Theorem: Necessary-condition obstruction to bare closure

Fix an observation test from Theorem thm:freewitness. Suppose a future admission has the following properties:

1. the leading classical symbol of $Q({\mathcal{B}}_f)$ is ${\mathcal{B}}_f{|}_{{\mathcal{C}}_{\mathrm{sol}}}$;

2. the leading differential on that symbol is the copied free action (eq:sB);

3. all further corrections have nonnegative formal powers of $\hslash$ and strictly positive powers of $\kappa$ in the differential beyond $s_0$.

Then it cannot satisfy

through coupling order ${\kappa }^2$ and quantum order ${\hslash }^0$. The same conclusion holds for the physically weighted bare dressing $\kappa Q({\mathcal{B}}_f)$ through coupling order ${\kappa }^2$.

Proof. For the unweighted insertion, the coefficient at ${\hslash }^0{\kappa }^0$ of the proposed identity is $s_0{\mathcal{B}}_f{|}_{{\mathcal{C}}_{\mathrm{sol}}}=0$. Theorem thm:freewitness contradicts it.

For the weighted insertion, the first coefficient is instead ${\hslash }^0{\kappa }^1$, and it is the same nonzero variation. Positive-order corrections to the differential cannot alter this coefficient. Nor can an order-${\kappa }^2$ tree correction alter it; its copied free variation is zero by (eq:sT).