# Generalized matroids

Consider the following definition of matroid.

A matroid over a set $$X$$ is a family $$\mathcal B\subseteq\mathcal P(X)$$ of subsets of $$X$$ (the set of bases) with the following properties:

1. $$\mathcal B$$ is non-empty.
2. (base exchange property) For any two $$A,B\in \mathcal B$$ and $$a\in A\setminus B$$, there is a $$b\in B\setminus A$$, so that $$A\setminus\{a\}\cup\{b\}\in \mathcal B\qquad\text{and}\qquad B\setminus\{b\}\cup\{a\}\in \mathcal B$$

I am interested in a generalization, lets call it $$k$$-matroids, in which I replace the second axiom by the following:

1. For any two $$A,B\in \mathcal B$$ and distinct $$a_1,...,a_k\in A\setminus B$$, there is are distinct $$b_1,...,b_k\in B\setminus A$$, so that $$A\setminus\{a_1,...,a_k\}\cup\{b_1,...,b_k\}\in \mathcal B\qquad\text{and}\qquad B\setminus\{b_1,...,b_k\}\cup\{a_1,...,a_k\}\in \mathcal B$$

Is this generalization known and if so, where to read about it. If not, are there some obvious connections to classical matroids (which are 1-matroids).

• Isn't your original definition of matroid the definition of a $\delta$-matroid, which is already a generalization of matroids? Aug 8, 2019 at 6:43
• @quarague I was under the impression that this was one of the equivalent definitions of matroids (from Wikipedia). What exactly is a $\delta$-matroid? Is it because the exchange property is symmetric, that is it makes a statement about $A\setminus\{a\}\cup\{b\}$ and $B\setminus \{b\}\cup\{a\}$ instead of just one? I learned that this is equivalent to just one direction. But actually, I do not need the symmetry. If this is the problem I would rather edit the question. Aug 8, 2019 at 7:01
• You are correct, you defined a matroid. Your definition is the one that is generalized when defining $\delta$-matroids, that's why I got confused. In a $\delta$-matroid one uses symmetric difference and not all bases need to have the same cardinality. See en.wikipedia.org/wiki/Delta-matroid Aug 8, 2019 at 7:12

• Note that $k$ is a fixed number, so for example, in a 2-matroid the usual base exchange property does not necessarily hold. But yes, every $k$-matroid is also a $dk$-matroid for every $d\ge 1$. Aug 20, 2021 at 9:20