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This is crossposted at stack exchange as http://math.stackexchange.com/questions/248391/dirac-operators-on-s1.

I am trying to understand the Dirac operators associated to the 2 spinor bundles on $S^1.$ I have been getting very confused about why one bundle has nontrivial harmonic spinors and the other doesn't.(Harmonic spinors are solutions $s$ to the equation $Ds = 0$ where $D$ is the Dirac operator and $s$ is a section.)

Here is my argument (which must be wrong somewhere). We have 2 spin structures given by the connected 2-fold covering and the disconnected 2-fold covering. Since the tangent bundle $TS^1$ is trivial, we can choose the trivial connection on it given by $f \rightarrow df.$ When considered as a connection on the principal bundle of frames (also isomorphic to $S^1$), i.e. as a Lie algebra valued one form on $S^1,$ it must be the zero form.

Ok, so now given either spin structure, the connection must lift to the $0$ connection. Furthermore, any complex line bundle over the circle is trivial, so both cases look exactly the same, and the Dirac operator appears to be $f \rightarrow i\frac{df}{dx}.$

However, I am told that in the case of the connected double cover we should have an additional condition on our $f,$ namely that it should satisfy $f(-x) = -f(x).$ With this extra condition, there cannot be harmonic spinors on the spinor bundle associated to the connected spin structure. Where have I gone wrong?

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Hopefully, my answer at stackexchange is answering your question. –  Kofi Dec 1 '12 at 19:29

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This one is tricky and extremely confusing. The bundle of spinors is a complex line bundle so it is trivializable. You detect the spin structure only if you look at the Dirac operator. For one spin structure the Dirac operator has a kernel, for the other, it does not.

For a more detailed discussion on the pathological case of spin structures on $S^1$ I recommend you to have a look at the discussion on spin structures on page 150-151 of these notes. There I discuss Milnor's point of view on this subject, but even Milnor in his nice little paper on this subject (see the precise reference in the above notes) is not very careful about this issue.

Addendum. Over $S^1$ we have two real line bundles, $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ the trivial line bundle $\underline{\bR}$ and the nontrivial line bundle $\tilde{\underline{\bR}}$ with nontrivial first Stieffel-Whitney class $w_1$. Note that $\tilde{\underline{\bR}}$ can be identified with the tautological (real) line bundle over the projective line $\bR\mathbb{P}^1$. These line bundles are equipped with natural metrics and compatible connections $\nabla$ and respectively $\tilde{\nabla}$. Something miraculous happens. Although these two real lines bundles are not isomorphic, the complex line bundles $\underline{\bR}\otimes\bC$ and $\tilde{\underline{\bR}}\otimes\bC$ are isomorphic as complex line bundles over $S^1$ (duh!). The connections $\nabla$ and $\tilde{\nabla}$ induce connections on the complexifications $\underline{\bR}\otimes\bC$ and $\tilde{\underline{\bR}}\otimes\bC$ and we obtain two Dirac operators $\newcommand{\ii}{\boldsymbol{i}}$ $D=-\ii\nabla_\theta$ and $-\ii\tilde{\nabla}_\theta$ on the same line bundle. However, the isomorphism $\Phi$ that maps $\underline{\bR}\otimes\bC$ to $\tilde{\underline{\bR}}\otimes\bC$ does not map the complexification of $\nabla$ to the complexification of $\tilde{\nabla}$ so that the operators $D$ and $\tilde{D}$ are not conjugate to each other. The operator $D$ is $-\ii\frac{d}{d\theta}$, and

$$ \Phi^{-1}\tilde{D}\Phi =-\ii\frac{d}{d\theta}+\frac{1}{2}= D+\frac{1}{2}. $$

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If the Dirac operator is simply expressed as $f \rightarrow -i\frac{df}{dx}$ how does it pick up any extra structure? If this is not the correct form for it, how do we derive that form from the local expression for the Dirac operator? See also mathoverflow.net/questions/116306/…. Thank you for your help. –  mkreisel Dec 14 '12 at 16:06
    
The two Dirac operators are $-\boldsymbol{i}\frac{d}{d\theta}$ and $-\boldsymbol{i}\frac{d}{d\theta}+\frac{1}{2}$, $\theta\in [0,2\pi]$. –  Liviu Nicolaescu Dec 14 '12 at 20:19
    
@mkreisel I've included an addendum to my answer that maybe will help clarify the above statement. –  Liviu Nicolaescu Dec 14 '12 at 20:59
    
Thanks, that helps a lot! –  mkreisel Dec 15 '12 at 3:58

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