In 1971, Aubrey Ingleton [1] showed that the rank function of representable matroids satisfies additional linear inequalities that do not follow from the usual rank axioms. The inequality he observed now bears his name. It states that, for all sets $A,B,C,D$ in a representable matroid $M$,

\begin{align*}

&r(A\cup B) + r(A \cup C) + r(A \cup D) + r(B \cup C) + r(B \cup D) \\

\ge & \ r(A) + r(B) + r(A \cup B \cup C) + r(A \cup B \cup D) + r(C \cup D)

\end{align*}

As difficult to typeset as it may be, this is an intriguing fact. The problem of characterizing which matroids are representable is difficult; various authors [2][3] have considered whether it is possible to extend one of the usual axiom systems in a natural way to capture precisely the representable matroids. Ingleton’s inequality suggests the kind of extra axiom that might need to be added to the rank axioms if thinking along these lines. Perhaps more importantly, the Ingleton inequality gives a concise way to certify that a matroid is not representable; if $(A,B,C,D)$ is a $4$-tuple in a matroid $M$ violating the inequality, then $M$ is non-representable. This has been useful to many authors that wish to construct non-representable matroids, in particular in a result of Mayhew, Newman, Welsh and Whittle [4] that constructs a very rich family of excluded minors for the real-representable matroid, all violating Ingleton’s inequality.

Finally, the Ingleton inequality is closely related to everyone’s favourite non-representable matroid, the Vámos matroid $V_8$. This rank-$4$ matroid has eight elements and ground set $E$ that is the disjoint union of four two-element sets $P_1,P_2,P_3,P_4$, in which the dependent four-element sets are precisely the pairs $P_i \cup P_j$ with $i < j$ and $(i,j) \ne (3,4)$. The tuple $(A,B,C,D) = (P_1,P_2,P_3,P_4)$ violates the inequality in this case: the left-hand and right-hand sides are $15$ and $16$ respectively. On the other hand, satisfying Ingleton’s inequality for all choices of $A,B,C,D$ is not enough to imply representability; for example, the non-Desargues matroid satisfies Ingleton’s inequality but is still non-representable, as is the direct sum of the Fano and non-Fano matroids.

**Counting**

This post is about a recent result [7] I’ve obtained with Jorn van der Pol that answers what we consider to be a natural question: how `close’ is the class of matroids that satisfy Ingleton’s inequality to the class of representable matroids? For convenience, we will call a matroid *Ingleton* if Ingleton’s inequality is satisfied for all choices of $(A,B,C,D)$. It might not be immediatley clear what the answer should be. Some (weak) positive evidence is the fact that the Ingleton matroids are closed under minors and duality (see [5], Lemmas 3.9 and 4.5) and that $V_8$ is non-Ingleton. However, I think the natural educated guess goes in the other direction; Ingleton’s inequality seems too coarse to capture, even approximately, a notion as intricate as linear representability In fact, Mayhew, Newman and Whittle [3] have in fact shown that it is impossible to define the class of representable matroids by adding *any* finite list of rank inequalities to the usual rank axioms, let alone a single inequality.

Our result confirms this suspicion.

**Theorem 1:** For all sufficiently large $n$, the number of Ingleton matroids with ground set $\{1,\dotsc,n\}$ is at least $2^{\frac{1.94 \log n}{n^2}\binom{n}{n/2}}$.

To view this result in context, the lower bound should be compared to the counts of representable matroids and of all matroids. On a fixed ground set $[n] = \{1,\dotsc, n\}$ The number of representable matroids is at most $2^{n^3/4}$ for all $n \ge 12$ [6], and the number of matroids is at least $2^{\tfrac{1}{n}\binom{n}{n/2}}$. (Both these upper and lower bounds are in fact correct counts up to a constant factor in the exponent). This first expression is singly exponential in $n$, having the form $2^{\mathrm{poly}(n)}$, while the second is doubly exponential, having the form $2^{2^n/\mathrm{poly}(n)}$. The lower bound in our theorem is of the second type, showing that the Ingleton matroids asymptoticlaly dwarf the representable matroids in number. In other words, knowing that a matroid is Ingleton tells you essentially nothing about whether it is representable. (In fact, our techniques show that the number of *rank-$4$* Ingleton matroids on $[n]$ is asymptotically larger than the class of all representable matroids on $[n]$, which seems surprising.) The ideas in the proof of the above theorem are simple and we obtain a nice excluded minor result on the way; I briefly sketch them below. For the full proof, see https://arxiv.org/abs/1710.01924 .

**Ingleton Sparse Paving Matroids**

Our proof actually constructs a large number of Ingleton matroids of a specific sort: *sparse paving*. These matroids, which have come up in previous matroidunion posts, play a very special role in the landscape of all matroids; they are matroids that, while having a somewhat trivial structure, are conjectured to comprise almost all matroids. For the definition, it is easier to talk about the nonbases of a matroid than its bases. Let $\binom{[n]}{r}$ denote the collection of $r$-element subsets of $[n]$. Given a rank-$r$ matroid on $[n]$, call a dependent set in $\binom{[n]}{r}$ a *nonbasis *of $M$.

A rank-$r$ matroid is *sparse paving* if for any two nonbases $U_1,U_2$ of $M$, we have $|U_1 \cap U_2| < r-1$. Equivalently, no two nonbases of $M$ differ by a single exchange, or every dependent set of $M$ is a circuit-hyperplane of $M$. In fact, this condition itself implies that the matroid axioms are satisfied; given any collection $\mathcal{K}$ of $r$-element subsets of $[n]$, if no two sets in $K$ intersect in exactly $r-1$ elements, then $\mathcal{K}$ is the set of nonbases of a sparse paving matroid on $[n]$. Thus, an easy way to guarantee that a set $\mathcal{K}$ is actually the set of nonbases of a matroid is to prove that no two of its members intersect in $r-1$ elements.

Our key lemma gives a simpler way to understand Ingleton’s inequality for sparse paving matroids. In general, it is very hard to mentally juggle the $A$’s, $B$’s, $C$’s and $D$’s while working with the inequality, but for sparse paving matroids, things are much simpler.

**Lemma 1:** If $M$ is a rank-$r$ sparse paving matroid, then $M$ is Ingleton if and only if there do not exist pairwise disjoint sets $P_1,P_2,P_3,P_4,K$ where $|P_1| = |P_2| = |P_3| = |P_4| = 2$ and $|K|=r-4$, such that the five $r$-element sets $K \cup P_i \cup P_j: i < j, (i,j) \ne (3,4)$ are nonbases, while the set $K \cup P_3 \cup B_4$ is a basis.

This statement may look technical, but it should also look familiar. If $P_1,P_2,P_3,P_4,K$ are sets as above, then the minor $N = (M / K)|(P_1\cup P_2 \cup P_3 \cup P_4)$ is an eight-element sparse paving matroid having a partition $(P_1,P_2,P_3,P_4)$ into two-element sets, where precisely five of the six sets $P_i \cup P_j$ are nonbases, and the last is a basis. This is a structure very similar to that of the Vámos matroid. Call an eight-element, rank-$4$ matroid $N$ having such a property* Vámos-like*. (Such an $N$ need not be precisely the Vámos matroid, as there may be four-element sets of other forms that are also nonbases of $N$). In *any* Vámos-like matroid, $(A,B,C,D) = (P_1,P_2,P_3,P_4)$ will violate Ingleton’s inequality. We can restate Lemma 1 as follows.

**Lemma 1 (simplified):** If $M$ is a sparse paving matroid, then $M$ is Ingleton if and only if $M$ has no Vámos-like minor.

There are evidently only finitely many Vámos-like matroids, since they have eight elements; in fact, thanks to Dillon Mayhew and Gordon Royle’s excellent computational work [8], we know all $39$ of them; as well as the Vámos matroid itself, they include the matroid AG$(3,2)^-$ obtained by relaxing a circuit-hyperplane of the rank-$4$ binary affine geometry. It is easy to show that the sparse paving matroids are themselves a minor-closed class with excluded minors $U_{1,1} \oplus U_{0,2}$ and $U_{0,1} \oplus U_{2,2}$. Combined with Lemma 1, this gives us a nice excluded minor theorem:

**Theorem 2:** There are precisely $41$ excluded minors for the class of Ingleton sparse paving matroids: the $39$ Vámos-like matroids, as well as $U_{1,1} \oplus U_{0,2}$ and $U_{0,1} \oplus U_{2,2}$.

**The Proof**

Armed with Lemma 1, we can now take a crack at proving our main theorem. For simplicity, we will prove a slightly weaker result, with a worse constant of $0.2$ and no logarithmic factor in the exponent. The stronger result is obtained by doing some counting tricks a bit more carefully. The good news is that the proof of the weaker theorem is short enough to fit completely in this post.

**Theorem 3:** For all sufficiently large $n$, there are at least $2^{0.2n^{-2}\binom{n}{n/2}}$ Ingleton sparse paving matroids with ground set $[n]$

**Proof**

Let $n$ be large and let $r = \left\lfloor n/2 \right\rfloor$ and $N = \binom{n}{r}$. Let $c = 0.4$. We will take a uniformly random subset $\mathcal{X}$ of $\left\lfloor c n^{-2}\binom{n}{r}\right\rfloor$ of the $\binom{n}{r}$ sets in $\binom{[n]}{r}$. We then hope that $\mathcal{X}$ is the set of nonbases of an Ingleton sparse paving matroid. If it is not, we remove some sets in $\mathcal{X}$ so that it is.

We consider two possibilities that, together, encompass both ways $\mathcal{X}$ can fail to be the set of nonbases of a sparse paving matroid. They are

- $\mathcal{X}$ is not the set of nonbases of a sparse paving matroid. (That is, there are sets $U_1,U_2 \in \mathcal{X}$ whose intersection has size $r-1$.)
- $\mathcal{X}$ is the set of nonbases of a sparse paving matroid, but this matroid fails to be Ingleton. (That is, there are pairwise disjoint subsets $P_1,P_2,P_3,P_4,K$ of $[n]$ where $|P_1| = |P_2| = |P_3| = |P_4| = 2$ and $|K|=r-4$, such that at least five of the six $r$-element sets $K \cup P_i \cup P_j: i < j, (i,j)$ are nonbases.)

Let $a(\mathcal{X}),b(\mathcal{X})$ denote the number of times each of these types of failure occurs. The condition in (2) is slightly coarser than required, for reasons we will see in a minute. So $a(X)$ is the number of pairs $(U_1,U_2)$ with $|U_1 \cap U_2| = r-1$ and $U_1,U_2 \in X$, and $b(X)$ is the number of $5$-tuples $(P_1,P_2,P_3,P_4,K)$ satisfying the condition in (2).

**Claim:** If $n$ is large, then $\mathbf{E}(a(\mathcal{X}) + 2b(\mathcal{X})) < \tfrac{1}{2}|\mathcal{X}|$.

**Proof of claim:** The number of pairs $U_1,U_2$ intersecting in $r-1$ elements is $\binom{n}{r}r(n-r) < n^2\binom{n}{r}$. For each such pair, the probability that $U_1,U_2 \in X$ is at most $(|\mathcal{X}|/\binom{n}{r})^2 \le c^2n^{-4}$. Therefore \[\mathbf{E}(a(\mathcal{X})) \le c^2n^{-2}\binom{n}{r} = (1+o(1))c|\mathcal{X}|.\]

The number of $5$-tuples $(K,P_1,P_2,P_3,P_4)$ pf disjoint sets where $|K| = r-4$ and $|P_i| = 2$ is at most $\binom{n}{2}^4\binom{n-8}{r-4} < \tfrac{n^8}{16}\binom{n}{r}$. The probability that, for some such tuple, at least five of the sets $K \cup P_i \cup P_j$ are in $X$ is at most $6(|\mathcal{X}|/\binom{n}{r})^5 \le 6c^5n^{-10}$. Therefore

\[\mathbf{E}(b(\mathcal{X})) \le \frac{n^8}{16}\binom{n}{r}\cdot 6c^5 n^{-10} = (1+o(1))\tfrac{3}{8}c^4|\mathcal{X}|.\]

By linearity of expectation, the claim now holds since $c = 0.4$ gives $c + \tfrac{3}{4}c^4 < \tfrac{1}{2}$.

Now, let $\mathcal{X}_0$ be a set of size $\left\lfloor c n^{-2}\binom{n}{r}\right\rfloor$ for which $a(\mathcal{X}) + 2b(\mathcal{X}) < \tfrac{1}{2}|\mathcal{X}_0|$. We now remove from $\mathcal{X}_0$ one of the two sets $U_1,U_2$ for each pair contributing to $a(\mathcal{X}_0)$, and two of the sets $K \cup P_i \cup P_j$ for each tuple $(K,P_1,P_2,P_3,P_4)$ contributing to $b(\mathcal{X}_0)$. This leaves a subset $\mathcal{X}_0’$ of $\mathcal{X}_0$ of size $\tfrac{1}{2}|\mathcal{X}_0| \approx \tfrac{1}{2}cn^{-2}\binom{n}{r} = 0.2n^{-2}\binom{n}{r}$. By construction and Lemma 1, $\mathcal{X}_0’$ is the set of nonbases of a rank-$r$ Ingleton sparse paving matroid on $[n]$. However (and this is where condition (2) needed to be strengthened slightly), so are all subsets of $\mathcal{X}_0’$. This puts the size of $\mathcal{X}_0’$ in the exponent; there are therefore at least $2^{0.2n^{-2}\binom{n}{n/2}}$ Ingleton sparse paving matroids on $[n]$, as required.

**References**

- A.W. Ingleton. Representation of matroids. In D.J.A Welsh, editor,
*Combinatorial mathematics and its applications (Proceedings of a conference held at the Mathematical Institute, Oxford, from 7-10 July, 1969).*Academic Press, 1971. - P. Vámos. The missing axiom of matroid theory is lost forever.
*J. London Math. Soc. 18 (1978), 403-408* - D. Mayhew, M. Newman and G. Whittle. Yes, the “missing axiom” of matroid theory is lost forever,
*arXiv:1412.8399* - D. Mayhew, M. Newman and G. Whittle. On excluded minors for real-representability.
*J. Combin. Theory Ser. B 66 (2009), 685-689.* - A. Cameron. Kinser inequalities and related matroids.
*Master’s Thesis, Victoria University of Wellington.*Also available at*arXiv:1401.0500* - P. Nelson. Almost all matroids are non-representable.
*arXiv:1605.04288* - P. Nelson and J. van der Pol. Doubly exponentially many Ingleton matroids.
*arXiv:1710.01924* - D. Mayhew and G.F. Royle. Matroids with nine elements.
*J. Combin. Theory Ser. B 98 (2008), 882-890.*

Great blog post, Peter!

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.)

Thanks for the correction.

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.