~~~\gamma \in \Gamma.$$

Answer 1

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.