When do the circuits of a matroid have a connected intersection graph?

Question

When does a matroid $M$ have a set of circuits $\mathcal{C}$ with a connected intersection graph i.e. when is the graph $G$ with$V(G)=\mathcal{C}$ and adjacencies $\{A,B\}\in E(G)\iff A\cap B\neq\emptyset$ connected?

This is equivalent to charactering the matroids with a partial ear-decomposition i.e. the matroids with circuits that can be indexed $C_1,\ldots C_n$ so we get that $\forall 0<i\leq n\exists j<i:C_i\cap C_j\neq\emptyset$ (where note this indexing is not necessarily injective i.e. there might exist $i\neq j$ with $C_i=C_j$)

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Suppose we call matroids with this property special now if two matroids $M_1$ and $M_2$ are special and some circuit in $M_1$ is not disjoint to some circuit in $M_2$ then $M_1\oplus M_2$ is also special, with that said then what do "special" matroids look like? Is there a simple way to characterise these?

Answer 1

This holds if and only if $M$ has at most one connected component which contains a circuit. Clearly, the intersection graph of circuits is disconnected if $M$ has two connected components which each contain a circuit. For the other direction, suppose that $M$ has at most one connected component $N$ which contains a circuit. If $M$ has at most one circuit, then clearly the intersection graph of circuits is connected. Otherwise, let $C_1$ and $C_2$ be distinct circuits of $M$. Note that $C_1$ and $C_2$ are circuits of $N$. Choose $e \in C_1$ and $f \in C_2$. Since $N$ is connected, there is a circuit $C_3$ of $N$ such that $\{e,f\} \subseteq C_3$. Thus, there is path of length $2$ between $C_1$ and $C_2$ in the intersection graph of circuits.

Answer 2

It seems that the question was edited while I typed. The second question I refer is when a matroid M has an ordering $C_1,\dots, C_n$ of its circuits such that for each $2\le i\le n$, there exists $j<i$ such that $C_i$ and $C_j$ intersect:

The questions are not equivalent. The answer to the second question (the one about the graph) is given by Tony Huynh: $M$ is connected except for coloops. This happens to be the answer to the first question too (the one about the circuit ordering).

We reduce the proof to the case that $M$ is coloopless. On one hand if $M$ has such an ordering for its circuits, then $M$ is connected by the answer for the other question.

The other implication is proved by induction on the number of elements. Suppose that $M$ is connected and smaller connected matroids than $M$ have such an ordering of its circuits. There is a result that says that $M$ has an element $e$ such that either $M\backslash e$ is connected or $e$ is in serial pair of $M$ and $M/e$ is connected. In the latter case, a desired ordering of the circuits of $M/e$ induces an ordering of the corresponding circuits in $M$. In the former case, one just has to add the circuits of $M$ containing $e$ to the end of a desired ordering of the circuits of $M\backslash e$.