# Does a compact Lie algebra come from a Compact Lie Group?

I came across this definition in the book by Knapp:

Definition: A Lie algebra $$\mathfrak{g}$$ is a compact Lie algebra if the subgroup $$\text{Int } \mathfrak{g}$$ of $$\text{Aut } \mathfrak{g}$$ with Lie algebra $$\text{ad } \mathfrak{g}$$ is compact.

I was wondering if it could imply that $$g$$ comes from a compact matrix Lie group. Or, is isomorphic to a Lie algebra which comes from a compact matrix Lie group. Any hints in this direction would be appreciated!

• Yes, it will be isomorphic to the Lie algebra of a compact (matrix) Lie group ( matrix is superfluous, any compact Lie group is iso to a matrix Lie group). Perhaps this appears later in the book. Hint: the definitions should be meaningful. Jun 11, 2021 at 5:17
• @orangeskid, I could not find it at first glance in the book. Perhaps you could give me another source for the proof? Jun 16, 2021 at 7:06

This is standard Lie groups and algebras theory, perhaps consult Bourbaki Lie Groups - the book on compact Lie groups.

A sketch of the proofs:

First, note that if $$\frak{g}$$ has center $$0$$, then $$\frak{g} \simeq$$ $$\text{ad}\frak{g}$$, so we are done. So we have to deal with the general case, when $$\frak{g}$$ has a nonzero center.

$$\text{Int} \frak{g}$$ being a compact subgroup of $$\text{gl}(\frak{g})$$ implies there exists a positive quadratic form $$\langle\ ,\rangle$$ on $$\frak{g}$$ invariant under $$\text{Int} \frak{g}$$. For this form we have $$\langle [X,Y], Z\rangle + \langle Y, [X,Z]\rangle = 0$$

From here we conclude that if $$\frak{i} \triangleleft \frak{g}$$ is an ideal of $$\frak{g}$$, then $$\frak{i}^{\perp}$$ is also an ideal and $$i \oplus \frak{i}^{\perp} = \frak{g}$$

Take $$\frak{i} = z$$, the center of $$\frak{g}$$. Then we $$\frak{g}$$ is a direct sum of $$\frak{z}$$ and $$\text{ad}\frak{g}$$. You are more or less done.

The following are equivalent:

• $$\frak{g}$$ is a compact Lie algebra

• there exists a positive quadratic form on $$\frak {g}$$ that is $$\frak{g}$$ invariant ( see the equality above)

• the Killing form on $$\frak{g}$$ is negative semidefinite

• $$\frak{g}$$ is isomorphic to a Lie subalgebra of $$so(n,\mathbb{R})$$ for some $$n$$.

In particular, a Lie subalgebra of a compact Lie algebra is also compact