Perfect matchings for the three-term Gale-Robinson sequences pdf

Throughout the '90s, proofs of integrality were known only for individual sp In the early '00s, Sergey Fomin and Andrei Zelevinsky proved Gale and Robinson's integrality They that genera

Perfect matchings for the three-term

Gale-Robinson sequences

Mireille Bousquet-Mélou (LaBRI), James Propp , Julian West

(Submitted on 17 Jun 2009)

In memoryof David Gale, 1921-2008

In 1991, David Gale and Raphael Robinson, building on explorations

out b hael Somos in the 1980s, intro a three-parameter family of

pro non-integral rational numbers Throughout the '90s, proofs of integrality

were known only for individual sp In the early '00s, Sergey Fomin and

Andrei Zelevinsky proved Gale and Robinson's integrality They

that generalize Gale-Robinson numbers are polynomials with integer

ef-ts However, their proof did not oer any enumerative interpretation of the

Gale-Robinsonnumbers/polynomials Hereweprovide haninterpretationinthe

settingof p hings ofgraphs, h makesintegrality/polynomiality

obvi-ous Moreover,this interpretationimpliesthatthe tsoftheGale-Robinson

polynomials arepositive,asFomin and Zelevinsky

Linear are ubiquitousin as partof abroad generalframework

that is well-studied and well-understood; in many binatorially-dened

possible to reverse-engineer a binatorialinterpretation for the (see [4℄ and

s(n) = (s(n − 1)s(n − 3) + s(n − 2) 2 )/s(n − 4),

hwe prefer to write inthe form

s(n)s(n − 4) = s(n − 1)s(n − 3) + s(n − 2) 2 ,

are tered far less often, and there is no simple general theory that es the

s(n)s(n − k) = s(n − 1)s(n − k + 1) + s(n − 2)s(n − k + 2) + · · ·+ s(n − ⌊k/2⌋)s(n − ⌈k/2⌉).

If one puts s(0) = s(1) = · · · = s(k − 1) = 1 and denes subsequent terms using the

are entries A006720 through A006723 in [24℄.) Although integer satisfying

http://jamespropp.org/somo s.ht ml.)

Inspired by the workof Somos,DavidGale and RaphaelRobinson[13,12℄

a(n)a(n − m) = a(n − i)a(n − j) + a(n − k)a(n − ℓ),

withinitial a(0) = a(1) = · · · = a(m − 1) = 1,wherem = i + j = k + ℓ Wethis the three-term Gale-Robinson re e

1 The Somos-4 and Somos-5

arethesp where(i, j, k, ℓ)isequalto(3, 1, 2, 2)and(4, 1, 3, 2)resp ely Galeand Robinson that for all integers i, j, k, ℓ > 0 with i + j = k + ℓ = m, the

a(0), a(1), determined by this has all itsterms given by integers.About ten years later,this wasproved inaninuentialpaperby Fominand

In this paper, we rst give a ombinatorial proof of the integrality of the three-term

P (n; i, j, k, ℓ) (n > 0) and prove in Theorem 9 that the nth graph in the has

diamondgraphs, h are the hings graphs for the Gale-Robinson

is shown in Figure1 All are subgraphs of the square grid

Figure 1: The P (25; 6, 2, 5, 3) Its hing numberis a(25), where a(n) is theGale-Robinson asso with (i, j, k, ℓ) = (6, 2, 5, 3)

method (seeFigure11and the surroundingtext) that the graph P (n; i, j, k, ℓ)

in terms of the smaller graphs P (n ′

; i, j, k, ℓ) with n ′

Formula(2)in 3)thatallowsoneto thegraphP (n; i, j, k, ℓ)immediately.The heart of our proof is the demonstration that if one denes a(n) as the number of

p hings of P (n) ≡ P (n; i, j, k, ℓ), the a(0), a(1), a(2), satises the

P (n)has a(n)p hingsforalln, hyieldsadditionallytheintegralityofa(n)

terms a(n 1 )a(n 2 ) of the Gale-Robinson

dened by

p(n)p(n − m) = w p(n − i)p(n − j) + z p(n − k)p(n − ℓ),

withi + j = k + ℓ = mandp(0) = p(1) = · · · = p(m − 1) = 1,isa ofpolynomials

spe horizontaledges(theexponentof the variableu)and the numberofv edges(the exponent of the variable v) The that p(n) is a polynomial with ts in

Zwasproved in[11℄, but no binatorialexplanation wasgiven and the non-negativity

A paper by one of these students, David Speyer [25℄,

Gale-Robinson and thusyieldsa binatorialproof ofthe integrality ofthe

polyno-mials mentioned abo e are indeed polynomials, and have non-negative ts One

se-Moreover, the of our graphs as subgraphs of the square grid looks

moreregular,and may beuseful tostudy limitshapesofrandomp hings

rather, the equivalentdomino tilings)

LetusmentionthatshortlyafterSpeyerdidhisworkonp hings,he andhis

work gives, as two sp binatorial proofs of the integrality of Somos-6 and

Somos-7

ThestrategiesthatledtoSpeyer's [25℄andtothepresent arenotentirely

to hers seeking toapply similar hniques to otherproblems; others may want to

skip the rest of the intro

singly-indexedGale-Robinsonnumbera(n)byatriply-indexed quantityA(n, p, q) satisfyingtheperturbed

A(n, p, q)A(n−m, p, q) = A(n−i, p−1, q)A(n−j, p+1, q)+A(n−k, p, q+1)A(n−ℓ, p, q−1).

(This of perturbation is not as sp as it looks: all that matters is that the

pairs (−1, 0), (1, 0), (0, 1), (0, −1)that e the perturbations of the and thirdordinates inthe fourindex-triples onthe right-handside, viewed aspointsintheplane,

parallelogramis tantamount to asimple re-indexing of the Ifwe

take as our initial A(n, p, q) = x n,p,q for all n between 0 and m − 1 and p, q

arbitrary, with (formal) indeterminates x n,p,q, then h A(n, p, q) with n > m beexpressed as a rational of these indeterminates It should be emphasized here

thatforalln, p, q, r, s,therational A(n, p, q)andA(n, r, s)arethesame

up to re-indexing of the indeterminates

Propp that hA(n, p, q)isaLaurentpolynomialinsome nitesubsetofthe(innitelymany)indeterminatesx n,r,s,withinteger ts;thatis, hA(n, p, q)

is an element of Z[x ± n,r,s 1 ] This was subsequently proved by Fomin and Zelevinsky [11℄.Note that if one sets all the indeterminates x n,r,s equal to 1, the Laurent polynomials

tin h hLaurentpolynomialispositive(a thatisnot proved byFomin

and Zelevinsky's method)and furthermore is equal to 1

Proppknewthatinthe i = j = k = ℓ = 1,the LaurentpolynomialsA(n, p, q)

beinterpretedasmultivariate hingpolynomialsofsuitablegraphs,namely,the

t of its relationto the study of determinants, and had shown (with Rumsey) [22℄

on this with determinants, see [5℄.) The work by Elkies, Kuperberg, Larsen,

and Propp [10℄ had shown that the monomialsin these Laurent polynomials ond

beextended tothe Gale-Robinson family of

It should be knowledged here that the idea behind the sp triply-indexed

per-turbation A(n, p, q) of the Gale-Robinson that proved so fruitful from an

of Zabrodin [28℄ that was brought to Propp's attention by k Kenyon This

What the REACHstudents wereable todo, afterdiligentexaminationof theLaurent

polynomialsA(n, p, q),is viewthose Laurent polynomialsas multivariate hingnomials of suitable graphs Bousquet-Mélou and West, independently, did the same for

poly-small values of n, untilthey were able to extrapolate these examplesto the form

There is a general strategy here for reverse-engineering binatorial interpretations

another ofthisreverse-engineeringmethod(inthe textofMarkonumbers

and frieze patterns), see [18℄

In this we dene a family of subgraphs of the square h we

the set of edges of E \ E ′

havingboth endpoints inV ′′

of E h that every vertex

We willsometimes omit the word p and

m(G), is the number of p hings of G More generally, we shall often

the set E of edges as a set of uting indeterminates, and asso with a (phing E ′

inthissum-of-pro byanon-negativeintegern e,thenthis expressionb a

there are n e edges joining the v x and y for all e = {x, y} in E (and no edgesjoiningx and y if {x, y} isnot in E) In if hn e is set equal to0or 1,then

mostregularofthemare the(Azte diamondgraphs, hare thedualsof the

diamonds, h were rst studied in detail in [10℄ A diamond graph of width

2k − 1 is obtained by taking e rows of squares, of length 1, 3, , 2k − 3, 2k −

1, 2k −3, , 3, 1and kingthemfromtoptobottom,withthemiddlesquaresinalltherows lining up v , asillustrated by Figure 2 Let A be a diamondgraph of width

2k − 1 LetA N bethe diamondgraphof width 2k − 3 obtained by deletingthe leftmostand rightmost squares of A as well as the two lowest squares of h of the remaining

South, West and East sub-diamonds of A, denoted by A S , A W and A E Finally, let A C

be the tral sub-diamond of A of width 2k − 5 (Figure 3) The following result is a

diamond graph A is related to the polynomials of its sub-diamondsby

M(A)M(A C ) = nsM(A W )M(A E ) + ewM(A N )M(A S ),

where n, s, w,ande denote respe thetop (resp bottom, westmost,eastmost)edge of

A (see Figure 2)

In if a(n) (with n > 2) denotes the hing number of a diamond graph ofwidth 2n − 3,then

a(n)a(n − 2) = 2a(n − 1) 2

for all n > 2, provided we adopt the initial a(0) = a(1) = 1 This shows that

implies a(n) = 2( n 2 )

spe of it,and this is the point of viewwe adopt in this paper The key idea is

Corollary 2 LetAb adiamondgraph,andletGb aspanningsubgraphofA, ontainingthe edges n, s, w and e Let G N = G ∩ A N, and dene G S , G W , G E and G C similarly.Then

M(G)M(G C ) = nsM(G W )M(G E ) + ewM(G N )M(G S ).

A E

A N

A S

Figure3: The ve sub-diamondsof a diamondgraph of width 9

a = 0 in M(A), for every edge a that belongs to A but not to G The same propertyrelates M(G N ) and M(A N ), and so on Consequently, Corollary 2 issimplyobtained bysetting a = 0in Theorem 1,for every edge a that belongsto A but not to G

A standard pine one of width 2k − 1 is a subgraph P = (V, E) of the square

satisfyingthe three following illustrated by Figure4.a:

1. Thehorizontaledges formi + j + 1segmentsof oddlength, startingfromthepoints

(0, 1), (1, 2) , (i − 1, i)and (0, 0), (1, −1), , (j, −j),forsomei > 1,j > 0

More-o er, if L m denotes the length of the segment lyingat ordinatem, then

L − j < · · · < L − 1 < L 0 = 2k − 1 = L 1 > L 2 > · · · > L i

the abo e horizontaledges

3. Lete = {(a, b), (a, b + 1)}beav edgeofthesquare joiningtwov

of V Ifa + b is even, then e belongs tothe set ofedges E, and we say that e is aneven edgeof P Otherwise, emay belong to E,or not (Figure4.a), and we e a(present orabsent) odd edge

The leftmostv ofa standard are always (0, 0)and (0, 1) However, times it is venient to graphsobtained by shifting h a graph toa dierent

In a transplanted the leftmost v are (a, b) and (a, b + 1), where a + b is

intend, we omit the modier and simplyuse the word

tains no vertex of degree 1 h a vertex only o at the right border of the

and o if the rightmost vertex of some horizontalsegment does not belong

to a v edge, as shown in Figure 4.b.) An diamond graph is an example of

tains 2k − 1 butmay tainnosquareatall Denotingbyℓ(resp.r)theleftmost (resp rightmost) of the longest row of P, we say that P is rooted on (ℓ, r).(If P is standard, then ℓ is the with (0, 0) as its lower-left We refer to thelongest row of a as row 0 The row abo e it (resp below) is row 1 (resp −1),and so on

hrowisa k square In this the rightmost ksquarein hrowisalsothe

position of its k squares Equivalently,it is determinedby the positionof

itsodd v edges Conversely, any nite set S of ksquares whoseleft v lie in the 90 degree wedge bounded by the rays y = x > 0 and −y = x > 0.Assume that S is monotone, in the following sense: the rows that tain at least onesquareof S are e(sayfromrow−j torowi, whererowm refersto lobetween ordinates mand m + 1)andform > 0(resp m < 0),the rightmost ksquare

lower-inrowm o tothe leftoftherightmost ksquare inrowm − 1(resp.m + 1) Then

weight 1

hingofP Sp ,ifv isavertex ofdegree1 inP,theninany p hing

tinuingthe pro of andforbiddingotheredges Anexampleof thisisshown

reader k (starting from the rightmost frontier of P and working

leftward) that h of the isolated edges is a edge (that is, it must be tained

a p hing of P ¯ together with the set of isolated edges shown at right In this

ofP We thisthe ore ofP anddenoteitbyP ¯ (IfP isnotastandard

dene P 0 asthe standard obtained by translating P by (−a, −b), and we denethe of P asthe of P 0 translated by (a, b) However, for the rest of this

P, let b 1 be the rightmost k square in row 1 of P that lies to the left of b 0,let b 2 be the rightmost k square in row 2 of P that lies tothe left of b 1, and

so on(pro upwards); likewise, let b − 1 be the rightmost k square in row −1of

P that lies to the leftof b 0, and so on(pro downwards) If atsome pointthere is no k square that satises the requirement, we leave b m undened Considerall the of P that lie in the same row as,and lieweakly tothe left of, one of one of

in row 0 of Q be no farther to the right than b 0, h implies that the rightmost

k square in row 1 of Q be no farther to the right than b 1, and likewise for

of the leftmostand rightmost v edgesin h row(and noother v

edges) along with some horizontal edges (Figure6) In the rightmost v

ely of horizontal edges (see Figure 5); let H be the edge set of this

of P by adjoining the edges in H, so m(P ) > m( ¯ P ) We now show that every phing ofP is obtained froma p hing of P ¯ inthis wa

Proposition 3 Let P b a pine one with ore P ¯ Then m( ¯ P ) = m(P )

degree1alongtherightboundaryofQ Letv = (a, b)beoneofthetherightmostv

of degree1 inQ Thenv isthe rightmost vertex inone of the rows of Q Assume for the

moment that v lies abo e the longestrowof Q (that is, b > 1) See the top part

of Figure7 for anillustration of the following argument Let u be the vertex to the left

while every other edge taining uis forbiddenfrom belongingto any p hing

obtained fromQ bydeletingu, v,and every edge twith

u or v has the same hing number as Q Furthermore, Q ′

vertex v 1 = (a − 1, b + 1)belongs toQ In this v 1 has degree 1 in Q ′

Let i be thelargest integer h that v j = (a − j, b + j) belongs to Q for all 0 6 j 6 i Applying thedeletion pro to the v v = v 0 , v 1 , , v i (in this order) yields a Q ∗

Assume now that b = 1 Applying the deletion pro to all the v of Q of theform (a − j, 1 + j) or (a − j, −j) yields again a Q ∗

(see Figure 7, bottom) By

symmetry,wehave vered allpossible values of b

,weremoved someedges

that belong to P ¯ The examination of Figure5 shows that we would have, in

b the removed edges were all or forbidden, whereas the rightmost edges of

¯

P are neither nor forbidden(Figure6)

To prove thatm( ¯ P ) = m(P ),take Q = P and use the operationrepeatedly

, Q ∗∗

, hthat m(P ) = m(Q) = m(Q ∗

) = m(Q ∗∗

) = · · · and P = ¯ ¯ Q = Q ∗ = Q ∗∗ = · · · Eventually we arrive at a

sub-of P whose is P ¯; that is, we arrive at P ¯ itself And h step of ourpreserves m(Q),we that m( ¯ P ) = m(P ),as

Let P be a with longest row of 2n + 1 squares Let A be thesmallestdiamondgraph tainingP (thelongestrowofA tains 2n + 1

Let G denote the spanning subgraph of A whose edge-set of all edges of P, allhorizontal edges of A, and all even v edges of A (Figure 8) Observe that G is a

P

G

Let usnow use the notationof Corollary 2 That is, G N = G ∩ A N,and soon Then

G N , G S , G W , G E andG C are (standardortransplanted) LetP N

,P S,P W,P Eand P C

,P S,P W, P Eand P C

the

An example is given in Figure9 Let ℓ 0 (resp r 0) be the leftmost (resp rightmost)

of the longest row R 0 of P Similarly,let r 1 (resp r − 1) denote the rightmost of therow just abo e (resp below) R 0 Observe that the r 0 , r 1 and r − 1 ond to

k squares of P Finally, let ℓ ′

0 be the k ell of R 0 following ℓ 0, and let r ′

0 the

k square of R 0 r 0 (if it exists) In light of the properties of the

Proposition 4 Let P b a d pine one With the above notation, P N

0) Finally, P C

is the largest d sub-pine one rooted on (ℓ ′

0 , r ′

0 )

, P S, P Eand P W

ℓ ′ 0

P N

Similarly,

for r < 0, row r of P with row r + 1 of P S

It thus remains to determine the

longest row of P This row is obtained by adding a 2-by-1 2

to the left of the

longest rowof P E

,and thensuperimposing the longest rowof P W

LetusnowapplyCorollary2tothegraphGobtainedby P intoaspanning

of A (Figure8) ByProposition3, P isthe of G,m(G) = m(P ),andsimilar identities relate the hing numbers of G N and P N

,

m(P )m(P C ) = m(P W )m(P E ) + m(P N )m(P S ).

numberof p hings ofthe diamondgraphofwidth2n − 3 isthenthterminthe2

a(n)a(n − 2) = a(n − 1)a(n − 1) + a(n − 1)a(n − 1),

with initial a(0) = a(1) = 1 More generally, the three-term Gale-Robinson

with initial a(n) = 1 for n = 0, 1, , m − 1 Here, i, j, k and ℓ are positiveintegers hthat i + j = k + ℓ = m, and weadopt the following(important) vention

j = min{i, j, k, ℓ}.

(P (n; i, j, k, ℓ)) n>0 for hsetofparameters{i, j, k, ℓ} hthati+j = k +ℓ = m,andto

its set of odd v edges, that is, v edges of the form {(a, b), (a, b + 1)} where

h will be used to determine the positions of the odd v edges in the

P (n; i, j, k, ℓ): forr > 0,let

U(n, r, c) = 2c + r − 3 − 2  mc + kr + i − n − 1

j

 ,

L(n, r, c) = 2c + r − 3 − 2  mc + ℓr + i − n − 1

j



(2)

Observe that the parameters k and ℓ play roles Also, U(n, 0, c) = L(n, 0, c)

and its South-West liesat ordinates (0, 0), asshown inFigure4

i + j = m and j 6 i) Retain those values U(n, r, c) that are larger than r, and a

and(U(n, r, c), r+1)(anoddedge, U(n, r, c)+r isodd) Therstrownot taining

propertyguaranteesthatiftherthrowisempty,thenallhigherrowsareemptytoo Italso

totherightoftherightmostv edgeinrowr +1 (Tosee this,notethatU(n, r, 0)−r

isalwaysanoddnumber.) SotheinequalityU(n, r + 1, 0) − (r + 1) < U(n, r, 0) − r implies

U(n, r + 1, 0) − (r + 1) 6 U(n, r, 0) − r − 2, orU(n, r + 1, 0) < U(n, r, 0) That is,the set

of odd edges (equivalently, of k squares) given by Formula(2) satisesthe top part

in rowr > 0, if itexists, liestothe leftof the rightmostodd edge inrow r − 1

Similarly, to lo the edges in row −r 6 0, the values of L(n, r, 0) > L(n, r, 1) > · · · and retain those larger than r For h, a v edge inrow −r

at L(n, r, c), that is, (L(n, r, c), −r) and (L(n, r, c), −r + 1) Observe

same whether it is determinedfrom U or fromL

The y properties satised by the positions of the odd edges imply that

horizontaledges from(r, r + 1) to(U(n, r, 0), r + 1) and from(−r, −r) to(L(n, r, 0), −r)

for allr > 0 Finally, allthe appropriateeven v edges these steps are

has been sp This pointof view simpliesthe exposition

Observe that P (n) is empty if and only if U(n, 0, 0) < 0, h is equivalent to

U(n, 0, 0) 6 −1 U(n, 0, 0) is odd), h is easily seen to be equivalent to n < m

(using the that m = i + j)

Example Take (i, j, k, ℓ) = (5, 2, 3, 4) and determine P (12) The abo e denition of U

U(n, r, c) = 2c + r − 3 − 2  7c + 3r − 8

2

 ,

L(n, r, c) = 2c + r − 3 − 2  7c + 4r − 8

2



Inrow0,wendodd edgeswith lowerv (5, 0)and (1, 0) Inrow1,thereisone oddedge at (4, 1) This is the top row of the diagram b U(12, 2, 0) = 1 < 2 Turning

to the lower portion of the diagram, there isone odd edge with lower vertex (2, −1) andnoneinrow−2orbelow Completingthe diagramisnowroutine,and givesthe

the Gale-Robinson a(n) asso with (5, 2, 3, 4)satises a(12) = 14

A larger example ispresented afterCorollary 10

(2) (4)

(0, 0)

hings (a stands for any of the two hings of asquare)

edges (or, equivalently, the k squares) in rows r and r + 1 are interleaved That is,between two ksquaresinrowr > 0,thereisa ksquareinrowr + 1,andsimilarly,between two ksquares inrowr 6 0, thereisa k square inrowr − 1 This beked onthe small example of Figure 10, but is more visibleon the bigger example of

Figure12

Lemma 6 (The interleaving property) For all values of n, r and c, the U

and L dened by (2) satisfy

U(n, r, c + 1) + 1 6 U(n, r + 1, c) 6 U(n, r, c) − 1

The three other inequalitiesare proved in asimilar manner

havejustdened Usingthe notationofTheorem 5,wewillverifythat,up totranslation,

P (n) W = P (n − i), P (n) E = P (n − j), P (n) N = P (n − k), P (n) S = P (n − ℓ) and

P (n) C = P (n − m) Theseequiv willfollowfromthe interleavingproperty andthe

Lemma 7 For any e of parameters (i, j, k, ℓ), the U and L dened by (2)satisfy:

U(n − k, r − 1, c) = U(n, r, c) − 1, L(n − ℓ, r − 1, c) = L(n, r, c) − 1, U(n − ℓ, r + 1, c − 1) = U(n, r, c) − 1, L(n − k, r + 1, c − 1) = L(n, r, c) − 1, U(n − m, r, c − 1) = U(n, r, c) − 2, L(n − m, r, c − 1) = L(n, r, c) − 2.

Proof The L-identities are to the U-identities upon hanging k and ℓ, sothat there are really5 identities to prove These all be veried by routine

manipulations Letus k for the fourth identity satisedby U:

Weleave itto the reader toverify the remaining 4identities

Wenow kthattheseidentitiesimplythatthe arerelatedtooneanother

These identities hold up to a translation

Proof Beginby king that P (n) W = P (n − i) Using the of P (n) W

given

inProposition 4,and the that the k squares of P (n) are interleaved, we see thatthe odd v edges in P (n) W

are those of P (n), that the rightmost odd edge

in h rowhas been removed (Ifthis was the onlyodd edge inthe row, then the entire

row disappears.) Therefore P (n) W

P (n), but beginning with c = 1, 2, instead of c = 0, 1, This means that in row