Editing Ordered exponential (section)

== Example ==

Given a manifold <math>M</math> where for a <math>e \in TM</math> with [[Group theory|group]] transformation <math>g: e \mapsto g e</math> it holds at a point <math>x \in M</math>:

: <math>de(x) + \operatorname{J}(x)e(x) = 0.</math>

Here, <math>d</math> denotes [[exterior differentiation]] and <math>\operatorname{J}(x)</math> is the connection operator (1-form field) acting on <math>e(x)</math>. When integrating above equation it holds (now, <math>\operatorname{J}(x)</math> is the connection operator expressed in a coordinate basis)

: <math>e(y) = \operatorname{P} \exp \left(- \int_x^y J(\gamma (t)) \gamma '(t) \, dt \right) e(x)</math>

with the path-ordering operator <math>\operatorname{P}</math> that orders factors in order of the path <math>\gamma(t) \in M</math>. For the special case that <math>\operatorname{J}(x)</math> is an [[Antisymmetry|antisymmetric]] operator and <math>\gamma</math> is an infinitesimal rectangle with edge lengths <math>|u|,|v|</math> and corners at points <math>x,x+u,x+u+v,x+v,</math> above expression simplifies as follows :

: <math>
\begin{align}
& \operatorname{OE}[- \operatorname{J}]e(x) \\[5pt]
= {} & \exp [- \operatorname{J}(x+v) (-v)] \exp [- \operatorname{J}(x+u+v) (-u)] \exp [- \operatorname{J}(x+u) v] \exp [- \operatorname{J}(x) u] e(x) \\[5pt]
= {} & [1 - \operatorname{J}(x+v) (-v)][1 - \operatorname{J}(x+u+v) (-u)][1 - \operatorname{J}(x+u) v][1 - \operatorname{J}(x) u] e(x).
\end{align}
</math>

Hence, it holds the group transformation identity <math>\operatorname{OE}[- \operatorname{J}] \mapsto g \operatorname{OE}[\operatorname{J}] g^{-1}</math>. If <math>- \operatorname{J}(x)</math> is a smooth connection, expanding above quantity to second order in infinitesimal quantities <math>|u|,|v|</math> one obtains for the ordered exponential the identity with a correction term that is proportional to the [[Riemann curvature tensor| curvature tensor]].