Littlewood Richardson polynomials

Transcription

1 Littlewood Richardson polynomials Alexander Molev University of Sydney

2 A diagram (or partition) is a sequence λ = (λ 1,..., λ n ) of integers λ i such that λ 1 λ n 0, depicted as an array of unit boxes.

3 A diagram (or partition) is a sequence λ = (λ 1,..., λ n ) of integers λ i such that λ 1 λ n 0, depicted as an array of unit boxes. Example. The diagram λ = (5, 5, 3) is

4 A diagram (or partition) is a sequence λ = (λ 1,..., λ n ) of integers λ i such that λ 1 λ n 0, depicted as an array of unit boxes. Example. The diagram λ = (5, 5, 3) is The number of boxes is the weight of the diagram, denoted λ. The number of nonzero rows is its length, denoted l(λ).

5 A diagram (or partition) is a sequence λ = (λ 1,..., λ n ) of integers λ i such that λ 1 λ n 0, depicted as an array of unit boxes. Example. The diagram λ = (5, 5, 3) is λ = 13 l(λ) = 3 The number of boxes is the weight of the diagram, denoted λ. The number of nonzero rows is its length, denoted l(λ).

6 Littlewood Richardson coefficients c ν λµ

7 Littlewood Richardson coefficients c ν λµ Let l(λ) n and let V λ denote the irreducible gl n -module with the highest weight λ.

8 Littlewood Richardson coefficients c ν λµ Let l(λ) n and let V λ denote the irreducible gl n -module with the highest weight λ. Then V λ V µ L = cλµ ν V ν. ν Here l(λ), l(µ), l(ν) n.

9 Let λ = k and let χ λ denote the corresponding irreducible character of the symmetric group S k.

10 Let λ = k and let χ λ denote the corresponding irreducible character of the symmetric group S k. Then Ind S k+l S k S l ( χ λ χ µ) = ν c ν λµ χν. Here λ = k, µ = l, ν = k + l.

11 Let λ = k and let χ λ denote the corresponding irreducible character of the symmetric group S k. Then Ind S k+l S k S l ( χ λ χ µ) = ν c ν λµ χν. Here λ = k, µ = l, ν = k + l. In particular, c ν λµ 0 = ν = λ + µ.

12 Let n and N be nonnegative integers with n N and let Gr n,n denote the Grassmannian of the n-dimensional vector subspaces of C N. The cohomology ring H (Gr n,n ) has a basis of the Schubert classes σ λ parameterized by all diagrams λ contained in the n m rectangle, m = N n.

13 Let n and N be nonnegative integers with n N and let Gr n,n denote the Grassmannian of the n-dimensional vector subspaces of C N. The cohomology ring H (Gr n,n ) has a basis of the Schubert classes σ λ parameterized by all diagrams λ contained in the n m rectangle, m = N n. We have σ λ σ µ = ν c ν λµ σ ν. Here λ, µ, ν are contained in the n m rectangle.

14 Let x = (x 1, x 2,... ) be an infinite set of variables.

15 Let x = (x 1, x 2,... ) be an infinite set of variables. Given any partition λ = (λ 1,..., λ n ), define the corresponding monomial symmetric function by m λ (x) = σ x λ 1 σ(1) x λ 2 σ(2)... x λn σ(n), summed over permutations σ of the x i which give distinct monomials.

16 Let x = (x 1, x 2,... ) be an infinite set of variables. Given any partition λ = (λ 1,..., λ n ), define the corresponding monomial symmetric function by m λ (x) = σ x λ 1 σ(1) x λ 2 σ(2)... x λn σ(n), summed over permutations σ of the x i which give distinct monomials. The algebra of symmetric functions Λ is defined as the Q-span of all monomial symmetric functions.

17 Examples: power sums symmetric functions p k (x) = m (k) (x) = i=1 x k i,

18 Examples: power sums symmetric functions p k (x) = m (k) (x) = i=1 x k i, elementary symmetric functions e k (x) = m (1 k ) (x) = x i1... x ik, i 1 > >i k 1

19 Examples: power sums symmetric functions p k (x) = m (k) (x) = i=1 x k i, elementary symmetric functions e k (x) = m (1 k ) (x) = x i1... x ik, i 1 > >i k 1 complete symmetric functions h k (x) = m λ (x) = x i1... x ik. λ =k i 1 i k 1

20 Schur functions

21 Schur functions Given a diagram λ, a reverse λ-tableau T is obtained by filling in the boxes of λ with the numbers 1, 2,... in such a way that the entries weakly decrease along the rows and strictly decrease down the columns. If α = (i, j) is a box of λ we let T (α) = T (i, j) denote the entry of T in the box α.

22 Schur functions Given a diagram λ, a reverse λ-tableau T is obtained by filling in the boxes of λ with the numbers 1, 2,... in such a way that the entries weakly decrease along the rows and strictly decrease down the columns. If α = (i, j) is a box of λ we let T (α) = T (i, j) denote the entry of T in the box α. Example. A reverse λ-tableau for λ = (5, 5, 3):

23 The Schur function s λ (x) corresponding to λ is defined by s λ (x) = T x T (α), α λ summed over the reverse λ-tableaux T.

24 The Schur function s λ (x) corresponding to λ is defined by s λ (x) = T x T (α), α λ summed over the reverse λ-tableaux T. Example. For λ = (2, 1) the reverse tableaux are i j with i j and i > k. k

25 The Schur function s λ (x) corresponding to λ is defined by s λ (x) = T x T (α), α λ summed over the reverse λ-tableaux T. Example. For λ = (2, 1) the reverse tableaux are i j with i j and i > k. k Hence s (2,1) (x) = x i x j x k. i j, i>k

26 Note also hk (x) = s(k ) (x), ek (x) = s(1k ) (x).

27 Note also h k (x) = s (k) (x), e k (x) = s (1 k ) (x). Tableaux i 1 i 2 i 1 i 2 i k. i k

28 Note also h k (x) = s (k) (x), e k (x) = s (1 k ) (x). Tableaux i 1 i 2 i 1 i 2 i k. i k Hence s (k) (x) = x i1... x ik, s (1 k ) (x) = i 1 i k 1 i 1 > >i k 1 x i1... x ik.

29 The Schur functions s λ (x) parameterized by all diagrams form a basis of the algebra of symmetric functions Λ.

30 The Schur functions s λ (x) parameterized by all diagrams form a basis of the algebra of symmetric functions Λ. The relation s λ (x) s µ (x) = ν c ν λµ s ν(x) defines the Littlewood Richardson coefficients c ν λµ.

31 History: D. E. Littlewood and A. R. Richardson (1934), (general formulation, a proof in the case l(µ) 2), G. de B. Robinson (1938, proof contains gaps).

32 History: D. E. Littlewood and A. R. Richardson (1934), (general formulation, a proof in the case l(µ) 2), G. de B. Robinson (1938, proof contains gaps). Complete proofs: G. P. Thomas (1974 PhD thesis, 1978 paper), M. P. Schützenberger (1977).

33 History: D. E. Littlewood and A. R. Richardson (1934), (general formulation, a proof in the case l(µ) 2), G. de B. Robinson (1938, proof contains gaps). Complete proofs: G. P. Thomas (1974 PhD thesis, 1978 paper), M. P. Schützenberger (1977). Now: A couple of dozens of versions of the LR rule, cλµ ν counts tableaux, trees, hives, honeycombs, cartons, puzzles,....

34 Knutson Tao Woodward puzzles Suppose that λ, µ, ν are contained in n m rectangle. Write each partition in the binary code of length n + m.

35 Knutson Tao Woodward puzzles Suppose that λ, µ, ν are contained in n m rectangle. Write each partition in the binary code of length n + m. Example. The diagram λ = (5, 5, 3) inside 4 7 rectangle is represented as follows:

36 Knutson Tao Woodward puzzles Suppose that λ, µ, ν are contained in n m rectangle. Write each partition in the binary code of length n + m. Example. The diagram λ = (5, 5, 3) inside 4 7 rectangle is represented as follows:

37 Write the sequences corresponding to λ, µ, ν around the border of an equilateral triangle of side length n + m as indicated:

38 Write the sequences corresponding to λ, µ, ν around the border of an equilateral triangle of side length n + m as indicated: λ µ ν

39 Write the sequences corresponding to λ, µ, ν around the border of an equilateral triangle of side length n + m as indicated: λ µ ν Theorem [KTW 03]. The Littlewood Richardson coefficient c ν λµ equals the number of triangular puzzles which can be obtained with the use of the following set of unit puzzle pieces.

40 Puzzle pieces

41 Example. Calculation of cλµ ν, λ = (2), µ = (1), ν = (2, 1).

42 Example. Calculation of cλµ ν, λ = (2), µ = (1), ν = (2, 1). Take n = m = 2 so that

43 Example. Calculation of cλµ ν, λ = (2), µ = (1), ν = (2, 1). Take n = m = 2 so that λ

44 Example. Calculation of cλµ ν, λ = (2), µ = (1), ν = (2, 1). Take n = m = 2 so that λ µ

45 Example. Calculation of cλµ ν, λ = (2), µ = (1), ν = (2, 1). Take n = m = 2 so that λ µ ν

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51 c 1 0 λµ ν =

52 A tableau version of the LR rule Let R denote a sequence of diagrams µ = ρ (0) ρ (1) ρ (l 1) ρ (l) = ν, ρ σ means σ is obtained from ρ by adding one box.

53 A tableau version of the LR rule Let R denote a sequence of diagrams µ = ρ (0) ρ (1) ρ (l 1) ρ (l) = ν, ρ σ means σ is obtained from ρ by adding one box. Let r i denote the row number of the box added to ρ (i 1). The sequence r 1 r 2... r l is the Yamanouchi symbol of R.

54 Example. Let R : (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1)

55 Example. Let R : (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) or, with diagrams,

56 Example. Let R : (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) or, with diagrams, the Yamanouchi symbol is

57 The column word of a tableau T is the sequence of all entries of T written in the column order: by reading the entries by columns from left to right and from bottom to top in each column.

58 The column word of a tableau T is the sequence of all entries of T written in the column order: by reading the entries by columns from left to right and from bottom to top in each column. Example. A reverse λ-tableau for λ = (5, 5, 3):

59 The column word of a tableau T is the sequence of all entries of T written in the column order: by reading the entries by columns from left to right and from bottom to top in each column. Example. A reverse λ-tableau for λ = (5, 5, 3): Its column word is

60 Let λ and ν be two diagrams.

61 Let λ and ν be two diagrams. A reverse λ-tableau T is called ν-bounded if the entries in the top row do not exceed the respective column lengths of ν: T (1, 1) ν 1, T (1, 2) ν 2,....

62 Let λ and ν be two diagrams. A reverse λ-tableau T is called ν-bounded if the entries in the top row do not exceed the respective column lengths of ν: T (1, 1) ν 1, T (1, 2) ν 2,.... Such tableaux exist only if λ ν.

63 Let λ and ν be two diagrams. A reverse λ-tableau T is called ν-bounded if the entries in the top row do not exceed the respective column lengths of ν: T (1, 1) ν 1, T (1, 2) ν 2,.... Such tableaux exist only if λ ν.

64 Let λ and ν be two diagrams. A reverse λ-tableau T is called ν-bounded if the entries in the top row do not exceed the respective column lengths of ν: T (1, 1) ν 1, T (1, 2) ν 2,.... Such tableaux exist only if λ ν. Maximal entries:

65 Theorem. The Littlewood Richardson coefficient c ν λµ equals the number of common elements in the two sets: { } column words of the ν-bounded reverse λ-tableaux and { } Yamanouchi symbols of the sequences from µ to ν.

66 Theorem. The Littlewood Richardson coefficient c ν λµ equals the number of common elements in the two sets: { } column words of the ν-bounded reverse λ-tableaux and { } Yamanouchi symbols of the sequences from µ to ν. Remarks. This is a particular case of a more general theorem.

67 Theorem. The Littlewood Richardson coefficient c ν λµ equals the number of common elements in the two sets: { } column words of the ν-bounded reverse λ-tableaux and { } Yamanouchi symbols of the sequences from µ to ν. Remarks. This is a particular case of a more general theorem. The theorem is equivalent to the puzzle rule (T. Tao).

68 Example. Calculation of cλµ ν, λ = µ = (2, 1), ν = (3, 2, 1).

69 Example. Calculation of cλµ ν, λ = µ = (2, 1), ν = (3, 2, 1). Here ν 1 = 3, ν 2 = 2, ν 3 = 1. The ν-bounded λ-tableaux are

70 Example. Calculation of cλµ ν, λ = µ = (2, 1), ν = (3, 2, 1). Here ν 1 = 3, ν 2 = 2, ν 3 = 1. The ν-bounded λ-tableaux are The set of column words is { } 2 3 2, 1 3 2, 2 3 1, 1 3 1, 1 2 2,

71 The sequences from (2, 1) to (3, 2, 1): (2, 1) (3, 1) (3, 2) (3, 2, 1) (2, 1) (3, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 2) (3, 2) (3, 2, 1) (2, 1) (2, 2) (2, 2, 1) (3, 2, 1) (2, 1) (2, 1, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 1, 1) (2, 2, 1) (3, 2, 1)

72 The sequences from (2, 1) to (3, 2, 1): (2, 1) (3, 1) (3, 2) (3, 2, 1) (2, 1) (3, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 2) (3, 2) (3, 2, 1) (2, 1) (2, 2) (2, 2, 1) (3, 2, 1) (2, 1) (2, 1, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 1, 1) (2, 2, 1) (3, 2, 1) The set of the Yamanouchi symbols is { } 1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2,

73 The sequences from (2, 1) to (3, 2, 1): (2, 1) (3, 1) (3, 2) (3, 2, 1) (2, 1) (3, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 2) (3, 2) (3, 2, 1) (2, 1) (2, 2) (2, 2, 1) (3, 2, 1) (2, 1) (2, 1, 1) (3, 1, 1) (3, 2, 1) (2, 1) (2, 1, 1) (2, 2, 1) (3, 2, 1) The set of the Yamanouchi symbols is { } 1 2 3, 1 3 2, 2 1 3, 2 3 1, 3 1 2, Hence c ν λµ = 2.

74 Double symmetric functions The elements of the algebra of symmetric functions Λ can be viewed as sequences of symmetric polynomials: i=1 x k i x k 1, x k 1 + x k 2,..., x k 1 + x k x k n,...

75 Double symmetric functions The elements of the algebra of symmetric functions Λ can be viewed as sequences of symmetric polynomials: i=1 x k i x k 1, x k 1 + x k 2,..., x k 1 + x k x k n,... The polynomials in such a sequence are compatible with the evaluation homomorphisms ϕ n : P(x 1,..., x n ) P(x 1,..., x n 1, 0).

76 Let a = (a i ), i Z, be a sequence of variables. Denote by Λ n the ring of symmetric polynomials in x 1,..., x n with coefficients in Q[a].

77 Let a = (a i ), i Z, be a sequence of variables. Denote by Λ n the ring of symmetric polynomials in x 1,..., x n with coefficients in Q[a]. Consider the sequences of symmetric polynomials compatible with the evaluation homomorphisms ϕ n : Λ n Λ n 1, P(x 1,..., x n ) P(x 1,..., x n 1, a n ).

78 Let a = (a i ), i Z, be a sequence of variables. Denote by Λ n the ring of symmetric polynomials in x 1,..., x n with coefficients in Q[a]. Consider the sequences of symmetric polynomials compatible with the evaluation homomorphisms ϕ n : Λ n Λ n 1, P(x 1,..., x n ) P(x 1,..., x n 1, a n ). The ring Λ a of double symmetric functions is formed by such sequences of polynomials. The sequences can also be regarded as formal series.

79 Examples. We have ϕ n : n i=1 (x k i n 1 ai k ) i=1 (x k i a k i ) hence p k (x a) = i=1 (x k i a k i ) Λ a, the double power sums symmetric function.

80 Examples. We have ϕ n : n i=1 (x k i n 1 ai k ) i=1 (x k i a k i ) hence p k (x a) = i=1 (x k i a k i ) Λ a, the double power sums symmetric function. Λ a is the ring of polynomials in p 1 (x a), p 2 (x a),.... with coefficients in Q[a].

81 Examples. We have ϕ n : n i=1 (x k i n 1 ai k ) i=1 (x k i a k i ) hence p k (x a) = i=1 (x k i a k i ) Λ a, the double power sums symmetric function. Λ a is the ring of polynomials in p 1 (x a), p 2 (x a),.... with coefficients in Q[a]. Note that Λ 0 = Λ.

82 Double Schur functions For any diagram λ define the double Schur function by s λ (x a) = T (x T (α) a T (α) c(α) ), α λ summed over the reverse λ-tableaux T, c(α) = j i is the content of the box α = (i, j).

83 Double Schur functions For any diagram λ define the double Schur function by s λ (x a) = T (x T (α) a T (α) c(α) ), α λ summed over the reverse λ-tableaux T, c(α) = j i is the content of the box α = (i, j). The double Schur functions form a basis of Λ a over Q[a].

84 Example. For λ = (2, 1) the reverse tableaux are i j with i j and i > k k

85 Example. For λ = (2, 1) the reverse tableaux are i j with i j and i > k k Hence s (2,1) (x a) = (x i a i )(x j a j 1 )(x k a k+1 ). i j, i>k

86 Set h k (x a) = s (k) (x a), e k (x a) = s (1 k )(x a).

87 Set h k (x a) = s (k) (x a), e k (x a) = s (1 k )(x a). Tableaux i 1 i 2 i 1 i 2 i k. i k

88 Set h k (x a) = s (k) (x a), e k (x a) = s (1 k )(x a). Tableaux i 1 i 2 i 1 i 2 i k. i k Double complete and elementary symmetric functions: h k (x a) = (x i1 a i1 )... (x ik a ik k+1), i 1 i k e k (x a) = (x i1 a i1 )... (x ik a ik +k 1). i 1 > >i k

89 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a).

90 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a). Properties. cλµ ν (a) 0 only if ν λ + µ.

91 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a). Properties. cλµ ν (a) 0 only if ν λ + µ. cλµ ν (a) is homogeneous of degree λ + µ ν.

92 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a). Properties. cλµ ν (a) 0 only if ν λ + µ. cλµ ν (a) is homogeneous of degree λ + µ ν. c ν λµ (a) = c ν λµ if λ + µ = ν or a = (0).

93 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a). Properties. cλµ ν (a) 0 only if ν λ + µ. cλµ ν (a) is homogeneous of degree λ + µ ν. c ν λµ (a) = c ν λµ if λ + µ = ν or a = (0). c ν λµ (a) = c ν µλ (a).

94 Define the Littlewood Richardson polynomials cλµ ν (a) Q[a] by s λ (x a) s µ (x a) = ν c ν λµ (a) s ν(x a). Properties. cλµ ν (a) 0 only if ν λ + µ. cλµ ν (a) is homogeneous of degree λ + µ ν. c ν λµ (a) = c ν λµ if λ + µ = ν or a = (0). c ν λµ (a) = c ν µλ (a). cλµ ν (a) 0 only if λ ν and µ ν.

95 Calculation of c ν λµ (a) Given a sequence R from µ to ν with the Yamanouchi symbol r 1 r 2... r l, introduce the set T (λ, R) of barred reverse λ-tableaux T with entries from {1, 2,... } such that T contains entries r 1, r 2,..., r l listed in the column order.

96 Calculation of c ν λµ (a) Given a sequence R from µ to ν with the Yamanouchi symbol r 1 r 2... r l, introduce the set T (λ, R) of barred reverse λ-tableaux T with entries from {1, 2,... } such that T contains entries r 1, r 2,..., r l listed in the column order. We will distinguish these entries by barring each of them.

97 Calculation of c ν λµ (a) Given a sequence R from µ to ν with the Yamanouchi symbol r 1 r 2... r l, introduce the set T (λ, R) of barred reverse λ-tableaux T with entries from {1, 2,... } such that T contains entries r 1, r 2,..., r l listed in the column order. We will distinguish these entries by barring each of them. An element T T (λ, R) is a pair consisting of a reverse λ-tableau and a sequence of barred entries compatible with R.

98 Example. Let R be the sequence (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) so that the Yamanouchi symbol is

99 Example. Let R be the sequence (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) so that the Yamanouchi symbol is Let λ = (5, 5, 3). The barred λ-tableau belongs to T (λ, R).

100 Example. Let R be the sequence (3, 1) (3, 2) (3, 2, 1) (3, 3, 1) (4, 3, 1) so that the Yamanouchi symbol is Let λ = (5, 5, 3). The barred λ-tableau belongs to T (λ, R).

101 Given a sequence of diagrams R : µ = ρ (0) ρ (1) ρ (l 1) ρ (l) = ν, set ρ(α) = ρ (i) for any box α occupied by an unbarred entry of T, between r i and r i+1 in column order.

102 Given a sequence of diagrams R : µ = ρ (0) ρ (1) ρ (l 1) ρ (l) = ν, set ρ(α) = ρ (i) for any box α occupied by an unbarred entry of T, between r i and r i+1 in column order. The barred entries r 1, r 2,..., r l of T divide the tableau into regions marked by the elements of the sequence R : r 1 r 2 r l ρ (l) ρ (0) ρ (1)

103 Theorem (Kreiman & M. 07, independently). We have c ν λµ (a) = R T α λ T (α) unbarred ( a T (α) ρ(α)t (α) a T (α) c(α) ), summed over all sequences R from µ to ν and all ν-bounded reverse λ-tableaux T T (λ, R).

104 Theorem (Kreiman & M. 07, independently). We have c ν λµ (a) = R T α λ T (α) unbarred ( a T (α) ρ(α)t (α) a T (α) c(α) ), summed over all sequences R from µ to ν and all ν-bounded reverse λ-tableaux T T (λ, R). Moreover, in each factor ρ(α) T (α) > c(α).

105 Theorem (Kreiman & M. 07, independently). We have c ν λµ (a) = R T α λ T (α) unbarred ( a T (α) ρ(α)t (α) a T (α) c(α) ), summed over all sequences R from µ to ν and all ν-bounded reverse λ-tableaux T T (λ, R). Moreover, in each factor ρ(α) T (α) > c(α). Remarks. If ν = λ + µ then this is a version of the LR rule.

106 Theorem (Kreiman & M. 07, independently). We have c ν λµ (a) = R T α λ T (α) unbarred ( a T (α) ρ(α)t (α) a T (α) c(α) ), summed over all sequences R from µ to ν and all ν-bounded reverse λ-tableaux T T (λ, R). Moreover, in each factor ρ(α) T (α) > c(α). Remarks. If ν = λ + µ then this is a version of the LR rule. cλµ ν (a) is Graham-positive: it is a polynomial in the differences a i a j, i < j, with positive integer coefficients.

107 Example. Calculation of c ν λµ (a), λ = (2, 1), µ = (3, 1), ν = (4, 1, 1).

108 Example. Calculation of c ν λµ (a), λ = (2, 1), µ = (3, 1), ν = (4, 1, 1). Here ν 1 = 3, ν 2 = 1, ν 3 = 1, ν 4 = 1. The ν-bounded λ-tableaux

109 Example. Calculation of c ν λµ (a), λ = (2, 1), µ = (3, 1), ν = (4, 1, 1). Here ν 1 = 3, ν 2 = 1, ν 3 = 1, ν 4 = 1. The ν-bounded λ-tableaux There are two sequences R 1 : (3, 1) (4, 1) (4, 1, 1) and R 2 : (3, 1) (3, 1, 1) (4, 1, 1) with the respective Yamanouchi symbols 1 3 and 3 1.

110 T (λ, R 1 ) contains one barred tableau 3 1 with T (α) = 1, ρ(α) = (4, 1, 1), c(α) = 1, 1 contributing a T (α) ρ(α)t (α) a T (α) c(α) = a 3 a 0.

111 T (λ, R 1 ) contains one barred tableau 3 1 with T (α) = 1, ρ(α) = (4, 1, 1), c(α) = 1, 1 contributing a T (α) ρ(α)t (α) a T (α) c(α) = a 3 a 0. T (λ, R 2 ) contains two barred tableaux with contributions 3 1 a 2 a 2, a 1 a 3. 2

112 T (λ, R 1 ) contains one barred tableau 3 1 with T (α) = 1, ρ(α) = (4, 1, 1), c(α) = 1, 1 contributing a T (α) ρ(α)t (α) a T (α) c(α) = a 3 a 0. T (λ, R 2 ) contains two barred tableaux with contributions 3 1 a 2 a 2, a 1 a 3. 2 Hence c ν λµ (a) = a 3 a 0 + a 2 a 2 + a 1 a 3.

113 Example. For the product of the double Schur functions s (2) (x a) and s (2,1) (x a) we have s (2) (x a) s (2,1) (x a) = s (4,1) (x a) + s (3,2) (x a) + s (3,1,1) (x a) + s (2,2,1) (x a) + ( ) a 1 a 0 s(2,1,1) (x a) + ( ) a 1 a 2 s(2,2) (x a) + ( ) a 1 a 2 + a 2 a 0 s(3,1) (x a) + ( ) ( ) a 1 a 2 a 1 a 0 s(2,1) (x a).

114 Example. For the product of the double Schur functions s (2) (x a) and s (2,1) (x a) we have s (2) (x a) s (2,1) (x a) = s (4,1) (x a) + s (3,2) (x a) + s (3,1,1) (x a) + s (2,2,1) (x a) + ( ) a 1 a 0 s(2,1,1) (x a) + ( ) a 1 a 2 s(2,2) (x a) + ( ) a 1 a 2 + a 2 a 0 s(3,1) (x a) + ( ) ( ) a 1 a 2 a 1 a 0 s(2,1) (x a).

115 Example. For the product of the double Schur functions s (2) (x a) and s (2,1) (x a) we have s (2) (x a) s (2,1) (x a) = s (4,1) (x a) + s (3,2) (x a) + s (3,1,1) (x a) + s (2,2,1) (x a) + ( ) a 1 a 0 s(2,1,1) (x a) + ( ) a 1 a 2 s(2,2) (x a) + ( ) a 1 a 2 + a 2 a 0 s(3,1) (x a) + ( ) ( ) a 1 a 2 a 1 a 0 s(2,1) (x a).

116 Example. For any diagram λ, c λ λλ (a) = (i, j) λ ( ai λi a λ j j+1 ).

117 Example. For any diagram λ, c λ λλ (a) = (i, j) λ ( ai λi a λ j j+1 ). Setting a i = i for all i gives the product of the hooks of λ.

118 Example. For any diagram λ, c λ λλ (a) = (i, j) λ ( ai λi a λ j j+1 ). Setting a i = i for all i gives the product of the hooks of λ. Proof of the theorem. Calculate cλµ ν (a) by induction on ν µ.

119 Example. For any diagram λ, c λ λλ (a) = (i, j) λ ( ai λi a λ j j+1 ). Setting a i = i for all i gives the product of the hooks of λ. Proof of the theorem. Calculate cλµ ν (a) by induction on ν µ. Starting point: the Vanishing Theorem (A. Okounkov, 96): s λ (a ρ a) = 0 unless λ ρ, s λ (a λ a) 0, where a ρ = (a 1 ρ1, a 2 ρ2,... ).

120 Hence, if R = {µ} is a one-term sequence, then c µ λµ (a) = s λ(a µ a), a µ = (a 1 µ1, a 2 µ2,... ),

121 Hence, if R = {µ} is a one-term sequence, then c µ λµ (a) = s λ(a µ a), a µ = (a 1 µ1, a 2 µ2,... ), and so c µ λµ (a) = T α λ ( a T (α) µt (α) a T (α) c(α) ).

122 Hence, if R = {µ} is a one-term sequence, then c µ λµ (a) = s λ(a µ a), a µ = (a 1 µ1, a 2 µ2,... ), and so c µ λµ (a) = T α λ ( a T (α) µt (α) a T (α) c(α) ). Then use the recurrence ( cλµ ν (a) = 1 cλµ ν a ν a µ +(a) cλµ ), ν (a) µ µ + ν ν where a ν a µ = ) i 1 ((a ν ) i (a µ ) i (M. & Sagan, 99).

123 Knutson Tao puzzles Write the binary sequences corresponding to λ, µ, ν around the border of an equilateral triangle:

124 Knutson Tao puzzles Write the binary sequences corresponding to λ, µ, ν around the border of an equilateral triangle: λ µ ν

125 Knutson Tao puzzles Write the binary sequences corresponding to λ, µ, ν around the border of an equilateral triangle: λ µ ν Theorem [KT 03]. The Littlewood Richardson polynomial cλµ ν (a) equals the sum of weights of triangular puzzles, where an additional puzzle piece can be used.

126 Additional puzzle piece

127 Additional puzzle piece Each occurrence of this puzzle piece contributes a factor by the rule: i j a i m a j m

128 Quantum immanants (Okounkov, 96) Let l(λ) n and k = λ. Consider the irreducible representation V λ of the symmetric group S k.

129 Quantum immanants (Okounkov, 96) Let l(λ) n and k = λ. Consider the irreducible representation V λ of the symmetric group S k. The orthonormal Young basis {v U } of V λ is parameterized by the standard λ-tableaux U.

130 Quantum immanants (Okounkov, 96) Let l(λ) n and k = λ. Consider the irreducible representation V λ of the symmetric group S k. The orthonormal Young basis {v U } of V λ is parameterized by the standard λ-tableaux U. Set φ U = (s v U, v U ) s 1 C[S k ]. s S k Consider the natural action of the symmetric group S k on the tensor product of k copies of C n. Denote by Φ U the image of φ U in (Mat n ) k.

131 Consider the Lie algebra gl n with its standard basis {E ab }, where a, b {1,..., n}.

132 Consider the Lie algebra gl n with its standard basis {E ab }, where a, b {1,..., n}. For any c C set n E c = e ab ( E ab δ ab c ) Mat n U(gl n ). a,b=1

133 Consider the Lie algebra gl n with its standard basis {E ab }, where a, b {1,..., n}. For any c C set n E c = e ab ( E ab δ ab c ) Mat n U(gl n ). a,b=1 The tensor products of the form (E c 1 )... (E c k ) are regarded as elements of the algebra Mat n... Mat n U(gl n ).

134 The quantum immanant S λ associated with λ is the element of U(gl n ) defined by S λ = 1 H λ tr (E c 1 )... (E c k ) Φ U, where the trace is taken over all matrix factors, U is a standard λ-tableau, c r = j i if r occupies the box (i, j) in U, H λ = k!/ dim V λ is the product of the hooks of λ.

135 Examples. Quantum minors (Capelli elements) S (1 k ) = a 1 < <a k p S k sgn p E a1,a p(1)... (E + k 1) ak,a p(k).

136 Examples. Quantum minors (Capelli elements) S (1 k ) = a 1 < <a k p S k sgn p E a1,a p(1)... (E + k 1) ak,a p(k). Quantum permanents S (k) = 1 α a 1 a 1!... α n! k p S k E a1,a p(1)... (E k + 1) ak,a p(k), where α i is the multiplicity of i in a 1,..., a k, each a r {1,..., n}.

137 The quantum immanants S λ with l(λ) n form a basis of the center of the universal enveloping algebra U(gl n ).

138 The quantum immanants S λ with l(λ) n form a basis of the center of the universal enveloping algebra U(gl n ). Define the coefficients fλµ ν by the expansion S λ S µ = ν f ν λµ S ν.

139 The quantum immanants S λ with l(λ) n form a basis of the center of the universal enveloping algebra U(gl n ). Define the coefficients fλµ ν by the expansion S λ S µ = ν f ν λµ S ν. Corollary. f ν λµ = c ν λµ (a) for the specialization a i = i for i Z.

140 The coefficient fλµ ν is zero unless λ, µ ν. If λ, µ ν then f ν λµ = R T α λ T (α) unbarred ( ) ρ(α) T (α) c(α), summed over all sequences R from µ to ν and all ν-bounded reverse λ-tableaux T T (λ, R). In particular, the f ν λµ are nonnegative integers.

141 Example. For any n 3 we have S (2) S (2,1) = S (4,1) + S (3,2) + S (3,1,1) + S (2,2,1) + S (2,1,1) + 5 S (3,1) + 3 S (2,2) + 3 S (2,1).

142 Example. For any n 3 we have S (2) S (2,1) = S (4,1) + S (3,2) + S (3,1,1) + S (2,2,1) + S (2,1,1) + 5 S (3,1) + 3 S (2,2) + 3 S (2,1). If n = 2 then S (2) S (2,1) = S (4,1) + S (3,2) + 5 S (3,1) + 3 S (2,2) + 3 S (2,1).

143 Equivariant Schubert calculus on the Grassmannian The torus T = (C ) N acts naturally on Gr n,n. The equivariant cohomology ring H T (Gr n,n) is a module over Z[t 1,..., t N ] = H T ({pt}).

144 Equivariant Schubert calculus on the Grassmannian The torus T = (C ) N acts naturally on Gr n,n. The equivariant cohomology ring H T (Gr n,n) is a module over Z[t 1,..., t N ] = H T ({pt}). It has a basis of the equivariant Schubert classes σ λ parameterized by all diagrams λ contained in the n m rectangle, m = N n.

145 Corollary. We have σ λ σ µ = ν d ν λµ σ ν, where dλµ ν = c λµ ν (a) with the sequence a specialized as follows: a m+1 = t 1,..., a n = t N, and a i = 0 for all remaining values of i.

146 Corollary. We have σ λ σ µ = ν d ν λµ σ ν, where dλµ ν = c λµ ν (a) with the sequence a specialized as follows: a m+1 = t 1,..., a n = t N, and a i = 0 for all remaining values of i. The d ν λµ are polynomials in the t i t j, i > j with positive integer coefficients (the positivity property, Graham 01).

147 The coefficients d ν λµ, regarded as polynomials in the a i, are independent of n and m, as soon as the inequalities n λ 1 + µ 1 and m λ 1 + µ 1 hold (the stability property).

148 The coefficients d ν λµ, regarded as polynomials in the a i, are independent of n and m, as soon as the inequalities n λ 1 + µ 1 and m λ 1 + µ 1 hold (the stability property). Remark. The puzzle rule of Knutson and Tao gives a manifestly positive formula for the dλµ ν, while the tableau rule is manifestly stable.

149 Example. For any n 3 and m 4 we have σ (2) σ (2,1) = σ (4,1) + σ (3,2) + σ (3,1,1) + σ (2,2,1) + (t m t m 1 ) σ (2,1,1) + (t m+2 t m 1 ) σ (2,2) + (t m+2 t m 1 + t m t m 2 ) σ (3,1) + (t m+2 t m 1 ) (t m t m 1 ) σ (2,1).

150 Dimensions of skew diagrams Let µ λ be two diagrams. The skew diagram θ = λ/µ is the set-theoretical difference of the diagrams λ and µ:

151 Dimensions of skew diagrams Let µ λ be two diagrams. The skew diagram θ = λ/µ is the set-theoretical difference of the diagrams λ and µ: Example. λ = (10, 8, 5, 4, 2) and µ = (6, 3): θ = λ/µ

152 If θ has n = θ boxes, then a standard θ-tableau is obtained by filling the boxes bijectively with the numbers {1, 2,..., n} in such a way that the entries increase along the rows and down the columns.

153 If θ has n = θ boxes, then a standard θ-tableau is obtained by filling the boxes bijectively with the numbers {1, 2,..., n} in such a way that the entries increase along the rows and down the columns. The dimension dim θ of a skew diagram θ is the number of the standard θ-tableaux.

154 If θ has n = θ boxes, then a standard θ-tableau is obtained by filling the boxes bijectively with the numbers {1, 2,..., n} in such a way that the entries increase along the rows and down the columns. The dimension dim θ of a skew diagram θ is the number of the standard θ-tableaux. Set H θ = θ! dim θ.

155 If θ is normal (nonskew), then H θ coincides with the product of the hooks of θ due to the hook formula.

156 If θ is normal (nonskew), then H θ coincides with the product of the hooks of θ due to the hook formula. Example. The hooks of θ = (4, 3, 1):

157 If θ is normal (nonskew), then H θ coincides with the product of the hooks of θ due to the hook formula. Example. The hooks of θ = (4, 3, 1): Hence H θ = = 576 and dim θ = 70.

158 If θ is normal (nonskew), then H θ coincides with the product of the hooks of θ due to the hook formula. Example. The hooks of θ = (4, 3, 1): Hence H θ = = 576 and dim θ = 70. If θ = θ 1 θ r, then H θ = H θ1... H θr.

159 Example. Let θ = (3, 2)/(1). The standard θ-tableaux are

160 Example. Let θ = (3, 2)/(1). The standard θ-tableaux are Hence dim θ = 5 and H θ = 24/5.

161 Example. Let θ = (3, 2)/(1). The standard θ-tableaux are Hence dim θ = 5 and H θ = 24/5. Corollary. We have c ν λµ = ρ ( 1) ν/ρ H ρ H ν/ρ H ρ/λ H ρ/µ, summed over the diagrams ρ which contain both λ and µ, and are contained in ν.

162 Example. Let λ = µ = (2, 1), ν = (3, 2, 1).

163 Example. Let λ = µ = (2, 1), ν = (3, 2, 1). Then ρ runs over the set of diagrams { } (2, 1), (3, 1), (2, 2), (2, 1, 1), (3, 2), (3, 1, 1), (2, 2, 1), (3, 2, 1).

164 Example. Let λ = µ = (2, 1), ν = (3, 2, 1). Then ρ runs over the set of diagrams { } (2, 1), (3, 1), (2, 2), (2, 1, 1), (3, 2), (3, 1, 1), (2, 2, 1), (3, 2, 1). Here H ν/ρ = H ρ/λ = H ρ/µ = 1 for all ρ.

165 Example. Let λ = µ = (2, 1), ν = (3, 2, 1). Then ρ runs over the set of diagrams { } (2, 1), (3, 1), (2, 2), (2, 1, 1), (3, 2), (3, 1, 1), (2, 2, 1), (3, 2, 1). Here H ν/ρ = H ρ/λ = H ρ/µ = 1 for all ρ. Hence c (3,2,1) (2,1)(2,1) = = 2.