What really is a path-ordered exponential?

Question

In some texts about gauge theories in Physics I've found one object called a path-ordered exponential which I'm not sure what it means. As I understood, the idea is as follows: let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $L_g : G\to G$ be the left translation by $g$, i.e. $L_g(g') = gg'$.

If $\gamma : I\subset \mathbb{R}\to G$ is a curve in $G$ then by virtue of left translation we have the following:

$$\gamma'(t) = (L_{\gamma(t)})_{\ast} \beta(t)$$

where $\beta : I\to \mathfrak{g}$ is defined by

$$\beta(t) = (L_{\gamma(t)^{-1}})_\ast \gamma'(t).$$

So given $\gamma$ we can find $\beta$. Now, if someone gives $\beta$ and wants to find $\gamma$, then $\gamma$ is the curve satisfying the differential equation

$$\gamma'(t) = (L_{\gamma(t)})_{\ast} \beta(t)$$

and then as I understand, the solution to this equation is the path-ordered exponential.

If $G$ is the multiplicative group of real numbers, then $\mathfrak{g}$ is also the real numbers. In that case $\gamma(t) = f(t)$ is a real function and so is $\beta(t) = g(t)$. The differential equation is then

$$f'(t) = f(t) g(t)$$

whose solution is just the usual exponential

$$f(t) = C\exp\left(\int g(t)dt\right).$$

Now I simply can't understand what is this path-ordered exponential. What this path-ordered exponential really is, and how can one show that it is the solution to that differential equation?

Answer 1

For non-commuting matrices, $e^{A}·e^B\ne e^{A+B}$. If one uses an approximation with infinitesimal time steps $dt$, then $$ γ(t+dt)\simeq γ(t)(1+β(t)·dt)\simeq γ(t)·e^{β(t)·dt} $$ so that $$ γ(t+N·dt)=γ(t)·e^{β(t)·dt}·e^{β(t+dt)·dt}·…·e^{β(t+(N-1)·dt)·dt} $$ However, since the matrices (if in a matrix representation) $β(t+k·dt)$ do not commute, you can not combine the exponentials into one exponential with a Riemann sum or integral.

Staying with the unexponentiated approximation, you get $$ γ(t+N·dt)=γ(t)·(1+β(t)·dt)·(1+β(t+dt)·dt)·(1+β(t+2dt)·dt)·…·(1+β(t+(N-1)·dt)·dt) $$ where the expansion leads to Riemann-like sums $$ γ(t+N·dt)=γ(t)+γ(t)·\sum_{i=0}^{N-1}β(t+i·dt)·dt\\+γ(t)·\sum_{0\le i<j<N}β(t+i·dt)·β(t+j·dt)·dt^2\\+γ(t)·\sum_{0\le i<j<k<N}β(t+i·dt)·β(t+j·dt)·β(t+k·dt)·dt^3+… $$ which shows where the "time ordered" originates.