Welcome to MathOverflow

Question

1

Let $f\in \mathbb{C}[x_1,\dots, x_n]$ be a reduced homogeneous polynomial of degree n.

Let $\mathfrak{g}=\{ \delta \in Der_{\mathbb{C}^n}|\delta(f)\in (f)\mathcal{O}_{\mathbb{C}^n} \text{ and weight}(\delta) =0 \}$ be a (reductive) complex Lie algebra with minimal system of generators $\langle \sigma_1, \dots, \sigma_s, \delta_1, \dots, \delta_r\rangle$ such that:

$\sigma_1, \dots, \sigma_s$ are simultaneously diagonalizable,
$\delta_1, \dots, \delta_r$ are nilpotent,
$[\sigma_i,\delta_j]\in \mathbb{Q} \cdot \delta_j$ for all i,j.

Is it true that the centre of $\mathfrak{g}$ is made only of diagonalizable elements?

edited May 5 2010 at 18:51

S. Carnahan♦
25.6k●41●90

asked May 4 2010 at 11:22

Michele Torielli
476●3●9

3

What do you mean by made of diagonal elements? Clearly, nothing in your formulation prevents $\delta_1$ to be in the centre, but then it is diagnosable itself – Bugs Bunny May 4 2010 at 11:37

I meant that the centre is a subset of $\langle \sigma_1, \dots, \sigma_s \rangle$. Moreover, in my case $\mathfrak{g}$ is a Lie algebra of weight zero vector fields. – Michele Torielli May 4 2010 at 13:08

Why can't you take $\mathfrak{g}$ to be abelian and $r\geq 1$? – José Figueroa-O'Farrill May 4 2010 at 13:41

2

@Michele: You can have $s\geq 1$ and still have an abelian algebra. This satisfies all 3 of your conditions. I suspect that there is an extra condition that is implicit. – José Figueroa-O'Farrill May 4 2010 at 13:50

1

@Michele: Then why is this information missing from the question? My guess is that you are hoping to prove a result about a particular Lie algebra by applying some result about general Lie algebras. (Your recent questions all suggest this.) It does not seem to me that you are spending enough time testing your conjectures, though. – José Figueroa-O'Farrill May 4 2010 at 21:21

show 6 more comments

Adam Gal · Answer 1 · 2010-05-04 21:06:00Z UTC

-2

If the algebra is reductive it has no center. The meaningful center is the center of the universal enveloping algebra. Even there, the elements are not only combinations of what you call diagonalizable elements - they are btw called a cartan subalgebra of g. There is the Harish-Chandra theorem that says that the center is isomorphic to symmetric functions on the cartan subalgebra.

answered May 4 2010 at 21:06

Adam Gal
203●1●8

1

Every Lie algebra $L$ has a centre. It consists of those $x\in L$ such that $[x,y]=0$ for all $y\in L$. Perhaps the centre is zero, perhaps it is not. A reductive Lie algebra is a direct sum of a semisimple Lie algebra and an abelian Lie algebra. If follows that the centre of a reductive Lie algebra $L$ is zero if and only if $L$ is semisimple. – Robin Chapman May 4 2010 at 21:12

You're right, I was thinking of semi-simple. – Adam Gal May 4 2010 at 21:26

Centre of a Lie algebra

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.