calculate the span of the Lie bracket of two lists of vectors in a Lie algebra, calculate the span of the matrix commutator of two lists of matrices - Maple Help

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LieAlgebras[BracketOfSubspaces] - calculate the span of the Lie bracket of two lists of vectors in a Lie algebra, calculate the span of the matrix commutator of two lists of matrices

Calling Sequences

BracketOfSubspaces(S1, S2)

BracketOfSubspaces(M1, M2)

Parameters

S1, S2   - two lists of vectors whose spans determine subspaces of a Lie algebra $\mathrm{𝔤}$

M1, M2   - two lists of square  matrices

Description

 • Let be a Lie algebra and let and be two subspaces (not necessarily subalgebras). Then denotes the span of all vectors of the form with and If span {and span {then

span{|  and  .

Likewise, if ${M}_{1}$ and are two subspaces of the Lie algebra of all matrices), then denotes the span of all matrices of form , with and .

 • The first calling sequence BracketOfSubspaces(S1, S2) calculates the subspace A list of linearly independent vectors defining a basis for  is returned. If (that is, all the vectors in commute with all the vectors in ), then an empty list is returned.
 • The second calling sequence BracketOfSubspaces(M1, M2) calculates the subspace . A list of linearly independent vectors defining a basis for  is returned. If (that is, all the matrices in in commute with all the matrices in ${M}_{2}$), then an empty list is returned.
 • The command BracketOfSubspaces is part of the DifferentialGeometry:-LieAlgebras package.  It can be used in the form BracketOfSubspaces(...) only after executing the commands with(DifferentialGeometry) and with(LieAlgebras), but can always be used by executing DifferentialGeometry:-LieAlgebras:-BracketOfSubspaces(...).

Examples

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

Example 1.

First we initialize a Lie algebra.

 > $\mathrm{L1}≔\mathrm{_DG}\left(\left[\left["LieAlgebra",\mathrm{Alg1},\left[4\right]\right],\left[\left[\left[2,4,1\right],1\right],\left[\left[3,4,3\right],1\right]\right]\right]\right)$
 ${\mathrm{L1}}{:=}\left[\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e1}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}\right]$ (2.1)
 > $\mathrm{DGsetup}\left(\mathrm{L1}\right):$

We bracket the subspaces span  and span {

 Alg1 > $\mathrm{S1}≔\left[\mathrm{e1},\mathrm{e2}\right]:$$\mathrm{S2}≔\left[\mathrm{e3},\mathrm{e4}\right]:$
 Alg1 > $\mathrm{BracketOfSubspaces}\left(\mathrm{S1},\mathrm{S2}\right)$
 $\left[{\mathrm{e1}}\right]$ (2.2)

We bracket the subspace span{ with itself.

 Alg1 > $\mathrm{S3}≔\left[\mathrm{e1},\mathrm{e2},\mathrm{e3}\right]:$
 Alg1 > $\mathrm{BracketOfSubspaces}\left(\mathrm{S3},\mathrm{S3}\right)$
 $\left[{}\right]$ (2.3)

Example 2.

The command also works with lists of matrices.

 > $\mathrm{M1}≔\left[\mathrm{Matrix}\left(\left[\left[0,1,1\right],\left[0,0,0\right],\left[0,0,0\right]\right]\right),\mathrm{Matrix}\left(\left[\left[0,1,0\right],\left[0,0,0\right],\left[1,0,0\right]\right]\right)\right]$
 ${\mathrm{M1}}{:=}\left[\left[\begin{array}{rrr}{0}& {1}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {1}& {0}\\ {0}& {0}& {0}\\ {1}& {0}& {0}\end{array}\right]\right]$ (2.4)
 > $\mathrm{M2}≔\left[\mathrm{Matrix}\left(\left[\left[0,0,1\right],\left[0,0,0\right],\left[0,0,0\right]\right]\right),\mathrm{Matrix}\left(\left[\left[0,0,0\right],\left[0,0,1\right],\left[0,0,0\right]\right]\right),\mathrm{Matrix}\left(\left[\left[0,0,0\right],\left[0,0,0\right],\left[0,0,1\right]\right]\right)\right]$
 ${\mathrm{M2}}{:=}\left[\left[\begin{array}{rrr}{0}& {0}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {1}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {0}\\ {0}& {0}& {1}\end{array}\right]\right]$ (2.5)
 > $\mathrm{BracketOfSubspaces}\left(\mathrm{M1},\mathrm{M2}\right)$
 $\left[\left[\begin{array}{rrr}{0}& {0}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{-}{1}& {0}& {0}\\ {0}& {0}& {0}\\ {0}& {0}& {1}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {1}\\ {-}{1}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {0}\\ {-}{1}& {0}& {0}\end{array}\right]\right]$ (2.6)