Let $M$ be a connected Hausdorff second countable topological space. I will call $M$ self-covering if it is its own $n$-fold cover for some $n>1$. For instance, the circle is its own double cover by $e^{2\imath\varphi}\mapsto e^{\imath\varphi}$.


  1. What would be a reasonable assumption on $M$ to ensure it is self-covering? Is every connected compact/closed topological manifold self-covering?

  2. Assume that $M$ is a differentiable manifold, and let $\phi:M\to M$ be the local diffeomorphism providing the $n$-fold cover for $n>1$. Is it true that the modulus of the Jacobian $|\det J(d\phi(x))|<1$ everywhere aside from countably many isolated points $x$ where it can be 1? In other words, is a covering map a volume contraction?

Please, note that I am not a topologist; more accessible language and references would be appreciated.

  • 5
    $\begingroup$ 1 is certanly false, for instance if M is simply connected it has no coverings; 2 makes no sense because volumes are not defined on a manifold. $\endgroup$ Mar 24, 2017 at 15:01
  • 2
    $\begingroup$ An obvious necessary condition is that $\pi_1(M)$ contains a subgroup of index $n$ abstractly isomorphic to itself. In particular, $\pi_1(M)$ must be infinite. $\endgroup$ Mar 24, 2017 at 15:10
  • 2
    $\begingroup$ Why should be such a quotient homeomorphic to $M$ itself? A priori, we only know that it has the same fundamental group. $\endgroup$ Mar 24, 2017 at 15:14
  • 4
    $\begingroup$ If $M$ has this property then 1) its Euler characteristic must be zero, and 2) all of its Pontrjagin numbers must be zero. If it is smooth this is very close to making $M$ be the boundary of another manifold. If your $n$ is even then 3) all of its Stiefel--Whitney numbers must be zero, in which case $M$ is the boundary of another manifold. $\endgroup$ Mar 24, 2017 at 15:23
  • 1
    $\begingroup$ @FrancescoPolizzi Yes in both cases I stand corrected. I think there are examples already for double covers with $\pi_1=\mathbb Z$ when one does get non-homotopy-equivalent manifolds this way. $\endgroup$ Mar 24, 2017 at 15:24

4 Answers 4


Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$. We define $M_f = F \times I \,/\, f$ to be the mapping torus with monodromy $f$. That is, form the product $F \times I$ and identify the two ends via $f$. Note that $M_f$ is an $F$-bundle over the circle.

Now, the $(k+1)$-fold cover of $M_f$, obtained by "unwrapping the circle direction" is the mapping torus for $f^{k+1} = f$. Thus the $(k+1)$-fold cover is homeomorphic to $M_f$.

As a concrete example, the trefoil knot complement $X_T$ is a once-punctured torus bundle over the circle, with monodromy of order six. Thus $X_T$ seven-fold covers itself. (And also five-fold covers itself.) This trick works for any torus knot complement.

  • $\begingroup$ In comments to the question @KevinCasto pointed out that for $M$ simply connected and $f$ with the property that all $f^k$ are pairwise non-homotopic we would obtain an example of $M$ with non-co-Hopfian fundamental group ($\mathbb Z$ in this case) that cannot cover itself nontrivially. However I could not find a rigorous argument for the reverse of your statement; more precisely, whether $M_f$ and $M_{f^k}$ are not homotopy equivalent if $f$ and $f^k$ are not homotopic. It might be not difficult to show this for some specifically chosen example, but is it true in general? $\endgroup$ Mar 25, 2017 at 8:02
  • $\begingroup$ (Reason of interest in such example lies in finding out generality to which sufficiency of non-co-Hopfianness of the fundamental group, as described in the answer by @NeilHoffman, could be extended) $\endgroup$ Mar 25, 2017 at 8:11
  • $\begingroup$ This is great use of torsion in the automorphism group! The delooping of the automorphism group of an object classifies bundles with that object as their regular fibre. What results relate the torsion of a group G to G-principal bundles? $\endgroup$ Jun 20, 2018 at 16:43

A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points out, a necessary condition is that the fundamental group is isomorphic to one of its proper finite-index subgroups, which is rather strong (in particular, implies that the group is not co-Hopfian). If such a group is also finitely presented, then it will be realized as the fundamental group of a 4-manifold (or $n$-manifold, for $n\geq 4$). However, it's not at all clear when the corresponding finite covers will be homeomorphic.

Examples comes from the non-co-Hopfian groups $BS(1,k)$. These are ascending extensions of cyclic groups by cyclic groups, and unwrapping the fiber gives isomorphic finite-index subgroups: $BS(1,k)\cong \langle a, b | aba^{-1}=b^k\rangle$, with index $k-1$ subgroup $\langle a, b^{k-1}\rangle \cong BS(1,k)$. 4-manifolds with such fundamental group were classified by Hambleton-Kreck-Teichner. Their classification is (essentially) in terms of the equivariant intersection form on $\pi_2(M)$. When this intersection form is trivial (which they prove is realized in Theorem B(i)), then the covers will be homeomorphic by Theorem A (this example came out of a discussion I had with van Limbeek and Teichner).


Adding further counter-examples to the discussion, compact hyperbolic manifolds in all dimensions are ruled out by the fact that Gromov's "simplicial volume" is nonzero (being proportional to hyperbolic volume with positive proportionality constant depending only on dimension), and that this invariant follows the same multiplication rule for covering maps that is followed by Euler characteristic.


EDIT: The following edits (in bold) are in response to Francesco Polizzi astute comment.

To address question 1, there are some obvious necessary conditions For example, the Euler characteristic of your manifold must be 0. We can find examples along these lines fairly easily: the circle, torus and $n$-torus.

However, Euler characteristic is not sufficient, meaning the second part of your question 1 is no, not even for manifolds with Euler characteristic 0. For example, consider any compact hyperbolic 3-manifold (or a simply-connected odd-dimensional manifold to appear to Bruno Martelli's comment above).

For connected, orientable, prime 3-manifolds, a sufficient condition is that the fundamental group of the manifold is not co-Hopfian. Recall a group $G$ is co-Hopfian if $G$ is not isomorphic to any of its proper subgroups. This is a little stronger than the necessary condition, which would be that $G$ is isomorphic to one of its finite index proper subgroups. (To observe that these to conditions are distinct consider free groups of rank $\geq2$.) However, I believe this requires invoking the affirmative proof of the Geometrization Conjecture, which is highly non-trivial.

This question points to many more examples of 3-manifold groups which are co-Hopfian: Hopfian and Co-Hopfian groups (examples)

  • 3
    $\begingroup$ Is it trivial to see that having not co-Hopfian fundamental group suffices? $\endgroup$ Mar 24, 2017 at 15:49
  • $\begingroup$ @Francesco Polizzi Thank you for pointing out! $\endgroup$ Mar 24, 2017 at 20:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.