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Is their a systematic way, based on a topology meterizable $(X, t)$ to define or compute some metric $d$ on $X$ such that the open balls in $(X, d)$ is a metric of $(X, t)$?

Christopher King
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  • i think what you are looking for is the "metrizability" of a topology. – akkkk Aug 24 '13 at 12:44
  • metrizability is based on existence. I want a general definition or algorithm. – Christopher King Aug 24 '13 at 12:51
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    Given how non-constructive pretty much all proofs of 'metrisability theorems' are, I think it is very ambitious to hope for an algorithm. Perhaps you'll be a little better off if you restrict yourself to special classes of metrisable spaces (for instance Riemannian manifolds). – Dan Rust Aug 24 '13 at 12:54
  • Even a metric defined some how based on the topology would be good (like counting the common open sets or something like that.) – Christopher King Aug 24 '13 at 13:02
  • Well the only way to define a metric, given only a metrisable topological space, would be via its topology, so you're not really asking anything easier here. – Dan Rust Aug 24 '13 at 13:05
  • What would be an example of function $d$ that is a metric for $(X, t)$ in terms of $X$ and $t$. – Christopher King Aug 24 '13 at 13:07
  • My point was that the only information we have about this set is that it has a topology defined on it, and that topology is metrisable. Seeing as we can't just use the set to define a metric, we'll have to use the topology defined on it in some way. That is, any answer will have to be given in terms of its open sets and any properties they satisfy given that they admit a metric. – Dan Rust Aug 24 '13 at 13:13
  • Oh, so we can't define $d$ based on $X$ and $t$ in general. Okay. – Christopher King Aug 24 '13 at 13:17
  • In many cases $d$ will not exist, but even if it does, at this time we don't know a general method to construct such a $d$ (and as said, once you've found one metrization, you'll have found infinitely many). – akkkk Aug 24 '13 at 19:46

2 Answers2

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First of all, Daniel Rust is absolutely right that in general there is no 'canonical' metric on the given topological space. Nevertheless, there are many situations in topology when it is still important to find one, namely, in the case of smooth compact manifolds. Typically, one looks for Einstein metrics or metrics of constant curvature (sectional or scalar) or Kähler-Einstein metrics, provided you have a fixed conformal or complex structure, depending on the setting. Some of the most remarkable results in topology and geometry are proofs of existence and uniqueness of such metrics. Examples are: uniformization of Riemann surfaces, Mostow rigidity theorem, Thurston-Perelman geometrization theorems for 3-dimensional manifolds, solution of Yamabe's problem (Trudinger, Aubin, Schoen), Yau's proof of Calabi's conjecture, and, most recently, theorems of Tian, Donaldson, Chen and Sun on Kahler-Einstein metrics on Fano manifolds.

Edit: Here is a couple of references: C.LeBrun "Optimal Metrics, Curvature Functionals, and ..." http://www.math.sunysb.edu/~claude/madrid.pdf

M.Berger "What is the Best Riemannian Metric on a Compact Manifold?" http://link.springer.com/chapter/10.1007%2F978-3-642-18245-7_11#page-1

Moishe Kohan
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  • Chen-Donaldson-Sun's work on Kahler-Einstein metrics is unrelated to their uniqueness. Uniqueness of KE metrics has been known for a long time, in general uniqueness of constant scalar curvature Kaehler metrics is due to (in varying levels of generality) Donaldson, Chen-Tian, Berman-Berndtsson. – user31559 Jul 07 '14 at 16:26
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    @user31559: very true but I was talking about BOTH existence and uniqueness results (which you can see by reading my answer more closely). – Moishe Kohan Jul 07 '14 at 16:50
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There is no canonical metric associated to a metrisable topological space because, if $d\colon X\times X\rightarrow\mathbb{R}$ is a metric, then so is $d_a$ defined by $d_a(x,y)=a.d(x,y)$ for some $a> 0$. If $a\neq 1$ then $d_a\neq d$ for all $X$ with more than one element.

In fact, it gets worse. We can't even define a canonical metric on $X$ 'up to a choice of constant multiple' because if $d$ is a metric, then so is $d^*$ defined by $d^*(x,y)=\min\{1,d(x,y)\}$ and so is $d'$ defined by $d'(x,y)=\dfrac{d(x,y)}{d(x,y)+1}$. Which are both bounded metrics which induce the same topology as $d$, and so if $d$ is not bounded, both $d'$ and $d^*$ are equivalent metrics which are not equal to $d$ or even equal 'up to a choice of constant multiple'.

Dan Rust
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