Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

All isometries of the n-torus

  1. Oct 21, 2006 #1
    How does one start this kind of question? I'm completely stumped.
     
  2. jcsd
  3. Oct 21, 2006 #2
    The set of all isometries on any riemannian manifold M forms a group called the isometry group on M. For R^n this is the set of all rigid motions each of which has the form f(x) = Ax + b where A is in O(n) and b is in R^n. For S^n this is the set of all orthagonal transformations, i.e. O(n). Now here is my guess. Since T^n = S^1 X ... X S^1 (n times) then perhaps it's isometry group is O(1) X ... X O(1) with the product group structure. If this fails perhaps you could use the fact that T^n is homeomorphic to R^n/Z^n. Good luck
     
  4. Oct 22, 2006 #3
    First of all, thanks a lot for taking the time.

    I'd already figured out that T^n being isometric to R^n/Z^n might be helpful. But it is true that in this case, every isometry of R^n/Z^n extends to an isometry of R^n?

    This is actually what's bothering me.

    I'll admit I hadn't thought of the S^1 X...X S^1 idea, but it looks like I have a similar problem in this case.
     
  5. Dec 1, 2006 #4

    Chris Hillman

    User Avatar
    Science Advisor

    Lifting from torus to plane

    "Lift", not "extend". The textbook by Boothby, An Introduction to Differentiable Manifolds and Riemannian Geometry should help.

    Chris Hillman
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: All isometries of the n-torus
Loading...