GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n

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1 GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n KANG LU FINITE DIMENSIONAL REPRESENTATIONS OF gl n Let e j,, j =,, n denote the standard bass of the general lnear Le algebra gl n over the feld of complex numbers The subalgebra gl n s spanned by the bass elements e j wth, j =,, n Denote by h = h n the dagonal Cartan subalgebra n gl n The elements e,, e nn form a bass of h Fnte-dmensonal rreducble representatons of gl n are n a one-to-one correspondence wth n-tuples of complex numbers λ = (λ,, λ n ) such that () λ λ + Z + for =,, n Such an n-tuple λ s called the hghest weght of the correspondng representaton whch we shall denote by L(λ) It contans unque, up to a multple, nonzero vector v + (the hghest vector) such that e v + = λ v + for =,, n and e j v + = 0 for < j n BGG RESOLUTION Last semnar, we talked about the BGG resoluton Now let us recall the BGG resoluton and deduce the Weyl character formula from t Denote by M λ the Verma module wth hghest weght λ Let W be the Weyl group and s be the smple refectons, =,, rank g and ρ s the Weyl vector Defne the shfted acton of the Weyl group on h by w λ = w(λ + ρ) ρ Denote P + and Φ + the set of all domnant ntegral weghts and the set of all postve roots of g Theorem ([BGG]) Let λ P + Then there exsts a lon exact sequence () 0 M w0 λ M s λ M λ L(λ) 0 w W,l(w)=k M w λ where l(w) s the length of an element w W, and w 0 s the longest element of the Weyl group Now we are gong to deduce the Weyl character formula from Theorem Note that the character of M λ s ch(m λ ) = q λ α Φ + q α = qλ ( + q α + q α + ) α Φ + Snce character s addtve, we can take the alternatng sum wth respect to character n () It follows that () ch(l(λ)) = ( ) w ch(m w λ ) = w W( ) w q w λ w W α Φ +( q α ) Let λ = 0, we have ( q α ) = α Φ + w W ( ) w q wρ ρ

2 KANG LU It follows that (3) ch(l(λ)) = w W( ) w q w(λ+ρ) w W ( ) w q wρ, whch s the celebrated Weyl character formula Snce we are nterested n the case g = gl n We know that W s the symmetrc group S n Also, we have ρ = (n,,, 0) If we wrte x for q (0,,,,0), where (0,,,, 0) has all components zero except for the -th component Then we can rewrte (3) as ch(l(λ)) = det ( x λ ) +n j,j n det ( x n j ) = S λ (x,, x n ),,j n where S λ s exactly the Schur polynomal Our next goal s to deduce the Branchng rule for the reducton gl n gl n from Schur polynomal Frst we need the followng proposton Proposton We have where µ runs over all parttons such that S λ (x,, x n, x n = ) = S µ (x,, x n ), µ λ µ λ µ λ n µ n λ n Proof We prove t only for the case n = 3 And from ths we can defntely get the dea to prove t for general n Let λ = (λ, λ, λ 3 ), λ Z + We have x λ + x λ + x λ 3 S λ (x, x, x 3 ) = x λ + (x x )(x x 3 )(x x 3 ) x λ + x λ 3 x λ + 3 x λ + 3 x λ 3 3 Settng x 3 =, we have x λ + x λ + x λ 3 S λ (x, x, ) = (x x )(x )(x ) x λ + x λ + x λ 3 Subtractng the nd column from the st, and then 3rd column from the nd, we get x λ + S λ (x, x, ) = x λ + x λ + x λ 3+ (x x )(x )(x ) x λ + x λ + x λ + x λ 3+ We see that the st row can be dvded by x, and the second one by x Note that for a, b Z + such that a > b: x a x b x = x a c:b c<a We see that by dvdng by x n the frst row n the frst place we get a sum whch can be wrtten as x µ+ = x µ+ λ + µ +<λ + λ µ λ

3 GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n 3 We choose to call the summaton ndex µ + so that we wll have the desred result Consderng three other matrx elements n a smlar way, we conclude the proof Corollary 3 We have where µ runs over all parttons such that S λ (x,, x n, x n ) = S µ (x,, x n )x λ µ n, µ λ µ λ µ λ n µ n λ n From Proposton, one can easly get the branchng rule for the reducton gl n gl n Theorem 4 The restrcton of L(λ) to the subalgebra gl n s somorphc to the drect sum of parwse nequvalent rreducble representatons L(λ) gln = summed over the hghest weghts µ satsfyng the betweenness condtons (4) λ µ Z + and µ λ + Z + for =,, n µ L µ 3 GELFAND-TSETLIN BASIS The next two sectons are coped from [M] The subsequent applcatons of the branchng rule to the subalgebras of the chan gl gl gl n gl n yeld a parameterzaton of bass vectors n L(λ) by the combnatoral objects called the Gelfand-Tsetln patterns Such a pattern Λ (assocate wth λ) s an array of row vectors λ n λ n λ n,n λ nn λ n, λ n,n λ λ λ where the upper row concdes wth λ and the followng condtons hold λ k λ k, Z +, λ k, λ k,+ Z +, =,, k for each k =,, n The Gelfand-Tsetln bass of L(λ) s provded by the followng theorem Let us set l k = λ k + Theorem 3 ([GT]) There exsts a bass {v Λ } n L(λ) parametrzed by all patterns Λ such that the acton of generators of gl n s gven by the formulas ( k k (3) e kk v Λ = λ k λ k, )v Λ, = = (3) e k,k+ v Λ = k = (l k l k+, ) (l k l k+,k+ ) (l k l k ) (l k l kk ) v Λ+δ k,

4 4 KANG LU (33) e k+,k v Λ = k (l k l k, ) (l k l k,k ) (l = k l k ) (l k l kk ) v Λ δ k The arrays Λ ± δ k are obtaned from Λ by replacng λ k by λ k ± It s supposed that v Λ = 0 f the array Λ s not a pattern; the symbol ndcates that the zero factor n the denomnator s skpped The Gelfand-Tsetln subalgebra of U(gl n ) s the subalgebra H n generated by the centers Z(gl ),, Z(gl n ) of the subalgebra chan U(gl ) U(gl ) U(gl n ) nduced by gl gl gl n Theorem 3 ([V]) The Gelfand-Tsetln subalgebra H n of U(gl n ) s maxmal commutatve 4 CONSTRUCTION OF THE BASIS: LOWERING AND RAISING OPERATORS For each =,, n ntroduce the followng elements of the unversal envelopng algebra U(gl n ) (4) z n = e e e s s e s n(h h j ) (h h jr ), > > > s (4) z n = e e e s s e ns (h h j ) (h h jr ), < < < s n where s runs over nonnegatve ntegers, h = e + and {j,, j r } s the complementary subset to {,, s } n the set {,, } or { +,, n }, respectvely For nstance, z 3 = e 3, z 3 = e 3 (e e ) + e e 3, z 3 = e 3, z 3 = e 3 (e e + ) + e e 3 Consder now the rreducble fnte-dmensonal representaton L(λ) of gl n wth the hghest weght λ = (λ,, λ n ) and the hghest vector v + Denote by L(λ) + the subspace of gl n -sngular vectors n L(λ): L(λ) + = {v L(λ) e j v = 0, < j n} Gven a gl n -weght µ = (µ,, µ n ) we denote by L(λ) + µ the correspondng weght subspace n L(λ) + : L(λ) + µ = {v L(λ) + e v = µ v, =,, n } The man property of the elements z n and z n s descrbed by the followng lemma Lemma 4 Let v L(λ) + µ Then for any =,, n we have z n v L(λ) + µ+δ and z n v L(λ) + µ δ, where the weght µ ± δ s obtaned from µ by replacng µ wth µ ± Ths result allows us to regard the elements z n and z n as operators n the space L(λ) + They are called the rasng and lowerng operators, respectvely By the branchng rule (Theorem 4) the space L(λ) + µ s one-dmensonal f the condtons (4) hold and t s zero otherwse

5 GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n 5 Lemma 4 Suppose that µ satsfes the betweenness condtons (4) Then the vector v µ = z λ µ n z λ n µ n n,n v + s nonzero Moreover, the space L(λ) + µ s spanned by v µ The U(gl n )-span of each nonzero vector v µ s a gl n -module somorphc to L (µ) Iteratng the constructon of the vectors v µ for each par of Le algebras gl k gl k we shall be able to get a bass n the entre space Λ(λ) Theorem 43 The bass vector v Λ of Theorem 3 can be gven by the formula (43) v Λ = ( λ z k, λ k, k k=,,n z λ k,k λ k,k ) k,k v +, where the factors n the product are ordered n accordance wth ncrease of the ndces Instead of gvng a proof frst, I wll check n Secton 5 that the constructon n Theorem 43 does satsfy the formulas n Theorem 3 for some smple cases The proof wll be gven n Secton 6 5 EXAMPLES In ths secton, we are gong to check that the vectors constructed from Theorem 43 does satsfy the formulas n Theorem 3 for n =, 3 The case for gl 3 s not complete 5 The case of gl Let us fx the hghest weght λ = (λ, λ ), then λ = λ, λ = λ Denote by µ = λ, then each pattern s determned by k = λ µ for k = 0,, λ λ By (43), the vector corresponds to k s gven by v k = e λ µ v + = e k v+ Now t s trval that e v k = v k+, whch s exactly (33)(note that µ corresponds to k + ) Snce for any vector v of weght (µ, µ ), the vector e v has weght (µ, µ + ) Hence v k = e k v+ has weght (λ k, λ + k) Therefore, we have e v k = (λ k)v k = µv k = λ v k, e v k = (λ + k)v k = (λ + λ λ )v k Formula (3) s checked Fnally, to check (3), we need the followng useful formula n U(gl ) (5) [e, e k ] = kek (k e + e ) Now (3) holds because of e v k =e e k v+ = e k e v + + [e, e k ]v+ = ke k (k e + e )v + = (λ µ)(λ µ λ + λ )v k = (λ λ )(λ λ + )v k = (l l )(l l )v k

6 6 KANG LU 5 The case of gl 3 Ths case s more nvolved Frst we fx the hghest weght λ = (λ, λ, λ 3 ) = (λ 3, λ 3, λ 33 ) Then each pattern s unque determned by λ, λ and λ We denote the vector constructed n Theorem 43 correspondng to ths patter by (λ, λ ; λ ), then It s easy to see (λ, λ ; λ ) = e λ λ (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + e (λ, λ ; λ ) =e λ λ + (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + =(λ, λ ; λ ) Snce for any vector v of weght (µ, µ, µ 3 ), the vectors e v, e 3 v and e 3 v have weght (µ, µ +, µ 3 ), (µ, µ, µ 3 + ) and (µ, µ, µ 3 + ) respectvely Moreover, e e + acts on v by scalar product One has that the weght of (λ, λ ; λ ) s Hence we get formulas (3) (λ, λ + λ λ, λ 3 + λ 3 + λ 33 λ λ ) e (λ, λ ; λ ) = λ (λ, λ ; λ ), e (λ, λ ; λ ) = (λ + λ λ )(λ, λ ; λ ), e 33 (λ, λ ; λ ) = (λ 3 + λ 3 + λ 33 λ λ )(λ, λ ; λ ) Now let us consder e (λ, λ ; λ ) By (5), we have e (λ, λ ; λ ) =e e λ λ (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + =e λ λ e (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + (λ λ )e λ λ (λ λ e + e )(e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + =e λ λ e (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + (λ λ )(λ λ )(λ, λ ; λ + ) For the last equalty, we use the fact that (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + has weght (λ, λ, λ 3 + λ 3 + λ 33 λ λ ) Now t s enough to check that (5) e λ λ e (e 3 (e e + ) + e e 3 ) λ 3 λ e λ 3 λ 3 v + = 0 We compute e (e 3 (e e + ) + e e 3 ) frst We have e (e 3 (e e + ) + e e 3 ) =(e 3 e e 3 )(e e + ) + e e e 3 + (e e )e 3 =e 3 e (e e ) + e 3 e e 3 (e e ) e 3 + e e 3 e + (e e )e 3 =e 3 (e e )e e 3 e + e 3 e + e 3 e 3 + e e 3 e =(e 3 (e e ) + e e 3 )e Now one has (5) snce v + s a sngular vector and [e, e 3 ] = 0

7 GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n 7 6 PROOF OF THEOREM 3 It suffces to show that wth the formulas gven by (3), (3) and (33) the vector space Λ v Λ dose gve us a smple representaton of gl n wth hghest weght λ Frst we need to check t gves us a representaton Let A be the Cartan matrx of sl n, and set e = e,+, f = e +, and h = e e +,+ for every =,, n Then the Serre relatons for sl n are [h, e j ] = a j e j, [h, f j ] = a j f j, [e, f j ] = δ j h, [h, h j ] = 0, Ad a j e e j = Ad a j f f j = 0, for every, j =,, n Hence t s enough to check that [h, e j ]v Λ = a j e j v Λ, [h, f j ]v Λ = a j f j v Λ, [e, f j ]v Λ = δ j h v Λ, [h, h j ]v Λ = 0, Ad a j e e j v Λ = Ad a j f f j v Λ = 0, for every, j =,, n and Gelfand-Tsetln pattern Λ To our convenence, for each gven pattern Λ assocated to λ, we wrte Λ (k) := = k λ k, and (6) E Λ k := (l k l k+, ) (l k l k+,k+ ) (l k l k ) (l k l kk ), FΛ k := (l k l k, ) (l k l k,k ) (l k l k ) (l k l kk ) for every k =,, n and =,, k Because h and h j dagonally act on Λ v Λ, hence [h, h j ]v Λ = 0 To prove [h, e j ]v Λ = a j e j v Λ, t suffces to show (6) [e, e j,j+ ]v Λ = δ,k e,j+ v Λ δ j+, e j v Λ Snce the patterns nvolved n e j,j+ v Λ are gven by ncreasng one of the entres n the j-th row of Λ by and the acton of e on v Λ only nvolves the sum of -th and ( )- th row of Λ, one can easly check (6) Hence we have [h, e j ]v Λ = a j e j v Λ and smlarly [h, f j ]v Λ = a j f j v Λ Now let us prove [e, f j ]v Λ = δ j h v Λ Wth the notaton n (6), we have e,+ e j+,j v Λ = e,+ e j+,j e,+ v Λ = e j,j+ j k= s= F Λ jk v Λ δ jk = E Λ s v Λ+δ s = j k= j k= s= s= F Λ jk EΛ δ jk s v Λ δjk +δ s, F Λ+δ s jk E Λ s v Λ δ jk +δ s Snce Fjk Λ only nvolves the j-th and (j )-th rows of Λ whle EΛ s only nvolves the -th and ( + )-th rows, we have (63) F Λ jk EΛ δ jk s for = j and = j If = j, then we have = F Λ+δ s jk Es Λ F Λ jk F Λ+δ s jk = l jk l j,s l jk l j,s, E Λ s E Λ δ jk s = l s l +,k l s l +,k + = l jk l j,s l jk l j,s,

8 8 KANG LU e (63) holds If = j and s = k, one can check (63) drectly Therefore [e, f j ]v Λ = δ j h v Λ s equvalent to ( (lk + l, ) (l k + l, ) (l k l +, ) (l k l +,+ ) (64) (l k + l ) (l k + l ) (l k l ) (l k l ) k= (l k l, ) (l k l, ) (l ) k l +, ) (l k l +,+ ) (l k l ) (l k l ) (l k l ) (l k l ) =(l + + l ) (l, + + l, ) (l +, + + l +,+ ) Denote the rght hand sde of (64) by M, then M s a symmetrc ratonal functon wth respect to l,, l Moreover, t has at most smple poles Note that a symmetrc functon cannot have smple poles, t follows that M s a polynomal n l,, l By consderng the asymptotc behavor, we know that M l as l tends to nfnty By symmetry, t follows that M (l + + l ) s ndependent of l,, l It s easy to see l +, s a lnear term n M Let us consder the coeffcent of l +, Ths coeffcent s k= ( (lk + l, ) (l k + l, ) (l k + l ) (l k + l ) (l k l, ) (l k l, ) (l k l ) (l k l ) (l k l +, ) (l k l +,+ ) (l k l ) (l k l ) (l k l +, ) (l k l +,+ ) (l k l ) (l k l ) Let ϕ(x) = (x + l, ) (x + l, )(x l +, ) (x l +,+ ), then deg ϕ = Set a k = l k and a k+ = l k for every k =,,, we have ϕ(a k ) (a j =k k a j ) =(l k + l, ) (l k + l, ) (l k + l ) (l k + l ) (l k l +, ) (l k l +,+ ) (l k l ) (l k l ), ϕ(a k+ ) (a j =k+ k a j ) = (l k l, ) (l k l, ) (l k l ) (l k l ) By Lagrange Interpolatng Polynomal, we have ϕ(a k ) k= j =k Consder the leadng coeffcent, t follows that ϕ(a k ) k= j =k (l k l +, ) (l k l +,+ ) (l k l ) (l k l ) (x a j ) (a k a j ) = ϕ(x) (a k a j ) = Hence the coeffcent of l +, s equal to By symmetry all the coeffcent of l +,k are equal to for every k =,, + Smlarly, all the coeffcent of l,k are equal to )

9 GELFAND-TSETLIN BASIS FOR THE REPRESENTATIONS OF gl n 9 for every k =,, Now t s easy to see M ( (l + + l ) (l, + + l, ) (l +, + + l +,+ ) ) s a constant Let l sk = k for every s =,, + and k =,, s, we can get ths constant, whch s exactly Hence we proved [e, f j ]v Λ = δ j h v Λ Remark 6 A smple proof of (64), as gven by Prof Tarasov, s the followng Let p j (u) = (u l j ) (u l jj ), then the left hand sde of (64) s equal to k= Res u=lk p (u)p + (u ) p (u)p (u ) and the rght hand sde of (64) s equal to Hence they must equal p Res (u)p + (u ) u= p (u)p (u ) p Res (u)p + (u ) u=lk + p k= (u)p (u ) Smlar to the frst half of the proof of [e, f j ]v Λ = δ j h v Λ, we can prove [e,+, e j,j+ ]v Λ = 0 f j = 0 ( = j s trval) Let us consder [e,+, [e,+, e +,+ ]]v Λ Frst, we have [x, [x, y]] = x y + yx xyx and e,+ e +,+v Λ = + k= s= e,+ e +,+ e,+ v Λ = e +,+ e,+ v Λ = + k= t= + k= s= E+,k Λ EΛ+δ +,k s E Λ+δ +,k+δ s t v Λ+δ+,k +δ s +δ t, s= t= t= E Λ+δ s +,k EΛ s EΛ+δ +,k+δ s t v Λ+δ+,k +δ s +δ t, E Λ+δ s+δ t +,k Es Λ EΛ+δ s t v Λ+δ+,k +δ s +δ t To prove [e,+, [e,+, e +,+ ]]v Λ = 0, t suffces to show the coeffcent of v Λ+δ+,k +δ s +δ t s zero If s = t, after takng the common factors, the coeffcent s a multple of (l s l +,k )(l t l +,k ) (l s l t )(l t l s ) + (l t l +,k )(l s l +,k ) (l t l s )(l s l t ) (l s l +,k )(l t l+,k ) (l s l t )(l t l s ) (l t l +,k )(l s l+,k ) (l t l s )(l s l t ) + (l s l +,k )(l t l +,k ) (l s l t )(l t l s ) + (l t l +,k )(l s l +,k ) (l t l s )(l s l t ) It s easy to see the frst lne of ths expresson s equal to and the second l s l t Hence the coeffcent s zero Smlarly, f s = t, then t suffces to check l t l s (l s l +,k )(l s l +,k ) (l s l +,k )(l s l +,k ) + (l s l +,k )(l s + l +,k ) = 0, whch s obvous Smlarly, we can prove Ad a j e e j v Λ = Ad a j f f j v Λ = 0 Let Λ 0 be the Gelfand-Testln pattern assocated to λ defned by λ k = λ for every k =,, n and =,, k Now by the formulas (3) and (3), we have e v L0 = λ v Λ0, e,+ v Λ0 = 0

10 0 KANG LU Hence v Λ0 s a sngular vector wth weght λ Note that by defnton, Λ v Λ s fntedmensonal Moreover, by Branchng rule (Theorem 4), the dmenson of L(λ) s equal to the number of Gelfand-Testln patterns assocated to λ It follows that L(λ) = Λ v Λ, and the proof of Theorem 3 s complete 7 FURTHER TOPICS We mght consder how the Gelfand-Tsetln algebra acts on ths bass To be contnued REFERENCES [BGG] [GT] [M] [V] IN Bernsten, I M Gelfand, S I Gelfand, Structure of representatons generated by vectors of hghest weght, Functonal Analyss and Its Applcatons 5 (97), no, pp -9 I M Gelfand and M L Tsetln, Fnte-dmensonal representatons of the group of unmodular matrces (n Russan), Doklady Akad Nauk SSSR (NS) 7 (950), Englsh translaton n: I M Gelfand Collected papers, vol II Sprnger-Verlag, Berln, 987, pp A Molev, Gelfand-Tsetln bases for classcal Le algebras, -65, mathrt/089v È B Vnberg, On certan commutatve subalgebras of a unversal envelopng algebra, Mathematcs of the USSR-Izvestya 36 (99), no, pp -3 KL: DEPARTMENT OF MATHEMATICS, INDIANA UNIVERSITY-PURDUE UNIVERSITY-INDIANAPOLIS, 40 NBLACKFORD ST, LD 70, INDIANAPOLIS, IN 460, USA E-mal address: lukang@upuedu