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This is from my lecture notes.

Let $\mathfrak{g}$ be a comple lie algebra. Let $x\in \mathfrak{g}$ and $\Sigma(x)\subseteq \mathbb{C}$ be the set of eigenvalues of $ad(x)\in \mathfrak{gl(g)}$. ($ad(x)$ here means adjoint of x, i.e. $ad(x)y=xy - yx$.) Since $ad(x)x=[x,x]=0$ for all $x$ we see that $0\in \Sigma(x)$ for all $x$. For $\lambda\in\Sigma(x)$ we denote by $\mathfrak{g}_{\lambda}(x)$ the generalized eigenspace of λ, that is,

$$\mathfrak{g}_{\lambda}(x) = \{y\in \mathfrak{g} | (ad(x) - λid_\mathfrak{g})^ny = 0 \text{ for some } n\in \mathbb{N}\}$$

How do you get from these definitions, the following equality $$\mathfrak{g}=\oplus_{\lambda\in\Sigma(x)}\mathfrak{g}_{\lambda}(x)$$ Here the direct sum is the direct sum of lie algebras.

coolpenguin
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1 Answers1

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This is quite literally the decomposition of $V:=\mathfrak g$ via the Jordan normal form of $ad(x) \in \mathrm{End}_\mathbb C (V)$, with the normalised eigenspace to $\lambda$ being the space spanned by those basis vectors giving the Jordan blocks with $\lambda$'s on the diagonal.

To be precise, linear algebra gives us that decomposition as vector spaces, and it's left to you to check that those generalized eigenspaces are stable under the Lie bracket.

(Note that in general it is not true though that $[\mathfrak g_{\lambda_1}(x), \mathfrak g_{\lambda_2}(x) ] = 0$, i.e. the direct sum is NOT a direct sum of Lie algebras in the standard meaning of the word; rather, it is a direct sum of vector spaces, and each summand is a Lie subalgebra (but not necessarily an ideal) of $\mathfrak g$.)

Torsten Schoeneberg
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