Denominator exponent and optimal T count for determinant-one unitaries

Reduce an exact word supplied by [F1] as in [F2], retaining its scalar ${\omega }^q$. When $h=0$, the matrix is monomial with unit-modulus nonzero entries, so [F5] gives $k=0$. Since its determinant is one, its diagonal exponent is even: its determinant is ${\omega }^{2q+a+4e}$. Hence the matrix uses no T gates, and neither does its conjugate by $T$, which changes its monomial exponent by $0$ or $2$. When $h=1$, every entry has modulus $1\mathrm{/}\sqrt{2}$, and [F4] gives exponent $1$ for every entry; thus $k=1$.

For $h\ge 2$, set $s=h+1>0$. All four squared moduli have exponent $s$: they are either $\mathrm{\mid }u{\mathrm{\mid }}^2$ or $1-\mathrm{\mid }u{\mathrm{\mid }}^2$, and the latter has primitive numerator whenever the former has positive least exponent, since subtracting from $(\sqrt{2}{)}^s$ cannot make its primitive numerator divisible by $\sqrt{2}$. By [F4], every entry therefore has exponent $\lceil s\mathrm{/}2\rceil$, so $k=\lceil (h+1)\mathrm{/}2\rceil$. Together with step 1.1, every case with $h\ge 1$ has $h\in \{2k-2,2k-1\}$, except that $h=0$ was treated separately.

Write $A=p(a)+p(c)\in \{0,1,2\}$. The exact reduced word has determinant ${\omega }^{2q+a+c+\sum b_i+4(h+e)}=1$, so $A+h-1$ is even. Its optimal T count is $h-1+A$ by [F3]. Conjugating by $T$ changes the endpoint exponents to $a+1$ and $c-(-1{)}^e$ and leaves every odd interior exponent unchanged; hence both endpoint parities flip. The conjugate is still reduced with the same $h$ and has optimal count $h-1+(2-A)$. If $h=2k-2$ is even, the determinant parity forces $A=1$, giving both counts $2k-2$. If $h=2k-1$ is odd, it forces $A=0$ or $2$, giving the two counts $2k-2$ and $2k$. This includes $h=1,k=1$. Together with the monomial case in step 1.1, all assertions follow.