2
$\begingroup$

I'm struggling to formally demonstrate the hypothesis that every flat of the dual matroid ($F \in \mathcal{F}(M^*)$) is the complement of some circuit or union of circuits of the original matroid $M$.

I tried to translate the following statements into a simpler language and I think I figured out the idea but not how to write it in a generic way:

  1. Hyperplanes are maximal proper flats
  2. Every flat is the intersection of some collection of hyperplanes of the matroid
  3. Circuits are minimal dependent sets
  4. Hyperplanes of $M^*$ are the complements of the circuits of $M$.

If all the flats of the matroid can be constructed by forming all possible intersections of various collections of hyperplanes and $E(M) - H^* = C$, then all the unions of various collections of circuits can be constructed by $E(M) - F^*$.

Summing up:

$F^* = H^*_1 \cap H^*_2 = {({H^*_1}^C \cup {H^*_2}^C)}^C = {(C_1 \cup C_2)}^C$

Could you help me see the best way to write this?

My main reference is the book Matroids: A Geometric Introduction, by Gary Gordon and Jennifer McNulty.

$\endgroup$

1 Answer 1

2
$\begingroup$

You seem to have all of the pieces in hand to prove the statement in general.

$F^*$ is a flat of $M^*$ if and only if there are hyperplanes $H^*_1, \dots, H^*_n$ of $M^*$ ($n\ge 1$) such that $F^* = \bigcap_{i=1}^nH^*_i$. But $H^*$ is a hyperplane of the dual if and only if there is a circuit $C$ of $M$ such that $E \setminus C= H^*$. So $F^*$ is a flat of $M^*$ if and only if $F^* = \bigcap_{i=1}^n(E \setminus C_i)$, where $C_i$ is the complement of $H_i$. It follows that $F^* = E \setminus \bigcup C_i$ since the intersection of complements of sets in $E$ is the complement of the union of the sets.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .