# Define a metric based on a topology

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 $(X, t)$?

• 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. – PyRulez Aug 24 '13 at 12:51
• 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.) – PyRulez 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

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

• 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 7 '14 at 16:26
• @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 7 '14 at 16:50

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'.

• Sorry, I meant some metric (I corrected the question.) – PyRulez Aug 24 '13 at 12:42