Determinant-one Clifford+T words are dense in SU(2)

By [F7], determinant is a homomorphism on $G$, and [F2] places its image in the finite group ${\mu }_8$. Therefore its kernel $G_1=\{W\in G:\det W=1\}$ has finite index in $G$. Let $\pi :U(2)\to PU(2)$ be the quotient map. Then $\pi (G_1)$ has finite index in $\pi (G)$, and [F1] makes $\pi (G)$ dense. Its closure $\overline{\pi (G_1)}$ is consequently a closed finite-index subgroup of $PU(2)$: the dense group $\pi (G)$ is a finite union of cosets of $\pi (G_1)$, so its closure is the same finite union of the corresponding closed cosets of $\overline{\pi (G_1)}$. Every projective class has an $SU(2)$ representative: write $\det Q=e^{i\varphi }$ using [F8] and multiply $Q$ by $\lambda =e^{-i\varphi \mathrm{/}2}$. Every $SU(2)$ matrix has eigenvalues $e^{i\alpha },e^{-i\alpha }$; [F4] diagonalizes it and the path $s\mapsto S\mathrm{diag}(e^{is\alpha },e^{-is\alpha })S^{\ast }$ joins it to $I$. Thus $SU(2)$, and hence its continuous image $PU(2)$, is connected. The complement of a finite-index closed subgroup is a finite union of its closed cosets, so that subgroup is also open. Since $PU(2)$ is connected, its only nonempty clopen subgroup is $PU(2)$ itself; hence $\overline{\pi (G_1)}=PU(2)$.

Choose $\delta >0$ and, by the density just proved, a word $W\in G_1$ and a scalar $\mathrm{\mid }\lambda \mathrm{\mid }=1$ with $\mathrm{\parallel }W-\lambda R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<\delta$. Write the columns of $W$ and $R$ as $w_1,w_2$ and $r_1,r_2$. Multilinearity [F5] gives $\det W-\det (\lambda R)=\det (w_1-\lambda r_1,w_2)+\det (\lambda r_1,w_2-\lambda r_2)$. For $x=(x_1,x_2{)}^T$ and $y=(y_1,y_2{)}^T$, expanding both sides gives $\mathrm{\mid }\det (x,y){\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2-\mathrm{\mid }⟨x,y⟩{\mathrm{\mid }}^2=\mathrm{\mid }{x}_1{\mathrm{\mid }}^2\mathrm{\mid }{y}_2{\mathrm{\mid }}^2+\mathrm{\mid }{x}_2{\mathrm{\mid }}^2\mathrm{\mid }{y}_1{\mathrm{\mid }}^2-2\mathrm{Re}(x_1{y}_2{\bar{x}}_2{\bar{y}}_1)\le \mathrm{\parallel }x{\mathrm{\parallel }}^2\mathrm{\parallel }y{\mathrm{\parallel }}^2$. Since the columns are unit vectors and each column difference has norm less than $\delta$ by [F6], $\mathrm{\mid }1-{\lambda }^2\mathrm{\mid }\le 2\delta$.

Put $a=\mathrm{\mid }\lambda -1\mathrm{\mid }$ and $b=\mathrm{\mid }\lambda +1\mathrm{\mid }$. Then $a^2+b^2=4$ and $ab=\mathrm{\mid }1-{\lambda }^2\mathrm{\mid }\le 2\delta$. Since $\max (a,b)\ge \sqrt{2}$ from $a^2+b^2=4$, we have $\min (a,b)=ab\mathrm{/}\max (a,b)\le \sqrt{2}\delta$. Choose $\sigma \in \{1,-1\}$ attaining this minimum. By the triangle inequality and $\mathrm{\parallel }R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$, $\mathrm{\parallel }W-\sigma R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<(1+\sqrt{2})\delta$. The scalar $-I$ is a word in $G$: direct multiplication gives $X=HZH=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),XZ=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right),(XZ{)}^2=-I\mathrm{.}$ Thus $\sigma W\in G_1$, and $\mathrm{\parallel }\sigma W-R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }W-\sigma R{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ is less than $(1+\sqrt{2})\delta$. Because $\det (-I)=1$ and $\det W=1$, this scalar correction leaves the determinant equal to $1$. Choosing $\delta <\epsilon \mathrm{/}(1+\sqrt{2})$ proves the claim.