Circle in $\mathbb{S}^3$ not mapping to a point in $\mathbb{S}^2$ under Hopf map

Question

The Hopf fibration is a mapping $h:\mathbb{S^3} \mapsto\mathbb{S}^2$ defined by $r\mapsto ri\bar{r}$ where $r$ is a unit quaternion in the form $r=a+bi+cj+dk $ where $a,b,c,d \in \mathbb{R}$ and $ijk=-1$. Explicitly, $$h(a,b,c,d)=(a^2+c^2-b^2+d^2,2(ad+bc),2(bd-ac))$$ Now, it is known that for a point in $\mathbb{S^2}$ the preimage $h^{-1}$ is a circle in $\mathbb{S^3}$. How can this be? If you consider the set of points $$C= \{(\cos(t),0,0,\sin(t)) \mid t\in\mathbb{R}\} \in \mathbb{S^3}$$ which can be written in terms of quaternions as $C=e^{k t}$ where $ t \in \mathbb{R}$. This clearly doesn't map to a single point in $\mathbb{S^2}$ under the Hopf fibration. It maps to the circle $i\cos(2t)+j\sin(2t)$. So I am not understanding what's going on here. Why is this circle on in 4 dimensional space not mapping to a point in 3 dimensional space?

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

As Cheerful Parsnip explained in his comment, the claim is not that every circle maps to a point, but that all preimages of points are circles.

A simple example illustrating this is the projection $p : S^2 \times S^1 \to S^2$. All preimages of points are $1$-spheres, but of course not every image of a $1$-sphere in $S^2 \times S^1$ is a point. Take for example the equator $E \subset S^2$. Then $E \times \{1\}$ is a $1$-sphere, but its image is $E$.

Answer 2

The unit quaternion $\hat{\Psi}$ in the SO(3) picture. \begin{equation} \hat{\Psi}\;=\; \begin{pmatrix} q_1^2+q_i^2-q_j^2-q_k^2 & 2(q_iq_j-q_1q_k) & 2(q_iq_k+q_1q_j) \\ 2(q_iq_j+q_1q_k) & q_1^2-q_i^2+q_j^2-q_k^2 & 2(q_jq_k-q_1q_i) \\ 2(q_iq_k-q_1q_j) & 2(q_jq_k+q_1q_i) & q_1^2-q_i^2-q_j^2+q_k^2 \end{pmatrix} \end{equation} We can label the elements as: \begin{equation*} \hat{\Psi}\;=\; \begin{pmatrix} \mathcal{I}^i_{l} & \mathcal{I}^j_{l} & \mathcal{I}^k_{l} \\ \mathcal{J}^i_{l} & \mathcal{J}^j_{l} & \mathcal{J}^k_{l} \\ \mathcal{K}^i_{l} & \mathcal{K}^j_{l} & \mathcal{K}^k_{l} \end{pmatrix} \qquad\qquad\qquad\qquad \hat{\Psi}\;=\; \begin{pmatrix} \mathcal{I}^i_{r} & \mathcal{J}^i_{r} & \mathcal{K}^i_{r} \\ \mathcal{I}^j_{r} & \mathcal{J}^j_{r} & \mathcal{K}^j_{r} \\ \mathcal{I}^k_{r} & \mathcal{J}^k_{r} & \mathcal{K}^k_{r} \end{pmatrix} \end{equation*}

The Lie algebra matrices defined by \begin{equation} \hat{\pi}_i\;\equiv\; \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\qquad \hat{\pi}_j\;\equiv\; \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\qquad \hat{\pi}_k\;\equiv\; \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \end{equation}
For a quaternion $\hat{\Psi}$ there are 6 possible mappings from $\mathbb{S}^3\mapsto\mathbb{S}^2$, given by \begin{align} \hat{\mathcal{I}}_l\;=\;\hat{\Psi}^t\hat{\pi}_i\hat{\Psi} \qquad\qquad\;\; \hat{\mathcal{J}}_l\;&=\;\hat{\Psi}^t\hat{\pi}_j\hat{\Psi} \qquad\qquad\; \hat{\mathcal{K}}_l\;=\;\hat{\Psi}^t\hat{\pi}_k\hat{\Psi} \\ \hat{\mathcal{I}}_r\;=\;\hat{\Psi}\hat{\pi}_i\hat{\Psi}^t \;\qquad\qquad \hat{\mathcal{J}}_r\;&=\;\hat{\Psi}\hat{\pi}_j\hat{\Psi}^t \qquad\qquad\; \hat{\mathcal{K}}_r\;=\;\hat{\Psi}\hat{\pi}_k\hat{\Psi}^t \end{align}

To rotate a quaternion in $\mathbb{R}^4$, consider the basis matrices:

The left Cayley matrices: \begin{equation} \hat{l}_i\;=\; \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\;\;\;\; \hat{l}_j\;=\; \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}\;\;\;\; \hat{l}_k\;=\; \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix} \end{equation} which satisfy the relation $$\hat{l}_i^2\;=\;\hat{l}_j^2\;=\;\hat{l}_k^2\;=\;\hat{l}_i\hat{l}_j\hat{l}_k\;=\;-\hat{l}_1$$ The right Cayley matrices: \begin{equation} \hat{r}_i\;=\; \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\;\;\;\; \hat{r}_j\;=\; \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}\;\;\;\; \hat{r}_k\;=\; \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix} \end{equation} which satisfy the relation $$\hat{r}_i^2\;=\;\hat{r}_j^2\;=\;\hat{r}_k^2\;=\;\hat{r}_i\hat{r}_j\hat{r}_k\;=\;-\hat{r}_1$$ and $\hat{l}_1,\hat{r}_1$ are the $4\times4$ identity matrices.

In vector form the quaternion is represented \begin{equation*} \vec{\Psi}\;=\; \begin{pmatrix} q_1 \\ q_i \\ q_j \\ q_k \end{pmatrix} \end{equation*} Rotate the quaternion $\vec{\Psi}$ in the left Cayley basis, through an angle $\varphi$ in the $\hat{l}_i$ axis. $$\vec{\Psi}'\;=\;\exp\left(\frac{\varphi}{2}\hat{l}_i\right)\vec{\Psi}$$ \begin{equation} \vec{\Psi}'\;=\; \begin{pmatrix} \cos\left(\tfrac{\varphi}{2}\right) & -\sin\left(\tfrac{\varphi}{2}\right) & 0 & 0 \\ \sin\left(\tfrac{\varphi}{2}\right) & \cos\left(\tfrac{\varphi}{2}\right) & 0 & 0 \\ 0 & 0 & \cos\left(\tfrac{\varphi}{2}\right) & -\sin\left(\tfrac{\varphi}{2}\right) \\ 0 & 0 & \sin\left(\tfrac{\varphi}{2}\right) & \cos\left(\tfrac{\varphi}{2}\right) \end{pmatrix} \begin{pmatrix} q_1 \\ q_i \\ q_j \\ q_k \end{pmatrix} \;=\; \begin{pmatrix} \cos\left(\tfrac{\varphi}{2}\right)q_1-q_i\sin\left(\tfrac{\varphi}{2}\right) \\ \cos\left(\tfrac{\varphi}{2}\right)q_i+q_1\sin\left(\tfrac{\varphi}{2}\right) \\ \cos\left(\tfrac{\varphi}{2}\right)q_j-q_k\sin\left(\tfrac{\varphi}{2}\right) \\ \cos\left(\tfrac{\varphi}{2}\right)q_k+q_j\sin\left(\tfrac{\varphi}{2}\right) \end{pmatrix} \end{equation} By extension - the quaternion can be rotated in a circle in $\mathbb{R}^4$ using the 6 rotation axes $\hat{l}_i,\hat{l}_j,\hat{l}_k,\hat{r}_i,\hat{r}_j,\hat{r}_k$.

$\vec{\Psi}'$ in the SO(3) picture: \begin{equation*} \hat{\Psi}'\;=\; \begin{pmatrix} \mathcal{I}^i_{l} & \mathcal{I}^j_{l} & \mathcal{I}^k_{l} \\ \mathcal{J}^i_{l}\cos(\varphi) - \mathcal{K}^i_{l}\sin(\varphi) & \mathcal{J}^j_{l}\cos(\varphi) - \mathcal{K}^j_{l}\sin(\varphi) & \mathcal{J}^k_{l}\cos(\varphi) - \mathcal{K}^k_{l}\sin(\varphi) \\ \mathcal{K}^i_{l}\cos(\varphi) + \mathcal{J}^i_{l}\sin(\varphi) & \mathcal{K}^j_{l}\cos(\varphi) + \mathcal{J}^j_{l}\sin(\varphi) & \mathcal{K}^k_{l}\cos(\varphi) + \mathcal{J}^k_{l}\sin(\varphi) \end{pmatrix}\\ \end{equation*} where $\varphi\in[0,4\pi]$.

Now to your question:

Why is this circle on in 4 dimensional space not mapping to a point in 3 dimensional space?

The circle in $\mathbb{R}^4$,is projected to a point in $\mathbb{R}^3$ via: \begin{equation*} \hat{\mathcal{I}}_l\;=\;\hat{\Psi}^t\hat{\pi}_i\hat{\Psi} \;=\;\hat{\Psi}'^t\hat{\pi}_i\hat{\Psi}' \end{equation*}

To view the fibre in the traditional form, consider the stereographic projection of the quaternion via the mapping: \begin{equation*} \mathbb{S}^3/(1,0,0,0)\;\mapsto\;\mathbb{R}^3 \end{equation*} given by \begin{equation} (q_1',q_i',q_j',q_k')\;\mapsto\;\left(\frac{q_i'}{1-q_1'},\frac{q_j'}{1-q_1'},\frac{q_k'}{1-q_1'}\right) \end{equation} with $\varphi\in[0,4\pi]$.