REPRESENTATIONS OF S n AND GL(n, C)

REPRESENTATIONS OF S n AND GL(n, C) SEAN MCAFEE 1 outline For a given finite group G, we have that the number of irreducible representations of G is equal to the number of conjugacy classes of G Although numerically equal, there is no general formula to map the conjugacy classes of a group to its irreducible representations For the case of the symmetric group S n, however, there is a remarkably simple correspondence; we will see that each irreducible representation of S n is determined uniquely by a combinatorial object called a Young diagram corresponding to a given conjugacy class In section 2, we will introduce these diagrams and show how they determine a representation V λ of S n In section 3, we will show that each such representation is irreducible and, in fact, the collection of V λ s accounts for every irreducible representation of S n Finally, in section 4, we will discuss how these representations can be used to give a large class of irreducible representations of the general linear group GL k For more details, please refer to Chapters 4, 6, and 15 of Representation Theory: A First Course by William Fulton and Joe Harris 2 Group algebras, Young diagrams, Young tableaux, and Young symmetrizers Our goal is to describe the irreducible representations of S n We first define the group algebra CS n to be the set of all finite formal sums of the form σ S n z σ e σ, z σ C, where the e σ are basis elements indexed by the σ S n The group algebra is equipped with a natural addition and multiplication which work as follows: i) z i e σi + y i e σi = (z i + y i )e σi ii) ( z i e σi ) ( ) y j e σj = z i y j e σiσ j For example, in the case of CS 3 we have and i,j 3e (12) + 7e (132) + 5e (12) + 2e (132) = 2e (12) + 9e (132), (2e 1 3e (12) )(e (123) + 5e (12) ) = 2e 1(123) + 10e 1(12) 3e (12)(123) 15e (12)(12) = 2e (123) + 10e (12) 3e (23) 15e 1 We can also view CS n as a complex vector space with basis elements indexed by the e σ s There is a natural linear action of CS n (the group algebra) on CS n 1

2 SEAN MCAFEE (the vector space) given by the multiplication rule above In other words, we have a representation ρ : S n GL(CS n ) = GL(C n! ), called the regular representation of S n The rest of this section will be devoted to describing certain elements c λ of CS n (the group algebra), called Young symmetrizers The images of these elements in CS n (the vector space) will be subspaces V λ of CS n that are invariant under the action of S n Thus we will arrive at representations ρ λ : S n GL(V λ ) which, we will see in section 3, are irreducible Recall that a partition λ = λ 1 + + λ k, (λ 1 λ k ) of an integer n gives rise to a collection of stacked squares called a Young diagram; where the length of each row is given by λ i For example, the partition 4 + 2 + 2 + 1 of n = 9 induces the Young diagram By filling in the boxes of a Young diagram in any way with one each of the numbers 1,, n, we get what is known as a Young tableau For our purposes, the arrangement of numbers will not matter (this is not obvious and takes some work to prove), so we will fill in our Young diagrams in the natural way, starting with 1 in the upper left and increasing by 1 as we move from left to right and top to bottom: Given such a tableau, we define 1 2 3 4 5 6 7 8 9 P λ = {σ S n σ preserves each row } Q λ = {σ S n σ preserves each column } We then use these sets to define elements a λ, b λ in the group algebra CS n : a λ := σ P λ e σ b λ := sgn(σ)e σ σ Q λ Finally, we define the Young symmetrizer: c λ := a λ b λ For an easy example, consider the partition 3 = 2 + 1; this gives us the following Young tableau: 1 2 3

3 so and We then have P λ = {1, (12)} and Q λ = {1, (13)}, a λ = e 1 + e (12) and b λ = e 1 e (13), c λ = (e 1 + e (12) )(e 1 e (13) ) = e 1 + e (12) e (13) e (132) As discussed above, c λ acts on the vector space CS 3 ; we take this action to be on the right The matrix of this action (using the ordered basis {1, (12), (13), (23), (123), (132)}) is 1 1 1 0 1 0 1 1 0 1 0 1 1 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 The first and third columns form a basis for the column space of this matrix, thus the image V λ = CS n c λ of c λ is the span of e 1 + e (12) e (13) e (132) and e 1 + e (13) e (23) + e (123) Assuming for the moment that we have proven each such V λ is irreducible, this gives us a two dimensional irreducible representation of S 3, ie V λ is equivalent to the standard representation As a final remark, notice that any element d CS n gives us a subspace CS n d of CS n which is invariant under the (left) action of CS n Thus the Young symmetrizers c λ are not special in this respect; what does make them special is that the CS n - invariant subspaces which they determine will turn out to be irreducible under the CS n -action (and, actually, unique for each partition λ) Exercise 21 Explicitly show that the representation above is equivalent to the standard representation of S 3, where S 3 acts on the subspace of R 3 spanned by the vectors e 1 e 2, e 2 e 3 by permuting indices Exercise 22 Show that the Young diagrams 1 2 3 and 1 2 3 give rise to the trivial and alternating representations of S 3, respectively Exercise 23 Show that, for S n, the partitions (n) and (1 + 1 + + 1) give rise to the trivial and alternating representations of S n, respectively Exercise 24 Show that, for arbitrary n, the partition ((n 1) + 1) gives rise to the standard representation of S n

4 SEAN MCAFEE We will prove the following: 3 proof that CS n c λ is irreducible Theorem 31 Given S n, let λ be a partition of n Let V λ be the subspace of CS n spanned by the Young symmetrizer c λ (so V λ = CS n c λ ) Then i) V λ is an irreducible CS n -module (ie V λ is an irreducible representation of S n ) ii) If λ, µ are distinct partitions of n, then V λ = Vµ iii) The V λ account for all irreducible representations of S n The objects involved are combinatorial in nature, so the proof relies on a couple unavoidably ugly lemmata The proofs of these will be omitted, but can be found in chapter 4 of Fulton and Harris Lemma 32 For all x CS n, c λ xc λ is a scalar multiple of c λ Lemma 33 If λ µ, then c λ CS n c µ = 0 Assuming the above have been shown, we are ready to prove the theorem: Proof of theorem 31: i) Let V λ = CS n c λ for a given Young symmetrizer c λ By Lemma 32, we have that c λ V λ Cc λ Let W be a nonzero subrepresentation of V λ We will show that W = V λ First, we claim that both c λ V λ, c λ W are nonzero To see this, suppose that c λ V λ = 0 Then V λ V λ = CS n (c λ V λ ) = 0 As vector spaces, there exists a projection π : CS n CS n c λ that commutes with the action of S n (this can be obtained by taking the projection induced by c λ and averaging over the action of S n ) This projection can be described as right multiplication on CS n (the group algebra) by an element x CS n by letting x := π(1) This x must lie in V λ, since x = 1 x By the definition of a projection, x = x 2 V λ V λ = 0, thus x = 0, a contradiction since the nonzero c λ itself lies in CS n c λ Thus we must have c λ V λ 0 Similarly, we show c λ W 0 So we have W is a subspace of V λ, c λ V λ Cc λ, and c λ W 0; thus we must have c λ W = Cc λ Therefore, V λ = CS n c λ = CS n (Cc λ ) = CS n (c λ W ) W, where the inclusion on the right follows from the fact that W is a subrepresentation of V λ, ie it is invariant under the action of CS n Thus, we have V λ = W, completing the proof that V λ is irreducible ii) Let λ, µ be distinct partitions of n, and let V λ, V µ their corresponding representations By the above, we have that c λ V λ = Cc λ 0 By Lemma 33, we have that c λ V µ = c λ CS n c µ = 0 Therefore, V λ and V µ cannot be isomorphic as CS n modules, proving ii)

5 iii) Each partition λ of n corresponds to a distinct conjugacy class of S n We have shown in ii) that the V λ determined by such partitions are all inequivalent Thus, since the number of conjugacy classes of a finite group is equal to the number of irreducible representations, we have accounted for all of the irreducible representations of S n 4 irreducible representations of GL n (C) Equipped with our construction of the irreducible representations of S n, it is actually fairly simple to describe all irreducible, finite-dimensional, rational representations of the general linear group GL n (C) By rational, we mean the following: given a finite dimensional representation ρ : GL(n, C) GL(m, C), we say that the representation is polynomial if, for each A GL(n, C), we have that ρ(a) has entries which are polynomial in the entries of A We say that the representation is rational if, for each A GL(n, C), we have that ρ(a) has entries which are rational polynomials in the entries of A Example 41 For any GL(n, C), we have the standard representation, given by ρ : GL(n, C) GL(n, C) A A This is clearly a polynomial (and thus rational) representation Exercise 42 Show that the standard representation of GL(n, C) is irreducible Exercise 43 Show that the representation ρ(g) = (g 1 ) T is a representation of GL(n, C) on C 2 which is rational, but not polynomial In studying the irreducible representations of S n, we started by observing that any σ S n acts (by right multiplication) on the group algebra CS n by permuting the basis elements We then used linearity to extend this action to a CS n action on CS n The image of this action under the element c λ ended up being an irreducible representation V λ of S n The process for studying the irreducible representations of GL(n, C) will be similar We will start by giving a sketch of the construction, then we will proceed to fill in some of the details Consider GL(n, C) acting on the vector space V = C n by matrix multiplication For any positive integer d, this multiplication can be extended to the dth tensor power of V, written V d For such a fixed d, we can look at a partition λ = (d 1,, d k ) and (just like in the S n case) construct the Young symmetrizer c λ This symmetrizer acts on the right of V d in the same way that it did on the group algebra CS n : by permuting elements The image V d c λ will turn out to be a GL(n, C)-invariant subspace of V d We write S λ V to denote this subspace, and call the map V S λ V the Schur functor These S λ V s will turn out to be precisely the irreducible polynomial representations of GL(n, C) Let s begin to unpack the above paragraph by looking at the action of GL(n, C) on V d For a basis vector v 1 v 2 v d of V d, and for g GL(n, C), we define g (v 1 v 2 v d ) := gv 1 gv 2 gv d By linearity, we can extend this to an action on any linear combination of such basis vectors

6 SEAN MCAFEE Exercise 44 Check that the above defines a representation of GL(n, C) on V d Now, let s analyze the right action of S d on V d For σ S d and v 1 v 2 v d a basis element of V d, define (v 1 v 2 v d ) σ := v σ(1) v σ(2) v σ(d) Again, by linearity we can extend this to a right action of the group algebra CS d on all of V d This action clearly commutes with the left action of GL(n, C) defined above We define two important subspaces of V d, called the symmetric tensors Sym d V and the alternating tensors Alt d V (this is often denoted d V, and its elements are written as v 1 v 2 v d ): { } Sym d V := v σ(1) v σ(2) v σ(d) σ S d { } Alt d V := sgn(σ)v σ(1) v σ(2) v σ(d) σ S d For example, if d = 2, then (for arbitrary vectors v 1, v 2 V ) and v 1 v 2 + v 2 v 1 Sym 2 V, v 1 v 2 v 2 v 1 Alt 2 V Intuitively, we have that Sym d V is the set of vectors in V d that are not changed by permuting their components, and Alt d V is the set of vectors in V d that change sign whenever two components are switched (this is due to S d being generated by transpositions, which have negative signature) Exercise 45 Verify that Sym d V and Alt d V are invariant subspaces of V d under the action of CS d Under the (right) action of CS d, we can redefine Sym d V and Alt d V as ( ) Sym d V := V d Alt d V := V d ( σ S d e σ σ S d sgn(σ)e σ Exercise 46 Check that the above definitions coincide with our original definitions of Sym d V and Alt d V Exercise 47 Check that Sym d V and Alt d V are invariant under the (left) action of GL(n, C) Exercise 47 gives two examples of a more general (and extremely useful) phenomenon: for any element c of the group algebra CS d, we have that V d c is invariant under the action of GL(n, C) Indeed, since the actions of GL(n, C) and CS d commute, then, for any g GL(n, C), v V d, g(vc) = (gv)c V d c )

7 This gives us an entire family of invariant (and potentially irreducible) subspaces of V d obtained by simply choosing an element at random from CS d As mentioned before, the only elements we need will be the Young symmetrizers c λ There is another thing to point out about Sym d V and Alt d V : they are the images V d c λ = S λ V of the Young symmetrizers corresponding to the partitions (d) and (1, 1,, 1), respectively As subspaces of V d, they have zero intersection, ie they are direct summands in the decomposition of V d : V d = Sym d V Alt d V (Sym d V Alt d V ) Notice the relation between this and the irreducible spaces V (d) and V (1,1,,1) in CS d in our representation of S d There is a very simple correspondence between the decomposition of CS d into irreducible representations V λ of S d and the decomposition of V d into irreducible representations S λ V of GL(n, C) The above discussion and more is summarized in the following theorem, which we will not prove here For details, refer to pages 84-87 of Fulton and Harris Theorem 48 i) Let k = dim V Then S λ V = 0 if λ k+1 0; that is, if the Young diagram corresponding to λ has more than k rows ii) Let m λ be the dimension of the irreducible representation V λ of S d corresponding to a partition λ of d Then V d = (S λ V ) m λ λ iii) Each S λ V is an irreducible polynomial representation of GL(n, C) iv) All rational representations of GL(n, C) can be described by a tensor product S λ V (det) k, where z Z >0 and det is the (one dimensional) determinant representation of GL(n, C) Example 49 Consider GL(2, C) with its natural representation on V = C 2 Let d = 3, and consider the partition λ = (1, 1, 1) Then we have Then, for v 1 v 2 v 3 V 3, c (1,1,1) = e 1 e (12) e (13) e (23) + e (123) + e (132) (v 1 v 2 v 3 ) c (1,1,1) = v 1 v 2 v 3 v 2 v 1 v 3 v 3 v 2 v 1 v 1 v 3 v 2 + v 3 v 1 v 2 + v 2 v 3 v 1 Since V is two dimensional, we may write v 3 = av 1 + bv 2 ; the above then becomes v 1 v 2 av 1 + v 1 v 2 bv 2 v 2 v 1 av 1 v 2 v 1 bv 2 av 1 v 2 v 1 bv 2 v 2 v 1 v 1 av 1 v 2 v 1 bv 2 v 2 + av 1 v 1 v 2 + bv 2 v 1 v 2 + v 2 av 1 v 1 + v 2 bv 2 v 1 If we use the properties of tensors to move the constants a, b to the front of each monomial, we see that the above sum collapses to 0, illustrating part i) of the theorem

8 SEAN MCAFEE Example 410 We use the setup of the previous example, this time with partition λ = (2, 1) Then we have Thus, for v 1 v 2 v 3 V 3, c (2,1) = 1 + e (12) e (13) e (132) (v 1 v 2 v 3 ) c (2,1) = v 1 v 2 v 3 + v 2 v 1 v 3 v 3 v 2 v 1 v 3 v 1 v 2 We have an embedding of Alt 2 V V into V V V given by (v 1 v 3 ) v 2 v 1 v 2 v 3 v 3 v 2 v 1 This realizes the image of c (2,1) as the subspace of Alt 2 V V spanned by vectors of the form (v 1 v 3 ) v 2 + (v 2 v 3 ) v 1 These vectors (check this!) are the kernel of the canonical map φ : Alt 2 V Alt 3 V, (v 1 v 2 ) v 3 v 1 v 2 v 3, thus S (2,1) V = Ker φ By part ii) of the theorem, and by our discussion of S 3 in section 2, we have the decomposition V V V = Sym 3 V Alt 3 V Ker φ Ker φ, where the two copies of Ker φ are due to the fact that CS 3 c (2,1) is a two-dimensional (irreducible) representation of S 3, ie m (2,1) = 2 To summarize this section: we can parametrize the irreducible rational representations of GL(n, C) by a triple (d, λ d, z), where d is any positive integer, λ d is a partition of d whose Young diagram has no more than n rows, and k is any positive integer The resulting representation is the tensor product S λd V (det) k There is much, much more to say about representations of GL(n, C) than is contained in this paper For instance, the use of Schur polynomials to describe the characters of S λ V, or the decomposition of the tensor product S λ V S µ V, or the connection between the representations of GL(n, C) and those of SL(n, C) and U(n) We encourage the reader to investigate these on their own