discrete mathematics – Proving that any arbitrary set of the rational numbers with a sum of leq 1 is not a matroid

Let $(X,I)$ be a matroid ($X$ is a finite set). We define $I subseteq 2^S$.

By definition $(X,I)$ is a matroid iff

a) $ Zin I$ and $Y subseteq Z implies Yin I$ (hereditary condition)

c) $Y,Z in I $ and $ |Y|lt |Z|$ then there exists an $z in Zsetminus Y$ such that $(Ycup z) in I$

Now, we consider finite sets of rational numbers $Q$, with the condition that the sum of the elements of any arbitrary $Q$ is less than or equal to 1.

That being said, if we define $Q$ so that $Q={-5,5}$, then it would mean that $Qin S$, however, according to part b) of our definition, a subset $Y$, say, in this case $Y={5}$, would also have to be in $S$, which is not the case. We conclude; $(X,I)$ can’t possibly be a matroid.

Is this proof permissible? I would really appreciate you guys’ thoughts 🙂

co.combinatorics – Dimension of Circuit Space of a Matroid

If $G$ is a graph with edge set $E$, let $W$ be the $mathbb{Z}/2$-vector space generated by the elements of $E$. If $A = {a_1, dots, a_n} subset E$, let $bar{A} = a_1 + dots + a_n in V$; then $bar{A}_1 + bar{A}_2 = overline{A_1 Delta A_2}$, where $Delta$ indicates symmetric difference.

I’ll define the cycle space of $G$ to be the subspace of $W$ generated by simple cycles of $G$. More precisely, the cycle space of $G$ is the subspace of $W$ generated by the set ${bar{C} mid C text{ is a simple cycle of } G}$. We could also view the cycle space as the first simplicial homology group of $G$ over $mathbb{Z}/2$. It is not difficult to show that the dimension of the cycle space of $G$ is the corank of the cycle matroid of $G$.

Given any matroid $M$ with ground set $E$, we could define the circuit space of $M$ in a completely analogous way, just using the word “circuit” instead of “simple cycle.” My question is: is it always true that the dimension of the circuit space of $M$ is the corank of $M$? If not, for what types of matroids is this true? Finally, can anyone recommend good resources that deal with this sort of thing?

Thanks!

co.combinatorics – What are the coloops of a hypergraphic matroid?

For a hypergraph $H=(V,E)$ call $Fsubseteq E$ a hyperforest in $H$ iff there is a forest graph $G$ and a bijection $smallphi:E(G)to F$ satisfying $smallforall ein E(G)(esubseteq phi(e))$ or equivalently $smallforall Isubseteq F(|I|+1leq |cup_{Sin I}S|)$, further we call the matroid $M(H)=(E,mathcal{I})$ such that $mathcal{I}={Isubseteq E:Itext{ is a hyperforest in }H}$ the hypergraphic matroid of $H$ so in particular if $H$ is an undirected graph then this coincides with the standard definition described here while analogously if $mathcal{E}in E$ then $M(Hsetminusmathcal{E})=M(H)setminusmathcal{E}$.

Now with all of that said, my question is given any hypergraph $H$ what are the coloops of $M(H)$?

I suspect they relate to the line graph $smallmathcal{L}(H)=(E(H),{{X,Y}subseteq E(H):Xneq Ytext{ and }Xcap Yneqemptyset})$.

What is the rank of a transversal matroid for a family of sets $mathcal{F}$ such that $T=(mathcal{F},subseteq)$ is a tree-order?

If we let $small r(mathcal{F})=min{|mathcal{F}setminus mathcal{G}|+|cup_{Xinmathcal{G}}X|:mathcal{G}subseteqmathcal{F}}=max{|text{img}(f)|:ftext{ is a choice function of }mathcal{F}}$and $small (mathcal{F},subseteq)text{ is a tree}iffforall X,Y,Zinmathcal{F}(Xsubseteq Ycap Zrightarrow Ycup Zin {Y,Z})$ can $small r(mathcal{F})$ be explicitly defined?

For example if $omega(mathcal{F})$ is the number of edges in the hasse diagram of $(mathcal{F},subseteq)$ (which is a tree graph) then $r(mathcal{F})leq omega(mathcal{F})+1$, in particular if $mathcal{F}$ is an inclusion chain (i.e. if the hasse diagram of $(mathcal{F},subseteq)$ is a path graph) then we know: $r(mathcal{F})=omega(mathcal{F})+1$. But what about in ​general? Is there a way to write $r(mathcal{F})$ in terms of a weighted sum of paths in the hasse diagram of $(mathcal{F},subseteq )$?

co.combinatorics – Is the smallest packing number of any port in a matroid $M$ equal to the maximum number of pairwise disjoint bases in $M$?

If for any matroid $M=(E,mathcal{I})$ we let $M_e={Ssubseteq Esetminus{e}:Scup{e}text{ is a circuit of }M}$ and for any family of non-empty sets $mathcal{F}neqemptyset$ we let $nu(mathcal{F})$ be the maximum number of pairwise disjoint sets in $mathcal{F}$ then is it true that $min{nu(M_e):ein E}=nu({B:Btext{ is a base of }M})$? I’m fairly sure this is correct though I just want to make sure.

Sorry if this is elementary, I know by NW that if $beta(M)=nu({B:Btext{ is a base of }M})$ then we have $smallbeta(M)=maxleft{frac{|E|-|X|}{r(E)-r(C)}:Xsubseteq Etext{ and }r(X)neq r(E)right}$ and $min{nu(M_e):ein E}$ is equal to either $beta(M)-1$ or $beta(M)$ so basically I want to verify this quantity always evaluates to the latter.

ag.algebraic geometry – What’s known about the matroid induced by the PlĂĽcker coordinates of the representation of a matroid?

Let $M$ be a linear matroid with ground set $E$ and independent subsets $mathcal I$, represented by $rho: E rightarrow V$.
This induces a map
$$
hatrho: mathcal I rightarrow mathbf P(Lambda V), {e_1,ldots,e_k} mapsto (rho(e_1) wedge cdots wedge rho(e_k)).
$$

The subset $operatorname{Im}(hatrho) subset mathbf P(Lambda V)$ in turn defines a linear matroid $hat M$ over $Lambda V$.

Is there anything known about this matroid $hat M$, e.g. how it depends on the choice of representation, and what its bases look like in relation to the bases of $M$? I’m interested in particular in the case where $M$ is uniform (for instance, I think the representation shouldn’t matter in that case) or arises combinatorially, and would like to obtain combinatorial descriptions of $hat M$. But any pointers would be greatly appreciated.

co.combinatorics – Does the non-extreme point operator of a closure space being idempotent, imply said space is a matroid (in flat lattice form)?

Given any set $X$ and some closure operator $text{cl}:2^Xto 2^X$ on $X$, we define $psi:2^Xto 2^X$ so that $psi(Q)={qin Q:qintext{cl}(Qsetminus{q})}$ for all $Qsubseteq X$, now if $forall Ssubseteq X(psi(psi(S))=psi(S))$ (i.e. if $psi$ is idempotent) must $text{cl}$ be the closure operator of some matroid?

Sorry if this is either trivially true or trivially false as I am still new to matroid theory.

I know that if $text{cl}$ is the closure operator of a matroid then $psi$ must be idempotent so the converse statement holds. Further I have been able to reduce the problem to proving $mathscr{C}={Csubseteq X:psi(C)neq emptysetlandbigcup_{Ssubset S}psi(C)=emptyset}$ is such that for all $C_1,C_2in mathscr{C}$ we have $ein C_1cap C_2implies exists C_3in mathscr{C}:C_3subseteq (C_1cup C_2)setminus {e}$ i.e. that the family of sets $mathscr{C}$ satisfies the matroid circuit axioms. This is because if $text{cl}$ is the closure operator of a matroid $M$ grounded on $X$, then $psi$ maps every $Ssubseteq X$ to the maximal cyclic set of $M$ in $S$ or in other words $psi(S)$ is the union of all circuits of $M$ contained in $S$, so in particular the closure operator $text{cl}^*:2^Xto 2^X$ of the dual matroid $M^*$ of $M$ satisfies $text{cl}^*(Q)=Xsetminuspsi(Xsetminus Q)$ for all $Qsubseteq X$.

Independent pair in matroid – MathOverflow

Given a matroid $M$ with ground set $E$ of size $2n$, suppose there exists $Asubseteq E$ of size $n$ such that both $A$ and $Esetminus A$ are independent. What is the minimum number of $Bsubseteq E$ such that both $B$ and $Esetminus B$ are independent?

With $n=2$, some casework shows that the answer is $4$: suppose ${1,2},{3,4}$ are independent. Using the augmentation property with ${1}$ and ${3,4}$, we get that wlog ${1,3}$ is independent. If ${2,4}$ is independent, we get four sets $B$, so using ${2}$ against ${3,4}$, it must be that ${2,3}$ is independent. But then using ${4}$ against ${1,2}$ gives us the claim. It is possible that the independent sets are $emptyset,{1},{2},{3},{4},{1,2},{3,4},{1,3},{2,4}$, giving the answer of $4$.

co.combinatorics – Is matroid realizability computable?

Contra my suspicions, the Internet is telling me that Vámos proved in “A necessary and sufficient condition for a matroid to be linear” (citation below) that it is decidable if a matroid is representable over a field. See quote on pg. 3 of Solving Rota’s Conjecture which says “The first exception is the algorithmic problem of determining when a given matroid is representable over an unspecified field, which was proved to be decidable by Vámos (28).”

Vamos, P., A necessary and sufficient condition for a matroid to be linear, Möbius Algebras, Conf. Proc. Waterloo 1971, 162-169 (1975). ZBL0374.05017.