 LeviDecomposition - Maple Help

Query[LeviDecomposition] - check that a pair of subalgebras define a Levi decomposition of a Lie algebra

Calling Sequences

Query([R, S], "LeviDecomposition")

Parameters

R     - a list of independent vectors in a Lie algebra $\mathrm{𝔤}$

S     - a list of independent vectors in a Lie algebra $\mathrm{𝔤}$ Description

 • A pair of subalgebras  in a Lie algebra define a Levi decomposition if is the radical of $\mathrm{𝔤}$, $S$ is a semi-simple subalgebra, and (vector space direct sum). Since the radical is an ideal we have , and The radical $R$ is unique, the semisimple subalgebra in a Levi decomposition is not.
 • Query([R, S], "LeviDecomposition") returns true if the pair R, S is a Levi decomposition of  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 three different Lie algebras and print their multiplication tables.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[3\right]\right],\left[\left[\left[1,3,1\right],1\right],\left[\left[2,3,1\right],1\right],\left[\left[2,3,2\right],1\right]\right]\right]\right):$
 > $\mathrm{L2}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg2},\left[3\right]\right],\left[\left[\left[1,2,1\right],1\right],\left[\left[1,3,2\right],-2\right],\left[\left[2,3,3\right],1\right]\right]\right]\right):$
 > $\mathrm{L3}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg3},\left[5\right]\right],\left[\left[\left[1,3,2\right],-1\right],\left[\left[1,4,1\right],-1\right],\left[\left[2,4,2\right],1\right],\left[\left[2,5,1\right],-1\right],\left[\left[3,4,3\right],2\right],\left[\left[3,5,4\right],-1\right],\left[\left[4,5,5\right],2\right]\right]\right]\right):$
 > $\mathrm{DGsetup}\left(\mathrm{L1},\left[x\right],\left[a\right]\right):$$\mathrm{DGsetup}\left(\mathrm{L2},\left[y\right],\left[b\right]\right):$$\mathrm{DGsetup}\left(\mathrm{L3},\left[z\right],\left[c\right]\right):$
 Alg3 > $\mathrm{print}\left(\mathrm{MultiplicationTable}\left(\mathrm{Alg1},"LieTable"\right),\mathrm{MultiplicationTable}\left(\mathrm{Alg2},"LieTable"\right),\mathrm{MultiplicationTable}\left(\mathrm{Alg3},"LieTable"\right)\right)$ Alg1 is solvable and therefore the radical is the entire algebra.

 Alg3 > $\mathrm{R1}≔\left[\mathrm{x1},\mathrm{x2},\mathrm{x3}\right]:$$\mathrm{S1}≔\left[\right]:$
 Alg3 > $\mathrm{Query}\left(\left[\mathrm{R1},\mathrm{S1}\right],"LeviDecomposition"\right)$
 ${\mathrm{true}}$ (2.1)

Alg2 is semisimple and therefore the radical is the zero subalgebra.

 Alg1 > $\mathrm{R2}≔\left[\right]:$$\mathrm{S2}≔\left[\mathrm{y1},\mathrm{y2},\mathrm{y3}\right]:$
 Alg1 > $\mathrm{Query}\left(\left[\mathrm{R2},\mathrm{S2}\right],"LeviDecomposition"\right)$
 ${\mathrm{true}}$ (2.2)

Alg3 has a non-trivial Levi decomposition.

 Alg1 > $\mathrm{R3}≔\left[\mathrm{z1},\mathrm{z2}\right]:$$\mathrm{S3}≔\left[\mathrm{z3},\mathrm{z4},\mathrm{z5}\right]:$
 Alg1 > $\mathrm{Query}\left(\left[\mathrm{R3},\mathrm{S3}\right],"LeviDecomposition"\right)$
 ${\mathrm{true}}$ (2.3)

It is easy to see that in this last example the Levi decomposition is not unique. First we find the general complement to the radical $\mathrm{R3}$ using the ComplementaryBasis program.

 Alg1 > $\mathrm{SS0}≔\mathrm{ComplementaryBasis}\left(\mathrm{R3},\left[\mathrm{z1},\mathrm{z2},\mathrm{z3},\mathrm{z4},\mathrm{z5}\right],k\right)$
 ${\mathrm{SS0}}{:=}\left[{\mathrm{k1}}{}{\mathrm{z1}}{+}{\mathrm{k2}}{}{\mathrm{z2}}{+}{\mathrm{z3}}{,}{\mathrm{k3}}{}{\mathrm{z1}}{+}{\mathrm{k4}}{}{\mathrm{z2}}{+}{\mathrm{z4}}{,}{\mathrm{k5}}{}{\mathrm{z1}}{+}{\mathrm{k6}}{}{\mathrm{z2}}{+}{\mathrm{z5}}\right]{,}\left\{{\mathrm{k1}}{,}{\mathrm{k2}}{,}{\mathrm{k3}}{,}{\mathrm{k4}}{,}{\mathrm{k5}}{,}{\mathrm{k6}}\right\}$ (2.4)



Next we determine for which values of the parameters for which he subspace SS0 is a Lie subalgebra.  We find that .

 Alg3 > $\mathrm{TF},\mathrm{Eq},\mathrm{SOL},\mathrm{SubAlgList}≔\mathrm{Query}\left(\mathrm{SS0},"Subalgebra"\right)$
 ${\mathrm{TF}}{,}{\mathrm{Eq}}{,}{\mathrm{SOL}}{,}{\mathrm{SubAlgList}}{:=}{\mathrm{true}}{,}\left\{{0}{,}{-}{3}{}{\mathrm{k1}}{,}{-}{3}{}{\mathrm{k6}}{,}{-}{\mathrm{k2}}{+}{\mathrm{k3}}{,}{\mathrm{k4}}{+}{\mathrm{k5}}{,}{-}{\mathrm{k5}}{-}{\mathrm{k4}}\right\}{,}\left[\left\{{\mathrm{k1}}{=}{0}{,}{\mathrm{k2}}{=}{\mathrm{k3}}{,}{\mathrm{k3}}{=}{\mathrm{k3}}{,}{\mathrm{k4}}{=}{-}{\mathrm{k5}}{,}{\mathrm{k5}}{=}{\mathrm{k5}}{,}{\mathrm{k6}}{=}{0}\right\}\right]{,}\left[\left[{\mathrm{k3}}{}{\mathrm{z2}}{+}{\mathrm{z3}}{,}{\mathrm{k3}}{}{\mathrm{z1}}{-}{\mathrm{k5}}{}{\mathrm{z2}}{+}{\mathrm{z4}}{,}{\mathrm{k5}}{}{\mathrm{z1}}{+}{\mathrm{z5}}\right]\right]$ (2.5)
 Alg3 > $\mathrm{S4}≔{\mathrm{SubAlgList}}_{1}$
 ${\mathrm{S4}}{:=}\left[{\mathrm{k3}}{}{\mathrm{z2}}{+}{\mathrm{z3}}{,}{\mathrm{k3}}{}{\mathrm{z1}}{-}{\mathrm{k5}}{}{\mathrm{z2}}{+}{\mathrm{z4}}{,}{\mathrm{k5}}{}{\mathrm{z1}}{+}{\mathrm{z5}}\right]$ (2.6)
 Alg3 > $\mathrm{Query}\left(\left[\mathrm{R3},\mathrm{S4}\right],"LeviDecomposition"\right)$
 ${\mathrm{true}}$ (2.7)