On crystal bases and Enright s completions

Transcription

1 Journal of Algebra 312 (2007) On crystal bases and Enrght s completons Djana Jakelć Department of Mathematcs, Unversty of Vrgna, Charlottesvlle, VA 22904, USA Receved 6 January 2006 Avalable onlne 2 February 2007 Communcated by Leonard L. Scott, Jr. Dedcated to the memory of Des Sheham Abstract We nvestgate the nterplay of crystal bases and completons n the sense of Enrght on certan nonntegrable representatons of quantum groups. We defne completons of crystal bases, show that ths noton of completon s compatble wth Enrght s completon of modules, prove that every module n our category has a crystal bass whch can be completed and that a completon of the crystal lattce s unque. Furthermore, we gve two constructons of the completon of a crystal lattce Elsever Inc. All rghts reserved. Keywords: Enrght s completon; Crystal bass; Crystal lattce; Nonntegrable representatons; Quantum groups Introducton In ths paper we are concerned wth brngng together two dstnct notons of the representaton theory of quantzed envelopng algebras those of crystal bases and completons. The theory of crystal and canoncal bases s one of the most remarkable developments n representaton theory. It was ntroduced ndependently by M. Kashwara [7,8] and G. Lusztg [11] n a combnatoral and a geometrc way, respectvely. In ths paper we follow Kashwara s approach. Roughly speakng, a crystal bass of an ntegrable representaton for the quantzed envelopng algebra U q (g) of a fnte-dmensonal complex semsmple Le algebra g (or more generally, of a symmetrzable Kac Moody Le algebra) s a par consstng of a lattce of the module, called a crystal lattce, and a vector space bass of a quotent of the crystal lattce, called a crystal. It s actually a certan parametrzaton of bases of the module wth a number E-mal address: dj7m@vrgna.edu /$ see front matter 2007 Elsever Inc. All rghts reserved. do: /j.jalgebra

2 112 D. Jakelć / Journal of Algebra 312 (2007) of desrable propertes and t encodes the ntrnsc structure of the module n a combnatoral way. One of the most mportant combnatoral realzatons of crystals s Lttelmann s path model [9,10]. T. Enrght [3] ntroduced a noton of completon wth respect to a smple root of g on the category I(g) of weght modules of g that are U (g)-torson free and U + (g)-fnte. Completon s an effectve process of obtanng new representatons from a gven one, contanng the orgnal one as a subrepresentaton. Enrght used the completon functor n order to algebracally construct the fundamental seres representatons. Soon after, V. Deodhar [2] realzed the completon functor va Ore localzaton gvng a concrete model of completon and used t to prove Enrght s unqueness conjecture arsng from consderng successve completons (stressng the sl 2 -nature of completons). A. Joseph [5] then generalzed ths functor to the Bernsten Gelfand Gelfand category O(g) and gave a refnement of the Jantzen conjecture by studyng the lftng of a contravarant form under the acton of the completon functor. Later, Y.M. Zou [15] extended some of the results obtaned by Enrght and Deodhar to the quantum groups settng. The startng motvaton of ths work was to study both crystal bases and completons of modules belongng to the category I(U q (g)) (thus nonntegrable ones) and examne how the two concepts relate to each other wth the am of ntroducng a noton of completon of crystal bases whch would be compatble wth Enrght s completon of modules. It s natural to expect that such a program may eventually lead to some potentally nterestng nterplay between the theory of crystal bases and that of fundamental seres representatons. Furthermore, snce all the Verma modules n a lattce of nclusons of Verma modules may be obtaned from the correspondng rreducble one by means of completons, one may expect that a successful crystal base theory related to completons could produce a combnatoral tool relevant for studyng Jantzen fltratons and some other Kazhdan Lusztg Theory related topcs. Besde defnng crystal bases of ntegrable representatons, Kashwara [8] also defned the crystal bass of the quantzaton Uq (g) of the unversal envelopng algebra of the nlpotent part of g by consderng Uq (g) as a module for Kashwara s algebra B q(g). Although the former was the man goal of [8], the latter provded a way to smultaneously consder the bases of all ntegrable representatons, and the nterplay of the two notons of crystal bases played a central role n the paper. Snce [8] there have been only a few papers concerned wth crystal bases or crystals of representatons whch are not necessarly ntegrable, ncludng [6,13,14]. Also, n a jont work wth V. Char and A. Moura [1] we consdered the problem of tensor product decompostons nto ndecomposables for nonntegrable modules n the BGG category O by ntroducng the combnatoral objects called branched crystals whch satsfy a relaxed axom for formal nvertblty of Kashwara s operators. In ths paper, we follow Kashwara s defnton of crystal bases, thus the Uq -torson free modules n queston are naturally vewed as B q (g)-modules. On the other hand, n regard to completons they are thought of as U q (g)-modules. Ths aspect makes the stuaton more nterestng and a synchronzaton of the two structures becomes essental. As the U q (sl 2 )-case s already qute ntrcate, we restrct ourselves to that case n ths paper leavng the consderaton on how to extend ths theory to the hgher rank case to a future work. We ntroduce a category Ĩ consstng of U q -modules n I = I(U q (sl 2 )) wth a compatble B q -structure and obtan a decomposton of every module n Ĩ nto a drect sum of ndecomposable U q -submodules whch are also B q -nvarant (cf. Theorem 2.4). Due to ths decomposton, we are able to ntroduce weght spaces nto our consderaton of crystal bases of modules n Ĩ even though these crystal bases arse from the B q -structures. Therefore, we get a settng resemblng the one of ntegrable U q -modules. We defne a noton of a complete crystal bass and a

3 D. Jakelć / Journal of Algebra 312 (2007) completon of a crystal bass. In the process, we take advantage of a symmetrc acton of the Kashwara operators ẽ and f on crystal bases. However, unlke the case of modules where the completon of a module contans the module, one cannot expect a completon of a crystal lattce to contan the crystal lattce; ths beng ndeed clear from the case of Verma modules. Our defnton of completon of bases nvolves a very natural connecton of crystal bases arsng from B q -structures wth the ones arsng from U q -structures. We prove that a crystal bass s complete f and only f ts correspondng module s. We also show that every module n Ĩ has a crystal bass whch can be completed and moreover a completon of the crystal lattce s unque (cf. Theorems 3.5 and 3.6). The paper s arranged as follows. In Secton 1, we set the notaton and revew relevant results on completons and crystal bases. We thnk t nstructve to have ths secton done for U q (g), where g s any smple Le algebra, as t s then more transparent how crucal a step the U q (sl 2 )- case s n solvng the problem. However, the remanng sectons treat the U q (sl 2 )-case, except where stated otherwse. In Secton 2, we consder some natural B q -structures on the ndecomposable U q -modules n I, collect the desrable propertes that a B q -structure should have wth respect to a U q -structure n order to defne the category Ĩ, and prove the smultaneous decomposton theorem for modules n Ĩ. In Secton 3, we look nto the crystal bases wth whch modules n Ĩ are naturally endowed va ther B q -structures, defne complete crystal bases, show they correspond to complete modules, defne a completon of a crystal bass, and prove the aforementoned Theorems 3.5 and 3.6. In Secton 4 we gve two constructons of the completon of a crystal lattce one obtaned by modfyng Deodhar s model of completon of modules and the other by applyng an operator used by Kashwara n [7] to construct the operators ẽ and f. 1. Prelmnares 1.1. Let (a j ) 1,j l be the Cartan matrx of a fnte-dmensonal complex smple Le algebra g and d unque postve ntegers such that gcd (d 1,...,d l ) = 1 and the matrx (d a j ) 1,j l s symmetrc. Let q be an ndetermnate and Q(q) the feld of ratonal functons of q wth coeffcents n Q. Set q = q d. The quantzed envelopng algebra U q (g) s the Q(q)-algebra wth generators e, f, t, t 1,1 l, and defnng relatons where t t 1 t e j t 1 = 1 = t 1 t, t t j = t j t, = q a j e j, t f j t 1 = q a j f j, t t 1 e f j f j e = δ j q q 1, 1 a j s=0 ( 1) s e (s) e j e (1 a j s) = 0, 1 a j s=0 ( 1) s f (s) f j f (1 a j s) = 0 ( j) e (n) = en [n]!, [n] = qn q n q q 1 f(n) = f n [n]!, [n] (!=[1] [2]...[n] n Z + ), and (n Z).

4 114 D. Jakelć / Journal of Algebra 312 (2007) Denote by U q + (g) (respectvely U q (g)) the subalgebra of U q(g) generated by e (respectvely f ), 1 l. LetUq 0(g) be the subalgebra of U q(g) generated by t and t 1,1 l. Multplcaton defnes an somorphsm of Q(q)-vector spaces Uq (g) U q 0(g) U q + (g) U q (g). Denote by U q (g) the subalgebra of U q (g) generated by e, f, t and t 1. Then U q (g) = U q (sl 2 ) Let h be a Cartan subalgebra of g and Φ the root system of (g, h).let{α } 1 l h be a set of smple roots of g and {h } 1 l h the correspondng set of coroots so that α j (h ) = a j. Let Q = l =1 Zα be the root lattce and P ={λ h λ(h ) Z, 1 l} the weght lattce for g. Also,setQ + = l =1 Z + α and P + ={λ h λ(h ) Z +,1 l}. Denote by W be the Weyl group of Φ and by s the reflecton wth respect to the smple root α (1 l). For a U q (g)-module M and λ P, we defne the λ-weght space of M as M λ ={m M t m = q λ(h ) m, 1 l}. M s a weght module f t s a drect sum of ts weght spaces. If n addton there exst λ P and a nonzero vector m M λ such that e m = 0 for all and M = U q (g) m, then M s a hghest weght module wth hghest weght λ and hghest weght vector m. The Verma module M(λ) s the U q (g)-module U q (g)/j (λ) where J(λ) s the left deal of U q (g) generated by {t q λ(h ),e 1 l}. M(λ) s the unversal hghest weght module of weght λ. The unque rreducble quotent V(λ)of M(λ) s fnte-dmensonal ff λ P We recall the defntons and some propertes of category I, completons, and T modules (cf. [2 4,15]). Let I(U q (g)) be the category of U q (g)-modules M satsfyng: () M s a weght module, () Uq (g)-acton on M s torson free, and () M s U q + (g)-fnte,.e., e acts locally nlpotently on M for all. Fx {1,...,l}. Set M n ={m M t m = q nm} for n Z, Me ={m M e m = 0}, and M e n f n+1 = M n M e. A module M n I(U q (g)) s sad to be complete wth respect to f : M e n M e n 2 s bjectve for all n Z+. A module N n I(U q (g)) s a completon of M wth respect to provded: () N s complete wth respect to, () M s embedded n N, and () N/M s f -fnte. Theorem. (Cf. [2,3,15].) () Every module M n I(U q (g)) has a completon C (M) wth respect to (1 l) and any two such completons are naturally somorphc. () Let w W and M I(U q (g)). For any two reduced expressons w = s 1...s k = s j1...s jk, there exsts an somorphsm F : C 1 (C 2...(C k (M))...) C j1 (C j2...(c jk (M))...) such that F M s the dentty. Thus the process of completon depends essentally on the U q (sl 2 )-representaton theory and so we pay specal attenton to the case g = sl For brevty, wrte U q = U q (sl 2 ) and I = I(U q (sl 2 )). Denote the generators of sl 2 by e, f, t ±1.

5 D. Jakelć / Journal of Algebra 312 (2007) Let n Z. The quantum Casmr element C = qt + q 1 t 1 (q q 1 + feacts on the Verma module ) 2 M(n) as multplcaton by scalar c n = qn+1 + q n 1 (q q 1 ) 2. Consder the left deal I(n) of U q defned by I(n)= U q {t q n 2,e n+2,(c c n ) 2 }.TheT module T(n)s defned as T(n)= U q / I(n). It s an ndecomposable module belongng to the category I and 0 M(n) T(n) M( n 2) 0 s exact. Theorem. (Cf. [3,15].) () The M(n) (n Z) and the T(n) (n Z + ) are precsely all the ndecomposable objects of the category I. Among these, the M(n) for n 1 and T(n)for n Z + are the complete ones. The completon of M( n 2) s M(n) for n 1. () Every module n I s a drect sum (not necessarly fnte) of ndecomposable ones In the subsectons that follow, we recall the man results on Kashwara s crystal bases (cf. [8]). Let M be a fnte-dmensonal U q (g)-module. Fx an ndex (1 l). By the representaton theory of U q (sl 2 ), M = λ P,0 k λ(h ) f (k) (Ker e M λ ). Hence, every element u M λ can be unquely wrtten as u = k 0 f (k) u k where u k Ker e M λ+kα. The Kashwara operators ẽ and f are defned by ẽ u = k 1 f (k 1) u k and f u = k 0 f (k+1) u k. It follows that ẽ M λ M λ+α and f M λ M λ α. Also, f u Ker e and n>0, then f n u = f (n) u. Let A be the subrng of Q(q) consstng of ratonal functons regular at q = 0. Defnton. (See [8].) A free A-submodule L of M s called a crystal lattce f: (a) M = Q(q) A L,(b)L = λ P L λ where L λ = L M λ, and (c) ẽ L L and f L L, 1 l. A crystal bass of M s a par (L, B) satsfyng the followng condtons: () L s a crystal lattce of M, () B s a Q-bass of L/qL, () B = λ P B λ where B λ = B (L λ /ql λ ), (v) ẽ B B {0} and f B B {0}, 1 l, and (v) for b,b B, f b = b f and only f b = ẽ b. For λ P +, consder the fnte-dmensonal rreducble U q (g)-module V(λ) wth hghest weght λ and hghest weght vector u λ.letl(λ) be the smallest A-submodule of V(λ)contanng u λ whch s stable under f s,.e., L(λ) s the A-span of the vectors of the form f 1... f r u λ, where 1 j l and r Z +. Set B(λ) ={b L(λ)/qL(λ) b = f 1... f r u λ mod ql(λ)}\{0}. Then (L(λ), B(λ)) s a crystal bass of V(λ)(cf. [8, Theorem 2]). Snce crystal bases are stable under drect sums, every fnte-dmensonal U q (g)-module M has a crystal bass Isomorphsm of crystal bases s defned as follows: Defnton. Let (L 1,B 1 ) and (L 2,B 2 ) be crystal bases of fnte-dmensonal U q (g)-modules M 1 and M 2, respectvely. We say that (L 1,B 1 ) = (L 2,B 2 ) f there exsts a U q (g)-somorphsm ϕ : M 1 M 2 whch nduces an somorphsm of A-lattces ϕ : L 1 L 2 such that ϕ(b 1 ) = B 2 where ϕ : L 1 /ql 1 L 2 /ql 2 s the nduced somorphsm of Q-vector spaces.

6 116 D. Jakelć / Journal of Algebra 312 (2007) If (L, B) s a crystal bass of a fnte-dmensonal U q (g)-module M, then there exsts an somorphsm M = j V(λ j ) by whch (L, B) s somorphc to j (L(λ j ), B(λ j )) (cf. [8, Theorem 3]) Kashwara s algebra B q (g) (cf. [8,12]) s the Q(q)-algebra generated by e and f,1 l, subject to relatons 1 a j s=0 e f j = q a j f j e + δ j, (1.1) ( 1) s( 1 a e ) (s) ( )( ) e j e (1 aj s) j = 0, ( 1) s f (s) f j f (1 a j s) = 0 ( j). (1.2) For the case g = sl 2, wrte B q = B q (sl 2 ). Then the generators e and f of B q satsfy s=0 e f = q 2 fe + 1. (1.3) For each P Uq (g), there exst unque R,Q U q (g) such that [e,p]= t Q t 1 R q q 1. Defne e End(U q (g)) by e (P ) = R. Then, U q (g) s a left B q(g)-module where e acts as descrbed and f acts by the left multplcaton. Moreover, Uq (g) s an rreducble B q(g)-module and Uq (g) O(B q(g)), where O(B q (g)) s the category of B q (g)-modules M satsfyng that for all u M, there exsts r 0 such that e 1 e 2...e r u = 0 for any 1 1,..., r l. Furthermore, O(B q (g)) s semsmple and Uq (g) s the unque rreducble object of O(B q(g)), upto somorphsm. Remark. For each, e (1) = 0 and e (f p ) = q (p 1) [p] f p 1 for p Let M belong to the category O(B q (g)). Fx an ndex (1 l). Then M = k 0 f (k) Ker e. (1.4) Defne Kashwara s operators ẽ, f End(M) by ẽ ( f (k) u ) = { f (k 1) u, k 1, 0, k= 0, ( (k) f f u ) = f (k+1) u, (1.5) for u Ker e and extend lnearly. Then ẽ f = 1. Also, f ẽ s the projecton onto f M wth respect to M = f M Ker e. Defnton. (See [8].) A free A-submodule L of M s called a crystal lattce f: (a) M = Q(q) A L, and (b) ẽ L L and f L L, 1 l. A crystal bass of M s a par (L, B) satsfyng the followng condtons: () L s a crystal lattce of M, () B s a Q-bass of L/qL, () ẽ B B {0} and f B B, 1 l, and (v) f b B such that ẽ b B, then b = f ẽ b.

7 D. Jakelć / Journal of Algebra 312 (2007) Isomorphsm of crystal bases of modules n O(B q (g)) s defned analogously to somorphsm of crystal bases of fnte-dmensonal modules replacng U q (g) by B q (g) n Defnton 1.6. It wll be clear from the context f we mean crystal bass n the sense of Defnton 1.5 or n the sense of Defnton 1.8. Otherwse, we wll make the dstncton. Let L( ) be the smallest A-submodule of Uq (g) contanng 1 that s stable under f s,.e., L( ) = A-span of { f 1... f r 1 1 j l; r Z + }. Set B( ) ={b L( )/ql( ) b = f 1... f r 1modqL( )}. Then (L( ), B( )) s a crystal bass of Uq (g) [8, Theorem 4]. Moreover, any crystal bass of Uq (g) concdes wth (L( ), B( )) up to a constant multple. The relaton of (L( ), B( )) to (L(λ), B(λ)) s gven by [8, Theorem 5]. 2. U q -modules wth B q -module structure In ths secton we are concerned wth U q -modules endowed wth B q -module structures whch allow decompostons that are smultaneously U q and B q -nvarant. In 2.1 we drop the assumpton g = sl 2, but we uphold t otherwse For λ P, we consder M(λ), the Verma module wth hghest weght λ. Ifm λ s a hghest weght vector of M(λ), then M(λ) = U q (g)m λ = U q (g)m λ. Snce U q (g) s a B q(g)- module, M(λ) has a natural B q (g)-structure va composton of Q(q)-algebra homomorphsms B q (g) End ( U q (g)) Θ End ( M(λ) ). Namely, f ϕ λ : Uq (g) M(λ) s the U q (g)-module somorphsm sendng 1 to m λ, then Θ(g) = ϕ λ gϕλ 1 for g End(Uq (g)). Hence, f f,e End(U q (g)), 1 l, are defned as n 1.7, then f,e End(M(λ)) are gven as follows: f um λ = Θ(f )(um λ ) = ϕ λ f (u) = f um λ and e um λ = Θ(e )(um λ) = ϕ λ e (u) = e (u)m λ for u Uq (g). Usng the same symbols for f and e n both cases should create no confuson. Remark. The followng conclusons are evdent. (1) e m λ = 0 for each. (2) f,e End(M(λ)) do not depend on the choce of a hghest weght vector. (3) ϕ λ : Uq (g) M(λ) s also a B q(g)-module somorphsm. (4) M(λ) O(B q (g)). (5) M(λ) s rreducble as a B q (g)-module. We emphasze Remark 2.1(5) at ths pont. For example, n sl 2 -case, M( n 2) s a submodule of M(n) (n Z + ) wth respect to the U q -structure, but as B q -modules M( n 2) = M(n) Next, we study the B q -structure on T modules and therefore we consder g = sl 2.Let n Z +. We am to extend the B q -acton on M(n) embedded n T(n)to the whole of T(n). Let z be a U q -generator of T(n)of weght n 2 and let v be a hghest weght vector of the Verma submodule of T(n)wth hghest weght n. Consder the U q -decomposton T(n)= Uq v U q z. (2.1)

8 118 D. Jakelć / Journal of Algebra 312 (2007) We defne f,e End(T (n)) usng (2.1) and f,e End(U q ) as follows. For u U q,set f uv = fuv, f uz = fuz, e uv = e (u)v, e uz = e (u)z. (2.2) The next asserton s easly seen. Lemma. Equatons (2.2) defne a B q -module structure on T(n). Remark. (1) Ker e = Q(q)v Q(q)z (cf. Remark 1.7). (2) T(n) O(B q ). We call the above standard B q -structures on Verma and T modules We now consder the category I and collect together desrable propertes that any B q - structure must have wth respect to the U q -structure so that a synchronzaton of the two would be plausble. Let M be n I and denote by M c the generalzed c-egenspace of the quantum Casmr element C on M. Snce M s U + q -fnte and C s central, then M = r Z Mc r where c r = qr+1 + q r 1 (q q 1 ) 2. Clearly, c r = c r 2. Defnton. Let Ĩ be the category wth objects all fntely generated U q -modules M n the category I whch are also equpped wth a B q -module structure satsfyng the followng condtons: (a) f B q acts the same as f U q. (b) If e m = 0 for a weght vector m M \ f M of weght n Z, then e m = 0. (c) If e m 0 for a weght vector m M c n \ f M of weght n 2forn Z +, then there exst a Q(q)-subspace T of M such that (1) T s both U q and B q -submodule of M, (2) T = T(n) both as U q and B q -module where T(n) s endowed wth standard B q - structure, (3) e m = e m for some U q -generator m of T of weght n 2. Morphsms are defned to be Q(q)-lnear maps that are both U q and B q -morphsms. The followng lemma and proposton are mmedate. Lemma. Let M be n Ĩ, m M \ f M, and p 1. Then e f p m = q 2p f p e m + 1 q 2p 1 q 2 f p 1 m. By the above lemma, the acton of e on M s determned by ts acton on M \ f M. Proposton. () The modules M(r), r Z, and T(n), n Z +, wth the standard B q -structures belong to Ĩ. () Fnte drect sums of modules n Ĩ arealsonĩ.

9 D. Jakelć / Journal of Algebra 312 (2007) We next am to show: Theorem. Let M be n Ĩ. Then M = M (1) M (s) for some s>0, where: () M (j) s a U q and B q -submodule of M; () M (j) s U q -somorphc ether to M(r),r Z, ort(n), n Z + ; () M (j) s B q -somorphc ether to M(r), r Z,orT(n), n Z +, wth standard B q -structures. Proof. If follows from Theorem 1.4 and fnte generaton of M that M = N (1) N (p) N (s) where, as U q -modules, N (j) = M(rj ) for some r j Z for 1 j p, and N (j) = T(n j ) for some n j Z + for p + 1 j s. For 1 j p, letm j be a hghest weght vector of N (j). It s clear that m j / f M and e m j = 0. So by Defnton 2.3(b), e m j = 0. It then follows from Lemma 2.3 that N (j) s stable under e,.e., N (j) s a B q -submodule as well, and moreover t s equpped wth the standard B q -structure. Now consder N (p+1) whch s U q -somorphc to T(n p+1 ) and for smplcty let n = n p+1 and N = N (p+1).letz be a U q -generator of N of weght n 2 and let v be a hghest weght vector of the Verma submodule of N wth hghest weght n. Note that z/ f M, e z 0, and z M c n. So by Defnton 2.3(c), there exst a Q(q)-subspace T of M whch s both U q and B q -somorphc to T(n)wth standard B q -acton and a generator z of T such that e z = e z. Set w = z z. Then e w = 0 and w s of weght n 2. Let R = s j=1,j p+1 N (j). Then M = R N. Hence w = w R + w N for some w R R, w N N, and e w R = 0 = e w N. Snce w N s of weght n 2, t s a lnear combnaton of z and f n+1 v and snce e z 0, then w N = αf n+1 v for some α Q(q). Consder z 0 = z w N = z αf n+1 v. Then, z 0 also generates N and z 0 = w R + z. Therefore N R + T and so M = R + T. It s easly seen by weght consderaton that the U q -submodule R T of T cannot be somorphc nether to T nor to Verma submodules of T snce the correspondng weght spaces of T and N have the same dmenson and R + T = R N. Hence R T ={0}, and so M = R T as U q -modules. Therefore, we can replace N (p+1) n the drect sum we started wth by T. Denote T by T (p+1). Now we replace n the same way one by one the U q -submodules N (p+2),...,n (s) by U q -submodules T (p+2),...,t (s) wth standard B q -module structure such that M = N (1) N (p) T (p+1) T (s). Ths proves the theorem. Corollary. Let M be n Ĩ. Then: () Ker e = r Z (Ker e M r ); () Ker e has a Q(q)-bass consstng of weght vectors of the form f k u where k Z + and u Ker e. 3. Completons of crystal bases of modules n the category Ĩ Now we consder crystal bases n the sense of Defnton 1.8 for modules n Ĩ wth any B q - structure. Snce every module n Ĩ has completon belongng to Ĩ, as well as a crystal bass, t s natural to examne the nterplay between these two concepts. We am to defne a noton of completon of crystal bases of modules n Ĩ that wll be compatble wth the noton of completon of modules.

10 120 D. Jakelć / Journal of Algebra 312 (2007) The frst two subsectons are wrtten for any smple Le algebra g.letm, n Z +. Recall (cf. [8]), f (n) f (m) = [ ] n + m f (n+m) n, (3.1) [n] q n+1 (1 + qa) for n 0, (3.2) [n]! q n(n 1) 2 (1 + qa), (3.3) [ m n ] q n(m n) (1 + qa) (3.4) where [ m n ] = [m] [m 1]...[m n + 1] [1] [2]...[n] for n>0 and [ ] m = 1. 0 Denote by A the unts n A. The next lemma s mmedate and t holds for crystal lattces both n the sense of Defnton 1.8 and n the sense of Defnton 1.5. Lemma. Let L be a crystal lattce of a module M and let L = q r al for some r Z and a A. Then L s also a crystal lattce of M. Ifr>0 then L L, whle f r<0 then L L. Ifr = 0 then L = L The followng proposton s an analogue of a statement for crystal bases of fntedmensonal modules,.e., n the sense of Defnton 1.5. We gve a proof for reader s convenence. Proposton. Let M and M be n O(B q (g)) and let ϕ : M M be a B q (g)-module somorphsm. If (L, B) s a crystal bass of M, then (ϕ(l), ϕ(b)) s a crystal bass of M, where ϕ s the nduced map L/qL ϕ(l)/qϕ(l). Proof. Consder the decompostons of both M and M as n (1.4) and ẽ and f defned as n (1.5). It s plan that Ker e M = ϕ(ker e M) and both ẽ and f commute wth ϕ. Thus, ϕ(l) s a crystal lattce of M. The restrcton ϕ L : L ϕ(l) s an A-module somorphsm, and so the nduced map ϕ : L/qL ϕ(l)/qϕ(l) s a Q-vector space somorphsm. Hence ϕ(b) s a Q-bass of ϕ(l)/qϕ(l). Clearly both ẽ and f commute wth ϕ. For λ P,letm λ be a hghest weght vector of the Verma module M(λ) and let ϕ λ : Uq (g) M(λ) be a B q (g)-module somorphsm such that ϕ λ (u) = um λ.letl (λ) be the A-span of { f 1... f p m λ 1 j l; p Z + } and B (λ) ={ f 1... f p m λ 1 j l; p Z + }, where m λ = m λ + ql (λ). Snce (L( ), B( )) s a crystal bass of Uq (g), t mmedately follows: Corollary. (L (λ),b (λ) ) s a crystal bass of the Verma module M(λ). Every crystal bass of M(λ) s a nonzero scalar multple of (L (λ),b (λ) ).

11 D. Jakelć / Journal of Algebra 312 (2007) Example. If g = sl 2 and r Z, then any crystal bass of the Verma module M(r) s of the form (L (r),b (r) ) where L (r) = k Z + Af (k) m r and B (r) ={f (k) m r k Z + } for a hghest weght vector m r From ths pont on, we focus exclusvely on the U q (sl 2 )-case. We consder T modules. Recall that (L( ) 2,B( ) 2 ) s a crystal bass of Uq U q, where L( ) 2 = L( ) L( ), B( ) 2 = (B( ) 0) (0 B( )) L( ) 2 /ql( ) 2, L( ) = k Z + Af (k) 1 and B( ) ={f (k) 1 k Z + } where 1 = 1 + ql( ). The next corollary follows mmedately from consderng the map ψ n : Uq U q T(n) gven by (u 1,u 2 ) u 1 v + u 2 z whch s a B q -module somorphsm where Uq U q s equpped wth the obvous B q -structure. Corollary. Let z be a U q -generator of T(n) of weght n 2 and let v be a hghest weght vector of Verma submodule of T(n)wth hghest weght n. Set ( ) ( ) L {n} = Af (k) v Af (k) z k Z + k Z + and B {n} = { f (k) v mod ql {n} k Z +} { f (k) z mod ql {n} k Z +}. Then (L {n},b {n} ) s a crystal bass of T(n). Moreover, f (L, B) s any crystal bass of T(n), then there exst z and v as above such that (L, B) s of the form (L {n},b {n} ). The followng theorem now easly follows from Theorem 2.4, Corollares 3.2 and 3.3. Theorem. Let M be a module n the category Ĩ. Then M belongs to the category O(B q ) and there exst unque ntegers r 1,...,r m and unque nonnegatve ntegers n 1,...,n t such that M has a crystal bass (L, B) whch s a drect sum of crystal bases of the form (L (r ),B (r ) ),for = 1,...,m, and crystal bases of the form (L {n j },B {n j } ),forj = 1,...,t. Furthermore, any crystal bass of M s somorphc to (L, B). Defnton. We call the crystal bass (L, B) from the prevous theorem a standard crystal bass of M. Evdently, the ntegers r 1,...,r m and n 1,...,n t depend only on M and not on (L, B). Every crystal bass of an ndecomposable module n Ĩ s obvously a standard crystal bass For M n Ĩ, recall the decomposton M = k 0 f (k) Ker e (see 1.8). By Corollary 2.4, Ker e = n Z (Ker e M n ). Thus we brng the weght spaces of M n the pcture and have: M = k 0 n Z f (k) (Ker e M n ). (3.5) For a crystal lattce L of M, setl n = L M n. Then L = n Z L n.also,setl e ={m L em = 0}=L M e and L e n = Le M n = L M e n.

12 122 D. Jakelć / Journal of Algebra 312 (2007) Lemma. Let M be n Ĩ and let L be a crystal lattce of M. Then f n+1 (M e n ) Me n 2 and f n+1 (L e n ) Le n 2 for n Z+. Proof. Let m M e n. By Corollary 2.4, m = k 0 f (k) u k where f (k) u k M e and u k Ker e M n+2k. Thus f n+1 m = k 0 f n+1 f (k) u k = k 0 f (n+1+k) u k M n 2. Furthermore, utlzng (3.1) and commutaton relatons, e f n+1 f (k) u k = 1 [ n+k+1 n+1 ]ef (n+1) f (k) u k = 1 [ n+k+1 n+1 ] (f (n+1) e + f (n) q n t q n t 1 q q 1 ) f (k) u k = 0 for all k 0. Therefore f n+1 m M e, and so f n+1 (Mn e) Me n 2. In addton, snce f(l) L, then f n+1 (L e n ) Le n 2. Defnton. Let L be a crystal lattce of a module M n Ĩ. WesayL s complete f f n+1 : L e n L e n 2 s bjectve for all n Z+. A crystal lattce L of the completon C(M) of M s sad to be a completon of a crystal lattce L of M provded () L e L and L L; () L M = L L; () L/( L L) s a crystal lattce of C(M)/M (to be precse, ts mage under L/( L L) C(M)/M). Remark. () Unlke the case of modules where C(M) contans M, one cannot expect L to contan L. If we consder crystal lattces of a reducble Verma module and ts rreducble Verma U q -submodule, denoted by L 1 and L 2 respectvely, although L 1 s somorphc to L 2, when talkng about completons we are nterested n ther exact relatonshp and ndeed L 2 L 1. Moreover, we observe that there s no p Z such that q p L 2 L 1. In fact, the actons of the Kashwara operators ẽ and f on L 1 and L 2 are dfferent,.e., ẽ M (ẽ C(M) ) M and f M ( f C(M) ) M. () Condton () n Defnton 3.4 s a very natural connecton to expect between crystal lattces and completons. Also, we note that C(M)/M s a fnte-dmensonal U q -module, thus condton () gves a connecton between crystal lattces arsng from the B q -structures,.e., n the sense of Defnton 1.8, wth crystal lattces arsng from the U q -structures,.e., n the sense of Defnton 1.5. () The condton L L,.e., L s not properly contaned n L, follows from condton () n the case that L s a crystal lattce of a module M that s not complete. Proposton. Let L be a crystal lattce of a module M n Ĩ and let L be a completon of L. Then L e n 2 = Le n 2 for all n Z+. Proof. Let n Z +. We note that snce C(M)/M s a fnte-dmensonal U q -module, then (C(M)/M) e n 2 = 0, and so C(M)e n 2 = Me n 2. Now, by Defnton 3.4(), L e n 2 = L C(M) e n 2 = L M n 2 e = ( L M) M n 2 e = ( L L) M n 2 e = L e n 2 Le n 2. Hence, L e n 2 Le n 2. The clam now follows from Defnton 3.4().

13 D. Jakelć / Journal of Algebra 312 (2007) Let (L, B) be a crystal bass of M. Set B n = L n /ql n B where L n /ql n = {m mod ql m L n }. Then B = n Z B n.alsosetb e n = (L n/ql n ) e B where (L n /ql n ) e = {m mod ql m L e n }. Lemma. Let M be n Ĩ and let (L, B) be a crystal bass of M. Then: () f n+1 (B e n ) Be n 2 for n Z+ ; () If L s complete, then f n+1 : B e n Be n 2 s bjectve for all n Z+. Proof. Part () of the lemma follows from Lemma 3.4 and from B beng nvarant under f.its easly seen from Corollary 2.4 that Bn e s a bass of (L n/ql n ) e, thus mplyng part (). be the nduced Q-epmorphsm from the canoncal A-ep- Let ϕ : L/q L L/(L L) q( L/(L L)) morphsm ϕ : L L/(L L). Defnton. A crystal bass (L, B) of M n Ĩ s sad to be complete f the crystal lattce L s complete. A crystal bass ( L, B) of C(M) s a completon of a crystal bass (L, B) f (1) L s a completon of L, (2) ( L/(L L), ϕ( B)) s a crystal bass of C(M)/M. The next theorem verfes that the above defnton of completeness for crystal bases s compatble wth the noton of completeness for modules. Theorem. Let (L, B) be a crystal bass of a module M n Ĩ. Then (L, B) s complete f and only f M s complete. Proof. Let M be complete and let n Z +. Then f n+1 : Mn e Me n 2 s bjectve. Snce M s n Ĩ, ts weght spaces are fnte-dmensonal. Hence, dm Q(q) M e n = dm Q(q) M e n 2. (3.6) L s a crystal lattce of M, so there s an A-bass of L that s also a Q(q)-bass of M. Ths bass s made up of weght vectors of the form f (k) u where k 0 and u Ker e. If a nonzero sum of such vectors s annhlated by e, then also each of them s (see Theorem 2.4 and Corollary 2.4). Thus, a Q(q)-bass of Mn e s an A-bass of Le n. By (3.6), rank A L e n = rank A L e n 2. Now, snce A has the nvarant dmenson property and snce ẽ f = 1onM, then f n+1 : L e n Le n 2 s bjectve. The converse s true snce f s njectve on M, and the above steps can be reversed for afxedn. Example. The modules T(n) for n Z + and M(n) for n 1 are complete, so ther crystal bases defned n 3.3 and 3.2, respectvely, are complete by the theorem. Ths can also be seen drectly from the Defntons 3.4 and 3.5.

14 124 D. Jakelć / Journal of Algebra 312 (2007) Corollary. A complete crystal bass (L, B) of a module M n Ĩ s a completon of tself. Furthermore, L s the only completon of tself. Proof. By the prevous theorem, (L, B) beng complete mples M = C(M). It s obvous that a crystal bass of a complete module satsfes Defntons 3.4 and 3.5 for beng a completon of tself. Now assume L s another completon of L. Then L = L C(M) = L M = L L, thus L L. The desred concluson follows from the condton L L The rest of the secton s devoted to the proof of the followng theorem. Theorem. A standard crystal bass (L, B) of any module M n Ĩ has a completon. Moreover, there exsts a unque standard crystal lattce whch s a completon of L. By Theorem 3.3, every module M n category Ĩ has a standard crystal bass, thus t has a crystal bass that can be completed as n the above theorem. We frst prove: Proposton. A crystal bass (L, B) of any ndecomposable module M n Ĩ has a completon. Furthermore, L has a unque completon. Proof. By Theorem 3.5 and Corollary 3.5, n order to show exstence, we need to consder only Verma modules n Ĩ whch are not complete. Let M be a Verma module wth hghest weght n 2forsomen Z + and let L be ts crystal lattce. Then L = k 0 Af (k) m 0 for some hghest weght vector m 0 of M and L s clearly not complete. Snce M has a completon C(M) and f n+1 : C(M) e n C(M)e n 2 = Me n 2 s bjectve, there exsts m n C(M)e n such that f (n+1) m = m 0.Let L = k 0 Af (k) m and B ={f (k) m + q L k Z + }. Then ( L, B) s a crystal bass of C(M) and we clam t s a completon of (L, B). It follows from Secton 3.1 that L = Af (k) f (n+1) m = [ n k A k k 0 k 0 ] f (n+1+k) m = k 0 Aq k(n+1) f (n+1+k) m. The last equalty s a consequence of the elements of 1 + qa beng unts n A. We have L e = Am 0 L, L L = k 0 Af (n+1+k) m, and L/( L L) = n k=1 Af (k) m where m = m + q( L L). Thus the condtons () () of Defnton 3.4 are satsfed. It s now easy to check condton (2) of Defnton 3.5. Unqueness follows from parts () and () of Defnton 3.4. Remark. () Consderng the lattce L from the proof of prevous proposton, notce that L M = k 0 Af (n+1+k) m whch s a proper A-submodule of the crystal lattce L = k 0 Aq k(n+1) f (n+1+k) m of M,.e., n order to complete L to a crystal lattce of C(M) that s nvarant under Kashwara operators ẽ C(M) and f C(M), L has to be made thnner as an A-lattce n a partcular way.

15 D. Jakelć / Journal of Algebra 312 (2007) () Usng the proof of the prevous proposton, we can also see the followng. Let n Z + and let M = M(n) be the Verma module wth hghest weght n and M ts unque Verma submodule. Let L be any crystal lattce of M and L any crystal lattce of M. Then, there exsts r Z + such that L/(L q r L ) s somorphc to a crystal lattce of the rreducble U q -module M/M Proposton. Drect sum of completons of crystal bases s a completon of drect sum of crystal bases. Proof. For = 1, 2, let (L,B ) be a crystal bass of M and ( L, B ) ts completon. We clam that ( L 1, B 1 ) ( L 2, B 2 ) s a completon of (L 1,B 1 ) (L 2,B 2 ). Snce both crystal bases and completons of modules n Ĩ respect drect sums, ( L 1, B 1 ) ( L 2, B 2 ) = ( L 1 L 2, B 1 B 2 ) s a crystal bass of C(M 1 ) C(M 2 ) = C(M 1 M 2 ). Condton () of Defnton 3.4 s evdent snce (L 1 L 2 ) e = L e 1 Le 2 L 1 L 2. Furthermore, L 1 L 2 L 1 L 2. By defnton, L M = L L. Clearly, for j, M L j = 0. Therefore, ( L 1 L 2 ) (M 1 M 2 ) = ( L 1 L 2 ) (L 1 L 2 ), so condton () s satsfed, as well. Next, ( L 1 L 2 )/ ( ( L 1 L 2 ) (L 1 L 2 ) ) = ( L 1 L 2 )/ ( ( L 1 L 1 ) ( L 2 L 2 ) ) On the other hand, = L 1 /( L 1 L 1 ) L 2 \ ( L 2 L 2 ). C(M 1 M 2 )/(M 1 M 2 ) = ( C(M 1 ) C(M 2 ) ) /(M 1 M 2 ) = C(M 1 )/M 1 C(M 2 )/M 2. Condton () now follows from crystal lattces respectng drect sums. Furthermore, t s easy to see that ϕ( B 1 B 2 ) = ϕ( B 1 ) ϕ( B 2 ) provng condton (2) of Defnton 3.5. Lemma. Let M be a module n Ĩ and suppose L,L, = 1,...,k, are crystal lattces of ndecomposable submodules of M such that L 1 L 2 L k = L 1 L 2 L k.if L, L are the completons of L,L, respectvely, then L 1 L 2 L k = L 1 L 2 L k. Proof. It suffces to consder the case when M s a generalzed egenspace of the Casmr element wth egenvalue c n. If all of the L are complete, then so s L 1 L 2 L k and we are done usng Proposton 3.7 and Corollary 3.5. To smplfy the notaton we consder only the case when k = 2, L 1 s a crystal lattce of a Verma submodule of M wth hghest weght n, and L 2 s a crystal lattce of a Verma submodule of M of hghest weght n 2. The proof of the general case s smlar, but wth more nvolved notaton. It s clear that ether L 1 or L 2 s a crystal lattce of a Verma submodule of M wth hghest weght n, so we assume t s L 1. Furthermore, t s then obvous from the condton L 1 L 2 = L 1 L 2 that we must have L 1 = L 1 = L 1 = L 1.Letv Ker e M n and u, u Ker e M n 2 be such that L 1,L 2,L 2 are the A-spans of f () v,f () u, f () u ( 0), respectvely. Clearly u = af (n+1) v + bu where a,b A. Snce u L 1 L 2, then b A.Letm, m be the unque elements n C(M) such that f (n+1) m = u and f (n+1) m = u. Then L 2 = 0 Af () m and L 2 = 0 Af () m by the

16 126 D. Jakelć / Journal of Algebra 312 (2007) proof of Proposton 3.6. Also, f (n+1) m = f (n+1) (av + bm). Snce M s Uq -torson free, then m = av + bm L 1 L 2 and m = b 1 (m av) L 2 L 1 provng the lemma n ths case It s convenent to have the followng defnton. Defnton. Let M 1, M 2 be modules n Ĩ and let (L 1,B 1 ), (L 2,B 2 ) be crystal bases of M 1,M 2, respectvely. We say that (L 1,B 1 ) s strongly somorphc to (L 2,B 2 ) f there exsts a B q - somorphsm ϕ : M 1 M 2 such that () ϕ nduces an somorphsm of the crystal bases (L 1,B 1 ) and (L 2,B 2 ); () ϕ s weght-preservng; () ϕ(ker e M1 ) = Ker e M2. For example, any two standard crystal bass of a module n Ĩ are strongly somorphc. Proof of Theorem 3.6. By Defnton 3.3, a standard crystal bass s a drect sum of crystal bases of the form (L (r),b (r) ),for = 1,...,m, and crystal bases of the form (L {n j },B {n j } ), for j = 1,...,t, where the ntegers r and n j depend on M,.e., these are the hghest weghts of Verma and T modules present n a decomposton of M. The frst clam of the theorem now follows from Propostons 3.6 and 3.7. We now prove the unqueness of completon of L. Snce L s a standard crystal lattce of M, then L = L 1 L k for some crystal lattces L, = 1,...,k, of ndecomposable U q -submodules M of M such that M = M 1 M k.if L denotes the completon of L, then L 1 L k s a completon of L. Let L denote another standard completon of L. Then L s a crystal lattce of C(M) = C(M 1 ) C(M k ), thus t s a drect sum of standard crystal lattces L, = 1,...,k,of ndecomposable submodules M of C(M) such that C(M) = M 1 M k and such that L s strongly somorphc to L, = 1,...,k (up to reorderng). Snce L,L 1,...,L k are standard crystal lattces and L 1 L k s a completon of L, t s easly seen that there exst crystal lattces L, = 1,...,k, strongly somorphc to L, = 1,...,k, respectvely, such that L s a completon of L and L = L 1 L k. The concluson now follows from the prevous lemma. 4. Constructons of completons of crystal lattces We now construct a completon of a crystal lattce modfyng Deodhar s constructon of completon for modules n Ĩ. By Theorems 2.4 and 3.5 and Proposton 3.7, t suffces to consder crystal bases of Verma modules M( n 2) for n Z We start by recallng Deodhar s constructon for U q (sl 2 ) (cf. [2,15]). Denote by A(U q ) the category of U q -weght modules that are U q -torson free. For M n A(U q), lets M be the set of formal symbols S M ={f r m r Z +,m M}. Defne an equvalence relaton on S M by f r m f k m ff f k m = f r m. Set D(M) = S M /. Then D(M) has a vector space structure. For z U q and r Z +, there exst u U q and s Z + such that f s z = uf r.now,for f r m D(M) and z U q, defne z f r m = f s um. Ths acton makes D(M) a U q -module. Any nonzero element v D(M) can be unquely expressed as f n m(n Z +,m M) where n s mnmal wth respect to ths property. Ths expresson s called the mnmal expresson for v.

17 D. Jakelć / Journal of Algebra 312 (2007) Clearly, f n m s a mnmal expresson ff m/ fm.also,v M ff n = 0 n the mnmal expresson for v. Set C(M) ={v D(M) e k v = 0forsomek>0}. Theorem. (Cf. [2,15].) Let M be a module n the category I. () C(M) s a completon of M. () Let v D(M) \ M be a weght vector such that tv = q a v for some a Z and let v = f n m be the mnmal expresson for v. Then v C(M) ff the followng two condtons are satsfed: (a) a = n j for some j>0, and (b) e k f k j m = 0 for some k j. The next corollary gves bass vectors for the completon C(M) of an rreducble Verma module M. Corollary. Let M be the Verma module wth hghest weght n 2 for n Z + and let m 0 be a hghest weght vector of M. Then the completon of M s C(M) = k n 1 Q(q)f k m 0. Proof. Applyng Deodhar s constructon on M = k 0 Q(q)f k m 0, we obtan D(M) = k Z Q(q)f k m 0 because f s njectve and f r f k m 0 = f k r m 0 for r, k Z +. Snce M C(M), we need to consder only f k m 0 for k>0. By Theorem 4.1(), f k m 0 C(M) ff (a) n 2 + 2k = k j for some j>0, and (b) e p f p j m 0 = 0forsomep j. Thss equvalent to k<n+ 2(p = j = n + 2 k) It s convenent to ntroduce the followng formal notaton. Set [ k]! = [ 1]...[ k] and f ( k) m = f k m for k>0 and m M. Then [ k]! = ( 1) k [k]! and f (k) m = f k m for [ k]! [k]! k Z and m M. For, k Z + and m M, f () f ( k) m = f []! f k m [ k]! = f k m [ k]! = []![ k]! []![ k]! f ( k) m. (4.1) Thus, C(M) = Q(q)f (k) m 0 = Q(q)f () f ( n 1) m 0, k n 1 0 where we follow the notaton of Corollary We now construct drectly from the crystal lattce of an rreducble Verma module a crystal lattce of ts completon. Lemma. 1 (a) If n, then 1 = 1 + q ( n 1) a for some a A. (b) If >n, then q n 1 1 = 1 + q ( n 1) a for some a A.

18 128 D. Jakelć / Journal of Algebra 312 (2007) Proof. (a) For n, 1+ q +n+1 A. 1 q n 1 (b) For >n, = 1 + q ( n 1) q n 1 + 1,soleta = 1 + q n 1 A. Proposton. Let M be the Verma module wth hghest weght n 2 for some n Z + and let m 0 be a hghest weght vector of M. Then 2q n( n 1) () the A-module L generated by { 1 + q ( n 1) f ( n 1) m 0 0} s a crystal lattce of C(M); () L s a completon of the crystal lattce L = k 0 Af (k) m 0 of M. Proof. (1) By Corollary 4.1, C(M) s a Verma module wth a hghest weght vector f ( n 1) m 0. Thus, a crystal lattce of C(M) s of the form L = 0 Ar(q)f() f ( n 1) m 0 for r(q) Q(q). Utlzng (3.3), we have the followng: f () f ( n 1) n+1 [ n 1]! m 0 = ( 1) []![n + 1]! f ( n 1) m 0 Case 1. n + 1. = ( 1) n+1 [ n 1]! f ( n 1) m q 1 0 for some a A. 2 [( 1)+n(n+1)] (1 + aq) f () f ( n 1) m 0 = ( 1) n+1 q 1 2 ( n 2)( n 1) (1 + a q) f ( n 1) m q [( 1)+n(n+1)] (1 + aq) = a q ( 1)(n+1) f ( n 1) m 0 for some a A. for some a A Case 2. 0 n. f () f ( n 1) m 0 = ( 1) n+1 ( 1) n+1 q 1 2 (n )(n+1 ) (1 + a q),f ( n 1) m q 1 0 for some a A 2 [( 1)+n(n+1)] (1 + aq) = a q n f ( n 1) m 0 for some a A. (4.2) Hence, for r(q) = q n(n+1), we have by Lemmas 3.1 and 4.3 L = 0 Aq n(n+1) f () f ( n 1) m 0 = ( n =0 ) ( Aq n( n 1) f ( n 1) m 0 qn( n 1) n+1 Aq (n+1)( n 1) f ( n 1) m 0 ) 2qn( n 1) = A 1 + q ( n 1) f ( n 1) m 0 = A 1 + q ( n 1) f ( n 1) m

19 D. Jakelć / Journal of Algebra 312 (2007) () Evdently, L e = Am 0 L, L L, and L M = j 0 Aqj(n+1) f (j) m 0 = L L. Next, L/( L L) = = n Aq n(n+1 ) f ( n 1+) m 0 =0 n Aq n(n+1) f () f ( n 1) m 0 by (4.2). =0 where m 0 = m 0 + L L Snce q n(n+1) f ( n 1) m 0 (n fact, ts mage under L/( L L) C(M)/M) s a hghest weght vector of C(M)/M, then L/( L L) s a crystal lattce of C(M)/M. Thus, L s a completon of L. Remark. () The above proposton gves the bass vectors of a crystal lattce of C(M) naclosedform. However, f we remove the A -multples from the gven bass vectors of L, they come down to { q n( n 1) f ( n 1) m 0 for 0 n, q (n+1)( n 1) f ( n 1) (4.3) m 0 for >n, and they generate the same crystal lattce L. () If we start wth the lattce from Proposton 4.3 and apply f (n+1) to ts hghest weght vector, we get: In other words, m 0 L. ( 2q f (n+1) n(n+1) ) 1 + q n+1 f ( n 1) m 0 = 2q n(n+1) [0]! 1 + q n+1 [n + 1]![ n 1]! m 0 = a m 0 for some a A In the last subsecton, we show how an operator defned by Kashwara n [7] can be appled to a certan A-lattce n order to deform t to a crystal lattce of the completon. We consder the A-lattces L n M = M( n 2) and L n C(M) where L = k 0 Af (k) m 0 and L = k n 1 Af (k) m 0. Note that L s a crystal lattce of M,butL s not ts completon. Actually, L s not even a crystal lattce snce ẽ(l ) L where ẽ denotes the Kashwara operator on C(M). However, we wll drectly transform L to a completon of L. Let Δ = qt + q 1 t 1 + (q q 1 ) 2 fe 2 be a central element of U q and consder the acton of qtδ on C(M). Fork>0, f k Δ = Δf k, thus Δf k m 0 = f k Δm 0. Hence Δf (k) m 0 = f (k) Δm 0 = (q n 1 + q n+1 2)f (k) m 0 for k n 1. Now, qtδf (k) m 0 = (q n 1 + q n+1 2)q n 1 2k f (k) m 0 =[(q n 1 1)q k ] 2 f (k) m 0. We defne (qtδ) 1 2 End(C(M)) as follows: (qtδ) 1 2 f (k) m 0 = q k (q n 1 1)f (k) m 0 for k n 1 (cf. [7]). Thus, (qtδ) 1 2 End(C(M)) s defned by (qtδ) 1 2 f (k) m 0 = q k( q n 1 1 ) 1 f (k) m 0 (4.4)

20 130 D. Jakelć / Journal of Algebra 312 (2007) for k n 1. Let S n End(C(M)) be gven by S n = { q n 1 (qtδ) 1 2, on C(M) l for l n 2, d, on C(M) l for l> n 2. (4.5) Proposton. Let L = k 0 Af (k) m 0 be a crystal lattce of the Verma module M wth hghest weght n 2, and let L = k n 1 Af (k) m 0. Then q n(n+1) S n (qtδ) n 2 L s a completon of L. Proof. We note that q n 1 1 = aq n 1 where a = 1 q n+1 A. It follows from (4.4) that (qtδ) n 2 L = k n 1 Aqnk (q n 1 ) n f (k) m 0 = k n 1 Aqn(k+n+1) f (k) m 0. Also, for k 0, Therefore, S n f (k) m 0 = q n 1 q k( q n 1 1 ) 1 f (k) m 0 = q k a 1 f (k) m 0. (4.6) q n(n+1) S n (qtδ) n 2 L = S n ( k n 1 Aq nk f (k) m 0 ) ( ) ( = Aq (n+1)k f (k) m 0 k 0 n 1 k<0 Aq nk f (k) m 0 ). Hence, q n(n+1) S n (qtδ) n 2 L s the A-lattce generated by the same vectors as n (4.3). Therefore, t s equal to the crystal lattce L from the Proposton 4.3 whch s a completon of L. Acknowledgments My sncere thanks are due to V. Deodhar for many nsprng and useful dscussons, and n partcular for hs contrbuton to Secton 2, and to A. Moura for readng through the manuscrpt and for valuable suggestons. The referee s comments led to a clearer exposton and are apprecated. I am also pleased to thank the Max-Planck-Insttut für Mathematk n Bonn for ts hosptalty. References [1] V. Char, D. Jakelć, A. Moura, Branched crystals and the category O, J. Algebra 294 (2005) [2] V. Deodhar, On a constructon of representatons and a problem of Enrght, Invent. Math. 57 (1980) [3] T.J. Enrght, On the fundamental seres of a real semsmple Le algebra: Ther rreducblty, resolutons and multplcty formulae, Ann. of Math. 110 (1979) [4] D. Jakelć, Structure of T modules and restrcted duals: The classcal and the quantum case, Comm. Algebra 31 (11) (2003) [5] A. Joseph, The Enrght functor on the Bernsten Gelfand Gelfand category O, Invent. Math. 67 (1982) [6] S.-J. Kang, M. Kashwara, K. Msra, Crystal bases of Verma modules for quantum affne Le algebras, Compos. Math. 92 (1994) [7] M. Kashwara, Crystalzng the q-analogue of unversal envelopng algebras, Comm. Math. Phys. 133 (1990) [8] M. Kashwara, On crystal bases of the q-analogue of unversal envelopng algebras, Duke Math. J. 63 (2) (1991)

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