On the canonical ideal of the Ehrhart ring of the chain polytope of a poset

TL;DR

This paper investigates the algebraic properties of the Ehrhart ring of the chain polytope of a poset, linking its canonical ideal's properties to the combinatorial structure of the poset.

Contribution

It characterizes when the Ehrhart ring of the chain polytope is level or anticanonical level based on poset structure and explores properties of its canonical ideal.

Findings

If K[C(P)] is level, then K[O(P)] is also level.

Symbolic and ordinary powers of the canonical ideal coincide.

Degrees of generators of canonical and anticanonical ideals are consecutive integers.

Abstract

Let P be a poset, O(P) the order polytope of P and C(P) the chain polytope of P. In this paper, we study the canonical ideal of the Ehrhart ring K[C(P)] of C(P) over a field K and characterize the level (resp. anticanonical level) property of K[C(P)] by a combinatorial structure of P. In particular, we show that if K[C(P)] is level (resp. anticanonical level), then so is K[O(P)]. We exhibit examples which show the converse does not hold. Moreover, we show that the symbolic powers of the canonical ideal of K[C(P)] are identical with ordinary ones and degrees of the generators of the canonical and anticanonical ideals are consecutive integers.

Equations

$$x=z_0<\cdot z_1<\cdot \cdots <\cdot z_t=y.$$

$${\xi }^{{+}^{\prime }\uparrow }(z)+\xi (z)+{\xi }^{{+}^{\prime }\downarrow }(z)={\xi }^{+\uparrow }(z)+{\xi }^{{+}^{\prime }\downarrow }(z)={\xi }^{{+}^{\prime }\uparrow }(z)+{\xi }^{+\downarrow }(z)\le M.$$

$${\xi }^{+\downarrow }(x)+q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\uparrow }(y)\le M,$$

$${\xi }^{{+}^{\prime }\downarrow }(y)\ge q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\downarrow }(x)\text{mbox}and$$

$${\xi }^{{+}^{\prime }\uparrow }(x)\ge q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\uparrow }(y).$$

$$z_1,\dots ,z_s,z_0^{\prime },\dots ,z_{s^{\prime }}^{\prime },z_1^{\prime \prime },\dots ,z_{s^{\prime \prime }}^{\prime \prime }$$

$${\xi }^{+\downarrow }(x)+q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\uparrow }(y)$$

$${\xi }^{{+}^{\prime }\downarrow }(y)\ge \sum _{\ell =1}^{s^{\prime }-1}\xi (z_{\ell }^{\prime })+{\xi }^{+\downarrow }(x)\ge q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\downarrow }(x)$$

$${\xi }^{{+}^{\prime }\uparrow }(x)\ge \sum _{\ell =1}^{s^{\prime }-1}\xi (z_{\ell }^{\prime })+{\xi }^{+\uparrow }(y)\ge q^{(n)}\mathrm{dist}(x,y)-n+{\xi }^{+\uparrow }(y)$$

$$\mathbb{K}[T^f\mid f\in {\mathbb{Z}}^{X^{-}},f(-\infty )>0,\frac{1}{f(-\infty )}f\in \mathcal{P}]$$

$$\underset{f\in {\mathbb{Z}}^{X^{-}},f(-\infty )>0,\frac{1}{f(-\infty )}f\in \mathrm{relint}\mathcal{P}}{⨁}\mathbb{K}{T}^f$$

$$E_{\mathbb{K}}[\mathcal{C}(P)]=\underset{\xi \in {\mathcal{S}}^{(0)}}{⨁}\mathbb{K}{T}^{\xi }\text{mbox}and\omega =\underset{\xi \in {\mathcal{S}}^{(1)}}{⨁}\mathbb{K}{T}^{\xi }.$$

$$\xi(z)\mathrel{:=}\begin{cases}n,&\mbox{if $z\in P$;}\\ {q^{(n)}{\mathrm{dist}}}(-\infty,\infty),&\mbox{if $z=-\infty$,}\end{cases}$$

$${\omega }^{(n)}=\underset{\xi \in {\mathcal{S}}^{(n)}}{⨁}\mathbb{K}{T}^{\xi }.$$

$$\zeta^{\prime}(z)=\begin{cases}0,&\mbox{if $z\in C$;}\\ -m,&\mbox{otherwise.}\end{cases}$$

$$\xi_{1}(w)=\begin{cases}\lfloor\frac{1}{n}\xi(w)\rfloor,&\vtop{\noindent if $w$ is a maximal element of $P$ or $w=-\infty$;}\\ \lfloor\frac{1}{n}\xi^{+\uparrow}(w)\rfloor-\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(w)\rfloor,&\mbox{otherwise}\end{cases}$$

$${\omega }^{(n)}={\omega }^n=\underset{\xi \in {\mathcal{S}}^{(n)}}{⨁}\mathbb{K}{T}^{\xi }$$

$${\omega }^{(-n)}=({\omega }^{(-1)}{)}^n=\underset{\xi \in {\mathcal{S}}^{(-n)}}{⨁}\mathbb{K}{T}^{\xi }.$$

$$\xi_{1}(z)\mathrel{:=}\begin{cases}1,&\mbox{if $z\in A\cap(\bigcup_{C\in C_{\xi}^{[d-\epsilon]}}C)$ or $z=-\infty$;}\\ 0,&\mbox{otherwise}\end{cases}$$

$$q^{(ϵ)}(w_0,z_0,\dots ,w_s,z_s):=\sum _{\ell =0}^s{q}^{(ϵ)}\mathrm{dist}(w_{\ell },z_{\ell })-\sum _{\ell =0}^{s-1}{q}^{(ϵ)}\mathrm{dist}(w_{\ell +1},z_{\ell }).$$

$$q^{(ϵ)}\mathrm{dist}(x_i,y_j)<q^{(ϵ)}(x_i,y_i,\dots ,x_j,y_j)$$

$$q^{(ϵ)}\mathrm{dist}(x_i,y_j)+\sum _{\ell =i}^{j-1}{q}^{(ϵ)}\mathrm{dist}(x_{\ell +1},y_{\ell })\ge \sum _{\ell =i}^j{q}^{(ϵ)}\mathrm{dist}(x_{\ell },y_{\ell }).$$

$$q^{(ϵ)}\mathrm{dist}(x_i,y_j)+\sum _{\ell =i}^{j-1}{q}^{(ϵ)}\mathrm{dist}(x_{\ell +1},y_{\ell })$$

$$\le$$

$$d(j-i+1)$$

$$\deg T^{\xi}\leq\max\left\{q^{(\epsilon)}(x_{0},y_{0},\ldots,x_{t},y_{t})\left|\vbox{\noindent$y_{0}$, $x_{1}$, \ldots, $y_{t-1}$, $x_{t}$ is a $q^{(\epsilon)}$-reduced\ {sequence with condition N}', where we set $x_{0}\mathrel{:=}-\infty$ and $y_{t}\mathrel{:=}\infty$.}\right.\right\}.$$

$$d-ϵ={\xi }^{+}(C_0)=(\#C_0)ϵ\le q^{(ϵ)}\mathrm{dist}(-\infty ,\infty )-ϵ.$$

$${\xi }^{+\downarrow }(x_i)=\sum _{c\in C_i\cap (-\infty ,x_i]}\xi (c)\text{mbox}and{\xi }^{{+}^{\prime }\uparrow }(x_i)=\sum _{c\in C_i\cap (x_i,\infty )}\xi (c)$$

$${\xi }^{{+}^{\prime }\downarrow }(y_i)=\sum _{c\in C_i\cap (-\infty ,y_i)}\xi (c)\text{mbox}and{\xi }^{+\uparrow }(y_i)=\sum _{c\in C_i\cap [y_i,\infty )}\xi (c)$$

$${\xi }^{{+}^{\prime }\uparrow }(x_i)-{\xi }^{+\uparrow }(y_i)=\sum _{c\in C_i\cap (x_i,y_i)}\xi (c)=(\#(C_i\cap (x_i,y_i)))ϵ\le q^{(ϵ)}\mathrm{dist}(x_i,y_i)-ϵ$$

$${\xi }^{{+}^{\prime }\downarrow }(y_0)\le q^{(ϵ)}\mathrm{dist}(-\infty ,y_0)-ϵ\text{mbox}and{\xi }^{{+}^{\prime }\uparrow }(x_t)\le q^{(ϵ)}\mathrm{dist}(x_t,\infty )-ϵ$$

$${\xi }^{{+}^{\prime }\uparrow }(x_{i+1})-{\xi }^{+\uparrow }(y_i)\ge q^{(ϵ)}\mathrm{dist}(x_{i+1},y_i)-ϵ$$

Full text

Moreover, we show that the symbolic powers of the canonical ideal of $\mathbb{K}[\mathscr{C}(P)]$ are identical with ordinary ones and degrees of the generators of the canonical and anticanonical ideals are consecutive integers.

Key Words:chain polytope, order polytope, Ehrhart ring, level ring

MSC:52B20, 13H10, 13F50, 06A11, 06A07

1 Introduction

Let $P$ be a finite partially ordered set (poset for short). Stanley [Sta3] investigated two convex polytopes associated with $P$, the order polytope $\mathscr{O}(P)$ and the chain polytope $\mathscr{C}(P)$. He showed a surprising connection between $\mathscr{O}(P)$ and $\mathscr{C}(P)$. For example the Ehrhart rings $\mathbb{K}[\mathscr{O}(P)]$ and $\mathbb{K}[\mathscr{C}(P)]$ of $\mathscr{O}(P)$ and $\mathscr{C}(P)$ have the same Hilbert series.

About the same time, Hibi [Hib] constructed and studied the ring ${\mathcal{R}}_{\mathbb{K}}[H]$, where $H$ is a finite distributive lattice, which nowadays called the Hibi ring, in the study of algebras with straightening law (ASL). It turned out that ${\mathcal{R}}_{\mathbb{K}}[H]$ is identical with the Ehrhart ring $\mathbb{K}[\mathscr{O}(P)]$ of the order polytope $\mathscr{O}(P)$, where $P$ is the set of join-irreducible elements of $H$.

After that, Hibi rings, i.e., the ring of the form $\mathbb{K}[\mathscr{O}(P)]$ for some poset $P$, are extensively studied by many researchers, including the present author. In our previous papers, we studied many aspects of Hibi rings, especially described generators of the canonical ideals of Hibi rings and characterized level and anticanonical level properties of Hibi rings [Miy1, Miy2]. The key notion to investigate the generators of the canonical ideal of a Hibi ring is “sequence with condition N” (see below).

In this paper, we investigate the canonical ideal of the Ehrhart ring $\mathbb{K}[\mathscr{C}(P)]$ of the chain polytope $\mathscr{C}(P)$ of a poset $P$. The key notion in this case is “sequence with condition N’ ”.

• Definition 4.1

Let $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$, …, $y_{t-1}$, $x_{t}$ be a sequence of elements of $P$. We say that $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$, …, $y_{t-1}$, $x_{t}$ satisfy condition N (resp. condition N’) if

1. (1)

$y_{0}>x_{1}<y_{1}>x_{2}<\cdots<y_{t-1}>x_{t}$** and** 2. (2)

if $i\leq j-2$, then $y_{i}\not\geq x_{j}$ (resp. $y_{i}\not>x_{j}$).

We define that an empty sequence (i.e., $t=0$) satisfy condition N (resp. condition N’).

In the case of Hibi rings, i.e., the case of $\mathbb{K}[\mathscr{O}(P)]$, a Laurent monomial associated to a map $\nu\colon P^{-}\to\mathbb{Z}$ is a generator of the canonical (resp. anticanonical) ideal if and only if there is a sequence with condition N with an appropriate relation with $\nu$, where $P^{-}=P\cup\{-\infty\}$. Moreover, the symbolic powers of the canonical ideal are identical with the ordinary ones and the degrees of the generators of the canonical and anticanonical ideals are consecutive integers, i.e., if there are generators of the canonical (resp. anticanonical) ideal of degrees $d_{1}$ and $d_{2}$ with $d_{1}<d_{2}$, then for any integer $d$ with $d_{1}<d<d_{2}$, there is a generator of the canonical (resp. anticanonical) ideal with degree $d$. Further, $\mathbb{K}[\mathscr{O}(P)]$ is level (resp. anticanonical level) if and only if “$q^{(1)}$-reduced” (resp. “$q^{(-1)}$-reduced”) sequence with condition N is the empty sequence only. See Definition 4.7 for the definition of the notion $q^{(1)}$-reduced (resp. $q^{(-1)}$-reduced).

In this paper, we show that the almost identical phenomena occur for $\mathbb{K}[\mathscr{C}(P)]$. A Laurent monomial associated to a map $\xi\colon P^{-}\to\mathbb{Z}$ is a generator of the canonical (resp. anticanonical) ideal if and only if there is a sequence with condition N’ with an appropriate relation with $\xi$ (which is different from the one in the case of $\mathbb{K}[\mathscr{O}(P)]$, see Proposition 4.10). Moreover, the symbolic powers of the canonical ideal are identical with the ordinary ones (Theorem 3.8) and the generators of the canonical and anticanonical ideals are consecutive integers (Theorem 5.4). Further, $\mathbb{K}[\mathscr{C}(P)]$ is level (resp. anticanonical level) if and only if $q^{(1)}$-reduced (resp. $q^{(-1)}$-reduced) sequence with condition N’ is the empty sequence only (Theorem 4.17). Since condition N’ is weaker than condition N, it follows that if $\mathbb{K}[\mathscr{C}(P)]$ is level (resp. anticanonical level), then so is $\mathbb{K}[\mathscr{O}(P)]$. We also exhibit examples which show that the converse does not hold. See Examples 4.19 and 4.20.

2 Preliminaries

In this paper, all rings and algebras are assumed to be commutative with identity element unless stated otherwise.

First we state notations about sets used in this paper. We denote by $\mathbb{N}$ the set of nonnegative integers, by $\mathbb{Z}$ the set of integers, by $\mathbb{Q}$ the set of rational numbers and by $\mathbb{R}$ the set of real numbers. We denote the cardinality of a set $X$ by $\#X$. For nonempty sets $X$ and $Y$, we denote the set of maps from $X$ to $Y$ by $Y^{X}$. If $X$ is a finite set, we identify $\mathbb{R}^{X}$ with the Euclidean space $\mathbb{R}^{\#X}$. Let $X$ be a set and $A$ a subset of $X$. We define the characteristic function $\chi_{A}\in\mathbb{R}^{X}$ by $\chi_{A}(x)=1$ for $x\in A$ and $\chi_{A}(x)=0$ for $x\in X\setminus A$.

Now we define a symbol which is frequently used in this paper.

Definition 2.1

Let $X$ be a finite set and $\xi\in\mathbb{R}^{X}$. For $B\subset X$, we set $\xi^{+}(B)\mathrel{:=}\sum_{b\in B}\xi(b)$.

Next we define operations of elements in $\mathbb{R}^{X}$.

Definition 2.2

Let $X$ be a set. For $\xi$, $\xi^{\prime}\in\mathbb{R}^{X}$ and $a\in\mathbb{R}$, we define maps $\xi\pm\xi^{\prime}$ and $a\xi$ by $(\xi\pm\xi^{\prime})(x)=\xi(x)\pm\xi^{\prime}(x)$ and $(a\xi)(x)=a(\xi(x))$ for $x\in X$.

Note that if $X$ is a finite set, $B$ is a subset of $X$ and $a\in\mathbb{R}$, then $(\xi\pm\xi^{\prime})^{+}(B)=\xi^{+}(B)\pm(\xi^{\prime})^{+}(B)$ and $(a\xi)^{+}(B)=a(\xi^{+}(B))$. We denote $f\in\mathbb{R}^{X}$ with $f(x)=0$ for any $x\in X$ by 0.

Next we recall some definitions concerning finite partially ordered sets (poset for short). Let $Q$ be a finite poset. We denote the set of maximal (resp. minimal) elements of $Q$ by $\max Q$ (resp. $\min Q$). If $\max Q$ (resp. $\min Q$) consists of one element $z$, we often abuse notation and write $z=\max Q$ (resp. $z=\min Q$). A chain in $Q$ is a totally ordered subset of $Q$. For a chain $X$ in $Q$, we define the length of $X$ as $\#X-1$. The maximum length of chains in $Q$ is called the rank of $Q$ and denoted $\mathrm{rank}Q$. A subset $A$ of $Q$ is an antichain in $Q$ if every pair of two elements in $A$ are incomparable by the order of $Q$. If $I\subset Q$ and $x\in I$, $y\in Q$, $y\leq x\Rightarrow y\in I$, then we say that $I$ is a poset ideal of $Q$.

Let $+\infty$ (resp. $-\infty$) be a new element which is not contained in $Q$. We define a new poset $Q^{+}$ (resp. $Q^{-}$) whose base set is $Q\cup\{+\infty\}$ (resp. $Q\cup\{-\infty\}$) and $x<y$ in $Q^{+}$ (resp. $Q^{-}$) if and only if $x$, $y\in Q$ and $x<y$ in $Q$ or $x\in Q$ and $y=+\infty$ (resp. $x=-\infty$ and $y\in Q$). We set $Q^{\pm}\mathrel{:=}(Q^{+})^{-}$.

Let $Q^{\prime}$ be an arbitrary poset. (We apply the following definition for $Q^{\prime}=Q$, $Q^{+}$, $Q^{-}$ or $Q^{\pm}$.) If $x$, $y\in Q^{\prime}$, $x<y$ and there is no $z\in Q^{\prime}$ with $x<z<y$, we say that $y$ covers $x$ and denote $x\mathrel{<\!\!\!\cdot}y$ or $y\mathrel{\cdot\!\!\!>}x$. For $x$, $y\in Q^{\prime}$ with $x\leq y$, we set $[x,y]_{Q^{\prime}}\mathrel{:=}\{z\in Q^{\prime}\mid x\leq z\leq y\}$. Further, for $x$, $y\in Q^{\prime}$ with $x<y$, we set $[x,y)_{Q^{\prime}}\mathrel{:=}\{z\in Q^{\prime}\mid x\leq z<y\}$, $(x,y]_{Q^{\prime}}\mathrel{:=}\{z\in Q^{\prime}\mid x<z\leq y\}$ and $(x,y)_{Q^{\prime}}\mathrel{:=}\{z\in Q^{\prime}\mid x<z<y\}$. We denote $[x,y]_{Q^{\prime}}$ (resp. $[x,y)_{Q^{\prime}}$, $(x,y]_{Q^{\prime}}$ or $(x,y)_{Q^{\prime}}$) as $[x,y]$ (resp. $[x,y)$, $(x,y]$ or $(x,y)$) if there is no fear of confusion.

Definition 2.3

Let $Q^{\prime}$ be an arbitrary finite poset and let $x$ and $y$ be elements of $Q^{\prime}$ with $x\leq y$. A saturated chain from $x$ to $y$ is a sequence of elements $z_{0}$, $z_{1}$, …, $z_{t}$ of $Q^{\prime}$ such that

[TABLE]

Note that the length of the chain $z_{0}$, $z_{1}$, …, $z_{t}$ is $t$.

Definition 2.4

Let $Q^{\prime}$, $x$ and $y$ be as above. We define ${\mathrm{dist}}(x,y)\mathrel{:=}\min\{t\mid$ there is a saturated chain from $x$ to $y$ with length $t.\}$ and call ${\mathrm{dist}}(x,y)$ the distance of $x$ and $y$. Further, for $n\in\mathbb{Z}$, we define ${q^{(n)}{\mathrm{dist}}}(x,y)\mathrel{:=}\max\{nt\mid$ there is a saturated chain from $x$ to $y$ with length $t.\}$ and call ${q^{(n)}{\mathrm{dist}}}(x,y)$ the $n$-th quasi-distance of $x$ and $y$.

Note that ${q^{(-1)}{\mathrm{dist}}}(x,y)=-{\mathrm{dist}}(x,y)$ and ${q^{(1)}{\mathrm{dist}}}(x,y)=\mathrm{rank}([x,y])$. Note also that ${q^{(n)}{\mathrm{dist}}}(x,z)+{q^{(n)}{\mathrm{dist}}}(z,y)\leq{q^{(n)}{\mathrm{dist}}}(x,y)$ for any $x$, $z$, $y$ with $x\leq z\leq y$. Further, ${q^{(n)}{\mathrm{dist}}}(x,y)=n$ if $x\mathrel{<\!\!\!\cdot}y$, ${q^{(n)}{\mathrm{dist}}}(x,x)=0$ and ${q^{(mn)}{\mathrm{dist}}}(x,y)=m{q^{(n)}{\mathrm{dist}}}(x,y)$ for any positive integer $m$.

Now we state the following.

Definition 2.5

Let $Q$ be a finite poset and $\xi\in\mathbb{R}^{Q}$. For $z\in Q$, we set $\xi^{+\uparrow}(z)\mathrel{:=}\max\{\sum_{\ell=0}^{s}\xi(z_{\ell})\mid$ there exist elements $z_{0}$, $z_{1}$, …, $z_{s}$ with $z=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}+\infty$ in $Q^{+}\}$. (resp. $\xi^{+^{\prime}\uparrow}(z)\mathrel{:=}\max\{\sum_{\ell=1}^{s}\xi(z_{\ell})\mid$ there exist elements $z_{0}$, $z_{1}$, …, $z_{s}$ with $z=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}+\infty$ in $Q^{+}\}$, where we define the empty sum to be 0). Similarly, we set $\xi^{+\downarrow}(z)\mathrel{:=}\max\{\sum_{\ell=0}^{s}\xi(z_{\ell})\mid$ there exist elements $z_{0}$, $z_{1}$, …, $z_{s}$ with $z=z_{0}\mathrel{\cdot\!\!\!>}z_{1}\mathrel{\cdot\!\!\!>}\cdots\mathrel{\cdot\!\!\!>}z_{s}\mathrel{\cdot\!\!\!>}-\infty$ in $Q^{-}\}$ (resp. $\xi^{+^{\prime}\downarrow}(z)\mathrel{:=}\max\{\sum_{\ell=1}^{s}\xi(z_{\ell})\mid$ there exist elements $z_{0}$, $z_{1}$, …, $z_{s}$ with $z=z_{0}\mathrel{\cdot\!\!\!>}z_{1}\mathrel{\cdot\!\!\!>}\cdots\mathrel{\cdot\!\!\!>}z_{s}\mathrel{\cdot\!\!\!>}-\infty$ in $Q^{-}\}$).

It is clear from the definition that $\xi^{+\uparrow}(z)=\xi(z)+\xi^{+^{\prime}\uparrow}(z)$, $\xi^{+\downarrow}(z)=\xi(z)+\xi^{+^{\prime}\downarrow}(z)$ and $\xi^{+\uparrow}(z)+\xi^{+^{\prime}\downarrow}(z)=\xi^{+^{\prime}\uparrow}(z)+\xi^{+\downarrow}(z)=\max\{\xi^{+}(C)\mid C$ is a maximal chain in $Q$ with $C\ni z\}$. Note that $\xi^{+\uparrow}_{1}(z)+\xi^{+\uparrow}_{2}(z)=(\xi_{1}+\xi_{2})^{+\uparrow}(z)$ does not hold in general.

The following lemma is easily proved.

Lemma 2.6

Let $Q$ be a finite poset and $\xi\in\mathbb{R}^{Q}$.

1. (1)

Suppose that $z\in Q$, $z=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}+\infty$ in $Q^{+}$ and $\xi^{+\uparrow}(z)=\sum_{\ell=0}^{s}\xi(z_{\ell})$ (resp. $z=z_{0}\mathrel{\cdot\!\!\!>}z_{1}\mathrel{\cdot\!\!\!>}\cdots\mathrel{\cdot\!\!\!>}z_{s}\mathrel{\cdot\!\!\!>}-\infty$ in $Q^{-}$ and $\xi^{+\downarrow}(z)=\sum_{\ell=0}^{s}\xi(z_{\ell})$). Then for any $k$ with $0\leq k\leq s$, $\xi^{+\uparrow}(z_{k})=\sum_{\ell=k}^{s}\xi(z_{\ell})$ and $\xi^{+^{\prime}\uparrow}(z_{k})=\sum_{\ell=k+1}^{s}\xi(z_{\ell})$ (resp. $\xi^{+\downarrow}(z_{k})=\sum_{\ell=k}^{s}\xi(z_{\ell})$ and $\xi^{+^{\prime}\downarrow}(z_{k})=\sum_{\ell=k+1}^{s}\xi(z_{\ell})$). 2. (2)

Let $z\in Q\setminus\max Q$ (resp. $z\in Q\setminus\min Q$). Then $\xi^{+^{\prime}\uparrow}(z)=\max\{\xi^{+\uparrow}(z^{\prime})\mid z^{\prime}\mathrel{\cdot\!\!\!>}z\}$ (resp. $\xi^{+^{\prime}\downarrow}(z)=\max\{\xi^{+\downarrow}(z^{\prime})\mid z^{\prime}\mathrel{<\!\!\!\cdot}z\}$). 3. (3)

Suppose that $w_{1}$, $w_{2}\in Q$, $w_{1}<w_{2}$ and $w_{1}=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}=w_{2}$. Then $\xi^{+^{\prime}\uparrow}(w_{1})\geq\sum_{\ell=1}^{s-1}\xi(z_{\ell})+\xi^{+\uparrow}(w_{2})$ and $\xi^{+^{\prime}\downarrow}(w_{2})\geq\sum_{\ell=1}^{s-1}\xi(z_{\ell})+\xi^{+\downarrow}(w_{1})$.

Moreover, we see the following fact.

Lemma 2.7

Let $Q$ be a finite poset, $z\in Q$ and $n$ an integer. Set $M\mathrel{:=}\max\{\xi^{+}(C)\mid C$ is a maximal chain in $Q\}$. Then

[TABLE]

Further, if $x$, $y\in Q$, $x<y$ and $\xi(w)\geq n$ for any $w\in(x,y)$, then

[TABLE]

• Proof

The first assertion is clear from the fact that $\xi^{+\uparrow}(z)+\xi^{+^{\prime}\downarrow}(z)=\max\{\xi^{+}(C^{\prime})\mid C^{\prime}$ is a maximal chain in $Q$ and $C^{\prime}\ni z\}$.

In order to prove the other assertions, take elements

[TABLE]

of $Q$ such that $-\infty\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}=x=z^{\prime}_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime}_{s^{\prime}}=y=z^{\prime\prime}_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime\prime}_{s^{\prime\prime}}\mathrel{<\!\!\!\cdot}+\infty$ in $Q^{\pm}$, $\xi^{+\downarrow}(x)=\sum_{\ell=1}^{s}\xi(z_{\ell})$, ${q^{(n)}{\mathrm{dist}}}(x,y)=s^{\prime}n$ and $\xi^{+\uparrow}(y)=\sum_{\ell=1}^{s^{\prime\prime}}\xi(z^{\prime\prime}_{\ell})$, Then, since $z_{1},\ldots,z_{s},z^{\prime}_{1},\ldots,z^{\prime}_{s^{\prime}-1},z^{\prime\prime}_{1},\ldots,z^{\prime\prime}_{s^{\prime\prime}}$ is a maximal chain in $Q$ and $\xi(z^{\prime}_{\ell})\geq n$ for $1\leq\ell\leq s^{\prime}-1$, we see that

[TABLE]

Further,

[TABLE]

and

[TABLE]

**by Lemma 2.6 (3).  **

Next we fix notation about Ehrhart rings. Let $X$ be a finite set with $-\infty\not\in X$ and $\mathscr{P}$ a rational convex polytope in $\mathbb{R}^{X}$, i.e., a convex polytope whose vertices are contained in $\mathbb{Q}^{X}$. Set $X^{-}\mathrel{:=}X\cup\{-\infty\}$ and let $\{T_{x}\}_{x\in X^{-}}$ be a family of indeterminates indexed by $X^{-}$. For $f\in\mathbb{Z}^{X^{-}}$, we define the Laurent monomial $T^{f}$ by $T^{f}\mathrel{:=}\prod_{x\in X^{-}}T_{x}^{f(x)}$. We set $\deg T_{x}=0$ for $x\in X$ and $\deg T_{-\infty}=1$. Then the Ehrhart ring of $\mathscr{P}$ over a field $\mathbb{K}$ is the $\mathbb{N}$-graded subring

[TABLE]

of the Laurent polynomial ring $\mathbb{K}[T_{x}^{\pm 1}\mid x\in X][T_{-\infty}]$. We denote the Ehrhart ring of $\mathscr{P}$ over $\mathbb{K}$ by $E_{\mathbb{K}}[\mathscr{P}]$. (We use $-\infty$ as the degree indicating element in order to be consistent with the case of Hibi ring.) If $E_{\mathbb{K}}[\mathscr{P}]$ is a standard graded algebra, i.e., generated as a $\mathbb{K}$-algebra by degree $1$ elements, we denote $E_{\mathbb{K}}[\mathscr{P}]$ as $\mathbb{K}[\mathscr{P}]$.

Note that $E_{\mathbb{K}}[\mathscr{P}]$ is Noetherian since $\mathscr{P}$ is rational. Therefore normal by the criterion of Hochster [Hoc]. Further, by the description of the canonical module of a normal affine semigroup ring by Stanley [Sta2, p. 82], we see that the ideal

[TABLE]

of $E_{\mathbb{K}}[\mathscr{P}]$ is the canonical module of $E_{\mathbb{K}}[\mathscr{P}]$, where ${\rm{relint}}\mathscr{P}$ denotes the interior of $\mathscr{P}$ in the affine span of $\mathscr{P}$. We call this ideal the canonical ideal of $E_{\mathbb{K}}[\mathscr{P}]$.

Now we recall the definitions of order and chain polytopes [Sta3]. Let $P$ be a finite poset. The order polytope $\mathscr{O}(P)$ and the chain polytope $\mathscr{C}(P)$ are defined as follows. $\mathscr{O}(P)\mathrel{:=}\{f\in\mathbb{R}^{P}\mid 0\leq f(x)\leq 1$ for any $x\in P$ and if $x<y$ in $P$, then $f(x)\geq f(y)\}$, $\mathscr{C}(P)\mathrel{:=}\{f\in\mathbb{R}^{P}\mid 0\leq f(x)$ for any $x\in P$ and $f^{+}(C)\leq 1$ for any chain in $P\}$. The Ehrhart ring $\mathbb{K}[\mathscr{O}(P)]$ of the order polytope of $P$ is identical with the ring considered by Hibi [Hib], which is nowadays called the Hibi ring. We studied in our previous papers [Miy1, Miy2] the canonical ideals of Hibi rings.

The main object of the study of this paper is $E_{\mathbb{K}}[\mathscr{C}(P)]$ and its canonical ideal. We denote the canonical ideal of $E_{\mathbb{K}}[\mathscr{C}(P)]$ as $\omega$. In order to study $E_{\mathbb{K}}[\mathscr{C}(P)]$, we first collect some basic facts on $E_{\mathbb{K}}[\mathscr{C}(P)]$.

First, the vertices of $\mathscr{C}(P)$ are of the form $\chi_{A}$, where $A$ is an antichain of $P$ (including the empty set). Further, it is easily seen that $\mathscr{C}(P)=\{f\in\mathbb{R}^{P}\mid f(x)\geq 0$ for any $x\in P$ and $f^{+}(C)\leq 1$ for any maximal chain in $P\}$. Thus, ${\rm{relint}}\mathscr{C}(P)=\{f\in\mathbb{R}^{P}\mid f(x)>0$ for any $x\in P$ and $f^{+}(C)<1$ for any maximal chain in $P\}$.

Here, we define the following notation.

Definition 2.8

For $n\in\mathbb{Z}$, we set ${\mathcal{S}}^{(n)}(P)\mathrel{:=}\{\xi\in\mathbb{Z}^{P^{-}}\mid\xi(x)\geq n$ for any $x\in P$ and $\xi(-\infty)\geq\xi^{+}(C)+n$ for any maximal chain $C$ in $P\}$.

In the following, we fix a finite poset $P$ and abbreviate ${\mathcal{S}}^{(n)}(P)$ as ${\mathcal{S}}^{(n)}$ and apply the above definitions by setting $Q=P$, $P^{+}$, $P^{-}$ or $P^{\pm}$.

By the above consideration, we can describe $E_{\mathbb{K}}[\mathscr{C}(P)]$ and $\omega$ by using this notation.

[TABLE]

Finally in this section, we note the following fact which seems to be a folklore.

Proposition 2.9

$E_{\mathbb{K}}[\mathscr{C}(P)]$* is a standard graded $\mathbb{K}$-algebra.*

• Proof

It is enough to show that if $\xi\in{\mathcal{S}}^{(0)}$ and $\xi(-\infty)\geq 2$, then there are $\xi_{1}$, $\xi_{2}\in{\mathcal{S}}^{(0)}$ such that $\xi_{1}(-\infty)=1$ and $\xi_{1}+\xi_{2}=\xi$.

If $\xi(x)=0$ for any $x\in P$, it is enough to set $\xi_{1}(x)=\xi_{2}(x)=0$, for any $x\in P$, $\xi_{1}(-\infty)=1$ and $\xi_{2}(-\infty)=\xi(-\infty)-1$. Suppose that $\xi(x)>0$ for some $x\in P$. Set $A\mathrel{:=}\max\{x\in P\mid\xi(x)>0\}\cup\{-\infty\}$, $\xi_{1}\mathrel{:=}\chi_{A}$ and $\xi_{2}\mathrel{:=}\xi-\xi_{1}$. First it is clear from the definition that $\xi_{1}(-\infty)=1$ and $\xi=\xi_{1}+\xi_{2}$. Let $x$ be an arbitrary element of $P$. If $x\in A$, then $\xi(x)>0$ and $\xi_{1}(x)=1$. Thus, $\xi_{2}(x)\geq 0$. If $x\not\in A$. then $\xi_{1}(x)=0$ and $\xi(x)\geq 0$. Thus, $\xi_{2}(x)\geq 0$.

Now let $C$ be an arbitrary maximal chain in $P$. We have to show that $\xi^{+}_{1}(C)\leq\xi_{1}(-\infty)$ and $\xi^{+}_{2}(C)\leq\xi_{2}(-\infty)$. If $C\cap A\neq\emptyset$, then there is a unique element $x\in C\cap A$. Therefore, $\xi^{+}_{1}(C)=1=\xi_{1}(-\infty)$ and $\xi^{+}_{2}(C)=\xi^{+}(C)-1\leq\xi(-\infty)-1=\xi_{2}(-\infty)$.

Suppose that $C\cap A=\emptyset$. If $\xi(c)=0$ for any $c\in C$, $\xi^{+}_{1}(C)=\xi^{+}(C)=0$. Thus, $\xi^{+}_{1}(C)\leq 1=\xi_{1}(-\infty)$ and $\xi^{+}_{2}(C)=\xi^{+}(C)=0\leq\xi_{2}(-\infty)$. Assume that $\xi(c)>0$ for some $c\in C$. Set $x\mathrel{:=}\max\{c\in C\mid\xi(c)>0\}$. Since $x\not\in A$, there exists $y\in A$ with $y>x$. Take a saturated chain $x=z_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}=y$ and $y=z^{\prime}_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime}_{s^{\prime}}\mathrel{<\!\!\!\cdot}+\infty$. Then $C^{\prime}\mathrel{:=}(C\cap(-\infty,x])\cup\{z_{1},\ldots,z_{s},z^{\prime}_{1},\ldots,z^{\prime}_{s^{\prime}}\}$ is a maximal chain in $P$. Further, $\xi^{+}(C^{\prime})>\xi^{+}(C)$ since $\xi(c)=0$ for any $c\in C$ with $c>x$ and $C^{\prime}\ni z_{s}=y\in A$. Therefore, $\xi^{+}(C)\leq\xi^{+}(C^{\prime})-1\leq\xi(-\infty)-1$. Since $C\cap A=\emptyset$, $\xi^{+}_{1}(C)=\chi^{+}_{A}(C)=0\leq 1=\xi_{1}(-\infty)$ and $\xi^{+}_{2}(C)=\xi^{+}(C)-\xi^{+}_{1}(C)=\xi^{+}(C)\leq\xi(-\infty)-1=\xi_{2}(-\infty)$.

**Thus, we see that $\xi_{1}$, $\xi_{2}\in{\mathcal{S}}^{(0)}$.  **

3 Symbolic powers of the canonical ideal

In this section, we consider the symbolic powers $\omega^{(n)}$ of the canonical ideal $\omega$ of $\mathbb{K}[\mathscr{C}(P)]$ and describe the Laurent monomial basis of $\omega^{(n)}$ as a vector space over $\mathbb{K}$. Further, we show that the symbolic powers of $\omega$ are equal to the ordinary ones.

Here we note the following fact. Let $R$ be a Noetherian normal domain and $I$ a fractional ideal. $I$ is said to be divisorial if $R:_{Q(R)}(R:_{Q(R)}I)=I$, i.e., $I$ is reflexive as an $R$-module, where $Q(R)$ is the fraction field of $R$. It is known that the set of divisorial ideals form a group, denoted ${\mathrm{Div}}(R)$, by the operation $I\cdot J\mathrel{:=}R:_{Q(R)}(R:_{Q(R)}IJ)$. We denote the $n$-th power of $I$ in this group $I^{(n)}$, where $n\in\mathbb{Z}$. Note that if $I\subsetneq R$, then $I^{(n)}$ is identical with the $n$-th symbolic power of $I$. Note also that the inverse element of $I$ in ${\mathrm{Div}}(R)$ is $R:_{Q(R)}I$.

Suppose further that $R$ is an affine semigroup ring generated by Laurent monomials in the Laurent polynomial ring $\mathbb{K}[X_{1}^{\pm 1},\ldots,X_{s}^{\pm 1}]$ over $\mathbb{K}$, where $\mathbb{K}$ is a field and $X_{1}$, …, $X_{s}$ are indeterminates. Let $I$ be a divisorial ideal generated by Laurent monomials $m_{1}$, …, $m_{\ell}$. Then $R:_{Q(R)}I=\bigcap_{i=1}^{\ell}Rm_{i}^{-1}$. Thus, $R:_{Q(R)}I$ is an $R$-submodule of of $\mathbb{K}[X_{1}^{\pm 1},\ldots,X_{s}^{\pm 1}]$ generated by Laurent monomials. Therefore, the set of divisorial ideals generated by Laurent monomials form a subgroup of ${\mathrm{Div}}(R)$. It is known that the canonical module $\omega$ is reflexive and isomorphic to an ideal. Therefore $\omega\in{\mathrm{Div}}(R)$. Thus, if the canonical module $\omega$ of $R$ is isomorphic to an ideal of $R$ generated by Laurent monomials, then the inverse element $\omega^{(-1)}$ of $\omega$ in ${\mathrm{Div}}(R)$ is also an $R$-submodule $\mathbb{K}[X_{1}^{\pm 1},\ldots,X_{s}^{\pm 1}]$ generated by Laurent monomials.

Since an $R$-submodule of $\mathbb{K}[X_{1}^{\pm 1},\ldots,X_{s}^{\pm 1}]$ generated by Laurent monomials has a unique system of generators consisting of Laurent monomials, we call an element of this system a generator of the $R$-submodule.

Taking into account of this fact, we first note the following fact.

Lemma 3.1

Let $n$ be an integer. Set $\xi\in\mathbb{Z}^{P^{-}}$ by

[TABLE]

then $\xi\in{\mathcal{S}}^{(n)}$.

• Proof

**It is clear that $\xi(z)\geq n$ for any $z\in P$. Let $C$ be an arbitrary maximal chain in $P$. Then $C\cup\{-\infty,\infty\}$ is a saturated chain from $-\infty$ to $\infty$ of length $\#(C\cup\{-\infty,\infty\})-1$. Thus, $n(\#(C\cup\{-\infty,\infty\})-1)\leq{q^{(n)}{\mathrm{dist}}}(-\infty,\infty)=\xi(-\infty)$. Therefore, $\xi^{+}(C)+n=n(\#C)+n\leq\xi(-\infty)$.  **

Next we state the following.

Proposition 3.2

Let $n$ be an integer and $\xi\in\mathbb{Z}^{P^{-}}$. Then $T^{\xi}\in\omega^{(n)}$ if and only if $\xi\in{\mathcal{S}}^{(n)}$. In particular,

[TABLE]

• Proof

The cases where $n=0$ and $1$ are noted in the previous section. Let $m$ be a positive integer.

We first show that $T^{\xi}\in\omega^{(-m)}$ if and only if $\xi\in{\mathcal{S}}^{(-m)}$. In order to prove the “only if” part, first let $\eta$ be a map from $P^{-}$ to $\mathbb{Z}$ such that $\eta(z)=1$ for any $z\in P$ and $\eta(-\infty)=q^{(1)}{\mathrm{dist}}(-\infty,\infty)$. Then $T^{\eta}\in\omega$ by Lemma 3.1. Thus, $T^{m\eta}\in\omega^{m}$. Since $T^{\xi}\in\omega^{(-m)}$, we see that $T^{\xi+m\eta}\in\mathbb{K}[\mathscr{C}(P)]$. Therefore, $\xi(z)+m\eta(z)\geq 0$ for any $z\in P$ and we see that $\xi(z)\geq-m$ since $\eta(z)=1$. Next, let $C$ be an arbitrary maximal chain in $P$, $N$ a huge integer ($N>q^{(1)}{\mathrm{dist}}(-\infty,\infty)$) and let $\eta^{\prime}$ be a map from $P^{-}$ to $\mathbb{Z}$ such that $\eta^{\prime}(z)=N$ for $z\in C$, $\eta^{\prime}(z)=1$ for $z\in P\setminus C$ and $\eta^{\prime}(-\infty)=N(\#C)+1$. Then $T^{\eta^{\prime}}\in\omega$ since $N$ is a huge integer and therefore $T^{m\eta^{\prime}}\in\omega^{m}$. Thus, $T^{\xi+m\eta^{\prime}}\in\mathbb{K}[\mathscr{C}(P)]$ since $T^{\xi}\in\omega^{(-m)}$. Therefore, $\xi(-\infty)+m\eta^{\prime}(-\infty)\geq\xi^{+}(C)+m((\eta^{\prime})^{+}(C))$. Since $\eta^{\prime}(-\infty)=N(\#C)+1$ and $(\eta^{\prime})^{+}(C)=N(\#C)$, we see that $\xi(-\infty)-\xi^{+}(C)\geq m((\eta^{\prime})^{+}(C))-m\eta^{\prime}(-\infty)=-m$, i.e., $\xi(-\infty)\geq\xi^{+}(C)-m$.

Next, we prove the “if” part. Let $T^{\eta_{1}}$, …, $T^{\eta_{m}}$ be arbitrary Laurent monomials in $\omega$, where $\eta_{i}\in\mathbb{Z}^{P^{-}}$ for $1\leq i\leq m$. Since $\eta_{i}(z)\geq 1$ and $\xi(z)\geq-m$ for any $z\in P$ and $i$, we see that $(\xi+\eta_{1}+\cdots+\eta_{m})(z)\geq 0$ for any $z\in P$. Next let $C$ be an arbitrary maximal chain in $P$. Since $\eta_{i}^{+}(C)+1\leq\eta_{i}(-\infty)$ for any $i$, we see that $(\eta_{1}+\cdots+\eta_{m})^{+}(C)+m\leq(\eta_{1}+\cdots+\eta_{m})(-\infty)$. Since $\xi(-\infty)\geq\xi^{+}(C)-m$ by assumption, we see that $(\xi+\eta_{1}+\cdots+\eta_{m})(-\infty)\geq(\xi+\eta_{1}+\cdots+\eta_{m})^{+}(C)$. Therefore, we see that $T^{\xi+\eta_{1}+\cdots+\eta_{m}}\in\mathbb{K}[\mathscr{C}(P)]$. Since $T^{\eta_{1}}$, …, $T^{\eta_{m}}$ are arbitrary Laurent monomials in $\omega$, we see that $T^{\xi}\in\omega^{(-m)}$.

We next show that for $\xi\in\mathbb{Z}^{P^{-}}$, $T^{\xi}\in\omega^{(m)}$ if and only if $\xi\in{\mathcal{S}}^{(m)}$. We first show the “only if” part. Set $\zeta\in\mathbb{Z}^{P^{-}}$ by $\zeta(z)=-m$ for any $z\in P$ and $\zeta(-\infty)={q^{(-m)}{\mathrm{dist}}}(-\infty,\infty)$. Then $\zeta\in{\mathcal{S}}^{(-m)}$ by Lemma 3.1 and therefore we see that $T^{\zeta}\in\omega^{(-m)}$ by the fact proved above. Since $T^{\xi+\zeta}\in\mathbb{K}[\mathscr{C}(P)]$ by assumption, we see that $\xi(z)+\zeta(z)\geq 0$ for any $z\in P$. Thus, $\xi(z)\geq-\zeta(z)=m$ for any $z\in P$. Next let $C$ be an arbitrary maximal chain in $P$. Set $\zeta^{\prime}\in\mathbb{Z}^{P^{-}}$ by

[TABLE]

Then $\zeta^{\prime}\in{\mathcal{S}}^{(-m)}$ and therefore by the criterion of a Laurent monomial to be an element of $\omega^{(-m)}$ proved above, we see that $T^{\zeta^{\prime}}\in\omega^{(-m)}$. Thus, $T^{\xi+\zeta^{\prime}}\in\mathbb{K}[\mathscr{C}(P)]$ by assumption and we see that $\xi^{+}(C)+(\zeta^{\prime})^{+}(C)\leq\xi(-\infty)+\zeta^{\prime}(-\infty)$. Therefore, $\xi(-\infty)\geq\xi^{+}(C)+(\zeta^{\prime})^{+}(C)-\zeta^{\prime}(-\infty)=\xi^{+}(C)+m$.

**Next we show the “if” part. Let $\xi$ be a map from $P^{-}$ to $\mathbb{Z}$ such that $\xi(z)\geq m$ for any $z\in P$ and $\xi(-\infty)\geq\xi^{+}(C)+m$ for any maximal chain $C$ in $P$. Let $T^{\zeta}$, $\zeta\in\mathbb{Z}^{P^{-}}$ be an arbitrary Laurent monomial in $\omega^{(-m)}$. Then by the first part of this proof, we see that $\zeta(z)\geq-m$ for any $z\in P$ and $\zeta(-\infty)\geq\zeta^{+}(C)-m$ for any maximal chain $C$ in $P$. Therefore, $(\xi+\zeta)(z)=\xi(z)+\zeta(z)\geq 0$ and $(\xi+\zeta)(-\infty)=\xi(-\infty)+\zeta(-\infty)\geq\xi^{+}(C)+m+\zeta^{+}(C)-m=(\xi+\zeta)^{+}(C)$ for any maximal chain $C$ in $P$. Thus, we see that $T^{\xi+\zeta}\in\mathbb{K}[\mathscr{C}(P)]$. Since $T^{\zeta}$ is an arbitrary Laurent monomial in $\omega^{(-m)}$, we see that $T^{\xi}\in\omega^{(m)}$.  **

As a corollary, we see the following fact.

Corollary 3.3

Let $n$ be an integer and let $\xi\colon P^{-}\to\mathbb{Z}$ be the map defined in Lemma 3.1. Then $T^{\xi}$ is an element of $\omega^{(n)}$ with $\deg T^{\xi}={q^{(n)}{\mathrm{dist}}}(-\infty,\infty)$ and for any map $\zeta\colon P^{-}\to\mathbb{Z}$ with $T^{\zeta}\in\omega^{(n)}$, $\deg T^{\zeta}\geq\deg T^{\xi}$.

• Proof

**By Lemma 3.1, we see that $\xi\in{\mathcal{S}}^{(n)}$. Thus, $T^{\xi}\in\omega^{(n)}$ by Proposition 3.2 and $\deg T^{\xi}=\xi(-\infty)={q^{(n)}{\mathrm{dist}}}(-\infty,+\infty)$. Also, by Proposition 3.2, we see that $\zeta\in{\mathcal{S}}^{(n)}$. Take a maximal chain $C$ in $P$ with ${q^{(n)}{\mathrm{dist}}}(-\infty,\infty)=n(\#(C\cup\{-\infty,\infty\})-1)$. Then ${q^{(n)}{\mathrm{dist}}}(-\infty,\infty)=n(\#C)+n$. Since $\zeta(c)\geq n$ for any $c\in C$, we see that $\zeta^{+}(C)\geq n(\#C)$. Thus, $\deg T^{\zeta}=\zeta(-\infty)\geq\zeta^{+}(C)+n={q^{(n)}{\mathrm{dist}}}(-\infty,\infty)=\deg T^{\xi}$.  **

Next, we show that the symbolic powers of $\omega$ are equal to ordinary ones. For $\alpha\in\mathbb{R}$, we denote by $\lfloor\alpha\rfloor$ the maximum integer less than or equals to $\alpha$ and by $\lceil\alpha\rceil$ the minimum integer larger than or equals to $\alpha$. Let $\epsilon$ be $1$ or $-1$, $n$ an integer with $n\geq 2$ and $T^{\xi}$ an arbitrary Laurent monomial in $\omega^{(n\epsilon)}$, where $\xi\in\mathbb{Z}^{P^{-}}$. In this setting, we define two maps $\xi_{1}$ and $\xi_{2}\in\mathbb{Z}^{P^{-}}$ as follows. Set $\xi_{1}$ by

[TABLE]

and $\xi_{2}=\xi-\xi_{1}$. We first show the following.

Lemma 3.4

In the above notation, $\xi^{+\uparrow}_{1}(z)=\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor$ for any $z\in P$ and $\max\{\xi^{+}_{1}(C)\mid C$ is a maximal chain in $P\}=\lfloor\frac{1}{n}\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}\rfloor$.

• Proof

First we show the former assertion by Noetherian induction on $z$. The case where $z$ is a maximal element of $P$ is trivial. Assume that $z$ is not maximal. Then by Lemma 2.6 (2), we see that $\xi^{+^{\prime}\uparrow}(z)=\max\{\xi^{+\uparrow}(z^{\prime})\mid z^{\prime}\mathrel{\cdot\!\!\!>}z\}$. Since $\lfloor\frac{1}{n}\max\{\xi^{+\uparrow}(z^{\prime})\mid z^{\prime}\mathrel{\cdot\!\!\!>}z\}\rfloor=\max\{\lfloor\frac{1}{n}\xi^{+\uparrow}(z^{\prime})\rfloor\mid z^{\prime}\mathrel{\cdot\!\!\!>}z\}$, we see by the induction hypothesis and Lemma 2.6 (2) that $\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor=\max\{\xi^{+\uparrow}_{1}(z^{\prime})\mid z^{\prime}\mathrel{\cdot\!\!\!>}z\}=\xi^{+^{\prime}\uparrow}_{1}(z)$. Therefore, by the definition of $\xi_{1}$, we see that $\xi^{+\uparrow}_{1}(z)=\xi_{1}(z)+\xi^{+^{\prime}\uparrow}_{1}(z)=\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor-\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor+\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor=\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor$.

**Next we show the latter assertion. Since $\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}=\max\{\xi^{+\uparrow}(z)\mid z$ is a minimal element of $P\}$ and the corresponding equation for $\xi_{1}$, we see that $\max\{\xi^{+}_{1}(C)\mid C$ is a maximal chain in $P\}=\max\{\xi^{+\uparrow}_{1}(z)\mid z$ is a minimal element of $P\}=\max\{\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor\mid z$ is a minimal element of $P\}=\lfloor\frac{1}{n}\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}\rfloor$.  **

Now we prove the following.

Lemma 3.5

In the above setting, $\xi_{1}\in{\mathcal{S}}^{(\epsilon)}$.

• Proof

It is enough to show that $\xi_{1}(z)\geq\epsilon$ for any $z\in P$ and $\xi_{1}(-\infty)\geq\max\{\xi^{+}_{1}(C)\mid C$ is a maximal chain in $P\}+\epsilon$.

Let $z$ be an arbitrary element of $P$. If $z$ is a maximal element of $P$, then $\xi_{1}(z)=\lfloor\frac{1}{n}\xi(z)\rfloor\geq\epsilon$ since $\xi(z)\geq n\epsilon$ by Proposition 3.2. Assume that $z$ is not a maximal element of $P$. Then $\xi^{+\uparrow}(z)=\xi(z)+\xi^{+^{\prime}\uparrow}(z)\geq n\epsilon+\xi^{+^{\prime}\uparrow}(z)$. Therefore, $\xi_{1}(z)=\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor-\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor\geq\lfloor\frac{1}{n}(n\epsilon+\xi^{+^{\prime}\uparrow}(z))\rfloor-\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor=\epsilon$.

**Further, since $T^{\xi}\in\omega^{(n\epsilon)}$, we see by Proposition 3.2 that $\xi(-\infty)\geq\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}+n\epsilon$. Thus, $\xi_{1}(-\infty)=\lfloor\frac{1}{n}\xi(-\infty)\rfloor\geq\lfloor\frac{1}{n}(\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}+n\epsilon)\rfloor=\max\{\xi^{+}_{1}(C)\mid C$ is a maximal chain in $P\}+\epsilon$ by Lemma 3.4.  **

Next we show that $\xi_{2}\in{\mathcal{S}}^{((n-1)\epsilon)}$. First we state the following.

Lemma 3.6

In the above setting, $\xi^{+\uparrow}_{2}(z)=\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil$ for any $z\in P$ and $\max\{\xi^{+}_{2}(C)\mid C$ is a maximal chain in $P\}=\lceil\frac{n-1}{n}\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}\rceil$.

• Proof

We first prove the former assertion by Noetherian induction on $z$. If $z$ is a maximal element of $P$, then $\xi^{+\uparrow}_{2}(z)=\xi_{2}(z)=\xi(z)-\xi_{1}(z)=\xi(z)-\lfloor\frac{1}{n}\xi(z)\rfloor=\lceil\frac{n-1}{n}\xi(z)\rceil=\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil$ by the definition of $\xi_{1}$ and $\xi_{2}$. Suppose that $z$ is not a maximal element of $P$. Take $z_{1}\in P$ with $z_{1}\mathrel{\cdot\!\!\!>}z$ and $\xi^{+\uparrow}(z_{1})=\max_{z^{\prime}\mathrel{\cdot\!\!\!>}z}\xi^{+\uparrow}(z^{\prime})$. Then since $\xi^{+\uparrow}_{2}(z)=\xi_{2}(z)+\max_{z^{\prime}\mathrel{\cdot\!\!\!>}z}\xi^{+\uparrow}_{2}(z^{\prime})$, we see by induction hypothesis that $\xi^{+\uparrow}_{2}(z)=\xi_{2}(z)+\max_{z^{\prime}\mathrel{\cdot\!\!\!>}z}\lceil\frac{n-1}{n}\xi^{+\uparrow}(z^{\prime})\rceil=\xi_{2}(z)+\lceil\frac{n-1}{n}\xi^{+\uparrow}(z_{1})\rceil$. Since $\xi_{2}(z)=\xi(z)-\xi_{1}(z)=\xi(z)-(\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor-\lfloor\frac{1}{n}\xi^{+^{\prime}\uparrow}(z)\rfloor)=\xi(z)-\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor+\lfloor\frac{1}{n}\xi^{+\uparrow}(z_{1})\rfloor$, we see that $\xi^{+\uparrow}_{2}(z)=\xi_{2}(z)+\lceil\frac{n-1}{n}\xi^{+\uparrow}(z_{1})\rceil=\xi(z)-\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor+\lfloor\frac{1}{n}\xi^{+\uparrow}(z_{1})\rfloor+\lceil\frac{n-1}{n}\xi^{+\uparrow}(z_{1})\rceil=\xi(z)+\xi^{+\uparrow}(z_{1})-\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor=\xi^{+\uparrow}(z)-\lfloor\frac{1}{n}\xi^{+\uparrow}(z)\rfloor=\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil$.

**Now we prove the latter assertion. By the fact $\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}=\max\{\xi^{+\uparrow}(z)\mid z$ is a minimal element of $P\}$ and the corresponding fact for $\xi_{2}$, we see that $\max\{\xi^{+}_{2}(C)\mid C$ is a maximal chain in $P\}=\max\{\xi^{+\uparrow}_{2}(z)\mid z$ is a minimal element of $P\}=\max\{\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil\mid z$ is a minimal element of $P\}=\lceil\frac{n-1}{n}\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}\rceil$.  **

Now we show the following.

Lemma 3.7

$\xi_{2}\in{\mathcal{S}}^{((n-1)\epsilon)}$.

• Proof

Let $z\in P$. If $z$ is a maximal element of $P$, then by Lemma 3.6 $\xi_{2}(z)=\xi^{+\uparrow}_{2}(z)=\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil=\lceil\frac{n-1}{n}\xi(z)\rceil\geq(n-1)\epsilon$, since $\xi(z)\geq n\epsilon$. Next suppose that $z$ is not a maximal element of $P$. Take $z_{1}\in P$ with $z_{1}\mathrel{\cdot\!\!\!>}z$ and $\xi^{+\uparrow}(z_{1})=\max_{z^{\prime}\mathrel{\cdot\!\!\!>}z}\xi^{+\uparrow}(z^{\prime})=\xi^{+^{\prime}\uparrow}(z)$. Then by Lemma 3.6, we see that $\xi^{+\uparrow}_{2}(z_{1})=\max_{z^{\prime}\mathrel{\cdot\!\!\!>}z}\xi^{+\uparrow}_{2}(z^{\prime})=\xi^{+^{\prime}\uparrow}_{2}(z)$. Therefore, by Lemma 3.6, we see that $\xi_{2}(z)=\xi^{+\uparrow}_{2}(z)-\xi^{+\uparrow}_{2}(z_{1})=\lceil\frac{n-1}{n}\xi^{+\uparrow}(z)\rceil-\lceil\frac{n-1}{n}\xi^{+\uparrow}(z_{1})\rceil\geq(n-1)\epsilon$, since $\xi^{+\uparrow}(z)=\xi(z)+\xi^{+\uparrow}(z_{1})\geq n\epsilon+\xi^{+\uparrow}(z_{1})$.

**Further, since $\xi(-\infty)\geq\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}+n\epsilon$, we see by Lemma 3.6 that $\xi_{2}(-\infty)=\xi(-\infty)-\xi_{1}(-\infty)=\xi(-\infty)-\lfloor\frac{1}{n}\xi(-\infty)\rfloor=\lceil\frac{n-1}{n}\xi(-\infty)\rceil\geq\lceil\frac{n-1}{n}(\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}+n\epsilon)\rceil=\max\{\xi^{+}_{2}(C)\mid C$ is a maximal chain in $P\}+(n-1)\epsilon$. Thus, we see that $\xi_{2}\in{\mathcal{S}}^{((n-1)\epsilon)}$.  **

By Proposition 3.2, Lemmas 3.5 and 3.7, and induction on $n$, we see the following.

Theorem 3.8

Let $n$ be a positive integer. Then

[TABLE]

and

[TABLE]

Remark 3.9

Let $P=\{x,y\}$ be a poset with $x<y$ and let $\xi\colon P^{-}\to\mathbb{Z}$ be a map with $\xi(x)=\xi(y)=3$ and $\xi(-\infty)=8$. Then $\xi\in{\mathcal{S}}^{(2)}$. If we set $\xi^{\prime}(w)=\lfloor\frac{1}{2}\xi(w)\rfloor$ for any $w\in P^{-}$, then $\xi^{\prime}\in{\mathcal{S}}^{(1)}$ but $\xi-\xi^{\prime}\not\in{\mathcal{S}}^{(1)}$. Thus, simply setting $\xi_{1}(w)=\lfloor\frac{1}{n}\xi(w)\rfloor$ for any $w\in P^{-}$ does not meet our demand.

Further, let $P_{1}=\{x_{1},x_{2},y\}$ be a poset with order relation $x_{1}<x_{2}$ and let $\xi\colon P^{-}\to\mathbb{Z}$ be the map with $\xi(x_{1})=\xi(x_{2})=2$, $\xi(y)=4$ and $\xi(-\infty)=6$. Then $\xi\in{\mathcal{S}}^{(2)}$. If we set $\xi^{\prime}(z)=1$ for $z\in P$ and $\xi^{\prime}(-\infty)=3$, then $\xi^{\prime}\in{\mathcal{S}}^{(1)}$ but $\xi-\xi^{\prime}\not\in{\mathcal{S}}^{(1)}$. Thus, setting $\xi_{1}(z)=1$ for any $z\in P$ and $\xi_{1}(-\infty)=q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$ does not meet our demand also.

For a poset $Q$, we define a graph $G(Q)$ whose vertex set is $Q$ and for $a$, $b\in Q$, $\{a,b\}$ is an edge of $G(Q)$ if and only if $a\neq b$ and $a$ and $b$ are comparable by the order of $Q$. A graph $G$ such that there is a poset $Q$ with $G=G(Q)$ is called a comparability graph. Recall that for a finite graph $G$ with vertex set $V$ and edge set $E$, a stable set of $G$ is a subset of $V$ pairwise nonadjacent. Further, the stable set polytope (vertex-packing polytope) of $G$ is the convex polytope in $\mathbb{R}^{V}$ which is the convex hull of $\{\chi_{A}\mid A$ is a stable set$\}$.

Since in our setting, $\mathscr{C}(P)$ is the convex hull of $\{\chi_{A}\mid A$ is an antichain of $P\}$ and a subset of $P$ is an antichain if and only if it is a stable set of $G(P)$, we see that $\mathscr{C}(P)$ is the stable set polytope of $G(P)$. Further, it is known the following fact.

Fact 3.10** ([GH, Theorem 1])**

Let $G$ be a graph. Then $G$ is a comparability graph of a poset if and only if each odd cycle of $G$ has at least one triangular chord.

By the consideration above, we see the following.

Corollary 3.11

Let $G$ be a finite graph. Suppose that each odd cycle of $G$ has a triangular chord. Denote the canonical ideal of the Ehrhart ring of the stable set polytope of $G$ as $\omega$. Then for any positive integer $n$, $\omega^{(n)}=\omega^{n}$ and $\omega^{(-n)}=(\omega^{(-1)})^{n}$. In particular, the symbolic power of the canonical ideal of the Ehrhart ring of the stable set polytope of a bipartite graph is identical with the ordinary power.

4 Characterizations of level and anticanonical level properties of $\mathbb{K}[\mathscr{C}(P)]$

In this section, we describe generators of $\omega$ (resp. $\omega^{(-1)}$) by “sequence with condition N’ ” in $P$. Recall that we [Miy1, Miy2] described generators of the canonical and anticanonical ideals of $\mathbb{K}[\mathscr{O}(P)]$ by “sequence with condition N” in $P$.

First we recall the definition of condition N and define condition N’ of a sequence of elements of $P$.

Definition 4.1

Let $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$, …, $y_{t-1}$, $x_{t}$ be a sequence of elements of $P$. We say that $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$, …, $y_{t-1}$, $x_{t}$ satisfy condition N (resp. condition N’) if

1. (1)

$y_{0}>x_{1}<y_{1}>x_{2}<\cdots<y_{t-1}>x_{t}$ and 2. (2)

if $i\leq j-2$, then $y_{i}\not\geq x_{j}$ (resp. $y_{i}\not>x_{j}$).

We define that an empty sequence (i.e., $t=0$) satisfy condition N (resp. condition N’).

When considering a sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N (resp. condition N’), we usually set $x_{0}=-\infty$ and $y_{t}=\infty$ and consider a sequence $x_{0}$, $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$, $y_{t}$ in $P^{\pm}$.

We note the following basic fact because it is used frequently.

Lemma 4.2

Let $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ be a sequence with condition N’. If $y_{i}\leq y_{j}$ (or $x_{i}\leq x_{j}$) and $i\neq j$, then $i<j$. In particular, $y_{0}$, $y_{1}$, …, $y_{t-1}$ (resp. $x_{1}$, $x_{2}$, …, $x_{t}$) are different elements of $P$. Further, if $y_{k}\leq x_{\ell}$, then $\ell\geq k+2$.

• Proof

**If $j<i$, then $y_{j}\geq y_{i}>x_{i+1}$ (or $y_{j-1}>x_{j}\geq x_{i}$) and $j\leq(i+1)-2$ (or $j-1\leq i-2$), contradicting condition N’. Similarly, if $\ell\leq k$, then $\ell-1\leq(k+1)-2$ and $y_{\ell-1}>x_{\ell}\geq y_{k}>x_{k+1}$ violating condition N’. Further, since $x_{k+1}<y_{k}$, we see $\ell\neq k+1$. Thus, $\ell\geq k+2$.  **

Here we recall the definition of level (resp. anticanonical level) property.

Definition 4.3** ([Sta1, Pag])**

Let $R$ be a standard graded Cohen-Macaulay algebra over a field. If the degree of all the generators of the canonical module $\omega$ of $R$ are the same, then we say that $R$ is level. Moreover, if $R$ is normal (thus, is a domain) and the degree of all the generators of $\omega^{(-1)}$ are the same, we say that $R$ is anticanonical level.

In the following, we characterize if a Laurent monomial in $\omega$ (resp. $\omega^{(-1)}$) is a generator by sequences with condition N’ in $P$ and also characterize level and anticanonical level properties of $\mathbb{K}[\mathscr{C}(P)]$ by sequences with condition N’ in $P$. Since the cases of $\omega$ and $\omega^{(-1)}$ are similar, we treat both cases simultaneously.

In the following, let $\epsilon$ be $1$ or $-1$. Further, we define the following notation.

Definition 4.4

Let $\xi\in\mathbb{Z}^{P}$, $\xi\in\mathbb{Z}^{P^{-}}$, $\xi\in\mathbb{Z}^{P^{+}}$ or $\xi\in\mathbb{Z}^{P^{\pm}}$ and $n\in\mathbb{Z}$. We set $C_{\xi}^{[n]}\mathrel{:=}\{C\mid C$ is a maximal chain in $P$ and $\xi^{+}(C)=n\}$.

Note that if $\xi\in{\mathcal{S}}^{(m)}$ and $\xi(-\infty)=d$, then $C_{\xi}^{[n]}=\emptyset$ for any $n$ with $n>d-m$.

Next we note the following.

Lemma 4.5

Suppose that $\xi\in{\mathcal{S}}^{(\epsilon)}$ and set $d\mathrel{:=}\xi(-\infty)$. Then $T^{\xi}$ is not a generator of $\omega^{(\epsilon)}$ if and only if there is an antichain $A$ (which may be an empty set) such that for any $C\in C_{\xi}^{[d-\epsilon]}$, $C\cap A\neq\emptyset$ and $\xi(a)>\epsilon$ for the unique element $a\in C\cap A$. (This condition is trivially satisfied if $C_{\xi}^{[d-\epsilon]}=\emptyset$.)

• Proof

We first prove the “if” part. Set $\xi_{1}\colon P^{-}\to\mathbb{Z}$ by

[TABLE]

and $\xi_{2}\mathrel{:=}\xi-\xi_{1}$. Then $\xi_{2}(z)\geq\epsilon$ for any $z\in P$ by assumption. Further, $C_{\xi_{2}}^{[m]}=\emptyset$ for any $m$ with $m\geq d-\epsilon$ since $A\cap C\neq\emptyset$ for any $C\in C_{\xi}^{[d-\epsilon]}$ and $C_{\xi}^{[n]}=\emptyset$ for any $n$ with $n>d-\epsilon$. Thus, $\xi^{+}_{2}(C)\leq d-\epsilon-1$ for any maximal chain $C$ in $P$. Therefore, $\xi_{2}\in{\mathcal{S}}^{(\epsilon)}$ since $\xi_{2}(-\infty)=d-1$ and we see that $T^{\xi_{2}}\in\omega^{(\epsilon)}$. Since $\xi_{1}\in{\mathcal{S}}^{(0)}$, we see that $T^{\xi_{1}}\in\mathbb{K}[\mathscr{C}(P)]$ and therefore $T^{\xi}$ is not a generator of $\omega^{(\epsilon)}$, since $T^{\xi}=T^{\xi_{1}}T^{\xi_{2}}$.

**Now we prove the “only if” part. Suppose that $T^{\xi}$ is not a generator of $\omega^{(\epsilon)}$. Then, since $\mathbb{K}[\mathscr{C}(P)]$ is a standard graded algebra, there are $\xi_{1}\in{\mathcal{S}}^{(0)}$ and $\xi_{2}\in{\mathcal{S}}^{(\epsilon)}$, such that $\xi=\xi_{1}+\xi_{2}$ and $\xi_{1}(-\infty)=1$. Set $A\mathrel{:=}\{a\in P\mid\xi_{1}(a)>0\}$. Then, since $\xi_{1}(-\infty)=1$ and $\xi_{1}\in{\mathcal{S}}^{(0)}$, we see that $A$ is an antichain. Let $C$ be an arbitrary element of $C_{\xi}^{[d-\epsilon]}$. Since $\xi_{2}(-\infty)=d-1$ and $\xi_{2}\in{\mathcal{S}}^{(\epsilon)}$, we see that $\xi^{+}_{2}(C)+\epsilon\leq d-1$. Since $\xi^{+}_{1}(C)+\xi^{+}_{2}(C)=\xi^{+}(C)=d-\epsilon$, we see that $\xi^{+}_{1}(C)\geq 1$. Thus, $C\cap A\neq\emptyset$ and for the unique element $a\in A\cap C$, $\xi(a)=\xi_{1}(a)+\xi_{2}(a)\geq 1+\epsilon>\epsilon$ since $\xi_{2}\in{\mathcal{S}}^{(\epsilon)}$.  **

By using this criterion of a Laurent monomial $T^{\xi}\in\omega^{(\epsilon)}$ to be a generator of $\omega^{(\epsilon)}$, we next state a sufficient condition for a Laurent monomial $T^{\xi}\in\omega^{(\epsilon)}$ to be a generator of $\omega^{(\epsilon)}$.

Lemma 4.6

Let $\xi\in{\mathcal{S}}^{(\epsilon)}$ and set $\xi(-\infty)=d$. If there are elements $z_{0}$, $w_{1}$, $z_{1}$, $w_{2}$, …, $z_{s-1}$, $w_{s}$ of $P$ with $z_{0}>w_{1}<z_{1}>w_{2}<\cdots<z_{s-1}>w_{s}$ and $C_{0}$, $C_{1}$, …, $C_{s}\in C_{\xi}^{[d-\epsilon]}$ such that

1. (1)

$s=0$* or $C_{0}\ni z_{0}$, $C_{s}\ni w_{s}$ and $C_{i}\ni w_{i}$, $z_{i}$ for $1\leq i\leq s-1$ (the latter part is trivially valid if $s=1$), and* 2. (2)

$\xi(z)=\epsilon$* for any $z\in C_{i}\cap(w_{i},z_{i})$ and for any $i$ with $0\leq i\leq s$, where we set $w_{0}=-\infty$ and $z_{s}=\infty$,*

then $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$

• Proof

Assume the contrary. Then by Lemma 4.5, we see that there is an antichain $A$ in $P$ such that for any $C\in C_{\xi}^{[d-\epsilon]}$, $A\cap C\neq\emptyset$ and $\xi(a)>\epsilon$ for the unique element $a\in A\cap C$.

**If $s=0$, then for $a\in A\cap C_{0}$, $\xi(a)=\epsilon$ by assumption and this is a contradiction. Thus, $s>0$. Set $A\cap C_{i}=\{a_{i}\}$ for $0\leq i\leq s$. Since $a_{s}\leq w_{s}$ and $a_{0}\geq z_{0}$ by assumption, we see that there exists $i$ with $1\leq i\leq s$, $a_{i}\leq w_{i}$ and $a_{i-1}\geq z_{i-1}$. Since $z_{i-1}>w_{i}$, this contradicts to the fact that $A$ is an antichain.  **

Next we show a strong converse of this lemma. We show that if $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$, then we can not only take a sequence $z_{0}$, $w_{1}$, …, $z_{s-1}$, $w_{s}$ satisfying the condition of Lemma 4.6, but also $z_{0}$, $w_{1}$, …, $z_{s-1}$, $w_{s}$ satisfy condition N’. Further, we show that we can take sequence which are “$q^{(\epsilon)}$-reduced”.

Definition 4.7

For a sequence of elements $w_{0}$, $z_{0}$, $w_{1}$, …, $z_{s-1}$, $w_{s}$, $z_{s}$ with $w_{0}<z_{0}>w_{1}<\cdots<z_{s-1}>w_{s}<z_{s}$ in $P^{\pm}$, we set

[TABLE]

Further, for a sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ of elements in $P$ with condition N’, we say that $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced if

[TABLE]

for any $x_{i}$, $y_{j}$ with $x_{i}<y_{j}$ and $0\leq i<j\leq t$, where we set $x_{0}=-\infty$ and $y_{t}=\infty$.

Example 4.8

If there is a part of the sequence of the following form

[TABLE]

then it is not $q^{(1)}$-reduced, since $q^{(1)}{\mathrm{dist}}(x_{i},y_{i+2})=1>0=q^{(1)}(x_{i},y_{i},x_{i+1},y_{i+1},x_{i+2},y_{i+2})$ in the former case and $q^{(1)}{\mathrm{dist}}(x_{i},y_{i+1})=1=q^{(1)}(x_{i},y_{i},x_{i+1},y_{i+1})$ in the latter case. Later, we seek a sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ such that $q^{(\epsilon)}(x_{0},y_{0},\ldots,x_{t},\infty)$ as large as possible. If there is a part of the first kind in the sequence with condition N’, we can replace it with $y_{0}$, $x_{1}$, …, $x_{i}$, $y_{i+2}$, $x_{i+2}$, …, $x_{t}$ and obtain a sequence with larger $q^{(1)}(x_{0},y_{0},\ldots)$. Further, if there is a part of the second kind in the sequence with condition N’, we apply the replacement above and remove redundancy.

Lemma 4.9

Let $\xi$ be an element of ${\mathcal{S}}^{(\epsilon)}$ with $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$. Set $d\mathrel{:=}\xi(-\infty)$. Then there exists a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ such that there are $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ with the following conditions.

1. (1)

$t=0$* or $C_{0}\ni y_{0}$, $C_{t}\ni x_{t}$ and $C_{i}\ni x_{i}$, $y_{i}$ for $1\leq i\leq t-1$, and* 2. (2)

$\xi(z)=\epsilon$* for any $z\in C_{i}\cap(x_{i},y_{i})$ and for any $i$ with $0\leq i\leq t$, where we set $x_{0}=-\infty$ and $y_{t}=\infty$.*

• Proof

First we note that $C_{\xi}^{[d-\epsilon]}\neq\emptyset$ by Lemma 4.5. Further, if there exists $C\in C_{\xi}^{[d-\epsilon]}$ with $\xi(z)=\epsilon$ for any $z\in C$, then the empty sequence satisfies the required condition. Thus, in the following of the proof, we assume that for any $C\in C_{\xi}^{[d-\epsilon]}$, there is $z\in C$ with $\xi(z)>\epsilon$.

Put $Y_{0}\mathrel{:=}\{y\in P\mid\exists C\in C_{\xi}^{[d-\epsilon]};y=\min\{c\in C\mid\xi(c)>\epsilon\}\}$, $X_{1}\mathrel{:=}\{x\in P\mid\exists y\in Y_{0};x<y\}$, $Y_{1}\mathrel{:=}Y_{0}\cup\{y\in P\mid\exists x\in X_{1},\exists C\in C_{\xi}^{[d-\epsilon]};x\in C,y=\min\{c\in C\mid c>x,\xi(c)>\epsilon\}\}$, $X_{2}\mathrel{:=}\{x\in P\mid\exists y\in Y_{1};x<y\}$, $Y_{2}\mathrel{:=}Y_{1}\cup\{y\in P\mid\exists x\in X_{2},\exists C\in C_{\xi}^{[d-\epsilon]};x\in C,y=\min\{c\in C\mid c>x,\xi(c)>\epsilon\}\}$, $X_{3}\mathrel{:=}\{x\in P\mid\exists y\in Y_{2};x<y\}$ and so on. Also put $X\mathrel{:=}\bigcup_{i=1}^{\infty}X_{i}$ and $Y\mathrel{:=}\bigcup_{i=0}^{\infty}Y_{i}$. Set $A\mathrel{:=}\max Y$. Then $A$ is an antichain in $P$. Moreover, $Y\setminus A\subset X$. In fact, if $y\in Y\setminus A$, there is $a\in A$ with $a>y$. Since $a\in A\subset Y$, we see that $y\in X$.

If $A\cap C\neq\emptyset$ for any $C\in C_{\xi}^{[d-\epsilon]}$, then by Lemma 4.5, we see that $T^{\xi}$ is not a generator of $\omega^{(\epsilon)}$, since $\xi(a)>\epsilon$ for any $a\in A$, contradicting the assumption. Therefore, there exists $C\in C_{\xi}^{[d-\epsilon]}$ such that $C\cap A=\emptyset$.

Take such $C$ and set $\{c\in C\mid\xi(c)>\epsilon\}=\{c_{1},c_{2},\ldots,c_{v}\}$, $c_{1}<c_{2}<\cdots<c_{v}$. By the definition of $Y_{0}$, we see that $c_{1}\in Y_{0}\subset Y$. Since $C\cap A=\emptyset$, we see that $c_{1}\in Y\setminus A\subset X$. Therefore, we see that $c_{2}\in Y$, and so on. Repeating this argument, we see that $c_{v}\in X$.

Set $t\mathrel{:=}\min\{\ell\mid c_{v}\in X_{\ell}\}$, $x_{t}\mathrel{:=}c_{v}$ and $C_{t}\mathrel{:=}C$. Since $x_{t}\in X_{t}$, we see that there is $y_{t-1}\in Y_{t-1}$ with $y_{t-1}>x_{t}$. Note that $y_{t-1}\not\in Y_{t-2}$ since $x_{t}\not\in X_{t-1}$. By the definition of $Y_{t-1}$, we see that there are $x_{t-1}\in X_{t-1}$ and $C_{t-1}\in C_{\xi}^{[d-\epsilon]}$ with $x_{t-1}\in C_{t-1}$ and $y_{t-1}=\min\{c\in C_{t-1}\mid\xi(c)>\epsilon,c>x_{t-1}\}$. Note that $x_{t-1}\not\in X_{t-2}$ since $y_{t-1}\not\in Y_{t-2}$. Since $x_{t-1}\in X_{t-1}$, there is $y_{t-2}\in Y_{t-2}$ with $y_{t-2}>x_{t-1}$. Note that $y_{t-2}\not\in Y_{t-3}$ since $x_{t-1}\not\in X_{t-2}$.

Continueing this argument, we see that there exist a sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ of elements of $P$ and $C_{1}$, $C_{2}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ such that $y_{0}>x_{1}<\ldots<y_{t-1}>x_{t}$, $x_{k}\in X_{k}$ for $1\leq k\leq t$, $x_{k}\not\in X_{k-1}$ for $2\leq k\leq t$, $y_{k}\in Y_{k}$ for $0\leq k\leq t-1$, $y_{k}\not\in Y_{k-1}$ for $1\leq k\leq t-1$, $y_{k}$, $x_{k}\in C_{k}$ and $\xi(z)=\epsilon$ for any $z\in C_{k}\cap(x_{k},y_{k})$ for $1\leq k\leq t-1$, $x_{t}\in C_{t}$ and $\xi(z)=\epsilon$ for any $z\in C_{t}\cap(x_{t},\infty)$. Since $x_{k}\not\in X_{k-1}$ for $2\leq k\leq t$, we see that if $i\leq j-2$, then $y_{i}\not>x_{j}$, i.e., $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ satisfies condition N’. Moreover, by the definition of $Y_{0}$, we can take $C_{0}\in C_{\xi}^{[d-\epsilon]}$ with $y_{0}=\min\{c\in C_{0}\mid\xi(c)>\epsilon\}$. Then $y_{0}\in C_{0}$ and for any $z\in C_{0}\cap(-\infty,y_{0})$, $\xi(z)=\epsilon$.

Next we show that $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced. Assume the contrary and suppose that there are $i$ and $j$ with $0\leq i<j\leq t$, $x_{i}<y_{j}$ and $q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{j})\geq\sum_{\ell=i}^{j}q^{(\epsilon)}{\mathrm{dist}}(x_{\ell},y_{\ell})-\sum_{\ell=i}^{j-1}q^{(\epsilon)}{\mathrm{dist}}(x_{\ell+1},y_{\ell})$. Then

[TABLE]

Take a saturated chain $x_{\ell+1}=z_{\ell,0}\mathrel{<\!\!\!\cdot}z_{\ell,1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{\ell,s_{\ell}}=y_{\ell}$ with $\epsilon s_{\ell}=q^{(\epsilon)}{\mathrm{dist}}(x_{\ell+1},y_{\ell})$ for $i\leq\ell\leq j-1$. Then, $C^{\prime}_{\ell}\mathrel{:=}(C_{\ell+1}\cap(-\infty,x_{\ell+1}])\cup\{z_{\ell,1},\ldots,z_{\ell,s_{\ell}-1}\}\cup(C_{\ell}\cap[y_{\ell},\infty))$ is a maximal chain in $P$ for $i\leq\ell\leq j-1$. Further, take a saturated chain $x_{i}=z^{\prime}_{0}\mathrel{<\!\!\!\cdot}z^{\prime}_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime}_{s^{\prime}}=y_{j}$ with $\epsilon s^{\prime}=q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{j})$. Then, $C^{\prime\prime}\mathrel{:=}(C_{i}\cap(-\infty,x_{i}])\cup\{z^{\prime}_{1},\ldots,z^{\prime}_{s^{\prime}-1}\}\cup(C_{j}\cap[y_{j},\infty))$ is also a maximal chain in $P$.

Therefore,

[TABLE]

since $C^{\prime}_{\ell}$ for $i\leq\ell\leq j-1$ and $C^{\prime\prime}$ are maximal chains in $P$ and $\xi\in{\mathcal{S}}^{(\epsilon)}$. On the other hand,

[TABLE]

since for any $i\leq\ell\leq j$, $C_{\ell}\in C_{\xi}^{[d-\epsilon]}$ and $\xi(z)=\epsilon$ for any $z\in C_{\ell}\cap(x_{\ell},y_{\ell})$. Thus, we see by inequation (4.1) that equalities hold in (Proof) and (4.3) and therefore $\xi(z^{\prime}_{m})=\epsilon$ for $1\leq m\leq s^{\prime}-1$ and $\xi^{+}(C^{\prime\prime})=d-\epsilon$. This means that $y_{j}\in Y_{i}$, contradicting the fact that $y_{j}\not\in Y_{j-1}$.

**Therefore, we see that $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced.  **

By Lemmas 4.6 and 4.9, we see the following.

Proposition 4.10

Let $\xi\in{\mathcal{S}}^{(\epsilon)}$ and set $d\mathrel{:=}\xi(-\infty)$. Then the following conditions are equivalent.

1. (1)

$T^{\xi}$* is a genenrator of $\omega^{(\epsilon)}$.* 2. (2)

There are elements $z_{0}$, $w_{1}$, …, $z_{s-1}$, $w_{s}$ of $P$ with $z_{0}>w_{1}<\cdots<z_{s-1}>w_{s}$ and $C_{0}$, $C_{1}$, …, $C_{s}\in C_{\xi}^{[d-\epsilon]}$ such that

1. (a)

$s=0$* or $C_{0}\ni z_{0}$, $C_{s}\ni w_{s}$ and $C_{i}\ni w_{i}$, $z_{i}$ for $1\leq i\leq s-1$, and* 2. (b)

$\xi(z)=\epsilon$* for any $z\in C_{i}\cap(w_{i},z_{i})$ and for any $i$ with $0\leq i\leq s$, where we set $w_{0}=-\infty$ and $z_{s}=\infty$,* 3. (3)

There exists a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ such that there are $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ with the following conditions.

1. (a)

$t=0$* or $C_{0}\ni y_{0}$, $C_{t}\ni x_{t}$ and $C_{i}\ni x_{i}$, $y_{i}$ for $1\leq i\leq t-1$, and* 2. (b)

$\xi(z)=\epsilon$* for any $z\in C_{i}\cap(x_{i},y_{i})$ and for any $i$ with $0\leq i\leq t$, where we set $x_{0}=-\infty$ and $y_{t}=\infty$.*

As a corollary, we obtain an upper bound of the degrees of generators of $\omega^{(\epsilon)}$.

Corollary 4.11

Let $T^{\xi}$, $\xi\in{\mathcal{S}}^{(\epsilon)}$, be a generator of $\omega^{(\epsilon)}$. Then

[TABLE]

(See Definition 4.7 for the notation.)

• Proof

Set $d\mathrel{:=}\deg T^{\xi}$. Then by Proposition 4.10, we see that there are a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ and $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ satisfying (3) of Proposition 4.10.

If $t=0$, then since $\xi(c)=\epsilon$ for any $c\in C_{0}$, we see that

[TABLE]

Therefore, $\deg T^{\xi}=d\leq q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)=q^{(\epsilon)}(x_{0},y_{0})$.

Now suppose that $t>0$. Since $\max\{\xi^{+}(C)\mid C$ is a maximal chain in $P\}=d-\epsilon$, we see that $d-\epsilon\geq\xi^{+\downarrow}(x_{i})+\xi^{+^{\prime}\uparrow}(x_{i})\geq\sum_{c\in C_{i}\cap(-\infty,x_{i}]}\xi(c)+\sum_{c\in C_{i}\cap(x_{i},\infty)}\xi(c)=\xi^{+}(C_{i})=d-\epsilon$. Therefore,

[TABLE]

for $1\leq i\leq t$. We see by the same way that

[TABLE]

for $0\leq i\leq t-1$. In particular,

[TABLE]

for $1\leq i\leq t-1$, since $\xi(c)=\epsilon$ for any $c\in C_{i}\cap(x_{i},y_{i})$. Moreover, we see that

[TABLE]

by the same way. On the other hand, we see by Lemma 2.7,

[TABLE]

for $0\leq i\leq t-1$. Thus,

[TABLE]

In fact, this upper bound is the least upper bound, see Proposition 4.16. Further, it is seen, by Corollary 3.3 that the minimum degree of the Laurent monomials in $\omega^{(\epsilon)}$ is $q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$.

The $q^{(\epsilon)}$-reduced sequence with condition N’ satisfying (3) of Proposition 4.10 is far from unique. However, for a given $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ we can construct an element $\xi\in{\mathcal{S}}^{(\epsilon)}$ such that $d=\xi(-\infty)=q^{(\epsilon)}(-\infty,y_{0},\ldots,x_{t},\infty)$ and there are $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ which satisfy conditions of (3) of Proposition 4.10. In order to do this task, suppose that a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ is given and fixed. We set $x_{0}\mathrel{:=}-\infty$, $y_{t}\mathrel{:=}\infty$ and $d\mathrel{:=}q^{(\epsilon)}(x_{0},y_{0},\ldots,x_{t},y_{t})$.

If $t=0$, then it is enough to set

[TABLE]

since $\xi\in{\mathcal{S}}^{(\epsilon)}$ by Lemma 3.1 and there is a maximal chain $C_{0}$ in $P$ with $\xi^{+}(C_{0})=\epsilon(\#C_{0})=q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)-\epsilon=d-\epsilon$. Thus, we assume that $t>0$ in the following.

We define two maps $\mu^{\prime}\colon\{y_{0},y_{1},\ldots,y_{t}\}\to\mathbb{Z}$ and $\mu^{\prime\prime}\colon\{x_{0},x_{1},\ldots,x_{t}\}\to\mathbb{Z}$ by $\mu^{\prime\prime}(x_{i})\mathrel{:=}q^{(\epsilon)}(x_{i},y_{i},\ldots,x_{t},y_{t})$ for $0\leq i\leq t$, $\mu^{\prime}(y_{i})\mathrel{:=}q^{(\epsilon)}(x_{i+1},y_{i+1},\ldots,x_{t},y_{t})-q^{(\epsilon)}{\mathrm{dist}}(x_{i+1},y_{i})$ for $0\leq i\leq t-1$ and $\mu^{\prime}(y_{t})\mathrel{:=}0$. Note that $\mu^{\prime\prime}(x_{0})=d$, $\mu^{\prime\prime}(x_{i})=\mu^{\prime}(y_{i})+q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i})$ for $0\leq i\leq t$, $\mu^{\prime\prime}(x_{i})=\mu^{\prime}(y_{i-1})+q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i-1})$ for $1\leq i\leq t$ and $\mu^{\prime\prime}(x_{i})-\mu^{\prime}(y_{j})=q^{(\epsilon)}(x_{i},y_{i},\ldots,x_{j},y_{j})$ for $0\leq i\leq j\leq t$.

Moreover, if $i<j$ and $x_{i}<y_{j}$, then $q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{j})<\mu^{\prime\prime}(x_{i})-\mu^{\prime}(y_{j})$, since $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced. Note that it may happen that $x_{i}=y_{j}$ for some $i$ and $j$, but this does not imply $\mu^{\prime\prime}(x_{i})=\mu^{\prime}(y_{j})$.

Example 4.12

Let $\epsilon=1$ and let $P$ and $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$ be as follows.

[TABLE]

Then $\mu^{\prime}(y_{2})=0$, $\mu^{\prime\prime}(x_{2})=3$, $\mu^{\prime}(y_{1})=2$, $\mu^{\prime\prime}(x_{1})=5$, $\mu^{\prime}(y_{0})=4$ and $\mu^{\prime\prime}(x_{0})=7$.

We first define the map $\xi_{0}\colon P\to\mathbb{Z}$ by

[TABLE]

Note that $y_{j}$ may be $y_{t}=\infty$ in the above definition, thus $\{\mu^{\prime}(y_{j})+q^{(\epsilon)}{\mathrm{dist}}(z,y_{j})\mid y_{j}>z$ in $P^{+}\}\neq\emptyset$, since $z\in P$. For $\xi_{0}$, we see the following fact.

Lemma 4.13

1. (1)

$\xi_{0}(z)\geq\epsilon$* for any $z\in P$.* 2. (2)

$\xi_{0}(z)=\epsilon$* if $z\not\in\{y_{0},\ldots,y_{t-1}\}$.* 3. (3)

$\xi^{+\uparrow}_{0}(y_{i})=\mu^{\prime}(y_{i})$* for $0\leq i\leq t-1$.* 4. (4)

$\xi^{+}_{0}(C)\leq d-\epsilon$* for any maximal chain $C$ in $P$.* 5. (5)

$q^{(\epsilon)}{\mathrm{dist}}(x_{0},y_{0})+\xi^{+\uparrow}_{0}(y_{0})=d$.

• Proof

(2)** is obvious from the definition. We next prove (1). The case where $z\not\in\{y_{0},\ldots,y_{t}\}$ follows from (2). Assume that $z=y_{i}$ and $0\leq i\leq t-1$. Take $j$ with $y_{j}>y_{i}$ and \xi_{0}(y_{i})=\mu^{\prime}(y_{i})-\big{(}\mu^{\prime}(y_{j})+q^{(\epsilon)}{\mathrm{dist}}(y_{i},y_{j})\big{)}+\epsilon. Then $i<j$ by Lemma 4.2 and $x_{i}<y_{j}$. Since $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced sequence with condition N’, we see that**

[TABLE]

On the other hand, since

[TABLE]

we see that

[TABLE]

Therefore,

[TABLE]

Next we prove (3) by backward induction on $i$. First consider the case where $i=t-1$. Since $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ satisfies condition N’, we see that $y_{j}>y_{t-1}$ implies $j=t$ by Lemma 4.2. Therefore,

[TABLE]

On the other hand, since $z\in P$, $z>y_{t-1}$ implies $z\not\in\{y_{0},\ldots,y_{t-1}\}$ by Lemma 4.2, we see that

[TABLE]

Therefore, we see that

[TABLE]

Next suppose that $0\leq i\leq t-2$. We first show that $\xi^{+\uparrow}_{0}(y_{i})\leq\mu^{\prime}(y_{i})$. Take $z_{0}$, $z_{1}$, …, $z_{s}$ with $y_{i}=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}\infty$ and $\sum_{\ell=0}^{s}\xi_{0}(z_{\ell})=\xi^{+\uparrow}(y_{i})$. First consider the case where $\{z_{1}$, …, $z_{s}\}\cap\{y_{0}$, …, $y_{t-1}\}=\emptyset$. Since $\xi_{0}(z_{\ell})=\epsilon$ for $1\leq\ell\leq s$, we see that

[TABLE]

If $\{z_{1},\ldots,z_{s}\}\cap\{y_{0},\ldots,y_{t-1}\}\neq\emptyset$, take minimal $u$ with $u\geq 1$ and $z_{u}\in\{y_{0},\ldots,y_{t-1}\}$ and set $z_{u}=y_{j}$. Then $j>i$ by Lemma 4.2. Thus, $\xi^{+\uparrow}_{0}(y_{j})=\mu^{\prime}(y_{j})$ by the induction hypothesis. Moreover, $\xi^{+\uparrow}_{0}(y_{j})=\sum_{\ell=u}^{s}\xi_{0}(z_{\ell})$ by Lemma 2.6 (1). On the other hand, since $\xi_{0}(z_{\ell})=\epsilon$ for $1\leq\ell\leq u-1$, we see that

[TABLE]

Therefore,

[TABLE]

Thus,

[TABLE]

Next we prove that $\xi^{+\uparrow}_{0}(y_{i})\geq\mu^{\prime}(y_{i})$. Take $j$ with $y_{j}>y_{i}$ and

[TABLE]

Then $j>i$ by Lemma 4.2. Take $z_{0}$, …, $z_{u}\in P^{+}$ with $y_{i}=z_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{u}=y_{j}$ and $u\epsilon=q^{(\epsilon)}{\mathrm{dist}}(y_{i},y_{j})$. Then by Lemma 2.6 (3) and the induction hypothesis

[TABLE]

where we set $\xi^{+\uparrow}_{0}(y_{t})\mathrel{:=}0$. Since $\xi_{0}(z_{\ell})\geq\epsilon$ for $1\leq\ell\leq u-1$ by (1), we see that

[TABLE]

by (4.4). Thus, we have proved (3).

(5)** follows from (3) and the fact $d=\mu^{\prime\prime}(x_{0})=\mu^{\prime}(y_{0})+q^{(\epsilon)}{\mathrm{dist}}(x_{0},y_{0})$.**

Finally, we prove (4). Let $C$ be a maximal chain in $P$. We set $C=\{c_{1},\ldots,c_{s}\}$, $-\infty\mathrel{<\!\!\!\cdot}c_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}c_{s}\mathrel{<\!\!\!\cdot}\infty$. First, consider the case where $C\cap\{y_{0},\ldots,y_{t-1}\}=\emptyset$. Then $\xi_{0}(c_{i})=\epsilon$ for any $i$ by (2). Thus,

[TABLE]

Since $y_{0}$, $x_{1}$, …$y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced, we see that

[TABLE]

Therefore,

[TABLE]

Next we consider the case where $C\cap\{y_{0},\ldots,y_{t-1}\}\neq\emptyset$. Take minimal $u$ with $c_{u}\in\{y_{0},\ldots,y_{t-1}\}$ and set $c_{u}=y_{j}$. Then

[TABLE]

Since $\xi_{0}(c_{\ell})=\epsilon$ for $1\leq\ell\leq u-1$, we see that

[TABLE]

since $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced. On the other hand, since $\xi^{+\uparrow}_{0}(y_{j})\geq\sum_{\ell=u}^{s}\xi_{0}(c_{\ell})$, we see that

[TABLE]

**by (3).  **

Next we state a lemma which is used in the induction argument.

Lemma 4.14

Let $k$ be an integer with $0\leq k\leq t-1$ and suppose that a map $\xi_{k}\colon P\to\mathbb{Z}$ is defined so that

• (H1)

$\xi_{k}(z)\geq\epsilon$** for any $z\in P$.**

• (H2)

$\xi_{k}(z)=\epsilon$** if $z\not\in\{x_{1},\ldots,x_{k},y_{1},\ldots,y_{t-1}\}$.**

• (H3)

$\xi^{+\uparrow}_{k}(y_{i})=\mu^{\prime}(y_{i})$** for $0\leq i\leq t-1$.**

• (H4)

$\xi^{+}_{k}(C)\leq d-\epsilon$** for any maximal chain $C$ in $P$.**

• (H5)

$\xi^{+\downarrow}_{k}(x_{i})+q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i})+\xi^{+\uparrow}_{k}(y_{i})=d$** for $0\leq i\leq k$.**

are satisfied, where we set $\xi^{+\downarrow}_{k}(x_{0})=0$.

Let $\xi_{k+1}\colon P\to\mathbb{Z}$ be a map such that

[TABLE]

Then

1. (1)

$\xi_{k+1}(z)\geq\xi_{k}(z)$* for any $z\in P$. In particular, $\xi_{k+1}(z)\geq\epsilon$ for any $z\in P$.* 2. (2)

$\xi_{k+1}(z)=\epsilon$* if $z\not\in\{x_{1},\ldots,x_{k+1},y_{0},\ldots,y_{t-1}\}$.* 3. (3)

$\xi^{+\uparrow}_{k+1}(y_{i})=\mu^{\prime}(y_{i})$* for $0\leq i\leq t-1$.* 4. (4)

$\xi^{+}_{k+1}(C)\leq d-\epsilon$* for any maximal chain $C$ in $P$.* 5. (5)

$\xi^{+\downarrow}_{k+1}(x_{i})+q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i})+\xi^{+\uparrow}_{k+1}(y_{i})=d$* for $0\leq i\leq k+1$, where we set $\xi^{+\downarrow}_{k+1}(x_{0})=\xi^{+\uparrow}_{k+1}(y_{t})=0$.*

• Proof

First note that we have to be careful because that it might happen $x_{k+1}\in\{y_{0},\ldots,y_{t-1}\}$.

(2)** is obvious from (H2) and the definition of $\xi_{k+1}$. Next we prove (1). Since**

[TABLE]

by (H4) and Lemma 2.7, we see that

[TABLE]

by (H1). Therefore (1) follows from the definition of $\xi_{k+1}$.

Next we prove (4). Let $C$ be an arbitrary maximal chain in $P$. If $C\not\ni x_{k+1}$, then by the definition of $\xi_{k+1}$, we see that

[TABLE]

by (H4). If $C\ni x_{k+1}$, then

[TABLE]

by the definition of $\xi_{k+1}$.

Now we prove (3). First, we see by (1) and (H3) that

[TABLE]

In order to prove the converse inequality, take a saturated chain $y_{i}=z_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}\infty$ from $y_{i}$ to $\infty$ with $\xi^{+\uparrow}_{k+1}(y_{i})=\sum_{\ell=0}^{s}\xi_{k+1}(z_{\ell})$. If $x_{k+1}\not\in\{z_{0},\ldots,z_{s}\}$, then

[TABLE]

by (H3) and the definition of $\xi_{k+1}$. Now suppose that $x_{k+1}\in\{z_{0},\ldots,z_{s}\}$ and set $x_{k+1}=z_{u}$. Then, since $y_{k}>x_{k+1}\geq y_{i}>x_{i+1}$, we see by condition N’ that $k\not\leq(i+1)-2$, i.e., $k\geq i$. (In fact, $i\leq k-1$, but we do not use this fact.) Since $\xi_{k}(z)\geq\epsilon$ for any $z\in P$, we see by Lemma 2.7 that

[TABLE]

Therefore, by (H5), (1) and (4), we see that

[TABLE]

Thus, we see that

[TABLE]

Finally, we prove (5). By (4) and Lemma 2.7, we see that

[TABLE]

Thus,

[TABLE]

for any $i$. If $i\leq k$, then by (H5), we see that

[TABLE]

In order to prove the case where $i=k+1$, take a saturated chain $x_{k+1}=z_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{s}\mathrel{<\!\!\!\cdot}z_{s+1}=\infty$ with $\xi^{+\uparrow}_{k+1}(x_{k+1})=\sum_{\ell=0}^{s}\xi_{k+1}(z_{\ell})$. Take minimal $u$ with $1\leq u\leq s+1$ and $z_{u}\in\{y_{0},\ldots,y_{t}\}$ and set $z_{u}=y_{j}$. Then

[TABLE]

by (3) and Lemma 2.6 (1). We claim here that $j=k$ or $j=k+1$. In fact, by condition N’, we see that $j\geq k$. If $j\geq k+2$, then, since $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced, we see that

[TABLE]

Thus, by (4.5), (3) and Lemma 2.7, we see that

[TABLE]

This is a contradiction. Therefore, $j=k$ or $j=k+1$ and

[TABLE]

Therefore, we see by (4.5) that

[TABLE]

by (3). Further, since $\xi_{k+1}(z)\geq\epsilon$ for any $z$, we see that $\xi^{+^{\prime}\uparrow}_{k+1}(x_{k+1})\geq\xi^{+\uparrow}_{k+1}(y_{k+1})+q^{(\epsilon)}{\mathrm{dist}}(x_{k+1},y_{k+1})-\epsilon$ by Lemma 2.6 (3). Thus,

[TABLE]

Since

[TABLE]

by the definition of $\xi_{k+1}$, we see that

[TABLE]

and therefore,

[TABLE]

By Lemmas 4.13 and 4.14 and induction, we see that $\xi_{t}$ is defined and satisfies

1. (1)

$\xi_{t}(z)\geq\epsilon$ for any $z\in P$, 2. (2)

$\xi_{t}(z)=\epsilon$ if $z\not\in\{x_{1},\ldots,x_{t},y_{0},\ldots,y_{t-1}\}$. 3. (3)

$\xi^{+}_{t}(C)\leq d-\epsilon$ for any maximal chain $C$ in $P$ and 4. (4)

$\xi^{+\downarrow}_{t}(x_{i})+q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i})+\xi^{+\uparrow}_{t}(y_{i})=d$ for $0\leq i\leq t$, where we set $\xi^{+\downarrow}_{t}(x_{0})=\xi^{+\uparrow}_{t}(y_{t})=0$.

Example 4.15

Let $\epsilon=1$ and let $P$ and $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$ be as follows.

[TABLE]

Then $d=6$,

[TABLE]

Let $\epsilon=1$ and let $P$ and $y_{0}$, $x_{1}$, $y_{1}$, $x_{2}$ be as follows.

[TABLE]

Then $d=7$,

[TABLE]

Let $i$ be an integer with $1\leq i\leq t-1$. Take a saturated chain $x_{i}=z_{0}\mathrel{<\!\!\!\cdot}z_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z_{u}=y_{i}$ with $u\epsilon=q^{(\epsilon)}{\mathrm{dist}}(x_{i},y_{i})$. Then $z_{1}$, …, $z_{u-1}\not\in\{x_{1},\ldots,x_{t},y_{0},\ldots,y_{t-1}\}$. In fact, if there is $z_{\ell}\in\{x_{1},\ldots,x_{t},y_{0},\ldots,y_{t-1}\}$ with $1\leq\ell\leq u-1$, then $z_{\ell}=x_{i+1}$ or $z_{\ell}=y_{i-1}$ by condition N’ and Lemma 4.2. If $z_{\ell}=x_{i+1}$, then $q^{(\epsilon)}{\mathrm{dist}}(x_{i},x_{i+1})=\ell\epsilon$ and $q^{(\epsilon)}{\mathrm{dist}}(x_{i+1},y_{i})=(u-\ell)\epsilon$ and therefore

[TABLE]

This contradicts to the assumption that $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is $q^{(\epsilon)}$-reduced.

Therefore, $z_{\ell}\neq x_{i+1}$. We see that $z_{\ell}\neq y_{i-1}$ by the same way. Thus, $z_{\ell}\not\in\{x_{1},\ldots,x_{t},y_{0},\ldots,y_{t-1}\}$ and therefore $\xi_{t}(z_{\ell})=\epsilon$ for $1\leq\ell\leq u-1$ by (2).

Take saturated chains $-\infty\mathrel{<\!\!\!\cdot}z^{\prime}_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime}_{u^{\prime}}=x_{i}$ and $y_{i}=z^{\prime\prime}_{0}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}z^{\prime\prime}_{u^{\prime}}\mathrel{<\!\!\!\cdot}\infty$ with $\xi^{+\downarrow}_{t}(x_{i})=\sum_{\ell=1}^{u^{\prime}}\xi_{t}(z^{\prime}_{\ell})$ and $\xi^{+\uparrow}_{t}(y_{i})=\sum_{\ell=0}^{u^{\prime\prime}}\xi_{t}(z^{\prime\prime}_{\ell})$. Then $C_{i}\mathrel{:=}\{z^{\prime}_{1},\ldots,z^{\prime}_{u^{\prime}},z_{1},\ldots,z_{u-1},z^{\prime\prime}_{0},\ldots,z^{\prime\prime}_{u^{\prime\prime}}\}$ is a maximal chain in $P$ with $x_{i}$, $y_{i}\in C_{i}$ and $\xi_{t}(z)=\epsilon$ for any $z\in C_{i}\cap(x_{i},y_{i})$. Further,

[TABLE]

by (5) above.

By a similar way, we can take maximal chains $C_{0}$ and $C_{t}$ with $\xi^{+}_{t}(C_{0})=\xi^{+}_{t}(C_{t})=d-\epsilon$ and for $\xi_{t}(z)=\epsilon$ for any $z\in(C_{0}\cap(-\infty,y_{0}))\cup(C_{t}\cap(y_{t},\infty))$.

Define $\xi\colon P^{-}\to\mathbb{Z}$ by

[TABLE]

Then by the facts above and Proposition 4.10, we see the following.

Proposition 4.16

Let $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ be a $q^{(\epsilon)}$-reduced sequence with condition N’ in $P$. Set $x_{0}\mathrel{:=}-\infty$ and $y_{t}\mathrel{:=}\infty$. Then there exists $\xi\in{\mathcal{S}}^{(\epsilon)}$ such that $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$ and

[TABLE]

Now we state characterizations of level and anticanonical level properties of $\mathbb{K}[\mathscr{C}(P)]$.

Theorem 4.17

The following conditions are equivalent.

1. (1)

$\mathbb{K}[\mathscr{C}(P)]$* is level (resp. anticanonical level).* 2. (2)

$q^{(1)}$-reduced (resp. $q^{(-1)}$-reduced) sequence with condition N’ is the empty sequence only. 3. (3)

For any sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’,

[TABLE]

(resp. $q^{(-1)}(x_{0},y_{0},\ldots,x_{t},y_{t})\leq q^{(-1)}{\mathrm{dist}}(x_{0},y_{t})$), where we set $x_{0}\mathrel{:=}-\infty$ and $y_{t}\mathrel{:=}\infty$.

• Proof

Let $\epsilon=1$ or $-1$. We prove the level case and the anticanonical level case simultaneously by setting $\epsilon=1$ when considering the level case and $\epsilon=-1$ when considering the anticanonical level case. The contraposition of (1)$\Rightarrow$(2) follows from the fact that $q^{(\epsilon)}(-\infty,y_{0},x_{1},\cdots,y_{t-1},x_{t},\infty)>q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$ for any nonempty $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’, since by Proposition 4.16 and Corollary 3.3, there are generators of $\omega^{(\epsilon)}$ with degrees $q^{(\epsilon)}(-\infty,y_{0},x_{1},\cdots,y_{t-1},x_{t},\infty)$ and $q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$ respectively.

We next show (2)$\Rightarrow$(1). Let $T^{\xi}$, $\xi\in{\mathcal{S}}^{(\epsilon)}$ be an arbitrary generator of $\omega^{(\epsilon)}$ and set $d\mathrel{:=}\deg T^{\xi}$. Then by Proposition 4.10, we see that there exists a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ and $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ satisfying conditions of (3) of Proposition 4.10. By assumption $t=0$. Therefore $\xi(z)=\epsilon$ for any $z\in C_{0}$. Thus, $\epsilon(\#C_{0})=\xi^{+}(C_{0})=d-\epsilon$. On the other hand, since $C_{0}\cup\{-\infty,\infty\}$ is a saturated chain from $-\infty$ to $\infty$, we see that $d=\epsilon(\#C_{0})+\epsilon=\epsilon(\#C_{0}+1)\leq q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$. Therefore, $d=q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$ by Corollary 3.3.

(3)$\Rightarrow$(2)**** is trivial. In order to prove (2)$\Rightarrow$(3), let $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ be an arbitrary sequence with condition N’. If there are $i$, $j$ with $i<j$ and $x_{i}<y_{j}$ and

[TABLE]

then by setting $x^{\prime}_{u}\mathrel{:=}x_{u}$ for $1\leq u\leq i$, $x^{\prime}_{u}\mathrel{:=}x_{u-j+i}$ for $i<u\leq t-j+i$, $y^{\prime}_{u}\mathrel{:=}y_{u}$ for $0\leq u<i$ and $y^{\prime}_{u}\mathrel{:=}y_{u-j+i}$ for $i\leq u<t-j+i$, we get a sequence $y^{\prime}_{0}$, $x^{\prime}_{1}$, …$y^{\prime}_{t-j+i-1}$, $x^{\prime}_{t-j+i}$ with condition N’ and

[TABLE]

where we set $x_{0}=x^{\prime}_{0}=-\infty$ and $y_{t}=y^{\prime}_{t-j+i}=\infty$.

Repeating this argument, we see that there is a $q^{(\epsilon)}$-reduced sequence $y^{\prime\prime}_{0}$, $x^{\prime\prime}_{1}$, …, $y^{\prime\prime}_{t^{\prime\prime}-1}$, $x^{\prime\prime}_{t^{\prime\prime}}$ with condition N’ such that

[TABLE]

where we set $x^{\prime\prime}_{0}\mathrel{:=}-\infty$ and $y^{\prime\prime}_{t^{\prime\prime}}\mathrel{:=}\infty$. Since $q^{(\epsilon)}$-reduced sequence with condition N’ is the empty sequence only by assumption, we see that $t^{\prime\prime}=0$, i.e.,

[TABLE]

Note that if one replaces condition N’ to condition N in (2) and (3) of Theorem 4.17, one obtains a criterion of level (resp. anticanonical level) property of $\mathbb{K}[\mathscr{O}(P)]$ [Miy1, Miy2]. Since condition N’ is a weaker condition than condition N, we see the following.

Corollary 4.18

If $\mathbb{K}[\mathscr{C}(P)]$ is level (resp. anticanonical level), then so is $\mathbb{K}[\mathscr{O}(P)]$.

The converse of this corollary does not hold.

Example 4.19

Let $n$, $m_{1}$, $m_{2}$ be integers with $n\geq 4$, $m_{1}$, $m_{2}\geq n-2$ and let $P=\{a_{1}$, …, $a_{n}$, $b_{1}$, …, $b_{m_{1}}$, $c_{1}$, …, $c_{m_{2}}$, $d\}$ be a poset with covering relations $a_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}a_{n}$, $b_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}b_{m_{1}}\mathrel{<\!\!\!\cdot}d\mathrel{<\!\!\!\cdot}c_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}c_{m_{2}}$ and $a_{1}\mathrel{<\!\!\!\cdot}d\mathrel{<\!\!\!\cdot}a_{n}$.

[TABLE]

Then $\mathbb{K}[\mathscr{O}(P)]$ is level and $\mathbb{K}[\mathscr{C}(P)]$ is not. In fact, it is easily verified that there is no $q^{(1)}$-reduced sequence with condition N except the empty sequence. Thus, $\mathbb{K}[\mathscr{O}(P)]$ is level. However, $d$, $a_{1}$, $a_{n}$, $d$ is a $q^{(1)}$-reduced sequence with condition N’. Further, if we define $\xi\colon P^{-}\to\mathbb{Z}$ by

[TABLE]

then $T^{\xi}$ is a generator of $\omega$ with $\deg T^{\xi}=m_{1}+m_{2}+n-1$, while the minimal degree of the generators of $\omega$ is $q^{(1)}{\mathrm{dist}}(-\infty,\infty)=m_{1}+m_{2}+2<m_{1}+m_{2}+n-1$, since $n\geq 4$.

Example 4.20

Let $n$ be a positive integer and let $P=\{a_{1}$, …, $a_{n+3}$, $b_{1}$, …, $b_{n}$, $c_{1}$, …, $c_{n}$, $d_{1}$, $d_{2}$, $d_{3}\}$ be a poset with covering relations $a_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}a_{n+3}$, $d_{1}\mathrel{<\!\!\!\cdot}d_{2}\mathrel{<\!\!\!\cdot}d_{3}$, $a_{1}\mathrel{<\!\!\!\cdot}b_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}b_{n}\mathrel{<\!\!\!\cdot}d_{2}\mathrel{<\!\!\!\cdot}c_{1}\mathrel{<\!\!\!\cdot}\cdots\mathrel{<\!\!\!\cdot}c_{n}\mathrel{<\!\!\!\cdot}a_{n+3}$.

[TABLE]

Then $\mathbb{K}[\mathscr{O}(P)]$ is anticanonical level and $\mathbb{K}[\mathscr{C}(P)]$ is not. In fact, it is easily verified that there is no $q^{(-1)}$-reduced sequence with condition N except the empty sequence. Thus, $\mathbb{K}[\mathscr{O}(P)]$ is anticanonical level. However, $d_{2}$, $a_{1}$, $a_{n+3}$, $d_{2}$ is a $q^{(-1)}$-reduced sequence with condition N’. Further, if we set $\xi\colon P^{-}\to\mathbb{Z}$ by

[TABLE]

then $T^{\xi}$ is a generator of $\omega^{(-1)}$ with degree $n-4$, while the minimal degree of the generators of $\omega^{(-1)}$ is $q^{(-1)}{\mathrm{dist}}(-\infty,\infty)=-4$.

5 Degrees of the generators of canonical and anticanonical ideals

In this section, we show that the degrees of generators of $\omega$ and $\omega^{(-1)}$ are consecutive integers. Since the cases of $\omega$ and $\omega^{(-1)}$ are similar, we set $\epsilon=1$ or $-1$ and treat the cases of $\omega$ and $\omega^{(-1)}$ simultaneously. Recall that the minimal degree of the generators of $\omega^{(\epsilon)}$ is $q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$ by Corollary 3.3.

We first note the following fact.

Lemma 5.1

Let $\xi\in{\mathcal{S}}^{(\epsilon)}$ and $d=\xi(-\infty)$. Suppose that $C_{1}$, $C_{2}\in C_{\xi}^{[d-\epsilon]}$ and $z\in C_{1}\cap C_{2}$. Then $(C_{1}\cap(-\infty,z])\cup(C_{2}\cap(z,\infty))\in C_{\xi}^{[d-\epsilon]}$.

• Proof

Set $C=(C_{1}\cap(-\infty,z])\cup(C_{2}\cap(z,\infty))$ and $C^{\prime}=(C_{2}\cap(-\infty,z])\cup(C_{1}\cap(z,\infty))$. Then, it is easily verified that $C$ and $C^{\prime}$ are maximal chains in $P$. Further,

[TABLE]

and

[TABLE]

Since $\sum_{c\in C_{1}\cap(-\infty,z]}\xi(w)+\sum_{c\in C_{1}\cap(z,\infty)}\xi(w)=\xi^{+}(C_{1})=d-\epsilon$ and $\sum_{c\in C_{2}\cap(-\infty,z]}\xi(w)+\sum_{c\in C_{2}\cap(z,\infty)}\xi(w)=\xi^{+}(C_{2})=d-\epsilon$, we see that

[TABLE]

**On the other hand, since $\xi\in{\mathcal{S}}^{(\epsilon)}$, we see that $\xi^{+}(C^{\prime\prime})\leq d-\epsilon$ for any maximal chain $C^{\prime\prime}$ in $P$. Thus, we see that $\xi^{+}(C)=\xi(C^{\prime})=d-\epsilon$.  **

Next we show the following fact.

Lemma 5.2

Let $\xi$ be an element of ${\mathcal{S}}^{(\epsilon)}$ with $d=\xi(-\infty)>q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$. Then for any $C\in C_{\xi}^{[d-\epsilon]}$, there exists $z\in C$ such that $\xi(z)>\epsilon$.

• Proof

**Assume the contrary and take $C\in C_{\xi}^{[d-\epsilon]}$ with $\xi(c)\leq\epsilon$ for any $c\in C$. Since $\xi\in{\mathcal{S}}^{(\epsilon)}$, we see that $\xi(c)=\epsilon$ for any $c\in C$. Therefore, $d-\epsilon=\xi^{+}(C)=\epsilon(\#C)=\epsilon(\#(C\cup\{-\infty,\infty\})-1)-\epsilon\leq q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)-\epsilon$. This contradicts to the assumption that $d>q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$.  **

Now we prove the following.

Lemma 5.3

Let $\xi$ be an element of ${\mathcal{S}}^{(\epsilon)}$ such that $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$ with degree $d>q^{(\epsilon)}{\mathrm{dist}}(-\infty,\infty)$. Define $\xi_{1}\colon P^{-}\to\mathbb{Z}$ by

[TABLE]

Then $\xi_{1}\in{\mathcal{S}}^{(\epsilon)}$ and $T^{\xi_{1}}$ is a generator of $\omega^{(\epsilon)}$ with degree $d-1$.

• Proof

First we show that $\xi_{1}\in{\mathcal{S}}^{(\epsilon)}$. It is clear that $\xi_{1}(z)\geq\epsilon$ for any $z\in P$ by the definition of $\xi_{1}$. Let $C$ be an arbitrary maximal chain in $P$. Then $\xi^{+}(C)\leq d-\epsilon$ since $\xi\in{\mathcal{S}}^{(\epsilon)}$. If $\xi^{+}(C)\leq d-\epsilon-1$, then $\xi^{+}_{1}(C)\leq d-\epsilon-1$, since $\xi_{1}(z)\leq\xi(z)$ for any $z\in P$. Suppose that $C\in C_{\xi}^{[d-\epsilon]}$. Then by Lemma 5.2, we see that there is $z\in C$ with $\xi(z)>\epsilon$. Therefore, if we set $z=\max\{c\in C\mid\xi(c)>\epsilon\}$, then $\xi_{1}(z)=\xi(z)-1$. Thus, we see that

[TABLE]

Now we show that $T^{\xi_{1}}$ is a generator of $\omega^{(\epsilon)}$. Since $T^{\xi}$ is a generator of $\omega^{(\epsilon)}$, we see by Proposition 4.10 that there is a $q^{(\epsilon)}$-reduced sequence $y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ with condition N’ and $C_{0}$, $C_{1}$, …, $C_{t}\in C_{\xi}^{[d-\epsilon]}$ which satisfy the conditions of (3) of Proposition 4.10.

If $t=0$, then $C_{0}\in C_{\xi}^{[d-\epsilon]}$ and $\xi(z)=\epsilon$ for any $z\in C_{0}$. This contradicts to Lemma 5.2. Thus, we see that $t>0$. Further, we see that $\{c\in C_{i}\mid\xi(c)>\epsilon\}\neq\emptyset$ for $0\leq i\leq t$. In particular, $\{c\in C_{i}\mid\xi_{1}(c)=\xi(c)-1\}\neq\emptyset$ for $0\leq i\leq t$.

Set $z_{i}\mathrel{:=}\min\{c\in C_{i}\mid\xi_{1}(c)=\xi(c)-1\}$ for $0\leq i\leq t$. Since $\xi(c)=\epsilon$ for any $c\in C_{t}\cap(x_{t},\infty)$, we see that $z_{t}\leq x_{t}$. Set $j\mathrel{:=}\min\{i\mid z_{i}\leq x_{i}\}$. Then $j>0$. Further, for any $i$ with $0\leq i<j$, $z_{i}\geq y_{i}$. For $i$ with $0\leq i\leq j$, take $C^{\prime}_{i}\in C_{\xi}^{[d-\epsilon]}$ with $z_{i}=\max\{c\in C^{\prime}_{i}\mid\xi(c)>\epsilon\}$. Then by Lemma 5.1, we see that

[TABLE]

and moreover, by the definition of $z_{i}$, and the choice of $C^{\prime}_{i}$, $\{c\in C^{\prime\prime}_{i}\mid\xi_{1}(c)\neq\xi(c)\}=\{z_{i}\}.$ In particular, $\xi^{+}_{1}(C^{\prime\prime}_{i})=d-\epsilon-1$ for $0\leq i\leq j$ since $\xi_{1}(z_{i})=\xi(z_{i})-1$.

**Since $y_{0}>x_{1}<y_{1}>x_{2}<\cdots<y_{j-1}>z_{j}$, $y_{0}\in C^{\prime\prime}_{0}$, $z_{j}\in C^{\prime\prime}_{j}$, $x_{i}$, $y_{i}\in C^{\prime\prime}_{i}$ for $1\leq i\leq j-1$ and $C^{\prime\prime}_{i}\in C_{\xi_{1}}^{[d-1-\epsilon]}$ for $0\leq i\leq j$ and $\xi_{1}(z)=\epsilon$ for any $z\in C^{\prime\prime}_{i}\cap(x_{i},y_{i})$ for $i$ with $0\leq i\leq j-1$ and $z\in C^{\prime\prime}_{j}\cap(z_{j},\infty)$, we see by Proposition 4.10 that $T^{\xi_{1}}$ is a generator of $\omega^{(\epsilon)}$.  **

Now we show the following.

Theorem 5.4

Let $P$ be a finite poset. Set $d_{0}\mathrel{:=}q^{(1)}{\mathrm{dist}}(-\infty,\infty)$ (resp. $d_{0}\mathrel{:=}q^{(-1)}{\mathrm{dist}}(-\infty,\infty)$) and ${d_{\max}}\mathrel{:=}\max\{\sum_{\ell=0}^{t}q^{(1)}(x_{0},y_{0},\ldots,x_{t},y_{t})\mid y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is a $q^{(1)}$-reduced sequence with condition N’, where $x_{0}=-\infty$ and $y_{t}=\infty\}$ (resp. ${d_{\max}}\mathrel{:=}\max\{\sum_{\ell=0}^{t}q^{(-1)}(x_{0},y_{0},\ldots,x_{t},y_{t})\mid y_{0}$, $x_{1}$, …, $y_{t-1}$, $x_{t}$ is a $q^{(-1)}$-reduced sequence with condition N’, where $x_{0}=-\infty$ and $y_{t}=\infty\}$). Then for any integer $d$ with $d_{0}\leq d\leq{d_{\max}}$, there exists a generator of $\omega$ (resp. $\omega^{(-1)}$) of degree $d$.

• Proof

**We prove the assertions of $\omega$ and $\omega^{(-1)}$ simultaneously by setting $\epsilon=1$ or $-1$. By Corollary 3.3, we see that arbitrary generator of $\omega^{(\epsilon)}$ has degree greater than or equals to $d_{0}$. Further, by Corollary 4.11, we see that any generator of $\omega^{(\epsilon)}$ has degree less than or equals to ${d_{\max}}$. On the other hand, we see by Proposition 4.16 that there is a generator of $\omega^{(\epsilon)}$ with degree ${d_{\max}}$. Thus, by Lemma 5.3 and the backward induction, we see the result.  **

By the same argument as the end of §3, we see the following fact.

Corollary 5.5

Let $G$ be a finite graph. Suppose that each odd cycle of $G$ has a triangular chord. Then the degree of the generators of the canonical and anticanonical ideals of the Ehrhart ring of the stable set polytope of $G$ are consecutive integers.