Yes, the "missing axiom" of matroid theory is lost forever
YYES, THE “MISSING AXIOM” OF MATROID THEORY ISLOST FOREVER
DILLON MAYHEW, MIKE NEWMAN, AND GEOFF WHITTLE
Abstract.
We prove there is no sentence in the monadic second-orderlanguage MS that characterises when a matroid is representable overat least one field, and no sentence that characterises when a matroid is K -representable, for any infinite field K . By way of contrast, becauseRota’s Conjecture is true, there is a sentence that characterises F -rep-resentable matroids, for any finite field F . Introduction
A matroid captures the notion of a discrete collection of points in space.Sometimes these points can be assigned coordinates in a consistent way, andsometimes they cannot. The problem of characterising when a matroid is representable has been the prime motivating force in matroid research sinceWhitney’s founding paper [13].Plenty of effort has been invested in characterising matroid representabil-ity via excluded minors. Less attention has been paid to the prospect of char-acterisating representability via axioms. Perhaps this is because of V´amos’swell-known article [12], which has been interpreted as stating that no suchcharacterisation exists (see [4]). In [9], we pointed out that the possibility ofcharacterising representable matroids in the language of Whitney’s axiomswas still open; that, in other words, we still did not know if “the missingaxiom of matroid theory is lost forever”, contra
V´amos’s title. We conjec-tured that in fact there was no such characterisation, and we made somepartial progress towards resolving the conjecture by showing that it was im-possible to characterise the class of representable matroids, or the class ofmatroids representable over an infinite field, using a logical language basedon the rank function. However, that language imposed quite strong con-straints on the form of quantification. In this article, we present a languagewith no such constraints, and we prove that it is impossible to characteriserepresentability or representability over an infinite field in this more naturallanguage. This is not to say that representability cannot be characterised instronger languages: indeed, any language will suffice if it is strong enoughto express the statement that the independent sets are in correspondencewith the linearly independent sets of columns in a matrix.
Date : October 11, 2018. a r X i v : . [ m a t h . C O ] M a r MAYHEW, NEWMAN, AND WHITTLE
The language that we develop here is a form of monadic second-order logicfor matroids (similar to that used by Hlinˇen´y [6]), which we denote MS . Aswe show in Section 2, MS is expressive enough to state the matroid axioms,and to state when a matroid contains an isomorphic copy of a fixed minor.This means that any minor-closed class of matroids can be characterisedwith an MS sentence, as long as it has a finite number of excluded minors.In particular, since Rota’s Conjecture has been positively resolved by Geelen,Gerards, and Whittle (see [5]), it follows that the class of F -representablematroids can be characterised by a sentence in MS whenever F is a finitefield. Our main results show that this is not the case for infinite fields.Nor is it possible to characterise the matroids that are representable overat least one field using an MS sentence. When we say that a matroid is representable we mean it is representable over at least one field. Theorem 1.1.
There is no sentence, ψ , in MS , such that a matroid isrepresentable if and only if it satisfies ψ . Theorem 1.2.
Let K be any infinite field. There is no sentence, ψ K , inMS , such that a matroid is K -representable if and only if it satisfies ψ K . These theorems may seem stronger than those in [9], but in fact theresults are independent of each other. The logical language used in [9] hadconstraints on quantification, unlike MS , but it also had access to the rankfunction, and to the arithmetic of the integers, while MS does not.Theorems 1.1 and 1.2 follow easily from two lemmas. Let k be a positiveinteger. Let M and M be matroids. We will say that a k -certificate for M and M is a pair, ( M (cid:48) , ψ ), where M (cid:48) is a matroid satisfying E ( M (cid:48) ) ∩ ( E ( M ) ∪ E ( M )) = ∅ , and ψ is a sentence in MS with k variables such that ψ is satisfied by exactly one of the direct sums M ⊕ M (cid:48) and M ⊕ M (cid:48) . Wedefine M and M to be k -equivalent if there is no k -certificate for M and M . This relation is obviously reflexive and symmetric. Assume that M is k -equivalent to M , and M is k -equivalent to M , but that ( M (cid:48) , ψ ) is a k -certificate for M and M . Relabelling the ground set of a matroid has noeffect on whether it satisfies a sentence in MS . Therefore we can assumethat E ( M (cid:48) ) is disjoint from E ( M ) ∪ E ( M ) ∪ E ( M ). Now ( M (cid:48) , ψ ) is a k -certificate for M and M , or for M and M , a contradiction. Therefore k -equivalence truly is an equivalence relation.If two matroids are k -equivalent, then no k -variable sentence can distin-guish them, even after adjoining an arbitrary matroid via a direct sum. Lemma 1.3.
Let k be a positive integer. There are only finitely manyequivalence classes of matroids under the relation of k -equivalence. In Section 3, we will find an explicit bound on the number of equivalenceclasses. By using Lemma 1.3, we can easily deduce Theorem 1.1.
Proof of
Theorem 1.1 . Assume that there is a sentence, ψ , in MS , thatcharacterises representable matroids. Let k be the number of variables in HE “MISSING AXIOM” OF MATROID THEORY 3 ψ . We apply Lemma 1.3. Because there are infinitely many prime numbers,we can assume that M and M are k -equivalent, where M ∼ = PG(2 , p ) and M ∼ = PG(2 , p (cid:48) ) for distinct primes, p and p (cid:48) . We choose M (cid:48) to be isomorphicto M , where E ( M (cid:48) ) ∩ ( E ( M ) ∪ E ( M )) = ∅ . Then ψ is satisfied by both M ⊕ M (cid:48) and M ⊕ M (cid:48) , or it is satisfied by neither. But M ⊕ M (cid:48) ∼ =PG(2 , p ) ⊕ PG(2 , p ) is representable over GF( p ) [11, Proposition 4.2.11].On the other hand, both PG(2 , p (cid:48) ) and PG(2 , p ) are isomorphic to minorsof M ⊕ M (cid:48) [11, 4.2.19], so it follows from [11, Proposition 3.2.4] and [1,Proposition 7.3] that if M ⊕ M (cid:48) is representable over a field, then thatfield must simultaneously have subfields isomorphic to GF( p ) and GF( p (cid:48) ),an impossibility. To summarise, M ⊕ M (cid:48) is representable, and M ⊕ M (cid:48) isnot, but ψ is satisfied by both, or by neither. Thus ψ certainly does notcharacterise representable matroids. (cid:3) The notion of k -equivalence is reminiscent of the Myhill-Nerode character-isation of regular languages (see [10] or [3, Section 6.1]). Lemma 1.3 is alsoa matroid analogue of the fact that a graph property definable in monadicsecond-order logic can be recognised by an automaton [2], and is therefore finite , in the sense of Lengauer and Egon [7]. By way of contrast, the theo-rem in [9] used a proof technique that was essentially an Ehrenfeucht-Fra¨ıss´egame (see [3, Section 2.2]). Note that if two matroids are k -equivalent, thenthey satisfy exactly the same k -variable sentences (since the empty matroidis not a k -certificate). This implies the known fact that there are only finitelymany rank- k -types (see [8, Section 3.4] for an explanation).Our second lemma will be used to prove Theorem 1.2. In this case, it willnot suffice to use direct sums, as the sum of two K -representable matroids isalso K -representable. Thus we use the notion of a proper amalgam (whichwill be precisely defined in Section 4). Let M (cid:96) be the set of matroidsthat contain a U , -restriction on the set (cid:96) = { a, b, x, y, z } . If M and M are matroids in M (cid:96) , and E ( M ) ∩ E ( M ) = (cid:96) , then the proper amalgamof M and M exists, and is denoted by Amal( M , M ). The ground setof Amal( M , M ) is E ( M ) ∪ E ( M ), and Amal( M , M ) | E ( M i ) = M i for i = 1 , k be a positive integer. Let M and M be matroids in M (cid:96) . A( k, (cid:96) ) -certificate is a pair, ( M (cid:48) , ψ ), where M (cid:48) ∈ M (cid:96) satisfies E ( M (cid:48) ) ∩ ( E ( M ) ∪ E ( M )) = (cid:96) , and ψ is a k -variable sentence that is satisfied byexactly one of Amal( M , M (cid:48) ) and Amal( M , M (cid:48) ). We say that M and M are ( k, (cid:96) ) -equivalent if there is no such certificate. Lemma 1.4.
Let k be a positive integer. There are only only finitely manyequivalence classes of M (cid:96) under the relation of ( k, (cid:96) ) -equivalence. Again, we will explicitly bound the number of equivalence classes.In Section 5, we will construct two families of matroids in M (cid:96) by using gain graphs . Loosely speaking, a gain graph is a graph equipped with edgelabels that come from a group. For each such graph, there is a correspondinggain-graphic matroid, whose ground set is the edge set of the graph. Let K MAYHEW, NEWMAN, AND WHITTLE be a field, let s, t ≥ α and β be elements in K − { } with orders greater than, respectively, s and 2 t ( t − K , s, α ) and ( K , t, β ), there are unique gain graphs, which we willdenote by Γ( K , α, s ) and ∆( K , β, t ). (We postpone the exact descriptionsuntil Section 5.) The edge labels of Γ( K , α, s ) and ∆( K , β, t ) come from themultiplicative group of K .Assume that M corresponds to the gain graph Γ( K , α, s ), and that M (cid:48) corresponds to ∆( K , β, t ). We also assume that E ( M ) ∩ E ( M (cid:48) ) = (cid:96) . In thecase that α = β , where the order of α is greater than max { s, t ( t − } , both M and M (cid:48) can be represented over K , but Amal( M, M (cid:48) ) can be representedover K if and only if s = t . This means that Lemma 1.4 quickly leads to aproof of Theorem 1.2, with the two families of gain-graphic matroids playingthe same role that projective planes did in the proof of Theorem 1.1. Detailsof the proof will be left until the end of the paper.In fact, Lemma 1.4 is sufficient to prove both Theorem 1.1 and Theo-rem 1.2, since, if Amal( M, M (cid:48) ) is not representable over the field K , thenit is not representable over any field (Lemma 5.3). However, we feel thatLemma 1.3 is more intuitive, and also interesting in its own right, so weprefer to prove that lemma, and then note the changes required to producea proof of Lemma 1.4.Lemma 1.4 also implies the following (unsurprising) facts: using MS tocharacterise increasingly large finite fields requires increasingly large sen-tences. Furthermore, it is not possible to axiomatise the class of matroidsrepresentable over a given characteristic. Corollary 1.5.
Let Q be the set of prime powers. For each q ∈ Q , let ψ q bean MS sentence such that a matroid is GF( q ) -representable if and only ifit satisfies ψ q . There is no integer, N , such that every sentence in { ψ q } q ∈Q has at most N variables. Corollary 1.6.
Let c be either or a prime number. There is no sentence, ψ c , in MS such that a matroid is representable over a field of characteristic c if and only if it satisfies ψ c . The paper is structured as follows: Section 2 introduces the MS languagefor matroids, and discusses its expressive power; Section 3 gives a proof ofLemma 1.3; in Section 4 we define the proper amalgam of matroids along a U , -restriction, and prove some of its properties; Section 5 introduces gain-graphic matroids, and defines the two special classes of matroids. Finally,in Section 6, we prove Lemma 1.4, and complete the proof of Theorem 1.2,and Corollaries 1.5 and 1.6. For all matroid essentials we refer to Oxley [11].2. Monadic second-order logic
In this section we give a formal definition of our monadic second-orderlanguage for matroids. The language MS includes a countably infinitesupply of variables, X , X , X , . . . along with the binary predicate, ⊆ , the HE “MISSING AXIOM” OF MATROID THEORY 5 unary predicates, Sing and Ind, as well as the standard connectives ∧ and ¬ , and the quantifier ∃ .We recursively define formulas in MS , and simultaneously define theirsets of variables. The following statements define expressions known as atomic formulas .(1) X i ⊆ X j is an atomic formula, for any variables X i and X j , andVar( X i ⊆ X j ) = { X i , X j } .(2) Sing( X i ) is an atomic formula, for any variable X i , andVar(Sing( X i )) = { X i } .(3) Ind( X i ) is an atomic formula, for any variable X i , andVar(Ind( X i )) = { X i } .A formula is an expression generated by a finite application of the follow-ing rules. Every formula has an associated set of variables and free variables :(1) Every atomic formula, ψ , is a formula, and Fr( ψ ) = Var( ψ ).(2) If ψ is a formula, then ¬ ψ is a formula, and Var( ¬ ψ ) = Var( ψ ),while Fr( ¬ ψ ) = Fr( ψ ).(3) If ψ and ψ are formulas, and Fr( ψ i ) ∩ (Var( ψ j ) − Fr( ψ j )) = ∅ for { i, j } = { , } , then ψ ∧ ψ is a formula, and Var( ψ ∧ ψ ) =Var( ψ ) ∪ Var( ψ ), while Fr( ψ ∧ ψ ) = Fr( ψ ) ∪ Fr( ψ ).(4) If ψ is a formula and X i ∈ Fr( ψ ), then ∃ X i ψ is a formula, andVar( ∃ X i ψ ) = Var( ψ ), while Fr( ∃ X i ψ ) = Fr( ψ ) − { X i } .A variable in Var( ψ ) is free if it is in Fr( ψ ), and bound otherwise. A formulais quantifier-free if all of its variables are free, and is a sentence if all itsvariables are bound. If ψ is a quantifier-free formula, then we will define the depth of ψ to be the number of applications of Rules (2) and (3) required toconstruct ψ . Rule (3) insists that no variable can be free in one of ψ and ψ and bound in the other, if ψ ∧ ψ is to be a formula. We can overcomethis constraint if necessary by renaming the bound variables in a formula.If ψ is a formula and X i ∈ Fr( ψ ), then we use ∀ X i ψ as a shorthand for ¬ ( ∃ X i ¬ ψ ). We also use the shorthand ψ ∨ ψ to mean ¬ (( ¬ ψ ) ∧ ( ¬ ψ ))and we use ψ → ψ to mean ( ¬ ψ ) ∨ ψ . Likewise, we use ψ ↔ ψ to mean( ψ → ψ ) ∧ ( ψ → ψ ). We use X (cid:42) Y to stand for ¬ ( X ⊆ Y ).Let ψ be a formula in MS . An interpretation of ψ is a pair ( M, τ ), where M = ( E, I ) consists of a set, E , and a collection, I , of subsets of E , and τ is a function from Fr( ψ ) into the power set of E . We will recursively definewhat it means for ( M, τ ) to satisfy ψ , starting with the case that ψ is atomic.If ψ is X i ⊆ X j , then ( M, τ ) satisfies ψ if and only if τ ( X i ) ⊆ τ ( X j ). If ψ isSing( X i ), then ( M, τ ) satisfies ψ if and only if | τ ( X i ) | = 1. Finally, if ψ isInd( X i ), then ( M, τ ) satisfies ψ if and only if τ ( X i ) is in I .Now we assume that ψ is not atomic. If ψ is ¬ φ for some formula φ , then( M, τ ) satisfies ψ is if and only if ( M, τ ) does not satisfy φ . Assume that ψ is φ ∧ φ . Then ( M, τ ) satisfies ψ if and only if ( M, τ (cid:22)
Fr( φ ) ) satisfies φ and( M, τ (cid:22)
Fr( φ ) ) satisfies φ . Finally, assume that ψ is ∃ X i φ , where X i is a freevariable in the formula φ . Then ( M, τ ) satisfies ψ if and only if there exists MAYHEW, NEWMAN, AND WHITTLE a subset, Y i ⊆ E , such that the interpretation ( M, τ ∪ { ( X i , Y i ) } ) satisfies φ . If ψ is an MS sentence, then we say that M = ( E, I ) satisfies ψ (or ψ is satisfied by M ) if the interpretation ( M, ∅ ) satisfies ψ .We will spend some time illustrating the expressive power of MS . It ispowerful enough to state the axioms for matroids, and to characterise whena matroid contains a fixed minor.If t ≥ t ( X i , . . . , X i t , X i t +1 ) as shorthand forthe formula ∀ X Sing( X ) → ( X ⊆ X i t +1 ↔ (cid:95) ≤ j ≤ t X ⊆ X i j ) . The variable X stands for some variable different from each of X i , . . . , X i t +1 .Clearly the formula Union t ( X i , . . . , X i t , X i t +1 ) is satisfied by the interpre-tation ( M, τ ) if and only if τ ( X i t +1 ) is equal to τ ( X i ) ∪ · · · ∪ τ ( X i t ).We let Max( X i ) stand for the formulaInd( X i ) ∧ ( ∀ X X i ⊆ X → ( X ⊆ X i ∨ ¬ Ind( X ))) . Therefore Max( X i ) is satisfied by τ in M = ( E, I ) if and only if τ ( X i ) is amaximal member of I .Let E be a finite set, and let I be a collection of subsets of E . Then I isthe family of independent sets of a matroid, M = ( E, I ), if and only if M satisfies the following sentences: I1. ∃ X Ind( X ) I2. ∀ X ∀ X (Ind( X ) ∧ ( X ⊆ X )) → Ind( X ) I3. ∀ X ∀ X (Max( X ) ∧ Ind( X ) ∧ ¬ Max( X )) →∃ X Sing( X ) ∧ ( X ⊆ X ) ∧ ( X (cid:42) X ) ∧∃ X (Union ( X , X , X ) ∧ Ind( X ))The sentence I3 declares that if X is a maximal set in I , and X is a non-maximal set, then there is an element x ∈ X − X such that X ∪ { x } isin I . It is not difficult to show that these axioms imply that the maximalmembers of I are equicardinal. From this it follows immediately that themaximal members of I obey the matroid basis axioms. Therefore I1 , I2 ,and I3 axiomatise matroids, as claimed.Next we let N be a fixed matroid on the ground set { , . . . , n } , with I as its collection of independent sets. Let D be the set of dependentsubsets of N . A matroid has a minor isomorphic to N if and only if itcontains distinct elements x , . . . , x n , and an independent set, X n +1 , suchthat { x , . . . , x n } ∩ X n +1 = ∅ , and { x i , . . . , x i t } ∪ X n +1 is independentprecisely when { i , . . . , i t } is an independent set of N . In this case, N isisomorphic to the minor produced by contracting X n +1 and restricting tothe set { x , . . . , x n } . Thus we see that a matroid has a minor isomorphic to HE “MISSING AXIOM” OF MATROID THEORY 7 N if and only if it satisfies the following sentence: ∃ X · · ·∃ X n ∃ X n +1 Ind( X n +1 ) ∧ (cid:94) ≤ i ≤ n (Sing( X i ) ∧ ( X i (cid:42) X n +1 )) ∧ (cid:94) ≤ i Let k be a positive integer. Define g ( k, 0) to be 2 k ( k +1) k , and recursivelydefine g ( k, n +1) to be 2 g ( k,n ) . Let f ( k ) be g ( k, k ). Our goal in this sectionis to prove Lemma 1.3. We restate the lemma here, with an explicit boundon the number of equivalence classes. Lemma 3.1. Let k be a positive integer. There are at most f ( k ) equivalenceclasses of matroids under the relation of k -equivalence.Proof. We define a registry to be a ( k + 2) × k matrix with rows indexed byInd, Sing, and X , . . . , X k , and columns indexed by X , . . . , X k . An entryin row Ind or in row X i must be ‘T’ or ‘F’. An entry in the row indexedby Sing is either ‘0’, ‘1’, or ‘ > ’. It follows that there are at most g ( k, depth- tree to be a registry. Recursively, a depth- ( n + 1) tree is a non-empty set of depth- n trees. An easy inductive argument shows thatthere are no more than g ( k, n + 1) depth-( n + 1) trees, and hence no morethan f ( k ) depth- k trees.A stacked matroid is a tuple M = ( M, Y , . . . , Y m ), where M is a matroid,and each Y i is a subset of E ( M ). We define ||M|| to be m . We can identifythe matroid M with the stacked matroid M = ( M ), and note that in thiscase, ||M|| = 0.To each stacked matroid, M , satisfying ||M|| ≤ k , we are going to as-sociate a tree, T ( M ), of depth k − ||M|| . We start by assuming that k − ||M|| = 0, so that T ( M ) is a depth-0 tree, which is to say, a reg-istry. Let M be ( M, Y , . . . , Y k ). For every j in { , . . . , k } , set the entry of T ( M ) in row Ind and column X j to be ‘T’ if Y j is independent in M , andotherwise set it to be ‘F’. Now, for every pair i, j ∈ { , . . . , k } , set the entryof T ( M ) in row X i and column X j to be ‘T’ if and only if Y i ⊆ Y j . Finally,for each j ∈ { , . . . , k } , set the entry of T ( M ) in row Sing and column X j MAYHEW, NEWMAN, AND WHITTLE to be ‘0’ if | Y j | = 0, set it to be ‘1’ if | Y i | = 1, and set it to ‘ > ’ otherwise.This defines T ( M ) in the case that k − ||M|| = 0.Now we make the inductive assumption that T ( M ) is defined when k − ||M|| ≤ n , where n is some integer in { , . . . , k − } . Let M =( M, Y , . . . , Y k − n − ) be a stacked matroid. Thus k − ||M|| = n + 1. Let Y k − n be any subset of E ( M ). If M (cid:48) = ( M, Y , . . . , Y k − n − , Y k − n ), then k − ||M (cid:48) || = n , so our inductive assumption means that T ( M (cid:48) ) is definedand is a depth- n tree. Since a depth-( n + 1) tree is a non-empty set ofdepth- n trees, we simply define T ( M ) to be the set {T ( M, Y , . . . , Y k − n − , Y k − n ) : Y k − n ⊆ E ( M ) } . We have now defined T ( M ) for each stacked matroid, M , that satisfies ||M|| ≤ k . Note that if M is a matroid, then the stacked matroid M = ( M )satisfies ||M|| = 0, and hence T ( M ) is a depth- k tree.Let ψ be a formula in MS such that either ψ is quantifier-free, orVar( ψ ) = { X , . . . , X k } . Let b ( ψ ) be the number of bound variables in ψ . We are going to define what it means for T and T (cid:48) to be ψ -compatible when T and T (cid:48) are depth- b ( ψ ) trees.In the first case, we assume that ψ is quantifier-free, so that b ( ψ ) = 0,and T and T (cid:48) are depth-0 trees; that is, registries. To start with, we assumethat ψ is an atomic formula. If ψ is X i ⊆ X j , then we define T and T (cid:48) to be ψ -compatible if and only if their entries in row X i and column X j are both‘T’. Similarly, if ψ is Ind( X j ), then we define T and T (cid:48) to be ψ -compatibleif and only both T and T (cid:48) have ‘T’ as their entries in row Ind and column X j . Next we assume that ψ is Sing( X j ). Let ω be the entry of T in rowSing and column X j . Let ω (cid:48) be the analogous entry of T (cid:48) . We define T and T (cid:48) to be ψ -compatible if and only if { ω, ω (cid:48) } = { ‘0’ , ‘1’ } .This defines ψ -compatibility in the case that ψ is atomic, so we will nowassume it is not atomic. Since it is quantifier-free, this means that ψ has theform ¬ φ or φ ∧ φ . First assume that ψ is ¬ φ , where φ is quantifier-free. Byinduction on the depth of quantifier-free formulas, we can determine whetheror not T and T (cid:48) are φ -compatible. We define T and T (cid:48) to be ψ -compatibleif and only if T and T (cid:48) are not φ -compatible. Next assume that ψ is φ ∧ φ .Again, φ and φ have no bound variables, and by induction on the depth ofquantifier-free formulas, we can determine whether T and T (cid:48) are compatiblerelative to φ and φ . We define T and T (cid:48) to be ψ -compatible if and only if T and T (cid:48) are both φ -compatible and φ -compatible. We have now defined ψ -compatibility in the case that ψ has no bound variables.Next we will assume that Var( ψ ) = { X , . . . , X k } . By the previous para-graphs, we can make the inductive assumption that ψ -compatibility is de-fined if b ( ψ ) ≤ n , where n is some integer in { , . . . , k − } . Let ψ be a formulawith Var( ψ ) = { X , . . . , X k } and assume that ψ has n + 1 bound variables.By renaming variables, we will assume that Fr( ψ ) = { X , . . . , X k − n − } , andthat X k − n , . . . , X k are the bound variables of ψ . By standard techniques, HE “MISSING AXIOM” OF MATROID THEORY 9 we can assume that ψ is in prenex normal form . That is, ψ = Q k − n X k − n · · · Q k X k ψ (cid:48) where each Q j is either ∃ or ∀ , and ψ (cid:48) is a quantifier-free formula in MS withVar( ψ (cid:48) ) = { X , . . . , X k } . Let φ be the formula Q k − n +1 X k − n +1 · · · Q k X k ψ (cid:48) obtained from ψ by removing the quantification of X k − n .Let T and T (cid:48) be trees of depth b ( ψ ) = n +1. Thus T and T (cid:48) are non-emptyset of depth- n trees. First consider the case that Q k − n = ∃ . The number ofbound variables in φ is n . If T is a depth- n tree contained in T , and T (cid:48) isa depth- n tree in T (cid:48) , then by the inductive hypothesis, φ -compatibility of T and T (cid:48) is defined. We define T and T (cid:48) to be ψ -compatible if and only ifthere exist trees, T ∈ T and T (cid:48) ∈ T (cid:48) that are φ -compatible.Similarly, if Q k − n = ∀ , we define T and T (cid:48) to be ψ -compatible if and onlyif T and T (cid:48) are φ -compatible for every tree T ∈ T and every tree T (cid:48) ∈ T (cid:48) .This completes the definition of ψ -compatibility.The following claim contains the heart of the proof of Lemma 3.1. Claim 3.1.1. Let ψ be an MS formula such that either ψ is quantifier-free, or Var( ψ ) = { X , . . . , X k } . If Var( ψ ) = { X , . . . , X k } , then let m be | Fr( ψ ) | and assume that Fr( ψ ) = { X , . . . , X m } . Otherwise, let m be k .Let M = ( M, Y , . . . , Y m ) and M (cid:48) = ( M (cid:48) , Y (cid:48) , . . . , Y (cid:48) m ) be stacked matroids,where E ( M ) ∩ E ( M (cid:48) ) = ∅ . Define τ to be the function that takes X i to Y i ∪ Y (cid:48) i , for each X i ∈ Fr( ψ ) . The interpretation ( M ⊕ M (cid:48) , τ ) satisfies ψ ifand only if the trees, T ( M ) and T ( M (cid:48) ) , are ψ -compatible.Proof. Let b ( ψ ) be the number of bound variables in ψ . We will prove theclaim by induction on b ( ψ ). Note that T ( M ) and T ( M (cid:48) ) both have depth k − m = b ( ψ ).For our base case, we assume that b ( ψ ) = 0, so that ψ is quantifier-free, ||M|| = ||M (cid:48) || = k , and both T ( M ) and T ( M (cid:48) ) are registries. Start byassuming that ψ is an atomic formula. Consider the case that ψ is X i ⊆ X j .Then ( M ⊕ M (cid:48) , τ ) satisfies ψ if and only if τ ( X i ) ⊆ τ ( X j ), which is true ifand only if Y i ⊆ Y j and Y (cid:48) i ⊆ Y (cid:48) j . But this is the case if and only if T ( M )and T ( M (cid:48) ) both contain ‘T’ in row X i and column X j . This is exactly whatit means for T ( M ) and T ( M (cid:48) ) to be ψ -compatible, so we are done in thiscase.In our next case, ψ is Ind( X j ). Then ( M ⊕ M (cid:48) , τ ) satisfies ψ if and only if τ ( X j ) is independent in M ⊕ M (cid:48) . This is true if and only if Y j is independentin M and Y (cid:48) j is independent in M (cid:48) . In turn, this is true if and only if T ( M )and T ( M (cid:48) ) both contain ‘T’ in row Ind and column X j , which is the case ifand only if T ( M ) and T ( M (cid:48) ) are ψ -compatible.Next, we assume that ψ is Sing( X j ). Then ( M ⊕ M (cid:48) , τ ) satisfies ψ if andonly if | τ ( X j ) | = 1, and this is true if and only if {| Y j | , | Y (cid:48) j |} = { , } . Thisholds if and only if the entries of T ( M ) and T ( M (cid:48) ) in row Sing and column X j are ‘0’ and ‘1’, in some order. Once again, this is true precisely when T ( M ) and T ( M (cid:48) ) are ψ -compatible. We have finished the case that ψ isatomic, so now we assume that ψ is not atomic.Since ψ is quantifier-free, it has the form ¬ φ or φ ∧ φ . Consider theformer case. By induction on the depth of quantifier-free formulas, we canconclude that T ( M ) and T ( M (cid:48) ) are φ -compatible if and only if ( M ⊕ M (cid:48) , τ )satisfies φ . The definition of compatibility means that T ( M ) and T ( M (cid:48) )are ψ -compatible if and only if they are not φ -compatible, which is the caseexactly when ( M ⊕ M (cid:48) , τ ) satisfies ψ .Next we assume that ψ is φ ∧ φ , where φ and φ have no boundvariables. Again, we use induction on the depth of quantifier-free formulas.We conclude that ( M ⊕ M (cid:48) , τ (cid:22) Fr( φ α ) ) satisfies φ α if and only if T ( M ) and T ( M (cid:48) ) are φ α -compatibile, for α = 1 , 2. This holds if and only if T ( M ) and T ( M (cid:48) ) are ψ -compatibile. Thus we have proved the claim in the case that b ( ψ ) = 0.We make the inductive assumption that the claim holds when the num-ber of bound variables is at most n , for some integer n ∈ { , . . . , k − } .Consider the case that b ( ψ ) = n + 1. We have assumed that Fr( ψ ) = { X , . . . , X k − n − } , and we can also assume that ψ = Q k − n X k − n · · · Q k X k ψ (cid:48) where each Q j is a quantifier, and ψ (cid:48) is quantifier-free and satisfies Var( ψ (cid:48) ) = { X , . . . , X k } . Let φ be Q k − n +1 X k − n +1 · · · Q k X k ψ (cid:48) .Consider the case that Q k − n = ∃ . Then ( M ⊕ M (cid:48) , τ ) satisfies ψ if andonly if there are subsets Y k − n ⊆ E ( M ) and Y (cid:48) k − n ⊆ E ( M (cid:48) ) such that( M ⊕ M (cid:48) , τ ∪ { ( X k − n , Y k − n ∪ Y (cid:48) k − n ) } )satisfies φ . By the inductive assumption, this holds if and only if T ( M, Y , . . . , Y k − n − , Y k − n ) and T ( M (cid:48) , Y (cid:48) , . . . , Y (cid:48) k − n − , Y (cid:48) k − n ) are φ -compat-ible. Now T ( M, Y , . . . , Y k − n − , Y k − n ) is a depth- n tree contained in thedepth-( n + 1) tree T ( M ), and T ( M (cid:48) , Y (cid:48) , . . . , Y (cid:48) k − n − , Y (cid:48) k − n ) is similarly con-tained in T ( M (cid:48) ). Thus the recursive definition of compatibility means that( M ⊕ M (cid:48) , τ ) satisfies ψ if and only if T ( M ) and T ( M (cid:48) ) are ψ -compatibile,exactly as desired.The case when Q k − n = ∀ is similar. In this case, ( M ⊕ M (cid:48) , τ ) satisfies ψ ifand only if ( M ⊕ M (cid:48) , τ ∪ { ( X k − n , Y k − n ∪ Y (cid:48) k − n ) } ) satisfies φ for every choiceof subsets Y k − n ⊆ E ( M ) and Y (cid:48) k − n ⊆ E ( M (cid:48) ). By induction, this is trueif and only if T ( M, Y , . . . , Y k − n − , Y k − n ) and T ( M (cid:48) , Y (cid:48) , . . . , Y (cid:48) k − n − , Y (cid:48) k − n )are φ -compatible, for every choice of Y k − n and Y (cid:48) k − n . This holds if and onlyif T and T (cid:48) are φ -compatible, for all trees T ∈ T ( M ) and T (cid:48) ∈ T ( M (cid:48) ).This is exactly what it means for T ( M ) and T ( M (cid:48) ) to be ψ -compatible, sothe proof is complete. (cid:3) Let M and M be two matroids, which we consider as stacked matroids M = ( M ) and M = ( M ). We complete the proof of Lemma 3.1 byshowing that if the trees T ( M ) and T ( M ) are equal, then M and M are k -equivalent. This will imply that the number of equivalence classes isat most the number of depth- k trees, and we will be done. Thus we assumethat T ( M ) = T ( M ).Let M (cid:48) be any matroid with E ( M (cid:48) ) ∩ ( E ( M ) ∪ E ( M )) = ∅ , and let M (cid:48) =( M (cid:48) ) be the corresponding stacked matroid. Let ψ be any MS sentence withVar( ψ ) = { X , . . . , X k } . Then Claim 3.1.1 implies that M ⊕ M (cid:48) satisfies ψ if and only if T ( M (cid:48) ) is ψ -compatible with T ( M ) = T ( M ), which holdsif and only if M ⊕ M (cid:48) satisfies ψ . Therefore no k -certificate exists for M and M , so they are k -equivalent, exactly as desired. (cid:3) Amalgams Let M and M be simple matroids with ground sets E and E , rankfunctions r and r , and closure operators cl and cl . Let (cid:96) be E ∩ E ,where we assume that M | (cid:96) = M | (cid:96) . A matroid, M , on the ground set E ∪ E is an amalgam of M and M if M | E = M and M | E = M . Amatroid is modular if r ( F ) + r ( F (cid:48) ) = r ( F ∩ F (cid:48) ) + r ( F ∪ F (cid:48) ) whenever F and F (cid:48) are flats. If M | (cid:96) is a modular matroid, then [11, Theorem 11.4.10]implies that(1) r ( X ) = min { r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) : X ⊆ Y ⊆ E ∪ E } is the rank function of an amalgam of M and M , known as the properamalgam . We denote this amalgam by Amal( M , M ). Every rank-2 ma-troid is modular. (To see this, note that either r ( F ) = r ( F (cid:48) ) = 1, or one of F and F (cid:48) is contained in the other. Neither of these cases lead to a violationof modularity.) Henceforth, we consider only the case that r ( (cid:96) ) = 2. Thismeans that M | (cid:96) is modular, so that Amal( M , M ) is defined. Proposition 4.1. Assume that M i is a simple matroid with ground set E i ,rank function r i , and closure operator cl i , for i = 1 , . Let (cid:96) = E ∩ E ,where M | (cid:96) = M | (cid:96) and r ( (cid:96) ) = 2 . Let X be a subset of E ∪ E . If X ∩ E is dependent in M , or if X ∩ E is dependent in M , then X is dependent in Amal( M , M ) . If X ∩ E is independent in M and X ∩ E is independentin M , then X is dependent in Amal( M , M ) if and only if (i) (cid:96) ⊆ cl ( X ∩ E ) and r (( X − E ) ∪ (cid:96) ) < r ( X − E ) + 2 , (ii) (cid:96) ⊆ cl ( X ∩ E ) and r (( X − E ) ∪ (cid:96) ) < r ( X − E ) + 2 , or (iii) there is an element y ∈ (cid:96) such that y ∈ cl ( X − E ) ∩ cl ( X − E ) .Proof. If X ∩ E is dependent in M , then X ∩ E is dependent inAmal( M , M ), since Amal( M , M ) | E = M . By symmetry, X is depen-dent in Amal( M , M ) if X ∩ E is dependent in M or if X ∩ E is dependentin M . Henceforth we assume that X ∩ E is independent in M and X ∩ E is independent in M . Assume statement (i) holds. Let Y be X ∪ (cid:96) . Then | X | = | X ∩ E | + | X − E | = r ( X ∩ E ) + r ( X − E ) > r ( X ∩ E ) + r (( X − E ) ∪ (cid:96) ) − r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) , so by (1), the rank of X in Amal( M , M ) is less than | X | , as desired. Bysymmetric arguments, we see that if (i) or (ii) holds, then X is dependentin Amal( M , M ).Next we assume that (iii) holds. Since X ∩ E contains no circuits of M it follows that y is not in X . If X ∩ (cid:96) contains distinct elements, u and v ,then by performing circuit elimination on { y, u, v } and a circuit contained in( X − E ) ∪ y that contains y , we obtain a circuit of M contained in X ∩ E .This contradiction means that | X ∩ (cid:96) | ∈ { , } . Let Y be X ∪ y . Then | X | = | X ∩ E | + | X ∩ E | − | X ∩ (cid:96) | = r ( X ∩ E ) + r ( X ∩ E ) − | X ∩ (cid:96) | = r ( Y ∩ E ) + r ( Y ∩ E ) − ( r ( Y ∩ (cid:96) ) − > r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) . Again we see that X is dependent in Amal( M , M ), and this completes theproof of the ‘if’ direction.For the ‘only if’ direction, we assume that X is dependent inAmal( M , M ). As X ∩ E is independent in M and X ∩ E is independentin M , it follows that X is contained in neither E nor E . There is some set Y such that X ⊆ Y ⊆ E ∪ E and | X | > r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ).Assume that amongst all such sets, Y has been chosen so that it is as smallas possible. If y is an element in Y − ( X ∪ E ), then we could replace Y with Y − y . Therefore no such element exists. By symmetry it follows that Y − X ⊆ (cid:96) . If Y = X , then Y ∩ E is independent in M and Y ∩ E is inde-pendent in M , so | X | > r ( Y ∩ E )+ r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) = | Y | = | X | . Thiscontradiction means that there is an element, y , in Y − X . The minimalityof Y means that r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) < r (( Y − y ) ∩ E ) + r (( Y − y ) ∩ E ) − r (( Y − y ) ∩ (cid:96) ) . It follows that y is in cl (( Y − y ) ∩ E ) and cl (( Y − y ) ∩ E ), but notcl (( Y − y ) ∩ (cid:96) ). We combine the observations in this paragraph to deducethat | X ∩ (cid:96) | < | Y ∩ (cid:96) | < | X ∩ (cid:96) | = 1, so that | Y ∩ (cid:96) | = 2 and Y = X ∪ y . Let x be theelement in X ∩ (cid:96) . Since cl ( X ∩ E ) = cl (( Y − y ) ∩ E ) contains x and y , itcontains (cid:96) . As y is in cl (( X − E ) ∪ x ), it follows that r (( X − E ) ∪ (cid:96) ) = r (( X − E ) ∪ x ) < r ( X − E ) + 2. Therefore statement (i) holds. HE “MISSING AXIOM” OF MATROID THEORY 13 Now we assume that | X ∩ (cid:96) | = 0. If Y ∩ (cid:96) = { y } , then Y = X ∪ y , and y is incl (( Y − y ) ∩ E ) ∩ cl (( Y − y ) ∩ E ) = cl ( X − E ) ∩ cl ( X − E ) , so statement (iii) holds. Therefore we assume that Y ∩ (cid:96) = { y, y (cid:48) } , and hence Y = X ∪ { y, y (cid:48) } . Earlier statements imply that y ∈ cl (( X ∩ E ) ∪ y (cid:48) ) ∩ cl (( X ∩ E ) ∪ y (cid:48) ) and y (cid:48) ∈ cl (( X ∩ E ) ∪ y ) ∩ cl (( X ∩ E ) ∪ y ) . If y is in neither cl ( X ∩ E ) nor cl ( X ∩ E ), then r ( Y ∩ E ) = r (( X ∩ E ) ∪ y ) = r ( X ∩ E ) + 1, and similarly, r ( Y ∩ E ) = r ( X ∩ E ) + 1. Butthis means that r ( Y ∩ E ) + r ( Y ∩ E ) − r ( Y ∩ (cid:96) ) = r ( X ∩ E ) + r ( X ∩ E ) = | X | , which is a contradiction. Hence, by using symmetry, we can assume that y isin cl ( X ∩ E ). This means that y (cid:48) , and hence (cid:96) , is contained in cl ( X ∩ E ).Also, r (( X − E ) ∪ (cid:96) ) = r (( X − E ) ∪ y ) < r ( X − E ) + 2so statement (i) holds, and the proof is complete. (cid:3) Gain-graphic matroids In this section we introduce two families of matroids via gain graphs. Let G be an undirected graph (possibly containing loops and multiple edges)with edge set E and vertex set V . Define A ( G ) to be the following subsetof E × V × V : { ( e, u, v ) : e is a non-loop edge joining u and v }∪ { ( e, u, u ) : e is a loop incident with u } . A gain graph (over the group H ) is a pair ( G, σ ), where G is a graph, and σ is a function from A ( G ) to H , such that σ ( e, u, v ) = σ ( e, v, u ) − for everynon-loop edge e with end-vertices u and v . We say that σ is a gain function .If C = v e v e · · · e t v t +1 is a cycle of G , where v = v t +1 , then σ ( C ) isdefined to be σ ( e , v , v ) σ ( e , v , v ) · · · σ ( e t , v t , v t +1 ) . Note that, in general, H may be nonabelian, and the value of σ ( C ) dependson the choice of starting point and orientation for C ; however, if σ ( C ) isequal to the identity of H , then this equality will hold no matter whichstarting point and orientation we choose. In this case, we say that C is balanced . A cycle that is not balanced is unbalanced .The gain-graphic matroid M ( G, σ ) has the edge set of G as its ground set.The circuits of M ( G, σ ) are exactly the edge sets of balanced cycles, alongwith the minimal edge sets that induce connected subgraphs containing atleast two unbalanced cycles and no balanced cycles. Any such subgraph iseither a theta graph , a loose handcuff , or a tight handcuff . A theta graph consists of two vertices joined by three internally-disjoint paths; a loosehandcuff consists of two vertex-disjoint cycles joined by a single path thatintersects the cycles only in its end-vertices; and a tight handcuff consists oftwo edge-disjoint cycles that share exactly one vertex.Assume that ( G, σ ) is a gain graph, where σ takes A ( G ) to the multiplica-tive group of a field, K . Let v , . . . , v m and e , . . . , e n be orderings of thevertex and edge sets of G . We define a matrix, D ( G, σ ), with entries from K . The columns of D ( G, σ ) are labelled by e , . . . , e n . Let b , . . . , b m be thestandard basis vectors. Assume that e i is incident with vertices v j and v k ,where j ≤ k . (If e i is a loop, then j = k .) The column labelled by e i is equalto b j − σ ( e i , v j , v k ) b k . Note that if e i is a balanced loop, then column e i isthe zero vector, and if e i is an unbalanced loop, then the column contains asingle non-zero entry. Proposition 5.1 (Theorem 2.1 of [14]) . Let ( G, σ ) be a gain graph overthe multiplicative group of the field K . The matrix D ( G, σ ) represents thematroid M ( G, σ ) over K . Next we construct two families of gain graphs. Let K be a field. The gainfunctions of the two families will be into the multiplicative group of K . Let s ≥ α be an element in K − { } with order greaterthan s . The gain graph Γ( K , s, α ) has vertex set { u , . . . , u s +1 } . Each vertex u i in { u , . . . , u s } is incident with a loop, a i . In addition, u is incident withthe loop a , and u s +1 is incident with the loop b . The parallel edges x i and y i join u i and u i +1 for each i in { , . . . , s } . Moreover, the edges x , y , and z join u and u s +1 . We define the gain function, σ , so that it takes each loop to α and each x i to 1. Furthermore, σ ( y i , u i , u i +1 ) = α for each i in { , . . . , s } ,while σ ( x, u , u s +1 ) = 1, σ ( y, u , u s +1 ) = α s − , and σ ( z, u , u s +1 ) = α s .Now let t ≥ β be an element in K − { } withorder greater than 2 t ( t − K , t, β ). It has { v , . . . , v t } as its vertex set. Each vertex v i ∈ { v , . . . , v t − } is incidentwith a loop, b i , while v is incident with the loop a and v t is incident withthe loop b . For each i ∈ { , . . . , t − } , the edges e i and f i join v i to v i +1 .The edges x , y , z , and g join the vertices v and v t . The gain function, σ , takes each loop to β , and each edge e i to 1. The triple ( f i , v i , v i +1 ) istaken to β t − when i ∈ { , . . . , t } , and to β t when i ∈ { t + 1 , . . . , t − } .Thus t of the edges f , . . . , f t − receive the label β t − , and the other t − β t . The values of σ ( x, v , v t ), σ ( y, v , v t ), σ ( z, v , v t ), and σ ( g, v , v t ) are 1, β t − , β t , and β t ( t − , respectively.Figure 1 shows Γ( K , s, α ) and ∆( K , t, β ). The edge labels of loops havebeen omitted. Every edge label corresponds to the orientation of the edgeshown in the drawing.Note that whenever M = M (Γ( K , s, α )) and M (cid:48) = M (∆( K , t, β )), then E ( M ) ∩ E ( M (cid:48) ) = { a, b, x, y, z } . Let (cid:96) be this intersection. Then M | (cid:96) and M (cid:48) | (cid:96) are both isomorphic to U , , so the discussion in Section 4 implies thatAmal( M, M (cid:48) ) is defined. HE “MISSING AXIOM” OF MATROID THEORY 15 u u u u u s − u s u s +1 v v v t v t +1 v t +2 v t − v t α α α α α β t − β t − β t β t α s − α s β t − β t β t ( t − ) Figure 1. The gain graphs Γ( K , s, α ) and ∆( K , t, β ).If G is a graph and X is a set of edges, then G [ X ] denotes the subgraphof G containing the edges in X and all vertices that are incident with atleast one edge in X . Lemma 5.2. Let K be a field, let s ≥ be an integer, and let α be an elementin K − { } with order greater than s ( s − . Let M be M (Γ( K , s, α )) andlet M (cid:48) be M (∆( K , s, α )) . Then Amal( M, M (cid:48) ) is K -representable.Proof. Let (cid:96) be { a, b, x, y, z } . Let ( G, σ ) stand for the signed graphΓ( K , s, α ), and let ( G (cid:48) , σ (cid:48) ) stand for ∆( K , s, α ). The lemma will follow fromProposition 5.1 if we can prove that Amal( M, M (cid:48) ) is gain-graphic over themultiplicative group of K . To this end, we construct a graph, H , by gluingtogether G and G (cid:48) . We identify the vertices u and v as the new vertex w , and identify u s +1 and v s as w (cid:48) . The edge-set of H is exactly the unionof the edge-sets of G and G (cid:48) . Any edge incident with u or v in G or G (cid:48) is incident with w in H , and any edge incident with u s +1 or v s is incidentwith w (cid:48) in H . All other incidences are exactly as in G or G (cid:48) . Let e be anedge of G or G (cid:48) , and let u and v be the vertices incident with e . (It may bethe case that u = v .) If u is u or v , then let ˆ u be w , and if u = u s +1 or v s , then let ˆ u be w (cid:48) . Otherwise define ˆ u to be u . We define ˆ v in exactly thesame way. We define the function θ so that θ ( e, ˆ u, ˆ v ) = σ ( e, u, v ) if e is anedge of G , and θ ( e, ˆ u, ˆ v ) = σ (cid:48) ( e, u, v ) if e is an edge of G (cid:48) . It is clear that θ is a well-defined gain function for H .Let N be the gain-graphic matroid M ( H, θ ). We can prove the lemma bychecking that N and Amal( M, M (cid:48) ) are equal. We do this by showing thata set, X , is dependent in N if and only if it is dependent in Amal( M, M (cid:48) ).Note that N is obviously an amalgam of M and M (cid:48) . For the first direction, we assume that X is a circuit in N . As N is anamalgam of M and M (cid:48) , we assume that X is contained in neither E ( M )nor E ( M (cid:48) ). We start by considering the case that X is a balanced cycle in( H, θ ). If X contains an edge joining w and w (cid:48) , then this edge is g , and H [ X ]contains a path with vertex sequence w, u , u , . . . , u s , w (cid:48) , for otherwise X is contained in E ( M ) or E ( M (cid:48) ). The product of edge labels along thispath is α j , where j ≤ s . We also require that α j = α s ( s − , since g islabelled with α s ( s − , and X is a balanced cycle. But α j = α s ( s − cannothold, as α has order greater than 2 s ( s − s ≥ s ( s − > s .Therefore we conclude that X does not contain any edge between w and w (cid:48) , and since X is not contained in E ( M ) or E ( M (cid:48) ), it follows that it isthe edge-set of a Hamiltonian cycle. Let α j be the product of edge labelsalong the path in H [ X ] with vertex sequence w, u , u , . . . , u s , w (cid:48) . Thus0 ≤ j ≤ s . Let α p ( s − qs be the product of edge labels along the pathin H [ X ] with vertex sequence w, v , v , . . . , v s − , w (cid:48) , where p and q arenon-negative integers satisfying 0 ≤ p ≤ s and 0 ≤ q ≤ s − 1. Thus0 ≤ p ( s − 1) + qs ≤ s ( s − 1) and α j = α p ( s − qs , as X is balanced. Asthe order of α is greater than 2 s ( s − j = p ( s − 1) + qs ,and hence j is equal to 0, s − 1, or s . In these three cases, x , y , or z isan element in cl M ( X − E ( M (cid:48) )) ∩ cl M (cid:48) ( X − E ( M )). Thus statement (iii) ofProposition 4.1 holds, so X is dependent in Amal( M, M (cid:48) ).Now we can assume that X does not contain a balanced cycle of ( H, θ ).Thus H [ X ] is a theta graph or a handcuff. Let { M , M } be { M, M (cid:48) } . As-sume that H [ X − E ( M )] is a path from w to w (cid:48) . None of the internalvertices of this path has degree three or more in H [ X ]. It follows that,regardless of whether H [ X ] is a theta graph or a handcuff, H [ X ∩ E ( M )]contains a unbalanced cycle joined by a path to the loop a , and an unbal-anced cycle joined by a path to the loop b . Therefore { a, b } (and hence allof (cid:96) ) is contained in cl M ( X ∩ E ( M )). Also, H [( X − E ( M )) ∪ { a, b } ] is ahandcuff, and hence ( X − E ( M )) ∪ { a, b } is a circuit of M that spans (cid:96) .This means that r M (( X − E ( M )) ∪ (cid:96) ) < r M ( X − E ( M )) + 2. Now (i) ofProposition 4.1 holds, so X is dependent in Amal( M, M (cid:48) ).We can now assume that neither H [ X − E ( M )] nor H [ X − E ( M (cid:48) )] is apath from w to w (cid:48) . Assume H [ X − E ( M )] is a forest. As H [ X ] has novertices of degree one, the forest must be a path, and its end-vertices mustbe w and w (cid:48) , contradicting our assumption. By symmetry, it follows thateach of H [ X − E ( M )] and H [ X − E ( M (cid:48) )] contains an unbalanced cycle.Since H [ X ] is connected, either w or w (cid:48) is on a path from one of these cyclesto the other. Let us assume the former, since the latter case is identical.Now a is in a handcuff in G [( X − E ( M (cid:48) )) ∪ a ], and hence in a circuit of M that is contained in ( X − E ( M (cid:48) )) ∪ a . By symmetry, a is also contained ina circuit of M (cid:48) that is contained in ( X − E ( M )) ∪ a . Thus statement (iii)of Proposition 4.1 holds, and X is dependent in Amal( M, M (cid:48) ). We haveproved that if X is dependent in N , it is dependent in Amal( M, M (cid:48) ). HE “MISSING AXIOM” OF MATROID THEORY 17 For the other direction, we assume that X is independent in N . Thismeans that H [ X ] contains no balanced cycles, and any connected componentof H [ X ] contains at most one cycle. Let us assume for a contradiction that X is dependent in Amal( M, M (cid:48) ). In fact, we can assume that X is a circuitof Amal( M, M (cid:48) ). As N is an amalgam of M and M (cid:48) , it follows that neither X ∩ E ( M ) nor X ∩ E ( M (cid:48) ) is dependent, so X is contained in neither E ( M )nor E ( M (cid:48) ). One of the three statements in Proposition 4.1 must hold.We prove the following statements for M and M (cid:48) simultaneously, by let-ting { M , M } be { M, M (cid:48) } . Claim 5.2.1. The subgraph H [ X − E ( M )] either contains a connected com-ponent that contains both w and w (cid:48) , or a connected component that containsa cycle and at least one of w and w (cid:48) .Proof. Assume the claim is false, so that any component of H [ X − E ( M )]contains at most one of w and w (cid:48) , and any component containing one ofthese vertices contains no cycle. This means that if p and q are distinctelements of (cid:96) , then H [( X − E ( M )) ∪ { p, q } ] contains no balanced cycles,and no theta graphs or handcuffs. From this it follows that cl M ( X − E ( M ))does not contain any element of (cid:96) , so statement (iii) of Proposition 4.1 doesnot hold. Moreover, r M (( X − E ( M )) ∪ (cid:96) ) = r M ( X − E ( M )) + 2. Asone of the three statements in Proposition 4.1 must hold, it follows that (cid:96) is in cl M ( X ∩ E ( M )) and r M (( X − E ( M )) ∪ (cid:96) ) < r M ( X − E ( M )) + 2.But this now means that X contains at least two elements of (cid:96) , or else (cid:96) (cid:42) cl M ( X ∩ E ( M )). Hence (cid:96) ⊆ cl M ( X ∩ (cid:96) ), so Proposition 4.1 impliesthat X ∩ E ( M ) is dependent in Amal( M, M (cid:48) ). Since we have assumed X is a circuit of Amal( M, M (cid:48) ), this means that X ⊆ E ( M ), contrary tohypothesis. Therefore Claim 5.2.1 holds. (cid:3) Claim 5.2.2. There is no connected component of H [ X − E ( M )] that con-tains both w and w (cid:48) .Proof. Assume that H is such a component. Then H is contained in aconnected component, H , of H [ X ∩ E ( M )]. If H contains a cycle, thenby applying Claim 5.2.1 to H [ X − E ( M )], we can deduce that the unionof H with a component of H [ X − E ( M )] contains a theta graph or ahandcuff. We have assumed that H [ X ] contains no such subgraph, so thisis a contradiction. Therefore H contains no cycle, from which we deducethat H [ X − E ( M )] is a path from w to w (cid:48) and X ∩ (cid:96) = ∅ . Note that H [( X − E ( M )) ∪ a ] contains no circuit of M , so (cid:96) (cid:42) cl M ( X − E ( M )).If there is a component of H [ X − E ( M )] that contains w and w (cid:48) , thenby the reasoning in the previous paragraph, H [ X − E ( M )] is a path from w to w (cid:48) , and (cid:96) (cid:42) cl M ( X − E ( M )). Therefore H [ X ] is a Hamiltonian cycle,and the only statement in Proposition 4.1 that can hold is statement (iii).We have noted that a (and by symmetry b ) is not in cl M ( X − E ( M )),so there is an edge, p , joining w and w (cid:48) , such that p is in both cl M ( X − E ( M )) and cl M ( X − E ( M )). This means that H [( X − E ( M )) ∪ p ] and H [( X − E ( M )) ∪ p ] are both balanced cycles. Thus the product of edgelabels on the path H [ X − E ( M )] is the inverse of the product on the path H [ X − E ( M )] (assuming that we travel in a consistent direction aroundthe Hamiltonian cycle H [ X ]). Hence X is a balanced cycle, a contradiction.Therefore no component of H [ X − E ( M )] contains w and w (cid:48) .Recall that X ∩ (cid:96) = ∅ and neither a nor b is in cl M ( X − E ( M )). As oneof the statements from Proposition 4.1 must hold, either there is an edgebetween w and w (cid:48) that is in both cl M ( X − E ( M )) and cl M ( X − E ( M )),or cl M ( X − E ( M )) contains (cid:96) . Therefore in either case we can let p bean edge between w and w (cid:48) that is in cl M ( X − E ( M )). Let C be a circuitof M contained in ( X − E ( M )) ∪ p that contains p . No component of H [ X − E ( M )] contains w and w (cid:48) so H [ C − p ] is not connected. It followsthat H [ C ] is a loose handcuff, and p is an edge in the path between the twocycles. Therefore H [ X − E ( M )] contains two distinct components, eachcontaining a cycle and one of w and w (cid:48) . As H [ X − E ( M )] is a path from w to w (cid:48) , it follows that H [ X ] is a handcuff, a contradiction. (cid:3) We have shown that neither H [ X − E ( M )] nor H [ X − E ( M (cid:48) )] contains acomponent that contains w and w (cid:48) . By using Claim 5.2.1, symmetry, and thefact that X contains no handcuffs, we can assume the following: there is acomponent of H [ X − E ( M )] that contains w and a cycle, and any componentthat contains w (cid:48) contains no cycle; similarly, there is a component of H [ X − E ( M )] that contains w (cid:48) and a cycle, and any component that contains w contains no cycle. It follows from this assumption (and the fact that H [ X ]contains no handcuffs) that X ∩ (cid:96) = ∅ . Notice that a is the only element incl M ( X − E ( M )) ∩ (cid:96) = cl M ( X ∩ E ( M )) ∩ (cid:96). Similarly, cl M ( X ∩ E ( M )) ∩ (cid:96) = { b } . Therefore none of the statements inProposition 4.1 can hold, so we have a contradiction.Now it follows that if X is independent in N it is also independent inAmal( M, M (cid:48) ), so N = Amal( M, M (cid:48) ), exactly as desired. This completesthe proof of Lemma 5.2. (cid:3) Lemma 5.3. Let K be a field and let s and t be distinct integers satis-fying s, t ≥ . Let α be an element in K − { } with order greater than max { s, t ( t − } . Let M be M (Γ( K , s, α )) and let M (cid:48) be M (∆( K , t, α )) .Then Amal( M, M (cid:48) ) is not representable over any field.Proof. Let us assume that the matrix D represents Amal( M, M (cid:48) ) over thefield L . Let B be the set { a , . . . , a s , a, b, b , . . . , b t − } . Thus B is the set ofall loops in Γ( K , s, α ) and ∆( K , t, α ). It is clear that B ∩ E ( M ) and B ∩ E ( M (cid:48) )are independent in M and M (cid:48) , and moreover, r M (( B − E ( M (cid:48) )) ∪ (cid:96) ) = r M ( B − E ( M (cid:48) )) + 2 and r M (cid:48) (( B − E ( M )) ∪ (cid:96) ) = r M (cid:48) ( B − E ( M )) + 2.Now it follows easily from Proposition 4.1 that B cannot be dependentin Amal( M, M (cid:48) ). If e is any element of the ground set of Amal( M, M (cid:48) )that is not in B , then B ∪ e contains a circuit of either M or M (cid:48) , andthis circuit has cardinality three. From this it follows that B is a basis HE “MISSING AXIOM” OF MATROID THEORY 19 of Amal( M, M (cid:48) ). We can assume that the columns of D labelled by theelements of B form an identity matrix. As the fundamental circuits relativeto B all have cardinality three, every column of D contains either one ortwo non-zero elements. By scaling, we can assume that the first non-zeroentry in each column is 1. Thus D = D ( G, θ ), for some gain graph ( G, θ )over the multiplicative group of L . By examining the fundamental circuitsrelative to B , we see that G is the graph in Figure 2. wu u u u s − u s w (cid:48) α α α α s − α s v v t v t +1 v t +2 v t − β β t β t +1 β t − δ γ (cid:15) ζ Figure 2. The gain graph ( G, θ ).By scaling rows of D , we can assume that θ ( x , w, u ) = θ ( x s , u s , w (cid:48) ) = θ ( e i , w, v ) = 1 . Moreover, we can also assume that θ ( x i , u i , u i +1 ) = 1 for each i = 2 , . . . , s − θ ( e i , v i , v i +1 ) = 1 for each i = 2 , . . . , t − 2. Note that { x , . . . , x s , x } is a balanced cycle in Γ( K , s, α ), and that { e , . . . , e t − , x } is a balanced cycle in ∆( K , t, α ). It now follows from Proposition 4.1 that { x , . . . , x s , e , . . . , e t − } is dependent in Amal( M, M (cid:48) ), and we deduce thatit is the edge-set of a balanced cycle in ( G, θ ). This in turn implies that θ ( e t − , v t − , w (cid:48) ) = 1.For i ∈ { , . . . , s − } , let α i be the value θ ( y i , u i , u i +1 ). Define α tobe θ ( y , w, u ) and α s to be θ ( y s , u s , w (cid:48) ). Similarly, for i ∈ { , . . . , t − } , let β i be θ ( f i , v i , v i +1 ). Define β to be θ ( f , w, v ), and let β t − be θ ( f t − , v t − , w (cid:48) ). Let γ , δ , (cid:15) , and ζ be θ ( x, w, w (cid:48) ), θ ( y, w, w (cid:48) ), θ ( z, w, w (cid:48) ),and θ ( g, w, w (cid:48) ), respectively.Because { x , . . . , x s , x } is a balanced cycle in Γ( K , s, α ), and hence a cir-cuit in Amal( M, M (cid:48) ), it follows that it is also a balanced cycle in ( G, θ ).This means that γ = 1. Next we notice that ( { y , . . . , y s } − y i ) ∪ { x i , y } is abalanced cycle of Γ( K , s, α ) and hence of ( G, θ ), for any i in { , . . . , s } . The product of edge labels on this cycle in ( G, θ ) is α · · · α s α − i δ − , which impliesthat α i = α · · · α s δ − for any i ∈ { , . . . , s } . Let α stand for α · · · α s δ − ,so that α i = α for any i ∈ { , . . . , s } , and δ = α s − . As { y , . . . , y s , z } is abalanced cycle, it follows that (cid:15) = α s .Next we observe that ( { e , . . . , e t − } − e i ) ∪ { f i , y } is a balanced cycle in∆( K , t, α ), and hence in ( G, θ ), for any i ∈ { , . . . , t } . Thus β i = δ = α s − for any such i . Similarly, ( { e , . . . , e t − } − e i ) ∪ { f i , z } is a balanced cyclefor any i ∈ { t + 1 , . . . , t − } , from which we deduce that β i = (cid:15) = α s .As { f , . . . , f t , e t +1 , . . . , e t − , g } and { e , . . . , e t , f t +1 , . . . , f t − , g } areboth balanced cycles in ∆( K , t, α ), it now follows that the products β · · · β t = ( α s − ) t and β t +1 · · · β t − = ( α s ) t − are both equal to ζ . Thus α st − t = α st − s , implying α s = α t . Let o be the order of α in L . Since s (cid:54) = t ,we know that o < max { s, t } . But if o < s , then { y , . . . , y o , x o +1 , . . . , x s , x } isa balanced cycle in ( G, θ ), although it is not a circuit in M . Therefore o < t .Now the product of edge labels on the cycle { f , . . . , f o , e o +1 , . . . , e t − , x } is ( α s − ) o = 1, so this is a balanced cycle in ( G, θ ), although not a circuitin M (cid:48) . This contradiction proves the lemma. (cid:3) Proof of Lemma 1.4 This section is dedicated to proving Lemma 1.4, which we restate with anexplicit bound. Let k be a positive integer. Define g ( k, 0) to be 2 k k k .Recursively define g ( k, n + 1) to be 2 g ( k,n ) , and let f ( k ) be g ( k, k ). Recallthat (cid:96) is the set { a, b, x, y, z } , and M (cid:96) is the class of matroids having a U , -restriction on (cid:96) . A pair of matroids is ( k, (cid:96) )-equivalent if they have no( k, (cid:96) )-certificate, as defined in the introduction. Lemma 6.1. Let k be a positive integer. There are at most f ( k ) equivalenceclasses of M (cid:96) under the relation of ( k, (cid:96) ) -equivalence.Proof. The main ideas required here are essentially identical to those inSection 3, so we omit many details. A registry is a ( k + 2) × k matrix withcolumns indexed by the variables X , . . . , X k , and rows indexed by Ind, Sing,and X , . . . , X k . As before, an entry in row X i is either ‘T’ or ‘F’, and anentry in row Sing is either ‘0’, ‘1’, or ‘ > ’. Let A be the set { D , S } ∪ { α : α ⊆ (cid:96), | α | ≤ } ∪ { ( α, β ) : α, β ⊆ (cid:96), | α | , | β | ≤ , α ∩ β = ∅} . A registry entry in row Ind must be a member of A . A simple calculationshows that |A| = 49. Therefore there are at most 2 k k k = g ( k, depth- tree is a registry, and a depth- ( n + 1) tree is anon-empty set of depth- n trees. Hence there are no more than f ( k ) depth- k trees.A stacked matroid is a tuple, M = ( M, Y , . . . , Y m ), where M is in M (cid:96) ,and each Y i is a subset of E ( M ). If ||M|| = m ≤ k , then we associate adepth-( k − ||M|| ) tree, T ( M ) to M . We give the definition of T ( M ) only inthe case that T ( M ) is a registry, because otherwise the definition is identicalto that in Lemma 3.1. Assume that M = ( M, Y , . . . , Y k ). The entry in row HE “MISSING AXIOM” OF MATROID THEORY 21 X i and column X j of the registry T ( M ) is ‘T’ if and only if Y i ⊆ Y j . Theentry in row Sing and column X i is ‘0’, ‘1’, or ‘ > ’, according to whether | Y i | is less than, equal to, or greater than one.The rules defining the entries in row Ind are more complicated. Let ω stand for the entry in row Ind and column X j . If Y j is dependent in M , thenwe set ω to be ‘D’. Now we assume that Y j is independent. Let π be theinteger r M ( Y j − (cid:96) ) − r M ( Y j ∪ (cid:96) ) + 2. This is known as the local connectivity of Y j − (cid:96) and (cid:96) . The submodularity of the rank function shows that π ≥ r M ( Y j − (cid:96) ) ≤ r M ( Y j ∪ (cid:96) ), it follows that π ≤ 2. If π = 2, then Y j − (cid:96) spans (cid:96) , and we set ω to be ‘S’. In the next case, we assume that π = 0. Certainly | Y j ∩ (cid:96) | ≤ 2, as we have assumed that Y j is independent in M . So Y j ∩ (cid:96) is in A , and we set ω to be Y j ∩ (cid:96) . Finally, we consider thecase that π = 1. Thus | Y j ∩ (cid:96) | ≤ 1, for otherwise r M ( Y j ) = r M ( Y j ∪ (cid:96) ) = r M ( Y j − (cid:96) ) − π + 2 = | Y j − (cid:96) | + 1 < | Y j | , which contradicts our assumption that Y j is independent in M . We let β bethe set Y j ∩ (cid:96) . Let α be cl M ( Y j − (cid:96) ) ∩ (cid:96) . Note that α ∩ β = ∅ , as otherwise Y j contains a circuit of M . Moreover, r M ( α ) ≤ r M (cl M ( Y j − (cid:96) )) + r M ( (cid:96) ) − r M (cl M ( Y j − (cid:96) ) ∪ (cid:96) ) = π = 1 , so | α | ≤ 1. Therefore ( α, β ) is in A , and we set ω to be ( α, β ).Let ψ be an MS formula such that either ψ is quantifier-free, or Var( ψ ) = { X , . . . , X k } . Let b ( ψ ) be the number of bound variables in ψ , and let T and T (cid:48) be depth- b ( ψ ) trees. We will define what it means for T and T (cid:48) to be ψ -compatible. We give the definition only in the case that b ( ψ ) = 0and ψ is the atomic formula Ind( X j ): otherwise the definition is identical tothat in Lemma 3.1. Let ω and ω (cid:48) be the entries of T and T (cid:48) in row Ind andcolumn X j . It easiest to define the rules that determine the ψ -compatibilityof T and T (cid:48) via a flowchart, which is exactly what we do in Figure 3. Whenfollowing this flowchart, we start in the shaded cell. A terminal node thatis hollow signifies that T and T (cid:48) are ψ -compatible. A filled terminal nodesignifies that they are not. Note that if ω is not ‘D’ or ‘S’, then it is eithera subset of (cid:96) , or a pair ( α, β ), where α and β are subsets of (cid:96) . The samecomment applies to ω (cid:48) . Claim 6.1.1. Let ψ be an MS formula such that either ψ is quantifier-free, or Var( ψ ) = { X , . . . , X k } . If Var( ψ ) = { X , . . . , X k } , then let m be | Fr( ψ ) | and assume that Fr( ψ ) = { X , . . . , X m } . Otherwise, let m be k . Let M and M (cid:48) be matroids in M (cid:96) satisfying E ( M ) ∩ E ( M (cid:48) ) = (cid:96) , andlet M = ( M, Y , . . . , Y m ) and M (cid:48) = ( M (cid:48) , Y (cid:48) , . . . , Y (cid:48) m ) be stacked matroids.Define τ to be the function that takes X i to Y i ∪ Y (cid:48) i , for each X i ∈ Fr( ψ ) . Theinterpretation (Amal( M, M (cid:48) ) , τ ) satisfies ψ if and only if the trees, T ( M ) and T ( M (cid:48) ) , are ψ -compatible.Proof. The proof of this claim differs from that of Claim 3.1.1 only in thebase case when ψ is the atomic formula Ind( X j ). Therefore we need only ω ⊆ (cid:96) and ω (cid:48) ⊆ (cid:96) ? ω (cid:42) (cid:96) and ω (cid:48) (cid:42) (cid:96) ? β ∪ β (cid:48) = ∅ ?Set { δ , δ } = { ω, ω (cid:48) } with δ ⊆ (cid:96) , δ = ( α, β ). δ ∩ α = ∅ ? | δ ∪ β | ≤ α = α (cid:48) and α (cid:54) = ∅ ? ω (cid:54) = D and ω (cid:48) (cid:54) = D? δ = ∅ ? NY NY NYY | ω ∪ ω (cid:48) | ≤ NYN Y NY NYYN NYN Set ( α, β ) = ω ,( α (cid:48) , β (cid:48) ) = ω (cid:48) . ω (cid:54) = S and ω (cid:48) (cid:54) = S? Set { δ , δ } = { ω, ω (cid:48) } with δ = S. Figure 3. Deciding whether T and T (cid:48) are ψ -compatible.consider this case. Let ψ be the formula Ind( X j ). Let ω be the entry in row HE “MISSING AXIOM” OF MATROID THEORY 23 Ind and column X j of the registry T ( M ), and let ω (cid:48) be the correspondingentry of T ( M (cid:48) ). We will trace all possible outcomes in the flowchart shownin Figure 3. We will prove that if T ( M ) and T ( M (cid:48) ) are ψ -compatible, then Y j ∪ Y (cid:48) j is independent in Amal( M, M (cid:48) ), whereas if they are not ψ -compatible,then Y j ∪ Y (cid:48) j is dependent. This will establish the claim. Let X be the set Y j ∪ Y (cid:48) j .If either ω or ω (cid:48) is ‘D’, then either Y j is dependent in M , or Y (cid:48) j is dependentin M (cid:48) . In this case T ( M ) and T ( M (cid:48) ) are not ψ -compatible, and X iscertainly dependent in Amal( M, M (cid:48) ). Therefore we will assume that ω (cid:54) = Dand ω (cid:48) (cid:54) = D, so Y j is independent in M and Y (cid:48) j is independent in M (cid:48) .In the next case, we assume that either ω or ω (cid:48) is ‘S’. By symmetry, we canassume that ω = S. Then Y j − (cid:96) spans (cid:96) in M . Since Y j is independent in M ,we observe that Y j − (cid:96) = Y j . Assume that ω (cid:48) (cid:54) = ∅ , so that T ( M ) and T ( M (cid:48) )are not ψ -compatible. If ω (cid:48) is a non-empty subset of (cid:96) , then ω (cid:48) = Y (cid:48) j ∩ (cid:96) ,and it follows that an element of Y (cid:48) j is in the closure of Y j − (cid:96) in M , so that X is dependent. If ω (cid:48) is not a subset of (cid:96) , then r M (cid:48) ( Y (cid:48) j − (cid:96) ) − r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) +2 > 0, meaning that r M (cid:48) (( X − E ( M )) ∪ (cid:96) ) < r M (cid:48) ( X − E ( M )) + 2. ThusProposition 4.1 implies that X is dependent in Amal( M, M (cid:48) ). On the otherhand, if ω (cid:48) = ∅ , then T ( M ) and T ( M (cid:48) ) are ψ -compatible. Furthermore, r M (cid:48) ( Y (cid:48) j − (cid:96) ) − r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) + 2 = 0 and Y (cid:48) j ∩ (cid:96) = ∅ , meaning that Y (cid:48) j − (cid:96) = Y (cid:48) j .Now we know that X ∩ (cid:96) = ∅ , so that X ∩ E ( M ) is independent in M and X ∩ E ( M (cid:48) ) is independent in M (cid:48) . The fact that r M (cid:48) ( Y (cid:48) j ) + 2 = r M (cid:48) ( Y (cid:48) j ∪ (cid:96) )implies that cl M (cid:48) ( Y (cid:48) j ) ∩ (cid:96) = ∅ . None of the statements in Proposition 4.1apply, so X is independent in Amal( M, M (cid:48) ).We now follow the branch of the flowchart in which ω (cid:54) = S and ω (cid:48) (cid:54) = S.This means that neither Y j − (cid:96) nor Y (cid:48) j − (cid:96) spans (cid:96) . Assume that ω and ω (cid:48) are both subsets of (cid:96) . This implies that r M ( Y j − (cid:96) ) − r M ( Y j ∪ (cid:96) ) + 2and r M (cid:48) ( Y (cid:48) j − (cid:96) ) − r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) + 2 are both zero. From this we deduce thatcl M ( Y j − (cid:96) ) ∩ (cid:96) and cl M (cid:48) ( Y (cid:48) j − (cid:96) ) ∩ (cid:96) are empty. Assume that | ω ∪ ω (cid:48) | > 2. Then T ( M ) and T ( M (cid:48) ) are not ψ -compatible. As (cid:96) is a rank-2 set, obviouslyit follows that X ∩ E ( M ) and X ∩ E ( M (cid:48) ) are dependent. Therefore weassume that | ω ∪ ω (cid:48) | ≤ 2, so that T ( M ) and T ( M (cid:48) ) are ψ -compatible.As r M ( Y j − (cid:96) ) = r M ( Y j ∪ (cid:96) ) − 2, and X ∩ (cid:96) = ω ∪ ω (cid:48) contains at mosttwo elements, we see that X ∩ E ( M ) is independent in M . By exactlythe same argument, X ∩ E ( M (cid:48) ) is independent in M (cid:48) . The information wehave assembled in this paragraph is enough to determine that none of thestatements in Proposition 4.1 apply, so X is independent in Amal( M, M (cid:48) ).Next we consider the branch where neither ω nor ω (cid:48) is a subset of (cid:96) . Thismeans that both r M ( Y j − (cid:96) ) − r M ( Y j ∪ (cid:96) ) + 2 and r M (cid:48) ( Y (cid:48) j − (cid:96) ) − r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) + 2are equal to one. Let ω be ( α, β ), where α and β are disjoint subsets of (cid:96) of size at most one, and similarly assume that ω (cid:48) = ( α (cid:48) , β (cid:48) ). Assume that α = α (cid:48) and that α (cid:54) = ∅ , so that T ( M ) and T ( M (cid:48) ) are not ψ -compatible. Thesingle element in α belongs to both cl M ( Y j − (cid:96) ) and cl M (cid:48) ( Y (cid:48) j − (cid:96) ). Statement (iii) of Proposition 4.1 now implies that X is dependent. Thus we assumethat either α (cid:54) = α (cid:48) , or α = α (cid:48) = ∅ . Assume that β ∪ β (cid:48) (cid:54) = ∅ , so that T ( M )and T ( M (cid:48) ) are not ψ -compatible. By symmetry, we will assume that β (cid:54) = ∅ ,and e is the single element in β . Then e is in Y j ∩ (cid:96) , but not in cl M ( Y j − (cid:96) ).Since r M ( Y j ∪ (cid:96) ) = r M ( Y j − (cid:96) ) + 1, we now see that Y j spans (cid:96) in M . As r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) = r M (cid:48) ( Y j − (cid:96) ) + 1, Proposition 4.1 tells us that X is dependent.On the other hand, if β ∪ β = ∅ , then T ( M ) and T ( M (cid:48) ) are ψ -compatibleand X ∩ (cid:96) is empty, which means that X ∩ E ( M ) is independent in M and X ∩ E ( M (cid:48) ) is independent in M (cid:48) . Earlier we followed the branch inwhich neither cl M ( Y j − (cid:96) ) nor cl M ( Y (cid:48) j − (cid:96) ) contains (cid:96) . It follows that neithercl M ( X ∩ E ( M )) nor cl M (cid:48) ( X ∩ E ( M (cid:48) )) contains (cid:96) . There is no element of (cid:96) in both cl M ( Y j − (cid:96) ) nor cl M ( Y (cid:48) j − (cid:96) ), since in that case the element wouldbe in α and α (cid:48) . Therefore Proposition 4.1 implies that X is independent.Finally we arrive at the branch of the flowchart where exactly one of ω and ω (cid:48) is a subset of (cid:96) . By symmetry, we will assume that ω (cid:48) ⊆ (cid:96) and ω = ( α, β ), where α and β are disjoint subsets of (cid:96) of size at most one. Ifthere is an element of ω (cid:48) in α , then this element is in ( Y (cid:48) j ∩ (cid:96) ) ∩ cl M ( Y j − (cid:96) ),which implies that X ∩ E ( M ) is dependent in M . As T ( M ) and T ( M (cid:48) ) arenot ψ -compatible in this branch, this is the desired outcome. Therefore weassume that ω (cid:48) ∩ α = ∅ . Assume that ω (cid:48) ∪ β contains distinct elements, e and f . This means that T ( M ) and T ( M (cid:48) ) are not ψ -compatible. We have justassumed that ω (cid:48) ∩ α = ∅ , from which it follows that e is not in cl M ( Y j − (cid:96) ).As r M ( Y j − (cid:96) ) = r M ( Y j ∪ (cid:96) ) − 1, we deduce that r M (( Y j − (cid:96) ) ∪ e ) = r M ( Y j ∪ (cid:96) ).Therefore ( Y j − (cid:96) ) ∪ e spans f in M , so X ∩ E ( M ) is dependent. Now weassume that ω (cid:48) ∪ β contains at most one element. Therefore T ( M ) and T ( M (cid:48) ) are ψ -compatible. Since ω (cid:48) ∩ α = ∅ , it follows easily that ( Y j − (cid:96) ) ∪ ( ω (cid:48) ∪ β ) = X ∩ E ( M ) is independent in M . Similarly, X ∩ E ( M (cid:48) ) = ( Y (cid:48) j − (cid:96) ) ∪ ( ω (cid:48) ∪ β ) is independent in M (cid:48) . Because r M (cid:48) ( Y (cid:48) j − (cid:96) ) − r M (cid:48) ( Y (cid:48) j ∪ (cid:96) )+2 = 0, thereis no element in cl M (cid:48) ( Y (cid:48) j − (cid:96) ) ∩ (cid:96) . Proposition 4.1 implies that the only way X can be dependent in Amal( M, M (cid:48) ) is if (cid:96) is contained in cl M (cid:48) ( X ∩ E ( M (cid:48) )).But this is impossible, as r M (cid:48) ( Y j − (cid:96) ) = r M (cid:48) ( Y (cid:48) j ∪ (cid:96) ) − 2, and there is at mostone element in X ∩ (cid:96) . Therefore X is independent in Amal( M, M (cid:48) ), exactlyas desired. (cid:3) We complete the proof of Lemma 6.1 by observing that Claim 6.1.1 impliesthat the number of ( k, (cid:96) )-equivalence classes is bounded above by the numberof depth- k trees. (cid:3) We can now prove Theorem 1.2 and Corollaries 1.5 and 1.6. Proof of Theorem 1.2 . Let K be an infinite field. Assume that ψ K is asentence in MS characterising K -representable matroids. Observe that K contains non-zero elements with arbitrarily large order: to see this, assumethat the order of every element in K −{ } is bounded above by the integer K .Then every element in K − { } is a root of the polynomial ( x K − x K − − HE “MISSING AXIOM” OF MATROID THEORY 25 · · · ( x − K is finite. Thiscontradiction proves our claim.Let k be | Var( ψ K ) | . We apply Lemma 6.1. Choose the element α ∈ K −{ } with high enough order so that there are at least f ( k ) + 1 integers, s ,such that s ≥ s ( s − 1) is less than the order of α . Then thereare two distinct integers, s and t , satisfying these constraints, such that M = M (Γ( K , s, α )) and M = M (Γ( K , t, α )) are ( k, (cid:96) )-equivalent. Welet M (cid:48) be M (∆( K , s, α )). Then ψ K is satisfied by both of Amal( M , M (cid:48) )and Amal( M , M (cid:48) ), or by neither. However, the first of these amalgams is K -representable by Lemma 5.2, and the second is not representable over anyfield at all, by Lemma 5.3. This contradiction completes the proof of thetheorem. (cid:3) Proof of Corollary 1.5 . Let { ψ q } q ∈Q be a set of sentences characteris-ing GF( q )-representability, and assume that N is an integer such that | Var( ψ q ) | ≤ N for all q ∈ Q . Recall that if q ∈ Q , then the multiplica-tive group of GF( q ) has an element of order q − 1. We apply Lemma 6.1.Choose q ∈ Q large enough so that there are least f ( N ) + 1 integers, s , satisfying s ≥ s ( s − < q − 1. Let α be a generator of themultiplicative group of GF( q ). Assume that ψ q contains k ≤ N variables.As f ( N ) + 1 ≥ f ( k ) + 1, there are distinct integers, s and t , such that s, t ≥ s ( s − , t ( t − < q − M = M (Γ( K , s, α )) and M (cid:48) = M (Γ( K , t, α )) are ( k, (cid:96) )-equivalent. Now we obtain a contradictionfrom Lemmas 5.2 and 5.3 exactly as before. (cid:3) Proof of Corollary 1.6 . If K is an infinite field with characteristic c , then K contains elements of arbitrarily high order, so all matroids of the form M = M (Γ( K , s, α )) and M = M (Γ( K , t, α )) are K -representable. Therefore theproof proceeds exactly as in Theorem 1.2. (cid:3) Acknowledgements We thank Noam Greenberg for helpful advice and the referees for theirconstructive feedback. The research in this article was supported by theRutherford Discovery Fellowship, the Marsden Fund of New Zealand, andNSERC Canada. References [1] M. Aigner. Combinatorial theory . Classics in Mathematics. Springer-Verlag, Berlin(1997).[2] B. Courcelle. 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School of Mathematics and Statistics, Victoria University of Wellington,New Zealand E-mail address : [email protected] Department of Mathematics and Statistics, University of Ottawa, Ottawa,Canada E-mail address : [email protected] School of Mathematics and Statistics, Victoria University of Wellington,New Zealand E-mail address ::