Proof of conjugacy of Cartan subalgebras

Question

I was looking at the proof of the conjugacy of Cartan subalgebras from Carter's Lie Algebra's of Finite and Affine Type. An important part of the proof is to show that every Cartan subalgebra $H$ has a regular element. To show this, the proof constructs a polynomial function $f$ such that $$f(x) = T_x(x)$$ where $T_x$ is in the group of inner automorphisms, $G$, and then shows that the range of $f$ is large. Is there a more explicit way to prove the presence of a regular element in $H$? Why not consider the set of all $G$ translates of $H$? I am also interested in knowing if there is a simpler proof if we allow conjugacy by automorphisms which are not inner.

Answer 1

For the proof of the conjugacy of Cartan subalgebras in $L$ over an algebraically closed field one needs to show that the regular elements form a non-empty Zariski open set in $L$, invariant under all automorphisms of $L$; and, that for a regular element $h_0$ the Lie algebra $H=L_0(h_0)$ is nilpotent, and equal to its own normaliser. Then the Theorem follows easily. Proving "Zariski-open" has to use that the condition $\dim L_0(h_0)>rank (L)$ can be expressed by the vanishing of polynomial functions as above, given the definition of regular.

If you want a more explicit proof of the conjugacy, then follow the proof given in the lecture notes Lie algebras - Harvard Mathematics Department by Sternberg, Theorem $10$, page $76$ till page $81$.