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?


2 Answers 2


I think there is a slight mistake in the formulation of the question. $\mathbb{CP}^n$ 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 $\mathbb{CP}^n$ (for odd $n$, there is unique spin structure on $\mathbb{CP}^n$, 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 $\mathbb{CP}^n$ 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 $\mathbb{C}^n$. 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 $\mathbb{CP}^n$ 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 $\mathbb{Z}/2$). 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 $q:G \to G'$, $pq=\operatorname{id}$. 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 $\mathbb{CP}^n$ is indeed $SU(n+1)$-equivariant, if it exists. Grassmannians and other homogeneous spaces can be dealt with in the same way.

  • $\begingroup$ 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? $\endgroup$ Oct 14, 2010 at 20:59
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    $\begingroup$ 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. $\endgroup$ Oct 15, 2010 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.

  • $\begingroup$ 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 $\Bbb{P}^n = Gr(1,n+1)$. $\endgroup$ Oct 13, 2010 at 19:47
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    $\begingroup$ 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. $\endgroup$ Dec 4, 2010 at 20:14

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