2
$\begingroup$

Assume that $G$ is a (finite) abelian group and $M$ is a matroid whose ground set is $G$. Let $X$ and $Y$ be subsets of $G$, and $H$ is the stabilizer of $X+Y$. That is $X+Y+H=X+Y$. We denote the rank function of $M$ by $r$. Then can we say that $r(X+Y)\geq$ $r(X)+r(Y)-r(H)$? Or under what condition this can be true? I am fine to assume $M$ to be a partition matroid(or even a uniform matroid) and also assume $G$ is cyclic of prime order. (A similar result holds for the cardinality of sumsets of a given group, well-known as Kneser's theorem)

Note: To make the statement plausible, adding the condition of $X+Y$ to be independent won't hurt.

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
+50
$\begingroup$

Without any condition on the matroid structure, there is really no reason for your inequality to hold.

For example, take $X=\{a\},Y=\{b\}$ so that $X+Y=\{a+b\}$ where $a,b$ are any nonzero elements of $G$. Then take any matroid structure such that $a+b$ is a loop but $a,b$ and $0$ are not. We have $0=r(X+Y)<r(X)+r(Y)-r(H)=1$.

Maybe you want to add some kind of compatibility condition between the group and matroid structures to make your question more interesting?

EDIT: Even in a very simple and and well-behaved setting (say $G$ cyclic and $M$ uniform), the inequality fails. For example, let $G=\mathbb{Z}/N\mathbb{Z}$ with the $n$-uniform matroid structure, where $n<N/2$. Let $X\subset [0, N/2)$ be any subset of size $n$. Then $r(X+X)=r(X)=r(Y)=n$ and $H$ is trivial so $r(H)=1$. Hence your inequality fails whenever $n>1$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thanks, Antonie. I just edited my question. I agree that assuming nothing on the group structure and no restriction on the matroid will make things very wild. $\endgroup$
    – Shahab
    Sep 4, 2021 at 16:55
  • $\begingroup$ Is it understood, in the first place, that the matroid is closed under taking pairwise sumsets? And that its structure is compatible with the subset relation of the sets in $G$? $\endgroup$
    – Jukka Kohonen
    Sep 4, 2021 at 21:10
  • $\begingroup$ The matroid is defined over a group, so sumsets are all belong to the ground set, $G$. The matroid itself may not be closed under taking pairwise sumsets. That is, if $X+Y$ is not an independent set, it is legit to discuss $r(X+Y)$ (which is equal to the cardinality of the largest subset of $X+Y$ that is independent in $M$) @Jukka Kohonen $\endgroup$
    – Shahab
    Sep 4, 2021 at 22:28
  • $\begingroup$ Another counterexample: Take a sum-free subset of $A$ of $G$, and let $M$ be a matroid whose vertices set $V(M)$ is $A$. Set $X=Y=A$. Let $r(M)=n$. Then we have: $r(X+Y)=0$ but $r(x)=r(Y)=n$, and $r(H)\leq n$. This violates the claimed inequality $\endgroup$
    – Shahab
    Sep 6, 2021 at 19:13
  • $\begingroup$ Yes @Antoine Labelle, you are right. I however hope that for certain subsets of probably a very small family of matroids it might be the case. Indeed, for situations that $X$ and $Y$ are not full-rank but $X+Y$ is close to being full-rank? $\endgroup$
    – Shahab
    Sep 8, 2021 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.