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Sur les éléments réguliers dans les algèbres de Lie réductives. (Regular elements of reductive Lie algebras). (French) Zbl 0681.17008
Let \({\mathfrak g}\) be a finite dimensional Lie algebra over a field k of characteristic 0 and \(f\in {\mathfrak g}^*\). Denote by \(B_ f\) the alternating bilinear form (x,y) \(\mapsto f([x,y])\) on \({\mathfrak g}\times {\mathfrak g}\) and by \({\mathfrak g}^ f\) the kernel of \(B_ f\). The index of \({\mathfrak g}\) is the integer \(m_{{\mathfrak g}}=Inf\{\dim {\mathfrak g}\); \(f\in {\mathfrak g}^*\}\) and f is called a regular linear form on \({\mathfrak g}\) if dim \({\mathfrak g}^ f=m_{{\mathfrak g}}\). The set of regular forms on \({\mathfrak g}\) is denoted by \({\mathfrak g}^*_ r\). It is known that for \(f\in {\mathfrak g}^*\), \({\mathfrak g}^ f\) is a commutative Lie algebra but there exists an example of a nilpotent Lie algebra which shows that the converse is inexact.
In this paper, the author shows that if \({\mathfrak g}\) is reductive, \(f\in {\mathfrak g}^*_ r\) if and only if [\({\mathfrak g}^ f,{\mathfrak g}^ f]=0\). The theorem follows from the fact that for a simple Lie algebra \({\mathfrak g}\), a distinguished element x of \({\mathfrak g}\) is regular if and only if the centralizer of x in \({\mathfrak g}\) is commutative. This fact is shown for each type of simple Lie algebra separately.
Reviewer: E.Abe

17B20 Simple, semisimple, reductive (super)algebras
20G15 Linear algebraic groups over arbitrary fields