HomomorphismSubalgebras - Maple Help

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(${\mathbf{φ}}$, keyword)

HomomorphismSubalgebras(${\mathbf{φ}}$,S, keyword)

Parameters

$\mathrm{φ}$         - a transformation mapping one Lie algebra 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 and $k$be Lie algebras and let be a Lie algebra homomorphism .The kernel of is the ideal of vectors ker(| The image of is the subalgebra of vectors im$\left(\mathrm{φ}\right)$ =  for some . If $S$ is a subalgebra of $k$, then the inverse image of with respect to is the subalgebra  .
 • HomomorphismSubalgebras(${\mathbf{φ}}$, "Kernel") calculates ker$\left(\mathrm{φ}\right).$  A list of independent vectors defining a basis for the kernel is returned.  If ker(then an empty list is returned.
 • HomomorphismSubalgebras(${\mathbf{φ}}$, "Image") calculates im($\mathrm{φ}$).  A list of independent vectors defining a basis for the image is returned. If  im(then an empty list is returned.
 • HomomorphismSubalgebras(${\mathbf{φ}}$, S, "InverseImage") calculates ${\mathrm{φ}}^{-1}\left(S\right)$. A list of independent vectors defining a basis for the inverse image is returned. If ${\mathrm{\phi }}^{-1}\left(S\right)$ = 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(...).

Examples

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

Example 1.

First we initialize a pair of Lie algebras and display the multiplication tables.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[3\right]\right],\left[\left[\left[2,3,1\right],1\right]\right]\right]\right):$
 > $\mathrm{DGsetup}\left(\mathrm{L1},\left[x\right],\left[\mathrm{α}\right]\right):$
 Alg1 > $\mathrm{L2}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg2},\left[4\right]\right],\left[\left[\left[1,4,2\right],1\right],\left[\left[3,4,3\right],1\right]\right]\right]\right):$
 Alg1   > $\mathrm{DGsetup}\left(\mathrm{L2},\left[y\right],\left[\mathrm{β}\right]\right):$
 Alg1 > $\mathrm{print}\left(\mathrm{MultiplicationTable}\left(\mathrm{Alg1},"LieBracket"\right),\mathrm{MultiplicationTable}\left(\mathrm{Alg2},"LieBracket"\right)\right)$
 $\left[\left[{\mathrm{x2}}{,}{\mathrm{x3}}\right]{=}{\mathrm{x1}}\right]{,}\left[\left[{\mathrm{y1}}{,}{\mathrm{y4}}\right]{=}{\mathrm{y2}}{,}\left[{\mathrm{y3}}{,}{\mathrm{y4}}\right]{=}{\mathrm{y3}}\right]$ (2.1)

We define a transformation Phi from Alg1 to Alg2 and check that it is a Lie algebra homomorphism.

 Alg2 > $\mathrm{Φ}≔\mathrm{Transformation}\left(\left[\left[\mathrm{x1},0&mult\mathrm{y1}\right],\left[\mathrm{x2},\mathrm{y2}\right],\left[\mathrm{x3},\mathrm{y3}\right]\right]\right)$
 ${\mathrm{Φ}}{≔}\left[\left[{\mathrm{x1}}{,}{0}{}{\mathrm{y1}}\right]{,}\left[{\mathrm{x2}}{,}{\mathrm{y2}}\right]{,}\left[{\mathrm{x3}}{,}{\mathrm{y3}}\right]\right]$ (2.2)
 Alg2 > $\mathrm{Query}\left(\mathrm{Alg1},\mathrm{Alg2},\mathrm{Φ},"Homomorphism"\right)$
 ${\mathrm{true}}$ (2.3)

We find the kernel of Phi.

 Alg2 > $\mathrm{HomomorphismSubalgebras}\left(\mathrm{Φ},"Kernel"\right)$
 $\left[{\mathrm{x1}}\right]$ (2.4)

We find the image of Phi.

 Alg1 > $\mathrm{HomomorphismSubalgebras}\left(\mathrm{Φ},"Image"\right)$
 $\left[{\mathrm{y2}}{,}{\mathrm{y3}}\right]$ (2.5)

We find the inverse image of the subalgebra spanned by   with respect to $\mathrm{φ}$.

 Alg2 > $\mathrm{S1}≔\left[\mathrm{y3},\mathrm{y4}\right]:$
 Alg2 > $\mathrm{HomomorphismSubalgebras}\left(\mathrm{Φ},\mathrm{S1},"InverseImage"\right)$
 $\left[{\mathrm{x1}}{,}{-}{\mathrm{x3}}\right]$ (2.6)