We identify $S^3$ with the unit quaternions and $S^2$ the unit pure quaternions, and the conjugate of $z=a+bi+cj+dk$ is defined as $z^*=a-bi-cj-dk$. Then we consider the map $$\pi:S^3\ni z\mapsto ziz^*\in S^2.$$ The reason why $\pi$ maps into $S^2$ is given in this answer, but I'm stuck as to how to show it is surjective (in fact, a surjective submersion)?

Thanks to @Xipan Xiao's comment now I understand why $\pi$ is surjective: in fact for each $p\in S^2$ we can pick $\tilde q=\cos(\theta/2)+q\sin(\theta/2)\in S^3,\, q=e_1\times p/\|e_1\times p\|\in S^2$ so that $p=\tilde qi\tilde q^*$.

However, I'm still quite at a loss how to show $\pi$ is a submersion? Can we possibly avoid explicitly expanding the quaternions and using the stereographical projection charts to compute the Jacobian?

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    $\begingroup$ let $a=\cos\frac{\phi}{2}$, and $z=\cos\frac{\phi}{2}+V\sin\frac{\phi}{2}$ then $zxz*$ is the rotation of $\frac{\phi}{2}$ along the axis $V$. So the map is to rotate $(1, 0, 0)$ and it's onto since we can set $V$ to any unit pure quaternion and $\phi$ to any value. $\endgroup$ – Xipan Xiao Apr 23 '17 at 5:48
  • $\begingroup$ @XipanXiao thanks for your comment, which clarifies. $\endgroup$ – Vim Apr 23 '17 at 7:11

You can compute the derivative directly by viewing the spheres as submanifolds in $\mathbb H$ and the space of purely imaginary quaternions, respectively. Indeed, the map $f(z)=ziz^*$ makes sense as a map from $\mathbb H$ to imaginary quaternions, so you can differentiate it as any map $\mathbb R^4\to\mathbb R^3$, and then think about the restriction to the unit spheres. In partiuclar, you can compute $Df(z)(w)$ as the derivative at $t=0$ of $f(z+tw)$. Oberserving that quaternionic multiplication is $\mathbb R$ bilinear, while conjugation is $\mathbb R$-linear, you get $Df(z)(w)=wiz^*+ziw^*$, and this is twice the imaginary part of $wiz^*$.

So what you actually have to prove is that if you have a purely imaginary quaternion $q$ perpendicular to $ziz^*$ (i.e. tangent to $S^2$ in that point), then you can write it as $wiz^*+ziw^*$, for some $w$ which is perpendicular to $z$ (and hence tangent to $S^3$ in $z$). Now this is easy for $z=1$. Here $q$ must be a linear combination of $j$ and $k$, while $w$ has to be purely imaginary, and you immediately write out an explicit solution. For general $z$, the orthogonal space to $ziz^*$ is spanned by $zjz^*$ and $zkz^*$ and you can similarly find an explicit solution.

Having verified that $\pi$ is a submersion, it follows that it is surjective. Since $d\pi(z)$ is surjective, the implicit function theorem shows that the image of $\pi$ contains an open nighborhood of $z$ in $S^2$ and since this works for each $z$, you see that $\pi(S^3)$ is open in $S^2$. On the other hand, continuity of $\pi$ shows that $\pi(S^3)$ is a compact subset of $S^2$ and thus closed. Since $S^2$ is connected, you get $\pi(S^3)=S^2$.


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