1 Irreducible Representations of S n

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1 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes REPRESENTATION THEORY OF S n Steven Byrnes Abstract We use Young tableaux and Young symmetrizers to classify the irreducible representations over C of the symmetric group on n letters, S n. We establish an isomorphism between a ring of all representations of finite symmetric groups and the ring of symmetric functions, and, as a corollary, prove the Frobenius formula for the characters of representations of S n. We use the theory thus developed to characterize the representations of the Lie algebra sl n. Irreducible Representations of S n. Young tableaux In any finite group, the number of conjugacy classes equals the number of irreps. But the symmetric group S n (i.e. the group of permutations of n objects, under composition) is one of the few where there is a natural one-to-one correspondence between the two. As a first step in defining this correspondence, we define a partition of a positive integer n to be a nonincreasing sequence of positive integers summing to n. That is, λ = (λ, λ 2,...,λ m ) is a partition of n iff λ λ 2 λ m and m i= λ i = n. When convenient, we will allow there to be 0s at the end of a partition, with the understanding that (λ,...,λ m, 0,..., 0) = (λ,...,λ m ). The partitions of n are in one-to-one correspondence with conjugacy classes of S n : the partition (λ,, λ m ) is associated with the conjugacy class of permutations which, written in cycle notation, contain cycles of length λ, λ 2,, λ m. If λ is a partition of n, we write λ n or λ = n. The next step is to associate each partition with a Young diagram. A Young diagram consists of rows of square boxes, with each row left-justified, and with each row no longer than the row above 2. The Young diagram associated with a partition λ = (λ,, λ m ) is the one with λ i boxes in the ith row from the top. Abusing notation, we use λ to denote the Young diagram in addition the partition, when there is no risk of confusion. Recall that the regular representation of a group G is the group algebra C[G], with For the purposes of this paper, an irrep denotes an irreducible representation over C. 2 This is the English convention; diagrams written using the French convention have the smallest row on top.

2 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 2 Figure : The Young diagram of (3, 2, 2) G-action defined by left-multiplication, and that each irrep of G appears as a subrepresentation of the regular representation. Thus each irrep of S n is associated with an S n -invariant subspace of C[S n ], and to define the irrep it suffices to define a map into one of its associated subspaces. Starting with a Young diagram λ n, number the boxes to n, left-to-right top-tobottom. Let S n permute the numbers in the boxes, and define the subgroups A λ, B λ S n to be the permutations preserving rows and the permutations preserving columns, respectively. Now define two elements of C[S n ] by a λ = g A λ g and b λ = g B λ (sign(g))g (where sign(g) is the determinant of the associated permutation matrix). The Young symmetrizer of λ is defined to be c λ = a λ b λ..2 Characterization of the irreps of S n The fundamental theorem is the following: Theorem. For λ n, the image of the map ϕ λ : C[S n ] C[S n ], x xc λ is an irrep of S n. As λ varies over partitions of n, each irrep of S n (up to isomorphism) is generated exactly once. Proof. We start by putting an ordering on the set of Young diagrams of size n. For λ = (λ,...,λ m ) and η = (η,...,η m ), we say λ > η iff there is a j with λ i = η i for i < j and λ j = η j. Next we define an Young tableau as follows: if λ n is a Young diagram, we make a Young tableau with shape λ by putting a number in each box from to n with no repeats. Note that S n operates on tableaux of a given size by permuting the numbers. Now we prove a useful lemma about Young tableaux. Lemma. Suppose that S and T are Young tableaux with shape σ and τ respectively (where σ = τ = n), and suppose σ τ. Then either we can find two distinct integers which are in the same row of S and the same column of T; or λ = λ and there is a p S n preserving rows of S and a q S n preserving columns of T, with ps = qt. Proof. Suppose that there are no two distinct integers in the same row of S and the same column of T. Then each of the σ numbers in the first row of S is in a different column of T. In particular, T must have at least σ different columns, and since σ > τ, T has exactly σ = τ columns, and there is a q S n preserving columns of T such that

3 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 3 the top row of q T has the same elements as the top row of S. Likewise, examining the second row of S, we see that σ 2 = τ 2 and there is a q 2 S n fixing the top row of T and preserving columns of T, such that the second row of q 2 q T has the same elements as the second row of S. Repeating this process, and noting that (q m q m q ) still preserves columns of T, we have proven the lemma. We define the canonical tableau of a given shape to be the one numbered in order left-to-right, top-to-bottom; this is the tableau we used implicitly when defining A λ and B λ. To complete the proof of Theorem, we need four more straightforward results. Lemma 2. For p A λ and q B λ, pa λ = a λ p = p and (sign(q)q)b λ = b λ (sign(q)q) = b λ. Proof. This follows from writing out the summation definition of a λ and b λ. Lemma 3. The set of elements x C[S n ] such that px(sign(q)q) = x for all p A λ, q B λ, is Cc λ. Proof. One inclusion follows from Lemma 2. For the opposite inclusion, suppose that x = g S n α g g satisfies px(sign(q)q) = x. Since the set of such x s is closed under addition and negatives, we may subtract a multiple of c λ to assume WLOG that α e = 0 (where e denotes the identity in S n ). Then for p A λ, q B λ, it follows from px(sign(q)q) = x that α pq = α e sign(q) = 0. And for g that cannot be written in the form pq, we have by Lemma that there are two elements in the same row of T and the same column of gt (where T is the canonical tableau with shape λ). Let p A λ be the transposition of those two elements and let q = g pg B λ. Then it follows from px(sign(q)q) = x that α g sign q = α pgq = α g, and since sign q = sign p =, we have α g = 0. So we have proven that α g = 0 for all g, so indeed only scalar multiples of c λ satisfy the given identity. For any tableau T with shape λ, we can define b T in a way analogous to b λ, but starting from the tableau T instead of the canonical tableau T λ (in particular, b λ = b Tλ ). Note that xb T x = b xt. Lemma 4. If λ > η, then c λ xc η = 0 for all x C[S n ]. Proof. First we note that for any S of shape λ and T of shape η, by Lemma, there are two elements in the same row of S and the same column of T. Letting s S n switch those two elements, we have a S b T = (a S s)(sb T ) = (a S )( b T ) by Lemma 2, so a S b T = 0. For x in the subgroup S n of C[S n ], we have a λ xb η = a Tλ b xtη x = 0x = 0. By taking linear combinations of these, we get a λ xb η = 0 for all x C[S n ]. In particular, a λ (b λ xa η )b η = 0, proving the lemma. Lemma 5. Suppose a finite group G acts on C[G] by left-multiplication, and suppose ν : C[G] C[G] is a G-invariant C-endomorphism. Then ν is equivalent to rightmultiplication by an element of C[G].

4 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 4 Proof. For g G, we have ν(g) = ν(g()) = g(ν()). Then by C-linearity of ν, we have ν(x) = xν() for general x C[G]. Now we can finish the proof of Theorem. For λ n, let S λ = C[S n ]c λ be the image of ϕ λ. Recall that S n acts on C[S n ] on the left, so S λ is certainly an S n -invariant subspace. By Lemmas 2 and 3, we have c λ xc λ Cc λ for all x C[S n ]. It follows from this that c λ S λ Cc λ. Now suppose for contradiction that c λ S λ = 0. Then S λ S λ = C[S n ]c λ S λ = 0. By basic representation theory, there is an S n -invariant projection from C[S n ] to S λ ; by Lemma 5 suppose the projection is right-multiplication by x C[S n ]. Then x = x S λ and by definition of a projection, x = x 2 S λ S λ = {0}, so S λ = 0, a contradiction. Thus c λ S λ 0, and since c λ S λ Cc λ, a -dimensional space, we get c λ S λ = Cc λ. Using the same reasoning, if W is a nonzero subrepresentation of S λ, then c λ W = Cc λ, so S λ = C[S n ]c λ = C[S n ]c λ W W, since W is an S n -invariant subspace, and thus W = S λ. So S λ is irreducible. Finally, assume λ η, and we will show S λ = Sη. WLOG λ > η, so by Lemma 4, c λ S η = {0}. But we saw above that c λ S λ {0}, so S λ and S η have distinct C[S n ]-actions, and thus they are not isomorphic as representations. Now we are done we have found a distinct irrep for each conjugacy class of S n (see Section.), so this exhausts the irreps (up to isomorphism)..3 Corollaries of the characterization theorem As practice working with these representations, we can look at the -dimensional representations of S n and their tensor products with other irreps. For n 2, there are exactly two -dimensional irreps of S n. To see this, note that all of S n is generated by the conjugacy class of two-letter transpositions, which in a -dimensional irrep is mapped to a square-root of, i.e. to ±. If it is mapped to, we get the trivial representation ρ(g) = ; and if it is mapped to, we get the alternating representation A, defined by ρ(g) = sign(g). It follows immediately from the construction of S λ that S (n) is the trivial representation, and S (,,...,) is the alternating representation. We define the conjugate partition λ of λ n by flipping the associated Young diagram about a 45 line. In terms of the partition λ = (λ,..., λ m ), we have λ i = j iff λ j i and λ j+ < i. Corollary. For n 2, let A be the alternating representation of S n, and suppose λ n. Then S λ = A Sλ. Proof. We have S λ = C[S n ]a λ b λ, and we let V = C[S n ]b λ a λ. First I will prove V = A S λ. We have that C[S n ] = A C[S n ] as S n -modules, by the automorphism taking g S n to (g sign(g)), and this automorphism maps b λ to a λ and a λ to b λ, and thus V to A S λ. Hence, V = A S λ. So it suffices to show that S λ = V. First note that both

5 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 5 S λ and V are irreps if V were reducible, then tensoring each component with A would give a decomposition of S λ, which is impossible. Now right-multiplication by a λ b λ gives an endomorphism of S λ, which by Schur s Lemma is multiplication by a scalar. I claim that the scalar is not 0; in fact, Lemma 6. c λ c λ = n! dim S λ c λ. Proof. By Lemmas 2 and 3, c λ c λ = αc λ for some α C. As before, let ϕ λ : x xc λ act on C[S n ]. Using the basis S n of C[S n ], we have that Tr(ϕ λ ) is n!, since it has only s on the diagonal (the g-component of gc λ is ). Now we compute Tr(ϕ λ ) in a different basis, consisting of a basis for ker ϕ λ, unioned with the preimage of a basis for Img(ϕ λ ) = S λ. For the latter, we can just pick a basis of S λ, since α 0 (if c 2 λ = 0 then its eigenvalues would all be 0, so ϕ λ would have trace 0 which we saw above is false). Then ϕ λ is multiplication by 0 on the basis elements in ker ϕ λ, and multiplication by α on the basis elements in S λ, so Tr(ϕ λ ) = α dim(s λ ). We thus have α dim S λ = Tr(ϕ λ ) = n!, proving the lemma. So indeed, right-multiplication by a λ b λ is an isomorphism of S λ. Then right-multiplication by a λ is an injective (hence nontrivial) homomorphism S λ V, so by Schur s Lemma S λ = V proving the corollary. 2 Symmetric Functions and Irreps of S n We will define two Z-graded rings the ring of homogeneous symmetric functions (a concept closely related to symmetric polynomials), and a ring of representations of all finite symmetric groups and show that they are isomorphic. The first step will be proving some basic facts about symmetric polynomials. 2. Symmetric polynomials We define the elementary symmetric polynomial e n in m variables by e n (x,...,x m ) = x i x in i < <i n m (in particular, if m > n, e n (x,..., x m ) = 0), and we define the complete symmetric polynomial h n in m variables by h n (x,...,x m ) = x i x in. i i n m

6 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 6 For a partition λ = (λ,...,λ j ), we define e λ and h λ by e λ = e λ e λj ; h λ = h λ h λj. Next we define the Schur function s λ (x,...,x m ). We will give two definitions, both of which we will need later (for example, cf. () and (7)). The first definition is in terms of a Young semi-standard tableau (abbreviated sst, plural ssts). An sst T of shape λ n is obtained by numbering the boxes of the Young diagram λ with positive integers in such a way that the numbers are non-strictly increasing along each row and strictly increasing along each column. For an sst T with entries t,..., t n (in some order), define (x,...,x m ) T = x t x tn, or (x,...,x m ) T = 0 in the case that some t i > m. Now we can give our first definition of Schur functions: for λ n, s λ (x,...,x m ) = (x,...,x m ) T. () T an sst of shape λ Our second definition is simpler to state: for λ n, we say s λ (x,...,x m ) = 0 if λ has more than m rows; otherwise we write λ = (λ,...,λ m ) (possibly with ending 0s), and then x λ +m x λ +m m.. x λm+m m x λm+m m m s λ (x,...,x m ) = x m x m. (2) m.. We will prove the equivalence by induction, by proving for both definitions the Pieri formula : s λ (x,...,x m )s (p) (x,...,x m ) = s η (x,..., x m ), (3) η where the sum is over all ssts η that are obtained from λ by adding p boxes, each in a different column. Lemma 7. Define s λ as in (). Then the Pieri formula (3) holds. Proof. We set up a one-to-one correspondence between a pair of ssts for λ and (p), and an sst for some such η containing the same total collection of box-labels. We do this by a process called row-bumping. Given an sst µ and a positive integer x, we join x to µ as follows: If x is at least as large as the right-most entry of the top row, we add a box onto the right of the top row and label it with x. Otherwise, we put x into its correct lexicographic place in the top row, shifting all the later entries one box to the right and bumping the right-most entry into the second row, where we repeat the process.

7 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes Figure 2: An example of row-bumping. Given ssts in shapes λ and (p), we join the entry of the leftmost box of (p) to λ, then the second-to-leftmost, and so on. Since the entries of (p) are non-strictly increasing, it is clear from the algorithm that each added box will be strictly to the right of the previous added box (in particular, the added boxes will all be in different columns), and the entries of the resulting ssts are clearly the union of the entries of the two we started with. Conversely, starting with an sst with shape η as in the lemma, we can take the rightmost added box and run the row-bumping algorithm in reverse, and repeat that process to get two ssts with shapes λ and (p). The definition of Schur polynomials () gives the desired result. Lemma 8. Define s λ as in (2). Then the Pieri formula (3) holds. Proof. Let f(n,...,n m ) be the determinant of the m m matrix with (i, j)th entry x n j i. So we need to prove that f(l + m,...,l m )h p = f(n + m,..., n m ), with the sum over n l n 2 l 2 n m l m with n i = p + l i (this expresses the condition that p squares were added, no two in the same column). Expressing this for all ps at once in a formal power series (and changing variables n i n i + m i, l i l i + m i), it suffices to show that f(l,...,l m ) m i= x i = f(n,...,n m ), (4) with the sum over all n l > n 2 l 2 > n m l m. We prove this by induction on m. The base case m = is trivial. And expanding the determinant along the top row, we see that (4) follows immediately from the induction hypothesis. So we have proven that Pieri s formula (3) holds, and now we can easily prove that the two definitions of Schur polynomials () and (2) are equivalent. Since s (p) = h p (under either definition), we can repeatedly apply Pieri s formula to get, for λ n, h λ = η n K ηλ s η, (5) where K ηλ counts the number of ways to build up η by adding successively λ,...,λ j boxes in such a way that no two are added to the same column in a given stage, and such that the result of a given stage is a valid Young diagram. Equivalently, K ηλ is the number of ssts of shape η with λ s, λ 2 2s, etc. (the equivalence of the definitions follows from labeling the λ boxes added at the first stage, at the second stage 2, etc.). The

8 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 8 K ηλ are known as Kostka numbers. If η > λ in the lexicographic ordering defined above, it is clear from the first definition that K ηλ = 0, and it is also clear that K ηη =. Thus, (K ηλ ) η,λ n is an upper triangular matrix with s on the diagonal. (6) In particular, we can invert the matrix K to get an expression for s λ in terms of h η. It follows that the two definitions for Schur functions are indeed equivalent. And it is clear from (2) that the Schur polynomials are symmetric. We define two more symmetric polynomials. For λ = (λ,..., λ j ) n, the monomial symmetric polynomial m λ (x,...,x m ) = x λ i x λ j i j, with the sum over sets {i,...,i j } of distinct elements in {,...,m}. And the Newton power sum p λ (x,...,x m ) = j p λi (x,...,x m ) = i= j i= (x λ i + + xλ i m ). We prove another lemma relating these functions: Lemma 9. The following equality holds for formal power series: m m = x j= i y j λ i= = λ = λ m λ (x,...,x m )h λ (y,...,y m ) s λ (x,...,x m )s λ (y,..., y m ) z(λ) p λ(x,...,x m )p λ (y,...,y m ), with the sum over all partitions λ of any integer n, where if (λ,..., λ j ) contains the integer n a n times, z(λ) := n nan n!. Proof. The first line follows from a simple expansion: m l m ( + x i y j + x 2 i y2 j + ) = x n i h n(y,...,y m ) i= j= i= n= = m λ (x,...,x m )h λ (y,..., y m ). λ The second line follows from the definition (2) of Schur polynomials. We first have sλ (x)s λ (y) (x) (y) = det ( ( x i y j ) ) i,j m, (7)

9 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 9 where (x) := det x j i i,j m : expanding the RHS, the coefficient of x l x lm m for l > > l m is det y l j i i,j n, and vice versa, so this equation holds. On the other hand, det ( ( x i y j ) ) i,j m = (x) (y) m i= j= m ( x i y j ), which follows from induction on m: subtract the first row from each other row, then factor out (x i x ) from the ith row (2 i m) and x y j from the jth column ( j m). Next, subtract the first column from each other column, then factor out (y j y ) from the jth column (2 j m) and x i y from the ith row (2 i m). The determinant of the new matrix, by expansion in the first row, equals a smaller matrix of the original form, so we can use the induction hypothesis to prove this formula. So we have proven the second line of the lemma. For the third line, we have a proof using formal power series: ( m m m ) ( (x i t) n m ) (x i t) n = exp( log( x i t)) = exp = exp x i t n n i= i= n= n= i= ( ) pn (x)t n ( ) pn (x)t n an = exp = n n a n= n= a n! = z(λ) p λ(x)t λ. n=0 λ i= If instead of (x,...,x m ), we plug in (x y, x y 2,...,x m y m ), and plug in t =, we get the third line of the lemma. 2.2 The ring Λ of symmetric functions We define a symmetric function p of degree d to be a family of polynomials p = (p n ) n Z + such that each p n is a symmetric homogeneous polynomial of degree d in n variables with integer coefficients, and such that for m > n, p m (x,...,x n, 0,..., 0) = p n (x,...,x n ). All the symmetric polynomials we have defined so far can be grouped into such families, so we can regard them all as symmetric functions for example, instead of having the Schur polynomial s λ in m variables for some m Z, we can speak simply of the Schur function s λ. Let Λ d be the set of all symmetric functions of degree d. We define Λ = d= Λ d (with Λ 0 := Z). Λ is a graded ring, under ordinary polynomial addition and multiplication. Lemma 0. {s λ λ d} is a basis of Λ d. Proof. In a nontrivial sum λ a λs λ = 0, let η be the first (lexicographically) λ with a λ 0, and then the coefficient of x η xη j j is a λ, a contradiction proving linear independence.

10 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 0 And for an arbitrary element f Λ d, write f in d variables, let η be the first lexicographic λ with a nonzero coefficient of x λ xλ j j, and subtract s λ times that coefficient from f. Repeating this process will express f as a linear combination of s λ (both written in d variables). For more than d variables, the same relation will hold, since each term of f (being degree d) is uniquely determined by its terms that contain just the variables x through x d. So {s λ } is a spanning set. It follows that {s λ λ d, d Z + } is a basis for Λ. Thus we can define an inner product, on Λ by making {s λ } orthonormal. A lemma about this inner product will be useful later. Lemma. h λ, m η = δ λη (i.e. if λ = η, 0 otherwise). Also, p λ, p η = z(λ)δ λη. Proof. Write h λ = ν x λνs ν and m η = ν y λνs ν. By Lemma 9, we get ( ( )) (s ν (x,...,x m )) s µ (y,...,y m ) x λν y λµ δ νµ = 0 ν µ for any m. Since the s λ (x, x 2,...) are linearly independent over Z, they are certainly linearly independent over Z[y, y 2,...], so we get µ s µ(y,...,y m ) ( λ x λνy λµ δ νµ ) = 0 for any m. Again, by the linear independence of s µ, we have λ x λνy λµ = δ νµ, and by definition of the inner product, we get h λ, m η = δ λη. The identity p λ, p η = z(λ)δ λη follows from Lemma 9 in an analogous way. λ 2.3 The ring R of representations of symmetric groups For n, we let R n be the free (additive) abelian group generated by the irreps of S n (or more precisely, the isomorphism classes of the irreps of S n ). Any representation of S n is embedded in this group, by identifying V W with V + W R n. We further define a multiplication map R n R m R m+n defined by V W = Ind S n+m S n S m (V W) (8) where V and W are irreps. For general elements, is defined to distribute over addition i.e. ( ) ( ) a λ S λ b η S η = a λ b η (S λ S η ). λ n η m Elementary properties of tensor products and induced representations imply that equation (8) holds for any representations V, W. We now define R = n=r n.

11 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes R is a graded ring. The irreps of all S n for n are certainly a basis for R, so we define an inner product, on R by making the irreps of all S n an orthonormal set. Note that, when restricted to R n, this inner product agrees with the regular inner product of representations, defined by V, W = n! g S n χ V (g)χ W (g), where χ V denotes the character of V. For λ n, we define H λ to be the representation of S n induced from the trivial representation of the subgroup A λ (where A λ is defined as in section.). An equivalent definition is H λ = C[S n ]a λ, where a λ is again defined as in section Isomorphism theorem We will show in this section that Λ and R are isomorphic and isometric. The following lemma is the first step in connecting these two rings: Lemma 2. For partitions λ, η n, define ξη λ so that p η = λ ξλ η m λ (i.e. ξη λ coefficient of x λ x λn n in the expansion of p η ). Then (I) h λ = η is the z(η) ξλ η p η, and (II) χ Hλ (C(η)) = ξ λ η, where C(η) is the conjugacy class of elements of S n with cycle structure η. Proof. (I) follows from Lemma : Taking the inner product with h λ gives h λ, p η = ξη λ. Thus h λ, p η = ν z(ν) ξλ νp ν, p η for all p η, and since {p η } span Λ (by Lemma 9) and the inner product is nondegenerate, (I) must hold. For (II), by the character formula for induced representations, χ Hλ (C(µ)) = [S n : A λ ] C(µ) A λ. C(µ) Let a k be the number of cycles of length k in µ. We have, by combinatorics, C(µ) = n! i ia ia i! (write an element of C(µ) in cycle notation, then remove the parentheses to get one of n! permutations; but this overcounts by a factor of the expression in the denominator). Moreover, A λ = λ! λ n!. Finally, we compute C(µ) A λ, or the number of permutations with cycle structure µ that preserve the rows of T λ. Given such a permutation, let r pq be the number of q-cycles that permute elements of the pth row of T λ. We have r p + 2r p2 + + nr pn = λ p, r q + + r nq = a q, (9) and given a set of numbers r pq satisfying (9), there are n λ p! p= rp r p! n rpn r pn! different permutations characterized by r pq (by the same counting method as above). Combining

12 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 2 these, we get χ Hλ (C(µ)) = n! i ia i a i! n λ p! λ! λ n! n! r p rp! n rpn r p= pn! = n q= a q! r q! r nq!, with the sum over all sets of nonnegative integers r pq satisfying (9). Meanwhile, for q n, we have (x q + + xq n = a q! )mq r q! r nq! xqr q x qrnq n (the sum again over r pq satisfying (9)). Taking the product from q = to n, we get that χ Hλ (C(µ)) is the coefficient of x λ x λn n in p λ, as desired. We can now prove the isomorphism theorem. Theorem 2. Define ϕ : Λ R to be an additive homomorphism satisfying ϕ(h λ ) = H λ. Then (I) ϕ is an isomorphism of graded rings respecting the inner product, (II) ϕ(s λ ) = S λ, and (III) ϕ has inverse ψ, satisfying, for V a representation of S n, ψ(v ) = µ n z(µ) χ V (C(µ))p µ. (Note that we need to pass to Λ Q to define ψ, but its image is nevertheless in Λ Λ Q.) Proof. First we note that ϕ is well-defined, since h λ is a basis for Λ by Lemma 0: the h λ span Λ by Lemma 9, and they are linearly independent since {h λ } and {s λ } have the same number of terms with any given degree. Next, we note that H λ H λj = H λ, as follows from transitivity of induced representations and the definitions of H k and. Any element of Λ can be written as a polynomial in {h n }, so ϕ is a homomorphism of rings. It trivially respects the Z-grading. It follows from the definition of ψ that it is an additive homomorphism into Λ Q (although we will soon see that its image is in Λ Λ Q). I will show that ψ ϕ is the identity map. For λ n, we have ψ(ϕ(h λ )) = ψ(h λ ) = µ n z(µ) χ H λ (C(µ))p µ. By parts (I) and (II) of Lemma 2, this sum is indeed h λ. So indeed ψ ϕ is the identity on h λ, and thus on all of Λ, and it follows that ψ is the inverse of ϕ on the image of ϕ, so that ϕ is an isomorphism into its image.

13 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 3 Next we show that ψ is an isometry. For V and W representations of S n, we have ψ(v ), ψ(w) = χ V (C(λ))χ W (C(η)) p λ, p η z(λ)z(η) λ,η n = χ V (C(λ))χ W (C(λ)) z(λ) λ n = C(λ) χ V (C(λ))χ W (C(λ)) n! λ n = V, W. Since the representations of S n linearly span R n, we have, for x, y R n, ψ(x), ψ(y) = x, y. Moreover, ψ respects the Z-grading, so if x R n, y R m, m n, then ψ(x), ψ(y) = 0 = x, y. This proves that ψ is an isometry. In particular, it is injective, so ϕ is surjective, so ϕ is an isomorphism. This completes the proof of (I) and (III). Finally, we prove (II). Fix an n. For λ n, we can write ϕ(s λ ) = µ n m µs µ for some m µ Z. Since = s λ, s λ = ϕ(s λ ), ϕ(s λ ) = m 2 µ, there is an η such that ϕ(s λ ) = ±S η. By equations (5) and (6), we can write h λ = s λ + η<λ K ηλs η. Now suppose inductively that ϕ(s η ) = S ν for all η < λ. Since ϕ is a homomorphism, we have H λ, S λ = ϕ(s λ ) + η<λ K ηλ S η, S λ = ϕ(s λ ), S λ. Now H λ, S λ, since we have a nonzero homomorphism from H λ to S λ, by writing H λ = C[S n ]a λ and S λ = C[S n ]a λ b λ, and using the surjection f : x xb λ. So ϕ(s λ ), S λ. But we showed above that ϕ(s λ ) = ±S η for some η, and it follows that ϕ(s λ ) = S λ. By induction, (II) holds. 2.5 Frobenius character formula A particularly important corollary of this theorem is the Frobenius character formula, which allows us to directly compute the character of S λ : Corollary 2. (Frobenius character formula.) Let χ λ (η) be the character of S λ on the conjugacy class C(η). Then: (I) s η = λ χ z(λ) η(λ)p λ ; (II) p η = λ χ λ(η)s λ ; and (III) χ λ (η) is the coefficient of x λ +m x λ 2+m 2 2 x λm m in the expansion of ( (x)p η (x)), where x = (x,...,x m ) and (x) = det x j i i,j m = i<j (x i x j ). Proof. (I) follows from the fact that ψ(s λ ) = s λ. (II) follows from (I) and Lemma, in an analogous way to part (I) of Lemma 2. Finally, (III) follows from (II) and the definition (2) of Schur polynomials.

14 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 4 3 Representations of sl m A direct application of what we have proven above is the classification of the irreps of the Lie algebra sl m, defined to be the C-algebra of traceless complex m m matrices, with the Lie bracket being matrix commutation. We will be using some elementary facts and terminology for simple Lie algebras and their representations for definitions and proofs, see [], especially chapter 4. We will first use the representation theory of S n to characterize the irreps of sl m, and then we will relate the representation ring of sl m to the rings Λ and R we defined in the previous section. 3. Characterization of the irreps of sl m From the definition of sl m, it has an m-dimensional defining representation V. Any tensor power V n of V is also a representation of sl m, by the usual Lie algebra construction: g(v v n ) = (g(v ) v 2 v n ) + + (v v n g(v n )). (0) But S n acts on V n as well, on the right, by permuting tensor components: (v v n )σ = v σ() v σ(n). These two actions clearly commute (justifying our notation), so we can break down V n into irreps with respect to S n, i.e. V λ = V n c λ, and we have that V λ is a representation of sl m. We will prove that V λ = 0 if λ has more than m rows, and otherwise V λ is an irrep; and moreover all irreps of sl m are isomorphic to V λ for some λ. Take the basis {e,...,e m } of V such that the representation V maps matrices in sl m to themselves. A numbering T of the λ with integers in [, m] corresponds to an element of the space V λ as follows: if a,..., a n are the numbers in the boxes, from left to right, top to bottom, T gets mapped to e T := (e a e an )c λ. {e T } is certainly a spanning set for V λ, but not linearly independent for n, m >. However, it turns out that {e T T is an sst with entries in [, m]} is a basis. The first step in proving this is: Lemma 3. {e T T is an sst with entries in [, m]} spans V λ. Proof. We want show that if T is not an sst, we can write e T as a linear combination of e T where T are ssts. We have two relations between the e T :

15 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 5 (I)e T is alternating under permutations within each column; (II)Fixing a j and k, we have e T = e T, with the sum over all numberings which result from switching the top k entries of the (j + )st row with some set of k entries in the jth row, while fixing the vertical order of the entries in each set. (I) follows directly from Lemma 2, but the proof to (II) is more involved. For a nonempty subset Y of the (j + )st column of T, we define γ Y (T) = e(s, T)e S, with the sum over all S that result from exchanging a subset of Y with a subset of the jth column, preserving the vertical order of each subset, and with e(s, T) being ( ) to the power of the number of entries in each subset. I claim γ Y (T) = 0. We let X be the set of entries exchanged with Y, let H be the subset of S n preserving the elements outside X Y, and let K be the subset of H taking X to itself (and Y to itself). Note that if σ, σ H differ by an element of K, then by (I), sign(σ)e σt = sign(σ 2 )e σ2 T. Thus γ Y (T) = so it suffices to show that 0 = h H σ H/K sign(h)e ht = sign(σ)e σt = q B T sign(q) ( X! Y! p A T h H sign(h)e ht, h H ) sign(h)(e t e tn )hp q. We fix a q, and pick an t H that transposes two elements in the same row of qt. Then, since (ht)p = hp and sign(ht) = sign(h), we can pair up terms to prove that the sum is in fact 0. To finish the proof of (II), we note that Y ( ) Y γ Y (T) = e T inclusion-exclusion principle, so indeed e T = e T. e T, by the Now we define an ordering on tableaux of shape λ with entries in [, m], by S > T if S has some entry bigger than T, but S and T agree in all entries directly below and all entries (non-directly) to the right. Take an arbitrary numbering T, and use (I) to assume WLOG that it is strictly increasing along columns. Suppose that T is not an sst. Then take the rightmost column that contains an entry greater than the one to its left, and the lowest such entry in that column say this is the kth from the top in the (j +)st column. We perform the exchange (II) with the top k entries of the jth column, and we get a sum of e S, with each S > T in the above ordering. We then repeat this process with each non-sst in our expression for T, and so on. Since there are only finitely many possible numberings, the process must terminate, with e T expressed as a linear combination of e S for ssts S. Next we can prove Lemma 4. {e T T is an sst with entries in [, m]} is a basis for V λ.

16 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 6 Proof. We count dimensions. We have C[S n ] = λ n (S λ ) dim S λ, so V n = λ n (V λ ) dim S λ. Taking the dimension of each side and using Lemma 3 and part (II) of Corollary 2, m n = λ n(dim S λ )(dim V λ ) λ n(χ λ ( n ))(# of ssts of shape λ with entries in [, m]) = λ n(χ λ (( n ))(s λ ( m )) () = p ( n )( m ) = m n (where the notation ( k ) denotes the k-tuple (,...,)). Hence the inequality is an equality, so the number of ssts of shape λ with entries in [, m] is equal to dim V λ. Combining this with Lemma 3, we are done. If λ has more than m rows, there are no ssts of shape λ with entries in [, m], so V λ = 0. But if V λ has at most m rows, V λ is nonzero and irreducible, and we prove this next. We choose a basis for sl m as follows: Let M ij be the m m matrix with at (i, j), 0 elsewhere. Now sl m is the linear span of: () H i = M i,i M i+,i+ for i m ; (2) E ij = M ij for i < j m; (3) F ij = M ij for j < i m. It is easy to show that {H i } spans a Cartan subalgebra and that we can say each E ij is a positive root vector and each F ij is a negative root vector. For any representation V, we say v V is a weight vector iff it is an eigenvector of each H i ; since the H i are already diagonal, the weight vectors are exactly the scalar multiples of e T for ssts T. We say a nonzero weight vector v is a highest weight vector iff E ij (v) = 0 for all i < j m. A basic fact, which we will not prove, is that a representation is irreducible iff it has a unique highest weight vector. Lemma 5. Suppose a partition λ has at most m rows. Then e T is a highest weight vector of V λ iff T is the sst of shape λ such that each entry in row i is i. Proof. Let T be the specified sst. First we write a formula for the action of sl m on a weight vector e T. Let T t...t n be the numbering of λ with entries t,...,t n, in left-to-right, top-to-bottom order, and let g ij be the (i, j)th entry of g sl m. I claim ge Tt...tn = n i= m g j,ti e. (2) Tt...t i jt i+...tn j=

17 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 7 Since the actions of sl m and S n commute, it suffices to prove that g(e t e tn ) = n m g j,ti (e t e ti e j e ti+ e tn ), i= j= which follows directly from equation (0). Now it follows easily that E pq e T = 0: If there were a nonzero term in (2), it would have j = p, t i = q, p < q. Hence the modified tableau e T = e Tt...t i jt i+ will have a...tn p entry in the qth row. Looking directly above that entry in the pth row, there will be another p entry, so by antisymmetry in the columns, e T = 0. Hence we have proven that E pq e T = 0. Next, suppose T T is an sst with entries in [, m], and write T = T t...t n. Let i be the lowest number such that the ith box is in the jth row, but j t i. By definition of an sst, j < t i, so we can let g = E j,ti. Then e Tt...t i jt i+ is an sst with entries in [, m],...tn so looking at (2), we see ge T 0. We have thus proven the lemma. It follows that each V λ is an irrep of sl m. We now prove that each irrep of sl m is of the form V λ. The proof is based purely on Lie algebra theory, so we will be brief. Lemma 6. Each representation of sl m is isomorphic to V λ for some partition λ. Proof. One computes the simple roots (0,...,0,,, 0,..., 0), and the fundamental weights (,...,, 0,..., 0). The pth fundamental weight w p is the sum of the p highest weights of V, and thus is the highest weight of p V = V ( p ). Thus, for any a,...,a m 0, we know that ( V ) a ( m V ) am V (a + +a m) has highest weight a w + + a m w m. Hence every irrep W is in a tensor power of V. We decomposed this tensor power into spaces of the form V λ, so W = V λ for some λ. The highest weight of V λ is the weight of e T (as defined in Lemma 5), which is (λ λ 2,...,λ m λ m ), so V λ = Vλ iff λ i λ i is constant for i m. Combining this with the above Lemmas, we have proven: Theorem 3. If λ has more than m rows, V λ = 0, and otherwise V λ is an irrep of sl m. All irreps of sl m are isomorphic to some V λ. If λ and λ have at most m rows, then V λ = Vλ iff λ λ = λ 2 λ 2 = = λ m λ m. 3.2 The representation ring of sl m We define the representation ring R m of sl m to be the free abelian group on the irreps of sl m (more precisely, on the isomorphism classes of irreps of sl m ). We embed all

18 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 8 representations of sl m into R m by writing U W = U +W, and then define multiplication so that U W = U W. We define the character of a representation W of sl m to be a function Char W (x,...,x m ) on the nonzero complex variables x,...,x m, defined as follows: Let γ : W U be the exponentiation map, so that U is a representation of SL m C. Then define Char W (x,...,x m ), for x x m =, to be the trace of the element D(x,...,x m ) SL m C on U, where D(x,...,x m ) = x... The map γ takes the Cartan subalgebra to the set of diagonal matrices, and thus takes weight vectors to eigenvectors of D(x,...,x m ). Moreover, we can compute x m. D(x,..., x m )γ(e T ) = γ(d(ln x,..., ln x m )(e T )) = γ(( lnx ti )e T ) i = γ((ln x,...,lnx m ) T e T ) = (x,...,x m ) T γ(e T ), and it follows from Lemma 4 that Char Vλ (x,...,x m ) = s λ (x,...,x m ). Note that the restriction x x m = is important for example, if λ i + = λ i for i m, then V λ = Vλ but Char Vλ = s λ (x,...,x m ) = x x m s λ (x,...,x m ) = x x m Char Vλ. It is easy to verify that Char U W = Char U + Char W and Char U W = Char U Char W. Char is thus a homomorphism from R m to Λ(m)/(x x m ), where Λ(m) Λ is the ring of symmetric polynomials in m variables. We also have an surjective additive homomorphism α : R R m, taking λ to V λ (which might be 0). Putting together Char, α, and the map ψ defined in Theorem 2, we have a composite map Λ R R m Λ(m). This map takes the symmetric function s λ Λ to the symmetric polynomial s λ (x,...,x m ) Λ(m)/(x x m ), and since the s λ are a basis for Λ, it takes any symmetric function f(x, x 2,...) to f(x,...,x m, 0, 0,...). By Theorem 3, if we let I be the ideal generated by s λ with λ having more than m rows, and s λ s λ, where λ and λ differ by one or more columns of m squares, then V λ = Λ/I. Put another way, Theorem 4. Let I be the ideal of Λ generated by the symmetric functions e m+ and (e m ). Then the Z/m-graded ring Λ/I is isomorphic to R m.

19 Math 250 Final Paper Jan. 4, 2005 Steven Byrnes 9 Proof. Note that both e m+ = 0 and e m = respect the Z/m-grading. The ideal generated by e m+ consists of all symmetric functions f with f(x,..., x m ) = 0, so Λ/e m+ = Λ(m). And in Λ(m), (em ) = (x x m ). So this theorem follows from the above discussion. Thus we can, for example, decompose tensor products of representations of sl m by doing the analogous operation on Schur polynomials. The Z/m-grading on R m is sometimes called m-ality, or triality in the case m = 3. As a closing note, the Weyl group of sl m is S m, and S m is a symmetry of R m = Λ(m)/I. This property holds in general. References [] W. Fulton, J. Harris, Representation Theory, New York: Springer-Verlag, 99. [I used the material in Chapters 4, 6, and Appendix A throughout the paper. Section was based especially closely on Chapter 4.] [2] W. Fulton, Young Tableaux, New York: Cambridge University Press, 997. [I used material throughout this book, in particular Chapters 2, 6, 7, and 8. Section 2 in particular was based mostly on this book.] [3] D. Knutson, λ-rings and the Representation Theory of the Symmetric Group, New York: Springer-Verlag, 973. [I did not use this explicitly, but read it through for reference.]