I have the following exterior differential system for one forms $\alpha, \beta, \gamma$, where the $\theta^i$ are a cotangent basis on $SO(3)$, i.e. they satisfy $d \theta^i = \epsilon\_{ijk} \theta^j \wedge \theta^k$ for the antisymmetric tensor $\epsilon$:


$\theta^2 \wedge \alpha = 0$


$\theta^2 \wedge \gamma = 0$


$d \alpha = - \theta^3 \wedge \beta$


$d \beta = \theta^3 \wedge \alpha - \theta^1 \wedge \gamma$


$d \gamma = \theta^1 \wedge \beta$


In addition there is a coordinate $\rho$, i.e. the one forms can be expanded as $\alpha = \alpha\_a \theta^a + \alpha\_\rho d\rho$ etc.


I guess it doesn't look that imposing (I don't know much of EDSses), but I'm having a seminar talk in a few days, and I've still got plenty to do... It would be awesome if I had a solution to the system above to show the audience! I'd be happy with a proof of existence (a la Frobenius or something I guess) and SOME info on the solution as well or pointers to maybe similar systems in the literature. I'm aware of the text books by Bryant et al.[(Google books link)](http://books.google.fi/books?id=pPmHMAEACAAJ&dq=exterior+differential+systems&hl=en&sa=X&ei=YOKWUM3zBoem4gTAjoCYBQ&ved=0CCsQ6AEwAA) and Ivey&Landsberg [(Google books link)](http://books.google.fi/books?id=8fxN1A7MLBQC&dq=exterior+differential+systems+ivey&hl=en&sa=X&ei=c-KWUKG4NOzb4QSt8oCgDA&ved=0CCgQ6AEwAA), but these are thick books and I'm almost outta time!!


I'll think about this myself too, of course, and I'll post here about it too, but now I need to get some sleep...



Exterior differential system on $SO(3;\mathbb R) \times \mathbb R$
The only solutions are:


$ \alpha = a \theta^2, \qquad \beta = -a \theta^1 - c \theta^3, \qquad \gamma = c \theta^2, $


where $a, c$ are constants. (No $d\rho$ terms appear.) Here's how the computation goes:


From your first two equations, we have


$ \alpha = a \theta^2, \qquad \gamma = c \theta^2$


for some functions $a, c$. We can also assume that


$ \beta = b\_1 \theta^1 + b\_2 \theta^2 + b\_3 \theta^3 + b\_\rho d\rho$


for some functions $b\_1, b\_2, b\_3, b\_\rho$. 


Now, substitute these expressions into your equation for $d\alpha$. Reducing modulo $\theta^2$ implies that $b\_\rho = 0$ and $b\_1 = -a$. Then the remaining terms imply that


$ da = a\_2 \theta^2 - b\_2 \theta^3$


for some function $a\_2$. Similarly, reducing the equation for $d\gamma$ modulo $\theta^2$ implies that $b\_3 = -c$, and then the remaining terms in $d\gamma$ imply that 


$dc = b\_2 \theta^1 + c\_2 \theta^2$


for some function $c\_2$. Now, substitute all this into your equation for $d\beta$. Reducing modulo $\theta^2$ implies that $b\_2=0$, and then the remaining terms imply that $a\_2 = c\_2 = 0$ as well. Therefore, $da = dc = 0$, and so $a$ and $c$ are constants, which leads to the solutions above. (It is straightforward to check that these are, in fact, solutions for any constants $a, c$.)


