Change of basis for \(V_7\)

In our work we use two different bases of the vector space \(V_7\), which is the half spin representation \(S^{+}\). The first basis is given by the vectors \(e_I\) where \(I\) is a subset of \(\{1,2,3,4,5\}\) with even cardinality, and \(e_I\) is the wedge product of the vectors \(e_{i}\) for \(i \in I\). The second basis is the one produced by de Graaf's algorithm where the term order used is the lex term order on the lowering operators in the order they are listed in the Macaulay2 package LieAlgebraRepresentations.

The LieAlgebraRepresentations package has the functionality required to construct both of these representations, as well as to compute an isomorphism between two irreducible representations.

Macaulay2, version 1.25.11
Type "help" to see useful commands

i1 : needsPackage("LieAlgebraRepresentations");

i2 : so10 = simpleLieAlgebra("D",5);

i3 : V00010 = irreducibleLieAlgebraModule({0,0,0,1,0},so10);

i4 : LAB = lieAlgebraBasis(so10);

i5 : L = halfspinRepresentationMatrices(5,0);

i6 : rho1 = lieAlgebraRepresentation(V00010,LAB,L);

i7 : rho2 = deGraafRepresentation({0,0,0,1,0},so10);
Warning:  F4 Algorithm not available over current coefficient ring or inhomogeneous ideal.
Converting to Naive algorithm.
Warning:  F4 Algorithm not available over current coefficient ring or inhomogeneous ideal.
Converting to Naive algorithm.
max-lev=17
Finished level 1. {#G,#B}={4, 2}
Finished level 2. {#G,#B}={8, 3}
Finished level 3. {#G,#B}={11, 5}
Finished level 4. {#G,#B}={15, 7}
Finished level 5. {#G,#B}={19, 9}
Finished level 6. {#G,#B}={25, 11}
Finished level 7. {#G,#B}={31, 13}
Finished level 8. {#G,#B}={39, 14}
Finished level 9. {#G,#B}={45, 15}
Finished level 10. {#G,#B}={51, 16}
Finished level 11. {#G,#B}={55, 16}
Finished level 12. {#G,#B}={58, 16}
Finished level 13. {#G,#B}={59, 16}
Finished level 14. {#G,#B}={60, 16}
Finished level 15. {#G,#B}={60, 16}
Finished level 16. {#G,#B}={60, 16}
Finished level 17. {#G,#B}={60, 16}
Compute rho(B_0)
Compute rho(B_1)
Compute rho(B_2)
Compute rho(B_3)
Compute rho(B_4)
Compute rho(B_5)
Compute rho(B_6)
Compute rho(B_7)
Compute rho(B_8)
Compute rho(B_9)
Compute rho(B_10)
Compute rho(B_11)
Compute rho(B_12)
Compute rho(B_13)
Compute rho(B_14)
Compute rho(B_15)
Compute rho(B_16)
Compute rho(B_17)
Compute rho(B_18)
Compute rho(B_19)
Compute rho(B_20)
Compute rho(B_21)
Compute rho(B_22)
Compute rho(B_23)
Compute rho(B_24)
Compute rho(B_25)
Compute rho(B_26)
Compute rho(B_27)
Compute rho(B_28)
Compute rho(B_29)
Compute rho(B_30)
Compute rho(B_31)
Compute rho(B_32)
Compute rho(B_33)
Compute rho(B_34)
Compute rho(B_35)
Compute rho(B_36)
Compute rho(B_37)
Compute rho(B_38)
Compute rho(B_39)
Compute rho(B_40)
Compute rho(B_41)
Compute rho(B_42)
Compute rho(B_43)
Compute rho(B_44)

i8 : P = isomorphismOfRepresentations(rho1,rho2)
Length 1 complete. 10 new words found
Length 2 complete. 5 new words found

o8 = | 0 0 0  0 0 0  0  0 0 0 0  0  0 0 0  1 |
     | 0 0 0  1 0 0  0  0 0 0 0  0  0 0 0  0 |
     | 0 0 0  0 0 -1 0  0 0 0 0  0  0 0 0  0 |
     | 0 0 0  0 0 0  0  1 0 0 0  0  0 0 0  0 |
     | 0 0 0  0 0 0  0  0 0 1 0  0  0 0 0  0 |
     | 0 0 0  0 0 0  0  0 1 0 0  0  0 0 0  0 |
     | 0 0 0  0 0 0  0  0 0 0 -1 0  0 0 0  0 |
     | 0 0 0  0 0 0  0  0 0 0 0  -1 0 0 0  0 |
     | 0 0 0  0 0 0  0  0 0 0 0  0  1 0 0  0 |
     | 0 0 0  0 0 0  0  0 0 0 0  0  0 1 0  0 |
     | 0 0 0  0 0 0  0  0 0 0 0  0  0 0 -1 0 |
     | 1 0 0  0 0 0  0  0 0 0 0  0  0 0 0  0 |
     | 0 1 0  0 0 0  0  0 0 0 0  0  0 0 0  0 |
     | 0 0 -1 0 0 0  0  0 0 0 0  0  0 0 0  0 |
     | 0 0 0  0 1 0  0  0 0 0 0  0  0 0 0  0 |
     | 0 0 0  0 0 0  -1 0 0 0 0  0  0 0 0  0 |

              16       16
o8 : Matrix QQ   <-- QQ

The documentation for isomorphismOfRepresentations indicates that the matrix \(P\) satisfies \( \rho_2(X) = P^{-1} \rho_1(X) P\) for each \(X \in \mathfrak{g}\). Thus, the matrix \(P\) shown above is the change of matrix from the de Graaf basis to the basis given by the even wedge products. In the functions we wrote to evaluate the invariant polynomials \(F_{7,1}\) and \(F_{7,2}\), we go in the opposite direction, and hence need the inverse of the matrix shown above.

i9 : inverse P

o9 = | 0 0 0  0 0 0 0  0  0 0 0  1 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 0 0  0 1 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 0 0  0 0 -1 0 0  |
     | 0 1 0  0 0 0 0  0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 0 0  0 0 0  1 0  |
     | 0 0 -1 0 0 0 0  0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 0 0  0 0 0  0 -1 |
     | 0 0 0  1 0 0 0  0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 1 0  0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 1 0 0  0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 -1 0  0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  -1 0 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  1 0 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 1 0  0 0 0  0 0  |
     | 0 0 0  0 0 0 0  0  0 0 -1 0 0 0  0 0  |
     | 1 0 0  0 0 0 0  0  0 0 0  0 0 0  0 0  |

              16       16
o9 : Matrix QQ   <-- QQ