Manuscript I-B-VP-1@1 · I-B-VP-1-v1.tex · proposition · 6d700be9

Proposition: Euler identity and scale dependence

Dilation residues and compatibility of presentations › The unavoidable scalar residue · lines 1071–1116 · proof included or linked

I-B-VP-1@1#prop:Euler, I-B-VP-1@1#eq:EulerU, I-B-VP-1@1#eq:scaleshift

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The generating finite part obeys

(E + 2) U_{L} = σ_{α} δ_{W}, L \frac{\partial U _{L}}{\partial L} = - σ_{α} δ_{W},

and hence

U_{L^{'}} - U_{L} = - σ_{α} lo g (L^{'} / L) δ_{W} .

For the transverse dilation $x \mapsto λ x$ ,

U_{L} (λ x) = λ^{- 2} (U_{L} (x) + σ_{α} lo g λ δ_{W})

as a pullback identity of generalized functions. No extension of $r^{- 2}$ differing from $U_{L}$ by a multiple of $δ_{W}$ can be strictly homogeneous.

Proof. For a test $f$ ,

⟨ (E + 2) U_{L}, f ⟩ = - ⟨ U_{L}, E f ⟩ .

The subtraction term vanishes because $E f (y, 0) = 0$ . Radial integration gives

ε ↓ 0 lim α \int_{r > ε} r^{- 2} E f d^{2} x = - σ_{α} f (y, 0)

after the longitudinal integration. This proves the first identity. Differentiating the defining subtraction proves the second, and integration in $L$ gives (eq:scaleshift). Integrating the Euler equation along dilations gives the pullback formula. Since $(E + 2) δ_{W} = 0$ , adding a scalar contact cannot remove the residue.

Displayed formulas in this passage

I-B-VP-1@1#eq:EulerU · line 1074
$(E + 2) U_{L} = σ_{α} δ_{W}, L \frac{\partial U _{L}}{\partial L} = - σ_{α} δ_{W},$
I-B-VP-1@1#eq:scaleshift · line 1081
$U_{L^{'}} - U_{L} = - σ_{α} lo g (L^{'} / L) δ_{W} .$
unlabeled · line 1086
$U_{L} (λ x) = λ^{- 2} (U_{L} (x) + σ_{α} lo g λ δ_{W})$
unlabeled · line 1099
$⟨ (E + 2) U_{L}, f ⟩ = - ⟨ U_{L}, E f ⟩ .$
unlabeled · line 1104
$ε ↓ 0 lim α \int_{r > ε} r^{- 2} E f d^{2} x = - σ_{α} f (y, 0)$

Proposition: Euler identity and scale dependence

Displayed formulas in this passage

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