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It's well known that if I have a Riemannian manifold $M$ and a Lie group of isometries $G$ that acts freely and properly on $M$, then the quotient $M/G$ is a manifold and inherits the Riemannian metric from $M$.

Given this, say I know the Riemannian distance metric on $M$, say $d(x,y)$. I want to compute the distance metric on $M/G$. My ansatz is:

$d_{M/G}([x],[y]) = \inf \{ d(x,y) ~|~ x \in [x], y \in [y]\}$

i.e., the smallest distance between any two points in the respective orbits $[x],[y]$.

For example, if we take $S^{2n-1} \subset \mathbb{C}^n$ to be the standard unit sphere in $\mathbb{C}^n$, the distance metric is given by:

$$d(x,y) = \cos^{-1}(\Re\{\langle x,y\rangle \})$$

If I form the quotient $S^{2n-1}/U(1)$ (where the group acts by phase shifts), my ansatz formula yields:

$$d([x],[y]) = \cos^{-1}(|\langle x,y\rangle|)$$

which is the distance metric induced by the Fubini-Study metric on $\mathbb{CP}^n$.

This seems intuitive, but I haven't been able to find an explicit reference. I tried to prove it, but ran into some technical issues related to lifting paths on $M/G$ to paths on $M$.

Is there something that prevents this working generally, or am I just not looking in the right place for a reference?

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  • $\begingroup$ I am not sure about the Riemannian case, but in general your formula does not satisfy the triangle inequality, hence not a metric. You need to take chains. Google quotient metric space. $\endgroup$
    – Asaf Shachar
    Commented Jul 10, 2016 at 3:18
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    $\begingroup$ The infimum formula works. The quotient map $M\to M/G$ is a Riemannian submersion, Any minimal geodesic in the quotient lifts to a minimal geodesic orthogonal to the two fibers (orbits) corresponding to the endpoints of the geodesic. $\endgroup$
    – Igor Belegradek
    Commented Jul 10, 2016 at 4:23
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    $\begingroup$ The first "well-known" statement is false: the Lie group of reals acts freely (and by isometries) on the 2-torus (translating along an irrational line), with dense orbits, so the quotient is not a manifold. $\endgroup$
    – YCor
    Commented Jul 10, 2016 at 11:16
  • $\begingroup$ @YCor I forgot to include "properly", which rules out that counter example. $\endgroup$
    – Jarred Barber
    Commented Jul 10, 2016 at 14:03
  • $\begingroup$ @IgorBelegradek Thanks, I think the piece that I was missing is that in this case the quotient becomes a principal G-bundle $\endgroup$
    – Jarred Barber
    Commented Jul 10, 2016 at 14:43

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