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Question: Show that tfae:

  1. The endomorphism of the torus $T_A:\mathbb{T}^n\to\mathbb{T}^n, [x]\mapsto[Ax]$, where $[x]:=\{x+y:y\in\mathbb{Z}^n\}$ and $A\in M_n(\mathbb{Z})$, is invertible.

  2. $x\in\mathbb{Z}^n \iff Ax\in\mathbb{Z}^n$.

  3. $|\det A|=1$.

Motivation: This question is from Barreira & Valls' Dynamical Systems: an Introduction (p. 25), which I am reading as a reading course.

I think we need to establish a relation between $|\det A|=1$ and that $A^{-1}\in M_n(\mathbb{Z})$. I could not prove this, but supposing it I could do $1\implies2, 3\implies2$ and $3\implies1$.

Alp Uzman
  • 10,742

2 Answers2

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Well, $\det(A\cdot B)=\det A\cdot\det B$, and also, $\det A\in\Bbb Z$ if the entries of $A$ are from $\Bbb Z$.
So, if $\exists A^{-1}\in M_n(\Bbb Z)$, then we have the following product of integers as $AA^{-1}=I$ and $\det(I)=1$: $$\det A\cdot \det(A^{-1})=1\,.$$

For $2\implies3$: if $A$ (any real matrix) is not invertible, then $0$ is an eigenvalue, so there is a whole line mapped to $0$, and a line definitely contains points out of $\Bbb Z^n$. So, assuming $2$, $A$ must be invertible, but the condition also implies that the preimages of the standard basis vectors are all in $\Bbb Z^n$, which are just the columns of $A^{-1}$.

Berci
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It took me a while but I think I got it now. Here is my take (which is a slight variant of Berci's answer):

First I'll expand and reformulate the problem. Fix $d\in\mathbb{Z}_{\geq1}$ as the dimension. Let $GL(d,\mathbb{R})$ denote the group of $d\times d$ invertible matrices with real coefficients and let $GL(d,\mathbb{Z})$ denote the group of $d\times d$ invertible matrices with integer coefficients whose inverse is also with integer coefficients (as in General linear group over integers). The determinant of a matrix with integer coefficients is also an integer, and by the multiplicativity of the determinant, one can also write more explicitly $GL(d,\mathbb{Z})=M_d(\mathbb{Z})\cap\{|\det|=1\}$, where $M_d(\mathbb{Z})$ is the algebra of $d\times d$ matrices with integer coefficients. For convenience we'll identify matrices with linear maps throughout.

The $d$-torus $\mathbb{T}^d=\mathbb{R}^d/\mathbb{Z}^d$ is an abelian Lie group with Lie algebra $\mathbb{R}^d$ (with zero bracket). Let us denote the points of $\mathbb{T}^d$ by $[v]=\{w\in\mathbb{R}^d\vert w-v\in\mathbb{Z}^d\}$, $v\in\mathbb{R}^d$, where $v\mapsto [v]$ is the universal covering map (= exponential map). If $f:\mathbb{T}^d\to \mathbb{T}^d$ is a Lie group endomorphism, the linear map $f'(e_{\mathbb{T}^d}):\mathbb{R}^d\to\mathbb{R}^d$ completely determines it ($e_{\mathbb{T}^d}=[0]$). Conversely, if $A:\mathbb{R}^d\to \mathbb{R}^d$ is a linear map and $\forall v\in\mathbb{R}^d: v\in\mathbb{Z}^d\implies Av\in\mathbb{Z}^d$ (which means $A\in M_d(\mathbb{Z})$), then $A$ factors through $v\mapsto [v]$. The resulting map is denoted by $T_A$ as above. Thus any Lie group endomorphism of $\mathbb{T}^d$ is of the form $T_A: [x]\mapsto [Ax]$ for some $A\in M_d(\mathbb{Z})$. Note that $[0]$ is a fixed point of any $T_A$.

The question is about which Lie group endomorphisms of $\mathbb{T}^d$ are actually Lie group automorphisms. It suffices to find those $A$s in $M_d(\mathbb{Z})$ for which $(T_A)^{-1}:\mathbb{T}^d\to\mathbb{T}^d$ exists.

Question, rephrased: Let $A\in M_d(\mathbb{Z})$. Then the following are equivalent:

  1. $T_A$ is a Lie group automorphism of $\mathbb{T}^d$.
  2. $\forall v\in\mathbb{R}^d: v\in\mathbb{Z}^d \iff Av\in\mathbb{Z}^d$.
  3. $A\in GL(d,\mathbb{Z})$.

Proof: (1.$\implies$2.) Necessity is automatic from $A\in M_d(\mathbb{Z})$. For sufficiency, let $v\in\mathbb{R}^d$ be such that $Av\in\mathbb{Z}^d$. Then $[Av]=[0]$. Since $T_A$ is invertible, $[v]=(T_A)^{-1}\circ T_A ([v])=(T_A)^{-1}([Av])=(T_A)^{-1}([0])=[0]$, so $v\in\mathbb{Z}^d$.

(2.$\implies$3.) Let $v\in \ker(A)$. Then $Av=0\in\mathbb{Z}^d$, whence by hypothesis $v\in\mathbb{Z}^d$, so that $\ker(A)\leq \mathbb{Z}^d$, and thus $\ker(A)=0$ and $A$ in invertible (i.e. $A\in GL(d,\mathbb{R})$). Now let $w\in\mathbb{R}^d$ and put $v=A^{-1}w$. Then the hypothesis reads $A^{-1}w\in\mathbb{Z}^d \iff w\in\mathbb{Z}^d$. In particular choosing the standard basis vectors $e_1=(1,0,...,0),..., e_d=(0,...,0,1)$ as $w$ gives that any column of $A^{-1}$ is a vector in $\mathbb{Z}^d$, so that $A^{-1}\in M_d(\mathbb{Z})$. Thus $A\in GL(d,\mathbb{Z})$.

(3.$\implies$1.) By hypothesis and our considerations above regarding Lie group endomorphisms of $\mathbb{T}^d$, both $A$ and $A^{-1}$ induce endomorphisms of $\mathbb{T}^d$. Then $(T_A)^{-1}\circ T_A([x])=[A^{-1}Ax]=[x]$ and similarly $T_A\circ(T_A)^{-1}=\operatorname{id}_{\mathbb{T}^d}$, so that $(T_A)^{-1}=T_{A^{-1}}$ and $T_A$ is a Lie group automorphism of $\mathbb{T}^d$.

Finally let me note that $SL(d,\mathbb{Z})\leq GL(d,\mathbb{Z})$. The smaller group consists of Lie group automorphisms of $\mathbb{T}^d$ that not only preserve volume (which is automatic) but also orientation.

Alp Uzman
  • 10,742