Halfway to Rota’s basis conjecture authored

by

Matija Bucić, Matthew Kwan, Alexey Pokrovskiy, and Benny Sudakov

Abstract: In 1989, Rota made the following conjecture. Given $n$ bases $B_{1},\dots,B_{n}$ in an $n$-dimensional vector space $V$, one can always find $n$ disjoint bases of $V$, each containing exactly one element from each $B_{i}$ (we call such bases transversal bases). Rota’s basis conjecture remains wide open despite its apparent simplicity and the efforts of many researchers (for example, the conjecture was recently the subject of the collaborative “Polymath” project). In this paper we prove that one can always find $\left(1/2-o\left(1\right)\right)n$ disjoint transversal bases, improving on the previous best bound of $\Omega\left(n/\log n\right)$. Our results also apply to the more general setting of matroids.

]]>Proposition: Let be a simple graph edge-coloured in such a way that the edges in each colour class form a spanning tree . Then there is a partition of into edge-disjoint rainbow trees.

Proof: Choose an arbitrary vertex of and consider the orientation of each corresponding to rooting $t_c$ at (i. e. every edge points away from .) Assign to each edge the vertex it points to. For each vertex , consider the subgraph consisting of the edges to which is assigned. Since each vertex is assigned to precisely one edge from each colour class, these graphs form a collection of edge-disjoint rainbow trees. ]]>

This is a marvelous result and a huge improvement over the previous bound! It is very tempting now to speculate that a combination of explicit construction and random construction (a la Keevash as it were) could be the path to the full conjecture.

]]>**Theorem 1**

Given disjoint bases in a rank- matroid, there are at least disjoint transversal bases.

Throughout the post, I will use the natural logarithm, and define .

Let . For each , let be disjoint -element subsets of chosen at random. Now for each , note that the sets are subsets of that are chosen independently and uniformly at random. If contains a transversal basis with probability at least 1/2, then by the linearity of expectation, the expected number of disjoint transversal bases of will be at least . It follows that there exist at least disjoint transversal bases.

Therefore, to prove Theorem 1, it suffices to prove the following:

**Theorem 2**

Let be disjoint bases for a rank- matroid. If are -element subsets chosen independently and uniformly at random, then the probability that contains a transversal basis is at least 1/2.

There is a very useful theorem by Rado which characterizes the existence of transversal basis.

**Rado’s Theorem**

Let be sets of elements in a rank- matroid. Then there is a transversal basis of if and only if for all .

In order to prove Theorem 2, we focus on the probability of failure of each of the conditions in Rado’s Theorem. Let be bases (not necessarily disjoint) of a rank- matroid. We let denote the probability that, when -element subsets are chosen independently and uniformly at random from respectively, we have .

Note that we do not require the sets to be disjoint. In fact, the case that is interesting and plays an important role in the proof. In this case we have , and hence the failure probability depends only on and ; we denote it by . It turns out this is an upper bound on all the other failure probabilities. We prove this by naively constructing an independent set within the randomly chosen , and showing that it has rank less than with probability *equal to* .

We then compute , which is closely related to the Coupon Collector’s Problem, as well as a bipartite matching problem considered by Erdős and Renyi. We get

Finally, by the union bound, the probability that *does not* contain a transversal basis is at most the sum of the failure probabilities of each of the conditions in Rado’s Theorem; hence, it is at most With some algebraic manipulation, we can show that this is bounded by 1/2, as required by our theorem.

I’d like to try to construct more counterexamples from graphs. As a start, can someone construct an graphical counterexample that (a) contains no doubled edges and (b) contains no induced subgraph isomorphic to ?

To recap, this means constructing a bi-tree together with an arrangement of its edges into two columns such that no set of row permutations produces spanning trees in both columns.

]]>I see. I was hoping that this might yield a different type of counterexample, but your construction here doesn’t seem to do that. Consider the four sets , where each two-digit number represents an edge connecting vertices and in a graph with five vertices labeled 1 to 5. This is an instance of your construction, and it is “almost” an counterexample, but unfortunately there is a unique way to make both columns into bases.

]]>It is a subgraph of a bi-tree, you can extend it into a bi-tree in several ways. Adding another edge won’t decrease the number in which the minors intersect.

]]>How is with a subdivided edge a bi-tree? It has an odd number of edges.

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