How to prove that two compact 1-dim Riemannian manifolds with same length must be isometric?

Question

How to prove that two compact 1-dim Riemannian manifolds with same length must be isometric ?

I know the compact 1-dim manifold must be homeomorphism to $S^1$ , but how to do a specific isometric ?

Answer 1

Here's a sketch: Fix a point $m$ in $(M, g)$ and a point $n$ in $(N, h)$, and unit tangent vectors $u$ at $m$ and $v$ at $n$. Let $\alpha$ denote the geodesic in $M$ satisfying $\alpha(0) = m$, $\alpha'(0) = u$, and let $\beta$ denote the geodesic in $N$ satisfying $\beta(0) = n$, $\beta'(0) = v$.

Check that:

• The geodesic $\alpha$ is a local isometry (as well as a covering map) from the Euclidean line to $M$; similarly for $\beta$ and $N$.

• Because $M$ and $N$ have the same length, say $\ell$, we have $\alpha(t + \ell) = \alpha(t)$ and $\beta(t + \ell) = \beta(t)$ for all real $t$.

• Show there exists a unique mapping $f:M \to N$ satisfying $f \circ \alpha = \beta$. Formally, $f(p) = \beta \circ \alpha^{-1}(p)$. Check that $f$ is bijective and a local isometry, hence diffeomorphism, hence an isometry.