2

Prop.2.29 on Hatcher's algebraic topology says $\mathbb Z_2$ is the only nontrivial group that can act freely on $S^n$ if $n$ is even.

If we have a unit sphere $S^2$ sitting in a standard coordinate, then let $f$ to be rotation of $S^2$ around $z$ axis by $\pi/2$ and let $g$ to be rotation of $S^2$ around $x$ axis by $\pi/2$. I think composition $h = g \circ f$ should be a homeomorphism with no fixed points. But $h^2$ is not the identity map. Where I get wrong ?

iefjkfdhfure
  • 425
  • Every rotation has an axis of rotation, where it fixes the points on the axis' intersection with $S^2$. One shows this by observing that every matrix in $SO(3)$ has an eigenvector with eigenvalue $1$. (actually it has an eigenvector since the characteristic polynomial is degree three, then use the fact that you are in $SO(3)$ to deduce the claim) – Andres Mejia May 09 '21 at 00:04

1 Answers1

3

But $h = g \circ f$ does have a fixed point. Let me assume that your rotations are positive (counterclockwise) with respect to the right hand rule using the positive orientations on the $x$ and $z$ axis. Consider the octant $x \ge 0$, $y \le 0$, $z \ge 0$. That octant intersects $S^2$ in a right angled spherical triangle. At the center of that triangle one finds the point $$P = \left(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\right) $$ which is fixed by $h = g \circ f$. The function $h$ is a rotation of $S^2$ around the line $\overline{OP}$ by $2\pi/3$.

Lee Mosher
  • 120,280