- #1
FreHam
- 10
- 0
Quick question:
Suppose I have a (transitive) R^n action on a manifold M. If the isotropy group of R^n is discrete, does that mean that it is automatically isomorphic to Z/kZ, with 0<=k<=n?
Basically, my discrete subgroup is a lattice then, right?
Thanks!
Suppose I have a (transitive) R^n action on a manifold M. If the isotropy group of R^n is discrete, does that mean that it is automatically isomorphic to Z/kZ, with 0<=k<=n?
Basically, my discrete subgroup is a lattice then, right?
Thanks!