Second Stiefel Whitney class of quotients of odd spheres

Question

I don't know much of algebraic topology so the following question could be very silly. Let $G$ a finite subgroup of $U(n)$ that acts linearly (the action induced by the action of $U(n)$ on $\mathbb{C}^{n}$) and freely on the unit sphere $S^{2n-1}\subset \mathbb{C}^{n}$. Let $X$ be the quotient manifold $$X:=S^{2n-1}/G$$
I want to compute $w_{2}(X)\in H^{2}(X,\mathbb{Z}_{2})$ i.e. the second Stiefel-Whitney class of the tangent bundle of $X$. More precisely i'd like to know if it is $0$ or not and under which hypotheses on $G$ it vanishes. Is this a standard problem? Is there some reference for this type of calculation or something similar?

Thank you in advance!

Answer 1

The second Stiefel-Whitney class of $TX$ will vanish if and only if $X$ admits a spin structure and this will be the case if and only if the action of $G$ on the oriented orthonormal frame bundle of $X$ lifts to an action on the (unique) spin bundle of the round sphere. The total space of the oriented orthonormal frame bundle of $S^{2n-1}$ is the Lie group $SO(2n)$ and the action of $G$ is given by left multiplication in $SO(2n)$ under the natural embedding $G < U(n) < SO(2n)$. Let $\theta: Spin(2n) \to SO(2n)$ denote the spin cover. Then $G$ lifts if and only if there is a subgroup $\hat G < Spin(2n)$ isomorphic to $G$ and which covers $G$ under $\theta$.

Now the above does not use the fact that $G < U(n)$ and it is the general situation for any subgroup of $SO(2n)$. In your case one should be able to say more.

For example, if $G< SU(n)$ then the quotient would be spin: in fact, $SU(n)$ acts faithfully on the spinor representations of $Spin(2n)$.

For $G<U(n)$ but not a subgroup of $SU(n)$, one way to explicitly determine the existence of the lift is the following. Exhibit $G$ in terms of generators and relations and find lifts to $Spin(2n)$ of the generators. There is usually a "sign" ambiguity in lifting each generator and then you have to decide whether these signs can be chosen in such a way that the relations are satisfied in $Spin(2n)$ on the nose and not just up to a sign.

An illustrative example can be found in §5.1.1 in this paper of Joan Simón and myself; although we do not look at finite groups $G$ but rather circle groups. But clearly you can choose finite cyclic subgroups of the circle group and the same calculation applies.

Answer 2

For cyclic groups, you can in fact go further and calculate all of the Stiefel-Whitney classes similarly to how you calculate them for real projective spaces. You can find the latter in standard texts (eg Milnor-Stasheff, Characteristic Classes). If G is a cyclic group of order $m$, acting by a standard linear action, the quotient is a high-dimensional lens space. As such, there is an $m$-fold covering map from a sphere to the lens space, which embeds in the unit sphere bundle of a complex line bundle $\lambda$. The basic idea is to show that the tangent bundle, stabilized by adding trivial line bundles, splits as a sum of powers of $\lambda$; this leads directly to a formula for all Stiefel-Whitney (and Pontrjagin) classes.

I would imagine that a similar argument works for more general G, but some representation theory probably intervenes. In general, it's probably easier to try to solve the lifting problem algebraically as in José's answer above.

References for the cyclic case: J. Folkman, Equivariant maps of spheres into the classical groups, Mem. Amer . Math. Soc. 95 (1971).

Kwak, Jin Ho Some cyclic group actions on a homotopy sphere and the parallelizability of its orbit spaces. Publ. Res. Inst. Math. Sci. 21 (1985), no. 4, 807–818.

Ewing, J., S. Moolgavkar, L. Smith and R. S. Stong, Stable parallelizability of lens spaces, J. of Pure and Applied Algebra, 10 (1977), 177-191.

Answer 3

Because $G \subset U(n)$, the stabilised tangent bundle of $X = S^{2n-1}/G$ has an obvious complex structure. The determinant bundle $\lambda$ is the quotient of $S^{2n-1} \times \mathbb{C}$ by $G$ acting as $(p, z) \mapsto (g(p), (\det g) z)$. Now $\det : G \to \mathbb{R}/\mathbb{Z}$ descends to a homomorphism $H_1(X) \to \mathbb{R}/\mathbb{Z}$, so defines an element $\alpha \in H^1(X; \mathbb{R}/\mathbb{Z})$. Then $c_1(\lambda) \in H^2(X; \mathbb{Z})$ is the image of $\alpha$ under the Bockstein map associated to $0 \to \mathbb{Z} \to \mathbb{R} \to \mathbb{R}/\mathbb{Z} \to 0$. Its mod 2 reduction equals $w_2(X) \in H^2(X; \mathbb{Z}/2\mathbb{Z})$.

Equivalently, $w_2(X)$ is the image of $\alpha$ under the Bockstein of $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{R}/2\mathbb{Z} \to \mathbb{R}/\mathbb{Z} \to 0$. Whether $\alpha$ is in the image of $H^1(X; \mathbb{R}/2\mathbb{Z})$ depends on whether one can make a consistent choice of square roots of $\det g$ for $g \in G$ (essentially the same lifting problem that appears in José's answer). I believe this is possible if and only if any $\bar g$ in the abelianisation of $G$ with $\det \bar g = -1$ has order divisible by 4 (or if you like: $g^2$ is not a commutator for any $g \in G$ with $\det g = -1$). That is then a criterion for $w_2(X)$ to vanish.