Polynomial Representations of the Lie superalgebra osp (1 | 2 n ) - PowerPoint PPT Presentation

Polynomial Representations of the Lie superalgebra osp (1 | 2 n ) Asmus Kjær Bisbo Department of Applied Mathematics, Computer Science and Statistics, Faculty of Science, Ghent University 18. June 2019 Joint work with Joris Van der Jeugt and Hendrik de Bie 1 / 16

Representation Theory of osp (1 | 2 n ) Classify the representations. Find character formulas. Construct bases. Calculate matrix elements. Formulate inner products. What do we know about osp (1 | 2 n ) representations?: Finite dimensional representations: Classified and characters are understood. [Kac; 1977] Infinite dimensional representations: Lowest weight representations: Classified and characters are mostly understood. [Dobrev, Zhang; 2006]. Paraboson Fock representations: Character formula, basis and matrix elements(”abstract”). [Lievens, Stoilova, Van der Jeugt; 2008] 2 / 16

Definition of osp (1 | 2 n ) Definition as a matrix algebra [Kac; 1977]. Definition as a superalgebra by means of generators and relations [Ganchev, Palev; 1980]: Odd generators b + i , b − i , i ∈ { 1 , . . . , n } , satisfying [ { b ξ i , b η j } , b ǫ l ] = ( ǫ − ξ ) δ i , l b η j + ( ǫ − η ) δ j , l b ξ i , for i , j , l ∈ { 1 , . . . , n } and η, ǫ, ξ ∈ { + , −} , to be interpreted as ± 1 in the algebraic relations. We can interpret b + and b − as parabosonic creation and i i annihilation operators. 3 / 16

Parabosonic Fock space Definition 1 For p ∈ N , the paraboson Fock space is an osp (1 | 2 n ) irrep. with unique vacuum | 0 � satisfying i | 0 � = 0 and 1 i }| 0 � = p b + 2 { b − i , b + 2 | 0 � , ( i ∈ { 1 , . . . , n } ) . C [ R np ] polynomials in n · p variables, x i , j , C ℓ p Clifford algebra generated by e j , satisfying { e i , e j } = 2 δ ij . Then osp (1 | 2 n ) acts on C [ R np ] ⊗ C ℓ p with, p p � � b + and b − i �→ X i = x i , j e j , i �→ D i = ∂ x i , j e j . � �� � � �� � j =1 j =1 Green’s ansatz Green’s ansatz 4 / 16

Polynomial Representation Let W ( µ + p 2 ) be the osp (1 | 2 n ) irrep. with a (not necessarily unique) vacuum vector | µ ; 0 � satisfying 1 i }| µ ; 0 � = ( µ + p 2 { b − i , b + 2) | µ ; 0 � , ( i ∈ { 1 , . . . , n } ) . Decomposition [Cheng, Kwong, Wang; 2010], [Salom; 2013]: 2 W ( µ + p � C [ R np ] ⊗ C ℓ p = 2) , m µ + p µ ∈P with m µ + p 2 being the multiplicities. The case µ = 0 gives a paraboson Fock space. Let | 0; 0 � �→ 1, then 1 2 { D i , X i } (1) = p 2 , ( i ∈ { 1 , . . . , n } ) , W ( p 2) = span C { X k 1 1 · · · X k n n (1) : k 1 , . . . , k n ∈ N 0 } . 5 / 16

Character Formula P Set of all partitions. ( λ 1 , . . . , λ k ) , λ 1 ≥ · · · ≥ λ k , k ∈ N 0 ℓ ( λ ) Length of λ ∈ P . λ ℓ ( λ ) > 0 , λ l = 0 , l > ℓ ( λ ) s λ Schur function of the partition λ ∈ P K λµ Kostka numbers for λ ∈ P and µ ∈ N n 0 Theorem (Lievens, Stoilova, Van der Jeugt; 2008) char W ( p � 2) = ( t 1 · · · t n ) p / 2 s λ ( t 1 , . . . , t n ) λ ∈P , ℓ ( λ ) ≤ p � � K λµ t µ 1 = ( t 1 · · · t n ) p / 2 1 · · · t µ n n µ ∈ N n λ ∈P ,ℓ ( λ ) ≤ p 0 6 / 16

Weight Spaces Definition char W ( p dim W ( p 2) = ( t 1 · · · t n ) p / 2 � 2 t µ 1 1 · · · t µ n 2) µ + p n . µ ∈ N n 0 So for all µ ∈ N n 0 dim W ( p � 2) µ + p 2 = K λµ λ ∈P ,ℓ ( λ ) ≤ p K λµ := # { Semistandard Young Tableaux of shape λ and weight µ } ≤ � �� � 1 1 2 4 � < , λ = (4 , 3 , 2), and µ = (2 , 2 , 1 , 2 , 2) 2 3 4 5 5 7 / 16

Young Tableaux and Basis � S.s. Young Tableaux of at most p rows, � Y ( p ) = and weight in N n 0 There exists a basis for W ( p 2 ): Consisting of vectors ν A , for A ∈ Y ( p ). Tableaux A ∈ Y ( p ) of weight µ , gives ν A ∈ W ( p 2) µ + p 2 . 8 / 16

Tableaux Vectors i 1 i 1 � i 2 i 2 → I = ( i 1 , . . . , i k ) → X I := sgn( σ ) X i σ (1) · · · X i σ ( k ) : σ ∈ S k i k Definition For A = ( A [1] , . . . , A [ l ]) ∈ Y ( p ), s.s. Young tableau with l columns, define ω A := X A [ l ] X A [ l − 1] · · · X A [1] (1) . Remark For each I, X I = 0 iff k > p. 9 / 16

Basis Vectors 1 1 2 4 A = �→ ω A = X 4 X (2 , 4) X (1 , 3 , 5) X (1 , 2 , 5) (1) . 2 3 4 5 5 Theorem W ( p 2 ) has basis { ω A ∈ W ( p 2) : A ∈ Y ( p ) } , with ω A ∈ W ( p 2 ) µ A + p 2 . Proof strategy: Construct a total order < on Y ( p ) such that ω A / ∈ span { ω B : B ∈ Y ( p ) , B < A } . 10 / 16

Monomial Expansion M n , p ( N 0 ) , n by p positive integer matrices γ ∈ M n , p ( N 0 ) p n � � µ γ = ( µ 1 , j , . . . , µ n , j ) , and η γ = ( µ i , 1 , . . . , µ i , p ) j =1 i =1 p n � � x γ = x γ i , j e η γ = e ( η γ ) 1 · · · e ( η γ ) p and . p i , j 1 i =1 j =1 Proposition For A ∈ Y ( p ) of shape λ A ∈ P , � ω A = ( λ A ) ′ 1 ! · · · ( λ A ) ′ c A ( γ ) x γ e η γ ( λ A ) 1 ! γ ∈ Mn , p ( N 0) µ A = µγ 11 / 16

Monomial Coefficients A k ∈ Y ( p ) k ’th subtableaux of A ∈ Y ( p ) , A = 1 1 2 4 ⇒ A 4 = 1 1 2 4 , A 3 = 1 1 2 , A 2 = 1 1 2 = , A 2 3 4 2 3 4 2 3 2 Theorem Let A ∈ Y ( p ) n ( p ) , σ ∈ S µ A and γ ∈ M n , p ( N 0 ) with µ γ = µ A . p 1 � � sgn( σ )( − 1) N A ( γ ) c A ( γ ) = sgn( L A ( σ, α )) ( η γ ) 1 ! · · · ( η γ ) p ! σ ∈ S µ A α =1 The values N A ( γ ) and L A ( σ, α ) being combinatorial expressions in γ ∈ M n , p ( N 0 ) , λ A 1 , . . . , λ A n ∈ P and σ ∈ S ( µ A ) 1 × · · · × S ( µ A ) n 12 / 16

Leading Monomial Proposition For A , B ∈ Y ( p ) , B < A, then c A ( γ ) ∈ Z for all γ ∈ M n , p ( N 0 ) and a) c A ( γ A ) � = 0 b) c B ( γ A ) = 0 Where ( γ A ) i , j = # { Number of i’s in the j’th row of A } = ( λ A i ) j − ( λ A i − 1 ) j d µ := dim W ( p 2) µ + p 2 . { A 1 , . . . , A d µ } ⊂ Y ( p ) , tableaux of weight µ s.t. A 1 < · · · < A d µ . 13 / 16

Action on Basis Elements Inner product on W ( p 2 ) ⊂ C [ R np ] ⊗ C ℓ p : � x γ e η , x γ ′ e η ′ � := δ γ,γ ′ δ η,η ′ . For v ∈ W ( p 2 ) and k , l ∈ { 1 , . . . , d µ } , f µ ( v ) l = � x γ Al e η γ Al , v � ( U µ ) k , l = c A k ( γ A l ) , and 1 ω B = ( λ B ) 1 ! · · · ( λ B ) n ! ω B . Proposition The matrix U µ is integer and upper triangular, and for any v ∈ dim( W n ( p )) µ + p 2 , d µ � ( U − 1 v = µ ( f µ ( v )) k ω A l . k =1 14 / 16

Action on Basis Elements X i ω A ∈ W n ( p ) µ A + ǫ i + p 2 , and D i ω A ∈ W n ( p ) µ A − ǫ i + p 2 . Proposition Let A , B ∈ Y ( p ) and i ∈ { 1 , . . . , n } . Then p α − 1 � x γ A e η γ A , X i ¯ � � ( − 1) ( λ A ) β c B ( γ A − ǫ i ,α ) ω B ( p ) � = α =1 β =1 p α − 1 � x γ A e η γ A , D i ¯ � � ( − 1) ( λ A ) β (( γ A ) i ,α + 1) c B ( γ A + ǫ i ,α ) . ω B ( p ) � = α =1 β =1 This determines the vector f µ ( X i ω B ) and f µ ( D i ω B ), and thus the action. 15 / 16

Example We calculate X i ω B for B = 2 3 , µ B = (0 , 1 , 1 , 1), i = 1: 4 1 1 4 1 3 A 1 = 2 , A 2 = , A 3 = , A 4 = 2 4 , A 5 = 1 3 1 3 4 , 2 2 3 2 3 4 4 1 2 1 2 1 2 4 1 2 3 1 2 3 4 . A 6 = , A 7 = 3 4 , A 8 = , A 9 = , A 10 = 3 3 4 4 1 − 1 1 1 − 1 − 1 − 1 1 − 1 1 − 1     0 1 0 − 1 1 0 1 − 1 0 − 1 − 1 0 0 1 1 − 1 0 0 0 − 1 1 1         0 0 0 1 − 1 0 0 0 1 1 0     0 0 0 0 1 0 0 0 0 − 1 − 1     U µ B + ǫ i = , f µ B + ǫ i = .     0 0 0 0 0 1 1 − 1 1 − 1 − 1     0 0 0 0 0 0 1 − 1 − 1 − 1 1         0 0 0 0 0 0 0 1 0 1 0     0 0 0 0 0 0 0 0 1 − 1 1 0 0 0 0 0 0 0 0 0 1 0 X 1 ω 2 3 = − 8 ω A 1 − 4 ω A 2 +3 ω A 3 − 2 ω A 4 − 1 ω A 5 − 4 ω A 6 +2 ω A 7 + ω A 9 . 4 16 / 16