Theorem: S 2 scaling residue

By Theorem thm:wpA-Sbv, $S=c(Q-i0{)}^{-1}$ on ${\mathbb{R}}^4\setminus \{0\}$ with $c=1/(4{\pi }^2)$ and $Q=\eta (z,z)$. Let $u=S^2=c^2(Q-i0{)}^{-2}\in {\mathcal{H}}_4$. Then, for every $\mu >0$ and every admissible $\varpi$ with $\varpi (0)=1$,

and $\mathcal{A}(S^2)=(E+4){\mathfrak{R}}_{\mathrm{L},\mu }(S^2)=r\delta$. The value of $r$ is independent of $\mu$ and of the radial profile of $\varpi$.

Proof. By Proposition prop:M6-dil, $r=-⟨S^2,E{\varpi }_{\mu }⟩$. In Euclidean polar coordinates $z=r\omega$ on the unit sphere $S^3$, the obstruction identity for a homogeneous density of degree $-4$ reduces $r$ to the spherical integral of the angular part of $S^2$. By Theorem thm:wpA-Sbv that angular part is $c^2(Q-i0{)}^{-2}$. Parametrizing $S^3$ by ${\omega }_0=\cos \chi$ and spatial radius $\sin \chi$, one has $Q=\cos (2\chi )$ and

The $u$-integral equals $-2i\pi$ by the residue at the interior double pole $z_m=i(e-\sqrt{e^2+1})$ of the unit-circle integrand $2i(z-1{)}^2/(z^2-2iez+1{)}^2$ after $e\downarrow 0$ (certificate: scaling_anomaly.py). Hence ${\int }_{S^3}(Q-i0{)}^{-2}\mathrm{d}\sigma =-2i{\pi }^2$ and $r=c^2(-2i{\pi }^2)$. Nonvanishing uses $c\ne 0$; with $c=1/(4{\pi }^2)$, $r=-2i{\pi }^2/(16{\pi }^4)=-i/(8{\pi }^2)$. The side $Q-i0$ fixes the sign of the angular integral; the counterfactual $Q+i0$ would conjugate it and flip $r$ (Remark rem:wpA-M6). Independence of the radial profile of $\varpi$ is the polar telescope ${{\int }_0}^{\infty }{\partial }_r{\varpi }_{\mu }(r\omega )\mathrm{d}r=-{\varpi }_{\mu }(0)=-1$.