Manuscript XVI-I@1 · XVI-I.tex · proposition · a7046a8d

Proposition: Diagonal polynomial and its derivatives

Classical kernel, domains, and exact matching › The classical kernel input · lines 397–432 · proof included or linked

XVI-I@1#prop:Ft, XVI-I@1#eq:Ft, XVI-I@1#eq:Dh

Define

F_{t} (h, ϕ) = ⟨ t_{D}, h \otimes h \otimes ϕ ⟩ .

Then, on the inherited real smooth completion domain,

F_{t} (h, ϕ) = C_{3, D} (f, g) [h, ϕ] .

For real smooth tensor directions $a, c$ ,

D_{h} F_{t} [a] = 2 ⟨ t_{D}, a \otimes h \otimes ϕ ⟩, D_{h}^{2} F_{t} [a, c] = 2 ⟨ t_{D}, a \otimes c \otimes ϕ ⟩ .

There is no additional factorial in $F_{t}$ .

Proof. Degree-two homogeneity of the diagonal completion gives

O_{pol} (h, h; ϕ) = \frac{1}{2} (C_{3, D} [2 h, ϕ] - 2 C_{3, D} [h, ϕ]) = C_{3, D} [h, ϕ] .

The kernel representation proves the first assertion. Differentiating the quadratic polynomial and using tensor-slot symmetry proves (eq:Dh).

Displayed formulas in this passage

XVI-I@1#eq:Ft · line 400
$F_{t} (h, ϕ) = ⟨ t_{D}, h \otimes h \otimes ϕ ⟩ .$
XVI-I@1#eq:Dh · line 410
$D_{h} F_{t} [a] = 2 ⟨ t_{D}, a \otimes h \otimes ϕ ⟩, D_{h}^{2} F_{t} [a, c] = 2 ⟨ t_{D}, a \otimes c \otimes ϕ ⟩ .$
unlabeled · line 406
$F_{t} (h, ϕ) = C_{3, D} (f, g) [h, ϕ] .$
unlabeled · line 423
$O_{pol} (h, h; ϕ) = \frac{1}{2} (C_{3, D} [2 h, ϕ] - 2 C_{3, D} [h, ϕ]) = C_{3, D} [h, ϕ] .$