Irreducible Representations of symmetric group S n Yin Su 045 Good references: ulton Young tableaux with applications to representation theory and geometry ulton Harris Representation thoery a first course James The representation theory of the symmetric groups Basic results from representation theory G finite group ground field GL(V ) = GL n () if dim V = n That is G acts on V A representation of G is a group homomorphism ϕ : G Representation of G: ϕ : G GL(V ) = GL n () Left modules of the group algebra G: ϕ : G End(V ) = M n () Important facts: G is semisimple: Every G-module is a direct sum of simple module If char = 0 then G is semisimple To understand the representation of S n it s enough to understand all the simple modules Goal: Assume char = 0 ind all irreducible representations /simple modules of S n
Idea: Module S n : spanned by all λ tableaux Permutation modules: spanned by all λ-tabloids Specht modules: spanned by all λ-polytabloids Specht modules are simple Tableaux Definition Suppose λ = (λ λ λ t ) is a partition of n Then a λ-tableau is called a Young tableau if it s filled with { n} without repetition Example 4 5 5 4 6 are Young tableaux but 4 is not Action of S n on Young tableaux: permuting numbers in the tableau Example ()(4) 4 = 4 5 5 Note: λ-tableaux - correspond to elements of S n 4 = (4) (4) 5 If we view tableaux as elements in S n then the action of S n is just composition Take the vector space generated by spanning all λ-tableaux This is a S n -module isomorphic to S n Example λ = ( ) S = = T T T T T T
Permutation module of S n or any λ-tableau T define its row stabilizer and column stabilizer as subgroups of S n : Row stabilizer: R T = {σ S n σ fixes all rows of T} Column stabilizer: C T = {σ S n σ fixes all columns of T} Equivalence relation on λ-diagrams: T T R T = R T T = σt for some σ R T Notation: {T} = the equivalence class of T Note: {T} is the R T -orbit of T {T} = R T = λ!λ! λ t! Example: T = 4 5 T = 4 5 R T = S {4} S {5} and C T = S {4} S {} 4 5 Notation: {T} = 4 5 4 5 but Example ()-tabloids: {T } = = {T } = = {T } = = Proposition and Definition(Permutation module) Let M λ be the -vector space spanned by all λ-tabloids Then M λ is an S n -module under the action σ {T} = {σ T} Example:
σ = ()(4) {T} = Then σ {T} = {σ T} = 4 5 4 5 Note: The action is well defined because the identity R σ T σ = σr T Choose another representative of {T} = Then σ {T} = 4 5 acts: ) M λ is a cyclic S n -module generated by any tabloid because S n acts on tabloids transitively ) dim M λ = #{λ tabloids} = S n n! = S λ λ! λ t! Example: ()-permutation module is M () = {T } {T } {T } Compare: S = T T T T T T Question: Is M λ irreducible? Answer: No in most cases Example: M Let x = {T } + {T } + {T } Then the action of S n fixes x So the span of x is a submodule of M () which is isomorphic to the trivial module Specht module Wilhelm Specht (907-985): German mathematician He introduced Specht modules in 95 Definition (Polytaloid) or any λ-tableau T define e T = κ T {T} M λ where κ T = σ C T ( ) σ σ Note e T depends on the tableau T not only on the tabloid {T} A different representative of {T} will give a different polytabloid Example T = Then C T = S {4} S {5} and κ T = e (4) (5) + (4)(5) Then the corresponding tabloid e T = κ T {T} = (e (4) (5) + (4)(5)) = 4 5 5 4 + 4
e T = κ T {T } = 4 5 5 4 + Proposition and definition (Specht module) S λ = e T : Tλ tabloid is a submodule of M λ Note The action of S n on S λ is induced from M λ which is σ e T = σ κ T {T} = ( ) τ στ{t} = κ σt {σt} = e σt τ C T To prove the action is well defined we need an identity C σt σ = σc T Note The Specht module S λ is a cyclic module generated by any polytabloid Example Specht module S () e T = κ T {T } = (e ()){T } = = {T } {T } e T = κ T {T } = (e ()){T } = e T = κ T {T } = (e ()){T } = e T = κ T {T } = (e ()){T } = e T = κ T {T } = (e ()){T } = e T = κ T {T } = (e ()){T } = Thus S () is spanned by all the above six elements: = {T } {T } = {T } {T } = {T } {T } = {T } {T } = {T } {T } S () = e T e T e T e T e T e T Compare: M () = {T } {T } {T } S = T T T T T T M () Theorem: When char = 0 S λ s are all the simple modules of S n Example Representations of S : trivial representation alternating representation -dim representation λ = () S () is the trivial module λ = ( ) S () is the -dim module λ = ( ) S () is the alternating module Note The simple modules correspond to the partitions of n 5
Representations of S 4 : Specht modules S (4) S () S () S () S () Question: What s the dimension of the Specht module? What s the basis? dim(s (λ) ) = S n = n! λ n Dimension and basis of Specht modules e T = {T } {T } e T = {T } {T } e T = {T } {T } e T = {T } {T } e T = {T } {T } e T = {T } {T } Thus S () is spanned by all the above six e T s: S () = e T e T e T e T e T e T M () But the dimension of S () is Possible basis is {e T e T } Example λ = ( ) Then 4 polytabloids {e T } will span the Specht module S () But actually the dimension is very small: dim S () = Basis: Let T = T = Then {e T e T } is a basis of S () 4 4 Write down the basis: e T = 4 4 4 + 4 ; e T = 4 4 4 + 4 Take T = 4 Then e T = 4 4 4 + 4 = e T + e T Theorem ) {e T : T is a standard λ tableau} is a basis of the Specht module S λ ) Dimension of Specht module (Hook formula): dim S λ = #{standard λ tableaux} = n! x T Hook(x) Example 6
λ = ( ) 5 There are 5 standard λ-tableaux: 4 5 5 4 4 5 5 4 The corresponding polytabloids form a basis of the Specht module S () and dim S () = 5 Hook formula: 5! dim S () = 4 = 5 Topic Decomposition of permutation modules Note Having found all irreducible representations of S n when char = 0 we can decompose any S n -module into direct sum of simple modules S n = (S (λ) ) dim Sλ eg S = S () (S () ) S () λ n Question: What is the decomposition of the permutation module M λ? One approach: character and orthogonal projection Note that characters of irreducible representations from an orthonormal basis of the space of class functions The character of S λ can be computed (computable but not fast) and then the number of occurrence of S µ in M λ is equal to the inner product S µ M λ Therefore M λ = (S (µ) ) Sµ M λ Another approach: Young s rule M λ = (S µ ) k µλ where k µλ = #{semistandard λ tableaux of type µ} µ λ µ Example M = S (5) S (4) S () Topic A taste of modular representation Question: What are the simple modules of S n if char = p 0? Note A representation of a group G in the ground field with characteristic p is called a modular representation Difficulty When char = p the Specht modules are not simple anymore Example λ = ( ) n = char = Our Specht module is spanned by e T e T T T standard Let x = {T } + {T } + {T } We claim x S () (Note If char = 0 x / S () ) e T = {T } {T } = {T } + {T } e T = {T } {T } = {T } + {T } Then x = e T + e T Obviously σ x = x Therefore x S () is a nontrivial submodule The Specht module S () is not simple! 7
How to fix this? Solution: Define an inner product on M λ and define a new module D λ = Sλ /(Sλ Sλ ) Theorem All the nonzero D λ s give all the simple modules of S n The dimension is computable by the Gram matrix (but we don t have a general formula) Difficulty If char = p n! then S n is not semisimple That is S n -modules can t be decomposed into direct sum of simple modules! or example the permutation module M λ Solution: Instead of decomposing the modules into direct sum we find the composition factors of this module The composition factors are among all the simple modules D (λ) But even the composition factors of S λ is not clear! Next week: Tentative list of topics Decomposition of M λ when char = 0 by using Young s rule Restriction and induction of Specht modules: S λ S m and S λ S n Simple S n -modules D λ when char = p Thank you! 8