# 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 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.