THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e (4,1) Ana Prlić University of Zagreb, Croatia
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1 GLASNIK MATEMATIČKI Vol. 5(7)(017), THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e (4,1) Ana Prlić University of Zagreb, Croatia Abstract. Let G be the Lie group SO e(4,1), with maximal compact subgroup K = S(O(4) O(1)) e = SO(4). Let g = so(5,c) be the complexification of the Lie algebra g 0 = so(4,1) of G, and let U(g) be the universal enveloping algebra of g. Let g = k p be the Cartan decomposition of g, and C(p) the Clifford algebra of p with respect to the trace form B(X,Y) = tr(xy) on p. In this paper we give explicit generators of the algebra (U(g) C(p)) K. 1. Introduction Let G be a connected real reductive Lie group with Cartan involution Θ, such that K = G Θ is a maximal compact subgroup of G. Let g be the complexification of the Lie algebra g 0 of G, and let U(g) be the universal enveloping algebra of g. Let g 0 = k 0 p 0 and g = k p be the Cartan decompositions of g 0 and g corresponding to Θ. Let B be a fixed nondegenerate invariant symmetric bilinear form on g 0, such that B is negative definite on k 0 and positive definite on p 0, and such that k 0 is orthogonal to p 0. We use the same letter B for the extension of B to g. In general, one can define B as a suitable extension of the Killing form, while for matrix examples one can take the trace form. Let C(p) be the Clifford algebra of p with respect to B. If X is an irreducible (g,k) module, then X is completely determined by the action of the centralizer of k in U(g) on any K isotypic component of X. This was proved by Harish-Chandra in []. The problem with this approach 010 Mathematics Subject Classification. E47, E46. Key words and phrases. Lie group, Lie algebra, representation, Dirac operator, Dirac cohomology. The author was supported by grant no of the Croatian Science Foundation and by the QuantiXLie Center of Excellence. 75
2 76 A. PRLIĆ to determination of all X is that the structure of the algebra U(g) K is very complicated to describe. There is a version of this theorem proved in [6] where one considers modules X S, where X is a (g,k) module, and S is the spin module for C(p). Let K be the spin double cover of K. Then X S is a (U(g) C(p), K) module and it is proved in [6] that an irreducible (U(g) C(p), K) module is determined by the action of the algebra (U(g) C(p)) K on any nontrivial K isotypic component of X S. The algebra(u(g) C(p)) K is completely determined forthe groupsu(, 1) in [8]. Except for the obvious generators from the center Z(k) of U(k) and C(p) K, there are two more generators: the Dirac operator, and its k version called k Dirac. In this paper we provethat a similar result holds for the group SO e (4,1). In this case k is equivalent to the product of two copies of sl(,c) and we prove that the centralizer of k in U(g) C(p) is generated by two copies of the Casimir operator of sl(,c), by one generator from C(p) K, by the Dirac operator and by the k Dirac. Among the side results we get that the algebra (U(g) C(p)) K is a free U(g) K module of rank 16. This is a special case of the recent result of Kralje vić [5, Corollary 1] which asserts that for any finite-dimensional complex K module V the space (U(g) V) K is a free U(g) K module of rank dimv. The algebra (U(g) C(p)) K is also important for the Dirac induction developed in [6] and further studied in [9]. For the reader s convenience, let us recall the basic definitions and results important in the setting of the Dirac operator. The Dirac operator corresponding to a real reductive Lie group was defined by Parthasarathy [7], where it was used to construct the discrete series representations. The final form of the construction was given by Atiyah and Schmid[1]. The algebraic version of the Dirac operator and the notion of Dirac cohomology were defined by Vogan in [10] and further studied by Huang and Pandžić [3,4]. Let b i be a basis of p and let d i be the dual basis with respect to B. Then D = b i d i U(g) C(p). i The sum does not depend of the choice of basis b i. If X is a (g,k) module, and S a spin module for C(p), then the Dirac cohomology of X is H D (X) = KerD/ImD KerD. It is a K module. In [6] the opposite construction was described, i.e. a construction of a (g, K) module X whose Dirac cohomology contains a given irreducible K module. It was proved in [6,9] that some discrete series representation can be constructed in that way. The final goal is to prove that the same construction works for all discrete series representation, and also for some other modules which may be difficult to construct otherwise.
3 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 77. The algebra (S(g) (p)) K We use the same strategy as in [8]: we study the algebra (S(g) (p)) K since the algebrass(g) (p) and U(g) C(p) are isomorphic as K modules, and the algebra S(g) (p) is easier to study. Let G be the Lie group SO(4,1) = {g GL(5,R) g τ γg = γ}, where γ = The (real) Lie algebra of G is g 0 = so(4,1) = {x gl(5,r) x τ = γxγ, trx = 0}. We use the following basis for the complexification g of g 0 : H 1 = ie 1 ie 1, H = ie 34 ie 43, E 1 = 1 (e 13 e 4 ie 3 ie 14 e 31 +e 4 +ie 3 +ie 41 ), E = 1 (e 13 +e 4 ie 3 +ie 14 e 31 e 4 +ie 3 ie 41 ), F 1 = 1 (e 13 e 4 +ie 3 +ie 14 e 31 +e 4 ie 3 ie 41 ), F = 1 (e 13 +e 4 +ie 3 ie 14 e 31 e 4 ie 3 +ie 41 ), E 3 = e 15 ie 5 +e 51 ie 5, E 4 = e 35 ie 45 +e 53 ie 54, F 3 = e 15 +ie 5 +e 51 +ie 5, F 4 = e 35 +ie 45 +e 53 +ie 54. where e ij denotes the matrix with the ij entry equal to 1 and all other entries equal to 0. The commutation relations are given by the following table:
4 78 A. PRLIĆ H E 1 E F 1 F E 3 E 4 F 3 F 4 H 1 0 E 1 E F 1 F E 3 0 F 3 0 H E 1 E F 1 F 0 E 4 0 F 4 E 1 0 H 1 +H E 4 E 3 E 0 H 1 H 0 E 3 F 4 0 F 1 0 F 4 F F E F 3 E 3 E 1 H 1 E E 4 F H F 3 F 1 Let g = k p be the Cartan decomposition of g corresponding to the Cartan involution θ(x) = X τ. Then k = span{h 1,H,E 1,E,F 1,F } = so(4,c), and p = span{e 3,E 4,F 3,F 4 }. We have k = k 1 k, where k 1 = span{h 1 +H,E 1,F 1 } sl(,c), k = span{h 1 H,E,F } sl(,c) with H 1 +H, E 1 and F 1 (resp. H 1 H, E and F ) corresponding to the standard basis of sl(,c). Algebras k 1 and k mutually commute. We set a 1 = (H 1 +H ) +4E 1 F 1 S(k 1 ) S(g) a = (H 1 H ) +4E F S(k ) S(g). Note that a 1 (respectively a ) symmetrizes to a multiple of the Casimir element of U(k 1 ) (respectively U(k )), and that they are both K invariant. From [8, Lemma.1] it follows that (.1) S(k 1 ) k1 = C[a 1 ] and S(k ) k = C[a ]. and also (.) S(k 1 ) = S(k 1 ) k1 H k1 and S(k ) = S(k ) k H k, with the spaces of harmonics H k1 and H k decomposing as H k1 = V (n,n) and H k = V (n, n). n Z + n Z + Here V (n,n) denotes the so(4,c) module with the highest weight vector E n 1, on which H 1 and H both act by n. Similarly, V (n, n) denotes the so(4,c) module with the highest weight vector E n, on which H 1 acts by n and H by n. Similarly as in the proof of [8, Lemma.3.], we have S n (p) = V (n,0) (E 3 F 3 +E 4 F 4 )S n (p),
5 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 79 where V (n,0) is the so(4,c) module with the highest weight vector E n 3. Let us denote b = E 3 F 3 +E 4 F 4. One can easily check that b is K invariant. Now we have (.3) S(p) = S(p) K H p = C[b] H p, where H p = n Z + V (n,0). From (.1), (.) and (.3) it follows that S(g) (p) = C[a 1,a,b] H k1 H k H p (p). The k-submodules of (p) are: 0 (p) = 4 (p) = V (0,0) 1 (p) = 3 (p) = V (1,0) span{e 3 E 4,E 3 F 3 +E 4 F 4,F 3 F 4 } = V (1,1) span{e 3 F 4,E 3 F 3 E 4 F 4,E 4 F 3 } = V (1, 1). It follows that (p) decomposes under k as (p) = 0 {}}{ V (0,0) 1 {}}{ V (1,0) {}}{{}}{ V (1,1) 3 {}}{ V (1,0) 4 {}}{ V (0,0), V (1, 1) where each of the numbers over braces denotes the degree in which the corresponding k-module is appearing. If V is a k 1 module and W is a k module, then we denote by V W the external tensor product of V and W. As a vector space, V W is equal to the direct sum of V and W, and the action of k = k 1 k module is defined so that k 1 acts on V and k acts on W. Then we have V (n,m) V n+m V n m as k modules,
6 80 A. PRLIĆ where V k denotes the sl(,c) module with highest weight k. It follows that H k1 H k = n Z + V (n,n) m Z + V (m, m) V n V 0 V 0 V m n Z + m Z + (V n V 0 ) (V 0 V m ) n,m Z + V n V m n,m Z + V (n+m,n m). n,m Z + We have seen that H p = k Z + V (k,0). On the other hand, (.4) V (n+m,n m) V (k,0) = (V n V m ) (V k V k ) = (V n V k ) (V m V k ) = ( V n+k V n+k V n k ) ( V m+k V m+k V m k ) = (V n+k V m+k ) (V n+k V m+k ) ( V n k V m k ). Furthermore, we have 0 {}}{ (p) = V 0 V 0 1 {}}{ V 1 V 1 {}}{{}}{ V V 0 V 0 V 3 {}}{ V 1 V 1 4 {}}{ V 0 V 0.
7 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 81 where as before, the numbers over braces denote the degrees in (p). In order to get an invariant, we can tensor V 0 V 0 only with V 0 V 0. In (.4), we have V 0 V 0 only if n k = m k = 0, that is, n = m = k. In that case, we have one invariant in degree 4n and one invariant in degree 4n+4, where n 0. Furthermore, V 1 V 1 can be tensored only with V 1 V 1, and in (.4) we have V 1 V 1 only for n k = m k = 1. We have several cases: If n = m,k = n+1, we have one invariant in degree 4n+ and one invariant in degree 4n+4, where n 0. If n = m,k = n 1, we have one invariant in degree 4n and one invariant in degree 4n+, where n 1. If m = n+1,k = n+1, we have one invariant in degree 4n+3 and one invariant in degree 4n+5, where n 0. If m = n 1,k = n 1, we have one invariant in degree 4n 1 and one invariant in degree 4n+1, where n 1. TheexteriorproductV V 0 canbetensoredonlywithv V 0. In(.4),we have V V 0 only for n k =, m k = 0 or n k +=, m k = 0. The cases are: If k = n, we have one invariant in degree 4n+, where n 1. If n k =, we have one invariant in degree 4n 1, where n 1. If n k =, we have one invariant in degree 4n+5, where n 0. Finally, V 0 V can be tensored only with V 0 V, and in (.4) we have V 0 V only for n k = 0, m k = or n k = 0, m k + =. The cases are: If m = n, we have one invariant in degree 4n+, where n 1. If m k =, we have one invariant in degree 4n+3, where n 0. If m k =, we have one invariant in degree 4n+1, where n 1. From all the above we conclude that we have the following table for the K invariants in H k1 H k H p (p): degree number of linearly independent invariants k, k 1 4 4k +1, k 1 4 4k +, k 1 4 4k +3, k 1 4 We have proved: Proposition.1. The algebra of K invariants in S(g) (p) can be written as (S(g) (p)) K = C[a 1,a,b] (H k1 H k H p (p)) K.
8 8 A. PRLIĆ The number of invariants in (H k1 H k H p (p)) K in each degree is given by the above table. The elementsa 1,a and bofs(g) canbe viewedaselementsofs(g) (p) by the identification S(g) = S(g) 1. We define further elements, which are all easily checked to be K-invariant: D = E 3 F 3 +E 4 F 4 +F 3 E 3 +F 4 E 4, ( the Dirac operator) c = ((E 1 E F 3 E 1F F 4 +F 1F E 3 F 1E E 4 ) (H 1 H )(E 1 F 3 F 4 +F 1 E 3 E 4 ) (H 1 +H )(F E 3 F 4 +E F 3 E 4 ) (H 1 H )(H 1 +H )(E 3 F 3 E 4 F 4 )) 1, d = E 1 F 3 F 4 (H 1 +H ) (E 3 F 3 +E 4 F 4 ) F 1 E 3 E 4, e = E E 4 F 3 +(H 1 H ) (E 3 F 3 E 4 F 4 )+F E 3 F 4, f = (E 1 F 3 F 4 E 1 F 4 F 3 ) (H 1 +H )(E 3 F 3 +E 4 F 4 F 3 E 3 F 4 E 4 ) (F 1 E 3 E 4 F 1 E 4 E 3 ), g = (E F 3 E 4 E E 4 F 3 ) +(H 1 H )(E 3 F 3 E 4 F 4 F 3 E 3 +F 4 E 4 ) +(F E 3 F 4 F F 4 E 3 ) h = (E 1 E F 3 F 3 E 1 F F 4 F 4 +F 1 F E 3 E 3 F 1 E E 4 E 4 ) (H 1 H )(E 1 F 3 F 4 +E 1 F 4 F 3 +F 1 E 3 E 4 +F 1 E 4 E 3 ) (H 1 +H )(F E 3 F 4 +F F 4 E 3 +E F 3 E 4 +E E 4 F 3 ) 1 (H 1 H )(H 1 +H )(E 3 F 3 +F 3 E 3 E 4 F 4 F 4 E 4 ), i = 1 E 3 E 4 F 3 F 4, j = E 3 E 4 F 3 F 4 E 4 E 3 F 3 F 4 +F 3 E 3 E 4 F 4 F 4 E 3 E 4 F 3. Proposition.. Let S and T be the following subsets of (S(g) (p)) K : S = 1 an bn3 c n4 n 1,n,n 3,n 4 N 0 } T = {1,D,d,e,f,g,h,i,j,Dd,De,Df,Dg,fg,Dh,dg}. Then the set S T of products of elements of S and T in the algebra S(g) (p) is a basis for (S(g) (p)) K. Proof. We first prove the linear independence of the set S T. Notice that it is enough to prove the linear independence of following sets: a) S, b) S {D} S {f} S {g} S {h}, c) S {d} S {Df} S {e} S {Dg} S {Dh} S {fg},
9 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 83 d) S {j} S {Dd} S {De} S {dg}. The rest of the independence then follows by considering just the second factors in the tensor products. i I a) Let i I λ i a n1,i 1 a n,i b n3,i c n4,i = 0. In the expansion of the summand a n1,i 1 a n,i b n3,i c n4,i we consider the terms without H 1 H, E 1, E 3, F 3. There is only one such term and it is (H 1 +H ) n1,i (4E F ) n,i (E 4 F 4 ) n3,i ( F 1 E E 4 )n4,i. We have λ i (H 1 +H ) n1,i (4E F ) n,i (E 4 F 4 ) n3,i ( F 1 E E4) n4,i = 0, i I that is, λ i 4 n,i ( ) n4,i (H 1 +H ) n1,i E n,i+n4,i E n3,i+n4,i 4 F n4,i 1 F n,i F n3,i 4 = 0. It follows from the Poincaré-Birkhoff-Witt theorem that λ i = 0 for all j {1,,l}. b) We consider summands of the form F 3. It is enough to show the linear independence of the set 1 an c bn3 n4 E 3 n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 ( E 1 F 4 (H 1 +H )E 3 ) n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 (E E 4 +(H 1 H )E 3 ) n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 (E 1 E F 3 (H 1 H )E 1 F 4 (H 1 +H )E E 4 1 (H 1 H )(H 1 +H )E 3 ) n 1,n,n 3,n 4 N 0 }. As in case a), we consider the terms without H 1 H, E 1, E 3, F 3. We have λ k (H 1 +H ) n 1,k (4E F ) n,k (E 4 F 4 ) n 3,k ( F 1 E E4 )n 4,k (E E 4 ) k K + l Lλ l (H 1 +H ) n 1,l (4E F ) n,l (E 4 F 4 ) n 3,l ( F 1 E E 4 )n 4,l ( (H 1 +H )E E 4 ) = 0. It follows that λ k (H 1 +H ) n 1,k (4E F ) n,k (E 4 F 4 ) n 3,k ( F 1 E E4 )n 4,k = 0 k K and λ l (H 1 +H ) n1,l+1 (4E F ) n,l (E 4 F 4 ) n 3,l ( F 1 E E4) n 4,l = 0. l L
10 84 A. PRLIĆ i I By the same arguments as in case a), we get λ k = 0 for all k K and λ l = 0 for all l L. Then we have i I j J λ i a n1,i 1 a n,i b n3,i c n4,i E 3 + λ j a n1,j 1 a n,j b n3,j c n4,j ( E 1 F 4 (H 1 +H )E 3 ) = 0 Now we consider the terms without H 1 +H, E, E 4, F 4. We have λ i (4E 1 F 1 ) n1,i (H 1 H ) n,i (E 3 F 3 ) n3,i (F 1 F E3) n4,i E 3 = 0, i I which implies λ i 4 n1,i n4,i (H 1 H ) n,i E n1,i 1 E n3,i+n4,i 3 F n1,i+n4,i 1 F n4,i F n3,i 3 = 0. Hence, λ i = 0 for all i I. So we have j J λ j a n1,j 1 a n,j b n3,j c n4,j = 0. Now a) implies that λ j = 0 for all j J. c) In this case we consider summands of the form E 3 F 3. It is enough to prove the linear independence of the set S ( (H 1 +H )) S ( E 1 F 4 F 3 (H 1 +H )E 3 F 3 (H 1 +H )E 3 F 3 F 1 E 3 E 4 ) S (H 1 H ) S (F 3 (E E 4 +(H 1 H )E 3 )+E 3 ((H 1 H )F 3 +F F 4 )) S (F 3 (E 1 E F 3 (H 1 H )E 1 F 4 (H 1 +H )E E 4 1 (H 1 H )(H 1 +H )E 3 )+E 3 ( F 1 F E 3 +(H 1 H )F 1 E 4 +(H 1 +H )F F (H 1 H )(H 1 +H )F 3 ) S ((H 1 +H )F 3 +F 1 E 4 )(E E 4 +(H 1 H )E 3 ) +(E 1 F 4 +(H 1 +H )E 3 )( (H 1 H )F 3 F F 4 )).
11 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 85 i I + j J Let λ i a n1,i 1 a n,i b n3,i c n4,i ( (H 1 +H )) λ j a n1,j 1 a n,j b n3,j c n4,j ( E 1 F 4 F 3 (H 1 +H )E 3 F 3 (H 1 +H )E 3 F 3 F 1 E 3 E 4 )+ k K + l L λ k a n 1,k 1 a n,k b n 3,k c n 4,k (H 1 H ) λ l a n 1,l 1 a n,l b n 3,l c n 4,l (F 3 (E E 4 +(H 1 H )E 3 ) +E 3 ((H 1 H )F 3 +F F 4 ))+ m M λ m a n1,m 1 a n,m b n3,m c n4,m (F 3 (E 1 E F 3 (H 1 H )E 1 F 4 (H 1 +H )E E 4 1 (H 1 H )(H 1 +H )E 3 ) +E 3 ( F 1 F E 3 +(H 1 H )F 1 E 4 +(H 1 +H )F F (H 1 H )(H 1 +H )F 3 ) + λ n a n1,n 1 a n,n b n3,n c n4,n ((H 1 +H )F 3 +F 1 E 4 )(E E 4 +(H 1 H )E 3 ) n N +(E 1 F 4 +(H 1 +H )E 3 )( (H 1 H )F 3 F F 4 )) = 0. We first consider the summands without H 1 H,E 1,E 3 and F 3 and we get, similarly as in case a), that λ i = 0 for all i I and λ n = 0 for all n N. Now we consider summands without H 1 +H,E,E 3 and F 3, and we get that λ k = 0 for all k K. Then we consider summands without H 1 +H,E,E 4 and F 4, and we get that λ l = 0 for all l L and λ m = 0 for all m M. From this we conclude that also λ j = 0 for all j J. d) This time we consider summands of the form E 4 F 3 F 4. It is enough to show linear independence of the set 1 an c bn3 n4 E 3 n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 ((H 1 +H )E 3 +E 1 F 4 ) n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 ((H 1 H )E 3 E E 4 ) n 1,n,n 3,n 4 N 0 } 1 an c bn3 n4 ( 4E 1 E F 3 +(H 1 H )E 1 F 4 +(H 1 +H )E E 4 + (H 1 +H )(H 1 H )E 3 ) n 1,n,n 3,n 4 N 0 }. We consider the summands without H 1 H, E 1, E 4, F 4 and we get λ i = 0 for all i I and λ j = 0 for all j J. Then we consider the summands without H 1 +H, E, E 3, F 3 and we get λ l = 0 for all l L and λ k = 0 for all k K.
12 86 A. PRLIĆ This finishes the proof of linear independence of the set S T. To prove that S T is also a spanning set, we consider the degrees of the elements of the set {c n4 } T. The degrees of these elements are respectively 4n 4,4n 4 +,4n 4 +3,4n 4 +3,4n 4 +3,4n 4 +3,4n 4 +4,4n 4 +4,4n n 4 +5,4n 4 +5,4n 4 +5,4n 4 +5,4n 4 +6,4n 4 +6,4n Considering the number of invariants in each degree of the set {c n4 n 4 Z + } T, we get the following table: degree number of invariants k, k 1 4 4k +1, k 1 4 4k +, k 1 4 4k +3, k 0 4 This finishes the proof, since we got the same table as in Proposition The set of generators for (U(g) C(p)) K Recall that the symmetrization map σ : S(g) U(g) given by σ(x 1 x x n ) = 1 x n! α(1) x α() x α(n), n N,x 1,,x n g α S n is an isomorphism of K modules. The Chevalley map τ : (p) C(p) given by τ(v 1 v n ) = 1 sgn(α)v α(1) v α() v α(n) n! α S n is also an isomorphism of K modules. Since for z 1,,z n g and α S n we have it follows that z 1 z n z α(1) z α(n) U n 1 (g), σ(z 1 z n ) = z 1 z n modulo U n 1 (g). Similarly, since for y 1,,y k p and α S k we have it follows that So we have y 1 y k sgn(α)y α(1) y α(k) C k 1 (p), τ(y 1 y k ) = y 1 y k modulo C k 1 (p). (3.1) σ(xy) = σ(x)σ(y) +U n+m 1 (g), x U n (g),y U m (g),
13 THE CENTRALIZER OF K IN U(g) C(p) FOR THE GROUP SO e(4,1) 87 and (3.) τ(zw) = τ(z)τ(w) +C k+l 1 (g), z C k (p),w C l (p). Let ρ = σ τ. Using (3.1), (3.) and Proposition. one shows by induction that the following lemma holds: Lemma 3.1. The algebra (U(g) C(p)) K is generated by elements ρ(a 1 ), ρ(a ), ρ(b), ρ(c), ρ(d), ρ(d), ρ(e), ρ(f), ρ(g), ρ(h), ρ(i) and ρ(j). We define the k-dirac operator D k U(g) C(p) mentioned in the introduction by D k = E 1 α(f 1 )+E α(f )+F 1 α(e )+F α(e ) +(H 1 H ) α(h 1 H )+(H 1 +H ) α(h 1 +H ). We can now reduce the set of generators for (U(g) C(p)) K given by Lemma 3.1, since we have ρ(b) = 1 D +D k, ρ(d) = D k 1 ) (D k ρ(i)+ρ(i) D k, ρ(e) = D k 1 ) (D k ρ(i)+ρ(i) D k, ρ(j) = 1 ( ) ρ(i) D D ρ(i), ρ(f) = 1 ( ) ρ(d) D D ρ(d) 3ρ(j), ρ(g) = 1 ( ) ρ(e) D D ρ(e) 3ρ(j), ρ(h) = 1 ( ρ(d) (ρ(g)+ 3 ) 4 D +ρ(e) ρ(c) = 1 4 (ρ(f) 3 ) D 3 ( ρ(f) ρ(g) D) ), 4 ( ( ρ(f) 3 ) D (ρ(g)+ 3 ) D + (ρ(g)+ 3 ) D (ρ(f) 3 ) D (ρ(a ) ρ(d)+ρ(d) ρ(a ))+(ρ(a 1 ) ρ(e)+ρ(e) ρ(a 1 )) 1 ( ) ρ(g) D D ρ(g) + 1 ( ) ρ(f) D D ρ(f) +(4 ρ(b)+5) (ρ(d) ρ(e))+6 ρ(b) 8 ρ(a 1 ) 8 ρ(a ) From this we conclude Theorem 3.. The algebra (U(g) C(p)) K is generated by the elements ρ(a 1 ), ρ(a ), ρ(i) C(p) K, D and D k. By the same argument as in [8, Corollary 4.5.], we get ).
14 88 A. PRLIĆ Corollary 3.3. The algebra (U(g) C(p)) K is a free module over U(g) K of rank 16. The last result is a special case of the more general result proved in Corollary [5, Corollary 1]. Acknowledgements. The author would like to thank Pavle Pandžić for valuable comments and suggestions which helped to improve the paper. References [1] M. Atiyah and W. Schmid, A geometric construction of the discrete series for semisimple Lie groups, Invent. Math. 4 (1977), 1 6. [] Harish-Chandra, Representations of semisimple Lie groups. II, Trans. Amer. Math. Soc. 76 (1954), [3] J.-S. Huang and P. Pandžić, Dirac cohomology, unitary representations and a proof of a conjecture of Vogan, J. Amer. Math. Soc. 15 (00), [4] J.-S. Huang and P. Pandžić, Dirac Operators in Representation Theory, Mathematics: Theory and Applications, Birkhäuser, Boston, MA 006. [5] H. Kraljević, K structure of U(g) for su(n, 1) and so(n, 1), arxiv: v1. [6] P. Pandžić and D. Renard, Dirac induction for Harish-Chandra modules, J. Lie Theory 0 (010), [7] R. Parthasarathy, Dirac operator and the discrete series, Ann. of Math. 96 (197), [8] A. Prlić, K invariants in the algebra U(g) C(p) for the group SU(,1), Glas. Mat. Ser. III 50 (015), [9] A. Prlić, Algebraic Dirac induction for nonholomorphic Discrete Series of SU(, 1), J. Lie Theory 6 (016), no. 3, [10] D.A. Vogan, Jr., Dirac operators and unitary representations, 3 talks at MIT Lie groups seminar, Fall A. Prlić Department of Mathematics University of Zagreb Zagreb Croatia anaprlic@math.hr Received:
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