# infinite order automorphism on torus

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

• 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 $\operatorname{ord}(A)\leq \operatorname{ord}(A_T)$. – Jason DeVito Aug 19 at 14:55

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.

• Very nice. Indeed, I was hoping to obtain the result on the eigenvalues. – klirk Aug 19 at 15:39
• 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. – Lee Mosher Aug 19 at 15:44