# Transforming the Dirac Operator on $S^1$

My goal is to understand as much as I can about the Dirac operator on $S^1$ where we give $S^1$ the spin structure given by the connected double cover of the frame bundle. The spinor bundle on $S^1$ can be constructed as follows. The bundle is trivial over $U_1 = S^1 \setminus \{i\}$ and $U_2 = S^1 \setminus \{-i\}$ and the transition function $T_{12}$ defined on $U_1 \cap U_2$ is given by $T_{12}(z) = 1$ when $\textbf{Re}z > 1$ and $T_{12}(z) = -1$ when $\textbf{Re}z < 1.$

Denote by $\Phi(x): \mathbb{R} \rightarrow U_1$ the inverse of stereographic projection given by $x \rightarrow \text{arccos}(\frac{x^2-1}{x^2+1})$ where elements of $U_1$ are given by an angle $\psi.$ Note that $\Phi^{-1}$ is given by $\Phi^{-1}(\psi) = \frac{\text{cos}\psi}{1-\text{sin}\psi}.$

Sections of this bundle can be thought of as pairs of functions $f_1, f_2: \mathbb{R} \rightarrow \mathbb{C}$ such that $$f_1(\frac{1}{x}) = \begin{cases} f_2(x) & \quad \text{if x>0}\\ -f_2(x) & \quad \text{if x<0} \end{cases}$$

This condition arises since the change of coordinates from $U_1$ to $U_2$ is given by $\frac{1}{x},$ and the minus sign comes from the transition functions of the bundle. Locally, the Dirac operator $D$ is given by $Df_1 = i\frac{df_1}{dx}, Df_2 = i\frac{df_2}{dx}.$

However, as explained in this post, sections of this bundle can also be identified with functions $g: S^1 \rightarrow \mathbb{C}$ satisfying $g(-\theta) = -g(\theta).$ I would like to compute what the Dirac operator does when we identify our sections in this way.

Since I only know how to compute the Dirac operator locally, I will need to start with a $g$ as above, convert it to $f_1, f_2$ and apply the Dirac operator locally, then convert that back to a function on the circle. By symmetry I really only need to apply this method to $f_1.$ What I hope to find (because many books tell me this is the right answer) is that $Dg = i\frac{dg}{d\theta}$ where the derivative is globally defined by lifting $g$ to the universal cover $\mathbb{R}$ and differentiating it there.

So, given $g: S^1 \rightarrow \mathbb{C}$ as above and following the method from the post, we can get $f_1$ by taking $$f_1(x) = g(\frac{\Phi(x)}{2}).$$

Now we apply the Dirac operator to get $$Df_1(x) = i\frac{df_1}{dx} = i\frac{dg}{d\theta}(\frac{\Phi(x)}{2}) \cdot \frac{1}{2}\frac{d\Phi}{dx} = i\frac{dg}{d\theta}(\frac{\Phi(x)}{2})\cdot \frac{-1}{x^2+1}.$$

To convert this back to a function on $S^1$ we set $x = \Phi^{-1}(2\theta)$ to get

$$i\frac{dg}{d\theta}(\theta) \cdot \frac{-1}{(\frac{\text{cos}2\theta}{1-\text{sin}2\theta})^2+1} =i\frac{dg}{d\theta}\frac{(-1)(1-\text{sin}2\theta)^2}{1-\text{sin}2\theta}=i\frac{dg}{d\theta}(1-\text{sin}2\theta).$$

So it looks like $Dg = i\frac{dg}{d\theta}(1-\text{sin}2\theta)...$ Can anyone see the mistake I've made? Or, if this method of calculation is entirely wrong, can you demonstrate how to compute the Dirac operator in this case?

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Anyone have any ideas? – mck Dec 7 '12 at 15:30