Theorem: Algebraic consistency criterion

There exists a complex-linear map ${\mathsf{T}}_2:{\mathcal{D}}_2\to {\mathcal{P}}_{\mathrm{adm}}{\otimes }_{\mathbb{C}}\mathbb{C}[\hslash ]/({\hslash }^4)$ satisfying all prescribed values, classical multiplication, symmetry, and the binary involution identities if and only if the following finite-relation condition holds:

for every finite $I_0\subset I$ and every complex choice of coefficients.

This is an algebraic theorem. It does not assert fixed-support continuity or identify a distributional time-ordering construction.

Proof. Write ${\mathsf{T}}_2=A+X$. Symmetry is equivalent to $X(\sigma d)=X(d)$, so $X$ factors through $Q$. The binary involution is equivalent to

For prescribed tensors this requires the assignments

Any linear map must respect their finite relations, proving necessity of (eq:relation-test).

Conversely, that condition defines a complex-linear map $x$ on

The space $U$ is conjugation-stable, and the assignments give $x(\gamma u)=-x(u{)}^{\ast }$. Set $y=ix$. Then

On the real fixed-point space $U_{\mathbb{R}}=\{u\in U:\gamma u=u\}$, the map $y$ takes values in the real fixed-point space of $\hslash {\mathcal{P}}_{\mathrm{adm}}{\otimes }_{\mathbb{C}}\mathbb{C}[\hslash ]/({\hslash }^4)$.

Choose a real vector-space complement of $U_{\mathbb{R}}$ in $Q_{\mathbb{R}}$, extend $y$ by zero there, and complexify. Every $q\in Q$ has the decomposition

into real fixed-point vectors. The extended $y$ therefore remains conjugation-equivariant. Define $X=-iy$, pulled back to ${\mathcal{D}}_2$, and ${\mathsf{T}}_2=A+X$.

The range of $X$ lies in $\hslash {\mathcal{P}}_{\mathrm{adm}}{\otimes }_{\mathbb{C}}\mathbb{C}[\hslash ]/({\hslash }^4)$, so the classical coefficient is unchanged. The construction gives symmetry, prescribed values, and (eq:X-involution). All outputs are finite linear combinations of admitted coefficient functionals. Linearity of the carrier conditions preserves their required descendants.