An orientable non-spin${}^c$ manifold with a spin${}^c$ covering space

Question

Is there a closed, smooth, orientable manifold which is not spin${}^c$ but has a finite cover which is spin${}^c$?

Such examples exist when spin${}^c$ is replaced by spin: an Enriques surface is not spin but it is double covered by a K3 surface which is spin.

Every orientable manifold of dimension at most four is spin${}^c$, so any such examples must have dimension at least five. The only non-spin${}^c$ manifold I know of is the Wu manifold $SU(3)/SO(3)$. Of course taking products or connected sums with other manifolds provides more examples.

Answer 1

This is probably overkill, but I couldn't resist advertising a preprint that Diarmuid Crowley and I recently posted to the arXiv: https://arxiv.org/abs/1802.01296

In the final section we discuss examples due to Teichner of closed $6$-manifolds with extraordinary cohomological properties. These examples are constructed as sphere bundles of rank $3$ vector bundles over closed $4$-manifolds. In particular, in Proposition 5.9 we consider the total space of a sphere bundle $S^2\to N:=S(E)\to M$ associated to a vector bundle $E$ whose base manifold has $$ \pi_1(M)=\mathbb{Z}/8\rtimes\mathbb{Z}/2. $$ The manifold $N$ is not spin$^c$. This follows from Theorem 1.3 and Proposition 5.9, since $N$ has a cohomology class $x\in H^2(N;\mathbb{Z}/2)$ such that $\beta(x^2)$ is not a multiple of $\beta(x)$ (here $\beta$ denotes the Bockstein from mod $2$ to integral cohomology).

Now, if we pull back $E$ to obtain a bundle $\widetilde{E}$ over the universal cover $\widetilde{M}$, then $\widetilde{N}:=S(\widetilde{E})$ will be a finite cover of $N$. It remains to show that $\widetilde{N}$ is spin$^c$.

Let $\pi:\widetilde{N}\to \widetilde{M}$ denote the bundle projection. A general argument about tangent bundles of sphere bundles gives a bundle isomorphism $$ T\widetilde{N}\oplus\mathbb{R}\cong \pi^*T\widetilde{M} \oplus \pi^*\widetilde{E}, $$ which (since $\widetilde{M}$ and $\widetilde{E}$ are orientable) gives $$w_2(\widetilde{N})= \pi^*w_2(\widetilde{M}) + \pi^*w_2(\widetilde{E}).$$ Therefore, $$\beta w_2(\widetilde{N})= \pi^*\beta w_2(\widetilde{M}) + \pi^*\beta w_2(\widetilde{E}).$$ The first term on the right vanishes since an orientable $4$-manifold is spin$^c$. The second term vanishes since $\widetilde{E}$ is an orientable bundle of rank $3$, and therefore $\beta w_2(\widetilde{E})=e(\widetilde{E})$ equals the Euler class, which pulls back to zero in the sphere bundle. Hence $\beta w_2(\widetilde{N})=0$, and $\widetilde{N}$ is spin$^c$ as claimed.

Answer 2

The Dold manifold $P(1,2)$ is a simple example. It is defined by quotienting the ${\rm Spin}^c$ 5-manifold $S^1\times \mathbb{CP}^2$ via the (fixed points free and orientation preserving) involution $(x,z) \mapsto (-x,\bar z)$. In general this construction works for every $P(m,n)=S^m \times \mathbb{CP}^n$ with $m$ odd and $n$ even.

See this paper of Plymen from 1986, Section 4.7