# Path ordered integral over simplices?

1. May 11, 2012

### ianhoolihan

Hi all,

I am trying to follow the derivation by Carroll of the parallel propagator in these notes, beginning page 66 or so.

My question is with the integral in equation 3.40.

1) why is it that this is an integral over a simplice, and not an n-cube (which the limits of integration seem to suggest)?

2) why is it that $\eta_n \geq \eta_{n-1}\geq \ldots \geq \eta_1$ so that path ordering must take place? (See between equations 3.40 and 3.41 for discussion.)

Cheers,

Ianhoolihan

2. May 11, 2012

### chiro

Hey ianhoolihan.

I can't answer your question specifically, but one thing that you may want to think about is the actual final map that your products of your individual A maps would do to your integral for each simplex.

My guess is that the orientation would be screwed up completely for one and this would affect the integral in a bad way and not give the right result.

Just out of curiosity, what does your linear map A represent?

3. May 11, 2012

### ianhoolihan

Hmmm, the A maps are (I'm guessing here) an infinitesimal "parallel propagation" in the direction tangent to the curve $\gamma$. As for the product of the maps, I'm not sure.

I wonder if the reason for $\eta_n \geq \eta_{n-1}\geq \ldots \geq \eta_1$ is to do with the iteration process...actually, I think that might be it. For those not looking at the pdf, the solution for the parallel propagator is

$${P^\mu}_\rho (\lambda,\lambda_0) = {\delta^\mu}_\rho + \int^\lambda_{\lambda_0} {A^\mu}_\sigma(\eta) {P^\sigma}_\rho(\eta,\lambda_0) d\eta$$

So solving by iteration,

$${P^\mu}_\rho (\lambda,\lambda_0) = {\delta^\mu}_\rho +\int^\lambda_{\lambda_0} {A^\mu}_\rho(\eta)d \eta+ \int^\lambda_{\lambda_0} \int^\eta_{\lambda_0} {A^\mu}_\sigma(\eta){A^\sigma}_\rho(\eta')d\eta d\eta' + \ldots$$

The point being that, in the first equation it is ${P^\sigma}_\rho(\eta,\lambda_0)$, so the next substitution must only range from $\eta$ to $\lambda_0$, if you understand the abuse of language.

Yup, I think that works.

Now, for the simplices...oh, maybe it follows from $\eta_n \geq \eta_{n-1}\geq \ldots \geq \eta_1$ quite obviously. For example, in the second term, the integral is over both $\eta,\eta'$ such that $\eta' \leq \eta$. Hence, simplices.

Thank you.

4. May 11, 2012

### chiro

I don't think I did much to warrant a thank-you, but I'm glad you got it in the end.