# Lie algebra, supersolvable

*triangular Lie algebra*

A finite-dimensional Lie algebra $\mathfrak g$ over a field $k$ for which the eigen values of the operators $\ad X$ of the adjoint representation belong to $k$ for all $X\in\mathfrak g$ (cf. Adjoint representation of a Lie group).

A supersolvable Lie algebra is solvable. The class of supersolvable Lie algebras contains the class of nilpotent Lie algebras and is contained in the class of exponential Lie algebras (cf. Lie algebra, nilpotent; Lie algebra, exponential). It is closed with respect to transition to subalgebras, quotient algebras and finite direct sums, but it is not closed with respect to extensions.

A supersolvable Lie algebra over a perfect field has many of the properties of solvable Lie algebras (cf. Lie algebra, solvable) over an algebraically closed field (Lie's theorem, the presence of a chain of ideals $\mathfrak g=\mathfrak g_0\supset\mathfrak g_1\supset\dots\supset\mathfrak g_n=\{0\}$ for which $\dim\mathfrak g_i=\dim\mathfrak g-i$, and others). In an arbitrary finite-dimensional Lie algebra $\mathfrak g$ there are maximal supersolvable subalgebras and they contain the nil radical. If $k=\mathbf R$ or $\mathbf C$, or if $k$ is perfect and $\mathfrak g$ is an algebraic linear Lie algebra, then all supersolvable subalgebras are conjugate. The Lie algebra $\mathfrak g$ over $k$ corresponding to a $k$-split algebraic group (cf. Split group) over a perfect field $k$ is a supersolvable Lie algebra.

Any supersolvable Lie algebra over a field of characteristic 0 can be isomorphically imbedded in the Lie algebra of upper-triangular matrices with coefficients from $k$ (which is itself supersolvable). The simplest example of a supersolvable Lie algebra that is not nilpotent is a two-dimensional Lie algebra with basis $X,Y$ and defining relation $[X,Y]=X$. For references see Lie group, supersolvable.

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Lie algebra, supersolvable.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Lie_algebra,_supersolvable&oldid=43553