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Here's a statement and a proof given in a Lie Algebra course (in the tutorial):

Let $L$ be a semisimple Lie algebra over a field $F$ with $\text{char} F=0$. Let $x\in L$ be a semisimple element. Prove that $x$ is regular if and only $x$ is an element of exactly one CSA.

Proof (this is just one direction: "not regular $\Rightarrow$ belongs to more than one CSA". The other direction is clear)

Suppose $x$ is not regular. So, $C_L(x)$ is not a maximal toral subalgebra of $L$. Using the adjoint representation, we can assume that $L$ is a matrix algebra with $x$ a diagonal matrix. $C_L(x)$ properly contains a maximal toral subalgebra $H$ with $x \in H$.

Now, $C_L(x)$ is not abelian and therefore we can take some $y \in C_L(x) \setminus H$ with $L_0(\text{ad}y)$ minimal. This $L_0(\text{ady})$ is minimal Engel and therefore a CSA. It is not $H$ because $y \in L_0(\text{ad}y)\setminus H$. So $x$ is an element of two different CSAs.

My confusion:

  1. I don't understand the reasoning in "Now, $C_L(x)$ is not abelian and therefore we can take some $y \in C_L(x) \setminus H$ with $L_0(\text{ad}y)$ minimal". The problem is with the therefore part.
  2. I don't see where viewing $L$ as a matrix algebra helps in the proof given.
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Gils
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    What does $L_0$ represent? And does the adjoint action mean $\operatorname{ad}y(z)$ for all $z\in L$ or $\operatorname{ad}\ell(y)$ for some $\ell\in L$? – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Dec 13 '20 at 21:04
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    @TheSimpliFire $L_0(ady)$ are the elements of $L$ which are eventually kiled by some power of $ad y$, i.e., $L_0(ad y) = {x \in L: (ad y)^k x = 0 \mbox{ for some k}}$. I think that he means $L\cong \mbox{ad (L)}\subseteq \mbox{End}(L)$ – user2345678 Dec 14 '20 at 09:10

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