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I have another exterior differential system for one forms $U^i$, 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 $\epsilon$. This is related to a previous question here.

\begin{array}{c} \text{dU}^1+\sqrt{3} \theta^1 \wedge U^3+\sqrt{3} \theta^2\wedge U^4 =0\\ \text{dU}^2+\theta^1 \wedge U^3-\theta^2 \wedge U^4+2 \cosh(\rho) \theta^3 \wedge U^5 =0\\ \text{dU}^3+\sinh(\rho)\theta^2\wedge U^5+\theta^3 \wedge U^4 =0\\ \text{dU}^4-\sinh(\rho)\theta^1\wedge U^5-\theta^3\wedge U^3 =0\\ -\sinh(\rho) \text{dU}^5 -\cosh(\rho) \text{d$\rho$}\wedge U^5-\theta^1\wedge U^4+\theta^2\wedge U^3 =0\\ \cosh(\rho) \text{dU}^5+\sinh(\rho) \text{d$\rho$}\wedge U^5+\theta^1\wedge U^4+\theta^2 \wedge U^3-2 \theta^3 \wedge U^2 =0\\ -\text{dU}^4+\cosh(\rho) \theta^1\wedge U^5+\sqrt{3} \theta^2\wedge U^1-\theta^2 \wedge U^2+\theta^3\wedge U^3 =0\\ -\text{dU}^3+\sqrt{3} \theta^1\wedge U^1+\theta^1\wedge U^2+\cosh(\rho) \theta^2\wedge U^5-\theta^3\wedge U^4=0 \end{array}

FYI: this problem is related to my work on asymptotic symmetries of certain noncompact homogeneous spaces, i.e. I want to find diffeomorphisms which preserve a certain metric tensor asymptotically.

It seems like it should be possible to solve systems like this (semi-) automatically with a computer algebra package. After all, the system reduces to an overdetermined system of first order partial differential equations, for which such tools already exist... but the EDS form is so much more convenient that I would hate to rewrite everything as 1st order PDEs!

EDIT: As R.B. suggested in the comments, I forgot to mention that as in the other question, there is also a coordinate $\rho$ (of course), i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc. Also, the summation convention is not implied in the expression for $d \theta^i$.

EDIT2: Of course a good start would be to add the third and last equations to get rid of $dU^3$ and similarly for $dU^4$ and $dU^5$. The number of unknowns will then be reduced from 20 to 7, but the system still seems quite difficult to solve... I've been hacking away at it with Mathematica and some exterior differential and wedge product functions, but it's still a mess!

3 added 7 characters in body

I have another exterior differential system for one forms $U^i$, 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 $\epsilon$. This is related to a previous question here.

\begin{array}{c} \text{dU}^1+\sqrt{3} \theta^1 \wedge U^3+\sqrt{3} \theta^2\wedge U^4 =0\\ \text{dU}^2+\theta^1 \wedge U^3-\theta^2 \wedge U^4+2 \cosh(\rho) \theta^3 \wedge U^5 =0\\ \text{dU}^3+\sinh(\rho)\theta^2\wedge U^5+\theta^3 \wedge U^4 =0\\ \text{dU}^4-\sinh(\rho)\theta^1\wedge U^5-\theta^3\wedge U^3 =0\\ -\sinh(\rho) \text{dU}^5 -\cosh(\rho) \text{d$\rho$}\wedge U^5-\theta^1\wedge U^4+\theta^2\wedge U^3 =0\\ \cosh(\rho) \text{dU}^5+\sinh(\rho) \text{d$\rho$}\wedge U^5+\theta^1\wedge U^4+\theta^2 \wedge U^3-2 \theta^3 \wedge U^2 =0\\ -\text{dU}^4+\cosh(\rho) \theta^1\wedge U^5+\sqrt{3} \theta^2\wedge U^1-\theta^2 \wedge U^2+\theta^3\wedge U^3 =0\\ -\text{dU}^3+\sqrt{3} \theta^1\wedge U^1+\theta^1\wedge U^2+\cosh(\rho) \theta^2\wedge U^5-\theta^3\wedge U^4=0 \end{array}

FYI: this problem is related to my work on asymptotic symmetries of certain noncompact homogeneous spaces, i.e. I want to find diffeomorphisms which preserve a certain metric tensor asymptotically.

It seems like it should be possible to solve systems like this (semi-) automatically with a computer algebra package. After all, the system reduces to an overdetermined system of first order partial differential equations, for which such tools already exist... but the EDS form is so much more convenient that I would hate to rewrite everything as 1st order PDEs!

EDIT: As R.B. suggested in the comments, I forgot to mention that as in the other question, there is also a coordinate $\rho$ (of course), i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc. Also, the summation convention is not implied in the expression for $d \theta^i$.

2 Clarifications as suggested by R.B.

I have another exterior differential system for one forms $U^i$, 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 $\epsilon$. This is related to a previous question here.

\begin{array}{c} \text{dU}^1+\sqrt{3} \theta^1 \wedge U^3+\sqrt{3} \theta^2\wedge U^4 =0\\ \text{dU}^2+\theta^1 \wedge U^3-\theta^2 \wedge U^4+2 \cosh(\rho) \theta^3 \wedge U^5 =0\\ \text{dU}^3+\sinh(\rho)\theta^2\wedge U^5+\theta^3 \wedge U^4 =0\\ \text{dU}^4-\sinh(\rho)\theta^1\wedge U^5-\theta^3\wedge U^3 =0\\ -\sinh(\rho) \text{dU}^5 -\cosh(\rho) \text{d$\rho$}\wedge U^5-\theta^1\wedge U^4+\theta^2\wedge U^3 =0\\ \cosh(\rho) \text{dU}^5+\sinh(\rho) \text{d$\rho$}\wedge U^5+\theta^1\wedge U^4+\theta^2 \wedge U^3-2 \theta^3 \wedge U^2 =0\\ -\text{dU}^4+\cosh(\rho) \theta^1\wedge U^5+\sqrt{3} \theta^2\wedge U^1-\theta^2 \wedge U^2+\theta^3\wedge U^3 =0\\ -\text{dU}^3+\sqrt{3} \theta^1\wedge U^1+\theta^1\wedge U^2+\cosh(\rho) \theta^2\wedge U^5-\theta^3\wedge U^4=0 \end{array}

FYI: this problem is related to my work on asymptotic symmetries of certain noncompact homogeneous spaces, i.e. I want to find diffeomorphisms which preserve a certain metric asymptotically.

It seems like it should be possible to solve systems like this (semi-) automatically with a computer algebra package. After all, the system reduces to an overdetermined system of first order partial differential equations, for which such tools already exist... but the EDS form is so much more convenient that I would hate to rewrite everything as 1st order PDEs!

EDIT: As R.B. suggested in the comments, I forgot to mention that as in the other question, there is also a coordinate $\rho$ (of course), i.e. the one forms can be expanded as $\alpha = \alpha_a \theta^a + \alpha_\rho d\rho$ etc. Also, the summation convention is not implied in the expression for $d \theta^i$.

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