I don’t know enough about orientable matroids to have proper intuition. I can give an answer to Q1 part (ii) (the answer is no) and for what it’s worth, I think the answer to Q2 will be yes.

Take AG(2,3), the ternary affine plane. The appendix in Oxley’s book tells us that it has no M(K_4)-minor. It is not real-representable, so no class in your list contains only real-representable matroids.

On the other hand, AG(2,3) is base-orderable, and therefore orientable. So I used Sage to construct ternary coextensions of AG(2,3) that do not have M(K_4), in the hope that some of them will be non-orientable. Here’s an example, represented over GF(3):

[1 2 1 2 1 2 1 2 1 0]

[0 0 1 1 2 2 1 1 0 0]

[1 1 1 1 1 1 0 0 0 0]

[0 0 1 1 0 1 2 0 2 1]

But now I’m stuck, because I don’t know how to quickly test whether this is non-orientable.

Cheers,

Dillon

EDIT: I got the order of subclass inclusion wrong, so of course AG(2,3)

tells us nothing about any of the classes other than Class (1), the

matroids with no M(K_4)-minor. This class certainly contains

matroids that are not real-representable.

I can now do better than that, and show that Class (1) contains

non-orientable matroids. In their paper *Orientability of
Matroids*, Bland and Las Vergnas call the following matroid the

matroids

projective geometry

configuration to Friedrich Levi, in

Konfigurationen

In any case, I entered the matroid into Sage with the following command:

M=Matroid(groundset=[‘a’,’b’,’c’,’v’,’x1′,’x2′,’y’,’z’],

circuit_closures=[(2,{‘a’,’b’,’c’}),(2,{‘a’,’v’,’x2′}),(2,{‘a’,’x1′,’y’}

),(2,{‘b’,’x1′,’x2′}),(2,{‘b’,’y’,’z’}),(2,{‘c’,’v’,’y’}),(2,{‘c’,’x2′,’

z’}),(2,{‘v’,’x1′,’z’}),(3,{‘a’,’b’,’c’,’v’,’x1′,’x2′,’y’,’z’})])

Bland and Las Vergnas state that M is an excluded minor for the class of

orientable matroids (p. 114). On the other hand, if I ask Sage:

M.has_minor(matroids.CompleteGraphic(4))

it reports

False.

So M is an example of a M(K_4)-free matroid that is non-orientable.

]]>Question for y’all,

Here is a partial list of complete matroid classes (closed under minors, duals, direct sums, principal extensions). Each class in the list is properly contained in its predecessors. (In fact, M(K_4) is a minimal excluded minor for each class.)

1. M(K_4)-free

2. Base orderable

3. k-base orderable (k=2,3,…)

4. Strongly base orderable

5. Gammoids

Now, every gammoid is orientable (indeed R-representable).

Q1. Does another class in this list consist entirely of orientable (or R-representable) matroids?

Q2. Is there a non-orientable matroid without M(K_4) minor?

I have not found much helpful literature so far.

I only saw your comment today (Jan 24)! I’m not a member of the Matroid Union and didn’t get an email notification that there was a comment… As for your comment itself, sounds great! Please send me the details when you’ve written this up. I tried to extend our theory to the non-commutative case using GP functions and failed; your idea to use Plucker ratios instead of GP functions is very cool.

All the best,

Matt

Defining a (left) weak matroid over a skew hyperfield is easy enough, you just use your weak circuit axioms. The key to proving duality etc. is to replace the Grassmann-Plucker functions with Plucker ratio’s, as follows.

For any matroid $N$, let

$$A_N:=\{(B,B’): B, B’\text{ are adjacent bases of }N\}.$$

Consider a (skew) hyperfield $H$. Then $[.]:A_N\rightarrow H$ is a * left Plucker ratio map* if

**(P0)** $[Fa, Fb] [Fb, Fa]=1$

**(P1)** $[Fac,Fbc] [Fab, Fac] [Fbc, Fab]=1$

**(P2)** $[Fa,Fb] [Fb, Fc] [Fc, Fa]=-1$

**(P3)** $[Fac, Fbc]=[Fad, Fbd]$ whenever $Fab$ is not a basis.

**(P4)** $1\in [Fbd, Fab] [Fac, Fcd]+[Fad, Fab] [Fbc, Fcd]$

In the presence of an underlying matroid $N$, these axioms are cryptomorphic to the left circuit axioms (C0)-(C3). From any left $H$-matroid $M$ with circuits $\mathcal{C}$, one can derive a Plucker ratio map by setting

$$[Fa, Fb]_M:= X(a)^{-1}X(b)$$

for any circuit $X\in\mathcal{C}$ with $a,b\in \underline{X}\subseteq Fab$.

In the converse direction, a Plucker ratio map allows the reconstruction of $H$-oriented circuits $\mathcal{C}$ that satisfy the weak circuit axioms (given the underlying matroid $N$).

The dual of a left Plucker ratio map for $N$ is a right Plucker ratio map for $N^*$ determined by

$$[B, B’]^* = -[E-B, E-B’].$$

This allows to construct the dual $M^*$ of a weak left $H$-matroid $M$ via the Plucker ratio map.

The *cross ratio* is defined as $cr(F,a,b,c,d):=[Fac, Fbc][Fbd, Fad]$.

Only axiom (P4) refers to the hyperfield addition, and although I have not written it up for tracts I’m quite confident that the whole thing works in that setting as well. In the proofs I only ever refer to the multiplicative group of $H$ and the collection $$\{(a,b) \in H^2: 1\in a+b\}.$$

]]>I don’t think our result quite shows that, since we only forbid $5$- and $6$-tuples of nonbases in a specific structure; potentially we still keep rank-$4$ minors on eight elements with $> 4$ nonbases that are not Vamos-like.

However, I’ve thought about similar questions, and I think it should be the case that there is some constant $\alpha > 0$ so that there are a lot of matroids whose $m$-element minors all have at most $\alpha m$ nonbases; for $s \approx m/2$ this is very few indeed.

In fact, I think we can show, by sampling from the Graham/Sloane construction instead of the whole Johnson graph, that for a worse value of $\alpha$, the same statement holds when $n^{-2}$ is replaced with $n^{-h}$ for any $h$ strictly greater than $1$.

This is certainly eroding my belief in the conjecture that for sparse paving $N$, almost every matroid contains $N$ as a minor.

]]>The proof of your final claim shows that there are very many sp matroids (at least $c n^{-2}\binom{n}{n/2}$ in the exponent) so that each minor on $m=8$ elements of rank $s=4$ has at most $k=4$ nonbases. What function $k=k(m,s)$ makes this true in general?

(Point 2 just above that claim seems to have swallowed some surrounding text.)

]]>https://en.wikipedia.org/wiki/Dickson%27s_lemma

and

https://en.wikipedia.org/wiki/Buchberger%27s_algorithm

Dicksons lemma follows by a straightforward induction argument.

So no need to apply Zorns Lemma here.

]]>I think your interpretation of the ultraproduct is exactly right.

]]>