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:

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

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

  • 3
    $\begingroup$ 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 $\endgroup$ – Bugs Bunny May 4 '10 at 11:37
  • $\begingroup$ 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. $\endgroup$ – Michele Torielli May 4 '10 at 13:08
  • $\begingroup$ Why can't you take $\mathfrak{g}$ to be abelian and $r\geq 1$? $\endgroup$ – José Figueroa-O'Farrill May 4 '10 at 13:41
  • 2
    $\begingroup$ @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. $\endgroup$ – José Figueroa-O'Farrill May 4 '10 at 13:50
  • 1
    $\begingroup$ @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. $\endgroup$ – José Figueroa-O'Farrill May 4 '10 at 21:21

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