It's not: the first set succeeds. The second set + k1*k0 + k2*k0 - k3*k0 - k4*k0 + 2*k1*k3 - k2^2 + k1*k5 + k3*k4 + k3*k5 - k1*k6 - k2*k6 + k3*k6 + k4*k6 - 2*k2*k4 + k3^2 - k2*k1 - k4*k5 - k5*k2, etc. is the one that fails; or if you prefer my version (I tested both), 4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5, etc. = constant. WFL On 8/11/14, Bill Gosper <billgosper@gmail.com> wrote:
Fred Lunnon>
Panic on again --- my contrived "explanation" for the Jacobian rank being 5 instead of 4 is utter tripe. The polynomial rank equals the dimension of the tangent space at a generic point of the variety, and is the same whether we are in projective space or Euclidean: the rank for any Steiner constraint basis has to equal n-1 no matter what.
So I contrived a (non-canonical) numerical example. The first relation set k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 etc. have rank 2, and checks out to 10 decimals; the second set FAILS !! Phew ...
The counterexample used above had [k1,...,k5,k6,k0] = [-19.62529364, -5.201811345, -27.16907840, -55.16907840, -50.50676305, -37.32621846, 4.034461316]; full details of numerical test cases can be provided on request.
A minor omission in my earlier theorem on opposite pairs for n even: proof should read << Modulo similarity, any n-ring D equals C transformed by some Moebius inversion X , and conversely.
<WFL
I'm confused. How is In[5]:= k1^2 + k2 k3 + k2 k4 + k3 k5 -> k1 k2 + k3^2 + k1 k4 + k2 k5 /. Thread[{k1, k2, k3, k4, k5, k6, k0} -> {-19.62529364, -5.201811345, -27.16907840, -55.16907840, -50.50676305, -37.32621846, 4.034461316}]
Out[5]=2185.6819136605263 -> 2185.681912691753 a failure?
--rwg
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