- #1
SmithWillSuffice
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I'm reading about bundes and connections but I cannot get past a little problem involving path-ordered exponentials. I hope someone can help me out. I'll try to state the problem as well as possible with plain text LaTeX. My question is just this: How does the the integral (of a general non-commuting matrix function A, such as a connection) like this,
\int_{t\ge t_1\ge \ldots \ge t_n \ge 0} A(t_1)\ldotsA(t_n) dt_n\ldots dt_1
become,
\frac{1}{n!} \int{t_i\in [0,1]} \mathcal{P}A(t_1)\ldotsA(t_n) dt_n\ldots dt_1
where \mathcalP} is the path-ordered product operator, and how does this path-ordered integral become,
\frac{1}{n!} \mathcal{P} \left( \int_0^t A(s) ds \right)
?
Background: the first integral arises when one constructs for example an iterated solution to a simple first-order matrix DE, du/dt = -A(t)u(t), whereby A(t1) need not commute with A(t2).
I can see how the latter conversion can be made if the multiple integral $\int{t_i\in [0,1]}...$ can be taken as a product of $n$ decoupled integrals. But I do not quite see how the $1/n!$ arises in the first conversion from the coupled multiple integral to the decoupled path-ordered multiple integral. I've tried integration by parts, Taylor expansion, and a few other tricks, including staring at the integrals for a long time! I'm missing a basic clue though.
I think I can take the final result and differentiate to prove that it solves the matrix DE that arises in the parallel transport equation for a vector say $u$ on a bundle fibre,
d u(t) / dt + A(t) u(t) = 0
where $A$ is the connection for the covariant derivative along a path. The solution is obviously,
u(t) = \mathcal{P} \exp^{-\int_0^t A(s) ds} u(0)
It's just bugging the heck out of me that I cannot figure out how to construct the formal path-ordered exponential solution without resorting to "guess the answer and back-substitute".
muchos gracias in advance for any hints/help.
---
NotAStudentButAPerpetualSeeker
\int_{t\ge t_1\ge \ldots \ge t_n \ge 0} A(t_1)\ldotsA(t_n) dt_n\ldots dt_1
become,
\frac{1}{n!} \int{t_i\in [0,1]} \mathcal{P}A(t_1)\ldotsA(t_n) dt_n\ldots dt_1
where \mathcalP} is the path-ordered product operator, and how does this path-ordered integral become,
\frac{1}{n!} \mathcal{P} \left( \int_0^t A(s) ds \right)
?
Background: the first integral arises when one constructs for example an iterated solution to a simple first-order matrix DE, du/dt = -A(t)u(t), whereby A(t1) need not commute with A(t2).
I can see how the latter conversion can be made if the multiple integral $\int{t_i\in [0,1]}...$ can be taken as a product of $n$ decoupled integrals. But I do not quite see how the $1/n!$ arises in the first conversion from the coupled multiple integral to the decoupled path-ordered multiple integral. I've tried integration by parts, Taylor expansion, and a few other tricks, including staring at the integrals for a long time! I'm missing a basic clue though.
I think I can take the final result and differentiate to prove that it solves the matrix DE that arises in the parallel transport equation for a vector say $u$ on a bundle fibre,
d u(t) / dt + A(t) u(t) = 0
where $A$ is the connection for the covariant derivative along a path. The solution is obviously,
u(t) = \mathcal{P} \exp^{-\int_0^t A(s) ds} u(0)
It's just bugging the heck out of me that I cannot figure out how to construct the formal path-ordered exponential solution without resorting to "guess the answer and back-substitute".
muchos gracias in advance for any hints/help.
---
NotAStudentButAPerpetualSeeker
