TITLE: Involutions of $S^2$
QUESTION [17 upvotes]: are there some complete results on the involutions of 2 sphere?
at least I have three involutions:
(let $\mathbb{Z}_2=\{1,g\}$,and $S^2=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=1\}$)
1.$g(x,y,z)=(-x,-y,-z)$(antipodal map) with null fixed point set,and orbit space $\mathbb{R}P^2$
actully,for free involution on $S^n$ with $n\leq3$,the orbit space is homeomorphic to real projective space (Livesay 1960)
2.$g(x,y,z)=(-x,-y,z)$ (rotation $\pi$ rad around $z$ axis)  with fixed point set $S^0$(the north pole and south pole) and orbit space $S^2$.
3.$g(x,y,z)=(x,y,-z)$(reflection along $z=0$) with fixed point set $S^1$ (the equator)and orbit space $D^2$
i want to know if there are some other involutions over 2-sphere.
here we take two involutions as equivalent if there are conjugate in the homeomorphism group of $S^2$

REPLY [3 votes]: Yes, your three examples are the only three up to conjugation by a homeomorphism.  In the orientation-preserving case, this is a theorem of Kerekjarto (Sur les groupes compacts de transformations topologiques des surfaces, 1941). A streamlined modern proof, including the orientation-reversing case, is offered by Boris Kolev (Note sur les sous-groupes compacts d'homeomorphismes de la sphere, 2006). Both of these references are in French.
From what I gather in skimming Kolev (but not poring through the proofs), the outline of the proof has two parts.  First, one shows that each fixed point has an invariant disc neighborhood.  Then, one uses the fact that any involution of a 2-disc is conjugate to the cone of an involution on a circle, the latter of which must conjugate to a linear one.