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.

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    $\begingroup$ Isn't it the same for general $n$. Aren't they just in the same orbit if and only if their entries have the same gcd? $\endgroup$ – Derek Holt Apr 4 '14 at 7:20
  • $\begingroup$ I suspect it is the same, but how can I even begin to explicitly show that? $\endgroup$ – Isabelle Apr 4 '14 at 13:59

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.

  • $\begingroup$ This is correct (except for $m=1$, where the sign is also an invariant...) $\endgroup$ – YCor Apr 7 '14 at 21:49
  • $\begingroup$ True: I was implicitly assuming that $m>1$. $\endgroup$ – Moishe Kohan Apr 7 '14 at 22:19

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