
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

Pick a point $\displaystyle q\in N$. There is an $\displaystyle \epsilon>0$ such that $\displaystyle \exp_q$ is a diffeomorphism of the ball $\displaystyle B(0,\epsilon)=B_{\epsilon}^N\subset T_qN$ (the tangent space) into $\displaystyle N$. Since $\displaystyle \phi$ is a homeomorphism, $\displaystyle \phi^{1}(\exp_q(B_{\epsilon}^N))$ is an open set in $\displaystyle N$ and $\displaystyle \exists \ p\in\phi^{1}({q})\bigcap\phi^{1}(\exp_q(B_{\epsilon}^N))$.
Now, since $\displaystyle \phi$ is an isometry, for all geodesics $\displaystyle \gamma:(\delta,\delta)\rightarrow N$ with $\displaystyle \gamma(0)=\phi(p)$, for the geodesic $\displaystyle \beta:(\eta,\eta)\rightarrow M$ with $\displaystyle \beta(0)=p$ and $\displaystyle {\rm d}\phi_p(\beta'(0))=\gamma'(0)$, we have that $\displaystyle \gamma=\phi\circ\beta.$
This means $\displaystyle ({\rm d}\phi_p)^{1}$ exists (for all of the tangent space, as $\displaystyle N$ is complete); and since $\displaystyle M$ is complete, we can consider the map $\displaystyle \psi:\exp_q(B_{\epsilon}^N)\rightarrow\exp_p(B_{ \epsilon}^M)$ defined by $\displaystyle \psi=\exp_p\circ({\rm d}\phi_p)^{1}\circ(\exp_q(B_{\epsilon}^N))^{1}.$ This is a diffeomorphism, and inverse to the (restricted map) $\displaystyle \phi:\exp_p(B_{\epsilon}^M)\rightarrow\exp_q(B_{ \epsilon}^N)$. So $\displaystyle \phi$ is a diffeomorphism.
Note. I do not see why this would not work if $\displaystyle \phi$ was only a local isometry.
Note 2. Thin Lizzy  Whiskey in the jar

Sorry, stupid me  use the argument about $\displaystyle ({\rm d}\phi)^{1}$ to get just $\displaystyle \phi$ 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 :D


What!! :eek:
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 :( :o :confused: