# Lattice automorphisms of finite order

Are there any known examples of lattice automorphisms of finite order in indefinite lattices being classified up to conjugacy?

Yes, there are such examples. You can refer to this question on MO where the automorphism goup of an indefinite lattice is described. In this example, torsion elements are of order 2, and an element in each conjugacy class is given. Another simple example is the case of the form $H\perp<2>$ ($H$ is the hyperbolic plane) whose orthogonal group is isomorphic to $\mathbf Z/2 \times \Gamma$, where $\Gamma$ is isomorphic to $\mathbf Z/4\star_{\mathbf Z/2}\mathbf Z/6$.
The very well known torsion elements are the reflections, whose conjugacy classes are known in small dimensions thanks to the work of Vinberg (and some other great mathematicians). When the group is generated by reflections (as in the case of $H\perp<2>$, you obtain by Vinberg algorithm a fundamental domain, and you just have to compute the stabilizers of the bounding elements (which are intersections of the hyperplanes relative to which the reflections act) to obtain a finite set of representants of all conjugacy classes of torsion elements. Then a little more work identify representants in the same class. This should work with the group $\mathrm{O}_{1,n}(\mathbf Z)$ for $n\leq 13$, for example (see Vinberg : on groups of unit elements of certain quadratic forms).
• @Bob : It should not be that hard to do the job with $H\perp H$ : a model is the module $\mathrm{M}_2(\mathbf Z)$, with the determinant. This gives you a morphism $\mathrm{GL}_2(\mathbf Z)\times_{det} \mathrm{GL}_2(\mathbf Z)\to \mathrm{SO}(H\perp H)$. The image is of small index (perharps of order 2, generated by $M\mapsto M^t$, this would need to be checked). – few_reps Sep 17 '13 at 9:44