AIX – MARSEILLE UNIVERSITY—————————∗ ∗ ∗—————————

MASTER THESIS

INTRODUCTION TO TORIC VARIETIES

Specialty : Fundamental Mathematics

Advisor : Prof. Anne PichonProf. Guillaume Rond

Student: Nguyen Nho

Marseille, June 2017

Contents

1 From combinatorial geometry to toric varieties 31.1 Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Monoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Affine toric varieties 172.1 Laurent polynomials . . . . . . . . . . . . . . . . . . . . . . . 172.2 Some basic results of algebraic geometry . . . . . . . . . . . . 182.3 Affine toric varieties . . . . . . . . . . . . . . . . . . . . . . . 20

3 Toric varieties 293.1 Fans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.2 Toric varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 The torus action and the orbits 344.1 The torus action . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.3 Compactness and smoothness . . . . . . . . . . . . . . . . . . 40

1

Introduction and acknowledgements

The main goal of this work is to study the basic theories of the toric varieties.The procedure of the construction of the toric varieties associates to a

cone σ in space Rn successively: the dual of cone σ, a monoid Sσ, a finitelygenerated C−algebra Rσ and finally an algebraic variety Xσ. We will describethe steps of the procedure:

σ → ∨σ → Sσ → Rσ → Xσ.

Consider the group action of the torus ((C∗)n) on the toric varieties, therelations between combinatorial geometry and algebraic geometry, the com-pactness and the smoothness of the toric varieties.

This stage is done at Luminy, Aix-Marseille University, under the super-vision of Professor Anne Pichon, Professor Guillaume Rond and ProfessorJean-Paul Brasselet. I would like to express my sincere appreciation andgratitude to my advisors for providing extensive support and valuable guid-ance.

I am very thankful to my professors at Aix-Marseille University for theirenthusiastic teaching in my courses.

Special thanks to Professor Le cong Trinh, Professor Ngo Lam Xuan Chauand Professor Lionel Nguyen Van The for giving me an opportunity to studyat Aix-Marseille University.

Special thanks to Le Van An, Jihane Alamedine, Pedro Javier Ortiz, VienVien and my friends for helping me to complete this stage.

Marseille, June 2017

2

1 From combinatorial geometry to toric va-

rieties

1.1 Cones

Definition 1.1.1 Let A = {v1, ..., vr} be a finite set of vectors in Rn, theset

σ = {x ∈ Rn|x = λ1v1 + λ2v2 + ...+ λrvr, λi ≥ 0, λi ∈ R}

is called a convex polyhedral cone. The vectors v1, ..., vr are called gen-erators of the cone σ.

Convention: If A = ∅ then σ = {0}.

In Rn, let ei be the i − th column of the identity matrix, so we havee1 = (1, 0, ...., 0), ...., en = (0, 0, ...., 1). Then (e1, ..., en) is a basis of Rn.

Example 1.1.2 In R2 with a basis (e1, e2),

O

σ1

O−e1 e1

e2e2

O−e1

e1 + e2e2 − e1 e2

σ2 σ3

Fig. 1 Examples of cones

• σ1 is generated by vector e2.• σ2 is generated by vectors (e1,−e1, e2).• σ3 is generated by vectors (−e1, e2−e1, e2, e1+e2), the vectors (−e1, e1+

e2) are also generators of σ3.

Definition 1.1.3 The dimension of a cone σ, denoted by dimσ, is thedimension of the smallest linear space containing σ.

Example 1.1.4 In Example 1.1.2, one has dimσ1 = 1, dimσ2 = dimσ3 = 2.In the following, N will denote by a fixed lattice N ∼= Zn ⊂ Rn.

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Definition 1.1.5 A cone σ is a lattice (or rational) cone if it admitsgenerators v1, ..., vr such that vi belongs to N for all i.

A cone is strongly convex if it does not contain any straight line goingthrough the origin (i.e σ

⋂(−σ) = {0}).

Example 1.1.6 In Example 1.1.2, the cones σ1 and σ3 are the stronglyconvex lattice cones. The cone σ2 is not strongly convex.

Definition 1.1.7 Let (Rn)∗ be the dual space of Rn and 〈, 〉 the dualpairing.

The dual of cone σ is the set

∨σ = {u ∈ (Rn)∗|〈u, v〉 ≥ 0,∀v ∈ σ}.

The set of vectors (e∗1, ..., e∗n) is a dual basis of (Rn)∗, where e∗i is the i−th

column of the identity matrix.Given a lattice N in Rn, we define the dual lattice by M = HomZ(N,Z) ∼=

Zn in (Rn)∗.

Lemma 1.1.8 Let σ be a convex polyhedral cone generated by the vectors

(v1, v2, ..., vr), then∨σ =

r⋂i=1

∨τi where τi is the ray, generated by the vector vi.

PROOF:For every vi, we have a cone τi = R≥0vi and

∨τi = {u ∈ (Rn)∗|〈u, vi〉 ≥ 0}.

Then

∨σ = {u ∈ (Rn)∗|〈u, vi〉 ≥ 0, i = 1, ..., r} =

r⋂i=1

{u ∈ Rn|〈u, vi〉 ≥ 0}.

So we have∨σ =

r⋂i=1

∨τi

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Lemma 1.1.9 Let v ∈ N and set τ = R≥0v. Then∨τ is a lattice cone.

PROOF:We set

v⊥ = {u = (u1, ..., un) ∈ (Rn)∗|〈u, v〉 = 0}.

If v = 0 then∨τ = (Rn)∗.

If v 6= 0, suppose that v =n∑i=1

λiei, then v∗ =n∑i=1

λie∗i , we have

〈v∗, v〉 =n∑i=1

λ2i > 0.

One has

∨τ = {u ∈ (Rn)∗|〈u, v〉 ≥ 0}.

For every u ∈ ∨τ , set t =〈u, v〉〈v∗, v〉

. Hence, one has 〈u − tv∗, v〉 = 0. This

means that u− tv∗ ∈ v⊥, then u belongs to R≥0v∗ + v⊥.

So we obtain

∨τ = R≥0v

∗ + v⊥.

Since v ∈ N , v∗ ∈M . We will show that v⊥ is a lattice cone.Suppose that v = (v1, ..., vn) ∈ Rn, set the matrix

A =[v1 ... vn

]∈ R1×n,

then

v⊥ = {u = (u1, ..., un) ∈ (Rn)∗|〈u, v〉 = u1v1 + ...+ unvn = 0} = KerA.

Since v1, ..., vn ∈ Z, we can get n− 1 vectors a1, ..., an−1in Zn such that

v⊥ =n−1∑i=1

Rai =n−1∑i=1

R≥0ai +n−1∑i=1

R≥0(−ai).

Therefore, v⊥ is a lattice cone. Hence,∨τ is a lattice cone.

5

Example 1.1.10 Let us denote by (e∗1, e∗2) the dual basis of (R2)∗.

O

τ

O−e1∗ e∗1

e2∗e2∗

∨τ

−e1

e1 + 2e2

σ

O O

−2e∗1 + e∗2 e∗2

∨σ

Fig. 2 Examples of dual cones.

Proposition 1.1.11

i) If σ is a convex polyhedral cone, then∨σ is a polyhedral convex cone.

ii) If σ is a lattice cone then∨σ is a lattice cone (relative to the lattice M).

PROOF:i) Suppose that σ is generated by (v1, ..., vr), then

σ = {r∑i=1

λivi|λi ≥ 0}

By Weyl-Minkowski’s Theorem for cones, there are vectors a1, ..., ak ∈Rn such that

σ = {x ∈ Rn|atix ≥ 0,∀i = 1, ..., k}.

Then

∨σ = {

r∑i=k

λia∗i |λi ≥ 0},

6

where a∗i := ai ∈ (Rn)∗. Hence∨σ is a convex polyhedral cone.

ii) See [7], page 7, Proposition 1.3.

Proposition 1.1.12 If σ is a convex polyhedral cone, then (∨σ)∨ = σ.

PROOF:Suppose that σ is generated by (v1, ..., vr). We have

(∨σ)∨ = {v ∈ Rn|〈u, v〉 ≥ 0,∀u ∈ ∨σ}.

Let v ∈ σ, then 〈u, v〉 ≥ 0 for all u ∈ ∨σ. Hence, we have σ ⊂ (∨σ)∨.

Conversely, suppose that v ∈ (∨σ)∨, v /∈ σ.

Since v /∈ σ, there do not exist λi ≥ 0 such thatr∑i=1

λivi = v. By Farkas’

Lemma, there is u ∈ ∨σ such that 〈u, v〉 < 0 . Contradiction with v ∈ (∨σ)∨.

Therefore (∨σ)∨ ⊂ σ and we obtain the result.

Definition 1.1.13 In fact, a convex polyhedral cone σ can also be definedas intersection of half-spaces. For each (co)vector u ∈ (Rn)∗, we define ahalf-space by

Hu = {v ∈ Rn|〈u, v〉 ≥ 0}.

If the cone∨σ is generated by the vectors (u1, u2, ..., ut) , then

σ = {v ∈ Rn|〈u1, v〉 ≥ 0, .., 〈ut, v〉 ≥ 0} =t⋂i=1

{v ∈ Rn|〈ui, v〉 ≥ 0}.

So one has

σ =t⋂i=1

Hui .

Notice that if σ is a strongly convex cone, then∨σ is not necessarily a strongly

convex cone (cone τ in Example 1.1.10 is an example).

7

Proposition 1.1.14 The sum of two convex polyhedral cones is also a con-vex polyhedral cone. The intersection of two convex polyhedral cones is alsoa convex polyhedral cone.

PROOF:Suppose σ1 and σ2 are two convex polyhedral cones. σ1 is generated by

the vectors (a1, .., ar), and σ2 is generated by the vectors (b1, ..bk).

σ1 + σ2 = {r∑i=1

λiai +k∑i=1

βibi|λi, βi ∈ R≥0}.

So σ1+σ2 is a convex polyhedral cone generated by the vectors (a1, ..ar, b1, ..bk).Suppose σ = σ1 + σ2,

∨σ = {u ∈ (Rn)∗|〈u, v〉 ≥ 0,∀v ∈ σ} =

2⋂i=1

{u ∈ (Rn)∗|〈u, v〉 ≥ 0,∀v ∈ σi}.

Therefore, we have

∨σ =

∨σ1 ∩

∨σ2.

Finally,

σ1 ∩ σ2 = (∨σ1)∨ ∩ (

∨σ2)∨ = (

∨σ1 +

∨σ2)∨

is a convex polyhedral cone, so we obtain the result.

1.2 Faces

Definition 1.2.1 Let σ be a cone and λ ∈ ∨σ⋂M , then

τ = σ⋂

λ⊥ = {v ∈ σ|〈λ, v〉 = 0}

is called a face of σ. We will denote by τ < σ.A cone is a face of itself, other faces are called proper faces.A one-dimensional face is called an edge.

Example 1.2.2 In Example 1.1.2, the cone σ1 has 4 faces σ1,τ1 = R≥0e1, τ2 = R≥0e2 and {0}.

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Property 1.2.3 Let σ be a cone generated by the vectors (v1, ..., vr), then

i) Every face τ of σ is a convex polyhedral cone, σ has a finite number offaces.

ii) Every intersection of faces of σ is a face of σ.

iii) Every face of a face is a face.

PROOF:i) There is λ ∈ ∨σ

⋂M such that

τ = σ⋂λ⊥ = {v ∈ σ|〈λ, v〉 = 0}.

Let {a1, ..., at} be the set of all elements vi belonging to A = {v1, ..., vr}such that 〈λ, vi〉 = 0.

Let v ∈ τ , suppose that v =r∑i=1

αivi. If 〈λ, vj〉 > 0 for some j, then

αj = 0 (because 〈λ, v〉 = 0).Hence, we have

τ =t∑i=1

R≥0ai.

This means that τ is a convex polyhedral cone.The set A has a finite number of subsets; therefore, σ has a finite number

of faces.

ii) Suppose that τ1, τ2 are two faces of σ, so there are λ1, λ2 ∈∨σ⋂M such

that τ1 = σ⋂λ⊥1 , τ2 = σ

⋂λ⊥2 . Show that τ1 ∩ τ2 is a face of σ.

Firstly, we will prove that

σ ∩ (λ⊥1 ∩ λ⊥2 ) = σ ∩ (λ1 + λ2)⊥.

If v ∈ σ ∩ (λ⊥1 ∩ λ⊥2 ), then 〈λ1, v〉 = 〈λ2, v〉 = 0, so 〈λ1 + λ2, v〉 = 0, hencev ∈ σ ∩ (λ1 + λ2)⊥.

Conversely, if v ∈ σ ∩ (λ1 + λ2)⊥ then 〈λ1 + λ2, v〉 = 〈λ1, v〉+ 〈λ2, v〉 = 0.

Since λ1, λ2 ∈∨σ⋂M , we have 〈λ1, v〉 ≥ 0 and 〈λ2, v〉 ≥ 0. Hence 〈λ1, v〉 =

〈λ2, v〉 = 0 and this implies that v ∈ λ⊥1 ∩ λ⊥2 .Finally, we have

τ1 ∩ τ2 = (σ ∩ λ⊥1 ) ∩ (σ ∩ λ⊥2 ) = σ ∩ (λ⊥1 ∩ λ⊥2 ) = σ ∩ (λ1 + λ2)⊥.

9

We have λ1 + λ2 ∈∨σ⋂M ; therefore, τ1 ∩ τ2 is a face of σ.

iii) Suppose that γ < τ < σ, τ = σ ∩ λ⊥, γ = τ ∩ α⊥ where λ ∈ ∨σ⋂M,α ∈

∨τ⋂M .We will prove that γ < σ.Firstly, we will prove that if vi ∈ σ \ τ , then there is k ∈ Z≥0 such that

〈α + kλ, vi〉 > 0.For every vi ∈ σ \ τ , we have 〈λ, vi〉 > 0, 〈α, vi〉 ∈ R, so there is ki ∈ Z≥0

such that 〈α + kiλ, vi〉 > 0. Choose k = max{ki}, then 〈α + kλ, vi〉 > 0 forall vi ∈ σ \ τ .

If vi ∈ τ then 〈α + kλ, vi〉 ≥ 0.Therefore, for every vi ∈ σ, we have 〈α + kλ, vi〉 ≥ 0, then α + kλ ∈ σ.Finally, we prove that

γ = σ ∩ (α + kλ)⊥.

If v ∈ γ then v ∈ τ and v ∈ τ , so 〈λ, v〉 = 0 , 〈α, v〉 = 0, hence〈α + kλ, v〉 = 0, this means that v ∈ σ ∩ (α + kλ)⊥.

Suppose that v =r∑i=1

βivi ∈ σ ∩ (α + kλ)⊥.

If vi /∈ τ then 〈α + kλ, vi〉 > 0, then βi = 0 (because 〈α + kλ, v〉 = 0).Hence, if βi 6= 0 then vi ∈ τ , so 〈λ, v〉 = 0. Since 〈α + kλ, v〉 = 0, one has〈α, v〉 = 0, then v ∈ γ.

Remark 1.2.4 Indeed, if τ < σ then∨σ ⊂ ∨τ .

Remark 1.2.5 Let τ be a face of σ. If x and y belong to σ and x + ybelongs to τ , then x and y belong to τ .

Proposition 1.2.6 Let τ = σ⋂λ⊥ be a face of σ (λ ∈ ∨σ ∩M), then

∨τ =

∨σ + R≥0(−λ).

PROOF:

• We show that σ ∩∨

(−λ) = σ ∩ λ⊥.

If v ∈ σ ∩∨

(−λ) then 〈−λ, v〉 ≥ 0 and 〈λ, v〉 ≥ 0, so we have 〈λ, v〉 = 0,hence v ∈ σ ∩ λ⊥.

If v ∈ σ ∩ λ⊥ then 〈λ, v〉 = 0, so v ∈ σ ∩∨

(−λ).• Consider the following relation

10

τ = σ⋂λ⊥ = σ ∩

∨(−λ) = (

∨σ + R≥0(−λ))∨.

So we have the result

∨τ =

∨σ + R≥0(−λ).

Example 1.2.7 Let us consider the following example.

O

O

e2

O

τ

∨τ

O

e2

e1 O

e∗2

λ = e∗1−λ

∨σσ

Fig. 3

Firstly, consider τ is a face of σ. One has τ = σ⋂λ⊥, and

∨τ =

∨σ + R≥0(−λ).

Secondly, consider the face {0}, we have∨{0} = (Rn)∗.

Set η = e∗1 + e∗2, we have η ∈ ∨σ⋂M , and then

(Rn)∗ =∨σ + R≥0(−η).

Definition 1.2.8 The relative interior of a cone σ is the interior of σas a subset of the space Rσ generated by σ. A point of the relative interioris obtained by taking a strictly positive linear combination of dimσ linearlyindependent vectors among the generators of σ.

For any vector v in σ, there is a face τ < σ such that v is in the relativeinterior of τ.

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Property 1.2.9 Let σ be a convex polyhedral cone σ and v ∈ σ. If v is in

the relative interior of σ then∨σ ∩ v⊥ = σ⊥.

PROOF:Suppose that σ is generated by the vectors (v1, ..., vr) and dimσ = d.Let the set A = {v1, ..., vr}.

Assume that v is in the relative interior of σ, then there are vectors a1, ..., adin A such that the set of vectors B = {a1, ..., ad} is linearly independent and

v =d∑i=1

λiai (λi > 0,∀i = 1, ..., d).

Since dim(σ) = d and B is linearly independent. For every vector x ∈ σ,we have

x ∈d∑i=1

Rai.

We have

v⊥ = {u ∈ (Rn)∗|〈u, v〉 = 0}.

If u ∈ ∨σ ∩ v⊥, then 〈u, vi〉 ≥ 0,∀i = 1, ..., r and 〈u, v〉 = 0. Hence

〈u, ai〉 = 0,∀i = 1, ..., d. Then 〈u, x〉 = 0 for all x ∈ σ, then u ∈ σ⊥.

If u ∈ σ⊥ then 〈u, vi〉 = 0 for i = 1, ..., r, hence u ∈ ∨σ ∩ v⊥.

Finally, we have∨σ ∩ v⊥ = σ⊥.

Proposition 1.2.10 Let τ be a face of σ, then∨σ ∩ τ⊥ is a face of

∨σ with

dim(τ) + dim(∨σ ∩ τ⊥) = n. This provides a one-to-one correspondence be-

tween faces of σ and faces of∨σ.

PROOF:• If τ < σ then

∨σ ∩ τ⊥ is a face of

∨σ.

Faces of∨σ are the cones

∨σ ∩ v⊥ with v ∈ (

∨σ)∨ ∩N = σ ∩N .

Let v be in the relative interior of τ . By Property 1.2.9, we have

∨σ ∩ τ⊥ =

∨σ ∩ (

∨τ ∩ v⊥) = (

∨σ ∩ ∨τ) ∩ v⊥ =

∨σ ∩ v⊥

is a face of∨σ.

• dim(τ) + dim(∨σ ∩ τ⊥) = n.

There is λ ∈ ∨σ ∩M such that τ = σ ∩ λ⊥, and we have

12

∨τ =

∨σ + R≥0(−λ).

Since v ∈ τ , we have 〈λ, v〉 = 0. Then one has R≥0(−λ) ⊂ v⊥, hence

τ⊥ =∨τ ∩ v⊥ = (

∨σ + R≥0(−λ)) ∩ v⊥ =

∨σ ∩ v⊥ + R≥0(−λ).

Since λ ∈ ∨σ ∩M , one has

dim(∨σ ∩ v⊥) = dim(

∨σ ∩ v⊥ + R≥0(−λ)) = dim(τ⊥).

Therefore,

dim(τ) + dim(∨σ ∩ τ⊥) = dim(τ) + dim(

∨σ ∩ v⊥) = dim(τ) + dim(τ⊥) = n.

• Provide a one-to-one correspondence between faces of σ and faces of∨σ.

Consider the finite sets of cones

A = {τ |τ < σ},B = {τ |τ < ∨

σ}.

Consider the mapping

Θ : A→ B

τ 7→ ∨σ ∩ τ⊥

and

Θ′ : B → A

τ 7→ σ ∩ τ⊥

Let γ be a face of∨σ. This means that γ =

∨σ ∩ v⊥ for some v ∈ σ ∩ N .

There is a face τ of σ such that v is in the relative interior of τ . ThenΘ(τ) = γ, hence Θ is surjective. Similarly, we have Θ′ is surjective.

We denote by the number of all elements of A by |A|. Since Θ,Θ′ aresurjective, then we have |A| ≥ |B|, |A| ≤ |B|, so |A| = |B|. Then Θ isbijective. Therefore, we have a one - to - one correspondence between faces

of σ and faces of∨σ.

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Remark 1.2.11 Let σ be a convex polyhedral cone. For every vector v inσ, there exists a unique face τ < σ such that v is in the relative interior of τ.

In fact, let τ1 and τ2 be the faces of σ.Suppose that v is in the relative interior of the faces τ1 and τ2, then we

have

∨σ ∩ τ⊥1 =

∨σ ∩ τ⊥2 =

∨σ ∩ v⊥.

This means that two faces∨σ ∩ τ⊥1 and

∨σ ∩ τ⊥2 are the same, so we have

τ1 = τ2. (By applying the one - to-one correspondence between faces of σ

and faces of∨σ).

Remark 1.2.12 Let σ be a strongly convex cone in Rn, then dim∨σ = n.

1.3 Monoids

Definition 1.3.1 A semi-group (i.e. a non empty set S with an associativeoperation + : S × S → S ) is called a monoid if it is commutative, has azero element (0 + s = s, ∀s ∈ S) and satisfies the simplification law, i.e:

s+ t = s′+ t⇒ s = s

′for s, s

′, t ∈ S.

Lemma 1.3.2 If σ is a cone then σ ∩M is a monoid.

PROOF:• Let x and y be in σ ∩M , then x+ y = y + x is in σ ∩M .• The zero element 0 ∈ σ ∩M , and v + 0 = v for all v ∈ σ ∩M .• Since σ ∩M ⊂M , the operation satisfies the simplification law.

Definition 1.3.3 A monoid S is finitely generated if there are elementsa1, a2, ..ak in S such that

∀s ∈ S, s = λ1a1 + ...+ λkak where λi ∈ Z≥0.

Such elements a1, a2, ..ak are called generators of the monoid.

Lemma 1.3.4 (Gordon’s Lemma). If σ is a convex polyhedral latticecone, then σ ∩N is a finitely generated monoid.

PROOF:Let σ be a convex polyhedral lattice cone, generated by the vectors

(v1, ..., vr) such that vi ∈ σ ∩N for all i = 1, ...r.

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Consider the map

f : Rr → R

t = (t1, ..., tr) 7→r∑i=1

tivi

Set K = f([0; 1]r), so K is compact.Let x ∈ Rr and set

B(x, 1) = {y ∈ Rr|d(y, x) < 1}.

One hasK ⊂⋃x∈K

B(x, 1), so there exists n0 in N such thatK ⊂n0⋃i=1

B(xi, 1).

Since B(xi, 1) ∩N is a finite set, one has K ∩N is a finite set.Let us show that K ∩N and v1, ..., vr generates σ ∩N .

Let v =r∑i=1

λivi be in σ ∩N , then

v =r∑i=1

(ni + ri)vi =r∑i=1

nivi +r∑i=1

rivi ∈ N

where ni ∈ Z≥0, 0 ≤ ri ≤ 1.

Therefore, u =r∑i=1

rivi ∈ K ∩N.

So v = n1v1 + ...+ nrvr + u, and we obtain the result.

Proposition 1.3.5 Let σ be a convex polyhedral lattice cone, then∨σ ∩M

is a finitely generated monoid. We will denote∨σ ∩M by Sσ.

PROOF:By Proposition 1.1.1,

∨σ is also a convex polyhedral lattice cone.

By Lemma 1.3.4, Sσ is a finitely generated monoid.

Example 1.3.6 In Rn, consider the 0-dimensional cone σ = {0}.We have

∨σ = (Rn)∗, then the monoid Sσ =

∨σ ∩M = M . Hence, Sσ is

generated by the vectors (e∗1, ...e∗n,−e∗1, ....,−e∗n).

15

Example 1.3.7 In R2, let σ be the cone generated by (−2e1 + e2, e2).A

u

Cv w

a

j

−e∗1

∨σ

O

e∗1 + 2e∗22e∗2−e∗1 + 2e∗2

−e∗1 + e∗2 e∗2

−e∗1O

e2−2e1 + e2 −e1 + e2

σ

Fig. 4

• σ ∩N is generated by the vectors (−2e1 + e2, e2,−e1 + 2e2).

• Sσ =∨σ ∩M is generated by the vectors (−e∗1, e∗2, e∗1 + 2e∗2).

Proposition 1.3.8 Let σ be a rational convex polyhedral cone and τ =σ ∩ λ⊥ is a face of σ, with λ ∈ Sσ ∩M, then

Sτ = Sσ + Z≥0.(−λ).

PROOF:According to Proposition 1.2.6, we have

∨τ =

∨σ + R≥0(−λ).

By taking the intersection of both sides by M , give us

∨τ ∩M = (

∨σ + R≥0(−λ)) ∩M.

Since R≥0(−λ) ⊂M , one gets

Sτ = Sσ + Z≥0.(−λ).

Example 1.3.9 In Example 1.2.7

∨τ =

∨σ + R≥0(−e∗1).

Sτ is generated by the vectors (e∗1, , e∗2,−e∗1).

Sσ is generated by the vectors (e∗1, e∗2).

Therefore, we have

Sτ = Sσ + Z≥0.(−e∗1).

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2 Affine toric varieties

2.1 Laurent polynomials

Let us denote by C[z, z−1] = C[z1, ...zn, z−11 , ..., z−1

n ] the ring of Laurentpolynomials. A Laurent monomial is written by λ.za = λzα1

1 ...zαnn , whereλ ∈ C∗ and a = (α1, ..., αn) ∈ Zn.

One of the important facts in the definition of toric varieties, and the keyof the second step, that is the mapping

θ : Zn → C[z, z−1]

a = (α1, ..., αn) 7→ za = zα11 ...zαnn ,

which is an isomorphism between the additive group Zn and the multiplica-tive group of monic Laurent monomials. Monic means that the coefficient ofthe monomal is 1.

Proposition 2.1.1 The mapping

θ : Zn → C[z, z−1]

a = (α1, ..., αn) 7→ za = zα11 ...zαnn

is an isomorphism between the additive group Zn and the multiplicative groupof monic Laurent monomials.

PROOF:• θ is an homomorphism:

Let a = (α1, ..., αn), b = (β1, ..., βn) ∈ Zn, then

θ(a+b) = θ(α1 +β1, ..., αn+βn) = zα1+β11 ...zα1+βn

n = zα11 ...zαnn zβ11 ...z

βnn = za.zb

Therefore, θ(a+ b) = θ(a)θ(b).• θ is injective:

Let a = (α1, ..., αn) ∈ Zn. If θ(a) = 1 then α1 = ... = αn = 0, so a = 0,this means that θ is injective.• θ is surjective:

Each monic za = zα11 ...zα1

n in C[z, z−1], choose a = (α1, ..., αn) and wehave θ(a) = za, so θ is surjective.

Finally, we have θ is an isomorphism.

Definition 2.1.2 The support of a Laurent polynomial f =∑

finite λaza

is defined by

supp(f) = {a ∈ Zn : λa 6= 0}.

17

Proposition 2.1.3 Let σ be a lattice cone, the ring

Rσ = {f ∈ C[z, z−1] : supp(f) ⊂ Sσ}

is a finitely generated monomial algebra (i.e. is a C-algebra generated byLaurent monomials).

PROOF:According to Proposition 1.3.5, Sσ is a finitely generated monoid. Sup-

pose that Sσ is generated by the vectors (a1, ..., at) with ai ∈ Zn for all i.We prove that Rσ is generated by the monomials (za1 , ..., zat).Take a monic Laurent monomial za ∈ Rσ, then a ∈ Sσ, so there are

λ1, ..., λt ∈ Z≥0 such that

a = λ1a1 + ...+ λtat.

Hence

za = (za1)λ1 ...(zat)λt .

So for every monic Laurent monomial inRσ is generated by the monomials(za1 , ..., zat). If f ∈ Rσ then all its monic Laurent monomials belong to Rσ;therefore, we have the result.

2.2 Some basic results of algebraic geometry

Let C[ξ] = C[ξ1, ..., ξk] be the ring of polynomials in k variables over C.

Definition 2.2.1 Let E = {f1, ..., fr} ⊂ C[ξ], then

V (E) = {x ∈ Ck : f1(x) = ... = fr(x) = 0}

is called the affine algebraic set defined by E. Let I be the ideal generatedby E, then V (I) = V (E).

Definition 2.2.2 Let X ⊂ Ck, then

I(X) = {f ∈ C[ξ] : f|X = 0}

is an ideal, called the vanishing ideal of X.

18

Proposition 2.2.3 For x = (x1, ..., xk) in Ck, let us consider E = {ξ1 −x1, ..., ξk−xk}. Then V (E) = {x} and I({x}) = C[ξ](ξ1−x1)+ ...+C[ξ](ξk−xk) is a maximal ideal. We will denote I({x}) by Mx.

PROOF:• V (E) = {x ∈ Ck : ξ1 − x1 = ... = ξk − xk = 0} = {x}.• Mx = {f ∈ C[ξ] : f(x) = 0} is the ideal generated by E, then we have

Mx = C[ξ](ξ1 − x1) + ...+ C[ξ](ξk − xk).• The mapping

φ : C[ξ]→ Cf 7→ f(x)

is a surjective homomorphism.We have Kerφ = I({x}), hence C[ξ]�I({x}) ∼= C. Since C is a field,

then Mx is a maximal ideal.

Theorem 2.2.4 (Weak version of the Nullstellensatz): Every max-imal ideal in C[ξ] can be written by Mx for a point x.

Corollary 2.2.5 The correspondence x←→Mx is a one - to - one corre-spondence between points in Ck and maximal ideals M of C[ξ].

Ck ←→ {M ⊂ C[ξ] :M is a maximal ideal} =: Spec(C[ξ]).

Lemma 2.2.6 Let I be an ideal of C[ξ], then

V (I) = {x ∈ Ck|I ⊂Mx}.PROOF:Let y ∈ V (I). For every f ∈ I, we have f(y) = 0 , hence f ∈ My. And

then I ⊂My. Therefore, y ∈ {x ∈ Ck|I ⊂Mx}Conversely, let x ∈ Ck such that I ⊂ Mx, this means that f(x) = 0 for

all f ∈ I, then x ∈ V (I).

Definition 2.2.7 Let us denote the vanishing ideal of V (I) by IV = I(V (I)),then RV = C[ξ]/IV is the coordinate ring of the affine algebraic set V (I).It is generated as a C - algebra by the classes ξj of the coordinate function ξj.

The generators ξj = ξj + IV of RV are the restrictions of coordinate func-tions to the affine algebraic set V.

If I = {0}, then V (I) = Ck and we have RV = Ck.

19

Corollary 2.2.8 There is a one - to - one correspondence

V ←→ {M ⊂ RV |M is a maximal ideal} =: Spec(RV ).

By considering the Zariski topology on each side, we obtain an homeomor-phism

V ∼= Spec(RV ).

2.3 Affine toric varieties

Definition 2.3.1 The affine toric variety corresponding to a rational,polyhedral, strongly convex cone σ is Xσ := Spec(Rσ).

Example 2.3.2 In Example 1.3.7, let a1 = −e∗1, a2 = e∗2 and a3 = e∗1 + 2e∗2be a system of generators of Sσ.

The isomorphism θ is given by

θ : Z2 → C[z1, z2, z−11 , z−1

2 ]

a1 7→ z−11 = u1

a2 7→ z2 = u2

a3 7→ z1.z22 = u3

By Proposition 2.1.3,

Rσ = C[u1, u2, u3].

The relation between a1, a2 and a3 is 2a2 = a1 + a3, this provides therelation u2

2 = u1.u3 between u1, u2, u3 .Consider the mapping

i : C[ξ1, ξ2, ξ3]→ C[u1, u2, u3]

ξ1 7→ u1

ξ2 7→ u2

ξ3 7→ u3

For every f ∈ C[ξ1, ξ2, ξ3], if i(f) = 0 then f ∈ Ker(i) = C[ξ](ξ22 − ξ1.ξ3).

Indeed, i is surjective, hence, we have

Rσ = C[u1, u2, u3] ∼= C[ξ1, ξ2, ξ3]/Ker(i).

Therefore,

Xσ = V (Ker(i)) = {ξ = (ξ1, ξ2, ξ3) ∈ C3|ξ22 = ξ1.ξ3}.

20

Exercise 2.3.3 In Rn, let σ be a rational, polyhedral, strictly convex cone.How to find the affine toric variety Xσ?

SOLUTION:Step 1: Find a system of generators of Sσ.By Proposition 1.3.5, suppose that Sσ is generated by (a1, ..., ak), where

ai = (α1i , ..., α

ni ).

Step 2: Find the relations between coordinates.By the isomorphism θ, we obtain monic Laurent monomials ui = zα

1i ...zα

ni ∈

C[z, z−1] for i = 1, ..., k. The C -algebra Rσ = C[u1, ..., uk] can be representedby

Rσ = C[ξ1, ..., ξk]/Iσ

for some ideal Iσ that we must determine.Find all linear relations between generators of Sσ. (The number of linear

relations between them is finite, see Exercise 2.3.4). Each linear relation canbe written as

k∑j=1

νjaj =k∑j=1

µjaj νj, µj ∈ Z≥0.

We obtain the relation between coordinates

uν11 ...uνkk = uµ11 ...u

µkk ,

and finally we have the binomial relation

ξν11 ...ξνkk = ξµ11 ...ξµkk .

Step 3: Iσ is generated by all binomial relations and Xσ = V (Iσ).

Exercise 2.3.4 Let v1, ..., vk ∈ Zn, find all α1, ..., αk ∈ R such that

α1v1 + ....+ αkvk = 0.

SOLUTION:Suppose that (e1, ..., en) is the basis of Rn, where e1 = (1, 0, .., 0), ..., en =

(0, ..., 0, 1) ∈ Zn.There are aji ∈ Z ( for j = 1, ..., n and i = 1, ..., k) such that

vi = a1ie1 + ...+ anien for i = 1, ..., k.

By

21

α1v1 + ....+ αkvk = 0,

we have

e1

k∑i=1

a1iαi + ...+ en

k∑i=1

aniαi = 0,

if and only if

k∑i=1

a1iαi = ... =k∑i=1

aniαi = 0,

if and only if (α1, ..., αk) ∈ KerA , where A is the matrix

A =

a11 ... a1k

. .

. .

. .an1 ... ank

.KerA is a linear space, dim(KerA) = n − rank(A) = d. We can get a

basis (u1, ..., ud) of KerA, where ui ∈ Zn for all i = 1, ..., d.

Theorem 2.3.5 Let σ be a lattice cone in Rn and A = (a1, ..., ak) a systemof generators of Sσ, the corresponding toric variety Xσ is represented by theaffine toric variety V (Iσ) ⊂ Ck where Iσ is an ideal of C[ξ1, ..., ξk] generatedby finitely many binomials corresponding to the relations between elements ofA.PROOF:

By Exercise 2.3.4, the number of binomial relations corresponding torelations between elements of A is finite. For each binomial relation, we haveone binomial.

In the rest of the proof, we show that every element of Iσ is a sum ofbinomials of the previous type. (See [4], Theorem VI.2.7.)

Example 2.3.6 Let us consider the cone σ = {0}, the dual cone is∨σ =

(Rn)∗. We can choose different systems of generators of Sσ, for example

A1 = (e∗1, ..., e∗n,−e∗1, ...,−e∗n),

A2 = (e∗1, ..., e∗n,−(e∗1 + ...+ e∗n)).

Let us take the first system of generators A1. The corresponding mono-mial C -algebra is

22

Rσ = C[z1, ..., zn, z−11 , ..., z−1

n ] = C[ξ1, ..., ξ2n]/Iσ,

where

Iσ = C[ξ](ξ1.ξn+1 − 1) + ...+ C[ξ](ξn.ξ2n − 1).

Therefore

Xσ = V (ξ1.ξn+1 − 1, ..., ξn.ξ2n − 1).

We set

T = {(x1, ..., xn) ∈ Cn|xi 6= 0, i = 1, ..., n} = (C∗)n.

Consider the projection

π : C2n → Cn

(x1, ..., x2n) 7→ (x1, ..., xn).

So Xσ is homeomorphic to T.For the second system of generators A2, we have

Rσ = C[z1, ..., zn, z−11 ...z−nn ] = C[ξ1, ..., ξn+1]/Iσ,

where

Iσ = C[ξ](ξ1...ξn+1 − 1).

In this case,

Xσ = V (ξ1...ξn+1 − 1) ⊂ Cn+1

Consider the projection

π : Cn+1 → Cn

(x1, ..., xn+1) 7→ (x1, ..., xn).

Hence Xσ is also homeomorphic to T.

Definition 2.3.7 The set T = (C∗)n is called the complex algebraicn-torus.

We have dim T = n.

Remark 2.3.8 T is a closed subset of C2n, but as a subspace of Cn, it isnot closed.

23

Proposition 2.3.9 Let σ be a lattice cone in Rn, then affine toric varietyXσ contains the torus T as a Zariski open dense subset.

PROOF:Let (a1, ..., ak) be a system of generators for the monoid Sσ and let

V (Iσ) ⊂ Ck be a representation of Xσ. Each ai is written by ai = (α1i , ..., α

ni )

with αji ∈ Z, and t = (t1, ..., tn) ∈ T = (C∗)n with ti 6= 0 for i = 1, ..., n.Consider the relation

k∑j=1

νjaj =k∑j=1

µjaj νj, µj ∈ Z≥0.

For every t ∈ T, let tai = tα1i

1 ...tαnin ∈ C∗, we have

(ta1)ν1 ...(tak)νk = t(ν1a1+...+νkak) = t(µ1a1+...+µkak) = (ta1)µ1 ...(tak)µk .

This means that (ta1 , ..., tak) satisfies the binomial relation

ξν11 ...ξνkk = ξµ11 ...ξµkk .

Consider the embedding

h : T→ Xσ

t = (t1, ..., tn) 7→ (ta1 , ..., tak)

We prove that h is bijective from T to Xσ ∩ (C∗)k.

Firstly, we need to prove that there is b ∈ Sσ such that all points b + e∗iare in Sσ.

If σ = {0}, then it is obvious.Assume that σ 6= {0} is generated by (v1, ..., vr).Since σ is strongly convex cone, there is b1 ∈ Sσ such that 〈b1, vj〉 > 0 for

all j = 1, ..., r.Let us fix i ∈ {1, ..., n}.If bi + e∗i ∈ Sσ then bi+1 = bi. We have 〈bi+1, vj〉 > 0 for all j = 1, ..., r.If bi + e∗i /∈ Sσ, for every vj such that 〈bi + e∗i , vj〉 < 0, then 〈bi, vj〉 > 0

and 〈e∗i , vj〉 < 0, then there exists kj ∈ Z>0 such that 〈kj.bi + e∗i , vj〉 > 0. Letni = max{kj}, then ni.bi+e∗i ∈ Sσ, and 〈ni.bi+e∗i , vj〉 > 0 for all j = 1, ..., r.In this case we let bi+1 = ni.bi + e∗i .

Choosing b = bn, we have the result.

The Laurent monomials zb = f0(u), zb+e∗i = fi(u) are in Rσ = C[u] ⊂

C[z, z−1].

24

Let h(t) = x be a point in Xσ ∩ (C∗)k, then fi(h(t)) = tif0(h(t)), we haveti = fi(h(t))/f0(h(t)). Therefore h is injective .

For every x ∈ Xσ ∩ (C∗)k, we have

x = h((f1(x)/f0(x), ..., fn(x)/f0(x)).

Hence h is surjective from T to Xσ ∩ (C∗)k.Finally, since Xσ ∩ (C∗)k is dense in Xσ, one gets h(T) is dense in Xσ.

Then affine toric variety Xσ contains the torus T as a Zariski open densesubset.

Remark 2.3.10 If σ is a rational, polyhedral, strictly convex cone in Rnthen dimCXσ = n.

By Proposition 2.3.9 and dimCXσ is finite, then dimCXσ = dimT = n.

Example 2.3.11 In the case of Example 1.3.7, let a1 = −e∗1, a2 = e∗2 anda3 = e∗1 + 2e∗2, then

h : T→ Xσ

t = (t1, t2) 7→ (t−11 , t2, t1t

22)

Example 2.3.12 Let σ be the cone in Rn generated by (e1, ..., ep) withp < n. Then Sσ is generated by (e∗1, ..., e

∗n,−(e∗p+1 + ...+ e∗n)).

Rσ = C[z1, ..., zn, z−1p+1...z

−1n ] = C[ξ1, ..., ξn+1]/Iσ,

where

Iσ = C[ξ](ξp+1...ξn+1 − 1).

Therefore

Xσ = V (Iσ) = Cp × (C∗)n−p.

If p = n then Xσ = Cn.Xσ is smooth for all p ≤ n.

Remark 2.3.13 Let us denote NR = N ⊗Z R. A lattice homomorphismϕ : N ′ → N defines a homomorphism of real vector spaces ϕR : N

′

R → NR.Assume that ϕR maps a ( polyhedral, rational, strictly convex) cone σ′ of N ′

to a (polyhedral, rational, strictly convex) cone σ of N. Then the dual map∨ϕ : M → M ′ provides a map Sσ → Sσ′ . It defines a map Rσ → Rσ′ and amap Xσ′ → Xσ.

We apply this remark to an example.

25

Example 2.3.14 This is the example of a 2-dimensional affine toric variety.Let us consider in R2 the cone generated by e2 and pe1− qe2, for integers

p, q ∈ Z>0 such that 0 < q < p and (p, q) = 1.Then Rσ = ⊕C(zi1z

j2), the sum over (i, j) with j ≤ p

qi. Let N ′ be the

sublattice of N generated by pe1 − qe2 and e2, i.e. by pe1 and e2. Let us callσ′ the cone σ considered in N ′ instead of N. Then σ′ is generated by twogenerators of the lattice N ′, so Xσ′ is C2.

In this situation, the inclusion N ′ ⊂ N provides a map Xσ′ → Xσ (Re-mark 2.3.13). The map can be made explicitly in the following way:

Let us denote by x and y the monomials corresponding to the generatorse∗1 and e∗2 of the dual lattice M. The dual lattice M ′ ⊃ M correspondingto N ′ is generated by 1

pe∗1 and e∗2. The monomials corresponding to these

generators are u and y such that up = x. The monoid Sσ′ is generated by1pe∗1 and q

pe∗1 + e∗2, then

Rσ′ = C[u, uqy] = C[u, v] with v = uqy.

On the other hand,

Rσ = ⊕C[xiyj] = ⊕C[upi−qjvj] , the sum over (i, j) with j ≤ pqi.

Consider the group of p− th roots of unity

Γp = {z ∈ C|zp − 1 = 0} ∼= Z/pZ

Consider the group action ϕ of Γp on Rσ = C[u, v] given by

ϕ : Γp × C[u, v]→ C[u, v]

(ζ, f) 7→ ζ.f = f(ζu, ζqv)

Set

C[u, v]Γp = {f ∈ C[u, v]|ζ.f = f, ∀ζ ∈ Γp}.

We have

(upi−qjvj)(ζu, ζqv) = ζpiupi−qjvj = upi−qjvj for all j ≤ pqi.

This shows that

Rσ ⊂ C[u, v]Γp .

For any utvk ∈ C[u, v], t, k ∈ Z≥0, we have

utvk(ζu, ζqv) = ζt+kqutvk.

26

Since (p, q) = 1, one has

t+ kq ≡ 0 mod(p) if and only if t = pi− kq for some i.

Since t ≥ 0, we have k ≤ pqi, hence utvk = upi−qkvk in Rσ.

This implies that

Rσ = C[u, v]Γp

Consider the group action ϕ of Γp on Xσ′ .

ϕ : Γp ×Xσ′ → Xσ′

(ζ, (u, v)) 7→ (ζu, ζqv)

Then Xσ′/Γp is the quotient of the action, that is the set of all orbits ofXσ′ under the action of Γp. The orbit of (u, v) is

Γp.(u, v) = {(ζu, ζqv)|ζ ∈ Γp}.

We will show that Xσ = Xσ′/Γp.The inclusion Rσ ⊂ Rσ′ induces a map

θ : Spec(Rσ′)→ Spec(Rσ)

M 7→M∩Rσ

which is surjective.For every (u0, v0), (u1, v1) ∈ C2, we have the maximal ideals M0,M1 ∈

Spec(Rσ′) corresponding to (u0, v0), (u1, v1). One has

M0 ∩Rσ = {f ∈ Rσ|f(u0, v0) = 0}

Let f = up − up0 ∈M0 ∩Rσ, g = up−qv − up−q0 v0 ∈M0 ∩Rσ.Assume that M0 ∩ Rσ = M1 ∩ Rσ, this means that f(u1, v1) = 0 and

g(u1, v1) = 0, hence

up1 = up0 and up−q1 v1 − up−q0 v0 = 0.

Therefore there is ζ ∈ Γp such that u1 = ζu0, and

ζp−qup−q0 v1 − up−q0 v0 = 0.

If u0 6= 0 then

ζp−qv1 = v0.

27

This shows that v1 = ζqv0.If u0 = 0, let h = vp − vp0 ∈ M0 ∩ Rσ, one has vp1 = vp0, then there is

ζ1 ∈ Γp such that v1 = ζp1v0.In two cases (u0 = 0 or u0 6= 0), we have (u1, v1) = (ζu0, ζ

qv0) for someζ ∈ Γp .

This implies that (u1, v1) is in the orbit of (u0, v0).Conversely, suppose that (u, v) and (ζu, ζqv) correspond to the maximal

ideals M, ζM∈ Spec(Rσ′). Since Rσ = C[u, v]Γp , we have

M∩Rσ = {f ∈ Rσ|f(u, v) = 0} = {f ∈ Rσ|f(ζu, ζqv) = 0} = ζM∩Rσ.

Finally we have the result Xσ∼= Xσ′/Γp = C2/Γp.

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3 Toric varieties

3.1 Fans

Definition 3.1.1 A fan ∆ in the space Rn is a finite union of cones suchthat

i) Every cone of ∆ is a strongly convex, polyhedral, rational cone.

ii) Every face of a cone of ∆ is a cone of ∆.

iii) If σ and σ′ are the cones of ∆, then σ∩σ′ is a common face of σ and σ′.

In the following, unless specified, all cones we will consider will be poly-hedral, rational cones.

Example 3.1.2 Example of fans

e2

−e1 − e2 e1 − e2

−e1 + e2 e1 + e2

e1

σ1

σ2

OO

σ3

σ4

σ5

Fig. 5

Example 3.1.3 Let us denote by (t0, t1, t2) the homogeneous coordinatesof the space P2. Let z1 = t1/t0, z2 = t2/t0.P2 has three coordinate chartsU0 = {(t0 : t1 : t2) ∈ P2|t0 6= 0} ∼= C2

(z1,z2), corresponding to the algebra

C[z1, z2],U1 = {(t0 : t1 : t2) ∈ P2|t1 6= 0} ∼= C2

(z−11 ,z−1

1 z2), corresponding to the

algebra C[z−11 , z−1

1 z2],U2 = {(t0 : t1 : t2) ∈ P2|t2 6= 0} ∼= C2

(z−12 ,z1z

−12 )

, corresponding to the

algebra C[z−12 , z1z

−12 ].

Let us consider in R2 the following fan:

29

e2

−e1 − e2

−e∗1 + e∗2

e∗1

σ1

e1

−e∗2

e∗2

e∗1 − e∗2

−e∗1

∨σ0

∨σ2

∨σ1

σ0

σ2

O O

Fig. 6

i) Sσ0 is generated by (e∗1, e∗2), hence Rσ0 = C[z1, z2], then Xσ0 = C2

(z1,z2) =U0.

ii) Sσ1 is generated by (−e∗1,−e∗1 + e∗2), hence Rσ0 = C[z−11 , z−1

1 z2], thenXσ1 = C2

(z−11 ,z−1

1 z2)= U1.

iii) Sσ2 is generated by (−e∗2, e∗1−e∗2), hence Rσ2 = C[z−12 , z1z

−12 ], then Xσ2 =

C2(z−1

2 ,z1z−12 )

= U2.

We have σ0 ∩ σ1 = τ , which is the cone generated by e2.Let us glue Xσ0 and Xσ1 along Xτ .We have

Sτ = Sσ0 + Z≥0(−e∗1) = Sσ1 + Z≥0(e∗1).

Then Xτ = C∗z1 ×Cz2 in Xσ0 and Xτ = C∗z−11

×Cz−11 z2

in Xσ1 . So we have

Xσ0�(z1 = 0) ∼= Xτ∼= Xσ1�(z−1

1 = 0).

One has

Xτ = {(t0 : t1 : t2) ∈ P2|t0 6= 0, t0 6= 0}

which is a subset of U0 and U1. Since U0 ∩ U1 = Xτ , the gluing Xσ0 and Xσ1

along Xτ is U0 ∪ U1 = P2 \ {(0 : 0 : 1)}.

30

3.2 Toric varieties

In a general way, let τ be a face of a cone σ, then∨τ =

∨σ + R≥0(−λ) where

λ ∈ ∨σ ∩M and τ = σ⋂λ⊥ (Proposition 1.2.6).

The monoid Sτ is thus obtained from Sσ by adding one generator −λ. Asλ can be chosen as an element of a system of generators (a1, ..., ak) for Sσ.We may assume that λ = ak is the last vector in the system of generators ofSσ and we denote ak+1 = −λ. In order to obtain the relationships betweenthe generators of Sτ , one has to consider previous relationships between thegenerators (a1, ..., ak) of Sσ and the supplementary relationship ak+ak+1 = 0.

This relationship corresponds to the multiplicative one ukuk+1 = 1 andthat is the only supplementary relationship we need in order to obtain Rτ

from Rσ. As the generators ui are precisely the coordinate function on thetoric varieties Xσ and Xτ , this mean that the projection

Ck+1 → Ck

(x1, ..., xk, xk+1) 7→ (x1, ..., xk)

identifies Xτ with the open subset of Xσ defined by xk 6= 0. That can bewritten as follows.

Lemma 3.2.1 There is a natural identification Xτ∼= Xσ \ (uk = 0).

Remark 3.2.2 Let us suppose that τ is the common face of two cones σand σ′. Lemma 3.2.1 allows us to glue together Xσ and Xσ′ along Xτ . Thisis performed in the following way:

Let us write (v1, ..., vl) the coordinates on Xσ′. By Lemma 3.2.1, there isa homeomorphism Xτ

∼= Xσ′ \ (ul = 0), we obtain the gluing map

ψσ,σ′ : Xσ \ (uk = 0)∼=→ Xτ

∼=→ Xσ′ \ (vl = 0).

Definition 3.2.3 (Toric varieties) Let ∆ be a fan in Rn. Consider the dis-

joint union∐σ∈∆

Xσ where two points x ∈ Xσ and x′ ∈ Xσ′ are identified if

ψσ,σ′(x) = x′. The resulting space X∆ is called a toric variety. It is topo-logical space endowed with an open covering by the affine toric varieties Xσ

for σ ∈ ∆. It is an algebraic variety whose charts are defined by binomial

31

relations.

In fact, we have shown that , for a face τ of a cone σ, one has inclusions:

τ ↪→ σ∨τ ←↩ ∨σ

Rτ ←↩ Rσ

Xτ ↪→ Xσ.

Proposition 3.2.4 Let ∆ be a fan in Rn. Consider the disjoint union∐σ∈∆

Xσ. We write x ∼ x′ if ψσ,σ′(x) = x′ for some σ, σ′ ∈ ∆. Then the

relation ∼ is an equivalence relation.

PROOF:• Reflexivity:

ψσ,σ : Xσ

∼=→ Xσ

∼=→ Xσ

So we have x ∼ x.• Symmetry: If x ∼ x′ ,then there is the map

ψσ,σ′ : Xσ \ (uk = 0)∼=→ Xτ

∼=→ Xσ′ \ (vl = 0)

such that ψσ,σ′(x) = x′. Therefore we have ψσ′,σ(x′) = x, then x′ ∼ x.• Transitivity: If x ∼ x′ and x′ ∼ x′′, there are the maps

ψσ,σ′ : Xσ \ (uk = 0)∼=→ Xτ

∼=→ Xσ′ \ (vl = 0)

ψσ′,σ′′ : Xσ′ \ (vm = 0)∼=→ Xτ ′

∼=→ Xσ′′ \ (sr = 0)

hence

ψτ,τ ′ : Xτ \ (vm = 0)∼=→ Xσ′ \ (vm = 0, vl = 0)

∼=→ Xτ ′ \ (vl = 0)

and ψτ,τ ′(x) = x′′.

Remark 3.2.5 X∆ =∐σ∈∆

Xσ/ ∼.

32

Proposition 3.2.6 Every n−dimensional toric variety contains the torusT = (C∗)n as an Zariski open dense subset.PROOF:

The torus T corresponds to the zero cone, which is a face of every σ ∈ ∆.(i.e. T = X{0} ). The embedding of the torus into every affine toric varietyXσ has been shown in Proposition 2.3.9. By the previous identifications, allthe tori corresponding to affine toric varieties Xσ in X∆ are identified as anopen dense subset in X∆.

Example 3.2.7 Consider the following fan:

O

e2

e1

−e2

−e1

O

e∗2

e∗1

−e∗1

−e∗2

σ0σ1

σ2σ3

∨σ1

∨σ0

∨σ2

∨σ3

Fig. 7We have that

Sσ0 is generated by (e∗1, e∗2), hence Rσ0 = C[z1, z2] and Xσ0 = C2

(z1,z2),

Sσ1 is generated by (−e∗1, e∗2), hence Rσ1 = C[z−11 , z2] and Xσ1 = C2

(z−11 ,z2)

,

Sσ2 is generated by (−e∗1,−e∗2), hence Rσ2 = C[z−11 , z−1

2 ] and Xσ2 = C2(z−1

1 ,z−12 )

,

Sσ3 is generated by (e∗1,−e∗2), hence Rσ3 = C[z1, z−12 ] and Xσ3 = C2

(z1,z−12 )

.

The face τ1 = σ0 ∩ σ1 , which is generated by e2, then Xτ1 = C∗z1 ×Cz2 inXσ0 and Xτ1 = C∗

z−11

× Cz2 in Xσ1 .

We have Xσ0 = U0 × Cz2 and Xσ1 = U1 × Cz2 where U0 = {(t0 : t1) ∈P1|t0 6= 0}, and U1 = {(t0 : t1) ∈ P1|t1 6= 0}.

Let U = {(t0 : t1) ∈ P1|t0 6= 0, t1 6= 0} ∼= C∗.One has Xτ1 = U × Cz2 = Xσ0 ∩ Xσ1 in P1 × Cz2 . Therefore the gluing

Xσ0 and Xσ1 along Xτ1 is P1 × Cz2 .The face τ2 = σ2 ∩ σ3. Similarly, we have the gluing Xσ2 and Xσ3 along

Xτ2 is P1 × Cz−12

.

Finally, we glue P1 × Cz2 and P1 × Cz−12

together along P1 × U, we have

P1 × P1.

33

4 The torus action and the orbits

4.1 The torus action

Definition 4.1.1 The torus T = (C∗)n is a group operating on itself bymultiplication. The action of the torus on each affine toric variety Xσ isdescribed as follows:

Let (a1, ..., ak) be a system of generators for the monoid Sσ. For theprevious coordinates of Rn, each ai is written by ai = (α1

i , ..., αni ) with α1

i ∈ Z,and t ∈ T is written by t = (t1, ..., tn) where tj ∈ C∗. A point x ∈ Xσ iswritten by x = (x1, ..., xk) ∈ Ck. The action of T on Xσ is given by

T×Xσ → Xσ

(t, x) 7→ t.x = (ta1x1, ..., takxk)

where tai = tα1i

1 ...tαnin ∈ C∗.

Let x = (x1, ..., xk) and y = (y1, ..., yk) in Ck and satisfy the binomialrelation

ξν11 ...ξνkk = ξµ11 ...ξµkk .

Since

(x1y1)ν1 ...(xkyk)νk = (x1)ν1 ...(xk)

νk(y1)ν1 ...(yk)νk =

(x1)µ1 ...(xk)µk(y1)µ1 ...(yk)

µk = (x1y1)µ1 ...(xkyk)µk ,

then xy = (x1y1, ..., xkyk) also satisfies this binomial relation.Therefore, if x and y are in Xσ then xy is also in Xσ.By Proposition 2.3.9, for every t = (t1, ..., tn) ∈ T, one has (ta1 , ..., tak) ∈

Xσ. Then t.x ∈ Xσ for all x ∈ Xσ.

Example 4.1.2 In the case of Example 1.3.7, let a1 = −e∗1, a2 = e∗2 anda3 = e∗1 + 2e∗2. The action of T on Xσ is the map

T×Xσ → Xσ

(t, x) = ((t1, t2), (x1, x2, x3)) 7→ (t−11 x1, t2x2, t1t

22x3).

Remark 4.1.3 Let τ be a face of σ. By Lemma 3.2.1, we can suppose thatSσ is generated by (a1, ..., ak) , Sτ is generated by (a1, ..., ak,−ak) and thereis a natural identification

p : Xτ

∼=→ Xσ \ (xk 6= 0)

(x1, ..., xk, xk+1) 7→ (x1, ..., xk).

34

The action of T on Xσ is the map

T×Xσ → Xσ

(t, x) 7→ t •σ x = (ta1x1, ..., takxk).

The action of T on Xτ is the map

T×Xτ → Xτ

(t, x) 7→ t •τ x = (ta1x1, ..., takxk, t

−akxk+1).

Hence, for every x = (x1, ..., xk) ∈ Xσ \ (xk 6= 0), one has p−1(x) =(x1, ..., xk, x

−1k ). Therefore

p−1(t •σ x) = (ta1x1, ..., takxk, t

−akx−1k ) = t •τ p−1(x).

Then t •σ x = p(t •τ p−1(x)).For every x = (x1, ..., xk, xk+1) ∈ Xτ , we have

p(t •τ x) = (ta1x1, ..., takxk) = t •σ p(x).

Theorem 4.1.4 Let ∆ be a fan in Rn, the torus action on the affine toricvarieties, for σ ∈ ∆, provide a torus action on the toric variety X∆.

PROOF:Suppose that x and x′ are identified, then there are σ and σ′ in ∆ such

that ψσ,σ′(x) = x′, where ψσ,σ′ is the gluing map. We also have

ψσ,σ′ : Xσ \ (uk = 0)p−11→ Xτ

p2→ Xσ′ \ (vl = 0), τ = σ ∩ σ′.

Then ψσ,σ′(x) = p2(p−11 (x)) = x′.

By Remark 4.1.3, we havet •σ′ x′ = t •σ′ (p2(p−1

1 (x))) = p2(t •τ p−11 (x)) = p−1

2 p1(t •σ x) = ψσ,σ′(t •σ x).This shows that t •σ′ x′ and t •σ x are identified in X∆.

4.2 Orbits

Let us consider the case ∆ = {0}, then X∆ = (C∗)n is the torus. There isonly one orbit which is the total space X∆ and is the orbit of the point whosecoordinates ui are (1, ..., 1) in Cn.

In the general case, the apex σ = {0} of ∆ provides an open dense orbitwhich is the embedded torus T = (C∗)n (Proposition 3.2.6). Let us describethe other orbits.

There is a correspondence (see Corollary 2.2.5)

35

Ck ←→ {M ⊂ C[ξ] :M maximal ideal} ←→ HomC−alg(C[ξ],C).

With this correspondence, the point x = (x1, ..., xk) corresponds to theideal Mx = C[ξ](ξ1 − x1) + ...+ C[ξ](ξk − xk) and to the homomorphismϕ : C[ξ]→ C such that Kerϕ =Mx, (i.e. ϕ(f) = f(x)).

If I is an ideal in C[ξ], then V = V (I) = {x ∈ Ck : I ⊂ Mx} andIV = I(V (I)). The set V is an affine algebraic set whose coordinate ring isRV = C/IV and we have the correspondence (see Corollary 2.2.8 )

V ←→ {M ⊂ RV :M maximal ideal} ←→ HomC−alg(RV ,C).

As a semi- group, the dual lattice M is generated by (e∗1, ..., e∗n,−e∗1, ..., e∗n)

and the Laurent polynomial ring C[M ] is generated by (z1, ..., zn, z−11 , ..., z−nn ).

We have identifications

T = Spec(C[M ]) ∼= Hom(M,C∗) ∼= (C∗)n

where N ∼= Hom(M,C) and Hom(M,C) are group homomorphisms.

All semi-groups Sσ =∨σ ∩M are semi-groups of the lattice M and C[Sσ]

is a sub - algebra of C[M ]. These sub-algebras are generated by monomialsin variables ui.

If Sσ is generated by (a1, ..., ak), then elements ui = zai , i = 1, ..., k, aregenerators of the C-sub-algebra C[Sσ], with multiplication zaza

′= za+a′ and

z0 = 1.

Remark 4.2.1 Points of Spec(C[Sσ]) correspond to homomorphisms ofsemi-groups of Sσ in C where C = C∗ ∪ {0} is an abelian semi-group viamultiplication:

Xσ = Spec(C[Sσ]) ∼= Homsg(Sσ,C)

(semi-group homomorphisms). If ϕ ∈ Homsg(Sσ,C), the point x correspond-ing to ϕ satisfies ϕ(a) = za(x) (evaluation in x) for all a ∈ Sσ. This meansthat ϕ(ai) is the i− th coordinate of x, i.e. x = (ϕ(a1), ..., ϕ(an)) ∈ Ck.

The action of T on Xσ can be interpreted in the following way:

t ∈ T is identified with the group homomorphism Mt→ C∗, and

x ∈ Xσ is identified with the group homomorphism Sσx→ C, then

t.x ∈ Xσ is identified with the group homomorphism Sσt.x→ C, u 7→ t(u).x(u).

Indeed, we have

36

t.x = (ta1x1, ..., takxk),

then (t.x)(ai) = taixi = t(ai).x(ai), (by t(ai) = tai and x(ai) = xi).

Hence, for every u =k∑i=1

λiai ∈ Sσ, where λi ∈ Z≥0 , one has

(t.x)(u) = (ta1x1)λ1 ...(takxk)λk = tλ1a1+...+λkak .x(λ1a1)...x(λkak) =

tux(λ1a1 + ...+ λkak) = t(u).x(u).

Definition 4.2.2 Distinguished points. Let σ be a cone and Xσ theassociated affine toric variety. We associate to each face τ of σ a distin-guished point xτ corresponding to the semi-group homomorphism defined ongenerators a of Sσ by

ϕτ (a) =

{1 if a ∈ τ⊥

0 in other cases.

Exercise 4.2.3 Prove that ϕτ : Sσ → C is a semi-group homomorphism.

SOLUTION:Let a, b ∈ Sσ, consider two cases:If a, b ∈ τ⊥ = {u ∈ (Rn)∗|〈u, v〉 = 0,∀v ∈ τ}, then (a + b) ∈ τ , by the

definition of ϕτ , we have

ϕτ (a+ b) = ϕτ (a) = ϕτ (b) = ϕτ (a)ϕτ (b) = 1.

If a /∈ τ⊥, then there is v ∈ τ such that 〈a, v〉 > 0, hence 〈a + b, v〉 > 0(since 〈b, v〉 ≥ 0) , this means that a+ b /∈ τ⊥. Then we have

ϕτ (a+ b) = ϕτ (a) = ϕτ (a)ϕτ (b) = 0.

Therefore, we have the result.

37

Example 4.2.4 In the case of Example 1.3.7, the generators of Sσ area1 = −e∗1, a2 = e∗2 and a3 = e∗1 + 2e∗2. A

u

Cv w

a

j

−e∗1

∨σ

O

e∗1 + 2e∗22e∗2−e∗1 + 2e∗2

−e∗1 + e∗2 e∗2

−e∗1O

τ1 a2

a1

a3

τ2

e2−2e1 + e2 −e1 + e2

σ

Fig. 8

The cone σ has four faces {0}, τ1 is generated by e2, τ2 is generated by−2e1 + e2, and σ.

We have {0}⊥ = (R2)∗, τ⊥1 = Ra1, τ⊥2 = Ra3, σ⊥ = {0}.

The distinguished points:

x{0} = (ϕ{0}(a1), ϕ{0}(a2), ϕ{0}(a3)) = (1, 1, 1),

xτ1 = (ϕτ1(a1), ϕτ1(a2), ϕτ1(a3)) = (1, 0, 0),

xτ2 = (ϕτ2(a1), ϕτ2(a2), ϕτ2(a3)) = (0, 0, 1),

xσ = (ϕτ2(a1), ϕτ2(a2), ϕτ2(a3)) = (0, 0, 0).

Definition 4.2.5 Let σ be a cone in Rn and τ a face of σ. The orbit of Tin Xσ corresponding to the face τ is the orbit of the distinguished point xτ ,we denote by Oτ .

Oτ = {t.xτ |t ∈ T = (C∗)n}.

Example 4.2.6 In Example 4.2.5, for each distinguished point, we haveO{0} = {t.x{0}|t = (t1, t2) ∈ T = (C∗)2} = {(t−1

1 , t2, t1t22)|(t1, t2) ∈ (C∗)2},

then

O{0} ∼= (C∗)2.

Oσ = {(0, 0, 0},Oτ1 = C∗ξ1 × {0} × {0},Oτ2 = {0} × {0} × C∗ξ3 .

Consider the disjoint union Oσ

∐Oτ1

∐Oτ2

∐O{0}, we have

38

Oσ

∐Oτ1

∐Oτ2

∐O{0} = Xσ = {(x1, x2, x3) ∈ C3|x1x3 = x2

2}.

Indeed, suppose x = (x1, x2, x3) ∈ Xσ.If x2 6= 0 then x ∈ O{0}.If x2 = 0 then x1 = 0 or x3 = 0, therefore x ∈ Oσ

∐Oτ1

∐Oτ2 .

Proposition 4.2.7 Let σ be a cone in Rn and Oσ be the orbit of the dis-tinguished point xσ. If dimσ = k then

Oσ∼= (C∗)n−k.

PROOF:If dimσ = n then σ⊥ = {0}. Hence xσ = (0, ..., 0) ∈ Xσ, then Oσ = {0}.If σ = {0} then xσ = (1, ..., 1) ∈ Xσ, then Oσ = T = TN.If dimσ < n, let Nσ be the sublattice of N , which is generated by σ ∩N ,

then

Nσ = (σ ∩N) + (−σ ∩N).

Then we have a decomposition

N = Nσ

⊕N ′, σ = σ′

⊕{0},

where σ′ is in a cone in Nσ ⊂ Rk. Then dimN ′R = n− k.The dual decomposition M = M ′⊕M ′′, then dimM ′′

R = n− k . One has

Sσ = ((σ′)∨ ∩M ′))⊕

M ′′,

Xσ = Xσ′ × TN ′ .

Consider two toric actions on Xσ′ and TN ′ :

TNσ ×Xσ′ → Xσ′ ,

TN ′ × TN ′ → TN ′ .

By T = TNσ × TN ′ , we have toric action on Xσ

T× (Xσ′ × TN ′)→ Xσ′ × TN ′ .

Since σ = σ′⊕{0}, then xσ = xσ′ × x{0} ∈ Xσ′ × TN ′ . Hence

Oσ = Oσ′ ×O{0} ⊂ Xσ′ × TN ′ .

We have dim(σ′) = dim(Nσ)R = k, then Oσ′ = {0} ⊂ Xσ′ . Then

Oσ∼= TN ′ = (C∗)n−k.

39

Remark 4.2.8 In Proposition 4.2.7, we can suppose that the lattice Nσ

is generated by (e1, ..., ek), then N ′ is generated by (ek+1, ..., en). Hence thelattice M ′′ is generated by (e∗k+1, ..., e

∗n), then

M ′′ = τ⊥ ∩M .

Hence,

Oσ∼= TN ′ = Homsg(τ

⊥ ∩M,C∗).

Theorem 4.2.9 Let σ be a cone in Rn, then

Xσ =∐τ<σ

Oτ ,

where Oτ is the orbit of the distinguished point xτ .

PROOF:We have Oτ ⊂ Xσ for all τ < σ, then

Xσ ⊃∐τ<σ

Oτ .

Conversely, for every x ∈ Xσ, x corresponds to a semi-homomorphismϕ : Sσ → C such that ϕ(a) = za(x). Then ϕ(0) = 1, this shows thatϕ−1(C∗) 6= ∅.

Let a, b ∈ Sσ such that (a + b) ∈ ϕ−1(C∗), so ϕ(a + b) ∈ C∗, thenϕ(a)ϕ(b) ∈ C∗, hence a and b ∈ ϕ−1(C∗). Then

ϕ−1(C∗) =∨σ ∩ τ⊥ ∩M for some τ < σ.

Hence ϕ ∈ Homsg(∨σ ∩ τ⊥ ∩M,C∗), this means that x ∈ Oτ .

4.3 Compactness and smoothness

For each k in Z, one has algebraic group homomorphism

C∗ → C∗

z 7→ zk

providing the isomorphism Homalg.gr(C∗,C∗) = Z.Let N be a lattice, with dual lattice M, one has

(A) T = TN = Hom(M,C∗)

40

and with the choice of a basis for N , one has isomorphisms

(B) Hom(C∗,T) ∼= Hom(Z, N) ∼= N .Every one-parameter sub-group λ : C∗ → T corresponds to an unique

v ∈ N. Let us denote by λv the one-parameter sub-group corresponding tov. One has

v = (v1, ..., vn) λv(z) = (zv1 , ..., zvn).

In a dual way, one has

Hom(T,C∗) ∼= Hom(N,Z) ∼= M.

Every character corresponding to a unique u ∈M. Let χu ∈ Hom(T,C∗)be the character corresponding to u = (u1, ..., un) ∈ M. For t = (t1, ..., tn) ∈T, then χu(t) = tu11 ...t

unn . We will denote also by χu the corresponding func-

tion in the coordinate ring C[M ].Let us recall that a basis of the complex vectorial space C[M ] is given

by the elements χu with u ∈ M .The generators ui ∈ M correspond to thegenerators χui for the C−algebra C[M ]. More precisely, if (e1, ..., en) is abasis for N , then (e∗1, ..., e

∗n) is a basis for N and χei = χi a basis for the ring

of Laurent polynomial with n variables over C[M ].If z ∈ C∗, then λv(z) ∈ T, and (by (A)), λv(z) corresponds to a group

homomorphism from M in C∗. More explicitly

λv(z)(u) = χu(λv(z)) = z〈u,v〉,

where 〈, 〉 is the dual pairing M⊗

N → Z, i.e.

u v 7→ 〈u, v〉M×N → Z

Hom(T,C∗)×Hom(C∗,T) →Hom(C∗,C∗)χ λ 7→z 7→ z〈u,v〉 = χu(λv(z))

In fact, let λv(z) = (t1, ..., tn) ∈ T and v = (v1, ..., vn) ∈ N then

λv(z)(u) = tu11 ...tunn = χu(t) = χu(λv(z)),

and

λv(z)(u) = (zv1 , ..., zvn)(u) = (zv1)u1 ...(zvn)un = zu1v1+unvn = z〈u,v〉.

41

Example 4.3.1 Let σ be a cone generated by a part (e1, ..., ep) of a basisof N , then Xσ = Cp × Cn−p (Example 2.3.12).

For v = (v1, ..., vn) ∈ Zn, then λv(z) = (zv1 , ..., zvn). The limit limz→0λv(z)exists and lies in Xσ if and only if all vi are nonnegative and vi = 0 for i > p.In other words, the limit exists in Xσ if and only if v ∈ σ. In that case, thelimit is (y1, ..., yn), where yi = 0 if vi > 0 and yi = 1 if vi = 0. The possiblelimits are the distinguished point xτ for some face τ of σ.

We denote the union of the cones of ∆ by |∆|.

Proposition 4.3.2 Let v be in |∆| and τ be a cone of ∆ containing v inits relative interior. If limz→0λv(z) exists then limz→0λv(z) = xτ .

PROOF:By Property 1.2.9, we have

∨τ ∩ v⊥ = τ⊥.

For every σ ∈ ∆ and containing τ , we work in Xσ. For z ∈ C∗, we haveλv(z) = (zv1 , ..., zvn) ∈ T = Hom(M,C∗), then one has the map

λv(z) : M −→ C∗

u 7−→ z〈u,v〉 = λv(z)(u).

Consider the restriction of λv(z) on Sσ = M ∩ ∨σ, we have

λv(z) |Sσ : Sσ −→ C∗

u 7−→ z〈u,v〉 = λv(z)(u).

Then λv(z) |Sσ∈ Hom(Sσ,C) = Xσ.For every u ∈ Sσ, we have 〈u, v〉 ≥ 0, and 〈u, v〉 = 0 if and only if

u ∈ ∨τ ∩ v⊥ = τ⊥. Then

limz→0(λv(z) |Sσ) : Sσ −→ C

u 7−→

{1 if u ∈ τ⊥

0 in other cases.

This shows that limz→0(λv(z) |Sσ) = xτ . Hence if limz→0λv(z) exists thenlimz→0λv(z) = xτ .

42

Proposition 4.3.3 If v does not belong to any cone of ∆, then limz→0λv(z)does not exist in X∆.

PROOF:Suppose that v /∈ σ, then there is u ∈ ∨

σ such that 〈u, v〉 < 0. There-fore limz→0z

〈u,v〉 = ∞. This shows that limz→0λv(z) does not exist inHom(M,C∗). Then limz→0λv(z) does not exist in X∆.

Definition 4.3.4 A cone σ defined by the set of vectors (v1, ..., vr) is asimplex if all the vectors vi are linearly independent. A fan ∆ is simplicialif all cones of ∆ are simplices.

Definition 4.3.5 A vector v ∈ Zn is primitive if its coordinate are co-prime. A cone is regular if the vectors (v1, ..., vr) spanning the cone are prim-itive and there exist primitive vectors (vr+1, ..., vn) such that det(v1, ..., vn) =±1. In other words, the vectors (v1, ..., vr) can be completed in a basis of thelattice N. A fan is regular if all its cones are regular ones.

Definition 4.3.6 A fan ∆ is complete if its cones cover Rn, i.e. |∆| = Rn.

Theorem 4.3.7 Let ∆ be a fan in Rn. If X∆ is compact, then ∆ is com-plete.

PROOF:If |∆| is not all Rn, then there is a vector v such that v does not belong

to any cone (∆ is finite). In that case, λv(z) does not have a limit in X∆

when z goes to 0. That gives a contradiction with the compacity. (If X iscompact, then every infinite subset of X has at least one limit point in X).

Remark 4.3.8 Torus T = (C∗)n is smooth.Let σ = {0} be a cone in Rn, by Example 2.3.6, we have

T ∼= Xσ = V (ξ1...ξn+1 − 1) ⊂ Cn+1.

Set F (ξ1, ..., ξn+1) = ξ1...ξn+1 − 1 ∈ C[ξ1, ..., ξn+1], then F is irreducible.The set of singular points of Xσ is

V (ξ1...ξn+1 − 1) ∩ V (∂F

∂ξ1

, ....,∂F

∂ξn+1

)

= V (ξ1...ξn+1 − 1) ∩ V (ξ2...ξn+1, ...., ξ1...ξn) = ∅.

Then T = (C∗)n is smooth.

43

Theorem 4.3.9 Let ∆ be a fan in Rn. If ∆ is regular, then X∆ is smooth.

PROOF:Let (e1, ..., en) be a basis of N.Suppose that σ0 is generated by (e1, ..., ep), then σ0 is regular. By Exam-

ple 2.3.12, we have Xσ0∼= Cp × (C∗)n−p. Hence Xσ0 is smooth.

If σ is regular, then σ is generated by the vectors (v1, ..., vr), and the setof the vectors (v1, ..., vr, vr+1, ..., vn) is a basis of N, for some r = 1, ..., n.

Hence, there is a matrix A ∈ M(n,Z) such that Av1 = e1, ..., Avn = en.Consider the mapping

A : N → N

v 7→ Av.

We have that A is an isomorphism ( because det(A) = ±1).One has thatAσ = σ′ is generated by (e1, ..., er). ThenXσ′

∼= Cr×(C∗)n−ris smooth. By σ ∼= σ′ then Xσ

∼= Xσ′ . Hence Xσ is smooth.

Theorem 4.3.10 Let ∆ be a fan in Rn. If the toric variety X∆ is smooth,then the fan ∆ is regular.

PROOF:For every cone σ ∈ ∆, we prove that σ is regular.Since X∆ is smooth, then Xσ is smooth for all σ ∈ ∆.

Firstly, suppose that dimσ = n. Then we have σ⊥ = {0}, so the distin-guished point xσ = (0, ..., 0) ∈ Xσ.

Assume that Sσ =∨σ ∩M is generated by the vectors (a1, ..., ak), one has

Rσ = C[Sσ] = C[za1 , ..., zak ] = C[ξ1, ..., ξk]/Iσ,

and

Xσ = V (Iσ) ⊂ Ck.

Let us denote by M the maximal ideal of Rσ corresponding to the pointxσ. Let ϕ ∈ Homsg(Sσ,C) correspond to xσ, then ϕ(a) = za(x) for a ∈ Sσ.

One has

M∼= Ker(ϕ).

44

Since xσ = (0, ..., 0) = (ϕ(a1), ..., ϕ(ak)) ∈ Xσ, then ϕ(a1) = ... = ϕ(ak) =0. This shows that Kerϕ = Sσ − {0} (because ϕ(0) = 1). Therefore, M isgenerated by all zu such that u ∈ Sσ − {0}. So M2 is generated by zu suchthat u is the sum of two elements of Sσ − {0}.

And M/M2 is identified with the cotangent space at xσ. Since Xσ issmooth, then Rσ is regular, so dimRσ = dimM/M2.

Since dimRσ = dim(Xσ) = dimT = n, then

dimM/M2 = n.

In other words, M/M2 has a basis the images of elements zu for u ∈Sσ − {0}, such that u is not the sum of two vectors in Sσ − {0}.

Let H be a set of vectors u ∈ Sσ − {0} such that u is not the sum of twovectors in Sσ − {0}, then H ⊂ {a1, ..., ak}, and H generates Sσ. Hence H isfinite. This shows that M/M2 and H have n elements.

One has dim∨σ = n, then the elements of H are linearly independent, and

Sσ+(−Sσ) = M . Then H is the basis of M . This implies that σ is generatedby a basis of N and Xσ = Cn.

Let us consider the general case, i.e. dimσ = k ≤ n.Consider the sub-lattice

Nσ = (σ ∩N) + (−σ ∩N).

Then we have a decomposition

N = Nσ

⊕N ′, σ = σ′

⊕{0},

where σ′ is in a cone in Nσ ⊂ Rk. Then dimN ′R = n− k.The dual decomposition M = M ′⊕M ′′, then dimM ′′

R = n− k . One has

Sσ = ((σ′)∨ ∩M ′))⊕

M ′′,

Xσ = Xσ′ × TN ′ .

Since Xσ is smooth, then Xσ′ is smooth.The toric variety Xσ′ corresponds to the cone σ′ in the lattice Nσ. Since

dimσ′ = dim(Nσ)R = k, then σ′ is regular. This shows that σ is regular.

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Example 4.3.11 Let us consider the following cone σ in R2.

O

e2

O

v1

v2

v3

v4

e1 − 3e2

σ

σ1

σ2∆

σ3

v5

σ4

Fig. 9

We have that Xσ is not smooth, because σ is not regular. Now, we willdecompose σ to get a regular fan ∆. So we will add the vectors between e2

and e1 − 3e2 to get the cones, which are regular.Set v1 = e2, v2 = e1, v3 = e1 − e2, v4 = e1 − 2e2, v5 = e1 − 3e3.Since det(vi, vi+1) = ±1, for i = 1, ..., 4. Then, for i = 1, ..., 4, the cone σi

is generated by (vi, vi+i) is regular. The fan ∆ = {σi, i = 1, ..., 4} is regular,then X∆ is smooth.

We can see that v2 = v1 + v3, 2v3 = v2 + v4.

Proposition 4.3.12 Let v1, v2, v3 be the vectors in R2. If det(v1, v2) =det(v2, v3) = 1, then there is α ∈ Z such that αv2 = v1 + v3.

PROOF:Suppose that v1 = (a1, a2), v2 = (b1, b2), v3 = (c1, c2). Then

det(v1, v2) = a1b2 − a2b1 = 1,

det(v2, v3) = b1c2 − b2c1 = 1.

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Hence, (b1, b2) is a solution of the following system of equations{a2x− a1y = −1

c2x− c1y = 1.

Then,

b1 =a1 + c1

a1c2 − a2c1

, b2 =a2 + c2

a1c2 − a2c1

.

Therefore, set α = a1c2 − a2c1, so we have αv2 = v1 + v3.

Proposition 4.3.13 Let v1, v2 be the vectors in R2. If det(v1, v2) = 1 andv3 = αv2 − v1 for some α ∈ Z , then det(v2, v3) = 1.

PROOF:Assume that v1 = (a1, a2), v2 = (b1, b2), then v3 = (αb1 − a1, αb2 − a2).

Hence,

det(v2, v3) = det

[b1 αb1 − a1

b2 αb2 − a2

]= det(a1b2 − a2b1) = det(v1, v2) = 1.

Example 4.3.14 Let us consider the cone σ generated by two vectors e2

and 7e1 − 3e2. We have that Xσ is not smooth. Then, let us decompose σto get a regular fan.

• Step 1: Consider the Hirzebruch-Jung fraction of7

3:

7

3= 3− 1

2− 12

• Step 2: Set v0 = (0, 1) = e2, v1 = (1, 0). Calculate

v2 = 3v1 − v0 = (3,−1),

v3 = 2v2 − v1 = (5,−2),

v4 = 2v3 − v2 = (7,−3) = 7e1 − 3e2.

• Step 3: Decompose σ by cones σi, generated by the vectors (vi, vi+1), fori = 0, ..., 3. Thus, we have a regular fan ∆, and then X∆ is smooth. X∆ is aresolution of singularities of Xσ.

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Proposition 4.3.15 Let σ ⊂ NR ∼= R2 be a 2−dimensional strongly convexlattice cone. Then there exists a basis (v1, v2) for N such that σ is generatedby the vectors (v2,mv1 − kv2) where (k,m) = 1 and 0 ≤ k < m.

PROOF:Suppose that σ is generated by (u1, u2), where u1, u2 are primitive vectors.Set v2 = u2, since u2 is a primitive vector, then we can take it as a part

of a basis of N. Hence we have a basis (v′1, v2) for some v′1 ∈ N. And then,we have

u1 = mv′1 + lv2 for some m 6= 0.

Then we can assume that m > 0 (if m < 0, then we can get v′′1 = −v′1).

There are integers s, k such that l = sm− k, where 0 ≤ k < m. Let us takethis integer s. Let v1 = v′1 + sv2, then (v1, v2) is a basis of N and

u2 = mv′1 + lv2 = m(v1 − sv2) + lv2 = mv1 + (l −ms)v2 = mv1 − kv2.

Remark 4.3.16 A 2−dimensional strongly convex lattice cone σ in R2

is isomorphic to a cone σ′, generated by the vectors (e2,me1 − ke2), where(k,m) = 1 and 0 ≤ k < m.

Example 4.3.17 Let σ be a cone in R2, generated by the vectors (v1 =4e1 − 3e2, v2 = e1 + e2). We have that σ is not regular. Consider the matrix

A =

[1 −10 1

],

one has detA = 1 and

Av1 =

[1 −10 1

] [4−3

]=

[7−3

],

Av2 =

[1 −10 1

] [11

]=

[01

].

Let σ′ be a cone generated by the vectors (u1 = 7e1−3e2, u2 = e2). Then,

σA∼= σ′, hence Xσ

∼= Xσ′ .By Example 2.3.14, Xσ′

∼= C2/Γ7.By Theorem 4.3.7, Xσ is not compact.Xσ is not smooth. By Example 4.3.14, we can decompose σ′ to get a

regular fan ∆. Then X∆ is smooth and is a resolution of singularities of Xσ.

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References

[1] J.-P Brasselet Introduction to Toric Varieties, IML Luminy Case 907,Marseille, France.

[2] L. Birbrair, J.-P Brasselet and A. Fernandes The separation setfor toric varieties. Preprint, Nov. 2013.

[3] M. Barthel, J.-P Brasselet et K.-H. Fieseler Classes de Chern desvarietes toriques singulieres. c.R.Acad.Sci. Paris 315 (1992), 187-192.

[4] W. Fulton Introduction to Toric Varieties. Annals of Math. Studies,Princeton Univ. Press 1993.

[5] G. Ewald Combinatorial convexity and Algebraic Geometry. GraduateTexts in Mathematics. 168. 1996.

[6] I.R. Shafarevich Basic Algebraic Geometry, Study Edition, SpringerVerlag, Berlin, 1977.

[7] T.Oda Convex Bodies and Algebraic Geometry. Ergebn.Math.Grenwgb. (3.Folge), Bd.15, Springer-Verlag, Berlin etc.,1988.

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