2
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ I'm not convinced that this is a research-level question, but certainly Victor has answered it. For a broader survey of regular nilpotent elements, you might want to look at my online notes (written shortly after Kostant's death): people.math.umass.edu/~jeh/pub/regular.pdf $\endgroup$
    – Jim Humphreys
    Oct 16, 2017 at 13:23

2 Answers 2

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks! I read from this blog the construction of Cartan subalgebra amathew.wordpress.com/2010/02/01/cartan-subalgebras. It looks like it does not need the assumption that the element is semisimple but proves it at last. Could you remind me what's the issue here? $\endgroup$
    – Dai
    Oct 16, 2017 at 18:02
  • $\begingroup$ Sorry, at the last, I realized my confusion is due to the subtlety that in the blog, the subalgebra is defined as elements that can be taken to be zero by finite time of adjoint action, Not the subalgebra spanned by centralizer, though they both reach the minimal dimension, rank $l$. So there's no contradiction here. In the first case , things are much stricter and elements we are seeking for at last turn out to be semisimple. $\endgroup$
    – Dai
    Oct 16, 2017 at 20:42
2
$\begingroup$

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.).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks for the clarification. A good way to avoid the confusion would be to write "regular-nilpotent element". $\endgroup$
    – YCor
    Jul 18, 2019 at 14:27
  • 2
    $\begingroup$ I think the usual modern definition of regular elements is those with minimal-dimensional centraliser, regardless of Jordan decomposition; and then one observes that the condition on a semisimple element can be reëxpressed in terms of the characteristic polynomial as you have done. That is, I think that it is the first definition (of 'regular semisimple') that should be taken as a special or exceptional case, not the second. In this way one gets the theorem that the nilpotent part of a regular element is regular in the centraliser of the semisimple part. $\endgroup$
    – LSpice
    Jul 18, 2019 at 15:49
  • 1
    $\begingroup$ See, for example, Steinberg - Regular elements of semisimple algebraic groups (MSN), where this definition is given in the first paragraph. $\endgroup$
    – LSpice
    Jul 18, 2019 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.