# Is every closed simply connected manifold a non-trivial covering space?

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?

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

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 $$k$$).

• 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. Sep 11, 2020 at 21:08
• Ah somehow I forgot that Qiaochu's answer also answers this question. Sep 11, 2020 at 21:09
• @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. 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.

• This is all really interesting, is there a standard reference I can look at regarding your point on classification of covered spaces? Sep 11, 2020 at 21:22
• 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 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 $$\mathbb{R}P^{10}$$.

• This was unexpected for me, how interesting! Sep 11, 2020 at 21:15
• 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. Sep 12, 2020 at 9:17
• @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 $\Sigma = -\Sigma$. Sep 12, 2020 at 16:06
• At first glance, I thought it is an obvious fact but by your comment seems that it is a bit deep!! Thanks a lot. Sep 12, 2020 at 16:14
• Wow...lovely idea! Sep 12, 2020 at 17:29