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\begin{document}

\title[Size, circumference and circuit removal in $3$--connected matroids]{On size, circumference and circuit removal in $3$--connected matroids}

\author{Manoel Lemos}
\address{Departamento de Matem\'atica\\
        Universidade Federal de Pernambuco\\
        Recife\\
        Pernambuco 50740-540\\ 
	Brazil}
\email{manoel@dmat.ufpe.br}

\author{James Oxley}
\address{Department of Mathematics\\
	Louisiana State University\\
	Baton Rouge\\ 
	Louisiana 70803--4918\\
	USA}
\email{oxley@math.lsu.edu}




\subjclass{05B35}
\date{}

\begin{abstract}
This paper proves several extremal results for $3$-connected matroids: a $3$-connected matroid $M$ of rank at least six has circumference $c(M)$ at least six and, provided $|E(M)| \ge 4r(M) -5$, has a circuit whose deletion leaves a $3$-connected matroid; if $M$ has a basis $B$ such that $M\ba e$ is not $3$-connected for all $e$ in $E(M) - B$, then $|E(M)| \le 3r(M) -4$; and if $M$ is minimally $3$-connected but not hamiltonian, then $|E(M)|\le 3r(M) -c(M)$.
\end{abstract}

\maketitle
\section{Introduction}


Let $M$ be a matroid and $A$ be a subset of $E(M)$. Lemos and Oxley \cite{Paper1} and Lemos, Oxley, and Reid \cite{Paper2} considered the problem of finding a sharp upper bound on $|E(M') -A|$ where $M'$ is a $3$-connected minor of $M$ that is minimal with the property that $M|A = M'|A$.  The following theorem, the main result of \cite{Paper1}, solves this problem in the case when $A$ spans $M$. 
Let $\lambda_1(A,M)$ denote  the number of connected components of $M|A$. Now $M|A$ can be constructed from a collection  $\Lambda_2(A,M)$ of $3$-connected matroids by using the operations of direct sum and 2-sum. It follows from work of Cunningham  and Edmonds \cite{cunn} that  $\Lambda_2(A,M)$ is unique up to isomorphism. We  denote by $\lambda_2(A,M)$ the number of matroids in $\Lambda_2(A,M)$ that are not isomorphic to $U_{1,3}$, the three-element cocircuit. 

\begin{theorem}
\label{Theorem 0} 
Let $M$ be a $3$--connected matroid other than $U_{1,3}$ and let $A$ be a non-empty spanning subset of $E(M)$. If $M$ has no proper $3$--connected minor $M'$ such that $M'|A=M|A$, then
$$|E(M)|\le |A|+\lambda_1(A,M)+\lambda_2(A,M)-2,$$
unless $A$ is a circuit of size at least four,  in which case,  
$$|E(M)|\le 2|A|-2.$$
\end{theorem}


It was also shown in \cite{Paper1} that, in a strong sense, the bound in this theorem is best-possible: given a simple matroid $N$ having at least one circuit, there is a minor-minimal 3-connected matroid $M$ for which  $M|E(N)=N$ such that $|E(M)|$ attains the bound in the theorem. Therefore the theorem is  best-possible for every restriction $M|A$ for which $r^*(M|A)\neq 0$. The first new result of this paper sharpens Theorem~\ref{Theorem 0} in the case that $A$ is a basis of $M$.

\begin{theorem}
\label{Theorem 2} Let $M$ be a $3$-connected matroid of rank at least four and  $B$ be a basis of $M$. If $M$ has no proper $3$-connected minor that contains $B$ and has $B$ as a basis, then $|E(M)|\le 3r(M)-4$.
\end{theorem}

A {\it minimally $3$-connected matroid} is a 3-connected matroid for which no single-element deletion is 3-connected.  Generalizing a result of Halin \cite[Satz 7.6]{halin} for graphs, Oxley \cite[Theorem 4.7]{qjm} proved the following bound on the size of a minimally 3-connected matroid  and characterized the matroids attaining equality in this bound.

\begin{theorem}
\label{ox} Let $M$ be a minimally $3$-connected matroid of rank at least three. Then 
$$|E(M)|\le\left\{
\begin{tabular}{ll}
$2r(M)$,&\hbox{ if $r(M) \le 6$;}\\
$3r(M)-6$,&\hbox{ if $r(M) \ge 7$.}\\
\end{tabular}\right.
$$
\end{theorem}

Our second theorem uses Theorem~\ref{Theorem 0} to derive a new bound on the size of a minimally 3-connected matroid.  The {\it circumference} $c(M)$ of a matroid that is not free is the maximum size of a circuit of $M$. 

\begin{theorem}
\label{Theorem 1} Let $M$ be a minimally $3$-connected matroid. Then
$$|E(M)|\le\left\{
\begin{tabular}{ll}
$2r(M)$,&\hbox{if $M$ has  a hamiltonian circuit;}\\
$3r(M)-c(M)$,&\hbox{otherwise.}\\
\end{tabular}\right.
$$
\end{theorem}

A comparison of the last two results prompts one to seek a lower bound on the circumference of a 3-connected matroid.  The following such bound is obtained in Section 3, and an immediate consequence of this bound is that Theorem~\ref{Theorem 1} sharpens Theorem~\ref{ox}.

\begin{theorem}
\label{Theorem 3} For a $3$-connected matroid $M$ of rank at least six,  $c(M)\ge 6$.
\end{theorem}

Mader \cite{mader} showed that every $k$-connected simple graph $G$ with minimum degree at least $k+2$ has a cycle $C$ such that $G\ba C$, the graph obtained from $G$ by deleting the edges of $C$, is $k$-connected. Jackson \cite{jack} extended this result for $2$-connected graphs.  In \cite{Rem}, Lemos and Oxley proved that if $M$ is a  2-connected matroid  for which 
$$|E(M)|\ge 2r(M) +2 + \max\{0, 1 + r(M) -c(M)\},$$
then $M$ must have a circuit $C$ such that $M\backslash C$ is 2-connected. 
Moreover, if $|E(M)| \ge 3r(M)$, they showed that this circuit  can be chosen so as to avoid some arbitrarily chosen basis $B$ of $M$.  In Section 4, we prove the corresponding results in the 3-connected case:

\begin{theorem}
\label{Theorem 4} 
Let $M$  be a $3$-connected matroid such that $r(M)\ge 6$. If
$$|E(M)|\ge 3r(M) +1 + \max\{0, r(M) -c(M)\},$$
then $M$ has a circuit $C$ such that $M\backslash C$ is $3$-connected.
\end{theorem}

\begin{theorem}
\label{Theorem 5} 
Suppose that $M$ is a $3$-connected matroid such that $r(M)\ge 4$ and let $B$ be a basis of $M$.
If $$|E(M)|\ge 4r(M)-3,$$
then $M$ has a circuit $C$ such that $M\backslash C$ is $3$-connected and $C\cap B=\emptyset$. 
\end{theorem}

The terminology used here will follow Oxley \cite{oxbook}.


\section{A bound to the size of a minimally 3-connected matroid}


In this section, we shall prove Theorems~\ref{Theorem 2} and \ref{Theorem 1}, a best-possible bound on the size of a minimal 3-connected matroid that maintains a fixed set as a basis, and a new bound on the size of a minimally 3-connected matroid as a function of its rank and circumference. Both of these results are consequences of Theorem~\ref{Theorem 0}. For a 3-connected matroid $M$ and a subset $A$ of its ground set, $(M,A)$ is called a {\it minimal pair} if $M$ has no 3-connected minor $M'$ for which $M'|A=M|A$.

The following lemma plays a key role in the proofs of both theorems. For a basis $B$ of a matroid $M$ and an element $e$ of $E(M) -B$, the unique circuit of $M$ that is contained in $B \cup e$ is denoted by $C(e,B)$.

\begin{lemma}
\label{Lemma 1} 
Suppose that $(M,B \cup  e )$ is a minimal pair and that $r(M) \neq 0$. If $B$ is a basis of $M$ and $e\in E(M)- B$, then 
$$|E(M)|\le\left\{
\begin{tabular}{ll}
$2r(M)$,&\hbox{if $C(e,B)$ is a hamiltonian circuit of $M$;}\\
$3r(M)-|C(e,B)|$,&\hbox{otherwise.}\\
\end{tabular}\right.
$$
\end{lemma}


\begin{proof} Suppose first that $C(e,B)$ is a hamiltonian circuit of $M$. If $r(M) \in \{1,2\}$, then $E(M) = B \cup e$ and the result holds. Thus we may assume that $r(M) \ge 3$.  Then, by applying Theorem~\ref{Theorem 0} to the minimal pair $(M,B\cup e )$, we get that $|E(M)|\le 2|C(e,B)|-2=2r(M)$, and the result follows. 

We may now  suppose that $C(e,B)$ is not a hamiltonian circuit of $M$. 
Observe that $M|(B\cup e )$ has $C(e,B)$ as a connected component and $B- C(e,B)$ as a non-empty set of coloops. Therefore 
$$\lambda_1(B\cup e ,M)=1+|B- C(e,B)|=1+(|B|-|C(e,B)|+1).$$
Thus $\lambda_1(B\cup e ,M)=r(M)+2-|C(e,B)|$. We also have that
\vspace{-0.7cm}
\begin{center}
\begin{eqnarray*}
\lambda_2(B\cup e ,M)&=&\lambda_2(C(e,B),M)+|B- C(e,B)|\\
&=&(|C(e,B)|-2)+(|B|-|C(e,B)|+1)=r(M)-1.
\end{eqnarray*}
\end{center}
Hence, by Theorem~\ref{Theorem 0}, 
\vspace{-0.7cm}
\begin{center}
\begin{eqnarray*}
|E(M)|&\le&|B\cup e |+\lambda_1(B\cup e ,M)+\lambda_2(B\cup e ,M)-2\\
&=&(r(M)+1)+(r(M)+2-|C(e,B)|)+(r(M)-1)-2\\
&=&3r(M)-|C(e,B)|.
\end{eqnarray*}
\end{center}
\vspace{-0.7cm}\end{proof}
\vspace{0.3cm}

Theorem~\ref{Theorem 1} is a straightforward consequence of this lemma.

\medskip

\noindent{\it Proof of Theorem~\ref{Theorem 1}.} Let $C$ be a circuit of $M$ such that $|C|=c(M)$. Choose an element $e$ of $C$ and a basis $B$ of $M$ such that $C- e \subseteq B$. Clearly $C=C(e,B)$. The result  follows immediately from Lemma~\ref{Lemma 1}.
\hfill $\square $

\medskip

Lemos and Oxley \cite{pack} proved the analogue of the last  theorem for minimally 2-connected matroids, namely, if $M$ is a minimally 2-connected matroid, then
$$|E(M)|\le 2r(M)+2-c(M).$$

A matroid is {\it hamiltonian} if it has a spanning circuit.  Evidently the circumference of such a matroid is one more than its rank. An interesting aspect of Theorem~\ref{Theorem 1} is that the wheels and whirls are hamiltonian matroids that are extremal examples for the theorem.  Moreover, for each $r$ and each $c$ such that $6\le c\le r$, we shall now describe an extremal example $M$ for this theorem that has rank $r$ and circumference $c$. Let $\{v_1,v_2,v_3\}$ be one of the vertex classes in the bipartition of $K_{3,r+3-c}$.  Let $G$ be obtained from $K_{3,r+3-c}$ as follows: add a path of length $c-4$ to $K_{3,r+3-c}$ that links $v_1$ and $v_3$ but is otherwise disjoint from $K_{3,r+3-c}$; then add a new edge joining $v_2$ to every vertex of this path other than the ends. It is not difficult to check that the cycle matroid $M$ of $G$ has circumference $c$, rank $r$, and has $3r-c$ elements.



Next we shall prove Theorem~\ref{Theorem 2} giving a best-possible upper bound on $|E(M)|$  for a minimal pair $(M,A)$ when $A$ is a basis of $M$.
In this theorem, observe that  the bound on the rank of $M$ in the hypothesis cannot be lowered: take $M$ to be the rank-3 wheel  and let $B$ be its set of  spokes. Then $|E(M)| = 6$, but $3r(M) - 4 = 5$.


\medskip

\noindent{\it Proof of Theorem~\ref{Theorem 2}.} First observe that $(M,B \cup  e )$ is a minimal pair for all $e$ in $E(M) - B$.  Suppose  that $C(e,B)$ is hamiltonian for some $e$ in $E(M) -B$.  Then, by Lemma~\ref{Lemma 1}, $|E(M)| \le 2r(M)$ and so, as $r(M) \ge 4$, we have $|E(M)| \le 3r(M) - 4$.  Thus we may assume that $C(e,B)$ is non-hamiltonian for all $e$.  In that case, by Lemma~\ref{Lemma 1} again, $|E(M)|\le 3r(M)-|C(e,B)|$ for all $e$.  Hence we may assume that 
$$\hbox{$|C(e,B)|=3$ for every element $e$ of $E(M)- B$.}\eqno{(1)}$$

Next  we shall prove that the theorem holds unless 
$$\hbox{$C(e,B)\cap C(f,B)\not=\emptyset$ whenever $e\not=f$.}\eqno{(2)}$$
Suppose that $e\not=f$ and that $C(e,B)\cap C(f,B)=\emptyset$. In this case, we shall consider the minimal pair $(M,B\cup\{e,f\})$. Observe that $M|(B\cup\{e,f\})$ has $C(e,B),C(f,B)$, and each individual element of $B- [C(e,B)\cup C(f,B)]$ as its connected components. Thus, by (1),  
$$\lambda_1(B\cup\{e,f\},M)=r(M)-2\hbox{ and }\lambda_2(B\cup\{e,f\},M)=r(M)-2.$$
Hence, by Theorem~\ref{Theorem 0}, 
$$|E(M)|\le |B\cup\{e,f\}|+(r(M)-2)+(r(M)-2)-2=3r(M)-4$$
and the result follows. Thus we may suppose that (2) holds.

Assume that there are distinct elements $e,f,g$ in $E(M)- B$ such that
$$C(e,B)=\{e,a,b\},C(f,B)=\{f,b,c\},C(g,B)=\{g,c,a\}.\eqno{(3)}$$
In this case, we shall obtain  a contradiction. We may assume that  $E(M)-(B\cup\{e,f,g\})$ contains an element $h$ otherwise  the theorem  certainly holds. By (2), $C(h,B)$ must intersect all of  $C(e,B),$ $C(f,B)$, and $C(g,B)$. Hence $C(h,B)- h\subseteq\{a,b,c\}$. Thus $\{a,b,c\}$ spans $E(M)- B$. Moreover, $B-\{a,b,c\}$ is non-empty since $r(M) \ge 4$. 
Hence $\{B-\{a,b,c\}, (E(M) - B) \cup \{a,b,c\}\}$ is a 1-separation of the 3-connected matroid $M$; a contradiction.  Thus (3) cannot occur.

We deduce that $B$ has an element $b$  such that $b\in C(e,B)$ for every $e$ in  $E(M)- B$. Let $b_e$ be the element of $C(e,B)-\{e,b\}$. If $b_e=b_{f}$ for $e\not=f$, then  $M|\{b,b_e,e,f\}$ is isomorphic to $U_{2,4}$ and hence $M\backslash e$ is 3-connected; a contradiction. Thus  $b_e\not=b_{f}$ whenever  $e\not=f$. Therefore 
$$|E(M)- B|\le |B- b|=r(M)-1$$
and the result follows. \hfill $\square $


\medskip

We now show that Theorem~\ref{Theorem 2} is  best-possible. Let $K_{3,n}''$ be the graph that is obtained from $K_{3,n}$ by adding a new edge from one of the degree-$n$ vertices of the latter to each of the other two degree-$n$ vertices.
Then equality is attained in the bound in Theorem~\ref{Theorem 2} if we take $M$ to be $M(K_{3,n}'')$ and $B$ to be the set of edges meeting  the vertex of degree $n+2$ in $K_{3,n}''$. 


\section{The circumference of a 3-connected matroid}

For a $k$-connected graph $G$,the minimum vertex degree is at least $k$. When $k\ge 2$, a well-known result of Dirac \cite[Theorem 4]{dirac}  implies that the circumference of $G$ is at least $2k$ provided that $|V(G)|\ge 2k$. Moreover, this result is  best-possible. Thus a 3-connected graph with at least six vertices has circumference  at least six. In this section, we shall prove Theorem~\ref{Theorem 3}, a generalization of this result to  3-connected matroids having rank at least six. 


Let $L$ be a subset of the ground set of a matroid $M$ and suppose that $L$ is the union of a set of circuits of $M$ and $r^*(M|L)=2$. Then $L$ is what Tutte \cite{wtt} has called a ``line'' of $M$.  We shall call $L$ a {\it Tutte-line} since the word ``line'' is also commonly used in matroid theory to mean a rank-2 flat.  It is not difficult to see that every Tutte-line $L$ of a matroid $M$ has a {\it canonical partition} $\{L_1,L_2,\dots,L_k\}$ such that a subset $C$ of $L$  is a circuit of $M|L$  if and only if $C=L- L_i$ for some $i$ in $\{1,2,\dots,k\}$. A Tutte-line $L$ is {\it connected} if  $M|L$ is a connected matroid.

In the next proof, we shall make frequent use of the next two lemmas. Both parts of the first of these are elementary consequences of {\it orthogonality}, the property of a matroid that a circuit and a cocircuit cannot have exactly one common element. The second lemma was proved by Oxley \cite[Theorem 2.5]{qjm}.



\begin{lemma}
\label{Lemma A} 
\begin{enumerate}
\item[(i)] If $C$ is a circuit of a $3$-connected matroid and $T_1$ and $T_2$ are distinct triads of  $M$ both of which meet $C$, then  $C\cap T_1\not=C\cap T_2$.
\item[(ii)] If $L$ is a Tutte-line of a matroid $M$ and $T$ is a triad of $M$, then $T$ meets an odd number  of sets in the canonical partition of $L$. 
\end{enumerate}
\end{lemma}

\begin{lemma}
\label{Lemma B} 
Let $C$ be a circuit of a minimally $3$-connected matroid $M$ and suppose that $|E(M)| \ge 4$. Then $M$ has at least two distinct triads intersecting $C$.
\end{lemma}


\noindent{\it Proof of Theorem~\ref{Theorem 3}.} Let $M$ be a counterexample to the theorem for which $|E(M)|$ is minimal.  Clearly $M$ must be  minimally 3-connected.  Let $C$ be a circuit of $M$ such that $|C|=c(M)$. Then $3\le |C|\le 5$.

Let ${\mathcal D}$ be the set of circuits $D$ of $M/C$ such that $D$ is not a circuit of $M$ and $|D|\ge 2$.


\begin{lemma}
\label{Lemma 1'} 
\begin{enumerate}
\item[(i)] $|D|=2$  for every $D$  in ${\mathcal D}$; 
\item[(ii)] $|C|\ge 4$; and 
\item[(iii)] for all $D$ in ${\mathcal D}$, the set $C\cup D$ is a connected Tutte-line of $M$ having  canonical partition $\{X_D,Y_D,D\}$ for some $X_D$ and $Y_D$ with  $|X_D|,|Y_D|$ in $\{2,3\}$.
\end{enumerate}
\end{lemma}

\begin{proof} Suppose that $D \in {\mathcal D}$. Then $C\cup D$ is a connected Tutte-line $L$ of $M$. Let $\{L_1,L_2,\dots,L_k\}$ be the canonical partition of $L$. As $L$ is connected, $k\ge 3$. Since $(C \cup D) - D$ is a circuit of $M|L$, we may assume, without loss of generality, that $D=L_1$. As $L- L_1 = C$, a maximum-sized circuit of $M$, we must have that $|L-L_1|\ge  |L-L_i|$ for all $i$.  But $|D| \ge 2$. Thus 
$$2\le |D| = |L_1| \le |L_i|\eqno{(4)}$$
for all $i$. As $\cup_{j=2}^{k}L_j = C$ and $|C| \le 5$, we deduce that $k \le 3$.  Hence $k = 3$. Moreover, $|L_2|, |L_3| \in \{2,3\}$ and $\min\{|L_2|,|L_3|\} = 2$.  Thus, by (4), $|D| = 2$. In addition, $|C| = |L_2| + |L_3| \ge 4$, and (iii) holds. 
\end{proof}
 
\begin{lemma}
\label{Lemma 2} $c(M/C)\le 2$.
\end{lemma}

\begin{proof}
Let $C'$ be a circuit of $M/C$ with at least three elements. By Lemma~\ref{Lemma 1'}, $C' \not \in {\mathcal D}$. Thus $C'$ is  a circuit of $M$. Observe that $M|(C\cup C')=(M|C)\oplus (M|C')$. Choose an element $d$ of $C'$. Then $M$ has   a circuit $C_d$ that contains $d$ and meets $C$. Clearly $C_d- C$ is a union of circuits of $M/C$. Take such a circuit $D$  that contains $d$.  Since $D$ and $C'$ are both circuits of $M/C$, we cannot have that $D = \{d\}$ otherwise $D$ is a proper subset of $C'$. Hence $|D| \ge 2$.  Lemma~\ref{Lemma 1'} now implies that $|D| = 2$, say $D = \{d,d'\}$.  Moreover, $M|(C \cup D)$ is connected. Since $M|C'$ is also connected, it follows that $M|(C \cup D \cup C')$, which equals $M|(C \cup C' \cup d')$, is connected.  Deleting $d'$ from the last matroid produces the disconnected matroid $M|(C \cup  C')$ with components $M|C$ and $M|C'$. Thus $M|(C \cup C' \cup d')$ is the series connection, with basepoint $d'$, of $[M|(C \cup C' \cup d')]/C'$ and $[M|(C \cup C' \cup d')]/C$. Now $M|(C \cup C' \cup d')$ has a circuit $D'$ that contains $d'$ and at least two elements of $C'$, otherwise every element of $C'$ is parallel to $d$, a contradiction to the fact that $C'$ is a circuit of $M$ of size at least three.  Then $D' = D'_1 \cup D'_2 \cup d'$ where $D'_1 \cup d'$ and $D'_2 \cup d'$ are  circuits of $[M|(C \cup C' \cup d')]/C'$ and $[M|(C \cup C' \cup d')]/C$, respectively.   Hence $D' - C = D'_2 \cup d'$ and so $D' -C$ is a circuit of $M/C$ that is not a circuit of $M$, and $|D' - C| \ge 3$. Thus $D' - C \in {\mathcal D}$ yet Lemma~\ref{Lemma 1'} fails for it; a contradiction. 
\end{proof}


By Lemma~\ref{Lemma 2}, the connected components of $M/C$ consist of loops and parallel classes.  But each parallel class of $M/C$ is a cocircuit of $M$ and therefore has at least three elements.  Let these rank-one components of $M/C$ be  $H_1,H_2,\dots,H_n$. Then $n \ge 1$ since $C$ does not span $M$.  Therefore, as $M$ is 3-connected and $|E(M) - E(H_i)| \ge |C| \ge 3$, we have that 
$$r(E(H_i)) + r(E(M)-E(H_i)) - r(M) \ge 2.$$
Thus, as $E(M) - E(H_i)$ is a hyperplane of $M$, it follows that 
$$r(E(H_i)) \ge 3 \hbox{  for all $i$.}\eqno{(5)}$$

\begin{lemma}
\label{Lemma 3} $C$ does not contain a triad of $M$.
\end{lemma}

\begin{proof} Suppose that $C$ contains a triad $T$ of $M$. Choose a subset $A$ of $E(H_1)$ such that $|A|=2$. Then  $A\in{\mathcal D}$ so, by Lemma~\ref{Lemma 1'},  $C\cup A$ is a connected Tutte-line of $M$ having canonical partition $\{A,X,Y\}$ with $|X|,|Y|\in\{2,3\}$. Since $X \cup Y = C$, it follows, by Lemma~\ref{Lemma A}(ii), that $\{X,Y\} = \{T,C-T\}$.  Hence, as $|C| \le 5$, we deduce that $|C-T| \le 2$.  Thus $|C-T| = 2$ and $|C| = 5$.  Moreover, for all 2-element subsets $A$ of $E(H_1)$, the set $A \cup (C - T)$ is a circuit $C_A$ of $M$.

As $M$ is minimally 3-connected, Lemma~\ref{Lemma B} implies that $M$ has distinct triads $T_1$ and $T_2$ both of which meet $C_A$.  By Lemma~\ref{Lemma A}(i), $T_1 \cap C_A \neq T_2 \cap C_A$. Thus, as $|A| = 2$, at least one of $T_1$ and $T_2$, say the former, meets $C-T$. 
Now, by Lemma~\ref{Lemma A}(ii), either (i) $T_1 \cap C \subseteq C-T$; or (ii) $T_1 = \{a,t,c\}$ for  some $a$ in $A$, some $t$ in $T$, and some $c$ in $C-T$. Let $A'$ be a subset of $E(H_1)- a$ such that $|A\cap A'| = 1.$  Then $C_{A'}$, which equals $A' \cup (C-T)$, cannot meet $T_1$ in a single element, so (ii) does not hold.  Hence  (i) holds.


Let $A'$ be a subset of $E(H_1)$ such that $|A' \cap A| = 1$.  By Lemma~\ref{Lemma A}(ii), $A' \cap T_1 = \emptyset$, since $\{A',T,C-T\}$ is the canonical partition of $A' \cup C$. 
Let $C - T = \{x,y\}$.  Then  $(C_A\cup C_{A'})- y$, which equals $A\cup A'\cup x$, contains a circuit  of $M$. As $|A\cup A'|=3$ and  $(A\cup A'\cup x)\cap T_1=\{x\}$, it follows that $A\cup A'$ is a triangle of $M$. As $A'$ was an arbitrarily chosen subset of $E(H_1)$ for which $|A \cap A'| = 1$, we deduce that $A$ spans $E(H_1)$ in $M$. Hence $r(E(H_1)) = 2$; a contradiction to (5). \end{proof}


\begin{lemma}
\label{Lemma 4} If $i \in \{1,2,\ldots,k\}$ and $\{a,b\}$ is a $2$-element subset of $E(H_i)$ that is contained in a triad of $M$, then $E(H_i)$ is a triad of $M$.
\end{lemma}

\begin{proof} Let $T$ be a triad of $M$ that contains $\{a,b\}$ and suppose that $T\cap E(H_i) = \{a,b\}$. By orthogonality, $T\cap C = \emptyset$.   Thus $\{a,b\}$ is a union of cocircuits of $M|(C \cup E(H_i))$.  But, since $E(H_i)$ is a parallel class of $M/C$ of size at least three, it follows that $r(C \cup E(H_i)) = r(C) + 1$ and $E(H_i)$ is a cocircuit of $M|(C \cup E(H_i))$.  This is a contradiction since $\{a,b\}$  is a proper subset of $E(H_i)$.  We conclude that $T \subseteq E(H_i)$. Since  both $T$ and $E(H_i)$ are cocircuits of $M$, equality must hold here. 
\end{proof}


\begin{lemma}
\label{Lemma 5} For all $i$, the matroid $M|(C \cup E(H_i))$ has at most one $2$-cocircuit  contained in $C$.
\end{lemma}


\begin{proof} Suppose that $M|(C \cup E(H_i))$ has at least two 2-cocircuits $W_1$ and $W_2$  contained in $C$. We now distinguish the following two cases:  (i) $|W_1 \cap W_2| = 1$; and (ii) $|W_1 \cap W_2| = 0$. In each case, we shall show that every 3-element subset of $E(H_i)$ is a triangle so $r(E(H_i))=2$; a contradiction to (5).

Assume that (i) holds. Then $W_1 \cup W_2$ is contained in a series class of  $M|(C \cup E(H_i))$. Let $\{a,b,c\}$ be an arbitrary 3-element subset of $E(H_i)$.  Then Lemma~\ref{Lemma 1'} implies that, for each $A$ in $\{\{a,b\}, \{b,c\}\}$, the set $C \cup A$ is a connected Tutte-line of $M$ having canonical partition $\{A,X_A,Y_A\}$ where each of $X_A$ and $Y_A$ has either two or three elements.  Since both $A \cup X_A$ and $A \cup Y_A$ are circuits of $M$, we must have that $W_1 \cup W_2$ is contained in and therefore equals $X_A$ or $Y_A$.  Thus both $W_1 \cup W_2 \cup \{a,b\}$ and $W_1 \cup W_2 \cup \{b,c\}$ are circuits of $M$.  Hence if $w \in W_1$, then $(W_1 \cup W_2 \cup \{a,b,c\}) - w$ contains a circuit $C_w$ of $M$. As $W_1 \cup W_2$ is contained in a series class of $M|(C \cup E(H_i))$ and $w \not \in C_w$, it follows that $C_w$ avoids $W_1 \cup W_2$.  Hence $C_w$ is contained in and therefore equals $\{a,b,c\}$.  It follows that $r(E(H_i))= 2$; a contradiction to (5).

We may now assume that (ii) holds. Let $\{a,b,c\}$ be an arbitrary 3-element subset of $E(H_i)$. Then, for each 2-element subset $A$ of $\{a,b,c\}$, Lemma~\ref{Lemma 1'} implies that $C\cup A$ is a connected Tutte-line.  Moreover, for the canonical partition $\{A,X_A,Y_A\}$ of $C\cup A$, we must have that each of $W_1$ and $W_2$ is contained in $X_A$ or $Y_A$.  The fact that each of $X_A$ and $Y_A$ has two or three elements, but their union has at most five elements implies that $A \cup W_1$ or $A \cup W_2$ is a circuit of $M$.  Hence there are distinct 2-element subsets $A$ and $A'$ of $\{a,b,c\}$ such that, for some $j$ in $\{1,2\}$, both $A \cup W_j$ and $A' \cup W_j$ are circuits of $M$.  Thus, if $w \in W_j$, then $M$ has a circuit $C_1$ contained in $(A \cup A' \cup W_j) - w$.  Since $W_j$ is a 2-cocircuit of $M|(C \cup E(H_i))$, it follows that $C_1$ is contained in and hence equals $A \cup A'$.  Therefore, as in case (i), we deduce that every 3-element subset of $E(H_i)$ is a triangle, so $r(E(H_i))= 2$; a contradiction to (5). 
\end{proof}


\begin{lemma}
\label{Lemma 6} Suppose that $i \in \{1,2,\ldots,k\}$ and  that $\{a,b,c\}$ is a $3$-element subset of $E(H_i)$. Then $\{a,b\}$ is contained in a $4$-circuit $D$ of $M$ where $|D \cap C| = 2$. Moreover,  either
\begin{enumerate}
\item[(i)] $M$ has a triad that meets  both $\{a,b\}$ and $D \cap C$; or 
\item[(ii)] $E(H_i)$ is a triad of $M$ and there is a triad of $M$ that contains $c$ and meets $C$.
\end{enumerate}
In both cases, $M$ has a triad that meets both $E(H_i)$ and $C$.
\end{lemma}

\begin{proof} The fact that $D$ exists follows immediately from Lemma~\ref{Lemma 1'}(iii) since $|C| \le 5$.  Let $D \cap C = \{\alpha,\beta\}$. 
Suppose that (i) does not occur. By Lemma~\ref{Lemma B}, there are triads $T$ and $T'$ of $M$ that meet $D$. By Lemma~\ref{Lemma A}(i), $T \cap D \neq T' \cap D$.  The canonical partition of the connected Tutte-line $\{a,b\} \cup C$ must be $\{\{a,b\}, \{\alpha,\beta\}, C-\{\alpha,\beta\}\}$.  Thus orthogonality and Lemma~\ref{Lemma A}(ii) imply that each of $|T \cap (C \cup \{a,b\})|$ and $|T' \cap (C \cup \{a,b\})|$ is 2.  Hence we may assume, without loss of generality, that $T \cap (C \cup \{a,b\}) = \{a,b\}$ and $T' \cap (C \cup \{a,b\}) = \{\alpha,\beta\}.$  Therefore, by Lemma~\ref{Lemma 4}, $E(H_i) = \{a,b,c\}$ and $E(H_i)$ is a triad of $M$. We may suppose that the element of $T' - (C \cup \{a,b\})$ is not in $C \cup E(H_i)$ otherwise this element is $c$ and (ii) follows.  

Applying Lemma~\ref{Lemma B} again, this time to the circuit $C$, we deduce that $M$ has a triad $T''$ that meets $C$ and is different from $T'$.  By Lemma~\ref{Lemma 3}, $|T'' \cap C| = 2$.  We show next  that $T'' \cap E(H_i)$ is empty. If not, then $c \in T''$, or, by Lemma \ref{Lemma A}(ii), $T''$ meets both $\{a,b\}$ and $D \cap C$. But, in both these cases, the lemma holds.  Thus   $T'' \cap E(H_i)$ is indeed empty. Hence $T'' \cap C$ is a 2-cocircuit of $M|(C \cup E(H_i))$. But $T' \cap C$ is also a 2-cocircuit of  $M|(C \cup E(H_i))$ and Lemma~\ref{Lemma A}(i) implies that $T' \cap C \neq T'' \cap C$.  Hence  $M|(C \cup E(H_i))$ has two 2-cocircuits contained in $C$.  This contradiction to Lemma~\ref{Lemma 5} completes the proof of Lemma~\ref{Lemma 6}. 
\end{proof}






\begin{lemma}
\label{Lemma 7} $M$ has exactly two rank-one components, $H_i$. Moreover, $|C|=5$ and $r(M)=6$.
\end{lemma}

\begin{proof}  Suppose that the number $n$ of rank-one components of $M/C$ is at least three. Then, by Lemma~\ref{Lemma 6}, for each $i$ in $\{1,2,\dots,n\}$, there is a triad $T_i$ such that both $T_i\cap C$ and $T_i\cap E(H_i)$ are non-empty. By Lemma~\ref{Lemma A}(ii), if $i$ and $j$ are distinct members of  $\{1,2,\dots,n\}$, then $T_i \cap C \neq T_j \cap C$. Thus $M|(C \cup E(H_1))$ has both $T_2 \cap C$ and $T_3 \cap C$ as 2-cocircuits contained in $C$, a contradiction to Lemma~\ref{Lemma 5}.  We conclude that $n \le 2$. Thus 
$$6\le r(M)=|C|-1+n\le |C|+1\le 6,$$
so $r(M) = 6$. Moreover, $n=2$ and  $|C|=5$. 
\end{proof}



We now work towards obtaining a final contradiction that will complete the proof of Theorem~\ref{Theorem 3}.  By Lemma~\ref{Lemma 6}, for each $i$ in $\{1,2\}$, there is a triad $T_i$ that meets both $E(H_i)$ and $C$.  If, for a fixed $i$, there are two such triads $T_{i,1}$ and $T_{i,2}$, then, for $j \neq i$, the sets $T_{i,1}\cap C$ and $T_{i,2} \cap C$ are 2-cocircuits of $M|(C \cup E(H_j))$. Thus, by Lemma~\ref{Lemma 5}, $T_{i,1}\cap C = T_{i,2}\cap C$, a contradiction to Lemma~\ref{Lemma A}(ii).  Hence $T_i$ is unique.

Suppose that, for some $i$ in $\{1,2\}$, the set $E(H_i)$ is not a triad of $M$. Then  Lemma~\ref{Lemma 6} implies that, for each 2-element subset $A$ of $E(H_i)$, there is a 4-element circuit, $D_A,$ that contains $A$ and meets $C$ in exactly two elements.  Thus, as (ii) of Lemma~\ref{Lemma 6} does not hold, (i) of that lemma implies that  there are at least two triads of $M$ that meet $C$ and $E(H_i)$. This contradiction to the uniqueness of $T_i$ implies that both $E(H_1)$ and $E(H_2)$ are triads of $M$.

Let $E(H_1)=\{a,b,c\}$ and $E(H_2)=\{a',b',c'\}$. Without loss of generality, we may suppose that $T_1\cap E(H_1) = \{c\}$ and $T_2\cap E(H_2) = \{c'\}$. 
Then, for some $\{\alpha, \beta\} \subseteq C$, there is a circuit 
 $\{\alpha,\beta,a,b\}$  of $M$. By orthogonality, either 

\begin{enumerate}
\item[(i)] $T_1=\{\alpha,\beta,c\}$; or 
\item[(ii)] $T_1\cap \{\alpha,\beta,a,b\}=\emptyset$. 
\end{enumerate}
In both cases, we shall show that 
\begin{enumerate}
\item[(iii)] $T_1\cap T_2=\emptyset$; and 
\item[(iv)] there are distinct elements $\mu$ and $\nu$ of $T_1 - c$ so that 
 $\{a,c,\varepsilon,\mu\}$ and $\{b,c,\varepsilon,\nu\}$ are circuits of $M$ where  $\varepsilon$ is the element of $C-(T_1\cup T_2)$.
\end{enumerate}

Suppose that (ii) occurs. Then,  by  Lemma~\ref{Lemma B}, $M$ has a triad $T$ that meets $\{\alpha,\beta,a,b\}$ and is different from $E(H_1)$. Since $E(H_1) \cap \{\alpha,\beta,a,b\} = \{a,b\}$, it follows, by Lemma~\ref{Lemma A}(ii), that $T \cap \{\alpha,\beta,a,b\} \neq \{a,b\}$. Thus either $T$ meets both $E(H_1)$ and $C$, or $T \cap \{\alpha,\beta,a,b\}  =\{\alpha,\beta\}$. Since $T_1$ is the only triad meeting both $E(H_1)$ and $C$, the first case implies that $T_1 = T$; a contradiction since $T_1$ avoids  $\{\alpha,\beta,a,b\}$ by (ii). We conclude that $T \cap \{\alpha,\beta,a,b\}  =\{\alpha,\beta\}$. If $T$ avoids $E(H_2)$, then $T_1 \cap C$ and $T \cap C$ are distinct 2-cocircuits of $M|(C \cup E(H_2))$ contained in $C$; a contradiction to Lemma~\ref{Lemma 5}.  Thus $T$ meets $E(H_2)$. Since $T$ also meets $C$, it follows that $T= T_2$ and hence that $T_2 = \{\alpha,\beta,c'\}$.  Hence $T_1 \cap T_2 = \emptyset$, that is, (iii) holds.  We may now assume that 
 $T_1=\{\gamma,\delta,c\}$ where $C= \{\alpha, \beta,\gamma,\delta,\varepsilon\}$. By Lemma~\ref{Lemma 1'} and the fact that $|C| = 5$, the connected Tutte-line $C \cup \{a,c\}$ contains a 5-element circuit 
$C_1$ of $M$ that contains $\{a,c\}$. By applying Lemma~\ref{Lemma A}(ii)  to $T_1$, we deduce that $|C_1\cap\{\gamma,\delta\}|=1$, say $\gamma\in C_1$. As $T_2 = \{\alpha,\beta,c'\}$, it follows that  $\{\alpha, \beta\}$ is a series class of $M|(C \cup \{a,c\})$ and hence that 
$C_1=\{\alpha,\beta,\gamma,a,c\}$. Thus  $\{a,c,\delta,\varepsilon\}$ is a circuit of $M$. 

Now use the connected Tutte-line $C \cup \{b,c\}$ to give that $M$ has a 5-element circuit $C_2$ that contains $\{b,c\}$. Then, arguing as for $C_1$, we deduce that  $\{b,c,\delta,\varepsilon\}$ or $\{b,c,\gamma,\varepsilon\}$ is a circuit of $M$. The first possibility is excluded since it implies that   $\{a,b,\delta,\varepsilon\}$ contains a circuit of $M$. This contradicts the fact that $\{\{a,b\}, \{\alpha, \beta\}, \linebreak \{\gamma,\delta,\varepsilon\}\}$ is the canonical partition of $C \cup \{a,b\}$. We conclude that if (ii) occurs, then (iii) and (iv) hold.

Now suppose that (i) holds. Then, by Lemma~\ref{Lemma A}(i) and (ii), $T_1\cap T_2=\emptyset$, that is, (iii) holds. Thus we may assume  that $T_2=\{\gamma,\delta,c'\}$. Let $C_3$ be a 5-element circuit  of $M|(C \cup \{a,c\})$ that contains $\{a,c\}$. Then, by applying Lemma~\ref{Lemma A}(ii) to the triad $\{\alpha,\beta,c\}$, we have that $|C_3\cap\{\alpha,\beta\}|=1$, say $\alpha\in C_3$. As $T_2 = \{\gamma,\delta,c'\}$, it follows that $\{\gamma,\delta\}$ is a cocircuit of $M|(C \cup \{a,c\})$. Thus the canonical partition of $C \cup \{a,c\}$ is $\{\{a,c\}, \{\alpha, \gamma,\delta\}, \{\beta,\varepsilon\}\}$. Hence 
 $\{a,c,\beta,\varepsilon\}$ is a circuit of $M$. Arguing similarly using a 5-element circuit $C_4$ of $M|(C \cup \{b,c\})$ that contains $\{b,c\}$, we deduce that  $\{b,c,\beta,\varepsilon\}$ or $\{b,c,\alpha,\varepsilon\}$ is a circuit of $M$. The first possibility cannot occur because it implies that 
 $\{a,b,\beta,\varepsilon\}$ contains a circuit of $M$ which contradicts the fact that $\{\{a,b\}, \{\alpha, \beta\}, \{\gamma,\delta,\varepsilon\}\}$ is the canonical partition of $C \cup \{a,b\}$. We conclude that $\{b,c,\alpha,\varepsilon\}$ is a circuit of $M$ and hence that, when (i) occurs, both (iii) and (iv) hold.

Since  $\{\{a,b\}, \{\alpha, \beta\}, \{\gamma,\delta,\varepsilon\}\}$ is the canonical partition of $C \cup \{a,b\}$ and $T_1\cap T_2=\emptyset$, Lemma~\ref{Lemma A}(ii) implies that either $T_1 = \{\alpha,\beta,c\}$ and  $T_2 = \{\gamma,\delta,c'\}$, or $T_1 = \{\gamma,\delta,c\}$ and $T_2 = \{\alpha,\beta,c'\}$. Using Lemma~\ref{Lemma A}(ii) again, it follows that, in each case, either $\{\{a',b'\}, \{\alpha, \beta,\varepsilon\}, \{\gamma,\delta\}\}$ or 
$\{\{a',b'\}, \{\alpha, \beta\}, \{\gamma,\delta,\varepsilon\}\}$ is the canonical partition of $C \cup \{a',b'\}$. Thus $\{a',b',\gamma,\delta\}$ or 
$\{a',b',\alpha,\beta\}$ is a circuit of $M$. Using this circuit in place of $\{a,b,\alpha,\beta\}$, and $E(H_2)$ and $T_2$ in place of $E(H_1)$ and $T_1$, we may now argue as in (i) and (ii) above to deduce that there are distinct elements $\mu'$ and $\nu'$ of $T_2-c'$ such that $\{a',c',\varepsilon,\mu'\}$ and $\{b',c',\varepsilon,\nu'\}$ are circuits of $M$.

By elimination,  $(\{a,c,\varepsilon,\mu\}\cup\{a',c',\varepsilon,\mu'\})-\varepsilon$ contains a circuit $C_5$ of $M$. The triads $\{a,b,c\}$ and $T_1$ and orthogonality imply that  either $C_5\cap\{a,c,\mu\}=\emptyset$, or $\{a,c,\mu\}\subseteq C_5$. Similarly, the triads $\{a',b',c'\}$ and $T_2$ imply that  $C_5\cap\{a',c',\mu'\}=\emptyset$, or $\{a',c',\mu'\}\subseteq C_5$. But $|C_5|\le 5$, so $C_5$ is either $\{a,c,\mu\}$ or $\{a',c',\mu'\}$. However, the connected Tutte-lines $C \cup \{a,c\}$ and $C \cup \{a',c'\}$ imply that both $\{a,c,\mu\}$ and $\{a',c',\mu'\}$ are
 properly contained in  circuits of $M$. This contradiction completes the proof of Theorem~\ref{Theorem 3}. \hfill $\square $

\medskip

To close this section, we shall present an example that shows that the lower bound on the rank in the hypothesis of Theorem~\ref{Theorem 3} is best-possible. Let $M$ be the tipless binary 5-spike, that is, $M$ is the matroid that is represented over $GF(2)$ by the matrix $[I_5|J_5 - I_5]$ where $J_5$ is the $5 \times 5$ matrix of all ones.  It is not difficult to show (see, for example, \cite[p.321]{oxbook}) that $M$ has circumference 5. 

\section{Removing circuits from matroids}

In this section, we shall prove Theorems~\ref{Theorem 4} and ~\ref{Theorem 5}. 

\medskip

\noindent{\it Proof of Theorem~\ref{Theorem 4}.}  By Lemma~\ref{Lemma 1}, there is a  3-connected matroid $N$ having the same rank and the same circumference as $M$ such that $N=M|E(N)$ and
$$|E(N)|\le 2r(M) + \max\{0, r(M) -c(M)\}.$$
Now 
\vspace{-0.7cm}
\begin{center}
\begin{eqnarray*}
|E(M)|-|E(N)|&\ge&[3r(M) +1 + \max\{0, r(M) -c(M)\}]\\
		& & \hspace{2cm}-[2r(M) + \max\{0, r(M) -c(M)\}]\\
&=&r(M)+1.
\end{eqnarray*}
\end{center}
Thus $E(M)- E(N)$ must be a dependent set of $M$ and so contains a circuit $C$ of $M$ avoiding $E(N)$. Observe that $M\backslash C$ is a 3-connected matroid, since it has $N$ as a minor and $E(N)$ spans $M$. 
\hfill $\square $

\medskip

On combining the last theorem with Theorem~\ref{Theorem 3}, we immediately obtain the following result.


\begin{corollary}
\label{Corollary} If $M$ is a $3$-connected matroid such that $r(M)\ge 6$ and
$$|E(M)|\ge 4r(M)-5,$$
then $M$ has a circuit $C$ such that $M\backslash C$ is $3$-connected.
\end{corollary}





The last corollary is best-possible as the next example shows.  
Let $M$ be the matroid that is obtained as follows.  Begin with a 3-point line $\{a,b,c\}$ and take the generalized parallel connection of $n$ copies, $N_1,N_2,\ldots,N_n$, of $M(K_4)$ across $\{a,b,c\}$.  Each $N_i$ has a unique 3-point line that meets $\{a,b,c\}$ at $a$.  Freely add a point $p_i$ on each such line.  Then $M$ is obtained by deleting $a$ from the resulting matroid.  Certainly $M$ is 3-connected and has rank $2+n$.  Moreover, for each $i$, the 4-element set $(E(N_i) \cup p_i) - \{a,b,c\}$ is a cocircuit of $M$ containing a triangle.  Using this fact and orthogonality, it is not difficult to see that $M$ has no circuit whose deletion leaves a 3-connected matroid.  But 
$$|E(M)|=4n+2=4(n+2)-6=4r(M)-6,$$
so the bound in the last corollary cannot be improved.
 

The proof of Theorem~\ref{Theorem 5} is obtained 
by making slight modifications to the proof of Theorem~\ref{Theorem 4} so that Theorem~\ref{Theorem 2} rather than Lemma~\ref{Lemma 1} can be used. We omit the straightforward details.

To obtain an example showing that Theorem~\ref{Theorem 5} is best-possible, we modify the previous example by freely adding two new points $q_1$ and $q_2$ on the line $\{b,c\}$.  Let the resulting matroid be $M'$.  Then $|E(M')| = 4n+4$ and $r(M') = n+2$, so $|E(M')| = 4r(M')-4$.  For each $i$, let $b_i$ be the coloop of $M'|[(E(N_i) \cup p_i) - \{a,b,c\}]$ and let $B = \{b,c,b_1,b_2,\ldots,b_n\}$.  Then $B$ is a basis for $M'$ and, arguing as for $M$, it is not difficult to see that $M'$ has no circuit $C$ that avoids $B$ such that $M' \backslash C$ is 3-connected.

\medskip

\noindent{\bf Acknowledgements}~~ The first author was partially supported  by CNPq, CAPES, FINEP and PRONEX 107/97. The second author was partially supported  by the National Security Agency. 

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\end{thebibliography}
\end{document}


Until the end of the proof, we shall fix the following notation: (i) $a,b$ and $c$ are different elements of $E(H_i)$; (ii) $C=\{\alpha,\beta,\gamma,\delta\}$ or $C=\{\alpha,\beta,\gamma,\delta\,\varepsilon\}$; (iii) $D=\{a,b,\alpha,\beta\}$ is a circuit of $M$; (iv) $Z=C\cup E(H_i)$.

