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All isometries of the n-torus

by Palindrom
Tags: isometries, ntorus
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Palindrom
#1
Oct21-06, 11:05 AM
P: 266
How does one start this kind of question? I'm completely stumped.
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mtiano
#2
Oct21-06, 03:35 PM
P: 6
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
Palindrom
#3
Oct22-06, 03:17 AM
P: 266
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.

Chris Hillman
#4
Dec1-06, 05:40 PM
Sci Advisor
P: 2,340
Post All isometries of the n-torus

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

Chris Hillman


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