change the basis for a representation, either in the Lie algebra or in the representation space - Maple Programming Help

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LieAlgebras[ChangeRepresentationBasis] - change the basis for a representation, either in the Lie algebra or in the representation space

Calling Sequences

ChangeRepresentationBasis(${\mathbf{ρ}}$, B, Fr)

ChangeRepresentationBasis(${\mathbf{ρ}}$, P, keyword, Fr)

Parameters

$\mathrm{ρ}$          - a representation of a Lie algebra on a vector space $V$

B          - a list of vectors defining a new basis B for either or $V$

Fr         - a Maple name or string, the name of the Lie algebra or vector space with the new basis B

P          - a change of basis matrix, the columns of which are the components of the new basis vectors Bwith respect to the original basis

keyword    - either "Domain" or "Range", indicating that the matrix $P$ is a change of basis matrix for the Lie algebra or the representation space

Description

 • Let  be a representation of a Lie algebra $\mathrm{𝔤}$ on a vector space $V$. Let be the given basis for $\mathrm{𝔤}$ and let be the given basis for $V$. Let , the matrix representing the linear transformation $\mathrm{ρ}\left({e}_{i}\right)$with respect to the basis $\left\{{E}_{r}\right\}$. If is another basis for the representation space $V$, then in this new basis where and is the change of basis matrix whose columns are the components of with respect to the basis If is another basis for then where and $Q$ is the change of basis matrix whose columns are the components of the with respect to the $\left\{{e}_{i}\right\}$, that is, .
 • If $B$ = $\left[{F}_{s}\right]$ is a list of vectors defining a basis for $V$, then ChangeReperesentationBasis(${\mathbf{\rho }}$ B, Fr) computes the matrices for the representation $\mathrm{ρ}$ with respect to the basis $\left[{F}_{s}\right]$. If is the change of basis matrix, then the calling sequence ChangeRepresentationBasis(${\mathbf{\rho }}$, P, "Range", Fr) produces the same result.
 • If is a list of vectors defining a basis for $\mathrm{𝔤}$, then ChangeRepresentationBasis(${\mathbf{\rho }}$, B, Fr) computes the matrices for the representation $\mathrm{ρ}$ with respect to the basis $\left[{f}_{j}\right]$. If Q is the change of basis matrix, then the command ChangeRepresentationBasis(${\mathbf{\rho }}$, Q, "Domain", Fr) produces the same result.

Examples

 > with(DifferentialGeometry): with(LieAlgebras):

Example 1.

We define a representation and make a change of basis for the representation space.

 > L := LieAlgebraData([[x1, x3] = x3, [x1, x4] = x4, [x2, x4] = x3], [x1, x2, x3, x4], Alg1);
 ${L}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e4}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{\mathrm{e3}}\right]$ (2.1)
 > DGsetup(L);
 ${\mathrm{Lie algebra: Alg1}}$ (2.2)
 Alg1 > DGsetup([x, y, z], V);
 ${\mathrm{frame name: V}}$ (2.3)
 V > M := [Matrix([[1, 0, 0], [0, 1, 0], [0, 0, 0]]), Matrix([[0, 1, 0], [0, 0, 0], [0, 0, 0]]), Matrix([[0, 0, 1], [0, 0, 0], [0, 0, 0]]), Matrix([[0, 0, 0], [0, 0, 1], [0, 0, 0]])]:
 V > rho := Representation(Alg1, V, M);
 ${\mathrm{ρ}}{:=}\left[\left[{\mathrm{e1}}{,}\left[\begin{array}{rrr}{1}& {0}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{e2}}{,}\left[\begin{array}{rrr}{0}& {1}& {0}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{e3}}{,}\left[\begin{array}{rrr}{0}& {0}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{e4}}{,}\left[\begin{array}{rrr}{0}& {0}& {0}\\ {0}& {0}& {1}\\ {0}& {0}& {0}\end{array}\right]\right]\right]$ (2.4)

Define the new basis for the representation space.

 Alg1 > B := evalDG([D_x + D_y + D_z, D_x - D_y, D_x + 2*D_y + D_z]);
 ${B}{:=}\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_x}}{+}{2}{}{\mathrm{D_y}}{+}{\mathrm{D_z}}\right]$ (2.5)

Compute the representation ${{\mathbf{φ}}}_{{1}}$in the basis B.

 V > phi1 := ChangeRepresentationBasis(rho, B, V);
 ${\mathrm{φ1}}{:=}\left[\left[{\mathrm{e1}}{,}\left[\begin{array}{rrr}{-}{2}& {0}& {-}{3}\\ {1}& {1}& {1}\\ {2}& {0}& {3}\end{array}\right]\right]{,}\left[{\mathrm{e2}}{,}\left[\begin{array}{rrr}{-}{1}& {1}& {-}{2}\\ {1}& {-}{1}& {2}\\ {1}& {-}{1}& {2}\end{array}\right]\right]{,}\left[{\mathrm{e3}}{,}\left[\begin{array}{rrr}{-}{1}& {0}& {-}{1}\\ {1}& {0}& {1}\\ {1}& {0}& {1}\end{array}\right]\right]{,}\left[{\mathrm{e4}}{,}\left[\begin{array}{rrr}{-}{1}& {0}& {-}{1}\\ {0}& {0}& {0}\\ {1}& {0}& {1}\end{array}\right]\right]\right]$ (2.6)

We can use the Query command to check that ${{\mathbf{\phi }}}_{{1}}$is a representation of Alg1.

 Alg1 > Query(phi1, "Representation");
 ${\mathrm{true}}$ (2.7)

Check, by example, that the matrices for are correct. We apply rho(e1) to Fr[1] and express the result as a linear combination of the vectors Fr. This should give the first column of the matrix for e1 in phi1.

 Alg1 > a := ApplyRepresentation(rho, e1, B[1]);
 ${a}{:=}{\mathrm{D_x}}{+}{\mathrm{D_y}}$ (2.8)
 V > GetComponents(a, B);
 $\left[{-}{2}{,}{1}{,}{2}\right]$ (2.9)

Example 2.

We obtain the same change of basis as in Example 1 using the other calling sequence for the procedure ChangeRepresentationBasis. We take the matrix A to be the matrix whose columns are the coefficients of the new basis in terms of the old.

 V > B;
 $\left[{\mathrm{D_x}}{+}{\mathrm{D_y}}{+}{\mathrm{D_z}}{,}{\mathrm{D_x}}{-}{\mathrm{D_y}}{,}{\mathrm{D_x}}{+}{2}{}{\mathrm{D_y}}{+}{\mathrm{D_z}}\right]$ (2.10)
 V > P := LinearAlgebra:-Transpose(Matrix(GetComponents(B, [D_x, D_y, D_z])));
 ${P}{:=}\left[\begin{array}{rrr}{1}& {1}& {1}\\ {1}& {-}{1}& {2}\\ {1}& {0}& {1}\end{array}\right]$ (2.11)
 V > phi2 := ChangeRepresentationBasis(rho, P, "Range", V);
 ${\mathrm{φ2}}{:=}\left[\left[{\mathrm{e1}}{,}\left[\begin{array}{rrr}{-}{2}& {0}& {-}{3}\\ {1}& {1}& {1}\\ {2}& {0}& {3}\end{array}\right]\right]{,}\left[{\mathrm{e2}}{,}\left[\begin{array}{rrr}{-}{1}& {1}& {-}{2}\\ {1}& {-}{1}& {2}\\ {1}& {-}{1}& {2}\end{array}\right]\right]{,}\left[{\mathrm{e3}}{,}\left[\begin{array}{rrr}{-}{1}& {0}& {-}{1}\\ {1}& {0}& {1}\\ {1}& {0}& {1}\end{array}\right]\right]{,}\left[{\mathrm{e4}}{,}\left[\begin{array}{rrr}{-}{1}& {0}& {-}{1}\\ {0}& {0}& {0}\\ {1}& {0}& {1}\end{array}\right]\right]\right]$ (2.12)

Example 3.

Now we make a change of basis in the Lie algebra. First we use the LieAlgebraData command to create the Lie algebra in the new basis.

 Alg1 > ChangeFrame(Alg1);
 ${\mathrm{Alg1}}$ (2.13)
 Alg1 > B := evalDG([e1 + e2, e3 - e2, e2 - 2*e3, e1 - e3 + e4]);
 ${B}{:=}\left[{\mathrm{e1}}{+}{\mathrm{e2}}{,}{-}{\mathrm{e2}}{+}{\mathrm{e3}}{,}{\mathrm{e2}}{-}{2}{}{\mathrm{e3}}{,}{\mathrm{e1}}{-}{\mathrm{e3}}{+}{\mathrm{e4}}\right]$ (2.14)
 Alg1 > L2 := LieAlgebraData(B, Alg2);
 ${\mathrm{L2}}{:=}\left[\left[{\mathrm{e1}}{,}{\mathrm{e2}}\right]{=}{-}{\mathrm{e2}}{-}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e3}}\right]{=}{2}{}{\mathrm{e2}}{+}{2}{}{\mathrm{e3}}{,}\left[{\mathrm{e1}}{,}{\mathrm{e4}}\right]{=}{-}{\mathrm{e1}}{-}{3}{}{\mathrm{e2}}{-}{2}{}{\mathrm{e3}}{+}{\mathrm{e4}}{,}\left[{\mathrm{e2}}{,}{\mathrm{e4}}\right]{=}{2}{}{\mathrm{e2}}{+}{2}{}{\mathrm{e3}}{,}\left[{\mathrm{e3}}{,}{\mathrm{e4}}\right]{=}{-}{3}{}{\mathrm{e2}}{-}{3}{}{\mathrm{e3}}\right]$ (2.15)
 Alg1 > DGsetup(L2, [f], [theta]);
 ${\mathrm{Lie algebra: Alg2}}$ (2.16)
 Alg2 > phi3 := ChangeRepresentationBasis(rho, B, Alg2);
 ${\mathrm{φ3}}{:=}\left[\left[{\mathrm{f1}}{,}\left[\begin{array}{rrr}{1}& {1}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f2}}{,}\left[\begin{array}{rrr}{0}& {-}{1}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f3}}{,}\left[\begin{array}{rrr}{0}& {1}& {-}{2}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f4}}{,}\left[\begin{array}{rrr}{1}& {0}& {-}{1}\\ {0}& {1}& {1}\\ {0}& {0}& {0}\end{array}\right]\right]\right]$ (2.17)
 Alg2 > Query(phi3, "Representation");
 ${\mathrm{true}}$ (2.18)

Example 4. We obtain the same change of basis as in Example 3 using the other calling sequence for the procedure ChangeRepresentationBasis. We take the matrix A to be the matrix whose columns are the coefficients of the new basis in terms of the old.

 Alg2 > B;
 $\left[{\mathrm{e1}}{+}{\mathrm{e2}}{,}{-}{\mathrm{e2}}{+}{\mathrm{e3}}{,}{\mathrm{e2}}{-}{2}{}{\mathrm{e3}}{,}{\mathrm{e1}}{-}{\mathrm{e3}}{+}{\mathrm{e4}}\right]$ (2.19)
 Alg1 > P := LinearAlgebra:-Transpose(Matrix(GetComponents(B, [e1, e2, e3, e4])));
 ${P}{:=}\left[\begin{array}{rrrr}{1}& {0}& {0}& {1}\\ {1}& {-}{1}& {1}& {0}\\ {0}& {1}& {-}{2}& {-}{1}\\ {0}& {0}& {0}& {1}\end{array}\right]$ (2.20)
 Alg1 > phi4 := ChangeRepresentationBasis(rho, P, "Domain", Alg2);
 ${\mathrm{φ4}}{:=}\left[\left[{\mathrm{f1}}{,}\left[\begin{array}{rrr}{1}& {1}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f2}}{,}\left[\begin{array}{rrr}{0}& {-}{1}& {1}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f3}}{,}\left[\begin{array}{rrr}{0}& {1}& {-}{2}\\ {0}& {0}& {0}\\ {0}& {0}& {0}\end{array}\right]\right]{,}\left[{\mathrm{f4}}{,}\left[\begin{array}{rrr}{1}& {0}& {-}{1}\\ {0}& {1}& {1}\\ {0}& {0}& {0}\end{array}\right]\right]\right]$ (2.21)