Cohen–Macaulayness of Bipartite Graphs, Revisited

Abstract

Cohen–Macaulayness of bipartite graphs is investigated by several mathematicians and has been characterized combinatorially. In this paper, we give some different combinatorial conditions for a bipartite graph which are equivalent to Cohen–Macaulayness of the graph. We prove that a bipartite graph is Cohen–Macaulay if and only if it is well covered and has a unique perfect matching. We also provide a fast algorithm to check Cohen–Macaulayness of a given bipartite graph.

1 Introduction an Preliminaries

Characterization and classification of Cohen–Macaulay graphs, specially bipartite graphs, have been extensively studied in the last decades (e.g., see [2–4, 6, 11]). A thorough background to the subject is provided in the above references and [9]. To make this note self-contained, we review some of the basic notions.

Throughout this paper, all graphs are finite and simple with no vertex of degree zero. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. For two adjacent vertices $v$ and $w$ in $G$, we write $v\sim w$. The set of all vertices of $G$ adjacent to a vertex $v$ is denoted by $N(v)$. We say that a set $F\subseteq V(G)$ is an independent set in $G$ if no two vertices of $F$ are adjacent. A set $P\subseteq E(G)$ is called a perfect matching if there is no pair of distinct edges in $P$ with a common vertex and any vertex in $G$ belongs to one of the edges in $P$. A graph $G$ is called bipartite if $V(G)=V\cup W$ such that $V\cap W=\varnothing$ and both $V$ and $W$ and independent sets in $G$. A bipartite graph is called complete bipartite if each vertex in $V$ is adjacent to each vertex of $W$.

Let $G$ be a graph on the vertex set $V(G) = \{v_1,\ldots ,v_n\}$. Let $K$ be a field and $S=K[x_1,\ldots ,x_n]$ the polynomial ring on $n$ variables with coefficients in $K$. The edge ideal $I(G)$ of $G$ is defined to be the ideal of $S$ generated by all square-free monomials $x_ix_j$, provided that $v_i\sim v_j$ in $G$. The quotient ring $R(G)=S/I(G)$ is called the edge ring of the graph $G$.

Let $[n] = \{1,2,\ldots ,n\}$. A (finite) simplicial complex $\Delta$ on $n$ vertices is a collection of subsets of $[n]$ such that the following conditions hold:

• (i) $\{i\}\in \Delta$ for each $i\in [n]$,

• (ii) if $E\in \Delta$ and $F\subseteq E$, then $F\in \Delta$.

An element of $\Delta$ is called a face, and a maximal face with respect to inclusion order is called a facet. The dimension of a face $F\in \Delta$ is defined to be $|F|-1$ and the dimension of $\Delta$ is the maximum of the dimensions of its faces. Faces with dimension 0 are called vertices.

Let $\Delta$ be a simplicial complex on $[n]$ and $S=K[x_1,\ldots ,x_n]$. Let $I_{\Delta }$ be the ideal of $S$ generated by all square-free monomials $x_{i_1}\cdots x_{i_s}$, provided that $\{i_1,\ldots ,i_s\}\not \in \Delta$. The quotient ring $K[\Delta ]=S/I_{\Delta }$ is called the Stanley–Reisner ring of the simplicial complex $\Delta$. Let $v_i$ be a vertex in a graph $G$ and $x_i$ be the corresponding variable in the polynomial ring. In this paper, we usually use $x_i$ instead of $v_i$. The same notation is used for vertices of a simplicial complex.

For a graph $G$, there is a simplicial complex which is called independence complex of $G$ and is defined by

$$\begin{aligned} \Delta _G = \{F \subseteq V : F \ \text{ is } \text{ an } \text{ independent } \text{ set } \text{ in } \ G\} . \end{aligned}$$

Let $R$ be a commutative ring with an identity. The depth of $R$, denoted by $\text{ depth }(R)$, is the largest integer $r$ such that there is a sequence $f_1, \ldots , f_r$ of elements of $R$ such that $f_i$ is not a zero-divisor in $S / (f_1, \ldots , f_{i-1})$ for all $1\le i \le r$, and $(f_1, \ldots , f_r)\ne R$. Such a sequence is called a regular sequence. The depth is an important invariant of a ring. It is bounded by another important invariant, the Krull dimension of the ring denoted by $\dim (R)$; the length of the longest chain of prime ideals in the ring. The ring $R$ is called Cohen–Macaulay if $\text{ depth }(R) = \dim (R)$. A graph $G$ (a simplicial complex $\Delta$, respectively) is called Cohen–Macaulay if the ring $R(G)$ (the ring $K[\Delta ]$, respectively) is Cohen–Macaulay. It is well known that $\dim (K[\Delta ])=\dim (\Delta )+1$ (see [5]).

A simplicial complex $\Delta$ is called pure if all its facets have the same cardinality. A graph $G$ is called well covered or unmixed if all maximal independent sets of vertices of $G$ have the same cardinality. It is clear that a graph $G$ is unmixed if and only if the simplicial complex $\Delta _G$ is pure. It is well known that a Cohen–Macaulay simplicial complex is pure ([5]), but the converse is not true, i.e., there are pure simplicial complexes which are not Cohen–Macaulay. Also it is known that if $G$ is a Cohen–Macaulay graph, then $\overline{G}$, the complement of $G$, is connected [5].

A pure simplicial complex $\Delta$ on vertex set $[n]$ is called completely balanced if there is a partition of $[n]$ as $C_1,\ldots ,C_r$ such that each facet of $\Delta$ has exactly one vertex in common with each $C_i$. Here, a partition means that $C_1\cup \cdots \cup C_r=[n]$, and for each $i\ne j, \, C_i\cap C_j=\varnothing$. Such simplicial complexes were studied by R. Stanley [8]. He proved that, in a completely balanced simplicial complex with partition $C_1, \ldots , C_r$, the elements $\theta _1,\ldots ,\theta _r$ form a homogeneous system of parameters, where

$$\begin{aligned} \theta _i=\sum _{j\in C_i}\bar{x}_j, \end{aligned}$$

where $\bar{x}_i$ denotes the image of $x_i$ in $K[\Delta ]$. By a homogeneous system of parameters in a standard graded ring $R$, we mean a set of homogeneous elements $\theta _1,\ldots ,\theta _r$ of nonzero degrees such that $\dim (R/(\theta _1,\ldots ,\theta _r))=0$.

2 Main Results

M. Estrada and R. H. Villarreal in [2] have proved that, for a bipartite graph $G$, Cohen–Macaulayness and shellability are equivalent, and if $G$ is Cohen–Macaulay, then there is a vertex $v$ in $G$ such that $G\setminus \{v\}$ is again Cohen–Macaulay.

Villarreal has proved in [12] that a bipartite graph $G$ with parts $V$ and $W$ is unmixed if and only if $|V|=|W|$ and there is an order on vertices of $V$ and $W$ as $x_1, \ldots , x_n$ and $y_1, \ldots , y_n$ respectively, such that

• (1) $x_i\sim y_i$ for $i=1, \ldots , n$,

• (2) for each $1\le i<j<k\le n$ if $x_i\sim y_j$ and $x_j\sim y_k$, then $x_i\sim y_k$. J. Herzog and T. Hibi in [4] have proved that a bipartite graph $G$ is Cohen–Macaulay if and only if it is unmixed (has properties (1) and (2) above), and

• (3) if $x_i\sim y_j$, then $i\le j$.

In these results, one needs to find an appropriate order on the vertices of $G$ for which the criterion holds. Therefore, checking the Cohen–Macaulayness of a given bipartite graph in practice is rather complicated. In this paper, we show that one can check Cohen–Macaulayness of a given bipartite graph in a quite easy way. We prove that a graph $G$ is Cohen–Macaulay if and only if it is unmixed and has a unique perfect matching. These kind of graphs have applications in computational biology. The unique maximum-weight perfect matching can be used to predict the folding structure of RNA molecules (see [10]). Our main theorem is the following.

Theorem 2.1

Let $G$ be a bipartite graph with parts $V$ and $W$. Then, $G$ is Cohen–Macaulay if and only if there is a perfect matching in $G$ as $\{x_1,y_1\}, \ldots , \{x_n,y_n\}$, such that $x_i\in V$ and $y_i\in W$ for $i=1,\ldots ,n$, and the following two conditions hold.

• (1) The induced subgraph on $N(x_i)\cup N(y_i)$ is a complete bipartite graph, for $i=1,\ldots ,n$.

• (2) If $x_i\sim y_j$ for $i\ne j$, then $x_j\not \sim y_i$.

Before proving the theorem, we prove some preliminary lemmas.

Lemma 2.1

Let $G$ be an unmixed bipartite graph with a perfect matching $\{x_1,y_1\}, \ldots ,$ $\{x_n,y_n\}$. Then, $G$ is Cohen–Macaulay if and only if the sequence $\bar{x}_1+\bar{y}_1, \ldots , \bar{x}_n+\bar{y}_n$ is a regular sequence in $R(G)$.

Proof

The sets $\{x_1,y_1\}, \ldots , \{x_n,y_n\}$ form a partition of vertices of $G$ and any maximal independent set intersects each of them in exactly one vertex. Thus, the simplicial complex $\Delta _{G}$ is completely balanced. By Corollary 4.2 and its remark in [8], the sequence $x_1+y_1, \ldots , x_n+y_n$ is a system of parameters in $R(G)$. By Theorem 17.4 in [7], $R(G)$ is Cohen–Macaulay if and only if every system of parameters is a regular sequence in $R(G)$. $\square$

Lemma 2.2

Let $I$ be an ideal of $S=K[x_1,\ldots ,x_n]$, generated by some quadratic monomials. Let for some $i,j, \, 1\le i<j\le n, \, x_i^2\not \in I$ and $x_j^2\not \in I$. Then, $\bar{x}_i+\bar{x}_j$ is a zero-divisor in $S/I$ if and only if one of the following conditions holds:

• (i) There is $x_k, \, k\not \in \{i,j\}$ such that $\bar{x}_k(\bar{x}_i+\bar{x}_j)=0$.

• (ii) There are integers $k,l, \, 1\le k<l\le n$, both distinct from $i$ and $j$, such that $x_kx_l\not \in I$ and $\bar{x}_k\bar{x}_l(\bar{x}_i + \bar{x}_j)=0$.

Here, $\bar{x}_i$ denotes the image of $x_i$ in $S/I$.

Proof

Without loss of generality, we may assume that $i=1$ and $j=2$. It is well known that a polynomial $f$ in $S$ belongs to a monomial ideal $I$ if and only if all monomials of $f$ belong to $I$. Let $\prec$ be the lexicographic order on monomials of $S$ induced by $x_1\succ x_2 \succ \cdots \succ x_n$. Let $\bar{x}_1+\bar{x}_2$ be a zero-divisor in $S/I$. Then, there is a polynomial $h$ in $S$ such that $\bar{h}$ is nonzero in $S/I$ and $\bar{h}(\bar{x}_1+\bar{x}_2)=0$ or equivalently, $f=h(x_1+x_2)\in I$. Let $h=h_1+h_2+\cdots + h_r$ such that $h_i$s are monomials and $h_1\succ h_2\succ \cdots \succ h_r$. We may assume that $h_1\not \in I$. Now, $h_1x_1$ is the greatest monomial of $f$ with respect to the order $\prec$ which cannot be canceled by other monomials. Therefore, $h_1x_1\in I$ and there is a quadratic monomial in the monomial generating set of $I$ which divides $h_1x_1$ and does not divide $h_1$. This monomial must be of the form $x_1x_k$ for some $k, \, 1< k\le n$. On the other hand, $x_1\not \mid h_1$ because $x_k|h_1$ and $h_1\not \in I$. Since the order is lexicographic, $x_1$ does not divide any other monomials in $h$. Therefore, in the polynomial $hx_1+hx_2$, there is no monomial of the summand $hx_2$, which is divisible by $x_1$. In this summand, $h_1x_2$ is the greatest monomial with respect to $\prec$ and can not be canceled by other monomials. This implies that $h_1x_2\in I$. Similar to the above argument, there is a quadratic monomial in the generating set of $I$ which divides $h_1x_2$ but not $h_1$. This monomial must be of the form $x_2x_l$ for some $2<l\le n$. We also have $x_2\not \mid h_1$ but $x_k|h_1$ and $x_l|h_1$. If $k=l$, then $x_k(x_1+x_2)\in I$ and if $k\ne l$, then $x_kx_l(x_1+x_2)\in I$. Note that, $x_kx_l\not \in I$ because $x_kx_l|h_1$ and $h_1\not \in I$. This completes the proof in one direction. The converse is trivial. $\square$

Proof of Theorem 2.1  We proceed the proof in 3 steps. First we prove that a bipartite graph $G$ is unmixed if and only if there is a perfect matching in $G$ satisfying condition 1. Then, in Step 2, we prove that for an unmixed bipartite graph, condition 2 is necessary for Cohen–Macaulayness. In Step 3 we prove that condition 2 is also sufficient for Cohen–Macaulayness of such a graph.

Step 1. Let $G$ be unmixed. There is no isolated vertex and any vertex in $V$ is adjacent to some vertices in $W$. Therefore, there is no vertex in $V$ independent to the set $W$. This means that $W$ is a maximal independent set in $G$. Similarly, $V$ is a maximal independent set. By unmixedness of $G, \, |V|=|W|$. Let $A\subseteq V$ be a nonempty set. Suppose $|N(A)|< |A|$. There is no edge between $A$ and $W\setminus N(A)$. Therefore, $A\cup (W\setminus N(A))$ is an independent set and its size is strictly greater than $|W|$, which is a contradiction. Therefore, $|N(A)| \ge |A|$ for each nonempty subset $A$ of $V$. Therefore, by the marriage theorem of Hall [1], there is a perfect matching between $V$ and $W$.

Now, let $V=\{x_1,\ldots ,x_n\}, \, W=\{y_1,\ldots ,y_n\}$ and $\{x_1,y_1\}, \ldots , \{x_n,y_n\}$ be a perfect matching in $G$. The graph $G$ is unmixed and any maximal independent set of vertices in $G$ has cardinality $n$. Therefore, any maximal independent set intersects each edge of the perfect matching in exactly one vertex. Suppose for some $j, \, 1\le j\le n$, the induced subgraph on $N(x_j)\cup N(y_j)$ is not a complete bipartite graph. Then, there are $x\in N(y_j)$ and $y\in N(x_j)$ such that $x\not \sim y$. The set $\{x,y\}$ is independent and so there is a maximal independent set containing it. But, this maximal independent set does not meet the edge $\{x_j,y_j\}$ which is a contradiction. Therefore, condition 1 holds.

Conversely, assume that there is a perfect matching $\{x_1,y_1\}, \ldots , \{x_n,y_n\}$ in $G$ which satisfies condition 1. Let $A$ be a maximal independent set in $G$. Then $A$ meets each edge in the perfect matching in at most one vertex. Suppose that for some $j, \, 1\le j\le n, \, A\cap \{x_j,y_j\}=\varnothing$. Then, neither $x_j$ nor $y_j$ is independent to $A$, and there are $x,y\in A$ such that $x\sim y_j$ and $y\sim x_j$. But, $x$ and $y$ are not adjacent and the induced subgraph on $N(x_j)\cup N(y_j)$ is not a complete bipartite graph, which is a contradiction. Therefore, $A$ meets any edge in the perfect matching and has cardinality $n$. It means that $G$ is unmixed.

Step 2. Let $G$ be a bipartite graph with a perfect matching which satisfies condition 1 but condition 2 fails. That is, for some $i$ and $j, \, 1\le i<j\le n$, we have $x_i\sim y_j$ and $x_j\sim y_i$. Then, in the quotient ring $R(G)/(x_i + y_i)$, the element $\bar{x}_i$ is nonzero and $\bar{x}_i(\bar{x}_j+\bar{y}_j)=0$ because $\bar{x}_i=-\bar{y}_i$. Therefore, $\bar{x}_j+\bar{y}_j$ is a zero-divisor in $R(G)/(x_i + y_i)$. This means that the sequence $\bar{x}_1+\bar{y}_1, \ldots , \bar{x}_n+\bar{y}_n$ is not a regular sequence in $R(G)$ and hence by Lemma 2.1, $R(G)$ is not Cohen–Macaulay.

Step 3. Let $G$ be a bipartite graph with a perfect matching satisfying condition 1. In this case, $\dim (R(G))=n$ and to prove that $R(G)$ is Cohen–Macaulay, it is enough to show that the sequence $\bar{x}_1+\bar{y}_1, \ldots , \bar{x}_n+\bar{y}_n$ is a regular sequence in $R(G)$ (Lemma 2.1). For an integer $i, \, 1\le i< n$, the ring $R(G)/(x_1+y_1,\ldots ,x_{i-1}+y_{i-1})$ can be viewed as the ring $R'(G)$ obtained by $R(G)$ identifying $x_j$ with $-y_j$ for $j=1,\ldots ,i-1$. By the Lemma 2.2 and its proof, the only possibility for $\bar{x}_i+\bar{y}_i$ to be zero-divisor in $R'(G)$ is that there is $j, \, 1\le j\le i-1$, such that $\bar{x}_j(\bar{x}_i+\bar{y}_i)=0$. Therefore, $\bar{x}_j\bar{y}_i=0$ and $\bar{x}_j\bar{x}_i=0$ or equivalently, $\bar{y}_j\bar{x}_i=0$. Therefore, $x_j\sim y_i$ and $y_j\sim x_i$. But, in this case, condition 2 fails. This completes the proof. $\square$

Proposition 2.1

Condition 1 in Theorem 1 which is equivalent to unmixedness of a bipartite graph is also equivalent to say that none of the polynomials $x_1+y_1, \ldots , x_n+y_n$ is a zero-divisor in $R(G)$.

Proof

The assertion is clear by the Lemma 2.2 and the Theorem 2.1. $\square$

Remark 2.1

Condition 2 in Theorem 1 is equivalent to say that, for each $i$ and $j, \, 1\le i<j\le n$, the induced subgraph on vertices $\{x_i,y_i,x_j,y_j\}$ has connected complement.

Corollary 2.1

Let $G$ be a bipartite Cohen–Macaulay graph and let $\{x_i,y_i\}$ be any edge in the perfect matching mentioned in Theorem 1. Then, $G\setminus \{x_i,y_i\}$ is again Cohen–Macaulay.

Proof

Here, by $G\setminus \{x_i,y_i\}$ we mean the graph obtained by deleting vertices $x_i$ and $y_i$ and all edges passing through one of these vertices. It is clear that if condition 1 or 2 in Theorem 1 holds for $G$, then it holds for $G\setminus \{x_i,y_i\}$ for each $i=1,\ldots ,n$. $\square$

Proposition 2.2

Let $G$ be a bipartite Cohen–Macaulay graph with parts $V$ and $W$. Then, there is at least one vertex of degree one in each part.

Proof

Let $y$ be a vertex in $W$ such that for any other vertex $y'\in W$, we have $\deg (y')\le \deg (y)$. Let $x\in V$ be the vertex such that $\{x,y\}$ is in a perfect matching in $G$. If $\deg (x)>1$, then there is a vertex $y'\in W\setminus \{y\}$ such that $x\sim y'$. Let $x'$ be a vertex in $V\setminus \{x\}$ such that $\{x',y'\}$ is in the perfect matching. Since $G$ is Cohen–Macaulay, the induced subgraph on $N(x)\cup N(y)$ is a complete bipartite graph and $x'\not \in N(y)$. Then, $y'$ is adjacent to each vertex in $N(y)\cup \{x'\}$. This means that $\deg (y')>\deg (y)$, which is a contradiction. Therefore, $\deg (x)=1$. $\square$

Let $G$ be a Cohen–Macaulay bipartite graph. There are some vertices in both parts with degree one. If we remove the vertex adjacent to a vertex of degree one, the edge consisting of these two vertices in a perfect matching will be removed and the remaining graph is also Cohen–Macaulay.

Corollary 2.2

Let $G$ be a Cohen–Macaulay bipartite graph. Then, there is a unique perfect matching in $G$.

Proof

Let $V$ and $W$ be two parts of $G$. By the Theorem 1, there is a perfect matching. Let $P$ be a perfect matching in $G$. By the above proposition, there is a vertex $x_1$ of degree one in $V$. Let $y_1\in W$ be the unique vertex adjacent to $x_1$. Then $\{x_1,y_1\}\in P$. The graph $G\setminus \{x_1,y_1\}$ is again Cohen–Macaulay and $V\setminus \{x_1\}$ has a vertex of degree one as $x_2$. Let $y_2\in W\setminus \{y_1\}$ be the unique vertex adjacent to $x_2$. Then, $\{x_2,y_2\}\in P$. Continuing this process, $P$ will be uniquely determined. $\square$

Corollary 2.3

Let $G$ be an unmixed bipartite graph. Then, the following conditions are equivalent.

• (i) $G$ is Cohen–Macaulay.

• (ii) There is a unique perfect matching in $G$.

• (iii) For each two edges $e_1, e_2$ in a perfect matching, the complement of the induced subgraph on vertices of $e_1$ and $e_2$ is connected.

• (iv) For a perfect matching $P$ in $G$, there is an order on edges of $P$ such that $P=\{\{x_1,y_1\},\ldots ,\{x_n,y_n\}\}$ and $x_i\sim y_j$ implies $i\le j$.

Proof

(i$\rightarrow$ii) This is proved in Corollary 2.2. Let $G$ be unmixed but not Cohen–Macaulay. Then, there is a perfect matching and two edges in the perfect matching as $\{x_i,y_i\}$ and $\{x_j,y_j\}$ such that $x_i\sim y_j$ and $x_j\sim y_i$. Substituting $\{x_i,y_i\}$ and $\{x_j,y_j\}$ by $\{x_i,y_j\}$ and $\{x_j,y_i\}$, we get a different perfect matching. This proves (ii$\rightarrow$i). The equivalence of (i) and (iii) is clear by Theorem 2.1 and Remark 2.1. To prove (i$\rightarrow$iv), let $x_1, \ldots , x_n$ be vertices of $V$ such that $\deg (x_i)\ge \deg (x_{i+1})$ for each $i=1,\ldots , n-1$. Then, $\deg (y_1)=1$. Remove $\{x_1,y_1\}$. In the remaining graph, $\deg (y_2)=1$ and continuing this process shows that there is no any edge between $x_i$ and $y_j$ if $i>j$. Finally, condition (iv) clearly implies condition (iii). $\square$

Now for a given bipartite graph $G$, we present a fast polynomial-time algorithm to check whether $G$ is Cohen–Macaulay.

Algorithm

Let $G$ be a given bipartite graph with parts $V$ and $W, \, |V|=|W|=n$.

Step 1. :

Take $i=0$.

Step 2. :

If there is no vertex with degree 1 in $V$, go to Step 6.

Step 3. :

Set $i=i+1$. Choose a vertex of degree one in $V$ and name it $x_i$. Name the vertex in $W$ adjacent to $x_i$ to be $y_i$. Take $V=V\setminus \{x_i\}, W=W\setminus \{y_i\}$. If $i<n$, go to Step 2.

Step 4. :

If there is $j, \, 2\le j\le n$ such that, a vertex in $N(x_j)$ and a vertex in $N(y_j)$ are not adjacent, then, go to Step 6.

Step 5. :

Write “G is Cohen–Macaulay” and end the algorithm.

Step 6. :

Write “G is not Cohen–Macaulay” and end the algorithm.