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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!

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Two questions: First, are you assuming that the $U^i$ are linear combinations of the $\theta^j$ and $d\rho$, as before? Second, in your structure equation for the $\theta$s, are you using the summation convention or not? I.e., *with* the summation convention, you'd have $d\theta^1 = 2\ \theta^2\wedge\theta^3$ while, without the summation convention, you'd have $d\theta^1 = \theta^2\wedge\theta^3$. –  Robert Bryant Nov 5 '12 at 20:51
    
Oops, should have been a bit more careful... edited the post accordingly! –  H. Arponen Nov 5 '12 at 21:44
    
I found an appropriate package for REDUCE: reduce-algebra.com/docs/eds.pdf Unfortunately I have virtually no experience with REDUCE... –  H. Arponen Nov 5 '12 at 21:48
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I did some not-very-pretty Maple computations, and if I'm doing it right, it looks like there's a family of solutions parametrized by 2 arbitrary functions of 2 variables. I don't see any good way to write them down explicitly, though - is it enough for you to know that solutions exist? –  Jeanne Clelland Nov 7 '12 at 23:01
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Jeanne: that is GREAT news, exactly what I was hoping!! Actually I also had a third equation which is a bit simpler than this one but more complicated than the other one, for which I managed to get solutions parametrized by one function of three variables. The result(s) mean that there's an infinite number of asymptotic symmetries of a certain noncompact homogeneous space. The problem is based on this one: arxiv.org/abs/1209.5597 –  H. Arponen Nov 8 '12 at 13:41

1 Answer 1

up vote 15 down vote accepted

Jeanne's calculations give the right answer, i.e., that the solutions depend on two arbitrary functions of 2 variables.

It turns out, though, that, with the right choice of variables, one can reduce the problem to an underdetermined system of $2$ linear equations for a pair of tensors of rank $2$ on the $2$-sphere. (This choice of variables is suggested by examining the characteristic variety and tableau of the EDS.)

First of all, if one solves for the coefficients of the $U^i$ as the OP suggests, one does get $7$ parameters. One way to do this is as follows: $$ \begin{align} U^1 &= -(\sqrt{3}/3)\bigl((v_4{-}(\cosh(\rho)-\sinh(\rho))v_1)\ \theta^1 -(v_2{+}(\cosh(\rho)+\sinh(\rho))v_3)\ \theta^2\bigr)\\\\ U^2 &= (\cosh(\rho)-\sinh(\rho))v_1\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_3\ \theta^2\\\\ U^3 &= (\cosh(\rho)-\sinh(\rho))v_5\ \theta^1 + v_6\ \theta^2 + 2\sinh(\rho)v_3\ \theta^3 +v_4\ d\rho\\\\ U^4 &= v_7\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_5\ \theta^2 + 2\sinh(\rho)v_1\ \theta^3 +v_2\ d\rho\\\\ U^5 &= (\cosh(\rho)-\sinh(\rho))v_2\ \theta^1-(\cosh(\rho)+\sinh(\rho))v_4\ \theta^2.\\\\ \end{align} $$ The $v_i$ are now $7$ unknowns, and substituting this into the above $2$-forms gives an involutive differential system for the $v_i$ that is generated by five $2$-forms and has characters $s_1=5$ and $s_2=2$, with $s_i=0$ for $i>2$.

However, examining the characteristic variety and tableau of this system suggests that one should reparametrize the first four $v_i$ as $$ \begin{align} v_1 &= -(2\sinh(\rho){+}\cosh(\rho))\ p_2-\cosh(\rho)\ q_2\\\\ v_2 &= (2\cosh(\rho)\sinh(\rho)+2\cosh^2(\rho)-1)(p_1{+}q_1)\\\\ v_3 &= +(2\sinh(\rho)-\cosh(\rho))\ p_1 + \cosh(\rho)\ q_1\\\\ v_4 &= (2\cosh(\rho)\sinh(\rho)-2\cosh^2(\rho)+1)(p_2{-}q_2).\\\\ \end{align} $$ for some functions $p_1,p_2,q_1,q_2$. When one does this, one finds that the differential equations imply $$ \begin{align} dp_1 &= (p_0{+}p_3)\ \theta^1 + (p_4{+}p_5)\ \theta^2-3p_2\ \theta^3\\\\ dp_2 &= (p_4{-}p_5)\ \theta^1 + (p_0{-}p_3)\ \theta^2+3p_1\ \theta^3\\\\ dq_1 &= (q_0{+}q_3)\ \theta^1 + (q_4{+}q_5)\ \theta^2-1q_2\ \theta^3\\\\ dq_2 &= (q_4{-}q_5)\ \theta^1 + (q_0{-}q_3)\ \theta^2+1p_1\ \theta^3\\\\ \end{align} $$ for some functions $p_0,p_3,p_4,p_5,q_0,q_3,q_4,q_5$. This implies that the (complex-valued) linear form $Q = (q_1{-}iq_2)(\theta^1{+}i\theta^2)$ and the (complex-valued) cubic form $P = (p_1{-}ip_2)(\theta^1{+}i\theta^2)^3$ are well-defined on the $2$-sphere $S^2 = \mathrm{SO}(3)/\mathrm{SO}(2)$, where the $\mathrm{SO}(2)$-subgroup of $\mathrm{SO}(3)$ is the one that is an integral of the forms $\theta^1$ and $\theta^2$. Thus, $P$ and $Q$ are sections of natural complex line bundles over the $2$-sphere.

One then finds that the integral manifolds of the differential system must satisfy $$ \begin{align} v_5 &= -\sinh(\rho)\ (p_4{+}q_5)-\cosh(\rho)\ (p_5{+}q_4)\\\\ v_6 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0 +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}1)\ p_3\\\\ &\quad +(\cosh(\rho)\sinh(\rho){-}\cosh^2(\rho){+}{\tfrac{1}{3}})\ q_0 +(\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho))\ q_3\\\\ v_7 &= (\cosh^2(\rho){-}\cosh(\rho)\sinh(\rho){-}2)\ p_0 +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}1)\ p_3\\\\ &\quad +(\cosh(\rho)\sinh(\rho){+}\cosh^2(\rho){-}{\tfrac{1}{3}})\ q_0 +(\cosh^2(\rho){+}\cosh(\rho)\sinh(\rho))\ q_3\ .\\\\ \end{align} $$

Finally, once these are substituted into the $2$-forms, one finds that the differential system reduces to a pair of second order linear differential equations on $S^2$. Using notation that can be found in a paper of mine (TAMS 290 (1985), 259–271), this second order equation for the tensors $P$ and $Q$ can be written in the form $$ Y(Y(P)) = \tfrac12\ Q + \tfrac23\ X(Y(Q)) - \tfrac13\ X(X({\overline{Q}})), $$ where $X$ and $Y$ are certain first-order differential operators that are invariant under $\mathrm{SO}(3)$ and that generalize $\partial$ and $\bar\partial$ to symmetric $(1,0)$-forms of arbitrary degree. This is an elliptic linear equation of second order that is underdetermined (it is $2$ equations for the $4$ components of $P$ and $Q$). It is invariant under the rotations of $S^2$ and equivalent to the original (overdetermined) EDS on $\mathrm{SO}(3)\times\mathbb{R}$.

It is conceivable that this equation admits an explicit solution in terms of a potential (which would be a section of a rank $2$ vector bundle over $S^2$), but I have not tried to check whether this is true or not.

Added Comment: I have now checked about the possibility of a potential and, miraculously, it turns out that there is a potential: One can show that the solutions of the above equation for $P$ and $Q$ are expressible in the form $$ \begin{align} P &= X(X(L))\\\\ Q &= L + (XY{+}YX)L + X(X({\overline{L}})\bigr)\\\\ \end{align} $$ where $L=(L_1{+}iL_2)(\theta^1{+}i\theta^2)$ is an arbitrary complex-valued $(1,0)$-form on $S^2$. Thus, components of $L$ are the two arbitrary functions of $2$ variables predicted by the theory for the general solution.

$L$ is not quite unique. It turns out that one can also add an expression of the form $X(a_1+ib_2)$ to $L$ where $a_1$ is the restriction to the $2$-sphere of a linear function in $\mathbb{R}^3$ and $b_2$ is the restriction to the $2$-sphere of a harmonic homogeneous quadratic function in $\mathbb{R}^3$. This describes the (local and global) ambiguity in the potential $L$ completely.

Thus, with this construction and the above formulae, we have the complete description of the (local and global) integral manifolds of the original EDS in terms of the second and third derivatives of $L$ (which is arbitrary).

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That is absolutely beautiful, thank you Robert!! That's something I need a while to digest, but I'll read your paper (seems it's open access). –  H. Arponen Nov 9 '12 at 20:14
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You are welcome; it was interesting and a little bit surprising that it's really a 2-dimensional problem in disguise. The only part of my TAMS paper that is relevant to this problem is the first two pages of Section 1 (pp. 260–261), where $X$ and $Y$ are defined and Proposition 1.1 is proved, so ignore Proposition 1.2 and everything after. I could have written the equations out without using this notation, but, while the formulae aren't complicated, one might not immediately see that the two equations are elliptic. –  Robert Bryant Nov 9 '12 at 21:00
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Dang, Robert, that's pretty awesome. :) –  Jeanne Clelland Nov 9 '12 at 23:33

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