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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?

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up vote 9 down vote accepted

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.

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Thanks alot for your answer. One question though, if the spinor bundle is $SU(N+1)$-equivarient then it must correspond to a represention of $U(N)$, right? What is this representation? – Jean Delinez Oct 14 '10 at 20:59
Yes, the spinor bundle will be an SU(N+1)-equivariant vector bundle on SU(N+1)/G, with G= SU(n+1) \cap (U(n) \times U(1)), which is the same information as a representation of G. How do you find out which one it is? Well, you look at a G-fixed point, see a representation of G, viewed as a map G \to SO(2N). You lift it to Spin(2N) and compose with (one of) the spin representation of the orthogonal group. – Johannes Ebert Oct 15 '10 at 9:37

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.

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Hi Charles, a couple suggested edits: (1) In complex projective space and Grassmannians, it makes sense to identify c_1, a class in $H^2$, with a number but only because $H^2$ is canonically identified with $\Bbb{Z}$. (2) I think you mean $\Bbb{P}^n = Gr(1,n+1)$. – Dave Anderson Oct 13 '10 at 19:47
One can compute the first Chern class of the complex Grassmannian of $k$-planes in $C^N$ as follows. For the purpose of computing its Chern character, the holomorphic tangent bundle $T$ is a product of the tautological rank $k$ bundle $V$ and (formally) $N−V$. This leads to the formula $c_1(T) = (N−2k)v$ where $v=c_1(V)$ is a generator of $H^2$. So it looks like the Grassmannian is spin iff $N$ is even. – Simon Salamon Dec 4 '10 at 20:14

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