Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

How can one get equation 117 on page 21 of the below document by employing Stokes' Theorem to solve the square of the line integral on the left such that the path is closed and infinitesimal, e.g., an infinitesimal rectangle?

Probably very related to that is how one can get the following identity:

$ S(C) - 1 = \Delta x\Delta y\left [ i\left [ \frac{\partial G_\mu}{\partial \nu} -\frac{\partial G_\nu}{\partial \mu} \right ] - \left [ G_\mu, G_\nu \right ]\right ] $

from the equation below by doing the line integrals also on an infinitesimal rectangle.

$ S(C)({\mathbf R}) = 1 + i\int_{0}^{\mathbf R}G({\mathbf R'})d{\mathbf R'} - \int_{0}^{\mathbf R}\left [ \int_{0}^{\mathbf R'}G({\mathbf R''})d{\mathbf R''}\right ]G({\mathbf R'})d{\mathbf R'} $

The part I cannot get is the one with the commutator. The other parts of the derived equation are fine.



share|cite|improve this question
Did you keep in mind that $G$'s are matrices and therefore do not commute? – Fabian May 5 '11 at 8:53
Yes, and it is still unclear to me how the commutator comes from the double integral... – Raphael R. May 5 '11 at 14:39

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.