LieAlgebras[HomomorphismSubalgebras] - find the kernel or image of a Lie algebra homomorphism; find the inverse image of a subalgebra with respect to a Lie algebra homomorphism
Calling Sequences
HomomorphismSubalgebras(φ, keyword)
HomomorphismSubalgebras(φ, S, keyword)
Parameters
φ - a transformation mapping one Lie algebra 𝔤 to another k
keyword - a keyword string, one of "Kernel", "Image", "InverseImage"
S - a list of vectors defining a basis for a subalgebra of k
Description
Examples
Let g and kbe Lie algebras and let φ:g →k be a Lie algebra homomorphism .The kernel of φ is the ideal of vectors ker(φ) = {x ∈ g | φx = 0}. The image of φ is the subalgebra of vectors imφ = {y ∈k | y = φx for some x ∈ g}. If S is a subalgebra of k, then the inverse image of S with respect to φ is the subalgebra φ−1S = {x ∈ g | φx ∈ S}.
HomomorphismSubalgebras(φ, "Kernel") calculates kerφ. A list of independent vectors defining a basis for the kernel is returned. If ker(φ) = 0,then an empty list is returned.
HomomorphismSubalgebras(φ, "Image") calculates im(φ). A list of independent vectors defining a basis for the image is returned. If im(φ) = 0,then an empty list is returned.
HomomorphismSubalgebras(φ, S, "InverseImage") calculates φ−1S. A list of independent vectors defining a basis for the inverse image is returned. If φ−1S = 0, then an empty list is returned.
The command HomomorphismSubalgebras is part of the DifferentialGeometry:-LieAlgebras package. It can be used in the form HomomorphismSubalgebras(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-HomomorphismSubalgebras(...).
with⁡DifferentialGeometry:with⁡LieAlgebras:
Example 1.
First we initialize a pair of Lie algebras and display the multiplication tables.
L1≔_DG⁡LieAlgebra,Alg1,3,2,3,1,1:
DGsetup⁡L1,x,α:
L2≔_DG⁡LieAlgebra,Alg2,4,1,4,2,1,3,4,3,1:
DGsetup⁡L2,y,β:
print⁡MultiplicationTable⁡Alg1,LieBracket,MultiplicationTable⁡Alg2,LieBracket
x2,x3=x1,y1,y4=y2,y3,y4=y3
We define a transformation Phi from Alg1 to Alg2 and check that it is a Lie algebra homomorphism.
Φ≔Transformation⁡x1,0&multy1,x2,y2,x3,y3
Φ ≔ x1,0⁢y1,x2,y2,x3,y3
Query⁡Alg1,Alg2,Φ,Homomorphism
true
We find the kernel of Phi.
HomomorphismSubalgebras⁡Φ,Kernel
x1
We find the image of Phi.
HomomorphismSubalgebras⁡Φ,Image
y2,y3
We find the inverse image of the subalgebra spanned by y3, y4 with respect to φ.
S1≔y3,y4:
HomomorphismSubalgebras⁡Φ,S1,InverseImage
x1,−x3
See Also
DifferentialGeometry
LieAlgebras
ApplyHomomorphism
MultiplicationTable
Query[Homomorphism]
Transformation
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