Arbitrary-size triple-encoded projective braid approximation

By [F6], the code consists of $n$ local two-path triple factors and the full spectator space. The fusion recurrence from [F9]–[F10] gives $m_2(1)=m_2(\tau )=1$, then $m_3(1)=m_2(\tau )=1$ and $m_3(\tau )=m_2(1)+m_2(\tau )=2$. Thus the local triple total-$\tau$ factor is two-dimensional and the total-$1$ factor is one-dimensional. If $U=I$, choose the empty target list and the empty braid proves the claim. For the rest of the proof suppose $U\mathrm{\ne }I$. For $n\ge 2$, [F1] gives a finite QSD list of one-qubit gates and CNOTs; for $n=1$, use $U$ as its one-qubit list. Replace each nonadjacent two-qubit target on wires $i<j$ by adjacent SWAPs carrying wire $j$ next to $i$, the target on that adjacent pair, and the reverse SWAPs. On each computational basis vector, conjugation by this wire permutation transfers the same two target coordinates and undoing the permutation restores the others; hence the replacement is exactly the original gate. The resulting finite list contains only one-qubit and adjacent two-qubit targets.

Let $M>0$ be the number of factors in the resulting list; positivity follows from $U\mathrm{\ne }I$ in step 1.1. By [F14] choose an integer $N>\max \{2M\mathrm{/}ϵ,2\mathrm{/}{\eta }_0\}$ and put $\delta =1\mathrm{/}N$. Then $\delta <ϵ\mathrm{/}(2M)$ and $\delta <{\eta }_0\mathrm{/}2$. For a one-qubit factor $G\in U(2)$ write $\det G=a+ib$; since $G$ is unitary, $a^2+b^2=1$. If $a=-1$, set $\alpha =i$; otherwise define $x=\sqrt{(1+a)\mathrm{/}2},y=\mathrm{sgn}(b)\sqrt{(1-a)\mathrm{/}2},\alpha =x+iy,$ with $\mathrm{sgn}(0)=1$. The roots exist by [F13], and direct squaring gives ${\alpha }^2=\det G$ and $\mathrm{\mid }\alpha \mathrm{\mid }=1$. Therefore $S={\alpha }^{-1}G$ is unitary with determinant one. By [F2] and [A1], compile $S$ to a normalized braid word within $\delta$. If its Artin exponent sum is $k$, the physical action on the local total-$\tau$ code sector is within $\delta$ of $(R_1{z}^{-1}{)}^k{\alpha }^{-1}G$.

For an adjacent two-qubit factor $G$, [F4] and [A1] give one normalized six-strand braid and a code-preserving ideal extension with full ambient operator error at most $\delta$, uniformly over all spectator fusion paths. If its exponent sum is $k$, [F8] multiplies both the physical word and ideal extension by the same scalar $R_1^k$; their norm error remains $\delta$ and the code target is $R_1^{k}G$. For a one-qubit word from step 2.1, use [F11]–[F12] to decompose the register by the selected triple's total charge and spectator fusion paths. On the two-dimensional total-$\tau$ sector define the ideal block to be the target $G$ times the phase $(R_1{z}^{-1}{)}^k{\alpha }^{-1}$; on the one-dimensional total-$1$ sector define it to equal the physical braid. These blocks form a unitary preserving the code. Its full ambient norm difference from the physical word is at most $\delta$: the error is at most $\delta$ on the total-$\tau$ block by step 2.1 and zero on total charge $1$, and tensoring with spectator identities and taking an orthogonal direct sum does not increase the maximum block norm.

Each ideal extension preserves the code and its restriction to the code is the corresponding QSD factor multiplied by its recorded scalar. Their product therefore restricts to $\omega (U\otimes I_{V({\tau }^n\to c)})$ for the product phase $\omega$. The physical/ideal factor errors are each at most $\delta$, so [F5] bounds the full product error by $M\delta <ϵ\mathrm{/}2$. Applying [F7] gives logical block error and leakage amplitude at most $M\delta <ϵ$. For certified Cauchy target names [F3] guarantees that the B6 subsearch terminates; its density proof supplies no useful length or time cutoff. The only choice axiom is [A1], inherited from the density suppliers; all remaining arguments are finite-dimensional.