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

  • $\begingroup$ 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. $\endgroup$ – Berci Jun 12 at 23:00
  • $\begingroup$ @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.. $\endgroup$ – Mariah Jun 12 at 23:04
  • $\begingroup$ There might be special cases or more advanced reasons where/why it holds.. $\endgroup$ – Berci Jun 12 at 23:16
  • $\begingroup$ 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? $\endgroup$ – Torsten Schoeneberg Jun 13 at 4:40
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    $\begingroup$ @TorstenSchoeneberg added a screen shot $\endgroup$ – 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.)

  • $\begingroup$ Ah, yes probably then.. $\endgroup$ – Mariah Jun 13 at 12:13

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