Here $I\_{p,q}$ is the unique-up-to-isometry unimodular lattice of signature $(p,q)$, whose Gram matrix is diagonal with $p$ 1s and $q$ -1s.


In his paper "ON GROUPS OF UNIT ELEMENTS OF CERTAIN QUADRATIC FORMS", Vinberg gives a description of the automorphism group of the lattice $I\_{p,1}$. It is a semi-direct product of the subgroup generated by reflections, which is a hyperbolic Coxeter group that can be effectively described, and a subgroup of the symmetries of the fundamental polyhedron for this Coxeter group.


Are there similar descriptions of the orthogonal group of $I\_{p,q}$? What about the special case $q=2$?



Orthogonal group of the lattice $I_{p,q}$?
The subgroup generated by reflections is normal, and therefore is finite-index
by the [Margulis normal subgroup theorem](http://groupprops.subwiki.org/wiki/Margulis%27_normal_subgroup_theorem) (as long as the rank is $\geq 2$, so $|p|\geq 2, |q|\geq 2$). 


**Addendum:** 


The conjugate of a reflection is a reflection. In fact, a reflection may be defined as a matrix element $A$ such that $I−A$ has rank 1. This is clearly conjugacy invariant. Also, if you conjugate a reflection in the vector $v$ by a matrix $B$, then you get a reflection in $Bv$. 


There's also the [congruence subgroup property](http://www.ams.org/mathscinet-getitem?mr=507030), so any finite-index subgroup is a congruence subgroup (in rank >1). What you can do (in principle) is start enumerating congruence subgroups (and use Reidemeister-Schreier to find generators), and start multiplying together reflections. Eventually, you will find generators of a finite-index congruence subgroup which are products of finitely many reflections. Take the normal subgroup generated by these (assuming we have included a conjugate of every reflection) in the finite quotient to determine the subgroup generated by reflections. 


