# finitely many orbits of integer solutions module unimodular groups

Let $Q$ be the Lorentzian matrix, that is $Q=\mathrm{diag}(I_n,-1)$, where $I_n$ denotes de $n\times n$ identity matrix. Let $\mathcal R$ the set of integer solutions $x\in\mathbb Z^{n+1}$ of $$Q[x]:=x^tQx=x_1^2+\dots+x_{n}^2-x_{n+1}^2=-1.$$ The set $\Gamma=\mathrm{O}(n,1)\cap \mathrm{M}(n+1,\mathbb Z)$ acts by left multiplication on $\mathcal R$. It's known that there are finitely many $\Gamma$-orbits.

I think (but I'm not completely sure) that $\Gamma\backslash \mathcal R$ has only one class if $2\leq n\leq 7$, and two if $n=8$ (since $(0,0,0,0,0,0,0,0,1)$ and $(1,1,1,1,1,1,1,1,3)$ leaves in different $\Gamma$-orbits).

What happens in the hermitian case?

More precisely, let $\Gamma=\mathrm{U}(n,1)\cap\mathrm M(n+1,\mathcal O)$ with $\mathcal O$ the ring of integer of a quadratic imaginary extension of $\mathbb Q$, and let $\mathcal R$ be the set of "integer" solutions $x\in\mathcal O^{n+1}$ of $$Q[x]=x^*Qx=|x_1|^2+\dots+|x_{n}|^2-|x_{n+1}|^2=-1.$$

How many $\Gamma$-orbits are there in the set $\mathcal R$?

[Edit: It's enough for $n=2,3$ and some particular ring $\mathcal O$, like $\mathbb Z[\sqrt{-1}]$.]

Let ${\bf Q}[x]=|x_0|^2+|x_1|^2+\cdots+ |x_n|^2-|x_{n+1}|^2$. A solution to $Q[x]=-1$ gives a solution to ${\bf Q}[x]=0$ by setting $x_0=1$. Orbits of solutions to ${\bf Q}[x]=0$ correspond to cusps (ends) of the orbifold $\mathbb{CH}^{n+1}/{\bf \Gamma}$, where ${\bf \Gamma} = U({\bf Q};\mathcal{O})$. Thus, Stover's theorem says that for any $k$, there are only finitely many $\mathcal{O}$ and dimensions $n$ such that ${\bf \Gamma}$ has less than $k$ cusps (I think it's fairly standard that for different $\mathcal{O}$ quadratic rings of integers, the lattices $U({\bf Q};\mathcal{O})$ will be non-commensurable).
I would like to say now that this should imply that there are only finitely many $n, \mathcal{O}$ that there are $\leq k$ solutions to $Q[x]=-1$ up to the action of $\Gamma$. The problem is that not every ${\bf \Gamma}$-orbit of solution to ${\bf Q}[x]=0$ necessarily has a representative with $x_0=1$. However, I would still conjecture this to be true, and one might study the techniques in Stover's paper to see if they carry over to your question.
Dear Agol, I'm just interested in the number of elements of $\Gamma\backslash \mathcal R$ for $n=2$ or $3$. – emiliocba Dec 28 '11 at 18:23