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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

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    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
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