Existence of toral subalgebra

by Mariah   Last Updated June 13, 2019 12:20 PM

If $L$ is a semi-simple Lie algebra every element $x \in L$ has a Jordan decomposition into $x = x_s + x_n$ where $[x_s,x_n] = 0$ and $ad(x_s)$ is semi-simple while $ad(x_n)$ is nilpotent.

Thus if $L$ isn't nilpotent there is an $x \in L$ such that $ad(x_s) \neq 0$ and so the sub-algebra span$\{x_s\}$ is a toral sub-algebra.

In Humphrey's book he says that span$\{x_s : x \in L, x_s \neq 0\}$ is also a toral sub-algebra.

However, if $x, y \in L$, does it necessarily follow that $x_s + y_s$ is an ad-semi-simple element? I can't find a reason, for instance, that $x_s + y_s = (x+y)_s$.

Why is span$\{x_s : x \in L, x_s \neq 0\}$ a toral sub-algebra?

Screen shot from Humphrey's book: enter image description here

Answers 1

As I read it, the text in the book just claims that the 1-dimensional subalgebra spanned by one nonzero $x_s$ is a non-zero subalgebra consisting of semisimple elements. (I agree the formulation is not completely clear.)

Andreas Cap
Andreas Cap
June 13, 2019 12:07 PM

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