On the structure of spikes

Spikes are an important class of 3-connected matroids. For an integer $r\geq 3$, there is a unique binary r-spike denoted by $Z_{r}$. When a circuit-hyperplane of $Z_{r}$ is relaxed, we obtain another spike and repeating this procedure will produce other non-binary spikes. The $es$-splitting operation on a binary spike of rank $r$, may not yield a spike. In this paper, we give a necessary and sufficient condition for the $es$-splitting operation to construct $Z_{r+1}$ directly from $Z_{r}$. Indeed, all binary spikes and many of non-binary spikes of each rank can be derived from the spike $Z_{3}$ by a sequence of The $es$-splitting operations and circuit-hyperplane relaxations.


Introduction
Azanchiler [1], [2] extended the notion of n-line splitting operation from graphs to binary matroids. He characterized the n-line splitting operation of graphs in terms of cycles of the respective graph and then extended this operation to binary matroids as follows. Let M be a binary matroid on a set E and let X be a subset of E with e ∈ X. Suppose A is a matrix that represents M over GF (2). Let A e X be a matrix obtained from A by adjoining an extra row to A with this row being zero everywhere except in the columns corresponding to the elements of X where it takes the value 1, and then adjoining two columns labeled α and γ to the resulting matrix such that the column labeled α is zero everywhere except in the last row where it takes the value 1, and γ is the sum of the two column vectors corresponding to the elements α and e. The vector matroid of the matrix A e X is denoted by M e X . The transition from M to M e X is called an es-splitting operation. We call the matroid M e X as es-splitting matroid.
Let M be a matroid and X ⊆ E(M), a circuit C of M is called an OXcircuit if C contains an odd number of elements of X, and C is an EX-circuit if C contains an even number of elements of X. The following proposition characterizes the circuits of the matroid M e X in terms of the circuits of the matroid M. Proposition 1. [1] Let M = (E, C) be a binary matroid together with the collection of circuits C. Suppose X ⊆ E, e ∈ X and α, γ / ∈ E. Then M e X = (E ∪ {α, γ}, C ′ ) where C ′ = (∪ 5 i=0 C i ) ∪ Λ with Λ = {e, α, γ} and C 0 = {C ∈ C : C is an EX-circuit}; C 1 = {C ∪ {α} : C ∈ C and C is an OX-circuit}; C 2 = {C ∪ {e, γ} : C ∈ C, e / ∈ C and C is an OX-circuit}; C 3 = {(C \ e) ∪ {γ} : C ∈ C, e ∈ C and C is an OX-circuit}; C 4 = {(C \ e) ∪ {α, γ} : C ∈ C, e ∈ C and C is an EX-circuit}; C 5 = The set of minimal members of {C 1 ∪ C 2 : C 1 , C 2 ∈ C, C 1 ∩ C 2 = ∅ and each of C 1 and C 2 is an OX-circuit}.
It is observed that the es-splitting of a 3-connected binary matroid may not yield a 3-connected binary matroid. The following result, provide a sufficient condition under which the es-splitting operation on a 3-connected binary matroid yields a 3-connected binary matroid.

Proposition 2. [4]
Let M be a 3-connected binary matroid, X ⊆ E(M) and e ∈ X. Suppose that M has an OX-circuit not containing e. Then M e X is a 3-connected binary matroid.
To define rank-r spikes, let E = {x 1 , x 2 , ..., x r , y 1 , y 2 ..., y r , t} for some r ≥ 3. Let The set of circuits of every spike on E includes C 1 ∪ C 2 . Let C 3 be a, possibly empty, subsets of {{z 1 , z 2 , ...z r } : z i is in {x i , y i } for all i} such that no two members of C 3 have more than r − 2 common elements. Finally, let C 4 be the collection of all (r + 1)-element subsets of E that contain no member of

Proposition 3. [3]
There is a rank-r matroid M on E whose collection C of circuits is The matroid M on E with collection C of circuits in the last proposition is called a rank-r spike with tip t and legs L 1 , L 2 , ...L r where L i = {t, x i , y i } for all i. In the construction of a spike, if C 3 is empty, the corresponding spike is called the rank-r free spike with tip t. In an arbitrary spike M, each circuit in C 3 is also a hyperplane of M. Evidently, when such a circuit-hyperplane is relaxed, we obtain another spike. Repeating this procedure until all of the circuit-hyperplanes in C 3 have been relaxed will produce the free spike. Now 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 labeled, 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 and denoted by Z r . Oxley [3] showed that all rank-r, 3-connected binary matroids without a 4-wheel minor can be obtained from a binary r-spike by deleting at most two elements.

Circuits of Z r
In this section, we characterize the collection of circuits of Z r . To do this, we use the next well-known theorem.

Theorem 4. [3]
A matroid M is binary if and only if for every two distinct circuits C 1 and C 2 of M, their symmetric difference, C 1 ∆C 2 , contains a circuit of M. Now let M = (E, C) be a binary matroid on the set E together with the set C of circuits where E = {x 1 , x 2 , ..., x r , y 1 , y 2 ..., y r , t} for some r ≥ 3. Suppose Y = {y 1 , y 2 ..., y r }. For k in {1, 2, 3, 4}, we define ϕ k as follows.
Proof. Let M be a matroid on the set E = {x 1 , x 2 , ..., x r , y 1 , y 2 ..., y r , t} such that Then, for every two distinct circuits C 1 and C 2 of ϕ 3 , we have C 1 ∩ Y = C 2 ∩ Y and |C j ∩ {x i , y i }| = 1 for all i and j with 1 ≤ i ≤ r and j ∈ {1, 2}. We conclude that there is at least one y i in C 1 such that y i / ∈ C 2 and so x i is in C 2 but it is not in C 1 . Thus, no two members of ϕ 3 have more than r − 2 common elements. It is clear that every member of ϕ 4 has (r + 1)-elements and contains no member of ϕ 1 ∪ ϕ 2 ∪ ϕ 3 . By Proposition 3, we conclude that M is a rank-r spike. It is straightforward to show that for every two distinct members of C, their symmetric difference contains a circuit of M . Thus, by Theorem 4, M is a binary spike.
It is not difficult to check that if r is odd, then the intersection of every two members of ϕ 3 has odd cardinality and the intersection of every two members of ϕ 4 has even cardinality and if r is even, then the intersection of every two members of ϕ 3 has even cardinality and the intersection of every two members of ϕ 4 has odd cardinality. Clearly, |ϕ 1 | = r, |ϕ 2 | = r(r−1) 2 and |ϕ 3 | = |ϕ 4 | = 2 r−1 . Therefore, every rank-r binary spike has 2 r + r(r+1)

The es-splitting operation on Z r
By applying the es-splitting operation on a given matroid with k elements, we obtain a matroid with k + 2 elements. In this section, our main goal is to give a necessary and sufficient condition for X ⊆ E(Z r ) with e ∈ X, to obtain Z r+1 by applying the es-splitting operation on X. Now suppose that M = Z r be a binary rank-r spike with the matrix representation [I r |J r −I r |1] over GF (2) whose columns are labeled, in order x 1 , x 2 , ..., x r , y 1 , y 2 , ..., y r , t. Suppose ϕ = ϕ 1 ∪ ϕ 2 ∪ ϕ 3 ∪ ϕ 4 be the collection of circuits of Z r defined in section 2. Let X 1 = {x 1 , x 2 , ..., x r } and Y 1 = {y 1 , y 2 , ..., y r } and let X be a subset of E(Z r ). By the following lemmas, we give six conditions for membership of X such that, for every element e of this set, (Z r ) e X is not the spike Z r+1 .
Lemma 6. If r ≥ 4 and t / ∈ X, then, for every element e of X, the matroid (Z r ) e X is not the spike Z r+1 .
Proof. Suppose that t / ∈ X. Without loss of generality, we may assume that there exist i in {1, 2, ..., r} such that x i ∈ X and e = x i . By Proposition 1, the set Λ = {x i , α, γ} is a circuit of (Z r ) e X . Now consider the leg L i = {t, x i , y i }, we have two following cases. (i) If y i ∈ X, then |L i ∩ X| is even. By Proposition 1, the leg L i is a circuit of (Z r ) e X . Now if all other legs of Z r have an odd number of elements of X, by Proposition 1, we observe that these legs transform to circuits of cardinality 4 and 5. So there are exactly two 3-circuit in (Z r ) e X . If not, there is a j = i such that L j is a 3-circuit of (Z r ) e X and (Λ ∩ L i ∩ L j ) = ∅, we conclude that in each case, for every element e of X, the matroid (Z r ) e X is not the spike Z r+1 . Since Z r+1 has r + 1 legs and the intersection of the legs of Z r+1 is non-empty.
(ii) If y i / ∈ X, then |L i ∩ X| is odd. By Proposition 1, (L i \ x i ) ∪ γ is a circuit of (Z r ) e X . Now if there is the other leg L j such that |L j ∩ X| is even, then L j is a circuit of (Z r ) e X . But (L j ∩ Λ ∩ ((L i \ x i ) ∪ γ)) = ∅, so (Z r ) e X is not the spike Z r+1 . We conclude that every leg L j with j = i has an odd number of elements of X. Since x i / ∈ L j , by Proposition 1 again, L j is not a 3-circuit in (Z r ) e X . Therefore, (Z r ) e X has only two 3-circuits and so, for every element e of X, the matroid (Z r ) e X is not the spike Z r+1 .
Lemma 7. If r ≥ 4 and e = t, then, for every element e of X − t, the matroid (Z r ) e X is not the spike Z r+1 . Proof. Suppose that e = t. Without loss of generality, we may assume that there exist i in {1, 2, ..., r} such that x i ∈ X and e = x i . By Proposition 1, the set Λ = {x i , α, γ} is a circuit of (Z r ) e X and by Lemma 6, to obtain Z r+1 , the element t is contained in X. Now consider the leg L i = {t, x i , y i }. We have two following cases.
is not the spike Z r+1 . We conclude that every leg L j with j = i has an odd number of elements of X. Since x i / ∈ L j , by Proposition 1 again, L j is not a 3-circuit in (Z r ) e X . Therefore (Z r ) e X has only two 3-circuit and so (Z r ) e X is not the spike Z r+1 . (ii) If y i / ∈ X, then |L i ∩ X| is even. So L i is a circuit of (Z r ) e X . By similar arguments in Lemma 6 (i), one can show that for every element e of X − t, the matroid (Z r ) e X is not the spike Z r+1 . Next by Lemmas 6 and 7, to obtain the spike Z r+1 , we take t in X and e = t.
Lemma 8. If r ≥ 4 and there is a circuit C of ϕ 3 such that |C ∩ X| is even, then the matroid (Z r ) t X is not the spike Z r+1 . Proof. Suppose that C is a circuit of Z r such that is a member of ϕ 3 and |C ∩ X| is even. Then, by Proposition 1, the circuit C is preserved under the es-splitting operation. So C is a circuit of (Z r ) t X . But |C| = r. Now if r > 4, then C cannot be a circuit of Z r+1 , since it has no r-circuit, and if r = 4, then, to preserve the members of ϕ 2 in Z 4 under the es-splitting operation and to have at least one member of ϕ 3 which has even number of elements of X, the set X must be E(Z r ) − t or t. But in each case (Z 4 ) t X has exactly fourteen 4-circuits, so it is not the spike Z 5 , since this spike has exactly ten 4-circuit. We conclude that the matroid (Z r ) t X is not the spike Z r+1 . Lemma 9. If r ≥ 4 and |X ∩ {x i , y i }| = 2, for i in {1, 2, ..., r}, then the matroid (Z r ) t X is not the spike Z r+1 unless r is odd and for all i, {x i , y i } ⊂ X, in which case Z r+1 has γ as a tip.
Proof. Suppose that {x i , y i } ⊂ X for i ∈ {1, 2, ..., r}. Since t ∈ X and e = t, after applying the es-splitting operation, the leg {t, x i , y i } turns into two circuits {t, x i , y i , α} and {x i , y i , γ}. Now consider the leg L j = {t, x j , y j } where j = i. If |L j ∩ X| is even (this means {x j , y j } X), then L j is a circuit of (Z r ) e X . But {x i , y i , γ} ∩ {t, x j , y j } = ∅ and this contradicts the fact that the intersection of the legs of a spike is not the empty set. So {x j , y j } must be a subset of X. We conclude that {x k , y k } ⊂ X for all k = i. Thus X = E(Z r ). But in this case, r cannot be even since every circuit in ϕ 3 has even cardinality and by Lemma 8, the matroid (Z r ) t X is not the spike Z r+1 . Now we show that if X = E(Z r ), and r is odd, then (Z r ) t X is the spike Z r+1 with tip γ. Clearly, every leg of Z r has an odd number of elements of X. Using Proposition 1, after applying the es-splitting operation, we have the following changes. For i in {1, 2, ..., r}, L i transforms to two circuits (L i \ t) ∪ γ and L i ∪ α, every member of ϕ 2 is preserved, and if C ∈ ϕ 3 , then C ∪ α and C ∪ {t, γ} are circuits of (Z r ) t X . Finally, if C ∈ ϕ 4 , then C and (C \ t) ∪ {α, γ} are circuits of (Z r ) t X . Note that, since X = E(M ) with e = t, there are no two disjoint OX-circuits in Z r such that their union be minimal. Therefore the collection C 5 in Proposition 1 is empty. Now suppose that α and t play the roles of x r+1 and y r+1 , respectively, and γ plays the role of tip. Then we have the spike Z r+1 with tip γ whose collection ψ of circuits is ψ 1 ∪ ψ 2 ∪ ψ 3 ∪ ψ 4 where In the following lemma, we shall use the well-known facts that if a matroid M is n-connected with E(M ) ≥ 2(n − 1), then all circuits and all cocircuits of M have at least n elements, and if A is a matrix that represents M over GF (2), then the cocircuit space of M equals the row space of A.
Lemma 10. If |X| ≤ r, then the matroid (Z r ) t X is not the spike Z r+1 .
the es-splitting operation L j transforms to {x j , y j , γ}. Now let L k = {t, x k , y k } be another leg of Z r . If |L k ∩ X| is even, then L k is a circuit of (Z r ) t X . But (L k ∩ Λ ∩ {x j , y j , γ}) = ∅. Hence, in this case, the matroid (Z r ) t X is not the spike Z r+1 . We may now assume that every other leg of Z r has an odd number of elements of X. Then, for all j = i, the elements x j and y j are not contained in X. We conclude that |X| = 1 and in the last row of the matrix that represents the matroid (Z r ) t X there are two entries 1 in the corresponding columns of t and α. Hence, (Z r ) t X has a 2-cocircuit and it is not the matroid Z r+1 since spikes are 3-connected matroids.
By Lemmas 9 and 10, we must check that if |x| = r + 1, then, by using the es-splitting operation, can we build the spike Z r+1 ?
Lemma 11. If r ≥ 4 and |X ∩ X 1 | be odd, then the matroid (Z r ) t X is not the spike Z r+1 .
Proof. Suppose that r is even and |X ∩ X 1 | is odd. Since t ∈ X and |X| = r + 1, so |X ∩ Y 1 | must be odd. Therefore the set X must be C ∪ t where C ∈ ϕ 3 . But |C ∩ X| is even and by Lemma 8, the matroid (Z r ) t X is not the spike Z r+1 . Now Suppose that r is odd, r ≥ 4 and |X ∩ X 1 | is odd. Then |X ∩ Y 1 | must be even and so X = C where C ∈ ϕ 4 . By definition of binary spikes, there is a circuit C ′ in ϕ 4 such that C ′ = C∆{x i , y i , x j , y j } for all i and j with 1 ≤ i < j ≤ r. Clearly, |(E − C ′ ) ∩ X| = 2. Since (E − C ′ ) is a circuit of Z r and is a member of ϕ 3 , by Lemma 8, the matroid (Z r ) t X is not the spike Z r+1 . Now suppose that M is a binary rank-r spike with tip t and r ≥ 4. Let X ⊆ E(M ) and e ∈ X and let E(M ) − E(M e X ) = {α, γ} such that {e, α, γ} is a circuit of M e X . Suppose ϕ = ∪ 4 i=0 ϕ i be the collection of circuits of M where ϕ i is defined in section 2. With these preliminaries, the next two theorems are the main results of this paper.
Theorem 12. Suppose that r is an even integer greater than three. Let M be a rank-r binary spike with tip t. Then M e X is a rank-(r + 1) binary spike if and only if X = C where C ∈ ϕ 4 and e = t.
Proof. Suppose that M = Z r and X ⊆ E(M ) and r is even. Then, by combining the last six lemmas, |X| = r + 1; and X contains an even number of elements of X 1 with t ∈ X. The only subsets of E(Z r ) with these properties are members of ϕ 4 . Therefore X = C where C ∈ ϕ 4 and by Lemma 7, e = t. Conversely, let X = C where C ∈ ϕ 4 . Then, by using Proposition 1, every leg of Z r is preserved under the es-splitting operation since they have an even number of elements of X.
Evidently, if C ∈ ϕ 4 , then |C ∩ X| is odd and by Proposition 1, C ∪ α and (C \ t) ∪ γ are circuits of M t X . Moreover, there are no two disjoint OX-circuit in ϕ. So the collection C 5 in Proposition 1 is empty. To complete the proof, suppose that α and γ play the roles of x r+1 and y r+1 , respectively, then we have the spike Z r+1 with collection of circuits ψ = ψ 1 ∪ ψ 2 ∪ ψ 3 ∪ ψ 4 where Theorem 13. Suppose that r is an odd integer greater than three. Let M be a rank-r binary spike with tip t. Then M e X is a rank-(r + 1) binary spike if and only if X = C ∪ t where C ∈ ϕ 3 or X = E(M ), and e = t.
Proof. Suppose that M = Z r and X ⊆ E(M ). Let X = E(M ). Then, by Lemma 9, the matroid M t X is the spike Z r+1 with tip γ. Now, by combining the last six lemmas., |X| = r + 1 and X contains an even number of elements of X 1 with t ∈ X. The only subsets of E(Z r ) with these properties are in {C ∪ t : C ∈ ϕ 3 }. Conversely, let X = C ∪ t where C ∈ ϕ 3 . Clearly, every member of ϕ 3 contains an odd number of elements of X. Now let C ′ be a member of ϕ 4 . If C ′ = E(Z r ) − C, then C ′ contains an odd number of elements of X. If C ′ = E(Z r )−C, then there is a C ′′ ∈ ϕ 3 such that C ′ = E(Z r ) − C ′′ . Therefore |C ∩ C ′ | = |C ∩ (E(Z r ) − C ′′ )| = |C − (C ∩ C ′′ )| and so |C ∩ C ′ | is even. So C ′ contains an odd number of elements of X and, by Proposition 1 again C ′ ∪ α and (C \ t) ∪ γ are circuits of M t X . Evidently, if C 1 and C 2 be disjoint OX-circuits of Z r , then one of C 1 and C 2 is in ϕ 3 and the other is in ϕ 4 where C 2 = E(Z r ) − C 1 , as C 1 ∪ C 2 is not minimal, it follows by Proposition 1 that C 5 is empty. Now if α and γ play the roles of x r+1 and y r+1 , respectively. Then M t X is the spike Z r+1 with collection of circuits ψ = ψ 1 ∪ ψ 2 ∪ ψ 3 ∪ ψ 4 where Remark 14. Note that the binary rank-3 spike is the Fano matroid denoted by F 7 . It is straightforward to check that any one of the seven elements of F 7 can be taken as the tip, and F 7 satisfies the conditions of Theorem 13 for any tip. So, there are exactly 35 subset X of E(F 7 ) such that (F 7 ) e X is the binary 4-spike where e is a tip of it. Therefore, by Theorem 13, these subsets are X = E(F 7 ) for every element e of X and C ∪ z for every element z in E(F 7 ) not contained in C with e = z where C is a 3-circuit of F 7 .