On the dimension of the sheets of a reductive Lie algebra
Résumé
Let $g$ be a complex finite dimensional Lie algebra and $G$ its adjoint group. For $f \in g^{*}$, we denote by $B_f$ the skew-symmetric bilinear form on $g x g$ defined by : $B_f(x,y)=f([x,y])$, for all $x,y \in g$. For $m \in \mathbb{N}$, the subset $$g_{m}^{*}: =\{f \in g^* | rank B_f = 2m\}$$ is a $G$-invariant locally closed subset of $g^{*}$. The study of the sets $g_{m}^{*}$ should give information about the coadjoint orbits of $g$, because the kernel of $B_f$ is nothing but the stabilizer of $f$ for the coadjoint action. Following a suggestion of A. A. Kirillov, we investigate the dimension of the sets $g_{m}^{*}$. We quickly realize that this problem is closely related to the problem of computing {\it the index of $g$}. Because computing the index for any Lie algebra is a very hard problem in general, it would be too ambitious to hope to get these dimensions for any Lie algebra. In this paper we focus on the reductive case. If $g$ is reductive, then $g$ is identified with its dual $g^{*}$, and the problem reduces to compute the dimension of the set $g^{(m)}=\{x \in g | \dim (G \cdot x) =2m \}$, for $m \in \mathbb{N}$, where $G \cdot x$ is the $G$-orbit of $x$, for $x \in g$. It is known that $g^{(m)}$ is nonempty if and only if $2m$ is the dimension of a nilpotent orbit of $g$, and that its irreducible components, called {\it sheets of $g$}, are parameterized by the pairs $(\mathfrak{l},\mathcal{O}_{\mathfrak{l}})$, up to $G$-conjugation class, consisting of a Levi subalgebra $\mathfrak{l}$ of $g$ and a {\it rigid} nilpotent orbit $\mathcal{O}_{\mathfrak{l}}$ in $\mathfrak{l}$. That's why our approach consists in computing the dimension of the sheets of $g$, whence we deduce the dimension of the subsets $g^{(m)}$.
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