Spreads: Decomposing projective geometries

Happy New Year from all of us at the Matroid Union!

Decomposition of objects into smaller or simpler objects is a central theme in combinatorics. Think about decomposing a graph into its connected components or blocks, or into forests. In today’s post, I discuss decomposing finite projective geometries into projective geometries of smaller rank.

Definition: Let $q$ be prime power, let $n \ge t \ge 1$ be integers and let $M = \text{PG}(n-1,q)$ be the projective geometry over the field with $q$ elements. A $t$-spread of $G$ is a collection of rank-$t$ flats of $M$ that partition the points of $M$—that is, these flats are pairwise disjoint and their union is the set of all points of $M$.

(Geometers often use dimension instead of rank; the dimension of a projective geometry is one less than its rank. Their $t$-spreads are collections of $t$-dimensional subspaces of an $n$-dimensional space. We will however stick to the terminology that is more familiar in the matroid theory setting.)

Every projective geometry has a 1-spread: the flats making up the spread are simply the points of the geometry. It is much more interesting to look into the existence of $t$-spreads when $t \ge 2$.

The Fano plane $F_7 = \text{PG}(2,2)$ does not admit a decomposition into triangles. The quickest argument to see this is based on counting the number of points: any matroid that can be decomposed into triangles necessarily has a number of points that is a multiple of 3, but the Fano plane has seven points. By contrast, the projective geometry $\text{PG}(3,2)$, which has fifteen points, can be partitioned into five triangles (as encoded using different colours in the following representation:

The projective geometry PG(3,2) can be decomposed into five triangles.
The projective geometry $\text{PG}(3,2)$ can be decomposed into five triangles (indicated here using five different colours).

The counting argument above can be generalised. In order for $M = \text{PG}(n-1,q)$ to have any hope of admitting a $t$-spread, the number of points in a projective geometry of rank $t$ should evenly divide the number of points in $M$, or $\frac{q^t-1}{q-1} \bigg| \frac{q^n-1}{q-1}$. It is a nice exercise to show that the latter happens precisely when $n$ is a multiple of $t$. So, if $\text{PG}(n-1,q)$ can be decomposed into rank-$t$ subgeometries, we must have $t|n$.

It turns out that this divisibility criterion is not only necessary—it is sufficient as well. We end this blog post with a slick proof of this fact. The proof can be found in many introductory texts on projective geometry.

Theorem: The projective geometry $M=\text{PG}(n-1,q)$ admits a $t$-spread if and only if $t|n$.

Proof: We already proved the implication from left to right. The implication in the other direction follows from a construction.

Suppose that $n = kt$ for some integer $k$. The points of $M$ can be identified with the 1-dimensional subspaces of the vector space $\mathbb{F}_q^n$. We will show that $\mathbb{F}_q^n$ can be decomposed into $t$-dimensional subpspaces that are pairwise disjoint (except that they all contain the zero-vector). As the 1-dimensional subspaces of $\mathbb{F}_q^t$ can be identified with the points of a $\text{PG}(t-1,q)$, this immediately implies the theorem.

Consider the $k$-dimensional $\mathbb{F}_{q^t}$-vector space $\mathbb{F}_{q^t}^k$. Its 1-dimensional subspaces intersect only in the zero-vector. These 1-dimensional subspaces give us our spread, as each such subspace can alternatively be viewed as a $t$-dimensional $\mathbb{F}_q$-vector space. $\Box$

Triangles, arcs, and ovals

Tuza’s Conjecture

One of my favourite open problems in Tuza’s Conjecture, which relates triangle packings and triangle hitting sets in graphs.

Let $G$ be a graph. A triangle packing is a set of triangles in $G$ no two of which have an edge in common. The maximum size of a triangle packing is called the triangle packing number of $G$, for which we write $\nu(G)$.

A triangle-hitting set is a subset $X \subseteq E(G)$ with the property that removing $X$ from $G$ results in a triangle-free graph. We write $\tau(G)$ for the size of the smallest triangle-hitting set in $G$.

A straightforward argument shows that $\nu(G) \le \tau(G) \le 3\nu(G)$. Tuza conjectured that the upper bound can be improved.

Conjecture (Tuza, 1981): $\tau(G) \le 2\nu(G)$ for all graphs $G$.

In other words, a graph without $t+1$ edge-disjoint triangles can be made triangle-free by removing at most $2t$ edges.

The factor 2 in the conjecture, if true, is best possible: $K_4$ and $K_5$ are examples for which $\tau(G) = 2\nu(G)$.

K_5 contains two edge-disjoint triangles (pink) and can be made triangle-free by removing the four dashes edges.
$K_5$ contains two edge-disjoint triangles (pink) and can be made triangle-free by removing the four dashed edges.

Geometric Tuza’s Conjecture

Tuza’s Conjecture, being formulated in terms in terms of edges and triangles only, can be naturally generalised to matroids. A triangle in a matroid $M$ is a restriction isomorphic to $U_{2,3}$. We define $\nu(M)$ to be the maximum number of disjoint triangles in $M$, and $\tau(M)$ to be the minimum size of a set $X$ such that $M\backslash X$ does not contain any triangles.

Unfortunately, the geometric version of Tuza’s Conjecture fails for matroids in general: For the Fano plane, $F_7 = \mathrm{PG}(2,2)$, we have $\tau(F_7) = 3$ and $\nu(F_7) = 1$.

The Fano plane does not contain two disjoint triangles and can be made triangle-free by removing the three empty elements.
$F_7$ does not contain two disjoint triangles and can be made triangle-free by removing the three empty elements.

Kazu Nomoto and I believe that, at least for binary matroids, the Fano plane is the only obstruction to a geometric version of Tuza’s Conjecture.

Conjecture (Nomoto–Van der Pol, 2021): Let $M$ be a simple binary matroid with no restriction isomorphic to $F_7$. Then $\tau(M) \le 2\nu(M)$.

As graphic matroids do not have restrictions isomorphic to $F_7$, the geometric version of Tuza’s conjecture implies the graphic version. We verified the conjecture for cographic matroids and several other classes of matroids.

Projective planes

Moving beyond binary matroids, one could ask if Tuza’s Conjecture holds for other classes of matroids. Here, we ask if it holds for projective planes. We’ve already seen that the conjecture fails for the Fano plane. It fails spectacularly for all other finite projective planes as well. This result is closely related to classical results in projective geometry.

A finite projective plane is a set system $\Pi = (E,\mathcal{L})$ — the elements of $E$ are called points, and the elements of $\mathcal{L}$ are called lines — with the following properties:

  1. Any pair of distinct points is contained in exactly one line;
  2. Any pair of lines intersects in exactly one point; and
  3. There are four points such that no line contains more than two of them.

For each finite projective plane $\Pi$ there is a number $q \ge 2$, the order of $\Pi$, such that every point is contained in $q+1$ lines, and every line contains $q+1$ points. It follows that a projective plane of order $q$ contains exactly $q^2 + q + 1$ points.

Well-known examples are the Desarguesian planes $\mathrm{PG}(2,q)$, whose points and lines are the 1- and 2-dimensional subspaces of $\mathbb{F}_q^3$, respectively, but there are other examples as well.

Arcs

The lines in $\mathcal{L}$ form the collection of hyperplanes of a rank-3 matroid on $E$, so it makes sense to talk of $\tau(\Pi)$ and $\nu(\Pi)$ for finite projective planes $\Pi$.

An arc in $\Pi$ is a set $A$ of points such that no line intersects $A$ in more than two points. In our terminology, $A$ is a triangle-free restriction of $\Pi$. (Confusingly, the term triangle is also used for an arc of size 3, which is the opposite of our use of the term.)

The following result is well known in projective geometry.

Lemma: Let $\Pi$ be a projective plane of order $q$ and let $A$ be an arc. Then $|A| \le q+2$. If $q$ is odd, then $|A| \le q+1$.

Proof. Let $p \in A$ be any point, and let $\ell_1, \ell_2, \ldots, \ell_{q+1}$ be an enumeration of the lines incident with $p$. Each point in $\Pi$ is on one of those $q+1$ lines, and each line contains at most one point from $A$ other than $p$, so $|A| \le 1 + (q+1)$.

To prove the second bound, suppose for the sake of contradiction that $q$ is odd and $|A| = q+2$. Every line of $\Pi$ intersects $A$ in either $0$ or $2$ points; this is in particular true for all lines through a fixed point $p \not\in A$. It follows that $|A|$ is twice the number of lines incident with $p$ that intersect $A$, so $|A|$ must be even: a contradiction, so $|A| < q+2$. $\Box$

Arcs of size $q+1$ are called ovals, while arcs of size $q+2$ are called hyperovals. Constructions for ovals and hyperovals exist for the planes $\mathrm{PG}(2,q)$; Segre proved that in $\mathrm{PG}(2,q)$ of odd order, all ovals are nondegenerate conic sections.

By the lemma, sets containing more than $q+2$ points must contain a triangle, so $\tau(\Pi)$ is large.

Lemma: Let $\Pi$ be a projective plane of order $q$. If $q$ is even, then $\tau(\Pi) \ge q^2 – 1$. If $q$ is odd, then $\tau(\Pi) \ge q^2$.

Trivially, $\nu(\Pi) \le \frac{1}{3}(q^2+q+1)$. This is sufficient to show that Tuza’s conjecture fails for projective planes of sufficiently large order, as $\tau(\Pi) \sim 3\nu(\Pi)$ as $q \to \infty$.

The trivial upper bound on $\nu$ is correct, as can be shown by a simple construction.

Lemma: Let $\Pi$ be a projective plane of order $q \ge 3$. Then $\Pi$ can be decomposed into disjoint triangles and at most one point. Consequently, $\nu(\Pi) = \lfloor\frac{1}{3}(q^2+q+1)\rfloor$.

Proof. The exact construction depends on the remainder of $q$ modulo 3. Here, we only give the construction for the case $q = 3k+1$; the case $q = 3k+2$ allows a similar construction, while the case $q=3k$ is easier. See the figure following the proof for an illustration in the case $q=4$.

Let $p \in \Pi$ be any point, and let $\ell_0, \ell_1, \ldots, \ell_q$ be an enumeration of the lines incident with $p$. Let $e \in \ell_0$ be a point distinct from $p$, and let $\ell’ = {e, e’_1, \ldots, e’_q}$ and $\ell” = {e, e”_1, \ldots, e”_q}$ be two distinct lines incident with $e$ such that $e’_i, e”_i \in \ell_i$ for $i = 1, 2, \ldots, q$.

Consider the $(q+2)/3$ disjoint triangles

$\{e’_2, e’_3, e’_4\}, \{e’_5, e’_6, e’_7\}, \ldots, \{e’_{q-2}, e’_{q-1}, e’_q\},\quad\text{and}\quad\{e,e”_1, e”_2\}$,

and let $Z$ be the set of all points in those triangles. $Z$ contains exactly one point on each of the lines $\ell_i$, except that it contains two points of $\ell_2$. It follows that each of the sets $\ell_i\setminus (Z \cup {e})$ (for $i \neq 2$) and $\ell_2\setminus Z$ can be partitioned into triangles. $\Box$

A decomposition of PG(2,4) into seven triangles.
A decomposition of $\mathrm{PG}(2,4)$ into seven triangles.

It follows from the above results that

$\frac{\tau(\Pi)}{\nu(\Pi)} \ge \frac{q^2-1}{\frac{1}{3}(q^2+q+1)} \ge \frac{15}{7} > 2$

whenever the order $q$ of $\Pi$ is at least 4. For the (unique) projective geometry $\Pi = \mathrm{PG}(2,3)$ of order $q = 3$, we have $\tau(\Pi) = 9$ and $\nu(\Pi) = 4$, while for the (unique) projective geometry $\Pi = \mathrm{PG}(2,2)$ or order $q=2$ we have $\tau(\Pi)=3$ and $\nu(\Pi) = 1$, so Tuza’s Conjecture fails for all finite projective planes.

Colouring matroid reductions

Covering a graph by forests

Given a loopless graph $G = (V,E)$ with at least one edge, what is the smallest possible number of forests needed to cover all edges? This number is known as the arboricity $a(G)$ of $G$.

For example, consider the following 6-vertex graph, which is obtained from the complete graph $K_5$ by removing one edge, and adding a new pendant edge to one of the vertices of degree 3.

A 6-vertex graph whose edges can be covered by three forests, but not by two.
A 6-vertex graph whose edges can be covered by three forests, but not by two.

As the graph is connected and has six vertices, each forest in $G$ contains at most five edges. As the graph has ten edges in total, we need at least two forests to cover all edges.

But try as we might, we’ll never be able to cover all edges with only two forests. The reason is that similar bounds must hold in every subgraph of $G$. If we remove the unique degree-1 vertex and the edge incident with it, we are left with a connected graph on five vertices and nine edges. As in this subgraph every forest has at most four edges, we need at least three forests to cover the whole graph. I’ll leave it as an exercise to show that the original graph can actually be covered by three forests.

So, each subgraph $G’$ of $G$ (with at least one edge) provides a lower bound for $a(G)$, as $a(G) \ge \left\lceil \frac{|E(G’)|}{|V(G’)|-c(G’)}\right\rceil$ (here, $c(G’)$ denotes the number of components of $G’$).

Nash-Williams showed that we don’t need more forests than suggested by the worst possible bound obtained in this way: $$a(G) = \max_{G’\subseteq G : E(G’) \neq \emptyset} \left\lceil \frac{|E(G’)|}{|V(G’)|-c(G’)} \right\rceil.$$

Colouring matroids by independent sets

Each of the definitions above can be generalised to matroids, in which case the corresponding question becomes: Given a matroid $M$, how many independent sets do we need to cover the entire ground set of $M$?

The minimum number of independent sets that we need is known as the colouring number (or covering number) of the matroid, and we write $\text{col}(M)$. As loops are not contained in any independent set, we will assume that $M$ has no loops. Edmonds showed (PDF) that the following generalisation of Nash-Williams’ formula holds: $$\text{col}(M) = \max_{E’ \subseteq E: r(E’) > 0} \left\lceil\frac{|E’|}{r_M(E’)}\right\rceil.$$

We say that $M$ is $k$-colourable if $\text{col}(M) \le k$. Note that $a(G) = \text{col}(M(G))$ for any graph $G$, so the colouring number is a proper generalisation of arboricity.

(I like to use “colouring number” rather than “covering number”, as the definition is analoguous to that of “chromatic number” in graphs, which, after all, is the smallest number of independent sets of vertices into which the graph can be partitioned.)

Colouring partition matroids

The colouring number of the uniform matroid $U_{r,n}$ is $\lceil n/r\rceil$. It is clear that this is a lower bound, as $U_{r,n}$ has $n$ elements, and each independent set contains at most $r$ elements.

It is also an upper bound, for we can arbitrarily partition $U_{r,n}$ into $\lfloor n/r\rfloor$ independent sets of size $r$ and, if $r$ does not divide $n$, one more independent set of size less than $r$.

It is an exercise to show that $\text{col}(M_1 \oplus M_2) = \max\{\text{col}(M_1), \text{col}(M_2)\}$.

A matroid is a partition matroid if it is a direct sum of uniform matroids. The following result follows immediately from the above observations:

Lemma.If $M = \bigoplus\limits_{i=1}^k U_{r_i, n_i}$, then $\text{col}(M) = \max\limits_{i=1}^k \lceil n_i / r_i \rceil$.

A 1-partition matroid is a partition matroid in which each of the summands $U_{r_i, n_i}$ has rank 1. The colouring number of a 1-partition matroid is just the size of its largest component.

While colouring 1-partition matroids sounds trivial, the problem occurs naturally in discrete optimisation. For example, edge-colouring bipartite graphs can be described as finding an optimal colouring of the intersection of two 1-partition matroids corresponding to the two sides of the bipartition:

An edge-colouring of the bipartite graph K(3,4).
Proper edge-colourings of the bipartite graph $K_{3,4}$ are common independent sets of two partition matroids corresponding to the two colour classes.

Weak maps and bounds

Suppose that $M$ is some complicated matroid and we want to know or bound $\text{col}(M)$. If $N$ is another matroid on the same ground set such that every independent set in $N$ is independent in $M$ (that is, $N$ is a weak map image of $M$), then $\text{col}(M) \le \text{col}(N)$.

This is particularly useful when $N$ is a 1-partition matroid, as in that case $\text{col}(M)$ is bounded by the size of the largest uniform summand in $N$.

Thus, we have the problem of finding a 1-partition reduction: given a matroid $M$, can we find a 1-partition matroid $N$ that is a weak map image of $M$ whose colouring number is not much larger than that of $M$?

In particular, does there exist a constant $c$ such that we can always find a 1-partition reduction $N$ such that $\text{col}(M) \le \text{col}(N) \le c\cdot\text{col}(M)$? We call a matroid $(1,c)$-decomposable if admits such a 1-partition reduction.

A conjecture

This question was asked by Bérczi, Schwartz and Yamaguchi, who conjectured that every matroid is (1,2)-decomposable; that is, for every $k$-colourable matroid $M$, there exists a partition $E(M) = X_1 \cup X_2 \cup \ldots \cup X_\ell$ such that each $X_i$ contains at most $2k$ elements, and each transversal of $\{X_1, X_2, \ldots, X_\ell\}$ is independent.

While this sounds like a very strong statement, it is true for a large number of natural classes of matroids, including graphic matroids and transversal matroids. (In those cases, an actual 1-partition reduction can be found in polynomial time, based on the natural representations of such matroids).

Unfortunately, the general conjecture was recently shown to fail for large binary projective geometries by Abdolazimi, Karlin, Klein and Oveis Gharan (in fact, their argument shows that there are matroids that are not $(1,c)$-decomposable for arbitrary $c$; a stronger result was later obtained by Leichter, Moseley and Pruhs). I was later able to show the slightly stronger result that almost every $\mathbb{F}_q$-representable matroid is not $(1,c)$-decomposable.

However, not all is lost. While there are counterexamples to the conjecture, the number of such counterexamples may be small; more precisely, the conjecture may still hold for almost all matroids.

Let $\mathcal{P}$ be a matroid property that is closed under isomorphism. We say that almost every matroid has property $\mathcal{P}$ if the fraction of matroids on ground set $[n]$ that fail $\mathcal{P}$ vanishes as $n$ tends to infinity, that is $$\lim_{n \to \infty} \frac{\#\{\text{matroids on $[n]$ with $\mathcal{P}$}\}}{\#\{\text{matroids on $[n]$}\}} = 1.$$

Bérczi, Schwarcz, and Yamaguchi proved that paving matroids are (1,2)-decomposable. It is widely believed that almost every matroid is paving; proving this conjecture would imply that almost every matroid is (1,2)-decomposable.

In fact, since we know a little more about random matroids, the factor 2 can be improved slightly.

First, we know that almost every matroid has rank close to half its size. More precisely, for $\varepsilon > 0$, almost every matroid satisfies $$\left(\frac{1}{2}-\varepsilon\right)|E(M)| \le r(M) \le \left(\frac{1}{2}+\varepsilon\right)|E(M)|.$$

Second, Bérczi, Schwarcz and Yamaguchi proved that a paving matroid of rank $r \approx n/2$ is 2- or 3-colourable, and that if it is $k$-colourable, it has a $\lceil\frac{r}{r-1}k\rceil$-colourable partition reduction. This leads to the following updated version of the conjecture, which may still be true:

Conjecture. Almost every matroid is (1, 3/2)-decomposable.

Matroids and the De Bruijn–Erdős Theorem

Finite linear spaces

finite linear space is a set $E$ of points together with a family $\mathcal{L}$ of lines with the property that

  • every pair of points is on a line, and
  • every line contains at least two points.

In matroidal terms, a finite linear space is a simple matroid of rank 3, given by its hyperplanes (unless all points are on a single line). We will drop the word “finite”, and simply talk about linear spaces instead.

How many lines must be in a linear space?

Let’s write $n = |E|$. As every pair of points defines a line, we immediately see that $|\mathcal{L}| \le \binom{n}{2}$. Equality holds when $\mathcal{L}$ is the family of all pairs in $E$; or equivalently, if $(E,\mathcal{L})$ is the rank-3 uniform matroid on $E$.

In $U_{3,n}$, all lines are as short as they can be. A more interesting question is: How many “long” lines can a linear space contain? Here, a line is long if it contains 3 or more points.

Proposition. A linear space on $n$ points contains at most $\frac{1}{3}\binom{n}{2}$ long lines.

Proof. Every pair of points is in at most one long line, and every long line contains at least three pairs of points. $\Box$

The proposition says that a simple rank-3 matroid on $n$ points has at most $\frac{1}{3}\binom{n}{2}$ long lines.

Equality holds if and only if each line contains exactly 3 points. In this case, $\mathcal{L}$ is the set of blocks of a Steiner triple system, and we necessarily have $n \equiv 1,3$ modulo 6.

In the other direction, we can ask: What is the smallest possible number of lines in a linear space? A reasonable guess is that this number is minimised when $\mathcal{L}$ consists one line containing all but one of the points, and $n-1$ two-point lines, each containing the remaining point of $E$. We know this as the matroid $U_{n-1,2} \oplus U_{1,1}$, but this linear space is also called a near-pencil.

Projective planes on $n$ points, when they exist, have $n$ lines as well. De Bruijn and Erdős showed that near-pencils and projective planes have the smallest number of lines among all linear spaces.

Theorem (De Bruijn–Erdős, 1948). A linear space on $n$ points, not all collinear, has at least $n$ lines. Equality holds if and only if the linear space is a near-pencil or a projective plane.

This translates to the following matroid version.

Theorem (De Bruijn–Erdős, matroid version). A simple rank-3 matroid $M$ on $n$ points has at least $n$ hyperplanes. Equality holds if and only if $M \cong U_{2,n-1} \oplus U_{1,1}$ or $M$ is a finite projective plane.

A slick proof due to Conway can be found in Jukna’s book Extremal Combinatorics. It was already noted by De Bruijn and Erdős that the Euclidean analogue of this theorem (i.e., when we assume that $M$ is real-representable) follows from the Sylvester–Gallai Theorem Jim Geelen discussed in a previous post.

Higher rank

Linear spaces can be generalised to higher dimensions. One such generalisation is known as a Hartmanis partition, which to the matroid theorist is probably better known as a paving matroid. A paving matroid is a matroid in which the only “interesting” flats are the hyperplanes: each flat of lower rank is an independent set (alternatively, each circuit has cardinality at least the rank of the matroid). For example, a matroid of rank 3 is paving if and only if it is simple, while the Vámos matroid is a paving matroid of rank 4.

The De Bruijn–Erdős Theorem is the base case of the following result.

Lemma. Let $3 \le r \le n$ and let $M$ be a paving matroid of rank $r$ on $n$ points. Then $M$ has at least $n$ hyperplanes.

Proof. For $r=3$, this is just the bound from the De Bruijn–Erdős Theorem. We proceed by induction on $r$. Suppose that the theorem holds for $r = s$, and let $M$ be a matroid of rank $s+1$. Let $e \in E(M)$. The hyperplanes of $M/e$ are in natural correspondence with the hyperplanes of $M$ that contain $e$. By induction, $M/e$ has at least $n-1$ hyperplanes, so $M$ has at least $n-1$ hyperplanes that contain $e$. Finally, as $e$ is not a loop, $M$ has at least one hyperplane that does not contain $e$. We conclude that $M$ has at least $n$ hyperplanes. $\Box$

It is not hard to construct rank-$r$ matroids with exactly $n$ hyperplanes—one example is $U_{2,n-r+2}\oplus U_{r-2,r-2}$. More generally, we can take any of the tight examples from the De Bruijn–Erdős Theorem on $n-r+3$ points and add $r-3$ coloops. However, the matroids obtained in this way are far from paving.

Question. For $r \ge 4$, is the bound on the number of hyperplanes in the previous lemma tight?

I suspect the answer is “no”. If I had to guess the correct number, it would be $1+\binom{n-1}{r-2}$, which is the number of hyperplanes of $U_{r-1,n-1}\oplus U_{1,1}$.

We now drop the assumption that $M$ is paving and write $f_k(r,n)$ for the minimal number of rank-$k$ flats in a simple matroid of rank $r$ on $n$ points. The De Bruijn–Erdős Theorem asserts that $f_2(3,n) = n$.

The following lemma shows that $f_k(r,n)$ is increasing in $r$. The truncation of a matroid is obtained by removing the hyperplanes from its lattice of flats; this operation decreases the rank of the matroid by 1.

Lemma. For all $2 \le k < r-1 < n$ we have $f_k(r,n) \ge f_k(r-1,n)$.

Proof. Let $M$ be a matroid of rank $r$ on $n$ points with $f_k(r,n)$ flats of rank $k$. The truncation of $M$ has rank $r-1$ and $f_k(r,n)$ flats of rank $k$. $\Box$

Let me conclude by asking about a higher-dimensional analogue of the De Bruijn–Erdős Theorem.

Question. For $2 \le k < r \le n$, what is $f_k(r,n)$?