# A lattice framework for generalizing shellable complexes and matroids

Rakhi Pratihar1 Inria Centre de Saclay, Campus Polytechnique, Palaiseau,
France  and  Tovohery H. Randrianarisoa2 and Klara Stokes2 Department of
Mathematics and Mathematical Statistics, Umeå University, Umeå, Sweden
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We introduce the notion of _power lattices_ that unifies and extends the
equicardinal geometric lattices, Cartesian products of subspace lattices, and
multiset subset lattices, among several others. The notions of shellability
for simplicial complexes, $q$-complexes, and multicomplexes are then unified
and extended to that of complexes in power lattices, which we name as
_$\mathcal{P}$ -complexes_. A nontrivial class of shellable
$\mathcal{P}$-complexes are obtained via $\mathcal{P}$-complexes of the
independent sets of a matroid in power lattice, which we introduce to
generalize matroids in Boolean lattices, $q$-matroids in subspace lattices,
and sum-matroids in Cartesian products of subspace lattices. We also prove
that shellable $\mathcal{P}$-complexes in a power lattice yield shellable
order complexes, extending the celebrated result of shellability of order
complexes of (equicardinal) geometric lattices by Björner and also, a recent
result on shellability of order complexes of lexicographically shellable
$q$-complexes. Finally, we provide a construction of matroids on the lattice
of multiset subsets from weighted graphs. We also consider a variation of
Stanley-Reisner rings associated with shellable multicomplexes than the one
considered by Herzog and Popescu and proved that these rings are sequentially
Cohen-Macaulay.

###### Key words and phrases:

Shellability, order complexes, matroids, sequential Cohen-Macaulay,
multicomplexes, power lattices

###### 2010 Mathematics Subject Classification:

13F55, 05E45, 06C10, 05B35

##  1\. Introduction

Shellability is an important notion having interesting applications in
combinatorics, commutative algebra, and algebraic topology. Historically, it
was introduced to prove the Euler-Poincaré formula for higher-dimensional
convex polytopes. The shellability property was first implicitly assumed by
Schläfli [20] in 1852 and was formally established after many decades by
Bruggesser and Mani [7] in 1971.

The original definition of shellability for polyhedral complexes takes the
following form for abstract simplicial complexes pioneered by Björner and
Wachs [4]. An abstract simplicial complex $\Delta$ is said to be shellable if
it is pure (i.e., all its facets or maximal faces have the same dimension) and
there is a linear ordering $F_{1},\ldots,F_{t}$ of its facets such that for
each $j=2,\ldots,t$, the complex
$\left<F_{j}\right>\cap\left<F_{1},\ldots,F_{j-1}\right>$ is generated by a
nonempty set of maximal proper faces of $F_{j}$. Here for $i=1,\ldots,t$, by
$\left<F_{1},\ldots,F_{i}\right>$ we denote the complex generated by
$F_{1},\ldots,F_{i}$, i.e., the smallest simplicial complex containing
$F_{1},\ldots,F_{i}$.

From the topological point of view, a shellable simplicial complex is weak
homotopy equivalent to a wedge sum of spheres, thus the homology groups are
well understood [15, §2.3]. Shellable simplicial complexes are important in
commutative algebra, partly because their “face rings” or Stanley-Reisner
rings (over any field) are Cohen-Macaulay. These rings were introduced
independently by Hochster and Stanley and they have nice properties. Since
Gröbner degenerations of the coordinate rings of many classes of algebraic
varieties turn out to be Stanley-Reisner rings of simplicial complexes,
shellability provides a tool to prove their Cohen-Macaulayness. Even if the
shellable simplicial complexes are not pure, the associated Stanley-Reisner
ring is still sequentially Cohen-Macaulay [27, Chapter 2, Section 2]. The
notion of shellability has been generalized to multicomplexes (e.g. [9,
Definition 3.4]). It was also shown that when a pure multicomplex is
shellable, the associated topological space is a wedge of spheres [9,
Corollary 3.10]. More recently, $q$-analogues of these results are obtained in
[12, 13]where (multi)complexes are replaced with $q$-complexes, introduced at
least by Rota [25] and then by Alder [1]. It was shown in [13] that when a
$q$-complex is shellable, the associated topological space is also a wedge of
spheres.

Motivated by Stanley’s work on R-labeling, Björner introduced the notions of
EL-labelling and CL-labelling that give efficient methods for establishing
shellability. Important known classes of shellable simplicial complexes
include the boundary complex of a convex polytope [7], the order complex of a
bounded, locally upper semimodular poset [11], and matroid complexes, i.e.,
complexes formed by the independent subsets of matroids [24]. Among different
constructions, one can obtain shellable multicomplexes and $q$-complexes as
independent sets of discrete polymatroids [16] and $q$-matroids [12],
respectively. For comprehensive background and the proofs supporting these
claims, we refer to the literature, including the books by Stanley [27] and
Bruns and Herzog [8], the compilation of lecture notes in [14], and Björner’s
survey article [4].

Furthermore, a more general notion of sum-matroids has been introduced in
[23], which is a direct generalization of the notions of matroids and
$q$-matroids. Therefore, it is natural to ask if the corresponding
“independent elements” have a “shellable” property and if the associated
topological space is also a wedge of spheres. With this primary motivation, we
introduce the notion of _power lattices_ and give an affirmative answer to
this question. Within this lattice, we define a general notion of
“$\mathcal{P}$-complexes” and their shellability. We will show that the
associated topological space is again a wedge of spheres. Our approach is
similar to the one in [13] where we show that the order complex associated to
a shellable $\mathcal{P}$-complex is a shellable simplicial complexes, and
then we use the homotopy equivalence between the associated topological
spaces. We obtain non-trivial classes of shellable $\mathcal{P}$-complexes by
considering the $\mathcal{P}$-complexes of independent sets of matroids in
power lattices, that we introduce to generalize the notion of matroids,
polymatroids, $q$-matroids.

In the remaining part of the paper, we focus on the Stanley-Reisner ring
associated with multicomplexes. It was shown in [17, Corollary 10.6] that the
corresponding quotient ring, which we denote by $R(\Gamma)$ is (sequentially)
Cohen-Macaulay when $\Gamma$ is a shellable multicomplex. However, as
mentioned in [9, Remark 1.5], the ideal $I(\Gamma)$ is not the same as the
Stanley-Reisner ideal $I_{\Gamma}$ when the multicomplex is a simplicial
complex. In [17], they consider the ideal $I(\Gamma)$ associated with a
multicomplex $\Gamma$ where $I(\Gamma)$ is the monomial ideals spanned by all
monomials which are not in $\Gamma$. To provide an appropriate generalization,
we consider a monomial ideal $I_{\Gamma}$ generated by monomials not in
$\Gamma$ but belonging to a power lattice of multiset subsets of some fixed
multiset. We justify this choice by showing that $I_{\Gamma}$ can be obtained
by using the same construction of the Stanley-Reisner ring using a section
ring of sheaves in [28]. Our result, therefore, states that the quotient ring
$R_{\Gamma}$ by the ideal $I_{\Gamma}$ is sequentially Cohen-Macaulay whenever
$\Gamma$ is a shellable multicomplex. Since, in general, a shellable
multicomplex can be obtained from the independent elements of a matroid on the
lattice of multiset subsets, we show how we can construct such matroids from
weighted graphs. This construction can be seen as a generalization of graphic
matroids.

The rest of this paper is organized as follows. In Section 2, we introduce the
notion of _power lattices_ and study their basic properties. We describe some
classes of lattices that it generalizes including the _equicardinal geometric
lattices_ and the _lattices of multiset subsets_. In Section 3, we introduce
the notion of complexes in power lattices that generalizes the notions of
simplicial complexes, $q$-complexes and multicomplexes. In this section, we
also introduce the notion of shellability for complexes in a power lattice
that extends and generalizes the existing notions of shellability for the
aforementioned complexes. The main result of this section is the shellability
of order complexes of a shellable complex in a power lattice. Then, we
introduce the notion of matroids in power lattices in Section 4 and prove that
the complexes induced by them are shellable complexes in power lattices. We
construct a matroid in the power lattice of multiset subsets from weighted
graphs. Finally, in Section 5, we associate Stanley-Reisner rings to
multicomplexes, and we establish the sequentially Cohen-Macaulayness of the
Stanley-Reisner rings of shellable multicomplexes.

##  2\. Power lattices

In this section, we start with recalling basic notions and definitions
regarding finite posets. Then we introduce power lattices that generalize and
extend the equicardinal geometric lattices, subset lattices, subspace lattices
and lattices of multiset subsets, among others. We give a total order on the
elements of the same rank of a power lattice. This total order will be useful
to study the complexes associated to the lattice in the subsequent sections.

######  Definition 1 (Posets).

A partially ordered set or poset is a set $P$ with a binary relation $\preceq$
defined on its elements, which satisfies

  1. P1:

(Reflexive) for all $x\in P$, $x\preceq x$,

  2. P2:

(Transitive) if $x\preceq y$ and $y\preceq z$, then $x\preceq z$,

  3. P3:

(Antisymmetric) if $x\preceq y$ and $y\preceq x$, then $x=y$.

Throughout the text, we will use $(P,\preceq)$ to denote a poset. If the
relation $\preceq$ is clear from the context, we simply use $P$ omitting the
symbol $\preceq$.

The symbol $\preceq$ reads “precedes” or “contained in” or “is less than or
equal to”. For $x,y\in P$, if $x\preceq y$ and $x\neq y$, then one writes
$x\prec y$. If $x\prec y$ and there exists no $z\in P$ such that $x\prec
z\prec y$, then one writes $x\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}y$ that
reads $y$ covers $x$ (or $x$ is covered by $y$). A poset $P$ is said to have a
lower (resp. upper) bound $x\in P$ if $\forall y\in P$, $x\preceq y$ (resp.
$y\preceq x$). A poset is _bounded_ if it admits both an upper and a lower
bound (which must be unique). In a bounded poset, the unique lower (resp.
upper) bound is denoted by $\hat{0}$ (resp. $\hat{1}$). A _subposet_
$(S,\preceq)$ of $(P,\preceq)$ is a subset $S\subset P$ with the partial order
in $S$ induced from the partial order $\preceq$ on $P$.

######  Definition 2.

Let $x,y$ be elements of the poset $(P,\preceq)$. The _join_ of $x$ and $y$
(if it exists) is an element $z\in P$ such that $x,y\preceq z$ and every $u\in
P$ with $x,\,y\preceq u$ satisfies $z\preceq u$. The join of $x$ and $y$ is
denoted by $x\vee y$. The _meet_ of $x$ and $y$ (if it exists) is an element
$z\in P$ such that $z\preceq x,\,y$ and every $u\in P$ with $u\preceq x,\,y$
satisfies $u\preceq z$. The meet of $x$ and $y$ is denoted by $x\wedge y$.

By definition, meet and join of two elements in a poset are unique. The meet
and join satisfy the following properties.

$x\preceq y\Longleftrightarrow x\vee y=y\Longleftrightarrow x\wedge y=x.$ (1)

######  Definition 3 (Lattices).

A _join-semilattice_ (resp. _meet-semilattice_) is a poset $P$ in which every
pair of elements admits a join (resp. meet). A poset is called a _lattice_ if
it is both a join-semilattice and a meet-semilattice.

######  Definition 4 (Ranked lattices).

A rank function on a poset $(P,\preceq)$ is a non-negative integer valued
function $\rho:P\rightarrow\mathbb{N}$ which satisfies

  1. (i)

if $x\prec y\in P$, then $\rho(x)<\rho(y)$,

  2. (ii)

if $x\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}y$, then $\rho(y)=\rho(x)+1$.

A lattice $(P,\preceq)$ with a rank function $\rho$ is called a _graded_ or
_ranked_ lattice, often denoted as $(P,\preceq,\rho)$ or simply as $(P,\rho)$.
We denote by $P(l)$ the elements of $P$ of rank $l$.

For a bounded ranked lattice $P$, the standard convention is to take
$\rho(\hat{0})=0$ and the rank of the lattice $P$ is defined to be
$\rho(P):=\rho(\hat{1})$.

######  Definition 5 (Semimodular lattice).

A semimodular lattice is a ranked lattice $(P,\rho)$ such that for all $x,y\in
P$,

$\rho(x\vee y)+\rho(x\wedge y)\leq\rho(x)+\rho(y).$ (2)

If equality holds in (2), the lattice is called modular.

The elements of $P(1)$ of a ranked lattice $(P,\rho)$ are called atoms of $P$.
These elements cover $\hat{0}$. We introduce the notion of multiplicity of an
element of $P$ with respect to an atom.

######  Definition 6 (Multiplicity in a lattice).

Let $(P,\rho)$ be a ranked lattice. For any element $x\in P$, we describe the
set of all atoms preceding $x$ by

${A}(x):=\\{w\in P(1)\colon w\preceq x\\}.$

In the case $A(x)=\\{w\\}$, we call $x\in P$ a _power_ of the atom $w$. The
set of all powers of an atom $w$ is denoted by $\mathrm{pow}(w):=\\{y\in
P\colon A(y)=w\\}$. In this case $v_{w}(y):=\rho(y)$ is called the _$w$
-multiplicity_ of $y$ (or the valuation of $y$ at $w$).

For an atom $w\preceq y$, the _$w$ -multiplicity_ of $y$ (or the valuation of
$y$ at $w$) is defined as

$v_{w}(y):=\max\\{v_{w}(z)\colon z\preceq y\text{ and
}z\in\mathrm{pow}(w)\\}.$

If an atom $w\npreceq y$, then we define $v_{w}(y)=0$.

######  Definition 7.

A power lattice is a finite, semimodular lattice $(P,\rho)$ which satisfies
the following properties:

  1. (i)

For any $w\in P(1)$ and integer $r$, there exists at most one element in
$\mathrm{pow}(w)$ of rank $r$.

  2. (ii)

For any $x,y\in P$, $\sum\limits_{w\in P(1)}v_{w}(x)=\sum\limits_{w\in
P(1)}v_{w}(y)\Longleftrightarrow\rho(x)=\rho(y)$.

Since by definition, a power lattice is finite, it is bounded. If $y$ is a
power of $x$, for some atom $x$, then we write $y=x^{\rho(y)}=x^{v_{x}(y)}$.

######  Lemma 8.

Given a power lattice $P$, if $x\preceq y$, then $v_{w}(x)\leq v_{w}(y)$ for
all $w\in P(1)$.

###### Proof.

Suppose $x\preceq y$, then for an atom $w$, $w^{v_{w}(x)}\preceq x\preceq y$
and therefore $v_{w}(x)\leq v_{w}(y)$. ∎

Later we will see that the converse of this lemma is also true.

######  Example 9.

The example on the right provides a finite ranked lattice $(P,\rho)$ which is
semimodular. Moreover, $3$ is the only element of rank 2 which is power of the
atom $6$. Notice that $\rho(3)=\rho(2)=2$, but $A(3)=\\{6\\}$ and
$A(2)=\\{4,5,6\\}$. Thus the total valuation of 3 is $\sum\limits_{w\in
P(1)}v_{w}(3)=2$, whereas $\sum\limits v_{w}(2)=3$. Hence it does not satisfy
Definition 7 $(ii)$ and thus $P$ is not a power lattice.

$\hat{1}$$2$ $3$$4$ $5$ $6$$\hat{0}$

Following the non-example, now we give a list of some lattices which are
included in the class of power lattices.

######  Example 10.

  1. (1)

For a finite set $A$, the Boolean lattice $2^{A}$ of all subsets of $A$ is a
power lattice $(2^{A},\subseteq)$, where $\vee$ is the union of two sets and
$\wedge$ is the intersection. The rank $\rho$ is given by the size of subsets.
For any subset $U\in 2^{A}$, $v_{\\{x\\}}(U)=1$ if $x\in U$ and
$v_{\\{x\\}}(U)=0$, otherwise. Thus for any atom $w$, the power set of $w$ is
$\mathrm{pow}(w)=\\{w\\}$. Also, the condition of equal rank elements having
equal cardinality in Definition 7 $(ii)$ is satisfied from the definition of
the rank function.

  2. (2)

Let $(\Sigma(\mathbb{F}_{q}^{n}),\subseteq)$ be the lattice of
$\mathbb{F}_{q}$-subspaces of the $n$-dimensional vector space
$\mathbb{F}_{q}^{n}$ over $\mathbb{F}_{q}$, where $\vee$ is the sum of two
spaces and $\wedge$ is the intersection. Then
$(\Sigma(\mathbb{F}_{q}^{n}),\subseteq)$ is power lattice with the rank
function $\rho$ given by dimension. Here the atoms are the one-dimensional
subspaces. For $U\in\Sigma(\mathbb{F}_{q}^{n})$, $v_{\left<x\right>}(U)=1$, if
$x\in U$ and $v_{\left<x\right>}(U)=0$, otherwise. The conditions $(i)$ and
$(ii)$ of Definition 7 are satisfied for similar reasons as in the previous
example.

  3. (3)

Let $n$ be an integer. The set $div(n)$ of all divisors of $n$ forms a power
lattice $(div(n),|)$ where $a|b$ if a divides $b$, $\vee$ is the least common
multiple and $\wedge$ is greatest common divisor of two numbers. The rank
function $\rho$ is the number of prime factors counted with multiplicity. If
$x\in div(n)$ and $x=p_{1}^{a_{1}}\dots p_{t}^{a_{t}}$ is the prime
factorization of $x$, then $v_{p_{i}}(x)=a_{i}$, if $p$ doesn’t divide $x$,
then $v_{p}(x)=0$. This is “equivalent” to the lattice of multiset we define
next and which we will also see in a later section.

  4. (4)

Let $M$ be a multiset. The set $2^{M}$ of all multiset subsets of $M$ forms a
power lattice $(2^{M},\subseteq)$ where $\vee$ is the multiset union of two
sets and $\wedge$ is the intersection. The rank function $\rho$ is defined by
the size of subsets (with multiplicity). Let $U\in 2^{M}$, if $x\in U$, then
$v_{\\{x\\}}(U)$ is the multiplicity of $x$ in $U$ and if $x\notin U$, then
$v_{\\{x\\}}(U)=0$.

  5. (5)

For $i=1,\dots,l$, let $n_{i}$ be an integer. The Cartesian product of
lattices
$\Sigma(\mathbb{F}_{q}^{n_{1}})\times\dots\times\Sigma(\mathbb{F}_{q}^{n_{l}})$
is a power lattice. $\vee$ is defined by the componentwise sums of subspaces
whereas $\wedge$ is defined by the componentwise intersections. The rank
function $\rho$ is the sum of the dimension of the components. Let
$U=(U_{1},\dots,U_{l})$ and let $u=(\langle 0\rangle,\dots,\langle
0\rangle,\langle x\rangle,\langle 0\rangle,\dots,\langle 0\rangle)$ where
$x\neq 0$ and $\left<x\right>$ is at the position $i$ for some $1\leq i\leq
l$. If $x\in U_{i}$ then $v_{u}(U)=1$. Otherwise, $v_{u}(U)=0$.

  6. (6)

If $G$ is a cyclic group, then the lattice of subgroups of $G$ forms a power
lattice $L(G)$ where $\vee$ is given by the subgroup generated by the union of
two subgroups and $\wedge$ is the intersection. Let $H\in L(G)$ such that
$|H|=\prod p_{i}^{l_{i}}$, where $p_{i}$’s are pairwise coprime. Then
$\rho(H)=\sum_{i}l_{i}$. If $C\in L(H)$ with $|C|=p_{i}$, then
$v_{C}(H)=l_{i}$, otherwise if $C\notin L(H)$, then $v_{C}(H)=0$.

  7. (7)

Let $G$ be a finite abelian group, not necessarily cyclic. By the fundamental
theorem of finite abelian groups, $G$ can be expressed as the direct sum of
cyclic groups of prime-power order say $G=\prod_{i=1}^{s}G_{i}$. So the
lattice of subgroups of $G$ can be defined as the Cartesian product of
lattices $L(G)=\prod_{i=1}^{s}L(G_{i})$. Similarly to the Cartesian product of
lattice of subspaces, we also get a power lattice of subgroups of abelian
groups.

  8. (8)

Equicardinal geometric lattices are power lattices. Indeed, a geometric
lattice is equicardinal if its elements of equal rank have equal cardinality.
Note that, the atomistic property of a geometric lattice implies that the
valuation of an element $x$ w.r.t. any atom $w$ is $v_{w}(x)=1$ if $w\in A(x)$
and 0, otherwise. Thus the property $(ii)$ of Definition 7 becomes equivalent
to the property of the same rank elements having the same cardinality. The
other properties follows from the definition of geometric lattices.

######  Remark 11.

  1. (a)

Following the characterization of geometric lattices as lattices of flats of
simple matroids, the class of equicardinal geometric (semi)lattices include
the equicardinal matroids. For more on this topic, one can refer to, e.g., the
article [2] by Alon, Babai, and Suzuki and the references mentioned there.

  2. (b)

There exists a solvable non-abelian group $G$ whose subgroup lattice do not
satisfy the property (i) in the definition of a power lattice. For example,
take the quaternion group $Q_{8}=\\{\pm 1,\pm i,\pm j,\pm k\\}$. Then
$\\{-1,1,i,-i\\}$ and $\\{-1,1,j,-j\\}$ are two powers of the atom
$\\{-1,1\\}$ and they have the same order.

######  Lemma 12.

Let $P$ be a ranked lattice with $p\in P(1)$. Then for $x,\,y\in P$,

  1. (1)

$v_{p}(x\wedge y)=\min\\{v_{p}(x),v_{p}(y)\\}$,

  2. (2)

$v_{p}(x\vee y)\geq\max\\{v_{p}(x),v_{p}(y)\\}$.

###### Proof.

Since $x\wedge y\preceq x$ and $x\wedge y\preceq y$, we have $v_{p}(x\wedge
y)\leq v_{p}(x)$ and $v_{p}(x\wedge y)\leq v_{p}(y)$. Therefore $v_{p}(x\wedge
y)\leq\min\\{v_{p}(x),v_{p}(y)\\}$. Now,
$p^{\min\\{v_{p}(x),v_{p}(y)\\}}\preceq x$ and
$p^{\min\\{v_{p}(x),v_{p}(y)\\}}\preceq y$. Therefore
$p^{\min\\{v_{p}(x),v_{p}(y)\\}}\preceq x\wedge y$. Hence
$\min\\{v_{p}(x),v_{p}(y)\\}\leq v_{p}(x\wedge y)$. Thus
$\min\\{v_{p}(x),v_{p}(y)\\}=v_{p}(x\wedge y)$. For the second point,
$p^{v_{p}(x\vee y)}\preceq x$ and $p^{v_{p}(x\vee y)}\preceq x$. Hence
$v_{p}(x\vee y)\geq v_{p}(x)$ and $v_{p}(x\vee y)\geq v_{p}(x)$. ∎

######  Remark 13.

Notice that in the second point of Lemma 12, strict inequality can happen. For
example if we have the lattice of subspaces, the sum of two subspaces can
produce a new vector which was not in the two original subspaces.

The rest of the section deals with defining a total order on the elements of
same rank of a power lattice. For that, first we will introduce the notion of
factorization of an element of $P$ into atoms. We consider an arbitrary but
fixed total order $\preceq_{1}$ on $P(1)$ i.e., the atoms of $P$.

######  Definition 14 (Factorization).

Let $x\in P$ and $A(x)=\\{x_{1},\dots,x_{l}\\}$ such that
$x_{1}\prec_{1}x_{2}\prec_{1}\cdots\prec_{1}x_{l}$. Then the factorization of
$x$ is defined as

$F(x)=(\overbrace{x_{1},\dots,x_{1}}^{v_{x_{1}}(x)\text{ times
}},\dots,\overbrace{x_{l},\dots,x_{l}}^{v_{x_{l}}(x)\text{ times }}).$

The length of the factorization of $x$ is defined as $\sum\limits_{x_{i}\in
A(x)}v_{x_{i}}(x)$. For brevity, we will denote $F(x)$ as

$F(x)=\prod_{i=1}^{l}x_{i}^{v_{x_{i}}(x)}.$

######  Remark 15.

Because of the property (ii) in Definition 7 of a power lattice, the
factorizations of two elements of the same rank must have the same length
i.e., the cardinalities are the same when counted with multiplicity.

We show next that factorization uniquely defines an element of a power
lattice.

######  Lemma 16.

Let $(P,\preceq)$ be a power lattice. For $x\in P$, if
$F(x)=\prod_{i=1}^{l}x_{i}^{v_{i}}$, then
$x=\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}$. Moreover, if $x,y\in P$ such that
$x\neq y$, then $F(x)\neq F(y)$.

###### Proof.

Suppose that $F(x)=\prod_{i=1}^{l}x_{i}^{v_{i}}$. We know that
$v_{i}=v_{x_{i}}(x)$ and from the definition of $v_{x_{i}}(x)$,
$x_{i}^{v_{i}}\preceq x$ and therefore, by definition of join,
$\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\preceq x$. By Lemma 8,
$v_{x_{i}}\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)\leq
v_{x_{i}}(x)$. On the other side, since
$x_{i}^{v_{i}}\preceq\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}$ and
$v_{i}=v_{x_{i}}(x)$, by Lemma 8, $v_{x_{i}}(x)\leq
v_{x_{i}}\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. Hence
$v_{x_{i}}(x)=v_{x_{i}}\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. It
is also clear that $A(x)\subseteq
A\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. Since
$\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\preceq x$, then
$A\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)\subseteq A(x)$. Thus
$A(x)=A\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. Therefore
$\sum\limits_{w\in P(1)}v_{w}(x)=\sum\limits_{w\in
P(1)}v_{w}\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. By the property
(ii) of a power lattice, this implies that
$\rho(x)=\rho\left(\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\right)$. Since
$\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}\preceq x$, we must have
$\bigvee\limits_{i=1}^{l}x_{i}^{v_{i}}=x$.

For the second part, if $F(x)=F(y)=\prod_{i=1}^{l}x_{i}^{v_{x_{i}}(x)}$, then
the first part of the lemma says that
$x=y=\bigvee\limits_{i=1}^{l}x_{i}^{v_{x_{i}}(x)}$. ∎

Now, we can show that the converse of Lemma 8 is also true.

######  Corollary 17.

Let $(P,\preceq)$ be a power lattice. For $x,y\in P$, $x\preceq y$ if and only
if $v_{w}(x)\leq v_{w}(y)$ for all $w\in P(1)$ (equivalently, $F(x)\subseteq
F(y)$).

###### Proof.

The first part of the proof is just Lemma 8. Conversely, suppose that for all
$w\in P(1)$, $v_{w}(x)\leq v_{w}(y)$. Therefore, $w^{v_{w}(x)}\preceq
w^{v_{w}(y)}\preceq\bigvee\limits_{w\in P(1)}w^{v_{w}(y)}=y$. The last
equality is a consequence of Lemma 16. This implies
$\bigvee\limits_{w}w^{v_{w}(x)}\preceq y$ and by Lemma 16, $x\preceq y$. ∎

######  Definition 18 (Total order on the elements of same rank).

Let $(P,\preceq,\rho)$ be a power lattice. For a positive integer $l$ with
$P(l)\neq\emptyset$, we define a total order on $P(l)$ as follows. Let $x,y$
be two distinct elements of $P(l)$ with the corresponding factorizations
$F(x)=(x_{1},\dots,x_{t})$ and $F(y)=(y_{1},\dots,y_{t})$.

We say that $x\preceq_{l}y$ if

$\begin{cases}x=y\quad\text{or}\\\ x\neq y\text{ and if }i\text{ is the
smallest integer such that }x_{i}\neq y_{i},\text{ then
}x_{i}\prec_{1}y_{i}.\end{cases}$

From Remark 15, $F(x)$ and $F(y)$ in the previous definition have the same
length. Moreover, Corollary 16 says that two distinct elements of $P(l)$ must
have distinct factorization. The relation $\preceq_{l}$ on the factorizations
is just the lexicographic ordering induced by $\preceq_{1}$, so it is clear
that we have a total order.

We give an equivalent definition of the total order of Definition 18, that
will be useful in the following sections.

######  Lemma 19.

Let $P$ be a power lattice and let $x\neq y$ be two elements of $P(l)$. Then
$x\prec_{l}y$ if and only if

$\min_{\preceq_{1}}F(x)\backslash
F(y)\prec_{1}\min_{\preceq_{1}}F(y)\backslash F(x),$

where the multiset difference is done by also considering the multiplicity.

###### Proof.

Since $F(x)$ and $F(y)$ have the same length and they are distinct, then
$F(x)\backslash F(y)$ and $F(y)\backslash F(x)$ are non-empty. Let
$F(x)=(x_{1},x_{2},\dots,x_{t})$ and $F(y)=(y_{1},y_{2}\dots,y_{t})$ be the
factorizations of $x$ and $y$ respectively. Suppose that $j$ is the smallest
integer such that $x_{j}\neq y_{j}$ and $x_{j}\prec_{1}y_{j}$. Then $x_{j}\in
F(x)\backslash F(y)$ and therefore $\min_{\preceq_{1}}F(x)\backslash
F(y)\preceq_{1}x_{j}$. Now $F(y)\backslash
F(x)\subset\\{y_{j},\dots,y_{t}\\}$. Therefore
$y_{j}\preceq_{1}\min_{\preceq_{1}}F(y)\backslash F(x)$. Since
$x_{j}\prec_{1}y_{j}$, we get the desired result.

Conversely, let $\min_{\preceq_{1}}F(x)\backslash
F(y)\prec_{1}\min_{\preceq_{1}}F(y)\backslash F(x)$. Suppose, $x\succ_{l}y$
and then it follows that $x_{j}\succ_{1}y_{j}$. This implies $y_{j}\in
F(y)\backslash F(x)$ and in fact, $y_{j}=\min_{\preceq_{1}}F(y)\backslash
F(x)$. Combining all these we get,

$\min_{\preceq_{1}}F(x)\backslash
F(y)\prec_{1}\min_{\preceq_{1}}F(y)\backslash F(x)=y_{j}\prec_{1}x_{j}.$ (3)

Since all $x_{i}$ for $i<j$, precedes $y$, $\min_{\preceq_{1}}F(x)\backslash
F(y)\prec_{1}x_{j}$ gives a contradiction. This proves $x\prec_{l}y$. ∎

######  Lemma 20.

Let $P$ be a power lattice. Let $x,y\in P$ be such that $x\in P(1)$ and
$A(y)=\\{x\\}$. Then there is a unique chain, i.e., totally ordered set
$x\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}x^{2}\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}\cdots\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}y=x^{\rho(y)}$.
The elements in this chain are all powers of $x$.

###### Proof.

For the existence of such a chain, we first note that any element $z\in P$
such that $x\prec z\prec y=x^{\rho(y)}$ has to be in $\mathrm{pow}(x)$.
Indeed, since Lemma 8 implies $\\{x\\}=A(x)\subseteq A(z)\subseteq
A(y)=\\{x\\}$ and thus $A(z)=\\{x\\}$. The existence of power of $x$ for each
$r$ with $1\leq r\leq\rho(y)$ follows from Definition 4 of ranked lattice. The
uniqueness follows from $(i)$ of Definition 7 of a power lattice. ∎

##  3\. Complexes in power lattices

In this section, we introduce the notion of complexes in power lattices, that
we call $\mathcal{P}$-complexes, to generalize and extend the notion of
abstract simplicial complexes, $q$-complexes, and multicomplexes. Then we
define the notion of $\mathcal{P}$-shellability of $\mathcal{P}$-complexes
extending the existing notions of shellability of the complexes mentioned
above. The main result of this section is the shellability of the order
complex of a $\mathcal{P}$-shellable $\mathcal{P}$-complex.

######  Definition 21 ($\mathcal{P}$-complexes).

Let $(P,\preceq)$ be a power lattice. A $\mathcal{P}$-complex $S$ in $P$ is a
non-empty subposet of $P$ such that $\forall x\in S$ and $y\in P$, $y\preceq
x$ implies $y\in S$.

In other words, a $\mathcal{P}$-complex is a subset of $P$ which is closed
under $\preceq$. For $Q\subseteq P$, the $\mathcal{P}$-complex generated by
$Q$ is the $\mathcal{P}$-complex defined by

$\langle Q\rangle:=\\{x\in P\colon x\preceq y\text{ for some }y\in Q\\}.$

######  Definition 22.

Let $(P,\rho)$ be a power lattice and let $S$ be a $\mathcal{P}$-complex in
$P$. The elements of $S$ are called the faces of $S$ and the maximum elements
of $S$ are called the facets of $S$. If all the facets have same rank, then
$S$ is called a _pure_ $\mathcal{P}$-complex. The _rank_ of a
$\mathcal{P}$-complex is defined to be the maximum of ranks of its faces.

######  Definition 23 ($\mathcal{P}$-shellable $\mathcal{P}$-complexes).

Let $(P,\preceq,\rho)$ be a power lattice. A $\mathcal{P}$-complex $S\subseteq
P$ of rank $r$ is called $\mathcal{P}$-shellable if $S$ is pure and there
exists an ordering $f_{1},f_{2},\ldots,f_{t}$ on the facets of $S$ such that
for all $1\leq i<j\leq t$, there exists $k<j$ for which $f_{i}\wedge
f_{j}\preceq f_{k}\wedge f_{j}$ and $\rho(f_{k}\wedge f_{j})=r-1$.

If we take a finite set $A$ as in Example 10, then we recover the notion of
shellable simplicial complex.

######  Definition 24.

A simplicial complex $S$ is a non-empty set of subsets of $2^{A}$ such that
$S$ is closed under inclusion. A shellable simplicial complex is a simplicial
complex $S$ such that its facets have the same cardinality $r$ and there
exists an ordering $F_{1},F_{2},\cdots,F_{t}$ of all the facets of $S$ such
that for all $i<j\leq t$, there exists $k<j$ such that $F_{i}\cap
F_{j}\subseteq F_{k}\cap F_{j}$ and $|F_{k}\cap F_{j}|=r-1$.

If we consider the power lattice $\Sigma(\mathbb{F}_{q}^{n})$, then we have
the $q$-shellable $q$-complexes

######  Definition 25 ([1, 12]).

A $q$-complex $S$ is a non-empty set of subsets of
$\Sigma(\mathbb{F}_{q}^{n})$ such that $S$ is closed under inclusion. A
$q$-shellable $q$-complex is a $q$-complex $S$ such that its facets have the
same dimension $r$ and there exists an ordering $F_{1},F_{2},\cdots,F_{t}$ of
all the facets of $S$ such that for all $i<j\leq t$, there exists $k<j$ such
that $F_{i}\cap F_{j}\subseteq F_{k}\cap F_{j}$ and
$\dim_{\mathbb{F}_{q}}F_{k}\cap F_{j}=r-1$.

######  Definition 26.

Let $P$ be a power lattice of rank $n$. For $l\leq n-1$, an $l$-sphere $S_{x}$
of $P$ is a subposet such that there exists $x\in P(l+1)$ such that
$S_{x}=\\{y\in P\colon y\prec x\\}$.

######  Remark 27.

When $P=2^{A}$ or $P=\Sigma(\mathbb{F}_{q}^{n})$, the $l$-spheres in $P$ are
$\mathcal{P}$-shellable.

For the lattice $2^{A}$ of subsets, the shellability helps to describe the
homology degrees of some topological space associated to the simplicial
complex [3, Appendix]. For the lattice of subspaces
$\Sigma(\mathbb{F}_{q}^{n})$, the shellability helped partially to describe
the homology degrees [12]. However, by considering the associated order
complex, the homology can be fully obtained [13]. We generalize this results
in the setting of $\mathcal{P}$-complexes in a power lattice.

######  Definition 28 (Order complex).

Let $P$ be a poset. A chain in $P$ is a totally ordered subset of $P$ and we
write a chain as a sequence $(x_{1}\prec x_{2}\prec\dots\prec x_{l})$. The
order complex of $\mathcal{K}(P)$ is the poset of all chains in $P$.

The order complex of a poset is known to be a simplicial complex. The next
theorem says that the order complex of an $l$-sphere is shellable as a
simplicial complex. For that we first need an ordering on the maximal chains.

######  Definition 29 (Reverse lexicographic ordering).

Let $(P,\preceq)$ be a power lattice and let $S$ be a pure
$\mathcal{P}$-complex of rank $r$. Suppose that $\preceq_{i}$ is the total
ordering on $P(i)$. The reverse ordering $\trianglelefteq$ on the facets of
$\mathcal{K}(S)$ is defined as follows: Let $X=(x_{0}\preceq\dots\preceq
x_{r})$ and $Y=(y_{0}\preceq\dots\preceq y_{r})$ be two maximal chains of $S$.
$X\trianglelefteq Y$ if $X=Y$ or if $X\neq Y$ and $x_{i}\prec_{i}y_{i}$ where
$i$ is the largest integer such that $x_{i}\neq y_{i}$.

It is clear that $\trianglelefteq$ is a total order, since this is just the
reverse lexicographic order.

######  Lemma 30.

Let $(P,\preceq,\rho)$ be a power lattice and let
$(x_{i_{1}}\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}\dots\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}x_{i_{s}})$
be a chain in $P$ with $\rho(x_{i_{j}})=i_{j}$. If for all $i_{1}<i<i_{s}$,
$x_{i}=\min_{\preceq_{i}}\\{a\in P(i)\colon x_{i-1}\prec a\prec x_{i+1}\\}$,
then for all $i_{1}<i\leq i_{s}$,

$\min_{\preceq_{1}}F(x_{i})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{i_{s}})\backslash F(x_{i-1}).$

###### Proof.

Let $i_{1}<i<i_{s}$. First we show that

$\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1}).$

Let $z=\min_{\preceq_{1}}F(x_{i+1})\backslash F(x_{i-1})$. Let $t$ be the
smallest positive integer such that $z^{t}\npreceq x_{i-1}$ and therefore
surely $z^{t}\preceq x_{i+1}$. So $x_{i-1}\prec x_{i-1}\vee z^{t}\preceq
x_{i+1}$ and by semimodularity of the lattice, we have $\rho(x_{i-1}\vee
z^{t})\leq\rho(x_{i-1})+\rho(z^{t})-\rho(x_{i-1}\wedge
z^{t})=(i-1)+t-(t-1)=i$. Since $\rho(x_{i+1})=i+1$, we have
$x_{i-1}\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}x_{i-1}\vee
z^{t}\prec\mathrel{\mkern-5.0mu}\mathrel{\cdot}x_{i+1}$ If $x_{i-1}\vee
z^{t}\neq x_{i}$, then $x_{i}\prec_{i}x_{i-1}\vee z^{t}$. Thus by Lemma 19,

$\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1}\vee
z^{t})\prec_{1}\min_{\preceq_{1}}F(x_{i-1}\vee z^{t})\backslash
F(x_{i})\preceq_{1}z.$ (4)

We get the second inequality because $z^{t}\npreceq x_{i}$ as otherwise
$x_{i-1}\vee z^{t}\preceq x_{i}$. And since they have the same rank, they
would be equal which contradicts our assumption. Since $F(x_{i})\backslash
F(x_{i-1}\vee z^{t})\subseteq F(x_{i})\backslash F(x_{i-1})\subseteq
F(x_{i+1})\backslash F(x_{i-1})$,

$z=\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})\preceq_{1}\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1}\vee
z^{t}).$

But this together with Equation (4) implies $z\prec_{1}z$, which is a
contradiction.

Therefore, $x_{i-1}\vee z^{t}=x_{i}$ and this implies $z^{t}\preceq x_{i}$.
Thus $z\in F(x_{i})\backslash F(x_{i-1})$. This implies that
$\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1})\preceq_{1}z$. On the other
side, we know that $z=\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})\preceq_{1}\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1})$. Hence
$z=\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1})$. Thus it proves that

$\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{i})\backslash F(x_{i-1}),\quad\forall
i,i_{1}<i<i_{s}.$

Let $i_{1}<i\leq i_{s}$. Now we show inductively that, for all $i<j\leq i_{s}$

$\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{j})\backslash F(x_{i-1}).$

The case $j=i_{s}$ will give the result of the Lemma. The statement is
obviously true for $j=i+1$. Suppose that it is true for $j>i$ i.e.

$\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{j})\backslash F(x_{i-1}).$ (5)

We want to show that

$\min_{\preceq_{1}}F(x_{i+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{j+1})\backslash F(x_{i-1}).$ (6)

Since $x_{j}\prec x_{j+1}$, then, using Equation (5)

$z:=\min_{\preceq_{1}}F(x_{j+1})\backslash
F(x_{i-1})\preceq_{1}\min_{\preceq_{1}}F(x_{j})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{i+1})\backslash F(x_{i-1}).$ (7)

We claim that $z\in F(x_{i+1})\backslash F(x_{i-1})$. This implies that
$\min_{\preceq_{1}}F(x_{i+1})\backslash F(x_{i-1})\preceq_{1}z$ and therefore
$\min_{\preceq_{1}}F(x_{j+1})\backslash
F(x_{i-1})=\min_{\preceq_{1}}F(x_{i+1})\backslash F(x_{i-1})$ which concludes
the proof.

Proof of claim: Suppose that $z\notin F(x_{i+1})\backslash F(x_{i-1})$. Thus
$v_{z}(x_{i-1})=v_{z}(x_{i})=v_{z}(x_{i+1})=r$. Since
$v_{z}(x_{j+1})>v_{z}(x_{i-1})$ and $i+1<j+1$, then we can find $j>k\geq i$
such that $v_{z}(x_{k-1})=v_{z}(x_{k})=v_{z}(x_{k+1})=r$ and
$v_{z}(x_{k+2})>r$.

Now, take $x_{k}\preceq x_{k}\vee z^{r+1}\preceq x_{k+2}$. Since
$z^{r+1}\npreceq x_{k}$, then $x_{k}\prec x_{k}\vee z^{r+1}$. Furthermore, by
semimodularity, $\rho(x_{k}\vee
z^{r+1})\leq\rho(x_{k})+\rho(z^{r+1})-\rho(x_{k}\wedge
z^{r+1})=k+1<\rho(x_{k+2})$. Hence $x_{k}\vee z^{r+1}\prec x_{k+2}$. Hence
$x_{k}\prec x_{k}\vee z^{r+1}\prec x_{k+2}$. By the minimality of $x_{k+1}$ in
this interval, $x_{k+1}\preceq_{k+1}x_{k}\vee z^{r+1}$. Since $z^{r+1}\npreceq
x_{k+1}$, then $x_{k+1}\prec_{k+1}x_{k}\vee z^{r+1}$. By Lemma 19,

$\min_{\preceq_{1}}F(x_{k+1})\backslash F(x_{k}\vee
z^{r+1})\prec_{1}\min_{\preceq_{1}}F(x_{k}\vee z^{r+1})\backslash F(x_{k+1}).$
(8)

Since $z^{r+1}\npreceq x_{k+1}$, then $z\in F(x_{k}\vee z^{r+1})\backslash
F(x_{k+1})$. Thus $\min_{\preceq_{1}}F(x_{k}\vee z^{r+1})\backslash
F(x_{k+1})\preceq_{1}z$. Combining this with Equation (8), we have

$\min_{\preceq_{1}}F(x_{k+1})\backslash F(x_{k}\vee z^{r+1})\prec_{1}z.$ (9)

Since $k<j$, $x_{k+1}\preceq x_{j}$. Therefore $F(x_{k+1})\subseteq F(x_{j})$.
This implies that $F(x_{k+1})\backslash F(x_{k}\vee z^{r+1})\subseteq
F(x_{j})\backslash F(x_{k}\vee z^{r+1})$. Thus

$\min_{\preceq_{1}}F(x_{j})\backslash F(x_{k}\vee
z^{r+1})\preceq_{1}\min_{\preceq_{1}}F(x_{k+1})\backslash F(x_{k}\vee
z^{r+1}).$

Together with Equation (9), this gives us

$\min_{\preceq_{1}}F(x_{j})\backslash F(x_{k}\vee z^{r+1})\prec_{1}z.$ (10)

Now, since $i\leq k$, then $x_{i-1}\preceq x_{i}\preceq x_{k}\preceq x_{k}\vee
z^{r+1}$. Thus $F(x_{j})\backslash F(x_{k}\vee z^{r+1})\subseteq
F(x_{j})\backslash F(x_{i-1})$. Thus $\min_{\preceq_{1}}F(x_{j})\backslash
F(x_{i-1})\preceq_{1}\min_{\preceq_{1}}F(x_{j})\backslash F(x_{k}\vee
z^{r+1})$. Together with Equation (7), this gives us
$z\preceq_{1}\min_{\preceq_{1}}F(x_{j})\backslash F(x_{k}\vee z^{r+1})$.
Combining this with Equation (10), we have $z\prec_{1}z$ which is impossible.
Hence the claim. ∎

If $P$ is the lattice of subsets or lattice of subspaces, the rank function
$\rho$ is modular. In this case, every sphere is shellable as $P$ is a modular
lattice. For general power lattices, we only have semimodularity of the rank
function $\rho$. So we cannot directly conclude whether the spheres are
shellable $\mathcal{P}$-complexes or not. However, we can directly show the
shellability of the associated order complexes.

######  Theorem 31 (Shellability of the order complex of a sphere).

Let $P$ be a power lattice of rank $n$ and let $l$ be an integer such that
$l\leq n-1$. For an $l$-sphere $S_{x}$ in $P$, the order complex
$\mathcal{K}(S_{x})$ is a shellable simplicial complex with respect to the
ordering $\trianglelefteq$ on the facets of $\mathcal{K}(S_{x})$.

###### Proof.

Suppose that $S_{x}=S_{z_{l+1}}$ where $z_{l+1}\in P(l+1)$. Let
$x_{l},y_{l}\in P(l)$ such that $x_{l}\prec z_{l+1}$ and $y_{l}\prec z_{l+1}$.
Suppose that $X=(x_{0}\prec\dots\prec x_{l})$ and $Y=(y_{0}\prec\dots\prec
y_{l})$ are two distinct maximal chains in $P$ such that $X\triangleleft Y$.
We want to find a chain $U\triangleleft Y$ such that $X\cap Y\subset U\cap Y$
and $|U\cap Y|=l$.

Let $x_{l+1}=y_{l+1}=z_{l+1}$. Let $s$ be the largest integer such that $s\leq
l$ and $x_{s}\neq y_{s}$. Therefore, $x_{s}\prec_{s}y_{s}$ and
$x_{s+1}=y_{s+1}$. Since $x_{0}=y_{0}$, we can find integer $r$ such that
$0\leq r<s$, $x_{r}=y_{r}$ and $x_{i}\neq y_{i}$ for $r<i\leq s$. If $r=s-1$,
then take the chain

$U=(y_{0}\prec\dots\prec y_{s-1}\prec x_{s}\prec y_{s+1}\prec\dots\prec
y_{l}).$

Otherwise, assume that $r\leq s-2$. We claim that
$y_{i}\neq\min_{\preceq_{i}}\\{a\in P\colon y_{i-1}\prec a\prec y_{i+1}\\}$
for some $r<i\leq s$. In this case we take the chain

$U=(y_{0}\prec\dots\prec y_{i-1}\prec\min_{\preceq_{i}}\\{a\in P\colon
y_{i-1}\prec a\prec y_{i+1}\\}\prec y_{i+1}\prec\dots\prec y_{l}).$

That would conclude the proof.

Proof of claim: Assume that for all $i$ such that $r<i\leq s$,
$y_{i}=\min_{\preceq_{i}}\\{a\in P\colon y_{i-1}\prec a\prec y_{i+1}\\}$.

Let $j$ be the largest integer such that $r\leq j<s$ and $y_{j}\preceq x_{s}$.
Such $j$ exists since $y_{r}=x_{r}\prec x_{s}$.

First, since $y_{j}\prec y_{s}\prec y_{s+1}$, then $F(y_{j})\subset
F(y_{s})\subset F(y_{s+1})$. Therefore

$\min_{\preceq_{1}}F(y_{s+1})\backslash
F(y_{j})\preceq_{1}\min_{\preceq_{1}}F(y_{s+1})\backslash F(y_{s})$ (11)

Now, since $x_{s}\prec x_{s+1}=y_{s+1}$, then $F(x_{s})\subset F(y_{s+1})$.
Therefore,

$\min_{\preceq_{1}}F(y_{s+1})\backslash
F(y_{s})\preceq_{1}\min_{\preceq_{1}}F(x_{s})\backslash F(y_{s})$ (12)

Equations (11) and (12) imply that

$\min_{\preceq_{1}}F(y_{s+1})\backslash
F(y_{j})\preceq_{1}\min_{\preceq_{1}}F(x_{s})\backslash F(y_{s})$ (13)

Now, since $x_{s}\prec_{s}y_{s}$, by Lemma 19,
$\min_{\preceq_{1}}F(x_{s})\backslash
F(y_{s})\prec_{1}\min_{\preceq_{1}}F(y_{s})\backslash F(x_{s})$ and together
with Equation (13), this implies that

$\min_{\preceq_{1}}F(y_{s+1})\backslash
F(y_{j})\prec_{1}\min_{\preceq_{1}}F(y_{s})\backslash F(x_{s})$ (14)

Now, by Lemma 30, we have $a:=\min_{\preceq_{1}}F(y_{s+1})\backslash
F(y_{j})=\min_{\preceq_{1}}F(y_{j+1})\backslash F(y_{j})$ and the second
equality implies that $v_{a}(y_{j})<v_{a}(y_{j+1})\leq v_{a}(y_{s})$. Hence
$a\in F(y_{j+1})\subseteq F(y_{s})$. If $a\in F(y_{s})\backslash F(x_{s})$,
then $\min_{\preceq_{1}}F(y_{s})\backslash F(x_{s})\preceq_{1}a$ i.e.

$\min_{\preceq_{1}}F(y_{s})\backslash
F(x_{s})\preceq_{1}\min_{\preceq_{1}}F(y_{s+1})\backslash F(y_{j}).$

Together with Equation (14), this gives $\min_{\preceq_{1}}F(y_{s})\backslash
F(x_{s})\prec_{1}\min_{\preceq_{1}}F(y_{s})\backslash F(x_{s})$. This is
impossible and therefore, $a\notin F(y_{s})\backslash F(x_{s})$. This implies
that $v_{a}(y_{s})\leq v_{a}(x_{s})$ and thus,
$v_{a}(y_{j})<v_{a}(y_{j+1})\leq v_{a}(y_{s})\leq v_{a}(x_{s})$. Since
$a\preceq y_{s}$, this also implies that $a\preceq x_{s}$. Let
$t=v_{a}(y_{j+1})$ and therefore, $a^{t}\npreceq y_{j}$, but $a^{t}\preceq
y_{j+1}$ and $a^{t}\preceq x_{s}$. Since $y_{j}\prec x_{s}$, then $a^{t}\vee
y_{j}\preceq x_{s}$. Now $y_{j}\preceq a^{t}\vee y_{j}\preceq y_{j+1}$. The
first inequality is strict because $a^{t}\npreceq y_{j}$ and thus comparing
the ranks, we must have $a^{t}\vee y_{j}=y_{j+1}$. Therefore, $y_{j+1}\preceq
x_{s}$ and by the property of $j$, we have $j+1\geq s$. Hence $y_{s}\preceq
y_{j+1}\preceq x_{s}$. Since $\rho(y_{s})=\rho(x_{s})$, then $x_{s}=y_{s}$
which is a contradiction. Hence we conclude that there exists $r<i\leq s$ such
that $y_{i}\neq\min_{\prec_{i}}\\{a\in P\colon y_{i-1}\preceq a\prec
y_{i+1}\\}$ and we are done. ∎

######  Definition 32.

Let $P$ be a power lattice and suppose $S$ is a shellable
$\mathcal{P}$-complex of rank $r$ with the linear ordering of facets given by
$\preceq_{r}$. We define a linear ordering $\ll$ on the maximal chains of $S$
as follows. Suppose $X=(x_{0}\prec x_{1}\prec\dots\prec x_{r})$ and
$Y=(y_{0}\prec y_{1}\prec\dots\prec y_{r})$ with $X\neq Y$. Define $X\ll Y$ if
either $x_{r}\prec_{l}y_{r}$ or

$x_{r}=y_{r},\quad\text{ and }\quad(x_{0}\preceq\dots\preceq
x_{r-1})\triangleleft(y_{0}\preceq\dots\preceq y_{r-1}).$

This is again a total order. Now, we show the main theorem of this section
relating the shellability of a $\mathcal{P}$-complex to the shellability of
the associated order complex.

######  Theorem 33.

Let $P$ be a power lattice. If $S$ is a shellable $\mathcal{P}$-complex with
the linear ordering of facets given by $\preceq_{r}$, then the order complex
$\mathcal{K}(S)$ is a shellable simplicial complex whose facets are ordered
with $\ll$.

###### Proof.

Let $X=(x_{0}\prec\dots\prec x_{r})$ and $Y=(y_{0}\prec\dots\prec y_{r})$ be
two distinct maximal chains in $P$ such that $X\ll Y$. Let $e$ be the largest
index such that $x_{e}\neq y_{e}$.

Case 1. Suppose that $e<r$ and thus $x_{e+1}=x_{e+1}$, $e\leq r$. Let $f$ be
the largest index smaller than $e$ such that $U_{f}=V_{f}$. Such $f$ exists
because $U_{0}=V_{0}$. Consider the two chains

$\displaystyle\overline{X}=(y_{0}\preceq\dots\preceq y_{f}\preceq
x_{f+1}\preceq\dots\preceq x_{e})$
$\displaystyle\overline{Y}=(y_{0}\preceq\dots\preceq y_{f}\preceq
y_{f+1}\preceq\dots\preceq y_{e})$

We see that $\rho(x_{e})=\rho(y_{e})=e$ and $x_{e}\preceq x_{e+1}$,
$y_{e}\preceq x_{e+1}$, with $\rho(x_{e+1})=e+1$. Applying Theorem 31, we can
find a chain $\overline{U}=(u_{0}\preceq\dots\preceq u_{e})$ such that
$\overline{X}\cap\overline{Y}\subset\overline{U}\cap\overline{Y}$,
$|\overline{U}\cap\overline{Y}|=e$ and
$\overline{U}\triangleleft\overline{Y}$. Take $U=(u_{0}\preceq\dots\preceq
u_{e}\preceq y_{e+1}\preceq\dots\preceq y_{r})$ and we have $X\cap Y\subset
U\cap Y$, $|U\cap Y|=r$ and $U\ll Y$.

Case 2. Suppose that $e=r$, we have $x_{r}\prec_{r}y_{r}$. Then let $f$ be the
largest index such that $x_{f}=y_{f}$. Again $f$ exists because $x_{0}=y_{0}$.
Now, by the shellability of $S$, there is $z_{r}$ such that
$z_{r}\prec_{r}y_{r}$, $\rho(z_{r}\wedge y_{r})=r-1$ and $x_{r}\wedge
y_{r}\subset z_{r}\wedge r_{r}$. Let $t$ be the largest index such that
$y_{t}\prec z_{r}$. Then $t\geq f$. Moreover, since $z_{r}\neq y_{r}$ and
$\rho(z_{r})=\rho(y_{r})$, then $y_{r}\npreceq z_{r}$ i.e. $t<r$. Now, take
the chains

$\displaystyle Z=(y_{0}\prec\dots\prec y_{f}\prec\dots\prec y_{t}\prec
z_{t+1}\prec\dots\prec z_{r}),$ $\displaystyle Y=(y_{0}\prec\dots\prec
y_{f}\prec\dots\prec y_{t}\prec y_{t+1}\prec\dots\prec y_{r}).$

Since $y_{r}\neq z_{r}$, then $y_{r}\prec y_{r}\vee z_{r}$. Therefore
$r=\rho(y_{r})<\rho(y_{r}\vee
z_{r})\leq\rho(y_{r})+\rho(z_{r})-\rho(y_{r}\wedge z_{r})$ (because of the
semimodularity of $P$). Since $\rho(z_{r})=\rho(y_{r})=r$ and
$\rho(z_{r}\wedge y_{r})=r-1$, then $r<\rho(y_{r}\vee z_{r})\leq r+1$.
Therefore $z_{r}$ and $y_{r}$ precede $y_{r}+z_{r}$ which is of rank $r+1$.
Thus we can apply Theorem 31 to find a chain $U=(u_{0}\preceq\dots\preceq
u_{e})$ such that $Z\cap Y\subset U\cap Y$, $|U\cap Y|=e$ and $U\ll Y$. This
concludes the proof. ∎

The main consequence of this theorem is that the homotopy type of a shellable
$\mathcal{P}$-complex is understood.

As topological spaces, a finite poset and its associated order complex are
weakly homotopy equivalent [21]. Moreover, a shellable simplicial complex has
the same homotopy type as a wedge of spheres, i.e., the reduced simplicial
homology is zero except at maximal dimension [3, Appendix]. Therefore, by
Theorem 33, we have the following corollary that generalizes [13] and [9,
Corollary 3.10].

######  Corollary 34.

Let $P$ be a power lattice. If $S$ is a shellable $\mathcal{P}$-complex of
rank $r$ with the linear ordering of facets given by $\preceq_{r}$, then it
has the homotopy type of a wedge of spheres.

As mentioned in the introduction, another way to obtain shellable simplicial
complexes is to start with EL-shellable or CL-shellable posets. The order
complexes associated to these posets give us a shellable simplicial complex
[3, 5]. Our condition is different as instead of using the EL/CL-shellability,
we are using the $\mathcal{P}$-shellability of the poset as a
$\mathcal{P}$-complex in a power lattice. Shellable simplicial complexes are
of interests as they have nice topological and algebraic properties and the
construction via the order complex provides new classes of shellable
simplicial complexes.

##  4\. Matroids in power lattices

In the classical theory of matroids, the independent sets of a matroids form a
shellable simplicial complex. To continue with the analogy, we want to provide
constructions of shellable $\mathcal{P}$-complexes via a generalization of
matroids. Then we show how to get shellable $\mathcal{P}$-complexes from the
matroids.

######  Definition 35.

Let $(P,\preceq)$ be a power lattice with rank function $\rho$. A matroid
$(P,\mathcal{I})$ on $P$ is defined by a subset $\mathcal{I}$ of $P$ such that

  1. (I1)

$\hat{0}\in\mathcal{I}$,

  2. (I2)

if $x\preceq y$ and $y\in\mathcal{I}$, then $x\in\mathcal{I}$,

  3. (I3)

For all $x,y\in\mathcal{I}$ such that $\rho(x)<\rho(y)$, there exists an atom
$a\in P$ such that $v_{a}(x)<v_{a}(y)$, $x\vee a^{v_{a}(x)+1}\in\mathcal{I}$.

######  Remark 36.

Notice that, this definition is more general to the definition of $q$-matroids
in the lattice of subspaces [19]. However, the axioms in the definition were
enough to get shellability of both for matroid and $q$-matroid [12, 4].

######  Example 37.

  1. (1)

Take $(P,P)$. This is a trivial matroid.

  2. (2)

More generally we can define the uniform matroids $U_{k}(P)$. Let $P$ be a
power lattice with $\rho(\hat{1})=n$. Suppose $k\leq n$ and let
$\mathcal{I}=U_{k}(P):=\\{x\in P\colon\rho(x)\leq k\\}$. Then (I1) and (I2)
are clear. To check (I3). Suppose that $\rho(x)<\rho(y)\leq k$ and hence
$\rho(x)\leq k-1$. There exists an atom $a\in P$ such that
$v_{a}(x)<v_{a}(y)$. Now take $x\vee a^{v_{a}(x)+1}$. By semimodularity, we
have

$\displaystyle\rho(x\vee a^{v_{a}(x)+1})$
$\displaystyle\leq\rho(x)+\rho(a^{v_{a}(x)+1})-\rho(x\wedge a^{v_{a}(x)+1})$
$\displaystyle\leq\rho(x)+\rho(a^{v_{a}(x)+1})-\rho(a^{v_{a}(x)})$
$\displaystyle\leq\rho(x)+1$ $\displaystyle\leq k$

Hence $x\vee a^{v_{a}(x)+1}\in\mathcal{I}$.

######  Example 38.

  1. (i)

The classical matroids on the lattice of subsets [22]. They consist of a set
$I$ of a subsets of subsets of a finite set. $I$ is closed under inclusion and
for two subsets $A,B\in I$ such that $|A|<|B|$, there exists $x\in B\setminus
A$ such that $A\cup\\{x\\}\in I$.

  2. (ii)

The relatively recent $q$-matroids on the lattice of subspaces [19]. They
consist of a set $I$ of subspaces of a vector space $\mathbb{F}_{q}^{n}$. $I$
is closed under inclusion and for two subspaces $A,B\in I$ such that $\dim
A<\dim B$, there exists $x\in B\setminus A$ such that $A\cup\\{x\\}\in I$.

  3. (iii)

The discrete polymatroids on the lattice of multiset subsets [16]. They
consist of a set $I$ of multiset subsets of a finite multiset. $I$ is closed
under inclusion and for two subsets $A,B\in I$ such that $|A|<|B|$, there
exists $x\in B\setminus A$ (the difference takes into account the
multiplicity) such that $A\cup\\{x\\}\in I$, the last union means that the
multiplicity of $x$ in $A$ is increased by $1$.

  4. (iv)

The sum-matroids on the Cartesian product of lattice of subspaces. Although in
[23], the sum matroids were defined by rank functions, one can show that by
taking the independent elements, i.e. the elements such that whose rank is the
same as the total dimension, we get a matroid over the Cartesian product of
lattice of subspaces.

######  Definition 39.

Let $(P,\preceq)$ be a power lattice with rank function $\rho$ and let
$(P,\mathcal{I})$ be a matroid in $P$, the elements of $\mathcal{I}$ are
called the independent elements of $(P,\mathcal{I})$. A maximum element in
$\mathcal{I}$ with respect to $\preceq$ is called a basis of
$(P,\mathcal{I})$. Let $\mathcal{B}$ denote the set of all bases of
$(P,\mathcal{I})$.

######  Lemma 40.

Let $(P,\preceq)$ be a power lattice with rank function $\rho$ and let
$(P,\mathcal{I})$ be a matroid in $P$. Let $\mathcal{B}$ be the set of bases
of the matroid. All elements of $\mathcal{B}$ have the same rank.

###### Proof.

Let $x,y\in\mathcal{B}$. Suppose $\rho(x)<\rho(y)$. By (I3), there exists an
atom $a\in P$ such that $v_{a}(x)<v_{a}(y)$, $x\vee
a^{v_{a}(x)+1}\in\mathcal{I}$. It is clear that $x\preceq x\vee
a^{v_{a}(x)+1}$ and by Lemma 12, $v_{a}(x)<v_{a}(x\vee a^{v_{a}(x)+1})$ and
this implies $x\neq x\vee a^{v_{a}(x)+1}$. This contradicts the maximality of
$x$ in $\mathcal{I}$. Therefore $\rho(x)=\rho(y)$. ∎

Similar to classical matroids and $q$-matroids, the bases of a matroid
satisfies the following nice properties.

######  Theorem 41.

Let $(P,\preceq)$ be a power lattice with rank function $\rho$ and let
$(P,\mathcal{I})$ be a matroid in $P$. Let $\mathcal{B}$ be the set of bases
of the matroid. Then

  1. (B1)

$\mathcal{B}\neq\emptyset$,

  2. (B2)

For all $x,y\in\mathcal{B}$, if $x\preceq y$, then $x=y$,

  3. (B3)

For all $x,y\in\mathcal{B}$, and for each $u\preceq x$ with
$\rho(u)=\rho(x)-1$ and $x\wedge y\preceq u$, there exists an atom $a$ such
that $v_{a}(u)<v_{a}(y)$ and $u\vee a^{v_{a}(u)+1}\in\mathcal{B}$.

###### Proof.

  1. (B1)

$\mathcal{I}$ is non-empty and it is finite so there must be a least one
maximal element.

  2. (B2)

Clear.

  3. (B3)

Let $x,y\in\mathcal{B}$, and let $u\preceq x$ with $\rho(u)=\rho(x)-1$ and
$x\wedge y\preceq u$. By (I2) $u\in\mathcal{I}$. Since $\rho(u)<\rho(x)$, by
Lemma 40, $\rho(u)<\rho(y)$. Therefore, by (I3), there exists an atom $a\in P$
such that $v_{a}(u)<v_{a}(y)$, $u\vee a^{v_{a}(u)+1}\in\mathcal{I}$. Now
$u\prec u\vee a^{v_{a}(u)+1}$ and $\rho(u)=\rho(x)-1$. Therefore $u\vee
a^{v_{a}(u)+1}$ is a basis.

∎

######  Remark 42.

Again, the axioms are more general than in the case of $q$-matroids.

Now we show the dual basis exchange property for matroids.

######  Theorem 43 (Dual basis exchange property).

Let $(P,\preceq)$ be a power lattice with rank function $\rho$ and let
$(P,\mathcal{I})$ be a matroid in $P$. Let $\mathcal{B}$ be the set of bases
of the matroid. Let $x,y\in\mathcal{B}$ such that $x\neq y$ and let $a$ be an
atom with $v_{a}(y)>v_{a}(x)$. Then there exist $u\in P$ with
$\rho(u)=\rho(x)-1$ and an atom $b$ such that $v_{b}(y)<v_{b}(x)$, $x\wedge
y\preceq u$, $x=u\vee b^{v_{b}(u)+1}$ and $u\vee a^{v_{a}(u)+1}$ is a basis.

###### Proof.

Let $r=\rho(x)$ and let $s=r-\rho(x\wedge y)$. If $s=1$, then $\rho(x\wedge
y)=r-1$. Take $u=x\wedge y$. Then we observe that

  1. (i)

$x\wedge y\preceq u$ is true.

  2. (ii)

Since $x\neq y$ and $\rho(x)=\rho(y)$, there exists $b$ such that
$v_{b}(y)<v_{b}(x)$. Moreover, we have $x\wedge y\preceq u\prec u\vee
b^{v_{b}(u)+1}$. Since $v_{b}(u)\leq v_{b}(y)$, $v_{b}(u)<v_{b}(x)$. Therefore
$b^{v_{b}(u)+1}\preceq x$. Hence $x\wedge y\preceq u\prec u\vee
b^{v_{b}(u)+1}\preceq x$. Since $\rho(x\wedge y)=r-1$ and $\rho(x)=r$, we have
$u\vee b^{v_{b}(u)+1}=x$.

  3. (iii)

Similarly, we can check that $u\vee a^{v_{b}(a)+1}=y$.

We proved that the statement holds for $s=1$. We now show the statement by
induction on $s$. Assume that the statement holds for $r-\rho(x\wedge y)<s$,
where $s>1$. We want to show that the statement holds for $r-\rho(x\wedge
y)=s$.

Suppose that $r-\rho(x\wedge y)=s\geq 2$. Therefore $x\wedge y\prec w\prec y$
for some $w$ such that $\rho(w)=r-1$ and $v_{a}(w)>v_{a}(x\wedge y)$. By (B3),
there is an atom $z$ such that $v_{z}(w)<v_{z}(y)$ and $t=w\vee
z^{v_{z}(w)+1}$ is a basis. Now, by Lemma 12, $v_{z}(x\wedge
y)=\min\\{v_{z}(x),v_{z}(y)\\}$. Hence $\min\\{v_{z}(x),v_{z}(y)\\}\leq
v_{z}(w)<v_{z}(y)$ and therefore $v_{z}(x\wedge y)=v_{z}(x)$. This gives us
$v_{z}(x)\leq v_{z}(w)<v_{z}(t)$. So $x\neq t$ and $x\wedge y\prec x\wedge t$.
Moreover, $v_{a}(y)\geq v_{a}(w)>v_{a}(x\wedge y)$. By Lemma 12,
$v_{a}(x\wedge y)=v_{a}(x)$. In addition $v_{a}(t)\geq v_{a}(w)>v_{a}(x\wedge
y)=v_{a}(x)$ so that $v_{a}(t)>v_{a}(x)$. By the induction hypothesis, there
exist $u$ with $\rho(u)=\rho(x)-1$ and an atom $b$ such that
$v_{b}(t)<v_{b}(x)$, $x\wedge t\preceq u$, $x=u\vee b^{v_{b}(u)+1}$ and $u\vee
a^{v_{a}(u)+1}$ is a basis. We easily check that $x\wedge y\preceq x\wedge
t\preceq u$. Finally, $x\wedge y\preceq x\wedge t$ implies that $v_{b}(x\wedge
y)\leq v_{b}(x\wedge t)$. Since $v_{b}(t)<v_{b}(x)$, then $v_{b}(x\wedge
y)\leq v_{b}(t)$. If $v_{b}(x)\leq v_{b}(y)$, then
$v_{b}(x)=\min\\{v_{b}(x),v_{b}(y)\\}=v_{b}(x\wedge y)\leq v_{b}(t)$ which is
a contradiction. Therefore $v_{b}(y)<v_{b}(x)$.

∎

######  Theorem 44.

Let $(P,\preceq)$ be a power lattice with rank function $\rho$ and let
$(P,\mathcal{I})$ be a matroid in $P$. Let $\mathcal{B}$ be the set of bases
of the matroid whose elements have rank $r$. Then $\mathcal{I}$ is shellable
where the elements of $\mathcal{B}$ are ordered with $\preceq_{r}$.

###### Proof.

Let $x$ and $y$ be two bases such that $x\prec_{r}y$. Let
$F(x)=(x_{1},\dots,x_{t})$ and $F(y)=(y_{1},\dots,y_{t})$ be the factorization
of $x$ and $y$ respectively. We also assume that these factorizations are
already written in such a way that $x_{i}\preceq_{1}x_{i+1}$ and
$y_{i}\preceq_{1}y_{i+1}$. Therefore

$\min_{\preceq_{1}}F(x)\setminus F(y)\prec_{1}\min_{\preceq_{1}}F(x)\setminus
F(y).$

Let $i_{0}$ be the smallest integer such that $x_{i_{0}}\neq y_{i_{0}}$. In
this case $x_{i_{0}}\prec_{1}y_{i_{0}}$ and
$v_{x_{i_{0}}}(x)>v_{x_{i_{0}}}(y)$. By Theorem 43, There exists $u$ with
$\rho(u)=\rho(x)-1$ and $y_{j}$ such that $v_{y_{j}}(x)<v_{y_{j}}(y)$,
$x\wedge y\preceq u$ and $y=u\vee y_{j}^{v_{y_{j}}(u)+1}$ and $z=u\vee
x_{i_{0}}^{v_{x_{i_{0}}}(u)+1}$ is a basis. We have $x\wedge y\preceq u\preceq
z\wedge y$. The next step is to show that $z\prec_{r}y$.

For clarity, write

$F(x)=\prod_{i=1}^{s}a_{i}^{v_{a_{i}}(x)},\quad a_{i}\prec a_{i+1}$

and

$F(y)=\prod_{i=1}^{s}b_{i}^{v_{b_{i}}(y)},\quad b_{i}\prec b_{i+1}$

and $i_{1}$ is the smallest integer such that $v_{a_{i}}>v_{b_{i}}$. Notice
that $x_{i_{0}}=a_{i_{1}}$. For $z$, we have
$(\bigvee_{i=1}^{i_{1}-1}b_{i}^{v_{b_{i}}(y)})\vee
b_{i_{1}}^{v_{b_{i_{1}}}+1}\prec z$. Moreover $z$ may contain a new atom $a$
which is not equal to $b_{i}$, $i=1,\dots,i_{1}-1$. But in any case, whether
$a\prec_{1}b_{i_{1}}$ or not, we always have $z\prec_{r}y$. This also implies
that $x\wedge y\prec z$ so that $\rho(x\wedge y)\leq\rho(u)\leq\rho(z\wedge
y)<\rho(z)$. Hence $\rho(z\wedge y)=r-1$. ∎

###  4.1. Weighted graphic matroids in the lattice of multiset subsets

In this subsection, we consider the particular power lattice of multiset
subsets of a finite multiset. We generalize the classical graphic to matroids
on the lattice of multiset subsets from weighted graphs.

Let $S$ be a finite multiset i.e., some elements can be repeated. For
simplicity, we write $S=\prod_{i=1}^{l}x_{i}^{n_{i}}$ where the elements of
$S$ are the $x_{i}$’s with the corresponding multiplicities denoted by the
$n_{i}$’s. These elements of $S$ can be thought as variables whereas $S$ can
be thought as a monomial. The lattice $P$ of subset of $S$ is the set of all
multiset subsets of $S$ (which can be thought as monomials) i.e.

$P=\left\\{\prod_{i=1}^{l}x_{i}^{a_{i}}\colon 0\leq a_{i}\leq n_{i}\right\\}.$

The atoms of $P$ are the variables $x_{i}$. The maximal element is $\hat{1}=S$
and the minimal element is $\hat{0}=1$. The valuation of a monomial $\prod
x_{j}^{a_{j}}$ at a variable $x_{i}$ is $v_{x_{i}}(\prod
x_{j}^{a_{j}})=a_{i}$. The meet and join between two monomials are given by

$(\prod x_{i}^{a_{i}})\cap(\prod x_{i}^{b_{i}})=\prod
x_{i}^{\min\\{a_{i},b_{i}\\}}\text{ and }(\prod x_{i}^{a_{i}})\cup(\prod
x_{i}^{b_{i}})=\prod x_{i}^{\max\\{a_{i},b_{i}\\}}.$

In fact the meet is just the greatest common divisors of two monomials whereas
the join is the least common multiple. The rank function $\rho$ is the same as
the sum of the valuation at every atoms. It is not difficult to show that we
have a power lattice with the partial ordering defined by multiset inclusion,
which we denote by $\subseteq$. In the remaining part of this paper, whenever
we talk about a lattice of multiset subsets, we always use the above
notations.

When $P$ is the lattice of multiset subsets of a finite multiset, the
$\mathcal{P}$-complexes in $P$ are known as multicomplexes [27].

######  Definition 45.

Let $P$ be the lattice of multiset subsets of
$S=\prod_{i=1}^{l}x_{i}^{n_{i}}$. A multicomplex $\Delta$ in $S$ is a subset
of $P$ such that

  1. (i)

$\Delta\neq\emptyset$,

  2. (ii)

if $x\in\Delta$ and $y|x$, then $y\in\Delta$.

Here we recall the definition of shellable multicomplexes from [9] that is
also obtained as the particular case of general shellable
$\mathcal{P}$-complexes as we defined in Definition 23.

######  Definition 46.

[9, Definition 3.4] A pure shellable multicomplex is a multicomplex $S$ such
that its facets have the same cardinality $r$ when counted with multiplicity
and there exists an ordering $F_{1},F_{2},\cdots,F_{t}$ of all the facets of
$S$ such that for all $i<j\leq t$, there exists $k<j$ such that $F_{i}\cap
F_{j}\subseteq F_{k}\cap F_{j}$ and $|F_{k}\cap F_{j}|=r-1$, where the
cardinality is counted with multiplicity.

As we have seen in a previous section, one way to get a shellable multicomplex
is to build it from a matroid. For clarity, let us recall the definition of
matroids in the lattice of multiset subsets.

######  Definition 47.

Let $P$ be a lattice of multiset subsets of $S=\prod_{i=1}^{l}x_{i}^{n_{i}}$
with rank function $\rho$. A matroid $(P,\mathcal{I})$ on $P$ is defined by a
subset $\mathcal{I}$ of $P$ such that

  1. (I1)

$\hat{0}\in\mathcal{I}$,

  2. (I2)

if $x\subseteq y$ and $y\in\mathcal{I}$, then $x\in\mathcal{I}$,

  3. (I3)

For all $x,y\in\mathcal{I}$ such that $\rho(x)<\rho(y)$, there exists $x_{i}$
such that $v_{x_{i}}(x)<v_{x_{i}}(y)$, $x\cup
x_{i}^{v_{x_{i}}(x)+1}\in\mathcal{I}$.

Matroids on multisets are already known as discrete polymatroids. One can have
a look at the paper [16] and the properties (D1) and (D2) in the Introduction.
Various constructions of matroids on multisets therefore already exist (see
also [6]). In the following, we present a construction similar to graphical
matroids, using weighted graphs.

######  Definition 48 (Weighted graphs).

A weighted graph $G=(V,E)$ is a set of vertices $V$ together with a subset
$E\subseteq V\times V$ such that to each $e\in E$ is attached a positive
integer $wt_{G}(e)$, called the weight of $e$.

######  Definition 49.

Let $G=(V,E)$ be a weighted graph. A subgraph of $G$ is a weighted graph
$G_{1}=(V_{1},E_{1})$ such that $V_{1}\subseteq V$, $E_{1}\subseteq E$ and for
$e\in E_{1}$, $0<wt_{G_{1}}(e)\leq wt_{G}(e)$. For a set of edges $E_{1}$ we
denote by $V(E_{1})$ the set of vertices which are either an origin or a tail
of an edge in $E_{1}$.

######  Definition 50 (Cycles).

Given a weighted graph $(V,E)$, a cycle of $(V,E)$ is a subgraph
$G_{1}=(V(E_{1}),E_{1})$ where $E_{1}=\\{e_{0},\dots,e_{n}\\}\subset E$ such
that $wt_{G_{1}}(e_{i})=wt_{G}(e_{i})$ and the tail of $e_{i-1\mod n}$ is the
origin of $e_{i\mod n}$ for any $i$.

In other words, a cycle is similar to the classical notion of cycle on simple
graph except that the weight of every edge in the cycle must be maximal, i.e.
the same as its weight in the original graph.

######  Definition 51 (Lattice of multiset subsets associated to a weighted
graph).

Let $G=(V,E)$ be a weighted graph. Let $S=\prod_{e\in E}x_{e}^{wt_{G}(e)}$ be
a multiset on the edges in $E$. Define $P$ be the lattice of multiset subsets
of $S$. For a multiset subset $H$ in $P$, the subgraph induced by $H$ is the
weighted graph $(V(H),E(H))$, where $E(H)$ is the set of elements of $H$, the
weight of an edge is its multiplicity in $H$ and $V(H):=V(E(H))$.

######  Definition 52.

Let $G=(V,E)$ be a weighted graph. The independent complex $\mathcal{I}_{G}$
associated to $G$ is a set of multiset subsets $H$ of edges where
$H\in\mathcal{I}_{G}$ if $(V(H),E(H))$ doesn’t contain a subgraph which is a
cycle of $G$. In addition, the empty set is added to $\mathcal{I}_{G}$.

Let $P(G)$ be the power lattice of multiset subsets of the multiset $S$
associated to a graph $G=(V,E)$. It is clear that the independent complex
$\mathcal{I}_{G}$ is a $P(G)$-complex i.e. it is not empty and it is closed
under inclusion which we denote by $\preceq$. Now we show that this form a
matroid $(P(G),\mathcal{I}_{G})$ in the lattice of multiset subsets $P(G)$.

######  Theorem 53.

Let $G=(V,E)$ be a weighted graph. Let $\mathcal{I}_{G}$ be the independent
complex associated to $G$ in the power lattice $P(G)$ associated to $G$. Then,
$(P(G),\mathcal{I}_{G})$ is a matroid and we call it the graphic matroid
associated to a weighted graph $G=(V,E)$.

###### Proof.

What remains to show is the property (I3) which we recall:

For all $x,y\in\mathcal{I}_{G}$ such that $\rho(x)<\rho(y)$, there exists an
atom $a\in P$ such that $v_{a}(x)<v_{a}(y)$, $x\cup
a^{v_{a}(x)+1}\in\mathcal{I}_{G}$.

For the proof, let $x,y\in\mathcal{I}_{G}$ such that $\rho(x)<\rho(y)$, then
there exist an edge $e$ such that $v_{e}(x)<v_{e}(y)$. Let $A=\\{e\in E\colon
v_{e}(x)<v_{e}(y)\\}$. Therefore $A$ is non-empty. Let $A^{C}=E\setminus A$
and therefore, for $e\in A^{C}$, $v_{e}(x)\geq v_{e}(y)$. We distinguish two
cases:

Case 1. Suppose there exists $e\in A$, such that $v_{e}(x)<w_{G}(e)-1$. Then
$z=x\vee e^{v_{e}(x)+1}\in\mathcal{I}$, since the only change done to $x$ to
obtain $z$ is to increase the weight of one edge but the weight is still small
and can’t be part of a cycle.

Case 2. Suppose that for all $e\in A$, we have $wt_{G}(e)-1\leq v_{e}(x)$.
Then $wt_{G}(e)-1\leq v_{e}(x)<v_{e}(y)\leq wt_{G}(e)$. Hence, for all $e\in
A$, $wt_{G}(e)=v_{e}(y)=v_{e}(x)+1$. Since $\rho(x)<\rho(y)$, $\sum_{e\in
E}v_{e}(x)<\sum_{e\in E}v_{e}(y)$ and therefore

$\sum_{e\in A^{C}}v_{e}(x)<\sum_{e\in A^{C}}v_{e}(y)+|A|.$ (15)

Let

$B=\\{e\in A^{C}\colon v_{e}(x)=v_{e}(y)=wt_{G}(e)\\}$

and

$C=\\{e\in A^{C}\colon v_{e}(x)>v_{e}(y)\text{ or }v_{e}(x)<wt_{G}(e)\\}.$

Hence $B\cup C=A^{C}$. Equation (15) implies

$\sum_{e\in B}v_{e}(x)+\sum_{e\in C}v_{e}(x)<\sum_{e\in B}v_{e}(y)+\sum_{e\in
C}v_{e}(y)+|A|.$

By definition of $B$, the previous inequality implies

$\sum_{e\in C}v_{e}(x)-\sum_{e\in C}v_{e}(y)<|A|.$

Let $D=\\{e\in A^{C}\colon v_{e}(x)>v_{e}(y)\\}$. We have $D\subseteq C$. Then

$\sum_{e\in D}v_{e}(x)-\sum_{e\in D}v_{e}(y)\leq\sum_{e\in
C}v_{e}(x)-\sum_{e\in C}v_{e}(y)<|A|.$

Hence

$|D|<|A|\quad\text{i.e. }|D|+|B|<|A|+|B|$ (16)

Notice that $B\cap D=B\cap A=\emptyset$. Consider the simple graph
$G_{x}=(V,E_{x})$ where $e\in E_{x}$ if $v_{e}(x)=wt_{G}(e)$. Similarly,
define $G_{y}=(V,E_{y})$ where $e\in E_{y}$ if $v_{e}(y)=wt_{G}(e)$. One can
check that $E_{x}\subseteq B\cup D$ and $A\cup B\subseteq E_{y}$. These
together with Equation (16) implies $|E_{x}|<|E_{y}|$. The number of connected
components in $G_{x}$ is $N_{x}=|V|-|E_{x}|$ and the number of connected
components in $G_{y}$ is $N_{y}=|V|-|E_{y}|$ and we have $N_{y}<N_{x}$. By the
pigeonhole principle, there is a component $z$ of $G_{y}$ which contains
vertices from two or more components of $G_{x}$. Take a path $z$ in $G_{y}$
which connects two components of $G_{x}$. This path contains an edge $a$ (i.e.
$a\preceq z$ which connect two components of $G_{x}$. Therefore adding this
edge to $G_{x}$ cannot produce a cycle when adding $a$ to the graph $G_{x}$.
More importantly, $x\vee a^{v_{a}(x)+1}$ does not contain a cycle and hence it
is independent.

∎

##  5\. Stanley-Reisner rings associated to multicomplexes

The face rings or Stanley-Reisner rings associated to shellable simplicial
complexes are Cohen-Macaulay rings [27]. Thus for a shellable
$\mathcal{P}$-complex in a power lattice, the Stanley-Reisner ring associated
with its order complex is Cohen-Macaulay. In this section, we consider a
quotient of polynomial ring associated to a multicomplex and show that the
ring is sequentially Cohen-Macaulay if the multicomplex is
$\mathcal{P}$-shellable. We also explain why the ring we consider
corresponding to a multicomplex is a generalization of Stanley-Reisner ring of
a simplicial complex. To do this, we adapt the construction of sheaf of rings
by Yuzvinsky in [28] for a multicomplex.

Let $P$ be the lattice of multiset subsets of
$\hat{1}=\prod_{i=1}^{l}x_{i}^{n_{i}}$ and suppose that $\Delta$ is a
multicomplex in $P$. In this section, we also assume that
$\hat{1}\notin\Delta$. To this multiset, we associate the ring
$R=\mathbb{F}[x_{1},\dots,x_{l}]$, where $\mathbb{F}$ is any arbitrary field.

Let $X=X(\Delta)$ be the set of all meets of facets of $\Delta$. Consider $X$
as a poset where for $A,B\in X$, $A\preceq B$ if $B\subseteq A$. The minimal
elements of the poset $X$ are therefore the elements of $B$.

Now, we construct a sheaf $\mathfrak{A}$ of rings on $X$. For each $\sigma\in
X$, let the stalk

$(\mathfrak{A})_{\sigma}=\mathbb{F}[x_{1},\dots,x_{l}]/(x_{1}^{v_{x_{1}}(\sigma)+1},\dots,x_{l}^{v_{x_{l}}(\sigma)+1})$

be the quotient ring modulo the ideal
$(x_{1}^{v_{x_{1}}(\sigma)+1},\dots,x_{l}^{v_{x_{l}}(\sigma)+1})$.

Let $\prod_{\sigma\in X}(\mathfrak{A})_{\sigma}$ be the Cartesian product of
the stalks associated to the elements of $X$. For $\sigma\preceq\tau$ in $X$
(i.e., $\tau\subseteq\sigma$), we define the ring homomorphism
$\rho_{\tau\sigma}:(\mathfrak{A})_{\sigma}\rightarrow(\mathfrak{A})_{\tau}$ as
the natural projection.

One can verify that these define a sheaf of rings $\mathfrak{A}$.

######  Definition 54.

Let $\Delta$ be a multicomplex and let $X=X(\Delta)$ be the set of all meets
of facets of $\Delta$. The ring of sections $\Gamma(\mathfrak{A})$ on $X$ is
defined by

$\Gamma(\mathfrak{A})=\\{(f_{\sigma})_{\sigma\in X}\in\prod_{\sigma\in
X}(\mathfrak{A})_{\sigma}\colon\rho_{\tau\sigma}(f_{\sigma})=f_{\tau},\sigma\preceq\tau,\sigma,\tau\in
X\\}.$ (17)

The next theorem shows that the ring of sections on $X$ is, in fact,
isomorphic to a quotient of $\mathbb{F}[x_{1},\dots,x_{l}]$.

######  Theorem 55.

Let $P$ be the lattice of multiset subsets of
$\hat{1}=\prod_{i=1}^{l}x_{i}^{n_{i}}$ and let $\Delta$ be a multicomplex in
$P$. Suppose $X=X(\Delta)$ is the set of all meets of facets of $\Delta$ and
$I_{\Delta}$ is the ideal of $\mathbb{F}[x_{1},\dots,x_{l}]$ generated by the
monomials corresponding to the elements in $P\setminus\Delta$. Then the ring
of sections on $X$ is isomorphic to the quotient of the polynomial ring by
$I_{\Delta}$, i.e.,

$\Gamma(\mathfrak{A})\simeq\mathbb{F}[x_{1},\dots,x_{l}]/I_{\Delta}.$

###### Proof.

Consider the homomorphism

$\displaystyle\phi:\mathbb{F}[x_{1},\dots,x_{l}]$
$\displaystyle\longrightarrow\Gamma(\mathfrak{A})$ $\displaystyle f$
$\displaystyle\longmapsto(f+(x_{1}^{v_{x_{1}}(\sigma)+1},\dots,x_{l}^{v_{x_{l}}(\sigma)+1}))_{\sigma\in
X}.$

The above map is a well-defined ring homomorphism. We first show that
$\ker\phi=I_{\Delta}$.

Let $f\in I_{\Delta}$ be a generator monomial. If $\phi(f)\neq 0$, then there
exists $\sigma\in X$ such that
$f\notin(x_{1}^{v_{x_{1}}(\sigma)+1},\ldots,x_{l}^{v_{x_{l}}(\sigma)+1})$.
This implies that $v_{x_{i}}(f)\leq v_{x_{i}}(\sigma)$ for all
$i\in\\{1,\ldots,l\\}$, which contradicts that $f$ is a monomial corresponding
to a nonface, i.e., elements in $P\backslash\Delta$. Thus $f\in\ker\phi$ and
therefore, $I_{\Delta}\subseteq\ker\phi$.

Conversely, let $f\in\ker\phi$, i.e.,
$f\in(x_{1}^{v_{x_{1}}(\sigma)+1},\dots,x_{l}^{v_{x_{l}}(\sigma)+1})$ for all
$\sigma\in X$. For for any $\sigma\in X$, we associate the ideal
$P_{\sigma}=(x_{1}^{v_{x_{1}}(\sigma)+1},\dots,x_{l}^{v_{x_{l}}(\sigma)+1})$.
Thus $f\in\ker\phi$ implies that $f\in\bigcap\limits_{\sigma\in X}P_{\sigma}$.
We claim that $I_{\Delta}=\bigcap\limits_{\sigma\in X}P_{\sigma}$. That would
imply that $f\in I_{\Delta}$ and therefore $\ker\phi\subseteq I_{\Delta}$.
Thus $\ker\phi=I_{\Delta}$. Finally, we note that the map $\phi$ is a
surjective map follows from Equation (17) since $\emptyset\in\Delta$ and
$(\mathfrak{A})_{\emptyset}=\mathbb{F}[x_{1},\ldots,x_{l}]$. This would
complete the proof of the Theorem.

Proof of the claim: It is well known that if $J$ is a monomial ideal of
$\mathbb{F}[x_{1},\ldots,x_{l}]$, then $(ab,J)=(a,J)\bigcap(b,J)$ where $a,b$
are relatively prime monomials. Thus the monomial ideal $I_{\Delta}$ can be
written as intersection of ideals of the form
$(x_{1}^{b_{1}},\ldots,x_{l}^{b_{l}})$ i.e.,

$I_{\Delta}=\bigcap_{b_{1},\dots,b_{l}}(x_{1}^{b_{1}},\ldots,x_{l}^{b_{l}}).$

Note that if $I_{\Delta}\subseteq(x_{1}^{b_{1}},\ldots,x_{l}^{b_{l}})$, then
the multiset $\prod_{i=1}^{l}x_{i}^{b_{i}-1}$ is a face of $\Delta$. Indeed,
otherwise, if it is a nonface, the corresponding monomial
$\prod_{i=1}^{l}x_{i}^{b_{i}-1}\in
I_{\Delta}\subseteq(x_{1}^{b_{1}}\ldots,x_{l}^{b_{l}})$, which is not
possible, considering the degrees. Thus the ideal $I_{\Delta}$ is of the
intersection of ideals of the form $P_{\sigma}$ for _some_ faces
$\sigma\in\Delta$ i.e.,

$I_{\Delta}=\bigcap_{\sigma\in\Gamma}P_{\sigma},$ (18)

where $\Gamma\subseteq\Delta$. On the other hand, we claim that for any facet
$\sigma$ of $\Delta$, $I_{\Delta}\subseteq P_{\sigma}$. Let
$f=\prod_{i=1}^{l}x_{i}^{b_{i}}$ be any generator of $I_{\Delta}$
corresponding to a nonface. That means for any facet $\sigma$,
$b_{i}>v_{x_{i}}(\sigma)$ for some $i$. This implies
$\prod_{i=1}^{l}x_{i}^{b_{i}}\in P_{\sigma}$ which proves the claim. Thus we
have

$I_{\Delta}\subseteq\bigcap\limits_{\sigma\text{ facet of }\Delta}P_{\sigma}.$
(19)

Moreover, if $\tau\subset\sigma$, then $P_{\sigma}\subset P_{\tau}$. Since any
facet $\sigma$ of $\Delta$ is an element of $X$, we have the following
equalities

$\bigcap_{\sigma\text{ facet of }\Delta}P_{\sigma}=\bigcap_{\sigma\in
X}P_{\sigma}=\bigcap_{\sigma\in\Delta}P_{\sigma}$ (20)

Now, since
$\bigcap\limits_{\sigma\in\Delta}P_{\sigma}\subseteq\bigcap\limits_{\sigma\in\Gamma}P_{\sigma}$,
combining equations (18), (19) and (20), we get
$I_{\Delta}\subseteq\bigcap\limits_{\sigma\in
X}P_{\sigma}=\bigcap\limits_{\sigma\in\Delta}P_{\sigma}\subseteq\bigcap\limits_{\sigma\in\Gamma}P_{\sigma}=I_{\Delta}$
and that implies $I_{\Delta}=\bigcap\limits_{\sigma\in X}P_{\sigma}$. This
completes the proof of the claim.

∎

In the last theorem, if we consider $\Delta$ to be a simplicial complex, then
the ring of sections becomes isomorphic to the Stanley-Reisner ring associated
with $\Delta$. This leads to the next definition.

######  Definition 56.

Let $P$ be the lattice of multiset subsets of
$\hat{1}=\prod\limits_{i=1}^{l}x_{i}^{n_{i}}$ and let $\Delta$ be a
multicomplex in $P$. The Stanley-Reisner ideal associated to $\Delta$ is the
ideal $I_{\Delta}$ generated by the monomials corresponding to the elements in
$P\setminus\Delta$. The quotient ring
$\mathbb{F}[x_{1},\dots,x_{n}]/I_{\Delta}$ is called the Stanley-Reisner ring
associated with $\Delta$.

######  Remark 57.

Notice that our ideal $I_{\Delta}$ is the monomial ideal generated by the
monomials in $P\setminus\Delta$. On the other hand, the ideal considered in
[17, 18, 26] is the one generated by _any_ monomials which are not in
$\Delta$, regardless of whether they are in $P$ or not. The difference is
demonstrated by an example in [9, Remark 1.5].

Our goal is to show that the shellability of $\Delta$ implies that
$\mathbb{F}[x_{1},\dots,x_{l}]/I_{\Delta}$ is sequentially Cohen-Macaulay. To
do this, we recall some notions in commutative algebra. For detail, one can
refer to, e.g., the ‘green book’ by Stanley [27].

######  Definition 58.

Let $R=\mathbb{F}[x_{1},\cdots,x_{l}]$ be a polynomial ring. Suppose
$M_{k}=x_{1}^{a_{k,1}}\dots x_{l}^{a_{k,l}}$ is a monomial in $R$. We define
the polarization of $M_{k}$ to be the square-free monomial

$pol(M_{k})=x_{1,1}x_{1,2}\dots x_{1,a_{k,1}}x_{2,1}x_{2,2}\dots
x_{2,a_{k,2}}\dots x_{l,1}x_{l,2}\dots x_{l,a_{k,l}}$

in the polynomial ring $S=\mathbb{F}[x_{i,j}\colon 1\leq i\leq l,1\leq j\leq
a_{k,i}]$.

If $I=(M_{1},\dots,M_{t})$ is an ideal of $R$, then the polarization of $I$ is
the ideal,

$pol(I)=(pol(M_{1}),\dots,pol(M_{t})),$

in the polynomial ring $\mathbb{F}[x_{i,j}:1\leq i\leq l,1\leq
j\leq\max_{k}a_{k,i}]$.

######  Definition 59.

Let $R$ be a commutative Noetherian local ring and let $M$ be an $R$-module.
An element $a\in M$ is said to be $M$-regular if $ax\neq 0$ for all $0\neq
x\in M$. A sequence $(a_{1},\dots,a_{r})$ of elements of $M$ is an
$M$-sequence if

  1. (a)

$a_{1}$ is $M$-regular and $a_{i+1}$ is $M/a_{i}M$-regular for every $i\geq
1$, and

  2. (b)

$M/(\sum_{i}a_{i}M)\neq 0$.

######  Definition 60.

Let $R$ be a commutative Noetherian local ring and let $M$ be a finitely
generated $R$-module. Every maximal $M$-sequences have the same length
$0pt(M)$, which is called the depth of $M$.

######  Definition 61.

Let $R$ be a commutative Noetherian local ring and let $M$ be a finitely
generated $R$-module. The Krull dimension $\dim M$ of the module $M$ is the
Krull dimension of the ring $R/(Ann_{R}(M))$, where

$Ann_{R}(M)=\\{x\in R\colon xy=0\text{ for all }y\in M\\},$

the annihilator of $M$.

######  Definition 62.

Let $R$ be a commutative Noetherian local ring and let $M$ be a finitely
generated $R$-module. We say that $M$ is a Cohen-Macaulay module if
$0pt(M)=\dim M$.

######  Definition 63.

Let $R$ be a commutative Noetherian local ring and let $M$ be a finitely
generated (graded) $R$-module. A finite filtration $0=M_{0}\subset
M_{1}\subset\dots\subset M_{l}=M$ of $M$ by submodules of $M$ is called a
$CM$-filtration if every $M_{i}/M_{i-1}$ is Cohen-Macaulay and

$\dim(M_{1}/M_{0})<\dim(M_{2}/M_{1})<\dots<\dim(M_{l}/M_{l-1}).$

$M$ is said to be sequentially Cohen-Macaulay if $M$ admits a $CM$-filtration.

For the rest of this paper, we fix an arbitrary multicomplex $\Delta$ in a
lattice $P$ of multiset subsets of
$\hat{1}=\prod\limits_{i=1}^{l}x_{i}^{n_{i}}$ and let $I_{\Delta}$ be the
monomial ideal generated by all the monomials which are in $P\setminus\Delta$
i.e., $I_{\Delta}=(M_{1},\dots,M_{|P\setminus\Delta|})$ where $M_{i}$’s are
the elements of $P\setminus\Delta$. To prove the sequential Cohen-Macaulayness
of $R/I_{\Delta}$ when $\Delta$ is shellable, we will use the following
relation between an ideal and its polarization.

######  Proposition 64 ([10, Proposition 4.11]).

Let $R$ and $S$ be defined as in Definition 58. Then $R/I_{\Delta}$ is
sequentially Cohen-Macaulay if and only if $S/pol(I_{\Delta})$ is sequentially
Cohen-Macaulay.

A set corresponds to a square-free monomial, when written multiplicatively. We
let $P_{1}$ be the lattice of subsets of
$\prod\limits_{i=1}^{l}\prod\limits_{j=1}^{n_{i}}x_{i,j}$. Let $J$ be the set
of all monomials of $S$ which are in $pol(I_{\Delta})$. Then
$\Delta_{1}=P_{1}\setminus J$ is a $\mathcal{P}$-complex and it is clearly a
simplicial complex. It is also clear by definition that the Stanley-Reisner
ring associated to $\Delta_{1}$ is given by $S/pol(I_{\Delta})$. The
simplicial complex $\Delta_{1}$ is called the polarized simplicial complex
associated to the multicomplex $\Delta$.

Here is another description of the polarized simplicial complex associated to
$\Delta$ that will be useful later.

######  Theorem 65.

Let $\Delta_{1}$ be the polarized simplicial complex associated to a
multicomplex $\Delta$. Then

$\Delta_{1}=\left\langle\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
a_{i}+1\end{subarray}}x_{i,j}\colon x_{1}^{a_{1}}\cdots
x_{l}^{a_{l}}\in\Delta\right\rangle.$

###### Proof.

Let $\tilde{\Delta}$ be the simplicial complex given by

$\tilde{\Delta}=\left\langle\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
a_{i}+1\end{subarray}}x_{i,j}\colon x_{1}^{a_{1}}\cdots
x_{l}^{a_{l}}\in\Delta\right\rangle.$

Our goal is to show that $\Delta_{1}=\tilde{\Delta}$.

First, we show that $\tilde{\Delta}\subseteq\Delta_{1}$. To do this, since
$\Delta_{1}$ is a simplicial complex, it is enough to show that
$\prod\limits_{1\leq i\leq l}\prod\limits_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq a_{i}+1\end{subarray}}x_{i,j}\in\Delta_{1}$ for every
$x_{1}^{a_{1}}\cdots x_{l}^{a_{l}}\in\Delta$. So, let $x_{1}^{a_{1}}\cdots
x_{l}^{a_{l}}\in\Delta$ and, by contradiction, suppose instead that

$\prod\limits_{1\leq i\leq l}\prod\limits_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq a_{i}+1\end{subarray}}x_{i,j}\notin\Delta_{1}.$

Hence $\prod\limits_{1\leq i\leq l}\prod\limits_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq a_{i}+1\end{subarray}}x_{i,j}\in J$. Since $J$ is a
monomial ideal, $\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
a_{i}+1\end{subarray}}x_{i,j}$ is divisible by a monomial which is a generator
in $pol(I_{\Delta})$. More precisely $\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
a_{i}+1\end{subarray}}x_{i,j}$ is divisible by $\prod\limits_{1\leq i\leq
l}\prod\limits_{1\leq j\leq b_{i}}x_{i,j}$, where $b_{i}\leq a_{i}$ and
$x_{1}^{b_{1}}\cdots x_{l}^{b_{l}}\notin\Delta$. But $x_{1}^{a_{1}}\cdots
x_{l}^{a_{l}}\in\Delta$ implies that $x_{1}^{b_{1}}\cdots
x_{l}^{b_{l}}\in\Delta$ and hence we have a contradiction. Therefore, by
contradiction, $\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
a_{i}+1\end{subarray}}x_{i,j}\in\Delta_{1}$ and we are done with the first
part of this proof.

Conversely, let $u=\prod\limits_{1\leq i\leq l}\prod\limits_{1\leq j\leq
n_{i}}x_{i,j}^{r_{i,j}}\in\Delta_{1}$, where $r_{i,j}\in\\{0,1\\}$. Since
$u\in\Delta_{1}$, then $u\notin pol(I_{\Delta})$. Let $b_{i}=\max\\{k\colon
r_{i,j}=1,\text{ for all }1\leq j\leq k\\}$ (if the set is empty, we take
$b_{i}=0)$. Now, $v=\prod\limits_{1\leq i\leq l}\prod\limits_{1\leq j\leq
b_{i}}x_{i,j}$ is not in $pol(I_{\Delta})$ (otherwise $u$ would be in
$pol(I_{\Delta})$ as $v$ divides $u$. Since $v\notin pol(I_{\Delta})$, then
$\prod\limits_{i=1}^{l}x_{i}^{b_{i}}\notin I_{\Delta}$. Therefore
$\prod\limits_{i=1}^{l}x_{i}^{b_{i}}\in\Delta$. By definition of
$\tilde{\Delta}$, $w=\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b_{i}+1\end{subarray}}x_{i,j}\in\tilde{\Delta}$. The only variables which are
not in $w$ are the variables in $\prod\limits_{i=1}^{l}x_{i,b_{i}+1}$. And by
the definition of the $b_{i}$’s, those are variables neither in $u$. Therefore
$u$ divides $w$ and since $\tilde{\Delta}$ is a simplicial complex and
$w\in\tilde{\Delta}$, we have $u\in\tilde{\Delta}$. Hence
$\Delta_{1}\in\tilde{\Delta}$.

∎

######  Remark 66.

Notice that the polarized simplicial complex $\Delta_{1}$ associated with
$\Delta$ is not necessarily pure. For example, consider the power lattice of
the multiset subsets of $\hat{1}=x_{1}^{3}x_{2}^{3}$. Let $\Delta$ be the
multicomplex with only two facets $F_{1}=x_{1}^{2}x_{2}^{2}$ and
$F_{2}=x_{1}x_{2}^{3}$. Then the polarized simplicial complex $\Delta_{1}$
associated to $\Delta$ has at least two facets $x_{1,1}x_{1,2}x_{2,1}x_{2,2}$
and $x_{1,1}x_{1,3}x_{2,1}x_{2,2}x_{2,3}$.

######  Corollary 67.

Let $\Delta_{1}$ be the polarised simplicial complex associated to the
multicomplex $\Delta$. Then $\Delta_{1}=\langle F\rangle$ where

$F=\left\\{\prod\limits_{1\leq i\leq l}\prod_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq b_{i}+1\end{subarray}}x_{i,j}\colon 0\leq b_{i}\leq
a_{i},\prod_{i=1}^{l}x_{i}^{a_{i}}\text{ is a facet of }\Delta\right\\}$

###### Proof.

This is clear. ∎

We recall the notion of shellability for non-pure simplicial complexes
introduced by Björner and Wachs.

######  Definition 68.

Let $\Delta_{1}$ be a (not necessarily pure) simplicial complex. $\Delta_{1}$
is said to be shellable if there exists an ordering of the facets of
$\Delta_{1}$, $F_{1},\dots,F_{l}$ such that for every $i<j$, there is $k$ such
that $F_{i}\cap F_{j}\subseteq F_{k}\cap F_{j}$ and $|F_{k}\cap
F_{j}|=|F_{j}|-1$.

######  Theorem 69.

If $\Delta$ is a shellable multicomplex, then the polarized simplicial complex
$\Delta_{1}$ associated to $\Delta$, is a (not necessarily pure) shellable
simplicial complex.

###### Proof.

Let us denote by $\prec_{\Delta}$ the ordering of the facet of $\Delta$ which
makes it shellable. For this proof, we need to define an ordering on the
facets of $\Delta_{1}$. Notice that an element of $F$ in Corollary 67 is not
necessarily a facet of $\Delta_{1}$. However, a facet of $\Delta_{1}$ must be
an element of $F$. Let
$u=\prod\limits_{i=1}^{l}(\prod\limits_{j=1}^{n_{i}}x_{i,j}\setminus
x_{i,b_{i}+1})$ be a facet of $\Delta_{1}$. To $u$, we associated the face
$U_{1}=\prod\limits_{i=1}^{l}x_{i}^{b_{i}}$ of $\Delta$ and the facet
$U=\prod\limits_{i=1}^{l}x_{i}^{a_{i}}$ such that if $U_{1}\subseteq
U^{\prime}$ for another facet $U^{\prime}$ of $\Delta$, then
$U\prec_{\Delta}U^{\prime}$. This means that $U$ is the smallest facet of
$\Delta$ which contains $U_{1}$. Similarly, let
$v=\prod\limits_{i=1}^{l}(\prod\limits_{j=1}^{n_{i}}x_{i,j}\setminus
x_{i,b^{\prime}_{i}+1})$, $V_{1}=\prod\limits_{i=1}^{l}x_{i}^{b^{\prime}_{i}}$
be the associated face of $\Delta$ and
$V=\prod\limits_{i=1}^{l}x_{i}^{a^{\prime}_{i}}$ be the associated facet. We
say that $u\preceq_{\Delta_{1}}v$ if either $u=v$ or $u\neq v$ and one of the
following is satisfied:

  1. (i)

$U=V$ and if $e$ is the smallest integer such that $b_{e}\neq b^{\prime}_{e}$,
then $b_{e}>b^{\prime}_{e}$.

  2. (ii)

$U\prec_{\Delta}V$.

This is a total ordering $\preceq_{\Delta_{1}}$ on the facets of $\Delta_{1}$.
We now show that this defines a shelling on $\Delta_{1}$.

So, let us write $u=\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b_{i}+1\end{subarray}}x_{i,j}$ and $v=\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}$ and suppose $u\prec_{\Delta_{1}}v$.
Assume that $u,v$ correspond, respectively, to
$U_{1}=\prod\limits_{i=1}^{l}x_{i}^{b_{i}}\subseteq\prod\limits_{i=1}^{l}x_{i}^{a_{i}}=U$
and
$V_{1}=\prod\limits_{i=1}^{l}x_{i}^{b^{\prime}_{i}}\subseteq\prod\limits_{i=1}^{l}x_{i}^{a^{\prime}_{i}}=V$.
There are two cases:

Case 1. $U=V$. Let $e$ be the smallest integer such that $b_{e}\neq
b^{\prime}_{e}$. In this case $b_{e}>b^{\prime}_{e}$. We can then take
$W_{1}=x_{e}^{b_{e}}\prod\limits_{\begin{subarray}{c}1\leq i\leq l\\\ i\neq
e\end{subarray}}x_{i}^{b^{\prime}_{i}}\subseteq U$. Since $V_{1}\subseteq
W_{1}$, then $U$ must also be the smallest (w.r.t $\preceq_{\Delta}$) which
contains $W_{1}$. This defines $w=\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}$, where $b^{*}_{i}=b^{\prime}_{i}$ for
$i\neq e$ and $b^{*}_{e}=b_{e}$. It is not hard to see that
$w\prec_{\Delta_{1}}v$. What remains is to show that $u\cap v\subseteq w\cap
v$ and $|w\cap v|=|v|-1$.

$\displaystyle u\cap v$ $\displaystyle=\left(\prod\limits_{1\leq i\leq
l}\prod_{\begin{subarray}{c}j=1\\\ j\neq
b_{i}+1\end{subarray}}^{n_{i}}x_{i,j}\right)\cap\left(\prod_{1\leq i\leq
l}\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$
$\displaystyle=\prod\limits_{1\leq i\leq
l}\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$
$\displaystyle=\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{e}\\\ j\neq
b_{e}+1\end{subarray}}x_{e,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{e}\\\ j\neq
b^{\prime}_{e}+1\end{subarray}}x_{e,j}\right)\cdot\prod\limits_{\begin{subarray}{c}1\leq
i\leq l\\\ i\neq e\end{subarray}}\left(\prod_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq
b_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$

But for $i\neq e$,

$\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)\subseteq\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)=\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right).$

Hence

$\displaystyle u\cap v$
$\displaystyle\subseteq\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{e}\\\
j\neq
b_{e}+1\end{subarray}}x_{e,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{e}\\\ j\neq
b^{\prime}_{e}+1\end{subarray}}x_{e,j}\right)\cdot\prod\limits_{\begin{subarray}{c}1\leq
i\leq l\\\ i\neq e\end{subarray}}\left(\prod_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$

Since $b_{e}=b^{*}_{e}$, then

$\displaystyle u\cap v$ $\displaystyle\subseteq\prod\limits_{1\leq i\leq
l}\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$
$\displaystyle\subseteq\left(\prod\limits_{1\leq i\leq
l}\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod\limits_{1\leq i\leq
l}\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)$ $\displaystyle\subseteq w\cap
v.$

Finally,

$\displaystyle|w\cap v|$ $\displaystyle=\left|\left(\prod\limits_{1\leq i\leq
l}\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod\limits_{1\leq i\leq
l}\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)\right|$
$\displaystyle=\left|\prod\limits_{1\leq i\leq
l}\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)\right|$
$\displaystyle=\left|\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{e}\\\
j\neq
b_{e}+1\end{subarray}}x_{e,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{e}\\\ j\neq
b^{\prime}_{e}+1\end{subarray}}x_{e,j}\right)\cdot\prod\limits_{\begin{subarray}{c}1\leq
i\leq l\\\ i\neq e\end{subarray}}\left(\prod_{\begin{subarray}{c}1\leq j\leq
n_{i}\\\ j\neq b^{\prime}_{i}+1\end{subarray}}x_{i,j}\right)\right|.$

But

$\left|\left(\prod_{\begin{subarray}{c}1\leq j\leq n_{e}\\\ j\neq
b_{e}+1\end{subarray}}x_{e,j}\right)\cap\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{e}\\\ j\neq
b^{\prime}_{e}+1\end{subarray}}x_{e,j}\right)\right|=\left|\left(\prod_{\begin{subarray}{c}1\leq
j\leq n_{e}\\\ j\neq b^{\prime}_{e}+1\end{subarray}}x_{e,j}\right)\right|-1.$

Hence $|w\cap v|=|v|-1$

Case 2. $U\prec V$. By shellability of the multicomplex, there is
$W\prec_{\Delta}V$ such that $U\cap V\subseteq W\cap V$ and $|W\cap V|=|V|-1$.
For $|W\cap V|=|V|-1$ to hold, $W$ must be of the form
$W=x_{m}^{a^{\prime}_{m}+1}x_{n}^{a^{\prime}_{n}-1}\prod\limits_{\begin{subarray}{c}1\leq
i\leq l\\\ i\neq m,n\end{subarray}}x_{i}^{a^{\prime}_{i}}$ for some $1\leq
m,n\leq l$. Since $W\prec_{\Delta}V$ and $V$ being the “smallest” to contain
$V_{1}$, we must have $V_{1}\nsubseteq W$. Therefore $b^{\prime}_{n}\geq
a^{\prime}_{n}$. By looking at the valuation at $x_{n}$, $v_{x_{n}}(W\cap
V)\leq v_{x_{n}}(W)=a^{\prime}_{n}-1$. Furthermore $U\cap V\subseteq W\cap V$,
therefore $v_{x_{n}}(U\cap V)\leq v_{x_{n}}(W\cap V)\leq a^{\prime}_{n}-1$.
Since $v_{x_{n}}(V)=a^{\prime}_{n}$ and $v_{x_{n}}(U\cap
V)=\min\\{v_{x_{n}}(U),v_{x_{n}}(V)\\}$, we must have $a_{n}=v_{x_{n}}(U)\leq
a^{\prime}_{n}-1$. Therefore $b_{n}\leq a_{n}\leq a^{\prime}_{n}-1$. So we can
take $W_{1}=x_{n}^{b_{n}}\prod\limits_{\begin{subarray}{c}1\leq i\leq l\\\
i\neq n\end{subarray}}x_{i}^{b^{\prime}_{i}}$ and we have $W_{1}\subseteq W$.
This gives us $w=\prod\limits_{1\leq i\leq
l}\prod\limits_{\begin{subarray}{c}1\leq j\leq n_{i}\\\ j\neq
b^{*}_{i}+1\end{subarray}}x_{i,j}$, where $b^{*}_{i}=b^{\prime}_{i}$ for
$i\neq n$ and $b^{*}_{n}=b_{n}$. Since $W\prec_{\Delta}V$, then
$w\prec_{\Delta_{1}}v$. Now $b_{n}\leq a^{\prime}_{n}-1<a^{\prime}_{n}\leq
b^{\prime}_{n}$ so that $b^{*}_{n}\neq b^{\prime}_{n}$ and similarly to the
previous case, one can show that $u\cap v\subseteq w\cap v$ and $|w\cap
v|=|v|-1$.

∎

We record here a result about sequential Cohen-Macaulayness of the Stanley-
Reisner ring associated to a non-pure shellable simplicial complex.

######  Proposition 70 ([27, Sec III.2]).

If $\Delta_{1}$ is a (not necessarily pure) shellable simplicial complex whose
associated Stanley-Reisner ideal in a polynomial ring $R_{1}$ is
$I_{\Delta_{1}}$, then $R_{1}/I_{\Delta_{1}}$ is sequentially Cohen-Macaulay.

As a consequence of Proposition 70, Theorem 69 and Proposition 64, we have the
main theorem of this section.

######  Theorem 71.

If $\Delta$ is a shellable multicomplex in the power lattice of multiset
subsets of $x_{1}^{n_{1}}\cdots x_{l}^{n_{l}}$, then the Stanley-Reisner ring
$\mathbb{F}[x_{1},\dots,x_{l}]/I_{\Delta}$ is a sequentially Cohen-Macaulay
ring.

#### Acknowledgement

We thank Ananthnarayan Hariharan and Neeraj Kumar for helpful discussion on
questions related to Stanley-Reisner rings.

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