Manuscript XVI-J@1 · XVI-J.tex · proposition · 6c1f306d

Proposition: Complete decorated coefficient

Independent polynomial proofs of the numerical tables › A second contraction method · lines 999–1049 · proof included or linked

XVI-J@1#prop:Ktable, XVI-J@1#eq:Ktable, XVI-J@1#eq:Kcompact

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For $h = h_{*}$ and a general symmetric jet $J$ , the tensor is

K^{00} K^{01} K^{11} K^{12} K^{22} K^{23} K^{33} = 14 J_{11} + 10 J_{22} + 10 J_{33}, = - 14 J_{01}, = 14 (J_{00} - J_{22} - J_{33}), = 14 J_{12}, = 10 J_{00} - 14 J_{11} - 10 J_{33}, = 10 J_{23}, = 10 J_{00} - 14 J_{11} - 10 J_{22} . K^{02} K^{13} = - 10 J_{02}, = 14 J_{13}, K^{03} = - 10 J_{03},

The remaining entries follow by symmetry.

Proof. Apply (eq:covdiag)–(eq:covoff) to (eq:Qmatrix). A compact form of the resulting identity is

E_{L} [Q (H; J)] = - 10 (η tr (η J) - η J η) + 4 (E tr (η J) + η J_{11} - E J η - η J E),

where $E$ is the contravariant matrix with $E^{11} = 1$ and all other entries zero. Both sides are linear in $J$ . Substitution of the ten symmetric component jets, using the complete quadratic contraction tables, proves this polynomial identity. Expanding its diagonal and off-diagonal entries gives (eq:Ktable).

For additional explicit arithmetic checks, the next lemma evaluates three component jets directly.

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XVI-J@1#eq:Ktable · line 1002
$K^{00} K^{01} K^{11} K^{12} K^{22} K^{23} K^{33} = 14 J_{11} + 10 J_{22} + 10 J_{33}, = - 14 J_{01}, = 14 (J_{00} - J_{22} - J_{33}), = 14 J_{12}, = 10 J_{00} - 14 J_{11} - 10 J_{33}, = 10 J_{23}, = 10 J_{00} - 14 J_{11} - 10 J_{22} . K^{02} K^{13} = - 10 J_{02}, = 14 J_{13}, K^{03} = - 10 J_{03},$
XVI-J@1#eq:Kcompact · line 1030
$E_{L} [Q (H; J)] = - 10 (η tr (η J) - η J η) + 4 (E tr (η J) + η J_{11} - E J η - η J E),$

Proposition: Complete decorated coefficient

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