The standard seven-sphere as the $(1,0)$ quaternionic Hopf sphere bundle

On the base chart $q_0\mathrm{\ne }0$, put $z=q_1{q}_0^{-1}$ and write $(q_0,q_1)=(1,z)\lambda \mathrm{/}\sqrt{1+\mathrm{\mid }z{\mathrm{\mid }}^2}$, with unit $\lambda =q_0\mathrm{/}\mathrm{\mid }{q}_0\mathrm{\mid }$. On $q_1\mathrm{\ne }0$, put $w=q_0{q}_1^{-1}=z^{-1}$ and write $(q_0,q_1)=(w,1){\lambda }^{\mathrm{\prime }}\mathrm{/}\sqrt{1+\mathrm{\mid }w{\mathrm{\mid }}^2}$, with ${\lambda }^{\mathrm{\prime }}=q_1\mathrm{/}\mathrm{\mid }{q}_1\mathrm{\mid }$. These formulas and their inverses are smooth. The base is the one-point compactification of $\mathbb{H}$; replacing the second coordinate $w$ by $\bar{w}$ gives the usual stereographic transition $z\mapsto z\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^2$, hence its smooth structure is that of $S^4$. On the equator $\mathrm{\mid }z\mathrm{\mid }=1$ set $a=z$; the fibre transition is ${\lambda }^{\mathrm{\prime }}=a\lambda$, since quaternionic multiplication has the displayed order. Thus the two hemisphere product charts of $S^7$ glue by precisely [F1], giving $M_{1,0}\cong S^7$.

By [F2], $q(W_{1,0})=4$ and $\sigma (W_{1,0})=1$. Therefore [F3] gives $\lambda (M_{1,0})=2\cdot 4-1=7\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}7)$, agreeing with the standard sphere's disk filling.