Spaced-support moment criterion for order-one AE correction

The two codewords are orthogonal because their supports are disjoint. For each $r=0,1,2$, the operator $J_z^r$ is diagonal in the magnetic basis, so its cross matrix element between the codewords is zero and its two diagonal matrix elements agree by the given moment equality. Thus the code satisfies the pair-product conditions for $\{I,J_z\}$.

Define the linear flip $S:V_J\otimes V_1\to V_1\otimes V_J$ on the product basis by $S(e_{J,m}\otimes e_{1,q})=e_{1,q}\otimes e_{J,m}$. It preserves the inner product because it sends an orthonormal basis bijectively to an orthonormal basis. Applying both sides to each product-basis vector gives $S{J}_a^{\mathrm{t}\mathrm{o}\mathrm{t},(J,1)}=J_a^{\mathrm{t}\mathrm{o}\mathrm{t},(1,J)}S$ for $a=z,+,-$. For any allowed output spin $J^{\mathrm{\prime }}$, the joint conditions of weight $J^{\mathrm{\prime }}$ and annihilation by the raising operator define a one-dimensional line in either ordered product: by F5 and the ladder formula, only the unique spin-$J^{\mathrm{\prime }}$ summand contributes to that joint space. Thus the flip sends the recorded unit highest-weight vector in order $(J,1)$ to ${\lambda }_{J^{\mathrm{\prime }}}$ times the unit highest-weight vector in order $(1,J)$, with $\mathrm{\mid }{\lambda }_{J^{\mathrm{\prime }}}\mathrm{\mid }=1$. Use the source's coefficient phases for the latter order. Commuting $S$ through every power of the lowering operator, with the identical normalization in F2, gives the same scalar for every lowered vector. Comparing product-basis coefficients yields $C_{J,m;1,q}^{J^{\mathrm{\prime }},M}={\lambda }_{J^{\mathrm{\prime }}}{C}_{1,q;J,m}^{J^{\mathrm{\prime }},M}$ for all $m,q,M$. In particular the proportionality is independent of both magnetic labels and component index.

For rank zero, F3 permits only the input block as target. On that block the rank-zero tensor equation says that its sole component commutes with $J_z$ and $J_{\pm }$. The distinct magnetic eigenvalues of $J_z$ force it to be diagonal, and commuting with $J_{+}$ forces adjacent diagonal entries to be equal because its ladder coefficient is nonzero between consecutive magnetic labels. Thus it is a scalar multiple of the identity on $V_J$, including the one-dimensional case. Extended to $H$, it is a scalar multiple of the input-block projection $Q={\iota }_J{\pi }_J$, and $QP=PQ=P$.

The spacing and step 1.1 meet the hypotheses of the explicit-operator criterion. Consequently the pair-product condition holds for the coherent Eq. (3) operators on the valid selected target branches, extended by zero outside the input block. Removing branches preserves that condition.

Fix a selected target block and a rank-one tensor family. By F3 each component is a common reduced scalar times the coherent component with coefficients $C_{J,m;1,q}^{J^{\mathrm{\prime }},m+q}$. If $J^{\mathrm{\prime }}$ is absent from the coupling list the family is zero. Otherwise step 1.2 converts these coefficients to the source's coupling order by one nonzero scalar. The nonzero component identification in F4 then makes each tensor component a scalar multiple of the corresponding Eq. (3) component; the scalar may depend on $q$, which does not affect membership in the linear span. Zero components need no additional generator.

Each such extended operator $K_q$ also has scalar compression $P{K}_{q}P$. If its target spin differs from $J$, this compression is zero because the target block is orthogonal to the code. In the same block, a component with $q=\pm 1$ has zero compression: a nonzero matrix element between supported magnetic labels would require their difference to be $q$, contrary to spacing. The remaining component is $L_0$, proportional to $J_z$, whose compression is scalar by step 1.1. Taking adjoints also gives scalar $P{K}_q^{\ast }P$. Since $P{I}_{H}P=P$, all pair products involving the ambient identity and these components are scalar, so adjoining $I_H$ to the family in step 2.1 preserves the pair-product condition.

The ambient identity and the valid Eq. (3) components satisfy the pair-product condition by step 3.1. Adjoining $Q$ preserves it: $P{Q}^{\ast }QP=P$, and for each operator $K$ of that family, $P{Q}^{\ast }KP=PKP$ and $P{K}^{\ast }QP=P{K}^{\ast }P$, which are scalar by pairing $K$ with $I_H$. Steps 2.2 and 1.3 place every generator of the stated tensor error span inside the span of this enlarged family. Sesquilinearity therefore gives $P{E}^{\ast }FP=c_{E,F}P$ for every $E,F$ in the full stated span.