We know that every universal covering space is simply connected. The converse is trivially true, every simply connected space is a covering space of itself. But I'm wondering what simply connected spaces, specifically manifolds, are covering spaces of another (connected) space other than itself.

I found this paper that shows that there exist certain contractible open simply connected manifolds that aren't non-trivial covering spaces of manifolds. If we restrict to closed simply connected manifolds, can we also find such examples, or can there always be a different manifold that it covers?

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    $\begingroup$ You might also be interested in orbifolds (which can, in some cases, be covered by manifolds that need not cover any other manifold). $\endgroup$ Sep 12, 2020 at 22:31

3 Answers 3


The manifold $\mathbb{CP}^2$ is not the universal covering space of any manifold other than itself.

One way to see this is to note that if $M \to N$ is a $d$-sheeted covering of closed manifolds, then $\chi(M) = d\chi(N)$. As $\chi(\mathbb{CP}^2) = 3$, we see that $d = 1$, in which case $N = \mathbb{CP}^2$, or $d = 3$. If a three-sheeted covering were to exist, the manifold $N$ would satisfy $\pi_1(N) \cong \mathbb{Z}_3$ (as this is the only group of order $3$). As $\mathbb{Z}_3$ has no index two subgroups, the manifold $N$ is orientable. But then the signature of $N$ satisfies $1 = \sigma(\mathbb{CP}^2) = 3\sigma(N)$ which is impossible, so the only manifold which is covered by $\mathbb{CP}^2$ is itself.

More generally, the connected sum of $k$ copies of $\mathbb{CP}^2$ does not cover any manifold other than itself. The argument above for orientability of the quotient doesn't apply when $k$ is even, instead you can use the argument in this answer (which works for any $k$).

  • $\begingroup$ Thanks! Both you and Qiaochu's answers are great. I wish I could accept both of them. By the way, could you explain a bit more why $\pi_1 (N) \cong \mathbb{Z}_3$? I don't know the relationship between the Euler characteristic and the fundamental group or the number of sheets. $\endgroup$ Sep 11, 2020 at 21:08
  • $\begingroup$ Ah somehow I forgot that Qiaochu's answer also answers this question. $\endgroup$ Sep 11, 2020 at 21:09
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    $\begingroup$ @Paul: if $\pi : X \to Y$ is a covering map and $X$ is simply connected then $X$ is the universal cover, so $Y$ must be obtained from the quotient of the action of $\pi_1(Y)$ on $X$. In other words, universal covers are Galois. If $\pi$ is an $n$-sheeted covering then the classification of covering spaces tells us that $\pi_1(X)$ must be a subgroup of $\pi_1(Y)$ of index $n$, which when $\pi_1(X)$ is trivial tells us that $\pi_1(Y)$ must be a finite group of order $n$. And when $n = 3$ there is exactly one such group. $\endgroup$ Sep 11, 2020 at 21:26

A closed simply connected $n$-manifold for $n = 2, 3$ is a sphere, so the smallest example occurs in dimension at least $4$. In dimension $4$ we can show that $\mathbb{CP}^2$ doesn't cover another manifold. There are two nice arguments given in the answers to this math.SE question, which I'll briefly sketch:

  1. Using the Lefschetz fixed point theorem, we can show that every diffeomorphism $f : \mathbb{CP}^2 \to \mathbb{CP}^2$ has a fixed point. It follows that $\mathbb{CP}^2$ does not admit a free action by any nontrivial group. This argument generalizes to all $\mathbb{CP}^{2k}$.

  2. $\chi(\mathbb{CP}^2) = 3$, so a nontrivial space that $\mathbb{CP}^2$ covers can only have $\chi(X) = 1$. This space must be a quotient by some action of $\mathbb{Z}/3$ and so has $\pi_1(X) \cong \mathbb{Z}/3$, which implies that $H^1(X, \mathbb{F}_2) = 0$ and hence that $X$ is orientable. But this implies $b_4 = 1$ so $\chi(X) \ge 2$; contradiction. This argument generalizes to $\mathbb{CP}^{2k}$ whenever $2k+1$ is prime.

It's worth noting the way in which both of these arguments fail for the spheres $S^n$ (which cover the real projective spaces $\mathbb{RP}^n$): 1) Lefschetz shows that a diffeomorphism can be fixed-point-free if it acts by $(-1)^{n+1}$ on $H_n$, which the antipode map does, and 2) $\chi(S^n) = 1 + (-1)^n$ so when $n$ is odd we only learn that a covered space also has $\chi = 0$ (true of e.g. the lens spaces) and when $n$ is even we learn that a nontrivial covered space has $\chi = 1$, $\pi_1 = \mathbb{Z}_2$ and can't be orientable, which is consistent, and true of $\mathbb{RP}^n$.

I also want to note that unlike the classification of covering spaces, which is purely homotopy-theoretic in that it only depends on $\pi_1$, the classification of covered spaces depends delicately on homeomorphism type. For example a point doesn't cover anything nontrivially but $\mathbb{R}$ does. In fact for every group $G$ and every simply connected space $X$ we can find a homotopy equivalent space $X'$ which covers a space with $\pi_1 \cong G$. So studying covered spaces is genuinely a point-set topological question.

  • $\begingroup$ This is all really interesting, is there a standard reference I can look at regarding your point on classification of covered spaces? $\endgroup$ Sep 11, 2020 at 21:22
  • $\begingroup$ Not that I'm aware of. I don't think people study this question at that level of generality, although the classification of space forms is a question like this: en.wikipedia.org/wiki/Space_form $\endgroup$ Sep 11, 2020 at 21:29

Below, I'll give an example which, as a topological manifold can non-trivially cover, but as a smooth manifold it cannot.

Recall that in each dimension, the set of diffeomorphism types of $S^n$ forms an abelian group under the connect sum operation, where the inverse to an element is given by switching the orientation.

Elements of order $2$ in this group are precisely the exotic spheres which admit an orientation reversing diffeomorphism.

In dimension $10$, the group of exotic spheres has order $6$. By Cauchy's theorem, there is an element $\Sigma$ of order $3$. In particular, such a $\Sigma$ does not admit an orientation reversing diffeomorphism.

It follows from an easy application of the Lefschetz fixed point formula that any orientation preserving diffeomorphism $f:\Sigma\rightarrow \Sigma$ must have a fixed point. Thus, $\Sigma$ cannot smoothly cover anything.

On the other hand, $\Sigma$, being homeomorphic to $S^{10}$ can topologically cover $\mathbb{R}P^{10}$.

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    $\begingroup$ This was unexpected for me, how interesting! $\endgroup$ Sep 11, 2020 at 21:15
  • $\begingroup$ Interesting different argument! The proof of second and third para is easy to check? In Milnor-Kervaire it has been proved for h-cobordisms classes. Do you have a good reference containing its proof. $\endgroup$
    – C.F.G
    Sep 12, 2020 at 9:17
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    $\begingroup$ @C.F.G: First, the second paragraph is slightly mistated: The set of oriented diffeomorphism types of $S^n$ forms an abelian group under the connect sum operation, except possibly in dimension $4$. This is outside my expertise, but the shortest proof I know is via Milnor-Kervaire and then Smale's H-cobordism theorem. Granting the second paragraph, the third is easy: If $\Sigma$ has an orientation reversing diffeo, then $\Sigma = -\Sigma$ in this group. Thus, $2\Sigma = 0$. Conversely, if $2\Sigma = 0$, then $\Sigma = -\Sigma$. $\endgroup$ Sep 12, 2020 at 16:06
  • $\begingroup$ At first glance, I thought it is an obvious fact but by your comment seems that it is a bit deep!! Thanks a lot. $\endgroup$
    – C.F.G
    Sep 12, 2020 at 16:14
  • $\begingroup$ Wow...lovely idea! $\endgroup$ Sep 12, 2020 at 17:29

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