Regular nilpotent element in complex simple Lie algebra In Hitchin's paper Lie group and Teichmuller space,
http://www.sciencedirect.com/science/article/pii/004093839290044I he mentioned in section 4 that a regular nilpotent element $e$ is an element that is nilpotent with respect to adjoint action and its dimension of centralizer is equal to the rank $l$ of the complex simple lie algebra $\mathfrak{g}^{c}$.

But I also read from the construction of Cartan subalgebra, a regular element $e$ whose dimension of centralizer is minimized, i.e. equal to the rank of lie algebra $l$ must lie in the Cartan subalgebra $\mathfrak{h}=\{v\in\mathfrak{g}^{c} | (ad_e)^kv=0, k\in \mathbb{N}\}$ it spans. In particular, it is semisimple.  My confusion is how can such an element be both semisimple and nilpotent with respect to adjoint action? Could someone remind me what I mess up here?
 A: There are two notions, unfortunately both called regular, which are completely different:
First there is the notion of a regular element in a Lie algebra $\mathfrak{g}$ (see for instance Serre's book "Complex semisimple Lie algebras, chaper III 2.). This is an element $x$ such that the characteristic polynomial of $ad(x)$ has 0 as eigenvalue with multiplicity the rank of $\mathfrak{g}$.
Second, there is the notion of a regular nilpotent element (also called principal nilpotent), which is a nilpotent element whose centralizer is of dimension the rank of $\mathfrak{g}$.
Since the characteristic polynomial of a nilpotent element is just a monomial, it is never regular (in the first sens).
In a semisimple Lie algebra, all regular elements are semisimple (see Serre's book; chapter III 5.).
A: Regular elements need not be semisimple! For example, in the Lie algebra $\frak{sl}_2$, every non-zero element is regular, with the centralizer spanned by the element itself. Among the elements of the standard basis, $e$ and $f$ are nilpotent, whereas $h$ is semisimple. Only $h$ spans a Cartan subalgebra; the subalgebras spanned by $e$ and $f$ are normalized by $h$, so they are not self-normalizing.
Your statement that a regular element belongs to a Cartan subalgebra is false, due to a missing crucial hypothesis: the element needs to be semisimple. Indeed, in a semisimple Lie algebra, a Cartan subalgebra is a maximal abelian subalgebra consisting of semisimple elements.
