Representable Matroid Lifts

Posted on September 7, 2026 by Daniel Bernstein

In this blog post, I’m going to discuss a part of a paper I wrote with Zach Walsh a few years back called “Matroid lifts and representability.” We’ll start by motivating the idea of a matroid lift.

Let $A \in \mathbb{F}^{r_1 \times n}$ and $B \in \mathbb{F}^{r_2 \times n}$ be matrices such that $A$ is a row-submatrix of $B$. Let $E$ denote the column set of $B$, which also naturally indexes the column set of $A$. Let $M_A$ and $M_B$ denote the matroids on ground set $E$ defined by linear independence in $A$ and $B$, respectively. Since $A$ is a row-submatrix of $B$, there is a special relationship between $M_A$ and $M_B$. In particular, every flat of $M_A$ is a flat of $M_B$. This motivates the following definition: given two matroids $M_1$ and $M_2$ on a common ground set $E$, one says that $M_2$ is a lift of $M_1$ if every flat of $M_1$ is a flat of $M_2$.

An Example

Recall that $U_{r,n}$ denotes the uniform matroid of rank $r$ on $n$ elements. One can check that $U_{2,3}$ is a lift of $U_{1,3}$. One can check this by noting that the flats of $U_{1,3}$ are the empty set, the full ground set, and each singleton set and that each of these are also flats of $U_{2,3}$. Alternatively, one can consider the following $\mathbb{Q}$-matrix

$\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix}$

and note that the matroid of the top row is $U_{1,3}$ and the matroid of the whole matrix is $U_{2,3}$. One interesting thing to note is that even though $U_{1,3}$ and $U_{2,3}$ are both $\mathbb{F}_2$-representable, there is no $2\times 3$ $\mathbb{F}_2$ matrix $B$ that represents $U_{2,3}$, whose top row represents $U_{1,3}$.

Notational Remark

For any matroid $Q$, I will use the notation $r_Q$ and ${\rm cl}_Q$ to denote the rank function and closure operator in $Q$.

Elementary lifts

If $M_2$ is a lift of $M_1$, then the rank of $M_2$ is greater than or equal to that of $M_1$. When $M_2$ is a lift of $M_1$ whose rank is exactly one greater than that of $M_1$, one says that $M_2$ is an elementary lift of $M_1$. I will now describe a classical theorem of Brylawski, the dual version of which was previously stated by Crapo, that gives an elegant combinatorial characterization of all elementary lifts of a matroid $M$.

Let $M$ be a matroid on ground set $E$. A linear class of $M$ is a collection $\mathcal{L}$ of circuits of $M$ such that if $C_1,C_2 \in \mathcal{L}$ satisfy

$|C_1 \cup C_2| – r_M(C_1 \cup C_2) = 2$

then if $C_3$ is a circuit of $M$ contained in $C_1 \cup C_2$, then $C_3 \in \mathcal{L}$. For each such linear class, one can define a function $r_\mathcal{L}: 2^{E} \rightarrow \mathbb{N}$ as follows

$r_\mathcal{L}(X) = \begin{cases}r_M(X) & \textnormal{if each circuit of } M|X \textnormal{ is in } \mathcal{L} \\ r_M(X) + 1 & {\rm otherwise}.\end{cases}$

For each linear class $\mathcal{L}$ of $M$, $r_\mathcal{L}$ is the rank function of a matroid $L$ such that $L$ is an elementary lift of $M$. What is even more interesting is that for every elementary lift $L$ of $M$, there exists a linear class $\mathcal{L}$ of circuits of $M$ such that the rank function of $L$ is $r_{\mathcal{L}}$.

Higher-rank lifts

I became aware of Zach’s existence when his paper “A new matroid lift construction and an application to group-labeled graphs” hit the arXiv. It gave a method for constructing non-elementary lifts of matroids in a way that generalized Brylawski’s construction. At the time, I was on the prowl for matroid theory tools that could be useful in rigidity theory, and matroid lifts had recently played a key role in a groundbreaking paper of Clinch, Jackson, and Tanigawa called “Abstract 3-Rigidity and Bivariate C12-Splines II: Combinatorial Characterization.” So I was very intrigued by Zach’s construction, which is given in the following theorem.

Theorem (Walsh): Let $M$ be a matroid on ground set $E$ with circuit set $\mathcal{C}$. Let $N$ be a matroid on ground set $\mathcal{C}$ such that if $|C_1 \cup C_2| – r_M(C_1 \cup C_2) = 2$, then each $C \in \mathcal{C}$ such that $C \subset C_1 \cup C_2$ satisfies $C \in {\rm cl}_N(\{C_1 \cup C_2\})$. Then the function $\rho_N: 2^E \rightarrow \mathbb{N}$ defined as follows is the rank function of a lift of $M$ of rank $r_M(E) + r_N(\mathcal{C})$

$\rho_N(X) = r_M(X) + r_N(\{C : C \textnormal{ is a circuit of } M|X\}).$

Zach’s construction generalizes Brylawski’s construction in the sense that when $r_N(\mathcal{C}) = 1$, the rank-zero flat of $N$ is a linear class $\mathcal{L}$ of circuits of $M$, and that $r_\mathcal{L} = \rho_N$. Zach had hoped that this construction would completely generalize Brylawski’s result in that every lift of a matroid $M$ would have rank function of the form $\rho_N$ for some matroid $N$ on the circuit set of $M$. As we will see, this is not the case. However, it does hold for matroid lifts that arise from taking the matroid of a matrix, and a row-submatrix. This gives us a way to test that a given lift is “representable” as a lift, and can even be turned into a test for matroid representability.

Theorem (B-Walsh): Let $A \in \mathbb{F}^{r_1 \times n}$ and $B \in \mathbb{F}^{r_2 \times n}$ such that $A$ is a row-submatrix of $B$. Let $M_A,M_B$ respectively denote the column matroids of $A$ and $B$ and let $\mathcal{C}$ denote the set of circuits of $A$. Then there exists a matroid $N$ on ground set $\mathcal{C}$ such that $\rho_N$ is the rank function of $M_B$.

Here is a sketch of the proof. Since $A$ is a row-submatrix of $B$ we may without loss of generality assume that $B$ has the following form for some matrix $C$

$B = \binom{A}{C}$.

For each circuit $C$ of $M_A$ let $x_C$ be a fixed element of $\mathbb{F}^n$ such that $Ax = 0$. Then $Bx_C$ can be written in the following form for some $x_C’ \in \mathbb{F}^{r_2-r_1}$

$Bx_C = \binom{0}{x_C’}$.

This enables us to define a matroid $N$ on the circuit set $\mathcal{C}$ of $A$ by asserting that $\{C_1,\dots,C_k\}$ is independent in $N$ if and only if the set of vectors $\{x_{C_1}’,\dots,x_{C_k}’\}$ is linearly independent in $\mathbb{F}^{r_2-r_1}$. Then, the rank function of $M_B$ is $\rho_N$.

Non-representable higher rank lifts

The proof of the above theorem heavily depends on access to a linear representation of the higher rank matroid, that has a row-submatrix representing the lower rank matroid. This dependence on a representation cannot be avoided, at least not in full generality. In particular, we can obtain a counterexample from the Vámos matroid.

Given any matroid $M$ on ground set $E$ and an independent subset $I \subset E$, $M\setminus I$ is always a lift of $M / I$ and the difference in ranks is $|I|$ and all lifts can be obtained in this way. Let $V_8$ be the matroid with ground set $\{1,2,\dots,8\}$ of rank $4$, such that every subset of size four or smaller is independent, with the exceptions of the following four-element subsets

$1234,3456,5678,1278,3478.$

This is known as the Vámos matroid, which is known to not be representable over any field. Then $V_8 \setminus \{7,8\}$ is a lift of $V_8 / \{7,8\}$, and as Zach and I show in our paper, there is no matroid $N$ of rank $2$ on the circuit set of $V_8 / \{7,8\}$ such that the rank function of $V_8 \setminus \{7,8\}$ is $\rho_N$. We generalize this construction to construct a two-parameter infinite antichain (in the minor partial order) of sparse paving matroids that are not representable over any field, that satisfy Ingleton’s inequality.

Concluding remarks

As with many papers I’ve written, the bulk of the time I spent “writing” this paper was spent chasing something I wound up being unable to prove. In particular, I was hoping to generalize the theorem about representable lifts of matroids being obtainable via Zach’s construction to the family of algebraic matroids. Alas, not only was I unable to prove that, I’m not so sure if it’s true anymore. But I really don’t have any reason to make a conjecture either way.

Direct sum of q-matroids

Posted on August 3, 2026 by Relinde Jurrius

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

Posted on July 21, 2026 by Guest Contributor

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:

explain the statement of the theorem; and
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 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

Posted on July 6, 2026 by Jorn van der Pol

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.

The Matroid Union

A blog for and by the matroid community