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Math Help - Riemannian geometry

  1. #1
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    Riemannian geometry

    Hello.
    Let M,N be a connected smooth riemannian manifolds.
    I define the metric as usuall, the infimum of lengths of curves between the two points.
    (the length is defined by the integral of the norm of the velocity vector of the curve).

    Suppose phi is a homeomorphism which is a metric isometry.
    I wish to prove phi is a diffeomorphism.

    Please, anyone who can help.
    Thanks in advance,

    Roey
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  2. #2
    Super Member Rebesques's Avatar
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    Pick a point q\in N. There is an \epsilon>0 such that \exp_q is a diffeomorphism of the ball B(0,\epsilon)=B_{\epsilon}^N\subset T_qN (the tangent space) into N. Since \phi is a homeomorphism, \phi^{-1}(\exp_q(B_{\epsilon}^N)) is an open set in N and \exists \ p\in\phi^{-1}({q})\bigcap\phi^{-1}(\exp_q(B_{\epsilon}^N)).

    Now, since \phi is an isometry, for all geodesics -\delta,\delta)\rightarrow N" alt="\gamma-\delta,\delta)\rightarrow N" /> with \gamma(0)=\phi(p), for the geodesic -\eta,\eta)\rightarrow M" alt="\beta-\eta,\eta)\rightarrow M" /> with \beta(0)=p and {\rm d}\phi_p(\beta'(0))=\gamma'(0), we have that \gamma=\phi\circ\beta.

    This means ({\rm d}\phi_p)^{-1} exists (for all of the tangent space, as N is complete); and since M is complete, we can consider the map \psi:\exp_q(B_{\epsilon}^N)\rightarrow\exp_p(B_{ \epsilon}^M) defined by \psi=\exp_p\circ({\rm d}\phi_p)^{-1}\circ(\exp_q(B_{\epsilon}^N))^{-1}. This is a diffeomorphism, and inverse to the (restricted map) \phi:\exp_p(B_{\epsilon}^M)\rightarrow\exp_q(B_{ \epsilon}^N). So \phi is a diffeomorphism.


    Note. I do not see why this would not work if \phi was only a local isometry.

    Note 2. Thin Lizzy - Whiskey in the jar
    Last edited by Rebesques; August 26th 2006 at 06:53 PM.
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  3. #3
    Super Member Rebesques's Avatar
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    Sorry, stupid me - use the argument about ({\rm d}\phi)^{-1} to get just \phi to be a diffeo. No need to make our lives harder by computing inverses. Sorry again, I was... checking my signature out and was carried away
    Last edited by Rebesques; August 26th 2006 at 10:35 PM.
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  4. #4
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    M,N are not complete
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  5. #5
    Super Member Rebesques's Avatar
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    What!!

    I can see how the hypotheses on M can be relaxed, by following the same argument. But no completeness for N, well... There is no way I can see it happen, and my guess is there is no simple way to do this
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