Almost-Toric Hypersurfaces Bo Lin University of California, - - PowerPoint PPT Presentation

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Almost-Toric Hypersurfaces Bo Lin University of California, - - PowerPoint PPT Presentation

Introduction and the Main Theorem An example Sketch of Proof Implementation Almost-Toric Hypersurfaces Bo Lin University of California, Berkeley April 18th, 2015 STAGS 2015, Brown University arXiv:1410.0776 Bo Lin Almost-Toric


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Introduction and the Main Theorem An example Sketch of Proof Implementation

Almost-Toric Hypersurfaces

Bo Lin

University of California, Berkeley

April 18th, 2015 STAGS 2015, Brown University arXiv:1410.0776

Bo Lin Almost-Toric Hypersurfaces

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Abstract

An almost-toric hypersurface is parametrized by monomials multiplied by polynomials in one extra variable. We determine the Newton polytope of such a hypersurface, and apply this to give an algorithm for computing the implicit equation.

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Toric Part of ZA,f

Let A be a n × (n + 2) full-rank matrix with integer entries. A =

  • a0

a1 . . . an+1

  • .

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Toric Part of ZA,f

Let A be a n × (n + 2) full-rank matrix with integer entries. A =

  • a0

a1 . . . an+1

  • .

If the entries of each vector ai have the same sum d, then ta0, ta1, . . . , tan+1 are Laurent monomials in n variables t1, . . . , tn of the same degree d, where tai = ta1,i

1

ta2,i

2

. . . tan,i

n

.

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Coefficient Part of ZA,f

We fix K as an algebraically closed field. Let f0, f1, . . . , fn+1 ∈ K[x] be univariate polynomials in another variable x.

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Almost-Toric Hypersurface ZA,f

Definition An n-dimensional almost-toric hypersurface ZA,f is a codimension 1 hypersurface that is the Zariski closure of following parametrization: {(ta0f0(x) : . . . : tan+1fn+1(x)) ∈ Pn+1|t ∈ (K∗)n, x ∈ K}. Remark We need some mild hypothesis of A, f to guarantee that ZA,f is a codimension 1 hypersurface. We will see the hypothesis in our main theorem.

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Newton Polytope

Suppose ZA,f is an almost-toric hypersurface, then its ideal is

  • principal. The Newton polytope of ZA,f is defined as the Newton

polytope of the generator (up to a scalar multiple) of this principal

  • ideal. This generator, denoted as p(u0, . . . , un+1), is called the

implicit equation of ZA,f. Proposition The Newton polytope of an almost-toric hypersurface ZA,f is at most 2-dimensional in Rn+2. Its vertices are non-negative lattice points.

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Newton Polygon

Proof. If we substitute all ui by taifi(x) in p, we get another polynomial in variables t1, . . . , tn, x. By definition this is the zero polynomial. After this substitution, each term in p becomes the product of a monomial in variables t1, t2, . . . , tn and a polynomial in x. Because p is the generator of a principle ideal, all such monomials in variables t1, t2, . . . , tn are the same, which gives n independent linear equations on the vertices of Newt(ZA,f) and we conclude that Newt(ZA,f) is at most 2-dimensional.

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Main Theorem

Theorem Suppose ZA,f is defined as before. It has a Pl¨ ucker matrix PA for its toric part and a valuation matrix Vf for its coefficients. (a) if rank(PA · Vf) = 0 then ZA,f is not a hypersurface; (b) if rank(PA · Vf) = 1 then ZA,f is a toric hypersurface; (c) if rank(PA · Vf) = 2 then ZA,f is a hypersurface but not

  • toric. The directed edges of the Newton polygon of ZA,f are

the nonzero column vectors of PA · Vf.

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Main Theorem

Theorem Suppose ZA,f is defined as before. It has a Pl¨ ucker matrix PA for its toric part and a valuation matrix Vf for its coefficients. (a) if rank(PA · Vf) = 0 then ZA,f is not a hypersurface; (b) if rank(PA · Vf) = 1 then ZA,f is a toric hypersurface; (c) if rank(PA · Vf) = 2 then ZA,f is a hypersurface but not

  • toric. The directed edges of the Newton polygon of ZA,f are

the nonzero column vectors of PA · Vf. Remark The sum of each row in PA · Vf is zero.

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Pl¨ ucker Matrices

The Pl¨ ucker matrix PA of A is a (n + 2) × (n + 2) square matrix whose row vectors span the kernel of A. The i, j-th entry of PA is pi,j =     

1 δ(−1)i+j det(A[i,j]),

i < j; −pj,i, i > j; 0, i = j. where A[i,j] is the submatrix obtained from A by deleting the i-th and j-th columns of A and δ is the greatest common divisor of all det(A[i,j]).

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Pl¨ ucker Matrices

Example: A = 3 2 1 1 2 3

  • .

Then δ = 3 and PA =     −1 2 −1 1 −3 2 −2 3 −1 1 −2 1     .

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Pl¨ ucker Matrices

Example: A = 3 2 1 1 2 3

  • .

Then δ = 3 and PA =     −1 2 −1 1 −3 2 −2 3 −1 1 −2 1     . Proposition PA is skew-symmetric. The rank of PA is 2. The entries in each row and column of PA sum to 0.

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Valuation Matrices

The valuation matrix of ZA,f is defined from the polynomials f0, f1, . . . , fn+1. Suppose g1, g2, . . . , gm are all irreducible factors of n+1

i=0 fi. Then

we define vectors uj = deg(gj) · (ordgjf0, ordgjf1, . . . , ordgjfn+1) ∈ Nn+2 for 1 ≤ j ≤ m. Now we need to simplify the set of these vectors. We combine the pairwise linearly dependent vectors, because they correspond to the same edge of the polygon.

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Valuation Matrices - continued

Now let S = {u1, u2, . . . , um}. If two vectors in S are linearly dependent, then we delete them and add their sum to the set. We repeat this procedure. After finite steps, we end up with another set without pairwise linearly dependent vectors. Finally we get S′ = {v1, v2, . . . , vl}. Definition The valuation matrix of ZA,f is Vf =

  • vT

1

vT

2

. . . vT

l

(− l

j=1 vj)T

. The last vector represents the valuation at ∞.

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Valuation Matrices - continued

Now let S = {u1, u2, . . . , um}. If two vectors in S are linearly dependent, then we delete them and add their sum to the set. We repeat this procedure. After finite steps, we end up with another set without pairwise linearly dependent vectors. Finally we get S′ = {v1, v2, . . . , vl}. Definition The valuation matrix of ZA,f is Vf =

  • vT

1

vT

2

. . . vT

l

(− l

j=1 vj)T

. The last vector represents the valuation at ∞. Proposition The sum of each row in Vf is zero.

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Valuation Matrices

Example: n = 2, f0 = x−1, f1 = (x−1)2(x+1), f2 = (x+1)x3, f3 = (x−2)x. Then the irreducible factors are x − 1, x, x + 1, x − 2.

1 x − 1 corresponds to (1, 2, 0, 0) 2 x corresponds to (0, 0, 3, 1) 3 x + 1 corresponds to (0, 1, 1, 0) 4 x − 2 corresponds to (0, 0, 0, 1)

These vectors are pairwise linearly independent, so the valuation matrix is Vf =     1 −1 2 1 −3 3 1 −4 1 1 −2     .

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3-dimensional hypersurface ZA,f

Let ZA,f admit the following parametrization over C:

(t2

1(x2+1) : t1t2x3(x−1) : t1t3x(x+1) : t2 2(x−2)(x2+1) : t2 3(x−1)2(x+1)).

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Pl¨ ucker Matrix of ZA,f

In this example A =   2 1 1 1 2 1 2   , d = δ = 2. Then PA =       −2 2 1 −1 2 −4 2 −2 4 −2 −1 2 −1 1 −2 1       .

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Valuation Matrix of ZA,f

In this example the irreducible factors are x, x − 1, x + i, x − i, x + 1, x − 2. But we can combine x ± i into x2 + 1. So the vectors are (0, 3, 1, 0, 0), (0, 1, 0, 0, 2), (2, 0, 0, 2, 0), (0, 0, 1, 0, 1), (0, 0, 0, 1, 0), (−2, −4, −2, −3, −3). The valuation matrix of ZA,f is Vf =       2 −2 3 1 −4 1 1 −2 2 1 −3 2 1 −3       .

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Result of The Main Theorem

The main theorem tells us that these directed edges are the column vectors of PA · Vf. The product is       −4 −4 2 1 1 4 −4 4 4 −2 −2 12 4 −8 −2 −6 2 −2 −2 1 1 −6 −2 4 1 3       .

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Implicit Equation of ZA,f

Using ideal elimination in Macaulay2 we can compute the implicit equation of ZA,f in variables u0, u1, u2, u3, u4: 16u4

1u16 2 u2 3 − 40u0u4 1u14 2 u2 3u4 + 8u2 0u2 1u14 2 u3 3u4 − 16u0u6 1u12 2 u3u2 4

+20u2

0u4 1u12 2 u2 3u2 4 + 159u3 0u2 1u12 2 u3 3u2 4 + u4 0u12 2 u4 3u2 4

+54u2

0u6 1u10 2 u3u3 4 − 77u3 0u4 1u10 2 u2 3u3 4 + 379u4 0u2 1u10 2 u3 3u3 4

+5u2

0u8 1u8 2u4 4 − 27u3 0u6 1u8 2u3u4 4 − 29u4 0u4 1u8 2u2 3u4 4

+163u5

0u2 1u8 2u3 3u4 4 − 12u3 0u8 1u6 2u5 4 − 35u4 0u6 1u6 2u3u5 4

−425u5

0u4 1u6 2u2 3u5 4 + 4u6 0u2 1u6 2u3 3u5 4 + 87u5 0u6 1u4 2u3u6 4

+717u6

0u4 1u4 2u2 3u6 4 + 103u6 0u6 1u2 2u3u7 4 − 115u7 0u4 1u2 2u2 3u7 4

+12u7

0u6 1u3u8 4 + 4u8 0u4 1u2 3u8 4.

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Newton Polytope of ZA,f

Using Sagemath we can find the vertices of Newton polytope of this defining polynomial are (0, 4, 16, 2, 0), (2, 8, 8, 0, 4), (3, 8, 0, 6, 5), (7, 6, 0, 1, 8), (8, 4, 0, 2, 8), (4, 0, 12, 4, 2). Then the directed edges are (2, 4, −8, −2, 4), (−4, 4, 4, −2, −2), (−4, −4, 12, 2, −6) (1, −2, 0, 1, 0), (4, −2, −6, 1, 3), (1, 0, −2, 0, 1).

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Tropical variety

Let K be a field with valuation and f(x1, . . . , xn) =

  • u∈Nn

cu

n

  • i=1

xui

i ∈ K[x1, . . . , xn]

be a polynomial.

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Tropical variety

Let K be a field with valuation and f(x1, . . . , xn) =

  • u∈Nn

cu

n

  • i=1

xui

i ∈ K[x1, . . . , xn]

be a polynomial. Then for w ∈ Rn, the tropical polynomial is defined as trop(f)(w) = min

u∈Nn (uiwi + val(cu)).

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Tropical variety

Let K be a field with valuation and f(x1, . . . , xn) =

  • u∈Nn

cu

n

  • i=1

xui

i ∈ K[x1, . . . , xn]

be a polynomial. Then for w ∈ Rn, the tropical polynomial is defined as trop(f)(w) = min

u∈Nn (uiwi + val(cu)).

The tropical hypersurface is defined as

trop(V (f)) = {w ∈ Rn|minimum is attained at least twice in trop(f)(w)}.

If V is a variety, the tropical variety trop(V ) is defined as trop(V ) =

  • f∈I(V )

trop(V (f)).

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XA and Yf

Definition XA is defined as the Zariski closure of the following parametrization: {(ta0 : ta1 : . . . : tan+1) ∈ Pn+1|t ∈ (K∗)n}. Then XA is a toric variety.

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XA and Yf

Definition XA is defined as the Zariski closure of the following parametrization: {(ta0 : ta1 : . . . : tan+1) ∈ Pn+1|t ∈ (K∗)n}. Then XA is a toric variety. Definition Yf is defined as the Zariski closure of the following parametrization: {(f0(x) : f1(x) : . . . : fn+1(x)) ∈ Pn+1|x ∈ K}. Then Yf is a curve.

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trop(ZA,f)

The tropicalization of XA and Yf are easy to find, and since ZA,f is the Hadamard Product of XA and Yf, trop(ZA,f) is the Minkowski sum of trop(XA) and trop(Yf).

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trop(ZA,f)

The tropicalization of XA and Yf are easy to find, and since ZA,f is the Hadamard Product of XA and Yf, trop(ZA,f) is the Minkowski sum of trop(XA) and trop(Yf). Proposition Let ZA,f be an almost-toric hypersurface, then

trop(ZA,f) = {u + λ · vT |u ∈ row(A), v ∈ col(Vf), λ ≥ 0}.

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Sketch of Proof

The proof of our main theorem consists of two parts:

1 Show that the edges of Newt(ZA,f) have the same directions

as the column vectors of PA · Vf.

2 Show that each edge has the same length as the

corresponding column vector of PA · Vf. For 1, the tropicalization of ZA,f has maximal cells corresponding to rays. These cells also correspond to the edges of Newt(ZA,f), in the following way such that every edge is orthogonal to the ray of the cell. (PA is skew-symmetric so for any column vector v we have vT · (PA · v) = 0) Then we determine the directions of the edges of Newt(ZA,f).

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Sketch of Proof - continued

For 2, we should be careful. We need the notion of multiplicity of maximal cells. It is defined using commutative algebra. However we use some results in [1] and find them using the index of lattices. Then we prove our main theorem. To get the correct length we need δ in the definition of the Pl¨ ucker matrix.

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Goal of Implementation

For almost-toric hypersurfaces, our goal is: given the parametrization of an almost-toric hypersurface, to find its implicit equation p(u0, u1, . . . , un+1).

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Goal of Implementation

For almost-toric hypersurfaces, our goal is: given the parametrization of an almost-toric hypersurface, to find its implicit equation p(u0, u1, . . . , un+1). Existing approaches include ideal elimination using Gr¨

  • bner bases,

which is inefficient when n is large.

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Algorithm

Based on the main theorem we have the following algorithm:

1 Compute PA from A, factorize f0, f1, . . . , fn+1 over Q into

irreducible factors to get Vf.

2 Compute PA · Vf and check that it has rank 2. 3 Find Newt(ZA,f) using our theorem. 4 Determine all possible monomials in variables u0, u1, . . . , un+1

that could appear in the implicit equation.

5 Use linear algebra to compute the coefficients of these

monomials.

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Step (1) and (2)

In step (1) we can use mathematical software to factor the polynomials in f0, . . . , fn+1.

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Step (3)

By our theorem, the set of directed edges are the column vectors

  • f PA · Vf. Then we need to arrange them in the correct order.

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Step (3)

By our theorem, the set of directed edges are the column vectors

  • f PA · Vf. Then we need to arrange them in the correct order.

We could project these vectors to a 2-dimensional space, by choosing two of the coordinates 1 ≤ c1 < c2 ≤ n + 2. There is still the problem of orientation: these directed edges admit two different arrangements. The correct orientation is determined by the sign of the (c1, c2)-th entry of PA.

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Step (3) - continued

Suppose we have arranged the column vectors of PA · Vf in correct

  • rder, say v1, . . . , vm. Then if we fix one point c in Zn+2, we can

add those vectors one by one to get all vertices of a polygon: c, c + v1, c + v1 + v2, . . . , c + v1 + v2 + . . . + vm−1, which is a translation of Newt(ZA,f).

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Step (3) - continued

Suppose we have arranged the column vectors of PA · Vf in correct

  • rder, say v1, . . . , vm. Then if we fix one point c in Zn+2, we can

add those vectors one by one to get all vertices of a polygon: c, c + v1, c + v1 + v2, . . . , c + v1 + v2 + . . . + vm−1, which is a translation of Newt(ZA,f). We claim that Newt(ZA,f) is uniquely determined. Note that each vertex of Newt(ZA,f) corresponds a monomial appeared in the implicit equation p(u0, u1, . . . , un+1) and its i-th coordinate is the exponent of ui−1 in the monomial. Since p is irreducible, for each i the minimum of these exponents must be 0. Then we know exactly how to translate our polygon to get Newt(ZA,f)

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Step (4)

From Newt(ZA,f) we find all lattice points inside or on the boundary of this polygon. Thus we find all monomials that may appear in p(u0, u1, . . . , un+1).

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Step (5)

Suppose we have undetermined coefficients for all possible monomials in p. After the substitution ui = taifi(x) we can cancel the unique monomial in variables t1, . . . , tn and get a univariate polynomial in x with undetermined coefficients. Since this polynomial is identically zero, the undetermined coefficients satisfy a system of homogeneous linear equations.

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Step (5) - continued

Next we use interpolation. Suppose there are k possible monomials in the implicit equation, then we plug in x by integers ranging from −r to r, where r = ⌊ k

2⌋. Each interpolation gives a linear equation

with k coefficients. Then we use the solve command in Maple 17 to solve these coefficients. Since this is a homogeneous linear system, the solution space should be 1-dimensional. This leads us to add another equation, for example a1 = 1 where a1 is one of the coefficients, to guarantee the uniqueness of solution. After getting the solution, if all coefficients are rational, we normalize them so that their content is 1.

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Efficiency Test

We then try some examples that both Macaulay2 and Sagemath cannot solve in a reasonable time. We generate some samples as the input using the following method: let n be the dimension of the torus, d the degree of the homogeneous monomials and k a positive integer. Then we choose n + 2 degree d monomials randomly from all possible n+d−1

d

  • choices. For the univariate

polynomials, we choose n + 2 polynomials of the form (x − 2)∗(x − 1)∗x∗(x + 1)∗(x + 2)∗, where each ∗ is a random integer between 0 and k.

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Efficiency Test

sample degree # of terms time cost 1 213 109 12.484s 2 109 80 1.594s 3 172 129 10.421s 4 474 275 156.969s 5 291 137 20.375s 6 179 97 8.110s 7 40 32 0.156s 8 27 14 0.140s 9 79 71 1.766s 10 281 148 20.719s

Table: n = 4, d = 4, k = 5

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Efficiency Test

The table shows that the time needed to find the implicit equation

  • f almost-toric hypersurfaces given by randomly generated inputs.

Our algorithm improves the efficiency of finding the implicit equation of almost-toric hypersurfaces.

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Reference

[1]

  • D. Maclagan and B. Sturmfels: Introduction to Tropical

Geometry, Graduate Studies in Mathematics 161, American Mathematical Society, Providence, RI, 2015.

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The End

Thank you!

Bo Lin Almost-Toric Hypersurfaces