# If the universal cover of a manifold is spin, must it admit a finite cover which is spin?

If $$M$$ is non-orientable, then it has a finite cover which is orientable (in particular, the orientable double cover).

If $$M$$ is non-spin, then it does not necessarily have a finite cover which is spin, e.g. $$M = \mathbb{CP}^2$$. As a cover of a spin manifold is spin, a necessary condition for $$M$$ to admit such a finite cover is that its universal cover is spin (which is not the case in the previous example). In analogy with the first sentence, note that the universal cover is always orientable.

Let $$M$$ be a closed smooth manifold whose universal cover is spin. Is there a finite cover of $$M$$ which is spin?

This question is partially motivated by the study of positive scalar curvature. The Dirac operator on a spin manifold can be used to obtain obstructions to positive scalar curvature à la Lichnerowicz, Hitchin, Gromov & Lawson, Rosenberg, etc. More generally, these techniques can be applied to manifolds which admit a spin cover. Things are generally more difficult in the non-compact case than in the compact case, so if the answer to my question were 'yes', we could just pass to a compact cover which is spin and apply the techniques there (as opposed to passing to the potentially non-compact universal cover).

• You can even get an example with contractible universal covering. I will write details later when I have more time, or someone else can do this using Davis' trick. – Moishe Kohan Sep 8 at 18:16
• @MoisheKohan: I would be very interested to see such examples. – Michael Albanese Sep 8 at 21:32

No, this is not true: for each dimension $$d \geq 4$$, there is a closed, oriented $$d$$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.

The reason is simply that there are finitely presented groups which have no nontrivial finite quotient. One example is Higman's group $$H$$, see https://en.wikipedia.org/wiki/Higman_group.

The key features of $$H$$ are:

1. $$H$$ is infinite,
2. $$H$$ does not admit a nontrivial finite quotient,
3. $$H$$ is acyclic,
4. $$H$$ has a classifying space $$BH$$ which is a finite $$2$$-dimensional CW-complex.

The proof of 1,2 can be found in Tao's blog https://terrytao.wordpress.com/2008/10/06/finite-subsets-of-groups-with-no-finite-models/, and the proof of 3,4 in ''The topology of discrete groups'' by Baumslag, Dyer, Heller).

Now pick an element $$1 \neq x \in H$$, which induces an injective homomorphism $$\mathbb{Z} \to H$$ and form the amalgamated product $$G=H \ast_{\mathbb{Z}} H$$. The group $$G$$ is infinite and has no nontrivial homomorphism to a finite group $$F$$, since any homomorphism $$G \to F$$ must vanish on the two copies of $$H$$.

The pushout $$BH \cup_{S^1} BH$$ is aspherical by Whiteheads asphericity theorem and hence a model for $$BG$$. I have designed things so that $$H_2(BG) \cong \mathbb{Z}$$ and all other homology groups are trivial. In particular, $$G$$ is perfect, and the Quillen plus construction $$BG^+$$ must be homotopy equivalent to $$S^2$$, so that there is a homology equivalence $$f:BG \to S^2$$. Now let $$V \to S^2$$ be the nontrivial oriented vector bundle of rank $$d$$, which has $$w_2 (V) \neq 0$$. It follows that the vector bundle $$f^\ast V \to BG$$ is not spin. $$BG$$ has no nontrivial cover, and $$BG$$ is aspherical, so the pullback of $$f^\ast V$$ to the universal cover is trivial.

Now there exists, when $$d \geq 4$$, a closed $$d$$-manifold $$M$$ with a $$2$$-connected map $$\ell: M \to BG$$ and a bundle isomorphism $$TM\oplus \mathbb{R}\cong \ell^\ast f^\ast V \oplus \mathbb{R}$$. This is achieved by surgery below the middle dimension. In particular, $$\pi_1 (M)\cong G$$. Hence $$\pi_1(M)$$ has no nontrivial finite index normal subgroup, and therefore no nontrivial finite index subgroup at all. It follows that $$M$$ does not have a nontrivial finite cover.

By construction, $$w_2 (TM) \neq 0$$, but the universal cover of $$M$$ is stably parallelizable.

• Thanks for your answer. I think I follow the construction up until the penultimate paragraph. I am trying to understand the construction of $M$ via Theorem 1.2 on page 22 here. Is there a more direct result I can apply to deduce the existence of $M$? – Michael Albanese Sep 9 at 15:16
• I don't think you can get it in a more direct way, because you also want to control the tangent/normal bundle along the process. An easier strategy to construct a manifold with fundamental group $G$ which is often mentioned is to embed the 2-skeleton of BG into R^5 (or some larger R^n), to take a regular neighborhood of the image and to take its boundary. The result would, however, be stably parallelizable. – Johannes Ebert Sep 9 at 16:01
• I know this is technique is relatively standard, but it's my first time seeing it. I don't see how the aforementioned theorem allows us to deduce that there is a $2$-connected map $\ell : M \to BG$ with $TM\cong\ell^*f^*V$. Rather, from the definition of normal map on page 19, it seems to me that one obtains a $2$-connected map $\ell : M \to BG$ with $TM\oplus\ell^*f^*V$ stably trivial. What am I missing? – Michael Albanese Sep 9 at 18:57
• One only gets an isomorphism $TM \oplus \mathbb{R}\cong \ell^\ast f^\ast V\oplus \mathbb{R}$; I have corrected the statement. To deduce my claim from the result in Wall's book, you pick a complementary bundle $V^\bot$ of large dimension, apply Wall's result to get $TM \oplus \ell^\ast f^\ast V^\bot$ stably trivial, add further trivial bundles to $V^\bot$ to make the sum actually trivial. Add a copy of $\ell^\ast f^\ast V$ to both sides of the equation and get that $TM \oplus \mathbb{R}^n \cong \ell^\ast f^\ast V \oplus \mathbb{R}^n$ for some large $n$. (ctd.) – Johannes Ebert Sep 10 at 7:33
• Thanks for fleshing out the details, I can now see why $w_2(TM) \neq 0$. Is the universal cover of $M$ parallelisable or only stably parallelisable? I can show the latter using the isomorphism $TM\oplus\mathbb{R} \cong \ell^*f^*V\oplus\mathbb{R}$. – Michael Albanese Sep 10 at 22:25

As promised, here is my solution based on the Davis trick. First, there is a very general construction of PL aspherical 4-manifolds (it also works in higher dimensions). Start with a finite aspherical 2-dimensional CW complex $$W$$. Up to homotopy, $$W$$ always embeds in the Euclidean 4-space $$E^4$$ (I think, this is due to Stallings). Take such an embedding and let $$N=N(W)$$ denote a regular neighborhood of $$W$$ in $$E^4$$. Now, apply "Davis trick" to $$N$$: Introduce a reflection orbifold structure on the boundary of $$N$$ such that the corresponding stratification of the boundary is dual to a triangulation of $$\partial N$$. The resulting orbifold $${\mathcal O}$$ is very good (admits a finite orientable manifold-covering $$M\to {\mathcal O}$$) and its universal covering (same for $$M$$ and for $${\mathcal O}$$) is contractible. As a bonus, $$\pi_1(W)$$ embeds in $$\pi_1(M)< \pi_1({\mathcal O})$$. For details see

Mess, Geoffrey, Examples of Poincaré duality groups, Proc. Am. Math. Soc. 110, No. 4, 1145-1146 (1990). ZBL0709.57025.

and, of course, the original paper by Mike Davis from 1983. (Actually, it was Bill Thurston who came up with this trick in the context of 3-manifolds: He used it for the proof of his hyperbolization theorem.) This construction allows one to embed 2-dimensional finitely presented groups with "exotic properties" in fundamental groups of closed aspherical PL manifolds.

I will use a relative version of this construction. Start with a closed connected oriented surface of genus $$\ge 1$$; I'll take the torus $$T^2$$. Let $$E\to T^2$$ be the 2-disk bundle over $$T^2$$ with the Euler number $$\pm 1$$. The boundary of the 4-manifold $$E$$ is a 3-dimensional Nil-manifold: The total space of a nontrivial circle bundle over the torus. The group $$\pi_1(\partial E)$$ has two generators $$a, b$$, and $$\pi_1(\partial E)$$ has the presentation $$\langle a, b| [a,b]=t, [a,t]=1, [b,t]=1\rangle.$$ Represent $$a, b$$ by simple disjoint loops $$\alpha, \beta$$ in $$\partial E$$. Now, take your favorite finite 2-dimensional aspherical complex $$W$$ whose fundamental group is nontrivial and has no proper finite index subgroups (I care only about the homotopy type of $$W$$). The standard example is the presentation complex of Higman group. But there are many other examples. As before, embed $$W$$ in $$E^4$$, take a regular neighborhood $$N$$ of $$W$$ in $$E^4$$. Then $$\pi_1(\partial N)$$ maps nontrivially to $$\pi_1(W)$$. Pick two simple loops $$\alpha', \beta'\subset \partial N$$ which map nontrivially to $$\pi_1(W)$$ (you can take the same loop).

Now, take two copies $$N_a, N_b$$ of $$N$$ and attach them to $$E$$ by identifying a regular neighborhood of $$\alpha'$$ to that of $$\alpha$$ for $$N_a$$ and identifying a regular neighborhood of $$\beta'$$ to that of $$\beta$$ for $$N_b$$. The result is a compact PL aspherical 4-manifold with boundary $$Z$$. The $$\pi_1(Z)$$ is an amalgam of $$\pi_1(E)\cong {\mathbb Z}^2$$ with two copies of $$\pi_1(N)$$ (along infinite cyclic subgroups). For each homomorphism to a finite group $$\phi: \pi_1(Z)\to \Phi$$ the subgroups $$\pi_1(N_a), \pi_1(N_b)$$ will have to map trivially. Hence, $$a$$ and $$b$$ will have to map trivially as well. Since $$a, b$$ generate $$\pi_1(E)$$, $$\pi_1(Z)$$ has no nontrivial homomorphisms to finite groups. Now, apply Davis trick to $$Z$$. The result is an orbifold $${\mathcal O}$$. Since $$Z$$ was aspherical, so is $${\mathcal O}$$ (i.e. it has contractible universal covering space).

Take a finite orientable manifold-covering $$M\to {\mathcal O}$$. Then $$M$$, of course, has contractible (hence, spin) universal covering. I claim that $$M$$ has no finite spin-covering spaces. Indeed, for each finite-sheeted covering $$p: M'\to M$$, the manifold $$int(Z)\subset M$$ has to lift trivially; more precisely, $$p$$ restricts to a trivial covering $$p^{-1}(int Z)\to int Z.$$ This is because $$\pi_1(Z)$$ has no nontrivial homomorphisms to finite groups. Thus, $$M'$$ contains a copy of $$E$$. In particular, $$M'$$ contains a 2-torus with odd self-intersection, i.e. the intersection form of $$M'$$ is not even, i.e. $$M'$$ is not spin.

I was working in the PL category but in dimension 4, PL is the same DIFF, so you get a smooth example as well.

Edit. Lemma. Let $$M$$ be a triangulated manifold, $$W\subset M$$ is a subcomplex and $$N=N(W)$$ is the regular neighborhood of $$W$$ in $$M$$. Then the inclusion map $$W\to N$$ is a homotopy-equivalence; if $$W$$ is connected and has codimension $$\ge 2$$ in $$M$$ then $$\partial N$$ is connected and the induced map $$\pi_1(\partial N)\to \pi_1(W)$$ is surjective.

Proof. The homotopy-equivalence part is standard and holds for general simplicial complexes $$M$$, not just for manifolds. Moreover, the inclusion map $$\partial N\to (N \setminus W)$$ is also a homotopy-equivalence. (Both are proven using "straight-line homotopy.")

I will prove the second part. Take an arc $$\alpha$$ in $$N$$ connecting two points $$x, y\in \partial N$$. Since $$W$$ has codimension $$\ge 2$$, taking $$\alpha$$ in general position, we see that it will be disjoint from $$W$$, hence, is homotopic relative to $$\{x, y\}$$ to an arc in $$\partial N$$. (I am using here and below the h.e. $$\partial N\to N-W$$.) Thus, $$\partial N$$ is connected. Next, let $$\alpha$$ be a loop in $$N$$ based at $$x\in \partial N$$. By the same argument, $$\alpha$$ is homotopic to a loop based at $$x$$ and contained in $$N-W$$, hence, to a loop in $$\partial N$$.

• I think you mean $\pi_1(\partial E)$ has three generators $a$, $b$, and $t$. In the following paragraph, how do you know that $\pi_1(\partial N)$ maps non-trivially to $\pi_1(W)$? Is the inclusion map $\partial N \hookrightarrow N$ followed by the deformation retraction $N \to W$ non-trivial on $\pi_1$? – Michael Albanese Sep 14 at 15:36
• @MichaelAlbanese If you prefer three, but two suffice, since [𝑎,𝑏]=𝑡. – Moishe Kohan Sep 14 at 18:54
• [Feel free to ignore this question] If you don't include $t$ as a generator, what does $[a, b] = t$ mean ($t$ is not a word in $a$ and $b$)? – Michael Albanese Sep 14 at 22:47
• Am I correct in saying that the reason you apply the Davis trick to $Z$ is to obtain a closed aspherical manifold $M$ (as opposed to $Z$ which is an aspherical manifold with boundary)? – Michael Albanese Sep 14 at 22:49
• @MichaelAlbanese: $t$ is a letter in a presentation: You can eliminate it if you like by writing down relators of the form $[a,[a,b]]=1$. The bottom line is that you need just two generators. – Moishe Kohan Sep 14 at 22:49