 antennense

2022-07-01

I am studying Helgason's Differential Geometry and Symmetric Spaces, trying to understand real forms of Lie Algebras.
My problem is related to Lemma 6.1 (1st edition):
Let ${K}_{0}$ be the Killing form of a Lie algebra $\mathfrak{g}$ over $\mathbb{R}$, and $K$ the Killing form of the complexification Lie algebra ${\mathfrak{g}}_{\mathbb{C}}$. Then

My problem:
Understanding the previous equation as $K\left(X,Y\right)\equiv K\left(X+i0,Y+i0\right)$ I would get twice the result stated in the book. This is due to the following facts.
1. ${\text{ad}}_{\mathbb{C}}\left(X\right)\in {\text{End}}_{\mathbb{C}}\left({\mathfrak{g}}_{\mathbb{C}}\right)$, acting as
${\text{ad}}_{\mathbb{C}}\left(X\right)\left[A+iB\right]=\left[X+i0,A+iB{\right]}_{\mathbb{C}}=\left[X,A\right]+i\left[X,B\right]$
therefore I can write it as a direct sum of the real adjoint map
${\text{ad}}_{\mathbb{C}}\left(X\right)=\text{ad}\left(X\right)\oplus \text{ad}\left(X\right)$
2. $\text{Trace}\left[{\text{ad}}_{\mathbb{C}}\left(X\right)\circ {\text{ad}}_{\mathbb{C}}\left(Y\right)\right]=\text{Trace}\left[\text{ad}\left(X\right)\circ \text{ad}\left(Y\right)\oplus \text{ad}\left(X\right)\circ \text{ad}\left(Y\right)\right]=2\text{Trace}\left[\text{ad}\left(X\right)\circ \text{ad}\left(Y\right)\right]$

Background:
1. Helgason's definition of complexification: ${\mathfrak{g}}_{\mathbb{C}}=\mathfrak{g}×\mathfrak{g}\simeq \mathfrak{g}\oplus \mathfrak{g}$ with the complex structure
$J:\left(X,Y\right)\equiv X+iY↦\left(-Y,X\right)\equiv -Y+iX$
extending the Lie bracket by $\mathbb{C}$-linearity:
$\left[X+iY,Z+iT{\right]}_{\mathbb{C}}=\left[X,Z\right]-\left[Y,T\right]+i\left[X,T\right]+i\left[Y,Z\right]$
2. Killing form of any Lie algebra over an arbitrary field $\mathbb{K}$:
$B\left(X,Y\right)=\text{Trace}\left[\text{ad}\left(X\right)\circ \text{ad}\left(Y\right)\right]$
where $\text{ad}\left(X\right)$ is the $\mathbb{K}$-linear map $\left[X,\cdot \right]$. Expert

$K\left(a,b\right)=Tr\left(a{d}_{a}\circ a{d}_{b}\right)$ a basis of $g$ induces a basis of ${g}_{\mathbb{C}}$ and $a{d}_{a}\circ a{d}_{b}$ have the same matrix in both basis.