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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 $\mathbb Z[\sqrt{-1}]$.]

Thank you in advance.-.

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1 Answer 1

I don't know the answer to your question (if there is one), but I just noticed a preprint by Matt Stover on the arxiv that I think is relevant to a related question. This preprint says that the number of commensurability classes of arithmetic lattices with a bounded number of ends is finite. Let's try to see how this might be related to your question.

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.

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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

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