Veit, You're welcome. It's an interesting problem. What's the URL of the website? Victor On Wed, Aug 24, 2011 at 5:13 PM, Veit Elser <ve10@cornell.edu> wrote:
Thanks Victor! The complex Hadamard matrices of form
{{1, 1, 1, 1, 1, 1, 1}, {1, -1 - b - c - d - e - f, b, c, d, e, f}, {1, b, b^2/d, b c, (b d)/f, (b e)/c, (b f)/e}, {1, c, b c, c^2/e, (c d)/b, (c e)/f, (c f)/d}, {1, d, (b d)/f, (c d)/b, d^2/e, d e, (d f)/c}, {1, e, (b e)/c, (c e)/f, d e, e^2/d, (e f)/b}, {1, f, (b f)/e, (c f)/d, (d f)/c, (e f)/b, f^2}}
are now completely classified. I took your univariate polynomials in b (same as c), d (same as e), and f and found all the roots that have modulus 1. I then brute-force evaluated the polynomials
{b c d e + b^2 c d f + b c^2 e f + b c d e f + b d^2 e f + c d e^2 f + b c d e f^2, b^2 c d e + b c d^2 f + b c e f + b c d e f + c^2 d e f + b c d e^2 f + b d e f^2, b c^2 d e + b c d f + b^2 d e f + b c d e f + b c d^2 e f + b c e^2 f + c d e f^2, b c d^2 e + b^2 c e f + c d e f + b c d e f + b c^2 d e f + b d e^2 f + b c d f^2, b c d e^2 + b c^2 d f + b d e f + b c d e f + b^2 c d e f + c d^2 e f + b c e f^2}
at all combinations of the modulus 1 roots and identified those combinations on which all five of the above vanished. This gave 48 solutions overall. There were no surprises, but at least now I have algebraic expressions for the exotic ones (roots of polynomials of degree 18 and 36).
There is a website run by polish academics that records these things and I will credit you for this contribution.
Veit
On Aug 24, 2011, at 1:00 PM, Victor Miller wrote:
Here's the factorization (over Q) of the groebner basis of the radical ideal. I've introduced extra variables I call bb,cc,dd,ee,ff which are the inverses of b,c,d,e,f respectively, enforced by equations like b*bb-1==0. The solutions are obtained by setting all of these polynomials to 0.
Victor
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