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

A second contraction method

Independent polynomial proofs of the numerical tables › A second contraction method · lines 930–998

XVI-J@1#eq:secondmethod, XVI-J@1#eq:covdiag, XVI-J@1#eq:covoff

The following auxiliary operation is finite-dimensional polynomial algebra, not a probability expectation:

E_{L} [p (a)] = \frac{1}{2} i, j \sum L_{ij} \partial_{a_{i}} \partial_{a_{j}} p (a)_{a = 0} .

Let

H (a) = i \sum a_{i} Z_{i} .

Then

K (J) = E_{L} [Q (H (a); J)] .

Indeed, the coefficient of $a_{i} a_{j}$ in the quadratic source is its polarization on $Z_{i}, Z_{j}$ .

The diagonal components of $H (a)$ are

H_{00} H_{11} H_{22} H_{33} = \frac{1}{2} - 1 - 1 - 1 - 1 - 1 - 1 11 - 1 1 - 1 1 - 1 11 - 1 a_{0} a_{4} a_{7} a_{9} .

Its off-diagonal components are

(H_{01}, H_{02}, H_{03}, H_{12}, H_{13}, H_{23}) = \frac{1}{2} (a_{1}, a_{2}, a_{3}, - a_{5}, - a_{6}, - a_{8}) .

Thus the diagonal quadratic contraction table, in order $00, 11, 22, 33$ , is

(E_{L} [H_{μμ} H_{ν ν}]) = - 3 - 3 11 - 3 - 3 33 13 - 3 - 1 13 - 1 - 3 .

The off-diagonal squares are

μν E_{L} [H_{μν}^{2}] 013021031 12 - 3 13 - 3 23 - 1

and all mixed contractions involving two different off-diagonal components, or one diagonal and one off-diagonal component, vanish.

These tables follow by multiplying the displayed rational change-of-basis matrix with (eq:Ltable). They provide a contraction calculation independent of the double polarization sum.

Displayed formulas in this passage

XVI-J@1#eq:secondmethod · line 944
$K (J) = E_{L} [Q (H (a); J)] .$
XVI-J@1#eq:covdiag · line 971
$(E_{L} [H_{μμ} H_{ν ν}]) = - 3 - 3 11 - 3 - 3 33 13 - 3 - 1 13 - 1 - 3 .$
XVI-J@1#eq:covoff · line 983
$μν E_{L} [H_{μν}^{2}] 013021031 12 - 3 13 - 3 23 - 1$
unlabeled · line 934
$E_{L} [p (a)] = \frac{1}{2} i, j \sum L_{ij} \partial_{a_{i}} \partial_{a_{j}} p (a)_{a = 0} .$
unlabeled · line 940
$H (a) = i \sum a_{i} Z_{i} .$
unlabeled · line 952
$H_{00} H_{11} H_{22} H_{33} = \frac{1}{2} - 1 - 1 - 1 - 1 - 1 - 1 11 - 1 1 - 1 1 - 1 11 - 1 a_{0} a_{4} a_{7} a_{9} .$
unlabeled · line 964
$(H_{01}, H_{02}, H_{03}, H_{12}, H_{13}, H_{23}) = \frac{1}{2} (a_{1}, a_{2}, a_{3}, - a_{5}, - a_{6}, - a_{8}) .$

A second contraction method

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