 SubalgebraNormalizer - Maple Help

LieAlgebras[SubalgebraNormalizer] - find the normalizer of a subalgebra

Calling Sequences

SubalgebraNormalizer(h, k)

Parameters

h         - a list of vectors defining a subalgebra $h$ in a Lie algebra $\mathrm{𝔤}$

k         - (optional) a list of vectors defining a subalgebra $k$ of $\mathrm{𝔤}$ containing the subalgebra $h$ Description

 • Let be a Lie algebra and let be subalgebras.The normalizer $n$ of $h$ in $k$ is the largest subalgebra $n$ of $k$ which contains $h$ as an ideal. The normalizer of $h$ always contains $h$ itself.
 • SubalgebraNormalizer(h, k) calculates the normalizer of $h$ in the subalgebra $k$. If the second argument $k$ is not specified, then the default is and the normalizer of in is calculated.
 • A list of vectors defining a basis for the normalizer of $h$ is returned.
 • The command SubalgebraNormalizer is part of the DifferentialGeometry:-LieAlgebras package.  It can be used in the form SubalgebraNormalizer(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-SubalgebraNormalizer(...). Examples

 > $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{LieAlgebras}\right):$

Example 1.

First initialize a Lie algebra and display the Lie bracket multiplication table.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[5\right]\right],\left[\left[\left[2,5,1\right],1\right],\left[\left[3,4,1\right],1\right],\left[\left[3,5,2\right],1\right]\right]\right]\right):$
 Alg1   > $\mathrm{DGsetup}\left(\mathrm{L1}\right):$
 > $\mathrm{MultiplicationTable}\left("LieBracket"\right)$
 $\left[\left[{\mathrm{e2}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e5}}\right]{=}{\mathrm{e2}}\right]$ (2.1)

Calculate the normalizer of span in span$\left\{{e}_{1},{e}_{3},{e}_{4}\right\}$.

 Alg1 > $\mathrm{S1}≔\left[\mathrm{e3}\right]:$$\mathrm{S2}≔\left[\mathrm{e1},\mathrm{e3},\mathrm{e4}\right]:$
 Alg1 > $\mathrm{SubalgebraNormalizer}\left(\mathrm{S1},\mathrm{S2}\right)$
 $\left[{\mathrm{e3}}{,}{\mathrm{e1}}\right]$ (2.2)

Calculate the normalizer of ${S}_{3}=$spanin .

 Alg1 > $\mathrm{S3}≔\left[\mathrm{e2},\mathrm{e4}\right]:$$\mathrm{S4}≔\left[\mathrm{e1},\mathrm{e2},\mathrm{e4},\mathrm{e5}\right]:$
 Alg1 > $\mathrm{SubalgebraNormalizer}\left(\mathrm{S3},\mathrm{S4}\right)$
 $\left[{\mathrm{e4}}{,}{\mathrm{e2}}{,}{\mathrm{e1}}\right]$ (2.3)

Calculate the normalizer of ${S}_{5}=$spanin the Lie algebra Alg1.

 Alg1 > $\mathrm{S5}≔\left[\mathrm{e1},\mathrm{e2}\right]:$
 Alg1 > $\mathrm{SubalgebraNormalizer}\left(\mathrm{S5}\right)$
 $\left[{\mathrm{e5}}{,}{\mathrm{e4}}{,}{\mathrm{e3}}{,}{\mathrm{e2}}{,}{\mathrm{e1}}\right]$ (2.4)