free group acting on the 3D sphere

Question

Let's say I have a 3 dimensional sphere $S^2$ and I define the following transformation $R_1$

• rotate the sphere around the north-south axis of $\frac{\pi}{2}$ (north and south are fixed points till now)
• rotate a single, predetermined meridian (so a circle containing north and south) of $\frac{\pi}{2}$ (no fixed points remain)

It seems to me that no point remains fixed and applying the mapping 4 times ${R_1}^4$ corresponds to the identity.

Why isn't the map $R_1$ - defined as the composition of the above two transformations - acting freely on the sphere as the quotient group $\mathbb{Z}/4\mathbb{Z}$?

I know that it contradicts a well known theorem according to which only the antipodal mapping acts freely on $S^{2n}$.

Answer 1

The theorem mentioned in the question is based on the invariant degree of a continuous mapping, but the second part of the mapping described in the question (i.e. rotating only a meridian of a sphere) is not a continuous function, indeed the following doesn't hold:

A function between two topological spaces $X$ and $Y$ is continuous if for every open set $V \subseteq Y$, the inverse image is an open subset of $X$

Answer 2

There is indeed a fixed point. One could demonstrate this with matrices, but I'll do a more geometric demonstration.

Let's imagine that the sphere is subdivided into three right-angled triangles by three great circles intersecting each other in six points, meeting at right angles. The six points are denoted $\pm A,\pm B,\pm C$ (think of $\pm A$ as the north-south poles, $\pm B$ as the east-west poles, and $\pm C$ as the front-back poles).

The first image on this page shows the subdivision that I have in mind.

Now I'll describe the rotations and their effect on the six poles.

The first rotation, around $\pm A$ by angle $\pi/2$, fixes $A$ and $-A$ and rotates the $B,C,-B,-C$ great circle, permuting those vertices in a 4-cycle $B \mapsto C \mapsto -B \mapsto -C \mapsto B$. Altogether the first rotation has the effect \begin{align*} A &\mapsto A \\ -A &\mapsto -A \\ B &\mapsto C \\ -B &\mapsto -C \\ C &\mapsto -B \\ -C &\mapsto B \end{align*} The second rotation, around $\pm B$ by angle $\pi/2$, fixes $B$ and $-B$ and rottes the $A,C,-A,-C$ great circle, permuting those 4 vertices in a 4-cycle $A \mapsto C \mapsto -A \mapsto -C \mapsto A$, for the effect \begin{align*} A &\mapsto C \\ -A &\mapsto -C \\ B & \mapsto B \\ -B &\mapsto -B \\ C &\mapsto -A \\ -C &\mapsto A \end{align*} Now let's compose the first rotation followed by the second. The effect on the four vertices is this: \begin{align*} A &\mapsto C \\ -A &\mapsto -C \\ B &\mapsto -A \\ -B &\mapsto A \\ C &\mapsto -B \\ -C &\mapsto B \end{align*} The three great circles subdivide the sphere into three right-angled triangles, each of which has one vertex chosen from $\pm A$, another from $\pm B$, and the third from $\pm C$.

Consider the right angled triangle with vertices $A,C,-B$. It's three vertices are permuted in a 3-cycle $A \mapsto C \mapsto -B \mapsto A$, and so that triangle is rotated by angle $2\pi/3$, fixing the center of the triangle. So yes, there's a fixed point.