We all know that the antipodal map is a free action of $\mathbb{Z}_2$ on $S^2$. Considering $\mathbb{Z}_2 = \{1, -1\}$, a free action may be viewed as a map $f : S^2 \rightarrow S^2$, i.e. the action of $-1$, with no fixed points, such that $f \circ f = Id_{S^2}$. We then readily see that $deg(f) = \pm 1$. Since $f$ does not have fixed points, it cannot have degree 1, and hence $f$ is homotopic to the antipodal map.

My question is: does this imply that the quotient of $S^2$ by this action is homeomorphic to $\mathbb{RP}^2$? Actually, I know this is true, because the quotient must be a closed surface with fundamental group $\mathbb{Z}_2$, so we can use the classification of closed surfaces. But I would like to be able to generalize this to spheres of any even dimension, so I wanted to see if I could get an answer that doesn't use that.


Unfortunately, the result is not true in dimensions of the form $4k$. That is, in each dimension of the form $4k$, there is a $Z_2$ action on $S^{4k}$ with quotient not homeomorphic to $\mathbb{R}P^{4k}$. (I don't know what happens in other dimensions. I suspect that Ricci flow techniques in dimension $3$ would show that the the only $Z_2$ quotients of $S^3$ is the standard $\mathbb{R}P^3$. I'm far from an expert in this area, though.)

More specifically, in

D. Ruberman. Invariant Knots of free involutions of S4, Top. Appl. 18 (1984), 217-224

Ruberman shows that there is a non-smoothable $4$-manifold $X$ which is homotopy equivalent, but not homeomorphic to $\mathbb{R}P^4$.

The universal cover $\tilde{X}$ of $X$ must therefore be homotopy equivalent to $S^4$. By Freedman's classification, we know $\tilde{X}$ is homeomorphic to $S^4$. Now, the deck group (which is isomorphic to $\pi_1(X)\cong \pi_1(\mathbb{R}P^2)\cong \mathbb{Z}_2$) acts freely on $\tilde{X}\cong S^4$.

Further, in

R. Fintushel and R. J. Stern, Smooth free involutions on homotopy 4k-spheres, Michigan Math. J. 30 (1983), no.1, 37–51,

Fintushel and Stern find examples of smooth $4k$ manifolds (with $k>1$) which are homotopy equivalent, but not homeomorphic to $\mathbb{R}P^{4k}$. One can repeat the argument above (using Smale's proof of the Poincare conjecture in high dimensions instead of Freedman's result) to find the involutions on $S^{4k}$.

  • $\begingroup$ Is the quotient always homotopically equivalent to $\mathbb{RP}^2$ though? $\endgroup$ – Pedro Oct 31 '14 at 8:44
  • $\begingroup$ For the $\mathbb{R}P^2$ case, you were exactly right: The quotient of $S^2$ by a free action of $\mathbb{Z}_2$ is always homeomorphic to $\mathbb{R}P^2$ (and your argument is the one I would have used). I believe that, in general, the quotient $X$ of $S^n$ by a free $\mathbb{Z}_2$ action is always homotopy equivalent to $\mathbb{R}P^n$. The proof I'm thinking of: First, show $H^1(X;\mathbb{Z}_2) \cong \mathbb{Z}_2$. Since $\mathbb{R}P^\infty$ is $K(\mathbb{Z}_2, 1)$, this gives a homotopically nontrivial map $X\rightarrow \mathbb{R}P^\infty$ which, by cellular approximation is... $\endgroup$ – Jason DeVito Oct 31 '14 at 13:53
  • $\begingroup$ homotopic to a map $X\rightarrow \mathbb{R}P^n$. This map should be an isomoprhism on $\pi_1$, and all other homotopy groups vanish for trivial reasons. Hence, Whitehead's theorem implies this map is a homotopy equivalence. $\endgroup$ – Jason DeVito Oct 31 '14 at 13:55
  • $\begingroup$ Jason: In dimension 3 this indeed follows from Perelman's theorem (although, I think, it was known earlier). By Perelman's theorem, every closed 3-manifold with finite fundamental group is homeomorphic to the quotient of $S^3$ by a finite subgroup of $O(4)$ acting freely. Now, the problem of uniqueness reduces to simple linear algebra. $\endgroup$ – Moishe Kohan Nov 5 '14 at 22:31
  • $\begingroup$ @studiosus: I had some vague idea that Perelman's theorem implied that a closed $3$-manifold with finite fundamental group was a quotient of $S^3$, but I didn't realize one could assume the action of the deck group is conjugate to a subgroup of $O(4)$. Granting that, I agree it's simple linear algebra from there. $\endgroup$ – Jason DeVito Nov 6 '14 at 3:27

The quotients of such actions are called "Fake real projective spaces" and you find their classification in http://www.map.mpim-bonn.mpg.de/Fake_real_projective_spaces


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