Symmetry, Integrability and Geometry: Methods and Applications

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1 Symmetry, Integrablty and Geometry: Methods and Applcatons Zero Acton on Perfect Crystals for U q G 1) 2 Kalash C. MISRA, Mahathr MOHAMAD and Masato OKADO Department of Mathematcs, North Carolna State Unversty, Ralegh, North Carolna , USA E-mal: msra@unty.ncsu.edu Department of Mathematcal Scence, Graduate School of Engneerng Scence, Osaka Unversty, Toyonaka, Osaka , Japan E-mal: mahathr75@yahoo.com, okado@sgmath.es.osaka-u.ac.jp SIGMA ), 022, 12 pages Receved November 13, 2009, n fnal form March 03, 2010; Publshed onlne March 09, 2010 do: /sigma ) Abstract. The actons of 0-Kashwara operators on the U q 1)) G 2 -crystal Bl n [Yamane S., J. Algebra ), ] are made explct by usng a smlarty technque from that of a U q 3)) D 4 -crystal. It s shown that {Bl } l 1 forms a coherent famly of perfect crystals. Key words: combnatoral representaton theory; quantum affne algebra; crystal bases 2010 Mathematcs Subject Classfcaton: 05E99; 17B37; 17B67; 81R10; 81R50 1 Introducton Let g be a symmetrzable Kac Moody algebra. Let I be ts ndex set for smple roots, P the weght lattce, α P a smple root I), and h P = HomP, Z)) a smple coroot I). To each I we assocate a postve nteger m and set α = m α, h = h /m. Suppose h, α j ),j I s a generalzed Cartan matrx for another symmetrzable Kac Moody algebra g. Then the subset P of P consstng of λ P such that h, λ s an nteger for any I can be consdered as the weght lattce of g. For a domnant ntegral weght λ let B g λ) be the hghest weght crystal wth hghest weght λ over U q g). Then, n [5] Kashwara showed the followng. The theorem n [5] s more general.) Theorem 1. Let λ be a domnant ntegral weght n P. Then, there exsts a unque njectve map S : B g λ) B g λ) such that wt Sb) = wt b, Se b) = e m Sb), Sf b) = f m Sb). In ths paper, we use ths theorem to examne the so-called Krllov Reshetkhn crystal. Let g be the affne algebra of type D 3) 4. The generalzed Cartan matrx h, α j ),j I I = {0, 1, 2}) s gven by Set m 0, m 1, m 2 ) = 3, 3, 1). Then, g defned above turns out to be the affne algebra of type G 1) 2. Ther Dynkn dagrams are depcted as follows D 3) 4 : G 1) 2 : Ths paper s a contrbuton to the Proceedngs of the Workshop Geometrc Aspects of Dscrete and Ultra- Dscrete Integrable Systems March 30 Aprl 3, 2009, Unversty of Glasgow, UK). The full collecton s avalable at

2 2 K.C. Msra, M. Mohamad and M. Okado For G 1) 2 a famly of perfect crystals {B l } l 1 was constructed n [7]. However, the crystal elements there were realzed n terms of tableaux gven n [2], and t was not easy to calculate the acton of 0-Kashwara operators on these tableaux. On the other hand, an explct acton of these operators was gven on perfect crystals { ˆB l } l 1 over U q 3)) D 4 n [6]. Hence, t s a natural dea to use Theorem 1 to obtan the explct acton of e 0, f 0 on B l from that on ˆB l wth sutable l. We remark that Krllov Reshetkhn crystals are parametrzed by a node of the Dynkn dagram except 0 and a postve nteger. Both B l and ˆB l correspond to the par 1, l). Our strategy to do ths s as follows. We defne V l as an approprate subset of ˆB3l that s closed under the acton of ê m, ˆf m where ê, ˆf stand for the Kashwara operators on ˆB 3l. Hence, we can regard V l as a U q 1)) 1)) G 2 -crystal. We next show that as a Uq G 2 {0,1} = U qa 2 ))-crystal 1)) and as a U q G 2 {1,2} = U qg 2 ))-crystal, V l has the same decomposton as B l. Then, we can conclude from Theorem 6.1 of [6] that V l s somorphc to the U q 1)) G 2 -crystal Bl constructed n [7] Theorem 2). The paper s organzed as follows. In Secton 2 we revew the U q 3)) D 4 -crystal ˆBl. We then construct a U q 1)) G 2 -crystal Vl n ˆB 3l wth the ad of Theorem 1 and see t concdes wth B l gven n [7] n Secton 3. Mnmal elements of B l are found and {B l } l 1 s shown to form a coherent famly of perfect crystals n Secton 4. The crystal graphs of B 1 and B 2 are ncluded n Secton 5. 2 Revew on U q 3)) D 4 -crystal ˆB l In ths secton we recall the perfect crystal for U q 3)) D 4 constructed n [6]. Snce we also consder U q 1)) G 2 -crystals later, we denote t by ˆBl. Kashwara operators e, f and ε, ϕ on ˆB l are denoted by ê, ˆf and ˆε, ˆϕ. Readers are warned that the coordnates x, x and steps by Kashwara operators n [6] are dvded by 3 here, snce t s more convenent for our purpose. As a set 3x ˆB l = b = x 1, x 2, x 3, x 3, x 2, x 1 ) Z 0 /3) x 3 mod 2), x + x ) + x 3 + x 3 )/2 l/3 In order to defne the actons of Kashwara operators ê and ˆf for = 0, 1, 2, we ntroduce some notatons and condtons. Set x) + = maxx, 0). For b = x 1, x 2, x 3, x 3, x 2, x 1 ) ˆB l we set =1,2. and sb) = x 1 + x 2 + x 3 + x x 2 + x 1, 2.1) z 1 = x 1 x 1, z 2 = x 2 x 3, z 3 = x 3 x 2, z 4 = x 3 x 3 )/2. 2.2) Now we defne condtons E 1 ) E 6 ) and F 1 ) F 6 ) as follows F 1 ) z 1 + z 2 + z 3 + 3z 4 0, z 1 + z 2 + 3z 4 0, z 1 + z 2 0, z 1 0, F 2 ) z 1 + z 2 + z 3 + 3z 4 0, z 2 + 3z 4 0, z 2 0, z 1 > 0, F 3 ) z 1 + z 3 + 3z 4 0, z 3 + 3z 4 0, z 4 0, z 2 > 0, z 1 + z 2 > 0, F 4 ) z 1 + z 2 + 3z 4 > 0, z 2 + 3z 4 > 0, z 4 > 0, z 3 0, z 1 + z 3 0, F 5 ) z 1 + z 2 + z 3 + 3z 4 > 0, z 3 + 3z 4 > 0, z 3 > 0, z 1 0, F 6 ) z 1 + z 2 + z 3 + 3z 4 > 0, z 1 + z 3 + 3z 4 > 0, z 1 + z 3 > 0, z 1 > )

3 1)) Zero Acton on Perfect Crystals for U q G 2 3 The condtons F 1 ) F 6 ) are dsjont and they exhaust all cases. E ) 1 6) s defned from F ) by replacng > resp. ) wth resp. <). We also defne A = 0, z 1, z 1 + z 2, z 1 + z 2 + 3z 4, z 1 + z 2 + z 3 + 3z 4, 2z 1 + z 2 + z 3 + 3z 4 ). 2.4) Then, for b = x 1, x 2, x 3, x 3, x 2, x 1 ) ˆB l, ê b, ˆf b, ˆε b), ˆϕ b) are gven as follows x 1 1/3,...) f E 1 ),..., x 3 1/3, x 3 1/3,..., x 1 + 1/3) f E 2 ),..., x 3 2/3,..., x 2 + 1/3,...) f E 3 ), ê 0 b =..., x 2 1/3,..., x 3 + 2/3,...) f E 4 ), x 1 1/3,..., x 3 + 1/3, x 3 + 1/3,...) f E 5 ),..., x 1 + 1/3) f E 6 ), x 1 + 1/3,...) f F 1 ),..., x 3 + 1/3, x 3 + 1/3,..., x 1 1/3) f F 2 ),..., x 3 + 2/3,..., x 2 1/3,...) f F 3 ), ˆf 0 b =..., x 2 + 1/3,..., x 3 2/3,...) f F 4 ), x 1 + 1/3,..., x 3 1/3, x 3 1/3,...) f F 5 ),..., x 1 1/3) f F 6 )...., x 2 + 1/3, x 1 1/3) f z 2 z 3 ) +, ê 1 b =..., x 3 + 1/3, x 3 1/3,...) f z 2 < 0 z 3, x 1 + 1/3, x 2 1/3,...) f z 2 ) + < z 3 ), x 1 1/3, x 2 + 1/3,...) f z 2 ) + z 3 ), ˆf 1 b =..., x 3 1/3, x 3 + 1/3,...) f z 2 0 < z 3,..., x 2 1/3, x 1 + 1/3) f z 2 > z 3 ) +, {..., x 3 + 2/3, x 2 1/3,...) f z 4 0, ê 2 b =..., x 2 + 1/3, x 3 2/3,...) f z 4 < 0, {..., x 2 1/3, x 3 + 2/3,...) f z 4 0, ˆf 2 b =..., x 3 2/3, x 2 + 1/3,...) f z 4 > 0, ˆε 0 b) = l 3sb) + 3 max A 32z 1 + z 2 + z 3 + 3z 4 ), ˆϕ 0 b) = l 3sb) + 3 max A, ˆε 1 b) = 3 x x 3 x 2 + x 2 x 3 ) + ) +, ˆϕ 1 b) = 3x 1 + 3x 3 x 2 + x 2 x 3 ) + ) +, ˆε 2 b) = 3 x x 3 x 3 ) +, ˆϕ 2 b) = 3x x 3 x 3 ) ) If ê b or ˆf b does not belong to ˆB l, namely, f x j or x j for some j becomes negatve or sb) exceeds l/3, we should understand t to be 0. Forgettng the 0-arrows, ˆB l l B G 2 jλ1 ), j=0 where B G 2 λ) s the hghest weght Uq G 2 )-crystal of hghest weght λ and G 2 stands for the smple Le algebra G 2 wth the reverse labelng of the ndces of the smple roots α 1 s the short

4 4 K.C. Msra, M. Mohamad and M. Okado root). Forgettng 2-arrows, ˆB l l 2 =0 j 0,j 1 l j 0,j 1 l mod 3) B A 2 j 0 Λ 0 + j 1 Λ 1 ), where B A 2 λ) s the hghest weght U q A 2 )-crystal wth ndces {0, 1}) of hghest weght λ. 3 U q 1)) G 2 -crystal In ths secton we defne a subset V l of ˆB 3l and see t s somorphc to the U q 1)) G 2 -crystal Bl. The set V l s defned as a subset of ˆB 3l satsfyng the followng condtons: x 1, x 1, x 2 x 3, x 3 x 2 Z. 3.1) For an element b = x 1, x 2, x 3, x 3, x 2, x 1 ) of V l we defne sb) as n 2.1). From 3.1) we see that sb) {0, 1,..., l}. Lemma 1. For 0 k l {b V l sb) = k} = 1 k + 1)k + 2)2k + 3)3k + 4)3k + 5). 120 Proof. We frst count the number of elements x 2, x 3, x 3, x 2 ) satsfyng the condtons of coordnates as an element of V l and x 2 + x 3 + x 3 )/2 + x 2 = m m = 0, 1,..., k). Accordng to a, b, c, d) a, d {0, 1/3, 2/3}, b, c {0, 1/3, 2/3, 1, 4/3, 5/3}) such that x 2 Z + a, x 3 2Z + b, x 3 2Z + c, x 2 Z + d, we dvde the cases nto the followng 18: ) 0, 0, 0, 0), ) 0, 0, 2/3, 2/3), ) 0, 0, 4/3, 1/3), v) 0, 1, 1/3, 1/3), v) 0, 1, 1, 0), v) 0, 1, 5/3, 2/3), v) 1/3, 1/3, 1/3, 1/3), v) 1/3, 1/3, 1, 0), x) 1/3, 1/3, 5/3, 2/3), x) 1/3, 4/3, 0, 0), x) 1/3, 4/3, 2/3, 2/3), x) 1/3, 4/3, 4/3, 1/3), x) 2/3, 2/3, 0, 0), xv) 2/3, 2/3, 2/3, 2/3), xv) 2/3, 2/3, 4/3, 1/3), xv) 2/3, 5/3, 1/3, 1/3), xv) 2/3, 5/3, 1, 0), xv) 2/3, 5/3, 5/3, 2/3). The number of elements x 2, x 3, x 3, x 2 ) n a case among the above such that a+b+c)/2+d = e e = 0, 1, 2, 3) s gven by fe) = ) m e+3 3. Snce there s one case wth e = 0 ) and e = 3 xv) and 8 cases wth e = 1 and e = 2, the number of x 2, x 3, x 3, x 2 ) such that x 2 +x 3 + x 3 )/2+ x 2 = m s gven by f0) + 8f1) + 8f2) + f3) = 1 2 2m + 1)3m2 + 3m + 2). For each x 2, x 3, x 3, x 2 ) such that x 2 +x 3 + x 3 )/2+ x 2 = m m = 0, 1,..., k) there are k m+1) cases for x 1, x 1 ), so the number of b V l such that sb) = k s gven by k m= m + 1)3m2 + 3m + 2)k m + 1). A drect calculaton leads to the desred result.

5 1)) Zero Acton on Perfect Crystals for U q G 2 5 We defne the acton of operators e, f = 0, 1, 2) on V l as follows. x 1 1,...) f E 1 ),..., x 3 1, x 3 1,..., x 1 + 1) f E 2 ),..., x2 2 3, x 3 2 3, x , x ,... ) f E 3 ) and z 4 = 1 3,..., x2 1 e 0 b = 3, x 3 4 3, x , x ,... ) f E 3 ) and z 4 = 2 3,..., x 3 2,..., x 2 + 1,...) f E 3 ) and z 4 1 3, 2 3,..., x 2 1,..., x 3 + 2,...) f E 4 ), x 1 1,..., x 3 + 1, x 3 + 1,...) f E 5 ),..., x 1 + 1) f E 6 ), x 1 + 1,...) f F 1 ),..., x 3 + 1, x 3 + 1,..., x 1 1) f F 2 ),..., x 3 + 2,..., x 2 1,...) f F 3 ),..., x2 + 1 f 0 b = 3, x , x 3 2 3, x 2 2 3,... ) f F 4 ) and z 4 = 1 3,..., x , x , x 3 4 3, x 2 1 3,... ) f F 4 ) and z 4 = 2 3,..., x 2 + 1,..., x 3 2,...) f F 4 ) and z 4 1 3, 2 3, x 1 + 1,..., x 3 1, x 3 1,...) f F 5 ),..., x 1 1) f F 6 ),..., x 2 + 1, x 1 1) f x 2 x 3 x 2 x 3 ) +, e 1 b =..., x 3 + 1, x 3 1,...) f x 2 x 3 < 0 x 3 x 2, x 1 + 1, x 2 1,...) f x 2 x 3 ) + < x 2 x 3, x 1 1, x 2 + 1,...) f x 2 x 3 ) + x 2 x 3, f 1 b =..., x 3 1, x 3 + 1,...) f x 2 x 3 0 < x 3 x 2,..., x 2 1, x 1 + 1) f x 2 x 3 > x 2 x 3 ) +, {..., x , x 2 1 3,... ) f x 3 x 3, e 2 b = f 2 b =..., x , x 3 2 3,... ) f x 3 < x 3, {..., x2 1 3, x ,... ) f x 3 x 3,..., x3 2 3, x ,... ) f x 3 > x 3. We now set m 0, m 1, m 2 ) = 3, 3, 1). Proposton 1. 1) For any b V l we have e b, f b V l {0} for = 0, 1, 2. 2) The equaltes e = ê m and f = ˆf m hold on V l for = 0, 1, 2. Proof. 1) can be checked easly. For 2) we only treat f. To prove the = 0 case consder the followng table F 1 ) F 2 ) F 3 ) F 4 ) F 5 ) F 6 ) z 1 1/3 1/ /3 1/3 z 2 0 1/3 1/3 2/3 1/3 0 z 3 0 1/3 2/3 1/3 1/3 0 z /3 1/3 0 0

6 6 K.C. Msra, M. Mohamad and M. Okado Ths table sgnfes the dfference z j for ˆf 0 b) z j for b) when b belongs to the case F ). Note that the left hand sdes of the nequaltes of each F ) 2.3) always decrease by 1/3. Snce z 1, z 2, z 3 Z, z 4 Z/3 for b V l, we see that f b belongs to F ), ˆf 0 b and ˆf 0 2b also belong to F ) except two cases: a) b F 4 ) and z 4 = 1/3, and b) b F 4 ) and z 4 = 2/3. If a) occurs, we have ˆf 0 b, ˆf 0 2b F 3). Hence, we obtan f 0 = ˆf 0 3 n ths case. If b) occurs, we have ˆf 0 b F 4 ), ˆf 0 2b F 3). Therefore, we obtan f 0 = ˆf 0 3 n ths case as well. In the = 1 case, f b belongs to one of the 3 cases, ˆf 1 b and ˆf 1 2 b also belong to the same case. Hence, we obtan f 1 = ˆf 1 3. For = 2 there s nothng to do. Proposton 1, together wth Theorem 1, shows that V l can be regarded as a U q 1)) G 2 -crystal wth operators e, f = 0, 1, 2). 1)) Proposton 2. As a U q G 2 {1,2} -crystal V l l B G 2 kλ 1 ), k=0 where B G 2 λ) s the hghest weght U q G 2 )-crystal of hghest weght λ. Proof. For a subset J of {0, 1, 2} we say b s J-hghest f e j b = 0 for any j J. Note from 2.5) that b k = k, 0, 0, 0, 0, 0) 0 k l) s {1, 2}-hghest of weght 3kΛ 1 n ˆB 3l. By settng g = G 2 = G 2 wth the reverse labelng of ndces), m 1, m 2 ) = 3, 1), g = G 2 n Theorem 1, we know that the connected component generated from b k by f 1 = ˆf 1 3 and f 2 = ˆf 2 s somorphc to B G 2 kλ 1 ). Hence by Proposton 1 1) we have l B G 2 kλ 1 ) V l. k=0 3.2) Now recall Weyl s formula to calculate the dmenson of the hghest weght representaton. In our case we obtan B G 2 kλ 1 ) = 1 k + 1)k + 2)2k + 3)3k + 4)3k + 5). 120 However, ths s equal to {b V l sb) = k} by Lemma 1. Therefore, n 3.2) should be =, and the proof s completed. 1)) Proposton 3. As a U q G 2 {0,1} -crystal V l l/2 =0 j 0,j 1 l B A 2 j 0 Λ 0 + j 1 Λ 1 ), where B A 2 λ) s the hghest weght U q A 2 )-crystal wth ndces {0, 1}) of hghest weght λ. Proof. For ntegers, j 0, j 1 such that 0 l/2, j 0, j 1 l, defne an element b,j0,j 1 of V l by { 0, y1, 3y b,j0,j 1 = 0 2y 1 +, y 0 +, y 0 + j 0, 0) f j 0 j 1, 0, y 0, y 0 +, 2y 1 y 0 +, 2y 0 y 1 + j 0, 0) f j 0 > j 1. Here we have set y a = l j a )/3 for a = 0, 1. From 2.5) one notces that b,j0,j 1 s {0, 1}- hghest of weght 3j 0 Λ 0 + 3j 1 Λ 1 n ˆB 3l. For nstance, ˆε 0 b,j0,j 1 ) = 0 and ˆϕ 0 b,j0,j 1 ) = 3j 0 snce

7 1)) Zero Acton on Perfect Crystals for U q G 2 7 sb,j0,j 1 ) = l and max A = 2z 1 + z 2 + z 3 + 3z 4 = j 0. By settng g = g = A 2, m 0, m 1 ) = 3, 3) n Theorem 1, the connected component generated from b,j0,j 1 by f = ˆf 3 = 0, 1) s somorphc to B A 2 j 0 Λ 0 + j 1 Λ 1 ). Hence, by Proposton 1 1) we have l/2 =0 j 0,j 1 l B A 2 j 0 Λ 0 + j 1 Λ 1 ) V l. However, from Proposton 2 one knows that V l = l B G 2 kλ 1 ). k=0 Moreover, t s already establshed n [7] that l B G 2 kλ 1 ) = k=0 l/2 =0 j 0,j 1 l B A 2 j 0 Λ 0 + j 1 Λ 1 ). Therefore, the proof s completed. Theorem 6.1 n [6] shows that f two U q 1)) G 2 -crystals decompose nto 0 k l BG 2 kλ 1 ) as U q G 2 )-crystals, then they are somorphc to each other. Therefore, we now have Theorem 2. V l agrees wth the U q 1)) G 2 -crystal Bl constructed n [7]. The values of ε, ϕ wth our representaton are gven by ε 0 b) = l sb) + max A 2z 1 + z 2 + z 3 + 3z 4 ), ϕ 0 b) = l sb) + max A, ε 1 b) = x 1 + x 3 x 2 + x 2 x 3 ) + ) +, ϕ 1 b) = x 1 + x 3 x 2 + x 2 x 3 ) + ) +, ε 2 b) = 3 x x 3 x 3 ) +, ϕ 2 b) = 3x x 3 x 3 ) ) 4 Mnmal elements and a coherent famly The noton of perfect crystals was ntroduced n [3] to construct the path realzaton of a hghest weght crystal of a quantum affne algebra. The crystal B l was shown to be perfect of level l n [7]. In ths secton we obtan all the mnmal elements of B l n the coordnate representaton and also show {B l } l 1 forms a coherent famly of perfect crystals. For the notatons such as P cl, P + cl ) l, see [3]. 4.1 Mnmal elements From 3.3) we have c, ϕb) = ϕ 0 b) + 2ϕ 1 b) + ϕ 2 b) = l + max A + 2z 3 + z 2 ) + ) + + 3z 4 ) + z 1 + z 2 + 2z 3 + 3z 4 ), where z j 1 j 4) are gven n 2.2) and A s gven n 2.4). The followng lemma was proven n [6], although Z s replaced wth Z/3 here. Lemma 2. For z 1, z 2, z 3, z 4 ) Z/3) 4 set ψz 1, z 2, z 3, z 4 ) = max A + 2z 3 + z 2 ) + ) + + 3z 4 ) + z 1 + z 2 + 2z 3 + 3z 4 ). Then we have ψz 1, z 2, z 3, z 4 ) 0 and ψz 1, z 2, z 3, z 4 ) = 0 f and only f z 1, z 2, z 3, z 4 ) = 0, 0, 0, 0).

8 8 K.C. Msra, M. Mohamad and M. Okado From ths lemma, we have c, ϕb) l = ψz 1, z 2, z 3, z 4 ) 0. Snce c, ϕb) εb) = 0, we also have c, εb) l. Suppose c, εb) = l. It mples ψ = 0. Hence from the lemma one can conclude that such element b = x 1, x 2, x 3, x 3, x 2, x 1 ) should satsfy x 1 = x 1, x 2 = x 3 = x 3 = x 2. Therefore the set of mnmal elements B l ) mn n B l s gven by B l ) mn = {α, β, β, β, β, α) α Z 0, β Z 0 )/3, 2α + 3β l}. For b = α, β, β, β, β, α) B l ) mn one calculates εb) = ϕb) = l 2α 3β)Λ 0 + αλ 1 + 3βΛ Coherent famly of perfect crystals The noton of a coherent famly of perfect crystals was ntroduced n [1]. Let {B l } l 1 be a famly of perfect crystals B l of level l and B l ) mn be the subset of mnmal elements of B l. Set J = {l, b) l Z >0, b B l ) mn }. Let σ denote the somorphsm of P + cl ) l defned by σ = ε ϕ 1. For λ P cl, T λ denotes a crystal wth a unque element t λ defned n [4]. For our purpose the followng facts are suffcent. For any P cl -weghted crystal B and λ, µ P cl consder the crystal T λ B T µ = {t λ b t µ b B}. The defnton of T λ and the tensor product rule of crystals mply ẽ t λ b t µ ) = t λ ẽ b t µ, f t λ b t µ ) = t λ f b t µ, ε t λ b t µ ) = ε b) h, λ, ϕ t λ b t µ ) = ϕ b) + h, µ, wt t λ b t µ ) = λ + µ + wt b. Defnton 1. A crystal B wth an element b s called a lmt of {B l } l 1 f t satsfes the followng condtons: wt b = 0, εb ) = ϕb ) = 0, for any l, b) J, there exsts an embeddng of crystals f l,b) : T εb) B l T ϕb) B sendng t εb) b t ϕb) to b, B = l,b) J Im f l,b). If a lmt exsts for the famly {B l }, we say that {B l } s a coherent famly of perfect crystals. Let us now consder the followng set { B = b = ν 1, ν 2, ν 3, ν 3, ν 2, ν 1 ) Z/3) 6 ν 1, ν 1, ν 2 ν 3, ν 3 ν 2 Z, 3ν 3 3 ν 3 mod 2) }, and set b = 0, 0, 0, 0, 0, 0). We ntroduce the crystal structure on B as follows. The actons of e, f = 0, 1, 2) are defned by the same rule as n Secton 3 wth x and x replaced wth ν and ν. The only dfference les n the fact that e b or f b never becomes 0, snce we allow a coordnate to be negatve and there s no restrcton for the sum sb) = 2 ν + ν )+ν 3 + ν 3 )/2. For ε, ϕ wth = 1, 2 we adopt the formulas n Secton 3. For ε 0, ϕ 0 we defne ε 0 b) = sb) + max A 2z 1 + z 2 + z 3 + 3z 4 ), ϕ 0 b) = sb) + max A, =1

9 1)) Zero Acton on Perfect Crystals for U q G 2 9 where A s gven n 2.4) and z 1, z 2, z 3, z 4 are gven n 2.2) wth x, x replaced by ν, ν. Note that wt b = 0 and ε b ) = ϕ b ) = 0 for = 0, 1, 2. Let b 0 = α, β, β, β, β, α) be an element of B l ) mn. Snce εb 0 ) = ϕb 0 ), one can set σ = d. Let λ = εb 0 ). For b = x 1, x 2, x 3, x 3, x 2, x 1 ) B l we defne a map f l,b0 ) : T λ B l T λ B by where f l,b0 )t λ b t λ ) = b = ν 1, ν 2, ν 3, ν 3, ν 2, ν 1 ), ν 1 = x 1 α, ν 1 = x 1 α, ν j = x j β, ν j = x j β j = 2, 3). Note that sb ) = sb) 2α + 3β). Then we have wt t λ b t λ ) = wt b = wt b, ϕ 0 t λ b t λ ) = ϕ 0 b) + h 0, λ = ϕ 0 b ) + l sb)) + sb ) l 2α 3β) = ϕ 0 b ), ϕ 1 t λ b t λ ) = ϕ 1 b) + h 1, λ = ϕ 1 b ) + α α = ϕ 1 b ), ϕ 2 t λ b t λ ) = ϕ 2 b) + h 2, λ = ϕ 2 b ) + 3β 3β = ϕ 2 b ). ε t λ b t λ ) = ε b ) = 0, 1, 2) also follows from the above calculatons. From the fact that z j for b) = z j for b ) t s straghtforward to check that f b, e b B l resp. b, f b B l ), then f l,b0 )e t λ b t λ )) = e f l,b0 )t λ b t λ ) resp. f l,b0 )f t λ b t λ )) = f f l,b0 )t λ b t λ )). Hence f l,b0 ) s a crystal embeddng. It s easy to see that f l,b0 )t λ b 0 t λ ) = b. We can also check B = l,b) J Im f l,b). Therefore we have shown that the famly of perfect crystals {B l } l 1 forms a coherent famly. 5 Crystal graphs of B 1 and B 2 In ths secton we present crystal graphs of the U q 1)) G 2 -crystals B1 and B 2 n Fgs. 1 and 2. In the graphs b b stands for b = f b. Mnmal elements are marked as. Recall that as a U q G 2 )-crystal B 1 B0) BΛ 1 ), B 2 B0) BΛ 1 ) B2Λ 1 ). We gve the table that relates the numbers n the crystal graphs to our representaton of elements accordng to whch U q G 2 )-components they belong to. B0) : φ = 0, 0, 0, 0, 0, 0) BΛ 1 ): 1 = 1, 0, 0, 0, 0, 0) 2 = 0, 1, 0, 0, 0, 0) 3 = 0, 2 3, 2 3, 0, 0, 0) 4 = 0, 1 3, 4 3, 0, 0, 0) 5 = 0, 1 3, 1 3, 1, 0, 0) 6 = 0, 1 3, 1 3, 1 3, 1 3, 0) 7 = 0, 0, 1, 1, 0, 0) 8 = 0, 0, 1, 1 3, 1 3, 0) 9 = 0, 0, 0, 4 3, 1 3, 0) 10 = 0, 0, 0, 2 3, 2 3, 0) 11 = 0, 0, 0, 0, 1, 0) 12 = 0, 0, 0, 0, 0, 1) 13 = 0, 0, 2, 0, 0, 0) 14 = 0, 0, 0, 2, 0, 0)

10 10 K.C. Msra, M. Mohamad and M. Okado Fgure 1. Crystal graph of B 1. s f 1 and s f 2. B2Λ 1 ): 15 = 2, 0, 0, 0, 0, 0) 16 = 1, 1, 0, 0, 0, 0) 17 = 1, 2 3, 2 3, 0, 0, 0) 18 = 1, 1 3, 4 3, 0, 0, 0) 19 = 1, 1 3, 1 3, 1, 0, 0) 20 = 1, 1 3, 1 3, 1 3, 1 3, 0) 21 = 1, 0, 1, 1, 0, 0) 22 = 1, 0, 1, 1 3, 1 3, 0) 23 = 1, 0, 0, 4 3, 1 3, 0) 24 = 1, 0, 0, 2 3, 2 3, 0) 25 = 1, 0, 0, 0, 1, 0) 26 = 1, 0, 0, 0, 0, 1) 27 = 1, 0, 2, 0, 0, 0) 28 = 1, 0, 0, 2, 0, 0) 29 = 0, 2, 0, 0, 0, 0) 30 = 0, 5 3, 2 3, 0, 0, 0) 31 = 0, 4 3, 4 3, 0, 0, 0) 32 = 0, 4 3, 1 3, 1, 0, 0) 33 = 0, 4 3, 1 3, 1 3, 1 3, 0) 34 = 0, 1, 1, 1, 0, 0) 35 = 0, 1, 1, 1 3, 1 3, 0) 36 = 0, 1, 0, 4 3, 1 3, 0) 37 = 0, 1, 0, 2 3, 2 3, 0) 38 = 0, 1, 0, 0, 1, 0) 39 = 0, 1, 0, 0, 0, 1) 40 = 0, 1, 2, 0, 0, 0) 41 = 0, 1, 0, 2, 0, 0) 42 = 0, 2 3, 2 3, 0, 1, 0) 43 = 0, 1 3, 4 3, 0, 1, 0) 44 = 0, 1 3, 1 3, 1, 1, 0) 45 = 0, 1 3, 1 3, 1 3, 4 3, 0) 46 = 0, 0, 1, 1, 1, 0) 47 = 0, 0, 1, 1 3, 4 3, 0) 48 = 0, 0, 0, 4 3, 4 3, 0) 49 = 0, 0, 0, 2 3, 5 3, 0) 50 = 0, 0, 0, 0, 2, 0) 51 = 0, 0, 0, 0, 1, 1) 52 = 0, 0, 2, 0, 1, 0) 53 = 0, 0, 0, 2, 1, 0) 54 = 0, 2 3, 2 3, 0, 0, 1) 55 = 0, 1 3, 4 3, 0, 0, 1) 56 = 0, 1 3, 1 3, 1, 0, 1) 57 = 0, 1 3, 1 3, 1 3, 1 3, 1) 58 = 0, 0, 1, 1, 0, 1) 59 = 0, 0, 1, 1 3, 1 3, 1) 60 = 0, 0, 0, 4 3, 1 3, 1) 61 = 0, 0, 0, 2 3, 2 3, 1) 62 = 0, 0, 0, 0, 0, 2) 63 = 0, 0, 2, 0, 0, 1) 64 = 0, 0, 0, 2, 0, 1) 65 = 0, 2 3, 8 3, 0, 0, 0) 66 = 0, 1 3, 10 3, 0, 0, 0) 67 = 0, 1 3, 7 3, 1, 0, 0) 68 = 0, 1 3, 7 3, 1 3, 1 3, 0) 69 = 0, 0, 3, 1, 0, 0) 70 = 0, 0, 3, 1 3, 1 3, 0) 71 = 0, 0, 2, 4 3, 1 3, 0) 72 = 0, 0, 2, 2 3, 2 3, 0) 73 = 0, 0, 4, 0, 0, 0) 74 = 0, 0, 2, 2, 0, 0) 75 = 0, 2 3, 2 3, 2, 0, 0) 76 = 0, 1 3, 4 3, 2, 0, 0) 77 = 0, 1 3, 1 3, 3, 0, 0) 78 = 0, 1 3, 1 3, 7 3, 1 3, 0) 79 = 0, 0, 1, 3, 0, 0) 80 = 0, 0, 1, 7 3, 1 3, 0) 81 = 0, 0, 0, 10 3, 1 3, 0) 82 = 0, 0, 0, 8 3, 2 3, 0) 83 = 0, 0, 0, 4, 0, 0) 84 = 0, 2 3, 5 3, 1, 0, 0) 85 = 0, 1 3, 4 3, 4 3, 1 3, 0) 86 = 0, 0, 1, 5 3, 2 3, 0) 87 = 0, 2 3, 5 3, 1 3, 1 3, 0) 88 = 0, 2 3, 2 3, 4 3, 1 3, 0) 89 = 0, 1 3, 1 3, 5 3, 2 3, 0) 90 = 0, 2 3, 2 3, 2 3, 2 3, 0) 91 = 0, 1 3, 4 3, 2 3, 2 3, 0) Comparng our crystal graphs wth those n [7] we found that some 2-arrows are mssng n Fg. 3 of [7].

11 1)) Zero Acton on Perfect Crystals for U q G 2 11 Fgure 2. Crystal graph of B 2. s f 0, s f 1 and others are f 2. Acknowledgements KCM thanks the faculty and staff of Osaka Unversty for ther hosptalty durng hs vst n August, 2009 and acknowledges partal support from NSA grant H MM would lke to thank Unverst Tun Hussen Onn Malaysa for supportng ths study. MO would lke to thank the organzers of the conference Geometrc Aspects of Dscrete and Ultra-Dscrete Integrable Systems held durng March 30 Aprl 3, 2009 at Glasgow for a warm hosptalty and acknowledges partal support from JSPS grant No

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