# charged spinors and spin^c structures

+ 4 like - 0 dislike
82 views

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?

+ 4 like - 0 dislike

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 (4,840 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 $S \otimes L^{1/2}$?

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.