Two-phase reduction for Clifford+T Rz synthesis up to phase

Write $\det U={\omega }^k$ by [F1], and set $V=e^{-ik\pi \mathrm{/}8}{R}^{\ast }U$. Since $\det R=1$, [F4] gives $\det V=e^{-ik\pi \mathrm{/}4}{\omega }^k=1$. The unitary $V$ is normal, so [F2] diagonalizes it. Its two eigenvalues have product $1$ and modulus $1$, hence are $e^{i\varphi },e^{-i\varphi }$ for some $\varphi$ by [F9]; consequently $\mathrm{tr}V=2\cos \varphi$ is real. Choose $\sigma \in \{1,-1\}$ such that $W=\sigma V$ has nonnegative trace. Since $-1=e^{i8\pi \mathrm{/}8}$, $W=e^{in\pi \mathrm{/}8}{R}^{\ast }U$ for an integer $n$, and $W$ is unitary with determinant $1$.

Since $W\in SU(2)$, its eigenvalues are a unit scalar and its conjugate; by [F9] write them $e^{it},e^{-it}$ and, swapping them if needed, choose $\alpha \in [0,\pi ]$ with eigenvalues $e^{i\alpha },e^{-i\alpha }$. Then $\mathrm{tr}W=2\cos \alpha \ge 0$; [F10,F11] imply $\alpha \le \pi \mathrm{/}2$. For any unit scalar $\mu$, choose its angle $t\in [0,2\pi )$ by [F9] and set $\psi =t$ if $t\le \pi$, or $\psi =t-2\pi$ otherwise. Thus $\psi \in [-\pi ,\pi ]$. By [F2] and [F3], $\mathrm{\parallel }I-W{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\mid }1-e^{i\alpha }\mathrm{\mid }$ and $\mathrm{\parallel }I-\mu W{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\max \{\mathrm{\mid }1-e^{i(\psi +\alpha )}\mathrm{\mid },\mathrm{\mid }1-e^{i(\psi -\alpha )}\mathrm{\mid }\}$. If $\psi \ge 0$, put $t=\psi +\alpha \in [\alpha ,3\pi \mathrm{/}2]$. For $t\le \pi$, $\sin (t\mathrm{/}2)\ge \sin (\alpha \mathrm{/}2)$ by [F10]; for $t\ge \pi$, sine decreases on $[\pi \mathrm{/}2,3\pi \mathrm{/}2]$, so $\sin (t\mathrm{/}2)\ge \sin (3\pi \mathrm{/}4)$. By [F11,F12], $\sin (3\pi \mathrm{/}4)=\sin (\pi \mathrm{/}4)=1\mathrm{/}\sqrt{2}$, and $\alpha \mathrm{/}2\le \pi \mathrm{/}4$ gives $\sin (\alpha \mathrm{/}2)\le 1\mathrm{/}\sqrt{2}$ by [F10]. If $\psi \le 0$, apply the same bound to $t=\mathrm{\mid }\psi \mathrm{\mid }+\alpha$ from the second eigenvalue. By [F9,F10], $\mathrm{\mid }1-e^{ix}\mathrm{\mid }=2\mathrm{\mid }\sin (x\mathrm{/}2)\mathrm{\mid }$ and these bounds imply $\mathrm{\parallel }I-W{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }I-\mu W{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$.

Left multiplication by $R$ preserves operator norm, since it preserves the norm of every vector in the maximum [F3]. As $W=e^{in\pi \mathrm{/}8}{R}^{\ast }U$, step 2.1 implies $\mathrm{\parallel }R-e^{in\pi \mathrm{/}8}U{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }I-W{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }R-\lambda U{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. Write $n=2q+r$ with $r\in \{0,1\}$. Then $e^{in\pi \mathrm{/}8}={\omega }^q{e}^{ir\pi \mathrm{/}8}$. Since ${\omega }^q$ is a free scalar Clifford phase in [F6], moving it between the explicit scalar and the representative leaves the projective T-count unchanged. Thus the two phase representatives are $e^{ir\pi \mathrm{/}8}\in \{1,e^{i\pi \mathrm{/}8}\}$, proving the reduction.

In the control-first basis [F8], direct block multiplication gives $C(\lambda U)=\mathrm{diag}(I,\lambda U)=(\mathrm{diag}(1,\lambda )\otimes I)C(U)$, in agreement with the supplier identity [F7]. The first factor is generally not a global scalar on the joint control-target system, so replacing $U$ by $\lambda U$ under control requires this relative phase correction.