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!



Second Stiefel Whitney class of quotients of odd spheres
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](http://uk.arxiv.org/abs/hep-th/0401206) 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. 


