Orthogonal group of the lattice \$I_{p,q}\$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T00:01:22Z http://mathoverflow.net/feeds/question/73088 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73088/orthogonal-group-of-the-lattice-i-p-q Orthogonal group of the lattice \$I_{p,q}\$? A. Pascal 2011-08-17T18:31:42Z 2011-08-18T16:10:50Z <p>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.</p> <p>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.</p> <p>Are there similar descriptions of the orthogonal group of \$I_{p,q}\$? What about the special case \$q=2\$?</p> http://mathoverflow.net/questions/73088/orthogonal-group-of-the-lattice-i-p-q/73116#73116 Answer by Agol for Orthogonal group of the lattice \$I_{p,q}\$? Agol 2011-08-18T02:47:14Z 2011-08-18T16:10:50Z <p>The subgroup generated by reflections is normal, and therefore is finite-index by the <a href="http://groupprops.subwiki.org/wiki/Margulis%27_normal_subgroup_theorem" rel="nofollow">Margulis normal subgroup theorem</a> (as long as the rank is \$\geq 2\$, so \$|p|\geq 2, |q|\geq 2\$). </p> <p><strong>Addendum:</strong> </p> <p>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\$. </p> <p>There's also the <a href="http://www.ams.org/mathscinet-getitem?mr=507030" rel="nofollow">congruence subgroup property</a>, 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. </p>