Let $S^n \rightarrow X$ be a two-fold covering. Is it true that $X$ is homeomorphic to $\mathbb{R}\mathrm{P}^n$? If so, how does one prove it?

  • $\begingroup$ You should check the book "Involutions on Manifolds" by Lopez de Medrano. I remember that he has a detailed discussion of free involutions on spheres of dimension $\ge 5$, but I do not remember in which category: Smooth or topological. If you work in the smooth category, there will be examples which are not diffeomorphic to the standard $RP^n$ (at least in dimension 4 - Fintushel-Stern and Capell-Shaneson, and, most likely, in all higher dimensions). $\endgroup$ – Moishe Kohan Mar 25 '17 at 16:54

There are different answers in the categories of homotopy type, topological, PL, and smooth manifolds. Let's start from the bottom and go up.

Suppose the finite $n$-dimensional CW complex $X$ has 2-fold cover homeomorphic to $S^n$. The map $X \to \Bbb{RP}^\infty$ classifying the fundamental group factors through $\Bbb{RP}^n$ by cellular approximation. Pulling back the universal bundle restricted to $\Bbb{RP}^n$ we still get the same bundle on $X$, but get an equivariant map $S^n \to S^n$ whose induced map on the quotient is the given map; note that the action on the sphere in both cases is free. If this map were of degree 1, we'd be finished: the map $f: X \to \Bbb{RP}^n$ clearly induces an isomorphism on all homotopy groups up to degree $(n-1)$, the degree statement implies it on degree $n$, and you can use a (particularly nasty) form of the Hurewicz theorem to conclude the map is a homotopy equivalence. Unfortunately I see no particular reason to believe it is degree 1; however, you can modify an "inverse map" from $\Bbb{RP}^{n-1}$ on the top degree cell so that it is a homotopy equivalence; the details are in Wall's "Surgery on compact manifolds". Conversely a manifold homotopy equivalent to $\Bbb{RP}^n$ is the quotient of a homotopy $S^n$ by some $\Bbb Z/2$ action.

In fact the classification of $n$-manifolds homotopy equivalent to $\Bbb{RP}^n$ but not (homeomorphic, PL homeomorphic, diffeomorphic) is the same as the classification of free $\Bbb Z/2$ actions on (possibly exotic) $S^n$ of the appropriate type up to the appropriate kind of equivalence (equivariant homeomorphism, diffeomorphism, etc).

TOP/PL/Smooth. These all agree in dimensions up to 3. It is classical then that there is only $\Bbb{RP}^2$ and a corollary of Perelman's proof of the elliptization conjecture says that there is only $\Bbb{RP}^3$ in dimension 3. In dimension 4, PL = Smooth, but there are non-smoothable 4-manifolds, and most topological 4-manifolds admit many distinct smooth structures. In dimension 4 there are exactly two manifolds homotopy equivalent to $\Bbb{RP}^4$ up to homeomorphism; one is not smoothable. See Hambleton-Kreck-Teichner Up to diffeomorphism (even when the universal cover is diffeomorphic to $S^4$) this is still open, but there are a few inequivalent smooth structures on $\Bbb{RP}^4$ with universal cover known to be standard. I don't know if there are infinitely many yet. Probably not.

In dimensions at least $5$ this is answered by surgery theory. Wall has a complete description in his book for PL $\Bbb{RP}^n$s; I suspect there is only minor change from the topological setting. (I do not really want to decode his notation and calculate the number of $\Bbb{RP}^n$s in each degree. It's finite except in dimensions equivalent to $3 \mod 4$.) His book will also provide a good reference to what's known smoothly in high dimensions and where to look.

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  • $\begingroup$ Actually in the second paragraph $X$ just has to be a finite CW complex with a double cover homotopy equivalent to $S^n$. $\endgroup$ – user98602 Mar 26 '17 at 3:57
  • $\begingroup$ Thanks for the detailed answer! I have two more question: Firstly, I did not quite understand how exactly the question if there are spaces homotopy equivalent but not isomorphic to $\mathbb{R}\mathrm{P}^n$ (in some category) relates to the original question of coverings by $S^n$. Secondly, do you see some way to get a homeomorphism (not just a homotopy equivalence) between $X$ and $\mathbb{R}\mathrm{P}^n$ in the CW-setting? $\endgroup$ – Robin Stoll Mar 29 '17 at 13:35
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    $\begingroup$ @RobinStoll No, it's just not true then. If $X \simeq \Bbb{RP}^n$ and $X$ is a manifold, then its universal cover is seen to be homotopy equivalent to $S^n$ and is a manifold, but thanks to the now-solved Poincare conjecture in all dimensions, the universal cover must be homeomorphic to $S^n$. All of these fake $\Bbb{RP}^n$s can be given CW structures. $\endgroup$ – user98602 Mar 29 '17 at 16:34

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