Representation Theory

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1 Representation Theory Lectures by Stuart Martin Notes by David Mehrle Cambridge University Mathematical Tripos Part III Lent 206 Contents Introduction The Okounkov-Vershik approach Coxeter generators acting on the Young basis Content vectors and tableaux Main result and its consequences Young s Seminormal and Orthogonal Forms Hook Length Formula A bijection that counts Extra Material Last updated May 20, 206.

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3 Lecture 5 January 206 Introduction Remark.. Welcome to rep theory. This is kind of a big audience, so I ll do my best to reduce it by half by at least Monday." The topics we re going to cover in this course are as follows: Overview of representations and characters of finite groups. Representations of symmetric groups: Young symmetrizers, Sþecht modules, branching rule, Gelfand-Tzetlin bases. (This is a very modern approach to representation theory of S n ). Young Tableaux, hook-length formula, RSK algorithm (and what Serre said was the most beautiful proof in all of mathematics.") All of this started with Young, who was actually a clergyman. References B.E. Sagan, The Symmetric Group: representations, combinatorial algorithms and symmetric functions (2nd edn), GTM 203, Springer 200. The classical approach to representation theory of S n. T. Ceccherini-Silberstein, F. Scarabotti and F. Tolli, Representation theory of the symmetric groups: the Okounkov-Vershik approach, character formulas and partition algebras, CUP 200. This has all of the modern stuff in it, including Gelfand-Tzetlin bases. R.P. Stanley, Enumerative Combinatorics, Volume 2 (Chapter 7), CUP 200. James and Kerber CUP 86 (You can tell who wrote which bits, because the stuff James wrote is all correct and everything Kerber wrote is wrong). A. Kleschchev, Linear and Projective Representations of Symmetric Groups. Very dense, and has more than what we need. W. Fulton, Young tableaux, Cambridge University Press, Basic Representation Theory What is representation theory all about? We have groups on one hand, and symmetries of some object on the other hand. It is the total opposite of geometry. In geometry, we have some object and try to figure out what groups describe it. In representation theory, we are given a group and we want to find the things that are described by the groups. GROUPS SYMMETRIES OF SOMETHING symmetric group S n ÐÑ finite set general linear group GL n pcq ÐÑ vector space V, dim V n 3

4 Lecture 5 January 206 For us, GL n pcq is the main continuous group, and S n is the main discrete group we will work with. Definition.2. Let G be a group. A (complex, finite dimensional, linear) representation of G is a homomorphism ρ : G Ñ GLpVq where V is some finitedimensional vector space over C. Equivalently, a representation is a homomorphism R : G Ñ GL n pcq, in which case we may think about matrices Rpgq instead of endomorphisms ρpgq. Note that we have Rpg g 2 q Rpg qrpg 2 q and Rpg q Rpgq. Example.3. Let C n be the finite cyclic group of order n generated by g: C n t, g,..., g n u such that g n. A representation of G on V defines an invertible endomorphism ρpgq P GLpVq with ρpq id V and ρpg k q ρpgq k. Therefore, all other images of ρ are determined by the single operator ρpgq. So what are all the representations of C n? The one dimensional representations R : C n Ñ GL pcq Cˆ are completely determined by Rpgq ζ with ζ n. Hence, ζ is an n-th root of unity. There are n non-isomorphic - dimensional representations of C n. Actually, this isn t an accident: Lemma.4. For any abelian group G, the number of one-dimensional representations is G. Example.5 (Continued from Example.3). What about the d-dimensional representations? Choose a basis of V, such that ρpgq corresponds to a matrix M Rpgq which takes Jordan Normal Form.» J J 2 M where the Jordan blocks J k are of the form» fi λ λ ffi ffi J k.... ffi.. ffi ffi ffi λ fl λ Impose the condition that M n id V. But M n is also block-diagonal, and the blocks of M n are just powers of the Jordan blocks.» M n J n J n 2... J n m... J m fi» ffi ffi ffi fl fi ffi ffi ffi fl... fi ffi ffi ffi fl 4

5 Lecture 2 8 January 206 Hence, for any Jordan block J k, we must have Jk n. Now let s compute: let N be the Jordan matrix with λ 0, so J k λ ` N. Hence, ˆn Jk n pλ ` Nqn λ n id ` λ n N `... ` N n But N p for any p is a matrix with zeros and ones only with the ones along a line in position pi, jq with i j ` p. So J n id only if λ n and N 0. Thus, each J k is a ˆ block and M must be diagonal with respect to this basis. We have just proved: Proposition.6. If V is a representation of C n, there is a basis of V for which the action of every element of C n is a diagonal matrix, with the n-th roots of on the diagonal. In particular, the d-dimensional representations of C n are classified up to isomorphism by unordered d-tuples of n-th roots of unity. Exercise.7. Do the same thing for representations of the infinite cyclic group pz, `q. Show that the d-dimensional representations of Z are in bijection with the conjugacy classes of GL d pcq. Exercise.8. If G is a finite abelian group, show that the d-dimensional isomorphism classes of representations of G are in bijection with unordered d-tuples of -dimensional representations. Definition.9. Two representations R, R 2 of G are equivalent if for each g P G, R pgq CR 2 pgqc for some fixed nonsingular matrix C. Definition.0 (Operations on Representations). Let ρ : G Ñ GLpV q and ρ 2 : G Ñ GLpV 2 q be two representations of G, with dim V k and dim V 2 k 2. Then (a) the direct sum of these representations is a representation ρ ρ ρ 2 : G Ñ GLpV V 2 q of dimension k ` k 2 such that j R pgq 0 Rpgq 0 R pgq (b) the tensor product of these representations is a representation ρ ρ b ρ 2 : G Ñ GLpV b Wq, of dimension k k 2. Last time we defined the tensor product and direct sum of representations. There are many more things we could do here: for any operations on vector spaces, there are similar operations on representations, such as symmetric and exterior powers of representations. Definition.. () A representation ρ is called decomposable if it is equivalent to a direct sum of two other representations, ρ ρ ρ 2, with ρ, ρ 2 both nontrivial, that is, dim ρ 2, ρ 2 ě. Otherwise, ρ is indecomposable. 5

6 Lecture 2 8 January 206 (2) A representation ρ : G Ñ GLpVq is reducible if there is a subspace W Ĺ V with W t0u such that all operators ρpgq preserve W (for all w P W, ρpgqpwq P W). Otherwise, ρ is irreducible (sometimes called simple). Clearly, irreducible implies indecomposable. Theorem.2 (Maschke s Theorem). Over C, for all finite groups G, a representation ρ is irreducible if and only if ρ is indecomposable. Moreover, any representation is a direct sum of irreducible representations, that is, ρ ρ ρ 2... ρ k with ρ i irreducible. In this case, we say ρ is completely reducible or semisimple. For any representation ρ of G, there is a decomposition ρ ρ a... ρ a k k where all the ρ i are distinct irreducible representations; the decomposition of ρ into a direct sum of these k-many factors is unique, as are the ρ i that occur and their multiplicities a i. This is called the isotypical decomposition of ρ. Finally, there are only finitely many irreducible representations. Remark.3. Henceforth, we will call an irreducible representation an irrep. Remark.4. Questions for rep theory () Classify (construct) the irreps, ρ,..., ρ l of G. (2) Decompose the tensor product of two representations ρ i and ρ j into irreps, since it s rarely irreducible. ρ i b ρ j pρ... ρ q pρ 2... ρ 2 q... ρ m ρ m How should we calculate the multiplicity m k m i,j,k of ρ k in ρ i b ρ j? This second question is still unsolved even in characteristic zero, although when we work with Hopf algebras and quantum groups and the like, there are many more techniques available to throw at it, so more is known in that case..2 Symmetric Groups, Young diagrams, and Partitions Definition.5. The elements w of the symmetric group S n are bijections w : t,..., nu Ñ t,..., nu. The operation is composition of maps, written from right to left w w 2. Remark.6. There are several ways to write elements of S n : 6

7 Lecture 3 20 January 206 one-line notation two-line notation wpq wp2q... wpnq ˆ 2... n wpq wp2q... wpnq cycle notation w pa,..., a n qpb,..., b l q... Example.7. In S 5, consider the permutation wpq 2, wp2q 3, wp3q, wp4q 4, wp5q 5. The two line notation is ˆ The cycle notation for w is p 2 3qp4qp5q. The product of the two elements u 2 3 p 2q ˆ 2 3 w 3 2 p2 3q of S is uw p 2 3q. ˆ 2 3 and 3 2 Example.8. Some representations of S n () The trivial representation w ÞÑ pq. (2) The sign / alternating representation w ÞÑ sgnpwq #` w even w odd (3) The defining / standard representation # R : w ÞÑ permutation matrix of w. wpiq j That is, Rpwq px ij q where x ij 0 otherwise. For example, if te, e 2, e 3 u is the standard basis of C 3, then S 3 acts in the standard representation by permuting coordinates: Rpwqe i e wpiq. This representation is not irreducible, because the subspace spanned by the sum of basis vectors is invariant. Exercise.9. Check that the trivial and alternating reps are all the -dimensional representations of S n. Exercise.20. Show that the standard representation R of S n decomposes as R R R 2, with R trivial and R 2 an pn q-dimensional irrep. Definition.2. Let n P N. A partition of n is a finite sequence λ pλ,..., λ k q such that λ i P N and λ ě λ 2 ě... ě λ k and λ ` λ 2 `... ` λ k n. Write λ $ n. The set of all partitions of n is Ppnq. Conjugacy classes of S n are parameterized by the partitions of n. The conjugacy class associated with the partition λ $ n consists of all permutations w P S n whose cycle decomposition is of the form w pa,..., a λ qpb,..., b λ2 q pc,..., c λk q. 7

8 Lecture 3 20 January 206 Remark.22. Irreps of S n correspond to partitions of n. We ve seen that conjugacy classes of S n are defined by cycle type, and cycle types correspond to partitions. Therefore partitions correspond to conjugacy classes, which correspond to irreps. Proposition.23. Given λ P Ppnq, let c λ P CS n be the sum of all the permutations in S n with cycle type λ. Then tc λ λ P Ppnqu is a basis for ZpCS n q. (General fact: dim ZpCS n q # of conjugacy classes.) Proof. First, note that each c µ is invariant under conjugation, since the conjugacy classes are by cycle type. Hence, each c µ P ZpCS n q. The c µ are linearly independent, since there s no way to add things of different cycle types and end up with zero. To show that the c µ are spanning, let f P ZpCS n q and let τ P S n. We have that τ f f τ since f lies in the center, so τ f τ f. So if we write f ÿ σps n a σ σ, then a σ a τστ. Hence, the a σ are constant on conjugacy classes; if σ, ρ have the same cycle type, then a σ a ρ. So we may write So the c µ are spanning. f ÿ λpppnq a λ c λ. We can conveniently talk about partitions using Young tableaux, but this will first require a long list of definitions. Definition.24. A paritition λ $ n is represented by a what is called alternatively a Young diagram/frame/ferrers diagram, e.g. the partition p4, 3, q $ 8 corresponds to the diagram Definition.25. A box is described by its coordinates px, yq where x goes down and y goes across. Example.26. For example, in the diagram g is in box p2, 3q. a b c d e f g h Definition.27. The content of a box is cpx, yq y x. 8

9 Lecture 3 20 January 206 Definition.28. A Young tableau of shape λ or λ-tableau is a bijection between boxes of the Young diagram of shape λ and t,..., nu, e.g Definition.29. A Young tableau is called standard if the numbers filled into boxes are increasing both along the rows (left-to-right) and along the columns (top-to-bottom).. Example.30. For example, is standard Definition.3. A box px, yq is removable if there is no box below (in position px `, yq) or to the right (in position px, y ` q). Precisely, px, yq is removable if and only if (x ă k & y λ x ą λ x` ) or (x k & y λ k ). Removing such a box produces a Young diagram associated with a partition of n. Similar for addable boxes. Example.32. For example, a b c d e f g h d, g, h are removable but none of the others are. Boxes are addable with coordinates p, 5q, p2, 4q, p3, 2q and p4, q, as indicated with s below. a b c d e f g h Definition.33. If both λ, µ $ n, then λ ă µ in lexicographic order if, for some i, λ j µ j for j ă i and λ i ă µ i. This defines a total order on Ppnq. Example.34. for example p 6 q ă p2, 4 q ă p2 2, 2 q ă... ă p5, q ă p6q. Definition.35. For λ $ n, Tabpλq SYTpλq standard λ-tableaux (. SYTpnq ď λ$n SYTpλq Definition.36. Let λ pλ,..., λ k q $ n. In a Young diagram for λ, there exists t : λ columns. The j-th column contains exactly λ j : ti : λ i ě ju boxes. The conjugate partition to λ is the parittion λ pλ,..., λ tq 9

10 Lecture 3 20 January 206 Example.37. Conjugating the partition p4, 4, 3,, q gives p5, 3, 3, 2q, which corresponds to flipping the corresponding Young diagram over the main diagonal: Notice that conjugating twice gives the original partition back. Definition.38. A diagram of λ is a hook if λ pn k, k q pn k,,,..., q for some 0 ď k ď n. k is the height of the hook. Example.39. is a hook of height 3. Definition.40. Suppose that λ pλ,..., λ k q and λ pλ,..., λ tq are conjugates. The hook length of a box of coordinate pi, jq is hpi, jq h ij : pλ i jq ` pλ j iq ` i pλ i jq is the arm of the hook: the number of boxes in the same row to the right; pλ j q is the leg of the hook: number of boxes in the same column below, and counts the box in the corner. Example.4. In the partition λ p7, 4, 3, 2q the hook length of the box p2, 2q is h 22 2 ` 2 ` 5. The hook is shaded in the diagram. Remark.42. The French justify the Young diagrams along the bottom instead of the top. They hate our convention because it violates Descartes s convention whereby the x-axis increases to the right and the y-axis increases up. The way we draw them is known as the English convention." Example.43. The partitions of 3 correspond to irreps of S 3. For example, is the trivial representation, 0

11 Lecture 4 22 January 206 is the sign representation, and is the 2-dimensional component of the regular representation. Definition.44. Write tv λ λ $ nu for the irreps of S n over C. These are called the Specht modules..3 Things we will prove later Remark.45. No sudden movements or sharp noises; please contain your excitement." Proposition.46. dim V λ # SYTpλq. Definition.47. f λ : # SYTpλq Proposition.48 (Frobenius-Young identity). ÿ fλ 2 n! λ$n Combinatorially, Proposition.48 states that #tpp, Qq: P, Q std tableaux of shape λ, λ $ nu #S n n! and the Robinson-Schensted-Knuth (RSK) correspondence gives the bijection. Theorem.49 (Hook-length formula). f λ n! ś hpi, jq. pi,jqpλ Example.50. If λ pn, nq, then p2nq! f λ pn!qp2 3 nqpn ` q ˆ2n. n ` n This is in Richard Stanley s book as part of an infamous exercise, in which he asks you to prove the equivalence of 66 different combinatorial expressions for the Catalan numbers. Remark.5 (Top Travel Tip). Bring Richard Stanley s book on combinatorics with you when you travel, and you ll never be bored. Let G be some finite group and let H ď G be a subgroup. There are two operations we can do on representations of G and H. Given a representation of G, we can restrict to a representation of H, and given a representation of H, we can induce a representation of G. These are linked by Frobenius reciprocity. Definition.52. If V is a representation of G, then Res G H V is the restriction of V to the subgroup H. Alternatively written VÓ H.

12 Lecture 4 22 January 206 Remark.53. If V is an irreducible G-module, then Res G H V need not be irreducible. For example, if G S n, the irreps are V λ for λ $ n. To consider Res S n S V n λ, we need to embed S n into S n, which we do in the standard way: those permutations of t, 2,..., nu leaving n fixed. Definition.54. The Bratelli diagram of representations of symmetric groups has vertices: irreducible representations of S n edges: two vertices V λ, V µ are joined by k edges from λ to µ if V µ is a representation of S n and V λ is a representation of S n, such that V µ is an irreducible component of Res S n S n V λ. Definition.55. The Young poset is the set Y tλ λ $ n, n P Nu of all partitions with the poset structure as follows. Let µ pµ,..., µ k q $ n and λ pλ,..., λ h q $ m be partitions in Y. Then we say that µ ď λ ðñ m ě n, h ě k and λ j ě µ k This simply means that µ is a subdiagram of λ. Example.56. If µ p3, 2, q and λ p4, 3, q, then µ ď λ. λ µ λ { µ We use the notation λ { µ to denote the squares that remain after removing µ from λ: these are the unshaded ones in the diagram on the right. Definition.57. If µ, λ P Y, then we say that λ covers (or is covered by) µ if µ ď λ and µ ď ν ď λ, ν P Y implies that ν µ or ν λ. Clearly λ covers µ if and only if µ ď λ and λ { µ consists of a single box. We write λ Ñ µ or µõλ. Definition.58. The Hasse diagram of Y or the Young (branching) graph is an oriented graph with vertex set Y and an arrow λ Ñ µ if and only if λ covers µ. We will eventually show that the Young graph Y is the same as the Bratelli diagram for the branching of representations upon restriction from S n to S n. Lemma.59 (Branching Rule). Res S n S n V λ à µ$pn q µďλ where the direct sum is over all Young diagrams µ obtained from λ by removing a single removable box. Note that if V µ occurs at all, it occurs with multiplicity one the branching is simple. V µ 2

13 Lecture 4 22 January 206 There is a corresponding result for induction, but we won t bother with it right now; see Corollary 5.5. Example.60. Consider the Young tableaux λ p5, 4, 4, 2q and the representation V λ is obtained by removing any one of the removable (shaded) boxes. Res S 5 S 4 V p5,4,4,2q V p4,4,4,2q V p5,4,3,2q V p5,4,4,q To prove Lemma.59, we need to think about it combinatorially. To that end, we can associate paths in the Young graph with standard Young tableaux. Definition.6. A path in the Young graph is a sequence π λ pq pnq Ñ λ pn q Ñ Ñ λ of partitions λ pjq $ j such that λ pjq covers λ pj q for j 2, 3,..., n. Notice that a path always ends at the trivial partition λ pq pq $. Let n lpπq be the length of the path π. Then Π n pyq is the set of all paths of length n in the Young graph. Further define ΠpYq : ď ně Π n pyq. Given λ $ n and a path π λ pnq Ñ λ pn q Ñ Ñ λ pq, there exists a corresponding standard tableaux T of shape λ obtained by placing the integer k P t,..., nu in the box λ pkq {λ pk q. 3

14 Lecture 5 25 January 206 Figure. Bottom of the Hasse diagram of Y Example.62. Consider the path π λ p8q p4, 3, q Ñ p4, 3q Ñ p3, 3q Ñ p3, 2q Ñ p2, 2q Ñ p2, q Ñ p2q Ñ pq λ pq. We read the path backwards from λ pq to λ p8q and add boxes one at a time to reconstruct the standard Young tableaux. H ù 2 ù 2 3 ù ù ù ù ù So the path π corresponds to the standard tableaux Fact.63. There is a natural bijection between Π n pyq and SYTpnq, which extends to a bijection ΠpYq Ø Ť ně SYTpnq. 4

15 Lecture 5 25 January 206 Once we know that the branching graph for S n corresponds to the Young graph, we can see deduce interesting results about the representation theory of S n. For instance, we can easily show that dim V λ # SYTpλq f λ. Recall the statement of the Branching Rule (Lemma.59). Res S n S n pv λ q à µ%pn q µďλ Slightly more generally, we could instead restrict to any m ă n. Res S n S m pv λ q Res S m` S m Res S m`2 S Res S n m` S pv n λ q, and at each step of the consecutive restrictions, the decomposition is simple (multiplicity-free) and it occurs according to the branching graph. Therefore, the multiplicity of V µ in the restriction Res S n S m V λ is the number of paths in Y that start at λ and end at µ. This is the number of ways in which you can obtain a diagram of shape λ from one of shape µ from adding successively pn mq addable boxes to the diagram of shape µ. Recall Proposition.46, which claims that V µ. f λ : dim V λ # SYTpλq. We can prove this by counting dimensions in the Branching Rule. We see that dim V λ ÿ dim V µ λñµ ÿ λñµñν dim V ν ÿ λ λ pnq Ñ Ñλ pq pq dim V pq # of paths from pλq to pq in Y. We ll construct a basis of V λ where each basis vector corresponds to a downward path in the Young graph from pλq to pq. As in the previous lecture, each such path corresponds to a standard Young tableaux of shape λ. Example.64. There are three different paths from p3, q down to pq in Y. One such path is the following ù 2 3 ù 2 ù.4 Back to basics: Young symmetrizers Let s go back to the problem of constructing these modules V λ. The classical method of constructing irreps of S n is via Young symmetrizers. While this approach is fast, it is inferior because it takes a lot of effort to prove things about representations of S n using Young symmetrizers. In contrast, the Okounkov- Vershik approach that we will take is more effort upfront, but it makes proving things about representations much easier. 5

16 Lecture 5 25 January 206 Definition.65. Let G tg,..., g r u be a finite group. The group algebra CG is the C-algebra whose elements are formal linear combinations for α i P C. Multiplication is given by α g ` ` α r g r pα g ` ` α r g r qpβ g ` β r g r q ÿ i,j Example.66. Let G S 3. Then pα i β j qg i g j rÿ ˆ k ÿ α i β j i,j g i g j g k CG CS 3 tα ` βp 2q ` γp 3q ` δp2 3q ` εp 2 3q ` ζp 3 2qu is a 6-dimensional C-algebra. g k Definition.67. The regular representation V CG has a left-action of G by multiplication with dim V G. gpα g ` ` α r g r q α pgg q ` ` α r pgg r q, G. Every irreducible representation is contained in the regular representation Theorem.68. Let V i be the irreducible representations of G, for i P I. Then CG à ipi pdim V i qv i Corollary.69. Example.70. ÿ pdim V i q 2 G ipi CS 3 V V V V V corresponds to the trivial representation, V corresponds to the alternating representation, and V is the 2-dimensional component of the standard representation. So how do we find V in CS 3? Well, it corresponds to a -dimensional subspace, so therefore we need a vector to span it. Try ř wps 3 w. Indeed, V C ÿ wps 3 w G. Similarly, V C ÿ wps 3 p q w w G. 6

17 Lecture 6 27 January 206 We write p q w sgnpwq. How do we find V V inside CS 3? Consider c p ` p 2qqp p 3qq P CS 3. Take CS 3 c Ď CS 3. The claim is that V CS 3 c, which is a special case of the construction called the Young symmetrizer. Definition.7. Pick any tableau T of shape λ $ n. Define P P λ σ P S n σ preserves each row of T ( Q Q λ σ P S n σ preserves each column of T ( Now let a λ ÿ wpp w P CS n b λ ÿ wpqp q w w Finally, c λ a λ b λ is the Young symmetrizer. The Specht module V λ is the ideal V λ : CS n c λ Ď CS n. Example.72. If λ p3, 2q, and the tableaux is then Theorem.73. T a λ p ` p 3q ` p 4q ` p3 4q ` p 3 4q ` p 4 3qq p ` p2 5qq b λ p p 2qqp p3 5qq () Some scalar multiple of c λ is an idempotent: c 2 λ n λc λ, for some n λ P C. (2) V λ is an irrep of S n. (3) Every irrep of S n can be obtained in this way for a unique partition λ $ n. Proof. See Fulton & Harris 4.2 / Example sheet. Remark.74. () We will ultimately see that each irrep can be defined over Q instead of over C. In fact, the scalar n λ in Theorem.73() is n λ n! { dim Vλ. (2) Any tableau gives an irrep, not just standard ones. They will be isomorphic to those of the standard ones, but in with things in a different order., Example.75. λ pnq, c λ a λ ř wps n w V λ CS n c λ CS n ÿ wps n w trivial rep 7

18 Lecture 6 27 January 206 µ p n q p,,..., q c µ b µ ř wps n p q w w V µ CS n c µ CS n ÿ p q w w alternating rep wps n λ p2, q $ 3 c p2,q p ` p2 qqp p 3qq ` p 2q p 3q p 3 2q P CS 3. V p2,q xc p2,q, p 3qc p2,q y Exercise.76. Prove V λ b sgn Vλ, where λ is the conjugate partition to λ..5 Coxeter Generators Definition.77. S n is generated by the adjoint transpositions s i pi, i ` q for ď i ď n with coxeter relations s 2 i s i s j s i s j s i s j if i j s i s j s j s i for i j ě 2 Definition.78. s i s ik is reduced if there is no shorter product. In this case k is the Coxeter length. How do these act on tableaux? If T is a tableau of shape λ, let σ P S n. We obtain a new tableau σt by replacing i with σpiq for each i. Example.79. Let σ p qp2 5q. T σt Notice that T is not standard, but σt is. Definition.80. If T is standard, say s i is admissible for T if s i T is still standard. So s i is admissible for T if and only if i, i ` belong to neither the same row nor the same column of T. Definition.8. For π P S n, an inversion for π is a pair pi, jq with i, j P t,..., nu such that i ă j ùñ πpiq ą πpjq. Ipπq tall inversions in πu. Let lpπq Ipπq. Theorem.82. The Coxeter length of π equals lpπq. Proof. First, notice that lpπs i q # lpπq πpiq ą πpi ` q lpπq ` πpi ` q ą πpiq 8

19 Lecture 6 27 January 206 This shows that, if π s i s i2 s ik is a minimal representation of π by Coxeter generators, then lpπq lps i s i2 s ik q ď lps i s i2 s ik q ` ď lps i s i2 s ik 2 q ` 2 ď... ď k. () So the inversion length is bounded above by the Coxeter length. It remains to show that the Coxeter length of π is bounded below by lpπq. Proof by induction on lpπq. If lpπq 0, then π is the identity, with Coxeter length zero. Assume now that lpπq ą 0. Now let j P t, 2,..., nu such that πpjq n. Define τ πs jn s jn` s n, (2) such that τpnq n. Then using () lpτq lpπs jn s jn` s n q lpπs jn s jn` s jn 2 q lpπs jn s jn` s jn 3 q 2. lpπq pn jq. Hence, lpτq ă lpπq, so by induction, lpτq is the Coxeter length of τ, so τ can be written as the product of lpτq-many Coxeter generators. (2) now shows us that π can be written as the product of lpτq ` pn jq lpπq many Coxeter generators. Hence, the Coxeter length of π is bounded above by lpπq. Definition.83. λ pλ,..., λ k q $ n. 2 λ T λ λ ` λ ` 2 λ `λ 2 λ k n where λ k λ `... ` λ k `. T λ is called the canonical tableau. Given T P SYTpλq, denote by σ T P S n the unique permutation such that σ T T T λ. Proposition.84. T P SYTpλq, l lpσ T q. Then there is a sequence of l admissible transpositions which transforms T into T λ. 9

20 Lecture 7 29 January 206 Proof. Let j be the number in the rightmost box of the last row of T. There are two cases: either j n or j n. If j n then, as this box is removable, we can consider standard tableau pt of shape p λ pλ,..., λ k q $ pn q obtained by removing that box. By induction applied to p T, there exists a sequence of p l lpσ pt q admissible transpositions which transforms p T into T pλ defined by p λ. The same sequence will transform T into T λ and l p l. If j n, then s j is an admissible transposition for T. Similarly, s j` is admissible for s j T, and so on, so s n is admissible for s n 2 s n 3 s j` s j T. Finally, s n s n 2 s j T has n in the rightmost box on the last row of T and we ve reduced to the previous case. Corollary.85. If T, S P SYTpλq, then S can be obtained from T by applying a sequence of admissible adjacent transpositions. 2 The Okounkov-Vershik approach This approach to the representation theory of S n is due to Andrei Okounkov and Anatoly Vershik, 996, Main Steps in the Okounkov-Vershik approach branching S n Ñ S n is multiplicity free, see Lemma.59 given an irreducible S n -module V λ, branching is simple so the decomposition of V λ into irreducible S n -modules depends only on the given partition and nothing else. Each module decomposes canonically into irreducible S n 2 -modules. Iterating we get a canonical decomposition of V λ into irreducible S -modules. So there exists a canonical basis of V λ determined modulo scalars, called the Gelfand-Tsetlin basis (abbreviated GZ-basis). Let Z n ZpCS n q be the center of the group algebra of S n. The Gelfand- Tsetlin algebra GZ n is a (commutative) subalgebra of CS n generated by Z Y... Y Z n. The next step in the Vershik-Okounkov approach is to show that GZ n consists of all elements of CS n that act diagonally in the GZ-basis in every irreducible representation. GZ n is a maximal commutative subalgebra of CS n with dimension equal to the sum of dimensions of the distinct irreducible S n -modules. Thus, any vector in the GZ-basis (in any irrep) is uniquely determined by the eigenvalues of the elements of the GZ-algebra on this vector. For i,..., n, let X i p, iq ` p2, iq `... ` pi, iq P CS n. These are called the Young-Jucys-Murphy elements (YJM-elements). We will show that these YJM elements generate the GZ algebra. 20

21 Lecture 7 29 January 206 To a GZ-vector v (meaning an element of the GZ-basis for some irrep), we associate a tuple αpvq pa, a 2,..., a n q, where a i is the eigenvalue of X i on the vector v, and let Specpnq tαpvq v is a GZ-vectoru. By a previous step, for GZ-vectors u, v, we have that u v if and only if αpuq αpvq. Hence, Specpnq is equal to the sum of dimensions of the distinct irreducible inequivalent representations of S n. The last step in the Vershik-Okounkov approach is to construct a bijection Specpnq Ñ SYTpnq such that tuples in Specpnq whose GZ-vectors belong to the same irrep go to standard Young tableaux of the same shape. This proceeds by induction, using relations s 2 i, X i X i` X i` X i, s i X i ` X i` s i where s i pi, i ` q. Definition 2.. A part of a partition µ pµ,..., µ k q is µ i for some i. Let P pnq be the set of all pairs pµ, iq where µ $ n is a partition and i is a part of µ. A part µ i of µ is non-trivial if µ i ě 2. Let #µ be the sum of its nontrivial parts. Recall by Theorem.82 that σ P S n can be written as a product of lpσq-many Coexter transpositions and cannot be written as a product of any fewer. Remark 2.2 (Conventions). All algebras are finite dimensional over C and unital. Subalgebras contain the unit; algebra homomorphisms preserve the unit. Given elements or subalgebras A,..., A n of an algebra A, denote by xa,..., A n y the subalgebra of A generated by A Y... Y A n. Definition 2.3. Let G be a group. Let tu G ď G 2 ď... ď G n ď... be an (inductive) chain of (finite) subgroups of G. Write G^n for the set of equivalence classes of finite dimensional complex representations of G n. V λ is the irreducible G n -module corresponding to λ P G^n. Definition 2.4. The branching multigraph / Bratelli diagram of a chain of groups tu G ď G 2 ď... ď G n ď... has vertices: the elements of the set š ně G^n edges: two vertices λ, µ are joined by k directed edges from λ to µ whenever µ P G^n and λ P G^n for some n, and the multiplicity of µ in the restriction of λ to G n is k. 2

22 Lecture 8 February 206 We call G^n as the n-th level of the Bratelli diagram. Write λ Ñ µ if pλ, µq is an edge of the diagram. Assume that the Bratelli diagram is a graph, i.e. all multiplicities of all restrictions are 0 or (multiplicity free or simple branching). Take a G n -module V λ with λ P G^n. By simple branching, the decomposition V λ À µ V µ where the sum over all µ P G^n with λ Ñ µ is canonical. Iterating the decomposition, we obtain a canonical decomposition of V λ into irreducible G -modules, that is, -dimensional subspaces V λ à T V T, (3) where the sum is over all possible chains λ pnq Ñ Ñ λ pq (4) with λ pnq P G^i and λ pnq λ, (or equivalently, the sum is over all possible tableaux of shape λ when G S n ). Definition 2.5. Choosing a nonzero vector v T in each one-dimensional space V T, we obtain a basis tv T u of V λ, called the Gelfand-Tsetlin basis (GZ-basis). By definition of v T, we have pcg i qv T V λ piq for each i, because v T P V λ piq, which is an irrep, and so the action of CG i on v T can recover the entirety of the irrep V λ piq. Note that the chains (4) are in bijection with directed paths in the Bratelli diagram from λ to the unique element λ pq of G^. We have a canonical basis (up to scalars) for V λ : the GZ-basis. Can we identify those elements of CG n that act diagonally in this basis, for every irrep? In other words, consider the algebra isomorphism φ : CG n ÀλPG^n End V λ g (5) g V λ ÝÑ Vλ : λ P G^n We know that φ is an isomorphism because, if φpxq φpyq, then x and y necessarily act the same on each irrep and hence on the regular representation. Therefore, x y. Counting dimensions establishes surjectivity. Definition 2.6. Let DpV λ q be the set of operators on V λ diagonal in the GZ-basis of V λ. What is the image under φ (5) of the subalgebra À λpg^n DpV λq of À λpg^n EndpV λq? Definition 2.7 (Notation). Let Z n ZpCG n q be the center of the group algebra. Definition 2.8. We can easily see that GZ n xz,..., Z n y is a commutative C-algebra of CG n. This is the Gelfand Tsetlin algebra of the inductive chain of subgroups. 22

23 Lecture 8 February 206 Theorem 2.9. GZ n is the image of À DpV λ q under the isomorphism (5). That is, GZ n consists of elements of CG n that act diagonally in the GZ-basis in every irreducible representation of G n. Thus, GZ n is a maximal commutative subalgebra of CG n and its dimension is dim GZ n ÿ dim V λ λpg^n Proof. Consider the chain T from (4). For each i,..., n, we will denote by p λ piq P Z i the central idempotent corresponding to the representation defined by λ piq P G^i. Define p T p λ pq p λ p2q p λ pnq P GZ n. The image of p T under (5) is p f µ : µ P G^nq where f µ 0 if µ λ and f λ is projection onto V T with respect to (3). Hence, the image of GZ n under (5) includes À λpg^n DpV λq, which is a commutative maximal subalgebra of À λpg^n EndpV λq. Since GZ n is itself commutative, the result follows. Definition 2.0. A GZ-vector of G n (modulo scalars) is an element v T of the GZ-basis for some irrep of G n. An immediate corollary of Theorem 2.9 says something interesting about these vectors. Corollary 2. (Corollary of Theorem 2.9). (i) Let v P V λ, λ P G^n. If v is an eigenvector (for the action) of every element of GZ n then (a scalar multiple of) v is a GZ-basis vector of V λ, that is, v is of the form v T for some path T. (ii) Let u, v be two GZ-vectors. If u, v have the same eigenvalues for every element of GZ n, then u v. Remark 2.2. Later we ll find an explicit set of generators for the GZ-algebras of the symmetric groups. 2.2 Symmetric groups have simple branching To prove this, we first need several preliminary theorems from the theory of semisimple algebras. We won t prove them, or even use them often. Theorem 2.3 (Artin-Wedderburn Theorem). If A is a semisimple C-algebra, then A decomposes as a direct sum of matrix algebras over C. Theorem 2.4 (Double Centralizer). Let A be a finite dimensional central simple algebra (central means ZpAq Cq, and B Ď A a simple subalgebra. Let C Z A pbq be the centralizer. Then C is simple, and Z A pcq B, and moreover dim C paq dim C pbq dim C pcq. 23

24 Lecture 8 February 206 Theorem 2.5. Let M be a finite dimensional semisimple complex algebra and let N be a semisimple subalgebra. Let ZpM, Nq Z M pnq be the centralizer of the pair pm, Nq, ZpM, Nq tm P M mn n P Nu. Then ZpM, Nq is semisimple and the following are equivalent: (a) the restriction of any finite dimensional complex irreducible representation of M to N is multiplicity free (b) ZpM, Nq is commutative. We ll apply this theorem in the case where M CS n and N CS n. Proof. Without loss of generality (by Artin-Wedderburn, Theorem 2.3) M À ti M i where each M i is some matrix algebra. Write elements of M as tuples pm,..., m k q where each m i P M i. Let N i be the image of N under the projection M Ñ M i. It s a homomorphic image of a semisimple algebra, and so N i is semisimple. Now ZpM, Nq À k i ZpM i, N i q. By the Double Centralizer Theorem (Theorem 2.4), each ZpM i, N i q is simple, and therefore ZpM, Nq is semisimple. Now let s establish the equivalence of (a) and (b). Let V i " pm,..., m k q P M ˇ m j 0 for i j and with all entries of m i not in the first column equal to zero The V i are all the distinct inequivalent irreducible M modules and the decomposition of V i into irreducible N-modules is identical to the decomposition of V i into irreducible N i -modules. Now notice that ZpM, Nq is commutative if and only if ZpM i, N i q is commutative for all i, which is true if and only if all irreducible representations of ZpM i, N i q have dimension. So it suffices to show that all irreps of ZpM i, N i q are have dimension if and only if the restriction of irreps of M i to N i is multiplicity free. First, assume that all irreps of ZpM i, N i q have dimension. Let U be an irrep of M i and V an irrep of N i. In particular, Hom Ni pv, Uq is an irrep of ZpM i, N i q, and so has dimension. By Schur s lemma, this is the multiplicity of V in Res M i N U, so the branching is multiplicity-free. i Conversely, assume ZpM i, N i q has an irrep of dimension ą. Let U be an irrep of M i and V an irrep of N i. We have by Schur s Lemma that End C puq M, so End Ni puq ZpM i, N i q. Hence, if Res M i N U À i j W j is the decomposition of U into simple N i -modules, then ZpM i, N i q End Ni puq à Hom Ni pw j, W k q j,k *. is a decomposition of ZpM i, N i q into irreducible representations. So if there is an irrep of ZpM i, N i q with dimension ą, then W j W k for some distinct j, k. This irrep of N i then occurs with multiplicity ą in Res M i N i U; branching is not simple. 24

25 Lecture 9 3 February Involutive algebras Definition 2.6. Let F R or F C. If F C, then for α P C denote by α the complex conjugate. If F R and α P R, then α α. An F-algebra A is involutive if it has a conjugate linear anti-automorphism of order 2, that is, a bijective map x ÞÑ x such that px ` yq x ` y pαxq αx pxyq y x px q x for all x, y P A, α P F. x is called the adjoint of x. An element x P A is normal if it commutes with it s own adjoint: xx x x. x is self adjoint or hermitian if x x. Definition 2.7. Let A be involutive over R. Then the -complexification of A is the algebra whose elements are pairs px, yq P A ˆ A, written x ` iy, and the operations are the obvious ones: px ` iy q ` px 2 ` iy 2 q px ` x 2 q ` ipy ` y 2 q αpx ` iyq pαxq ` ipαyq px ` iy qpx 2 ` iy 2 q px x 2 y y 2 q ` ipx y 2 ` x 2 y q px ` iyq x iy A real element of the -complexification is an element of the form x ` i0 for some x P A. Example 2.8. If F R or C, and G is a finite group, FG is involutive under ÿ α i g i ÿ α g i i. (6) i i Recall that we defined Z n ZpCG n q and the Gelfand-Tsetlin algebra GZ n xz,..., Z n y. Recall also that CG n à EndpV λ q. λpg^n Remark 2.9. Here is what we know about the Gelfand-Tsetlin algebra so far, from Theorem 2.9 (0) GZ n is commutative () GZ n is the algebra of diagonal matrices with respect to the GZ-basis (2) GZ n is a maximal commutative subalgera of CG n (3) v P V λ is in the GZ-basis if and only if v is a common eigenvector of elements of GZ n 25

26 Lecture 9 3 February 206 (4) Each basis element is uniquely determined by the eigenvalues of elements of GZ n. If A ě B, then ZpA, Bq ta P A ab b P Bu. Remark We proved the following in Theorem 2.5. Let H be a subgroup of G. Then the following are equivalent: () Res G H is multiplicity-free; (2) ZpCG, CHq is commutative. Here s some more stuff about involutive algebras. Theorem 2.2. Let A be an involutive C-algebra. Then (i) An element x P A is normal if and only if x y ` iz for some self-adjoint y, z P A that commute. (ii) A is commutative if and only if every element of A is normal. (iii) If A is a -complexification of a real involutive algebra, then A is commutative if every real element of A is self-adjoint. Proof. Proof of (i). Assume that x is normal. Then xx x x. Define y 2 px ` x q and z i 2 px xq. Then we have that y 2 px ` x q px 2 ` xq y z p 2 i px xqq i2 px xq i2 px x q pix 2 ixq z yz 4 i px ` x qpx xq 4 i pxx x 2 ` px q 2 ` x xq 4 i ppx q 2 x 2 q zy 4 i px xqpx ` x q 4 i px x ` px q 2 x 2 xx q 4 i ppx q 2 x 2 q y ` iz 2 px ` x q ` i2 px 2 xq 2 px ` x x ` xq x Conversely, if x y ` iz for self-adjoint y, z P A that commute, then xx py ` izqpy ` izq py ` izqpy iz q py ` izqpy izq y 2 ` z 2 x x py ` izq py ` izq py iz qpy ` izq py izqpy ` izq y 2 ` z 2 Proof of (ii). Let x, y P A. By part piq, write x x ` ix 2 for some x, x 2 that are self-adjoint and commute. Likewise, write y y ` iy 2 for some y, y 2 that are self-adjoint and commute. Note that, since x x and y y, and x y P A is normal, then px y q x y ùñ y x x y ùñ y x x y, 26

27 Lecture 9 3 February 206 and likewise x commutes with y 2 and y commutes with x 2 and y 2, x 2 commute. Therefore, xy px ` ix 2 qpy ` iy 2 q x y ` ix y 2 ` ix 2 y x 2 y 2 y x ` iy 2 x ` iy x 2 y 2 x 2 py ` iy 2 qpx ` ix 2 q yx Proof of (iii). Every element of A can be written as x ` iy for some real elements x, y. It suffices by part (ii) to show that every element of A is normal, and to show that, it suffices to show by part (i) that every element of A is of the form x ` iy for x, y self-adjoint and commuting. We know that x, y are self adjoint because they are real elements. We know further that xy is a real element, and therefore xy pxyq y x yx. Hence, x and y are self-adjoint and commute, so we are done. This concludes the proof of Theorem 2.5. Theorem The centralizer ZpCS n, CS n q is commutative. Proof. Claim that the involutive subalgebra ZpCS n, CS n q is the -complexification of ZpRS n, RS n q, and therefore commutative by Theorem 2.2. So it s enough to show that every element of ZpRS n, RS n q is self-adjoint by Theorem 2.2. To that end, let f ř πps n α π π for α π P R be an element of ZpRS n, RS n q. Fix σ P S n. S n is ambivalent (meaning that every element is conjugate to its inverse) since σ and σ are of the same cycle type. To produce a permutation τ in S n conjugating σ to σ : write the permutation σ in cycle form and write down σ in cycle form below such that the lengths correspond to each other. The permutation in S n taking an element of the top row to the corresponding element in the bottom row cojugates σ to σ. For example, in S 9, write σ p24qp35qp6879q σ p42qp35qp6978q and then τ p24qp89q. Moreover, we want this τ to represent an element of S n. We can always choose a conjugating τ that fixes any of the numbers that σ moves, that is, there is some τ such that τpnq n and τστ σ. We can do this by permuting the cycle of σ containing n until n lines up in both the cycles of σ and σ. Continuing the previous example, we can write p6978q p7869q, so σ p24qp35qp6879q σ p42qp35qp7869q 27

28 Lecture 9 3 February 206 and then τ p24qp67q conjugates σ to σ. Therefore, we can choose τ P S n such that τστ σ. Since τ P S n and f P ZpRS n, RS n q, we see that τ f f τ ùñ f τ f τ. Hence, f τ f τ ÿ α pτπτ π q, πps n so α σ are constant on conjugacy classes. Therefore, α σ α σ. Since σ was arbitrary in S n, we see that f f (see (6)). Denote the centralizer ZpCS n, CS n q by Z pn,q. We ll see a second proof of commutativity in Theorem Young-Jucys-Murphy elements (YJM elements) Henceforth G n S n, so chains in the Bratelli diagram refer to chains in the Bratelli diagram of the symmetric groups. Definition For i 2,..., n, define Y i to be the sum of all i-cycles in S n. By convention, Y n 0. Define Y i as the sum of all i-cycles in S n containing n. For pµ, iq P P pnq, (see Definition 2.), let c pµ,iq P CS n be the sum of permutations π in S n such that the type of π is µ and the size of the cycle of π containing n is i. Remark Each of Y 2,..., Y n, Y2,..., Y n equals c pµ,iq for suitable µ and i. In particular, Y j c pµ,q for µ pj,,..., q, and Yj c pµ,jq for µ pj,,..., q. Lemma (i) tc pµ,iq pµ, iq P P pnqu is a basis of Z pn,q. Therefore, we have that xy 2,..., Y n, Y 2,..., Y ny Ď Z pn,q. (ii) c pµ,iq P xy 2,..., Y k, Y2,..., Y ky for k #µ. (iii) Z pn,q xy 2,..., Y n, Y 2,..., Y ny (iv) Z n xy 2,..., Y n y. Proof. (i) The first bit is an exercise, similar to the proof Proposition.23 that tc µ µ P Ppnqu is a basis of Z n. The second bit follows from Remark (ii) Induction on #µ. If #µ 0, then c pµ,iq is the identity permutation, which lies in the subalgebra xy 2,..., Y k, Y2,..., Y ky. So now assume true when #µ ď k. Consider pµ, iq P P pnq with #µ k `. Let the nontrivial parts of µ be µ,..., µ l in some order. There are several cases (a) First, prove it for i. Consider the product Y µ Y µl By (i), Y µ Y µl α pµ,q c pµ,q ` ÿ pτ,q α pτ,q c pτ,q. 28

29 Lecture 0 5 February 206 where α pµ,q 0 and the sum is over all pτ, q with #τ ă #µ. Then we are done by induction. (b) If i ą, without loss of generality assume that i µ. Consider the product Yµ Y µ2 Y µl. By (i), we see that Yµ Y µ2 Y µl α pµ,iq c pµ,iq ` ÿ α pτ,jq c pτ,jq. pτ,jq where α pµ,iq 0 and the sum is over all pτ, jq with #τ ă #µ. Then we are done by induction. (iii) Follows from piq and piiq. (iv) Similar to piiiq. Definition For ď i ď n, define the Young-Jucys-Murphy elements (YJM elements) By convention, X 0. X i p, iq ` p2, iq `... ` pi, iq P CS n. This is equal to the sum of all the 2-cycles in S i minus the sum of all 2-cycles in S i. Note that X i is the difference of an element of Z i and an element of Z i. Therefore X i R Z i for all ď i ď n. X i and pi, q don t commute, for example. Theorem 2.27 (Okounkov-Vershnik, 2004). (i) Z pn,q xz n, X n y (ii) GZ n xx,..., X n y. Proof. (i) Evidently, xz n, X n y Ď Z pn,q because X n Y2 and then apply Lemma 2.25(iii). Conversely, we already know that Y k P Z n, so it s enough to show that Y 2,..., Y n P xz n, X n y. Since Y 2 X n, then Y 2 P xz n, X n y. This forms the base case for induction. Now assume that Y2,..., Y k` P xz n, X n y. We aim to show for an inductive step that Yk`2 P xz n, X n y. We ll sink to computing with elements and just hit this theorem with a club until it dies. Write Yk` as Yk` ÿ pi,..., i k, nq i,...,i k summed over all distinct i,..., i k P t, 2,..., nu. Consider now Yk` X n P xz n, X n y. Yk` X n ÿ n ÿ pi,..., i k, nq pi, nq (7) i,...,i k i 29

30 Lecture 0 5 February 206 and take a typical element pi,..., i k, nqpi, nq of this product. There are two possibilities: either i i j for any j,..., k, or i i j for some j. If i i j for any j,..., k, then the product is pi, i,..., i k, nq. If i i j for some j, then the product is pi,..., i j qpi j`,..., nq. Hence, (7) becomes ÿ i,i,...,i k pi, i,..., i k, nq ` ÿ i,...,i k j kÿ pi,..., i j qpi j`,..., i k, nq (8) The first sum is over all distinct i, i,..., i k P t,..., n u and the sum on the right is over all distinct i,..., i k P t, 2,..., n u. Rewrite (8) as Yk`2 ` ÿ α pµ,iq c pµ,iq pµ,iq where the sum is over all pµ, iq such that #µ ď k `. Now by induction and Lemma 2.25(ii), we have Y k`2 P xz n, X n y. (ii) Induction on n. The cases for n and n 2 are trivial. Now assume GZ n is generated by xx, X 2,..., X n y. We want to show that GZ n xgz n, X n y. Clearly, GZ n Ě xgz n, X n y, since X n is the difference of an element of Z n and an element of Z n. So we just have to check that GZ n Ď xgz n, X n y. To show this, it s enough to show that Z n Ď xgz n, X n y. But this is clear by (i), since Z n Ď Z pn,q Ď xz n, X n y Ď xgz n, X n y. Remark Theorem 2.27(i) implies that Z pn,q is commutative, because Z pn,q xz n, X n y and X n commutes with every element in Z n. This gives another proof of the fact that Z pn,q is commutative. Definition The GZ-basis for G S n is called the Young basis. By Corollary 2.(i), the Young / GZ-vectors are common eigenvectors for GZ n. Definition Let v be a Young vector for S n. αpvq pa,..., a n q P C n, where a i is the eigenvalue of X i on v. Call αpvq the weight of v. Note that a 0 since X 0. Definition 2.3. Let Specpnq tαpvq v is a Young vectoru. This is the spectrum of YJM-elements. By Corollary 2.(ii), Specpnq dim GZ n ÿ dim λ. λps^n 30

31 Lecture 8 February 206 By definition, Specpnq is in natural bijection with chains T, as in (4). Given α P Specpnq, denote by v α the Young vector with weight α and T α the corresponding chain in the Bratelli diagram. Given a chain T as in (4), we denote the corresponding weight vector αpv T q by αptq. Hence, we have a one-to-one correspondence T ÞÑ αptq; α ÞÑ T α between chains in the Bratelli diagram and Specpnq. Moreover, there is a natural equivalence relation on Specpnq defined as follows. Definition Let α, β P Specpnq, α β ðñ v α, v β belong to the same irreducible module for S n ðñ T α, T β start at the same vertex. Clearly, #pspecpnq{ q is the number of paths in the Bratelli diagram from level n to level 0, which (although we haven t proved it yet) is equal to # SYTpλq #S^n. So this gives some circumstantial evidence that looking at Specpnq is interesting and relevant to representations of S n. We want to describe the set Specpnq describe the relation calculate the matrix elements in the Young basis calculate the characters of irreducible representations of S n Remark The book by Curtis-Reiner from 962 is a good reference for Artin-Wedderburn theory. Here s the story so far: each irrep V λ has a nice basis called the GZ-basis tv T u, each v T corresponding to some chain λ pnq Ñ λ pn q Ñ Ñ λ pq pq. The YJM elements are X k ř k i pi, kq P CS n for ď k ď n. The GZ-algebra GZ n is generated by the YJM elements. GZ n is a maximal commutative subalgebra of CS n by Theorem The GZ-basis is the unique basis such that the basis elements are common eigenvectors of the X k. X i v T a i v T. Note that a i depends on T as well as i. αptq pa,..., a n q P C n We re looking at the spectrum Specpnq tαptq T is a pathu. 3