I was seeking for the solution of following partial differential equation for two unknowns $\vec{u}(s,t), \vec{w}(s,t)$
$$\partial\_t \vec{u} = \partial\_s \vec{w} - [\vec{w} \times \vec{u}].$$
Using computed-based Lie symmetries method I achieved to get its general solution
$$\vec{u} = \partial\_s \vec{a} + \frac{a - \sin a}{a^3} (\vec{a} (\vec{a} \cdot \partial\_s \vec{a}) - a^2 \partial\_s \vec{a}) - \frac{1 - \cos a}{a^2} [\vec{a} \times \partial\_s \vec{a}],$$
$$\vec{w} = \partial\_t \vec{a} + \frac{a - \sin a}{a^3} (\vec{a} (\vec{a} \cdot \partial\_t \vec{a}) - a^2 \partial\_t \vec{a}) - \frac{1 - \cos a}{a^2} [\vec{a} \times \partial\_t \vec{a}].$$
Actually, it was obtained by purely algebraic methods, not involving its intrinsic geometric nature. It would be a great pleasure for me, if it could be derived by methods of differential geometry. 


Do you have any ideas?



Partial differential equation parametrization
This is a standard geometric formula, disguised because of the old-fashioned notation and the mixture of the PDE with the formula for exponentiation in the orthogonal group. 


Rewrite it this way: Set 
$$
\hat u = \begin{pmatrix}0 & u\_3 & - u\_2\\ -u\_3 & 0 & u\_1\\ u\_2 & -u\_1 & 0\end{pmatrix}
\quad\text{and}\quad
\hat w = \begin{pmatrix}0 & w\_3 & - w\_2\\ -w\_3 & 0 & w\_1\\ w\_2 & -w\_1 & 0\end{pmatrix}.
$$
and set $\alpha = \hat u\,\mathrm{d}s+\hat w\,\mathrm{d}t$. Then the PDE becomes
$$
\mathrm{d}\alpha + \alpha\wedge\alpha = 0.\tag1
$$ 
(You may need to check signs and replace $u$ and $w$ by their negatives to get this. I'll leave the details to you.) Of course, $(1)$ is a famous equation and it is known that the general solution is given by 
$$
\alpha = A^{-1}\,\mathrm{d}A,\tag2
$$
where $A(s,t)$ is an orthogonal matrix. Now write $A(s,t) = \exp\bigl(a(s,t)\bigr)$, where $a(s,t)$ is a skew symmetric matrix (which can always be done). In fact, letting $\lambda$ be the square root of $-\tfrac12$ the trace of $a^2$, one has $a^3 + \lambda^2 a = 0$ (because this is the characteristic polynomial of $a$), then one can compute the exponential of $a$ as
$$
\exp(a) = I + \frac{\sin\lambda}{\lambda} a + \frac{1-\cos\lambda}{\lambda^2} a^2.\tag3
$$
Putting all this together gives the explicit formula for the general solution.


