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Given a metric space $X$ and an equivalence relation $\sim$, the quotient (pseudo-)metric on $X/\sim$ is defined as follows:

  • $d'([x],[y]) = \inf \left \{ d(p_1,q_1) + d(p_2,q_2) + ... + d(p_n,q_n)\right \}$

where $[p_1] = [x], [q_n] = [y],$ and $[q_i] = [p_{i+1}]$.

Under what circumstances does this reduce to the much simpler equation

  • $d'([x],[y]) = \inf \left \{ d(p,q): [p] = [x], [q] = [y] \right \}$

?

It appears to work for $\mathbb{R}^n/\mathbb{Z}^m$ for $m \leq n$, but I'm curious about the general case.

Mike Battaglia
  • 7,878
  • Always by the triangle inequality? – Gunnar Þór Magnússon Feb 06 '14 at 22:54
  • No, you usually need to consider all chains for it to work out. – Mike Battaglia Feb 06 '14 at 23:09
  • The first metric is less than the second metric @gunnar. The triangle inequality does not work, as $p_{n+1}$ may be distinct from $ q_ n$ – Stefan Hamcke Feb 06 '14 at 23:18
  • For instance, consider the real line with the absolute value as a metric, but where the numbers 0 and 1 are identified and nothing else. By the simpler expression, the distance between -1 and 2 is still |2--1|=3, but by the more complicated one involving chains, you can consider the chain (-1, 0, 1, 2), for which the sum of successive distances is 1+0+1=2. – Mike Battaglia Feb 06 '14 at 23:20

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