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Introduction to amoebas and tropical geometry Milo Bogaard 17-2-2015 Master Thesis Supervisor: dr. Raf Bocklandt KdV Instituut voor wiskunde Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

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Page 1: Introduction to amoebas and tropical geometry · 2020-02-08 · Introduction to amoebas and tropical geometry Milo Bogaard 17-2-2015 Master Thesis Supervisor: dr. Raf Bocklandt KdV

Introduction to amoebas and tropicalgeometry

Milo Bogaard

17-2-2015

Master Thesis

Supervisor: dr. Raf Bocklandt

KdV Instituut voor wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica

Universiteit van Amsterdam

Page 2: Introduction to amoebas and tropical geometry · 2020-02-08 · Introduction to amoebas and tropical geometry Milo Bogaard 17-2-2015 Master Thesis Supervisor: dr. Raf Bocklandt KdV

AbstractIn this thesis we give an introduction to the theory of tropicalgeometry and its applications to amoebas. We treat Kapranov’stheorem and Mikhalkin’s limit construction for amoebas.We also compute a number of examples.

InformationTitle: Introduction to amoebas and tropical geometryAuthor: Milo Bogaard, [email protected], 5743117Supervisor: dr. Raf BocklandtSecond reviewer: Hessel PosthumaSubmitted: 17-2-2015

Korteweg de Vries Instituut voor WiskundeUniversiteit van AmsterdamScience Park 904, 1098 XH Amsterdamhttp://www.science.uva.nl/math

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Contents

Introduction 3

1 Polyhedral complexes and tropical curves 61.1 Polyhedral geometry . . . . . . . . . . . . . . . . . . . . . . . 61.2 The Newton polytope and its subdivisions . . . . . . . . . . . 81.3 Tropical algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4 Tropical hypersurfaces . . . . . . . . . . . . . . . . . . . . . . 131.5 The Newton polytope of a tropical curve . . . . . . . . . . . . 15

2 The amoeba of a planar curve 192.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . 192.2 Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3 The amoeba and the Newton polytope . . . . . . . . . . . . . 232.4 The components of the complement of the amoeba . . . . . . 26

3 Puiseux series and Kapranov’s theorem 273.1 Valuations and Puiseux series . . . . . . . . . . . . . . . . . . 273.2 Tropical polynomials . . . . . . . . . . . . . . . . . . . . . . . 293.3 Kapranov’s theorem . . . . . . . . . . . . . . . . . . . . . . . 32

4 A limit of amoebas 364.1 The construction of the limit . . . . . . . . . . . . . . . . . . . 364.2 The Hausdorff distance . . . . . . . . . . . . . . . . . . . . . . 384.3 A neighborhood of the tropical curve . . . . . . . . . . . . . . 384.4 The limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5 Appendix I 445.1 Endomorphisms of (C∗)2 . . . . . . . . . . . . . . . . . . . . . 445.2 The transformation of the amoeba . . . . . . . . . . . . . . . . 47

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6 Appendix II 496.1 Computing the amoeba of a planar curve . . . . . . . . . . . . 496.2 Critical points . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Populaire samenvatting(Dutch) 54

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Introduction

In this thesis we give an introduction to the related subjects of amoebas andtropical geometry.

If X is an algebraic variety in the algebraic torus (K∗)n over a field Kwith a norm | · |K the amoeba is the image of X under the map definedby LogK(z1, ..., zn) = (log(|z1|K), ...., log(|z1|K)). This construction was firstused in 1971 by George Bergman to proof a theorem about subgroups ofGL(n,Z) in [2]. We briefly explain the relation between amoebas and thegroup GL(n,Z) in appendix I.

The name ’amoeba’ was introduced by Gelfand, Kapranov and Zelevinskywho rediscovered the concept when studying the combinatorics of discrim-inants of polynomials in [8]. The name is very appropriate considering forexample figure 1.

Figure 1: An example of an amoeba of a curve in (C∗)2.

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Tropical geometry is a concept from computer science and is named afterthe Brazilian mathematician and computer scientist Imre Simon. A tropicalcurve in R2 is the set of points where a finite collection of linear functionsdoes not have a unique maximum, for example figure 2.

Figure 2: An example of a tropical curve.

It turns out that many results from algebraic geometry also hold for trop-ical curves, for example Bezout’s theorem for the number of intersections oftwo curves. Many results of this type can be found in [21] and [16].

For fields with a non-Archimedean absolute value the relation betweenamoebas and tropical curves is very direct, the amoeba of a hypersurfaceis given by a tropical hypersurface. This is Kapranov’s theorem, which weprove in chapter 3.

The relation between curves over the complex numbers and amoebas ismore complicated. Using complex analysis Mikhail Passare and Hans Rull-gard constructed a tropical curve which is a deformation retract of a givenamoeba. This construction can be found in [18].

In the last chapter of this thesis we study Mikhalkin’s theorem on thelimit of a sequence of scaled amoebas from the article [14].

A nice application of this theorem is the construction of amoebas with aprescribed number of components. For example the theorem shows that thescaled amoeba of f = u3 +w3 +u2w4 +u4w3 +ru3w+r2u2w2 +ruw3 +ru3w3

converges to figure 2 for r →∞.

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If we pick r = 4 we already get the four bounded components we wanted.We can change the coefficients somewhat without changing the topologyof the amoeba and figure 1 is the amoeba of the curve in (C∗)2 given byu3 + w3 + u2w4 + u4w3 + 3.8u3w + 12.9u2w2 + 4.55uw3 + 4.6u3w3.

A much more serious application of amoebas is to study the topology andanalytic structure of real and complex varieties as in [20], [14] and [15] byViro and Mikhalkin.

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Chapter 1

Polyhedral complexes andtropical curves

In this chapter we give a brief introduction to tropical geometry. The goalis to show the connection between tropical hypersurfaces and subdivisions ofpolytopes, which is theorem 1.28.

1.1 Polyhedral geometry

In this section we state a number of definitions of polyhedral geometry. Mostof this section is derived from the excellent reference work [26].

A subset X of Rn is called convex if for every pair of points x,y ∈ X theline segment connecting x and y is contained in X. The convex hull of a setA ⊂ Rn is the intersection of all convex sets which contain A and is denotedby conv(A).

Definition 1.1. A set P ⊂ Rn is called a polytope if it is the convex hull afinite set of points.

The following concrete description of the set of points of a polytope issometimes useful.

Proposition 1.2. If A ⊂ Rn is a finite set then conv(A) consists of all pointsof the form

∑v∈B λvv with B ⊆ A, λv ∈ (0, 1] and

∑v∈B λv = 1

Sums of this form are called convex combinations of the points in A.We denote the space of linear functionals on Rn by (Rn)∨. If ϕ is a

nonzero functional then any level set is a hyperplane and for c ∈ R the set{x ∈ Rn|ϕ(x) ≤ c} is called a half space.

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Definition 1.3. The intersection of a finite number of half spaces is calleda polyhedron.

The previous two definitions are related by the following theorem.

Theorem 1.4. A subset P ⊂ Rn is a polytope if and only if it is a boundedpolyhedron.

The proof can be found in [26] in chapter 1 and is harder than one wouldexpect. It is now immediately clear that the intersection of two polytopes isagain a polytope, which is not easy to prove directly from the definition.

If we would define a polytope as a bounded polyhedron then it would behard to prove that the Minkowski sum(defined in the following section) oftwo polytopes would be a polytope, so theorem 1.4 is fundamental. 1

A subset F of a polyhedron P is called a face of P if F = ∅ or thereis a functional ϕ ∈ (Rn)∨ such that F = {x ∈ P |ϕ(x) = M}, where Mis the maximum of ϕ on P . The face F is called the supporting face of ϕand is denoted with Pϕ. Because M is the maximum of ϕ on P we haveF = P ∩ {x ∈ Rn|ϕ(x) ≥M}, so F itself is a polyhedron. If P is a polytopeit follows from theorem 1.4 that a face is again a polytope.

A face of dimension 0 is called a vertex and a face of dimension dim(P )−1is called a facet. A vertex contains only one point and we can identify thevertex with the point it contains. The set of vertices is denoted with vert(P ).

The set of all faces of a polyhedron has the following structure.

Definition 1.5. A polyhedral complex is a collection C of polyhedra suchthat

1. If P ∈ C then every face of P is an element of C.

2. If P,Q ∈ C then P ∩Q is a face of P and of Q.

The support |C| of C is the union over all elements of C. We call C apolyhedral subdivision of |C|. Notice that |C| is not necessarily a polyhedron.

Sometimes we refer to the elements of C as the cells of the complex. Thecells are partially ordered by inclusion. If all maximal cells have the samedimension n we say that C is pure of dimension n and denote dim(C) = n.The vertices of the cells of C are cells of C by property 1. and are called thevertices of C. If C is pure of dimension n the cells of dimension n − 1 arecalled the facets of C.

1This version of theorem 1.4 gives no easy proof for the fact that the Minkowski sumof two polyhedra is again polyhedron, which follows directly from a more precise version,see page 30 of [26].

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An isomorphism ϕ between complexes C and D is a bijection ϕ : |C| → |D|which is affine on all elements of C and induces a bijection C → D.

Now we give an example of a polyhedral complex we encounter later.

Example 1.6. Let P ⊂ R× Rn be a polytope and let Pmax ⊂ P be the setof all (λ,x) ∈ P such that if (t,x) ∈ P for some t ∈ R then t ≤ λ. Thus Pmax

consists of all points of P which are maximal on a line of the form R× {x}with x ∈ Rn. The set of all faces of P contained in Pmax is called the upperenvelope of P . We can define the set Pmin as the set of points which areminimal on lines of the form R × {x}, then the set of all faces contained inPmin is called the lower envelope of P .

Lemma 1.7. The upper envelope of a polytope P is a polyhedral subdivisionof Pmax and the lower envelope is a polyhedral subdivision of Pmin.

Proof. It follows directly from the definitions that the upper envelope isa polyhedral complex. Now let p = (λ,x) ∈ Pmax an let H1, ..., Hk bethe half spaces defining P . By definition of Pmax the must be an i suchthat (λ + ε,x) /∈ Hi for every ε ∈ R>0. Now p is contained in the face Fdefined by Hi. If L is the boundary of Hi then F = L ∩ P . If w ∈ L thenw + (ε, 0, ..., 0) /∈ Hi for any ε ∈ R>0, so F is an upper face of P .

The case of the lower envelope is identical.

1.2 The Newton polytope and its subdivi-

sions

We will regularly see the following class of polytopes throughout this thesis.The importance of the Newton polytope for tropical geometry is explainedby theorem 1.28 If K is a field the ring of Laurent polynomials is the ringK[z1, ..., zn, z

−11 , ..., z−1n ]. Thus a Laurent polynomial is a finite sum f =∑

v∈Zk avzv, with av ∈ K and zv = zv11 · · · zvnn . This gives the following

polytope.

Definition 1.8. If f =∑

v∈Zn avzv is a Laurent polynomial then the Newton

polytope Newt(f) is the convex hull of the set {v ∈ Zn|av 6= 0}.

We can immediately verify that Newt(f) behaves nicely with respect tomultiplication. The Minkovski sum A+B of two subsets A,B ⊂ Rn is definedby A+B = {a+ b|a ∈ A, b ∈ B}.

Lemma 1.9. If f and g are Laurent polynomials over a field, then Newt(fg) =Newt(f) + Newt(g).

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Proof. If f =∑

i aizi and g =

∑i biz

i, then fg =∑

i cizi, where ci =∑

j,k:j+k=i ajbk. If ci 6= 0 then there are j, k ∈ Zn such that aj 6= 0 andbk 6= 0, so i = j + k with j ∈ Newt(f) and k ∈ Newt(g). Thus Newt(fg) ⊂Newt(f) + Newt(g).

To show the converse it is enough to show that all vertices of Newt(f) +Newt(g) are contained in Newt(fg). If k is a vertex of Newt(f) + Newt(g),then k is a sum of unique vertices i ∈ Newt(f) and j ∈ Newt(g). Thus thecoefficient ck = aibj 6= 0, so k ∈ Newt(fg).

The Newton polytope is an example of a lattice polytope, which is apolytope such that all vertices are integral points. A polyhedral subdivisionof a lattice polytope is called a lattice subdivision if all of its elements arelattice polytopes. It is clear that any lattice polytope can be realized as theNewton polytope of a polynomial.

We now describe a way to make lattice subdivisions of a lattice polytope.This method works similarly for arbitrary polytopes and can be found inlecture 5 of [26].

Let Q ⊂ Rn be a lattice polytope and let A ⊂ Q be a set of integralpoints containing all vertices of Q and let v : A → R be any function. Inthe applications of this construction Q will be the Newton polytope of apolynomial f and v will depend on the coefficients of f .

We define a polytope P ⊂ Rn+1 by P = conv{(v(i), i)|i ∈ A}. We have anatural projection π : P → Q.

Lemma 1.10. The restriction π : Pmax → Q is a bijection.

Proof. If x ∈ P then we can write x =∑

i∈A λi(v(i), i) with λi ∈ [0, 1] and∑i∈A λi = 1, so π(x) =

∑i∈A λii ∈ Q. Thus π(P ) ⊂ Q. It is clear that

π is injective. Now let x ∈ Q. We have Q = conv(A) because A containsall vertices of Q. Thus we can write x =

∑i∈A λii with λi ∈ [0, 1] and∑

i∈A λi = 1. Now y =∑

i∈A λi(v(i), i) ∈ P . Thus Pmax must contain anelement y such that y1 = y1,...,yn = yn. Clearly π(y) = π(y) = x so π issurjective.

If p is a vertex of a face contained in Pmax then p is also a vertex ofP so p is of the form (v(i), i1, ..., in) for some i ∈ A, so π(p) = (i1, ..., in)is an integral point. Thus the projection of the upper envelope defines alattice subdivision of Newt(f). A subdivision of a polytope which can beconstructed in this way is called a regular subdivision.

We can immediately construct an interesting lattice subdivision.

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Proposition 1.11. If Q ⊂ Rn is a lattice polytope there is a function v : Q∩Zn → Q such that the set of vertices of the regular subdivision induced by vis Q ∩ Zn.

Proof. After a translation we can assume Q is contained in the box [0, N ]n

for some N ∈ N. Let f : [0, N ]→ R be a concave function which is nowherelinear, for example f(t) = N2 − t2. Now we define v : Q ∩ Zn → Q byv(i) =

∑nj=1 f(ij) and note that the image indeed lies in Q.

We must show that (v(k), k) is an upper vertex of the polytope conv({(v(i), i)|i ∈Q ∩ Zn}) in R× Rn for every k ∈ Q ∩ Zn.

Suppose that k =∑

i∈A λii with A ⊂ Q ∩ Zn and λi ∈ (0, 1].We get

v(k) =n∑j=1

f(∑i∈A

λiij) ≥n∑j=1

∑i∈A

λif(ij)

=∑i∈A

λi

n∑j=1

f(ij) =∑i∈A

λiv(i)

by applying Jensen’s inequality to each term of the first sum. This showsv(k) is maximal, so (v(k), k) lies in the upper envelope.

Because f is not linear we have equality if and only if ij = kj for all i ∈ A,which can only happen if A = {k}. This shows (v(k), k) is not contained inconv({(v(i), i)|i ∈ Q ∩ Zn − {k}}), so it is a vertex.

The function used in the proof of the previous lemma is generally not themost convenient when constructing a subdivision. In the following examplewe use a simpler one.

Example 1.12. Consider the ring C(t)[u,w] of polynomials in two variablesover the field of rational functions over C and let

f = u3 + w3 + 1 + u2w3 + u3w2 + tu+ tu2 + tw + tw2 + tu3w

+tuw3 + t2uw + t2u2w + t2uw2 + t2u2w2,

then Newt(f) = conv({(0, 0), (0, 3), (3, 0), (3, 2), (2, 3)}). We define v : Newt(f)∩Z2 → R by v(i) = valt(ai) where ai is the coefficient of f at ui1wi2 . It is easyto check that figure 1.1 is the associated subdivision of Newt(f).

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Figure 1.1: The subdivision of Newt(f) of example 1.12.

Not every lattice subdivision of a lattice polytope is regular, see [26] page132 for a simple counterexample. Many results about regular subdivisionscan be found in [8]. More counterexamples can be found in [12].

1.3 Tropical algebra

Tropical varieties are also derived from algebra, so we give the necessary def-initions of tropical algebra in this section. Similarly to algebraic geometrythere are two ways to define tropical polynomials, as abstract sums of mono-mials or as polynomial functions. As in the case of finite fields a distinctformal sums can define the same functions.

In this thesis we only consider tropical hypersurfaces which are easy todefine directly using a function on Rn and in this chapter this concrete defini-tion would be sufficient. In chapter 3.2 briefly consider the algebra of tropicalpolynomials.

Definition 1.13. A semiring is a set R with addition ⊕ and multiplication� such that

• (R,⊕) is a commutative monoid with identity element e⊕

• (R,�) is a monoid with identity element e�

and for all a, b, c ∈ R the operations must satisfy

• a� (b⊕ c) = a� b⊕ a� c

• (b⊕ c)� a = b� a⊕ c� a

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• e⊕ � a = e⊕.

The axioms are the same as those of a ring except for the lack of anadditive inverse and the fact that e⊕�a = e⊕ does not follow from the otheraxioms.

A standard example is R≥0 with addition and multiplication. The follow-ing example is less intuitive.

Example 1.14. For t ∈ R>1 let Rt = R ∪ {−∞}, let x ⊕t y = logt(tx + ty)

and x�t y = x+ y, where we conveniently put t−∞ = 0. We have logt(x)⊕tlogt(y) = logt(x + y) and logt(x) �t logt(y) = logt(xy) for every x, y ∈ R≥0,so the operation on Rt is just the operation of R≥0 transfered by the maplogt. Thus Rt is a semiring and Rt

∼= R≥0.

Definition 1.15. The tropical semiring is the set T = R ∪ {−∞} withoperations x⊕ y = max{x, y} and x� y = x+ y.

It is not hard to check that T satisfies all the axioms of a semiring.We can also get T as a limit of the semirings Rt given in the previous

example. We have max{x, y} = limt→∞ logt(tx + ty). To see this let x =

max{x, y}, so log(1 + ty−x) ≤ log(2) is bounded. Now limt→∞(tx + ty) =

x+limt→∞log(1+ty−x)

log(t)= x. This gives an immediate proof that T is a semiring.

The process of taking the limit of the semirings Rt is called Maslov de-quantization. It is explained briefly on page 28 of [14].

Definition 1.16. A tropical polynomial is a formal sum⊕

i∈I ai�xi11 �...�xinn

where I ⊂ Zn is a finite index set, ai ∈ T and x1, ..., xn are real variables.

If x = (x1, ..., xn) and i = (i1, ..., in) ∈ Zn we use xi instead of xi11 � ...�xinnand i · x = i1x1 + ...+ inxn, where · is the standard inner product.

Using the definition of ⊕ and � a tropical polynomial f =⊕

i∈I ai � xi

defines a function on Rn given by f(x1, ..., xn) = maxi∈I{ai + i · x}. We saytwo tropical polynomials are equivalent if they define the same function.

Example 1.17. Two tropical polynomials with different coefficients can beequal. For example 0 ⊕ x ⊕ x2 and 0 ⊕ x2 both define the function f(x) =max{0, 2x} on R.

For every equivalence class of tropical polynomials we can define a pickcanonical polynomial with the least possible number of terms as follows. If fis given by maxi∈I{ai+i·x} and k ∈ I define Ck = {x ∈ Rn|ak+k ·x = f(x)}.On Ck the function f is given by ak + k · x.

Put J = {i ∈ I|C◦i 6= ∅} and fred = maxi∈J{ai+i·x} then f and fred definethe same function on Rn. It is clear that all tropical polynomials equivalentto f have the same reduced polynomial.

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1.4 Tropical hypersurfaces

Analogous to algebraic geometry we want to assign a hypersurface to a trop-ical polynomial. The following definition is natural.

Definition 1.18. If f is a tropical polynomial in n variables with at least 2terms then the hypersurface V (f) of f is the set of all points of Rn where fis not locally linear.

It is clear that equivalent tropical polynomials define the same tropicalhypersurface. A tropical hypersurface in R2 is called a tropical curve.

Lemma 1.19. For a tropical polynomial f = maxi∈I{ai + i ·x} the followingsets are equal

1. The hypersurface V (f).

2. The set of all p ∈ Rn where f achieves its maximum at least twice.

3. The intersection⋂k∈I(C

◦k)c.

Proof. Suppose f is not locally linear in p ∈ Rn, then there are distinctk1, k2 ∈ I such that f is equal to ai + ki · x on points arbitrarily close to p.Because f is continuous it follows that f(p) = a1 + k1 · p = a2 + k2 · p, sothe first set is contained in the second.

If f achieves its maximum for both k1, k2 ∈ I it is clear that p cannot liein C◦k for any k ∈ I, so the second set is contained in the third.

Suppose p ∈⋂k∈I(C

◦k)c, then there must be distinct k1, k2 ∈ I such that

a1 + k1 · p = a2 + k2 · p = f(p). If p /∈ V (f) then a1 + k1 · x = a2 + k2 · x onan open neighborhood of p, so k1 = k2. This shows the third set is containedin the first.

Example 1.20. Figure 1.20 is the tropical curve defined by max{0, 3y, 3x, 3x+2y, 2x+ 3y, x+ 1, 2x+ 1, y + 1, 2y + 1, 3x+ y + 1, x+ 3y + 1, x+ y + 2, x+2y + 2, 2x + y + 2, 2x + 2y + 2}. The edges which cross the axes extend toinfinity. If we consider figure 1.20 and 1.1 as graphs they are dual. We willsee that when we put the structure of a polyhedral complex on V (f) theycan be seen as dual polyhedral complexes.

We now construct a polyhedral subdivision of Rn such that the union overall facets is V (f). In the case of figure 1.2 it is easy to see that this is possible.This way we also get a polyhedral subdivision of V (f).

For A ⊆ I let CA = {x ∈ Rn|ai + i · x = f(x) for all i ∈ A}. Notice thatC{k} = Ck and CA = ∩k∈ACk. Let V(f) = {CA|A ⊂ I and #A ≥ 1} ∪ {∅}.

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Figure 1.2: The tropical curve of example 1.20.

Proposition 1.21. For a tropical polynomial f = maxi∈I{ai + i · x} thecollection V(f) is a polyhedral complex and V(f) = V(fred).

Proof. If fred = maxi∈J{ai+ i ·x}, then for any A ⊆ J the set CA is the samein V(f) as in V(fred) as f and fred define the same function on Rn. ThusV(fred) ⊆ V(f).

For all A ⊆ I have CA = ∩k∈ACk and Ck = {x ∈ Rn|ak + k · x = f(x)} isa polyhedron, so all sets in V(f) are polyhedra.

We now want to show that CA ∩ CB = CA∪B is a face of CA and CBfor any nonempty A,B ⊂ I. Let k, j be distinct elements of I then theequation ak + k · x = aj + j · x defines a hypersurface H such that Ck andCj lie in different half spaces defined by H. By definition of H we haveH ∩ Ck = H ∩ Cj and Cj ∩ Ck ⊂ H, so Cj ∩ Ck = H ∩ Ck = H ∩ Cj isa face of both Cj and Ck. For any nonempty A ⊂ I and k ∈ A we haveCA∩Ck = ∩i∈ACi∩Ck, so CA is an intersection of faces of Ck, so CA is also aface of Ck. By the same argument CA ∩CB = CA∪B is a face of Ck for somek ∈ A, so CA ∩ CB is also a face of CA.

Now let j ∈ J , so C◦k 6= ∅ and dim(Cj) = n. The boundary of Cj iscontained in the union of the hypersurfaces Hi defined by aj +j ·x = ai+ i ·xfor j 6= k. Each Hj defines a face Hi∩Cj = C{i,j} of Cj of dimension at mostn = 1 so these faces are proper.

If F is a facet of Cj then dim(F ) = n − 1 and F = ∪i 6=jHi ∩ F . Wehave dim(Hi ∩ F ) ≤ dim(F ) and the union is finite so there is a k 6= jwith dim(Hk ∩ F ) = dim(F ). As Hk ∩ F = (Hk ∩ Cj) ∩ F it follows that

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Hj ∩ F is a face of F so F = F ∩ Hj. Therefore F ⊆ Hk ∩ Cj, so F isalso a face of Hk ∩ Cj = C{k,j}. As dim(F ) = dim(Hk ∩ F ) this impliesF = C{k,j} ∈ V(fred).

It now follows directly that any face of Cj is also in V(fred) because it isthe intersection of all facets containing it. The same statement also followsfor faces of general CA ∈ V(fred), because a face of CA is also a face of Ck forany k ∈ A. This shows V(fred) is a polyhedral complex.

Now pick a general CA ∈ V(f) then CA = ∪j∈JCA ∩ Cj. As CA ∩ Cj is aface of CA it follows by comparing the dimensions that the is a j ∈ J suchthat CA = CA ∩ Cj. As CA ∩ Cj is a face of Cj and V(fred) is a polyhedralcomplex it follows that CA ∈ V(fred), so it follows that V(f) = V(fred).

Corollary. The complex V(f) is pure of dimension n and any nonemptyelement is the intersection of maximal elements. The tropical curve V (f) isthe union over all facets of V(f).

Proof. The first statement is clear for V(fred), so it also holds for V(f). Thesecond part follows directly from lemma 1.19.

1.5 The Newton polytope of a tropical curve

Now we give a different construction of V(f). This construction allows us tomake the connection with subdivisions mentioned in example 1.20. To makethis subdivision we need to define what the Newton polytope of a tropicalpolynomial is and we need to introduce normal cones.

Definition 1.22. For a polytope P with face F the normal cone NF (P ) ofP at F is the subset of (Rn)∨ defined by

NF (P ) = {ϕ ∈ (Rn)∨|ϕ(z) ≤ ϕ(y) for all z ∈ P and y ∈ F}.

This means that NF (P ) consists of all functions ϕ such that the maximumof ϕ on P is attained on all of F . If ϕ is a functional such that Pϕ = F , thenϕ ∈ NG(P ) for any face G with G ⊆ F and F is the maximal face for whichϕ is contained in the normal cone.

It is usual to identify Rn with its dual (Rn)∨ via the map given by x 7→(y 7→ x · y), where · is the usual inner product. With this identification thenormal cone is given by

NF (P ) = {x ∈ Rn|x · z ≤ x · y for all z ∈ P and y ∈ F}.

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We only need to check maximality of ϕ on vertices so we have

NF (P ) = {x ∈ Rn|x · z ≤ x · y for all z ∈ A and y ∈ vert(F )}, (1.5.1)

where A is any subset of P which contains vert(P ).The following properties of normal cones can be found in [26]. A polyghe-

dron C is a cone if λx ∈ C for all x ∈ C and λ ∈ R≥0.

Proposition 1.23. Let P ⊂ Rn be a polytope and let F and G be nonemptyfaces of P .

1. The normal cone NF (P ) is a polyhedron and a cone.

2. We have NF (P ) ⊂ NG(P ) if and only if G ⊂ F and equality if andonly if F = G.

3. The normal cone satisfies dim(F ) + dim(NP (F )) = n.

4. If F1, ..., Fn are faces of P then there is a face K such that ∩ni=1NFi(P ) =NK(P ).

We can use normal cones to find the upper faces of a polytope. LetP ⊂ R× Rn be a polytope and let H1 = {1} × Rn.

Proposition 1.24. A face F of P is an upper face if and only if NF (P ) ∩H1 6= ∅.

Proof. If we consider (t,x) ∈ R × Rn as a functional it defines a face P byF = {(s,y) ∈ R×Rn|ts+ x ·y = M}, where M is the maximum of (t,x) onP . If t > 0 and (s,y) ∈ P (t,x) then (s,y) is a maximal point of P on the lineR× y as M is the maximum of (t,x) on P . Thus P (t,x) is an upper face.

If NF (P ) ∩ H1 6= ∅ pick (1,x) ∈ NF (P ) ∩ H1, then F ⊆ P (1,x) which isa upper face, so F is an upper face as well. This shows one direction of theequivalence.

Let H1, ..., Hn be the half spaces defining P and let L1, ...Ln be the cor-responding hyperspaces, then P ∩ Li is a face of P for i = 1, ..., n and theboundary of P is contained in ∪ni=1Li.

Let F be a nonempty upper face of P then F = ∪ni=1F ∩ Li and F ∩ Liis a face of F . We reorder the half spaces such that L1 ∩ F, ..., Lk ∩ F areproper faces of F and Lk ∩F, ..., Ln ∩F are equal to F . As the dimension ofa proper face of F is smaller that the dimension of F it follows that k < n.

Let z be a point of F which is not contained in any proper face of F .If Li ∩ F is a proper face of F then z ∈ H◦i . Thus z is contained in the

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interior of ∩ki=1Li, so there is an ε > 0 such that z + (ε, 0, ..., 0) ∈ ∩ki=1Li. Asz + (ε, 0, ..., 0) /∈ P there must be a j > k such that z + (ε, 0, ..., 0) /∈ Hj.

The half space Hj is defined by a functional (t,x) and a constant M ∈ Rvia Hj = {(s,y) ∈ R × Rn|ts + x · y ≤ M}. As w · (t,x) = M and(z + (ε, 0, ..., 0)) · (t,x) > M it follows that t > 0. Now (1, 1

tx) also defines

Hj, so (1, 1tx) is maximal on F . Thus (1, 1

tx) is the required element of

NF (P ) ∩H1.

We will later combine the previous proposition with the following generalfact about cones.

Lemma 1.25. Let C ⊂ Rn be a polyhedron which is a cone of dimensiond and let H be an affine hyperspace which does not contain the origin. IfC ∩H 6= ∅ then dim(C ∩H) = d− 1.

Proof. The smallest linear subspace L of Rn which contains C has dimensiond and H∩L is a hyperspace in L, so without loss of generality we can assumedim(C) = n.

After a linear transformation we can assume H is defined by x1 = 1. If allinterior points of C satisfy x1 ≤ 0 then all points of C satisfy x1 ≤ 0, whichis a contradiction because C ∩H = ∅ . Thus we can pick a a point x ∈ C◦.After multiplication with a positive constant it follows that H must containan interior point of C, so C ∩ H contains a disc of dimension d − 1, whichproves the lemma.

Now we can define the Newton polytope of a tropical polynomial andcheck that this definition makes sense.

Definition 1.26. If f = maxi∈I{ai+i·x} is a tropical polynomial Newt(f) =conv(I).

Lemma 1.27. The Newton polytope of f only depends on the function definedby f on Rn

Proof. It is enough to show Newt(f) = Newt(fred) and it is clear thatNewt(fred) ⊆ Newt(f).

Let k be a vertex of Newt(f), then Nk(Newt(f)) has an interior point x.This point satisfies i · x < k · x for every i ∈ I − {k}, so for a sufficientlylarge λ ∈ R we have i · λx + ai < k · λx + ak for every i ∈ I − {k}. Thusf is equal to k · x + ak on λx + Nk(Newt(f)). As λx + Nk(Newt(f)) is atranslated cone of dimension n it follows that C◦k 6= ∅, so k · x + ak is a termof fred and k ∈ Newt(fred). This shows Newt(f) = Newt(fred).

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We can now use the construction from section 2 to define a lattice subdivi-sion of Newt(f). We have seen that the function v : I → R given by v(i) = aidefines a polytope P = conv{(ai, i)|i ∈ I} ⊂ R×Rn and the projection of theupper hull of P is lattice subdivision of Newt(f). We denote this subdivisionwith S(f).

The following elementary proof is based on lecture notes by Bernd Sturm-fels which can be found at [24] A proof using the Legendre transform can befound in [14].

Theorem 1.28. If f is a tropical polynomial there is an inclusion reversingbijection between the nonempty cells of V(f) and the nonempty cells of S(f).

Proof. By definition we have an inclusion preserving bijection between S(f)and the upper hull of P . We can include V(f) in R × Rn as {1} × V(f), soit suffices to give an inclusion reversing bijection Φ from the upper hull of Pto {1} × V(f).

For an upper face F of P define Φ(F ) = NF (P )∩H1, where H1 = |{1} ×V(f)| = {1} × Rn. Because any vertex of P is of the form (ai, i) with i ∈ Iformula 1.5.1 shows the normal cone of a face F is given by

NF (P ) = {(t,x) ∈ R× Rn|tai + x · i ≤ taj + x · j for all i ∈ I, j ∈ A},where A is the set of all j ∈ I such that (aj, j) is a vertex of F . ThusNF (P )∩H1 = {1}×CA which is a cell of {1}×V(f). In particular N(ai,i) ={1} × Ci for all i ∈ I such that (ai, i) is a vertex.

The function Φ is inclusion reversing by lemma 1.23.Let k ∈ I such that C◦k 6= ∅, pick x ∈ C◦k and let F = P (1,x) then (1,x) ∈

NF (P ) so F is a upper face by lemma 1.24. Because NF (P ) ∩H1 is a cell of{1}×V(f) and {1}×Ck is maximal it follows that NF (P )∩H1 = {1}×Ck.Now lemma 1.25 shows that F is a vertex of P and it is clear that F canonly be (ak, k).

Let {1} × CA be any nonempty cell of {1} × V(f) then we can chooseA such that C◦k 6= ∅ for all k ∈ A by corollary 1.4. Now {1} × CA =∩k∈AN(ak,k)(P )∩H1 = (∩k∈AN(ak,k)(P ))∩H1. By lemma 1.23 there is a faceF of P such that NF (P ) = ∩k∈AN(ak,k)(P ) and F is an upper face by lemma1.24. Thus Φ is surjective.

If F and G are upper faces of P and NF (P ) ∩ H1 = NG(P ) ∩ H1 then(NF (P )∩NG(P ))∩H1 = NF (P )∩H1 By proposition 1.23 there is a face K ofP with NF (P )∩NG(P ) = NK(P ) which contains NF (P ) and NG(P ) and bylemma 1.24 this is an upper face. By lemma 1.25 the dimensions of NF (P ),NG(P ) and NK(P ) are equal, so by proposition 1.23 the dimensions of F , Gand K are equal. As F and G are faces of K it follows that F = G = K.

This shows the assignment Φ is an inclusion reversing bijection.

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Chapter 2

The amoeba of a planar curve

In this chapter we give the definition of an amoeba which was first introducedin [8]. The idea of applying a logarithmic function to an algebraic variety isnot a completely unexpected one. Applying the logarithm to a real algebraiccurve was explored by Oleg Viro, see for example [19]

Also a basic result of multivariable complex analysis is that there is aconvergent Laurent series on a set of complex numbers if and only if thelogarithm of the absolute values of this set is a convex set. This result isused to derive basic results about amoebas and we cite it as proposition 2.5.

We will give some elementary properties and examples of amoebas. Thepicture are drawn with the algorithm in appendix 6.

2.1 Definitions and examples

In this chapter we work with Laurent polynomials, which are elements ofK[z1, ..., zn, z

−11 , ..., z−1n ], where K is a field. In most of the section C is the

base field, n = 2 and we use u and w instead of z1 and z2. Throughout thischapter f(z) = f(z1, ..., zn) denotes a Laurent polynomial, and Cf denotesthe hypersurface defined by f in (C∗)n.

If K is a field then an absolute value on K is a function | · |K : K → R≥0such that for all a, b ∈ K we have

1. |a|K = 0 if and only if a = 0

2. |ab|K = |a|K |b|K

3. |a+ b|K ≤ |a|K + |b|K

If we have the stronger condition

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3’ |a+ b|K ≤ max{|a|K , |b|K}

then | · |K is called non-archimedean.If K has an absolute value then we have a map LogK : (K∗)n → Rn given

by LogK(x1, ..., xn) = (log(|x1|K), ..., log(|xn|K)). If K = C we use | · | insteadof | · |C and Log instead of LogC. In this chapter we only consider amoebasover the complex numbers and in the next chapter we look at amoebas overa field with a non-archimedean absolute value.

Definition 2.1. If X ⊂ (K∗)n is an algebraic variety then the amoeba of Xis the set LogK(X).

We can immediately note the following basic properties.

Proposition 2.2. If X is an irreducible complex variety then the amoeba ofX is closed and connected.

Proof. The absolute value map is a proper map by the Heine-Borel theoremand (C∗)n is locally compact so the absolute value map is a closed map. Thusthe logC map is also closed, so the amoeba of X is closed.

An irreducible complex variety is always connected(see for example section7.2 of [22]), so the amoeba is connected as well.

It is not immediately clear that this definition gives an interesting set.However the following picture, which shows the amoeba of the curve givenby f(u,w) = u3 + w3 + 2uw + 1 clearly shows that the amoeba is an objectwith non-trivial geometry.

Figure 2.1: The amoeba of the curve defined by f(u,w) = u3 +w3 +2uw+1.

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Example 2.3. The amoeba of an affine hyperplane. In (C∗)2 such a hy-perplane is given by an equation of the form f(u,w) = au + bw + c witha, b, c ∈ C where at least two of a, b, c are not zero.

If a = 0 then w = −cb

so the amoeba is the set {(x, log(|c|)− log(|b|)) : x ∈R}, which is the line defined by y = log(|c|)− log(|b|) in R2. The case b = 0is similar.

If c = 0 then the amoeba of f is the line given by y = x+log(|a|)−log(|b|).

Figure 2.2: The boundary of the amoeba given by f(u,w) = u+ w + 1.

In the case a, b, c are not zero, then we can consider the case a = b = c = 1,as other values give a translation of this amoeba by (log(c)− log(a), log(c)−log(b)). In this case Cf = {(u,−u− 1) : u ∈ C∗}. If |u| > 1 then | − u− 1| =|u + 1| can assume any value in [|u| − 1, |u| + 1], if |u| = 1, then |u + 1| canassume any value in (0, 2] and if 0 < |u| < 1 then |u + 1| can assume anyvalue in [1− |u|, |u|+ 1]. Thus the absolute values of the points in Cf is thepart of R2

>0 bounded by the lines {(t, t+ 1)|t ∈ (0,∞)}, {(t, 1− t)|t ∈ (0, 1)}and {(t, t− 1)|t ∈ (1,∞)}. We get the amoeba by applying the logarithm.

Notice that in the cases where the area of the amoeba of a line is 0 theNewton polytope of f is not of full dimension.

2.2 Laurent series

From now on we only consider the case of the amoeba of a hypersurface de-fined over the complex numbers by a Laurent polynomial f . To understand

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the complement of the amoeba we need to look at the Laurent series expan-sion of 1

f. A Laurent series S centered at 0 is a formal sum

∑v∈Zn avz

v, whereav ∈ C and zv = zv11 · · · zvnn . Because Zn has no natural order a Laurent se-ries is said to converge in a point if it converges absolutely. The domain ofconvergence is the interior of the set of all point where S converges.

Example 2.4. Let f(u,w) = u + w + 1 be the polynomial form example2.3 and consider the Laurent series S =

∑i,j∈Z2(−1)i+j

(i+jj

)uiwj, where we

note that(i+jj

)= 0 if i < 0 or j < 0. If S converges in (u0, w0) then we

have S(u0, w0) =∑∞

k=0(−1)k(u0 + w0)k by the binomial theorem. This is a

geometric series so it follows that S is a Laurent expansion of 1f. .

Applying the binomial theorem to the absolute value gives∑i,j∈Z2

|(−1)i+j(i+ j

j

)uiwj| =

∑i,j∈Z2

(i+ j

j

)|u|i|w|j =

∞∑n=0

(|u|+ |w|)n,

so this series converges absolutely if and only if |u|+|w| < 1. Thus the domainof convergence of S is the set {(u,w) ∈ (C∗)2||u|+ |w| < 1} and applying theLog function to this domain gives one component of the complement of theamoeba of f .

Using Pascals identity for binomial coefficients and similar arguments tothe previous example one can show that∑

i,j∈Z2

(−1)−i+1

(−i+ 1

j

)uiwj and

∑i,j∈Z2

(−1)−j+1

(−j + 1

i

)uiwj

are Laurent expansions of 1f

and that the domains of convergence are {(u,w) ∈(C∗)2||w| < |u|− 1} and {(u,w) ∈ (C∗)2||u| < |w|− 1}. Thus the domains ofconvergence give the other components of the complement of the amoeba. Aconsistent way to find certain Laurent expansions of 1

fis given in the remark

at the end of this section.

The correspondence between components of amoebas and Laurent seriesfollows from a general result in multivariable complex analysis. A slightlydifferent version of this proposition is cited in [8]. See [17], theorem 2, for aproof of part (b) and [11] theorem 2.3.2 for part (a).

Proposition 2.5. (a) If S(z) =∑

v∈Zn avzv is a Laurent series centered at

0 the domain of convergence is of the form Log−1(B) where B ⊂ Rn is aconvex open subset. 1

1In multivaliable complex analysis such a domain is referred to as a logarithmicallyconvex Reinhardt domain.

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(b) If ϕ(z) is a holomorphic function on a domain D of the form Log−1(B)with B ⊂ Rn open and connected then there is a unique Laurent series cen-tered at 0 which converges to ϕ(z) on D.

We can directly apply this to amoebas.

Corollary. If f is a Laurent polynomial then the components of Rn−Log(Cf )are convex and the components correspond bijectively to the Laurent seriesexpansions of 1

fcentered at 0.

Proof. Let A be a component of Rn − Log(Cf ) then 1f

is a holomorphic

function defined on Log−1(A). By part (b) of proposition 2.5 there is aunique Laurent series S which converges to 1

fon Log−1(A). By part (a) of

the same proposition the domain of convergence of S is of the form Log−1(B)with B ⊂ Rn open and convex. We clearly have A ⊂ B and as B is convexit is also connected, so A = B.

Thus the components of the complement of the amoeba are convex andwe can associate a unique Laurent series to each.

If S ′ is another Laurent series of 1f

centered at 0, then it has a domain of

convergence of the form Log−1(B) where B ⊂ Rn is a convex open subset.As B cannot intersect the amoeba there is a component A of the complementsuch that A ∩ B 6= ∅. If S is the power series expansion corresponding to Aby the first part of the proof then S ′ = S by part (b) of proposition 2.5.

Remark. It is possible compute some of the Laurent series of 1f

directly. If

γ is a vertex of Newt(f) then we can write f(z) = aγzγ(1 + g(z)), where

g(z) =∑

v∈Zk,v 6=γavaγ

zv−γ. Using the geometric series we get a formal identity

1

f(z)= a−1γ z−γ

∞∑i=0

(−1)ig(z)i. (2.2.1)

Expanding the terms gives a Laurent series which converges on a componentof the complement of the amoeba. The details can be found in [8] or [7].

2.3 The amoeba and the Newton polytope

In this section we show some elementary properties of amoebas. In particularall amoebas will look somewhat similar to figure 2.1, with a finite number oftentacles. Furthermore the directions of the tentacles are prescribed by theNewton polytope.

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For proposition 2.7 we only consider curves in (C∗)2. This eliminates theneed to use Laurent series. The proofs in [8] use the Laurent series of 1

fand

work in all dimensions.For a vertex γ of Q = Newt(f) the normal cone Nγ(Q) can be identified

with a subset of Rn by formula 1.5.1, so Nγ(Q) = {a ∈ Rn|a · (v − γ) ≤0 for every v ∈ Q}.

Proposition 2.6. There is a vector b ∈ Nγ(Q) such that b+Nγ(Q) does notintersect the amoeba of f .

Proof. For anyM ∈ R we can find a vector b ∈ Nγ(Q) such that b·(v−γ) < Mfor every v ∈ Q∩Zn different from γ. Now this condition also holds for everyelement of b+Nγ(Q).

For any v ∈ (Q−{γ})∩Zn and z ∈ (C∗)n such that Log(z) ∈ b+Nγ(Q),we have Log(|zv−γ|) = Log(z) · (v − γ) < M . This gives |zv| < |zγ|eM . Letk + 1 be the number of nonzero coefficients of f , then

|∑v 6=γ

avzv| <

∑v 6=γ

|av||zγ|eM ≤ kmax{|av| : v 6= 0}eM |zγ|.

This shows that if we pick M such that kmax{av : v 6= γ}eM ≤ |aγ| we get|∑

v 6=γ avzv| < |aγ||zγ|. Thus f cannot have a zero in z, so a does not lie in

the amoeba.

Remark. As stated in the proof it is enough to find b such that (b, ω − γ) <M for every integral ω ∈ Q different from γ, where M ∈ R is such thatkmax{|av| : v 6= 0}eM ≤ |aγ| where k+1 is the number of nonzero coefficientsof f .

Remark. If we compute the cones in the example 2.3 we can use M = log(12)

so we can pick b(0,0) = (log(12), log(1

2)), b(1,0) = and b(0,1) = (0, log(1

2)).

Proposition 2.7. The translated cones b+Nγ(Q) in the previous propositionlie in distinct components of the complement of the amoeba.

Proof. Let γ and δ be vertices of Newt(f) and let bγ +Nγ(Q) and bδ +Nδ(Q)be the cones which do not intersect the amoeba of f . Also let Aγ and Aδ bethe components containing bγ +Nγ(Q) and bδ +Nδ(Q).

First we assume γ and δ are adjacent vertices. We can assume that γ andδ lie on the y-axis, Q lies in R2

≥0 and Q meets the x-axis. By proposition5.6 this can be achieved by an automorphism of C[u,w, u−1, w−1] defined inequation 5.1.1 and by noting that Q can be moved by integer vectors withoutchanging the curve in (C∗)2. Now f does not contain negative exponents andis not divisible by u and w in C[u,w]. In this case f also defines a curve in

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Figure 2.3: The boundary of the amoeba given by f(u,w) = u+w + 1 withthe translated normal cones.

C2 which restricts to Cf in (C∗)2. On the w-axis of C2 the polynomial f isgiven by f(0, w) which has more than one term, so f has a zero a = (0, a2)with a2 6= 0.

As curves have no isolated points there are points {(aj1, aj2)}j∈N ∈ Cf suchthat limj→∞(aj1, aj2) = a. Now limj→∞ log(|(aj1)|) = −∞ and limj→∞ log(|aj2|) =log(|(a2)|). Thus for sufficiently large j the point (aj1, aj2) lies betweenbγ+Nγ(Q) and bδ+Nδ(Q). As components are convex this shows bγ+Nγ(Q)and bδ +Nδ(Q) cannot lie in the same component so Aγ 6= Aδ.

Now we assume γ and δ are arbitrary vertices with Aγ = Aδ and we showthat we can find an adjacent vertex η of γ with Aγ = Aη. We note that byconvexity Aγ contains bγ + bδ +Nγ(Q) +Nδ(Q).

We can write Nγ(Q) = H1∩H2 for open half planes H1 and H2. If Nδ(Q)∩H1 = ∅ = Nδ(Q)∩H2 then Nγ(Q) +Nδ(Q) = R2, which is impossible, so wecan assume Nδ(Q)∩H1 6= ∅. If η is the adjacent vertex of γ for which Nη(Q)lies in H1 then any translation of Nη(Q) intersects bγ + bδ +Nγ(Q) +Nδ(Q).

Thus Aγ ∩Aη 6= ∅ and therefore Aγ = Aη, which completes the proof.

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2.4 The components of the complement of

the amoeba

Using the integral formula 2 for the coefficients of the Laurent expansionof 1

fit is possible to derive more information about the complement of the

amoeba. In [7] the authors construct a distinct point in Newt(f) ∩ Zn forevery component of Acf . The same function was constructed by Mikhalkinin [15] using the topological linking number.

The results can be stated as follows.

Theorem 2.8 (Fosberg, Passare, Tsikh). The order of a component of Acfis an element of Newt(f) with integer coefficients and the orders of distinctcomponents are distinct.

Corollary. For a Laurent polynomial f the number of distinct Laurent ex-pansions of 1

fis at most the number points in Newt(f) ∩ Zn and at least the

number of vertices.

The following property of the order will be useful later on.

Proposition 2.9. Let k ∈ Zn∩Newt(f) and let z ∈ (C∗)n such that Log(z) /∈Af and |akzk| > |

∑j 6=k ajz

j|, then the order of the complement of Acf thatcontains Log(z) is k.

2See for example [11] theorem 2.7.1

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Chapter 3

Puiseux series and Kapranov’stheorem

The aim of this chapter is to give a proof of Kapranov’s theorem. This theo-rem describes the amoeba of a hypersurface over an algebraically closed fieldwith a non archimedean absolute value in terms of tropical geometry. It wasproved by Mikhail Kapranov in an unpublished manuscript [10]. Kapranov,Einslieder and Lind later published a significantly more abstract version ofthe theorem in [4].

The original result is quoted in [14] and a more general result is provedin the forthcoming book [13] by Maclagan and Sturmfels. The proof in thischapter is based on course notes [24] by Bernd Stumfels.

3.1 Valuations and Puiseux series

Let K be a field with absolute value | · |K . The absolute value | · |K is callednon-Archimedean if we have |a + b|K ≤ max{|a|K , |b|K} for every a, b ∈ K.This is a stronger version of the normal triangle inequality. It is often intuitiveto define non-Archimedean absolute values in terms of valuations.

Definition 3.1. A valuation on a field K is a function val : K → R ∪ {∞}such that for all a, b ∈ K we have

1. val(a) =∞ if and only if a = 0

2. val(ab) = val(a) + val(b)

3. val(a+ b) ≥ min{val(a), val(b)}

It follows directly from the seconds axiom that val(1) = 0 and val(a−1) =− val(a). We also note the following.

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Lemma 3.2. If val(a) 6= val(b) then val(a+ b) = min{val(a), val(b)}.

Proof. We can assume val(a) < val(b), then val(a+b) ≥ val(a). Suppose thatval(a+ b) > val(a), then val(a) = val((a+ b)− b) ≥ min{val(a+ b), val(b)} >val(a) which is impossible, so val(a+ b) ≤ val(a).

We can switch between non-Archimedean absolute values and valuationsby the following proposition.

Proposition 3.3. If | · |K is a non-Archimedean absolute value on a fieldK then val(a) = − log |a|K defines a valuation on K. Conversely if valis a valuation on K the function defined by |a|K = e− val(a) defines a non-Archimedean absolute value on K.

Now we construct the field of Puiseux series, which is an extension of thefield of Laurent series in which more exponents are allowed.

The valuation on the field C((t)) of formal Laurent series over C definedby for p =

∑i∈I ait

i by val(p) = min{i ∈ I|ai 6= 0} is an example of anon-Archimedean valuation. For k > 0 the field C((t1/k)) is an extension ofC((t)) of degree k and the valuation can be extended by the same definition.

A Puiseux series over C is a series of the form∑

i∈I aiti, where ai ∈ C

and I ⊂ Q is a set of fractions with a common denominator and a minimalelement. We denote the set of Puiseux series over C with C{{t}}. If k is acommon denominator for I then

∑i∈I ait

i ∈ C((t1/k)). This shows we have

C((t1k )) ⊂ C{{t}} for all k ∈ Z>0 and C{{t}} =

⋃∞k=1C((t

1k )). Thus C{{t}}

is a field extending C((t)).If p =

∑i∈I ait

i is a Puiseux series we put val(p) = min{J}. It is easy toverify that val is a non-Archimedean valuation on C{{t}}.

The following theorem is due to Puiseux but was also known to Newton.In the language of modern algebra this theorem can be proved easily byapplying Hensel’s lemma to the ring C[[t]] as is done in [6].

Theorem 3.4 (Puiseux). The field C{{t}} is the algebraic closure of C((t)).

We can now take a look at amoebas defined over K = C{{t}}. Theabsolute value on K is given by |p|K = e− val(p), so the logK function on(K∗)n is given by logK(p1, ..., pn) = (− val(p1), ...,− val(pn)).

We now compute a very simple example of an amoeba which is similar tothe general case.

Example 3.5. Let Cf be the curve in (K∗)2 defined by f = u + w + t andlet T be the tropical curve V (x⊕ y ⊕−1). We claim LogK(Cf ) = T ∩Q2.

A point of Cf is of the form (p,−p− t) for some p ∈ K∗. If − val(p) 6= −1then val(−p−t) = min{val(p), 1}, so (− val(p),− val(−p−t)) = (− val(p),−1)

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if− val(p) < −1 and (− val(p),− val(−p−t)) = (− val(p),− val(p)) if− val(p) >−1. In both cases (− val(p),− val(−p − t)) lies in T . If − val(p) = −1 wehave − val(−p− t) ≤ −min{val(p), 1} = −1, so (− val(p),− val(−p− t)) liesin T . As val(K) = Q we get LogK(Cf ) ⊂ T ∩Q2.

We have

T ∩Q2 = {(−q,−1)|q ∈ Q≥1} ∪ {(−1,−q)|q ∈ Q≥1} ∪ {(q, q)|q ∈ Q≥−1}.

For q ∈ Q<−1 we have (t−q,−t−q−t) ∈ Cf and LogK(t−q,−t−q−t) = (q,−1).For q ∈ Q>−1 we have (t−q,−t−q − t) ∈ Cf and LogK(t−q,−t−q − t) = (q, q).For q ∈ Q≤−1 we have (−t−q−t, t−q) ∈ Cf and LogK(t−q,−t−q−t) = (−1, q).Thus LogK(Cf ) = T ∩Q2.

3.2 Tropical polynomials

Now we look at the algebra of tropical polynomials extending section 1.3.We have delayed this section because it is necessary to use the notion of atropical hypersurface.

Let T[x1, ..., xn] be the set of tropical polynomials in the variables x1, ..., xn.If f = ⊕i∈Iai � xi11 � · · · � xinn and g = ⊕j∈Jbj � xj11 � · · · � xjnn we putf ⊕ g = ⊕k∈K(ak ⊕ bk) � xk11 � · · · � xknn where K = I ∪ J and ak = −∞ ifk /∈ I. We also put f � g = ⊕k∈Kck � xk11 � · · · � xknn where K = I + J andck = ⊕i,j;i+j=kai � bj.

Example 3.6. Let f = x⊕ y ⊕−1 and g = −x⊕−y ⊕−1, then

f � g = −1�−x⊕−1�−y ⊕ 0⊕ x�−y ⊕ y �−x⊕−1� x⊕−1� y.

The tropical curve associated to f � g can be seen in figure 3.1. As expectedit is the union of the tropical curves of f and g.

The proof that a polynomial ring over a semiring is a semiring is identicalto the proof that a polynomial ring is a ring so we can note the followingfact.

Proposition 3.7. The set T[x1, ..., xn] is a commutative semiring for theoperations ⊕ and �.

In order to use addition and multiplication of tropical polynomials weneed to look at the functions the define on Rn. In case all coefficients arezero the statement about the union of the tropical hypersurfaces is essentiallyproposition 7.12 in [26].

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Figure 3.1: The tropical curve defined by (x⊕ y ⊕−1)� (−x⊕−y ⊕−1).

Proposition 3.8. If f and g are tropical polynomials and v ∈ Rn thenf ⊕ g(v) = max{f(v), g(v)} and f � g(v) = f(v) + g(v). FurthermoreV (f � g) = V (f) ∪ V (g).

Proof. Pick v ∈ Rn then

max{f(v), g(v)} = max{maxi∈I{ai + i · v},max

j∈J{aj + j · v}}

= maxk∈I∪J

{max{ak, bk}+ k · v} = (f ⊕ g)(v),

which proves the first assertion.Suppose ak + k · v is a maximal term in f(v) and al + l · v is a maximal

term in g(v). Thus we have ai + i · v + bj + j · v ≤ ak + k · v + bl + l · vfor all i ∈ I and j ∈ J . If i + j = k + j this gives ai + bj ≤ ak + bl so thecoefficient ck+j = maxi,j:i+j=k+j{ai + bj} of f � g at k + j is equal to ak + bl.From the same inequality it follows that ak + bl + (k + l) · v is a maximalterm of f � g(v). This proves the second assertion.

If v ∈ V (f) then there are distinct k, k′ ∈ I such that ak + k · v andak′ + k′ · v are maximal. If bl + l · v is a maximal term of g at v thenak + bl + (k+ l)v and ak′ + bl + (k′ + l)v are maximal terms of f � g at v sov ∈ V (f �g). The case v ∈ V (g) is identical. If v /∈ V (f)∪V (g) then f andg are linear on a neighborhood of v so f ⊕ g is as well, so we have provedthe third assertion.

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The following fact will be used together with lemma 3.2.

Lemma 3.9. Let f and g be tropical polynomials and let ck = maxi,j;i+j=k{ai+bj} be the coefficient of f � g at xk. If ck + k · x is maximal at a pointv /∈ V (f � g) then the maximum in ck = maxi,j;i+j=k{ai + bj} is unique.

Proof. By lemma 3.8 we have v /∈ V (f) and v /∈ V (g) so we can pick i ∈ Iand j ∈ J such that ai′ + i′ · x < ai + i · x for every i′ ∈ I distinct from i andaj′ + j′ · x < aj + j · x for every j′ ∈ J distinct from j. As in the proof oflemma 3.8 it follows that i+ j = k and ck = ai + bj.

If ck = ai′ + bj′ for i′ ∈ I − {i} and j′ ∈ J − {j} then

ck + k · v = ai′ + bj′ + (i′ + j′) · v < ai + bj + (i+ j) · v = ck + k · v,

so ck is indeed unique.

Definition 3.10. If K is a field with a valuation, f =∑

i∈I aizi with I ⊂ Zn

a Laurent polynomial then

trop(f) =⊕i∈I

− val(ai)� xi.

It is necessary to use − val because if ai = 0 then − val(ai) = −∞, so theterm − val(ai)� xi does not contribute to trop(f) either.

The following example shows that we have to look at functions on Rn

rather than tropical polynomials.

Example 3.11. Let f = z + t and g = z + (−t + t2), then trop(fg) =x2⊕−2� x⊕−2 and trop(f)� trop(g) = x2⊕−1� x⊕−2. However bothtrop(fg) and trop(f)� trop(g) define the function max{−2, 2x} on R2.

The following proposition explains why tropicalization is a meaningfulconstruction.

Proposition 3.12. If f and g are polynomials over a field with a valuationthen trop(fg) defines the same function on Rn as trop(f) � trop(g). Inparticular we have V (trop(fg)) = V (trop(f)) ∪ V (trop(g)).

Proof. Let f =∑

i∈I aizi and g =

∑i∈J bjz

i with I, J ⊂ Zn, then fg =∑k∈K ckz

k with K = I + J and ck =∑

i,j;i+j=k aibj. Thus trop(fg) =⊕k∈K − val(ck)�xk. We also have trop(f)� trop(g) =

⊕k∈K ck�xk, where

ck = maxi,j;i+j=k

{− val(ai)− val(bj)} = maxi,j;i+j=k

{− val(aibj)}.

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By lemma 3.2 we get

− val(ck) ≤ − mini,j;i+j=k

{val(aibj)} = maxi,j;i+j=k

{− val(aibj)} = ck.

Thus trop(fg)(v) ≤ trop(f)� trop(g)(v) for every v ∈ Rn.Let v /∈ V (trop(f) � trop(g)) and let ck + k · v be the maximal term of

trop(f) � trop(g) at v then the maximum in ck = maxi,j;i+j=k{− val(ai) −val(bj) is unique by lemma 3.9. Thus by lemma 3.2 we have − val(ck) = ck,so trop(fg)(v) ≥ − val(ck) + k · v = trop(f)� trop(g)(v).

This shows trop(fg) and trop(f)�trop(g) are equal on the complement ofV (trop(f)� trop(g)). As both functions are continuous they must be equal.

This proves the first assertion and the second one follows directly.

The same type of statement is false for the tropical sum of polynomials.

Example 3.13. Let f = z + t + t2 and g = z − t then trop(f)⊕ trop(g) =(x⊕−1)⊕(x⊕−1) = (x⊕−1) while trop(f+g) = x⊕−2, so V (trop(f+g)) ={−2} and V (trop(f)⊕ trop(g)) = {−1}.

3.3 Kapranov’s theorem

Now we can sharpen Puiseux’s theorem in two steps. The second step willbe Kapranov’s theorem. For a polynomial f over the field of Puiseux seriesit gives a complete description of the amoeba Af in terms of a tropicalhypersurface.

The first step is basically Kapranov’s theorem in one dimension. It is aneasy consequence of Puiseux’s theorem and lemma 3.12.

Lemma 3.14. Let f ∈ C{{t}}[z] be a polynomial in one variable over thefield of Puiseux series and suppose v ∈ V (trop(f)) then f has a root u suchthat − val(u) = v.

Proof. By lemma 3.12 we can assume f is monic. We prove the lemma byinduction on deg(f). If deg(f) = 1 then f = z−u and trop(f) = x⊕− val(u),so the result is clear, even in the case u = 0.

Suppose the lemma holds for polynomials of degree at most n. Letdeg(f) = n + 1 and let v ∈ V (trop(f)). By Puiseux theorem we have aroot w of f . If − val(w) = v we are done. Otherwise we write f = (z−w)f ′.By the case n = 1 we have V (z − w) = {− val(w)} and by lemma 3.12 itfollows that V (trop(f)) = V (trop(f ′)) ∪ {− val(w)}. Thus v ∈ V (trop(f ′))so by induction we get a root u of f ′ such that − val(u) = v. As u is also aroot of f we are done.

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For products of linear polynomials we could repeat the proof of lemma 3.14with a suitable version of example 3.5, but this does not work for irreduciblepolynomials.

The solution from [24] is to use an inclusion of K∗ in (K∗)n and applylemma 3.14 to get a root of the pullback of the polynomial.

However the inclusion used in [24] does not always work because tropical-ization does not always behave nicely with respect to pullbacks as we can seein example 3.16. This can be fixed by requiring an additional assumption onthe inclusion.

Theorem 3.15 (Kapranov). If f =∑

i∈I aizi is a polynomial over the field

of Puiseux series in n variables then Af = V (trop(f)) ∩Qn.

Proof. Let f =∑

i∈I aizi with I ⊂ Zn.

For p ∈ Cf let v = LogK(p) = (− val(p1), ...,− val(pn)) then− val(aipi) =

− val(ai) + i · v for all i ∈ I. We have

trop(f)(v) = maxi∈I{− val(ai) + i · v}. (3.3.1)

and this maximum is attained for only one index in I if and only if theminimum mini∈I{val(aip

k)} is attained by one index in I. If that is the casewe get val(

∑i∈I aip

i) = mini∈I{val(aipk)} by lemma 3.2. This is impossible

as ∞ = val(0) = val(∑

i∈I aipi) and val(aip

k) <∞ for some i ∈ I. Thus wehave Af ⊆ V (trop(f)) ∩Qn.

Now let v ∈ V (trop(f))∩Qn. We construct an inclusion of K∗ into (K∗)n

to apply lemma 3.14.For any nonzero α ∈ Zn with coprime coefficients the function ϕ(s) =

(t−v1sα1 , ..., t−vnsαn) defines an inclusion ϕ : K∗ → (K∗)n. If we define ψ : R→Rn by ψ(λ) = v + λα then the diagram

K∗− val //

ϕ

��

Rψ��

(K∗)nLogK // Rn

(3.3.2)

is commutative and ψ(0) = v. Thus it suffices to find an α such thattrop(ϕ∗f) has a root at 0.

To ensure this we pick an α ∈ Zn such that α · i 6= α · j for every pairof distinct i, j ∈ I. As each of these equations defines the complement ofhypersurface it is possible to pick such an α. Now the pullback of f(z)is the polynomial g = ϕ∗f given by g(z) = f(t−v1zα1 , ..., t−vnzαn) on K∗.To get the tropicalization of g we expand it so g(z) =

∑k∈K bkz

k, where

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bk =∑

i∈I:α·i=k ait−i·v. By the choice of α each sum bk has only one term so

− val(bk) = − val(ai) + i · v if there is a i ∈ I with α · i = k and bk = −∞otherwise.

The restriction of trop(f) to the image of ψ is given by trop(f)(v +λα) =maxi∈I{− val(ai) + i · (v + λα)} = maxi∈I{− val(ai) + i · v + (i · α)λ}.

Thus trop(g) is equal to the restriction of trop(f) and trop(g) is not locallylinear at 0, so g has a zero s with − val(s) = 0. Now ϕ(s) is a root of fbecause g is the pullback of f and LogK(ϕ(s)) = v because diagram 3.3.2commutes.

The following example shows that tropicalization does not always com-mute with pullbacks.

Example 3.16. Let f = tuw2+(−t+t2)u2w+tu5, then trop(f) = max{−1+x+ 2y,−1 + 2x+ y,−1 + 5x} and V (f) is the curve in figure 3.2

Figure 3.2: The tropical curve defined by max{−1+x+2y,−1+2x+y,−1+5x}.

Pick α = (1, 1) and define ϕ and ψ as in diagram 3.3.2. Now α does notsatisfy the conditions required in the proof of Kapranov’s theorem.

The pullback ϕ∗f of f toK∗ is given by ϕ∗f(z) = t2z3+tz5, so trop(ϕ∗f) =max{−2 + 3x,−1 + 5x} and the restriction of trop(f) to R is given bymax{−1 + 3x,−1 + 5x}.

Thus trop(ϕ∗f) is locally linear at 0 and the root is at x = −12. Thus with

lemma 3.14 we get a root p of f with LogK(p) = (−12,−1

2). In figure 3.2 we

can see that the root has shifted along one of the branches of V (trop(f)).

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Even if trop commutes with the pullback lemma 3.14 can still give thewrong root if the inclusion of R in Rn does not intersect V (f) transversally.

Example 3.17. Let f = u + w + t, then trop(f) = max{x, y,−1} andv = (0, 0) ∈ V (trop(f)). Let α = (1, 1) and define ϕ and ψ as in diagram3.3.2. Now ϕ∗f(z) = 2z + t, so trop(ϕ∗f) = max{λ,−1}. Also trop(f) =max{x, y,−1}, so the restriction to R is given by max{λ, λ,−1} which isequal to trop(ϕ∗f).

The tropical root of trop(ϕ∗f) is λ = −1, so the root p of f given by lemma3.14 satisfies LogK(p) = (−1,−1), so it is not the root we were looking for.

Remark. Kapranov’s theorem holds for any algebraically closed field witha non-Archimedean absolute value. The proof given here also works in thegeneral case. We proved 3.12 for arbitrary fields with a valuation, so the proofof lemma 3.14 is identical in the general case. In the proof of Kapranov’stheorem the elements t−v1 , ..., t−vn have to be replaced with elements of theright absolute value. This is possible in an algebraically closed field.

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Chapter 4

A limit of amoebas

In this chapter we study the the limit of a collection of amoebas. We followan article [14] by Grigory Mikhalkin. This approach makes it possible toconstruct algebraic curves which have certain topological properties.

This kind of construction was first used by Oleg Viro to construct realalgebraic curves with a particular topology in [19]. In [14] it is used to findthe topology of an arbitrary smooth hypersurface in (K∗)n.

4.1 The construction of the limit

Let f =∑

j∈Q ajzj be a Laurent polynomial, where Q ⊂ Zn is the set

{j ∈ Zn|aj 6= 0}. We make a limit of amoebas by changing both the baseof the logarithm and the coefficients of f . Let Cf be the curve defined by fand let Af be the amoeba of f . For r ∈ R>1 we define Logr : Cn → Rn byLogr(z1, ..., zn) = (logr |z1|, ..., logr |zn|), then Logr(Cf ) = Af ·M , where M isthe diagonal matrix with entries 1

log(r). Thus Logr(Cf ) is homeomorphic with

Af . The coefficients of f are changed by the formula fr =∑

j∈Q ajrv(j)zj for

some function v : Q → Q. If Xr is the hypersurface defined by fr, then wedefine Ar = Logr(Xr).

We can also look at the tropical polynomial ft(z) =∑

j∈Q ajt−v(j)zj ∈

C{{t}}[z±]. The fact that the minus sign appears in the exponent is anunfortunate consequence of our conventions in the Log map and the definitionof the tropicalization of a polynomial.

The polynomial ft is called the patchworking polynomial. As C{{t}}is a field with a non-archimedean norm, the polynomial ft defines a non-archimedean amoeba. By Kapranov’s theorem the closure of this amoeba isthe tropical curve defined by trop(ft) = maxj∈Q{v(j) + j · x}, where x =(x1, ..., xn) is a real variable. We denote this tropical curve with AK .

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The goal of this chapter is to prove that for r → ∞ the amoebas Arconverge to AK .

Example 4.1. We can already compute an easy example. Let f = u+w+1and v(0, 0) = v(1, 0) = v(0, 1) = 0, then Ar = M · Af , where M is thediagonal matrix with entries 1

log(r). It is easy to see these set converge to the

union of the three lines given by {((−t, 0)|t ∈ R≥0)} ∪ {((0, t)|t ∈ R≥0)} ∪{((−t,−t)|t ∈ R≥0)}, which is the tropical curve given by x⊕y⊕0 = trop(f).

Figure 4.1: The set Ar with r = e7 in blue and the tropical curve of x⊕y⊕1in red.

In case f is a Laurent polynomial in two variables we have vol(Af ) ≤π2 vol(Newt(f)) by [18]. Thus we get vol(Logr(Cf )) ≤ π2

log(r)nvol(Newt(f)),

so the volume of Logr(Cf ) tends to zero when r goes to∞. This also happenswhen we change the coefficients of f as the newton polytope remains thesame. Thus we know that if limr→∞Ar exists it has zero volume.

Furthermore proposition 2.6 and theorem 1.28 together with the precedingexample already suggest a strong relation between the limit and a tropicalcurve.

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4.2 The Hausdorff distance

We need to make precise what it means for a sequence of sets to converge,this is expressed by the Hausdorff distance.

Definition 4.2. The Hausdorff distance between two closed sets A,B ⊂ Rn

is given bymax{sup

a∈A(d(a,B)), sup

b∈B(d(b, A))}.

The Hausdorff distance does not give a metric on the set of closed subsetsof Rn because it is not always finite. However it does satisfy the triangleinequality. For a collection {Ar}r∈R>1 we can define the limit as the closedset B ⊂ Rn such that limr→∞ dH(Ar, B) = 0 if such a set exists. This set isunique by the following lemma.

Lemma 4.3. If {Ar}r∈R>1 converges to B then B consists of all b ∈ Rn forwhich there are ar ∈ Ar such that limr→∞ ar = b.

Proof. By definition of dH for every b ∈ B we can find a ar ∈ Ar withd(ar, b) ≤ dH(Ar, B), so limr→∞ ar = b.

If b /∈ B then d(b, B) > ε for some ε ∈ R>0. By the convergence of theAr we have an M ∈ R>0 such that dH(Ar, B) ≤ ε

2for all r > M . Thus also

supa∈Ar d(a,B) ≤ ε2. By the triangle inequality we get that d(b, Ar) >

ε2

forr > M , so b is not the limit of points in the sets Ar.

The converse of the lemma does not hold, for example if Ar = [0, r] thenthe set of limit points is [0,∞), but dH([0, r], [0,∞)) =∞ for every r ∈ R>0.

4.3 A neighborhood of the tropical curve

Now we can construct a collection of decreasing neighborhoods of AK whichcontain Ar. Let M ∈ R>1 and let ArK be the subset of Rn described by theinequalities

ck + k · x ≤ maxj∈Q−{k}

{cj + j · x}+ logr(M),

where k ranges over Q and cj = v(j) + logr |aj|. To ensure both AK ⊂ ArKand Ar ⊂ ArK we pick M = max{N,M2

f }, where N = #Q − 1 and Mf =maxj∈Q{|aj|, |aj|−1}.

Lemma 4.4. The set ArK is a neighborhood of AK for every r ∈ R>1 andlimr→∞ dH(ArK ,AK) = 0.

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Proof. The tropical curve AK is given by the inequalities

v(k) + k · x ≤ maxj∈Q−{k}

{v(j) + j · x},

so if y ∈ AK we get ck + k · y ≤ maxj∈Q−{k}{v(j) + j · y} + logr |ak|. Wehave logr |ak| ≤ logr(Mf ) and maxj∈Q−{k}{v(j) + j ·y} ≤ maxj∈Q−{k}{v(j) +logr |aj| + j · y} + logr(Mf ). This shows AK ⊂ ArK for every r ∈ R>0 andthis also implies supy∈AK (d(y, ArK)) = 0.

Now we estimate d(y,AK) for y ∈ ArK . If y ∈ AK , then d(y,AK) = 0.Thus we can assume y is a the component of the complement of AK definedby v(k) +k ·x > maxj∈Q−{k}{v(j) + j ·x} for some k ∈ Q. Pick j1 ∈ Q−{k}such that v(j1) + j1 · y = maxj∈Q−{k}{v(j) + j · y}.

By definition of ArK we have v(k) + logr |ak|+ k · y ≤ maxj∈Q−{k}{v(j) +logr |aj|+ j ·y}+logr(M), so v(k)+k ·y ≤ v(j1)+ j1 ·y+3 logr(M). Now let

y′ = y− 3 logr(M)||k−j1||2 (k−j1). We have v(k)+k·y′ ≤ v(j1)+j1 ·y′, so y′ does not lie

in the same component of AcK as y. Thus d(y,AK) ≤ d(y,y′) = 3 logr(M).This shows supy∈ArK

d(y,AK) ≤ 3 logr(M), which proves the lemma.

We only need the sets ArK to show the convergence of the Ar and thefollowing two properties are used in the proof of theorem 4.7.

Lemma 4.5. For every r ∈ R>0 we have Ar ⊂ ArK.

Proof. If x ∈ Ar we can pick z ∈ (C∗)n with fr(z) = 0 and Logr(z) = x.We have

∑j∈Q ajr

v(j)zj = 0, so |akrv(k)zk| = |∑

j∈Q−{k} ajrv(j)zj| for every

k ∈ Q. Applying logr and the triangle inequality gives

ck + k · Logr(z) = logr(|∑

j∈Q−{k}

ajrv(j)zj|)

≤ logr(∑

j∈Q−{k}

|ajrv(j)zj|)

= logr(∑

j∈Q−{k}

|rcj ||zj|)

≤ logr(N maxj∈Q−{k}

|rcj ||zj|)

= maxj∈Q−{k}

{cj + j · Logr(z)}+ logr(N),

so x = Logr(z) ∈ ArK .

Lemma 4.6. If the components of the complement of ArK given by cki+ki·x >maxj∈Q−{ki}{cj + j · x} for k1, k2 ∈ Q and k1 6= k2 are nonempty they lie indistinct components of the complement of Ar.

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Proof. Let x be in the component given by cki + ki · x > maxj∈Q−{ki}{cj +j · x} + logr(M) an let Logr(z) = x. We have rcj+j·x = rv(j)|aj||zj|, so weget rv(ki)|aki ||zki | > M maxj∈Q−{ki}{rv(j)|aj||zj|} > |

∑j∈Q−{ki} r

v(j)ajzj| by

the choice of M . Now we can apply theorem 2.9, so x lies in the componentof Ar corresponding to ki. By the same theorem the components of Arcorresponding to k1 and k2 are distinct which proves the lemma.

4.4 The limit

Now we can prove the statement of the introduction and construct somecurves with nice amoebas.

Theorem 4.7. When r →∞ the amoebas Ar converge to AK for the Haus-dorff distance.

Proof. By lemma 4.4 and 4.5 we have limt→∞ supy∈Ar d(y,AK) = 0.Let ε > 0. Because AK is a finite collection of subsets of hypersurfaces

and limr→∞ dH(ArK ,AK) = 0 we can find a T > 0 such that ArK does notcontain any ball of diameter ε.

Now let x ∈ AK . Because the components of the complement of ArKare convex we can find a ball B of diameter ε which intersects two distinctcomponents C1 and C2 of the complement of ArK . If B contains no point ofAr for some r > T we get that C1 and C2 are connected in the complementof Ar. This is impossible by lemma 4.6. Thus d(x,Ar) < ε.

We now consider an example.

Example 4.8. Consider the curve defined by f(u,w) = u3 +w3 + λuw + 1.If λ = 0 then Af is the image of the amoeba u + w + 1 under a linear mapby corollary 5.2 and therefore the complement has no bounded components.In figure 4.2 we can see that the amoeba also has no holes for small values ofλ. 1 Nu we pick the function v by v(0, 0) = v(3, 0) = v(0, 3) and v(1, 1) = 1and note that Ar is the amoeba of u3 + w3 + ruw + 1 multiplied coordinatewise with 1

log(r). Now theorem 4.7 shows that Acf has a bounded component

for sufficiently large values of λ. The case of λ = 2 is figure 2.1.

We can also use theorem 4.7 to show that the maximum number of com-ponents given by corollary 2.4 is achieved by some curves.

Theorem 4.9. For every finite set Q ⊂ Zn there is a Laurent polynomialg with Newt(g) = conv(Q) such that the number of components of Ag is# Newt(g) ∩ Zn.

1This could in principle be proved by looking at the critical points of Log map.

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Figure 4.2: The amoeba of u3 + w3 + λuw + 1 for λ = 1.

Proof. Let I = Newt(g) ∩ Zn and let f((z)) =∑

i∈I zi. By proposition 1.11there is a function v : I → Z such that the subdivision of Newt(f) = Newt(g)induced by the tropical curve V given by maxi∈I{ai + i · x} has I as set ofvertices. By theorem 1.28 the components of the complement of V correspondbijectively to I. Now we apply theorem 4.7 and note AK = V .

Thus for sufficiently large r it follows that Acr has at least #I components.The amoeba of g((z)) =

∑i∈I r

v(i)zi is homeomorphic to Acr so it follows thatAcg has exactly #I components.

Now we give an example of an amoeba constructed this way.

Example 4.10. We want to construct a polynomial with Newton polytopeN = conv{(0, 0), (3, 0), (0, 3), (2, 3), (3, 2)} and maximal number of compo-nents. LetQ = N∩Z2 and let f =

∑j∈Q u

j1wj2 . Now we must pick a functionv : Q → Q such that trop(ft) has a maximal number of components. If wepick v as in example example 1.12 then trop(ft) will be the tropical polyno-mial from example 1.20 which indeed has the maximum possible number ofcomponents.

The convergence of the modified amoebas Ar to the tropical curve can beseen in figure 4.3.

For sufficiently large r the complement of Ar has a component for everycomponent of the complement of the tropical curve of ft by lemma 4.6.

As Ar is the amoeba of fr multiplied with a constant we can use any fr

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(a) r=2 (b) r=3

(c) r=5 (d) r=100

Figure 4.3: Converging amoebas.

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with sufficiently large r, for example f14.

f14 = u3 + w3 + 1 + u2w3 + u3w2 + 14u+ 14u2 + 14w + 14w2 + 14u3w

+14uw3 + 196uw + 196u2w + 196uw2 + 196u2w2

Figure 4.4: The amoeba of f14.

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Chapter 5

Appendix I

In this appendix we consider the effect of an endomorphism of (C∗)2 on theamoeba of a curve. This is the special case n = d = 2 of section 4 of [25],although the formulation is slightly different.

We are interested in the endomorphisms of (C∗)2 with the structure ofa maximal ideal spectrum and not of an algebraic group, so multiplicationwith a constant is an endomorphism.

5.1 Endomorphisms of (C∗)2

The points of (C∗)2 correspond to the closed points of spec(C[u,w, u−1, w−1]),which are by definition the maximal ideals of C[u,w, u−1, w−1]. By Hilbert’snullstellensatz and elementary properties of localization the set of maximalideals of C[u,w, u−1, w−1] is {(u − λ1, w − λ2)|λ1, λ2 ∈ C∗}. The correspon-dence between maximal ideals and points is given by mapping a maximalideal m to the single element of {(λ1, λ2) ∈ (C∗)2|f(λ1, λ2) = 0 for all f ∈ m},thus (u− λ1, w − λ2) 7→ (λ1, λ2).

For M ∈ Mat2(Z) and α = (α1, α2) ∈ (C∗)2 we define

(α1, α2)M = (αm11

1 αm212 , αm12

1 αm222 ).

By definition the endomorphisms of (C∗)2 correspond the endomorphismsof C[u,w, u−1, w−1] as C-algebra and we can use commutative algebra todescribe these.

Lemma 5.1. The set End(C[u,w, u−1, w−1]) consists of all maps defined by

f(z) 7→ f(α · zM),

with M ∈ Mat2(Z) and α = (α1, α2) ∈ (C∗)2, where f(z) = f(u,w) ∈C[u,w, u−1, w−1]. Furthermore all these maps are distinct.

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Proof. Denote A = C[u,w, u−1, w−1]. By the universal property of local-ization, homomorphisms ϕ : A → A are the unique extensions of homomor-phisms ϕ : C[u,w]→ A such that ϕ(u) and ϕ(w) are units. The set of unitsis {αijuiwj|αij ∈ C∗ and i, j ∈ Z}. Thus a homomorphism ϕ : C[u,w] → Aextends to an endomorphism of A if and only if ϕ(u) = α1u

m11wm21 andϕ(w) = α2u

m12wm22 with α1, α2 ∈ C∗ and m11,m21,m12,m22 ∈ Z. In thiscase ϕ and its extension ϕ : A → A are given by f(z) 7→ f(α · zM). Thisproves the first statement in the lemma.

If ϕ is the endomorphism given by M ∈ Mat2(Z) and α = (α1, α2) ∈(C∗)2 then ϕ(u) = α1u

m11wm21 and ϕ(w) = α2um12wm22 , so M and α are

unique.

We can use this lemma to show the group of automorphisms is given by asemidirect product. It is easy to check (αβ)M = αMβM and αMN = (αN)M

for all α, β ∈ (C∗)2 and M,N ∈ Mat2(Z). This way we get a homomorphismGL(2,Z) → AutGrp((C∗)2) and thus a semidirect product (C∗)2 o GL(2,Z)of groups.

Proposition 5.2. The map Ψ: (C∗)2 o GL(2,Z)→ Aut(C[u,w, u−1, w−1]),defined for (α,M) ∈ (C∗)2oGL(2,Z) and f(z) = f(u,w) ∈ C[u,w, u−1, w−1]by

Ψ(α,M)(f(z)) = f(α · zM), (5.1.1)

is an isomorphism.

Proof. By the previous lemma we know that formula 5.1.1 defines an endo-morphism of C[u,w, u−1, w−1]. It is clear that Ψ maps the unit element of(C∗)2 o GL(2,Z) to the identity map, so if Ψ is multiplicative all endomor-phisms defined by formula 5.1.1 are automorphisms.

Let ψ and ϕ be the endomorphisms defined by (β,N) and (α,M) withβ, α ∈ (C∗)2 and N,M ∈ Mat2(Z) then we have

ψ ◦ ϕ(u) = α1βm111 βm21

2 um11n11+m21n12wm11n21+m21n22 (5.1.2)

and

ψ ◦ ϕ(w) = α2βm121 βm22

2 um12n11+m22n12wm12n21+m22n22 . (5.1.3)

This shows ψ ◦ϕ is the endomorphism defined by (αβM , NM). In particularwe have Ψ(β,N)◦Ψ(α,M) = Ψ(αβM , NM) = Ψ((β,N)(α,M)), which showsthat Ψ is well defined and a homomorphism.

If ϕ is an endomorphism and ψ = ϕ−1 it follows by equation 5.1.2 and5.1.3 that NM is the unit matrix, so N ∈ GL(2,Z) and ϕ ∈ Im(Ψ). FinallyΨ is surjective by the uniqueness of (α,M) in the previous lemma.

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Equations 5.1.2 and 5.1.3 also show that the composition of endomor-phisms is also well behaved with respect to matrix multiplication, so wouldalso be possible to give a description of End(C[u,w, u−1, w−1]) in terms ofthe semigroups (C∗)2 and Mat2(Z).

For every endomorphism of C[u,w, u−1, w−1] we have an endomorphismof the maximal spectrum. The following proposition shows how this map be-haves with respect to the identification of the maximal spectrum with (C∗)2.Proposition 5.3. If ϕ is the endomorphism of C[u,w, u−1, w−1] defined byα ∈ (C∗)2 and M ∈ Mat2(Z) then the corresponding endomorphism ϕ∗ of(C∗)2 is given by ϕ∗(λ) = αλM .

Proof. Denote A = C[u,w, u−1w−1] and let msp(A) be the maximal idealspectrum ofA. If ϕ in an endomorphism ofA the canonical map ϕ : msp(A)→msp(A) is given by ϕ(m) = ϕ−1(m) for maximal ideals m of A.

We have identified (C∗)2 with msp(A) by the map Z : msp(A) → (C∗)2which maps a maximal ideal to its set of zeros. Thus we need to show thatthe diagram

msp(A) Z //

ϕ

��

(C∗)2

ϕ∗

��msp(A) Z // (C∗)2

commutes. Let m ∈ msp(A), then m = (u−λ1, w−λ2) for some λ1, λ2 ∈ C∗,so Z(m) = (λ1, λ2) = λ and ϕ∗Z(m) = αλM . If f ∈ ϕ−1(m) then f(αλM) =ϕ(f)(λ), so αλM is the zero of ϕ−1(m).

Proposition 5.4. Let ϕ be the endomorphism of C[u,w, u−1, w−1] definedby α ∈ (C∗)2 and M ∈ Mat2(Z). If det(M) 6= 0 then the correspondingendomorphism ϕ∗ of (C∗)2 is surjective.

Proof. It is enough to to prove this with α = (1, 1). By the Smith normalform theorem we can find A,B,∈ GL(2,Z) and D ∈ Mat2(Z) such thatAMB = D and D is a diagonal matrix. From the assumptions it followsdirectly that det(D) 6= 0, so D defines a surjective endomorphism of (C∗)2.

The equation AMB = D also holds when we consider the maps definedby A,M,B,D, so M is also surjective.

The above argument can easily be turned into an equivalence.

Proposition 5.5. Let ϕ∗ be a surjective endomorphism of (C∗)2 as in theprevious proposition. If Cf ⊂ (C∗)2 is the curve defined by f ∈ C[u,w, u−1, w−1]then the restriction ϕ∗|Cϕ(f) : Cϕ(f) → Cf is surjective.

Proof. We have f(ϕ∗(λ)) = ϕ(f)(λ) for every λ ∈ (C∗)2, so λ ∈ Cϕ(f) if andonly if ϕ∗(λ) lies in Cf . The corollary follows because ϕ∗ is surjective.

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5.2 The transformation of the amoeba

It turns out that the surjective maps of curves in the previous section givelinear bijections between the amoebas of the curves.

First we give the action on the Newton polytope.

Proposition 5.6. Let ϕ is the endomorphism of C[u,w, u−1, w−1] defined byα ∈ (C∗)2 and M ∈ Mat2(Z), then Newt(ϕ(f)) = M Newt(f)

Proof. The monomial uiwj occurs in f if and only if the monomial ϕ(uiwj) =un11i+n12jwn21i+n22j occurs in ϕ(f). As a linear map maps the convex hull ofa set to the convex hull of the image of the set, the proposition follows.

The map ϕ∗|Cϕ(f) : Cϕ(f) → Cf from corollary 5.5 translates directly to amap between the amoebas of f and ϕ(f). Because ϕ∗ is surjective this mapbecomes a bijection.

Corollary. Let Cf ⊂ (C∗)2 be the curve defined by f ∈ C[u,w, u−1, w−1]and let ϕ be the endomorphism of C[u,w, u−1, w−1] defined by α ∈ (C∗)2 andM ∈ Mat2(Z) with det(M) 6= 0, then

Af = MAϕ(f) − Log(α).

Proof. By proposition 5.5 the map ϕ∗|Cϕ(f) : Cϕ(f) → Cf is surjective and

by proposition 5.3 we have ϕ∗(λ) = αλM . Let T : R2 → R2 be defined by

T

(xy

)= M ·

(xy

)−(

log |α1|log |α2|

), then it is easy to check that the diagram

Cϕ(f)

Log��

ϕ∗ // Cf

Log��

R2 T // R2

commutes, which shows the sets Log(Cf ) and T Log(Cϕ(f)) are equal.

Example 5.7. Corollary 5.2 makes it possible to draw some amoebas in two

different ways. Consider f = u + w + 1 and M =

(3 57 2

), then ϕ(f) =

u3w7 + u5w2 + 1.In figure 5.1 the amoeba of f has been translated by M , however the

distribution of the computed points has become rather uneven as only a fewpoints of the left tentacle have been computed.

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Figure 5.1: The amoeba of f translated by M .

Directly computing points of the amoeba of u3w7 + u5w2 + 1 is slowerbecause of the higher degree, but the points are distributed evenly as can beseen in figure 5.2.

Figure 5.2: The amoeba of u3w7 + u5w2 + 1 computed directly.

The difference in scale is because both the amoeba of f and ϕ(f) werecomputed in the square defined by −4 ≤ x, y ≤ 4.

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Chapter 6

Appendix II

In this appendix we give the Sage [23] code used to draw the amoebas in thisthesis. It is very similar to the matlab code available at [1].

From a practical point of view the use of Sage it convenient because it hasa number built in operations on polynomials. Furthermore it is open sourceand free to use.

6.1 Computing the amoeba of a planar curve

The following code can be put directly into a worksheet, except for an extralinebreak in line 9.

In the first box we declare the box [xmin, xmax]× [ymin, ymax] in whichwe want to draw the amoeba, the variables, the number of points drawn andthe polynomial f .

1 var ( ’ xmin ’ , ’ xmax ’ , ’ ymin ’ , ’ ymax ’ )2 xmin=−33 xmax=34 ymin=−35 ymax=36 var ( ’ ang l e s t ep ’ , ’ g r i d s t e p ’ )7 g r i d s t e p =2508 a n g l e s t e p =1209 var ( ’ u ’ , ’w’ )

10 f =4∗u∗w−uˆ2∗w−wˆ2∗u−1

The following box computes points along vertical lines in the chosen box.Because we only look at polynomials with real coefficients the angle of thepossible solutions can be taken in [0, π).

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1 x g r i d s t e p s i z e =(xmax−xmin )/ g r i d s t e p2 L i s t x =[ ]3 f o r j in range (1 , g r i d s t e p ) :4 r=xmin+j ∗ x g r i d s t e p s i z e5 f o r k in range (0 , a n g l e s t e p ) :6 theta=(k/ a n g l e s t e p )∗ pi ∗ I7 z1=eˆ r∗e ˆ( theta )8 g=f . s u b s t i t u t e (u=z1 )9 S=g . r oo t s (w, r i ng=ComplexField ( prec =10)

10 , m u l t i p l i c i t i e s=False )11 f o r p2 in S :12 logp2=log ( abs ( p2 ) )13 i f logp2>ymin and logp2<ymax :14 L i s t x . append ( ( r , logp2 ) )

We can now plot the list of points with the following command.

1 l i s t p l o t (Lx , s i z e =2, a s p e c t r a t i o =1)

If we only compute points along vertical lines any tentacles of the amoebawhich are vertical will be drawn badly because the become thinner than thespace between the lines on which the points are computed. This can be seenin the next example.

Example 6.1. If we run the algorithm for f = 1 +u3 +w3 + 2uw in the box[−3, 3]× [−3, 3] with grid step = 250 and angle step = 120 we get figure 6.1

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Figure 6.1: Incomplete amoeba for f = 1 + u3 + w3 + 2uw.

Thus we have to add a second list of points, now computed on horizontallines. Then we get the full picture. We can plot both list with the followingcommand.

1 l i s t p l o t ( L i s t x )+ l i s t p l o t ( L i s t y )

The complete amoeba of example 6.1 is figure 2.1

6.2 Critical points

If we compute too few point to fill up the amoeba we get a picture where wecan see some indication as to how the curve is folded onto the amoeba. Forexample in figure 6.2 we can see that the boundary of the hole in the amoebais the union of three foldlines which continue in the amoeba.

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Figure 6.2: Some points on the amoeba of f = 1 + u3 + w3 + 1.25uw.

These lines are critical points of the Log map on the curve. In [25] analgorithm to compute these lines is given.

In the following amoeba the boundary of the hole in the amoeba seems tobe smooth.

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Figure 6.3: Some points on the amoeba of f = 4uw − u2w − w2u− 1.

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Populaire samenvatting

Een vlakke kromme is de verzameling van punten in het vlak die aan eenbepaald soort vergelijking voldoen. In figuur 6.4(a) staan bijvoorbeeld deoplossingen van de vergelijking y − x2 − 2 = 0. In dit geval is de krommeeen parabool. 1 In het algemeen bestaat een kromme uit een aantal lijnenen cirkels.

In figuur 6.4(b) staat een geschaalde versie2 van dezelfde kromme. Vanuitdit perspectief lijkt de kromme uit twee rechte lijnen te bestaan.

(a) parabool (b) geschaalde parabool

Figure 6.4

Om te zien met welke rechte lijnen we te maken hebben bekijken we eentropische kromme. Deze naam heeft niets met een regenwoud te makenmaar met het feit dat dit soort meetkunde vernoemd is naar de Braziliaansewiskundige Imre Simon.

In figuur 6.5 is aangegeven in welk deel van het vlak de functies 2x, y enlog(2) maximaal zijn. De punten waar meerdere van deze functies maximaalzijn vormen de drie rode lijnstukken. Dit is de tropische kromme van 2x, yen log(2). We zien dat twee van de lijnen figuur 6.4(b) goed benaderen.

1Om technische redenen kijken we alleen naar oplossingen die niet op de coordinaatassenliggen.

2We sturen een punt (x, y) naar (log |x|, log |y|).

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Figure 6.5: Een tropisch kromme.

Nu gaan we uitzoeken waar de derde lijn vandaan komt.In figuur 6.4(a) hebben we oplossingen van y−x2−2 = 0 met reele getallen

bepaald en dit geeft een kromme in het vlak. We kunnen ook zoeken naaroplossingen met complexe getallen. Als we dit doen geven de oplossingeneen oppervlak in een vierdimensionale ruimte, waar we dus geen plaatje vankunnen tekeken.

Wel kunnen we weer dezelfde soort schaling toepassen als we afbeelding6.4(b) gedaan hebben. De vierdimensionale ruimte wordt nu naar een twee-dimensionale ruimte afgebeeld. Het beeld van verzameling van oplossingenis nu te zien in figuur 6.6. We kunnen nu opmerken dat de lijn in figuur6.4(b) de bovenrand van figuur 6.6 is en dat de we de tak gevonden hebbendie hoort bij de derde lijn van de tropische kromme.

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Figure 6.6: De amoeba van de vergelijking y − x2 − 2 = 0.

De figuur die we krijgen als we de procedure van afbeelding 6.6 toepassenop de complexe oplossingen van een vergelijking heet de amoeba van devergelijking. De gelijkenis met de eencellige organismes met dezelfde naamwordt duidelijk als we naar een ander voorbeeld van een amoeba kijken,bijvoorbeeld afbeelding 6.7.

Figure 6.7: De amoeba van de vergelijking x3 + y3 + x2y4 + x4y3 + 3.8x3y +12.9x2y2 + 4.55xy3 + 4.6x3y3 = 0.

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In deze sciptie hebben we een constructie van Grigory Mikhalkin bekekenwaar een amoeba steeds verder vervormd wordt tot die steeds meer op eentropische kromme gaat lijken. Het effect van deze constructie is duidelijk tezien in afbeelding 6.8.

(a) r=2 (b) r=3

(c) r=5 (d) r=100

Figure 6.8: Het vervormen van een amoeba.

Ook hebben we het geval bekeken waar we niet kijken naar oplossingenvan vergelijkingen met complexe getallen, maar met oplossingen in een andergetalsyteem. Dit geval wordt geheel beschreven door de stelling van Kapra-nov.

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