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Lecture 25 —Dimensions and Characters of Semisimple Lie Algebras Prof

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Lecture 25 —Dimensions and Characters of Semisimple Lie Algebras Prof 18.745 Introduction to Lie Algebras Dec 9, 2010 Lecture 25 |Dimensions and Characters of Semisimple Lie Algebras Prof. Victor Kac Scribe: Wenzhe Wei Let g be as in the last lecture - finite dimesional semisimple lie algebra. Let h be a Cartan subalgebra, and Π = fα1; : : : ; αrg ⊂ ∆+ ⊂ ∆, as before, a system of simple roots. We have the triangular decomposition g = n− + h + n+, b = h+ + n+, with b - a Borel subalgebra, and [b; b] = n+. Let (·; ·) 1 be a nondegenerate invariant symmetric bilinear form on g, let ρ = 2 Σα2∆+ α. Let fEi;Hi;Fig be −1 2ν (αi) the Chevalley generators satisfying Hi = , Ei 2 gα, Fi 2 g−α and such that < Ei;Hi;Fi > (αi,αi) ∗ form the standard basis of sl2(F). Define the subset P+ = fλ 2 h jλ(Hi) 2 Z+ for all i = 1; : : : rg Theorem 25.1. (Cartan) The g-modules fL(Λ)gΛ2P+ are, up to isomorphism, all irreducible finite- dimensional g-modules. (Recall from previous lectures that L(Λ) is the irreducible heigest weight module with heighest weight λ.) Q (Λ+ρ,α) Theorem 25.2. (H.Weyl Dimension formula) If Λ 2 P+, then dimL(Λ) = (ρ,α) α2∆+ ∗ Example 25.1. g = sl2(F) =< E; F; H >. Then all Verma modules M(Λ), where Λ 2 h = F ∗ J since h = FH, have basis F vΛ,j 2 Z+. By the key sl2 lemma, M(Λ) is irreducible unless Λ+1 Λ+1 Λ = Λ(H) 2 Z+ In the latter case (by the same lemma) EF vΛ = 0, hence F vΛ is a singular Λ+1 j vector. So M(Λ) is not irreducible. But L(Λ) = M(Λ)=U(sl2(F)F vΛ is irreducible since F vΛ, 0 ≤ j ≤ m = Λ(H) are independent. So by fundamental sl2 lemma, dim(L(Λ)) = m + 1 Proof of Theorem 25.1. Recall that σi =< Ei;Fi;Gi >' sl2(F ) and vΛ 2 L(Λ) satisfies Ei(vΛ) = 0;Hi(vΛ) = Λ(Hi)vΛ. Hence by fundamental sl2 lemma, Λ(Hi) 2 Z+. So dimL(Λ) < 1 ) Λ 2 P+. Coversely, if Λ 2 P+, then by Theorem 2, dim(Λ) < 1. 1 P Lemma 25.1. Recall ρ = 2 α. Then ρ(Hi) = 1 α2∆+ Proof. Consider the reflection from the Weyl group corresponding to αi: 1 1 X r ρ = r ( α + α) = ρ − α αi αi 2 i 2 i α2∆nfαig But rαi (λ) = λ − λ(Hi)αi hence ρ(Hi) = 1. Example 25.2. g = sl2, Then Λ(H) = m means that Λ = mρ. So (m + 1)(α; α) dimL(Λ) = = m + 1 (α; α) 2 −1 Example 25.3. g = sl . We have Π = fα ; α g,(α ; α ) = , ρ = α + α , ρ(α ) = 1, 3 1 2 i j −1 2 1 2 i Λ = m1Λ1 + m2Λ2, where (Λi; αj) = δij. By Cartan's theorem, dimL(Λ) < 1 iff m1; m2 2 Z+. We 1 compute (Λ + ρ, α1) = m1 + 1, (Λ + ρ, α2) = m2 + 1, and (Λ + ρ, α1 + α2) = m1 + m2 + 1, so we have (m + 1)(m + 1)(m + m + 1) dimL(Λ) = 1 2 1 2 2 In general, we may write Λ = ΣikiΛi, where Λi(Hj) = δij. Then dimL(Λ) < 1 iff ki 2 Z+. These m1 m2 ki are called labels of the heighest weight. They are depicted on the Dynkin diagram: We'll deduce Weyl's Dimensional formula from the Weyl character formula i i L Definition 25.1. Let M be a g-module which is h-diagonalizable, let M = Mλ, then λ2h∗ X λ ch(M) = (dimMλ)e λ Here eλeµ = eλ+µ; e0 = 1 Q −α Theorem 25.3. (Weyl Character Formula) Let R = (1 − e ). if Λ 2 P+ then α2∆+ X eρRchL(Λ) = (detw)ew(Λ+ρ) (∗) w2W We'll first derive Theorem2 from Theorem3, then prove Theorem3. Given µ 2 h∗, consider the following linear map from linear comibnation of the eλ to functions of λ t(λ,µ) t that Fµ(e ) = e .Apply Fµ to both sides of (*), we have t(ρ,ρ) Y −t(ρ,α) X t(w(Λ+ρ);w−1(ρ)) e (1 − e )FρchL(Λ) = (detw)e w2W X = (detw)et(w(Λ+ρ);w(ρ)) w2W X w(ρ) = FΛ+ρ (detw)e w2W Now we can value the sum.Note that L(0) is the trivial 1-dim g-module, hence chL(0) = 1. There- fore (*) implies X eρR = (detw)ew(ρ) (∗∗) w2W The above equality becomes ρ −t(Λ+ρ,α) FΛ+ρe R Y (1 − e ) et(ρ,ρ)(F ch(L(Λ))) = = ρ Q (1 − e−t(ρ,α)) (1 − e−t(ρ,α)) α2∆+ α2∆+ As t ! 0, et(ρ,ρ) ! 1. hence X t(ρ,Λ) LHS = Fρch(L(Λ)) = dimL(Λ)e = dimL(Λ) λ And by L'Hospitals rule, Y (λ + ρ, α) e−t(λ+ρ,α) lim RHS = lim = RHS of Weyl dimension fomula t!0 t!0 (ρ, α) e−t(ρ,α) α2∆+ 2 Proof of the Weyl's character formula. Lemma 25.2. If Λ(Hi) 2 Z+, then chL(Λ) is ri-invariant. Λ(Hi)+1 Λ(Hi)+1 Proof. By the key sl2 lemma, EiFi vΛ = 0, EjFi vΛ = 0 for j 6= i since Ej and Fi Λ(Hi)+1 commute. So Fi vΛ is a singular vector of L(Λ) and since L(Λ) is irreducible, so it has no Λ(Hi)+1 singular weights other than Λ, we conclude that Fi vΛ = 0. N But L(Λ) = U(g)vΛ, hence for v 2 L(Λ) we conclude that Fi v = 0 for N 0 Also obviously N Ei v = 0 for N 0. It follows that any v 2 L(Λ) lies in a sl2-invariant finite dimensional subspace. Hence by Weyl's complete reducibility theorem L(Λ) is a direct sum of irreducible sl2-modules. So it suffices to prove that the character of a finite dimensional irreducible sl2-modules is rαi -invariant. Note that chL(mρ) = emρ + e(m−2)ρ + ::: + e−mρ Since rαi (ρ) = ρ − rαi , the character is rαi -invariant. Hence the lemma holds for L(Λ) as well. Lemma 25.3. RchM(Λ) = eΛ Proof. We know from last lecture that vectors Em1 :::EmN form a basis of M(Λ). Hence −β1 −βN X chM(Λ) = eΛ−m1β1:::−mN βN N (m1;:::mN )2Z+ X = eΛ eΛ−m1β1:::−mN βN N (m1;:::mN )2Z+ eΛ = (1) Q (1 − e−α) α2∆+ by geometric progression. Lemma 25.4. w(eρR) = (detw)eρR for any w 2 W ρ ρ Proof. Since W is generated by rαi , it suffices to check rαi (e R) = −e R. Indeed, we can rewrite R as Y R = (1 − eαi ) (1 − e−α) α2∆+nαi Q −α Note that (1 − e ) is rαi -invariant, so by Lemma 1 α2∆+nαi ρ ρ−αi αi Y −α rαi (e R) = e (1 − e ) (1 − e ) α2∆+nαi Y = eρ(e−αi − 1) (1 − e−α) α2∆+nαi = −eρR (2) as wanted. 3 Lemma 25.5. Let Λ 2 h∗ and let V be a highest weight module with highest weight Λ. Let R D(Λ) = fΛ − Σkiαi; ki 2 Z+g. Then chV = Σλ2B(Λ)aλ chL(λ); where B(Λ) = fλ 2 D(Λ)j(Λ + ρ, Λ + ρ) = (λ + ρ, λ + ρ) and aΛ = 1; aλ 2 Z+g: P Proof. Proof is by induction on dimV = dimVλ < 1 due to Theorem2(h) from last lecture λ2B(Λ) that jB(Λ)j < 1 is finite. If P = 1 then Λ is the only singular weight hence by Theorem2(c) λ2B(Λ) from last lecture that V = L(Λ) so chV = chL(Λ). If there another singular vector vλ; λ 6= Λ, let U = U(g)vλ and consider the following exact sequence of g-modules 0 ! U ! V ! V=U ! 0 Then ch(V ) = ch(U) + ch(U=V ), now we apply the induction assumption to each of the terms. Lemma 25.6. In the assumptions of Lemma 5 and V = L(Λ) is irreducible, we have chV = Σλ2B(λ)bλchMλ, where bΛ = 1 and bλ 2 Z. P Proof. By Lemma 5 we have for any µ 2 B(Λ): chM(µ) = aλ,µchL(λ). Let B(Λ) = fΛ = λ2B(µ) P λ1; : : : ; λrg. Order them in such a way that λi −λj 2= f kiαijki 2 Z+g if i > j. We get a system of i P equations chMλj = aijchL(λi), where aii = 1; aij = 0 for i > j. So the matrix aij of this system i is upper triangular matrix of integers with 1's on the diagonal and so its inverse, which expresses chL(Λ)’s in terms of chM(µ)'s for µ 2 B(Λ) is a matrix of integers with ones on the diagonal as well, and we are done. P Proof. Proof of Theorem 3. With Lemma 6, chL(Λ) = bλchM(λ), where bΛ = 1. Multiply λ2B(λ) both sides by eρR we get from Lemma 3 that ρ X λ+ρ e RchL(Λ) = bλe (!) λ2B(Λ) By Lemma 2 L(Λ) is W -invariant, hence by Lemma 4, eρRchL(Λ) is W -anti-invariant (i.e. multi- plied by the determinant). Hence the left hand side of the equation is anti-invariant, and therefore so is the right hand side. Hence using any simple transitivity of W on weyl chambers we can rewrite (!) as follows: ρ X w(Λ+ρ) X X w(λ+ρ) e RchL(Λ) = (detw)e + bλ (detw)e w2W λ2B(Λ)nfΛg,λ+ρ2P+ w2W So it remains to show that the second term in this sum is 0, i.e.
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