Six-anyon braid target extends uniformly to fusion spectators

Each triple has fusion factor $V({\tau }^3\to \tau )\cong {\mathbb{C}}^2$. Associating the $n$ triple charges first gives the code factorization $({\mathbb{C}}^2{)}^{\otimes n}\otimes V({\tau }^n\to c)$; the second factor records all spectator fusion paths. This is an orthogonal fusion-tree decomposition by [F1, F3].

Let the chosen adjacent pair of triples occupy strands $3j-2,\dots ,3j+3$, and put $p=3(j-1)$ and $q=3(n-j-1)$. Reassociate the ordered input only, retaining the prefix of $p$ strands, the selected six-strand block, and the suffix of $q$ strands in that order; no permutation of strands is used. The fusion-tree basis gives the orthogonal decomposition $V({\tau }^{3n}\to c)\cong {⨁}_{\alpha ,a,e,d}V({\tau }^p\to \alpha )\otimes V({\tau }^6\to a)\otimes V_{\alpha a}^e\otimes V({\tau }^q\to d)\otimes V_{ed}^c,$ where the sums run over simple charges and zero-dimensional summands are omitted (with ${\tau }^0=\mathbf{1}$). Every factor other than $V({\tau }^6\to a)$ records a prefix, suffix, or fusion-multiplicity spectator. By [F3], a normalized braid word $\beta$ supported on the selected six consecutive strands acts on each summand as the identity on all other factors and ${\tilde{\rho }}_a(\beta )$ on $V({\tau }^6\to a)$.

For each local total charge $a$, let $T_a$ be the target extension in [F1]. On the decomposition in step 1.2 define $\mathcal{W}$ on each summand by the identity on every factor other than $V({\tau }^6\to a)$ and by $T_a$ on that block. This is a unitary direct sum; it preserves the triple code and acts there as $U$ on the selected adjacent pair and identity on all other logical and spectator factors. Joint density [F1] gives one normalized six-strand braid word $\beta$ with $\mathrm{\parallel }{\tilde{\rho }}_a(\beta )-T_a{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \delta$ for both $a=1,\tau$.

In a summand of step 1.2, group the factors before $V({\tau }^6\to a)$ as $S_L$ and those after it as $S_R$. Choose orthonormal bases $(e_k)$ and $(f_{\mathrm{\ell }})$ of $S_L,S_R$ by [F4] and write $\xi ={\sum }_{k,\mathrm{\ell }}{e}_k\otimes {\xi }_{k\mathrm{\ell }}\otimes f_{\mathrm{\ell }}$. Then $\mathrm{\parallel }(I\otimes A\otimes I)\xi {\mathrm{\parallel }}^2={\sum }_{k,\mathrm{\ell }}\mathrm{\parallel }A{\xi }_{k\mathrm{\ell }}{\mathrm{\parallel }}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2{\sum }_{k,\mathrm{\ell }}\mathrm{\parallel }{\xi }_{k\mathrm{\ell }}{\mathrm{\parallel }}^2$, with equality on a maximizing vector in the middle factor tensored with basis vectors; hence the operator norm on each tensor summand equals $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. For an orthogonal direct sum of blocks, squared norms add, so its operator norm is the maximum block norm (equality follows by testing a maximizing vector in a maximizing block). The decomposition in step 1.2 and the bounds in step 2.1 therefore give $\mathrm{\parallel }\tilde{\rho }(\beta )-\mathcal{W}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \delta$ on the whole ambient space.

Apply [F2] to $\tilde{\rho }(\beta )$ and $\mathcal{W}$. Their compression to the code differs from the target by at most $\delta$, and $(I-P)\mathcal{W}P=0$, so the leakage amplitude is also at most $\delta$. The norm estimate is on the whole ambient space, so it remains valid for inputs with prior leakage. The physical categorical word has the common phase of The braid phase is a common scalar character, which is harmless only after phase matching or when reported projectively.