Periodicity of irreducible modular and quantum characters
aa r X i v : . [ m a t h . R T ] F e b PERIODICITY OF IRREDUCIBLE MODULAR ANDQUANTUM CHARACTERS
PETER FIEBIG
Abstract.
For a root system R , a field K and an invertible element q in K let U ( K ,q ) ( R ) be the associated quantum group, defined via Lusztig’s dividedpowers construction. We study the irreducible characters of this algebra withintegral (but not necessarily dominant) highest weight. If σ l ( q ) = 0, where σ l is the l -th cyclotomic polynomial, then these characters exhibit a certain l -periodicity. Introduction
Let R be a root system, K a field and q ∈ K an invertible element. We denoteby U K = U ( K ,q ) ( R ) the corresponding quantum group that is obtained via basechange from Lusztig’s integral form U Z [ v,v − ] . This is a triangularized algebra, i.e.we have a decomposition U K = U − K ⊗ U K ⊗ U + K . Let X be the weight lattice of R .Any λ ∈ X gives rise to a character U K → K (of “type 1”), and the standardconstruction (via Verma modules) yields the irreducible U K -module L K ( λ ) ofhighest weight λ . It has a weight decomposition L K ( λ ) = L µ ∈ X L K ( λ ) µ . For l > σ l ∈ Z [ v ] the l -th cyclotomic polynomial. The main result of thisarticle is the following. Theorem 1.
Let λ, µ ∈ X and assume that µ ≤ λ . Suppose that l ≥ is strictlybigger than any coefficient in the expansion of λ − µ as a sum of simple roots,and that σ l ( q ) = 0 in K . Then dim L K ( λ ) µ = dim L K ( λ + lγ ) µ + lγ for all γ ∈ X . Suppose that q = 1. If char K = p >
0, then σ l (1) = 0 in K if and only if l = p r for some r ≥
0. So in this case the above result specializes to the following.
Theorem 2.
Suppose that char K = p > and q = 1 . Let λ, µ ∈ X andsuppose that µ ≤ λ . Suppose that r ≥ is such that p r is strictly bigger than anycoefficient in an expansion of λ − µ in terms of simple roots. Then dim L K ( λ ) µ = dim L K ( λ + p r γ ) µ + p r γ for all γ ∈ X . Note that in this case the periodicity has a certain fractal nature. Note alsothat for r = 1 one can find the statement of Theorem 2 somewhat hidden in theproof of the theorem in the appendix of [J1]. Jantzen’s arguments resemble thearguments that we present in this article in some respects.For q = 1 and arbitrary K there is a surjective homomorphism U K → D K ,where D K is the hyperalgebra of the split semisimple, simply connected algebraicgroup G K with root system R . For a dominant weight λ we obtain L K ( λ ) byrestriction from the irreducible rational representation of G K with highest weight λ . For dominant weights λ and λ + p r γ the statement of Theorem 2 is impliedby Steinberg’s tensor product theorem for irreducible representations of G K .Given arbitrary λ, µ we can find r ≫ γ such that λ + p r γ is dominant. Sothe characters of all L K ( λ ) can be deduced from the set of dominant charac-ters. Steinberg’s tensor product theorem shows that the dominant characterscan be deduced from the (finite) set of restricted dominant characters. So evenin this more general setup, the restricted irreducible characters contain sufficientinformation to determine all irreducible characters.The main idea of the proof of Theorem 1 is to realize L K ( λ ) as the quotient ofan auxiliary object P K ( λ ), which is an X -graded vector space freely generated bythe set of simple root paths starting at λ . The quotient is taken by the radical of abilinear form b λ . Note that a pair ( K , q ) as above as a Z = Z [ v, v − ]-algebra K that is a field. The objects P K ( λ ) and b λ can be obtained from integral versionsover Z by base change. But the radical, of course, depends heavily on K and q . We study the Z -matrix of b λ in the basis of simple root paths. Each entry isgiven by evaluating a quantum polynomial function X → Z [ v, v − ] at λ . Thesefunctions are polynomials in the quantized linear functions [ f ( · )] : X → Z , butwith coefficients in the quotient field Q := Q ( v ). It is not difficult to controlthe denominators, and a simple arithmetic statement about the periodicity ofquantum polynomial functions then yields the periodicity of Theorem 1. Acknowledgement:
This material is based upon work supported by a grantof the Institute for Advanced Study School of Mathematics.2.
The quantum group U K Let R be a root system and Π ⊂ R a basis. For any α ∈ R we denote by α ∨ its coroot. Set a α,β = h α, β ∨ i for α, β ∈ Π. Then there is a vector d = ( d α )with entries d α ∈ { , , } such that ( d α a α,β ) is a symmetric matrix and eachirreducible component of R contains an α with d α = 1. The goal of this sectionis to define the quantum group U ( K ,q ) and its category O ( K ,q ) of highest weightrepresentations for any field K and any non-zero element q in K . It has atriangular decomposition and a category O of representations. The (integral) quantum group.
Set Z := Z [ v, v − ]. For n ∈ Z define[ n ] := v n − v − n v − v − = , if n = 0, v n − + v n − + · · · + v − n +1 , if n > − v − n − − v − n − − · · · − v n +1 , if n < . For n ≥ n ] ! = [1] · [2] · · · [ n ], and for n ∈ Z and r ≥ h nr i := [ n ] · [ n − · · · [ n − r + 1][1] · [2] · · · [ r ] . For α ∈ Π set v α := v d α and[ n ] α := [ n ]( v α ) = v nα − v − nα v α − v − α . Define [ n ] ! α , (cid:2) nr (cid:3) α ∈ Z likewise. All of the above are elements in Z .Set Q := Q ( v ) = Quot ( Z ). The quantum group U Q = U Q ( R ) associated with R is the associative, unital Q -algebra with set of generators { e α , f α , k α , k − α | α ∈ Π } subject to the following relations (cf. [L, Section 1.1]): k α k β = k β k α , k α k − α = 1 = k − α k α ,k α e β = v a αβ α e β k α , k α f β = v a αβ α f β k α , [ e α , f β ] = 0 ( α = β ) , [ e α , f α ] = k α − k − α v α − v − α and the Serre-relations X r + s =1 − a αβ ( − s (cid:20) − a αβ s (cid:21) α e rα e β e sα for α = β, X r + s =1 − a αβ ( − s (cid:20) − a αβ s (cid:21) α f rα f β f sα for α = β. We endow U Q with an X -grading such that deg( e α ) = α , deg( f α ) = − α anddeg( k α ) = deg( k − α ) = 0.For α ∈ Π and n ≥ e [ n ] α := e nα / [ n ] ! α , f [ n ] α := f nα / [ n ] ! α , (cid:20) k α n (cid:21) α := n Y s =1 k α v − s +1 α − k − α v s − α v sα − v − sα . Following ideas of Kostant, Lusztig defined the integral version U Z of U Q as the Z -subalgebra generated by the set { e [ n ] α , f [ n ] α | α ∈ Π , n ≥ } ∪ { k α , k − α | α ∈ Π } .Then U + Z , U − Z and U Z are defined as the Z -subalgebras in U Z generated by e [ n ] α ’s,the f [ n ] α ’s, and the set { k α , k − α , (cid:2) k α n (cid:3) α | α ∈ Π , n ≥ } , resp. ( cf. [L, Section 1.3& 1.4]). All the algebras above inherit an X -grading from U Q . In [L, Section 6.5] the elements e [ n ] α and f [ n ] α are defined for each positive root α ∈ R + and n ≥
0. Then we have the following PBW-type theorem.
Theorem 2.1 ([L, Theorem 6.7]) . (1) The algebra U + Z is free as a Z -module,and the elements Q α ∈ R + e [ n α ] α with n α ∈ Z ≥ form a basis. (2) The algebra U − Z is free as a Z -module, and the elements Q α ∈ R + f [ n α ] α with n α ∈ Z ≥ form a basis. (3) The algebra U Z is free as a Z -module, and the elements Q α ∈ Π k δ α α h k α r α i α with δ α ∈ { , } and r α ∈ Z ≥ form a basis. (4) The multiplication defines an isomorphism U − Z ⊗ U Z ⊗ U + Z ∼ −→ U Z of Z -modules. Note that the products in parts (1) and (2) in the theorem above have to betaken with respect to certain orders on the set of positive roots, cf. Theorem [L,Theorem 6.7].Fix a field K and an invertible element q in K . Then K acquires an Z -algebra structure such that v ∈ Z acts as multiplication with q . We set U K = U ( K ,q ) := U Z ⊗ Z K and define U + K , U − K and U K likewise. Theorem 2.1 implies that these are sub-algebras in U K and that the multiplication map U − K ⊗ U K ⊗ U + K → U K is anisomorphism.2.2. Representation theory of U K . We denote by X the weight lattice asso-ciated with R . By [APW, Lemma 1.1], every µ ∈ X yields a character (denotedby the same letter) µ : U Z → Z k ± α v ±h µ,α ∨ i α (cid:20) k α r (cid:21) α (cid:20) h µ, α ∨ i r (cid:21) α ( α ∈ Π, r ≥ . A U K -module M is called a weight module if M = L µ ∈ X M µ , where M µ := { m ∈ M | H.m = µ ( H ) m for all H ∈ U K } . By definition, all our weight modules are “of type 1” (cf. [J3, Section 5.1]).
Remark . Suppose that K is a field and q = 1. There is a surjective homo-morphism U K → D ( G K ), where G K is the split semisimple, simply connectedand connected algebraic group with root system R over K , and D ( G K ) is thealgebra of distributions of G K . The homomorphism maps e α and f α to the Serregenerators of D ( G K ), (cid:2) k α r (cid:3) α to (cid:0) h α r (cid:1) for all α ∈ Π, while k α is mapped to 1 (cf. Theorem 2.1, [L, Section 8.15]). By [J2, Section II, 1.20], a finite dimensionalrepresentation of U K is the same as a rational representation of G K -module.The following (well-known) relation is a cornerstone of the approach towardsrepresentations of quantum groups chosen in this article. Lemma 2.3.
Let M be a U K -module. For all α, β ∈ Π and v ∈ M µ we thenhave e [ m ] α f [ n ] β ( v ) = ( f [ n ] β e [ m ] α ( v ) , if α = β, P min( m,n ) r =0 h h µ,α ∨ i + m − nr i α f [ n − r ] β e [ m − r ] α ( v ) , if α = β. Proof.
The case α = β follows directly from the fact that e α and f β commute.By [L, Section 6.5] we have the following relation in the algebra U Z : e [ m ] α f [ n ] β = min( m,n ) X r =0 f [ n − r ] α (cid:20) k α ; 2 r − m − nr (cid:21) α e [ m − r ] α , where (cid:20) k α ; cr (cid:21) α = r Y s =1 k α v c − ( s − α − k − α v − ( c − ( s − α v sα − v − sα . This is an element in U K that acts as multiplication with r Y s =1 v h ν,α ∨ i + c − ( s − α − v − ( h ν,α ∨ i + c − ( s − α v sα − v − sα . on a vector of weight ν . For v ∈ M µ we obtain e [ m ] α f [ n ] β ( v ) = min( m,n ) X r =0 f [ n − r ] α r Y s =1 v ζ − ( s − α − v − ( ζ − ( s − α v sα − v − sα e [ m − r ] α ( v ) , where ζ = h µ + ( m − r ) α, α ∨ i + 2 r − m − n = h µ, α ∨ i + m − n . It remains toshow that (cid:20) ζr (cid:21) α = r Y s =1 v ζ − ( s − α − v − ( ζ − ( s − α v sα − v − sα , which is (almost) immediate from the definition. (cid:3) Category O . Given the triangular decomposition of U K , the following is astandard definition. Definition 2.4.
We denote by O K = O ( K ,q ) the full subcategory of the categoryof U K -modules that contains all M that are weight modules and have the propertythat for any m ∈ M the K -vector space U + K .m is finite dimensional. Clearly, O K is an abelian category. Note that in the case char K = 0 theabove category is studied in [AM].Set U ≥ K := U K U + K ⊂ U K . For λ ∈ X we define the Verma module with highestweight λ by ∆ K ( λ ) := U K ⊗ U ≥ K K λ , where K λ is the one-dimensional K -vector space, endowed with the U ≥ K -actionsuch that U K acts via the character λ , and e [ n ] α ∈ U + K acts trivially for n > K ( λ ) is anobject in O K . Standard arguments also show that the following holds (cf. [J3,Section 5.5]). Lemma 2.5.
For all λ ∈ X , the Verma module ∆ K ( λ ) has a unique irreduciblequotient L K ( λ ) in O K . The L K ( λ ) with λ in X form a full set of representativesof the irreducible objects in O K . A model for the semisimple objects in O K In this section we want to study a category that resembles the category O K ,but “ignores the Serre-relations”. We will see later that it is equivalent to thefull subcategory of O K that contains all semisimple objects.We consider X -graded K -vector spaces M = L µ ∈ X M µ endowed with K -linear operators E α,n , F α,n : M → M for all α ∈ Π and n > nα and − nα , resp. We will always set E α, = F α, = id M µ .There are three conditions that we will assume on the above data. The first twoare easy to formulate.(C1) There exists some γ ∈ X such that M µ = 0 implies µ ≤ γ .(C2) For all µ ∈ X , α, β ∈ Π, m, n >
0, and v ∈ M µ , E α,m F β,n ( v ) = ( F β,n E α,m ( v ) , if α = β , P min( m,n ) r =0 h h µ,α ∨ i + m − nr i α F α,n − r E α,m − r ( v ) , if α = β (cf. Lemma 2.3).(cf. Lemma 2.3.) A remark on property (C1): Note that for any finite subset T of X there exists some γ ∈ X such that µ ≤ γ for all µ ∈ T .In order to formulate the third condition, we need some definitions. For any µ ∈ X define M δµ := M α ∈ Π ,n> M µ + nα and let E µ : M µ → M δµ ,F µ : M δµ → M µ be the column and the row vector with entries E α,n | M µ and F α,n | M µ + nα , resp. Hereis the third condition.(C3) For all µ ∈ X we have M µ = ker E µ ⊕ im F µ .Now we can define the model category. Definition 3.1.
The category C K = C ( K ,q ) is defined as follows. Its objects arethe X -graded K -vector spaces M = L µ ∈ X M µ endowed with K -linear endomo-morphisms E α,n , F α,n : M → M of degree + nα and − nα , resp., for all α ∈ Π and n >
0, for which the conditions (C1), (C2) and (C3) are satisfied. A morphism f : M → N in C K is a degree 0 homogeneous K -linear map that commutes withall E ’s and F ’s.For a morphism f : M → N in C K we denote by f µ : M µ → N µ its µ -component. Let us already formulate one of the main results of this article. Theorem 3.2. (1)
Each object M in C K carries a unique U K -module struc-ture such that the following holds. (a) The X -grading on M is the weight space decomposition. (b) For α ∈ Π and n > , the homomorphisms E α,n and F α,n are theaction maps of e [ n ] α and f [ n ] α , resp. (2) With the U K -module structure defined by (1), each object M of C K is asemisimple object in O K of finite rank. (3) We obtain an equivalence of C K with the full subcategory of O K thatcontains all semisimple ojects of finite rank. Extending morphisms.
Let us call a subset J of X open if λ ∈ J and λ ≤ µ imply µ ∈ J . A useful property of the category C K is that “partialmorphisms”, i.e. morphisms that are defined only on open subsets, extend. Lemma 3.3.
Let M and N be objects in C K , and let J be an open subset of X .Suppose that for all ν ∈ J we have a K -linear homomorphism f ( ν ) : M ν → N ν such that the diagrams M ν + nαf ( ν + nα ) / / F α,n (cid:15) (cid:15) N ν + nαF α,n (cid:15) (cid:15) M ν f ( ν ) / / N ν M ν + nαf ( ν + nα ) / / N ν + nα M νE α,n O O f ( ν ) / / N νE α,n O O commute for all ν ∈ J , α ∈ Π and n > . Then there exists a morphism f : M → N in C K such that f ν = f ( ν ) for all ν ∈ J .Proof. By definition there is some γ such that M µ = 0 or N µ = 0 implies µ ≤ γ .We can hence enlarge J by adding all weights ν γ and setting f ( ν ) = 0 forthose ν . After that we can use an inductive argument and need to show thefollowing. If µ ∈ X is such that µ
6∈ J and J ′ := J ∪ { µ } is open again, thenthere exists a homomorphism f ( µ ) : M µ → N µ such that the diagrams M µ + nαf ( µ + nα ) / / F α,n (cid:15) (cid:15) N µ + nαF α,n (cid:15) (cid:15) M µ f ( µ ) / / N µ M µ + nαf ( µ + nα ) / / N µ + nα M µE α,n O O f ( µ ) / / N µE α,n O O commute for all α ∈ Π and n > α, β ∈ Π and m, n >
0. We now show that the diagram M µ + nβf ( µ + nβ ) (cid:15) (cid:15) F β,n / / M µ E α,m / / M µ + mαf ( µ + mα ) (cid:15) (cid:15) N µ + nβ F β,n / / N µ E α,m / / N µ commutes. Let v ∈ M µ + nβ . Using (C2) we obtain some scalars c r ∈ K such that f ( µ + mα ) E α,m F β,n ( v ) = f ( µ + mα ) min( m,n ) X r c r F β,n − r E α,m − r ( v )= min( m,n ) X r c r f ( µ + mα ) F β,n − r E α,m − r ( v )= min( m,n ) X r c r F β,n − r f ( µ + mα +( n − r ) β ) E α,m − r ( v )= min( m,n ) X r c r F β,n − r E α,m − r f ( µ + nβ ) ( v )= E α,m F β,n f ( µ + nα ) ( v )(note that f ( µ + nβ ) ( v ) also has weight µ + nβ ). It now follows that the diagram M δµf ( δµ ) (cid:15) (cid:15) F µ / / M µ E µ / / M δµf ( δµ ) (cid:15) (cid:15) N δµ F µ / / N µ E µ / / N δµ commutes. By (C3), the homomorphisms E µ are injective on the image of F µ , sowe obtain a unique induced homomorphism ˜ f ( µ ) such that the diagram M δµf ( δµ ) (cid:15) (cid:15) F µ / / im F µ ˜ f ( µ ) (cid:15) (cid:15) E µ / / M δµf ( δµ ) (cid:15) (cid:15) N δµ F µ / / im F µ E µ / / N δµ commutes. By property (C3) we have M µ = ker E µ ⊕ im F µ and N µ = ker E µ ⊕ im F µ . One now checks easily that f ( µ ) := (cid:18) f ( µ ) (cid:19) : M µ → N µ serves ourpurpose. (cid:3) Root paths and bilinear forms
The goal of this section is to construct for any λ ∈ X an object S K ( λ ) in C K .We obtain S K ( λ ) as the quotient of an object P K ( λ ) with E - and F -operatorsthat is freely generated by root paths by the radical of a bilinear form. Note that,in general, P K ( λ ) is not an object in C K , it won’t satisfy condition (C3).4.1. Simple root paths. A simple root path , or simply a root path , is a sequence δ := ( δ , . . . , δ l ) of (not necessarily distinct) simple roots δ , . . . , δ l ∈ Π with l ≥ l ( δ ) = l its length and by ht( δ ) := δ + · · · + δ l its height . For i ∈ { , . . . , l } we set δ ( i ) := ( δ , . . . , b δ i , . . . , δ l )(delete the i -th entry). This is a path of length l − δ ( i ) ) =ht( δ ) − δ i . We denote by ∅ the path of length 0.Fix λ ∈ X . For µ ∈ X we denote by P Q ( λ ) µ the Q -vector space with basis { δ | ht( δ ) = λ − µ } , and we set P Q ( λ ) := L µ ∈ X P Q ( λ ) µ . Let α ∈ Π. We nowdefine Q -linear homogeneous operators ǫ α , ϕ α on P Q ( λ ) of degree + α and − α .For a path δ = ( δ , . . . , δ l ) set ϕ α ( δ ) := ( α, δ , . . . , δ l ) ,ǫ α ( δ , . . . , δ l ) := X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α δ ( i ) . Here, and in the following, we simplify notation slightly and write [ ν, α ∨ ] α insteadof [ h ν, α ∨ i ] α . We define, for n ≥ ǫ [ n ] α := ǫ nα / [ n ] ! α , ϕ [ n ] α := ϕ nα / [ n ] ! α . These are K -linear operators on P Q ( λ ) of degree + nα and − nα , resp.4.2. Commutation relations.
We now show that the operators defined abovesatisfy the relation in assumption (C2).
Lemma 4.1.
Let α, β ∈ Π and m, n > . For all µ ∈ X and v ∈ P Q ( λ ) µ wehave ǫ [ m ] α ϕ [ n ] β ( v ) = ( ϕ [ n ] β ǫ [ m ] α ( v ) , if α = β P min( m,n ) r =0 h h µ,α ∨ i + m − nr i α ϕ [ n − r ] α ǫ [ m − r ] α ( v ) , if α = β. Proof.
First suppose that α = β . In this case it suffices to consider the case m = n = 1. Let δ = ( δ , . . . , δ l ) be a path. Then ǫ α ϕ β ( δ ) = ǫ α ( β, δ , . . . , δ l )= X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α ( β, δ , . . . , b δ i , . . . , δ l )= X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α ϕ β ( δ ( i ) )= ϕ β ( X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α δ ( i ) )= ϕ β ǫ α ( δ ) . Now suppose that α = β . We first consider the case m = n = 1. For a path δ of height λ − µ we have ǫ α ϕ α ( δ ) = ǫ α ( α, δ , . . . , δ l )= [ λ − ( λ − µ ) , α ∨ ] α ( δ , . . . , δ l )++ X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α ( α, δ , . . . , b δ i , . . . , δ l )= [ µ, α ∨ ] α δ + ϕ α ( X i,δ i = α [ λ − δ i +1 − · · · − δ l , α ∨ ] α ( δ , . . . , b δ i , . . . , δ l ))= [ µ, α ∨ ] α δ + ϕ α ǫ α ( δ ) . This is the claimed equation in the case m = n = 1.Now let M = L l ∈ Z M l be the Z -graded Q -vector space with M l := M µ ∈ X, h µ,α ∨ i = l P Q ( λ ) µ . Then ǫ α and ϕ α induce linear operators e, f : M → M that are homogeneous ofdegree +2 and −
2, resp. By what we have shown above, we have ( ef − f e ) | M l =[ l ] α id M l . We are hence in the situation of Lemma 6.1 in the appendix (replacethe variable v in Lemma 6.1 by the variable v α ). From this lemma we deduce ǫ mα ϕ nα ( v ) = min( m,n ) X r =0 [ m ] ! α [ n ] ! α [ m − r ] ! α [ n − r ] ! α (cid:20) h µ, α ∨ i + m − nr (cid:21) α ϕ n − rα ǫ m + rα ( v ) , or ǫ [ m ] α ϕ [ n ] α ( v ) = min( m,n ) X r =0 (cid:20) h µ, α ∨ i + m − nr (cid:21) α ϕ [ n − r ] α ǫ [ m − r ] α ( v )for all v ∈ M µ and m, n > (cid:3) A bilinear form.
For an arbitrary path γ = ( γ , . . . , γ l ) define ǫ γ := ǫ γ ◦ · · · ◦ ǫ γ l ,ϕ γ := ϕ γ ◦ · · · ◦ ϕ γ l . These are Q -linear operators on P Q ( λ ) of degree +ht( γ ) and − ht( γ ), resp.For a path δ = ( δ , . . . , δ l ) let δ r := ( δ l , δ l − , . . . , δ ) be the reversed path. Then l ( δ r ) = l ( δ ) and ht( δ r ) = ht( δ ). If δ and γ have the same height, then ǫ δ r ( γ ) isa Q -multiple of the empty path. We identify Q = Q ∅ , so that the followingdefinition makes sense. Definition 4.2.
We denote by b λ : P Q ( λ ) × P Q ( λ ) → Q the bilinear form definedby b λ ( δ, γ ) := ( , if ht( δ ) = ht( γ ) ,ǫ δ r ( γ ) , if ht( δ ) = ht( γ )for paths δ , γ . Proposition 4.3. (1)
The bilinear form b λ is symmetric. (2) For µ = µ ′ , the weight spaces P Q ( λ ) µ and P Q ( λ ) µ ′ are b λ -orthogonal. (3) For each path δ , the ǫ δ is biadjoint to the operator ϕ δ r .Proof. Statement (2) is clear from the definition. Statement (1) is the quantumanalogue of Lemma 3.9 in [F] and is proven in an analogous way. In order toprove (3), by induction we only need to consider the case δ = ( α ) with α ∈ Π. Again, Lemma 3.9 in [F] contains the corresponding statement in the non-quantum world. The arguments in the quantum world are analogous. (cid:3)
A lattice inside P Q ( λ ) . For any path δ define a scalar δ ! ∈ Z \ { } byinduction on the length. We set ∅ ! = 1. Suppose that l ≥ δ =( δ , . . . , δ l ). Set α := δ and s := max { i ∈ { , . . . , l } | α = δ = δ = · · · = δ i } .Now set δ ! := [ s ] ! α · δ ′ ! , where δ ′ = ( δ s +1 , . . . , δ l ). For example, for δ = ( α, α, β, γ, β, β, β ) with α = β and β = γ we have δ ! = [2] ! α [1] ! β [1] ! γ [3] ! β . Note that δ ! = δ r ! . Now set h δ i := 1 δ ! δ ∈ P Q ( λ ) . Let P Z ( λ ) be the Z -module generated inside P Q ( λ ) by the elements h δ i for allpaths δ . Then P Z ( λ ) inherits an X -grading P Z ( λ ) = L µ ∈ X P Z ( λ ) µ , and thepaths h δ i with ht( δ ) = λ − µ form a basis of P Z ( λ ) µ as a Z -module. Lemma 4.4.
Let α ∈ Π and n ≥ . Then the operators ǫ [ n ] α and ϕ [ n ] α stabilize P Z ( λ ) . Proof.
Let δ = ( δ , . . . , δ l ) be a path. We want to show that ǫ [ n ] α ( h δ i ) and ϕ [ n ] α ( h δ i )are contained in P Z ( λ ). This is clear in the case δ = ∅ , so we can assume l ≥ β := δ and let s ≥ β = δ = · · · = δ s = δ s +1 . Hence h δ i = ϕ [ s ] β ( h δ ′ i ) with δ ′ = ( δ s +1 , . . . , δ l ).We begin with showing the ϕ [ n ] α ( h δ i ) ∈ P Z ( λ ). First, suppose that α = β .Then ϕ [ n ] α ( h δ i ) = h ϕ nα ( h δ i ) i is contained in P Z ( λ ). If α = β , then ϕ [ n ] α ( h δ i ) = ϕ [ n ] α ϕ [ s ] α ( h δ ′ i )= [ n + s ] ! α [ n ] ! α [ s ] ! α ϕ [ n + s ] α ( h δ ′ i ) . As [ n + s ] ! α [ n ] ! α [ s ] ! α ∈ Z and l ( δ ′ ) < l ( δ ), we can use induction over the length and deducethat ϕ [ n ] α ( h δ i ) ∈ P Z ( λ ).Now we want to show that ǫ [ n ] α ( h δ i ) ∈ P Z ( λ ). We have ǫ [ n ] α ( h δ i ) = ǫ [ n ] α ϕ [ s ] β ( h δ ′ i )= X r ≥ c r ϕ [ s − r ] β ǫ [ n − r ] α ( h δ ′ i )for some c r ∈ Z by Lemma 4.1. By induction we can assume ǫ [ n − r ] α ( h δ ′ i ) ∈ P Z ( λ ) .So ϕ [ s − r ] β ǫ [ n − r ] α ( h δ ′ i ) ∈ P Z ( λ ) by what we have already shown above. It followsthat ǫ [ n ] α ( h δ i ) ∈ P Z ( λ ). (cid:3) For a path γ set ϕ h γ i = γ ! ϕ γ and ǫ h γ i = γ ! ǫ γ . Note that these operators arecompositions of various ϕ [ n ] α ’s and ǫ [ n ] α ’s, hence they stabilize P Z ( λ ) ⊂ P Q ( λ ). Lemma 4.5.
The restriction of b λ to P Z ( λ ) × P Z ( λ ) takes values in Z .Proof. For paths δ and γ of the same height we have( h δ i , h γ i ) = δ ! − γ ! − ( δ, γ )= δ ! − γ ! − ǫ δ r ( γ )= ǫ h δ r i ( h γ i ) ∈ Z ∅ , as P Z ( λ ) is stable under the operator ǫ h δ r i . (cid:3) Base change and the radical of b λ . Now let K again be a field, and q ∈ K × an invertible element. As before we view K as a Z -algebra. Set P K ( λ ) := P Z ( λ ) ⊗ Z K . For any path γ we denote by the same symbols ǫ h γ i and ϕ h γ i the induced K -linearoperators on P K ( λ ), and we denote by b λ the bilinear form on P K ( λ ) inducedby ( · , · ) Z . Let rad K ( λ ) ⊂ P K ( λ ) be the radical of the bilinear form b λ . Thenrad K ( λ ) is a graded subspace in P K ( λ ), i.e. rad K ( λ ) = L µ ∈ X rad K ( λ ) µ . Lemma 4.6.
The radical rad K ( λ ) is stable under the operators ǫ h γ i and ϕ h γ i forall paths γ .Proof. This follows immediately from the fact that ǫ h γ i and ϕ h γ i are biadjointwith respect to b λ . (cid:3) We set S K ( λ ) := P K ( λ ) / rad K ( λ ) . This is an X -graded space with induced operators ǫ h γ i and ϕ h γ i for all paths γ .Moreover, it carries a symmetric bilinear form b λ that is non-degenerate. Wedenote the linear operators induced by ǫ [ m ] α and ϕ [ n ] β by E α,m and F β,n , resp. Proposition 4.7.
The K -vector space S K ( λ ) together with the induced X -grading and the set of operators E α,n , F α,n with α ∈ Π , n > is an objectin C K . It has the following properties. (1) S K ( λ ) λ is one dimensional, and S K ( λ ) µ = 0 implies µ ≤ λ . (2) End C K ( S K ( λ )) = K · id S K ( λ ) . In particular, an endomorphism on S K ( λ ) is an automorphism if and only if its λ -component is an automorphism.Proof. In order to show that S K ( λ ) is an object in C K , we need to show that itsatisfies properties (C1), (C2) and (C3). By construction, S K ( λ ) µ = 0 implies µ ≤ λ , so property (C1) holds. Property (C2) follows from the commutationrelations that we have proven in Lemma 4.1. So we need to verify (C3). As λ is the highest weight of S K ( λ ) we deduce im F λ = 0 and ker E λ = S K ( λ ) λ , so S K ( λ ) λ = im F λ ⊕ ker E λ . Now suppose µ < λ . As the elements ϕ h γ i ( ∅ ), where γ runs over all paths, form a basis of P K ( λ ), and hence a generating set for S K ( λ ),we have im F µ = S K ( λ ) µ . So it remains to show that ker E µ = 0. So let v ∈ ker E µ . Then b λ ( v, F α,m ( w )) = b λ ( E α,m ( v ) , w ) = 0 for all α ∈ Π, m > w ∈ S K ( λ ) µ + mα . Hence ker E µ is b λ -orthogonal to im F µ . As im F µ = S K ( λ ) µ and as b λ is non-degenerate and the weight space decomposition is orthogonal, we deduceker E µ = 0. So also property (C3) is satisfied. Hence S K ( λ ) is indeed an objectin C K . The properties stated in (1) follow immediately from the construction.So let us show that property (2) holds. Let f be an endomorphism on S K ( λ ).As S K ( λ ) λ is one-dimensional, we can identify f λ with an element in K . We set g := f − f λ id S K ( λ ) . This now is an endomorphism that vanishes on S K ( λ ) λ . Butthen g ( ϕ h γ i ( ∅ )) = ϕ h γ i g ( ∅ ) = 0. As S K ( λ ) is generated as a K -vector space bythe elements ϕ h γ i ( ∅ ) we deduce g = 0, hence f = f λ · id S K ( λ ) . (cid:3) Semisimplicity of C K . Now we show that, essentially, we have alreadyconstructed all objects in C K . Proposition 4.8.
Let M be an object in C K and suppose that M µ is finite di-mensional for all µ ∈ X . Then there exists an index set I and weights λ i ∈ X for i ∈ I such that M ∼ = S K ( λ ) ⊕ · · · ⊕ S K ( λ l ) . Proof.
By property (C1) there is a maximal element λ ∈ X such that M λ = 0.We set J := { µ ∈ X | λ ≤ µ } . Note that this is an open subset of X . Set n = dim M λ . Then we can choose a vector space isomorphism f ( λ ) : S K ( λ ) ⊕ nλ ∼ −→ M λ . We denote by f µ : S K ( λ ) µ → M µ the zero homomorphism for all µ ∈ J , µ = λ (both sides are 0 anyways). Then we can apply Lemma 3.3 and deducethat there is a morphism f : S K ( λ ) ⊕ n → M in C K with f λ = f ( λ ) . In the sameway we obtain an extension ˜ f : M → S K ( λ ) ⊕ n of f − λ ) : M λ ∼ −→ S K ( λ ) ⊕ nλ . Thecomposition ˜ f ◦ f is an endomorphism of S K ( λ ) ⊕ n that restricts to the identityon the highest weight space. Proposition 4.7 implies that ˜ f ◦ f is the identity.Hence S K ( λ ) ⊕ n is a direct summand of M , and a direct complement N satisfies N λ = 0. From here we can proceed by induction. (cid:3) Connection to representation theory.
The following shows that wedealt with X -graded representations of U K all along. Theorem 4.9.
There exists a unique U K -module structure on S K ( λ ) such that E α,n and F α,n are the action maps of e [ n ] α and f [ n ] α , resp., and k α , k − α act on S K ( λ ) µ via the character µ , for all α ∈ Π , n > and µ ∈ X . This makes S K ( λ ) into an object in O K that isomorphic to L K ( λ ) .Proof. We actually prove a reverse statement. Consider L K ( λ ) as an X -gradedspace with operators by letting E α,n and F α,n be the action maps of e [ n ] α and f [ n ] α , resp. Then property (C1) is obviously satisfied by this data, and Lemma2.3 implies (C2). As λ is the maximal weight, we have L K ( λ ) = ker E λ andim F λ = 0. As L K ( λ ) µ has no non-trivial primitive vectors for µ = λ , we deduce(C3). So L K ( λ ) can be viewed as an object in C K . Proposition 4.8 implies nowthat L K ( λ ) is isomorphic to a direct sum of copies of various S K ( µ )’s. As L K ( λ )is generated (over the F α,n -maps) by a non-zero vector of weight λ , we deduce L K ( λ ) ∼ = S K ( λ ) in C K . This module structure is unique, as U K is generated bythe elements e [ n ] α , f [ n ] α with α ∈ Π, k α , k − α for α ∈ Π and n ≥ (cid:3) Cyclotomic polynomials and periodicity
The aim of this section is to study the matrix of the bilinear forms b λ in termsof the basis of simple root paths. It turns out that we can view the entries of thismatrix as quantum polynomial functions .5.1. Quantum polynomial functions associated to paths.
Let f : X → Z be a group homomorphism, and α ∈ Π. We denote by [ f ( · )] α : X → Z the map λ [ f ( λ )] α = v f ( λ ) α − v − f ( λ ) α v α − v − . Definition 5.1.
We denote by Z [ X ] the Z -subalgebra inside the algebra of allmaps from X to Z that is generated by the maps of the form [ f ( · )] α for all grouphomomorphisms f : X → Z and α ∈ Π. We set Q [ X ] := Z [ X ] ⊗ Z Q . For paths δ, γ of the same height define a function ˜ a δ,γ : X → Z by˜ a δ,γ ( λ ) = b λ ( δ, γ ) . Then the following is immediate.
Lemma 5.2. (1)
We have ˜ a ∅ , ∅ ( λ ) = 1 for all λ ∈ X . (2) We have ˜ a δ,γ = ˜ a γ,δ for all paths δ , γ of the same height. (3) We have ˜ a δ,γ ( λ ) = X i,δ l = γ i [ λ − γ i +1 − · · · − γ l , δ ∨ ] δ ˜ a δ ( l ) γ ( i ) ( λ ) for all paths δ , γ of the same height and all λ ∈ X . It follows from (1) and (3) by induction that ˜ a δ,γ ∈ Z [ X ]. Now set a δ,γ := 1 δ ! γ ! ˜ a δ,γ . Then a δ,γ ∈ Q [ X ]. For λ ∈ X we have, however, a δ,γ ( λ ) = δ ! γ ! b λ ( δ, γ ) = b λ ( h δ i , h γ i ) ∈ Z , by Lemma 4.5, i.e. a δ,γ is a function in Q [ X ] with a δ,γ ( X ) ⊂ Z .For ν ≥ A ν := (cid:16) a δγ (cid:17) ht( δ )=ht( γ )= ν with entries in Q [ X ]. For λ ∈ X we denote by A ν ( λ ) the matrix obtained byevaluating each entry at λ . So A ν ( λ ) is a symmetric matrix with entries in Z .For ( K , q ) as above we denote by A µ ( λ ) K the matrix with entries in K obtainedby base change Z → K . It represents the restriction of the bilinear form b λ tothe λ − ν weight space of P K ( λ ) with respect to the basis {h δ i} ht( δ )= ν . Proposition 5.3.
For all λ, µ ∈ X we have dim K L K ( λ ) µ = rk K A λ − µ ( λ ) K . Proof.
We have dim K L K ( λ ) µ = dim K S K ( λ ) µ by Theorem 4.9. By definition, S K ( λ ) µ = P K ( λ ) µ / rad K ( λ ) µ and hence dim K S K ( λ ) µ = rk K A λ − µ ( λ ) K . (cid:3) Cyclotomic polynomials.
For l ≥ σ l ∈ Z [ v ] the l -th cy-clotomic polynomial (so σ = v − σ = v + 1, σ = v + v + 1,...). Then v n − Q l ≥ ,l | n σ l . So [ n ] = v n − v − n v − v − = v − n +1 v n − v − v − n +1 Y l ≥ ,l | n σ l . Lemma 5.4.
Let F be an element in Z [ X ] . Then F ( λ + lγ ) ≡ F ( λ ) mod σ l for all λ, γ ∈ X . Proof.
It is sufficient to show the statement for all F of the form [ f ( · )] α for somegroup homomorphism f : X → Z and some α ∈ Π. For these, we calculate asfollows. [ f ( λ + lγ )] α = [ f ( λ ) + lf ( γ )] α = v f ( λ )+ lf ( γ ) α − v − f ( λ ) − lf ( γ ) α v α − v − α = v f ( λ ) α ( v lα ) f ( γ ) − v − f ( λ ) α ( v − lα ) f ( γ ) v α − v − α ≡ v f ( λ ) α − v − f ( λ ) α v α − v − α = [ f ( λ )] α mod σ l as v lα = v d α l = 1 mod σ l for all α ∈ Π. (cid:3) For ν ≥ c α ( ν ) ∈ Z ≥ by ν = P α ∈ Π c α ( ν ) α . Lemma 5.5.
Let ν ≥ and l ∈ N be such that l > c α ( ν ) for all α ∈ Π . Then A ν ( λ + lγ ) ≡ A ν ( λ ) mod σ l for all λ, γ ∈ X . Let ( K , q ) be as before. If σ l ( q ) = 0, then the above Lemma means A ν ( λ + lγ ) K = A ν ( λ ) K . Proof.
Let s ∈ Q [ X ] be the entry in A ν at row δ and column γ . Then ˜ s := δ ! γ ! s ∈ Z [ X ]. Lemma 5.4 implies that ˜ s ( λ + lγ ) − ˜ s ( λ ) = δ ! γ ! ( s ( λ + lγ ) − s ( λ ))is divisible by σ l for all λ, γ ∈ X . Note that the height of δ and γ is ν , so ourassumption on l implies that each simple root α occurs in either path at most l − σ l divides neither δ ! nor γ ! . So we deduce that s ( λ + lγ ) − s ( λ ) is divisible by σ l . (cid:3) Let ( K , q ) be as before. Proposition 5.3 and Lemma 5.5 immediately implythe following, which is the main result of this article. Theorem 5.6.
Suppose that λ, µ ∈ X and l ∈ N are such that µ ≤ λ and l > c α ( λ − µ ) for all α ∈ Π . Suppose furthermore that σ l ( q ) = 0 . Then dim K L K ( λ ) µ = dim K L K ( λ + lν ) µ + lν for all ν ∈ X . Appendix
This section contains a lengthy calculation that is used in the main body of thearticle. Let M = L l ∈ Z M l be a Z -graded Q = Q ( v )-module. Let e, f : M → M be Q -linear endomorphisms of degree +2 and −
2, resp. Suppose that for all l ∈ Z we have ( ef − f e ) | M l = [ l ] · id M l = v l − v − l v − v − id M l . Lemma 6.1.
For all m, n > and l ∈ Z we have e m f n | M l = min( m,n ) X r =0 [ m ] ! [ n ] ! [ m − r ] ! [ n − r ] ! (cid:20) l + m − nr (cid:21) f n − r e m − r | M l . Proof.
We first prove this in the case m = 1. Then the claim reads ef n ( v ) = f n e ( v ) + [ n ][ l + 1 − n ] f n − ( v )for all v ∈ M l . For n = 1 this is the assumed commutation relation between theoperators e and f . So suppose the claim holds for some n . Let v ∈ M l . Then ef n +1 ( v ) = ( ef n ) f ( v )= f n ef ( v ) + [ n ][ l − − n ] f n ( v ) (as f ( v ) ∈ M l − )= f n +1 e ( v ) + [ l ] f n ( v ) + [ n ][ l − − n ] f n ( v ) (as [ e, f ]( v ) = [ l ] v ) , and the claim boils down to showing that[ l ] + [ n ][ l − − n ] = [ n + 1][ l − n ] , which follows from Lemma 6.2 (set a = n + 1, b = l − n , c = 1). So we haveproven the claim in the case m = 1 and arbitrary n > n and prove the formula by induction on m . For m = 1 this issettled already. So suppose the above equation holds for some m . Let v ∈ M l .We calculate, using the induction hypothesis and setting ζ := l + m − n , e m +1 f n ( v ) = e ( e m f n )( v )= e ( min( m,n ) X r =0 [ m ] ! [ n ] ! [ m − r ] ! [ n − r ] ! (cid:20) ζr (cid:21) f n − r e m − r ( v ))= min( m,n ) X r =0 [ m ] ! [ n ] ! [ m − r ] ! [ n − r ] ! (cid:20) ζr (cid:21) ef n − r e m − r ( v ) . Now fix, for a moment, a number r with 0 ≤ r ≤ min( m, n ). Set w r := e m − r ( v ).This is an element in M l +2( m − r ) . Then the already proven case m = 1 yields ef n − r ( w r ) = f n − r e ( w r ) + [ n − r ][ l + 2( m − r ) + 1 − ( n − r )] f n − r − ( w r )= f n − r e ( w r ) + [ n − r ][ ζ + m − r + 1] f n − r − ( w r ) . We obtain e m +1 f n ( v ) = P ≤ s ≤ min( m +1 ,n ) c s f n − s e m +1 − s ( v ) with c s = [ m ] ! [ n ] ! [ m − s ] ! [ n − s ] ! (cid:20) ζs (cid:21) ++ [ m ] ! [ n ] ! [ n − ( s − ζ + m − ( s −
1) + 1][ m − ( s − ! [ n − ( s − ! (cid:20) ζs − (cid:21) for 0 ≤ s ≤ min( m, n ) and, in the case n ≥ m + 1, c m +1 = [ m ] ! [ n ] ! [ n − m ][ ζ + 1][ n − m ] ! (cid:20) ζm (cid:21) = [ m + 1] ! [ n ] ! [ n − ( m + 1)] ! (cid:20) ζ + 1 m + 1 (cid:21) . For c m +1 we immediately see that this equals the coefficient of f n − ( m +1) on theleft hand side of the equation that we want to prove.We now fix s with 0 ≤ s ≤ min( m, n ) and write c s = [ m ] ! [ n ] ! [ m − s ] ! [ n − s ] ! d s with d s = (cid:20) ζs (cid:21) + [ n − ( s − ζ + m − s + 2][ m − s + 1][ n − s + 1] (cid:20) ζs − (cid:21) = (cid:20) ζs (cid:21) + [ ζ + m − s + 2][ m − s + 1] (cid:20) ζs − (cid:21) . Now the claim is equivalent to showing that c s = [ m +1] ! [ n ] ! [ m +1 − s ] ! [ n − s ] ! (cid:2) ζ +1 s (cid:3) or d s = [ m +1][ m − s +1] (cid:2) ζ +1 s (cid:3) , i.e. (cid:20) ζs (cid:21) + [ ζ + m − s + 2][ m − s + 1] (cid:20) ζs − (cid:21) = [ m + 1][ m − s + 1] (cid:20) ζ + 1 s (cid:21) or [ m − s + 1] (cid:20) ζs (cid:21) + [ ζ + m − s + 2] (cid:20) ζs − (cid:21) = [ m + 1] (cid:20) ζ + 1 s (cid:21) which follows from Lemma 6.2 ( a = m + 1, b = ζ + 1, c = s ). (cid:3) Lemma 6.2.
Let a, b ∈ Z and c ≥ . Then we have [ a ] (cid:20) bc (cid:21) = [ a − c ] (cid:20) b − c (cid:21) + [ a + b − c ] (cid:20) b − c − (cid:21) . Proof.
We multiply both sides with [ c ] ! and arrive at the equivalent equation[ a ][ b ] · · · [ b − c + 1] = [ a − c ][ b − · · · [ b − c ] + [ a + b − c ][ c ][ b − · · · [ b − c + 1] . Both sides are divisible by [ b − · · · [ b − c + 1], hence the above follows from theequation [ a ][ b ] = [ a − c ][ b − c ] + [ a + b − c ][ c ] . Multiplying this equation by ( v − v − ) we obtain the equivalent equation( v a − v − a )( v b − v − b ) = ( v a − c − v − ( a − c ) )( v b − c − v − ( b − c ) )+( v a + b − c − v − ( a + b − c ) )( v c − v − c )The right hand side is v a + b − c − v a − b − v − a + b + v − a − b +2 c + v a + b − v a + b − c − v − a − b +2 c + v − a − b , which simplifies to − v a − b − v − a + b + v a + b + v − a − b . This equals the left handside. (cid:3) References [AM] Andersen, Henning Haahr; Mazorchuk, Volodymyr,
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