Exact eight-dimensional Fibonacci B6 density

The fusion rule and $F\mathrm{/}R$ convention in [F8], together with recurrence [F7], give $\dim V({\tau }^3\to 1)=1$, $\dim V({\tau }^3\to \tau )=2$, and $\dim V({\tau }^4\to \tau )=3$. The eight paths in ${\mathcal{P}}_{\tau }$ thus decompose by the first-three-charge label: charge $1$ contributes $1\cdot 2=2$ dimensions, and charge $\tau$ contributes $V({\tau }^3\to \tau )\otimes V(\tau \otimes {\tau }^3\to \tau )$ of dimension $2\cdot 3=6$. The operator $C:=B_1{B}_2^{-1}$ is identity on the first summand. On the two-dimensional first-three charge-$\tau$ factor it has determinant $1$ and trace ${\varphi }^{-2}$, so the full spectrum of $C$ is $1$ with multiplicity $2$ and $e^{i\theta },e^{-i\theta }$ each with multiplicity $3$, where $2\cos \theta ={\varphi }^{-2}$.

The minimal polynomial of ${\varphi }^{-2}$ is $x^2-3x+1$, whose other root is ${\varphi }^2>2$. If $e^{i\theta }$ were a root of unity, every algebraic conjugate of its trace $e^{i\theta }+e^{-i\theta }$ would be a sum of two roots of unity and have absolute value at most $2$, contradicting the conjugate ${\varphi }^2$. Thus $\theta \mathrm{/}(2\pi )$ is irrational.

By [F3], powers of $C$ are dense in the circle $T=\mathrm{diag}(I_2,e^{it}{I}_3,e^{-it}{I}_3)$ after choosing an eigenbasis. Let $H$ be the closure of the normalized B6 image in $U(8)$. Then $T\subseteq H$, so the nonzero tangent $X=C-C^{-1}$ lies in $\mathrm{Lie}(H)$.

Starting with $Z_0=X$, construct a list as follows, all over ${\mathbb{F}}_{31}$ for the independence tests: repeatedly scan ${\mathrm{Ad}}_i(Z_j)=B_i{Z}_j{B}_i^{-1}$ in current-list order $j$ and generator order $i=1,\dots ,5$, appending a word exactly when its specialized row is independent of previous rows. Using the resulting list as a snapshot, scan commutators $[Z_j,Z_i]$ in lexicographic order $j>i$, again appending exactly the independent rows. Repeat full rounds until none is appended. Each accepted word is an adjoint or bracket applied to earlier Lie algebra elements, so it lifts to $\mathrm{Lie}(H)$. Unitary conjugation and commutators preserve traceless skew-Hermitian matrices.

[F2] makes reduction of every coefficient and denominator in these words well defined. The exact integer-arithmetic certificate research/scaffold-audit-2026-10-16/topological-quantum-computing/b6-su8-lie-rank-certificate.py selects 63 words by the procedure of step 4.1 (the script labels them $Y_0,\dots ,Y_{62}$; this proof renames them $Z_0,\dots ,Z_{62}$) and returns determinant $22$ for the $63\times 63$ minor formed from the first $63$ row-major coordinates of these matrices. A zero algebraic determinant would map to zero in ${\mathbb{F}}_{31}$; since $22\mathrm{\ne }0$, the exact determinant is nonzero. The omitted coordinate is the final diagonal entry, determined by trace zero. Thus the selected traceless skew-Hermitian matrices are linearly independent over $\mathbb{C}$.

Each $B_i$ has determinant $r^{8-3}=r^5$ by [F1], hence $\det (H)\subseteq ⟨r^5⟩$ is finite. Since $H^0$ is connected, its determinant is $1$, so $H^0\subseteq SU(8)$. A connected coordinate neighborhood of the identity in the Lie group $H$ lies in $H^0$; translating it shows that $H^0$ is open in $H$, so $\mathrm{Lie}(H^0)=\mathrm{Lie}(H)\subseteq \mathfrak{s}\mathfrak{u}(8)$. By [F5], $\mathfrak{s}\mathfrak{u}(8)$ has real dimension $63$. Step 5.1 supplies $63$ independent elements of $\mathrm{Lie}(H)$, so $\mathrm{Lie}(H)=\mathfrak{s}\mathfrak{u}(8)$. Thus the inclusion $H^0\hookrightarrow SU(8)$ has an isomorphism as its differential at the identity, and [F9] makes it a local diffeomorphism. Its image contains an open neighborhood of the identity and is an open subgroup of $SU(8)$. By [F6], $SU(8)$ is connected, so this subgroup equals $SU(8)$. Hence the B6 closure contains $SU(8)$.