aa r X i v : . [ m a t h . C O ] A p r Plabic R-matrices
Sunita Chepuri
Abstract
Postnikov’s plabic graphs in a disk are used to parametrize totally positive Grass-mannians. In recent years plabic graphs have found numerous applications in mathand physics. One of the key features of the theory is the fact that if a plabic graph isreduced, the face weights can be uniquely recovered from boundary measurements. Onsurfaces more complicated than a disk this property is lost. In this paper we under-take a comprehensive study of a certain semi-local transformation of weights for plabicnetworks on a cylinder that preserve boundary measurements. We call this a plabicR-matrix. We show that plabic R-matrices have underlying cluster algebra structure,generalizing recent work of Inoue-Lam-Pylyavskyy. Special cases of transformationswe consider include geometric R-matrices appearing in Berenstein-Kazhdan theory ofgeometric crystals, and also certain transformations appearing in a recent work ofGoncharov-Shen.
The relationship between total positivity and networks has been studied extensively (see [4],[6], [8]). In his groundbreaking paper [21], Postnikov develops a theory of plabic networksfor studying the connection between the totally nonnegative Grassmannian and planardirected networks in a disk. Plabic graphs have since been found to have many additionalapplications. They have been used by Kodama and Williams to study soliton solutionsto the KP equation [16, 17], by Arkani-Hamed, et. al., to study scattering amplitudes for N = 4 supersymmetric Yang-Mills [1, 2, 3], and by Gekhtman, Shapiro, and Vainshtein tostudy Poisson geometry [9, 10].Postnikov defines a set of local moves and reductions so that the boundary measurementmap gives a bijection between move-reduction equivalence classes for plabic networks ina disk and the totally nonnegative Grassmannian. However, there are plabic networkson a cylinder that are not move-reduction equivalent and yet have the same boundarymeasurements. In particular, we define a semi-local transformation on weights for plabicnetworks on a cylinder that preserves boundary measurements. We call this a plabic R-matrix . Plabic R-matrices are different from Postnikov’s moves and reductions in that theydo not alter the underlying graph. 1n the simplest case (see Example 4.14), we recover the geometric R-matrix. Thegeometric R-matrix arises in Berenstein and Kazhdan’s theory of geometric crystals [5]. Itis studied by Etingof in exploring set-theoretical solutions to the Yang-Baxter equation [7],and by Kajiwara, Nouni, and Yamada in the context of representations of affine Weylgroups [15]. Kashiwara, Nakashima, and Okado give a thorough survey constructing thegeometric R-matrix in different types [14]. The geometric R-matrix also appears in Lamand Pylyavskyy’s investigation of total positivity in loop groups [18].Another special case of the plabic R-matrix (see Examples 4.20, 6.12, and 7.4) is atransformation used by Goncharov and Shen to study Donaldson-Thomas invariants [12].Given a surface S with punctures, there is a birational Weyl group action on the modulispace X PGLm , S for each puncture in S . Goncharov and Shen show that this action is given bycluster Poisson transformations. We can obtain the plabic R-matrix from a generalizationof these transformations (see Sections 6 and 7), and in specific cases recover the Weyl groupaction. Section 2.
This section provides background on planar directed networks. In Sec-tion 2.1, we review Postinikov’s boundary measurement map (Section 4 of [21]). Section 2.2follows Gekhtman, Shapiro, and Vainshtein’s lifting of this construction to planar directednetworks on a cylinder [11].
Section 3.
Here we introduce plabic networks on a cylinder. Section 3.1 details howto obtain face and trail weights from edge weights and how to calculate the weight of apath in a face weighted network. In Section 3.2, we expose a major difficulty of lifting thetheory of plabic networks to the cylinder: Postinikov’s results (Section 10 of [21]) regardingchanging the orientation of edges of a plabic network do not hold. We do however show inTheorem 3.18 that plabic networks on a cylinder contain enough information that we candefine involutions on the level of plabic networks, rather than planar directed networks.
Section 4.
We begin by reviewing Postnikov’s moves and reductions for plabic net-works (Section 12 of [21]) and Postnikov diagrams, also known as alternating strand di-agrams (Section 14 of [21]). The definition of Postnikov diagrams can be generalized sothat it lifts to a cylinder. We prove in Theorem 4.5 that Postnikov diagrams on a cylinderare in bijection with leafless reduced plabic graphs on a cylinder with no unicolored edges,an analogue of Corollary 14.2 of [21]. Using alternating stand diagrams, we introduce afamily of plabic networks on a cylinder called cylindric k -loop plabic networks. We de-fine an transformation called a plabic R-matrix on weights for such networks, in both anedge-weighted and face-weighted setting. In Theorems 4.13 and 4.18, we show that plabicR-matrices preserve boundary measurements, are an involutions, give the only choices ofweights that preserve the boundary measurements, and satisfy the braid relation. Section 5.
This section gives a brief background on cluster algebras from quivers,including both x - and y -dynamics. Section 6.
In this section, we define spider web quivers and a mutation sequence τ for2hese quivers. Proposition 6.2 shows τ preserves the original quiver. We give formulas forboth the x and y variables after applying τ to a spider web quiver. In Theorems 6.6, 6.7, 6.9,and 6.10, we show that τ is an involution and satisfies the braid relation for both x and y variables. Section 7.
We show in Theorem 7.2 that for a cylindric k -loop plabic network theface weighted plabic R-matrix is realized by the y -dynamics of the dual quiver, under theinvolution τ from Section 6.The rest of the sections provide proofs of the main theorems. Section 8 proves some use-ful facts about plabic graphs on a cylinder using Postnikov diagrams, Section 9 gives proofsabout plabic R-matrices, and Section 10 contains proofs of the theorems from Section 6. Acknowledgements.
I would like to thank my adviser Pavlo Pylyavskyy for intro-ducing me to this topic and for the many helpful conversations and suggestions. I am alsothankful for the support of RTG NSF grant DMS-1148634.
Definition 2.1.
We will assume a planar directed graph on a surface with boundary,considered up to homotopy, has n vertices on the boundary, b , ..., b n . We will call these boundary vertices , and all other vertices internal vertices . Additionally, we will assume thatall boundary vertices are sources or sinks. A planar directed network is a planar directedgraph with a weight x e ∈ R > assigned to each edge. Definition 2.2.
The source set of a planar directed graph or network is the set I = { i ∈ [ n ] | b i is a source } . The sink set is I = [ n ] \ I = { j ∈ [ n ] | b j is a sink } . Definition 2.3.
For any path P in a planar directed network, the weight of P is x P = Y e ∈ P x e . The material in this section can be found in Section 4 of [21].For a planar directed graph or network in a disk, we will label the boundary vertices b , ..., b n in clockwise order. Definition 2.4.
For a path P in a planar directed graph or network in a disk from b i to b j , we define its winding index , wind ( P ). First, we smooth any corners in P and makethe tangent vector of P at b j have the same direction as the tangent vector at b i . Then wind ( P ) is the full number of counterclockwise 360 ◦ turns the tangent vector makes from3 i to b j , counting clockwise turns as negative. For C a closed curve, we can define wind ( C )similarly. See Lemma 4.2 in [21] for a recursive formula for wind ( P ) and wind ( C ).Figure 1: A path P with wind ( P ) = − Definition 2.5.
Let b i be a source and b j be a sink in a planar directed network in a diskwith graph G . Let the edge weights be the formal variables x e . Then the formal boundarymeasurement M form ij is the formal power series M form ij := X paths P from b i to b j ( − wind ( P ) x P . Lemma 2.6.
The formal power series M form ij sum to subtraction-free rational expressionsin the variables x e . Thus, M form ij is well-defined function on R | E ( G ) |≥ , where E ( G ) is the setof edges in the graph G . Definition 2.7.
The boundary measurements M ij for a planar directed network in a diskare nonnegative real numbers obtained by writing the formal boundary measurements M form ij as subtraction-free rational expressions, and then specializing them by assigning x e the real weight of the edge e . Example 2.8.
Suppose we have the following network: b b x = 1 x = 2 x = 1 x = 1 M form12 = x x x − x x x x x + x x x x x x x − ... = x x x ∞ X i =0 ( − x x ) i = x x x x x x e ’s, we find M = . Definition 2.9.
For 0 ≤ k ≤ n , the Grassmannian Gr kn is the manifold of k -dimensionalsubspaces of R n .We can associate any full-rank real k × n matrix to a point in Gr kn by taking the spanof its rows. Let M at ∗ kn be the space of full-rank real k × n matrices. Since left-multiplying amatrix by an element of the general linear group is equivalent to performing row operationsand row operations do not change the row-span of a matrix, we can think of Gr kn as thequotient GL k \ M at ∗ kn . Definition 2.10.
For a k × n matrix M , a maximal minor is ∆( M ) S where S ⊆ [ n ] and | S | = k . ∆( M ) S is the determinant of the submatrix of M obtained by taking only thecolumns indexed by S . Definition 2.11. If N et kn is the set of planar directed networks in a disk with k boundarysources and n − k boundary sinks, then we can define the boundary measurement map M eas : N et kn → Gr kn . M eas ( N ) is the point in Gr kn represented by the matrix A ( N ),which is defined as follows:(1) A ( N ) I , the submatrix of A ( N ) containing only the columns in the source set, is theidentity Id k .(2) For I = { i < ... < i k } , r ∈ [ k ], and b j ∈ I , we define a rj = ( − s M i r ,j , where s is thenumber of elements of I strictly between i r and j .Note that the map M eas is constructed so that M ij = ∆( A ( N )) ( I \{ i } ) ∪{ j } . Example 2.12.
Consider the network from the pervious example: b b x = 1 x = 2 x = 1 x = 1 In this case I = { } , so we put Id in the first column of A ( N ).We compute a = ( − M = M = .So, we have A ( N ) = (cid:2) (cid:3) . Throughout this paper, we will draw a cylinder as a fundamental domain of its universalcover such that it is a rectangle with boundary components on the left and right (seeFigure 2). 5igure 2: A cylinder, as we will represent them in this paper.The constructions in this section may be found in [11]. They are based on Postnikov’stheory of planar directed networks in a disk [21], as seen in 2.1.For a planar directed graph or network on a cylinder, we will label the boundary vertices b , ..., b n from the top of the left boundary component to the bottom and then from thebottom of the right boundary component to the top. Definition 2.13. A cut γ is an oriented non-self-intersecting curve from one boundarycomponent to another, considered up to homotopy. The endpoints of the cut are basepoints . We will always assume the cut is disjoint from the set of vertices of the graph andthat it corresponds to the top and bottom of our rectangle when we draw a cylinder. Thecut is denoted by a directed dashed line. Definition 2.14.
For a path P , the intersection number , int ( P ), is the number of times P crosses γ from the right minus the number where P crosses γ from the left. Definition 2.15. If P is a path from b to b ′ where b, b ′ are on the same boundary compo-nent, then C P is the closed loop created from following the path P and then going downalong the boundary from b ′ to b . If P is a path from b to b ′ where b, b ′ are on the differentboundary components, then C P is the closed loop created from following the path P goingdown on the boundary from b ′ to the base point of the cut, following the cut (or its reverse),and then down on the boundary from base point of the cut to b . Definition 2.16.
We can glue together the top and bottom of our rectangle, which rep-resents a cylinder, in the plane to form an annulus. Do this such that going up along theboundary of the rectangle corresponds to going clockwise around the boundary of the an-nulus (see Figure 3). Then for a path P , the winding index of P is defined to be wind ( C P ),when C P is drawn on this annulus and wind ( C P ) is calculated as in Definition 2.4.6igure 3: Turning a cylinder into an annulus. Example 2.17.
Here we have the cylinder depicted as an annulus. The black dashed line is the cut. Apath P is shown in blue. P crosses the cut once from left to right, so int ( P ) = −
1. Theextension of P to C P is shown in red. We can see wind ( C P ) = − Definition 2.18.
Let b i be a source and b j be a sink in a planar directed network on acylinder with graph G . Let the edge weights be the formal variables x e . Then the formalboundary measurement M form ij is the formal power series M form ij := X paths P from b i to b j ( − wind ( C P ) − ζ int ( P ) x P . Lemma 2.19 (Corollary 2.3 of [11]) . If N is a planar network on a cylinder, then theformal power series M form ij sum to rational expressions in the variables x e and ζ . Definition 2.20.
The boundary measurements M ij for a planar directed network on acylinder are rational functions in ζ obtained by writing the formal boundary measurements M form ij as rational expressions, and then specializing them by assigning x e the real weightof the edge e . 7 xample 2.21. Suppose we have the following network: b b b x = 1 x = 1 x = 1 x = 1 x = 2 x = 3 M form12 = x x x ζ − x x x x x x ζ + x x x x x x x x x ζ ... = x x x ζ ∞ X i =0 ( − x x x ζ ) i = x x x ζ x x x ζM form13 = − x x x x ζ + x x x x x x x ζ − x x x x x x x x x x ζ ... = − x x x x ζ ∞ X i =0 ( − x x x ζ ) i = − x x x x ζ x x x ζ Substituting our values for the x e ’s, we find M = ζ ζ and M = − ζ ζ . Definition 2.22.
The space of Grassmannian loops , LGr kn is the space of rational func-tions X : R → Gr kn . Elements in LGr kn can be represented as a full-rank k × n matrixwhere the entries are functions of a parameter ζ . Definition 2.23. If N et
Ckn is the set of planar directed networks on a cylinder with k boundary sources and n − k boundary sinks, we can define the boundary measurement map as M eas C : N et
Ckn → LG kn where M eas C ( N ) is represented by the matrix A ( N ) suchthat:(1) A ( N ) I is the identity Id k .(2) For I = { i < ... < i k } , r ∈ [ k ], and b j ∈ I , we define a rj = ( − s M i r ,j , where s is thenumber of elements of I strictly between i r and j .Note that the map M eas C is constructed so that M ij = ∆( A ( N )) ( I \{ i } ) ∪{ j } . Example 2.24.
Consider the network from the previous example:8 b b x = 1 x = 1 x = 1 x = 1 x = 2 x = 3 In this case I = { } , so we put Id in the first column of A ( N ).We compute a = ( − M = M = ζ ζ and a = ( − M = M = − ζ ζ .So, we have A ( N ) = h ζ ζ − ζ ζ i . Definition 3.1. A planar bicolored graph , or plabic graph , on a surface with boundary isa planar undirected graph such that each boundary vertex has degree 1 and each internalvertex is colored black or white. Definition 3.2 (Definition 11.5 of [21]) . A plabic network in a disk is a plabic graph witha weight y f ∈ R > assigned to each face.Postnikov [21] solves the inverse boundary problem for planar directed networks ina disk by turning them into plabic networks. We will approach the problem for planardirected networks on a cylinder in the same way. Definition 3.3 (Section 4 of [21]) . A gauge transformation is a rescaling of edge weightsin a planar directed network so that all incoming edges of a particular vertex are multipliedby a positive real number c and all outgoing edges of that vertex are divided by c .It is clear that gauge transformations preserve the boundary measurements, as theypreserve the weight of each path. This means that we can only ever hope to solve theinverse boundary problem up to gauge transformations. To this end, we introduce thespace of face and trail weights, which eliminates gauge transformations. This space wasintroduced by Gekhtman, Shapiro, and Vainshtein in [11]. Here we present their resultsand also state explicitly how to obtain boundary measurements from a face weighted planardirected network on a cylinder. 9 x x x x x f Figure 4: A face with y f = x x − x − x − x x − x − = x x x x x . Definition 3.4 (Section 11 of [21]) . For a face in a planar directed network, define the face weight to be y f := Y e i ∈ I + f x e i Y e j ∈ I − f x e j where I + f is the set of edges on the outer boundary of f oriented clockwise and edges onthe inner boundary (if f is not simply connected) oriented counterclockwise, and I − f is theset of edges on the outer boundary of f oriented counterclockwise and edges on the innerboundary (if f is not simply connected) oriented clockwise.We can see that the product of weights of all the faces is 1, as each edge is countedonce going clockwise and once going counterclockwise. Definition 3.5 (Section 2.3 of [11]) . A trail in a planar directed network on a cylinder N is a sequence of vertices v , ..., v m +1 where v , v m +1 are boundary vertices on differentboundary components and for each i , either ( v i , v i +1 ) or ( v i +1 , v i ) is an edge in N . The weight of a trail is t = Y ( v i ,v i +1 ) an edge in N x ( v i ,v i +1 ) Y ( v i +1 ,v i ) an edge in N x − v i +1 ,v i ) . Notice that the face and trail weights are invariant under gauge transformations.10 heorem 3.6 (Section 2.3 of [11]) . For a planar directed graph G on a cylinder with edgeset E and face set F , R E> / { gauge transformations } = ( R F − > if there is no trail, R F − > ⊕ R > otherwise.where R F − > is generated by the face weights under the relation that their product be equalto 1 and R > in the second case is generated by the weight of a trail. Having a network that falls under the first case is equivalent to having a network ina disk. In this case, we recover Postnikov’s face weight construction for planar directednetworks in a disk.
Definition 3.7.
We will define wt ( P, y, t ) in three cases:(1) For a path P that begins and ends on the same boundary component, draw enoughcopies of the fundamental domain that we can draw P as a connected curve. If P ispath from b i to b j , then P along with a segment of the boundary between b i and b j form a closed shape P b on the universal cover. When P b is to the right of P , wt ( P, y, t )is the product of the weights of the faces in the interior of P b . When P b is to the leftof P , wt ( P, y, t ) is the inverse of this product.(2) For a path P that begins on the same boundary component as the trail and ends on theother boundary component, draw enough copies of the fundamental domain we thatwe can draw P as a connected curve and that there is at least one copy of the trail thatlies completely to the right of P . Then wt ( P, y, t ) is the product of the weights of thefaces that lie to the right of P and to the left of a copy of the trail that is completelyto the right of P times the weight of the trail.(3) For a path P that begins on the boundary component where the trail ends and endson the other boundary component, draw enough copies of the fundamental domain wethat we can draw P as a connected curve and that there is at least one copy of the trailthat lies completely to the right of P . Then wt ( P, y, t ) is the product of the weightsof the faces that lie to the right of P and to the right of a copy of the trail that iscompletely to the right of P times the inverse of the weight of the trail.In Cases 2 and 3, wt ( P, y, t ) is well-defined because if we choose two trails that liecompletely to the right of P , the product of the weights of faces between the trails is 1. Example 3.8.
Suppose we have the following network with face and trail weights: abc abcd d t The trail and trail weight appear inblue, and the trail is oriented fromright to left.11 bc abcd dabc abcd d For the path P shown in red, theinterior of P b is to the left of P . So, wt ( P, y, t ) = ab . abc abcd dabc abcd d For the path P shown in red, P isgoing in the opposite direction of thetrail. So, wt ( P, y, t ) = dtb . Theorem 3.9.
For a path P in a planar directed network on a cylinder, wt ( P, y, t ) = x P . Proof.
We can see this by counting how many times the weight of an edge and its inverse arein the product wt ( P, y, t ) when the edge is in P , when it’s is between P and the boundarycomponent P that makes up part of P b (in Case 1), and when it’s between P and the trail(in Case 2 and 3). Definition 3.10 (Definition 9.2 of [21]) . A perfect network is a planar directed networkin which each boundary vertex has degree 1, and each internal vertex either has exactlyone edge incoming (and all others outgoing) or exactly one edge outgoing (and all othersincoming). Proposition 3.11 (Proposition 9.3 of [21]) . Any planar directed network in a disk can betransformed into a perfect network without changing the boundary measurements.
Proposition 3.12.
Any planar directed network on a cylinder can be transformed into aperfect network without changing the boundary measurements. x x x x x x x x x x x x x x x x x x x x x x x x x x x x x Figure 5: A planar directed network in a disk transformed into a perfect network.
Proof.
The proof for Proposition 3.11 holds for planar directed networks on a cylinder.
Definition 3.13 (Section 9 of [21]) . For an internal vertex v in a perfect network, definethe color of v , col ( v ), to be black if v has exactly one outgoing edge and white if v hasexactly one incoming edge. Theorem 3.14.
Let
N, N ′ be two perfect networks in a disk such that:(1) The underlying graphs G and G ′ are isomorphic as undirected graphs.(2) Each internal vertex of degree = 2 has col N ( v ) = col N ′ ( v ) .(3) If the undirected edge e is directed in the same way in N and N ′ , then x e = x ′ e . If theedge e has opposite direction in N and N ′ , the x e = ( x ′ e ) − .Then M eas ( N ) = M eas ( N ′ ) ∈ Gr kn . Example 3.15.
Let
N, N ′ be as below: N = b b A ( N ) = (cid:2) x x ( x + x ) (cid:3) x x x x N ′ = b b A ( N ′ ) = h ( x x x ) − x x − i = h x x ( x + x ) i x − x x − x − A ( N ′ ) by [ x x ( x + x )] ∈ GL gives A ( N ), these twomatrices represent the same point in the Grassmannian.Notice that in our example, N ′ could be obtained from N by reversing a path fromone boundary vertex to another. In fact, for any two networks N and N ′ satisfying theconditions of Theorem 3.14, N ′ can be obtained from N by reversing a set of paths betweenboundary vertices and a set of cycles. Thus, the theorem can be proven by showing thatreversing paths between boundary vertices and reversing cycles preserve the boundarymeasurement map.As the edge reversal described in Theorem 3.14 does not affect face weights, we may usethis theorem in conjunction with our definition of face weights to obtain a plabic networkfrom any planar directed network on a disk. Thus, plabic networks identify certain directedplanar networks that have the same image under the boundary measurement map.Unfortunately, the statement analogous to Theorem 3.14 for planar directed graphs ona cylinder does not hold. Theorem 3.16 (Theorem 4.1 of [11]) . Let P be a path with no self-intersections from b i to b j in a planar directed network on a cylinder N such that M ij and P does not intersectthe cut. Create N ′ from N by reversing the direction of all the edges in P and invertingtheir weights. Then ( M eas C ( N ′ ))( ζ ) = ( ( M eas C ( N ))( ζ ) b i , b j are on the same boundary component , ( M eas C ( N ))( − ζ ) otherwise . Since we cannot necessarily reverse paths that begin and end on different boundarycomponents without changing the image of the network under the boundary measurementmap, we cannot turn planar directed networks on a cylinder into plabic networks. Inparticular, we have to keep track of the orientation of the edges. However, as path reversalchanges the boundary measurements in a predictable way, plabic networks will still proveuseful to us (see Theorem 3.18).
Proposition 3.17 (Proposition 2.1 of [11]) . Let
N, N ′ be two networks with the samegraph and weights, where N has cut γ and N ′ has cut γ ′ obtained by interchanging one ofthe base points with b , the first boundary vertex below the base point. Then ( − wind ( C ′ P ) − ζ int ( P ′ ) x P ′ = (( − α ( P ) ζ ) β ( b,P ) ( − wind ( C P ) − ζ int ( P ) x P , here x P is as in Definition 2.3 and α ( P ) = ( if the endpoints of P are on the same boundary component, otherwise, β ( b, P ) = if b is the sink of P, − if b is the source of P, otherwise. Theorem 3.18.
Let
N, N ′ be two perfect networks on a cylinder such that:(1) The underlying graphs G and G ′ are isomorphic as undirected graphs.(2) Each internal vertex of degree = 2 has col N ( v ) = col N ′ ( v ) .(3) If the undirected edge e is directed in the same way in N and N ′ , then x e = x ′ e . If theedge e has opposite direction in N and N ′ , the x e = ( x ′ e ) − .Given an involution on the edge weights of N that preserves the boundary measurementmap, then there is a canonical way to define an involution of the edge weights of N ′ .Proof. As with planar directed networks on a disk, we can always obtain N ′ from N byreversing a set of paths and cycles. Therefore, we only need to show the conclusion showsfor N ′ equal to N with a cycle (with no self-intersections) reversed or N with a path (withno self-intersections) reversed.First consider a cycle C with no self-intersections. If C is a contractible loop, thenthe proof that reversing cycles on a disk does not change the boundary measurements stillholds (Lemma 10.5 of [21]).If C is not a contractible loop, a similar proof holds, except that the winding numberis more complicated. We consider what happens for paths that have edges in C . First, forpaths that begin and end on the same boundary component, the winding numbers behavethe same as for networks on a disk. So, the boundary measurements for pairs of verticeson the same boundary component remain the same. Now consider paths that begin andend on different components. For vertices v i and v j in the cycle, going from v i to v j in onedirection around the cycle, with as many loops as desired, crosses the cut the same numberof times from each side and going from v i to v j in the other direction crosses the cut onemore time from one side than from the other. Crossing the cut an additional time adds orremoves a loop in C P as we trace along the cut. So, we get an extra factor of − N ′ with no self-intersections, we can reverse the cycle, apply ourinvolution, and reverse the cycle again. Any boundary measurements that change when15e reverse the cycle change only by a factor of −
1, and they change again by the samefactor when we reverse the cycle a second time. So, for any cycle C , we have the followingcommutative diagram, where f is defined to be the map that makes this diagram commute: N e N N ′ f N ′ cycle reversalinvolution cycle reversal f f is an involution that preserves the boundary measurements.Now consider a path P with no self-intersections. If P does not intersect the cut,Theorem 3.16 says we can reverse P , possibly at the expense of replacing ζ with − ζ . If P does intersect the cut, we can move the cut so that P no longer intersects it. Movingthe cut changes the weight of each path by a power of ζ and a power of −
1. These powersdepend only on the source and sink of the path, so it changes the weight of each boundarymeasurement by a power of ζ and a power of −
1. Then reversing the path, since it no longerintersects the cut, either keeps the boundary measurement map the same, or replaces ζ with − ζ . Finally, we can move the cut back, which will cause each boundary measurementto again pick up a power of − ζ . This process gives us the followingcommutative diagram, where f is defined to be the map that makes the diagram commuteand g is the composition of functions: N e N M f M M ′ f M ′ N ′ f N ′ cut changeinvolution cut change path reversalpath reversal cut changecut change gg f It is clear f is an involution, so we just need to check that M eas C ( N ′ ) = M eas C ( f N ′ ).Given a matrix representing the image of a network under the boundary measurementmap, g is equivalent to possibly switching ζ for − ζ and multiplying each entry by a power16f − ζ . Since M eas C ( N ) = M eas C ( e N ), we can pick the same matrixrepresentative for them, and we can see M eas C ( g ( N )) = M eas C ( g ( e N ))Since the plabic network structure is useful to us, but also we can’t eliminate orientation,we will be working with directed plabic networks. Definition 3.19. A directed plabic graph on a cylinder is a planar directed graph on acylinder such that each boundary vertex has degree 1 and each internal vertex is coloredblack of white. A directed plabic network on a cylinder is a directed plabic graph with aweight y f ∈ R > assigned to each face and a specified trail with weight t . Postnikov solves the inverse boundary problem for plabic networks up to a set of localtransformations which do not alter the boundary measurements (Theorem 12.1 of [21]).These are as follows:(M1) Square move. y y y y y ↔ y − y y − y (1 + y ) y y − y (1 + y )(M2) Unicolored edge contraction/uncontraction. y y y y y ↔ y y y y y The unicolored edge may be white (as pictured) or black and there may be anynumber of edges on each of the vertices. All of the face weights remain unchanged.(M3) Middle vertex insertion/removal y y ↔ y y Vertex insertion/removal may be done with a vertex of either color.17R1) Parallel edge reduction y y y → y y − y (1 + y )(R2) Leaf reduction y y y → y y y The vertex with degree 1, which we call a leaf, may be any color, and the vertexconnected to the leaf (which is of the opposite color) may have any degree ≥ y → y All of these transformations specialize to directed edge weighted versions. From here,we will be using these transformations freely, and considering (directed) plabic networksthat differ by them as equivalent. Our goal in this section is to define and explore asemi-local transformation for planar directed networks on a cylinder.
Definition 4.1.
The transformations (M1) - (M3) are called moves and the transforma-tions (R1) - (R3) are called reductions . Two plabic graphs or networks are move-equivalent if they can be transformed into the same graph or network by moves.
Definition 4.2.
A plabic graph or network is reduced if it has no isolated connectedcomponents and it is not move-equivalent to any graph or network to which we can apply(R1) or (R2). A plabic graph or network is leafless if it has no non-boundary leaves.
Definition 4.3. A Postnikov diagram , also known as an alternating strand diagram, on asurface with boundary is a set of directed curves, called strands , such that when we drawthe strands on the universal cover of the surface we have the following:(1) Each strand begins and ends at a boundary vertex or is infinite.182) There is exactly one strand that enters and one strand that leaves each boundaryvertex.(3) No three strands intersect at the same point.(4) All intersections are transverse (the tangent vectors are independent).(5) There is a finite number of intersections in each fundamental domain.(6) Along any strand, the strands that cross it alternate crossing from the left and crossingfrom the right.(7) Strands do not have self-intersections, except in the case where a strand is a loopattached to a boundary vertex. Notice that this excludes the possibility of a closedcycle.(8) If two strands intersect at u and v , then one strand is oriented from u to v and onestrand is oriented from v to u .Postnikov diagrams are considered up to homotopy. We can obtain a plabic graph froma Postnikov diagram as follows:(1) Place a black vertex in every face oriented counterclockwise and a white vertex in everyface oriented clockwise.(2) If two oriented faces share a corner, connect the vertices in these two faces.Figure 6: A plabic graph and its Postnikov diagram.19otice that if our surface is a disk, we recover Definition 14.1 of [21] for a Postnikovdiagram. Theorem 4.4 (Corollary 14.2 of [21]) . Postnikov diagrams in a disk are in bijection withleafless reduced plabic graphs in a disk with no unicolored edges.
Theorem 4.5.
Postnikov diagrams on a cylinder are in bijection with leafless reducedplabic graphs on a cylinder with no unicolored edges.
See Section 8.1 for proof.
Definition 4.6.
A cylindric k -loop plabic graph is a plabic graph on a cylinder that arisesfrom a Postnikov diagram where exactly k of the strands are loops around the cylinderwith the same orientation.Cylindric k -loop plabic graphs have k strings of vertices around the cylinder. Thosestrings alternate black and white vertices, and the black vertices only have additional edgeson the left of the strand while the white vertices only have additional edges to the right ofthe strand (see Figure 7).plabic graph Postnikov diagramFigure 7: A cylindric 2-loop plabic graph and its Postnikov diagram. Definition 4.7.
For a cylindric k -loop plabic graph, any vertices that are not on one ofthe strings of vertices defined by the k loops and lie between two of these strings are called interior vertices . Theorem 4.8.
Any cylindric k -loop plabic graph can be transformed by moves to one thathas no interior vertices. k -loop plabic graph. By Theorem 8.9, we can choose two adjacentstrings and assume there are no interior vertices between them. We will be describing aninvolution on the edge weights of these two adjacent strings and the edges between them,so we’ll ignore the rest of the graph. That is, we’ll assume we have a cylindric 2-loop plabicgraph with no vertices other than those on the 2 strings. Definition 4.9.
The canonical orientation of a cylindric 2-loop plabic graph is the ori-entation where the edges on the strings are oriented up and the other edges are orientedfrom left to right.Let us choose the edges from white vertices to black vertices to be variables, and setall the other edges to have weight 1. We do this to have a canonical way to kill the gaugetransformations on our network (notice the number of variables is the number of faces, orthe dimension of R E / { gauge transformations } ). In our diagrams, any edges that are notlabeled are assumed to have weight 1. We now have a directed plabic network. Definition 4.10.
We can expand the directed plabic network by splitting each vertex thathas multiple edges to the other string into that many vertices, and inserting vertices of theopposite color between them. Let any new edges created have weight 1. Thus, we havea new network that is equivalent to the old one, but all the interior faces are hexagonswhere the colors of the vertices alternate and where there are 2 white vertices on the leftstring and 2 black vertices on the right string. We’ll call this the expanded directed plabicnetwork .Choose a white vertex on the left string of a cylindric 2-loop expanded directed plabicnetwork. Call the weight of the edge from this vertex to the black vertex above it on thestring x . For the white vertex on the left string that is part of the same face as thesetwo vertices, call the weight of the edge from this vertex to the black vertex above it onthe string y . Call the weight of the edge that makes up the upper boundary of the facecontaining these vertices z . Moving up the left string, give the next white to black edgethe weight x , and so on. Do the same on the right. Moving in the same direction, labelall the edges between the strings with weights z , z , etc. For a particular network, someof these values might be set to 1, because we created the edges when we expanded thenetwork. Let x = { x , x , ..., x n } , y = { y , y , ..., y n } , z = { z , z , ..., z n } . We will considerall of these indices to be modular.Consider the network on the universal cover of the cylinder. Choose a fundamentaldomain. Label all the edge weights in the fundamental domain with a superscript (1),so the weights are x (1)1 , y (1)1 , z (1)1 , x (1)2 , etc. Label all the edge weights in the copy of thefundamental domain that lies above with a superscript (2), and so on. Define λ i ( x, y, z ) =sum of the path weights from vertex a to vertex b , where vertex a is a the highest vertex on21 bcd efghijk ℓmn directed plabic network abcd efghijkℓmn expanded directed plabic networkFigure 8: A directed plabic network and the corresponding expanded directed plabic net-work.the left string of the interior face that has an edge labeled x (1) i and b is the lowest vertex onthe right string of the interior face that has an edge labeled x (2) i . Define A i = { i − j | j ≥ , x i − = ... = x i − j = 1 } and α i = | A i | . Define, B i = { i + j | j ≥ , y i +1 = ... = y i + j = 1 } and β i = | B i | . 22 xample 4.11. Consider the network below: x (1)1 x (2)1 x (1)2 x (2)2 x (1)3 x (2)3 x (1)4 x (1)5 x (1)6 x (0)6 y (1)6 y (2)6 y (1)1 y (2)1 y (1)2 y (2)2 y (1)3 y (1)4 y (1)5 y (0)5 z (1)6 z (1)6 z (1)1 z (1)1 z (1)2 z (1)2 z (1)3 z (1)4 z (1)5 z (0)5 λ ( x, y, z ) = paths from the vertexat the bottom of x (1)2 to the vertexat the top of y (2)6 , so λ ( x, y, z ) = z y y y y y + x z y y y y + x x z y y y + x x x z y y + x x x x z y + x x x x x z . A = ∅ because there are no weightsdirectly below x set to 1 from ex-panding the network. B = ∅ because there are no weightsdirectly above y set to 1 from ex-panding the network. α = β = 0. λ ( x, y, z ) = z y y y y y + x z y y y y + x x z y y y + x x x z y y + x x x x z y + x x x x x z . A = ∅ because there are no weightsdirectly below x set to 1 from ex-panding the network. B = { } because y , which is di-rectly above y is set to 1 from ex-panding the network, but y is notset to 1. α = 0 , β = 1.23 efinition 4.12. Define T e to be the transformation on edge weights from ( x, y, z ) to( x ′ , y ′ , z ′ ) where x ′ i = x i set to 1 , ( Q αik =0 y i − k − ) λ i − αi − ( x,y,z ) λ i ( x,y,z ) otherwise, y ′ i = y i set to 1 , (cid:16)Q βik =0 x i + k +1 (cid:17) λ i + βi +1 ( x,y,z ) λ i ( x,y,z ) otherwise, z ′ i = z i (cid:0)Q α i k =1 y i − k − (cid:1) λ i − α i − ( x, y, z ) (cid:16)Q β i k =1 x i + k +1 (cid:17) λ i + β i +1 ( x, y, z ) λ i − ( x, y, z ) λ i +1 ( x, y, z ) . We call T e the edge weighted plabic R-matrix . Theorem 4.13. T e has the following properties:1. It preserves the boundary measurements.2. It is an involution.3. ( x, y, z ) and ( x ′ , y ′ , z ′ ) are the only choices of weights on a fixed cylindric 2-loop plabicgraph that preserve the boundary measurements.4. It satisfies the braid relation. See Section 9 for proof.
Example 4.14.
Consider the network below: x x x x y y y y z z z z e gives us the following values: x ′ = y λ ( x, y, z ) λ ( x, y, z ) = y ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y y x ′ = y λ ( x, y, z ) λ ( x, y, z ) = y ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y x ′ = y λ ( x, y, z ) λ ( x, y, z ) = y ( x x x z + x x z y + x z y y + z y y ) x x x z + x x z y + x z y y + z y y y x ′ = y λ ( x, y, z ) λ ( x, y, z ) = y ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y y y ′ = x λ ( x, y, z ) λ ( x, y, z ) = x ( x x x z + x x z y + x z y y + z y y ) x x x z + x x z y + x z y y + z y y y y ′ = x λ ( x, y, z ) λ ( x, y, z ) = x ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y y ′ = x λ ( x, y, z ) λ ( x, y, z ) = x ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y y y ′ = x λ ( x, y, z ) λ ( x, y, z ) = x ( x x x z + x x z y + x z y y + z y y y ) x x x z + x x z y + x z y y + z y y y z ′ = z z ′ = z z ′ = z z ′ = z Notice that in this example, the directed plabic network is the same as the expandeddirected plabic network. These types of networks are of particular interest to us, bothbecause of their simplicity and because they correspond to the wiring diagrams studied inSection 6 of [19] where the horizontal wires are all in the same direction and either all thewire cycles are whirls or all the wire cycles are curls. Thus, if we let all of our z variablesequal 1, we have recovered the whurl relation of [19], which is also the geometric R-matrix.25 xample 4.15. Consider the network below: x x x y = 1 y y = 1 z z z x x x y y = 1 y z z z T e gives us the following values: x ′ = y λ ( x, y, z ) λ ( x, y, z )= x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y x ′ = y λ ( x, y, z ) λ ( x, y, z )= y ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y x ′ = y λ ( x, y, z ) λ ( x, y, z )= x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ′ = y λ ( x, y, z ) λ ( x, y, z )= y ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y x ′ = y λ ( x, y, z ) λ ( x, y, z )= x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y x x x x x z + x x x x z + x x x z y + x x z y + x x z y + x z y y + z y y x ′ = y λ ( x, y, z ) λ ( x, y, z )= y ( x x x x x z + x x x x z + x x x z y + x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y y ′ = x x λ ( x, y, z ) λ ( x, y, z )= x x ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y y ′ = 1 y ′ = x x λ ( x, y, z ) λ ( x, y, z )= x x ( x x x x x z + x x x x z + x x x z y + x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y y ′ = 1 y ′ = x x λ ( x, y, z ) λ ( x, y, z )= x x ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z + x x x z y + x x z y + x x z y + x z y y + z y y y ′ = 1 z ′ = z x λ ( x, y, z ) λ ( x, y, z )= z x ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y z ′ = z z ′ = z x λ ( x, y, z ) λ ( x, y, z )= z x ( x x x x x z + x x x x z + x x x z y + x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y z ′ = z ′ = z x λ ( x, y, z ) λ ( x, y, z )= z x ( x x x x x z + x x x x z + x x x z y + x x z y + x z y y + z y y ) x x x x x z + x x x x z y + x x x z y + x x z y y + x z y y + z y y y z ′ = z We will now rewrite T e in terms of face and trail weights. We begin with a cylindric2-loop plabic graph with the canonical orientation. Choose an edge adjacent to the leftboundary. Follow this edge, and then go up the left string. At the first opportunity, makea right to cross to the right string. Follow the right string up and at the first opportunitymake a right to the right boundary. If there is more than one edge on the right stringin this path, then the path passes a face between strings that has no edge to the rightboundary. This face must have an edge to the left boundary. Choose this edge to theleft boundary to start with and repeat. When this process yields a path that has only 5edges, we will select that to be our trail (going from the left to right boundary). Let thetrail weight be t . Label the faces on the left a , a , ..., a ℓ starting above the trail and goingup. In the same way, label the faces on the right b , b , ..., b m and label the faces in thecenter c , c , ..., c n − , c n = a a ...a ℓ b b ...b m c c ...c n − . We will consider all of these indices tobe modular.We will say a j is associated to i if the highest edge on the left string bordering the facelabeled a j also borders the face c i . Similarly, b j is associated to i if the lowest edge on theright string bordering the face labeled b j also borders the face c i . Lemma 4.16.
Suppose we have an expanded cylindric 2-loop plabic graph with the face andtrail weights as above. We can turn this into a directed graph with the canonical orientationand the following edge weights: • For a face labeled a j , give the edge from a white vertex to a black vertex on the leftstring that is highest on the face the weight a − j . If a j is associated to i , this shouldbe the edge that borders both the face labeled a j and the face labeled c i . • For a face labeled b j , give the edge from a white vertex to a black vertex on the leftstring that is lowest on the face the weight b j . If b j is associated to i , this should bethe edge that borders both the face labeled b j and the face labeled c i . • Give all other edges on the strings weight 1. • Give the edge between the two strings that is part of the trail the weight t . • Give each edge between the two strings the weight of the edge below it multiplied bythe weight of the face in between and the weight of the edges of the face on the right tring, and then divided by the weight of the edges of the face on the left string. Thatis, if an edge is between the faces labeled c i and c i +1 , give it the weight t Y j where a j isassociated to k ≤ i a j Y j where b j isassociated to k ≤ i b j Y j ≤ i c j . Proof.
Clear by computation.Using this, we can define b λ i ( a, b, c ) to be λ i ( z, y, z ), where x, y, z are defined from a, b, c as in Lemma 4.16 and we choose our indexing so that y = b . a a a a b b b b c c c c c c t x = a − x = a − x = a − x = 1 x = 1 x = a − y = b y = 1 y = b y = 1 y = b y = b z = a a a b b b b c c c c c t z = t z = a b c t z = a a b c c t z = a a b b c c c t z = a a b b b c c c c t Figure 9: Changing an expanded cylindric 2-loop plabic network with face and trail weightsinto one with edge weights as in Lemma 4.16.29 efinition 4.17.
Define T f to be the transformation on face weights from ( a, b, c ) to( a ′ , b ′ , c ′ ) where a ′ i = b λ j ( a, b, c ) b λ p ( a, b, c ) (cid:18)Q b r associated to s where p ≤ s Example 4.19. We revisit Example 4.14, but with face variables this time. a a a a b b b b c c c c t f gives us the following values: a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a a a a b b b b ( c + c c + c c c ) + 1 a a a b b b b (1 + c + c c + c c c ) a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a ( a a a a b b b b ( c c + c c c ) + 1 + c ) a a a a b b b b ( c + c c + c c c ) + 1 a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a ( a a a a b b b b c c c + 1 + c + c c ) a a a a b b b b ( c c + c c c ) + 1 + c a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a a a a b b b b (1 + c + c c + c c c ) a a a a b b b b c c c + 1 + c + c c b ′ = b λ ( a, b, c ) a b λ ( a, b, c ) = b ( a a a a b b b b ( c c + c c c ) + 1 + c a a a a b b b b ( c + c c + c c c ) + 1 b ′ = b λ ( a, b, c ) a b λ ( a, b, c ) = b ( a a a a b b b b c c c + 1 + c + c c ) a a a a b b b b ( c c + c c c ) + 1 + c b ′ = b λ ( a, b, c ) a b λ ( a, b, c ) = b (1 + c + c c + c c c ) a a a a b b b b c c c + 1 + c + c c b ′ = b λ ( a, b, c ) a b λ ( a, b, c ) = a a a a b b b b ( c + c c + c c c ) + 1 a a a a b b b (1 + c + c c + c c c ) c ′ = a a b b c b λ ( a, b, c ) b λ ( a, b, c ) = a a a a b b b b c (1 + c + c c + c c c ) a a a a b b b b ( c c + c c c ) + 1 + c c ′ = a a b b c b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a b b b b ( c + c c + c c c ) + 1) a a a a b b b b c c c + 1 + c + c c c ′ = a a b b c b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a b b b b ( c c + c c c ) + 1 + c )1 + c + c c + c c c c ′ = a a b b c b λ ( a, b, c ) b λ ( a, b, c ) = a a a a b b b b c ( a a a a b b b b c c c + 1 + c + c c ) a a a a b b b b ( c + c c + c c c ) + 131 xample 4.20. We revisit Example 4.15, but with face variables this time. a a a a a a b b b c c c c c c tT f gives us the following values: a ′ = b λ ( a, b, c ) b λ ( a, b, c ) = a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1 a a a a a b b b (1 + c + c c + c c c + c c c c + c c c c c ) a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a ( a a a a a a b b b ( c c + c c c + c c c c + c c c c c ) + 1 + c ) a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1 a ′ = b λ ( a, b, c ) b λ ( a, b, c ) = a ( a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c ) a a a a a a b b b ( c c + c c c + c c c c + c c c c c ) + 1 + c a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a ( a a a a a a b b b ( c c c c + c c c c c ) + 1 + c + c c + c c c ) a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c a ′ = b λ ( a, b, c ) b λ ( a, b, c ) = a ( a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c ) a a a a a a b b b ( c c c c + c c c c c ) + 1 + c + c c + c c c a ′ = b λ ( a, b, c ) b b λ ( a, b, c ) = a (1 + c + c c + c c c + c c c c + c c c c c ) a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c ′ = b λ ( a, b, c ) a a b λ ( a, b, c ) = b ( a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c ) a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1 b ′ = b λ ( a, b, c ) a a b λ ( a, b, c ) = b ( a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c ) a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c b ′ = b λ ( a, b, c ) a a b λ ( a, b, c )= b ( a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1) a a a a a a b b b ( a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c ) c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c ) = c a a a a a a b b b (1 + c + c c + c c c + c c c c + c c c c c ) a a a a a a b b b ( c c + c c c + c c c c + c c c c c ) + 1 + c c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1) a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a a a b b b ( c c + c c c + c c c c + c c c c c ) + 1 + c ) a a a a a a b b b ( c c c c + c c c c c ) + 1 + c + c c + c c c c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a a a b b b ( c c c + c c c c + c c c c c ) + 1 + c + c c ) a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c ) = c ( a a a a a a b b b ( c c c c + c c c c c ) + 1 + c + c c + c c c )1 + c + c c + c c c + c c c c + c c c c c c ′ = c a a b b λ ( a, b, c ) b λ ( a, b, c )= a a a a a a b b b c c c c c + 1 + c + c c + c c c + c c c c c c c c c ( a a a a a a b b b ( c + c c + c c c + c c c c + c c c c c ) + 1) Definition 5.1. A quiver Q is a directed graph with vertices labeled 1 , ..., n and no loopsor 2-cycles. Definition 5.2. If k is a vertex in a quiver Q , a quiver mutation at k , µ k ( Q ) is definedfrom Q as follows:(1) for each pair of edges i → k and k → j , add a new edge i → j ,(2) reverse any edges incident to k ,(3) remove any 2-cycles. 33 efinition 5.3. A seed is a pair ( Q, x ) where Q is a quiver and x = ( x , ..., x n ) with n the number of vertices of Q . Definition 5.4. If k is a vertex in Q , we can define a seed mutation of ( Q, x ) at k , µ k ( Q, x ) = ( Q ′ , x ′ ) by Q ′ = µ k ( Q ) and x ′ = ( x ′ , ..., x ′ n ) where x ′ i = ( x i i = k, Q nj =1 x { edges k → j in Q } j + Q nj =1 x { edges j → k in Q } j i = k. Recall that Q has no 2-cycles, so x j shows up in at most one of the products in the formulafor x ′ k . Definition 5.5. A seed with coefficients is a triple ( Q, x , y ) where Q is a quiver, x =( x , ..., x n ), and y = ( y , ..., y n ) with n the number of vertices of Q . Definition 5.6. If k is a vertex in Q , we can define a seed mutation of ( Q, x , y ) at k , µ k ( Q, x , y ) = ( Q ′ , x ′ , y ′ ) by Q ′ = µ k ( Q ) , x ′ = ( x ′ , ..., x ′ n ), and y ′ = ( y ′ , ..., y ′ n ) where x ′ i = ( x i i = k, Q nj =1 x { edges k → j in Q } j + Q nj =1 x { edges j → k in Q } j i = k,y ′ i = y − k i = ky i (1 + y − k ) − { edges k → i in Q } i = k, { edges k → i in Q } ≥ ,y i (1 + y k ) { edges i → k in Q } i = k, { edges i → k in Q } ≥ . Definition 5.7. A y -seed , ( Q, y ) and its mutations are defined as above, but without the x variables.Frequently the y -dynamics above will be defined over a semifield that may have adifferent addition. However, for our purposes, we will use these less general definitions. In [13], Inoue, Lam, and Pylyavskyy obtain the geometric R-matrix from a sequence ofcluster mutations of a triangular grid quiver. Triangular grid quivers are exactly the dualquivers to the plabic graph associated to certain wiring diagrams, such as the plabic graphin Example 4.14. In [12], Goncharov and Shen show that in another family of quivers, thesame mutation sequence gives a Weyl group action. The rest of this section will explorehow these cases generalize. 34 efinition 6.1. A spider web quiver is a quiver constructed as follows. We begin withthree or more concentric circles. Place as many vertices as desired (at least 2 so as to avoidloops) on each circle. Orient the edges of each circle counter clockwise. Then add edgesin the diagram between vertices on adjacent circles so that each face is oriented, containstwo edges between circles, and has at least 3 sides.For the purposes of this paper, we will assume a spider web quiver has 3 circles, unlessstated otherwise. Label the vertices of the middle circle 1 , ..., n following the arrows aroundthe circle. If the lowest index vertex in the middle circle with an edge to the outer circle is i , label the vertex in the outer circle which has an arrow to i as 1 − . Continue labeling thevertices of the outer circle 2 − , − , ... following the arrows around the circle. If the lowestindex vertex in the middle circle with an edge to the inner circle is j , label the vertex inthe inner circle which has an arrow to j as 1 + . Continue labeling the vertices of the outercircle 2 + , + , ... following the arrows around the circle. See Figure 10 for examples.12 3 41 + + + + − − − − − − − − − − 12 3 4 5 61 + + + Figure 10: Two spider web quivers. The quiver on the left is a triangular grid quiver, asin [13]. The quiver on the right is a 3-triangulation with 3 ideal triangles, as in Section 8of [12].We will always let n be the number of vertices on the middle circle of a spider webquiver. We’ll be considering the transformation τ := µ µ ...µ n − s n − ,n µ n µ n − ...µ where s n − ,n is the operation that transposes vertices n and n − Proposition 6.2. If Q is a spider web quiver, then τ ( Q ) = Q. For proof, see Section 10.1. 35or a spider web quiver, let α be the largest vertex on the outer circle connected to inmiddle circle and β be the largest vertex on the inner circle connected to the middle circle.Let x [ i ] − := x a if there is at least one vertex on the outer circle with an edge to anelement of { , ..., i } , and the largest such vertex is a, . Let x [ i ] + := x b if there is at least one vertex on the inner circle with an edge to anelement of { , ..., i } , and the largest such vertex is b, . Let x [ i ] −∗ := x a if there is at least one vertex on the outer circle with an edge to anelement of { , ..., i } , and the largest such vertex is a,x α otherwise . Let x [ i ] + ∗ := x b if there is at least one vertex on the inner circle with an edge to anelement of { , ..., i } , and the largest such vertex is b,x β otherwise . Theorem 6.3. For a spider web quiver Q , let x = P n − j =1 (cid:16)(cid:16)Q j − k =1 x k (cid:17) (cid:16)Q nk = j +2 x k (cid:17) x [ j ] −∗ x [ j ] + ∗ (cid:17) + (cid:16)Q n − k =2 x k (cid:17) x [ n ] − x [ n ] + Q nk =1 x k . Then, τ ( x i ) = x i x. For proof, see Section 10.2. Example 6.4. Consider the quiver on the left of Figure 10. In this case, we have x = x x x [1] −∗ x [1] + ∗ + x x x [2] −∗ x [2] + ∗ + x x x [3] −∗ x [3] + ∗ + x x x [4] − x [4] + x x x x = x x x − x + + x x x − x + + x x x − x + + x x x − x + x x x x So, the x -variables after applying τ are as follows: x ′ = x x x − x + + x x x − x + + x x x − x + + x x x − x + x x x x ′ = x x x − x + + x x x − x + + x x x − x + + x x x − x + x x x x ′ = x x x − x + + x x x − x + + x x x − x + + x x x − x + x x x x ′ = x x x − x + + x x x − x + + x x x − x + + x x x − x + x x x x -variables remain the same because there are no mutations are the corre-sponding vertices. Example 6.5. Consider the quiver on the right of Figure 10. In this case, we have x = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x So, the x -variables after applying τ are as follows: x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x x ′ = x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x + x x x x x − x x x x x x All other x -variables remain the same because there are no mutations are the corre-sponding vertices. Theorem 6.6. For a spider web quiver, τ is an involution on the x -variables.Proof. We know that for each k , µ k is an involution and also s n − ,n is an involution. Thismeans that for each i , µ µ ...µ n s n − ,n µ n − µ n − ...µ µ µ ...µ n − s n − ,n µ n µ n − ...µ ( x i ) = x i . Since µ n s n − ,n = s n − ,n µ n − and µ n − s n − ,n = s n − ,n µ n , we have( µ µ ...µ n − s n − ,n µ n µ n − ...µ ) ( x i ) = x i . Theorem 6.7. Suppose we have a spider web quiver with any number of concentric circles.Choose 2 adjacent circles. If we apply τ to one circle, then the other, then the first again,we obtain the same x -variables as if we had applied τ to the second circle, then the first,then the second again.Proof. Suppose the first circle has vertices labeled with variables a , ..., a n and the secondcircle has vertices labeled with b , ..., b m . Without loss of generality, we can assume thefirst circle is outside the second. If we apply the involution to the first circle, the onlyvariables that change are a , ..., a n , which get replaced with aa , ..., aa n . We denote thenew variables with primes, that is a ′ i = aa i and b ′ i = b i .37ext we apply the involution to the second circle. This means b ′ , ..., b ′ m get replacedwith b ′ b ′ , ..., b ′ b ′ m . Let’s compute b ′ . b ′ = P m − j =1 (cid:16)(cid:16)Q j − k =1 b ′ k (cid:17) (cid:16)Q mk = j +2 b ′ k (cid:17) b ′ [ j ] −∗ b ′ [ j ] + ∗ (cid:17) + (cid:16)Q m − k =2 b ′ k (cid:17) b ′ [ m ] − b ′ [ m ] + Q mk =1 b ′ k = P m − j =1 (cid:16)(cid:16)Q j − k =1 b k (cid:17) (cid:16)Q mk = j +2 b k (cid:17) ab [ j ] −∗ b [ j ] + ∗ (cid:17) + (cid:16)Q m − k =2 b k (cid:17) ab [ m ] − b [ m ] + Q mk =1 b k = ab So, b ′ b ′ i = abb i . We denote the new variables with additional primes, that is a ′′ i = aa i and b ′′ i = abb i .Finally, we apply the involution to the first circle again. This means a ′′ , ..., a ′′ n getreplaced with a ′′ a ′′ , ..., a ′′ a ′′ n . Let’s compute a ′′ . a ′′ = P m − j =1 (cid:16)(cid:16)Q j − k =1 a ′′ k (cid:17) (cid:16)Q mk = j +2 a ′′ k (cid:17) a ′′ [ j ] −∗ a ′′ [ j ] + ∗ (cid:17) + (cid:16)Q m − k =2 a ′′ k (cid:17) a ′′ [ m ] − a ′′ [ m ] + Q mk =1 a ′′ k = P m − j =1 (cid:16)(cid:16)Q j − k =1 aa k (cid:17) (cid:16)Q mk = j +2 aa k (cid:17) a [ j ] −∗ aba [ j ] + ∗ (cid:17) + (cid:16)Q m − k =2 aa k (cid:17) a [ m ] − aba [ m ] + Q mk =1 aa k = P m − j =1 (cid:16)(cid:16)Q j − k =1 a k (cid:17) (cid:16)Q mk = j +2 a k (cid:17) a [ j ] −∗ ba [ j ] + ∗ (cid:17) + (cid:16)Q m − k =2 a k (cid:17) a [ m ] − ba [ m ] + a Q mk =1 a k = b So, a ′′ a ′′ i = aba i .Thus, our variables after applying the involution three times are aba , ..., aba n and abb , ..., abb m . By symmetry, we can see that if we had applied the involution to the secondcircle, then the first, then the second again, we would have the same variables.Now we will discuss the y -dynamics of the quiver. Define y := 1 for ease of notation. Theorem 6.8. For any vertex i − connected to the middle circle, let d i be the minimal j so that i − is connected to j and e i be the maximal j so that i − is connected to j . Similarly,For any vertex i + connected to the middle circle, let f i be the minimal j so that i + isconnected to j and g i be the maximal j so that i + is connected to j . If we apply τ to aspider web quiver Q , we obtain the following y -variables:(1) If ≤ i ≤ n − , y ′ i = P nj = i Q j − k = i y k + ( Q nk = i y k ) (cid:16)P i − j =0 Q jk =0 y k (cid:17)P n − j = i +1 Q jk = i +1 y k + (cid:0)Q nk = i +1 y k (cid:1) (cid:16)P ij =0 Q jk =0 y k (cid:17) y ′ n = 1 + y n P n − j =0 Q jk =0 y k P n − j =1 Q jk =1 y k + Q nk =1 y k (3) If there are no edges between i − and the middle circle, then y ′ i − = y i − (4) If i is maximal so that there are edges between i − and the middle circle, then y ′ i − = y i − (cid:0)Q nk = e i +1 y k (cid:1) (cid:16)P n − j = d Q jk =0 y k + ( Q nk =0 y k ) (cid:16)P d − j =0 Q jk =0 y k (cid:17)(cid:17)P nj = e i +1 Q j − k = e i +1 y k + (cid:0)Q nk = e i +1 y k (cid:1) (cid:16)P e i − j =0 Q jk =0 y k (cid:17) (5) Otherwise, y ′ i − = y i − P n − j = e i Q jk = d i +1 y k + (cid:0)Q nk = d i +1 y k (cid:1) (cid:16)P e i − j =0 Q jk =0 y k (cid:17)P nj = d i +1 Q j − k = d i +1 y k + (cid:0)Q nk = d i +1 y k (cid:1) (cid:16)P d i − j =0 Q jk =0 y k (cid:17) (6) If there are no edges between i + and the middle circle, then y ′ i + = y i + (7) If i is maximal so that there are edges between i + and the middle circle, then y ′ i + = y i + (cid:16)Q nk = g i +1 y k (cid:17) (cid:16)P n − j = f Q jk =0 y k + ( Q nk =0 y k ) (cid:16)P f − j =0 Q jk =0 y k (cid:17)(cid:17)P nj = g i +1 Q j − k = g i +1 y k + (cid:16)Q nk = g i +1 y k (cid:17) (cid:16)P g i − j =0 Q jk =0 y k (cid:17) (8) Otherwise, y ′ i + = y i + P n − j = g i Q jk = f i +1 y k + (cid:16)Q nk = f i +1 y k (cid:17) (cid:16)P g i − j =0 Q jk =0 y k (cid:17)P nj = f i +1 Q j − k = f i +1 y k + (cid:16)Q nk = f i +1 y k (cid:17) (cid:16)P f i − j =0 Q jk =0 y k (cid:17) For proof, see Section 10.3. Theorem 6.9. For a spider web quiver, τ is an involution on the y -variables.Proof. Theorem 4.1 of [20] tells us that periodicity of x - and y -variables are equivalent.So, this follows from Theorem 6.6. Theorem 6.10. Suppose we have a spider web quiver with any number of concentriccircles. Choose 2 adjacent circles. If we apply τ to one circle, then the other, then the firstagain, we obtain the same y -variables as if we had applied τ to the second circle, then thefirst, then the second again. roof. By Theorem 4.1 of [20], this follows from Theorem 6.7. Example 6.11. Consider the quiver on the left of Figure 10. In this case, we have: y ′ = 1 + y + y y + y y y y + y y + y y y + y y y y y ′ = 1 + y + y y + y y y y + y y + y y y + y y y y y ′ = 1 + y + y y + y y y y + y y + y y y + y y y y y ′ = 1 + y + y y + y y y y + y y + y y y + y y y y y ′ − = y − ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ − = y − ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ − = y − ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ − = y − ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ + = y + ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ + = y + ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ + = y + ( y + y y + y y y + y y y y )1 + y + y y + y y y y ′ + = y + ( y + y y + y y y + y y y y )1 + y + y y + y y y Example 6.12. Consider the quiver on the right of Figure 10. In this case, we have: y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y ′ = 1 + y + y y + y y y + y y y y + y y y y y y + y y + y y y + y y y y + y y y y y + y y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ − = y − ( y + y y + y y y + y y y y + y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ + = y + ( y y + y y y + y y y y + y y y y y + y y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ + = y + ( y y + y y y + y y y y + y y y y y + y y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y y ′ + = y + ( y y + y y y + y y y y + y y y y y + y y y y y y + y y y y y y )1 + y + y y + y y y + y y y y + y y y y y Recall in Section 4 that we labeled the faces in the left column of a cylindric 2-loop plabicnetwork a , ..., a ℓ starting above the trail and going up. We labeled the faces on the right b , ..., b m and the faces in the center c , ..., c n − , c n = a ...a ℓ b ...b m c ...c n − . Proposition 7.1. Let Q be the dual quiver to a cylindric 2-loop plabic network. Label theface and trail weights of the plabic network as in Section 4. Let c = 1 for ease of notation.Setting each y -variable equal to the corresponding face weight and applying τ yields thefollowing y -variables:(1) For i ≤ n , y i = c i (cid:16)Q i − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = i Q j − k = i c k (cid:17) + P i − j =0 Q jk =0 c k (cid:16)Q i +1 k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = i +2 Q j − k = i +2 c k (cid:17) + P ij =0 Q jk =0 c k y n = (cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) + P n − j =0 Q jk =0 c k (cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j =1 Q jk =1 c k (cid:17) + Q kk =1 c k (3) If i is maximal, y i − = P d − j =0 Q jk =0 c k + (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j = d Q jk =0 c k (cid:17)(cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j =0 Q jk =0 c k (cid:17) (4) If i is second-largest, y i − = P n − j =0 Q jk =0 c k (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = d i +1 Q j − k =1 c k (cid:17) + P d i − j =0 Q jk =1 c k (5) For other i , y i − = (cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = e i +1 Q j − k = d i +1 c k (cid:17) + (cid:16)Q n − k = d i +1 c k (cid:17) (cid:16)P e i − j =0 Q jk =0 c k (cid:17)(cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = d i +1 Q j − k = d i +1 c k (cid:17) + (cid:16)Q n − k = d i +1 c k (cid:17) (cid:16)P nj =0 Q jk = d i − c k (cid:17) (6) If i is maximal and n ∈ B , y i + = 1 + (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j =1 Q jk =0 c k (cid:17)(cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j =0 Q jk =0 c k (cid:17) (7) If i is second largest and n ∈ B , y i + = P n − j =0 Q sk =0 c k (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = f i +1 Q j − k =1 c k (cid:17) + P f i − j =0 Q sk =1 c k (8) If i is maximal and n B , y i + = 1 + (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j =1 Q jk =0 c k (cid:17)(cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)(cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P n − j = g i Q jk =0 c k (cid:17) + P g i − j =0 Q jk =0 c k (cid:17) (9) For other i , y i + = (cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = g i +1 Q j − k = f i +1 c k (cid:17) + (cid:16)Q n − k = f i +1 c k (cid:17) (cid:16)P g i − j =0 Q jk =0 c k (cid:17)(cid:16)Q n − k =1 c k (cid:17) (cid:16)Q ℓk =1 a k (cid:17) ( Q mk =1 b k ) (cid:16)P nj = f i +1 Q j − k = f i +1 c k (cid:17) + (cid:16)Q n − j = f i +1 c k (cid:17) (cid:16)P nj =0 Q jk = f i − c k (cid:17) roof. Notice that because of the way the faces were numbered, n ∈ A and f = 1 forany such quiver. Then we can find the formulas for the y -variables by computation andTheorem 6.8. Theorem 7.2. Let Q be the dual quiver to a cylindric 2-loop plabic network. Label theface and trail weights of the plabic network as in Section 4. If we set each y -variable equalto the corresponding face weight and apply τ , the y -variables we obtain are the same as theface variables with the transformation T f applied to them.Proof. Let’s investigate b λ i ( a, b, c ). There are n terms in this sum; each one crosses fromthe left string to the right string at a different edge. If we first calculate the term wherewe go across as soon as possible, then our first term is i Y k =1 c k ! m Y k =1 b k ! Y b k assoc. to ji a k (cid:17) Substituting this into our expressions for the face variables under T f and using the previousproposition proves the theorem. Example 7.3. Notice that the quiver in Example 6.11 is dual to the plabic graph inExample 4.19. We can see that the formulas we have calculated are the same if we makethe following substitutions: y = c , y = c , y = c , y = a a a a b b b b c c c y − = a , y − = a , y − = a , y − = a y + = b , y + = b , y + = b , y + = b Example 7.4. Notice that the quiver in Example 6.12 is dual to the plabic graph inExample 4.20. We can see that the formulas we have calculated are the same if we makethe following substitutions: 43 = c , y = c , y = c , y = c , y = c , y = a a a a a a b b b c c c c c y − = a , y − = a , y − = a , y − = a , y − = a , y − = a y + = b , y + = b , y + = b Definition 8.1 (Section 13 of [21]) . For a plabic graph G , a trip is a walk in G that turnsright at each black vertex and left at each white vertex. Definition 8.2 (Section 13 of [21]) . A trip in a plabic graph is a round trip if it is a closedcycle. Definition 8.3 (Section 13 of [21]) . Two trips in a plabic graph have an essential inter-section if there is an edge e with vertices of different colors such that the two trips passthrough e in different directions. A trip in a plabic graph has an essential self-intersection if there is an edge e with vertices of different colors such that the trip passes through e indifferent directions. Definition 8.4 (Section 13 of [21]) . Two trips in a plabic graph have an bad double crossing if they have essential intersections at edges e and e where both trips are directed from e to e . Theorem 8.5. Let G be a leafless reduced plabic graph on a cylinder without isolatedcomponents. We will consider e G to be G drawn on the universal cover of the cylinder.Then G is reduced if and only if the following are true:(1) There are no round trips in e G .(2) e G has no trips with essential self-intersections.(3) There are no pairs of trips in e G with a bad double crossing.(4) If a trip begins and ends at the same boundary vertex, then either e G has a boundaryleaf at that vertex. The above theorem is analogous to Theorem 13.2 of [21]. Proof. Notice that G is reduced if and only if e G is reduced. The proof for Theorem 13.2of [21] holds to show that e G is reduced if and only if conditions (1) - (4) hold.Now we can prove Theorem 4.5. 44 roof. Notice that the trips in a plabic graph follow the same paths as the strands in theassociated Postnikov diagram. The conditions from Theorem 8.5 correspond exactly to theconditions we require in the definition of a Postnikov diagram. Lemma 8.6. Suppose a cylindric 2-loop plabic graph has an interior vertex that has oneedge to a vertex on a string and one edge to a different vertex on the same string, such thatthere is only one vertex on the string between these two vertices, and the square formedby these four vertices is the boundary of a single face. Then, we can reduce the numberof strand crossings in between the two loops in the associated Postnikov diagram using thesquare move.Proof. Without loss of generality, assume the interior vertex is black. Then we can applytransformations to our plabic graph as follows:leftloop = ... ... ......... ↔ ... ... ...... ... ↔ ... ... ...... ... ↔ ... ... ...... ... = leftloopThe Postnikov diagram on the left has two crossings between the left and the rightloop, in addition to those we can’t see in the picture. The Postnikov diagram on the righthas one crossing between the left and the right loop, aside from those we can’t see in thepicture. So, we have reduced the number of crossings. Lemma 8.7. Suppose a cylindric 2-loop plabic graph has at least one interior vertex.Assume no vertices in the plabic graph have degree two. Then at least one interior vertexmust have multiple edges to vertices on a string. roof. Note that only white vertices on the left string and black vertices on the right stringcan have edges to interior vertices. As the graph is bipartite, this means an interior vertexcannot have edges to vertices on both the left and right string.Since there are interior vertices, there must be a vertex on a string attached to aninterior vertex. Without loss of generality, assume there’s such a vertex on the left string.Expand this vertex so that we have a vertex on the left string that is attached only to oneinterior vertex and two vertices on the left string. We allow some vertices of degree 2 tobe created to keep the graph bipartite. Now we have an interior vertex must arise from apart of the alternating strand diagram that looks as follows:leftloop s s s Call the strands s , s , and s , as denoted in the diagram. The black vertex in thediagram is not on the right string, so s cannot be the right loop. Suppose s starts andends on the right boundary of the cylinder. If s does not make a turn and head downward,then s and the right loop would intersect in two places on the universal cover, and bothwould be oriented in the same direction (from one crossing to the other). So, s must turndownwards at some point. If s turns to the right to do this, it will have a self-crossing,which is not allowed. If s turns to the left to do this, then s = s . Then s and s intersect twice, and both are oriented from the crossing on the right to the crossing on theleft. This is not allowed, so s must cross the left loop.Suppose s crosses the left loop from left to right (the argument is very similar for s crossing from right to left). There must be a strand oriented from right to left that crossesthe left loop just above s . Call this strand s . Either s must cross s or s = s . Supposeit’s the former. The face above s and to the right of the left loop corresponds to a whitevertex on the left string, which must be connected to an interior vertex. Suppose there isanother strand that crosses s between the left loop and s in the same direction as s .Since this strand did not cross s in the same direction as s between the left loop and s , it must cross s somewhere above s . However, this strand would also have to cross s below s , as nothing crosses the left loop between s and s . This would introduce twocrossings of s and the additional upward pointing strand, and both of the strands wouldbe oriented in the same direction. This cannot happen, so we have this section of thealternating strand diagram that looks as follows:46eftloop s s Now suppose we are in the second case: s = s . We get a sequence of connectedvertices beginning with the white vertex in the face bounded by s , s , and the left loopand ending with the white vertex in the face bounded by s and the left loop such thatvertices alternate being in faces to the left and right of s . This sequence and the sectionof the string bounded below by s and above by s form a closed cycle, where there mayor may not be vertices shared by sequence zig-zagging across s and the left string:leftloop s s ... ...If there are no interior vertices inside this cycle, then we are done, as any black vertexin this cycle not on the left string, of which there is at least one, is adjacent to multiplevertices on the left string. Otherwise, we can repeat our process from the beginning, thistime choosing an interior vertex that is inside the cycle. We will obtain an s ′ , s ′ , s ′ , and s ′ . If s ′ = s ′ , then we get a contradiction, as above. Otherwise the cycle of vertices weobtain is inside the cycle we obtained from s , s and s . Since the graph is finite, thisprocess must eventually terminate, and we will have found an interior vertex with multiplenon-parallel edges to vertices on the left string.47 emma 8.8. Suppose a cylindric 2-loop plabic graph with no vertices of degree 2 has aninterior vertex that has (at least) two edges to vertices on a string. If there are any othervertices of the same color on the string between these two vertices, they must also haveedges to the same interior vertex.Proof. Assume for contradiction we have a plabic graph with an interior vertex that has(at least) two edges to vertices on a string and the vertices of the same color on the stringbetween these two vertices do not have edges to the same interior vertex. Without loss ofgenerality assume the interior vertex is a black vertex. We know this part of the Postnikovdiagram looks as shown, where the interior of the darker box is unknown:leftloop...If any of the strands on the left exit the shaded face out the top or bottom and don’treturn, they’ll disconnect the original edges from the interior vertex to the two vertices onthe string. If they exit out the strand on the right and don’t return, they’ll split the oneinterior vertex into multiple. Thus, all the strands crossing the left loop from the left mustalso be the same strands that cross from the right. All strands like this must be orientedthe opposite direction of the loop. The only way to pair up these strands in that way isfor each strand going to the right is paired with the strand immediately below it. There isno way to do this without self-crossings and without creating vertices of degree 2. So, wehave a contradiction. Theorem 8.9. Any cylindric 2-loop plabic graph can be transformed by moves to one thathas no interior vertices.Proof. If there are interior vertices, then by Lemmas 8.7 and 8.8, we are always in asituation where, by Lemma 8.6, we can reduce the number of edges until we get to a plabic48raph where there are no interior vertices. Lemma 8.10. If there are interior vertices in a cylindric 2-loop plabic graph, then theremust be a black interior vertex that is adjacent to the left string.Proof. Assume for contradiction we have a cylindric 2-loop plabic graph with interiorvertices, but no black interior vertex connected to the left string. Then each white vertexon the left string is connected only to two black vertices on left string and some positivenumber of black vertices on the right string. If we look at the area of the graph betweentwo consecutive edges that connect the strands, we have three possibilities: ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... We can draw in parts of some strands based on what we know of the graph: ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... For each area enclosed by edges in these figures, the strands drawn above must be theonly strands that enter or exit the area. If we test all the ways to connect the enteringstrands to the exiting strands, we find that in each case, there is only one way to do thiswhile maintaining the rules of Postnikov diagrams and without creating additional edgesbetween the vertices on the strands: 49 .. ... ... ... ... ... ... ... ... ... ... ... These Postnikov diagrams don’t have any interior vertices, which is a contradiction.Now we are ready to prove Theorem 4.8. Proof. First, we’ll just consider the first and second farthest right loops. If there areinterior vertices, then by Lemma 8.10, there must be a black interior vertex adjacent to theleft string. The proof of Lemma 8.7 tells us that there must be a black vertex connectedby multiple nonparallel edges to the left string. Then, by Lemmas 8.8 and 8.6, we cando a square move to reduce the number of crossings between the loops in the associatedalternating strange diagram. Since we are doing this square move on a black vertex, weare pushing these crossings to the left (as pictured in the proof of Lemma 8.6). We cancontinue this process until there are no interior vertices between the first and second farthestright loops. Now we’ll look at the second and third farthest right loops and, using the samemethod, remove all interior vertices here. Notice that we do not create any interior verticesbetween the first and second farthest right loops, as we are always pushing crossings to theleft. Repeating this process with each pair of loops, going from right to left, removes allinterior vertices. Lemma 9.1. We have the following relation between λ j and λ j +1 : λ j +1 ( x, y, z ) = y j λ j ( x, y, z ) + z j ( Q ni =1 x i − Q ni =1 y i ) x j +1 Proof. If we take a path that gives a term in λ j and does not start with z j , we can add twoadditional edges at the end on the right string and remove two edges at the beginning onthe left string, to get a path that gives a term in λ j +1 . This process gives us all the termsin λ j +1 except the one that comes from the path that stays on the left string and crossesover at the last moment on z j . 50e can choose one black vertex on the left string of a cylindric 2-loop extended plabicnetwork and the white vertex on the right string that is part of the same interior face.Adding one edge of weight p and one edge of weight − p from the white vertex to the blackvertex does not change any of the boundary measurements, because every time there isa path that goes through the edge with weight p , there is another path exactly the sameexcept it goes through the edge of weight − p . The weights of these paths will cancel outwhen computing the boundary measurements. We can expand these two vertices into 3vertices each, as seen in Figure 11, where the top white vertex on the right and bottomblack vertex on the left may or may not have a third edge, depending on whether or notthe original white and black vertices were of degree 2 or 3 before adding the edges weighted p and − p . a bdc − pp a bdc − pp Figure 11: Adding the edges p and − p to a cylindric 2-loop directed plabic network.Notice that what was a hexagon is now a pentagon on top of a square on top of apentagon. We can turn the upper pentagon into a square and perform the square move.This will allow us to perform the square move on the face above that, and so on. Whenwe change original edge weights x i +1 , y i , and z i by a square move, we will denote the newweights by g x i +1 , e y i , and e z i , as in Figure 12. We can continue doing square moves in thisway until we end up with a graph that looks the same as the one we started with. Nowthe edge weighted − p will be the edge at the top of the square and the edge originallyweighted p , which has a a new weight, has been pushed around the cylinder so that it isunder the edge weighted − p . Notice that each original edge weight is changed by a squaremove exactly once, so ignoring the two edges oriented from right to left, we end up with anew graph with edge weights e x, e y , and e z .Each square move looks as depicted in Figure 12. From the figure, we see that as weperform the sequence of square moves, there is always one edge, besides the fixed edgeweighted − p , oriented from right to left. Lemma 9.2. There is at most one value of p for which the weight of the bottom edge of y i z i x i +1 ↔ e z i g x i +1 W e y i Figure 12: A square move in the sequence of moves that push the edge weighted p aroundthe cylinder. the square in the network, after the square move has propagated all the way around thecylinder, is equal to p .Proof. If we are preforming the move where the edge oriented from right to left is pushedfrom between g z i − and z i to between e z i and z i +1 , then suppose this edge is weighted w = apbp + c where a, b, c don’t depend on p . Then the weight of the edge oriented fromright to left after the square move is x i +1 z i (1+ yiziw ) = ( ax i +1 ) p ( az i + by i ) p + cy i . So, by induction, we findthat the weight of the bottom edge of the square in the network, after the square movehas propagated all the way around the cylinder, is a constant times p divided by a linearfunction of p . Setting this equal to p , we get a linear equation, so there is at most onesolution. Theorem 9.3. If we add the edges weighted by p and − p such that the the face aboveedge p has the edge labeled y j , and the face below edge − p has the edge labeled x j , thenlet p = p j = Q ni =1 x i − Q ni =1 y i λ j ( x, y, z ) and perform the sequence of square moves described above.Then when the edge directed from the right string to the left string that is not the edgeweighted by − p is just above the face with the edge labeled f x k , the weight of this edge is p k = Q ni =1 x i − Q ni =1 y i λ k ( x, y, z ) . After doing all the square moves, we have the following values of e x i , e y i , and e z i : e x i = y i − λ i − ( x, y, z ) λ i ( x, y, z ) , e y i = x i +1 λ i +1 ( x, y, z ) λ i ( x, y, z ) , e z i = z i . roof. We only need to show this for one square move. e p j = x j +1 z j (cid:16) y j z j p j (cid:17) = x j +1 (cid:16) Q ni =1 x i − Q ni =1 y i λ j ( x,y,z ) (cid:17) z j (cid:16) Q ni =1 x i − Q ni =1 y i λ j ( x,y,z ) (cid:17) + y j = x j +1 ( Q ni =1 x i − Q ni =1 y i ) z j ( Q ni =1 x i − Q ni =1 y i ) + y j λ j ( x, y, z )= ( Q ni =1 x i − Q ni =1 y i ) λ j +1 ( x, y, z ) , by Lemma 9.1 g x j +1 = x j +1 z j p j y j = x j +1 z j ( Q ni =1 x i − Q ni =1 y i ) y j λ j ( x,y,z ) = x j +1 y j λ j ( x, y, z ) y j λ j ( x, y, z ) + z j ( Q ni =1 x i − Q ni =1 y i )= y j λ j ( x, y, z ) λ j +1 ( x, y, z ) , by Lemma 9.1 e y j = y j (cid:18) z j p j y j (cid:19) = y j + z j ( Q ni =1 x i − Q ni =1 y i ) λ j ( x, y, z )= y j λ j ( x, y, z ) + z j ( Q ni =1 x i − Q ni =1 y i ) λ j ( x, y, z )= x j +1 λ j +1 ( x, y, z ) λ j ( x, y, z ) , by Lemma 9.1 e z j = e y j p j (cid:16) y j z j p j (cid:17) = x j +1 λ j +1 ( x,y,z ) λ j ( x,y,z ) p j + y j z j = x j +1 z j λ j +1 ( x, y, z ) z j p j λ j ( x, y, z ) + y j λ j ( x, y, z )= x j +1 z j λ j +1 ( x, y, z ) z j ( Q ni =1 x i − Q ni =1 y i ) + y j λ j ( x, y, z )= z j , by Lemma 9.153 heorem 9.4. After pushing the square move through every face, we can remove the edgesweighted p and − p . Then we can use gauge transformations to force all the edge weightsthat were originally set to 1 equal to 1 again. We obtain the new edge weights x ′ i , y ′ i , z ′ i : x ′ i = x i set to , ( Q αik =0 y i − k − ) λ i − αi − ( x,y,z ) λ i ( x,y,z ) otherwise, y ′ i = y i set to , (cid:16)Q βik =0 x i + k +1 (cid:17) λ i + βi +1 ( x,y,z ) λ i ( x,y,z ) otherwise, z ′ i = z i Y ℓ ∈ A i e x ℓ Y m ∈ B i f y m = z i (cid:0)Q α i k =1 y i − k − (cid:1) λ i − α i − ( x, y, z ) (cid:16)Q β i k =1 x i + k +1 (cid:17) λ i + β i +1 ( x, y, z ) λ i − ( x, y, z ) λ i +1 ( x, y, z ) . Proof. Clear by computation.We can now prove the first part of Theorem 4.13. Proof. T e is the transformation we obtain when we do the process from Theorem 9.3 andthen Theorem 9.4. Since adding the p and − p edges, applying square moves, removing the p and − p edges, and applying gauge transformations all preserve the boundary measure-ments, T e preserves the boundary measurements. Lemma 9.5. For any i, λ i ( x, y, z ) = λ i ( e x, e y, e z ) .Proof. First suppose that we are in the case where the expanded plabic graph has novertices of degree 2. Then λ i ( x, y, z ) is exactly the coefficient of a power of ζ in one of theboundary measurements. So λ i ( x, y, z ) = λ i ( x ′ y ′ z ′ ), as the transformation does not changethe boundary measurements. But in this case, ( x ′ y ′ z ′ ) = ( e x, e y, e z ), so λ i ( x, y, z ) = λ i ( e x, e y, e z ).Now suppose we are in the general case. The lack of a third edge on some verticesdoes not affect the square move in any way. The transformation from ( x, y, z ) to ( e x, e y, e z )is the same as if all the vertices were degree 3, and thus, we still have that λ i ( x, y, z ) = λ i ( e x, e y, e z ).Since Q ni =1 x i = Q ni =1 e y i , Q ni =1 y i = Q ni =1 e x i , and λ i ( x, y, z ) = λ i ( e x, e y, e z ), − p j from ouroriginal network is equal to p j from the network with edge weights ( e x, e y, e z ). Thus, we can54nd ( ee x, ee y, ee z ) by taking our original network, moving the edge p j around by square moves,and then moving the edge − p j around by square moves. Lemma 9.6. After applying the gauge transformations to the network to get from ( e x, e y, e z ) to ( x ′ , y ′ , z ′ ) , the edges labeled by p j , − p j in the ( e x, e y, e z ) network are replaced with − p ′ j and p ′ j = Q ni =1 x ′ i − Q ni =1 y ′ i λ j ( x ′ , y ′ , z ′ ) , respectively.Proof. In the network with edges ( e x, e y, e z ), consider p j λ j ( e x, e y, e z ) = Q ni =1 x i − Q ni =1 y i =a sum of weights of some cycles beginning and ending at the vertex on the bottom rightof the face between the two strings with edge e x j . Since gauge transformations don’taffect the weight of cycles, p j λ j ( e x, e y, e z ) = − p ′ j λ j ( x ′ , y ′ , z ′ ) where − p ′ j is the weight of theedge previously weighted p j after applying gauge transformations to obtain the networkwith edge weights ( x ′ , y ′ , z ′ ) from the one with edge weights ( e x, e y, e z ). So, p ′ must be Q ni =1 y i − Q ni =1 x i λ ( x ′ , y ′ , z ′ ) = Q ni =1 x ′ i − Q ni =1 y ′ i λ ( x ′ , y ′ , z ′ ) . A similar argument holds for − p . Lemma 9.7. Gauge transformations commute with the square move.Proof. Since our formulas for the square move come from the face weighted square move,and gauge transformations don’t affect face weights, gauge transformations commute withthe square move. Theorem 9.8. We have the following commutative diagram. ( x, y, z ) ( e x, e y, e z ) ( x ′ , y ′ , z ′ )( ee x, ee y, ee z ) ( e x ′ , e y ′ , e z ′ )(( e x ) ′ , ( e y ) ′ , ( e z ) ′ ) = ( x ′′ , y ′′ , z ′′ ) square moves gauge trans.square moves square movesgauge trans. gauge trans.gauge trans. Proof. By Lemma 9.6 we can get from ( x ′ , y ′ , z ′ ) to ( e x ′ , e y ′ , e z ′ ) by square moves. ByLemma 9.5, we can find ( ee x, ee y, ee z ) from ( e x, e y, e z ) by doing square moves. Lemma 9.7 tells uswe have the bottom arrow of the square in the diagram such that the square commutes.Both (( e x ) ′ , ( e y ) ′ , ( e z ) ′ ) and ( x ′′ , y ′′ , z ′′ ) can be obtained from ( ee x, ee y, ee z ) by gauge transforma-tions. Since (( e x ) ′ , ( e y ) ′ , ( e z ) ′ ) and ( x ′′ , y ′′ , z ′′ ) both have the same edge weights set equal to1 and the number of edge weights not set to be 1 is the same as the dimension of the spaceof face and trail weights, (( e x ) ′ , ( e y ) ′ , ( e z ) ′ ) = ( x ′′ , y ′′ , z ′′ ).We can now prove the second part of Theorem 4.13.55 roof. This is a matter of checking that ( x ′′ , y ′′ , z ′′ ) = ( x, y, z ). We will to this using theleft branch of our commutative diagram in Theorem 9.8.Notice that rather than moving the edge p all the way around and then moving theedge − p all the way around, we can do one square move with p and one square movewith − p . Computations very similar to those in the proof of Theorem 9.3 show that( ee x, ee y, ee z ) = ( x, y, z ). This means we don’t have to do any gauge transformations to get to( x ′′ , y ′′ , z ′′ ) and in fact ( x ′′ , y ′′ , z ′′ ) = ( ee x, ee y, ee z ) = ( x, y, z ).Suppose we have a cylindric 2-loop plabic network with no interior vertices. Considerthe expanded graph on its universal cover. Let y i be a weight in the expanded plabic graphthat has not been set to 1. Let j be minimal such that j > i and x j is not set to 1. Let k, ℓ be maximal such that k ≤ i, ℓ < i and x k , y ℓ are not set to 1. Let a be the source below x k , b be the source between x j and x k , c be the source above x j , d be the sink between y i and y ℓ , and e be the sink below y ℓ . Let n ( w ) = the number of times the edge w crosses thecut from the right − the number of times the edge w crosses the cut from the left. Thenwe have the following relations: M bd = x j ( − ζ ) n ( x j ) M cd + y ℓ ( − ζ ) n ( y ℓ ) M be − x j y ℓ ( − ζ ) n ( x j )+ n ( y ℓ ) M ce M ad = x k ( − ζ ) n ( x k ) M bd + y ℓ ( − ζ ) n ( y ℓ ) M ae − x k y ℓ ( − ζ ) n ( x k )+ n ( y ℓ ) M be Thus, we have relations between y ℓ and x j and between y ℓ and x k . In particular, wehave: x j = M bd − y ℓ ( − ζ ) n ( y ℓ ) M be ( − ζ ) n ( x j ) M cd − y ℓ ( − ζ ) n ( x j )+ n ( y ℓ ) M ce y ℓ = M ad − x k ( − ζ ) n ( x k ) M bd ( − ζ ) n ( y ℓ ) M ae − ( − ζ ) n ( x k )+ n ( y ℓ ) x k M be Combining these, we get a relation between x j and x k . x j = M bd M ae − M ad M be ( − ζ ) n ( x j ) ( M ae M cd − M ad M ce ) + x k ( − ζ ) n ( x j )+ n ( x k ) ( M bd M ce − M be M cd ) Lemma 9.9. If we know the boundary measurements of a network, then repeated sub-stitution with the above formula to get x j as a function of x n will always yield a linearexpression in x n divided by another linear expression in x n .Proof. We only need to check one step. Suppose we have x j = A + Bx m C + Dx m and x m = EF + Gx n ,coming from the formula above. Then x j = ( AF + BE )+ AGx n ( CF + DE )+ CGx n .Now we prove the third part of Theorem 4.13.56 roof. Choose an edge on the left string that is not set to 1. Call it x j . Let x k be the nextedge below x j on the left string that is not set to one. Suppose we do a transformationthat preserves the boundary measurements and replaces x j with x ′ j , x k with x ′ k , and so on.Since the transformation does not change the boundary measurements, the above relationbetween x j and x k holds for x ′ j and x ′ k . Similarly, if x ′ ℓ is the next edge weight on the leftstring that is not set to 1, x ′ k and x ′ ℓ have the same relationship. So, substitution gives usa formula for x ′ j in terms of x ′ ℓ . Repeating this process all the way around the cylinder, weget an expression for x ′ j in terms of itself. By Lemma 9.9, clearing the denominator gives aquadratic equation in x ′ j . Quadratic equations have at most two solutions, so there are atmost 2 possibilities for x ′ j . We have identified two solutions in our involution, so we knowthese are the only two solutions.Suppose now that we have a cylindric 3-loop plabic graph. By Theorem 4.8, we canassume without loss of generality that there are no interior vertices between the left andmiddle strings or the middle and right strings.We will label the left, middle, and right strings 1, 2, and 3, respectively. Then T ( i,i +1) e will be applying T e to the strings i and i + 1. Taking the expanded directed plabic networkfor this graph, we can add the edges p, − p between strings 1 and 2, q, − q between strings2 and 3, and r, − r between strings 1 and 2 so that moving p around the cylinder by squaremoves until it is below − p is equivalent to applying T (1 , e , modulo gauge transformations,then moving q around the cylinder by square moves until it’s below p is equivalent toapplying T (2 , e , modulo gauge transformations, and finally moving r around the cylinder bysquare moves until it’s below q is equivalent to applying T (1 , e a second time, modulo gaugetransformations. Since gauge transformations commute with square moves (Lemma 9.7),we can apply all of our gauge transformations at the end, and get T (1 , e ◦ T (2 , e ◦ T (1 , e . Lemma 9.10. Starting with a plabic network that looks like the top half of a face with p, q , and r added, there is an 8-cycle of square moves. That is, there is a sequence of 8square moves where the networks we obtain after each of the first 7 square moves are alldifferent from each other and from the original network, but the network we obtain afterthe 8th square move is the same as the original network.Proof. Since this 8-cycle is on a portion of the cylinder isomorphic to the disk, we willuse an undirected plabic network with face weights to prove this. This gives us a moregeneral statement, as we could choose any orientations and sets of edge weights that givethe plabic networks in the following figures, and the proof would hold true. Thus, we willconsider that we are starting with a plabic network that looks as follows:57 c b de hf g − p − rrp − qq Figure 13: A face of a network with the edges p, − p, q, − q, r , and − r added as describedabove. The area above the red line is the top half to the face, as referenced in Lemma 9.10. g hcbf iae jd Let A = 1 + c + ac and B = 1 + b + bc + abc = 1 + bA . Applying unicolored edgecontractions shows that the above plabic network is the same as the network in the upperleft corner of the following 8-cycle: 58 bc defgh i ↔ ac c b (1 + c ) c defg hc c i (1 + c ) ↔ aA bA cac dA caceA fg hc c i (1 + c ) →← aA bA (1+ c ) Aac dA caceAfbAB gB hcb (1+ c ) B i (1 + c ) ↔ Aa bA Bc (1+ a ) d ( + a ) aceAfbAB gB hcb (1+ a ) B ai a ↔ Ba (1+ b ) bbc (1+ a ) c (1+ a ) B d ( + a ) ae (1+ b )1+ afb b gB hcb (1+ a ) B ai a →← a b bc (1+ a )1+ b d ( + a ) ae (1+ b )1+ afb b g ( + b ) h ai a ↔ a b bc b de (1 + b ) fb b g ( + b ) h i ↔ abc defgh i Finally, we can prove the last part of Theorem 4.13. Proof. Consider one face in our expanded plabic network. We can either add p, − p, q, − q, r ,and − r as described above, or we can add P, − P, Q, − Q, R , and − R where P, − P are59etween strings 2 and 3, Q, − Q are between strings 1 and 2, and R, − R are betweenstrings 2 and 3 so that moving P around the cylinder by square moves until it’s below − P is equivalent to applying T (2 , e , modulo gauge transformations, then moving Q around thecylinder by square moves until it’s below P is equivalent to applying T (1 , e , modulo gaugetransformations, and finally moving R around the cylinder by square moves until it’s below Q is equivalent to applying T (2 , e a second time, modulo gauge transformations. If we add P, Q, R to our network, to a face, the face looks as follows: a c b de hf g − P − RRP − QQ Let’s look at just the upper half of this picture. Notice this is the same as the secondstep in the 8-cycle in Lemma 9.10. Starting with p, q , and r inserted into a face, the 8-cycletells us that using a square move to change to a network where the upper half of that facelooks like the one with P, Q , and R , although we do not know that the edges have the rightweights, and then pushing each of those edges up one face on the cylinder, is the same aspushing p, q , and r up one face on the cylinder, and then using a square move to changeto a network that looks like one with P, Q , and R pushed up one face on the cylinder.Suppose we move p, q , and r all the way around the cylinder (performing T (1 , e ◦ T (2 , e ◦ T (1 , e , modulo gauge transformations) and then do a square move to get to a network thatlooks like one with P, Q , and R pushed around the cylinder. The previous paragraph tellsus that instead, we could push p, q , and r around the cylinder except for one face, then doa square move to get to a network that looks like one with P, Q , and R pushed around the60ylinder except for one face, then push the three edges up the last face. We can keep doingthis square move to get a face that looks like one with P, Q , and R begin pushed aroundthe cylinder one face earlier until it’s the first thing we do. Doing this move first gives usvalues on the edges that are the same as those after we push them around the cylinder, asthe values of p, q , and r are the same before and after being pushed around the cylinder.By Lemma 9.2 this means that the values are the P, Q , and R defined above, and further,that T (1 , e ◦ T (2 , e ◦ T (1 , e = T (2 , e ◦ T (1 , e ◦ T (2 , e .We now turn our attention to T f and Theorem 4.18. Theorem 9.11. Applying T e to the edge weights defined in Lemma 4.16, then turning thegraph back into a face weighted graph gives the following weights: a ′ i = b λ j ( a, b, c ) b λ p ( a, b, c ) (cid:18)Q b r associated to s where p ≤ s The formulas for a ′ i , b ′ i , and c ′ i follow from Theorem 9.3. Note there is no need touse the formulas for x ′ i , y ′ i , and z ′ , as these arise from the formulas for e x i , e y i by gaugetransformations, which do not affect the face weights. Because of the way we chose ourtrail, none of the gauge transformations will affect the weight of the trail. Thus the trailweight remains the same.Now we prove Theorem 4.18. Proof. By Theorem 9.11, these properties are inherited from Theorem 4.13. 10 Spider Web Quiver Proofs Let Q r = µ r µ r − ...µ ( Q ). Lemmas 10.1 through 10.8 may be proven by induction.61 emma 10.1. For r < n − , the edges of Q r between vertices in the middle circle are:(1) a possibly empty directed path → → ... → r ,(2) a directed path r + 1 → r + 2 → ... → n ,(3) an oriented triangle r → n → r + 1 → r . Let A be the set of vertices in the middle circle that have edges to the outer circle. Let A ′ = { vertices i | i + 1 ∈ A } . Lemma 10.2. Suppose ∈ A . Then for r < n − the edges between the middle and outercircle in Q r are:(1) (the largest vertex on the outer circle originally connected to a vertex on the middlecircle) → → − ,(2) r → (the largest vertex on the outer circle that originally had an edge to a vertexbetween 1 and r ) → (the largest vertex strictly between 1 and r in A ′ ) → (the nextlargest vertex along the outer circle originally connected to a vertex on the middlecircle) → (the next largest vertex strictly between 1 and r in A ′ ) → ... → (the smallestvertex strictly between 1 and r in A ′ ) → (the next largest vertex along the outer circleoriginally connected to a vertex on the middle circle) → ,(3) n → (the largest vertex on the outer circle that originally had an edge to a vertexbetween 1 and n − ) → (the largest vertex in A less than n ) → (the next largest vertexalong the outer circle originally connected to a vertex on the middle circle) → (thenext largest vertex in A ) → ... → (the largest vertex of A greater than r + 1 ) → (thenext largest vertex along the inner circle originally connected to a vertex on the middlecircle) → r + 1 . Lemma 10.3. Suppose A . For r < n − , if r is smaller that the smallest vertex of A , then the edges between the middle and outer circle in Q r are:(1) the same edges as in the original quiver, that is, (the largest vertex on the outer circleoriginally connected to a vertex on the middle circle) → (the largest vertex in A ) → (the next largest vertex along the outer circle originally connected to a vertex on themiddle circle) → (the next largest vertex in A ) → (the next largest vertex along theouter circle originally connected to a vertex on the middle circle) → ... → (the smallestvertex in A ) → (the largest vertex along the outer circle originally connected to a vertexon the middle circle).Otherwise the edges between the middle and outer circle in Q r are: 1) (the largest vertex on the outer circle originally connected to a vertex on the middlecircle) → (the smallest vertex in A along the middle circle),(2) r → (the largest vertex on the outer circle that originally had an edge to a vertexbetween 1 and r ) → (the largest vertex strictly between 1 and r in A ′ ) → (the nextlargest vertex along the outer circle originally connected to a vertex on the middlecircle) → (the next largest vertex strictly between 1 and r in A ′ ) → ... → (the smallestvertex marked in A ′ ),(3) n → (the largest vertex on the outer circle that originally had an edge to a vertexbetween 1 and n − ) → (the largest vertex in A less than n ) → (the next largest vertexalong the outer circle originally connected to a vertex on the middle circle) → (thenext largest vertex in A ) → ... → (the largest vertex of A greater than r + 1 ) → (thenext largest vertex along the inner circle originally connected to a vertex on the middlecircle) → r + 1 . Let B be the set of vertices in the middle circle that have edges to the inner circle. Let B ′ = { vertices i | i + 1 ∈ B } . Lemma 10.4. Suppose ∈ B . Then for r < n − the edges between the middle and innercircle in Q r are:(1) (the largest vertex on the inner circle originally connected to a vertex on the middlecircle) → → + ,(2) r → (the largest vertex on the inner circle that originally had an edge to a vertexbetween 1 and r ) → (the largest vertex strictly between 1 and r in B ′ ) → (the nextlargest vertex along the inner circle originally connected to a vertex on the middlecircle) → (the next largest vertex strictly between 1 and r in B ′ ) → ... → (the smallestvertex strictly between 1 and r in B ′ ) → (the next largest vertex along the inner circleoriginally connected to a vertex on the middle circle) → ,(3) n → (the largest vertex on the inner circle that originally had an edge to a vertexbetween 1 and n − ) → (the largest vertex in B less than n ) → (the next largest vertexalong the inner circle originally connected to a vertex on the middle circle) → (thenext largest vertex in B ) → ... → (the largest vertex of B greater than r + 1 ) → (thenext largest vertex along the inner circle originally connected to a vertex on the middlecircle) → r + 1 . Lemma 10.5. Suppose B . For r < n − , if r is smaller that the smallest vertex of B , then the edges between the middle and inner circle in Q r are:(1) the same edges as in the original quiver, that is, (the largest vertex on the inner circleoriginally connected to a vertex on the middle circle) → (the largest vertex in B ) → the next largest vertex along the inner circle originally connected to a vertex on themiddle circle) → (the next largest vertex in B ) → (the next largest vertex along theinner circle originally connected to a vertex on the middle circle) → ... → (the smallestvertex in B ) → (the largest vertex along the inner circle originally connected to a vertexon the middle circle).Otherwise the edges between the middle and inner circle in Q r are:(1) (the largest vertex on the inner circle originally connected to a vertex on the middlecircle) → (the smallest vertex in B along the middle circle),(2) r → (the largest vertex on the inner circle that originally had an edge to a vertexbetween 1 and r ) → (the largest vertex strictly between 1 and r in B ′ ) → (the nextlargest vertex along the inner circle originally connected to a vertex on the middlecircle) → (the next largest vertex strictly between 1 and r in B ′ ) → ... → (the smallestvertex marked in B ′ ),(3) n → (the largest vertex on the inner circle that originally had an edge to a vertexbetween 1 and n − ) → (the largest vertex in B less than n ) → (the next largest vertexalong the inner circle originally connected to a vertex on the middle circle) → (thenext largest vertex in B ) → ... → (the largest vertex of B greater than r + 1 ) → (thenext largest vertex along the inner circle originally connected to a vertex on the middlecircle) → r + 1 . Lemma 10.6. For r < n − , the edges of Q r between vertices in the outer circle are:(1) The directed circle − → − → ... → n − → − ,(2) The edge − → (the largest vertex on the outer circle originally connected to a vertexon the middle circle), if r ≥ (the smallest vertex in A ). Lemma 10.7. For r < n − , the edges of Q r between vertices in the inner circle are:(1) The directed circle + → + → ... → n + → + .(2) The edge + → (the largest vertex on the inner circle originally connected to a vertexon the middle circle), if r ≥ (the smallest vertex in B ). Lemma 10.8. For r < n − , the edges of Q r between vertices between the outer and innercircle are:(1) − → (the largest vertex on the inner circle originally connected to a vertex on themiddle circle) and + → (the largest vertex on the outer circle originally connected toa vertex on the middle circle), if the smallest vertex in B is the smallest vertex in A , − → (the largest vertex on the inner circle originally connected to a vertex on themiddle circle), if the smallest vertex in A is less than the smallest vertex in B ,(3) + → (the largest vertex on the outer circle originally connected to a vertex on themiddle circle), if the smallest vertex in B is less than the smallest vertex in A . Lemma 10.9. If s n − ,n is the transposes vertices n − and n , then s n − ,n µ n µ n − ( Q n − ) = Q n − .Proof. We can check all the cases.The proof of Proposition 6.2 follows easily. Proof. µ µ ...µ n − s n − ,n µ n µ n − ...µ ( Q ) = µ µ ...µ n − s n − ,n µ n µ n − ( Q n − )= µ µ ...µ n − ( Q n − ) by Lemma 10.9= Q as µ µ ...µ n − is ( µ n − µ n − ...µ ) − Recall the definitions of A and B from 10.1.Let x i,A := ( x α if i is the smallest element of A, . Let x i,B := ( x β if i is the smallest element of B, . We will define x := x n for ease of notation. Proposition 10.10. For a quiver Q as described at the beginning of the section, µ n − ...µ ( x i ) = P ij =0 x [ j ] − x [ j ] + (cid:16)Q j − k =0 x k (cid:17) (cid:16)Q ik = j +1 x k +1 x k,A x k,B (cid:17)Q ik =1 x k , for ≤ i ≤ n − .Proof. Notice that µ n − ...µ ( x i ) = µ i ...µ ( x i ). Using this and Lemmas 10.1 through 10.5,we see µ n − ...µ ( x ) = x n x [1] − x [1] + + x x ,A x ,B x , n − ...µ ( x i ) = x n x [ i ] − x [ i ] + + µ n − ...µ ( x i − ) x i +1 x i,A x i,B x i . Then the result follows by induction.We now have the proof of Theorem 6.3. Proof. First consider i = n, n − 1. In this case, µ µ ...µ n − s n − ,n µ n µ n − ...µ ( x i ) = s n − ,n µ n µ n − ...µ ( x i ). We can compute this using Lemmas 10.1 through 10.5 and Propo-sition 10.10. We have the following: τ ( x n − ) = x [ n − − x [ n − + + µ n − ...µ ( x n − ) x n − ,A x n − ,B x n τ ( x n ) = x [ n − − x [ n − + + µ n − ...µ ( x n − ) x n − ,A x n − ,B x n − Notice that for any i , x [ i ] −∗ = x [ i ] − (cid:16)Q jk = i +1 x k,A (cid:17) , as long as there is an element of A in { , ..., j } . Similarly, x [ i ] + ∗ = x [ i ] + (cid:16)Q jk = i +1 x k,B (cid:17) , as long as there is an element of B in { , ..., j } . Substituting for µ n − ...µ ( x n − ) in each of the above cases, we have what wewant.Now consider i < n − 2. Because mutation is an involution, we know µ i ...µ µ µ ...µ n − s n − ,n µ n µ n − ...µ ( x i ) = µ i +1 ...µ n − s n − ,n µ n µ n − ...µ ( x i ) . Since mutating at a vertex only affects the variable at that vertex, this is the same as µ i µ i − ...µ ( x i ). From Proposition 10.10, we know the formula for this expression. Thus,all we need to do is show, starting with the cluster variables e x i = x i x , mutating at 1through i gives us back the formulas in Proposition 10.10. µ i ...µ ( e x i ) = P ij =0 x [ j ] − x [ j ] + (cid:16)Q j − k =0 x k x (cid:17) (cid:16)Q ik = j +1 x k +1 xx k,A x k,B (cid:17)Q ik =1 x k x = P ij =0 x [ j ] − x [ j ] + (cid:16)Q j − k =0 x k (cid:17) ( x ) j (cid:16)Q ik = j +1 x k +1 x k,A x k,B (cid:17) ( x ) i − j (cid:16)Q ik =1 x k (cid:17) ( x ) i = P ij =0 x [ j ] − x [ j ] + (cid:16)Q j − k =0 x k (cid:17) (cid:16)Q ik = j +1 x k +1 x k,A x k,B (cid:17)(cid:16)Q ik =1 x k (cid:17) Lemma 10.11. For < r ≤ n − , the y -variables for the vertices in the middle circle ofthe quiver after r mutations are:(1) If i < r , then y ′ i = y i +1 P i − j =0 Q jk =0 y k P i +1 j =0 Q jk =0 y k (2) y ′ r = P r − j =0 Q jk =0 y k Q rk =0 y k (3) y ′ r +1 = Q r +1 k =0 y k P rj =0 Q jk =0 y k (4) If r + 1 < m < n , then y ′ m = y m (5) y ′ n = y n r X j =0 j Y k =0 y k Proof. This can be shown by induction. Lemma 10.12. After mutating all n vertices and performing the transposition s n − ,n , the y -variables for the vertices in the middle circle of the quiver are:(1) If i < n − , then y ′ i = y i +1 P i − j =0 Q jk =0 y k P i +1 j =0 Q jk =0 y k (2) y ′ n − = y n − (cid:16)P n − j =0 Q jk =0 y k (cid:17) (cid:16) y n P n − j =0 Q jk =0 y k (cid:17)P n − j =0 Q jk =0 y k (3) y ′ n − = 1 y n P n − j =0 Q jk =0 y k (4) y ′ n = P n − j =0 Q jk =0 y k Q n − k =0 y k Proof. These values can be easily computed. Lemma 10.13. If < r ≤ n − , then after mutating vertex r for the second time, the y -variables in the middle circle of the quiver are: 1) If ℓ < i − , then y ℓ = y ℓ +1 P ℓ − j =0 Q jk =0 y k P ℓ +1 j =0 Q jk =0 y k (2) y r − = (cid:16) y r P r − j =0 Q jk =0 y k (cid:17) (cid:16)(cid:16)P nj = r +1 Q j − k = r +1 y k (cid:17) + (cid:0)Q nk = r +1 y k (cid:1) (cid:16)P r − j =0 Q jk =0 y k (cid:17)(cid:17)P n − j =0 Q jk =0 y k (3) y r = P n − j =0 Q jk =0 y k (cid:16) y r +1 P r − j =0 Q jk =0 y k (cid:17) (cid:16)(cid:16)P nj = r +2 Q j − k = r +2 y k (cid:17) + (cid:0)Q nk = r +2 y k (cid:1) (cid:16)P rj =0 Q jk =0 y k (cid:17)(cid:17) (4) If i < m < n , then y m = P nj = m Q j − k = m y k + ( Q nk = m y k ) (cid:16)P m − j =0 Q jk =0 y k (cid:17) y m +1 (cid:16)P nj = m +2 Q j − k = m +2 y k + (cid:0)Q nk = m +2 y k (cid:1) (cid:16)P mj =0 Q jk =0 y k (cid:17)(cid:17) (5) y n = (cid:16)P r − j =0 Q jk =0 y k (cid:17) (cid:16) y n P n − j =0 Q jk =0 y k (cid:17) ( Q rk =0 y k ) (cid:16)P nj = r +1 Q j − k = r +1 y k + (cid:0)Q nk = r +1 y k (cid:1) (cid:16)P r − j =0 Q jk =0 y k (cid:17)(cid:17) Proof. This can be shown by induction.For 1 ≤ i ≤ n − 2, let e y i = Q ik =0 y k P i − j =0 Q jk =0 y k , b y i = y i +1 (cid:16)P i − j =0 Q jk =0 y k (cid:17) (cid:16)(cid:16)P nj = i +2 Q j − k = i +2 y k (cid:17) + (cid:0)Q ni +2 y k (cid:1) (cid:16)P ij =0 Q jk =0 y k (cid:17)(cid:17)P n − j =0 Q jk =0 y k . That is, e y i is the value of y ′ i before mutating at i for the first time, and b y i is the value of y ′ i before mutating at i for the second time. Let y ∗ n − = Q n − k =0 y k P n − j =0 Q jk =0 y k , y ∗ n = y n n − X j =0 j Y k =0 y k , so y ∗ n − is the value of y ′ n − before mutating at n − y ∗ n is the value of y ′ n beforemutating at n . Lemma 10.14. Suppose n ∈ A . Then after completing the mutation sequence, we havethe following y -variables for the outer circle. 1) If there are no edges between i − and the middle circle, then y ′ i − = y i − .(2) If i is maximal so that there are edges between i − and the middle circle, then y i − = y i − (1 + c y d ) (cid:16) f y d − (cid:17) . (3) If i is second largest so that there are edges between i − and the middle circle, then y i − = y i − (1 + y ∗ n − ) Q n − k = d i (1 + e y k )(1 + y ∗− n ) Q n − k = d i (cid:0) b y k − (cid:1) . (4) Otherwise, y i − = y i − Q e i − k = d i (1 + e y k ) Q e i − k = d i (cid:0) b y k − (cid:1) . Proof. We can prove this by using Lemmas 10.2 and 10.3 to know the arrows in Q r andapplying the y -mutation rules. Lemma 10.15. Suppose n A . Then after completing the mutation sequence, we havethe following y -variables for the outer circle.(1) If there are no edges between i − and the middle circle, then y ′ i − = y i − .(2) If i is maximal so that there are edges between i − and the middle circle, then y i − = y i − (1 + y ∗ n − ) (1 + c y d ) Q n − k = e i (1 + e y k )(1 + y ∗− n ) (cid:16) f y d − (cid:17) Q n − k = e i (cid:0) b y k − (cid:1) . (3) Otherwise, y i − = y i − Q e i − k = d i (1 + e y k ) Q e i − k = d i (cid:0) b y k − (cid:1) . Proof. We can prove this by using Lemmas 10.2 and 10.3 to know the arrows in Q r andapplying the y -mutation rules. Lemma 10.16. Suppose n ∈ B . Then after completing the mutation sequence, we havethe following y -variables for the inner circle.(1) If there are no edges between i + and the middle circle, then y ′ i + = y i + . 2) If i is maximal so that there are edges between i + and the middle circle, then y i + = y i + (1 + c y f ) (cid:16) f y f − (cid:17) . (3) If i is second largest so that there are edges between i + and the middle circle, then y i + = y i + (1 + y ∗ n − ) Q n − k = f i (1 + e y k )(1 + y ∗− n ) Q n − k = f i (cid:0) b y k − (cid:1) . (4) Otherwise, y i + = y i + Q g i − k = f i (1 + e y k ) Q g i − k = f i (cid:0) b y k − (cid:1) . Proof. We can prove this by using Lemmas 10.4 and 10.5 to know the arrows in Q r andapplying the y -mutation rules. Lemma 10.17. Suppose n B . Then after completing the mutation sequence, we havethe following y -variables for the inner circle.(1) If there are no edges between i + and the middle circle, then y ′ i + = y i + .(2) If i is maximal so that there are edges between i + and the middle circle, then y i + = y i + (1 + y ∗ n − ) (1 + c y f ) Q n − k = g i (1 + e y k )(1 + y ∗− n ) (cid:16) f y f − (cid:17) Q n − k = g i (cid:0) b y k − (cid:1) . (3) Otherwise, y i + = y i + Q g i − k = f i (1 + e y k ) Q g i − k = f i (cid:0) b y k − (cid:1) . Proof. 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