A compact space may not be 2nd countable, e.g. an uncountable space with finite-complement topology. If a compact hausdorff space is 2nd countable, then it is a Polish space, namely, a seperable complete metric space. All Polish spaces are Borel isomorphic. In particular, they are isomorphic to the unit intevel. Thus every compact space admits a probability measure (actually every measurable space admits a probability measure, simply take the Dirac measure).

Every compact group admits a suitabily regular bi-invariant probability measure, namely a **Haar measure.**~~ (a wrong proof: take a probability measure obtained above, translate it from both sides and then take the average. this is wrong and the arguement above is irrelevant.~~ To see this it’s enough to show that there exists a left-invariant measure (which one proves quite nontrivially using Riesz representation theorem). Since compactness allows us to show that the modular function is as .

A group is called unimodular if its modular function is trivial. A group is unimodular if and only if its left Haar measure and right Haar measure conincide. Examples of unimodular groups are abelian groups, compact groups, discrete groups, semisimple Lie groups and connected nilpotent Lie groups. An example of a non-unimodular group is the *ax* + *b* group of transformations of the form. this example shows that a solvable Lie group need not be unimodular.

Every compact Lie group (which is very well understood) admits a **bi-invariant metric **making it into a space of nonnegative sectional curvature (which are also well understood). In general noncompact Lie groups do not have a bi-invariant metric, though all connected semisimple (or reductive) Lie groups do. The existence of a bi-invariant metric implies that the Lie algebra is the Lie algebra of a compact Lie group;

Thanks to Haar measure, every (continuous) representation on a compact group is **unitarizable**. in general, for non-compact groups, it is a more serious question which representations are unitarizable.

However, bi-**invariant vector field** is a naive question with nagetive answer.

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