The smallest matroids with no large independent flat

We show that a simple rank-$r$ matroid with no $(t+1)$-element independent flat has at least as many elements as the matroid $M_{r,t}$ defined as the direct sum of $t$ binary projective geometries whose ranks pairwise differ by at most $1$. We also show for $r \ge 2t$ that $M_{r,t}$ is the unique example for which equality holds.


Introduction
Call a set S in a matroid M a claw of M if S is both a flat and an independent set of M. A k-claw is a claw of size k. These objects were introduced by Bonamy et al. [2] and studied by Nelson and Nomoto [3]; both of these papers consider the structure of 3-claw-free binary matroids.
Here we deal with general matroids, and address the simple extremal question of determining the smallest simple rank-r matroids omitting a given claw; we solve this problem and characterize the tight examples.
Theorem 1.8 of [3] shows that, for r ≥ 4, the unique smallest simple rank-r binary matroid with no 3-claw is the direct sum of two binary projective geometries of ranks ⌊r/2⌋ and ⌈r/2⌉. We show that, perhaps surprisingly, the exact same construction is also extremal for general matroids, and that its natural generalization is still extremal for excluding larger claws. For integers r ≥ 1 and t ≥ 1, let M r,t denote the matroid that is the direct sum of t (possibly empty) binary projective geometries, whose ranks sum to r and pairwise differ by at most 1. We prove the following, which was conjectured for the special case of binary matroids in [3]. Theorem 1.1. Let r, t ≥ 1 be integers. If M is a simple rank-r matroid with no (t + 1)-claw, then |M| ≥ |M r,t |. If equality holds and r ≥ 2t, then M ∼ = M r,t .
Note that for r ≤ t, the matroid M r,t is free and therefore the theorem is trivial. For t < r < 2t, there is a rather tame family of exceptional tight examples, which we describe in Theorem 3.3. One can also ask a similar question with 'simple' relaxed to 'loopless'. (One must still insist that M is loopless, since any matroid with a loop has no claw.) In this case the answer is much less interesting; the direct sum of r − t parallel pairs and t coloops has 2r−t elements and is the unique smallest loopless rank-r matroid with no (t + 1)-claw, this is a consequence of Lemma 3.2 below.
Graph Theory. The study of the structure of 3-claw-free binary matroids in [2,3] was motivated by structural results in graph theory. The notions of induced subgraphs, cliques, chromatic number and forests can be naturally extended from graphs to binary matroids, e.g. cliques correspond to projective geometries, while claws correspond to induced forests. A graph-theoretic analogue of Theorem 1.1 using this correspondence would characterize graphs on r vertices with minimum number of edges and no induced forests of given size. From the matroidal point of view, the natural measure of the size of a forest is the number of edges, but there seem to exist no direct analogue of Theorem 1.1 using this measure. Defining the size of a forest as the number of its vertices works much better, as follows.
Let G n,t denote the graph on n vertices which is a disjoint union of t complete subgraphs, whose sizes pairwise differ by at most one. Turán's classical theorem [5] is equivalent to the statement that |E(G)| ≥ |E(G n,t )| for every graph G on n vertices with no independent set of size t + 1. This observation implies that the following graph-theoretic analogue of Theorem 1.1 generalizes Turán's theorem for n ≥ 3t. Theorem 1.2. Let n, t ≥ 1 be integers such that n ≥ 3t. If G is a graph on n vertices having no forest on 2t + 1 vertices as an induced subgraph, then |E(G)| ≥ |E(G n,t )|.
We give a short proof of Theorem 1.2, obtained by adapting one of the standard proofs of Turan's theorem, in Section 4.
Triangle-free matroids. The extremal examples in Theorem 1.1 have many triangles, and our proof techniques analyze triangles closely. It seems plausible that if M is required to be triangle-free, then the sparsest examples, instead of projective geometries, come from binary affine geometries, which are triangle-free and have 2-claws but no 3-claws. (An affine geometry AG(r − 1, 2) is obtained from a projective geometry PG(r − 1, 2) by deleting a hyperplane.) This leads us to conjecture the following. Conjecture 1.3. Let t, r be integers with t ≥ 1 and t|r. If M is a simple triangle-free matroid with no (2t + 1)-claw, then |M| ≥ t2 r/t−1 .
This conjectured bound holds with equality when M is the direct sum of t copies of a rank-(r/t) binary affine geometry; these should be the only cases where equality holds. We prove this in the easy case where t = 1; see Lemma 3.4. In what follows, we use the notation of Oxley [4]; flats of a matroid of rank 1 and 2 are points and lines respectively. We additionally write |M| for E(M). A simplification of M is any matroid obtained from M by deleting all loops and all but one element from each parallel class. All such matroids are clearly isomorphic; we write si(M) for a generic matroid isomorphic to a simplification of M, and write ε(M) for | si(M)|, the number of points of M.

The Bound
In this section we give the easy proof of the lower bound in Theorem 1.1. Our first lemma shows that the property of being (t + 1)claw-free is essentially closed under contraction; if F is a k-claw of some simplification of M, call F a k-pseudoclaw of M.
Proof. Let M ′ = (M/X)\P be a simplification of M/X, and suppose that M ′ has a k-claw F . Since F is independent in M/X, it is independent in M, and is skew to X in M. Since F is a flat of M/X, we have The sets cl M (F ) and X are skew in M, so cl we also have e ∈ cl M/X (F ) and so e ∈ cl M ′ (F ) = F , contrary to the choice of e. It follows that cl M (F ) − F intersects neither X nor P so is empty; therefore F is a k-claw of M.
Let f (r, t) = |M r,t |. Since a rank-n projective geometry has 2 n − 1 elements, we clearly have f (r, t) = (t − a)2 ⌊r/t⌋ + a2 ⌈r/t⌉ − t, where a ∈ {0, . . . , t − 1} is the integer with a ≡ r (mod t). More importantly for our purposes, we can define f recursively; it is easy to check that f (r, t) = r for all 0 ≤ r ≤ t and f (r, t) = 2f (r − t, t) + t for r > t. We use this recurrence and the previous lemma to prove the lower bound in our main theorem.
Theorem 2.2. If t ≥ 1 is an integer and M is a simple rank-r matroid with no (t + 1)-claw, then |M| ≥ f (r, t).
Proof. Let M be a counterexample for which r + |M| is minimized. If M is a free matroid then clearly r ≤ t, in which case f (r, t) = r ≤ |M| so M is not a counterexample. Therefore M has a non-coloop e.
Since |M\e| < |M| ≤ f (r, t) but M\e is not a counterexample, there must be a (t + 1)-claw S ′ in M\e. Now the matroid M| cl M (S ′ ) has rank t + 1 and has at most t + 2 elements, so has at most one circuit. There is thus a t-element subset S of cl M (S ′ ) containing at most |C|−2 elements of each C circuit of M; this set S is a t-claw.
If there is some rank-(t+1) flat F containing S for which |F −S| = 1, then F is a (t + 1)-claw. Therefore every such flat satisfies |F − S| ≥ 2; since S is a flat, it follows that every parallel class of M/S has size at least 2. Moreover, si(M/S) is a rank-(r −t) matroid that by Lemma 2.1 has no (t + 1)-claw; inductively we have | si(M/S)| ≥ f (r − t, t). Now as required.

Equality
In this section we characterize matroids for which the bound in Theorem 2.2 holds with equality. We require two lemmas; the first (which uses Tutte's characterization of binary matroids as those with no U 2,4minor [6]) corresponds to the case t = 1 of Theorem 2.2. Proof. Let M be a minor-minimal counterexample. Clearly r(M) ≥ 3. Let e ∈ E(M) and let H be a hyperplane of M not containing e. Since M|H has no 2-claw, we have |H| ≥ 2 r−1 − 1 by the minimality of M.
For each x ∈ H, the line spanned by x and e contains an element of E(M) − {e, x}, and these lines pairwise intersect only in e, so we see that |M| ≥ 2|H| + 1 ≥ 2 r − 1 as required.
Note that if t < r < 2t then |M r,t | = 2r − t. In this range, the matroids M r,t are not the only ones satisfying the bound in Theorem 2.2 with equality. The other examples include direct sums of circuits and coloops, and the matroid M r,t is the special case where all these circuits are triangles. The following lemma shows that these are the only examples. It also implies the characterization of the smallest (t + 1)claw-free matroids that are not required to be simple that was claimed in the introduction. 2 (|M| − t) ≤ r − t. Therefore equality holds in the latter, so |C| = r − t and |M| = 2r − t, giving the required structure.
We may therefore assume that |C 0 | ≥ 3. Let B be a basis of M * containing all but one element of C 0 . Since M * has no coloops, for each x ∈ B, there is a circuit C x of M * for which x ∈ C and |C ∩B| = |C|−1; choose the C x so that C x = C 0 for each x ∈ C 0 . Let X = ∪ x∈X C x . Since each C x contains only one element outside B and the element of C 0 − B is chosen at least twice, we have |X| < 2r(M * ).
By construction, the set X contains a basis and, since X is a union of circuits of M * , the matroid M * |X has no coloops. Let Y = E(M) − X; by construction the set Y is independent in M, and M/Y has no loops, so Y is a flat of M. By hypothesis, we therefore have |Y | ≤ t and so |X| ≥ |M| − t = 2(r − t). By our upper bound on |X|, this gives 2(r − t) < 2r(M * ) = 2(|M| − r) ≤ 2(r − t), a contradiction.
We are now ready to prove the main theorem. Proof. Consider a counterexample M for which |M| + r is minimized. Clearly r > t, as otherwise |M| ≥ r = f (r, t) and there is nothing to prove. Therefore |M| ≥ f (r, t) > r, so M is not a free matroid.

M has a t-claw.
Subproof: Let e be a non-coloop of M; by the minimality of M, the matroid M\e has a (t + 1)-claw F , now M| cl M (F ) has rank at least t + 1 and has at most t + 2 elements, so has at most one circuit. There is thus a t-element subset of cl M (F ) that contains at most |C| − 2 elements of each circuit C of M| cl M (F ); this set is a t-claw of M. Subproof: The matroid M/S has rank r − t and, by Lemma 2.1, has no near-(t + 1)-claw. Therefore si(M/S) has no (t + 1)-claw, so ε(M/S) ≥ f (r − t, t). Moreover, if some parallel class Y of M/S has size 1, then S ∪ Y is a (t + 1)-claw of M, so every parallel class of M/S has size at least 2, giving |M/S| ≥ 2ε(M/S). Therefore f (r, t) ≥ |M| ≥ 2ε(M/S) + |S| ≥ 2f (r − t, t) + t = f (r, t).
Equality holds throughout, which gives the claim.
The matroid si(M/S) has no (t+1)-claw and has f (r −t, t) elements, so inductively satisfies one of the conclusions of the theorem. For each component N of M/S, the matroid si(N) is either a circuit or a binary projective geometry. Since every component of si(M/S) is a circuit or projective geometry, given any pseudoclaw K of M and any e, e ′ in the same component of M/S for which e ∈ K and e ′ / ∈ K, the set (K − e) ∪ {e ′ } is also a pseudoclaw. Since M/S has at least one t-pseudoclaw, both conclusions of the claim easily follow.
The above claim implies in particular that every element of M/S is in a t-pseudoclaw.

3.3.4.
For each t-pseudoclaw U of M/S, there is a bijection ψ U from U to S so that for each e ∈ U, the flat T e = cl M (e, ψ U (e)) is a triangle of M, and so that M| cl M (S ∪ U) = ⊕ e∈U (M|T e ).
Subproof: Since the closure of U in M/S is obtained from U by extending each element of U once in parallel, we have | cl M (S ∪ U)| = |S| + 2|U| = 3t. By Lemma 3.2, it follows that the simple rank-2t matroid M ′ = M| cl M (S ∪ U) is the direct sum of t circuits and some set of coloops, and therefore that is precisely the direct sum of t triangles. Since S is a t-claw of M ′ and U is a t-pseudoclaw of M ′ /S, both S and U must be transversals of this set of triangles. The claim follows.
Every element e of M/S is contained in a t-pseudoclaw, so the above claim implies that each such e is in exactly one triangle that intersects S. Write ψ(e) for the unique element of S for which e and ψ(e) are contained in a triangle; we have ψ U (e) = ψ(e) for each t-pseudoclaw U of M/S containing e.
Since M|S is a claw and no rank-1 flat of M/S has more than two elements, no line of M that intersects S has more than three elements. Moreover, each e ∈ E(M/S) is in exactly one triangle of M that intersects S, so the number of triangles of M that intersect S is exactly 1 2 |M/S| = 1 2 (2ε(M/S)) = f (r − t, t). Therefore S is generic. It follows from the choice of S that every t-claw of M is generic. , there is a bijection ψ i = ψ U ∪{e i } from U ∪ {e i } to S, and moreover for each e ∈ U we have ψ 1 (e) = ψ(e) = ψ 2 (e). Therefore ψ 1 and ψ 2 agree on all t − 1 elements of U; thus ψ 1 (e 1 ) = ψ 1 (e 2 ) and so ψ(e 1 ) = ψ(e 2 ).
Suppose now that e 1 and e 2 are in different components of M/S. By 3.3.3 there is a t-pseudoclaw U containing e 1 and e 2 . Since ψ U is a bijection we have ψ(e 1 ) = ψ U (e 1 ) = ψ U (e 2 ) = ψ(e 2 ), as required.
It follows from the first part that the image of ψ has size equal to the number of components of M/S. But clearly the image of ψ contains the image of ψ U , which is equal to S, for each t-pseudoclaw U. Therefore ψ has image S, so M/S has exactly |S| = t components. Subproof: Suppose not, so there is some e ∈ E(N) for which σ(e) > 0. Let U be a t-pseudoclaw of M/S containing e. Since U is a generic t-claw of M, we have u∈U τ (u) = f (r − t, t), so Since |M| = f (r, t) and σ(e) > 0, this is a contradiction.
Let N ∈ N . It is clear, since N is obtained from N/ψ(N) by t − 1 successive extension-contraction operations, that r(N) ≤ r( N) − 1. The matroid si(N) is a circuit or a binary projective geometry, so so equality holds throughout, and the sets E( N) are mutually skew in M. Thus M is the direct sum of t nonempty binary projective geometries. If M has components of ranks r 1 , r 2 with r 2 ≥ r 1 + 2, then deleting both and replacing them with projective geometries of rank r 2 − 1 and r 1 + 1 respectively gives a matroid M ′ with no (t + 1)-claw satisfying which contradicts the minimality of |M|. It follows that no two components of M have ranks differing by more than 1, so M ∼ = M r,t , contrary to the choice of M as a counterexample.
Finally, we prove the t = 1 case of Conjecture 1.3 as promised.
Proof. We may assume that r ≥ 3. We first show that every triple of distinct elements of M is contained in a four-element circuit; indeed, given such a triple I, since I is not a triangle or a 3-claw, we have r M (I) = 3 and cl M (I) = I. Thus there is some x ∈ cl M (I) − I. Since M is triangle-free, no pair of elements of I spans x, so I ∪ {x} is a 4-element circuit.
Let e ∈ E(M). Since M is triangle-free, the matroid M/e is simple. If M/e has a 2-claw I, then I ∪ {e} is clearly a 3-claw of M; therefore M/e is 2-claw-free and so |M/e| ≥ 2 r−1 − 1 by Lemma 3.1. It follows that |M| ≥ 2 r−1 as required.
If equality holds, then M/e ∼ = PG(r − 2, 2) so M/e is binary. This holds for arbitrary e ∈ E(M); it follows (since r ≥ 3) that M has no U 2,4 -minor so is also binary. A simple rank-r triangle-free binary matroid has at most 2 r−1 elements and equality holds only for binary affine geometries (see [1], for example); therefore M ∼ = AG(r−1, 2).
Theorem 4.1. Let n, t ≥ 1 be integers. Let G be a simple graph on n vertices such that no forest on 2t + 1 vertices is an induced subgraph of G. Then |E(G)| ≥ g(n, t).
Proof. We prove the theorem by induction on |V (G)|. The base case is trivial. For the induction step we can clearly assume that n ≥ 2t + 1.
Suppose first that n ≤ 3t. If there exists v ∈ V (G) such that deg(v) ≥ 3 then the desired inequality immediately follows by applying the induction hypothesis to G \ v. Thus we assume that deg(v) ≤ 2 for every v ∈ V (G), and so every component of G is a tree or a cycle. Let S be the set of vertices of cycles of G. As every cycle on k vertices contains an induced forest on at least 2k/3 vertices, it follows that G contains an induced forest on n − |S|/3 vertices. Thus n − |S|/3 ≤ 2t, so |E(G)| ≥ |S| ≥ 3(n − 2t) = g(n, t), as desired.
It remains to consider the case where n > 3t. Let X ⊆ V (G) be chosen maximal so that G[X], the subgraph of G induced by X, is a forest. Then |X| ≤ 2t. Let Z be the set of non-isolated vertices of G[X]. As G[X ∪ {v}] contains a cycle for every v ∈ V (G) \ X, every such v has at least two neighbors in Z. Thus z∈Z deg(z) ≥ |Z| + 2|V (G) − X| ≥ |Z| + 2(n − 2t).