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I am interested in matroids of rank two and would like to understand how interesting/big this class of matroids is.

I know that the 2-uniform matroid on (k+2) elements is not representable over any field with at most k elements. (see Oxley, p203).

Does there exist any survey on matroids of rank two?

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2 Answers 2

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Up to simplification (suppressing loops and parallel elements), every rank two matroid is just a rank two uniform matroid.

Note that the vectors $(1, a_1), \dots, (1, a_n)$ represent the uniform matroid $U_{2,n}$, provided that $a_1, \dots, a_n$ are all distinct. Thus, the rank two matroids are all representable over each infinite field.

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  • $\begingroup$ This is correct, at least for finite matroids; for finite matroids, one of the many ways to justify this, and an interestingly 'roundabout' way, is to appeal to Theorem 6.1.3, equivalence (i)$\leftrightarrow$(ii), in Oxley, Matroid Theory, 2nd edition, which for $r=2$ degenerates into the statement that the matroid is a projective geometry in dimension 1, which of course is a uniform matroid, as you say. $\endgroup$
    – Peter Heinig
    Commented Oct 11, 2017 at 13:14
  • $\begingroup$ Indeed, this can also be seen with the partition of Fedor Petrov (other answer). But in my particular problem parallel elements are important. $\endgroup$
    – Quentin Fortier
    Commented Oct 11, 2017 at 13:27
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At first, neglect all loops (elements of rank 0). Now call two elements $a,b$ similar, if $\{a,b\}$ is not a base. It follows that if $\{a,c\}$ is a base, than $\{b,c\}$ is also a base (else the set $\{a,b,c\}$ would have inclusion-maximal independent subsets $\{b\}$ and $\{a,c\}$ of different size.) In other words, similarity is an equivalence relation. So, all elements may be partitioned onto groups so that any two elements of different groups form a base, and from the same group not. Of course such a matroid is representable over sufficiently large field. Namely, if we have $m$ groups, we should have $m$ different directions of the lines on the plane, that is, the field must contain at least $m-1$ elements (and this suffices).

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