Orthogonal group of the lattice $I_{p,q}$?

2011-08-17T18:31:42Z

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 $q=2$?

Answer by Agol for Orthogonal group of the lattice $I_{p,q}$?

2011-08-18T02:47:14Z

The subgroup generated by reflections is normal, and therefore is finite-index by the Margulis 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, 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.

Orthogonal group of the lattice $I_{p,q}$? - MathOverflow

Orthogonal group of the lattice $I_{p,q}$?

Answer by Agol for Orthogonal group of the lattice $I_{p,q}$?