Two-qubit determinant obstruction to controlled T

By [F6], the two one-wire embeddings are $H\otimes I_2,I_2\otimes H$ and the corresponding tensor embeddings for $T,T^{†},S,S^{†},X,Y,Z$ and scalar phases. Direct diagonal/permutation calculations give $\det (H\otimes I_2)=\det (I_2\otimes H)=1$, $\det (T\otimes I_2)=\det (I_2\otimes T)={\omega }^2$, and $\det (T^{†}\otimes I_2)=\det (I_2\otimes T^{†})={\omega }^{-2}$. The same calculation gives determinant ${\omega }^4$ for embedded $S,S^{†}$, determinant $1$ for embedded $X,Y,Z$, and ${\omega }^{4j}$ for the embedded scalar phase ${\omega }^j{I}_2$. CNOT is a single transposition of $\mathrm{\mid }10⟩$ and $\mathrm{\mid }11⟩$, so its determinant is $-1={\omega }^4$. Thus every primitive determinant is an even power of $\omega$; [F4,F7] make the determinant of a finite circuit product an even power as well.

The target has determinant $\det C(T)=\omega$. Exact equality contradicts step 1.1. If a two-wire circuit $V$ equals $\lambda C(T)$ for a unit scalar $\lambda$ as in [F8], then its $(0,0)$ entry is $\lambda$, so [F1] gives $\lambda \in D$; unitarity gives $\lambda \bar{\lambda }=1$, so [F2] gives $\lambda ={\omega }^j$. By [F5], $\det V={\lambda }^4\det C(T)={\omega }^{4j+1}$, an odd power, again contradicting step 1.1.