If $G$ is a graph with edge set $E$, let $W$ be the $\mathbb{Z}/2$-vector space generated by the elements of $E$. If $A = \{a\_1, \dots, a\_n\} \subset E$, let $\bar{A} = a\_1 + \dots + a\_n \in V$; then $\bar{A}\_1 + \bar{A}\_2 = \overline{A\_1 \Delta A\_2}$, where $\Delta$ indicates symmetric difference.


I'll define the *cycle space* of $G$ to be the subspace of $W$ generated by simple cycles of $G$. More precisely, the *cycle space* of $G$ is the subspace of $W$ generated by the set $\{\bar{C} \mid C \text{ is a simple cycle of } G\}$. We could also view the cycle space as the first simplicial homology group of $G$ over $\mathbb{Z}/2$. It is not difficult to show that the dimension of the cycle space of $G$ is the corank of the cycle matroid of $G$.


Given any matroid $M$ with ground set $E$, we could define the *circuit space* of $M$ in a completely analogous way, just using the word "circuit" instead of "simple cycle." My question is: is it always true that the dimension of the circuit space of $M$ is the corank of $M$? If not, for what types of matroids is this true? Finally, can anyone recommend good resources that deal with this sort of thing?


Thanks!



Dimension of circuit space of a matroid
The dimension of the circuit space of a matroid $M$ is the corank of $M$ if and only if $M$ is binary. Here is a proof. Given a basis $B$ and $e \notin B$, we let $C(e,B)$ be the unique circuit contained in $B \cup \{e\}$. We will use the following well-known characterization of binary matroids (Theorem 9.1.2 in Oxley's *Matroid Theory* text).


**Theorem.** A matroid $M$ is binary if and only if, for all bases $B$ of $M$ and all circuits $C$ of $M$, $C= \triangle\_{e \in C-B} C(e,B)$.


Here, $\triangle$ means symmetric difference, or equivalently $\sum$, if we view sets as vectors over $\mathbb{F}\_2$.  

Let $\dim(M)$ denote the dimension of the circuit space of $M$ and $r^\*(M)$ denote the corank of $M$. We now prove that $\dim(M)=r^\*(M)$ if and only if $M$ is binary.


*Proof.* For each basis $B$ of $M$, let $\mathcal{C}\_B:=\{C(e,B) : e \notin B\}$. The circuits in $\mathcal{C}\_B$ are linearly independent since for each $e \notin B$, there is a unique circuit in $\mathcal{C}\_B$ containing $e$. Thus, $\dim(M) \geq r^\*(M)$, for every matroid $M$.


If $M$ is binary, then by the above theorem, $C= \triangle\_{e \in C-B} C(e,B)$, for every circuit $C$. Thus, $\dim(M)=r^\*(M)$.


If $M$ is not binary, then by the above theorem, there exists a basis $B'$ and a circuit $C'$ such that $C' \neq \triangle\_{e \in C-B'} C(e,B')$. In particular, $C'$ is not a sum of the vectors in $\mathcal{C}\_{B'}$. Thus, $\dim(M)>r^\*(M)$.


