Lecture 14 - Isomorphism Theorem of Harish-Chandra

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1 Lecture 14 - Isomorphsm Theorem of Harsh-Chandra March 11, 2013 Ths lectures shall be focused on central characters and what they can tell us about the unversal envelopng algebra of a semsmple Le algebra. 1 Invarant Polynomals In our dscusson last tme of formal characters, we showed that the formal character of a representaton s W-nvarant, and further that any W-nvarant character was a sum of characters of representatons. The same statement holds true for central characters. Obvously we must frst determne how a central character s obtaned from a representaton. Ths leads us to the study of nvarant polynomals. A polynomal on g s an element of S(g ). Gven a monomal P = g 1... g k, t acts wth homogenety k on g by (g 1... g k )(X) = g 1 (X)... g k (X). If P S(g ) s a homogeneous polynomal (of order k, say), we can polarze (symmetrze) t to obtan a functon of k varables. If P = g 1... g k, t acts on k g by P (X 1,..., X k ) = 1 g 1(X π1 )... gk(x πk ) (1) for X 1,..., X n g. Note that P (X) = P (X,..., X). Notce that many non-trval k- homogeneous polynomals P have homegety k acton beng precsely zero: P (X) = 0 for all X, although P 0. Nevertheless, polynomals wth non-zero homogenety-k actons play a dstngushed role n or study. Gven any homogenety-k polynomal P S(g ) and g g, we have the contragredent acton of g on P, gven by (g.p )(X 1,..., X k ) = P (X 1,..., [g, X ],..., X k ) (2) A polynomal P for whch g.p = 0, all g g s called nvarant. The algebra of nvarant polynomals s denoted S(g ) g. 1

2 Let τ : g V be a fnte-dmensonal representaton, and consder the homogenety k map Polarzng, we obtan P τ,k (X) = tr (τ(x)) k. (3) P τ,k (X 1,..., X k ) = 1 tr (τ(x π1 )... τ(x πk )) (4) for X 1,..., X k g. Breakng apart ths sum, consder a sngle monomal F, where say F (X 1,..., X k ) = tr(τ(x 1 )... τ(x k )). (5) Wth τ([g, X]) = τ(g)τ(x) τ(x)τ(g) we obtan F (X 1,..., [g, X ], X k ) = tr(τ(x 1 )... τ(g)τ(x )τ(x k )) tr(τ(x 1 )... τ(x )τ(g)τ(x k )) (6) = tr(τ(g)τ(x 1 )... τ(x k )) tr(τ(x 1 )... τ(x k )τ(g)). whch equals zero by the cyclc property of traces. Thus polynomals of the form P τ,k are nvarant. Now an nvarant polynomal on g restrcts to h, where n fact t s W-nvarant; the algebra of W-nvarant polynomals on h s denoted S(h ) W. To see ths, smply note that P τ s nvarant under σ α = e adx e ady e adx where span C {x, y, h} s the sl 2 -subspace of g correspondng to the root α. We have already see that the contragredent acton of σ α on h s the Weyl reflecton n α. Thus we have a restrcton map ρ : S(g ) g S(h ) W. (7) Proposton 1.1 The restrcton map ρ s an somorphsm. Indeed S(h ) W s spanned by polynomals of the form P τ,k as τ ranges over all fnte dmensonal rreducble representatons and k over all non-negatve ntegers. Pf. If λ s a weght, note that λ k S(h ). We can average λ k over the acton of W to obtan Aλ k S(h ) W. Further, we can assume λ s domnant. Let M λ be the set of domnant weghts λ so that λ < λ. Now f τ has hghest weght λ, then P τ,k s a polynomal n terms of Aλ k and lower weghts. In fact P τ,k Aλ k s a combnaton of the A(λ ) k for λ M λ. Now an nducton argument can proceed. We make a fnal note before movng on. The kllng form κ produces an somorphsm g g, whch extends to S(g ) S(g). Due to the assocatvty of κ, we have n fact κ : S(g ) W S(g) W 2

3 Now we have a vector-space somorphsm whch, among other possbltes, can be gven by B : S(g) U(g) (8) X 1 X k 1 X π1... X πk (9) Now f P S(g) g, then n fact B(P ) s central. Ths s clear from the commutator relaton B([g, X 1 X k ]) = [g, B(X 1 X k )] (10) and that the left sde equals zero. Lastly, f a monomal X 1... X k U(g) s central, t s g-nvarant, meanng 0 = [g, X 1... X k ] = [g, X 1 ]X 2... X k X 1... X k 1 [g, X k ] = B ([g, X 1 ], X 2,..., X k ) B(X 1,..., X k 1 [g, X k ]) mod S k 1 (g) (11) Smlarly for a sum of monomals n U(g). The lower-order terms are W-nvarant as well, so we see that s a vector space somorphsm. B : S(g) g U(g) g Z (12) 2 The Harsh-Chandra map 2.1 Evaluaton of characters Gven a central character χ, how do we determne ts value at a pont λ L? As usual, set n + = g α α Φ + n = (13) α Φ g α Consder the decomposton of the envelopng algebra Ug = Uh ( U(g)n + + n U(g) ) (14) whch we know s possble due to Poncare-Brkhoff-Wtt. Notce that U(g)n + and n U(g) are not dsjont. Now f z Z s central, then n fact z Uh ( U(g)n + n U(g) ). (15) 3

4 Defne the map γ to be the projecton onto the frst factor. Note that h s commutatve, so U(h) = S(h). One easly sees γ s an algebra homomorphsm. We therefore have the algebra homomorphsm γ : Z S(h). (16) How exactly does ths help? Consder the rreducble hghest-weght module V Λ of hghest weght Λ, and suppose v + s a hghest-weght module. Snce n + klls v +, we have z.v + = γ(z).v + = Λ(γ(z))v + (17) where the weght Λ acts on the polynomal γ(z) n the obvous way. Consderng S(h) to be the polynomal algebra on h, we can wrte Λ(γ(z)) = γ(z)(λ), whch s probably more natural. We have shown that χ Λ (z) = γ(z)(λ). (18) 2.2 Twsted Chracters Ths s all very nce, but, as t happens, unsymmetrcal. Consder the twsted character χ Λ defned by where δ s the Weyl vector. χ Λ = χ Λ δ (19) Consder also the twst map τ : S(h) S(h) gven on h U(h) by τ(h) = h δ(h)1. Defne the Harsh-Chandra map γ : Z S(h) by Then we have χ Λ (z) = γ(z)(λ). γ = τ γ. (20) χ Λ (z) = χ Λ δ (z) = γ(z)(λ δ) = τ γ(z)(λ) = γ(z)(λ) (21) Proposton 2.1 The twsted character s Weyl-nvarant: χ σλ = χ Λ for any σ W. Pf. Consder the Verma module V Λ δ. Recall that ths s an nfnte dmensonal module wth a maxmal submodule so that the quotent s the rreducble module of hghest weght Λ δ. Let v + be a hghest weght vector, so that γ(z)(λ)v + = χ Λ (z)v + = χ Λ δ (z)v + = λ(z).v + (22) 4

5 Now consder the sl 2 subalgebra {x, y, h} correspondng to the smple root α, and let r be the Dynkn coeffcent r = α, Λ. Note that the th Dynkn coeffcent of Λ δ s r 1. Therefore x.y r.v + = 0 (23) and because [x j, y] = 0 for any other x j n the root-space g αj (smple root α j ), we have that x j.y r.v + = y r.x j.v + = 0. Thus v + s klled by n +. Ths means that z.y r.v + = γ(z).y r.v + = γ(z)(λ r α δ)y r.z = γ(z)(σ Λ)y r.z (24) where σ s the Weyl reflecton n α. But z.y r.v + = y r.z.v + = y r.γ(z).v + = γ(z)(λ)v + (25) Therefore γ(z)(σ Λ) = γ(z)λ, whch means χ Λ = χ σλ. (26) Corollary 2.2 The Harsh-Chandra map γ s a homomorphsm γ : Z S(h) W. 3 The Harsh-Chandra Isomorphsm To see the bjectvty of γ, frst recall that we have the followng vector space somorphsms and vector space maps S(h) W U(h) W S(g) g U(g) g (27) S(h) W S(g) g Z γ U(h). (28) The un-twsted map γ : Z U(h) s a projecton, and whle not an algebra homomorphsm, t has no kernel on elements of U(h) W. Snce τ s a vector space somorphsm and γ = σ γ s an algebra homomorphsm on U(h) W, we get a composton of vector space somorphsms so that s an algebra somorphsm. S(h) W S(g) g Z γ U(h) W. (29) U(g) g Z γ U(h) W. (30) 5