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?