Manuscript XVI-H@1 · XVI-H.tex · proposition · af895076

Proposition: Response and completion homogeneity

Degree audit and real polarization › Displayed switch-correction coefficients · lines 659–694 · proof included or linked

XVI-H@1#prop:Chom, XVI-H@1#eq:Cscale, XVI-H@1#eq:Bresp

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Under Input input:completion,

u_{D} [s γ] = s^{2} u_{D} [γ], C_{3, D} (f, g) [s γ, ϕ] = s^{2} C_{3, D} (f, g) [γ, ϕ] .

The response polarization

B_{resp} (a, c) = \frac{1}{2} (u_{D} [a + c] - u_{D} [a] - u_{D} [c])

is symmetric and real bilinear.

Proof. The two fields $u_{D} [s γ]$ and $s^{2} u_{D} [γ]$ satisfy the same linear equation and zero initial data by (eq:Jscale). Smooth zero-data uniqueness proves their equality. Linearity of $A_{1}$ in the response proves the second identity in (eq:Cscale).

The field $B_{resp} (a, c)$ solves

L B_{resp} (a, c) = J_{D} (a, c)

with zero initial data. Each bilinearity defect solves the homogeneous zero-data equation, since the right-hand side is bilinear. Uniqueness makes every defect zero.

Displayed formulas in this passage

XVI-H@1#eq:Cscale · line 662
$u_{D} [s γ] = s^{2} u_{D} [γ], C_{3, D} (f, g) [s γ, ϕ] = s^{2} C_{3, D} (f, g) [γ, ϕ] .$
XVI-H@1#eq:Bresp · line 670
$B_{resp} (a, c) = \frac{1}{2} (u_{D} [a + c] - u_{D} [a] - u_{D} [c])$
unlabeled · line 688
$L B_{resp} (a, c) = J_{D} (a, c)$

Proposition: Response and completion homogeneity

Displayed formulas in this passage

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