MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is related to my question on stack exchange which has not yet received an answer. For the purposes of this question, fix the spin structure over $S^1$ to be given by the connected double cover.

Sections of the bundle of spinors on $S^1$ can be thought of in two ways. First, by trivializing the bundle over $U_1 = S^1 \setminus \{i\}$ and $U_2 = S^1 \setminus \{-i\},$ sections can be considered as pairs of real valued functions $f_1, f_2$ defined on $\mathbb{R}$ such that $f_1(\frac{1}{x}) = f_2(x)$ when $x> 0,$ and $f_1(\frac{1}{x}) = -f_2(x)$ when $x< 0.$ Following Lawson/Michelson's "Spin Geometry," we can compute the Dirac operator locally. For example, $ f_1 \rightarrow i\frac{df_1}{dx}.$

We can also identify the sections as function $f: S^1 \rightarrow \mathbb{C}$ such that $f(-\theta) = -f(\theta)$ by the procedure described here When we identify sections this way, I have seen the Dirac operator expressed as $f \rightarrow -i\frac{df}{d\theta}.$ However I have never seen a derivation of this fact from the local expression given above. My attempt at the computation is contained in the stack exchange link at the top of the page.

Can anyone tell me how to translate between these 2 points of view?

share|cite|improve this question
Even a comment outlining how you would do the calculation, even if it doesn't show the calculation in full, would be helpful. – mkreisel Dec 23 '12 at 17:28
In the first parametrization you identify $z\inS^1\setminux{\xi}$ with $x\in\mathbb R$ for example, for $\xi={-1}$ this is given by the Caley transform $x\mapsto -\frac{x-i}{x+i}$ with inverse $z\mapsto -i\frac{z-1}{z+1}$ in the second you identify $S^1$ with the double cover locally by the map $z\mapsto z^2$, the rest should be just using some chain rule, or do I misunderstand you question? – Marcel Bischoff Dec 24 '12 at 22:18
Could you explain what trivialisation you use when obtaining the local expression $f_1 \rightarrow i\frac{df_1}{dx}$ for the Dirac operator? The transition function $f_1(\frac{1}{x}) = f_2(x)$ suggests the trivialising sections of the spinor bundle is intended to have constant norm. But then the local expression looks to me like the Dirac operator of the Euclidean metric on $\mathbb{R}$, rather than with respect to the metric pulled back from the circle by stereographic projection. – Johannes Nordström Dec 25 '12 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.