Matroid lifts and representability

A 1965 result of Crapo shows that every elementary lift of a matroid $M$ can be constructed from a linear class of circuits of $M$. In a recent paper, Walsh generalized this construction by defining a rank-$k$ lift of a matroid $M$ given a rank-$k$ matroid $N$ on the set of circuits of $M$, and conjectured that all matroid lifts can be obtained in this way. In this sequel paper we simplify Walsh's construction and show that this conjecture is true for representable matroids but is false in general. This gives a new way to certify that a particular matroid is non-representable, which we use to construct new classes of non-representable matroids. Walsh also applied the new matroid lift construction to gain graphs over the additive group of a non-prime finite field, generalizing a construction of Zaslavsky for these special groups. He conjectured that this construction is possible on three or more vertices only for the additive group of a non-prime finite field. We show that this conjecture holds for four or more vertices, but fails for exactly three.


Introduction and preliminaries
Given matroids M and L on a common ground set E, L is a lift of M if there exists a matroid K on ground set E ∪ F such that M = K/F and L = K \ F .If L is a lift of M, then the rank of L is at least the rank of M. We say that L is a rank-k lift of M if the rank of L is k greater than that of M. Rank-1 lifts, called elementary lifts, are well understood.Indeed, a classical theorem of Brylawski [2], which was previously stated in the dual by Crapo [3], says that the elementary lifts of a matroid M are in bijection with the set of linear classes of circuits of M, where a linear class is a set C of circuits satisfying the following: We can state this bijection between linear classes of circuits and elementary lifts as follows.
Theorem 1 ( [2,3]).Let M be a matroid on ground set E and let C be a linear class of circuits of M. Then the function r M ′ : 2 E → Z defined, for all X ⊆ E, by is the rank function of an elementary lift M ′ of M.Moreover, every elementary lift of M can be obtained in this way.
This raises the question of whether a similar characterization of higher-rank lifts is possible.Walsh [12, Theorem 2] described a procedure that constructs a rank-k lift of a matroid M from a rank-k matroid N on the circuit set of M. For this construction to work, N must satisfy a particular constraint.When k = 1, this constraint precisely says that the loops of N are a linear class of circuits of M, so Walsh's construction generalizes Theorem 1.Our first main result, stated below using the notation cl N for the closure operator of a matroid N, is a simplification of Walsh's original construction in which we only require N to satisfy a condition concerning pairs of circuits of M. Theorem 2. Let M be a matroid on ground set E and let N be a matroid whose ground set is the circuit set of M. Assume that if C 1 , C 2 are circuits of M for which Then the function r : 2 E → Z defined, for all X ⊆ E, by r(X) = r M (X) + r N ({C : C is a circuit of M|X}) is the rank function of a rank-r(N) lift of M.
Given a matroid M and another matroid N on the circuits of N satisfying the hypothesis of Theorem 2, we write M N for the lift constructed in Theorem 2. There are natural choices for a matroid N satisfying the hypothesis of Theorem 2, such as the derived matroids [11,10,7], matroids from gain graphs over certain groups [12], and rank-2 uniform matroids.
It was conjectured in [12] that every lift of M is isomorphic to M N for some matroid N on the circuits of M. We prove that this true for representable matroids but false in general.We use these facts to derive a new certificate for non-representability which we then use to generate new families of non-representable matroids.In particular, the following is our second main result.Theorem 3.For each integer r ≥ 5 there is a rank-r non-representable sparse paving matroid K with a two-element set X so that there is no matroid N on the set of circuits of K/X for which (K/X) N ∼ = K \X.This family of matroids may be of independent interest: they form an infinite antichain of non-representable sparse paving matroids that do not violate Ingleton's inequality.
Part of the motivation of [12] was to generalize Zaslavsky's application of Theorem 1 to gain graphs.A gain graph is a pair (G, φ) where G is a graph and φ is a gain function that orientably labels the edges of G by elements of a group Γ. Zaslavsky [15] famously applied Theorem 1 to gain graphs by showing that for each gain function on a graph G one can naturally construct a linear class B of circuits of M(G), the graphic matroid of G.The circuits in B are the balanced cycles of G with respect to the gain function, and the pair (G, B) is a biased graph.The elementary lift M(G, φ) of M(G) obtained from applying Theorem 1 with the linear class B is the lift matroid of (G, B), and it follows from Theorem 1 that a cycle of G is a circuit of M if and only if it is balanced.We direct the reader to [4,12,15] for more background on gain graphs and lift matroids.
For certain groups Walsh [12] generalized this construction by defining a matroid N on the circuits of M(G) that satisfies the hypothesis of Theorem 2. To avoid technicalities, we only consider the full Γ-gain graph graph (K Γ n , φ Γ n ) for a finite group Γ, where K Γ n has vertex set [n] and edge set [n]  2 × Γ, and the gain function φ Γ n orients edge ({i, j}, γ) from i to j when i < j and assigns the label γ.The previous version of Theorem 2 was applied to gain graphs to get the following, where Z j p denotes the direct sum of j copies of the cyclic group of order p.
Theorem 4 ([12, Theorem 3]).Let p be a prime, and let n ≥ 3 and j ≥ 2 be integers.For each integer i with 1 ≤ i ≤ j, there is a rank-i lift M of M(K Surprisingly, it was shown that for finite abelian groups such a construction is only possible for groups of this form; namely, the additive group of a non-prime finite field. Theorem 5 ([12,Theorem 4]).Let Γ be a nontrivial finite abelian group, and let n ≥ 3 be an integer.Let M be a lift of M(K Γ n ) so that a cycle of K Γ n is a circuit of M if and only if it is balanced.Then either Γ ∼ = Z j p for some prime p and integer j ≥ 2, or M is an elementary lift of M(K Γ n ).It was conjectured in [12,Conj. 25] that this holds more generally for every nontrivial finite group.For our third and final main result we show that this conjecture is true when n ≥ 4 and is false when n = 3.A group partition of a group Γ is a partition of the non-identity elements of Γ into sets A 1 , . . ., A k such that each A i ∪ {ǫ} is a subgroup of Γ for all i ∈ [k], where ǫ is the identity element of Γ.The partition is nontrivial if it has more than one part.Theorem 6.Let Γ be a nontrivial finite group and let n ≥ 3. Let M n,Γ be the class of lifts M of M(K Γ n ) so that a cycle of K Γ n is a circuit of M if and only if it is balanced.Then M n,Γ contains a non-elementary lift in precisely the following cases: 1. n = 3 and Γ has a nontrivial partition, or 2. n ≥ 4 and Γ = Z j p for some prime p and j ≥ 2.
We prove Theorem 2 in Section 2 and Theorem 3 in Section 3. One direction of Theorem 6 was given in [12] as Lemma 21; we prove the converse direction in Section 4. We follow the notation and terminology of Oxley [9].

A simplified construction
We now state the original construction from [12] and then show that it is equivalent to Theorem 2. For a collection X of sets we write ∪X for and no circuit in C ′ is contained in the union of the others.Equivalently, C ′ is contained in the collection of fundamental circuits with respect to a basis of M, because the set obtained from ∪C ′ by deleting one element from each circuit that is not in any other circuit in In order to prove Theorem 2, we need to recall the lift construction from [12].
Theorem 7 ([12, Theorem 2]).Let M be a matroid on ground set E, and let N be a matroid on the set of circuits of M so that Then the function r : 2 E → Z defined, for all X ⊆ E, by is the rank function of a rank-r(N) lift of M.
Theorem 2 follows from Theorem 7 and the following proposition, which shows that (7 * ) and the hypothesis from Theorem 2 are equivalent.Proposition 8. Let M be a matroid and let N be a matroid on the circuits of M. The following are equivalent Proof.The implication ( * ) =⇒ ( * ′ ) is immediate.We now show ( * ′ ) =⇒ ( * ).Assume ( * ′ ) and let C ′ be minimal so that ( * ) is false for Every subset of C ′ is a perfect collection of circuits of M; we will make repeated tacit use of this fact.If there is some Since M|(∪C ′ ) is connected and has corank at least two, there is a circuit ) has corank one and thus contains a unique circuit C 1 (where We will finish the proof by using the minimality of One advantage of ( * ′ ) over ( * ) is that it is a local condition rather than a global condition, and therefore may be easier to verify for certain choices of N. For example, when M is graphic it suffices to check condition ( * ′ ) only when C 1 and C 2 are in a common theta subgraph.

The converse
It was conjectured in [12,Conj. 1.6] that the converse of Theorem 2 holds: for every matroid K with a set X, there is a matroid N on the circuits of K/X so that (K/X) N ∼ = K \ X.We show that this is true if K is representable but false in general, even when |X| = 2. Proposition 9. Let K be an F-representable matroid and let X be a subset of its ground set.Then there exists an F-representable matroid N on the circuits of K/X such that (K/X) N ∼ = K \X.
Proof.Denote M := K/X and L := K \ X and let E denote the ground set of M and L. Without loss of generality we may assume that X is independent in K. Therefore there exists a matrix A whose column-matroid is K such that the columns corresponding to elements of X are distinct standard basis vectors.Thus one obtains an F-representation A M of M by deleting the columns corresponding to X, as well as the rows corresponding to the nonzero entries of these columns.One obtains an F-representation A L of L by deleting the X columns.From this, one obtains an F-representation A M of M by deleting the rows where the columns corresponding to X have their nonzero entries.
For each circuit C of M, let x C be an element of the kernel of A M whose support is C and let B be a matrix whose column-set is the following.
Let N be the column matroid of B, which we can view as a matroid whose ground set is the circuit set of M. Let C 1 and C 2 be circuits of  Now assume Y is dependent in L and let y be such that A L y = 0. Then A M y = 0. Define k := |Y | − r M (Y ) and note that this is the dimension of the kernel of the matrix D obtained from A L by restricting to columns corresponding to Y .Since the kernel of every matrix is spanned by its support-minimal elements, there exists a set of circuits {C 1 , . . ., C k } of M so that {x C 1 , . . ., x C k } is a basis of the nullspace of D, modulo adding/removing entries corresponding to elements of E \ Y .This gives us scalars λ 1 , . . ., λ k so that

Multiplying both sides of the above on the left by
If M and L are representable over a field F and L is a lift of M, this does not imply that there is an F-representable matroid K so that L = K \X and M = K/X for some set X ⊆ E(K).Consider the following example.Given nonnegative integers r ≤ n, the uniform matroid of rank r on n elements is denoted U r,n .Let M = U 1,3 and let L = U 2,3 .Then both M and L are representable over F 2 .Let K be a matroid on ground set E ∪{e 0 }.
We next construct an infinite family of matroids for which the converse of Theorem 2 does not hold.For certain values of r and t, we will define a set of r-element subsets C(r, t) of [2t + 2] and then show that they are the circuit-hyperplanes of a matroid of rank r on ground set [2t + 2] which we will denote K(r, t).We will then show that there is no matroid N on the circuit set of K(r, t)/{2t + 1, 2t + 2} such that K(r, t) \ {2t + 1, 2t + 2} = (K(r, t)/{2t + 1, 2t + 2}) N .Proposition 9 will then imply that K(r, t) is not representable over any field.Definition 10.Let r ≥ 4 and t ≥ 3 be integers satisfying r ≤ 2t − 2. For i = 1, . . ., t, let C i ⊆ [2t] be defined as with indices taken cyclically modulo 2t.Then define: 1. X := {2t + 1, 2t + 2}, 2. C ′ (r, t) := {C i ∪ X : i ∈ [t]}, and Define C(r, t) to be the set of subsets of [2t + 2] containing C ′ (r, t) ∪ C ′′ (r, t) and all (r + 1)-element subsets that do not contain an element of C ′ (r, t) or C ′′ (r, t).
We will soon see that C(r, t) is the circuit set of a matroid, but before doing this, we look at the case of r = 4 and t = 3.Here we have In particular, C ′ (4, 3) ∪ C ′′ (4, 3) is the set of circuit-hyperplanes of the Vámos matroid V 8 .So K(r, t) is a generalization that captures the cyclic nature of the set of circuithyperplanes of V 8 .Recall that a matroid of rank r is sparse paving if every r-element subset is either a basis or a circuit-hyperplane.Proposition 11.Let r ≥ 4 and t ≥ 3 be integers satisfying r ≤ 2t − 2. Then C(r, t) is the circuit set of a rank-r sparse paving matroid K(r, t) on ground set [2t + 2].
Proof.It suffices to show that no two sets in C ′ (r, t) ∪ C ′′ (r, t) intersect in r − 1 elements.We have three cases to consider.
Case 1: Let C ∈ C ′ (r, t) and The following implies Theorem 3.
Proof.Denote K := K(r, t) and X := {2t + 1, 2t + 2}.Let M := K/X and L := M \X, so L is a rank-2 lift of M. For each i ∈ [t], the set C i is a circuit of M and is independent in L. So for each i ∈ [t] the set C i is a non-loop of N. We will argue that the following statements hold for M and L: Now suppose there is a matroid N on the circuits of M so that M N ∼ = L. Then (a) and (c) together imply that C i and C i+1 are parallel in N for all i ∈ [t], which implies that C 1 and C t are parallel in N.But (b) and (d) together imply that C 1 and C t are independent in N, a contradiction.It now follows from Proposition 9 that K is not representable.
As illustrated by Theorem 12, the fact that the converse of Theorem 2 is false in general but true for representable matroids gives a new way to certify non-representability.
We hope that this method is in fact 'new' and that K(r, t) cannot be certified as nonrepresentable using existing means.However, there are many ways to certify non-representability, and we make no attempt to test K(r, t) against them all.We merely show that K(r, t) does not violate the most well-known certificate for non-representability: Ingleton's inequality.Ingleton [5] proved that if a matroid has sets A, B, C, D so that then it is not representable.We show that this cannot be used to certify non-representability of K(r, t) when r ≥ 5 and r ≤ 2t − 3. Following [8], we say that a matroid is Ingleton if any choice of four subsets satisfies the above inequality.
Proof.Nelson and van der Pol [8,Lemma 3.1] showed that a rank-r sparse paving matroid is Ingleton if and only if there are no pairwise disjoint subsets I, P 1 , P 2 , P 3 , P 4 so that |I| = r − 4 and |P i | = 2 for all i ∈ {1, 2, 3, 4}, while I ∪ P i ∪ P j is a circuit of for all {i, j} = {3, 4} and I ∪ P 3 ∪ P 4 is a basis.Suppose that such sets exist for K(r, t).Then I is an (r − 4)-element set contained in at least five circuit-hyperplanes of K(r, t), and Y = I ∪ P 1 ∪ P 2 ∪ P 3 ∪ P 4 is an (r + 4)-element set that contains at least five circuithyperplanes of K(r, t).Now let X = {2t + 1, 2t + 2}.
If I = C i ∩ C i+1 for some i (indices taken modulo t), then I is in at most one circuithyperplanes of the form C j ∪ X, and at most three of the form C j ∪ C j+1 , a contradiction.So I = C i ∩C i+1 for some i.Then I is contained in exactly five circuit-hyperplanes, namely , and C i+1 ∪C i+2 , which means that each of these sets is contained in Y .However, r ≤ 2t − 3 implies that It follows from Theorem 12 and results of Nelson and van der Pol [8] that K(r, t) also cannot be certified as non-representable via a small non-representable minor.Following [8], a rank-4 sparse paving matroid M is Vámos-like if it has a partition (P 1 , P 2 , P 3 , P 4 ) such that exactly five of the six pairs P i ∪ P j form circuits of M.There are 39 Vámos-like matroids, one of which is the Vámos matroid itself, and none are representable.They prove that a sparse paving matroid is Ingleton if and only if it has no Vámos-like minor.So, Theorem 12 implies that K(r, t) has no Vámos-like minor when r ≥ 5 and r ≤ 2t − 3.
While K(r, t) is non-representable, we conjecture that it is very close to being representable, in the following sense.Conjecture 14.For all integers r and t with r ≥ 4, t ≥ 3 and 2t+2 ≥ r +4, any matroid obtained from K(r, t) by relaxing a circuit-hyperplane into a basis is representable.
When r ≤ 2t − 3, K(r, t) has no element in every circuit-hyperplane, so Conjecture 14 would imply that K(r, t) is an excluded minor for the class of representable matroids.We prove one more interesting property of K(r, t).Proposition 15.Let r ≥ 4 and t ≥ 3 be integers satisfying r ≤ 2t − 3 and let M be a proper minor of K(r, t).Then M is not isomorphic to K(r ′ , t ′ ) for any integers r ′ ≥ 4 and t ′ ≥ 3 satisfying r ′ ≤ 2t ′ − 3.
Proof.Suppose K(r ′ , t ′ ) is a minor of K(r, t) and (r ′ , t ′ ) = (r, t).Then r ′ < r.Let A ⊆ E(K(r, t)) be independent so that K(r, t)/A has a spanning K(r ′ , t ′ )-restriction, so It is straightforward to show that the intersection of any h of the sets C i has size at most r−2h.
So {K(r, t) : r ≥ 4, t ≥ 3, r ≤ 2t − 3} is an infinite antichain of non-representable matroids that all satisfy Ingleton's inequality, and if Conjecture 14 is true then each is also an excluded minor for representability.We comment that while K(r, t) is constructed using rank-2 lifts, related constructions using rank-t lifts with t > 2 are likely possible as well.We conclude this section with the following question.
Question 16.Which matroids K have the property that for every set X ⊆ E(K) there is a matroid N on the circuits of K/X such that (K/X) N ∼ = K \X?
For example, if every algebraic matroid has this property then K(r, t) would be nonalgebraic for all r ≥ 4 and t ≥ 3.This may be a promising direction since the Vámos matroid is non-algebraic [6] and isomorphic to K(4, 3).

Gain graphs
Recall that (K Γ n , φ Γ n ) is the gain graph over a finite group Γ where K Γ n has vertex set [n] and edge set [n]  2 × Γ, and the gain function φ Γ n orients edge ({i, j}, α) from i to j when i < j and assigns the label α.We write α ij for the edge ({i, j}, α), for convenience.For each α ∈ Γ we write E α for {({i, j}, α) : 1 ≤ i < j ≤ n}; these are the edges labeled by α.For a set A ⊆ Γ we write E A for ∪ α∈A E α .A cycle of (K Γ n , φ Γ n ) is balanced if an oriented product of its edge labels is equal to the identity element of Γ.For the remainder of this section we shall refer to balanced and unbalanced cycles of K Γ n , with the gain function φ Γ n implicit.
We can use Γ to define special automorphisms of the graph K Γ n , as follows.Given an integer k ∈ [n] and an element β ∈ Γ, define an automorphism For each edge e of K Γ n we say that f β (e) is obtained from e by switching at vertex k with value β.If a set X of edges of K Γ n can be obtained from a set Y via a sequence of switching operations we say that X and Y are switching equivalent.It is straightforward to check that switching maps balanced cycles to balanced cycles and unbalanced cycles to unbalanced cycles.We comment that switching is typically an operation on gain functions, and our application of this operation to define a graph automorphism is nonstandard.
For each nontrivial finite group Γ and integer n ≥ 3, we define M n,Γ to be the class of lifts of M(K Γ n ) for which a cycle of K Γ n is a circuit of M if and only if it is balanced.Each matroid in M n,Γ is simple, since each 2-element cycle of M(K Γ n ) is unbalanced.We now generalize Theorem 5 in the case that n ≥ 4.
Theorem 17.Let n ≥ 4 be an integer, let Γ be a finite group, and let then there is a prime p and an integer j ≥ 2 so that Γ ∼ = Z j p .Proof.Let ǫ denote the identity element of Γ.It was proved in [12,Lemma 19]  Proof.Let α ∈ Γ, and let A ∈ A contain α.Since A ∪ ǫ is a proper subgroup of Γ, we have |A| < |Γ|/2, and so |Γ − A| > |Γ|/2.The centralizer of α contains Γ − A, and thus contains more than |Γ|/2 elements.Since the centralizer of A is a subgroup of Γ, it follows that it is equal to Γ. Thus, α commutes with every element of Γ.Since the same argument applies to every element of Γ, it follows that Γ is abelian.Now that Γ is abelian the theorem statement follows from Theorem 5.
Let |Γ| be minimal so that the theorem is false.We first show that Γ has a 2-element generating set.We may assume that there are elements a 1 ∈ A 1 and a 2 ∈ A 2 that do not commute, or else the theorem statement holds by Claim 17.1.Let Γ ′ be the subgroup of Γ generated by {a 1 , a 2 }, and let n is a circuit of M ′ if and only if it is balanced.If Γ ′ = Γ, then the minimality of |Γ| implies that Γ ′ is abelian.But then a 1 and a 2 commute, a contradiction.So {a 1 , a 2 } is a 2-element generating set of Γ.It follows from [12,Lemma 20] that E {a 1 ,a 2 ,ǫ} spans M, and it follows from the submodularity of r M that r M (E {a 1 ,a 2 ,ǫ} ) = n + 1.So M is a rank-2 lift of M(K Γ n ).Suppose for a contradiction that there is some A ∈ A so that A ∪ ǫ is not a normal subgroup of Γ.We will define a pair of hyperplanes of M that violate the hyperplane axioms.Let H 1 = E A∪ǫ .Note that H 1 is a hyperplane of M, by the definition of ∼.Let {V, V ′ } be a partition of [n] with |V |, |V ′ | ≥ 2, and let H 2 be the set of edges with both ends in the same part.Then H 2 is a flat of M(K Γ n ), and is therefore a flat of M because M is a lift of M(K Γ n ) (see [9,Prop. 7.3.6]).By adding an edge labeled by ǫ with one end in V and the other in V ′ and repeatedly adding the third edge of a balanced cycle we obtain all edges.So r M (H 2 ) ≥ r(M) − 1 and therefore H 2 is a hyperplane of M. Since A ∪ ǫ is not normal, there is some element b ∈ Γ so that b , where e is an edge with one end in V and the other in V ′ , with label b, and directed from V to V ′ .By the hyperplane axioms, B is contained in a hyperplane of M.However, we claim that B spans M. Let B ′ be obtained from B by switching with value b at each vertex in V .In B ′ , the edge e is labeled by ǫ, and the set of labels of edges between each pair of vertices in V is b −1 • (A ∪ ǫ) • b.Note that B ′ has a spanning tree of edges all labeled by ǫ.Then the set obtained from B ′ by taking closure under balanced cycles contains E {ǫ,a,c} , which spans M because a ≁ c.Since switching preserves balanced cycles, this implies that B also spans M, a contradiction.
We have shown that A ∪ ǫ is a normal subgroup of Γ for each A ∈ A. Since normal subgroups that intersect in the identity commute, it follows that each α ∈ A commutes with every element in Γ − A. Thus, the theorem holds by Claim 17.1.Surprisingly, Theorem 5 does not generalize to non-abelian groups when n = 3. Walsh showed that if there exists M ∈ M n,Γ whose rank is at least two greater than that of M(K Γ n ), then Γ has a nontrivial partition [12,Lemma 21].Theorem 19 establishes the converse for n = 3. Recall that a nontrivial partition of a group Γ with identity ǫ is a partition A of Γ − {ǫ} so that A ∪ ǫ is a subgroup of Γ for all A ∈ A, and |A| ≥ 2. For example, Z j p has a nontrivial partition into p j −1 p−1 copies of the cyclic group Z p , and the dihedral group has a nontrivial partition where each reflection is a part and the nontrivial rotations form a part.We direct the reader to [14] for background on group partitions.
A group may have multiple nontrivial partitions.That said, each finite group has a canonical partition called the primitive partition, first described in [13].It is universal in the following sense: if A is the primitive partition of a finite group G and B is another partition, then for each B ∈ B, the following is a partition of B ∪ ǫ {A ∈ A : A ∪ ǫ is a subgroup of B ∪ ǫ}.
For our purposes, the most important property of the primitive partition is the following.
which allows us to apply the construction given in Theorem 2. It remains to prove that L = M N .Let Y ⊆ E, let T ⊆ Y be a basis of M|Y , and for each e ∈ Y \ T let C e denote the unique circuit of M in T ∪ {e}.If Y is dependent in M N , then T is a proper subset of Y and there exists a nontrivial linear dependence of the following form e∈Y \T λ e A L x Ce = 0. Then e∈Y \T λ e x Ce lies in the kernel of A L .It is nonzero since x Ce is zero at all f ∈ Y \ (T ∪ {e}) and nonzero at e.So Y is also dependent in L.
and let C be a circuit of M with C ⊆ C 1 ∪C 2 .Consider the matrix obtained by restricting A M to the columns indexed by C 1 ∪ C 2 .It has a two-dimensional kernel with a basis obtained from {x C 1 , x C 2 } by deleting entries corresponding to elements of