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:
- $0\leq r(A)\leq\dim(A)$.
- If $A\subseteq B$ then $r(A)\leq r(B)$.
- $r(A+B)+r(A\cap B)\leq r(A)+r(B)$.
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:
- $\mathcal{I}\neq\emptyset$.
- If $J\in\mathcal{I}$ and $I\subseteq J$, then $I\in\mathcal{I}$.
- 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:
- 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.
References
[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.
