Quotient Space Metric with Nice Equivalence Classes

Question

Quotient Space Metric: The quotient metric for arbitrary quotient spaces is defined as

If $M$ is a metric space with metric $d$, and $\sim$ is an equivalence relation on $M$, then we can endow the quotient set $M/{\sim}$ with the following (pseudo)metric. Given two equivalence classes $[x]$ and $[y]$, we define $$ d'([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} $$ where the infimum is taken over all finite sequences $(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots, q_n)$ with $[p_1]=[x], [q_n]=[y], [q_i]=[p_{i+1}], i=1,2,\dots, n-1$.

Often we encounter "nice" spaces where the quotient metric can actually be defined as $d([x],[y]) = \inf\{d(p,q)|p\in[x], q\in [y]\}$ (i.e. the shortest path between equivalence classes $[x],[y]$ on the quotient space). This is not true in general, as dicussed here: Why are quotient metric spaces defined this way?

However, it seems that there are many useful quotient spaces where this holds true. Examples:

1. $R^2$ under the identification of vertical lines ($x,y \in \mathbb{R}^2$, $x \sim y$ if $x-y \in \{0\} \times \mathbb{R}$).
2. $R^2$ under the identification of circles ($x,y \in \mathbb{R}^2$, $x \sim y$ if $\|x\|_2^2=\|y\|_2^2$).
3. Special Euclidean Group $SE(2)$ under identification of $SO(2)$ ($x,y \in SE(2)$, $x \sim y$ if $x-y \in \{0,0\} \times SO(2)$).
4. Other spaces like $SE(3) / SO(3)$

That begs the question: What are the conditions under which $d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$ actually holds true?

Answer 1

Here is a sufficient condition that covers all your examples. If the quotient $M/{\sim}$ is the quotient of $M$ by a group $G$ of isometries $M\to M$, then the quotient metric is given by $d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$.

To prove this, it suffices to show that the definition $$d([x],[y]) = \inf\{d(p,q) | p\in[x], q\in [y]\}$$ satisfies the triangle inequality (since that implies the distance along any finite chain is bounded below by this simple distance). So, let $x,y,z\in M$, and suppose $p\in[x],q,q'\in[y],$ and $r\in [z]$. We wish to show that $d([x],[y])$ as defined above is less than or equal to $d(p,q)+d(q',r)$. To prove this, note that there is some $g\in G$ such that $q'=gq$. We then have $d(q',r)=d(gq,r)=d(q,g^{-1}r)$ since $g$ is an isometry. Thus $$d(p,q)+d(q',r)=d(p,q)+d(q,g^{-1}r)\geq d(p,g^{-1}r).$$ Since $g^{-1}r\in [z]$, $d(p,g^{-1}r)\geq d([x],[y])$, as desired.

For citations: This result can be quotient from Theorem 2.1 of this paper.