Cyclic Flats, Sticky Matroids, and Intertwines

Cyclic Flats, Sticky Matroids, and Intertwines Joseph E. Bonin The George Washington University

Part I Essential Matroid Background and Cyclic Flats

Matroids via Flats There are many equivalent formulations of the notion of a matroid. One formulation from flats... A matroid is a pair (S, F) where S is a finite set and F 2 S with 1. S F, 2. if X,Y F, then X Y F, and 3. if X F and y S X, then there is a (unique) Y F with a. X y Y and b. if Z F with X Z Y, then Z = Y. The sets in F are called flats.

Matroids via Flats The flats form a geometric (i.e., atomic and semimodular) lattice. {a, b, c, d, e, f} c {a, b, c} {b, d} {c, d, e} {a, e} {a, d, f} {c, f} {b, e, f} b d a f M e {a} {b} {c} {d} {e} {f} The lattice of flats of M 1. S F, 2. if X,Y F, then X Y F, and 3. if X F and y S X, then there is a (unique) Y F with a. X y Y and b. if Z F with X Z Y, then Z = Y.

Matroids via Rank A matroid is a pair (S,r) where S is a finite set and r is a function from 2 S to Z with 1. 0 r(x) X for all X S, 2. if X Y S, then r(x) r(y ), and 3. if X,Y S, then r(x) + r(y ) r(x Y ) + r(x Y ). We call r(x) the rank of the set X. The rank r(m) of a matroid M = (S,r) is r(s). How (S, F) gives (S,r)... Intersections of flats are flats, so, given (S, F), for any X S there is a least flat cl(x) (the closure of X) with X cl(x); r(x) is the rank of cl(x) in the geometric lattice of flats.

Flats to Rank {a, b, c, d, e, f} c {a, b, c} {b, d} {c, d, e} {a, e} {a, d, f} {c, f} {b, e, f} b d a f M e {a} {b} {c} {d} {e} {f} The lattice of flats of M E.g., cl(b,d,f ) = {a,b,c,d,e,f }, so r(b,d,f ) = 3; cl(a,b) = {a,b,c}, so r(a,b) = 2; cl(c,d,e) = {c,d,e}, so r(c,d,e) = 2.

Rank to Flats The description of getting ranks from flats shows that flats are the sets for which no proper superset has the same rank, so... Given (S,r), let F consist of the sets X with r(x y) > r(x) for all y S X.

Other Fundamental Matroid Concepts Hyperplanes are flats of rank r(m) 1. E.g., {a,b,c}, {c,f }. b c d Each flat is an intersection of hyperplanes. a f e A set X is independent if r(x) = X. E.g.,, {a}, {a,e,f }. Circuits are minimal dependent sets. E.g., {a,b,c}, {a,b,d,e}.

Deletion, Contraction, and Minors Consider a matroid M = (S, F) and X S. The deletion M\X, or restriction M (S X), is the matroid (S X, {F X : F F}). c c b d b d c d a f M e f M\a e b f M/a e The contraction M/X is (S X, {F X : F F, X F }). Combining deletion and contraction gives the minors of M.

Direct Sums Given M 1 = (S 1, F 1 ) and M 2 = (S 2, F 2 ) with S 1 S 2 =, the direct sum M 1 M 2 is the matroid (S 1 S 2, {F 1 F 2 : F 1 F 1, F 2 F 2 }).. The direct sum of a 3-point line and a 4-point line, which has rank 4.

Cyclic Flats A set in M is cyclic if it is a (possibly empty) union of circuits. An element x in a flat F of M is an isthmus of F if F x is a flat. Cyclic flats are the flats that have no isthmuses. Let Z(M) be the set of cyclic flats of M. Ordered by containment, Z(M) is a lattice.. {a, b, c, d, e, f, g, h} c a d b g e h f {a, b, c, d} {e, f, g, h} e a f b g c h d

Cyclic Flats and their Ranks A matroid is determined by its cyclic flats and their ranks. (Brylawski, 1975). {a, b, c, d, e, f, g, h} c a d b M g e h f {a, b, c, d} 3 2 4 3 {e, f, g, h} 3 2 e a g f c b N h d

Axioms for Matroids via Cyclic Flats Theorem For Z 2 S and r : Z Z, there is a matroid M on S with Z = Z(M) and r = r M Z iff (Z0) Z is a lattice under inclusion, (Z1) r(0 Z ) = 0, where 0 Z is the least element of Z, (Z2) 0 < r(y ) r(x) < Y X for all X,Y Z with X Y, (Z3) for all pairs of incomparable sets X,Y Z, r(x) + r(y ) r(x Y ) + r(x Y ) + (X Y ) (X Y ). (Sims, 1980; Bonin and de Mier, 2008)

The Aim of this Talk The point of this talk is to demonstrate how this axiom scheme can be useful in certain matroid construction problems. Illustrations: 1. progress on the sticky matroid conjecture and 2. a construction of infinitely many intertwines for most pairs of matroids. The moral is that cyclic flats let you effectively deal with certain key geometric information.

Part II The Sticky Matroid Conjecture

Amalgams An amalgam of matroids M 1 on S 1 and M 2 on S 2 is a matroid M on S 1 S 2 with M S 1 = M 1 and M S 2 = M 2. M 1 c b d f a e M 2 c d y x e z b f a c y d x e z b f a c y d x e z b a f c y d x z e rank 4 rank 3 rank 3 A necessary (but not sufficient) condition for having an amalgam: M 1 S 1 S 2 = M 2 S 1 S 2.

Amalgams Need Not Exist The matroid M 1 below is the Vámos matroid; each set {x,x,y,y } except {a,a,b,b } is coplanar. a a b b d d d d z d d c c M 1 c c M 2 c c M 1 S 1 S 2 = M 2 S 1 S 2 A pair M 1, M 2 with M 1 S 1 S 2 = M 2 S 1 S 2, yet with no amalgam.

Amalgams Need Not Exist d d c c z

Amalgams Need Not Exist a a b b d d d d z d d c c M 1 c c M 2 c c M 1 S 1 S 2 = M 2 S 1 S 2 A pair M 1, M 2 with M 1 S 1 S 2 = M 2 S 1 S 2, yet with no amalgam.

Amalgams Need Not Exist a a b b d d d d z d d c c M 1 c c M 2 c c M 1 S 1 S 2 = M 2 S 1 S 2 A pair M 1, M 2 with M 1 S 1 S 2 = M 2 S 1 S 2, yet with no amalgam. A matroid N is sticky if, whenever M 1 on S 1 and M 2 on S 2 satisfy M 1 S 1 S 2 = M 2 S 1 S 2 N, then M 1 and M 2 have an amalgam.

Modular Matroids A flat X in a matroid M is modular if, for all flats Y of M, r(x) + r(y ) = r(x Y ) + r(x Y ). A matroid is modular if all of its flats are modular d d. c c No lines are modular. All flats are modular. Theorem Modular matroids are precisely the direct sums of projective geometries.

The Sticky Matroid Conjecture Theorem Modular matroids are sticky. (Ingleton, 1970s) The Sticky Matroid Conjecture Sticky matroids are modular. (Poljak and Turzík, 1982)

More on Modular Flats Theorem Intersections of modular flats are modular. (Brylawski, 1975) Corollary A matroid is modular iff each of its hyperplanes is modular. Theorem A hyperplane is modular iff it intersects each line (rank-2 flat) in at least a point (rank-1 flat).

Earlier Progress on the Conjecture The conjecture holds in rank 3. Theorem A rank-3 matroid that has a pair of disjoint lines is not sticky. (Poljak and Turzík, 1982) Theorem Contractions of sticky matroids are sticky. (Bachem and Kern, 1988) Corollary A matroid of rank r 3 is not sticky if it has a pair of hyperplanes whose intersection has rank r 3. Theorem The sticky matroid conjecture is true if and only if it is true for rank-4 matroids. (Bachem and Kern, 1988)

Part of the Recent Progress Theorem A rank-4 matroid that has a pair of disjoint planes is not sticky. (Bonin, 2009) The rank-4 case is of most interest, but the counterpart holds for all ranks three or greater.

A Proof of the Rank-3 Case A rank-3 matroid N that has disjoint lines L 1,L 2 is not sticky.. L 1 L 2 N N L 1 L 2 M1 a a b b S A B L 1 A L 1 B S L 2 A L 2 B L 1 L 2 } 4 } 3 } 2. N L 1 L 2 Z(N) B = {b, b } A = {a, a } M 2

Sketch of the Proof for the Rank-4 Case A rank-4 matroid N that has disjoint planes P 1,P 2 is not sticky. Extend N, on S, to M 1 by adding independent elements p and q only to P 1, P 2, and S.

Sketch of the Proof for the Rank-4 Case A rank-4 matroid N that has disjoint planes P 1,P 2 is not sticky. Extend N, on S, to M 1 by adding independent elements p and q only to P 1, P 2, and S. P 1 A P 1 B S A B S P 1 P 2 Z(N). P 2 A P 2 B } } } 5 4 3 WLOG, assume P 1 and P 2 are cyclic. Let A and B be disjoint 3-sets, also disjoint from S. Modify Z(N) as shown. Check (Z0) (Z3). This gives M 2. If M were an amalgam of M 1 and M 2, then p,q are in cl M (P 1 A) and cl M (P 2 A) (both of rank 4); this yields p,q cl M (A). Similarly, p,q cl M (B). Semimodularity applied to A {p, q} and B {p, q} gives r(p,q) 1, contrary to {p,q} being independent in M 1.

Additional Recent Progress Related ideas can be used to prove a result that gives the following theorem of Bachem and Kern (1988). (Their proof was flawed.) Theorem The sticky matroid conjecture holds for rank-4 matroids that satisfy the bundle condition:. given four lines in rank 4 with no three coplanar, if five of the six pairs of lines are coplanar, then so is the sixth pair....

Status of the Conjecture It now suffices to prove the conjecture in rank 4 when (a) each pair of planes intersects in a line and (b) the bundle condition fails. It is conjectured that no rank-4 matroids satisfy (a) and (b). This would follow from the rank-4 case of Kantor s conjecture: for sufficiently large r, if, in a finite rank-r matroid M, each pair of hyperplanes intersects in a flat of rank r 2, then some extension of M is modular. (Having a modular extension implies that the bundle condition holds.)

Part III Intertwines

Motivation and Definition If classes C and C of matroids are minor-closed, then so is C C. Any excluded minor M for C C has a minor that is an excluded minor for C and another that is an excluded minor for C. Also, no proper minor of M has two such minors. A matroid M is an intertwine of matroids M 1 and M 2 if M has a minor isomorphic to M 1 and a minor isomorphic to M 2, and no proper minor of M has both M 1 - and M 2 -minors. We may assume the ground sets of M 1 and M 2 are disjoint.

Examples An intertwine is a minor-minimal matroid having both M 1- and M 2-minors. Some minors of M 1 M 2 and their (various) parallel connections are intertwines of M 1 and M 2.. M 1.. M 2 The rank-3 intertwines of M 1 and M 2. (Contract the blue point to get M 1.)

The Basic Problem of Intertwines Problem Can two matroids have infinitely many intertwines? (Brylawski, 1986; Robertson; Welsh; Oxley, Problem 14.4.6)

The Basic Problem of Intertwines Problem Can two matroids have infinitely many intertwines? (Brylawski, 1986; Robertson; Welsh; Oxley, Problem 14.4.6) Yes. (Vertigan, mid-1990 s, unpublished)

The Basic Problem of Intertwines Problem Can two matroids have infinitely many intertwines? (Brylawski, 1986; Robertson; Welsh; Oxley, Problem 14.4.6) Yes. (Vertigan, mid-1990 s, unpublished) Our goal: construct and explore infinite sets of intertwines.

Background Free Extension and Free Coextension Fix a set X that is disjoint from the ground set S of M. The free extension of M by X is the matroid M + X on S X whose cyclic flats and ranks are 1. the proper cyclic flats F of M, with rank r M (F), and 2. S X, of rank r(m). The free coextension of M by X is the matroid M X on S X whose cyclic flats and ranks are 1. the sets F X, of rank r M (F) + X, for F Z(M) with F, and 2. the empty set, of rank 0.

Examples of Free Extension and Free Coextension a b c f M 1 d e a a b b M 2 (rank 2) c c a b b a c p M 2 p (rank 3) c p b c d q a q b b c a f M 1 + {p, q} e a p c M 2 {p, q} (rank 4)

Free Coextension a b c f M d e {a, b, c} 2 {a, b, c, d, e, f} 3 {c, d, e} 2 0 {a, d, f} 2 Z(M) {b, e, f} 2 {a, b, c, d, e, f} X 3 + X {a, b, c} X {c, d, e} X {a, d, f} X {b, e, f} X 2 + X 2 + X 2 + X 2 + X Z(M X) 0 Free coextension preserves the nullity of corresponding proper, nonempty cyclic flats: η M (F) = F r M (F) = η M X (F X).

The Construction of Intertwines We will use cyclic flats to define a particular amalgam of certain free coextensions of M 1 and M 2. Let Z (M) be the set of proper, nonempty cyclic flats of M. Let the ground sets of the matroids M 1 and M 2 be S 1 and S 2, respectively, with S 1 S 2 =.

The Construction of Intertwines Let Z = Z ( M 1 (T 1 S 2) ) Z ( M 2 (T 2 S 1) ) {,S 1 S 2 T 1 T 2 } for some S 1 S 1 and S 2 S 2, and where, for some integer k r(m 1 ) + η 1 (S 1 ) + r(m 2) + η 2 (S 2 ), T 1 and T 2 are sets with k = r(m 1 ) + S 2 + T 1 and k = r(m 2 ) + S 1 + T 2 with T 1, T 2, and S 1 S 2 disjoint. Define r : Z Z by 1. r( ) = 0, 2. r(s 1 S 2 T 1 T 2 ) = k, 3. r(f T 1 S 2 ) = r 1(F) + T 1 + S 2 for F Z (M 1 ), and 4. r(f T 2 S 1 ) = r 2(F) + T 2 + S 1 for F Z (M 2 ).

An Example S 1 u v w M 1 y x S 1 S 2 T 1 T 2 k b c d {x, y, a} T 1 k 1 {a, b, c, u, v, w} T 2 {c, d, e, u, v, w} T 2 k 1 k 1 S 2 a M 2 e 0 T 1 = k 2 1 = k 3, T 2 = k 3 3 = k 6

This Defines a Matroid Theorem The pair (Z,r) satisfies axioms (Z0)-(Z3). The resulting rank-k matroid M on S 1 S 2 T 1 T 2 (i) is an amalgam of M 1 (T 1 S 2 ) and M 2 (T 2 S 1 ), and (ii) η 1 (F) = η M (F T 1 S 2 ) for F Z (M 1 ) and η 2 (F) = η M (F T 2 S 1 ) for F Z (M 2 ).

The Main Result Theorem Assume neither M 1 nor M 2 arises, up to isomorphism, from the other by any combination of minors, free extension, and free coextension. For i {1,2}, fix S i with FI(M i ) S i S i FI(M i ). If k 4max{ S 1, S 2 }, then the matroid M defined above is an intertwine of M 1 and M 2. FI(M): the set of elements in no proper cyclic flat of M. (free elements and isthmuses) FI(M ): the intersection of all nonempty cyclic flats of M. (cofree elements and loops)

A Sketch of the Proof Duality simplifies the proof. Theorem With k = S 1 + S 2 + T 1 + T 2 k, we have ( Mk (M 1,S 1,T 1;M 2,S 2,T 2) ) = Mk (M 1,S 1 S 1,T 2;M 2,S 2 S 2,T 1). It suffices to show that no single-element deletion M\a has both an M 1 - and an M 2 -minor. By symmetry, it suffices to treat a S 1 T 1.

A Sketch of the Proof (continued) The inequality on k implies that if M\a\X/Y M i, then either X T 1 η(m 1 ) or Y T 1 r(m 2 ).

A Sketch of the Proof (continued) The inequality on k implies that if M\a\X/Y M i, then either X T 1 η(m 1 ) or Y T 1 r(m 2 ). The first implies (1) M\(X T 1 ) = ( M 2 (T 2 S 1 )) + ( (S 1 S 1 ) (T 1 X) ), (2) M\(X T 1 ) has no M 1 -minor, and (3) η 2 (F) = η M\(X T1 )(F T 2 S 1 ) for F Z (M 2 ). The second implies (1 ) M/(Y T 1 ) = ( M 1 ((T 1 Y ) S 2 )) + ( (S 2 S 2 ) T 2), (2 ) M/(Y T 1 ) has no M 2 -minor, and (3 ) η 1 (F) = η M/(Y T1 )(F (T 1 Y ) S 2 ) for F Z (M 1 ).

A Sketch of the Proof (continued) For a (S 1 S 1 ) T 1, assume M\a has a minor M\a\X/Y M 1.

A Sketch of the Proof (continued) For a (S 1 S 1 ) T 1, assume M\a has a minor M\a\X/Y M 1. Either (1) M\(X T 1 ) = ( M 2 (T 2 S 1 )) + ( (S 1 S 1 ) (T 1 X) ), (2) M\(X T 1 ) has no M 1 -minor, and X (3) η 2 (F) = η M\(X T1 )(F T 2 S 1 ) for F Z (M 2 ) or (1 ) M/(Y T 1 ) = ( M 1 ((T 1 Y ) S 2 )) + ( (S 2 S 2 ) T 2), (2 ) M/(Y T 1 ) has no M 2 -minor, and (3 ) η 1 (F) = η M/(Y T1 )(F (T 1 Y ) S 2 ) for F Z (M 1 ).

A Sketch of the Proof (last step) For a S 1, assume M\a has a minor M\a\X/Y M 2. Either (1) M\(X T 1 ) = ( M 2 (T 2 S 1 )) + ( (S 1 S 1 ) (T 1 X) ), (2) M\(X T 1 ) has no M 1 -minor, and (3) η 2 (F) = η M\(X T1 )(F T 2 S 1 ) for F Z (M 2 ) or (1 ) M/(Y T 1 ) = ( M 1 ((T 1 Y ) S 2 )) + ( (S 2 S 2 ) T 2), (2 ) M/(Y T 1 ) has no M 2 -minor, and X (3 ) η 1 (F) = η M/(Y T1 )(F (T 1 Y ) S 2 ) for F Z (M 1 ). By (3), since a is in proper cyclic flats, deleting a makes the sum of the nullities of the proper cyclic flats less than that sum in M 2, so M\a\X/Y M 2.

Knowing More About M 1 and M 2 Lets Us Say More If neither M 1 nor M 2 has circuit-hyperplanes, then we can alter the construction, adding any collection of k-subsets of T 1 T 2 (no two intersecting in k 1 elements) as circuit-hyperplanes; the result is also an intertwine. Thus, in this case, the number of intertwines of M 1 and M 2 grows rapidly (at least exponentially) as a function of the rank.

Corollaries of the Construction Theorem The matroid M is transversal if and only if M 1 and M 2 are. Corollary If M 1 and M 2 satisfy the conditions in the main theorem and are transversal, then infinitely many of their intertwines are transversal. The same holds for cotransversal matroids, bitransversal matroids, and gammoids. Corollary If M 1 and M 2 satisfy the conditions in the theorem, then, for any integer n, infinitely many of their intertwines 1. are n-connected and 2. have U n,2n -minors.

Corollaries of the Construction If S 1 = = S 2, then M = T k( (M 1 T 1 ) (M 2 T 2 ) ) where T k is the k-fold truncation. If S 1 = S 1 and S 2 = S 2, then M = L j( (M 1 + T 1 ) (M 2 + T 2 ) ) where L j is the (k r(m 1 ) r(m 2 ))-fold lift. Corollary Assume a class C of matroids is closed under direct sums, free extensions, free coextensions, truncations, and lifts. Assume M 1,M 2 C satisfy the hypotheses of the main theorem and either FI(M 1 ) = = FI(M 2 ) or FI(M 1 ) = = FI(M 2 ). Then C contains infinitely many intertwines of M 1 and M 2. This applies to the class of matroids representable over a given infinite field or the class representable over a given characteristic.

Open Problems Open Problem Under precisely what conditions do M 1 and M 2 have infinitely many intertwines? Open Problem Can a matroid and a uniform matroid have infinitely many intertwines? (Geelen, Some Open Problems on Excluding a Uniform Matroid) Open Problem What can be said about i(k + 1;M 1,M 2 ) i(k;m 1,M 2 ) where i(k;m 1,M 2 ) is the number of rank-k intertwines of M 1 and M 2?