Rank Connectivity and Pivot-Minors of Graphs

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For one of its applications, we present a result on the size of excluded pivot-minors for graphs of rank-width k. Rank-width is a width parameter of graphs introduced by Oum and Seymour [12]. An excluded pivot-minor for graphs of rank-width at most k is a pivot-minor-minimal graph having rank-width larger than k. As a corollary of Theorem 4.1, we prove the following in Section 7. • Every excluded pivot-minor for graphs of rank-width at most k has at most 7 k+1 ( /12 · 6 − 1)/5 vertices for k ≥ 2. • Every excluded pivot-minor for graphs of rank-width at most 2 has at most 16 vertices. These are improvements of a previous result [10], showing that such graphs have at most (6k+1 − 1)/5 vertices. For graphs of rank-width 1, the list of excluded pivot-minors and vertex-minors are known exactly, see Oum [11], but the list is not known for graphs of rank-width 2. Now it may be possible to search all graphs up to 16 vertices to determine the list of excluded pivot-minors for the class of graphs of rank-width at most 2. There are other chain theorems known in other contexts. One of the well-known chain theorems is Tutte’s Wheels and Whirls Theorem for matroids. Allys [1] proved the following chain theorem on 2-rank-connected graphs, which can be considered as a generalization of Tutte’s Wheels and Whirls Theorem in a certain sense. (See Geelen [7] for an alternative proof.) Theorem 1.1 (Allys [1, Theorem 4.3]). Let G be a prime graph with at least 5 vertices. Then G has a prime pivot-minor H such that |V (H)| = |V (G)|− 1, unless G is pivot- equivalent to a cycle. Our goal is to explore chain theorems to higher rank connectivity. Our main theorem is motived by the following theorem on internally 4-connected matroids by Hall [8]. For the definitions on matroids, we refer to the book of Oxley [13]. Theorem 1.2 (Hall [8, Theorem 3.1]). Let M be an internally 4-connected matroid, and let {a, b, c} be a triangle of M. Then at least one of the following hold. (1) At least one of M \ a, M \ b and M \ c is 4-connected up to 3-separators of size 4. (2) At least two of M \ a, M \ b and M \ c are 4-connected up to 3-separators of size 5. The above theorem has been used later in several papers [5, 6] to prove stronger theorems for internally 4-connected matroids. Why is it possible to find theorems on the rank connectivity of graphs analogous to those on the matroid connectivity? We do not need matroids for the proofs in the paper, but there are interesting connections between bipartite graphs and binary matroids, fully discussed in [10]. Let M be the binary matroid on a finite set E = X ∪ Y with a binary representation X Y 1 1   X . A . ..    1    Then in the matroid M, the rank rM (S) of a set S is the dimension of the vector space spanned by column vectors indexed by S. The connectivity function λM (S) of M is RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 3 defined as λM (S) = rM (S)+ rM (E − S) − rM (E) for all S ⊆ E. A matroid M is n-connected if for all integers k

2. Preliminaries: Submodularity and Pivoting For an X × Y matrix M and subsets U of X and V of Y , we write M[U, V ] to denote the U × V submatrix of M. The following proposition is well known; for instance, see [9, Proposition 2.1.9], [17, Lemma 2.3.11], or [16].

Proposition 2.1. Let M be an X × Y matrix. Let X1, X2 ⊆ X and Y1,Y2 ⊆ Y . Then rank(M[X1,Y1])+rank(M[X2,Y2]) ≥ rank(M[X1∩X2,Y1∪Y2])+rank(M[X1∪X2,Y1∩Y2]).

All graphs in this paper are simple. For a graph G = (V, E), Let AG be the adjacency matrix of a graph G over the binary ﬁeld GF(2), that is a V × V matrix such that the (i, j) entry is 1 if and only if i is adjacent to j in G. For subsets X and Y of V , Let ρG(X,Y ) = rank(AG[X,Y ]). The cut-rank function ρG(X) is ρG(X,V (G) − X). We will sometimes write ρ for ρG if it is not ambiguous. A partition (A, B) of the vertex set of a graph G is called a split if ρG(A) ≤ 1 and min(|A|, |B|) ≥ 2. By Proposition 2.1, one can easily show the following two lemmas, which will be very useful later. Lemma 2.2. Let G be a graph and let a, b be distinct vertices. Let A ⊆ V (G) − {a} and B ⊆ V (G) − {b}. If b∈ / A and a∈ / B, then

(1) ρG(A ∩ B)+ ρG\a\b(A ∪ B) ≤ ρG\a(A)+ ρG\b(B). If b ∈ A and a∈ / B, then

(2) ρG\b(A ∩ B)+ ρG\a(A ∪ B) ≤ ρG\a(A)+ ρG\b(B). If b ∈ A and a ∈ B, then

(3) ρG\a\b(A ∩ B)+ ρG(A ∪ B) ≤ ρG\a(A)+ ρG\b(B). 4 SANG-IL OUM

Lemma 2.3. Let G be a graph and let x be a vertex of G. Let X,Y ⊆ V (G) − {x}. Then

(4) ρG\x(X ∩ Y )+ ρG(X ∪ Y ∪ {x}) ≤ ρG\x(X)+ ρG(Y ∪ {x}),

(5) ρG(X ∩ Y )+ ρG\x(X ∪ Y ) ≤ ρG\x(X)+ ρG(Y ). We deﬁned k+ℓ-rank-connected graphs in Section 1. Proposition 2.4. Let k be a positive integer. Let G = (V, E) be a k+0-rank-connected graph with at least 2k vertices. Let v ∈ V . Then both G and G \ v are (k − 1)+0-rank- connected. Consequently if G is k-rank-connected and |V | ≥ 2k, then G \ v is (k − 1)-rank- connected for all v ∈ V . Proof. Let us ﬁrst prove that G\v is (k−1)+0-rank-connected for each v ∈ V . Let (X,Y ) be a partition of V − {v} such that |X|, |Y | ≥ k − 1 and ρG\v(X) < k − 1. We deduce that |X| < k because ρ(X,Y ∪ {v}) < k. Similarly |Y | < k because ρ(X ∪ {v},Y ) < k. Then |V | = |X| + |Y | + 1 < 2k, a contradiction. Now let us assume that G is not (k − 1)+0-rank-connected. Suppose that (X,Y ) be a partition of V such that |X|, |Y | ≥ k − 1 and ρG(X) < k − 1. We may assume that |X| ≤ |Y |. Since |V | ≥ 2k, |Y | ≥ k. We pick v ∈ Y . Then |Y − {v}| ≥ k − 1 and +0 ρG\v(X) < k − 1; a contradiction because G \ v is (k − 1) -rank-connected. For a vertex v of a graph G, let N(v) be the set of neighbors of v and let N[v] = N(v) ∪ {v}. Pivoting an edge vw of a graph G is an operation to obtain a new graph denoted by G ∧ vw from G by the following procedures: (i) For each (x,y) ∈ (N(v) − N(w) − {w}) × (N(w) − N(v) − {v}), (x,y) ∈ (N(v) − N(w) − {w}) × (N(w) ∩ N(v)), or (x,y) ∈ (N(v) ∩ N(w)) × (N(w) − N(v) − {v}), we remove the edge xy if x, y are adjacent in G and we add a new edge xy if x, y are non-adjacent. (ii) Swap the labels of v and w; in other words, make NG(w) the set of neighbors of v in G ∧ vw, and make NG(v) the set of neighbors of w in G ∧ vw. A graph H is pivot-equivalent to G if H is obtained from G by a sequence of pivots. A graph H is a pivot-minor of G if H is an induced subgraph of a graph pivot-equivalent to G. Pivots preserve the cut-rank functions (see [10, Proposition 2.6]); ρG(X)= ρH (X) for all sets X if G and H are pivot-equivalent. It is easy to verify that if y1,y2 ∈ N(x), then G ∧ xy1 = (G ∧ xy2) ∧ y1y2; see [10, Proposition 2.5]. Since (G ∧ xy1) \ x is pivot-equivalent to (G ∧ xy2) \ x, we often write G/x to denote (G ∧ xy) \ x for some neighbor y of x, when there is no confusion. (If x has no neighbors, we let G/x = G \ x.) We have to pay attention to cause no confusions; for instance, if y, z ∈ N(x), then G/x \ y is somewhat ambiguous because (G ∧ xy) \ x \ y and (G ∧ xz) \ x \ y are not always pivot-equivalent. The following lemma is in [10].

Lemma 2.5 ([10, Lemma 4.4]). Let G be a graph. Let v be a vertex in G. Let (X1, X2) and (Y1,Y2) be partitions of V (G) − {v}. Then

ρG\v(X1)+ ρG/v(Y1) ≥ ρG(X1 ∩ Y1)+ ρG(X2 ∩ Y2) − 1. RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 5

Proof. In [10, Proposition 4.3], it is shown that ρG/v(Y1)= ρG(Y1 ∪ {v},Y2 ∪ {v}) − 1 if v has at least one neighbor. So if v is not an isolated vertex, then we obtain the desired inequality from Proposition 2.1. If v is an isolated vertex, then G/v = G\v and therefore ρG\v(X1)+ ρG\v(Y1) ≥ ρG\v(X1 ∩ Y1)+ ρG\v(X2 ∩ Y2) > ρG(X1 ∩ Y1)+ ρG(X2 ∪ Y2) − 1 by Proposition 2.1. By Lemma 2.5, we deduce the following proposition, motivated by Bixby’s lemma [2]. Proposition 2.6. Let G be a k+ℓ-rank-connected graph and let v be a vertex in G. Then either G \ v or G/v is k+(2ℓ+k−1)-rank-connected. Proof. Suppose that both G \ v and G/v are not k+(2ℓ+k−1)-rank-connected. There exist partitions (X1, X2) and (Y1,Y2) of V (G) − {v} such that ρG\v(X1) < k, ρG/v(Y1) < k and |X1|, |X2|, |Y1|, |Y2| ≥ 2k + 2ℓ − 1. By swapping Y1 and Y2 if necessary, we may +ℓ assume that |X1 ∩ Y1| ≥ k + ℓ. Since G is k -rank-connected, ρG(X1 ∩ Y1) ≥ k. By +ℓ Lemma 2.5, ρG(X2 ∩ Y2) < k. Since G is k -rank-connected, |X2 ∩ Y2| < k + ℓ. By applying Lemma 2.5 with (X1, X2) and (Y2,Y1), we deduce that either ρG(X1 ∩ Y2) < k or ρG(X2 ∩ Y1) < k. If ρG(X1 ∩ Y2) < k, then |X1 ∩ Y2| < k + ℓ, and therefore |Y2| = |X1 ∩ Y2| + |X2 ∩ Y2| < 2k + 2ℓ − 1. Similarly, if ρG(X2 ∩ Y1) < k, then |X2 ∩ Y1| < k + ℓ, and therefore |X2| < 2k + 2ℓ − 1. This is a contradiction. 3. Generating prime pivot-minors

A set X of vertices of a graph G is fully closed if ρG(X ∪ {v}) > ρG(X) for all v ∈ V (G) − X. Our goal in this section is to prove the following proposition. Proposition 3.1. Let G = (V, E) be a prime graph with |V |≥ 8. Suppose that G has a fully closed set A such that ρG(A) ≥ 2. Then A has a vertex v such that either G \ v or G/v is prime. We can not omit the condition that A is fully closed; for instance, if G is a cycle of length at least 6, then neither G \ v nor G/v is prime for a vertex v of G. Let us discuss how to prove Proposition 3.1. By Proposition 2.6, for each vertex v of G, either G \ v or G/v is 2+1-rank-connected. The rest of this section is devoted to ﬁnd a vertex v so that G \ v or G/v is prime. Lemma 3.2. Let G = (V, E) be a prime graph with |V | ≥ 7 and let a ∈ V . Suppose that G \ a is 2+1-rank-connected and let b, c be distinct vertices of G \ a such that +1 ρG\a({b, c}) ≤ 1. Suppose that G \ b is 2 -rank-connected but not prime. (i) If c has a neighbor other than a and b, then G \ a \ b is prime. (ii) If |V | > 7 and c is adjacent to only a and b, then G \ c is prime. (iii) If |V | = 7, c is adjacent to only a and b, and G \ c is not prime, then |A| = |B| = 3 for every split (A, B) of G \ c, and (G ∧ ac) \ a \ c is prime. Proof. Let us ﬁrst prove (i). Since G \ b is not prime, there exists a set X such that ρG\b(X) ≤ 1 and |X|≥ 2, |V |− 1 − |X|≥ 2. We may assume that c ∈ X. If a ∈ X, then By (3) of Lemma 2.2 with ρG\a({b, c}) = 1, ρG\b(X) ≤ 1, we deduce that ρG(X∪{b}) ≤ 1; a contradiction because G is prime. So a∈ / X. By (2) of Lemma 2.2, we deduce that +1 ρG\a(X ∪ {b}) ≤ 1. Since G \ a is 2 -rank-connected, |V − X| = 2. Let V − X = {a, d}. So ρG\b({a, d}) ≤ 1. 6 SANG-IL OUM

To prove that G \ a \ b is prime, let us assume on the contrary that G \ a \ b has a split (A, B). We may assume that c ∈ A. From (4) of Lemma 2.3, we deduce that ρG\a\b({c})+ ρG\a(A ∪ {b}) ≤ ρG\a\b(A)+ ρG\a({b, c}) ≤ 2. Since c has a neighbor other than a and b, we have ρG\a(A ∪ {b}) ≤ 1. Since G \ a is 2+1-rank-connected, we have |B| = 2. If d ∈ A, then ρG\a\b({d})+ ρG(A ∪ {a, b}) ≤ ρG\a(A ∪ {b})+ ρG\b({a, d}) ≤ 2. Since |A| ≥ 2 and |B| ≥ 2, we have ρG(A ∪ {a, b}) ≥ 2 and therefore ρG\a\b({d}) = 0. Then by (4) of Lemma 2.3, ρG\a({b, c, d}) ≤ ρG\a\b({d})+ ρG\a({b, c}) ≤ 1. This contracts to the assumption that G \ a is 2+1-rank-connected. Therefore we may assume that d∈ / A. We know that ρG\a(B) = ρG\a(A ∪ {b}) ≤ 1. By (2) of Lemma 2.2, ρG\a({d})+ ρG\b(B ∪ {a}) ≤ ρG\a(B)+ ρG\b({a, d}) ≤ 2. Since |B ∪ {a}| ≥ 3 and G \ b is 2+1-rank-connected, we deduce that |A| = 2. Then, |V | ≤ |A| + |B| + 2 ≤ 6, a contradiction. This proves that if c has a neighbor other than a and b, then G \ a \ b is prime. To prove (ii) and (iii), let us assume that c has no neighbors other than a and b. Suppose that G \ c is not prime. Let A, B be subsets of V (G \ c) such that B = V − (A ∪ {c}), |A| ≥ 2, |B| ≥ 2, and ρG\c(A) ≤ 1. We may assume that a ∈ A. Since ρG(A ∪ {a}) ≥ 2, we have b ∈ B. From (2) of Lemma 2.2, ρG\c({b})+ ρG\a(B ∪ {c}) ≤ ρG\a({b, c})+ ρG\c(B) ≤ 2. We +1 deduce that ρG\a(B ∪ {c}) ≤ 1. Since G \ a is 2 -rank-connected, |A − {a}| ≤ 2 and therefore |A|≤ 3. We also know that ρG\b({a, c}) = 1 because the degree of c is 2. From (2) of Lemma 2.2, ρG\c({a}) + ρG\b(A ∪ {c}) ≤ ρG\b({a, c}) + ρG\c(A) ≤ 2. We conclude that |B − {b}| ≤ 2 because G \ b is 2+1-rank-connected. Then |V | = |A| + |B| + 1 ≤ 7. So we may assume that |V | = 7, G \ c is not prime, and every split (A, B) of G satisﬁes |A| = |B| = 3. We omit the remaining ﬁnite case checking to see why (G ∧ ac) \ a \ c is prime. We will show that reducing by one vertex is enough when there is a good fully closed set. Lemma 3.3. Let G = (V, E) be a prime graph. Let A be a fully closed set and let v ∈ A. If X ⊆ V − {v}, |X| = 2, and ρG\v(X) ≤ 1, then |X ∩ A|= 6 1.

Proof. By the submodular inequality, we obtain ρG\v(X ∩ A)+ ρG(X ∪ A) ≤ ρG\v(X)+ ρG(A) ≤ ρG(A) + 1 by (4) of Lemma 2.3. Suppose that |X ∩ A| = 1. Since A is fully closed, ρG(X ∪ A) > ρG(A). Thus, ρG\v(X ∩ A) = 0, contradictory to the fact that G \ v is connected. Lemma 3.4. Let G = (V, E) be a prime graph with |V | ≥ 4. Let A be a fully closed set with |A|≥ 2 and let v ∈ A. Suppose that G \ v is 2+1-rank-connected but not prime. Then there is a subset X of A − {v} such that |X| = 2 and ρG\v(X) ≤ 1. Proof. If |V | = 4, then it is trivial because every set X of 2 vertices of G\v would satisfy +0 ρG\v(X) ≤ 1. Therefore we may assume that |V |≥ 5. Since G \ v is 1 -rank-connected by Proposition 2.4, G\v is not 2+0-rank-connected. So there exists a subset X of V −{v} such that ρG\v(X) ≤ 1 and |X| = 2. By Lemma 3.3, we may assume that X ∩ A = ∅. Then ρG(A) ≥ 2, because G is prime and |V − A| ≥ |A|≥ 2. RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 7

′ We claim that we may assume that ρG(A) = 2. Suppose not. Let A be a maximal ′ ′ ′ subset of A such that ρG(A ) = 2. Let w ∈ V − A . If w ∈ A, then ρG(A ∪ {w}) > 2 ′ ′ because G is prime and A is chosen maximally. If w ∈ V −A, then ρG(A)+ρG(A ∪{w}) ≥ ′ ′ ′ ρG(A ∪ {w})+ ρG(A ) and ρG(A ∪ {w}) > ρG(A ) because A is fully closed. Therefore ′ ′ A is a fully closed subset of A. So if ρG(A) > 2, then we may replace A with A and assume that ρG(A) = 2. Let B = V − A. By Lemma 2.3,

ρG\v(X)+ ρG(B) ≥ ρG(X)+ ρG\v(B).

Since G is prime, we have ρG(X) = 2 and therefore 2 = ρG(A) ≥ ρG\v(A−{v})+1. Since +1 G\v is 2 -rank-connected, |A| = 3. Therefore ρG\v(A−{v}) ≤ 1 and |A−{v}| = 2.

Proof of Proposition 3.1. Suppose A has no such vertex. Let v1 ∈ A. We may assume +1 ′ that G \ v1 is 2 -rank-connected, because otherwise G = G ∧ v1w1 for a neighbor w1 +1 of v1 is 2 -rank-connected. Since G \ v1 is not prime, there exists a set X such that

|X| = 2, X ⊆ A − {v1}, and ρG\v1 (X) ≤ 1 by Lemma 3.4. Let X = {v0, v2}. Since v0 has at least two neighbors, let w0 be a neighbor of v0 such that w0 6= v1. Then one of +1 +1 the graphs G \ v0 or G ∧ v0w0 \ v0 is 2 -rank-connected. If G \ v0 is 2 -rank-connected, then a sequence v0, v1 in a graph G has the desired property that both G \ v0 and G \ v1 +1 +1 are 2 -rank-connected. If (G ∧ v0w0) \ v0 is 2 -rank-connected, then a sequence v0, v1 in a graph G ∧ v0w0 has the desired property. So we may assume that both G \ v0 and +1 G \ v1 are 2 -rank-connected. By (ii) of Lemma 3.2, v2 has a neighbor other than v0 +1 and v1. Then either G \ v2 or (G ∧ v2w2) \ v2 is 2 -rank-connected. So we may assume +1 that G \ v0, G \ v1, and G \ v2 are 2 -rank-connected and ρG\v1 ({v0, v2}) ≤ 1. Therefore we can choose a sequence v0, v1, . . . , vk ∈ A of distinct vertices of a graph ′ ′ ′ G pivot-equivalent to G such that ρG \vi ({vi−1, vi+1}) ≤ 1 for all 1 ≤ i < k and G \ v0, ′ ′ +1 G \ v1, . . ., G \ vk are 2 -rank-connected and k is maximum. We may assume that G = G′. If vi−1 is adjacent to only vi and vi+1, then G \ vi−1 is prime by Lemma 3.2. So vi−1 has at least one neighbor other than vi and vi+1. Similarly vi+1 has at least one neighbor other than vi and vi−1. Thus vi−1 and vi+1 are twins in G \ vi. (Two vertices x,y of G are called twins if no vertex in V (G) − {x,y} is adjacent to only one of x and y.)

Since G \ vk is not prime, there is a set X such that ρG\vk (X) ≤ 1 and |X| = 2. By Lemma 3.4, we may assume that X ⊆ A. We claim that X ∩ {vk−2, vk−1} = {vk−1}. If X ∩ {vk−2, vk−1} = {vk−2}, then +1 ρG\vk−1 (X ∪ {vk}) ≥ 2 because G \ vk−1 is 2 -rank-connected. Then

2 ≥ ρG\vk−1 ({vk−2, vk})+ ρG\vk (X) ≥ ρG\vk ({vk−2})+ ρG\vk−1 (X ∪ {vk}) ≥ 3; a contradiction. If X ∩ {vk−2, vk−1} = ∅ or X ∩ {vk−2, vk−1} = {vk−2, vk−1}, then let Z = X or

Z = V (G \ vk) \ X so that Z ∩ {vk−2, vk} = {vk−2} and ρG\vk (Z) ≤ 1. Then

2 ≥ ρG\vk−1 ({vk−2, vk})+ ρG\vk (Z) ≥ ρG\vk\vk−1 ({vk−2})+ ρG(Z ∪ {vk}).

Since G is prime and vk−2 has a neighbor other than vk−1 and vk, we have ρG(Z∪{vk}) ≥

2 and ρG\vk\vk−1 ({vk−2}) ≥ 1; a contradiction. Thus we conclude that X∩{vk−2, vk−1} = {vk−1}. 8 SANG-IL OUM

Let vk+1 be a vertex in A such that X = {vk−1, vk+1}. We claim that vk+1 is distinct from v0, v1, . . ., vk. We already proved that vk+1 6= vk−1, vk. Suppose that vk+1 = vi for some 0 ≤ i < k such that i ≡ k + 1 (mod 2). Then for all w∈ / vi, vi+1, . . . , vk−1, vi and vk−1 are twins in G \ vi+1 \ vi+2 \···\ vk−2, because vi+2j and vi+2j+2 are twins in G \ vi+2j+1 for j = 0, 1,..., (k − i − 3)/2. In particular if w = vk, then we deduce that vk is adjacent to both vk−1 and vi or vk is nonadjacent to both vk−1 and vi. If vi has no neighbors other than vk−1 and vk, then by Lemma 3.2, G \ vi is prime, a contradiction. Thus vi has some neighbors other than vk−1 and vk. Similarly vk−1 has some neighbors other than vi and vk. Therefore we deduce that vi and vk−1 are twins in G \ vk. But if so, then ρG({vi, vk−1) ≤ 1, a contradiction to the assumption that G is prime. Now suppose that vk+1 = vi for some 0 ≤ i < k with i ≡ k (mod 2). Then v0, v1, . . ., vk have the same set of neighbors in V (G) − {v0, v1, . . . , vk}. This implies ρG({v0, v1, . . . , vk}) ≤ 1, a contradiction to the assumption that G is prime. This completes the proof of the claim that vk+1 is distinct from v0, v1, . . . , vk. Let X = {v0, v1, . . . , vk, vk+1}. Since G \ vk+1 is not prime, vk+1 has a neighbor other than vk and vk−1 by Lemma 3.2 and therefore vk+1 and vk−1 are twins in G\vk. Thus we can argue that v0, v2,... have the same set N0 of neighbors in V (G) − X and v1, v3,... also have the same set N1 of neighbors in V (G) − X. Since ρG(X) ≥ 2, both N0 and N1 are nonempty. This implies that vk+1 has a neighbor w in V (G) − X. +1 We know that G\vk+1 is not 2 -rank-connected, because k is chosen as the maximum. ′ ′ +1 Let G = G ∧ vk+1w. Then G \ vk+1 is 2 -rank-connected and it is easy to verify that v0, v1, . . . , vk+1 is a longer sequence satisfying all the conditions, which leads to a contradiction. Therefore there must exist a vertex x ∈ A such that G \ x or G/x is prime.

4. Generating prime 3+3-rank-connected pivot-minors Here is our main theorem which provides a proper prime 3+3-rank-connected pivot- minor of a 3+2-rank-connected graph. Theorem 4.1. If G is a prime 3+2-rank-connected graph with at least 10 vertices, then G has a prime 3+3-rank-connected pivot-minor H with one fewer vertices. We consider three following cases: (1) G is 3-rank-connected, (2) G is prime 3+1-rank-connected but not 3-rank-connected, (3) G is prime 3+2-rank-connected but not 3+1-rank-connected. The cases (1) and (3) are easier than the case (2). We will treat those two easy cases in this section. Lemma 4.2. Let G be a 3-rank-connected graph with at least 6 vertices. For each vertex v of G, either G \ v or G/v is prime 3+2-rank-connected. Proof. This follows from Propositions 2.4 and 2.6. Lemma 4.3. Let G be a prime 3+2-rank-connected graph with at least 10 vertices. If G is not 3+1-rank-connected, then there exists a vertex x of G such that G \ x or G/x is prime 3+3-rank-connected. RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 9

Proof. Since G is not 3+1-rank-connected, there exists a subset X = {a, b, c, d} of V (G) +2 such that ρG({a, b, c, d}) = 2. Since |V (G)| ≥ 10 and G is 3 -rank-connected, ρG(X ∪ {v}) > 2 for all v ∈ V (G) − X. Therefore X is fully closed. By Proposition 3.1, we may assume that G \ a is prime. Since G \ a is prime, ρG\a({b, c, d}) ≥ 2. Because ρG\a({b, c, d}) ≤ ρG(X) = 2, we have ρG\a({b, c, d}) = 2. Let us consider a partition (A, B) of V (G\a) such that ρG\a(A) ≤ 2. We may assume that |A ∩ X| ≥ |B ∩ X|. (1) First suppose that b, c, d ∈ A. We claim that |A|≤ 3 or |B|≤ 4. By Lemma 2.3, ρG\a(A)+ ρG(X) ≥ ρG\a(A ∩ X)+ ρG(A ∪ X). Since ρG(X)=2 and ρG\a({b, c, d}) = 2, +2 we have ρG(A ∪ {a}) ≤ ρG\a(A). Since G is 3 -rank-connected, we have |A| ≤ 3 or |B|≤ 4. (2) Now consider the case when |A ∩ X| = 2. We may assume that A ∩ X = {b, c}. By the submodular inequality, ρG\a({b, c, d})+ρG\a (A) ≥ ρG\a(A∪{d})+ρG\a({b, c}). Since G\a is prime, we have ρG\a({b, c}) = 2. Thus, we deduce that ρG\a(A∪{d}) ≤ ρG\a(A). By (1), either |A| + 1 ≤ 3 or |B|− 1 ≤ 4. By (1) and (2), G \ a is 3+3-rank-connected.

5. Triplets To prove Theorem 4.1, we may assume that a graph G is prime 3+1-rank-connected but not 3-rank-connected by Lemmas 4.2 and 4.3. Therefore G must have a set T such +3 that ρG(T ) = 2. Our aim is to prove that G \ x is prime 3 -rank-connected for some x ∈ T . However, we can not achieve this if G \ x is not prime. Therefore we wish to make sure that ρG\x(T − {x}) = 2 for some x ∈ T . It turns out that we can achieve a stronger statement; there exists a graph G′ pivot- equivalent to G such that ρG′\x(T −{x})=2for all x ∈ T . In this section, we will prove this. In a graph G, Let us call a set T of vertices a triplet if |T | = 3, ρG(T ) = 2 and ρG\x(T − {x}) = 2 for each x ∈ T .

Lemma 5.1. Let G be a prime graph and ρG({a, b, c}) = 2. Then there exists a graph G′ pivot-equivalent to G such that {a, b, c} is a triplet of G′. Proof. A graphical proof is possible, but we present a proof using Lemma 2.5. Suppose that {a, b, c} is not a triplet. Let us assume that ρG\c({a, b}) = 1. Let x be a neighbor of c, other than a and b. Such x should exist, because ρG({a, b, c}) = 2. Then by Lemma 2.5,

ρG∧cx\c({a, b})+ ρG\c({a, b}) ≥ ρG({a, b})+ ρG({a, b, c}) − 1 = 3. ′ So ρG∧cx\c({a, b}) = 2. Suppose that {a, b, c} is not a triplet of G = G ∧ cx. We may assume that ρG′\b({a, c}) = 1. We deduce that ρG\b({a, c}) = 1 and therefore ρG\a({b, c}) = 2, because if a matrix with 3 rows has rank 2, then there are two linearly independent rows. Let y be a neighbor of b in G′, other than a and c. Similarly to the above argument, we deduce that ρG′∧by\b({a, c}) = 2. In addition, ρG′∧by\c({a, b}) = ρG′\c({a, b})=2 and ρG′∧by\a({b, c})= ρG′\a({b, c})= ρG\a({b, c}) = 2. 10 SANG-IL OUM

Lemma 5.2. Let G = (V, E) be a prime 3+1-rank-connected graph. If {a, b, c} is a triplet of G, then G \ a, G \ b, and G \ c are prime.

Proof. Suppose that (X,Y ) is a split of G \ a. Since ρG(X ∪ {a}) ≤ 2 and ρG(X) ≤ 2, we have |X| ≤ 2 or |Y | ≤ 2. We may assume that |X| = 2. Since X 6= {b, c}, we may assume that b ∈ Y . If c ∈ X, then ρG\b(X ∪ {a}) + ρG({a, b, c}) ≥ ρG\b({a, c}) + ρG(X ∪ {a, b}) by +1 Lemma 2.3. Since G is 3 -rank-connected, ρG(X ∪ {a, b}) ≥ 3. Therefore ρG\b(X ∪ {a}) ≥ 3. We obtain a contradiction from Lemma 2.2 as follows:

ρG\a(X)+ ρG\b({a, c}) = 3 ≥ ρG\a({c})+ ρG\b(X ∪ {a}) ≥ 4.

Now we assume that c∈ / X. Then ρG\a(Y )+ρG({a, b, c}) ≥ ρG\a({b, c})+ρG(Y ∪{a}) by Lemma 2.3 and therefore ρG(Y ∪ {a}) ≤ 1, contradictory to the assumption that G is prime. Lemma 5.3. Let G = (V, E) be a prime 3+1-rank-connected graph. Let {a, b, c} be a triplet of G. Let (X,Y ) be a partition of V (G\a) such that ρG\a(X) = 2 and |X|, |Y |≥ 5. Then |X ∩ {b, c}| = 1 and ρ(X − {b, c},Y ∪ {a})= ρ(X ∪ {a},Y − {b, c}) = 3.

Proof. If {b, c} ⊆ X, then ρG\a(X) + ρG({a, b, c}) ≥ ρG\a({b, c}) + ρG(X ∪ {a}) by Lemma 2.3 and therefore ρG(X ∪ {a}) ≤ 2, contradictory to the assumption that G is 3+1-rank-connected. Therefore {b, c} 6⊆ X. Similarly {b, c} 6⊆ Y by the symmetry between X and Y . We conclude that |X ∩ {b, c}| = 1. Then ρG(X ∪ {a, b, c}) ≥ 3 because G is 3+1-rank-connected and |Y − {b, c}| ≥ 4. By symmetry, we may assume that b ∈ X and c ∈ Y . Since ρ({a, b, c},Y − {c}) = ρ({a, b},Y − {c}), we have ρ(X ∪ {a},Y − {c}) = ρ(X ∪ {a, c},Y − {c}) ≥ 3, and therefore ρ(X ∪ {a},Y − {c})=3as ρ(X ∪ {a},Y − {c}) ≤ ρ(X,Y − {c}) + 1 ≤ ρ(X,Y ) + 1 = 3. By symmetry between X and Y , ρ(X − {b},Y ∪ {a}) = 3.

6. Prime 3+1-rank-connected graphs To prove the remaining case (2) of Theorem 4.1, we will prove the following. Proposition 6.1. Let G = (V, E) be a prime 3+1-rank-connected graph with a triplet {a, b, c}. Then (i) at least one of G \ a, G \ b, and G \ c is prime 3+2-rank-connected, or (ii) at least two of G \ a, G \ b, and G \ c are prime 3+3-rank-connected. In the remaining of this section we are going to prove Proposition 6.1. We follow the outline of the proof of Hall [8, Theorem 3.1] on internally 4-connected matroids. Let G = (V, E) be a prime 3+1-rank-connected graph with a triplet {a, b, c} that does not satisfy (i). Then there exist partitions (Ab, Ac), (Ba,Bc), and (Ca,Cb) of V (G \ a), V (G \ b), V (G \ c), respectively, such that

• ρ(Ab, Ac)= ρ(Ba,Bc)= ρ(Ca,Cb) = 2, • |Ab|, |Ac|, |Ba|, |Bc|, |Ca|, |Cb|≥ 5, • b ∈ Ab, c ∈ Ac, a ∈ Ba, c ∈ Bc, a ∈ Ca, and b ∈ Cb. Lemma 6.2.

(i) ρ(Ab ∩ Bc, Ac ∪ Ba − {a})= ρ(Ab ∩ Bc). RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 11

(ii) ρ(Ac ∩ Bc, Ab ∪ Ba − {a, b})= ρ(Ac ∩ Bc). (iii) ρ(Ac ∩ Ba, Ab ∪ Bc − {b})= ρ(Ac ∩ Ba). Proof.

ρ(Ab ∩ Bc, Ac ∪ Ba − {a})

= ρ(Ab ∩ Bc, Ac ∪ Ba) because ρ(Bc,Ba − {a})= ρ(Bc,Ba),

= ρ(Ab ∩ Bc, Ac ∪ Ba ∪ {b}) because ρ({a, c},V − {a, b, c})= ρ({a, b, c},V − {a, b, c}).

ρ. (Ac ∩ Bc, Ab ∪ Ba − {a, b})

= ρ(Ac ∩ Bc, Ab ∪ Ba − {b}) because ρ(Bc,Ba − a)= ρ(Bc,Ba),

= ρ(Ac ∩ Bc, Ab ∪ Ba) because ρ(Ac, Ab − b)= ρ(Ac, Ab).

ρ(Ac ∩ Ba, Ab ∪ Bc − {b})

= ρ(Ac ∩ Ba, Ab ∪ Bc) because ρ(Ac, Ab − {b})= ρ(Ac, Ab),

= ρ(Ac ∩ Ba, Ab ∪ Bc ∪ {a}) because ρ({b, c},V − {a, b, c})= ρ({a, b, c},V − {a, b, c}). Lemma 6.3.

Proof. (i) Since G is prime, ρ(Ab ∩Bc) ≥ 2 and by Lemma 6.2, ρ(Ab ∩Bc, Ac ∪Ba −{a}) ≥ 2. By the submodular inequality,

ρ(Ab − {b}, Ac)+ ρ(Bc,Ba) ≥ ρ(Ab ∩ Bc, Ac ∪ Ba)+ ρ(Ab ∪ Bc − {b}, Ac ∩ Ba).

Since ρ(Ab, Ac)=2 and ρ(Ab ∩ Bc, Ac ∪ Ba) ≥ ρ(Ab ∩ Bc, Ac ∪ Ba − {a}) ≥ 2, we deduce that ρ(Ac ∩ Ba, Ab ∪ Bc − {b}) ≤ 2. By Lemma 6.2, ρ(Ac ∩ Ba) ≤ 2 and |Ac ∩ Ba|≤ 3 because G is 3+1-rank-connected. (ii) This is equivalent to (i) after swapping a, A with b, B respectively. (iii) By the submodular inequality,

ρ(Ab − {b}, Ac)+ ρ(Ba − {a},Bc) ≥ ρ(Ab ∩ Ba, Ac ∪ Bc)+ ρ(Ab ∪ Ba − {a, b}, Ac ∩ Bc).

Since ρ(Ab, Ac)= ρ(Ba,Bc)=2and ρ(Ac ∩Bc, Ab ∪Ba −{a, b})= ρ(Ac ∩Bc), we deduce +1 that ρ(Ac ∩ Bc) ≤ 2. Since G is 2 -rank-connected, |Ac ∩ Bc|≤ 3. (iv) From the assumption, ρ(Ab ∩ Ba, Ac ∪ Bc ∪ {a}) ≤ 2. Then ρ(Ab ∩ Ba, Ac ∪ Bc ∪ {a})+ ρ(V − {a, b, c}, {a, b, c}) ≥ ρ(Ab ∩ Ba, Ac ∪ Bc ∪ {a, b})+ ρ(V − {a, b, c}, {a, c}) +1 and we deduce that ρ(Ab ∩ Ba) ≤ 2. Since G is 3 -rank-connected, |Ab ∩ Ba|≤ 3. +3 Lemma 6.4. If |Ab ∩ Ba| = 3 and ρ(Ab ∩ Ba, Ac ∪ Bc) ≤ 1, then G \ c is prime 3 - rank-connected.

To check the ﬁrst case, let us assume that Ab ∩ Ba ⊆ Ca. From the submodular inequality,

ρ((Ab ∩Ba)∪{b}, Ac ∪Bc)+ρ(Ca,Cb) ≥ ρ(Ca ∪{b},Cb −{b})+ρ(Ab ∩Ba, Ac ∪Bc ∪{b}).

We know that ρ(Ca,Cb)=2and ρ((Ab ∩Ba)∪{b}, Ac ∪Bc) ≤ 1+ρ(Ab ∩Ba, Ac ∪Bc) ≤ 2. Since ρ({b, c},V −{a, b, c})= ρ({a, b, c},V −{a, b, c}), we have ρ(Ab ∩Ba, Ac ∪Bc∪{b})= ρ(Ab ∩Ba). Since G is prime, ρ(Ab ∩Ba) ≥ 2. So we deduce that ρ(Ca ∪{b},Cb −{b}) ≤ 2. Since ρ({a, b},V −{a, b, c})= ρ({a, b, c},V −{a, b, c}), we deduce that ρ(Ca ∪{b, c},Cb − +1 {b})= ρ(Ca ∪ {b},Cb − {b}) ≤ 2. Since G is 3 -rank-connected, we have |Cb − {b}| ≤ 3. So |Cb|≤ 4. Now we need to check the remaining case in which |(Ab ∩ Ba) ∩ Ca| = 2. Let x be the unique element in Ab ∩ Ba ∩ Cb. Let Ab ∩ Ba ∩ Ca = {y, z}. Since x ∈ Ba, x 6= b. From the submodular inequality,

ρ(Ca,Cb)+ ρ({x,y,z},V − {x,y,z}) ≥ ρ({y, z},V − {y, z})+ ρ(Ca ∪ {x},Cb − {x}), and we deduce that ρ(Ca ∪ {x},Cb − {x}) ≤ 2. Again we use the submodular inequality, and obtain that

ρ({x, y, z, b},V − {x, y, z, b})+ ρ(Ca ∪ {x},Cb − {x})

≥ ρ(Ca ∪ {x, b},Cb − {b, x})+ ρ({x,y,z},V − {x,y,z}).

We have ρ({x, y, z, b}) ≤ 1+ρ(Ab∩Ba, Ac∪Bc) = 2 because b∈ / Ac∪Bc. So it follows that ρ(Ca ∪ {x, b},Cb − {b, x}) ≤ 2. Since ρ({a, b},V − {a, b, c}) = ρ({a, b, c},V − {a, b, c}), we have ρ(Ca ∪ {x, b, c},Cb − {b, x}) = ρ(Ca ∪ {x, b},Cb − {b, x}) ≤ 2. Since G is +1 3 -rank-connected and ρ(Cb − {b, x}) ≤ 2, we have |Cb|− 2 ≤ 3.

+1 Lemma 6.5. Let G be a prime 3 -rank-connected graph. Suppose that (Ab, Ac), (Ba,Bc), (Ca,Cb) are chosen so that min(|Ab|, |Ac|), min(|Ba|, |Bc|), min(|Ca|, |Cb|) are maximum. If |Ab ∩ Ba|≤ 1, |Ab ∩ Bc|≤ 1, |Ac ∩ Ba|≤ 1, or |Ac ∩ Bc| ≤ 2, then at least two of G \ a, G \ b, and G \ c are 3+3-rank-connected.

For the simplicity, we let

λ = |Ab ∩ Bc ∩ Ca|, µ = |Ac ∩ Ba ∩ Cb|,

λ1 = |Ac ∩ Bc ∩ Ca|, µ1 = |Ab ∩ Ba ∩ Cb|,

λ2 = |Ab ∩ Ba ∩ Ca|, µ2 = |Ac ∩ Bc ∩ Cb|,

λ3 = |Ab ∩ Bc ∩ Cb|, µ3 = |Ac ∩ Ba ∩ Ca|. Above 8 sets and {a, b, c} partition V = V (G) and therefore 3 λ + µ + (λi + µi)= |V |− 3. Xi=1

From now on, we assume that (Ab, Ac), (Ba,Bc), (Ca,Cb) are chosen so that each of min(|Ab|, |Ac|), min(|Ba|, |Bc|), min(|Ca|, |Cb|) is maximized. Then by Lemma 6.5 and Lemma 6.3, we may assume the following.

(6) 2 ≤ λ + λi ≤ 3, 2 ≤ µ + µi ≤ 3 for all i = 1, 2, 3,

(7) λ1 + µ2 = |Ac ∩ Bc|− 1 ≥ 2, λ2 + µ1 = |Ab ∩ Ba|≥ 2,

(8) λ2 + µ3 = |Ba ∩ Ca|− 1 ≥ 2, λ3 + µ2 = |Bc ∩ Cb|≥ 2,

(9) λ3 + µ1 = |Ab ∩ Cb|− 1 ≥ 2, λ1 + µ3 = |Ac ∩ Ca|≥ 2. Since all graphs up to 11 vertices are 3+3-rank-connected, we may assume that |V |≥ 13. By (iii) and (iv) of Lemma 6.3, we have the following.

(10) Either (a) λ1 + µ2 = 2 or (b) ρ(Ab ∩ Ba) ≤ 2, λ2 + µ1 ≤ 3.

(11) Either (a) λ2 + µ3 = 2 or (b) ρ(Bc ∩ Cb) ≤ 2, λ3 + µ2 ≤ 3.

(12) Either (a) λ3 + µ1 = 2 or (b) ρ(Ac ∩ Ca) ≤ 2, λ1 + µ3 ≤ 3. Lemma 6.6. +3 (i) If λ1 + µ2 ≥ 3 and λ2 + µ1 = 3, then G \ c is 3 -rank-connected. +3 (ii) If λ2 + µ3 ≥ 3 and λ3 + µ2 = 3, then G \ a is 3 -rank-connected. +3 (iii) If λ3 + µ1 ≥ 3 and λ1 + µ2 = 3, then G \ b is 3 -rank-connected. Proof. This is a restatement of Lemma 6.4. Lemma 6.7. If λ + µ ≤ 1, then G \ a, G \ b, and G \ c are 3+3-rank-connected.

Proof. By symmetry we may assume that λ = 0 and µ ≤ 1. Then λi ≥ 2 and µi ≥ 1 for all i ∈ {1, 2, 3}. Then by (10–12), λi = 2 and µi = 1 for all i. Then the conclusion follows Lemma 6.6. Lemma 6.8. If λ = µ = 1, then at least two of G \ a, G \ b, and G \ c are 3+3-rank- connected.

Proof. By (6), λi,µi ∈ {1, 2} for all i ∈ {1, 2, 3}. Since |V | ≥ 13, we may assume that µ3 = |Ac ∩ Bc ∩ Cb| = 2. Since G is prime, ρ(Ac ∩ Bc ∩ Cb) = 2. By Lemma 6.3, ρ(Ac ∩ Ba) ≤ 2. By the submodular inequality,

ρ(Ac ∩ Ba)+ ρ(Ac ∩ Ca) ≥ ρ(Ac ∩ Bc ∩ Cb)+ ρ((Ac ∩ Ba) ∪ (Ac ∩ Ca)). 14 SANG-IL OUM

Since |(Ac ∩ Ba) ∪ (Ac ∩ Ca)| = µ + µ3 + λ1 ≥ 4, ρ((Ac ∩ Ba) ∪ (Ac ∩ Ca)) > 2, and therefore ρ(Ac ∩ Ca) > 2. By (12), λ3 = µ1 = 1. Since λ2 + µ3 > 2, ρ(Bc ∩ Cb) ≤ 2 by (11). By the submodular inequality,

ρ(Ac ∩ Cb)+ ρ(Bc ∩ Cb) ≥ ρ(Ac ∩ Bc ∩ Cb)+ ρ((Ac ∩ Cb) ∪ (Bc ∩ Cb)).

Lemma 6.3 conﬁrms that ρ(Ac∩Cb) ≤ 2. If µ2 = |Ac∩Bc∩Cb| = 2, then ρ(Ac∩Bc∩Cb) ≥ 2. Moreover, |(Ac ∩Cb)∪(Bc ∩Cb)| = µ+µ2 +λ3 ≥ 4 and ρ((Ac ∩Cb)∪(Bc ∩Cb)) ≤ 2 by the above inequality. But this contracts to the assumption that G is 3+1-rank-connected. Therefore we conclude that µ2 = 1. +3 Then |Cb| = 1+ µ + µ1 + µ2 + λ3 = 5 and therefore G \ c is 3 -rank-connected, because (Ca,Cb) was chosen so that min(|Ca|, |Cb|) is maximum. Suppose that λ1 = λ2 = 2. Since λ1 + µ2 > 2, we have ρ(Ab ∩ Ba) ≤ 2 by (10). By the submodular inequality,

ρ(Ab ∩ Ba)+ ρ(Ab ∩ Ca) ≥ ρ(Ab ∩ Ba ∩ Ca)+ ρ((Ab ∩ Ba) ∪ (Ab ∩ Ca)). +1 Since λ2 = 2, ρ(Ab ∩ Ba ∩ Ca) ≥ 2. Since G is 3 -rank-connected and |(Ab ∩ Ba) ∪ (Ab ∩ Ca)| = λ + λ2 + µ1 ≥ 4, ρ((Ab ∩ Ba) ∪ (Ab ∩ Ca)) > 2. This contradicts to Lemma 6.3 stating that ρ(Ab ∩ Ca) ≤ 2. Therefore either λ1 =1 or λ2 = 1. +3 If λ2 = 1, then |Ab| =1+ λ + µ1 + λ2 + λ3 = 5 and so G \ a is 3 -rank-connected. If +3 λ1 = 1, then |Bc| =1+ λ + λ1 + µ2 + λ3 =5 and so G \ b is 3 -rank-connected.

Lemma 6.9. If λ ≥ 2, then λ + λ1 + λ2 + λ3 ≤ 3. If µ ≥ 2, then µ + µ1 + µ2 + µ3 ≤ 3. It follows that either λ ≤ 1 or µ ≤ 1.

Proof. We may assume that λ = 2, because otherwise λi = 0 for all i. By Lemma 6.3, ρ(Ab ∩ Bc) ≤ 2 and ρ(Bc ∩ Ca) ≤ 2. By the submodular inequality,

ρ(Ab ∩ Bc)+ ρ(Bc ∩ Ca) ≥ ρ(Ab ∩ Bc ∩ Ca)+ ρ((Ab ∩ Bc) ∪ (Bc ∩ Ca)).

Since G is prime and 2 ≤ λ ≤ 3, ρ(Ab ∩Bc ∩Ca) ≥ 2. Therefore ρ((Ab ∩Bc)∪(Bc ∩Ca)) ≥ 2. Now, |(Ab ∩ Bc) ∪ (Bc ∩ Ca)| = (λ + λ3) + (λ + λ1) − λ = λ + λ1 + λ3 ≤ 4. Since G is +1 3 -rank-connected, λ1 + λ3 ≤ 1. Similarly we deduce that λ2 + λ3 ≤ 1 and λ1 + λ2 ≤ 1. So at most one of λ1, λ2, λ3 is 1. So λ + λ1 + λ2 + λ3 ≤ 3. By symmetry, if µ ≥ 2, then µ + µ1 + µ2 + µ3 ≤ 3. If λ, µ ≥ 2, then |V |≤ 9, a contradiction. So either λ ≤ 1 or µ ≤ 1.

Lemma 6.10. If λ2 = λ3 = 0, µ1 ≥ 2 and ρ(Ab ∩ Ba, Ac ∪ Bc) ≤ 1, then µ + µ1 = 2. In particular, if λ2 = λ3 = 0 and µ1 ≥ 2, then λ1 + µ2 = 2 or µ + µ1 = 2.

Proof. Suppose that λ2 = λ3 = 0, µ1 ≥ 2, and ρ(Ab∩Ba, Ac∪Bc) ≤ 1. Then ρ((Ab∩Ba)∪ {b}, Ac ∪Bc) ≤ 2. Since λ2 = λ3 = 0, Ab ∩Ba = Ab ∩Ba ∩Cb = Ab ∩Cb −{b}. Thus ρ(Ab ∩ Cb, Ac ∪Ca −{a}) ≤ 2. By Lemma 6.2, ρ(Ab ∩Cb)= ρ(Ab ∩Cb, Ac ∪Ca −{a}) ≤ 2. By the submodular inequality, ρ(Ab ∩Cb)+ρ(Ba ∩Cb) ≥ ρ(Ab ∩Ba ∩Cb)+ρ((Ab ∩Cb)∪(Ba ∩Cb)). By Lemma 6.3, ρ(Ba ∩Cb) ≤ 2. Since G is prime and µ1 ≥ 2, we have ρ(Ab ∩Ba∩Cb) ≥ 2. Therefore ρ((Ab ∩ Cb) ∪ (Ba ∩ Cb)) ≤ 2. Then |(Ab ∩ Cb) ∪ (Ba ∩ Cb)| =1+ µ1 + µ ≤ 3, +1 because G is 3 -rank-connected. We conclude that µ + µ1 ≤ 2. If λ1+µ2 ≥ 3, then |Ac∩Bc| = 1+λ1+µ2 ≥ 4. This implies that ρ(Ab∩Ba, Ac∪Bc) ≤ 1 by Lemma 6.3.

Lemma 6.11. µ + λ3 > 0. RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 15

Proof. We claim that if µ = λ3 = 0, then λ2 = 0 and µ3 = 2. Suppose that λ2 6= 0 or µ3 6= 2. Since µ3 ≥ 2, we have λ2 + µ3 ≥ 3. By Lemma 6.3, ρ(Bc ∩ Cb,Ba ∪ Ca) ≤ 1. Since λ3 = µ = 0, Bc ∩ Cb = Ac ∩ Bc ∩ Cb = Ac ∩ Cb. So ρ(Bc ∩ Cb,Ba ∪ Ca)= ρ(Ac ∩ Cb, Ab ∪Ca −{b}) ≤ 1. Since ρ(Ac, Ab)= ρ(Ac, Ab −{b}), we have ρ(Ac ∩Cb, Ab ∪Ca) ≤ 1. Since G \ c is prime, |Ac ∩ Cb| = µ2 ≤ 1. However, this contradicts to the condition that µ + µ2 ≥ 2. By symmetry, the above proof can be modiﬁed to prove the following claims:

• If µ = λ2 = 0, then λ1 = 0 and µ2 = 2. • If µ = λ1 = 0, then λ3 = 0 and µ1 = 2.

Now we conclude that λ1 = λ2 = 0 and µ1 = µ2 = µ3 = 2. Then |V |≤ 3+λ+2+2+2 ≤ 12, a contradiction to our assumption.

Finally we are ready to ﬁnish the proof of Proposition 6.1. By Lemma 6.9, we may assume that µ ≤ 1. We may assume that λ ≥ 2 by Lemma 6.7 and Lemma 6.8. By Lemma 6.9, λ + λ1 + λ2 + λ3 ≤ 3. We may assume that λ2 = λ3 = 0. By Lemma 6.11, µ = 1. (1) Suppose that λ = 2. Since |Ab| =1+ λ + µ1 ≥ 5 and µ + µ1 ≤ 3, we have µ1 = 2 and |Ab| = 5. By Lemma 6.10, λ1 + µ2 = 2. Then |Bc| =1+ λ + λ1 + µ2 + λ3 = 5 and therefore both G \ a and G \ b are 3+3-rank-connected. (2) Suppose that λ = 3. By Lemma 6.9, λ1 = 0. Since |V | ≥ 13, we have µ1 = µ2 = µ3 = 2. By Lemma 6.10, ρ(Ab ∩ Ba, Ac ∪ Bc) ≥ 2 and therefore Lemma 6.3 implies that ρ(Ac ∩ Bc) ≤ 2. By the submodular inequality, ρ(Ac ∩ Bc)+ ρ(Ac ∩ Cb) ≥ ρ(Ac ∩ Bc ∩ Cb)+ ρ((Ac ∩ Bc) ∪ (Ac ∩ Cb)). By Lemma 6.3, ρ(Ac ∩ Cb) ≤ 2. Since G is prime and µ2 = 2, we have ρ(Ac ∩ Bc ∩ Cb)= 2. Then ρ((Ac ∩ Bc) ∪ (Ac ∩ Cb)) ≤ 2, but |(Ac ∩ Bc) ∪ (Ac ∩ Cb)| =1+ µ + µ2 ≥ 4, a contradiction to the assumption that G is 3+1-rank-connected. This completes the proof of Proposition 6.1 and Theorem 4.1.

7. Excluded pivot-minors for rank-width k As an application, we will discuss excluded pivot-minors of graphs of rank-width at most k, which are pivot-minor-minimal graphs whose rank-width is larger than k. First, we review the definition of rank-width, defined by Oum and Seymour [12]. A tree is subcubic if every vertex has degree 1 or 3. A rank-decomposition of a graph G is a pair (T,µ) of a subcubic tree T and a bijection µ : V (G) → {x : x is a leaf of T }. For a rank-decomposition (T,µ) and each edge e of T , we pick a component Ce of T \ e and we let Ae = {v ∈ V (G) : µ(v) ∈ V (Ce)}. Then the width of an edge e of T in a rank-decomposition (T,µ) is defined as ρG(Ae). Since T \ e has exactly two components and ρG(Ae) = ρG(V (G) − Ae), the width of e is well defined. We define the width of a rank-decomposition (T,µ) as the maximum width of an edge e over all edges e of T . The rank-width rwd(G) of a graph G is the minimum width of (T,µ) over all rank-decompositions (T,µ) of the graph. See [11] for a survey on rank-width. We can define k-branched sets recursively as follows: A set B of vertices is k-branched if and only if ρG(B) ≤ k and either |B| = 1 or there exists a proper nonempty subset B′ of B such that both B′ and B − B′ are k-branched. A set A of vertices of a graph G is titanic if for every partition (A1, A2, A3) of A, there exists i ∈ {1, 2, 3} such that 16 SANG-IL OUM

ρG(Ai) ≥ ρG(A). The following lemma originates from Robertson and Seymour [14]. A proof can be found in [10, Lemma 5.1].

Lemma 7.1. Let G be a graph. If A is a titanic set such that ρG(A) ≤ k = rwd(G), then V (G) − A is k-branched. The following lemma is a restatement of [10, Lemma 5.3] and is a consequence of Proposition 2.6 and Lemma 7.1. We provide a proof. Lemma 7.2. Let m be an integer. Let G be a graph such that rwd(G) > m and rwd(G \ v) < rwd(G) and rwd(G/v) < rwd(G) for every vertex v of G. If G is m+ℓ- rank-connected, then G is (m + 1)+(6ℓ+5m−5)-rank-connected. Proof. Let k = rwd(G) − 1. Suppose that G is not (m + 1)+(6ℓ+5m−5)-rank-connected. There exists a partition (A, B) of V (G) such that ρG(A) < m +1 and |A|, |B| ≥ (6ℓ + 5m − 5) + (m + 1). We may assume that B is not k-branched. Let v ∈ A. By Proposition 2.6, either G \ v or G/v is m+(2ℓ+m−1)-rank-connected. We may assume that G \ v is m+(2ℓ+m−1)-rank-connected. In every partition (X1, X2, X3) of A − {v}, there exists i ∈ {1, 2, 3} such that |Xi|≥ (|A|−1) +(2ℓ+m−1) /3. We may assume that |X1| ≥ 2ℓ + 2m − 1. Since G \ v is m -rank- connected, ρG\v(X1) ≥ m ≥ ρG\v(A − {v}). Therefore A − {v} is a titanic set of G \ v. By Lemma 7.1, B is k-branched in G \ v. Since B is not k-branched in G, there must exist a subset X of B such that ρG(X) = ρG\v(X) + 1. By Lemma 2.3, ρG\v(X)+ ρG(B) ≥ ρG(X)+ ρG\v(B) and therefore ρG(B) = ρG\v(B) + 1. But this is +(2ℓ+m−1) a contradiction because G \ v is m -rank-connected and ρG\v(B) 2. If rwd(G \ v) < rwd(G) and rwd(G/v) < rwd(G) for every vertex v of G, then G is prime 3+2-rank-connected. Proof. Let k = rwd(G) − 1 ≥ 2. It is easy to see that G is connected and therefore G is 1+0-rank-connected. By Lemma 7.2, G is 2+0-rank-connected. So G is prime. Suppose that G is not 3+2-rank-connected. Let (A, B) be a partition of V (G) such that |A|, |B|≥ 5 and ρG(A) ≤ 2. If both A and B were k-branched, then G would have rank-width k. Therefore, we may assume that B is not k-branched. We may assume that B is minimal with such properties. ′ ′ If ρG(A ∪ {w}) ≤ 2 for some w ∈ B, then (A ,B ) = (A ∪ {w},B − {w}) is another partition. By assumption, either |B| = 5 or B − {w} is k-branched. If |B| = 5, then B − {w} is k-branched. If B − {w} is k-branched, then B is k-branched, contrary to the assumption that B is not. Therefore ρG(A ∪ {w}) > 2 for all w ∈ B. So ρG(A)=2 and A is fully closed. Then by Proposition 3.1, there exists v ∈ A such that either G\v or G/v is prime. We may assume that G \ v is prime, because otherwise we replace G with G ∧ vw for some neighbor w of v. Since |A|≥ 5, in every 3-partition (X1, X2, X3) of A − {v}, there exists i such that |Xi|≥ 2. Since G \ v is prime, ρG\v(Xi) ≥ 2 ≥ ρG\v(A − {v}) and therefore RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 17

A−{v} is titanic in G\v0. Since G\v has rank-width k, B is k-branched in G\v. Since B is not k-branched in G, there exists a subset X of B such that ρG\v(X)= ρG(X)−1. By Lemma 2.3, ρG\v(X)+ ρG(B) ≥ ρG(X)+ ρG\v(B) and therefore ρG(B)= ρG\v(B) + 1. But this is a contradiction because G \ v is prime but ρG\v(B) = 1. By combining Lemma 7.2 with Proposition 7.3, we obtain the following corollary, which is an improvement over the previous theorem [10, Theorem 5.4] stating that such k − +( 6 1 −k) graphs are (k + 1) 5 -rank-connected for every k ∈ {2, 3,..., rwd(G) − 1}. Corollary 7.4. Let G be a graph such that rwd(G) > 2. If rwd(G \ v) < rwd(G) · k − +( (7/12) 6 1 −k) and rwd(G/v) < rwd(G) for every vertex v of G, then G is (k + 1) 5 -rank- connected for every k ∈ {2, 3, 4,..., rwd(G) − 1}. Proof. This is a direct corollary of Lemma 7.2 and Proposition 7.3, as (7/12) · 6k − 1 (7/12)6k+1 − 1 6 − k + 5(k + 1) − 5= − (k + 1). 5 5 Thanks to the improvement by Corollary 7.4, we can now obtain an upper bound on the maximum number of vertices in such graphs, which is better than the previous upper bound (6k+1 − 1)/5 in [10, Theorem 5.4]. Theorem 7.5. Let k ≥ 2. If a graph G has rank-width larger than k but all of its proper (7/12)·6k+1−1 pivot-minors have rank-width at most k, then |V (G)|≤ 5 .

(7/12)·6k −1 +N Proof. Let n = |V (G)|. Let N = 5 − k. By Corollary 7.4, G is (k + 1) - rank-connected. Let v be a vertex of G. We may assume, by Proposition 2.6, that G \ v is (k + 1)+(2N+k)-rank-connected. Since G \ v has rank-width at most k, there exists a partition (A, B) of V (G \ v) such that ρG\v(A) ≤ k and |A|, |B| ≥ (n − 1)/3. We may assume that |A| ≤ |B|. Since G \ v is (k + 1)+(2N+k)-rank-connected, |A| < (7/12)·6k+1−1 (2N + k) + (k + 1). Therefore n ≤ 6(N + k)+1= 5 . When k = 2, we can get a better bound 16 by using Theorem 4.1 on 3+2-rank- connected graphs. Previous bounds are 43 by [10, Theorem 5.4], and 25 by Theorem 7.5. The author tried to compute the complete list of excluded pivot-minors for the class of graphs of rank-width at most 2 by searching graphs up to 16 vertices, but was not able to ﬁnish computation. Theorem 7.6. If a graph G has rank-width 3, but all of its proper pivot-minors have rank-width at most 2, then |V (G)|≤ 16. Proof. By Proposition 7.3, G is prime 3+2-rank-connected. Suppose that |V (G)| ≥ 17. By Theorem 4.1, G has a 3+3-rank-connected pivot-minor H such that |V (H)| = |V (G)|− 1 ≥ 16. Since the rank-width of H is 2, there is a partition (A, B) of V (H) such that |A|, |B| ≥ |V (H)|/3 > 5 and ρH (A) ≤ 2, contradicting the assumption that G is 3+3-rank-connected. If we had a stronger connectivity condition on G, then we could obtain a stronger bound from Proposition 6.1. 18 SANG-IL OUM

Proposition 7.7. Suppose that a graph G has rank-width 3, but all of its proper pivot- minors have rank-width at most 2. If G is 3+1-rank-connected, then |V (G)|≤ 14. Proof. Suppose that there is a vertex x ∈ G such that G\x or G/x is 3+2-rank-connected. We may assume that G \ x is 3+2-rank-connected because otherwise we can replace G with G ∧ xy for some neighbor y of x. Since G \ x has rank-width 2, there is a partition (A, B) of V (G) − {x} such that ρ(A, B)=2 and |A|, |B|≥ (|V (G)|− 1)/3. Since G \ x is 3+2-rank-connected, min(|A|, |B|) ≤ 4. Therefore |V (G)| ≤ 13. We may now assume that G/x are G \ x are not 3+2-rank-connected. By Lemma 4.2, G is not 3-rank-connected and therefore there is a set X = {a, b, c} of vertices such that ρ(X) = 2. By Lemma 5.1, we may assume that X is a triplet of G, because otherwise we can apply pivoting. By Proposition 6.1, we may assume that G \ a is 3+3-rank-connected. Let (T,µ)bea rank-decomposition of G \ a having width 2. We orient every edge uv of T from u to v if the partition given by T \ uv has at most 5 leaves in the component containing u. Since T has more vertices than edges, there is a vertex v of T such that all edges incident to v is outgoing. If v is a leaf, then |V (G)|≤ 6. So we may assume that v is not a leaf and therefore there exists a partition (A1, A2, A3) of vertices of G \ a such that |Ai|≤ 5 and ρ(Ai) ≤ 2 for all i ∈ {1, 2, 3}. If there exists i such that b, c ∈ Ai, then ρ(Ai ∪{a}) ≤ 2 because ρ({b, c},V (G)−X)= +1 ρ(X,V (G)−X). Since G is 3 -rank-connected, either |Ai|+1 ≤ 3 or |V (G)−Ai|−1 ≤ 3. If |Ai|≤ 2, then |V (G)|≤ 1+5+5+2= 13. If |V (G)−Ai|≤ 4, then |V (G)|≤ 4+5 = 9. We conclude that if there exists i such that b, c ∈ Ai, then |V (G)|≤ 13. Now suppose that no Ai contains both b and c. We may assume that b ∈ A1 and c ∈ A2. Then ρ(A1 ∪ A2, A3) ≤ 2 and therefore ρ(A1 ∪ A2 ∪ {a}, A3) ≤ 2. Since G +1 is 3 -rank-connected, either |A1 ∪ A2| + 1 ≤ 3 or |A3| ≤ 3. If |A1 ∪ A2| ≤ 2, then |V (G)|≤ 1+2+5=8. If |A3|≤ 3, then |V (G)|≤ 1+5+5+3=14.

References [1] L. Allys. Minimally 3-connected isotropic systems. Combinatorica, 14(3):247–262, 1994. [2] R. E. Bixby. A simple theorem on 3-connectivity. Linear Algebra Appl., 45:123–126, 1982. [3] A. Bouchet. Reducing prime graphs and recognizing circle graphs. Combinatorica, 7(3):243–254, 1987. [4] W. H. Cunningham. Decomposition of directed graphs. SIAM J. Algebraic Discrete Methods, 3(2):214–228, 1982. [5] J. Geelen and X. Zhou. A splitter theorem for internally 4-connected binary matroids. SIAM J. Discrete Math., 20(3):578–587 (electronic), 2006. [6] J. Geelen and X. Zhou. Generating weakly 4-connected matroids. J. Combin. Theory Ser. B, 98(3):538–557, 2008. [7] J. F. Geelen. Matchings, matroids and unimodular matrices. PhD thesis, University of Waterloo, 1996. [8] R. Hall. A chain theorem for 4-connected matroids. J. Combin. Theory Ser. B, 93(1):45–66, 2005. [9] K. Murota. Matrices and matroids for systems analysis, volume 20 of Algorithms and Combinatorics. Springer-Verlag, Berlin, 2000. [10] S. Oum. Rank-width and vertex-minors. J. Combin. Theory Ser. B, 95(1):79–100, 2005. [11] S. Oum. Rank-width: algorithmic and structural results. Discrete Appl. Math., 231:15–24, 2017. [12] S. Oum and P. Seymour. Approximating clique-width and branch-width. J. Combin. Theory Ser. B, 96(4):514–528, 2006. [13] J. G. Oxley. Matroid theory. Oxford University Press, New York, 1992. RANKCONNECTIVITYANDPIVOT-MINORSOFGRAPHS 19

[14] N. Robertson and P. Seymour. Graph minors. X. Obstructions to tree-decomposition. J. Combin. Theory Ser. B, 52(2):153–190, 1991. [15] J. Spinrad. Prime testing for the split decomposition of a graph. SIAM J. Discrete Math., 2(4):590– 599, 1989. [16] K. Truemper. A decomposition theory for matroids. I. General results. J. Combin. Theory Ser. B, 39(1):43–76, 1985. [17] K. Truemper. Matroid decomposition. Academic Press Inc., Boston, MA, 1992.

Discrete Mathematics Group, Institute for Basic Science (IBS), Daejeon, South Ko- rea. Department of Mathematical Sciences, KAIST, Daejeon, South Korea. Email address: [email protected]