Representation Theory

Andrew Kobin Fall 2014 Contents Contents

Contents

1 Introduction 1 1.1 Group Theory Review ...... 1 1.2 Partitions ...... 2

2 Group Representations 4 2.1 Representations ...... 4 2.2 G-homomorphisms ...... 9 2.3 Schur’s Lemma ...... 12 2.4 The Commutant and Endomorphism Algebras ...... 13 2.5 Characters ...... 17 2.6 Tensor Products ...... 24 2.7 Restricted and Induced Characters ...... 25

3 Representations of Sn 28 3.1 Young Subgroups of Sn ...... 28 3.2 Specht Modules ...... 33 3.3 The Decomposition of M µ ...... 44 3.4 The Hook Length Formula ...... 48 3.5 Application: The RSK Algorithm ...... 49

4 Symmetric Functions 52 4.1 Generating Functions ...... 52 4.2 Symmetric Functions ...... 53 4.3 Schur Functions ...... 57 4.4 Symmetric Functions and Character Representations ...... 60

i 1 Introduction

1 Introduction

The following are notes from a course in representation theory taught by Dr. Frank Moore at Wake Forest University in the fall of 2014. The main topics covered are: group representations, characters, the representation theory of Sn, Young tableaux and tabloids, Specht modules, the RSK algorithm and some further applications to combinatorics.

1.1 Group Theory Review Deﬁnition. A group is a nonempty set G with a binary operation “·00 : G×G → G satisfying (1) (Associativity) a(bc) = (ab)c for all a, b, c ∈ G.

(2) (Identity) There exists an identity element e ∈ G such that for every a ∈ G, ae = ea = a.

(3) (Inverses) For every a ∈ G there is some b ∈ G such that ab = ba = e. Examples.

1 The groups of real- or complex-valued n × n matrices, denoted GLn(R) and GLn(C), respectively, are groups under matrix multiplication. The identity in each case is the 1 0 ··· 0 0 1 ··· 0   identity matrix I = . . .. .. . . . . 0 0 ··· 1

2 For a vector space V over R (resp. C), the general linear group of V , denoted GL(V ), is defined to be the group of invertible linear transformations V → V . If V ∼ is finite dimensional over R (resp. C) then GL(V ) = GLn(R) (resp. GLn(C)) where n = dim V . The general linear group is a group under function composition, with the identity function id : V → V defined by v 7→ v and inverses defined by invertibility of the elements of GL(V ).

3 The most important example in these notes is the symmetric group on n symbols, Sn. Explicitly, Sn is deﬁned to be the group of bijective functions π : {1, . . . , n} → {1, . . . , n}. The group law here is function composition and the identity and inverses are deﬁned in the same manner as in 2 . There are three types of notation for an element π ∈ Sn which will be used interchangeably: 1 2 3 ··· n Two-line notation: π(1) π(2) π(3) ··· π(n) One-line notation: π(1) π(2) ··· π(n) – this is just the bottom row of the two-line notation Cycle notation, e.g. (1 3)(2 4 5)(6 7)

1 1.2 Partitions 1 Introduction

Deﬁnition. The cycle type of a permutation π ∈ Sn is the number and length of disjoint cycles into which π decomposes. The following theorem from group theory guarantees that the cycle type is well-deﬁned for any permutation.

Theorem 1.1.1. For all π ∈ Sn, π can be written as the product of disjoint cycles. Example 1.1.2. Consider the permutation π = (2 4 3). In one-line notation, this can be written π = 1 4 2 3. The cycle type of π is 31, whereas the cycle type of σ = (1 4 6 3)(2 5)(7 8) 1 2 in S8 is 4 , 2 .

1.2 Partitions

Deﬁnition. A partition of a natural number n is a sequence λ = (λ1, . . . , λm) such that

(1) The sequence of λi is weakly decreasing, that is, λ1 ≥ λ2 ≥ · · · ≥ λm. Pm (2) i=1 λi = n. A partition will often be denoted λ ` n. Deﬁnition. Let G be any group. Two elements g, h ∈ G are said to be conjugate in G if there is a k ∈ G such that g = khk−1, or equivalently if gk = kh. The conjugacy class of −1 an element g is Kg = {h ∈ G | g = khk for some k ∈ G}. Conjugacy is an equivalence relation. In general, the equivalence classes of an equivalence relation on a set S form a partition of S, as is the case with the symmetric group: Proposition 1.2.1. The conjugacy classes of G form a partition of G, that is, they are disjoint and contain every element of G.

Example 1.2.2. Let G = GLn(C), the group of n × n matrices with complex entries. Suppose M and N are conjugate, n × n matrices, i.e. there exists a matrix K ∈ G such that M = KNK−1. This corresponds to a commutative diagram

n K n CK Cstd

N M

n n CK Cstd K

(The subscripts on the vector spaces denote diﬀerent bases, as K may be thought of as a change of basis matrix.) This diagram in fact shows that M and N are also similar matrices. In general, conjugacy and similarity are equivalent in GLn(C). (For more, see the section on Jordan canonical form in any linear algebra text.)

2 1.2 Partitions 1 Introduction

Let π ∈ Sn and suppose π = (i1 i2 ··· i`) ··· (im im+1 ··· in) is the cycle decomposition −1 −1 of π. Let σ be the permutation sending ij 7→ j, so that σ sends j 7→ ij. Then σπσ = (1 2 3 ··· `) ··· (m m + 1 ··· n) has the same cycle type as π. This is the main ingredient in the proof of the following theorem.

Theorem 1.2.3. In Sn, all permutations of the same cycle type are conjugate. Proposition 1.2.4. There is a one-to-one correspondence between the set of conjugacy classes of Sn and partitions λ of n. Definition. The generating function for a partition is a formal series given by ∞ ∞ ∞ ∞ ! X Y 1 Y X p tn = = tni . n 1 − ti n=0 i=1 i=1 n=i Definition. We say a group G acts on a set X provided there is some mapping “·00 : G × X → X such that (1) For all g, h ∈ G and x ∈ X, g · (h · x) = gh · x. (2) For the identity element e ∈ G, e · x = x for every x ∈ X. The group action axioms easily imply that g−1 · (g · x) = x = g · (g−1 · x). Definition. Given a group action G X, the orbit and stabilizer of an element x ∈ X are defined by

OrbG(x) = Ox = {g · x | g ∈ G}

and StabG(x) = Gx = {g ∈ G | g · x = x}.

Remark. The stabilizer Gx is a subgroup of G for every x ∈ X. An easy consequence of this fact and Lagrange’s theorem is the following result. Theorem 1.2.5 (Orbit-Stabilizer Theorem). Suppose G acts on X. Then there is a bijection between orbits and quotient groups of G by the stabilizers given by

Ox ←→ G/Gx

g · x 7−→ gGx.

In particular, if G has ﬁnite order then for any x ∈ X, |G| = |Ox| |Gx|. Examples. 1 Any group G acts on itself via conjugation: g · h = ghg−1. The orbits of this action are precisely the conjugacy classes of G: Oh = Kh. On the other hand, the stabilizers, −1 called centralizers, are of the form Zh = {g ∈ G | ghg = h}. The Orbit-Stabilizer Theorem then gives the important relationship |G| = |Kh| |Zh| for any h ∈ G.

m1 m2 mn 2 Let λ = (1 , 2 , . . . , n ) be a partition of n and take g ∈ Sn with cycle type λ. De- m1 m2 mn note the order of the centralizer by zλ = |Zg|, so that zλ = 1 (m1!)2 (m2!) ··· n (mn!). The Orbit-Stabilizer Theorem then gives us n! |Kλ| = . zλ

3 2 Group Representations

2 Group Representations

2.1 Representations

We begin with several additional concepts which are fundamental to group theory.

Deﬁnition. A group homomorphism is a map ϕ : G → H between two groups G and H such that

(1) ϕ(gg0) = ϕ(g)ϕ(g0) for all g, g0 ∈ G.

(2) ϕ(eG) = eH . Remark. It is an easy consequence of the two homomorphism axioms that inverses are preserved, so a homomorphism may be thought of as a map that transfers group structure.

There are two important subgroups induced by a homomorphism.

Definition. The kernel of a homomorphism ϕ : G → H is the subgroup of G defined by ker ϕ = {g ∈ G | ϕ(g) = eH }. On the other hand, the image of ϕ is the subgroup of H defined by im ϕ = {h ∈ H | ϕ(g) = h for some g ∈ G}.

We now introduce the central concept in representation theory.

Deﬁnition. For a group G, a matrix representation of G is a group homomorphism X : G → GLd(C), for some integer d, called the degree of the representation.

Deﬁnition. We say a representation X is faithful if ker X = {eG}. Examples. There are several important examples of representations which we will repeat- edly use as examples of new concepts.

1 For an arbitrary group G, the trivial representation is a one-dimensional representation ∗ X : G → GL1(C) = C which sends each g ∈ G to 1 ∈ C.

2 The sign representation of Sn is another one-dimensional representation deﬁned by

∗ X : Sn −→ C ( 1 π is even π 7−→ −1 π is odd.

3 The deﬁning representation of Sn is

X : Sn −→ GLn(C) ( 1 π(j) = i π 7−→ (xij) = 0 otherwise.

4 2.1 Representations 2 Group Representations

When n = 3, the deﬁning representation of S3 looks like

1 0 0 0 1 0 0 0 1 X((1)) = 0 1 0 X((1 2)) = 1 0 0 X((1 3)) = 0 1 0 etc. 0 0 1 0 0 1 1 0 0

These can realized as permutations of the rows of the matrix X((1)) = I3 – or indeed in general, In.

4 We will determine all one-dimensional representations of Cn, the cyclic group of order ∗ n. Let X : Cn → C be a representation. Since Cn is generated by an element g, X is completely determined by X(g). Denote this image by c = X(g). Since X is a homomorphism and gn = e, we see that cn = X(g)n = X(gn) = X(e) = 1, so c is an nth root of unity.

There are several main goals of representation theory:

Understand all ﬁnite degree representations of a group G. We will do this for Sn in Chapter 3.

Understand how representations are “built” out of smaller ones.

Deﬁne what makes a representation irreducible.

Classify irreducible representations of G. We will also do this for Sn in Chapter 3. Deﬁnition. Let G be a group. A vector space V is a G-module if there is a group homomorphism ρ : G → GL(V ). Equivalently, V is a G-module if there is a G-action on V satisfying

(1) gv ∈ V for all g ∈ G and v ∈ V .

(2) g(cv + dw) = c(gv) + d(gw) for any g ∈ G, v, w ∈ V and c, d ∈ C. (3) (gh)v = g(hv) for all g, h ∈ G and v ∈ V .

(4) eGv = v for all v ∈ V .

If G acts on a (ﬁnite) set X = {x1, . . . , xn}, one can easily deﬁne an associated G-module by taking V = CX, the vector space whose basis is X. The action G V is given by

g(c1x1 + ... + cnxn) = c1(gx1) + ... + cn(gxn).

Notice the sets {x1,..., xn} and {gx1,..., gxn} are equal for any g, giving rise to the term ‘permutation representation’:

Deﬁnition. When a group G acts on a set X, the module CX is called the permutation representation of G associated to X.

5 2.1 Representations 2 Group Representations

Examples.

5 Sn acts naturally on the set S = {1, . . . , n}. Here the permutation representation associated to S is CS = {c11 + c22 + ... + cnn |i∈ C}. The Sn-action is given by π : Sn × S → S, π(c11 + ... + cnn) = c1π(1) + ... + cnπ(n). Since CS is an Sn-module, for each choice of basis of GL(CS) the homomorphism Sn → GL(CS) determines a matrix representation X : Sn → GLn(C). If n = 3 for example, then the standard basis {1, 2, 3} induces the defining representation X : S3 → GL3(C) – one may check that all of the matrices defined this way coincide with the defining representation described in 3 :

0 1 0 (1 2) 7−→ 1 0 0 0 0 1

However, diﬀerent bases give rise to diﬀerent (matrix) representations. For instance, given the basis {1 + 2 + 3, 1 + 2, 1} we have

1 0 0  (1 2) 7−→ 0 1 1  0 0 −1

6 Any group G acts on itself via left multiplication: g · h = gh. The corresponding G-module C[G] = {c1g1 + ... + cngn | ci ∈ C, gi ∈ G} is called the group algebra of G or alternatively the left regular representation of G (depending on whether C[G] is thought of as a module or a representation). The multiplicative structure of this algebra is deﬁned as g1 · g2 = g1g2 for group elements and extended by linearity in C[G].

In G = S3, this group algebra structure may be seen through the following computation:

(1(1) + 2(1 2) + 3(1 2 3))(2(1 3) − 2(1 3 2)) = 1 · 2(1)(1 3) + 2 · 2(1 2)(1 3) + 3 · 2(1 2 3)(1 3) − 2 · 1(1)(1 3 2) − 2 · 2(1 2)(1 3 2) − 3 · 2(1 2 3)(1 3 2) = 2(1 3) + 4(1 3 2) + 6(2 3) − 2(1 3 2) − 4(1 3) − 6(1) = −6(1) − 2(1 3) + 6(2 3) + 2(1 3 2).

Deﬁnition. Let V be a G-module. A subspace W ⊂ V is a G-submodule of V if g ·w ∈ W for every g ∈ G and w ∈ W . Such a subspace is often called a G-invariant subspace of V .

Examples.

7 Every G-module V has the trivial submodules {0} and V .

6 2.1 Representations 2 Group Representations

8 Let V = C{1,..., n}, G = Sn and consider the subspace W = C{1 + ... + n}. We will see that W is already known to us. Take π ∈ Sn. Then by commutativity, we can write π(1 + ... + n) = π(1) + ... + π(n) = 1 + ... + n. W is therefore the trivial representation: ∼ ∗ X : Sn −→ GL(W ) = C ∗ π 7−→ id ↔ 1 ∈ C .

So it turns out that the trivial representation of Sn may be realized as a nontrivial submodule of the deﬁning representation for Sn.

9 We can do the same with the sign representation. Let V = C[Sn] and take the submodule W = P sgn(σ)σ . In S ,(1) − (1 2) − (1 3) − (2 3) + (1 2 3) + (1 3 2) C σ∈Sn 3 generates W . Another element of the Sn-module C[W] is ! X X (1 2) sgn(σ)σ = sgn(σ)(1 2)σ

σ∈S3 σ∈S3 = (1 2) − (1) − (1 3 2) − (1 2 3) + (2 3) + (1 3) X = − sgn(σ)σ.

σ∈S3 And in general it is easy to see that π P sgn(σ)σ = sgn(π) P sgn(σ)σ, σ∈S3 σ∈S3 so W is indeed the sign representation of S3. Notice that in these two examples, our two favourite one-dimensional representations of Sn correspond to one-dimensional submodules. Deﬁnition. A G-module V is reducible if it contains a nontrivial submodule, 0 ( W ( V . Otherwise V is said to be irreducible. Equivalently, V is reducible if there is a basis B of V such that for every g ∈ G, X(g) has the form A(g) B(g) X(g) = 0 C(g) where A(g) and C(g) are square matrices. It would be even nicer to be able to write X(g) down as above with B(g) = 0. Such a block diagonal matrix representation coincides with the idea of direct product, which we will deﬁne in a moment. Examples.

10 In the deﬁning representation for S3 and the basis B = {1 + 2 + 3, 2, 3}, we have

 1 1 0  X((1 2)) =  0 -1 0  0 -1 1

The obstacle in making X((1 2)) block diagonal is the fact that W = C{2, 3} is not a G-module, since for example (1 2)2 = 1 6∈ W .

7 2.1 Representations 2 Group Representations

We deﬁne an inner product h·, ·i on V by hi, ji = δij, the Kronecker symbol for i and j, and make the inner product linear in the ﬁrst slot and conjugate-linear in the second slot. In particular, for v = a1+b2+c3 and w = x1+y2+z3, hv, wi = ax+by+cz. Also notice that 0 0 0 h·, ·i is S3-invariant: for all v, v ∈ S3 we have hπv, πv i = hv, v i. Now given a submodule W ⊂ V , we may form its orthogonal complement W ⊥ = {v ∈ V | hv, wi = 0 for all w ∈ W }. Proposition 2.1.1. If W is a submodule of V then W ⊥ is also submodule of V . Proof. First, W ⊥ is always a subspace of V . Let u ∈ W ⊥; we must show that πu ∈ W ⊥ for all π ∈ S3. Well for all w ∈ W , we have

−1 −1 hπu, wi = hπ πu, π wi by S3-invariance = hu, π−1wi by the ring action = 0 since u ∈ W ⊥ and π−1w ∈ W.

So this shows that πu ∈ W ⊥ too, proving W ⊥ is a submodule of V .

Let’s apply this new submodule construction to example 10 . Let W = C{1 + 2 + 3} and consider W ⊥. It’s easy to compute that the kernel of the matrix 1 1 1 is spanned −1 −1 by  0  and  1 , which correspond to 2 − 1 and 3 − 1. So {2 − 1, 3 − 1} is a basis 1 0 for W ⊥ and under this basis we have the matrix representation

1 0 0 1 0 0  X((1)) = 0 1 0 X((1 2)) = 0 −1 −1 0 0 1 0 0 1

1 0 0  1 0 0 X((1 3)) = 0 1 0  X((2 3)) = 0 0 1 0 −1 −1 0 1 0

1 0 0  1 0 0  X((1 2 3)) = 0 −1 −1 X((1 3 2)) = 0 0 1  0 1 0 0 −1 −1

We have thus managed to find a submodule of V for which each X(g) has a block diagonal structure as a matrix. Maschke’s Theorem says that we can do this for every G-module when G is a finite group. Theorem 2.1.2 (Maschke). Let G be a finite group and let V be a nonzero G-module. Then V may be written as a direct sum, V = W (1) ⊕ W (2) ⊕ · · · ⊕ W (k), where each W (i) is an irreducible submodule of V .

Proof. For simplicity, we will assume the ground ﬁeld is C but the theorem holds whenever |G| is a unit of the ground ﬁeld. Suppose V has a nontrivial submodule – otherwise the proof is trivial. Call this submodule W . Pick a basis B for V that contains a basis for W ;

8 2.2 G-homomorphisms 2 Group Representations

say B = {w1, . . . , w`, v`+1, . . . , vd} where {w1, . . . , w`} is the basis for W and d = dim V . Let h·, ·i denote the standard linear/conjugate-linear inner product on this basis – note that 0 P 0 h·, ·i may not necessarily be G-invariant. Deﬁne hhv, v ii = g∈Ghgv, gv i which is an inner product on V (this is easy to check). For any g ∈ G, we see that X X hhgv, gv0ii = hhgv, hgv0i = hgv, gv0i = hhv, v0ii h∈G g∈G

since G acts transitively on itself by left multiplication. Therefore hh·, ·ii is G-invariant. The process described in the above example allows us to write V = W ⊕ W ⊥, where W ⊥ is the perpendicular subspace of W with respect to hh·, ·ii. Since W is a nontrivial submodule of V , dim W < dim V and dim W ⊥ < dim V so induction proves the desired decomposition in Maschke’s Theorem.

Corollary 2.1.3 (Matrix Version). Let G be a ﬁnite group and let X be some matrix representation of G. Then there is a ﬁxed matrix T such that for all g ∈ G,

 (1)  X (g) 0  X(2)(g)  −1   TX(g)T =  ..   0 .  X(k)(g)

where each X(i) is an irreducible matrix representation of G.

2.2 G-homomorphisms

Deﬁnition. Let V and W be G-modules. A G-homomorphism is a linear transformation θ : V → W such that θ(g · v) = g · θ(v) for all v ∈ V .

In matrix language, pick bases B and C of V and W , respectively, and let T = [θ]B→C be the matrix form of θ in the bases B and C. Let X and Y be the matrix representations corresponding to (V, B) and (W, C), respectively, i.e. X : G → GLn(C) and Y : G → GLm(C). Then for all g ∈ G, TX(g) = Y (g)T and it suﬃces to check this property to show that θ is a G-homomorphism.

Example 2.2.1. Let G = Sn and consider the G-modules V = C{v} (the trivial representation, via π · v = v) and W = C{1, 2,..., n} (the deﬁning representation via π · i = π(i)). Deﬁne a map

θ : V −→ W v 7−→ 1 + 2 + ... + n.

Take any permutation π ∈ Sn. Then

θ(π · v) = θ(v) = 1 + 2 + ... + n = π · (1 + 2 + ... + n) = π · θ(v).

9 2.2 G-homomorphisms 2 Group Representations

So θ is a G-homomorphism. An important case of this is the homomorphism

θ : V −→ C[G] X v 7−→ g. g∈G

Example 2.2.2. For another example, let G = Sn but this time consider the G-modules C{v} (the sign representation via π · v = sgn(π)v) and W = C[Sn] (the group ring via π · σ = πσ). We claim that

θ : V −→ W X v 7−→ sgn(σ)σ

σ∈Sn is an Sn-homomorphism. To see this, take σ ∈ Sn and v ∈ V . Then X θ(σ · v) = θ(sgn(σ)v) = sgn(σ) sgn(π)π

π∈Sn X = sgn(σ−1) sgn(π)π since sgn(σ−1) = sgn(σ)

π∈Sn X = sgn(σ−1π)π since sgn is a homomorphism.

π∈Sn

Note that if the coeﬃcient of π is sgn(σ−1π) then the coeﬃcient of σπ must be sgn(σ−1(σπ)) = sgn(π). So we have X X θ(σ · v) = sgn(π)σπ = σ sgn(π)π = σ · θ(v).

π∈Sn π∈Sn

Thus θ is an Sn-homomorphism. Deﬁnition. We say two G-modules V and W are G-equivalent, or G-isomorphic, if there exists a bijective G-homomorphism between them.

In the matrix version of the deﬁnition, there exists an invertible matrix T such that we can write Y (g) = TX(g)T −1 for all g ∈ G. In other words, X(g) and Y (g) are similar and there is a single change-of-basis matrix T taking one set to the other. For the rest of the section, we will use the extended example of G = S3 with the subgroup H = h(2 3)i = S{1} × S{2,3}. The cosets look like G/H = {H, (1 2)H, (1 3)H} and G acts on this quotient group via π · (σH) = πσH. A useful object for thinking about these subgroups is called the Young tabloid.

Deﬁnition. Let λ = (λ1, . . . , λ`) be a partition of n. Then a Young tabloid of shape λ is an array with ` rows such that row i contains λi integers between 1 and n. The order of the integers doesn’t matter and there may be repeats of any of the integers.

10 2.2 G-homomorphisms 2 Group Representations

In our example, λ = (2, 1) is a partition of n = 3 and a Young tabloid of this shape is

2 3 3 2 1 3 which is equal to but not 1 1 2

There is a natural action of Sn on the set of tabloids of shape λ that permutes the entries according to the permutation of the set {1, . . . , n}. For example, in S9,

3 1 4 1 3 2 1 2 (1 2 3 4) · 5 9 = 5 9 2 3

Returning to λ = (2, 1) and S3, let S be the set of all tabloids of shape λ. Then CS is an S3-module.

Proposition 2.2.3. The following S3-modules are S3-isomorphic: CS, C{1, 2, 3} viewed as the deﬁning representation and CG/H, the coset representation.

Proof. Deﬁne a map θ : CG/H → CS by

2 3 1 3 1 2 H 7−→ (1 2)H 7−→ and (1 3)H 7−→ 1 2 3

We need to check θ(π · σH) = π · θ(σH) for each π ∈ S3 and each coset σH. For π = (1 3) and σ = (1 2),

1 3 1 3 θ((1 3) · (1 2)H) = θ((1 2 3)H) = θ((1 2)H) = = (1 3) · 2 2

so the property holds. The rest are easy to verify as well. Thus θ is an isomorphism on bases of CG/H and CS, which is suﬃcient to prove that these are S3-isomorphic. Now deﬁne η : C{1, 2, 3} → CS by

2 3 1 3 1 2 1 7−→ 2 7−→ and 3 7−→ 1 2 3

It is easy to check that η is another S3-isomorphism, so we have shown that CS, C{1, 2, 3} and CG/H are all isomorphic as S3-modules. In the future, given a partition λ ` n and the set S of all tabloids of shape λ, we will write M λ = CS. We will prove (1,...,1) ∼ Theorem 2.2.4. M = C[Sn] and this module contains all irreducible representations of Sn.

11 2.3 Schur’s Lemma 2 Group Representations

2.3 Schur’s Lemma

We begin with a simple fact about G-homomorphisms.

Lemma 2.3.1. If θ : V → W is a G-homomorphism then ker θ and im θ are G-submodules of V and W , respectively.

Proof. Let v ∈ ker θ. Then for any g ∈ G, θ(g · v) = g · θ(v) = g · 0 = 0. So g · v ∈ ker θ, proving ker θ is a G-submodule of V . This immediately gives us the simple but important Schur’s Lemma.

Lemma 2.3.2 (Schur’s Lemma). Let V and W be irreducible G-modules and let θ : V → W be a G-homomorphism. Then either θ is the zero map or θ is invertible.

Lemma 2.3.3 (Schur’s Lemma, Matrix Version). Let X and Y be irreducible matrix representations of G and suppose T is a matrix such that TX(g) = Y (g)T for all g ∈ G. Then either T = 0 or T is invertible.

Deﬁnition. For any two G-modules V and W , we deﬁne the vector space

HomG = {θ : V → W | θ is a G-homomorphism}.

Note that Hom is always a vector space but not necessarily a G-module.

Corollary 2.3.4. Let V and W be G-modules and suppose V is irreducible. Then HomG(V,W ) = 0 if and only if W does not contain any submodules isomorphic to V .

L (i) (i) Proof. By Maschke’s Theorem (2.1.2), we may write W = i∈I W , where each W is irreducible. Then ! M (i) M (i) HomG(V,W ) = HomG V, W = HomG(V,W ). i∈I i∈I

(i) Then by Schur’s Lemma (2.3.2), HomG(V,W ) is nonzero exactly when V is isomorphic to one of the W (i). An important consequence of this is that if T is a matrix satisfying TX(g) = X(g)T for all g ∈ G, that is T ∈ HomG(V,V ), then (T − cI)X(g) = X(g)(T − cI) for any scalar c ∈ C, where I is the appropriate identity matrix. In particular, take c to be an eigenvalue of T , so that T − cI is not invertible. By Schur’s Lemma (2.3.2), this means T − cI = 0, which can be written T = cI. This proves the following theorem.

Theorem 2.3.5. If X is an irreducible matrix representation of G over C, then the only matrices T for which TX(g) = X(g)T are scalar multiples of the identity matrix. In particular, dim HomG(V,V ) = 1. Note that this does not necessarily hold when the ground ﬁeld is changed to something other than C.

12 2.4 The Commutant and Endomorphism Algebras 2 Group Representations

2.4 The Commutant and Endomorphism Algebras

Deﬁnition. Given X a matrix representation of G and V a corresponding G-module, we deﬁne the commutant algebra

Com X = {T ∈ Matd | TX(g) = X(g)T for all g ∈ G} and the endomorphism ring

EndG(V ) = HomG(V,V ).

Example 2.4.1. This example will illustrate how the commutant algebra interacts with direct products of representations. Consider a representation

X(1) 0 X = = X(1) ⊕ X(2), 0 X(2) where X(1) and X(2) are non-isomorphic, irreducible matrix representations of a group G. By (i) Theorem 2.3.5, we know that Com X = {cIdi | c ∈ C} for each i = 1, 2, where di = deg Xi. Take T T T = 11 12 ∈ Com X. T21 T22 Then for all g ∈ G,

(1) (1) T11 T12 X (g) 0 X (g) 0 T11 T12 (2) = (2) T21 T22 0 X (g) 0 X (g) T21 T22 (1) (2) (1) (1) T11X (g) T12X (g) X (g)T11 X (g)T12 =⇒ (1) (2) = (2) (2) . T21X (g) T22X (g) X (g)T21 X (g)T22

(1) (2) From this we see that T11 ∈ Com X and T22 ∈ Com X . Moreover, T12 corresponds (2) (1) (1) with some element of HomG(X ,X ) so by Schur’s Lemma (2.3.2), T12 = 0 since X is irreducible. By similar logic, T21 = 0. Therefore we have T 0 T = 11 0 T22

(i) where Tii ∈ Com X for each i = 1, 2. Finally, this means that the commutant algebra for X looks like c1Id1 0 Com X = : c1, c2 ∈ C . 0 c2Id2 If instead we have X = X(1) ⊕ X(1), which we will denote 2X(1), then the above work shows that every element of Com X looks like

c I c I T = 11 d1 12 d1 . c21Id1 c22Id1 This matrix form motivates the introduction of the tensor product.

13 2.4 The Commutant and Endomorphism Algebras 2 Group Representations

Deﬁnition. Given two matrices X = (xij) and Y of any type, their tensor product (sometimes also called the Kronecker product) is deﬁned as   x11Y x12Y ··· x1nY  x Y x Y ··· x Y   21 22 2n  X ⊗ Y =  . . .. .   . . . .  xm1Y xm2Y ··· xmnY

If X is m × n and Y is m0 × n0, then X ⊗ Y is mm0 × nn0. In the language of tensors, the commutant algebra in the last example consists of matrices c11 c12 of the form ⊗ Id1 . This extends to mX = X ⊕ X ⊕ · · · ⊕ X: c21 c22 Proposition 2.4.2. Suppose X is an irreducible matrix representation. Then

Com(mX) = {C ⊗ Id | C ∈ Matm, d = deg X} and dim Com(mX) = m2. Proof. Similar to the previous example. For a general representation X, we can use Maschke’s Theorem (2.1.2) to decompose X:

(1) (2) (k) X = m1X ⊕ m2X ⊕ · · · ⊕ mkX , where each X(i) is irreducible and X(i) 6∼= X(j) for each i 6= j. By Schur’s Lemma (2.3.2) and Proposition 2.4.2, we have

k ( k ) M (i) M (i) Com X = Com(miX ) = Mmi ⊗ Idi : Mmi ∈ Matmi , di = deg X . i=1 i=1 Example 2.4.3. Let X = 2X(1) ⊕ X(2), where X(1) and X(2) are distinct, irreducible representations, deg X(1) = 3 and deg X(2) = 4. Then dim X = 10 and by Proposition 2.4.2, an element of Com X is a 10 × 10 block matrix:  a 0 b 0   a b     0 a 0 b   0   c 0 d 0     c d     0 c 0 d     e   0   0 e   e   0  e Accordingly, dim Com X = 5. In general, the dimension of the commutant algebra is given by the following formula.

14 2.4 The Commutant and Endomorphism Algebras 2 Group Representations

(1) (2) (k) (i) Proposition 2.4.4. Given X = m1X ⊕m2X ⊕· · ·⊕mkX , where the X are distinct, Pk 2 irreducible representations, dim Com X = i=1 mi . Proof. Apply Proposition 2.4.2. Note that Com X is in general not a commutative algebra, despite its name. Thus in many cases Z(Com X) = {T ∈ Com X | TS = ST for all S ∈ Com X} is nontrivial. Our goal is to compute the center in the general case; however let’s begin with the simplest case imaginable, Matd. Let Eii represent the d × d matrix with the iith entry as a 1 and 0’s elsewhere. Then for any C ∈ Z(Matd), the identity CEii = EiiC looks like     c1i 0  c   2i    0 . 0 = ci1 ci2 ··· cim  .    cni 0 This holds for any i, so the oﬀ-diagonal entries are all zero:   c1 0  c   2  C =  ..   0 .  cn

Further, for any i 6= j, the identity C(Eij + Eji) = (Eij + Eji)C, when written out, implies that ci = cj. Therefore all diagonal entries of C are equal, so we see that the center in this case is one-dimensional: Z(Matd) = {cId | c ∈ C}. To prove the general case, we need the following properties of direct sums and tensors.

Lemma 2.4.5. Let A, X ∈ Matd and B,Y ∈ Mate. Then (1) (A ⊕ B)(X ⊕ Y ) = AX ⊕ BY . (2) (A ⊗ B)(X ⊗ Y ) = AX ⊗ BY . Proof omitted. Lk Now let C ∈ Z(Com X), which can be written C = i=1(Cmi ⊗ Idi ), and take T = Lk i=1(Tmi ⊗ Idi ) ∈ Com X. Using Lemma 2.4.5, we compute k ! k ! k M M M CT = (Cmi ⊗ Idi ) (Tmi ⊗ Idi ) = ((Cmi Tmi ) ⊗ Idi ) i=1 i=1 i=1 k M and TC = ((Tmi Cmi ) ⊗ Idi ). i=1 Therefore the center consists of ( k ) ( k ) M M Z(Com X) = ((ciImi ) ⊗ Idi ): ci ∈ C = ciImidi : ci ∈ C . i=1 i=1 We summarize our ﬁndings in the next two theorems.

15 2.4 The Commutant and Endomorphism Algebras 2 Group Representations

(1) (k) Theorem 2.4.6. Let X be a representation and write X = m1X ⊕ · · · ⊕ mkX , where (i) each X is irreducible of degree di. Then

k X (1) deg X = midi. i=1

( k ) M (2) Com X = (Mmi ⊗ Idi ): Mmi ∈ Matmi . i=1

k X 2 (3) dim Com X = mi . i=1

( k ) M (4) Z(Com X) = ciImidi : ci ∈ C . i=1 (5) dim Z(Com X) = k.

(1) Theorem 2.4.7 (G-module Version). Let V be a G-module and write V = m1V ⊕ · · · ⊕ (k) (i) mkV , where each V is irreducible. Then

k X (1) dim V = midi. i=1

k ∼ M (2) HomG(V,V ) = Matmi . i=1

k X 2 (3) dim End V = mi . i=1   c 0   1  ∼  ..  (4) Z(End V ) =  .  : ci ∈ C .    0 ck 

(5) dim Z(End V ) = k.

Another important observation is contained in the next proposition.

Proposition 2.4.8. If V and W are G-modules and V is irreducible, the the dimension of HomG(V,W ) is the number of isomorphic copies of V in an irreducible decomposition of W .

Lk (i) (i) Proof. Suppose W = i=1 W for distinct irreducibles W . Then HomG(V,W ) can be written k ! k M (i) ∼ M (i) HomG(V,W ) = HomG V, W = mi HomG(V,W ). i=1 i=1

16 2.5 Characters 2 Group Representations

(i) ∼ (i) We proved that HomG(V,W ) = 0 when V 6= W , so ﬁnally

( ∼ (i) mi if V = W for some i dim HomG(V,W ) = 0 otherwise.

2.5 Characters

Deﬁnition. Given X, a matrix represention of G, the character of X is the homomorphism

X tr χ : G −→ GLd(C) −→ C. If X and Y are equivalent representations of G with respective characters χ and ψ, then there exists a matrix T such that Y (g) = TX(g)T −1 for all g ∈ G, and therefore

ψ(g) = tr ◦Y (g) = tr(TX(g)T −1) = tr ◦X(g) = χ(g).

So equivalent matrix representations have the same character, which allows us to deﬁne the character of a G-module.

Deﬁnition. The character of a G-module V , denoted χV , is the character of any matrix representation of V . Examples.

1 Let G = S3 with the deﬁning representation V = C{1, 2, 3}. Then we can compute the traces of each matrix in example 10 from Section 2.1 to produce the following table of character values:

g (1) (1 2) (1 3) (2 3) (1 2 3) (1 3 2) χV (g) 3 1 1 1 0 0

There are a couple things to notice. First, the character value of the identity in G is always equal to dim V – here, χV ((1)) = 3, the dimension of the deﬁning representation of S3. In addition, χV is constant across conjugacy classes of S3.

2 Let V = C[G] be the left regular representation (i.e. G acts on the group algebra by left multiplication). Let X be the matrix representation of V in the basis G of V . Then for each g ∈ G, χ(g) = tr(X(g)) is exactly the number of 1’s on the diagonal of X(g), or equivalently the number of ﬁxed points of the G-action. Since the identity eG is the only element of G that ﬁxes every element, we have ( |G| g = e χ(g) = G 0 g 6= eG.

17 2.5 Characters 2 Group Representations

Proposition 2.5.1. Let X be a matrix representation of G with character χ. Then

(1) χ(eG) = deg X. (2) If x and y are conjugate elements of G then χ(x) = χ(y).

(3) If Y is an equivalent representation, then the character ψ of Y is equal to χ.

Proof. (1) follows from the same logic as in example 2 above and (3) was proven at the start of the section. To show (2), suppose x = zyz−1 for some z ∈ G. Then

χ(x) = tr(X(x)) = tr(X(zyz−1)) = tr(X(z)X(y)X(z−1)) by homomorphism axioms = tr(X(y)) since trace is invariant under commutativity = χ(y).

Deﬁnition. Given a group G, we deﬁne the class function space

R(G) = {f : G → C | f is constant on conjugacy classes}.

R(G) is a C-vector space whose dimension equals the number of conjugacy classes of G. Remark. While χ(gh) = χ(hg) for any character χ, in general χ(gh) 6= χ(g)χ(h) except in the case when χ is the character of a one-dimensional representation. Deﬁnition. The character table of a group G is the array K

χ χK

where χ are the inequivalent, irreducible characters of G and K are the distinct conjugacy classes of G.

So far our character table for S3 looks like: (1 2) (1 3 2) (1) (1 3) (1 3 2) (2 3) χtriv 1 1 1 χsign 1 −1 1 ? ? ? ? Our goal will be to piece together the last row of the table with the information for what turns out to be the ﬁnal irreducible representation of S3.

18 2.5 Characters 2 Group Representations

Deﬁnition. Given two characters χ and ψ, we deﬁne their inner product by

1 X hχ, ψi = χ(g)ψ(g), |G| g∈G

where ψ(g) denotes the complex conjugate of ψ(g).

Consider the function space Func(G, C) = {functions f : G → C}, which is a G-module under the action (g · f)(h) = f(gh). The standard inner product with respect to the basis {1g}g∈G deﬁned by

1g : G −→ C g 7−→ 1 h 7−→ 0 if h 6= g

0 P 0 0 is hf, f i = g∈G f(g)f (g). Notice that hf, f i is G-invariant: for any h ∈ G, we have X X X hh · f, h · f 0i = (h · f)(g)(h · f 0)(g) = f(hg)f 0(hg) = f(g)f 0(g) = hf, f 0i g∈G g∈G g∈G

1 (after reordering the sum). The |G| coeﬃcient in the deﬁnition of our inner product makes the basis orthonormal, which has several desirable properties. To represent the G-module Func(G, C), let Y be the matrix representation of G using 0 the orthonormal basis {1G}g∈G. For all f, f ∈ Func(G, C), hf, f 0i = hg · f, g · f 0i = hhY (g)f, Y (g)f 0ii,

where the latter is the standard inner product on R|G|. This shows Y (g) is a unitary matrix for all g ∈ G. Now from linear algebra, we know that when Y (g) is unitary,

T Y (g−1) = Y (g)−1 = Y (g) .

If ψ is the character of the matrix representation Y and Y (g) is unitary for all g ∈ G, then

T ψ(g−1) = tr(Y (g−1)) = tr(Y (g)−1) = tr(Y (g) ) = tr(Y (g)) = ψ(g).

So we could write our inner product instead as

1 X hχ, ψi = χ(g)ψ(g−1). |G| g∈G

Theorem 2.5.2. If χ and ψ are irreducible characters of a group G then hχ, ψi = δχ,ψ, that is, the rows of the character table for G, suitably weighted, are orthonormal.

19 2.5 Characters 2 Group Representations

Proof. Let χ be the character of A, a representation of degree d, and let ψ be the character of B, a representation of degree f. Also let X = (xij) be a d × f matrix of indeterminants 1 P −1 and set Y = (yij) = |G| g∈G A(g)XB(g) . For any h ∈ G,

1 X 1 X A(h)YB(h)−1 = A(h)A(g)XB(g)−1B(h)−1 = g ∈ GA(h)A(g)XB(g−1)B(h−1) |G| |G| g∈G 1 X 1 X = A(hg)XB(g−1h−1) = g ∈ GA(hg)XB((hg)−1) |G| |G| g∈G 1 X = A(g)XB(g) after reordering |G| g∈G = Y.

Therefore by Theorem 2.3.5, we have ( cI if χ = ψ Y = d 0 if χ 6= ψ.

First assume χ 6= ψ. We know yij = 0 for all i, j. Let aij(g) and bij(g) denote respectively the ijth positions of A(g) and B(g). Then

1 X X y = a (g)x b (g−1). ij |G| ik k` `j k,` g∈G

So for all i, j, 1 X X a (g)b (g−1) = 0 |G| ik `j k,` g∈G

since yij is just a polynomial in xk` (and the coeﬃcients have to sum to 0). Recall that −1 −1 −1 −1 χ(g) = tr ◦A(g) = a11(g) + ... + add(g) and ψ(g ) = tr ◦B(g ) = b11(g ) + ... + bff (g ). Then 1 X hχ, ψi = (a (g) + ... + a (g))(b (g−1) + ... + b (g−1)). |G| 11 dd 11 ff g∈G Choosing k = i and ` = j, we see that

1 X X a (g)b (g−1) = 0. |G| ii jj i,j g∈G

It follows that hχ, ψi = 0. On the other hand, suppose χ = ψ – we may assume that A = B. Here we have ( 1 X X −1 c if i = j aik(g)a`j(g ) = |G| 0 if i 6= j. k,` g∈G

20 2.5 Characters 2 Group Representations

1 P P −1 It suﬃces to compute |G| i g∈G aii(g)aii(g ). Recall that when χ = ψ,

1 X Y = A(g)XA(g−1) = cI |G| d g∈G

for some constant c. Taking the trace of both sides gives

1 X tr(cI ) = tr(A(g)XA(g−1)) d |G| g∈G 1 X |G| cd = tr X = (x + ... + x ) = x + ... + x . |G| |G| 11 dd 11 dd g∈G

1 This can be rewritten as cd = tr X, or c = d tr X = yii, or even 1 1 X X y = (x + ... + x ) = a (g)x a (g−1). ii d 11 dd |G| ik k` `j k,` g∈G

Then ( 1 1 X −1 d if k = ` aik(g)a`i(g ) = |G| 0 if k 6= `. g∈G 1 That is, the k` coeﬃcient is just d δk,`. So we can let k = ` = i and conclude that

d 1 X X X 1 a (g)a (g−1) = = 1. |G| ii ii d i g∈G i=1

This ﬁnishes the proof.

Corollary 2.5.3. Let X be a matrix representation of G with character χ. Write X = (1) (k) (i) (i) m1X ⊕ · · · ⊕ mkX for distinct, irreducible representations X having character χ . Then

(1) (k) (1) χ = m1χ + ... + mkχ .

(i) (2) For each i = 1, . . . , k, hχ, χ i = mi.

Pk 2 (3) hχ, χi = i=1 mi . (4) χ is irreducible if and only if hχ, χi = 1.

(5) X and Y are equivalent if and only if χ = ψ.

Proof. (1) – (4) follow direction from Theorem 2.5.2, and (5, =⇒ ) is obvious. For the other Lk (i) (i) (i) (i) direction of (5), write Y = i=1 niX . Then hχ, χ i = hψ, χ i. By (2), hχ, χ i = mi (i) and hψ, χ i = ni. Hence Y is equivalent to X.

21 2.5 Characters 2 Group Representations

Examples.

3 In Section 2.1 we saw that V = C{1 + 2 + 3, 2 − 1, 3 − 1} is a representation of S3. Denote its character by χ(2,1). We will now show that it is irreducible and distinct from the trivial and sign representations and is therefore the ﬁnal entry in the character table for S3. First, we compute

χ(2,1)((1)) = 2, χ(2,1)((1 2)) = 0 and χ(2,1)((1 2 3)) = −1

(write down the matrices and compute their traces). Then we have hχ(2,1), χ(2,1)i = 1 2 2 2 6 (2,1) 6 (2 + 0 + 2 · (−1) ) = 6 = 1. So by (4) of Corollary 2.5.3, χ is indeed irreducible. We will see that the superscript (2, 1) corresponds to the partition λ = (2, 1) and more importantly that the following is the complete character table for S3:

(1) (1 2) (1 2 3) χtriv 1 1 1 χsign 1 −1 1 χ(2,1) 2 0 −1

L (i) 4 Let V = C[G], the group algebra. By Maschke’s Theorem (2.1.2), C[G] = i miV where the V (i) are all the irreducible G-modules (at this point there may be infinitely many). Note that since C[G] is finite dimensional, there can only be finitely many mi that are nonzero. Let χ be the character of C[G] and let χ(i) be the character of V (i). Then for each i, 1 X hχ, χ(i)i = m = χ(g)χ(i)(g−1). i |G| g∈G Recall that there are no fixed points of the action of any nonidentity element in the left regular representation, so the character for the group algebra is ( |G| if g = e χ(g) = G 0 otherwise.

1 (i) 1 (i) (i) Then mi = |G| χ(eG)χ (eG) = |G| |G|χ (eG) = χ (eG). This proves the following theorem.

Theorem 2.5.4. Every irreducible G-module appears in C[G] and the multiplicity of each irreducible piece is its dimension, i.e.

M (i) (i) C[G] = (dim V )V . i

Moreover, there are only ﬁnitely many V (i).

Theorem 2.5.5. For any group G, the number of pairwise-inequivalent, irreducible G- modules is equal to the number of conjugacy classes of G.

22 2.5 Characters 2 Group Representations

∼ Proof sketch. First, C[G] = EndG(C[G]) as vector spaces. The map

θ : C[G] −→ EndG(C[G]) V 7−→ φV : C[G] → C[G] w 7→ w · v

is an anti-isomorphism of algebras, i.e. θ(fg) = θ(g)θ(f). This induces an anti-isomorphism ∼ Z(C[G]) = Z(EndG(C[G])). Recall that we proved that dim Z(EndG(C[G])) is equal to the number of irreducible representations of V . Let z = c1g1 + ... + cngn ∈ Z(C[G]) and take −1 −1 −1 h ∈ G. Then z = hzh =⇒ c1g1 + ... + cngn = c1hg1h + ... + cnhgnh . This implies that if gi and gj are conjugate, ci = cj. So for each conjugacy class K, we get X zK := g ∈ Z(C[G]). g∈K

Since the conjugacy classes partition G, the collection {zK } forms a basis of Z(C[G]). Hence the number of conjugacy classes equals dim Z(C[G]) = dim Z(EndG(C[G])) which in turn equals the number of distinct, irreducible representations. Corollary 2.5.6. There are orthogonality relations on the columns of the character table for G given by X |G| χ χ = δ K L |K| K,L χ irreducible for each pair of conjugacy classes K and L. hp i Proof. Let U = |K|/|G| χK which is a square matrix. The rows of U are orthogonal (in the usual sense). Then

D E X |K| p|K|/|G| χ , p|K|/|G| ψ = χ ψ K K |G| K K K 1 X = |K|χ ψ |G| K K K 1 X = χ(g)ψ(g) = hχ, ψi = δ . |G| χ,ψ g∈G So the rows of U are in fact orthonormal. For any two conjugacy classes K and L, Dp p E δK,L = |K|/|G| χK , |L|/|G| χL X p p = |K|/|G| χK |L|/|G| χL χ p |K| |L| X = χ χ . |G| K K χ The proves the desired formulas.

23 2.6 Tensor Products 2 Group Representations

If we didn’t already know the ﬁnal row of the character table in example 3 , we could use these orthogonality relations to determine the entries of the table:

X |G| χ(1)χ(1) = = 6 =⇒ 12 + 12 + (χ(2,1)(1))2 = 6 =⇒ χ(2,1)(1) = 2; |(1)| χ X |G| χ(2 3)χ(2 3) = = 2 =⇒ 12 + (−1)2 + (χ(2,1)(2 3))2 = 2 =⇒ χ(2,1)(2 3) = 0; |(2 3)| χ and χ(2,1)(2 3 1) = −1 follows from earlier formulas.

2.6 Tensor Products

In this section we introduce and brieﬂy discuss tensor products of representations.

Definition. Let G and H be finite groups with representations X of degree d and Y of degree f, respectively. The tensor product of G and H is the element of GLdf (C) defined by (X ⊗Y )(g, h) = X(g)⊗Y (h), where the latter is the Kronecker product (or matrix tensor).

Theorem 2.6.1. Let G and H be ﬁnite groups, X and Y be representations of these groups with deg X = d and deg Y = f. Then

(1) X ⊗ Y is a representation of the direct product group G × H.

(2) If χ and ψ are characters of X and Y , respectively, then the character χ⊗ψ is deﬁned by the relation (χ ⊗ ψ)(g, h) = χ(g)ψ(h).

(3) If X and Y are irreducible then so is X ⊗ Y .

(4) All irreducible representations of G × H are obtained in this way.

Proof. (1) Let eG be the identity in G and eH be the identity in H. Then

(X ⊗ Y )(eG, eH ) = X(eG) ⊗ Y (eH ) = Id ⊗ If = Idf .

Now let g, g0 ∈ G and h, h0 ∈ H and consider

(X ⊗ Y )(gg0, hh0) = X(gg0) ⊗ Y (hh0) = X(g)X(g0) ⊗ Y (h)Y (h0) = (X(g) ⊗ Y (h))(X(g0) ⊗ Y (h0)) = (X ⊗ Y )(g, h)(X ⊗ Y )(g0, h0).

Thus X ⊗ Y is a representation of G × H. (2) The key property here is that trace splits up tensors in the following way: tr(A⊗B) = (tr A)(tr B) (this is easy to see since trace just takes the sum along diagonals of triangular matrices, and by Maschke’s Theorem (2.1.2) allows us to write any representation this way). Using this property, we have

tr(X ⊗ Y )(g, h) = tr(X(g) ⊗ Y (g)) = (tr X(g))(tr Y (g)) = χ(g)ψ(g).

24 2.7 Restricted and Induced Characters 2 Group Representations

(3) Take characters χ of X and ψ of Y . Then

1 X hχ ⊗ ψ, χ ⊗ ψi = (χ ⊗ ψ)(g, h)(χ ⊗ ψ)(g−1, h−1) |G| |H| g∈G,h∈H 1 X X = χ(g)χ(g−1)ψ(h)ψ(h−1) |G| |H| g∈G h∈H ! ! 1 X 1 X = χ(g)χ(g−1) ψ(h)ψ(h−1) |G| |H| g∈G h∈H = hχ, χihψ, ψi.

If χ and ψ are both irreducible, then hχ, χ, ihψ, ψi = 1 · 1 = 1 so χ ⊗ ψ is irreducible as well. (4) Let p be the number of conjugacy classes of G and q be the number of conjugacy classes of H. The same calculation as in (3) shows that if χ0 is another character of X and ψ0 0 0 is another character of Y , then hχ ⊗ ψ, χ ⊗ ψ i = δ(χ,ψ),(χ0,ψ0). This tells us there are at least pq pairwise inequivalent, irreducible representations of (G × H)-modules of the form X ⊗ Y . But the number of conjugacy classes in G × H is precisely pq, so this is all of them.

2.7 Restricted and Induced Characters

Given H a subgroup of G, are there natural associations between G-modules and H-modules? Of course! The assignment of G-modules down to H-modules is called restriction, while the assignment of H-modules up to G-modules is called induction; we deﬁne both formally here.

G Deﬁnition. Let H be a subgroup of G and take V a G-module. The restriction V ↓H is the ρ H-module obtained by the composition H,→ G −→V GL(V ), that is, the H-action is simply the G-action of the elements of H on V .

Deﬁnition. Let H be a subgroup of G with {t1, . . . , t`} a transversal of H in G. For Y a G matrix representation of H, we deﬁne the induction Y ↑H by

 −1 −1 −1  Y (t1gt1 ) Y (t1gt2 ) ··· Y (t1gt` ) Y (t gt−1) Y (t gt−1) ··· Y (t gt−1) G  2 1 2 2 2 `  Y ↑H (g) =  . . .. .  ∈ GLd`(C)  . . . .  −1 −1 −1 Y (t`gt1 ) Y (t`gt2 ) ··· Y (t`gt` )

where Y (g) = 0 if g 6= H and d = deg Y .

Remark. G-modules are in correspondence with left C[G]-modules and this induces a ring homomorphism C[H] ,→ C[G] which makes every C[G]-module into a C[H]-module via f · m 7→ im(f) · m.

Remark. Notice that we have deﬁned restrictions on G-modules, whereas inductions are deﬁned in terms of matrix representations. Given a C[H]-module W , there is a parallel G notion of induction known as extension of scalars: W↑H = C[G] ⊗C[H] W .

25 2.7 Restricted and Induced Characters 2 Group Representations

G Proposition 2.7.1. (1) Y ↑H is a well-deﬁned representation of G.

G (2) Y ↑H is invariant across any choice of transversal of H ≤ G. (3) (Frobenius Reciprocity) Let X be a G-representation with character χ and let Y be G G an H-representation with character φ. Then hφ↑H , χi = hφ, χ↓H i. This says that restriction and induction are an adjoint pair.

G H G (4) Let K ≤ H ≤ G and take X a representation of K. Then X↑K = X↑K ↑H .

(5) If H has transversal {t1, . . . , t`} in G with cosets H = {t1H, . . . , t`H}, the matrices in G the representation 1↑H are equal to the matrices of the representation corresponding to the coset action of G on H (with H as a basis).

Proof. (1), (2) and (4) are routine to check. We will prove Frobenius reciprocity and (5). 1 X 1 X X (3) Consider hφ↑G , χi = φ↑G (g)χ(g−1) = φ(x−1gx)χ(g−1). Let H |G| H |G| |H| g∈G g∈G x∈G y = x−1gx, so that g−1 = xy−1x−1. We then have

1 X X hφ↑G , χi = φ(y)χ(xy−1x−1) conjugation is an automorphism H |G| |H| y∈G x∈G 1 X X = φ(y)χ(y−1) since χ is constant on conjugacy classes |G| |H| y∈G x∈G 1 X = φ(y)χ(y−1) |H| y∈G 1 X = φ(y)χ(y−1) since φ(y) = 0 when y 6 inH |H| y∈H G = hφ, χ↓H i.

G (5) Let X = 1↑H = (xij). Also let Z be the representation of the action G CH via g · (giH) = (gti)H = tjH. Note that X and Z are both ` × ` matrices consisting of 1’s and −1 0’s. Then xij(g) = 1 ⇐⇒ ti gtj ∈ H ⇐⇒ gtjH = tiH ⇐⇒ zij(g) = 1, so X = Z. This ﬁnishes what we set out to prove.

G Given an H-representation Y , we know Y ↑H is a G-representation. Given a chacter φ of G Y , is there a natural character of Y ↑H associated to φ? Recall that for any g ∈ G,

 −1  Y (t1 gt1) Y ↑G (g) =  ..  H  .  −1 Y (t` gt`)

26 2.7 Restricted and Induced Characters 2 Group Representations

G Accordingly, we can prove a nice formula for φ↑H :

` G X −1 φ↑H (g) = tr Y (ti gti) i=1 ` 1 X X = tr Y (h−1t−1gt h) since t−1gt ∈ H ⇐⇒ h−1t−1gt h ∈ H |H| i i i i i i i=1 h∈H 1 X = tr Y (x−1gx) since cosets partition G. |H| x∈G

G 1 P −1 Therefore our formula for the induced character of φ is φ↑H (g) = |H| x∈G φ(x gx). Remark. (Hom-tensor adjointness) Let R → S be a ring homomorphism – this should be thought of as an analog to the inclusion C[H] ,→ C[G]. Let W be a right R-module and V a right S-module. Hom-tensor adjointness says that there is an isomorphism of abelian groups ∼ HomS(W ⊗R S,V ) = HomR(W, HomS(S,V )), ∼ and in fact HomR(W, HomS(S,V )) = HomR(W, V ). These two are analogs of restriction (HomR(W, V )) and extension of scalars (HomS(W ⊗R S,V )). In our setting, ∼ HomG(W ⊗C[H] C[G],V ) = HomH (W, V ). This is equivalent to Frobenius reciprocity.

27 3 Representations of Sn

3 Representations of Sn

3.1 Young Subgroups of Sn

Recall that λ ` n denotes a sequence (λ1, . . . , λn) of nonincreasing integers summing to n. A Ferrers diagram is a way of representing λ visually:

(4, 4, 3, 2) ←→

Given Sn, we will write subgroups Sk, k ≤ n, as a symmetric group on a subset of {1, . . . , n} to denote which symbols they permute. For example, S{1,3,4} is a subgroup of S4 which is isomorphic to S3. In this way we can associate such subgroups to partitions of n.

Deﬁnition. Given a partition λ = (λ1, . . . , λ`) of n, the Young subgroup of shape λ is the subgroup

Sλ := S{1,...,λ1} × S{λ1+1,...,λ1+λ2} × · · · × S{λ1+...+λk−1+1,...,n} ≤ Sn.

Example 3.1.1. For n = 13, a Young subgroup is S{1,2,3,4} × S{5,6,7,8} × S{9,10,11} × S{12,13}. This corresponds to the partition λ = (4, 4, 3, 2).

Given a Young subgroup Sλ, consider the trivial representation:

∗ 1 : Sλ −→ GL1(C) = C .

We can induce this up to Sn: 1↑Sn = S = S /S . Sλ C λ C n λ

In order to do this, we will pick a transversal of Sλ in Sn and understand what this looks like in Sn. Deﬁnition. A Young tableau t of shape λ – sometimes denoted tλ – is a Ferrers diagram of shape λ ﬁlled in with the numbers 1, . . . , n bijectively. We say t is standard if the entries increase along every row and every column. Below are two Young tableaux. The one on the right is standard, while the one on the left is not. 4 1 2 1 2 4 3 5 3 5

Clearly the second is just a rearrangement of the ﬁrst. In fact, Sn acts on the set of all Young tableaux of shape λ:

4 1 2 1 2 4 (1 2 4) · = 3 5 3 5

28 3.1 Young Subgroups of Sn 3 Representations of Sn

Deﬁnition. We say two Young tableaux tλ and sλ are (row) equivalent if the entries in each row of tλ and sλ are the same up to reordering. This equivalence relation is denoted tλ ∼ sλ. The equivalence class of a tableau t is called the tabloid of t, denoted {t}.

As in Section 2.2, a tabloid is denoted with bars between rows:

1 2 4 4 1 2 = 3 5 3 5

λ Recall that we denote by M the Sn-module generated by the set of λ-tabloids under the action π ·{t} = {πt}. Our goals in this chapter are as follows:

λ 1) Show that the irreducible Sn-module corresponding to λ appears in M with multiplicity 1.

2) Describe a partial ordering on partitions, called the dominance ordering, such that µ D λ if and only if the irreducible Sn-module corresponding to µ is a submodule of M λ.

3) Find a combinatorial rule for determining the multiplicities of each of these submodules.

Examples.

(n) n o 1 Consider M = C 1 2 3 ··· n . For any π ∈ Sn, π · 1 2 3 ··· n = 1 2 3 ··· n so M (n) is the trivial representation.

(( )n ) 1 2 ··· ˆı ··· n 2 Next consider M (n−1,1) = C . Note that each tabloid is i i=1 identiﬁed by i, the entry in the second row, so M (n−1,1) = C{1, 2,..., n}, the deﬁning (2,1) (2,1) (2,1) (3) representation. For S3, we saw that M could be written M = S ⊕ S , the direct sum of the irreducible (2, 1) representation and the trivial representation.

3 Let (1n) denote the partition (1, 1,..., 1) ` n. Consider   i  1    (1n)  i2  M = C . : {i1, . . . , in} = {1, . . . , n} . .      in 

This is the left regular representation of Sn on the group algebra C[Sn]. It also turns out that S(1n) is the sign representation.

Deﬁnition. We say a G-module V is cyclic if V = C[G] · ~v for some vector ~v ∈ V .

29 3.1 Young Subgroups of Sn 3 Representations of Sn

Proposition 3.1.2. M λ is cyclic for all partitions λ of n. Moreover, M λ is cyclically generated by any tabloid {t} of shape λ and its dimension is given by the formula

λ n! dimC M = . λ1!λ2! ··· λ`! Proof. Take a λ-tabloid {t}. Clearly {t} generates M λ because given any other tabloid {s}, there is some (maybe even many) permutations π such that π ·{t} = {s}. Let Sn act on the set of λ-tabloids. By the ﬁrst statement, this action is transitive so by the Orbit-Stabilizer Theorem, λ dimC M = | OrbSn ({t})| = [Sn : StabSn ({t})].

But | StabSn ({t})| = λ1!λ2! ··· λ`! = λ! which implies the desired formula.

Theorem 3.1.3. Let λ be a partition of n, Sλ the Young subgroup of shape λ and

1 2 3 ··· λ1

λ1 + 1 λ1 + 2 ··· λ1 + λ2

{tλ} = . .

n − λ` + 1 ··· n

λ ({tλ} is sometimes known as the superstandard tabloid of shape λ). Then V := λ CSn/Sλ is isomorphic to M as Sn-modules. λ λ Proof. Let π1, . . . , πk be a transversal for Sλ ≤ Sn. We have a map θ : V → M given n! λ λ by πiSλ 7→ {πitλ}. Since k = λ! by Proposition 3.1.2, it is clear that dim V = dim M . λ λ Moreover, {πiSλ} ↔ {{πitλ}} is a bijection on the natural bases of V and M , respectively, so θ is a vector space isomorphism. To see that the Sn-action is preserved, consider

θ(π(πiSλ)) = θ((ππi)Sλ)

= θ(πjSλ) for some πj in the transversal

= {πjtλ}

= {ππitλ} since Sλ is the stabilizer

= π ·{πitλ}

= πθ(πiSλ).

Hence θ is an Sn-module isomorphism. Definition. A set A is called a poset (short for partially ordered set) if there is a relation ≤ on A that satisfies the following properties for all a, b, c ∈ A: (1) (Reflexivity) a ≤ a

30 3.1 Young Subgroups of Sn 3 Representations of Sn

(2) (Antisymmetry) If a ≤ b and b ≤ a then a = b

(3) (Transitivity) If a ≤ b and b ≤ c then a ≤ c.

Example 3.1.4. R with the usual ordering ≤ is a poset; in fact it is a totally ordered poset since every x, y ∈ R satisﬁes either x ≤ y, y ≤ x or x = y.

Example 3.1.5. The Boolean algebra Bn = (P{1, . . . , n}, ⊆) is a poset. An easy way to visualize B3 for example is in the form of a Hasse diagram:

{1, 2, 3}

{1, 2} {1, 3} {2, 3}

{1} {2} {3}

∅

Each line indicates that the lower subset is contained in the upper subset; the diagram is also transitive, so that not every containment is shown.

Deﬁnition. Suppose λ = (λ1, . . . , λ`) and µ = (µ1, . . . , µm) are partitions of n. We say λ dominates µ, written λ D µ, if for each i = 1, . . . , n there are inequalities

i i X X λj ≥ µj. j=1 j=1

Proposition 3.1.6. The dominance order D is a partial order on the set of partitions of n. Proof. Easy to check.

Deﬁnition. Given a partition λ, its conjugate λC is found by reversing the rows and columns of a Ferrers diagram of shape λ.

Example 3.1.7. The conjugate of (4, 3, 2) is (3, 3, 2, 1):

←→

31 3.1 Young Subgroups of Sn 3 Representations of Sn

Example 3.1.8. The Hasse diagram for the dominance ordering on partitions of n = 6 is shown below. Notice that the diagram is reversible, that is, one can ﬂip it upside-down and it looks the same. (6)

(5,1)

(4,2)

(3,3) (4,1,1)

(3,2,1)

(3,1,1,1) (2,2,2)

(2,2,1,1)

(2,1,1,1,1)

(1,1,1,1,1,1)

There is another useful partial ordering of partitions which we deﬁne here.

Deﬁnition. The lexicographic order is deﬁned on partitions of n by λ < µ if there exists an i between 1 and n such that λj = µj for all j < i and λi < µi.

Proposition 3.1.9. Lexicographic order is a reﬁnement of dominance order, that is if λ D µ then λ ≥ µ.

32 3.2 Specht Modules 3 Representations of Sn

Pk Proof. Take λ 6= µ and let i be the first index where they differ. If λ D µ then j=1 λj = k i Pi λj=1µj for k = 1, . . . , i − 1 and λj=1λj > j=1 µj. So λj = µj for j < i and λi > µi which shows this choice of i suffices to show λ > µ under the lexicographic order.

3.2 Specht Modules

Deﬁnition. For a tableau t of shape λ ` n, the row stabilizer of t is deﬁned as

Rt = SR1 × SR2 × · · · × SR` ≤ Sn, where Ri is the ith row of t. Similarly, the column stabilizer of t is

Ct = SC1 × SC2 × · · · × SCm ≤ Sn, where Cj is the jth column of t.

4 1 2 Example 3.2.1. For t = , the stabilizers are Rt = S × S and Ct = 3 5 {1,2,4} {3,5}

S{3,4} × S{1,5}.

Deﬁnition. Given a subset H ⊂ Sn, deﬁne the following elements of C[Sn]: X X H+ := σ and H− := sgn(σ)σ. σ∈H σ∈H

− Using this notation with the column stabilizers of a tableau, we set Kt = Ct and deﬁne:

Deﬁnition. The polytabloid associated to a tableau t is et := Kt{t}. Example 3.2.2. For the same tableau t as above, we have

Ct = {(1), (1 5), (3 4), (3 4)(1 5)}

Kt = {(1) − (1 5) − (3 4) + (3 4)(1 5)}.

Then the polytabloid for t is calculated to be

4 1 2 et = Kt{t} = [(1) − (1 5) − (3 4) + (3 4)(1 5)] 3 5 4 1 2 4 5 2 3 1 2 3 5 2 = − − + . 3 5 3 1 4 5 4 1

λ Deﬁnition. For a partition λ ` n, deﬁne the Specht module S to be the span of et over all tableaux t of shape λ. Notice that Sλ is a submodule of M λ.

Lemma 3.2.3. Let t be a tableau and take π ∈ Sn. Then

−1 (1) Rπt = πRtπ .

33 3.2 Specht Modules 3 Representations of Sn

−1 (2) Cπt = πCtπ .

−1 (3) Kπt = πKtπ .

(4) eπt = πet. Proof. To prove (1), we have the following string of equivalent statements:

σ ∈ Rπt ⇐⇒ σ{πt} = {πt} ⇐⇒ π−1σπ{t} = {t} −1 ⇐⇒ π σπ ∈ Rt −1 ⇐⇒ σ ∈ πRtπ .

(2) is proven in a similar fashion and (2) implies (3). For (4), we have:

−1 −1 eπt = Kπt{πt} = πKtπ {πt} = πKtπ π{t} = πKt{t} = πet.

Notice further that (4) implies each Sλ is a cyclic submodule of M λ.

Examples.

λ − 1 Let λ = (n). Then S = Span e1 2 3 ··· n and C1 2 ··· n· 1 2 3 ··· n = 1 2 3 ··· n (this is the only polytabloid of shape λ. So we see that S(n) is a one-dimensional representation and in fact S(n) = M (n), the trivial representation.

n λ 2 Now let λ = (1 ), so that S = Span eπt, where

π(1) 1 π(2) 2 πt = . for the standard tableau t = . . . . π(n) n

Then K = P sgn(σ)σ and e = P sgn(σ)σ ·{t}. We see that by (4) of t σ∈Sn t σ∈Sn Lemma 3.2.3, X X π · et = eπt = sgn(σ)σ{πt} = sgn(σ)σπ{t}

σ∈Sn σ∈Sn X X = sgn(σ0) sgn(π)σ0{t} = sgn(π) sgn(σ)σ{t} 0 σ ∈Sn σ∈Sn

= sgn(π)et.

(1n) So π acts on et by applying sgn(π) to et. Thus S is the sign representation of Sn.

34 3.2 Specht Modules 3 Representations of Sn

3 Let λ = (n − 1, 1). We will denote the tableau

1 2 ··· ˆi ··· n i

by î, with tabloid {ti}. In total we have n tabloids of shape (n−1, 1) and for each i ≥ 2, ˆ ˆ ˆ ˆ ˆ eti = Kti {ti} = (e − (1 i)) · i = i − 1. The polytabloid for i = 1 is simply 1 − 2, so we see that there is a set of n − 1 polytabloids that span S(n−1,1): {2ˆ − 1ˆ, 3ˆ − 1ˆ,..., nˆ − 1ˆ}. Consider the defining representation, M (n−1,1) = S(n−1,1) ⊕S(n). We haven’t yet proven that M λ must be the inner product of these irreducibles, but we will show this in the future. Note that for any i ≥ 2, π(î − 1)ˆ = πd(i) − πd(1) = (πd(i) − 1)ˆ − (πd(1) − 1).ˆ If π(i) = i then then π(î − 1)ˆ = (î − 1)ˆ − (πd(1) − 1)ˆ so this contributes a 1 to the character. On the other hand, if π(1) = j, then π(ˆj − 1)ˆ = (πd(j) − 1)ˆ − (ˆj − 1)ˆ and this contributes a −1 to the character; the latter case only occurs once. Thus the character for λ = (n − 1, 1) is equal to the number of fixed points minus 1:   (n−1,1) X χ (π) =  1 − 1. i:π(i)=i

It turns out that this is an irreducible character, and computing the dimension of S(n−1,1) will show that there is only one of these representations of dimension n − 1. Let µ be a partition of n. The Submodule Theorem characterizes submodules of M µ in terms of Specht modules: Theorem 3.2.4 (Submodule Theorem). Let U be a submodule of M µ. Then either Sµ ⊆ U or U ⊂ (Sµ)⊥. To make sense of the statement of the theorem, we need to define an inner product with respect to which to take the perpendicular space (Sµ)⊥. Definition. For two tabloids {t} and {s} of shape µ, define their inner product to be h{t}, {s}i = δ{t},{s}, where δ is the Kronecker delta symbol. Remark. In a characteristic zero environment, any space W always satisfies W ∩W ⊥ = {0} but this is not true in general for positive characteristic. For example, take   1   1   p W = Span{~v} = Span . ∈ Fp. .    1 

Then ~v is a nonzero vector satisfying h~v,~vi = 0 so ~v ∈ W ∩ W ⊥. Lemma 3.2.5. Let h·, ·i be the inner product deﬁned above on M µ. In what follows, if − − H = {π} we will denote H by π . Let H ≤ Sn. Then

35 3.2 Specht Modules 3 Representations of Sn

(1) If π ∈ H then πH− = H−π = (sgn π)H−1.

(2) For every u, v ∈ M µ, hH−u, vi = hu, H−vi.

− (3) If (b c) ∈ H where (b c) is a transposition in Sn, then H = k(e − (b c)) for some k ∈ Sn. (4) If t is a tableau with b and c in the same row and (b c) ∈ H, then H−{t} = 0.

Proof. (1) For any π ∈ H, X X X X π (sgn σ)σ = (sgn σ)πσ = (sgn σ0π−1)σ0 = (sgn π) (sgn σ0)σ0. σ∈H σ∈H σ0∈H σ0∈H

Hence πH− = (sgn π)H− and the other equality is proven similarly. (2) Let u, v ∈ M µ. Then X hH−u, vi = h(sgn σ)σu, vi σ∈H X −1 = hu, (sgn σ)σ vi by bilinearity, Sn-invariance σ∈H = hu, H−vi after reindexing.

(3) Since {(1), (b c)} =: K is a subgroup of H, take a transversal k1, k2, . . . , k` ∈ H of K. S` Then H = i=1 kiK is a disjoint union and

` ! ` ` X X X − (sgn ki)ki (e − (b c)) = (sgn ki)ki + sgn(ki(b c))ki(b c) = H . i=1 i=1 i=1

P` Choose k = i=1(sgn ki)ki and we have proven (3). (4) The key is that since b and c are in the same row, (b c){t} = {t}. Then

H−{t} = k(e − (b c)){t} by (3) = k({t} − (b c){t}) = k(0) = 0.

The following lemma gives a criterion for when a partition λ dominates µ.

Lemma 3.2.6 (Dominance). Let tλ and sµ be tableaux of shape λ and µ, respectively. If for µ λ each i the elements of row i of s are in different columns of t then λ D µ. Proof. By hypothesis, we can sort entries in each column of tλ so that the rows 1, . . . , i of µ λ s all occur in the first i rows of t . Now consider λ1 + ... + λi; this equals the number of elements in the first i rows of tλ, which is just the number of elements of sµ occurring in the λ first i rows of t , that is, µ1 + µ2 + ... + µi. This holds for all i, and therefore we conclude that λ D µ.

36 3.2 Specht Modules 3 Representations of Sn

We immediately obtain the following corollaries. There will be many more (sometimes surprising) consequences of the Dominance Lemma to come.

Corollary 3.2.7. Let t be a λ-tableau and s be a µ-tableau. If Kt{s}= 6 0 then λ D µ, and furthermore if λ = µ then Kt{s} = ±et, where the sign is sgn π for the permutation π such that {s} = {πt}.

Proof. By (4) of Lemma 3.2.5, for every b, c in the same row of s,(b c) 6∈ Ct so b and c must be in diﬀerent columns of t. Hence by the Dominance Lemma λ D µ. Now suppose µ = λ. Then {s} = {πt} for some π ∈ Sn and we see that

Kt{s} = Kt{πt} = Ktπ{t} = (sgn π)Kt{t} = (sgn π)et.

λ Corollary 3.2.8. If u ∈ M and t is a µ-tableau, then Ktu = cet for a scalar c.

λ P Proof. Take u ∈ M and write u = ci{si} for scalars ci and λ-tabloids {si}. Then by the Dominance Lemma (3.2.6), X X X X Ktu = Kt ci{si} = ciKt{si} = ±ciet = ±ci et

P for some choice of signs on the ci. Setting c = ±ci for this choice of signs, we see that Ktu = cet. Theorem (3.2.4). Let U ⊆ M µ be a submodule. Then either Sµ ⊆ U or U ⊆ (Sµ)⊥.

Proof. Take u ∈ U and let t be a µ-tableau. Then by Corollary 3.2.8, Ktu = fet for some 1 µ element f of the base ﬁeld. If f 6= 0 then et = f Ktu ∈ U. Since S is cyclically generated by µ any et for a µ-tableau t, this implies in our case that S ⊆ U. On the other hand, suppose for every u ∈ U and µ-tableau t, that Ktu = 0. Then by (2) of Lemma 3.2.5,

hu, eti = hu, Kt{t}i = hKtu, {t}i = h0, {t}i = 0.

This holds for any u ∈ U, so we conclude that U ⊆ (Sµ)⊥.

Corollary 3.2.9. If char F = 0 or char F = p > n, the Specht modules Sλ ⊂ Sn are all irreducible.

Proof. Under these assumptions, if V is a vector space and h·, ·i is an inner product on V , for any subspace W ⊆ V it is always true that V = W ⊕W ⊥. So in particular M λ = Sλ ⊕(Sλ)⊥, and as a consequence Sλ ∩ (Sλ)⊥ = 0. If U ⊆ Sλ is a submodule, it is also a submodule of M λ. By the Submodule Theorem (3.2.4), either U ⊇ Sλ or U ⊆ (Sλ)⊥. If U ⊇ Sλ then U = Sλ. On the other hand, if U ⊆ (Sλ)⊥ then U ⊆ Sλ ∩ (Sλ)⊥ = 0. Hence Sλ is irreducible.

λ µ Proposition 3.2.10. Suppose char F = 0 and θ ∈ Hom(S ,M ) is nonzero. Then λ D µ. Moreover, if λ = µ then θ is multiplication by a constant.

37 3.2 Specht Modules 3 Representations of Sn

λ λ λ λ ⊥ Proof. Pick et ∈ S such that θ(et) 6= 0. Since M = S ⊕ (S ) in characteristic zero, we can extend θ to a map θ˜ : M λ → M µ, (s, s0) 7→ θ(s). Then θ˜ is also nonzero, and ˜ ˜ ˜ X 0 6= θ(et) = θ(Kt{t}) = Ktθ({t}) = Kt ci{si}

where {si} are µ-tableaux. So we see that Kt{si} 6= 0 for some i, and by Corollary 3.2.7, this implies λ D µ. For the second statement, suppose λ = µ. Then by Corollary 3.2.7, ˜ X X X θ(et) = Kt ci{ti} = ciKt{ti} = ±ci et. ˜ Hence θ(et) is a constant multiple of et, and this implies that θ is a multiplication map. Theorem 3.2.11. The Specht modules Sλ for partitions λ ` n form a complete list of irreducible Sn-modules. Proof. It suﬃces to show that the Specht modules are all distinct, since they are indexed by the conjugacy classes of Sn and this is exactly the number of irreducible Sn-modules. Suppose Sλ ∼= Sµ. Then Hom(Sλ,Sµ) ⊆ Hom(Sλ,M µ). Since Hom(Sλ,Sµ) contains an isomorphism, λ µ it is nonempty, so Hom(S ,M ) is nonempty as well. Then by Proposition 3.2.10, λ D µ. Reversing the roles of λ and µ, we have that µ D λ as well, so λ = µ. Hence the Specht modules over all distinct partitions of n are distinct. Deﬁnition. Let µ ` n and write

µ M λ M = KλµS . λ`n

The constants Kλµ are called the Kostka numbers.

Remark. Notice that Kµµ = 1, and for any partitions λ, µ ` n, their Kostka number satisﬁes λ µ Kλµ = dim Hom(S ,M ) = hχSλ , χM µ i. Recall that a tableau t is standard if the rows and columns of t are (strictly) increasing. The main structure theorem of Specht modules is:

Theorem 3.2.12. For any partition λ ` n, the set of polytabloids {et | t is a standard λ-tableau} is a basis of Sλ. To prove this theorem, we need some more general machinery of combinatorics, called compositions. These generalize partitions of n. Deﬁnition. A composition of a natural number n is an ordered sequence of nonnegative P` integers λ = (λ1, . . . , λ`) such that i=1 λi = n. Note that partitions are compositions, but compositions are in fact more general since they need not be weakly decreasing sequences. We can deﬁne the dominance order on compositions in the same way as we did for partitions. Given a tabloid {t} of shape λ, construct the following combinatorial objects for each 1 ≤ i ≤ n: {ti} : the tabloid formed by all elements ≤ i in {t} λi : the composition that is the shape of {ti}. The sequence λi is called the composition sequence of {t}.

38 3.2 Specht Modules 3 Representations of Sn

4 1 2 Example 3.2.13. For the tabloid {t} = we have 3 5

i {ti} λi

1 1 (1, 0) ∅

1 2 2 (2, 0) ∅

1 2 3 (2, 1) 3

1 2 4 4 (3, 1) 3

1 2 4 5 (3, 2) = λ 3 5

Deﬁnition. Given λ-tabloids {s} and {t} with composition sequences λi and µi, respectively, i i we say {s} dominates {t}, denoted {s} D {t}, if λ D µ with the usual dominance order for all 1 ≤ i ≤ n.

This gives a poset structure on the set of tabloids of a given shape.

Theorem 3.2.14 (Dominance Lemma for Tabloids). If k < ` and k appears in a lower row than ` in a tabloid {t}, then {t} / (k `){t}.

Proof. Let λi and µi be the composition sequences of {t} and (k `){t}, respectively. Then for i < k or i ≥ `, we know that λi = µi. Let k ≤ i < `. If r and q are the rows of {t} in which k and ` appear, then λi is equal to µi with the qth part decreased by one and the rth part increased by one.

r k r `

(k `) q ` q k

λi µi

Pr i Pr i Pm i Then the column sums satisfy j=1 λj < j=1 µj, and for all other m > r, j=1 λj ≤ Pm i i i j=1 µj. Hence λ / µ for all i, so {t} / (k `){t}.

39 3.2 Specht Modules 3 Representations of Sn

Corollary 3.2.15. If t is a standard tableau and {s} appears in et then {t} D {s}.

Proof. Suppose s = πt where π ∈ Ct. Induct on the number of column inversions of s, i.e. pairs k, ` with k < ` and ` above k in the same column. For any such k, `, {s} / (k `){s} by the Dominance Lemma (3.2.6), and (k `){s} has one less column inversion. Since k, ` appear in the same column of s,(k `) ∈ Ct so (k `){s} is also in et. By induction, {s} / (k `){s} E {t}.

Deﬁnition. For a poset (A, ≤), we say an element b ∈ A is maximal if a ≥ b implies a = b for any a ∈ A. Further, b is a maximum of A if b ≥ c for all c ∈ A.

In this language, Corollary 3.2.15 says that for any standard tableau t, {t} is maximum in et with respect to the dominance order.

1 2 Example 3.2.16. Consider t = . Since t is standard, we can compute 3 4

Kt = e − (1 3) − (2 4) + (1 3)(2 4) 1 2 2 3 3 4 1 4 and et = − − + . 3 4 1 4 1 2 2 3

Then by Corollary 3.2.15, {t} is maximum in et. Similarly, for another standard tableau 1 3 s = we have 2 4

Ks = e − (1 2) − (3 4) + (1 2)(3 4) 1 3 2 3 1 4 2 4 and es = − − + . 2 4 1 4 2 3 1 3

As a result of Corollary 3.2.15, we see that any standard tableau does not show up in the polytabloid for a diﬀerent standard tableau.

µ Lemma 3.2.17. Let v1, . . . , vm ∈ M and suppose for each vi we may choose a tabloid {ti} appearing in {vi} such that {ti} is maximum in vi and the {ti} are all distinct. Then {v1, . . . , vm} is linearly independent.

Proof. Label the polytabloids so that {t1} is maximal among the {ti}. If {t1} occurs in vi then {t1}/{ti} so {t1} may only occur in v1. Thus a linear combination c1v1 +...+cmvm = 0 forces c1 = 0. We may induct on m to obtain ci = 0, 1 ≤ i ≤ m for any m. We have thus established the linear independence part of Theorem 3.2.12. To prove that the polytabloids span Sλ, choose a tableau t and assume the columns of t are decreasing. Suppose we can ﬁnd a sequence of permutations {π} such that

In each tableau πt, a row descent has been eliminated.

40 3.2 Specht Modules 3 Representations of Sn

P The element g = π(sgn π)π satisﬁes get = 0. P Then we will have et = − π(sgn π)eπt. The elements g have a special name. Deﬁnition. Let A and B be disjoint subsets of {1, . . . , n} and choose a transversal {π} of S the subgroup SA × SB ≤ SA∪B, written SA∪B = π π(SA × SB). Then the Garnir element associated to the pair A, B is X gA,B = (sgn π)π. π

How can we pick a transversal for such a subgroup SA × SB? Let SA∪B act on the set {(A0,B0) ⊆ {1, . . . , n}2 : |A0| = |A|, |B0| = |B|,A0 ∪ B0 = A ∪ B}. We illustrate a choice of transversal with an example:

Example 3.2.18. Let A = {5, 6} and B = {2, 4}, which we will denote by A = 56 and B = 24. Then the set of (A0,B0) in {1,..., 6}2 satisfying the above properties is shown 0 0 below, along with the permutations in S{2,4,5,6} that take (A, B) to (A ,B ). (A0,B0) π (56, 24) e (46, 25) (4 5) (26, 45) (2 5) (45, 26) (4 6) (25, 46) (2 6) (24, 56) (2 5)(4 6)

Then {π} is a transversal of S{5,6} × S{2,4}, but of course it is not unique. For a tableau t, let A and B be subsets of the jth and (j +1)st columns of t, respectively. The transversal we pick to deﬁne gA,B is chosen so that the entries in πt for any π in the transversal increase when moving down along A or B.

1 2 3 Example 3.2.19. Let t = 5 4 which is not standard because of the row descent 6

5 > 4. Let A = {5, 6} and B = {2, 4} as above. We compute a transversal of SA × SB as follows:

1 2 3 1 2 3 1 4 3 1 2 3 1 4 3 1 5 3 5 4 4 5 2 5 4 6 2 6 2 6 6 6 6 5 5 4

π = e (4 5) (2 4 5) (4 6 5) (2 4 6 5) (2 5)(4 6)

Then the Garnir element for A, B is computed to be gA,B = e − (4 5) + (2 4 5) + (4 6 5) − (2 4 6 5) + (2 5)(4 6). One can verify that gA,Bet = 0.

41 3.2 Specht Modules 3 Representations of Sn

Proposition 3.2.20. Let t be a tableau with decreasing columns and take A, B to be subsets of the jth and (j + 1)st columns of t, respectively. Let gA,B be a Garnir element for A, B. If |A ∪ B| is greater than the number of elements in column j, then gA,Bet = 0.

− Proof. We first claim that SA∪Bet = 0. To see this, let σ ∈ Ct. By hypothesis, there are elements a, b ∈ A ∪ B such that a and b are in the same row of σt (since |A ∪ B| is larger − than the size of column j). Then (a b) ∈ SA∪B and by (4) of Lemma 3.2.5, SA∪B{σt} = 0. By definition e = P so we see that S− e = 0. Now for the transversal {π} chosen t σ∈Ct A∪B t S − − to define gA,B, we have SA∪B = π π(SA × SB) so in turn, SA∪B = gA,B(SA × SB) . Then − gA,B(SA × SB) = 0 by the above work. Note that since SA × SB ⊂ Ct, if σ ∈ SA × SB then

− − (sgn σ)σet = (sgn σ)σCt {t} = Ct {t} = et.

− − Thus (SA × SB) et = |SA × SB|et. This implies gA,B(SA × SB) et = |SA × SB|gA,Bet = 0, but since A, B are nonempty, we must have gA,Bet = 0. We next deﬁne an analog of a tabloid for columns of a tableau t.

Deﬁnition. For a tableau t, the column tabloid of t is deﬁned to be the orbit of the action of Ct on t, denoted [t].

1 2 Example 3.2.21. For the tableau t = we use a similar bar notation to write the 3 column tabloid [t]: 1 2 1 2 1 2 3 2 = = 3 3 3 , 1 We deﬁne a dominance order on column tabloids in the same fashion as with tabloids, replacing ‘row’ with ‘column’ in the deﬁnition. A version of the Dominance Lemma (3.2.6) holds for column tabloids:

Theorem 3.2.22 (Dominance Lemma for Column Tabloids). If k < ` and k appears in a column to the right of `, then [t] / (k `)[t].

Proof. Analagous to the proof of the Dominance Lemma (3.2.6) for (row) tabloids. We are now ready to prove the main structure theorem for Specht modules.

λ Theorem 3.2.23 (Peel). The set {et | t is a standard λ-tableau} is a basis for S . Proof. Linear independence follows from Lemma 3.2.17. If t is a tableau of shape λ whose columns are not increasing, pick σ ∈ Ct so that σt has increasing columns. Then et = (sgn σ)eσt so we may assume t has increasing columns. The maximum element of the poset of column tabloids of shape λ is [t0], where t0 is the λ-tableau formed by ﬁlling in the numbers 1, . . . , n from the top left, increasing down each column, and moving to the next column to the right when the current column is exhausted. Then t0 is standard so t0 lies in the span of {et | t is standard}. To induct, assume for all s such that [t] / [s] that es ∈ Span{et | t is standard}. If t itself is standard, we’re done. Otherwise, there’s a row

42 3.2 Specht Modules 3 Representations of Sn

descent, say ai > bi in row i of t. Let A = {ai, . . . , ap be a subset of the jth column, where the length of the column is p, and let B = {b1, . . . , bi} be a subset of the (j + 1)st column. Choose a transversal π of SA × SB as before. Then notice that |A ∪ B| = p + 1 > p so P by Proposition 3.2.20, gA,Bet = 0. Thus et = − π6=e(sgn π)eπt. Now each nonidentity element π in the transversal exchanges some element of B for some element of A. Since b1 < b2 < . . . < bi < ai < . . . < ap, we see by deﬁnition of the dominance order on column tabloids that [πt] . [t]. Hence by induction, et is in the span of {et | t is standard}. This λ proves {et | t is standard} is indeed a basis of S . Corollary 3.2.24. For any partition λ ` n,

(1) dim Sλ = f λ, where f λ is the number of standard λ-tableaux. X (2) (f λ)2 = n!. λ`n The equation in (2) leads one to believe there is a correspondence

[ standard tableau standard tableau S ←→ × n of shape λ of shape λ λ`n π 7−→ (P,Q)

Where P and Q are both standard Young tableaux whose shape is the cycle type of π. This remarkable fact is called the Robinson-Schensted correspondence. We will study a more sophisticated version of this correspondence, called the RSK algorithm, in Section 3.5.

Example 3.2.25. The partitions of n = 4, along with the Ferrers diagram and f λ for each, are shown in the table below. λ t f λ (4) 1

(3, 1) 3

(2, 2) 2

(2, 1, 1) 3

(1, 1, 1, 1) 1

X Then by (2) of Corollary 3.2.24, (f λ)2 = 1 + 9 + 4 + 9 + 1 = 24. λ`4

43 µ 3.3 The Decomposition of M 3 Representations of Sn

4 2 Example 3.2.26. Let’s illustrate Theorem 3.2.23 with an example. Let t = . First 3 1 3 1 apply the permutation (1 2)(3 4) to change t into t0 = and note that sgn(1 2)(3 4) = 1, 4 2

so that et = et0 . Let A = {3, 4} and B = {1}. The transversal of S34 × S1 ≤ S{1,3,4} is determined by: 3 1 1 3 1 4 4 2 4 2 3 2

π = e (1 3) (1 4 3)

0 0 0 0 Then gA,B = e−(1 3)+(1 4 3) and gA,Bet = gA,Bet = 0 so et = e(1 3)t −e(1 4 3)t = es1 −es2 , 1 3 1 4 where s1 = and s2 = . Note that s1 and s2 are not standard tableaux, 4 2 3 2 but they are closer in the dominance order to being standard than t0. We can repeat the algorithm on s1 and s2, multiple times if necessary, to get et as a linear combination of standard polytabloids.

3.3 The Decomposition of M µ

µ We have shown that for a given partition µ ` n, the Sn-module M can be decomposed into Specht modules: µ M λ M = KλµS , λDµ

where Kλµ are the Kostka numbers and Kµµ = 1. We would like to determine a combinatorial way of computing the other Kλµ. This is done by generalizing our existing theory of Young tableaux. Deﬁnition. For a partition λ ` n, a generalized Young tableau of shape λ is any labelling of a Ferrers diagram of shape λ with positive integers. A generalized Young tableau T is determined by its shape (λ) and its type, which is an ordered sequence of integers (a1, . . . , an) such that ai counts the number of instances of i in T . Example 3.3.1. The generalized tableau

4 1 4 T = 1 3

has shape (3, 2) and type (2, 0, 1, 2).

λ µ Recall that Kλµ = dim Hom(S ,M ). The idea is for each generalized tableau T , to λ µ deﬁne a map θT : S → M such that T has shape λ and type µ. Deﬁnition. A Young tableau t is semi-standard if t is weakly increasing along rows and down columns.

44 µ 3.3 The Decomposition of M 3 Representations of Sn

λ µ It turns out that a basis of all such maps θT : S → M consists of θT such that T is a semi-standard Young tableau. Deﬁnition. If T is a generalized tableau of shape λ, we set T (i) to be the entry of T in the same box as i in the standard tableau t having the same shape as T . Example 3.3.2. For n = 5 and λ = (3, 2), any generalized tableau T of shape λ is of the form T (1) T (2) T (3) 1 2 3 T = where t = T (4) T (5) 4 5

Let Tλ,µ denote the set of all λ-tableaux of content µ. Deﬁne a function µ θ : M −→ C[Tλ,µ] {s} 7−→ T, where T is the generalized tableau such that T (i) is the number of the row in s in which i appears. 1 2 3 Example 3.3.3. Let t = which is a standard tableau of shape λ = (3, 2). Let 4 5 µ = (2, 2, 1). Consider the action of θ on a tabloid: 2 3 2 1 1 {s} = 1 5 7−→ T = 3 2 4 Also deﬁne an inverse map

−1 µ θ : C[Tλ,µ] −→ M T 7−→ {s},

µ where s has i in the row corresponding to T (i). This establishes a bijection M ↔ C[Tλ,µ]. We want θ to be an Sn-homomorphism. Suppose θ({s}) = T and take π ∈ Sn. Then (πT )(i) is the row number of i in π{s}, which by the action of Sn on tabloids is equal to the row number of π−1(i) in {s}. This is precisely equal to T (π−1i), so we see that θ is a homomorphism of Sn-modules. We can define an action of Sn on generalized λ-tableaux by having each element simply −1 permute the boxes themselves. This makes θ an Sn-homomorphism as well, so θ is an isomorphism. Thus to compute Kλµ, it will be sufficient to compute a basis for C[Tλ,µ]. For each T ∈ Tλ,µ, define the homomorphism λ µ θT : M −→ M = C[Tλ,µ] X {t} 7−→ S, S∈{T }

λ where {T } is the orbit of T under the action of Rt on generalized tableaux. Note that M is cyclically generated by {t} so the image of {t} determines the whole map.

45 µ 3.3 The Decomposition of M 3 Representations of Sn

1 2 3 2 1 1 (3,2) Example 3.3.4. As above, take t = and T = . Then θT : M → 4 5 3 2 M (2,2,1) has the following images:

2 1 1 1 2 1 1 1 2 2 1 1 1 2 1 1 1 2 θT ({t}) = + + + + + 3 2 3 2 3 2 2 3 2 3 2 3

θT ((1 2 4){t}) = (1 2 4)θT ({t}) 3 2 1 3 1 1 3 1 2 2 2 1 2 1 1 2 1 2 = + + + + + 1 2 2 2 1 2 1 3 2 3 1 3

λ µ ¯ λ µ Given θ : M → M , denote by θT the restriction θT |Sλ : S → M , so that the following diagram commutes:

θT λ µ M M = C[Tλ,µ]

¯ θT Sλ

For a tableau t, we have

¯ X θT (et) = θT (et) = θT (Kt{t}) = KtθT ({t}) = Kt S. S∈{T }

The problem is that some of these terms are zero, so we need a way of knowing when this happens.

Lemma 3.3.5. If t is a λ-tableau and T ∈ Tλ,µ then KtT = 0 if and only if T has two equal elements in the same column.

Proof. ( =⇒ ) If KtT = 0 then X T + (sgn π)(πT ) = 0.

π∈Ctr{e}

So there must be a permutation π ∈ Ct with sgn π = −1 and πT = T . If we write π as a product of disjoint, nontrivial cycles, the elements of T that this cycle permutes must be equal. Also, π ∈ Ct so T has two equal elements in at least one of its columns. ( ⇒ = ) Suppose T (i) and T (j) are in the same column and equal. Then (e − (i j))T = 0 and (i j) ∈ Ct so Kt = g(e − (i j)). This implies that KtT = 0. ¯ P Lemma 3.3.5 allows us to reduce the expression θT (et) = Kt S∈{T } S to the case where the rows of T are weakly increasing – this only depends on {T } – and the sum is over all generalized tableaux in {T } that don’t have a repeated column entry anywhere. Multiplying by an appropriate element of Ct, we may also assume the columns of T weakly increase, but then by Lemma 3.3.5 these will be strictly increasing.

46 µ 3.3 The Decomposition of M 3 Representations of Sn

Deﬁnition. A generalized tableau T is semi-standard if it is weakly increasing across rows and strictly increasing down columns. The set of all semi-standard Young tableaux of shape 0 λ and content µ is denoted Tλ,µ. ¯ 0 λ µ Theorem 3.3.6. The set {θT | T ∈ Tλ,µ} forms a basis of Hom(S ,M ). In particular, Kλµ is equal to the number of semi-standard Young tableaux of shape λ and content µ. Proof. This closely follows the proof for Sλ using Young tableaux (not generalized).

(3,2,1) Example 3.3.7. In S6, let’s decompose M into Specht modules and determine the Kostka numbers; write (3,2,1) M λ M = Kλ(3,2,1)S . λD(3,2,1) (3,2,1) We’ve shown that K(3,2,1)(3,2,1) = 1 so S shows up once in this decomposition. Using the Hasse diagram for partitions of n = 6 from Example 3.1.8, we list the partitions (up to duality) and the semi-standard generalized tableaux to the right.

λ semi-std. gen. tab. Kλ(3,2,1)

1 1 1 (3, 3) 1 2 2 3

1 1 1 2 (4, 1, 1) 2 1 3

1 1 1 2 1 1 1 3 (4, 2) 2 2 3 2 2

1 1 1 2 3 1 1 1 2 2 (5, 1) 2 2 3

(6) 1 1 1 2 2 3 1

Therefore M (3,2,1) = S(3,2,1) ⊕ S(3,3) ⊕ S(4,1,1) ⊕ (S(4,2) ⊕ S(4,2)) ⊕ (S(5,1) ⊕ S(5,1)) ⊕ S(6). n (1n) ∼ L λ λ Example 3.3.8. For λ = (1 ) = (1, 1,..., 1), we know M = C[Sn] = λ`n(dim S )S . So the Kostka numbers count the number of standard Young tableaux of each shape par- λ titioning n, since Kλ(1n) = f for each λ ` n. This makes sense, since any semi-standard tableau of content (1n) is standard. Thus the remaining piece of the puzzle required in fully describing the representation λ theory of Sn is to ﬁnd a formula for f , the number of standard Young tableaux of shape λ. In the next section we describe such a formula.

47 3.4 The Hook Length Formula 3 Representations of Sn

3.4 The Hook Length Formula

In this section we complete the description of the representation theory of Sn by describing the hook formula for f λ, the number of standard tableaux of shape λ, which in turn allows us to calculate dim Sλ.

Deﬁnition. Let λ be a Ferrers diagram. If v = (i, j) is a node in the diagram, the hook of v is the set 0 0 0 0 0 0 Hv = {(i , j ) | i ≥ i and j = j, or j ≥ j and i = i}.

The size of this set, denoted h(i,j) = |H(i,j)|, is called the hook length of v = (i, j). Theorem 3.4.1 (Hook Length Formula; Frame-Robinson-Thrall). If λ is a partition of n then a formula for f λ is λ n! f = Q . (i,j)∈λ h(i,j)

λ Q Proof. (Sketch) We will prove that n! = f (i,j)∈λ h(i,j). On the left, n! is the number of all λ-tableaux, so the hook length formula is proven by constructing a bijection

standard arrays of shape λ {λ-tableaux} ←→ × λ-tableaux where (i, j) has h(i,j) values

t 7−→ (tstd,J),

where tstd is the standard tableaux in the equivalence class {t}, and J keeps track of the “slides” used to standardize t.

Example 3.4.2. For n = 5 and λ = (2, 2, 1), we have written the Ferrers diagram corresponding to λ with the hook lengths ﬁlled in at each (i, j), and computed the hook length formula to the right:

4 2 5! 3 1 f λ = = 5. 4 · 3 · 2 · 1 · 1 1

Thus there are 5 standard λ-tableaux, and dim Sλ = 5. Since n is rather small in this example, it’s easy to write down all the standard tableaux of shape (2, 2, 1):

1 4 1 3 1 3 1 2 1 2 2 5 2 5 2 4 3 5 3 4 3 4 5 4 5

48 3.5 Application: The RSK Algorithm 3 Representations of Sn

3.5 Application: The RSK Algorithm

Given a permutation π ∈ Sn, combinatorialists are often interested in ﬁnding the longest increasing subsequence of π.

Example 3.5.1. Let π = (6 2 5 3 1 4 8 7) ∈ S8. Examples of subsequences of π include (6 5 4), (6 1 8 7), etc. The subsequence (2 4) is increasing, but we could make it longer by adding 8 to obtain (2 4 8). This cannot be made into a longer increasing subsequence. In fact there are two subsequences of length 4: (2 3 4 8) and (2 3 4 7), and these are the longest possible.

The RSK algorithm gives us a quick way to answer this combinatorial question, and many more.

Theorem 3.5.2 (Robinson-Schensted-Knuth). There exists a bijection between matrices with nonnegative integer entries, with ﬁnitely many nonzero entries, and pairs of semi- standard Young tableaux of the same shape.

The Robinson-Schensted (RS) correspondence describes this relation for permutation matrices and standard Young tableaux. This correspondence is part of a three-step process, by which one takes a matrix, constructs a permutation array and then generates a tableaux pair:    PQ  a b c ··· −→ −→   x y z ···   matrix permutation array tableaux pair

Deﬁnition. Schensted insertion is a process to insert a number into a semi-standard Young tableau. Speciﬁcally, to insert x into a semi-standard tableau T ,

(1) Let x0 = x. Find the smallest number in the ﬁrst row that’s greater than x0 and call this entry x1. Replace x1 with x0.

(2) Find the smallest number in the second row bigger than x1 and call it x2. Replace x2 with x1.

(3) Continue recursively deﬁning xi and replacing xi−1 with xi.

(4) When xi is at least as large as every element in row i − 1, place xi at the end of the row and stop.

1 1 3 4 8

Example 3.5.3. Let T = 2 5 6 . To insert x0 = 3, we obtain the sequence 6 8

49 3.5 Application: The RSK Algorithm 3 Representations of Sn

x0 = 3, x1 = 4, x2 = 5, x3 = 6 and stop with x3. This produces the following tableau:

1 1 3 3 8 2 4 6 5 8 6

On the other hand, inserting x0 = 9 yields

1 1 3 4 8 9 2 5 6 6 8

Schensted insertion allows us to prove multiplication rules for Schur functions (Sec- tion 4.3) like the Littlewood-Richardson rule. The content of the RSK theorem above is really the RSK algorithm, which associates a permutation to a pair (P,Q) of semi-standard Young tableaux of the same shape.

RSK Algorithm. Let π ∈ Sn. (1) Insert the ﬁrst letter of the bottom row of π (in two-line notation) into an empty Ferrers diagram using Schensted insertion. This tableau will be P .

(2) Record where the new entry was placed in P using the recording tableau Q, placing the corresponding letter from the top row of π in the entry of Q where Schensted insertion places the bottom entry in P .

(3) Repeat, inserting each letter one by one from the bottom row of π into P and recording the top row entry in Q.

Deﬁnition. Given integers x, y, z, if x < y < z we say yxz ∼ yzx and xzy ∼ zxy, called Knuth equivalence.

Theorem 3.5.4. Two permutations π and σ are Knuth equivalent if and only if their insertion tableaux are the same, i.e. π 7→ (P,Q) and σ 7→ (P,R).

Proposition 3.5.5. Given a permutation π ∈ Sn and its corresponding tableaux pair (P,Q), the length of the longest increasing subsequence of π is λ1, where λ = (λ1, λ2, . . . , λ`) is the shape of P and Q.

The following is a corollary to the RSK theorem.

P λ 2 Corollary 3.5.6. For an integer n ≥ 1, n! = λ`n(f ) . Proof. This follows easily from RSK when we notice that n! is the number of permutations of n elements and for each λ ` n,(f λ)2 is the number of pairs of semi-standard Young tableaux of shape λ.

50 3.5 Application: The RSK Algorithm 3 Representations of Sn

1 2 3 4 5 6 7 8 Example 3.5.7. Consider the permutation π = . A few steps of 6 2 5 3 1 4 8 7 the RSK algorithm for π are shown below. P (insertion) Q (recording)

6 1

2 1 6 2

2 1 3 5 2 6

. . . .

1 3 4 7 1 3 6 7 2 8 2 8 5 4 6 5 Then the shape of P and Q is (4, 2, 1, 1) so by Proposition 3.5.5, this tells us that the longest increasing subsequence of π has length 4, as we saw at the beginning of the section. Example 3.5.8. What happens when you apply the RSK algorithm to the inverse of a given 1 2 3 4 5 6 7 8 permutation? For the same π as in the above example, we have π−1 = . 5 2 4 6 3 1 8 7 The RSK algorithm yields: 1 3 6 7 1 3 4 7 2 8 2 8 P = and Q = 4 5 5 6 So it appears that if π 7→ (P,Q) then π−1 7→ (Q, P ).

Proposition 3.5.9. If, under the correspondence in the RSK theorem, π ∈ Sn corresponds to (P,Q) then π−1 corresponds to (Q, P ). Corollary 3.5.10. The set of symmetric matrices is in bijective correspondence with semi- standard Young tableaux. Proof. Suppose A = AT . Form the permutation array σ from A, which must satisfy σ−1 = σ. Then by Proposition 3.5.5, the pair (P,Q) corresponding to σ must satisfy (P,Q) = (Q, P ), i.e. Q = P . Thus the correspondence may be written A ↔ P .

51 4 Symmetric Functions

4 Symmetric Functions

4.1 Generating Functions

Deﬁnition. Given a set S, a set I and a function

wt : S −→ C[[{xi}i∈I ]] into the ring of formal complex power series in the variables {xi}i∈I , the generating function for this data is the formal sum X fS(x) = wt(s). s∈S Example 4.1.1. Let S be the set of all partitions and let I = {1} just be a singleton. Deﬁne a weight function

wt : S −→ C[[x]] λ 7−→ x|λ| where |λ| = λ1 + ... + λ`. Then the corresponding generating function is

X |λ| X n fS(x) = x = p(n)x λ n≥0 where p(n) is the number of partitions of n. Consider the product formula (due to Euler):

1 1+1 1+1+1 2 2+2 3 3+3 fS(x) = (1 + x + x + x + ...)(1 + x + x + ...)(1 + x + x + ...) ··· 1 1 1 = ··· 1 − x 1 − x2 1 − x3 Y 1 = . 1 − xn n≥1

This infinite product converges to fS(x) in the following sense. Q Definition. A formal product of functions i≥1 fi(x) converges to f(x) if for every n ≥ 1, there exist an N ≥ 1 such that the coefficient of xn in f(x), denoted [xn]f(x), is equal to its counterpart in the product truncated at N, i.e.

N ! n n Y [x ]f(x) = [x ] fi(x) . i=1

Example 4.1.2. Let S0 be the set of partitions λ with only odd parts. For example, (3, 3, 1) |λ| is an element of S0. As in Example 4.1.1, let wt : S0 → C[[x]] be the function wt(λ) = x . Then this time the generating function is X Y 1 f (x) = p (n)xn = . S0 0 1 − x2k−1 n≥0 k≥1 n odd

52 4.2 Symmetric Functions 4 Symmetric Functions

Example 4.1.3. Let Sd be the set of partitions λ with distinct parts, e.g. (7, 4, 2, 1) but not (3, 3, 1). Again let wt(λ) = x|λ|. Then

X n Y n fSd (x) = pd(n)x = (1 + x ). n≥0 n≥1

Theorem 4.1.4 (Euler). For all n, p0(n) = pd(n). That is, the number of odd partitions of n is equal to the number of partitions of n into distinct parts.

Proof. It suﬃces to show the generating functions for S0 and Sd are equal. Consider

Y Y (1 + xn)(1 − xn) f (x) = (1 + xn) = Sd 1 − xn n≥1 n≥1 Y 1 − x2n Y 1 = = 1 − xn 1 − x2n−1 n≥1 n≥1 X n = p0(n)x = fS0 (x). n≥0

Therefore p0(n) = pd(n) for every n.

4.2 Symmetric Functions

Denote byx = {x1, x2, x3,...} a set of variables indexed by the natural numbers. Then C[[x]] will denote the set of formal complex power series in these variables. Note that for any n, the symmetric group Sn acts on the ring C[[x]] via σ · xi = xσ(i) for the components with i ≤ n and σ · xi = xi if i > n. Usually, when a group G acts on a ring R, an important object of study will be RG, the set of ﬁxed points of the action. In this case, we build up such a set degree-by-degree.

Deﬁnition. We say a monomial xλ1 ··· xλ` in [[x]] has degree d = P` λ . A polynomial i1 i` C i=1 ` f ∈ C[[x]] is called homogeneous of degree d if every monomial in f has degree d.

Let C[[x]]d denote the C-vector space of all homogeneous elements of C[[x]] of degree d. Then Sn acts on C[[x]]d for all n, d ≥ 1. Deﬁne

d Λ = {f ∈ C[[x]]d | σ · f = f for all σ ∈ Sn, n ≥ 1}.

Alternatively, Λd can be deﬁned as the C-span of all monomials X m = xλ1 ··· xλ` λ i1 i` I⊆N |I|=`(λ) where λ ` d, `(λ) is the length of λ and I are the ordered subsets of N.

Deﬁnition. mλ is called a monomial symmetric function.

53 4.2 Symmetric Functions 4 Symmetric Functions

Example 4.2.1. For λ = (2, 1),

2 2 2 2 2 2 m(2,1) = x1x2 + x1x2 + x1x3 + x1x3 + x2x3 + x2x3 + ...

Note that {1, 2} is considered to be distinct from the subset {2, 1} (since they are ordered), 2 2 so x1x2 and x1x2 show up separately.

d Lemma 4.2.2. For all d, Λ = SpanC{mλ | λ ` d}. d Proof. Let L = SpanC{mλ | λ ` d}. Clearly Sn fixes L for every n, so L ⊆ Λ . On the other hand, take f ∈ Λd. Then every term of f has degree at most d. Suppose xλ1 ··· xλ` is a i1 i` monomial appearing in f with coefficient c, such that λ1 +...+λ` = d. Then for any σ ∈ Sn, λ1 λ` the coefficient of x ··· x must also be c. This means we can write f = cmλ + g where σ(i1) σ(i`) the sorted exponent vectors of monomials appearing in g do not contain λ. Since there are finitely many partitions λ ` d, we know that f is the finite sum X f = cλmλ λ`d

where cλ is the coefficient corresponding to λ. Hence f ∈ L and we are done. L Definition. A graded ring is a ring R with decomposition R = i∈I Ri such that RiRj ⊆ Ri+j for all i, j ∈ I. Definition. The ring of symmetric functions is the graded ring

M d Λ = Λ = SpanC{mλ | λ is a partition}. d≥0 P Remark. The infinite sum λ mλ over all partitions λ is fixed by Sn for all n, but it does not belong to Λ since mλ 6= 0 for infinitely many λ. Informally, symmetric functions are the finite polynomials fixed by every symmetric group.

Lemma 4.2.3. For each d ≥ 0, dim Λd = p(d), the number of partitions of d.

This is immediate from Lemma 4.2.2. We next describe some other bases of Λd.

Deﬁnition. For n ≥ 1, deﬁne:

n n n pn = m(n) = x1 + x2 + x3 + ..., the nth power sum. X n en = m(1 ) = xi1 ··· xin , the nth elementary symmetric function.

i1<···

2 2 Example 4.2.4. e3 = x1x2x3+x1x2x4+x2x3x4+... and h2 = x1+x1x2+x2+x1x3+x2x3+...

54 4.2 Symmetric Functions 4 Symmetric Functions

` Y Deﬁnition. For a partition λ ` n and for each f ∈ {p, e, h}, deﬁne fλ = fλ` . i=1 Theorem 4.2.5. The following are bases of Λn for each n:

(1) {mλ | λ ` n}

(2) {pλ | λ ` n}

(3) {eλ | λ ` n}

(4) {hλ | λ ` n}. Proof. (1) was proven in Lemma 4.2.2 and we will prove (4) at the end of the section. (2) Note that for any λ ` n,

2 λ1 λ1 λ λ2 λ` λ` pλ = (x1 + x2 + ...)(x1 + x2 + ...)(··· (x1 + x2 + ...).

µ1 µm If x1 ··· xm appears in pλ for some µ ` n, then µ is a “coarsening” of λ in the sense that µ D λ. Then X X pλ = cµλmµ = cλλmλ + cµλmµ µDλ µ.λ

and cλλ 6= 0 (since it is equal to the factorial of the content of λ). Thus we can order the partitions lexicographically, which implies that the change-of-basis matrix between {mλ} and {pλ} is upper (or lower) triangular with all diagonal entries nonzero. Hence {pλ} is a basis. 0 0 0 (3) Let λ = (λ1, . . . , λ`) and let λ = (λ1, . . . , λ`) be the transpose of λ. We claim that X eλ0 = mλ + dλµmµ. µ.λ

Lexicographically, the largest term in eλ is the lead term, which is the product of the lead terms of each e 0 : λi

λ1 λ` (x ··· x 0 )(x ··· x 0 ) ··· (x ··· x 0 ) = x ··· x . 1 λ` 1 λ2 1 λ` 1 ` If we take any other term, it’s like moving the “bits” in the partition to later parts of the partition. Clearly (λ1, . . . , λ`) . (λ1, . . . , λi − 1, . . . , λj + 1, . . . , λ`), proving the claim. So d the eλ0 are a basis for Λ as well, and we can just list each λ as the transpose of another partition, showing {eλ} is a basis.

Example 4.2.6. Let n ≥ 1, set Sn = {λ | λ is a partition with `(λ) = n} and deﬁne a weight function wt : Sn → C[[x]] by λ 7→ xλ1 xλ2 ··· xλn . Then the generating function for Sn is X X fSn (x) = wt(s) = xλ1 xλ2 ··· xλn = en(x).

s∈Sn λ∈Sn

55 4.2 Symmetric Functions 4 Symmetric Functions

Example 4.2.7. Let Sd be the set of partitions with distinct parts, as in Example 4.1.3, n and deﬁne wt : Sd → C[[x, t]] by λxλ1 xλ2 ··· xλn t , where n = `(λ). Then the generating function for this data is X n E(x, t) := fSd (x, t) = en(x)t . n≥0 Example 4.2.8. Similarly, for the set S of all partitions and the weight function wt : S → C[[x, t]] given by λ 7→ xc(λ)t`(λ), where c(λ) is the content of λ, the generating function is

X n H(x, t) := fS(x, t) = hn(x)t . n≥0

Y Y 1 Proposition 4.2.9. E(x, t) = (1 + x t) and H(x, t) = . i 1 − x t i≥1 i≥1 i Proof. Note that these products converge in the sense of Section 4.1. The coefficient of tn in (1 + x1t)(1 + x2t) ··· is en(x) because a term in the coefficient is determined by a choice of the second term in n of the factors in this infinite product. Similarly, by geometric series, n 2 2 2 2 the coefficient of t in (1 + x1t + x1t + ...)(1 + x2t + x2t + ...) ··· has terms that are determined by a choice of partition of length n and the weights of such partitions give all monomials in hn(x). Remark. A useful property of the E and H functions is:

Y Y 1 E(x, t)H(x, −t) = (1 + x t) = 1. i 1 + x t i≥1 i≥1 i

This is a manifestation of Koszul duality in commutative algebra. X tn Proposition 4.2.10. p (x) = ln H(x, t). n n n≥1

Proof. By Proposition 4.2.9, we can write H(x, t) = Q 1 . Then i≥1 1−xit

Y 1 X 1 ln = ln 1 − x t 1 − x t i≥1 i i≥1 i n n X X (xit) X X (xit) = = n n i≥1 n≥1 n≥1 i≥1 X tn X X tn = xn = p (x) . n i n n n≥1 i≥1 n≥1

Now we can prove (4) of Theorem 4.2.5.

56 4.3 Schur Functions 4 Symmetric Functions

Proof. By the above remark, E(x, t)H(x, −t) = 1 and equating the coeﬃcients of tn in this equality gives n X n−r (−1) hn−r(x)er(x) = 0. r=0 n d So en = h1en−1 − h2en−2 + ... + (−1) hn. We prove that {hλ} is a basis of Λ by induction. The base case is e0 = h0. Now assume n ≥ 1. Then by induction, we can write ej for j < n as a linear combination of the hλ for λ ` j. So en ∈ Span{hλ} as desired. Moreover, since n #{hλ | λ ` n} = dim Λ , we are done.

4.3 Schur Functions

µ µ1 µ2 µ` Given a composition µ = (µ1, µ2, . . . , µ`), we will writex = x1 x2 ··· x` . Also, for a generalized tableau T , write

T Y X x = xTi,j and |T | = Ti,j. (i,j)∈T (i,j)∈T

4 1 1 T 2 Example 4.3.1. For T = we havex = x x2x3x4 and |T | = 11. 3 2 1

In this section, we will consider the set S of semistandard generalized tableaux T of shape λ. For this set, deﬁne a weight function wt : S → C[[x, t]] mapping T 7→ xT t|T |. This has generating function X n fS(x, t) = ssλ(n)t n≥0

where ssλ(n) is the generating function of semistandard generalized λ-tableaux of total weight n.

Deﬁnition. The Schur function corresponding to a partition λ is sλ(x) = fS(x, 1). Example 4.3.2. Let λ = (2, 1). The ﬁrst few semistandard generalized Young tableaux of shape (2, 1) are:

1 1 1 2 1 1 1 3 1 2 1 3 2 3 2 2 ··· 2 2 3 3 3 2 3 3

2 2 2 2 2 2 Then s(2,1)(x) = x1x2 + x1x2 + x1x3 + x1x3 + 2x1x2x3 + x2x3 + x2x3 + .... Notice that this is our ﬁrst example of a symmetric function with coeﬃcients other than 1. Example 4.3.3. For λ = (n), the semistandard generalized tableaux all have the form

| {z } n

Thus s(n)(x) = hn(x).

57 4.3 Schur Functions 4 Symmetric Functions

n Example 4.3.4. For λ = (1 ), s(1n)(x) = en(x).

Note that for any λ, the x1 ··· xn coeﬃcient in sλ(x) is equal to the number of standard λ-tableaux, that is, fλ.

Proposition 4.3.5. For any partition λ, the Schur function sλ(x) is a symmetric function.

P 0 µ Proof. By definition, sλ(x) = µ |Tλµ|x , where the sum is over all compositions µ and 0 Tλµ is equal to the set of semistandard Young tableaux of shape λ and content µ. (Note 0 that when µ is a partition, |Tλµ| = Kλµ, the Kostka number for λ, µ.) It’s sufficient to show 0 0 µ |Tλµ| = |Tλµ˜| for any rearrangementµ ˜ of µ. Going back to the definition of M in Section 2.2, we can define this module on compositions as well, and as in Theorem 3.3.6, the formula λ µ 0 µ ∼ µ˜ dim Hom(S ,M ) = |Tλµ| holds even when µ is a composition. Obviously, M = M for any rearrangementµ ˜ of µ, so their dimensions are equal as required.

Corollary 4.3.6. For any partition λ, X sλ(x) = Kλµmµ. partitions µ

n Theorem 4.3.7. The set {sλ | λ ` n} is a basis of Λ . P Proof. By Corollary 4.3.6, sλ(x) = µ Kλµmµ. Moreover, it follows from Proposition 3.2.10 that Kλλ = 1 and Kλµ = 0 if λ D6 µ. So X sλ(x) = Kλµmµ λDµ which implies the change-of-basis matrix from {mλ} to {sλ} is (upper or lower) triangular n with nonzero diagonal entries and hence {sλ} is a basis of Λ .

n The above proof shows that {sλ} is a basis of Λ over Z, which is even stronger than being a C-basis.

Theorem 4.3.8 (Jacobi-Trudi Determinants). Let λ = (λ1, . . . , λ`). Then sλ(x) = det(hλi−i+j) 0 0 and if λ is the transpose of λ, then sλ (x) = det(eλi−i+j). Example 4.3.9. For λ = (2, 1), we have

hλ1−1+1 hλ1−1+2 h2 h3 s(2,1)(x) = = = h1h2 − h3 hλ2−2+1 hλ2−2+2 h0 h1 2 2 3 2 = (x1 + x2 + x3 + ...)(x1 + x1x2 + x2 + x1x3 + x2x3 + ...) − (x1 + x1x2 + ...) ··· 2 2 = x1x2 + x1x2 + ... + 2x1x2x3 + ...

= m(2,1) + 2m(1,1,1).

Compare this to the computation in Example 4.2.1.

58 4.3 Schur Functions 4 Symmetric Functions

For ` ≥ 1, deﬁne Λ` to be the subspace of Λ of symmetric polynomials in ` variables. Then

Λ` = {f(x1, . . . , x`, 0, 0,...) | f ∈ Λ}

S` = C[x1, . . . , x`] = {g ∈ C[x1, . . . , x`] | σ · g = g for all σ ∈ S`}.

Definition. A polynomial f ∈ C[x1, . . . , x`] is called skew-symmetric if σ · f = (sgn σ)f for all σ ∈ S`. Definition. Given a composition µ of length `, define the alternant of µ to be

X µ aµ(x1, . . . , x`) = (sgn π)π · x .

π∈S` More generally, given a ﬁnite group G and a one-dimensional character χ of G, we can deﬁne the weighted Reynolds operator

χ ρ : C[x1, . . . , x`] −→ C[x1, . . . , x`] 1 X f 7−→ χ(g−1)g · f. |G| g∈G

χ The key point of this operator is that im ρ = {f ∈ C[x1, . . . , x`] | π·f = χ(π)f}. This image is called the χ-weight submodule of C[x1, . . . , x`]. In this context, we can write an alternant χ µ as aµ(x1, . . . , x`) = n!ρ (x ) where χ is the sign representation. Note that the alternant may also be realized as a determinant:

µj aµ(x1, . . . , x`) = det(xi ).

Example 4.3.10. For λ = (4, 2, 1), we have

4 2 4 2 2 4 2 4 2 4 4 2 a(4,2,1)(x1, x2, x3) = x1x2x3 + x1x2x3 + x1x2x3 − x1x2x3 − x1x2x3 − x1x2x3 4 2 x1 x1 x1 4 2 = x2 x2 x2 . 4 2 x3 x3 x3

Example 4.3.11. For any ` ≥ 1, set δ = (` − 1, ` − 2,..., 2, 1, 0),

`−1 `−2 2 x1 x1 ··· x1 x1 1 `−1 `−2 2 x2 x2 ··· x2 x2 1 Y aδ = ...... = (xi − xj)...... i

(This may be recognizable as a Vandermonde determinant.)

59 4.4 Symmetric Functions and Character Representations 4 Symmetric Functions

Remark. If µ has a repeated part, then aµ = 0 because there are two columns of the corresponding matrix that are equal. However, if λ is a partition and δ is as above, the componentwise sum λ + δ is a partition with no repeated parts so we can look at aλ+δ. Also, xi − xj is a factor of aλ+δ for all λ since in the quotient C[x1, . . . , x`]/(xi − xj), this matrix has two equal rows. By this, we see that aλ+δ is a symmetric polynomial. aδ

aλ+δ Theorem 4.3.12. For any λ, sλ = . aδ

Lemma 4.3.13. Let µ = (µ1, . . . , µ`) be a composition and consider the ` × ` matrices

µi `−i (j) Aµ = (xj ) Hµ = (hµi−`+j) E = ((−1) e`−i)

(j) where all entries are polynomials in ` variables and en is the elementary symmetric function in x1, . . . , xj−1, xj+1, . . . , xn. Then Aµ = HµE. Proof. Define (j) X (j) n Y E (x, t) = en (x)t = (1 + xit) n≥0 i6=j (the second equality comes from Proposition 4.2.9). Also by Proposition 4.2.9, we know P n Q 1 (j) H(x, t) = hn(x)t = and by the remark following that proposition, H(x, t)E (x, −t) = n≥0 i≥j 1−xit µ 1 . Consider the coefficient of tµi on each side. On the right, it is equal to x i by geometric 1−xj t j series, whereas on the left, the coefficient is

` X `−k (j) hµi−`+k(−1) e`−k. k=1

µi Now compare the ijth entries of Aµ and HµE. For Aµ it’s xj but this is equal to the above sum, while for HµE, the entry is the sum of the products of the ikth entries of Hµ and the P` `−k (j) kjth entries of E, which is precisely k=1 hµi−`+k(−1) e`−k. Hence the matrices agree in every entry so they are equal.

4.4 Symmetric Functions and Character Representations

µ µ Theorem 4.4.1. If φλ is the value of the character of M on the conjugacy class corresponding to λ, then X µ pλ = φλmµ. µDλ

Proof. Let λ = (λ1, . . . , λ`). Then

` Y λi λi X pλ = (x1 + x2 + ...) = cλµmµ. i=1 µDλ

µ µ1 µm We prove the theorem by comparing the coeﬃcients ofx = x1 ··· xm on each side. On the right, it’s cλµ. On the left, it’s the number of ways to distribute the parts of λ into

60 4.4 Symmetric Functions and Character Representations 4 Symmetric Functions

subpartitions λ1, . . . , λm such that S λi = λ and λi ` µi for all i, where we consider parts of equal size as distinct for counting purposes. For any π ∈ Sn of cycle type λ, we have

µ µ φλ = φ (π) = #{{t} | t is a µ-tableau and π{t} = {t}}.

If π = π1π2 ··· π` where the πi are disjoint cycles of length λi, then this is the same distri- µ bution as above. Hence cλµ = φλ.

Example 4.4.2. To demonstrate this, we compute the pλ for partitions of n = 3:

3 3 p(3) = x1 + x2 + ... = m(3) 2 2 p(2,1) = (x1 + x2 + ...)(x1 + x2 + ...) 3 2 2 = (x1 + x2 + ...) + (x1x2 + x1x2 + ...)

= m(3) + m(2,1) 3 p(1,1,1) = (x1 + x2 + ...) 3 3 2 2 = (x1 + x2 + ...) + (3x1x2 + 3x1x3 + ...) + (6x1x2x3 + ...)

= m(3) + 3m(2,1) + 6m(1,1,1).

Putting these in a table,

p(3) p(2,1) p(1,1,1) m(3) 1 1 1 m(2,1) 0 1 3 m(1,1,1) 0 0 6 we are also able to compute the decomposition of M µ for µ ` 3 as we would have been able to in Section 3.3: (3) (2, 1) (1, 1, 1) M (3) 1 1 1 M (2,1) 0 1 3 M (1,1,1) 0 0 6 P µ P We have shown that pλ = µ φλmµ and sλ = µ Kλµmµ, but the Kostka numbers form an invertible matrix, so there are inverse Kostka numbers Keλµ such that X mµ = Keλµsλ. λ Then we can write ! X ν X ν X X X ν pµ = φµmν = φµ Keλνsλ = Keλνφµsλ. ν ν λ λ ν

n n So the change-of-basis matrix (Λ , {pµ}) → (Λ , {sλ}) is given by KAe , where Ke is the matrix µ of inverse Kostka numbers and A = (φλ)µ,λ.

61 4.4 Symmetric Functions and Character Representations 4 Symmetric Functions

Example 4.4.3. For n = 3, we have

1 1 1 1 0 0 A = 0 1 3 and Kλµ = 1 1 0 0 0 6 1 2 1

from Example 4.4.2, so the inverse Kostka matrix is

 1 0 0 Keλµ = −1 1 0 1 −2 1 and therefore the change-of-basis matrix from {pλ} to {sλ} is

 1 1 1 KeλµA = −1 0 2 . 1 −1 1

Notice that this is precisely the character table for S3 that we computed in Example 3 of Section 2.5.