Expansion and contraction functors on matriods

TL;DR

This paper investigates how expansion and contraction functors affect matroid properties, showing that certain properties are preserved under expansions and introducing contraction as a new tool to analyze White's conjecture.

Contribution

It introduces the contraction functor for matroids, establishes the equivalence of properties under expansions, and links White's conjecture to expansions and contractions.

Findings

Expansion preserves graphic, binary, and transversal properties.

White's conjecture holds for a matroid if and only if it holds for all its expansions.

Some classes of matroids are identified that satisfy White's conjecture.

Abstract

Let $M$ be a matroid. We study the expansions of $M$ mainly to see how the combinatorial properties of $M$ and its expansions are related to each other. It is shown that $M$ is a graphic, binary or a transversal matroid if and only if an arbitrary expansion of $M$ has the same property. Then we introduce a new functor, called contraction, which acts in contrast to expansion functor. As a main result of paper, we prove that a matroid $M$ satisfies White's conjecture if and only if an arbitrary expansion of $M$ does. It follows that it suffices to focus on the contraction of a given matroid for checking whether the matroid satisfies White's conjecture. Finally, some classes of matroids satisfying White's conjecture are presented.

Equations

$$A^{\alpha }=\{x_{i_{1}1},\dots ,x_{i_1{k}_{i_1}},\dots ,x_{i_{r}1},\dots ,x_{i_r{k}_{i_r}}\}.$$

$$\{A{\}}^{\alpha }=\{\{x_{i_1{j}_1},\dots ,x_{i_r{j}_r}\}:1\le j_l\le k_{i_l}\text{mbox}foralll\}.$$

$${\mathcal{A}}^{\alpha }=\bigcup _{A\in \mathcal{A}}\{A{\}}^{\alpha }.$$

$${\mathcal{B}}_{M_X}=\{B\cap X:B\in {\mathcal{B}}_M\}.$$

$$[n{]}^{\alpha }=\{x_{11},\dots ,x_{1k_1},\dots ,x_{i1},\dots ,x_{i{k}_i},x_{i(k_i+1)},\dots ,x_{n1},\dots ,x_{n{k}_n}\}$$

$$([n{]}^{\beta }{)}^{1+{\epsilon }_{i{k}_i}}=\{x_{111},\dots ,x_{1k_{1}1},\dots ,x_{i11},\dots ,x_{i{k}_{i}1},x_{i{k}_{i}2},\dots ,x_{n11},\dots ,x_{n{k}_{n}1}\}.$$

$$\varphi(x_{rst})=\left\{\begin{array}[]{ll}x_{rs}&t=1\\ x_{i(k_{i}+1)}&t=2.\end{array}\right.$$

$$\begin{array}[]{cr}\theta:&(\mathcal{A}^{\beta})^{\mathbf{1}+\varepsilon_{ik_{i}}}\rightarrow\mathcal{A}^{\alpha}.\\ &F\mapsto\varphi(F)\end{array}$$

$${\mathcal{B}}_M=\{\{x_1,x_2,x_3\},\{x_1,x_2,x_4\},\{x_1,x_3,x_4\},\{x_1,x_3,x_5\},$$

$$\{x_1,x_4,x_5\},\{x_2,x_3,x_4\},\{x_2,x_3,x_5\},\{x_2,x_4,x_5\}\}.$$

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$$|\bigcup A_i|\ge |K|.$$

$$\{x_1{y}_2,x_2{y}_1\},\{x_1{y}_1,x_3{y}_2\},\{x_2{y}_1,x_3{y}_2\}$$

$${\mathcal{B}}_{M[\mathcal{A}]}=\{\{x_1,x_2\},\{x_1,x_3\},\{x_2,x_3\}\}.$$

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}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}\end{array}$$

$${\mathcal{B}}_{M(\mathcal{P})}=\{U\subset A:|U\cap A_i|\le 1\text{mbox}foralli\text{mbox}and|U|=t\}.$$

$$x_i\sim x_j⟺\{A\backslash x_i:A\in \mathcal{B},x_i\in A\}=\{A\backslash x_j:A\in \mathcal{B},x_j\in A\}.$$

$$x_{i}\sim x_{j}\Longleftrightarrow\mbox{for all}\ A\in\mathcal{B}\left\{\begin{array}[]{ll}\mbox{if}\ x_{i}\in A\ \mbox{then}\ (A\backslash x_{i})\cup x_{j}\in\mathcal{B}\\ \mbox{if}\ x_{j}\in A\ \mbox{then}\ (A\backslash x_{j})\cup x_{i}\in\mathcal{B}.\end{array}\right.$$

$$E(x;A_r,A_s)=\{y\in A_s:(A_r\backslash x)\cup y,(A_s\backslash y)\cup x\in {\mathcal{B}}_M\}.$$

$$(A_1,\dots ,A_{r-1},(A_r\backslash x)\cup y,A_{r+1},\dots ,A_{s-1},(A_s\backslash y)\cup x,A_{s+1},\dots ,A_m)$$

$$E(U;A_r,A_s)=\{V\subset A_s:(A_r\backslash U)\cup V,(A_s\backslash V)\cup U\in {\mathcal{B}}_M\}.$$

$$(A_1,\dots ,A_m){\sim }_M^i(C_{11},\dots ,C_{1m}){\sim }_M^i\dots {\sim }_M^i(C_{t1},\dots ,C_{tm}){\sim }_M^i(B_1,\dots ,B_m).$$

$$M\in \mathrm{TE}(i)\iff M^{\alpha }\in \mathrm{TE}(i).$$

$$(B_1,B_2){\sim }_M^1(C_{11},C_{12}){\sim }_M^1\dots {\sim }_M^1(C_{t1},C_{t2}){\sim }_M^1(B_2,B_1).$$

$${\mathcal{B}}_{M_1\oplus M_2}=\{B_1\cup B_2:B_1\in {\mathcal{B}}_{M_1},B_2\in {\mathcal{B}}_{M_2}\}.$$

$$\mathcal{B}=\{A\cup ({\cup }_{k=1}^t{z}_{i_k{j}_k}):A\in {\mathcal{B}}_M,1\le i_1<\dots <i_t\le r,j_k\in [s]\}$$

$${\mathcal{B}}_M=\{\{x_1,x_2,x_3,x_5\},\{x_1,x_2,x_3,x_7\},\{x_1,x_2,x_5,x_6\},\{x_1,x_2,x_5,x_7\},\{x_1,x_2,x_6,x_7\},$$

$$\{x_1,x_3,x_4,x_5\},\{x_1,x_3,x_4,x_7\},\{x_1,x_4,x_5,x_6\},\{x_1,x_4,x_5,x_7\},\{x_1,x_4,x_6,x_7\},$$

$$\{x_2,x_3,x_5,x_7\},\{x_2,x_5,x_6,x_7\},\{x_3,x_4,x_5,x_7\},\{x_4,x_5,x_6,x_7\}\}.$$

$${\mathcal{B}}_{\overline{M}}=\{\{x_1,x_2,x_3,x_5\},\{x_1,x_2,x_3,x_7\},\{x_1,x_2,x_5,x_7\},\{x_2,x_3,x_5,x_7\}\}.$$

Full text

Expansion and contraction functors on matriods ††

• Corresponding Author.

2010 Mathematics Subject Classification: Primary 05B35; Secondary 52B40.

Key words and phrases: expansion functor, contraction functor, White’s conjecture.

Rahim Rahmati-Asghar∗

*Department of Mathematics, Faculty of Basic Sciences, University of Maragheh, P. O. Box 55181-83111, Maragheh, Iran.

e-mail* : rahmatiasghar.r@gmail.com

Abstract. Let $M$ be a matroid. We study the expansions of $M$ mainly to see how the combinatorial properties of $M$ and its expansions are related to each other. It is shown that $M$ is a graphic, binary or a transversal matroid if and only if an arbitrary expansion of $M$ has the same property. Then we introduce a new functor, called contraction, which acts in contrast to expansion functor. As a main result of paper, we prove that a matroid $M$ satisfies White’s conjecture if and only if an arbitrary expansion of $M$ does. It follows that it suffices to focus on the contraction of a given matroid for checking whether the matroid satisfies White’s conjecture. Finally, some classes of matroids satisfying White’s conjecture are presented.

Introduction

Matroids are abstract combinatorial structures that capture the notion of independence that is common to a surprisingly large number of mathematical entities. They were introduced by Whitney in 1935 as a common generalization of independence in linear algebra and independence in graph theory [21]. Matroid theory is one of the most fascinating research areas in combinatorics. It was linked to projective geometry by Mac Lane [12], and have found a great many applications in several branches of mathematics [20]. In this regard studying the structural properties of matroids from different point views have been considered by many researchers. Also classifying matroids with a desired property or making modifications to a matroid so that it satisfies a special property is the subject of many research papers, see for example [1, 7, 9, 11, 15, 17, 20].

The notion of expansion is a known notion appeared in different terminologies as parallelization or duplication in combinatorics [5, 4, 13, 16]. In [15], the authors studied behaviors of expansion functor on some algebraic structures associated to discrete polymatroids. It was shown that a nonempty finite set is a discrete polymatroid if and only if its an arbitrary expansion is a discrete polymatroid (c.f. [15, Theorem 1.2.]). The discrete polymatroid is a multiset analogue of the matroid. Moreover, there are several classes of matroids so that the study of each of them is interesting in its own right. This motivates to focus on the behaviors of the expansion functor on some classes of matroids and to investigate some structural properties of them in this paper. Our goal in this paper is to investigate more relations between the exchange property of bases of a matroid and those of its expansions. It turns out that the exchange properties of bases of a matroid are preserved under taking the expansion functor and so this construction is a very good tool to make new matroids with a desired property. Moreover, by taking another functor, contraction functor, which is the opposite to the expansion functor we will able to construct a new matroid, with possibly smaller ground set, from a given matroid and check a desired property on new matroid instead of primary one.

White in 1980 proposed a conjecture about the bases of a matroid [20]. This conjecture has received much attention in recent years and has some algebraic and combinatorial variants, all of which are open problems. Up to now, several mathematicians confirmed only some variants of this conjecture for special classes of matroids (see for example [1, 2, 7, 9, 11, 15, 17, 18]).

White [20] defined three classes $\mathrm{TE}(1)$, $\mathrm{TE}(2)$ and $\mathrm{TE}(3)$ of matroids and conjectured that $\mathrm{TE}(1)=\mathrm{TE}(2)=\mathrm{TE}(3)=$the class of all matroids.

We investigate the effect of the expansion functor on the exchange property for bases of matroids and conclude that White’s conjecture is preserved under taking the expansion or contraction functor.

The paper is organized as follows. In Section 1, we review some preliminaries which are needed in the sequel. In Section 2, we investigate the expansion of some classes of matroids. We show that a matroid is graphic, binary or transversal if and only if its an arbitrary expansion has such a property (see Theorems 2.1, 2.4 and 2.7). Also, we prove that the expansion of an uniform matroid is a partition matroid and, conversely, every partition matroid is an expansion of an uniform matroid (see Theorem 2.9). In Section 3, we introduce the contraction functor which acts in contrast to expansion functor. The last section is devoted to the study of unique exchange property. After recalling some notions and notations from [20], we formulate White’s conjecture [20, Conjecture 12]. As one of the main results, we show that a matroid $M$ satisfies White’s conjecture if and only if an arbitrary expansion of $M$ does (see Theorem 3.6). This concludes that $M$ satisfies White’s conjecture if and only if its contraction does (see Corollary 3.11). On the other hand, since the class of contracted matroids is very smaller than the class of all matroids, it follows from Corollary 3.11 that to test White’s conjecture for a given class of matroids it suffices to turn our attention to their contractions. Finally, we give some classes of matroids which satisfy White’s conjecture.

1 Preliminaries

A matroid $M$ is a pair $(\mathcal{E}_{M},\mathcal{B}_{M})$ consisting of a finite set $\mathcal{E}_{M}$ and a non-empty family $\mathcal{B}_{M}$ of subsets of $\mathcal{E}_{M}$ such that no set in $\mathcal{B}_{M}$ properly contains another set in $\mathcal{B}_{M}$ and, moreover, $\mathcal{B}_{M}$ satisfies the following exchange property:

for every $B_{1},B_{2}\in\mathcal{B}_{M}$ and $x\in B_{1}\backslash B_{2}$ there exists $y\in B_{2}\backslash B_{1}$, such that $(B_{1}\backslash x)\cup y\in\mathcal{B}_{M}$.

$\mathcal{E}_{M}$ and $\mathcal{B}_{M}$ are, respectively, called the ground set and the basis set of $M$. The background from matroid theory which we use may be obtained from [14] or [19].

Recall from [15] the concept of expansion functor on a family of subsets of $[n]=\{x_{1},\ldots,x_{n}\}$. Let $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{N}^{n}$. For $A=\{x_{i_{1}},\ldots,x_{i_{r}}\}\subseteq[n]$, the expansion of $A$ is defined

[TABLE]

Let $\mathcal{A}$ be a family of subsets of $[n]$ and $A=\{x_{i_{1}},\ldots,x_{i_{r}}\}\in\mathcal{A}$. Set $[n]^{\alpha}=\{x_{ij}:1\leq i\leq n,1\leq j\leq k_{i}\}$. The expansion of the singleton family $\{A\}$ with respect to $\alpha$ is denoted by $\{A\}^{\alpha}$ and it is a family of subsets of $[n]^{\alpha}$ defined as follows:

[TABLE]

Also, the expansion of $\mathcal{A}$ with respect to $\alpha$ is denoted by $\mathcal{A}^{\alpha}$ and it is defined

[TABLE]

Let $2^{[n]}$ denote the set of all subsets of $[n]$ and let $\alpha\in\mathbb{N}^{n}$. We define the map $\pi:2^{[n]^{\alpha}}\rightarrow 2^{[n]}$ by setting $\pi(\{x_{i_{1}j_{1}},\ldots,x_{i_{r}j_{r}}\})=\{x_{i_{1}},\ldots,x_{i_{r}}\}$.

The following theorem is a direct consequence of [15, Theorem 1.2]:

Theorem 1.1**.**

Let $M$ be a nonempty family of subsets of $[n]$ and let $\alpha\in\mathbb{N}^{n}$. Then $M$ is a matroid if and only if $M^{\alpha}$ is.

The restriction of a matroid $(\mathcal{E}_{M},\mathcal{B}_{M})$ to $X\subseteq\mathcal{E}_{M}$ is denoted by $M_{X}$ and it is a matroid with the ground set $\mathcal{E}_{M_{X}}=\mathcal{E}_{M}\cap X$ and the basis set

[TABLE]

2 The expansion of some classes of matroids

Let $G=(V(G),E(G))$ be an undirected graph (with possibly loops or parallel edges). A spanning subgraph of a graph $G$ is a subgraph whose vertex set is the entire vertex set of $G$. If this spanning subgraph is a tree, it is called a spanning tree of the graph.

Let $\mathcal{E}=E(G)$ and $\mathcal{B}=\{B\subset E(G):B\ \mbox{is a spanning tree of}\ G\}$. Then $M=(\mathcal{E},\mathcal{B})$ is a matroid called a graphic matroid. The graphic matroid associated with the graph $G$ is denoted by $M(G)$.

Theorem 2.1**.**

Let $M$ be a matroid on $[n]$ and $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{N}^{n}$. Then $M$ is graphic if and only if $M^{\alpha}$ is graphic.

We need some notations and an auxiliary lemma:

For any $1\leq i\leq n$, let $\varepsilon_{i}=(a_{1},\ldots,a_{n})\in\mathbb{N}^{n}$ be defined as a_{j}=\left\{\begin{array}[]{ll}0&\hbox{if}\ j\neq i\\ 1&\hbox{if}\ j=i.\end{array}\right. Set $\mathbf{1}=(1,1,\ldots,1)\in\mathbb{N}^{n}$.

Lemma 2.2**.**

Let $\mathcal{A}$ be a family of subsets of $[n]$, $\beta=(k_{1},\ldots,k_{n})\in\mathbb{N}^{n}$ and $\alpha=\beta+\delta_{i}$. Then $(\mathcal{A}^{\beta})^{\mathbf{1}+\varepsilon_{ik_{i}}}\cong\mathcal{A}^{\alpha}$.

Proof.

Note that

[TABLE]

and

[TABLE]

Define $\varphi:([n]^{\beta})^{\mathbf{1}+\varepsilon_{ik_{i}}}\rightarrow[n]^{\alpha}$ given by

[TABLE]

Then $\varphi$ induces the bijection

[TABLE]

∎

Now we prove Theorem 2.1:

Proof.

Let $G$ be a graph with $E(G)=\{x_{1},\ldots,x_{n}\}$ and $M\cong M(G)$. We use induction on $\alpha$.

First, suppose that $k_{1}=2$ and $k_{i}=1$ for all $i>1$. Let $G^{\prime}$ be a graph with the vertex set $V(G)$ and the edge set $E(G^{\prime})=\{x_{i1},x_{12}:i=1,\ldots,n\}$ where $x_{i1}=x_{i}$ for all $i$ and $x_{11}$ and $e_{12}$ are parallel. It is easy to check that $M(G)^{\alpha}\cong M(G^{\prime})$.

Now, suppose that $k_{1}>1$ and $\beta\in\mathbb{N}^{n}$ with $\beta(i)=k_{i}$ if $2\leq i\leq n$ and $\beta(1)=k_{1}-1$. Assume that $M(G)^{\beta}\cong M(H)$ is graphic. Then it follows from induction hypothesis and Lemma 2.2 that $M(G)^{\alpha}\cong(M(G)^{\beta})^{\varepsilon_{1}}\cong M(G^{\prime})$ where $G^{\prime}$ is obtained from $H$ by adding a parallel edge.

Conversely, suppose $M^{\alpha}$ is graphic and $M^{\alpha}\cong M(G^{\prime})$. Set $X=\{x_{i1}:1\leq i\leq n\}$. Then $(M^{\alpha})_{X}\cong M$ is graphic by [8, page 842]. ∎

Example 2.3*.*

Consider the matroid $M$ on $[6]$ with basis set

[TABLE]

[TABLE]

$M$ is a graphic matroid associated with the graph $G$ shown in Figure 1. Let $\alpha=(1,1,1,1,2,2)\in\mathbb{N}^{6}$. Then $M^{\alpha}$ is a graphic matroid and $M^{\alpha}$ is associated with $G^{\prime}$ (see Figure 1), obtained from $G$ by adding parallel edges to $x_{5}$ and $x_{6}$.

A subset of the ground set of a matroid $M$ that is contained in no bases of $M$ is called dependent. A circuit in $M$ is a minimal dependent subset (with respect to inclusion) of $\mathcal{E}_{M}$ and the set of circuits of $M$ is denoted by $\mathcal{C}(M)$. A matroid $M$ is binary if and only if for every pair of circuits of $M$, their symmetric difference contains another circuit. See [14, Theorem 9.1.2] for other equivalent definitions of binary matroids.

Theorem 2.4**.**

Let $M$ be a matroid on $[n]$ and $\alpha\in\mathbb{N}^{n}$. Then $M$ is binary if and only if $M^{\alpha}$ is.

Proof.

Let $M$ be binary and let $C^{\prime}_{1},C^{\prime}_{2}\in\mathcal{C}(M^{\alpha})$. If $\pi(C^{\prime}_{1})=\pi(C^{\prime}_{2})$ then there exist $x_{ij}\in C^{\prime}_{1}\backslash C^{\prime}_{2}$ and $x_{ij^{\prime}}\in C^{\prime}_{2}\backslash C^{\prime}_{1}$ with $j\neq j^{\prime}$. Set $C^{\prime}=\{x_{ij},x_{ij^{\prime}}\}$. Then $C^{\prime}\subset C^{\prime}_{1}\triangle C^{\prime}_{2}$ and $C^{\prime}\in\mathcal{C}(M^{\alpha})$, and hence the assertion is completed. So suppose that $\pi(C^{\prime}_{1})\neq\pi(C^{\prime}_{2})$.

If $|\pi(C^{\prime}_{1})|=|C^{\prime}_{1}|$ and $|\pi(C^{\prime}_{2})|=|C_{2}|$ then since $\pi(C^{\prime}_{1}),\pi(C^{\prime}_{2})\in\mathcal{C}(M)$ there exists $C\in\mathcal{C}(M)$ such that $B\subseteq\pi(C^{\prime}_{1})\triangle\pi(C^{\prime}_{2})$. Since $\pi(C^{\prime}_{1})\triangle\pi(C^{\prime}_{2})\subseteq\pi(C^{\prime}_{1}\triangle C^{\prime}_{2})$ we have $B\subseteq\pi(C^{\prime}_{1}\triangle C^{\prime}_{2})$ and so it follows that $C^{\prime}\subseteq C^{\prime}_{1}\triangle C^{\prime}_{2}$ for some $C^{\prime}\in\mathcal{C}(\mathcal{M}^{\alpha})$ with $\pi(C^{\prime})=C$.

Consider $|\pi(C^{\prime}_{1})|\neq|C^{\prime}_{1}|$ or $|\pi(C^{\prime}_{2})|\neq|C^{\prime}_{2}|$. Let, for example, $|\pi(C^{\prime}_{1})|\neq|C^{\prime}_{1}|$. Then $\{x_{il},x_{im}\}\subset C^{\prime}_{1}$ for some $l,m$. Since $\{x_{il},x_{im}\}\in\mathcal{C}(M^{\alpha})$, thus it should be $C^{\prime}_{1}=\{x_{il},x_{im}\}$. Consider the following cases:

• •

$|\pi(C^{\prime}_{2})|\neq|C^{\prime}_{2}|$: Then $C^{\prime}_{2}=\{x_{jl^{\prime}},x_{jm^{\prime}}\}$ for some $j,l^{\prime}$ and $m^{\prime}$. Since $\pi(C^{\prime}_{1})\neq\pi(C^{\prime}_{2})$ we have $i\neq j$. Set $C^{\prime}=\{x_{il},x_{im}\}$.

• •

$|\pi(C^{\prime}_{2})|=|C^{\prime}_{2}|$: Then $C^{\prime}_{2}=\{x_{i_{1}j_{1}},\ldots,x_{i_{r}j_{r}}\}$ and $C^{\prime}_{1}\triangle C^{\prime}_{2}=\{x_{i_{1}j_{1}},\ldots,x_{i_{r}j_{r}},x_{il},x_{im}\}$ or $C^{\prime}_{1}\triangle C^{\prime}_{2}=\{x_{i_{1}j_{1}},\ldots,x_{i_{r-1}j_{r-1}},x_{il}\}$ where $x_{i_{r}j_{r}}=x_{im}$. At the first case, set $C^{\prime}=\{x_{il},x_{im}\}$ and at the second one, set $C^{\prime}=\{x_{i_{1}j_{1}},\ldots,x_{i_{r-1}j_{r-1}},x_{il}\}$.

Thus $C^{\prime}\in\mathcal{C}(M^{\alpha})$ and $C^{\prime}\subset C^{\prime}_{1}\triangle C^{\prime}_{2}$. Therefore $M^{\alpha}$ is binary.

Conversely, suppose that $M^{\alpha}$ is binary and $C_{1},C_{2}\in\mathcal{C}(M)$. Let $C_{1}=\{x_{i_{1}},\ldots,x_{i_{r}}\}$ and $C_{2}=\{x_{j_{1}},\ldots,x_{j_{s}}\}$. Set $C^{\prime}_{1}=\{x_{i_{1}1},\ldots,x_{i_{r}1}\}$ and $C^{\prime}_{2}=\{x_{j_{1}1},\ldots,x_{j_{s}1}\}$. Then $C^{\prime}_{1},C^{\prime}_{2}\in\mathcal{C}(M^{\alpha})$ and so there exists $C^{\prime}\in\mathcal{C}(M^{\alpha})$ with $C^{\prime}\subset C^{\prime}_{1}\triangle C^{\prime}_{2}$. It follows that $\pi(C^{\prime})\in\mathcal{C}(M)$ and $\pi(C^{\prime})\subset C_{1}\triangle C_{2}$. Therefore $M$ is binary. ∎

We recall from [14, page 46.] the definition of a transversal matroid. Let $S\subset[n]$. A set system $(S,\mathcal{A})$ is a set $S$ along with a multiset $\mathcal{A}=(A_{j}:j\in J)$ of (not necessarily distinct) subsets of $S$. If $J=\{1,\ldots,m\}$ then we may denote $\mathcal{A}$ by $\mathcal{A}=(A_{1},\ldots,A_{m})$. A transversal of $\mathcal{A}=(A_{j}:j\in J)$ is a subset $T$ of $S$ for which there is a bijection $\varphi:J\rightarrow T$ with $\varphi(j)=A_{j}$ for all $j\in J$.

Theorem 2.5**.**

[6]** A finite set system $(S,(A_{j}:j\in J))$ has a transversal if and only if, for all $K\subset J$,

[TABLE]

If $X\subset S$, then $X$ is a partial transversal of $\mathcal{A}=(A_{j}:j\in J)$ if, for some subset $K$ of $J$, $X$ is a transversal of $\mathcal{A}$. The partial transversals of a $\mathcal{A}$ are the independent sets of a matroid. We call such a matroid a transversal matroid and denote it by $M[\mathcal{A}]$. $\mathcal{A}$ is called a presentation of $M[\mathcal{A}]$.

Theorem 2.6**.**

[3]** Let $M$ be a transversal matroid on $[n]$. Then so is $M_{X}$ for each $X\subset[n]$. If $(A_{1},\ldots,A_{m})$ is a presentation of $M$, then $(A_{1}\cap X,\ldots,A_{m}\cap X)$ is a presentation of $M_{X}$.

If $\mathcal{A}=(A_{j}:j\in J)$ is a family of subsets of $S\subset[n]$ then the bipartite graph associated with $\mathcal{A}$, denoted by $G[\mathcal{A}]$, has the vertex set $S\cup\{A_{j}:j\in J\}$ and the edge set $\{x_{i}A_{j}:j\in J\ \mbox{and}\ x_{i}\in A_{j}\}$.

A matching in a graph $G$ is the set of edges in $G$ no two of which have a common endpoint. A subset $X$ of $S$ is a partial transversal of $\mathcal{A}$ if and only if there is a matching in $G[\mathcal{A}]$ which every edge has one endpoint in $X$.

Theorem 2.7**.**

Let $M$ be a matroid on $[n]$ and let $\alpha\in\mathbb{N}^{n}$. Then $M$ is transversal if and only if $M^{\alpha}$ is.

Proof.

“Only if part”: Let $M$ be a transversal matroid with $M\cong M[\mathcal{A}]$ where $\mathcal{A}=(A_{1},A_{2},\ldots,A_{m})$ and $A_{i}\subset[n]$. Set $\mathcal{A}^{\alpha}=(A^{\alpha}_{1},A^{\alpha}_{2},\ldots,A^{\alpha}_{m})$. We claim that $\mathcal{A}^{\alpha}$ is a presentation of $M^{\alpha}$. We associate to $A_{j}$ and $A^{\alpha}_{j}$, respectively, the vertices $y_{j}$ and $y^{\prime}_{j}$.

Let $B^{\prime}\in\mathcal{B}_{M[\mathcal{A}]^{\alpha}}$. We may assume that $B^{\prime}=\{x_{1i_{1}},\ldots,x_{ri_{r}}\}$. So there exists the maximal matching $\{x_{1}y_{j_{1}},\ldots,x_{r}y_{j_{r}}\}$ in the bipartite graph $G[\mathcal{A}]$ with the partition $[n]\dot{\cup}\{y_{1},\ldots,y_{m}\}$. It is clear that $B^{\prime\prime}=\{x_{1i_{1}}y^{\prime}_{j_{1}},\ldots,x_{ri_{r}}y^{\prime}_{j_{r}}\}$ is a matching in $G[\mathcal{A}^{\alpha}]$. Suppose, on the contrary, that $B^{\prime\prime}$ is not maximal. So for some matching $C$ in $G[\mathcal{A}^{\alpha}]$ we have $B^{\prime\prime}\subset C$. Let $x_{st}y^{\prime}_{l}\in C\backslash B^{\prime\prime}$. Since $\{x_{1}y_{j_{1}},\ldots,x_{r}y_{j_{r}}\}$ is maximal, we have $s\in\{1,\ldots,r\}$. Moreover, it is clear that $l\not\in\{j_{1},\ldots,j_{r}\}$. Let $s=1$ and let $X=\{x_{1},\ldots,x_{r}\}$. By Theorem 2.6, $M_{X}$ is transversal with presentation $(A_{1}\cap X,\ldots,A_{m}\cap X)$. But $|\cup_{k\in\{j_{1},\ldots,j_{r},l\}}(A_{k}\cap X)|=r<r+1=|\{j_{1},\ldots,j_{r},l\}|$. This contradicts Theorem 2.5. Therefore $B^{\prime\prime}$ is a maximal matching in $G[\mathcal{A}^{\alpha}]$ and so $B^{\prime}\in\mathcal{B}_{M[\mathcal{A}^{\alpha}]}$.

In a similar argument we show that $\mathcal{B}_{M[\mathcal{A}^{\alpha}]}\subseteq\mathcal{B}_{M[\mathcal{A}]^{\alpha}}$. Therefore $M[\mathcal{A}]^{\alpha}\cong M[\mathcal{A}^{\alpha}]$, as desired.

“If part”: Let $M^{\alpha}$ be transversal with the presentation $\mathcal{B}=(B_{1},\ldots,B_{m})$. One may suppose that $[n]:=\{x_{i1}:1\leq i\leq n\}$. By Theorem 2.6, $(M^{\alpha})_{[n]}$ is a transversal matroid with the presentation $\mathcal{B}^{\prime}=(B_{1}\cap[n],\ldots,B_{m}\cap[n])$. Since $(M^{\alpha})_{[n]}=M$, the assertion is completed.

∎

Example 2.8*.*

Let $S=\{x_{1},x_{2},x_{3},x_{4}\}\subset[n]$ and $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{N}^{n}$. Let $\mathcal{A}=(\{x_{1},x_{2}\},\{x_{1},x_{3}\},\{x_{3}\})$. Then the matchings in $G[\mathcal{A}]$ are

[TABLE]

and so

[TABLE]

$G[\mathcal{A}]$ and $G[\mathcal{A}^{\alpha}]$ are shown in Figure 2.

A matroid on $[n]$ of rank $t\leq n$ is an uniform matroid if all $t$-element subsets of $[n]$ are bases and it is denoted by $U_{t,n}$.

A partition matroid of rank $t$ [10] is a matroid $M(\mathcal{P})$ associated with a partition $\mathcal{P}=\{A_{1},\ldots,A_{m}\}$ of $\mathcal{E}_{M(\mathcal{P})}$ and the basis set

[TABLE]

Theorem 2.9**.**

The expansion of every uniform matroid is a partition matroid. Conversely, every partition matroid is the expansion of an uniform matroid.

Proof.

Let $U_{t,n}$ be an uniform matroid on $[n]$ of rank $t$ and let $\alpha=(k_{1},\ldots,k_{n})\in\mathbb{N}^{n}$. Set $A_{i}=\{x_{i1},\ldots,x_{ik_{i}}\}$ for all $i$. Then it is easy to see that $U^{\alpha}_{t,n}=M(\mathcal{P})$ where $\mathcal{P}=\{A_{1},\ldots,A_{n}\}$.

Conversely, let $M(\mathcal{P})$ be a partition matroid of rank $t$ with $\mathcal{P}=\{A_{1},\ldots,A_{m}\}$. Set $A^{\prime}_{i}:=\{x_{i}\}$ for all $i$. Then $M(\mathcal{P})\cong U^{\alpha}_{t,m}$ where $\alpha=(k_{1},\ldots,k_{m})$ and $k_{i}=|A_{i}|$ for all $i$. ∎

3 The contraction functor

Definition 3.1**.**

Let $\mathcal{A}$ be a family of subsets of $[n]$ and let $\mathcal{B}$ denote the maximal elements of $\mathcal{A}$ (with respect to inclusion). We define the relation $``\sim$” on $[n]$ in the following form:

[TABLE]

In other words,

[TABLE]

It is easily shown that $\sim$ is an equivalence relation. Let $[m]=\{y_{1},\ldots,y_{m}\}$ be the set of equivalence classes under $\sim$.

Let $y_{i}=\{x_{i1},\ldots,x_{ia_{i}}\}$. Set $\alpha=(a_{1},\ldots,a_{m})$. For $A\in\mathcal{B}$, define $\overline{A}=\{y_{i}:y_{i}\cap A\neq\emptyset\}$ and $\overline{\mathcal{A}}$ a family of subsets of $[m]$ with the set $\{\overline{A}:A\in\mathcal{B}\}$ of maximal elements. We call $\overline{\mathcal{A}}$ the contraction of $\mathcal{A}$ by $\alpha$. Clearly, every family of subsets of $[n]$ has an unique contraction.

A family $\mathcal{A}$ of subsets of $[n]$ is called contracted if $\mathcal{A}$ and $\overline{\mathcal{A}}$ coincide up to a relabeling.

Remark 3.2*.*

Note that the contraction functor behaves exactly the opposite to expansion functor. Actually, if $\mathcal{A}$ is a family of subsets of $[n]$ and $\overline{\mathcal{A}}$ is the contraction of $\mathcal{A}$ by $\alpha$, then $(\overline{\mathcal{A}})^{\alpha}$ and $\mathcal{A}$ coincide up to a relabeling of $[n]$. Also, for every $\alpha\in\mathbb{N}^{n}$, two families $\overline{\mathcal{A}^{\alpha}}$ and $\overline{\mathcal{A}}$ coincide. Therefore $\mathcal{A}$ is a matroid if and only if $(\overline{\mathcal{A}})^{\alpha}$ is a matroid. Equivalently, by Theorem 1.1, $\overline{\mathcal{A}}$ is a matroid.

Remark 3.3*.*

All of uniform matroids of rank $t>1$ are contracted.

Corollary 3.4**.**

The contraction of a partition matroid is an uniform matroid.

Proof.

Let $M$ be a partition matroid on $[n]$. By Theorem 2.9, $M$ is the expansion of an uniform matroid $U_{t,n}$ with respect to some $\alpha\in\mathbb{N}^{n}$. It follows from Remarks 3.2 and 3.3 that $\overline{M}=\overline{U^{\alpha}_{t,n}}=\overline{U_{t,n}}=U_{t,n}$. ∎

In view of Theorems 2.1, 2.4, 2.7 and 2.9 we have the following:

Corollary 3.5**.**

The contraction of a graphic (resp. binary, transversal) matroid is graphic (resp. binary, transversal).

4. Unique exchange property for bases with a view towards White’s conjecture

In this section, we investigate the preservation of White’s conjecture under taking the expansion and contraction functors. First, we recall some notions and notations from [20].

Let $M$ be a matroid. Two sequences of bases $(A_{1},\ldots,A_{m})$ and $(B_{1},\ldots,B_{m})$ are compatible if $A_{1}\cup\ldots\cup A_{m}=B_{1}\cup\ldots\cup B_{m}$. Let $(A_{1},\ldots,A_{m})$ be a sequence of bases of a matroid $M$ and let $1\leq r<s\leq m$ and $x\in A_{r}$. Set

[TABLE]

Let $y\in E(x;A_{r},A_{s})$. Then we say

[TABLE]

is obtained from $(A_{1},\ldots,A_{m})$ by a symmetric exchange. For two sequences of bases of a matroid $M$, $(A_{1},\ldots,A_{m})\sim^{1}_{M}(B_{1},\ldots,B_{m})$ means that $(B_{1},\ldots,B_{m})$ is obtained from $(A_{1},\ldots,A_{m})$ by a symmetric exchange. Also, $(A_{1},\ldots,A_{m})\sim^{2}_{M}(B_{1},\ldots,B_{m})$ implies to $(B_{1},\ldots,B_{m})$ is obtained from $(A_{1},\ldots,A_{m})$ by a symmetric exchange or a permutation of the order of the bases.

If $U\subset A_{r}$, we let

[TABLE]

Let $V\in E(U;A_{r},A_{s})$. Then we say $(A_{1},\ldots,A_{r-1},(A_{r}\backslash U)\cup V,A_{r+1},\ldots,A_{s-1},(A_{s}\backslash V)\cup U,A_{s+1},\ldots,A_{m})$ is obtained from $(A_{1},\ldots,A_{m})$ by a symmetric subset exchange. We write $(A_{1},\ldots,A_{m})\sim^{3}_{M}(B_{1},\ldots,B_{m})$ if $(B_{1},\ldots,B_{m})$ is obtained from $(A_{1},\ldots,A_{m})$ by a symmetric subset exchange.

For $i=1,2,3$, set $\mathrm{TE}(i)$ the class of matroids with the property that for every matroid $M\in\mathrm{TE}(i)$ and every two compatible sequences $(A_{1},\ldots,A_{m})$ and $(B_{1},\ldots,B_{m})$ of bases of $M$, there exist the sequences $(C_{j1},\ldots,C_{jm})$, for $j=1,\ldots,t$, of bases of $M$ such that

[TABLE]

It is easy to check that $\mathrm{TE}(1)\subseteq\mathrm{TE}(2)\subseteq\mathrm{TE}(3)$. White conjectured that theses classes are equal to the class of all matroids [20, Conjecture 12]. We will say that a matroid $M$ satisfies White’s conjecture if $M\in\mathrm{TE}(i)$, for all $i=1,2,3$.

Theorem 3.6**.**

For a matroid $M$ on $[n]$, $\alpha\in\mathbb{N}^{n}$ and $i=1,2,3$ we have

[TABLE]

Before proving Theorem 3.6, we present some auxiliary results.

Lemma 3.7**.**

Let $M$ be a matroid on $[n]$ and $\alpha\in\mathbb{N}^{n}$.

1. (i)

For any two compatible sequences $(A_{1},\ldots,A_{m})$ and $(B_{1},\ldots,B_{m})$ of bases of $M$, there exist two compatible sequences $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ and $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ of bases of $M^{\alpha}$ such that $\pi(A^{\prime}_{i})=A_{i}$ and $\pi(B^{\prime}_{i})=B_{i}$; 2. (ii)

If $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ and $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ are two compatible sequences of bases of $M^{\alpha}$ then $(\pi(A^{\prime}_{1}),\ldots,\pi(A^{\prime}_{m}))$ and $(\pi(B^{\prime}_{1}),\ldots,\pi(B^{\prime}_{m}))$ are two compatible sequences of bases of $M$.

Lemma 3.8**.**

Let $M$ be a matroid on $[n]$ and let $\alpha\in\mathbb{N}^{n}$. For $B^{\prime}_{1},B^{\prime}_{2}\in\mathcal{B}_{M^{\alpha}}$, if $(B^{\prime}_{1},B^{\prime}_{2})$ is obtained from $(A^{\prime}_{1},A^{\prime}_{2})$ by a symmetric subset exchange then

1. (i)

$\pi(B^{\prime}_{i})=\pi(A^{\prime}_{i})$* for $i=1,2$ or* 2. (ii)

$(\pi(B^{\prime}_{1}),\pi(B^{\prime}_{2}))$* is obtained from $(\pi(A^{\prime}_{1}),\pi(A^{\prime}_{2}))$ by a symmetric subset exchange.*

Lemma 3.9**.**

Let $M$ be a matroid on $[n]$ and let $\alpha\in\mathbb{N}^{n}$. Suppose that $(B_{1},B_{2})$ is obtained from $(A_{1},A_{2})$ by a symmetric subset exchange. Let $(A^{\prime}_{1},A^{\prime}_{2})$ and $(B^{\prime}_{1},B^{\prime}_{2})$ be two compatible sequences of bases of $M^{\alpha}$ with $\pi(A^{\prime}_{i})=A_{i}$, $\pi(B^{\prime}_{i})=B_{i}$. Then $(B^{\prime}_{1},B^{\prime}_{2})$ is obtained from $(A^{\prime}_{1},A^{\prime}_{2})$ by a symmetric subset exchange.

Now we prove Theorem 3.6 in three parts:

Membership in $\mathrm{TE}(3)$:

Assume that $M\in\mathrm{TE}(3)$ and let $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ and $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ be two compatible sequences of bases of $M^{\alpha}$. Let $A_{i}=\pi(A^{\prime}_{i})$ and $B_{i}=\pi(B^{\prime}_{i})$ for $i=1,\ldots,m$. By Lemma 3.7(ii), $(A_{1},\ldots,A_{m})$ and $(B_{1},\ldots,B_{m})$ are compatible and, by the assumption, $(B_{1},\ldots,B_{m})$ is obtained from $(A_{1},\ldots,A_{m})$ by a composition of symmetric subset exchanges. It follows from Lemma 3.9 that $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ is obtained from $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ by a composition of symmetric subset exchanges. Therefore $M^{\alpha}\in\mathrm{TE}(3)$.

Similarly, the converse direction follows from Lemmas 3.7(i) and 3.8.

Membership in $\mathrm{TE}(2)$:

Let $M\in\mathrm{TE}(2)$. Let $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ and $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ be two compatible sequences of $M^{\alpha}$. By Lemma 3.7(ii), $(\pi(A^{\prime}_{1}),\ldots,\pi(A^{\prime}_{m}))$ and $(\pi(B^{\prime}_{1}),\ldots,\pi(B^{\prime}_{m}))$ are two compatible sequences of bases of $M$ and so, by the assumption, $(\pi(B^{\prime}_{1}),\ldots,\pi(B^{\prime}_{m}))$ is obtained from $(\pi(A^{\prime}_{1}),\ldots,\pi(A^{\prime}_{m}))$ by a composition of symmetric exchanges and permutations of the order of the bases. It follows from Lemma 3.9 that $(B^{\prime}_{1},\ldots,B^{\prime}_{m})$ is obtained from $(A^{\prime}_{1},\ldots,A^{\prime}_{m})$ by a composition of symmetric exchanges and permutations of the order of the bases.

The converse direction obtains in a similar argument by using Lemmas 3.7(i) and 3.8.

Membership in $\mathrm{TE}(1)$:

Lemma 3.10**.**

([11]) Let $M$ be a matroid. Then $M\in\mathrm{TE}(1)$ if and only if $M\in\mathrm{TE}(2)$ and any pair $(A_{2},A_{1})$ of bases of $M$ is obtained from $(A_{1},A_{2})$ by a composition of symmetric exchanges.

Since $M\in\mathrm{TE}(2)$ if and only if $M^{\alpha}\in\mathrm{TE}(2)$, it suffices to show that any pair $(A_{2},A_{1})$ of bases of $M$ is obtained from $(A_{1},A_{2})$ by a composition of symmetric exchanges if and only if any pair of bases of $M^{\alpha}$ has this property.

Suppose that any pair $(A_{2},A_{1})$ of bases of $M$ is obtained from $(A_{1},A_{2})$ by a composition of symmetric exchanges. Consider a pair $(B^{\prime}_{2},B^{\prime}_{1})$ of bases of $M^{\alpha}$. Let $B_{i}=\pi(B^{\prime}_{i})$. By the assumption, $(B_{2},B_{1})$ is obtained from $(B_{1},B_{2})$ by a composition of symmetric exchanges. Therefore

[TABLE]

By Lemma 3.7(i), one can choose the bases $C^{\prime}_{ij}\in\mathcal{B}_{M^{\alpha}}$ with $\pi(C^{\prime}_{ij})=C_{ij}$ such that $(B^{\prime}_{1},B^{\prime}_{2}),(C^{\prime}_{11},C^{\prime}_{12}),\ldots,(C^{\prime}_{t1},C^{\prime}_{t2})$ are pairwise compatible. It follows from Lemma 3.9 that $(B^{\prime}_{2},B^{\prime}_{1})$ is obtained from $(B^{\prime}_{1},B^{\prime}_{2})$ by a composition of symmetric exchanges.

The converse follows from Lemmas 3.7(ii) and 3.8 in a similar argument.

Corollary 3.11**.**

Let $M$ be a matroid on $[n]$ and let $\alpha\in\mathbb{N}^{n}$. Then $M$ satisfies White’s conjecture if and only if the contraction of $M$ does.

Remark 3.12*.*

Note that the class of contracted matroids is very smaller than the class of all matroids. It follows from Corollary 3.11 that to test White’s conjecture for a class of matroids it suffices to turn our attention to their contractions.

Corollary 3.13**.**

Every partition matroid satisfies White’s conjecture.

Proof.

Let $M$ be a partition matroid. By Corollary 3.4, the contraction of $M$ is an uniform matroid. In view of Corollary 3.11, if we show that $U_{t,n}\in\mathrm{TE}(1)$ then the assertion is completed. It was shown in [18] that the toric ideal of any uniform matroid is generated by quadratic binomials corresponding to symmetric exchanges and this is an algebraic meaning of the property $\mathrm{TE}(2)$. Hence $U_{t,n}\in\mathrm{TE}(2)$. Now suppose that $(A_{2},A_{1})$ is a pair of bases of $U_{t,n}$. Let $A_{1}=\{x_{i_{1}},\ldots,x_{i_{t}}\}$ and $A_{2}=\{x_{j_{1}},\ldots,x_{j_{t}}\}$. Let $i_{s+1}=j_{s+1},\ldots,i_{t}=j_{t}$ and $i_{p}\neq j_{q}$ for $p,q\leq s$. Then

$(A_{1},A_{2})\sim^{1}_{U_{t,n}}(\{x_{j_{1}},x_{i_{2}},\ldots,x_{i_{t}}\},\{x_{i_{1}},x_{j_{2}},\ldots,x_{j_{t}}\})\sim^{1}_{U_{t,n}}\ldots$

$\sim^{1}_{U_{t,n}}(\{x_{j_{1}},x_{j_{2}},\ldots,x_{j_{s-1}},x_{i_{s}},\ldots,x_{i_{t}}\},\{x_{i_{1}},x_{j_{2}},\ldots,x_{i_{s-1}},x_{j_{s}},\ldots,x_{j_{t}}\})\sim^{1}_{U_{t,n}}(A_{2},A_{1}).$

It follows from Lemma 3.10 that $M\in\mathrm{TE}(1)$, as desired. ∎

Let $M_{1}$ and $M_{2}$ be matroids on disjoint ground sets. The direct sum of $M_{1}$ and $M_{2}$ is denoted by $M_{1}\oplus M_{2}$ and it is a matroid, by [14, Proposition 4.2.12], on $\mathcal{E}_{M_{1}}\cup\mathcal{E}_{M_{2}}$ with the basis set

[TABLE]

Lemma 3.14**.**

Let $M_{1}$ and $M_{2}$ be matroids on disjoint ground sets. If $M_{1}$ and $M_{2}$ satisfy White’s conjecture then $M_{1}\oplus M_{2}$ does, too.

The proofs of above lemma is easy and we leave them to the reader.

Remark 3.15*.*

(i) Let $M$ be a matroid on $[n]=\{x_{1},\ldots,x_{n}\}$ and let $z$ be disjoint from $x_{i}$’s. Then $M$ satisfies White’s conjecture if and only if the matroid $N$ with the basis set $\{A\cup z:A\in\mathcal{B}_{M}\}$ does.

(ii) Let $M$ be a matroid on $[n]$. Consider $\{z_{ij}:i=1,\ldots,r,j=1,\ldots,s\}$ a set disjoint from $[n]$. Let $t\leq r,s$. Then the set

[TABLE]

is the basis set of a matroid $N$. In fact, $N=M\oplus M(\mathcal{P})$ where $\mathcal{P}=\{A_{1},\ldots,A_{r}\}$ and $A_{i}=\{z_{i1},\ldots,z_{is}\}$ for all $i$. Especially, if $M$ satisfies White’s conjecture then it follows from Lemma 3.14 and Corollary 3.13 that $N$ satisfies White’s conjecture, too.

Combining Theorem 2.4 with Theorem 3.6 we obtain

Corollary 3.16**.**

A binary matroid satisfies White’s conjecture if and only if its contraction does.

Example 3.17*.*

Consider the matroid $M$ on $[7]$ with the basis set

[TABLE]

[TABLE]

[TABLE]

It is easy to check that $M$ is binary. To see that $M$ satisfies White’s conjecture, we first contract $M$. The contraction of $M$ is a binary matroid on $\{x_{1},x_{2},x_{3},x_{5},x_{7}\}$ with the basis set

[TABLE]

On the other hand, $\mathcal{B}_{\overline{M}}$ can be obtained by adding $x_{2}$ to all bases of the uniform matroid $U_{3,4}$ with the ground set $\{x_{1},x_{3},x_{5},x_{7}\}$. It follows from Remark 3.15(i) that $\overline{M}$ satisfies White’s conjecture and so $M$ satisfies White’s conjecture, too.