The combinatorial derived matroid

This post is co-authored by Ragnar Freij-Hollanti.

As has been mentioned on this blog before, an idée fixe of the late Henry Crapo was to naturally define a matroid $\delta M$ whose ground set is the collection $\mathcal{C}(M)$ of circuits (or the collection $\mathcal{C}(M^*)$ of cocircuits) of an underlying matroid $M$. Similar questions have also been asked by Gian-Carlo Rota. We will call such a construction a derived matroid of $M$, partly in analogy with the construction of derived codes in coding theory. Throughout this blogpost, $n$ will denote the size and $k$ will denote the rank of $M$.

The derived matroid could be seen as a combinatorialization of the notion of syzygies and resolutions in commutative algebra, regarding circuits as combinatorial “relations”, which generate a space in which we can again study “relations”, and so on. From this viewpoint, it is natural to try to define a matroid structure on the set of circuits. On the other hand, it seems like a desirable feature that the nullity of $M$ would give, or at least bound, the number of “independent” circuits (i.e. relations) in $M$, or in other words that the number of independent cocircuits would equal (or be bounded by) the nullity of $M^*$, which is the rank of $M$. In this light studying the sequence of repeated derivations of a matroids $M$ would be more natural if $\delta M$ had as its ground set the cocircuits of $M$. Since the cocircuits are in natural bijection with the hyperplanes of $M$, a derived matroid with ground set $\mathcal{C}(M^*)$ would model the hyperplane arrangement described by the point set in the representable case. There are thus good arguments both in favour of grounding the derived matroid on $\mathcal{C}(M)$ and on $\mathcal{C}(M^*)$. In our paper, we choose the first alternative, but the last word has hardly been said in this debate, and of course, one definition can be obtained from the other by dualizing.

Let us first consider the case for represented matroids, and let $G$ be a $k\times n$ matrix of full rank representing $M$ over a field $\mathbb{F}$. Then, every circuit $C\in\mathcal{C}(M)$ is the inclusion-minimal support of a vector $q_C\in\mathbb{F}^n$ such that $G q_C=\mathbf{0}$, and this vector is unique up to scalar multiple. The collection of vectors $\{q_C\}_{C\in\mathcal{C}(M)}$ represent some matroid with ground set $\mathcal{C}(M)$, and we will denote this matroid by $\delta_{OW}(G)$, where the subscript refers to the authors Oxley and Wang, who studied this construction [OW19].

It is not difficult to see that $\delta_{OW}(G)$ has rank $n-k$, because the circuit vectors together generate the null space of the matrix $G$. While this is certainly a desirable feature, it is an unfortunate fact that the construction $\delta_{OW}$ depends on $G$ rather than only on $M$. The geometrically easiest illustration of this feature may be when $M$ is the uniform matroid $U_{3,6}$, represented by a configuration $P$ of six points in general position in the projective plane. Then, the hyperplane arrangement of the $\binom{6}{2}=15$ lines (through pairs of $P$) can be of two different kinds, as shown in the picture below (not all lines are drawn):

Two possibilities for the derived matroid of $U_{3,6}$

Indeed, in addition the five lines intersecting in each point of $P$, it will generically be the case that all other intersections of lines are distinct. However, it is also possible that three of the lines, that do not share a point in $P$, might meet in a point in the projective plane. In this latter case, the cocircuits corresponding to the three lines will be “dependent”; in the former case they will be “independent” in $\delta_{OW}(Q^*)$.

It is fair to say that the dependence between the three lines in the previous example is “accidental”, in that it is not forced by the matroid structure of $U_{3,6}$ itself. If we are to define “the” derived matroid of the matroid $U_{3,6}$ combinatorially, we in particular want the complements of these lines to be independent. Following this line of thought, we consider independence to be the generic state of things, while dependence is forced by constraints coming from the underlying matroid. Our construction of $\delta M$ is thus twofold. First we identify which collections of circuits “have to” be dependent in any “reasonably defined” derived matroid of $M$. After this, we add just enough other dependent sets, to guarantee that $\delta M$ is indeed a matroid, and we do it in such a way as to not introduce unnecessary dependent sets, and in particular not in low rank.

Let us illustrate this procedure with the infamously non-representable Vámos matroid $M$. All sets of three or fewer elements are independent and among the sets of four elements only the five depicted as grey rectangles in the figure are circuits. All sets of five or more elements are dependent.

The Vámos matroid

The ground set of the derived matroid $\delta M$ has 41 elements, the circuits of $M$. If we consider the three planes $abcd$, $adef$ and $bcef$, we intuitively want this to be a dependent set in $\delta M$, because it “looks like” these planes make a “circuit”. To make this mathematically more precise, we can say that this set of circuits is dependent because its size (3) is larger than the nullity in $M$ of the union of all its elements (these six elements have nullity 2 in $M$). That is to say, we want that a set of circuits is dependent in $M$ if it belongs to the family
$$\mathcal{A}_0:=\{A\subseteq\mathcal{C}:|A|>n(\cup_{C\in A}C)\}.$$
It is readily checked that in the previous example of $U_{6,3}$, the three discussed lines are not in this family, because $3\not>3$.

The above family $\mathcal{A}_0$ gives us a precise description of which collections of circuits “have to” be dependent in $\delta M$. Note that this is a combinatorial description: it does not depend on a representation of the matroid. What we want to do now, is make sure that we find all dependent sets of the derived matroid. To see how to get to this, consider the axioms for the collection $\mathcal{D}$ of dependent sets of a matroid.

  1. $\emptyset\notin\mathcal{D}$;
  2. if $D\in\mathcal{D}$ and $D\subseteq D’$ then $D’\in\mathcal{D}$;
  3. if $D_1,D_2\in\mathcal{D}$ and $D_1\cap D_2\notin\mathcal{D}$, then $(D_1\cup D_2)\setminus\{e\}\in\mathcal{D}$ for all $e\in D_1\cap D_2$.

Our collection $\mathcal{A}_0$ satisfies the first two axioms, but in general, it does not satisfy the third. So we want to add extra elements in order to get a collection that does satisfy (D3). There are two ways to do this, as we see from a logical rewriting of the axiom:

  1. If $D_1, D_2\in\mathcal{D}$, then $D_1\cap D_2\in\mathcal{D}$ or $(D_1\cup D_2)\setminus\{e\}\in\mathcal{D}$ for all $e\in D_1\cap D_2$.

This means that for any $A_1,A_2\in\mathcal{A}_0$, we can either decide that $A_1\cap A_2$ is dependent, or that $(A_1\cup A_2)\setminus{C}$ is dependent for all $C\in A_1\cap A_2$ (or both). If we decide on the first, we will get a lot of small dependent sets, and often this leads to a derived matroid of rank 0. We therefore decide to define the following operations.

Definition. Let $\mathcal{C}$ be the set of circuits of some matroid, and let $\mathcal{A}\subseteq \mathcal{C}$. Then we define the collections
$$\epsilon(\mathcal{A})=\mathcal{A}\cup\left\{(A_1 \cup A_2) \setminus {C} : A_1, A_2\in \mathcal{A}, A_1\cap A_2\not\in\mathcal{A}, C\in A_1\cap A_2\right\}$$ and $${\uparrow}\mathcal{A}=\{A\subseteq \mathcal{C}: \exists A’\in\mathcal{A}: A’\subseteq A\}.$$

Observe that, by definition, $\mathcal{A}\subseteq {{\uparrow} \mathcal{A}}$ and $\mathcal{A}\subseteq \epsilon(\mathcal{A})$ for every $\mathcal{A}\subseteq2^\mathcal{C}$. The operations ${\uparrow}$ and $\epsilon$ are designed to guarantee properties (D2) and (D3) in the matroid axioms.

Definition. Let $M$ be a matroid, and $\mathcal{C}=\mathcal{C}(M)$ its collection of circuits. Define the collection
$$ \mathcal{A}_0:=\{A \subseteq \mathcal{C}: |A|> n(\cup_{C\in A} C)\}. $$
Inductively, we let $\mathcal{A}_{i+1}={\uparrow}\epsilon(\mathcal{A}_i)$ for $i\geq 1$, and
$$\mathcal{A}=\bigcup_{i\geq 0} \mathcal{A}_i.$$

Note that the sequence $\mathcal{A}_i$ is both increasing and contained in the finite set $2^\mathcal{C}$. Hence, we have that $\mathcal{A}_0\subseteq\mathcal{A}$ and $\mathcal{A}=\mathcal{A}_n$ for some $n\geq 0$.

Definition (combinatorial derived matroid). Let $M$ be a matroid with circuits $\mathcal{C}$. Then the combinatorial derived matroid $\delta M$ is a matroid with ground set $\mathcal{C}(M)$ and dependent sets $\mathcal{A}$.

By carefully checking the dependent axioms, we can prove that this definition gives indeed a matroid. In fact, we can prove two more ways to construct this derived matroid, by means of its circuits: we refer the interested reader to the full paper [FJK23].

Theorem. For any matroid $M=(E,\mathcal{C})$ the collection $\mathcal{A}$ is the collection of dependent sets of some matroid with ground set $\mathcal{C}$.

This is good news! We now have a definition of a derived matroid that is purely combinatorial: it does not depend on a representation, and hence it exists for any matroid. As far as we know, this is the first definition of this kind. We also managed to show that there is more good news: this definition behaves well with connectedness. That is, the derived matroid of a direct sum is the direct sum of the derived matroids of the connected components.

Unfortunately, it is not all good news. It is not difficult to show that the rank of $\delta M$ is bounded by $n-k$, as desired, but equality does not always hold. Computer calculations by Knutsen [Knu23] show that the rank of the derived Vámos matroid is 3, and not 4. However, this news is not as bad as it sounds, since the Vámos matroid was shown to not have an adjoint by Cheung [Che74]. The construction of the adjoint of a matroid is, in some sense, the dual of that of the derived matroid: it has the set of cocircuits of $M$ as its ground set.

Many questions are still open regarding this construction of $\delta M$. Most prominently: how does this construction relate to earlier constructions of the derived matroid? In particularly, that of Oxley and Wang, but also to the adjoint of a matroid. For more questions, we refer to the full paper [FJK23]. We hope this post inspired you to think more about derived matroids!


[Che74] Cheung, A.L.C. (1974). Adjoints of a geometry. Canadian Mathematical Bulletin, 17(3), 363-365.
[FJK23] Freij-Hollanti, R. & Jurrius, R.P.M.J. & Kuznetsova, O. (2023). Combinatorial Derived Matroids. Electronic Journal of Combinatorics, 30(2), P2.8.
[Knu23] Knutsen, T.D. (2023). Codes, matroids and derived matroids. Master thesis, UiT Arctic University of Norway.
[OW19] Oxley, J & Wang, S. (2019). Dependencies among dependencies in matroids. Electronic Journal of Combinatorics, 26(3), P3.46.

The $q$-analogue of a matroid

The goal of this post is to give you some intuition and background about the q-analogue of a matroid. To keep this a gentle introduction, I’ll leave out most technical details. If you are interested, you can read more in [JP2018] and [BCJ2017].

Before discussing what a q-analogue is, let me start by giving the point of view of matroids that will help us with this. As an example, consider the following matroid.

We will describe the rank function of this matroid via a colouring of the Boolean lattice. The matroid is on four points, so we look at the lattice of all subsets of the set of four elements. We colour all covers in this lattice: we make them red if the rank goes up and green if it stays the same. So our matroid becomes:

For every element in the lattice, we can find its rank by counting the number of red lines in a chain from the bottom to this element. For example, $\{a,c\}$ has rank 2 and $\{b,c\}$ has rank 1. The whole matroid has rank 3: no matter what path we take from $\emptyset$ to $\{a,b,c,d\}$, we always use 3 red lines.

In this way we can represent any matroid on $n$ points as a colouring of the Boolean lattice of height $n$. We can ask ourselves the opposite question as well: suppose I want to colour a Boolean lattice, how do I do that in a way that gives a matroid? We call such a colouring matroidial. It turns out it is enough to look at the intervals of height two, that I will call diamonds. A colouring is matroidial if and only if every diamond is of one of the following types:

(The mixed diamond can of course be mirrored, this depends on how you draw the Boolean lattice.) Note that both paths from the bottom to the top of the diamond have the same number of red edges: we need this for the rank function to be well defined. Also, we cannot have a diamond where the bottom edges are green and the top are red, because this would violate the semimodularity of the rank function. It is equivalent to saying that the sum of two loops can never have rank 1.

If you want to familiarise yourself more with this way of looking at matroids, you can ask yourself how to recognise certain matroid properties in this picture. For example, an independent set is an element of the Boolean lattice such that the whole interval between the bottom (the empty set) and this element only has red lines. The closure of a subset is found by going up from the element via green lines only as far as possible. Can you see how to characterise flats, bases, hyperplanes, and circuits in this picture? Also, some slightly more difficult questions: how would duality work? And taking minors? (You can find the answers to all these questions in the first pages of [BCJ2017].)

Now that we have this image of a matroid in mind, let’s talk about the q-analogue. The concept of a q-analogue in combinatorics started to get attention in the 1980’s. The idea is to ‘generalise’ from a finite set to a finite dimensional vectorspace. This space was originally over a finite field, because then we can count things. I will keep that point of view here, but note that as long as the dimension is finite, the underlying field can be infinite.

Here is a first example of a q-analogue. We know that we can count the number of subsets of size $k$ of a set of size $n$ with the binomial ${n\choose k}$. The q-analogue of this is the Gaussian binomial, that counts the number of $k$-dimensional subspaces of an $n$-dimensional space over $\mathbb{F}_q$. It is given by

\[ \genfrac{[}{]}{0pt}{}{n}{k}_q=\frac{(q^n-1)(q^{n-1}-1)\cdots(q^{n-k+1}-1)}{(q^k-1)(q^{k-1}-1)\cdots(q-1)}. \]

In essence, what you do by taking a q-analogue is going from the Boolean lattice to the subspace lattice, the lattice of a vectorspace and all its subspaces. If you want to go back from the q-analogue to the set-case, you let $q\to1$. One could say that set theory is just geometry over the field of one element. If that sounds intriguing to you, I recommend the very accessible paper [Coh2004] on this topic.

During the last decade, q-analogues became in fashion again because of their application to network coding. Together with Ruud Pellikaan, who had been my PhD supervisor, I started thinking about the q-analogue of a matroid around 2013. (Don’t worry, I’ll tell you what that is in a minute.) Our motivation came from coding theory: in my thesis, I had studied the relation between the weight enumerator of a code and the Tutte polynomial of a matroid, and we wanted to do the same thing for the q-analogue of codes, that are used in network coding. To achieve this we needed the q-analogue of a matroid and its Tutte polynomial. That looked easy enough at first, but the Tutte polynomial for q-matroids is still not well-defined, as far as I know.

A few years later I attended a matroid workshop in Eindhoven, were I was going to speak about q-matroids. Already in the first coffee break I was introduced to Henry Crapo, who had read my abstract and showed great interested in my work. At that time I was a postdoc and you can imagine I was a bit nervous about such a big name showing interest in what I was doing. It turned out that Henry developed already the idea of a q-matroid in his PhD thesis in 1964, but it became somewhat forgotten. His point of view was quite different from mine and did not have anything to do with vectorspaces at all. But our two approaches turned out to complement each other very well and this was the start of a great collaboration. Unfortunately, we did not get the time to finish some of our big questions.

Now, back to the mathematics. The q-analogue of a matroid, that we call a q-matroid, is a colouring of the subspace lattice. Here is an example:

And here is another one:

In both cases, the underlying vectorspace is $\mathbb{F}_2^3$ and the elements of the lattice are give by their Plücker coordinates. We see that the first q-matroid has rank 1, where the second one has rank 2.

Just as with the colouring of the Boolean lattice, we can say when we find such a colouring to be matroidial. We look again at diamonds, the intervals of heigh two. Now there are not 2 elements in the middle of a diamond, but $q+1$ (the number of points on a projective line). The following diamonds are allowed:

The interesting case here is the mixed diamond. It can have only one green line at the bottom. If there were two, the whole diamond has to be green: the sum of two loops can not have rank 1.

We can define a q-matroid also by its rank function. Then it is a pair $(E,r)$ where $E$ is a finite dimensional vectorspace and $r$ an integer valued function defined on the subspaces of $E$. For all subspaces $A$ and $B$ of $E$ the rank function satisfies:

  1. $0\leq r(A)\leq\dim(A)$.
  2. If $A\subseteq B$ then $r(A)\leq r(B)$.
  3. $r(A+B)+r(A\cap B)\leq r(A)+r(B)$.

These axioms are a direct q-analogue of the rank axioms for classical matroids. In the first axiom, cardinality of a set is replaced by dimension of a subspace. Both are the height of the element in the underlying lattice. In the last axiom the union of two subsets is replaced by the direct sum of subspaces: both are the join of elements in the underlying lattice.

Inspired by the above relatively straightforward q-analogues of two cryptomorphic definitions of a matroid, it is a logical thing to ask if we can do this for more cryptomorphisms. Define an independent space of a q-matroid as a subspace whose rank is equal to its dimension. If we denote the family of all independent spaces by $\mathcal{I}$, we could write down the following axioms for independent spaces:

  1. $\mathcal{I}\neq\emptyset$.
  2. If $J\in\mathcal{I}$ and $I\subseteq J$, then $I\in\mathcal{I}$.
  3. If $I,J\in\mathcal{I}$ with $\dim I<\dim J$, then there is some 1-dimensional subspace $x\subseteq J$, $x\not\subseteq I$ with $I+x\in\mathcal{I}$.

But here is an interesting thing: these properties follow from the rank axioms, but they do not completely define a q-matroid! (This object is studied in the work of Terwilliger [Ter1996].) You can check that the following diamond is not excluded by the three independence axioms:

So, we need an extra axiom for independent spaces. This one works:

  1. Let $A,B\subseteq E$ and let $I,J$ be maximal independent subspaces of $A$ and $B$, respectively. Then there is a maximal independent subspace of $A+B$ that is contained in $I+J$.

(If you find this axiom not very appealing, I agree. But it is exactly what goes wrong if you try to prove semimodularity from the independence axioms. It might be that there is a ‘prettier’ way to solve the need for a fourth axiom — maybe by changing the third axiom? I’m open to suggestions!)

Note that the similar statement for the classical case follows from the three standard independence axioms. This is a good example of something that makes q-analogues difficult: if you add two subspaces $A$ and $B$ together, you get not only all 1-dimensional subspaces that were already in $A$ or $B$, but a whole lot more. It can be hard to control what is going on with these extra 1-dimensional spaces.

Another difficulty in q-analogues is the notion of a complement. For sets, if $A$ is a subset of $E$, then the complement of $A$ in $E$ is just the set of all elements that are in $E$ but not in $A$. For spaces, this is a lot more difficult. If $A$ is a subspace of $E$, we could say that we want its complement to contain all 1-dimensional subspaces of $E$ that are not in $A$. But these don’t form a subspace. We could also take the orthogonal complement $A^\perp$ of $A$ in $B$: this is well-defined and it has the dimension we want. But it can be that $A\cap A^\perp$ is nontrivial (remember we are working over finite fields) and this gives problems. Another option is to take a subspace $A^c$ such that $A\oplus A^c=E$, but then $A^c$ is not unique. All these options might be the type of complement you need in a proof of a q-analogue, but it is not always clear which one.

As an illustration, think about the proof that the complements of bases of a matroid form the family of bases of the dual matroid. This involves proving that the (strong) base exchange axiom holds for all complements of bases:

  • If $B_1$ and $B_2$ are bases and $x\in B_1-B_2$, then there is and element $y\in B_2-B_1$ such that $B_1-x\cup y$ and $B_2-y\cup x$ are both bases.

For classical matroids, this all works very smoothly: removing element $x$ and adding element $y$ form a basis is like removing $y$ and adding $x$ to its complement. For q-analogues, this proof is an absolute disaster, no matter which definition of complement you try. So in order to define the dual of a q-matroid, we have to come up with something else that does not use bases. Such as: “Take the coloured lattice, put it upside down, and interchange red and green lines.” I’ll leave it to you to convince yourself that this gives again a q-matroid. (Hint: apply this procedure to the diamonds of a matroidial colouring.)

The theory of q-matroids is still in its early stages of development. There are many, many concepts in matroid theory out there waiting for a q-analogue. Like the Tutte polynomial, as already mentioned. But there are more basic questions to answer first. Take for example the fact that a circuit and a cocircuit can never have one element in common. Is it the case in q-matroids that a circuit and a cocircuit can never intersect in a 1-dimensional subspace? Or, how about a q-analogue of a graph that generalises to a q-matroid? Inspiration for open questions can be found in the last section of [JP2018], or you can simply start q-ifying your own favourite matroid result.


[BCJ2017] Bollen, G. & Crapo, H. & Jurrius, R.P.M.J. (2017). The Tutte q-polynomial. Eternal preprint. arXiv

[Coh2004] Cohn, H. (2004). Projective geometry over $\mathbb{F}_1$ and the Gaussian binomial coefficients. American Mathematical Monthly, 111:487–495. pdf

[JP2018] Jurrius, R.P.M.J. & Pellikaan, R. (2018). Defining the q-analogue of a matroid. Electronic Journal of Combinatorics, 25(3), P3.2. doi

[Ter1996] Terwilliger, P. (1996). Quantum matroids. In: Bannai, E. & Munemasa, A., editors, Progress in Algebraic Combinatorics, volume 24 of Advanced Studies in Pure Mathematics. Mathematical Society of Japan, Tokyo, 1996.