check if a Lie algebra is Abelian - Maple Programming Help

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Query[Abelian] - check if a Lie algebra is Abelian

Calling Sequences

Query(Alg, "Abelian")

Query(S, "Abelian")

Parameters

Alg     - (optional) name or string, the name of an initialized Lie algebra $\mathrm{𝔤}$

S       - a list of vectors defining a basis for a subalgebra

Description

 • A Lie algebra  is Abelian if for all .
 • Query(Alg, "Abelian") returns true if Alg is an Abelian Lie algebra and false otherwise.  If the algebra is unspecified, then Query is applied to the current algebra.
 • Query(S, "Abelian") returns true if the subalgebra S is an Abelian Lie algebra and false otherwise.
 • The command Query is part of the DifferentialGeometry:-LieAlgebras package.  It can be used in the form Query(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-Query(...).

Examples

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

Example 1.

We initialize a pair of 3-dimensional Lie algebras.The first algebra is not Abelian and the second is, as confirmed by a call to Query.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[3\right]\right],\left[\left[\left[1,2,1\right],1\right]\right]\right]\right)$
 ${\mathrm{L1}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{\mathrm{e1}}\right]$ (2.1)
 > $\mathrm{L2}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg2},\left[3\right]\right],\left[\right]\right]\right)$
 ${\mathrm{L2}}{:=}\left[{}\right]$ (2.2)
 > $\mathrm{DGsetup}\left(\mathrm{L1},\left[x\right],\left[a\right]\right):$$\mathrm{DGsetup}\left(\mathrm{L2},\left[y\right],\left[b\right]\right):$
 Alg2 > $\mathrm{Query}\left(\mathrm{Alg1},"Abelian"\right)$
 ${\mathrm{false}}$ (2.3)
 Alg1 > $\mathrm{ChangeLieAlgebraTo}\left(\mathrm{Alg2}\right):$
 Alg2 > $\mathrm{Query}\left("Abelian"\right)$
 ${\mathrm{true}}$ (2.4)

The subalgebra in Alg1 is not Abelian while the subalgebra  is Abelian.

 Alg2 > $\mathrm{S1}≔\left[\mathrm{x1},\mathrm{x2}\right]:$$\mathrm{S2}≔\left[\mathrm{x2},\mathrm{x3}\right]:$
 Alg2 > $\mathrm{Query}\left(\mathrm{S1},"Abelian"\right)$
 ${\mathrm{false}}$ (2.5)
 Alg1 > $\mathrm{Query}\left(\mathrm{S2},"Abelian"\right)$
 ${\mathrm{true}}$ (2.6)
 See Also

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