Compatible algebras with straightening laws on distributive lattices

TL;DR

This paper characterizes finite distributive lattices that admit a unique compatible algebra with straightening laws, providing a clear understanding of their algebraic structure.

Contribution

It introduces a characterization of finite distributive lattices with a unique compatible algebra with straightening laws, advancing the theoretical understanding of these algebraic structures.

Findings

Identifies conditions for the existence of a unique compatible algebra with straightening laws.

Provides a classification of finite distributive lattices based on their algebraic compatibility.

Enhances the theoretical framework connecting distributive lattices and straightening laws.

Abstract

We characterize the finite distributive lattices on which there exists a unique compatible algebra with straightening laws.

Equations

$$\mathcal{O}(P)=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n:0\le x_i\le 1,1\le i\le n,x_i\ge x_j\text{ if }{p}_i\le p_j\text{ in }P\}.$$

$$\mathcal{C}(P)=\{x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n:x_i\ge 0,1\le i\le n,$$

$$\text{ and }{x}_{i_1}+\cdots +x_{i_r}\le 1\text{ if }{p}_{i_1}<\cdots <p_{i_r}\text{is a maximal chain in }P\}.$$

$$\psi (\alpha )\psi (\beta )=\psi (\alpha \ast \beta )\psi (\alpha \cup \beta )$$

$$\delta (\alpha )\delta (\beta )=\delta (\alpha \cap \beta )\delta (\alpha \circ \beta )$$

$$\alpha \cap \beta =\alpha \ast \beta \text{ and }\overline{\alpha }\cap \overline{\beta }=\overline{\alpha \circ \beta }$$

$$k=n-(\mathrm{rank}(\alpha \cup {\alpha }^{\prime })-\mathrm{rank}(\alpha \cap {\alpha }^{\prime })).$$

$${\omega }_{\alpha }{\omega }_{{\alpha }^{\prime }}={\omega }_{\beta }{\omega }_{{\beta }^{\prime }},\text{ with }\beta \subseteq \alpha \cap {\alpha }^{\prime }\text{ and }{\beta }^{\prime }⊋\alpha \cup {\alpha }^{\prime }.$$

$${\alpha }_1\subseteq {\beta }^{\prime }\text{ and }\alpha \cap {\alpha }_1=\alpha \cap ({\alpha }^{\prime }\cup \{q\})=\alpha \cap {\alpha }^{\prime }.$$

$$\mathrm{rank}(\alpha \cup {\alpha }_1)-\mathrm{rank}(\alpha \cap {\alpha }_1)=\mathrm{rank}(\alpha \cup {\alpha }^{\prime }\cup \{q\})-\mathrm{rank}(\alpha \cap {\alpha }^{\prime })=$$

$$=\mathrm{rank}(\alpha \cup {\alpha }^{\prime })+1-\mathrm{rank}(\alpha \cap {\alpha }^{\prime })=n-k.$$

$${\omega }_{\alpha }{\omega }_{{\alpha }^{\prime }}={\omega }_{\beta }{\omega }_{{\beta }^{\prime }}\text{ and }{\omega }_{\alpha }{\omega }_{{\alpha }_1}={\omega }_{\alpha \cap {\alpha }^{\prime }}{\omega }_{\alpha \cup {\alpha }_1}.$$

$${\omega }_{\alpha \cap {\alpha }^{\prime }}{\omega }_{{\alpha }^{\prime }}{\omega }_{\alpha \cup {\alpha }_1}={\omega }_{\beta }{\omega }_{{\alpha }_1}{\omega }_{{\beta }^{\prime }}.$$

$$\alpha \cap {\alpha }^{\prime }\subset {\alpha }^{\prime }\subset \alpha \cup {\alpha }_1\text{ and }\beta \subset \alpha \cap {\alpha }^{\prime }=\alpha \cap {\alpha }_1\subset {\alpha }_1\subset {\beta }^{\prime }.$$

Full text

Introduction

Let $P$ be a finite partial ordered set (poset for short) and ${\mathcal{I}}(P)$ the distributive lattice of the poset ideals of $P.$ By a famous theorem of Birkhoff [1], for every finite distributive lattice $L$, there exists a unique subposet $P$ of $L$ such that $L={\mathcal{I}}(P).$ The order polytope ${\mathcal{O}}(P)$ and the chain polytope ${\mathcal{C}}(P)$ were introduced in [7]. In [3] it was shown that the toric ring $K[{\mathcal{O}}(P)]$ over a field $K$ is an algebra with straightening laws (ASL in brief) on the distributive lattice ${\mathcal{I}}(P)$ over the field $K.$ In [5] it was shown that the ring $K[{\mathcal{C}}(P)]$ associated with the chain polytope shares the same property.

Let $S=K[x_{1},\ldots,x_{n},t]$ be the polynomial ring over a field $K$ and $\{w_{\alpha}\}_{\alpha\in{\mathcal{I}}(P)}$ be a set of monomials in $x_{1},\ldots,x_{n}$ indexed by ${\mathcal{I}}(P)$. Let $K[\Omega]\subset S$ be the toric ring generated over $K$ by the set of monomials $\Omega=\{\omega_{\alpha}\}_{\alpha\in{\mathcal{I}}(P)}$ where $\omega_{\alpha}=w_{\alpha}t$ for all $\alpha\in{\mathcal{I}}(P).$ Clearly, $K[\Omega]$ is a graded algebra if we set $\deg(\omega_{\alpha})=1$ for all $\alpha\in{\mathcal{I}}(P).$ Let $\varphi:{\mathcal{I}}(P)\rightarrow K[\Omega]$ be the injective map defined by $\varphi(\alpha)=\omega_{\alpha}$ for all $\alpha\in{\mathcal{I}}(P).$ Assume that $K[\Omega]$ is an ASL on ${\mathcal{I}}(P)$ over $K.$ According to [5], $K[\Omega]$ is a *compatible * ASL if each of its straightening relations is of the form $\varphi(\alpha)\varphi(\alpha^{\prime})=\varphi(\beta)\varphi(\beta^{\prime})$ with $\beta\subseteq\alpha\cap\alpha^{\prime}$ and $\beta^{\prime}\supseteq\alpha\cup\alpha^{\prime},$ where $\alpha,\alpha^{\prime}$ are incomparable elements in ${\mathcal{I}}(P).$ If $K[\Omega]$ and $K[\Omega^{\prime}]$ are compatible ASL on ${\mathcal{I}}(P)$ over $K,$ we identify them if they have the same straightening relations. In this case, we write $K[\Omega]\equiv K[\Omega^{\prime}].$

In [5, Question 5.1 (b)], Hibi and Li asked the following question:* For which posets $P$, does there exists a unique compatible ASL on ${\mathcal{I}}(P)$ over $K?$*

In this note, we give a complete answer to this question. Namely, we prove the following.

Theorem 1**.**

Let $P$ be a finite poset. Then, the following statements are equivalent:

• (i)

There exists a unique compatible ASL on ${\mathcal{I}}(P)$ over $K.$

• (ii)

$K[{\mathcal{O}}(P)]\equiv K[{\mathcal{C}}(P)]\equiv K[{\mathcal{C}}(P^{\ast})],$* where $P^{\ast}$ denotes the dual poset of $P.$*

• (iii)

Each connected component of $P$ is a chain, that is, $P$ is a direct sum of chains.

1. Order polytopes, chain polytopes, and their associated toric rings

Let $P=\{p_{1},\ldots,p_{n}\}$ be a finite poset. For the basic terminology regarding posets which is used in this paper we refer to [1] and [8, Chapter 3]. The order polytope ${\mathcal{O}}(P)$ is defined as

[TABLE]

In [7, Corollay 1.3] it was shown that the vertices of ${\mathcal{O}}(P)$ are $\sum_{p_{i}\in\alpha}{\mathbb{e}}_{i},\alpha\in{\mathcal{I}}(P).$ Here ${\mathbb{e}}_{i}$ denotes the unit coordinate vector in ${\mathbb{R}}^{n}.$ If $\alpha=\emptyset,$ then the corresponding vertex in ${\mathcal{O}}(P)$ is the origin of ${\mathbb{R}}^{n}.$

The chain polytope ${\mathcal{C}}(P)$ is defined as

[TABLE]

[TABLE]

In [7, Theorem 2.2], it was proved that the vertices of ${\mathcal{C}}(P)$ are $\sum_{p_{i}\in A}{\mathbb{e}}_{i}$ where $A$ is an antichain in $P.$ Recall that an antichain in $P$ is a subset of $P$ such that any two distinct elements in the subset are incomparable. Since every poset ideal is uniquely determined by its antichain of maximal elements, it follows that ${\mathcal{O}}(P)$ and ${\mathcal{C}}(P)$ have the same number of vertices. However, as it was observed in [7], in general, these two polytopes are not combinatorially equivalent, that is, ${\mathcal{O}}(P)$ and ${\mathcal{C}}(P)$ need not have the same number of $i$-dimensional faces for $i>0.$ Combinatorially equivalence of order and chain polytopes is studied in [6].

The toric rings $K[{\mathcal{O}}(P)]$ and $K[{\mathcal{C}}(P)]$

To each subset $W\subset P$ we attach the squarefree monomial $u_{W}\in K[x_{1},\ldots,x_{n}],$ $u_{W}=\prod_{p_{i}\in W}x_{i}.$ If $W=\emptyset,$ then $u_{W}=1.$ The toric ring $K[{\mathcal{O}}(P)]$, known as the Hibi ring associated with the distributive lattice ${\mathcal{I}}(P),$ is generated over $K$ by all the monomials $u_{\alpha}t\in S$ where $\alpha\in{\mathcal{I}}(P).$ The toric ring $K[{\mathcal{C}}(P)]$ is generated by all the monomials $u_{A}t$ where $A$ is an antichain in $P.$ Also, as we have already mentioned in Introduction, both rings are algebras with straightening laws on ${\mathcal{I}}(P)$ over $K.$

We recall the definition of an ASL. For a quick introduction to this topic we refer to [2] and [4, Chapter XIII]. Let $K$ be a field, $R=\oplus_{i\geq 0}R_{i}$ with $R_{0}=K$ be a graded $K$-algebra, $H$ a finite poset, and $\varphi:H\rightarrow R$ an injective map which maps each $\alpha\in H$ to a homogeneous element $\varphi(\alpha)\in R$ with $\deg\varphi(\alpha)\geq 1$. A standard monomial in $R$ is a product $\varphi(\alpha_{1})\varphi(\alpha_{2})\cdots\varphi(\alpha_{k})$ where $\alpha_{1}\leq\alpha_{2}\leq\cdots\leq\alpha_{k}$ in $H.$

Definition 2**.**

The $K$-algebra $R$ is called an algebra with straightening laws on $H$ over $K$ if the following conditions hold:

• (1)

The set of standard monomials is a $K$–basis of $R;$

• (2)

If $\alpha,\beta\in H$ are incomparable and if $\alpha\beta=\sum c_{i}\gamma_{i1}\ldots\gamma_{ik_{i}},$ where $c_{i}\in K\setminus\{0\}$ and $\gamma_{i1}\leq\ldots\leq\gamma_{ik_{i}},$ is the unique expression of $\alpha\beta$ as a linear combination of standard monomials, then $\gamma_{i1}\leq\alpha,\beta$ for all $i.$

The above relations $\alpha\beta=\sum c_{i}\gamma_{i1}\ldots\gamma_{ik_{i}}$ are called the straightening relations of $R$ and they generate the defining ideal of $R.$

Let us go back to the toric rings $K[{\mathcal{O}}(P)]$ and $K[{\mathcal{C}}(P)].$

One considers $\varphi:{\mathcal{I}}(P)\rightarrow K[{\mathcal{O}}(P)]$ defined by $\varphi(\alpha)=u_{\alpha}t$ for every $\alpha\in{\mathcal{I}}(P).$ As it was proved by Hibi in [3], $K[{\mathcal{O}}(P)]$ is an ASL on ${\mathcal{I}}(P)$ over $K$ with the straightening relations $\varphi(\alpha)\varphi(\beta)=\varphi(\alpha\cap\beta)\varphi(\alpha\cup\beta)$ where $\alpha,\beta$ are incomparable elements in ${\mathcal{I}}(P).$

On the other hand, one defines $\psi:{\mathcal{I}}(P)\rightarrow K[{\mathcal{C}}(P)]$ by setting $\psi(\alpha)=u_{\max\alpha}t$ for all $\alpha\in{\mathcal{I}}(P)$ where $\max\alpha$ denotes the set of the maximal elements in $\alpha.$ Note that, for every $\alpha\in{\mathcal{I}}(P),$ $\max\alpha$ is an antichain in $P$ and each antichain $A\subset P$ determines a unique ideal $\alpha\in{\mathcal{I}}(P),$ namely, the poset ideal generated by $A.$ Therefore, $\psi$ is an injective well defined map and by [5, Theorem 3.1], the ring $K[{\mathcal{C}}(P)]$ is an ASL on ${\mathcal{I}}(P)$ over $K$ with the straightening relations

[TABLE]

where $\alpha\ast\beta$ is the poset ideal of $P$ generated by $\max(\alpha\cap\beta)\cap(\max\alpha\cup\max\beta).$

We observe that one may also consider $K[{\mathcal{C}}(P^{\ast})]$ as an ASL on ${\mathcal{I}}(P)$, where $P^{\ast}$ is the dual poset of $P.$ We may define $\delta:{\mathcal{I}}(P)\rightarrow K[C(P^{\ast})]$ by $\delta(\alpha)=u_{\min\overline{\alpha}}t$ for $\alpha\in{\mathcal{I}}(P),$ where $\min\overline{\alpha}$ is the set of minimal elements in $\overline{\alpha}$ and $\overline{\alpha}$ is the filter $P\setminus\alpha$ of $P.$ We recall that a filter $\gamma$ in $P$ (or dual order ideal) is a subset of $P$ with the property that for every $p\in\gamma$ and every $q\in P$ with $q\geq p,$ we have $q\in\gamma.$ Thus, a filter in $P$ is simply a poset ideal in the dual poset $P^{\ast}.$ The ring $K[C(P^{\ast})]$ is an ASL on ${\mathcal{I}}(P)$ over $K$ as well with the straightening relations

[TABLE]

for incomparable elements $\alpha,\beta\in{\mathcal{I}}(P),$ where $\alpha\circ\beta$ is the poset ideal of $P$ which is the complement in $P$ of the filter generated by $\min(\overline{\alpha}\cap\overline{\beta})\cap(\min\overline{\alpha}\cup\min\overline{\beta}).$ Let us also observe that all the algebras $K[{\mathcal{O}}(P)],K[{\mathcal{C}}(P)]$, and $K[{\mathcal{C}}(P^{\ast})]$ are compatible algebras with straightening laws.

2. Proof of Theorem 1

We clearly have (i) $\Rightarrow$ (ii). Let us now prove (ii) $\Rightarrow$ (iii). By hypothesis, the straightening relations of $K[{\mathcal{O}}(P)],K[{\mathcal{C}}(P)],$ and $K[{\mathcal{C}}(P^{\ast})]$ coincide. Therefore, we must have

[TABLE]

for all $\alpha,\beta$ incomparable elements in ${\mathcal{I}}(P).$ From the second equality in (1), it follows that $\overline{\alpha}\cap\overline{\beta}$ is the filter of $P$ generated by $\min(\overline{\alpha}\cap\overline{\beta})\cap(\min\overline{\alpha}\cup\min\overline{\beta}).$ Assume that there exists two incomparable elements $p,p^{\prime}\in P$ such that there exists $q\in P$ with $q>p$ and $q>p^{\prime}.$ Consider $\overline{\alpha}$ the filter generated by $p$ and $\overline{\beta}$ the filter generated by $p^{\prime}.$ Then $\min(\overline{\alpha}\cap\overline{\beta})\cap(\min\overline{\alpha}\cup\min\overline{\beta})=\emptyset$, but obviously, $\overline{\alpha}\cap\overline{\beta}\neq\emptyset.$ This shows that, for any two incomparable elements $p,p^{\prime}\in P,$ there is no element $q\in P$ with $q>p,q>p^{\prime}.$

Similarly, by using the first equality in (1), we derive that, for any two incomparable elements $p,p^{\prime}\in P,$ there is no element $q\in P$ with $q<p,q<p^{\prime}$. This shows that every connected component of the poset $P$ is a chain.

Finally, we prove (iii) $\Rightarrow$ (i). Let $P$ be a poset such that all its connected components are chains and assume that the cardinality of $P$ is equal to $n.$ Let $\{\omega_{\alpha}\}_{\alpha\in{\mathcal{I}}(P)}$ be the generators of $K[\Omega]\subset S$ and assume that the straightening relations of $K[\Omega]$ are $\varphi(\alpha)\varphi(\alpha^{\prime})=\varphi(\beta)\varphi(\beta^{\prime})$ where $\beta\subseteq\alpha\cap\alpha^{\prime},$ $\beta^{\prime}\supseteq\alpha\cup\alpha^{\prime},$ and $\alpha,\alpha^{\prime}$ are incomparable elements in ${\mathcal{I}}(P).$ We have to show that, for all $\alpha,\alpha^{\prime}$ incomparable elements in ${\mathcal{I}}(P),$ we have $\beta=\alpha\cap\alpha^{\prime}$ and $\beta^{\prime}=\alpha\cup\alpha^{\prime}.$

We proceed by induction on

[TABLE]

If $k=0,$ that is, $\operatorname{rank}(\alpha\cup\alpha^{\prime})-\operatorname{rank}(\alpha\cap\alpha^{\prime})=n$, then $\alpha\cup\alpha^{\prime}=P$ and $\alpha\cap\alpha^{\prime}=\emptyset,$ thus $\beta=\alpha\cap\alpha^{\prime}$ and $\beta^{\prime}=\alpha\cup\alpha^{\prime}.$ Assume that the desired conclusion is true for $\operatorname{rank}(\alpha\cup\alpha^{\prime})-\operatorname{rank}(\alpha\cap\alpha^{\prime})=n-k$ with $k\geq 0.$ Let us choose now $\alpha,\alpha^{\prime}$ incomparable in ${\mathcal{I}}(P)$ such that $\operatorname{rank}(\alpha\cup\alpha^{\prime})-\operatorname{rank}(\alpha\cap\alpha^{\prime})=n-k-1$ and assume that we have a straightening relation $\varphi(\alpha)\varphi(\alpha^{\prime})=\varphi(\beta)\varphi(\beta^{\prime})$ with $\beta\subsetneq\alpha\cap\alpha^{\prime}$ or $\beta^{\prime}\supsetneq\alpha\cup\alpha^{\prime}.$ By duality, we may reduce to considering $\beta^{\prime}\supsetneq\alpha\cup\alpha^{\prime}.$ In other words, in $K[\Omega]$ we have

[TABLE]

As $P$ is a direct sum of chains, we may find $p\in\max(\alpha\cup\alpha^{\prime})$ and $q\in\beta^{\prime}\setminus(\alpha\cup\alpha^{\prime})$ such that $q$ covers $p$ in $P$, that is, there is no other element $q^{\prime}$ in $P$ with $q>q^{\prime}>p.$ Without loss of generality, we may assume that $p\in\alpha^{\prime}.$ Let $\alpha_{1}$ be the poset ideal of $P$ generated by $\alpha^{\prime}\cup\{q\}.$ As all the connected components of $P$ are chains, we have $\alpha_{1}=\alpha^{\prime}\cup\{q\}$ since there are no other elements in $P$ which are smaller than $q$ except those which are on the same chain as $p$ and $q$ which are in $\alpha^{\prime}.$ Moreover, by the choice of $q,$ we have

[TABLE]

On the other hand,

[TABLE]

[TABLE]

By the inductive hypothesis, it follows that $\varphi(\alpha)\varphi(\alpha_{1})=\varphi(\alpha\cap\alpha_{1})\varphi(\alpha\cup\alpha_{1}),$ or, equivalently, in $K[\Omega]$ we have the equality $\omega_{\alpha}\omega_{\alpha_{1}}=\omega_{\alpha\cap\alpha_{1}}\omega_{\alpha\cup\alpha_{1}}.$ Thus, we have obtained the following equalities in $K[\Omega]:$

[TABLE]

This implies that

[TABLE]

In addition, we have:

[TABLE]

This implies that both monomials in (2) are standard monomials in $K[\Omega]$ which is in contradiction to the condition that the standard monomials form a $K$-nasis in $K[\Omega].$ Therefore, our proof is completed.