Clifford+T Is Dense Projectively on One Qubit

From [F1] and [F9], $X,Z$ are Hermitian and traceless, and $X^2=Z^2=I$, $XZ=-iY$, $ZX=iY$. Hence $H^{\ast }=H$, $\mathrm{tr}H=0$ and $H^2=(X^2+XZ+ZX+Z^2)\mathrm{/}2=I$. Directly, $H=2^{-1\mathrm{/}2}\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)$ has determinant $-1$, so $h=iH$ is unitary with determinant $1$. By [F6], $t=e^{-i\pi \mathrm{/}8}T=\mathrm{diag}(e^{-i\pi \mathrm{/}8},e^{i\pi \mathrm{/}8})$ is also in $SU(2)$. Put $V=ht$, $s=\sin (\pi \mathrm{/}8)$ and $c=\cos (\pi \mathrm{/}8)$. Multiplying the displayed Pauli matrices gives $V=(s\mathrm{/}\sqrt{2})I+i(cX-sY+cZ)\mathrm{/}\sqrt{2}$; hence $\mathrm{tr}V=\sqrt{2}s=:x$. By [F12], $0=\cos (\pi \mathrm{/}2)=2{\cos }^2(\pi \mathrm{/}4)-1$. Since $0<\pi \mathrm{/}4<\pi \mathrm{/}2$ and cosine is strictly decreasing by [F11], $\cos (\pi \mathrm{/}4)>\cos (\pi \mathrm{/}2)=0$, hence $\cos (\pi \mathrm{/}4)=1\mathrm{/}\sqrt{2}=\sqrt{2}\mathrm{/}2$. Applying the other double-angle form in [F12] at $u=\pi \mathrm{/}8$ gives $\sqrt{2}\mathrm{/}2=1-2s^2$. Therefore $x^2=2s^2=1-\sqrt{2}\mathrm{/}2$.

The number $x$ is a root of the monic rational quartic $P(X)=X^4-2X^2+1\mathrm{/}2$. It has no rational root: if $r\in \mathbb{Q}$ were a root, then $r^2$ would solve $z^2-2z+1\mathrm{/}2=0$, whose roots $1\pm \sqrt{2}\mathrm{/}2$ are irrational. Indeed, if $\sqrt{2}=a\mathrm{/}b$ in lowest terms, then $a^2=2b^2$ is even, so $a$ is even because an odd integer has odd square. Substituting $a=2c$ gives $b^2=2c^2$ even, so $b$ is even too, a contradiction. If $P$ factored over $\mathbb{Q}$, its lack of a rational root forces a factorization into two monic rational quadratics; write $P=(X^2+aX+b)(X^2-aX+d)$. Comparing coefficients gives $a(d-b)=0$, $b+d-a^2=-2$, and $bd=1\mathrm{/}2$. If $a=0$, then $b,d$ would be rational roots of $Z^2+2Z+1\mathrm{/}2$, whose discriminant is $2$, not a rational square. If $d=b$, then $b^2=1\mathrm{/}2$, also impossible in $\mathbb{Q}$. Thus $P$ is irreducible and is the monic minimal polynomial of $x$. Every power of $x$ reduces using $P(x)=0$ to a rational linear combination of $1,x,x^2,x^3$, while these four powers are linearly independent over $\mathbb{Q}$ by irreducibility of $P$. Hence $[\mathbb{Q}(x):\mathbb{Q}]=4$, so $\mathbb{Q}(x)$ is a number field by [F3]. Finally [F2] shows $x$ is not an algebraic integer because $P\mathrm{\notin }\mathbb{Z}[X]$.

Write the eigenvalues of $V$ as $e^{i\alpha },e^{-i\alpha }$, so $x=2\cos \alpha$. If $\alpha \mathrm{/}\pi$ were rational, these eigenvalues would be roots of unity. For each positive integer $m$, the polynomials $S_0=2$, $S_1=X$, $S_{m+1}=X{S}_m-S_{m-1}$ are monic integer polynomials for $m\ge 1$ and satisfy $S_m(\lambda +{\lambda }^{-1})={\lambda }^m+{\lambda }^{-m}$. For a root of unity of order $m$, this makes $x$ a root of the monic integer polynomial $S_m(X)-2$, contradicting step 2.1. Hence $\alpha \mathrm{/}\pi$ is irrational.

By [F7], $V\in SU(2)$ has eigenvalues $e^{i\alpha },e^{-i\alpha }$. Step 3.1 makes $\alpha \mathrm{/}(2\pi )$ irrational, so [F4] makes its integer multiples dense modulo one. Diagonalizing $V$ shows its powers are dense in the torus $\{Q\mathrm{diag}(e^{it},e^{-it})Q^{\ast }:t\in \mathbb{R}\}$, because the parametrization is continuous and $2\pi$-periodic by [F6,F8]. The Pauli-vector part of the explicit matrix in step 1.1 is proportional to $v=(c,-s,c)$, where $c=\cos (\pi \mathrm{/}8)$ and $s=\sin (\pi \mathrm{/}8)$. Since $x=\sqrt{2}s$ and $x^2=1-\sqrt{2}\mathrm{/}2$, the Pythagorean identity gives $c^2=1-s^2=1-x^2\mathrm{/}2>1\mathrm{/}2$, so both the $Z$ coordinate $c$ and the $XY$ component $(c,-s)$ of $v$ are nonzero. The subgroup generated by $h,t$ contains $tV{t}^{-1}$, so its closure also contains the conjugate torus. Conjugation by $t=\exp (-i\pi Z\mathrm{/}8)$ rotates Pauli vectors about the $Z$ axis by $\pi \mathrm{/}4$: direct multiplication gives $tX{t}^{\ast }=\cos (\pi \mathrm{/}4)X+\sin (\pi \mathrm{/}4)Y$ and $tY{t}^{\ast }=-\sin (\pi \mathrm{/}4)X+\cos (\pi \mathrm{/}4)Y$, while $tZ{t}^{\ast }=Z$. If the rotated vector were parallel to $v$, the nonzero $Z$ coordinate would force the scalar to be $1$, so the nonzero vector $(c,-s)$ would be fixed by the planar rotation $R_{\pi \mathrm{/}4}$. But $\det (R_{\pi \mathrm{/}4}-I)=2(1-\cos (\pi \mathrm{/}4))>0$ by [F11], so no nonzero planar vector is fixed. Hence the two axes are nonparallel. Normalize them as $a=v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$ and $b=R_z(\pi \mathrm{/}4)v\mathrm{/}\mathrm{\mid }v\mathrm{\mid }$; they are nonparallel unit vectors because planar rotations preserve Euclidean length. Thus the closure contains the one-parameter subgroups about two nonparallel unit axes.

By [F5] the closure of the subgroup generated by $h,t$ is $SU(2)$. The quotient map from $SU(2)$ to $PU(2)$ is continuous by the quotient topology. It is onto: for any $Q\in U(2)$, unitarity gives $\mathrm{\mid }\det Q\mathrm{\mid }=1$. By [F10] and Euler's formula [F6], write $\det Q=e^{i\theta }$ and set $\lambda =e^{-i\theta \mathrm{/}2}$. Then $\mathrm{\mid }\lambda \mathrm{\mid }=1$, ${\lambda }^2=(\det Q{)}^{-1}$, and $\lambda Q\in SU(2)$ has the same projective class as $Q$. A continuous image of a dense subset is dense in the image, so the projective classes of words in $h,t$ are dense in $PU(2)$. Finally $h=iH$ and $t=e^{-i\pi \mathrm{/}8}T$, so these classes equal those of $H,T$.