Alternate characterizations of the Hopf circles in $S^3$

Question

The Hopf fibration is a continuous map $S^3 \to S^2$ whose fibers are all circles on $S^3$.

1. Is every one of these fiber circles a great circle of $S^3$? (The Wikipedia page implies this at one point when discussing one explicit construction, but I just want to check that this is true in general.)

2. The Hopf fibration does not map every great circle of $S^3$ to a single point on $S^2$, but only the great circles $$x(a, b, c, d) + y(-b, a, -d, c) \cap S^3,$$ where $a,b,c,d$ are fixed real numbers (not all zero) and $x$ and $y$ are variable real numbers. (These are just the real and imaginary parts of the unimodular points on the complex line $\lambda (z_1, z_2)$ in $\mathbb{C}^2$.) Is there any other simple characterization (e.g. geometric) of these "Hopf circle" fibers in $S^3$?

3. To what subset of $S^2$ does a non-Hopf great circle in $S^3$ map under the Hopf fibration?

Answer 1

(2) The group $S^3$ acts on $S^2$ by conjugation (where $S^2\subset\mathbb{R}^3$ is interpreted as being comprised of pure imaginary, or pure vector, quaternions) transitively. The stabilizer of $i$ is $S^1\subset\mathbb{C}\subset\mathbb{H}$, the group of unit complex numbers, and hence by orbit-stabilizer the fibers of the map $S^3\to S^2$ ($q\mapsto qiq^{-1}$) are precisely the cosets of $S^1$.

(1) No, only those in "complex lines" in $\mathbb{H}\cong\mathbb{C}^2$. So, for instance, the great circle in the span of $\{1,j\}$ is not one of the fibers. (Unless one uses a different conjugation map $S^3\to S^2$, i.e. $q\mapsto qjq^{-1}$.) Note that the moduli space of all (oriented) 2D subspaces of 4D space is the (oriented) Grassmanian $\widehat{\mathrm{Gr}}(4,2)\simeq S^2\times S^2$ and the collection of those 2D subspaces which are complex 1D subspaces forms a $\{\ast\}\times S^2$ within.

(3) Any great circle is of the form $\cos\theta p+\sin\theta q$ where $p\perp q$, hence $q=\mathbf{u}p$ for some pure imaginary quaternion $\mathbf{u}$, i.e. of the form $e^{\theta\mathbf{u}}p$, and the image is then $pe^{\theta\mathbf{u}}\mathbf{i}e^{-\theta\mathbf{u}}p^{-1}$ which is the effect of first rotating the vector $\mathbf{i}$ around $\mathbf{u}$ by all possible angles, then rotating the resulting 2D circle around according to conjugation by whatever the quaternion $p$ is. So, the image is a circle.