A notion of minor-based matroid connectivity

TL;DR

This paper characterizes specific matroids based on their connectivity properties, identifying unique conditions under which matroids maintain minors after element deletion or contraction, and linking connectivity to clonal classes.

Contribution

It establishes that only $U_{1,2}$ and $M( ext{W}_2)$ are special connected matroids with certain minor-preservation properties, and relates $U_{0,1} igoplus U_{1,1}$-connectivity to trivial clonal classes.

Findings

$U_{1,2}$ is the only connected matroid with minor-preservation under deletion/contraction.

$U_{1,2}$ and $M( ext{W}_2)$ are the only connected matroids with a specific minor-intersection property.

$U_{0,1} igoplus U_{1,1}$-connectivity characterizes trivial clonal classes.

Abstract

For a matroid $N$, a matroid $M$ is $N$-connected if every two elements of $M$ are in an $N$-minor together. Thus a matroid is connected if and only if it is $U_{1,2}$-connected. This paper proves that $U_{1,2}$ is the only connected matroid $N$ such that if $M$ is $N$-connected with $|E(M)| > |E(N)|$, then $M \backslash e$ or $M / e$ is $N$-connected for all elements $e$. Moreover, we show that $U_{1,2}$ and $M(\mathcal{W}_2)$ are the only connected matroids $N$ such that, whenever a matroid has an $N$-minor using $\{e,f\}$ and an $N$-minor using $\{f,g\}$, it also has an $N$-minor using $\{e,g\}$. Finally, we show that $M$ is $U_{0,1} \oplus U_{1,1}$-connected if and only if every clonal class of $M$ is trivial.

Full text

1. Introduction

Our terminology follows Oxley [8]. We say that a matroid $M$ uses an element $e$ or a set $Z$ of elements if $e\in E(M)$ or $Z\subseteq E(M)$. Let $N$ be a matroid. A matroid $M$ with $|E(M)|\geq 2$ is $N$-connected if, for every pair of distinct elements $e,f$ of $E(M)$, there is a minor of $M$ that is isomorphic to $N$ and uses $\{e,f\}$.

We will assume, unless otherwise stated, that the matroids discussed here have at least two elements. Note that $U_{1,2}$-connectivity coincides with the usual notion of connectivity for matroids. Hence, relying on a well-known inductive property of matroid connectivity [13], we have that if $M$ is $U_{1,2}$-connected, $e\in E(M)$, and $|E(M)|\geq 3$, then $M\backslash e$ or $M/e$ is $U_{1,2}$-connected. Our first theorem shows that $U_{1,2}$ is the only connected matroid with this property.

Theorem 1.1**.**

Let $N$ be a matroid. If, for every $N$-connected matroid $M$ with $|E(M)|>|E(N)|$ and, for every $e$ in $E(M)$, at least one of $M\backslash e$ or $M/e$ is $N$-connected, then $N$ is isomorphic to one of $U_{1,2}$, $U_{0,2}$, or $U_{2,2}$.

One attractive property of matroid connectivity is that elements can be assigned to components. We say that a matroid $N$ has the transitivity property if, for every matroid $M$ and every triple $\{e,f,g\}\subseteq E(M)$, if $e$ is in an $N$-minor with $f$, and $f$ is in an $N$-minor with $g$, then $e$ is in an $N$-minor with $g$. Let $M(\mathcal{W}_{2})$ be the rank-$2$ wheel. In Section 6, we prove the following result.

Theorem 1.2**.**

The only matroids with the transitivity property are $U_{1,2}$ and $M(\mathcal{W}_{2})$.

On combining the last two theorems, we get the following result, which indicates how special the usual matroid connectivity is.

Corollary 1.3**.**

Let $N$ be a matroid with the transitivity property such that whenever $M$ is an $N$-connected matroid, $e\in E(M)$, and $|E(M)|>|E(N)|$, at least one of $M\backslash e$ and $M/e$ is $N$-connected. Then $N\cong U_{1,2}$.

The concept of $N$-connectivity can also convey interesting information when $N$ is disconnected, as the next result indicates.

Theorem 1.4**.**

A matroid $M$ is $U_{0,1}\oplus U_{1,1}$-connected if and only if every clonal class of $M$ is trivial.

The paper is structured as follows. In the next section, we recall Cunningham and Edmonds’s decomposition theorem for connected matroids that are not $3$-connected, which is a basic tool in our proofs. Sections 3, 4, and 5 treat the cases of $N$-connected matroids when $N$ is $3$-connected, connected, and disconnected, respectively. In particular, we prove Theorems 1.1, and 1.2 in Section 6 and Theorem 1.4 in Section 5. Finally, in Section 7, we consider what can be said when every set of three elements occurs in some minor. Moss [6] showed that $3$-connected matroids can be characterized as those in which every set of four elements is contained in a minor isomorphic to a member of $\{\mathcal{W}^{2},\mathcal{W}^{3},\mathcal{W}^{4},M(\mathcal{W}_{3}),M(\mathcal{W}_{4}),Q_{6}\}$.

2. Preliminaries

The concept of $N$-connectivity is closely related to roundedness, which is exemplified by Bixby’s [1] result that if $e$ is an element of a $2$-connected non-binary matroid $M$, then $M$ has a $U_{2,4}$-minor using $e$. Formally, let $t$ be a positive integer and let $\mathcal{N}$ be a class of matroids. A matroid $M$ has an $\mathcal{N}$-minor if $M$ has a minor isomorphic to a member of $\mathcal{N}$. Seymour [11] defined $\mathcal{N}$ to be $t$-rounded if, for every $(t+1)$-connected matroid $M$ with an $\mathcal{N}$-minor and every subset $X$ of $E(M)$ with at most $t$ elements, $M$ has an $\mathcal{N}$-minor using $X$. Thus Bixby’s result shows that $\{U_{2,4}\}$ is $1$-rounded. Seymour [10] extended this result as follows.

Theorem 2.1**.**

Let $M$ be a $3$-connected matroid having a $U_{2,4}$-minor, and let $e$ and $f$ be elements of $M$. Then $M$ has a $U_{2,4}$-minor using $\{e,f\}$.

The connectivity function $\lambda_{M}$ of a matroid $M$ is defined for every subset $X$ of $E(M)$ by $\lambda_{M}(X)=r(X)+r(E(M)-X)-r(M)$; equivalently, $\lambda_{M}(X)=r(X)+r^{*}(X)-|X|.$ For disjoint subsets $A,B$ of $E(M)$, define $\kappa_{M}(A,B)=\min\{\lambda_{M}(X):A\subseteq X\subseteq E(M)-B\}$.

Lemma 2.2**.**

If $N$ is a minor of $M$ and $A,B$ are disjoint subsets of $E(N)$, then $\kappa_{N}(A,B)\leq\kappa_{M}(A,B)$.

Next we give a brief outline of Cunningham and Edmonds’s decomposition [4] of matroids that are $2$-connected but not $3$-connected. More complete details can be found in [8, Section 8.3]. First recall that when $(X,Y)$ is a $2$-separation of a connected matroid $M$, we can write $M$ as $M_{X}\oplus_{2}M_{Y}$ where $M_{X}$ and $M_{Y}$ have ground sets $X\cup p$ and $Y\cup p$. A matroid-labeled tree is a tree $T$ with vertex set $\{M_{1},M_{2},\ldots,M_{n}\}$ such that each $M_{i}$ is a matroid and, for distinct vertices $M_{j}$ and $M_{k}$, the sets $E(M_{j})$ and $E(M_{k})$ are disjoint if $M_{j}$ and $M_{k}$ are non-adjacent, whereas if $M_{j}$ and $M_{k}$ are joined by an edge $e$, then $E(M_{j})\cap E(M_{k})=\{e\}$, and $\{e\}$ is not a separator in either $M_{j}$ or $M_{k}$.

When $f$ is an edge of a matroid-labeled tree $T$ joining vertices $M_{i}$ and $M_{j}$, if we contract the edge $f$, we obtain a new matroid-labeled tree $T/f$ by relabeling the composite vertex that results from this contraction as $M_{i}\oplus_{2}M_{j}$, with every other vertex retaining its original label.

A tree decomposition of a $2$-connected matroid $M$ is a matroid-labeled tree $T$ such that if $V(T)=\{M_{1},M_{2},\ldots,M_{n}\}$ and $E(T)=\{e_{1},e_{2},\ldots,e_{n-1}\}$, then

1. (i)

$E(M)=(E(M_{1})\cup E(M_{2})\cup\cdots\cup E(M_{n}))-\{e_{1},e_{2},\ldots,e_{n-1}\}$; 2. (ii)

$|E(M_{i})|\geq 3$ for all $i$ unless $|E(M)|<3$, in which case, $n=1$ and $M_{1}=M$; and 3. (iii)

the label of the single vertex of $T/\{e_{1},e_{2},\ldots,e_{n-1}\}$ is $M$.

We call the members of $\{e_{1},e_{2},\ldots,e_{n-1}\}$ basepoints since each member of this set is the basepoint of a $2$-sum when we construct $M$. Cunningham and Edmonds (in [4]) proved the following (see also [8, Theorem 8.3.10]).

Theorem 2.3**.**

Let $M$ be a $2$-connected matroid. Then $M$ has a tree decomposition $T$ in which every vertex label that is not a circuit or a cocircuit is $3$-connected, and there are no adjacent vertices that are both labeled by circuits or are both labeled by cocircuits. Moreover, $T$ is unique up to relabeling of its edges.

The tree decomposition $T$ whose existence is guaranteed by the last theorem is called the canonical tree decomposition of $M$. Although circuits and cocircuits with at most three elements are $3$-connected matroids, when we refer to a $3$-connected vertex, we shall mean one with at least four elements. Clearly, for each edge $p$ of $T$, the graph $T\backslash p$ has two components. Thus $p$ induces a partition of $V(T)$ and a corresponding partition $(X_{p},Y_{p})$ of $E(M)$. The latter partition is a $2$-separation of $M$; we say that it is displayed by the edge $p$. Moreover, $M=M_{X_{p}}\oplus_{2}M_{Y_{p}}$ where $M_{X_{p}}$ and $M_{Y_{p}}$ have ground sets $X_{p}\cup p$ and $Y_{p}\cup p$, respectively. We shall refer to this 2-sum decomposition of $M$ as having been induced by the edge $p$ of $T$.

We shall frequently use the following well-known result, which appears, for example, as [9, Lemma 2.15].

Lemma 2.4**.**

Let $M_{1}$ and $M_{2}$ label distinct vertices in a tree decomposition $T$ of a connected matroid $M$. Let $P$ be the path in $T$ joining $M_{1}$ and $M_{2}$, and let $p_{1}$ and $p_{2}$ be the edges of $P$ meeting $M_{1}$ and $M_{2}$, respectively. Then $M$ has a minor that uses $(E(M_{1})\cup E(M_{2}))\cap E(M)$ and is isomorphic to the $2$-sum of $M_{1}$ and $M_{2}$, with respect to the basepoints $p_{1}$ and $p_{2}$.

We will often use the next result, another consequence of Theorem 2.3.

Lemma 2.5**.**

Let $(X,Y)$ be a $2$-separation displayed by an edge $p$ in a $2$-connected matroid $M$. Suppose $y\in Y$. Then $M$ has, as a minor, the matroid $M_{X}(y)$ that is obtained from $M_{X}$ be relabeling $p$ by $y$. In particular, let $N$ be a $3$-connected minor of $M$ with $|E(N)|\geq 4$ and $|E(N)\cap Y|\leq 1$. If $|E(N)\cap Y|=1$, let $y\in E(N)\cap Y$; otherwise let $y$ be an arbitrary element of $Y$. Then $M_{X}(y)$ has $N$ as a minor.

Let $T$ be the canonical tree decomposition of a $2$-connected matroid $M$, and let $M_{0}$ label a vertex of $T$. Let $p_{1},p_{2},\ldots,p_{d}$ be the edges of $T$ that meet $M_{0}$. For each $p_{i}$, let $(X_{i},Y_{i})$ be the $2$-separation of $M$ displayed by $p_{i}$, where $M_{0}$ is on the $X_{i}$-side of the $2$-separation. For each $i$, let $y_{i}\in Y_{i}$. Then, by repeated application of Lemma 2.5, we deduce that $M$ has, as a minor, the matroid that is obtained from $M_{0}$ by relabeling $p_{i}$ by $y_{i}$ for all $i$ in $\{1,2,\ldots,d\}$. We denote this matroid by $M_{0}(y_{1},y_{2},\ldots,y_{d})$ and call it a specially relabeled $M_{0}$-minor of $M$.

The following result, which is straightforward to prove by repeated application of Lemma 2.2, is well known.

Lemma 2.6**.**

Let $N$ be a $3$-connected matroid with $|E(N)|\geq 3$. Let $M$ be a $2$-connected matroid with canonical tree decomposition $T$. Then there is a unique vertex $M^{\prime}$ of $T$ such that, for each edge $p$ of $T$, the partition of $V(T)$ induced by $p$ has the vertex $M^{\prime}$ on the same side as at least $|E(N)|-1$ elements of $N$. Moreover, there is a specially relabeled $M^{\prime}$-minor of $M$ that has $N$ as a minor.

3. $3$-connected matroids

Let $\mathcal{N}$ be a set of matroids. A matroid $M$ is $\mathcal{N}$-connected if, for every two distinct elements $e$ and $f$ of $M$, there is an $N$-minor of $M$ that uses $\{e,f\}$ for some $N$ in $\mathcal{N}$. A consequence of [8, Proposition 4.3.6] is that a matroid with at least three elements is $\{U_{1,3},U_{2,3}\}$-connected if and only if it is connected. The first result in this section characterizes $U_{2,3}$-connected matroids. One may hope for a characterization of $3$-connectivity in terms of $\mathcal{N}$-connectivity, but no such characterization exists. To see this, note that if $M$ is $\mathcal{N}$-connected, then so is $M\oplus_{2}M$. A characterization of $3$-connectivity in terms of minors containing $4$-element sets, as opposed to the $2$-element sets currently under consideration, is given in [6].

Proposition 3.1**.**

A matroid $M$ is $U_{2,3}$-connected if and only if $M$ is connected and simple.

Proof.

Suppose $M$ is $U_{2,3}$-connected. Clearly $M$ is connected and simple. Conversely, if $M$ is connected and simple, and $e$ and $f$ are distinct elements of $M$, then $M$ has a circuit $C$ containing $\{e,f\}$ and $|C|\geq 3$. Hence $M$ has a $U_{2,3}$-minor using $\{e,f\}$, so $M$ is $U_{2,3}$-connected. ∎

Corollary 3.2**.**

A matroid $M$ is $U_{1,3}$-connected if and only if $M$ is connected and cosimple.

We will describe $N$-connectivity for a $3$-connected matroid $N$ by first considering the case when $N$ is $U_{2,4}$. We will refer to binary and non-binary matroids that label vertices of a canonical tree decomposition as binary and non-binary vertices.

Theorem 3.3**.**

A matroid $M$ is $U_{2,4}$-connected if and only if $M$ is connected and non-binary, and, in the canonical tree decomposition of $M$,

1. (i)

every binary vertex has at most one element that is not a basepoint; and 2. (ii)

on every path between two binary vertices that each contain a unique element of $E(M)$, there is a non-binary vertex.

Proof.

Suppose $M$ is non-binary and connected, and the canonical tree decomposition $T$ of $M$ satisfies the above conditions. Suppose $e$ and $f$ are distinct elements of $M$. If $e$ and $f$ are in the same $3$-connected vertex $M_{0}$ of $T$, then, by (i), $M_{0}$ is non-binary. Thus, by Theorem 2.1, $M$ has a $U_{2,4}$-minor using $\{e,f\}$.

Next suppose $e$ belongs to a binary vertex $M_{1}$ of $T$, and $f$ belongs to a non-binary vertex $M_{0}$ of degree $d$. By Lemma 2.5, $M$ contains a specially labeled $M_{0}$-minor $M_{0}(e,y_{2},y_{3},\ldots,y_{d})$ using $\{e,f\}$. Similarly, let $e$ and $f$ belong to binary vertices $M_{1}$ and $M_{2}$, and let $M_{0}$ be a non-binary vertex on the path between them in $T$. Then $M$ contains a specially labeled $M_{0}$-minor $M_{0}(e,f,y_{3},y_{4},\ldots,y_{d})$. Thus, by Theorem 2.1, $M$ has a $U_{2,4}$-minor using $\{e,f\}$.

Suppose now that $M$ is $U_{2,4}$-connected. Clearly $M$ is non-binary and connected. If a binary vertex $M_{1}$ in $T$ contains two non-basepoints $e$ and $f$, then, by Lemma 2.6, a $U_{2,4}$-minor of $M$ using $\{e,f\}$ must be a minor of $M_{1}$; a contradiction.

Now suppose $e$ and $f$ are the unique non-basepoints of binary vertices $M_{1}$ and $M_{2}$, respectively, in $T$, and let $N$ be a $U_{2,4}$-minor of $M$ using $\{e,f\}$. By Lemma 2.6, $T$ has a nonbinary vertex $M_{0}$ such that, for every edge $p$ of $T$, the partition of $V(T)$ induced by $p$ has $M_{0}$ on the same side as at least $|E(N)|-1$ elements of $N$. Let $p_{1}$ be the edge incident with $M_{0}$ such that $M_{1}$ and $M_{0}$ are on opposite sides of the induced partition of $V(T)$. Then $M_{2}$ must be on the same side of this partition as $M_{0}$. Hence $M_{0}$ lies on the path in $T$ between $M_{1}$ and $M_{2}$. ∎

The last theorem can be generalized as follows.

Theorem 3.4**.**

Let $N$ be a $3$-connected matroid with at least four elements. A matroid $M$ is $N$-connected if and only if $M$ is connected, has $N$ as a minor, and, in the canonical tree decomposition of $M$,

1. (i)

every vertex that is not $N$-connected has at most one element that is not a basepoint; and 2. (ii)

on every path between two vertices that are not $N$-connected and that each have unique non-basepoints, there is an $N$-connected vertex.

4. Connected matroids

In this section, we consider $N$-connected matroids when $N$ is connected but not $3$-connected.

Theorem 4.1**.**

A matroid $M$ is $M(\mathcal{W}_{2})$-connected if and only $M$ is connected and non-uniform.

Proof.

If $M$ is $M(\mathcal{W}_{2})$-connected, then it is clearly both connected and non-uniform. To prove the converse, suppose $M$ is connected and non-uniform. We argue by induction that $M$ is $M(\mathcal{W}_{2})$-connected. This is immediate if $|E(M)|=4$, since $M(\mathcal{W}_{2})$ is the unique $4$-element connected, non-uniform matroid. Assume it holds for $|E(M)|<n$ and let $|E(M)|=n>4$. Distinguish two elements $x$ and $y$ of $E(M)$.

Suppose there is an element $e$ of $E(M)-\{x,y\}$ such that $M/e$ is disconnected. Then $M$ is the parallel connection, with basepoint $e$, of two matroids $M_{1}$ and $M_{2}$. Now $M\backslash e$ is connected. We may assume that it is uniform; otherwise, by the induction assumption, $M\backslash e$ and hence $M$ has an $M(\mathcal{W}_{2})$-minor using $\{x,y\}$. Now $r(E(M_{1})-e)+r(E(M_{2})-e)-r(M\backslash e)=1$. Suppose each of $|E(M_{1})-e|$ and $|E(M_{2})-e|$ has at least two elements. Then $M\backslash e$ has a $2$-separation. Since $M\backslash e$ is uniform, it follows that $M\backslash e$ is a circuit or a cocircuit. In the latter case, $M$ is also a cocircuit; a contradiction. If $M\backslash e$ is a circuit, then $M$ is the parallel connection of two circuits, and $M$ is easily seen to have an $M(\mathcal{W}_{2})$-minor using $\{x,y\}$.

Now suppose that $|E(M_{1})-e|=1$. Thus $M$ has a circuit, $\{e,f\}$ say, containing $e$. As $M\backslash e$ is uniform but $M$ is not, $r(M)\geq 2$, so $M\backslash e$ has a circuit containing $\{f,x,y\}$. It follows that $M$ has an $M(\mathcal{W}_{2})$-minor with ground set $\{e,f,x,y\}$.

We may now assume that $M/e$ is connected for all $e$ in $E(M)-\{x,y\}$. Moreover, by replacing $M$ with $M^{*}$ in the argument above, we may also assume that $M\backslash e$ is connected for all such $e$. If $M\backslash e$ or $M/e$ is non-uniform, then, by the induction assumption, $M$ has an $M(\mathcal{W}_{2})$-minor using $\{x,y\}$. Thus both $M\backslash e$ and $M/e$ are uniform. Let $r(M\backslash e)=r$. Then every circuit of $M\backslash e$ has $r+1$ elements. Since $M$ is not uniform, it has a circuit containing $e$ that has at most $r$ elements. Contracting $e$ from $M$ produces a rank-$(r-1)$ matroid having a circuit with at most $r-1$ elements. Since $M/e$ is uniform, this is a contradiction. ∎

We omit the straightforward proof of the next result.

Lemma 4.2**.**

If $M$, $N$, and $N^{\prime}$ are matroids such that $M$ is $N$-connected and $N$ is $N^{\prime}$-connected, then $M$ is $N^{\prime}$-connected.

If we wish to describe the class of $N$-connected matroids for a $3$-connected matroid $N$, it suffices to describe the $N$-connected matroids that are $3$-connected and then apply Theorem 3.4. If $N$ is not $3$-connected, the task of describing $N$-connected matroids becomes harder, and we omit any attempt to provide a general theorem for $N$-connectivity in this case. We will instead give characterizations for two specific matroids that are not $3$-connected, namely $U_{1,4}$ and its dual $U_{3.4}$. We will use the following theorem of Oxley [7].

Theorem 4.3**.**

Let $M$ be a $3$-connected matroid having rank and corank at least three, and suppose that $\{x,y,z\}\subseteq E(M)$. Then $M$ has a minor isomorphic to one of $U_{3,6},P_{6},Q_{6},\mathcal{W}^{3}$, or $M(K_{4})$ that uses $\{x,y,z\}$.

Proposition 4.4**.**

A $3$-connected matroid $M$ is $U_{1,4}$-connected if and only if either $M\cong U_{2,n}$ for some $n\geq 5$, or $M$ has rank and corank at least three.

Proof.

Clearly if $n\geq 5$, then $U_{2,n}$ is $U_{1,4}$-connected. Now assume that $r(M)\geq 3$ and $r^{*}(M)\geq 3$. Suppose $\{x,y\}\subseteq E(M)$. Then, by Theorem 4.3, $M$ has an $\mathcal{N}$-minor using $\{x,y\}$ where $\mathcal{N}$ is $\{U_{3,6},P_{6},Q_{6},\mathcal{W}^{3},M(K_{4})\}$. One easily checks that each member of $\mathcal{N}$ is $U_{1,4}$-connected. Hence, by Lemma 4.2, $M$ is $U_{1,4}$-connected.

To prove the converse, assume that $M$ is $U_{1,4}$-connected. Since $r^{*}(U_{1,4})=3$, it follows that $r^{*}(M)\geq 3$. The required result holds if $r(M)\geq 3$. But, since $M$ is $3$-connected and $U_{1,4}$-connected, $r(M)\geq 2$. Moreover, if $r(M)=2$, then $M\cong U_{2,n}$ for some $n\geq 5$. ∎

Duality gives a corresponding result for $U_{3,4}$-connectivity.

Corollary 4.5**.**

A $3$-connected matroid $M$ is $U_{3,4}$-connected if and only if either $M\cong U_{n-2,n}$ where $n\geq 5$, or $M$ has rank and corank at least $3$.

Observe that this fails to fully characterize $U_{3,4}$-connectivity for if we let $M=M(K_{2,3})$, then $M$ is $U_{3,4}$-connected but none of the matroids in its canonical tree decomposition is $U_{3,4}$-connected. We can instead describe $U_{3,4}$-connectivity in terms of forbidden configurations of matroids in the canonical tree decomposition.

Proposition 4.6**.**

Suppose $M$ is not $3$-connected. Then $M$ is $U_{3,4}$-connected if and only if $M$ is connected and simple, and, in the canonical tree decomposition $T$ of $M$, there is no vertex of degree at most two that is labeled by some $U_{2,n}$ such that its only neighbors in $T$ are cocircuits that use elements of $E(M)$.

Proof.

Let $T$ be the canonical tree decomposition of $M$. Assume $M$ is $U_{3,4}$-connected. Then, by Lemma 4.2, $M$ is $U_{2,3}$-connected, so $M$ is connected and simple. Suppose that $T$ has a vertex $M_{0}$ whose degree $d$ is at most two such that $M_{0}$ is labeled by some $U_{2,n}$ and has its only neighbors $M_{1},\dots,M_{d}$ labeled by cocircuits that use elements of $E(M)$. For each $i$ in $\{1,\dots,d\}$, suppose $f_{i}\in E(M_{i})\cap E(M)$. Then $M$ can be obtained from a copy of $U_{2,n}$ using $\{f_{1},\dots,f_{d}\}$ by, for each $i$, adjoining some matroid via parallel connection across the basepoint $f_{i}$. If $d=1$, let $f_{2}$ be an element of $M_{0}$ other than $f_{1}$. Clearly $M$ has no circuit using $\{f_{1},f_{2}\}$ that has more than three elements.

Now assume that $M$ is connected and simple and that $T$ satisfies the specified conditions. Let $\{e,f\}$ be a subset of $E(M)$ that is not contained in a $U_{3,4}$-minor. Assume first that $e$ and $f$ belong to the same vertex $M_{1}$ of $T$. As $M$ is simple, $M_{1}$ is not a cocircuit. Now $M$ has a specially relabeled $M_{1}$-minor using $\{e,f\}$. Thus, by Corollary 4.5, $M_{1}\cong U_{2,n}$ for some $n\geq 3$. Let $p$ be an edge of $T$ that meets $M_{1}$. Consider the $2$-sum $N_{1}\oplus_{2}N_{2}$ induced by $p$ where $\{e,f\}\subseteq E(N_{1})$. Certainly $N_{1}$ has a circuit containing $\{e,f,p\}$, and $N_{2}$ has a circuit of size at least three containing $p$. Thus $M$ has a $U_{3,4}$-minor containing $\{e,f\}$; a contradiction.

We now know that $e$ and $f$ belong to distinct vertices $M_{1}$ and $M_{2}$ of $T$. Each edge $p$ of the path $P$ in $T$ joining $M_{1}$ and $M_{2}$ induces a 2-sum decomposition of $M$ into two matroids, $N_{1p}$ and $N_{2p}$. Moreover, an element $x_{i}$ of $E(N_{ip})$ is in a circuit of $N_{ip}$ of size at least three containing $p$ unless $x_{i}$ is parallel to $p$ in $N_{ip}$. Thus $e$ or $f$ is parallel to $p$ in $N_{1p}$ or $N_{2p}$, respectively. Let the edges of $P$, in order, be $p_{1},p_{2},\dots,p_{k}$ where $p_{1}$ meets $M_{1}$. We may assume that $e$ is parallel to $p_{1}$ in $N_{1p_{1}}$. Then the vertex $M_{1}$ of $T$ containing $e$ is a cocircuit.

Suppose $k\geq 3$. As no two adjacent vertices of $T$ are cocircuits, neither $e$ nor $f$ is parallel to $p_{2}$ in $N_{1p_{2}}$ or $N_{2p_{2}}$. Hence $M$ has a $U_{3,4}$-minor using $\{e,f\}$. This contradiction implies that $k\in\{1,2\}$. Suppose $k=2$. Then $f$ is parallel to $p_{2}$ in $N_{2p_{2}}$. Thus $M_{2}$ is a cocircuit. Since $M$ has no $U_{3,4}$-minor using $\{e,f\}$, the vertex $M_{3}$ of $T$ that is adjacent to both $M_{1}$ and $M_{2}$ is isomorphic to some $U_{2,n}$. By assumption, $M_{3}$ must have another neighbor in $T$ to which it is joined by the edge $q$, say. Then, for the 2-sum decomposition $Q_{1}\oplus_{2}Q_{2}$ of $M$ induced by $q$, there is a circuit of $Q_{1}$ containing $\{e,f,q\}$ and a circuit of $Q_{2}$ of size at least three containing $q$. Thus $M$ has a $U_{3,4}$-minor using $\{e,f\}$. This contradiction implies that $k=1$. Then $M=N_{1p_{1}}\oplus_{2}N_{2p_{1}}$. Thus the specially relabeled minor $N_{2p_{1}}(e)$ uses $\{e,f\}$. Now the canonical tree decomposition $T^{\prime}$ of $N_{2p_{1}}(e)$ can be obtained from the component of $T\backslash p_{1}$ using $N_{2p_{1}}$ by replacing $M_{2}$ by $M_{2}(e)$. As $e$ and $f$ are contained in the same vertex of $T^{\prime}$, we deduce from the second paragraph that $N_{2p_{1}}(e)$, and hence $M$, has a $U_{3,4}$-minor using $\{e,f\}$; a contradiction. ∎

5. Disconnected matroids

We now turn our attention to $N$-connectivity where $N$ is disconnected. The following is essentially immediate.

Proposition 5.1**.**

Let $n$ be an integer exceeding one. A matroid $M$ is $U_{n,n}$-connected if and only if $M$ is simple with rank at least $n$.

Recall that elements $x$ and $y$ of a matroid $M$ are clones if the bijection on $E(M)$ that interchanges $x$ and $y$ but fixes every other element yields the same matroid. Next we prove Theorem 1.4, showing that a matroid is $U_{0,1}\oplus U_{1,1}$-connected if and only if no element has a clone. The proof will use the well-known fact (see, for example, [2]) that two elements in a matroid are clones if and only if they are in precisely the same cyclic flats.

Proof of Theorem 1.4.

Suppose every clonal class of $M$ is trivial and let $x$ and $y$ be distinct elements of $M$. Then $M$ has a cyclic flat $F$ that contains exactly one of $x$ and $y$, say $x$. In $M/(F-x)$, the element $x$ is a loop but $y$ is not. Thus $M$ has a $U_{0,1}\oplus U_{1,1}$-minor using $\{x,y\}$, so $M$ is $U_{0,1}\oplus U_{1,1}$-connected.

Conversely, assume $M$ is $U_{0,1}\oplus U_{1,1}$-connected, but $M$ has elements $x$ and $y$ that are in the same cyclic flats. Suppose that $M/C\backslash D\cong U_{0,1}\oplus U_{1,1}$ and $E(M/C\backslash D)=\{x,y\}$. Let $x$ be the loop of $M/C\backslash D$. Then $x\in\mathrm{cl}_{M}(C)$. Thus $y\in\mathrm{cl}_{M}(C)$, so $y$ is a loop in $M/C\backslash D$; a contradiction. ∎

Recall, for the next result, that an element is free in a matroid if it is not a coloop and every circuit that contains it is spanning.

Theorem 5.2**.**

A matroid $M$ is $U_{1,2}\oplus U_{1,1}$-connected if and only if $M$ is loopless, has at most one coloop, and has at most one free element.

Proof.

Clearly if $M$ is $U_{1,2}\oplus U_{1,1}$-connected, then it obeys the specified conditions. Conversely, suppose $M$ is loopless, has at most one coloop, and has at most one free element. Let $e$ and $f$ be elements of $M$. Suppose first that $M$ is disconnected. If $e$ and $f$ are in the same component, then they are in a $U_{1,2}$-minor of that component, so $M$ has a $U_{1,2}\oplus U_{1.1}$-minor using $\{e,f\}$. If $e$ and $f$ are in different components, then one of these components is not a coloop. That component has a $U_{1,2}$-minor using $e$ or $f$. It follows that $M$ has a $U_{1,2}\oplus U_{1,1}$-minor using $\{e,f\}$.

Now suppose $M$ is connected. Suppose that $e$ is free in $M$. Then $f$ is in some non-spanning circuit, $C_{f}$. Choose $g$ in $C_{f}-f$. Contracting $C_{f}-\{f,g\}$ and deleting every other element of $M$ yields a $U_{1,2}\oplus U_{1,1}$-minor of $M$ using $\{e,f\}$.

Suppose neither $e$ nor $f$ is free in $M$. If there is a non-spanning circuit $C$ containing $\{e,f\}$, we can find a $U_{1,2}\oplus U_{1,1}$-minor by contracting every element of $C$ except $e$ and $f$, and deleting every other element except for one. Now suppose every circuit containing $\{e,f\}$ is spanning. Since $e$ is not free, there is a non-spanning circuit $C$ containing $e$. Clearly $f\not\in\mathrm{cl}(C)$ otherwise $M|{\mathrm{cl}(C)}$ is a connected matroid of rank less than $r(M)$ so it contains a circuit containing $\{e,f\}$; a contradiction. Therefore, after we contract all of $C$ except for $e$ and one other element, we see that $f$ will not be a loop. Thus we can find a $U_{1,2}\oplus U_{1,1}$-minor using $\{e,f\}$. ∎

Corollary 5.3**.**

A matroid $M$ is $U_{1,2}\oplus U_{0,1}$-connected if and only if $M$ is coloopless and has at most one element that is in every dependent flat.

6. $N$-connectivity as compared to connectivity

Before proving Theorem 1.1, we state and prove its converse.

Proposition 6.1**.**

If $N\in\{U_{1,2},U_{0,2},U_{2,2}\}$, then, for every $N$-connected matroid $M$ with $|E(M)|\geq 3$ and for every $e$ in $E(M)$, at least one of $M\backslash e$ or $M/e$ is $N$-connected.

Proof.

The result is immediate if $N\cong U_{1,2}$. By duality, it suffices to deal with the case when $N\cong U_{2,2}$. Suppose $M$ is $U_{2,2}$-connected, and $|E(M)|\geq 3$. By Proposition 5.1, $M$ is simple with rank at least two. Therefore if $M$ is $U_{2,2}$-connected and $r(M)>2$, we can delete any element $e$ of $M$ and still have an $N$-connected matroid. Observe that if $r(M)=2$, then $M$ must be connected since it is simple. Therefore $M$ has no coloops, so $r(M\backslash e)=2$ for all $e$ of $E(M)$. Thus $M\backslash e$ is $U_{2,2}$-connected. ∎

Proof of Theorem 1.1.

First we consider the case when $N$ is connected. Then $N$ is $U_{1,2}$-connected. Thus, by Lemma 4.2, every $N$-connected matroid is $U_{1,2}$-connected and so is connected. Suppose $M$ is an $N$-connected matroid with $|E(M)|>|E(N)|$.

Assume $N$ is simple. Then, by Proposition 3.1 and Lemma 4.2, $N$, and hence $M$, is $U_{2,3}$-connected. Let $M_{1}$ and $M_{2}$ be isomorphic copies of $M$ with disjoint ground sets. Pick arbitrary elements $g_{1}$ and $g_{2}$ in $M_{1}$ and $M_{2}$, and let $M_{3}$ be the parallel connection of $M_{1}$ and $M_{2}$ with respect to the basepoints $g_{1}$ and $g_{2}$, which we relabel as $g$ in $M_{3}$. Then one easily sees that $M_{3}$ is $N$-connected. Let $e,f\in E(M_{1})-g$. By assumption, we can remove all the elements of $E(M_{1})-\{e,f,g\}$ from $M_{3}$ via deletion or contraction to obtain a matroid $M_{4}$ that is still $N$-connected. Since $M_{4}$ is $U_{2,3}$-connnected, it follows that $\{e,f,g\}$ is a triangle in $M_{4}$. Moreover, $\{e,f\}$ is a series pair in $M_{4}$. However, neither $M_{4}\backslash e$ nor $M_{4}/e$ is $U_{2,3}$-connected since $M_{4}\backslash e$ is disconnected, and $M_{4}/e$ has $f$ and $g$ in parallel. We deduce that $N$ is not simple. Dually, $N$ is not cosimple. The only uniform matroid that is neither simple nor cosimple is $U_{1,2}$, so either $N\cong U_{1,2}$, or $N$ is non-uniform.

Next we show that $N$ cannot be non-uniform. Suppose, instead, that $N$ is non-uniform. Then, as $N$ is connected, by Theorem 4.1, $N$ is $M(\mathcal{W}_{2})$-connected.

Recall that $M$ is $N$-connected with $|E(M)|>|E(N)|$. Let $n=|E(N)|+1$ and distinguish elements $e,f$ of $E(M)$. Let each of $M_{1},M_{2},\ldots,M_{n}$ be a copy of $M$ and let $e_{i}$ and $f_{i}$ be the elements of $M_{i}$ corresponding to $e$ and $f$. Let $M^{\prime}$ be the parallel connection of $M_{1},M_{2},\ldots,M_{n}$ with respect to the basepoints $e_{1},e_{2},\ldots,e_{n}$ where these elements are relabeled as $e$ in $M^{\prime}$. By assumption, for each $M_{i}$, we can remove $E(M_{i})-\{e,f_{i}\}$ from $M^{\prime}$ in such a way that the resulting matroid $M^{\prime\prime}$ is $N$-connected. Since $M^{\prime\prime}$ is connected, it must be isomorphic to $U_{1,n+1}$, which is clearly not $M(\mathcal{W}_{2})$-connected; a contradiction. We conclude that $N$ cannot be non-uniform, and hence the theorem holds when $N$ is connected.

Next we consider the case when $N$ is disconnected, first showing the following.

6.2.1**.**

If each element of $N$ is a loop or a coloop, then $N\cong U_{0,2}$ or $U_{2,2}$.

Suppose $n\geq 3$ and let $N\cong U_{n,n}$. Let $M=U_{2,3}\oplus U_{n-2,n-2}$. Then $M$ is $N$-connected, but if $e$ is a coloop of $M$, then neither $M\backslash e$ nor $M/e$ has a $U_{n,n}$-minor. Therefore $N\not\cong U_{n,n}$; dually, $N\not\cong U_{0,n}$.

If $N=U_{0,1}\oplus U_{1,1}$, then let $M=M(K_{4})$. By Theorem 1.4, $M$ is $N$-connected, but, for every $e$ of $E(M)$, both $M\backslash e$ and $M/e$ have nontrivial clonal classes and are therefore not $N$-connected. Now assume $N\cong U_{0,n}\oplus U_{m,m}$ for some $n\geq 2$ and $m\geq 1$. Then $U_{0,n+1}\oplus U_{m,m}$ is an $N$-connected matroid, say $M$. But if $e$ is a coloop, then neither $M\backslash e$ nor $M/e$ has an $N$-minor. On combining this contradiction with duality, we conclude that 6.2.1 holds.

Now assume that $N$ has $k+s$ components $N_{1},N_{2},\ldots,N_{k+s}$ where those with at least two elements are $N_{1},N_{2},\ldots,N_{k}$. Then $k\geq 1$. For each $i$ in $\{1,2,\ldots,k\}$, choose an element $e_{i}$ of $N_{i}$ and relabel it as $p$. Let $M^{\prime}$ be the parallel connection of $N_{1},N_{2},\ldots,N_{k}$ with respect to the basepoint $p$ where we take $M^{\prime}=N_{1}$ if $k=1$. Let $N^{\prime}$ be a copy of $N$ whose ground set is disjoint from $E(N)$, and let $n^{\prime}_{i}$ be the component of $N^{\prime}$ corresponding to $N_{i}$. Let $M_{1}=N^{\prime}\oplus M^{\prime}$. We show next that

6.2.2**.**

$M_{1}$* is $N$-connected.*

Suppose $\{e,f\}\subseteq E(M_{1})$. Certainly $M_{1}$ has an $N$-minor using $\{e,f\}$ if $\{e,f\}\subseteq E(N^{\prime})$. Next suppose that $e\in E(M^{\prime})$. Then, since $M^{\prime}$ is a connected parallel connection, we see that, for each $i$ in $\{1,2,\ldots,k+s\}$, there is an $N_{i}$-minor of $M^{\prime}$ using $e$. Thus, if $f\in E(N^{\prime})$, say $f\in E(N^{\prime}_{j})$, then we can choose $i\neq j$ and get an $N$-minor of $M_{1}$ using $\{e,f\}$ unless $k=1=j$. In the exceptional case, $M^{\prime}$ has an $N_{2}$-minor with ground set $\{e\}$ and again we get an $N$-minor of $M_{1}$ using $\{e,f\}$. We may now assume that $f\in E(M^{\prime})$, say $f\in E(N_{j})$. Then $M^{\prime}$ has an $N_{j}$-minor using $\{e,f\}$, so $M_{1}$ has an $N$-minor using $\{e,f\}$. Thus 6.2.2 holds.

Since $M_{1}$ is $N$-connected, by assumption, we may delete or contract elements of $M_{1}$ until we obtain an $N$-connected matroid $M_{2}$ with $|E(M_{2})|=|E(N)|+1$. In particular, we may remove elements from $M^{\prime}$ in $M_{1}$ until a single element $g$ remains. Now choose $e$ in $E(N^{\prime}_{1})$. Then $M_{2}\backslash e$ or $M_{2}/e$ is isomorphic to $N$. But both $M_{2}\backslash e$ and $M_{2}/e$ have more one-element components than $N^{\prime}$; a contradiction. ∎

Recall that we say that a matroid $N$ has the transitivity property if, for every matroid $M$ and every triple $\{e,f,g\}\subseteq E(M)$, if $e$ is in an $N$-minor with $f$, and $f$ is in an $N$-minor with $g$, then $e$ is in an $N$-minor with $g$. Clearly $N$ has the transitivity property if and only if $N^{*}$ has the transitivity property.

Lemma 6.3**.**

Suppose $N$ is a matroid having the transitivity property. Let $N^{\prime}$ be obtained from $N$ by adding an element $f$ in parallel to a non-loop element $e$ of $N$. Then there is an element $g$ of $E(N^{\prime})$ such that $N^{\prime}\backslash g$ is isomorphic to $N$ and has $\{e,f\}$ as a $2$-circuit. Moreover, $g$ is in a $2$-circuit in $N$.

Proof.

The transitivity property implies that $\{e,f\}$ is in an $N$-minor of $N^{\prime}$. Since $r^{*}(N^{\prime})>r^{*}(N)$, there must be an element $g$ of $E(N^{\prime})-\{e,f\}$ such that $N^{\prime}\backslash g\cong N$. Since we have introduced a new $2$-circuit in constructing $N^{\prime}$, when we delete $g$, we must destroy a $2$-circuit. ∎

By the last lemma and duality, we obtain the following result.

Corollary 6.4**.**

If $N$ is a matroid having the transitivity property, then $N$ has a component with more than one element.

The following elementary observation and its dual will be used repeatedly in the proof of Theorem 1.2.

Lemma 6.5**.**

Suppose $N$ is a matroid with the transitivity property. Let $N_{0}$ be a component of $N$ with the largest number of elements. Suppose $f$ is added in parallel to an element $e$ of $N_{0}$. Let $N_{0}^{\prime}$ and $N^{\prime}$ be the resulting extensions of $N_{0}$ and $N$, respectively. Suppose $g\in E(N^{\prime})$ such that $N^{\prime}\backslash g\cong N$. Then $g\in E(N_{0}^{\prime})$.

Recall that a set $S$ of elements of a matroid $M$ is a fan if $|S|\geq 3$ and there is an ordering $(s_{1},s_{2},\ldots,s_{n})$ of the elements of $S$ such that, for all $i$ in $\{1,2,\ldots,n-2\}$,

1. (i)

$\{s_{i},s_{i+1},s_{i+2}\}$ is a triangle or a triad; and 2. (ii)

when $\{s_{i},s_{i+1},s_{i+2}\}$ is a triangle, $\{s_{i+1},s_{i+2},s_{i+3}\}$ is a triad; and when $\{s_{i},s_{i+1},s_{i+2}\}$ is a triad, $\{s_{i+1},s_{i+2},s_{i+3}\}$ is a triangle.

Note that the above extends the definition given in [8] by eliminating the requirement that $M$ be simple and cosimple. We shall follow the familiar practice here of blurring the distinction between a fan and a fan ordering.

Lemma 6.6**.**

Let $(s_{1},s_{2},\ldots,s_{n})$ be a fan $X$ in a matroid $M$ such that each of $\{s_{1},s_{2}\}$ and $\{s_{n-1},s_{n}\}$ is a circuit or a cocircuit. Then $X$ is a component of $M$.

Proof.

By switching to the dual if necessary, we may assume that $\{s_{1},s_{2},s_{3}\}$ is a triangle of $M$. Thus $\{s_{1},s_{2}\}$ is a cocircuit. Observe that $\{s_{i}:i\mathrm{~{}is~{}odd}\}$ spans $X$. If $n$ is odd, this is immediate, and if $n$ is even, it follows from the fact that $\{s_{n-1},s_{n}\}$ is a circuit in this case. By duality, $\{s_{i}:i\mathrm{~{}is~{}even}\}$ spans $X$ in $M^{*}$. Hence $r(X)+r^{*}(X)\leq|X|$; that is, $\lambda(X)\leq 0$, so $X$ is a component of $M$. ∎

We define a special fan to be a fan $(s_{1},s_{2},\ldots s_{k})$ such that $\{s_{1},s_{2}\}$ is a cocircuit of $M$. We will now show that $U_{1,2}$ and $M(\mathcal{W}_{2})$ are the only connected matroids with the transitivity property.

Proof of Theorem 1.2.

It is clear that $U_{1,2}$ has the transitivity property. By Theorem 4.1, two elements of $M$ are in an $M(\mathcal{W}_{2})$-minor together if and only if they are in a connected, non-uniform component together. It follows that $M(\mathcal{W}_{2})$ has the transitivity property.

Suppose that $N$ has the transitivity property. Assume that $N$ is not isomorphic to $U_{1,2}$ or $M(\mathcal{W}_{2})$. Next we show the following.

6.7.1**.**

Let $N_{0}$ be a largest component of $N$. Then $N_{0}$ is isomorphic to $U_{1,2}$ or $M({\mathcal{W}}_{2})$.

Assume that this assertion fails. Then, by Corollary 6.4, $N_{0}$ has at least two, and hence at least three, elements. Take an element $e$ of $N_{0}$ and add an element $f$ in series with it. Let the resulting coextensions of $N_{0}$ and $N$ be $N^{\prime}_{0}$ and $N^{\prime}$, respectively. Then, by the transitivity property, $N^{\prime}/a\cong N$ for some element $a$ of $E(N^{\prime})-\{e,f\}$. Furthermore, by the dual of Lemma 6.5, $a\in E(N_{0})$. We deduce that $N_{0}$ has a 2-cocircuit, say $\{a,b\}$. In $N_{0}$, add an element $c$ in parallel to $a$ to get $N_{1}$. Then, by transitivity and Lemma 6.5, there is an element $s_{1}$ of $E(N_{1})-\{a,c\}$ such that $N_{1}\backslash s_{1}\cong N_{0}$. Since $N_{1}\backslash b$ has $\{a,c\}$ as a component, the component sizes of $N_{1}\backslash b$ and $N_{0}$ do not match, so $s_{1}\neq b$. Thus $s_{1}\in E(N)-\{a,b,c\}$, so $N_{1}\backslash s_{1}$ has $\{c,a,b\}$ as a cocircuit. Next add an element $d$ to $N_{1}\backslash s_{1}$, putting it in series with $c$. Let the resulting matroid be $N_{2}$. By the dual of Lemma 6.5, there is an element $s_{2}$ of $E(N_{2})-\{c,d\}$ such that $N_{2}/d\cong N_{0}$. Moreover, $s_{2}$ must be in a $2$-cocircuit of $N_{2}$, and $s_{2}$ is in a triangle in $N_{2}$ as $N_{2}/s_{2}$ must have a $2$-circuit that is not present in $N_{2}$ since adding $d$ destroyed the 2-circuit $\{a,c\}$. Now $s_{2}\neq a$ since $N_{2}/a$ has $\{c,d\}$ as a component.

Suppose $s_{2}=b$. Then $b$ is in a $2$-cocircuit $\{b,e\}$ in $N_{2}$. Moreover, $N_{2}$ has a triangle $T$ containing $b$. By orthogonality, $T=\{b,e,a\}$. Then $(d,c,a,e)$ is a fan $X$ in $N_{2}/b$ having $\{c,d\}$ as a cocircuit and $\{a,e\}$ as a circuit. By Lemma 6.6, $X=E(N_{2}/b)$, so $N_{0}\cong N_{2}/b\cong M(\mathcal{W}_{2})$; a contradiction.

We now know that $s_{2}\neq b$, so $s_{2}\not\in\{a,b,c,d\}$. Thus $N_{0}$ has $(d,c,a,b)$ as a special fan. Among all the special fans of $N_{0}$ and $N_{0}^{*}$, take one, $(a_{1},a_{2},\ldots,a_{k})$, with the maximum number of elements. Then $k\geq 4$. First assume $\{a_{k-2},a_{k-1},a_{k}\}$ is a triad. Suppose $\{a_{k-1},a_{k}\}$ is a $2$-circuit of $N_{0}$. Then, by Lemma 6.6, the special fan is the whole component $N_{0}$. As $N_{0}\not\cong M({\mathcal{W}}_{2})$, we see that $k\geq 6$. Add an element $f$ in parallel to $a_{3}$ to form a new matroid $N_{0}^{\prime}$. Then $\{a_{1},a_{3}\}$ is in an $N_{0}$-minor of $N_{0}^{\prime}$, and so is $\{a_{1},f\}$. By the transitivity property, $N_{0}^{\prime}$ has $\{a_{3},f\}$ in an $N_{0}$-minor. Since $N_{0}^{\prime}$ has $\{a_{3},f\}$ and $\{a_{k-1},a_{k}\}$ as its only 2-circuits, while $N_{0}$ has a single 2-circuit, we deduce that $N_{0}^{\prime}\backslash a_{k}\cong N_{0}$. But every element of $N_{0}$ is in a cocircuit of size at most three, yet $f$ is in no such cocircuit of $N_{0}^{\prime}\backslash a_{k}$; a contradiction.

It remains to deal with the cases when, in $N_{0}$, either $\{a_{k-2},a_{k-1},a_{k}\}$ is a triad and $\{a_{k-1},a_{k}\}$ is not a circuit, or $\{a_{k-2},a_{k-1},a_{k}\}$ is a triangle. In these cases, add $a_{0}$ in parallel with $a_{1}$ to produce $N_{3}$. To obtain an $N_{0}$-minor of $N_{3}$ using $\{a_{0},a_{1}\}$, we must delete an element $z$ of $N_{3}$ that belongs to a $2$-circuit. Now $z$ is not in $\{a_{2},a_{3},\ldots,a_{k}\}$ as none of these elements is in a $2$-circuit, so $N_{3}\backslash z$ is isomorphic to $N_{0}$ and has $(a_{0},a_{1},\ldots,a_{k})$ as a special fan. This contradicts our assumption that a special fan in $N_{0}$ or $N_{0}^{*}$ has at most $k$ elements. We conclude that 6.7.1 holds.

6.7.2**.**

$N$* has no single-element component.*

To see this, let $N_{0}$ be a largest component of $N$. By 6.7.1, $N_{0}$ is isomorphic to $U_{1,2}$ or $M({\mathcal{W}}_{2})$. Assume that $N$ has a single-element component $N_{1}$ with $E(N_{1})=\{a\}$. By replacing $N$ by its dual if necessary, we may assume that $a$ is a coloop of $N$. Let $c$ be an element that is in a 2-cocircuit of $N_{0}$. Now let $N^{\prime}$ be obtained from $N$ by adding an element $b$ so that $N^{\prime}$ has $\{a,b,c\}$ as a triangle and $\{a,b\}$ as a cocircuit. Then, by the transitivity property, $N^{\prime}\backslash g\cong N$ for some element $g$ not in $\{a,b\}$. By the choice of $N_{0}$, we deduce that $g$ must be in the same component $N_{0}^{\prime}$ of $N^{\prime}$ as $\{a,b,c\}$. Moreover, $g$ must be in a 2-cocircuit of $N^{\prime}_{0}$. But $N^{\prime}_{0}$ contains no such element. Hence 6.7.2 holds.

6.7.3**.**

$N$* has a single component of maximum size.*

Assume that this fails, and let $N_{0}$ and $N_{1}$ be components of $N$ of maximum size. Let $\{a_{i},b_{i}\}$ be a 2-circuit of $N_{i}$. Let $N^{\prime}_{i}$ be obtained from $N_{i}$ by adding $c_{i}$ in series with $b_{i}$. Now take a copy of $U_{2,3}$ with ground set $\{c_{0},z,c_{1}\}$ and adjoin $N^{\prime}_{0}$ and $N^{\prime}_{1}$ via parallel connection across $c_{0}$ and $c_{1}$, respectively. Truncate the resulting matroid to get $N_{01}$. Then $r(N_{01})=r(N_{0})+r(N_{1})+1$. Let $N^{\prime}$ be obtained from $N$ by replacing $N_{0}\oplus N_{1}$ by $N_{01}$. Now $N_{01}/c_{0}$ and $N_{01}/c_{1}$ have $(N_{0}\oplus N_{1})$-minors using $\{z,c_{1}\}$ and $\{z,c_{0}\}$, respectively. Hence $N^{\prime}/c_{0}$ and $N^{\prime}/c_{1}$ have $N$-minors using $\{z,c_{1}\}$ and $\{z,c_{0}\}$. Thus, by transitivity, $N^{\prime}$ has an $N$-minor $\tilde{N}$ using $\{c_{0},c_{1}\}$. As $r(N^{\prime})=r(N)+1$, there are elements $e$, $f$, and $g$ of $E(N^{\prime})-\{c_{0},c_{1}\}$ such that $\tilde{N}=N^{\prime}/e\backslash f,g$. Now $N^{\prime}/e$ must have two disjoint 2-circuits that are not in $N^{\prime}$. Thus $e\in E(N_{01})$. As $e\not\in\{c_{0},c_{1}\}$, it follows that $N_{0}\cong M({\mathcal{W}}_{2})\cong N_{1}$ and, by symmetry, we may assume that $e=a_{0}$. But $N_{01}/a_{0}$ does not have an $(M({\mathcal{W}}_{2})\oplus M({\mathcal{W}}_{2}))$-minor. Thus 6.7.3 holds.

By 6.7.1 and 6.7.3, $N$ has a single largest component $N_{0}$ and it is isomorphic to $M({\mathcal{W}}_{2})$. As $N$ is disconnected, we may assume by duality that $N$ has a component $N_{1}$ that is isomorphic to $U_{1,k}$ for some $k$ in $\{2,3\}$. Now take a copy of $U_{2,3}$ with ground set $\{c_{0},z,c_{1}\}$ and adjoin copies of $U_{2,k+1}$ via parallel connection across $c_{0}$ and $c_{1}$, letting the resulting matroid be $N_{01}$. Replacing $N_{0}\oplus N_{1}$ by $N_{01}$ in $N$ to give $N^{\prime}$, we see that $r(N^{\prime})=r(N)+1$. Moreover, $N^{\prime}/c_{0}$ and $N^{\prime}/c_{1}$ have $N$-minors using $\{c_{1},z\}$ and $\{c_{0},z\}$, respectively. But $c_{0}$ and $c_{1}$ are the only elements $e$ of $N^{\prime}$ such that $N^{\prime}/e$ has two disjoint 2-circuits that are not in $N^{\prime}$. Thus $N^{\prime}$ has no $N$-minor using $\{c_{0},c_{1}\}$. This contradiction completes the proof of the theorem. ∎

We conclude this section by proving Corollary 1.3, which demonstrates how two of the basic properties of matroid connectivity are enough to characterize it.

Proof of Corollary 1.3.

Assume that $N\not\cong U_{1,2}$. Then, by Theorem 1.1 and duality, we may assume that $N\cong U_{2,2}$. But $U_{2,2}$ does not have the transitivity property as the matroid $U_{1,2}\oplus U_{1,1}$ shows. ∎

7. Three-element sets

The notion of $N$-connectivity defined here relies on sets of two elements. Sets of size three have already been an object of some study. Seymour asked whether every $3$-element set in a $4$-connected non-binary matroid belongs to a $U_{2,4}$-minor but Kahn [5] and Coullard [3] answered this question negatively. Seymour [12] characterized the internally $4$-connected binary matroids that are $U_{2,3}$-connected, but the problem of completely characterizing when every triple of elements in an internally $4$-connected matroid is in a $U_{2,3}$-minor remains open [8, Problem 15.9.7].

For a $3$-connected binary matroid $M$ having rank and corank at least three, Theorem 4.3 shows that every triple of elements of $M$ is in an $M(K_{4})$-minor. The next result extends this theorem to connected binary matroids. As the proof, which is based on Lemma 2.6, is so similar to those appearing earlier, we omit the details.

Proposition 7.1**.**

Let $M$ be a connected binary matroid. For every triple $\{x,y,z\}\subseteq E(M)$, there is an $M(K_{4})$-minor using $\{x,y,z\}$ if and only if every matroid in the canonical tree decomposition of $M$ has rank and corank at least $3$.