Partition models for the crystal of the basic U q ( sl ^ n ) -module
aa r X i v : . [ m a t h . C O ] F e b Partition models for the crystal of the basic U q ( b sl n )-module Matthew Fayers
Queen Mary, University of London, Mile End Road, London E1 4NS, U.K. [email protected]
Abstract
For each n >
3, we construct an uncountable family of models of the crystal of the basic U q ( b sl n )-module. These models are all based on partitions, and include the usual n -regular and n -restricted models, as well as Berg’s ladder crystal, as special cases. Let n >
2, and let b sl n denote the Kac–Moody algebra of type A (1) n − , and U q ( b sl n ) its quantisedenveloping algebra. A key object in the combinatorial representation theory of a quantised Kac–Moody algebra U is a crystal basis for a module M , and the associated crystal graph . When U = U q ( b sl n )and M is the irreducible highest-weight module with highest weight Λ , this crystal graph plays ar ˆole in the representation theory of Iwahori–Hecke algebras of type A at an n th root of unity: thereis a correspondence between the vertices of this crystal graph and simple modules for these Heckealgebras, with the arrows in the crystal graph corresponding to certain functors refining inductionand restriction between these algebras.There is particular interest in modelling crystal graphs, i.e. realising their vertex sets as sets ofwell-understood combinatorial objects. Of particular prominence in the case of b sl n -crystals is themodelling of crystals of highest-weight representations by multipartitions (where the number ofcomponents in each multipartition equals the level of the highest weight). In this paper we restrictattention to the better-understood case of level 1, where crystals are modelled by partitions. Thereare two well-established models of the highest-weight crystal with highest weight Λ , in termsof n -regular and n -restricted partitions. These models come directly from Fock space models for(modules containing) the corresponding highest-weight module V ( Λ ).More recently, Berg [B] found a realisation of B ( Λ ) in terms of a new class of partitions, inthe case where n >
3. Berg’s construction is purely combinatorial – his proof involves an explicitisomorphism of crystals between his crystal and the n -regular crystal – and it is unclear whetherthere is more algebraic structure behind the scenes.In this paper we extend Berg’s work, by showing that his crystal model is one of an uncountablefamily of models of the crystal B ( Λ ), for each n >
3. Again, our construction and proof are purely1 MatthewFayerscombinatorial. The proof is rather awkward – it is not clear how to give an explicit isomorphismfrom one of our crystals to a known model of B ( Λ ), and so we use a result of Stembridge, whichenables recognition of crystals of highest-weight modules for simply-laced algebras in terms oflocal properties. In fact, it is too di ffi cult for this author even to use Stembridge’s results directly,and we use an alternative route due to Danilov, Karzanov and Koshevoy. In some ways this proofis unsatisfactory; it would be nice to have an algebraic understanding – perhaps via an analogueof the Fock space – for why these models exist.We now indicate the layout of this paper. In the next section we give enough definitions toenable a statement of the main theorem. In Section 3, we give the necessary background informationon crystals. Section 4 gives a brief review of ± -sequences, which are at the heart of our crystalconstructions. Section 5 contains an abstract construction of crystals which turn out to be crystalsof highest-weight sl -modules; these crystals are used in Section 6 to complete the proof of themain theorem. Finally in Section 7 we show that the various models we have created for B ( Λ ) fora given value of n are distinct, so that we really do get an uncountable family. Acknowledgements.
The author’s thanks are due to Chris Berg, who made an early version of hispaper [B] available. The author enjoyed fruitful discussions with Chris (as well as Monica Vaziraniand Brant Jones) at MSRI Berkeley in March 2008, during the concurrent MSRI programmes‘Combinatorial representation theory’ and ‘Representation theory of finite groups and relatedtopics’. The author would like to thank the organisers of these programmes, as well as MSRI forsome financial support.This work was begun while the author was a visiting Postdoctoral Fellow at MassachusettsInstitute of Technology, with the aid of a Research Fellowship from the Royal Commission for theExhibition of 1851. The author is very grateful to M.I.T. for its hospitality, and the 1851 Commissionfor its generous support.
In this section we state our main theorem. In order to get to the result as quickly as possible, wegive only minimal definitions here; we assume the reader has some familiarity with representationsof Kac–Moody algebras and quantum groups. Further definitions are given in later sections. b sl n In this paper we fix an integer n >
3, and consider the Kac–Moody algebra b sl n . This has simpleroots indexed by the set I = Z / n Z , and the Cartan matrix is given by a ij = δ ij − δ i ( j + − δ i ( j − for i , j ∈ I . We may abuse notation and label elements of Z / n Z by the integers 0 , . . . , n −
1; of course,the subscripts in the above formula should then be read modulo n .The usual realisation of this Cartan matrix is given as follows. The Cartan subalgebra h is an( n + C with basis { h i | i ∈ Z / n Z } ∪ { D } . artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 3The dual space h ∗ has basis { Λ i | i ∈ Z / n Z } ∪ { δ } , and the weight lattice P is the Z -span of this basis. The two bases are paired via h h i , Λ j i = δ ij , h D , δ i = , h h i , δ i = h D , Λ j i = , and the simple roots α i ( i ∈ Z / n Z ) are given by α i = Λ i − Λ i − − Λ i + + δ i δ. In this paper, we shall be concerned with the highest-weight module V ( Λ ) for the quantumenveloping algebra U q ( b sl n ); as well as being a prototype for highest-weight modules of a ffi ne Kac–Moody algebras, this has important applications in the representation theory of Iwahori–Heckealgebras; for example, see [G]. As usual, a partition is a weakly decreasing sequence λ = ( λ , λ , . . . ) of non-negative integerssuch that the sum λ + λ + . . . is finite. Partitions are often written with equal parts groupedtogether and zeroes omitted, and the partition (0 , , . . . ) is written as ∅ . A partition λ is oftenidentified with its Young diagram , which is the set[ λ ] = n ( a , c ) ∈ N (cid:12)(cid:12)(cid:12) c λ a o . We adopt the English convention for drawing Young diagrams, in which a increases down the pageand c increases from left to right.We refer to elements of N as nodes , and elements of [ λ ] as nodes of λ . A node ( a , c ) of λ is removable if [ λ ] \ { ( a , c ) } is the Young diagram of a partition (i.e. if c = λ a > λ a + ), while a node ( a , c )not in λ is an addable node of λ if [ λ ] ∪ { ( a , c ) } is the Young diagram of a partition.If λ is a partition, the conjugate partition λ ′ is defined by λ ′ a = (cid:12)(cid:12)(cid:12) { c | λ c > a } (cid:12)(cid:12)(cid:12) . Now suppose we have fixed n >
2. We define the residue of a node ( a , c ) to be c − a + n Z . If ( a , c )has residue i ∈ Z / n Z , we refer to it as an i-node . B ( Λ ) Now we describe the two usual partition models for the basic crystal B ( Λ ) for U q ( b sl n ); seeSection 3 for an introduction to crystals.Say that a partition λ is n-restricted if λ a − λ a + < n for all a . We write Rest n for the setof n -restricted partitions, and we define crystal operators ˜ e i , ˜ f i for i ∈ Z / n Z as follows. Given λ ∈ Rest n , let ( a , c ) , . . . , ( a r , c r ) be the list of all addable and removable i -nodes of λ , ordered so that c < · · · < c r . Now write a sequence of signs π = π . . . π r , where π j equals + if ( a j , c j ) is an addablenode, and − otherwise. MatthewFayersIf there is no good position in this sequence (see § e i λ =
0. Otherwise, if position j is the good position, we define ˜ e i λ to be the partition obtainedfrom λ by removing the node ( a j , c j ). Similarly, if there is no cogood position in π , then we define˜ f i λ =
0, and otherwise we define ˜ f i λ to be the partition obtained by adding the node correspondingto the cogood position in π .We define a function wt on partitions bywt( λ ) = Λ − X i ∈ Z / n Z c i α i , where c i is the number of i -nodes of λ . Then we have the following. Theorem 2.1. [MM, Theorem 4.7]
The operators ˜ e i , ˜ f i and the weight function wt endow Rest n with thestructure of a crystal for b sl n . This crystal is isomorphic to B ( Λ ) . Now we describe the other well-known realisation of the basic U q ( b sl n )-crystal in terms ofpartitions. Say that a partition λ is n-regular if there is no a for which λ a = λ a + n − >
0; in otherwords, λ is n -regular if λ ′ is n -restricted. We may define crystal operators ˜ e i , ˜ f i on the set of n -regular partitions; this is done exactly as for n -restricted partitions above, except that we replace c < · · · < c r above with c > · · · > c r . Then we get a counterpart to Theorem 2.1. An explicitisomorphism between these two crystals may be given in terms of the Mullineux map [M].The object of the present paper is to describe, for each n >
3, an uncountable family of realisationsof the crystal B ( Λ ) in terms of partitions. The n -regular and n -restricted versions described abovearise as special cases, as does Berg’s ‘ladder crystal’ [B], which was the starting point for the presentwork. Since the first version of this paper was written, Tingley [T] has shown that another modelfrom the literature – Nakajima’s monomial crystal – occurs as a special case too.We now fix n >
3, and describe our crystals. The rest of the paper is devoted to proving thatthese crystals are all isomorphic to the crystal B ( Λ ).Each of our crystals will be indexed by an arm sequence , which is defined to be a sequence A = A , A , . . . of integers such that • t − A t ( n − t for all t >
1, and • A t + u ∈ { A t + A u , A t + A u + } for all t , u > λ is any partition and ( a , c ) is a node of λ , then the( a , c ) -hook of λ is the set of nodes of λ directly to the right of or directly below ( a , c ), including ( a , c )itself. The ( a , c ) -hook length of λ is the number of nodes in the ( a , c )-hook, i.e. the number λ a − c + λ ′ c − a + , while the ( a , c ) -arm length is just the number of nodes strictly to the right of ( a , c ), i.e. λ a − c .Now suppose A is an arm sequence. If λ is a partition, we say that λ is A-regular if there is nonode ( a , c ) of λ such that the ( a , c )-hook length of λ equals nt and the ( a , c )-arm length equals A t , forany t .We refer to a hook with length nt and arm length A t as an illegal hook ; so an A -regular partitionis one with no illegal hooks.artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 5 Examples.
1. Suppose A t = ( n − t for all t . Then a partition λ is A -regular if and only if it is n -restricted.To see this, suppose first that λ is not n -restricted. Then we have λ a − λ a + > n for some a , sothe ( a , λ a − n + λ has length n and arm length n −
1. Conversely, suppose that forsome node ( a , c ) of λ and t > a , c )-hook length of λ equals nt while the ( a , c )-arm lengthequals ( n − t . Then we have λ a = c + ( n − t and λ a + t < c , so λ a − λ a + t > ( n − t , and hencewe must have λ b − λ b + > n for some b ∈ { a , a + , . . . , a + t − } ; so λ is not n -restricted.2. Suppose A t = t for each t ; then a partition is A -regular if and only if it does not possess a hookwhose length is exactly n times its arm length. Hence by [B, Theorem 13.2.3] the A -regularpartitions are precisely the partitions appearing in Berg’s ladder crystal. This characterisationof these partitions in terms of illegal hooks was the starting point for the present work.Write R A for the set of A -regular partitions; this will be the underlying set of our crystal. Inorder to define the crystal operators ˜ e i , ˜ f i on R A , we need to introduce a total order on the set of allnodes of a given residue. If ( a , c ) and ( b , d ) are distinct nodes with the same residue, then the axialdistance b − a + c − d equals nt for some integer t ; by interchanging ( a , c ) and ( b , d ) if necessary, wesuppose t >
0. Now we set ( a , c ) ≻ ( b , d ) if c − d > A t , and ( b , d ) ≻ ( a , c ) otherwise; for this purpose,we read A as 0.It easy to check, using the definition of an arm sequence, that ≻ is transitive, and is thereforea total order on the set of nodes of a given residue. Now we can define our crystal operators.Suppose λ ∈ R A and i ∈ Z / n Z . Let ( a , c ) , . . . , ( a r , c r ) be the list of addable and removable i -nodesof λ , ordered so that ( a , c ) ≻ · · · ≻ ( a r , c r ). Define the sequence of signs π = π . . . π r by putting π j = + if ( a j , c j ) is an addable node, and π j = − otherwise. If π has no good position, then set˜ e i λ =
0. Otherwise, let j be the good position, and say that ( a j , c j ) is the i -good node for λ ; define˜ e i λ by removing this node from λ . Similarly, if there is no cogood position in π , then set ˜ f i λ = k be the cogood position and say that ( a k , c k ) is the i -cogood node for λ ; define ˜ f i λ by adding this node to λ .Now we can state our main theorem. Theorem 2.2.
Suppose n > and A is an arm sequence. Then the crystal operators ˜ e i , ˜ f i and the weightfunction wt endow R A with the structure of an b sl n -crystal. This crystal is isomorphic to B ( Λ ) . Example.
Suppose n =
3, and A satisfies A = A = A =
4. Part of the crystal R A is shownin Figure 1. In this section, we give the definitions and basic results we shall need concerning crystals. Anindispensable introduction to this subject is Kashiwara’s book [K2].
Suppose I is a finite set. We define an I-crystal to be a finite set B , equipped with functions˜ e i , ˜ f i : B −→ B ⊔ { } MatthewFayers ∅ ✚✚✚✚✚❃ ✚✚✚✚✚❃ ✲ ✚✚✚✚❃ ✲ ✚✚✚✚✚❃ ✲ ✚✚✚✚❃ ❩❩❩❩⑦ ❩❩❩❩⑦ ✚✚✚✚✚❃ ✲ ✚✚✚✚❃ ❩❩❩❩⑦ ✚✚✚❃ ✚✚✚✚✚❃ ❩❩❩❩⑦ ✚✚✚✚❃ ✲ ✚✚✚✚✚❃ ❩❩❩❩⑦ ✲ ✡✡✡✡✡✡✡✣ ✡✡✡✡✡✡✡✣ ✲ ✡✡✡✡✡✡✡✣ ❩❩❩⑦ ✚✚✚✚✚❃ ✚✚✚✚❃ ❩❩❩❩⑦ ✚✚✚❃ ✲ ✲ ✚✚✚✚✚❃ ❩❩❩❩❩⑦ ✲ ✲ ✚✚✚✚❃ ✲ ❩❩❩❩⑦ ✚✚✚✚❃ ❩❩❩❩⑦ ❩❩❩❩❩⑦ ❩❩❩❩❩⑦ ✚✚✚✚❃ ❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ ✚✚✚✚✚❃ ✲ ❏❏❏❏❏❏❏❫ ❏❏❏❏❏❏❏❫ ✲ ❏❏❏❏❏❏❫ ✚✚✚✚❃ ✚✚✚✚✚❃ ❩❩❩❩❩⑦ ❩❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ ✚✚✚✚✚❃ ❩❩❩❩⑦ ❩❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ ✚✚✚✚❃ ✚✚✚✚✚❃ ❩❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ ✲ ❩❩❩❩❩⑦ Figure 1artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 7for each i ∈ I , satisfying the following axioms.C1. If b , b ′ ∈ B , then b = ˜ e i b ′ ⇐⇒ b ′ = ˜ f i b . C2. If we set ǫ i ( b ) = max n ǫ (cid:12)(cid:12)(cid:12) ˜ e ǫ i ( b ) , o , φ i ( b ) = max n φ (cid:12)(cid:12)(cid:12) ˜ f φ i ( b ) , o , then ǫ i ( b ) , φ i ( b ) are finite.Here 0 is what Kashiwara calls a ‘ghost element’. In other words, ˜ e i and ˜ f i may be regarded asmutually inverse injective partial functions from B to itself. To an I -crystal is associated a crystalgraph : this has vertex set B , with an arrow labelled with i ∈ I from b to ˜ f i b whenever ˜ f i b ,
0. An I -crystal is often identified with its crystal graph, and accordingly one may speak of a crystal beingconnected, or talk of the connected components of a crystal. Of course, each connected componentis itself a crystal.Now let g be a symmetrisable Kac–Moody algebra. This is defined by a Cartan matrix A = ( a ij ) i , j ∈ I , where I is an indexing set, which we assume to be finite. For each i ∈ I , we let α i denotethe corresponding simple root, and h i the simple coroot. P denotes the lattice of integral weights.A g - crystal is defined to be an I -crystal, equipped with a functionwt : B −→ P such that the following additional axioms are satisfied for each i ∈ I .C3. If b ∈ B and ˜ e i b ,
0, then wt(˜ e i b ) = wt( b ) + α i . C4. For each b ∈ B , h h i , wt( b ) i = φ i ( b ) − ǫ i ( b ) . Remarks.
1. In fact, the definition we have given above is the definition of a semi-normal crystal; in general,a more liberal definition of ǫ i , φ i is permitted. Since in this paper we shall only be concernedwith semi-normal crystals, we use the term ‘crystal’ as defined above.2. Note that if the Cartan matrix is non-singular (in particular, if g is of finite type), then byaxiom C4 the functions ˜ e i , ˜ f i determine the weight function wt. Hence in this case we mayspecify a crystal simply by giving the functions ˜ e i , ˜ f i for each i ; the assertion that we have acrystal is then the assertion that these functions, together with the implied weight function,satisfy the axioms. Subcrystals
Given an I -crystal B and a subset J of I , one obtains a J -crystal B J simply by forgetting thefunctions ˜ e i , ˜ f i for i ∈ I \ J ; we call this a subcrystal of B . If B is a g -crystal with g having indexing set I , then B J is a g J -crystal, where g J is the subalgebra of g corresponding to J ; the weight function on B J is induced from the weight function on B .We shall be particularly interested in the case where | J | =
2; in this case we refer to B J as a rank subcrystal of B . MatthewFayers Tensor products If B , B ′ are I -crystals, the tensor product B ⊗ B ′ is an I -crystal defined as follows. The underlyingset is the Cartesian product B × B ′ , whose elements we write in the form b ⊗ b ′ , for b ∈ B and b ′ ∈ B ′ .For i ∈ I the functions ˜ e i , ˜ f i are given by˜ e i ( b ⊗ b ′ ) = (˜ e i b ) ⊗ b ′ ( φ ( b ) > ǫ ( b ′ )) b ⊗ (˜ e i b ′ ) ( φ ( b ) < ǫ ( b ′ )) , ˜ f i ( b ⊗ b ′ ) = ( ˜ f i b ) ⊗ b ′ ( φ ( b ) > ǫ ( b ′ )) b ⊗ ( ˜ f i b ′ ) ( φ ( b ) ǫ ( b ′ ));in these definitions, b ⊗ ⊗ b ′ should both be taken to equal 0.If B , B ′ are g -crystals, then B ⊗ B ′ is also a g -crystal, with weight functionwt( b ⊗ b ′ ) = wt( b ) + wt( b ′ ) . The point of this definition is that if B , B ′ are crystals arising from crystal bases of U q ( g )-modules V , V ′ , then B ⊗ B ′ is the crystal given by the associated crystal basis of V ⊗ V ′ (which is a U q ( g )-moduleonce an appropriate coproduct on U q ( g ) is chosen). Of particular interest in the study of g -crystals are highest-weight crystals. For each dominantintegral weight Λ of g , Kashiwara [K1] proved that the irreducible highest-weight module V ( Λ )has an essentially unique crystal basis; we write B ( Λ ) for the corresponding g -crystal. FollowingDanilov et al. [DKK], we say that a g -crystal is regular if it is isomorphic to B ( Λ ) for some dominantintegral weight Λ .The aim of this paper is to give a family of constructions of one particular regular crystal B ( Λ ).Our approach to this is to define an abstract crystal, and then to prove that it is regular and reado ff the highest weight. In order to do this, we shall need to use the following results. Proposition 3.1.
Suppose B , B ′ are regular g -crystals. Then every connected component of B ⊗ B ′ is aregular crystal. Proof.
This follows from the fact that a tensor product of two irreducible highest-weight modulesdecomposes as a direct sum of irreducible highest-weight modules. (cid:3)
Now say that an element b of an I -crystal B is a source if ˜ e i b = i ∈ I . Theorem 3.2. [KMN, Proposition 2.4.4]
Suppose B is a g -crystal. Then B is regular if and only if B hasa unique source and every connected component of every rank subcrystal of B is regular. Once we know that a crystal is regular, it is straightforward to find its highest weight; this isjust the weight of the unique source.In view of Theorem 3.2, it will be useful to have a characterisation of regular crystals in thecase where g has rank 2. In fact, the only cases which will interest us are those which arise as rank2 subalgebras of b sl n for n >
3, namely sl ⊕ sl and sl . These cases were studied by Stembridge[S]; he gave a list of ‘local’ axioms which characterise highest-weight crystals in these cases. Inartitionmodelsforthecrystalofthebasic U q ( b sl n )-module 9fact, his results are stronger: starting from an I -crystal B where | I | = a priori a crystal forany particular g ) his axioms characterise whether B is a regular crystal for sl ⊕ sl or sl (with theimplied weight function – see Remark 2 in § g is simply-laced. Highest-weight crystals for sl ⊕ sl Suppose | I | =
2, say I = { i , j } , and that the Cartan matrix A is given by a ii = a jj = , a ij = a ji = sl ⊕ sl , and we identify it with the latter.The following result is a special case of the results in [S]. Proposition 3.3.
Suppose B is a connected I-crystal, where I = { i , j } . Then B is a regular sl ⊕ sl -crystal ifand only if ˜ e i , ˜ f i both commute with ˜ e j , ˜ f j . What this says is that B is a regular sl ⊕ sl -crystal if and only if its crystal graph is ‘rectangular’,as in the following diagram: • • • . . . •• • • . . . • ... ... ... ... • • • . . . • ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ✲ i ❄ j ❄ j ❄ j ❄ j ❄ j ✲ i ✲ i ✲ i ✲ i . Highest-weight crystals for sl If I = { i , j } , with a ii = a jj = , a ij = a ji = − , then we shall identify g with sl . Stembridge’s axioms to characterise regular sl -crystals arecomplicated, and it seems to be di ffi cult to verify these directly for the sl -crystals in this paper.Accordingly, we shall use a di ff erent version, due to Danilov, Karzanov and Koshevoy.Suppose we are given an I -crystal B , and define the functions ǫ i , φ i , ǫ j , φ j on B as in § b ∈ B and ˜ e i ( b ) ,
0, then either ǫ j (˜ e i b ) = ǫ j ( b ) + , φ j (˜ e i b ) = φ j ( b )0 MatthewFayersor ǫ j (˜ e i b ) = ǫ j ( b ) , φ j (˜ e i b ) = φ j ( b ) − . (b) If in addition ˜ f i b ,
0, then either ǫ j (˜ e i b ) > ǫ j ( b ) or φ j ( ˜ f i b ) > φ j ( b ) . The same statements hold with i and j interchanged.A3. (a) If b ∈ B , ˜ e i b ,
0, ˜ e j b , ǫ j (˜ e i b ) = ǫ j ( b ), then˜ e i ˜ e j b = ˜ e j ˜ e i b , ǫ i (˜ e j b ) > ǫ i ( b ) . (b) If b ∈ B , ˜ f i b ,
0, ˜ f j b , φ j ( ˜ f i b ) = φ j ( b ), then˜ f i ˜ f j b = ˜ f j ˜ f i b , φ i ( ˜ f j b ) > φ i ( b ) . The same statements hold with i and j interchanged.There is also an axiom A4 in [DKK], but we shall not need this, since we use the followingtheorem. Proposition 3.4. [DKK, Proposition 5.3]
Suppose B is an { i , j } -crystal satisfying axioms A2 and A3, andsuppose that B has a unique source. Then B is a regular sl -crystal. ± -sequences In this section, we give some basic properties of ± -sequences; these are essential to the definitionsof our crystals.We define a ± -sequence to be a finite word π = π . . . π m from the alphabet { + , , −} . We shallperform arithmetic with these signs, interpreting + as + − as − π . . . π m is a ± -sequence, we define h i ( π ) = π i + · · · + π m for i = , . . . , m +
1. We define φ ( π ) = max n h i ( π ) (cid:12)(cid:12)(cid:12) i m + o , and if φ ( π ) > cogood position in π to bemax n i (cid:12)(cid:12)(cid:12) h i ( π ) = φ ( π ) o . Similarly, we define g i ( π ) = − ( π + · · · + π i )artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 11for i = , . . . , m , and we let ǫ ( π ) = max n g i ( π ) (cid:12)(cid:12)(cid:12) i m o ;if ǫ ( π ) > good position in π to bemin n i (cid:12)(cid:12)(cid:12) g i ( π ) = ǫ ( π ) o . Example.
Suppose π = − + − − + + − − + + − . Then the values of the functions h and g are as follows. h = − − − − − − − g = . Hence ǫ ( π ) = φ ( π ) =
1, with the good and cogood positions being 5 and 14 respectively.
Remarks.
1. The reader will observe that the zeroes in a ± -sequence π have no e ff ect on φ ( π ) or ǫ ( π ).That is, if we define a new sequence π ′ by removing some or all of the zeroes, then we have φ ( π ) = φ ( π ′ ) and ǫ ( π ) = ǫ ( π ′ ), and the good and cogood positions in π ′ correspond to thosein π . It is slightly more useful to us in this paper to allow zeroes.2. The reader may prefer an alternative way to work out ǫ ( π ) and φ ( π ) and the good and cogoodpositions. If π is a ± -sequence, then we define the reduction of π by repeatedly replacing anysegment of the form + . . . − with a segment 00 . . . + preceding a − ). Then ǫ ( π ) is the number of − signsin the reduction of π , and the good position is the position of the last such sign; φ ( π ) is thenumber of + signs in the reduction of π , and the cogood position is the position of the first + .For instance, the reduction of the sequence π given in the above example is the following: − − + . We now collect some basic properties of ± -sequences. We omit the proofs in the hope that thereader will find it more instructive to prove the results himself. Lemma 4.1.
Suppose π = π . . . π m is a ± -sequence.1. φ ( π ) − ǫ ( π ) = π + · · · + π m .2. If ǫ ( π ) > and i is the good position in π , then π i = − .3. If φ ( π ) > and j is the cogood position in π , then π j = + .4. If ǫ ( π ) , φ ( π ) > and i , j are the good and cogood positions respectively, then i < j.5. If i is the good position in π and j > i with π j = − , then there exists k such that i < k < j and π k = + .
6. If j is the cogood position in π and i < j with π i = + , then there exists k such that i < k < j and π k = − . Now we consider the relationship between two ± -sequences. Lemma 4.2.
Suppose π = π . . . π m and ρ = ρ . . . ρ m are two ± -sequences, and i ∈ { , . . . , m } is such that π i = − , ρ i = + , π j = ρ j for all j , i . Then position i is the good position in π if and only if it is the cogood position in ρ . If this is the case, then ǫ ( ρ ) = ǫ ( π ) − , φ ( ρ ) = φ ( π ) + . Lemma 4.3.
Suppose π = π . . . π m and ρ = ρ . . . ρ m are two ± -sequences, and i ∈ { , . . . , m } is such that ρ i = π i − , π j = ρ j for all j , i . Then we have the following.1. If φ ( π ) > and the cogood position in π is at or before position i, then ǫ ( ρ ) = ǫ ( π ) , φ ( ρ ) = φ ( π ) − . If in addition ǫ ( π ) > , then π and ρ have the same good position.2. If φ ( π ) = or the cogood position in π is strictly after position i, then ǫ ( ρ ) = ǫ ( π ) + , φ ( ρ ) = φ ( π ) .
3. If ǫ ( π ) > and the good position in π is after position i, then π and ρ have the same good position. Finally, we need to consider the concatenation of two ± -sequences. Lemma 4.4.
Suppose π = π . . . π l and ρ = ρ . . . ρ m are two ± -sequences, and let π ∗ ρ = π . . . π l ρ . . . ρ m denote their concatenation. • Suppose φ ( π ) > ǫ ( ρ ) . Then ǫ ( π ∗ ρ ) = ǫ ( π ) , and if this is positive, then the good position in π ∗ ρ coincides with the good position in π . • Suppose φ ( π ) < ǫ ( ρ ) . Then ǫ ( π ∗ ρ ) = ǫ ( π ) + ǫ ( ρ ) − φ ( π ) , and the good position in π ∗ ρ correspondsto the good position in ρ . • Suppose φ ( π ) ǫ ( ρ ) . Then φ ( π ∗ ρ ) = φ ( ρ ) , and if this is positive, then the cogood position in π ∗ ρ corresponds to the cogood position in ρ . • Suppose φ ( π ) > ǫ ( ρ ) . Then φ ( π ∗ ρ ) = φ ( π ) + φ ( ρ ) − ǫ ( ρ ) , and the cogood position in π ∗ ρ coincideswith the cogood position in π . artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 13 sl -crystals In this section, we give a construction of rank 2 crystals, and use Proposition 3.4 to provethat these crystals are regular sl -crystals. The crystals defined in this way will arise as rank 2subcrystals of our b sl n -crystals, but it will be helpful to describe them in a more abstract way here.Throughout this section, i and j should be regarded as abstract symbols; in Section 6 they willtake values in Z / n Z . Definition.
Define a biorder to be a finite set S , equipped with two total orders > i and > j and afunction Γ : S → { i , j } , satisfying the following condition: if s , t ∈ S are such that s > i t > j s , then Γ ( s ) = j and Γ ( t ) = i .Given a biorder S and given s , t ∈ S , we define s ≫ t if s > i t and s > j t . We also define s ⋗ t ifeither s > i t or s > j t , and we define ⊲ to be the transitive closure of ⋗ . Then ≫ is a partial order on S , while ⊲ is a preorder. Example.
Let S = { p , q , r , s , t } , with p > i q > i r > i s > i t , r > j p > j q > j t > j s . Then we must have Γ ( p ) = Γ ( q ) = Γ ( s ) = j , Γ ( r ) = Γ ( t ) = i . The partial order ≫ is given by the Hasse diagram pq rs t ❅❅❅(cid:0)(cid:0)(cid:0)(cid:0) , while the preorder ⊲ is defined by x ⊲ y if and only if x ∈ { p , q , r } or y ∈ { s , t } .If S is a biorder, we define a configuration for S to be a function a : S → { , , } ; we write C( S )for the set of configurations of S . Given a ∈ C( S ), we shall find it helpful to abuse notation andincorporate the function Γ into a ; for example, we may write a ( s ) = i to mean that a ( s ) = Γ ( s ) = i .We shall define functions ˜ e i , ˜ f i , ˜ e j , ˜ f j which will make C( S ) into an { i , j } -crystal. Take a ∈ C( S ),and for s ∈ S define π i ( a , s ) = + (if a ( s ) = i or 1 j ) − (if a ( s ) = i or 2 j )0 (otherwise) . i-signature of a to be the ± -sequence π i ( a , s ) . . . π i ( a , s m ), where s , . . . , s m are theelements of S arranged so that s > i · · · > i s m .If there is no good position in the i -signature of a , then define ˜ e i a =
0; otherwise, let k be thegood position, and define ˜ e i a to be the configuration with(˜ e i a )( s ) = a ( s ) − δ ss k . We shall say that s k is the i -good element of S for a .Similarly, if there is no cogood position in the i -signature, then we define ˜ f i a =
0; otherwise, welet l be the cogood position, and define ˜ f i a to be the configuration with( ˜ f i a )( s ) = a ( s ) + δ ss l ;we say that s l is the i -cogood element of S for a .By interchanging the symbols i and j throughout the above definition, we obtain the definitionof the j -signature of a configuration and functions ˜ e j and ˜ f j . The next lemma shows that thesefunctions make C( S ) into an { i , j } -crystal. Lemma 5.1.
Suppose S is a biorder, and h ∈ { i , j } .1. If a , b ∈ C( S ) , then ˜ e h a = b if and only if ˜ f h b = a.2. If a ∈ C( S ) and we define ǫ h ( a ) = max n ǫ (cid:12)(cid:12)(cid:12) ˜ e ǫ h a , o , φ h ( g ) = max n φ (cid:12)(cid:12)(cid:12) ˜ f ǫ h a , o , then ǫ h ( a ) = ǫ ( π ) and φ h ( a ) = φ ( π ) , where π is the h-signature of a. Proof.
1. We shall suppose that b = ˜ e h a and prove that a = ˜ f h b ; the other direction is similar. We have b ( s ) = a ( s ) − δ ss ′ for each s ∈ S , where s ′ is the h -good element of S for a . By Lemma 4.1(2), we musthave π h ( a , s ′ ) = − , and therefore (from the definition of π h ) we have π h ( b , s ′ ) = + ; clearly π h ( b , s ) = π h ( a , s ) for all s , s ′ . Now by Lemma 4.2 s ′ is the h -cogood element of S for b , andtherefore ˜ f h b = a .2. We prove that ǫ h ( a ) = ǫ ( π ) by induction on ǫ ( π ). If ǫ ( π ) =
0, then by definition ˜ e h a =
0, so that ǫ h ( a ) =
0. If ǫ ( π ) >
0, then c has a good position, and so ˜ e h a ,
0. We write b = ˜ e h a . From above,the h -signature ρ of b is obtained from π by replacing − with + in the good position. Henceby Lemma 4.2 we have ǫ ( ρ ) = ǫ ( π ) −
1; by induction ǫ h ( b ) = ǫ ( ρ ), and the result follows.Proving that φ h ( a ) = φ ( π ) is very similar. (cid:3) Example.
Let S = { p , q , r } , with r > i p ≫ q > j r (so that necessarily Γ ( p ) = Γ ( q ) = i and Γ ( r ) = j ).Then the crystal graph of C( S ) is given in Figure 2 (where we write a configuration a in the form( a ( p ) , a ( q ) , a ( r ))).artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 15 ◗◗◗◗◗◗◗s j ✑✑✑✑✑✑✑✰ i ◗◗◗◗◗◗◗s j ✑✑✑✑✑✑✑✰ i ◗◗◗◗◗◗◗s j ✑✑✑✑✑✑✑✰ i ❍❍❍❍❍❍❥ j ❍❍❍❍❍❍❥ j ✟✟✟✟✟✟✙ i ❍❍❍❍❍❍❥ j ✟✟✟✟✟✟✙ i ✟✟✟✟✟✟✙ i ❍❍❍❍❍❍❥ j ❍❍❍❍❍❍❥ j ✟✟✟✟✟✟✙ i ❍❍❍❍❍❍❥ j ✟✟✟✟✟✟✙ i ✟✟✟✟✟✟✙ i (2 , , , , , , , ,
0) (2 , , , ,
0) (2 , , , , , , , , , , , ,
0) (2 , , , , , , ❍❍❍❍❍❍❥ j (0 , , ❍❍❍❍❍❍❥ j (1 , , ❍❍❍❍❍❍❥ j (0 , , ✟✟✟✟✟✟✙ i (0 , , ✟✟✟✟✟✟✙ i (0 , , ✟✟✟✟✟✟✙ i ◗◗◗◗◗◗◗s j (0 , , ✑✑✑✑✑✑✑✰ i (0 , , , ,
2) (1 , , ❍❍❍❍❍❍❥ j ✟✟✟✟✟✟✙ i (1 , ,
1) (1 , , , , We see from Figure 2 that not every component of C( S ) is a regular sl -crystal: the secondcomponent pictured fails to satisfy axiom A3, and does not have a unique source. So we need torestrict attention to certain components of C( S ). Definition.
Suppose S is a biorder, and a ∈ C( S ). We say that a is good for S if it satisfies the followingconditions.G1. There do not exist s , t ∈ S such that s > i t > j s and a ( s ) = a ( t ) = s , t ∈ S such that s ⋗ t ⊲ s , Γ ( s ) = Γ ( t ) and a ( s ) < a ( t ).G3. There do not exist q , r , s , t ∈ S such that: • q ⋗ r ≫ s ⋗ t , • r ≫ t ⋗ q ≫ s , • a ( q ) = a ( s ) = a ( r ) = a ( t ) = S ) for the set of good configurations for S . We want to show that GC( S ) is a unionof connected components of the crystal C( S ). That is, we prove the following. Proposition 5.2.
Suppose S is a biorder, h ∈ { i , j } and a , b are configurations for S with ˜ f h a = b. Then a isgood if and only if b is good. First we need a lemma concerning axiom G2; it says that if a configuration contains a coun-terexample to G2, then it contains a counterexample of a particular form.
Lemma 5.3.
Suppose S is a biorder, and a is a configuration for S which does not satisfy axiom G2. Thenthere exist s , t , r ∈ S such that s ≫ t ⋗ r ⋗ s , Γ ( s ) = Γ ( t ) , a ( s ) < a ( t ) . Proof.
By hypothesis, we can find m > s , t , r , . . . , r m ∈ S such that s ⋗ t ⋗ r ⋗ · · · ⋗ r m ⋗ s , Γ ( s ) = Γ ( t ) , a ( s ) < a ( t ) . All we need to do is show that we can make such a choice with m =
1; the conditions s ⋗ t and Γ ( s ) = Γ ( t ) guarantee that s ≫ t .So suppose we have chosen s , t , r , . . . , r m as above with m as small as possible, and supposefor a contradiction that m >
1. Note first that r ≫ t ; for if not, then s ⋗ t ⋗ r ⋗ r ⋗ · · · ⋗ r m ⋗ s ,contradicting the minimality of m . So we have r ⋗ r ≫ t ⋗ r , and this implies t ⋗ r ⋗ t and r ⋗ r ⋗ r . By the definition of a biorder, we then have Γ ( t ) , Γ ( r ) , Γ ( r ), so that Γ ( r ) = Γ ( t ).In a similar way, we find that Γ ( r m ) , Γ ( s ). Since Γ ( r ) = Γ ( t ) = Γ ( s ), we see in particular that m >
2. This then implies that s ≫ r ; for if r ⋗ s , then we have s ⋗ t ⋗ r ⋗ r ⋗ s , contradicting theminimality of m .artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 17Since a ( s ) < a ( t ), we must have either a ( s ) < a ( r ) or a ( r ) < a ( t ). In the first case, we get s ≫ r ⋗ r ⋗ · · · ⋗ r m ⋗ s , Γ ( s ) = Γ ( r ) , a ( s ) < a ( r ) , while in the second case we get r ≫ t ⋗ r ⋗ r , Γ ( r ) = Γ ( t ) , a ( r ) < a ( t );either way, we have a contradiction to the minimality of m . (cid:3) Proof of Proposition 5.2.
We assume that h = i ; the case where h = j then follows from thesymmetry of the definitions. We assume that a is not a good configuration, and prove that b is notgood; the reverse direction follows in a very similar way.Let s ′ be the i -cogood element for a . Then b is given by b ( s ) = a ( s ) + δ ss ′ for s ∈ S , and a ( s ′ ) equals either 0 i or 1 j .The assumption that a < GC( S ) means that a violates one of the axioms G1–3. We consider eachof the possibilities in turn. a does not satisfy G1 In this case there are s , t ∈ S such that s > i t > j s and a ( s ) = a ( t ) =
1. Fromthe definition of a biorder, we must have Γ ( s ) = j and Γ ( t ) = i ; hence we cannot have s ′ = t . If s ′ , s , then obviously b does not satisfy G1, so we suppose that s ′ = s .We have π i ( b , s ) = π i ( b , t ) = − . Since s > i t , the − corresponding to t occurs after the − corresponding to s in the i -signature of b . But the − corresponding to s is in the good position,so by Lemma 4.1(5) there must be a + between these two positions. That is, there is r ∈ S suchthat s > i r > i t and b ( r ) equals either 0 i or 1 j .If b ( r ) = i , then we have r ⋗ t ⋗ s ⋗ r , Γ ( r ) = Γ ( t ) , b ( r ) < b ( t ) , so b violates G2. On the other hand, if b ( r ) = j , then, we have s ≫ r by the definition of abiorder, whence r > i t > j r , b ( r ) = b ( t ) = , so b violates G1. Either way, we find b < GC( S ). a does not satisfy G2 In this case, we apply Lemma 5.3, and we find that there are r , s , t ∈ S suchthat s ≫ t ⋗ r ⋗ s , Γ ( s ) = Γ ( t ) and a ( s ) < a ( t ). If the same r , s , t do not yield a violation of G2 in b , we must be in one of the following two situations:1. s ′ = s , a ( s ) = i , a ( t ) = i , Γ ( r ) = j , t > j r > i s ;2. s ′ = s , a ( s ) = j , a ( t ) = j , Γ ( r ) = i , t > i r > j s .As above, the fact π i ( b , s ) = π i ( b , t ) = − while s > i t means that there must be some q ∈ S suchthat s > i q > i t and b ( q ) equals either 0 i or 1 j .8 MatthewFayers • Suppose we are in case 1. If b ( q ) = i then b violates G2, since q ⋗ t ⊲ q , Γ ( q ) = Γ ( t ) , b ( q ) < b ( t ) . So suppose instead that b ( q ) = j . Then we have r > i q , and since Γ ( q ) = j , this meansthat r > j q too. Hence t > j q , and so we have q > i t > j q , b ( q ) = b ( t ) = , so b violates G1. • Next suppose we are in case 2. In this case, if we have b ( q ) = j , then b violates G2 (viathe pair ( q , t )), so we suppose instead that b ( q ) = i . Now we have q > i r , so q ⋗ r ⊲ q , Γ ( q ) = Γ ( r ) , b ( q ) = , so that if b ( r ) > b violates G2. So let us suppose that b ( r ) =
0. Then we find that s ⋗ q ≫ t ⋗ r , q ≫ r ⋗ s ≫ t , b ( s ) = b ( t ) = , b ( q ) = b ( r ) = , and b violates G3. a does not satisfy G3 Suppose q , r , s , t are as in G3. Note that the axioms for a biorder imply that Γ ( q ) , Γ ( r ) , Γ ( s ) , Γ ( t ). Obviously if s ′ < { q , r , s , t } , then b violates G3, so suppose otherwise. s ′ cannot equal either q or s . If s ′ = t , then b violates G2, because r ⋗ t ⊲ r , Γ ( r ) = Γ ( t ) , b ( r ) < b ( t );so suppose s ′ = r . This means that Γ ( r ) = i and hence Γ ( s ) = j , so that π i ( b , r ) = π i ( b , s ) = − .Arguing as in previous cases, we find that there must be some p ∈ S such that r > i p > i s and b ( p ) equals either 0 i or 1 j .If b ( p ) = j , then b violates G2, since we have p > i s ⊲ p , Γ ( p ) = Γ ( s ) , b ( p ) < b ( s );so suppose instead that b ( p ) = i . Since p > i s and Γ ( p ) = i , the definition of a biorder impliesthat p ≫ s , so we have q ⋗ p ≫ s ⋗ t , p ≫ t ⋗ q ≫ s , b ( q ) = b ( s ) = , b ( p ) = b ( t ) = , and b violates G3. (cid:3) Given Proposition 5.2, it makes sense to refer to a connected component of C( S ) as good if everyvertex is labelled by a good configuration, and to view GC( S ) as a crystal. Our aim is to show thatevery component of GC( S ) is a regular sl -crystal.artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 19 GC( S ) First we note the following simple lemma, which follows from the definitions.
Lemma 5.4.
Suppose a , b ∈ C( S ) with b = ˜ e i a. Then the j-signature of b is obtained from the j-signature ofa either by replacing a with a − or by replacing a + with a . This lemma, together with Lemmata 4.3 and 5.1, implies that the first part of axiom A2 holds inC( S ) (and in particular, in GC( S )). In fact, we can make a more precise statement. Lemma 5.5.
Suppose a , b ∈ C( S ) with b = ˜ e i a. If φ j ( a ) = , then we have ǫ j ( b ) = ǫ j ( a ) + , φ j ( b ) = φ j ( a ) . If not, then let t be the i-good element of S for a, and let s be the j-cogood element for a. • If t > j s, then ǫ j ( b ) = ǫ j ( a ) + , φ j ( b ) = φ j ( a ) . • If s > j t, then ǫ j ( b ) = ǫ j ( a ) , φ j ( b ) = φ j ( a ) − , and if in addition ǫ j ( a ) > then the j-good element for b is the same as the j-good element for a. Proof.
This follows from Lemmata 4.3, 5.1 and 5.4. (cid:3)
Of course, the same result holds with i and j interchanged. Now to complete the proof thataxiom A2 holds in GC( S ), we need to show the following. Lemma 5.6.
There is no a ∈ GC( S ) such that ˜ e i a , ˜ f i a , , ǫ j (˜ e i a ) = ǫ j ( a ) , φ j ( ˜ f i a ) = φ j ( a ) . Proof.
Suppose for a contradiction that we can find such a configuration a . Then obviously the i -signature of a has both a good and a cogood position. By Lemma 5.5 the j -signature of a has acogood position, and similarly the j -signature must also have a good position. We let q , r , s , t be the i -cogood, j -good, j -cogood and i -good elements for a , respectively.We have t > i q and r > j s , by Lemma 4.1(4). We also have s > j t , by Lemma 5.5, and in asymmetrical way we have q > j r .So we have t > i q > j t , and hence Γ ( q ) = i , Γ ( t ) = j . Since π i ( a , q ) = + and π i ( a , t ) = − , we musttherefore have a ( q ) = i and a ( t ) = j .Now consider the value of Γ ( r ). If Γ ( r ) = i , then (since π j ( a , r ) = − ) we have a ( r ) = i . But thenwe have q ≫ r (since q > j r and Γ ( r ) = i ), with Γ ( q ) = Γ ( r ), r ⊲ q and a ( q ) < a ( r ), and this contradictsaxiom G2. On the other hand, if we have Γ ( r ) = j , then a ( r ) = j . But then we have r > j t , Γ ( r ) = Γ ( t ), t ⊲ r and a ( r ) < a ( t ), which again contradicts axiom G2. (cid:3) Of course, the same result holds with i and j interchanged, and we see that axiom A2 holds inGC( S ).0 MatthewFayers GC( S ) Lemma 5.7.
Suppose a ∈ GC( S ) with ˜ e i a , ˜ e j a , and ǫ j (˜ e i a ) = ǫ j ( a ) . Then ˜ e i ˜ e j a = ˜ e j ˜ e i a , , and ǫ i (˜ e j a ) = ǫ i ( a ) + . Proof.
Let r and t be the j -good and i -good elements for a respectively. By Lemma 5.5, r is also the j -good element for ˜ e i a . So in order to show that ˜ e i ˜ e j a = ˜ e j ˜ e i a ,
0, we need to show that t is also the i -good element for ˜ e j a ; for then we shall have ˜ e i ˜ e j a = ˜ e j ˜ e i a = b , where b ( s ) = a ( s ) − δ sr − δ st . Let s be the j -cogood element for a . (Note that there must be such an element, by Lemma 5.5;moreover, that lemma implies that s > j t . We also have r > j s by Lemma 4.1(4).)The i -signature of ˜ e j a is obtained from the i -signature of a by subtracting 1 from the entrycorresponding to r . If r > i t , then by Lemma 4.3 the position corresponding to t will still be good inthe i -signature of ˜ e j a , which is what we require; so suppose otherwise, i.e. t > i r .Now we have r > j s > j t > i r ; hence r > j t > i r , which implies that Γ ( r ) = i and Γ ( t ) = j . Since π j ( a , r ) = π i ( a , t ) = − , we then have a ( r ) = a ( t ) = s . Because π j ( a , s ) = + , we have either a ( s ) = i or a ( s ) = j . In the latter case wewould have s ⋗ t ⊲ s , with Γ ( s ) = Γ ( t ) and a ( s ) < a ( t ), and this contradicts G2. So instead we musthave a ( s ) = i . Hence π i ( a , s ) = − , with t > i s ; since t is the i -good element for a , there must thereforebe some q ∈ S such that t > i q > i s and π i ( a , q ) = + . We consider the two possibilities for a ( q ).If a ( q ) = j , then we have t ≫ q , so that s > j q > i s ; but then we have a contradiction to G1. Onthe other hand, if a ( q ) = i , then we have q ⋗ s ⊲ q with Γ ( q ) = Γ ( s ) and a ( q ) < a ( s ), and we have acontradiction to G2.It remains to show that ǫ i (˜ e j a ) = ǫ i ( a ) +
1. If this is not the case, then by Lemma 5.5 (with i and j interchanged) we have φ i ( a ) >
0, and p > i r , where p is the i -cogood element of S . So we have r > j s > j t > i p > i r ; in particular, r > j t > i r , so that Γ ( r ) = i and Γ ( t ) = j . Hence a ( r ) = a ( t ) = a ( p ) = i or 1 j . If a ( p ) = i , then p ⋗ r ⊲ p with Γ ( p ) = Γ ( r ) and a ( p ) < a ( r ), contradictingaxiom G2. So instead a ( p ) = j . Similarly a ( s ) = i . Now we have Γ ( t ) = Γ ( p ) and t ⋗ p , whichimplies t ≫ p , and hence s > j p . We also have r ≫ s , which implies p > i s . So we have p > i s > j p and a ( p ) = a ( s ) =
1, but this contradicts G1. (cid:3)
The same result holds with i and j interchanged, so part (a) of axiom A3 holds in GC( S ). Part(b) is proved in the same way. Now suppose that the biorder S is such that s ⊲ t for every s , t ∈ S ; we shall say that S is transitive if this is the case. By Proposition 3.4, in order to show that every component of GC( S ) isregular, it is enough to show that any such component has only one source. In fact, we shall provethat under the assumption of transitivity GC( S ) has only one source; this implies in particular thatit has only one component. Proposition 5.8.
Suppose S is a transitive biorder. Then
GC( S ) has only one source. artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 21 Proof.
It is clear that the configuration a with a ( s ) = s ∈ S is good and is a source. So wemust show that there is no source b which is good and has b ( s ) > s .Suppose b is such a configuration, and suppose first that b ( s ) =
1. Without loss of generality,we assume Γ ( s ) = i , and in fact we assume that s is maximal with respect to the order > i such that b ( s ) = i .The i -signature of b contains a − corresponding to s . Since b is a source, this signature cannotcontain a good position, and therefore there must be a + preceding this − in the i -signature. Choosesuch a + , and let t be the corresponding element of S . Then b ( t ) equals either 0 i or 1 j . In the firstcase, we have t > i s , Γ ( t ) = Γ ( s ), s ⊲ t (because S is transitive) and b ( t ) < b ( s ), and this contradictsaxiom G2. So instead we must have b ( t ) = j . If s > j t , then b violates axiom G1, so we must have t ≫ s . Repeating the above argument with t in place of s and with i and j interchanged, we findthat there is some t ′ ∈ S such that t ′ ≫ t and b ( t ′ ) = i . This implies that t ′ ≫ s , but this contradictsthe choice of s .So we cannot have b ( s ) = s . We choose some s such that b ( s ) =
2. Without loss ofgenerality, we assume that Γ ( s ) = j , and we assume that s is maximal with respect to the order > i such that b ( s ) = j . Arguing as above, there must be some t ∈ S such that t > i s and there is a + corresponding to t in the i -signature of b . The assumption that b ( t ) , b ( t ) = i .Now the definition of a biorder gives t ≫ s .Since S is transitive, we can find r , . . . , r m ∈ S such that s ⋗ r ⋗ · · · ⋗ r m ⋗ t . We make such achoice with m as small as possible. We note first that m must be greater than 1. For if m =
1, then wehave s ⋗ r ⋗ s , which means that Γ ( r ) = i ; but we also have r ⋗ t ⋗ r , which implies that Γ ( r ) = j .We observe that t ≫ r m − . Indeed, if not, then s ⋗ r ⋗ · · · ⋗ r m − ⋗ t , which contradicts theminimality of m . Now the fact that r m − ⋗ r m implies that t ⋗ r m . So t ⋗ r m ⋗ t , and we therefore have Γ ( r m ) = j , and r m > i t > j r m . This gives r m > i s , and now the choice of s implies that b ( r m ) =
0. Butthen we have r m ≫ s ⊲ r m , Γ ( r m ) = Γ ( s ) and b ( r m ) < b ( s ), and this contradicts axiom G2. (cid:3) Now Proposition 3.4 implies the following.
Corollary 5.9.
Suppose S is a transitive biorder. Then
GC( S ) has exactly one component, which is a regular sl -crystal. GC( S ) is a regular crystal Now we complete the proof of the main result of this section by proving that if S is a (possiblyintransitive) biorder, then every component of GC( S ) is a regular sl -crystal. In order to do this, weuse tensor products and exploit Proposition 3.1. Proposition 5.10.
Suppose S is a biorder, and suppose that we can decompose S as S ⊔ S in such a waythat for all s ∈ S and t ∈ S we have s ≫ t. Then as crystals we have C( S ) (cid:27) C( S ) ⊗ C( S ) . Furthermore,under this isomorphism, every good component of C( S ) arises as a component of D ⊗ D , where D is a goodcomponent of C( S ) and D is a good component of C( S ) . Proof.
Given a configuration a for S and given g ∈ { , } , we define a g to be the restriction of a to2 MatthewFayers S g ; so a g is a configuration for S g . This defines a bijection χ : C( S ) −→ C( S ) × C( S ) a ( a , a ) , and we claim that this bijection gives an isomorphism of crystals, i.e. for a ∈ C( S ) and h ∈ { i , j } χ (˜ e h a ) = (˜ e h a , a ) ( φ h ( a ) > ǫ h ( a ))( a , ˜ e h a ) ( φ h ( a ) < ǫ h ( a )) ,χ ( ˜ f h a ) = ( ˜ f h a , a ) ( φ h ( a ) > ǫ h ( a ))( a , ˜ f h a ) ( φ h ( a ) ǫ h ( a ))(where we interpret ( a ,
0) and (0 , a ) as 0). To see this, we note that the h -signature of a consists ofthe h -signature of a followed by the h -signature of a , and then apply Lemma 4.4.For the second part of the proposition, we simply observe that if a is a good configuration for S , then a and a are good configurations for S and S respectively; this is immediate from thedefinition of a good configuration. (cid:3) Finally we can prove the main result of this section.
Theorem 5.11.
Suppose S is a biorder. Then every component of
GC( S ) is a regular sl -crystal. Proof.
We proceed by induction on | S | . If S is transitive, then the result follows from Corollary5.9, so suppose S is not transitive. This means that we can write S = S ⊔ S , where S , S arenon-empty and s ≫ t for all s ∈ S and t ∈ S (for example, choose s , t ∈ S such that t S s , and let S = { s ∈ S | s Q s } ). By induction every good component of C( S ) or C( S ) is a regular sl -crystal;by Proposition 5.10, every good component of C( S ) is isomorphic to a connected component of thetensor product of a good component of C( S ) and a good component of C( S ), and so by Proposition3.1 is a regular sl -crystal. (cid:3) Now we can return to our crystal R A for a given arm sequence A , and prove Theorem 2.2. Fromnow on, we write I for the set Z / n Z . R A is an b sl n -crystal The first thing we need to do in order to prove Theorem 2.2 is to show that the operators ˜ e i , ˜ f i actually map R A to R A ⊔ { } . That is, we need to prove the following proposition. Proposition 6.1.
Suppose A is an arm sequence, λ ∈ R A and i ∈ Z / n Z .1. If ˜ f i λ , , then ˜ f i λ ∈ R A . artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 23
2. If ˜ e i λ , , then ˜ e i λ ∈ R A . Proof.
We prove (1); the proof of (2) is similar. Let µ = ˜ f i λ , and suppose µ is obtained from λ byadding the i -node ( a , c ). We suppose for a contradiction that µ has an illegal hook. Since λ has noillegal hooks, an illegal hook of µ must be either the ( b , c )-hook for some b < a , or the ( a , d )-hook forsome d < c . We assume the latter; the other case follows in a similar way (or indeed by the fact thatthe definitions have a symmetry corresponding to conjugation of partitions).So we assume that the ( a , d )-hook of µ has length nt and arm length c − d = A t , for some t . Thismeans that λ ′ d = a + nt − A t − . Let b = λ ′ d +
1. Then the node ( b , d ) is not a node of λ , but the node ( b − , d ) is. And in fact ( b , d )must be an addable node of λ ; for if not, then d > b , d −
1) does not lie in λ . But thisthen means that the ( a , d − λ is illegal (with length nt and arm length A t ).Since c − d = A t , we have ( b , d ) ≻ ( a , c ). So λ has addable i -nodes ( b , d ) ≻ ( a , c ), and ( a , c ) isthe i -cogood node. By Lemma 4.1(6), this means that there is a removable i -node ( f , g ) of λ with( b , d ) ≻ ( f , g ) ≻ ( a , c ). We now consider three cases. • First suppose f < a . The fact that ( f , g ) and ( a , c ) have the same residue implies that g − c + a − f = nu for some positive integer u . Since ( f , g ) ≻ ( a , c ), we have g − c > A u . Combining this withthe fact that c − d = A t , we get g − d > A t + A u , whence g − d > A t + u . On the other hand the factthat ( b , d ) ≻ ( f , g ) means that g − d A t + u . So g − d = A t + u , which means that the ( f , d )-hookof λ has length n ( t + u ) and arm length A t + u , a contradiction. • Next suppose a < f < b . Now we have g − d + b − f = nu for some positive u , and we claimthat g − d = A u , which means that the ( f , d )-hook of λ is illegal.Now the ordering ( b , d ) ≻ ( f , g ) ≻ ( a , c ) gives g − d A u , and c − g A t − u ; using the fact that c − d = A t , we get g − d > A t − A t − u > A u . So g − d = A u . • Finally, we suppose f > b . Now we have d − g + f − b = nu for some positive u , and we willshow that d − − g = A u , which means that the ( b , g )-hook of λ is illegal.The ordering of the nodes gives d − g > A u and c − g A t + u . The latter yields d − g A t + u − A u ,which is at most A u +
1. So d − g = A u +
1, as required. (cid:3)
Now we can show that R A is an I -crystal, by checking axiom C1. Lemma 6.2.
Suppose A is an arm sequence, λ, µ ∈ R A and i ∈ I. Then ˜ e i λ = µ if and only if ˜ f i µ = λ . Proof.
This follows from the definitions, together with Lemma 4.2. (cid:3)
Next we need to check that R A is an b sl n -crystal, with the weight function wt given in § λ ∈ R A and i ∈ I , and let π be the ± -sequence corresponding to the addable and removable i -nodes of λ . Then by Lemma 4.2, we have ǫ i ( λ ) = ǫ ( π ) and φ i ( λ ) = φ ( π ) (cf. the proof of Lemma4 MatthewFayers5.1). Hence by Lemma 4.1(1), φ i ( λ ) − ǫ i ( λ ) equals the number of addable i -nodes of λ minus thenumber of removable i -nodes. It is well-known (and easy to prove by induction on | λ | ) that thisequals h h i , wt( λ ) i . So axiom C4 is satisfied, and R A is an b sl n -crystal. R A has a unique source We have seen that R A is an b sl n -crystal; to show that it is isomorphic to the highest-weight crystal B ( Λ ), we verify the hypotheses of Theorem 3.2. The first thing we need to check is that the crystal R A has a unique source. Proposition 6.3.
Suppose A is an arm sequence. If λ ∈ R A and λ , ∅ , then λ has at least one good node. Proof.
For this proof, we introduce a partial order on N : we put ( a , c ) > ( b , d ) if and only if thereare γ, δ > a , c ) < ( b + γ, d + δ ). It is easy to check that > is indeed a partial order, whichrestricts to the order < on the set of nodes of a given residue.Now let ( a , c ) be a node of λ which is maximal with respect to the order > , and let i be theresidue of ( a , c ); then we claim that λ has a good node of residue i . Certainly ( a , c ) is removable,since we have ( a + , c ) > ( a , c ) and ( a , c + > ( a , c ). Therefore if there is no good i -node, then theremust be some addable i -node ( b , d ) of λ such that ( b , d ) ≻ ( a , c ). We shall show that this forces λ tohave an illegal hook, which gives a contradiction.We assume that b > a ; the other case is very similar. Since ( b , d ) and ( a , c ) have the same residue,we can write b − a + c − d = nt with t a positive integer. Then the fact that ( b , d ) ≻ ( a , c ) implies that c − d A t . The node ( b − , d )lies in λ (because ( b , d ) is an addable node of λ ), so by maximality we have ( b − , d ) (cid:11) ( a , c ). Hence( b − , d ) (cid:11) ( a , c + a , c + ≻ ( b − , d ) (since these two nodes have the same residue).This means that c + − d > A t ; so c − d = A t . So the ( a , d )-hook of λ has arm length A t ; it has length c − d + b − a = nt , and therefore is an illegal hook. (cid:3) So we know that the empty partition ∅ is the unique source of R A ; since wt( ∅ ) = Λ , all thatremains is to check the final condition of Theorem 3.2, namely that every rank 2 subcrystal of R A isregular. sl ⊕ sl -subcrystals Suppose i , j ∈ I with i , j ±
1, and consider the subcrystal of R A given by just the i - and j -arrows.This subcrystal is an sl ⊕ sl -crystal, so by Proposition 3.3, all we need to check is the following. Proposition 6.4.
Suppose i , j are distinct elements of Z / n Z with j , i ± . Then the operators ˜ e i , ˜ f i on R A commute with ˜ e j , ˜ f j . Proof.
This is easy to see from the definitions: since j , i ±
1, two nodes of residues i , j can-not be adjacent. Therefore applying ˜ e i or ˜ f i , which involves adding or removing an i -node, cannota ff ect the set of addable and removable j -nodes of a partition. Hence ˜ e i , ˜ f i will commute with ˜ e j , ˜ f j . (cid:3) artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 25 sl -subcrystals Now we consider the rather more awkward case of sl -subcrystals. Throughout this subsection,we fix i ∈ Z / n Z , and we set j = i +
1. We consider the subcrystal R iA of R A obtained by deleting allarrows other than those labelled i or j . R iA is an { i , j } -crystal; we must prove that each componentof this crystal is a regular sl -crystal.Given a connected component C of R iA , our aim is to construct a biorder S and an isomorphism ψ : C −→ D , where D is a component of GC( S ); then by Theorem 5.11 we shall know that C isregular.Let λ be a partition in C , and define two partitions C ∨ and C ∧ as follows: C ∨ is defined byrepeatedly removing removable nodes of residues i and j until there are no more; and C ∧ is definedby repeatedly adding addable nodes of residues i and j until there are no more. Note that ingeneral, C ∧ and C ∨ need not lie in C . However, since C is connected, the definition of C ∨ and C ∧ does not depend on the choice of λ . Example.
Suppose n = i = λ = (3 , , ). Then we have C ∨ = (3 ,
1) and C ∧ = (5 , , ).Now define an i-domino in N to be a pair of adjacent nodes of which one has residue i and theother has residue j . Since j = i +
1, the node of residue j in an i -domino must lie either immediatelyabove or immediately to the right of the node of residue i ; we say that the domino is vertical in thefirst case, and horizontal in the second case.With C ∧ and C ∨ defined as above, the set [ C ∧ ] \ [ C ∨ ] is a disjoint union of dominoes. We nowdefine a biorder S C : as a set, this is the set of dominoes in [ C ∧ ] \ [ C ∨ ]. The order > i is defined bytaking the i -nodes in the dominoes, and using the order ≻ on these; that is, for s , t ∈ S C , s > i t ⇐⇒ (the i -node in s ) ≻ (the i -node in t );the order > j is defined in exactly the same way, using the nodes of residue j . The function Γ isdefined by Γ ( s ) = i (if s is horizontal) j (if s is vertical) . It is straightforward to check that S C satisfies the axioms for a biorder.Now we define a map ψ : C → C( S C ). Given a partition λ in C , we define ψ ( λ ) by ψ ( λ )( s ) = (cid:12)(cid:12)(cid:12) [ λ ] ∩ s (cid:12)(cid:12)(cid:12) for each s ∈ S C . Example.
Continuing the last example, [ C ∧ ] \ [ C ∨ ] consists of three dominoes: • a horizontal domino p = { (1 , , (1 , } ; • a horizontal domino q = { (2 , , (2 , } ; • a vertical domino r = { (3 , , (4 , } .6 MatthewFayersIf A is an arm sequence with A = A =
3, then we find that(1 , ≻ (2 , ≻ (3 , , (4 , ≻ (1 , ≻ (2 , , so that S C is precisely the biorder given in the example in § C of R iA is givenin Figure 3 (which the reader should compare with the first diagram in Figure 2); in the Youngdiagrams, we have marked with × the nodes belonging to C ∨ , so that the reader can more easilyidentify the added nodes. Lemma 6.5.
Suppose λ ∈ C . Then ψ ( λ ) is a good configuration. Proof.
We use proof by contradiction, showing that if ψ ( λ ) violates one of axioms G1–3, then λ possesses an illegal hook, so does not lie in R A . For brevity, let us write a for ψ ( λ ). a does not satisfy G1 In this case, we have s , t ∈ S C such that s > i t > j s and a ( s ) = a ( t ) =
1. Thisimplies that s is a vertical domino, while t is a horizontal domino. Let ( b , c ) be the i -node in s ,and ( d , e ) the i -node in t . Then we have ( b , c ) ≻ ( d , e ), but ( d , e + ≻ ( b − , c ). Since ( b , c ) and( d , e ) have the same residue, we have | b − d + e − c | = nu for some positive integer u .Suppose first that b > d . Now ( b , c ) ≻ ( d , e ) implies that e − c A u , while ( d , e + ≻ ( b − , c )implies that e + − c > A u , so we have e − c = A u . Now consider the ( d , c )-hook of λ . Since a ( t ) =
1, we have ( d , e ) ∈ λ = ( d , e + d , c )-hook is e − c = A u .Since a ( s ) =
1, we have ( b , c ) < λ ∋ ( b − , c ), and we find that the length of the ( d , c )-hook is b − d + e − c = nu . So λ possesses an illegal hook.Alternatively, suppose d > b . Now we have c − e − = A u , and we claim that the ( b , e )-hook of λ is illegal. Since a ( s ) = b , c ) < λ . On the other hand, the definition of C ∧ impliesthat ( b , c − ∈ λ ; for otherwise ( b , c ) could not lie in C ∧ . So the arm length of the ( b , e )-hook of λ is c − − e = A u . To compute the length of the ( b , e )-hook, we observe that ( d , e ) ∈ λ (because a ( t ) = d + , e ) < λ (by the definition of C ∨ ). So we find that the length of the ( b , e )-hookis d − b + c − e = nu , as required. a does not satisfy G2 Applying Lemma 5.3, there are r , s , t ∈ S C such that s ≫ t ⋗ r ⋗ s , Γ ( s ) = Γ ( t )and a ( s ) < a ( t ). We let ( b , c ) be the i -node in s and ( d , e )-the i -node in t . Now we have( b , c ) ≻ ( d , e ), and we claim that ( d + , e + ≻ ( b , c ). To show this, we’ll assume Γ ( s ) = i (theother case being very similar). In this case, if we let ( f , g ) be the i -node in r , then t ⋗ j r gives( d , e + ≻ ( f − , g ), and hence ( d + , e + ≻ ( f , g ). But we also have ( f , g ) ≻ ( b , c ) since r ⋗ i s ,so ( d + , e + ≻ ( b , c ). As above, we have | b − d + e − c | = nu for some positive u .First suppose b > d . Then we compute e − c = A u . This enables us to find an illegal hook in λ ,but there are various cases. If s and t are horizontal dominoes, then the ( d , c )-hook is illegal if a ( s ) = a ( t ) =
1, while the ( d , c + a ( s ) < a ( t ) =
2. On the other hand if s and t are vertical, then the ( d , c )-hook is illegal if a ( s ) = a ( t ) =
2, while the ( d − , c )-hookis illegal if a ( s ) = < a ( t ).The case where d > b is similar.artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 27 ◗◗◗◗◗◗◗◗◗◗s ✑✑✑✑✑✑✑✑✑✑✰ ◗◗◗◗◗◗◗◗◗◗s ✑✑✑✑✑✑✑✑✑✑✰ ◗◗◗◗◗◗◗◗◗◗s ✑✑✑✑✑✑✑✑✑✑✰ ❍❍❍❍❍❍❍❍❍❍❥ ❍❍❍❍❍❍❍❍❍❍❥ ✟✟✟✟✟✟✟✟✟✟✙ ❍❍❍❍❍❍❍❍❍❍❥ ✟✟✟✟✟✟✟✟✟✟✙ ✟✟✟✟✟✟✟✟✟✟✙ ❍❍❍❍❍❍❍❍❍❍❥ ❍❍❍❍❍❍❍❍❍❍❥ ✟✟✟✟✟✟✟✟✟✟✙ ❍❍❍❍❍❍❍❍❍❍❥ ✟✟✟✟✟✟✟✟✟✟✙ ✟✟✟✟✟✟✟✟✟✟✙ ×××××××××××××××× ×××××××× ×××××××××××××××××××××××× ×××××××××××× Figure 38 MatthewFayers a does not satisfy G3 Suppose q , r , s , t ∈ S C are as in G3. Then Γ ( q ) , Γ ( r ) , Γ ( s ) , Γ ( t ), and weconsider the case where Γ ( q ) = j ; the case where Γ ( q ) = i is very similar. Let ( b , c ) be the i -node in r , and ( d , e ) the i -node in s . Then we have ( b , c ) ≻ ( d , e ), but (since s > i t > j q > i r )( d + , e + ≻ ( b , c ). Now, using similar arguments to the previous cases, we find that the( d , c )-hook of λ is illegal if b > d , while the ( b , e )-hook is illegal if d > b . (cid:3) Finally, we note the following.
Lemma 6.6. ψ commutes with ˜ e i , ˜ f i , ˜ e j , ˜ f j . Proof.
We consider ˜ e i and ˜ f i ; the proof for ˜ e j and ˜ f j is similar. Suppose λ ∈ C , and let AR i ( λ ) denotethe set of addable and removable i -nodes of λ . The definitions of C ∧ and C ∨ imply that each nodein AR i ( λ ) is contained in [ C ∧ ] \ [ C ∨ ], and so is contained in some domino in [ C ∧ ] \ [ C ∨ ]. So wehave a map ∂ : AR i ( λ ) → S C , given by mapping a node to the domino that contains it; since eachdomino contains a unique i -node, ∂ is injective; furthermore, ∂ is order-preserving in the sense thatif ( a , c ) ≻ ( b , d ) in AR i ( λ ), then ∂ (( a , c )) > i ∂ (( b , d )). Now let a = ψ ( λ ), and for s ∈ S C define π i ( a , s ) asin § π i ( a , s ) = + (if s = ∂ (( a , c )), with (( a , c )) an addable node of λ ) − (if s = ∂ (( a , c )), with (( a , c )) a removable node of λ )0 (if s does not lie in the image of ∂ ) . This, together with the fact that ∂ is order-preserving, means that the ± -sequence obtained fromAR i ( λ ) is precisely the i -signature of ψ ( λ ) with the 0s removed. Moreover, if π i ( a , s ) = + , thenincreasing a ( s ) by 1 corresponds to adding the i -node in s to λ , while if π i ( a , s ) = − , then decreasing a ( s ) by 1 corresponds to removing the i -node in s from λ . In view of Remark 1 in Section 4, we seethat the definitions of ˜ e i and ˜ f i on λ and on a are essentially the same. (cid:3) Since C is a connected component of R A , Lemma 6.6 implies that the image of ψ is a connectedcomponent of C( S ); this component is good, by Lemma 6.5. It is easy to see that ψ is injective, andso C is isomorphic to a component of GC( S ), and so by Theorem 5.11 C is a regular sl -crystal.So we have verified all the hypotheses of Theorem 3.2 for the crystal R A , and the proof ofTheorem 2.2 is complete. ff erent arm sequences are di ff erent For each n >
3, we have defined a family of crystals for U q ( b sl n ); or rather, a family of models forthe same crystal. However, it is not clear that di ff erent arm sequences give di ff erent models. Theaim of this section is to prove this statement, by showing the following. Proposition 7.1.
Suppose n > and A , A ′ are distinct arm sequences. Then the sets R A and R A ′ aredistinct. artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 29We begin by singling out two particular arm sequences. Call the two sequences (0 , , , . . . ) and( n − , n − , n − , . . . ) the extreme sequences. Lemma 7.2.
Suppose A is a non-extreme arm sequence.1. For each t > , we have t A t ( n − t − .2. Let u > and let λ be the partition ( A u + , nu − A u − ) . Then λ possesses an illegal hook of length nu,but does not possess an illegal hook of any other length. Proof.
1. This is simple to check.2. The (1 , nu . Now suppose the ( a , c )-hook is an illegal hook of length nt for some t < u . Then either a = c A u +
1, or c = a nu − A u . In thefirst case we find nt = A t +
1, which contradicts the fact that A t ( n − t −
1, while in thesecond case we find that A t =
0, which also contradicts the first part of this lemma. (cid:3)
Proof of Proposition 7.1.
Suppose first that A is the extreme sequence ( n − , n − , . . . ). Then A = n −
1, while A ′ < n −
1, so the partition ( n ) is A ′ -regular but not A -regular. A similar argumentapplies in the case where A = (0 , , , . . . ), using the partition (1 n ).Now assume that neither A nor A ′ is extreme, and let u be minimal such that A u , A ′ u . Considerthe partition λ = ( A u + , nu − A u − ). By Lemma 7.2(2), λ does not lie in R A ; nor does it have a hookwith length nt and arm length A t = A ′ t , for any t < u . The only hook of length nt with t > u is the(1 , nu and arm length A u , A ′ u , and therefore λ does lie in R A ′ . (cid:3) Remarks.
1. The proof of Proposition 7.1 relies on the restriction t − A t ( n − t . In fact, if we broadenthe definition of an arm sequence to allow 0 A t nt −
1, then Theorem 2.2 still holds. Butthe crystals R A and R A ′ will be identical whenever A = A ′ = A = A ′ = n − A -regular ones). Then di ff erent arm sequences give di ff erent crystals, even withthe broader definition above. However, except in very special cases, the components of thiscrystal other than the component R A are not regular. It seems that some modification of thedefinitions is appropriate to make these components regular; the author hopes to address thisin the future.We end the paper by giving an alternative parametrisation of our crystals; this shows in partic-ular that we have uncountably many arm sequences. Lemma 7.3.
Suppose A is an arm sequence. Then the sequence (cid:18) A t t (cid:19) converges to some y A ∈ [1 , n − . Proof.
We claim that for any t , u we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) A t t − A u u (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < { t , u } , uA t A tu < uA t + u and similarly tA u A tu < tA u + t . Hence A t t − A u u < A tu tu − A tu − ttu = u , A u u − A t t < A tu tu − A tu − utu = t . (cid:3) We shall abuse terminology by saying that the arm sequence
A tends to y if y = lim t →∞ A t t . Lemma 7.4.
Suppose y ∈ [1 , n − .1. If y is irrational, then the sequence A y given byA yt = ⌊ yt ⌋ is an arm sequence. Furthermore, A y is the unique arm sequence that tends to y.2. If y is rational, then the two sequences A y , + and A y , − given byA y , + t = ⌊ yt ⌋ , A y , − t = ⌈ yt − ⌉ are arm sequences. They are the only arm sequences that tend to y. Proof.
It is straightforward to verify that the given sequences are arm sequences tending to y . Nowsuppose A is an arm sequence tending to y . First we claim that A t yt for each t . If not, then forsome t we have A t = yt + δ for δ >
0. For any N > A Nt > NA t = yNt + N δ , and hence A Nt Nt > y + δ t , so A does not tend to y ; contradiction.In a similar way, using the inequality A Nt NA t + N −
1, we get A t > yt − t . Since each A t is an integer, this specifies each A t uniquely, except when yt is an integer. So the proof of (1)is complete. To complete the proof of (2), we must show that either A t = yt whenever yt ∈ N , or A t = yt − yt ∈ N ; in other words, it is not possible to find t and u such that A t = yt − A u = yu . But if we do have such t , u , then the inequality at the end of the proof of Lemma 7.3is violated. (cid:3) These results imply that we may parametrise arm sequences by real numbers in the interval[1 , n − B ( Λ )), and that there is a natural total order on arm sequences. In the notation of Lemma 7.4, theextreme sequences (0 , , , . . . ) and ( n − , n − , . . . ) are the sequences A , − and A n − , + respectively,and the sequence (1 , , , . . . ) defining Berg’s ladder crystal is A , + .artitionmodelsforthecrystalofthebasic U q ( b sl n )-module 31 References [B] C. Berg, ‘( l , b sl l ’,arXiv:math.CO / A -crystals’, J. Algebra (2007), 218–34.[G] I. Grojnowski, ‘A ffi ne b sl p controls the representation theory of the symmetric group and relatedHecke algebras’, arXiv:math / ffi necrystals and vertex models’, Internat. J. Modern Phys. Supp 1A (1992), 449–84.[K1] M. Kashiwara, ‘On crystal bases of the q -analogue of universal enveloping algebras’, DukeMath. J. (1991), 465–516.[K2] M. Kashiwara, Bases cristallines des groupes quantiques , Cours sp´ecialis´ees , Soci´et´eMath´ematique de France, Paris, 2002.[MM] K. Misra & T. Miwa, ‘Crystal base for the basic representation of U q ( b sl ( n ))’, Comm. Math.Phys. (1990), 79–88.[M] G. Mullineux, ‘Bijections on p -regular partitions and p -modular irreducibles of the symmetricgroups’, J. London Math. Soc. (2) (1979), 60–6.[S] J. Stembridge, ‘A local characterization of simply-laced crystals’, Trans. Amer. Math. Soc.355