A generalization of cancellative dimer algebras to hyperbolic surfaces
aa r X i v : . [ m a t h . R A ] J a n A GENERALIZATION OF CANCELLATIVE DIMER ALGEBRASTO HYPERBOLIC SURFACES
KARIN BAUR AND CHARLIE BEIL
Abstract.
Cancellative dimer algebras on a torus are noncommutative crepantresolutions, and in particular have many nice algebraic and homological proper-ties. All of these properties disappear, however, for dimer algebras on higher genussurfaces. We consider a new class of quiver algebras on surfaces, called ‘geodesicghor algebras’, that reduce to cancellative dimer algebras on a torus, yet continueto possess nice properties on higher genus surfaces. We show that noetherian lo-calizations of these algebras are endomorphism rings of modules over their centers,and establish a rich interplay between their central geometry and the topology ofthe surface in which they are embedded. Introduction
Cancellative dimer algebras on a torus have been extensively studied in the con-texts of noncommutative resolutions, Calabi-Yau algebras, and stability conditions,e.g., [Br, D, B2]. It is well known that every cancellative dimer algebra on a torus is anoncommutative crepant resolution, and every three dimensional affine toric Goren-stein singularity admits a noncommutative crepant resolution given by such a dimeralgebra. However, if the dimer algebra is on a surface of genus g ≥
2, then these niceproperties disappear: the center of such a dimer algebra is simply the polynomialring in one variable, and so there can be no interesting interactions beetween thetopology of the surface and the algebras central geometry and representation theory.In this article we consider special quotients of dimer algebras, called ‘ghor algebras’.A ghor algebra is a quiver algebra whose quiver embeds in a surface, with relationsdetermined by the perfect matchings of its quiver (the precise definition is given inSection 2). Ghor algebras were introduced in [B1, B4] to study nonnoetherian dimeralgebras on a torus. In Section 2 we introduce a special property that certain ghoralgebras possess, called ‘geodesic’. On a torus, a ghor algebra is geodesic if and onlyif it is a cancellative dimer algebra. On higher genus surfaces, certain localizations
Mathematics Subject Classification.
Key words and phrases.
Dimer algebra, hyperbolic surface, non-noetherian ring, noncommutativealgebraic geometry. In [B4], we called ghor algebras ‘homotopy algebras’ because their relations are homotopy relationson the paths in the quiver when the surface is a torus. However, in the higher genus case homologouscycles are also identified (see Theorem 3.11 below), and therefore the name ‘homotopy’ is less suitablefor general surfaces. The word ‘ghor’ is Klingon for surface. of geodesic ghor algebras remain endomorphism rings of modules over their centers,but new features arise. The purpose of this article is to show that geodesic ghoralgebras exhibit a rich interplay between their algebraic properties, central geometry,and the topology of the surface in which they are embedded. Our main theorem isthe following.
Theorem.
Suppose A = kQ/ ker η is a geodesic ghor algebra on a surface Σ obtainedfrom a regular N -gon P by identifying the opposite sides, and vertices, of P . Set R = k [ ∩ i ∈ Q ¯ τ ( e i Ae i )] and S = k [ ∪ i ∈ Q ¯ τ ( e i Ae i )] ; then R is isomorphic to the center of A . Furthermore, the following holds.(1) If there is a cycle p such that p n R for each n ≥ , then A and R arenonnoetherian. In this case, R is depicted by the cycle algebra S .(2) The center R and cycle algebra S have Krull dimension dim R = dim S = N + 1 . In particular, if Σ is a smooth genus g ≥ surface, then dim R = rank H (Σ) + 1 = 2 g + 1 . (3) At each point m ∈ Max R for which the localization R m is noetherian, thelocalization A m := A ⊗ R R m is an endomorphism ring over its center: for each i ∈ Q , we have A m ∼ = End R m ( A m e i ) . The locus of such points lifts to an open dense subset of the algebraic variety
Max S . Preliminary definitions
Notation 2.1.
Throughout, k is an uncountable algebraically closed field. We de-note by Spec S and Max S the prime ideal spectrum (or scheme) and maximal idealspectrum (or affine variety) of S , respectively. Given a quiver Q , we denote by kQ thepath algebra of Q ; by Q ℓ the paths of length ℓ ; by t , h : Q → Q the tail and headmaps; and by e i the idempotent at vertex i ∈ Q . By cyclic subpath of a path p , wemean a subpath of p that is a nontrivial cycle. We denote by [ n ] the set { , , . . . , n } .In this article we consider surfaces Σ that are obtained from a regular convex2 N -gon P , N ≥
2, by identifying the opposite sides, and vertices, of P . This classof surfaces includes all smooth orientable compact closed connected genus g ≥ • if P is a 4 g -gon, then Σ is a smooth genus g surface; and • if P is a 2(2 g + 1)-gon, then Σ is a genus g surface with a pinched point. GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 3 ( i ) ( ii ) ( iii ) Figure 1.
Examples of geodesic ghor algebras. Opposite sides of thepolygons are identified. The ghor algebra (i) is on a pinched torus, andthe ghor algebras (ii) and (iii) are on a smooth genus 2 surface. Thecenters of (i) and (ii) are given explicitly in [BB, Sections 3.2, 3.3]. y y yx x x x x x x x x x x x x x x x x x x x x x yyx x x x yx Figure 2.
The polynomial ghor algebras A = k [ x , x , y ], A = k [ x , x , x , y ], and A = k [ x , x , x , x , y ].The polygon P is then a fundamental polygon for Σ.If N = 2, then Σ is a torus, and the covering space of Σ is the plane R . For N ≥
3, the covering space of Σ is the hyperbolic plane H . The hyperbolic planemay be represented by the interior of the unit disc in R , where straight lines in H are segments of circles that meet the boundary of the disc orthogonally. In thecovering, the hyperbolic plane is tiled with regular 2 N -gons, with 2 N such polygonsmeeting at each vertex. In this case, Σ is said to be a hyperbolic surface. Definition 2.2. • A dimer quiver on Σ is a quiver Q whose underlying graph Q embeds in Σ, such KARIN BAUR AND CHARLIE BEIL
11 11 111 1 122 22 2 22 22 3 33 33 34 44
Figure 3.
A generalization of the conifold dimer algebra on a torus(shown on the left) to geodesic ghor algebras on genus g surfaces. Thegrey polygon region is the fundamental domain P of the surface. Thesegeodesic ghor algebras are noetherian, satisfy R = S and S = P , andare the only known noetherian ghor algebras with more than one vertexin the case g ≥ \ Q is simply connected and bounded by anoriented cycle, called a unit cycle .- A perfect matching of a dimer quiver Q is a set of arrows D ⊂ Q such thateach unit cycle contains precisely one arrow in D .- A perfect matching D is called simple if Q \ D contains a cycle that passesthrough each vertex of Q (equivalently, Q \ D supports a simple kQ -moduleof dimension vector (1 , , . . . , P and S the set of perfect and simple matchings of Q , respectively.We will consider the polynomial rings k [ P ] and k [ S ] generated by these matchings. Throughout, we assume that each arrow of Q is contained in a perfect matching. • Denote by e ij ∈ M n ( k ) the n × n matrix with a 1 in the ij -th slot and zeroselsewhere. Consider the two algebra homomorphisms η : kQ → M | Q | ( k [ P ]) and τ : kQ → M | Q | ( k [ S ])defined on the vertices i ∈ Q and arrows a ∈ Q by η ( e i ) = e ii , η ( a ) = e h( a ) , t( a ) Q x ∈P : x ∋ a x,τ ( e i ) = e ii , τ ( a ) = e h( a ) , t( a ) Q x ∈S : x ∋ a x, and extended multiplicatively and k -linearly to kQ . We call the quotient A := kQ/ ker η the ghor algebra of Q . GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 5 • The dimer algebra of Q is the quotient of kQ by the ideal I = h p − q | ∃ a ∈ Q such that pa, qa are unit cycles i ⊂ kQ, where p, q are paths.A ghor algebra A = kQ/ ker η is the quotient of the dimer algebra kQ/I since I ⊆ ker η : if pa, qa are unit cycles with a ∈ Q , then η ( p ) = e h( p ) , t( p ) Y x ∈P : x a x = η ( q ) . Dimer algebras on non-torus surfaces have been considered in the context of, forexample, cluster categories [BKM, K], Belyi maps [BGH], and gauge theories [FGU,FH].
Notation 2.3.
Let π : Σ + → Σ be the projection from the covering space Σ + (here, R or the hyperbolic plane H ) to the surface Σ. Denote by Q + := π − ( Q ) ⊂ Σ + the(infinite) covering quiver of Q , and by p + the lift of a path p to Q + . Definition 2.4.
We say a homotopy H : Q × [0 , → Σ of the underlying graph Q of Q is dimer-preserving if for all t ∈ [0 , H ( − , t ) : Q → Σ is an embedding, andeach connected component of Σ \ H ( Q, t ) is simply connected and bounded by unitcycle. We will often omit the overline and simply write H : Q × [0 , → Σ.Fix a tiling of the covering space of Σ by fundamental polygons P , and label thesides 1 , , . . . , N of P in counterclockwise order, with indices taken modulo 2 N . Let H : Q × [0 , → Σ be a dimer-preserving homotopy for which all the vertices of H ( Q,
1) lift to vertices that lie in the interior of P , and no arrow intersects a cornervertex v of P :(1) H ( Q , ∩ π ( ∂P ) = ∅ , H ( Q , ∩ π ( v ) = ∅ . We say a path p transversely intersects side k of P with respect to H if p intersects k transversely in H ( Q, c , we define the class of c to be[ c ] := X k ∈ [ N ] ( n k − n k + N )( δ kℓ ) ℓ ∈ Z N , where n k , k ∈ [2 N ], is the number of times c transversely intersects side k of P . If Σ issmooth (that is, if N is even), then [ c ] is the homology class of c in H (Σ) := H (Σ , Z ).We introduce the following special class of ghor algebras which generalizes can-cellative dimer algebras on a torus. Definition 2.5. • A cycle p ∈ A is geodesic if the lift to Q + of each cyclic permutation of eachrepresentative of p does not have a cyclic subpath. • Two cycles are parallel if they do not transversely intersect.
KARIN BAUR AND CHARLIE BEIL • A ghor algebra is geodesic if for each k ∈ [2 N ] there is a geodesic cycle γ k with class [ γ k ] = ( δ kℓ − δ k + N,ℓ ) ℓ ∈ [ N ] ∈ Z N , with indices modulo 2 N , and a set of pairwise parallel geodesic cycles { c i ∈ e i kQe i } i ∈ Q such that c t( γ k ) = γ k . Remark 2.6.
We note that if Σ is hyperbolic, then for fixed k ∈ [ N ], the parallelgeodesic cycles c i will in general be in different homology classes; in particular, [ c i ]need not equal [ γ k ]. However, if Σ is flat (that is, N = 2), then for fixed k ∈ [2], theparallel geodesic cycles c i may be chosen to be in the same homology class. Remark 2.7.
Consider the ghor algebra A with quiver Q on a genus 2 surfacegiven in Figure 1.ii. Label the sides of P by 1 , . . . ,
8, starting with the top side andcontinuing counterclockwise around P . Observe that A is geodesic. However, thereis no geodesic cycle at the vertex in the center of P that intersects both sides 1 and7 transversely. Thus, it is too restrictive to require that there is a geodesic cycle ateach vertex in each homology class of Σ; see also Remark 3.12.For i, j ∈ Q , consider the k -linear maps¯ η : e j kQe i → k [ P ] and ¯ τ : e j kQe i → k [ S ]defined by sending p ∈ e j kQe i to the single nonzero matrix entry of η ( p ) and τ ( p )respectively; that is, η ( p ) = ¯ η ( p ) e ji and τ ( p ) = ¯ τ ( p ) e ji . These maps define multiplicative labelings of the paths of Q . We will often write p for ¯ η ( p ) or ¯ τ ( p ).An important monomial is the ¯ η - and ¯ τ -images of each unit cycle in Q , namely σ P := Y x ∈P x and σ S := Y x ∈S x. We will omit the subscript P or S if it is clear from the context.Since unit cycles σ i are contractible curves on Σ, the topology of Σ is more closelyreflected in the quotient rings(2) k [ P ] / ( σ P −
1) and k [ S ] / ( σ S − . If polynomials g, h are equal in the quotient, that is, if there is an ℓ ∈ Z such that g = hσ ℓ , then we will write g σ = h. Notation 2.8.
Given paths p, q , we write ¯ η ( p ) | ¯ η ( q ) (resp. ¯ τ ( p ) | ¯ τ ( p )) if ¯ η ( p ) divides¯ η ( q ) in k [ P ] (resp. ¯ τ ( p ) divides ¯ τ ( q ) in k [ S ]). GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 7 A bridge from topology to algebra: subdivisions and simplematchings
Let A = kQ/ ker η be a ghor algebra. Lemma 3.1. (i) If p + is a cycle in Q + , that is, p = π ( p + ) is a contractible cycle, then ¯ η ( p ) σ = 1 and ¯ τ ( p ) σ = 1 . (ii) If p + , q + are paths in Q + satisfying (3) t( p + ) = t( q + ) and h( p + ) = h( q + ) , then ¯ η ( p ) σ = ¯ η ( q ) and ¯ τ ( p ) σ = ¯ τ ( q ) . Proof. (i) We proceed by induction on the number of faces contained in the region R p bounded by p + . Factor p into a minimum number of subpaths p = p m · · · p p , where each p j is a subpath of a unit cycle. For each j ∈ [ m ], let r j be the path forwhich r j p j is a unit cycle and r + j lies in R p . Since m is minimum, the concatenation r = r · · · r m − r m is a cycle whose lift r + lies in the region R p .Without loss of generality, we may assume that at least one p j is not a unit cycle,hence at least one r j is not a vertex. Thus, by the induction hypothesis, there is an n ≥ η ( r ) = σ n P . Furthermore, ¯ η ( r )¯ η ( p ) = Y j ¯ η ( r j ) Y j ¯ η ( p j ) = Y j ¯ η ( r j p j ) = σ m P . Therefore ¯ η ( p ) = σ m − n P . Similarly, ¯ τ ( p ) = σ m − n S .(ii) Suppose p + , q + are paths in Q + satisfying (3). Let r + be a path in Q + fromh( p + ) to t( p + ). Then by Claim (i), there is an m, n ≥ η ( r )¯ η ( p ) = ¯ η ( rp ) = σ m P and ¯ η ( r )¯ η ( q ) = ¯ η ( rq ) = σ n P . Whence ¯ η ( p ) = ¯ η ( q ) σ m − n P . Similarly, ¯ τ ( p ) = ¯ τ ( q ) σ m − n S . (cid:3) KARIN BAUR AND CHARLIE BEIL
Lemma 3.2.
In the quotient rings (2), every path p ∈ e j Ae i has an inverse q ∈ e i Ae j : pq σ = qp σ = 1 . We will write p − := q .Proof. Fix a path p , and let q be any path satisfyingt( q + ) = h( p + ) and h( q + ) = t( p + ) . Then ( pq ) + is a cycle in Q + . Consequently, pq = pq σ = 1 by Lemma 3.1.i. (cid:3) In the following we describe an algebraic feature of ghor algebras that is a conse-quence of the curvature of Σ.
Remark 3.3. If A is a geodesic ghor algebra on a torus, and i, j ∈ Q +0 are distinctvertices, then there is always a path p + from i to j in Q + such that σ ∤ p [B1,Proposition 4.20.iii]. However, this implication no longer holds if the surface Σ is notflat. Indeed, in this case there are always distinct vertices i, j ∈ Q +0 for which everypath p + from i to j satisfies p = σ ℓ for some ℓ ≥
1; see [BB, Remark 2.7].
Proposition 3.4.
Suppose p, q are cycles in the same class, [ p ] = [ q ] . Then p σ = q .Proof. Fix a tiling of the covering space by fundamental polygons P , and a dimer-preserving homotopy H ( Q, ֒ → Σ satisfying (1). For a cycle p in Q (not moduloker η ), denote by ord( p ) the sequence of sides of P intersected by the arrow subpathsof p in H ( Q,
1) in order, ord( p ) := ( j (1) , j (2) , . . . , j ( m )) . If p and q both do not intersect the boundary π ( ∂P ), then by Lemma 3.1.ii, p σ = 1 σ = q. So suppose p (and thus q ) intersects π ( ∂P ). Consider the ordering of p and q withrespect to H ( Q, p ) = ( j (1) , j (2) , . . . , j ( m )) and ord( q ) = ( j ′ (1) , j ′ (2) , . . . , j ′ ( n )) . Without loss of generality, we may assume m ≥ n .Factor p into paths p = p ap , where p has minimal length such that a is an arrowthat intersects the boundary π ( ∂P ). Consider the cyclic permutation p ′ := ap p .Then p ′ = p . Furthermore, since p has minimal length, we have ord( p ) = ord( p ′ ).It therefore suffices to assume that the leftmost arrow subpath of p intersects π ( ∂P ).Under this assumption, we may factor p into paths p = a m p m · · · a p a p , where each a k ∈ Q is an arrow that intersects π ( ∂P ), and each p k ∈ Q ≥ is a paththat does not intersect π ( ∂P ). GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 9
Fix i ∈ Q . For each k ∈ [ m ] (modulo m ), consider a path s k − from h( a k − ) to i , and a path t k from i to t( a k ), both of which lift to paths that lie in the interior of P ; see Figure 4. By Lemma 3.1.ii, we have t k s k − σ = p k . Thus,(4) p = m Y k =1 a k p k σ = m Y k =1 a k t k s k − . Set r k := s k a k t k ∈ e i kQe i . Since p and q are in the same class u := [ p ] = [ q ] ∈ Z N and m ≥ n , we can pair off m − n of the m r k cycles, say r ℓ , r ℓ ′ , such that a ℓ and a ℓ ′ intersect the same side of P but in opposite directions. Consequently, the lift of theconcatenation r ℓ r ℓ ′ is a cycle in Q + . Thus r ℓ r ℓ ′ σ = 1, by Lemma 3.1.i. Therefore, upto a factor of σ , we can omit these m − n cycles from p : p σ = m Y k =1 r k σ = n Y κ =1 r k ( κ ) . It thus suffices to suppose m = n .Consider a permutation ϕ of [ m −
1] for which j ( ϕ ( k )) = j ′ ( k ). The cycle(5) r := t m r ϕ ( m − r ϕ ( m − · · · r ϕ (2) r ϕ (1) s m then satisfies(6) ord( r ) = ord( q ) . By (4) and (5), we have(7) r σ = p. Let α be path from t( q ) to t( p ) = t( r ), and let β be a path from t( p ) to t( q ), suchthat the lift ( αβ ) + lies in a single fundamental polygon P . In particular, ( αβ ) + is acycle in Q + . Thus αβ σ = 1, by Lemma 3.1.i. Furthermore, (6) implies thatord( r ) = ord( q ) = ord( αqβ ) . Whence, t(( αqβ ) + ) = t( r + ) and h(( αqβ ) + ) = h( r + ) . Thus(8) r σ = αqβ = qαβ σ = q, by Lemma 3.1.ii. Therefore, by (7) and (8), p σ = r σ = q, which is what we wanted to show. (cid:3) a k − s k − p k t k a k i Figure 4.
Setup for Proposition 3.4. · · · · · · · · · ·· · · · · · · · · · / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ a column subquiver · · · · ·· · · · · · · / / / / / / / / (cid:21) (cid:21) - - / / / / / / I I U U (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ X X ✶✶✶✶✶ (cid:6) (cid:6) ✌✌✌✌✌ m m a pillar subquiver Figure 5.
A column and pillar of a dimer quiver Q . The black interiorarrows are arrows of Q ; the blue and red boundary arrows are paths oflength at least one in Q ; and each interior cycle is a unit cycle of Q . Theleftmost and rightmost brown arrows of the column are identified. Notethat the blue and red bounding paths of the pillar are equal moduloker η .The following definitions connect the topology of the surface Σ with the simplematchings of Q , and thus with the algebraic structure of the ghor algebra A . Definition 3.5.
We call the subquiver given in Figure 5.i a column , and the subquivergiven in Figure 5.ii a pillar . A subdivision of Q is a set F of columns and pillars suchthat each arrow either (i) lies in the interior of at most one column or pillar; or (ii)belongs to the boundary of a column or pillar.The following lemma is immediate. Lemma 3.6.
Suppose A = kQ/ ker η is geodesic. Then for each k ∈ [2 N ] , the set ofparallel geodesic cycles { c i } i ∈ Q containing γ k determines a subdivision of Q . By ‘path’, we always mean a possibly trivial path unless stated otherwise.
GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 11
Figure 6.
The four minimal subdivisions of the geodesic ghor algebrain Figure 1.ii. x x x x y y y y z z z z Figure 7.
The simple matchings of the geodesic ghor algebra in Figure 1.ii.
Lemma 3.7.
Let p, q be paths in Q whose lifts p + , q + have no cyclic subpaths andbound 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 + ) = t( p + ) , then p ≡ q . Figure 8.
The minimal subdivisions and simple matchings of the geo-desic ghor algebra in Figure 1.iii. There are three minimal subdivisions(drawn in green) that each yield four simple matchings (drawn in red).Each minimal subdivision consists only of columns in the respectivedirections horizontal, vertical, and diagonal. Each pair of minimalsubdivisions yield a common simple matching, and so there are a totalof nine simple matchings.
Consequently, each subdivision yields a simple matching of Q .Proof. Follows from [B1, Lemmas 4.12, 4.14, 4.15], with the covering space R re-placed by the hyperbolic plane H if Σ is not flat. (cid:3) Example 3.8.
The geodesic ghor algebra in Figure 1.ii embeds in a genus 2 surface,and admits four minimal subdivisions (that is, subdivisions consisting of a minimalnumber of columns or pillars) shown in Figure 6. These four subdivisions yield thetwelve simple matchings shown in Figure 7.The geodesic ghor algebra in Figure 1.iii also embeds in a genus 2 surface, and ad-mits three minimal subdivisions shown in Figure 8. These three subdivisions consistonly of columns, and yield nine distinct simple matchings.
Proposition 3.9. If A is geodesic, then each arrow of A is contained in a simplematching.Proof. Suppose A is geodesic, and assume to the contrary that there is an arrow a ∈ Q that is not contained in any simple matching. GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 13
Let ˜ P ⊂ H be a fundamental polygon in the cover of Σ, and let ι : ˜ P → R bean embedding such that P := ι ( ˜ P ) is a regular 2 N -gon with respect to the standardmetric on R . In the following, we will define a dimer-preserving homotopy H : Q × [0 , N ] → Σthat rotates the arrow a by 2 π . However, since Σ is a compact surface, such ahomotopy cannot exist, and thus the arrow a cannot exist, contrary to assumption.We define H inductively. Let H ( − ,
0) : Q → Σ be a dimer embedding (that is,each connected component of Σ \ H ( Q,
0) is simply connected and bounded by a unitcycle), and suppose H : Q × [0 , k ] → Σhas been defined. To define H : Q × [ k, k + 1] → Σ, set H + ( − , t ) := ι ( ˜ P ∩ π − ( H ( − , t ))) . Then H + ( Q, t ) ⊂ P ⊂ R . We do not require the image H + ( b, t ) of an arrow b to bea line segment in R , that is, we allow H to ‘bend’ arrows.Denote by ~ε k the unit vector in R based at the midpoint of side k of P , orthogonalto side k , and pointing to the center of P . Construct a subdivision F from a set ofparallel geodesic cycles { c i } i ∈ Q in the direction ~ε k , as in Lemma 3.6. We proceed inthree steps.First, let H + : Q × [ k, k + ] → P be a dimer-preserving homotopy that straightens the boundaries of the columns inthe subdivision F so that they are parallel to ~ε k (in R ). Specifically, let q be aboundary path of a column in F . Then H + ( − , t ), t ∈ [ k, k + ], ‘straightens’ q sothat each vertex subpath of q lies on a line L ⊂ R that is parallel to ~ε k , by shrinkingthe (images of the) arrow subpaths of q that deviate from L , and then rotating andtranslating these arrows so that they lie along L .If a is a subpath of a column boundary path of F , then the arrow image H + ( a, k + )is parallel to ~ε j . In this case, set H ( Q, t ) = H ( Q, k + ) , for t ∈ [ k + , k + 1] . If a is an interior arrow of a column, then a belongs to a simple matching, contraryto assumption. So suppose a is a subpath of either a pillar boundary in F , or an evenarrow in the interior of a pillar.Let L t( a ) , L h( a ) be lines in R that are parallel to ~ε k , and contain the respectivevertices t( a ), h( a ). Since the columns in F have all been straightened and run parallelto L , we may form paths p and q from paths that are • subpaths of unit cycles that L t( a ) and L h( a ) intersect, respectively; and • subpaths of cycles in { c i } i ∈ Q . Since p and q are constructed from subpaths of parallel geodesic cycles, p and q arethemselves geodesic cycles. Thus we may define a homotopy H + : Q × [ k + , k + ] → P that straightens both p and q so that all of their arrow subpaths lie on L t( a ) and L h( a ) respectively, while leaving all other vertices of Q fixed.Finally, define a homotopy H + : Q × [ k + , k + 1] → P by first translating the path q along L h( a ) in the direction ~ε k , and the path p along L t( a ) in the direction − ~ε k , so that the angle between ~ε k and the vector ~a from t( a )to h( a ) is less than π . After this is accomplished, bring the two lines L t( a ) and L h( a ) sufficiently close together so that ~a is pointing in the direction ~ε k , to within somedesired approximation. Then the image H + ( a, k + 1) of a is nearly parallel to ~ε k .Consequently, the homotopy H ( − , t ) rotates a by 2 π as t runs from 0 to 2 N . Butthis is not possible since Σ is a compact surface, a contradiction. (cid:3) Proposition 3.10.
Suppose A is geodesic. A cycle p is contractible if and only if itsatisfies p σ = 1 .Proof. If p is contractible, then p σ = 1 by Lemma 3.1.i.So assume to the contrary that there is a cycle p for which p σ = 1, but p is notcontractible, [ p ] = 0. Denote by ℓ ( p ) the number of times p transversely intersectsthe boundary ∂P of the fundamental polygon P .(i) We first claim that ℓ ( p ) ≥
2. Indeed, assume to the contrary that ℓ ( p ) = 1.Then, since A is geodesic, there is a geodesic cycle γ k for which [ γ k ] = [ p ]. Let s + and t + be paths in Q + from t( p + ) to t( γ + k ), resp. h( γ + k ) to h( p + ). Then tsγ k = tγ k s σ = p, by Lemma 3.1.ii. Furthermore, ts σ = 1 since ts is contractible, by Lemma 3.1.i.Whence γ k σ = p. Since A is geodesic, there is a subdivision F in the direction k which supports γ k (that is, γ k is formed from subpaths of boundary paths of columns and pillars in F ).Moreover, F yields a simple matching x ∈ S by Lemma 3.6. Thus, x ∤ γ k . However, γ k = 1 since each arrow is contained in a simple matching, by Proposition 3.9. Butthen p σ = γ k σ = 1, contrary to our choice of p .(ii) Since ℓ ( p ) ≥
2, choose p so that ℓ ( p ) is minimal among all such noncontractiblecycles. Set n := ℓ ( p ).By Lemma 3.1.i, there is a cycle q that satisfies q σ = p , and factors into cycles q = q n · · · q q , such that for each j ∈ [ n ], ℓ ( q j ) = 1 and t( q j ) = t( p ). GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 15
Since A is geodesic, there is a geodesic cycle γ k for which [ γ k ] = [ q ]. Let r and r be paths wholly contained in P from t( p ) to t( γ k ), resp. t( γ k ) to t( p ). Then byLemma 3.1,(9) q σ = r γ k r = r r · γ k σ = γ k . Let γ k + N be the complementary geodesic cycle to γ k ; then γ k + N γ k is a contractiblecycle. Whence, γ k + N σ = γ − k .Set r := q n · · · q . By (9) we have rγ k σ = rq = q σ = p σ = 1 . Thus, r σ = γ k + N . Consequently, there is a j ∈ [2 , n ] such that q j transversely intersects side ν j of P where ν j ∈ { ( k + N ) − , k + N, ( k + N ) + 1 } , by Lemma 3.6. By possibly permuting the factors of β , we may assume j = 2. Thusthere is a path t + in Q + from h( r + ) to t( q +1 ), with ℓ ( t ) minimal, that runs backwardsalong r + and avoids the fundamental polygon containing h( q +1 ). Consequently, ℓ ( t ) ≤ ℓ ( p ) − . Furthermore, ( pt ) + is a cycle in Q + , and so by Lemma 3.1.i, t σ = tp = tp σ = 1 . We also have that [ t ] = 0 since [ p ] = 0. But then we arrive at a contradiction to theminimality of ℓ ( p ) since ℓ ( t ) < ℓ ( p ). (cid:3) The following theorem shows that a ghor algebra reflects the topology of the surfacein which it is embedded.
Theorem 3.11.
Let p and q be cycles in Q . Then [ p ] = [ q ] if and only if p σ = q. In particular, if Σ is smooth, then p and q are homologous if and only if p σ = q .Proof. The forward implication was shown in Proposition 3.4.So suppose [ p ] = [ q ]. Set i := t( p + ) and j := t( q + ). Let s + , t + , q ′ + be paths in Q + respectively from i to j ; j to i ; and h( q + ) to j . Then t σ = s − and q ′ σ = q − , byLemma 3.1.i. Furthermore, since [ p ] = [ q ], the cycle sptq ′ is not contractible. Thus pq − σ = stpq − σ = sptq ′ σ = 1 , where the last inequality holds by Proposition 3.10. Therefore p σ = q . (cid:3) Remark 3.12.
If Σ is smooth and p, q are homologous cycles, then there is some ℓ ∈ Z such that p = qσ ℓ , by Theorem 3.11. It may then be asked whether there isany significance to the exponent ℓ . We expect that the exponent is a sort of discretemeasure of curvature of the surface Σ. If this is the case, then the ghor algebrawould reveal aspects of Σ that are invisible to homologous cycles alone. We leave aninvestigation of this possibility for future work. Proposition 3.13.
Suppose A is geodesic. Let p, q ∈ Q ≥ be paths for which t( p ) = t( q ) and h( p ) = h( q ) . Then ¯ η ( p ) = ¯ η ( q ) if and only if ¯ τ ( p ) = ¯ τ ( q ) . Proof. (i) First suppose ¯ η ( p ) = ¯ η ( q ). Then¯ τ ( p ) = ¯ η ( p ) | x =1: x = ¯ η ( q ) | x =1: x = ¯ τ ( q ) . (ii) Now suppose ¯ τ ( p ) = ¯ τ ( q ), and assume to the contrary that ¯ η ( p ) = ¯ η ( q ). Let r be a path from h( p ) to t( p ), and let u, v ∈ Z N be the classes of the cycles rp and rq .First suppose u = v . Since ¯ η ( p ) = ¯ η ( q ), there is some x ∈ P \ S and m ≥ x m divides only one of ¯ η ( p ), ¯ η ( q ); say x m | ¯ η ( p ) and x m ∤ ¯ η ( q ). Thus, since u = v ,¯ η ( rp ) = ¯ η ( rq ) σ n P for some n ≥
1, by Proposition 3.4. Therefore,¯ τ ( r )¯ τ ( p ) = ¯ τ ( rp ) = ¯ η ( rp ) | x =1: x = ¯ η ( rq ) σ m P | x =1: x = ¯ τ ( rq ) σ n S = ¯ τ ( r )¯ τ ( q ) σ n S . But this contradicts our assumption that ¯ τ ( p ) = ¯ τ ( q ) since n ≥ u = v . Let s ∈ e h( p ) kQe h( p ) be a cycle with class v − u = 0. Then byProposition 3.10,(10) ¯ τ ( s ) σ = 1 . Since the cycles rsp and rq both have class v , we have(11) ¯ η ( rsp ) σ = ¯ η ( rq ) , by Proposition 3.4. Whence¯ τ ( s )¯ τ ( rq ) = ¯ τ ( s )¯ τ ( rp ) = ¯ τ ( rsp ) = ¯ η ( rsp ) | x ′ =1: x ′ σ = ¯ η ( rq ) | x ′ =1: x ′ = ¯ τ ( rq ) . Consequently, ¯ τ ( s ) σ = 1 since k [ S ] is an integral domain, contrary to (10). (cid:3) Corollary 3.14. If A is geodesic, then A := kQ/ ker η ∼ = kQ/ ker τ. In particular, it suffices to only consider the simple matchings of Q to determine therelations of A . GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 17
Proof.
Follows from Proposition 3.13. (cid:3)
Corollary 3.15. If A is geodesic, then the algebra homomorphism τ : kQ → M | Q | ( k [ S ]) induces an injective algebra homomorphism on A , τ : A ֒ → M | Q | ( k [ S ]) . Consequently, R is isomorphic to the center of A .Proof. By the definition of A , two paths p, q ∈ e j Ae i satisfy ¯ η ( p ) = ¯ η ( q ) if and onlyif p = q . Furthermore, if A is geodesic, then ¯ τ ( p ) = ¯ τ ( q ) if and only if p = q , byCorollary 3.14. (cid:3) Central geometry and endomorphism ring structure
Let A = kQ/ ker η be a ghor algebra. Recall that R is the center of A by Corollary3.15, and S is the cycle algebra of A . Throughout this section, set p := ¯ τ ( p ) for p ∈ e i kQe j , and σ := σ S = Q x ∈S x .4.1. Nonnoetherian central geometry.
Throughout this section, the ground field k is algebraically closed and uncountable.In contrast to the torus case, the centers of geodesic ghor algebras on hyperbolicsurfaces are usually nonnoetherian. We can nevertheless view such a center as thecoordinate ring on a geometric space, using the framework of nonnoetherian geom-etry introduced in [B3] (see also [B5]). In short, the geometry of a nonnoetheriancoordinate ring of finite Krull dimension looks just like a finite type algebraic variety,except that it has some positive dimensional closed points. Definition 4.1. A depiction of an integral domain k -algebra R is a finitely generatedoverring S such that the morphismSpec S → Spec R, q q ∩ R, is surjective, and(12) U S/R := { n ∈ Max S | R n ∩ R = S n } = { n ∈ Max S | R n ∩ R is noetherian } 6 = ∅ . For example, the algebra S = k [ x, y ] is a depiction of its nonnoetherian subalgebra R = k + xS . We thus view Max R as the variety Max S = A k , except that the line { x = 0 } is identified as a 1-dimensional (closed) point of Max R . In particular, thecomplement { x = 0 } ⊂ A k is the ‘noetherian locus’ U S/ ( k + xS ) [B3, Proposition 2.8].Let A be a ghor algebra with center R and cycle algebra S . Lemma 4.2.
The cycle algebra S is a finitely generated k -algebra.Proof. The dimer quiver Q is finite, and so there are only a finite number of cyclesin Q that do not have cyclic proper subpaths. (cid:3) Lemma 4.3.
Let A be a ghor algebra with center R . Then R is noetherian if andonly if for each monomial g ∈ S , there is an n ≥ such that g n is in R . Proof. ⇒ : First suppose there is a monomial g ∈ S such that g n R for each n ≥
1. ByLemma 3.1.i, there is an m ≥ n ≥ g n σ m is in R . Consider theinfinite ascending chain of ideals0 ⊆ σ m R ⊆ (1 , g ) σ m R ⊆ (1 , g, g ) σ m R ⊆ (1 , g, g , g ) σ m R ⊆ · · · . We claim that each inclusion is strict. Assume to the contrary that there is some n ≥ r , . . . , r n − ∈ R for which g n σ m = n − X i =0 r i g i σ m . Then, since k [ S ] is an integral domain, g n − n − X i =1 r i g i = r ∈ R. Furthermore, R is generated by monomials in the polynomial ring k [ S ], and so eachmonomial summand of the polynomial r is in R . Thus, since g n R , there is some1 ≤ i ≤ n − g n − i is a monomial summand of r i . But then r i is not in R ,a contradiction. Therefore R is nonnoetherian. ⇐ : Now suppose that for each monomial g ∈ S , there is an n ≥ g n isin R . By Lemma 4.2, S is generated by a finite number of cycles s j ,(13) S = k [ s j | j ∈ [1 , ℓ ]] . By assumption, for each j ∈ [0 , ℓ ], there is a minimum n j ≥ s n j j is in R .Consider the two sets of cycles X := { cycles in R that contain at most one of each vertex } ,Y := ( ℓ Y j =1 s m j j ∈ S \ R | m j ∈ [0 , n j − ) . We claim that R = k [ X, XY ∩ R ] . Indeed, let r ∈ X , s , s ∈ Y , and suppose ( rs ) s ∈ ( XY ∩ R ) Y ∩ R . Then( rs ) s = r ( s s ) ∈ ( XY ∩ R if s s Rk [ X ] if s s ∈ R Consequently, ( XY ∩ R ) Y ∩ R ⊆ k [ X, XY ∩ R ] , proving our claim. But | Y | < ∞ , and | X | < ∞ since Q has a finite number ofvertices. Whence | XY ∩ R | ≤ | X || Y | < ∞ . Thus R = k [ X, XY ∩ R ] is a finitelygenerated k -algebra, and therefore noetherian. (cid:3) GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 19
If Σ is a torus, then a ghor algebra A is geodesic if and only if it is noetherian, if andonly if its center R is noetherian, if and only if A is a finitely generated R -module [B2,Theorem 1.1]. If Σ is hyperbolic, then only one direction of the implication survives: Lemma 4.4. If R is noetherian, then(1) A is a finitely generated R -module;(2) A is noetherian; and(3) A is geodesic.Proof. Suppose R is noetherian.(1): Recall the finite generating set (13) of S . By Lemma 4.3, there are minimumintegers n i ≥ s n i i ∈ R . Set m := max { n , . . . , n ℓ } . Let L be the length of the longest path in Q with no cyclic proper subpath. Theneach path of length ≥ ( L + 1) ℓm will contain a cyclic subpath whose ¯ τ -image is in R . Therefore A is generated as an R -module by the set of all paths in Q of length atmost ( L + 1) ℓm .(2): Follows from (1) and the assumption that R is noetherian.(3): Suppose A is not geodesic. Then there is a vertex i ∈ Q and direction k ∈ [2 N ] for which every cycle at i parallel to k is not geodesic. Let p be a cycle at i parallel to k , whose lift p + to the cover Q + contains no cyclic subpaths modulo ker η (though some cyclic permutation of p + necessarily contains a cyclic subpath since p isnot geodesic). Let q + be formed from a cyclic permutation of p + by removing at leastone cyclic subpath. Then q n is in S \ R for all n ≥
1, by Theorem 3.11. Therefore R is nonnoetherian by Lemma 4.3. (cid:3) Lemma 4.5.
The morphisms κ S/ ˆ Z : Max S → Max R, n n ∩ R,ι
S/R : Spec S → Spec R, q q ∩ R, are well-defined and surjective.Proof. (i) We first claim that the map κ S/R is well-defined. Indeed, let n be inMax S . By Lemma 4.2, S is of finite type, and by assumption k is algebraicallyclosed. Therefore the intersection n ∩ R is a maximal ideal of R (e.g., [B3, Lemma2.1]).(ii) We claim that κ S/R is surjective. Fix m ∈ max R . Then S m is a proper idealof S since S is a subalgebra of the polynomial ring k [ S ]. Thus, since S is noetherian,there is a maximal ideal n ∈ Max S containing S m . Whence, m ⊆ S m ∩ R ⊆ n ∩ R. But n ∩ R is a maximal ideal of R by Claim (i). Therefore m = n ∩ R . q q s = q p q p q ( i ) ( ii ) ( iii ) Figure 9. (i): Setup for Lemma 4.6 (with irrelevant paths omitted).(ii): A specific example of (i), described in Example 4.7. (iii): Asubdivision consisting of two columns (drawn in green).(iii) It is clear that the map ι S/R is well-defined. Finally, we claim that ι S/R is sur-jective. By [B3, Lemma 3.6], if D is a finitely generated algebra over an uncountablefield k , and C ⊆ D is a subalgebra, then ι D/C : Spec D → Spec C is surjective if andonly if κ D/C : Max D → Max C is surjective. Therefore, ι S/R is surjective by Claim(ii). (cid:3)
Lemma 4.6.
Let n ∈ Max S be a maximal ideal. Suppose that each cycle p + ker η that passes through each vertex of Q satisfies p ∈ n . Then the localization R n ∩ R isnonnoetherian.Proof. Let p be a cycle for which p is in R \ n . Then p m n for each m ≥
1. Thus, byassumption, p m + ker η does not pass through each vertex of Q . In particular, thereis a vertex j ∈ Q such that e j is not a subpath of p m + ker η for m ≥ p is in R , there is a cycle q ∈ e j kQe j such that q = p . By Theorem 3.11, wehave [ p ] = [ q ]. It thus suffices to suppose that q has a cyclic proper subpath s thatruns along one of the edges of the fundamental polygon P ; see Figure 9.i.Let ℓ ≥ σ ℓ t( s ) + ker η passes through each vertex of Q . Then s n σ ℓ isin R for each n ≥
1. However, there is no cycle that intersects p with monomial s n for any n ≥
1, since such a cycle would necessarily be in the same class as s n , againby Theorem 3.11, and p + ker η and q + ker η do not intersect. Thus s n is not in R n ∩ R for each n ≥
1. But then the ascending chain of ideals of R n ∩ R , sσ ℓ ⊂ ( s, s ) σ ℓ ⊂ ( s, s , s ) σ ℓ ⊂ · · · , does not stabilize. Therefore R n ∩ R is nonnoetherian. (cid:3) Example 4.7.
In Lemma 4.6 we considered a ghor algebra that contains a cycle p whose monomial p is in R , yet p m + ker η does not pass through each vertex of Q for all m ≥
1. A priori it is unclear whether such ghor algebras exist; however,
GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 21 an example with this property is given in Figure 9.ii. This particular example maybe regarded as a two-vertex generalization of the polynomial ghor algebra given inExample 2.
Theorem 4.8.
If the center R of a ghor algebra A is nonnoetherian, then the cyclealgebra S of A is a depiction of R .Proof. Suppose that R is nonnoetherian.(i) We first claim that the locus U S/R is nonempty.Choose a maximal ideal n of S for which σ n (that is, for each simple matching x ∈ S , there is a nonzero constant c ∈ k ∗ such that ( x − c ) k [ S ] ∩ S ⊂ n ). Let m ≥ σ mi + ker η passes through each vertex of Q . Let s be any cycle in Q .Then sσ m = sσ m t( p ) ∈ R . Whence s = sσ m σ m ∈ R n ∩ R . Thus S ⊂ R n ∩ R , and therefore S n ⊆ R n ∩ R ⊆ S n . Consequently, n ∈ U S/R .(ii) Let n ∈ Max S , and suppose the localization R n ∩ R is noetherian. We claimthat R n ∩ R = S n .Fix a cycle s . Since R n ∩ R is noetherian, there is a cycle p which passes througheach vertex of Q such that p n , by Lemma 4.6. Since p passes through each vertexof Q , the monomials p and sp are both in R . Thus s = spp ∈ R n ∩ R . Whence S ⊂ R n ∩ R . Therefore R n ∩ R = S n .(iii) Finally, the morphism Spec S → Spec R is surjective by Lemma 4.5. (cid:3) The Krull dimension of the center.
Suppose A is geodesic. Denote by T thesubalgebra of S generated by σ and the ¯ τ -images of the 2 N geodesic cycles γ , . . . , γ N in Definition 2.5; without loss of generality we may assume(14) γ k γ k + N σ = 1 , by Lemma 3.1.i. Lemma 4.9.
The following inclusions hold: (i) S ⊂ T [ σ − ] . (ii) S ⊂ R [ σ − ] . (iii) R [ σ − ] = T [ σ − ] = S [ σ − ] .In particular, the algebraic varieties Max R , Max T , and Max S are birationally equiv-alent.Proof. (i) We first claim that S ⊂ T [ σ − ]. Let s be a cycle. Let p + be a path lyingin the fundamental polygon P from t( s + ) to a vertex i +0 ∈ π − ( i ), and let q + be a path lying in P from a vertex i +1 ∈ π − ( i ) to t( s + ). Then there is an ℓ ≥ pq = σ ℓ , by Lemma 3.1.i.Construct a path t + in Q + by concatenating lifts of cycles in T from i +0 to i +1 . Thent( t + ) = t(( psq ) + ) and h( t + ) = h(( psq ) + ) . Thus there is an m ∈ Z such that ¯ t = psqσ m , by Lemma 3.1.ii. But then s = σ − ℓ psq = ¯ tσ − ℓ − m ∈ T [ σ − ] . (ii) We now claim that S ⊂ R [ σ − ]. Let s be a cycle. Let p + be a cycle in Q + suchthat p = π ( p + ) passes through each vertex of Q . By Lemma 3.1.i, there is an ℓ ≥ p = σ ℓ . Therefore ¯ s = σ − ℓ ps ∈ R [ σ − ] . (iii) We have R [ σ − ] = S [ σ − ] (i) ⊆ T [ σ − ] ⊆ S [ σ − ] (ii) ⊆ R [ σ − ] , where ( i ) holds by Claim (i), and ( ii ) holds by Claim (ii). (cid:3) Lemma 4.10.
The Krull dimension of T is N + 1 .Proof. For j ∈ [ N ], set p j := ( σ, γ , . . . , γ N , γ N +1 , . . . , γ j + N ) T, and consider the chain of ideals of T ,(15) 0 ⊆ p ⊆ p ⊆ · · · ⊆ p N . (i) We first claim that each p j is prime.Indeed, let p, q be cycles for which p, q ∈ T , and suppose pq ∈ p j . Then there is a k ∈ [ j + N ] and a monomial t ∈ T such that pq = tγ k . By Theorem 3.11, the cycles pq and tγ k are in the same class. But then γ k is a factorof p or a factor q , whence p ∈ p j or q ∈ p j .(ii) We now claim that the inclusions in (15) are strict.Fix j ∈ [ N −
1] and k ∈ [ j + 1 , N ]. Assume to the contrary that γ k + N ∈ p j . Byassumption, σ ∤ γ k + N . Whence γ k + N σT . Thus, since T is toric, there is some i ∈ [ j + N ] and a monomial t ∈ T such that γ k + N = tγ i . But then by Theorem 3.11, γ k + N and tγ i are in the same class, a contradiction. Therefore the chain (15) is strict.(iii) Finally, the chain (15) is maximal: Let p be a prime ideal of T generated bymonomials, and let k ∈ [ N ]. Then σ is in p if and only if γ k or γ k + N is in p by(14). (cid:3) Proposition 4.11.
The Krull dimensions of R , S , and T are equal. GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 23
Proof. If B is a nonnoetherian integral domain depicted by C , then dim B = dim C ,by [B3, ]. Furthermore, if an integral domain is finitely generated over k , then itsKrull dimension and transcendence degree are equal. Thereforedim R (i) = dim S (ii) = trdeg k (Frac S ) = trdeg k (Frac S [ σ − ]) (iii) = trdeg k (Frac T [ σ − ]) = trdeg k (Frac T ) (iv) = dim T, where ( i ) holds since S is a depiction of R by Theorem 4.8; ( iii ) holds by Lemma4.9; and ( ii ) and ( iv ) hold since S [ σ − ] and T [ σ − ] are finitely generated over k byLemma 4.2. (cid:3) Theorem 4.12.
The Krull dimensions of R and S satisfy (16) dim R = dim S = N + 1 . In particular, if Σ is a smooth genus g ≥ surface, then dim R = rank H (Σ) + 1 = 2 g + 1 . Proof.
We have dim R = dim S = dim T by Proposition 4.11. The equalities (16)therefore hold by Lemma 4.10. (cid:3) Endomorphism ring structure.
Let A be a ghor algebra with center R andcycle algebra S . In the following, set p := ¯ τ ( p ) for p ∈ e j kQe i . Lemma 4.13.
Suppose A is geodesic. Fix i, j ∈ Q , and let f ∈ End R ( Ae i ) . Thenthere is some g ∈ k [ S ] such that f ( p ) = gp for all p ∈ e j Ae i .Proof. Fix i, j ∈ Q , and let f ∈ End R ( Ae i ). Consider a nontrivial path p ∈ e j kQe i .Since A is geodesic, there is a simple matching x ∈ S such that x | p , by Proposition3.9. Let m ≥ x m | p and x m +1 ∤ p . Since x is simple, there is a path q ∈ e j kQe i for which x ∤ q .Let r + be a path in Q + from h( p + ) to t( p + ). Let ℓ ≥ prσ ℓ and qrσ ℓ are both in R . Then, since f is an R -module homomorphism, we have( pσ ℓi r ) f ( q ) = f ( pσ ℓi rq ) = f ( qσ ℓi rp ) = ( qσ ℓi r ) f ( p ) . Thus, since k [ S ] is an integral domain,(17) f ( q ) q = f ( p ) p . But x m | p and x ∤ q , and so (17) implies x m | f ( p ). Therefore, since x ∈ S was anarbitrary simple matching for which x | p , we have p | f ( p ) . Set(18) g := f ( p ) p ∈ k [ S ] . Then f ( p ) = gp for all p ∈ e j Ae i , again by (17). (cid:3) Proposition 4.14.
Suppose A is geodesic. Then the topological ring A σ := A ⊗ R R [ σ − ] is an endomorphism ring over its center: for each i ∈ Q , we have A σ ∼ = End R σ ( A σ e i ) . Proof.
Fix vertices i, j ∈ Q , and a left R -module endomorphism f ∈ End R ( Ae i ). Bythe linearity of f , we may assume that the polynomial g ∈ k [ S ] corresponding to f and j , as defined in Lemma 4.13, is a monomial. Let p ∈ e j kQe i be a path; then q := f ( p ) is also a path since q = gp is a monomial. Let t + be a path in the cover Q + from h( p + ) to h( q + ). Thus, by Lemma 3.1.ii, there is an ℓ ∈ Z such that tp = qσ ℓ . But then the path tσ − ℓ ∈ e k A σ e j satisfies tσ − ℓ = g . Therefore f acts on e j A σ e i by left multiplication by tσ − ℓ .Conversely, every element of A σ defines a left R σ -module endomorphism of A σ e i by left multiplication. (cid:3) Lemma 4.15.
Suppose A is geodesic. If p, q ∈ kQe j are paths satisfying p = q , then h( p + ) = h( q + ) .Proof. Suppose p, q ∈ kQe j satisfy p = q . Let r + be a path in the cover Q + fromh( p + ) to h( q + ). Then rp = qσ ℓ for some ℓ ∈ Z , by Lemma 3.1.ii. Whence r = σ ℓ since p = q . But then t( r + ) = h( r + ) by Proposition 3.10. (cid:3) Recall the noetherian locus U S/R defined in (12), which is an open dense subset ofthe algebraic variety Max S of the cycle algebra S . Theorem 4.16.
Suppose A is geodesic. At each point m ∈ Max R which lifts to thenoetherian locus U S/R ⊆ Max S , that is, for which the localization R m is noetherian,the localization A m := A ⊗ R R m is an endomorphism ring over its center: for each i ∈ Q , we have A m ∼ = End R m ( A m e i ) . Proof.
Fix m ∈ Max R for which R m is noetherian; vertices i, j ∈ Q ; and a left R -module endomorphism f ∈ End R ( Ae i ). By Theorem 4.8, S is a depiction of R ,and thus there is an n ∈ Max S such that n ∩ R = m and S n = R m .By the linearity of f , we may assume that the polynomial g ∈ k [ S ] correspondingto f and j , as defined in Lemma 4.13, is a monomial. Thus, by Lemma 4.15 it sufficesto show that there is a path t ∈ kQe j and cycle s with s n , for which g = ts − ∈ ¯ τ ( e h( t ) Ae j ) ⊗ R R m . GENERAL. OF CANCELLATIVE DIMER ALGEBRAS TO HYPERBOLIC SURFACES 25
Let m ≥ σ m | g and σ m +1 ∤ g . If g = σ m , then we maytake t = σ mj by Proposition 3.10. So suppose there is a simple matching x ∈ S suchthat x ∤ gσ − m . Since x is simple, there is a path p ∈ e j kQe i such that x ∤ p . Set q := f ( p ) and k := h( q ); then q = f ( p ) = gp .Again since x is simple, there are paths r ∈ e i kQe k and t ∈ e k kQe j for which x ∤ r and x ∤ t . Then t pr ∈ e k kQe k is a cycle satisfying(19) x ∤ t pr. Furthermore, x ∤ gprσ − m = qrσ − m . Thus, by Theorem 3.11 and the fact that pr | qr , we may choose t so that eachcomponent of the class [ t pr ] ∈ Z N satisfies(20) sign([ t pr ] ν ) = sign([ qr ] ν ) and | [ t pr ] ν | ≤ | [ qr ] ν | , with ν ∈ [ N ].Choose a cycle t (at any vertex of Q ) such that(i) [ t ] = [ qr ] − [ t pr ], and(ii) t is minimally divisible by σ such that (i) holds.Set ℓ := t( t ).Since R n ∩ R is noetherian, there is a cycle t t = t e ℓ t ∈ e k kQe k that passes through ℓ for which t t n , by Lemma 4.6. By (i) and Theorem 3.11 we have qr = t · t prσ n for some n ∈ Z . Furthermore, by (ii), (19), and (20), we have n ≥
0. Therefore,setting t := t t t t , we find g = qp = t t t t pt t · p = tt t ∈ ¯ τ ( e k Ae j ) ⊗ R R m . Consequently, g is the ¯ τ -image of the cycle tt t ∈ e k A m e j . Conversely, every element of A m defines a left R m -module endomorphism of A m e i by left multiplication. (cid:3) Remark 4.17.
Theorem 4.16 is a generalization of the well known fact that can-cellative dimer algebras A on a torus are endomorphism rings: for each i ∈ Q , thereis an algebra isomorphism A ∼ = End R ( Ae i ). Indeed, a cancellative dimer algebra ona torus is a geodesic ghor algebra for which U S/R = Max S = Max R (noting that,in this case, R = S ). Thus Theorem 4.16 implies that for each m ∈ Max R , thelocalization A m is an endomorphism ring, A m ∼ = End R m ( A m e i ). This in turn impliesthe isomorphism A ∼ = End R ( Ae i ). Acknowledgments.
The authors were supported by the Austrian Science Fund(FWF) grant P 30549-N26. The first author was also supported by the AustrianScience Fund (FWF) grant W 1230 and by a Royal Society Wolfson Fellowship.
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