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LieAlgebras[HomomorphismSubalgebras] - find the kernel, image of a Lie algebra homomorphism; find the inverse image of a subalgebra with respect to a Lie algebra homomorphism
Calling Sequences
HomomorphismSubalgebras(Phi, keyword)
HomomorphismSubalgebras(Phi, S, keyword)
Parameters
Phi - 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
Let Phi: g -> k be a Lie algebra homomorphism. The kernel of Phi is the ideal of vectors x in g such that Phi(x) =0. The image of Phi is the subalgebra of vectors y in k such that y = Phi(x) for some x in g. If S is a subalgebra of k, then the inverse image of S with respect to Phi is the subalgebra of vectors x in g such that Phi(x) in S.
HomomorphismSubalgebras(Phi, "Kernel") calculates the kernel of Phi. A list of independent vectors defining a basis for the kernel is returned. If the kernel is trivial (that is, consists solely of the zero), then an empty list is returned.
HomomorphismSubalgebras(Phi, "Image") calculates the image of Phi. A list of independent vectors defining a basis for the image is returned. If the image is trivial (that is, consists solely of the zero), then an empty list is returned.
HomomorphismSubalgebras(Phi, S, "InverseImage") calculates the inverse image of S in the domain algebra g. A list of independent vectors defining a basis for the inverse image is returned. If the inverse image is trivial (that is, consists solely of the zero), 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(...).
Examples
Example 1.
First we initialize a pair of Lie algebras and display the multiplication tables.
We define a transformation Phi from Alg1 to Alg2 and check that it is a Lie algebra homomorphism.
We find the kernel of Phi.
We find the image of Phi.
We find the inverse image of the subalgebra S1 = [y3, y4] with respect to Phi.
See Also
DifferentialGeometry, LieAlgebras, ApplyHomomorphism, MultiplicationTable, Query[Homomorphism], Transformation
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