Maximal Dominant Weights for Affine Lie Algebra Representations - PowerPoint PPT Presentation

Maximal Dominant Weights for Affine Lie Algebra Representations Suzanne Crifo Advisor: Kailash Misra Department of Mathematics North Carolina State University June 5, 2018 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 1 / 15

Outline 1 Definitions 2 Integrable Highest Weight Modules B (1) 3 n , V ( k Λ 0 ) n = 5 , k = 3 Results for arbitrary n and k Crifo (NCSU) Maximal Dominant Weights June 5, 2018 2 / 15

Definitions g = B (1) n Consider B (1) with Cartan matrix and Dynkin diagram: n   2 0 − 1 0 0 . . . 0 0 2 − 1 0 0 0 . . .     − 1 − 1 2 − 1 0 . . . 0     0 0 − 1 2 − 1 0 . . . A =    . .  ... . .   . .     0 . . . 0 0 − 1 2 − 1   0 0 0 0 − 2 2 . . . 1 . . . ⇒ 1 2 2 2 2 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 3 / 15

Definitions Definitions Let h = span { h 0 , h 1 , . . . , h n , d } be the Cartan subalgebra Π = { α 0 , α 1 , . . . , α n } ⊂ h ∗ is the set of simple roots. Note that α j ( h i ) = a ij . Let δ = α 0 + α 1 + 2 α 2 + · · · + 2 α n be the null root c = h 0 + h 1 + 2 h 2 + · · · + 2 h n − 1 + h n is the canonical central element The set { Λ 0 , Λ 1 , . . . , Λ n } is the set of fundamental weights such that Λ j ( h i ) = δ ij and Λ j ( d ) = 0. Crifo (NCSU) Maximal Dominant Weights June 5, 2018 4 / 15

Integrable Highest Weight Modules Integrable Highest Weight Modules A weight λ is dominant integral if λ ( h i ) ∈ Z ≥ 0 for i ∈ I . Let P + be the set of dominant integral weights. Crifo (NCSU) Maximal Dominant Weights June 5, 2018 5 / 15

Integrable Highest Weight Modules Integrable Highest Weight Modules A weight λ is dominant integral if λ ( h i ) ∈ Z ≥ 0 for i ∈ I . Let P + be the set of dominant integral weights. For Λ ∈ P + there exists an unique (up to isomorphism) irreducible, integrable highest weight module V (Λ) generated by a highest weight vector v Λ . Crifo (NCSU) Maximal Dominant Weights June 5, 2018 5 / 15

Integrable Highest Weight Modules V (Λ) Let P (Λ) be the set of all weights of V (Λ). If λ ∈ P (Λ) then λ = Λ − � n i =0 b i α i where b i ∈ Z ≥ 0 . Crifo (NCSU) Maximal Dominant Weights June 5, 2018 6 / 15

Integrable Highest Weight Modules V (Λ) Let P (Λ) be the set of all weights of V (Λ). If λ ∈ P (Λ) then λ = Λ − � n i =0 b i α i where b i ∈ Z ≥ 0 . Definition A weight λ ∈ P (Λ) is called maximal if λ + δ �∈ P (Λ). Denote the set of all maximal weights as max(Λ). Crifo (NCSU) Maximal Dominant Weights June 5, 2018 6 / 15

Integrable Highest Weight Modules V (Λ) Let P (Λ) be the set of all weights of V (Λ). If λ ∈ P (Λ) then λ = Λ − � n i =0 b i α i where b i ∈ Z ≥ 0 . Definition A weight λ ∈ P (Λ) is called maximal if λ + δ �∈ P (Λ). Denote the set of all maximal weights as max(Λ). Then � P (Λ) = { λ − n δ | n ∈ Z ≥ 0 } . λ ∈ max(Λ) Any λ ∈ max(Λ) is W -conjugate to some µ ∈ max(Λ) ∩ P + , which is known to be a finite set. However, only partial results for an explicit description of this set are known. Crifo (NCSU) Maximal Dominant Weights June 5, 2018 6 / 15

Integrable Highest Weight Modules Find max(Λ) ∩ P + θ = δ − a 0 α 0 λ = λ − λ ( c )Λ 0 − ( λ | Λ 0 ) δ kC af ∩ (Λ + Q ) = { λ ∈ ˚ R | λ ( h i ) ≥ 0 for all i ∈ ˚ h ∗ I , ( λ | θ ) ≤ k } Proposition (Kac) The map λ �→ λ defines a bijection from max(Λ) ∩ P + onto kC af ∩ (Λ + Q ) . In particular, the set of dominant maximal weights of V (Λ) is finite. We want to determine the maximal dominant weights for the integrable B (1) n -module V ( k Λ 0 ). Crifo (NCSU) Maximal Dominant Weights June 5, 2018 7 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) ⇒ λ = 3Λ 0 − � 5 λ ∈ max(3Λ 0 ) ∩ P + = i =0 b i α i Crifo (NCSU) Maximal Dominant Weights June 5, 2018 8 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) ⇒ λ = 3Λ 0 − � 5 λ ∈ max(3Λ 0 ) ∩ P + = i =0 b i α i Then 5 5 5 � � � λ = 3Λ 0 − b i α i − (3Λ 0 − b i α i )( c )Λ 0 − (3Λ 0 − b i α i | Λ 0 ) δ i =0 i =0 i =0 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 8 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) ⇒ λ = 3Λ 0 − � 5 λ ∈ max(3Λ 0 ) ∩ P + = i =0 b i α i Then 5 5 5 � � � λ = 3Λ 0 − b i α i − (3Λ 0 − b i α i )( c )Λ 0 − (3Λ 0 − b i α i | Λ 0 ) δ i =0 i =0 i =0 5 � = − b i α i + b 0 δ i =0 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 8 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) ⇒ λ = 3Λ 0 − � 5 λ ∈ max(3Λ 0 ) ∩ P + = i =0 b i α i Then 5 5 5 � � � λ = 3Λ 0 − b i α i − (3Λ 0 − b i α i )( c )Λ 0 − (3Λ 0 − b i α i | Λ 0 ) δ i =0 i =0 i =0 5 � = − b i α i + b 0 δ i =0 5 � = − b i α i + b 0 ( α 0 + α 1 + 2 α 2 + 2 α 3 + 2 α 4 + 2 α 5 ) i =0 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 8 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) ⇒ λ = 3Λ 0 − � 5 λ ∈ max(3Λ 0 ) ∩ P + = i =0 b i α i Then 5 5 5 � � � λ = 3Λ 0 − b i α i − (3Λ 0 − b i α i )( c )Λ 0 − (3Λ 0 − b i α i | Λ 0 ) δ i =0 i =0 i =0 5 � = − b i α i + b 0 δ i =0 5 � = − b i α i + b 0 ( α 0 + α 1 + 2 α 2 + 2 α 3 + 2 α 4 + 2 α 5 ) i =0 = ( b 0 − b 1 ) α 1 + (2 b 0 − b 2 ) α 2 + (2 b 0 − b 3 ) α 3 + (2 b 0 − b 4 ) α 4 + (2 b 0 − b 5 ) α 5 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 8 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) Let x 1 = b 0 − b 1 , x i = 2 b 0 − b i for i = 2 , . . . , 5. Crifo (NCSU) Maximal Dominant Weights June 5, 2018 9 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n Example: B (1) 5 , V (3Λ 0 ) Let x 1 = b 0 − b 1 , x i = 2 b 0 − b i for i = 2 , . . . , 5. ( � 5 i =1 x i α i )( h 1 ) = 2 x 1 − x 2 ≥ 0 (0 , 0 , 0 , 0 , 0) ( � 5 i =1 x i α i )( h 2 ) = − x 1 + 2 x 2 − x 3 ≥ 0 (1 , 1 , 1 , 1 , 1) ( � 5 i =1 x i α i )( h 3 ) = − x 2 + 2 x 3 − x 4 ≥ 0 (1 , 2 , 2 , 2 , 2) ( � 5 i =1 x i α i )( h 4 ) = − x 3 + 2 x 4 − x 5 ≥ 0 = ⇒ (1 , 2 , 3 , 3 , 3) ( � 5 i =1 x i α i )( h 5 ) = − 2 x 4 + 2 x 5 ≥ 0 (1 , 2 , 3 , 4 , 4) ( � 5 i =1 x i α i | θ ) = x 2 ≤ 3 (1 , 2 , 3 , 4 , 5) (2 , 2 , 2 , 2 , 2) (2 , 3 , 3 , 3 , 3) (2 , 3 , 4 , 4 , 4) (2 , 3 , 4 , 5 , 5) (2 , 3 , 4 , 5 , 6) (3 , 3 , 3 , 3 , 3) Crifo (NCSU) Maximal Dominant Weights June 5, 2018 9 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n What are the weights? Recall λ = 3Λ 0 − � 5 i =0 b i α i . Given an element of kC af ∩ (Λ + Q ), ( x 1 , x 2 , x 3 , x 4 , x 5 ), we need to find the corresponding λ . Crifo (NCSU) Maximal Dominant Weights June 5, 2018 10 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n What are the weights? Recall λ = 3Λ 0 − � 5 i =0 b i α i . Given an element of kC af ∩ (Λ + Q ), ( x 1 , x 2 , x 3 , x 4 , x 5 ), we need to find the corresponding λ . Example ( x = (2 , 3 , 4 , 5 , 5)) b 0 − b 1 = 2 = ⇒ b 0 ≥ 2 b 0 ≥ 3 2 b 0 − b 2 = 3 = ⇒ 2 2 b 0 − b 3 = 4 = ⇒ b 0 ≥ 2 = ⇒ b 0 ≥ 3 b 0 ≥ 5 2 b 0 − b 4 = 5 = ⇒ 2 b 0 ≥ 5 2 b 0 − b 5 = 5 = ⇒ 2 Assume b 0 = 3 + r for r ∈ Z > 0 . Then λ + δ =3Λ 0 − (3 + r − 1) α 0 − (1 + r − 1) α 1 − (3 + 2 r − 2) α 2 − (2 + 2 r − 2) α 3 − (1 + 2 r − 2) α 4 − (1 + 2 r − 2) α 5 Therefore, b 0 = 3 and λ = 3Λ 0 − 3 α 0 − α 1 − 3 α 2 − 2 α 3 − α 4 − α 5 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 10 / 15

B (1) , V ( k Λ 0 ) n = 5 , k = 3 n max(3Λ 0 ) ∩ P + Element of max(3Λ 0 ) ∩ P + x vector x 2 0 (0 , 0 , 0 , 0 , 0) 3Λ 0 1 (1 , 1 , 1 , 1 , 1) 3Λ 0 − α 0 − α 2 − α 3 − α 4 − α 5 2 (1 , 2 , 2 , 2 , 2) 3Λ 0 − α 0 2 (1 , 2 , 3 , 3 , 3) 3Λ 0 − 2 α 0 − α 1 − 2 α 2 − α 3 − α 4 − α 5 2 (1 , 2 , 3 , 4 , 4) 3Λ 0 − 2 α 0 − α 1 − 2 α 2 − α 3 2 (1 , 2 , 3 , 4 , 5) 3Λ 0 − 3 α 0 − 2 α 1 − 4 α 2 − 3 α 3 − 2 α 4 − α 5 2 (2 , 2 , 2 , 2 , 2) 3Λ 0 − 2 α 0 − 2 α 2 − 2 α 3 − 2 α 4 − 2 α 5 3 (2 , 3 , 3 , 3 , 3) 3Λ 0 − 2 α 0 − α 2 − α 3 − α 4 − α 5 3 (2 , 3 , 4 , 4 , 4) 3Λ 0 − 2 α 0 − α 2 3 (2 , 3 , 4 , 5 , 5) 3Λ 0 − 3 α 0 − α 1 − 3 α 2 − 2 α 3 − α 4 − α 5 3 (2 , 3 , 4 , 5 , 6) 3Λ 0 − 3 α 0 − α 1 − 3 α 2 − 2 α 3 − α 4 3 (3 , 3 , 3 , 3 , 3) 3Λ 0 − 3 α 0 − 3 α 2 − 3 α 3 − 3 α 4 − 3 α 5 Crifo (NCSU) Maximal Dominant Weights June 5, 2018 11 / 15

B (1) , V ( k Λ 0 ) Results for arbitrary n and k n Elements of kC af ∩ (Λ + Q ) Theorem Given ˚ A the Cartan matrix for type B n of finite type, the set of solutions to � ˚ A x ≥ 0 x 2 ≤ k � x 2 is { x = 0 } ∪ { x = ( x 1 , x 2 , . . . , x n ) ∈ Z n � ≥ 0 | 1 ≤ x 2 ≤ k , x 1 = + l 1 , 0 ≤ 2 � x 2 � x 2 , x i = x 2 + � i � � l 1 ≤ j =3 l j , 0 ≤ l 3 ≤ − l 1 , 0 ≤ l n ≤ l n − 1 ≤ l n − 2 ≤ 2 2 · · · ≤ l 4 ≤ l 3 for all 2 ≤ i ≤ n } . Crifo (NCSU) Maximal Dominant Weights June 5, 2018 12 / 15