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A manifold is called prime if whenever it is homeomorphic to a connected sum, one of the two factors is homeomorphic to a sphere.

Is there an example of a finite covering $\pi : N \to M$ of closed orientable manifolds where $M$ is prime and $N$ is not?

There are no examples in dimensions two or three. If one is willing to forgo the orientability requirement, then there are examples in dimension three. In this paper, Row constructs infinitely many topologically distinct, irreducible (and hence prime), closed 3-manifolds with the property that none of their orientable covering spaces are prime.

There are examples where $N$ is prime and $M$ is not, such as the double covering $\pi : S^1\times S^2 \to \mathbb{RP}^3\#\mathbb{RP}^3$.

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There are examples analogous to Row's in dimensions $n>2$ which are orientable when $n$ is even. I'll give a bit of motivation for the example at the end.

Consider the action of the group $G= \mathbb{Z}^n\rtimes \{\pm I\}=\{ x \mapsto \pm x+ m, m\in \mathbb{Z}^n\}$ on $\mathbb{R}^n$. The subgroup $G_{m/2}=\{x,-x+m\}, m\in \mathbb{Z}^n $ is the stabilizer of $m/2\in \frac12\mathbb{Z}^n$. Remove open balls of radius $r<1/4$ about the lattice points $\frac12\mathbb{Z}^n$ to get the simply-connected manifold $V= \mathbb{R}^n -\mathcal{N}_r(\frac12\mathbb{Z}^n)$. When $n$ is even, $V$ admits an orientation which is $G$-invariant. This induces an orientation on $\partial V$. Since $G_{m/2}$ acts as the antipodal map on the sphere of radius $r$ about $m/2$, the quotient $W'=V/G$ will be a manifold with $2^n$ boundary components (corresponding to $\frac12\mathbb{Z}^n/\mathbb{Z}^n \cong (\mathbb{Z}/2\mathbb{Z})^n$) each of which is homeomorphic to $\mathbb{RP}^{n-1}$. The fundamental group of each boundary component will correspond to some $G_{m/2}$ up to conjugacy. $W'$ has a 2-fold cover $V/\mathbb{Z}^n$ which is homeomorphic to $T^n$ punctured at $2^n$ balls.

Take the $2^n$ boundary components of $W'$, and glue them together in pairs, so that when $n$ is even, the induced orientations get reversed, to get a manifold $W$. For concreteness, let's say that we identify the boundary components corresponding to $m/2+\mathbb{Z}^n$ and $m/2+\frac12^n +\mathbb{Z}^n$, inducing a homomorphism $\alpha_m:G_{m/2}\to G_{m/2+\frac12^n}$. In even dimensions, $W$ will be orientable. Since $\pi_1W'= G$, and the subgroup of a boundary component corresponding to the coset $m/2+\mathbb{Z}^n$ will be conjugate to $G_{m/2}$, we see that $\pi_1 W = G \ast_{m\in 0\times\{0,1\}^{n-1}} \alpha_m$ is a multiple HNN extension by the isomorphisms pairing the subgroups. Each HNN extension will introduce a new group element $t_m$ along with a relation of the form $t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}$. So we can give a relative presentation for the fundamental group as $$\pi_1 W \cong \langle G, t_m | t_mG_{m/2}t_m^{-1}=G_{m/2+\frac12^n}, m\in 0\times\{0,1\}^{n-1}\rangle.$$ Note that there is some choice here of subgroup representative up to conjugacy which does not affect the overall group isomorphism type.

The claim is that $\pi_1 W$ does not split as a free product. This follows from the Kurosh subgroup theorem, and will be proved below.

Now suppose that $W$ is a non-trivial connect sum $W= W_1 \# W_2$. Then $\pi_1(W)=\pi_1(W_1)\ast \pi_1(W_2)$ by the Seifert-van Kampen theorem. Since $\pi_1(W)$ is not a non-trivial free product, that means that $\pi_1(W_1)=1$ (possibly after reindexing).

We need to show that $W_1'=W_1\backslash D^n$ is homeomorphic to the $n$-ball, and hence $W_1$ is the $n$-sphere. $W_1'$ lifts to the double cover of $W$ coming from the homomorphism $\pi_1(W)\to \mathbb{Z}/2\mathbb{Z}$, which is homeomorphic to $T^n \#( S^{n-1}\times S^1)^{\# 2^{n-1}}$ (a non-prime manifold). In turn, $W_1'$ lifts to the universal cover of this manifold which is a submanifold of $\mathbb{R}^n$ (because it is an infinite connect sum of $\mathbb{R}^n$s). Hence $W_1'$ is an $n$-ball by the Schoenflies Theorem, and we see that $W$ is irreducible and orientable when $n>2$ is even.

Now let's see why $\pi_1 W$ is freely indecomposable. Suppose that $\pi_1 W=A\ast B$. Since $ G < \pi_1 W$ is freely indecomposable, by the Kurosh subgroup theorem $G$ is conjugate to a subgroup of $A$ or $B$, let's say $A$. Moreover, the group $H=\pi_1 W/ \ll \mathbb{Z}^n \gg$ obtained by killing $\mathbb{Z}^n$ will be isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ since $G/\mathbb{Z}^n\cong \mathbb{Z}/2\mathbb{Z}$. Thus we see that the image $\overline{A}$ of $A$ in $H$ will contain $\mathbb{Z}/2\mathbb{Z}$, and hence will be non-trivial. Moreover, the quotient $H$ will split as a free product $\overline{A}\ast B$. However, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}^{\ast 2^{n-1}}$ is not a free product, since it has a non-trivial center, a contradiction.

Motivation

If a finitely generated group $G$ splits as a free product, then any Cayley graph for $G$ (associated to a finite generating set) has more than one end. If $G< G'$ is finite index, then the Cayley graphs of $G$ and $G'$ are almost equivalent (quasi-isometric), in fact a Cayley graph for $G$ can be obtained from one for $G'$ by collapsing some finite trees equivariantly (this is basically the Reidemeister-Schreier method).
Hence if $G$ has more than one end, so does $G'$.

Now a theorem of Stallings implies that if a group $G'$ has more than one end, then $G'$ is a graph of groups with finite edge groups. Thus, in this example, we found a manifold whose fundamental group is an HNN extension over $\mathbb{Z}/2\mathbb{Z}$ subgroups, but itself is not a free product. But it has an index 2 subgroup that does split as a free product.

If an $n$-manifold $M$ is a connect sum, and $\pi_k(M)=0$ for $k < n-1$, then a similar argument shows that $\pi_1(M)=A\ast B$ is a non-trivial free product. So any manifold finitely covered by such an $M$ will have fundamental group splitting over a finite group. One can probably find many more examples with such properties. I don't know how to find an example which is a connect sum with simply-connected summands, but finitely covers a prime manifold.

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  • $\begingroup$ Thanks for your answer, it will take me some time to fully digest it. First of all, is $\pi_1(\mathcal{O})$ the orbifold fundamental group or the regular fundamental group? Second, I am not that familiar with HNN extensions, so I apologise if this is trivial, but what is the normal subgroup $\mathbb{Z}/2\mathbb{Z}$? Does the subgroup $\mathbb{Z}/2\mathbb{Z}$ in $\pi_1(\mathcal{O})$ become normal in the iterated HNN extension? $\endgroup$ – Michael Albanese Jul 2 at 15:12
  • $\begingroup$ @MichaelAlbanese: Your first question is correct about the orbifold fundamental group - I got rid of the intermediate orbifold which is a useful trick to come up with such examples, but does not play a role in the end. Also, the subgroup is not normal: the group is a semidirect product, but there are many lifts of the {±𝐼} group. So I've modified the argument to first kill $\mathbb{Z}^n$ before obtaining a similar contradiction. Let me know if you need further clarification or see other mistakes. $\endgroup$ – Ian Agol Jul 2 at 18:51
  • $\begingroup$ Thanks a lot for taking the time to expand on your answer, I appreciate it. It's not clear to me why the universal cover of $T^n\#(S^{n−1}\times S^1)^{\# 2^{n−1}}$ is a submanifold of $\mathbb{R}^n$. Even the description as an infinite connected sum of $\mathbb{R}^n$'s is mysterious to me (in part because the universal cover of $S^{n-1}\times S^1$ is $\mathbb{R}^n\setminus\{0\}$). $\endgroup$ – Michael Albanese Jul 3 at 12:55
  • $\begingroup$ @MichaelAlbanese $R^n-0 = R^n\# R^n$. So this holds for $S^{n-1}\times S^1$. Actually, any compact submanifold in the universal cover will live inside a connect sum of finitely many $R^n$s which is just $R^n$ punctured at some points. $\endgroup$ – Ian Agol Jul 3 at 13:46
  • $\begingroup$ Please let me know if my interpretation of your use of the Schoenflies Theorem is correct. As $W_1$ is simply connected, the manifold $W_1\setminus D^n$ is a submanifold of the universal cover (here $D^n$ denotes the open $n$-ball). As you mentioned above, it follows that $W_1\setminus D^n$ embeds in $\mathbb{R}^n$, and hence there is an embedding $S^{n-1} = \partial(W_1\setminus D^n) \to \mathbb{R}^n$. By the Schoenflies Theorem, $S^{n-1}$ bounds a ball, so $W_1\setminus D^n$ is a closed ball and $W_1 = S^n$. Is that what you had in mind? $\endgroup$ – Michael Albanese Jul 3 at 18:23

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