# Compact real form of Semisimple Lie algebra

I am reading B.Hall's book "Lie groups, Lie algebras, and Representations". You can find there a definition for semisimple Lie algebra.

DEF1:

A complex Lie algebra $$\mathfrak{g}$$ is reductive if there exists a compact matrix Lie group K such that: $$\mathfrak{g} \cong \mathfrak{t}_{\mathbb{C}}.$$ Where $$\mathfrak{t}$$ is the Lie algebra of K. Lie algebra $$\mathfrak{g}$$ is semisimple if it is reductive and its centre is trivial.

$$\square$$

And the definition for compact real form.

DEF2:

A real subalgebra $$\mathfrak{t}$$ of complex semisimple Lie algebra $$\mathfrak{g}$$ is a compact real form of $$\mathfrak{g}$$ if:

• $$\mathfrak{t}$$ isomorphic to Lie algebra of some compact Lie group.
• Any element $$Z \in \mathfrak{g}$$ can be written uniquely as $$Z=X+iY$$ with $$X,Y \in \mathfrak{t}$$

$$\square$$

The question is, how can I see that any semisimple Lie algebra has such a compact real form? As far as I understand it is not true that for any $$\mathfrak{t}$$ elements of $$\mathfrak{t}_{\mathbb{C}}$$ can be uniquely written as $$X+iY, \ X,Y\in \mathfrak{t}$$. As an example if some $$X$$ belongs to $$\mathfrak{t}$$ together with $$iX$$ then $$X$$ as an element of $$\mathfrak{t}_{\mathbb{C}}$$ can be written as $$X=X$$ or as $$X=i(-iX)$$.

It seems like using the Killing form, one can show that $$X$$ and $$iX$$ can't belong to the algebra of compact group simultaneously. It would contradict the fact that for a compact group, the Killing form is negative definite. However, is there a method without using those?

To address your points at the end it is not true that $$X$$ and $$iX$$ are both in $$\mathfrak{t}$$. This is a real subalgebra so it is not closed under multiplication by $$i$$, indeed $$\mathfrak{g} = \mathfrak{t}^\mathbb{C} = \mathfrak{t} \oplus i\mathfrak{t}$$. Using the Killing form we can confirm this since $$\mathfrak{t}$$ is negative definite so $$(X,X) \leq 0$$ and $$(iX,iX) = -(X,X) \geq 0$$ so $$iX$$ cannot be in $$\mathfrak{t}$$.
To show that any semisimple Lie algebra (from the usual definition) admits a compact real form we can proceed as follows. Take any real form $$\mathfrak{g}_0 \leq \mathfrak{g}$$ and pick a Cartan decomposition: $$\mathfrak{g}_0 = \mathfrak{k}_0 \oplus\mathfrak{p}_0$$ i.e. $$\mathfrak{k}_0$$ is a maximal compact subalgebra and $$\mathfrak{p}_0$$ is its orthocomplement. Then $$\mathfrak{u}_0 = \mathfrak{k}_0 \oplus i\mathfrak{p}_0$$ is a compact real form of $$\mathfrak{g}$$. Obviously you need to see that we can always find at least one real form and that such a thing admits a Cartan decomposition but this isn't too hard.
• @NikolayEbel That is true but we have made a much stronger assumption here. We have already assumed that $\mathfrak{g} = \mathfrak{t}^\mathbb{C} = \mathfrak{t} \oplus i\mathfrak{t}$ for some compact, real $\mathfrak{t}$ in Def 1. By Def 2 this immediately means $\mathfrak{t}$ is a compact real form (although we can make others). Note that by writing $\mathfrak{t}^\mathbb{C}$ we are assuming that $\mathfrak{t}$ is a real form (and thus $i\mathfrak{t}$ is complementary to $\mathfrak{t}$) although we can deduce it must be via the Killing form as well. Jul 20, 2021 at 9:22
• $\mathfrak{t}^\mathbb{C}$ precisely consists of unique linear combinations of the form $X = iY$ for $X,Y \in \mathfrak{t}$. $X$ and $i(-iX)$ are not different combinations since $i(-i) = 1$ Jul 20, 2021 at 9:28