# 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:

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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
June 13, 2019 12:07 PM

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