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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 $q=2$?

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Thanks. I have been looking at Allcock's papers, which are very interesting, but haven't found anything besided the hyperbolic/Lorentzian case. Do you have a particular paper in mind? –  A. Pascal Aug 17 '11 at 21:19
Yes, it's annoying that the Russian version is free but you have to pay for the translation. At least I hope some of that money goes to support Russian mathematics. –  A. Pascal Aug 17 '11 at 21:23
My campus math library has an article copying service. If they have it (at a separate storage facility a few miles away) one can request a scanned pdf, although I suppose they would refuse for something book length or too recent, and they do have translated Mat. Sbornik. The one he recommended for me was "new complex- and quaternion-hyperbolic reflection groups", but actually his email was more informative. This refers to my several MO questions on covering radius and class number for positive forms/lattices. Even the standard, Neumaier and Siedel 1983, is just Lorentzian. –  Will Jagy Aug 17 '11 at 22:38
Did you read his Fall 2011 class description, ma.utexas.edu/users/allcock/teaching/Coxeter.html where he recommends Humphries, "Reflection Groups and Coxeter Groups". –  Will Jagy Aug 17 '11 at 22:40
I see, he mistyped, the book is by Jim Humphreys, an MO user. –  Will Jagy Aug 18 '11 at 1:48
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1 Answer

up vote 3 down vote accepted

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 $|p|\geq 2, |q|\geq 2$).


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.

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Why is it obvious that the subgroup is normal? –  Will Jagy Aug 18 '11 at 4:10
I think that the group generated by reflections must be normal since the set of generators is stable under the adjoint action. –  A. Pascal Aug 18 '11 at 7:50
Thanks. I didn't know this theorem. So this reduces the problem to describing the group generated by reflections. In the hyperbolic case, Vinberg gives an algorithm for doing so. What about in other signatures? –  A. Pascal Aug 18 '11 at 7:52
Agol, thanks for the additional explanation. –  Will Jagy Aug 18 '11 at 20:25
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