Theorem: A faithful generated Lorentz representation need not provide quantum transmission

There is a datum in the frozen class whose component order has a faithful generated representation with cone ${\mathcal{L}}_3$, but whose designated $A$-to-$B$ connection is classical and cannot transmit an unknown qubit with half-diamond error below $1/2$ using local preprocessing and receiver decoding without an additional communication resource.

The representation in this example is an independently chosen mathematical witness, not an operational extraction satisfying XXII-A's non-circularity requirement.

Proof. For every experiment, use the definite-order circuit that initializes a qubit at $A$, connects $A$'s output to $B$'s input through

and discards $B$'s output. Party $C$ receives an independent initialized qubit and has its output discarded. Definite-order circuits give valid process matrices OCB.

Preparing $|0⟩$ or $|1⟩$ at $A$ and measuring at $B$ gives ${\Delta }_e(A,B)=1$. There is no reverse edge and no edge incident to $C$. The extracted order is therefore the disjoint union of

Let $u_e$ be a countable dense sequence in the unit circle of ${\mathbb{R}}^2$. Define, only for this mathematical example,

The comparable differences are exactly

The $A_e,B_e$ pair is future-null. Within one experiment, the spatial distance from $C_e$ to $B_e$ is at least three, exceeding the temporal difference one. Across experiments, the spatial separation of events at different time levels is at least nine, again exceeding their temporal difference. Distinct events at the same time level are also incomparable. The embedding is therefore injective and faithful for ${\mathcal{L}}_3$.

The closure of the cone generated by $(1,u_e)$ is ${\mathcal{L}}_3$: density supplies every unit-circle boundary generator, and the closed unit disk is the convex hull of that circle. These differences span ${\mathbb{R}}^3$. Thus all three representation equations hold.

Now introduce a qubit message at $A$ and permit a quantum receiver output at $B$, but add no new communication channel or shared entanglement resource. Every influence reaching the receiver through (eq:classical-channel) is a classical measurement outcome. Local source processing and receiver decoding leave the resulting message-to-receiver map in measure-and-prepare form. Proposition prop:eb-floor with $d=2$ gives the error floor $1/2$.

The spatial placements and dense directions above were chosen independently of the operational data. They prove a nonimplication from representation equations and cone shape to transport capacity. They do not define an admissible data-derived selector.