I am trying to show that a diffeomorphism $\phi : S \rightarrow \overline{S}$ is an isometry if and only if the arc length of any parametrized curve in S is equal to the arc length of the image curve by $\phi$.

$"\Rightarrow"$ We know that if the diffeomorphism $\phi$ is an isometry, then $$\left< \alpha '(t) , \alpha'(t) \right>_p = \left <d\phi_p(\alpha'(t)), d\phi_p(\alpha'(t)) \right>_{\phi(p)},$$ and since $\lvert \alpha'(t) \rvert = \left< \alpha'(t), \alpha'(t) \right >_p = \left <d\phi_p(\alpha'(t)), d\phi_p(\alpha'(t)) \right>_{\phi(p)}=\lvert d\phi_p (\alpha'(t)) \rvert = \lvert (\phi \circ \alpha)'(t) \rvert$, we conclude that arc length of $\alpha\left(t\right)$, $s(t) = \int_{0}^{t_0} \mid \alpha'(t) \mid dt = \int_{0}^{t_0} \lvert(\phi \circ \alpha)'\left(t\right)\rvert dt$ = length of image curve by $\phi$.

$\therefore$ If $\phi$ is an isometry, then the arclength under $\phi$ is preserved.

"$\Leftarrow$" NTS: If arclength under the map $\phi$ is preserved, then $\phi$ is an isometry.

Since arc length of $\alpha$ is preserved under the diffeomorphism $\phi$, $$s(t) = \int_{0}^{t_0} \lvert \alpha'(t) \rvert dt = \int_{0}^{t_0} \lvert (\phi \circ \alpha)'(t) \rvert dt.$$ Differentiating both sides, we get: $\frac{ds}{dt} = \lvert \alpha'(t) \rvert = \lvert (\phi \circ \alpha)'(t) \rvert$. Hence $$\left< \alpha'(t) , \alpha'(t) \right>_p = \left< d\phi (\alpha'(t), d\phi(\alpha'(t)) \right>_{\phi(p)}.$$

Since $\phi$ is a diffeomorphic map, and the differential $d\phi$ preserves the inner product, we conclude that $\phi : S \rightarrow \overline{S} $ is an isometry.

$\therefore$ diffeomorphism $\phi$ is an isometry if and only if the arc length of any parametrised curve in $S$ is equal to the arc length of the image curve by $\phi$.

Is this correct? Thanks


It is good. If you want to nitpick, note that you have proven that $\langle d\phi_p(v), d\phi_p(v)\rangle_{\phi(p)} = \langle v,v\rangle_p$ for all $v \in T_pS$, and it follows that $\langle d\phi_p(v), d\phi_p(\color{red}{w})\rangle_{\phi(p)} = \langle v,\color{red}{w}\rangle_p$ for all $v,w \in T_pS$ by polarization (you might mention that word somewhere).

The same argument shows that a diffeomorphism between pseudo-Riemannian manifolds is an isometry if and only if it preserves the energy of all curves, where by energy of a curve $\alpha: I \to M$ in $(M,\langle\cdot,\cdot\rangle)$ I mean $$E[\alpha] = \frac{1}{2} \int_I \langle \alpha'(t),\alpha'(t)\rangle_{\alpha(t)}\,{\rm d}t.$$

  • $\begingroup$ I see what you mean, so I must show the differential of the map preserves the inner product of all pairs of vectors on the tangent plane. Is there a way to show this without using polarization? I have not yet learnt what this is. $\endgroup$
    – T J. Kim
    May 3 '18 at 2:53
  • $\begingroup$ Here en.m.wikipedia.org/wiki/Polarization_identity $\endgroup$
    – Ivo Terek
    May 3 '18 at 2:55
  • 1
    $\begingroup$ Does that mean $\left<d\phi_p(v), d\phi_p(v)\right>_{\phi(p)} = \left<v, v \right>_p $ for all $v \in T_pS$ if and only if $\left<d\phi_p(v), d\phi_p(w)\right>_{\phi(p)} = \left<v, w \right>_p $ for all $v,w \in T_pS$ ,by polarization? $\endgroup$
    – T J. Kim
    May 3 '18 at 3:02

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