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How to convince myself (imagine) that $\Bbb S^1$-action on $\Bbb S^3$ fixes a circle of sphere?

Due to this comment of Jason DeVito, it is easy to see that action of $\Bbb S^1$ on $\Bbb S^3\subset \Bbb C^2$ defined by $z*(w_1,w_2)=(zw_1,w_2)$ fixes the entire circle $\{(0,w):|w|=1\}\subset\Bbb S^3\subset \Bbb C^2$. But I can't imagine it, because the common picture of action in my mind is that an circle action is a kind of rotation, so it has a rotation axis and spinning around this axis can fix at most 2 point. Is it possible that the axis of rotation is not a line?

Now, how can I think about this action geometrically? $z*(w_1,w_2)=(zw_1,\bar zw_2)$.

Edit: My understanding of last action is that: one side of $\Bbb S^3$ is spinning clockwise and other side is spinning counterclockwise (in different plane from first action) and these actions effect on the middle of sphere and it become scarious and kink in middle, Like cylinder if we spin the boundaries of it in different directions it become kink in middle like screw.

C.F.G
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  • Looks like, similar to that one on the comment, fixes two circles just in one of the cases also reverses orientation. – astro Aug 16 '20 at 13:47
  • @astro: what two circles? – C.F.G Aug 16 '20 at 14:56
  • ${ (0,\omega):|\omega|=1}$ is fixed with reversed orientation and ${ (\omega,0):|\omega|=1}$ is fixed with the same orientation (under the action described by yourself at the bottom of the question). EDIT: Not totally fixed, I guess the right term is $\textit{invariant}$. – astro Aug 16 '20 at 18:36
  • aha, I thought you meant the first action. – C.F.G Aug 16 '20 at 18:42
  • I think that the confusion comes actually from the wrong use of the therm $\textit{fixed}$ since your remark about the common points having to be fixed would be true if the circles were actually fixed, but that does not happen in this context since the invariant circles for the action don't exhibit global fixed points in any of the two cases. – astro Aug 16 '20 at 18:49
  • First you want to understand the Hopf fibration $S^1\to S^3\to S^2$, particularly why the circular fibers are parametrized by points on a $2$-sphere and how this is visualized in $\Bbb R^3$ under stereographic projection. Then multiplying by phasors amounts to simultaneously rotating all of the circles by an angle. (Also, viz Cartan-Dieudonne, one should think of rotations not in terms of axes of rotation but 2D planes of rotation, and a possible fixed axis in odd dimension.) Would you want an answer elaborating on this? – anon Aug 18 '20 at 09:12

2 Answers2

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One would not expect a rotation in $\mathbb C^2 \approx \mathbb R^4$ to have an "axis of rotation" which is a line, i.e. something of real dimension $1$. On the other hand, one would expect the "axis of rotation" to have real codimension $2$, which it does: the entire plane $w_1=0$ is fixed. And when you intersect that plane with $S^3$ you get a circle that is fixed.

If you want to visualize this example, it can be done using the fact that $S^3$ is the one-point compactification of $\mathbb R^3$, which I'll write as $S^3 = \mathbb R^3 \cup \{\infty\}$. In this model, one can visualize the circle of fixed points as the unit circle in the $x,y$-plane: $$\{(x,y,0) \mid x^2 + y^2 = 1\} $$ Outside of this circle of fixed points, every other orbit of the action is a circle, and one can visualize these circle orbits in $\mathbb R^3 \cup \{\infty\}$ using $(r,\theta,z)$ cylindrical coordinates, as follows. One of the circle orbits is $\text{$z$-axis} \cup \{\infty\}$. Then, for each constant angle $\theta_0$, the half plane $\theta = \theta_0$ pierces the fixed circle at the single point $P(\theta_0)$ with coordinates $(r,\theta,z)=(1,\theta_0,0)$, the boundary edge of that half plane is the $z$-axis which is an orbit, and the rest of the half-plane is foliated by a family of circle orbits which approach that single point in one direction getting smaller and smaller, and which approach the $z$-axis in the other direction getting larger and larger (in the hyperbolic metric $\frac{dr^2+dz^2}{r^2}$ on this half-plane, these are the concentric circles centered on $P(\theta_0)$).

C.F.G
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Lee Mosher
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For me, the way I think about rotations is a consequence of the maximal torus theorem for $\mathrm{SO}(n)$. Namely, given any $A\in \mathrm{SO}(n)$ (i.e., a rotation of $\mathbb{R}^n$ which fixes $0$), there is some basis of $\mathbb{R}^n$ with the property that in this basis, $A$ consists of a bunch of regular $2$-dimensional rotation blocks.

More precisely, writing $R(\theta) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$ for the standard counterclockwise rotation matrix, there is always an orthonormal basis of $\mathbb{R}^n$ in which $A$ takes the block diagonal form $$A=\begin{cases} \operatorname{diag}\Big(R(\theta_1), R(\theta_2),..., R(\theta_{n/2})\Big) & n \text{ even}\\ \operatorname{diag}\Big(R(\theta_1), R(\theta_2),..., R(\theta_{(n-1)/2},1)\Big) & n \text{ odd}\end{cases}.$$

This indicates that rotations are fundamentally two-dimensional ideas which are then bootstrapped to higher dimensions. In fact, it gives a recipe for constructing all rotations of $\mathbb{R}^n$: Pick any $2$-plane and rotate it a bit. In the orthogonal complement, pick any $2$-plane and rotate it. In the orthogonal complement of these two $2$-planes, pick any $2$-plane and rotate it, etc.

Thinking about $\mathbb{R}^3$ for a moment, a rotation in the $xy$-plane doesn't change the distance from a point in the $xy$ plane to any point in the $z$-axis. In fact, a rotation in the $xy$ plane has no effect on the $z$ axis. The above decomposition indicates that this idea propagates to higher dimensions. For example, in $\mathbb{R}^4$ (with coordinates, say, $(x,y,z,t)$) a rotation in the $xy$ plane doesn't change the distance from a point in the $xy$ plane to a point in the $zt$ plane.

This is why, for example, your action on $\Bbb S^3$ can rotate two things in opposite directions. It's hard to visualize, but a rotation in the $xy$-plane has no effect on the $zt$-plane, so no "twisting" of $\Bbb S^3$ occurs in your action.

On the other hand, for your cylinder action, note that your action is not a rotation of $\mathbb{R}^3$ restricted to the cylinder, so none of the above applies. In fact, I wouldn't call your action on the cylinder a rotation. It is a rotation on each boundary component, but who knows what it is in between!

C.F.G
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Jason DeVito - on hiatus
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  • Your example $\Bbb R^3$ is about whole of it or it is correct for its subspaces? – C.F.G Sep 02 '20 at 13:54
  • So I should accept it as a postulate to imagination rules. Although it is not to hard to show it by some algebra and calculus. – C.F.G Sep 02 '20 at 14:10
  • @C.F.G: My $\mathbb{R}^3$ example is about a rotation of all of $\mathbb{R}^3$. It also works for all of its vector subspaces, but not necessarily for all of it's topological subspaces. I don't know if I'd say I "accept it" - it's just an attempt at an intuitive picture, gained from the correct, rigorous algebra and calculus – Jason DeVito - on hiatus Sep 02 '20 at 17:11
  • You wrote that "In the orthogonal complement of these two 2-planes, pick any 2-plane and rotate it, etc." Is that possible that some rotations involve 3-space instead of 2-plane? I don't know my words is correct or not but something like complex dimensional rotation. e.g. $R(ix+y=\theta)$? – C.F.G Sep 04 '20 at 13:03
  • @C.F.G: A 3-d rotation may look like something you write, but you can always repick your basis to see a rotating 2-plane and a fixed vector. You already see this in $\mathbb{R}^3$: Every rotation has a fixed axis, so is really just rotating the orthogonal complement to the fixed axis. – Jason DeVito - on hiatus Sep 04 '20 at 14:33
  • Is this explanation correct? $z*(a,b):=(a,zb\bar{z})$ action first rotate sphere in zw-plane counterclockwise then rotate the result clockwise in the same plane. so it is zero rotation. – C.F.G Sep 05 '20 at 16:37
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    @C.F.G: Assuming $(a,b)\in \mathbb{C}^2$, yes, that is one way of thinking about it. Alternatively, note that if $b\in \mathbb{Z}$, then $zb\overline{z} = z\overline{z}b = |z|^2 b = b$. – Jason DeVito - on hiatus Sep 05 '20 at 20:49
  • It seems that you meant in $\Bbb C$ (because it is commutative). by knowing this, why you wrote $z^k q \overline{z}^k$ in this MO answer? By the above argument it is equal to $q$. What exactly this redundant factor do? – C.F.G Sep 06 '20 at 04:22
  • Oops, yes: I meant $b\in\mathbb{C}$ above. In the MO answer, I was treating $S^3\subseteq\mathbb{H}$, where $\mathbb{H}$ refers to the unit quaternions. These are non-commutative. If we write $q=a +bj$ with $(a,b)\in \mathbb{C}^2$, then $z^s q \overline{z}^t$ is the same action as $(z^{s-t} a, z^{s+t} b)$. – Jason DeVito - on hiatus Sep 06 '20 at 13:36
  • Ah-ha. Sorry for my comments, I really appreciate your help. Could you please correct this one: "in $\mathbb{H}$, $ij=k$ means that if we rotate $i$ in direction of $j$ we get $k$" which is wrong obviously. But for $\Bbb C$ it is correct: if we turn $1=(1,0)$ in direction of $i$ we get $i$. i.e. $1.i=i$. What is wrong? – C.F.G Sep 06 '20 at 13:56
  • @C.F.G. I am not exactly sure how to answer the question. Multiplying by $i$ causes a $\pi/2$ rotation both in the ${1,i}$ plane and also in the ${j,k}$ plane. – Jason DeVito - on hiatus Sep 06 '20 at 15:16