Consider the change of coordinates in $\mathbb{R}^3$ from cartesian $(x,y,z)$ to cylindrical polars $(r,\phi,u)$ i.e. $x = r \, cos \phi$, $y = r \, sin \phi$ and $z = u $. This transformation is obviously an isometry of $\mathbb{R}^3$.

Since a local chart is a local change of coordinates in a manifold, then it is correct to say that a local diffeomorphism $\psi: M \rightarrow \mathbb{R}^n$ of a Riemannian manifold $M$ is a local isometry?

Thanks in advance.

  • $\begingroup$ why is this an isometry? And what is obvious about that? $\endgroup$ – Thomas Feb 9 '17 at 10:28
  • $\begingroup$ Because it preserves the Euclidean space, or I am wrong? $\endgroup$ – jaogye Feb 9 '17 at 10:37

A diffeomorphism $f:M\to N$ is a differentiable bijection with a differentiable inverse. In other words, $f$ is smooth (or $\mathcal{C}^\infty$) and in addition $f$ is bijective and has a smooth inverse map.

Given a diffeomorphism $\phi:M\to N$ (where $N$ can be $\mathbb{R}^n$ for your question) we say that $\phi$ is an isometry if $$<u, v>_p = <d\phi_p(u),d\phi_p(v)>_{\phi(p)},$$ for all $p\in M$, $u,v\in T_pM$.

The notion of local isometry is based on applying the previous definition to a neighborhood $U\subset M$ of $p$ where $f$ is a local diffeomorphism over $f(U)$.

It is clear by the definitions that not every diffeomorphism (global or local) is an isometry.

  • $\begingroup$ Many thanjs for your answer. $\endgroup$ – jaogye Feb 9 '17 at 13:42
  • $\begingroup$ You're welcome @JuanAlvarado $\endgroup$ – Edu Feb 9 '17 at 13:46

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