On Density-Critical Matroids

TL;DR

This paper characterizes density-critical matroids, identifies ten minimal obstructions for covering by two independent sets, and explores the structure of such matroids, especially those with density less than 2.

Contribution

It provides a complete classification of certain density-critical matroids and explicitly solves the case for density at most 9/4.

Findings

Ten minimal obstructions for covering by two independent sets.

Density-critical matroids with density less than 2 are series-parallel networks.

Solved the classification problem for density at most 9/4.

Abstract

For a matroid $M$ having $m$ rank-one flats, the density $d(M)$ is $\tfrac{m}{r(M)}$ unless $m = 0$, in which case $d(M)= 0$. A matroid is density-critical if all of its proper minors of non-zero rank have lower density. By a 1965 theorem of Edmonds, a matroid that is minor-minimal among simple matroids that cannot be covered by $k$ independent sets is density-critical. It is straightforward to show that $U_{1,k+1}$ is the only minor-minimal loopless matroid with no covering by $k$ independent sets. We prove that there are exactly ten minor-minimal simple obstructions to a matroid being able to be covered by two independent sets. These ten matroids are precisely the density-critical matroids $M$ such that $d(M) > 2$ but $d(N) \le 2$ for all proper minors $N$ of $M$. All density-critical matroids of density less than $2$ are series-parallel networks. For $k \ge 2$, although finding all density-critical matroids of density at most $k$ does not seem straightforward, we do solve this problem for $k=\tfrac{9}{4}$.

Equations

$$kr(A)\ge |A|.$$

$$kr(M)=|E(M)|-1.$$

$$|E(M)|>kr(M)\ge kr(M\setminus x)\ge |E(M\setminus x)|=|E(M)|-1.$$

$$|E(M)|-\epsilon (M/S)>d(M)r(S).$$

$$\frac{\epsilon (M/S)}{r(M/S)}<\frac{\epsilon (M)}{r(M)}=\frac{|E(M)|}{r(M)}.$$

$$r(M)d(M)r(S)=|E(M)|r(S)<r(M)\left(|E(M)|-\epsilon (M/S)\right).$$

$$r(W\cup Q)=r(W\cup (Q-\{{\beta }_{m+1}\}))=n+m-1.$$

$$W\cup (Q-\{{\beta }_{m+1}\})=E(M).$$

$${\beta }_{m+1}\in W\cup (Q-\{{\beta }_{m+1}\}).$$

$$|E(M)|\le 2r(M)-1.$$

$$d(M_1\oplus M_2)\le \max \{d(M_1),d(M_2)\}.$$

$$\frac{n_1+n_2-2}{r_1+r_2-1}>\frac{n_1}{r_1}.$$

$$r_1{n}_2-2r_1>r_2{n}_1-n_1.$$

$$r_2{n}_1-2r_2>r_1{n}_2-n_2.$$

$$r(\{x,y,z\})$$

$$=r(N)-1+3-r(N)=2.$$

$$\frac{9}{4}\ge d(N|(X\cup Y))\ge \frac{17-k}{8-k}.$$

Full text

1. Introduction

Our notation and terminology follow Oxley [7]. For a positive integer $k$, let $\mathcal{M}_{k}$ be the class of matroids $M$ for which $E(M)$ is the union of $k$ independent sets. We say such a matroid can be covered by $k$ independent sets. Edmonds [3] gave the following characterization of the members of $\mathcal{M}_{k}$.

Theorem 1.1**.**

A matroid $M$ has $k$ independent sets whose union is $E(M)$ if and only if, for every subset $A$ of $E(M)$,

[TABLE]

Clearly, $\mathcal{M}_{k}$ is closed under deletion. However, $\mathcal{M}_{k}$ is not closed under contraction. For example, the $6$-element rank-$3$ uniform matroid $U_{3,6}$ can be covered by two independent sets, yet contracting a point of this matroid gives $U_{2,5}$, which cannot. For all $k$, the loop is the unique minor-minimal matroid not in $\mathcal{M}_{k}$. On that account, we limit the types of obstructions we consider. We first examine the minor-minimal loopless matroids that are not in $\mathcal{M}_{k}$. We find the following result.

Proposition 1.2**.**

The unique minor-minimal loopless matroid that cannot be covered by $k$ independent sets is $U_{1,k+1}$.

Restricting attention to minor-minimal simple matroids not in $\mathcal{M}_{k}$, we find much more structure. We have the following collection of ten matroids for the case when $k$ is two. In this result, $P(M_{1},M_{2})$ denotes the parallel connection of matroids $M_{1}$ and $M_{2}$, this matroid being unique when both $M_{1}$ and $M_{2}$ have transitive automorphism groups. Geometric representations of the nine of these ten matroids of rank at most four are shown in Figure 1. A diagram representing the tenth matroid, $P(M(K_{4}),M(K_{4}))$ is also given where we note that this matroid has rank five.

Theorem 1.3**.**

The minor-minimal simple matroids that cannot be covered by two independent sets are $U_{2,5}$, $P(U_{2,4},U_{2,4})$, $O_{7}$, $P_{7}$, $F_{7}^{-}$, $F_{7}$, $P(U_{2,4},M(K_{4}))$, $M(K_{5}\setminus e)$, $M^{*}(K_{3,3})$, and $P(M(K_{4}),M(K_{4}))$.

The following consequence of Theorem 1.1 will be helpful.

Lemma 1.4**.**

Let $M$ be a minor-minimal matroid that cannot be covered by $k$ independent sets. Then

[TABLE]

Moreover, $M$ has no coloops.

For a matroid $M$, we write $\varepsilon(M)$ for $|E(\operatorname{si}(M))|$, the number of rank-one flats of $M$. The density $d(M)$ of $M$ is $\tfrac{\varepsilon(M)}{r(M)}$ unless $r(M)=0$. In the exceptional case, $\varepsilon(M)=0$ and we define $d(M)=0$. We say that $M$ is density-critical when $d(N)<d(M)$ for all proper minors $N$ of $M$. Note that all density-critical matroids are simple. By Lemma 1.4 and Theorem 1.1, $M$ is a minor-minimal simple matroid that cannot be covered by $k$ independent sets if and only if $d(M)>k$ but $d(N)\leq k$ for all proper minors $N$ of $M$. Such matroids are strictly $k$-density-critical where, for $t\geq 0$, we say a matroid is strictly $t$-density-critical when its density is strictly greater than $t$ while all its proper minors have density at most $t$. Thus Theorem 1.3 explicitly determines all ten strictly $2$-density-critical matroids.

We propose the following.

Conjecture 1.5**.**

For all positive integers $k$, there are finitely many minor-minimal simple matroids that cannot be covered by $k$ independent sets.

More generally, we make the following conjectures. For $t>0$, we say a matroid is $t$-density-critical when its density is at least $t$ while all of its proper minors have density strictly less than $t$.

Conjecture 1.6**.**

For all $t\geq 0$, there are finitely many strictly $t$-density-critical matroids.

Conjecture 1.7**.**

For all $t>0$, there are finitely many $t$-density-critical matroids.

We also propose the following weakening of the last conjecture.

Conjecture 1.8**.**

For all $t\geq 0$, there are finitely many density-critical matroids with density exactly $t$.

We note that these conjectures hold over any class of matroids that is well-quasi-ordered with respect to minors. In particular, by a result announced by Geelen, Gerards, and Whittle (see, for example, [4]), these conjectures hold within the class of matroids representable over a fixed finite field.

Because the two excluded minors for series-parallel networks, $U_{2,4}$ and $M(K_{4})$, have density exactly two, for $k<2$, all density-critical matroids of density at most $k$ are series-parallel networks. For $k>2$, finding all density-critical matroids of density at most $k$ does not seem straightforward. However, we were able to solve this problem when $k=\tfrac{9}{4}$. For all $n\geq 2$, we denote by $P_{n}$ any matroid that can be constructed from $n$ copies of $M(K_{3})$ via a sequence of $n-1$ parallel connections. In particular, $P_{2}\cong M(K_{4}\backslash e)$. There are two choices for $P_{3}$ depending on which element of $M(K_{4}\backslash e)$ is used as the basepoint of the parallel connection with the third copy of $M(K_{3})$. We denote by $M_{18}$ the $18$-element matroid that is obtained by attaching, via parallel connection, a copy of $M(K_{4})$ at each element of an $M(K_{3})$.

Theorem 1.9**.**

*The following is a list of all pairs $(M,d)$ where $M$ is a density-critical matroid of density $d$ and $d\leq\tfrac{9}{4}$:

$(U_{1,1},1)$, $(U_{2,3},\tfrac{3}{2})$, $(M\left(P_{n}\right),\tfrac{2n+1}{n+1})$ for all $n\geq 2$, $(U_{2,4},2)$, $(M(K_{4}),2)$, $(P(M(K_{4}),M(K_{4})),\tfrac{11}{5})$, $(P(U_{2,4},M(K_{4})),\tfrac{9}{4})$, $(M(K_{5}\backslash e),\tfrac{9}{4})$, $(M^{*}(K_{3,3}),\tfrac{9}{4})$, $(M_{18},\tfrac{9}{4})$.*

2. Preliminaries

This section proves some preliminary results beginning with two that were stated in the introduction.

Proof of Proposition 1.2..

Clearly, $U_{1,k+1}$ is a minor-minimal loopless matroid that cannot be covered by $k$ independent sets. Conversely, suppose that $M$ is a minor-minimal loopless matroid that cannot be covered by $k$ independent sets. Certainly, $M$ contains some element $e$. Let $P\cup\{e\}$ be the parallel class of $M$ that contains $e$ where $P=\{e_{1},e_{2},\dots,e_{\ell}\}$ and $e\not\in P$. Now $M/e\setminus P$ is loopless, so, by minimality, $M/e\setminus P$ can be covered by $k$ independent sets $\{A_{1},A_{2},\dots,A_{k}\}$. Note that each $A_{i}\cup\{e\}$ is independent in $M$, so if $|P|=\ell\leq k-1$, then $\{A_{1}\cup\{e_{1}\},A_{2}\cup\{e_{2}\},\dots,A_{\ell}\cup\{e_{\ell}\},A_{\ell+1}\cup\{e\},\dots,A_{k}\cup\{e\}\}$ is a set of $k$ independent sets that covers $M$. Thus $|P|\geq k$, and so $M\cong U_{1,k+1}$. ∎

Since $U_{1,k+1}$ is a $(k+1)$-element cocircuit, the matroids having no $U_{1,k+1}$-minor are precisely the matroids for which every cocircuit has at most $k$ elements.

Proof of Lemma 1.4..

Take $x$ in $E(M)$. Then $M\setminus x$ can be covered by $k$ independent sets. Thus, by Theorem 1.1,

[TABLE]

We deduce that $kr(M)=|E(M)|-1$ and $r(M)=r(M\setminus x)$ so $M$ has no coloops. ∎

Lemma 2.1**.**

Let $M$ be a density-critical matroid of rank at least two. For each subset $S$ of $E(M)$,

[TABLE]

In particular, every element of $M$ is in a triangle and is in at least two triangles when $d(M)\geq 2$.

Proof.

Since $M$ is density-critical and therefore simple,

[TABLE]

Hence $r(M)\varepsilon(M/S)<|E(M)|(r(M)-r(S))$, so

[TABLE]

Thus $d(M)r(S)<|E(M)|-\varepsilon(M/S)$. In particular, $d(M)<|E(M)|-\varepsilon(M/e)$ for all $e$ in $E(M)$. Hence every such element $e$ is in at least one triangle, and $e$ is in at least two triangles when $d(M)\geq 2$. ∎

The next result will be useful in the proof of Theorem 1.3.

Lemma 2.2**.**

In a $3$-connected matroid $M$, let $F$ be a $2k$-element set $\{b_{1},a_{1},b_{2},a_{2},\dots,b_{k},a_{k}\}$. Suppose $\{b_{1},b_{2},\dots,b_{k}\}$ is independent and $\{b_{i},a_{i},b_{i+1}\}$ is a circuit for all $i$, where $b_{k+1}=b_{1}$. Then $M|F$ is a wheel of rank at least three or a whirl of rank at least two.

Proof.

Since $M$ is $3$-connected with at least four elements, it is simple. Now $M|F$ has $\{a_{i},b_{i+1},a_{i+1}\}$ as a triad, where $a_{k+1}=a_{1}$. By a result of Seymour [8] (see also [7, Lemma 8.8.5(ii)]), $M|F$ is a wheel or a whirl of rank $k$. ∎

3. The matroids that cannot be covered by two independent sets

In this section, we prove Theorem 1.3 thereby specifying all of the minor-minimal simple matroids that cannot be covered by two independent sets.

Proof of Theorem 1.3..

It is straightforward to check that each of the matroids listed is a minor-minimal simple matroid that cannot be covered by two independent sets. Now let $M$ be such a matroid. The next two assertions are immediate consequences of Lemmas 1.4, 2.1, and 1.1. However, we include proofs independent of Edmonds’s result for completeness.

3.1.1**.**

Every element of $M$ is contained in at least two triangles.

Let $e$ be an element of $M$ and let $M^{\prime}=\operatorname{si}(M/e)$. By minimality, $M^{\prime}$ has a partition into two independent sets $A$ and $B$. Suppose $e$ is not in a triangle. Then $E(M^{\prime})=E(M)-\{e\}$ and we have $r_{M}(A\cup\{e\})=r_{M^{\prime}}(A)+1=|A|+1$ and $r_{M}(B\cup\{e\})=|B|+1$, so $M$ is covered by the independent sets $A\cup\{e\}$ and $B\cup\{e\}$, which is a contradiction.

Now suppose $e$ is in exactly one triangle $\{e,c,d\}$ of $M$. We may assume that $M^{\prime}=M/e\backslash c$ and that $d\in A$. Then $r_{M}(A\cup\{c\})=r_{M}(A\cup\{c,e\})=r_{M^{\prime}}(A)+1=|A|+1$ and $r_{M}(B\cup\{e\})=r_{M^{\prime}}(B)+1=|B|+1$, so $M$ is covered by the independent sets $A\cup\{c\}$ and $B\cup\{e\}$. This contradiction implies that 3.1.1 holds.

3.1.2**.**

$|E(M)|\leq 2r(M)+1$* and $|A|\leq 2r(A)$ for every proper subset $A$ of $E(M)$.*

Suppose $A$ is a proper subset of $E(M)$. By the minimality of $M$, we can cover $M|A$ by two independent sets, and so $|A|\leq 2r(A)$. It follows easily that $|E(M)|\leq 2r(M)+1$. Thus 3.1.2 holds.

We construct a simple auxiliary graph $G$ from $M$, the vertices of which are the elements of $M$; two such vertices are adjacent exactly when they share a triangle in $M$. Next, we show the following.

3.1.3**.**

Let $Z$ be the vertex set of a component of $G$. Then $M|Z$ has a wheel or a whirl as a restriction.

We may assume that $M|Z$ has no line with four or more points otherwise $M$ has a rank-$2$ whirl as a restriction. For $b_{1}$ in $Z$, by 3.1.1, we can construct a maximal sequence $b_{1},a_{1},b_{2},a_{2},\dots,b_{n}$ of distinct elements such that $\{b_{1},b_{2},\dots,b_{n}\}$ is independent and $\{b_{i},a_{i},b_{i+1}\}$ is a triangle for all $i$ in $\{1,2,\dots,n-1\}$. Then $n\geq 3$.

Now $M$ has a triangles $\{b_{n},a_{n},b_{n+1}\}$ and $\{b_{0},a_{0},b_{1}\}$ that differ from $\{b_{n-1},a_{n-1},b_{n}\}$ and $\{b_{1},a_{1},b_{2}\}$, respectively. Let $A^{\prime}=\{b_{1},a_{1},b_{2},a_{2},\dots,b_{n-1},a_{n-1},b_{n}\}$. Assume that both $\{a_{n},b_{n+1}\}$ and $\{a_{0},b_{0}\}$ avoid $A^{\prime}$. Then $|A^{\prime}\cup\{a_{n},b_{n+1}\}|=2n+1=2r(A^{\prime}\cup\{a_{n},b_{n+1}\})+1$. Thus, by 3.1.2, $A^{\prime}\cup\{a_{n},b_{n+1}\}=E(M)$. By symmetry, $A^{\prime}\cup\{a_{0},b_{0}\}=E(M)$. Hence $\{a_{n},b_{n+1}\}=\{b_{0},a_{0}\}$, so $\{b_{n},a_{n},b_{n+1},b_{1}\}$ is a $4$-point line, a contradiction.

We may now assume that $b_{n+1}$ is a member $c_{i}$ of $\{b_{i},a_{i}\}$ for some $i$ with $1\leq i\leq n-1$. Then $\{c_{i},b_{i+1},b_{i+2},\dots,b_{n}\}$ is an independent set in $M|Z$ such that every two consecutive elements in the given cyclic order are in a triangle. Thus, by Lemma 2.2, $M|Z$ has a wheel or whirl of rank $n-i+1$ as a restriction. Hence 3.1.3 holds.

3.1.4**.**

For some component of $G$ having vertex set $Z$, the matroid $M|Z$ is not a wheel or a whirl.

Assume that this fails. Then, by 3.1.1, the only components of $G$ are rank-$2$ whirls or rank-$3$ wheels. Assume there are $s$ of the former and $t$ of the latter. Then $|E(M)|=4s+6t=2(2s+3t)$. Clearly $r(M)\leq 2s+3t$. By 3.1.2, equality must hold here. Hence each component of $G$ corresponds to a wheel or whirl component of $M$. As each wheel and each whirl can be covered by two independent sets, so too can $M$, a contradiction. Thus 3.1.4 holds.

Now take a component of $G$ having vertex set $Z$ such that $M|Z$ is not a wheel or a whirl. By 3.1.3, consider a wheel or whirl restriction of $M|Z$ with basis $B=\{b_{1},b_{2},\dots,b_{n}\}$ and ground set $W=\{b_{1},a_{1},b_{2},a_{2},\dots,b_{n},a_{n}\}$. Let $\{b_{i},a_{i},b_{i+1}\}$ be a triangle for all $i$ where $b_{n+1}=b_{1}$. As $W\neq Z$, there is a point $\beta_{1}$ in $W$ that is contained in a triangle $\{\beta_{1},\alpha_{1},\beta_{2}\}$ that is not a triangle of $M|W$. If $M|W$ is a rank-$2$ whirl or a rank-$3$ wheel, then, by symmetry, we may assume that $\beta_{1}=a_{1}$. If, instead, $M|W$ is neither a rank-$2$ whirl nor a rank-$3$ wheel, then 3.1.1 guarantees that such a triangle $\{\beta_{1},\alpha_{1},\beta_{2}\}$ exists with $\beta_{1}=a_{1}$. By repeatedly using 3.1.1, we can construct a sequence $\beta_{1},\alpha_{1},\dots,\beta_{m+1}$ where $\{\beta_{i},\alpha_{i},\beta_{i+1}\}$ is a triangle for all $i$ in $\{1,2,\dots,m\}$ and $B\cup\{\beta_{2},\dots,\beta_{m+1}\}$ is dependent but $B\cup\{\beta_{2},\dots,\beta_{m}\}$ is independent. By potentially interchanging $\alpha_{m}$ and $\beta_{m+1}$, we may assume that $\alpha_{m}\notin W$. Let $Q=\{\beta_{1},\alpha_{1},\dots,\beta_{m+1}\}$. Then

[TABLE]

As $|W\cup(Q-\{\beta_{m+1}\})|=2(n+m-1)+1=2r(W\cup(Q-\{\beta_{m+1}\}))+1$, we deduce, by 3.1.2, that

[TABLE]

Hence

[TABLE]

Assume that the theorem fails. We now show that

3.1.5**.**

$M|Z$* has no wheel-restriction of rank exceeding three and no whirl-restriction of rank exceeding two.*

Assume that this fails. Then we may assume that $M|W$ is a wheel of rank at least four or a whirl of rank at least three. Now $r(W)=n$ and $r(Q)\leq m+1$. By (1) and submodularity, $r(\operatorname{cl}(W)\cap\operatorname{cl}(Q))\leq 2$. Assume $W$ does not span $M$. Then, by (1) and (2), we see that $m>1$ and the only possible elements of $W$ that can lie in triangles with elements of $Q-W$ are $\beta_{1}$ and $\beta_{m+1}$. But a wheel of rank at least four and a whirl of rank at least three have at least three elements that are in unique triangles. Hence one of these elements will violate 3.1.1.

We now know that $W$ spans $M$, so the unique element of $Q-W$ is $\alpha_{1}$. Each of $a_{1},a_{2},\dots,a_{n}$ must be in a triangle with $\alpha_{1}$, the other element of which is in $W$. Assume both $\{a_{1},\alpha_{1},a_{3}\}$ and $\{a_{1},\alpha_{1},a_{n-1}\}$ are triangles. Then $n=4$. Suppose $\{a_{2},\alpha_{1},a_{4}\}$ is also a triangle. Then, by Lemma 2.2, for each $i$ in $\{2,4\}$, deleting $a_{i}$ from $M|(W\cup Q)$ gives a wheel or whirl of rank four. As $\{b_{1},b_{4},\alpha_{1},a_{2}\}$ and $\{b_{2},b_{3},\alpha_{1},a_{4}\}$ are circuits, both of these deletions are wheels. It follows that $M|(W\cup Q)\cong M^{*}(K_{3,3})$, so $M\cong M^{*}(K_{3,3})$, a contradiction. Thus, we may assume that $\{a_{2},\alpha_{1},a_{4}\}$ is not a triangle. Since $\alpha_{1}\not\in\operatorname{cl}(\{b_{1},b_{2},b_{3}\})\cup\operatorname{cl}(\{b_{2},b_{3},b_{4}\})$, there is no triangle containing $\{a_{2},\alpha_{1}\}$, a contradiction.

We may now assume that $\{a_{1},\alpha_{1},a_{3}\}$ is not a triangle. Then, by 3.1.1, $W$ has distinct elements $x$ and $y$ such that $\{a_{1},\alpha_{1},x\}$ and $\{a_{3},\alpha_{1},y\}$ are triangles. Thus $\{a_{1},a_{3},x,y\}$ contains a circuit. Now $\{a_{1},a_{3}\}$ is not in a triangle of $M|W$. Moreover, if $\{a_{1},x,y\}$ is a triangle, then $\{x,y\}=\{b_{1},b_{2}\}$. Using the triangles, $\{a_{1},\alpha_{1},x\}$ and $\{a_{3},\alpha_{1},y\}$, we deduce that $a_{3}\in\operatorname{cl}(\{b_{1},b_{2}\})$, a contradiction. It follows that $\{a_{1},a_{3},x,y\}$ is a circuit of $M$. Thus $M|W$ is either a rank-$3$ whirl or a rank-$4$ wheel.

Suppose $M|W$ is a rank-$3$ whirl. Then $M$ is an extension of this matroid by $\alpha_{1}$ in which every element is in at least two triangles. If $\{a_{1},a_{2},\alpha_{1}\}$ or $\{a_{2},a_{3},\alpha_{1}\}$ is a triangle, then one easily checks that $M\cong O_{7}$ or $M\cong P_{7}$, a contradiction. Hence we may assume that none of $\{a_{1},a_{2},\alpha_{1}\}$, $\{a_{2},a_{3},\alpha_{1}\}$, or $\{a_{3},a_{1},\alpha_{1}\}$ is a triangle. Then, to avoid having $U_{2,5}$ as a minor of $M$, we must have $\{a_{1},b_{3},\alpha_{1}\}$, $\{a_{2},b_{1},\alpha_{1}\}$, and $\{a_{3},b_{2},\alpha_{1}\}$ as triangles, that is, $M\cong F^{-}_{7}$, a contradiction.

We are left with the possibility that $M|W$ is a rank-$4$ wheel. Since it has $\{a_{1},a_{3},x,y\}$ as a circuit, it follows that $\{x,y\}=\{a_{2},a_{4}\}$. Then $M$ has either $\{a_{1},a_{2},\alpha_{1}\}$ and $\{a_{3},a_{4},\alpha_{1}\}$ as triangles or $\{a_{1},a_{4},\alpha_{1}\}$ and $\{a_{2},a_{3},\alpha_{1}\}$ as triangles. By symmetry, we may assume that we are in the second case. Then, by submodularity using the sets $\{b_{1},b_{2},a_{1},a_{4},b_{4},\alpha_{1}\}$ and $\{b_{2},b_{3},a_{2},a_{3},b_{4},\alpha_{1}\}$, we deduce that $r(\{b_{1},b_{4},\alpha_{1}\})=2$. It follows that $M\cong M(K_{5}\backslash e)$, a contradiction. We conclude that 3.1.5 holds.

Now suppose that $W$ spans $Z$. If $M|W$ is a rank-$2$ whirl, then $M|Z\cong U_{2,5}$, a contradiction. If $M|W$ is a rank-$3$ wheel, then one easily checks that $M|Z$ is isomorphic to one of $O_{7}$, $F_{7}^{-}$, or $F_{7}$, a contradiction.

We may now assume that $W$ does not span $Z$. Then $m>1$. By (3), $\beta_{m+1}\in W\cup(Q-\{\beta_{m+1}\})$. We will first suppose that $\beta_{m+1}=\beta_{i}$ for some $i$ in $\{1,2,\dots,m\}$. Then $\{\beta_{i},\beta_{i+1},\dots,\beta_{m}\}$ is an independent set and $\{\beta_{j},\alpha_{j},\beta_{j+1}\}$ is a triangle for all $j$ in $\{i,i+1,\dots,m\}$. By 3.1.5 and Lemma 2.2, for $R=\{\beta_{i},\alpha_{i},\beta_{i+1},\alpha_{i+1},\dots,\beta_{m},\alpha_{m}\}$, the matroid $M|R$ is a rank-$3$ wheel or a rank-$2$ whirl. Then the matroid obtained from $M|Z$ by contracting $\{\alpha_{2},\alpha_{3},\dots,\alpha_{i-1}\}$ and simplifying is the parallel connection of $M|W$ and $M|R$, that is, $M|Z$ has as a minor one of $P(U_{2,4},U_{2,4})$, $P(U_{2,4},M(K_{4}))$, and $P(M(K_{4}),M(K_{4}))$, a contradiction.

Finally, suppose that $\beta_{m+1}\not\in\{\beta_{1},\beta_{2},\dots,\beta_{m}\}$. Then $\beta_{m+1}$ is $\alpha_{i}$ for some $i\geq 1$, or $\beta_{m+1}\in W$. Consider the first case and take $\alpha_{m+1}=\beta_{i}$. Then, by 3.1.5 and Lemma 2.2, with $R=\{\beta_{i+1},\alpha_{i+1},\dots,\beta_{m+1},\alpha_{m+1}\}$, we have that $M|R$ is a rank-3 wheel or a rank-2 whirl. Contracting $\{\alpha_{2},\alpha_{3},\dots,\alpha_{i-1}\}$ from $M|Z$ and simplifying, we obtain one of $P(U_{2,4},U_{2,4})$, $P(U_{2,4},M(K_{4}))$, and $P(M(K_{4}),M(K_{4}))$, a contradiction. In the second case, when $\beta_{m+1}\in W$, we recall that $\beta_{1}=a_{1}$. Suppose that $\{\beta_{1},\beta_{m+1}\}$ is not in a triangle of $M|W$. Then $M|W\cong M(K_{4})$ and $\beta_{m+1}=b_{3}$. By assumption, $\{b_{1},b_{2},b_{3}\}\cup\{\beta_{2},\dots,\beta_{m}\}$ is independent. By Lemma 2.2, the triangles $\{b_{1},b_{2},a_{1}\}$, $\{a_{1},\alpha_{1},\beta_{2}\},\dots,\{\beta_{m},\alpha_{m},b_{3}\}$, $\{b_{3},a_{3},b_{1}\}$ imply that $M|Z$ has a wheel or whirl of rank at least four as a restriction, a contradiction. We deduce that $\{\beta_{1},\beta_{m+1}\}$ is in a triangle of $M|W$. Then, by symmetry, we may assume that $\beta_{m+1}=b_{1}$. We let $\alpha_{m+1}=b_{2}$. Then, for $R=\{\beta_{1},\alpha_{1},\dots,\beta_{m+1},\alpha_{m+1}\}$, we have that $M|R$ is a rank-3 wheel or a rank-2 whirl. But $\alpha_{1}\not\in\operatorname{cl}(W)$, so $M|R$ is a rank-3 wheel. If $M|W$ is a rank-$2$ whirl, then $O_{7}$ is a restriction of $M|Z$, a contradiction. If $M|W$ is a rank-$3$ wheel, then $M|(W\cup R)$ has rank four and consists of two copies of $M(K_{4})$ sharing a triangle. This matroid is $M(K_{5}\backslash e)$, a contradiction. ∎

4. The density-critical matroids of small density

In this section, we prove Theorem 1.9. The following result [6] (see also [7, Lemma 4.3.10]) will be used repeatedly in this proof.

Lemma 4.1**.**

Let $M$ be a connected matroid having at least two elements and let $\{e_{1},e_{2},\dots,e_{m}\}$ be a cocircuit of $M$ such that $M/e_{i}$ is disconnected for all $i$ in $\{1,2,\dots,m-1\}$. Then $\{e_{1},e_{2},\dots,e_{m-1}\}$ contains a $2$-circuit of $M$.

We shall make repeated use of the following consequence of this lemma.

Corollary 4.2**.**

Let $M$ be a simple connected matroid and $Z$ be a non-empty non-spanning subset of $E(M)$. Then $M$ has a simple connected minor N such that $N|Z=M|Z$ and $r(N)=r_{M}(Z)$.

Proof.

Let $C^{*}$ be a cocircuit of $M$ that is disjoint from $\operatorname{cl}(Z)$. As $M$ is simple, it follows by Lemma 4.1 that there is an element $e$ of $C^{*}$ such that $M/e$ is connected. Since $e\not\in\operatorname{cl}(Z)$, we see that $(M/e)|Z=M|Z$. Clearly we can label $\operatorname{si}(M/e)$ so that its ground set contains $Z$. If $r(M)-r(Z)=1$, then we take $N=\operatorname{si}(M/e)$. Otherwise we repeat the above process using $\operatorname{si}(M/e)$ in place of $M$. After $r(M)-r(Z)$ applications of this process, we obtain the desired minor $N$. ∎

The next result, which was proved by Dirac [2], follows easily by induction after recalling that a connected matroid with no minor isomorphic to $U_{2,4}$ or $M(K_{4})$ is isomorphic to the cycle matroid of a series-parallel network.

Lemma 4.3**.**

Let $M$ be a simple matroid having no minor isomorphic to $U_{2,4}$ or $M(K_{4})$. Then

[TABLE]

We omit the elementary proof of the next result a consequence of which is that every density-critical matroid is connected.

Lemma 4.4**.**

Let $M_{1}$ and $M_{2}$ be matroids of rank at least one. Then

[TABLE]

Moreover, equality holds here if and only if $d(M_{1})=d(M_{2})$.

The next result will be useful in identifying the density-critical matroids of density at most two.

Lemma 4.5**.**

Let $M$ be a density-critical matroid with $d(M)\leq 2$. If $(X_{1},X_{2})$ is a $2$-separation of $M$, then there is an element $p$ in $\operatorname{cl}(X_{1})\cap\operatorname{cl}(X_{2})$, and $M=P(M|(X_{1}\cup\{p\}),M|(X_{2}\cup\{p\}))$.

Proof.

As $(X_{1},X_{2})$ is a $2$-separation of $M$, for some element $q$ not in $E(M)$, we can write $M$ as $M_{1}\oplus_{2}M_{2}$ where each $M_{i}$ has ground set $X_{i}\cup\{q\}$. Let $|E(M_{i})|=n_{i}$ and $r(M_{i})=r_{i}$. Assume that both $M_{1}$ and $M_{2}$ are simple. Then $\tfrac{|E(M)|}{r(M)}>\tfrac{|E(M_{1})|}{r(M_{1})}$, so

[TABLE]

Hence

[TABLE]

By symmetry,

[TABLE]

Adding the last two inequalities gives $n_{1}+n_{2}>2(r_{1}+r_{2}),$ so $n_{i}>2r_{i}$ for some $i$. Thus $d(M_{i})>2$. Since $M$ is density-critical with density at most two, this is a contradiction. We conclude that $M_{1}$ or $M_{2}$, say $M_{1}$, is non-simple. Thus it has an element $p$ in parallel with the basepoint $q$ of the $2$-sum. Hence $M=P(M|(X_{1}\cup\{p\}),M|(X_{2}\cup\{p\}))$. ∎

Lemma 4.6**.**

Let $N$ be a simple connected matroid in which all but at most one element is in at least two triangles. Then $N$ has no $2$-cocircuits. Moreover, if $N$ has $\{a,b,c\}$ as a triad, then either

• (i)

$\{a,b,c\}$* is contained in a $4$-point line and $N=P(U_{2,4},N\setminus\{a,b,c\})$; or*

• (ii)

$N$* has a triangle $\{x,y,z\}$ such that $N|\{a,b,c,x,y,z\}\cong M(K_{4})$ and $N$ is the generalized parallel connection of $N|\{a,b,c,x,y,z\}$ and $N\setminus\{a,b,c\}$ across the triangle $\{x,y,z\}$.*

Proof.

As $N$ has at most one element that is not in at least two triangles, $N$ has no $2$-cocircuits. Suppose $\{a,b,c\}$ is a triad of $N$. If $\{a,b,c\}$ is also a triangle, then $\{a,b,c\}$ is $2$-separating in $N$. Moreover, $\{a,b,c\}$ is contained in a $4$-point line $\{a,b,c,d\}$ and (i) holds.

We may now assume that $\{a,b,c\}$ is not a triad of $N$. Then, because at least two of $a,b,$ and $c$ are in at least two triangles, the hyperplane $E(N)-\{a,b,c\}$ of $N$ contains distinct elements $x$, $y$, and $z$ such that $\{a,b,z\}$, $\{a,y,c\}$, and $\{x,b,c\}$ are triangles. Now

[TABLE]

Thus $\{x,y,z\}$ is a triangle of $N$ and $N|\{a,b,c,x,y,z\}\cong M(K_{4})$. It follows by a result of Brylawski [1] (see also [7, Proposition 11.4.15]) that (ii) holds. ∎

Corollary 4.7**.**

Let $N$ be a simple connected matroid in which all but at most one element is in at least two triangles and $d(N)\leq\tfrac{9}{4}$. If $r(N)=2$, then $N\cong U_{2,4}$. If $r(N)=3$, then $N\cong M(K_{4})$. If $r(N)=4$, then $N\cong P(U_{2,4},M(K_{4})),M(K_{5}\setminus e)$, or $M^{*}(K_{3,3})$.

Proof.

We omit the straightforward proof for the case when $r(N)\in\{2,3\}$. Assume $r(N)=4$. By Lemma 4.6, $N$ has no $2$-cocircuits. Now suppose $N$ has $\{a,b,c\}$ as a triad. If (i) of Lemma 4.6 holds, then $N=P(U_{2,4},N\setminus\{a,b,c\})$. By the result in the rank-$3$ case, $N\setminus\{a,b,c\}\cong M(K_{4})$, so $N\cong P(U_{2,4},M(K_{4}))$. If, instead, (ii) of Lemma 4.6 holds, then $N$ is the generalized parallel connection across a triangle $\{x,y,z\}$ of $M(K_{4})$ and $N\setminus\{a,b,c\}$. In the latter, $E(N\setminus\{a,b,c,x,y,z\})$ must be a triad of $N$, so $N\setminus\{a,b,c\}\cong M(K_{4})$. Hence $N$ is the generalized parallel connection across a triangle of two copies of $M(K_{4})$, so $N\cong M(K_{5}\setminus e)$.

We may now assume that $N$ has no triads. Then every cocircuit of $N$ has at least four elements. As $N$ certainly has a plane that contains two intersecting triangles, $\{x,f_{1},g_{1}\}$ and $\{x,f_{2},g_{2}\}$, we deduce that $|E(N)|\geq 9$, so $|E(N)|=9$. Let $\{a,b,c,d\}$ be the cocircuit $E(N)-\{x,f_{1},f_{2},g_{1},g_{2}\}$. Because $N$ has no plane with more than five points and has all but at most one element in two triangles, we may assume that $\{a,b,g_{1}\}$ and $\{a,c,g_{2}\}$ are triangles of $N$. Then $N\setminus d$ has $\{x,f_{1},g_{1}\},\{g_{1},b,a\},\{a,c,g_{2}\},\{g_{2},f_{2},x\}$ as triangles. By Lemma 2.2, $N\setminus d$ is a rank-$4$ wheel or whirl. In this matroid, $f_{1}$, $b$, $c$, and $f_{2}$ are in unique triangles. It follows that $N$ must have $\{d,f_{1},c\}$ and $\{d,b,f_{2}\}$ as triangles. Thus $N\setminus d$ is a rank-$4$ wheel. Likewise, $N\setminus f_{1}$ and $N\setminus c$ are also rank-$4$ wheels, so $N\cong M^{*}(K_{3,3})$. ∎

Lemma 4.8**.**

Let $N$ be a simple matroid of rank at least three in which every element is in at least two triangles. Suppose $e\in E(N)$. Then

• (i)

$e$* is in a plane of $N$ having at least seven points; or*

• (ii)

every element of $\operatorname{si}(N/e)$ is in at least two triangles; or

• (iii)

$N$* has a $U_{2,4}$- or $M(K_{4})$-restriction using $e$.*

Proof.

Assume that neither (i) nor (iii) holds. We show that every element of $\operatorname{si}(N/e)$ is in at least two triangles. First consider a triangle $\{e,c_{1},c_{2}\}$ of $N$ containing $e$. Let $\{c_{1},d_{1},f_{1}\}$ and $\{c_{2},d_{2},f_{2}\}$ be triangles of $N$ where neither contains $e$. If $r(\{e,c_{1},d_{1},f_{1},c_{2},d_{2},f_{2}\})=4$, then, in $\operatorname{si}(N/e)$, the element $c$ corresponding to $c_{1}$ and $c_{2}$ is in at least two triangles. Now suppose $r(\{e,c_{1},d_{1},f_{1},c_{2},d_{2},f_{2}\})=3$. Since $N$ has no plane with more than six points, we may assume that $f_{1}=f_{2}$. Rename this element $f$. If $\{e,d_{1},d_{2}\}$ is not a triangle, then $\operatorname{si}(N/e)$ has a $4$-point line containing $c$, so $c$ is in at least two triangles of this matroid. If $\{e,d_{1},d_{2}\}$ is a triangle of $N$, then $N|\{e,c_{1},c_{2},d_{1},d_{2},f\})\cong M(K_{4})$, a contradiction.

Now let $f$ be an element of $N$ that is not in a triangle with $e$. Let $\{f,g_{1},h_{1}\}$ and $\{f,g_{2},h_{2}\}$ be triangles of $N$. Then $\operatorname{si}(N/e)$ has at least two triangles containing $f$ otherwise $N|\{e,f,g_{1},g_{2},h_{1},h_{2}\})\cong M(K_{4})$, a contradiction. ∎

Recall that $M_{18}$ is the $18$-element matroid that is obtained by attaching, via parallel connection, a copy of $M(K_{4})$ at each element of an $M(K_{3})$.

Lemma 4.9**.**

Let $N$ be a simple connected non-empty matroid in which every element is in a $U_{2,4}$- or $M(K_{4})$-restriction. Assume that $d(N)\leq\tfrac{9}{4}$ but $d(N^{\prime})<\tfrac{9}{4}$ for all proper minors $N^{\prime}$ of $N$. Then $N$ is isomorphic to $U_{2,4},M(K_{4}),P(U_{2,4},M(K_{4})),P(M(K_{4}),M(K_{4})),M(K_{5}\setminus e)$, or $M_{18}$.

Proof.

Since $d(N^{\prime})\leq\tfrac{9}{4}$ for all minors $N^{\prime}$ of $N$, we see that, in any such $N^{\prime}$, no line has more than four points and no plane has more than six points. Next we show the following.

4.9.1**.**

If $N$ has a $4$-point line, then $N$ is isomorphic to $U_{2,4}$ or $P(U_{2,4},M(K_{4}))$.

This is immediate if $r(N)=2$. Because $N$ has no plane with more than six points, $r(N)\neq 3$. Let $L$ be a $4$-point line of $N$ and let $Z$ be a subset of $E(N)$ not containing $L$ such that $N|Z$ is isomorphic to $U_{2,4}$ or $M(K_{4})$. If $L\cap Z\neq\emptyset$, then again, since $N$ has no plane with more than six points, we deduce that $N\cong P(U_{2,4},M(K_{4}))$. We may now assume that $L\cap Z=\emptyset$. If $r(L\cup Z)\leq r(Z)+1$, then $N$ has a rank-$3$ or rank-$4$ restriction of density exceeding $\tfrac{9}{4}$, a contradiction. We deduce that $r(L\cup Z)=r(Z)+2$.

By Corollary 4.2, $N$ has a simple connected minor $N^{\prime}$ such that $N^{\prime}|(L\cup Z)=N|(L\cup Z)$ and $r(N^{\prime})=r(Z)+2$. As $N^{\prime}$ is connected, it has an element $x^{\prime}$ that is not in the closure of $L$ or of $Z$. Then $N^{\prime}/x^{\prime}$ has $N|L$ and $N|Z$ as restrictions and has rank $r(Z)+1$. Thus $\operatorname{si}(N^{\prime}/x^{\prime})$ has either a plane with more than six points or has $P(U_{2,4},M(K_{4}))$ as a restriction. Each possibility yields a contradiction, so 4.9.1 holds.

We may now assume that every element of $N$ is in an $M(K_{4})$-restriction. We may also assume that $N$ is not isomorphic to $M(K_{4})$ or $P(M(K_{4}),M(K_{4}))$. Next we show the following.

4.9.2**.**

Let $X$ and $Y$ be distinct subsets of $E(N)$ such that both $N|X$ and $N|Y$ are isomorphic to $M(K_{4})$. If $|X\cap Y|\geq 2$, then $N\cong M(K_{5}\setminus e)$.

Since $N$ has no plane with more than six points, $r(X\cup Y)>3$. As $|X\cap Y|\geq 2$, it follows by submodularity that $r(X\cup Y)=4$ and $r(X\cap Y)=2$. As $d(N|(X\cup Y))\leq\tfrac{9}{4}$, we deduce that $|X\cup Y|=9$, so $|X\cap Y|=3$ and $N=N|(X\cup Y)$. Moreover, $N|X$ and $N|Y$ meet in a triangle $\Delta$. By Lemma 4.6, $N$ is the generalized parallel connection of $N|X$ and $N|Y$ across $\Delta$. Thus $N\cong M(K_{5}\setminus e)$ as each of $N|X$ and $N|Y$ is isomorphic to $M(K_{4})$, so 4.9.2 holds.

We may now assume that $E(N)$ has at least three distinct subsets $X$ with $N|X\cong M(K_{4})$ and that no two such subsets meet in more than one element.

4.9.3**.**

$N$* does not have $P(M(K_{4}),M(K_{4}))$ as a restriction.*

Assume that $N|X\cong P(M(K_{4}),M(K_{4}))$ and $N|Y\cong M(K_{4})$ where $Y\not\subseteq X$. Suppose $|X\cap Y|=k$ where $k\in\{1,2\}$. Then $r(X\cup Y)\leq 8-k$ and $|X\cup Y|=17-k$, so

[TABLE]

Simplifying we obtain the contradiction that $4\geq 5k\geq 5$. We deduce using 4.9.2 that $|X\cap Y|=0$. Then $r(X\cup Y)=8$ otherwise $d(N|(X\cup Y))>\tfrac{9}{4}$.

By Corollary 4.2, $N$ has a simple connected minor $N^{\prime}$ such that $N^{\prime}|(X\cup Y)=N|(X\cup Y)$ and $r(N^{\prime})=8$. As $N|(X\cup Y)$ is disconnected, $N^{\prime}$ must contain an element that is not in $X\cup Y$. Hence $|E(N^{\prime})|\geq 18$, so $d(N^{\prime})\geq\tfrac{9}{4}$. Thus $N^{\prime}=N$ and $|E(N)|=18$, so $N$ has a single element $z$ that is not in $X\cup Y$. The $M(K_{4})$-restriction of $N$ that contains $z$ is forced to have more than one element in common with $Y$ or one of the $M(K_{4})$-restrictions of $N|X$. This contradiction to 4.9.2 completes the proof of 4.9.3.

We now know that any two $M(K_{4})$-restrictions of $N$ have disjoint ground sets. Let $X$, $Y$, and $Z$ be distinct subsets of $E(N)$ such that each of $N|X$, $N|Y$, and $N|Z$ is isomorphic to $M(K_{4})$. Next we show the following.

4.9.4**.**

$r(X\cup Y)=6$. Moreover, $r(X\cup Y\cup Z)=9$ unless $N\cong M_{18}$.

As $|X\cup Y|=12$ and $d(N|(X\cup Y))<\tfrac{9}{4}$, we deduce that $r(X\cup Y)=6$. The density constraint also means that $r(X\cup Y\cup Z)\geq 8$. Suppose $r(X\cup Y\cup Z)=8$. Then $d(N|(X\cup Y\cup Z))=\tfrac{9}{4}$, so $N=N|(X\cup Y\cup Z)$. Now $r(N/Z)=5$. As $\tfrac{12}{5}>\tfrac{9}{4}$, we must have some parallel elements in $N/Z$. As $Z$ is skew to each of $X$ and $Y$, we know that $(N/Z)|X=N|X$ and $(N/Z)|Y=N|Y$. Thus there must be elements $x$ of $X$ and $y$ of $Y$ that are parallel in $N/Z$. If there is a second such parallel pair, then $r(N/Z)\leq 4$, a contradiction. In $N$, we see that $r(Z\cup\{x,y\})=4$. Hence, in $N/x$, we obtain a $7$-point plane $Z\cup y$ unless $\{x,y,z\}$ is a triangle of $N$ for some $z$ in $Z$. Observe that each of $N/x$, $N/y$, and $N/z$ is disconnected, so $N$ is obtained from $M(K_{3})$ by attaching a copy of $M(K_{4})$ via parallel connection at each element. Thus $N\cong M_{18}$ and 4.9.4 holds.

By Corollary 4.2, $N$ has a simple connected minor $N^{\prime}$ of rank $9$ such that $N^{\prime}|(X\cup Y\cup Z)=N|(X\cup Y\cup Z)$. As $N^{\prime}$ is connected, there is an element $g$ of $E(N^{\prime})-(X\cup Y\cup Z)$. Since $N^{\prime}$ has no plane with more than six points, $g$ is not in the closure of any of $X$, $Y$, or $Z$ in $N^{\prime}$. As $N^{\prime}/g$ has rank $8$ but has density less than $\tfrac{9}{4}$, the eighteen elements of $X\cup Y\cup Z$ cannot all be in distinct parallel classes of $N^{\prime}/g$. Thus $N^{\prime}$ has a triangle $\{x,y,g\}$ where we may assume that $x\in X$ and $y\in Y$. Since $N^{\prime}|(X\cup Y\cup Z\cup g)$ has $Z$ as a component, there is an element $h$ of $E(N^{\prime})$ that is in neither $\operatorname{cl}_{N^{\prime}}(X\cup Y)$ nor $\operatorname{cl}_{N^{\prime}}(Z)$. As above, $N^{\prime}$ has a triangle $\{h,z,t\}$ where $t\in X\cup Y$ and $z\in Z$. Contracting $g$ and $h$ from $N^{\prime}|(X\cup Y\cup Z\cup\{g,h\})$ and simplifying, we get a rank-$7$ matroid with $16$ elements. As $\tfrac{16}{7}>\tfrac{9}{4}$, we have a contradiction that completes the proof of Lemma 4.9. ∎

Lemma 4.10**.**

Let $N$ be a simple connected matroid having an element $z$ such that each of $N$ and $\operatorname{si}(N/z)$ has every element in at least two triangles. If $d(N)\leq\tfrac{9}{4}$ and $d(N^{\prime})<\tfrac{9}{4}$ for all proper minors $N^{\prime}$ of $N$, then $N$ is isomorphic to $P(U_{2,4},M(K_{4}))$, $M(K_{5}\setminus e)$, or $M^{*}(K_{3,3})$.

Proof.

We argue by induction on $r(N)$, which must be at least three. Suppose it is exactly three. Since $\operatorname{si}(N/z)$ has density less than $\tfrac{9}{4}$, it is isomorphic to $U_{2,4}$. As $d(N)\leq\tfrac{9}{4}$, we see that $|E(N)|\leq 6$. By Lemma 4.6, $N$ has no $2$-cocircuits. Thus $N$ has a triangle whose complement is a triad. By Lemma 4.6 again, $N\cong M(K_{4})$ and we get a contradiction. Hence $r(N)\geq 4$. If $r(N)=4$, then, by Corollary 4.7, $N$ is isomorphic to $P(U_{2,4},M(K_{4})),M(K_{5}\setminus e)$, or $M^{*}(K_{3,3})$.

Now assume the result holds for $r(N)<k$ and let $r(N)=k\geq 5$. Let $N_{1}=\operatorname{si}(N/z)$. Every element of $N_{1}$ is in at least two triangles. Let $N_{2}$ be a component of $N_{1}$. By Lemma 4.8, either every element of $N_{2}$ is in a $U_{2,4}$- or $M(K_{4})$-restriction, or $N_{2}$ has an element $z_{2}$ such that every element of $\operatorname{si}(N_{2}/z_{2})$ is in at least two triangles. If the latter occurs, then, by the induction assumption, $N_{2}$ is isomorphic to $P(U_{2,4},M(K_{4})),M(K_{5}\setminus e)$, or $M^{*}(K_{3,3})$. Each of these matroids has density $\tfrac{9}{4}$, a contradiction. Thus every element of $N_{2}$ is in a $U_{2,4}$- or $M(K_{4})$-restriction. As $d(N_{2})<\tfrac{9}{4}$, Lemma 4.9 implies that $N_{2}$, and hence each component of $N_{1}$, is isomorphic to one of $U_{2,4}$, $M(K_{4})$, or $P(M(K_{4}),M(K_{4}))$.

Suppose that $N_{2}=N_{1}$. Then, as $r(N)\geq 5$, we deduce that $N_{1}\cong P(M(K_{4}),M(K_{4}))$. As $N_{1}=\operatorname{si}(N/z)$, we see that $r(N)=6$. Because $d(N)\leq\tfrac{9}{4}$, it follows that $|E(N)|\leq 13$. Since $z$ is in at least two triangles of $N$, we deduce that $|E(N)|\geq|E(N_{1})|+3=14$, a contradiction.

We may now assume that $N_{1}$ has more than one component. Hence, for some $k\geq 2$, there is a collection $N^{1},N^{2},\dots,N^{k}$ of connected matroids such that $E(N^{i})\cap E(N^{j})=\{z\}$ for all $i\neq j$, the matroid $N^{i}/z$ is connected for all $i$, and $N$ is the parallel connection of $N^{1},N^{2},\dots,N^{k}$ across the common basepoint $z$. As noted above, each $\operatorname{si}(N^{i}/z)$ is isomorphic to one of $U_{2,4}$, $M(K_{4})$, or $P(M(K_{4}),M(K_{4}))$. As every element of $N$ is in at least two triangles, every element of each $N^{i}$ except possibly $z$ is in at least two triangles of $N^{i}$. Thus, by Corollary 4.7, $N^{i}\cong M(K_{4})$; or $r(N^{i})=4$ and $|E(N^{i})|=9$; or $r(N^{i})>4$. In the first case, $\operatorname{si}(N^{i}/z)\not\cong U_{2,4}$; in the second case, $d(N^{i})=\tfrac{9}{4}$. Both of these possibilities give contradictions, so $\operatorname{si}(N^{i}/z)\cong P(M(K_{4}),M(K_{4}))$ for each $i$. As $z$ is in at least two triangles of $N$, we may assume the elements of two such triangles lie in $E(N^{1})\cup E(N^{2})$. As $|E(\operatorname{si}(N^{i}/z))|=11$ and $r(N^{i}/z)=5$, we see that $|E(N^{1})\cup E(N^{2})|\geq 25$ and $r(E(N^{1})\cup E(N^{2}))=11$. But $\tfrac{25}{11}>\tfrac{9}{4}$, a contradiction. ∎

We conclude the paper by proving Theorem 1.9. In this proof, we will make extensive use of the Cunningham-Edmonds canonical tree decomposition of a connected matroid. The definition and properties of this decomposition may be found in [7, Section 8.3]. In brief, associated with each connected matroid $M$, there is a tree $T$ that is unique up to the labelling of its edges. Each vertex of $T$ is labelled by a circuit, a cocircuit, or a $3$-connected matroid with at least four elements. Moreover, no two adjacent vertices of $T$ are labelled by circuits and no two adjacent vertices are labelled by cocircuits. For an edge $e$ of $T$ whose endpoints are labelled by matroids $M_{1}$ and $M_{2}$, the ground sets of these two matroids meet in $\{e\}$. When we contract $e$ from $T$, the composite vertex that results by identifying the endpoints of $e$ is labelled by the $2$-sum of $M_{1}$ and $M_{2}$. By repeating this process, contracting all of the remaining edges of $T$ one by one, we eventually obtain a single-vertex tree. Its vertex is labelled by $M$.

Each edge $f$ of $T$ induces a partition of $E(M)$. This partition is a $2$-separation of $M$ displayed by $f$. The remaining $2$-separations of $M$ coincide with those that are displayed by those vertices of $T$ that are labelled by circuits or cocircuits. For such a vertex $v$ having label $N$, there is a partition $\{X_{1},X_{2},\dots,X_{k}\}$ of $E(M)-E(N)$ induced by the components of $T-v$. A partition $(X,Y)$ of $E(M)$ is displayed by the vertex $v$ if each $X_{i}$ is contained in $X$ or $Y$. Every such partition of $E(M)$ with both $X$ and $Y$ having at least two elements is a $2$-separation of $M$ and these $2$-separations along with those displayed by the edges of $T$ are all of the $2$-separations of $M$. Recall that, for all $n\geq 2$, we denote by $P_{n}$ any matroid that can be constructed from $n$ copies of $M(K_{3})$ via a sequence of parallel connections.

Proof of Theorem 1.9..

Let $M$ be a density-critical matroid with $d(M)\leq\tfrac{9}{4}$. Suppose $d(M)\geq 2$. By Lemma 2.1, every element of $M$ is in at least two triangles. By Corollary 4.7, if $r(M)\in\{2,3\}$, then $M$ is $U_{2,4}$ or $M(K_{4})$. We may now assume that $r(M)\geq 4$. By Lemma 4.8, either every element of $M$ is in a $U_{2,4}$- or $M(K_{4})$-restriction, or, for some element $z$ of $M$, every element of $\operatorname{si}(M/z)$ is in at least two triangles. In the first case, by Lemma 4.9, $M$ is isomorphic to $P(U_{2,4},M(K_{4})),P(M(K_{4}),M(K_{4})),M(K_{5}\setminus e)$, or $M_{18}$. In the second case, by Lemma 4.10, $M$ is isomorphic to $P(U_{2,4},M(K_{4})),M(K_{5}\setminus e)$, or $M^{*}(K_{3,3})$. Thus the theorem identifies all possible density-critical matroids with density in $[2,\tfrac{9}{4}]$ and one easily checks that each of the matroids identified is indeed density-critical.

Now suppose that $d(M)<2$. By Lemma 4.4, $M$ is connected. Clearly, if $r(M)$ is $1$ or $2$, then $M$ is isomorphic to $U_{1,1}$ or $U_{2,3}$. As $U_{2,4}$ and $M(K_{4})$ both have density $2$, $M$ is a series-parallel network (see, for example, [7, Corollary 12.2.14]). Thus, in the Cunningham-Edmonds canonical tree decomposition $T$ of $M$, every vertex is labelled by a circuit or a cocircuit. Since $M$ is simple, for every vertex of $T$ that is labelled by a cocircuit $C^{*}$, at most one element of $C^{*}$ is in $E(M)$. Let $e$ be an edge of $T$ that meets the vertex labelled by $C^{*}$. Then, for the $2$-separation $(X,Y)$ of $M$ that is displayed by $e$, Lemma 4.5 implies that $M$ has an element $p$ in $\operatorname{cl}(X)\cap\operatorname{cl}(Y)$. Thus $p\in C^{*}$, so $C^{*}$ contains exactly one element of $M$.

Now take a vertex of $T$ that is labelled by a circuit $C$ where $C=\{e_{1},e_{2},\dots,e_{k}\}$ and suppose that $k\geq 4$. Suppose $e_{1}\in E(M)$. Then $M/e_{1}$ is simple having rank $r(M)-1$. As $\tfrac{|E(M)|-1}{r(M)-1}<\tfrac{|E(M)|}{r(M)},$ we obtain the contradiction that $|E(M)|<r(M)$. We deduce that $C\cap E(M)=\emptyset$. Now $T\setminus e_{1},e_{2}$ has exactly three components. Let $T^{\prime}$ be the one containing $e_{3}$ and let $X$ be the subset of $E(M)$ corresponding to $T^{\prime}$. Then $(X,E(M)-X)$ is a $2$-separation of $M$. By Lemma 4.5, there is an element $p$ of $M$ that is in $\operatorname{cl}(X)\cap\operatorname{cl}(E(M)-X)$. But the tree decomposition implies that there is no such element. We deduce that $C$ has exactly three elements. Thus every vertex of $T$ that is labelled by a circuit is labelled by a triangle. Since every vertex of $T$ that is labelled by a cocircuit has exactly one element of $E(M)$ in that cocircuit, a straightforward induction argument establishes that, for some $n\geq 2$, the matroid $M$ is obtained from $n$ copies of $M(K_{3})$ by a sequence of $n-1$ parallel connections. Thus $M\cong P_{n}$.

Finally, we show by induction that $P_{n}$ is density-critical. This is true for $n=1$. Assume it true for $n<m$ and let $n=m\geq 2$. Take $x$ in $P_{n}$. Assume first that $x$ is in exactly one triangle $\{x,y,z\}$. Then $\operatorname{si}(P_{n}/x)\cong P_{n}/x\setminus z$. As the last matroid is easily seen to be isomorphic to the density-critical matroid $P_{n-1}$ and $d(P_{n-1})<d(P_{n})$, we deduce that every minor of $P_{n}/x$ has density less that $d(P_{n})$. Now assume that $x$ is in at least two triangles of $P_{n}$. Then $\operatorname{si}(P_{n}/x)$ is easily seen to be the direct sum of a collection of matroids each of which is isomorphic to some $P_{k}$ with $k<n$. By Lemma 4.4 and the induction assumption, every minor of $P_{n}/x$ has density less that $d(P_{n})$. We conclude that $P_{n}$ is density-critical, so the theorem is proved. ∎

Acknowledgements

The work for this paper was done at the Tutte Centenary Retreat (https://www.matrix-inst.org.au/events/tutte-centenary-retreat/) held at the MAThematical Research Institute (MATRIx), Creswick, Victoria, Australia, 26 November – 2 December, 2017. The authors are very grateful for the support provided by MATRIx. The first, second, and fourth authors were supported, respectively, by NSERC Scholarship PGSD3-489418-2016, National Science Foundation Grant 1500343, and the New Zealand Marsden Fund.