Base-orderable matroid

In mathematics, a base-orderable matroid is a matroid that has the following additional property, related to the bases of the matroid.

For any two bases

$A$

and

$B$

there exists a feasible exchange bijection, defined as a bijection

$f$

from

$A$

to

$B$

, such that for every

$a\in A\setminus B$

, both

$(A\setminus \{a\})\cup \{f(a)\}$

and

$(B\setminus \{f(a)\})\cup \{a\}$

are bases.

The property was introduced by Brualdi and Scrimger. A strongly-base-orderable matroid has the following stronger property:

For any two bases

$A$

and

$B$

, there is a strong feasible exchange bijection, defined as a bijection

$f$

from

$A$

to

$B$

, such that for every

$X\subseteq A$

, both

$(A\setminus X)\cup f(X)$

and

$(B\setminus f(X))\cup X$

are bases.

The property in context

Base-orderability imposes two requirements on the function $f$:

1. It should be a bijection;
2. For every $a\in A\setminus B$, both $(A\setminus \{a\})\cup \{f(a)\}$ and $(B\setminus \{f(a)\})\cup \{a\}$ should be bases.

Each of these properties alone is easy to satisfy:

1. All bases of a given matroid have the same cardinality, so there are n! bijections between them (where n is the common size of the bases). But it is not guaranteed that one of these bijections satisfies property 2.
2. All bases $A$ and $B$ of a matroid satisfy the symmetric basis exchange property, which is that for every $a\in A\setminus B$, there exists some $f(a)\in B\setminus A$, such that both $(A\setminus \{a\})\cup \{f(a)\}$ and $(B\setminus \{f(a)\})\cup \{a\}$ are bases. However, it is not guaranteed that the resulting function f be a bijection - it is possible that several $a$ are matched to the same $f(a)$.

Matroids that are base-orderable

Every partition matroid is strongly base-orderable. Recall that a partition matroid is defined by a finite collection of categories, where each category $C_i$ has a capacity denoted by an integer $d_i$ with $0\le d_i\le |C_i|$. A basis of this matroid is a set which contains exactly $d_i$ elements of each category $C_i$. For any two bases $A$ and $B$, every bijection mapping the $d_i$ elements of $C_i\cap A$ to the $d_i$ elements of $C_i\cap B$ is a strong feasible exchange bijection.

Every transversal matroid is strongly base-orderable.

Matroids that are not base-orderable

Some matroids are not base-orderable. A notable example is the graphic matroid on the graph K₄, i.e., the matroid whose bases are the spanning trees of the clique on 4 vertices. Denote the vertices of K₄ by 1,2,3,4, and its edges by 12,13,14,23,24,34. Note that the bases are:

• {12,13,14}, {12,13,24}, {12,13,34}; {12,14,23}, {12,14,34}; {12,23,24}, {12,23,34}; {12,24,34};
• {13,14,23}, {13,14,24}; {13,23,24}, {13,23,34}; {13,24,34};
• {14,23,24}, {14,23,34}; {14,24,34}.

Consider the two bases A = {12,23,34} and B = {13,14,24}, and suppose that there is a function f satisfying the exchange property (property 2 above). Then:

• f(12) must equal 14: it cannot be 24, since A \ {12} + {24} = {23,24,34} which is not a basis; it cannot be 13, since B \ {13} + {12} = {12,14,24} which is not a basis.
• f(34) must equal 14: it cannot be 24, since B \ {24} + {34} = {13,14,34} which is not a basis; it cannot be 13, since A \ {34} + {13} = {12,13,23} which is not a basis.

Then f is not a bijection - it maps two elements of A to the same element of B.

There are matroids that are base-orderable but not strongly-base-orderable.

Properties

In base-orderable matroids, a feasible exchange bijection exists not only between bases but also between any two independent sets of the same cardinality, i.e., any two independent sets $A$ and $B$ such that $|A|=|B|$.

This can be proved by induction on the difference between the size of the sets and the size of a basis (recall that all bases of a matroid have the same size). If the difference is 0 then the sets are actually bases, and the property follows from the definition of base-orderable matroids. Otherwise by the augmentation property of a matroid, we can augment $A$ to an independent set $A\cup \{x\}$ and augment $B$ to an independent set $B\cup \{y\}$. Then, by the induction assumption there exists a feasible exchange bijection $f$ between $A\cup \{x\}$ and $B\cup \{y\}$. If $f(x)=y$, then the restriction of $f$ to $A$ and $B$ is a feasible exchange bijection. Otherwise, $f^{-1}(y)\in A$ and $f(x)\in B$, so $f$ can be modified by setting: $f(f^{-1}(y)):=f(x)$. Then, the restriction of the modified function to $A$ and $B$ is a feasible exchange bijection.

Completeness

The class of base-orderable matroids is complete. This means that it is closed under the operations of minors, duals, direct sums, truncations, and induction by directed graphs.Template:Rp It is also closed under restriction, union and truncation.Template:Rp

The same is true for the class of strongly-base-orderable matroids.