# Crystal Bases as Tuples of Integer Sequences

Authors: Deniz Kus
Journal: ISRN Combinatorics
 (2013)

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## Abstract

We describe a setℛ∞ consisting of tuples of integer sequences and provide certain explicit maps on it. We show that this defines a semiregular crystal for 𝔰𝔩n+1 and 𝔰𝔭2n, respectively. Furthermore, we define for any dominant integral weight λ a connected subcrystal ℛ(λ) in ℛ∞, such that this crystal is isomorphic to the crystal graph B(λ). Finally, we provide an explicit description of these connected crystals ℛ(λ).


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## Body

## 1. Introduction

Let𝔤 be a symmetrizable Kac-Moody algebra and let 𝔘q(𝔤) be the corresponding quantum algebra. For these quantum algebras, Kashiwara developed the crystal bases theory for integrable modules in [1] and thus provided a remarkable combinatorial tool for studying these modules. In particular crystal bases can be viewed as bases at q=0 and they contain structures of edge-colored oriented graphs satisfying a set of axioms, called the crystal graphs. These crystal graphs have certain nice properties; for instance, characters of 𝔘q(𝔤)-modules can be computed and the decomposition of tensor products of modules into irreducible ones can also be determined from the crystal graphs, to name but a few. It is thus an important problem to have explicit realizations of crystal graphs.There are many such realizations, combinatorial and geometrical, worked out during the last decades; for instance, we refer to [2–5]. In [2], the authors give a tableaux realization of crystal graphs for irreducible modules over the quantum algebra for all classical Lie algebras, which is a purely combinatorial model. Another significant combinatorial model for any symmetrizable Kac-Moody algebra is provided in [3], called Littelmann's path model. The underlying set here is a set of piecewise linear maps, and the crystal graph of an irreducible module of any dominant integral highest weight λ can be generated by an algorithm using the straight path connecting 0 and λ.A geometrical realization of crystals is also known and is provided by Nakajima [5] by showing that there exists a crystal structure on the set of irreducible components of a lagrangian subvariety of the quiver variety ℳ. This realization can be translated into a purely combinatorial model, the set of Nakajima monomials, where the action of the Kashiwara operators can be understood as a multiplication with monomials. Moreover, it is shown in [6] that the connected component of any highest weight monomial of highest weight λ is isomorphic to the crystal graph B(λ) obtained from Kashiwara's crystal bases theory. For special highest weight monomials these connected components are explicitly characterized for 𝔰𝔩n+1 in [7] and for the other classical Lie algebras in [8]. A combinatorial isomorphism from connected components corresponding to arbitrary highest weight monomials of highest weight λ and those in [7, 8] is provided in [9] for the types A and C and in [10] for the types B and D.In this paper we introduce a setℛ∞ consisting of tuples of integer sequences; that is, a typical element in ℛ∞ is given by
(1)x=(x1,x2,…)∈ℛ∞,where each component xj consists of certain ordered pairs of integers, xj=(i1,i1′)⋯(is,is′) (see Definition 3). Furthermore, the number of nonzero components is finite. We provide certain maps on ℛ∞, the Kashiwara operators e~l, f~l, and maps ϵl, φl for all l=1,…,n and prove that ℛ∞ is a semiregular crystal if 𝔤 is 𝔰𝔩n+1 or 𝔰𝔭2n (see Definition 3 and Proposition 9).Moreover, we introduce for any dominant integral weightλ a subcrystal ℛ(λ) as the connected component of ℛ∞ containing a highest weight element rλ and prove the following theorem.Theorem 1.
Letλ be a dominant integral weight, then there exists a crystal isomorphism
(2)ℛ(λ)⟶B(λ),
mapping rλ to the highest weight element bλ∈B(λ).Therefore, similar to the setting of Nakajima monomials, a natural question arises; namely, can one characterize for each dominant integral weightλ explicitly the sequences appearing in ℛ(λ)? We answer this question by describing explicitly these connected components (for the special linear Lie algebra in Theorem 18 and the symplectic Lie algebra in Theorem 19).Our paper is organized as follows: in Section2 we fix some notations and review briefly the crystal theory. In Section 3 we present the main definitions, especially the definition of ℛ∞ and we equip our main object with a crystal structure. In Section 4 Nakajima monomials are recalled. In Section 5 we introduce for any dominant integral weight λ the subcrystals ℛ(λ) and describe them explicitly. Finally, in Section 6 we prove that they are isomorphic to B(λ).


## 2. Notations and a Review of Crystal Theory

Let𝔤 be a complex simple Lie algebra of rank n with index set I={1,…,n} and fix a Cartan subalgebra 𝔥 in 𝔤 and a Borel subalgebra 𝔟⊇𝔥. We denote by Φ⊆𝔥* the root system of the Lie algebra, and corresponding to the choice of 𝔟 let Φ+ be the subset of positive roots. Further, we denote by Π={α1,…,αn} the corresponding basis of Φ and the basis of the dual root system Φ∨⊆𝔥 is denoted by Π∨={α1∨,…,αn∨}. Let 𝔤=𝔫+⊕𝔥⊕𝔫- be a Cartan decomposition and for a given root α∈Φ let 𝔤α be the corresponding root space. For a dominant integral weight λ we denote by V(λ) the irreducible 𝔤-module with a highest weight λ. Fix a highest weight vector vλ∈V(λ), then V(λ)=𝔘(𝔫-)vλ, where 𝔘(𝔫-) denotes the universal enveloping algebra of 𝔫-. For an indetermined element q we denote by 𝔘q(𝔤) the corresponding quantum algebra. The theory of studying modules of quantum algebras is quite parallel to that of Kac-Moody algebras and the irreducible modules are classified again in terms of highest weights (see [11]). Using the crystal bases theory, introduced by Kashiwara in [1], we can compute the character of an integrable module M in the category 𝒪q as follows:
(3)chM=∑μ‍♯(Bμ)eμ,
whereby (L,B) is the crystal bases of M (see [11]). The crystal graph associated with the irreducible module of highest weight λ is denoted by B(λ). So finding expressions for the characters can be achieved by finding explicit combinatorial description of crystal bases. For some examples we refer to [2–4].From now on we assume that𝔤 is a classical Lie algebra of type An or Cn. Note that the positive roots are all of the following form:
(4)TypeAn:αi,j=αi+αi+1+⋯+αj,for1≤i≤j≤n,TypeCn:αi,j=αi+αi+1+⋯+αj,for1≤i≤j≤n,αi,j¯=αi+αi+1+⋯+αn+αn-1+⋯+αj,for1≤i≤j≤n.
Furthermore, let P=⨁i∈Iℤωi be the set of classical integral weights and P+=⨁i∈Iℤ+ωi be the set of classical dominant integral weights. Before we discuss the crystal bases theory in detail we review first the notion of abstract crystals.


### 2.1. Abstract Crystals

Crystal bases of integrable𝔘q(𝔤)-modules in the category 𝒪q are characterized by certain maps satisfying some properties. One can define the abstract notion of crystals associated with a Cartan datum as follows.Definition 2.
LetI be a finite index set and let A=(ai,j)i,j∈I be a generalized Cartan matrix with the Cartan datum (A,Π,Π∨,P,P∨). A crystal associated with the Cartan datum (A,Π,Π∨,P,P∨) is a set B together with maps wt:B→P, e~l,f~l:B→B∪{0}, and ϵl,φl:B→ℤ∪{-∞} satisfying the following properties for all l∈I:(1)
φl(b)=ϵl(b)+〈αl∨,wt(b)〉,(2)
wt(e~lb)=wt(b)+αl if e~lb∈B,(3)
wt(f~lb)=wt(b)-αl if f~lb∈B,(4)
ϵl(e~lb)=ϵl(b)-1, φl(e~lb)=φl(b)+1 if e~lb∈B,(5)
ϵl(f~lb)=ϵl(b)+1, φl(f~lb)=φl(b)-1 if f~lb∈B,(6)
f~lb=b′ if and only if e~lb′=b for b,b′∈B,(7)
ifφl(b)=-∞ for b∈B, then f~lb=e~lb = 0. Furthermore, a crystal B is said to be semiregular if the equalities
(5)ϵl(b)=max{k≥0∣e~lkb≠0},φl(b)=max{k≥0∣f~lkb≠0}
hold.The mapse~l and f~l are called Kashiwara's crystal operators and the map wt is called the weight function. So, on the one hand, one can associate with any integrable 𝔘q(𝔤)-module a set B satisfying the properties from Definition 3, and, on the other hand, one can study the notion of abstract crystals. A natural question which arises at this point is therefore the following: can one determine whether an abstract crystal is the crystal of a module? Stembridge [12] gave a set of local axioms to characterize the set of crystals of module in the class of all crystals when 𝔤 is simply laced and a list of local axioms for B2-crystals is provided in [13]. In the following sections we define our underlying set and realize the crystal obtained from Kashiwara's crystal bases theory for the types An and Cn. We start by equipping our underlying set with an abstract crystal structure and later we prove that this crystal is the crystal of a module.


## 2.1. Abstract Crystals

Crystal bases of integrable𝔘q(𝔤)-modules in the category 𝒪q are characterized by certain maps satisfying some properties. One can define the abstract notion of crystals associated with a Cartan datum as follows.Definition 2.
LetI be a finite index set and let A=(ai,j)i,j∈I be a generalized Cartan matrix with the Cartan datum (A,Π,Π∨,P,P∨). A crystal associated with the Cartan datum (A,Π,Π∨,P,P∨) is a set B together with maps wt:B→P, e~l,f~l:B→B∪{0}, and ϵl,φl:B→ℤ∪{-∞} satisfying the following properties for all l∈I:(1)
φl(b)=ϵl(b)+〈αl∨,wt(b)〉,(2)
wt(e~lb)=wt(b)+αl if e~lb∈B,(3)
wt(f~lb)=wt(b)-αl if f~lb∈B,(4)
ϵl(e~lb)=ϵl(b)-1, φl(e~lb)=φl(b)+1 if e~lb∈B,(5)
ϵl(f~lb)=ϵl(b)+1, φl(f~lb)=φl(b)-1 if f~lb∈B,(6)
f~lb=b′ if and only if e~lb′=b for b,b′∈B,(7)
ifφl(b)=-∞ for b∈B, then f~lb=e~lb = 0. Furthermore, a crystal B is said to be semiregular if the equalities
(5)ϵl(b)=max{k≥0∣e~lkb≠0},φl(b)=max{k≥0∣f~lkb≠0}
hold.The mapse~l and f~l are called Kashiwara's crystal operators and the map wt is called the weight function. So, on the one hand, one can associate with any integrable 𝔘q(𝔤)-module a set B satisfying the properties from Definition 3, and, on the other hand, one can study the notion of abstract crystals. A natural question which arises at this point is therefore the following: can one determine whether an abstract crystal is the crystal of a module? Stembridge [12] gave a set of local axioms to characterize the set of crystals of module in the class of all crystals when 𝔤 is simply laced and a list of local axioms for B2-crystals is provided in [13]. In the following sections we define our underlying set and realize the crystal obtained from Kashiwara's crystal bases theory for the types An and Cn. We start by equipping our underlying set with an abstract crystal structure and later we prove that this crystal is the crystal of a module.


## 3. Tuples of Integer Sequences as Crystals

In this section we introduce a setℛ∞ consisting of tuples of integer sequences (see Definition 16) and a crystal structure on it in the sense of Definition 3. Our purpose is to identify for any dominant integral weight λ certain subcrystals ℛ(λ); that is, ⋃λ∈P+‍ℛ(λ)⊆ℛ∞, such that ℛ(λ) has a strong connection to the crystal graph B(λ) (see Corollary 21).


### 3.1. Set of Tuples of Integer Sequences

In order to defineℛ∞ we consider a total order on I={1,…,n} if 𝔤 is of type An and a total order on I={1,…,n,n-1¯,…,1¯} if 𝔤 is of type Cn, namely,
(6)1<2<⋯<n,1<2<⋯<n<n-1¯<⋯<1¯,
respectively. Furthermore, especially in Section 5, we need for type Cn the following bijective map:
(7):I-⟶I,n↦n,i¯↦i,i↦i¯,fori∈{1,…,n-1}.
For 1≤i≤n,s∈ℤ≥0 we set ℛsi to be the set of all sequences (i1,i1′)⋯(is,is′) with ij,ij′∈I, such that
(8)1≤is<is-1<⋯<i1≤i≤i1′<i2′<⋯<is′≤maxI,ij′≤ij¯,j=1,…,s,
where maxI denotes the maximal element in I with respect to <. We denote by ∅i the unique element in ℛ0i.Definition 3.
We defineℛ∞ to be the set of all infinite sequences x=(x1,x2,x3,…) where each component xj is contained in ℛ=⋃s,i‍ℛsi∪{0} and only finitely many components are nonzero. We identify ℛk with the sequences of the form (x1,…,xk,0,0,…).Before we mention the crystal structure onℛ∞ we will initially introduce a list of properties. We need these to define the Kashiwara operators. Let x=(i1,i1′)⋯(is,is′)∈ℛsi be an arbitrary element and fix l∈I:(a)
(9)l∉{i1,…,is,i1′,…,is′},l+1∈{i1,…,is},ifl<i,l-1∈{i1′,…,is′},ifl>i,/,ifl=i,(l+1¯∈{i1′,…,is′}∨l¯∉{i1′,…,is′}),(a′)
replace in (a)∨ by ∧,(b)
(10)l¯∉{i1′,…,is′},l+1¯∈{i1′,…,is′},l+1∉{i1,…,is}∨l∈{i1,…,is},ifl<i,l-1∉{i1′,…,is′}∨l∈{i1′,…,is′},ifl>i,l=i1∨l=i1′,ifl=i,(c)
(11)l∈{i1,…,is,i1′,…,is′},l+1∉{i1,…,is},ifl<i,l-1∉{i1′,…,is′},ifl>i,(i1,i1′)=(l,l),ifl=i,(l+1¯∈{i1′,…,is′}∨l¯∉{i1′,…,is′}),(d)
(12)l¯∈{i1′,…,is′},l+1¯∉{i1′,…,is′},l+1∉{i1,…,is}∨l∈{i1,…,is},ifl<i,l-1∉{i1′,…,is′}∨l∈{i1′,…,is′},ifl>i,l=i1∨l=i1′,ifl=i,(d′)
replace everywhere in (d)∨ by ∧.Let us consider an example.Example 4.
(
1
) Let 𝔤=A5, l=2, and
(13)x1=(3,4)(1,5)∈ℛ23,x2=(2,3)∈ℛ12,
then x1 satisfies (a) while x2 violates (a).
(
2
) Let 𝔤=C3, l=1, and
(14)x=(1,2¯)∈ℛ12,
then x satisfies (c).Remark 5.
Ifx satisfies (a′) and (d′), respectively, then it satisfies also (a) and (d), respectively. If 𝔤 is further of type An, these properties can be simplified. In particular, the properties (a′), (b), (d), and (d′) are superfluous.Henceforth we define a crystal structure onℛ∞, such that the semiregularity holds. For this let x=(x1,x2,x3,…) be such a sequence with finitely many components different from zero; recall that each component is a sequence as in (8). The weight function is given by
(15)wt(x)=∑i=1n‍ciωi-∑j‍wt(xj),
where
(16)wt(xj)={∑j=1sαij,ij′ifxj=(i1,i1′)⋯(is,is′),0ifxj=0orxj∈{∅1,…,∅n}
and ci=♯{xj≠0∣xj∈⋃s‍ℛsi}. Suppose that the nonzero components in x are given by xq1∈ℛs1j1,…,xqk∈ℛskjk. For fixed l∈I we define the following maps.(i) For2≤j≤k+1, let σlj:ℛ∞→ℤ≥0 be the map given by
(17)σlj(x)=alj(x)+blj(x),
where
(18)alj(x)=♯{xqp∣1≤p≤j-1,xqpsatisfies(a)or(b)},blj(x)=♯{xqp∣1≤p≤j-1,xqpsatisfies(a′)}.
Furthermore, we define θl(xqp) to be the sequence which arises out of xqp by
(19)replacingl+1byl,ifl<jp,replacingl-1byl,ifl>jp,adding(l,l),ifl=jp,
if xqp satisfies (a). If xqp satisfies (b), let θl(xqp) be the sequence which arises out of xqp by replacing l+1¯ by l¯. If neither (a) nor (b) is fulfilled, we set θl(xqp)=0.(ii) For2≤j≤k, let τlj:ℛ∞→ℤ≥0 be the map given by
(20)τlj(x)=clj(x)+dlj(x),
where
(21)clj(x)=♯{xqp∣2≤p≤j,xqpsatisfies(c)or(d)},dlj(x)=♯{xqp∣2≤p≤j,xqpsatisfies(d′)}.
We define ρl(xqp) to be the sequence which arises out of xqp by
(22)replacinglbyl+1,ifl<jp,replacinglbyl-1,ifl>jp,erasing(l,l),ifl=jp,
if xqp satisfies (c). If xqp satisfies (d), let ρl(xqp) be the sequence which arises out of xqp by replacing l¯ by l+1¯. If neither (c) nor (d) is fulfilled, we set ρl(xqp)=0.Remark 6.
(
1
) For j=1, we set σlj=τlj=0.
(
2
) Note that the image of x∈⋃s‍ℛsi under the maps  θl and ρl, respectively, is contained in ⋃s‍ℛsi∪{0}, that is,
(23)θl:⋃s‍ℛsi⟶⋃s‍ℛsi∪{0},ρl:⋃s‍ℛsi⟶⋃s‍ℛsi∪{0}.One important fact about these maps is described in the next lemma.Lemma 7.
Letx,x~ be nonzero sequences as in (8), then one has
(24)θl(x)=x~iffρl(x~)=x.Proof.
One can easily show thatx satisfies (a) if and only if x~ satisfies (c). Hence, we can suppose that x does not fulfill all properties enumerated in (a). By observing the action we see that x~ arises from x by replacing l+1¯ by l¯, which means that (c) is violated. In particular l+1¯ does not appear in x~ and l¯ appears in x~, which means that the properties in (d) hold. Hence, ρ(x~)=x. The arguments for the reverse direction are the same.Let(25)fl(x)≔max{1≤p≤k∣σlp(x)-τlp(x)=min{σlj(x)-τlj(x)∣1≤j≤k}},(26)el(x)≔min{1≤p≤k∣σlp(x)-τlp(x)=min{σlj(x)-τlj(x)∣1≤j≤k}}.
Now we are able to define the Kashiwara operators,
(27)f~lx={0,ifθl(xqfl(x))=0,(…,θl(xqfl(x)),…),else,e~lx={0,ifρl(xqel(x))=0,(…,ρl(xqel(x)),…),else.
Let us consider an example.Example 8.
(
1
) Let 𝔤=A4 and x=(x1,x2,x3,0,0,…) with
(28)x1=(2,2)(1,3)∈ℛ22,x2=(2,4)∈ℛ14,x3=(3,3)(2,4)∈ℛ23.
For l=2, we get σ21(x)=σ22(x)=σ23(x)=τ21(x)=0,τ22(x)=1,τ23(x)=2 and hence
(29)f~2x=0.
(
2
)  Let 𝔤=C3 and x=(x1,x2,0,0,…) with
(30)x1=(1,2¯)∈ℛ12,x2=(3,3)(1,2¯)∈ℛ23,
then σ32(x)=0,τ32(x)=1, and
(31)e~3x=(x1,x1,0,0⋯).It remains to define the mapsφl and ϵl. These maps are given by the next formula:
(32)ϵl(x)=τl2((∅1,xq1,0,0⋯))-min{σlj(x)-τlj(x)∣1≤j≤k},φl(x)=σlk+1(x)-τlk(x)-min{σlj(x)-τlj(x)∣1≤j≤k}.Proposition 9.
The setℛ∞ becomes a semiregular crystal.Proof.
In Lemma10 we proof the semiregularity of ℛ∞, which ensures that (4) and (5) from Definition 3 hold. So, to verify the proposition, it is sufficient to prove (1), (2), (3), and (6), where (2) and (3) are easily checked with the help of Remark 6. Let us start by proving (1); so let x∈ℛ∞ be arbitrary with finitely many nonzero components, say, xq1∈ℛs1j1,…,xqk∈ℛskjk. Then we order these components in a way such that the first components are contained in ⋃s‍⋃m<l‍ℛsm followed by components in ⋃s‍⋃m>l‍ℛsm and the last ones are contained in ⋃s‍ℛsl. So we can write the set of nonzero components of x as a disjoint union of three subsets A<l∪A>l∪A=l. Further let x1 be the element in ℛ∞ obtained from x by replacing all components not belonging to A<l by 0. And x2 and x3, respectively, are similarly defined using A>l and A=l, respectively. Then we get
(33)φl(x)-ϵl(x)=σlk+1(x)-τlk(x)-τl2((∅1,xq1,0,0⋯))=σl|A<l|+1(x1)-τl|A<l|(x1)+σl|A>l|+1(x2)-τl|A>l|(x2)+σl|A=l|+1(x3)-τl|A=l|(x3)-∑i=13‍τl2((∅1,yi,0,0⋯)),
where the sequence yi is the first nonzero element in xi. A short calculation shows
(34)σl|A<l|+1(x1)-τl|A<l|(x1)-τl2((∅1,y1,0,0⋯))=-∑1≤i≤k:xqi∈A<l‍〈αl∨,wt(xqi)〉,σl|A>l|+1(x2)-τl|A>l|(x2)-τl2((∅1,y2,0,0⋯))=-∑1≤i≤k:xqi∈A>l‍〈αl∨,wt(xqi)〉,σl|A=l|+1(x3)-τl|A=l|(x3)-τl2((∅1,y3,0,0⋯))=cl-∑1≤i≤k:xqi∈A=l‍〈αl∨,wt(xqi)〉.
Now we proceed to prove (6). By the definition of the Kashiwara operators and Lemma 7, it is enough to show that  fl(e~lx)=el(x) and el(f~lx)=fl(x). Since the proofs are similar, we prove only the latter equation. Assume that f~lx≠0 and fl(x)≠1, then
(35)σlj(f~lx)={σlj(x),ifj≤fl(x),σlj(x)-1,ifj>fl(x),τlj(f~lx)={τlj(x),ifj≤fl(x)-1,τlj(x)+1,ifj≥fl(x).
Subsequently we have
(36)el(f~lx)=min{1≤p≤k∣σlp(f~lx)-τlp(f~lx)=min{σlj(f~lx)-τlj(f~lx)∣1≤j≤k}}=min{τlfl(x)(x)1≤p≤k∣σlp(f~lx)-τlp(f~lx)=σlfl(x)(x)-τlfl(x)(x)-1}=fl(x).
In the case where fl(x)=1, we obtain
(37)el(f~lx)=min{1≤p≤k∣σlp(f~lx)-τlp(f~lx)=0}=1.Hence we have shown that the setℛ∞ is an abstract crystal provided that the semiregularity is shown. Thus, our aim now is to verify that the maps φl and ϵl, respectively, determine how often one can act with f~l and e~l, respectively. The semiregularity is a necessary condition of a crystal B, if one wants to identify it with the crystal graph B(λ).Lemma 10.
Letϵl and φl  be as in (32). For a given element x∈ℛ∞ one obtains
(38)ϵl(x)=max{k≥0∣e~lkx≠0},φl(x)=max{k≥0∣f~lkx≠0}.Proof.
We proof the statement by induction onz:=max{k≥0∣e~lkx≠0}; so let z=0 and suppose first that el(x)≠1. By the definition of el(x) and ρl we have
(39)σlel(x)(x)-τlel(x)(x)<σlel(x)-1(x)-τlel(x)-1(x)⟺σlel(x)(x)<σlel(x)-1(x),
which is a contradiction to el(x)≠1. Hence, we have el(x)=1 and as a consequence we obtain ϵl(x)=τl2((∅1,xq1,0,0,⋯))=0, which proves the initial step. Now assume that z>0 and consider the element e~lx, where we again presume initially el(x)≠1. By applying the induction hypothesis and using
(40)σlj(e~lx)={σlj(x),ifj≤el(x),σlj(x)+1,ifj>el(x),τlj(e~lx)={τlj(x),ifj≤el(x)-1,τlj(x)-1,ifj≥el(x),
we arrive at
(41)max{k≥0∣e~lk+1x≠0}=ϵl(e~lx)=τl2((∅1,xq1,0,0⋯))-(σle(x)(x)-τle(x)(x)+1)=ϵl(x)-1.
If el(x)=1,
(42)ϵl(e~l(x))=τl2((∅1,ρl(xq1),0,0⋯))+σle(x)(x)-τle(x)(x)=ϵl(x)-1.
The proof of the remaining equality φl(x)=max{k≥0∣f~lkx≠0} is quite parallel.Before we introduce the subcrystalsℛ(λ) we need some facts about the theory of tensor products of crystals. The tensor product rule is a very nice combinatorial feature and important to realize the crystal bases of a tensor product of two 𝔘q(𝔤)-modules.


## 3.1. Set of Tuples of Integer Sequences

In order to defineℛ∞ we consider a total order on I={1,…,n} if 𝔤 is of type An and a total order on I={1,…,n,n-1¯,…,1¯} if 𝔤 is of type Cn, namely,
(6)1<2<⋯<n,1<2<⋯<n<n-1¯<⋯<1¯,
respectively. Furthermore, especially in Section 5, we need for type Cn the following bijective map:
(7):I-⟶I,n↦n,i¯↦i,i↦i¯,fori∈{1,…,n-1}.
For 1≤i≤n,s∈ℤ≥0 we set ℛsi to be the set of all sequences (i1,i1′)⋯(is,is′) with ij,ij′∈I, such that
(8)1≤is<is-1<⋯<i1≤i≤i1′<i2′<⋯<is′≤maxI,ij′≤ij¯,j=1,…,s,
where maxI denotes the maximal element in I with respect to <. We denote by ∅i the unique element in ℛ0i.Definition 3.
We defineℛ∞ to be the set of all infinite sequences x=(x1,x2,x3,…) where each component xj is contained in ℛ=⋃s,i‍ℛsi∪{0} and only finitely many components are nonzero. We identify ℛk with the sequences of the form (x1,…,xk,0,0,…).Before we mention the crystal structure onℛ∞ we will initially introduce a list of properties. We need these to define the Kashiwara operators. Let x=(i1,i1′)⋯(is,is′)∈ℛsi be an arbitrary element and fix l∈I:(a)
(9)l∉{i1,…,is,i1′,…,is′},l+1∈{i1,…,is},ifl<i,l-1∈{i1′,…,is′},ifl>i,/,ifl=i,(l+1¯∈{i1′,…,is′}∨l¯∉{i1′,…,is′}),(a′)
replace in (a)∨ by ∧,(b)
(10)l¯∉{i1′,…,is′},l+1¯∈{i1′,…,is′},l+1∉{i1,…,is}∨l∈{i1,…,is},ifl<i,l-1∉{i1′,…,is′}∨l∈{i1′,…,is′},ifl>i,l=i1∨l=i1′,ifl=i,(c)
(11)l∈{i1,…,is,i1′,…,is′},l+1∉{i1,…,is},ifl<i,l-1∉{i1′,…,is′},ifl>i,(i1,i1′)=(l,l),ifl=i,(l+1¯∈{i1′,…,is′}∨l¯∉{i1′,…,is′}),(d)
(12)l¯∈{i1′,…,is′},l+1¯∉{i1′,…,is′},l+1∉{i1,…,is}∨l∈{i1,…,is},ifl<i,l-1∉{i1′,…,is′}∨l∈{i1′,…,is′},ifl>i,l=i1∨l=i1′,ifl=i,(d′)
replace everywhere in (d)∨ by ∧.Let us consider an example.Example 4.
(
1
) Let 𝔤=A5, l=2, and
(13)x1=(3,4)(1,5)∈ℛ23,x2=(2,3)∈ℛ12,
then x1 satisfies (a) while x2 violates (a).
(
2
) Let 𝔤=C3, l=1, and
(14)x=(1,2¯)∈ℛ12,
then x satisfies (c).Remark 5.
Ifx satisfies (a′) and (d′), respectively, then it satisfies also (a) and (d), respectively. If 𝔤 is further of type An, these properties can be simplified. In particular, the properties (a′), (b), (d), and (d′) are superfluous.Henceforth we define a crystal structure onℛ∞, such that the semiregularity holds. For this let x=(x1,x2,x3,…) be such a sequence with finitely many components different from zero; recall that each component is a sequence as in (8). The weight function is given by
(15)wt(x)=∑i=1n‍ciωi-∑j‍wt(xj),
where
(16)wt(xj)={∑j=1sαij,ij′ifxj=(i1,i1′)⋯(is,is′),0ifxj=0orxj∈{∅1,…,∅n}
and ci=♯{xj≠0∣xj∈⋃s‍ℛsi}. Suppose that the nonzero components in x are given by xq1∈ℛs1j1,…,xqk∈ℛskjk. For fixed l∈I we define the following maps.(i) For2≤j≤k+1, let σlj:ℛ∞→ℤ≥0 be the map given by
(17)σlj(x)=alj(x)+blj(x),
where
(18)alj(x)=♯{xqp∣1≤p≤j-1,xqpsatisfies(a)or(b)},blj(x)=♯{xqp∣1≤p≤j-1,xqpsatisfies(a′)}.
Furthermore, we define θl(xqp) to be the sequence which arises out of xqp by
(19)replacingl+1byl,ifl<jp,replacingl-1byl,ifl>jp,adding(l,l),ifl=jp,
if xqp satisfies (a). If xqp satisfies (b), let θl(xqp) be the sequence which arises out of xqp by replacing l+1¯ by l¯. If neither (a) nor (b) is fulfilled, we set θl(xqp)=0.(ii) For2≤j≤k, let τlj:ℛ∞→ℤ≥0 be the map given by
(20)τlj(x)=clj(x)+dlj(x),
where
(21)clj(x)=♯{xqp∣2≤p≤j,xqpsatisfies(c)or(d)},dlj(x)=♯{xqp∣2≤p≤j,xqpsatisfies(d′)}.
We define ρl(xqp) to be the sequence which arises out of xqp by
(22)replacinglbyl+1,ifl<jp,replacinglbyl-1,ifl>jp,erasing(l,l),ifl=jp,
if xqp satisfies (c). If xqp satisfies (d), let ρl(xqp) be the sequence which arises out of xqp by replacing l¯ by l+1¯. If neither (c) nor (d) is fulfilled, we set ρl(xqp)=0.Remark 6.
(
1
) For j=1, we set σlj=τlj=0.
(
2
) Note that the image of x∈⋃s‍ℛsi under the maps  θl and ρl, respectively, is contained in ⋃s‍ℛsi∪{0}, that is,
(23)θl:⋃s‍ℛsi⟶⋃s‍ℛsi∪{0},ρl:⋃s‍ℛsi⟶⋃s‍ℛsi∪{0}.One important fact about these maps is described in the next lemma.Lemma 7.
Letx,x~ be nonzero sequences as in (8), then one has
(24)θl(x)=x~iffρl(x~)=x.Proof.
One can easily show thatx satisfies (a) if and only if x~ satisfies (c). Hence, we can suppose that x does not fulfill all properties enumerated in (a). By observing the action we see that x~ arises from x by replacing l+1¯ by l¯, which means that (c) is violated. In particular l+1¯ does not appear in x~ and l¯ appears in x~, which means that the properties in (d) hold. Hence, ρ(x~)=x. The arguments for the reverse direction are the same.Let(25)fl(x)≔max{1≤p≤k∣σlp(x)-τlp(x)=min{σlj(x)-τlj(x)∣1≤j≤k}},(26)el(x)≔min{1≤p≤k∣σlp(x)-τlp(x)=min{σlj(x)-τlj(x)∣1≤j≤k}}.
Now we are able to define the Kashiwara operators,
(27)f~lx={0,ifθl(xqfl(x))=0,(…,θl(xqfl(x)),…),else,e~lx={0,ifρl(xqel(x))=0,(…,ρl(xqel(x)),…),else.
Let us consider an example.Example 8.
(
1
) Let 𝔤=A4 and x=(x1,x2,x3,0,0,…) with
(28)x1=(2,2)(1,3)∈ℛ22,x2=(2,4)∈ℛ14,x3=(3,3)(2,4)∈ℛ23.
For l=2, we get σ21(x)=σ22(x)=σ23(x)=τ21(x)=0,τ22(x)=1,τ23(x)=2 and hence
(29)f~2x=0.
(
2
)  Let 𝔤=C3 and x=(x1,x2,0,0,…) with
(30)x1=(1,2¯)∈ℛ12,x2=(3,3)(1,2¯)∈ℛ23,
then σ32(x)=0,τ32(x)=1, and
(31)e~3x=(x1,x1,0,0⋯).It remains to define the mapsφl and ϵl. These maps are given by the next formula:
(32)ϵl(x)=τl2((∅1,xq1,0,0⋯))-min{σlj(x)-τlj(x)∣1≤j≤k},φl(x)=σlk+1(x)-τlk(x)-min{σlj(x)-τlj(x)∣1≤j≤k}.Proposition 9.
The setℛ∞ becomes a semiregular crystal.Proof.
In Lemma10 we proof the semiregularity of ℛ∞, which ensures that (4) and (5) from Definition 3 hold. So, to verify the proposition, it is sufficient to prove (1), (2), (3), and (6), where (2) and (3) are easily checked with the help of Remark 6. Let us start by proving (1); so let x∈ℛ∞ be arbitrary with finitely many nonzero components, say, xq1∈ℛs1j1,…,xqk∈ℛskjk. Then we order these components in a way such that the first components are contained in ⋃s‍⋃m<l‍ℛsm followed by components in ⋃s‍⋃m>l‍ℛsm and the last ones are contained in ⋃s‍ℛsl. So we can write the set of nonzero components of x as a disjoint union of three subsets A<l∪A>l∪A=l. Further let x1 be the element in ℛ∞ obtained from x by replacing all components not belonging to A<l by 0. And x2 and x3, respectively, are similarly defined using A>l and A=l, respectively. Then we get
(33)φl(x)-ϵl(x)=σlk+1(x)-τlk(x)-τl2((∅1,xq1,0,0⋯))=σl|A<l|+1(x1)-τl|A<l|(x1)+σl|A>l|+1(x2)-τl|A>l|(x2)+σl|A=l|+1(x3)-τl|A=l|(x3)-∑i=13‍τl2((∅1,yi,0,0⋯)),
where the sequence yi is the first nonzero element in xi. A short calculation shows
(34)σl|A<l|+1(x1)-τl|A<l|(x1)-τl2((∅1,y1,0,0⋯))=-∑1≤i≤k:xqi∈A<l‍〈αl∨,wt(xqi)〉,σl|A>l|+1(x2)-τl|A>l|(x2)-τl2((∅1,y2,0,0⋯))=-∑1≤i≤k:xqi∈A>l‍〈αl∨,wt(xqi)〉,σl|A=l|+1(x3)-τl|A=l|(x3)-τl2((∅1,y3,0,0⋯))=cl-∑1≤i≤k:xqi∈A=l‍〈αl∨,wt(xqi)〉.
Now we proceed to prove (6). By the definition of the Kashiwara operators and Lemma 7, it is enough to show that  fl(e~lx)=el(x) and el(f~lx)=fl(x). Since the proofs are similar, we prove only the latter equation. Assume that f~lx≠0 and fl(x)≠1, then
(35)σlj(f~lx)={σlj(x),ifj≤fl(x),σlj(x)-1,ifj>fl(x),τlj(f~lx)={τlj(x),ifj≤fl(x)-1,τlj(x)+1,ifj≥fl(x).
Subsequently we have
(36)el(f~lx)=min{1≤p≤k∣σlp(f~lx)-τlp(f~lx)=min{σlj(f~lx)-τlj(f~lx)∣1≤j≤k}}=min{τlfl(x)(x)1≤p≤k∣σlp(f~lx)-τlp(f~lx)=σlfl(x)(x)-τlfl(x)(x)-1}=fl(x).
In the case where fl(x)=1, we obtain
(37)el(f~lx)=min{1≤p≤k∣σlp(f~lx)-τlp(f~lx)=0}=1.Hence we have shown that the setℛ∞ is an abstract crystal provided that the semiregularity is shown. Thus, our aim now is to verify that the maps φl and ϵl, respectively, determine how often one can act with f~l and e~l, respectively. The semiregularity is a necessary condition of a crystal B, if one wants to identify it with the crystal graph B(λ).Lemma 10.
Letϵl and φl  be as in (32). For a given element x∈ℛ∞ one obtains
(38)ϵl(x)=max{k≥0∣e~lkx≠0},φl(x)=max{k≥0∣f~lkx≠0}.Proof.
We proof the statement by induction onz:=max{k≥0∣e~lkx≠0}; so let z=0 and suppose first that el(x)≠1. By the definition of el(x) and ρl we have
(39)σlel(x)(x)-τlel(x)(x)<σlel(x)-1(x)-τlel(x)-1(x)⟺σlel(x)(x)<σlel(x)-1(x),
which is a contradiction to el(x)≠1. Hence, we have el(x)=1 and as a consequence we obtain ϵl(x)=τl2((∅1,xq1,0,0,⋯))=0, which proves the initial step. Now assume that z>0 and consider the element e~lx, where we again presume initially el(x)≠1. By applying the induction hypothesis and using
(40)σlj(e~lx)={σlj(x),ifj≤el(x),σlj(x)+1,ifj>el(x),τlj(e~lx)={τlj(x),ifj≤el(x)-1,τlj(x)-1,ifj≥el(x),
we arrive at
(41)max{k≥0∣e~lk+1x≠0}=ϵl(e~lx)=τl2((∅1,xq1,0,0⋯))-(σle(x)(x)-τle(x)(x)+1)=ϵl(x)-1.
If el(x)=1,
(42)ϵl(e~l(x))=τl2((∅1,ρl(xq1),0,0⋯))+σle(x)(x)-τle(x)(x)=ϵl(x)-1.
The proof of the remaining equality φl(x)=max{k≥0∣f~lkx≠0} is quite parallel.Before we introduce the subcrystalsℛ(λ) we need some facts about the theory of tensor products of crystals. The tensor product rule is a very nice combinatorial feature and important to realize the crystal bases of a tensor product of two 𝔘q(𝔤)-modules.


## 4. Tensor Products and Nakajima Monomials

In this section, we want to recall tensor products of crystals and investigate the action of Kashiwara operators on tensor products. With the aim of having a different realization ofB(λ) from our approach, we want to introduce the set of all Nakajima monomials, such that we can think of B(λ) in terms of certain monomials. This theory is discovered by Nakajima [14] and generalized by Kashiwara [6].


### 4.1. Tensor Product of Crystals

Suppose that we have two abstract crystalsB1, B2 in the sense of Definition 3, then we can construct a new crystal which is as a set B1×B2. This crystal is denoted by B1⊗B2 and the Kashiwara operators are given as follows:
(43)f~l(b1⊗b2)={(f~lb1)⊗b2,ifφl(b1)>ϵl(b2),b1⊗(f~lb2),ifφl(b1)≤ϵl(b2),e~l(b1⊗b2)={(e~lb1)⊗b2,ifφl(b1)≥ϵl(b2),b1⊗(e~lb2),ifφl(b1)<ϵl(b2).
Furthermore, one can describe explicitly the maps wt, φl, and ϵl on B1⊗B2, namely,
(44)wt(b1⊗b2)=wt(b1)+wt(b2),φl(b1⊗b2)=max{φl(b2),φl(b1)+φl(b2)-ϵl(b2)},ϵl(b1⊗b2)=max{ϵl(b1),ϵl(b1)+ϵl(b2)-φl(b1)}.
One of the most important interpretations of the tensor product rule is the following theorem (for more details see [11]).Theorem 11.
LetMj be an integrable module in the category 𝒪q and let (Lj,Bj)  be a crystal  bases of Mj,(j=1,2). Set L=L1⊗L2 and B=B1⊗B2. Then (L,B)  is a crystal  bases of M1⊗M2.


### 4.2. Nakajima Monomials

Fori∈I and n∈ℤ, we consider monomials in the variables Yi(n); that is, we obtain the set of Nakajima monomials ℳ as follows:
(45)ℳ≔{∏i∈I,n∈ℤ‍Yi(n)yi(n)∣yi(n)∈ℤvanishexceptforfinitelymany(i,n)∏i∈I,n∈ℤ‍}.
With the goal of defining a crystal structure on ℳ, we take some integers c=(ci,j)i≠j such that ci,j+cj,i=1. Let now M=∏i∈I,n∈ℤ‍Yi(n)yi(n) be an arbitrary monomial in ℳ and l∈I; then we set.
(46)wt(M)=∑i‍(∑n‍yi(n))ωi,φl(M)=max{∑k≤n‍yl(k)∣n∈ℤ},ϵl(M)=max{-∑k>n‍yl(k)∣n∈ℤ},nfl=min{n∣φl(M)=∑k≤n‍yl(k)},nel=max{n∣ϵl(M)=-∑k>n‍yl(k)}.
The Kashiwara operators are defined as follows:
(47)f~lM={Al(nfl)-1M,ifφl(M)>0,0,ifφl(M)=0,e~lM={Al(nel)M,ifϵl(M)>0,0,ifϵl(M)=0,
whereby
(48)Al(n):=Yl(n)Yl(n+1)∏i≠l‍Yi(n+ci,l)〈αi∨,αl〉.
The following two results are shown by Kashiwara [6].Proposition 12.
With the mapswt, φl, ϵl, f~l, e~l, l∈I, the set ℳ becomes a semiregular crystal.Remark 13.
Apriori the crystal structure depends on c; hence, we will denote this crystal by ℳc. But it is easy to see that the isomorphism class of ℳc does not depend on this choice. In the literature c is often chosen as
(49)ci,j={0,ifi>j,1,elseorci,j={0,ifi<j,1,else.Proposition 14.
LetM be a monomial in ℳ, such that e~lM=0 for all l∈I. Then the connected component of ℳ containing M is isomorphic to B(wt(M)).According to the latter proposition, it is of great interest to describe these connected components explicitly. This is worked out for special highest weight monomials for all classical Lie algebras in [7, 8] and for the affine Lie algebra An(1) in [15]. We recall the results here only for type Cn stated originally in [8].Proposition 15.
Letλ=∑i=1n‍miωi be a dominant integral weight and consider the highest weight monomial M=Y1(1)m1⋯Yn(1)mn. Then the connected component of ℳ containing M is characterized as the set of monomials of the form
(50)Xt1,1(1)⋯Xt1,α1(1)⋯Xtn,1(n)⋯Xtn,αn(n),tr,s∈{1<⋯<n<n¯<⋯<1¯},
satisfying  (1)
αj=mj+⋯+mn  for all j=1,…,n,(2)
tj,1≥⋯≥tj,αj  for all j=1,…,n,(3)
tj-1,k>tj,k  for all j=2,…,n and k=1,…,αj, where
(51)Xi(m)=Yi-1(m+1)-1Yi(m),Xi¯(m)=Yi-1(m+(n-i+1))Yi(m+(n-i+1))-1.Summarized, we have a semiregular crystalℳand for each dominant integral weight λ certain connected subcrystals contained in ℳ. These are isomorphic to B(λ) and an explicit description of these components is worked out for the classical simple Lie algebras and An(1). In the remaining parts of this paper we prove a similar result as Propositions 14 and 15, whereby our “big” semiregular crystal is ℛ∞.


## 4.1. Tensor Product of Crystals

Suppose that we have two abstract crystalsB1, B2 in the sense of Definition 3, then we can construct a new crystal which is as a set B1×B2. This crystal is denoted by B1⊗B2 and the Kashiwara operators are given as follows:
(43)f~l(b1⊗b2)={(f~lb1)⊗b2,ifφl(b1)>ϵl(b2),b1⊗(f~lb2),ifφl(b1)≤ϵl(b2),e~l(b1⊗b2)={(e~lb1)⊗b2,ifφl(b1)≥ϵl(b2),b1⊗(e~lb2),ifφl(b1)<ϵl(b2).
Furthermore, one can describe explicitly the maps wt, φl, and ϵl on B1⊗B2, namely,
(44)wt(b1⊗b2)=wt(b1)+wt(b2),φl(b1⊗b2)=max{φl(b2),φl(b1)+φl(b2)-ϵl(b2)},ϵl(b1⊗b2)=max{ϵl(b1),ϵl(b1)+ϵl(b2)-φl(b1)}.
One of the most important interpretations of the tensor product rule is the following theorem (for more details see [11]).Theorem 11.
LetMj be an integrable module in the category 𝒪q and let (Lj,Bj)  be a crystal  bases of Mj,(j=1,2). Set L=L1⊗L2 and B=B1⊗B2. Then (L,B)  is a crystal  bases of M1⊗M2.


## 4.2. Nakajima Monomials

Fori∈I and n∈ℤ, we consider monomials in the variables Yi(n); that is, we obtain the set of Nakajima monomials ℳ as follows:
(45)ℳ≔{∏i∈I,n∈ℤ‍Yi(n)yi(n)∣yi(n)∈ℤvanishexceptforfinitelymany(i,n)∏i∈I,n∈ℤ‍}.
With the goal of defining a crystal structure on ℳ, we take some integers c=(ci,j)i≠j such that ci,j+cj,i=1. Let now M=∏i∈I,n∈ℤ‍Yi(n)yi(n) be an arbitrary monomial in ℳ and l∈I; then we set.
(46)wt(M)=∑i‍(∑n‍yi(n))ωi,φl(M)=max{∑k≤n‍yl(k)∣n∈ℤ},ϵl(M)=max{-∑k>n‍yl(k)∣n∈ℤ},nfl=min{n∣φl(M)=∑k≤n‍yl(k)},nel=max{n∣ϵl(M)=-∑k>n‍yl(k)}.
The Kashiwara operators are defined as follows:
(47)f~lM={Al(nfl)-1M,ifφl(M)>0,0,ifφl(M)=0,e~lM={Al(nel)M,ifϵl(M)>0,0,ifϵl(M)=0,
whereby
(48)Al(n):=Yl(n)Yl(n+1)∏i≠l‍Yi(n+ci,l)〈αi∨,αl〉.
The following two results are shown by Kashiwara [6].Proposition 12.
With the mapswt, φl, ϵl, f~l, e~l, l∈I, the set ℳ becomes a semiregular crystal.Remark 13.
Apriori the crystal structure depends on c; hence, we will denote this crystal by ℳc. But it is easy to see that the isomorphism class of ℳc does not depend on this choice. In the literature c is often chosen as
(49)ci,j={0,ifi>j,1,elseorci,j={0,ifi<j,1,else.Proposition 14.
LetM be a monomial in ℳ, such that e~lM=0 for all l∈I. Then the connected component of ℳ containing M is isomorphic to B(wt(M)).According to the latter proposition, it is of great interest to describe these connected components explicitly. This is worked out for special highest weight monomials for all classical Lie algebras in [7, 8] and for the affine Lie algebra An(1) in [15]. We recall the results here only for type Cn stated originally in [8].Proposition 15.
Letλ=∑i=1n‍miωi be a dominant integral weight and consider the highest weight monomial M=Y1(1)m1⋯Yn(1)mn. Then the connected component of ℳ containing M is characterized as the set of monomials of the form
(50)Xt1,1(1)⋯Xt1,α1(1)⋯Xtn,1(n)⋯Xtn,αn(n),tr,s∈{1<⋯<n<n¯<⋯<1¯},
satisfying  (1)
αj=mj+⋯+mn  for all j=1,…,n,(2)
tj,1≥⋯≥tj,αj  for all j=1,…,n,(3)
tj-1,k>tj,k  for all j=2,…,n and k=1,…,αj, where
(51)Xi(m)=Yi-1(m+1)-1Yi(m),Xi¯(m)=Yi-1(m+(n-i+1))Yi(m+(n-i+1))-1.Summarized, we have a semiregular crystalℳand for each dominant integral weight λ certain connected subcrystals contained in ℳ. These are isomorphic to B(λ) and an explicit description of these components is worked out for the classical simple Lie algebras and An(1). In the remaining parts of this paper we prove a similar result as Propositions 14 and 15, whereby our “big” semiregular crystal is ℛ∞.


## 5. Explicit Description of the Connected Components

In this section we define for the dominant integral weightsλ=∑i=1n‍miωi certain connected subcrystals ℛ(λ)⊆ℛ∞. Furthermore, we provide an explicit description of these crystals in Theorems 18 and 19, respectively; that is, we give a set of conditions describing ℛ(λ).Definition 16.
For a dominant integral weightλ=∑i=1n‍miωi, let ℛ(λ) be the connected component of ℛ∞ containing
(52)rλ=(∅1,…,∅1︸m1,…,∅n,…,∅n︸mn,0,0⋯).Note that the weight ofrλ is precisely λ. Furthermore, by definition, ℛ(λ) is connected and for λ=ωi we can immediately provide a description of ℛ(ωi). To be more accurate we prove as a first step the following proposition:Proposition 17.
Letλ=ωi; then one obtains
(53)ℛ(λ)=(⋃s‍Rsi,0,0,…)=⋃s‍Rsi.Proof.
Since(⋃s‍Rsi,0,0,…) is stable under the Kashiwara operators f~l and e~l (Remark 6), it is enough to prove that ∅i is the unique highest weight element in ⋃s‍Rsi, that is,
(54)e~l(x,0,0,…)=0∀l∈I⟹x=(∅i,0,0⋯).
Assume x=(i1,i1′)⋯(is,is′) and e~l(x,0,0,…)=0foralll∈I. If i1<i, we have e~i1(x,0,0,…)≠0 and if i<i1′≤n, we have e~i1′(x,0,0,…)≠0. So let n-1¯≤i1′ (this case cannot appear if 𝔤=An); then since i1′∈{i1′,…,is′}, i1′¯+1¯∉{i1′,…,is′} and i1′≤i1¯=i¯, we obtain that i1′¯∈{i,…,n-1}. Thus, x satisfies (d) and hence e~i1′¯(x,0,0,…)≠0. According to these calculations, the pair (i,i) must appear in x. This implies e~i(x,0,0,…)≠0.
For generalλ’s, we can describe the connected components and refer to the following two subsections.


### 5.1. Explicit Description ofℛ(λ) in Type An

In this subsection we give an explicit characterization ofℛ(λ) if 𝔤 is the special linear Lie algebra; that is, we give conditions whether a sequence x is contained in ℛ(λ) or not (Theorem 18). Initially we note that ℛ(λ) lives in ℛk=∑i‍mi and the first k components are nonzero. For simplicity we set el(x)=e(x), fl(x)=f(x), θl=θ, and ρl=ρ.Theorem 18.
The crystalℛ(λ) consists of all sequences
(55)x=(x1,…,xk,0,0,…),
such that (1)
for all pairs(xq≠∅i,xq+1≠∅j), say,  xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj one has
(i)
♯{ik∣1≤ik≤jt-p}≥p+1, for all 0≤p≤u:=max{0≤r≤t-1∣jt-r≤i},(ii)
jt-p≤ijt-p-p+s-i′, for all p∈{u+1≤r≤t-1∣jt-r≤r+i},(iii)
jp′≤ij-i+s-t+p′, for all 1≤p≤i-j+t,(2)
there is no pair(xq,xq+1) of the form (∅i,xq+1≠∅j) with i≥jt.Proof.
First we note that the elementrλ is contained in ℛ(λ) and is a highest weight element. Furthermore, we claim that rλ is the unique highest weight element. So suppose that we have another element x=(x1,…,xk,0,0,…) satisfying e~lx=0 for all l∈I. Let z be the lowest integer which appears in one of the sequences x1,…,xk and let p be the minimal integer such that z appears in xp, say,xp=(j1,j1′)⋯(jt=z,jt′)∈ℛtj. In the case where such a z does not exist, we have x=rλ. We remark that
(56)e~l(xp,0,0⋯)=0∀l≥z
would imply the claim, whereby the reason is the following: assume that (56) holds and let r=max{1≤r≤t∣jr-1≠jr+1}. If j1=j and r=1, we set i=j1′+1 and i=jr+1else. In either case we obtain ϵi-1(xp,0,0,…)≠0, which is a contradiction to (56). So if suffices to show (56).
Primarily we claim thatx1,…,xp-1∈{∅1,…,∅n}. Assume that the element xp-1=(i1,i1′)⋯(is,is′)∈ℛsi is not in the aforementioned set; then the property (1)(i) is violated by the choice of z and p because
(57)0=♯{ik∣1≤ik≤z}≥1.
Hence, there exists a pair (∅j,xp) which forces j<z and thus, again by the choice of z, the remaining sequences are again contained in the set {∅1,…,∅n}. As a consequence, a short calculation by using Lemma 10 shows the uniqueness of the a highest weight element, namely,
(58)ϵl(x)=0∀l∈I⟹ϵl((xp,…,xk,0,0⋯))=0∀l≥z⟹ϵl((xp,0,0⋯))=0∀l≥z.
In order to obtain a connected crystal it remains to show that e~lx,f~lx∈ℛ(λ)∪{0}. Assume that f~lx≠0, say,(59)f~lx=(…,xf(x)-1,θ(xf(x)),xf(x)+1,…),
where we set for simplicity f(x)=q. Our goal here is to show that the properties (1) and (2) hold for f~lx, where we start by proving (2).
It is easy to see that(2) can only be violated if one of the following two cases occurs:(i)
(xq-1,xq)=(∅i,∅i)andl=i,(ii)
xq=(i1,i1′)⋯(is,is′)∈ℛsi,xq-1=∅is-1andl=is-1<i. In either case we obtain σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is a contradiction to the choice of q. The proof of the fact that (1) holds will proceed in several cases. In the remaining parts we denote the entries of θ(xq) by θ(ik),θ(ik′), that is,
(60)θ(xq)=(θ(i1),θ(i1′))⋯(θ(is+δl,i),θ(is+δl,i′)).
Case1.1 ((1) is violated for the pair (θ(xq),xq+1)). Let xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj. We first consider the case l<i, which means that we replace l+1 by l. Since the entries ip′ stay unchanged, only property (1)(i) can be violated. However, property (1)(i) is still fulfilled because
(61)♯{ik∣1≤ik≤jt-p}≤♯{θ(ik)∣1≤θ(ik)≤jt-p},∀0≤p≤u.
If l>i we replace l-1 by l and hence (1) is obviously fulfilled. So suppose that l=i, which means that we add the entry (l,l). The equality
(62)♯{ik∣1≤ik≤jt-p}=♯{θ(ik)∣1≤θ(ik)≤jt-p},∀0≤p≤u-1
and inequality
(63)♯{ik∣1≤ik≤jt-u}≤♯{θ(ik)∣1≤θ(ik)≤jt-u}
imply that property (1)(i) is still fulfilled. Furthermore, since θ(ik′)=ik-1′ for k>1, θ(i1′)=i, and θ(xq)∈ℛs+1i, we obtain
(64)jt-p≤ijt-p-p+s-i′=θ(ijt-p-p+s+1-i′),∀p∈{u+1≤r≤t-1∣jt-r≤r+i},jp′≤ij-i+s-t+p′=θ(ij-i+s+1-t+p′),∀1≤p≤i-j+t,
which particularly means that Case 1.1 can never appear.
Case   1.2 ((1)(i) is violated for the pair (xq-1,θ(xq))). For simplicity, we set xq-1=(i1,i1′)⋯(is,is′)∈ℛsi and xq=(j1,j1′)⋯(jt,jt′)∈ℛtj. In that case there exists at least one 0≤p≤θ(u)=max{r∣θ(jt-r)≤i}, such that
(65)♯{ik∣1≤ik≤θ(jt-p)}<p+1.
Necessarily we must have l≤i and in the case where l<j (65) implies
(66)♯{ik∣1≤ik≤l+1}>♯{ik∣1≤ik≤l},♯{ik∣1≤ik≤l-1}=♯{ik∣1≤ik≤l}.
As a consequence we get that l+1 appears in xq-1 while l does not appear and thus σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is a contradiction to the choice of q.
Eventually ifl=j=i, we obtain in a similar way
(67)t≥♯{ik∣1≤ik≤θ(j1)=j}=s≥♯{ik∣1≤ik≤j1}≥t,
which forces on the one hand s=t and on the other hand that j does not appear in xq-1. To be more precise, we can conclude the latter statement with the help of (1)(iii) and (1)(i), namely,
(68)j<j1′≤is-t+1′=i1′,s≥♯{ik∣1≤ik<i}≥♯{ik∣1≤ik≤j1}≥t=s.
Thus, we get again σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is once more a contradiction to the choice of q.
Case  1.3 ((1)(ii) is violated for the pair (xq-1,θ(xq))). Here we have l≤j and if l=j, we must have j>i because otherwise (1)(ii) would not be violated. We first consider the case where l=j and notice that the only possible violation is given by the following inequality:
(69)j>ij-t+s-i′.
We can conclude that j does not occur in xq-1 because either j=t+i and hence
(70)is′<j
or j+1≤t+i and (1)(iii) is applicable, which yields j<j1′≤ij-t+s-i+1′. To obtain a contradiction we have to show that j-1 appears in xq-1. If j1>i, this follows by the subsequent calculation:
(71)ij-t+s-i′≤j-1≤j1+(j-1-j1)≤ij1-t+1+s-i′+(j-1-j1)≤ij-t+s-i′.
If j1≤i, we obtain with property (1)(i) that s≥{ik∣1≤ik≤j1}≥t. In particular we actually have s=t because otherwise we would get
(72)j≤j-t+s-1≤ij-t+s-i′.
Eventually we can conclude again that j-1 must appear in xq-1(73)ij-i′≤j-1≤ij-i′.
Now we suppose l<j. Then an easy consideration shows that (1)(ii) can only be violated if l>i and thus we obtain similarly as before that the only violation which can occur is the following:
(74)l>il-r+s-i′,
where we expect jt-r=l+1. We would like to show as before that l does not appear in xq-1 while l-1 appears. We either have l=r+i and thus l>is′ or l+1≤r+i. In the latter case we apply property (1)(ii) and obtain
(75)l<il-r+s-i+1′.
In either case we notice that l does not occur in xq-1.
In order to prove the remaining part we consider the elementjt-(r-1) which is considerable, since
(76)l-r+s-1≤il-r+s-i′≤l-1⟹1≤s≤r.
In the case where jt-(r-1) is greater than i, we obtain
(77)il-r+s-i′≤l-1≤jt-(r-1)+(l-1-jt-(r-1))≤ijt-(r-1)-r+1+s-i′+(l-1-jt-(r-1))≤il-r+s-i′
and otherwise by using property (1)(i) we get r=s. Thus,
(78)l-1≥il-i′≥l-1.
Case  1.4 ((1)(iii) is violated for the pair (xq-1,θ(xq))). We suppose that (1)(iii) is violated, which forces l≥j. In the case where l>j, we have
(79)∃p:jp′=l-1=ij-i+s-t+p′.
It follows that l-1 appears and l does not appear in xq-1 because l=ij-i+s-t+p+1′ would imply
(80)l=ij-i+s-t+p+1′≥jp+1′>jp′=l-1⟹jp+1′=l.
Consequently, σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x) and we obtain as usual a contradiction to the choice of q.
If the remaining casel=j occurs, then one of the inequalities
(81)θ(jp′)≤ij-i+s-t-1+p′,1≤p≤i-j+t+1
must be violated. Clearly, the only possibility is that j≤ij-i+s-t′ does not hold. If t=j-i, we get
(82)j>is′
and else we can assume j-i+1≤t so that (1)(iii) is applicable, which yields
(83)i≤ij-i+s-t′<j<j1′≤ij-i+s-t+1′.
Therefore, j does not appear in xq-1. In what follows, we finish our proof by showing that j-1 appears in xq-1. If j1>i, we can apply (1)(ii) and get
(84)ij-i+s-t′≤j1+(j-1-j1)≤ij1-t+1+s-i′+(j-1-j1)≤ij-i+s-t′.
If j1≤i, we can verify with (1)(i) that s≥t, but the assumption s-1≥t  yields in a contradiction, namely,
(85)ij-i+s-t′≥j+s-t-1≥j.
So s=t and we obtain the required equality j-1≥ij-i′≥j-1.
The proof ofe~lx∈ℛ(λ)∪{0} is similar, which completes the proof.


### 5.2. Explicit Description ofℛ(λ) in Type Cn

In this subsection we would like to give an explicit characterization ofℛ(λ) if 𝔤 is a symplectic Lie algebra. In order to state the main theorem we fix some notation.For an arbitrary subsetA⊆I, n∈{0,1}, and y∈I, we set
(86)δy,A={1,ify∈A,0,ify∉A,nA={A,ifn=1,∅,ifn=0.
The analogue result to Theorem 18 for type Cn is the following.Theorem 19.
The crystalℛ(λ) consists of all sequences
(87)x=(x1,…,xk,0,0,…),
such that  (1)
for all pairs(xq≠∅i,xq+1≠∅j), say,   xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj one has
(i)
♯{ik∣1≤ik≤jt-p}≥p+1, for all 0≤p≤u:=max{0≤r≤t-1∣jt-r≤i},(ii)
jt-p≤ijt-p-p+s-i′, for all p∈{u+1≤r≤t-1∣jt-r≤r+i},(iii)
jp′≤ij-i+s-t+p′, for all 1≤p≤i-j+t,(2)
there is no pair(xq,xq+1) of the form (∅i,xq+1≠∅j) with i≥jt,(3)
there is no pair(xq≠∅i,xq+1) with the following property: there exists ip′≥ir′≥n with
(a)
ir′¯-δir′¯,{i+1,…,n}∉(1-δir′¯,{i+1,…,n}){i1,…,is}∪δir′¯,{i+1,…,n}(I-{i1′,…,is′}),(b)
ip′¯-δip′¯,{j+1,…,n}∉(1-δip′¯,{j+1,…,n}){j1,…,jt}∪δip′¯,{j+1,…,n}(I-{j1′,…,jt′}),(c)
(p-r)+♯{jk∣jk<ip′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<ip′¯} ≥ max{0,ir′¯-i}-max{0,ip′¯-j},(4)
there is no pair(xq≠∅i,xq+1≠∅j) with the following property: there exists ip′≥jr′≥n with
(a)
ip′¯-δip′¯,{j+1,…,n}∉(1-δip′¯,{j+1,…,n}){j1,…,jt}∪δip′¯,{j+1,…,n}(I-{j1′,…,jt′}),(b)
jr′¯-δjr′¯,{j+1,…,n}∉(1-δjr′¯,{j+1,…,n}){j1,…,jt}∪δjr′¯,{j+1,…,n}(I-{j1′,…,jt′}),(c)
(i-j)+(t-s)+(p-r)+♯{jk′∣ip′¯≤jk′<jr′¯}-♯{jk∣ip′¯≤jk<jr′¯} ≥ max{0,jr′¯-j}-max{0,ip′¯-j}.Proof.
First of all we note as in theAn case that the element rλ is contained in ℛ(λ) and is a highest weight element. In order to prove that rλ is the unique highest weight element in ℛ(λ) we assume x=(x1,…,xk,0,0,…) to be another one. Let z be the lowest integer which appears in one of the sequences x1,…,xk and let p be the minimal integer such that z appears in xp, say,xp=(i1,i1′)⋯(is=z,is′)∈ℛsi. We can prove similar to Theorem 18 that the elements x1,…,xp are contained in {∅1,…,∅n}. Accordingly we get once more with Lemma 10 (similar to the An case)
(88)e~lx=0∀l∈I⟹e~l((xp,…,xk,0,0⋯))=0∀l≥z⟹e~l((xp,0,0⋯))=0∀l≥z.
Our aim is again to prove the impossibility of e~l((xp,0,0⋯))=0foralll≥z. Let r=max{1≤r≤s∣ir-1≠ir+1}. If i1=i, r=1, and i1′∉{n-1¯,…,1¯}, we set j=i1′+1 and if i1=i,r=1 is not satisfied, we set j=ir+1 and obtain similar to Theorem 18 that ϵj-1(xp,0,0,…)≠0. Thus, the only remaining case which can appear is when i1=i, r=1, and i1′∈{n-1¯,…,1¯}. In this particular case we set j=i1′¯ and claim ϵj(xp,0,0,…)≠0. The latter claim is true because on the one hand we have
(89)i1′∈{i1′,…,is′},i′1¯+1¯∉{i1′,…,is′}
and on the other hand we got j¯=i1′≤i1¯=i¯⇒j≥i, which verifies the properties listed in (d). To be more precise, if j>i, we have j-1∉{i1′,…,is′} and if j=i, we have j=i=i1.
In order to finish the theorem it remains to show thatℛ(λ) is stable under the Kashiwara operators, that is, e~lx,f~lx∈ℛ(λ)∪{0}. Assume that f~lx≠0, say,(90)f~lx=(…,xq-1,θ(xq),xq+1,…).
Our aim here is to prove that the properties (1)–(4) hold for f~lx, whereby the verification of the first and the second properties proceeds almost similar to Theorem 18. Nevertheless we will demonstrate some parts of it in Case 1. In the remaining parts of our proof, we set xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj, whenever they are contained in ℛ-{∅1,…,∅n}. We will divide our proof into several cases.
Case  1 (l±1↦l or +(l,l)). Here we assume that the action of the Kashiwara operator on x is given in a way such that xq satisfies property (a). It means that we either replace l±1 by l or add the pair (l,l) as described in Section 3. Here we consider again several cases, where each case assumes that a condition described in Theorem 19 is violated.
Subcase1.1 ((1)(i) is violated for the pair (xq-1,θ(xq)). Since (1)(i) is violated, there must exist an element θ(jt+δl,j-r)=l such that
(91)♯{ik∣1≤ik≤l}≤r.
Further, as in Subcase 1.2 of Theorem 18, one can verify that l does not appear in xq-1 and if in addition l≤i<j holds, then l+1 occurs in xq-1. Consequently l¯ must be contained in {i1′,…,is′} because otherwise we would obtain a contradiction to the choice of q. Hence, if we take ip′=ir′=l¯, it follows immediately that the properties (a), (b), and (c) in (3) hold for the pair (xq-1,xq), which is impossible. For instance (c) is fulfilled with (91), since
(92)♯{ik∣1≤ik≤l}=♯{ik∣ik<l}≤r≤♯{jk∣jk<l}.
The other violations of the properties in (1) or (2) can be proven similarly, so that we consider as a next step the following case.
Subcase1.2 ((3) is violated for the pair (θ(xq),xq+1)). A simple case-by-case observation shows that this case can only occur if there exists θ(ip′)≥θ(ir′)=l+1¯, such that (a), (b) and (c) are satisfied. Suppose that there exists an element m¯ in the set {θ(ir+1′),…,θ(ip′)}(m¯=θ(ir+u′)), such that
(93)m-δm,{i+1,…,n}∉(1-δm,{i+1,…,n}){i1,…,is}∪δm,{i+1,…,n}(I-{i1′,…,is′})
and m¯ is minimal with this property. In the case where such an element does not exist we set m¯=θ(ip′). We claim the following.
Claim. Let m¯ be as described before; then
(94)u+min{l,i}-min{m,i}-♯{ik∣m≤ik<l}+♯{ik′∣m≤ik′<l}≤l+1-m.
Proof of the Claim.We consider again various cases, starting with (i)
l≤i: the minimality of m¯ implies θ(ir+1′)¯,…,θ(ir+u-1′)¯∈{i1,…,is}, which yields
(95)♯{ik′∣θ(ir′)<ik′<m¯}︸=u-1≤♯{ik∣m≤ik<l}⟺♯{ik′∣m≤ik′<l}-♯{ik∣m≤ik<l}≤-u+1,(ii)
l>i and m≤i: using the minimality of m¯ we obtain similarly θ(ir+u-1′)¯,…,θ(ir+x′)¯∈{i1,…,is} and θ(ir+1′)¯-1,…,θ(ir+x-1′)¯-1∉{i1′,…,is′}, for some integer x. Therefore, the following calculation implies (94):
(96)i-♯{ik∣m≤ik}+♯{ik′∣ik′<l}≤l-(x-1)-♯{ik∣m≤ik}≤l-♯{ik′∣θ(ir′)<ik′<m¯},(iii)
m>i: the minimality of m¯ provides θ(ir+1′)¯-1,…,θ(ir+u-1′)¯-1∉{i1′,…,is′}. Hence,
(97)♯{ik′∣m≤ik′<l}≤l-m-♯{ik′∣θ(ir′)<ik′<m¯},
which finishes the proof of the claim.
As a consequence of (94) we obtain
(98)♯{jk∣jk<θ(ip′)¯}-♯{θ(ik)∣θ(ik)<θ(ir′)¯}+♯{θ(ik′)∣θ(ik′)<θ(ir′)¯}-♯{jk′∣jk′<θ(ip′)¯}≤♯{jk∣jk<ip-δl,i′¯}-♯{ik∣m≤ik<l}-♯{ik∣ik<m}+♯{ik′∣m≤ik′<l}+♯{ik′∣ik′<m}-♯{jk′∣jk′<ip-δl,i′¯}<l+1-m-min{l,i}+min{m,i}+max{0,m-i}-max{0,ip-δl,i′¯-j}-(p-r)=l+1-min{l,i}-max{0,ip-δl,i′¯-j}-(p-r)≤max{0,l+1-i}-max{0,ip-δl,i′¯-j}-(p-r)+δl,{1,…,i-1},
where the first estimation is strict if l<i. Hence, we have a contradiction to the assumption that (c) holds.
Subcase1.3  ((3) is violated for the pair (xq-1,θ(xq))). In that case there exists ip′=l+1¯≥ir′, such that (a), (b), and (c) is satisfied. Therefore, we must have l¯=ip+1′ because otherwise we obtain a contradiction to the choice of q. It follows that
(99)(p-r)+♯{θ(jk)∣θ(jk)<ip′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{θ(jk′)∣θ(jk′)<ip′¯}≤(p+1-r)+♯{jk∣jk<ip+1′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<ip+1′¯}<max{0,ir′¯-i}-max{0,l-j}≤max{0,ir′¯-i}-max{0,l+1-j}+δl,{j,…,n},
where the first estimation is strict whenever l≥j and thus provides a contradiction to (c).
Subcase1.4 ((4) is violated for the pair (θ(xq),xq+1)). It is easy to see that this case can never appear.
Subcase1.5 ((4) is violated for the pair (xq-1,θ(xq))). In that case we have two possibilities, where we start by supposing that there is ip′=l+1¯≥θ(jr′), such that the properties (a), (b), and (c) are fulfilled. Similar to Subcase 1.3, we must have an element l¯=ip+1′ because otherwise we would obtain a contradiction to the choice of q. Subsequently we get
(100)(i-j)+(t+δl,j-s)+(p-r)+♯{θ(jk′)∣ip′¯≤θ(jk′)<θ(jr′)¯}-♯{θ(jk)∣ip′¯≤θ(jk)<θ(jr′)¯}≤(i-j)+(t+δl,j-s)+(p+1-r)+♯{jk′∣l<jk′≤θ(jr′)¯}-♯{jk∣l≤jk<θ(jr′)¯}<max{0,θ(jr′)¯-j}-max{0,l-j}≤max{0,θ(jr′)¯-j}-max{0,l+1-j}+δl,{j,…,n},
where the first estimation is strict provided thatl≥j meaning that this calculation contradicts once more property (c).
The last and second possibility which can occur is that there existsip′≥θ(jr′)=l+1¯, such that the properties (a), (b), and (c) are fulfilled.
Then we make a similar construction as in Subcase 1.2; namely, we suppose that(101)m¯∈{θ(jr+1′),…,θ(jt+δl,j′)},say,m¯=θ(ju′)=ju-δl,j′,
such that
(102)m-δm,{j+1,…,n}∉(1-δm,{j+1,…,n}){j1,…,jt}∪δm,{j+1,…,n}(I-{j1′,…,jt′})
and m≥ip′¯. If such an element exists we choose m maximal with this property and otherwise we set m¯=ip′. Using the maximality, we can verify similar to (94) the correctness of
(103)(u-r)-♯{jk∣m≤jk<l}+♯{jk′∣m≤jk′<l}+min{l,j}-min{m,j}≤l+1-m.
As a corollary we obtain as usual a contradiction to property (c) (recall that θ(xq)∈ℛt+δl,jj)
(104)(i-j)+(t+δl,j-s)+(p-r)+♯{θ(jk′)∣ip′¯≤θ(jk′)<θ(jr′)¯}-♯{θ(jk)∣ip′¯≤θ(jk)<θ(jr′)¯}≤(i-j)+(t+δl,j-s)+(p-u)+(u-r)+♯{jk′∣ip′¯≤jk′<m}+♯{jk′∣m≤jk′<l}-♯{jk∣ip′¯≤jk<m}-♯{jk∣m≤jk<l}<max{0,m-j}-max{0,ip′¯-j}+l+1-m-min{l,j}+min{m,j}≤l+1-min{l,j}-max{0,ip′¯-j}≤max{0,l+1-j}-max{0,ip′¯-j}+δl,{1,…,j-1},
where the first estimation is strict whenever l<j is satisfied.
Case2 (l+1¯↦l¯). Now we assume that the action of the Kashiwara operator on x is given in a way such that xq satisfies property (b) while (a) is violated, which in particular means that we replace the entry l+1¯ by l¯. Since the proofs are similar to Case 1, we do not give them in full details. We only consider the case where we presume that property (3) is violated, that is, one has the following.
Subcase2.1 ((3) is violated for the pair (xq-1,θ(xq))). It is easy to see that this case can never appear.
Subcase2.2 ((3) is violated for the pair (θ(xq),xq+1)). The first possibility which can occur is that there exists θ(ip′)=ip′≥θ(ir′)=l¯, such that the properties (a), (b), and (c) are fulfilled. Because of (a) and ir′=l+1¯, we must have
(105)l+1-δl+1,{i+1,…,n}∉(1-δl+1,{i+1,…,n}){i1,…,is}∪δl+1,{i+1,…,n}(I-{i1′,…,is′}),
since otherwise we would obtain that xq satisfies (a) (from Section 3) and thus the Kashiwara operator would act as in Case 1. Accordingly we can apply our assumptions to ip′≥ir′=l+1¯ and obtain a contradiction to (c), namely,
(106)♯{jk∣jk<ip′¯}-♯{θ(ik)∣θ(ik)<l}+♯{θ(ik′)∣θ(ik′)<l}-♯{jk′∣jk′<ip′¯}=♯{jk∣jk<ip′¯}+min{l,i}-min{l,i}-♯{ik∣ik<l}+♯{ik′∣ik′<l}-♯{jk′∣jk′<ip′¯}<♯{jk∣jk<ip′¯}+min{l+1,i}-min{l,i}-♯{ik∣ik<l+1}+♯{ik′∣ik′<l+1}-♯{jk′∣jk′<ip′¯}<min{l+1,i}-min{l,i}+max{0,l+1-i}-max{0,ip′¯-j}-(p-r)=1+max{0,l-i}-max{0,ip′¯-j}-(p-r).
The second and last possibility which can occur in that case isthat there exists θ(ip′)=l¯≥θ(ir′)=ir′, such that the properties (a), (b), and (c) are fulfilled. As before we can assume ip′=l+1¯. Suppose that there exists m¯∈{ir′,…,ip′} (m¯=ir+u′), such that
(107)m-δm,{j+1,…,n}∉(1-δm,{j+1,…,n}){j1,…,jt}∪δm,{j+1,…,n}(I-{j1′,…,jt′})
and m is minimal with this property. If such an element does not exist, we set m¯=ir′. Similar to (94) one gets
(108)p-(r+u)-min{l,j}+min{m,j}-♯{jk∣l≤jk<m}+♯{jk′∣l≤jk′<m}≤m-l.
Using this inequality we arrive once more at a contradiction, namely,
(109)♯{jk∣jk<l}-♯{θ(ik)∣θ(ik)<ir′¯}+♯{θ(ik′)∣θ(ik′)<ir′¯}-♯{jk′∣jk′<l}=♯{jk∣jk<m}-♯{jk∣l≤jk<m}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<m}+♯{jk′∣l≤jk′<m}<max{0,ir′¯-i}-max{0,m-j}-(p-r)-min{m,j}+min{l,j}+m-l=max{0,ir′¯-i}-max{0,l-j}-(p-r).
The proof of e~lx∈ℛ(λ)∪{0} is similar, which completes the proof.


## 5.1. Explicit Description ofℛ(λ) in Type An

In this subsection we give an explicit characterization ofℛ(λ) if 𝔤 is the special linear Lie algebra; that is, we give conditions whether a sequence x is contained in ℛ(λ) or not (Theorem 18). Initially we note that ℛ(λ) lives in ℛk=∑i‍mi and the first k components are nonzero. For simplicity we set el(x)=e(x), fl(x)=f(x), θl=θ, and ρl=ρ.Theorem 18.
The crystalℛ(λ) consists of all sequences
(55)x=(x1,…,xk,0,0,…),
such that (1)
for all pairs(xq≠∅i,xq+1≠∅j), say,  xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj one has
(i)
♯{ik∣1≤ik≤jt-p}≥p+1, for all 0≤p≤u:=max{0≤r≤t-1∣jt-r≤i},(ii)
jt-p≤ijt-p-p+s-i′, for all p∈{u+1≤r≤t-1∣jt-r≤r+i},(iii)
jp′≤ij-i+s-t+p′, for all 1≤p≤i-j+t,(2)
there is no pair(xq,xq+1) of the form (∅i,xq+1≠∅j) with i≥jt.Proof.
First we note that the elementrλ is contained in ℛ(λ) and is a highest weight element. Furthermore, we claim that rλ is the unique highest weight element. So suppose that we have another element x=(x1,…,xk,0,0,…) satisfying e~lx=0 for all l∈I. Let z be the lowest integer which appears in one of the sequences x1,…,xk and let p be the minimal integer such that z appears in xp, say,xp=(j1,j1′)⋯(jt=z,jt′)∈ℛtj. In the case where such a z does not exist, we have x=rλ. We remark that
(56)e~l(xp,0,0⋯)=0∀l≥z
would imply the claim, whereby the reason is the following: assume that (56) holds and let r=max{1≤r≤t∣jr-1≠jr+1}. If j1=j and r=1, we set i=j1′+1 and i=jr+1else. In either case we obtain ϵi-1(xp,0,0,…)≠0, which is a contradiction to (56). So if suffices to show (56).
Primarily we claim thatx1,…,xp-1∈{∅1,…,∅n}. Assume that the element xp-1=(i1,i1′)⋯(is,is′)∈ℛsi is not in the aforementioned set; then the property (1)(i) is violated by the choice of z and p because
(57)0=♯{ik∣1≤ik≤z}≥1.
Hence, there exists a pair (∅j,xp) which forces j<z and thus, again by the choice of z, the remaining sequences are again contained in the set {∅1,…,∅n}. As a consequence, a short calculation by using Lemma 10 shows the uniqueness of the a highest weight element, namely,
(58)ϵl(x)=0∀l∈I⟹ϵl((xp,…,xk,0,0⋯))=0∀l≥z⟹ϵl((xp,0,0⋯))=0∀l≥z.
In order to obtain a connected crystal it remains to show that e~lx,f~lx∈ℛ(λ)∪{0}. Assume that f~lx≠0, say,(59)f~lx=(…,xf(x)-1,θ(xf(x)),xf(x)+1,…),
where we set for simplicity f(x)=q. Our goal here is to show that the properties (1) and (2) hold for f~lx, where we start by proving (2).
It is easy to see that(2) can only be violated if one of the following two cases occurs:(i)
(xq-1,xq)=(∅i,∅i)andl=i,(ii)
xq=(i1,i1′)⋯(is,is′)∈ℛsi,xq-1=∅is-1andl=is-1<i. In either case we obtain σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is a contradiction to the choice of q. The proof of the fact that (1) holds will proceed in several cases. In the remaining parts we denote the entries of θ(xq) by θ(ik),θ(ik′), that is,
(60)θ(xq)=(θ(i1),θ(i1′))⋯(θ(is+δl,i),θ(is+δl,i′)).
Case1.1 ((1) is violated for the pair (θ(xq),xq+1)). Let xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj. We first consider the case l<i, which means that we replace l+1 by l. Since the entries ip′ stay unchanged, only property (1)(i) can be violated. However, property (1)(i) is still fulfilled because
(61)♯{ik∣1≤ik≤jt-p}≤♯{θ(ik)∣1≤θ(ik)≤jt-p},∀0≤p≤u.
If l>i we replace l-1 by l and hence (1) is obviously fulfilled. So suppose that l=i, which means that we add the entry (l,l). The equality
(62)♯{ik∣1≤ik≤jt-p}=♯{θ(ik)∣1≤θ(ik)≤jt-p},∀0≤p≤u-1
and inequality
(63)♯{ik∣1≤ik≤jt-u}≤♯{θ(ik)∣1≤θ(ik)≤jt-u}
imply that property (1)(i) is still fulfilled. Furthermore, since θ(ik′)=ik-1′ for k>1, θ(i1′)=i, and θ(xq)∈ℛs+1i, we obtain
(64)jt-p≤ijt-p-p+s-i′=θ(ijt-p-p+s+1-i′),∀p∈{u+1≤r≤t-1∣jt-r≤r+i},jp′≤ij-i+s-t+p′=θ(ij-i+s+1-t+p′),∀1≤p≤i-j+t,
which particularly means that Case 1.1 can never appear.
Case   1.2 ((1)(i) is violated for the pair (xq-1,θ(xq))). For simplicity, we set xq-1=(i1,i1′)⋯(is,is′)∈ℛsi and xq=(j1,j1′)⋯(jt,jt′)∈ℛtj. In that case there exists at least one 0≤p≤θ(u)=max{r∣θ(jt-r)≤i}, such that
(65)♯{ik∣1≤ik≤θ(jt-p)}<p+1.
Necessarily we must have l≤i and in the case where l<j (65) implies
(66)♯{ik∣1≤ik≤l+1}>♯{ik∣1≤ik≤l},♯{ik∣1≤ik≤l-1}=♯{ik∣1≤ik≤l}.
As a consequence we get that l+1 appears in xq-1 while l does not appear and thus σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is a contradiction to the choice of q.
Eventually ifl=j=i, we obtain in a similar way
(67)t≥♯{ik∣1≤ik≤θ(j1)=j}=s≥♯{ik∣1≤ik≤j1}≥t,
which forces on the one hand s=t and on the other hand that j does not appear in xq-1. To be more precise, we can conclude the latter statement with the help of (1)(iii) and (1)(i), namely,
(68)j<j1′≤is-t+1′=i1′,s≥♯{ik∣1≤ik<i}≥♯{ik∣1≤ik≤j1}≥t=s.
Thus, we get again σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x), which is once more a contradiction to the choice of q.
Case  1.3 ((1)(ii) is violated for the pair (xq-1,θ(xq))). Here we have l≤j and if l=j, we must have j>i because otherwise (1)(ii) would not be violated. We first consider the case where l=j and notice that the only possible violation is given by the following inequality:
(69)j>ij-t+s-i′.
We can conclude that j does not occur in xq-1 because either j=t+i and hence
(70)is′<j
or j+1≤t+i and (1)(iii) is applicable, which yields j<j1′≤ij-t+s-i+1′. To obtain a contradiction we have to show that j-1 appears in xq-1. If j1>i, this follows by the subsequent calculation:
(71)ij-t+s-i′≤j-1≤j1+(j-1-j1)≤ij1-t+1+s-i′+(j-1-j1)≤ij-t+s-i′.
If j1≤i, we obtain with property (1)(i) that s≥{ik∣1≤ik≤j1}≥t. In particular we actually have s=t because otherwise we would get
(72)j≤j-t+s-1≤ij-t+s-i′.
Eventually we can conclude again that j-1 must appear in xq-1(73)ij-i′≤j-1≤ij-i′.
Now we suppose l<j. Then an easy consideration shows that (1)(ii) can only be violated if l>i and thus we obtain similarly as before that the only violation which can occur is the following:
(74)l>il-r+s-i′,
where we expect jt-r=l+1. We would like to show as before that l does not appear in xq-1 while l-1 appears. We either have l=r+i and thus l>is′ or l+1≤r+i. In the latter case we apply property (1)(ii) and obtain
(75)l<il-r+s-i+1′.
In either case we notice that l does not occur in xq-1.
In order to prove the remaining part we consider the elementjt-(r-1) which is considerable, since
(76)l-r+s-1≤il-r+s-i′≤l-1⟹1≤s≤r.
In the case where jt-(r-1) is greater than i, we obtain
(77)il-r+s-i′≤l-1≤jt-(r-1)+(l-1-jt-(r-1))≤ijt-(r-1)-r+1+s-i′+(l-1-jt-(r-1))≤il-r+s-i′
and otherwise by using property (1)(i) we get r=s. Thus,
(78)l-1≥il-i′≥l-1.
Case  1.4 ((1)(iii) is violated for the pair (xq-1,θ(xq))). We suppose that (1)(iii) is violated, which forces l≥j. In the case where l>j, we have
(79)∃p:jp′=l-1=ij-i+s-t+p′.
It follows that l-1 appears and l does not appear in xq-1 because l=ij-i+s-t+p+1′ would imply
(80)l=ij-i+s-t+p+1′≥jp+1′>jp′=l-1⟹jp+1′=l.
Consequently, σlq-1(x)-τlq-1(x)<σlq(x)-τlq(x) and we obtain as usual a contradiction to the choice of q.
If the remaining casel=j occurs, then one of the inequalities
(81)θ(jp′)≤ij-i+s-t-1+p′,1≤p≤i-j+t+1
must be violated. Clearly, the only possibility is that j≤ij-i+s-t′ does not hold. If t=j-i, we get
(82)j>is′
and else we can assume j-i+1≤t so that (1)(iii) is applicable, which yields
(83)i≤ij-i+s-t′<j<j1′≤ij-i+s-t+1′.
Therefore, j does not appear in xq-1. In what follows, we finish our proof by showing that j-1 appears in xq-1. If j1>i, we can apply (1)(ii) and get
(84)ij-i+s-t′≤j1+(j-1-j1)≤ij1-t+1+s-i′+(j-1-j1)≤ij-i+s-t′.
If j1≤i, we can verify with (1)(i) that s≥t, but the assumption s-1≥t  yields in a contradiction, namely,
(85)ij-i+s-t′≥j+s-t-1≥j.
So s=t and we obtain the required equality j-1≥ij-i′≥j-1.
The proof ofe~lx∈ℛ(λ)∪{0} is similar, which completes the proof.


## 5.2. Explicit Description ofℛ(λ) in Type Cn

In this subsection we would like to give an explicit characterization ofℛ(λ) if 𝔤 is a symplectic Lie algebra. In order to state the main theorem we fix some notation.For an arbitrary subsetA⊆I, n∈{0,1}, and y∈I, we set
(86)δy,A={1,ify∈A,0,ify∉A,nA={A,ifn=1,∅,ifn=0.
The analogue result to Theorem 18 for type Cn is the following.Theorem 19.
The crystalℛ(λ) consists of all sequences
(87)x=(x1,…,xk,0,0,…),
such that  (1)
for all pairs(xq≠∅i,xq+1≠∅j), say,   xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj one has
(i)
♯{ik∣1≤ik≤jt-p}≥p+1, for all 0≤p≤u:=max{0≤r≤t-1∣jt-r≤i},(ii)
jt-p≤ijt-p-p+s-i′, for all p∈{u+1≤r≤t-1∣jt-r≤r+i},(iii)
jp′≤ij-i+s-t+p′, for all 1≤p≤i-j+t,(2)
there is no pair(xq,xq+1) of the form (∅i,xq+1≠∅j) with i≥jt,(3)
there is no pair(xq≠∅i,xq+1) with the following property: there exists ip′≥ir′≥n with
(a)
ir′¯-δir′¯,{i+1,…,n}∉(1-δir′¯,{i+1,…,n}){i1,…,is}∪δir′¯,{i+1,…,n}(I-{i1′,…,is′}),(b)
ip′¯-δip′¯,{j+1,…,n}∉(1-δip′¯,{j+1,…,n}){j1,…,jt}∪δip′¯,{j+1,…,n}(I-{j1′,…,jt′}),(c)
(p-r)+♯{jk∣jk<ip′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<ip′¯} ≥ max{0,ir′¯-i}-max{0,ip′¯-j},(4)
there is no pair(xq≠∅i,xq+1≠∅j) with the following property: there exists ip′≥jr′≥n with
(a)
ip′¯-δip′¯,{j+1,…,n}∉(1-δip′¯,{j+1,…,n}){j1,…,jt}∪δip′¯,{j+1,…,n}(I-{j1′,…,jt′}),(b)
jr′¯-δjr′¯,{j+1,…,n}∉(1-δjr′¯,{j+1,…,n}){j1,…,jt}∪δjr′¯,{j+1,…,n}(I-{j1′,…,jt′}),(c)
(i-j)+(t-s)+(p-r)+♯{jk′∣ip′¯≤jk′<jr′¯}-♯{jk∣ip′¯≤jk<jr′¯} ≥ max{0,jr′¯-j}-max{0,ip′¯-j}.Proof.
First of all we note as in theAn case that the element rλ is contained in ℛ(λ) and is a highest weight element. In order to prove that rλ is the unique highest weight element in ℛ(λ) we assume x=(x1,…,xk,0,0,…) to be another one. Let z be the lowest integer which appears in one of the sequences x1,…,xk and let p be the minimal integer such that z appears in xp, say,xp=(i1,i1′)⋯(is=z,is′)∈ℛsi. We can prove similar to Theorem 18 that the elements x1,…,xp are contained in {∅1,…,∅n}. Accordingly we get once more with Lemma 10 (similar to the An case)
(88)e~lx=0∀l∈I⟹e~l((xp,…,xk,0,0⋯))=0∀l≥z⟹e~l((xp,0,0⋯))=0∀l≥z.
Our aim is again to prove the impossibility of e~l((xp,0,0⋯))=0foralll≥z. Let r=max{1≤r≤s∣ir-1≠ir+1}. If i1=i, r=1, and i1′∉{n-1¯,…,1¯}, we set j=i1′+1 and if i1=i,r=1 is not satisfied, we set j=ir+1 and obtain similar to Theorem 18 that ϵj-1(xp,0,0,…)≠0. Thus, the only remaining case which can appear is when i1=i, r=1, and i1′∈{n-1¯,…,1¯}. In this particular case we set j=i1′¯ and claim ϵj(xp,0,0,…)≠0. The latter claim is true because on the one hand we have
(89)i1′∈{i1′,…,is′},i′1¯+1¯∉{i1′,…,is′}
and on the other hand we got j¯=i1′≤i1¯=i¯⇒j≥i, which verifies the properties listed in (d). To be more precise, if j>i, we have j-1∉{i1′,…,is′} and if j=i, we have j=i=i1.
In order to finish the theorem it remains to show thatℛ(λ) is stable under the Kashiwara operators, that is, e~lx,f~lx∈ℛ(λ)∪{0}. Assume that f~lx≠0, say,(90)f~lx=(…,xq-1,θ(xq),xq+1,…).
Our aim here is to prove that the properties (1)–(4) hold for f~lx, whereby the verification of the first and the second properties proceeds almost similar to Theorem 18. Nevertheless we will demonstrate some parts of it in Case 1. In the remaining parts of our proof, we set xq=(i1,i1′)⋯(is,is′)∈ℛsi and xq+1=(j1,j1′)⋯(jt,jt′)∈ℛtj, whenever they are contained in ℛ-{∅1,…,∅n}. We will divide our proof into several cases.
Case  1 (l±1↦l or +(l,l)). Here we assume that the action of the Kashiwara operator on x is given in a way such that xq satisfies property (a). It means that we either replace l±1 by l or add the pair (l,l) as described in Section 3. Here we consider again several cases, where each case assumes that a condition described in Theorem 19 is violated.
Subcase1.1 ((1)(i) is violated for the pair (xq-1,θ(xq)). Since (1)(i) is violated, there must exist an element θ(jt+δl,j-r)=l such that
(91)♯{ik∣1≤ik≤l}≤r.
Further, as in Subcase 1.2 of Theorem 18, one can verify that l does not appear in xq-1 and if in addition l≤i<j holds, then l+1 occurs in xq-1. Consequently l¯ must be contained in {i1′,…,is′} because otherwise we would obtain a contradiction to the choice of q. Hence, if we take ip′=ir′=l¯, it follows immediately that the properties (a), (b), and (c) in (3) hold for the pair (xq-1,xq), which is impossible. For instance (c) is fulfilled with (91), since
(92)♯{ik∣1≤ik≤l}=♯{ik∣ik<l}≤r≤♯{jk∣jk<l}.
The other violations of the properties in (1) or (2) can be proven similarly, so that we consider as a next step the following case.
Subcase1.2 ((3) is violated for the pair (θ(xq),xq+1)). A simple case-by-case observation shows that this case can only occur if there exists θ(ip′)≥θ(ir′)=l+1¯, such that (a), (b) and (c) are satisfied. Suppose that there exists an element m¯ in the set {θ(ir+1′),…,θ(ip′)}(m¯=θ(ir+u′)), such that
(93)m-δm,{i+1,…,n}∉(1-δm,{i+1,…,n}){i1,…,is}∪δm,{i+1,…,n}(I-{i1′,…,is′})
and m¯ is minimal with this property. In the case where such an element does not exist we set m¯=θ(ip′). We claim the following.
Claim. Let m¯ be as described before; then
(94)u+min{l,i}-min{m,i}-♯{ik∣m≤ik<l}+♯{ik′∣m≤ik′<l}≤l+1-m.
Proof of the Claim.We consider again various cases, starting with (i)
l≤i: the minimality of m¯ implies θ(ir+1′)¯,…,θ(ir+u-1′)¯∈{i1,…,is}, which yields
(95)♯{ik′∣θ(ir′)<ik′<m¯}︸=u-1≤♯{ik∣m≤ik<l}⟺♯{ik′∣m≤ik′<l}-♯{ik∣m≤ik<l}≤-u+1,(ii)
l>i and m≤i: using the minimality of m¯ we obtain similarly θ(ir+u-1′)¯,…,θ(ir+x′)¯∈{i1,…,is} and θ(ir+1′)¯-1,…,θ(ir+x-1′)¯-1∉{i1′,…,is′}, for some integer x. Therefore, the following calculation implies (94):
(96)i-♯{ik∣m≤ik}+♯{ik′∣ik′<l}≤l-(x-1)-♯{ik∣m≤ik}≤l-♯{ik′∣θ(ir′)<ik′<m¯},(iii)
m>i: the minimality of m¯ provides θ(ir+1′)¯-1,…,θ(ir+u-1′)¯-1∉{i1′,…,is′}. Hence,
(97)♯{ik′∣m≤ik′<l}≤l-m-♯{ik′∣θ(ir′)<ik′<m¯},
which finishes the proof of the claim.
As a consequence of (94) we obtain
(98)♯{jk∣jk<θ(ip′)¯}-♯{θ(ik)∣θ(ik)<θ(ir′)¯}+♯{θ(ik′)∣θ(ik′)<θ(ir′)¯}-♯{jk′∣jk′<θ(ip′)¯}≤♯{jk∣jk<ip-δl,i′¯}-♯{ik∣m≤ik<l}-♯{ik∣ik<m}+♯{ik′∣m≤ik′<l}+♯{ik′∣ik′<m}-♯{jk′∣jk′<ip-δl,i′¯}<l+1-m-min{l,i}+min{m,i}+max{0,m-i}-max{0,ip-δl,i′¯-j}-(p-r)=l+1-min{l,i}-max{0,ip-δl,i′¯-j}-(p-r)≤max{0,l+1-i}-max{0,ip-δl,i′¯-j}-(p-r)+δl,{1,…,i-1},
where the first estimation is strict if l<i. Hence, we have a contradiction to the assumption that (c) holds.
Subcase1.3  ((3) is violated for the pair (xq-1,θ(xq))). In that case there exists ip′=l+1¯≥ir′, such that (a), (b), and (c) is satisfied. Therefore, we must have l¯=ip+1′ because otherwise we obtain a contradiction to the choice of q. It follows that
(99)(p-r)+♯{θ(jk)∣θ(jk)<ip′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{θ(jk′)∣θ(jk′)<ip′¯}≤(p+1-r)+♯{jk∣jk<ip+1′¯}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<ip+1′¯}<max{0,ir′¯-i}-max{0,l-j}≤max{0,ir′¯-i}-max{0,l+1-j}+δl,{j,…,n},
where the first estimation is strict whenever l≥j and thus provides a contradiction to (c).
Subcase1.4 ((4) is violated for the pair (θ(xq),xq+1)). It is easy to see that this case can never appear.
Subcase1.5 ((4) is violated for the pair (xq-1,θ(xq))). In that case we have two possibilities, where we start by supposing that there is ip′=l+1¯≥θ(jr′), such that the properties (a), (b), and (c) are fulfilled. Similar to Subcase 1.3, we must have an element l¯=ip+1′ because otherwise we would obtain a contradiction to the choice of q. Subsequently we get
(100)(i-j)+(t+δl,j-s)+(p-r)+♯{θ(jk′)∣ip′¯≤θ(jk′)<θ(jr′)¯}-♯{θ(jk)∣ip′¯≤θ(jk)<θ(jr′)¯}≤(i-j)+(t+δl,j-s)+(p+1-r)+♯{jk′∣l<jk′≤θ(jr′)¯}-♯{jk∣l≤jk<θ(jr′)¯}<max{0,θ(jr′)¯-j}-max{0,l-j}≤max{0,θ(jr′)¯-j}-max{0,l+1-j}+δl,{j,…,n},
where the first estimation is strict provided thatl≥j meaning that this calculation contradicts once more property (c).
The last and second possibility which can occur is that there existsip′≥θ(jr′)=l+1¯, such that the properties (a), (b), and (c) are fulfilled.
Then we make a similar construction as in Subcase 1.2; namely, we suppose that(101)m¯∈{θ(jr+1′),…,θ(jt+δl,j′)},say,m¯=θ(ju′)=ju-δl,j′,
such that
(102)m-δm,{j+1,…,n}∉(1-δm,{j+1,…,n}){j1,…,jt}∪δm,{j+1,…,n}(I-{j1′,…,jt′})
and m≥ip′¯. If such an element exists we choose m maximal with this property and otherwise we set m¯=ip′. Using the maximality, we can verify similar to (94) the correctness of
(103)(u-r)-♯{jk∣m≤jk<l}+♯{jk′∣m≤jk′<l}+min{l,j}-min{m,j}≤l+1-m.
As a corollary we obtain as usual a contradiction to property (c) (recall that θ(xq)∈ℛt+δl,jj)
(104)(i-j)+(t+δl,j-s)+(p-r)+♯{θ(jk′)∣ip′¯≤θ(jk′)<θ(jr′)¯}-♯{θ(jk)∣ip′¯≤θ(jk)<θ(jr′)¯}≤(i-j)+(t+δl,j-s)+(p-u)+(u-r)+♯{jk′∣ip′¯≤jk′<m}+♯{jk′∣m≤jk′<l}-♯{jk∣ip′¯≤jk<m}-♯{jk∣m≤jk<l}<max{0,m-j}-max{0,ip′¯-j}+l+1-m-min{l,j}+min{m,j}≤l+1-min{l,j}-max{0,ip′¯-j}≤max{0,l+1-j}-max{0,ip′¯-j}+δl,{1,…,j-1},
where the first estimation is strict whenever l<j is satisfied.
Case2 (l+1¯↦l¯). Now we assume that the action of the Kashiwara operator on x is given in a way such that xq satisfies property (b) while (a) is violated, which in particular means that we replace the entry l+1¯ by l¯. Since the proofs are similar to Case 1, we do not give them in full details. We only consider the case where we presume that property (3) is violated, that is, one has the following.
Subcase2.1 ((3) is violated for the pair (xq-1,θ(xq))). It is easy to see that this case can never appear.
Subcase2.2 ((3) is violated for the pair (θ(xq),xq+1)). The first possibility which can occur is that there exists θ(ip′)=ip′≥θ(ir′)=l¯, such that the properties (a), (b), and (c) are fulfilled. Because of (a) and ir′=l+1¯, we must have
(105)l+1-δl+1,{i+1,…,n}∉(1-δl+1,{i+1,…,n}){i1,…,is}∪δl+1,{i+1,…,n}(I-{i1′,…,is′}),
since otherwise we would obtain that xq satisfies (a) (from Section 3) and thus the Kashiwara operator would act as in Case 1. Accordingly we can apply our assumptions to ip′≥ir′=l+1¯ and obtain a contradiction to (c), namely,
(106)♯{jk∣jk<ip′¯}-♯{θ(ik)∣θ(ik)<l}+♯{θ(ik′)∣θ(ik′)<l}-♯{jk′∣jk′<ip′¯}=♯{jk∣jk<ip′¯}+min{l,i}-min{l,i}-♯{ik∣ik<l}+♯{ik′∣ik′<l}-♯{jk′∣jk′<ip′¯}<♯{jk∣jk<ip′¯}+min{l+1,i}-min{l,i}-♯{ik∣ik<l+1}+♯{ik′∣ik′<l+1}-♯{jk′∣jk′<ip′¯}<min{l+1,i}-min{l,i}+max{0,l+1-i}-max{0,ip′¯-j}-(p-r)=1+max{0,l-i}-max{0,ip′¯-j}-(p-r).
The second and last possibility which can occur in that case isthat there exists θ(ip′)=l¯≥θ(ir′)=ir′, such that the properties (a), (b), and (c) are fulfilled. As before we can assume ip′=l+1¯. Suppose that there exists m¯∈{ir′,…,ip′} (m¯=ir+u′), such that
(107)m-δm,{j+1,…,n}∉(1-δm,{j+1,…,n}){j1,…,jt}∪δm,{j+1,…,n}(I-{j1′,…,jt′})
and m is minimal with this property. If such an element does not exist, we set m¯=ir′. Similar to (94) one gets
(108)p-(r+u)-min{l,j}+min{m,j}-♯{jk∣l≤jk<m}+♯{jk′∣l≤jk′<m}≤m-l.
Using this inequality we arrive once more at a contradiction, namely,
(109)♯{jk∣jk<l}-♯{θ(ik)∣θ(ik)<ir′¯}+♯{θ(ik′)∣θ(ik′)<ir′¯}-♯{jk′∣jk′<l}=♯{jk∣jk<m}-♯{jk∣l≤jk<m}-♯{ik∣ik<ir′¯}+♯{ik′∣ik′<ir′¯}-♯{jk′∣jk′<m}+♯{jk′∣l≤jk′<m}<max{0,ir′¯-i}-max{0,m-j}-(p-r)-min{m,j}+min{l,j}+m-l=max{0,ir′¯-i}-max{0,l-j}-(p-r).
The proof of e~lx∈ℛ(λ)∪{0} is similar, which completes the proof.


## 6. Crystal Bases as Tuples of Integer Sequences

In this section we will verify with Theorem20 that the crystal ℛ(λ) can be identified with the crystal graph B(λ) obtained from Kashiwara's crystal bases theory. Our strategy here is to show that there exists an isomorphism of ℛ(λ) onto the connected component of ⨂i=1nB(ωi)⊗mi containing ⨂i=1nbi⊗mi, where bi denotes the highest weight element in B(ωi). For the proofs in type An we will need a result stated in [16], where the affine type A Kirillov-Reshetikhin crystals are realized via polytopes. We will need the realization of level 1 KR-crystals especially, since they are as classical crystals isomorphic to B(ωi). In type Cn we will use a short induction argument to prove our results.Theorem 20.
Letλ be an arbitrary dominant integral weight and set k=∑i‍mi as before. Thenone has the following.(1)
Ifk=1, one has an isomorphism of crystals ℛ(λ)→~B(λ).(2)
Ifk>1 and j is the maximal integer such that mj≠0, one obtains a strict crystal morphism
(110)ϕ:ℛ(λ)⟶ℛ(λ-ωj)⊗ℛ(ωj)
mapping x=(x1,…,xk,0,0⋯) to the tensor product xk-1⊗xk, where xk-1=(x1,…,xk-1,0,0⋯).Proof.
With the help of Proposition17, the crystal ℛ(ωi) is characterized as ⋃s‍ℛsi. In the case where 𝔤 is of type An, we will consider the map
(111)x=(i1,i1′)⋯(is,is′)↦X,
whereby X is a pattern as in [16, Definition 2], with filling
(112)ap,q={1,if(p,q)=(ir,ir′)forsomer0,else.
By an inspection of the crystal structure on the KR-crystal B1,i≅B(ωi) ([16, Section 3.2]), it is easy to see that this map becomes an isomorphism of crystals.
If𝔤 is of type Cn, the proof of part (1) will proceed by upward induction on i. An observation of the crystal graph of R(ω1)(113)∅1⟶1(1,1)⟶2⋯⟶n-1(1,n-1)⟶n(1,n)⟶n-1(1,n-1¯)⟶n-2⋯⟶1(1,1¯)
proves the initial step. Now we assume the correctness of the claim for all integers less than i, especially we have ℛ(ωi-1)≅B(ωi-1). For the purpose of completing the induction we consider the injective map
(114)η:ℛ(ωi)⟶ℛ(ωi-1)⊗ℛ(ω1)
given by
(115)η(x)={∅i-1⊗(1,i-1),ifs=0∅i-1⊗(1,i1′),ifs=1,i1=i(i1,i-1)(1-δi1,i)(i2,i1′)⋯(is,is-1′)⊗(1,is′),else.
If η would be a strict crystal morphism, we would get ℛ(ωi)≅ℐm(η)≅B(ωi), which finishes the induction. Therefore, we prove that η is a strict crystal morphism, where we consider the cases s=0 and s=1,i1=i separately. In the separated cases we draw a part of the crystal graph in order to see that the properties of a crystal morphism hold. If s=0, we obtain
(116)∅i⟶i(i,i),∅i-1⊗(1,i-1)=η(∅i)⟶i∅i-1⊗(1,i)=η((i,i))
and if s=1,i1=i, we get(1)
ifi1′≤n-1  (see Figure 1),(2)
ifi1′≥n and i1′≠i¯  (see Figure 2),(3)
ifi1′≥n and i1′=i¯  (see Figure 3).Thus, from now on we can presume that s>1 or s=1,i1≠i. Let l∈I be an arbitrary integer. We set for convenience η(x)=x1⊗x2; then
(117)φl(η(x))=max{φl(x2),φl(x1)+φl(x2)-ϵl(x2)}={φl(x1)+φl(x2),ifl≠is′,is′¯max{0,φl(x1)-1},else.
The proof of φl(η(x))=φl(x) is an intensive investigation of the properties (a)–(d) listed in Section 3. To avoid confusion with indices we consider only the following case:(i)
l=is′ or l=is′¯: the case l=is′ is very simple because x satisfies neither (a) nor (b), which yields φl(x)=0. Furthermore, since l+1¯=is′+1¯∉{i1′,…,is′}, we get φl(x1)≤1. Consequently φl(η(x))=0. So it remains to consider the case l=is′¯. Here we claim the following: let (a~) be the property which arises from (a) by erasing the condition l¯∉{i1′,…,is′}, then x satisfies (a~) if and only if x1 satisfies (a~).
First we observe that the propertyl+1¯∉{i1′,…,is′} is equivalent to l+1¯∉{i1′,…,is-1′}, so that we can ignore it. We start the proof by assuming that x1 satisfies (a~) and x does not satisfy (a~), for example, l∈{i1,…,is,i1′,…,is′}. This case is only possible if l=i1=i and thus l-1=i-1∈{i1′,…,is-1′}, which is a contradiction. Since l-1∈{i1′,…,is′} is always fulfilled if l>i>i-1, we can assume that l<i and additionally l+1∉{i1,…,is}. Without loss of generality we set l=i-1 because else we have l+1∈{i1(1-δi1,i),i2,…,is}. As a consequence we get
(118)l+1=i∉{i1,…,is}⟹i1≤i-1⟹l≠i-1,
which is again a contradiction.
According to these calculations,x must satisfy (a~). To show the other direction, let x1 violate one of the properties in (a~), for example, l∈{(i1,i-1)(1-δi1,i),i2,…,is,i1′,…,is-1′}. Then we must have l=i-1 and i1<i which violates l+1∈{i1,…,is}. The only additional possibility which can occur, such that x1 does not satisfy (a~), is l<i-1 and l+1∉{i1(1-δi1,i),i2,…,is}. Then we get l+1=i1=i, which is a contradiction to l<i-1 and so the reverse direction is also completed.
According to this we haveφl(x)=1⇒φl(x1)=2 and φl(x)=0⇒φl(x1)≤1.
To show the existence of a morphism of crystals we have to show (among others) the weight invariance ofη, which is proven by the following calculation:
(119)wt(η(x))=ωi-1+ω1-∑j=1s-1‍αij+1,ij′-α1,is′-(1-δi1,i)αi1,i-1=ωi-1+ω1-α1,i-1-∑j=1s‍αij,ij′=ωi-∑j=1s‍αij,ij′.
The verification of ϵl(η(x))=ϵl(x) is therefore proven with Definition 3(1) and (119). Suppose now that f~lη(x)≠0 and f~l acts on the second tensor. A short investigation of the crystal graph (113) yields
(120)l={is′+1,ifis′≤n-1is′¯-1,else.
Since φl(η(x))=φl(x)=1≠0, we obtain that
(121)f~lx={(i1,i1′)⋯(is,is′+1),ifl=is′+1(i1,i1′)⋯(is,is′¯-1¯),l=is′¯-1
because if l=is′¯-1 and x would satisfy (a), it would automatically follow that x satisfies (a′) which is impossible since φl(x)=1≠2. Thus, f~lη(x)=η(f~lx).
Iff~lη(x)≠0 and f~l acts on the first tensor, we get with the tensor product rule φl(x1)>ϵl(x2). Note that the operation with the Kashiwara operator f~l would change the entry is′ in x if and only if either l=is′+1 or l=is′¯-1 and x is not subject to (a). Our goal is to show here that f~l has no effect on is′. If l=is′+1, we would get φl(x1)=0 and thus a contradiction to φl(x1)>ϵl(x2). In the case where l=is′¯-1, we have that (b) is not fulfilled for x1. Therefore, x1 must be subject to (a). Consequently we obtain that x must fulfill (a) as well because otherwise we would end in a contradiction; namely, the only property in (a) which can be violated is l∉{i1,…,is,i1′,…,is′}. So if l is contained in the aforementioned set, we get by the definition of η that l=i1=i. Hence, l-1=i-1 must be contained in the set {i≤i1′,…,is-1′}, which is impossible.
Thus, the entryis′ stays unchanged in f~lx which provides f~lη(x)=η(f~lx). The proof of e~lη(x)=η(e~lx) is similar, which completes the proof of (1).
In order to prove(2) we will check as in (1) step for step the properties of a morphism of crystals. We get
(122)φl(xk-1⊗xk)=max{φl(xk),φl(xk-1)+φl(xk)-ϵl(xk)}=max{φl(xk),σlk(xk-1)-τlk-1(xk-1)-min{σlj(xk-1)-τlj(xk-1)∣1≤j≤k-1}+φl(xk)-ϵl(xk)}=max{φl(xk),φl(xk)+σlk(x)-τlk(x)-min{σlj(x)-τlj(x)∣1≤j≤k-1}}=σlk+1(x)-τlk(x)-min{σlj(x)-τlj(x)∣1≤j≤k}=φl(x).
The same is trivially fulfilled for ϵl because of Definition 3(1) and the weight invariance of ϕ. Now suppose that fl(x)=p is as in (25). If we apply the Kashiwara operator f~l to the tensor product xk-1⊗xk, we obtain with the above calculations(123)f~l(xk-1⊗xk)={f~lxk-1⊗xk,ifφl(xk-1)>ϵl(xk)xk-1⊗f~lxk,ifφl(xk-1)≤ϵl(xk)={f~lxk-1⊗xk,ifσlk(x)-τlk(x)>σlp(x)-τlp(x)xk-1⊗f~lxk,ifσlk(x)-τlk(x)≤σlp(x)-τlp(x).
According to this we get that p∈{1,…,k-1} if f~l acts on the first tensor and p=k if f~l acts on the second tensor. The proof for the Kashiwara operator e~l works similarly.Figure 1Figure 2Figure 3Corollary 21.
One has an isomorphism of crystals(124)ℛ(λ)≅B(λ).Proof.
The proof will proceed by induction onk, where the initial step is exactly part (1) of Theorem 20. If k>1 and j is the maximal integer where mj is non zero, we can assume with the induction hypothesis that ℛ(λ-ωj)≅B(λ-ωj) and ℛ(ωj)≅B(ωj). The rest of the proof is done with part (2) of Theorem 20, since the map ϕ is injective and the image is a connected crystal containing the highest weight element rλ-ωj⊗rωj.



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