Theorem: Background gauge identity of the one-tree observable

Let $\xi ,\zeta \in {\mathfrak{g}}_U$, and let the classical backgrounds transform by the complete two-jet law

with $\hat{\varphi }$ and $\hat{h}$ held fixed. Then, modulo $\mathfrak{J}$ and modulo ${\kappa }^3$,

In particular the one-tree observable is invariant through ${\kappa }^1$; the tree ${\mathsf{T}}_{\mathrm{tree}}$ contributes nothing to the variation at any order; and the ${\kappa }^2$ defect is bilinear in $(\hat{h},\hat{\varphi })$, independent of $h_{\mathrm{cl}},k_{\mathrm{cl}}$ and of $\zeta$.

Proof. ${\mathsf{T}}_{\mathrm{tree}}$ depends on $f,g,\epsilon ,\hat{\varphi }$ only, so it is invariant. Write $h=h_{\mathrm{cl}}+\hat{h}$ and ${k^{\prime }}_{\mathrm{full}}=k_{\mathrm{cl}}+{\mathcal{L}}_{\xi }h+K\zeta +\frac{1}{2}{\mathcal{L}}_{\xi }K\xi$. X-C's fixed-regulator identity, in X-D's explicit form, is a polynomial identity in the smooth tensor variables $(h,k)$:

$X_{\xi }={\xi }^{\mu }{\partial }_{\mu }$, $T_2=X_{\zeta }+\frac{1}{2}{X_{\xi }}^2$, with all functions inside $P_m$ compactly supported. Substituting $\hat{h}$ for the quantum part of $h$ and Wick ordering preserves the identity of kernels, which are polynomial in $h$ with distributional coefficients. By Lemma lem:ideal(i) the right sides are in $\mathfrak{J}$. This gives invariance at ${\kappa }^1$, since ${\hat{A}}_1[h^{\prime }]-{\hat{A}}_1[h]=\hat{\varphi }(d_1[K\xi ])$, and at ${\kappa }^2$ for the full transform $(h^{\prime },{k^{\prime }}_{\mathrm{full}})$.

The frozen transform has ${k^{\prime }}_{\mathrm{cl}}={k^{\prime }}_{\mathrm{full}}-{\mathcal{L}}_{\xi }\hat{h}$. By Lemma lem:affine, $d_2$ is affine in $k$ with slope $d_1$, so $d_2[h^{\prime },{k^{\prime }}_{\mathrm{cl}}]=d_2[h^{\prime },{k^{\prime }}_{\mathrm{full}}]-d_1[{\mathcal{L}}_{\xi }\hat{h}]$. Wick ordering and (eq:one-tree) give (eq:defect).