PBW filtration and bases for irreducible modules in type A n
aa r X i v : . [ m a t h . R T ] A p r PBW FILTRATION AND BASES FOR IRREDUCIBLEMODULES IN TYPE A n EVGENY FEIGIN, GHISLAIN FOURIER AND PETER LITTELMANN
Abstract.
We study the PBW filtration on the highest weight rep-resentations V ( λ ) of sl n +1 . This filtration is induced by the standarddegree filtration on U ( n − ). We give a description of the associatedgraded S ( n − )-module grV ( λ ) in terms of generators and relations. Wealso construct a basis of grV ( λ ). As an application we derive a gradedcombinatorial character formula for V ( λ ), and we obtain a new class ofbases of the modules V ( λ ) conjectured by Vinberg in 2005. Introduction
Let g be a simple Lie algebra and let g = n + ⊕ h ⊕ n − be a Cartan de-composition. For a dominant integral λ we denote by V ( λ ) the irreducible g -module with highest weight λ . Fix a highest weight vector v λ ∈ V ( λ ).Then V ( λ ) = U( n − ) v λ , where U( n − ) denotes the universal enveloping alge-bra of n − . The degree filtration U( n − ) s on U( n − ) is defined by:U( n − ) s = span { x . . . x l : x i ∈ n − , l ≤ s } . In particular, U( n − ) = C and gr U( n − ) ≃ S ( n − ), where S ( n − ) denotes thesymmetric algebra over n − . The filtration U( n − ) s induces a filtration V ( λ ) s on V ( λ ): V ( λ ) s = U( n − ) s v λ . We call this filtration the PBW filtration. In this paper we study the asso-ciated graded space grV ( λ ) for g of type A n .So from now on we fix g = sl n +1 . Note that grV ( λ ) = S ( n − ) v λ is a cyclic S ( n − )-module, so we can write grV ( λ ) ≃ S ( n − ) /I ( λ ) , for some ideal I ( λ ) ⊂ S ( n − ). For example, for any positive root α the power f ( λ,α )+1 α of a root vector f α ∈ n −− α belongs to I ( λ ) since f ( λ,α )+1 α v λ = 0in V ( λ ). To describe I ( λ ) explicitly, consider the action of the oppositesubalgebra n + on V ( λ ). It is easy to see that n + V ( λ ) s ֒ → V ( λ ) s , so weobtain the structure of an U ( n + )-module on grV ( λ ) as well. We show: Theorem A. I ( λ ) = S ( n − ) (cid:16) U( n + ) ◦ span { f ( λ,α )+1 α , α > } (cid:17) . Theorem A should be understood as a commutative analogue of the well-known description of V ( λ ) as the quotient V ( λ ) ≃ U( n − ) / h f ( λ,α )+1 α , α > i (see for example [H]).Our second problem (closely related to the first one) is to construct amonomial basis of grV ( λ ). The elements Q α> f s α α v λ with s α ≥ grV ( λ ) (recall that the order in Q α> f s α α is not important since f α are considered as elements of S ( n − )). For each λ we construct a set S ( λ ) ofmulti-exponents s = { s α } α> such that the elements f s v λ = Y α> f s α α v λ , s ∈ S ( λ )form a basis of grV ( λ ). To give a precise definition of S ( λ ) we need tointroduce the notion of a Dyck path , which occurs already in Vinberg’s con-jecture:Let α , . . . , α n be the set of simple roots for sl n +1 . Then all positive rootsare of the form α p,q = α p + · · · + α q for some 1 ≤ p ≤ q ≤ n . We call asequence p = ( β (0) , β (1) , . . . , β ( k )) , k ≥ , of positive roots a Dyck path (or simply a path) if it satisfies the followingconditions: either k = 0, and then p = ( α i ) for some simple root α i , or k ≥
1, and then β (0) = α i , β ( k ) = α j for some 1 ≤ i < j ≤ n and theelements in between obey the following recursion rule:if β ( s ) = α p,q then β ( s + 1) = α p,q +1 or β ( s + 1) = α p +1 ,q . Denote by D the set of all Dyck paths. For a dominant weight λ = P ni =1 m i ω i let P ( λ ) ⊂ R n ( n +1) ≥ be the polytope(0.1) P ( λ ) := (cid:26) ( r α ) α> | ∀ p ∈ D : If β (0) = α i , β ( k ) = α j , then r β (0) + · · · + r β ( k ) ≤ m i + · · · + m j (cid:27) , and let S ( λ ) be the set of integral points in P ( λ ). We show: Theorem B . The set of elements f s v λ , s ∈ S ( λ ) , forms a basis of grV ( λ ) . For s ∈ S ( λ ) define the weightwt( s ) := X ≤ j ≤ k ≤ n s j,k α j,k . As an important application we obtain
Corollary. i) For each s ∈ S ( λ ) fix an arbitrary order of factors f α in the product Q α> f s α α . Let f s = Q α> f s α α be the ordered product. Then theelements f s v λ , s ∈ S ( λ ) , form a basis of V ( λ ) . ii) dim V ( λ ) = ♯S ( λ ) . iii) charV ( λ ) = P s ∈ S ( λ ) e λ − wt( s ) . BW FILTRATION AND BASES 3
We note that the order in the corollary above is important since we areback to the action of the (in general) not commutative enveloping algebra.We thus obtain a family of bases for irreducible sl n +1 -modules. Motivatedby a different background, the existence of these bases (with the same in-dexing set) was conjectured by Vinberg (see [V]). Using completely differentarguments, he proved the conjecture for sl , for sp and G . Note also thatthe data labeling the basis vectors is similar to that for the Gelfand-Tsetlinpatterns (see [GT]). However, these bases are very different from the GTbasis. Example 0.1.
For g = sl ( C ), there are only three Dyck paths, the two oflength 1 corresponding to the simple roots, and the path which involves allpositive roots. In the following we write the elements of P ( λ ) in a triangularform, where we put r = r α and r = r α in the first row and r = r α + α in the second row. For λ = m ω + m ω the associated polytope is P ( λ ) = (cid:26) r r r ∈ R ≥ | ≤ r ≤ m , ≤ r ≤ m ,r + r + r ≤ m + m (cid:27) , For the set of integral points we get for example S ( ω ) = (cid:26) , , (cid:27) , S ( ω ) = (cid:26) , , (cid:27) . and S (2 ω + ω ) = (cid:26) , , , , , , , , , , , , , , (cid:27) . We finish the introduction with several remarks. The PBW filtration forrepresentations of affine Kac-Moody algebras was considered in [FFJMT],[F1], [F2]. It was shown that it has important applications in the represen-tation theory of current and affine algebras and in mathematical physics.There exist special representations V ( λ ) such that the operators f s consistonly of mutually commuting root vectors, even before passing to grV ( λ ).These modules can be described via the theory of abelian radicals and turnedout to be important in the theory of vertex operator algebras (see [GG],[FFL], [FL]).Let V w ( λ ) ֒ → V ( λ ) be a Demazure module for some element w from theWeyl group. For special choices of w there exists a basis of V w ( λ ) similar tothe one given in Theorem B . We conjecture that this should be true for all w ∈ W and we will discuss this elsewhere.Finally we note that grV ( λ ) carries an additional grading on each weightspace V ( λ ) µ of V ( λ ): grV ( λ ) µ = M s ≥ gr s V ( λ ) µ = M s ≥ V ( λ ) µs /V ( λ ) µs − . EVGENY FEIGIN, GHISLAIN FOURIER AND PETER LITTELMANN
The graded character of the weight space is the polynomial p λ,µ ( q ) := X s ≥ (dim V ( λ ) µs /V ( λ ) µs − ) q s . Define the degree deg( s ) := P ≤ j ≤ k ≤ n s j,k for s ∈ S ( λ ), and let S ( λ ) µ bethe subset of elements such that µ = λ − wt( s ). Then Corollary. p λ,µ ( q ) = P s ∈ S ( λ ) µ q deg s . We note that our filtration is different from the Brylinski-Kostant filtra-tion (see [Br], [K]).Our paper is organized as follows:In Section 1 we introduce notations and state the problems. Sections 2 and3 are devoted to the proof of Theorem B (see Theorem 1.5). In Section 2we prove the spanning property and in Section 3 the linear independence.In Section 4 we summarize our constructions and prove Theorem A (seeTheorem 4.5). 1. Definitions
Let R + be the set of positive roots of sl n +1 . Let α i , ω i i = 1 , . . . , n bethe simple roots and the fundamental weights. All roots of sl n +1 are of theform α p + α p +1 + · · · + α q for some 1 ≤ p ≤ q ≤ n . In what follows we denotesuch a root by α p,q , for example α i = α i,i .Let sl n +1 = n + ⊕ h ⊕ n − be the Cartan decomposition. Consider theincreasing degree filtration on the universal enveloping algebra of U( n − ):(1.1) U( n − ) s = span { x . . . x l : x i ∈ n − , l ≤ s } , for example, U( n − ) = C · λ = m ω + · · · + m n ω n let V ( λ ) be thecorresponding irreducible highest weight sl n +1 -module with a highest weightvector v λ . Since V ( λ ) = U( n − ) v λ , the filtration (1.1) induces an increasingfiltration V ( λ ) s on V ( λ ): V ( λ ) s = U( n − ) s v λ . We call this filtration the PBW filtration and study the associated gradedspace grV ( λ ). In the following lemma we describe some operators acting on grV ( λ ). Let S ( n − ) denotes the symmetric algebra of n − . Lemma 1.1.
The action of U( n − ) on V ( λ ) induces the structure of a S ( n − ) -module on grV ( λ ) and gr ( V ( λ )) = S ( n − ) v λ . The action of U( n + ) on V ( λ ) induces the structure of a U ( n + ) -module on grV ( λ ) .Proof. The first statement is obviously true by the definition of the filtrationsU( n − ) s and V ( λ ) s . The inclusions U( n + ) V ( λ ) s ֒ → V ( λ ) s imply the secondstatement. (cid:3) BW FILTRATION AND BASES 5
Our aims are: • to describe grV ( λ ) as an S ( n − )-module, i.e. describe the ideal I ( λ ) ֒ → S ( n − ) such that grV ( λ ) ≃ S ( n − ) /I ( λ ); • to find a basis of grV ( λ ).The description of the ideal will be given in the last section. To describe thebasis we recall the definition of the Dyck paths: Definition . A Dyck path (or simply a path ) is a sequence p = ( β (0) , β (1) , . . . , β ( k )) , k ≥ i) If k = 0, then p is of the form p = ( α i ) for some simple root α i ; ii) If k ≥
1, then a) the first and last elements are simple roots. More precisely, β (0) = α i and β ( k ) = α j for some 1 ≤ i < j ≤ n ; b) the elements in between obey the following recursion rule: If β ( s ) = α p,q then the next element in the sequence is of theform either β ( s + 1) = α p,q +1 or β ( s + 1) = α p +1 ,q . Example 1.3.
Here is an example for a path for sl : p = ( α , α + α , α + α + α , α + α , α , α + α , α ) . For a multi-exponent s = { s β } β> , s β ∈ Z ≥ , let f s be the element f s = Y β ∈ R + f s β β ∈ S ( n − ) . Definition . For an integral dominant sl n +1 -weight λ = P ni =1 m i ω i let S ( λ ) be the set of all multi-exponents s = ( s β ) β ∈ R + ∈ Z R + ≥ such that for allDyck paths p = ( β (0) , . . . , β ( k ))(1.2) s β (0) + s β (1) + · · · + s β ( k ) ≤ m i + m i +1 + · · · + m j , where β (0) = α i and β ( k ) = α j .In the next two sections we prove the following theorem. Theorem 1.5.
The set f s v λ , s ∈ S ( λ ) , form a basis of grV ( λ ) .Proof. In Section 2 we show that the elements f s v λ , s ∈ S ( λ ), span grV ( λ ),see Theorem 2.4. In Section 3 we show that the number ♯S ( λ ) is smallerthan or equal to dim V ( λ ) (see Theorem 3.11), which finishes the proof ofTheorem 1.5. (cid:3) The spanning property
The space grV ( λ ) is endowed with the structure of a cyclic S ( n − )-module,i.e. grV ( λ ) = S ( n − ) v λ and hence grV ( λ ) = S ( n − ) /I ( λ ), where I ( λ ) is someideal in S ( n − ). Our goal in this section is to prove that the elements f s v λ , s ∈ S ( λ ), span grV ( λ ). EVGENY FEIGIN, GHISLAIN FOURIER AND PETER LITTELMANN
Let λ = m ω + · · · + m n ω n . The strategy is as follows: f ( λ,α )+1 α v λ = 0in V ( λ ) for all positive roots α , so for α = α i + · · · + α j , i ≤ j we have therelation f m i + ··· + m j +1 α i + ··· + α j ∈ I ( λ ) . In addition we have the operators e α acting on grV ( λ ), and I ( λ ) is stablewith respect to e α . By applying the operators e α to f m i + ··· + m j +1 α i + ··· + α j , we obtainnew relations. We prove that these relations are enough to rewrite anyvector f t v λ as a linear combination of f s v λ with s ∈ S ( λ ).We start with some notations. For 1 ≤ i < j ≤ n set α i,j = α i + · · · + α j , s i,j = s α i,j , f i,j = f α i,j , and for convenience we set α i,i = α i , s i,i = s α i and f i,i = f α i .By the degree deg s of a multi-exponent we mean the degree of the corre-sponding monomial in S ( n − ), i.e. deg s = P s i,j .We are going to define a monomial order on S ( n − ). To begin with, wedefine a total order on the set of generators f i,j , 1 ≤ i ≤ j ≤ n . We say that( i, j ) ≻ ( k, l ) if i > k or if i = k and j > l . Correspondingly we say that f i,j ≻ f k,l if ( i, j ) ≻ ( k, l ), so f n,n ≻ f n − ,n ≻ f n − ,n − ≻ f n − ,n ≻ . . . ≻ f , ≻ f , ≻ f ,n ≻ . . . ≻ f , . We use the associated homogeneous lexicographic ordering on the set ofmonomials in these generators of S ( n − ).We use the “same” total order on the set of multi-exponents, i.e. s ≻ t if and only if f s ≻ f t . More explicitly: for two multi-exponents s and t wewrite s ≻ t : • if deg s > deg t , • if deg s = deg t and there exist 1 ≤ i ≤ j ≤ n such that s i j >t i j and for i > i and ( i = i and j > j ) we have s i,j = t i,j . Proposition 2.1.
Let p = ( p (0) , . . . , p ( k )) be a Dyck path with p (0) = α i and p ( k ) = α j . Let s be a multi-exponent supported on p , i.e. s α = 0 for α / ∈ p . Assume further that k X l =0 s p ( l ) > m i + · · · + m j . Then there exist some constants c t labeled by multi-exponents t such that (2.1) f s + X t < s c t f t ∈ I ( λ ) ( t does not have to be supported on p ).Remark . We refer to (2.1) as a straightening law because it implies f s = − X t < s c t f t in S ( n − ) /I ( λ ) ≃ grV ( λ ) . BW FILTRATION AND BASES 7
Proof.
We start with the case p (0) = α and p ( k ) = α n (so, k = 2 n − p with p (0) = α i , p ( k ) = α j and one would start with the relation f m i + ··· + m j +1 i,j ∈ I ( λ ) instead of the relation f m + ··· + m n +11 ,n ∈ I ( λ ) below.So from now on we assume without loss of generality that p (0) = α and p ( k ) = α n . Let S + ( h ⊕ n + ) ⊂ S ( h ⊕ n + ) be the maximal homogeneous idealof polynomials without constant term. The adjoint action of U ( n + ) on g induces an action of U ( n + ) on S ( g ) and hence on S ( n − ) ≃ S ( g ) /S ( n − ) S + ( h ⊕ n + ) . In the following we use the differential operators ∂ α defined by ∂ α f β = ( f β − α , if β − α ∈ △ + , , otherwise . The operators ∂ α satisfy the property ∂ α f β = c α,β (ad e α )( f β ) , where c α,β are some non-zero constants. In the following we use very oftenthe following consequence: if f β . . . f β l ∈ I ( λ ), then for any α , . . . , α s ∂ α . . . ∂ α s f β . . . f β l ∈ I ( λ ) . Since f m + ··· + m n +11 ,n v λ = 0 in grV ( λ ), it follows that f s p (0) + ··· + s p ( k ) ,n ∈ I ( λ ) . Write ∂ i,j for ∂ α i,j . For instance, we have(2.2) ∂ ,i f ,j = f i +1 ,j , ∂ j,n f i,n = f i,j − for 1 ≤ i < j ≤ n. For i, j = 1 , . . . , n set s • ,j = j X i =1 s i,j , s i, • = n X j = i s i,j . We consider first the vector(2.3) ∂ s • ,n − n,n ∂ s • ,n − n − ,n . . . ∂ s • , ,n f s p (0) + ··· + s p ( k ) ,n ∈ I ( λ ) . Because of the formulas in (2.2) we get: ∂ s • , n f s p (0) + ··· + s p ( k ) ,n = c f s p (0) + ··· + s p ( k ) − s • , ,n f s • , , for some nonzero constant c , and ∂ s • , n ∂ s • , n f s p (0) + ··· + s p ( k ) ,n = c f s p (0) + ··· + s p ( k ) − s • − s • ,n f s • , f s • , for some nonzero constant c etc. Summarizing, the vector (2.3) is propor-tional (with a nonzero constant) to f s • , , f s • , , . . . f s • ,n ,n ∈ I ( λ ) . EVGENY FEIGIN, GHISLAIN FOURIER AND PETER LITTELMANN
To prove the proposition, we apply more differential operators to the mono-mial f s • , , f s • , , . . . f s • ,n ,n . Consider the following element in I ( λ ) ⊂ S ( n − ):(2.4) A = ∂ s , • , ∂ s , • , . . . ∂ s n, • ,n − f s • , , f s • , , . . . f s • ,n ,n . We claim: (2.5) A = X t ≤ s c t f t for some c s = 0 . Now A ∈ I ( λ ) by construction, so the claim proves the proposition. Proof of the claim:
In order to prove the claim we need to introduce somemore notation. For j = 1 , . . . , n − A j = ∂ s j +1 , • ,j ∂ s j +2 , • ,j +1 . . . ∂ s n, • ,n − f s • , , f s • , , . . . f s • ,n ,n , so A = A . To start an inductive procedure, we begin with A n − : A n − = ∂ s n, • ,n − f s • , , f s • , , . . . f s • ,n ,n . Now s n, • = s n,n and ∂ ,n − f ,j = 0 except for j = n , so(2.7) A n − = cf s • , , f s • , , . . . f s • ,n − s n,n ,n f s n,n n,n , for some nonzero constant c .The proof will now proceed by decreasing induction. Since the inductionprocedure is quite involved and the initial step does not reflect the prob-lems occurring in the procedure, we discuss for convenience the case A n − separately.Consider A n − , up to a nonzero constant we have: A n − = ∂ s n − , • ,n − f s • , , f s • , , . . . f s • ,n − s n,n ,n f s n,n n,n . Now ∂ ,n − f ,j = 0 except for j = n − , n , and ∂ ,n − f n,n = 0, so A n − = s n − , • X ℓ =0 c ℓ f s • , , f s • , , . . . f s • ,n − − s n − , • + ℓ ,n − f s • ,n − s n,n − ℓ ,n f s n − , • − ℓn − ,n − f ℓn − ,n f s n,n n,n . We need to control which powers f ℓn − ,n can occur. Recall that s has supportin p . If α n − p , then s n − ,n − = 0 and s n − , • = s n − ,n , so f s n − ,n n − ,n is thehighest power occurring in the sum. Next suppose α n − ∈ p . This implies α j,n p unless j = n − n . Since s has support in p , this implies s • ,n = s ,n + . . . + s n − ,n + s n,n = s n − ,n + s n,n , and hence again the highest power of f n − ,n which can occur is f s n − ,n n − ,n , andthe coefficient is nonzero. So we can write(2.8) A n − = s n − ,n X ℓ =0 c ℓ f s • , , . . . f s • ,n − − s n − , • + ℓ ,n − f s • ,n − s n,n − ℓ ,n f s n − , • − ℓn − ,n − f ℓn − ,n f s n,n n,n . For the inductive procedure we make the following assumption:
BW FILTRATION AND BASES 9 A j is a sum of monomials of the form(2.9) f s • , , . . . f s • ,j ,j f s • ,j +1 −∗ ,j +1 . . . f s • ,n −∗ ,n | {z } X f t j +1 ,j +1 j +1 ,j +1 f t j +1 ,j +2 j +1 ,j +2 . . . f t n − ,n n − ,n f t n,n n,n | {z } Y having the following properties: i) With respect to the homogeneous lexicographic ordering, all themulti-exponents of the summands, except one, are strictly smallerthan s . ii) More precisely, there exists a pair ( k , ℓ ) such that k ≥ j + 1, s k ℓ > t k ℓ and s kℓ = t kℓ for all k > k and all pairs ( k .ℓ ) suchthat ℓ > ℓ . iii) The only exception is the summand such that t ℓ,m = s ℓ,m for all ℓ ≥ j + 1 and all m .The calculations above show that this assumption holds for A n − and A n − .We come now to the induction procedure and we consider A j − = ∂ s j, • ,j − A j .Note that ∂ ,j − f ,ℓ = 0 except for ℓ ≥ j , and in this case we have ∂ ,j − f ,ℓ = f j,ℓ . Furthermore, ∂ ,j − f k,ℓ = 0 for k ≥ j + 1, so applying ∂ ,j − to a sum-mand of the form in (2.9) does not change the Y -part in (2.9). Summarizing,applying ∂ s j, • ,j − to a summand of the form in (2.9) gives a sum of monomialsof the form(2.10) f s • , , . . . f s • ,j − ,j − f s • ,j −∗ ,j . . . f s • ,n −∗ ,n | {z } X ′ f t j,j j,j . . . f t j,n j,n | {z } Z f t j +1 ,j +1 j +1 ,j +1 f t j +1 ,j +2 j +1 ,j +2 . . . f t n,n n,n | {z } Y . We have to show that these summands satisfy again the conditions i )– iii )above (but now for the index ( j − i) and ii) hold for theindex j , then the same holds obviously also for the index ( j −
1) for allsummands in (2.10) because the Y -part remains unchanged.So it remains to investigate the summands of the form (2.10) obtained byapplying ∂ s j, • j − to the only summand in (2.9) satisfying iii) .To formalize the arguments used in the calculation for A n − we need thefollowing notation. Let 1 ≤ k ≤ k ≤ · · · ≤ k n ≤ n be numbers defined by k i = max { j : α i,j ∈ p } . For convenience we set k = 1. Example 2.3.
For p = ( α , α , . . . , α n , α n , . . . , α n,n ) we have k i = n forall i = 1 , . . . , n . Since s is supported on p we have(2.11) s i, • = k i X ℓ = k i − s i,ℓ , s • ,ℓ = X i : k i − ≤ ℓ ≤ k i s i,ℓ . Suppose now that we have a summand of the form in (2.10) obtained byapplying ∂ s j, • j − to the only summand in (2.9) satisfying iii) . Since the Y -partremains unchanged, this implies already t n,n = s n,n , . . . , t j +1 ,j +1 = s j +1 ,j +1 .Assume that we have already shown t j,n = s j,n , . . . , t j,ℓ +1 = s j,ℓ +1 , thenwe have to show that t j,ℓ ≤ s j,ℓ .We consider five cases: • ℓ > k j . In this case the root α j,ℓ is not in the support of p andhence s j,ℓ = 0. Since ℓ > k j ≥ k j − ≥ . . . ≥ k , for the samereason we have s i,ℓ = 0 for i ≤ j . Recall that the power of f ,ℓ in A j − in (2.6) is equal to s • ,ℓ . Now s • ,ℓ = P i>j s i,ℓ by thediscussion above, and hence f s • ,ℓ ,ℓ has already been transformedcompletely by the operators ∂ ,i , i > j , and hence t j,ℓ = 0 = s j,ℓ . • k j − < ℓ ≤ k j . Since ℓ > k j − ≥ . . . ≥ k , for the same reasonas above we have s i,ℓ = 0 for i < j , so s • ,ℓ = P i ≥ j s i,ℓ . Thesame arguments as above show that for the operator ∂ ,j − onlythe power f s j,ℓ ,ℓ is left to be transformed into a power of f j,ℓ , sonecessarily t j,ℓ ≤ s j,ℓ . • k j − = ℓ = k j . In this case s j, • = s j,ℓ and thus the operator ∂ s j, • ,j − = ∂ s j,ℓ ,j − can transform a power f ∗ ,ℓ in A j only into a power f qj,ℓ with q at most s j,ℓ . • k j − = ℓ < k j . In this case s j, • = s j,ℓ + s j,ℓ +1 + . . . + s j,k j .Applying ∂ s j, • ,j − to the only summand in (2.9) satisfying iii) , theassumption t j,n = s j,n , . . . , t j,ℓ +1 = s j,ℓ +1 implies that one has toapply ∂ s j,kj ,j − to f ∗ ,k j and ∂ s j,kj − ,j − to f ∗ ,k j − etc. to get the demandedpowers of the root vectors. So for f ∗ ,ℓ only the operator ∂ s j,ℓ ,j − isleft for transformations into a power of f j,ℓ and hence t j,ℓ ≤ s j,ℓ . • ℓ < k j − . In this case s j,ℓ = 0 because the root is not in thesupport. Since t j,ℓ = s j,ℓ for ℓ > ℓ and s j,ℓ = 0 for ℓ ≤ ℓ (samereason as above) we obtain ∂ s j, • ,j − = ∂ P ℓ>ℓ s j,ℓ ,j − . But by assumption we know that ∂ s j,ℓ ,j − is needed to transform thepower f s j,ℓ ,ℓ into f s j,ℓ j,ℓ for all ℓ > ℓ , so no power of ∂ ,j − is left andthus t j,ℓ = 0 = s j,ℓ .It follows that all summands except one satisfy the conditions i),ii) above.The only exception is the term where the powers of the operator ∂ s j, • ,j − aredistributed as follows: f s • , , ...f s • ,j − ,j − ( ∂ s j,j ,j − f s • ,j ,j )( ∂ s j,j +1 ,j − f s • ,j +1 −∗ ,j +1 ) ... ( ∂ s j,n ,j − f s • ,n −∗ ,n ) f s j +1 ,j +1 j +1 ,j +1 ...f s n,n n,n . By construction, this term is nonzero and satisfies the condition iii) , whichfinishes the proof of the proposition. (cid:3)
Theorem 2.4.
The elements f s v λ with s ∈ S ( λ ) span the module grV ( λ ) . BW FILTRATION AND BASES 11
Proof.
The elements f s v λ , s arbitrary multi-exponent, span S ( n − ) /I ( λ ) ≃ grV ( λ ). We use now the equation (2.1) in Proposition 2.1 as a straighteningalgorithm to express f t v λ , t arbitrary, as a linear combination of elements f s v λ such that s ∈ S ( λ ).Let λ = P ni =1 m i ω i and suppose s / ∈ S ( λ ), then there exists a Dyck path p = ( p (0) , . . . , p ( k )) with p (0) = α i , p ( k ) = α j such that k X l =0 s p ( l ) > m i + · · · + m j . We define a new multi-exponent s ′ by setting s ′ α = ( s α , α ∈ p , , otherwise. For the new multi-exponent s ′ we still have k X l =0 s ′ p ( l ) > m i + · · · + m j . We can now apply Proposition 2.1 to s ′ and conclude f s ′ = X s ′ > t ′ c t ′ f t ′ in S ( n − ) /I ( λ ) . We get f s back as f s = f s ′ Q β / ∈ p f s β β . For a multi-exponent t ′ occurring inthe sum with c t ′ = 0 set f t = f t ′ Q β / ∈ p f s β β and c t = c t ′ . Since we have amonomial order it follows:(2.12) f s = f s ′ Y β / ∈ p f s β β = X s > t c t f t in S ( n − ) /I ( λ ) . The equation (2.12) provides an algorithm to express f s in S ( n − ) /I ( λ ) asa sum of elements of the desired form: if some of the t are not elements of S ( λ ), then we can repeat the procedure and express the f t in S ( n − ) /I ( λ )as a sum of f r with r < t . For the chosen ordering any strictly decreasingsequence of multi-exponents is finite, so after a finite number of steps oneobtains an expression of the form f s = P c r f r in S ( n − ) /I ( λ ) such that r ∈ S ( λ ) for all r . (cid:3) The linear independence
In the following let R i denote the subset R i = { α ∈ R + | ( ω i , α ) = 1 } . We define for a dominant weight λ ∈ P + R λ = { α ∈ R + | ( λ, α ) > } . Recall that we use α i,j as an abbreviation for α i + α i +1 + . . . + α j (see Section2). The set R i can then be described as R i = { α j,k | ≤ j ≤ i ≤ k ≤ n } . We say a path p has color i if ∃ j s.t. β ( j ) ∈ R i . Note that a path can haveseveral different colors.To simplify the notation we often just write ( j, k ) for the root α j,k (if noconfusion is possible).Let λ = P nj =1 m j ω j and let i be minimal such that m i = 0. For s ∈ S ( λ ),we denote R s i = { ( j, k ) ∈ R i | s j,k = 0 } . We define two different orders on R , a partial order “ ≤ ”:( j , k ) ≤ ( j , k ) ⇔ ( j ≤ j ∧ k ≤ k ) , and a total order “ ≪ ”:( j , k ) ≪ ( j , k ) ⇔ if ( k < k ) or ( k = k ∧ j < j ) . By definition, “ ≪ ” covers “ ≤ ”. Example 3.1.
For g = sl and i = 2, the minimal element of R i with respectto both orders is (1 ,
2) = α , = α + α . Note that α + α + α ≪ α , butthe two are not comparable with respect to “ ≤ ”.A tuple s ∈ S ( λ ) will be considered as an ordered tuple with respect tothe order “ ≪ ”: s = ( s , , s , , s , , s , , s , , s , , . . . , s n,n ) . The induced lexicographic order on S ( λ ) is a total order which we againdenote by “ ≪ ”. Remark . The total order ≪ is different from the order ≺ used in Sec-tion 2. Example 3.3.
For g = sl let s be defined by s = 1 , s = 1 and s j,k = 0 otherwise,and let t be defined by t = 1 , t = 1 and t j,k = 0 otherwise.Then s = (0 , , , , ,
0) and t = (0 , , , , , s ≪ t . Definition . For s ∈ S ( λ ) denote by M s i the set of minimal elements in R s i with respect to ≤ . We denote by m s i the tuple m j,k = 1 if ( j, k ) ∈ M s i and m j,k = 0 otherwise. Example 3.5.
1) If R s i = R i , then M s i = { α ,i } .2) If R s i = { α i,i , α i − ,i +1 , . . . , α i − ℓ,i + ℓ } for some ℓ ≤ i , then M s i = R s i . BW FILTRATION AND BASES 13
Remark . s we have M s i = { α j l ,k l | l = 1 , . . . , m } for some m , and the indices have the property1 ≤ j < j < . . . < j m ≤ i ≤ k m < · · · < k < k ≤ n. If s ∈ S ( ω i ), then for the associated tuple m s i we get: m s i = s .2). The sets M s i satisfy the following important property: any Dyck pathcontains at most one element of M s i , because the elements of a Dyck pathare linearly ordered with respect to “ ≥ ”. Proposition 3.7.
For s ∈ S ( λ ) let M s i be the minimal set. Then m s i ∈ S ( ω i ) , and if s ′ is such that s = s ′ + m s i , then s ′ ∈ S ( λ − ω i ) .Proof. Note that m s i ∈ S ( ω i ) by Remark 3.6. Let s ′ be such that s = s ′ + m s i .We claim that s ′ ∈ S ( λ − ω i ). Let λ = P nj = i m j ω j . For a Dyck path p let q λ p = P j color of p m j be the upper bound for the defining inequality (1.2) of S ( λ ) associated to p .If p is a Dyck path such that i is not a color, then q λ p = q λ − ω i p and s β = s ′ β for β R i , so s ′ satisfies the defining inequality for S ( λ − ω i ) given by p .Let p be a Dyck path of color i , so q λ − ω i p = q λ p −
1. If p ∩ M s i = ∅ , then P ( j,k ) ∈ p s ′ j,k = P ( j,k ) ∈ p s j,k − ≤ q λ p − q λ − ω i p , so s ′ satisfies the defininginequality for S ( λ − ω i ) given by p .Suppose now that p is a Dyck path of color i but p ∩ M s i = ∅ . Recallthat the elements in supp p are linearly ordered. Let α l,m be the minimalelement in R s i ∩ supp p . Since i is minimal such that m i >
0, note that s β = 0 for all β ∈ supp p be such that β < α l,m . By assumption, α l,m M s i ,so let α r,t ∈ M s i such that α r,t < α l,m . Let ˜ p be the Dyck path( α r,r , α r,r +1 , . . . , α r,t , α r,t +1 , . . . , α r,m , α r +1 ,m , . . . , α l,m , β , . . . , β N ) , where { β , . . . , β N } are the elements in supp p such that β j > α l,m . Since α r,t ∈ supp ˜ p we know: X ( j,k ) ∈ p s j,k < X ( j,k ) ∈ ˜ p s j,k ≤ q λ p and hence P ( j,k ) ∈ p s j,k = P ( j,k ) ∈ p s ′ j,k ≤ q λ p − q λ − ω i p . (cid:3) For s ∈ S ( λ ) we define a mutation of s as follows: Definition . Let β = X ( j,k ) ∈ R s j,k α j,k and suppose β = X ( j,k ) ∈ R t j,k α j,k where t j,k = 0 if ( j, k ) / ∈ R λ ; t j,k ≥ j, k ) ∈ R λ , for some t = ( t j,k ) / ∈ S ( λ ). Then we call t a mutation of s . Example 3.9.
Let g = sl and λ = ω . Define s by s , = 1 , s , = 1 and s i,j = 0 else,and t by t , = 1 , t , = 1 and t i,j = 0 else.Then t is a mutation of s . Proposition 3.10.
For s ∈ S ( λ ) let M s i be the minimal set. If t is amutation of m s i , t = t + t ∈ S ( λ ) and t j,k ≥ , then m s i ≪ m t i .Proof. Recall (see Remark 3.6) that M s i = { ( j l , k l ) | l = 1 , . . . , m } with1 ≤ j < · · · < j m ≤ i ≤ k m < · · · < k . Let t be a mutation of m s i , so t j,k = 0 for ( j, k ) / ∈ R i . Then there exists σ ∈ S m \{ id } such that if t p,q = 0, then ( p, q ) = ( j l , k σ ( l ) ) for some 1 ≤ l ≤ m .We can even assume that σ ( l ) = l for all l , because otherwise ( j l , k l ) is notmutated and appears in m s i and t .It is clear that m t i ≪ m t i (or equal), so it suffices to show that m s i ≪ m t i .Let x = σ − ( m ), we claim that M t i ⊂ { ( j , k σ (1) ) , . . . , ( j x , k σ ( x ) ) } . Let l > x ,then j x < j l and k m > k σ ( l ) (since σ ( l ) = m ). So ( j x , k m ) < ( j l , k σ ( l ) ) for all l > x . (cid:3) Theorem 3.11.
Let λ = P j m j ω j ∈ P + . For each s ∈ S ( λ ) fix an arbitraryorder of factors f α in the product Q α> f s α α . Let f s = Q α> f s α α be theordered product in U ( n − ) . Then the elements f s v λ , s ∈ S ( λ ) , form a basisof V ( λ ) .Proof. We will prove the claim by induction on m = P nj =1 m j . By Theo-rem 2.4 we know that the f s v λ span the representation V ( λ ), so dim V ( λ ) ≥ ♯S ( λ ). For the initial step m = 1 the description of S ( ω i ) in Remark 3.6shows that the tuples have all different weights and hence the f s v ω i are alsolinearly independent, which proves the claim for the fundamental represen-tations.We assume that the claim holds for λ , we want to prove it for λ + ω i .We may assume again that i is minimal such that m i = 0. The highestweight vector v λ ⊗ v ω i generates V ( λ + ω i ) ⊂ V ( λ ) ⊗ V ( ω i ). We assumein the following that the roots are ordered in such a way that the f α with α ∈ R i are at the beginning. Every element s ∈ S ( λ + ω i ) defines a vectorof f s ( v λ ⊗ v ω i ) ∈ V ( λ + ω i ). We want to show that these vectors are linearlyindependent, so we have to show(3.1) X s ∈ S ( λ + ω i ) a s f s ( v λ ⊗ v ω i ) = 0 ⇒ a s = 0 ∀ s ∈ S ( λ + ω i ) . BW FILTRATION AND BASES 15
We may assume without loss of generality that all s have the same weight,say s ∈ S ( λ + ω i ) µ . By Proposition 3.7 we can split an element in S ( λ + ω i )such that s = s + m s i , where s ∈ S ( λ ). Assume that we have a non-triviallinear dependence relation in (3.1). Fix ¯ s ∈ S ( λ + ω i ) µ such that a ¯ s = 0in this relation and a t = 0 for all t such that m ¯ s i ≪ m t i . Consider first¯ s = ¯ s + m ¯ s i , so we have(3.2) f ¯ s ( v λ ⊗ v ω i ) = c m ¯ s i f ¯ s v λ ⊗ f m ¯ s i v ω i + other terms, where c m ¯ s i is a nonzero constant (product of binomial coefficients).All the terms occurring in the linear dependence relation (3.1) can berewritten as sums of terms of the form f r v λ ⊗ f r v ω i . So in order to provethat necessarily a s = 0 for all terms in (3.1), it is sufficient to show that thatthe terms f r v λ ⊗ f r v ω i satisfying wt ( r ) = wt (¯ s ) and wt ( r ) = wt ( m ¯ s i )are linearly independent.Let us first consider the possible terms in (3.2) occurring among the otherterms . It is a sum of elements f r v λ ⊗ f r v ω i , where r + r = ¯ s and r = m ¯ s i .If wt( r ) = wt( m ¯ s i ), then either r ∈ S ( ω i ), but then r = m ¯ s i for weightreasons, or r S ( ω i ). In the latter case the entries in r are zero for all α k,ℓ R i because of the special choice of the ordering, and hence r hasto be a mutation of m ¯ s i . Then by Proposition 3.10, m ¯ s i ≪ m r + r i = m ¯ s i which is a contradiction. So the other terms consist only of tensors of theform f r v λ ⊗ f r v ω i , where wt( r ) = wt(¯ s ) and wt( r ) = wt( m ¯ s i ), hence forproving linear independence we can neglect these terms.To obtain a non-trivial linear combination such that a t = 0 for some t = ¯ s , one needs an element t ∈ S ( λ + ω i ) µ which can be splitted t = t + t such that wt( t ) = wt(¯ s ), wt( t ) = wt( m ¯ s i ), and f t v λ = 0 , f t v ω i = 0.Suppose that one has such a t = t + t and t / ∈ S ( ω i ). By the samearguments as above, t is a mutation of m ¯ s i and hence by Proposition 3.10, m ¯ s i ≪ m t i . But in this case we have by assumption a t = 0, contradictingthe fact a t = 0.It follows t ∈ S ( ω i ) and hence, by weight arguments, t = m ¯ s i and t = t + m s i , where t = s .So if a term of the form f t v λ ⊗ f t v ω i wt( t ) = wt(¯ s ), wt( t ) = wt( m ¯ s i )occurs in the linear dependence relation (3.1), then necessarily t = m ¯ s i .Hence, by Proposition 3.7, t ∈ S ( λ ). Since the possible t are differentfrom ¯ s and by induction the terms { f t v λ ⊗ f m ¯ s i v ω i | t ∈ S ( λ ) } are linearlyindependent, it follows a ¯ s = 0, contradicting the assumption a ¯ s = 0.Summarizing, we have shown that for the order fixed at the beginningof the proof the f s v λ + ω i , s ∈ S ( λ + ω i ), are linearly independent and forma basis. This implies in particular that ♯S ( λ + ω i ) = dim V ( λ + ω i ). Nowby Theorem 2.4 we know that the f s v λ + ω i , s ∈ S ( λ + ω i ), span V ( λ + ω i )for any chosen total order. So, for dimension reason, they also have to belinearly independent for any chosen order. (cid:3) Proof of Theorem A and applications
In this section we collect some immediate consequences of the construc-tions in Sections 2 and 3. The proof of Theorem 3.11 shows:
Corollary 4.1. dim V ( λ ) = S ( λ ) = number of integral points in the polytope P ( λ ) . By the defining inequalities (see 0.1) for the polytope P ( λ ) it is obviousthat for two dominant integral weights λ, µ we have P ( λ )+ P ( µ ) ⊆ P ( λ + µ ),and hence for the integral points we have S ( λ ) + S ( µ ) ⊆ S ( λ + µ ), too. Infact, the reverse implication is also true: Proposition 4.2. S ( λ ) + S ( µ ) = S ( λ + µ ) .Proof. Set ν = λ + µ and write ν = P k i ω i as a sum of fundamental weights.Proposition 3.7 provides an inductive procedure to write an element s in S ( ν ) as a sum s = P ni =1 P k i j =1 m i,j such that m i,j ∈ S ( ω i ) for all 1 ≤ i ≤ n ,1 ≤ j ≤ k i . This sum can be reordered in such a way that s = s + s , s ∈ S ( λ ), s ∈ S ( µ ), so s ∈ S ( λ ) + S ( µ ). (cid:3) As an interesting application we obtain a combinatorial character formulafor the representation V ( λ ). Let P be the weight lattice and for s ∈ S ( λ )define the weight wt( s ) := X ≤ j ≤ k ≤ n s j,k α j,k . Let S ( λ ) µ be the subset of elements such that µ = λ − wt( s ) and let S ( λ ) µ := { s ∈ S ( λ ) | µ = λ − wt( s ) } be the number of elements of this set. Weobtain as a consequence of Theorem 1.5: Proposition 4.3. charV ( λ ) = X µ ∈ P S ( λ ) µ e µ . The big advantage of our approach is that it provides also a combinatorialformula for the graded character. Recall that grV ( λ ) carries an additionalgrading on each weight space V ( λ ) µ of V ( λ ): grV ( λ ) µ = M s ≥ gr s V ( λ ) µ = M s ≥ V ( λ ) µs /V ( λ ) µs − . The graded character of the weight space is the polynomial p λ,µ ( q ) := X s ≥ (dim V ( λ ) µs /V ( λ ) µs − ) q s and the graded character of V ( λ ) is char q ( V ( λ )) = X µ ∈ P p λ,µ ( q ) e µ . We have a natural notion of a degree for the multi-exponents:
BW FILTRATION AND BASES 17
Definition . deg( s ) := X ≤ j ≤ k ≤ n s j,k . As an immediate consequence of Theorem 1.5 we get
Corollary. p λ,µ ( q ) = P s ∈ S ( λ ) µ q deg s and char q ( V ( λ )) = X s ∈ S ( λ ) e λ − wt( s ) q deg( s ) . Finally, we note that the results of Sections 2 and 3 imply the descriptionof the annihilating ideal I ( λ ). Theorem 4.5. (4.1) I ( λ ) = S ( n − ) (cid:16) U( n + ) ◦ span { f ( λ,α )+1 α , α > } (cid:17) . Proof.
Since f ( λ,α )+1 α v λ = 0 in V ( λ ) for all positive roots α , the right handside of (4.1) belongs to I ( λ ). Section 2 shows that the relations in the RHSof (4.1) are enough to rewrite any element of grV ( λ ) in terms of the basiselement f s v λ , s ∈ S ( λ ). This proves our theorem. (cid:3) Acknowledgements
The work of Evgeny Feigin was partially supported by the Russian Presi-dent Grant MK-281.2009.1, the RFBR Grants 09-01-00058, 07-02-00799 andNSh-3472.2008.2, by Pierre Deligne fund based on his 2004 Balzan prize inmathematics and by Alexander von Humboldt Fellowship. The work of Ghis-lain Fourier was partially supported by the DFG project “KombinatorischeBeschreibung von Macdonald und Kostka-Foulkes Polynomen“. The workof Peter Littelmann was partially supported by the priority program SPP1388 of the German Science Foundation.
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Evgeny Feigin:Tamm Department of Theoretical Physics, Lebedev Physics Institute,Leninsky prospect, 53, 119991, Moscow, Russia, and
French-Russian Poncelet Laboratory, Independent University of Moscow
E-mail address : [email protected] Ghislain Fourier:Mathematisches Institut, Universit¨at zu K¨oln,Weyertal 86-90, D-50931 K¨oln,Germany
E-mail address : [email protected] Peter Littelmann:Mathematisches Institut, Universit¨at zu K¨oln,Weyertal 86-90, D-50931 K¨oln,Germany
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