GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN. 1. Introduction

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1 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN KYU-HWAN LEE AND KYUNGYONG LEE Abstract. We propose a combnatoral/geometrc model and formulate several conectures to descrbe the c-matrces of an arbtrary skew-symmetrzable matrx. In partcular, we ntroduce real Lösungen as an analogue of real roots and conecture that c-vectors are real Lösungen gven by non-self-crossng curves on a Remann surface. We nvestgate the rank 3 quvers whch are not mutaton equvalent to any acyclc quver and show that all our conectures are true for these quvers. 1. Introducton Let Q be a quver wth n vertces and no orented cycles of length 2. The most basc nvarant of a representaton of Q s ts dmenson vector. By Kac s Theorem [14], the dmenson vectors of ndecomposable representatons of Q are postve roots of the Kac Moody algebra g Q assocated to the quver Q. When Q s acyclc, a representaton M of Q s called rgd f Ext 1 (M, M) = 0, and the dmenson vectors of ndecomposable rgd representatons are called real Schur roots as they are ndeed real roots of g Q. In the category of representatons of Q, rgd obects are foundatonal. Therefore an explct descrpton of real Schur roots s essental for the study of the category, and there have been varous results related to descrpton of real Schur roots of an acyclc quver ([3, 12, 13, 22, 23, 26]). In a prevous paper [16], the authors conectured a correspondence between real Schur roots of an acyclc quver and non-self-crossng curves on a marked Remann surface and hence proposed a new combnatoral/geometrc descrpton. Recently, Felkson and Tumarkn [8] completed a proof of the conecture for all 2-complete acyclc quvers. (An acyclc quver s called 2-complete f t has multple edges between any par of vertces.) Now, when Q s general, t s natural to consder the c-vectors of Q as dmenson vectors of rgd obects. Indeed, when Q s acyclc, the set of postve c-vectors s dentcal wth the set of real Schur roots [18]. For an arbtrary quver Q, a postve c-vector s the dmenson vector of a rgd ndecomposable representaton of a quotent of the completed path algebra. The quotent was ntroduced by Derksen, Weyman and Zelevnksy [6], and s called a Jacoban algebra. Thus Ths work was partally supported by a grant from the Smons Foundaton (#318706). Ths work was partally supported by NSF grant DMS , the Unversty of Nebraska Lncoln, and Korea Insttute for Advanced Study. 1

2 2 K.-H. LEE AND K. LEE c-vectors naturally generalze real Schur roots n ths sense, though they are not necessarly real roots of the correspondng Kac Moody algebra. Orgnally, c-vectors (and c-matrces) were defned n the theory of cluster algebras [9], and together wth ther companons, g-vectors (and g-matrces), played fundamental roles n the study of cluster algebras (for nstance, see [6, 10, 11, 17, 20]). As a cluster algebra s defned not only for a skew-symmetrc matrx (.e. a quver) but also for an arbtrary skew-symmetrzable matrx, one can ask: Can we have a combnatoral/geometrc descrpton of the c-vectors (and c-matrces) of a cluster algebra assocated wth an arbtrary skew-symmetrzable matrx? In ths paper, we propose a conectural, combnatoral/geometrc model for c-matrces assocated to an arbtrary skew-symmetrzable matrx, whch extends our model from the acyclc case [16, 8]. Our proposal s based on the followng observaton. Even n the non-acyclc case, the c-vectors behave the same way as real Schur roots do f we ntroduce a broader class of generalzed Cartan matrces. We wll ntroduce the noton of real Lösungen as an analogue of real roots to explan ths property of the c-vectors systematcally. When a skew-symmetrzable matrx s acyclc, t s natural to consder the correspondng generalzed Cartan matrx. For a general skew-symmetrzable matrx, we need to consder a broader class of matrces. Generalzed ntersecton matrces (GIMs) 1 were ntroduced by Slodowy [25, 24]. A GIM s a square matrx A = (a ) wth ntegral entres such that (1) for dagonal entres, a = 2; (2) a > 0 f and only f a > 0; (3) a < 0 f and only f a Snce we are more nterested n cluster algebras assocated wth skew-symmetrzable matrces, we restrct ourselves to the class of symmetrzable GIMs. Ths class contans the collecton of all symmetrzable generalzed Cartan matrces as a specal subclass. Defnton 1.1. Let A = (a ) be an n n symmetrzable GIM, and D = dag(d 1,..., d n ) be the symmetrzer,.e. the dagonal matrx such that d Z >0, gcd(d 1,..., d n ) = 1 and AD s symmetrc. Let Γ = n =1 Zα be the lattce generated by the formal symbols α 1,, α n. (1) An element γ = m α Γ s called a Lösung f (1.2) d a m m = 2d k for some k = 1,..., n. 1, n (2) A Lösung s postve f m 0 for all. Each α s called a smple Lösung. 1 Some authors call them quas-cartan matrces. For example, see [1].

3 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 3 (3) For each = 1,..., n, defne s GL Z (Γ) by s (α ) = α a α, = 1,..., n, and let G be the group generated by s 1,..., s n. (4) A Lösung γ s real f γ = gα for some g G and some = 1,..., n. (5) When A s symmetrc, a postve real Lösung s Schur f (m 1,..., m n ) s the dmenson vector of some rgd ndecomposable module M of the Jacoban algebra J(Q(A), W ) of the quver Q(A) over a feld k wth generc potental W such that End J(Q(A),W ) (M) = k. If A s a generalzed Cartan matrx, then real Lösungen (resp. real Schur Lösungen) are the same as real roots (resp. real Schur roots) of the Kac Moody algebra assocated wth A. We expect that, for each symmetrzable GIM, there exsts a Le algebra for whch real roots can be defned and are compatble wth real Lösungen, but we do not yet know whch Le algebra would be adequate. Some related works can be found n [1, 2, 4, 5, 21, 24, 25, 27]. Fx an n n skew-symmetrzable matrx B and let D = dag(d 1,..., d n ) be ts symmetrzer such [ that ] BD s symmetrc, d Z >0 and gcd(d 1,..., d n ) = 1. ] Consder the n 2n matrx B I. After a sequence w of mutatons, we obtan [B w C w. The matrx C w s called the c-matrx and ts row vectors the c-vectors. There s a set G B of at most 2n! symmetrzable GIMs assocated wth B. For each sequence of mutatons and every symmetrzable GIM n G B, we defne (see Defnton 2.6) (1) an n-tuple R of reflectons; (2) an n-tuple of l-vectors, each of whch s a real Lösung; the resultng matrx s called the l-matrx ; (3) and an n-tuple U of admssble curves whch are n one-to-one correspondence wth the l-vectors. Then we conecture as follows: Conecture 1.3. (1) the product of reflectons n R (n some order) s equal to s 1...s n ; (2) the l-matrx equals the c-matrx for some symmetrzable GIM A G B ; (3) the admssble curves n U have no (nether self nor parwse) ntersectons; (4) and f B s skew-symmetrc (equvalently D s the dentty matrx), then the postve l-vectors are real Schur Lösungen for the same GIM A. In partcular, we conecture that C B A G B L A

4 4 K.-H. LEE AND K. LEE where C B s the set of (postve) c-vectors of B and L A s the set of real Lösungen of A gven by non-self-crossng admssble curves. By constructon, admssble curves do not dstngush postve c-vectors from negatve ones snce the dfference s merely an sotopy. Ths conecture extends our conecture n [16] from acyclc quvers to arbtrary skew-symmetrzable matrces. In ths paper we prove Conecture 1.3 for the rank 3 quvers whch are not mutaton equvalent to any acyclc quver. The proof can be modfed to work for a bgger class of rank 3 skew-symmetrzable matrces, but we do not attempt to do so n ths paper for smplcty of notatons. Remark 1.4. Conecture 1.3 (4) follows from Conecture 1.3 (2). See Proposton Example. To llustrate our conecture, we present an example. Consder the symmetrzable matrx B = wth the symmetrzer D = dag(3, 2, 2), and the sequence of consecutve mutatons at ndces 2, 3, 2, 1, 2: [ ] [2,3,2,1,2] B I Thus we have obtaned three c-vectors (5, 18, 15), ( 2, 7, 6) and (0, 2, 1). We take a GIM A wth the same symmetrzer D and compute AD as follows: A = 2 2 2, AD = In accordance wth (1.2), defne a quadratc form by Then we have q(x, y, z) = 6x 2 + 4y 2 + 4z 2 12xy 8yz + 12zx. q(5, 18, 15) = 6, q( 2, 7, 6) = 4, q(0, 2, 1) = 4. Thus all three c-vectors are Lösungen for A. From Defnton 2.6, we obtan three reflectons s v 1 = s 3 s 2 s 1 s 2 s 3 s 2 s 3 s 2 s 1 s 2 s 3 s 2 s 3 s 2 s 1 s 2 s 3, s v 2 = s 3 s 2 s 1 s 2 s 3 s 2 s 3 s 2 s 1 s 2 s 3, s v 3 = s 2 s 3 s 2, where v s the mutaton sequence [2, 3, 2, 1, 2]. Snce s v 3 sv 2 sv 1 = s 1s 2 s 3, the part (1) of Conecture 1.3 s verfed for ths example. The correspondng l-vectors are, by Defnton 2.6, λ v 1 = s 3 s 2 s 1 s 2 s 3 s 2 s 3 s 2 (α 1 ) = (5, 18, 15), λ v 2 = s 3 s 2 s 1 s 2 s 3 (α 2 ) = ( 2, 7, 6), λ v 3 = s 2 (α 3 ) = (0, 2, 1).

5 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN Thus the l-matrx equals the c-matrx, and the part (2) of Conecture 1.3 holds Fnally, we draw the correspondng admssble curves on the unversal cover of a trangulated torus to see that they have no ntersectons. The shortest curve corresponds to s v 3, and the longest one to s v 1. Ths verfes the part (3) of Conecture 1.3 for ths example Organzaton of the paper. In Secton 2, our conecture wll be presented n a more refned way after all necessary defntons are made, and some propertes of l-vectors wll be proven. In the followng secton, the settngs wll be specalzed to the case of rank 3 quvers and the man theorem wll be stated. Sectons 4-8 are devoted to a proof of the man theorem. The proof wll be acheved through nducton. Secton 4 s concerned wth the ntal case, and Secton 5 wll present all possble stuatons for the general case after the ntal case. Secton 6 wll cover the base nducton step for the general case, and Secton 7 wll establsh the man nducton step. The proof wll be completed n Secton 8 by collectng consequences of the results. 2. Conecture In ths secton, we present our conecture n a more precse way after makng necessary defntons. For a nonzero vector c = (c 1,..., c n ) Z n, we wrte c > 0 f all c are non-negatve, and c < 0 f all c are non-postve. Defne c = ( c 1,..., c n ), and set α 1 = (1, 0,..., 0), α 2 = (0, 1, 0,..., 0),..., α n = (0,..., 0, 1). Assume that M = (m ) be an n 2n matrx of ntegers. For w = [, 2,..., l ], {1, 2,..., n}, we defne the matrx M w = (m w ) nductvely: the ntal matrx s M for w = [ ], and assumng we have M w, defne the matrx M w[k] = (m w[k] ) for k {1, 2,..., n} wth w[k] := [, 2,..., l, k] by (2.1) m w[k] = m f = k or = k, m + sgn(mw k ) max(mw k mw k, 0) otherwse,

6 6 K.-H. LEE AND K. LEE where sgn(a) {1, 0, 1} s the sgnature of a. The matrx M w[k] s called the mutaton of M w at the ndex k. Let B = (b ) be an n n skew-symmetrzable matrx and D = dag(d 1,..., d n ) be ts symmetrzer[ such ] that BD s symmetrc, d Z >0 and gcd(d 1,..., d n ) = 1. Consder the n 2n matrx B I and a mutaton sequence w = [ 1,..., k ]. After the mutatons at the ndces ] 1,..., k consecutvely, we obtan [B w C w. Wrte ther entres as (2.2) B w = [ ] [ ] b w, C w = c w = cw1. c w n, where c w are the row vectors. Defnton 2.3. The row vectors c w are called c-vectors of B for any and w. It s well-known that the vector c w s non-zero for each, and ether c w > 0 or c w < 0 due to sgn coherence of c-vectors ([7, 10]). For σ S n and ε = 1 or 1, we defne a symmetrzable GIM A(σ, ε) = (a ) by εb f σ 1 () > σ 1 (), (2.4) a = 2 f =, εb f σ 1 () < σ 1 (). Clearly the symmetrzer of A s the same as that of B. From Defnton 1.1, we have Lösungen and smple reflectons s, for = 1,..., n, and the group G generated by s s. To ntroduce our geometrc model 2 for c-vectors, we need a Remann surface Σ equpped wth n labeled curves as below. Let P 1 and P 2 be two dentcal copes of a regular n-gon. For σ S n, label the edges of each of the two n-gons by T σ(1), T σ(2),..., T σ(n) counter-clockwse. On P ( = 1, 2), let L be the lne segment from the center of P to the common endpont of T σ(n) and T σ(1). Later, these lne segments wll only be used to desgnate the end ponts of admssble curves and wll not be used elsewhere. Fx the orentaton of every edge of P 1 (resp. P 2 ) to be counter-clockwse (resp. clockwse) as n the followng pcture. 2 An alternatve geometrc model can be found n [8].

7 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 7 T σ(3) T σ(2) T σ(n) T σ(n 1) L 2. T σ(1). L 1 T σ(n 1) T σ(n) T σ(2) T σ(3) Let Σ be the Remann surface of genus n 1 2 obtaned by glung together the two n-gons wth all the edges of the same label dentfed accordng to ther orentatons. The edges of the n-gons become n dfferent curves n Σ. If n s odd, all the vertces of the two n-gons are dentfed to become one pont n Σ and the curves obtaned from the edges become loops. If n s even, two dstnct vertces are shared by all curves. Let T = T 1 T n Σ, and V be the set of the vertex (or vertces) on T. Let W be the unversal Coxeter group of rank n, whch s by defnton somorphc to the free product of n-copes of Z/2Z, and let R be the set of reflectons n W. We wll denote an element of W as a word from the alphabet {1, 2,..., n}. In partcular, an element w of R can be wrtten as w = 1 2 k such that k s an odd nteger and = k+1 for all {1,..., k}. Defnton 2.5. An admssble curve s a contnuous functon η : [0, 1] Σ such that 1) η(x) V f and only f x {0, 1}; 2) there exsts ɛ > 0 such that η([0, ɛ]) L 1 and η([1 ɛ, 1]) L 2 ; 3) f η(x) T \ V then η([x ɛ, x + ɛ]) meets T transversally for suffcently small ɛ > 0; 4) υ(η) R, where υ(η) := 1 k W s gven by {x (0, 1) : η(x) T } = {x 1 < < x k } and η(x l ) T l for l {1,..., k}. We consder curves up to sotopy. When p = p+1, 1 p k 1, for υ(η) = 1 k, the curve η s sotopc to a curve η 1 wth υ(η 1 ) = 1 p 1 p+2 k. If η 1 and η 2 are curves wth υ(η 1 ) = 1 k and υ(η 2 ) = 1 l, defne ther concatenaton η 1 η 2 to be a curve such that υ(η 1 η 2 ) = 1 k 1 l. For w = 1 2 k W, defne s(w) = s 1...s k G and η(w) to be the curve such that υ(η(w))) = w. We wrte s(η) = s(υ(η)) for a curve η. For sequences w = [ 1,..., k ] and v = [ 1,..., l ], defne wv to be ther concatenaton [ 1,..., k, 1,..., l ]. Defnton 2.6. For each mutaton sequence w, defne (e w, tw ) W := {±1} W nductvely wth the ntal elements (e, t ) = (+1, ), = 1,..., n, as follows: (e w, tw k tw tw k ) f bw k cw kk > 0, (2.7) (e w[k], t w[k] ) = ( e w, tw ) f = k, (e w, tw ) otherwse,

8 8 K.-H. LEE AND K. LEE where b w k and cw kk are the entres of Bw and C w n (2.2), respectvely. Next, for a mutaton sequence w and for = 1,..., n, defne the reflectons and the admssble curves by (2.8) s w = s(t w ) G and η w = η(t w ) Σ, respectvely, and defne the l-vectors by (2.9) λ w = e w s 1 s p (α ) Z n, where we wrte t w = 1 p p 1. The matrx L w = The l-vectors satsfy the nductve rule gven by s w k λw f b w k cw kk > 0, (2.10) λ w[k] = λ w f = k, otherwse. It s clear that We also see that λ w s w = s (η w ). (2.11) (e w[k,k], t w[k,k] ) = (e w, t w ). Indeed, note that b w k cw kk (e w[k,k] λw1. wll be called the l-matrx. λ w n > 0 f and only f bw[k] k c w[k] kk > 0, and n ths case, we have, t w[k,k] ) = (e w[k], t w[k] k t w[k] t w[k] k ) = (e w, t w k (tw k tw t w k )tw k ) = (ew, t w ). The other cases are clear. Consequently, we obtan the followng lemma. Lemma For any w and for any, k, s w[k,k] = s w, λ w[k,k] = λ w and η w[k,k] = η w. Notce that admssble curves do not dstngush postve l-vectors from negatve ones snce the dfference s merely an sotopy. Now we state Conecture 1.3 n a more refned way. Conecture For each mutaton sequence w, we can choose an element σ S n and ε = +1 or 1 to determne a GIM A(σ, ε) such that the followng holds for the reflectons and l-vectors defned through A(σ, ε): (C1) C w = L w ;

9 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 9 (C2) (2.14) s w ( λ w ) = λ w, s w ( λ w ) = λ w ± b w λ w for, where the sgn + or s to be chosen for ±b w ; (C3) It s possble to order the l-vectors λ w so that the negatve vectors precede the postve ones, and the product of the reflectons s w correspondng to these vectors, taken n ths order, equals s 1 s n ; (C4) Each of the admssble curves η w, = 1,..., n, s a non-self-crossng admssble curve, and they form a famly of parwse non-crossng curves. Here t s mportant to notce that dfferent mutaton sequences may well determne dfferent GIMs as llustrated n the followng example Example As n Secton 1.1, consder the matrx B = wth the symmetrzer D = dag(3, 2, 2), and recall that we chose d S 3 and the GIM A(d, 1) = for the mutaton sequence [2, 3, 2, 1, 2] to check the conecture. Now, for the mutaton sequence w = [3, 1, 2], we choose the GIM A(σ, 1) = 2 2 2, where σ S 3 s gven by σ(1) = 2, σ(2) = 3, σ(3) = 1. Wth ths choce of a GIM, one can check Conecture 2.13 for w = [3, 1, 2]. In partcular, we have [ ] w B I , and the l-vector λ w 1 s gven by λ w 1 = s 1 s 3 s 2 s 3 (α 1 ) = (5, 3, 6), whch s equal to the c-vector c w 1. As for another example v = [1, 3, 2], we choose the GIM A( σ, 1) = 2 2 2, where σ S 3 s gven by σ(1) = 3, σ(2) = 1, σ(3) = 2. Then we have

10 10 K.-H. LEE AND K. LEE and the l-vector λ v 1 [ B ] I s gven by whch s equal to the c-vector c v 1. v , λ v 1 = s 2 s 1 s 3 (α 1 ) = (5, 9, 3), The followng two propostons show that the l-vectors (are expected to) form a specal subfamly of the Lösungen. Proposton The l-vectors are real Lösungen. Proof. If an l-vector s a Lösung, then t s a real Lösung by defnton. blnear form on Γ = n =1 Zα by (α, α ) = a d Defne a symmetrc and by extendng t through lnearty. To show (1.2) for l-vectors, t suffces to show that ths form s G-nvarant. In turn, t amounts to showng that for,, l {1, 2,..., n}. Indeed, we have (s (α ), s (α l )) = (α, α l ) = a l d l (s (α ), s (α l )) = (α a α, α l a l α ) = (α, α l ) a (α, α l ) a l (α, α ) + a a l (α, α ) = a l d l a a l d l a l a d + 2a a l d = a l d l, snce a l d l = a l d. Proposton Assume that B s skew-symmetrc, and suppose that Conecture 2.13 (C1) s true. Then the postve l-vectors are real Schur Lösungen for the GIM A(σ, ε). Proof. It s known [18, Theorem 15] that a postve c-vector s the dmenson vector of a rgd ndecomposable module M of the Jacoban algebra J(Q(A), W ) over k wth generc potental W such that End J(Q(A),W ) (M) = k. If Conecture 2.13 (C1) s true, the l-vectors are c-vectors, and they are real Schur Lösungen by defnton. Before we close ths secton, we prove the frst dentty n (C2) of Conecture 2.13 and establsh symmetres among the remanng denttes when B s skew-symmetrc. Proposton For a mutaton sequence w, fx a GIM A(σ, ε), σ S n, ε = +1 or 1. (1) For {1, 2,..., n}, we have (2.19) s w ( λ w ) = λ w.

11 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 11 (2) Further, assume that B s skew-symmetrc, and wrte for and x R. Then we have s w ( λ w ) = λ w + x λ w (2.20) s w ( λ w ) = λ w + x λ w. Proof. (1) We use nducton on the length of w. If w s empty, we have s ( λ ) = s (α ) = α = λ. Assume that (2.19) s true for w, and consder w[k]. There are two cases. Frst, assume that s w[k] = s w and λ w[k] = ±λ w. Then we obtan s w[k] λ w[k] Next, assume that s w[k] = s w k sw sw k 1. Then we see s w[k] λ w[k] = s w λ w = λ w = λ w[k]. and λw[k] = s w k λw. Wrte sw k λw = ɛ 1s w k λw for ɛ 1 = 1 or = s w k sw s w k sw k λw = ɛ 1 s w k sw λ w = ɛ 1 s w k λw = s w k λw = λ w[k]. (2) Snce s w s a reflecton, t follows from part (1) that we can wrte s w ( λw ) = λw + x λw for and for some x R. We wll prove (2.20) by nducton. If w s empty, t follows form the defntons. Assume that (2.20) s true for w, and consder w[k]. There are four cases to consder: (a) s w[k] = s w and s w[k] = s w, (b) s w[k] = s w k sw sw k and sw[k] = s w k sw sw k, (c) s w[k] = s w and s w[k] = s w k sw sw k. (d) s w[k] = s w k sw sw k In case (a), wrte Snce λ w[k] = λ w By nducton, we have and sw[k] = s w. s w[k] ( λ w[k] and λw[k] = λ w, we obtan ) = λ w[k] + x λ w[k]. s w ( λ w ) = λ w + x λ w. s w ( λ w ) = λ w + x λ w, and hence s w[k] In case (b), we have λ w[k] = s w k λw ( λ w[k] ) = λ w[k] + x λ w[k]. and λw[k] = s w k λw. Wrte, for some ɛ 1, ɛ 2 {1, 1}, s w k λw = ɛ 1 s w k λw and s w k λw = ɛ 2 s w k λw. Assume that s w[k] ( λ w[k] ) = λ w[k] + x λ w[k].

12 12 K.-H. LEE AND K. LEE Then we have Comparng both sdes, By nducton, we get Now we obtan s w[k] ( λ w[k] ) = s w k sw s w k ( sw k λw ) = ɛ 2 s w k sw λ w, λ w[k] + x λ w[k] = s w k λw + x s w k λw = ɛ 2 s w k λw + ɛ 1 x s w k λw. s w λ w = λ w + ɛ 1 ɛ 2 x λ w. s w λ w = λ w + ɛ 1 ɛ 2 x λ w. s w[k] ( λ w[k] ) = s w k sw s w k ( sw k λw ) = ɛ 1 s w k sw λ w = ɛ 1 s w k ( λw + ɛ 1 ɛ 2 x λ w ) = λ w[k] + x λ w[k]. Thus (2.20) s true for w[k] n ths case. In case (c), we have λ w[k] = λ w and λw[k] = s w k λw. Wrte, for some ɛ 2 {1, 1} and x 1, x 2, x 3 R, Then we have s w k λw = ɛ 2 s w k λw, s w k ( λw ) = λ w + x 1 λ w k s w ( λ w ) = λ w + x 2 λ w, s w ( λ w k ) = λw k + x 3 λ w. On the other hand, by nducton, s w[k] ( λ w[k] ) = s w s w k λw = ɛ 2 s w ( λ w + x 1 λ w k ) = ɛ 2 ( λ w + x 2 λ w ) + x 1 ɛ 2 ( λ w k + x 3 λ w ) = λ w[k] + ɛ 2 (x 2 + x 1 x 3 ) λ w[k]. s w[k] ( λ w[k] ) = s w k sw s w k λw = s w k sw ( λ w + x 3 λ w k ) = s w k ( λw + x 2 λ w ) + x 3 s w k ( λw k + x 1 λ w ) = λ w + x 2 ( λ w + x 1 λ w k ) + x 1x 3 ( λ w + x 1 λ w k ) = λ w[k] + ɛ 2 (x 2 + x 1 x 3 ) λ w[k], as desred. The case (d) can be proven smlarly. Ths completes the proof.

13 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN Man Result In the rest of ths paper, we prove Conecture 2.13 for the class of rank 3 quvers that are not mutaton-equvalent to any acyclc quver. We call a quver 2-complete f each quver n ts mutaton class has multple edges between any par of vertces. Notce that f a rank 3 quver s not mutaton-equvalent to any acyclc quver then t s automatcally 2-complete. We fx notatons for the quvers we deal wth. Wthout loss of generalty, assume that b 12 2, b 23 2 and b Then the quver Q can be depcted by (3.1) 2 b 12 b 23 b We also assume that a mutaton sequence starts wth 2 and does not have two consecutve mutatons at the same vertex. We do not lose generalty due to symmetry of the quver and Lemma Moreover, we take σ = d S 3 and ε = 1 throughout. For any mutaton sequence (startng wth 2), the symmetrc GIM s to be 2 b 12 b 31 A(d, 1) = b 12 2 b 23, b 31 b 23 2 and the smple reflectons s 1, s 2, s 3 are determned; n partcular, s 1 (α 2 ) = α 2 + b 12 α 1, s 2 (α 3 ) = α 3 + b 23 α 2, s 1 (α 3 ) = α 3 b 31 α 1. We obtan a trangulaton of the Remann surface Σ (whch s a torus n ths case) wth edges ndexed by 1, 2, 3 as follows: 3 2 We wll consder the followng statement for each mutaton sequence w[k]. (S0) If λ w[k] = s w k λw then λ w[k] and λ w k 1 are both postve or both negatve. Theorem 3.2. Let Q be a rank 3 quver whch s not mutaton-equvalent to any acyclc quver. Then Conecture 2.13 holds true for Q. Moreover, (S0) s true for any mutaton sequence w[k]. The above theorem can be extended to a larger class of rank 3 skew-symmetrzable matrces. However, for smplcty of notatons, we are content n ths paper to consder the quver (.e. skew-symmetrc) case only.

14 14 K.-H. LEE AND K. LEE Our proof of Theorem 3.2 wll be acheved by nducton, checkng n each step the statements (C1)-(C3) of Conecture 2.13 and (S0) gven above. The statement (C4) wll be proven n the last secton (Proposton 8.2) usng (S0). As for (C2), t suffces by Proposton 2.18 to check (3.3) (C2) for Theorem 3.2: for a fxed {1, 2, 3}, s w ( λ w 1 ) = λ w 1 ± b w 1, λ w, s w ( λ w +1 ) = λ w +1 ± b w +1, λ w, s w 1( λ w +1 ) = λ w +1 ± b w +1, 1 λ w 1, where ± s to be determned and the ndces are read modulo 3. Remark 3.4. The group G generated by s s s not always a Coxeter group. In partcular, when b 12 = b 23 = b 31 = 2, we have s 3 s 2 s 1 s 3 s 2 s 1 = d. It s ntrgung that the same relaton appears for the double affne Weyl group assocated to SL 2 (F ), where F s a two-dmensonal local feld such as Q p ((t)) or F q ((t 1 ))((t 2 )). See [15, 19]. 4. Intal case: untl the frst mutaton at Base step. The ntal c-vectors are α 1, α 2 and α 3. They are also l-vectors. Thus (C1)-(C3) and (S0) are true trvally Mutatons at 2 and 1 only. Recall our assumpton that a mutaton sequence starts wth 2. From the condtons on our quver, we have b 31 + b 12 b 23 > 0. After mutaton at vertex 2, we obtan From the defnton, (C1) We have 0 b 12 b 31 + b 12 b 23 1 b 12 0 B [2] = b 12 0 b 23, C [2] = b 31 b 12 b 23 b s [2] 1 = s 2s 1 s 2, s [2] 2 = s 2, s [2] 3 = s 3, λ [2] 1 = s 2α 1, λ [2] 2 = α 2, λ [2] 3 = α 3. (4.1) λ [2] 1 = s 2α 1 = α 1 + b 12 α 2 = (1, b 12, 0) = c [2] (4.2) λ [2] 2 = α 2 = c [2] 2, λ[2] 3 = α 3 = c [2] 3. 1,

15 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 15 (C2) By (3.3), t suffces to check (4.3) s [2] s [2] 2 ( λ[2] 1 ) = s 2s 2 α 1 = α 1 = α 1 + b 12 α 2 b 12 α 2 = λ [2] 1 + b[2] 2 ( λ[2] 3 ) = s 2α 3 = α 3 + b 23 α 2 = λ [2] 3 + b[2] 32 λ[2] 2, 12 λ[2] s [2] 3 ( λ[2] 1 ) = s 3s 2 α 1 = s 2 α 1 + (b 12 b 23 b 31 )α 3 = λ [2] 1 + b[2] 13 λ[2] 3. Note that b [2] 12 < 0, b[2] 32 > 0 and b[2] 13 > 0. (C3) We see that s [2] 2 s[2] 1 s[2] 3 = s 2s 2 s 1 s 2 s 3 = s 1 s 2 s 3. (S0) We have λ [2] 1 = s 2λ 1, and both λ [2] 1 and λ 2 are postve. Defne a sequence {g n } by For n 1, we obtan from mutatons g n = b 12 g n 1 g n 2, g 0 = 0, g 1 = 1. b [(21)n ] 12 = b 12, b [(21)n ] 13 = b [(21)n 1 2] 13, b [(21)n ] 23 = b [(21)n 1 2] 23 + b 12 b [(21)n 1 2] 13, b [(21)n 2] 12 = b 12, b [(21)n 2] 13 = b [(21)n ] 13 + b 12 b [(21)n ] 23, b [(21)n 2] 23 = b [(21)n ] 23, and g 2n 1 g 2n 0 g 2n+1 g 2n+2 0 (4.4) C [(21)n] = g 2n g 2n+1 0, C [(21)n2] = g 2n g 2n Lemma 4.5. We have b [(21)n 2] 13 = b 13 g 2n+1 + b 23 g 2n+2, b [(21)n+1 ] 23 = b 13 g 2n+2 + b 23 g 2n+3. Proof. We use nducton on n. The case n = 0 can be checked easly. We obtan b [(21)n+1 2] 13 = b [(21)n+1 ] 13 + b 12 b [(21)n+1 ] 23 = b [(21)n 2] 13 + b 12 (b 13 g 2n+2 + b 23 g 2n+3 ) = b 13 g 2n+1 b 23 g 2n+2 + b 12 b 13 g 2n+2 + b 12 b 23 g 2n+3 = b 13 (b 12 g 2n+2 g 2n+1 ) + b 23 (b 12 g 2n+3 g 2n+2 ) = b 13 g 2n+3 + b 23 g 2n+4. 2, Smlarly, we get b [(21)n+2 ] 23 = b 13 g 2n+4 + b 23 g 2n+5.

16 16 K.-H. LEE AND K. LEE From the defntons (2.7) and (2.8), we obtan s [(21)n ] 1 = (s 2 s 1 ) n 1 s 2 s 1 s 2 (s 1 s 2 ) n 1, s [(21)n 2] 1 = (s 2 s 1 ) n s 2 s 1 s 2 (s 1 s 2 ) n, s [(21)n ] 2 = (s 2 s 1 ) n s 2 (s 1 s 2 ) n, s [(21)n 2] 2 = (s 2 s 1 ) n s 2 (s 1 s 2 ) n, s [(21)n ] 3 = s 3, s [(21)n 2] 3 = s 3. Lemma 4.6 (C1). We have λ [(21)n ] 1 = (s 2 s 1 ) n 1 s 2 α 1 = c [(21)n ] 1, λ [(21)n 2] 1 = (s 2 s 1 ) n s 2 α 1 = c [(21)n 2] 1, λ [(21)n ] 2 = (s 2 s 1 ) n α 2 = c [(21)n ] 2, λ [(21)n 2] 2 = (s 2 s 1 ) n α 2 = c [(21)n 2] 2, λ [(21)n ] 3 = α 3 = c [(21)n ] 3, λ [(21)n 2] 3 = α 3 = c [(21)n 2] 3. Proof. As all the other cases are smlar, we wll only prove λ [(21)n 2] 1 = (s 2 s 1 ) n s 2 α 1 = c [(21)n 2] 1. When n = 0, t s checked n (4.1). By nducton, assume λ [(21)n 2] 1 = c [(21)n 2] 1, and we obtan from (4.4) λ [(21)n+1 2] 1 = (s 2 s 1 ) n+1 s 2 α 1 = s 2 s 1 λ [(21)n 2] 1 = s 2 s 1 c [(21)n 2] 1 = ( g 2n+1 + b 12 g 2n+2, b 12 g 2n+1 + (b )g 2n+2, 0 ) = (g 2n+3, g 2n+4, 0) = c [(21)n+1 2] 1. As for (C2), t suffces by (3.3) to check the denttes n the followng lemma. Lemma 4.7 (C2). For n 1, we have and b [(21)n ] 21 < 0, b [(21)n ] 23 > 0 and b [(21)n ] 31 > 0. For n 0, we have and b [(21)n 2] 12 < 0, b [(21)n 2] 32 > 0 and b [(21)n 2] 13 > 0. s [(21)n ] 2 ( λ [(21)n ] 1 ) = λ [(21)n ] 1 + b [(21)n ] 21 λ [(21)n ] 2, s [(21)n ] 2 ( λ [(21)n ] 3 ) = λ [(21)n ] 3 + b [(21)n ] 23 λ [(21)n ] 2, s [(21)n ] 1 ( λ [(21)n ] 3 ) = λ [(21)n ] 3 + b [(21)n ] 31 λ [(21)n ] 1, s [(21)n 2] 2 ( λ [(21)n 2] 1 ) = λ [(21)n 2] 1 + b [(21)n 2] 12 λ [(21)n 2] 2, s [(21)n 2] 2 ( λ [(21)n 2] 3 ) = λ [(21)n 2] 3 + b [(21)n 2] 32 λ [(21)n 2] 2, s [(21)n 2] 1 ( λ [(21)n 2] 3 ) = λ [(21)n 2] 3 + b [(21)n 2] 13 λ [(21)n 2] 1,

17 Proof. We only prove GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 17 s [(21)n 2] 2 ( λ [(21)n 2] 1 ) = λ [(21)n 2] 1 + b [(21)n 2] 12 λ [(21)n 2] 2. If n = 0, t s checked n (4.3). By Lemma 4.6 and the nducton hypothess, we have ( ) s [(21)n+1 2] 2 ( λ [(21)n+1 2] 1 ) = (s 2 s 1 s [(21)n 2] 2 s 1 s 2 ) s 2 s 1 ( λ [(21)n 2] 1 ) = (s 2 s 1 )s [(21)n 2] 2 ( λ [(21)n 2] 1 ) { } = s 2 s 1 λ [(21)n 2] 1 + b [(21)n 2] 12 λ [(21)n 2] 2 = s 2 s 1 λ [(21)n 2] 1 + b [(21)n 2] 12 s 2 s 1 λ [(21)n 2] 2 = λ [(21)n+1 2] 1 + b [(21)n+1 2] 12 λ [(21)n+1 2] 2. We also have b [(21)n+1 2] 12 = b 12 (C3) We see that (4.8) s [(21)n ] 1 s [(21)n ] 2 s [(21)n ] 3 = s 1 s 2 s 3 and s [(21)n 2] 2 s [(21)n 2] 1 s [(21)n 2] 3 = s 1 s 2 s 3. (S0) We have λ [(21)n ] 2 = s [(21)n 1 2] 1 λ [(21)n 1 2] 2, and λ [(21)n ] 2 and λ [(21)n 1 2] 1 are both postve. Smlarly, λ [(21)n 2] 1 = s [(21)n ] 2 λ [(21)n ] 1, and λ [(21)n 2] 1 and λ [(21)n ] 2 are both postve Frst mutaton at 3 after 2 and 1 only. Wrte v = [(21) n 2], n 0. Then we have b v[3] 12 = b 12 b v 13b v 23 > 0, b v[3] 13 = b v 13 < 0, b v[3] 23 = b v 23 > 0, and For v[13], we have g 2n+1 g 2n+2 b v C v[3] 13 = g 2n g 2n b v[13] 12 = b 12 + b v[1] 13 bv[1] 23 < 0, b v[13] 13 = b v[1] 13 > 0, b v[13] 23 = b v[1] 23 < 0, and g 2n+1 g 2n+2 0 C v[13] = g 2n+2 g 2n+3 b v[1] On the other hand, from the defnton (2.8), we obtan s v[3] 1 = s v 3 s v 1 s v 3 = s 3 s v 1 s 3, s v[13] 1 = s v[1] 1, s v[3] 2 = s v 2, s v[13] 2 = s v[1] 3 s v[1] 2 s v[1] 3 = s 3 s v[1] 2 s 3, s v[3] 3 = s v 3 = s 3, s v[13] 3 = s v[1] 3 = s 3.

18 18 K.-H. LEE AND K. LEE Lemma 4.9 (C1). We have λ v[3] 1 = s 3 λ v 1 = c v[3] 1, λ v[13] 1 = λ v[1] 1 = c v[13] 1, λ v[3] 2 = λ v 2 = c v[3] 2, λ v[13] 2 = s 3 λ v[1] 2 = c v[13] 2, λ v[3] 3 = λ v 3 = c v[3] 3, λ v[13] 3 = λ v[1] 3 = c v[13] 2. Proof. Snce the other denttes are clear from (4.4), we have only to check It follows from Lemma 4.5 that λ v[3] 1 = s 3 λ v 1 = c v[3] 1 and λ v[13] 2 = s 3 λ v[1] 2 = c v[13] 2. s 3 λ v 1 = (g 2n+1, g 2n+2, b 31 g 2n+1 + b 23 g 2n+2 ) = (g 2n+1, g 2n+2, b v 13) = c v[3] 1. Smlarly, by Lemma 4.5, s 3 λ v[1] 2 = (g 2n+2, g 2n+3, b 31 g 2n+2 + b 23 g 2n+3 ) = (g 2n+2, g 2n+3, b v[1] 23 ) = cv[13] 2. By (3.3), t s enough to check the denttes n the followng lemma for (C2). Lemma 4.10 (C2). We have where b v[3] 12 > 0, b v[3] 23 > 0, b v[3] 13 < 0, and where b v[13] 21 > 0, b v[13] 23 < 0 and b v[13] 13 > 0. s v[3] 2 ( λ v[3] 1 ) = λ v[3] 1 + b v[3] 12 λv[3] 2, s v[3] 2 ( λ v[3] 3 ) = λ v[3] 3 + b v[3] 23 λv[3] 2, s v[3] 1 ( λ v[3] 3 ) = λ v[3] 3 + b v[3] 13 λv[3] 1, s v[13] 2 ( λ v[13] 1 ) = λ v[13] 1 + b v[13] 21 λ v[13] 2, s v[13] 2 ( λ v[13] 3 ) = λ v[13] 3 + b v[13] 23 λ v[13] 2, s v[13] 1 ( λ v[13] 3 ) = λ v[13] 3 + b v[13] 13 λ v[13] 1, Proof. We only prove the frst and the thrd dentty for v[3]. By Lemma 4.7, the left-hand sde of the frst dentty becomes s v[3] 2 ( λ v[3] 1 ) = s v 2 s v 3 ( λ v 1 ) = s v 2 { λ v 1 + b v 13 λ v 3 } = λ v 1 + b v 12 λ v 2 + b v 13{ λ v 3 + b v 32 λ v 2 } and the rght-hand sde s = λ v 1 + b v 12 λ v 2 + b v 13 λ v 3 + b v 13b v 32 λ v 2, λ v[3] 1 + b v[3] 12 λv[3] 2 = s 3 λ v 1 + b v[3] 12 λv 2 = λ v 1 + b v 13 λ v 3 + ( b 12 b v 13b v 23) λ v 2.

19 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 19 Thus the frst dentty s verfed. Agan, by Lemma 4.7, the left-hand sde of the thrd dentty s and the rght-hand sde s s v[3] 1 ( λ v[3] 3 ) = s v 3 s v 1 s v 3 ( λ v 3 ) = s v 3 s v 1 ( λ v 3 ) = s v 3 ( λ v 3 + b v 13 λ v 1 ) = λ v 3 b v 13s v 3 λ v 1, λ v[3] 3 + b v[3] 13 λv[3] 1 = λ v 3 b v 13 s v 3 λ v 1 = λ v 3 b v 13 s v 3 λ v 1 = λ v 3 b v 13s v 3 λ v 1, snce s v 3 λv 1 > 0. Ths verfes the thrd dentty. (C3) From (4.8) we obtan (4.11) (4.12) s v[3] 2 s v[3] 3 s v[3] 1 = s v 2 s v 3 s v 3 s v 1 s v 3 = s v 2 s v 1 s v 3 = s 1 s 2 s 3, s v[13] 1 s v[13] 3 s v[13] 2 = s v[1] 1 s v[1] 3 s v[1] 3 s v[1] 2 s v[1] 3 = s v[1] 1 s v[1] 2 s v[1] 3 = s 1 s 2 s 3. (S0) For v[3], we see that λ v[3] 1 > 0 and λ v 3 > 0. For v[13], we have λv[13] 2 > 0 and λ v[1] 3 > General case: after the ntal case After the ntal case n the prevous secton, we clam that there are only four possble cases (I)-(IV) lsted below. In the next two sectons, we prove that the cases (I)-(IV) are ndeed the only possbltes. Note that (C2) and (C3) are ncluded n each of the cases (I)-(IV). (It follows from (3.3) that we need to check only three denttes for (C2).) Suppose that w s non-empty. In what follows, we read ndces modulo 3. I) The length of w s odd and w[ + 1, ] [(21) n 3] for any {1, 2, 3}. s w[+1,] λ w[+1,] λ w[+1,] ( λ w[+1,] 1 + b w[+1,], 1 λ w[+1,] 1 > λ w[+1,] 1 > 0, λ w[+1,] > 0, + b w[+1,] +1, λ w[+1,] +1 < λ w[+1,] +1 ) = λ w[+1,] + b w[+1,] 1, 1 s w[+1,] ( λ w[+1,] +1 ) = λ w[+1,] +1 + b w[+1,],+1 λ w[+1,] (b w[+1,], 1 > 0), λ w[+1,] (b w[+1,],+1 < 0), s w[+1,] 1 ( λ w[+1,] +1 ) = λ w[+1,] +1 + b w[+1,] 1,+1 λw[+1,] 1 (b w[+1,] 1,+1 > 0). s w[+1,] +1 s w[+1,] s w[+1,] 1 = s 1 s 2 s 3. II) The length of w s odd and w[ 1, ] [(21) n 2] for any {1, 2, 3}. s w[ 1,] λ w[ 1,] λ w[ 1,] ( λ w[ 1,] 1 + b w[ 1,], 1 λ w[ 1,] 1 > λ w[ 1,] 1 > 0, λ w[ 1,] < 0, + b w[ 1,] +1, λ w[ 1,] +1 < λ w[ 1,] +1 ) = λ w[ 1,] + b w[ 1,] 1 1, λ w[ 1,] (b w[ 1,] 1, < 0),

20 20 K.-H. LEE AND K. LEE s w[ 1,] ( λ w[ 1,] +1 ) = λ w[ 1,] +1 + b w[ 1,] +1, λ w[ 1,] (b w[ 1,] +1, > 0), s w[ 1,] 1 ( λ w[ 1,] +1 ) = λ w[ 1,] +1 + b w[ 1,] 1,+1 λw[ 1,] 1 (b w[ 1,] 1,+1 > 0). s w[ 1,] +1 s w[ 1,] s w[ 1,] 1 = s 1 s 2 s 3. III) The length of w s even. Assume w[ + 1, ] [(21) n ] for any {1, 2, 3}. s w[+1,] λ w[+1,] λ w[+1,] ( λ w[+1,] 1 + b w[+1,] 1, λ w[+1,] 1 < λ w[+1,] 1 < 0, λ w[+1,] < 0, + b w[+1,],+1 λ w[+1,] +1 > λ w[+1,] +1 > 0. ) = λ w[+1,] + b w[+1,] 1 1, s w[+1,] ( λ w[+1,] +1 ) = λ w[+1,] +1 + b w[+1,] +1, λ w[+1,] (b w[+1,] 1, > 0), λ w[+1,] (b w[+1,] +1, < 0), s w[+1,] 1 ( λ w[+1,] +1 ) = λ w[+1,] +1 + b w[+1,] +1, 1 λw[+1,] 1 (b w[+1,] +1, 1 > 0). s w[+1,] 1 s w[+1,] s w[+1,] +1 = s 1 s 2 s 3. IV) The length of w s even and w[ 1, ] [(21) n 23] for any {1, 2, 3}. s w[ 1,] λ w[ 1,] λ w[ 1,] ( λ w[ 1,] 1 + b w[ 1,] 1, λ w[ 1,] 1 < λ w[ 1,] 1 < 0, λ w[ 1,] > 0, + b w[ 1,],+1 λ w[ 1,] +1 > λ w[ 1,] +1 > 0. ) = λ w[ 1,] + b w[ 1,] 1, 1 s w[ 1,] ( λ w[ 1,] +1 ) = λ w[ 1,] +1 + b w[ 1,],+1 λ w[ 1,] (b w[ 1,], 1 < 0), λ w[ 1,] (b w[ 1,],+1 > 0), s w[ 1,] 1 ( λ w[ 1,] +1 ) = λ w[ 1,] +1 + b w[ 1,] +1, 1 λw[ 1,] 1 (b w[ 1,] +1, 1 > 0). s w[ 1,] 1 s w[ 1,] s w[ 1,] +1 = s 1 s 2 s 3. Proposton 5.1. Let Q be a rank 3 quver whch s not mutaton equvalent to any acyclc quver. Then (I)-(IV) are the only cases for mutaton sequences not equal to any of [(21) n 2], [(21) n 3], [(21) n ] and [(21) n 23]. The proof of ths proposton wll be acheved n the next two sectons, and wll use nducton on the length of a mutaton sequence.

21 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 21 Wrte v = [(21) n 2], n 0. We consder 6. Proof of Proposton 5.1: Base step ) v[32], ) v[31], ) v[132], v) v[131]. We wll show that ) satsfes the condtons of I) and ) those of II), ) those of III) and v) those of IV). At the end of ths secton, (S0) wll be checked for all the cases ()-(v). and ) v[32] We have b v[32] 12 = b v[3] 12 < 0, b v[32] 13 = b v[3] 13 + b v[3] 12 bv[3] 23 > 0, b v[32] 23 = b v[3] 23 < 0, From the defntons, we obtan C v[32] = g 2n+1 g 2n+2 b v 13 g 2n g 2n+1 0. b v[3] 23 g 2n b v[3] 23 g 2n+1 1 s v[32] 3 = s v[3] 2 s v[3] 3 s v[3] 2 = s v 2 s 3 s v 2, λ v[32] 3 = s v[3] 2 (λ v[3] 3 ) = s v 2 (α 3 ) = (s 2 s 1 ) n s 2 (s 1 s 2 ) n (α 3 ). Lemma 6.1. We have λ v[32] 3 = λ v[3] 3 + b v[3] 23 λv[3] 2. Proof. By Lemma 4.10, we get λ v[32] 3 = s v[3] 2 (λ v[3] 3 ) = s v[3] 2 ( λ v[3] 3 ) = λ v[3] 3 b v[3] 23 λv[3] 2 = λ v[3] 3 + b v[3] 23 λv[3] 2. We check the condtons of (I). Clearly, we see that We need to check λ v[32] 1 > 0, λ v[32] 2 > 0, λ v[32] 3 λ v[32] 2 + (b v[32] 21 1)λ v[32] 1 > 0, λ v[32] 2 + (b v[32] 32 1)λ v[32] 3 The former s mmedate snce b v[32] For the latter, we have b v[32] 32 = b v[3] λ v[32] 2 + (b v[32] 32 1)λ v[32] 3 = (g 2n, g 2n+1, 0) + (b v[32] 32 1) 23 2 and get ) < 0, ( b v[3] 23 g 2n, b v[3] 23 g 2n+1, 1 snce 1 + (b v[32] 32 1)( b v[3] 23 ) = 1 + (1 bv[3] 23 )bv[3] 23 < 0 and (b v[32] 32 1)

22 22 K.-H. LEE AND K. LEE Lemma 6.2. and s v[32] 2 ( λ v[32] 1 ) = λ v[32] 1 + b v[32] 21 λ v[32] 2, s v[32] 2 ( λ v[32] 3 ) = λ v[32] 3 + b v[32] 23 λ v[32] 2, s v[32] 1 ( λ v[32] 3 ) = λ v[32] 3 + b v[32] 13 λ v[32] 1, s v[32] 3 s v[32] 2 s v[32] 1 = s 1 s 2 s 3. Proof. We only prove the second dentty among the frst three snce the others are smlar. From Lemmas 6.1 and 4.10, we obtan and LHS = s v[32] 2 ( λ v[32] 3 ) = s v[3] 2 ( λ v[3] 3 ) + b v[3] 23 sv[3] 2 ( λ v[3] 2 ) = λ v[3] 3 + b v[3] 23 λv[3] 2 b v[3] 23 λv[3] 2 = λ v[3] 3, RHS = λ v[32] 3 + b v[32] 23 λ v[32] 2 = λ v[3] 3 b v[3] 23 λv[3] 2 + b v[3] 23 λv[3] 2 = λ v[3] 3 = λ v[3] 3. For the last dentty, t follows from (4.11) that ) v[31] We have s v[32] 3 s v[32] 2 s v[32] 1 = s v[3] 2 s v[3] 3 s v[3] 2 s v[3] 2 s v[3] 1 = s v[3] 2 s v[3] 3 s v[3] 1 = s 1 s 2 s 3. b v[31] 12 = b v[3] 12 < 0, b v[31] 13 = b v[3] 31 > 0, b v[31] 23 = b v[3] 23 + b v[3] 12 bv[3] 13 < 0, C v[31] = From the defntons, we obtan g 2n+1 g 2n+2 b v 13 g 2n g 2n+1 0 b v[3] 31 g 2n+1 b v[3] 31 g 2n+2 b v[3] 31 bv 13 1 s v[31] 3 = s v[3] 1 s v[3] 3 s v[3] 1 = s 3 s v 1 s 3 s v 1 s 3,. λ v[31] 3 = s 3 s v 1 (α 3 ) = s 3 (s 2 s 1 ) n s 2 s 1 s 2 (s 1 s 2 ) n (α 3 ). Lemma 6.3. λ v[31] 3 = λ v[3] 3 + b v[3] 31 λv[3] 1. Proof. It follows from Lemma 4.10 that λ v[31] 3 = s v[3] 1 λ v[3] 3 = s v[3] 1 λ v[3] 3 = λ v[3] 3 b v[3] 13 λv[3] 1 = λ v[3] 3 + b v[3] 31 λv[3] 1.

23 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 23 We check the condtons of (II). Immedately, we see that We need to check λ v[31] 1 < 0, λ v[31] 2 < 0, λ v[31] 3 > 0. λ v[31] 1 + (b v[31] 13 1)λ v[31] 3 > 0, λ v[31] 1 + (b v[31] 21 1)λ v[31] 2 The latter s mmedate snce b v[31] For the former, we have b v 13 = bv[3] 31 = b v[31] 13 2 and get λ v[31] 1 + (b v[31] 13 1)λ v[31] 3 = ( g 2n+1, g 2n+2, b v 13) + (b v[31] 13 1) snce 1 + (b v[31] 13 1)b v[3] 31 = b v 13 (bv 13 1) 1 > 0 and ( ) b v[3] 31 g 2n+1, b v[3] 31 g 2n+2, b v[3] 31 bv 13 1 > 0, b v 13 + (b v[31] 13 1)(b v[3] 31 bv 13 1) = b v 13(b v )(b v 13 2) + 1 > 0. Lemma 6.4. and s v[31] 2 ( λ v[31] 1 ) = λ v[31] 1 + b v[31] 21 λ v[31] 2, s v[31] 2 ( λ v[31] 3 ) = λ v[31] 3 + b v[31] 32 λ v[31] 2, s v[31] 1 ( λ v[31] 3 ) = λ v[31] 3 + b v[31] 31 λ v[31] 1, s v[31] 2 s v[31] 1 s v[31] 3 = s 1 s 2 s 3. Proof. We only prove the thrd dentty among the frst three snce the others are smlar. By Lemma 6.3, we have LHS = s v[31] 1 ( λ v[31] 3 ) = s v[3] 1 s v[3] 1 λ v[3] 3 = λ v[3] 3, RHS = λ v[31] 3 + b v[31] 31 λ v[31] 1 = λ v[3] 3 + b v[3] 31 λv[3] 1 b v[3] 31 λv[3] 1 = λ v[3] 3. As for the last dentty, t follows from (4.11) that ) v[132] We have s v[31] 2 s v[31] 1 s v[31] 3 = s v[3] 2 s v[3] 1 s v[3] 1 s v[3] 3 s v[3] 1 = s v[3] 2 s v[3] 3 s v[3] 1 = s 1 s 2 s 3. C v[132] = g 2n+1 g 2n+2 0 g 2n+2 g 2n+3 b v[1] 23 b v[13] 32 g 2n+2 b v[13] 32 g 2n+3 b v[13] 32 b v[1] Let us check the condtons of (III). It s obvous that 23 1 λ v[132] 1 < 0, λ v[132] 2 < 0, λ v[132] 3 > 0..

24 24 K.-H. LEE AND K. LEE We need to check λ v[132] 2 + (b v[132] 12 1)λ v[132] 1 < 0, λ v[132] 2 + (b v[132] 23 1)λ v[132] 3 > 0. The former s mmedate snce b v[132] For the latter, we have b v[1] 23 = b v[13] 32 = b v[132] 23 2 and λ v[132] 2 + (b v[132] 23 1)λ v[132] 3 = ( g 2n+2, g 2n+3, b v[1] 23 ) + (bv[132] 23 1) snce 1 + (b v[132] 23 1)b v[13] 32 = b v[13] 32 (b v[13] 32 1) 1 > 0 and ( ) b v[13] 32 g 2n+2, b v[13] 32 g 2n+3, b v[13] 32 b v[1] 23 1 > 0, b v[1] 23 + (b v[132] 23 1)(b v[13] 32 b v[1] 23 1) = b v[1] 23 (bv[1] )(b v[1] 23 2) + 1 > 0. Lemma 6.5. The followng denttes hold: and s v[132] 2 ( λ v[132] 1 ) = λ v[132] 1 + b v[132] 12 λ v[132] 2, s v[132] 2 ( λ v[132] 3 ) = λ v[132] 3 + b v[132] 32 λ v[132] 2, s v[132] 1 ( λ v[132] 3 ) = λ v[132] 3 + b v[132] 31 λ v[132] 1, s v[132] 1 s v[132] 2 s v[132] 3 = s 1 s 2 s 3. Proof. The frst three denttes can be checked smlarly as n the prevous cases. For the last dentty, t follows from (4.12) that s v[132] 1 s v[132] 2 s v[132] 3 = s v[13] 1 s v[13] 2 s v[13] 2 s v[13] 3 s v[13] 2 = s v[13] 1 s v[13] 3 s v[13] 2 = s 1 s 2 s 3. v) v[131] We have g 2n+1 g 2n+2 0 C v[131] = g 2n+2 g 2n+3 b v[1] 23. b v[13] 13 g 2n+1 b v[13] 13 g 2n+2 1 We check the condtons of (IV). Immedately, we see that We need to check λ v[131] 1 > 0, λ v[131] 2 > 0, λ v[131] 3 λ v[131] 1 + (b v[131] 31 1)λ v[131] 3 < 0, λ v[131] 1 + (b v[131] 12 1)λ v[131] 2 > 0. The latter s mmedate snce b v[131] For the former, we have b v[131] 31 = b v[13] 13 2 and get λ v[131] 1 + (b v[131] 31 1)λ v[131] 3 = (g 2n+1, g 2n+2, 0) + (b v[131] 31 1) ( ) b v[13] 13 g 2n+1, b v[13] 13 g 2n+2, 1 < 0,

25 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 25 snce 1 + (b v[131] 31 1)( b v[13] 13 ) = 1 b v[13] 13 (b v[13] 13 1) < 0 and (b v[131] 31 1) Lemma 6.6. We have the followng denttes: s v[131] 2 ( λ v[131] 1 ) = λ v[131] 1 + b v[131] 12 λ v[131] 2, s v[131] 2 ( λ v[131] 3 ) = λ v[131] 3 + b v[131] 23 λ v[131] 2, s v[131] 1 ( λ v[131] 3 ) = λ v[131] 3 + b v[131] 13 λ v[131] 1, and s v[131] 3 s v[131] 1 s v[131] 2 = s 1 s 2 s 3. Proof. The frst three denttes can be checked smlarly as n the prevous cases. For the last dentty, t follows from (4.12) that s v[131] 3 s v[131] 1 s v[131] 2 = s v[13] 1 s v[13] 3 s v[13] 1 s v[13] 1 s v[13] 2 = s v[13] 1 s v[13] 3 s v[13] 2 = s 1 s 2 s 3. Fnally, we check (S0) for all the cases ()-(v). (S0) For () v[32], we have λ v[32] 3 < 0 and λ v[3] 2 < 0; for () v[31], we see λ v[31] 3 > 0 and λ v[3] 1 > 0; for () v[132], observe λ v[132] 3 > 0 and λ v[13] 2 > 0; for (v) v[131], we get λ v[131] 3 < 0 and λ v[13] 1 7. Proof of Proposton 5.1: Man Inducton Step In order for our nducton to be completed, there are 8 cases to be consdered. We present all the statements n each case. However, snce there are symmetres comng from permutatons of ndces and the computatons are all smlar, we prove the frst two cases and omt detals for the other cases. We also check that (S0) s true for each nducton step (I) (III). We obtan from the defnton and nducton We need to show λ w[+1,, 1] 1 λ w[+1,, 1] = λ w[+1,] 1 < 0, λ w[+1,, 1] +1 = λ w[+1,] +1 < 0, = s w[+1,] 1 λ w[+1,] = λ w[+1,] λ w[+1,, 1] 1 λ w[+1,, 1] 1 + (b w[+1,, 1] 1, + (b w[+1,, 1] +1, 1 + b w[+1,], 1 λ w[+1,] 1 > 0. 1)λ w[+1,, 1] > 0 1)λ w[+1,, 1] +1

26 26 K.-H. LEE AND K. LEE Proof. Frst, snce b w[+1,], 1 λ w[+1,, 1] 1 = λ w[+1,] 1 = (b w[+1,], 1 Second, we get 2, we have + (b w[+1,, 1] 1, + (b w[+1,], 1 1)(λ w[+1,] λ w[+1,, 1] 1 1)λ w[+1,, 1] 1)(λ w[+1,] + (b w[+1,], 1 = λ w[+1,] 1 + b w[+1,], 1 λ w[+1,] 1 ) 1)λ w[+1,] 1 ) + (b w[+1,], 1 2)λ w[+1,] 1 > 0. + (b w[+1,, 1] +1, 1 1)λ w[+1,, 1] +1 + (b w[+1,] 1,+1 1)λw[+1,] +1 Now we show s w[+1,, 1] ( λ w[+1,, 1] 1 ) = λ w[+1,, 1] 1 s w[+1,, 1] ( λ w[+1,, 1] +1 ) = λ w[+1,, 1] s w[+1,, 1] 1 ( λ w[+1,, 1] +1 ) = λ w[+1,, 1] + b w[+1,, 1], 1 λ w[+1,, 1], +1 + b w[+1,, 1], b w[+1,, 1] +1, 1 Proof. To ease the notaton, n ths proof, we wrte v = w[ + 1, ]. For the frst dentty, we have LHS = s w[+1,, 1] λ w[+1,, 1], λ w[+1,, 1] 1. ( λ w[+1,, 1] 1 ) = s v 1s v s v 1( λ v 1 ) = s v 1s v ( λ v 1 ) = s v 1( λ v 1 + b v, 1 λ v ) = λ v 1 b v, 1s v 1( λ v ) = λ v 1 + b v 1,(λ v + b v, 1λ v 1) = λ w[+1,, 1] 1 For the second, we have + b w[+1,, 1], 1 λ w[+1,, 1] = RHS. LHS = s w[+1,, 1] ( λ w[+1,, 1] +1 ) = s v 1s v s v 1( λ v +1 ) = s v 1s v ( λ v +1 + b v 1,+1 λ v 1 ) = s v ( 1 λ v +1 + b v,+1 λ v + b v 1,+1( λ v 1 + b v, 1 λ v ) ) = λ v +1 + b v 1,+1 λ v 1 + b v,+1( λ v + b v, 1 λ v 1 ) + b v ( 1,+1 λ v 1 + b v, 1( λ v + b v, 1 λ v 1 ) ) = λ v +1 + (b v,+1 + b v 1,+1b v, 1) λ v + b v, 1(b v,+1 + b v, 1b v 1,+1) λ v 1 = λ v +1 + (b v,+1 + b v, 1b v 1,+1)( λ v + b v, 1 λ v 1 ) = λ w[+1,, 1] +1 + b w[+1,, 1],+1 λ w[+1,, 1] = RHS.

27 For the thrd, we get GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 27 LHS = s w[+1,, 1] 1 ( λ w[+1,, 1] +1 ) = s v 1( λ v +1 ) = λ v +1 + b v 1,+1 λ v 1 = λ w[+1,, 1] +1 + b w[+1,, 1] +1, 1 λ w[+1,, 1] 1 = RHS. We have, by nducton, s w[+1,, 1] +1 s w[+1,, 1] 1 s w[+1,, 1] (S0) We see that λ w[+1,, 1] = s w[+1,] +1 s w[+1,] 1 s w[+1,] 1 s w[+1,] s w[+1,] 1 = s w[+1,] > 0 and λ w[+1,] 1 > s w[+1,] s w[+1,] 7.2. (I) (IV). We obtan from the defnton and nducton We need to show λ w[+1,,+1] 1 λ w[+1,,+1] Proof. Frst, snce b w[+1,] +1, 1 = s 1 s 2 s 3. = λ w[+1,] 1 > 0, λ w[+1,,+1] +1 = λ w[+1,] +1 > 0, = s w[+1,] +1 λ w[+1,] = λ w[+1,] + b w[+1,] +1, λ w[+1,] +1 λ w[+1,,+1] +1 + (b w[+1,,+1],+1 1)λ w[+1,,+1] < 0 λ w[+1,,+1] +1 + (b w[+1,,+1] +1, 1 1)λ w[+1,,+1] 1 > 0. 2, we have λ w[+1,, 1] +1 + (b w[+1,, 1],+1 1)λ w[+1,, 1] = λ w[+1,] +1 + (b w[+1,] +1, 1)(λ w[+1,] = (b w[+1,] +1, Second, we get 1)(λ w[+1,] + (b w[+1,] +1, + b w[+1,] +1, λ w[+1,] +1 ) 1)λ w[+1,] +1 ) + (b w[+1,] +1, 2)λ w[+1,] +1 λ w[+1,, 1] +1 + (b w[+1,, 1] +1, 1 1)λ w[+1,, 1] 1 = λ w[+1,] +1 + (b w[+1,] 1,+1 1)λw[+1,] 1 > 0. We need to prove s w[+1,,+1] ( λ w[+1,,+1] 1 ) = λ w[+1,,+1] 1 s w[+1,,+1] ( λ w[+1,,+1] +1 ) = λ w[+1,,+1] s w[+1,,+1] 1 ( λ w[+1,,+1] +1 ) = λ w[+1,,+1] + b w[+1,,+1] 1, λ w[+1,,+1], +1 + b w[+1,,+1] +1, +1 + b w[+1,,+1] +1, 1 λ w[+1,,+1], λ w[+1,,+1] 1.

28 28 K.-H. LEE AND K. LEE Proof. To ease the notaton, n ths proof, we wrte v = w[ + 1, ]. For the frst dentty, we have LHS = s w[+1,,+1] ( λ w[+1,,+1] 1 ) = s v +1s v s v +1( λ v 1 ) = s v +1s v ( λ v 1 + b v 1,+1 λ v +1 ) = s v ( +1 λ v 1 + b v, 1 λ v + b v 1,+1( λ v +1 + b v,+1 λ v ) ) = λ v 1 + b v 1,+1 λ v +1 + b v, 1( λ v + b v,+1 λ v +1 ) + b v ( 1,+1 λ v +1 + b v,+1( λ v + b v,+1 λ v +1 ) ) = λ v 1 + (b v, 1 + b v 1,+1b v,+1) λ v + b v,+1(b v, 1 + b v,+1b v 1,+1) λ v +1 = λ v 1 (b v 1, + b v 1,+1b v +1,) λ v b v,+1(b v 1, + b v +1,b v 1,+1) λ v +1 = λ v 1 + (b v 1, + b v 1,+1b v +1,)( λ v b v,+1 λ v +1 ) = λ w[+1,,+1] +1 + b w[+1,,+1] +1, λ w[+1,,+1] = RHS. For the second, we have LHS = s w[+1,,+1] ( λ w[+1,,+1] +1 ) = s v +1s v s v +1( λ v +1 ) = s v +1s v ( λ v +1 ) For the thrd, we get = s v +1( λ v +1 + b v,+1 λ v ) = λ v +1 b v,+1( λ v + b v,+1 λ v +1 ) = λ v +1 + b v,+1( λ v b v +1,λ v +1) = λ w[+1,,+1] +1 + b w[+1,,+1] +1, λ w[+1,,+1] = RHS. LHS = s w[+1,,+1] 1 ( λ w[+1,,+1] +1 ) = s v 1( λ v +1 ) = λ v +1 + b v 1,+1 λ v 1 = λ w[+1,,+1] +1 + b w[+1,,+1] +1, 1 λ w[+1,,+1] 1 = RHS. By nducton, s w[+1,,+1] s w[+1,,+1] +1 s w[+1,,+1] 1 = s w[+1,] +1 s w[+1,] s w[+1,] +1 s w[+1,] +1 s w[+1,] 1 = s w[+1,] (S0) It s straghtforward to see that λ w[+1,,+1] +1 s w[+1,] s w[+1,] 1 = s 1 s 2 s 3. < 0 and λ w[+1,] (II) (III). We obtan from the defnton and nducton λ w[ 1,, 1] 1 λ w[ 1,, 1] = λ w[ 1,] 1 < 0, λ w[ 1,, 1] +1 = λ w[ 1,] +1 < 0, = s w[ 1,] 1 λ w[ 1,] = λ w[ 1,] + b w[ 1,], 1 λ w[ 1,] 1 > 0.

29 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN 29 We see that λ w[ 1,, 1] 1 λ w[ 1,, 1] 1 + (b w[ 1,, 1] 1, + (b w[ 1,, 1] 1, 1 1)λ w[ 1,, 1] > 0 1)λ w[ 1,, 1] +1 One can show s w[ 1,, 1] ( λ w[ 1,, 1] 1 ) = λ w[ 1,, 1] 1 s w[ 1,, 1] ( λ w[ 1,, 1] +1 ) = λ w[ 1,, 1] s w[ 1,, 1] 1 ( λ w[ 1,, 1] +1 ) = λ w[ 1,, 1] Usng nducton, one can check (S0) We see that λ w[ 1,, 1] s w[ 1,, 1] +1 s w[ 1,, 1] 1 > 0 and λ w[ 1,] 1 > (II) (IV). We obtan from the defnton We see that We obtan λ w[ 1,,+1] 1 λ w[ 1,,+1] s w[ 1,,+1] + b w[ 1,, 1], 1 λ w[ 1,, 1], +1 + b w[ 1,, 1], b w[ 1,, 1] +1, 1 s w[ 1,, 1] = s 1 s 2 s 3. = λ w[ 1,] 1 > 0, λ w[ 1,,+1] +1 = λ w[ 1,] +1 > 0, = s w[ 1,] +1 λ w[ 1,] = λ w[ 1,] λ w[ 1,, 1], λ w[ 1,, 1] 1. + b w[ 1,] +1, λ w[ 1,] +1 λ w[ 1,,+1] +1 + (b w[ 1,,+1],+1 1)λ w[ 1,,+1] < 0 λ w[ 1,,+1] +1 + (b w[ 1,,+1] +1, 1 1)λ w[ 1,,+1] 1 > 0. ( λ w[ 1,,+1] 1 ) = λ w[ 1,,+1] 1 s w[ 1,,+1] ( λ w[ 1,,+1] +1 ) = λ w[ 1,,+1] s w[ 1,,+1] 1 ( λ w[ 1,,+1] +1 ) = λ w[ 1,,+1] Usng nducton, one sees that (S0) We check λ w[ 1,,+1] + b w[ 1,,+1] 1, λ w[ 1,,+1], +1 + b w[ 1,,+1] +1, +1 + b w[ 1,,+1] +1, 1 s w[ 1,,+1] s w[ 1,,+1] +1 s w[ 1,,+1] 1 = s 1 s 2 s 3. < 0 and λ w[ 1,] +1 λ w[ 1,,+1], λ w[ 1,,+1] 1.

30 30 K.-H. LEE AND K. LEE 7.5. (III) (I). We obtan from the defnton One can check One can show By nducton, λ w[+1,, 1] 1 λ w[+1,, 1] s w[+1,, 1] = λ w[+1,] 1 > 0, λ w[+1,, 1] +1 = λ w[+1,] +1 > 0, = s w[+1,] 1 λ w[+1,] = λ w[+1,] λ w[+1,, 1] 1 λ w[+1,, 1] 1 ( λ w[+1,, 1] 1 + (b w[+1,, 1] 1,+1 + (b w[+1,, 1], 1 ) = λ w[+1,, 1] 1 s w[+1,, 1] ( λ w[+1,, 1] +1 ) = λ w[+1,, 1] s w[+1,, 1] 1 ( λ w[+1,, 1] +1 ) = λ w[+1,, 1] (S0) We obtan λ w[+1,, 1] s w[+1,, 1] + b w[+1,] 1, λ w[+1,] 1 1)λ w[+1,, 1] +1 > 0 1)λ w[+1,, 1] + b w[+1,, 1] 1, λ w[+1,, 1], +1 + b w[+1,, 1] +1, +1 + b w[+1,, 1] 1,+1 s w[+1,, 1] 1 s w[+1,, 1] +1 = s 1 s 2 s 3. < 0 and λ w[+1,] (III) (II). We obtan from the defnton and nducton One can check One can show λ w[+1,,+1] 1 λ w[+1,,+1] s w[+1,,+1] = λ w[+1,] 1 < 0, λ w[+1,,+1] +1 = λ w[+1,] +1 < 0, = s w[+1,] +1 λ w[+1,] = λ w[+1,] λ w[+1,, 1], λ w[+1,, 1] 1. + b w[+1,],+1 λ w[+1,] +1 > 0. λ w[+1,,+1] +1 + (b w[+1,,+1] +1, 1)λ w[+1,,+1] > 0 λ w[+1,,+1] +1 + (b w[+1,,+1] 1,+1 1)λ w[+1,,+1] 1 ( λ w[+1,,+1] 1 ) = λ w[+1,,+1] 1 s w[+1,,+1] ( λ w[+1,,+1] +1 ) = λ w[+1,,+1] s w[+1,,+1] 1 ( λ w[+1,,+1] +1 ) = λ w[+1,,+1] We obtan by nducton (S0) We see that λ w[+1,,+1] + b w[+1,,+1], 1 λ w[+1,,+1], +1 + b w[+1,,+1], b w[+1,,+1] 1,+1 s w[+1,,+1] 1 s w[+1,,+1] +1 s w[+1,,+1] = s 1 s 2 s 3. > 0 and λ w[+1,] +1 > 0. λ w[+1,,+1], λ w[+1,,+1] 1.

31 GEOMETRIC DESCRIPTION OF C-VECTORS AND REAL LÖSUNGEN (IV) (I). We obtan from the defnton and nducton λ w[ 1,, 1] 1 λ w[ 1,, 1] It can be shown that It can be checked that By nducton, s w[ 1,, 1] = λ w[ 1,] 1 > 0, λ w[ 1,, 1] +1 = λ w[ 1,] +1 > 0, = s w[ 1,] 1 λ w[ 1,] = λ w[ 1,] λ w[ 1,, 1] 1 λ w[ 1,, 1] 1 ( λ w[ 1,, 1] 1 + (b w[ 1,, 1] 1,+1 + (b w[ 1,, 1], 1 ) = λ w[ 1,, 1] 1 s w[ 1,, 1] ( λ w[ 1,, 1] +1 ) = λ w[ 1,, 1] s w[ 1,, 1] 1 ( λ w[ 1,, 1] +1 ) = λ w[ 1,, 1] (S0) We check λ w[ 1,, 1] s w[ 1,, 1] + b w[ 1,] 1, λ w[ 1,] 1 1)λ w[ 1,, 1] +1 > 0 1)λ w[ 1,, 1] + b w[ 1,, 1] 1, λ w[ 1,, 1], +1 + b w[ 1,, 1] +1, +1 + b w[ 1,, 1] 1,+1 s w[ 1,, 1] 1 s w[ 1,, 1] +1 = s 1 s 2 s 3. < 0 and λ w[ 1,] (IV) (II). We obtan from the defnton and nducton λ w[ 1,,+1] 1 λ w[ 1,,+1] It can be shown that It can be shown that s w[ 1,,+1] = λ w[ 1,] 1 < 0, λ w[ 1,,+1] +1 = λ w[ 1,] +1 < 0, = s w[ 1,] +1 λ w[ 1,] = λ w[ 1,] λ w[ 1,, 1], λ w[ 1,, 1] 1. + b w[ 1,],+1 λ w[ 1,] +1 > 0. λ w[ 1,,+1] +1 + (b w[ 1,,+1] +1, 1)λ w[ 1,,+1] > 0 λ w[ 1,,+1] +1 + (b w[ 1,,+1] 1,+1 1)λ w[ 1,,+1] 1 ( λ w[ 1,,+1] 1 ) = λ w[ 1,,+1] 1 s w[ 1,,+1] ( λ w[ 1,,+1] +1 ) = λ w[ 1,,+1] s w[ 1,,+1] 1 ( λ w[ 1,,+1] +1 ) = λ w[ 1,,+1] We obtan, by nducton, (S0) We see that λ w[ 1,,+1] + b w[ 1,,+1], 1 λ w[ 1,,+1] b w[ 1,,+1], b w[ 1,,+1] 1,+1 s w[ 1,,+1] 1 s w[ 1,,+1] +1 s w[ 1,,+1] = s 1 s 2 s 3. > 0 and λ w[ 1,] +1 > 0. λ w[ 1,,+1]. λ w[ 1,,+1] 1. Ths completes a proof of Proposton 5.1.