Does locally isometric imply locally diffeomorphic for regular surfaces

229 Views Asked by At

Let $\phi: S\rightarrow \tilde{S}$ be a differential map between two regular surfaces. Then $d\phi_p: T_p(S)\rightarrow T_{\phi(p)}(\tilde{S})$. Suppose $(v_1,v_2)=(d\phi_P(v_1),d\phi_p(v_2))$ for any $v_1, v_2\in T_p(S)$. Can we prove that $S$ and $\tilde{S}$ is locally diffeomorphism at $p$


I want to prove that by inverse function theorem. Since $(v_1,v_2)=(d\phi_P(v_1),d\phi_p(v_2))$, we can say $d\phi_{p}$ is an isomohphism. Then by inverse function theorem, it is locally diffeomorphic at $p$. Is it correct.


Another related question is that there is a theorem any two regular surfaces are locally conformal。 Can we prove by the same argument saying any two regular surfaces are locally diffeomorphic?