Every Rz epsilon region contains an exact norm-solvable grid point

By [F2,F3], the entries of $T,T^{†}$ lie in $D$. Also $\sqrt{2}=\omega -{\omega }^3$ and $1\mathrm{/}\sqrt{2}=(\omega -{\omega }^3)\mathrm{/}2$, so every entry of $H$ lies in $D$. The ring $D$ is closed under complex conjugation because $\bar{\omega }={\omega }^{-1}={\omega }^7\in D$. Apply [F1] to $R_z(\theta )\in SU(2)$ and choose an exact word $U$ with $\det U=1$ and $\mathrm{\parallel }{R}_z(\theta )-U{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<\epsilon$. Every entry of $U$ lies in $D$ because each generator has entries in $D$ and products of matrices over a subring remain over that subring. Let the first column be $(u,t{)}^T$. It is a unit vector; its orthogonal unit vector $(-\bar{t},\bar{u}{)}^T$ is therefore the second column up to a unit scalar $\lambda$. The determinant of the resulting matrix is $\lambda$, so $\det U=1$ forces $\lambda =1$. Hence $U=\left(\begin{array}{cc}u& -\bar{t}\\ t& \bar{u}\end{array}\right)$ and $t^{†}t=1-u^{†}u$.

Applying $\bullet$ to $U^{\ast }U=I$ preserves products and fixes $I$; [F4] also makes it commute with the adjoint. Thus $U^{\bullet }$ is unitary, and its first-column norm gives $\mathrm{\mid }{u}^{\bullet }{\mathrm{\mid }}^2+\mathrm{\mid }{t}^{\bullet }{\mathrm{\mid }}^2=1$. The original first column similarly gives $\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }t{\mathrm{\mid }}^2=1$, so $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }{u}^{\bullet }\mathrm{\mid }\le 1$.

The strict approximation in step 1.1 and [F5] imply $\mathrm{Re}(e^{i\theta \mathrm{/}2}u)>1-{\epsilon }^2\mathrm{/}2$, hence the required weak inequality. Together with step 1.1 this gives a grid candidate and an exact solution of its norm equation.