Let be a matroid on an -element set that is a disjoint union of independent sets of size . Assume that there exists another matroid on the same ground set with the following properties:

(1) is strongly base orderable.

(2) for all , where is the rank function of .

(3) All circuits of satisfying remain dependent in .

Then there is an grid whose th row comprises and whose columns are independent in .

That all strongly base-orderable matroids satisfy Rota’s Basis Conjecture follows at once if we take since condition (2) is automatic by pigeonhole. However, there is a lot more that we might extract out of Lemma 6. Wild asks whether a suitable *always* exists. As Wild recognizes, this is probably too optimistic, but he doesn’t have a counterexample. Maybe a suitable always exists for graphic matroids?

This sounds like an interesting idea, but can you provide some motivation for the condition that you state? I’m having a hard time understanding why one would expect this to be true.

]]>Oops, right, somewhere halfway through my argument I forgot that was not the total number of elements.

]]>I’m afraid I don’t understand your sentence, “One of these independent sets will have at most elements.” Of course, by the pigeonhole principle, one of these independent sets will have at most elements, but how are you getting ? In fact if we look at our favorite example of with , then we see that cannot be achieved.

]]>But what happens to the elements’ partitioning to two bases? You forget about that condition? I might be really misunderstanding something, but consider taking the original matrix ordered such that the first column forms a basis. The second column can be covered by two independent sets (the respective parts of the two original bases). One of these independent sets will have at most elements. Why can’t we pull these out?

]]>By “fixing it” I do mean converting it to something that’s not a counterexample. And yes, when I pull out an edge , I first delete , and then I add a new independent edge to the graph. If originally appeared in the two-element independent set , then replaces in . Another way to put it is that I alter the matroid by making all sets consisting of an independent set plus the given edge independent, and keeping all the other independent sets. Basically I want to delete the edge from the graph (because I’m imagining that this is just an inductive step in the ordinary Rota conjecture, and the edge is appearing in some other column) but technically I have to leave it in there because the two-column version of the conjecture involves exactly sets of size exactly 2.

]]>Could you be more precise? By putting out, you really want to add an independent edge to the graph? What is “fixing it?” Does it mean converting it to something that’s not a counterexample?

]]>This is a nice observation. Thinking about where to go from here, I was led to the following thought. Given a graph, let us define the operation of “pulling out an edge” to mean deleting the edge, and then adding two new vertices to the graph with a new edge between them. Question: Given an graphic counterexample to the two-column version of Rota’s Basis Conjecture, can we always fix it by pulling out at most edges?

I expect that the answer is probably no, but the counterexamples should be more interesting, because I don’t think you can get them just by stringing together disjoint copies of and “uncontracting” edges as in Luke Pebody’s example.

Should the answer unexpectedly be yes, then that gives hope that we can control the copies of . Pulling out edges can be thought of as a proxy for making sure that those edges get used in other columns, in some kind of induction argument.

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