Thanks to Gene and Bill for their solutions. It's not immediately obvious what Mathematica's Reduce[] function does, but presumably a spec for is not hard to dig up. As Gene pointed out, the space of solutions is 1-D (not 3-D) --- I should have called them "rank-3". These equations arose during the investigation of a geometric algebra for 3-space line geometry, using the Clifford algebra Cl(3,3). The spinors cover collineations / correlelations, so include proper (orientation preserving) Euclidean isometries, but in a rather strange fashion: Euclidean rotation is represented by a grade-4 spinor, instead of the expected grade-2. Hoping to find a more natural construction for this hefty (31-component) object, I set out to factorise it as a (unique) product of two orthogonal grade-2 spinors --- this is always possible, and there's an algorithm to hack it, unless the spinor happens to be isoclinic [such as Euclidean 4-space product of rotations through equal angles about completely orthogonal planes]. The algorithm didn't seem to work. In line geometry, Euclidean rotation turns out to be isoclinic. This inconvenient fact took some time to sink in; once it had done so, I went off to try some easy special cases. [In general, special cases have probability zero and so never occur; in particular, any interesting or useful case invariably proves to be special.] Rotation about a line through the origin has a simpler formula (as usual), for which a canonical factorisation proves easy to guess; the same is true of translation, once finished tangling ones shoelaces in the fact that it is not only isotropic (the axis lies at infinity, and has magnitude zero), but degenerates unexpectedly to a single parabolic grade-2 spinor, so is already factorised --- duh! In both cases the components are linear monomials in the components of the axis line; but strangely, much time spent trying to glue the two together to fudge a general formula delivered naught but frustration ... I was now reduced to translating the rotation axis away from the origin via the formula I'd guessed, which required in turn solving the linear equations posted earlier, modulo the Grassmann constraint on the Pluecker coordinates of a line. By this stage, I was tempted to hurl the laptop across the room when it insisted said equations were insoluble! After a few more beginner-level blunders, the factorisation finally emerged in all its glory --- most of the 15 grade-2 components of each factor are degree-6 polynomials with 15 terms --- hardly surprising that guessing them failed. Choosing a simpler, asymmetric solution yields less grisly cubic factorisations, but still nothing intuitively suggestive. The 1-D space of solutions to the line (moment) translation equations corresponds to the 1-D subspace of bivector axes: any projective orthogonal 2-basis for this subspace yields an orthogonal factorisation of the isoclinic grade-4 spinor. I've spent a solid week on this infuriatingly messy exercise. Nothing is simple. Fred Lunnon (23/02/11) On 2/20/11, Fred lunnon <fred.lunnon@gmail.com> wrote:
Everybody can solve these, can't they --- just like quadratic equations ... Here's a simple example: solve for x,y,z, b z - c y = f , c x - a z = g , a y - b x = h .
Oops --- the determinant vanishes: | c 0 -a | | -b a 0 | = 0 . | 0 -c b | No solution, says our CAS equation solver, firmly.
Except of course, if the RHS also vanishes there is obviously a 1-dimensional homogeneous solution; but simply substituting f,g,h -> 0 should encourage our CAS to disgorge some version of that too.
End of story? Not quite: you see, adding a equn1 + b equn2 + c equn3 eliminates all variables, leaving a f + b g + c h = 0 ; and in my particular application, this constraint happens to be satisfied by the coefficients. Now some head-scratching unearths a 3-dimensional (in general) solution spanned by the symmetric linear basis [ x -> -h/b, y -> 0, z -> +f/b ] , [ x -> +g/c, y -> -f/c, z -> 0 ] , [ x -> 0, y -> +h/a, z -> -g/a ] .
In particular, summing eventually delivers the symmetric solution x -> (g/c - h/b)/3 , y -> (h/a - f/c)/3 , z -> (f/b - g/a)/3 , which I knew ought to exist by the symmetry of the equations.
I have encountered this type of problem --- solving SLE's on a manifold --- on a number of occasions, without ever finding an effective solution algorithm. Surely it must have been considered before --- any pointers, anyone?
Is it reasonable to expect that the next generation of CAS solvers might inform us about constraints on our coefficients under which such equations might possess extra solutions?
Fred Lunnon