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Let $M$ be a matroid on (finite) ground set $E$ with $\mathcal B(M)$ as its set of bases and $\mathcal C(M)$ as its set of circuits.

If we consider the uniform matroid $U_{m,n}$ for $m,n \in \mathbb N$ with $m < n$, then we see that $\mathcal C(U_{m,n}) = \mathcal B(U_{m+1,n})$.

Question: Is there any other pair $(M_1, M_2)$ of matroids with $\mathcal C(M_1) = \mathcal B(M_2)$ that are not uniform?

user2520938
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Moritz
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2 Answers2

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Yes there are more pairs like that. We know we can specify a matroid by its set of circuits and bases.

So now let $$B=C=\{X\subset\{1,2,3\}| |X|=2\}$$ Now consider the matroids $$M_1=(\{1,2,3,4\},\text{bases}=B)$$ $$M_2=(\{1,2,3,4\},\text{circuits}=C)$$ So $M_1$ and $M_2$ both have underlying set $\{1,2,3,4\}$, and there bases resp circuits specified as above. These will not be uniform matroids, but by definition have the required property.

user2520938
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2

Claim: Given two loop-free matroids $M_1$ and $M_2$, $\mathcal{B}(M_{1}) =\mathcal{C}(M_2)$ if and only if $M_1$ and $M_2$ are both partition matroids with $M_1 = \bigoplus_{i = 1} ^nU_{r_i,n_i}$ and $M_2 =\bigoplus_{i = 1} ^n U_{r_i-1,n_i}$.

Since bases (respectively, circuits) of a direct sum of matroids are just the disjoint union of bases (respectively, circuits) of the summands, it's enough to prove the above claim in the case when $M_2$ is connected.

Claim: Let $M_2$ be connected. Then $\mathcal{B}(M_1)=\mathcal{C}(M_2)$ if and only if there exist $1<r<n$ such that $M_1 = U_{r,n}$ and $M_2 = U_{r-1,n}$.

The "if" direction follows directly from the definition of uniform matroids. To prove the "only if" direction first let $r$ be the rank of $M_1$. Then every circuit of $M_2$ has cardinality $r$ and so by connectivity $M_2$ has rank $r-1$. So by Exercise 1.3.4 in Oxley's Matroid Theory, $M_2$ is uniform. Say $M_2=U_{r-1,n}$. Then $B(M_1)$ is the set of all $r$ element sets of an $n$ element set and so $M_1=U_{r,n}$, proving the claim.

Aaron Dall
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  • My original question was indeed inspired by Oxley's book. I have one remark: Let $r=2$. If I follow your inference, then the circuits of $M_2$ all have cardinality $2$. Consider a partition matroid (Oxley's book: p. 18, example 1.2.8): there all circuits also have cardinality $2$ but the rank of the partition matroid can be arbitrary large. Do you understand what I mean? – Moritz Nov 10 '16 at 17:57
  • I don't have my copy of Oxley on hand at the moment to check, but if I remember correctly a partition matroid is the direct sum of uniform matroids. Is that correct? – Aaron Dall Nov 10 '16 at 18:44
  • @Moritz: You're exactly right. I was assuming that the matroids in question were connected. I've edited the answer to take care of the disconnected case. – Aaron Dall Nov 10 '16 at 19:39