Let $M = \Gamma \backslash G/K$ be a Riemannian locally symmetric space, where $G$ is a connected semisimple Lie group of rank at least $2$, $K$ its maximal compact subgroup and $\Gamma < G$ an irreducible lattice.

Question: Does $M$ admit a global geodesic symmetry?

By a geodesic symmetry I mean an isometry of $M$ fixing a point $x\in M$ and reversing all geodesics through $x$. If the answer is NO then are there related general results?

  • $\begingroup$ If $\Gamma$ is a maximal arithmetic subgroup then the answer is NO. Indeed, the symmetry (or any isometry of $M$) belongs to the normalizer of $\Gamma$ in $G$ but then (by the maximality) belongs to $\Gamma$. But the $\Gamma$ action is free so the symmetry can not belong to $\Gamma$. $\endgroup$ – Holonomia Jan 7 '17 at 12:34
  • $\begingroup$ @Holonomia: There is no reason for $\Gamma$ to be torsion-free. In fact, if $\Gamma_0$ is a lattice with torsion, every maximal lattice $\Gamma$ containing $\Gamma_0$ will again contain torsion. $\endgroup$ – Moishe Kohan Aug 5 '17 at 3:40
  • $\begingroup$ @MoisheCohen : if $M$ is a manifold then the action of the latice must be free, isnt'it? $\endgroup$ – Holonomia Aug 5 '17 at 8:16
  • $\begingroup$ @Holonomia: Sure, but why would a maximal arithmetic lattice yield a manifold? (It does not, in general.) I know that in rank 1 there are maximal lattices which are torsion-free, but I do not know why this would be the case in higher rank. $\endgroup$ – Moishe Kohan Aug 5 '17 at 8:59
  • $\begingroup$ @Moishe Cohen: the OP question assume $M$ to be a manifold, actually a locally symmetric space, isnt'it? $\endgroup$ – Holonomia Aug 5 '17 at 9:03

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