Question

I am trying to understand spin structures and am looking at the specific case of complex projective space (viewed as the quotient $SU(N)/U(N-1)$) and more generally the Grassmannians (viewed as the quotient $SU(N)/(U(N-k) \times U(k))$. My questions are as follows:

(1) For what values of $N$ does complex projective $N$-space have a spin structure?

(2) For these values, is the canonical spinor bundle equivarient with respect to the $U(N-1)$ action?

(3) If so, what is the associated representation of $U(N-1)$?

(4) All of the above for the Grassmannians?

Answer 1

I think there is a slight mistake in the formulation of the question. CP^n is the homogeneous space U(n+1)/(U(n) \times U(1))=SU(n+1) /G with G= SU(n+1) \cap (U(n) \times U(1)). The right formulation of question (2) is: is the spin structure on CP^n (for odd n, there is unique spin structure on CP^n, see Charles Siegel's answer) U(n+1)-equivariant?

The answer is no, for a very elementary reason: if the spin structure were U(n+1)-equivariant, then it certainly were U(n)-equivariant, where U(n) embeds into the product in the standard way. But the U(n)-action on CP^n has a fixed point and it is not too hard to see that the U(n)-representation on the tangent space to that fixed point is isomorphic to the standard representation of U(n) on C^n. So if the spin structure were equivariant, then the fixed-point representation has to be spin, which is of course wrong.

You can ask the same question for spheres (is the spin structure on S^n SO(n+1)-equivariant), and the answer is again no. But the spin structure on S^n is Spin(n+1)-equivariant; likewise the spin structure on CP^n will be equivariant under the double cover of U(n).

What you can guess from these two examples is that the question has something to do with double covers (alias central extensions of your group by Z/2). Here is the precise relation: M a spin manifold, s a spin structure (viewed as a double cover of the frame bundle of M), G a topological group acting on M by diffeomorphisms. The spin structure defines a new group G' and a surjective homomorphism p:G' \to G with kernel. G' consists of pairs (f,t), f \in G and t is an isomorphism of spin structures f^* s \to s. The spin structure is equivariant under G', and it is G-equivariant iff there is q:G \to G', pq=id. If G is a simply-connected topological group, this is always the case, but otherwise not in general.

This discussion implies that the spin structure on CP^n is indeed SU(n+1)-equivariant, if it exists. Grassmannians and other homogeneous spaces can be dealt with in the same way.

Answer 2

As far as when do spin structures exist, a manifold is spin if and only if the 2nd Stiefel Whitney class $w_2(X)=0$, which is the same as $c_1(X)\mod 2=0$. So we must calculate $c_1(T_X)$.

Let $R$ be the universal subbundle and $Q$ the universal quotient bundle. Then we have $0\to R\to \mathbb{C}^n\to Q\to 0$ for $Gr(k,n)$, and the total Chern classes satisfy $c(R)c(Q)=1$ Thus, $c_1(R)+c_1(Q)=0$, and we can show that $c_1(R)=-1$ and thus $c_1(Q)=1$. But the tangent bundle is $\hom(R,Q)=R^*\otimes Q$, which means that $c_1(T)=n$, and completely ignores $k$.

So in particular, $\mathbb{P}^n=Gr(1,n+1)$ has Chern class $n+1$, and so will be spin if and only if $n$ is odd.

I don't know the answers to 2 and 3.

Note: as Dave pointed out in the comment, I've identified $H^2$ with the integers for Grassmannians because there is a unique Schubert class $\sigma_1$ which is an ample generator, and so we do have a canonical identification. This is trickier for other spaces, of course.