Exact five-dimensional Fibonacci B6 density

Write $u={\varphi }^{-1}$ and $r=R_{\tau }\mathrm{/}{R}_1=e^{-3\pi i\mathrm{/}5}$ by [F8]. The fusion-tree recurrence [F7] gives $\dim V({\tau }^2\to 1)=\dim V({\tau }^2\to \tau )=1$, $\dim V({\tau }^3\to 1)=1$, $\dim V({\tau }^3\to \tau )=2$, and $\dim V({\tau }^4\to 1)=2$. The five paths in ${\mathcal{P}}_1$ therefore decompose by the charge of the first three strands as one dimension from charge $1$, and $2\cdot 2=4$ dimensions from charge $\tau$: the latter are two copies of the same two-dimensional three-strand block. On the first summand $C:=B_1{B}_2^{-1}$ is $1$. On each two-dimensional summand, $B_1=\mathrm{diag}(1,r)$ and $B_2=F{B}_{1}F$ with $F^2=I$, so direct multiplication gives $\det C=1$ and $\mathrm{tr}C=r+r^{-1}+(2-r-r^{-1})u^2=u^2={\varphi }^{-2}$; the last equality uses $u^2+u=1$, since $(2+u)u^2=1$. Consequently the full spectrum is $1,e^{i\theta },e^{-i\theta }$, with the last two eigenvalues each repeated twice and $2\cos \theta ={\varphi }^{-2}$.

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

Let $H$ be the closure of the normalized B6 image in $U(5)$. The powers of $C$ contain a dense subgroup of the circle $\mathrm{diag}(1,e^{it}{I}_2,e^{-it}{I}_2)$, by [F3] and step 2.1. Therefore the nonzero tangent $X=C-C^{-1}$ belongs to $\mathrm{Lie}(H)$.

Define $Y_0=X$ and, with ${\mathrm{Ad}}_i(Y)=B_{i}Y{B}_i^{-1}$, define the remaining $23$ words in order by $Y_1={\mathrm{Ad}}_1{Y}_0,Y_2={\mathrm{Ad}}_2{Y}_0,Y_3={\mathrm{Ad}}_3{Y}_0,Y_4=[Y_3,Y_0],Y_5=[Y_3,Y_1],Y_6=[Y_3,Y_2],Y_7={\mathrm{Ad}}_3{Y}_1,Y_8={\mathrm{Ad}}_3{Y}_2,Y_9={\mathrm{Ad}}_2{Y}_3,Y_{10}={\mathrm{Ad}}_4{Y}_3,$ $Y_{11}={\mathrm{Ad}}_2{Y}_4,Y_{12}={\mathrm{Ad}}_4{Y}_4,Y_{13}={\mathrm{Ad}}_4{Y}_5,Y_{14}={\mathrm{Ad}}_4{Y}_6,Y_{15}=[Y_{10},Y_3],Y_{16}=[Y_{10},Y_4],Y_{17}=[Y_{10},Y_5],Y_{18}=[Y_{10},Y_6],Y_{19}=[Y_{10},Y_7],Y_{20}=[Y_{10},Y_8],Y_{21}=[Y_{10},Y_9],Y_{22}=[Y_{12},Y_0],Y_{23}=[Y_{12},Y_1]\mathrm{.}$ Conjugation by an element of $H$ preserves $\mathrm{Lie}(H)$, and [F4] makes that Lie algebra bracket closed, so every $Y_j$ lies in it. Each is traceless and skew-Hermitian: this holds for $X$ because $C$ is unitary with spectrum invariant under inversion, and is preserved by unitary conjugation and commutators.

By [F2], every coefficient of the $Y_j$ matrices lies in $\mathbb{Z}[r,\sqrt{\varphi }]$ with all denominators units. Reduce coefficients modulo $31$ using $r\mapsto 15$, $r^{-1}\mapsto 29$, $\varphi \mapsto 19$, $\sqrt{\varphi }\mapsto 9$; for conjugated coefficients first replace $r$ by $r^{-1}$ in the number field. The exact integer-arithmetic certificate research/scaffold-audit-2026-10-16/topological-quantum-computing/b6-lie-rank-certificate.py forms the $24\times 24$ matrix whose rows are the first $24$ row-major coordinates of $Y_0,\dots ,Y_{23}$ and returns determinant $17$ in ${\mathbb{F}}_{31}$. Since the reduction is a ring homomorphism, a zero algebraic determinant would reduce to zero; the determinant is therefore nonzero over $\mathbb{C}$. Tracelessness lets the omitted coordinate be recovered from the other $24$, so the $Y_j$ are linearly independent.

The determinant of each $B_i$ is $r^3$ by [F1], so $\det (H)\subseteq ⟨r^3⟩$ is finite. Since $H^0$ is connected, its determinant is $1$, and hence $H^0\subseteq SU(5)$. 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 $\mathrm{Lie}(H^0)=\mathrm{Lie}(H)$. Step 4.1 and [F5] give $\mathrm{Lie}(H^0)=\mathfrak{s}\mathfrak{u}(5)$. Thus the inclusion $H^0\hookrightarrow SU(5)$ has an isomorphism as its differential at the identity, and [F9] makes it a local diffeomorphism. Its image is an open subgroup of $SU(5)$. By [F6], $SU(5)$ is connected, so this subgroup is all of $SU(5)$. Therefore the closure contains $SU(5)$.