# Is the topology on compact connected Lie groups metrizable?

The question is in the title, really.

Suppose $$G$$ is a compact connected Lie group. Is there a metric on $$G$$ which induces the underlying topology? (so in particular $$G$$ is compact and connected wrt this metric).

I am not really from this area (my differential geometry is slack, to say the least). I am purely interested in the existence of such a metric.

Thanks.

From a general topology viewpoint: a Lie group $$G$$ is locally Euclidean, so locally metrisable. Also $$G$$ is compact and hence paracompact. It's $$T_1$$ (from locally Euclidean) and as a topological group this means $$G$$ is Tychonoff, Hausdorff etc.

So several metrisation theorems do apply: Bing-Nagata-Smirnov, Smirnov, Urysohn even. I don't think connectedness is necessary : $$G$$ can have only finitely many components (being compact and locally connected) and a finite sum of metrisable spaces still is metrisable. So $$G$$ is certainly metrisable.

• Thanks. I think connectedness is not required for the existence of a metric, but is required to say that this metric is left invariant (i.e. $d(gx,gy)=d(x,y)$. See for example en.wikipedia.org/wiki/Locally_compact_group and mathoverflow.net/questions/262298/…. Feel free to chime in with words of (dis)agreement :) Commented Mar 22, 2020 at 17:37
• @Artur the Wikipedia page states that any locally compact topological group that is metrisable has a left invariant metric. No connectedness needed there. Commented Mar 22, 2020 at 17:48
• It says, and I quote, "Every locally compact group which is second-countable is metrizable as a topological group (i.e. can be given a left-invariant metric compatible with the topology) ". Now this confuses me further - is left invariance part of the definition of being metrizable as a topological group? Or perhaps this follows from something which I am failing to see? Commented Mar 22, 2020 at 17:53
• @Artur your $G$ must be second countable, connected or not. Commented Mar 22, 2020 at 17:54
• @Artur it’s Birkhoff’s theorem. Any good topological group text book should cover it. Commented Mar 22, 2020 at 18:02

The short answer is yes, I believe. The metric that you would be looking for would be the one induced by the Killing form. There is a really good discussion on the construction of the Killing form and its properties here:

http://scipp.ucsc.edu/~haber/ph251/KillingForm.pdf

• correct me if I am wrong, but is this metric $G$-invariant? i.e., $d(gx,gy)=d(x,y)$ for $x,y,g\in G$? Commented Mar 22, 2020 at 15:10
• Since you are using a compact Lie group, yes it is. In general, the Killing form forms the unique G-invariant Riemannian metric on symmetric spaces G/K for K a maximal compact subgroup. Commented Mar 22, 2020 at 19:33
• No, if $G$ is not semisimple, then the Killing form has a nontrivial kernel (if $G$ is abelian, the Killing form is zero). Still $G$ admits a bi-invariant Riemannian metric. If $G$ is semisimple then the Killing form indeed yields a bi-invariant metric, and the other ones are obtained by rescaling on each simple factor.
– YCor
Commented Mar 24, 2020 at 23:14