# Involutions of $S^2$

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 $S^2$

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All finite groups acting effectively on $S^2$ are conjugate to subgroups of $O_3$, the group of linear isometries of $S^2$. So your problem reduces to linear algebra, checking eigenvalues, and you have found representatives of all the conjugacy classes.

I'm not sure who is the first to prove the result I'm quoting but nowadays it follows immediately from 2-dimensional manifold+orbifold geometrization.

I think for involutions on $S^2$ the proof isn't so hard. If there's no fixed points the quotient is projective space and you're done. If there's fixed points use an equivariant tubularneighbourhood of the fixed point set and you've decomposed your manifold into either a circle + two discs (your reflection action) or two discs and an annulus (your rotation action). Either way, you're done. So the main ingredients in the argument are knowing 1) fixed points sets of finite group actions on manifolds are manifolds and have equivariant tubular neighbourhoods. 2) the classification of 2-manifolds. There are some combinatorial arguments I'm skipping like how you can rule out more than one circle as fixed point set, or anything other than two points, etc.

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The orbifold viewpoint seems to assume the actions are locally nice, i.e., locally linear. What about dropping this assumption? Presumably the result still holds, but what's a reference for this? I seem to recall that Bing constructed an involution on the 3-sphere with a fixed point set that was a wildly-embedded 2-sphere, though I don't know a reference for this either. – Allen Hatcher Apr 9 '10 at 18:42
I think that the original result that Ryan is referring to is due to Brouwer (Uber die periodischen transforationen der kugel, 1919), Eilenberg (Sur les transformations periodiques de la surface de sphere, 1934), and Kerekjarto (Uber die periodischen derkreisscheibe und der kugelflache, 1919). It is a consequence of the Jordan Curve theorem. – Dan Margalit Apr 9 '10 at 22:00
Right, both of my arguments are presuming smooth group actions. For discontinuous group actions it's certainly false. Continuous actions I don't know an argument but it sounds like Dan perhaps has the right references. I'm at a conference now and don't have easy MathSciNet access. – Ryan Budney Apr 10 '10 at 4:52
I believe that the argument for the continuous case is in Section 2.2 of Kulkarni's "Riemann surfaces admitting large automorphism groups." – Dan Margalit Apr 10 '10 at 17:51

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.

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