Representability of Matroids

Representability of matroids

Greg McGlynn March 9, 2011

1 Overview Definition 2. A matroid is binary if it is has a representation over the field GF(2). There are several interesting problems in the Definition 3. U 2 is the matroid with a field of matroid representation theory. Nice 4 ground set of cardinality 4 in which every 2- results include the characterization of bi- element set is a basis. nary, ternary, and quaternary matroids by 2 excluded minors. It remains an open ques- Theorem 1. U4 is not binary tion whether you can characterize the ma- Proof. In the past we have argued that a matroids representable over a given finite field troid of rank n only requires n rows in a by a list of excluded minors. 2 representation. U4 has rank 2. But there Here we give the proof that the binary ma- are only 3 distinct non-zero 2-element vec- troids are exactly those that exclude the mi- tors over GF(2)–not enough to represent the 2 nor U4 and also an example of a matroid that 4 elements of the ground set. has no representation over any field. Definition 4. A minor of a matroid M is a smaller matroid obtained by a sequence of 2 Binary matroids deletions and contractions in M. It is not too hard to show that if M has We start with a review of some stufffrom a representation over a field F, then so does class. any minor of M.

Definition 1. A representation of a matroid Theorem 2. A matroid M is binary if and 2 M is a matrix A with entries from some only if it does not have U4 as a minor. field F such that there is a one-to-one corre- Clearly if M is binary, then it cannot have spondence between the columns of A and the 2 U4 as a minor, because that would imply that ground set of M, and a set of columns in A is 2 U4 is binary also. It remains to show that if linearly independent (as vectors) iffthe cor- 2 M is not binary, then it contains a U4 minor. responding set is independent in M. The strategy is:

1 Construct a binary matroid N with a Definition 5. Suppose M and N are matroids • specific relationship to the non-binary over a common ground set S. Call X S M. wrong if it is a basis of only one of M,⊂ N. Call B S a far basis if it is a common Show that this relationship implies that basis of⊂ both M and N and there is no Y S • 4 ⊂ either M or N contains a U4 minor. such that Y is wrong and B Y = 2. | # | Conclude that M contains a U 2 minor. Lemma 1. Suppose M, N are distinct ma- • 4 troids over a common ground set S that have Here is the proof. a far basis B. Then at least one of M, N con- 2 Proof. Suppose M is a non-binary matroid tains a U4 minor. with ground set S. We construct a related bi- Proof. By contradiction. Suppose there is a nary matroid N, represented over GF(2) by a 2 far basis B, yet neither M nor N has a U4 set of column vectors x s S . { s | ∈ } minor. Choose B a far basis, and X wrong, to min- First pick a basis B of M. Make B a basis • imize B X (which must be 4). of N also by choosing xb b B to be | # | ≥ a set of linearly independent{ | vectors∈ } over Now if S (B X) = then we can • \ ∪ ' ∅ GF(2). delete S (B X). In the deletions, B is still a\ far∪ basis, so M (S (B X)) For each s S B there is a unique and N (S (B X)) are counterexample\ \ ∪ • C B + s that∈ is\ a circuit in M. Define \ \ ∪ s ⊂ matroids. xs = b Cs s xb. By construction, Cs is a cicruit∈ in− N as well. Similarly, if B X = then we can con- ! • tract B X and∩ get' even∅ smaller coun- ∩ Now xs defines a binary matroid N. terexamples. Claim:{ }There is no Y S such that ⊂ By repeating this procedure we can get B Y = 2 and Y is a basis of exactly one of • M,| # N. | some smaller counterexamples M, N (minors of the original matrices) with B Proof of claim: If B Y = 2 then Y = ∩ B b+s for some b B|, s# S| B. In both M X = and B X = S. − ∈ ∈ \ ∅ ∪ and N, B+s contains the same unique circuit Without loss of generality suppose X is a ba- Cs. Then B + s b is a basis in M or N iff sis of M only. Note that X is now the only − b Cs so that removing b breaks the circuit. wrong set of S. If there were a different wrong ∈ So Y cannot be a basis of only one of M, N. set Y, then B Y < B X , violating the This is the relationship between the non- assumed minimality| # | of | B# X| . binary M and the binary N outlined in the We now find a basis D| of# M with| B D = strategy. Now proving the following lemma 2. Pick an x X. B + x has a unique| # cir-| will complete the theorem. cuit in M. Remove∈ one of the elements of the

2 circuit that is a member of B, say b. Now By DC on b, c , c is independent in D = B + x b is a basis of M because it con- • N. { } { } tains no circuit.− D = X (since B D = 2, ' | # | while B X 4). Therefore D is common Use the exchange property on c and | # |≥ basis of M and N. • a, d in N. Either: a, c is independent{ } Note that D X < B X . We chose in{ N,} but not in M, which{ } contradicts the | # | | # | B X to be minimum such that definition of B as a far basis since they | # | B is a far basis and have symmetric difference 2; or c, d = • X is independent in N, contradicting{ } the X is wrong. definition of X as wrong. • D and X threaten to break this minimum. It must be that D is not a far basis. Thus there So we reach a contradiction from assuming 2 is some wrong Y with D Y = 2. Since X that M = U4 – that is, from the assump- ' is the only wrong thing| around,# | we conclude tion that neither of the original matroids con- 2 that D X = 2. tained a U4 minor. So one of them must have. Now| # from| B X = 0, B D = This completes the proof that the binary ma- | ∩ | | # | 2 2, D X = 2, we conclude that S = troids are exactly those that have no U4 mi- B | X#=| 4 and B = X = 2. | | nor. | Concretely,# | let| |S |=| a, b, c, d ,B = a, b ,X = c, d . We have{ assumed} that { } 2 { } M = U4 , so there is some 2-element subset 3 A non-representable of S' that is dependent in M. Without loss of generality suppose it is a, c . matroid Now some elementary{ reasoning} that leads to a contradiction: Next we look at representations of some specific matroids and construct a matroid that By DC on B, a is independent in M. • { } is not representable over any field. The fig- By EX on a and X, a, d is indepen- ures represent two matroids: the non-Fano • dent in M.{ } { } matroid and the Fano matroid. In the figures, the vertices represent the elements of By DC on X, c is independent in M. the ground set. The bases of the matroids • { } are all 3-element sets except those with a line By EX on c and B, b, c is indepen- through them. A simple check of the ex- • dent in M. { } { } change property verifies that these rules and Both a, d and b, c must be indepen- the figures define two matroids. • dent in{ N also,} or{ else} they would wrong Below is a representation of the Fano ma- and have a symmetric difference 2 with troid F7 over GF(2). Interestingly, it is also B, a far basis. a representation of the non-Fano matroid F7−

3 to 1. Next scale rows 2 and 3 to set the last column equal to all 1s. Finally scale columsn 2, 3, and 6 again to set their ﬁrst nonzero entries equal to 1. We are left with only 3 undetermined entries:

1 0 0 1 1 0 1 Figure 1: On the left, the non-Fano matroid. 0 1 0 a 0 1 1 On the right, the Fano matroid. (taken from   0 0 1 0 bc1 [3])   Finally the values of a, b, and c are con- if the field is taken to be R. strained to all be 1 by the fact that 3, 4, 7 , 2, 5, 7 , and 1, 6, 7 are all dependent.{ So} if 1 0 0 1 1 0 1 { } { } A = 0 1 0 1 0 1 1 there is any representation, then our original   0 0 1 0 1 1 1 matrix A is one.   Interestingly, the above reasoning never re- In fact, we can show that if M F ,F− lied on whether 4, 5, 6 was independent or 7 7 { } is representable over some field F,∈{ then the} dependent. So the above applies equally well above matrix is a representation over F. As- to F7 and F7−. What this tells us is that there suming M is representable over our field is no field over which there is a representa-

F, let’s build a representation of M. B = tion of both F7 and F7−. If there were, both 1, 2, 3 is a basis, so we can make the first matroids would be represented by A on that 3{ columns} the orthogonal unit vectors, as in field. But then they would have to be the A. Next we look at the unique circuits con- same matroid. Since they are different ma- tained in B + s, s 4, 5, 6, 7 . For example, troids, there can be no such field. ∈{ } 1, 2, 4 is a minimal dependent set, so the This makes it easy to create a matroid with { } 4th column must have non-zero entries in the no representation over any field. Define the first and second row, and a zero entry in the direct sum of matroids M =(S1,I1) and N = third. Carrying on in this way we find that (S2,I2) defined over disjoint ground sets S1, our representation must have non-zero entries S2 by distributed as below: M N =(S S , X Y X I ,Y I ). 1 0 0 0 ⊕ 1∪ 2 { ∪ | ∈ 1 ∈ 2} 0 1 0 ∗∗0 ∗   0 0 1∗ 0 ∗∗ Then M and N are both minors of M N, ∗ ∗ ∗ for instance M =(M N) N. Therefore⊕ if   We can scale rows and columns to set most M N is representable⊕ over\ some field F, so ⊕ of these entries to 1. Scale the columns to set are M and N. From this it follows that F7 F7− the first nonzero entry in each column equal cannot be represented over any field. ⊕

4 4 Problems

1. Show that if M is a matroid representable over a ﬁeld F, then any minor of M is representable over F (hint: it’s easy to see that a deletion of M is representable over F; it only remains to show that the dual M ! is too. A contraction is just the operation dual + deletion + dual).

5 References

(1) Michel Goemans’s lecture • notes at http://math.mit.edu/ goemans/18438/lec9.pdf

(2) A. Schrijver. Combinatorial Opti- • mization: Polyhedra and Eﬃciency, vol- ume B. Springer, 2003.

(3) J. Oxley. What is a matroid? Cubo, • 5:179218, 2003.

(4) J.G. Oxley. Matroid Theory. Oxford • University Press, 1992.