# Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?

There are many closed manifolds with universal cover homotopy equivalent to $$\mathbb{R}^n$$, they are precisely the closed aspherical manifolds. There are also many closed smooth manifolds with universal cover diffeomorphic to $$\mathbb{R}^n$$, e.g. those which admit a metric of non-positive curvature. If one weakens diffeomorphic to homeomorphic, then the only additional examples one could possibly obtain would be four-dimensional, but no such examples are known to exist, see this question.

If one considers $$\mathbb{R}^n\setminus\{x\}$$ with $$n > 2$$ instead as a universal cover, there are plenty of examples. Any quotient of $$S^{n-1}$$ will have universal cover $$S^{n-1}$$ which is homotopy equivalent to $$\mathbb{R}^n\setminus\{x\}$$. If one upgrades to diffeomorphism, then the product of a smooth quotient of $$S^{n-1}$$ with $$S^1$$ yields a suitable manifold. Unlike the case of $$\mathbb{R}^n$$, one can construct smooth manifolds with universal cover homeomorphic but not diffeomorphic to $$\mathbb{R}^n\setminus\{x\}$$. For example, for any exotic $$(n-1)$$-sphere $$\Sigma$$, the universal cover of $$\Sigma\times S^1$$ is diffeomorphic to $$\Sigma\times\mathbb{R}$$ which is homeomorphic to $$S^{n-1}\times\mathbb{R}$$, and hence $$\mathbb{R}^n\setminus\{x\}$$, but is not diffeomorphic to it, see these comments by Igor Belegradek.

What if we remove more than one point from $$\mathbb{R}^n$$?

Is there a closed manifold whose universal cover is homotopy equivalent/homeomorphic/diffeomorphic to $$\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$$ for some $$k > 1$$?

There are no such manifolds in dimension one or two, but I don't even know if such examples can arise in dimension three.

The space $$\mathbb{R}^n\setminus\{x_1,\dots, x_k\}$$ is homotopy equivalent to $$\bigvee_{i=1}^kS^{n-1}$$. If $$M$$ is a closed manifold with the given universal cover, one might hope that an analysis of the natural $$\pi_1(M)$$-action on $$\pi_{n-1}(M) \cong \mathbb{Z}^k$$ could provide some insight.

• For homeomorphic/diffeomorphic the answer should be no, by the Stallings theorem on ends of groups. The fundamental group is finitely generated; the assumption implies it has $k+1>1$ ends; Stallings' theorem tells you it has either $1$ or infinitely many. (I am sorry for answering in a comment. It is late, and I don't have time to check that I am not making a stupid error or to write anything in more detail.) Mar 30, 2022 at 0:18
• The theorem that the number of ends is $0,1,2$ or $\infty$ is due to Freudenthal and Hopf, decades before Stallings. By the way, this "ends" result says that the space of ends, if infinite, is a Cantor. Hence, for every $n\ge 3$ and every totally disconnected compact space $K$ with at least 3 elements and at least one isolated point, $S^n-K$ is not homeomorphic to the universal cover of any compact space.
– YCor
Mar 30, 2022 at 8:21
• @YCor Thanks! I did forget the case of 2 ends last night. The correct form is just now typed in an answer... But I did not remember it is due to Freudenthal. Mar 30, 2022 at 8:22
• 2 ends can be achieved (universal cover of $\mathbf{S}^{n-1}\times \mathbf{S}^1$). References for the Freudenthal-Hopf theorem can be found on Wikipedia: en.wikipedia.org/wiki/Stallings_theorem_about_ends_of_groups And of course 1 end (universal covering $\mathbf{R}^n$) and 0 end (universal covering $\mathbf{S}^n$) can be achieved too. I'm not sure about $\mathbf{S}^n$ minus Cantor (by the way this is not uniquely defined up to homeomorphism — let's say, minus a Cantor that fits inside a line).
– YCor
Mar 30, 2022 at 8:24
• @DavidESpeyer Actually there exists a Cantor subset in $\mathbf{S}^3$ whose complement is not simply connected (Antoine necklace). I don't know what can be done if the complement is required to be simply connected. (At the positive side all Cantor subsets in the line or plane are topologically equivalent.)
– YCor
Mar 30, 2022 at 13:38

If we demand that the universal cover is homeomorphic / diffeomorphic to $$\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$$ with $$k>1$$ the answer is no, there are no such closed manifolds. Each missing point (together with the "infinity" of the one-point compactification of $$\mathbb{R}^n$$) is an end of the covering space, and these are all the ends. Therefore the universal cover has $$k+1 \ge 3$$ ends.
Freudenthal and Hopf proved that a finitely generated group has $$0$$, $$1$$, $$2$$, or infinitely many ends. The ends of the fundamental group biject with the ends of the universal cover, so the theorem contradicts our assumption.
I suspect (but am really not sure) that it may be possible to extend the argument to closed $$n$$-manifolds with universal cover only homotopy equivalent to $$\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$$ with $$k>1$$. Perhaps it can be shown that the universal cover must have $$k+1$$ ends by thinking of the separation properties of representatives of $$H_{n-1}(\bigvee_i S^{n-1})$$.