The first commutator requires quadratic composites

Let $Q_{\epsilon }=L_{\epsilon }{G_{\mathrm{gr}}}^{\mathrm{R}}$ denote the coordinate response kernel, with its tensor indices written ${Q_{\epsilon }}^{\mu }{}_{ab}(x,y)$. Then

For a compact scalar test $w$ put $\chi ={\Delta }_{\varphi \varphi }w=-{\delta }_{m}w$, where ${\delta }_m={{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{R}}-{{\mathsf{E}}_{\mathsf{m}}}^{\mathrm{A}}$. The formal first commutator with a linear generator is

It has the explicit classical coefficient

The first summand is a local quadratic stress polynomial with smooth compact coefficient

Smoothness follows by transposing the composed retarded operators onto the smooth observation test.

Such local quadratic representatives are not smooth finite-rank kernels. More generally, a nonnegative local square $\int q{\varphi }^2$, with $q$ positive on an open set, has infinite rank on free scalar solutions: its quadratic form is positive on the infinite-dimensional space of solution restrictions to that set. A quadratic polynomial in finitely many linear smearings has finite rank.

Thus a finite-generator presentation does not automatically contain the composites demanded by (eq:comm). This is a domain warning, not the asserted impossibility of every bimodule extension. Every algebra can be placed in larger algebraic modules; leaving ${\mathcal{A}}_0$ is not a proof to the contrary. Nor does one summand of (eq:comm), without analysis of the whole coefficient, establish every possible on-shell nonrepresentation claim.