# Do all spaces doubly covered by $S^{2n}$ have the homeomorphism type of $\mathbb{P}^{2n}_{\mathbb{R}}$?

For reference, my motivation: It's of interest to classify free actions of groups on spheres of positive even dimension. Establishing such a classification up to homotopy is not too difficult: Every free group action on a sphere of even dimension is homotopic to either the trivial action by the trivial group or the antipodal action by $$\mathbb{Z}/2$$. The question in the title is, by the reduction that follows, equivalent to whether this classification is conserved if the "homotopic" in the above sentence is strengthened to "homeomorphic".

Suppose that every space doubly covered by $$S^{2n}$$ has the homeomorphism type of $$\mathbb{P}^{2n}_{\mathbb{R}}$$ and let $$\tau\ \colon S^{2n}\to S^{2n}$$ be some continuous involution lacking fixed points. Then:

1. By the compactness of $$S^{2n}$$, $$x\mapsto\text{dist}_{\text{standard subspace Euclidean metric on }S^{2n}}\left(x,\tau\left(x\right)\right)\colon S^{2n}\to\mathbb{R}_{\geq 0}$$ attains a nonzero mminimum on its domain.

2. By (1), the projection map $$\gamma\ \colon S^{2n}\to\text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)$$ is a covering map.

3. By (2), there exists an isomorphism $$\psi\ \colon \text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)\to\mathbb{P}_{\mathbb{R}}^{2n}.$$

4. By (2) and the lifting theorem for covering spaces, the $$\psi$$ of (3) lifts to an isomorphism $$\tilde{\psi}\ \colon S^{2n}\to S^{2n}$$ such that $$\text{anti}\circ\tau = \tau\circ \tilde{\psi}$$ (with $$\text{anti}\ \colon S^{2n}\to S^{2n}$$ the antipodal involution), precisely the desideratum.

• shouldn't it be "doubly covered"?
– YCor
Oct 30, 2021 at 6:59
• what do you mean by "nontrivially doubly covered by $S^{2n}$"? The double cover can't be trivial since $S^{2n}$ is connected.
– YCor
Oct 30, 2021 at 7:00
• Yes to both of these. Oct 30, 2021 at 8:28

One approach to a homeomorphism classification of closed manifolds simply homotopy equivalent to a closed manifold $$X$$ of dimension $$>4$$ is to compute the topological structure set $$\mathcal S^s_\text{TOP}(X)$$ and the group of homotopy classes of simple homotopy equivalences $$\text{Aut}_s(X)$$. Then $$\text{Aut}_s(X)$$ acts on $$\mathcal S^s_\text{TOP}(X)$$ by composition and the quotient is the desired set of homeomorphism classes of closed manifolds simply homotopy equivalent to $$X$$.
For $$X=\mathbb RP^n$$ the topological structure set is known, and the structure set always has at least 4 elements, and it is finite unless $$n-3$$ is divisible by $$4$$.
The group $$\text{Aut}_s(\mathbb RP^n)$$ is trivial if $$n$$ is even and has order $$2$$ if $$n$$ is odd. See Corollary 6 in "Coverings of fibrations" by Becker and Gottlieb.
Thus there are lots of fake even-dimensional real projective spaces in dimensions $$>4$$.
By Freedman's work this can be extended to dimension $$4$$, see Invariant knots of free involutions on $$S^4$$ by Ruberman for examples of fake $$\mathbb RP^4$$.
• Awesome, thanks! So if I'm understanding correctly, the idea is that for $n\geq 3$, there is a large ($\geq 2$) number of homeomorphism classes of manifolds simply homotopy equivalent to real projective space of dimension $2n$, and the Poincaré conjecture for topological manifolds of dimension $2n$ implies that any of these have a (universally covering) double cover homeomorphic to $S^{2n}$, together demonstrating the existence of the desired counterexample(s)? Oct 30, 2021 at 8:42
• @RafayAshary: what you stated is correct. Also in this case any homotopy equivalence is simple because the Whitehead group of $\mathbb Z_2$ is trivial. Oct 30, 2021 at 12:25