Determining a Binary Matroid from its Small Circuits

It is well known that a rank-r matroid M is uniquely determined by its circuits of size at most r. This paper proves that if M is binary and r > 3, then M is uniquely determined by its circuits of size at most r − 1 unless M is a binary spike or a special restriction thereof. In the exceptional cases, M is determined up to


Introduction
A matroid M uses an element e or a set X if e ∈ E(M ) or X ⊆ E(M ).Suppose M is non-binary.Bixby [2] showed that if M is 2-connected and and e ∈ E(M ), then M has a U 2,4 -minor using e.Later, Seymour [7] showed that if M is 3-connected and e, f ∈ E(M ), then M has a U 2,4 -minor using {e, f }.In addition, he conjectured that if M is 4-connected and e, f, g ∈ E(M ), then M has a U 2,4 -minor using {e, f, g}.Kahn [4] and Coullard [3] gave counterexamples to this conjecture leaving open the problem of characterising all 4-connected non-binary matroids that have a 3-element set that is not used by any U 2,4 -minor (see [6,Problem 15.9.7]).
A rich class of counterexamples to Seymour's conjecture is provided by frame matroids with at least three joints e, f , and g.It is readily checked that, in this case, no circuit contains e, f , and g, and hence, such matroids have no U 2,4 -minor using {e, f, g}.A refinement of Seymour's conjecture is to conjecture that if a matroid M is 4-connected, e, f, g ∈ E(M ), and M has a circuit containing e, f , and g, then M has a U 2,4 -minor using {e, f, g}.However, this conjecture is also false.Counterexamples are given by Kahn [4] and Coullard [3].Their counterexamples are obtained from binary matroids by relaxing circuit-hyperplanes and, indeed, the only known counterexamples to the modified version of Seymour's conjecture are obtained from binary matroids by relaxing circuithyperplanes; Kahn's counterexample relaxes a single circuit-hyperplane; Coullard's example relaxes two circuit-hyperplanes.As a possible approach to solving the modified version of Seymour's conjecture, this paper considers the problem of whether a binary matroid M is uniquely determined by a matroid obtained from M by a sequence of circuit-hyperplane relaxations.
Let J r and 1 be the r × r and r × 1 matrices of all ones.For r 3, let A r be the r × (2r + 1) matrix [I r |J r − I r |1] over GF (2) whose columns are labelled, in order, x 1 , x 2 , . . ., x r , y 1 , y 2 , . . ., y r , t.The vector matroid M [A r ] of this matrix is called the rank-r binary spike with tip t.For each i in {1, 2, . . ., r}, the set {t, \t is called the rank-r tipless binary spike.Its legs are the sets {x 1 , y 1 }, {x 2 , y 2 }, . . ., {x r , y r }.Throughout this paper, we will use the term binary spike to include binary spikes with tips as well as tipless binary spikes.
The following is the main result of the paper.As we shall see, most of the effort in proving this result is devoted to verifying the last sentence.
Theorem 1.For r 3, let M be a rank-r binary matroid on a given ground set, and suppose that si(M ) is not isomorphic to U r−1,r ⊕ U 1,1 or U r,r+1 .Then M is uniquely determined by its circuits of size at most r − 1 unless si(M ) is isomorphic to M (K 3,2 ) or can be obtained from a binary spike by deleting at most r − 1 elements no two of which belong to the same leg.In the exceptional cases, M is uniquely determined up to isomorphism.
Let M be a rank-r binary spike with r in {3, 4}.If r = 3 and M has a tip, then any set of three lines through a common point can be chosen as the legs of the spike.If r = 4 and M is tipless, then M ∼ = AG(3, 2) and again there are seven different choices for the sets of legs.In these, the only two, cases where there are choices for the sets of legs, the assertion in the theorem that the deleted elements can all be chosen from different legs means that there is a choice of legs for which this is true rather than that this is true for all choices of the legs.
For arbitrary r exceeding 2, consider M [A r ], the binary spike with tip.Let i be an element of {1, 2, . . ., r} and let {u i , v i } = {x i , y i }.It is well known [5, p.66] and easily checked that the dual of M [A r ]\u i , t is isomorphic to the rank-(r − 1) binary spike with tip v i .In each of M [A r ]\t, u i and M [A r ]\u i , we call v i the cotip.The last two matroids, which are well known to be unique up to isomorphism, are called, respectively, the rank-r binary spike with a cotip and no tip, and the rank-r binary spike with a tip and a cotip.
Theorem 1 is a strengthening, for binary matroids, of the well-known fact that an arbitrary matroid is uniquely determined by its non-spanning circuits.Observe that, unless n = 1 or n = 2, an n-element rank-1 or rank-2 binary matroid is not uniquely determined by its set of circuits of size zero or size at most one, respectively.Also, for r 3, the sets of circuits of size at most r − 1 are the same in U r,r+1 and U r−1,r ⊕ U 1,1 .Moreover, for all r 4, no rank-r binary spike is uniquely determined by its circuits of size at most r − 1.To see this, fix r 4, and let M 1 and M 2 be two rank-r binary spikes with tip t and legs {t, x 1 , y 1 }, {t, x 2 , y 2 }, . . ., {t, x r , y r } with the property that {x 1 , x 2 , . . ., x r } is a basis of M 1 and {x 1 , x 2 , . . ., x r−1 , y r } is a basis of M 2 .Since M 1 and M 2 are binary, {x 1 , x 2 , . . ., x r−1 , y r } is a circuit-hyperplane of M 1 , and {x 1 , x 2 , . . ., x r } is a circuit-hyperplane of M 2 .Hence M 1 = M 2 , but M 1 and M 2 have the same sets of circuits of size at most r − 1.Note that, for all r 4, we see the same phenomenon when M 1 and M 2 are both spikes without tips, or are both spikes with tips and cotips, or are both tipless spikes with cotips.However, these exceptions can be eliminated when r 5 if we know at least one circuit of size r or r + 1.
Theorem 2. For r 5, let M be a rank-r binary matroid on a given ground set.Let C + be a fixed circuit of M choosing |C + | r if possible.Then M is uniquely determined by the collection Note that M (K 4 ) and M (K 3,2 ) show that Theorem 2 cannot be extended to allow r to be in {3, 4}.Neither matroid is uniquely determined by any one of its 4-circuits.
The proofs of Theorems 1 and 2 are constructive and rely on the preliminary results in the next section.Indeed, the proof of Theorem 1 is essentially no more than a packaging of these results.Section 3 consists of the proofs of the two theorems.Throughout the paper, notation and terminology follows [6].We shall also freely use the properties of spikes noted there (see, in particular, pp.41, 73, 74, and 111) as well as the well-known fact that if C 1 and C 2 are circuits of a binary matroid, then their symmetric difference, C 1 C 2 , is a disjoint union of circuits.For convenience, whenever we write "determined", we mean "uniquely determined".

Preliminaries
This section consists of five preliminary results.The first, due to Acketa [1], lists all binary paving matroids.We denote by M (K − 4 ) the cycle matroid of the graph obtained from K 4 by deleting an edge.Let e l denote the column of D corresponding to f .Then e l is determined.If |C(e l , B)| = r + 1, then, for k = l, the unique zero in column e k is in row i if and only if {e i , e k , e l } is a circuit.Since r(M ) 5, we can decide if {e i , e k , e l } is a circuit, and so we can determine e k , and thus determine M .Now suppose that |C(e l , B)| = r.Then the column e l has exactly one zero, say in row i.For k = l, the column e k has no zeros if and only if {e i , e k , e l } is a circuit.Furthermore, if the column e k has exactly one zero, then it is in row j if and only if {e i , e j , e k , e l } is a circuit, where i = j.Since r(M ) 5, we can decide if {e i , e k , e l } and {e i , e j , e k , e l } are circuits, and so we can determine e k .Hence M is determined.

Theorem 3 .
An n-element binary matroid is paving if and only if it is isomorphic to one of the following matroids: a loopless rank-2 matroid with at most three parallel classes, the electronic journal of combinatorics 23(1) (2016), #P1.26 Proof of Theorem 2. It follows by Theorem 1 that we may assume M has a circuit C + with |C + | r.Let f ∈ C + .We next determine a basis B of M with C + − {f } ⊆ B. If |C + | = r + 1, then choose B to be C + − {f }.On the other hand, if |C + | = r, then, by Lemma 6, we can find a basis of M containing C + − {f }, in which case, choose B to be this basis.Let B = {e 1 , e 2 , . . ., e r }, and construct a binary representation [I r |D] of M with columns labelled, in order, e 1 , e 2 , . . ., e r , e r+1 , . . ., e n , where n = |E(M )|.We complete the proof by determining the columns of D. Let k ∈ {r + 1, r + 2, . . ., n}.If the fundamental circuit C(e k , B) has size at most r − 1, then the column e k is determined.Observing that such a column has at least two ones and at most r − 2 ones, we see that the columns e k that are not immediately determined have either r − 1 ones or r ones.Since M is binary and simple, there is at most one column e k of D with |C(e k , B)| = r + 1.