Tropical Approach to the Cyclic n-roots Problem
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1 Tropical Approach to the Cyclic n-roots Problem Danko Adrovic and Jan Verschelde University of Illinois at Chicago Joint Mathematics Meetings - AMS Session on Commutative Algebra - San Diego, California Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 / 7
2 Cyclic n-roots Problem x + x + + x n = x x + x x + + x n x n + x n x = C n (x) = n j+i i = 3, 4,..., n : x k mod n = j= x x x x n =. benchmark problem in the field of computer algebra (pop. by J. Davenport) extremely hard to solve for n 8 square systems we expect isolated solutions we find positive dimensional solution sets Lemma (Backelin) If m divides n, then the dimension of the cyclic n-roots polynomial system is at least m. J. Backelin: Square multiples n give infinitely many cyclic n-roots. Reports, Matematiska Institutionen, Stockholms Universitet, 989. J. Davenport. Looking at a set of equations. Technical Report 87-6, Bath Computer Science, 987. k=j Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 / 7
3 Our Approach a new polyhedral method for square systems and systems with more equations than unknowns a symbolic-numeric approach with an origin in polyhedral homotopies Bernshtein s Theorem A & B to solve polynomial systems with Puiseux series we aim to generalize polyhedral homotopies to develop positive dimensional solution sets our approach is inspired by the constructive proof of the fundamental theorem of tropical algebraic geometry Theorem (Fundamental Theorem of Tropical Algebraic Geometry) ω Trop(I ) Q n p V (I ) : val(p) = ω Q n. Anders Nedergaard Jensen, Hannah Markwig, Thomas Markwig: An Algorithm for Lifting Points in a Tropical Variety. Collect. Math. vol. 59, no., pages 9 65, 8. rephrasing the theorem rational vector in the tropical variety corresponds to the leading powers of a Puiseux series, converging to a point in the algebraic variety. we understand the fundamental theorem via polyhedral homotopies we see it as a generalization of Bernshtein s Theorem B Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 3 / 7
4 General Definitions Definition (Polynomial System) F (x) = f (x) = f (x) =. f n (x) = Definition (Laurent Polynomial) f (x) = a A c a x a, c a C \ {}, x a = x ±a x ±a x ±an n Definition (Support Set) The set of exponents A i is called the support set of f i. Definition (Newton Polytope) Let A i be the support set of the polynomial f i F(x) =. Then, the Newton polytope of f i is the convex hull of A i, denoted P i. equivalent representation of P i (or any polytope) in R n convex hull of finite set of points, i.e. V-representation intersection of finitely many closed half-spaces, i.e. H-representation Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 4 / 7
5 General Definitions Definition (Initial Form) Let f (x) = c a x a be a Laurent polynomial, v Z n a non-zero vector and let, a A denote the usual inner product. Then, the initial form with respect to v is given by in v (f (x)) = c a x a m = min { a, v a A} a A, m= a,v Definition (Initial Form System) For a system of polynomials F(x) =, the initial form system is defined by in v (F(x)) = (in v (f ), in v (f ),..., in v (f n )) =. Definition (Pretropism) A pretropism v Z n is a vector, which leads to an initial form system. Definition (Tropism) A tropism is a pretropism, which is the leading exponent vector of a Puiseux series expansion for a curve, expanded about t. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 5 / 7
6 Tropisms and d-dimensional Surfaces Definition (Cone of Tropisms) A cone of tropisms is a polyhedral cone, spanned by tropisms. v, v,..., v d span a d-dimensional cone of tropisms dimension of the cone is the dimension of the solution set we obtain the tropisms by using the Cayley trick. Let v = (v (,), v (,),..., v (,n ) ), v = (v (,), v (,),..., v (,n ) ),..., v d = (v (d,), v (d,),..., v (d,n ) ) be d tropisms. Let r, r,..., r n be the solutions of the initial form system in v (in v ( in vd (F ) ))(x) =. d tropisms generate a Puiseux series expansion of a d-dimensional surface x = t v (,) t v (,) t v (d,) d (r + c (,) t w (,) + c (,) t w (,) +... ) x = t v (,) t v (,) t v (d,) d (r + c (,) t w (,) + c (,) t w (,) +... ) x = t v (,) t v (,) t v (d,) d (r + c (,) t w (,) + c (,) t w (,) +... ). x n = t v (,n ) t v (,n ) t v (d,n ) d (r n + c (,n ) t w (,n ) + c (,n ) t w (,n ) +... ) Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 6 / 7
7 Cyclic 4-roots problem Cyclic 4-Root Polynomial System x + x + x + x 3 = x x + x x 3 + x x + x x 3 = C 4 (x) = x x x + x x x 3 + x x x 3 + x x x 3 = x x x x 3 = The only pretropism is (,,, ) cyclic 4-roots initial form system in direction (,,, ) x + x 3 = x x + x x 3 + x x + x x 3 = in (,,, ) (C 4 )(x) = x x x 3 + x x x 3 = x x x x 3 = Using M to transform in (,,, ) (C 4 ): M = x = z ; x = z z ; x = z z ; x 3 = z 3 z in (,,, ) (F )(z) = z /z + z 3 /z = z z + z z 3 + z + z 3 = z z z 3 /z + z z 3 /z = z z z 3 = Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 7 / 7
8 Cyclic 4-roots problem cyclic 4-root polynomial system transformed z + z 3 = z z + z z 3 + z + z 3 = in (,,, ) (C 4 )(z) = z z z 3 + z z 3 = z z z 3 = Solutions of the transformed initial form system are (z =, z =, z 3 = ) and (z =, z =, z 3 = ). Letting z = t and returning solutions to original coordinates with x = z ; x = z z ; x = z z ; x 3 = z 3 z For cyclic 4-roots, the initial terms of the series are exact solutions x = t x = t x = t x = t x = t and x = t x = t x = t Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 8 / 7
9 Cyclic 4,8,-roots problem cyclic 4-roots: tropism: (,-,,-) x = t, x = t, x = t, x 3 = t cyclic 8-roots: tropism: (,-,,-,,-,,-) x = t, x = t, x = it, x 3 = it, x 4 = t, x 5 = t, x 6 = it, x 7 = it cyclic -roots: tropism: (,-,,-,,-,,-,,-,,-) x = t, x = t, x = ( + )t, x 3 = ( + )t, x 4 = ( + )t, x 5 = ( + )t, x 6 = t, x 7 = t, x 8 = ( )t, x 9 = ( )t, x = ( )t, x = ( )t Observing structure among tropism coefficients numerical solver PHCpack was used we recognize the coefficients as n -roots of unity Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 9 / 7
10 Cyclic n-roots problem: n=4m case Proposition For n = 4m, there is a one-dimensional set of cyclic n-roots, represented exactly as x k = u k t x k+ = u k t for k =,..., n and u k = e iπk n = e i4πk n. taking random linear combination of the solutions α t + α t + α t + α 3 t + + α n t + α n t =, α j C and simplifying β t + β =, β j C we see that all space curves are quadrics. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 / 7
11 Cyclic 9-Roots Polynomial System The cone of pretropisms for the cyclic 9-roots polynomial system was generated by pretropism sequence v = (,,,,,,,, ) v = (,,,,,,,, ). x + x 5 + x 8 = x x 8 + x x 3 + x 5 x 6 = x x x + x x x 8 + x x 7 x 8 + x x x 3 + x x 3 x 4 + x 3 x 4 x 5 +x 4 x 5 x 6 + x 5 x 6 x 7 + x 6 x 7 x 8 = x x x x 8 + x x 3 x 4 x 5 + x 5 x 6 x 7 x 8 = x x x x 3 x 8 + x x 5 x 6 x 7 x 8 + x x 3 x 4 x 5 x 6 = In v (In v (C 9 ))(x) = x x x x 3 x 4 x 5 + x x x x 3 x 4 x 8 + x x x x 3 x 7 x 8 +x x x x 6 x 7 x 8 + x x x 5 x 6 x 7 x 8 + x x 4 x 5 x 6 x 7 x 8 + x x x 3 x 4 x 5 x 6 +x x 3 x 4 x 5 x 6 x 7 + x 3 x 4 x 5 x 6 x 7 x 8 = x x x x 3 x 4 x 5 x 8 + x x x x 5 x 6 x 7 x 8 + x x 3 x 4 x 5 x 6 x 7 x 8 = x x x x 3 x 4 x 5 x 6 x 8 + x x x x 3 x 5 x 6 x 7 x 8 + x x x 3 x 4 x 5 x 6 x 7 x 8 = x x x x 3 x 4 x 5 x 6 x 7 x 8 = For one of the first solutions of the cyclic 9-roots polynomial system, we refer to J. C. Faugère, A new efficient algorithm for computing Gröbner bases (F 4 ). Journal of Pure and Applied Algebra, Vol. 39, Number -3, Pages 6-88, Year 999. Proceedings of MEGA 98, 7 June 998, Saint-Malo, France. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 / 7
12 Cyclic 9-Roots Polynomial System Cont. We use the coordinate change to transform the initial form system and the original cyclic 9-roots system. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 / 7 v =(,, -,,, -,,, - ) v =(,, -,,, -,,, - ) The unimodular coordinate transformation x = z M acts on the exponents. The new coordinates are given by M = x = z x = z z x = z z z x 3 = z z 3 x 4 = z z z 4 x 5 = z z z 5 x 6 = z z 6 x 7 = z z z 7 x 8 = z z z 8
13 Cyclic 9-Roots Polynomial System Cont. The transformed initial form system in v (in v (C 9 ))(z) is given by z + z 5 + z 8 = z z 3 + z 5 z 6 + z 8 = z z 3 z 4 + z 3 z 4 z 5 + z 4 z 5 z 6 + z 5 z 6 z 7 + z 6 z 7 z 8 + z z 3 + z 7 z 8 + z + z 8 = z z 3 z 4 z 5 + z 5 z 6 z 7 z 8 + z z 8 = z z 3 z 4 z 5 z 6 + z 5 z 6 z 7 z 8 + z z 3 z 8 = z z 3 z 4 z 5 z 6 z 7 + z 3 z 4 z 5 z 6 z 7 z 8 + z z 3 z 4 z 5 z 6 + z 4 z 5 z 6 z 7 z 8 + z z 3 z 4 z 5 + z z 3 z 4 z 8 +z z 3 z 7 z 8 + z z 6 z 7 z 8 + z 5 z 6 z 7 z 8 = z 3 z 4 z 6 z 7 + z 3 z 4 + z 6 z 7 = z 4 z 7 + z 4 + z 7 = z z 3 z 4 z 5 z 6 z 7 z 8 = Its solution is z =, z 3 = +, z 4 = +, z 5 =, z 6 =, z 7 =, z 8 = +, where i =. While we used a numerical solver PHCpack, we recognized the solution as the 3 rd roots of unity. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 3 / 7
14 Cyclic 9-Roots Polynomial System Cont. The following assignment satisfies cyclic 9-roots polynomial system entirely. z = t x = t z = t z = z 3 = + z 4 = + z 5 = z 6 = z 7 = z 8 = + x = z x = z z x = z z z x 3 = z z 3 x 4 = z z z 4 x 5 = z z z 5 x 6 = z z 6 x 7 = z z z 7 x 8 = z z z 8 x = t t x = t t ( ) x 3 = t ( + ) x 4 = t t ( + ) x 5 = t t x 6 = t ( ) x 7 = t t ( ) x 8 = t t ( + ) Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 4 / 7
15 Cyclic 9-Roots Polynomial System Cont. Letting u = e πi 3 and y = t, y = t t, y = t t u we can rewrite the exact solution as x = t x 3 = t u x 6 = t u x = t t x 4 = t t u x 7 = t t u x = t u x 5 = t x 8 = t u x = y x 3 = y u x 6 = y u x = y x 4 = y u x 7 = y u x = y x 5 = y u x 8 = y u and put it in the same format as in the proof of Backelin s Lemma, given in J. C. Faugère, Finding all the solutions of Cyclic 9 using Gröbner basis techniques. In Computer Mathematics: Proceedings of the Fifth Asian Symposium (ASCM), pages -. World Scientific,. degree of the solution component Simplifying, the system becomes α t + α t t + α 3 t t = α 4 t + α 5 t t + α 6 t t = α i C t t β = t β = forwards : [, u, u ] [u, u, ] [u,, u] backwards : [u, u, ] [u,, u ] [, u, u] As the simplified system has 3 solutions, the cyclic 9 solution component is a cubic surface. With the cyclic permutation, we obtain an orbit of 6 cubic surfaces, which satisfy the cyclic 9-roots system. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 5 / 7
16 Cyclic 6-Roots Polynomial System This 3-dimensional cyclic 6-root solution component is a quartic surface. Using cyclic permutation, we obtain 4 = 8 components of degree 4. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 6 / 7 Extending the pattern we observed among tropisms of the cyclic 9-roots, v = (,,,,,,,, ) v = (,,,,,,,, ) we can get the correct cone of tropisms for the cyclic 6-roots. v = (,,, 3,,,, 3,,,, 3,,,, 3) v = (,,,,,,,,,,,,,,, ) v = (,,,,,,,,,,,,,,, ) Extending the solutions at infinity pattern, cyclic 9-roots: u = e πi 3 cyclic 6-roots: u = e πi 4 The 3-dimensional solution component of the cyclic 6-roots is given by: x = t x 4 = ut x 8 = u t x = u 3 t x = t t x 5 = ut t x 9 = u t t x 3 = u 3 t t x = t t t x 3 = t 3 t t x 6 = ut t t x 7 = ut 3 t t x = u t t t x = u t 3 t t x 4 = u 3 t t t x 5 = u 3 t 3 t t
17 Cyclic n-roots Polynomial System Summary We now generalize the previous results for the cyclic n-roots systems. Proposition For n = m, there is a (m )-dimensional set of cyclic n-roots, represented exactly as x km+ = u k t x km+ = u k t t x km+ = u k t t t. x km+m = u k t t t t m x km+m = u k t m+ t m+ tm 3 m for k =,,,..., m and u k = e ikπ/m. Proposition The (m )-dimensional solutions set has degree equal to m. Applying cyclic permutation, we can find m components of degree m. Danko Adrovic (UIC) the cyclic n-roots problem January th, 3 7 / 7
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