Math 128: Lecture 20. May 12, 2014

Transcription

1 Math 128: Lecture 20 May 12, 2014

2 Weigh space multiplicities: We re trying to calculate m λ µ, the dimension of L(λ) µ in L(λ), with λ P + = Z 0 {ω 1,..., ω r }. 1. First solution: Freudenthal s multiplicity formula. m λ 2 µ = µ + iα, α m λ λ, λ + 2ρ µ, µ + 2ρ µ+iα. α R + i=1 2. Second solution: Weyl character formula. The character of a finite-dimensional g-module V is ch(v ) = λ P dim(v λ )X λ. For irreducible modules, the character is given by ch(l(λ)) = a λ+ρ where a λ+ρ = det(w)x w(λ+ρ). a ρ w W 3. Third solution: Path model.

3 Littelmann path model h α1 h α1 +α 2 h α2

4 Littelmann path model p h α1 h α1 +α 2 h α2

5 Littelmann path model p h α 1 + h α1 +α 2 h α2

6 Littelmann path model h α 1 h α 1 i p + i α 1 /2-shifts

7 Littelmann path model p h α 1 + α 1 (a + 1) α 1 a

8 Littelmann path model p h α 1 + α 1 (a + 1) α 1 a

9 Littelmann path model p h α 1 +

10 Littelmann path model h α 1 f i p +

11 Littelmann path model e i f i p h α 1 +

12 Littelmann path model e i f i p h α 1 + A crystal B is a set of paths closed under {f i, e i i = 1,..., r}. The crystal graph has vertices p B and edges p i f i p.

13 Working example Fix g = A 2 with base B = {β 1, β 2 β i = ε i ε i+1}. Calculate m ρ

14 Highest weight crystals A highest weight path is a path p satisfying e i p = 0 for all i, which is equivalent to p(1) P + and p(t) C ρ for all t [0, 1].

15 Highest weight crystals A highest weight path is a path p satisfying e i p = 0 for all i, which is equivalent to p(1) P + and p(t) C ρ for all t [0, 1]. The weight of any path p is wt(p) = p(1). Prop. The crystals generated highest weight paths of the same weight are isomorphic. Let B(λ) be the crystal generated by any h.w. path of weight λ P +.

16 Highest weight crystals A highest weight path is a path p satisfying e i p = 0 for all i, which is equivalent to p(1) P + and p(t) C ρ for all t [0, 1]. The weight of any path p is wt(p) = p(1). Prop. The crystals generated highest weight paths of the same weight are isomorphic. Let B(λ) be the crystal generated by any h.w. path of weight λ P +. The character of a crystal is ch(b) = p B X wt(p). Theorem For λ P +, ch(b(λ)) = ch(l(λ)).

17 Highest weight crystals The character of a crystal is ch(b) = p B X wt(p). Theorem For λ P +, ch(b(λ)) = ch(l(λ)). Proposition Let B, B be finite crystals. 1. ch(b) = ch(b ) if and only if B = B. 2. The union B B is a crystal, and 3. ch(b) = p B p is highest weight ch(b B ) = ch(b) + ch(b ). ch(b(wt(p))).

18 Tensor product rules The concatenation of two paths p, p is defined by { pp p(2t) 0 t 1/2, = p(1) + p (2(t 1/2)) 1/2 t 1. Note that wt(pp ) = wt(p) + wt(p ).

19 Tensor product rules The concatenation of two paths p, p is defined by { pp p(2t) 0 t 1/2, = p(1) + p (2(t 1/2)) 1/2 t 1. Note that wt(pp ) = wt(p) + wt(p ). Theorem 1. For finite-dimensional g-modules V, V, B(V V ) = {pp p B(V ), p B(V )}.

20 Tensor product rules The concatenation of two paths p, p is defined by { pp p(2t) 0 t 1/2, = p(1) + p (2(t 1/2)) 1/2 t 1. Note that wt(pp ) = wt(p) + wt(p ). Theorem 1. For finite-dimensional g-modules V, V, B(V V ) = {pp p B(V ), p B(V )}. 2. With λ, µ P +, and p + λ ch(l(λ) L(µ)) = highest weight in B(λ), ch(l(λ + wt(q))). q B(µ) p + λ q highest weight

21 B(L(ω 1 ) L(ω 1 )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =.

22 B(L(ω 1 ) L(ω 1 )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =.

23 B(L(ω 1 ) L(ω 1 )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =. Highest weight paths: p 2 1 with weight 2ω 1 p 1 p 2 with weight ω 2

24 B(L(ω 1 ) L(ω 1 )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =. Highest weight paths: p 2 1 with weight 2ω 1 p 1 p 2 with weight ω 2 So ch(l(ω 1 ) L(ω 1 )) = ch(l(2ω 1 )) + ch(l(ω 2 )),

25 B(L(ω 1 ) L(ω 1 )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =. Highest weight paths: p 2 1 with weight 2ω 1 p 1 p 2 with weight ω 2 So ch(l(ω 1 ) L(ω 1 )) = ch(l(2ω 1 )) + ch(l(ω 2 )), implying L(ω 1 ) L(ω 1 ) = L(2ω 1 ) L(ω 2 )

26 B(L(ω 1 ) L(ω 1 )) = B(L( ) L( )) is the set containing p 2 1 =, p 1p 2 =, p 1 p 3 =, p 2 p 1 =, p 2 2 =, p 2p 3 =, p 3 p 1 =, p 3 p 2 =, p 2 3 =. Highest weight paths: p 2 1 with weight 2ω 1 = p 1 p 2 with weight ω 2 = So ch(l(ω 1 ) L(ω 1 )) = ch(l(2ω 1 )) + ch(l(ω 2 )), implying L( ) L( ) = L(ω 1 ) L(ω 1 ) = L(2ω 1 ) L(ω 2 ) = L( ) L ( )

27 Back to tableaux B(ω 1 ) = {p i i = 1,..., r + 1} where p i is the straight-line path to ε i 1 r+1 (ε 1 + ε ε r+1 ). p 1 = p 2 = and p 3 =.

28 Back to tableaux B(ω 1 ) = {p i i = 1,..., r + 1} where p i is the straight-line path to ε i 1 r+1 (ε 1 + ε ε r+1 ). p 1 = p 2 = and p 3 =. Generate B(λ) with the path p + λ = pλ 1 1 pλ 2 2 pλr r.

29 Back to tableaux B(ω 1 ) = {p i i = 1,..., r + 1} where p i is the straight-line path to ε i 1 r+1 (ε 1 + ε ε r+1 ). p 1 = p 2 = and p 3 =. Generate B(λ) with the path p + λ = pλ 1 1 pλ 2 2 pλr r. Is this ok? If p = p i1 p i2 p in, then p(t) C ρ t iff every initial path p i1 p ij has weight wt(p i1 p ij ) = j (ω ik ω ik 1) P + k=1 (iff its weight is the sum of ω i s).

30 Back to tableaux B(ω 1 ) = {p i i = 1,..., r + 1} where p i is the straight-line path to ε i 1 r+1 (ε 1 + ε ε r+1 ). p 1 = p 2 = and p 3 =. Generate B(λ) with the path p + λ = pλ 1 1 pλ 2 2 pλr r. Is this ok? If p = p i1 p i2 p in, then p(t) C ρ t iff every initial path p i1 p ij has weight wt(p i1 p ij ) = j (ω ik ω ik 1) P + k=1 (iff its weight is the sum of ω i s). Another way: define the reading word of p to be i 1 i 2 i n.

31 Back to tableaux B(ω 1 ) = {p i i = 1,..., r + 1} where p i is the straight-line path to ε i 1 r+1 (ε 1 + ε ε r+1 ). p 1 = p 2 = and p 3 =. Generate B(λ) with the path p + λ = pλ 1 1 pλ 2 2 pλr r. Is this ok? If p = p i1 p i2 p in, then p(t) C ρ t iff every initial path p i1 p ij has weight wt(p i1 p ij ) = j (ω ik ω ik 1) P + k=1 (iff its weight is the sum of ω i s). Another way: define the reading word of p to be i 1 i 2 i n. Then p(t) C ρ t if any only if every initial subword i 1 i 2 i j of the reading word of p has the property that is contains more 1 s than 2 s, more 2 s than 3 s, and so on.

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