Direct sum of q-matroids

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One of the most straightforward operations you can do when you want to make a new matroid out of old ones, is taking the direct sum. The direct sum of the matroids $M_1=(E_1,r_1)$ and $M_2=(E_2,r_2)$ is the matroid $M$ on ground set $E=E_1\sqcup E_2$ with for all $A\subseteq E$,
\[ r(A)=r_1(A\cap E_1)+r_2(A\cap E_2). \]Alternatively, its independent sets are the unions of an independent set in $M_1$ and an independent set in $M_2$. The third equivalent way to define the direct sum is by saying that $M=M_1\oplus M_2$ is precisely the matroid such that $M|_{E_1}=M/E_2=M_1$ and $M|_{E_2}=M/E_1=M_2$.

In this post, we will discuss the q-analogue of this concept. The q-analogues of matroids, called q-matroids, have been introduced here. Instead of considering a matroid on a finite ground set, where every subset has a rank that satisfies certain axioms, we take a q-matroid on a finite dimensional vector space where every subspace has a rank, again satisfying certain axioms. A good intuition is to think about matroids as a bicolouring of the Boolean lattice, and q-matroids as a bicolouring of the subspace lattice.

If we take the direct sum of two q-matroids $M_1=(E_1,r_1)$ and $M_2=(E_2,r_2)$, it makes sense for $M=M_1\oplus M_2$ to have as a ground space the direct sum of vector spaces $E_1\oplus E_2$. It is also reasonable to ask that $M|_{E_1}=M/E_2=M_1$ and $M|_{E_2}=M/E_1=M_2$. But then it gets difficult. Where for sets, we can write any subset $A\subseteq E_1\sqcup E_2$ as $(A\cap E_1)\sqcup(A\cap E_2)$, such a thing is not true for vector spaces. Suppose for example $E_1=\langle100\rangle$ and $E_2=\langle010,001\rangle$ (over my favourite finite field $\mathbb{F}_2$) such that $E_1\oplus E_2$ is a 3-dimensional space. Then the space $A=\langle111\rangle$ does not intersect $E_1$ nor $E_2$. What is its rank going to be?

One could hope that maybe the rank axioms of q-matroids take care of this. Look for example at the direct sum of $U_{1,1}$ and $U_{1,2}$ over $\mathbb{F}_2$.

We see that the interval between $0$ and $\langle100\rangle$ and the interval between $\langle010,001\rangle$ and $E$ are coloured as $U_{1,1}$, showing $M|_{E_1}=M/E_2=M_1$. On the other hand, the interval between $0$ and $\langle010,001\rangle$ and the interval between $\langle100\rangle$ and $E$ are coloured as $U_{1,2}$, showing that $M|_{E_2}=M/E_1=M_2$. There is now only one way to colour the rest of the lattice:

Unfortunately, this construction becomes not unique already in dimension 4. If we try to make the direct sum $U_{1,2}\oplus U_{1,2}$, semimodularity of the rank function gives that all 1-dimensional spaces have rank 1 and all 3-dimensional spaces have rank 2. Any 2-dimensional space intersecting $E_1$ or $E_2$ is a basis. But there are 2-dimensional spaces that intersect neither $E_1$ nor $E_2$ (over $\mathbb{F}_2$, there are three of them) and those can be either a basis or a circuit. Any choice will produce another q-matroid.

The solution to this problem is to write the direct sum in a very convoluted way that is not helpful for matroids over sets, but that does have a nice q-analogue. This is the following. As mentioned, let $E=E_1\oplus E_2$. Make a matroid $M_1’$ by adding loops to $M_1$ until its groud space is $E$: $M_1’$ is the q-matroid on $E$ such that $M_1’|_{E_1}=M_1$ and $M_1’|_{E_2}$ consists of only loops. Similarly, let $M_2’$ be the q-matroid on $E$ such that $M_2’|_{E_2}=M_2$ and $M_2’|_{E_1}$ consists of only loops. Now the direct sum $M_1\oplus M_2$ is defined as the matroid union $M_1’\vee M_2’$.

It turns out that adding loops is, inductively, possible in the q-analogue. Also matroid union (as an honour to the name of this blog, surely!) has a well-defined q-analogue: this operation is defined for matroids on the same ground set, and can thus be generalised to q-matroids with the same ground space without the troubles that come with the direct sum of vector spaces. The rank function of the direct sum of two q-matroids is \[ r_{M_1\oplus M_2}(A)=\min_{X\subseteq A}\{r_{M_1′}(X)+r_{M_2′}(X)+\dim A-\dim X\}. \]

An equivalent way to phrase this definition of the direct sum of q-matroids, is by requiring the direct sum to to be the “most independent” q-matroid satisfying $M|_{E_1}=M/E_2=M_1$ and $M|_{E_2}=M/E_1=M_2$. This can be made precise in category theory language.

It is unclear what the independent spaces of the direct sum of two q-matroids are. The vector space sum of an independent space in $M_1$ and an independent space in $M_2$ is independent in the direct sum. But there are many more independent spaces. Many other properties of the direct sum of matroids fail to have a nice q-analogue. What does work in the q-analogue, is that the cyclic flats of the direct sum are exactly the sums of the cyclic flats of $M_1$ and $M_2$.

You might have guessed that if $M_1$ and $M_2$ are q-matroids represented by matrices $G_1$ and $G_2$ (I’m going to use this notion without a formal definition), then their direct sum is not necessarily represented by the matrix \[ \left[ \begin{array}{cc} G_1 & 0 \\ 0 & G_2 \end{array} \right]. \] Interestingly, the question about representability can be translated to the language of linear sets in finite geometry, making it possible to use several results in finite geometry for q-matroids.

References

[AJNZ26] Gianira Alfarano, Relinde Jurrius, Alessandro Neri and Ferdinando Zullo. Representability of the direct sum of uniform q-matroids. Combinatorial Theory, 6 (2026).

[CJ24] Michela Ceria and Relinde Jurrius. The direct sum of q-matroids. Journal of Algebraic Combinatorics, 59, pp. 291-330 (2024).

[GLJ23] Heide Gluesing-Luerssen and Benjamin Jany. Coproducts in categories of q-matroids. European Journal of Combinatorics 112, 103733 (2023).

[GLJ24] Heide Gluesing-Luerssen and Benjamin Jany. Decomposition of q-matroids using cyclic flats. SIAM Journal on Discrete Mathematics 38, pp 2940–2970 (2024).

[GLJ25] Heide Gluesing-Luerssen and Benjamin Jany. Representability of the direct sum of q-matroids. Journal of Algebraic Combinatorics 61, 51 (2025).

The cycle double cover theorem

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This is a guest post by Johannes Carmesin.

You have probably all heard that OpenAI announced a fully automated proof of the cycle double cover conjecture, which was conjectured independently by Tutte, Itai and Rodeh, Szekeres, and Seymour about fifty years ago.

First, to answer the most immediate question: yes, the proof is correct.

Here, I would like to do two things:

  1. explain the statement of the theorem; and
  2. give an overview of how the proof has been verified.

Statement of the theorem

The theorem has a particularly appealing topological formulation: every bridgeless multigraph can be embedded in a pseudosurface in such a way that every face is a disc. In particular, every edge occurs exactly twice among the boundary walks of the faces. Here, a pseudosurface is a topological space obtained from a closed surface (possibly disconnected) by identifying finitely many points.

The use of pseudosurfaces is very natural in this context: embeddings of different blocks can be glued together at their common cut-vertices. An example that pseudosurfaces are genuinely necessary is given below.

The bowtie graph, consisting of two triangles glued together at a single vertex. It has a unique cycle double cover. It contains each triangle twice. The corresponding embedding has four disc faces and lies naturally in two spheres identified at the common vertex. This shows that pseudosurfaces are genuinely necessary in the topological formulation of the problem.
The bowtie graph, consisting of two triangles glued together at a single vertex.
It has a unique cycle double cover. It contains each triangle twice.
The corresponding embedding has four disc faces and lies naturally in two spheres identified at the common vertex. This shows that pseudosurfaces are genuinely necessary in the topological formulation of the problem.

The algebraic counterpart of this topological statement is the following equivalent formulation: every bridgeless multigraph $G$ has a cycle double cover. That is, there exists a family $\{C_i\mid i\in I\}$ of cycles of $G$ such that every edge belongs to exactly two of the cycles $C_i$.

To see the connection, suppose first that such a cycle double cover is given. For every cycle $C_i$, take a disc and glue its boundary to the corresponding cycle in the graph. Since every edge occurs in exactly two cycles of the family, every edge is incident with two such discs. The resulting space is a pseudosurface in which the original graph is embedded and the added discs are precisely its faces.

For a simple example showing that pseudosurfaces are genuinely necessary in the above topological formulation of the cycle double cover conjecture, see the figure below.

Conversely, given such an embedding in a pseudosurface, the family $\{C_i\mid i\in I\}$ consists of the boundary walks $C_i$ of the faces of the embedding, which are cycles. Taken over all faces, these cycles form a cycle double cover, because every edge occurs exactly twice among the boundary walks of the faces.

The assumption that the graph is bridgeless is necessary. Indeed, a bridge is a cocircuit-singleton and thus cannot belong to any cycle and therefore cannot be covered even once by a cycle, let alone exactly twice.

About the proof

The proof by OpenAI establishes the second, algebraic formulation. In a nutshell, it begins with a nowhere-zero $\mathbb{F}_2^3$-flow $f$ on $G$; that is, an assignment of a nonzero vector in $\mathbb{F}_2^3$ to each edge such that, at every vertex, the values on the incident edges sum to zero. The existence of this flow follows from Seymour’s nowhere-zero six flow theorem (see also the recent short proof by DeVos and Nurse).

Roughly speaking, this flow is not quite a cycle double cover. The paper studies how one locally needs to modify the flow at each vertex so that it becomes a cycle double cover, and then there is a compatibility condition that needs to hold between adjacent vertices.

The compatibility conditions between the local modifications, and the equations for the local modifications give rise to a system of linear equations over $\mathbb{F}_2$, with one vector variable in $\mathbb{F}_2^3$ for each vertex and one scalar variable in $\mathbb{F}_2$ for each edge. OpenAI then uses duality to prove that this system always has a solution. The entire proof occupies only two pages.

Sang-il Oum gave an excellent talk on the proof and provides a much more detailed explanation. Jim Geelen (link) and Sang-il Oum (link) independently wrote short notes giving the complete proof in a different presentation, which I both find particularly pleasant to read and the latter additionally contains open questions.

Finally, the proof has already been formalised independently several times in Lean 4. Krystal Guo produced one formalisation of the core argument (personal communication), while Vaibhav Bajpai, Utku Okur, and I produced another, and one by OpenAI (see also this discussion in the AI-authored projects channel of the Lean Zulip community).

Matroid varieties

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Universal models for graphic and representable matroids

Graphs have the useful property that each of them is a restriction of a complete graph on the same set of vertices. This property makes it easy, for example, to generate a random graph: simply flip a coin for each edge in the complete graph, and include the edge if the coin turns up heads. In matroidal terms, the property is that every rank-$n$ simple graphic matroid is a restriction of $M(K_{n+1})$.

There is another well known class of matroids for which a similar observation holds: those representable over a fixed finite field, in which case the projective geometry of a given rank serves as a “universal model”, in the sense that every simple rank-$n$ matroid that is representable over the finite field $\mathbb{F}_q$ can be obtained from $\text{PG}(n-1,\mathbb{F}_q)$ by restriction.

On the other hand, the same is not true for the class of all matroids. Even for matroids of rank 2, there is no universal model $M$ such that each such matroid is a restriction of $M$.

Dowling geometries

The class of graphic matroids is only one member of a family of classes with the same property: the Dowling matroids. These have been discussed before on the blog in the context of biased graphs, but let’s briefly recall their definition.

The rank-$n$ Dowling geometry $\text{DG}(n, \Gamma)$ is determined by a finite group $\Gamma$, similar to how projective geometries are determined by a field, and a Dowling matroid is simply any matroid that can be obtained by restricting $\text{DG}(n, \Gamma)$. Let’s assume that the operation in $\Gamma$ is multiplication. A $\Gamma$-gain graph is a graph $G$, together with an orientation of the edges of $G$ and a gain function $\varphi\colon E(G)\to\Gamma$. For a cycle $C$ of $G$, we pick a starting point an an orientation of $C$, which gives us an order of the edges of $C$, say $C = e_1e_2\ldots e_k$. Define $\varphi(C) = \varphi(e_1)^{s_1} \varphi(e_2)^{s_2} \ldots \varphi(e_k)^{s_k}$, where $s_i = 1$ if the orientation of $e_i$ agrees with the orientation of $C$, and $s_i = -1$ otherwise. The cycle $C$ is called “balanced” if $\varphi(C)$ is the identity in the group; it is a straightforward exercise to check that balance of $C$ does not depend on the orientation or starting point of $C$.

Two $\mathbb{Z}_3$-gain graphs. In the theta-graph on the left, all cycles are balanced; in the theta-graph on the right, only one cycle is balanced. Group elements are written additively.

A $\Gamma$-gain graph $G$ gives rise to a matroid on $E(G)$ whose circuits are the balanced cycles of $G$, as well as the theta-subgraphs with three unbalanced cycles, and subgraphs formed by two edge-disjoint unbalanced cycles connected by a (possibly empty) path that is disjoint from the two cycles except for its first and last vertex.

We can now define the Dowling geometry $\text{DG}(n,\Gamma)$ as the matroid obtained from the biased graph $K_n^\Gamma$ whose vertices are labelled $1, 2, \ldots, n$, which has one (unbalanced) loop attached at each vertex, and which has $|\Gamma|$ edges between vertices $i$ and $j$ (directed in towards the largest label, say), each labelled with a different group element.

If every (non-loop) edge is labelled by the identity in $\Gamma$, then every cycle is balanced. In particular, $\text{DG}(n,\langle 1\rangle) \cong M(K_{n+1})$, where we write $\langle 1\rangle$ for the trivial group. So, graphic matroids are Dowling matroids.

Varieties

Apart from graphic matroids, matroids representable over finite fields, and Dowling matroids, are there any other natural classes of matroids that have a sequence of universal models? It turns out that we have to be a bit careful with our definitions, but a beautiful result by Kahn and Kung from 1982 states that the answer is, essentially, no.

First, we need to define precisely what we mean by a universal model. Let $\mathcal{M}$ be a class of matroids. A sequence $M_1, M_2, M_3, \ldots$ of matroids is called a sequence of universal models for $\mathcal{M}$ if (i) for every $n$, $M_n$ is a rank-$n$ matroid, and (ii) for every $n$, every simple rank-$n$ matroid in $\mathcal{M}$ is isomorphic to a restriction of $M_n$. Thus, $\text{PG}(0,\mathbb{F}_q), \text{PG}(1,\mathbb{F}_q), \text{PG}(2, \mathbb{F}_q), \ldots$ is a sequence of universal models for the $\mathbb{F}_q$-representable matroids, and $\text{DG}(1,\Gamma), \text{DG}(2,\Gamma), \text{DG}(3,\Gamma), \ldots$ is a sequence of universal models for the Dowling matroids over the group $\Gamma$.

Second, we need to be careful about our definition of “natural” class of matroids. A class $\mathcal{M}$ is called a hereditary class if it is closed under isomorphism, as well as under taking minors and direct sums; so, if $M,N \in \mathcal{M}$ and $e$ is an element of $M$, then each of $M\backslash e$, $M/e$, and $M\oplus N$ are in $\mathcal{M}$ as well.

A variety of matroids is a hereditary class with a sequence of universal models. We are now ready to state Kahn and Kung’s result.

Theorem (Kahn–Kung, 1982). If $\mathcal{M}$ is a variety of matroids, then $\mathcal{M}$ is one of the following classes:

  • Matroids representable over a finite field;
  • Dowling matroids over a finite group; or
  • Matchstick geometries or Origami geometries.

The classes of matchstick and origami geometries have low connectivity. The universal models for matchstick geometries are $U_{2,n+1}^{\oplus k}$ and $U_{2,n+1}^{\oplus k} \oplus U_{1,1}$, depending on the parity of the rank, while the universal models for origami matroids are obtained from a basis ${b_1, …, b_r}$ by adding $n$ points freely to each of the lines spanned by pairs $\{b_i, b_{i+1}\}$.

Each of the assumptions (that the class of matroids be minor-closed, closed under direct sum, and have a sequence of universal models) in Kahn and Kung’s theorem is necessary for its conclusion. It is an amusing exercise to come up with classes of matroids that satisfy only a subset of these assumptions but not the others.

Trouble with infinite oriented matroids

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After many years, this series of posts on infinite matroids is finally coming to an end. We started with the basic definitions and axioms, and building on those we have seen how a great deal of matroid theory can be lifted to infinite matroids, from representability and graphic matroids through to excluded minor characterisations and matroid intersection. Today, however, we are going to look at an instructive case of ideas from finite matroid theory which do not generalise to infinite matroids.

This is an old project, which I have been thinking about on-and-off for years together with Winfried Hochstättler and Stefan Kaspar. Probably Stefan is the one who has put most into this project and who has kept it alive and interesting for so long. Thanks Stefan! It’s about trying and failing to sensibly define infinite oriented matroids.

Before we get to what goes wrong in the infinite world, however, we should take a moment to remind ourselves how finite oriented matroids are defined. Although there are lots of different cryptomorphic axiomatisations, for our purposes it will be enough to look at the axiomatisation in terms of signed circuits.

A signing $X$ of a set $\underline X$ is a partition of $\underline X$ into a pair of sets $(X^+, X^-)$. There are lots of examples of matroids that come with a natural signing of each of their circuits. For example, let $D$ be any directed graph, and let $M$ be the cycle matroid of its underlying undirected graph. Then any circuit of $M$ is a cycle, and we can naturally partition its edges according to which way they point around the cycle. More generally, if $M$ is any matroid represented over the reals, then any circuit of $M$ is necessarily linearly dependent, and the linear combination witnessing this is unique up to (nonzero) scaling. So we can partition the circuit into those elements with a positive and those with a negative coefficient.

Notice that in each case, when we took a signing $X$ we could equally have taken the opposite signing $-X$ with $(-X)^+ = X^-$ and $(-X)^- = X^+$. So it is normal to include both options, giving a collection of signed sets which is symmetric, in the sense of being closed under taking opposites. To capture these and many more general examples, we can define an oriented matroid to be a pair $(E, \mathcal{C})$ where $E$ is a (finite) set and $\mathcal{C}$ is a set of signed subsets of $E$ satisfying the following axioms:

(OC1) $\emptyset \notin \mathcal{C}$

(OC2) $\mathcal{C}$ is symmetric

(OC3) For any $X$ and $Y$ in $\mathcal{C}$ with $\underline X \subseteq \underline Y$ we have $X = \pm Y$

(OC4) For any $X$ and $Y$ in $\mathcal{C}$, any $e \in X^+ \cap Y^-$, there is some $Z \in \mathcal{C}$ with $Z^+ \subseteq (X^+ \cup Y^+) – e$ and $Z^- \subseteq (X^- \cup Y^-) – e$

It’s clear that the underlying sets of these signed circuits must then be the circuits of a matroid, and we call $\mathcal{C}$ an orientation of that matroid. There is no space here to go into the rich theory of finite oriented matroids, but a good introduction is [1]. All we will need is that, as for ordinary matroids, we can define minors and duality for oriented matroids.

To understand the duality, we need the notion of orthogonality. If $X$ and $Y$ are signed sets then we say they are orthogonal if they are disjoint or if there is both at least one element where they agree on the sign and also at least one element where they disagree on the sign. Then for any orientation of a matroid there is a unique orientation of its dual all of whose signed sets are orthogonal to the signed circuits of the original matroid. Having defined duality of oriented matroids in this way, we can go on to define minors, since contraction is dual to restriction and it is clear how restriction should be defined for oriented matroids.

How could we generalise these notions to infinite matroids? It’s clear how to define finitary oriented matroids: they should be pairs $(E, \mathcal{C})$ such that $E$ is a set and $\mathcal{C}$ is a set of finite signed subsets of $E$ satisfying (OC1)-(OC4). But how can we go beyond this? There are a few essential properties that any definition should satisfy:

  • Every infinite oriented matroid should satisfy at least the axioms (OC1)-(OC4), though there could be additional restrictions
  • Infinite oriented matroids should be closed under both minors and duality
  • Natural examples such as finitary oriented matroids and infinite regular matroids (defined earlier in this series) should be included.

Unfortunately, there is no way to satisfy all of these requirements simultaneously. To understand why, we need to closely examine a particularly simple class of finitary orientable matroids. Suppose that we have any set $E$ of nonzero vectors in $\mathbb{R}^3$, such that no three of them lie in a common plane through the origin. Then the corresponding vector matroid $M(E)$ is simply the uniform matroid of rank 3 on $E$. Since this matroid is real representable, we can orient it in the way discussed earlier. This gives us a finitary oriented matroid.

What about its dual $M^*(E)$? That should be an orientation of the uniform matroid on $E$ of corank 3, whose circuits are all the sets of the form $E \setminus \{a,b\}$ with $a$ and $b$ distinct elements of $E$. To see how such a set should be oriented, consider the plane $H(a,b)$ through $a$, $b$ and the origin. We should partition the elements of $E \setminus \{a,b\}$ according to which side of this hyperplane they lie on. It isn’t hard to check that this signed set is orthogonal to every signed circuit, since no linear combination of points on the same side of the plane with all coefficients positive (or all negative) can ever evaluate to zero. So this must be the signing of $E \setminus \{a,b\}$ in the dual matroid.

Finally, let’s consider what the axiom (OC4) applied to $M^*(E)$ says about the set $E$. Suppose that we have two signed circuits $X$ and $Y$ of $M^*(E)$, orienting $E \setminus \{a,b\}$ and $E \setminus \{c,d\}$ respectively, and let $e$ be any element of $X^+ \cap Y^-$. Thus $e$ is different from $a$, $b$, $c$ and $d$. The signed circuit $Z$ given by (OC4) cannot contain $e$, so it must be a signing of $E \setminus \{e,f\}$ for some $f$. Furthermore, the region of $\mathbb{R}^3$ on the positive side of both $H(a,b)$ and $H(c,d)$ but on the negative side of $H(e,f)$ cannot meet $E$. If $E$ is finite then we have some wiggle room and we can always find a suitable choice of $f$ achieving this. But at the other end of the spectrum, if $E$ is dense in $\mathbb{R}^3$, this is a very strong condition. It forces $H(e,f)$ to contain the intersection of $H(a,b)$ and $H(c,d)$.

Putting all of this together, let’s fix a countable dense set $E$ of nonzero vectors in $\mathbb{R}^3$ such that no three of them lie on a common plane through the origin and there are no six of them $a,b,c,d,e,f$ such that $H(a,b)$, $H(c,d)$ and $H(e,f)$ all meet in a common line. It is straightforward to recursively build such a set. Then the dual to the natural orientation of $M(E)$ doesn’t satisfy (OC4). So there is no notion of infinite oriented matroid satisfying all three of the conditions above.

Where does this leave us? If we want to define infinite oriented matroids then we face a trilemma: we must give up on one of the three conditions.

Firstly, we could give up on the familiar circuit axioms. This is the approach taken by Wenzel in [2], as part of a larger project of defining infinite matroids over arbitrary fuzzy rings. Although it gives a very general and flexible definition in that context, in the particular case of oriented matroids we would lose many of the familiar properties and most of the theory of finite oriented matroids would have no analogue in this very general context.

Secondly, we could give up on duality but hope to preserve minors. From a matroid theoretic point of view, this would be a radical step, and it is hard to find an appropriate place to draw the line which is more useful than simply taking the union of the classes of finitary and regular matroids that we wanted to include. As an example of what can go wrong with this approach, those who remember the circuit axioms for infinite matroids might be tempted to work with signings of the circuits of infinite matroids satisfying (OC1)-(OC3) together with the following strengthened version of (OC4):

(OC4′) Suppose that $X \in \mathcal{C}$, $I \subseteq \underline X$ and $(Y_e)_{e \in I}$ is a family of elements of $\mathcal{C}$ such that $\underline Y_e \cap I = \{e\}$ and if $e$ is in $X^+$ then it is in $Y_e^-$ and if it is in $X^-$ then it is in $Y_e^+$. Then for every $f \in X^+ \setminus \bigcup_{e \in I} Y_e^-$ there is some $Z \in \mathcal{C}$ such that $f \in Z^+ \subseteq (X^+ \cup \bigcup_{e \in I} Y_e^+) \setminus I$ and $Z^- \subseteq (X^- \cup \bigcup_{e \in I} Y_e^-) \setminus I$

Unfortunately the class of such oriented matroids is not even closed under minors. Returning to our earlier counterexample, we can recursively extend our earlier set $E$ to a larger countable set $E’$ such that for any five distinct elements $a,b,c,d$ and $e$ of $E’$ there is a further element $f$ such that $H(a,b)$, $H(c,d)$ and $H(e,f)$ do intersect in a common line. So $M^*(E’)$ does satisfy (OC4), and it can even be shown to satisfy (OC4′). But its contraction minor $M^*(E)$ does not.

The final alternative is to give up on including all finitary oriented matroids. We could still hope to include all regular infinite matroids, since that class is already closed under duality and minors. The most promising option here involves the Farkas condition, which states that every element of $E$ must be contained in a positive circuit $X$ (one with $X^+ = \underline X$) or a positive cocircuit. Imposing this condition on the matroid and all its minors already forces (OC4′). However, this is a very restrictive definition, and for example it remains open whether it even includes all regular infinite matroids.

All of these issues and many more are explored in far more depth in our preprint [3]. Even though we didn’t find a satisfying definition of infinite oriented matroids, we ran into a number of subtle and interesting problems along the way, and the exploration was well worth the effort.

—

**References**

[1] A. Björner, M. Las Vergnas, B. Sturmfels, N. White, and G. M. Ziegler, *Oriented Matroids*, Encyclopedia of Mathematics and its Applications, vol. 46, 2nd ed., Cambridge University Press, 1999.

[2] W. Wenzel, *Dual Pairs of Matroids with Coefficients in Finitary Fuzzy Rings of Arbitrary Rank*, Mathematica Pannonica **27/NS1** (2021), no. 1, 48-61.

[3] N. Bowler, W. Hochstättler, and S. Kaspar, *On the Possibilities of Defining Infinite Oriented Matroids*, preprint (2026), arXiv:2603.15843.