LeviDecomposition - Maple Help
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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 𝔤

     S     - a list of independent vectors in a Lie algebra 𝔤

 

Description

Examples

Description

• 

A pair of subalgebras R, S  in a Lie algebra 𝔤 define a Levi decomposition if R is the radical of 𝔤, S is a semi-simple subalgebra, and g = R  S (vector space direct sum). Since the radical is an ideal we have R,.R R, R, S R, and S, S S. The radical R is unique, the semisimple subalgebra S in a Levi decomposition is not.

• 

Query([R, S], "LeviDecomposition") returns true if the pair R, S is a Levi decomposition of 𝔤  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

withDifferentialGeometry:withLieAlgebras:

 

Example 1.

We initialize three different Lie algebras and print their multiplication tables.

L1_DGLieAlgebra,Alg1,3,1,3,1,1,2,3,1,1,2,3,2,1:

L2_DGLieAlgebra,Alg2,3,1,2,1,1,1,3,2,2,2,3,3,1:

L3_DGLieAlgebra,Alg3,5,1,3,2,1,1,4,1,1,2,4,2,1,2,5,1,1,3,4,3,2,3,5,4,1,4,5,5,2:

DGsetupL1,x,a:DGsetupL2,y,b:DGsetupL3,z,c:

Alg3 > 

printMultiplicationTableAlg1,LieTable,MultiplicationTableAlg2,LieTable,MultiplicationTableAlg3,LieTable

 

Alg1 is solvable and therefore the radical is the entire algebra.

Alg3 > 

R1x1,x2,x3:S1:

Alg3 > 

QueryR1,S1,LeviDecomposition

true

(2.1)

 

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

Alg1 > 

R2:S2y1,y2,y3:

Alg1 > 

QueryR2,S2,LeviDecomposition

true

(2.2)

 

Alg3 has a non-trivial Levi decomposition.

Alg1 > 

R3z1,z2:S3z3,z4,z5:

Alg1 > 

QueryR3,S3,LeviDecomposition

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 R3 using the ComplementaryBasis program.

Alg1 > 

SS0ComplementaryBasisR3,z1,z2,z3,z4,z5,k

SS0:=k1z1+k2z2+z3,k3z1+k4z2+z4,k5z1+k6z2+z5,k1,k2,k3,k4,k5,k6

(2.4)

Next we determine for which values of the parameters k1, k2, k3, k4, k5, k6 for which he subspace SS0 is a Lie subalgebra.  We find that k1=0,  k2= k3,  k4 = k5,  k6 =0. 

Alg3 > 

TF,Eq,SOL,SubAlgListQuerySS0,Subalgebra

TF,Eq,SOL,SubAlgList:=true,0,3k1,3k6,k2+k3,k4+k5,k5k4,k1=0,k2=k3,k3=k3,k4=k5,k5=k5,k6=0,k3z2+z3,k3z1k5z2+z4,k5z1+z5

(2.5)
Alg3 > 

S4SubAlgList1

S4:=k3z2+z3,k3z1k5z2+z4,k5z1+z5

(2.6)
Alg3 > 

QueryR3,S4,LeviDecomposition

true

(2.7)

See Also

DifferentialGeometry

LieAlgebras

ComplementaryBasis

LeviDecomposition

MultiplicationTable

Query