FORMULAS FOR MULTIPLICITIES OF sl(3) MODULES

Transcript

1 FORMULAS FOR MULTIPLICITIES OF sl() MODULES AMY BARKER AND LAUREN VOGELSTEIN Let g = sl(), = {α, α } and α = ε - ε, α = ε - ε, let λ be a dominant integral weight such that λ C +, let be the outer shell of WD(λ) and let µ WD(λ) C Figure. Weight diagram for λ = (6, ) Definitions Definition. If λ = (aω, bω ), then define λ θ as the line in C + such that λ θ = {λ qθ : 0 q min{a, b}}. Let C +I be the region in C + that contains all µ above λ θ and C +II is the region in C + that contains all µ below λ θ. Note: If µ C +I then there exist m, n Z + such that µ = λ mα nθ. If µ C +II then there exist l, k Z + such that µ = λ lα kθ. Definition. The distance between µ and λ is Date: June 8, 0. d(µ, λ) = min{t : µ = λ t α ij, α ij Φ + }. j=

2 AMY BARKER AND LAUREN VOGELSTEIN Figure. Weight diagram for λ = (6, ) with λ θ Definition. Let the distance between weight µ and outer shell be defined as follows: d(µ, ) = min{i : µ + iα / W D(λ), i Z +, α Φ + }. Barker Vogelstein Lemma Lemma. For all i in Z +, for all α Φ + either: d(µ + iα, λ) < d(µ, λ) or d(µ + iα, ) < d(µ, ) or both. Proof. Case A: Suppose µ C +I and µ + iα C +I for some i Z +. Then, by definition, d(µ, λ) = d(λ mα nθ, λ) = m + n. d(µ + iα, λ) = d(λ mα nθ + iα, λ) = d(λ (m i)α nθ, λ) = m i + n < m + n. Case B: Suppose µ C +I and µ + iα C +I where i Z +. Then, by definition, d(µ, δ) = d(λ mα nθ, δ) = n. Therefore, µ + iα = λ mα nθ + iα = λ mα nθ + i(θ α ) = λ (m + i)α (n ). d(µ + iα, δ) = d(λ (m + i)α (n )θ, δ) = n i < n.

3 FORMULAS FOR MULTIPLICITIES OF sl() MODULES Case C: Suppose µ C +I and µ + iθ C +I where i Z +. Then, by definition, d(µ, λ) = d(λ mα nθ, λ) = m + n. d(µ + iθ, λ) = d(λ mα (n i)θ, λ) = m + n i < m + n. Case A: Suppose µ C +II and µ + iα C +II where i Z +. Then, by definition, d(µ, δ) = d(λ lα kθ, δ) = k. Therefore, µ + iα = λ lα kθ + iα = λ lα kθ + i(θ α ) = λ (l + i)α (k i)θ. d(µ + iα, δ) = d(λ (l + i)α (k i)θ, δ) = k i < k. Case B: Suppose µ C +II then µ + iα C +II where i Z +. Then, by definition, d(µ, λ) = d(λ lα kθ, λ) = l + k. d(λ lα kθ + iα, λ) = d(λ (l i)α kθ, λ) = l + k i < l + k. Case C: Suppose µ C +II and µ + iθ C +II where i Z +. Case C is similar to Case C. Case A: Suppose µ λ θ and µ + iα λ θ where i Z +. Then, by definition, d(µ, δ) = d(λ qθ, δ) = q. Therefore, µ + iα = λ qθ + iα = λ qθ + i(θ α ) = λ iα (q i)θ. d(µ + iα ) = d(λ iα (q i)θ, δ) = q i < q. Case B: Suppose µ λ θ and µ + iα λ θ where i Z +. Then, by definition, d(µ, δ) = d(λ qθ, δ) = q. Therefore, µ + iα = λ qθ + iα = λ qθ + i(θ α ) = λ iα (q i)θ. d(µ + iα ) = d(λ iα (q i)θ, δ) = q i < q. Case C: Suppose µ λ θ andµ + iθ λ θ where i Z +. Case C is similar to Case C.

4 AMY BARKER AND LAUREN VOGELSTEIN Therefore, in C + either d(µ + iα, λ) < d(µ, λ) or d(µ + iα, δ) < d(µ, δ). This is true in all of WD(λ) by symmetry. Theorem (Freudenthal). [FH9,.] The multiplicity of µ W D(λ): Antoine-Speiser Formula (6) ((λ + ρ, λ + ρ) (µ + ρ, µ + ρ))m(µ) = α Φ + m(µ + iα)(µ + iα, α). Proposition 7 (Antoine-Speiser). Let g = sl(), then: () The multiplicity of any weight on the outer shell is. () The multiplicity increases by each time you move to any inner hexagonal shell. () Once the shells become triangular, the multiplicity remains constant. Proof. Let λ = (cω + dω ) = (c, d). First note that we can simplify the two sides of Freudenthal s Formula (6) as follows: On the left hand side: ((λ + ρ, λ + ρ) (µ + ρ, µ + ρ))m(µ) On the right hand side: = ((λ + θ, λ + θ) (µ + θ, µ + θ))m(µ) = ((λ, λ) + (λ, θ) (µ, µ) (µ, θ))m(µ). α Φ + m(µ + iα)(µ + iα, α) = [m(µ + iα )(µ + iα, α ) + m(µ + iα )(µ + iα, α ) +m(µ + iθ)(µ + iθ, θ)]. Our proof is a double induction. First we induct on distance to the outer shell and then we induct on distance to the vertex. We will do this first for the hexagonal shells and then for triangular shells. Base case for first induction on hexagonal shells: Suppose µ C + is on the outer shell, i.e. d(µ, ) =. We need to consider when µ is on the top edge of WD(λ) and when µ is on the right edge of WD(λ). Along the top edge: If µ is on the top edge, then µ = λ rα for some r Z 0.

5 FORMULAS FOR MULTIPLICITIES OF sl() MODULES The left hand side of Freudenthal s formula becomes ((λ, λ) + (λ, θ) (µ, µ) (µ, θ))m(µ) = ((λ, λ) + (λ, θ) (λ rα, λ rα ) (λ rα, θ))m(µ) = ((λ, λ) + (λ, θ) (λ, λ) + r(λ, α ) r (α, α ) (λ, θ) + r(α, θ))m(µ) = (r(cω + dω, ω ω ) r + r(ω ω, ω + ω ))m(µ) = (r(c(ω, ω ) c(ω, ω ) + d(ω, ω ) d(ω, ω )) r + r((ω, ω ) + (ω, ω ) (ω, ω ) (ω, ω ))m(µ) = (r(c c + d + d ) r + r( + ))m(µ) = (rc r + r)m(µ). The right hand side of Freudenthal s formula becomes α Φ + m(µ + iα)(µ + iα, α) = m(µ + iα )(µ + iα, α ) because µ + iα / WD(λ), µ + iθ / WD(λ). In order to simplify this further, we will now prove inductively that m(µ + iα ) =. Base Case: γ = λ kα, where k = 0 γ = λ therefore m(γ) = m(λ) =. Inductive Step: Assume m(γ) = for γ = λ kα such that k r. Then, because µ = λ rα m(µ + iα ) = m(λ rα + iα ) = m(λ (r i)α ) = m(λ kα ) =. Therefore m(µ + iα ) =. Returning to our simplification of the right hand side, we can now write: ()(µ + iα, α ) = r ((µ, α ) + i) = (µ, α ) r () + r i r(r + ) = r(µ, α ) + = r(λ rα, α ) + r(r + ) = r((λ, α ) r(α, α )) + r + r = r(λ, α ) r + r + r = r(cω + dω, ω ω ) r + r = r(c(ω, ω ) c(ω, ω ) + d(ω, ω ) d(ω, ω )) r + r = r(c c + d d ) r + r (8) = rc r + r.

6 6 AMY BARKER AND LAUREN VOGELSTEIN Combining the left and right hand sides (8), we have (rc r + r)m(µ) = rc r + r. Therefore, m(µ) =. Along the right edge: If µ is on the right edge, then µ = λ sα for some s Z 0. The left hand side of Freudenthal s formula becomes: ((λ, λ) + (λ, θ) (λ sα, λ sα ) (λ sα, θ))m(µ) = ((λ, λ) + (λ, θ) (λ, λ) + s(λ, α ) + s(λ, α ) s (α, α ) (λ, θ) + s(α, θ))m(µ) = (s(cω + dω, ω + ω ) s + s( ω + ω, ω + ω ))m(µ) = (s( c(ω, ω ) + c(ω, ω ) d(ω, ω ) + d(ω, ω )) s + s( (ω, ω ) (ω, ω ) + (ω, ω ) + (ω, ω )))m(µ) = (s( c + c d + d) s + s( + + )m(µ) = (sd s + s)m(µ). The right hand side of Freudenthal s formula becomes α Φ + m(µ + iα)(µ + iα, α) = m(µ + iα )(µ + iα, α ) because µ + iα / WD(λ), µ + iθ / WD(λ). Similarly, we can inductively prove that m(µ + iα ) =. Returning to our simplification of the right hand side, we can now write: (µ + iα, α ) = r [(µ, α ) + i] = (µ, α ) r () + r i s(s + ) = s(λ sα, α ) + = s(λ, α ) s (α, α ) + s(s + ) = s(cω + dω, ω + ω ) s + s + s = s( c(ω, ω ) + c(ω, ω ) d(ω, ω ) + d(ω, ω )) s + s = s( c + c d + d) s + s = sd s + s. Combining the left and right hand sides, we have Therefore, m(µ) =. (sd s + s)m(µ) = sd s + s.

7 FORMULAS FOR MULTIPLICITIES OF sl() MODULES 7 We have shown that for any µ on the top or right edge of the outer shell, m(µ) =. This is true on all of the outer shell by the symmetry of WD(λ). Inductive Assumption: We have proven that that the multiplicity of weights on the first shell is. Now we will assume that the multiplicity of the first k shells is m(µ S u ) = u, where S u is the u th shell and u {,,,..., k }. Base Case: vertex of the kth shell = µ = λ (k )θ = λ kθ + θ. The left hand side of Freudenthal s formula becomes: ((λ, λ) + (λ, θ) (λ kθ + θ, λ kθ + θ) (λ kθ + θ, θ))m(µ) = ((λ, λ) + (λ, θ) (λ, λ) + k(λ, θ) (λ, θ) + k(λ, θ) k (θ, θ) + k(θ, θ) (λ, θ) + k(θ, θ) (θ, θ) (λ, θ) + k(θ, θ) (θ, θ))m(µ) = ((k )(λ, θ) k + k + k + k )m(µ) = ((k )(λ, θ) k + 8k 6)m(µ). The right hand side of Freudenthal s formula is simplified by m(µ + iα) = k i for all α Φ + and becomes: α Φ + = m(µ + iα)(µ + iα, α) α Φ + = α Φ + = α Φ + = α Φ + (k i)(λ kθ + θ + iα, α) (k i)[(λ, α) k(θ, α) + (θ, α) + i] (λ, α)(k i) k(θ, α)(k i) + (θ, α)(k i) + i(k i) (λ, α)(k ) k term α Φ + (θ, α)(k i) + term α Φ + (θ, α)(k i) + term To simplify this further, we will break it down term by term. Term : (λ, α ) (k i) + (λ, α ) (k i) + (λ, θ) (k i) = ((λ, α ) + (λ, α ) + (λ, θ)) = (λ, θ)k(k ) = (k k)(λ, θ) k(k ) α Φ + i(k i). term

8 8 AMY BARKER AND LAUREN VOGELSTEIN Term : Term : Term : k(θ, α ) (k i) k(θ, α ) (k i) k(θ, θ) (k i) k(k ) k(k ) k(k ) = k k k = k (k ) k (k ) k (k ) = k (k ) (θ, α ) (k i) + (θ, α ) (k i) + (θ, θ) (k i) k(k ) = = k(k ) k(k ) + + (ki i ) = k i k(k ) k(k + ) k(k + )(k + ) = k 6 = 6k (k + ) k(k + )(k + ) i Term + Term + Term + Term Therefore, = (k k)(λ, θ) k (k ) + k(k ) + 6k (k + ) k(k + )(k + ) = (k k)(λ, θ) k + 8k 6k ((k )(λ, θ) k + 8k 6)m(µ) = (k k)(λ, θ) k + 8k 6k. Thus, m(µ) = k. This proves that the multiplicity of the vertex of the k th shell is k. Now, we will perform the second proof by induction. We will prove that the multiplicity of every weight on the k th shell is k by inducting on distance from the vertex. We will again do this along the top edge and the right edge of the k th hexagonal shell. Base for the Second Inductionon Hexagonal Shells: For µ on the k th shell, m(µ) = k. Second Inductive Assumption (Top Edge): Assume the multiplicity remains constant for (r ) steps away from the vertex of the k th shell in the α direction. Let µ be the weight r steps away from the k th vertex in the α direction such that µ = λ (k )θ + rα = λ kθ + θ + rα.

9 FORMULAS FOR MULTIPLICITIES OF sl() MODULES 9 The left hand side of Freudenthal s formula becomes ((λ, λ) (λ, θ) (λ kθ + θ + rα, λ kθ + θ + rα ) (λ kθ + θ + rα, θ))m(µ) = ((λ, λ) (λ, θ) (λ, λ) + k(λ, θ) (λ, θ) + r(λ, α ) + k(λ, θ) k (θ, θ) + k(θ, θ) kr(θ, α ) (λ, θ) + k(θ, θ) (θ, θ)r(θ, α ) rk(θ, α ) + r(θ, α ) r (α, α ) (λ, θ) + k(θ, θ) (θ, θ) + r(θ, α ))m(µ = ((k )(λ, θ) + r(λ, α ) k + k rk + k + r rk + r r + k + r)m(µ) = ((k )(λ, θ) + r(λ, α ) rk k + 8k + r 6 r )m(µ). The right hand side of Freudenthal s formula becomes m(µ + iα)(µ + iα, α) α Φ + = (λ kθ + θ rα + iα) = (λ, α) k(θ, α) + (θ, α) r(α, α) + i = m(µ + iα)[(λ, α) k(θ, α) + (θ, α) r(α, α) + i] α Φ + = α Φ + (λ, α)m(µ + iα) k }{{}}{{} r α Φ + term α Φ + (α, α)m(µ + iα) + term Term : α Φ + (λ, α)m(µ + iα) (θ, α)m(µ + iα) + term α Φ + r+k = (λ, α ) m(µ + iα ) +(λ, α ) (a) im(µ + iα). term α Φ + k m(µ + iα ) +(λ, θ) (b) (a): [ r+k r (λ, α ) m(µ + iα ) = (λ, α ) k + = (λ, α ) = (λ, α ) [ r+k k r+k i=r+ r+k i=r+ [ k(r + k) (k i + r) i + = (k k)(λ, α ) + rk(λ, α ) r=k i=r+ (θ, α)m(µ + iα) term k m(µ + iθ) (c) r ] ] (r + k)(r + k + ) r(r + ) ] + rk

10 0 AMY BARKER AND LAUREN VOGELSTEIN (b): (c): (λ, α ) (λ, θ) k m(µ + iα ) = (λ, α ) k (k i) = (k k)(λ, α ) k m(µ + iθ) = (λ, θ) Combing parts (a), (b), and (c) we see that Term : k (k i) = (k k)(λ, θ) Term = (k k)(λ, α ) + rk(λ, α ) + (k k)(λ, α ) + (k k)(λ, θ) k α Φ + = (k k)(λ, α + α + θ) + rk(λ, α ) = (k k)(λ, θ) + rk(λ, α ) = (rk k)(λ, θ) + rk(λ, α ) (θ, α)m(µ + iα) [ r+k = k (θ, α ) m(µ + iα ) + (θ, α ) k m(µ + iα ) + (θ, θ) ] k m(µ + iθ) However, by (a) we know r+k m(µ + iα ) = (k k + kr). Similarly, by (b)(c) we know k m(µ + iα ) = k m(µ + iθ) = (k k). Using this to simplify, we have [ Term = k (k k + kr) + ] (k k) + (k k) = k + k k r Term : (Note that Term = Term.) k α Φ + (θ, α)m(µ + iα) [ r+k = k(/k) (θ, α ) m(µ + iα ) + (θ, α ) = k + k kr k m(µ + iα ) + (θ, θ) ] k m(µ + iθ)

11 Term : r α Φ + FORMULAS FOR MULTIPLICITIES OF sl() MODULES (α, α)m(µ + iα) r+k = r(α, α ) m(µ + iα ) r(α, α ) r+k = r m(µ + iα ) + r r+k = r m(µ + iα ) k (k ) r k m(µ + iα ) r(α, θ) k (k ) k m(µ + iθ) From (a) we know r+k m(µ + iα ) = (k k + kr). Using this to simplify, we have Term : r+k r m(µ + iα ) = r( (k k + rk)) r+k im(µ + iα ) + = = r(k k + rk). k im(µ + iα ) + r r+k ki + i(k i + r) + r+k = k i r+ r+k i=r+ i + r = r k + rk + k k r+k i=r+ k im(µ + iθ) k i(k i) + i + 8k k i 8 k k i(k i) i Term + Term + Term + Term + Term = (k k)(λ, θ) + rk(λ, α ) rk k + k + rk + k k r k rk + rk + r k + rk + k k = k((k )(λ, θ) + r(λ, α ) rk k + 8k + r 6 r ) Therefore, ((k )(λ, θ) + r(θ, α ) k + 8k rk r + r 6)m(µ) = k((k )(λ, θ) + r(λ, α ) rk k + 8k + r 6 r ) Thus, m(µ) = k. This proves that every µ on the top edge of the k th hexagonal shell has a multiplicity of k. We can prove the same for weights on the right edge of the k th hexagonal shell. Second Inductive Assumption (Right Edge):

12 AMY BARKER AND LAUREN VOGELSTEIN Assume the multiplicity remains constant for (s ) steps away from the vertex of the k th shell in the α direction. Let µ be the weight s steps away from the k th vertex in the α direction such that µ = λ (k )θ + sα = λ kθ + θ + sα. Similarly, the left hand side of Freudenthal s formula becomes ((k )(λ, θ) + s(λ, α ) k + 8k sk 6 + s s )m(µ) and the right hand side becomes k((k )(λ, θ) + s(λ, α ) k + 8k sk 6 + s s ). Therefore, m(µ) = k. This proves that the multiplicity of any weight on the k th hexagonal shell is k. Now we will go through a similar process for triangular shells. We will prove that once the shells become triangular, the multiplicity of the weights is constant. This is done by first inductively proving that the multiplicity of the weights on the vertex of each triangular shell are equal. Then, by a second inductive proof, we will show that the multiplicity of weights along the top and right edge of each triangular shell equal the multiplicity of the weight on the vertex. Base Case for the First Induction on Triangular Shells: The multiplicity of the outermost triangular shell of W D(λ), where λ = (aω, bω ) is min(a, b) +. Inductive Assumption: (to prove that every triangular shell will have weights of equal multiplicity): Assume the multiplicity of the weights for the first (t ) th triangular shells is min(a, b)+. Now, we will examine the t th triangular vertex for the case when (a > b). In this case, µ = λ bθ t(θ + α ) = λ (b + t)θ tα. The left hand side of Freudenthal s formula can be simplified as follows ((λ, λ) + (λ, θ) (λ bθ tθ tα, λ bθ tθ tα ) (λ bθ tθ tα, θ))m(µ) = ((b + t)(λ, θ) + t(λ, α ) b 6bt 6t + b + 6t)m(µ). Note: (λ, θ) = (aω + bω, ω + ω ) = a + b (λ, α ) = (aω + bω, ω ω ) = a. With this, we can further reduce the left hand side. = ((b + t)(a + b) + t(a) b 6bt 6t + b + 6t)m(µ) = (ab + at bt 6t + b + 6t)m(µ) The right hand side of Freudenthal s formula can be simplified as follows

13 FORMULAS FOR MULTIPLICITIES OF sl() MODULES Note: m(µ + iα)(µ + iα, α) α Φ = (λ bθ tθ tα + iα, α) = (λ, α) b(θ, α) t(θ, α) t(α, α) + i = m(µ + iα)[(λ, α) (b + t)(θ, α) t(α, α) + i] α Φ = m(µ + iα)(λ, α) (b + t) m(µ + iα)(θ, α) α Φ α Φ }{{}}{{} term term t m(µ + iα)(α, α) + m(µ + iα)i. α Φ α Φ }{{}}{{} term term m(µ + iα ) = m(µ + iα ) = m(µ + iθ) = Therefore, m(µ + iα ) = m(µ + iα ) = m(µ + iθ) = { b + i t b + + t i t i b + t + t b+t+ (b + ) + (b + t + i) = b + bt + b + t. i=t+ We will use this to simplify Terms - below. Term : m(µ + iα)(λ, α) α Φ = (λ, α )(b + bt + b + t) + (λ, α )(b + bt + b + t) + (λ, θ)(b + bt + b + t) = [(λ, α ) + (λ, α ) + (λ, θ)](b + bt + b + t) = (λ, θ)(b + bt + b + t) = (λ, θ)(b + bt + b + t) = (a + b)(b + bt + b + t) = ab + abt + ab + at + b + b t + b + bt Term : (b + t) m(µ + iα)(θ, α) α Φ = [ (b + t)(θ, α ) (b + t)(θ, α ) (b + t)(θ, θ)](b + bt + b + t) = (b + t)( + + )(b + bt + b + t) = b b t b 8t bt 8bt

14 AMY BARKER AND LAUREN VOGELSTEIN Term : Term : m(µ + iα)i α Φ ( t = i(b + ) + = = ( ( (b + ) (b + ) t m(µ + iα)(α, α α Φ = [ t(α, α ) t(α, α ) t(α, θ)](b + bt + b + t) = [ t + t t](bt + t + b + b) = bt t b t bt t b+t+ b+t+ i=t+ i(b + + t i) ) b+t+ b+t+ i + (b + ) i + t i b+t+ i + t i=t+ i=t+ i b+t+ i=t+ i=t+ i ) ( ( ) (b + t + )(b + t + ) = (b + ) ( (b + t + )(b + t + )(b + t + ) 6 = 6bt + 6b t + bt + b + 6b + b + 6t + 6t b+t+ i=t+ i ) ( (b + t + )(b + t + ) + t )) t(t + )(t + ) 6 ) t(t + ) Term + Term + Term + Term = abt + ab + ab + at + b t + b + b + bt b t b bt 8bt 8t b t bt bt t + 6b t + b + 6b + bt + 6bt + 6t + b + 6t = (b + )(ab + at bt 6t + b + t). Now, setting the left hand side equal to the right hand side we have (ab + at bt 6t + b + 6t)m(µ) = (b + )(ab + at bt 6t + b + 6t). Thus, m(µ) = b +. Similarly, we can prove that m(µ) = a + for µ on the vertex of the t th triangular shell in the case where (a < b). Also, note that where (a = b) there will be no traingular shells. This proves that the weight on the vertex of every triangular shell will have the same multiplicity, namely min(a, b) +. Base Case for the Second Induction Step on Triangular Shells: For µ on the vertex of the t th triangular shell, m(µ) = min(a, b) + for any t. Inductive Assumption:

15 FORMULAS FOR MULTIPLICITIES OF sl() MODULES For a > b, assume that for the first (p-) steps from the t th triangular vertex in the α direction, these weights have a multiplicity of b + and µ = λ bθ tθ tα pα = λ (b + t)θ (p + t)α. The left hand side of Freudenthal s formula becomes ((λ, λ) + (λ, θ) (λ (b + t)θ (p + t)α, λ (b + t)θ (p + t)α ) (λ (b + t)θ (p + t)α, θ))m(µ) = (ab + at bt + ap 6t bp 6pt p + b + 6t + p)m(µ). The right hand side of Freudenthal s formula becomes m(µ + iα)(µ + iα, α) α Φ = (λ (b + t)θ (p + t)α + iα, α) = (λ, α) (b + t)(θ, α) (p + t)(α, α) + i = m(µ + iα)[(λ, α) (b + t)(θ, α) (p + t)(α, α) + i] α Φ = m(µ + iα)(λ, α) (b + t) m(µ + iα)(θ, α) (p + t) m(µ + iα)(α, α) α Φ α Φ α Φ }{{}}{{}}{{} term term term + m(µ + iα)i. α Φ }{{} term Note: { b + i t + p m(µ + iα ) = b + + p + t i p + t i b + p + t + { b + i t m(µ + iα ) = m(µ + iθ) = b + + t i t i b + t + Therefore, and α Φ α Φ b+p+t+ b+t+ m(µ + iα ) = α Φ b+t+ t+p m(µ + iα ) = (b + ) + m(µ + iθ) = b + bt + b + t b+p+t+ i=p+t+ b+p+t+ = (b + ) () + (p + t) = b + bp + bt + b + p + t. We will use this to simplify Terms - below. (b + + t + p i) b+p+t+ i=p+t+ () b+p+t+ i=p+t+ i

16 6 AMY BARKER AND LAUREN VOGELSTEIN Term : α Φ m(µ + iα)(λ, α) b+p+t+ = (λ, α ) b+t+ m(µ + iα ) + (λ, α ) b+t+ m(µ + iα ) + (λ, θ) m(µ + iθ = (λ, α )(b + bp + bt + b + p + t) + (λ, α )(b + bt + b + t) + (λ, θ)(b + bt + b + t) Term : (b + t) α Φ m(µ + iα)(θ, α) b+p+t+ = (b + t)(θ, α ) b+t+ m(µ + iα ) (b + t)(θ, α ) = (b + t)(b + bp + bt + b + p + t) (b + t)(b + bt + b + t) b+t+ m(µ + iα ) (b + t)(θ, θ) m(µ + iθ) Term : (p + t) α Φ m(µ + iα)(α, α) b+p+t+ b+t+ = (p + t)(α, α ) m(µ + iα ) (p + t)(α, α ) m(µ + iα ) b+t+ (p + t)(α, θ) m(µ + iθ) = (p + t)(b + bp + bt + b + p + t) Term : α Φ b+p+t+ m(µ + iα)i = m(µ + iα )i (a) b+t+ + b+t+ m(µ + iα )i + m(µ + iθ)i (b)

17 FORMULAS FOR MULTIPLICITIES OF sl() MODULES 7 (a): b+p+t+ m(µ + iα )i p+t b+p+t+ = i(b + ) + i(b + p + t + i) b+p+t+ = (b + ) i=p+t+ b+p+t+ i + (p + t) i=p+t+ b+p+t+ i i=p+t+ = b + b p + b t + b + bp + bpt + bp + bt + bt + b + p + pt + t + t + p i (b): b+t+ b+t+ m(µ + iα )i + b+t+ = 8 m(µ + iα )i m(µ + iθ)i = (6bt + 6b t + bt + b + 6b + b + 6t + 6t) Now, we have Term = (a) + (b) = b + b p + 6b t + 6b + bp + bpt + bp + 6bt + bt + b + p + pt + 6t + 6t + p. Combining Terms - will give us the complete and simplified right hand side of Freudenthal s formula. Term + Term + Term + Term = ((λ, α )(b + bp + bt + b + p + t) + (λ, α )(b + bt + b + t) + (λ, θ)(b + bt + b + t)) + ( (b + t)(b + bp + bt + b + p + t) (b + t)(b + bt + b + t)) + ( (p + t)(b + bp + bt + b + p + t)) + (b + b p + 6b t + 6b + bp + bpt + bp + 6bt + bt + b + p + pt + 6t + 6t + p) = (b + )(ab + at bt + ap 6t bp 6pt p + b + 6t + p) Setting the left hand side equal to the right hand side gives us: (ab + at bt + ap 6t bp 6pt p + b + 6t + p)m(µ) = (b + )(ab + at bt + ap 6t bp 6pt p + b + 6t + p) Thus, m(µ) = b +

18 8 AMY BARKER AND LAUREN VOGELSTEIN This proves that for a > b the multiplicity of any µ W D(λ) on a triangular shell when λ = (aω, bω ) is b +. Similarly, it can be proved that for b > a the multiplicity of any µ W D(λ) on a triangular shell when λ = (aω, bω ) is a +. Combining these conclusions, we see that the multiplicity of any µ on a triangular shell in W D(λ) for λ = (aω, bω )i is min(a, b) +. References [FH9] William Fulton and Joe Harris, Representation theory, Graduate Texts in Mathematics, vol. 9, Springer-Verlag, New York, 99. A first course; Readings in Mathematics. MR9 (9a:0069)