Báo cáo toán học: "Combinatorial Interpretations for Rank-Two Cluster Algebras of Affine Type." docx

We define the weight of a perfect matching S to be the product of the weights of its constituent edges, e∈S we... E.g., for n = 3, the graph G3 = H2 has two perfect matchings with respec

Combinatorial Interpretations for Rank-Two Cluster

Algebras of Affine Type

Gregg Musiker

Department of MathematicsUniversity of California, San Diego

La Jolla, CA 92093-0112gmusiker@math.ucsd.edu

James Propp

Department of Mathematical SciencesUniversity of Massachusetts Lowell

Lowell, MA 01854propp@jamespropp.org.ignorethisSubmitted: Feb 20, 2006; Accepted: Jan 11, 2007; Published: Jan 19, 2007

Mathematics Subject Classification: 05A99, 05C70

AbstractFomin and Zelevinsky [6] show that a certain two-parameter family of rationalrecurrence relations, here called the (b, c) family, possesses the Laurentness property:for all b, c, each term of the (b, c) sequence can be expressed as a Laurent polyno-mial in the two initial terms In the case where the positive integers b, c satisfy

bc <4, the recurrence is related to the root systems of finite-dimensional rank 2 Liealgebras; when bc > 4, the recurrence is related to Kac-Moody rank 2 Lie algebras

of general type [9] Here we investigate the borderline cases bc = 4, corresponding

to Kac-Moody Lie algebras of affine type In these cases, we show that the rent polynomials arising from the recurence can be viewed as generating functionsthat enumerate the perfect matchings of certain graphs By providing combinato-rial interpretations of the individual coefficients of these Laurent polynomials, weestablish their positivity

Lau-1 Introduction

In [5, 6], Fomin and Zelevinsky prove that for all positive integers b and c, the sequence

of rational functions xn (n ≥ 0) satisfying the “(b, c)-recurrence”

xn= (xb

n−1+ 1)/xn−2 for n odd(xc

n−1+ 1)/xn−2 for n even

is a sequence of Laurent polynomial in the variables x1 and x2; that is, for all n ≥ 2, xn

can be written as a sum of Laurent monomials of the form axi

1xj2, where the coefficient

a is an integer and i and j are (not necessarily positive) integers In fact, Fomin andZelevinsky conjecture that the coefficients are always positive integers

It is worth mentioning that variants of this recurrence typically lead to rational tions that are not Laurent polynomials For instance, if one initializes with x1, x2 anddefines rational functions

n−1+ 1)/xn−2 for n = 3(xd

n−1+ 1)/xn−2 for n = 4(xe

n−1+ 1)/xn−2 for n = 5with b, c, d, e all integers larger than 1, then it appears that x5 is not a Laurent polynomial(in x1 and x2) unless b = d and that x6 is not a Laurent polynomial unless b = d and

c = e (This has been checked by computer in the cases where b, c, d, e are all between 2and 5.)

One reason for studying (b, c)-recurrences is their relationship with root systems sociated to rank two Kac-Moody Lie algebras Furthermore, algebras generated by asequence of elements satisfying a (b, c)-recurrence provide examples of rank two clusteralgebras, as defined in [6, 7] by Fomin and Zelevinsky The property of being a sequence

as-of Laurent polynomials, Laurentness, is in fact proven for all cluster algebras [6] as well

as a class of examples going beyond cluster algebras [5] In this context, the positivity ofthe coefficients is no mere curiosity, but is related to important (albeit still conjectural)total-positivity properties of dual canonical bases [13]

The cases bc < 4 correspond to finite-dimensional Lie algebras (that is, semisimpleLie algebras), and these cases have been treated in great detail by Fomin and Zelevinsky[6, 12] For example, the cases (1, 1), (1, 2), and (1, 3) correspond respectively to theLie algebras A2, B2, and G2 In these cases, the sequence of Laurent polynomials xn isperiodic More specifically, the sequence repeats with period 5 when (b, c) = (1, 1), withperiod 6 when (b, c) = (1, 2) or (2, 1), and with period 8 when (b, c) = (1, 3) or (3, 1) Foreach of these cases, one can check that each xn has positive integer coefficients

Very little is known about the cases bc > 4, which should correspond to Kac-MoodyLie algebras of general type It can be shown that for these cases, the sequence of Laurentpolynomials xn is non-periodic

This article gives a combinatorial approach to the intermediate cases (2, 2), (1, 4)and (4, 1), corresponding to Kac-Moody Lie algebras of affine type; specifically algebras

of types A(1)1 and A(2)2 Work of Sherman and Zelevinsky [12] has also focused on therank two affine case In fact, they are able to prove positivity of the (2, 2)-, (1, 4)- and(4, 1)-cases, as well as a complete description of the positive cone They prove both casessimultaneously by utilizing a more general recurrence which specializes to either case Byusing Newton polygons, root systems and algebraic methods analogous to those used inthe finite type case [7], they are able to construct the dual canonical bases for these clusteralgebras explicitly

Our method is intended as a complement to the purely algebraic method of Shermanand Zelevinsky [12] In each of the cases (2, 2), (1, 4) and (4, 1) we show that the positivityconjecture of Fomin and Zelevinsky is true by providing (and proving) a combinatorialinterpretation of all the coefficients of xn That is, we show that the coefficient of xi

1xj2

in xn is actually the cardinality of a certain set of combinatorial objects, namely, theset of those perfect matchings of a particular graph that contain a specified number of

“x1-edges” and a specified number of “x2-edges” This combinatorial description provides

a different way of understanding the cluster variables, one where the binomial exchangerelations are visible geometrically

The reader may already have guessed that the cases (1, 4) and (4, 1) are closely related.One way to think about this relationship is to observe that the formulas

xn = (xbn−1+ 1)/xn−2 for n oddand

xn = (xc

n−1+ 1)/xn−2 for n evencan be re-written as

xn−2 = (xb

n−1+ 1)/xn for n oddand

Our approach to the (1, 4) case will be the same as our approach to the simpler (2, 2)case: in both cases, we will utilize perfect matchings of graphs as studied in [2, 10, 11, etal.]

Definition 1 For a graph G = (V, E), which has an assignment of weights w(e) to itsedges e ∈ E, a perfect matching of G is a subset S ⊂ E of the edges of G such that eachvertex v ∈ V belongs to exactly one edge in S We define the weight of a perfect matching

S to be the product of the weights of its constituent edges,

e∈S

w(e)

With this definition in mind, the main result of this paper is the construction of afamily of graphs {Gn} indexed by n ∈ Z \ {1, 2} with weights on their edges such thatthe terms of the (2, 2)- (resp (1, 4)-) recurrence, xn, satisfy xn = pn(x1, x2)/mn(x1, x2);where pn(x1, x2) is the polynomial

Thus pn(x1, x2) may be considered a two-variable generating function for the perfectmatchings of Gn, and xn(x1, x2) may be considered a generating function as well, with aslightly different definition of the weight that includes “global” factors (associated withthe structure of Gn) as well as “local” factors (associated with the edges of a particularperfect matching) We note that the families Gn with n > 0 and Gn with n < 0 given

in this paper are just one possible pair of families of graphs with the property that the

xn(x1, x2)’s serve as their generating functions

The plan of this article is as follows In Section 2, we treat the case (2, 2); it is simplerthan (1, 4), and makes a good warm-up In Section 3, we treat the case (1, 4) (whichsubsumes the case (4, 1), since we allow n to be negative) Section 4 gives comments andopen problems arising from this work

2 The (2, 2) case

Here we study the sequence of Laurent polynomials x1, x2, x3 = (x2

2 + 1)/x1, etc If

we let x1 = x2 = 1, then the first few terms of sequence {xn(1, 1)} for n ≥ 3 are

2, 5, 13, 34, 89, It is not too hard to guess that this sequence consists of every otherFibonacci number (and indeed this fact follows readily from Lemma 1 given on the nextpage)

For all n ≥ 1, let Hn be the (edge-weighted) graph shown below for the case n = 6

x

x x

x1, the two horizontal edges adjoining them having weight x2, and so on (ending at theright with two edges of weight x2 when n is even and with two edges of weight x1 when

n is odd) Let Gn = H2n−4 (so that for example the above picture shows G5), andlet pn(x1, x2) be the sum of the weights of all the perfect matchings of Gn Also let

mn(x1, x2) = xn−21 xn−32 for n ≥ 3 We note the following combinatorial interpretation

of this monomial: mn(x1, x2) = xi

1xj2 where i is the number of square cells of Gn withhorizontal edges having weight x2 and j is the number of square cells with horizontaledges having weight x1 Using these definitions we obtain

Theorem 1 For the case (b, c) = (2, 2), the Laurent polynomials xn satisfy

xn(x1, x2) = pn(x1, x2)/mn(x1, x2) for n 6= 1, 2where pn and mn are given combinatorially as in the preceding paragraph

E.g., for n = 3, the graph G3 = H2 has two perfect matchings with respective weights

Proof We will have proved the claim if we can show that the Laurent polynomials

pn(x1, x2)/mn(x1, x2) satisfy the same quadratic recurrence as the Laurent polynomials

of the graph Hn, so that pn(x1, x2) = q2n−4(x1, x2) for n ≥ 3 Each perfect matching of

Hnis either a perfect matching of Hn−1 with an extra vertical edge at the right (of weight1) or a perfect matching of Hn−2 with two extra horizontal edges at the right (of weight

x1 or weight x2, according to whether n is odd or even, respectively) We thus have

+ x41x22 = p24+ x41x22

so for induction we assume that

pn−1pn−3 = p2n−2+ x2n−81 x2n−102 for n ≥ 5 (8)Using Lemma 1 and (8) we are able to verify that polynomials pn satisfy the quadraticrecurrence relation (3):

pnpn−2 =(x21+ x22 + 1)pn−1pn−2− x21x22p2n−2 =(x21+ x22+ 1)pn−1pn−2− x21x22(pn−1pn−3− x2n−81 x2n−102 ) =

as the xn’s, and we have proven Theorem 1

An explicit formula has recently been found for the xn(x1, x2)’s by Caldero and sky using the geometry of quiver representations:

Zelevin-Theorem 2 [4, Zelevin-Theorem 4.1], [14, Zelevin-Theorem 2.2]

n − qr

They also present expressions (Equations (5.16) of [4]) for the xn’s in terms of bonacci polynomials, as defined in [8], which can easily seen to be equivalent to the com-binatorial interpretation of Theorem 1 Subsequently, Zelevinsky has obtained a shortelementary proof of these two results [14]

Fi-2.1 Direct combinatorial proof of Theorem 2

Here we provide yet a third proof of Theorem 2: instead of using induction as in sky’s elementary proof, we use a direct bijection This proof was found after Zelevinsky’sresult came to our attention First we make precise the connection between the combina-torial interpretation of [4] and our own

Zelevin-Lemma 2 The number of ways to choose a perfect matching of Hm with 2q horizontaledges labeled x2 and 2r horizontal edges labeled x1 is the number of ways to choose asubset S ⊂ {1, 2, , m − 1} such that S contains q odd elements, r even elements, and

no consecutive elements

Notice that in the case m = 2n + 2, this number is the coefficient of x2r−n−11 x2q−n2 in

xn+3 (for n ≥ 0), and when m = 2n + 1, this number is the coefficient of x2r−n1 x2q−n2 in

sn, as defined in [4, 12, 14]

Proof There is a bijection between perfect matchings of Hm and subsets S ⊂ {1, 2, ,

m − 1}, with no two elements consecutive We label the top row of edges of Hm from 1 to

m − 1 and map a horizontal edge in the top row to the label of that edge Since horizontaledges come in parallel pairs and span precisely two vertices, we have an inverse map aswell

With this formulation we now prove

Theorem 3 The number of ways to choose a subset S ⊂ {1, 2, , N } such that Scontains q odd elements, r even elements, and no consecutive elements is

n + 1 − rq

n − qr



if N = 2n + 1 and

n − rq

n − qr



if N = 2n

Proof List the parts of S in order of size and reduce the smallest by 0, the next smallest

by 2, the next smallest by 4, and so on (so that the largest number gets reduced by2(q + r − 1))

This will yield a multiset consisting of q not necessarily distinct odd numbers between

1 and 2n + 1 − 2(q + r − 1) = 2(n − q − r + 1) + 1 if N = 2n + 1, and between 1 and2(n − q − r + 1) if N = 2n, as well as r not necessarily distinct even numbers between 1and 2(n − q − r + 1), regardless of whether N is 2n + 1 or 2n

Conversely, every such multiset, when you apply the bijection in reverse, you get a setconsisting of q odd numbers and r even numbers in {1, 2, , 2n} (resp {1, 2, , 2n+1}),

no two of which differ by less than 2

The number of such multisets is clearly n−q−r+1q

× n−q−r+1r

in the first case and

, the claim follows

Once we know this interpretation for the coefficients of xn+3 (sn), we obtain a proof

of the formula for the entire sum, i.e Theorem 2 and Theorem 5.2 of [4] (Theorem 2.2 of[14])

It is worth remarking that the extra terms x2n+21 and x2n+22 in Theorem 2 correspond

to the extreme case in which one’s subset of {1, 2, , N } consists of all the odd numbers

in that range

The recurrence of Lemma 1 can also be proven bijectively by computing in two differentways the sum of the weights of the perfect matchings of the graph Gnt H3 (the disjointunion of Gn and H3, which has all the vertices and edges of graphs Gn and H3 and noidentifications) We provide this proof since this method will be used later on in the (1, 4)case

On the one hand, the sum of the weights of the perfect matchings of Gn is the nomial pn and the sum of the weights of the perfect matchings of H3 is x2

poly-1 + x2

2 + 1, sothe sum of the weights of the perfect matchings of Gnt H3 is (x2

On the other hand, observe that the graph Gn+1 can be obtained from Gn t H3

by identifying the rightmost vertical edge of Gn with the leftmost vertical edge of H3.Furthermore, there is a weight-preserving bijection φ between the set of perfect matchings

of the graph Gn+1 and the set of perfect matchings of Gnt H3 that do not simultaneouslycontain the two rightmost horizontal edges of Gn and the two leftmost horizontal edges

of H3 (a set that can also be described as the set of perfect matchings of Gnt H3 thatcontain either the rightmost vertical edge of Gn or the leftmost vertical edge of H3 orboth) It is slightly easier to describe the inverse bijection φ−1: given a perfect matching

of Gnt H3 that contains either the rightmost vertical edge of Gn or the leftmost verticaledge of H3 or both, view the matching as a set of edges and push it forward by the gluing

map from Gnt H3 to Gn+1 We obtain a multiset of edges of Gn+1 that contains either 1

or 2 copies of the third vertical edge from the right, and then delete 1 copy of this edge,obtaining a set of edges that contains either 0 or 1 copies of that edge It is not hard tosee that this set of edges is a perfect matching of Gn+1, and that every perfect matching

of Gn+1 arises from this operation in a unique fashion Furthermore, since the verticaledge that we have deleted has weight 1, the operation is weight-preserving

The perfect matchings of Gnt H3 that are not in the range of the bijection φ are thosethat consist of a perfect matching of Gn that contains the two rightmost horizontal edges

of Gn and a perfect matching of H3 that contains the two leftmost horizontal edges of

H3 Removing these edges yields a perfect matching of Gn−1 and a perfect matching of

H1 Moreover, every pair consisting of a perfect matching of Gn−1 and a perfect matching

of H1 occurs in this fashion Since the four removed edges have weights that multiply to

Remark 1 We can also give a bijective proof of the quadratic recurrence relation (3)

by using a technqiue known as graphical condensation which was developed by Eric Kuo[10] He even gives the unweighted version of this example in his write-up

Remark 2 As we showed via equation (1), there is a reciprocity that allows us to relatethe cluster algebras for the (b, c)- and (c, b)-cases by running the recurrence backwards.For the (2, 2)-case, b = c so we do not get anything new when we run it backwards; weonly switch the roles of x1 and x2 This reciprocity is a special case of the reciprocity thatoccurs not just for 2-by-n grid graphs, but more generally in the problem of enumerating(not necessarily perfect) matchings of m-by-n grid graphs, as seen in [1] and [11] For the(1, 4)-case, we will also encounter a type of reciprocity

Remark 3 We have seen that the sequence of polynomials qn(x1, x2) satisfies the relation

q2n−4q2n−8 = q2n−62 + x2n−61 x2n−82

It is worth mentioning that the odd-indexed terms of the sequence satisfy an analogousrelation

q2n−3q2n−7 = q22n−5− x2n−61 x2n−82 This relation can be proven via Theorem 2.3 of [10] In fact the sequence of Laurentpolynomials {q2n+1/xn

1xn

2 : n ≥ 0} are the collection of elements of the semicanoncialbasis which are not cluster monomials, i.e not of the form xp

nxqn+1 for p, q ≥ 0 Theseare denoted as sn in [4] and [14] and are defined as Sn(s1) where Sn(x) is the normalizedChebyshev polynomial of the second kind, Sn(x/2), and s1 = (x2

1 + x2

2 + 1)/x1x2 Weare thankful to Andrei Zelevinsky for alerting us to this fact We describe an analogouscombinatorial interpretation for the sn’s in the (1, 4)-case in subsection 3.4

3 The (1, 4) case

In this case we let

xn−2 for n evenfor n ≥ 3 If we let x1 = x2 = 1, the first few terms of {xn(1, 1)} for n ≥ 3 are:

As in the (2, 2)-case, it turns out that this sequence {xn(1, 1)} (respectively {xn}) has

a combinatorial interpretation as the number (sum of the weights) of perfect matchings

in a sequence of graphs We prove that these graphs, which we again denote as Gn,have the xn’s as their generating functions in the later subsections We first give theunweighted version of these graphs where graph Gn contains xn(1, 1) perfect matchings

We describe how to assign weights to yield the appropriate Laurent polynomials xn inthe next subsection, deferring proof of correctness until the ensuing two subsections Theproof of two recurrences, in sections 3.2 and 3.3, will conclude the proof of Theorem 4 Thefinal subsection provides a combinatorial interpretation for elements of the semicanonicalbasis that are distinct from cluster monomials

Definition 2 We will have four types of graphs Gn, one for each of the above foursequences (i.e for an, bn, cn, and dn) Graphs in all four families are built up fromsquares (consisting of two horizontal and two vertical edges) and octagons (consisting oftwo horizontal, two vertical, and four diagonal edges), along with some extra arcs Wedescribe each family of graphs by type

Firstly, G3 (a1) is a single square, and G5 (a2) is an octagon surrounded by threesquares While the orientation of this graph will not affect the number of perfect match-ings, for convenience of describing the rest of the sequence G2n+3, we assume the three

squares of G5 are attached along the eastern, southern, and western edges of the octagonand identify G3 with the eastern square For n ≥ 3 the graph associated to an+1, G2n+3,can be inductively built from the graph for an, G2n+1 by attaching a complex consisting

of one octagon with two squares attached at its western edge and northern/southern edge(depending on parity) We attach this complex to the western edge of G2n+1, and ad-ditionally adjoin one arc between the northeast (resp southeast) corner of the southern(resp northern) square of G2n+3 \ G2n+1 and the southeast (resp northeast) corner ofthe northern (resp southern) square of G2n+1\ G2n−1

We can inductively build up the sequence of graphs corresponding to the dn’s, G−2n−1,analogously Here G−1, consists of a single octagon with a single square attached alongits northern edge We attach the same complex (one octagon and two squares) exceptthis time we orient it so that the squares are along the northern/southern and easternedges of the octagon We then attach the complex so that the eastern square attaches to

G−2n+1 Lastly we adjoin one arc between the northwest (resp southwest) corner of thesouthern (resp northern) square of G−2n−1\ G−2n+1 and the southwest (resp northwest)corner of the northern (resp southern) square of G−2n+1\ G−2n+3

The graph corresponding to b1, G4, is one octagon surrounded by four squares whilethe graph corresponding to c1, G0, consists of a single octagon As in the case of an or

dn, for n ≥ 1, the graphs for bn+1 (G2n+4) and cn+1 (G−2n) are constructed from bn and

cn (resp.), but this time we add a complex of an octagon, two squares and an arc on bothsides Note that this gives these graphs symmetry with respect to rotation by 180◦.The graphs G2n+2 (bn) consist of a structure of octagons and squares such that thereare squares on the two ends, four squares around the center octagon, and additional arcsshifted towards the center (the vertices joined by an arc lie on the vertical edges closest

to the central octagon) On the other hand, the graphs G−2n+2 (cn) have a structure ofoctagons and squares such that the central and two end octagons have only two squaressurrounding them, and the additional arcs are shifted towards the outside

For the reader’s convenience we illustrate these graphs for small n In our pictures,the extra arcs are curved, and the other edges are line segments It should be noted thatall these graphs are planar, even though it is more convenient to draw them in such a waythat the (curved) extra arcs cross the other (straight) edges