3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $S^{4k}$.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.