7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.