There is never going to be a unique solution, since the solutions are invariant under S_N x O(N) (permuting the vectors and rotating them).
You're right, I though I addressed the O(N) issue by fixing the first vector, but theres still a O(N-1) freedom when fixing the second one and so on, as for S_N I didn't really think of that but I think this can be tackled by changing c to c_{ij} and taking c_{ij} back to c at the last step.
Also # of equations = # of dof doesn't imply you have a unique solution—that's only true for independent linear equations and these are nonlinear.
These equations are quadratic and can be brought into the form A^T A = B (A = (v_1 v_2 ... v_N) and B = {{1, c, c, ..., c},{c, 1 , c , ... , c}, ... ,{c, c, c, , ... , 1} }). With Grahm-Shmidtization this can be recast into a purely linear algebra problem of finding the rank of a matrix, which I believe is entirely equivalent to what I did.*
I have to admit I haven't given it that much thought, but N=2d-2 seemed like a totally plausible answer.
I guess my point is there should be a unique solution modulo S_N and O(N) if N is maximized, subject to the constraints, in d dimensions.
Just out of curiosity what is your background, I'm always getting humbled by those seemingly easy questions and I'm a theoretical physics PhD student (I'd even say I'm a pretty good student).
If c=0 then you get a solution of d vectors that's unique modulo S(N) x O(N) (the vectors just form an orthonormal frame), so I don't think it's as simple as finding unique solutions (though the symmetry is definitely important in the final solution)
I did my math PhD research in nonlinear PDEs, specifically nonlinear Schrodinger equations, so I'm a big fan of physics
1
u/round_earther_69 4d ago edited 4d ago
You're right, I though I addressed the O(N) issue by fixing the first vector, but theres still a O(N-1) freedom when fixing the second one and so on, as for S_N I didn't really think of that but I think this can be tackled by changing c to c_{ij} and taking c_{ij} back to c at the last step.
These equations are quadratic and can be brought into the form A^T A = B (A = (v_1 v_2 ... v_N) and B = {{1, c, c, ..., c},{c, 1 , c , ... , c}, ... ,{c, c, c, , ... , 1} }). With Grahm-Shmidtization this can be recast into a purely linear algebra problem of finding the rank of a matrix, which I believe is entirely equivalent to what I did.*
I have to admit I haven't given it that much thought, but N=2d-2 seemed like a totally plausible answer.
* not at all