Quick question:(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Discrete abelian subgroups

Loading...

Similar Threads - Discrete abelian subgroups | Date |
---|---|

A Non-Abelian Stokes theorem and variation of the EL action | May 31, 2016 |

Computing a discrete surface integral of a scalar function | Oct 24, 2013 |

Plausibility of a Discrete Point Manifold | Jul 3, 2013 |

Will limit of discrete steps give Pythagoras theorem? | Apr 3, 2011 |

Discrete quotient group from closed subgroup | Mar 28, 2011 |

**Physics Forums - The Fusion of Science and Community**