Fixed scalar extension u μ R (o70-sync)

. One analytic finite-part scheme:

Frozen: $\eta =\mathrm{diag}(+---)$; $Q-i0$; principal logarithm; $\mu >0$ (inverse length); finite part $=$ Laurent constant term; no additional finite $c\delta$. Proof identifier: lem:analytic_FP_u_mu_R_o70 (equivalent FP prescription to o68/o69; constructive holomorphy proof added).

(distributional analytic continuation).

1. $u_{\zeta }$ admits a distributional meromorphic continuation near $\zeta =0$ against arbitrary compact smooth tests, with a simple pole.

2. Holomorphy on tests supported in $U$: the pairing equals $⟨(Q-i0{)}^{-2},f⟩$ at $\zeta =0$.

3. The pole part is local: $\frac{rf(0)}{\zeta }$ with $r=-i{\pi }^2$, derived below — not by naming analytic regularization.

4. ${u_{\mu }}^R$ is Lorentz covariant and even under $z\mapsto -z$.

5. $\mathrm{sd}({u_{\mu }}^R)\le 4$; residual $c\delta$ freedom frozen by excluding any added contact (XVI-J prop:extension = freedom only).

integrability near the cone. Near a nonzero null point with local coordinate $t=Q$, ordinary absolute local integrability of $(Q-i0{)}^{-2+\zeta }$ requires ${{\int }_{-\epsilon }}^{\epsilon }|t{|}^{-2+\Re \zeta }dt<\infty$, hence $\Re \zeta >1$. At $\zeta =0$ the $U$-supported pairing is obtained by boundary-value / analytic-family continuation, not by an ordinary integral.

{Proof that ${\varphi }_{-1+\zeta }$ is holomorphic near $\zeta =0$.} Do not argue from the zero of $4\zeta (\zeta -1)$ alone.

1. Let $F(\zeta )=(Q-i0{)}^{\zeta }$. This is a holomorphic family of distributions near $\zeta =0$: instantiate Dütsch–Fredenhagen–Keller–Rejzner, arXiv:1311.5424, Definition 4.2 (analytic regularization) and 4.1–4.2 (complex powers / meromorphic two-point families), with hypotheses $Q-i0$, principal logarithm, tests in $\mathcal{D}({\mathbb{R}}^4)$, and Gate 0 normalization. (Classical support: Gelfand–Shilov / Hörmander complex powers $(x\pm i0{)}^{\lambda }$ are entire in $\lambda$.)

2. Const-coeff differentiation under the pairing gives the recurrence $\square {\varphi }_{\lambda }=4\lambda (\lambda +1){\varphi }_{\lambda -1}$, hence

   $$\square F(\zeta )=4\zeta (\zeta +1){\varphi }_{-1+\zeta }.$$

3. $F(0)=(Q-i0{)}^0=1$, so $\square F(0)=0$. Holomorphy of $F$ yields $\square F(\zeta )=\zeta G(\zeta )$ with $G$ holomorphic in ${\mathcal{D}}^{\prime }$.

4. Therefore

   $${\varphi }_{-1+\zeta }=\frac{\square F(\zeta )}{4\zeta (\zeta +1)}=\frac{G(\zeta )}{4(\zeta +1)}$$

   is holomorphic near $\zeta =0$.