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Mar 3, 2014 at 22:45 vote accept user47719
Mar 3, 2014 at 22:08 comment added Robert Bryant Because $G$ is compact, it has a bi-invariant metric and the geodesics of such a metric are the 1-parameter subgroups. Since the exponential map from a point is onto when the manifold is compact, and since the diameter of $G$ is bounded, there is a upper bound $D$ on the length of a minimizing geodesic. Now let $K$ be the set of vectors in $\frak{g}$ of length at most $D/(2\pi)$.
Mar 3, 2014 at 19:42 comment added user47719 Thanks! So what I was missing is your last statement. I know it is true but how can I prove the existence of the compact set $K$?
Mar 3, 2014 at 12:18 history answered Robert Bryant CC BY-SA 3.0