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Let $\gamma:[0,1] \to \mathbb R^2$ be a path describing a rectangle with vertices $x$, $x+u$, $x+u+v$, $x+v$, where $x, u, v \in \mathbb R^2$ ($u, v$ linearly independent).

Let $J:\mathbb R^2 \to \mathfrak{so}(n) $ be a smooth map (a family of antisymmetric matrices, $\mathfrak{so}(n)$ is the Lie algebra of SO(n)).

Then, the ordered exponential $OE[J]:[0,1]\to SO(n)$ (along $\gamma$) is the solution to $$\frac{d}{dt} OE[J](t) = J(\gamma(t))OE[J](t)$$ with $OE[J](0)= 1$.

In the last section of Wikipedia it is written that the following holds:

$$OE[- {J}] = \exp [- {J}(x+v) (-v)] \exp [- {J}(x+u+v) (-u)] \exp [- J(x+u) v] \exp [- {J}(x) u] $$ $$= [1 - {J}(x+v) (-v)][1 -{J}(x+u+v) (-u)][1 - \operatorname{J}(x+u) v][1 - {J}(x) u].$$

Why this is true? What puzzles me is that $OE[-J]$ in general is not an exponential, but in this case we can express as composition of exponentials. Moreover somehow in the second line there are no exponentials at all. How can we prove this?

Some notes on the answer by Qmechanic. We have that (see e.g. wikipedia) $$OE[J](t) = \sum_{n=0}^{+\infty} \frac 1 {n!} \int_{[0,t]^n} \mathcal{T}\{J(t_1)J(t_2)\cdots J(t_n)\}dt_1dt_2\cdots dt_n.$$

Now for $t\to 0$ $$\int_{[0,t]^n} \mathcal{T}\{J(t_1)J(t_2)\cdots J(t_n)\}dt_1dt_2\cdots dt_n = J(0)^n t^n + O(t^{n+1})$$ (this is obtained for n=2 by $\int_{[0,t]^2}f(s_1)f(s_2)ds_1ds_2 = \int_{[0,t]^2}(f(0)+O(s_1))(f(0)+O(s_2))ds_1ds_2$)

thus $$OE[J](t) = \sum_{n=0}^{+\infty} \frac 1 n \int_{[0,t]^n} \mathcal{T}\{J(t_1)J(t_2)\cdots J(t_n)\}dt_1dt_2\cdots dt_n = e^{J(0)t}+ o(t),$$

then we use $e^{J(0)t} = (1+J(0)t)+ o(t)$

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OP is presumably missing that Wikipedia mentions that $\gamma$ is an infinitesimal rectangle. Each of the 4 contour integrals of the 4 sides of $OE[-J]$ are replaced to with the initial value of $J$ times the infinitesimal side length to lowest order$^1$. The exponentials are Taylor expanded to first order in the infinitesimal side length.

 x+v        x+u+v
  -----<----- 
  |    3    |
  |         |
v v 4       ^ 2
  |         |
  |    1    |
  ----->-----
 x     u     x+u

$\uparrow$ Fig. 1. A counterclockwise infinitesimal rectangular loop.

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$^1$ This is similar to the estimate $$\int_0^{\epsilon}\! dx~f(x)~=~f(0)\epsilon + {\cal O}(\epsilon^2).$$

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