Toric weak Fano varieties associated to building sets

TL;DR

This paper characterizes when the toric varieties linked to building sets are weak Fano, providing a precise criterion based on the structure of the building set.

Contribution

It offers a complete characterization of weak Fano conditions for toric varieties derived from building sets, filling a gap in the understanding of their geometric properties.

Findings

Provides a necessary and sufficient condition for weak Fano property.

Connects combinatorial building set structures to geometric Fano conditions.

Enhances classification of toric varieties associated with building sets.

Abstract

We give a necessary and sufficient condition for the nonsingular projective toric variety associated to a building set to be weak Fano in terms of the building set.

Equations

$$B=\{\{1\},\{2\},\{3\},\{4\},\{5\},\{1,2,3,4\},\{2,3,4,5\},\{1,2,3,4,5\}\}.$$

$$\{\{1\},\{2\},\{3\},\{1,2,3,4\}\},\{\{1\},\{2\},\{4\},\{1,2,3,4\}\},$$

$$\{\{1\},\{3\},\{4\},\{1,2,3,4\}\},\{\{2\},\{3\},\{4\},\{1,2,3,4\}\},$$

$$\{\{2\},\{3\},\{4\},\{2,3,4,5\}\},\{\{2\},\{3\},\{5\},\{2,3,4,5\}\},$$

$$\{\{2\},\{4\},\{5\},\{2,3,4,5\}\},\{\{3\},\{4\},\{5\},\{2,3,4,5\}\},$$

$$\{\{1\},\{2\},\{3\},\{5\}\},\{\{1\},\{2\},\{4\},\{5\}\},\{\{1\},\{3\},\{4\},\{5\}\}$$

$$N_1=\{\{2\},\{3\},\{4\},\{1,2,3,4\}\},N_2=\{\{2\},\{3\},\{4\},\{2,3,4,5\}\}.$$

$${\mathbb{R}}_{\ge 0}{N}_1$$

$${\mathbb{R}}_{\ge 0}{N}_2$$

$$V(G)=\{1,2,3,4\},A(G)=\{(1,2),(2,3),(3,1),(1,4),(4,3)\}.$$

$$\{\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,2,3,4\}\},$$

$$\{\{1\},\{2\},\{3\},\{4\},\{1,2,3\},\{1,2,3,4\}\},$$

$$\{\{1\},\{2\},\{3\},\{1,2,3\},\{4\},\{5\},\{4,5\}\}.$$

$$\mathcal{N}(B{)}_C=\{N\subset (B\setminus B_{\max })\setminus \{C\}\mid N\cup \{C\}\in \mathcal{N}(B)\}$$

$$\{I\in (B\setminus B_{\max })\setminus \{C\}\mid \{I,C\}\in \mathcal{N}(B)\}.$$

$$C\setminus B=\{I\subset S\setminus C\mid I\ne \varnothing ;I\in B\text{mbox}orC\cup I\in B\}$$

$$i_1\in I_1\setminus I_2,i_2\in I_2\setminus I_1,$$

$$N\in \mathcal{N}(B{|}_{I_1\cap I_2}{)}_{\max },N^{\prime }\in \mathcal{N}(B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$\{I_k\}\cup N\cup (B{|}_{I_1\cap I_2}{)}_{\max }\cup N^{\prime }\cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$K\in \{I_1\}\cup N\cup (B{|}_{I_1\cap I_2}{)}_{\max },L\in N^{\prime }\cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$K_1,\dots ,K_r\in N\cup (B{|}_{I_1\cap I_2}{)}_{\max },L_1,\dots ,L_s\in N^{\prime }\cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$L_1,\dots ,L_s\in N^{\prime }\cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$J_1,J_2\in B,j_1\in J_1\setminus J_2,j_2\in J_2\setminus J_1,$$

$$N\in \mathcal{N}(B{|}_{J_1\cap J_2}{)}_{\max },N^{\prime }\in \mathcal{N}(B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$\{J_k\}\cup N\cup (B{|}_{J_1\cap J_2}{)}_{\max }\cup N^{\prime }\cup (B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$\{I_k\}\cup N\cup (B{|}_{I_1\cap I_2}{)}_{\max }\cup N^{\prime }\cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

$$J_1,J_2\in B,j_1\in J_1\setminus J_2,j_2\in J_2\setminus J_1,$$

$$N\in \mathcal{N}(B{|}_{J_1\cap J_2}{)}_{\max },N^{\prime }\in \mathcal{N}(B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$\{J_k\}\cup N\cup (B{|}_{J_1\cap J_2}{)}_{\max }\cup N^{\prime }\cup (B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$J_1,J_2\in B,j_1\in J_1\setminus J_2,j_2\in J_2\setminus J_1,$$

$$N\in \mathcal{N}(B{|}_{J_1\cap J_2}{)}_{\max },N^{\prime }\in \mathcal{N}(B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$\{J_k\}\cup N\cup (B{|}_{J_1\cap J_2}{)}_{\max }\cup N^{\prime }\cup (B{|}_{(J_1△J_2)\setminus \{j_1,j_2\}}{)}_{\max }$$

$$B$$

$$\{1,2,3,4,5\},\{2,3,4,5,6\},\{1,2,3,4,5,6\}\}.$$

$$B{|}_{I_1\cap I_2}=\{\{3\},\{4\}\},B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}=\{\{2\},\{5\}\}.$$

$$\{I_1\}\cup \varnothing \cup (B{|}_{I_1\cap I_2}{)}_{\max }\cup \varnothing \cup (B{|}_{(I_1△I_2)\setminus \{i_1,i_2\}}{)}_{\max }$$

Full text

1. Introduction

A toric Fano variety is a nonsingular projective toric variety over $\mathbb{C}$ whose anticanonical divisor is ample. It is known that there are a finite number of isomorphism classes of toric Fano varieties in any given dimension. The classification problem of toric Fano varieties has been studied by many researchers. In particular, Øbro [2] gave an explicit algorithm that classifies all toric Fano varieties for any dimension.

A nonsingular projective algebraic variety is said to be weak Fano if its anticanonical divisor is nef and big. Sato [5] classified toric weak Fano 3-folds that are not Fano but are deformed to Fano under a small deformation, which are called toric weakened Fano 3-folds.

We can construct a nonsingular projective toric variety from a building set. Since a finite simple graph defines a building set, which is called the graphical building set, we can also associate to the graph a toric variety (see, for example [8]). The author [6, 7] characterized finite simple graphs whose associated toric varieties are Fano or weak Fano, and building sets whose associated toric varieties are Fano. In this paper, we characterize building sets whose associated toric varieties are weak Fano (see Theorem 2.4). Our theorem is proved combinatorially by using the fact that the intersection number of the anticanonical divisor with a torus-invariant curve can be computed in terms of the building set.

A toric weak Fano variety defines a reflexive polytope. Higashitani [1] constructed integral convex polytopes from finite directed graphs and gave a necessary and sufficient condition for the polytope to be terminal and reflexive. We also discuss a difference between the class of reflexive polytopes defined by toric weak Fano varieties associated to building sets, and that of reflexive polytopes associated to finite directed graphs.

The structure of the paper is as follows: In Section 2, we review the construction of a toric variety from a building set and state the characterization of building sets whose associated toric varieties are weak Fano. In Section 3, we consider reflexive polytopes associated to building sets and finite directed graphs. In Section 4, we give a proof of the main theorem.

Acknowledgment*.*

This work was supported by Grant-in-Aid for JSPS Fellows 15J01000. The author wishes to thank his supervisor, Professor Mikiya Masuda, for his valuable advice, and Professor Akihiro Higashitani for his useful comments.

2. The main result

Let $S$ be a nonempty finite set. A building set on $S$ is a finite set $B$ of nonempty subsets of $S$ satisfying the following conditions:

1. (1)

$I,J\in B$ with $I\cap J\neq\emptyset$ implies $I\cup J\in B$. 2. (2)

$\{i\}\in B$ for every $i\in S$.

We denote by $B_{\rm max}$ the set of all maximal (by inclusion) elements of $B$. An element of $B_{\rm max}$ is called a $B$-component and $B$ is said to be connected if $B_{\rm max}=\{S\}$. For a nonempty subset $C$ of $S$, we call $B|_{C}=\{I\in B\mid I\subset C\}$ the restriction of $B$ to $C$. $B|_{C}$ is a building set on $C$. Note that $B|_{C}$ is connected if and only if $C\in B$. For any building set $B$, we have $B=\bigsqcup_{C\in B_{\rm max}}B|_{C}$. In particular, any building set is a disjoint union of connected building sets.

Definition 2.1*.*

Let $B$ be a building set. A nested set of $B$ is a subset $N$ of $B\setminus B_{\rm max}$ satisfying the following conditions:

1. (1)

If $I,J\in N$, then we have either $I\subset J$ or $J\subset I$ or $I\cap J=\emptyset$. 2. (2)

For any integer $k\geq 2$ and for any pairwise disjoint $I_{1},\ldots,I_{k}\in N$, the union $I_{1}\cup\cdots\cup I_{k}$ is not in $B$.

Note that the empty set is a nested set for any building set. The set $\mathcal{N}(B)$ of all nested sets of $B$ is called the nested complex. $\mathcal{N}(B)$ is in fact an abstract simplicial complex on $B\setminus B_{\rm max}$.

Proposition 2.2** ([8, Proposition 4.1]).**

Let $B$ be a building set on $S$. Then every maximal (by inclusion) nested set of $B$ has the same cardinality $|S|-|B_{\rm max}|$. In particular, if $B$ is connected, then the cardinality of every maximal nested set of $B$ is $|S|-1$.

We are now ready to construct a toric variety from a building set. First, suppose that $B$ is connected and $S=\{1,\ldots,n+1\}$. We denote by $e_{1},\ldots,e_{n}$ the standard basis for $\mathbb{R}^{n}$ and we put $e_{n+1}=-e_{1}-\cdots-e_{n}$. For a nonempty subset $I$ of $S$, we denote $e_{I}=\sum_{i\in I}e_{i}$. Note that $e_{S}=0$. For $N\in\mathcal{N}(B)\setminus\{\emptyset\}$, we denote by $\mathbb{R}_{\geq 0}N$ the $|N|$-dimensional cone $\sum_{I\in N}\mathbb{R}_{\geq 0}e_{I}$ in $\mathbb{R}^{n}$, where $\mathbb{R}_{\geq 0}$ is the set of nonnegative real numbers, and we define $\mathbb{R}_{\geq 0}\emptyset$ to be $\{0\}\subset\mathbb{R}^{n}$. Then $\Delta(B)=\{\mathbb{R}_{\geq 0}N\mid N\in\mathcal{N}(B)\}$ forms a fan in $\mathbb{R}^{n}$ and thus we have an $n$-dimensional toric variety $X(\Delta(B))$. If $B$ is not connected, then we define $X(\Delta(B))=\prod_{C\in B_{\rm max}}X(\Delta(B|_{C}))$.

Theorem 2.3** ([8, Corollary 5.2 and Theorem 6.1]).**

Let $B$ be a building set. Then the associated toric variety $X(\Delta(B))$ is nonsingular and projective.

The following is our main result:

Theorem 2.4**.**

Let $B$ be a building set. Then the following are equivalent:

1. (1)

The associated toric variety $X(\Delta(B))$ is weak Fano. 2. (2)

For any $B$-component $C$ and for any $I_{1},I_{2}\in B|_{C}$ such that $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2}$ and $I_{2}\not\subset I_{1}$, we have at least one of the following:

• (i)

$I_{1}\cap I_{2}\in B|_{C}$.

• (ii)

$I_{1}\cup I_{2}=C$* and $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\leq 2$.*

Remark 2.5*.*

In a previous paper [6], we proved that the toric variety associated to a graphical building set is weak Fano if and only if every connected component of the graph does not have a cycle graph of length $\geq 4$ or a diamond graph as a proper induced subgraph. However, it is unclear whether this result can be obtained from Theorem 2.4.

Example 2.6*.*

Theorem 2.4 implies that if $|S|\leq 4$, then the toric variety $X(\Delta(B))$ is weak Fano for any connected building set $B$ on $S$. Any building set is a disjoint union of connected building sets, and the disjoint union corresponds to the product of toric varieties associated to the connected building sets. Since the product of toric weak Fano varieties is also weak Fano, it follows that all toric varieties of dimension $\leq 3$ associated to building sets are weak Fano.

We recall a description of the intersection number of the anticanonical divisor with a torus-invariant curve, see [3] for details. Let $\Delta$ be a nonsingular complete fan in $\mathbb{R}^{n}$ and let $X(\Delta)$ be the associated toric variety. For $0\leq r\leq n$, we denote by $\Delta(r)$ the set of $r$-dimensional cones in $\Delta$. For $\tau\in\Delta(n-1)$, the intersection number of the anticanonical divisor $-K_{X(\Delta)}$ with the torus-invariant curve $V(\tau)$ corresponding to $\tau$ can be computed as follows:

Proposition 2.7**.**

Let $\Delta$ be a nonsingular complete fan in $\mathbb{R}^{n}$ and $\tau=\mathbb{R}_{\geq 0}v_{1}+\cdots+\mathbb{R}_{\geq 0}v_{n-1}\in\Delta(n-1)$, where $v_{1},\ldots,v_{n-1}$ are primitive vectors in $\mathbb{Z}^{n}$. Let $v$ and $v^{\prime}$ be the distinct primitive vectors in $\mathbb{Z}^{n}$ such that $\tau+\mathbb{R}_{\geq 0}v$ and $\tau+\mathbb{R}_{\geq 0}v^{\prime}$ are in $\Delta(n)$. Then there exist unique integers $a_{1},\ldots,a_{n-1}$ such that $v+v^{\prime}+a_{1}v_{1}+\cdots+a_{n-1}v_{n-1}=0$. Furthermore, the intersection number $(-K_{X(\Delta)}.V(\tau))$ is equal to $2+a_{1}+\cdots+a_{n-1}$.

Proposition 2.8** ([4, Proposition 6.17]).**

Let $X(\Delta)$ be an $n$-dimensional nonsingular projective toric variety. Then $X(\Delta)$ is weak Fano if and only if $(-K_{X(\Delta)}.V(\tau))$ is nonnegative for every $(n-1)$-dimensional cone $\tau$ in $\Delta$.

Example 2.9*.*

Let $S=\{1,2,3,4,5\}$ and

[TABLE]

Then the nested complex $\mathcal{N}(B)$ consists of

[TABLE]

and their subsets. The pair $I_{1}=\{1,2,3,4\}$ and $I_{2}=\{2,3,4,5\}$ does not satisfy the condition (2) in Theorem 2.4. Hence the 4-dimensional toric variety $X(\Delta(B))$ is not weak Fano. In fact, there exists a 3-dimensional cone $\tau$ in $\Delta(B)$ such that $(-K_{X(\Delta(B))}.V(\tau))\leq-1$. Let

[TABLE]

Then we have

[TABLE]

Let us consider $\tau=\mathbb{R}_{\geq 0}N_{1}\cap\mathbb{R}_{\geq 0}N_{2}=\mathbb{R}_{\geq 0}e_{2}+\mathbb{R}_{\geq 0}e_{3}+\mathbb{R}_{\geq 0}e_{4}$. Since $(e_{1}+e_{2}+e_{3}+e_{4})+(-e_{1})-e_{2}-e_{3}-e_{4}=0$, Proposition 2.7 gives $(-K_{X(\Delta(B))}.V(\tau))=2-3=-1$. Therefore $X(\Delta(B))$ is not weak Fano by Proposition 2.8.

3. Reflexive polytopes associated to building sets

An $n$-dimensional integral convex polytope $P\subset\mathbb{R}^{n}$ is said to be reflexive if [math] is in the interior of $P$ and the dual $P^{*}=\{u\in\mathbb{R}^{n}\mid\langle u,v\rangle\geq-1\mbox{ for any }v\in P\}$ is also an integral convex polytope, where $\langle\cdot,\cdot\rangle$ denotes the standard inner product in $\mathbb{R}^{n}$. Let $\Delta$ be a nonsingular complete fan in $\mathbb{R}^{n}$. If the associated toric variety $X(\Delta)$ is weak Fano, then the convex hull of primitive generators of rays in $\Delta(1)$ is a reflexive polytope. For a building set $B$ such that the associated toric variety $X(\Delta(B))$ is weak Fano, we denote by $P_{B}$ the corresponding reflexive polytope.

Higashitani [1] gave a construction of integral convex polytopes from finite directed graphs (with no loops and no multiple arrows). We describe his construction briefly. Let $G$ be a finite directed graph whose node set is $V(G)=\{1,\ldots,n+1\}$ and whose arrow set is $A(G)\subset V(G)\times V(G)$. For $\overrightarrow{e}=(i,j)\in A(G)$, we define $\rho(\overrightarrow{e})\in\mathbb{R}^{n+1}$ to be $e_{i}-e_{j}$. We define $P_{G}$ to be the convex hull of $\{\rho(\overrightarrow{e})\mid\overrightarrow{e}\in A(G)\}$ in $\mathbb{R}^{n+1}$. $P_{G}$ is an integral convex polytope in the hyperplane $H=\{(x_{1},\ldots,x_{n+1})\in\mathbb{R}^{n+1}\mid x_{1}+\cdots+x_{n+1}=0\}$. In a previous paper we proved that if $X(\Delta(B))$ is Fano, then $P_{B}$ can be obtained from a finite directed graph:

Theorem 3.1** ([7, Theorem 4.1]).**

Let $B$ be a building set. If the associated toric variety $X(\Delta(B))$ is Fano, then there exists a finite directed graph $G$ such that $P_{B}$ is equivalent to $P_{G}$, that is, there exists a linear isomorphism $f:\mathbb{R}^{n}\rightarrow H$ such that $f(\mathbb{Z}^{n})=H\cap\mathbb{Z}^{n+1}$ and $f(P_{B})=P_{G}$.

However, there exist infinitely many reflexive polytopes associated to building sets that cannot be obtained from finite directed graphs. The following proposition provides such examples:

Proposition 3.2**.**

Let $S=\{1,\ldots,n+1\}$ and $B=2^{S}\setminus\{\emptyset\}$. Then $X(\Delta(B))$ is weak Fano by Theorem 2.4 but the reflexive polytope $P_{B}$ cannot be obtained from any finite directed graph for $n\geq 3$.

Proof.

Suppose that there exists a finite directed graph $G$ such that $P_{B}$ is equivalent to $P_{G}$. Since $0\in P_{G}$, there exists a nonempty subset $A^{\prime}$ of $A(G)$ and positive real numbers $a_{\overrightarrow{e}}$ for $\overrightarrow{e}\in A^{\prime}$ such that $\sum_{\overrightarrow{e}\in A^{\prime}}a_{\overrightarrow{e}}\rho(\overrightarrow{e})=0$. If $(i_{1},i_{2})\in A^{\prime}$, then we must have $(i_{2},i_{3})\in A^{\prime}$ for some $i_{3}\in V(G)$. Continuing this process, eventually we obtain a directed cycle of $G$. In general, if $G$ has a nonhomogeneous cycle (a directed cycle is a nonhomogeneous cycle), then the dimension of $P_{G}$ is $|V(G)|-1$ (see [1, Proposition 1.3]). Hence we have $|V(G)|=n+1$. Since $G$ has at most $n(n+1)$ arrows, $P_{G}$ has at most $n(n+1)$ vertices. On the other hand, $P_{B}$ has $2^{n+1}-2$ vertices. Thus we have the inequality $2^{n+1}-2\leq n(n+1)$, but this inequality does not hold for $n\geq 3$. This is a contradiction. Thus we proved the proposition. ∎

Example 3.3*.*

There also exists a reflexive polytope associated to a finite directed graph that cannot be obtained from any building set. Let $G$ be the finite directed graph defined by

[TABLE]

Then $P_{G}$ cannot be obtained from any building set. $P_{G}$ is a reflexive 3-polytope with six lattice points. However, there are only three types of reflexive 3-polytopes with six lattice points that are obtained from building sets. They are realized by the following building sets:

[TABLE]

All the building sets yield reflexive polytopes not equivalent to $P_{G}$.

4. Proof of Theorem 2.4

First we introduce some notation.

Definition 4.1*.*

Let $B$ be a building set on $S$.

1. (1)

We denote by $\mathcal{N}(B)_{\rm max}$ the set of all maximal (by inclusion) nested sets of $B$. $\mathcal{N}(B)_{\rm max}$ is a subset of $\mathcal{N}(B)$. 2. (2)

For $C\in B\setminus B_{\rm max}$, we call

[TABLE]

the link of $C$ in $\mathcal{N}(B)$. $\mathcal{N}(B)_{C}$ is an abstract simplicial complex on

[TABLE] 3. (3)

For a nonempty proper subset $C$ of $S$, we call

[TABLE]

the contraction of $C$ from $B$. $C\setminus B$ is a building set on $S\setminus C$.

The symmetric difference of two sets $X$ and $Y$ is defined by $X\triangle Y=(X\cup Y)\setminus(X\cap Y)$.

Lemma 4.2**.**

Let $B$ be a connected building set on $S$ and let $I_{1},I_{2}\in B$ with $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2},I_{2}\not\subset I_{1}$ and $|I_{1}\triangle I_{2}|\geq 3$. Suppose that

[TABLE]

such that

[TABLE]

is not a nested set of $B$ for some $k=1,2$. Then there exist $I^{\prime}_{1},I^{\prime}_{2}\in B$ such that $I^{\prime}_{1}\supset I_{1},I^{\prime}_{2}\supset I_{2},i_{1}\in I^{\prime}_{1}\setminus I^{\prime}_{2},i_{2}\in I^{\prime}_{2}\setminus I^{\prime}_{1},I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$ and $I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$.

Proof.

The proof is similar to a part of the proof of [7, Lemma 3.4 (1)].

Without loss of generality, we may assume $k=1$. Note that $\{I_{1}\}\cup N\cup(B|_{I_{1}\cap I_{2}})_{\rm max}$ and $N^{\prime}\cup(B|_{(I_{1}\triangle I_{2})\setminus\{i_{1},i_{2}\}})_{\rm max}$ are nested sets of $B$. Thus (4.1) falls into the following three cases.

Case 1. Suppose that (4.1) does not satisfy the condition (1) in Definition 2.1. Then there exist

[TABLE]

such that $K\not\subset L,L\not\subset K$ and $K\cap L\neq\emptyset$. If $K\in N\cup(B|_{I_{1}\cap I_{2}})_{\rm max}$, then $K\cap L=\emptyset$, a contradiction. Thus we must have $K=I_{1}$. Then $I_{1}\cup L\in B$. We put $I^{\prime}_{1}=I_{1}\cup L$ and $I^{\prime}_{2}=I_{2}$. Since $L\subset I_{1}\triangle I_{2}$, it follows that $L\setminus I_{1}\subset(I^{\prime}_{1}\cap I^{\prime}_{2})\setminus(I_{1}\cap I_{2})$. Thus $I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$.

Case 2. Suppose that (4.1) does not satisfy the condition (2) in Definition 2.1, and there exist

[TABLE]

for $r,s\geq 1$ such that $K_{1},\ldots,K_{r},L_{1},\ldots,L_{s}$ are pairwise disjoint and $K_{1}\cup\cdots\cup K_{r}\cup L_{1}\cup\cdots\cup L_{s}\in B$. Then we have $I_{k}\cup L_{1}\cup\cdots\cup L_{s}\in B$ for each $k=1,2$. We put $I^{\prime}_{k}=I_{k}\cup L_{1}\cup\cdots\cup L_{s}$ for $k=1,2$. $L_{1}\cup\cdots\cup L_{s}\subset I_{1}\triangle I_{2}$ implies $I_{k}\subsetneq I^{\prime}_{k}$ for some $k=1,2$. Since $I^{\prime}_{k}\setminus I_{k}\subset(I^{\prime}_{1}\cap I^{\prime}_{2})\setminus(I_{1}\cap I_{2})$, we have $I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$.

Case 3. Suppose that (4.1) does not satisfy the condition (2) in Definition 2.1, and there exist

[TABLE]

such that $I_{1},L_{1},\ldots,L_{s}$ are pairwise disjoint and $I_{1}\cup L_{1}\cup\cdots\cup L_{s}\in B$. We put $I^{\prime}_{1}=I_{1}\cup L_{1}\cup\cdots\cup L_{s}$ and $I^{\prime}_{2}=I_{2}$. Since $L_{1}\cup\cdots\cup L_{s}\subset I_{2}$, it follows that $L_{1}\cup\cdots\cup L_{s}\subset(I^{\prime}_{1}\cap I^{\prime}_{2})\setminus(I_{1}\cap I_{2})$. Thus $I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$.

In every case, we have $i_{1}\in I^{\prime}_{1}\setminus I^{\prime}_{2},i_{2}\in I^{\prime}_{2}\setminus I^{\prime}_{1}$ and $I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$. This completes the proof. ∎

Lemmas 4.3 and 4.5 play key roles in the proof of Theorem 2.4.

Lemma 4.3**.**

Let $B$ be a connected building set on $S$ and let $I_{1},I_{2}\in B$ with $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2},I_{2}\not\subset I_{1}$ and $I_{1}\cap I_{2}\notin B$. Then there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B,J_{1}\cup J_{2}\subset I_{1}\cup I_{2}$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$. If $J_{1}\triangle J_{2}=\{j_{1},j_{2}\}$, then $N^{\prime}$ and $(B|_{(J_{1}\triangle J_{2})\setminus\{j_{1},j_{2}\}})_{\rm max}$ are understood to be empty.

Proof.

We use induction on $|I_{1}\triangle I_{2}|$. We have $|I_{1}\triangle I_{2}|\geq 2$. Suppose $|I_{1}\triangle I_{2}|=2$. We put $J_{1}=I_{1}$ and $J_{2}=I_{2}$. Then $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B$ and $J_{1}\cup J_{2}=I_{1}\cup I_{2}$. We pick $N\in\mathcal{N}(B|_{J_{1}\cap J_{2}})_{\rm max}$. Then $\{J_{k}\}\cup N\cup(B|_{J_{1}\cap J_{2}})_{\rm max}$ is a nested set of $B$ for each $k=1,2$.

Suppose $|I_{1}\triangle I_{2}|\geq 3$. We pick $i_{1}\in I_{1}\setminus I_{2},i_{2}\in I_{2}\setminus I_{1},N\in\mathcal{N}(B|_{I_{1}\cap I_{2}})_{\rm max}$ and $N^{\prime}\in\mathcal{N}(B|_{(I_{1}\triangle I_{2})\setminus\{i_{1},i_{2}\}})_{\rm max}$. If

[TABLE]

is a nested set of $B$ for each $k=1,2$, then there is nothing to prove. Otherwise, by Lemma 4.2, there exist $I^{\prime}_{1},I^{\prime}_{2}\in B$ such that $I^{\prime}_{1}\supset I_{1},I^{\prime}_{2}\supset I_{2},i_{1}\in I^{\prime}_{1}\setminus I^{\prime}_{2},i_{2}\in I^{\prime}_{2}\setminus I^{\prime}_{1},I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$ and $I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$.

Case 1. Suppose $I^{\prime}_{1}\cap I^{\prime}_{2}\notin B$. We have $|I^{\prime}_{1}\triangle I^{\prime}_{2}|=|I^{\prime}_{1}\cup I^{\prime}_{2}|-|I^{\prime}_{1}\cap I^{\prime}_{2}|<|I_{1}\cup I_{2}|-|I_{1}\cap I_{2}|=|I_{1}\triangle I_{2}|$. By the hypothesis of induction, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B,J_{1}\cup J_{2}\subset I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$.

Case 2. Suppose $I^{\prime}_{1}\cap I^{\prime}_{2}\in B$. We may assume that $I_{1}\subsetneq I^{\prime}_{1}$.

Subcase 2.1. Suppose $I^{\prime}_{1}\cap I_{2}\in B$. We put $I^{\prime\prime}_{1}=I_{1}$ and $I^{\prime\prime}_{2}=I^{\prime}_{1}\cap I_{2}$. Then we have $I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}=I_{1}\cap I_{2}\notin B,i_{1}\in I^{\prime\prime}_{1}\setminus I^{\prime\prime}_{2}$ and $I^{\prime}_{1}\setminus I_{1}\subset I^{\prime\prime}_{2}\setminus I^{\prime\prime}_{1}$. Since $i_{2}\in(I_{1}\cup I_{2})\setminus(I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2})$, we have $I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}\subsetneq I_{1}\cup I_{2}$.

Subcase 2.2. Suppose $I^{\prime}_{1}\cap I_{2}\notin B$. We put $I^{\prime\prime}_{1}=I^{\prime}_{1}$ and $I^{\prime\prime}_{2}=I_{2}$. Then we have $I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}=I^{\prime}_{1}\cap I_{2}\notin B,i_{1}\in I^{\prime\prime}_{1}\setminus I^{\prime\prime}_{2},i_{2}\in I^{\prime\prime}_{2}\setminus I^{\prime\prime}_{1}$ and $I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}=I^{\prime}_{1}\cup I_{2}=I_{1}\cup I_{2}$. Since $I^{\prime}_{1}\setminus I_{1}\subset(I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2})\setminus(I_{1}\cap I_{2})$, we have $I_{1}\cap I_{2}\subsetneq I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}$.

In every subcase, we have $|I^{\prime\prime}_{1}\triangle I^{\prime\prime}_{2}|=|I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}|-|I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}|<|I_{1}\cup I_{2}|-|I_{1}\cap I_{2}|=|I_{1}\triangle I_{2}|$. By the hypothesis of induction, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B,J_{1}\cup J_{2}\subset I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}\subset I_{1}\cup I_{2}$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$.

Therefore the assertion holds for $|I_{1}\triangle I_{2}|$. ∎

Example 4.4*.*

Let $S=\{1,2,3,4,5,6\}$ and

[TABLE]

Let us consider $I_{1}=\{1,2,3,4\}$ and $I_{2}=\{3,4,5,6\}$. We pick $i_{1}=1$ and $i_{2}=6$. Then

[TABLE]

The only maximal nested set of each is the empty set. However,

[TABLE]

is not a nested set because $\{3\}\cup\{4\}\cup\{2\}\cup\{5\}=\{2,3,4,5\}\in B$ (Lemma 4.2, Case 2). Thus we put

[TABLE]

We have $I^{(1)}_{1}\cap I^{(1)}_{2}=\{2,3,4,5\}\in B$ (Lemma 4.3, Case 2) and $I_{1}\subsetneq I^{(1)}_{1}$. Since $I^{(1)}_{1}\cap I_{2}=\{3,4,5\}\notin B$ (Subcase 2.2), we put

[TABLE]

We pick $i^{(2)}_{1}=1$ and $i^{(2)}_{2}=6$. Then

[TABLE]

The only maximal nested set of each is the empty set.

[TABLE]

is not a nested set because $\{3\}\cup\{4\}\cup\{5\}\cup\{2\}=\{2,3,4,5\}\in B$ (Lemma 4.2, Case 2). Thus we put

[TABLE]

We have $I^{(3)}_{1}\cap I^{(3)}_{2}=\{2,3,4,5\}\in B$ (Lemma 4.3, Case 2) and $I^{(2)}_{2}\subsetneq I^{(3)}_{2}$. Since $I^{(2)}_{1}\cap I^{(3)}_{2}=\{2,3,4,5\}\in B$ (Subcase 2.1), we put

[TABLE]

Then

[TABLE]

The only maximal nested set of $B|_{I^{(4)}_{1}\cap I^{(4)}_{2}}$ is the empty set and

[TABLE]

are nested sets of $B$.

Lemma 4.5**.**

Let $B$ be a connected building set on $S$ and let $I_{1},I_{2}\in B$ with $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2},I_{2}\not\subset I_{1}$ and $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\geq 3$. Then there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$. Furthermore, we have $J_{1}\cup J_{2}\subsetneq I_{1}\cup I_{2}$ or $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 3$. If $J_{1}\triangle J_{2}=\{j_{1},j_{2}\}$, then $N^{\prime}$ and $(B|_{(J_{1}\triangle J_{2})\setminus\{j_{1},j_{2}\}})_{\rm max}$ are understood to be empty.

Proof.

We use induction on $|I_{1}\triangle I_{2}|$. We have $|I_{1}\triangle I_{2}|\geq 2$. Suppose $|I_{1}\triangle I_{2}|=2$. We put $J_{1}=I_{1}$ and $J_{2}=I_{2}$. Then $J_{1}\cap J_{2}\neq\emptyset$ and $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 3$. We pick $N\in\mathcal{N}(B|_{J_{1}\cap J_{2}})_{\rm max}$. Then $\{J_{k}\}\cup N\cup(B|_{J_{1}\cap J_{2}})_{\rm max}$ is a nested set of $B$ for each $k=1,2$.

Suppose $|I_{1}\triangle I_{2}|\geq 3$. We pick $i_{1}\in I_{1}\setminus I_{2},i_{2}\in I_{2}\setminus I_{1},N\in\mathcal{N}(B|_{I_{1}\cap I_{2}})_{\rm max}$ and $N^{\prime}\in\mathcal{N}(B|_{(I_{1}\triangle I_{2})\setminus\{i_{1},i_{2}\}})_{\rm max}$. If

[TABLE]

is a nested set of $B$ for each $k=1,2$, then there is nothing to prove. Otherwise, by Lemma 4.2, there exist $I^{\prime}_{1},I^{\prime}_{2}\in B$ such that $I^{\prime}_{1}\supset I_{1},I^{\prime}_{2}\supset I_{2},i_{1}\in I^{\prime}_{1}\setminus I^{\prime}_{2},i_{2}\in I^{\prime}_{2}\setminus I^{\prime}_{1},I^{\prime}_{1}\cap I^{\prime}_{2}\supsetneq I_{1}\cap I_{2}$ and $I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$.

Case 1. Suppose $|(B|_{I^{\prime}_{1}\cap I^{\prime}_{2}})_{\rm max}|\geq 3$. We have $|I^{\prime}_{1}\triangle I^{\prime}_{2}|=|I^{\prime}_{1}\cup I^{\prime}_{2}|-|I^{\prime}_{1}\cap I^{\prime}_{2}|<|I_{1}\cup I_{2}|-|I_{1}\cap I_{2}|=|I_{1}\triangle I_{2}|$. By the hypothesis of induction, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$. Furthermore, we have $J_{1}\cup J_{2}\subsetneq I^{\prime}_{1}\cup I^{\prime}_{2}=I_{1}\cup I_{2}$ or $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 3$.

Case 2. Suppose $|(B|_{I^{\prime}_{1}\cap I^{\prime}_{2}})_{\rm max}|\leq 2$. For any $K\in(B|_{I_{1}\cap I_{2}})_{\rm max}$, there exists unique $L_{K}\in(B|_{I^{\prime}_{1}\cap I^{\prime}_{2}})_{\rm max}$ such that $K\subset L_{K}$. Hence there exists $L\in(B|_{I^{\prime}_{1}\cap I^{\prime}_{2}})_{\rm max}$ that contains more than one element of $(B|_{I_{1}\cap I_{2}})_{\rm max}$. Let $K_{1},\ldots,K_{r}$ be all elements of $(B|_{I_{1}\cap I_{2}})_{\rm max}$ contained in $L$. Note that $I_{1}\cap I_{2}\cap L$ is the disjoint union of $K_{1},\ldots,K_{r}$. If $L\subset I_{1}\cap I_{2}$, then $B\ni L=I_{1}\cap I_{2}\cap L=K_{1}\cup\cdots\cup K_{r}\notin B$, a contradiction. Thus $L\not\subset I_{1}\cap I_{2}$. We may assume $L\not\subset I_{1}$.

Subcase 2.1. Suppose $I_{1}\cap L\in B$. If $L\subset I_{2}$, then $B\ni I_{1}\cap L=I_{1}\cap I_{2}\cap L=K_{1}\cup\cdots\cup K_{r}\notin B$, a contradiction. Thus $L\not\subset I_{2}$. We put $I^{\prime\prime}_{1}=I_{1}\cap L$ and $I^{\prime\prime}_{2}=I_{2}$. Then we have $I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}=I_{1}\cap I_{2}\cap L\notin B,L\setminus I_{2}\subset I^{\prime\prime}_{1}\setminus I^{\prime\prime}_{2}$ and $i_{2}\in I^{\prime\prime}_{2}\setminus I^{\prime\prime}_{1}$. Since $i_{1}\in(I_{1}\cup I_{2})\setminus(I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2})$, we have $I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}\subsetneq I_{1}\cup I_{2}$.

Subcase 2.2. Suppose $I_{1}\cap L\notin B$. We put $I^{\prime\prime}_{1}=I_{1}$ and $I^{\prime\prime}_{2}=L$. Then we have $I^{\prime\prime}_{1}\cap I^{\prime\prime}_{2}=I_{1}\cap L\notin B,i_{1}\in I^{\prime\prime}_{1}\setminus I^{\prime\prime}_{2}$ and $I^{\prime\prime}_{2}\setminus I^{\prime\prime}_{1}=L\setminus I_{1}\neq\emptyset$. Since $i_{2}\in(I_{1}\cup I_{2})\setminus(I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2})$, we have $I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}\subsetneq I_{1}\cup I_{2}$.

In every subcase, by Lemma 4.3, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B,J_{1}\cup J_{2}\subset I^{\prime\prime}_{1}\cup I^{\prime\prime}_{2}\subsetneq I_{1}\cup I_{2}$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$.

Therefore the assertion holds for $|I_{1}\triangle I_{2}|$. ∎

Example 4.6*.*

Let $S=\{1,2,3,4,5,6,7\}$ and

[TABLE]

Let us consider $I_{1}=\{1,2,3,4,5\}$ and $I_{2}=\{3,4,5,6,7\}$. We pick $i_{1}=1$ and $i_{2}=7$. Then

[TABLE]

The only maximal nested set of each is the empty set. However,

[TABLE]

is not a nested set because $\{4\}\cup\{2\}\cup\{6\}=\{2,4,6\}\in B$ (Lemma 4.2, Case 2). Thus we put

[TABLE]

We have $I^{(1)}_{1}\cap I^{(1)}_{2}=\{2,3,4,5,6\}\in B$ (Lemma 4.5, Case 2) and $L=\{2,3,4,5,6\}$. Since $L\not\subset I_{1}$ and $I_{1}\cap L=\{2,3,4,5\}\in B$ (Subcase 2.1), we put

[TABLE]

We pick $i^{(2)}_{1}=2$ and $i^{(2)}_{2}=7$. Then

[TABLE]

The only maximal nested set of each is the empty set.

[TABLE]

is not a nested set because $\{2,3,4,5\}\cup\{6\}=\{2,3,4,5,6\}\in B$ (Lemma 4.2, Case 3). Thus we put

[TABLE]

We have $I^{(3)}_{1}\cap I^{(3)}_{2}=\{3,4,5,6\}\notin B$ (Lemma 4.3, Case 1) and

[TABLE]

The only maximal nested set of $B|_{I^{(3)}_{1}\cap I^{(3)}_{2}}$ is the empty set and

[TABLE]

are nested sets of $B$. Furthermore, we have $I^{(3)}_{1}\cup I^{(3)}_{2}\subsetneq I_{1}\cup I_{2}$.

Example 4.7*.*

Let $S=\{1,2,3,4,5,6,7\}$ and

[TABLE]

Let us consider $I_{1}=\{1,2,3,4,5\}$ and $I_{2}=\{3,4,5,6,7\}$. We pick $i_{1}=1$ and $i_{2}=7$. Then

[TABLE]

The only maximal nested set of $B|_{I_{1}\cap I_{2}}$ is the empty set. We choose $N^{\prime}=\{\{2\}\}\in\mathcal{N}(B|_{(I_{1}\triangle I_{2})\setminus\{i_{1},i_{2}\}})_{\rm max}$.

[TABLE]

is not a nested set because $\{1,2,3,4,5\}\cap\{2,6\}=\{2\}\neq\emptyset$ (Lemma 4.2, Case 1). Thus we put

[TABLE]

We have $I^{(1)}_{1}\cap I^{(1)}_{2}=\{3,4,5,6\}=\{3\}\cup\{4,5,6\}$ (Lemma 4.5, Case 2) and $L=\{4,5,6\}$. Since $L\not\subset I_{1}$ and $I_{1}\cap L=\{4,5\}\notin B$ (Subcase 2.2), we put

[TABLE]

We pick $i^{(2)}_{1}=1$ and $i^{(2)}_{2}=6$. Then

[TABLE]

The only maximal nested set of each is the empty set.

[TABLE]

is not a nested set because $\{4,5,6\}\cup\{2\}\cup\{3\}=\{2,3,4,5,6\}\in B$ (Lemma 4.2, Case 3). Thus we put

[TABLE]

We have $I^{(3)}_{1}\cap I^{(3)}_{2}=\{2,3,4,5\}\notin B$ (Lemma 4.3, Case 1) and

[TABLE]

The only maximal nested set of $B|_{I^{(3)}_{1}\cap I^{(3)}_{2}}$ is the empty set and

[TABLE]

are nested sets of $B$. Furthermore, we have $I^{(3)}_{1}\cup I^{(3)}_{2}\subsetneq I_{1}\cup I_{2}$.

Proposition 4.8** ([8, Proposition 3.2]).**

Let $B$ be a building set on $S$ and let $C\in B\setminus B_{\rm max}$. Then the correspondence

[TABLE]

induces an isomorphism $\mathcal{N}(B)_{C}\rightarrow\mathcal{N}(B|_{C}\cup(C\setminus B))$ of simplicial complexes.

Lemma 4.9**.**

Let $J_{1},J_{2}\in B$ with $J_{1}\cap J_{2}\neq\emptyset,J_{1}\not\subset J_{2},J_{2}\not\subset J_{1}$ and $J_{1}\cup J_{2}\subsetneq S$. Let $N^{\prime\prime}\in\mathcal{N}(B|_{J_{1}\cup J_{2}})$ such that $\{J_{k}\}\cup N^{\prime\prime}\in\mathcal{N}(B|_{J_{1}\cup J_{2}})_{\rm max}$ for each $k=1,2$. Then there exists $M\in\mathcal{N}(B)$ such that $\{J_{k},J_{1}\cup J_{2}\}\cup N^{\prime\prime}\cup M\in\mathcal{N}(B)_{\rm max}$ for each $k=1,2$.

Proof.

We pick $M^{\prime}\in\mathcal{N}((J_{1}\cup J_{2})\setminus B)_{\rm max}$. Then

[TABLE]

for each $k=1,2$. Hence by Proposition 4.8, there exists $M\in\mathcal{N}(B)$ such that $\{J_{k}\}\cup N^{\prime\prime}\cup M$ are maximal simplices of $\mathcal{N}(B)_{J_{1}\cup J_{2}}$. Hence $\{J_{k},J_{1}\cup J_{2}\}\cup N^{\prime\prime}\cup M\in\mathcal{N}(B)_{\rm max}$ for each $k=1,2$. ∎

Proposition 4.10** ([8, Proposition 4.5]).**

Let $B$ be a building set on $S$ and let $I_{1},I_{2}\in B$ with $I_{1}\neq I_{2}$ and $N\in\mathcal{N}(B)$ such that $N\cup\{I_{1}\},N\cup\{I_{2}\}\in\mathcal{N}(B)_{\rm max}$. Then the following hold:

1. (1)

We have $I_{1}\not\subset I_{2}$ and $I_{2}\not\subset I_{1}$. 2. (2)

If $I_{1}\cap I_{2}\neq\emptyset$, then $(B|_{I_{1}\cap I_{2}})_{\rm max}\subset N$. 3. (3)

There exists $\{I_{3},\ldots,I_{k}\}\subset N$ such that $I_{1}\cup I_{2},I_{3},\ldots,I_{k}$ are pairwise disjoint and $I_{1}\cup\cdots\cup I_{k}\in N\cup B_{\rm max}$ ($\{I_{3},\ldots,I_{k}\}$* can be empty).*

We are now ready to prove Theorem 2.4.

Proof of Theorem 2.4.

The disjoint union of connected building sets yields the product of toric varieties associated to the connected building sets. Since the product of nonsingular projective toric varieties is weak Fano if and only if every factor is weak Fano, it suffices to show that, for any connected building set $B$ on $S=\{1,\ldots,n+1\}$, the following are equivalent:

• $(1^{\prime})$

The associated toric variety $X(\Delta(B))$ is weak Fano.

• $(2^{\prime})$

For any $I_{1},I_{2}\in B$ such that $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2}$ and $I_{2}\not\subset I_{1}$, we have at least one of the following:

• $({\rm i}^{\prime})$

$I_{1}\cap I_{2}\in B$.

• $({\rm ii}^{\prime})$

$I_{1}\cup I_{2}=S$ and $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\leq 2$.

$(1^{\prime})\Rightarrow(2^{\prime})$: Let $I_{1},I_{2}\in B$ such that $I_{1}\cap I_{2}\neq\emptyset,I_{1}\not\subset I_{2},I_{2}\not\subset I_{1}$ and $I_{1}\cap I_{2}\notin B$. We show that if $I_{1}\cup I_{2}\subsetneq S$ or $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\geq 3$, then the toric variety $X(\Delta(B))$ is not weak Fano.

Case 1. Suppose $I_{1}\cup I_{2}\subsetneq S$. By Lemma 4.3, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B,J_{1}\cup J_{2}\subset I_{1}\cup I_{2}\subsetneq S$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$. Since the cardinality of (4.2) is $|J_{1}\cup J_{2}|-1$, (4.2) is a maximal nested set of $B|_{J_{1}\cup J_{2}}$. By Lemma 4.9, there exists $M\in\mathcal{N}(B)$ such that

[TABLE]

for $k=1,2$. Let

[TABLE]

Clearly

[TABLE]

Since $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 2$, Proposition 2.7 gives

[TABLE]

Therefore $X(\Delta(B))$ is not weak Fano by Proposition 2.8.

Case 2. Suppose that $I_{1}\cup I_{2}=S$ and $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\geq 3$. By Lemma 4.5, there exist

[TABLE]

such that $J_{1}\cap J_{2}\neq\emptyset,J_{1}\cap J_{2}\notin B$ and

[TABLE]

is a nested set of $B$ for each $k=1,2$. Furthermore, we have $J_{1}\cup J_{2}\subsetneq I_{1}\cup I_{2}=S$ or $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 3$. If $J_{1}\cup J_{2}\subsetneq S$, then a similar augment shows that $X(\Delta(B))$ is not weak Fano. Suppose that $J_{1}\cup J_{2}=S$ and $|(B|_{J_{1}\cap J_{2}})_{\rm max}|\geq 3$. Then (4.3) is a maximal nested set of $B$. Let

[TABLE]

Since $e_{J_{1}\cup J_{2}}=e_{S}=0$, it follows that

[TABLE]

Proposition 2.7 gives

[TABLE]

Therefore $X(\Delta(B))$ is not weak Fano by Proposition 2.8.

$(2^{\prime})\Rightarrow(1^{\prime})$: Let $I_{1},I_{2}\in B$ with $I_{1}\neq I_{2}$ and $N\in\mathcal{N}(B)$ such that $N\cup\{I_{1}\},N\cup\{I_{2}\}\in\mathcal{N}(B)_{\rm max}$. We need to show that $(-K_{X(\Delta(B))}.V(\mathbb{R}_{\geq 0}N))\geq 0$.

Case 1. Suppose $I_{1}\cap I_{2}=\emptyset$. By Proposition 4.10 (3), there exists $\{I_{3},\ldots,I_{k}\}\subset N$ such that $I_{1}\cup I_{2},I_{3},\ldots,I_{k}$ are pairwise disjoint and $I_{1}\cup\cdots\cup I_{k}\in N\cup B_{\rm max}=N\cup\{S\}$. Since

[TABLE]

Proposition 2.7 gives

[TABLE]

Hence $(-K_{X(\Delta(B))}.V(\mathbb{R}_{\geq 0}N))\geq 1$.

Case 2. Suppose $I_{1}\cap I_{2}\neq\emptyset$. By Proposition 4.10 (1), we have $I_{1}\not\subset I_{2}$ and $I_{2}\not\subset I_{1}$.

$({\rm i}^{\prime})$ Suppose $I_{1}\cap I_{2}\in B$. By Proposition 4.10 (2), we have $\{I_{1}\cap I_{2}\}=(B|_{I_{1}\cap I_{2}})_{\rm max}\subset N$. By Proposition 4.10 (3), there exists $\{I_{3},\ldots,I_{k}\}\subset N$ such that $I_{1}\cup I_{2},I_{3},\ldots,I_{k}$ are pairwise disjoint and $I_{1}\cup\cdots\cup I_{k}\in N\cup B_{\rm max}=N\cup\{S\}$. Since

[TABLE]

Proposition 2.7 gives

[TABLE]

Hence $(-K_{X(\Delta(B))}.V(\mathbb{R}_{\geq 0}N))\geq 0$.

$({\rm ii}^{\prime})$ Suppose that $I_{1}\cup I_{2}=S$ and $|(B|_{I_{1}\cap I_{2}})_{\rm max}|\leq 2$. By Proposition 4.10 (2), we have $(B|_{I_{1}\cap I_{2}})_{\rm max}\subset N$. Since $e_{I_{1}\cup I_{2}}=e_{S}=0$, it follows that

[TABLE]

Proposition 2.7 gives

[TABLE]

Therefore $X(\Delta(B))$ is weak Fano by Proposition 2.8. This completes the proof of Theorem 2.4. ∎