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With respect to answer, I am not able to see the following:

" Flat Metric Is Unique Up To Diffeomorpshim " What i meant can be seen by clicking the link.

Is it trivial? Can someone help me out.

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1 Answer 1

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You have to be a little careful in the interpretation here. Let $(M,g)$ be a complete, simply connected Riemannian manifold of constant curvature $K$, and let $(V^K, \rho)$ be Euclidean $\mathbb R^n$ if $K = 0$, $S^n$ if $K > 0$, and $H^n$ if $K < 0$ where $S^n$ is equipped with the round metric of curvature $K$, and $H^n$ is the hyperbolic plane equipped with the standard hyperbolic metric of curvature $K$. Then it is a classical result that there exists a diffeomorphism $$F: (M,g) \to (V^K, \rho)$$ so that $g = F^*\rho$ i.e. $M$ is isometric to $V^k$. As a corollary of this, if you have two flat metrics on the same simply connected, complete manifold, then they are related to one another by a diffeomorphism. On the other hand, you may of course have non-diffeomorhpic spaces with "the same" flat metrics, e.g. the cylinder and the plane.

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Thanks for the en-lighting answer....+1 for this.. –  zapkm Mar 16 '12 at 15:13
    
Can i get the reference of the theorem, you mentioned. –  zapkm Mar 17 '12 at 3:29
    
One reference: see proposition 4.2 at page 242 (roughly) of RW Sharpe's Differential Geometry: Cartan's Generalization of Klein's Erlangen Program. It is probably also somewhere in JM Lee's Riemannian Manifolds –  Willie Wong Apr 9 '12 at 15:58
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