Spin structures on finite covers

In the last couple of posts we have discussed spin structures and Stiefel-Whitney classes. It is well-known that any manifold admits a finite cover which is orientable. This is equivalent to saying that there is a cover where the first Stiefel-Whitney class vanishes, so one could reasonably ask whether it is always possible to arrange for the higher Stiefel-Whitney classes to vanish in finite covers. Of course, one could end up with simply connected manifolds which aren't spin, such as $\mathbb{CP}^2$, so one could impose that the universal cover is spin. Even with this hypothesis, however, this isn't always possible to arrange. Ebert gives a nice construction in this MO post, using properties of Higman's group (previously discussed here). However, the obstruction is still a little silly, in that the manifold has no finite covers at all! In a better behaved case, we have

Theorem 1 (Sullivan): Let $M$ be a finite volume hyperbolic manifold. Then $M$ has a finite cover which is spin.

The goal of this post is to describe a proof, due to Long--Reid, which is much easier than Sullivan's original argument. Since every compact oriented manifold of dimension at most 3 is spin (in fact has stably trivial tangent bundle), we assume that $M$ has dimension \(n \geq 4\) throughout. Many things about hyperbolic manifolds are relatively explicit and well-behaved, and in this case we make use of an explicit description of the frame bundle.

Frame bundles of hyperbolic manifolds

The oriented orthonormal frame bundle of $\mathbb{H}^n$ is the set ${\rm F}(\mathbb{H}^n)$ of all ordered $(n+1)$-tuples $(v_1,\ldots,v_n,x)$  in $(\mathbb{R}^{n,1})^{n+1}$, with the subspace topology, such that $x \in \mathbb{H}^n$,  and $\{v_1,\ldots,v_n\}$ is an orthonormal basis for ${\rm T}_x(\mathbb{H}^n)$, and ${v_1,\ldots, v_n, x}$ is a positively oriented basis of $\mathbb{R}^{n+1}$. We have the projection map $\pi: {\rm F}(\mathbb{H}^n) \to \mathbb{H}^n$ defined by $\pi(v_1,\ldots,v_n,x) = x$. 

Define $\xi: {\rm F}(\mathbb{H}^n) \to {\rm SO}^+(n,1)$ by $\xi(v_1,\ldots,v_n,x) = A$ where $A$ is the matrix whose columns vectors are $v_1, \ldots, v_n, x$. Then $\xi$ is a diffeomorphism. 

Let $e_1, \ldots, e_{n+1}$ be the standard basis of $\mathbb{R}^{n+1}$. Then 

$${\rm F}(\mathbb{H}^n) = \{(Ae_1,\ldots, Ae_{n+1}): A \in {\rm SO}^+(n,1)\}.$$

Let $\epsilon: {\rm SO}^+(n,1) \to \mathbb{H}^n$ be the evaluation map at $e_{n+1}$. Then $\epsilon \xi = \pi$. 

Now ${\rm SO}^+(n,1)$ is a principal ${\rm SO}(n)$-bundle over $\mathbb{H}^n$ with projection map $\epsilon$ and $B \in {\rm SO}(n)$ acting freely on the right of ${\rm SO}^+(n,1)$ by right multiplication by $\hat B$ where $\hat B \in {\rm SO}^+(n,1)$ is the block diagonal matrix with blocks $B$ and $(1)$. 

Moreover ${\rm SO}^+(n,1)$ is a trivial principal ${\rm SO}(n)$-bundle over $\mathbb{H}^n$. The group ${\rm SO}(n)$ acts freely on the right of $F(\mathbb{H}^n)$ by 

$$(Ae_1,\ldots, Ae_{n+1})B = (A\hat Be_1, \ldots, A\hat B e_{n+1})$$

making ${\rm F}(\mathbb{H}^n)$ into a principal ${\rm SO}(n)$-bundle over $\mathbb{H}^n$ equivalent to  ${\rm SO}^+(n,1)$ viat he diffeomorphism $\xi$. 

Let $\Gamma\backslash \mathbb{H}^n$ be an orientable hyperbolic space-form. Then $\Gamma$ acts diagonally on the left of ${\rm F}(\mathbb{H}^n)$. Moreover, $\Gamma$ acts freely and discontinuously on ${\rm F}(\mathbb{H}^n)$. 
The orthonormal frame bundle ${\rm F}(\Gamma\backslash \mathbb{H}^n)$ of $\Gamma\backslash \mathbb{H}^n$ is the orbit space $\Gamma\backslash {\rm F}(\mathbb{H}^n)$. 

The left action of $\Gamma$ on ${\rm F}(\mathbb{H}^n)$ corresponds to the left action of $\Gamma$ on ${\rm SO}^+(n,1)$  by group multiplication. We will identify ${\rm F}(\mathbb{H}^n)$ with ${\rm SO}^+(n,1)$ and ${\rm F}(\Gamma\backslash \mathbb{H}^n)$ with $\Gamma\backslash {\rm SO}^+(n,1)$. 
We have that $\Gamma\backslash {\rm SO}^+(n,1)$ is a principal ${\rm SO}(n)$-bundle over $\Gamma\backslash \mathbb{H}^n$ with right action of ${\rm SO}(n)$ induced by the right action of ${\rm SO}(n)$ on ${\rm SO}^+(n,1)$ and bundle map $\varepsilon: \Gamma\backslash {\rm SO}^+(n,1) \to \Gamma\backslash \mathbb{H}^n$ defined by $\varepsilon(\Gamma A) = \Gamma \epsilon(A)$. 

We will first reduce the proof of Theorem 1 to showing that certain central extensions are residually finite.

As with \(\mathrm{SO}(n)\), the group \(\mathrm{SO^{\circ}}(n,1)\) has a universal 2-fold cover, which we denote by \(\mathrm{Spin^+}(n,1)\) (with covering map \(\phi\)) and, as above, there is an exact sequence
\[
1 \to \{\pm 1\} \to \mathrm{Spin^+}(n,1) \to \mathrm{SO^{\circ}}(n,1) \to 1.
\]
Using this and the identification of the \(\mathrm{SO}(n)\)-principal bundle of oriented orthonormal frames on \(TM\) with \(\Gamma\backslash\mathrm{SO^{\circ}}(n,1)\), we can construct an extension \(\overline{\Gamma} < \mathrm{Spin^+}(n,1)\) with \(\phi(\overline{\Gamma}) = \Gamma\). Note that \(\overline{\Gamma}\backslash\mathrm{Spin^+}(n,1) \cong \Gamma\backslash\mathrm{SO^{\circ}}(n,1)\) since
\[
\Gamma\backslash\mathrm{SO^{\circ}}(n,1) \cong (\{\pm 1\}\backslash\overline{\Gamma}) \backslash (\{\pm 1\}\backslash\mathrm{Spin^+}(n,1)) \cong \overline{\Gamma}\backslash\mathrm{Spin^+}(n,1).
\]

Lemma 2:  Let \(M = \mathbb{H}^n/\Gamma\) be an orientable finite-volume hyperbolic manifold with \(n \geq 4\).

1. Suppose that there is a subgroup \(H < \overline{\Gamma}\) of index 2 so that \(\phi\) maps \(H\) isomorphically onto \(\Gamma\) (or, equivalently, \(H \cap \{\pm 1\} = 1\)). Then \(M\) is spinnable.
2. Let \(D < \overline{\Gamma}\) be of finite index containing an index 2 subgroup \(D_0\) such that \(D_0 \cap \{\pm 1\} = 1\). Let \(\Delta = \phi(D) < \Gamma\). Then \(\mathbb{H}^n/\Delta\) is a finite cover of \(M\) that is spinnable.

Proof sketch: For the first part, we have an exact sequence
\[
1 \to \{\pm 1\} \to \overline{\Gamma} \to \Gamma \to 1.
\]
If \(H\) is a subgroup of index 2, as claimed, then \(H\backslash\mathrm{Spin^+}(n,1)\) is a principal \(\mathrm{Spin}(n)\)-bundle that double covers \(\overline{\Gamma}\backslash\mathrm{Spin^+}(n,1)\), which, from the discussion above, is \(\cong \Gamma\backslash\mathrm{SO^{\circ}}(n,1)\). Moreover, the right action of \(\mathrm{SO}(n)\) lifts to the right action of \(\mathrm{Spin}(n)\): that is, \(M\) is spinnable. I think this is sort of believable and find the details a bit annoying. See also Theorem 2.1 of this paper. 

The second part follows from the first part on noting that \(\mathbb{H}^n/\Delta\) is a finite cover of \(M\). $\blacksquare$

A proof of Sullivan's theorem

Theorem 1 will follow quickly from the next proposition.

Proposition 3: Let \(M^n = \mathbb{H}^n/\Gamma\) be a finite-volume orientable hyperbolic \(n\)-manifold and let \(\overline{\Gamma} < \mathrm{Spin^+}(n,1)\) with \(\phi(\overline{\Gamma}) = \Gamma\). Then \(\overline{\Gamma}\) is residually finite.

Proof: Recall the following from the last post. Let \(V\) be an \(m\)-dimensional vector space over \(\mathbb{R}\) and let \(q\) be a non-degenerate quadratic form on \(V\). The Clifford algebra \(\mathcal{CL}(V,q)\) associated to \((V,q)\) is the associative algebra with 1 obtained from the free tensor algebra on \(V\) by adding relations \(v \otimes v = -q(v)1\) for each \(v \in V\). Note that \(V\) embeds naturally into \(\mathcal{CL}(V,q)\), and \(\mathcal{CL}(V,q)\) has the structure of a real vector space of dimension \(2^m\) with a basis \(B\) constructed naturally from \(V\). 

Let \(P(V,q)\) denote the multiplicative group of \(\mathcal{CL}(V,q)\) generated by all \(v \in V\) such that \(q(v) \neq 0\). Then the spin group of \((V,q)\) is the subgroup of \(P(V,q)\) defined as
\[
\mathrm{Spin}(V,q) = \{v_1 \cdots v_k : v_i \in V, q(v_i) = \pm 1 \text{ for each } i, \text{ and } k \text{ even}\}.
\]
In the case when \(q = J_n\), \(\mathcal{CL}(V,q)\) is denoted by \(\mathcal{CL}(n,1)\), the group \(P(V,q)\) is denoted by \(P(n,1)\) and \(\mathrm{Spin}(V,q) = \mathrm{Spin}(n,1)\). The group \(\mathrm{Spin^+}(n,1)\) is the connected component of the identity in \(\mathrm{Spin}(n,1)\).

Now the group \(\mathrm{Spin}(n,1)\) acts on the vector space \(\mathcal{CL}(n,1)\) by left multiplication (on the basis \(B\)) thereby determining a faithful linear representation \(L: \mathrm{Spin}(n,1) \to \mathrm{GL}(2^{n+1}, \mathbb{R})\).

Since \(L(\overline{\Gamma})\) is a finitely generated subgroup of \(\mathrm{GL}(2^{n+1}, \mathbb{R})\) and hence is residually finite by Malcev's theorem. $\blacksquare$

Proof of Theorem 1: Residual finiteness implies that there exists a finite quotient \(\psi: \overline{\Gamma} \to Q\) so that \(\psi\) is injective on \(\{\pm 1\}\). Let \(D < \overline{\Gamma}\) be the subgroup of finite index given by \(\psi^{-1}(\psi(\{\pm 1\}))\). Then \(D\) contains a subgroup \(D_0\) of index 2 such that \(D_0 \cap \{\pm 1\} = 1\). Let \(\Delta = \phi(D)\). Then \(\Delta\) is a finite index subgroup of \(\Gamma\) and \(\mathbb{H}^n/\Delta\) is spinnable by Lemma 2.$\blacksquare$

Remark: Note that what is really important in Proposition 2 is that \(\overline{\Gamma}\) is finitely generated. However, it is important that the extensions considered above are of arithmetic groups in \(\mathrm{SO^{\circ}}(n,1)\) since Millson constructs arithmetic groups in \(\mathrm{SL}(n, \mathbb{R}) \times \mathrm{SL}(n, \mathbb{R})\) that have extensions by \(\mathbb{Z}/2\mathbb{Z}\) that are not residually finite, and Deligne constructed central extensions of integer symplectic groups which aren't residually finite.

As mentioned before, all compact oriented 3-manifolds admit a spin structure. To check that Theorem 1 isn't vacuous, we should exhibit the existence of non-spin finite volume hyperbolic manifolds. This, however, turns out to be rather hard. It was only in 2020 that Martelli--Riolo--Slavich constructed the first example of a compact orientable hyperbolic 4-manifold $M$ that does not admit any spin structure. They construct this via a completely explicit gluing of cells, and keep track of this carefully enough that they can show there is a ($\pi_1$-injective, embedded) genus 3 surface $S \subset M$ with self-intersection 1. Since 4-manifolds are spin if and only if the intersection form is even, this implies $M$ isn't spin. Furthermore, $M$ is arithmetic of first/simplest type because of the tessellation into 120-cells. It is also the first example of a 4-manifold where the second homology classes can't be generated by geodesically immersed surfaces.

Using this and an embedding theorem of Kolpakov--Reid--Slavich, they show that, for all $n \geq 4$ there exists a nonspin, compact, orientable, arithmetic, hyperbolic $n$-manifold, and hence non-stably parallelisable ones. This last point is most interesting for odd $n$, since non-stably parallelisable implies vanishing Euler characteristic but for even $n$ Chern--Gauss--Bonnet says the Euler characteristic is non-vanishing.

While unrelated to spin structures, it is interesting to note that Martelli--Riolo--Slavich also show that $\pi_1(S)$ determines a cover $\tilde{M}$ which is geometrically finite and diffeomorphic to a rank 2 bundle over $S$ with Euler number 1.That there are nontrivial bundles over surfaces that cover some compact, hyperbolic 4-manifolds was first shown by Gromov--Lawson--Thurston. It is still open whether there is a surface-by-surface bundle which is hyperbolic in the sense of either groups or manifolds, although I'm sure everyone is staring at Kent--Leininger's breakthrough construction of purely pseudo-Anosov subgroups of the mapping class group with baited breath (and most people think that the bundles can be made at least word hyperbolic).