TITLE: Orbits of action of $SL_m(\mathbb{Z})$ on $\mathbb{Z}^m$
QUESTION [7 upvotes]: I'm considering the action of $SL_m(\mathbb{Z})$ on $\mathbb{Z}^m$: if $A\in SL_m(\mathbb{Z})$ and $v\in\mathbb{Z}^m$, then $Av\in\mathbb{Z}^m$.
My question is: what are the orbits of this action?  I'm especially interested in the case $m=3$.
For $m=2$, we have the following:
If $a$ and $b$ are (positive) relatively prime integers, then you can always find integers $c$ and $d$ so that $ad-bc=1$, so that
$\begin{pmatrix}
a&c \\
b&d
\end{pmatrix}$
is in $SL_2(\mathbb{Z})$.
That means that
$\begin{pmatrix}
a \\
b
\end{pmatrix}$
is in the orbit of
$\begin{pmatrix}
1 \\
0
\end{pmatrix}$
under this action. Conversely it's easy to see that nothing else can be in that orbit. More generally, the orbits of this action are in bijection with the nonnegative integers: the orbit corresponding to $n>0$ consists mainly of the lattice points
$\begin{pmatrix}
a \\
b
\end{pmatrix}$
with $\gcd(|a|,|b|)=n$. And if $n=0$, then the orbit consists of
$\begin{pmatrix}
0 \\
0
\end{pmatrix}$
only.

REPLY [0 votes]: Apply Euclid's algorithm to the set of entries of a nonzero integer vector $v$. The operations of the algorithm amount to multiplying by elementary integer matrices (with all 1's on the diagonal and $\pm 1$ in one off-diagonal entry), which are, therefore, in $GL(m,Z)$. After finitely many steps (which will amount to acting by a product of elementary matrices), you convert $v$ to a vector of the form $(d, 0,...,0)$, where $d$ is the gcd of the entries of $v$. In order to get a matrix in $SL(m,Z)$, just multiply by a matrix of the form $Diag(1,-1,1...)$ if needed. The conclusion is that there is only one orbit of the action for each absolute value of $gcd$ of the vector entries.
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