Tropical Algebraic Geometry 3

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties MCS 563 Lecture 41 Analytic Symbolic Computation Jan Verschelde, 25 April 2014 Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 1 / 32

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 2 / 32

an example Consider as an example the binomial system { x 2 0 x 1 x 4 2 x 3 3 1 = 0 x 0 x 1 x 2 x 3 1 = 0 A = [ 2 1 4 3 1 1 1 1 ] T c = [ 1 1 ]. As basis of the null space of A we can for example take u = ( 3, 2, 1, 0) and v = ( 2, 1, 0, 1). The vectors u and v are tropisms for a two dimensional algebraic set. Placing u and v in the first two rows of a matrix M, extended so det(m) = 1, we obtain a coordinate transformation, x = y M : M = 3 2 1 0 2 1 0 1 1 0 0 0 0 1 0 0 x 0 = y 3 0 y 2 1 y 2 x 1 = y 2 0 y 1y 3 x 2 = y 0 x 3 = y 1. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 3 / 32

monomial transformations By construction, as Au = 0 and Av = 0: MA = 3 2 1 0 2 1 0 1 1 0 0 0 0 1 0 0 2 1 1 1 4 1 3 1 = 0 0 0 0 2 1 1 1 = B. The corresponding monomial transformation x = y M performed on x A = c yields y MA = y B = c, eliminating the first two variables: { y 2 2 y 3 1 = 0 y 2 y 3 1 = 0. Solving this reduced system gives values z 2 and z 3 for y 2 and y 3. Leaving y 0 and y 1 as parameters t 0 and t 1 we find as solution (x 0 = z 2 t 3 0 t 2 1, x 1 = z 3 t 2 0 t 1, x 2 = t 0, x 3 = t 1 ). Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 4 / 32

1 2 3 computation of the degree To compute the degree of (x 0 = z 2 t 3 0 t 2 1, x 1 = z 3 t0 2t 1, x 2 = t 0, x 3 = t 1 ) we use two random linear equations: 2 { c10 x 0 + c 11 x 1 + c 12 x 2 + c 13 x 3 + c 14 = 0 1 c 20 x 0 + c 21 x 1 + c 22 x 2 + c 23 x 3 + c 24 = 0 after substitution: { c 10 t 3 0 t 2 c 20 t 3 0 t 2 1 + c 11 t2 0 t 1 + c 12 t 0 + c 13 t 1 + c 14 = 0 1 + c 21 t2 0 t 1 + c 22 t 0 + c 23 t 1 + c 24 = 0 0 4 3 2 1 0 1 2 3 Theorem (Koushnirenko s Theorem) If all n polynomials in f share the same Newton polytope P, then the number of isolated solutions of f(x) = 0 in (C ) n the volume of P. As the area of the Newton polygon equals 8, the surface has degree 8. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 5 / 32

monomial maps Solutions of binomial systems have monomial parametrizations. A monomial map of dimension d in n-space is defined by an integer matrix A Z d n and a coefficient vector c (C ) n : (C ) d (C ) n : t = (t 0, t 1,...,t d 1 ) x = ct A where (x 0 = c 0 t a 0, x 1 = c 1 t a 1,..., x n 1 = c n 1 t a n 1). The n columns of A define the exponents of t in the map. For (x 0 = z 2 t 3 0 t 2 1, x 1 = z 3 t0 2t 1, x 2 = t 0, x 3 = t 1 ), the matrix is A = [ 3 2 1 0 2 1 0 1 By Kushnirenko s theorem, the degree of a monomial map equals the volume of the polytope spanned by the columns in A. ]. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 6 / 32

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 7 / 32

an illustrative example for a numerical irreducible decomposition (x 2 x1 2)(x 1 2 + x 2 2 + x 3 2 1)(x 1 0.5) = 0 f(x 1, x 2, x 3 ) = (x 3 x1 3 )(x 1 2 + x 2 2 + x 3 2 1)(x 2 0.5) = 0 (x 2 x1 2)(x 3 x1 3)(x 1 2 + x 2 2 + x 3 2 1)(x 3 0.5) = 0 f 1 (0) = Z = Z 2 Z 1 Z 0 = {Z 21 } {Z 11 Z 12 Z 13 Z 14 } {Z 01 } 1 Z 21 is the sphere x 2 1 + x 2 2 + x 2 3 1 = 0, 2 Z 11 is the line (x 1 = 0.5, x 3 = 0.5 3 ), 3 Z 12 is the line (x 1 = 0.5, x 2 = 0.5), 4 Z 13 is the line (x 1 = 0.5, x 2 = 0.5), 5 Z 14 is the twisted cubic (x 2 x 2 1 = 0, x 3 x 3 1 = 0), 6 Z 01 is the point (x 1 = 0.5, x 2 = 0.5, x 3 = 0.5). Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 8 / 32

the illustrative example numerically computing positive dimensional solution sets Used in two papers in numerical algebraic geometry: first cascade of homotopies: 197 paths A.J. Sommese, J. Verschelde, and C.W. Wampler: Numerical decomposition of the solution sets of polynomial systems into irreducible components. SIAM J. Numer. Anal. 38(6):2022 2046, 2001. equation-by-equation solver: 13 paths A.J. Sommese, J. Verschelde, and C.W. Wampler: Solving polynomial systems equation by equation. In Algorithms in Algebraic Geometry, Volume 146 of The IMA Volumes in Mathematics and Its Applications, pages 133 152, Springer-Verlag, 2008. The mixed volume of the Newton polytopes of this system is 124. By theorem A of Bernshteǐn, the mixed volume is an upper bound on the number of isolated solutions. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 9 / 32

three Newton polytopes (x 2 x1 2)(x 1 2 + x 2 2 + x 3 2 1)(x 1 0.5) = 0 f(x 1, x 2, x 3 ) = (x 3 x1 3 )(x 1 2 + x 2 2 + x 3 2 1)(x 2 0.5) = 0 (x 2 x1 2)(x 3 x1 3)(x 1 2 + x 2 2 + x 3 2 1)(x 3 0.5) = 0 Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 10 / 32

looking for solution curves The twisted cubic is (x 1 = t, x 2 = t 2, x 3 = t 3 ). We look for solutions of the form x 1 = t v 1, v 1 > 0, x 2 = c 2 t v 2, c 2 C, x 3 = c 3 t v 3, c 3 C. Substitute x 1 = t, x 2 = c 2 t 2, x 3 = c 3 t 3 into f (0.5c 2 0.5)t 2 + O(t 3 ) = 0 f(x 1 = t, x 2 = c 2 t 2, x 3 = c 3 t 3 ) = (0.5c 3 0.5)t 3 + O(t 5 ) = 0 0.5(c 2 1.0)(c 3 1.0)t 5 + O(t 7 ) conditions on c 2 and c 3. How to find (v 1, v 2, v 3 ) = (1, 2, 3)? Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 11 / 32

faces of Newton polytopes Looking at the Newton polytopes in the direction v = (1, 2, 3): Selecting those monomials supported on the faces 0.5x 2 0.5x1 2 = 0 v f(x 1, x 2, x 3 ) = 0.5x 3 0.5x1 3 = 0 0.5x 2 x1 3 0.5x 3x1 2 + 0.5x 3x 2 + 0.5x1 5 = 0 Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 12 / 32

degenerating the sphere (x 2 x1 2)(x 1 2 + x 2 2 + x 3 2 1)(x 1 0.5) = 0 f(x 1, x 2, x 3 ) = (x 3 x1 3 )(x 1 2 + x 2 2 + x 3 2 1)(x 2 0.5) = 0 (x 2 x1 2)(x 3 x1 3)(x 1 2 + x 2 2 + x 3 2 1)(x 3 0.5) = 0 As x 1 = t 0: x 2 (x2 2 + x 3 2 1)( 0.5) = 0 (1,0,0) f(x 1, x 2, x 3 ) x 3 (x2 2 + x 3 2 1)(x 2 0.5) = 0 x 2 x 3 (x2 2 + x 3 2 1)(x 3 0.5) = 0 As x 2 = s 0: x 1 2(x 1 2 + x 3 2 1)(x 1 0.5) = 0 (0,1,0) f(x 1, x 2, x 3 ) (x 3 x1 3 )(x 1 2 + x 3 2 1)( 0.5) = 0 x1 2(x 3 x1 3)(x 1 2 + x 3 2 1)(x 3 0.5) = 0 Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 13 / 32

more faces of Newton polytopes Looking at the Newton polytopes along v = (1,0,0) and v = (0, 1, 0): (1,0,0) f(x 1, x 2, x 3 ) = x 2 (x2 2 + x 3 2 1)( 0.5) x 3 (x2 2 + x 3 2 1)(x 2 0.5) x 2 x 3 (x2 2 + x 3 2 1)(x 3 0.5) (0,1,0) f(x 1, x 2, x 3 ) = x1 2(x 1 2 + x 3 2 1)(x 1 0.5) (x 3 x1 3 )(x 1 2 + x 3 2 1)( 0.5) x1 2(x 3 x1 3)(x 1 2 + x 3 2 1)(x 3 0.5) Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 14 / 32

faces of faces The sphere degenerates to circles at the coordinate planes. (1,0,0) f(x 1, x 2, x 3 ) = x 2 (x2 2 + x 3 2 1)( 0.5) x 3 (x2 2 + x 3 2 1)(x 2 0.5) x 2 x 3 (x2 2 + x 3 2 1)(x 3 0.5) (0,1,0) f(x 1, x 2, x 3 ) = x1 2(x 1 2 + x 3 2 1)(x 1 0.5) (x 3 x1 3 )(x 1 2 + x 3 2 1)( 0.5) x1 2(x 3 x1 3)(x 1 2 + x 3 2 1)(x 3 0.5) Degenerating even more: x 2 (x3 2 1)( 0.5) (0,1,0) (1,0,0) f(x 1, x 2, x 3 ) = x 3 (x3 2 1)( 0.5) x 2 x 3 (x3 2 1)(x 3 0.5) The factor x 2 3 1 is shared with (1,0,0) (0,1,0) f(x 1, x 2, x 3 ). Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 15 / 32

representing a solution surface The sphere is two dimensional, x 1 and x 2 are free: x 1 = t 1 x 2 = t 2 x 3 = 1 + c 1 t1 2 + c 2t2 2. For t 1 = 0 and t 2 = 0, x 3 = 1 is a solution of x 3 1 = 0. Substituting (x 1 = t 1, x 2 = t 2, x 3 = 1 + c 1 t 2 1 + c 2t 2 2 ) into the original system gives linear conditions on the coefficients of the second term: c 1 = 0.5 and c 2 = 0.5. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 16 / 32

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 17 / 32

an example Consider f C[x ±1, y ±1 ], a Laurent polynomial: f = c 1,2 xy 2 + c 0,2 y 2 + c 2,1 x 2 y + c 1,1 xy + c 1,0 y + c 4,0 x 4 + c 2,0 x 2 + c 0,0 with its Newton polygon and its normal fan: (0,2) (0,1) (0,0) (1,2) (1,1) (2,1) (4,0) (2,0) (0,1) (1,0) (0, 1) ( 2, 3) Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 18 / 32

a balanced fan Definition Let Σ be a rational, pure d-dimensional fan in R n. Fix weights m(σ) N n for all d-dimensional cones σ Σ. For a (d 1)-dimensional cone τ Σ, let L be the linear space parallel to τ, dim(l) = d 1. The abelian group L Z = L Z n is free of rank d 1, with N τ = Z n /L Z = Z n d+1. For each σ Σ with τ σ, the set (σ + L)/L is a one dimensional cone in N r R. Let u σ be the first lattice point on this ray. The fan Σ is balanced at τ if σ τ m(σ)u σ = 0. The fan Σ is balanced if it is balanced at each τ Σ, dim(τ) = d 1. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 19 / 32

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 20 / 32

the structure theorem Definition A pure d-dimensional polyhedral complex Σ in R n is connected through codimension one if for any two d-dimensional cells P, Q Σ there is a chain P = P 1, P 2,...,P s = Q for which P i and P i+1 share a common facet F i, for 1 i < s. Since P i are facets of Σ and F i are ridges, we call this a facet-ridge path connecting P and Q. Theorem (Structure Theorem for Tropical Varieties) Let X be an irreducible d-dimensional variety of T n. Then trop(x) is the support of a balanced weighted Γ val -rational polyhedral complex pure of dimension d. Moreover, the polyhedral complex is connected through codimension 1. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 21 / 32

computing multiplicities Lemma (Lemma 3.4.6) Let f = c a x a K[x ±1 1, x ±1 ±1 2,..., x n ] and a A let be a regular subdivision of the Newton polytope of f, induced by val(c a ); and let Σ be the polyhedral complex supported on trop(v(f)) that is dual to. The multiplicity of a maximal cell σ Σ is the lattice length of the edge e(σ) of dual to σ. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 22 / 32

the example revisited Assume val( ) does not triangulate the Newton polygon of f. (0,1) 2 (1,0) (0, 1) 1 ( 2, 3) 1 ( ) ( 0 1 4 + 2 1 0 4 (0,2) (0,1) (0,0) ) ( 0 + 1 1 (1,2) (1,1) (2,1) (4,0) (2,0) ) ( 2 + 1 3 ) = ( 0 0 ) Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 23 / 32

applying a unimodular coordinate transformation in ( 2, 3) (f)(x, y) = x 4 + xy 2 (1,2) (0,2) ( ) 2 3 U =, det(u) = 1 1 1 (0,1) (1,1) (2,1) ( ) ( ) 4 1 8 8 U = (0,0) (4,0) 0 2 4 3 (2,0) in ( 2, 3) (f)(x = X 2 Y 1, y = X 3 Y 1 ) ( 2, 3) = (X 2 Y 1 ) 4 + (X 2 Y 1 )(X 3 Y 1 ) 2 1 = X 8 Y 4 + X 8 Y 3 = X 8 Y 3 (Y + 1) Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 24 / 32

proof of Lemma 3.4.6 Pick w in the relative interior of σ. The initial ideal in w ( f ) is generated by in w (f) = t val(ca) c a x a. a e(σ) Since dim(e(σ)) = 1, a b for a, b e(σ) is unique up to scaling, so take v = a b of minimal length. in w (f) is then a monomial times g K[x ±1 ] in the variable y = x v. We may multiply f with a monomial so in w (f) is a polynomial (without negative exponents) with nonzero constant term. deg(g) equals the lattice length of e(σ), which equals the multiplicity of σ. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 25 / 32

Tropical Algebraic Geometry 3 1 Monomial Maps solutions of binomial systems an illustrative example 2 The Balancing Condition balancing a polyhedral fan the structure theorem 3 The Fundamental Theorem formal power series and tropical varieties Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 26 / 32

formal power series The discrete valuation ring of formal power series in t 1/N is { } R N = K[[t 1/N ]] = c α t α/n c α K, for N > 0, α=0 with discrete valuation on s R N : α } val(s) = ord t (s) = min{ N c α 0 1 N Z { }. If N divides M, then R N is a subring of R M. Denote the quotient field of R N as L N. The direct limit of L N gives the field of fractional power (or Puiseux) series L = K{{t}} = lim L N = L N. N>0 Puiseux: if K is algebraically closed, then so is K{{t}}. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 27 / 32

tropical varieties For an ideal J in the polynomial ring K{{t}}[x], there are two ways to define the tropical variety of J, denoted by Trop(J): 1 using t-initial ideals: Trop(J) = { ω R n t in ω (J) is monomial free }, or where t in ω uses the weight ( 1, ω) on monomials in K[t, x]; 2 Trop(J) is the image of val(p) for all p V(J) of J in K{{t}} n. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 28 / 32

the main theorem The main theorem of tropical algebraic geometry states that both ways to defined tropical varieties are equivalent, or: Theorem (the fundamental theorem of tropical algebraic geometry) ω Trop(J) Q n p V(J) : val(p) = ω Q n. We can rephrase the theorem as follows: every rational vector in the tropical variety corresponds to the leading powers of a Puiseux series converging to a point in the variety. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 29 / 32

t-initial ideals A t-initial ideal of an ideal J is not generated by taking the t-initial forms of the given generators of J. Consider for example J = tx + y, x + t as an ideal in K{{t}}[x, y]. We have that y t 2 J. Let ω = (1, 1), so y = t in ω (y t 2 ) belongs to the t-initial ideal. However t in ω (tx + y), t in ω (x + t) = x and y x. To compute t-initial ideals, standard bases are used. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 30 / 32

prime ideals For pure dimensional prime ideals, the situation is better: Theorem Let I and ideal in C{{t}}[x]. If I is prime of dimension d, then Trop(I) is a pure d-dimensional polyhedral fan. A finite intersection of tropical hypersurufaces is a tropical prevariety. Every ideal I has a fine generating set {f 1, f 2,..., f r } such that Trop(I) = Trop(f 1 ) Trop(f 2 ) Trop(f r ). This set of generators is a tropical basis for the ideal. Every tropical variety is a tropical prevariety, but the converse does not hold. An algorithm to compute the tropical prevariety uses the common refinement of the polyhedral fans of hypersurfaces. Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 31 / 32

Summary + Exercises Tropical algebraic geometry provides a framework to apply polyhedral methods to solve polynomial systems. Exercises: 1 Consider the amoeba of the product of two linear equations. How many tentacles to you see in each direction? For which choices of the coefficients do you see holes in the amoeba? 2 For ideals I and J show that I J Trop(I) Trop(J); Trop(I J) = Trop(I) Trop(J); and Trop(I + J) Trop(I) Trop(J). Analytic Symbolic Computation (MCS 563) Tropical Algebraic Geometry 3 L-41 25 April 2014 32 / 32