A bound for the growth rate of binary matroids having no PG(k-1,2)

Bounds for binary matroids having no P G(k − 1, 2)−minor Keywords: matroid, projective geometry, growth-rate AMS subject classification (2012): 05D99, 05B35

i

A bound for the growth rate of binary matroids having no P G(k − 1, 2)-minor Sean McGuinness Thompson Rivers University Kamloops, BC V2C5N3 Canada

Abstract Let M be a simple matroid having no U2,l+2 -minor. We show that ) 3 2k ( if |E(M )| ≥ l3k 2(l+1) r(M2)+1 , then M contains a simple minor N where r(N ) ≥ k and |E(N )| ≥ 2r(N ) −1. In particular, if M is a simple ) 2k 3( binary matroid where |E(M )| ≥ 23 +3k r(M2)+1 , then M contains a P G(k − 1, 2)-minor. Keywords : Matroid, projective geometry, growth-rate. AMS Subject Classifications (2012) : 05D99,05B35.

1

Introduction

For a matroid M , we let ε(M ) denote the number of rank-1 flats in M , that is, the number of elements in si(M ), the simplification of M . For a subset X ⊆ E(M ), we shall write ε(X) to mean ε(M X). A line is a rank-2 flat, and a long line is a line containing at least three rank-1 flats. Let U(l) be the class of matroids containing no U2,l+2 -minor. Let M ⊆ U(l) be a minor-closed class of matroids. The growth rate function for M is the function hM : N → N where hM (n) = max{ε(M ) M ∈ M, r(M ) ≤ n}. The growth rate theorem given in [1] shows that hM (n) is either linear, quadratic, or exponential.

1

1.1 Theorem ( Geelen, Kung, Whittle, 2008 ) There is an integer c such that one of following is true: (i) hM (n) ≤ cn for all n ≥ 0, ( ) (ii) n+1 ≤ hM (n) ≤ cn2 for all n ≥ 0 and M contains all graphic 2 matroids, or −1 ≤ hM (n) ≤ cq n for all n ≥ 0 (iii) there is a prime power q such that qq−1 and M contains all GF (q)-representable matroids. n

By the above theorem, it follows that if M ∈ U(l) is a simple binary matroid having no P G(k − 1, 2)-minor, (then)there is a constant c, depending only on k and l such that |E(M )| ≤ c n+1 2 . Unfortunately, the constant c implied by the proof of Theorem 1.1 in [1] is astronomical, involves nested Ramsey functions, and is difficult to calculate. Furthermore, it depends on the linear bound hN (n) ≤ cn for the class of matroids N which contain no M (Kn )-minor. Even here (see [2]), the constant c is astronomical and difficult to calculate. For small values of k, good bounds are possible for binary matroids having no P G(k − 1, 2)-minor. For example, Heller [3] and Murty [5] showed that a simple ( )binary matroid M of rank n with no P G(2, 2)-minor can have at most n+1 elements, and Kung [4] showed that if 2 (n+1) elements. instead M has no P G(3, 2)-minor, then it can have at most 15 2 2 In this paper, we shall derive an explicit bound for hM (n), for the class M of binary matroids having no P G(k − 1, 2)-minor. While our bound is still extremely large, it is nonetheless a vastly better bound than that implied by the proof in [1]. We shall prove the following more general theorem: 1.2 Theorem ) 3 2k ( Let M ∈ U(l) and let k ∈ Z+ . If ε(M ) > l3k 2(l+1) r(M2)+1 , then M contains a minor N such that r(N ) ≥ k and ε(N ) ≥ 2r(N ) − 1. As an immediate consequence of the above, we obtain the following: 1.3 Corollary 2k +3k 3

Let M be a binary matroid and let k ∈ Z+ . If ε(M ) > 23 then M has a P G(k − 1, 2)-minor.

(r(M )+1) 2

,

The proof of the above theorem borrows heavily from concepts and ideas introduced in [1]. However, it differs significantly in the manner in which it 2

constructs a minor N where ε(N ) ≥ 2r(N ) − 1. Specifically, the construction in [1] depends on finding a minor isomorphic to M (Kn ), where n is very large. In this paper, we avoid this construction altogether and use instead a more straightforward idea involving a k-line product, defined in Section 3.

2

Round Minors and Books

A matroid N is round if every cocircuit of N is spanning. Equivalently, N is round if E(N ) ̸= F1 ∪ F2 for any proper flats F1 and F2 of N. Note that if N is round, then N/e is round for all elements e ∈ E(N ). We say that a flat F of N is round if N F is round. The following lemma is given in [1]. 2.1 Lemma Let M be a matroid, let F1 and F2 be round flats of M such that r(F1 ) = r(F2 ) = k and r(F1 ∪ F2 ) = k + 1, and let F be the flat spanned by F1 ∪ F2 . If F ̸= F1 ∪ F2 , then F is round. Let F be a set of round rank-(k − 1) flats of a matroid M. A rank-k flat F is said to be F-constructed if there exist two flats F1 , F2 ∈ F such that F = clM (F1 ∪ F2 ) and F ̸= F1 ∪ F2 . Let F + denote the set of all F-constructed flats. By Lemma 2.1, all the flats in F + are round. We shall use the following lemma from [1]. 2.2 Lemma Let M ⊂ U (l) be a minor-closed class of matroids. There exists an integervalued function f (k, α, l) such that for all integers k (≥ 2 and ) α ≥ 1, if M ∈ M is a simple matroid where |E(M )| > f (k, α, l) r(M2)+1 , then there exists a minor N of M and a set F of round rank-(k − 1) flats of N such that |F + | > αr(N )|F|. In the proof of this lemma, given in [1], the function f (k, α, l) is defined 2 recursively where, f (2, α, l) = α(l + 1)2 and f (k + 1, α, l) = f (k, l(k+1) α + lk , l), for k = 2, 3, 4, . . . . Using this definition, one obtains (by straightforward calculations) a bound for f (k, α, l), namely f (k, α, l) ≤ (l + 1)2 l

k(k+1)(2k+1) 6

(α + lk ).

(1)

Following the definitions given in [1], a sequence (F0 , F1 , . . . , Ft ) of round flats in a matroid M is called a k-book if r(F0 ) = k and F1 , F2 , . . . , Ft are distinct rank-(k + 1) flats containing F0 . In [1], it is shown that for simple )| matroid M ∈ U(l), one can find a large k-book minor provided |E(M is r(M )2 large enough. More specifically, we have the following: 3

2.3 Lemma ( ) Let M ∈ U(l). If ε(M ) > f (k + 1, α, l) r(M2)+1 , then there is a minor of N of M containing a k-book (F0 , F1 , . . . , Ft ) where t ≥ αr(N ). Proof. We may assume that M is simple. Then by Lemma 2.2, there is a minor N and a collection F of rank-k flats of N such that |F + | > αr(N )|F|. Given that each flat of F + contains flats from F, there must be at least one flat F ∈ F contained in at least αr(N ) flats of F + . Thus there is a k-book (F0 , F1 , . . . , Ft ) in N where t ≥ αr(N ).

3

Line Products

For integers k ≥ 1, we define a k-line product in a recursive manner. First, we define a 1-line product to be a long line. For k ≥ 2, M is a kline product if one can partition E(M ) into long lines L1 , L2 , . . . , Ln having elements ei ∈ Li , i = 1, 2, . . . , n for which N = M {e1 , . . . , en } is a (k − 1)line product, and r(M ) = r(N ) + n. In this section, we show how one can construct k-line products in certain matroids. 3.1 Lemma Let M ∈ U(l) where M is a k-line product. Then |E(M )| ≤ (l + 1)k and k r(M ) ≤ 1 + (l+1)l −1 . Proof. We shall use induction on k. When k = 1, we have r(M ) = 2 and |E(M )| ≤ l + 1 and the bounds are seen to hold. Suppose the bounds holds for all (k − 1)-line products for some fixed k ≥ 2. Let M be a k-line product where M ∈ U(l). By definition, there exists a partition of E(M ) into disjoint long lines L1 , . . . , Ln for which there exist elements ei ∈ Li , i = 1, . . . , n such that N = M {e1 , . . . , en } is a (k − 1)-line product and r(M ) = r(N ) + n. By k−1 −1

assumption, we have that r(N ) ≤ 1+ (l+1) l

and n = |E(N )| ≤ (l+1)k−1 .

Thus r(M ) ≤ 1 + (l+1) l −1 + (l + 1)k−1 = 1 + (l+1)l −1 . We also have that |E(M )| ≤ n(l + 1) ≤ (l + 1)k−1 (l + 1) = (l + 1)k . Thus the bounds hold for M . The proof now follows by induction. k−1

k

In what follows, we show that if every element in a matroid M ∈ U(l) belongs to enough long lines, then M contains a k-line product. First, we need the following lemma.

4

3.2 Lemma Let M ∈ U(l) where M is simple. Let C1 , . . . , Cn be circuits and let xi , i = 1, 2, . . . , n be elements such that Ci \Cj = {xi } for all i ̸= j. Then n ≤ l − 1. Proof. Let X = ∩ni=1 Ci . Given that Ci \Cj = {xi }, for all j ̸= i, it follows that X = Ci \{xi }, i = 1, . . . , n. Since |Ci | ≥ 3, i = 1, . . . , n, it follows that |X| ≥ 2. Let X ′ ⊆ X where |X ′ | = 2. Then (M C1 ∪ · · · ∪ Cn )/(X\X ′ ) ≃ U2,n+2 . Thus n ≤ l − 1. 3.3 Proposition Let M ∈ U (l) where M is simple and each element of M belongs to at least k l · 2(l+1) long lines. Then there exists T ⊆ E(M ) such that M T is a k-line product. Proof. By induction on k. For k = 1, the proposition is seen to hold. Suppose that for some k ≥ 2, the proposition holds for k − 1. Let M ∈ U(l) be a k simple matroid where each element of M is contained in at least l2 (l+1) lines. By assumption, there is a subset T ′ ⊆ E(M ) for which N ′ = M T ′ is a (k−1)-line product. Let S ⊆ T ′ (S possibly empty) and let Le , for e ∈ S, be ∪ disjoint long lines such that for N ′′ = M (T ′ ∪ e∈S Le ), r(N ′′ ) = r(N ′ )+|S|. Suppose that S is maximal with respect to the above. If S = T ′ , then N ′′ is seen to be a k-line product. Suppose that S ⊂ T ′ . Let f ∈ T ′ \S. We shall show that there is a line Lf in M containing f which is disjoint from the lines Le , e ∈ S, such that r(N ′′ ∪ Lf ) = r(N ′ ) + 1. Let I be a maximum independent set in N ′′ where f ∈ I. Let L1 , L2 , . . . , Ln be the long lines of M containing f , where Li = clM ({f ∪ gi }), i = 1, . . . , n. Here, we may assume that gi ̸∈ I, i = 1, . . . , n. If I ∪ {gi } is independent for some i, then taking Lf = Li will suffice. Suppose that for i = 1, . . . , n, I ∪ {gi } contains a circuit Ci (so gi ∈ Ci ). Let Ci′ = Ci \{gi }, i = 1, . . . , n (noting that Ci′ ⊆ I). Suppose we fix a subset I ′ ⊆ I. By Lemma 3.2, there are at most l − 1 circuits Ci such that Ci′ = I ′ . Given that I contains exactly ′

(l+1)k−1 −1

l 2|I| = 2r(N )+|S| ≤ 21+ Lemma 3.1), it follows that

+|S|

subsets (since r(N ′ ) ≤ 1 +

n ≤ (l − 1)21+

(l+1)k−1 −1 +|S| l

(l+1)k−1 −1 l

by

.

Given that |S| < |E(N ′ )| ≤ (l + 1)k−1 (by Lemma 3.1), the above bound implies that n ≤ (l − 1)(21+

(l+1)k−1 −1 +(l+1)k−1 l

5

k

) < l2(l+1) .

k

However, by assumption n ≥ l2(l+1) , giving a contradiction. Thus it follows that I ∪ {gi } is independent for some i, and taking Lf = Li will suffice. Now we can use S ′ = S ∪ {f } in place of S, observing that ∪ r(M (T ′ ∪ Li ∪ {Lf })) = r(N ′ ) + |S ′ |. e∈S

However, S ′ contradicts the maximality of S. We conclude that S = T ′ , and this completes the proof.

4

Contracting a k-book into a flat

ε(M ) The density of a matroid M is defined to be δ(M ) = r(M ) . In this section, we shall show that if M is a k-book (F0 , F1 , . . . , Fn ), where si(M/F0 ) k+2 is a long line, then either M has density at least 2 k+2−1 , or there is an element a ∈ E(M )\Fn for which M/a has density which is strictly greater than the minimum density of the flats Fi , i = 1, . . . , n. This will play a key role in the proof of Theorem 1.2. The ideas in this section owe their conception to a simple observation of Kung [4]. One specific observation (among other things) derived from that paper is that if M is a binary 2book (F0 , F1 , F2 , F3 ) where Fi ≃ M (K4 ), i = 1, 2, 3 and M/F0 is a triangle, then there is an element a ∈ E(M )\F3 for which si(M/a) ≃ F7 . To give a short explanation for why this works, think of M as imbedded in a matroid M ′ ≃ P G(3, 2). There is an element a′ ∈ clM ′ (F3 )\F3 , and a′ belongs to four triangles in M ′ which do not intersect F3 . Since F0 has cardinality two or three, it follows that |E(M )\F3 | ≥ 6. and hence at least one of the triangles contains two elements of E(M )\F3 . Any one of these elements, say a, is seen to have the property that si(M/a) ≃ F7 . Here we shall expand on this observation and prove something more general. In the proof below, we shall use the (easily proven) fact that if M is a simple, rank-k matroid where any two elements belong to a long line, then |E(M )| ≥ 2k − 1.

4.1 Proposition Let M be a k-book (F0 , F1 , . . . , Fn ). Suppose si(M/F0 ) is a long line. Then either ε(F0 ) ≥ 2k − 1 and ε(M ) ≥ 2k+2 − 1 or there exists a ∈ E(M )\Fn for which ε(M/a) > mini ε(Fi ). Proof. For convenience, we may assume that M is simple. Let a ∈ F1 \F0 and let M ′ = M/a. We observe that since si(M/F0 ) is a long line, it follows that n ≥ 3, r(M ) = k + 2, and r(M ′ ) = k + 1. If the cardinalities of the 6

sets Fi , i = 2, . . . , n are not all equal, then ε(M ′ ) > min1≤i≤n |Fi |, since r(M ′ ) = k + 1 and Fi ⊆ E(M ′ ), i = 2, . . . , n. The same would apply if a ∈ F2 \F0 , and the sets Fi , i = 1, 3, . . . , n had different cardinalities. Thus we may assume that for some integer m ≥ 2, |Fi \F0 | = m + 1, i = 1, . . . , n. Suppose that for some a′ ∈ (F2 ∪ · · · Fn )\F0 , no long line contains a and a′ . Given that n ≥ 3, we have that a′ ∈ / Fj \F0 for some j ∈ {2, . . . , n} and we see that Fj ∪ {a′ } ⊂ E(M ′ ) and a′ is parallel to no element of Fj . Consequently, ε(M ′ ) ≥ |Fj | + 1, and thus ε(M ′ ) > mini |Fi |. Thus we may assume that for each element a′ ∈ (F2 ∪ · · · Fn )\F0 , there is a long line containing a and a′ . Furthermore, the above also shows that we may assume that a long line containing a and a′ must intersect each of the sets Fi \F0 , i = 1, 2, . . . , n. Let L0 = {a1 , a2 , . . . , an } be one such long line where ai ∈ Fi \F0 , i = 1, 2, . . . , n and a1 = a. We observe that Fi ⊆ E(M ′ ), i = 2, . . . , n. If for some a′ ∈ F1 \(F0 ∪ {a}), there is no long line containing a and a′ , then a′ is parallel to no element of F0 in M ′ . Noting that a′ ∈ clM ′ (F0 ), in this case, ε(clM ′ (F0 )) > |F0 | and hence ε(M ′ ) > mini |Fi |, i = 2, . . . , n. Thus we may assume that for any pair of elements a′ , a′′ ∈ F1 \F0 , there is a long line containing a′ and a′′ . Such a line must clearly intersect F0 in exactly one element. More generally, we may assume that for i = 1, . . . , n − 1, for any pair of elements a′ , a′′ ∈ Fi \F0 , there is a long line containing a′ and a′′ (and this line intersects F0 ). We show that the same holds for Fn . For i = 1, . . . , n, let Fi \F0 = {ai0 , ai1 , . . . , aim }, where ai = ai0 . For 1 ≤ i < j ≤ m, we show there is a long line in Fn containing ani and anj (which intersects F0 ). By the above, we may assume that Li = {a1 , a2i , . . . , ani }, i = 0, 1, . . . , m are long lines. The plane P = clM ({a, a2i , a2j }) intersects Fn in a line, say L′ , and L′ contains ani and anj (because Li and Lj are contained in P ). Furthermore, since the long line containing a2i and a2j (which is also in P ) intersects F0 at a point, say a′ , it follows that a′ ∈ L′ . Thus any two elements of Fn \F0 belong to a long line intersecting F0 . For i = 1, . . . , n and j = 0, . . . , m, let S(aij ) denote the set of elements of F0 that belong to a long line in Fi containing aij . We first show that the sets S(aij ), i = 1, . . . , n; j = 0, . . . , m are all equal. Consider S(a2 ) and S(a3 ). Let a′ ∈ S(a2 ), where a′ belongs to a long line, say L, containing a2 and a2j , for some j ∈ {1, . . . , m}. The plane P = clM ({a1 , a2 , a2j }) intersects F3 in a line, say L′ . Since L, L0 and Lj are all contained in P , it follows that a3 , a3j , and a′ all belong to L′ . Thus a′ ∈ S(a3 ) and given that a′ was arbitrarily chosen, it follows that S(a2 ) ⊆ S(a3 ). Switching the roles of a2 and a3 , we also obtain that S(a3 ) ⊆ S(a2 ) and consequently, S(a2 ) = S(a3 ). However, essentially the same arguments can be used to show that S(ai ) = S(ai′ ), for any i, i′ , where 1 < i < i′ ≤ n. Thus S(a2 ) = 7

S(a3 ) = · · · = S(an ). If we now let a2 play the role of a1 , then by similar reasoning, we obtain that S(a1 ) = S(a3 ) = · · · = S(an ). From this it follows that S(a1 ) = S(a2 ) = · · · = S(an ). Repeating the above arguments with the line Li in place of L0 , for i = 1, . . . , m, one obtains that S(a1 ) = S(a2i ) = · · · = S(ani ), i = 0, 1, . . . , m. Switching the roles of a1 and a2 , one obtains that S(a2 ) = S(a1i ) = S(a3i ) = · · · = S(ani ), i = 0, 1, . . . , m. Thus it follows that the sets S(aij ), i = 1, . . . , n; j = 0, . . . , m are all equal. Let S = S(a1 ). We now show that any two elements of S belong to a triangle contained in S. Let a, a′ ∈ S. There exist long lines L and L′ in F1 containing a1 that contain a and a′ , respectively. Let j, j ′ ∈ {1, . . . , m}, be such that a1j and a1j ′ belong to L and L′ , respectively. Let L′′ be a long line containing a1j and a1j ′ . Let a′′ ∈ F0 be the point of L′′ which intersects F0 (so a′′ ∈ S(a1j )). The plane P = clM ({a1 , a1j , a1j ′ }) intersects F0 in a line L′′′ . Given that L, L′ , and L′′ are all contained in P , it follows that a, a′ , and a′′ all belong to L′′′ . Since S = S(a1 ) = S(a1j ), it follows that a′′ ∈ S, and {a, a′ , a′′ } is a triangle contained in S. Given that a and a′ were arbitrarily chosen elements of S, it follows that any two elements of S belong to a triangle contained in S. Let N = M S. By the above, any two elements of N belong to a long line in N , and as such |S| = |E(N )| ≥ 2r(N ) −1. However, since r(F1 \F0 ) = k +1 (because F1 is round), it follows that r(N ) = k and hence |S| ≥ 2k − 1. Consequently, |F0 | ≥ 2k − 1. Given that each element of S belongs to a long line of Fi containing ai , we deduce that |Fi \F0 | ≥ 2k , i = 1, . . . , n. Thus ε(M ) = |E(M )| ≥ n(2k ) + 2k − 1 = (n + 1)2k − 1 ≥ 2k+2 − 1, since n ≥ 3. This completes the proof.

5

Proof of the Main Theorem

In this section, we shall complete the proof of Theorem 1.2. The main idea is to construct a certain large k-book with the property that when one contracts a certain set of elements, one obtains another large k-book where the minimum density of its rank-(k + 1) flats is strictly larger than the previous k-book. One then continues creating new k-books whose minimum density of its rank-(k + 1) flats is always increasing, until one has the desired minor. This process is embodied in the next two lemmas. 5.1 Lemma Let M be a k-book (F0 , F1 , . . . , Ft ) where t ≥ 3, mini ε(Fi ) ≥ δ, and si(M/F0 )

8

is a (2k+1 − δ)-line product. Then M contains a minor N where either r(N ) = k and ε(N ) ≥ 2k − 1 or r(N ) = k + 2 and ε(N ) ≥ 2k+2 − 1. Proof. We shall use induction on 2k+1 − δ. Suppose 2k+1 − δ = 1. Then δ = 2k+1 − 1 and r(M ) = k + 2. If ε(F0 ) ≥ 2k − 1, then N = F0 will suffice. Suppose ε(F0 ) = 2k −λ for some λ ≥ 2. Then given that ε(Fi ) ≥ δ = 2k+1 −1, i = 1, . . . , t, we have that ε(Fi \F0 ) ≥ 2k+1 − 1 − (2k − λ) = 2k + λ − 1, i = 1, . . . , t. It now follows that ε(M ) ≥ 2k − λ + t(2k + λ − 1) ≥ 2k+2 . Given that r(M ) = k + 2, taking N = M will suffice in this case. Suppose M is a k-book (F0 , F1 , . . . , Ft ) where mini ε(Fi ) ≥ δ and L = si(M/F0 ) is a (2k+1 − δ)-line product, where 2k+1 − δ > 1. Note that the elements of si(M/F0 ) correspond to the sets Fi \F0 , i = 1, . . . .t Assume that for any matroid M ′ which is a k-book (F0′ , F1′ , . . . , Ft′′ ) where mini ε(Fi′ ) ≥ δ ′ , if si(M/F0′ ) is a (2k+1 − δ ′ )-line product and δ ′ > δ, then there is a minor N ′ of M ′ where either r(N ′ ) = k and ε(N ′ ) ≥ 2k − 1, or r(N ′ ) = k + 2 and ε(N ′ ) ≥ 2k+2 − 1. We show that M has the desired minor N , completing the induction step. We may assume that ε(F0 ) < 2k − 1, for otherwise, taking N = F0 will suffice. Let L1 , L2 , . . . , Lm be the disjoint lines of L k+1 − δ)-line product and let e ∈ L , i = 1, . . . , m be such that in the (2 i i L′ = L {e1 , . . . , em } is a (2k+1 − δ − 1)-line product. For i = 1, . . . , m, let Fi1 , Fi2 , . . . , Fiti denote the rank-(k + 1) flats in the k-book (F0 , F1 , . . . , Ft ) containing elements of Li . Note that ti ≥ 3 for all i, since L1 , . . . , Lm are long lines. Here, we may assume that ei ∈ Fiti , i = 1, . . . , m. For i = 1, . . . , m, let F i = Fi1 ∪ Fi2 ∪ · · · ∪ Fiti and let Mi = M F i . Since F i = clM (Fi1 ∪ Fi2 ) ̸= Fi1 ∪ Fi2 , it follows by Lemma 2.1 that F i is a round rank(k + 2) flat. Since ε(F0 ) < 2k − 1, it follows by Proposition 4.1, that for i = 1, . . . , m, there exists ai ∈ F i \Fiti such that ε(Mi /ai ) > min1≤j≤ti ε(Fij ). Note that {a1 , . . . , am } is independent in M . Let M ′ = M/{a1 , . . . , am } and for i = 1, . . . , m, let Fi′ = clM ′ (F i \ai ). Let F0′ = clM ′ (F0 ). We claim that ′ ) is a k-book. To see this, let j ∈ {1, . . . , t − 1} be such that (F0′ , F1′ , . . . , Fm 1 a1 ∈ F1j . Then clM/a1 (F0 ) = F1j \a1 , and hence clM/a1 (F0 ) is a round rank-k flat in M/a1 since F1j is round in M . Furthermore, F 1 \a1 is seen to be a round rank-(k + 1) flat in M/a1 , since F 1 is round in M. Since clM/a1 (F0 ) is round, we also see that clM/a1 (F i ) is a round rank-(k + 2) flat in M/a1 , for i = 2, . . . , m. If one continues, contracting a2 , . . . , am , then one sees that ′ ) is a k-book where Fi′ , i = 0, . . . , m are round flats of M ′ and (F0′ , F1′ , . . . , Fm ′ ′ ′ ′ k+1 ε(Fi ) ≥ δ + 1. Moreover, si(M /F0 ) = L is a (2 − δ − 1)-line product. It ′ now follows from our assumptions that M contains a minor N ′ where either 9

r(N ′ ) = k and ε(N ′ ) ≥ 2k − 1, or r(N ′ ) = k + 2 and ε(N ′ ) ≥ 2k+2 − 1. The proof now follows by induction. 5.2 Lemma Let M ∈ U (l) be a k-book (F0 , F1 , . . . , Ft ) where M is simple and t ≥ 2k+1

l2 · 2(l+1) r(M ). Then M contains a minor N such that r(N ) ≥ k and r(N ) ε(N ) ≥ 2 − 1. Proof. By induction on r(M ). The lemma is certainly true when r(M ) < k+1 (l + 1)2 . Assume that for any simple matroid M ′ ∈ U(l), if M ′ is a k-book 2k+1

(F0′ , F1′ , . . . , Ft′′ ) where r(M ′ ) < r(M ) and t′ ≥ l2 · 2(l+1) r(M ′ ), then M ′ ′) ′ ′ ′ r(N has a minor N where r(N ) ≥ k and ε(N ) ≥ 2 − 1. Let L = si(M/F0 ). Observe that the elements of L correspond to the sets Fi \F0 , i = 1, . . . , t. Thus |E(L)| = t. Let e ∈ L and let M ′ = si(M/e) and L′ = si(L/e). Suppose that e belongs to fewer than l · 2(l+1) |E(L′ )| > t − l2 2(l+1)

long lines in L. Then

2k+1

2k+1

≥ l2 · 2(l+1)

2k+1

(r(M ) − 1) = l2 · 2(l+1)

2k+1

r(M ′ ).

Thus M ′ is seen to be a k-book (F0′ , F1′ , . . . , Ft′′ ) where t′ = |E(L′ )| > l2 · 2(l+1)

2k+1

r(M ′ ).

It follows by our assumptions that there exists a minor N ′ of M ′ where ′ r(N ′ ) ≥ k and ε(N ′ ) ≥ 2r(N ) − 1. Therefore, we may assume that for each 2k+1

e ∈ L, there are at least l · 2(l+1) long lines in L which contain e. It now follows by Proposition 3.3 that there exists T ⊆ E(L) such that L T is a 2k+1 -line product. Let F1′′ , F2′′ , . . . , Ft′′′′ be the rank-(k + 1) flats in the kbook (F0 , F1 , . . . , Ft ) which correspond to the elements of T. Let M ′′ be the ′′ ′′ ′′ k-book (F0 , F1 , . . . , Ft′′ ). Since si(M /F0 ) ≃ L T is a 2k+1 -line product, it now follows by Lemma 5.1 that M ′′ contains a minor N where either r(N ) = k and ε(N ) ≥ 2k − 1, or r(N ) = k + 2 and ε(N ) ≥ 2k+2 − 1. This completes the proof. 3

Proof of Theorem 1.2: Let M ∈ U(l) where ε(M ) > l3k 2(l+1) By straightforward calculations, the reader can verify that l (l + 1)2 l

(k+1)(k+2)(2k+3) 6

k

2 (l2 2(l+1)

+ lk+1 ). 10

2k

3k3

(r(M )+1)

2 2k (l+1) 2

.

>

It now follows by the above and (1)

3

2k

that l3k 2(l+1)

2k

> f (k + 1, l2 2(l+1) , l). By Lemma 2.3, M has a k-book 2k

minor (F0 , F1 , . . . , Ft ) where t ≥ l2 2(l+1) r(M ). It now follows by Lemma 5.2 that M contains a minor N where r(N ) ≥ k and ε(N ) ≥ 2r(N ) − 1. Acknowledgement The author wishes to thank the referee who pointed out an error in the original proof of Proposition 4.1 and also suggested some shorter arguments in that proof which simplified the exposition. I have also used the referees short proof of Kung’s observation in the discussion prior to Proposition 4.1.

References [1] Geelen, J., Kung, J., Whittle, G., Growth rates of minor-closed classes of matroids, J. Comb. Theory, Series B 99 (2009) 420-427. [2] Geelen, J., Gerards, A.M.H., Whittle, G., Disjoint cocircuits in matroids with large rank, J. Comb. Theory, Series B 87 (2003) 270-279. [3] Heller, I., On linear systems with integral-valued solutions, Pacific J. Math. 7 (1957) 1351-1364. [4] Kung, J., Binary matroids with no P G(3, 2)-minor, Algebra univers. 59 (2008) 111-116. [5] Murty, U.S.R., Extremal matroids with forbidden restrictions and minor, Proceedings of the Seventh Southeastern Conference on Combinatorics, Graph Theory, and Computing, pp. 463-468, Utilitas Matematica, Winnipeg, Man. (1976).

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