
#1
Oct2106, 11:05 AM

P: 266

How does one start this kind of question? I'm completely stumped.




#2
Oct2106, 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




#3
Oct2206, 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. 



#4
Dec106, 05:40 PM

Sci Advisor
P: 2,341

All isometries of the ntorusChris Hillman 


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