Representation theory

Representation theory Dr. Stuart Martin 2. Chapter 2: The Okounkov-Vershik approach These guys are Andrei Okounkov and Anatoly Vershik. The two papers appeared in 96 and 05. Here are the main steps: branching from S n to S n 1 is multiplicity-free given irreducible S n -module V λ, branching is simple, so the decomposition of V λ into irreducible S n 1 -modules is canonical. Each module decomposes canonically into irreducible S n 2 -modules, and then iterating this we get a canonical decomposition of V λ into irreducible S 1 -modules, which are just trivial reps. So there exists a canonical basis of V λ determined modulo scalars, called the Gelfand-Tsetlin basis (GZ-basis) Let Z n = Z(CS n ) 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 1... Z n. The idea is to show that GZ n consists of all the element of the group algebra CS n that act diagonally in the Gelfand-Tsetlin basis that we defined above in every irreducible representation. GZ n is a maximal commutative subalgebra of CS n and its dimension is the sum of the dimensions of the distinct inequivalent irreducible S n -modules. Thus, any vector in the GZ-basis (in any irreducible representation) is uniquely determined by the eigenvalues of the elements of the GZ-algebra on this vector. For i = 1,..., n, let X i = (1, i) + (2, i) +... + (i 1, i) CS n, known as the Young- Jucys-Murphy (YJM) elements. We ll show these generate the GZ algebra. (around 74, Jucys produced a paper on symmetric functions, in Russian, and kind of got buried. And then Murphy came around in 81 interpreting a special basis for symmetric functions). To a Gelfand-Tsetlin vector v (meaning an element of the GZ-basis which lies in some irreducible representation) we associate a tuple α(v) = (a 1, a 2,..., a n ) where a i is the eigenvalue of X i on v, and let spec(n) = { α(v) v is a GZ-vector. By the previous bullet point, for GZ-vectors u, v, u = v α(u) = α(v), hence spec(n) = sum of the dimensions of the distinct irreducible inequivalent representations of S n. Construct a bijection spec(n) SYT(n) such that tuples in spec(n) whose GZ-vectors belong to the same irreducible representation go to standard Young tableaux of the same shape. We ll proceed by induction, using s 2 i = 1, X i X i+1 = X i+1 X i and s i X i + 1 = X i+1 s i where s i = (i, i + 1) are the Coxeter transpositions. 1

2 DR. STUART MARTIN Some extra notation! Let P 1 (n) be the set of all pairs (µ, i) where µ n and i is a part of µ. A part is non-trivial if it s 2, and we let #µ be the sum of the sizes of it s non-trivial parts. Also recall that σ S n can be written as product of l(σ) Coxeter transpositions and can t be written in any fewer. Conventions all algebras are finite dimensional over C and they have units subalgebras contain the unit, and algebra homomorphisms preserve units. given elements or subalgebras A 1,..., A n of an algebra A, denote by A 1,..., A n be the subalgebra of A generated by A 1... A n. Let G be a group. Let {1 = G 1 G 2... G n... be an (inductive) chain of (finite) subgroups of G. Let G n = set of equivalence classes of f.d. C-irreps of G n and let V λ = irred G n -module corresponding to λ G n Definition 2.1. The branching multigraph (the Bratelli diagram) of has vertices elements of the set n 1 G n (disjoint union), and λ and µ are joined by k directed edges from λ to µ whenever µ G n 1 and λ G n for some n and the multiplicity of µ in the restriction of λ to G n 1 is k. Call G n the n-th level of the Bratelli diagram. Write λ µ if (λ, µ) is an edge of the diagram. Assume that the Bratelli diagram is a graph, i.e., all multiplicities of all restrictions are 0 or 1 (multiplicity-free, or simple branching). Take a G n -module V λ, with λ G n. By simple branching, the decomposition of V λ is V λ = µ V µ where the sum is over all µ G n 1 with λ µ is canonical. Iterating the decomposition, we obtain a canonical decomposition of V λ into irreducible G 1 -modules, i.e. 1-dimensional subspaces: V λ = V T T where the sum is over all possible chains of connections λ (n)... λ (1) with λ (n) Gî and λ(n) = λ. Choosing a non-zero vector v T in each 1-dimensional space V T, we obtain a basis {v T of V λ, and this is called the Gelfand-Tsetlin (GZ) basis in this case. By definition of v T, we have (CG i ) v T = V λ (i) for 1 i n. Then chains are in bijection with directed paths in the Bratelli diagram from λ to the unique element λ (1) of G 1. We have a canonical basis (up to scalars), the GZ-basis, in each irreducible representation of G n. Can we identify those elements in CG n that act diagonally in this basis (in every irrep)? In other words, consider the algebra isomorphism CG n = EndV λ given by g λg n ( ) g V λ V λ : λ G n, g G n Let D(V λ ) be the operators on V λ which are diagonal in the GZ-basis of V λ.

REPRESENTATION THEORY 3 Question: what is the image under the isomorphism of the subalgebra D(V λ ) of λ G n EndV λ? λ G n Notation Let Z n = Z(CG n ). Easily GZ n = Z 1,..., Z n is a commutative subalgebra of CG n, the Gelfand-Tsetlin algebra associated with the inductive chain of subgoups. Theorem 2.2. GZ n is indeed the image of D(V λ ) under the isomorphism, i.e. 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 λ G n dim λ. Proof. Consider a chain T from. For i = 1,..., n, let p λ (i) Z i denote the central idempotent corresponding to the representation defined by λ (i) Gî (a corollary of Wedderburn s theorem). Define p T = p λ (1)... p λ (n) GZ n The image of p T under is (f µ : µ G n) where f µ = 0 if µ λ and otherwise f λ = projection on V T with respect to the decomposition. Hence the image of GZ n under includes D(V λ), λ G n which is a commutative maximal subalgebra of EndV λ. since GZ n is itself commutative, λ G n the result follows. Definition A GZ-vector of G n, modulo scalars, is a vector in the GZ-basis corresponding to some irreducible representation of G n. An immediate corollary of?? is Corollary 2.3. (1) Let v V λ, λ G n. Then if v is an eigenvector (for the action) of every element of GZ n, then (a scalar multiple of) v belongs to the GZ-basis. (2) Let u, v be two GZ-vectors. If u, v have the same eigenvalues for every element of GZ n, they re equal. Remark Later we will find an explicit set of generators for the GZ-algebras of the symmetric groups. Symmetric groups have simple branching. The idea is to give a straightforward criterion for simple branching, and use some C -algebra stuff on involutions. It comes from the following result, which is not very well-known, and also kind of a consequence of Wedderburn s theorem Theorem 2.4. Let M be a finite dimensional semisimple complex algebra and let N be a semisimple subalgebra. Let Z(M, N) be the centralizer of the pair (M, N) consisting of all elements of M that commute with N. Then Z(M, N) is semisimple, and the following are equivalent: (1) The restrictions of any finite dimensional complex irreducible representation of M to N is multiplicity-free. (2) The centralizer Z(M, N) is commutative. Proof. Without loss of generality (Wedderburn) M = k i=1 M i where each M i is some matrix algebra. Write elements of M as tuples (m 1,..., m k ) with m i M i. Let N i be the image of N under the projection M M i. It s a homomorphic image of a semisimple algebra, so N i is semisimple (omomorphic images of semisimple things are semisimple). Now Z(M, N) = k i=1 Z(M i, N i ). By the double centralizer theorem (DCT), Z(M, N) is semisimple. Digression: If B A, B C A (C A (B)) is always true, and the DCT gives you conditions under which there s equality in this inclusion. Let V i = { (m 1,..., m k ) M mj = 0 for j i and with all entries of m i not in the 1st column equal to 0

4 DR. STUART MARTIN. Then the V i for 1 i k 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. Again, by DCT, V i is multiplicity free as an N i -module for all i iff all irreducible modules of Z(M i, N i ) have dimension 1 for all i iff Z(M i, N i ) is abelian for all i iff Z(M, N) is abelian. Let F = R or C. If F = C, and α C, we ll denote the conjugate by α. If α R, then α = α. An F -algebra is involutive if it has a conjugate linear anti-automorphism (reversing the order of products) of order 2, i.e. there exists a bijective map x x such that (x + y) = x + y (αx) = αx (xy) = y x (x ) = x for all x, y A and α F. The element x is called the adjoint of x. We say x is normal if xx = x x; and self-adjoint (also Hermitian) if x = x. Let A be involutive over R. Define an involutive C-algebra, the *-complexification of A, as follows: The elements are (x, y) A A, written x + iy. The operations are the obvious ones: and (x + iy) = x iy. (x 1 + iy 1 ) + (x 2 + iy 2 ) = (x 1 + x 2 ) + i(y 1 + y 2 ), etc. A real element of the complexification is an element of the form x + i0 for some x A. Example If F = R or C and G is finite, then F G is involutive under ( i α ig i ) = i α ig 1 i and CG is the complexification of RG. First examples sheet will be uploaded on the website. *** Recap: we let Z n = Z(CG n ) and the GZ-algebra is Z 1,..., Z n, and CG n = λ G n EndV λ. We proved that: (1) GZ n is commutative because everything in the i-th term commutes with everything before it, (2) GZ n is an algebra of diagonal matrices with respect to the GZ-basis in each V λ. (3) GZ n is a maximal commutative subalgebra in CG n (4) v V λ is in GZ-basis iff v is a common eigenvector of the elements of GZ n. (5) Each basis element is uniquely determined by eigenvalues of elements of GZ n. If B A is a subalgebra, we defined the centralizer Z(A, B) = { a A ab = ba b B. We proved lemma 2.4., which said that if H G then the following are equivalent: (1) Res G H is multiplicity-free (2) Z(CG, CH) is commutative. Exercise (mostly definition chasing) Let A be an involutive C-algebra. Then: (1) An element of A is normal iff it s of the form x = y + iz for some self-adjoint y, z A such that yz = zy.

REPRESENTATION THEORY 5 (2) A is commutative iff every element of A is normal. (3) If A is the -complexification of some real involutive algebra, then A is commutative if every real element of A is self-adjoint. Theorem 2.5. The centralizer Z(CS n, CS n 1 ) is commutative Proof. The involutive subalgebra Z(CS n, CS n 1 ) is the -complexification of Z(RS n, RS n 1 ) (needs a bit of proof). So it s enough to show that every element of (RS n, RS n 1 ) is self-adjoint. Let f = π S n α π π, α π R, be an element of the centralizer Z(RS n, RS n 1 ). Fix some σ S n. Recall that S n is ambivalent (meaning that σ and σ 1 are conjugate, which is true in the case of S n because they have the same cycle type). To produce a permutation in S n conjugating σ to σ 1 : write the permutation σ in cycle form, and below it σ 1 in cycle form so that the lengths correspond to each other. Then the permutation in S n taking an element of the top row to the element below it on the bottom row conjugates σ to σ 1. We can always choose a conjugating τ that fixes any one of the numbers that σ moves, i.e. there exists a τ such that τ(n) = n and τστ 1 = σ 1. Thus we can choose a τ S n 1 such that τστ 1 = σ 1. Since τ S n 1, τf = fτ, or f = τfτ 1 = π S n α π (τπτ 1 ), hence α σ = α σ 1. Since σ is arbitrary, we have verified that f = f, hence the complexification is commutative by the above exercise. We will denote the centralizer Z(CS n, CS n 1 ) by Z (n 1,1). We ll give a second proof of commutativity in (3.8). 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. For i = 2,..., n, define Y i = sum of all i-cycles in S n 1. By convention, Y n = 0. Define Y i = sum of all i-cycles in S n containing n. For (µ, i) P 1 (n), let c (µ,i) CS n be the sum of permutations π S n such that So: the cycle type of π is µ the size of the cycle of π containing n is i. So each of Y 2,..., Y n 1, Y 2,..., Y n equals c (µ,i) for some µ and some i. Lemma 2.6. We have the following many things: (i) { c (µ,i) (µ, i) P1 (n) is a basis of Z (n 1,1). Hence Y 2,..., Y n 1, Y 2,..., Y n Z (n 1,1). (ii) c (µ,i) Y 2,..., Y k, Y 2,..., Y k for k = #µ. (iii) Z (n 1,1) = Y 2,..., Y n 1, Y 2,..., Y n (iv) Z n 1 = Y 2,..., Y n 1 Proof. First bit is an exercise on the example sheet; similar to the proof that { c µ µ P (n) is a basis for Z n. The second bit follows from. The rest are very similar. For (ii), we ll use induction on #µ. If #µ = 0: c (µ,i) is the identity permutation. Now assume it s true when #µ k. Consider (µ, i) P 1 (n) with #µ = k + 1. Let the non-trivial parts of µ be µ 1,..., µ l in some order. ***

6 DR. STUART MARTIN There are two cases: (a) i = 1. Consider the product Y µ1... Y µl. By (i), we know that Y µ1... Y µl = α (µ,1) c (µ,1) + (τ,1) α (τ,1) c (τ,1) where α (µ,1) 0 and the sum is over all (τ, 1) where τ < µ. Then we re done by induction. (b) i > 1. WLOG, assume that i corresponds to µ 1. Consider the product By (i), Y µ 1 Y µ2... Y µl Y µ 1 Y µ2... Y µl = α (µ,i) c (µ,i) + (τ,j) α (τ,j) c (τ,j) where α (µ,i) 0 and the sum is over all (τ, j) with τ < µ, and again we re done by induction. Finally, (iii) comes from (i) and (ii), and (iv) is very similar to (iii). Definition 2.7. For each 1 i n, define X i = (1, i) + (2, i) +... + (i 1, i) CS n the sum of all the 2-cycles in S i - sum of 2-cycles in S i 1. By convention, X 1 = 0. Note that X i is the difference of an element of Z i and an element of Z i 1. These are the Young-Lucys-Murphy elts (YJM). Note that X i / Z i (1 i n). Theorem 2.8 (Okounkov-Vershik, 04). We have: (i) Z (n 1,1) = Z n 1, X n (ii) GZ n = X 1,..., X n. Proof. We have (i) Evidently, Z n 1, X n Z (n 1,1) because X n = Y 2 and we apply the previous result. Conversely, it s enough to show that Y 2,..., Y n Z n 1, X n by the previous result. Since Y 2 = X n, so Y 2 Z n 1, X n. Assume inductively that Y 2,..., Y k+1 Z n 1, X n ; we aim to show that Y k+2 Z n 1, X n. Write Y k+1 as Y k+1 = (i 1, i 2,..., i k, n) i 1,i 2,...,i k summed over distinct i j. Consider Y k+1 X n Z n 1, X n : it is ( n 1 ) (i 1,..., i k, n) (i, n) (5) i 1,i 2,...,i k i=1 and take a typical element in the product, which looks like (i 1,..., i k, n)(i, n). There are two cases: If i i j for j = 1,..., k, the product is (i, i 1,..., i k, n), and if i = i j for some j, the product is (i 1,..., i j )(i j+1,..., n). Hence (5) is k (i, i 1,..., i k, n) + (i 1,..., i j )(i j+1,..., i k, n) (6) i,i 1,i 2,...,i k j=1 i 1,...,i k

REPRESENTATION THEORY 7 where the first sum is over all distinct i,..., i k {1,..., n 1 and the second over all distinct i 1,..., i k {1,..., n 1. Rewrite (6) as Y k+2 + (µ,i) α (µ,i) c (µ,i) with the sum over all (µ, i) such that µ k + 1. So by induction and part (ii) of??, we have Y k+2 Z n 1, X n. (ii) Induction on n. The case n = 1, 2 are trivial. Assume GZ n 1 = X 1, X 2,..., X n 1. We have to show that GZ n = GZ n 1, X n. Clearly LHS RHS, so we need to check LHS RHS, and for this it s enough to show that Z n GZ n 1, X n, which is clear by (i), since Z n Z (n 1,1). Remark Part (i) implies that Z (n 1,1) is commutative because Z (n 1,1) = Z n 1, X n and X n commutes with everything in Z n 1, so we get the simple branching in a different way. The GZ-basis in the case G = S n is called the Young basis. By (3.3)(i), the Young/GZ vectors are common eigenvectors for the GZ n. Let v be a Young vector (an element of the Young basis) for S n. Definition 2.9. Given such a v, let α(v) = (a 1,..., a n ) C n where a i is the eigenvalue of X i acting on v. We call α(v) the weight of v. Note that a 1 = 0 because X 1 = 0. Let be the spectrum of the YJM-elements. spec(n) = { α(v) v is a Young vector By (3.3)(ii), we know that dim GZ n = spec(n) = λ S n dim λ By definition, spec(n) is in natural bijection with chains T as in (3). Explicitly, given α spec(n), denote by v α the Young vector with weight α, i.e. α(v α ) = α, and by T α the corresponding chain in the Bratelli diagram. Given a chain T as in (3), we denote the corresponding weight vector α(v T ) by α(t ). Hence, we have a 1-1 correspondence associating T α(t ), or going the other way α T α between chains (3) and spec(n). And there s a natural equivalence relation on spec(n): for α, β spec(n), we say α β v α, v β belong to the same irreducible S n -module T α, T β start at the same vertex. Clearly, spec(n)/ = S n. We now want to describe the set spec(n) describe the relation calculate the matrix elements in the Young basis could calculate the characters of the irreducible representations ***

8 DR. STUART MARTIN References on general Wederburn theory: Curtis-Reiner 62, 81 Story so far: Each representation V λ has nice basis: GZ-basis {v T where (T ) : λ (n)... λ (1) = (1) YJM elements: X k = k 1 i=1 (i, k) CS n for 1 k n generate the GZ-algebra, a maximal commutative subalgebra of CS n GZ-basis is unique basis such that basis elements are common eigenvectors of the X k : We defined α(t ) = (a 1,..., a n ) C n X i v T = a i v T We re interested in spec(n) = { α(t ) paths T.