NNONLOCALITY AND THE CENTRAL GEOMETRY OF DIMERALGEBRAS
CHARLIE BEIL
Abstract.
Let A be a dimer algebra and Z its center. It is well known that if A is cancellative, then A and Z are noetherian and A is a finitely generated Z -module. Here we show the converse: if A is non-cancellative (as almost all dimeralgebras are), then A and Z are nonnoetherian and A is an infinitely generated Z -module. Although Z is nonnoetherian, we show that it has Krull dimension 3and is generically noetherian. Furthermore, we show that the reduced center is thecoordinate ring for a Gorenstein algebraic variety with the strange property thatit contains precisely one ‘smeared-out’ point of positive geometric dimension. Inour proofs we introduce formalized notions of Higgsing and the mesonic chiral ringfrom quiver gauge theory. Contents
1. Introduction 21.1. Overview 21.2. Preliminaries 62. Cycle structure 83. Cancellative dimer algebras 264. Non-cancellative dimer algebras and their homotopy algebras 314.1. Cyclic contractions 314.2. Reduced and homotopy centers 384.3. Homotopy dimer algebras 494.4. Non-annihilating paths 504.5. Nonnoetherian and nonlocal 544.6. Integral closure 69Appendix A. A brief account of Higgsing with quivers 72References 74
Mathematics Subject Classification.
Key words and phrases.
Dimer model, dimer algebra, brane tiling, superpotential algebra, quiverwith potential, noncommutative algebraic geometry, nonnoetherian geometry. a r X i v : . [ m a t h . AG ] M a y CHARLIE BEIL Introduction
Overview.
We begin by recalling the definition of a dimer algebra, which is atype of quiver with potential whose quiver is dual to a dimer model.
Definition 1.1. • Let Q be a finite quiver whose underlying graph Q embeds into a two-dimensional real torus T , such that each connected component of T \ Q is simply connected and bounded by an oriented cycle of length at least 2,called a unit cycle . The dimer algebra of Q is the quiver algebra A = kQ/I with relations(1) I := (cid:104) p − q | ∃ a ∈ Q such that pa and qa are unit cycles (cid:105) ⊂ kQ, where p and q are paths. • A and Q are called cancellative if for all paths p, q, r ∈ A , p = q whenever pr = qr (cid:54) = 0 or rp = rq (cid:54) = 0 . Cancellative dimer algebras are now well understood: they are 3-Calabi-Yau al-gebras and noncommutative crepant resolutions of 3-dimensional toric Gorensteinsingularities (e.g., [Bo, Theorem 10.2], [Br], [D, Theorem 4.3], [MR, Theorem 6.3]).
Non-cancellative dimer algebras, on the other hand, are much less understood. How-ever, almost all dimer algebras are non-cancellative, and so it is of great interest tounderstand them.Our main results apply to non-cancellative dimer algebras that admit cyclic con-tractions, which is a new notion we introduce. Before stating our results, we definethis notion.Let A = kQ/I be a (possibly non-cancellative) dimer algebra. Fix a subset ofarrows Q ∗ ⊂ Q , and consider the quiver Q (cid:48) obtained by contracting each arrow in Q ∗ to a vertex. This contraction defines a k -linear map of path algebras ψ : kQ → kQ (cid:48) . We call ψ a contraction of dimer algebras if Q (cid:48) is also a dimer quiver, and ψ inducesa k -linear map of dimer algebras ψ : A = kQ/I → A (cid:48) = kQ (cid:48) /I (cid:48) , More generally, dimer algebras may be defined where Q embeds into any compact surface; seefor example [Bo, Theorems 11.2 and 11.3] and [BKM]. However, here we will only consider dimeralgebras for which Q embeds into a torus. A ring A which is a finitely-generated module over a central normal Gorenstein subdomain R is Calabi-Yau of dimension n if (i) gl . dim A = dim R = n ; (ii) A is a maximal Cohen-Macaulay moduleover R ; and (iii) Hom R ( A, R ) ∼ = A , as A -bimodules [Bra, Introduction]. Let A be a dimer algebra such that each arrow is contained in a perfect matching. Then A iscancellative if and only if A satisfies a combinatorial ‘consistency condition’ [IU, Theorem 1.1], [Bo,Theorem 6.2]. Cancellative dimer algebras are thus also called ‘consistent’. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 3 that is, ψ ( I ) ⊆ I (cid:48) . An example is given in Figure 1.We now describe the structure we wish to be preserved under a contraction. Definition 1.2.
Let A = kQ/I be a dimer algebra. • A perfect matching D ⊂ Q is a set of arrows such that each unit cyclecontains precisely one arrow in D . • A simple matching D ⊂ Q is a perfect matching such that Q \ D supports asimple A -module of dimension 1 Q (that is, Q \ D contains a cycle that passesthrough each vertex of Q ). Denote by S the set of simple matchings of A .Let ψ : A → A (cid:48) be a contraction to a cancellative dimer algebra A (cid:48) . Denote by B := k [ x D | D ∈ S (cid:48) ]the polynomial ring generated by the simple matchings of A (cid:48) . Denote by E ij a squarematrix with a 1 in the ij -th slot and zeros elsewhere. Since A (cid:48) is cancellative, thereis an algebra monomorphism(2) τ : A (cid:48) (cid:44) → M | Q (cid:48) | ( B )defined on i ∈ Q (cid:48) and a ∈ Q (cid:48) by τ ( e i ) := E ii , τ ( a ) := E h( a ) , t( a ) (cid:89) a ∈ D ∈S (cid:48) x D , and extended multiplicatively and k -linearly to A (cid:48) (Theorem 3.5). For p ∈ e j A (cid:48) e i ,denote by ¯ τ ( p ) ∈ B the single nonzero matrix entry of τ ( p ), that is, τ ( p ) = ¯ τ ( p ) E ji .If S := k [ ∪ i ∈ Q ¯ τ ψ ( e i Ae i )] = k (cid:2) ∪ i ∈ Q (cid:48) ¯ τ ( e i A (cid:48) e i ) (cid:3) , then we say ψ is cyclic , and call S the cycle algebra of A .Cyclic contractions are universal localizations that preserve the cycle algebra. Fur-thermore, they formalize the notion of Higgsing in abelian quiver gauge theories,where (morally) the mesonic chiral ring is preserved (Appendix A). The cycle alge-bra is isomorphic to the center Z (cid:48) of A (cid:48) , and contains the reduced centerˆ Z := Z/ nil Z of A as a subalgebra (Theorems 3.5 and 4.27). Moreover, the cycle algebra is uniquelydetermined by A : it is independent of the choice of ψ . Indeed, S is isomorphic tothe GL-invariant functions on the Zariski-closure of the variety of simple A -modulesof dimension 1 Q [B3, Theorem 3.13].Our first main theorem is the following. CHARLIE BEIL2 1 21 2 1 · (cid:15) (cid:15) (cid:63) (cid:63) (cid:95) (cid:95) (cid:47) (cid:47) (cid:127) (cid:127) (cid:47) (cid:47) (cid:95) (cid:95) (cid:15) (cid:15) (cid:31) (cid:31) ψ −→ (cid:15) (cid:15) (cid:47) (cid:47) (cid:95) (cid:95) (cid:47) (cid:47) (cid:15) (cid:15) (cid:47) (cid:47) (cid:95) (cid:95) (cid:47) (cid:47) (cid:15) (cid:15) Q Q (cid:48)
Figure 1.
The non-cancellative dimer algebra A = kQ/I cyclicallycontracts to the cancellative dimer algebra A (cid:48) = kQ (cid:48) /I (cid:48) . Both quiversare drawn on a torus, and the contracted arrow is drawn in green. Here,the cycle algebra of A is S = k [ x , y , xy, z ] ⊂ B = k [ x, y, z ], and thehomotopy center of A is R = k + ( x , y , xy ) S . Theorem 1.3.
Let A be a dimer algebra with center Z , and suppose A admits acyclic contraction. Then the following are equivalent (Theorems 4.41 and 4.50):(1) A is cancellative.(2) A is noetherian.(3) Z is noetherian.(4) A is a finitely generated Z -module.(5) The vertex corner rings e i Ae i are pairwise isomorphic algebras.(6) Each vertex corner ring e i Ae i is isomorphic to Z .(7) Each arrow annihilates a simple A -module of dimension Q .(8) Each arrow is contained in a simple matching.(9) If ψ : A → A (cid:48) is a cyclic contraction, then S = k [ ∩ i ∈ Q ¯ τ ψ ( e i Ae i )] .(10) If ψ : A → A (cid:48) is a cyclic contraction, then ψ is trivial. We then use the notion of depiction and geometric dimension, introduced in [B2],to make sense of the central geometry of non-cancellative dimer algebras. A depictionof a nonnoetherian integral domain R is a closely related noetherian overring S thatprovides a way of visualizing the geometry of R (Definition 1.9). The underlyingidea is that nonnoetherian geometry is the geometry of nonlocal algebraic varietiesand schemes. In this framework, (closed) points can be ‘smeared-out’, and thus havepositive dimension. Such points are therefore inherently nonlocal. We use the term‘nonlocal’ in the sense that two widely separated points may somehow be very nearto each other.Our second main theorem characterizes the central geometry of non-cancellativedimer algebras. Theorem 1.4.
Let A be a non-cancellative dimer algebra which admits a cyclic con-traction. Then(1) Z and ˆ Z := Z/ nil Z each have Krull dimension 3 (Theorem 4.66).(2) ˆ Z is a nonnoetherian integral domain depicted by the cycle algebra of A (Corol-lary 4.28 and Theorem 4.68). ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 5 (3) The reduced induced scheme structure of
Spec Z is birational to a noetherianaffine scheme, and contains precisely one closed point of positive geometricdimension. Furthermore, the maximal ideal spectrum Max ˆ Z may be viewedas a Gorenstein algebraic variety X with one positive dimensional subvariety Y ⊂ X identified as a single ‘smeared-out’ point (Theorem 4.68). In the course of proving Theorems 1.3 and 1.4, we also introduce the homotopydimer algebra of A (Definition 4.33),˜ A := A/ (cid:104) p − q | ∃ path r such that pr = qr (cid:54) = 0 or rp = rq (cid:54) = 0 (cid:105) , and the homotopy center of A (Definitions 4.3 and 4.26), R := k [ ∩ i ∈ Q ¯ τ ψ ( e i Ae i )] . We show that the center of the homotopy dimer algebra ˜ A is isomorphic to thehomotopy center R of A (Theorem 4.35). Furthermore, the reduced center ˆ Z of A isa subalgebra of its homotopy center R (Theorem 4.27). In particular, the kernel of ψ , restricted to Z , coincides with the nilradical of Z (Theorem 4.24),ker ψ ∩ Z = nil Z. In general, the containment ˆ Z ⊆ R may be proper (Theorem 4.31). Even so, ˆ Z and R determine the same nonlocal variety (Theorem 4.68). Furthermore, their integralclosures coincide (Theorem 4.75). Finally, we give necessary and sufficient conditionsfor R to be normal; for instance, R is normal if and only if there is an ideal J of S such that R = k + J (Theorem 4.74). Conventions:
Throughout, k is an uncountable algebraically closed field of char-acteristic zero. Let R be an integral domain and a k -algebra. We will denote bydim R the Krull dimension of R ; by Frac R the ring of fractions of R ; by Max R theset of maximal ideals of R ; by Spec R either the set of prime ideals of R or the affine k -scheme with global sections R ; by R p the localization of R at p ∈ Spec R ; and by Z ( a ) the closed set { m ∈ Max R | m ⊇ a } of Max R defined by the subset a ⊂ R .We will denote by Q = ( Q , Q , t , h) a quiver with vertex set Q , arrow set Q ,and head and tail maps h , t : Q → Q . We will denote by kQ the path algebra of Q , and by e i the idempotent corresponding to vertex i ∈ Q . Multiplication of pathsis read right to left, following the composition of maps. We say a ring is noetherianif it is both left and right noetherian. By module we mean left module. By infinitelygenerated R -module, we mean an R -module that is not finitely generated. By non-constant monomial, we mean a monomial that is not in k . 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. CHARLIE BEIL
Preliminaries.
Algebra homomorphisms from perfect matchings.
In this subsection, A = kQ/I is a dimer algebra. Denote by P and S the sets of perfect and simple matchings of A respectively. The following lemma is clear (see e.g., [MR, Section 4]). Lemma 1.5.
Let A = kQ/I be a dimer algebra. If σ i , σ (cid:48) i are two unit cycles at i ∈ Q , then σ i = σ (cid:48) i . Furthermore, the element (cid:80) i ∈ Q σ i is in the center of A . We will denote by σ i ∈ A the unique unit cycle at vertex i . Lemma 1.6.
Consider the maps (3) τ : A → M | Q | ( k [ x D | D ∈ S ]) and η : A → M | Q | ( k [ x D | D ∈ P ]) defined on i ∈ Q and a ∈ Q by (4) τ ( e i ) := E ii , τ ( a ) := E h( a ) , t( a ) (cid:81) a ∈ D ∈S x D ,η ( e i ) := E ii , η ( a ) := E h( a ) , t( a ) (cid:81) a ∈ D ∈P x D , and extended multiplicatively and k -linearly to A . Then τ and η are algebra homo-morphisms. Furthermore, each unit cycle σ i ∈ e i Ae i satisfies (5) τ ( σ i ) = E ii (cid:89) D ∈S x D and η ( σ i ) = E ii (cid:89) D ∈P x D . Proof. If a ∈ Q and pa , qa are unit cycles, then τ ( p ) = E t( a ) , h( a ) (cid:89) a (cid:54)∈ D ∈S x D = τ ( q ) .τ is therefore well-defined by the relations I given in (1), and thus clearly an algebrahomomorphism. η is similarly an algebra homomorphism.By definition, each perfect matching contains precisely one arrow in each unit cycle.Therefore (5) holds. (cid:3) Notation 1.7.
For each i, j ∈ Q , denote by¯ τ : e j Ae i → B := k [ x D | D ∈ S ] and ¯ η : e j Ae i → k [ x D | D ∈ 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 . That is, ¯ τ ( p ) and ¯ η ( p ) are the single nonzero matrix entries of τ ( p ) and η ( p ) respec-tively. In Sections 2 and 3, we will set(6) p := ¯ τ ( p ) and σ := (cid:89) D ∈S x D (or occasionally, σ := (cid:89) D ∈P x D ). ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 7
In Section 4, we will consider a cyclic contraction ψ : A → A (cid:48) . Denote by S (cid:48) the setof simple matchings of A (cid:48) . For p ∈ e j Ae i and q ∈ e (cid:96) A (cid:48) e k , we will set(7) p := ¯ τ ψ ( p ) , q := ¯ τ ( q ) , and σ := (cid:89) D ∈S (cid:48) x D . Impressions.
The following definition, introduced in [B], captures a useful ma-trix ring embedding.
Definition 1.8. [B, Definition 2.1] An impression ( τ, B ) of a finitely generated non-commutative algebra A is a finitely generated commutative algebra B and an algebramonomorphism τ : A (cid:44) → M d ( B ) such that • for generic b ∈ Max B , the composition(8) A τ −→ M d ( B ) (cid:15) b −→ M d ( B/ b ) ∼ = M d ( k )is surjective; and • the morphism Max B → Max τ ( Z ), b (cid:55)→ b ∩ τ ( Z ), is surjective.An impression determines the center of A explicitly [B, Lemma 2.1], Z ∼ = { f ∈ B | f d ∈ im τ } ⊆ B. Furthermore, if A is a finitely generated module over its center, then its impressionclassifies all simple A -module isoclasses of maximal k -dimension [B, Proposition 2.5].Specifically, for each such module V , there is some b ∈ Max B such that V ∼ = ( B/ b ) d , where av := (cid:15) b ( τ ( a )) v for each a ∈ A , v ∈ V . If A is nonnoetherian, then itsimpression may characterize the central geometry of A using depictions [B2, Section3].Now let A be a dimer algebra. We will show that ( τ, k [ x D | D ∈ S ]) in (3) isan impression of A if A is cancellative (Theorem 3.5). Furthermore, we will showthat A is cancellative if and only if A admits an impression ( τ, B ), where B is anintegral domain and τ ( e i ) = E ii for each i ∈ Q (Corollary 3.7). However, if A is non-cancellative, then its ‘homotopy algebra’ admits such an impression (Definition 4.33and Theorem 4.35). We will then use this impression to understand the nonnoetheriancentral geometry of a large class of non-cancellative dimer algebras in Section 4.5.1.2.3. Depictions.
The following definition, introduced in [B2], aims to capture thegeometry of nonnoetherian algebras with finite Krull dimension.
Definition 1.9. [B2, Definition 2.11.] Let S be an integral domain and a noetherian k -algebra. Let R be a nonnoetherian subalgebra of S which contains S in its fractionfield, and suppose there is a point m ∈ Max R such that R m is noetherian.(1) We say R is depicted by S if(a) the morphism ι R,S : Spec S → Spec R , q (cid:55)→ q ∩ R , is surjective, and CHARLIE BEIL (b) for each n ∈ Max S , if R n ∩ R is noetherian, then R n ∩ R = S n .(2) The geometric codimension or geometric height of p ∈ Spec R is the infimumght( p ) := inf (cid:8) ht( q ) | q ∈ ι − R,S ( p ) , S a depiction of R (cid:9) . The geometric dimension of p is the differencegdim p := dim R − ght( p ) . We will consider the following subsets of the variety Max S ,(9) U ∗ R,S := { n ∈ Max S | R n ∩ R is noetherian } U R,S := { n ∈ Max S | R n ∩ R = S n } . The locus U R,S specifies where Max R ‘looks like’ the variety Max S . Furthermore,condition (1.b) in Definition 1.9 is equivalent to U ∗ R,S = U R,S . Example 1.10.
Let S = k [ x, y ], and consider its nonnoetherian subalgebra R = k [ x, xy, xy , . . . ] = k + xS.R is then depicted by S , and the closed point xS ∈ Max R has geometric dimension1 [B2, Proposition 2.8 and Example 2.20]. Furthermore, U R,S is the complement ofthe line Z ( x ) = { x = 0 } ⊂ Max S. In particular, Max R may be viewed as 2-dimensional affine space A k = Max S withthe line Z ( x ) identified as a single ‘smeared-out’ point. From this perspective, xS isa positive dimensional, hence nonlocal, point of Max R .In Section 4.5, we will show that both the reduced and homotopy centers of A ,namely ˆ Z and R , are depicted by its cycle algebra S . Furthermore, we will find thatˆ Z and R determine the the same nonlocal variety.2. Cycle structure
Unless stated otherwise, A = kQ/I is a dimer algebra with at least one perfectmatching (Definition 1.2). Furthermore, unless stated otherwise, by path or cyclewe mean path or cycle modulo I . Throughout, we use the notation (6). Notation 2.1.
Let π : R → T be a covering map such that for some i ∈ Q , π (cid:0) Z (cid:1) = i ∈ Q . Denote by Q + := π − ( Q ) ⊂ R the covering quiver of Q . For each path p in Q ,denote by p + the unique path in Q + with tail in the unit square [0 , × [0 , ⊂ R satisfying π ( p + ) = p . An example of a dimer algebra with no perfect matchings is given in [B3, Example 3.15].
ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 9
Furthermore, for paths p , q satisfying(10) t( p + ) = t( q + ) and h( p + ) = h( q + ) , denote by R p,q the compact region in R ⊃ Q + bounded by (representatives of) p + and q + . Finally, denote by R ◦ p,q the interior of R p,q . Definition 2.2. • We say say two paths p, q ∈ A are a non-cancellative pair if p (cid:54) = q , and thereis a path r ∈ A such that rp = rq (cid:54) = 0 or pr = qr (cid:54) = 0 . • We say a non-cancellative pair p, q is minimal if for each non-cancellative pair s, t , R s,t ⊆ R p,q implies { s, t } = { p, q } . Lemma 2.3.
Let p, q ∈ e j Ae i be distinct paths such that (11) t( p + ) = t( q + ) and h( p + ) = h( q + ) . Then(1) pσ mi = qσ ni for some m, n ≥ .(2) ¯ τ ( p ) = ¯ τ ( q ) σ m for some m ∈ Z .(3) ¯ η ( p ) = ¯ η ( q ) if and only if p, q is a non-cancellative pair.Proof. (1) Suppose p, q ∈ e j Ae i satisfy (11). We proceed by induction on the region R p,q ⊂ R bounded by p + and q + , with respect to inclusion.If there are unit cycles sa and ta with a ∈ Q , and R p,q = R s,t , then the claim isclear. So suppose the claim holds for all pairs of paths s, t such thatt( s + ) = t( t + ) , h( s + ) = h( t + ) , and R s,t ⊂ R p,q . Factor p and q into subpaths that are maximal length subpaths of unit cycles, p = p m p m − · · · p and q = q n q n − · · · q . Then for each 1 ≤ i ≤ m and 1 ≤ j ≤ n , there are paths p (cid:48) i and q (cid:48) j such that p (cid:48) i p i and q (cid:48) j q j are unit cycles, and p + i and q + j lie in the region R p,q . See Figure 2. Note that if p j isitself a unit cycle, then p (cid:48) j is the vertex t( p j ), and similarly for q j .Consider the paths p (cid:48) := p (cid:48) p (cid:48) · · · p (cid:48) m and q (cid:48) := q (cid:48) q (cid:48) · · · q (cid:48) n . Then by Lemma 1.5 there is some c, d ≥ p (cid:48) p = σ ci and q (cid:48) q = σ di . In Lemma 2.12.1, we will not require (10) to hold.
Now t( p (cid:48) + ) = t( q (cid:48) + ) , h( p (cid:48) + ) = h( q (cid:48) + ) , and R p (cid:48) ,q (cid:48) ⊂ R p,q . Thus by induction there is some c (cid:48) , d (cid:48) ≥ p (cid:48) σ c (cid:48) i = q (cid:48) σ d (cid:48) i . Therefore by Lemma 1.5, pσ d + d (cid:48) i = pq (cid:48) qσ d (cid:48) i = pp (cid:48) qσ c (cid:48) i = qσ c + c (cid:48) i , proving our claim.(2) By Claim (1) there is some m, n ≥ pσ mi = qσ ni . Thus by Lemma 1.6,(14) ¯ τ ( p ) σ m = ¯ τ ( pσ mi ) = ¯ τ ( qσ ni ) = ¯ τ ( q ) σ n ∈ B. Claim (2) then follows since B is an integral domain.(3.i) First suppose ¯ η ( p ) = ¯ η ( q ). Set σ := (cid:81) D ∈P x D . Then (14) implies σ m = σ n since B is an integral domain (with ¯ η in place of ¯ τ ). Recall our standing assumption, P (cid:54) = ∅ . In particular, σ (cid:54) = 1. Therefore m = n . Consequently, the path r = σ mj satisfies rp = rq (cid:54) = 0 by (13) and Lemma 1.5.(3.ii) Conversely, suppose p, q is a non-cancellative pair. Then there is a path r such that rp = rq (cid:54) = 0 or pr = qr (cid:54) = 0;say rp = rq . Whence ¯ η ( r )¯ η ( p ) = ¯ η ( rp ) = ¯ η ( rq ) = ¯ η ( r )¯ η ( q ) , by Lemma 1.6. Therefore ¯ η ( p ) = ¯ η ( q ) since B is an integral domain. (cid:3) Lemma 2.4. (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) Suppose paths p, q are equal modulo I . If i + is a vertex in R p,q , then there isa path r + from t( p + ) to h( p + ) that is contained in R p,q , passes through i + ,and satisfies (15) p = r = q (modulo I ) . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 11 ···· · · · ··· ... ... p (cid:82) (cid:82) p (cid:79) (cid:79) p m − (cid:54) (cid:54) p m (cid:52) (cid:52) q (cid:76) (cid:76) q (cid:79) (cid:79) q n − (cid:104) (cid:104) q n (cid:106) (cid:106) p (cid:48) (cid:18) (cid:18) p (cid:48) (cid:15) (cid:15) p (cid:48) m − (cid:118) (cid:118) p (cid:48) m (cid:116) (cid:116) q (cid:48) (cid:12) (cid:12) q (cid:48) (cid:15) (cid:15) q (cid:48) n − (cid:40) (cid:40) q (cid:48) n (cid:42) (cid:42) Figure 2.
Setup for Lemma 2.3.1. The paths p = p m · · · p , q = q n · · · q , p (cid:48) = p (cid:48) · · · p (cid:48) m , and q (cid:48) = q (cid:48) · · · q (cid:48) n are drawn in red, blue, purple,and violet respectively. Each product p (cid:48) i p i and q (cid:48) j q j is a unit cycle. Notethat the region R p (cid:48) ,q (cid:48) is properly contained in the region R p,q . Proof. (1.i) First suppose p, q are equal modulo I . The relations generated by theideal I in (1) lift to homotopy relations on the paths of Q + . Thus t( p + ) = t( q + ) andh( p + ) = h( q + ). Therefore p + and q + bound a compact region in R .(1.ii) Now suppose p, q is a non-cancellative pair. Then there is a path r such that rp = rq (cid:54) = 0, say. In particular, t(( rp ) + ) = t(( rq ) + ) and h(( rp ) + ) = h(( rq ) + ) byClaim (1.i.). Therefore t( p + ) = t( q + ) and h( p + ) = h( q + ) as well.(2) Recall that I is generated by relations of the form s − t , where there is anarrow a such that sa and ta are unit cycles. The claim then follows since each vertexsubpath of the unit cycle sa (resp. ta ) is a vertex subpath of s (resp. t ). (cid:3) Definition 2.5.
A unit cycle σ i ∈ A of length 2 is a . A 2-cycle is removable if the two arrows it is composed of are redundant generators for A , and otherwise the2-cycle is permanent . Lemma 2.6.
There are precisely two types of permanent 2-cycles. These are givenin Figures 3.ii and 3.iii.Proof.
Let ab be a permanent 2-cycle, with a, b ∈ Q . Let σ t( a ) = sa and σ (cid:48) t( b ) = tb be the complementary unit cycles to ab containing a and b respectively. Since ab ispermanent, b is a subpath of s , or a is a subpath of t .Suppose b is a subpath of s . If b is a leftmost subpath of s , that is, s = bp for somepath p , then we have the setup given in Figure 3.ii. Otherwise there is a non-vertexpath q such that s = qbp , in which case we have the setup given in Figure 3.iii. (cid:3) · · p (cid:15) (cid:15) q (cid:79) (cid:79) a (cid:116) (cid:116) b (cid:52) (cid:52) = · · p (cid:15) (cid:15) q (cid:79) (cid:79) · · p (cid:79) (cid:79) a (cid:116) (cid:116) b (cid:52) (cid:52) i ji j p (cid:79) (cid:79) q (cid:15) (cid:15) a (cid:116) (cid:116) b (cid:52) (cid:52) a (cid:116) (cid:116) ( i ) ( ii ) ( iii ) Figure 3.
Cases for Lemma 2.6. In each case, a and b are arrows,and p and q are non-vertex paths. In case (i) ap , bq , and ab are unitcycles; in case (ii) abp and ab are unit cycles; in case (iii) qbpa and ab are unit cycles, and p and q are cycles. In case (i) ab is a removable2-cycle, and in cases (ii) and (iii) ab is a permanent 2-cycle. Notation 2.7.
By a cyclic subpath of a path p , we mean a proper subpath of p thatis a non-vertex cycle. Consider the following sets of cycles in A : • Let C be the set of cycles in A (i.e., cycles in Q modulo I ). • For u ∈ Z , let C u be the set of cycles p ∈ C such thath( p + ) = t( p + ) + u ∈ Q +0 . • For i ∈ Q , let C i be the set of cycles in the vertex corner ring e i Ae i . • Let ˆ C be the set of cycles p ∈ C such that ( p ) + does not have a cyclic subpath;or equivalently, the lift of each cyclic permutation of p does not have a cyclicsubpath.We denote the intersection ˆ C ∩ C u ∩ C i , for example, by ˆ C ui . Note that C is the setof cycles whose lifts are cycles in Q + . In particular, ˆ C = ∅ . Furthermore, the lift ofany cycle p in ˆ C has no cyclic subpaths, although p itself may have cyclic subpaths. Lemma 2.8.
Suppose A does not contain a non-cancellative pair where one of thepaths is a vertex. Let p be a non-vertex cycle.(1) If p ∈ C , then p = σ m for some m ≥ .(2) If p ∈ C and A is cancellative, then p = σ mi for some m ≥ .(3) If p ∈ C \ ˆ C , then σ | p .(4) If p is a path for which σ (cid:45) p , then p is a subpath of a cycle in ˆ C .Proof. (1) Suppose p ∈ C , that is, p + is a cycle in Q + . Set i := t( p ). By Lemma2.3.1, there is some m, n ≥ pσ mi = σ ni . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 13 j · i q (cid:79) (cid:79) c (cid:121) (cid:121) d (cid:118) (cid:118) a (cid:57) (cid:57) b (cid:55) (cid:55) Figure 4.
Setup for Remark 2.9. Let m ≥
1. Then the path p := baq m +1 dc satisfies pσ mi = σ i .If m ≥ n , then the paths pσ m − ni and e t( p ) form a non-cancellative pair. But this iscontrary to assumption. Thus n − m ≥ τ is an algebra homomorphism on e i Ae i by Lemma 1.6. In particular,(16) implies pσ m = σ n . Therefore p = σ n − m since B is an integral domain.(2) If A is cancellative, then (16) implies p = σ n − mi .(3) Suppose p ∈ C \ ˆ C . Then there is a cyclic permutation of p + which contains acyclic subpath q + . In particular, q | p . Furthermore, since q ∈ C , Claim (1) implies q = σ m for some m ≥
1. Therefore σ | p .(4) Let p be a path for which σ (cid:45) p . Then there is a simple matching D ∈ S suchthat x D (cid:45) p . In particular, p is supported on Q \ D . Since D is simple, p is a subpathof a cycle q supported on Q \ D . Whence x D (cid:45) q . Thus σ (cid:45) q . Therefore q is in ˆ C bythe contrapositive of Claim (3). (cid:3) Remark 2.9.
In Lemma 2.8 we assumed that A does not contain a non-cancellativepair where one of the paths is a vertex. Such non-cancellative pairs exist: if A containsa permanent 2-cycle as in Figure 3.ii, then pσ t( p ) = σ t( p ) . In particular, p, e t( p ) is anon-cancellative pair.Furthermore, it is possible for m > n in (16). Consider a dimer algebra with thesubquiver given in Figure 4. Here a, b, c, d are arrows, q is a non-vertex path, and ad , bc , qdcba are unit cycles. Let m ≥
1, and set p := baq m +1 dc . Then pσ mi = baq m +1 dcσ mi (i) = ba ( qσ j ) m qdc = b ( aqd ) m +1 c = b ( aqdc ) = bc = σ i , where ( i ) holds by Lemma 1.5. Note, however, that such a dimer algebra does notadmit a perfect matching.By the definition of ˆ C , each cycle p ∈ C ui \ ˆ C has a representative ˜ p that factors intosubpaths ˜ p = ˜ p ˜ p ˜ p , where(17) p p = ˜ p ˜ p + I ∈ C and p = ˜ p + I ∈ ˆ C u . Proposition 2.10. If ˆ C ui = ∅ for some u ∈ Z \ and i ∈ Q , then A is non-cancellative.Proof. In the following, we will use the notation (17). Set σ := (cid:81) D ∈P x D .Suppose ˆ C ui = ∅ . Let p, q ∈ C ui be cycles such that the region R ˜ p ˜ p ˜ p , ˜ q ˜ q ˜ q contains the vertex i + + u ∈ Q +0 . Furthermore, suppose p and q admit representatives˜ p (cid:48) and ˜ q (cid:48) (possibly distinct from ˜ p and ˜ q ) such that the region R ˜ p (cid:48) , ˜ q (cid:48) has minimal areaamong all such pairs of cycles p, q . See Figure 5.By Lemmas 1.6 and 2.3.1, there is some m ∈ Z such that¯ η ( p p p ) = ¯ η ( q q q ) σ m . Suppose m ≥
0. Set s := p p p and t := q q q σ mi . Then(18) ¯ η ( s ) = ¯ η ( t ) . Assume to the contrary that A is cancellative. Then s = t by Lemma 2.3.3.Furthermore, there is a path r + in R ˜ s, ˜ t which passes through the vertex i + + u ∈ Q +0 and is homotopic to s + (by the relations I ), by Lemma 2.4.2. In particular, r factorsinto paths r = r e i r = r r , where r , r ∈ C ui . But p and q were chosen so that the area of R ˜ p (cid:48) , ˜ q (cid:48) is minimal. Thus there is some (cid:96) , (cid:96) ≥ r = ˜ p (cid:48) σ (cid:96) i and ˜ r = ˜ p (cid:48) σ (cid:96) i (modulo I ) . Set (cid:96) := (cid:96) + (cid:96) . Then(19) r = r r = p σ (cid:96) + (cid:96) i = p σ (cid:96)i . Since A is cancellative, the ¯ η -image of any non-vertex cycle in Q + is a positivepower of σ by Lemma 2.8.1 (with ¯ η in place of ¯ τ ). In particular, since ( p p ) + is anon-vertex cycle, there is an n ≥ η ( p p ) = σ n . Hence¯ η ( p )¯ η ( p ) (i) = ¯ η ( s ) = ¯ η ( r ) (ii) = ¯ η ( p ) σ (cid:96) (iii) = ¯ η ( p )¯ η ( p )¯ η ( p p ) σ (cid:96) (iv) = ¯ η ( p )¯ η ( p ) σ n + (cid:96) . Indeed, ( i ) and ( iii ) hold by Lemma 1.6, ( ii ) holds by (19), and ( iv ) holds by (20).Furthermore, the image of ¯ η is in the integral domain k [ x D | D ∈ P ]. Thus σ n + (cid:96) = 1 . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 15 · ·· ·· ···· q (cid:79) (cid:79) q (cid:79) (cid:79) p (cid:79) (cid:79) p (cid:79) (cid:79) q (cid:39) (cid:39) p (cid:119) (cid:119) q (cid:39) (cid:39) p (cid:119) (cid:119) q (cid:103) (cid:103) p (cid:55) (cid:55) q (cid:103) (cid:103) p (cid:55) (cid:55) i + i + + ui + +2 u ˜ q (cid:48) (cid:100) (cid:100) ˜ p (cid:48) (cid:58) (cid:58) ˜ q (cid:48) (cid:100) (cid:100) ˜ p (cid:48) (cid:58) (cid:58) Figure 5.
Setup for Proposition 2.10. The (lifts of the) cycles s = p p p and q q q are drawn in red and blue respectively. Therepresentatives ˜ p (cid:48) and ˜ q (cid:48) of p and q , such that the region R ˜ p (cid:48) , ˜ q (cid:48) hasminimal area, are drawn in green.But n ≥ (cid:96) ≥
0. Whence σ = 1. Therefore Q has no perfect matchings, contraryto our standing assumption that Q has at least one perfect matching. (cid:3) Definition 2.11.
We call the subquiver given in Figure 6.i a column , and the sub-quiver given in Figure 6.ii a pillar . In the latter case, h( a (cid:96) ) and t( a ) are either vertexsubpaths of q (cid:96) and p respectively, or p (cid:96) and q respectively. Lemma 2.12.
Suppose paths p + , q + have no cyclic subpaths, and bound a region R p,q which contains no vertices in its interior.(1) If p and q do not intersect, then p + and q + bound a column.(2) Otherwise p + and q + bound a union of pillars. In particular, if t( p + ) = t( q + ) and h( p + ) = h( q + ) (cid:54) = t( p + ) , then p = q (modulo I ).Proof. Suppose the hypotheses hold. 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 ofsubquivers given by the four cases in Figure 6.In case (i), p = p (cid:96) · · · p and q = q (cid:96) · · · q . In cases (ii) - (iv), the paths p (cid:96) · · · p and q (cid:96) · · · q are not-necessarily-proper subpathsof p and q respectively. In cases (ii) and (iv), h( a (cid:96) ) and t( a ) are either vertexsubpaths of q and p respectively, or p and q respectively. In case (iii), h( a (cid:96) ) andt( a ) are either both vertex subpaths of q , or both vertex subpaths of p . In all cases, ·· · ... ·· ·· p (cid:79) (cid:79) a (cid:39) (cid:39) q (cid:79) (cid:79) b (cid:119) (cid:119) (cid:79) (cid:79) (cid:79) (cid:79) q (cid:96) (cid:79) (cid:79) b (cid:96) (cid:119) (cid:119) p (cid:79) (cid:79) a (cid:39) (cid:39) q (cid:79) (cid:79) a (cid:96) (cid:39) (cid:39) ·· · ... ·· ·· p (cid:83) (cid:83) a (cid:19) (cid:19) q (cid:79) (cid:79) b (cid:119) (cid:119) (cid:79) (cid:79) (cid:79) (cid:79) q (cid:96) − (cid:79) (cid:79) b (cid:96) − (cid:119) (cid:119) p (cid:96) (cid:50) (cid:50) a (cid:96) (cid:44) (cid:44) q (cid:96) (cid:108) (cid:108) a (cid:96) − (cid:39) (cid:39) · ... ·· ·· ·· p b (cid:11) (cid:11) q (cid:75) (cid:75) (cid:79) (cid:79) a (cid:39) (cid:39) (cid:79) (cid:79) (cid:79) (cid:79) q (cid:96) − (cid:79) (cid:79) b (cid:96) − (cid:119) (cid:119) p (cid:96) (cid:50) (cid:50) a (cid:96) (cid:44) (cid:44) q (cid:96) (cid:108) (cid:108) a (cid:96) − (cid:39) (cid:39) ·· · ... ·· ·· · p (cid:83) (cid:83) a (cid:19) (cid:19) q (cid:79) (cid:79) b (cid:119) (cid:119) (cid:79) (cid:79) (cid:79) (cid:79) q (cid:96) − (cid:79) (cid:79) q (cid:48) (cid:96) − (cid:79) (cid:79) (cid:29) (cid:29) (cid:23) (cid:23) (cid:60) (cid:60) b (cid:96) − (cid:119) (cid:119) p (cid:96) (cid:50) (cid:50) a (cid:96) (cid:42) (cid:42) q (cid:96) (cid:108) (cid:108) a (cid:96) − (cid:39) (cid:39) ( i ) ( ii ) ( iii ) ( iv ) Figure 6.
Cases for Lemmas 2.12 and 2.15. The subquiver in case (i)is a ‘column’, and the subquiver in case (ii) is a ‘pillar’. The paths a , . . . , a (cid:96) , b , . . . , b (cid:96) are arrows, and the paths p (cid:96) · · · p and q (cid:96) · · · q ,drawn in red and blue respectively, are not-necessarily-proper subpathsof p and q . In case (iv), the green paths are also arrows. In cases (i),(ii), (iii), the cycles q j a j b j and a j p j b j − are unit cycles. Note that q + has a cyclic subpath in cases (iii) and (iv), contrary to assumption.Furthermore, pb (cid:96) = b (cid:96) q in case (i), and p (cid:96) · · · p = q (cid:96) · · · q in case (ii). ···· q j (cid:79) (cid:79) b j (cid:119) (cid:119) (cid:3) (cid:3) a j (cid:39) (cid:39) p j (cid:79) (cid:79) p (cid:48) j (cid:79) (cid:79) p j +1 (cid:79) (cid:79) · p j (cid:79) (cid:79) p j +1 (cid:79) (cid:79) ····· ... b j (cid:122) (cid:122) a j (cid:36) (cid:36) (cid:63) (cid:63) (cid:27) (cid:27) (cid:27) (cid:27) (cid:23) (cid:23) q j (cid:79) (cid:79) q j (cid:79) (cid:79) q jm (cid:79) (cid:79) ( i ) ( ii ) Figure 7.
Generalizations of the setup given in Figure 6.iv. In bothcases, the green paths are arrows. Case ( i ) shows that if t( a j ) (cid:54) = h( b j )(resp. h( a j ) (cid:54) = t( b j − )), then p + (resp. q + ) has a cyclic subpath. In case( ii ), q j = q jm · · · q j q j . This case shows that if q j a j b j (resp. a j p j b j − )is not a unit cycle, then q + (resp. p + ) has a cyclic subpath. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 17 i ·· i jj p (cid:79) (cid:79) p (cid:79) (cid:79) p (cid:79) (cid:79) q (cid:111) (cid:111) q (cid:56) (cid:56) q (cid:77) (cid:77) Figure 8.
Setup for Lemma 2.14.each cycle which bounds a region with no arrows in its interior is a unit cycle. Inparticular, in cases (i) - (iii), each path a j p j b j − , 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 7. i and 7. ii . Consequently, p + and q + bound either a column or a union of pillars. (cid:3) Notation 2.13.
For u ∈ Z , denote by (cid:98) C u ⊆ ˆ C u a maximal subset of cycles inˆ C u whose lifts do not intersect transversely in R (though they may share commonsubpaths).In the following two lemmas, fix u ∈ Z \ (cid:98) C u ⊆ ˆ C u . Lemma 2.14.
Suppose ˆ C ui (cid:54) = ∅ for each i ∈ Q . Then (cid:98) C ui (cid:54) = ∅ for each i ∈ Q .Proof. Suppose p, q ∈ ˆ C u intersect transversely at k ∈ Q . 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 p p and q = q q q , wheret( p +2 ) = t( q +2 ) and h( p +2 ) = h( q +2 ) , and k + is the tail or head of p +2 . See Figure 8.Since ˆ C ui (cid:54) = ∅ for each i ∈ Q , we may suppose that R p ,q contains no vertices inits interior. Since p and q are in ˆ C , p +2 and q +2 do not have cyclic subpaths. Thus p = q (modulo I ) by Lemma 2.12.2. Therefore the paths s := q p q and t := p q p equal q and p (modulo I ) respectively. In particular, s and t are in ˆ C u . The lemmathen follows since s + and t + do not intersect transversely at the vertices t( p +2 ) orh( p +2 ). (cid:3) Lemma 2.15.
Suppose ˆ C ui (cid:54) = ∅ for each i ∈ Q . Then there is a simple matching D ∈ S such that Q \ D supports each cycle in (cid:98) C u .Furthermore, if A contains a column, then there are two simple matchings D , D ∈S such that Q \ ( D ∪ D ) supports each cycle in (cid:98) C u . Proof.
Suppose the hypotheses hold. By Lemma 2.14, (cid:98) C ui (cid:54) = ∅ for each i ∈ Q sinceˆ C ui (cid:54) = ∅ for each i ∈ Q . Thus we may consider cycles p, q ∈ (cid:98) C u for which π − ( p ) and π − ( q ) bound a region R p,q with no vertices in its interior.Recall Figure 6. Since (cid:98) C ui (cid:54) = ∅ for each i ∈ Q , we may partition Q into columnsand pillars, by Lemma 2.12. Consider the subset of arrows D (resp. D ) consistingof all the a j arrows in each pillar of Q , and all the a j arrows (resp. b j arrows) in eachcolumn of Q . Note that D consists of all the right-pointing arrows in each column,and D consists of all the left-pointing arrows in each column. Furthermore, if Q does not contain a column, then D = D .Observe that each unit cycle in each column and pillar contains precisely onearrow in D , and one arrow in D . (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 asubpath of p or q . Therefore D and D are perfect matchings of Q .Recall that a simple A -module of dimension 1 Q is characterized by the propertythat there is a non-annihilating path between any two vertices of Q . Clearly Q \ D and Q \ D each support a path that passes through each vertex of Q . Therefore D and D are simple matchings. (cid:3) Lemma 2.16. If A is cancellative, then A has at least one simple matching.Proof. Follows from Proposition 2.10 and Lemma 2.15. (cid:3)
Lemma 2.16 will be superseded by Theorems 2.24 and 4.41, where we will showthat A is cancellative if and only if each arrow in Q is contained in a simple matching. Lemma 2.17.
Let u ∈ Z \ , and suppose ˆ C ui (cid:54) = ∅ for each i ∈ Q . Then A does notcontain 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 a path r such that r = pr (cid:54) = 0. Whencet( r + ) = t(( pr ) + ) and h( r + ) = h(( pr ) + ) , by Lemma 2.4.1. In particular, r = σ m r for some m ∈ Z , by Lemma 2.3.2. Thus σ = 1 since B is an integral domain.Therefore A has no simple matchings, S = ∅ . Consequently, for each u ∈ Z \ i ∈ Q such that ˆ C ui = ∅ , by Lemma 2.15. (cid:3) Lemma 2.18.
Let u ∈ Z \ . If p, q ∈ C u , then there is some n ∈ Z such that p = qσ n . In particular, if σ (cid:45) p and σ (cid:45) q , then p = q .Proof. Suppose the hypotheses hold. Since Q is a dimer quiver, there is a path r from t( p ) to t( q ). Thus there is some m, n ≥ rpσ m t( p ) = qrσ n t( p ) , ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 19 by Lemma 2.3.1. Furthermore, τ is an algebra homomorphism by Lemma 1.6. Thus rpσ m = ¯ τ (cid:0) rpσ n t( p ) (cid:1) = ¯ τ (cid:0) qrσ n t( p ) (cid:1) = qrσ n . Therefore p = qσ n − m since B is an integral domain. (cid:3) Lemma 2.19.
Suppose A is cancellative.(1) If a ∈ Q , p ∈ ˆ C u t( a ) , and q ∈ ˆ C u h( a ) , then ap = qa .(2) Each vertex corner ring e i Ae i is commutative.Proof. (1) Suppose a ∈ Q , p ∈ ˆ C u t( a ) , and q ∈ ˆ C u h( a ) . Thent (cid:0) ( ap ) + (cid:1) = t (cid:0) ( qa ) + (cid:1) and h (cid:0) ( ap ) + (cid:1) = h (cid:0) ( qa ) + (cid:1) . Let r + be a path in Q + from h (( ap ) + ) to t (( ap ) + ). Then by Lemma 2.8.2, there issome m, n ≥ rap = σ m t( a ) 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. Then qσ h( a ) = qab = apσ n − m t( a ) b = apbσ n − m h( a ) . Thus, since A is cancellative and n − m ≥ q = apbσ n − m − a ) . Furthermore, ( ba ) + is cycle in Q + since ba is a unit cycle. But this is a contradictionsince q is in ˆ C u . Whence m = n . Therefore qa = ap .(2) Since I is generated by binomials, it suffices to show that if p, q ∈ e i Ae i arecycles, then qp = pq . Let r + be a path in Q + from h (( qp ) + ) to t (( qp ) + ). Set r := π ( r + ). Then ( rqp ) + and ( rpq ) + are cycles. In particular, there is some m, n ≥ rqp = σ mi and rpq = σ ni , by Lemma 2.8.2. Thus, since τ is an algebra homomorphism, σ m = rqp = rqp = rpq = rpq = σ n . Furthermore, σ (cid:54) = 1 by Lemma 2.16. Whence m = n since B is an integral domain.Thus rqp = σ mi = rpq . Therefore qp = pq since A is cancellative. (cid:3) Proposition 2.20.
Let u ∈ Z \ . Suppose (i) A is cancellative, or (ii) ˆ C ui (cid:54) = ∅ foreach i ∈ Q .(1) If p ∈ ˆ C u , then σ (cid:45) p .(2) If p, q ∈ ˆ C u , then p = q .(3) If a cycle p is formed from subpaths of cycles in ˆ C u , then p ∈ ˆ C .(4) If p, q ∈ ˆ C ui , then p = q . Note that in Claim (2) the cycles p and q are based at the same vertex i , whereasin Claim (3) p and q may be based at different vertices. Proof. If A is cancellative, then ˆ C ui (cid:54) = ∅ for each i ∈ Q , by Proposition 2.10. Thereforeassumption (i) implies assumption (ii), and so it suffices to suppose (ii) holds.(1) By Lemma 2.14, it suffices to consider p ∈ ˆ C u \ (cid:98) C u . Then there are cycles s, t ∈ (cid:98) C u such that s = q p q , t = p q p , p = p p p , as in Figure 8. In particular, σ (cid:45) s and σ (cid:45) t by Lemma 2.15. Thus(21) s = t by Lemma 2.18. Set q := q q q .Assume to the contrary that σ | p . Then σ | pq = p p p q q q = st (i) = s , where ( i ) holds by (21). Therefore σ | s since σ = (cid:81) D ∈S x D . But this is a contradic-tion to Lemma 2.15 since s ∈ (cid:98) C u .(2) Follows from Claim (1) and Lemma 2.18.A direct proof assuming A is cancellative: Suppose p, q ∈ ˆ C u . Let r be a path fromt( p ) to t( q ). Then rp = qr by Lemma 2.19.1. Thus rp = rp = qr = qr = rq. Therefore p = q since B is an integral domain.(3) Let p = p (cid:96) · · · p ∈ C u be a cycle formed from subpaths p j of cycles q j in ˆ C u .Then g := q = · · · = q (cid:96) by Claim (2). Furthermore, σ (cid:45) g by Claim (1). In particular, there is a simplematching D ∈ S for which x D (cid:45) g . Whence x D (cid:45) p j for each 1 ≤ j ≤ (cid:96) . Thus x D (cid:45) p .Therefore σ (cid:45) p . Consequently, p ∈ ˆ C by the contrapositive of Lemma 2.8.3, withLemma 2.17.(4) Follows from Claim (3) and Figure 6.ii.A direct proof assuming A is cancellative: Suppose p, q ∈ ˆ C ui . Let r + be a path in Q + from h ( p + ) to t ( p + ). Then there is some m, n ≥ rp = σ mi and rq = σ ni , by Lemma 2.8.2. Suppose m ≤ n . Then rpσ n − mi = σ ni = rq . Thus pσ n − mi = q since A is cancellative. But then Claim (1) implies m = n since by assumption p and q arein ˆ C u . Therefore p = q . (cid:3) Lemma 2.21.
Suppose ˆ C ui (cid:54) = ∅ for each u ∈ Z \ and i ∈ Q . Let ε , ε ∈ Z . Thereis an n (cid:29) such that if p ∈ ˆ C ( ε , i and q ∈ ˆ C ( nε ,ε ) i , ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 21 then σ (cid:45) pq .Proof. Fix ε , ε ∈ Z . For each n ≥
0, consider the cycles p ∈ ˆ C ( ε , i and q n ∈ ˆ C ( nε ,ε ) i . (1) We claim that for each n ≥ q + n lies in the region R p q n − ,q n +1 (modulo I ).This setup is shown in Figure 9.i. Indeed, suppose q + n intersects q + n − as shown inFigure 9.ii. Then q n − and q n factor into paths q n − = s s s and q n = t t t , where t( s +2 ) = t( t +2 ) and h( s +2 ) = h( t +2 ). In particular, there is some m ∈ Z such that s = t σ m , by Lemma 2.3.2. Set r := t s t . Since q n − and q n are in ˆ C , we have(22) σ (cid:45) q n − and σ (cid:45) q n , by Proposition 2.20.1. Whence σ (cid:45) s and σ (cid:45) t . Thus m = 0. Therefore r = t s t = t t t = q n . In particular, σ (cid:45) r by (22). Thus the cycle r is in ˆ C by Lemmas 2.8.4 and 2.17.Furthermore, r is in C ( nε ,ε ) i by construction. Whence r is in ˆ C ( nε ,ε ) i . Therefore r = q n (modulo I ) by Proposition 2.20.4. This proves our claim.(2) By Claim (1), there is a cycle r ∈ ˆ C ( ε , such that the area of the region R s n q (cid:48) n ,q (cid:48) n +1 bounded by a rightmost subpath q (cid:48) + n of q + n , a rightmost subpath q (cid:48) + n +1 of q + n +1 , and asubpath s + n of a path in π − ( r ∞ ), tends to zero (modulo I ) as n → ∞ . See Figure9.iii. (The case r = p is shown in Figure 9.i.) Since Q is finite, there is some N (cid:29) n ≥ N , then(23) q (cid:48) n +1 = s n q (cid:48) n (modulo I ) . Now choose m ≥ r is a subpath of s := s N + m − · · · s N +1 s N . By iterating (23) m − q (cid:48) N + m = sq (cid:48) N . Furthermore, there is a simple matching D ∈ S such that x D (cid:45) q N + m , by Proposition2.20.1. Whence x D (cid:45) q (cid:48) N + m = sq (cid:48) N . ( i ) · · · · · · · ·· · · · · · · · q q q q · · · q n q n +1 (cid:79) (cid:79) p (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:71) (cid:71) (cid:63) (cid:63) (cid:58) (cid:58) (cid:55) (cid:55) (cid:53) (cid:53) (cid:52) (cid:52) (cid:52) (cid:52) p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) ( ii ) · · · ·· · ·· · q n − q n q n +1 p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) (cid:52) (cid:52) t (cid:79) (cid:79) t (cid:46) (cid:46) t (cid:55) (cid:55) s (cid:63) (cid:63) s (cid:63) (cid:63) s (cid:63) (cid:63) ( iii ) · · · · · · · ·· · · · · · · ·· · · · · · · q q q q · · · q n q n +1 · (cid:79) (cid:79) (cid:79) (cid:79) p (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:66) (cid:66) (cid:57) (cid:57) (cid:54) (cid:54) (cid:52) (cid:52) (cid:51) (cid:51) (cid:50) (cid:50) (cid:50) (cid:50) p (cid:47) (cid:47) p (cid:47) (cid:47) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) s (cid:47) (cid:47) s (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) s n (cid:47) (cid:47) p (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) s (cid:47) (cid:47) Figure 9.
Setup for Lemma 2.21. In (ii): q + n − and q + n intersect, andthus factor into paths q n − = s s s and q n = t t t , drawn in purpleand blue respectively. In (iii): the area of the region R s n q (cid:48) n ,q (cid:48) n +1 tends tozero as n → ∞ . Here, an infinite path in π − ( r ∞ ) is drawn in orange,each lift of the cycle p is drawn in red, and q + n and q + n +1 are drawn inblue.In particular,(24) x D (cid:45) s. Moreover, r | s since r is a subpath of s , by our choice of m . Whence x D (cid:45) r by(24). But p = r since p and r are both in ˆ C ( ε , , by Proposition 2.20.2. Thus x D (cid:45) p .Whence x D (cid:45) pq N + m . Therefore σ (cid:45) pq N + m . (cid:3) Consider the subset of arrows(25) Q S := { a ∈ Q | a (cid:54)∈ D for each D ∈ S} , ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 23 where S is the set of simple matchings of A .We will show in Theorem 2.24 below that the two assumptions considered in thefollowing lemma, namely that Q S (cid:54) = ∅ and ˆ C ui (cid:54) = ∅ for each u ∈ Z \ i ∈ Q ,can never both hold. Lemma 2.22.
Suppose ˆ C ui (cid:54) = ∅ for each u ∈ Z \ and i ∈ Q . Let δ ∈ Q S . Thenthere is a cycle p ∈ ˆ C t( δ ) such that δ is not a rightmost arrow subpath of p (modulo I ).Proof. Suppose ˆ C ui (cid:54) = ∅ for each u ∈ Z \ i ∈ Q , and let δ ∈ Q S . Assume to thecontrary that there is an arrow δ which is a rightmost arrow subpath of each cycle inˆ C t( δ ) .(i) We first claim that there is some u ∈ Z \ δ + ) lies in the interior R ◦ s,t of the region bounded by the lifts of representatives s, t ∈ kQ of the cycle inˆ C u h( δ ) . (There is precisely one cycle in ˆ C u h( δ ) by Proposition 2.20.4.)Indeed, by Lemma 2.21, there are vectors u = u n +1 , u , u , . . . , u n ∈ Z \ ≤ m ≤ n , the cycles p m = p (cid:48) m δ ∈ ˆ C u m h( δ ) satisfy(26) σ (cid:45) p m +1 p m . Assume to the contrary that δ + is not contained in each region R p m +1 p m ,p m p m +1 . Since p m is in ˆ C , its lift p + m does not have a cyclic subpath. In particular, the tail of δ + is only a vertex subpath of p + m at its tail. We therefore have the setup given inFigure 10.Set p := p and q := p n . Then there is a leftmost subpath δp (cid:48) of p and a rightmostsubpath q (cid:48) of q such that ( δp (cid:48) q (cid:48) ) + is a cycle in Q + . In particular, δp (cid:48) q (cid:48) = σ (cid:96) for some (cid:96) ≥
1, by Lemmas 2.8.1 and 2.17. Whence σ | pq . But this is a contradictionto (26).Thus there is some 0 ≤ m ≤ n such that δ + lies in R p m +1 p m ,p m p m +1 . Further-more, since t( δ + ) is not a vertex subpath of p + m or p + m +1 , t( δ + ) lies in the interior of R p m +1 p m ,p m p m +1 . The claim then follows by setting s + I = p m +1 p m and t + I = p m p m +1 . (ii) Let s and t be as in Claim (i). In particular, s = t . Assume to the contrarythat there is a simple matching D ∈ S for which x D (cid:45) s = t. ··· · δ (cid:79) (cid:79) (cid:25) (cid:25) (cid:21) (cid:29) q (cid:48) (cid:117) (cid:117) (cid:83) (cid:83) (cid:51) (cid:51) p (cid:48) (cid:43) (cid:43) p = p (cid:27) (cid:27) p (cid:8) (cid:8) p (cid:123) (cid:123) p (cid:104) (cid:104) (cid:91) (cid:91) (cid:72) (cid:72) p n − (cid:59) (cid:59) q = p n (cid:40) (cid:40) Figure 10.
Setup for Claim (i) in the proof of Lemma 2.22. Here, p, q ∈ ˆ C h( δ ) are drawn in red and blue respectively. Furthermore, δp (cid:48) isa leftmost subpath of p ; q (cid:48) is a rightmost subpath of q ; and ( δp (cid:48) q (cid:48) ) + isa cycle in Q + .Let r + be a path in R s,t from a vertex on the boundary of R s,t to t( δ + ). It sufficesto suppose the tail of r is a vertex subpath of s , in which case s factors into paths s = s e t( r ) s . See Figure 11. Then ( rs δ ) + is a cycle in Q + . Whence rs δ = σ (cid:96) for some (cid:96) ≥
1, by Lemmas 2.8.1 and 2.17. In particular, x D | rs δ = rs δ. Furthermore, by assumption, x D (cid:45) s and x D (cid:45) δ. Whence x D | r . Thus, since D ∈ S and r are arbitrary, σ divides the ¯ τ -image of eachpath in R s,t from the boundary of R s,t to t( δ + ). But t( δ + ) lies in the interior R ◦ s,t .Thus the vertex t( δ ) is a source in Q \ D . Therefore D is not simple, contrary toassumption. It follows that x D | s for each D ∈ S .(iii) By Claim (ii), σ | s . But this is a contradiction since s is in ˆ C , by Lemmas2.8.3 and 2.17. Therefore there is a cycle in ˆ C t( δ ) whose rightmost arrow subpath isnot δ (modulo I ). (cid:3) Remark 2.23.
There are dimer algebras with an arrow δ ∈ Q that is a rightmostarrow subpath of each cycle in ˆ C t( δ ) ; see Figure 12. Furthermore, if A has center Z ,admits a cyclic contraction (Definition 4.3), and δ ∈ Q S , then δ is a rightmost arrowsubpath of each cycle p ∈ Ze t( δ ) for which σ (cid:45) p , by [B3, Lemma 2.4]. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 25 ··· · t (cid:105) (cid:105) s (cid:83) (cid:83) s (cid:53) (cid:53) r (cid:47) (cid:47) δ (cid:15) (cid:15) Figure 11.
Setup for Claim (ii) in the proof of Lemma 2.22. · · · ··· δ (cid:111) (cid:111) (cid:47) (cid:47) (cid:111) (cid:111) (cid:7) (cid:7) (cid:87) (cid:87) (cid:7) (cid:7) (cid:87) (cid:87) (cid:47) (cid:47) (cid:47) (cid:47) (cid:72) (cid:72) (cid:22) (cid:22) · · · ·· ···· ··· (cid:43) (cid:43) (cid:51) (cid:51) δ (cid:111) (cid:111) (cid:47) (cid:47) (cid:47) (cid:47) (cid:75) (cid:75) (cid:19) (cid:19) (cid:47) (cid:47) (cid:111) (cid:111) (cid:15) (cid:15) (cid:79) (cid:79) (cid:95) (cid:95) (cid:127) (cid:127) (cid:63) (cid:63) (cid:31) (cid:31) (cid:127) (cid:127) (cid:47) (cid:47) (cid:119) (cid:119) (cid:95) (cid:95) (cid:47) (cid:47) (cid:103) (cid:103) (cid:79) (cid:79) (cid:79) (cid:79) (cid:15) (cid:15) (cid:15) (cid:15) · · · ···· · ····· δ (cid:111) (cid:111) (cid:38) (cid:38) (cid:47) (cid:47) (cid:77) (cid:77) (cid:56) (cid:56) (cid:47) (cid:47) (cid:17) (cid:17) (cid:15) (cid:15) (cid:79) (cid:79) (cid:127) (cid:127) (cid:95) (cid:95) (cid:116) (cid:116) (cid:106) (cid:106) (cid:79) (cid:79) (cid:15) (cid:15) (cid:127) (cid:127) (cid:95) (cid:95) (cid:95) (cid:95) (cid:127) (cid:127) (cid:4) (cid:4) (cid:90) (cid:90) (cid:79) (cid:79) (cid:15) (cid:15) (cid:111) (cid:111) (cid:47) (cid:47) (cid:47) (cid:47) (cid:23) (cid:23) (cid:71) (cid:71) (cid:36) (cid:36) (cid:58) (cid:58) Figure 12.
In each example, the arrow δ is a rightmost arrow subpathof 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 non-vertex paths in Q + , and the blackarrows are arrows in Q + .The following will be generalized in Theorem 4.41 below. Theorem 2.24.
Suppose (i) A is cancellative, or (ii) ˆ C ui (cid:54) = ∅ for each u ∈ Z \ and i ∈ Q . Then Q S = ∅ , that is, each arrow annihilates a simple A -module ofdimension Q .Proof. Recall that assumption (i) implies assumption (ii), by Proposition 2.10. Sosuppose (ii) holds, and assume to the contrary that there is an arrow δ in Q S .By Lemma 2.22, there is a cycle p ∈ ˆ C t( δ ) whose rightmost arrow subpath is not δ (modulo I ). Let u ∈ Z be such that p ∈ C u . By assumption, there is a cycle q inˆ C u h( δ ) . We thus have one of the three cases given in Figure 13.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 2.15, the brownarrows belong to a simple matching D ∈ S . In particular, δ is in D , contrary toassumption. ( i ) · · · ·· · · · ··· · · (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) δ (cid:79) (cid:79) (cid:127) (cid:127) (cid:95) (cid:95) (cid:127) (cid:127) (cid:95) (cid:95) (cid:127) (cid:127) δ (cid:79) (cid:79) (cid:47) (cid:47) (cid:127) (cid:127) (cid:47) (cid:47) ( ii ) · · · ·· · · ·· · · (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:28) (cid:28) (cid:47) (cid:47) (cid:66) (cid:66) (cid:46) (cid:46) (cid:47) (cid:47) δ (cid:79) (cid:79) (cid:127) (cid:127) (cid:110) (cid:110) (cid:92) (cid:92) (cid:127) (cid:127) δ (cid:79) (cid:79) ( iii ) · · ·· · ·· ·· ·· · · (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:28) (cid:28) (cid:48) (cid:48) (cid:47) (cid:47) (cid:47) (cid:47) (cid:66) (cid:66) (cid:47) (cid:47) (cid:47) (cid:47) δ (cid:79) (cid:79) (cid:7) (cid:7) (cid:87) (cid:87) (cid:112) (cid:112) (cid:2) (cid:2) (cid:87) (cid:87) (cid:7) (cid:7) δ (cid:79) (cid:79) Figure 13.
Setup for Theorem 2.24. The red and blue paths are the(lifts of the) cycles p and q , respectively. The red and blue arrows arepaths 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 ), againcontrary 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 2.15, the brown arrows belong to asimple matching D ∈ S . In particular, in case (ii) δ is in D , 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 . (cid:3) Cancellative dimer algebras
In this section, we give an explicit impression of all cancellative dimer algebras.Throughout, A = kQ/I is a cancellative dimer algebra, and we use the notation (6). Lemma 3.1.
Let p, q ∈ e j Ae i be paths such that t( p + ) = t( q + ) and h( p + ) = h( q + ) . If p = q , then p = q .Proof. Since A is cancellative, A has at least one simple matching by Lemma 2.16.In particular, σ (cid:54) = 1. Thus we may apply the proof of Lemma 2.3.3, with ¯ τ in placeof ¯ η . (cid:3) Lemma 3.2.
For each i ∈ Q , the corner ring e i Ae i is generated by σ i and ˆ C i .Proof. Since I is generated by binomials, e i Ae i is generated by C i . It thus suffices toshow that C i is generated by σ i and ˆ C i .Let u ∈ Z and p ∈ C ui . If u = 0, then p = σ mi for some m ≥ u (cid:54) = 0. Then there is a cycle q in ˆ C ui by Proposition 2.10. In particular, ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 27 p = qσ m for some m ∈ Z by Lemma 2.18. Furthermore, m ≥ p = qσ mi by Lemma 3.1. (cid:3) It is well known that if A is cancellative, then A is a 3-Calabi-Yau algebra [D, MR].In particular, the center Z of A is noetherian and reduced, and A is a finitely generated Z -module. In the following, we give independent proofs of these facts so that Theorem3.5 may be self-contained. Furthermore, we prove the converse for dimer algebraswhich admit cyclic contractions in Theorem 4.50 below. Theorem 3.3.
Suppose A is a cancellative dimer algebra, and let i, j ∈ Q . Then(1) e i Ae i = Ze i ∼ = Z .(2) ¯ τ ( e i Ae i ) = ¯ τ ( e j Ae j ) .(3) A is a finitely generated Z -module, and Z is a finitely generated k -algebra.(4) Z is reduced.Proof. (1) For each i ∈ Q and u ∈ Z \
0, there exists a unique cycle c ui ∈ ˆ C ui (modulo I ) by Propositions 2.20.4 and 2.10. Thus the sum (cid:88) i ∈ Q c ui ∈ (cid:77) i ∈ Q e i Ae i is in Z , by Lemma 2.19.1. Whence e i Ae i ⊆ Ze i by Lemma 3.2. Furthermore, Ze i = Ze i = e i Ze i ⊆ e i Ae i . Therefore Ze i = e i Ae i .We now claim that there is an algebra isomorphism Z ∼ = Ze i for each i ∈ Q .Indeed, fix i ∈ Q and suppose z ∈ Z is nonzero. Then there is some j ∈ Q suchthat ze j (cid:54) = 0. Furthermore, since Q is a dimer quiver, there is a path p from i to j .Assume to the contrary that ze j p = 0. Thus, since I is generated by binomials,it suffices to suppose ze j = c − c where c and c are paths. But since A iscancellative, ze j p = 0 implies c = c . Whence ze j = 0, a contradiction. Therefore ze j p (cid:54) = 0. Consequently, pe i z = pz = zp = ze j p (cid:54) = 0 . Whence ze i (cid:54) = 0. Thus the algebra homomorphism Z → Ze i , z (cid:55)→ ze i , is injective,hence an isomorphism. This proves our claim.(2) Follows from Proposition 2.20.2 and and Lemma 3.2.(3) A is generated as a Z -module by all paths of length at most | Q | by Claim (1)and [B, second paragraph of proof of Theorem 2.11]. Thus A is a finitely generated Z -module. Furthermore, A is a finitely generated k -algebra since Q is finite. Therefore Z is also a finitely generated k -algebra [McR, 1.1.3].(4) Suppose z ∈ Z and z n = 0. Fix i ∈ Q . Then( ze i ) n = z n e i = 0 . Furthermore, we may write ze i = (cid:96) (cid:88) j =1 p j ∈ e i Ae i , where each p j is a cycle with a scalar coefficient. For 1 ≤ j ≤ (cid:96) , let u j ∈ Z besuch that p j ∈ C u j . Then the path summand p j n · · · p j of ( ze i ) n is in C (cid:80) nk =1 u jk . Inparticular, if u k , 1 ≤ k ≤ n , has maximal length, then p nk is the only path summandof ( ze i ) n which is in C nu k . But no path is zero modulo I , and so z = 0. Therefore Z is reduced. (cid:3) Lemma 3.4.
Let p ∈ C u . Then u = 0 if and only if p = σ m for some m ≥ .Proof. ( ⇒ ) Lemma 2.8.1.( ⇐ ) Let u ∈ Z \
0, and assume to the contrary that p ∈ C u satisfies p = σ m forsome m ≥
0. Since A is cancellative, there is a cycle q ∈ ˆ C u t( p ) by Proposition 2.10.Furthermore, σ (cid:45) q by Proposition 2.20.1. Thus there is some n ≥ p = qσ n , by Lemma 2.3.2. Whence σ m = qσ n . Furthermore, σ (cid:54) = 1 by Lemma 2.16. Therefore m = n and(27) q = 1 , since B is a polynomial ring. But each arrow in Q is contained in a simple matchingby Theorem 2.24. Therefore q (cid:54) = 1, contrary to (27). (cid:3) We now prove our main theorem for this section.
Theorem 3.5.
Suppose A = kQ/I is a cancellative dimer algebra. Then the algebrahomomorphism τ : A → M | Q | ( B ) , defined in (4), is an impression of A . Therefore τ classifies all simple A -moduleisoclasses of maximal k -dimension. Furthermore, the following holds: (28) Z ∼ = k [ ∩ i ∈ Q ¯ τ ( e i Ae i )] = k [ ∪ i ∈ Q ¯ τ ( e i Ae i )] . Proof. (i) We first show that τ is injective.(i.a) We claim that τ is injective on the vertex corner rings e i Ae i , i ∈ Q . Fix avertex i ∈ Q and let p, q ∈ e i Ae i be cycles satisfying p = q . Let r be a path suchthat r + is path from h( p + ) to t( p + ). Then rp ∈ C . Thus there is some m ≥ rp = σ mi , by Lemma 2.8.2. Whence rq = rq = rp = rp = σ mi = σ mi = σ m . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 29
Thus rq ∈ C by Lemma 3.4. Hence rq = σ mi = rp by Lemma 3.1. Therefore p = q since A is cancellative.(i.b) 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.a). Therefore p = q since A is cancellative.Since A is generated by paths and τ is injective on paths, it follows that τ isinjective.(ii) For each b ∈ Z ( σ ) c ⊂ Max B = A |S| k , the composition (cid:15) b τ defined in (8) is a representation of A where, when viewed as avector space diagram on Q , each arrow is represented by a non-zero scalar. Thus (cid:15) b τ is simple since there is an oriented path between any two vertices in Q . Therefore (cid:15) b τ is surjective.(iii) We claim that the morphismMax B → Max τ ( Z ) , b (cid:55)→ b | Q | ∩ τ ( Z ) , is surjective. Indeed, for any m ∈ Max R , B m is a (nonzero) proper ideal of B .Thus there is a maximal ideal b ∈ Max B containing B m since B is noetherian.Furthermore, since B is a finitely generated k -algebra and k is algebraically closed,the intersection b ∩ R =: m (cid:48) is a maximal ideal of R . Whence m ⊆ B m ∩ R ⊆ b ∩ R = m (cid:48) . But m and m (cid:48) are both maximal ideals of R . Thus m = m (cid:48) . Therefore b ∩ R = m ,proving our claim.Claims (i), (ii), (iii) imply that ( τ, B ) is an impression of A .(iv) Since ( τ, B ) is an impression of A , Z is isomorphic to R by [B, Lemma 2.1(2)]. Furthermore, R is equal to S by Theorem 3.3.2. Therefore (28) holds. (cid:3) Remark 3.6.
The labeling of arrows we obtain, namely (4), agrees with the labelingof 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 differentimplications than the toric construction of [CQ].
Corollary 3.7.
A dimer algebra A is cancellative if and only if it admits an impres-sion ( τ, B ) where B is an integral domain and τ ( e i ) = E ii for each i ∈ Q .Proof. Suppose A admits an impression ( τ, B ) such that B is an integral domain and τ ( e i ) = E ii for each i ∈ Q . Consider paths p, q, r satisfying pr = qr (cid:54) = 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 ) . · · (cid:47) (cid:47) · · (cid:111) (cid:111) ·· (cid:79) (cid:79) ·· (cid:15) (cid:15) · · (cid:63) (cid:63) · · (cid:127) (cid:127) · · (cid:31) (cid:31) · · (cid:95) (cid:95) x y z w xz yw xw yz Figure 14.
A labeling of arrows in the quiver of a square dimer alge-bra that specifies an impression.Therefore p = q by the injectivity of τ .The converse follows from Theorem 3.5. (cid:3) Remark 3.8.
Although non-cancellative dimer algebras do not admit impressions( τ, B ) with B an integral domain by Corollary 3.7, we will find that their homotopyalgebras do; see Definition 4.33 and Theorem 4.35. Example 3.9.
A dimer algebra A = kQ/I is square if the underlying graph of itscover Q + is a square grid graph with vertex set Z × Z , and with at most one diagonaledge in each unit square. Examples of square dimer algebras are given in Figure 15(i.b, ii.b, iii.c, iv.c).By [B, Theorem 3.7], any square dimer algebra A admits an impression ( τ, B = k [ x, y, z, w ]), where for each arrow a ∈ Q +1 , ¯ τ ( a ) is the monomial corresponding tothe orientation of a given in Figure 14. Specifically, τ : A → M | Q | ( B ) is the algebrahomomorphism defined by τ ( e i ) = E ii and τ ( a ) = ¯ τ ( a ) E h( a ) , t( a ) for each i ∈ Q and a ∈ Q . If Q only possesses three arrow orientations, say up, left,and right-down, then we may label the respective arrows by x , y , and z , and obtainan impression ( τ, k [ x, y, z ]). In either case, A is cancellative by Corollary 3.7. Proposition 3.10.
Suppose A is cancellative, and let p ∈ C be a non-vertex cycle.Then p ∈ ˆ C if and only if σ (cid:45) p . In particular, τ ( Z ) ⊆ B is generated over k by σ and a set of monomials in B not divisible by σ .Proof. ( ⇒ ) Proposition 2.20.1.( ⇐ ) Lemma 2.8.3. (cid:3) Recall that an ideal p in a (possibly noncommutative) ring R is prime if for allideals I, J (cid:47) R , IJ ⊆ p implies I ⊆ p or J ⊆ p . Moreover, R is prime if for all a, b ∈ R , aRb = (0) implies a = 0 or b = 0, that is, the zero ideal is a prime ideal. Proposition 3.11.
Cancellative dimer algebras are prime.Proof.
We claim that for nonzero elements p, q ∈ A , we have qAp (cid:54) = 0 . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 31
It suffices to suppose that p ∈ e j Ae i and q ∈ e (cid:96) Ae k . Since Q is strongly connected, there is a path r from j to k . Furthermore, thepolynomials p , q , and r are nonzero since τ is injective, by Theorem 3.5. Thus theproduct qrp = qrp ∈ B is nonzero since B is an integral domain. Therefore qrp isnonzero. (cid:3) Non-cancellative dimer algebras and their homotopy algebras
Throughout, we use the notation (7).4.1.
Cyclic contractions.
In this section, we introduce a new method for studyingnon-cancellative dimer algebras that is based on the notion of Higgsing, or moregenerally symmetry breaking, in physics. Using this strategy, we gain informationabout non-cancellative dimer algebras by relating them to cancellative dimer algebraswith certain similar structure. Throughout, A = kQ/I is a dimer algebra, usuallynon-cancellative.In the following, we introduce a k -linear map induced from the operation of edgecontraction in graph theory. Definition 4.1.
Let Q = ( Q , Q , t , h) be a dimer quiver, let Q ∗ ⊂ Q be a subset ofarrows, and let Q (cid:48) = ( Q (cid:48) , Q (cid:48) , t (cid:48) , h (cid:48) ) be the quiver obtained by contracting each arrowin Q ∗ to a vertex. Specifically, Q (cid:48) := Q / { h( a ) ∼ t( a ) | a ∈ Q ∗ } , Q (cid:48) = Q \ Q ∗ , and h (cid:48) ( a ) = h( a ), t (cid:48) ( a ) = t( a ) for each a ∈ Q (cid:48) . Then there is a k -linear map of pathalgebras ψ : kQ → kQ (cid:48) defined by ψ ( a ) = (cid:40) a if a ∈ Q ∪ Q \ Q ∗ e t( a ) if a ∈ Q ∗ and extended multiplicatively to nonzero paths and k -linearly to kQ . If ψ induces a k -linear map of dimer algebras ψ : A = kQ/I → A (cid:48) = kQ (cid:48) /I (cid:48) , that is, ψ ( I ) ⊆ I (cid:48) , then we call ψ a contraction of dimer algebras . Remark 4.2.
The containment ψ ( I ) ⊆ I (cid:48) may be proper. Indeed, ψ ( I ) (cid:54) = I (cid:48) when-ever ψ contracts a unit cycle to a removable 2-cycle.We now describe the structure we wish to preserve under a contraction. To specifythis structure, we introduce the following commutative algebras. Definition 4.3.
Let ψ : A → A (cid:48) be a contraction to a cancellative dimer algebra,and let ( τ, B ) be an impression of A (cid:48) . If S := k [ ∪ i ∈ Q ¯ τ ψ ( e i Ae i )] = k (cid:2) ∪ i ∈ Q (cid:48) ¯ τ ( e i A (cid:48) e i ) (cid:3) =: S (cid:48) , then we say ψ is cyclic , and call S the cycle algebra of A .The cycle algebra is independent of the choice of ψ and τ by [B3, Theorem 3.13].Henceforth we will consider cyclic contractions ψ : A → A (cid:48) . Notation 4.4.
For p ∈ e j Ae i and q ∈ e (cid:96) A (cid:48) e k , set p := ¯ τ ψ ( p ) ∈ B and q := ¯ τ ( q ) ∈ B. An immediate question is whether all non-cancellative dimer algebras admit cycliccontractions. We will show that dimer algebras exist which do not admit contrac-tions (cyclic or not) to cancellative dimer algebras. For example, dimer algebrasthat contain permanent 2-cycles cannot be contracted to cancellative dimer alge-bras (Proposition 4.43). Furthermore, we will show that if ψ : A → A (cid:48) is a cycliccontraction and A is cancellative, then ψ is necessarily trivial (Theorem 4.41). Example 4.5.
Consider the three examples of contractions ψ : A → A (cid:48) given inFigure 15. In each example, ψ is cyclic; Q (cid:48) is a square dimer (Example 3.9) withan impression given by the arrow orientations in Figure 14; and B = k [ x, y, z, w ].In particular, quiver (a) is non-cancellative, quivers (b) and (c) are cancellative, andquiver (c) is obtained by deleting the removable 2-cycles in quiver (b). The non-cancellative quivers (a) first appeared respectively in [FHPR, Section 4], [FKR]; [Bo,Example 3.2]; and [DHP, Table 6, 2.6]. Their respective cycle algebras S are: (i): S = k [ xz, xw, yz, yw ] R = k + ( x zw, xyzw, y zw ) S (ii): S = k [ xz, yz, xw, yw ] R = k + ( xz, yz ) S (iii): S = k [ xz, yw, x w , y z ] R = k + ( yw, x w , y z ) S Notation 4.6.
For g, h ∈ B , by g | h we mean g divides h in B , even if g or h isassumed to be in S . Lemma 4.7.
Suppose ψ : A → A (cid:48) is a contraction of dimer algebras, and A (cid:48) has aperfect 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 (cid:48) . Fix a unit cycle σ i (cid:48) ∈ A (cid:48) . Since ψ is surjective on Q (cid:48) , there is a vertex i ∈ Q such that ψ ( i ) = i (cid:48) . Let p ∈ A be a path from i to j , and set p (cid:48) := ψ ( p ). Then(29) p (cid:48) σ i (cid:48) (i) = ψ ( pσ i ) (ii) = ψ ( σ j p ) (iii) = ψ ( σ j ) p (cid:48) = e ψ ( j ) p (cid:48) = p (cid:48) . Indeed, ( i ) and ( iii ) hold by Definition 4.1, and ( ii ) holds by Lemma 1.5. The algebra R will be introduced in Definition 4.26 below. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 332 1 21 2 1 · · · (cid:111) (cid:111) (cid:15) (cid:15) (cid:47) (cid:47) (cid:79) (cid:79) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:15) (cid:15) (cid:15) (cid:15) (cid:79) (cid:79) ψ −→ (cid:111) (cid:111) (cid:15) (cid:15) (cid:47) (cid:47) (cid:79) (cid:79) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) ( i.a ) ( i.b ) · (cid:47) (cid:47) (cid:15) (cid:15) (cid:103) (cid:103) (cid:31) (cid:31) (cid:111) (cid:111) (cid:79) (cid:79) (cid:47) (cid:47) (cid:79) (cid:79) (cid:127) (cid:127) (cid:55) (cid:55) (cid:111) (cid:111) (cid:79) (cid:79) ψ −→ (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:47) (cid:47) (cid:79) (cid:79) (cid:111) (cid:111) (cid:37) (cid:37) (cid:101) (cid:101) (cid:57) (cid:57) (cid:121) (cid:121) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:47) (cid:47) (cid:79) (cid:79) (cid:111) (cid:111) ( ii.a ) ( ii.b ) ( ii.c ) ·
11 3 · (cid:47) (cid:47) (cid:23) (cid:23) (cid:111) (cid:111) (cid:79) (cid:79) (cid:79) (cid:79) (cid:31) (cid:31) (cid:87) (cid:87) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:111) (cid:111) ψ −→ ·
11 3 (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:102) (cid:102) (cid:37) (cid:37) ·
11 3 (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) ( iii.a ) ( iii.b ) ( iii.c ) Figure 15.
Some non-cancellative dimer algebras that cyclically con-tract to square dimer algebras. Each quiver is drawn on a torus, andthe contracted arrows are drawn in green. In (ii.a), the red and bluearrows generate a free subalgebra of A ; see Example 4.48 below. · ·· (cid:101) (cid:101) (cid:121) (cid:121) (cid:95) (cid:95) (cid:41) (cid:41) (cid:127) (cid:127) (cid:53) (cid:53) (cid:23) (cid:23) (cid:111) (cid:111) (cid:71) (cid:71) ψ −→ · (cid:95) (cid:95) (cid:127) (cid:127) (cid:95) (cid:95) (cid:127) (cid:127) (cid:47) (cid:47) (cid:47) (cid:47) Figure 16.
The contraction of a unit cycle (drawn in green) to a ver-tex. Such a contraction cannot induce a contraction of dimer algebras ψ : A → A (cid:48) if A (cid:48) has a perfect matching, by Lemma 4.7.Denote by P (cid:48) is the set of perfect matchings of A (cid:48) . Set σ := (cid:81) D ∈P (cid:48) x D . Then (29)implies ¯ η ( p (cid:48) ) σ = ¯ η ( p (cid:48) σ i (cid:48) ) = ¯ η ( p (cid:48) ) ∈ k [ x D | D ∈ P (cid:48) ] , by Lemma 1.6. Whence σ = 1. But this contradicts our assumption that P (cid:48) (cid:54) = ∅ . (cid:3) An example where a unit cycle is contracted to a vertex is given in Figure 16.
Lemma 4.8.
Suppose ψ : A → A (cid:48) is a contraction of dimer algebras, and A (cid:48) has aperfect matching. Then ψ cannot contract a non-vertex cycle in the underlying graph Q of Q to a vertex. In particular,(1) ψ cannot contract a non-vertex cycle in Q to a vertex;(2) if p ∈ C \ C , then ψ ( p ) ∈ C (cid:48) \ C (cid:48) ; 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 (cid:48) of Q and Q (cid:48) are given by V = | Q | , E = | Q | , F = T \ Q,V (cid:48) = | Q (cid:48) | , E (cid:48) = | Q (cid:48) | , F (cid:48) = T \ Q (cid:48) . Since Q and Q (cid:48) each embed into a two-torus, their respective Euler characteristicsvanish:(30) V − E + F = 0 , V (cid:48) − E (cid:48) + F (cid:48) = 0 . Assume to the contrary that ψ contracts the cycles p , . . . , p (cid:96) in Q to vertices in Q (cid:48) . Denote by n and n the respective number of vertices and arrows in Q whichare subpaths of some p i , 1 ≤ i ≤ (cid:96) . Denote by m the number of vertices in Q (cid:48) of theform ψ ( p i ) for some 1 ≤ i ≤ (cid:96) . By assumption, m ≥ n ≥ n . Whence 0 = F (cid:48) − E (cid:48) + V (cid:48) = F (cid:48) − ( E − n ) + ( V − n + m )= F (cid:48) + ( − E + V ) + ( n − n ) + m ≥ F (cid:48) − F + m. Thus, since m ≥ F (cid:48) < F. Therefore ψ contracts a face of Q to a vertex. In particular, some unit cycle in Q iscontracted to a vertex. But this is a contradiction by Lemma 4.7. (cid:3) Remark 4.9.
Suppose ψ : A → A (cid:48) is a cyclic contraction. Then A (cid:48) is cancellativeby definition, and thus has a perfect matching by Lemma 2.16. Therefore Lemmas4.7 and 4.8 hold in the case ψ is cyclic. Remark 4.10.
Let ψ : A → A (cid:48) be a contraction. Consider a path p = a n · · · a a ∈ kQ with a , . . . , a n ∈ Q ∪ Q . If p (cid:54) = 0, then by definition ψ ( p ) = ψ ( a n ) · · · ψ ( a ) ∈ kQ (cid:48) . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 35
However, we claim that if ψ is non-trivial, then it is not an algebra homomorphism.Indeed, let δ ∈ Q ∗ . Consider a path a δa (cid:54) = 0 in A . By Lemma 4.8.1, δ is not acycle. In particular, h( a ) (cid:54) = t( a ). Whence ψ ( a a ) = ψ (0) = 0 (cid:54) = ψ ( a ) ψ ( a ) , proving our claim. We note, however, that the restriction ψ : (cid:15) A(cid:15) → A (cid:48) where (cid:15) := 1 A − (cid:88) δ ∈ Q ∗ e h( δ ) , is an algebra homomorphism [B3, Proposition 2.13.1]. Lemma 4.11.
Let p be a non-vertex cycle.(1) If p ∈ C , then p = σ m for some m ≥ .(2) If p ∈ C \ ˆ C , then σ | p .Proof. If p + is a cycle (resp. has a cyclic subpath) in Q + , then ψ ( p ) + is a cycle (resp.has a cyclic subpath) in Q (cid:48) + . Furthermore, A (cid:48) is cancellative. Claims (1) and (2)therefore hold by Lemmas 2.8.1 and 2.8.3 respectively. (cid:3) The following strengthens Lemma 2.3.3 for dimer algebras that admit cyclic con-tractions (specifically, the head and tail of the lifts p + and q + are not required tocoincide). Lemma 4.12.
Let p, q ∈ e j Ae i be distinct paths. Then the following are equivalent:(1) ψ ( p ) = ψ ( q ) .(2) p = q .(3) p, q is a non-cancellative pair.Proof. (1) ⇒ (3): Suppose ψ ( p ) = ψ ( q ). Consider lifts p + and q + for whicht( p + ) = t( q + ) . Let r + be a path from h( p + ) to h( q + ). Then ψ ( r + ) is a cycle in Q (cid:48) + since ψ ( p ) = ψ ( q ),by Lemma 2.4.1. Thus r + is also a cycle by the contrapositive of Lemma 4.8.2.Whence h( p + ) = t( q + ). Furthermore, σ (cid:54) = 1 since A is cancellative, by Lemma 2.16.Therefore p, q is a non-cancellative pair, by Claim (3.i) in the proof of Lemma 2.3(with ¯ τ ψ in place of ¯ η ).(3) ⇒ (2): Holds similar to Claim (3.ii) in the proof of Lemma 2.3.(2) ⇒ (1): Holds since ¯ τ : e ψ ( j ) A (cid:48) e ψ ( i ) → B is injective, by Theorem 3.5. (cid:3) The following is a converse to Lemma 2.18.
Lemma 4.13.
Let u, v ∈ Z \ . If p ∈ C u and q ∈ C v are cycles for which p = q , then u = v . Proof.
Suppose to the contrary that u (cid:54) = v . Then p and q intersect at some vertex i since u and v are both nonzero. Let p i and q i be the respective cyclic permutationsof p and q with tails at i . By assumption, p i = p = q = q i . Thus p i , q i is a non-cancellative pair Lemma 4.12. Therefore u = v by Lemma2.4.1. (cid:3) Recall that if p and q are paths satisfyingt( p + ) = t( q + ) and h( p + ) = h( q + ) , then representatives ˜ p + and ˜ q + of their lifts bound a compact region R ˜ p, ˜ q in R . Ifonly one pair of representatives is considered, then by abuse of notation we denote R ˜ p, ˜ q by R p,q . Furthermore, we denote the interior of R p,q by R ◦ p,q . Lemma 4.14.
Let p, q be a non-cancellative pair.(1) Suppose p and q do not have proper subpaths (modulo I ) which form a non-cancellative pair. If r is a path of minimal length such that rp = rq (cid:54) = 0 (resp. pr = qr (cid:54) = 0 ), then each rightmost (resp. leftmost) arrow subpath of r + lies inthe region R p,q (modulo I ).(2) If the non-cancellative pair p, q in Claim (2) is minimal, then r + lies in theregion R p,q , and its head h( r + ) (resp. tail t( r + ) ) lies in the interior R ◦ p,q .Proof. (1) Suppose the hypotheses hold, with r a path of minimal length such that rp = rq (cid:54) = 0. Assume to the contrary that a rightmost arrow subpath of r + does notlie in the interior of R p,q ; see Figure 17.By assumption, p and q do not have proper subpaths which form a non-cancellativepair. Thus there is a rightmost non-vertex subpath p of p (or q ) and a leftmost non-vertex subpath r of r (modulo I ) such that for some arrow s , r p s is a unit cycle.Let ts be the complementary unit cycle containing s .Since rp homotopes to rq , the path t + intersects the interior of R p,q . We thus havethe setup given in Figure 17.ii. But then q + intersects the interiors of the unit cycles( r p s ) + and ( ts ) + since h( q + ) = h( p +2 ), a contradiction. Therefore r +1 lies in theinterior of R p,q .(2) Suppose r + intersects p + ; see Figure 18. Here p and r factor into (possiblyvertex) paths p = p p and r = r r , with h( p ) = h( r ). We claim that the non-cancellative pair p, q is not minimal.Indeed, each rightmost arrow subpath of r +1 lies in R p,q by Claim (1). Thus thepath r is not a vertex. Whence r p ∈ C is not a vertex. Therefore there is some m ≥ r p = σ m by Lemma 4.11.1. In particular, the paths p σ mi and r q satisfy p σ m = r q. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 37 · · · ·· p (cid:46) (cid:46) p (cid:24) (cid:24) q (cid:70) (cid:70) r (cid:47) (cid:47) r (cid:47) (cid:47) s (cid:110) (cid:110) t (cid:20) (cid:20) · · · ·· t p (cid:46) (cid:46) p (cid:24) (cid:24) q (cid:70) (cid:70) r (cid:47) (cid:47) r (cid:47) (cid:47) s (cid:84) (cid:84) (cid:62) (cid:62) ( i ) ( ii ) Figure 17.
Setup for Lemma 4.14.1. In both cases, the paths p = p p , q , and r = r r are drawn in red, blue, and green respectively.The orange path s is an arrow, and the cycles st and sr p are unitcycles. This leads to a contradiction in case (ii) since q passes throughthe interior of these unit cycles. · · ·· r (cid:79) (cid:79) p (cid:45) (cid:45) p (cid:24) (cid:24) q (cid:70) (cid:70) r (cid:80) (cid:80) Figure 18.
Setup for Lemma 4.14.2. The paths p = p p p , q , and r are drawn in red, blue, and green respectively. This shows that thenon-cancellative pair p, q is not minimal.Thus p σ m t( p ) and r q form a non-cancellative pair by Lemma 4.12. But by choosinga representative of the unit cycle σ +t( p ) which lies in R p,q , it follows that R p σ m t( p ) ,r q is properly contained in R p,q . Therefore the non-cancellative pair p, q is not minimal. (cid:3) Remark 4.15.
The assumption in Lemma 4.14.1 that p and q do not have propersubpaths p (cid:48) and q (cid:48) such that ψ ( p (cid:48) ) = ψ ( q (cid:48) ) is necessary. Indeed, consider the subquivergiven in Figure 19. Here, ψ contracts the two arrows whose tails have indegree 1.Observe that rp = rq (cid:54) = 0, and r + does not intersect the interior of R p,q . However, p and q have proper subpaths p (cid:48) and q (cid:48) modulo I satisfying ψ ( p (cid:48) ) = ψ ( q (cid:48) ). · · i · ·· · · · ·· · j · ·· ···· · (cid:111) (cid:111) (cid:111) (cid:111) (cid:47) (cid:47) (cid:47) (cid:47) (cid:111) (cid:111) (cid:111) (cid:111) (cid:111) (cid:111) (cid:47) (cid:47) (cid:47) (cid:47) (cid:111) (cid:111) (cid:111) (cid:111) (cid:47) (cid:47) (cid:111) (cid:111) (cid:47) (cid:47) r (cid:111) (cid:111) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) p (cid:79) (cid:79) (cid:79) (cid:79) (cid:15) (cid:15) (cid:79) (cid:79) q (cid:79) (cid:79) (cid:15) (cid:15) (cid:79) (cid:79) (cid:31) (cid:31) (cid:31) (cid:31) (cid:31) (cid:31) (cid:127) (cid:127) (cid:127) (cid:127) (cid:127) (cid:127) (cid:31) (cid:31) (cid:127) (cid:127) (cid:63) (cid:63) Figure 19.
Setup for Remark 4.15. The paths p , q , r are drawn inred, blue, and green respectively.The following generalizes Lemma 2.19.1 and Proposition 2.20.2 to the case where A is non-cancellative. Lemma 4.16.
Let u ∈ Z \ . Suppose a ∈ Q , p ∈ ˆ C u t( a ) , and q ∈ ˆ C u h( a ) . If R ◦ ap,qa contains no vertices, then ap = qa . Consequently, p = q .Proof. Suppose the hypotheses hold. If ( ap ) + and ( qa ) + have no cyclic subpaths(modulo I ), then ap = qa by Lemma 2.12.2.So suppose ( qa ) + contains a cyclic subpath. The path q + has no cyclic subpathssince q is in ˆ C . Thus q factors into paths q = q q , where ( q a ) + is a cycle. Inparticular, t( p + ) = t(( q q ) + ) and h( p + ) = h(( q q ) + ) . Whence p and q q bound a compact region R p,q q . Furthermore, its interior R ◦ p,q q contains no vertices since R ◦ ap,qa contains no vertices.The path ( q ) + has no cyclic subpaths, again since q is in ˆ C . Thus ( q q ) + also hasno cyclic subpaths. Furthermore, p + has no cyclic subpaths since p is in ˆ C . Therefore p = q q by Lemma 2.12.2.Since there are no vertices in R ◦ ap,qa , there are also no vertices in the interior of theregion bounded by the cycle ( aq ) + . Thus there is some (cid:96) ≥ aq = σ (cid:96) h( a ) and q a = σ (cid:96) t( a ) . Therefore ap = aq q = σ (cid:96) h( a ) q (i) = q σ (cid:96) t( a ) = q q a = qa, where ( i ) holds by Lemma 1.5.It thus follows that p = q by the proof of Proposition 2.20.2 with r = a . (cid:3) Reduced and homotopy centers.
Throughout, A is a non-cancellative dimeralgebra and ψ : A → A (cid:48) is a cyclic contraction. We denote by Z and Z (cid:48) the respectivecenters of A and A (cid:48) . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 392 1 21 2 1 · (cid:15) (cid:15) (cid:63) (cid:63) (cid:95) (cid:95) (cid:47) (cid:47) (cid:127) (cid:127) (cid:47) (cid:47) (cid:95) (cid:95) (cid:15) (cid:15) (cid:31) (cid:31) · · · ·· ·· · · ··· · · ·· ·· · · · (cid:47) (cid:47) (cid:47) (cid:47) (cid:15) (cid:15) (cid:95) (cid:95) (cid:31) (cid:31) (cid:15) (cid:15) (cid:95) (cid:95) (cid:31) (cid:31) (cid:127) (cid:127) (cid:63) (cid:63) (cid:47) (cid:47) (cid:15) (cid:15) (cid:63) (cid:63) (cid:95) (cid:95) (cid:31) (cid:31) (cid:15) (cid:15) (cid:47) (cid:47) (cid:15) (cid:15) (cid:63) (cid:63) (cid:95) (cid:95) (cid:31) (cid:31) (cid:127) (cid:127) (cid:63) (cid:63) (cid:127) (cid:127) (cid:31) (cid:31) (cid:47) (cid:47) (cid:95) (cid:95) (cid:95) (cid:95) (cid:127) (cid:127) (cid:95) (cid:95) (cid:15) (cid:15) (cid:95) (cid:95) (cid:95) (cid:95) (cid:63) (cid:63) (cid:127) (cid:127) Figure 20.
A dimer algebra A for which nil Z (cid:54) = 0. A fundamentaldomain of Q is shown on the left, and a larger region of Q + is shownon the right. The paths p , q , a are drawn in red, blue, and greenrespectively. The element ( p − q ) a + a ( p − q ) is central and squares tozero.4.2.1. The central nilradical.
Recall that cancellative dimer algebras are prime (Propo-sition 3.11), and in particular have reduced centers (Theorem 3.3.4). In the following,we show that these properties do not necessarily hold in the non-cancellative case.
Theorem 4.17.
Dimer algebras exist with non-vanishing central nilradical. Conse-quently, dimer algebras exist which are not prime.Proof.
Consider the non-cancellative dimer algebra A with quiver Q given in Figure20. (A cyclic contraction of A is given in Figure 1.) The paths p , q , a satisfy z := ( p − q ) a + a ( p − q ) ∈ nil Z. In particular, nil Z (cid:54) = 0. A is therefore not prime since zAz = z A = 0 . ( A also contains non-central elements a, b with the property that aAb = 0; for exam-ple, ( p − q ) Ae = 0.) (cid:3) Question 4.18.
Is there a necessary and/or sufficient combinatorial condition forthe central nilradical of a non-cancellative dimer algebra to vanish?In the following, we characterize the central nilradical in terms of the cyclic con-traction ψ , and show that it is a prime ideal of Z . Lemma 4.19.
Suppose p + is a cycle in Q + . If p is not equal to a product of unitcycles (modulo I ), then p (cid:54)∈ Ze t( p ) .Proof. Suppose p ∈ C i satisfies p (cid:54) = σ ni for all n ≥
1. Two examples of such a cycleare given by the red cycles in Figures 21.i and 21.ii.Let r be a path with tail at i whose lift r + does not intersect the interior of theregion bounded by p + in R , modulo I . Suppose q is a cycle satisfying rp = qr . Then ( i ) Q · ·· · (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:79) (cid:79) (cid:111) (cid:111) (cid:47) (cid:47) (cid:95) (cid:95) (cid:63) (cid:63) (cid:127) (cid:127) (cid:31) (cid:31) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:23) (cid:23) (cid:119) (cid:119) (cid:87) (cid:87) (cid:55) (cid:55) ψ −→ (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:79) (cid:79) (cid:79) (cid:79) (cid:111) (cid:111) (cid:47) (cid:47) Q (cid:48) ( ii ) Q · ··· (cid:47) (cid:47) (cid:47) (cid:47) (cid:127) (cid:127) (cid:127) (cid:127) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:79) (cid:79) (cid:127) (cid:127) (cid:127) (cid:127) (cid:79) (cid:79) (cid:47) (cid:47) (cid:127) (cid:127) ψ −→ · (cid:47) (cid:47) (cid:47) (cid:47) (cid:127) (cid:127) (cid:127) (cid:127) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:79) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:47) (cid:79) (cid:79) (cid:127) (cid:127) (cid:127) (cid:127) Q (cid:48) Figure 21.
Examples for Remarks 4.20, 4.22, and Proposition 4.72.The quivers are drawn on a torus, the contracted arrows are drawn ingreen, and the 2-cycles have been removed from Q (cid:48) . In each example,the cycle in Q formed from the red arrows is not equal to a product ofunit cycles (modulo I ). However, in example (i) this cycle is mappedto a unit cycle in Q (cid:48) under ψ .the region bounded by q + in R contains p + . Furthermore, rp = qr implies p = q since B is an integral domain.But for a sufficiently long path r + , the ¯ τ ψ -image of any cycle at h( r ) whose liftcontains p + in its interior will clearly not equal p . Thus for such a path r , rp (cid:54) = qr for all cycles q . Therefore p is not in Ze i . (cid:3) Remark 4.20.
It is possible for two cycles in Q + , one of which is properly containedin the region bounded by the other, to have equal ¯ τ ψ -images. Indeed, consider Figure21.i: the red cycle and the unit cycle in its interior both have ¯ τ ψ -image σ . Lemma 4.21.
Let u ∈ Z \ . Suppose p, q ∈ C ui are cycles such that pq = qp and ψ ( p ) = ψ ( q ) . Then p = qp = q . Consequently ( p − q ) = 0 . Proof.
For brevity, if sa and ta are unit cycles with a ∈ Q , then we refer to s as anarc and t as its complementary arc.Suppose the hypotheses hold. If p = q , then the lemma trivially holds, so suppose p (cid:54) = q . Since pq = qp , there are subpaths q (cid:48) , q (cid:48)(cid:48) , a, c of q and a path b such that pq = p ( cq (cid:48) ) = ( pc ) q (cid:48) = ( q (cid:48)(cid:48) b ) q (cid:48) = q (cid:48)(cid:48) ( bq (cid:48) ) = q (cid:48)(cid:48) ( ap ) = ( q (cid:48)(cid:48) a ) p = qp. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 41
In particular, q = cq (cid:48) = q (cid:48)(cid:48) a, ap = bq (cid:48) , pc = q (cid:48)(cid:48) b. See Figure 22, where p is drawn in red, q is drawn in blue, and their common subpathsare drawn in violet. In case (i) a is a rightmost subpath of q alone, and in case (ii) a is a rightmost subpath of both p and q .(i) First suppose a is a rightmost subpath of q alone.We claim that p + and q + have no cycle subpaths. Indeed, if p + has a unit cyclesubpath σ + i , then σ + i can be homotoped to a unit cycle at h( p ) which contains a byLemma 1.5, yielding case (ii). Otherwise, if p + contains a cyclic subpath r + which isnot equal to a product of unit cycles, then pq (cid:54) = qp by Lemma 4.19. This proves ourclaim.Now suppose to the contrary that ac (cid:54) = b . Then, since p (cid:54) = q and pc = q (cid:48)(cid:48) b ,there is an arc subpath of pc (drawn in green in Figure 22) containing a leftmostarrow subpath of c and rightmost arrow subpath of p , whose complementary arc(also drawn in green) lifts to a path that lies in the region bounded by ( pc ) + and( q (cid:48)(cid:48) b ) + . In particular, the two cycles formed from the green paths and the orangearrow are unit cycles. But this is not possible since t( a ) = t( p ), a contradiction.Thus ac = b , yielding(31) qp = q (cid:48)(cid:48) ap = q (cid:48)(cid:48) bq (cid:48) = q (cid:48)(cid:48) acq (cid:48) = q . Similarly pq = p .(ii) Now suppose a is a rightmost subpath of both p and q . Then the relation pc = q (cid:48)(cid:48) b implies ac = b by the contrapositive of Lemma 4.14.1. Therefore (31)holds. (cid:3) Remark 4.22.
The assumption in Lemma 4.21 that p + and q + are not cycles isnecessary. Indeed, if p, q ∈ C , then it is possible for pq = qp , ψ ( p ) = ψ ( q ), and( p − q ) (cid:54) = 0. For example, consider Figure 21.ii: the cycle p ∈ e Ae formed fromthe red arrows satisfies pσ = σ p and ψ ( p ) = ψ ( σ ) . However, ( p − σ ) m (cid:54) = 0 for each m ≥ Lemma 4.23.
Consider a central element z = (cid:88) i ∈ Q ( p i − q i ) , where p i , q i are elements in e i Ae i . Then for each i ∈ Q , p i q i = q i p i . Proof.
For each i ∈ Q we have p i − p i q i = p i ( p i − q i ) = p i z = zp i = ( p i − q i ) p i = p i − q i p i . Whence p i q i = q i p i . (cid:3) ( i ) i i i ·· · · p (cid:26) (cid:26) (cid:59) (cid:59) (cid:55) (cid:63) p (cid:26) (cid:26) q (cid:48) (cid:45) (cid:45) c (cid:54) (cid:54) (cid:68) (cid:68) (cid:64) (cid:72) a (cid:45) (cid:45) q (cid:48)(cid:48) (cid:68) (cid:68) b (cid:52) (cid:52) (cid:114) (cid:114) (cid:82) (cid:82) ( ii ) i i i · · ·· (cid:47) (cid:47) p (cid:26) (cid:26) a (cid:47) (cid:47) (cid:47) (cid:47) p (cid:26) (cid:26) q (cid:48) (cid:47) (cid:47) c (cid:68) (cid:68) q (cid:48)(cid:48) (cid:68) (cid:68) b (cid:64) (cid:64) Figure 22.
Cases for Lemma 4.21.
Theorem 4.24.
Let A be a non-cancellative dimer algebra and ψ : A → A (cid:48) a cycliccontraction. Then the central elements in the kernel of ψ are precisely the nilpotentcentral elements of A , Z ∩ ker ψ = nil Z. Proof. (i) We first claim that if z ∈ Z ∩ ker ψ , then z = 0, and in particular z ∈ nil Z .Consider a central element z in ker ψ . Since z is central it commutes with thevertex idempotents, and so z is a k -linear combination of cycles. Therefore, since ψ sends paths to paths and I (cid:48) is generated by differences of paths, it suffices to suppose z is of the form z = (cid:88) i ∈ Q ( p i − q i ) , where p i , q i are cycles in e i Ae i with equal ψ -images modulo I (cid:48) . Note that there maybe vertices i ∈ Q for which p i = q i = 0.By Lemma 4.23, for each i ∈ Q we have p i q i = q i p i . Furthermore, by Lemmas 1.5 and 4.19 it suffices to suppose that the lifts p + i , q + i arenot cycles in Q + . Therefore by Lemma 4.21, z = ( (cid:88) i ∈ Q ( p i − q i )) = (cid:88) i ∈ Q ( p i − q i ) = 0 . Whence z ∈ nil Z .(ii) We now claim that if z ∈ nil Z , then z ∈ ker ψ .Suppose z n = 0. Then for each i ∈ Q ,¯ τ ψ ( ze i ) n (i) = ¯ τ ψ (( ze i ) n ) (ii) = ¯ τ ψ ( z n e i ) = 0 , ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 43 where ( i ) holds since ¯ τ ψ is an algebra homomorphism on e i Ae i , and ( ii ) holds since z is central. But ¯ τ ψ ( e i Ae i ) is contained in the integral domain B . Whence¯ τ ψ ( ze i ) = 0 . Therefore ψ ( ze i ) = 0 since ¯ τ is injective, by Theorem 3.5. But then ψ ( z ) (i) = ψ (cid:32) z (cid:88) i ∈ Q e i (cid:33) (ii) = (cid:88) i ∈ Q ψ ( ze i ) = 0 , where ( i ) holds since the vertex idempotents form a complete set, and ( ii ) holds since ψ is a k -linear map. Therefore ψ ( z ) = 0. (cid:3) The reduced center as a subalgebra.
Let ψ : A → A (cid:48) be a cyclic contraction.Consider the map ˜ τ : A → M | Q | ( B )defined on p ∈ e j Ae i by(32) ˜ τ ( p ) = pE ji = ¯ τ ψ ( p ) E ji , and extended k -linearly to A . Lemma 4.25.
The map ˜ τ : A → M | Q | ( B ) is an algebra homomorphism.Proof. By Lemma 1.6, τ : A (cid:48) → M | Q (cid:48) | ( B ) is an algebra homomorphism. Furthermore, ψ is a k -linear map, and an algebra homomorphism when restricted to each vertexcorner ring e i Ae i . (cid:3) In addition to the cycle algebra S := k [ ∪ i ∈ Q ¯ τ ψ ( e i Ae i )], which is generated bythe union of the ¯ τ ψ -images of the vertex corner rings of A , it will also be useful toconsider the algebra generated by their intersection: Definition 4.26.
The homotopy center of A is the algebra R := k [ ∩ i ∈ Q ¯ τ ψ ( e i Ae i )] = (cid:92) i ∈ Q ¯ τ ψ ( e i Ae i ) . Theorem 4.27.
Let A be a non-cancellative dimer algebra and ψ : A → A (cid:48) a cycliccontraction. Then there is an exact sequence (33) 0 −→ nil Z (cid:44) −→ Z ¯ ψ −→ R, where ¯ ψ is an algebra homomorphism. Therefore the reduction ˆ Z := Z/ nil Z of Z isisomorphic to a subalgebra of R .Proof. (i) We first claim that the composition τ ψ : A → A (cid:48) → M | Q (cid:48) | ( B )induces an algebra homomorphism(34) ¯ ψ : Z → R, z (cid:55)→ ze i , where i ∈ Q is any vertex.Consider a central element z ∈ Z and vertices j, k ∈ Q . By the construction of Q , there is a path p from j to k . For i ∈ Q , set z i := ze i ∈ e i Ae i . By Lemma 4.25,¯ τ ψ is an algebra homomorphism on each vertex corner ring e i Ae i . Thus pz j = pz j = pz = zp = z k p = z k p ∈ B. But the image p is nonzero since τ is an impression of A (cid:48) and the ψ -image of anypath is nonzero. Thus, since B is an integral domain,(35) z j = z k . Therefore, since j, k ∈ Q were arbitrary, z j ∈ k [ ∩ i ∈ Q ¯ τ ψ ( e i Ae i )] = R. (ii.a) Let z ∈ Z and i ∈ Q . We claim that ψ ( ze i ) = 0 implies ψ ( z ) = 0. For each j ∈ Q (cid:48) , denote by c j := (cid:12)(cid:12) ψ − ( j ) ∩ Q (cid:12)(cid:12) the number of vertices in ψ − ( j ). Since ψ maps Q surjectively onto Q (cid:48) , we have c j ≥
1. Furthermore, if k ∈ ψ − ( j ), then(36) ψ ( z ) e j = c j ψ ( ze k ) . Set z (cid:48) j := c − j ψ ( z ) e j . Then the sum z (cid:48) := (cid:88) j ∈ Q (cid:48) z (cid:48) j is in the center Z (cid:48) of A (cid:48) by (35) and (28) in Theorem 3.5. Therefore¯ τ ( z (cid:48) j ) = ¯ τ ( z (cid:48) e j ) = ¯ τ ( z (cid:48) e ψ ( i ) ) = ¯ τ ( z (cid:48) ψ ( i ) ) = ¯ τ ( c − ψ ( i ) ψ ( z ) e ψ ( i ) ) (i) = ¯ τ ( ψ ( ze i )) = 0 , where ( i ) holds by (36). Thus z (cid:48) j = 0 since ¯ τ is injective. Whence ψ ( z ) e j = c j z (cid:48) j = 0 . But this holds for each j ∈ Q (cid:48) . Therefore ψ ( z ) = 0, proving our claim.(ii.b) We now claim that the homomorphism (34) can be extended to the exactsequence (33). Indeed, ¯ ψ : Z → R factors into the homomorphisms Z · e i −→ Ze i ψ −→ ψ ( Ze i ) ¯ τ −→ R. Suppose z ∈ Z is in the kernel of ¯ ψ ,¯ τ ψ ( ze i ) = ¯ ψ ( z ) = 0 . Note that ψ ( z ) is not in Z (cid:48) if there are vertices i, j ∈ Q (cid:48) for which c i (cid:54) = c j . Therefore in general ψ ( Z ) is not contained in Z (cid:48) . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 45
By Theorem 3.5, ¯ τ is injective on each vertex corner ring e i Ae i . Thus ψ ( ze i ) = 0.Whence ψ ( z ) = 0 by Claim (ii.a). Therefore z ∈ nil Z by Theorem 4.24. This provesour claim. (cid:3) Corollary 4.28.
The algebras ˆ Z , R , and S are integral domains. Therefore thecentral nilradical nil Z of A is a prime ideal of Z , and the schemes Spec Z and Spec ˆ Z are irreducible.Proof. R and S are domains since they are subalgebras of the domain B . Thereforeˆ Z is a domain since it isomorphic to a subalgebra of R by Theorem 4.27. (cid:3) Lemma 4.29. If g is a monomial in B and gσ is in S , then g is also in S .Proof. Suppose the hypotheses hold. Then there is a cycle p ∈ A (cid:48) such that p = gσ .Let u ∈ Z be such that p ∈ C (cid:48) u . Since A (cid:48) is cancelative, there is a cycle q ∈ ˆ C (cid:48) u byProposition 2.10. Furthermore, σ (cid:45) q by Proposition 3.10. Thus there is some m ≥ qσ m = p = gσ, by Lemma 2.18. Therefore g = ( gσ ) σ − = qσ m − ∈ S (cid:48) (i) = S, where ( i ) holds since ψ is cyclic. (cid:3) Proposition 4.30. (1) If g ∈ R and σ (cid:45) g , then g ∈ ¯ ψ ( Z ) .(2) If g ∈ S , then there is some N ≥ such that for each n ≥ , g n σ N ∈ ¯ ψ ( Z ) .(3) If g ∈ R , then there is some N ≥ such that g N ∈ ¯ ψ ( Z ) .Proof. Since R is generated by monomials, it suffices to consider a non-constantmonomial g ∈ R . Then for each i ∈ Q , there is a cycle c i ∈ e i Ae i satisfying c i = g .(1) Suppose σ (cid:45) g . Fix a ∈ Q , and set p := c t( a ) and q := c h( a ) . See Figure 23. We claim that ap = qa .Let u, v ∈ Z be such that p ∈ C u and q ∈ C v . By assumption, σ (cid:45) g = p = q . Then u and v are both nonzero by Lemma 4.11.1.Thus u = v by Lemma 4.13. Therefore ( ap ) + and ( qa ) + bound a compact region R ap,qa in R .We proceed by induction on the number of vertices in the interior R ◦ ap,qa .First suppose there are no vertices in R ◦ ap,qa . Since σ (cid:45) g = p = q , p and q are in ˆ C by Lemma 4.11.2. Therefore ap = qa by Lemma 4.16. So suppose R ◦ ap,qa contains at least one vertex i + . Let w ∈ Z be such that c i ∈ C w .Then w = u = v , again by Lemma 4.13. Therefore c i intersects p at least twice or q at least twice. Suppose c i intersects p at vertices j and k . Then p factors into paths p = p e k te j p = p tp . Let s + be the subpath of ( c i ) + from j + to k + . Thent( s + ) = t( t + ) and h( s + ) = h( t + ) . In particular, s + and t + bound a compact region R s,t .Since we are free to choose the vertex i + in R ◦ ap,qa , we may suppose R ◦ s,t containsno vertices. Furthermore, c + i and p + have no cyclic subpaths since σ (cid:45) g , again byLemma 4.11.2. Thus their respective subpaths s + and t + have no cyclic subpaths.Whence s = t by Lemma 2.12.2.Furthermore, since R ◦ ap sp ,qa contains less vertices than R ◦ ap,qa , it follows by induc-tion that ap sp = qa. Therefore ap = a ( p tp ) = a ( p sp ) = qa. Since a ∈ Q was arbitrary, the sum (cid:80) i ∈ Q c i is central in A .(2) Fix an arrow a ∈ Q . Set i := t( a ) and j := h( a ). Let r + be a path in Q + fromh (( ac i ) + ) to t (( ac i ) + ). Then by Lemma 2.3.1, there is some (cid:96), m, n ≥ σ mi = rc j aσ (cid:96)i and ac i rσ (cid:96)j = σ nj . Thus σ m = ¯ τ ψ (cid:0) rc j aσ (cid:96)i (cid:1) = rc j aσ (cid:96) = ac i rσ (cid:96) = ¯ τ ψ (cid:0) ac i rσ (cid:96)j (cid:1) = σ n . Furthermore, σ (cid:54) = 1 since ¯ τ is injective. Whence n = m since B is an integral domain.Therefore(37) a ( c i σ ni ) = ac i (cid:0) rc j aσ (cid:96)i (cid:1) (i) = (cid:0) ac i rσ (cid:96)j (cid:1) c j a = σ nj c j a (ii) = ( c j σ ni ) a, where ( i ) and ( ii ) hold by Lemma 1.5.For each a ∈ Q there is an n = n ( a ) such that (37) holds. Set N := max { n ( a ) | a ∈ Q } . Then (37) implies that the element (cid:80) i ∈ Q c i σ Ni is central.Now fix n ≥ a ∈ Q . Again set i := t( a ) and j := h( a ). Then ac ni σ Ni = ac n − i (cid:0) c i σ Ni (cid:1) = (cid:0) c j σ Nj (cid:1) ac n − i (i) = c j ac n − i σ Ni = · · · = c nj σ Nh a, where ( i ) holds by Lemma 1.5. Therefore, for each n ≥
1, the element (cid:88) i ∈ Q c ni σ Ni is central. But its ¯ τ ψ -image is g n σ N , proving Claim (2). ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 47 · · j ik · · a (cid:47) (cid:47) a (cid:47) (cid:47) p (cid:79) (cid:79) t (cid:79) (cid:79) p (cid:79) (cid:79) q (cid:79) (cid:79) (cid:58) (cid:58) (cid:71) (cid:71) s (cid:108) (cid:108) c i (cid:80) (cid:80) Figure 23.
Setup for Proposition 4.30.1. The cycles p = p tp , q , c i ,are drawn in red, blue, and green respectively. The path a is an arrow.(3) By Claim (1), it suffices to suppose σ | g . Then there is a monomial h ∈ B such that g = hσ. By Lemma 4.29, h is in S . Therefore by Claim (2), there is some N ≥ g N = h N σ N ∈ ¯ ψ ( Z ) . (cid:3) The following theorem shows that it is possible for the reduced center ˆ Z to beproperly contained in the homotopy center R . However, we will show that theydetermine the same nonlocal variety (Theorem 4.68), and that their integral closuresare isomorphic (Theorem 4.75). Theorem 4.31.
There exists dimer algebras for which the containment ˆ Z (cid:44) → R isproper.Proof. Consider the contraction given in Figure 24. This contraction is cyclic: S = k [ x , xy, y , z ] = S (cid:48) . We claim that the reduced center ˆ Z of A = kQ/I is not isomorphic to R . By theexact sequence (33), it suffices to show that the homomorphism ¯ ψ : Z (cid:44) → R is notsurjective.We claim that the monomial zσ is in R , but is not in the image ¯ ψ ( Z ). It is clearthat zσ is in R from the ¯ τ ψ labeling of arrows given in Figure 24.Assume to the contrary that zσ ∈ ¯ ψ ( Z ). Then by (35), for each j ∈ Q there is anelement c j in Ze j whose ¯ τ ψ -image is zσ . Consider the vertex i ∈ Q shown in Figure · · i ··· · ··· · i ···· (cid:5) (cid:5) y (cid:89) (cid:89) xz (cid:47) (cid:47) (cid:111) (cid:111) x (cid:47) (cid:47) (cid:47) (cid:47) z (cid:79) (cid:79) x (cid:42) (cid:42) (cid:23) (cid:23) y (cid:116) (cid:116) z (cid:79) (cid:79) y (cid:116) (cid:116) x (cid:42) (cid:42) (cid:79) (cid:79) (cid:23) (cid:23) (cid:79) (cid:79) (cid:111) (cid:111) xz (cid:47) (cid:47) yz (cid:5) (cid:5) (cid:89) (cid:89) z (cid:79) (cid:79) x (cid:47) (cid:47) (cid:47) (cid:47) ψ −→ · · ·· · · y (cid:111) (cid:111) x (cid:47) (cid:47) z (cid:79) (cid:79) x (cid:23) (cid:23) z (cid:79) (cid:79) y (cid:7) (cid:7) z (cid:79) (cid:79) y (cid:111) (cid:111) x (cid:47) (cid:47) Q Q (cid:48)
Figure 24.
A cyclic contraction ψ : A → A (cid:48) for which ˆ Z (cid:40) R . Q and Q (cid:48) are drawn on a torus, and the contracted arrows are drawn ingreen. The arrows drawn in blue form removable 2-cycles under ψ .The arrows in Q are labeled by their ¯ τ ψ -images, and the arrows in Q (cid:48) are labeled by their ¯ τ -images.24. The set of cycles in e i Ae i with ¯ τ ψ -image zσ are as follows: p := δ a a ( δ a ) δ b δ ,p := δ b ( δ a ) δ b ,p := δ b ( δ a ) δ a δ δ p := δ a a ( δ a ) δ b ,p := δ a a ( δ a ) δ a δ δ . Thus for some coefficients α , . . . , α ∈ k , c i = (cid:88) (cid:96) =1 α (cid:96) p (cid:96) . Since c i = zσ , there is some (cid:96) for which α (cid:96) is nonzero. In particular, there are cycles p (cid:48) (cid:96) and p (cid:48)(cid:48) (cid:96) satisfying b p (cid:96) = p (cid:48) (cid:96) b and δ p (cid:96) = p (cid:48)(cid:48) (cid:96) δ . However, there are no cycles p (cid:48) , p (cid:48)(cid:48) , p (cid:48) , p (cid:48)(cid:48) , p (cid:48) , for which b p = p (cid:48) b , δ p = p (cid:48)(cid:48) δ , b p = p (cid:48) b , δ p = p (cid:48)(cid:48) δ , b p = p (cid:48) b . Thus no such element c i ∈ Ze i can exist, a contradiction. Therefore ˆ Z (cid:54)∼ = R . (cid:3) ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 49
Theorem 4.31 raises the following question.
Question 4.32.
Are there necessary and sufficient conditions for the isomorphismˆ Z ∼ = R to hold?4.3. Homotopy dimer algebras.
We introduce the following class of algebras. Re-call the definition of a non-cancellative pair (Definition 2.2).
Definition 4.33.
Let A be a dimer algebra. We call the quotient algebra˜ A := A/ (cid:104) p − q | p, q is a non-cancellative pair (cid:105) the homotopy (dimer) algebra of A .Note that a dimer algebra A equals its homotopy algebra ˜ A if and only if A iscancellative.Let ψ : A → A (cid:48) be a cyclic contraction. Recall the algebra homomorphism˜ τ : A → M | Q | ( B )defined in (32) (Lemma 4.25). This homomorphism induces an algebra monomor-phism on the quotient ˜ A , ˜ τ : ˜ A → M | Q | ( B ) , by Lemma 4.12. Remark 4.34.
The ideal (cid:104) p − q | p, q is a non-cancellative pair (cid:105) ⊂ A is contained in the kernel of ψ , but not conversely. Indeed, if ψ contracts an arrow δ , then δ − e t( δ ) is in the kernel of ψ , but δ and e t( δ ) do not form a non-cancellativepair. Theorem 4.35.
The algebra monomorphism ˜ τ : ˜ A → M | Q | ( B ) is an impression ofthe homotopy dimer algebra ˜ A . Consequently, the center of ˜ A is isomorphic to thehomotopy center R of A .Proof. Denote by ˜ Z the center of ˜ A .(i) For generic b ∈ Max B , the composition (cid:15) b ˜ τ (defined in (8)) is surjective byClaim (ii) in the proof of Theorem 3.5. Furthermore, the morphismMax B → Max τ ( ˜ Z ) , b (cid:55)→ b | Q | ∩ τ ( ˜ Z ) , is surjective by Claim (iii) in the proof of Theorem 3.5. Therefore (˜ τ , B ) is animpression of ˜ A .(ii) Since (˜ τ , B ) is an impression of ˜ A , ˜ Z is isomorphic to R [B, Lemma 2.1 (2)]. (cid:3) Recall that non-cancellative dimer algebras exist which are not prime, by Theorem4.17. However, their homotopy algebras will always be prime:
Proposition 4.36.
Homotopy dimer algebras are prime.
Proof.
Follows similar to Proposition 3.11, since ˜ τ : ˜ A → M | Q | ( B ) is injective byLemma 4.12. (cid:3) Non-annihilating paths.
Throughout, ψ : A → A (cid:48) is a cyclic contraction.Recall the subset of arrows Q S ⊆ Q defined in (25), and the definition of a minimalnon-cancellative pair (Definition 2.2). Proposition 4.37.
Let p, q be a minimal non-cancellative pair. If r is a path ofminimal length satisfying rp = rq (cid:54) = 0 or pr = qr (cid:54) = 0 , then each arrow subpath of r is in Q S . In particular, if A is non-cancellative, then Q S (cid:54) = ∅ .Proof. Set σ := (cid:81) D ∈P x D , and recall the algebra homomorphism η defined in (3).Assume to the contrary that there is an arrow subpath r of r = r r r which iscontained in a simple matching D ; here r and r are paths, possibly vertices.Since D is a simple matching, there is a path t (cid:48) whose arrow subpaths are notcontained in D , from t( p ) to h( r ). By Lemma 4.14.2, the path r + lies in R p,q , andthe vertex h( r + ) lies in the interior R ◦ p,q . Thus the vertex h( r +2 ) also lies in R ◦ p,q .Therefore there is a leftmost non-vertex subpath t + of t (cid:48) + contained in R p,q with tailon p or q ; suppose t has tail on p and has minimal length. Then p factors into paths p = p p , where h( p +1 ) = t( t + ), as shown in Figure 25.Consider the path s such that ( r s ) + is a unit cycle that lies in the region R t,r r p .(Note that s and r p may share arrows.) Then x D (cid:45) ¯ η ( s ) since r ∈ D . Whence(38) σ (cid:45) ¯ η ( s ) . Furthermore, x D (cid:45) ¯ η ( t ) since no arrow subpath of t is contained in D . Thus x D (cid:45) ¯ η ( s )¯ η ( t ) = ¯ η ( st ) . Therefore(39) σ (cid:45) ¯ η ( st ) . Since t + lies in R p,q and h( t + ) lies in R ◦ p,q , we have R r p ,st (cid:40) R p,q and R stp ,r q (cid:40) R p,q . Furthermore, by (39) and Lemma 2.3.2, there is some m ≥ σ m ¯ η ( st ) = ¯ η ( r p ) . Thus, by Lemma 2.3.3 and the minimality of the pair p, q , σ m h( s ) st = r p . Therefore(40) σ m h( s ) stp = r p. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 51 ··· ·· q (cid:105) (cid:105) p (cid:83) (cid:83) p (cid:53) (cid:53) t (cid:59) (cid:59) r (cid:15) (cid:15) s (cid:55) (cid:55) r (cid:15) (cid:15) Figure 25.
Setup for Proposition 4.37.Since p, q is a non-cancellative pair, ¯ η ( p ) = ¯ η ( q ) by Lemma 2.3.3. Whence (40)implies σ m ¯ η ( stp ) = ¯ η ( r p ) = ¯ η ( r )¯ η ( p ) = ¯ η ( r )¯ η ( q ) = ¯ η ( r q ) . Therefore, again by Lemma 2.3.3 and the minimality of the pair p, q ,(41) σ m h( s ) stp = r q. Consequently, r p (40) = σ m h( s ) stp = r q. But r is an arrow. Therefore r does not have minimal length such that rp = rq ,contrary to our choice of r . (cid:3) Theorem 4.38.
Suppose ψ : A → A (cid:48) is a cyclic contraction. Then Q ∗ ⊆ Q S . Proof.
Suppose the hypotheses hold, and assume to the contrary that there is anarrow δ ∈ Q ∗ \ Q S .Let s be a path for which sδ is a unit cycle. Since δ is contracted to a vertex, ψ ( s )is a unit cycle in Q (cid:48) . Whence(42) ¯ τ ψ ( s ) = σ. Since D is simple, there is a cycle p ∈ A supported on Q \ D which contains s as asubpath, as well as each arrow in Q \ D . Since s is a subpath of p , ψ ( s ) is a subpathof ψ ( p ). In particular, ¯ τ ψ ( s ) | ¯ τ ψ ( p ) = p. Therefore (42) implies(43) σ | p. Since A (cid:48) is cancellative, S (cid:48) is generated by σ and a set of monomials in B notdivisible by σ , by Proposition 3.10. Thus (43) implies that the monomial p/σ is in S (cid:48) . Therefore, since S = S (cid:48) , there is a cycle q ∈ A such that(44) q = p/σ. Set i := t( q ). Since p contains each arrow in Q \ D and D is a perfect matching, p also contains each vertex in Q . Thus we may assume that t( p ) = i , by possiblycyclically permuting the arrow subpaths of p . Furthermore, ¯ τ : e i Ae i → B is injectiveby Theorem 3.5. Therefore (44) implies(45) ψ ( p ) = ψ ( qσ i ) . Since p contains each vertex in Q , p is in R . Thus there is some n ≥ p n ∈ Ze i by Proposition 4.30.3. Whence p n ( qσ i ) n = ( qσ i ) n p n . Furthermore, ψ ( p n ) = ψ ( q n σ ni ) by (45). Therefore by Lemma 4.21,(46) p n = q n σ ni . Now let V be an A -module with support Q \ D . Then p n does not annihilate V since p n is supported on Q \ D . However, σ i contains an arrow in D since D is aperfect matching. Thus q n σ ni annihilates V . But this is a contradiction to (46). (cid:3) Lemma 4.39.
If an arrow annihilates a simple A -module of dimension Q , then itis contained in a simple matching of A .Proof. Let V ρ be a simple A -module of dimension 1 Q , and suppose ρ ( a ) = 0. Let i ∈ Q . Since V ρ is simple of dimension 1 Q , there is a path p from t( a ) to i such that ρ ( p ) (cid:54) = 0. Furthermore, σ i p = pσ t( a ) by Lemma 1.5. Thus, since a is a subpath σ t( a ) (modulo I ), we have ρ ( σ i ) ρ ( p ) = ρ ( σ i p ) = ρ ( pσ t( a ) ) = 0 . Whence ρ ( σ i ) = 0 . Thus each unit cycle contains at least one arrow that annihilates V ρ . Therefore thereare perfect matchings D , . . . , D m ∈ P such that V ρ is supported on Q \ ( D ∪· · ·∪ D m ).Moreover, since ρ ( a ) = 0, there is some 1 ≤ (cid:96) ≤ m such that D (cid:96) contains a . Since V ρ is simple, there is a path r supported on Q \ ( D ∪ · · · ∪ D m ) which passes througheach vertex of Q . In particular, r is supported on Q \ D (cid:96) . Therefore D (cid:96) is a simplematching containing a . (cid:3) Lemma 4.40.
Let ψ : A → A (cid:48) be a cyclic contraction. Then R = S if and only if ˆ C ui (cid:54) = ∅ for each u ∈ Z \ and i ∈ Q . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 53
Proof. ( ⇒ ) Suppose R = S . Let u ∈ Z \
0. Since A (cid:48) is cancellative, there is a cycle q ∈ ˆ C (cid:48) u , by Proposition 2.10. In particular, σ (cid:45) q by Proposition 2.20.1.Since R = S , for each i ∈ Q there is a cycle p ∈ C i such that p = q . Furthermore,since A admits a cyclic contraction, A does not contain a non-cancellative pair whereone of the paths is a vertex, by Lemma 4.8.3. Thus p ∈ ˆ C i since σ (cid:45) p , by Lemma 2.8.3. Moreover, ψ ( p ) ∈ C (cid:48) u since ¯ τ ( ψ ( p )) = ¯ τ ( q ), by Lemma2.18. Whence p ∈ C u , by Lemma 4.8.1. Therefore p ∈ ˆ C ui . In particular, ˆ C ui (cid:54) = ∅ . The claim then followssince u ∈ Z \ i ∈ Q were arbitrary.( ⇐ ) Follows from Proposition 2.20.2. (cid:3) Theorem 4.41.
Let A = kQ/I be a dimer algebra. The following are equivalent:(1) A is cancellative.(2) Each arrow annihilates a simple A -module of dimension Q .(3) Each arrow is contained in a simple matching, Q S = ∅ .(4) If ψ : A → A (cid:48) is a cyclic contraction, then R = S .(5) If ψ : A → A (cid:48) is a cyclic contraction, then ψ is trivial, Q ∗ = ∅ .Proof. (2) ⇔ (3): Lemma 4.39.(3) ⇒ (1): Proposition 4.37.(4) ⇒ (3): Suppose R = S . Then ˆ C ui (cid:54) = ∅ for each u ∈ Z \ i ∈ Q , byLemma 4.40. Therefore Q S = ∅ by Theorem 2.24.(1) ⇒ (4): If A is cancellative, then R = S by Theorem 3.5.(1) ⇒ (5): Suppose A is cancellative, and ψ : A → A (cid:48) is a cyclic contraction. Then Q ∗ (i) ⊆ Q S (ii) = ∅ , where ( i) holds by Theorem 4.38, and ( ii) holds by the implication (1) ⇒ (2). There-fore Q ∗ = ∅ .(5) ⇒ (4): Clear. (cid:3) Non-cancellative dimer algebras do not necessarily admit contractions to cancella-tive dimer algebras, cyclic or not. In the following, we use Theorem 4.41 to showthat if a dimer algebra has a permanent 2-cycle (Definition 2.5), then it cannot becontracted to a cancellative dimer algebra.
Lemma 4.42.
Suppose ψ : A → A (cid:48) is a contraction of dimer algebras, and A (cid:48) hasa perfect matching. Further suppose ab is a 2-cycle in both A and A (cid:48) . Then ab isremovable in A if and only if ab is removable in A (cid:48) . Proof.
Suppose ab is a permanent 2-cycle. Then ab is given in Figure 3.ii or Figure3.iii, by Lemma 2.6. In either case, ψ ( ab ) is also a permanent 2-cycle by Lemma4.8.1. (cid:3) Proposition 4.43.
Let A be a dimer algebra with a permanent 2-cycle. Then A isnon-cancellative, and does not admit a contraction (cyclic or not) to a cancellativedimer algebra.Proof. (i) We first claim that A is non-cancellative. By Lemma 2.6, there are twotypes of permanent 2-cycles, and these are given in Figures 3.ii and 3.iii. Let D bea perfect matching of A . Then either a or b is contained in D . Therefore no arrowsubpath of p (resp. p or q ) in Figure 3.ii (resp. Figure 3.iii) is contained in D . Inparticular, no arrow subpath of p (resp. p or q ) is contained in a simple matching.Whence Q S (cid:54) = ∅ . Therefore A is non-cancellative by Theorem 4.41.(ii) By Claim (i) and Lemma 4.42, any contraction ψ : A → A (cid:48) to a cancellativedimer algebra necessarily contracts the unit cycle ab to a vertex. However, this is notpossible by Lemma 4.7. (cid:3) Nonnoetherian and nonlocal.
Throughout, A is a dimer algebra, non-cancellativeunless stated otherwise, and ψ : A → A (cid:48) is a cyclic contraction. Let˜ A := A/ (cid:104) p − q | p, q is a non-cancellative pair (cid:105) and R = Z ( ˜ A ) be the homotopy algebra and homotopy center of A , respectively. Lemma 4.44. If p is a cycle such that p (cid:54)∈ R and σ (cid:45) p , then for each n ≥ , p n (cid:54)∈ R. Furthermore, if A is non-cancellative, then such a cycle exists.Proof. (i) Assume to the contrary that there is a cycle p ∈ e i Ae i such that p (cid:54)∈ R , σ (cid:45) p , and p n ∈ R for some n ≥
2. Let u ∈ Z be such that p ∈ C ui . Since p is not in R , there is a vertex j ∈ Q such that(47) p (cid:54)∈ ¯ τ ψ ( e j Ae j ) . Furthermore, since p n is in R , p n homotopes to a cycle q ∈ e i Ae i that passes through j ,(48) q = p n (modulo I ) . For v ∈ Z , denote by q + v ∈ π − ( q ) the preimage with tailt( q + v ) = t( q + ) + v ∈ Q +0 . It is clear (see Figure 26) that there is a path r + from j + to j + + u , constructedfrom subpaths of q + , q + u , and possibly q + mu for some m ∈ Z . In particular, the cycle r := π ( r + ) ∈ e j Ae j is in C uj . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 55 i i i i i i · · · p n = q (cid:26) (cid:26) (cid:26) (cid:26) q (cid:26) (cid:26) (cid:26) (cid:26) q (cid:26) (cid:26) (cid:26) (cid:26) p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) p (cid:47) (cid:47) (cid:26) (cid:26) r (cid:68) (cid:68) (cid:26) (cid:26) j j j Figure 26.
Setup for Lemma 4.44. The path r + is drawn in red. Itsprojection r = π ( r + ) to Q is a cycle at j .Furthermore, since σ (cid:45) p , there is a simple matching D such that x D (cid:45) p . Whence x D (cid:45) q by (48). Thus x D (cid:45) r , and so σ (cid:45) r . In particular, σ (cid:45) p, σ (cid:45) r, and p, r ∈ C u . Therefore r = p by Lemma 2.18. But then p = r ∈ ¯ τ ψ ( e j Ae j ) , contrary to (47).(ii) Now suppose A is non-cancellative. We claim that there exists a cycle p asin Claim (i). Indeed, R (cid:54) = S by Theorem 4.41. Assume to the contrary that foreach cycle p satisfying p (cid:54)∈ R , we have σ | p . Then by the contrapositive of thisassumption, for each cycle q satisfying σ (cid:45) q , we have q ∈ R . But S is generated by σ and a set of monomials in B not divisible by σ , by Proposition 3.10. Therefore S ⊆ R since σ ∈ R . Whence S = R , contrary to assumption. (cid:3) Theorem 4.45.
Let A be a non-cancellative dimer algebra that admits a cyclic con-traction. Then each algebra A, Z, ˆ Z, R, is nonnoetherian.Proof.
Let ψ : A → A (cid:48) be a cyclic contraction.(i) We first claim that R is nonnoetherian. Indeed, since A is non-cancellative,there is a cycle p ∈ A such that for each n ≥ p n (cid:54)∈ R by Lemma 4.44. Whence there is some N ≥ n ≥ p n σ N ∈ ¯ ψ ( Z )by Proposition 4.30.2. Therefore p n σ N ∈ R by Theorem 4.27. Thus there is an ascending chain of ideals of R ,(50) (cid:0) pσ N (cid:1) ⊆ (cid:0) pσ N , p σ N (cid:1) ⊆ (cid:0) pσ N , p σ N , p σ N (cid:1) ⊆ · · · Assume to the contrary that for some (cid:96) ≥
2, there are elements g , . . . , g (cid:96) − ∈ R such that p (cid:96) σ N = (cid:96) − (cid:88) n =1 g n p n σ N . Then since R is a domain, p (cid:96) = (cid:96) − (cid:88) n =1 g n p n . Furthermore, since R is generated by monomials in the polynomial ring B , there issome 1 ≤ n ≤ (cid:96) − g n is a monomial and p (cid:96) = g n p n . Whence g n = p (cid:96) − n , again since R is a domain. But then p (cid:96) − n is in R , a contradiction.Thus each inclusion in the chain (50) is proper. Therefore R is nonnoetherian, provingour claim.(ii) We now claim that A and Z are nonnoetherian. Again consider the elements(49) in τ ( Z ). Denote by z n ∈ Z the central element whose ¯ ψ -image is p n σ N . Considerthe ascending chain of (two-sided) ideals of A ,(51) (cid:104) z (cid:105) ⊆ (cid:104) z , z (cid:105) ⊆ (cid:104) z , z , z (cid:105) ⊆ · · · . Assume to the contrary that for some (cid:96) ≥
2, there are elements a , . . . , a (cid:96) − ∈ A such that(52) z (cid:96) = (cid:96) − (cid:88) n =1 a n z n . Since p (cid:54)∈ R , there is a vertex i ∈ Q such that p is not in ¯ τ ψ ( e i Ae i ). From (52) weobtain p (cid:96) σ N = ¯ τ ψ ( z (cid:96) e i ) = (cid:88) ¯ τ ψ ( a n e i ) p n σ N . Whence(53) p (cid:96) = (cid:88) ¯ τ ψ ( a n e i ) p n . Furthermore, since each z n is central, (cid:88) ( e i a n )( z n e i ) = e i z (cid:96) = z (cid:96) e i = (cid:88) ( a n e i )( z n e i ) . Thus each product a n e i is in the corner ring e i Ae i . Therefore each image ¯ τ ψ ( a n e i ) isin ¯ τ ψ ( e i Ae i ). In particular, ¯ τ ψ ( a n e i ) cannot equal p (cid:96) − n for any (cid:96) − n ≥ i . But R is generated by monomials in the polynomial ring B . It followsthat (53) cannot hold. Thus each inclusion in the chain (51) is proper. Therefore A is nonnoetherian. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 57
Since the elements z , z , z , . . . are in Z , we may consider the ascending chain ofideals of Z , ( z ) ⊆ ( z , z ) ⊆ ( z , z , z ) ⊆ · · · . A similar argument then shows that Z is also nonnoetherian.(iii) Finally, we claim that ˆ Z is nonnoetherian. For each i ∈ Q , there are no cyclesin (nil Z ) e i by Theorem 4.24. Thus the cycle z n e i is not in (nil Z ) e i . Whence z n isnot in nil Z . The claim then follows similar to Claim (ii). (cid:3) Although ˆ Z and R are nonnoetherian, we will show that they each have Krulldimension 3 and are generically noetherian (Theorem 4.66). Furthermore, we willshow that the homotopy algebra ˜ A of A is also nonnoetherian (Theorem 4.65). Theorem 4.46.
Let A be a non-cancellative dimer algebra that admits a cyclic con-traction. Then A is an infinitely generated Z -module.Proof. Let ψ : A → A (cid:48) be a cyclic contraction. Assume to the contrary that thereare elements a , . . . , a N ∈ A such that A = N (cid:88) n =1 Za n . It suffices to suppose that each a n is in some corner ring e j Ae i , for otherwise we couldinstead consider the finite set { e j a n e i | i, j ∈ Q , ≤ n ≤ N } as a generating set for A as a Z -module. For each cycle p ∈ C , there is thus a subset J p ⊆ { , . . . , N } and nonzero central elements z n ∈ Z such that(54) p = (cid:88) n ∈ J p z n a n . Therefore by Theorem 4.27,(55) p = (cid:88) n ∈ J p ¯ ψ ( z n ) a n ∈ (cid:88) n ∈ J p Ra n . Set J := (cid:91) p ∈C J p ⊆ { , . . . , N } . Then (55) implies that(56) S ⊆ (cid:88) n ∈ J Ra n . Again consider (54), and suppose n ∈ J p . Then a n is in e i Ae i since p is in e i Ae i and z n is central. Whence a n is in S . Thus S ⊇ (cid:88) n ∈ J Ra n . Therefore, together with (56), we obtain(57) S = (cid:88) n ∈ J Ra n . In particular, S is a finitely generated R -module.But R is an infinitely generated k -algebra by Theorem 4.45. Furthermore, S is afinitely generated k -algebra by Theorems 3.3 and 3.5. Therefore S is an infinitelygenerated R -module by the Artin-Tate lemma, in contradiction to (57). (cid:3) We will show that the homotopy algebra ˜ A is also an infinitely generated moduleover its center R (Theorem 4.65).Following Theorem 4.46, an immediate question is whether non-cancellative dimeralgebras satisfy a polynomial identity. Proposition 4.47.
Suppose ψ : A → A (cid:48) is a cyclic contraction. If the head (or tail)of each arrow in Q ∗ has indegree 1, then A contains a free subalgebra. In particular, A is not PI.Proof. Let a ∈ Q ∗ ; then the indegree of h( a ) is 1 by assumption. Consider the paths p, q and the path b of maximal length such that bap and baq are unit cycles. Let b (cid:48) ∈ Q (cid:48) be a leftmost arrow subpath of b . Since b has maximal length, the indegree ofh( b (cid:48) ) = h( b ) is at least 2. In particular, b (cid:48) is not contracted. Furthermore, no vertexin Q (cid:48) has indegree 1 since A (cid:48) is cancellative. Whence ψ ( bap ) = b (cid:48) p and ψ ( baq ) = b (cid:48) q are unit cycles in Q (cid:48) . Therefore p, q is a non-cancellative pair and p = q .Since A (cid:48) is cancellative, there is a simple matching D ∈ S (cid:48) which contains b (cid:48) , byTheorem 4.41. Furthermore, since D is a simple matching, there is a path s in Q (cid:48) from h( p ) to t( p ) which is supported on Q (cid:48) \ D . In particular, x D (cid:45) ¯ τ ( sp ) = ¯ τ ( sq ) . Since the indegree of the head of each contracted arrow is 1, ψ is surjective. Thusthere is path r in Q from h( p ) to t( p ) satisfying ψ ( r ) = s . Whence x D (cid:45) ¯ τ ( sp ) = rp = rq. Thanks to Paul Smith and Toby Stafford for inquiring whether non-cancellative dimer algebras arePI.
ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 59
But then b is not a subpath of rp or rq (modulo I ) since x D | b (cid:48) . Consequently, thereare no relations between the cycles rp and rq . Therefore k (cid:104) rp, rq (cid:105) is a free subalgebra of A . (cid:3) Example 4.48.
Consider the dimer algebras A and A with quivers Q and Q given in Figures 15.i.a and 15.ii.a respectively. Both A and A have free subalgebras,although only A satisfies the hypotheses of Proposition 4.47. Indeed, let a and b bethe red and blue arrows in Q . Then k (cid:104) a, b (cid:105) is a free subalgebra of A . Question 4.49.
Is there a non-cancellative dimer algebra that satisfies a polynomialidentity?
Theorem 4.50.
Let A be a dimer algebra which admits a cyclic contraction. Thenthe following are equivalent:(1) A is cancellative.(2) A is noetherian.(3) Z is noetherian.(4) A is a finitely generated Z -module.(5) The vertex corner rings e i Ae i are pairwise isomorphic algebras.(6) Each vertex corner ring e i Ae i is isomorphic to Z .Proof. First suppose A is cancellative. Then Z is noetherian and A is a finitelygenerated Z -module by Theorem 3.3.3. Therefore A is noetherian. Furthermore, thevertex corner rings are pairwise isomorphic and isomorphic to Z by Theorem 3.3.1.Conversely, suppose A is non-cancellative. Then A and Z are nonnoetherian byTheorem 4.45; A is an infinitely generated Z -module by Theorem 4.46; and the vertexcorner rings are not all isomorphic by Theorem 4.41. (cid:3) Notation 4.51.
In the remainder of this section, we will identify ˆ Z with its isomor-phic ¯ ψ -image in R (Theorem 4.27), and thus write ˆ Z ⊆ R . Lemma 4.52.
The cycle algebra S is a finitely generated k -algebra and a normaldomain of Krull dimension 3. Furthermore, Max S is a toric Gorenstein singularity.Proof. Since A (cid:48) is a cancellative dimer algebra, it is well known that its center Z (cid:48) is afinitely generated k -algebra (Theorem 3.3.3), and a normal toric Gorenstein domain(Corollary 4.28) of Krull dimension 3. Furthermore, Z (cid:48) (i) ∼ = S (cid:48) (ii) = S, where ( i ) holds by Theorem 3.5, and ( ii ) holds by our assumption that ψ is cyclic. (cid:3) Lemma 4.53.
The morphisms (58) κ Z : Max S → Max ˆ Z, n → n ∩ ˆ Z,κ R : Max S → Max R, n → n ∩ R, and ι Z : Spec S → Spec ˆ Z, q → q ∩ ˆ Z,ι R : Spec S → Spec R, q → q ∩ R, are well-defined and surjective.Proof. Let n ∈ Max S . By Lemma 4.52, S is a finitely generated k -algebra, and byassumption k is an algebraically closed field. Therefore the intersections n ∩ ˆ Z and n ∩ R are maximal ideals of ˆ Z and R respectively.Surjectivity of κ Z (resp. κ R ) follows from Claim (iii) in the proof of Theorem 3.5,with S in place of B , and ˆ Z (resp. R ) in place of τ ( Z ).By assumption k is also uncountable. Surjectivity of ι Z (resp. ι R ) then follows fromthe surjectivity of κ Z (resp. κ R ), by [B2, Lemma 2.15]. (cid:3) Lemma 4.54. If p ∈ Spec ˆ Z contains a monomial, then p contains σ .Proof. Suppose p contains a monomial g . Since p is a proper ideal, g is not in k .Thus there is a non-vertex cycle p such that p = g .Let q + be a path from h( p + ) to t( p + ). Then ( pq ) + is a cycle in Q + . Thus there issome n ≥ pq = σ n by Lemma 4.11.1.By Lemma 4.53, there is a prime ideal q ∈ Max S such that q ∩ R = p . Then pq = σ n is in n since q ∈ S and p = g ∈ p . Thus σ is also in q since q is a prime ideal.Therefore by Lemma 1.5, σ ∈ q ∩ ˆ Z = p . (cid:3) Denote the origin of Max S by n := ( s ∈ S | s a non-vertex cycle) S ∈ Max S. Consider the maximal ideals of ˆ Z and R respectively, z := n ∩ ˆ Z and m := n ∩ R. Lemma 4.55.
The localizations ˆ Z z and R m are nonnoetherian.Proof. Let p ∈ S \ ˆ Z be as in Claim (i) in the proof of Theorem 4.45. Since ˆ Z isgenerated by monomials in the polynomial ring B , the monomial p n is not in thelocalization ˆ Z z for any n ≥
1. Whence the chain (50) does not terminate in ˆ Z z .Therefore ˆ Z z is nonnoetherian. Similarly R m is nonnoetherian. (cid:3) ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 612 1 21 2 1 · · ·· ·· · (cid:111) (cid:111) (cid:111) (cid:111) (cid:15) (cid:15) (cid:47) (cid:47) (cid:47) (cid:47) (cid:79) (cid:79) (cid:47) (cid:47) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) (cid:111) (cid:111) (cid:15) (cid:15) (cid:15) (cid:15) (cid:79) (cid:79) ψ −→ (cid:111) (cid:111) (cid:15) (cid:15) (cid:47) (cid:47) (cid:79) (cid:79) (cid:47) (cid:47) (cid:15) (cid:15) (cid:111) (cid:111) Figure 27.
An example where σ divides all non-constant monomialsin R . The quivers are drawn on a torus, and the contracted arrows aredrawn in green. Lemma 4.56.
Suppose that each non-constant monomial in ˆ Z is divisible (in B ) by σ . If p ∈ Spec ˆ Z contains a monomial, then p = z .Proof. Suppose p ∈ Spec ˆ Z contains a monomial. Then σ is in p by Lemma 4.54.Furthermore, there is some q ∈ Spec S such that q ∩ ˆ Z = p by Lemma 4.53.Suppose f is a non-constant monomial in ˆ Z . Then by assumption, there is amonomial h in B such that f = σh . By Lemma 4.29, h is also in S . Therefore f = σh ∈ q since σ ∈ p ⊆ q . But f ∈ ˆ Z . Whence f ∈ q ∩ ˆ Z = p . Since f was arbitrary, p contains all non-constant monomials in ˆ Z . (cid:3) Remark 4.57.
In Lemma 4.56, we assumed σ divides all non-constant monomialsin ˆ Z . An example of a dimer algebra with this property is the dimer algebra withquiver given in Figure 27. In this example, R = k + σS and S = k [ xz, xw, yz, yw ](with impression given by the arrow orientations in Figure 14). Lemma 4.58.
Suppose that there is a non-constant monomial in ˆ Z which is notdivisible (in B ) by σ . Let m ∈ Max ˆ Z \ { z } . Then there is a non-constant monomial g ∈ ˆ Z such that g (cid:54)∈ m and σ (cid:45) g. Proof.
Let m ∈ Max ˆ Z \ { z } .(i) We first claim that there is a non-constant monomial in ˆ Z which is not in m .Assume otherwise. Then n ∩ ˆ Z ⊆ m . But n ∩ ˆ Z is a maximal ideal by Lemma 4.53. Thus z = n ∩ ˆ Z = m , a contradiction.(ii) We now claim that there is a non-constant monomial in ˆ Z \ m which is notdivisible by σ . Assume otherwise; that is, assume that every non-constant monomialin ˆ Z , which is not divisible by σ , is in m . By assumption, σ does not divide allnon-constant monomials in ˆ Z . Thus there is at least one monomial in m . Therefore σ is in m by Lemma 4.54. By Lemma 4.53, there is an n ∈ Max S such that n ∩ ˆ Z = m . Then σ ∈ n since σ ∈ m . Suppose σ divides the monomial g ∈ ˆ Z ; say g = σh for some monomial h ∈ B . Then h ∈ S by Lemma 4.29. Thus g = σh ∈ n . Whence g ∈ n ∩ ˆ Z = m . Thus every non-constant monomial in ˆ Z , which is divisible by σ , is also in m . There-fore every non-constant monomial in ˆ Z is in m . But this contradicts our choice of m by Claim (i). (cid:3) Recall the subsets (9) of Max S and the morphisms (58). For brevity, we will write U Z , U ∗ Z , U R , U ∗ R for the respective subsets U ˆ Z,S , U ∗ ˆ Z,S , U R,S , U ∗ R,S . Furthermore, wewill denote their respective complements with a superscript c . Proposition 4.59.
Let n ∈ Max S . Then (59) n ∩ ˆ Z (cid:54) = z if and only if ˆ Z n ∩ ˆ Z = S n . Consequently, κ Z ( U Z ) = Max ˆ Z \ { z } . Furthermore, κ R ( U R ) = Max R \ { m } . Proof. (i) Let n ∈ Max S and set m := n ∩ ˆ Z . We first want to show that if m (cid:54) = z ,then ˆ Z m = S n .Consider g ∈ S \ ˆ Z . It suffices to show that g is in the localization ˆ Z m . But S isgenerated by σ and a set of monomials in B not divisible by σ , by Proposition 3.10.Furthermore, σ is in ˆ Z by Lemma 1.5. Thus it suffices to suppose g is a non-constantmonomial which is not divisible by σ . Let u ∈ Z and p ∈ C u be such that p = g .We claim that u (cid:54) = 0. Indeed, suppose otherwise. Then p + is a cycle in Q + . Thus p = σ n for some n ≥ σ n is in ˆ Z . Consequently, p = g is inˆ Z , contrary to our choice of g . Therefore u (cid:54) = 0.(i.a) First suppose σ does not divide all non-constant monomials in ˆ Z . Fix i ∈ Q .By Lemma 4.58, there is a non-vertex cycle q ∈ e i Ae i such that q ∈ ˆ Z \ m and σ (cid:45) q. Let v ∈ Z be such that q ∈ C v . Then v (cid:54) = 0 since σ (cid:45) q .We claim that u (cid:54) = v . Assume to the contrary that u = v . Then by Lemma 2.3.1, p = q since σ (cid:45) p and σ (cid:45) q . But q is in ˆ Z , whereas p is not, a contradiction. Therefore u (cid:54) = v .Since u (cid:54) = v are nonzero, the paths p + and q + are transverse in Q + . Thus p + and q + intersect at a vertex j + . Therefore p and q factor into the product of paths p = p e j p and q = q e j q . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 63
Consider the cycle r := q p p q ∈ e i Ae i . Let (cid:96) ≥ σ (cid:96) | r and σ (cid:96) +1 (cid:45) r . Since σ (cid:45) p and σ (cid:45) q , there is a cyclicsubpath r +1 of ( p p q ) + that is not a subpath of ( p p ) + or q +1 , or a cyclic subpath r +2 of ( q p p ) + that is not a subpath of q +2 or ( p p ) + , such that r = σ m and r = σ m , where m + m = (cid:96) . Consider the cycle t obtained from r by removing the cyclicsubpaths r and r , modulo I . Then σ (cid:45) t .Since i ∈ Q was arbitrary, we have t ∈ R . But σ (cid:45) t . Thus by Proposition 4.30.1., t ∈ ˆ Z. Therefore, since q ∈ ˆ Z \ m , g = p = rq − = σ (cid:96) tq − ∈ ˆ Z m . But g was an arbitrary non-constant monomial. Thus, since S is generated by mono-mials, S ⊆ ˆ Z m . Therefore S n = ˆ Z m . (i.b) Now suppose σ divides all non-constant monomials in ˆ Z . Further suppose m (cid:54) = z . Then by Lemma 4.56, m does not contain any monomials. In particular, σ (cid:54)∈ m . By Lemma 4.30.2, there is an N ≥ gσ N ∈ ˆ Z . Thus g = ( gσ N ) σ − N ∈ ˆ Z m . But g was an arbitrary non-constant monomial. Therefore S ⊆ ˆ Z m . It follows that S n = ˆ Z m . Therefore, in either case (a) or (b), Claim (i) holds.(ii) Now suppose n ∈ Max S satisfies n ∩ R (cid:54) = m . We claim that R n ∩ R = S n .Since n ∩ R (cid:54) = m , there is a monomial g ∈ S \ n . By Proposition 4.30.3, there issome N ≥ g N ∈ ˆ Z . But g N (cid:54)∈ n since n is a prime ideal. Thus g N ∈ ˆ Z \ ( n ∩ ˆ Z ) . Whence n ∩ ˆ Z (cid:54) = z . Denote by ˜ m := m ˆ Z m the maximal ideal of ˆ Z m . Then, since ˆ Z ⊂ S ,ˆ Z m = ˆ Z ˜ m ∩ ˆ Z ⊆ S ˜ m ∩ S ⊆ ( ˆ Z m ) ˜ m ∩ ˆ Z m = ˆ Z m . Therefore S n (i) = ˆ Z n ∩ ˆ Z (ii) ⊆ R n ∩ R ⊆ S n , where ( i ) holds by Claim (i), and ( ii ) follows from Theorem 4.27. Consequently R n ∩ R = S n , proving our claim.(iii) Finally, we claim thatˆ Z z (cid:54) = S n and R m (cid:54) = S n . These inequalities hold since the local algebras ˆ Z z and R m are nonnoetherian byLemma 4.55, whereas S n is noetherian by Lemma 4.52. (cid:3) Lemma 4.60.
Let q and q (cid:48) be prime ideals of S . Then q ∩ ˆ Z = q (cid:48) ∩ ˆ Z if and only if q ∩ R = q (cid:48) ∩ R. Proof. (i) Suppose q ∩ ˆ Z = q (cid:48) ∩ ˆ Z , and let s ∈ n ∩ R . Then s ∈ R . Whence there issome n ≥ s n ∈ ˆ Z by Lemma 4.30.3. Thus s n ∈ q ∩ ˆ Z = q (cid:48) ∩ ˆ Z. Therefore s n ∈ n (cid:48) . Thus s ∈ q (cid:48) since q (cid:48) is prime. Consequently s ∈ q (cid:48) ∩ R . Therefore q ∩ R ⊆ q (cid:48) ∩ R . Similarly q ∩ R ⊇ q (cid:48) ∩ R .(ii) Now suppose q ∩ R = q (cid:48) ∩ R , and let s ∈ q ∩ ˆ Z . Then s ∈ ˆ Z ⊆ R . Thus s ∈ q ∩ R = q (cid:48) ∩ R. Whence s ∈ q (cid:48) ∩ ˆ Z . Therefore q ∩ ˆ Z ⊆ q (cid:48) ∩ ˆ Z . Similarly q ∩ ˆ Z ⊇ q (cid:48) ∩ ˆ Z . (cid:3) Proposition 4.61.
The subsets U Z and U R of Max S coincide, U Z = U R . Proof. (i) We first claim that U Z ⊆ U R . Indeed, suppose n ∈ U Z . Then since ˆ Z ⊆ R ⊂ S , S n = ˆ Z n ∩ ˆ Z ⊆ R n ∩ R ⊆ S n . Thus R n ∩ R = S n . Therefore n ∈ U R , proving our claim.(ii) We now claim that U R ⊆ U Z . Let n ∈ U R . Then R n ∩ R = S n . Thus by Proposition 4.59, n ∩ R (cid:54) = n ∩ R. Therefore by Lemma 4.60, n ∩ ˆ Z (cid:54) = n ∩ ˆ Z. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 65
But then again by Proposition 4.59, ˆ Z n ∩ ˆ Z = S n . Whence n ∈ U Z , proving our claim. (cid:3) Definition 4.62.
We say an integral domain R is generically noetherian if there isan open dense set W ⊂ Max R such that for each m ∈ W , the localization R m isnoetherian. Theorem 4.63.
The following subsets of
Max S are nonempty and coincide: (60) U ∗ Z = U Z = U ∗ R = U R = κ − Z (Max ˆ Z \ { z } ) = κ − R (Max R \ { m } ) . In particular, ˆ Z and R are isolated nonnoetherian singularities and generically noe-therian.Proof. The equalities (60) hold since: • By Proposition 4.61, U R = U Z . • By (59) in Proposition 4.59, U Z = κ − Z (cid:16) Max ˆ Z \ { z } (cid:17) and U R = κ − R (Max R \ { m } ) . • By Lemma 4.52, the localization S n is noetherian for each n ∈ Max S . There-fore again by (59), U ∗ Z = U Z and U ∗ R = U R . Furthermore, κ − R (Max R \ { m } ) is nonempty since there is a maximal ideal of R distinct from m , and κ R is surjective by Lemma 4.53. (cid:3) Proposition 4.64.
The locus U R ⊂ Max S is an open set.Proof. We claim that the complement of U R ⊂ Max S is the closed subvariety U cR = { n ∈ Max S | n ⊇ m S } =: Z ( m S ) . Indeed, let n ∈ Max S . First suppose m S ⊆ n . Then(61) m ⊆ m S ∩ R ⊆ n ∩ R. Whence n ∩ R = m since m is a maximal ideal of R . Thus n (cid:54)∈ U R by Theorem4.63. Therefore U cR ⊇ Z ( m S ).Conversely, suppose n (cid:54)∈ U R . Assume to the contrary that m S ⊆ n . Then (61)implies that n ∩ R = m , contrary to Theorem 4.63. Therefore U cR ⊆ Z ( m S ). (cid:3) Theorem 4.65.
The homotopy algebra ˜ A of A is nonnoetherian and an infinitelygenerated module over its center R .Proof. The following hold: • (˜ τ , B ) is an impression of ˜ A by Theorem 4.35. • U ∗ R = U R by Theorem 4.63. • R (cid:54) = S by Theorem 4.41.Therefore ˜ A is nonnoetherian and an infinitely generated R -module by [B2, Theorem3.2.2]. (cid:3) Theorem 4.66.
Let A be a non-cancellative dimer algebra and ψ : A → A (cid:48) a cycliccontraction. Then the center Z , reduced center ˆ Z , and homotopy center R each haveKrull dimension 3, dim Z = dim ˆ Z = dim R = dim S = 3 . Furthermore, the fraction fields of ˆ Z , R , and S coincide, (62) Frac ˆ Z = Frac R = Frac S. Proof.
Recall that ˆ Z , R , and S are domains by Corollary 4.28. Furthermore, thesubsets U Z and U R of Max S are nonempty by Theorem 4.63.Since U Z and U R are nonempty, the equalities (62) hold by [B2, Lemma 2.4].Furthermore, dim ˆ Z = dim S = dim R by [B2, Theorem 2.5.4]. Therefore ˆ Z and R each have Krull dimension 3 by Lemma4.52. Finally, let p ∈ Spec Z . Then nil Z ⊆ p . Therefore dim Z = dim ˆ Z . (cid:3) Lemma 4.67.
Let g, h ∈ S be non-constant monomials such that σ (cid:45) gh . If gh (cid:54)∈ R ,then g (cid:54)∈ R and h (cid:54)∈ R .Proof. Suppose the hypotheses hold, and assume to the contrary that g ∈ R . Fix i ∈ Q . Since g is a non-constant monomial in R and h is a non-constant monomialin S , there is some u, v ∈ Z and cycles p ∈ C ui and q ∈ C v such that p = g and q = h. By assumption σ (cid:45) gh = pq . Thus σ (cid:45) p and σ (cid:45) q . Therefore u (cid:54) = 0 and v (cid:54) = 0 byLemma 4.11.1.If u = v , then p = q by Lemma 2.18. Whence q ∈ R since p ∈ R . But then pq ∈ R ,contrary to assumption. Therefore u (cid:54) = v . It follows that the lifts p + and q + aretransverse cycles in Q + . Thus there is a vertex j + ∈ Q +0 where p + and q + intersect.We may therefore write p and q as products of cycles p = p e j p and q = q e j q . Consider the cycle r = p q q p ∈ e i Ae i . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 67
Then r = pq = gh. Therefore, since i ∈ Q was arbitrary, r = gh is in R , a contradiction. (cid:3) Recall that the reduction X red of a scheme X , that is, its reduced induced schemestructure, is the closed subspace of X associated to the sheaf of ideals I , where foreach open set U ⊂ X , I ( U ) := { f ∈ O X ( U ) | f ( p ) = 0 for all p ∈ U } .X red is the unique reduced scheme whose underlying topological space equals that of X . If R := O X ( X ), then O X red ( X red ) = R/ nil R , where nil R is the nilradical of R (that is, the radical of the zero ideal of R ). Theorem 4.68.
Let A be a non-cancellative dimer algebra, ψ : A → A (cid:48) a cycliccontraction, and ˜ A the homotopy algebra of A .(1) The reduced center ˆ Z and homotopy center R of A are both depicted by thecenter Z (cid:48) ∼ = S of A (cid:48) .(2) The reduced induced scheme structure of Spec Z and the scheme Spec R arebirational to the noetherian scheme Spec S , and each contain precisely oneclosed point of positive geometric dimension.(3) The spectra Max ˆ Z and Max R may both be viewed as the Gorenstein algebraicvariety Max S with the subvariety U cR identified as a single ‘smeared-out’ point.Proof. (1) We first claim that ˆ Z and R are depicted by S . By Theorem 4.63, U ∗ Z = U Z (cid:54) = ∅ and U ∗ R = U R (cid:54) = ∅ . Furthermore, by Lemma 4.53, the morphisms ι Z and ι R are surjective. (2.i) Claim (1) and [B2, Theorem 2.5.3] together imply that the schemes Spec ˆ Z andSpec R are birational to Spec S , and are isomorphic on U Z = U R . By Lemma 4.52, S is a normal toric Gorenstein domain. By Theorem 4.63, Max ˆ Z and Max R eachcontain precisely one point where the localizations of ˆ Z and R are nonnoetherian,namely z and m .(2.ii) We claim that the closed points z ∈ Spec ˆ Z and m ∈ Spec R have positivegeometric dimension.Indeed, since A is non-cancellative, there is a cycle p such that σ (cid:45) p and p n ∈ S \ R for each n ≥
1, by Lemma 4.44. In particular, p is not a product p = gh , where g ∈ R or h ∈ R , by Lemma 4.67. Therefore p (cid:54)∈ m S. Thus for each c ∈ k , there is a maximal ideal n c ∈ Max S such that( p − c, m ) S ⊆ n c . The fact that S is a depiction of R also follows from [B2, Theorem 3.2.1], since (˜ τ , B ) is animpression of ˜ A by Theorem 4.35. Consequently, m ⊆ ( p − c, m ) S ∩ R ⊆ n c ∩ R. Whence n c ∩ R = m since m is maximal. Therefore by Theorem 4.63, n c ∈ U cR . Set q := (cid:92) c ∈ k n c . Then q is a radical ideal since it is the intersection of radical ideals. Thus, since S is noetherian, the Lasker-Noether theorem implies that there are minimal primes q , . . . , q (cid:96) ∈ Spec S over q such that q = q ∩ · · · ∩ q (cid:96) . Since (cid:96) < ∞ , at least one q i is a non-maximal prime, say q . Then m = (cid:92) c ∈ k ( n c ∩ R ) = (cid:92) c ∈ k n c ∩ R = q ∩ R ⊆ q ∩ R. Whence q ∩ R = m since m is maximal.Since q is a non-maximal prime ideal of S ,ht( q ) < dim S. Furthermore, S is a depiction of R by Claim (1). Thusght( m ) ≤ ht( q ) < dim S (i) = dim R, where ( i ) holds by Theorem 4.66. Thereforegdim m = dim R − ght( m ) ≥ , proving our claim.(3) Follows from Claims (1), (2), and Theorem 4.63. The locus U cR is a (closed)subvariety by Proposition 4.64. (cid:3) Remark 4.69.
Although ˆ Z and R determine the same nonlocal variety using depic-tions, their associated affine schemes (cid:16) Spec ˆ Z, O ˆ Z (cid:17) and (Spec R, O R )will not be isomorphic if their rings of global sections, ˆ Z and R , are not isomorphic. ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 69
Integral closure.
It is well known that the center of a cancellative dimeralgebra is normal. In this section, we characterize the central normality of non-cancellative dimer algebras.Throughout, A is a non-cancellative dimer algebra and ψ : A → A (cid:48) is a cycliccontraction. Let ˜ A and R = Z ( ˜ A ) be the homotopy algebra and homotopy center of A , respectively. We denote by ˆ Z c and R c the respective integral closures of ˆ Z and R . Proposition 4.70.
The homotopy center R is normal if and only if σS ⊂ R .Proof. (1) First suppose σS ⊂ R .(1.i) By Lemma 4.52, S is normal. Therefore, since R is a subalgebra of S ,(63) R c ⊆ S. (1.ii) Now let s ∈ S \ R . We claim that s is not in R c . Indeed, assume otherwise.Since S is generated by monomials in the polynomial ring B , there are monomials s , . . . , s (cid:96) ∈ S such that s = s + · · · + s (cid:96) . Since s (cid:54)∈ R , there is some 1 ≤ k ≤ (cid:96) such that s k (cid:54)∈ R . Choose s k to have maximaldegree among the subset of monomials in { s , . . . , s (cid:96) } which are not in R .By assumption, σS ⊂ R . Thus σ (cid:45) s k . Since s ∈ R c , there is some n ≥ r , . . . , r n − ∈ R such that(64) s n + r n − s n − + · · · + r s = − r ∈ R. By Lemma 4.44, the summand s nk of s n is not in R since σ (cid:45) s k . Thus − s nk is asummand of the left-hand side of (64). In particular, for some 1 ≤ m ≤ n , there aremonomial summands r (cid:48) of r m and s (cid:48) = s j · · · s j m of s m , and a nonzero scalar c ∈ k ,such that r (cid:48) s (cid:48) = cs nk . By Lemma 4.67, r (cid:48) is a nonzero scalar since r (cid:48) ∈ R , s nk (cid:54)∈ R , and σ (cid:45) s nk . Further-more, s (cid:48) is a non-constant monomial since r (cid:48) ∈ R and s nk (cid:54)∈ R . Therefore s j · · · s j m = s (cid:48) = ( c/r (cid:48) ) s nk . By Lemma 4.67, each monomial factor s j , . . . , s j m is not in R . But the monomial s k was chosen to have maximal degree, a contradiction. Therefore(65) R c ∩ S = R. It follows from (63) and (65) that R c = R c ∩ S = R. (2) Now suppose σS (cid:54)⊂ R . Then there are monomials s ∈ S \ R and t ∈ S such that s = tσ . Let n ≥ n unit cycles σ ni ∈ A isequal (modulo I ) to a cycle that contains each vertex in Q . Then σ n S ⊂ R. In particular, the product s n = t n σ n is in R . But then s ∈ Frac S \ R is a root of themonic polynomial x n − s n ∈ R [ x ] . Thus s is in R c \ R . Therefore R is not normal. (cid:3) Corollary 4.71. (1) If the head or tail of each contracted arrow has indegree 1, then R is normal.(2) If ψ contracts precisely one arrow, then R is normal.Proof. In both cases (1) and (2), clearly σS ⊂ R . (cid:3) Proposition 4.72.
For each n ≥ , there are dimer algebras for which σ n S (cid:54)⊂ R and σ n +1 S ⊂ R. Consequently, there are dimer algebras for which R is not normal.Proof. Recall the conifold quiver Q with one nested square given in Figure 21.i.Clearly σS ⊂ R . More generally, the conifold quiver with n ≥ σ n − S (cid:54)⊂ R and σ n S ⊂ R. The corresponding homotopy center R is therefore not normal for n ≥ (cid:3) Let ˜ m ⊂ m be the ideal of R generated by all non-constant monomials in R whichare not powers of σ . Proposition 4.73.
Let n ≥ , and suppose σ n S (cid:54)⊂ R and σ n +1 S ⊂ R . Then (66) R = k [ σ ] + ( ˜ m , σ n +1 ) S. Proof.
Suppose p ∈ C u is a non-vertex cycle such that for each n ≥ p (cid:54) = σ n . Then u (cid:54) = 0 by Lemma 4.11.1. The equality (66) then follows similar to the proof ofLemma 4.67. (cid:3) Theorem 4.74.
The following are equivalent:(1) R is normal.(2) σS ⊂ R .(3) R = k + m S .(4) R = k + J for some ideal J in S . ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 71
Proof.
We have • (1) ⇔ (2) holds by Proposition 4.70. • (2) ⇔ (3) holds by Proposition 4.73. • (3) ⇔ (4) holds again by Proposition 4.73. (cid:3) Theorem 4.75.
The integral closures of the reduced and homotopy centers are iso-morphic, ˆ Z c ∼ = R c . Proof.
For brevity, we identify ˆ Z with its isomorphic ¯ ψ -image in R (Theorem 4.27),and thus write ˆ Z ⊆ R . This inclusion implies(67) ˆ Z c ⊆ R c . To show the reverse inclusion, recall that by Theorem 4.66.3,Frac ˆ Z = Frac R = Frac S. (i) First suppose r ∈ R . Then there is some n ≥ r n ∈ ˆ Z by Proposition4.30.3. Whence r is a root of the monic polynomial x n − r n ∈ ˆ Z [ x ] . Thus r is in ˆ Z c . Therefore(68) R ⊆ ˆ Z c . (ii) Now suppose s ∈ R c \ R . Since R c is generated by certain monomials in S , itsuffices to suppose s is a monomial. Then by Claim (2) in the proof of Theorem 4.70,there is a monomial t ∈ S such that s = tσ . Furthermore, by Proposition 4.30.2,there is some N ≥ m ≥ t m σ N ∈ ˆ Z. In particular, s N = t N σ N ∈ ˆ Z. Whence s is a root of the monic polynomial x N − s N ∈ ˆ Z [ x ] . Thus s is in ˆ Z c . Therefore, together with (68), we obtain(69) R c ⊆ ˆ Z c . The theorem then follows from (67) and (69). (cid:3)
Proposition 4.76.
The integral closures R c ∼ = ˆ Z c are nonnoetherian and properlycontained in the cycle algebra S . Proof.
Since A is non-cancellative, there is some s ∈ S such that σ (cid:45) s and s n (cid:54)∈ R foreach n ≥
1, by Lemma 4.44. In particular, σ (cid:45) s n for each n ≥ σ = (cid:81) D ∈S x D .Thus s is not the root of a monic binomial in R [ x ] by Lemma 4.67. Therefore s (cid:54)∈ R c since R is generated by monomials in the polynomial ring B .Similarly, s m (cid:54)∈ R c for each m ≥
1. It follows that R c is nonnoetherian by Claim(i) in the proof of Theorem 4.45. (cid:3) Appendix A. A brief account of Higgsing with quivers
Quiver gauge theories
According to string theory, our universe is 10 dimensional.
In many string theo-ries our universe has a product structure M × Y , where M is our usual 4-dimensionalspace-time and Y is a 6-dimensional compact Calabi-Yau variety.Let us consider a special class of gauge theories called ‘quiver gauge theories’, whichcan often be realized in string theory. The input for such a theory is a quiver Q , asuperpotential W , a dimension vector d ∈ N Q , and a stability parameter θ ∈ R Q .Let I be the ideal in C Q generated by the partial derivatives of W with respect tothe arrows in Q . These relations (called ‘F-term relations’) are classical equations ofmotion from a supersymmetric Lagrangian with superpotential W . Denote by A the quiver algebra C Q/I .According to these theories, the space X of θ -stable representation isoclasses ofdimension d is an affine chart on the compact Calabi-Yau variety Y . The ‘gaugegroup’ of the theory is the isomorphism group (i.e., change of basis) for representationsof A .Physicists view the elements of A as fields on X . More precisely, A may be viewedas a noncommutative ring of functions on X , where the evaluation of a function f ∈ A at a point p ∈ X (i.e., representation p ) is the matrix f ( p ) := p ( f ) (up toisomorphism). Vacuum expectation values
Given a path f ∈ A and a representation p ∈ X , denote by f (¯ p ) the matrixrepresenting f in the vector space diagram on Q associated to p .A field f ∈ A is ‘gauge-invariant’ if f ( p ) = f ( p (cid:48) ) whenever p and p (cid:48) are isomorphicrepresentations (i.e., they differ by a ‘gauge transformation’). If f is a path, then f will necessarily be a cycle in Q .The ‘vacuum expectation value’ of a field is its expected (average) energy in thevacuum (similar to rest mass), and is abbreviated ‘vev’. In our case, the vev of a Thanks to physicists Francesco Benini, Mike Douglas, Peng Gao, Mauricio Romo, and JamesSparks for discussions on the physics of non-cancellative dimers. More correctly, weakly coupled superstring theory requires 10 dimensions. Here we are considering theories with N = 1 supersymmetry. More correctly, the F-term relations plus the D-term relations imply the equations of motion.
ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 73 path f ∈ A at a point p ∈ X is the matrix f (¯ p ), which is just the expected energyof f in M × { p } . Higgsing
Spontaneous symmetry breaking is a process where the symmetry of a physicalsystem decreases, and a new property (typically mass) emerges.For example, suppose a magnet is heated to a high temperature. Then all of itsmolecules, which are each themselves tiny magnets, jostle and wiggle about randomly.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 theneighboring molecules settle down, they align themselves with the first molecule, untilall the molecules settle down in alignment with the first. The orientation of thefirst settled molecule then determines the direction of magnetization for the wholematerial, and the material no longer has rotational symmetry. One says that therotational symmetry of the heated magnet was spontaneously broken as it cooled,and a global magnetic field emerged. Higgsing is a way of using spontaneous symmetry breaking to turn a quantumfield theory with a massless field and more symmetry into a theory with a massivefield and less symmetry. Here mass (vev’s) takes the place of magnetization, gaugesymmetry (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 anotherexample of Higgsing. Higgsing in quiver gauge theories
We now give our main example. Suppose an arrow a in a quiver gauge theory withdimension 1 Q 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 becomeidentified as the single vertex e ;(2) a has zero vev at any representation where a is represented by zero, while e cannever have zero vev since it is a vertex, and X only consists of representationisoclasses with dimension 1 Q .We therefore see that contracting an arrow to a vertex is a form of Higgsing in quivergauge theories with dimension 1 Q . More precisely, there are domains of magnetization. This is an example of ‘global’ symmetry breaking, meaning the symmetry is physically observable. This is an example ‘gauge’ symmetry breaking, meaning the symmetry is not an actual observablesymmetry of a physical system, but only an artifact of the math used to describe it (like a choice ofbasis for the matrix of a linear transformation). This is another example of gauge symmetry breaking.
In the context of a 4-dimensional N = 1 quiver gauge theory with quiver Q , theHiggsing we consider in this paper is related to RG flow. We start with a non-superconformal (strongly coupled) quiver theory Q which admits a low energy effec-tive description, give nonzero vev’s to a set of bifundamental fields Q ∗ , and obtain anew theory Q (cid:48) that lies at a superconformal fixed point. The mesonic chiral ring and the cycle algebra
The cycle algebra S we introduce in this paper is similar to the mesonic chiral ringin 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. Ifthe gauge group is abelian, then the dimension vector is 1 Q . In the case of a dimertheory with abelian gauge group, two disjoint cycles may share the same ¯ τ ψ -image,but take different values on a point of the vacuum moduli space. These cycles wouldthen be distinct elements in the mesonic chiral ring, although they would be identifiedin the cycle algebra S ; see [B3, Remark 3.17]. Acknowledgments.
Part of this paper is based on work supported by the SimonsFoundation while the author was a postdoc at the Simons Center for Geometry andPhysics at Stony Brook University.
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ONLOCALITY AND THE CENTRAL GEOMETRY OF DIMER ALGEBRAS 75
Heilbronn Institute for Mathematical Research, School of Mathematics, HowardHouse, The University of Bristol, Bristol, BS8 1SN, United Kingdom.
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