# Existence of toral subalgebra

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:

• Good question. In general we don't have $x_s+y_s$ semisimple: e.g. $\pmatrix{1&0\\0&0}$ and $\pmatrix{0&0\\1&1}$ are both diagonalizable, but their sum is not. – Berci Jun 12 at 23:00
• @Berci right, thanks.. Probably just a mistake in the text.. He doesn't try to prove this, just states this in the introduction to the Root Space Decomposition chapter.. – Mariah Jun 12 at 23:04
• There might be special cases or more advanced reasons where/why it holds.. – Berci Jun 12 at 23:16
• As stated, this is wrong; for every semisimple complex Lie algebra $L$, that span is actually all of $L$. Are you sure you quote Humphreys correctly? What is the exact phrasing and context in the book? – Torsten Schoeneberg Jun 13 at 4:40
• @TorstenSchoeneberg added a screen shot – Mariah Jun 13 at 7:33

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.)