Red-faces all round alert: the claims made in my previous post concerning dimension and basis of the constraint set are unconditionally withdrawn. The only useful fact to emerge from my Gröbner explorations is that RWG's longer relation is equivalent to the more elegant and succinct (4*k5^2 - 6*k4*k1 + 2*k2*k3 - k0*k5 + k6*k5) = (4*k3^2 - 6*k2*k4 + 2*k5*k1 - k0*k3 + k6*k3) . However, even this minuscule improvement was called into question when it occurred to me to compute the dimension of the relation set via the rank of its Jacobian matrix [[d f_i / d k_j]] , where f_i and k_j range over the 10 relations and 7 variables resp. Disaster! The Jacobian rank is 5 ; but Steiner rings have dimension 3 modulo isometry, and there are only n+2 = 7 variables. Since 3+5 > 7 , I conclude that the relations cannot possibly hold for all Steiner 5-rings. Fred Lunnon On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
For n = 5 --- assuming their validity, which currently remains unproved --- RWG's experimental quadratic relations constitute a polynomial basis for constraints on Steiner ring curvatures, with dimension 4. Also there are no linear constraints. [Established using Gröbner bases.]
The constraints comprise cyclic shifts (fixing k0,k6 ) of
k1^2 + k3*k2 + k3*k5 + k2*k4 = k3^2 + k1*k2 + k1*k4 + k2*k5 ;
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 ;
there are only 5 (not 10) in each orbit, since symmetry causes reflections to be equivalent to shifts. One relation in each orbit is redundant.
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG:
Wait, you can do those weirdly unsymmetrical n=5's?
In principle, but it's a more complicated on account of involving cyclotomy / trigonometry (as well as quadratic polynomials).
Yeah, but in the States you need a prescription for that.
In the end I resorted to pentacyclic (Lie-sphere) rather than tetracyclic (Moebius) coordinates, mostly because the geometric algebra machinery was to hand. Guess you need security clearance for that ...
WFL
On 8/10/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
RWG: << Apparently for composite cycle length n = j h, break the list of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1). >>
All curvatures are determined by any three consecutive; hence there are n-1 independent constraints. Does this polynomial ideal have a linear basis?
For n = 4 , 2 out of 3 are given by RWG's linear relations (k_0 + k_5) - (k_1 + k_3) = (k_0 + k_5) - (k_2 + k_4) = 0 ; what is some third independent relation?
WFL
On 8/10/14, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 8:39 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 3:35 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Aug 8, 2014 at 7:05 AM, Bill Gosper <billgosper@gmail.com> wrote:
> If the four curvatures are k1, k2, k3, k4, and the bounding circles > are > k0 and k5, > then empirically, k1 + k3 = k2 + k4 = k0 + k5 for some assignment of > signs to k0 and k5. > --rwg > Found nothing for n=5,
k1^2 + k2 k3 + k2 k4 + k3 k5 == k1 k2 + k3^2 + k1 k4 + k2 k5 plus nine more, closing over D_5. Huh?
And apparently ten more like 0 == k0 k1 + k0 k2 - k1 k2 - k2^2 - k0 k3 + 2 k1 k3 + k3^2 - k0 k4 - 2 k2 k4 + k3 k4 + k1 k5 - k2 k5 + k3 k5 - k4 k5 - k1 k6 - k2 k6 + k3 k6 + k4 k6, where k0,k6 are the annulus.
And there seems to be a very weird one involving k0 and k6, which may well fail under scrutiny. --rwg
Failed. I've been finding these numerically (69D) using assorted transcendental homographic coefficients, figuring if I get a false positive, I'll be famous for finding a relation among π,e^-2, and EulerGamma, e.g. But I'm getting dozens of false positives. Boy am I confused.
but for n=6, {k1,...,k6} bounded by k0 and k7,
k1+k4=k2+k5=k3+k6 (opposite pairs)=3(k0-k7), *and* k1+k3+k5=k2+k4+k6. (Four relations.) --rwg
Apparently for composite cycle length n = j h, break the list
of n curvatures into j bursts of length h. Weight them by a(1),a(2),...,a(h-1),a(h), with sum_i a(i) = 0. Then sum(a(i) k_i)=0 But there are likely other relations involving as well the annulus a(0),a(n+1).
WFL> I can prove this via plane tetracyclic coordinates.
The later conjectures can undoubtedly be decided in the same fashion, for small fixed n .
Fred Lunnon
Yeah, but in the States you need a prescription for that.
Wait, you can do those weirdly unsymmetrical n=5's?
How about n=7? I'm getting dozens of unlikely results like k3^2 + 2 k1 k4 + 2 k4 k5 == 2 k3 k4 + k4^2 + 2 k1 k5 (no k0,k8,k2,k6,k7!) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun