Fibonacci single-qubit projective density

Put $C=B_1{B}_2^{-1}$. Using $F^2=I$, $v^2=u$, and $r+r^{-1}=-u$, direct multiplication gives $\det C=1$ and $\mathrm{tr}C=r+r^{-1}+(2-r-r^{-1})u^2=u^2$; the last equality uses $u^2+u=1$, since $(2+u)u^2=1$. Thus the eigenvalues of the unitary $C$ are $e^{i\theta }$ and $e^{-i\theta }$, with $2\cos \theta =u^2={\varphi }^{-2}$.

The 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 $e^{i\theta }+e^{-i\theta }$ would have all algebraic conjugates of absolute value at most $2$, since every such conjugate is a sum of two roots of unity. The conjugate ${\varphi }^2>2$ contradicts this. Hence $\theta \mathrm{/}(2\pi )$ is irrational.

Let $\bar{\rho }:PU(2)\to SO(3)$ be the topological group isomorphism supplied by [F4], and transport the standard smooth Lie-group structure of $SO(3)$ to $PU(2)$ along $\bar{\rho }$. This transported structure has exactly the quotient-metric topology. The covering map $\rho :SU(2)\to SO(3)$ from [F6] satisfies $\rho =\bar{\rho }\circ q$ and is a smooth local diffeomorphism; hence $q={\bar{\rho }}^{-1}\circ \rho$ is a smooth local diffeomorphism too, so $d{q}_I$ is an isomorphism on tangent spaces. Since $\theta \mathrm{/}\pi$ is irrational by step 2.1, [F2] shows that the projective powers of $C$ are dense in the smooth circle $T=\{[\mathrm{diag}(e^{it},e^{-it})]:t\in \mathbb{R}\}\subset PU(2)$ after choosing an eigenbasis; under $\bar{\rho }$ this is the ordinary rotation circle about an axis. Thus $T\subseteq H$, the closure of the projective braid image, and $H$ is closed in the transported Lie group. By [A1] and [F3], $H$ is an embedded Lie subgroup. In the eigenbasis, $C-C^{-1}=\mathrm{diag}(2i\sin \theta ,-2i\sin \theta )$ is a nonzero tangent vector to the corresponding circle in $SU(2)$, so $d{q}_I(C-C^{-1})$ is a nonzero tangent to $T$ and lies in $\mathrm{Lie}(H)$.

Write $X=dq(C-C^{-1})\in \mathrm{Lie}(H)$. Conjugation by $[B_1]$ sends $T$ to a circle with tangent ${\mathrm{Ad}}_{[B_1]}X$. If these circles had the same axis, the rotation in $SO(3)$ induced by $[B_1]$ would either preserve or reverse that axis. A reversing rotation is a half-turn, but this projective generator has angle $3\pi \mathrm{/}5\mathrm{\ne }\pi$. If it preserved the oriented axis, it would commute projectively with $C$. Choose a determinant-one lift of $[B_1]$ to $SU(2)$, using [F4]. Its conjugate of $C$ would then be either $C$ or $-C$, because the kernel of $q$ is $\{\pm I\}$. The second case contradicts invariance of trace and $\mathrm{tr}C={\varphi }^{-2}\mathrm{\ne }0$, so $B_1$ would commute with $C$, and hence with $B_2$, contrary to [F1]. Therefore the two axes are nonparallel. The real Lie algebra $\mathfrak{s}\mathfrak{o}(3)$ is the three-dimensional space of skew-symmetric $3\times 3$ matrices. Under its standard identification with ${\mathbb{R}}^3$, the Lie bracket is the cross product up to a nonzero scalar; the bracket of two nonparallel axes is nonzero and independent of both. Thus $\mathrm{Lie}(H)=\mathfrak{s}\mathfrak{o}(3)$ by [F3].

A connected coordinate neighborhood of the identity in the embedded Lie group $H$ lies in $H^0$; translating it shows that $H^0$ is open in $H$, so $\mathrm{Lie}(H^0)=\mathrm{Lie}(H)$. By step 4.1, the inclusion $H^0\hookrightarrow SO(3)$ has an isomorphism as its differential at the identity. The inverse-function theorem [F7] makes this inclusion a local diffeomorphism, so $H^0$ contains an open neighborhood of the identity and is an open subgroup of $SO(3)$. Since $SO(3)$ is connected, $H^0=SO(3)$ and $H=SO(3)$. By [F4], the projective braid image is dense in $PU(2)$. The only choice assumption is [A1], and the conclusion is projective; no efficient braid-length bound follows.