# index of a Lie algebra

Let $\mathfrak{q}$ be a Lie algebra over $\mathbb{K}$ and $\mathfrak{q}^{*}$ its vector space dual. For $\xi\in\mathfrak{q}^{*}$ let $\mathfrak{q}_{\xi}$ denote the stabilizer of $\xi$ with respect to the co-adjoint representation.

The index of $\mathfrak{q}$ is defined to be

 $\operatorname{ind\,}\mathfrak{q}:=\min\limits_{\xi\in\mathfrak{g}^{*}}\dim% \mathfrak{q}_{\xi}$

## Examples

1. 1.

If $\mathfrak{q}$ is reductive then $\operatorname{ind\,}\mathfrak{q}=\operatorname{rank\,}\mathfrak{q}$. Indeed, $\mathfrak{q}$ and $\mathfrak{q}^{*}$ are isomorphic as representations for $\mathfrak{q}$ and so the index is the minimal dimension among stabilizers of elements in $\mathfrak{q}$. In particular the minimum is realized in the stabilizer of any regular element of $\mathfrak{q}$. These elemtents have stabilizer dimension equal to the rank of $\mathfrak{q}$.

2. 2.

If $\operatorname{ind\,}\mathfrak{q}=0$ then $\mathfrak{q}$ is called a Frobenius Lie algebra. This is equivalent to condition that the Kirillov form $K_{\xi}\colon\mathfrak{q}\times\mathfrak{q}\to\mathbb{K}$ given by $(X,Y)\mapsto\xi([X,Y])$ is non-singular for some $\xi\in\mathfrak{q}^{*}$. Another equivalent condition when $\mathfrak{q}$ is the Lie algebra of an algebraic group $Q$ is that $\mathfrak{q}$ is Frobenius if and only if $Q$ has an open orbit on $\mathfrak{q}^{*}$.

Title index of a Lie algebra IndexOfALieAlgebra 2013-03-22 15:30:47 2013-03-22 15:30:47 benjaminfjones (879) benjaminfjones (879) 6 benjaminfjones (879) Definition msc 17B05 index of a Lie algebra Frobenius Lie algebra Kirillov form