Quantum generalized Harish-Chandra isomorphisms

Journal of Algebra 401 (2014) 144–160

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Journal of Algebra www.elsevier.com/locate/jalgebra

Quantum generalized Harish-Chandra isomorphisms ✩ Polyxeni Lamprou Weizmann Institute of Science, 76100 Rehovot, Israel

a r t i c l e

i n f o

Article history: Received 12 June 2013 Available online 3 January 2014 Communicated by J.T. Stafford

a b s t r a c t We give a different proof of generalized Harish-Chandra isomorphisms proven by Khoroshkin, Nazarov and Vinberg [9] and Joseph [6,7] as well as of their analogues in the quantum case. © 2013 Elsevier Inc. All rights reserved.

Keywords: Harish-Chandra isomorphism Quantized enveloping algebra

1. Introduction Throughout this paper, the ground field k is algebraically closed of characteristic 0. 1.1. Let g be a semisimple Lie algebra, Φ its root system with respect to a fixed Cartan subalgebra h, π a basis in Φ and W its Weyl group. Let Λ be the weight lattice of g and for any β ∈ Φ , let hβ denote the corresponding coroot. Let ( , ) be the W -invariant symmetric bilinear form on Λ. For all −n

q − qα [n]α = α , 1 qα − q− α n

α ∈ π , set qα := q(α ,α )/2 and

  [n]α ! = [1]α [2]α · · · [n]α ,

[n]α ! n = . k α [n − k]α ![k]α !

1.2. Let U q (g), or simply U q , be the quantized enveloping algebra of g of simply connected type. Then U q is the algebra generated over k(q) by symbols { E α , F α | α ∈ π } and {qλ | λ ∈ Λ} subject to the following relations.

q λ = 1,

for λ = 0.

qλ qμ = qμ qλ = qλ+μ ,

for all λ, μ ∈ Λ.

qλ E α q−λ = q(α ,λ) E α = qα

λ(hα )

✩

E α , for all λ ∈ Λ, α ∈ π .

Work supported in part by the Israel Science Foundation Grant No. 882/10.

0021-8693/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.11.009

P. Lamprou / Journal of Algebra 401 (2014) 144–160

−λ(h )

qλ F α q−λ = q−(α ,λ) F α = qα α F α , q α − q −α E α F β − F β E α = δα ,β , 1 qα − q− α



1−β(hα )



(−1)r

r =0



1−β(hα )



r

(−1)

r =0

145

for all λ ∈ Λ, α ∈ π . for all α , β ∈ π .



1 − β(hα ) 1−β(hα )−r Eα E β E rα = 0, r α

for all α , β ∈ π ,

α = β.

for all α , β ∈ π ,

α = β.



1 − β(hα ) 1−β(hα )−r Fα F β F αr = 0, r α

The qλ , λ ∈ Λ form a multiplicative group T with q−λ = (qλ )−1 . Denote by U q0 := k(q)[ T ] the group algebra of T . Let U q+ (resp. U q− ) be the subalgebra of U q generated by the E α (resp. the F α ), for all α ∈ π. 1.3. The algebra U q has a Hopf algebra structure, with : U q → U q ⊗ U q , ε : U q → k, S : U q → U q the co-product, co-unit and antipode respectively defined on the generators as follows. For all α ∈ π , λ ∈ Λ,

( E α ) = E α ⊗ q−α + 1 ⊗ E α ,

S ( E α ) = − E α qα ,

( F α ) = F α ⊗ 1 + qα ⊗ F α ,  λ

q

S ( F α ) = −q

 λ

=q ⊗q , λ

λ

S q

=q

−α

−λ

ε ( E α ) = 0, ε ( F α ) = 0,

Fα,

 

ε q λ = 1.

,

+ − 1.4. Let U + := {u ∈ U q+ | ε (u ) = 0}, U + := {u ∈ U q− | ε (u ) = 0} denote the augmentation ideals of + − U q , U q respectively. There is a vector space decomposition of U q as follows:





− + . U q = U q0 ⊕ U + Uq + UqU+

With respect to this decomposition one defines a Harish-Chandra map p : U q → U q0 as projection onto the first factor. It is known that the restriction of this map on the center Z (U q ) of U q is an W. , where (U q0 )ev is the subalgebra of U q0 generated by qλ , λ ∈ 2Λ and W . isomorphism onto (U q0 )ev denotes the translated action of the Weyl group W on the torus T , given by [4, Section 7.1.17]:

w .qλ = q w λ q(ρ , w λ−λ) ,

(1)

for all λ ∈ Λ, where ρ is the half sum of positive roots in Φ . Moreover, it is known that the center W. ) is a polynomial algebra. Z (U q ) (and hence, (U q0 )ev 1.5. Let V be a finite-dimensional simple U q -module of type 1 (i.e. its highest weight is qλ for some λ ∈ Λ). For all μ ∈ Λ set







V μ := v ∈ V  qλ v = q(λ,μ) v , for all λ ∈ Λ , the weight space of V corresponding to μ. We consider a left and a right action of U q on V ⊗ U q as follows. For all x, a ∈ U q , v ∈ V we set

x( v ⊗ a) :=

 i

xi v ⊗ xi a

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where (x) =

i

xi ⊗ xi and xi a denotes multiplication in U q . We also set

( v ⊗ a)x := v ⊗ ax, where ax denotes multiplication in U q .

The adjoint action of U q on itself is given by (ad x)(a) = i xi aS (xi ) for all x, a ∈ U q with (x) =

i i xi ⊗ x . One defines the adjoint action of U q on V ⊗ U q in a similar manner using the left and right action on U q on this module:

(ad x)( v ⊗ a) =



 

xi v ⊗ aS xi ,

i



for all x, a ∈ U q , v ∈ V with (x) = i xi ⊗ xi . Then, the diagonal action of U q on V ⊗ U q , if we view U q as an ad U q -module, coincides with the adjoint action of U q on V ⊗ U q . 1.6.

It is easy to check that there is a vector space (and a U q0 -bimodule) decomposition









V ⊗ U q = V ⊗ U q0 ⊕ U q− ( V ⊗ U q ) + ( V ⊗ U q )U q+ . We consider two maps;

Pˆ : V ⊗ U q → V ⊗ U q0 , defined as projection onto the first factor with respect to the above decomposition and

P := 1 ⊗ p : V ⊗ U q → V ⊗ U q0 , where p is the quantum Harish-Chandra map defined in Section 1.4. We call Pˆ the quantum generalized Harish-Chandra map. Our aim is to describe the image of ( V ⊗ U q )ad U q under Pˆ and P . 1.7. The problem we are considering here was firstly addressed by Khoroshkin, Nazarov and Vinberg [9] in the non-quantum case. They showed that (the classical version of) Pˆ maps the space of g-invariants in V ⊗ U (g) isomorphically to the invariants under the Zhelobenko operators in V 0 ⊗ U (h), where V is a simple finite-dimensional U (g)-module and V 0 its zero weight space. Later, Joseph [6] gave a different proof of this result; the key point of his proof was the computation of determinants analogous to the Parthasarathy–Ranga Rao–Varadarajan (PRV) determinants [11]. Shortly after that, he computed the image of ( V ⊗ U (g))g under the map 1 ⊗ p (where p is the usual HarishChandra map) [7]; it equals the space of invariants under certain operators, which he called the analogue Zhelobenko operators, in V 0 ⊗ U (h). The method of his proof in [7], used earlier to compute PRV determinants for quantum affine Lie algebras [8], is what we use in this paper, thus getting a straightforward proof of the Khoroshkin–Nazarov–Vinberg result (Section 2) which carries over nicely to the quantum case (Section 3). Our proof involves simple sl2 computations; moreover, one only needs to define the Zhelobenko operators on V 0 ⊗ U (h) (resp. V 0 ⊗ U q0 ) rather than V ⊗ U (g) (resp. V ⊗ U (g)). A related result  in the quantum case is recently published in [1]; the author describes the image under Pˆ of V ⊗ ( λ∈Λ+ V (λ)∗ ⊗ V (λ)) where Λ+ is the set of dominant weights, V (λ) is the highest weight U q -module of type 1 of highest weight qλ , λ ∈ Λ+ and V (λ)∗ is the left dual of V (λ). In the classical case the main result of [1] is equivalent to the Khoroshkin–Nazarov–Vinberg theorem modified to the setting of algebraic groups.

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2. Generalized Harish-Chandra isomorphisms in the non-quantum case 2.1. Preliminaries 2.1.1. Recall paragraph 1.2 and let g = n− ⊕ h ⊕ n be the triangular decomposition of g corresponding to h. Let U (g) (resp. S (g)) be the universal enveloping algebra (resp. symmetric algebra) of g; S (g) identifies with the algebra of polynomials on the dual algebra g∗ of g. We have a vector space decomposition





U (g) = U (h) ⊕ n− U (g) + U (g)n . Let p be the Harish-Chandra homomorphism, that is the projection of U (g) onto U (h) with respect to the above decomposition:





p : U (g) → U (h)  U (g)/ n− U (g) + U (g)n . Since h is commutative, U (h) identifies with S (h). It is known that p restricted to the center Z (g) of U (g) is an isomorphism onto S (h) W . . Here W . denotes the translated action of the Weyl group W on S (h), namely, for any polynomial q ∈ S (h) and any λ ∈ h∗ ,









w .q(λ) := q w −1 .λ = q w −1 (λ + ρ ) − ρ . 2.1.2. Let V be a simple finite-dimensional U (g)-module. We define a left and a right action on V ⊗ U (g) as follows. For all x, a ∈ U (g) and all v ∈ V

x( v ⊗ a) = xv ⊗ a + v ⊗ xa, where xa denotes multiplication in U (g). We also set

( v ⊗ a)x := v ⊗ ax, where ax denotes multiplication in U (g). Then, the diagonal action of U (g) on V ⊗ U (g), if we view U (g) as an ad U (g)-module, coincides with the adjoint action of U (g) on V ⊗ U (g):

(ad x)( v ⊗ a) := x( v ⊗ a) − ( v ⊗ a)x = xv ⊗ a + v ⊗ [x, a], for all x, a ∈ U (g), v ∈ V . Remark. The left and right action defined above are analogous to those in 1.5, since U (g) is a Hopf algebra with co-product (x) = x ⊗ 1 + 1 ⊗ x, for all x ∈ g (extended to U (g) as an algebra homomorphism). We have a decomposition of vector spaces (and U (h)-bimodules) as follows:













 

V ⊗ U (g) = V ⊗ U (h) ⊕ n− V ⊗ U (g) + V ⊗ U (g) n , thus we may identify the space V ⊗ U (h) with ( V ⊗ U (g))/(n− ( V ⊗ U (g)) + ( V ⊗ U (g))n). Denote by Pˆ the projection of V ⊗ U (g) onto V ⊗ U (h) with respect to this decomposition:



 

Pˆ : V ⊗ U (g) → V ⊗ U (h)  V ⊗ U (g)

     n− V ⊗ U (g) + V ⊗ U (g) n .

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We call Pˆ the generalized Harish-Chandra map. We also consider the map

P := 1 ⊗ p : V ⊗ U (g) → V ⊗ U (h), where recall that p is the Harish-Chandra homomorphism. In the following paragraphs of this section we will describe the image of ( V ⊗ U (g))g under Pˆ and P . 2.2. Invariants under the map Pˆ 2.2.1. Let α ∈ π and let pα be the minimal parabolic subalgebra of g corresponding to α . Let rα be the Levi factor of pα and m± α the nilradical and opposed nilradical of pα . The algebra rα is reductive and is equal to rα = h + sα = sα ⊕ h , where sα = k e α , hα , f α   sl2 is the sl2 subalgebra of g corresponding to α and h is the orthogonal complement of khα in h. We have U (h)  S (h )[hα ]. Then one easily checks that we have a vector space (and a U (h)-bimodule) decomposition

















V ⊗ U (g) = V ⊗ U (rα ) ⊕ m− α V ⊗ U (g) + V ⊗ U (g) mα , thus we may identify V ⊗ U (rα )  V ⊗ U (g)/(m− α ( V ⊗ U (g)) + ( V ⊗ U (g))mα ). 2.2.2.  For all λ ∈ h∗ we denote by V λ the corresponding weight space of V of weight λ. Let V Zα := n∈Z V nα . We consider the projection Pˆ α of V ⊗ U (g) onto V ⊗ U (rα ). Since the map Pˆ α is rα -equivariant, one has that



rα

V ⊗ U (g)

r Pˆ Pˆ α  −→ V Zα ⊗ U (rα ) α −→ V 0 ⊗ U (h).

Of course Pˆ = Pˆ Pˆ α . By Kostant’s separation of variables theorem [10], U (rα ) is free over its center, namely

U (rα )  H ⊗ Z (rα ), where H is the space of harmonic polynomials in U (rα ) and the isomorphism map is given by multiplication. Moreover, for any rα -module M, there are dim M 0 copies of M in H. Hence ( V ⊗ U (rα ))rα = ( V ⊗ H)rα ⊗ Z (rα ) and its dimension is equal to dim V 0 . Now V is the direct sum of simple rα (equivalently, sα )-modules. Let V (n) be the (2n + 1)dimensional simple rα -module, V (n)∗ its dual. All the weight spaces—and in particular the zero weight space—of V (n), V (n)∗ are one-dimensional; hence there is a unique copy of V (n)∗ in H and so dim( V (n) ⊗ H)rα = dim V (n)0 = 1. Then Pˆ (( V (n) ⊗ H)rα ⊗ Z (rα )) = U (h)sα . Pˆ (( V (n) ⊗ H)rα ), where sα denotes the simple reflection in W corresponding to the simple root α . We will compute Pˆ (( V (n) ⊗ H)rα ). 2.2.3. Recall that sα := k{e α , hα , f α }. Let v 0 be a non-zero highest weight vector in V (n); then { v k = f αk v 0 | k = 0, 1, . . . , 2n} is a basis for V (n). For the copy of V (n)∗ in H we may choose the basis {(ad f α )m enα | m = 0, 1, . . . , 2n}. A non-zero invariant in ( V (n) ⊗ H)rα then is:

x=

2n 

(−1)m f α2n−m v 0 ⊗ (ad f α )m enα .

m =0

Now

(ad f α )m enα =

m m   f m−r n f αr f m−r enα f αr (−1)r m! α eα (−1)r α = m! . (m − r )! r ! r !(m − r )! r =0

r =0

(2)

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149

Then, using the identity min{n,m} n

m

e α f α = n!m!

 k =0

f αm−k

(m − k)!

hα − n − m + 2k k



enα−k

(n − k)!

Eq. (2) becomes

(ad f α )m enα = m!

 n−k min{n,r } m   f m−r f αr −k eα hα − n − r + 2k (−1)r α n! k (m − r )! (r − k)! (n − k)! r =0

= m!n!

k =0

m 

(−1)r

r =0

min{n,r }



1

(m − r )!

k =0

f αm−k

(r − k)!

hα − n − r + 2k k



enα−k

(n − k)!

.

Then we have

x = n!

2n 

(−1)m f α2n−m v 0 ⊗

m  (−1)r

m =0

r =0

min{n,r }



m!

(m − r )!

k =0

f αm−k

(r − k)!

hα − n − r + 2k k



enα−k

(n − k)!

.

If r < n, then each summand in the last sum of the above formula for x is 0 modulo ( V ⊗ U (g))n. Hence for m < n, the term in the right-hand side of the tensor in the formula for x is 0 modulo ( V ⊗ U (g))n. Hence x is equivalent modulo ( V ⊗ U (g))n to

x ≡ n!

2n 

m!

m=n

m  (−1)m f α2n−m v 0 ⊗ r =n

(−1)r f αm−n (m − r )!(r − n)!

hα + n − r n

 .

But for all a ∈ U (g) and all v ∈ V one has

v ⊗ f αs a ≡ (−1)s f αs v ⊗ a,





mod n− V ⊗ U (rα )

and so

x ≡ (−1)n n!

2n  m=n

m!

m  r =n

f αn v 0 ⊗

(−1)r (m − r )!(r − n)!

hα + n − r n



or

x ≡ (−1)n

2n 

m!

m=n

m  r =n

vn ⊗

(−1)r (hα + n − r )(hα + n − r − 1) · · · (hα − r + 1). (m − r )!(r − n)!

One has that 2n  m  m=n r =n

=

(−1)r m! (hα + n − r )(hα + n − r − 1) · · · (hα − r + 1) (m − r )!(r − n)!

2n  2n  r =n m=r

(−1)r m! (hα + n − r )(hα + n − r − 1) · · · (hα − r + 1) (m − r )!(r − n)!

(3)

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P. Lamprou / Journal of Algebra 401 (2014) 144–160

=

2n 2n    (−1)r m r! (hα + n − r )(hα + n − r − 1) · · · (hα − r + 1) ( r − n )! r r =n m=r

=



2n  (−1)r 2n + 1 (hα + n − r )(hα + n − r − 1) · · · (hα − r + 1)r ! (r − n)! r+1 r =n

 n  r (r + n)! 2n + 1 = (−1) (hα − r )(hα − r − 1) · · · (hα − r − n + 1). (−1) r! n−r n

r =0

Thus Eq. (3) becomes

x≡

 n  (r + n)! 2n + 1 (−1)r v n ⊗ (hα − r )(hα − r − 1) · · · (hα − r − n + 1). r! n−r r =0

Lemma. One has that

 n  (r + n)! 2n + 1 (hα − r )(hα − r − 1) · · · (hα − r − n + 1) (−1)r r! n−r r =0

= n!(hα + 2)(hα + 3) · · · (hα + n + 1). Proof. We first compute the coefficient of hnα of the polynomial in the LHS of the above equation.



  n n   2n + 1 r (r + n)! 2n + 1 r n+r = n! (−1) (−1) r! n−r r n−r r =0

r =0

= n!

  n  −n − 1 2n + 1 r =0

r

n−r

= n!

by the Chu–Vandermonde identity. So the polynomial at the LHS of the equation in the statement of the lemma has degree n and the leading coefficient is n!. We will show that its roots are the −2, −3, . . . , −n − 1. That is, we have to show that for k = 2, 3, . . . , n + 1

 n  r +n (r + n)! 2n + 1 (k + r )(k + r + 1) · · · (k + r + n − 1) = 0 (−1) r! n−r

(4)

r =0

or n  (−1)r r =0

Consider P (x) = n + 1. Then

1 r !(n − r )!(r + n + 1)

(k + r )(k + r + 1) · · · (k + r + n − 1) = 0.

(x+k)(x+k+1)···(x+k+n−1) . This is a polynomial of degree n − 1 for any k x+n+1

 n  n (−1)r P (r ) = 0, r =0

hence the assertion.

2

r

(5)

= 2, 3, . . . ,

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151

2.2.4. Set ψα ,n := (hα + 2)(hα + 3) · · · (hα + n + 1) and recall that sα ∈ W is the simple reflection corresponding to α . Then Pˆ (( V (n) ⊗ U (rα ))rα ) = U (h)sα . v n ⊗ ψα ,n . One has that sα .ψα ,n = (−1)n hα (hα − 1) · · · (hα − n + 1). Observe that the roots of the polynomials ψα ,n and sα .ψα ,n are distinct, that is these two polynomials are relatively prime. For all α ∈ π , we define a homomorphism

ξˆα : V (n)0 ⊗ U (h) → V (n)0 ⊗ Fract U (h) by:

ξˆα ( v ⊗ a) := v ⊗

ψα ,n sα .ψα ,n

sα .a,

for all v ∈ V (n)0 , a ∈ U (h). Then an element v ⊗ a ∈ V (n)0 ⊗ U (h) is fixed by ξˆα if and only if ψα ,n divides a and the quotient ψa is sα -invariant, that is α ,n

sα .a sα .ψα ,n

=

a

ψα ,n

.

Extend ξˆα linearly on V 0 , which is the direct sum of the zero weight spaces of simple sα -modules, and consider the monoid Ξˆ := ξˆα | α ∈ π . The homomorphisms ξˆα ∈ Ξˆ are called the Zhelobenko operators. ˆ Theorem. (See [9,5].) The map Pˆ is an isomorphism from ( V ⊗ U (g))g onto ( V 0 ⊗ U (h))Ξ . ˆ Proof. By the above one has that Pˆ (( V Zα ⊗ U (rα ))rα ) = ( V 0 ⊗ U (h))ξα for all ˆ U (h))Ξ . A straightforward computation shows that for all b

α ∈ π and so Pˆ (( V ⊗

⊂ (V 0 ⊗ ∈ S (h) and all v ⊗ a ∈ V 0 ⊗ S (h) one has that ξˆα (b( v ⊗ a)) = (sα .b)ξˆα ( v ⊗ a). Then injectivity follows by [6, Lemma 2.6] and surjectivity as in [6, Section 3]. 2 U (g))g )

2.3. Invariants under the map P In this section we will briefly recall Joseph’s result [7]. Notation is as in Section 2.2. Define a homomorphism

ξα : V (n)0 ⊗ U (h) → V (n)0 ⊗ Fract U (h) by

ξα ( v 0 ⊗ a) := v 0 ⊗

sα .ψα ,n

ψα ,n

sα .a.

Extend ξα by linearity to V 0 and let Ξ = ξα | α ∈ π . Then by [7, Theorem 2.4], one has that

P



g 

V ⊗ U (g)

 Ξ = V 0 ⊗ U (h) .

The homomorphisms ξα in Ξ are called the analogue Zhelobenko operators [7]. 3. Quantum generalized Harish-Chandra isomorphisms The notation in this section is as in 1.2–1.6.

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3.1. Invariants under the map Pˆ 3.1.1. Recall paragraph 2.2.1. Fix α ∈ π and let U qα := U q (rα ) be the subalgebra of U q generated − + − over k(q) by E α , F α and T . Then U q (m+ α ) (resp. U q (mα )) is the subalgebra of U q (resp. U q ) generated ± by the E β , β ∈ π \ {α } (resp. the F β , β ∈ π \ {α }). Let U q (mα )+ denote the augmentation ideals of U q (m± α ). One checks that we have a vector space decomposition:













 

+ V ⊗ U q = V ⊗ U qα ⊕ U q m− α + ( V ⊗ U q ) + ( V ⊗ U q ) U q mα + .

3.1.2.

(6)

For any Hopf algebra A denote by F ( A ) its locally finite part, that is







F ( A ) := a ∈ A  dim(ad A )a < ∞ . For any Hopf subalgebra B of A let







Fˆ ( B ) = b ∈ B  dim(ad A )b < ∞ = F ( A ) ∩ B ⊂ F ( B ). We have ( V ⊗ U q )ad U q  HomU q ( V ∗ , U q ) and V ∗ lives in the locally finite part F (U q ) of U q . Hence it is equivalent to consider ad U q -invariants in V ⊗ F (U q ). Denote by Λ+ the set of dominant weights in Λ. The following lemma is proven in [4, Lemma 7.1.3]. Lemma. One has that qλ ∈ F (U q ) if and only if λ ∈ −2Λ+ . 3.1.3.

The decomposition in (6) for the locally finite part of U q becomes





V ⊗ F (U q ) = V ⊗ Fˆ U qα



        +  ⊕ U q m− α + V ⊗ F ( U q ) + V ⊗ F ( U q ) U q mα + .

Consider the projection Pˆ α : V ⊗ F (U q ) → V ⊗ Fˆ (U qα ) with respect to the above decomposition and  set V Zα := n∈Z V nα . By Lemma 3.1.2 and since the map Pˆ α is U qα -equivariant, we have that



ad U qα

V ⊗ F (U q )

 ad U qα Pˆ   Pˆ α  −→ V Zα ⊗ Fˆ U qα −→ V 0 ⊗ U q0 ev .

Note that Pˆ = Pˆ Pˆ α . An analogue of the Kostant separation theorem is proven to hold for F (U qα ) in [4], namely









F U qα  Hα ⊗ Z U qα , where Hα is an ad U qα -stable subspace of F (U qα ). Moreover, the number of copies of any module M

in Hα is equal to dim M 0 , that is [Hα : M ] = dim M 0 . Of course Fˆ ( Z (U qα )) = F ( Z (U qα )) = Z (U qα ) and









ˆ α ⊗ Z U qα , Fˆ U qα  H ˆ α : V Zα ] = dim V 0 . ˆ α = {x ∈ Hα | dim(ad U q )x < ∞}. Also [H where H α Now V is the direct sum of simple U q -modules. Let V (n) be the simple (2n + 1)-dimensional U qα -module of type 1; all its weight spaces are one-dimensional, hence there is a unique copy of V (n)

sα . ˆ ˆ α )ad U q ). Below we compute Pˆ (( V (n) ⊗ ˆ α . Then Pˆ (( V (n) ⊗ Fˆ (U qα ))ad U q ) = (U q0 )ev in H P (( V (n) ⊗ H α

ˆ α) H

ad U α q

).

α

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3.1.4.

153

Let v 0 be a non-zero highest weight vector of V (n); a basis for V (n) then is { F αk v 0 |

ˆ α we choose the basis {(ad F α )m E nα | m = 0, 1, . . . , 2n}. k = 0, 1, . . . , 2n}. For the copy of V (n)∗ in H α

A non-zero invariant in ( V Zα ⊗ U qα )ad U q is

x=

2n 

(−1)m q(αn−m)(n−m+1) F α2n−m v 0 ⊗ (ad F α )m E nα .

m =0

3.1.5.

For all s ∈ Z set s s  α  qα qα − q −α q − α q ; s := . −1 qα − qα

One may show by induction that for all n, m ∈ N: min{n,m}  n



m

Eα Fα =

k =0

 

n k α







k   α  m [k]α ! F αm−k q ; k + j − (n + m) E nα−k . k α j =1

(7)

Also by induction one shows that for all u ∈ U q , m ∈ N

  m  r −r (m−1) m (ad F α ) u = (−1) qα F αm−r qr α uq−r α F αr . m

r

r =0

(8)

α

By Eqs. (8) and (7) one obtains:

  m  r −r (m−1) m (ad F α ) E α = (−1) qα F αm−r qr α E nα q−r α F αr m n

r

r =0

=

m 

−r (m−1)+2nr

α



(−1)r qα

r =0

=



m F m−r E nα F αr r α α

  m  −r (m−1)+2nr m (−1)r qα F αm−r r α

r =0

×

k α

k =0

=

m 

−r (m−1)+2nr

r =0 min{n,r } 

 k =0

r −k

k α

(−1)r qα

×



min{n,r }      n r

[k]α ! F α





 qα ; k + j − (n + r ) E nα−k

j =1



m r α

  

n k α

k  





k   α  r [k]α ! F αm−k q ; k + j − (n + r ) E nα−k . k α j =1

(9)

Let u be an element in U q and v ∈ V . Then, by the right action of E α on V ⊗ U q we have that:

v ⊗ u E α = ( v ⊗ u ) E α ≡ 0,

+ mod( V ⊗ U q )U + .

(10)

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P. Lamprou / Journal of Algebra 401 (2014) 144–160

+ By Eqs. (9) and (10) we have that mod( V ⊗ U q )U +

x≡

2n 

(n−m)(n−m+1) 2n−m

(−1)m qα

Fα

m=n

v0

 n      m    r r −r (m−1)+2nr m m−n α ⊗ (−1) qα [n]α ! F α q ; j−r . r n α

r =n

α

j =1

Also, if the v is a weight vector in V of weight wt v by the left action of F α on V ⊗ U q we get: wt v (hα )

F α ( v ⊗ u ) = F α v ⊗ u + qα − wt v (hα )

thus v ⊗ F α u ≡ −qα

v ⊗ F α u,

− F α v ⊗ u , mod U + ( V ⊗ U q ) and by induction one shows that

−k wt v (hα ) k(k−1) k

v ⊗ F αk u ≡ (−1)k qα

− mod U + ( V ⊗ U q ).

F α v ⊗ u,

qα

(11)

− + So, mod(U + ( V ⊗ U q ) + ( V ⊗ U q )U + ),

n

x ≡ (−1)

m 2n  

r −2(m−n)−r (m−1)+2nr

(−1) qα

m=n r =n

3.1.6.

 n    [m]α ! α vn ⊗ q ; j −r . [m − r ]α ![r − n]α ! j =1

Now 2n  m  2(m−n)−r (m−1)+2nr (−1)r q− α m=n r =n

=

2n  2n 

 n    [m]α ! α q ; j −r [m − r ]α ![r − n]α ! j =1

r −2(m−n)−r (m−1)+2nr

(−1) qα

r =n m=r

j =1

2n  +r +2nr (−1)r q2n = α r =n

2n  = (−1)r r =n

 n    [m]α ! α q ; j −r [m − r ]α ![r − n]α !

[r ]α ! [r − n]α !



1

[r − n]α ! 

n   j =1



 qα ; j − r α [r ]α !

2n 

−(2+r )m

qα

m=r

   n  α  q ; j−r ,

2n + 1 r+1 α

j =1

by the identity: 2n  m=r

−m(r +2)

qα





m 2n(r +1)−r = q− α r α





2n + 1 . r+1 α





m r α

(12)

P. Lamprou / Journal of Algebra 401 (2014) 144–160

155

Then 2n  x ≡ (−1) v n ⊗ (−1)r n

r =n

≡ vn ⊗

[r ]α ! [r − n]α !

  n 

2n + 1 r+1 α



 qα ; j − r

j =1

   n  α  2n + 1 r [n + r ]α ! (−1) q ; j −n−r . n+r +1 α [r ]α ! 

n  r =0



j =1

We will show that up to a scalar in k(q) one has that:

    n n   α       2n + 1 r [n + r ]α ! (−1) q ; j − n − r = qα ; 2 qα ; 3 · · · qα ; n + 1 . n+r +1 α [r ]α ! r =0

j =1

This is equivalent to proving the following lemma. Lemma. Up to a scalar in k(q) n  (−1)r r =0

1

[r ]α ![n − r ]α ![r + n + 1]α

[α − r ]α [α − r − 1]α · · · [α − r − n + 1]α

= [α + 2]α [α + 3]α · · · [α + n + 1]α . The proof of the above lemma, which was communicated to me by Pee Choon Toh, can be found in Appendix A. 3.1.7. Quantum Zhelobenko operators Set ψˆ α ,n := [qα ; 2][qα ; 3] · · · [qα ; n + 1]. Recall the translated action of W on U q0 given by (1); one

2 −α = qα q2 . Then has that sα .qα = q−α q− α and sα .q α











sα .ψˆ α ,n = (−1)n qα ; 0 qα ; −1 · · · qα ; −n + 1 . Observe that the roots for qα in ψˆ α ,n and sα .ψˆ α ,n are pairwise distinct. We define an operator ξˆα : V (n)0 ⊗ (U q0 )ev → V (n)0 ⊗ Fract(U q0 )ev by

ξˆα ( v ⊗ a) := v ⊗

ψˆ α ,n sα .a. sα .ψˆ α ,n

Since for any α ∈ π , the module V is the direct sum of such V (n), n ∈ N, we may extend ξˆα linearly on V 0 ⊗ (U q0 )ev . Let Ξˆ = ξˆα | α ∈ π . We call ξˆ ∈ Ξˆ the quantum Zhelobenko operators. 3.1.8.

α

s . Recall that Pˆ (( V (n) ⊗ (U q0 )ev )ad U q ) = (U qα )evα ( v n ⊗ ψˆ α ,n ). By the definition of ξˆα one has that

ˆ

u ⊗ a ∈ ( V (n)0 ⊗ (U q0 )ev )ξα if and only if ψˆ α ,n divides a and the quotient a/ψˆ α ,n lies in (U q0 )evα . s .

ad U α

By the above we have that Pˆ (( V (n) ⊗ (U q0 )ev )

q

ˆ

) = ( V (n)0 ⊗ (U q0 )ev )ξα for all α ∈ π and so

ˆ Pˆ (( V ⊗ U q )ad U q ) ⊂ ( V 0 ⊗ (U q0 )ev )Ξ . We will show that this inclusion is in fact an equality. ˆ Theorem. The map Pˆ is an isomorphism from ( V ⊗ U q )ad U q onto ( V 0 ⊗ (U q0 )ev )Ξ .

We will prove the above theorem in the sections below.

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P. Lamprou / Journal of Algebra 401 (2014) 144–160

3.1.9. Proof of injectivity View U q as an ad U q -module and let φ : V → U q be a U q -module homomorphism. Define Φ : V ⊗ U q → U q by Φ( v ⊗ a) = φ( v )a for all v ∈ V and all a ∈ U q . Then Φ is a U q -bimodule map (for the left and right action of U q on V ⊗ U q defined in paragraph 1.5, and the left and right action of U q on itself defined by left and right multiplication respectively). Then one has that

 



p Φ Pˆ (b)

  = p Φ(b) ,

(13)

for all b ∈ V ⊗ U q , where recall that p denotes the Harish-Chandra projection. Lemma. The restriction of Pˆ on ( V ⊗ U q )ad U q is injective. Proof. We first remark that an analogue of Moeglin’s theorem, namely that every two-sided ideal in U (g) meets its center, holds for U q . The proof is identical to M. Gorelik’s proof of Moeglin’s theorem for U (g) (for g semisimple), which may be found in [5, Section 7.1.12]. Let b = v ⊗ a ∈ ( V ⊗ U q )ad U q ; then b ∈ ( V ⊗ V ∗ )ad U q . By Moeglin’s theorem, there exists a φ ∈ HomU q ( V , U q ) such that (φ( V ) V ∗ )ad U q = 0. Now Φ(b) ∈ Z (U q ) and p is injective on Z (U q ). But p (Φ( Pˆ (b))) = p (Φ(b)) = 0 by (13), hence Pˆ (b) = 0.

2

W. 3.1.10. By the injectivity of the map Pˆ , the space I := Pˆ (( V ⊗ U q )ad U q ) is a free (U q0 )ev -module of rank r := dim V 0 . Let J 1 , J 2 , . . . , J r be a set of free generators. We will first show that the space

ˆ

W. Z := ( V 0 ⊗ (U q0 )ev )Ξ is also a free (U q0 )ev -module of rank r. We need the following lemma.

Lemma. For all b ∈ (U q0 )ev and all v ⊗ a ∈ V 0 ⊗ (U q0 )ev one has that for all α ∈ π ,

  ξˆα b( v ⊗ a) = (sα .b)ξˆα ( v ⊗ a). Proof. It is a straightforward computation—we will verify it for v ∈ V (n)0 and b = qλ , λ ∈ Λ. Notice that

q λ ( v ⊗ a) = q λ v ⊗ q λ a = v ⊗ q λ a since v ∈ V 0 . Then ξˆα ( v ⊗ qλ a) = v ⊗

(sα .qλ )ξˆα ( v ⊗ a) = q−λ(hα ) q sα λ ( v ⊗ 3.1.11.

ψˆ α ,n sα .ψˆ α ,n

−λ(hα )

sα .(qλ a) = qα

ψˆ α ,n −λ(h ) sα .a) = qα α v sα .ψˆ α ,n

⊗

v⊗

ψˆ α ,n q sα λ sα .a. sα .ψˆ α ,n

On the other hand,

ψˆ α ,n q sα λ sα .a, hence the assertion. sα .ψˆ α ,n

2

Now we are able to prove the freeness of the module Z . ˆ

W. -module of rank r := dim V 0 . Lemma. The space Z = ( V 0 ⊗ (U q0 )ev )Ξ is a free (U q0 )ev W Proof. By [3] and [12] one knows that (U q0 )ev is a free (U q0 )ev -module of rank | W |, that is



U q0

 ev

 W = U q0 ev ⊗ H ,

for a | W |-dimensional vector space H (an exposition of this result can be found in [4, 7.5.8, A.1.20–A.1.23]). Consider the automorphism θ : (U q0 )ev → (U q0 )ev defined on qλ by θ(qλ )(μ) = q(λ,μ+ρ ) for all μ ∈ Λ. Then w .θ(qλ )(μ) = θ(qλ )( w −1 .μ) = θ(qλ )( w −1 (μ + W. W θ( wqλ )(μ) and so (U q0 )ev = θ((U q0 )ev ).

ρ ) − ρ ) = q(λ, w

−1 (μ+ρ ))

=

P. Lamprou / Journal of Algebra 401 (2014) 144–160

157

W. W. Hence (U q0 )ev = (U q0 )ev ⊗ θ( H ) and (U q0 )ev is a free (U q0 )ev -module. By Lemma 3.1.10, one has that

ˆ W. W. Z is a free (U q0 )ev -module over ( V 0 ⊗ θ( H ))Ξ . Let { ˆJ 1 , ˆJ 2 , . . . , ˆJ s } be a basis of Z over (U q0 )ev . The 0 W. rank r free (U q )ev -module I is contained in Z , hence s  r. On the other hand, one shows exactly as in [6, Lemma 2.7] that the ˆJ 1 , ˆJ 2 , . . . , ˆJ s are linearly independent over Fract(U q0 )ev . Thus s  r and so s = r, hence the assertion. 2

3.1.12.

Set again r := dim V 0 and let { v i }ri=1 be a basis for V 0 .

Then for all i with 1  i  r, we may write J i =

r

j =1

v j ⊗ ai , j , and ˆJ i =

r

j =1

v j ⊗ aˆ i , j for

ˆ = (ˆai , j )i , j . We will compute the determinants det A and det Aˆ ai , j , aˆ i , j ∈ (U q0 )ev . Set A := (ai , j )i , j , A and show thatthey are equal up to a scalar in k(q). Set D α := n∈N [qα ; n + 1]dim V nα . The lemma below follows exactly as in [6, Lemma 3.2]. ˆ and the quotient is sα -invariant. Lemma. For all α ∈ π , D α divides det A and det A 3.1.13.

Set D :=



n∈N



β∈Φ + [q

β ; ρ (h

β ) + n]

dim V nβ

.

ˆ Lemma. The product D divides det A and det A. Proof. First, notice that since

ρ (hα ) = 1 one has that

D/Dα =







qβ ; ρ (hβ ) + n

dim V nβ

.

n∈N β∈Φ + \{α }

For β =  α by a straightforward computation one has that sα .[qβ ; ρ (hβ ) + n] = [q sα β ; ρ (sα β) + n]. Recall that sα (Φ + \ {α }) = Φ + \ {α } and dim V nβ = dim V nsα β ; we conclude that sα .( D / D α ) = D / D α .

ˆ for all Now by the previous lemma D α divides det A and det A

α ∈ π . Since Φ = W π and U q0 is a

ˆ unique factorization domain, it follows that D also divides det A and det A.

2

3.1.14. Proof of theorem Define a filtration in U q by setting the F α to have degree 0, the E α degree 1 and the qα degree −1

for all α ∈ π . This

filtration is ad-invariant. By [4, Lemma 7.1.25], for all λ ∈ 2Λ, one has that λ = α ∈π kα α with α ∈π kα ∈ Z. One has that

deg D =

 

ht β dim V nβ ,

(14)

n∈N β∈Φ +

where ht β denotes the height of the positive root β , that is if β =

α ∈π k α ∈ N . On the other hand,

deg det A 



k[Hk : V ],



α ∈π kα α , then ht β =

(15)

k∈N

where Hk denotes the homogeneous subspace of H of degree k. Then, by [4, Corollary 8.1.11], the right-hand sides of (14) and (15) are equal. Since D divides det A, one has that det A = D up to a non-zero scalar in k(q). Then Theorem 3.1.8 follows as in [6, Theorem 3.5].

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P. Lamprou / Journal of Algebra 401 (2014) 144–160

3.2. Invariants under the map P 3.2.1.

We keep the notation of Section 3.1. Fix

α ∈ π and recall that P = 1 ⊗ p : V ⊗ U q → V ⊗ U q0 ,

where p is the Harish-Chandra homomorphism. Our aim is to compute the image of ( V ⊗ U q )ad U q under P . One has a vector space decomposition:









 

+ U q = U qα ⊕ U q m− α + U q + U q U q mα + .

As before, it is equivalent to consider ad U q -invariants in V ⊗ F (U q ); the above decomposition for the locally finite part of U q becomes













 

+ F (U q ) = Fˆ U qα ⊕ U q m− α + F ( U q ) + F ( U q ) U q mα + .

Consider the projection p α : F (U q ) → Fˆ (U qα ) with respect to the above decomposition and set P α = 1 ⊗ p α . By Lemma 3.1.2 and since the map P α is U qα -equivariant, we have that



ad U qα

V ⊗ F (U q )

 ad U qα P   Pα  −→ V Zα ⊗ Fˆ U qα −→ V 0 ⊗ U q0 ev .

One has that P = P P α . Recall 3.1.3 that









ˆ α ⊗ Z U qα , Fˆ U qα  H ˆ α : U ] = dim U 0 . ˆ α is equal to dim U 0 , that is [H and the number of copies of any module U in H 3.2.2. The U q -module V is the direct sum of simple U qα -modules. Let V (n) be the simple (2n + 1)-dimensional U qα -module; all its weight spaces are one-dimensional, hence there is a unique

ˆ α . One has copy of V (n) in H P





V (n) ⊗ Fˆ U qα

ad U qα 

 s .   α ˆ α ad U q . = U q0 evα P V (n) ⊗ H

In [4, Theorem 8.2.10] it is shown that up to a scalar in k(q)

P



ad U qα 

ˆ V (n) ⊗ H

= det P V (n) =

n  



qα − q(m−1)(α ,α ) q−α .

m =1

Here quantum PRV determinant. Notice that the product above is equal to n P Vα (n) denotes the α ; 0][qα ; −1] · · · [qα ; −n + 1] up to a scalar in k(q). Recall that the latter equals [ q ; − m + 1 ] = [ q m=1 sα .ψˆ α ,n of paragraph 3.1.7, and set ψα ,n := sα .ψˆ α ,n ; then sα .ψα ,n = ψˆ α ,n . Thus

P





V (n) ⊗ Fˆ U qα

ad U qα 

 s . = U q0 evα v n ⊗ ψα ,n .

P. Lamprou / Journal of Algebra 401 (2014) 144–160

159

3.2.3. We define an operator ξα : V (n)0 ⊗ (U q0 )ev → V (n)0 ⊗ Fract(U q0 )ev by

ξα ( v ⊗ a) := v ⊗

ψα ,n sα .ψα ,n

sα .a.

Since for any α ∈ π , the module V is the direct sum of such V (n), n ∈ N, we may extend ξα linearly on V 0 ⊗ (U q )ev . The ξα are the quantum versions of the analogue Zhelobenko operators. Let Ξ = ξα | α ∈ π . Then as in Lemma 3.1.10, one has that for all b ∈ (U q0 )ev , v ⊗ a ∈ V 0 ⊗ (U q0 )ev and α ∈ π,

  ξα b( v ⊗ a) = (sα .b)ξα ( v ⊗ a). 3.2.4. Using the same arguments as before, we obtain P (( V ⊗ U q )ad U q ) ⊂ ( V 0 ⊗ (U q0 )ev )Ξ . In fact, this inclusion is an equality. Theorem. The map P is an isomorphism from ( V ⊗ U q )ad U q onto ( V 0 ⊗ (U q0 )ev )Ξ . Proof. Injectivity follows exactly as in Lemma 3.1.9. Surjectivity follows as for the map Pˆ with D replaced by the quantum PRV determinant, namely,

det P V :=

  

 ∨ qβ − q(m−β (ρ ))(β,β) q−β .

2

m∈N+ β∈Φ +

Acknowledgment I would like to thank A. Joseph for suggesting this problem to me and for many valuable discussions. Appendix A A.1.

Set

(n)q := Set (a; q)k =

1 − qn 1−q

 (n)q ! := (1)q (2)q · · · (n)q and

,

k−1

j =0 (1 − aq

j

n k

= q

(n)q ! . (k)q !(n − k)q !

), the q-Pochhammer symbol. One has that

(q; q)k = (1 − q)k (k)q !, 

qs ; q





= (1 − q)k

k

q −s ; q

 k

(s + k − 1)q ! , (s − 1)q !

= (−1)k q−(2s−k+1)k/2 (1 − q)k

(s)q ! k  s. (s − k)q !

A.2. The q-Saalschutz sum [2] is given by n  −n    n −1  (c /a; q)n (c /b; q)n (q−n ; q)r (a; q)r (b; q)r qr ; := φ q , a , b ; c , ( ab )/ cq q ; q = , 3 2 n − 1 (c ; q)n (c /(ab); q)n (c ; q)r ((ab)/(cq ); q)r (q; q)r r =0

where φ is the q-hypergeometric function.

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P. Lamprou / Journal of Algebra 401 (2014) 144–160

A.3. Set a = q2(n+1) , b = q−2(x−n) , c = q2(n+2) and q = q2 on the q-Saalschutz sum. Then n  (q−2n ; q2 )r (q2(n+1) ; q2 )r (q−2(x−n) ; q2 )r r =0

(q2(n+2) ; q2 )r (q−2x ; q2 )r

q2r

(q2 ; q2 )r

=

(q2 ; q2 )n (q2(x+2) ; q2 )n , 2 ( (q n+2) ; q2 )n (q2(x−n+1) ; q2 )n

which in turn gives n  (−1)r qr (r +1) r =0

=

(n)q2 ! (2n + 1)q2 !

1

(r )q2 !(n − r )q2 !(n + r + 1)q2

(x − r )q2 (x − r − 1)q2 · · · (x − r − n + 1)q2

(x + 2)q2 (x + 3)q2 · · · (x + n + 1)q2 .

(16)

A.4. Recall paragraph 1.2; one has that

[n] = q−n+1 (n)q2

and

[n]! = q−n(n−1)/2 (n)q2 !.

A.5. The sum in (16) becomes n 

(−1)r

r =0

=

1

[r ]![n − r ]![n + r + 1]

[x − r ][x − r − 1] · · · [x − r − n + 1]

[n]! [x + 2][x + 3] · · · [x + n + 1], [2n + 1]!

hence Lemma 3.1.6 holds. References [1] M. Balagovic, Chevalley restriction theorem for vector-valued functions on quantum groups, Represent. Theory 15 (2011) 617–645. [2] G. Gasper, M. Rahman, Basic Hypergeometric Series, Cambridge University Press, Cambridge, England, 1990. [3] S.G. Hulsurkar, Proof of Verma’s conjecture on Weyl’s dimension polynomial, Invent. Math. 27 (1974) 45–52. [4] A. Joseph, Quantum Groups and Their Primitive Ideals, Springer-Verlag, 1995. [5] A. Joseph, An algebraic slice in the coadjoint space of the Borel and the Coxeter element, Adv. Math. 227 (1) (2011) 522–585. [6] A. Joseph, A direct proof of a generalized Harish-Chandra isomorphism, Transform. Groups 17 (2) (2012) 513–521. [7] A. Joseph, Analogue Zhelobenko invariants and the Kostant Clifford algebra conjecture, Transform. Groups 17 (3) (2012) 781–821. [8] A. Joseph, G. Letzter, Evaluation of the quantum affine PRV determinants, Math. Res. Lett. 9 (2002) 307–322. [9] S. Khoroshkin, M. Nazarov, E. Vinberg, A generalized Harish-Chandra isomorphism, Adv. Math. 226 (2011) 1168–1180. [10] B. Kostant, Lie group representations on polynomial rings, Amer. J. Math. 85 (1963) 327–404. [11] K.R. Parthasarathy, R. Ranga Rao, V.S. Varadarajan, Representations of complex semisimple Lie groups and Lie algebras, Ann. of Math. (2) 85 (1967) 383–429. [12] R. Steinberg, On a theorem of Pittie, Topology 14 (1975) 173–177.