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A polymatroid is a finite set $X$ and a rank function $d : P(X) \to {\mathbb N}$ such that

1) $d(\varnothing)=0$,

2) $A \subset B$ implies $d(A) \leq d(B)$, and

3) $d(A \cap B) + d(A \cup B) \leq d(A) + d(B)$ for all $A,B \in P(X)$.

A polymatroid is said to be representable over $GF(2)$ (the field with two elements), if there exists a collection of subvectorspaces $\lbrace V_x \mid x \in X \rbrace$ of $GF(2)^{\oplus d(X)}$, such that

$$d(A) = \dim_{GF(2)} \bigvee_{x \in A} V_x, \quad \forall A \in P(X).$$

(It is clear that any function $d$, which is defined this way is indeed a polymatroid.)

A polymatroid is called matroid if $d(\lbrace x\rbrace)=1$ for all $x \in X$. Tutte proved that a matroid is representable over $GF(2)$ if and only if the matroid $U_{2,4}$ does not appear as a minor. Here, $U_{2,4}$ is the matroid formed by four points that lie on one line, i.e. the underlying set is $\lbrace x_1,\dots,x_4 \rbrace$ and $d(\lbrace x_i,x_j\rbrace) = 2$ for $i \neq j$ and $d(\lbrace x_1,\dots,x_4\rbrace)=2$. (The necessity of this condition is obvious, since $GF(2)^{\oplus 2}$ has only $3$ non-zero elements.)

Question: Is there any useful characterization of the representability of a polymatroid over $GF(2)$ ? What is known about this question ?

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Minor correction. A matroid is binary if and only if it does not contain a 4-point line as a minor. Note that a matroid may have a 4-point line as a minor, but not as a restriction, e.g. $U_{3,5}$. – Tony Huynh Sep 2 '10 at 22:54
Thanks, I have corrected the question. – Andreas Thom Sep 3 '10 at 5:10
up vote 3 down vote accepted

Here's a construction due to Stefan van Zwam.

Let $U_{2,4}$ be a 4-point line with ground set $[4]$, and rank function $r$. For each $A \subset [4]$, let $\chi(A)$ be 1, if $1 \in A$, and 0 if $1 \notin A$. For each $k \in \mathbb{N}$, let $S_k$ be the polymatroid with ground set $[4]$ and rank function $r+k\chi$.

Lemma. $S_k$ is not representable over $\mathbb{F}_2$, for every $k$.

Proof. Let $(V_i : i \in [4])$ be a representation of $S_k$ over $\mathbb{F}_2$. Choose a basis $B_i$ for each $V_i$. Since {1} has rank $k+1$ and {1,2,3,4} has rank $k+2$, we may assume that $B_1$ consists of the first $k+1$ standard basis vectors in $\mathbb{F}_2^{k+2}$. Now, since {1,2}, {1,3}, and {1,4} all have rank $k+2$, it follows that $B_2, B_3$, and $B_4$ all must have last coordinate equal to 1. In particular, $B_2 + B_3 +B_4 \neq 0$. Also, since every two element subset of {2,3,4} has rank 2, it follows that $B_2, B_3$ and $B_4$ are distinct. But now {2,3,4} must have rank 3 in $S_k$, a contradiction. $\square$

Moreover, it is easy to check that every minor of $S_k$ is representable over $\mathbb{F}_2$, where minors of polymatroids are defined in the obvious way. Therefore, the set of binary polymatroids has an infinite set of excluded minors.

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