Dimer algebras, ghor algebras, and cyclic contractions

A ghor algebra is the path algebra of a dimer quiver on a surface, modulo relations that come from the perfect matchings of its quiver. Such algebras arise from abelian quiver gauge theories in physics. We show that a ghor algebra $\Lambda$ on a torus is a dimer algebra (a quiver with potential) if and only if it is noetherian, and otherwise $\Lambda$ is the quotient of a dimer algebra by homotopy relations. Furthermore, we classify the simple $\Lambda$-modules of maximal dimension and give an explicit description of the center of $\Lambda$ using a special subset of perfect matchings. In our proofs we introduce formalized notions of Higgsing and the mesonic chiral ring from quiver gauge theory.


Introduction
Let k be an algebraically closed field.A dimer quiver Q is a quiver that embeds in a compact surface Σ such that each connected component of Σ \ Q is simply connected and bounded by an oriented cycle, called a unit cycle.A perfect matching x of Q is a subset of arrows such that each unit cycle contains precisely one arrow in x.We will often assume that Q is nondegenerate, that is, each arrow is contained in at least one perfect matching.One can imagine associating a different color to each perfect matching, and coloring each arrow a by all the perfect matchings that contain a.In this article we introduce the ghor algebra of Q, which is a quotient of the path algebra kQ where paths with coincident heads and coincident tails are deemed equal if they have the same coloring.
More precisely, let P be the set of perfect matchings of Q, and let n := |Q 0 | be the number of vertices of Q.Consider the algebra homomorphism from kQ to the n × n matrix ring over the polynomial ring generated by P, η : kQ → M n (k[P]) , defined on the vertices e i ∈ Q 0 and arrows a ∈ Q 1 by (1) e i → e ii and a → e h(a),t(a) x∈P: x∋a x, and extended multiplicatively and k-linearly to kQ.The ghor algebra of Q is the quotient Λ := kQ/ ker η.Λ may be viewed as a tiled matrix ring by identifying it with its induced image under η.In companion articles we will find that ghor algebras, although often nonnoetherian, have rich geometric, algebraic, and homological properties [B2, B3, B5, B6]. 1hor algebras generalize cancellative dimer algebras.The dimer algebra of a dimer quiver Q is the superpotential algebra A = kQ/I, where I is the ideal (2) I := ⟨p − q | ∃a ∈ Q 1 s.t.ap, aq are unit cycles⟩ ⊂ kQ.
Dimer algebras originated in string theory [HK, F-W], and have found wide application to many areas of mathematics (e.g., [BKM, Br, IU, F-V, H, IN]).Although dimer algebras describe a class of abelian superconformal quiver gauge theories, it is really ghor algebras-and not dimer algebras-that describe abelian quiver gauge theories which are not superconformal.As we will show, ghor algebras are quotients of superpotential algebras, and the additional relations arise from the assumption that the gauge group is abelian.We consider the following questions for dimer quivers Q on a torus: • What is a minimal subset of perfect matchings P 0 ⊆ P with the property that the algebra homomorphism τ : kQ → M n (k[P 0 ]) , defined by (1) with P 0 in place of P, satisfies kQ/ ker η ∼ = kQ/ ker τ ?• How is the ghor algebra Λ related to the dimer algebra A?
• How is the center and representation theory of Λ related to the perfect matchings in P 0 ?Typically Q contains non-cancellative pairs of paths, which are distinct paths p, q in A with the property that there is some path r satisfying pr = qr ̸ = 0 or rp = rq ̸ = 0.
If A has a non-cancellative pair, then A and Q are said to be non-cancellative, and otherwise A and Q are cancellative.A fundamental characterization of this property is that A is cancellative if and only if A is noetherian [B3,Theorem 1.1].
To address our questions, we use the operation of cyclic contractions introduced in [B2], which formalizes and augments the operation of Higgsing in quiver gauge theories.A cyclic contraction is a k-linear map of dimer algebras where Q ′ is cancellative and is obtained by contracting a set of arrows in Q such that the cycles in Q are suitably preserved (Definition 3.2).Cyclic contractions are useful because they allow non-cancellative dimer algebras to be studied by relating them to well-understood cancellative dimer algebras that share similar structure.Moreover, cyclic contractions exist for all nondegenerate dimer algebras [B1,Theorem 1.1].Examples of cyclic contractions are given in Figure 1.
A perfect matching x ∈ P is called simple if there is a cycle in Q \ x that passes through each vertex of Q; in this case, Q \ x supports a simple Λ-module (and simple A-module) of dimension vector 1 Q 0 .Let S be the set of simple matchings of Q.We will show that if Q is cancellative, then we may take P 0 to be the simple matchings S.However, in general P 0 cannot equal S since there are (nondegenerate) dimer quivers for which S = ∅; see Figure 1.i.One may ask if P 0 is simply a minimal set of perfect matchings such that each arrow of Q is contained in some x ∈ P 0 .To the contrary: for our choice of P 0 , there will always be arrows which are not contained in any x ∈ P 0 whenever Q is non-cancellative.
Our main theorem is the following.
Theorem 1.1.(Theorems 5.6 and 5.9.)Let Q be a nondegenerate dimer quiver on a torus, and fix a cyclic contraction ψ : A → A ′ .Let P 0 ⊆ P be the perfect matchings x of Q for which ψ(x) is a simple matching of Q ′ .Set B := k[P 0 ].
(2) Suppose k is uncountable.The induced homomorphism τ ψ : Λ → M n (B) classifies all simple Λ-module isoclasses of maximal k-dimension: for each such module V , there is a maximal ideal b ∈ Max B such that where av := τ ψ (a)v for each a ∈ Λ, v ∈ (B/b) n .(3) The centers of Λ and Λ ′ ∼ = A ′ are given by the intersection and union of the vertex corner rings of Λ, To define the ghor algebra Λ it thus suffices to keep only those perfect matchings x ∈ P for which ψ(x) is a simple matching of Q ′ .
Ghor algebras are contrasted with Broomhead's construction of toric algebras [Br,5.1],which are also based on dimer quivers, in Remark 2.5.

Preliminaries
Let R be an integral domain and a k-algebra.We will denote by Max R the set of maximal ideals of R, and by Z(a) the closed set {m ∈ Max R | m ⊇ a} of Max R defined by the subset a ⊂ R.
. Some cyclic contractions.Each quiver is drawn on a torus, and ψ contracts the green arrows.
We will denote by Q = (Q 0 , Q 1 , t, h) the quiver with vertex set Q 0 , arrow set Q 1 , and head and tail maps h, t : Q 1 → Q 0 .We will denote by kQ the path algebra of Q, and by e i the idempotent corresponding to vertex i ∈ Q 0 .Multiplication of paths is read right to left, following the composition of maps.By module we mean left module.Finally, we will denote by e ij ∈ M d (k) the d × d matrix with a 1 in the ij-th slot and zeros elsewhere.
2.1.Algebra homomorphisms from perfect matchings.Let A = kQ/I be a dimer algebra on a torus.Denote by P and S the sets of perfect and simple matchings of Q respectively.
Consider the algebra homomorphisms Lemma 2.1.The ideal I ⊂ kQ given in ( 2) is contained in the kernels of τ and η.Therefore τ and η induce algebra homomorphisms on the dimer algebra A, Proof.Let p − q be a generator for I as given in (2); that is, there is an arrow a ∈ Q 1 such that pa and qa are unit cycles.Then Similarly, η(p) = η(q).Therefor p − q is in the kernels of τ and η. □ The following lemma is clear.
Furthermore, the sum i∈Q 0 σ i is in the center of A.
We will denote by σ i ∈ A the unique unit cycle at vertex i.In Section 5, given a cyclic contraction ψ : A → A ′ and elements p ∈ e j Ae i , q ∈ e ℓ A ′ e k , we will set p := τψ (p) := τ ψ(p), q := τ (q), and σ := where S ′ is the set of simple matchings of A ′ .
Remark 2.5.Let A = kQ/I be a dimer algebra.In [Br,5.1],Broomhead introduced the 'toric algebra' T of Q: where the maps defined by taking a face to the sum of its boundary arrows, and an arrow a to the difference h(a) − t(a).We remark that this algebra is almost never isomorphic to the ghor algebra Λ of Q.
Indeed, the toric and ghor algebras of Then T is isomorphic to the tiled matrix algebra where S appears in each diagonal component.The ghor algebra Λ of Q, on the other hand, is the tiled matrix algebra These two algebras are rarely isomorphic: on a torus, the two algebras coincide if and only if A is cancellative, in which case they are also isomorphic to A (see Theorem 30 below and [Br, Bo]).On higher genus surfaces, there are only two known exceptional families of dimer quivers whose toric and ghor algebras are isomorphic, given in [BB,Figures 2 and 3].However, in general the vertex corner rings η(e i kQe i ) are not all equal, and therefore the two algebras are not isomorphic.Furthermore, the two algebras have very different algebraic and geometric properties; for example, toric algebras are generically noetherian whereas ghor algebras are not [B3, B6], [BB,Section 4.1].

Impressions.
The following definition, introduced in [B7], captures a useful matrix ring embedding.
Definition 2.6.[B7, Definition 2.1] Let A be a finitely generated k-algebra and let Z be its center.An impression of A is an algebra monomorphism τ : A → M d (B) to a matrix ring over a commutative finitely generated k-algebra B, such that is surjective; and Surjectivity of (5) implies that the center Z is given by ( 6) If in addition A is a finitely generated module over its center, then τ classifies all simple A-module isoclasses of maximal k-dimension [B7, Proposition 2.5].Specifically, for each such module V , there is some b ∈ Max B such that where av := τ (a)v for each a ∈ A, v ∈ (B/b) d .If A is nonnoetherian, then τ may characterize the central geometry of A using the framework of depictions [B4, Section 3]; this relationship is used to study the central geometry of nonnoetherian ghor and dimer algebras in [B6].
We will show that for a ghor algebra Λ, the homomorphism τ ψ : Λ → M n (B) is an impression (Theorem 5.9.1).Furthermore, a dimer algebra admits an impression if and only if it equals its ghor algebra (Corollary 5.11).

Cyclic contractions
In this section, we describe a new method for studying non-cancellative dimer algebras, first introduced in [B2], that is based on the notion of Higgsing, or more generally symmetry breaking, in physics.Using this strategy, we gain information about non-cancellative dimer algebras by relating them to cancellative dimer algebras with similar structure.Throughout, A = kQ/I is a dimer algebra, typically noncancellative.
Definition 3.1.Let Q be a dimer quiver, let Q * 1 ⊂ Q 1 be a subset of arrows, and let Q ′ be the quiver obtained by contracting each arrow in Q * 1 .Specifically, Q ′ is formed by simultaneously removing each arrow a in Q * 1 , and merging together the head and tail vertices of a.This operation defines a k-linear map of path algebras and extended multiplicatively to paths and k-linearly to kQ.If ψ induces a k-linear map of dimer algebras ψ : A = kQ/I → A ′ = kQ ′ /I ′ , that is, ψ(I) ⊆ I ′ , then we call ψ a contraction of dimer algebras.
We now describe the structure we wish to preserve under a contraction.To specify this structure, we introduce the following commutative algebras.Definition 3.2.Let ψ : A → A ′ be a contraction to a cancellative dimer algebra.If , then we say ψ is cyclic, and call S the cycle algebra of A.
The cycle algebra is independent of the choice of ψ by [B2,Theorem 3.13].
Example 3.3.Four cyclic contractions are given in Figure 1.The non-cancellative quivers Q in (ii) and (iv) have appeared in the physics literature (e.g., [F-R, Section 4], [FKR]; and [DHP,Table 6,2.6]).The unit 2-cycles in (iii) and (iv) consist of arrows that are redundant generators for A ′ = kQ ′ /I ′ , and so may be removed from Q ′ .In (iii), let a, b, c be the respective red, blue, and green arrows in Q. Observe that in A = kQ/I, we have ab ̸ = ba and cab = cba.
Thus the pair ab, ba is non-cancellative (in fact, a and b generate a free subalgebra of A).In (i), Q has no simple matchings.These examples are considered in more detail in Example 5.13.Notation 3.4.For g, h ∈ B, by g | h we mean g divides h in B, even if g or h is assumed to be in S.
Lemma 3.5.Suppose ψ : A → A ′ is a contraction of dimer algebras, and A ′ has a perfect matching.Then ψ cannot contract a unit cycle of A to a vertex.
Proof.Assume to the contrary that ψ contracts the unit cycle σ j ∈ A to the vertex e ψ(j) ∈ A ′ .Fix a unit cycle Indeed, (i) and (iii) hold by Definition 3.1, and (ii) holds by Lemma 2.2.Denote by P ′ the set of perfect matchings of , by Lemma 2.1.Whence σ = 1.But this contradicts our assumption that The contraction of a unit cycle (drawn in green) to a vertex.Such a contraction cannot induce a contraction of dimer algebras ψ : A → A ′ if A ′ has a perfect matching, by Lemma 3.5.
An example where a unit cycle is contracted to a vertex is given in Figure 2.
Lemma 3.6.Suppose ψ : A → A ′ is a contraction of dimer algebras, and A ′ has a perfect matching.Then ψ cannot contract a cycle in the underlying graph Q of Q to a vertex.In particular, (1) ψ cannot contract a cycle in Q to a vertex; (2) if p is a cycle in Q that is not null-homotopic as a loop on the torus, then ψ(p) is a cycle in Q ′ that is also not null-homotopic; and (3) A does not have a non-cancellative pair where one of the paths is a vertex.
Proof.The number of vertices, edges, and faces in the underlying graphs Q and Q ′ of Q and Q ′ are given by Since Q and Q ′ each embed into a two-torus, their respective Euler characteristics vanish: V − E + F = 0, V ′ − E ′ + F ′ = 0. Assume to the contrary that ψ contracts the cycles p 1 , . . ., p ℓ in Q to vertices in Q ′ .Denote by n 0 and n 1 the respective number of vertices and arrows in Q which are subpaths of some p i , 1 ≤ i ≤ ℓ.Denote by m the number of vertices in Q ′ 0 of the form ψ(p i ) for some 1 ≤ i ≤ ℓ.By assumption, m ≥ 1.
In any cycle, the number of trivial subpaths equals the number of arrow subpaths.Furthermore, if two cycles share a common edge, then they also share a common vertex.Therefore Thus F ′ < F since m ≥ 1. Therefore ψ contracts a face of Q to a vertex.In particular, some unit cycle in Q is contracted to a vertex.But this is not possible by Lemma 3.5.□ Remark 3.7.Suppose ψ : A → A ′ is a cyclic contraction.We will show in Lemma 4.17 below that A ′ , being cancellative, necessarily has a perfect matching.Therefore Lemmas 3.5 and 3.6 hold in the case ψ is cyclic.
An immediate question is whether all non-cancellative dimer algebras admit cyclic contractions.If Q is nondegenerate, then A admits a cyclic contraction [B1, Theorem 1.1].However, there are degenerate dimer algebras that do not admit contractions (cyclic or not) to cancellative dimer algebras.For example, dimer algebras that contain permanent 2-cycles (Definition 4.5) cannot contract to cancellative dimer algebras, by Lemma 3.6.1.
A similar question is whether cyclic contractions can exist between two different cancellative dimer algebras.The answer to this question is negative: if However, we claim that if ψ is nontrivial, then it is not an algebra homomorphism.Indeed, let δ ∈ Q * 1 .Consider a path a 2 δa 1 ̸ = 0 in A. By Lemma 3.6.1,δ is not a cycle.In particular, h(a 1 ) ̸ = t(a 2 ).Whence We note, however, that the restriction is an algebra homomorphism [B2, Proposition 2.12.1].
Remark 3.9.The notion of Higgsing in quiver gauge theories (that is, contracting a set of arrows to vertices) was established in the early 2000's, before dimer algebras appeared (e.g., [F-H]).Higgsing was first introduced in the context of Morita equivalences by the author in 2013 in [B2,arXiv v1], and a couple of months later by Ishii and Ueda in [IU,arXiv v2; Morita equivalences did not appear in v1].The latter was based on Gulotta's 'inverse algorithm' from 2008 [G].Gulotta's algorithm produces cancellative dimer quivers for arbitrary polygons and uses Higgsing in an essential way.In contrast, both the cycle algebra and the operation of Higgsing that preserves the cycle algebra, that is, cyclic contractions, were introduced in [B2] to study non-cancellative dimer algebras.Cyclic contractions have subsequently played a fundamental role in the articles [B3, B5, B6].

Cycle structure
Let A = kQ/I be a dimer algebra, possibly with no perfect matchings.Unless stated otherwise, by path or cycle we mean path or cycle modulo I. Throughout, we use the notation (4).
Notation 4.1.Let π : R2 → T 2 be a covering map such that for some For paths p, q satisfying (9) t(p + ) = t(q + ) and h(p + ) = h(q + ), denote by R p,q the compact region in R 2 ⊃ Q + bounded by representatives p+ , q+ of p + , q + . 2 If the representatives are fixed (or arbitrary), we will write R p,q for R p,q .
Definition 4.2.We say a non-cancellative pair p, q ∈ A is minimal if for each noncancellative pair s, t ∈ A, we have {s, t} = {p, q} whenever there exists representatives p, q, s, t satisfying R s, t ⊆ R p,q .
The following hold.
(1) We proceed by induction on the region R p,q ⊂ R 2 bounded by p + and q + , with respect to inclusion.
If there are unit cycles sa and ta with a ∈ Q 1 , and R p,q = R s,t , then the claim is clear.So suppose the claim holds for all pairs of paths s, t such that Factor p and q into subpaths, such that for each 1 ≤ α ≤ m and 1 ≤ β ≤ n, there are paths p ′ α and q ′ β for which p ′ α p α and q ′ β q β are unit cycles, and p ′ α + and q ′ β + lie in the region R p,q .See Figure 3.Note that if p α is itself a unit cycle, then p ′ α is the vertex t(p α ), and similarly for q β .Consider the paths Then by Lemma 2.2 there is some c, d ≥ 0 such that ), and R p ′ ,q ′ ⊂ R p,q .Thus, by induction, there is some c ′ , d ′ ≥ 0 such that proving our claim.
(2) By Claim (1) there is some m, n ≥ 0 such that (10) 2) then follows since B is an integral domain.A similar result holds for τ in place of η by setting each non-simple perfect matching x ∈ P equal to 1 in (11).
(3) Suppose p, q is a non-cancellative pair.Then there is a path r such that by Lemma 2.1.Therefore η(p) = η(q) since B is an integral domain.Similarly, τ (p) = τ (q).(4) Finally, suppose P ̸ = ∅ and η(p) = η(q).Set σ := x∈P x.Then (11) implies since B is an integral domain (with η in place of τ ).By assumption, P ̸ = ∅.Whence σ ̸ = 1.Therefore m = n.Consequently, the path are drawn in red, blue, purple, and violet respectively.Each product p ′ α p α and q ′ β q β is a unit cycle.Note that the region R p ′ ,q ′ is properly contained in the region R p,q .
(1) Suppose paths p, q are either equal modulo I, or form a non-cancellative pair.
Then their lifts p + and q + bound a compact region R p,q in R 2 .(2) Suppose paths p, q are equal modulo I.If i + is a vertex in R p,q , then there is a path r + from t(p + ) to h(p + ) that is contained in R p,q , passes through i + , and satisfies p = r = q (modulo I).
Proof.(1.i) First suppose p, q are equal modulo I.The relations generated by I in (2) lift to homotopy relations on the paths of Q + .Thus t(p + ) = t(q + ) and h(p + ) = h(q + ).Therefore p + and q + bound a compact region in R 2 .
(2) The ideal I is generated by relations of the form s − t, where there is an arrow a such that sa and ta are unit cycles.The claim then follows since each trivial subpath of the unit cycle sa (resp.ta) is a trivial subpath of s (resp.t).□ Definition 4.5.A unit cycle σ i ∈ A of length 2 is a removable 2-cycle if the two arrows it is composed of are redundant generators for A, and otherwise σ i is a permanent 2-cycle.
Lemma 4.6.There are precisely two types of permanent 2-cycles, given in Figures 4.ii and 4.iii.Consequently, if a dimer algebra A has a permanent 2-cycle, then A is degenerate.
Figure 4. Cases for Lemma 4.6.In each case, a and b are arrows, and p and q are paths.In case (i) ap, bq, ab are unit cycles; in cases (ii) and (iii) qbpa, ab are unit cycles, and p, q are cycles.(In case (ii), q may be a vertex, and q need not be a unit cycle, that is, q + may contain arrows in its interior.)In case (i) ab is a removable 2-cycle, and in cases (ii) and (iii) ab is a permanent 2-cycle.
Proof.Let ab be a permanent 2-cycle, with a, b ∈ Q 1 .Let σ t(a) = sa and σ ′ t(b) = tb be the complementary unit cycles to ab containing a and b respectively.Since ab is permanent, b is a subpath of s, or a is a subpath of t.
Suppose b is a subpath of s.Then there are (possibly trivial) paths p, q such that s = qbp.We therefore have either case (ii) (if q is null-homotopic as a loop) or case (iii) (if q is not null-homotopic) shown in Figure 4. □ Notation 4.7.By a cyclic subpath of a path p, we mean a subpath of p that is a nontrivial cycle.Consider the following sets of cycles in A: be the set of cycles in the vertex corner ring e i Ae i .
• Let Ĉ be the set of cycles p ∈ C such that the lift of each cyclic permutation of each representative of p does not have a cyclic subpath.
We decorate C so as to specify a set of cycles; e.g., Ĉu 0) is the set of cycles whose lifts are cycles in Q + .In particular, Ĉ0 = Q 0 .Furthermore, the lift of any cycle p in Ĉ has no cyclic subpaths, although p itself may have cyclic subpaths.
Lemma 4.8.Suppose A does not have a non-cancellative pair where one of the paths is a vertex.Let p be a nontrivial cycle.
(2) If p ∈ C 0 and A is cancellative, then p = σ m t(p) for some m ≥ 1.Furthermore, τ is an algebra homomorphism on e i Ae i by Lemma 2.1.In particular, (12) implies pσ m = σ n .Therefore p = σ n−m since B is an integral domain. ( Then there is a cyclic permutation of p + which contains a cyclic subpath q + .In particular, q | p.Furthermore, since q ∈ C 0 , Claim (1) implies q = σ m for some m ≥ 1. Therefore σ | p.
(4) Let p be a path for which σ ∤ p. Then there is a simple matching x ∈ S such that x ∤ p.In particular, p is supported on Q \ x.Since x is simple, p is a subpath of a cycle q supported on Q \ x.Whence x ∤ q.Thus σ ∤ q.Therefore q is in Ĉ by the contrapositive of Claim (3).□ Remark 4.9.In Lemma 4.8 we assumed that A does not have a non-cancellative pair where one of the paths is a vertex.Such non-cancellative pairs exist: consider the permanent 2-cycle in Figure 4.ii, and suppose q is a trivial cycle.Then pσ t(p) = pab = σ t(p) .In particular, the cycles p, e t(p) form a non-cancellative pair.Furthermore, it is possible for m > n in (12).Consider a dimer algebra with the subquiver given in Figure 5.Here a, b, c, d are arrows, q is a nontrivial path, and ad, bc, qdcba are unit cycles.Let m ≥ 1, and set p := baq m+1 dc.Then Indeed, (i) holds by Lemma 2.2; (ii) holds since σ j = da; and (iii) holds since b = baqd in A. Note, however, that such a dimer quiver does not admit a perfect matching.
Remark 4.10.Some of the following results in this section that relate simple matchings, homotopy of paths, and cancellativity can be obtained by combining certain results of several articles [MR, D, Br, Bo, IU].The proofs we give here, however, are new, independent, and self-contained, and are based on different methods.In particular, it was shown by Broomhead in [Br,Proposition 6.2] that if A is a 'geometrically consistent' dimer algebra, then for each i, j ∈ Q + 0 there is a path p + from i to j such that, in our notation, p is not divisible by some 'extremal' (or 'corner') perfect matching x.This result was proceeded by similar results by Mozgovoy and Reineke in [MR,, and subsequently extended to cancellative dimer algebras by Bocklandt in [Bo,Theorem 8.7].The latter built on the work of Davison [D, Definition 2.5], which introduced cancellativity but did not consider perfect matchings.It was then shown by Ishii and Ueda in [IU,Proposition 8.2], a few years later, that simple matchings and extremal matchings coincide.Putting these three results together we obtain Proposition 4.11 below.
Similarly, a few of the lemmas in this section that involve simple matchings, namely Lemmas 4. 19,4.26,4.27,4.29,and Proposition 4.21 (see [Br,Proposition 1.2]), also follow from combining various results of these articles.We emphasize, however, that simple matchings are defined quite differently from extremal matchings, and were not considered in [Br], [Bo], [MR], or [D].These lemmas therefore do not follow directly from any of one these articles.
For the remainder of this section, we assume that Q has at least one perfect matching, P ̸ = ∅, unless stated otherwise.By the definition of Ĉ, each cycle p ∈ C u i \ Ĉ has a representative p that factors into subpaths p = p3 p2 p1 , where (13) Proof.In the following, we will use the notation (13).Set σ := x∈P x.

Suppose Ĉu
contains the vertex i + + u ∈ Q + 0 .Furthermore, suppose p and q admit representatives p′ and q′ (possibly distinct from p and q) such that the region R p′ ,q ′ has minimal area among all such pairs of cycles p, q.See Figure 6.
Assume to the contrary that A is cancellative.If s ̸ = t, then s, t would be a noncancellative pair by Lemma 4.3.4.Therefore s = t.Furthermore, there is a path r + in R s, t which passes through the vertex and is homotopic to s + (by the relations I), by Lemma 4.4.2.In particular, r factors into paths r = r 2 e i r 1 = r 2 r 1 , where r 1 , r 2 ∈ C u i .But p and q were chosen so that the area of R p′ ,q ′ is minimal.Thus there is some ℓ 1 , ℓ 2 ≥ 0 such that r1 = p′ σ ℓ 1 i and r2 = p′ σ ℓ 2 i (modulo I).
Since A is cancellative, the η-image of any nontrivial cycle in Q + is a positive power of σ by Lemma 4.8.1 (with η in place of τ ).In particular, since (p 1 p 3 ) + is a nontrivial cycle, there is an n ≥ 1 such that (15) η(p Indeed, (i) and (iii) hold by Lemma 2.1, (ii) holds by ( 14), and (iv) holds by (15).Thus, since k[P] is an integral domain, we have But n ≥ 1 and ℓ ≥ 0. Whence σ = 1.Therefore Q has no perfect matchings, contrary to our standing assumption that Q has at least one perfect matching.□ . Setup for Proposition 4.11, drawn on the cover Q + .The (lifts of the) cycles s = p 3 p 2 2 p 1 and q 3 q 2 2 q 1 are drawn in red and blue respectively.The representatives p′ and q′ of p and q, such that the region R p′ ,q ′ has minimal area, are drawn in green.
Definition 4.12.We call the subquiver given in Figure 7.i a column, and the subquiver given in Figure 7.ii a pillar.In the latter case, h(a ℓ ) and t(a 1 ) are either trivial subpaths of q ℓ and p 1 respectively, or p ℓ and q 1 respectively.Lemma 4.13.Suppose paths p + , q + have no cyclic subpaths (modulo I), and bound a region R p,q that contains no vertices in its interior.
(1) If p and q do not intersect, then p + and q + bound a column.
Proof.Since R p,q contains no vertices in its interior, each path that intersects its interior is an arrow.Thus p + and q + bound a union of subquivers given by the four cases in Figure 7.
In case (i), In cases (ii) -(iv), the paths p ℓ • • • p 1 and q ℓ • • • q 1 are not-necessarily-proper subpaths of p and q respectively.In cases (ii) and (iv), h(a ℓ ) and t(a 1 ) are either trivial subpaths of q and p respectively, or p and q respectively.In case (iii), h(a ℓ ) and t(a 1 ) are either both trivial subpaths of q, or both trivial subpaths of p.In all cases, each cycle which bounds a region with no arrows in its interior is a unit cycle.In particular, in cases (i) -(iii), each path a j p j b j−1 , b j q j a j is a unit cycle.Observe that in cases (iii) and (iv), q + has a cyclic subpath, contrary to assumption.Generalizations of case (iv) are considered in Figures 8.i and 8.ii.Consequently, p + and q + bound either a column or a union of pillars.□ Notation 4.14.For u ∈ Z 2 , denote by C u a maximal set of representatives of cycles in Ĉu whose lifts do not intersect transversely in R 2 (though they may share common subpaths).
In the following two lemmas, fix u ∈ Z 2 and a subset C u .
Lemma 4.15.Suppose Ĉu So suppose u ∈ Z 2 \ 0. Let p, q be representatives of cycles p, q in Ĉu that intersect transversely at k ∈ Q 0 .Then their lifts p+ and q+ intersect at least twice since p and q are both in C u .Thus p and q factor into paths p = p 3 p 2 p 1 and q = q 3 q 2 q 1 , where t(p + 2 ) = t(q + 2 ) and h(p + 2 ) = h(q + 2 ), and k + is the tail or head of p + 2 .See Figure 9. Since Ĉu i ̸ = ∅ for each i ∈ Q 0 , we may suppose that R p 2 ,q 2 contains no vertices in its interior.Since p and q are in Ĉ, p + 2 and q + 2 do not have cyclic subpaths.Thus p 2 = q 2 (modulo I) by Lemma 4.13.2.Therefore the paths s := p 3 q 2 p 1 and t := q 3 p 2 q 1 equal p and q (modulo I) respectively.In particular, s and t are in Ĉu .The lemma then follows since s+ and t+ do not intersect transversely at the vertices t(p + 2 ) or h(p + 2 ).□ Lemma 4.16.Let u ∈ Z 2 \ 0, and suppose Ĉu i ̸ = ∅ for each i ∈ Q 0 .Then there is a simple matching x ∈ S such that Q \ x supports each cycle in C u .Furthermore, if A contains a column, then there are two simple matchings Thus we may consider cycles p, q ∈ C u for which π −1 (p) and π −1 (q) bound a region R p,q with no vertices in its interior.
Recall Figure 7. Since C u i ̸ = ∅ for each i ∈ Q 0 , we may partition Q into columns and pillars, by Lemma 4.13.Consider the subset of arrows x 1 (resp.x 2 ) consisting of all the a j arrows in each pillar of Q, and all the a j arrows (resp.b j arrows) in each column of Q.Note that x 1 consists of all the right-pointing arrows in each column, and x 2 consists of all the left-pointing arrows in each column.Furthermore, if Q does not contain a column, then Observe that each unit cycle in each column and pillar contains precisely one arrow in x 1 , and one arrow in x 2 .(Note that this is not true for cases (iii) and (iv).)Furthermore, no such arrow occurs on the boundary of these regions, that is, as a subpath of p or q.Therefore x 1 and x 2 are perfect matchings of Q.
Recall that a simple A-module V of dimension 1 Q 0 is characterized by the property that there is a cycle which passes through each vertex of Q that does not annihilate V .Clearly Q \ x 1 and Q \ x 2 each contain a cycle that passes through each vertex of Q. Therefore x 1 and x 2 are simple matchings.□ Lemma 4.17.If A is cancellative, then Q has at least one simple matching.
Proof.Follows from Proposition 4.11 and Lemma 4.16.□ Lemma 4.17 will be superseded by Theorem 4.25 below.
Lemma 4.18.Let u ∈ Z 2 \ 0, and suppose Ĉu i ̸ = ∅ for each i ∈ Q 0 .Then A does not have a non-cancellative pair where one of the paths is a vertex.
Proof.Suppose e i , p is a non-cancellative pair.Then there is some m, n ≥ 0 such that In particular, if σ ∤ p and σ ∤ q, then p = q.
Proof.Consider cycles p, q ∈ C u .Since Q is a dimer quiver, there is a path r from t(p) to t(q).Thus there is some m, n ≥ 0 such that by Lemma 4.3.1.Furthermore, τ is an algebra homomorphism by Lemma 2.1.Thus , and q ∈ Ĉu h(a) , then ap = qa.
(2) Each vertex corner ring e i Ae i is commutative.
Let r + be a path in Q + from h ((ap) + ) to t ((ap) + ).Then by Lemma 4.8.2, there is some m, n ≥ 1 such that and rqa = σ n t(a) .Assume to the contrary that m < n.Then qa = apσ n−m t(a) since A is cancellative.Let b be a path such that ab is a unit cycle.By Lemma 2.2, we have . Furthermore, (ba) + is cycle in Q + since ba is a unit cycle.But this is a contradiction since q is in Ĉu .Whence m = n.Therefore qa = ap.
(3) If a cycle p is formed from subpaths of cycles in Ĉu , then p ∈ Ĉ.
Note that in Claim (4) the cycles p and q are based at the same vertex i, whereas in Claim (2) p and q may be based at different vertices.
Proof.If A is cancellative, then Ĉu i ̸ = ∅ for each i ∈ Q 0 , by Proposition 4.11.Therefore assumption (i) implies assumption (ii), and so it suffices to suppose (ii) holds.
Fix a maximal set C u of representatives of cycles in Ĉu whose lifts do not intersect transversely in R 2 .Consider a cycle p ∈ Ĉu .If p has a representative p in C u , then σ ∤ p, by Lemma 4.16.It thus suffices to suppose that p does not have a representative belonging to C u .In particular, for any representative p of p, there are cycles s, t ∈ C u such that s = p 3 q 2 p 1 , t = q 3 p 2 q 1 , p = p 3 p 2 p 1 , as in Figure 9.
Since s, t ∈ C u , we have σ ∤ s and σ ∤ t.Thus ( 16) s = t by Lemma 4.19.Set q := q 3 q 2 q 1 .Assume to the contrary that σ | p. Then σ | pq = p 3 p 2 p 1 q 3 q 2 q 1 = st where (i) holds by ( 16).Therefore σ | s since σ = x∈S x.But this is not possible since s ∈ C u .
Therefore p = q since B is an integral domain.
(4) Follows from Claim (3) and Figure 7.ii.A direct proof assuming A is cancellative: Suppose p, q ∈ Ĉu i .Let r + be a path in Q + from h (p + ) to t (p + ).Then there is some m, n ≥ 1 such that rp = σ m i and rq = σ n i , by Lemma 4.8.2.Suppose m ≤ n.Then rpσ n−m i = σ n i = rq.Thus pσ n−m i = q since A is cancellative.But then Claim (1) implies m = n since by assumption p and q are in Ĉu .Therefore p = q.□ Lemma 4.22.Suppose Ĉu with n ≥ 0.
(i) We claim that for each n ≥ 1, q + n lies in the region R p 2 q n−1 ,q n+1 (modulo I).This is shown in Figure 10.i.Indeed, suppose a representative q+ n of q + n intersects a representative q+ n−1 of q + n−1 , as shown in Figure 10.ii.Then q n−1 and q n factor into paths q n−1 = s 3 s 2 s 1 and q n = t 3 t 2 t 1 , where t(s + 2 ) = t(t + 2 ) and h(s + 2 ) = h(t + 2 ).In particular, there is some m ∈ Z such that s 2 = t 2 σ m , by Lemma 4.3.2.Set r := t 3 s 2 t 1 .Since q n−1 and q n are in Ĉ, we have (17) σ ∤ q n−1 and σ ∤ q n , by Proposition 4.21.1.Hence σ ∤ s 2 and σ ∤ t 2 .Thus m = 0. Therefore In particular, σ ∤ r by ( 17).Thus the cycle r is in Ĉ by Lemmas 4.8.4 and 4.18.Furthermore, r is in . Therefore r = q n (modulo I) by Proposition 4.21.4.This proves our claim.
(ii) By Claim (i), there is a cycle s ∈ Ĉ(ε 1 ,0) such that for each n ≥ 1, the area of the region R sq ′ n ,q ′ n+1 , bounded by a rightmost subpath q ′ + n of q + n , a rightmost subpath q ′ + n+1 of q + n+1 , and s + , tends to zero (modulo I) as n → ∞.See Figure 10.iii.(The case r = p is shown in Figure 10.i.) Since Q is finite, there is some N ≫ 1 such that if n ≥ N , then Fix n ≥ N .There is a simple matching x ∈ S such that x ∤ q n , by Proposition 4.21.1.Whence x ∤ q ′ n+1 = sq ′ n . (i) Figure 10.Setups for Lemma 4.22, drawn on the cover Q + .(In each case, q j labels the path from the lower line of p's to the upper line of p's with head at the corresponding vertex.)In (ii): q + n−1 and q + n intersect, and thus factor into paths q n−1 = s 3 s 2 s 1 and q n = t 3 t 2 t 1 , drawn in purple and blue respectively.In (iii): the area of the region R snq ′ n ,q ′ n+1 tends to zero as n → ∞.Here, an infinite path in π −1 (r ∞ ) is drawn in orange, each lift of the cycle p is drawn in red, and q + n and q + n+1 are drawn in blue.
In particular, x ∤ s.But p = s since p and s are both in Ĉ(ε 1 ,0) , by Proposition 4.21.2.Thus x ∤ p. Therefore x ∤ pq n , and so σ ∤ pq n .□ Consider the subset of arrows where S is the set of simple matchings of A.
We will show in Theorem 4.25 below that the two assumptions considered in the following lemma, namely that Q S 1 ̸ = ∅ and Ĉu i ̸ = ∅ for each u ∈ Z 2 and i ∈ Q 0 , can never both hold.
Lemma 4.23.Suppose Ĉu There is a cycle p ∈ Ĉt(δ) such that δ is not the rightmost arrow subpath of any representative of p.
Proof.Assume to the contrary that there is an arrow δ which is a rightmost arrow subpath of some representative of each cycle in Ĉt(δ) .
(i) We first claim that there is some u ∈ Z 2 \ 0 such that t(δ + ) lies in the interior of the region R s, t bounded by the lifts of two representatives s, t of the (unique) cycle in Ĉu h(δ) .(There is precisely one cycle in Ĉu h(δ) by Proposition 4.21.4.)Indeed, by Lemma 4.22, there are counter-clockwise ordered vectors

and cycles
Assume to the contrary that for each m, δ + is not contained in the region R p m+1 pm,pmp m+1 .
See Figure 11.i.Since the convex cone of u 0 , . . ., u n is R 2 , there is some m for which p m contains a leftmost subpath δr, and p m+1 contains a rightmost subpath r ′ , such that (δrr ′ ) + is a cycle in Q + .See Figure 12.Whence, for some ℓ ≥ 1, by Lemmas 4.8.1 and 4.18.Consequently, σ | p m+1 p m .But this is a contradiction to (18).Thus there is some 0 ≤ m ≤ n such that δ + lies in R p m+1 pm,pmp m+1 ; see Figure 11.ii.Since p m , p m+1 are in Ĉ, their lifts p + m , p + m+1 do not have cyclic subpaths.Thus, δ + only meets p + m and p + m+1 at t(p + m ) = h(δ + ).Consequently, t(δ + ) lies in the interior of R p m+1 pm,pmp m+1 .The claim then follows by setting s = p m+1 p m and t = p m p m+1 .
(ii) Let s and t be as in Claim (i).In particular, s = t.Assume to the contrary that there is a simple matching x ∈ S for which Figure 11.Setups for Lemma 4.23, drawn on the cover Q + .In both cases, δ + is the green δ arrow on the left.The black δ arrow on the right is the last arrow subpath of both p m p m+1 and p m+1 p m .In (i), δ + is not contained in the region R p m+1 pm,pmp m+1 , whereas in (ii) δ + is contained in this region.
Let r + be a path in R s, t from a vertex on the boundary of R s, t to t(δ + ).It suffices to suppose the tail of r is a trivial subpath of s, in which case s factors into paths s = s2 e t(r) s1 .
Whence x | r.Thus, since r is arbitrary, x divides the τ -image of each path in R s, t from the boundary of R s, t to t(δ + ).But t(δ + ) lies in the interior of R s, t.Thus the vertex t(δ) is a source in Q \ x.Therefore x is not simple, contrary to assumption.It follows that x | s for each x ∈ S.
(iii) By Claim (ii), σ | s.But this is a contradiction since s is in Ĉ, by Lemmas 4.8.3 and 4.18.Therefore there is a cycle p in Ĉt(δ) such that δ is not the rightmost arrow subpath of any representative of p. □ Remark 4.24.There are dimer algebras that have an arrow δ ∈ Q 1 which is a rightmost arrow subpath of each cycle in Ĉt(δ) ; see Figure 13.Furthermore, if A has center Z, admits a cyclic contraction, and δ ∈ Q S 1 , then δ is a rightmost arrow subpath of each cycle p ∈ Ze t(δ) for which σ ∤ p, by [B2,Lemma 2.4].
The following was shown in [IU,Corollary 11.4] for consistent dimer quivers; here we give a new and independent proof using columns and pillars.
In each example, the arrow δ is a rightmost arrow subpath of each path from t(δ) to a vertex on the boundary of the region R p,q bounded by the paths p + and q + , drawn in red and blue respectively.The red and blue arrows are nontrivial paths in Q + , and the black arrows are arrows in Q + .Theorem 4.25.Suppose (i) A is cancellative, or (ii Proof.Recall that assumption (i) implies assumption (ii), by Proposition 4.11.So suppose (ii) holds, and assume to the contrary that there is an arrow δ in Q S 1 .By Lemma 4.23, there is a cycle p ∈ Ĉt(δ) whose rightmost arrow subpath is not δ (modulo I).Let u ∈ Z 2 be such that p ∈ C u .By assumption, there is a cycle q in Ĉu h(δ) .By Lemma 4.15, we may choose representatives p, q of p, q such that R p,q contains no vertices in its interior.We thus have one of the three cases given in Figure 14.
Figure 14.Setup for Theorem 4.25.The red and blue paths are the (lifts of the) cycles p and q, respectively.The red and blue arrows are paths in Q + , and the black and brown arrows are arrows in Q + .In cases (i) and (ii), δ belongs to a simple matching, contrary to assumption.In case (iii), δ is a rightmost arrow subpath of p (modulo I), again contrary to assumption.
First suppose p+ and q+ do not intersect (that is, do not share a common vertex), as shown in case (i).Then p+ and q+ bound a column.By Lemma 4.16, the brown arrows belong to a simple matching x ∈ S. In particular, δ is in x, contrary to assumption.
So suppose p+ and q+ intersect, as shown in cases (ii) and (iii).Then p+ and q+ bound a union of pillars.Again by Lemma 4.16, the brown arrows belong to a simple matching x ∈ S. In particular, in case (ii) δ is in x, contrary to assumption.Therefore case (iii) holds.But then δ is a rightmost arrow subpath of p (modulo I), contrary to our choice of p. □ In [B3,Theorem 1.1] we show that the converse of Theorem 4.25 also holds: A is cancellative if and only if each arrow of Q is contained in a simple matching.
If p = q, then p = q.
Proof.Since A is cancellative, Q has at least one simple matching by Lemma 4.17.In particular, σ ̸ = 1.Thus we may apply the proof of Lemma 4.3.3,with τ in place of η. □ Lemma 4.27.Suppose A is cancellative.For each i ∈ Q 0 , the corner ring e i Ae i is generated by σ i and Ĉi .
Proof.Since I is generated by binomials, e i Ae i is generated by C i .It thus suffices to show that C i is generated by σ i and Ĉi .
(⇐) Let u ∈ Z 2 \ 0, and assume to the contrary that p ∈ C u satisfies p = σ m for some m ≥ 0. Since A is cancellative, there is a cycle q ∈ Ĉu t(p) by Proposition 4.11.Furthermore, σ ∤ q by Proposition 4.21.1.Thus there is some n ≥ 0 such that p = qσ n , by Lemma 4.3.2.Whence σ m = qσ n .Furthermore, σ ̸ = 1 by Lemma 4.17.Therefore m = n and ( 19) q = 1, since B is a polynomial ring.But each arrow in Q is contained in a simple matching by Theorem 4.25.Therefore q ̸ = 1, contrary to (19).□ Recall that B := k[S] is the polynomial ring generated by the simple matchings S.
Proof.(i) We fist claim that τ is injective on the vertex corner rings e i Ae i , i ∈ Q 0 .Fix a vertex i ∈ Q 0 and let p, q ∈ e i Ae i be cycles satisfying p = q.Let r be a path such that r + is path from h(p + ) to t(p + ).Then rp ∈ C 0 .Thus there is some m ≥ 0 such that rp = σ m i , by Lemma 4.8.2.Whence rq = rq = rp = rp = σ m i = σ m i = σ m .Thus rq ∈ C 0 by Lemma 4.29.Hence rq = σ m i = rp by Lemma 4.26.Therefore p = q since A is cancellative.
(ii) We now claim that τ is injective on paths.Let p, q ∈ e j Ae i be paths satisfying p = q.Let r be a path from j to i.The two cycles pr and qr at j then satisfy pr = qr.Thus pr = qr since τ is injective on the corner ring e j Ae j by Claim (i).Therefore p = q since A is cancellative.
(iii) Since A is generated by paths and τ is injective on paths, it follows that τ is injective.

Proof of main theorem
Throughout, let A = kQ/I be a dimer algebra, and let ψ : A → A ′ be a cyclic contraction to a cancellative dimer algebra A ′ = kQ ′ /I ′ .If A is cancellative, then we may take ψ to be the identity map.
Let τ : A ′ → M |Q ′ 0 | (B) be the algebra homomorphism defined in Lemma 2.1, with B the polynomial ring generated by the simple matchings of A ′ , To prove Theorem 1.1, we will use the composition of ψ with τ .Specifically, let be the k-linear map defined for each i, j ∈ Q 0 and p ∈ e j Ae i by Denote by p := τ ψ (p) := τ ψ(p) the single nonzero matrix entry of τ ψ (p).
Lemma 5.1.The map τ is an algebra homomorphism.Furthermore, ψ is a k-linear map, and an algebra homomorphism when restricted to each vertex corner ring e i Ae i .□ Lemma 5.2.Let p be a nontrivial cycle.
Proof.Suppose ψ(p) = ψ(q).Consider lifts p + and q + for which t(p + ) = t(q + ).Let r + be a path from h(p + ) to h(q + ).Then ψ(r + ) is a cycle in Q ′+ since ψ(p) = ψ(q), by Lemma 4.4.1.Thus r + is also a cycle by the contrapositive of Lemma 3.6.2.Whence h(p + ) = h(q + ).□ Denote by P and P ′ the set of perfect matchings of Q and Q ′ respectively.Consider the algebra homomorphisms defined in (1), . By Lemma 3.5, ψ cannot contract a unit cycle to a vertex.Thus, if y is a perfect matching of Q ′ , then ψ −1 (y) is a perfect matching of Q.We may therefore view k[P ′ ] as a subalgebra of k[P] under the identification y = ψ −1 (y) for each y ∈ P ′ .For g ∈ e j kQe i , we then have Proposition 5.4.The k-linear map ψ : kQ → kQ ′ induces a k-linear map of ghor algebras ψ : Λ → Λ ′ .Furthermore, for each i, j ∈ Q 0 , the restriction is injective.
(2) Suppose that either k is uncountable, or Q is cancellative.Then τ ψ classifies all simple Λ-module isoclasses of maximal k-dimension: for each such module V , there is some b ∈ Max B such that ) The centers of Λ and Λ ′ are given by the intersection and union of the vertex corner rings of Λ, ( 25) Proof.
(1) We first show that τ ψ and τ are impressions of Λ and Λ ′ . ( is a simple representation of Λ.Indeed, when viewed as a vector space diagram on Q of dimension vector 1 Q 0 , each arrow is represented by a nonzero scalar.Thus, since Q has a cycle containing each vertex, the representation is simple.It follows that the composition (26) is surjective.
(1.iii) Set R := τ ψ (Z(Λ)); Claims (1.i) and (1.ii), together with (6), imply that each element of R is product of a polynomial in B and the identity matrix For brevity, we will omit 1 |Q 0 | in our expressions.
We claim that the morphism (1.iv)By Claims (1.i), (1.ii), and (1.iii), τ ψ is an impression of Λ.It follows that τ itself is an impression of Λ ′ by letting ψ : A = A ′ → A ′ be the trivial cyclic contraction given by the identity map.
(2.ii) Now suppose k is uncountable.Using Claim (2.i), it was shown in [B2,Proposition 3.10 and Theorem 3.11] that, irrespective of whether Q is cancellative or non-cancellative, τ ψ classifies all simple Λ-modules (and A-modules) of dimension vector 1 Q 0 , up to isomorphism.
We claim that the dimension vector of the simple Λ-modules of maximal k-dimension is 1 Q 0 .Let V be a simple Λ-module.Then for each i ∈ Q 0 , e i V is a simple e i Λe imodule.But e i Λe i is a commutative countably generated k-algebra, and k is an uncountable algebraically closed field.Thus, e i V = 0 or e i V ∼ = k.Whence dim k e i V ≤ 1.Furthermore, there is a representation of Λ where each arrow is represented by 1 ∈ k since we may set each x equal to 1; this representation is simple of dimension 1 Q 0 since Q has a cycle containing each vertex.In particular, there is a simple Λ-module of dimension 1 Q 0 , proving our claim.
(i) and (iv) hold by Claims (1.i) and (1.ii), together with ( 6).(ii) holds since the contraction ψ is cyclic, and since A ′ = Λ ′ .Finally, (iii) holds by Proposition 4.28.2.Therefore (25) holds.□ Remark 5.10.In the case of cancellative dimer algebras, the labeling of arrows given in (3) agrees with the labeling of arrows in the toric construction of [CQ,Proposition 5.3].We note, however, that impressions are defined more generally for non-toric algebras and have different implications than the toric construction of [CQ].
Corollary 5.11.A dimer algebra A is cancellative if and only if it admits an impression τ : A → M |Q 0 | (B), where B is an integral domain and τ (e i ) = e ii for each i ∈ Q 0 .
Proof.Suppose A admits an impression τ : A → M |Q 0 | (B), where B is an integral domain and τ (e i ) = e ii for each i ∈ Q 0 .Consider paths p, q, r satisfying pr = qr ̸ = 0. Then pr = pr = qr = qr.
Thus p = q since B is an integral domain.Whence τ (p) = pe h(p),h(r) = qe h(p),h(r) = τ (q).Therefore p = q by the injectivity of τ .The converse holds by Theorem 5.9.□ Recall that an algebra A is prime if for all a, b ∈ A, aAb = 0 implies a = 0 or b = 0, that is, the zero ideal is a prime ideal.
In this heated state the material has rotational symmetry and no net magnet field.However, as the material cools, one molecule happens to settle down first.As the neighboring molecules settle down, they align themselves with the first molecule, until all the molecules settle down in alignment with the first. 7The orientation of the first settled molecule then determines the direction of magnetization for the whole material, and the material no longer has rotational symmetry.One says that the rotational symmetry of the heated magnet was spontaneously broken as it cooled, and a global magnetic field emerged. 8iggsing is a way of using spontaneous symmetry breaking to turn a quantum field theory with a massless field and more symmetry into a theory with a massive field and less symmetry.Here mass (vev's) takes the place of magnetization, gauge symmetry (or the rank of the gauge group) takes the place of rotational symmetry, and energy scale (RG flow) takes the place of temperature.
The recent discovery of the Higgs boson at the Large Hadron Collider is another example of Higgsing.9 Higgsing in quiver gauge theories We now give our main example.Suppose an arrow a in a quiver gauge theory with dimension 1 Q 0 is contracted to a vertex e.We make two observations: (1) the rank of the gauge group drops by one since the head and tail of a become identified as the single vertex e; (2) a has zero vev at any representation where a is represented by zero, while e can never have zero vev since it is a vertex, and X only consists of representation isoclasses with dimension 1 Q 0 .We therefore see that contracting an arrow to a vertex is a form of Higgsing in quiver gauge theories with dimension 1 Q 0 .10 In the context of a 4-dimensional N = 1 quiver gauge theory with quiver Q, the Higgsing we consider in this article is related to RG flow.We start with a non-superconformal (strongly coupled) quiver theory Q which admits a low energy effective description, give nonzero vev's to a set of bifundamental fields Q * 1 , and obtain a new theory Q ′ that lies at a superconformal fixed point.
The mesonic chiral ring and the cycle algebra The cycle algebra S, introduced in [B2], is similar to the mesonic chiral ring in the corresponding quiver gauge theory.In such a theory, the mesonic operators, which are the gauge invariant operators, are generated by the cycles in the quiver.If the gauge group is abelian, then the dimension vector is 1 Q 0 .In the case of a dimer theory a) = e h(a),t(a) and k-linearly to kQ.
Lemma 2.3.Each unit cycle σ i ∈ e i Ae i satisfies τ (σ i ) = e ii x∈S x and η (σ i ) = e ii x∈P x.Proof.Each perfect matching contains precisely one arrow in each unit cycle.□ Notation 2.4.For each i, j ∈ Q 0 , denote by τ : e j Ae i → B := k[S] and η : e j Ae i → k[P] the respective k-linear maps defined on p ∈ e j Ae i by τ (p) = τ (p)e ji and η(p) = η(p)e ji .In particular, τ (p) and η(p) are the single nonzero matrix entries of τ (p) and η(p).In Section 4, we will set (4) p := τ (p) and σ := x∈S x (or occasionally, σ := x∈P x).
then the paths pσ m−n i and e t(p) form a non-cancellative pair.But this is contrary to assumption.Thus n − m ≥ 1.
is surjective.Indeed, for any m ∈ Max R, Bm is a proper ideal of B. Thus there is a maximal ideal b ∈ Max B containing Bm since B is noetherian.Furthermore, since B is a finitely generated k-algebra and k is algebraically closed, the intersectionb ∩ R =: m ′ is a maximal ideal of R. Whence m ⊆ Bm ∩ R ⊆ b ∩ R = m ′ .But m and m ′ are both maximal ideals of R. Thus m = m ′ .Therefore b ∩ R = m, proving our claim.