antennense

Answered

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

${K}_{0}(X,Y)=K(X,Y)\text{}\text{}\mathrm{\forall}X,Y\in \mathfrak{g}$

My problem:

Understanding the previous equation as $K(X,Y)\equiv K(X+i0,Y+i0)$ I would get twice the result stated in the book. This is due to the following facts.

1. ${\text{ad}}_{\mathbb{C}}(X)\in {\text{End}}_{\mathbb{C}}({\mathfrak{g}}_{\mathbb{C}})$, acting as

${\text{ad}}_{\mathbb{C}}(X)[A+iB]=[X+i0,A+iB{]}_{\mathbb{C}}=[X,A]+i[X,B]$

therefore I can write it as a direct sum of the real adjoint map

${\text{ad}}_{\mathbb{C}}(X)=\text{ad}(X)\oplus \text{ad}(X)$

2. $\text{Trace}[{\text{ad}}_{\mathbb{C}}(X)\circ {\text{ad}}_{\mathbb{C}}(Y)]=\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)\oplus \text{ad}(X)\circ \text{ad}(Y)]=2\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)]$

Background:

1. Helgason's definition of complexification: ${\mathfrak{g}}_{\mathbb{C}}=\mathfrak{g}\times \mathfrak{g}\simeq \mathfrak{g}\oplus \mathfrak{g}$ with the complex structure

$J:(X,Y)\equiv X+iY\mapsto (-Y,X)\equiv -Y+iX$

extending the Lie bracket by $\mathbb{C}$-linearity:

$[X+iY,Z+iT{]}_{\mathbb{C}}=[X,Z]-[Y,T]+i[X,T]+i[Y,Z]$

2. Killing form of any Lie algebra over an arbitrary field $\mathbb{K}$:

$B(X,Y)=\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)]$

where $\text{ad}(X)$ is the $\mathbb{K}$-linear map $[X,\cdot ]$.

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

${K}_{0}(X,Y)=K(X,Y)\text{}\text{}\mathrm{\forall}X,Y\in \mathfrak{g}$

My problem:

Understanding the previous equation as $K(X,Y)\equiv K(X+i0,Y+i0)$ I would get twice the result stated in the book. This is due to the following facts.

1. ${\text{ad}}_{\mathbb{C}}(X)\in {\text{End}}_{\mathbb{C}}({\mathfrak{g}}_{\mathbb{C}})$, acting as

${\text{ad}}_{\mathbb{C}}(X)[A+iB]=[X+i0,A+iB{]}_{\mathbb{C}}=[X,A]+i[X,B]$

therefore I can write it as a direct sum of the real adjoint map

${\text{ad}}_{\mathbb{C}}(X)=\text{ad}(X)\oplus \text{ad}(X)$

2. $\text{Trace}[{\text{ad}}_{\mathbb{C}}(X)\circ {\text{ad}}_{\mathbb{C}}(Y)]=\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)\oplus \text{ad}(X)\circ \text{ad}(Y)]=2\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)]$

Background:

1. Helgason's definition of complexification: ${\mathfrak{g}}_{\mathbb{C}}=\mathfrak{g}\times \mathfrak{g}\simeq \mathfrak{g}\oplus \mathfrak{g}$ with the complex structure

$J:(X,Y)\equiv X+iY\mapsto (-Y,X)\equiv -Y+iX$

extending the Lie bracket by $\mathbb{C}$-linearity:

$[X+iY,Z+iT{]}_{\mathbb{C}}=[X,Z]-[Y,T]+i[X,T]+i[Y,Z]$

2. Killing form of any Lie algebra over an arbitrary field $\mathbb{K}$:

$B(X,Y)=\text{Trace}[\text{ad}(X)\circ \text{ad}(Y)]$

where $\text{ad}(X)$ is the $\mathbb{K}$-linear map $[X,\cdot ]$.

Answer & Explanation

enfeinadag0

Expert

2022-07-02Added 16 answers

$K(a,b)=Tr(a{d}_{a}\circ a{d}_{b})$ 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.

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