charged spinors and spin^c structures

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I have heard that charged spinors are sections of a spin^c bundle. In particular that one does not need a spin structure if there is a $U(1)$ symmetry. I'd like to understand this.

Normally I would think of a charged spinor by picking a spin structure, making a complex spinor bundle $S$ and forming the tensor product (over $\mathbb{C}$) of this bundle with a complex line bundle associated to the $U(1)$ gauge bundle according to the charge of the fermion.

Given a spin^c structure, there is the determinant line. How is this related to the gauge bundle?

asked May 2, 2014

1 Answer

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Let X be a orientable smooth manifold. The obstruction to the existence of a spin structure on X is given by the second Stiefel-Whitney class of the tangent bundle of X :  $w_{2}(TX) \in H^{2}(X, \mathbb{Z}/2)$, there exists a spin structure if and only if this class is zero (this comes from the long exact sequence in cohomology $H^{1}(X,Spin) \rightarrow H^{1}(X,SO) \rightarrow H^{2}(X, \mathbb{Z/2})$ which comes from the short exact sequence $0 \rightarrow \mathbb{Z}/2 \rightarrow Spin \rightarrow SO \rightarrow 1$).

A $Spin^{c}$ structure on X exists if and only if there exists a complex line bundle $L$ such that $c_{1}(L) mod 2 = w_{2}(X)$. Then we can define a bundle "$S \otimes L^{1/2}$": if X is not spin, the spin bundle $S$ do not exist, there is some obstruction related to the non-vanishing of $w_{2}(TX)$ and the complex line bundle $L^{1/2}$ do not exist for the same reason, but these two obstructions compensate each other and there exists a bundle $E$ that would be $S \otimes L^{1/2}$ if S and $L^{1/2}$ would exist. A charged spinor is a section of $E$ and the gauge bundle does not exist in general, it is $L^{1/2}$ if $L^{1/2}$ exists (and this happens only if there is in fact a spin structure on X).

answered May 2, 2014 by (5,140 points)

Great, thanks! Now, a follow-up question: a spin^c structure with determinant line $L$ is the same as a spin structure on $TX \oplus L$. If I take $L$ to be some other bundle, real maybe and perhaps with rank other than 1, does my spinor bundle still look like $S \otimes L^{1/2}$?

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