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Background

Let $M$ be a spin manifold and let $\Gamma$ be a finite group acting freely and isometrically on $M$ in such a way that $M/\Gamma$ is a smooth riemannian manifold. The quotient will be spin if and only if the action of $\Gamma$ on $M$ lifts to the spin bundle.

For reasons having to do with $11 = 7 + 4$, I got interested in $M=S^7$ with the round metric. There is a unique spin structure on $S^7$ and the spin bundle is $$\mathrm{Spin}(7) \to \mathrm{Spin}(8) \to S^7.$$

A while back, together with one of my students, we investigated which smooth quotients $S^7/\Gamma$ are spin and how many inequivalent spin structures they admit. This boils down to determining the isomorphic lifts of $\Gamma \subset \mathrm{SO}(8)$ to $\mathrm{Spin}(8)$.

There are lots of finite subgroups $\Gamma \subset \mathrm{SO}(8)$ acting freely on $S^7$, which are listed in Wolf's Spaces of constant curvature and to our surprise (this does not happen with $S^5$, say) we found that all quotients $S^7/\Gamma$ are spin; although they do not all have the same number of spin structures. Our results were obtained by a case-by-case analysis, but we always remained with the sneaky suspicion that there ought to be a simple topological explanation.

Question

Is there one? Perhaps based on the parallelizability of $S^7$?

Thanks in advance.

Edit

I'm answering Chris's questions in the first comment below.

The problem is indeed the existence of a subgroup $\Gamma' \subset \mathrm{Spin}(8)$ such that obvious square commutes: $$\Gamma' \to \Gamma \to \mathrm{SO}(8) = \Gamma' \to \mathrm{Spin}(8) \to \mathrm{SO}(8)$$ and where the first map $\Gamma' \to \Gamma$ is an isomorphism. This is the same as lifting $\Gamma \to \mathrm{SO}(8)$ via the spin double cover.

The simplest counterexample for $S^5$ is to take any freely acting cyclic subgroup $\Gamma \subset \mathrm{SO}(6)$ of even order.

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  • $\begingroup$ Two questions: (1) is this equivalent to asking whether the homomorphism $\Gamma \to SO(8)$ lifts to Spin(8)? (2) What is the counter example in the case of the five-sphere $S^5$. I have a sketch of a proof, but it seems to work too generally so something must be wrong. $\endgroup$
    – Chris Schommer-Pries
    Dec 9, 2009 at 17:02
  • $\begingroup$ I will answer this as an edit to the question, if that's alright. $\endgroup$
    – José Figueroa-O'Farrill
    Dec 9, 2009 at 17:37

2 Answers 2

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Here is a partial answer. If the order of $\Gamma$ is odd, then this is a trivial application of transfer maps. You have described your manifold as a quotient $\pi:S^7 \to M = S^7/\Gamma$, and hence $S^7$ is a covering space of $M$. The transfer map is a wrong way map in cohomology:

$ \tau^* : H^* (S^7) \to H^* (M) $

which exists for cohomology in, say, $\mathbb{Z}/2$-coefficients. The composition $\tau^* \pi^*$ is multiplication by the order of $\Gamma$, which in this case is an isomorphism when the order of $\Gamma$ is odd. But since the cohomology of $S^7$ vanishes in degrees 1 and 2, this proves that these groups also vanish for $M$ and hence $M$ has a spin structure and it is unique.

The more interesting case is when $\Gamma$ is 2-primary. For example why does $\mathbb{R}P^7$ have a spin structure? I suspect that your intuition is spot on and that it has to do the framing of $S^7$.

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  • $\begingroup$ Thanks! Nice argument. Most relevant groups, though, have even order :) In fact, the only freely acting subgroups of SO(8) which have odd order are those which are an extension of an odd-order cyclic by another odd-order cyclic (the orders satisfying some arithmetic condition). Can you point me to a reference for the transfer maps, by the way? $\endgroup$
    – José Figueroa-O'Farrill
    Dec 7, 2009 at 21:19
  • $\begingroup$ You don't really need the transfer for this: the map $S^7/\Gamma \to B\Gamma$ classifying the covering space is 6-connected, so if the group has odd order $S^7/\Gamma$ has no $\mathbb{F}_2$-cohomology to support Stiefel--Whitney classes. $\endgroup$
    – Oscar Randal-Williams
    Dec 7, 2009 at 22:26
  • $\begingroup$ @Oscar: You still need to know that Z/2 cohomology of $B \Gamma$ vanishes, and this is usually proven with transfers. @Jose: Most standard references have a chapter on transfers. The standard reference for me is Allen Hatcher Algebraic Topology which is available online from his website and has a chapter on transfers. (I think he does it for integral cohomology, but the Z/2 case is exactly the same). $\endgroup$
    – Chris Schommer-Pries
    Dec 8, 2009 at 0:15
  • $\begingroup$ On the otherhand, since we only need to know that the low dimensional cohomology groups (1 and 2) of $B\Gamma$ vanish, we can probably avoid transfers by looking at explicit interpretations of group cohomology. $ H^1 (\Gamma, Z/2) = Hom(\Gamma, Z/2)$ vanishes if $\Gamma$ is of odd order, so we just need to calculate that there are no non-trivial Z/2 central extensions of $\Gamma$ to show that $H^2$ vanishes. This is probably a standard calculation. $\endgroup$
    – Chris Schommer-Pries
    Dec 8, 2009 at 12:18
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I would suspect triality is involved. The two spinor representations of spin(8) of spin(8) have the same dimension as the fundamental vector representation of spin(8) for all other spinor groups the representations don't have the same dimension and I believe this is related to triality. Here is an article on triality:

http://en.wikipedia.org/wiki/Triality

It has references to more material. Also see this article on SO(8):

http://en.wikipedia.org/wiki/SO(8)

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  • $\begingroup$ Triality is indeed an automorphism of $\Spin(8)$ which relates the vector and the two half-spinor representations. I am not sure how to use it to lift $\Gamma$, though. Since $\Gamma$ is a subgroup of $\mathrm{SO}(V)$ where $V$ is the vector representation, then triality maps it isomorphically to a subgroup of $\mathrm{SO}(\Delta_\pm)$, where $\Delta_\pm$ are the half-spinor reps. Is it clear that either of these two subgroups lift to $\mathrm{Spin}(8)$? I'll have to think about it some more, but it looks promising, so thanks! You've got my +1. $\endgroup$
    – José Figueroa-O'Farrill
    Dec 7, 2009 at 20:11

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