Mixed tensor products and Capelli-type determinants
aa r X i v : . [ m a t h . R T ] F e b MIXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS
DIMITAR GRANTCHAROV AND LUKE ROBITAILLE
Abstract.
In this paper we study properties of a homomorphism ρ from the universal en-veloping algebra U = U ( gl ( n + 1)) to a tensor product of an algebra D ′ ( n ) of differentialoperators and U ( gl ( n )). We find a formula for the image of the Capelli determinant of gl ( n + 1)under ρ , and, in particular, of the images under ρ of the Gelfand generators of the center Z ( gl ( n + 1)) of U . This formula is proven by relating ρ to the corresponding Harish-Chandraisomorphisms, and, alternatively, by using a purely computational approach. Furthermore, wedefine a homomorphism from D ′ ( n ) ⊗ U ( gl ( n )) to an algebra containing U as a subalgebra, sothat σ ( ρ ( u )) − u ∈ G U , for all u ∈ U , where G = P ni =0 E ii .Keywords and phrases: Lie algebra, Capelli identities, Capelli determinants, tensor modules.MSC 2010: 17B10, 17B35 Introduction
Capelli-type determinants are a powerful tool in invariant theory. One fundamental resultis that the Capelli determinant C N ( T ) corresponding (up to a diagonal shift) to the N × N matrix E whose ( i, j )th entry is the elementary matrix E ij is a polynomial in T whose coefficients C k are central elements in the universal enveloping algebra U ( gl ( N )) of the Lie algebra gl ( N ).Moreover, the coefficients C k are generators of the center Z ( gl ( N )) of U ( gl ( N )), and are usuallyreferred as Capelli generators. On the other hand the elements G k = tr( E k ) for k = 1 , . . . , N also form a system of generators of Z ( gl ( N )), sometimes known as the Gelfand generators.There is a nice transition formula between the Gelfand generators and the Capelli generators,[11], and this formula can be considered as a noncommutative version of the Newton identities.The applications of Capelli-type determinants extend well beyond relations and properties ofelements in Z ( gl ( N )). There are direct applications to classical invariant theory (see for example[6]), and a more general treatment of the subject using the theory of Yangians (see [9] and thereferences therein).In this paper we relate Capelli-type determinants to another classical construction in rep-resentation theory - the mixed tensor type modules. These modules are modules over tensorproducts of an algebra of differential operators and a universal enveloping algebra U ( gl ( n )).Modules of mixed tensor type, also known as tensor modules, or modules of Shen and Larson(see [10] and [8]), can be defined over the Lie algebras sl ( n + 1) after considering a suitablehomomorphism. In this paper we define a gl ( n + 1)-version of this homomorphism, namely amap ρ : U ( gl ( n + 1)) → D ′ ( n ) ⊗ U ( gl ( n )), where D ′ ( n ) is the algebra of polynomial differentialoperators on C [ t ± , t , . . . , t n ] generated by t i /t , t ∂ j , i > j ≥
0. One of the main results ofthe paper is an explicit description of the image of the Capelli determinant C n +1 ( T ) of gl ( n + 1)under ρ . It turns out that the result is especially pleasant: up to a shift, ρ ( C n +1 ( T )) is a product of C n ( T ) and a linear factor. This leads to explicit formulas for the images ρ ( G gl ( n +1) k )of the Gelfand generators G gl ( n +1) k for gl ( n + 1) under ρ . As a corollary, we also obtain themixed tensor version of the noncommutative Cayley-Hamilton identities; see Corollary 5.3. Toprove the formula for ρ ( C n +1 ( T )) we relate ρ to the Harish-Chandra isomorphisms of gl ( n )and gl ( n + 1). We expect that the explicit formulas for ρ ( C n +1 ( T )) and ρ ( G gl ( n +1) k ) to haveapplications in the representation theory of the tensor modules, as one can easily and explicitlycompute central characters.Another result of the paper is finding a pseudo left inverse σ of ρ in the following sense. Thehomomorphism σ maps D ′ ( n ) ⊗ U ( gl ( n )) to an algebra U ′′ that contains U = U ( gl ( n + 1)) as asubalgebra is such that σ ( ρ ( u )) − u ∈ G U for every u ∈ U , where G = G gl ( n +1)1 = P ni =0 E ii . Inparticular, we obtain that the kernel of ρ is the ideal ( G ) in U , while the kernel of the sl ( n + 1)version of ρ is trivial.It is worth noting that the results concerning the images of the generators of Z ( gl ( n + 1))under ρ can be obtained with long and direct computation. For this (alternative) approach weprove some more general identities that are included in the Appendix of this paper. We believethat these identities may be of independent interest.The organization of the paper is as follows. In Section 3 we define the homomorphism ρ .In Section 4 we relate ρ with the corresponding Harish-Chandra isomoprhisms. The results ofSection 4 are applied in the next section where the image of the Capelli-type determinant under ρ is computed. Using the latter results we derive the formulas for the images of the Gelfandgenerators under ρ . In Section 6 we define a pseudo-left inverse of ρ and find the kernel of ρ .The Appendix contains some explicit formulas for the images under ρ of a set of homogeneouselements of U ( gl ( n + 1)). All the Gelfand generators for gl ( n + 1) can be written as sums ofelements from this set, so we obtain an alternative proof of the results in Sections 4 and 5. Acknowledgements:
The first author is partly supported by Simons Collaboration Grant358245. He also would like to thank the Max Planck Institute in Bonn (where part of this workwas completed) for the excellent working conditions. The second author thanks his family fortheir support. We thank Vera Serganova for the useful discussion.2.
Notation and conventions
Our base field is C . All vector spaces, tensor products, and associative algebras will beconsidered over C unless otherwise specified. By δ kl we denote the Kronecker delta function,which equals 1 if k = l and 0 otherwise. Throughout the paper we fix a positive integer n .For a Lie algebra a , by U ( a ) we denote the universal enveloping algebra of a and by Z ( a ) thecenter of U ( a ). By gl ( N ) (respectively, sl ( N )) we denote the Lie algebra of all (respectively,traceless) N × N matrices. We write I N for the identity matrix of gl ( N ). For an N × N matrix A , the entries A ij will be indexed by 1 ≤ i, j ≤ n if N = n , and by 0 ≤ i, j ≤ n for N = n + 1. In the latter case, we will refer to the top left entry as the (0 , gl ( n + 1) will be written as ( n + 1)-tuples ( µ , . . . , µ n ), while those of gl ( n ) as n -tuples ( µ , . . . , µ n ).We will write E ij for the ( i, j )th elementary matrix of gl ( N ), and E N for the N × N matrixwhose ( i, j )th entry is E ij . Henceforth, we fix the Borel subalgebra b N and the Cartan subal-gebra h N of gl ( N ) to be the ones spanned by E ij ( i ≤ j ) and E kk (all k ), respectively. By n + gl ( N ) IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 3 (respectively, n − gl ( N ) ) we denote the nilradical (respectively, the opposite nilradical) of b N . Inparticular, b N = h ⊕ n + gl ( N ) and gl ( N ) = b N ⊕ n − gl ( N ) . We will use these conventions both for N = n and N = n + 1.By default, determinants will be column determinants. Namely, for an n × n matrix A withentries A ij in an associative algebra,det( A ) := X σ ∈ S n A σ (1)1 A σ (2)2 . . . A σ ( n ) n , where S n is the n th symmetric group. The determinant of an ( n + 1) × ( n + 1) matrix is definedanalogously.For a square matrix A and variable or a constant v , the expression A + v (and v + A ) shouldbe understood as the sum of A and the scalar matrix of the same size as A having v on thediagonal.By D ( n ) we denote the algebra of polynomial differential operators on C n . The algebra D ′ ( n )is the subalgebra of differential operators on C [ t ± , t , . . . , t n ] generated by t i t for i = 1 , . . . , n and t ∂ j for j = 0 , . . . , n . Here ∂ a stands for ∂∂t a . We set E := P ni =0 t i ∂ i .We finish this section with some conventions that we will use throughout the paper. Weassume that P si = r x i = 0 whenever r > s . For a subset S of a ring R , by ( S ) we denote thetwo-sided ideal of R generated by S . For associative unital algebras A and A and elements a i ∈ A i , we often write a and a for the elements a ⊗ ⊗ a , respectively.3. The homomorphism ρ In this section we define the homomorphism ρ : U ( gl ( n + 1)) → D ′ ( n ) ⊗ U ( gl ( n )) that playsimportant role in this paper. This homomorphism can be considered as the gl ( n + 1)-versionof a homomorphism ρ s : U ( sl ( n + 1)) → D ′ ( n ) ⊗ U ( gl ( n )), which can be defined as follows, inbrief. For a finite-dimensional gl ( n )-module V consider the corresponding trivial vector bundle V on P n . Then there is a natural map from sl ( n + 1) to the algebra of differential operatorson the space of sections of V over the open subset U = { t = 0 } of P n . This map leads tothe homomorphism ρ s . For details we refer the reader to, for example, § ρ s (given in local coordinates) are listed in the proof of [5, Lemma 2.3].Introduce the following elements of D ′ ( n ) ⊗ U ( gl ( n )): R = − n + 1 (cid:16) E ⊗ ⊗ G gl ( n )1 (cid:17) , R = ( E + n ) ⊗ , where G gl ( n )1 = P ni =1 E ii . Note that R and R are central in D ′ ( n ) ⊗ U ( gl ( n )).We next introduce three natural homomorphisms: the natural embedding ι s : U ( sl ( n + 1)) → U ( gl ( n + 1)), π g : U ( gl ( n + 1)) → U ( sl ( n + 1)), and ι g : U ( gl ( n )) → U ( sl ( n + 1)), as follows: π g ( B ) = B − n + 1 tr( B ) I n +1 , ι g ( C ) = C − tr( C ) E . Define ρ = ρ s π g . The fact that ρ s is a homomorphism and the explicit formulas for ρ s implythe following. DIMITAR GRANTCHAROV AND LUKE ROBITAILLE
Lemma 3.1.
The following correspondence E ab t a ∂ b ⊗ ⊗ E ab + δ ab R for a, b > ,E a t a ∂ ⊗ − X i> t i t ⊗ E ai for a > ,E b t ∂ b ⊗ for b > ,E t ∂ ⊗ R . extends to the homomorphism ρ : U ( gl ( n +1)) → D ′ ( n ) ⊗ U ( gl ( n )) of associative unital algebras. We note that ρ ( G gl ( n +1)1 ) = 0 and ρι s = ρ s . Let us also define γ : U ( sl ( n + 1)) → D ′ ( n ) ⊗ U ( sl ( n + 1))by the identity γ = (1 ⊗ ι g ) ρι s . We finish this section by collecting the identities for thehomomorphisms that we introduced in this section. Proposition 3.2.
We have γ = (1 ⊗ ι g ) ρ s , π g ι s = Id , and ρ s π g = ρ , and all other relationsthat directly follow from these three; in that sense, the following diagram is commutative. U ( gl ( n + 1)) π g (cid:15) (cid:15) ρ / / D ′ ( n ) ⊗ U ( gl ( n )) ⊗ ι g (cid:15) (cid:15) U ( sl ( n + 1)) ι s O O γ / / ρ s ❦❦❦❦❦❦❦❦❦❦❦❦❦❦ D ′ ( n ) ⊗ U ( sl ( n + 1))4. Images under Harish-Chandra isomorphisms
In this section we relate the restriction of ρ on the center Z ( gl ( n + 1)) of U ( gl ( n + 1)) withthe Harish-Chandra isomorphisms.We first recall the definition of the Harish-Chandra isomorphism and define an isomorphismof Harish-Chandra type with domain C [ E ] ⊗ Z ( gl ( n )).For a weight λ = ( λ , . . . , λ n ) of gl ( n ) denote by M n ( λ ) and L n ( λ ) the Verma module ofhighest weight λ and its unique simple quotient, respectively. We similarly define the gl ( n + 1)-modules M n +1 ( µ ) and L n +1 ( µ ) for a weight µ = ( µ , µ , . . . , µ n ) of gl ( n + 1). In particular,if λ = ( λ , . . . , λ n ), then there is a (highest weight) vector v of M n ( λ ) such that M n ( λ ) = U ( n − gl ( n ) ) ⊗ C v as vector spaces, E ab v = 0 for 1 ≤ a < b ≤ n , and E ii v = λ i v for i = 1 , . . . , n .Henceforth we set δ N = (0 , − , . . . , − N + 1). If ( b , . . . , b N ) ∈ C N , then the evaluationhomomorphism ev b ,...,b N : C [ x , . . . , x N ] → C is defined by ev b ,...,b N ( p ) = p ( b , . . . , b N ). Every z ′ ∈ Z ( gl ( n )) acts on L ( λ ) as χ λ ( z ′ )Id, where χ λ ( z ′ ) = ev λ + δ n ( χ n ( z ′ )) and χ n : Z ( gl ( n )) → C [ ℓ , . . . , ℓ n ] S n is the Harish-Chandra isomorphism. We similarly define χ n +1 : Z ( gl ( n + 1)) → C [ ℓ , . . . , ℓ n ] S n +1 using the action of any element of z ∈ Z ( gl ( n + 1)) on a simple highest weightmodule of gl ( n + 1).Next, define χ ,n : C [ E ] ⊗ Z ( gl ( n )) → C [ ℓ ] ⊗ C [ ℓ , . . . , ℓ n ] S n by χ ,n ( P i E i ⊗ z i ) = P i ℓ i ⊗ χ n ( z i ). Note that χ ,n is an isomorphism and that ℓ = χ ,n ( R ), where ℓ := − n +1 (cid:16) ℓ + P ni =1 ℓ i + n ( n − (cid:17) . The latter follows from the fact that χ n ( G gl ( n )1 ) = P ni =1 ℓ i + n ( n − (see Example 7.3.4 in [9]). IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 5
Recall that a weight µ of gl ( n + 1) is antidominant if µ i − µ i +1 / ∈ Z ≥ for all i = 0 , . . . , n .A well-known fact is that M n +1 ( µ ) is simple if and only if µ is anti-dominant. Also, by aTheorem of Duflo, the annihilator of a simple Verma module M n +1 ( µ ) is generated by z − χ µ ( z ), z ∈ Z ( gl ( n + 1)); see for example § F a = Span { t a − k −···− k n t k . . . t k n n | k , . . . , k n ∈ Z ≥ } and consider F a as a D ′ ( n )-module. Note that F a = D ′ ( n )( t a ) and that E = a Id on F a . Lemma 4.1.
Let AW n denote the set of all anti-dominant gl ( n ) -weights λ such that λ / ∈ Z .Then the modules L a ∈ Z F a and L λ ∈AW n M n ( λ ) are faithful over D ′ ( n ) and U ( gl ( n )) , respec-tively.Proof. The fact that L λ ∈AW n M n ( λ ) is faithful follows from the Theorem of Duflo and the factthat the annihilator of the direct sum of modules is the intersection of their annihilators.We now prove that the annihilator of L a ∈ Z F a is trivial. Suppose for the sake of contradic-tion that x ∈ D ′ ( n ) annihilates L a ∈ Z F a . Next, choose a monomial x m = t a . . . t a n n ∂ b . . . ∂ b n n inthe sum expansion of x that is ∂ -lexicographically maximal (i.e. relative to the ∂ -degree, ∂ -degree, . . . , ∂ n -degree) among all monomials in x . Then, as e , . . . , e n vary in a suitable set, thecoefficient of t e + a − b . . . t e n + a n − b n n in x ( t e . . . t e n n ) is equal to some polynomial in e , . . . , e n . Fur-thermore, the leximaximal property of t a . . . t a n n ∂ b . . . ∂ b n n yields that the coefficient of e b . . . e b n n in this polynomial is nonzero. Thus this polynomial is nonzero, and for suitable e , . . . , e n itevaluates to a nonzero number, contradicting the fact that x annihilates D ′ ( n ). (cid:3) Theorem 4.2.
Let τ be the endomorphism on C [ ℓ , ℓ , . . . , ℓ n ] defined by τ ( p ( ℓ , ℓ , . . . , ℓ n )) = p ( ℓ + ℓ, ℓ + ℓ − , . . . , ℓ n + ℓ − . Then we have that ρ | Z ( gl ( n +1)) = χ − ,n τ χ n +1 . In particular, ρ ( Z ( gl ( n + 1))) is a subalgebra of C [ E ] ⊗ Z ( gl ( n )) , and the following diagram iscommutative: Z ( gl ( n + 1)) ρ −−−→ C [ E ] ⊗ Z ( gl ( n )) χ n +1 y y χ ,n C [ ℓ , ℓ , . . . , ℓ n ] S n +1 τ −−−→ C [ ℓ ] ⊗ C [ ℓ , . . . , ℓ n ] S n Proof.
Let λ = ( λ , . . . , λ n ) be in AW n and a ∈ Z . Also, let M ( a, λ ) = F a ⊗ M n ( λ ). We firstnote that, in order to prove the identity in the theorem, it is sufficient to check that ρ ( z ) = χ − ,n τ χ n +1 ( z ) for all z ∈ Z ( gl ( n + 1)), as an identity of endomorphisms of M ( a, λ ). Indeed, byLemma 4.1 and by the fact that the tensor product of faithful modules is a faithful module (see,for example, [1]), the module L a ∈ Z ,λ ∈AW n M ( a, λ ) is a faithful module over D ′ ( n ) ⊗ U ( gl ( n )).We next observe that if M ( a, λ ) is considered as a gl ( n +1)-module through ρ , then M ( a, λ ) ∼ = M n +1 (˜ λ ), where ˜ λ = ( a + r , λ + r , . . . , λ n + r ) and r = − n +1 ( a + P ni =1 λ i ). To provethis, let us fix a highest weight vector v λ of M n ( λ ). Then it is straightforward to check that E ab ( t a ⊗ v λ ) = 0 for a < b and that the weight of t a ⊗ v λ is ˜ λ . On the other hand, since a ∈ Z and λ ∈ AW n , ˜ λ is anti-dominant (since we know λ / ∈ Z ). Hence, M n +1 (˜ λ ) is simple. To concludethe proof of M ( a, λ ) ∼ = M n +1 (˜ λ ) we show that both modules have the same formal characters. DIMITAR GRANTCHAROV AND LUKE ROBITAILLE
Indeed, observe that for a monomial u ∈ U ( n − gl ( n ) ), the vector t a ( t /t ) k . . . ( t n /t ) k n ⊗ uv λ in M ( a, λ ), and the vector E k . . . E k n n uv ˜ λ in M n +1 (˜ λ ), have the same weights. Therefore,ch ( F a ⊗ M n ( λ )) = ch( t a C [ t /t , . . . , t n /t ] ⊗ U ( n − gl ( n ) ) v λ ) = ch( U ( n − gl ( n +1) ) v ˜ λ ) . Since M ( a, λ ) ∼ = M n +1 (˜ λ ), we have that every z ∈ Z ( gl ( n + 1)) acts on M ( a, λ ) as χ ˜ λ ( z )Id.Recall that by definition, χ λ ( z ′ ) = ev λ + δ n ( χ n ( z ′ )) , χ ˜ λ ( z ) = ev ˜ λ + δ n +1 ( χ n +1 ( z )) . for every z ′ ∈ Z ( gl ( n )) and z ∈ Z ( gl ( n + 1)).On the other hand, if ξ = χ − ,n τ χ n +1 , then ξ ( z ) acts on M ( a, λ ) as χ a,λ ( ξ ( z ))Id, where χ a,λ ( P i E i ⊗ z ′ i ) = P i a i χ λ ( z ′ i ), for z ′ i ∈ Z ( gl ( n )). Let p = χ n +1 ( z ). It remains to prove that χ a,λ χ − ,n τ ( p ) = ev ˜ λ + δ n +1 p . This follows from the fact that χ a,λ ( E i ⊗ z ′ ) = ev a,λ + δ n ( χ ,n ( E i ⊗ z ′ ))for every z ′ ∈ Z ( gl ( n )) and that ev a,λ + δ n τ = ev ˜ λ + δ n +1 . (cid:3) Capelli-type determinants
For any formal variable T , we define the Capelli determinant of gl ( n + 1) and gl ( n ) as follows: C n +1 ( T ) = det E − T E · · · E n E E − T − · · · E n ... ... . . . ... E n E n · · · E nn − T − n and C n ( T ) = det E − T E · · · E n E E − T − · · · E n ... ... . . . ... E n E n · · · E nn − T − n + 1 , respectively. Note that T is a formal variable that commutes with all E ij , and C n ( T ) and C n +1 ( T ) will be treated as polynomials in this formal variable. We also note that the polyno-mials C n +1 ( T ) appear in [9] and [11] with a slight change. Namely, the polynomial C Mn +1 ( T )defined in § C n +1 ( T ) via the identity C n +1 ( T ) = C Mn +1 ( − T ). On the other hand, the polynomial C Un +1 ( T ) defined in [11] satisfies therelation C n +1 ( T ) = C Un +1 ( T + n ).Define C ρ ( T ) via the identity C ρ ( ρ ( T )) = ρ ( C n +1 ( T )). For convenience we will write T for ρ ( T ). Theorem 5.1.
The following identity holds: C ρ ( T + R ) = ( E − T ) C n ( T + 1) . In particular, C ρ ( E + R ) = 0 .Proof. The identity is equivalent to the following: ρ ( C Mn +1 ( T )) = ( E + T + R ) C Mn ( T + R − . IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 7
To prove the latter we use that χ n ( C Mn ( T )) = ( T + ℓ ) . . . ( T + ℓ n ) . (Theorem 7.1.1 in [9]) and the corresponding formula for χ n +1 ( C Mn +1 ( T )). To complete the proofwe use Theorem 4.2 and that χ ,n ( R ) = ℓ . (cid:3) Example 5.2.
In the case n = 2 , the identity in Theorem 5.1 is the following: det t ∂ − T t ∂ t ∂ t ∂ − t t E − t t E t ∂ + E − T − t ∂ + E t ∂ − t t E − t t E t ∂ + E t ∂ + E − T − = ( E − T ) C ( T + 1) = ( E − T ) det (cid:20) E − T − E E E − T − (cid:21) . Corollary 5.3.
The following identities hold: C ρ ( E n + R − n ) = 0 , C ρ ( E tn − R ) = 0 Proof.
The identities follow from Theorem 5.1 and the noncommutative version of the Cayley-Hamilton theorem: C Mn ( − E n + n −
1) = C Mn ( − E tn ) = 0 . (This is Theorem 7.2.1 in [9].) (cid:3) We conclude this section by giving explicit formulas of the images of the Gelfand invariantsof Z ( gl ( n + 1)) under ρ . To define these invariants we first introduce some special elements in U ( gl ( n + 1)). Set r gl ( n +1)0 ( a, b ) = δ ab , and let(1) r gl ( n +1) k +1 ( a, b ) = X i ,...,i k E ai E i i . . . E i k b where the sum runs over all (not necessarily distinct) 0 ≤ i , . . . , i k ≤ n . Then G gl ( n +1) k = n X i =0 r gl ( n +1) k ( i, i )is the Gelfand invariant of degree k of gl ( n + 1). In other words, G gl ( n +1) k = tr( E kn +1 ). Wedefine r gl ( n ) k ( a, b ) and G gl ( n ) k for gl ( n ) analogously. It is well-known fact that Z ( gl ( n + 1)) is apolynomial algebra in G gl ( n +1) k for k = 1 , , . . . , n + 1. Theorem 5.4.
The following formula holds for all positive integers k : ρ ( G gl ( n +1) k ) = k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( E ⊗
1) + ( n + 1) R k + k − X g =0 (cid:18) kg (cid:19) R g (cid:16) ⊗ G gl ( n ) k − g (cid:17) − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! (cid:16) ⊗ G gl ( n ) m − (cid:17) . DIMITAR GRANTCHAROV AND LUKE ROBITAILLE
Proof.
Let us first recall the following identity: C N ( T − N + 1) − C N ( T − N ) C N ( T − N + 1) = ∞ X k =0 G gl ( N ) k T − − k . This is sometimes referred as the noncommutative analogue of the classical Newton formula; seeTheorem 4 in [11] (see also Theorem 7.1.3 in [9]). Note that this identity should be consideredover the ring of formal Laurent series with coefficients in Z ( gl ( N )).On the other hand, Theorem 5.1 implies that(2) C ρ ( T − n − C ρ ( T − n ) = (cid:18) − T − R − R (cid:19) C n ( T − R − n ) C n ( T − R − n + 1)in C [ E ] ⊗ Z ( gl ( n )). Applying ρ to the Newton formula for N = n + 1, we obtain C ρ ( T − n − C ρ ( T − n ) = 1 − X k ≥ ρ ( G gl ( n +1) k ) T − − k = 1 − ( n + 1) T − − T − X k ≥ ρ ( G gl ( n +1) k ) T − k . (recall that we identify ρ ( T ) with T ). Analogously, we can express the right hand side of (2)as a power series in T − . We complete the proof by comparing and computing the coefficientsof T − k − on both sides. (cid:3) Example 5.5.
Theorem 5.4 applied for k = 3 gives the following formula: ρ ( G gl ( n +1)3 ) = ( R + 3 R R + 3 R )( E ⊗
1) + ( n + 1) R + (1 ⊗ G gl ( n )3 ) + 3 R (1 ⊗ G gl ( n )2 )+3 R (1 ⊗ G gl ( n )1 ) − (1 ⊗ G gl ( n )2 ) − ( R + 3 R )(1 ⊗ G gl ( n )1 ) . Pseudo left inverse of ρ In this subsection we prove that the kernel of ρ : U ( gl ( n + 1)) → D ′ ( n ) ⊗ U ( gl ( n )) is ( G )and define a family of homomorphisms σ , such that σ ( ρ ( t )) − t ∈ G U for all t ∈ U . Here andhenceforth U = U ( gl ( n + 1)) and G = G gl ( n +1)1 = P ni =0 E ii . The map σ is pseudo left inverseof ρ in the sense that σρ = Id mod ( G ).We first define the domain of σ . From now on, we set for simplicity U = U ( gl ( n + 1)). Let U ′ be the extension of U defined by U ′ = U h X i / ([ U, X ] , C n +1 ( X )). Equivalently, we define U ′ by considering a trivial central extension g Xn +1 = gl ( n + 1) ⊕ C X of gl ( n + 1), and letting U ′ = U ( g Xn +1 ) / ( C n +1 ( X )). The following lemmais standard. Lemma 6.1.
Every element in U ′ can be written uniquely in the form u = P ni =0 w i X i for some w i ∈ U . In particular, C n ( X ) and C n ( X + 1) are nonzero in U ′ . Lemma 6.2. U ′ is a domain.Proof. Let U [ X ] = U ( g Xn +1 ). To prove that the ring U ′ = U [ X ] / ( C n +1 ( X )) is a domain, itis enough to show that the associated graded ring gr U ′ is a domain. Let I = ( C n +1 ( X )).Then we have that gr U ′ = gr U [ X ] / gr I (see for example, § U [ X ] isa polynomial ring, and hence, a unique factorization domain, it is enough to show that thepolynomial C n +1 ( − X ) is irreducible in gr U [ X ].Let C n +1 ( − X ) = X n +1 + c X n + · · · + c n +1 . We know that the center Z ( gl ( n + 1)) of U ( gl ( n + 1)) is the polynomial algebra C [ c , . . . , c n +1 ]. Assume that C n +1 ( − X ) = ( X k + · · · + IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 9 a k − X + a k )( X m + · · · + b m − X + b m ) for some a i , b i in gr U of graded degree i . In particular, a k b m = c n +1 . Note that c n +1 is the determinant of E and hence the ideal generated by c n +1 ingr U is a determinantal ideal. It is well-known that any such ideal is a prime ideal (equivalently,that gr U/ ( c n +1 ) is an irreducible variety); see, for example, Proposition 1.1 in [2]. Therefore, c n +1 is irreducible in gr U . We thus may assume that a k is a constant, and that b m is a constantmultiple of c n +1 . Then k = 0 and m = n + 1, and hence, C n +1 ( − X ) is irreducible. (cid:3) Since U ′ is a quotient of the universal enveloping algebra of g Xn +1 , U ′ is a right (and left)Noetherian domain. Then by a theorem of Goldie, [4], U ′ is also a right Ore domain. Let U ′′ denote the right quotient ring of U ′ , i.e. the right Ore localization of U ′ relative to all nonzeroelements of U ′ . We have natural embeddings ι : U → U ′ and ι ′ : U ′ → U ′′ that will allow us toconsider the elements of U as elements of U ′ , and those of U ′ as elements of U ′′ . Denote by Y the left and right inverse of C n ( X ) in U ′′ , i.e. Y C n ( X ) = C n ( X ) Y = 1. Lemma 6.3. Y commutes with X and E ij whenever i, j > .Proof. To prove the result we observe that Y commutes with all elements that commute with C n ( X ). (cid:3) In order to define the homomorphism σ : D ′ ( n ) ⊗ U ( gl ( n )) → U ′′ , we next introduce somedistinguished elements of U ′′ . We treat again T as a formal variable that commute with all E ij .We first define the n × n matrix M n ( T ), whose ( i, j )th entry is M n ( T ) ij = det E − T − E · · · · · · E n E E − T − · · · · · · E n ... ... . . . 0 . . . ... E i E i · · · · · · E in ... ... . . . 0 . . . ... E n E n · · · · · · E nn − T − n + 1 where the entry of 1 in the determinant above is the ( i, j )th entry, while the ( k, k )th entryequals E kk − T − k for 0 < k < j and E kk − T − k + 1 for k > j .Similarly, we define M n +1 ( T ) ij = det E − T − E · · · · · · E n E E − T − · · · · · · E n ... ... . . . 0 . . . ... E i E i · · · · · · E in ... ... . . . 0 . . . ... E n E n · · · · · · E nn − T − n . The next lemma is standard but for reader’s convenience we include a short proof. We willuse the result both for gl ( n ) and gl ( n + 1). Lemma 6.4.
Let V and W be n × n matrices with entries in U ( gl ( n ))[ T ] and let P be a nonzeroelement of Z ( gl ( n ))[ T ] such that V W = P I n . Then W V = P I n .Proof. Since P is central in U ( gl ( n ))[ T ], we can localize U ( gl ( n ))[ T ] relative to its multiplicativesubset generated by P . Denote by L the corresponding localized ring. Let V ′ = P − V . We then have n × n matrices V ′ and W with entries in L such that V ′ W = I n . We next use that U ( gl ( n ))[ T ] is Noetherian, so L is Noetherian, and hence the ring of n × n matrices with entriesin L is Noetherian as well. And since V ′ W = I n in the Noetherian ring L , we have W V ′ = I n ,[7]. Thus W V = P I n , as desired. (cid:3) Lemma 6.5.
The following identities hold: ( E tn − T ) M n ( T ) = M n ( T )( E tn − T ) = C n ( T ); M n ( T ) t ( E n − T − n ) = ( E n − T − n ) M n ( T ) t = C n ( T +1) . Also, ( E tn +1 − T ) M n +1 ( T ) = C n +1 ( T ) , etc.Proof. First we show ( E tn − T ) M n ( T ) = C n ( T ), or, equivalently, P k ( E ki − δ ki T ) M n ( T ) kj = δ ij C n ( T ). We adopt the reasoning used in Section 1 of [11]. For this we work over the algebraΛ n ⊗ U ( gl ( n )), where Λ n is the exterior algebra with generators e , . . . , e n . Let F m = P l E lm e l .Then M n ( T ) kj e . . . e n = ( F − ( T + 1) e ) . . . e k . . . ( F n − ( T + n − e n ) , where e k is the j th term in the product. We have that X k ( E ki − δ ki T ) M n ( T ) kj e . . . e n = ( − j − X k ( E ki − δ ki T ) e k ! ( F − ( T + 1) e ) . . .F j − − ( T + j − e j − )(( F j +1 − ( T + j ) e j ) . . . ( F n − ( T + n − e n )= ( − j − ( F i − T e i )( F − ( T + 1) e ) . . . ( F j − − ( T + j − e j − )( F j +1 − ( T + j ) e j ) . . . ( F n − ( T + n − e n ) . We next prove an identity analogous to Lemma 1 in [11]. Specifically, we claim that(3) ( F r − ( T + l ) e r )( F s − ( T + l + 1) e s ) = − ( F s − ( T + l ) e s )( F r − ( T + l + 1) e r ) , or, equivalently, X p,q ( E pr e p − ( T + l ) e r )( E qs e q − ( T + l + 1) e s ) + X p,q ( E qs e q − ( T + l ) e s )( E pr e p − ( T + l + 1) e r ) = 0 . The left hand side equals X p,q [ E pr , E qs ] e p e q + X q E qs e r e q + X p E pr e s e p = X p,q δ qr E ps e p e q − X p,q δ ps E qr e p e q + X q E qs e r e q + X p E pr e s e p = X p E ps e p e r + X q E qs e r e q ! + X p E pr e s e p − X q E qr e s e q ! = 0 , as claimed. Now, applying (3) repeatedly, we obtain X k ( E ki − δ ki T ) M n ( T ) kj e . . . e n = ( F − T e ) . . . ( F i − ( T + j − e i ) . . . ( F n − ( T + n − e n ) , where the F i − ( T + j − e i is the j th term. If j = i , this yields the desired. Otherwise, notethat if r = s in (3) then ( F r − ( T + l ) e r )( F r − ( T + l + 1) e r ) = 0. Applying (3) repeatedly and IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 11 combining it with the last identity (considering separately the cases i < j and i > j ), we obtain( F − T e ) . . . ( F i − ( T + j − e i ) . . . ( F n − ( T + n − e n ) = 0 . Thus, ( E tn − T ) M n ( T ) = C n ( T ), as desired.The identity M n ( T ) t ( E n − T − n ) = C n ( T + 1) follows directly from Proposition 2 of [11].The remaining statements for the n × n matrices follow from Lemma 6.4. The proofs of thestatements involving the ( n + 1) × ( n + 1) matrices are analogous. (cid:3) We next define some elements in U ′′ that will also be used to define the pseudo-left inverse σ of ρ . Let u = 1, and let u , . . . , u n be defined via the matrix equation[ u , u , . . . , u n ] = − [ E , E , . . . , E n ] M n ( X ) Y. Some properties of u i are listed in the following lemmas. Lemma 6.6. P na =0 u a E ia = u i X for i = 0 , , . . . , n .Proof. We have [ u , u , . . . , u n ]( E tn − X ) = − [ E , E , . . . , E n ] M n ( X ) Y ( E tn − X ). By Lemma6.3, Y commutes with E tn and with X and thus with E tn − X . Hence, M n ( X ) Y ( E tn − X ) = M n ( X )( E tn − X ) Y = C n ( X ) Y = I n by Lemma 6.5. Thus, [ u , u , . . . , u n ]( E tn − X ) = − [ E , E , . . . , E n ]. Since u = 1, this yieldsthe desired result for i = 1 , , . . . , n . It remains to prove it for i = 0.We have [ u , u , . . . , u n ]( E tn +1 − X ) = [ q, , , . . .
0] for q = P na =0 u a E a − u X , and need toverify that q = 0. After multiplying by M n +1 ( X ), we obtain0 = [ u , u , . . . , u n ] C n +1 ( X ) = [ u , u , . . . , u n ]( E tn +1 − X ) M n +1 ( X ) = [ q, , . . . , M n +1 ( X )by Lemma 6.5 and using that C n +1 ( X ) = 0. Thus, q multiplied by any entry of the leftmostcolumn of M n +1 ( X ) equals 0. In particular, qM n +1 ( X ) = 0. This implies q = 0 because M n +1 ( X ) = C n ( X + 1), because C n ( X + 1) = 0 from Lemma 6.1, and because U ′′ is a domain.This completes the proof of the Lemma. (cid:3) Lemma 6.7. If v , v , . . . , v n are such that P na =0 v a E ia = v i X for i = 0 , , . . . , n , then v i = v u i for i = 0 , , . . . , n .Proof. We need to show [ v , . . . , v n ] = v [ u , . . . , u n ]. The identities given in the statementimply that [ v , . . . , v n ]( E tn − X ) = − v [ E , E , . . . , E n ]. We multiply this matrix identity by M n ( X ) Y on the right. Then, using the definition of u , . . . , u n , Lemma 6.5, and the identity C n ( X ) Y = 1, we obtain [ v , . . . , v n ] = v [ u , . . . , u n ] as needed. (cid:3) Lemma 6.8. [ u i , E jk ] = δ k u j u i − δ ki u j for i, j, k from to n .Proof. Fix j and k and set v i = [ u i , E jk ] + δ ki u j for i = 0 , , . . . , n . We will show that v , . . . , v n satisfy the hypothesis of Lemma 6.7. Note first that v = [ u , E jk ] + δ k u j = δ k u j as u = 1.Now, for i = 0 , , . . . , n , by Lemma 6.6 we have that n X a =0 [ u a , E jk ] E ia + n X a =0 u a ( δ aj E ik − δ ki E ja ) = n X a =0 ([ u a , E jk ] E ia + u a [ E ia , E jk ])= n X a =0 [ u a E ia , E jk ] = [ n X a =0 u a E ia , E jk ]= [ u i X, E jk ] = [ u i , E jk ] X. On the other hand, n X a =0 u a ( δ aj E ik − δ ki E ja ) = n X a =0 u a δ aj E ik − n X a =0 u a δ ki E ja = u j E ik − δ ki u j X by Lemma 6.6. Thus ( P na =0 [ u a , E jk ] E ia ) + u j E ik − δ ki u j X = [ u i , E jk ] X . Then n X a =0 v a E ia = n X a =0 [ u a , E jk ] E ia ! + u j E ik = [ u i , E jk ] X + δ ki u j X = v i X. Lemma 6.7 implies that v i = v u i = δ k u j u i . Thus [ u i , E jk ] = v i − δ ki u j = δ k u j u i − δ ki u j , asclaimed. (cid:3) Lemma 6.9. [ u i , u j ] = 0 for ≤ i, j ≤ n .Proof. Fix j and let v i = [ u i , u j ]. In particular, v = 0. Lemmas 6.6 and 6.8 imply that P na =0 v a E ia = v i ( X + 1) for all i . Since v = 0, we can write the last set of identities in thefollowing matrix form: [ v , . . . , v n ]( E tn − ( X + 1)) = [0 , . . . , M n ( X + 1)on the right and using Lemma 6.5, we obtain v i C n ( X + 1) = 0. Since C n ( X + 1) = 0 (by Lemma6.1) and U ′′ is a domain, v i = 0 for all i = 1 , . . . , n . Thus we have the desired. (cid:3) Lemma 6.10.
The correspondence w π g ( w ) , w ∈ U , X X − n +1 G extends to a homo-morphism π ′ g : U ′ → U ′ of associative unital algebras. Furthermore, ker π ′ g = ( G ) in U ′ .Proof. We can see that this correspondence yields a well-defined homomorphism U ( g Xn +1 ) → U ( g Xn +1 ). To see that it yields a well-defined map U ′ → U ′ , note that the determinant definitionof C n +1 ( T ) easily implies the identity π ′ g ( C n +1 ( T )) = C n +1 (cid:0) π ′ g ( T ) + n +1 G (cid:1) , so we will have π ′ g ( C n +1 ( X )) = C n +1 ( X ); thus we have a well-defined homomorphism π ′ g : U ′ → U ′ . For thekernel of π ′ g , we use that ( P ni =0 w i X i ) − π ′ g ( P ni =0 w i X i ) ∈ ( G ) for every P ni =0 w i X i ∈ U ′ . (cid:3) Let U ′ s be the image of π ′ g . By Lemma 6.10, U ′ s ≃ U ′ / ( G ). We have a natural embedding ι ′ s : U ′ s → U ′ such that π ′ g ι ′ s = Id.Recall C ρ ( E + R ) = 0 by Theorem 5.1. Thus we may define a homomorphism ρ ′ : U ′ →D ′ ( n ) ⊗ U ( gl ( n )) of associative unital algebras by the identities ρ ′ ( w ) = ρ ( w ) for all w ∈ U ( gl ( n + 1)) and ρ ′ ( X ) = E + R . Also, define ρ ′ s : U ′ s → D ′ ( n ) ⊗ U ( gl ( n )) by ρ ′ s = ρ ′ ι ′ s . Onehence has the following diagram. U π g (cid:15) (cid:15) ι / / U ′ π ′ g (cid:15) (cid:15) ρ ′ / / D ′ ( n ) ⊗ U ( gl ( n )) σ (cid:15) (cid:15) U sι s O O ι / / U ′ sι ′ s O O ι ′ / / ρ ′ s ♣♣♣♣♣♣♣♣♣♣♣♣♣ U ′′ (We will define σ momentarily.) Note again that ρ and ρ s are obtained by taking the composi-tions of ρ ′ and ρ ′ s , respectively, with the natural embedding ι : U → U ′ . IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 13
Proposition 6.11.
Let S ∈ U ′′ be such that [ S, u i ] = 0 for i > , [ S, E i ] = 0 for i > , and [ S, E ab − δ ab E ] = 0 for a, b > . The correspondence t i t u i , for i > ,t ∂ j E j − δ j S, for j ≥ ,E ab E ab − u a E b − δ ab S, for all a, b > . extends to a homomorphism σ : D ′ ( n ) ⊗ U ( gl ( n )) → U ′′ of associative unital algebras. Further-more, σρ ′ = ι ′ π ′ g and σρ = ι ′ ιπ g . Remark 6.12.
We can see such S exists and thus such σ exists; e.g., we could take S = 0 .Proof. Note that [
S, E i + u i E ] = 0 by Lemma 6.6. To check that the correspondence extendsto a homomorphism, we verify that σ ([ x, y ]) = [ σ ( x ) , σ ( y )] whenever x, y equal one of thegenerators t i t , t ∂ i , E ab .By Lemma 6.9, we have σ ([ t i t , t j t ]) = σ (0) = 0 = [ u i , u j ] = [ σ ( t i t ) , σ ( t j t )], as desired.By Lemma 6.8, we have σ ([ t i t , t ∂ j ]) = σ ( δ j t i t − δ ij ) = δ j u i − δ ij = [ u i , E j ] = [ σ ( t i t ) , σ ( t ∂ j )]as u = 1 and [ u i , S ] = 0.By Lemma 6.8 and Lemma 6.9, σ ([ t i t , E ab ]) = σ (0) = 0 = δ b u a u i − δ bi u a − u a ( δ b u i − δ bi ) =[ u i , E ab ] − [ u i , u a ] E b − u a [ u i , E b ] = [ u i , E ab − u a E b − δ ab S ] = [ σ ( t i t ) , σ ( E ab )], as u = 1 and[ u i , u a ] = 0 and [ u i , S ] = 0.We have σ ([ t ∂ i , t ∂ j ]) = σ ( δ i t ∂ j − δ j t ∂ i ) = δ i ( E j − δ j S ) − δ j ( E i − δ i S ) = δ i E j − δ j E i = [ E i , E j ] = [ σ ( t ∂ i ) , σ ( t ∂ j )], by the definition of S .By Lemma 6.8 we have σ ([ t ∂ i , E ab ]) = 0 = [ E i − δ i S, E ab − u a E b − δ ab S ] = [ σ ( t ∂ i ) , σ ( E ab )])Finally, using again Lemma 6.8 we have σ ([ E ab , E cd ]) = σ ( δ bc E ad − δ ad E cb ) = δ bc ( E ad − u a E d − δ ad S ) − δ ad ( E cb − u c E b − δ cb S )= δ bc E ad − δ ad E cb − δ bc u a E d + δ ad u c E b = [ E ab , E cd ] + [ E ab , − u c E d ] + [ − u a E b , E cd ] + [ − u a E b , − u c E d ]= [ E ab − u a E b , E cd − u c E d ] + [ E ab , − δ cd S ] + [ − δ ab S, E cd ]= [ E ab − u a E b − δ ab S, E cd − u c E d − δ cd S ]= [ σ ( E ab ) , σ ( E cd )] . In the above sequence of identities we used that [
S, δ cd E ab − δ ab E cd ] = [ S, δ cd ( E ab − δ ab E ) − δ ab ( E cd − δ cd E )] = 0 and that [ E ab , − u c E d ] + [ − u a E b , E cd ] + [ − u a E b , − u c E d ] = − δ bc u a E d + δ ad u c E b .Thus σ is indeed a homomorphism of associative unital algebras.Note that σ ( E ) = X − S and σ ( R ) = S − n +1 G . Using the definitions of σ and ρ ′ , it iseasy to verify that σρ ′ ( X ) = X − n +1 G , σρ ′ ( E ij ) = E ij for i = j , and σρ ′ ( E ii ) = E ii − n +1 G .Hence, σρ ′ = ι ′ π ′ g . The identity σρ = ι ′ ιπ g follows from ρ = ρ ′ ι and π ′ g ι = ιπ g . (cid:3) Theorem 6.13.
We have the following: (i) ker ρ ′ = ( G ) in U ′ , and ker ρ = ( G ) in U . (ii) ker ρ ′ s = (0) in U ′ s , and ker ρ s = (0) in U s . Proof.
For part (i) we first note that ρ ′ ( G ) = ρ ( G ) = 0. To complete the proof we use Lemma6.10 along with Proposition 6.11 and the fact that the kernel of π g is ( G ) in U . As for part(ii), note that π ′ s π ′ s = π ′ s ; then, if t ∈ ker ρ ′ s = ( G ) ∩ U s , then we have t = π ′ s ( t ) = 0. Thusker ρ ′ s = { } . Then ker ρ s ⊂ ker ρ ′ s , so ker ρ s = { } . (cid:3) IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 15
Appendix. Formulas for the images of certain elements under ρ In this appendix we provide explicit formulas for the images under ρ of r gl ( n +1) k ( a, b ). Forthe definition of the latter see (1). The proofs of these formulas are independent of the resultsin the previous sections of the paper. In this way we have an alternative proof of Theorem5.4. It is interesting to note that one can then go “backwards” and prove Theorem 5.1 andthen Theorem 4.2 purely computationally. Thus, this appendix leads to an alternative (morecomputational) approach of the results established in Sections 4 and 5.Note that r gl ( N ) k +1 ( a, b ) = P i r gl ( N ) k ( a, i ) E ib for all nonnegative integers k , for N = n or n + 1,and for all a, b .For a positive integer m and for a and b with 0 ≤ a ≤ n and 1 ≤ b ≤ n , define f m ( a, b ) = n X i =1 t a ∂ i ⊗ r gl ( n ) m − ( i, b ) . Theorem.
For all positive integers k , a , b , such that a, b ≤ n , ρ ( r gl ( n +1) k ( a, b )) = k X m =1 f m ( a, b ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! + k X g =0 (cid:18) kg (cid:19) R g (cid:16) ⊗ r gl ( n ) k − g ( a, b ) (cid:17) ,ρ ( r gl ( n +1) k ( a, k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t a ∂ ⊗ − k − X g =0 (cid:18) kg (cid:19) R g X j> t j t ⊗ r gl ( n ) k − g ( a, j ) ! − (cid:18) t a t ⊗ (cid:19) k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! ,ρ ( r gl ( n +1) k (0 , b )) = k X m =1 f m (0 , b ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! ,ρ ( r gl ( n +1) k (0 , k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t ∂ ⊗
1) + R k − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! . Proof.
We prove all four statements simultaneously by induction on k . The base case k = 1follows from the definition of ρ . Suppose the formulas in the statement of the Theorem are truefor some positive integer k . Let us prove them for k + 1.First, we consider the value of ρ ( r gl ( n +1) k ( a, b )) for a, b >
0. We have ρ ( r gl ( n +1) k +1 ( a, b )) = ρ ( r gl ( n +1) k ( a, E b + X i> r gl ( n +1) k ( a, i ) E ib )= ρ ( r gl ( n +1) k ( a, ρ ( E b ) + X i> ρ ( r gl ( n +1) k ( a, i )) ρ ( E ib )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ( t a ∂ ⊗ − k − X g =0 (cid:18) kg (cid:19) R g X j> t j t ⊗ r gl ( n ) k − g ( a, j ) − (cid:18) t a t ⊗ (cid:19) k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! ( t ∂ b ⊗ X i> k X m =1 f m ( a, i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g + k X g =0 (cid:18) kg (cid:19) R g (cid:16) ⊗ r gl ( n ) k − g ( a, i ) (cid:17)! ( t i ∂ b ⊗ ⊗ E ib + δ ib R )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ( t a ∂ t ∂ b ⊗ − k − X g =0 (cid:18) kg (cid:19) R g X j> t j ∂ b ⊗ r gl ( n ) k − g ( a, j ) − (cid:18) t a t ⊗ (cid:19) k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ( t ∂ b ⊗ X i> k X m =1 f m ( a, i )( t i ∂ b ⊗ k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g + X i> k X g =0 (cid:18) kg (cid:19) R g (cid:16) t i ∂ b ⊗ r gl ( n ) k − g ( a, i ) (cid:17) + X i> k X m =1 f m ( a, i )(1 ⊗ E ib ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g + X i> k X g =0 (cid:18) kg (cid:19) R g (cid:16) ⊗ r gl ( n ) k − g ( a, i ) E ib (cid:17) + k X m =1 f m ( a, b ) k − m X g =0 (cid:18) kg (cid:19) R g +11 R k − m − g + k X g =0 (cid:18) kg (cid:19) R g +11 (cid:16) ⊗ r gl ( n ) k − g ( a, b ) (cid:17) . Write the last expression in the form X − X − X + X + X + X + X + X + X .Then X − X = P i> P kg =0 (cid:0) kg (cid:1) R g (cid:16) t i ∂ b ⊗ r gl ( n ) k − g ( a, i ) (cid:17) − P k − g =0 (cid:16)(cid:0) kg (cid:1) R g P j> t j ∂ b ⊗ r gl ( n ) k − g ( a, j ) (cid:17) = R k ( t a ∂ b ⊗ . On the other hand, X = P km =1 (( P j,i> t a ∂ j t i ∂ b ⊗ r gl ( n ) m − ( j, i ))( P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g )) . Also, X = P km =2 (( P j,i> t a ∂ j t i ∂ b ⊗ r gl ( n ) m − ( j, i ))( P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g )) . Then X − X = ( P j,i> t a ∂ j t i ∂ b ⊗ r gl ( n )0 ( j, i ))( P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) = ( P i> t a ∂ i t i ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) , so X + X − X =( P i ≥ t a ∂ i t i ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) = ( t a ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − g ) , as P i ≥ t a ∂ i t i ∂ b ⊗ t a ∂ b ⊗ R . Therefore we have ( X + X − X ) + ( X − X ) = ( t a ∂ b ⊗ P kg =0 (cid:0) kg (cid:1) R g R k − g ) . IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 17
Furthermore, since P i> f m ( a, i )(1 ⊗ E ib ) = f m +1 ( a, b ), we have X = P km =1 ( f m +1 ( a, b ) P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g ) = P k +1 m =2 f m ( a, b )( P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g ).Thus X + (( X + X − X ) + ( X − X )) = P k +1 m =1 f m ( a, b )( P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g ) . But X = P km =1 ( f m ( a, b ) P k − m +1 g =1 (cid:0) kg − (cid:1) R g R k − m − g +12 ) . Hence,( X + (( X + X − X ) + ( X − X ))) + X = P k +1 m =1 ( f m ( a, b ) P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g ) . Now using that P i> r gl ( n ) k − g ( a, i ) E ib = r gl ( n ) k +1 − g ( a, b ), we find X = P kg =0 (cid:0) kg (cid:1) R g (1 ⊗ r gl ( n ) k +1 − g ( a, b )).Also X = P k +1 g =1 (cid:0) kg − (cid:1) R g (1 ⊗ r gl ( n ) k +1 − g ( a, b )) implies that X + X = P k +1 g =0 (cid:0) k +1 g (cid:1) R g (1 ⊗ r gl ( n ) k +1 − g ( a, b )) . Therefore, ρ ( r gl ( n +1) k +1 ( a, b )) = (( X + (( X + X − X ) + ( X − X ))) + X ) + ( X + X ) = P k +1 m =1 ( f m ( a, b ) P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g + P k +1 g =0 (cid:0) k +1 g (cid:1) R g (1 ⊗ r gl ( n ) k +1 − g ( a, b )) . This completes theproof of the inductive step for ρ ( r gl ( n +1) k +1 ( a, b )) for a, b > ρ ( r gl ( n +1) k ( a, a >
0. We have ρ ( r gl ( n +1) k +1 ( a, ρ ( r gl ( n +1) k ( a, E + X i> r gl ( n +1) k ( a, i ) E i )= ρ ( r gl ( n +1) k ( a, ρ ( E ) + X i> ρ ( r gl ( n +1) k ( a, i )) ρ ( E i )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t a ∂ ⊗ − k − X g =0 (cid:18) kg (cid:19) R g X j> t j t ⊗ r gl ( n ) k − g ( a, j ) ! − (cid:18) t a t ⊗ (cid:19) k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) !! ( t ∂ ⊗ R )+ X i> k X m =1 f m ( a, i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! + k X g =0 (cid:18) kg (cid:19) R g (cid:16) ⊗ r gl ( n ) k − g ( a, i ) (cid:17)! ( t i ∂ ⊗ − X j> t j t ⊗ E ij )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t a ∂ t ∂ ⊗ − k − X g =0 (cid:18) kg (cid:19) R g X j> t j ∂ ⊗ r gl ( n ) k − g ( a, j ) ! − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> t a ∂ i t j ∂ ⊗ r gl ( n ) m − ( i, j ) ! + k − X g =0 (cid:18) kg (cid:19) R g +11 R k − − g ! ( t a ∂ ⊗ − k − X g =0 (cid:18) kg (cid:19) R g +11 X j> t j t ⊗ r gl ( n ) k − g ( a, j ) ! − (cid:18) t a t ⊗ (cid:19) k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g +11 R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! + k X m =1 X i> f m ( a, i )( t i ∂ ⊗ ! k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! + k X g =0 (cid:18) kg (cid:19) R g X i> t i ∂ ⊗ r gl ( n ) k − g ( a, i ) ! − k X m =1 X i,j> f m ( a, i ) (cid:18) t j t ⊗ E ij (cid:19)! k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! − k X g =0 (cid:18) kg (cid:19) R g X i,j> t j t ⊗ r gl ( n ) k − g ( a, i ) E ij ! . We write the last expression as X − X − X + X − X − X + X + X − X − X .Then X − X = P kg =0 (cid:0) kg (cid:1) R g (cid:16)P i> t i ∂ ⊗ r gl ( n ) k − g ( a, i ) (cid:17) − P k − g =0 (cid:0) kg (cid:1) R g (cid:16)P i> t i ∂ ⊗ r gl ( n ) k − g ( a, i ) (cid:17) = R k ( t a ∂ ⊗ X + X = (cid:16)P kg =1 (cid:0) kg − (cid:1) R g P j> t j t ⊗ r gl ( n ) k +1 − g ( a, j ) (cid:17) + (cid:16)P kg =0 (cid:0) kg (cid:1) R g P j> t j t ⊗ r gl ( n ) k +1 − g ( a, j ) (cid:17) = (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g P j> t j t ⊗ r gl ( n ) k +1 − g ( a, j ) (cid:17) .We next have X = P km =1 P i,j> f m ( a, i ) (cid:16) t j t ⊗ E ij (cid:17)! (cid:16)P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g (cid:17) . Now, for anypositive integer m , P i,j> f m ( a, i ) (cid:16) t j t ⊗ E ij (cid:17) = P i,j> (cid:16)P ℓ> t a ∂ ℓ ⊗ r gl ( n ) m − ( ℓ, i ) (cid:17) (cid:16) t j t ⊗ E ij (cid:17) = P j,ℓ> t a ∂ ℓ t j t − ⊗ (cid:16)P i> r gl ( n ) m − ( ℓ, i ) E ij (cid:17) = P j,ℓ> t a ∂ ℓ t j t − ⊗ r gl ( n ) m ( ℓ, j ) = (cid:16) t a t ⊗ (cid:17) P j,ℓ> ∂ ℓ t j ⊗ r gl ( n ) m ( ℓ, j ), which is the same as (cid:16) t a t ⊗ (cid:17) P i,j> ∂ i t j ⊗ r gl ( n ) m ( i, j ). Then X = (cid:16) t a t ⊗ (cid:17) P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) . Also, X = (cid:16) t a t ⊗ (cid:17) P km =2 (cid:16)P k +1 − mg =1 (cid:0) kg − (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) . Therefore, X + X = (cid:16) t a t ⊗ (cid:17) P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) .Next, for any positive integer m , P i> f m ( a, i )( t i ∂ ⊗
1) = P i,j> t a ∂ j t i ∂ ⊗ r gl ( n ) m − ( j, i ) = P i,j> t a ∂ i t j ∂ ⊗ r gl ( n ) m − ( i, j ). Then X = P km =1 (cid:16)P i,j> t a ∂ i t j ∂ ⊗ r gl ( n ) m − ( i, j ) (cid:17) (cid:16)P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g (cid:17) .On the other hand, X = P km =2 (cid:16)P i,j> t a ∂ i t j ∂ ⊗ r gl ( n ) m − ( i, j ) (cid:17) (cid:16)P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g (cid:17) . Thus X − X = ( P i,j> t a ∂ i t j ∂ ⊗ δ ij ) (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − − g (cid:17) = (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − − g (cid:17) ( t a ∂ ⊗ R − ( t ∂ ⊗ X − X + X = (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − − g (cid:17) ( t a ∂ ⊗ R = (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − g (cid:17) ( t a ∂ ⊗ X − X + X + X − X = (cid:16)P kg =0 (cid:0) kg (cid:1) R g R k − g (cid:17) ( t a ∂ ⊗ X = (cid:16)P kg =1 (cid:0) kg − (cid:1) R g R k − g (cid:17) ( t a ∂ ⊗ X − X + X + X − X + X = (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g R k − g (cid:17) ( t a ∂ ⊗ IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 19
Hence, ρ ( r gl ( n +1) k +1 ( a, X − X + X + X − X + X ) − ( X + X ) − ( X + X ) = (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g R k − g (cid:17) ( t a ∂ ⊗ − (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g P j> t j t ⊗ r gl ( n ) k +1 − g ( a, j ) (cid:17) − (cid:16) t a t ⊗ (cid:17) P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) , as desired. This com-pletes the inductive step for the value of ρ ( r gl ( n +1) k +1 ( a, ρ ( r gl ( n +1) k (0 , b )) for b > ρ ( r gl ( n +1) k +1 (0 , b )) = ρ ( r gl ( n +1) k (0 , E b + X i> r gl ( n +1) k (0 , i ) E ib )= ρ ( r gl ( n +1) k (0 , ρ ( E b ) + X i> ρ ( r gl ( n +1) k (0 , i )) ρ ( E ib )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t ∂ ⊗
1) + R k − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) !! ( t ∂ b ⊗ X i> k X m =1 f m (0 , i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g !! ( t i ∂ b ⊗ ⊗ E ib + δ ib R )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t ∂ t ∂ b ⊗
1) + R k ( t ∂ b ⊗ − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! ( t ∂ b ⊗ X i> k X m =1 f m (0 , i )( t i ∂ b ⊗ k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! + X i> k X m =1 f m (0 , i )(1 ⊗ E ib ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! + k X m =1 f m (0 , b ) k − m X g =0 (cid:18) kg (cid:19) R g +11 R k − m − g ! . We write the last expression as X + X − X + X + X + X .Now X = P km =1 (( P j,i> t ∂ j t i ∂ b ⊗ r gl ( n ) m − ( j, i ))( P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g )) . Also, X = P km =2 (( P j,i> t ∂ j t i ∂ b ⊗ r gl ( n ) m − ( j, i ))( P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g )) . Then X − X = ( P j,i> t a ∂ j t i ∂ b ⊗ r gl ( n )0 ( j, i ))( P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) = ( P i> t ∂ i t i ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) , so X + X − X = ( P i ≥ t ∂ i t i ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − − g ) = ( t ∂ b ⊗ P k − g =0 (cid:0) kg (cid:1) R g R k − g ) , as P i ≥ t ∂ i t i ∂ b ⊗ t ∂ b ⊗ R . Then X + X + X − X = ( t ∂ b ⊗ P kg =0 (cid:0) kg (cid:1) R g R k − g ) . Since P i> f m (0 , i )(1 ⊗ E ib ) = f m +1 (0 , b ), X = P km =1 ( f m +1 (0 , b ) P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g ) = P k +1 m =2 ( f m (0 , b ) P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g ). Then X + ( X + X + X − X ) = P k +1 m =1 ( f m (0 , b ) P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g ). We also have X = P km =1 ( f m (0 , b ) P k +1 − mg =1 (cid:0) kg − (cid:1) R g R k +1 − m − g ) . Therefore, ρ ( r gl ( n +1) k +1 (0 , b )) = ( X +( X + X + X − X ))+ X = P km =1 ( f m (0 , b ) P k +1 − mg =1 (cid:0) k +1 g (cid:1) R g R k +1 − m − g ),which completes the inductive step for the value of ρ ( r gl ( n +1) k +1 (0 , b )) . Finally, we consider the value of ρ ( r gl ( n +1) k (0 , ρ ( r gl ( n +1) k +1 (0 , ρ ( r gl ( n +1) k (0 , E + X i> r gl ( n +1) k (0 , i ) E i )= ρ ( r gl ( n +1) k (0 , ρ ( E ) + X i> ρ ( r gl ( n +1) k (0 , i )) ρ ( E i )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t ∂ ⊗
1) + R k − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) !! ( t ∂ ⊗ R )+ X i> k X m =1 f m (0 , i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g !! ( t i ∂ ⊗ − X j> t j t ⊗ E ij )= k − X g =0 (cid:18) kg (cid:19) R g R k − − g ! ( t ∂ t ∂ ⊗
1) + R k ( t ∂ ⊗ − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! ( t ∂ ⊗ k − X g =0 (cid:18) kg (cid:19) R g +11 R k − − g ! ( t ∂ ⊗
1) + R k +11 − k X m =2 k − m X g =0 (cid:18) kg (cid:19) R g +11 R k − m − g ! X i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) ! + X i> k X m =1 f m (0 , i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g !! ( t i ∂ ⊗ − X i> k X m =1 f m (0 , i ) k − m X g =0 (cid:18) kg (cid:19) R g R k − m − g !! X j> t j t ⊗ E ij ! . IXED TENSOR PRODUCTS AND CAPELLI-TYPE DETERMINANTS 21
We write the last expression as X + X − X + X + X − X + X − X .Since X = P km =1 (cid:16)P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g (cid:17) (cid:16)P i,j> ∂ j t i ⊗ r gl ( n ) m − ( j, i ) (cid:17) ( t ∂ ⊗ X − X = (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − − g (cid:17) (cid:0)P i> ∂ i t i ⊗ (cid:1) ( t ∂ ⊗
1) = (cid:16)P k − g =0 (cid:0) kg (cid:1) R g R k − − g (cid:17) ( R − ( t ∂ ⊗ t ∂ ⊗ X − X + X + X = (cid:16)P kg =0 (cid:0) kg (cid:1) R g R k − g (cid:17) ( t ∂ ⊗ X = (cid:16)P kg =1 (cid:0) kg − (cid:1) R g R k − g (cid:17) ( t ∂ ⊗ X − X + X + X + X = (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g R k − g (cid:17) ( t ∂ ⊗ X = P km =1 P i,j> f m (0 , i ) (cid:16) t j t ⊗ E ij (cid:17)! (cid:16)P k − mg =0 (cid:0) kg (cid:1) R g R k − m − g (cid:17) . For anypositive integer m , P i,j> f m (0 , i ) (cid:16) t j t ⊗ E ij (cid:17) = P i,j> (cid:16)P ℓ> t ∂ ℓ ⊗ r gl ( n ) m − ( ℓ, i ) (cid:17) (cid:16) t j t ⊗ E ij (cid:17) = P j,ℓ> ∂ ℓ t j ⊗ (cid:16)P i> r gl ( n ) m − ( ℓ, i ) E ij (cid:17) = P j,ℓ> ∂ ℓ t j ⊗ r gl ( n ) m ( ℓ, j ). Then we have X = P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) kg (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) . Also, X = P km =2 (cid:16)P k +1 − mg =1 (cid:0) kg − (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) , and therefore X + X = P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) .Combining the above, ρ ( r gl ( n +1) k +1 (0 , X − X + X + X + X ) + X − ( X + X ) = (cid:16)P kg =0 (cid:0) k +1 g (cid:1) R g R k − g (cid:17) ( t ∂ ⊗ R k +11 − P k +1 m =2 (cid:16)P k +1 − mg =0 (cid:0) k +1 g (cid:1) R g R k +1 − m − g (cid:17) (cid:16)P i,j> ∂ i t j ⊗ r gl ( n ) m − ( i, j ) (cid:17) . This completes the induction step for ρ ( r gl ( n +1) k +1 (0 , (cid:3) References [1] G. Bergman, Tensor products of faithful modules, arXiv:1610.05178.[2] W. Bruns, U. Vetter, Determinantal rings,
Springer Lecture Notes , Springer-Verlag, Berlin, Heidel-berg, New York, 1988, vii+236 pp.[3] J. Dixmier, Enveloping Algebras, Graduate studies in mathematics, , Translations of MathematicalMonographs, American Mathematical Society, 1996.[4] A. Goldie, The structure of prime rings under ascending chain conditions, Proc. London Math. Soc. (1958), 589–608.[5] D. Grantcharov, V. Serganova, Cuspidal representations of sl ( n + 1), Adv. Math. (2010) 1517–1547.[6] R. Howe, Remarks on classical invariant theory,
Trans. Amer. Math. Soc. (1989), 539–570; Erratum,
Trans. Amer. Math. Soc. (1990), 823.[7] N. Jacobson, Some remarks on one-sided inverses,
Proc. Amer. Math. Soc. (1950), 352–355.[8] T. A. Larsson, Conformal fields: A class of representations of V ect ( N ), Int. J. Mod. Phys. A (1992),6493–6508.[9] A. Molev, Yangians and Classical Lie Algebras, Mathematical Surveys and Monographs , , AmericanMathematical Society, Providence, RI, 2007.[10] G. Shen, Graded modules of graded Lie algebras of Cartan type. I. Mixed products of modules, Sci. SinicaSer. A (1986), 570–581.[11] T. Umeda, Newton’s formula for gl ( n ), Proc. Amer. Math. Soc. , (1998), 3169–3175.[12] D. Zhelobenko, Compact Lie groups and their representations, Transl. Math. Monographs , , AmericanMathematical Society, 1974. Department of Mathematics, University of Texas at Arlington, Arlington, TX 76019
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