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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) and Ivey&Landsberg (Google books link), 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...

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1 Answer

up vote 5 down vote accepted

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 $a, c$.)

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Thanks a lot Jeanne!! I guess rather than overthinking it, sometimes it's better to just go ahead and SOLVE that damn thing already! ;) Thanks also for the speed lesson in exterior differentials systems! –  H. Arponen Nov 5 '12 at 13:37
    
You're welcome! And yes - sometimes direct computation works and sometimes it doesn't, but it's always worth a try. :) –  Jeanne Clelland Nov 5 '12 at 13:49
    
Jeanne: umm... you don't happen to be able to say something about another (more difficult) problem here? mathoverflow.net/questions/111578/… It seems like this kind of systems would be best solved by a CAS package. I already found you Cartan package for Maple, but I guess it can't handle systems like this? –  H. Arponen Nov 5 '12 at 20:37
    
Well, I can take a look! I'm kind of swamped this week, but I'll give it a try as soon as I get a chance. –  Jeanne Clelland Nov 5 '12 at 22:29
    
By the way, I fould your Cartan package for Maple. Will it be able to handle/solve this type of EDSes? I don't have access to Maple right now, but it could be useful in the future... –  H. Arponen Nov 6 '12 at 8:49
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