I am wondering if the following statement is true:
Let $\pi: \widetilde{M} \to M$ be a surjective Riemannian submersion and $\tilde{d}, d$ the Riemannian distances on $\widetilde{M}, M$ respectively. Then,
$$d(p,q) = \inf_{\tilde{p} \in \pi^{-1}(p), \\ \tilde{q} \in \pi^{-1}(q)} \tilde{d}(\tilde{p}, \tilde{q}).$$
I know that it is true in the case when $M = \widetilde{M}/G$ and $G$ a compact Lie group of isometries acting freely and properly on $\tilde{M}$. In this case one can use the possibility of lifting curves from $M$ to $\widetilde{M}$ to prove the statement. However, as explained here: Lifting curves from Riemannian submersions: reconciling claims from two books, not all curves can be lifted for general Riemannian submersions.
Proof ideas
- Can one always lift a minimal geodesic?
- According to proposition 2.109 in Gallot, Hullin and Lafontaine (Riemannian geometry, 2004) geodesics can always be lifted locally. This implies that the two distances agree locally. Can this be extended to a global statement?
Best Answer
I'm going to use $f$ for the projection to avoid a notation clash.
Lee's answer in the linked question can be turned into a counterexample for this question. Specifically, take $\tilde{M}$ to be the interval $(0,2\pi+\epsilon)$ for some very small number $\epsilon$ and take $M=S^1$, both with usual metrics. Finally take $f$ to be $f(t)=(\cos( t), \sin( t))$.
Set $p=2\epsilon$ and $q=2\pi-\epsilon$, so both $p,q\in \tilde{M}$. Moreover $p$ is the unique preimage of $f(p)$, and similarly for $q$. So, the distance between the preimages is $2\pi-3\epsilon$. Taking $\epsilon$ small enough, we can make this distance larger than $\pi$, which is the maximum distance between points in $M$