Math-1900 · mathematics · Combinatorics

Math-1900

Scrambled Rota conjecture

Status Partially solved · unsolvedmath.comLLM status Open unverified

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Statement

Definitions and mathematical statement

Let $M=(S,\mathcal I)$ be a finite loopless matroid of rank $k\ge2$. Independent sets form a hereditary family satisfying exchange; bases are independent sets of size $k$. Looplessness means every singleton is independent. Suppose the ground set can be partitioned into $k$ bases, and hence has $k^2$ elements.

Now choose an arbitrary partition $S=A_1\sqcup\cdots\sqcup A_k$ with $|A_i|=k$. These new classes need not be independent. The scrambled Rota conjecture asks for $k-1$ pairwise disjoint bases $T_1,\ldots,T_{k-1}$ satisfying \[ |T_j\cap A_i|=1 \quad(1\le i\le k,\ 1\le j\le k-1). \] Each requested basis is thus a full transversal of the new partition.

The original basis partition only certifies that the matroid has enough disjoint bases; it need not have any alignment with the new classes $A_i$. The $k-1$ transversals together use all but one element of each class. The remaining $k$ elements are not required to be a basis. Requesting $k$ disjoint transversal bases for arbitrary new classes would be a stronger assertion than this conjecture. No assumption of representability or additional independence of the scrambled classes is included. The omitted case $k=1$ is vacuous because it asks for zero transversals.

Short English statement

After arbitrarily regrouping the elements of a matroid that splits into $k$ bases, must there still be $k-1$ disjoint transversal bases?

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