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?

Question

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?

Answer 1

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

Answer 2

There is already a good answer by Tsemo Aristide. Here we want to focus on OP's title question (v2).

1. Given a finite-dimensional real Lie algebra $\mathfrak{g}$, then Lie's third theorem (more precisely Lie-Cartan's theorem) guarantees the existence of a corresponding connected & simply-connected Lie group $G$, such that its Lie algebra is $\mathfrak{g}$. In a neighborhood of the identity, the Lie group is reconstructed by the exponential map $$ \exp(\mathfrak{g})~\subseteq~ G .$$

2. Under the additional assumption that $\mathfrak{g}$ is semisimple, then the Killing form is negative definite iff $G$ is compact, cf. Theorem 2.28 in Ref. 1. This answers OP's title question (v2).

References:

1. M.M. Alexandrino & R.G. Bettiol, Introduction to Lie groups, isometric and adjoint actions and some generalizations, arXiv:0901.2374; Theorem 2.28.

Answer 3

For a less formal but more conceptual reason, think about it like this:

There is a standard adjoint action of a Lie group $G$ on its Lie algebra $\mathfrak g$, and the Killing form on $\mathfrak g$ is invariant under that action. So if the Killing form is negative definite, its negative is positive definite i.e. a scalar product on $\mathfrak g$. So the adjoint action maps

$$Ad: G \rightarrow SO(\mathfrak g),$$

$SO$ being defined w.r.t. the (negative of) the Killing form. Since the real Lie groups $SO(V)$, and a fortiori their closed subgroups, are compact, this means that at least $G/ker(Ad) = G/Z(G)$ is compact.