In the definition of the Compact Lie algebra, it asserts if the Killing form of a Lie algebra is negative definite, then the Lie algebra is the Lie algebra of a compact semisimple Lie group. I am wondering whether this statement only asserts the existence of such a compact semisimple Lie group.
Another question is if we let $\exp: \mathfrak{g} \rightarrow G$ be the exponential map. And Lie algebra $\mathfrak{g}$ has a negative definite Killing form, is $\exp(\mathfrak{g})$ a compact Lie group?
Best Answer
The fact that every finite dimensional Lie algebra is associated to a Lie group is the Lie third theorem. In the 1 reference below, it is quote the Cartan theorem which establishes a correspondence between the Lie algebra and 1-connected Lie groups. This implies if the Lie agebra of $G$ and $H$ are isomorphic, their universal cover $\tilde{G}$ and $\tilde{H}$ are isomorphic. Suppose that $G$ is compact and semi-simple, its fundamental group is finite 2. This implies that $\tilde{G}$ is a finite cover of $G$ and is also compact.
1 https://en.wikipedia.org/wiki/Lie%27s_third_theorem
2 https://mathoverflow.net/questions/95637/connected-compact-semisimple-lie-group-finite-fundamental-group
If $G$ is a compact Lie group and ${\cal G}$ its Lie algebra then $exp:{\cal G}\rightarrow G$ is surjective so its image is compact.
https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Surjectivity_of_the_exponential