1 Construction of Haar Measure

Deﬁnition 1.1. A family G of linear transformations on a linear topological space X is said to be equicontinuous on a subset K of X if for every neighborhood V of the origin in X there is a neighborhood U of the origin such that the following condition holds

if k1, k2 ∈ K and k1 − k2 ∈ U, then G(k1 − k2) ⊆ V that is T (k1 − k2) ∈ V for all T ∈ G.

1 Theorem 1.2 (Kakutani). Let K be a compact, convex subset of a locally convex linear topological space X, and let G be a group of linear mappings which is equicontinuous on K and such that G(K) ⊆ K. Then there exists a point p ∈ K such that

T (p) = p ∀T ∈ G

Proof. By Zorn’s lemma, K contains a minimal non-void compact convex subset K1 such that G(K1) ⊆ K1. If K1 contains just one point then the proof is complete. If this is not the case, the compact set

K1 − K1 contains some point other than the origin.

2 Thus, there exists a neighborhood V of the origin such that

V¯ 6⊇ K1 − K1.

There is a convex neighborhood V1 of the origin such that αV1 ⊆ V for |α| ≤ 1.

By the equicontinuity of G on the set K1, there is a neighborhood U1 of the origin such that if k1, k2 ∈ K1 and k1 − k2 ∈ U1 then

G(k1 − k2) ⊆ V1.

3 Because each T ∈ G is invertible, T maps open sets to open sets (open mapping theorem) and T (A ∩ B) = TA ∩ TB for any sets A, B. Since T is linear,

T convex-hull(A) = convex-hullT (A) for any set A. Because G is a group, G(GA) = GA for any set A.

4 Thus

U2 := convex-hull(GU1 ∩ (K1 − K1))

= convex-hull(G(U1 ∩ (K1 − K1))) ⊆ V1 is relatively open in K1 − K1 and satisﬁes GU2 = U2 6⊇ K1 − K1. By continuity, GU¯2 = U¯2. Deﬁne

∞ > δ := inf{a : a > 0, aU2 ⊇ K1 − K1} ≥ 1 and U := δU2. For each 0 < < 1,

(1 + )U ⊇ K1 − K1 6⊆ (1 − )U.¯

5 −1 The family of relatively open sets {2 U + k}, k ∈ K1, is a covering of −1 −1 K1. Let {2 U + k1,..., 2 U + kn} be a ﬁnite sub-covering and let −1 p = (k1 + . . . kn)/n. If k is any point in K1, then ki − k ∈ 2 U for some 1 ≤ i ≤ n. Since ki − k ∈ (1 + )U for all i and all > 0, we have 1 p ∈ 2−1U + (n − 1) · (1 + )U + k. n 1 1 For = 4(n−1) , we have p ∈ (1 − 4n )U + k for each k ∈ K1. Let

\ 1 K = K ∩ (1 − )U¯ + k 6= ∅. 2 1 4n k∈K1

6 1 ¯ Because (1 − 4n )U 6⊇ K1 − K1, we have K2 6= K1. The closed set K2 is clearly convex. Further since T (aU¯) ⊆ aU¯ for T ∈ G, we have

T (aU¯ + k) ⊆ aU¯ + T k for all T ∈ G, k ∈ K1.

Recalling TK1 = K1 for T ∈ G, we ﬁnd that GK2 ⊆ K2, which contradicts the minimality of K1.

7 Theorem 1.3 (Haar Measure). Let G be a compact group. Let C(G) be the space of continuous maps from G to C. Then, there is a unique linear form m : C(G) −→ C having the following properties: 1. m(f) ≥ 0 for f ≥ 0 (m is positive). 2. m(11)= 1 (m is normalized).

3. m(sf) = m(f) where sf is deﬁned as the function

−1 sf(g) = f(s g) s, g ∈ G

(m is left invariant).

4. m(fs) = m(f) where fs(g) = f(gs) for s, g ∈ G (m is right invariant).

8 Proof. For f ∈ C(G), let Cf denote the convex hull of all left translates of f. The elements of Cf are ﬁnite sums of the form: X X g(x) = aif(six) ai > 0, ai = 1 finite finite Clearly ||g|| = max{|g(x)| : x ∈ G} ≤ ||f||

Thus all sets Cf (x) = {g(x): g ∈ Cf } are bounded and relatively compact in C. Since G is compact, f is uniformly continuous, namely for all > 0, ∃ a neighborhood V = V of the identity element e ∈ G such that: y−1x ∈ V ⇒ |f(x) − f(y)| <

9 Since (s−1y)−1s−1x = y−1x, we also have

−1 |sf(y) − sf(x)| < whenever y x ∈ V

Since the functions g are convex combinations of functions of the form sf, |g(y) − g(x)| < whenever y−1x ∈ V

Thus the set Cf is equicontinuous. By Ascoli’s theorem Cf is relatively compact in C(G). Define the compact convex set Kf = C¯f in C(G). The compact group G acts by left translations (isometrically) on C(G) and leaves Cf and hence Kf invariant. By Kakutani’s Theorem 1.2, there is a fixed point g of this action G in Kf . Such a fixed point satisfies by definition

−1 sg = g (∀s ∈ G) ⇒ g(s ) = sg(e) = c (∀s ∈ G) for some constant c.

10 By the deﬁnition of the set Kf , given any > 0 there exists a ﬁnite set

{s1, . . . sn} in G and ai > 0 such that

n n X X ai = 1 and c − aif(six) < (∀x ∈ G) (1.1) 1 1

We ﬁrst show that there is only one constant function Kf . Start the same construction as above, only now using right translations of f (e.g. we can apply the preceding construction to the opposite group G0 of G , 0 −1 0 or the function f = f(x )), obtaining a relatively compact set Cf with 0 0 compact convex closure Kf containing a constant function c . It will be 0 enough to show c = c .(all constants c in Kf must be equal to one 0 0 chosen constant c of Kf and conversely.)

11 There is certainly a ﬁnite combination of right translates which is close to c0 namely

0 X X |c − bjf(xtj)| < ( for some tj ∈ G, bj > 0 with bj = 1)

Let us multiply this inequality by ai and put x = si to get

0 X |c ai − aibjf(sitj)| < ai (1.2)

Summing over i, we obtain

0 X X X |c ai − aibjf(sitj)| < ai = (1.3) i,j

12 Operating symmetrically on Equation (1.1) (multiplying by bj, putting x = tj and summing over j), we ﬁnd: X |c − aibjf(sitj)| < (1.4) i,j Subtracting (or adding) Equation (1.3) from (1.4) we get |c − c0| < 2. Since was arbitrary this completes the proof.

From now on the constant c in Kf will be denoted by m(f). It is the only constant function which can be approximated arbitrarily close with convex combinations of left or right translates of f.

13 The following properties are obvious:

• m(11)= 1 since Kf = {1} if f = 1. • m(f) ≥ 0 if f ≥ 0.

• m(af) = am(f) for any a ∈ C (since Kaf = Kf ).

• m(sf) = m(f) = m(fs) (by uniqueness) The proof will be complete if we show that m is additive (hence linear). Let us take f, g ∈ C(G) and start with Equation (1.1) above with c = m(f). Further let X h(x) = aig(six)

Since h ∈ Cg, we certainly have Ch ⊆ Cg whence Kh ⊆ Kg. But the set

Kg contains only one constant: m(h) = m(g).

14 We can write X |m(h) − bjh(tjx)| < P for ﬁnitely many suitable tj ∈ G and bj > 0 with bj = 1. Using the deﬁnition of h and m(h) = m(g), this implies X |m(g) − aibjg(sitjx)| < (1.5) i,j

However multiplying Equation (1.1) by bj and replacing x by tjx and summing over j we ﬁnd X |m(f) − aibjf(sitjx)| < (1.6) i,j

15 Adding Equation (1.5) and (1.6), this implies X |m(f) + m(g) − aibj(f + g)(sitjx)| < 2 i,j

Thus the constant m(f) + m(g) is in Kf+g. However note that the only constant in this compact convex set is m(f + g). This completes the proof.

16 1.1 Exercises

Exercise 1.4. Let m be the normalized Haar measure of a compact group G. For f ∈ C(G) or L1(G) show that m(f) = m(f˜) where the function f˜ is deﬁned as the function f˜(x) = f(x−1). This equality is usually written as Z Z f(x)dx = f(x−1)dx G G Hint: Observe that f → m(f˜) is a Haar measure on G and use the uniqueness part of the Theorem on Haar measures, Theorem 1.3

17 Before stating the next exercise we need a deﬁnition Deﬁnition 1.5 (Semidirect products). Let L be a group and assume it contains a normal subgroup G and a subgroup H such that GH = L and G ∩ H = {e}. That is, suppose one can select exactly one element h from each coset of G so that {h} forms a subgroup H. If H is also normal then L is isomorphic with the direct product G × H. If H fails to be normal, we can still reconstruct L if we know how the inner automorphisms ρh behave on G. Namely for xj ∈ G and hj ∈ H (j = 1, 2), we have:

−1 (x1h1)(x2h2) = x1h1x2h1 h1h2 = (x1ρh1 (x2)) h1h2

18 The construction just given can be cast in an abstract form. Let G and

H be groups and suppose there is a homomorphism h → τh which carries H onto a group of automorphisms of G, namely τh ◦ τh0 = τhh0 0 for h, h ∈ H. Let GsH denote the cartesian product of G and H. For 0 0 (x, h) and (x , h ) in GsH, deﬁne:

0 0 0 0 (x, h)(x , h ) = (x (τh(x )) , hh )

Then GsH is a group; it is called a semidirect product of G and H. Its identity is (e1, e2) where e1 and e2 are the identities of G and H −1 −1 respectively. The inverse of (x, h) is (τh−1 (x ), h ). Let

G1 := {(x, e2): x ∈ G} and

H1 := {(e1, h): h ∈ H}

19 Then G1 is a normal subgroup of GsH and H1 is a subgroup. Since

−1 (e1, h) · (x, e2) · (e1, h) = (τh(x), e2)

the inner automorphism ρ(e1,h) for (e1, h) ∈ H1 reproduces the action τh on G. Thus every semidirect product is obtained by the process described in the previous paragraphs.

20 Exercise 1.6. Let G and H be compact groups and let GsHbe a semidirect product of G and H . Suppose also that the mapping

(x, h) → τh(x) is a continuous mapping of G × H onto G . In particular, each τh is a homeomorphism of G onto itself. Show that the semidirect product GsHwith the product topology is a compact group. What is the Haar measure on GsHin terms of the Haar measures on G and H ?

21 Exercise 1.7. Let On(R) be the group of n × n orthogonal matrices. Suppose that Zij, 1 ≤ i ≤ j ≤ n are i.i.d. standard normal random variables. Let U be the random orthogonal matrix with rows obtained by applying the Gram-Scmidt process to the vectors (Z11,...,Z1n), ...,

(Zn1,...Znn). Show that U is distributed according to the Haar measure on On(R).

22 2 Representations, General Constructions

For E , a complex Banach space, let Gl(E) denote the group of continuous isomorphisms of E onto itself. A representation π of a compact group G in E is a homomorphism π:

π : G −→ Gl(E) for which all the maps G → E deﬁned as s → π(s)v (v ∈ E) are continuous. The space E = Eπ in which the representation takes place is called the representation space of π. A representation π of a group G in a vector space E canonically deﬁnes an action (also denoted by π)

π : G × E −→ E (s, v) −→ π(s)v

23 The deﬁnition requires this action to be separately continuous. The action is then automatically globally continuous. We say that a representation π is unitary when E =H , is a Hilbert space and each operator π(s) (s ∈ G) is a unitary operator (i.e. each π(s) is isometric and surjective). Thus π is unitary when E =H is a Hilbert space and

π(s)∗ = π(s)−1 = π(s−1)(s ∈ G)

The representation π of G in E is said to be irreducible when E and {0} are distinct and are the only two closed invariant subspaces under all operators π(s) (s ∈ G)(topological irreducibility).

24 Two representations π and π0 of the same group G are called equivalent when the two spaces over which they act are G -isomorphic, namely there exists a continuous isomorphism A : E → E0 of their respective spaces with

A(π(s)v) = π0(s)Av (s ∈ G, v ∈ E)

More generally, continuous linear operators A : E → E0 satisfying all commutation relations A(π(s)) = π0(s)A for all s ∈ G are called intertwining operators or G -morphisms (from π to π0) and their set is a vector space denoted either by

0 0 HomG(E,E ) or Hom(π, π )

25 Proposition 2.8. Let π be a unitary representation of G in the Hilbert space H . If H1 is an invariant subspace of H (with respect to all ⊥ operators π(s), s ∈ G), then the orthogonal space H2 = H1 of H1 in H is also invariant.

Proof. We need to show that if v ∈ H, v ⊥ H1 then π(s)v is also orthogonal to H1 for all s in G . For any x ∈ H1,

hx, π(s)vi = hπ(s)∗x, vi = hπ(s−1)x, vi = 0

−1 since by assumption π(s )x also lies in H1.

26 Proposition 2.9. Let π be a representation of a compact group G in a Hilbert space H . Then there exists a positive deﬁnite hermitian form ϕ which is invariant under the G -action, and which deﬁnes the same topological structure on H .

Proof. By continuity of the mappings s → π(s)v, the mappings

s −→ hπ(s)v, π(s)wi (v, w ∈ H) are also continuous (by continuity of scalar product in H × H). We can thus deﬁne Z ϕ(v, w) = hπ(s)v, π(s)wids G using the Haar integral.

27 It is clear that ϕ is hermitian and positive. Let us show that it is non-degenerate and deﬁnes the same topology on H . Since G is compact, π(G) is also compact in Gl(H) (with the strong topology). In particular, π(G) is simply bounded and thus uniformly bounded (uniform boundedness principle ≡ Banach-Steinhaus theorem). Thus, there exists a positive constant M > 0 with

||π(s)v|| ≤ M||v|| (∀s ∈ G, v ∈ H)

This implies

||v|| = ||π(s−1)π(s)v|| ≤ M||π(s)v|| ≤ M 2||v||

Thus M −1||v|| ≤ ||π(s)v|| ≤ M||v||

28 Squaring and Integrating over G , we ﬁnd

M −2||v||2 ≤ ϕ(v, v) ≤ M 2||v||2

Thus ϕ(v, v) = 0 implies ||v|| = 0 and v = 0. Thus ϕ and || · ||2 induce equivalent topologies (equivalent norms) on H . Invariance of ϕ comes from the invariance of the Haar measure. Z Z ϕ(π(t)v, π(t)w) = hπ(st)v, π(st)wids = f(st)ds G G Z Z = ft(s)ds = f(s)ds = ϕ(v, w) G G This shows that π is ϕ-unitary as desired. These propositions imply any representation of a compact group in a Hilbert space is equivalent to a unitary one, and any ﬁnite dimensional representation (the dimension of a representation is the dimension of its rep. space) is completely reducible (direct sum of irreducible ones.)

29 Deﬁnition 2.10 (left translations). In any space of functions on G , deﬁne the left translations by [l(s)f](x) = f(s−1x)

(If we do not want to identify elements of Lp(G) with functions or classes of functions, we can simply extend translations from C(G) to Lp(G) by continuity). Thus we have l(s) ◦ l(t) = l(st) and we get homomorphisms l : G → Gl(E), s → l(s) with any E = Lp(G), 1 ≤ p < ∞. Exercise 2.11. Check that these homomorphisms are continuous in the representation sense.

30 The above were the left regular representations of G. The right regular representations of G in the Banach space Lp(G) are deﬁned similarly with

p [r(s)f](x) = f(xs)(f ∈ L (G)) With this definition, one has r(s) ◦ r(t) = r(st). One can also consider the biregular representations of l × r of G × G in Lp(G) defined as −1 p [l × r(s, t)f](x) = f(s xt)(f ∈ L (G)) and its restriction to the diagonal G → G × G, s → (s, s) which is the adjoint representation of G . It is defined as

−1 p [Ad(s)f](x) = f(s xs)(f ∈ L (G)) The regular representations are faithful, i.e π(s) = 11 ⇔ s = e

31 Let π : G → Gl(E) and π0 : G0 → Gl(E0) be two representations. We can deﬁne the external direct sum representation of G × G0 in E ⊕ E0 by π ⊕ π0(s, s0) = π(s) ⊕ π0(s0)(s ∈ G, s0 ∈ G0) When G = G0, we can restrict this external direct sum to the diagonal G of G × G, obtaining the usual direct sum of πand π0

π ⊕ π0 : G → Gl(E ⊕ E0) s → π(s) ⊕ π0(s)

The external tensor product π ⊗ π0 as a representation of G × G0 in E × E0is deﬁned as

π ⊗ π0(s, s0) = π(s) ⊗ π0(s0)(s ∈ G, s0 ∈ G0)

32 We assume the two spaces E ,E0 are ﬁnite dimensional, thus this algebraic tensor product is complete; in general some completion has to be devised. The usual tensor product of two representations of the same group G is the restriction to the diagonal of the external tensor product (G = G0) and is given by

π ⊗ π0(s) = π(s) ⊗ π0(s)(s ∈ G)

33 For a given ﬁnite dimensional representation π : G → Gl(E), deﬁne the contragredient representation πˇ. This representation acts in the dual E0 of E (namely the space of linear forms on E) and

πˇ(s) = tπ(s−1)(s ∈ G)

Since transposition reverses the order of composition of mappings, namely t(AB) = tBtA, it is necessary to reverse the operations by taking the inverse in the group. The above construction allows us to conclude that πˇ(st) = πˇ(s)πˇ(t) as is required for a representation.

34 Conjugate representation π: When E =H is a Hilbert space the conjugate π¯ of π is a representation acting on the conjugate H¯ of H . Recall that H¯ has the same underlying additive group as H , but with the scalar multiplication in H¯ twisted by complex conjugation, namely the external operation of scalars is given by

(a, v) −→ a · v =av ¯ ( we use a dot in H¯ )

The inner product h·, ·i− of H¯ is deﬁned as

hv, wi− = hv, wi = hw, vi

This suggests that an element v ∈ H is written as v¯ when we consider it as an element of the dual Hilbert space H¯ . With this notation we have:

− av = a · v (a ∈ C) and hv,¯ w¯i = hv, wi

35 The identity map H → H¯ , v → v¯ is an anti-isomorphism. The conjugate of π is deﬁned as π¯(s) = π(s) in H¯ . Since the (complex vector) subspaces of H and H¯ are the same by deﬁnition, π and π¯ are reducible or irreducible simultaneously. However it is important to distinguish these two representations (in particular they are not always equivalent). Any orthonormal basis (ei) of H is also an orthonormal P basis of H¯ , but a decomposition v = viei in H gives rise to the decomposition X v¯ = v¯ie¯i (complex conjugate components in H¯ )

Thus the matrix representations associated with π and π¯ are complex conjugate to one another. Exercise 2.12. Show that when π is unitary and ﬁnite dimensional, the contragredient πˇ and the conjugate π¯ of π are equivalent.

36 2.1 Exercises

Exercise 2.13. Show that the left and right representations l and r of a group G (in any Lp(G) space) are equivalent. Exercise 2.14. If πand π0 are two representations of the same group G (acting in respective Hilbert spaces H and H0), show that the matrix 0 0 0 coeﬃcients of π ⊗ π (with respect to bases (ei) in H and (ej) in H 0 0 0 and ei ⊗ ej in H ⊗ H ) are products of matrix coeﬃcients of πand π (Kronecker product of matrices).

Exercise 2.15. Let 11n denote the identity representation of the group G in dimension n (the space of this identity representation is thus Cn

n and 11n(s) = idC for all s ∈ G). Show that for any representation πof G ,

π ⊗ 11n is equivalent to π ⊕ π ⊕ · · · ⊕ π (n terms)

37 Exercise 2.16. (Schur’s lemma) Let k be an algebraically closed field, V a finite dimensional vector space over k and Φ any irreducible set of operators in V (the only invariant subspaces, relatively to all operators belonging to Φ are V and {0}). Then, if an operator A commutes with all operators in Φ, A is a multiple of the identity operator (i.e. A is a scalar operator). Hint: Take an eigenvalue a in the algebraically closed field k and consider A − a · I, which still commutes with all operators of Φ. Show that the Ker(A − a · I)(6= {0}) is an invariant subspace.

38 3 Finite dimensional representations of compact groups (Peter-Weyl theorem)

Theorem 3.17 (Peter-Weyl). Let G be a compact group. for any s 6= e in G , there exists a ﬁnite dimensional irreducible representation π of G such that π(s) 6= 11.

Proof. We start with two Lemmas.

Lemma 3.18. Let G be a compact group, k : G × G → C a continuous function and K : L2 → C(G) the operator with kernel k, namely: Z (Kf)(x) = k(x, y)f(y)dy G Then K is a compact operator. Moreover if k(x, y) = k(y, x) identically on G × G, K is a Hermitian as an operator from L2(G) to C(G).

39 Lemma 3.19. Let K be a compact Hermitian operator (in some Hilbert space H ). Then the spectrum S of K consists of eigenvalues. Each eigenspace Hλ with respect to a non-zero eigenvalue λ ∈ S is ﬁnite dimensional and the number of eigenvalues outside any neighborhood of 0 is ﬁnite. Moreover, S ⊆ R and ||K|| = sup{|λ| : λ ∈ S} and the eigenspaces associated to distinct eigenvalues are orthogonal, i.e

Hλ ⊥ Hµ for λ 6= µ in S

40 Proof of Theorem 3.17: Assume that s 6= e in G and take an open symmetric neighborhood V = V −1 of e in G such that s∈ / V 2. There exists a positive continuous function f such that

f(e) > 0 , f(x) = f(x−1) = fˇ(x) , Supp(f) ⊆ V where Supp(f) denotes the support of f, namely the complement of the largest open set on which f vanishes. Consider the function ϕ = f ∗ f deﬁned as Z ϕ(x) = f(y)f(y−1x)dy G The support of ϕ is contained in V 2 and

ϕ(s) = 0 (s∈ / V 2) , ϕ(e) = ||f||2 > 0.

41 We also see that l(s)ϕ 6= ϕ. But the operator K with kernel k(x, y) = f(y−1x) is compact (see Lemma 3.18) and the convergence in quadratic mean of X f = f0 + fi , fi ∈ Ker(K − λi) = Hi (λi ∈ Spec(K)) implies that X X ϕ = Kf = Kfi = λifi where 1 fi = Kfi ∈ Im(K) ⊆ C(G) λi where we have uniform convergence holding in the series above. Since l(s)ϕ 6= ϕ, we must have l(s)fi 6= fi for at least one index i. However the deﬁnition of the kernel k shows that

k(sx, sy) = k(x, y) = f(y−1x)(s, x, y ∈ G)

42 The consequence of these identities is the translation invariance of all the eigenspaces Hi of K. The left regular representation restricted to a suitable finite dimensional subspace Hi (for any i, with l(s)fi 6= fi) will furnish an example of a finite dimensional representation π with π(s) 6= e. The corollaries of this theorem are numerous and important. Corollary 3.20. A compact group is commutative if and only if all its finite dimensional irreducible representations have dimension 1.

Proof. Exercise.

43 Corollary 3.21 (Peter-Weyl). Any continuous function on a compact group is a uniform limit of (ﬁnite) linear combinations of coeﬃcients of irreducible representations.

Proof. Let π be a (finite dimensional) irreducible representation of the compact group G and take a basis in the representation space of π in order to be able to identify in π : G → Gln(C), the coefficients of π being the continuous functions on G defined as

i i cj : g −→ cj(g) = hei, π(g)eji

More generally if u and v are elements of H, we can deﬁne the u (function) coeﬃcient cv of π on G by

u g −→ cv (g) = hu, π(g)vi

44 These functions are obviously finite linear combinations of the previously i defined matrix coefficients cj. Introduce the subspace V (π) of C(G) i u spanned by the cj, or equivalently by all cv for v, u ∈ Hπ. Observe that the subspaces of C(G) attached in this way to two equivalent representations π and π0 coincide namely, V (π) = V (π0). Thus we can form the algebraic sum (a priori this algebraic sum is not a direct sum) M AG = Vπ ⊆ C(G) where the summation index π runs over all (classes of) finite dimensional irreducible representations of G . The corollary can be restated in the following form:

AG is a dense subspace of the Banach space C(G) in the uniform norm

45 But this algebraic sum AG is a subalgebra of C(G) (the product of two continuous functions being the usual pointwise product). The product of the coefficients u s cv of π and γt of σ is a coefficient of the representation π ⊗ σ (the coefficient of this representation with respect to the two vectors u ⊗ s and v ⊗ t). Taking π and σ to be finite dimensional representations of G , π ⊗ σ will be finite dimensional, hence completely reducible and all its coefficients (in v s particular the product of cu and γt ) are finite linear combinations of coefficients of (finite dimensional) irreducible representations of G .

This subalgebra AG of C(G) contains the constants, is stable under complex conjugation (because π is irreducible precisely when π¯ is irreducible) and separates points of G by the main Theorem 3.17. By the Stone-Weistrass theorem the proof is complete.

46 3.1 Exercises

Exercise 3.22. Let G be a compact totally discontinuous group. Show that AG is the algebra of all locally constant functions on G . (Observe that a locally constant function on G is uniformly locally constant, hence can be identiﬁed with a function on a quotient G/H where H is some open subgroup of G . Conversely any ﬁnite dimensional representation of G must be trivial on an open subgroup H of G .)

Exercise 3.23. Let G be any compact group. Show that AG consists of the continuous functions f on G for which the left and right translates of f generate a ﬁnite dimensional subspace of C(G). In particular if G1 and G2 are two compact groups, any continuous homomorphism h : G1 → G2 has a transpose th : A2 → A1 where Ai = AGi , deﬁned by th(f) = f ◦ h. A priori this transpose is a linear mapping th : C(G2) → C(G1).

47 Exercise 3.24. Let G = Un(C) with its canonical representation π in V = Cn. Since π is unitary, we can identify π¯ with the contragredient of π : it acts in the dual V ∗ of V . p (a) Let Aq denote the space of linear combinations of coeﬃcients of the representation

p ⊗p ⊗q ∗ ⊗p ⊗q p πq = π¯ ⊗ π in (V ) ⊗ V = Tq (V )

p Prove that the sum of the subspaces Aq of C(G)is an algebra A (show p r pr p q that Aq As ⊆ Aqs), stable under conjugation (show that Aq = Ap), which separates the points of G . Using the Stone-Weierstrass theorem, conclude that A is dense in C(G).

(b) Show that A = AG. (use part(a) to prove that any irreducible p representation of G appears as a subrepresentation of some πq , or in other words can be realized on a space of mixed tensors.)

48 Exercise 3.25. Let G be a closed subgroup of Un(C). Using the fact that any ﬁnite dimensional representation of G appears in the restriction of some ﬁnite dimensional representation of Un(C) (this is a consequence of the theory of induced representations), show that G is a real algebraic subvariety of Un(C). (The transpose of the embedding G,→ Un(C) is the operation of restriction on polynomial functions and is surjective. Hence AG is a quotient of the polynomial algebra A of Un(C). By the exercise 3.24, A is generated by the coordinate functions

i i Un(C) −→ C , x = (xj) 7−→ xj and their conjugates. )

49 4 Decomposition of the regular representation

Lemma 4.1. Let V ⊆ L2(G) be a ﬁnite dimensional subspace which is invariant under the right regular representation of G. Then V consists of (classes of) continuous functions and each f ∈ V can be written as

f(x) = Tr(Aπ(x)) for some A ∈ End(V )

Here π denotes the restriction of the right regular representation to V .

50 i Proof. Take an orthonormal basis (χi) of V and recall the coeﬃcients cj of π deﬁned as

X j π(x)χi = ci (x)χj, x ∈ G j

P i For f = i a χi, we have

X i X i j π(x)f = a π(x)χi = a ci (x)χj i i,j Hence

X j i i i f(x) = [r(x)f](e) = ci (x)aj (with aj = a χj(e)) i,j Thus, f(x) = Tr(Aπ(x)) as claimed.

51 Let (π,V ) be any finite dimensional representation of the compact group G. For any endomorphism A ∈ End(V), we define the corresponding coefficient cA of π by cA(x) = T r(A · π(x)). The right translates of these coefficients are easily identified as

[r(s)cA](x) = cA(xs) = T r(Aπ(x)π(s))

= T r(π(s) · Aπ(x)) = T r(Bπ(x)) = cB(x) where B = π(s) · A. Consider the representation of G in End(V ) deﬁned by

lπ(s)A = π(s) · A (s ∈ G, A ∈ End(V ))

The above computations show that A → cA is a G-morphism

2 c : End(V ) −→ C(G) ⊆ L (G) (intertwining lπ and r.)

52 Suppose now that (π,V ) is an irreducible ﬁnite dimensional representation of the compact group G.

Write L2(G, π) for the image of End(V ) under the map c. Note that the vector space L2(G, π) only depends on the equivalence class of π. The representation (lπ, End(V )) is equivalent to π ⊕ · · · ⊕ π (n times, where n = dim(V )) and L2(G, π) is r-invariant, so the restriction of r to L2(G, π) is equivalent to π ⊕ · · · ⊕ π (m times for some m ≤ n). Thus, L2(G, π) has dimension mn ≤ n2. If V 0 is a r-invariant subspace of L2(G) such that the restriction of r to V 0 is equivalent to π, then V 0 is a subspace of L2(G, π) by Lemma 4.1. Hence, L2(G, π) is the sum of all subrepresentations of (L2(G), r) which are equivalent to π.

53 Definition 4.2. Let π be a finite dimensional irreducible representation of a compact group G. The space L2(G, π) consisting of the sum of all subspaces of the right regular representation which are equivalent to π is called the isotypical component of π in L2(G). Note that a function f ∈ L2(G) belongs to L2(G, π) precisely when the right translates of f generate an invariant subspace (of the right regular representation) equivalent to a finite multiple of π (that is, a finite direct sum of subrepresentations equivalent to π).

54 We shall now prove that the dimension of an isotypical component L2(G, π) is exactly (dim π)2.

55 2 Recall the G-morphism c : End(V) → L (G) , A 7→ cA := T r(Aπ). The fact that cA 6= 0 for A 6= 0 will be deduced from a computation of the quadratic norm of these coeﬃcient functions. It is easier to start with the case of rank ≤ 1 linear mappings. We use the isomorphisms

Vˇ ⊗ V −→ End(V )(Vˇ = dual of V ) deﬁned as follows: If u ∈ Vˇ and v ∈ V , the operator corresponding to u ⊗ v is u ⊗ v : x → u(x)v = hu, xiv

56 The image of u ⊗ v consists of multiples of v, and u ⊗ v has rank 1 when u and v are non-zero (quite generally, decomposable tensors corresponding to operators of rank ≤ 1). The coefficient cA with respect to the operator A = u ⊗ v coincides with the previously defined coefficient u cv = hu, π(x)vi = cu⊗v(x)

57 Lemma 4.3. Let π and σ be two representations of a compact group G and A : Vπ → Vσ be a linear mapping. Then Z A\ = σ(g)Aπ(g)−1dg G

\ is a G-morphism from Vπ to Vσ, namely A ∈ HomG(Vπ,Vσ).

Proof. We have Z Z A\π(s) = σ(g)Aπ(g)−1π(s)dg = σ(g)Aπ(s−1g)−1dg and replacing g by sg (i.e. s−1g by g) Z A\π(s) = σ(sg)Aπ(g)−1dg = σ(s)A\

58 Thus the averaging operation (given by the Haar integral) of Lemma 4.3 leads to a projector

\ \ : Hom(Vπ,Vσ) → HomG(Vπ,Vσ) ,A → A

In particular when π and σ are disjoint, i.e HomG(Vπ,Vσ) = 0, we must have A\ = 0. This is certainly the case when π and σ are non-equivalent irreducible representations (Schur’s lemma). Another case of special interest is π = σ, ﬁnite dimensional and irreducible. Schur’s lemma gives HomG(Vπ,Vσ) = C and thus A = λAid is a scalar operator.

59 Proposition 4.4. If π is a ﬁnite dimensional irreducible representation of the compact group G in V , the projector

\ End(V ) → EndG(V ) = C id, A → A = λA id, is given explicitly by the following formula: Z \ −1 T r(A) A = π(g)Aπ(g) dg = idV G dimV

Proof. Since we know a priori that the operator A\ is a scalar operator

λAid, we can determine the value of the scalar by taking traces in the deﬁning equalities Z Z −1 −1 λAT r(idV ) = T r π(g)Aπ(g) dg = T r π(g)Aπ(g) dg G G Z = T r(A)dg = T r(A) G

60 Theorem 4.5 (Schur’s orthogonality relations). Let G be a compact group and π, σ be a two ﬁnite dimensional irreducible representations of G. Assume that π and σ are unitary. Then (a) If π and σ are non-equivalent, L2(G, π) and L2(G, σ) are orthogonal in L2(G). (b) If π and σ are equivalent, L2(G, π) = L2(G, σ) and the inner product of the two coeﬃcients of this space is given by Z u x hcv , cy i = hu, π(g)vihx, π(g)yidg = hu, xihv, yi/dimV G

(c) More generally in the case π = σ, the inner product of general coeﬃcients is given by Z ∗ hcA, cBi = T r(Aπ(g))T r(Bπ(g))dg = T r(A B)/dimV G

61 Proof. (a) follows from Lemma 4.3 and (b) follows similarly from Proposition 4.4. It will be enough to show how (b) is derived. For this purpose, we consider the particular operators v¯ ⊗ y (v¯ ∈ V¯π, y ∈ Vπ) and apply the result of the proposition Z −1 T r(¯v ⊗ y) hv, yi π(g)(¯v ⊗ y)π(g) dg = idV = idV G dim V dimV Let us apply this operator to the vector u and take the inner product with the vector x Z hv, yi hu, xihv, yi hx, π(g)(¯v ⊗ y)π(g)−1u dgi = hx, ui = G dimV dimV

62 But we have π(g)(¯v ⊗ y)π(g)−1u = π(g)hv, π(g−1)uiy = hπ(g)v, uiπ(g)y = hu, π(g)viπ(g)y hence Z Z −1 u x hx, π(g)(¯v ⊗ y)π(g) u dgi = hu, π(g)vihx, π(g)yi dg = hcv , cy i G G as expected. Finally (c) follows from (b) by linearity since the operators v¯ ⊗ y (of rank ≤ 1) generate End(V ).

In particular we see that if 0 6= A ∈ End(V ),

2 ∗ ||cA|| = T r(A A)/ dim V 6= 0 and the mapping c : End(V ) → L2(G, π) is one-to-one and onto. The dimension of this isotypical component is thus (dim V )2.

63 Corollary 4.6. The Hilbert space L2(G) is the Hilbert sum of all isotypical components

2 M 2 L (G) = dL (G, π) The summation index π runs over equivalence classes of ﬁnite dimensional irreducible representations of the compact group G.

64 Proof. We have already seen that the isotypical subspaces L2(G, π) are mutually orthogonal to each other. Thus our corollary will be proved if we show that the algebraic sum

M 2 AG = L (G, π) ⊆ C(G) is dense in the Hilbert space L2(G).

But AG consists of coefficients of finite dimensional representations of G (we have proved that all finite dimensional representations are completely reducible), and the Peter-Weyl theorem 3.17 has shown that

AG is dense in C(G) for the uniform norm. A fortiori AG will be dense in L2(G) for the quadratic norm.

65 Corollary 4.7. Any (continuous, topologically) irreducible representation of a compact group G in a Banach space is ﬁnite dimensional.

Proof. Let σ : G → Gl(E) be such a representation and let E0 denote the (topological) dual of E, namely E0 is the Banach space of continuous linear forms on E. By the Hahn-Banach theorem, for each 0 6= x ∈ E, there is a continuous linear form x0 ∈ E0 with x0(x) 6= 0. For u ∈ E0 and v ∈ E we can consider the corresponding coeﬃcient of σ

u u cv ∈ C(G): g → cv (g) = hu, σ(g)vi

66 Letting v vary in E, we get a linear mapping

2 u Q : E → C(G) ⊆ L (G) , v → cv Since G is compact and the mappings g → σ(g)v are continuous (by the deﬁnition of continuous representations), the sets σ(G)v are compact, hence bounded in E (for each v ∈ E). By the uniform boundedness principle (Banach-Steinhaus theorem), the set σ(G) of operators in E is equicontinuous and bounded

sup ||σ(g)|| = M < ∞ g∈G Hence

u ||Qv|| = sup |cv (g)| = sup |hu, σ(g)vi| g∈G g∈G

≤ ||u||E0 sup ||σ(g)v||E ≤ M||u||E0 ||v||E g∈G

67 This proves that Q is continuous from E into C(G) (equipped with the uniform norm); it’s kernel is a proper and closed subspace F 6= E if u 6= 0 (in this case u(v) 6= 0 for some v ∈ E and thus u cv (e) = hu, vi = u(v) 6= 0). Take v ∈ E with Q(v) 6= 0, and apply the orthogonal Hilbert sum decomposition of the preceding corollary to Q(v). X Pπ(Qv) = Qv 6= 0 with

Pπ = orthogonal projector from L2(G) onto L2(G, π)

68 This implies that there is a π (ﬁnite dimensional irreducible representation of G) with PπQv 6= 0. For this π, we consider the composite Q 2 Pπ 2 E −→ L (G) −→ L (G, π) and its kernel which is a proper closed subspace Fπ 6= E. But Q is a G-morphism (intertwining σ and the right regular representation)

u u cσ(x)v(g) = hu, σ(g)σ(x)vi = cv (gx) which implies that Q(σ(x)v) = ρ(x)Q(v). Since Pπ is also a G-morphism, the kernel Fπ of the composite PπQ must be an invariant subspace of E. However E is irreducible by assumption so that Fπ = {0}, and the composite PπQ is one-to-one (into):

2 2 dim E ≤ dim L (G, π) = (dim V )

69 Definition 4.8. The dual Gb of a compact group G is the set of equivalence classes of irreducible representations of G. Let π be an irreducible representation of the compact group G and, $ = [π] its equivalence class ($ ∈ Gb). We say that π is a model of $ in this case i.e. when π ∈ $. The dimension of $ is defined as dim (π) = dim (Vπ) independently from the model chosen. Similarly the isotypical component of $ (in the right regular representation) is defined as L2(G, $) = L2(G, π) independently from the model π chosen for $. By the finiteness of the dimension of the irreducible representations of the compact group G namely Corollaries 4.6 and 4.7 M 2(G) = d 2(G, $) L $∈Gˆ L

Instead of π ∈ $ ∈ Gb we shall write more simply π ∈ Gb.

70 Proposition 4.9. Let G be a compact group. Then the following properties are equivalent (i) The dual Gb is countable. (ii) L2(G) is separable. (iii) G is metrizable.

Proof. The equivalence between (i) and (ii) is obvious since L2(G) is the Hilbert sum of the ﬁnite dimensional isotypical components L2(G, $) over the index set Gb. Moreover G can always be embedded in a product Y ∼ U(π) with U(π) = Udim (π)(C) (metrizable group) Gb

71 Since any countable product of metrizable topological spaces is metrizable, we see that (i)⇒(iii). Finally, the implication (iii)⇒(ii) is a classical application of the Stone-Weirstrass theorem.

72 4.1 Exercises

Exercise 4.10. Let V be a finite dimensional vector space and Vˇ be its dual and for u ∈ Vˇ , u ∈ V denote by u ⊗ v the operator x → u(x)v as defined in the beginning of this Section. Show (a) T r(u ⊗ v) = hu, vi = u(v) (intrinsic definition of T r), (b) (u ⊗ v) · (x ⊗ y) = hu, yix ⊗ v, (c) tAu ⊗ Bv = B · (u ⊗ v) · A for A, B ∈ End(V ). Moreover, identifying the dual of Vˇ ⊗ V with the dual of V ⊗ Vˇ in the obvious canonical way show (d) t(u ⊗ v) = v ⊗ u Assume now that V is now a representation space for a group G. Using (a) and (c) prove u (e) cu⊗v = cv ( i.e. T r(π(x) · u ⊗ v) = hu, π(x)vi).

73 Exercise 4.11. Let G be a compact group and (π,V ) be a ﬁnite dimensional representation of G. We denote by V G the subspace of invariants of V :

V G = {v ∈ V : π(g)v = v ∀g ∈ G} R (a) Check that the operator P = G π(g)dg is a projector from V onto V G. If π is unitary, P is the orthogonal projector on this subspace. (b) For two ﬁnite dimensional representations π and σ of G, consider Hom(Vπ,Vσ) as a representation space of G via the action

g · A = π(g) · Aσ(g)−1

Observe that G Hom(Vπ,Vσ) = HomG(Vπ,Vσ) and deduce a proof of the Lemma 4.3 from this observation.

74 (c) Using the G-isomorphism

Vˇπ ⊗ Vσ −→ Hom(Vπ,Vσ) with Vˇπ ⊗ Vσ being equipped with the representation πˇ ⊗ σ, conclude that the projector \ : Vˇπ ⊗ Vσ → (Vˇπ ⊗ Vσ)G is given by R G(πˇ ⊗ σ)(g)dg

75 5 Convolution, Plancherel formula and Fourier Inversion

Deﬁnition 5.1 (Convolution). On a compact group G, the convolution of two continuous functions f and g is deﬁned as Z f ∗ g(x) = f(y)g(y−1x)dy G

Deﬁning f ∗(x) = f(x−1), we can also write Z Z f ∗ g(x) = f(xy)g(y−1)dy = f(xy−1)g(y)dy G G Z = f ∗(yx−1)g(y)dy = hr(x−1)f ∗, gi G

76 The Cauchy-Schwartz inequality gives:

∗ |f ∗ g(x)| ≤ ||f ||2||g||2 = ||f||2||g||2 whence

||f ∗ g||∞ = sup |f ∗ g(x)| ≤ ||f||2||g||2 x∈G Consequently the convolution product can be extended by continuity from C(G) to L2(G) and by deﬁnition

2 2 ∗ : L (G) × L (G) → C(G),, (f, g) → f ∗ g is a continuous bilinear mapping still satisfying the above inequality. On the other hand the preceding formulas show:

∗ 2 ∗ hf, gi = f ∗ g(e) , ||f||2 = f ∗ f(e)

77 The convolution product for functions in L1(G) can be deﬁned by the same integral formula, but this will not converge for every x ∈ G and the result will no longer be continuous in general. To see what happens, take f, g ∈ L1(G). By Fubini’s theorem, Z Z Z |f ∗ g(x)|dx ≤ dx dy|f(y)g(y−1x)| Z Z = dy|f(y)| dx|g(y−1x)| Z = ||g||1 |f(y)|dy = ||f||1||g||1 < ∞

78 These inequalities prove that R f(y)g(y−1x)dy converges absolutely for almost all x ∈ G and the result f ∗ g ∈ L1(G) satisﬁes:

||f ∗ g||1 ≤ ||f||1||g||1

This convolution product is associative and L1(G) is an algebra for convolution. This algebra has no unit element in general (more precisely, it has no unit when G is not discrete i.e. G is not ﬁnite). We shall see that this algebra L1(G) is commutative exactly when G is commutative.

79 5.1 Integration of representations

Assume that π is a unitary representation of the compact group G in a Hilbert space H. We can “extend” π to a representation of L1(G) by the formula Z 1 1 π (f) = f(x)π(x)dx (f ∈ L (G)) G These integrals converge absolutely in norm: ||π(x)|| = 1 implies Z Z ||f(x)π(x)||dx = |f(x)|dx = ||f||1

Thus, 1 1 ||π (f)|| ≤ ||f||1 (f ∈ L (G))

80 Although G is not really embedded in L1(G) (when G is inﬁnite), we consider π1 as an extension of π. Later on we shall even drop the index 1, writing π instead of π1. The association

1 1 π : G → Gl(H) ; π : L (G) → End(H) can even be made when π is a representation in a Banach space since π(G) is always a bounded set of End(H): being weakly compact, it is uniformly bounded.

81 5.2 Comparison of several norms

Let A ∈ End(V ) be any endomorphism in V . Take any orthonormal i basis (ei) of V and assume that A is represented by the matrix (aj) in the basis (ei). Obviously

2 X i 2 ||A||HS = |aj| i,j deﬁnes a norm on End(V ) called the Hilbert-Schmidt norm (a priori this norm depends on the choice of the orthonormal basis (ei)).

82 i If B is another endomorphism, represented by the matrix (bj) (in the same basis), a small computation shows that

∗ X i i T r(A B) = a¯jbj ij This shows that the Hilbert-Schmidt norm is derived from the Hilbert-Schmidt inner product

X i i ∗ hA, BiHS = a¯jbj = T r(A B) i,j on End(V ), and is in particular independent from the choice of the orthonormal basis (ei) of V .

83 Return to a compact group G and a unitary irreducible representation π ∈ Gb in some ﬁnite dimensional Hilbert space V = Vπ. The spaces

2 Vˇ ⊗ V, End(V ) , L (G, π) are G-isomorphic. We shall give explicit isomorphisms between them keeping track of the various norms involved.

84 We have introduced the coeﬃcients

u ˇ cv (x) = hu, π(x)vi (u ∈ V , v ∈ V ) and the more general coeﬃcients (linear combinations of the preceding ones)

cA(x) = T r(Aπ(x)) (A ∈ End(V )) with the relation u cA = cv for A = u ⊗ v

85 For ﬁxed u ∈ V , the linear mapping

u 2 u c : V → L (G, π) , v → cv is a G-morphism form π to r, the right regular representation.

86 Similarly if v ∈ V is ﬁxed,

u u −1 −1 l(s)cv (x) = cv (s x) = hu, π(s )π(x)vi t −1 πˇ (s)u = h π(s )u, π(x)vi = cv (x) shows that the linear mapping

ˇ 2 u cv : V → L (G, π) , u → cv is a G-morphism from πˇ to l (the left regular representation). Summing up, ˇ 2 u c : V ⊗ V → L (G, π) , u ⊗ v → cv is a πˇ ⊗ π to l × r (biregular representation) G × G-morphism.

87 Note that

−1 −1 [l × r(s, t)]cA(x) = cA(s xt) = T r(Aπ(s )π(x)π(t)) −1 = T r(π(t)Aπ(s) π(x)) = cπ(t)Aπ(s)−1 (x)

So, the corresponding action of G × G on End(V ) is deﬁned by

(s, t) · A = π(t)Aπ(s)−1

88 In the following diagram of G-morphisms, Vˇ ⊗ V is equipped with the inner product hu ⊗ v, x ⊗ yi = hu, xihv, yi (We use the Riesz isomorphism between Vˇ and V¯ ). This inner product corresponds to the Hilbert-Schmidt norm

hu ⊗ v, x ⊗ yi = T r((u ⊗ v)∗x ⊗ y)

89 90 Schur’s orthogonality relations (Theorem 4.5) say 1 hcu, cxi = hu, xihv, yi (dim π = dim V ), v y dim π and hence √ u −1 c : u ⊗ v → cv is dim π × an isometry map

91 The inverse of c is nearly the extension π1 of π. Let us compute 1 u π (cv ). For this purpose, we apply this operator to a vector y and compute the inner product of the result with a vector x Z 1 u u u x hx, π (cv )yi = cv (s)hx, π(s)yids = hc¯v , cy i

This quantity will vanish when π¯ is not equivalent to π.

92 However, Schur’s relations give

1 u u x −1 hx, π (¯cv )yi = hcv , cy i = (dim π) hu, xihv, yi = (dim π)−1hx, hv, yiui = hx, (dim π)−1v ⊗ u(y)i

This implies 1 u −1 π (¯cv ) = (dim π) v ⊗ u and by linearity 1 −1 ∗ π (¯cA) = (dim π) A

93 2 Since the c¯A are the coeﬃcients of π¯, we see that on L (G, π¯), f → π1(f) is of the form √ 1 −1 π 2 ¯ = dim π × an isometry map L (G,π) The composite: conj End(V ) → L2(G, π) −→ L2(G, π¯) → End(Vˇ ) 1 A → cA c¯A = f → π (f) can be identiﬁed with

(dim π)−1 · (A → A∗) = (dim π)−1 × an isometry map

94 NOTE: From now on, we write π1(f) as simply π(f). Theorem 5.2 (Plancherel theorem). Let G be a compact group. For π ∈ Gˆ denote by Pπ : L2(G) → L2(G, π), the orthogonal projector on the isotypical component of π, and for f ∈ L2(G), let fπ = Pπ(f) so P 2 that the series Gˆ fπ converges to f in L (G). Then (a) fπ(x) = dim π · T r(π(fˇ)π(x)). Here fˇ(x) = f(x−1) (b) ||f ||2 = dim π · ||π(fˇ)||2 (Hilbert-Schmidt norm on RHS). π 2 HS (c) ||f||2 = P dim π · ||π(f)||2 (Parseval equality) 2 π∈Gˆ 2

95 Proof. The orthogonal projection fπ of f in L2(G, π), is the element cA of this space having the same inner product with all elements cB of L2(G, π). Let us determine A as a function of f. We must have

−1 ∗ hcB, fi = hcB, fπi = hcB, cAi = (dim π) T r(B A)

But Z Z hcB, fi = cB(x)f(x)dx = T r(Bπ(x))f(x)dx Z Z = T r(π(x−1)B∗)f(x)dx = T r(B∗ π(x−1)f(x)dx)

= T r(B∗π(fˇ))

Comparing the two results obtained for all B, we indeed ﬁnd

A = (dim π) · π(fˇ)

96 This gives ˇ fπ(x) = cA(x) = T r(Aπ(x)) = (dim π) · T r(π(f)π(x)) as asserted in part (a). Moreover Schur’s relations show that

||f ||2 = ||c ||2 = (dim π)−1||A||2 = (dim π) · ||π(fˇ)||2 π 2 A 2 HS HS Thus (b) is proved and (c) follows from the observation that f and fˇ have the same L2(G): we can interchange f and fˇ. Also observe that the dimensions of π and πˇ are the same.

97 5.3 Exercises

Exercise 5.3. Let G be a compact group, f, g ∈ L1(G). For any representation σ of G (in a Banach space), prove

σ(f ∗ g) = σ(f) · σ(g)

If σ is unitary, prove also σ(f) = σ(f ∗). (recall that f ∗(x) = f(x−1)) Exercise 5.4. Show that the “extensions” of the regular representations of a compact group G are given by

l(f)(ϕ) = f ∗ ϕ , r(g)(ϕ) = ϕ ∗ gˇ where f, g ∈ L1(G) and ϕ ∈ L2(G) (recall that gˇ(x) = g(x−1)). Conclude from this that

||f ∗ ϕ||2 ≤ ||f||1 ||ϕ||2

98 Moreover using exercise 5.3 deduce the associativity (f ∗ g) ∗ ϕ = f ∗ (g ∗ ϕ)

Here f, g ∈ L1(G) and ϕ ∈ L2(G) or all three functions in C(G). Also check that for any representation σ of G σ(l(x)f) = σ(x)σ(f) , σ(r(x)f) = σ(f)σ(x−1) Exercise 5.5. Let G be a compact group and denote by 1 1 1 Linv = Linv(G) the closure of L (G) of the subspace of continuous functions f satisfying f(xy) = f(yx) (or equivalently f(y−1xy) = f(x)) 1 1 for all x, y ∈ G. Show that Linv is contained in the center of L (G) (as 1 convolution algebra: prove that f ∗ g = g ∗ f for f, g ∈ Linv). For any irreducible representation π : G → Gl(V ) prove that

−1 1 π(f) = (dim π) hχ, fi1V (f ∈ Linv) where χ(g) = T r(π(g)) and hχ, fi = R χ(g)f(g)dg.

99 Hint: Use Schur’s lemma to prove that π(f) is a scalar operator and then take traces to determine the value of the constant in this multiple of the identity. Exercise 5.6. Let σ : G → Gl(V ) be a unitary representation of a compact group G. Check that σ(1) = P (here 1 is the constant function 1 in L1(G)) is the orthogonal projector V → V G on the subspace of G-invariants of V .(Hint: show that 1 ∗ 1 = 1 and 1∗ = 1 in 1 R L (G); more generally 1 ∗ f = f ∗ 1 is the constant function f(x)dx.) Exercise 5.7. Show that the “extended” left regular representation

1 1 2 l = l : L (G) → End(L (G)) has trivial kernel {0}.(Hint: Let 0 6= f ∈ L1(G) and construct a R sequence (gn) ⊆ C(G) with gn ≥ 0, gn(x)dx = 1 and 1 l(f)(gn) = f ∗ gn → f 6= 0). Conclude that if 0 6= f ∈ L (G), there exists a π ∈ Gˆ such that π(f) 6= 0. (continued on the next page)

100 Finally prove that

1 L (G) commutative ⇐⇒ G commutative Exercise 5.8. Let G be a compact group, π ∈ Gˆ and consider the adjoint representation of G in End(V ) (V = Vπ) deﬁned by the following composition Ad: G −→ G × G −→ End(V ) s −→ (s,s) −→ (A → π(s)Aπ(s)−1) (s,t) −→ (A → π(t)Aπ(s)−1) Prove that the multiplicity of the identity representation in this adjoint representation is 1. (This identity representation acts on the subspace of scalar operators: Schur’s lemma).

101 Exercise 5.9. The decomposition of the biregular representation of a compact group G in L2(G) gives the decomposition of the left (resp. right) regular representation simply by the composition with

i1 : G → G × G (resp. i2 : G → G × G s → (s, e) s → (e, s)) Conclude that l =∼ ⊕ˆ πˇ ⊗ 1 =∼ ⊕ˆ dim π · πˇ = dim π · π r =∼ ⊕ˆ 1 ⊗ π =∼ ⊕ˆ dim π · π

102 6 Characters and Group algebras

Let (π,V ) be a ﬁnite dimensional representation of a compact group G. The character χ = χπ of π is the (complex valued) continuous function on G deﬁned by χ(x) = T r(π(x))

Note that this is the function cA for A = idV ∈ End(V ). When dim (V ) = 1, χ and π can be identiﬁed. In this case χ is a homomorphism. Quite generally, since the trace satisﬁes the identity T r(AB) = T r(BA), we see that the characters of two equivalent representations are equal.

103 Moreover, characters satisfy

χ(xy) = χ(yx) as χ(y−1xy) = χ(x)(x, y ∈ G)

Thus characters are invariant functions

−1 χ ∈ Cinv = {f ∈ C(G): f(y xy) = f(x), x and y ∈ G}

Observe that

1 1 Linv = closure of C(G) in L (G) 2 2 Linv = closure of C(G) in L (G) Invariant functions are also called central functions, they belong to the center L1(G) with respect to convolution.

104 Still quite generally, the character of the contragredient πˇ of π is given by

t −1 −1 −1 χπˇ (x) = T r(πˇ(x)) = T r( π(x )) = T r(π(x )) = χ(x ) hence χπˇ =χ ˇ When π is unitary, π(x−1) = π(x)∗ (πˇ is equivalent to π¯) and χˇ is the complex conjugate of χ.

105 One can also check without diﬃculty that for two ﬁnite dimensional representations π and σ of G

χπ⊕σ = χπ + χσ , χπ⊗σ = χπ · χσ

106 When π is irreducible, Schur’s lemma shows that the elements z in the center Z of G are mapped to scalar operators by π: π(z) = λzidV so −1 that χ(z) = λzdim (V ). Thus the restriction of (dim V ) χ to the center Z is a homomorphism

× λ : Z −→ C This is the central character of π: it is independent of the special model chosen in the equivalence class of π. In particular if λ(z) is not contained in {±1} , π and π¯ are not equivalent: their central characters are diﬀerent. Also observe that χ(e) = dim (V ) (= dim π)

107 Proposition 6.1. Any continuous central function f ∈ Cinv on a compact group G is a uniform limit of linear combinations of characters of irreducible representations of G.

108 Proof. Let > 0. By the Peter-Weyl theorem 3.17, we know that there is a ﬁnite dimensional representation of (σ,V ) and a A ∈ End(V ) with

|f(x) − T r(Aσ(x))| < (x ∈ G)

In this representation, replace x by one of its conjugates yxy−1:

|f(x) − T r(Aσ(yxy−1))| = |f(x) − T r(σ(y−1)Aσ(y)σ(x))| <

Integrating over y, we have Z |f(x) − T r(Bσ(x))| < where B = σ(y−1)Aσ(y)dy

By the invariance of the Haar measure, the operator B commutes with all the operators σ(x).

109 Hence, if we decompose σ into isotypical components ∼ M ∼ M σ = nππ = π ⊗ 1nπ π π the operator B will have the form M B = iddim π ⊗ Bπ and M Bσ(x) = σ(x)B =∼ π ⊗ Bπ X T r(Bσ(x)) = aπχπ(x)(aπ = T r(Bπ))

This shows that X |f(x) − aπχπ(x)| < ﬁnite

110 Theorem 6.2. Let π and σ be two ﬁnite dimensional representations of a compact group G with respective characters χπ and χσ. Then

hχπ, χσi = dim HomG(Vπ,Vσ)

111 Proof. By Lemma 4.3, we know that the integral Z πˇ(x) ⊗ σ(x)dx is an expression for the projector

\ : Vˇπ ⊗ Vσ −→ (Vˇπ ⊗ Vσ)G

Hom(Vπ,Vσ) −→ HomG(Vπ,Vσ)

The dimension of the image space is the trace of this projector. Thus Z hχπ, χσi = χπ(x)χσ(x)dx = dim HomG(Vπ,Vσ)

112 Corollary 6.3. Let π be a ﬁnite dimensional representation of G. Then p π is irreducible ⇐⇒ ||χπ||2 = hχπ, χπi = 1

Corollary 6.4. Let π, σ ∈ Gˆ. Then

hχπ, χσi = δπσ (= 1 if π is equivalent to σ, 0 otherwise)

113 Corollary 6.5. Let σ be a ﬁnite dimensional representation of G and L ˆ σ = I nππ (summation over a ﬁnite subset I ⊆ G) be a decomposition into irreducible components. Then (a) nπ = hχπ, χσi is well determined 2 P 2 (b) ||χσ|| = I nπ

L n Proof. Let Vσ = (Vτ ⊗ C τ ). Since each G-morphism Vσ → Vπ n must vanish on all isotypical components Vτ ⊗ C τ where τ is not equivalent to π, we have

nπ HomG(Vσ,Vπ) = HomG(Vπ ⊗ C ,Vπ) nπ nπ = C ⊗ HomG(Vπ,Vπ) = C This proves assertion (a). Moreover (b) follows from Pythagoras theorem and Corollary 6.3

114 Corollary 6.6. The set of characters (χπ)π∈Gˆ is an orthonormal basis 2 2 of the Hilbert space L (G)inv. Every function f ∈ L (G)inv can be expanded in the series

X 2 f = hχπ, fiχπ (convergence in L (G)) Gˆ

115 Theorem 6.7. Let G be a compact group. For π ∈ Gˆ, let Pπ denote the projector L2(G) → L2(G, π) onto the isotypical component of π (in the right regular representation). Then Pπ is given by the convolution with the normalized character ϑπ = dim π · χπ Pπ : f → fπ = Pπf = f ∗ ϑπ

Proof. We have already seen that

fπ(x) = dim π · T r(π(fˇ)π(x)) Thus Z fπ(x) = dim π · T r f(y−1)π(y)π(x)dy Z = dim π · f(y)T r(π(y−1x))dy = f ∗ ϑπ(x)

116 Exercise 6.8. Let H1 and H2 be two Hilbert spaces. Prove that for any operator A in H1 ⊗ H2 which commutes to all operators T ⊗ 1

T ∈ End(H1) can be written in the form 1 ⊗ B for some B ∈ End (H2).

(Hint: Introduce an orthonormal basis (ei) of H1 and write A as an matrix of blocks with respect to this basis

X i i A(ej ⊗ x) = ei ⊗ Ajx (Aj ∈ End(H2)). i

Using the commutations (Pj ⊗ 1)A = A(Pj ⊗ 1) where Pj is the i orthogonal projector on Cej, conclude that Aj = 0 for i 6= j. Finally, using the commutation relations of A with the operators Uji ⊗ 1

Uji(ei) = ej ,Uji(ek) = 0 for k 6= i,

i conclude that Ai = B ∈ End(H2) is independent of i.)

117 Exercise 6.9. Check that the formula X f = hχπ, fiχπ coincides with the Fourier inversion formula.

118 7 Induced representations and Frobenius-Weil reciprocity

Suppose that K is a closed (hence compact) subgroup of G. Recall that K\G is the space of right cosets Kg, g ∈ G.

119 Suppose for the moment that K\G is ﬁnite, i.e.

K\G = {Kg1, . . . , Kgn} for some n so that G is the disjoint union of

Kg1, . . . , Kgn. For each s ∈ G we have an associated permutation π(s) of {1, . . . , n} −1 that sends i to the unique j with Kgis = Kgj. We can deﬁne an representation ρ of G on Cn by

ρ(s)(a1, . . . , an) := (aπ−1(1), . . . , aπ−1(n))

120 Equivalently, if we think of Cn as the space of functions from K\G into C, then, for s ∈ G and a coset L ∈ K\G, ρ(s)f(L) = f(Ls)

The space of functions from K\G into C can be identiﬁed with the space of functions from G to C that are constant on right cosets of K, that is, with the space of functions f : G → C such that f(kx) = f(x), k ∈ K, x ∈ G

Note that ρ is just the right regular representation for G restricted to this subspace. Can we build other representations of G by similar constructions?

121 We return to the case where K is an arbitrary closed subgroup of G (so that K\G is not necessarily ﬁnite).

122 p The canonical projection G → K\G pushes the Haar measure ds on G forward to a measure dx˙ on K\G characterized by Z Z f(x ˙)dx˙ = f(p(s))ds (f ∈ C(K\G)) K\G G Lemma 7.1. Negligible sets in K\G (relative to the measure dx˙) are those sets N for which p−1(N) is negligible in G (relative to the Haar measure ds of G). Moreover, for any f ∈ C(G) (or by extension for any f ∈ L1(G)) Z Z Z f(kx) dk dx˙ = f(x)dx K\G K G In particular, the measure dx˙ is invariant under left translations from G in K\G.

Proof. Exercise.

123 Now let (σ,Vσ) be a unitary representation of K. We deﬁne the Hilbert space L2(G, Vσ) as the completion of the space C(G, Vσ) of continuous functions G → Vσ with the norm Z ||f||2 = ||f(x)||2dx G The norm under the integral sign is computed in Vσ. Note that L2(G, Vσ) is a Hilbert space with inner product Z hf, gi = hf(x), g(x)idx G The inner product under the integral sign is computed in Vσ.

124 The elements

2 f ∈ L (G, Vσ) such that f(kx) = σ(k)f(x) for all k ∈ K constitute a subspace H ⊆ L2(G, Vσ). Since ||f(x)|| only depends on the coset Kx of x for f ∈ H (σ is assumed to be unitary), we can take the norm and inner product on H to be deﬁned by Z 2 2 ||f||H = ||f(x ˙)|| dx˙ K\G Z hf, giH = hf(x ˙), g(x ˙)i dx˙ K\G

As before, the norm and the inner product under the integral sign are computed in Vσ.

125 Note that if f ∈ H, r is the analogue of the right regular representation of G on L2(G, Vσ) (that is, (r(s)h)(x) = h(xs) for h ∈ L2(G, Vσ) and s, x ∈ G) and k ∈ K, then

(r(s)f)(kx) = f(kxs) = σ(k)f(xs) = σ(k)(r(s)f)(x), so that r(s)f ∈ H also. That is, H is r-invariant. G The induced representation ρ = IndK (σ) is by deﬁnition the “right regular representation” r of G restricted to H ⊆ L2(G, Vσ). The induced representation is unitary. For example, if σ is the identity representation of K (in dimension 1), Vσ = C, L2(G, Vσ) = L2(G), then H is simply L2(K \ G) and we get the construction considered at the start of this section.

126 Write HG for the G-ﬁxed elements in H (with respect to the action given by ρ).

K As before, write Vσ for the K-ﬁxed elements of Vσ (with respect to the action given by σ). G K Proposition 7.2. (1) The linear map H → Vσ given by f → f(e) is an isomorphism of vector spaces. (2) Let (π,Hπ) be a unitary representation of G. Then there is an equivalence

G ∼ G ˆ π ⊗ IndK (σ) → IndK (π|K ⊗ σ): Hπ⊗H → He given by v ⊗ f → ϕ with ϕ(x) = [π(x)v] ⊗ f(x)

127 Proof. The elements of HG are certainly functions f : G → Vσ which are (equal nearly everywhere to a) constant

f(x) = f(ex) = r(x)f(e) = f(e)

In particular, f(k) = f(e), k ∈ K By deﬁnition of H, f ∈ H implies

f(k) = f(ke) = σ(k)f(e), k ∈ K

K Thus, f(e) = σ(k)f(e) for all k ∈ K and so f(e) ∈ Vσ , giving part (1) of the proposition.

128 To check part (2), let us ﬁrst show that the functions ϕ (as deﬁned in the proposition) belong to the space of the induced representation G IndK (π|K ⊗ σ) ϕ(kx) = [π(kx)v] ⊗ f(kx) = [π(k)π(x)v] ⊗ [σ(k)f(x)]

= [[π|K ⊗ σ](k)](ϕ(x))

129 Next we show that v ⊗ f → ϕ is a G-morphism (intertwining π ⊗ ρ) and G ρ˜ = IndK (π|K ⊗ σ), which we recall is the restriction of the the right regular representation to H˜ . Note that

[π ⊗ ρ](s)(v ⊗ f) = [π(s)v] ⊗ [ρ(s)f] is mapped to the function on G given by

x 7→ [π(x)π(s)v] ⊗ [(ρ(s)f)(x)] = [π(xs)v] ⊗ f(xs) = (˜ρ(s)ϕ)(x) as desired.

130 Now we check that v ⊗ f → ϕ is isometric and hence injective. If (ei) is an orthonormal basis of Hπ, every element of Hπ⊗ˆ H can be written P P 2 uniquely as ei ⊗ fi with ||fi|| < ∞, and such an element has its P image the function ϕ = ϕi given by X x 7→ π(x)ei ⊗ fi(x)

The norm of ϕ is Z Z 2 2 X 2 ||ϕ||H˜ = ||ϕ(x)|| dx˙ = || π(x)ei ⊗ fi(x)|| dx˙ K\G K\G Z Z X 2 X 2 = ||fi(x)|| dx˙ = ||fi(x)|| dx˙ K\G K\G X 2 X 2 = ||fi||H = || ei ⊗ fi||

The third equality is justiﬁed by the fact that π(x)ei is also an orthonormal basis of Hπ, since π is unitary.

131 Finally to see that v ⊗ f → ϕ is onto, it is enough to see that all continuous functions Φ ∈ H˜ belong to the image. The function Φ has a unique expansion in the orthonormal basis (π(x)ei) of Hπ: X Φ(x) = π(x)ei ⊗ fi(x)(fi(x) ∈ V )

2 X 2 ||Φ(x)|| = ||fi(x)||

Therefore X π(kx)ei ⊗ fi(kx) = Φ(kx) = [π(k) ⊗ σ(k)]Φ(x) X = [π(k)π(x)ei] ⊗ [σ(k)fi(x)]

The uniqueness of the decomposition gives fi(kx) = σ(k)fi(x). Thus, fi ∈ H, as required.

132 Note that if we consider C as a trivial G- or K- space, then ∼ G HomG(C,H) = H and ∼ K HomK (C,Vσ) = Vσ

(We identify A ∈ HomG(C,H) with the image h ∈ H of 1 ∈ C and observe that the assumption that A intertwines with the induced representation H is equivalent to the assumption that ρ(g)h = h for all g ∈ G. A similar comment holds for HomK (C,Vσ).) Therefore, the isomorphism of part (1) of the preceding proposition can be written as ∼ HomG(C,H) → HomK (C,Vσ) This form admits the following generalization.

133 Theorem 7.3 (Frobenius-Weil). Let (π,Hπ) be a unitary representation of a compact group group G and (σ,Vσ) a unitary G representation of one of its closed subgroups K. Put ρ = IndK (σ) and H = Hρ. Then there is a canonical isomorphism

∼ MorG(Hπ,Hρ) → MorK (Hπ,Vσ) where we take for morphisms between two representations spaces, the Hilbert-Schmidt morphisms. (If our spaces are ﬁnite dimensional, then

MorG is what we have written before as HomG and, similarly, MorK is just HomK .)

134 Proof. We have seen in ﬁnite dimensions that in the identiﬁcation ˇ ∼ Hπ ⊗ Hρ → Hom(Hπ,Hρ) the representation πˇ ⊗ ρ is transformed into the representation

−1 A → ρ(s)Aπ(s) (s ∈ G, A ∈ Hom(Hπ,Hρ))

Much the same thing happens for inﬁnite dimensional unitary representations , we complete the algebraic tensor product of Hilbert spaces and get an isomorphism with the space of Hilbert-Schmidt operators. ˇ ∼ Hπ⊗ˆ Hρ → MorHS(Hπ,Hρ)

135 ˇ Thus G-morphisms Hπ → Hρ correspond to G-invariants in Hπ⊗ˆ Hρ. In other words, ∼ ˇ G MorG(Hπ,Hρ) = (Hπ⊗ˆ Hρ) ˇ ˜ Since Hπ⊗ˆ Hρ = H can be identiﬁed with the space of the representation of G induced from the representation π|K ⊗ σ of K (part (2) of Proposition 7.2), we infer from part (1) of Proposition 7.2 that

ˇ G ∼ ˜ G ∼ ˇ K (Hπ⊗ˆ Hρ) = H = (Hπ⊗ˆ Vσ)

The conclusion follows from the identity

ˇ K ∼ (Hπ⊗ˆ Vσ) = MorK (Hπ,Vσ)

136 Corollary 7.4. Consider (π,Hπ) ∈ Gˆ and (σ,Vσ) ∈ Kˆ . Then the G multiplicity of π in IndK (σ) is the same as the multiplicity of σ in π|K

Proof. Denote the (inﬁnite dimensional in general) space of G ρ = IndK (σ) by Hρ.

Any G-morphism Hπ → Hρ must send Hπ into the isotypical component π in Hρ: this isotypical component is isomorphic to a ⊕I Hπ whence M M MorG(Hπ, ⊕I Hπ) = MorG(Hπ,Hπ) = C I I by Schur’s lemma.

137 On the other hand, every K-morphism Hπ → Vσ must vanish oﬀ of those components of Hπ into a direct sum of irreducibles (for K) which are not equivalent to (σ,Vσ). Let us write the direct sum of the copies of (σ,Vσ) in Hπ as Vσ ⊗ Cm. Then

m ∼ m MorK (Hπ,Vσ) = MorK (Vσ ⊗ C ,Vσ) → Mor(C , MorK (Vσ,Vσ))

138 To see the isomorphism

m ∼ m MorK (Vσ ⊗ C ,Vσ) → Mor(C , MorK (Vσ,Vσ)),

m m note that if (ei) is a basis for C we can write any element of Vσ ⊗ C P m as i vi ⊗ ei. Any map A ∈ MorK (Vσ ⊗ C ,Vσ), is speciﬁed by the m maps v 7→ A(v ⊗ ei) = Ai belonging to MorK (Vσ,Vσ), and we can m think of these m maps as a single map from C to MorK (Vσ,Vσ) deﬁned by ei 7→ Ai.

139 Since MorK (Vσ,Vσ) = CidV (Schur’s lemma), we have ∼ m MorK (Hπ,Vσ) = dual of C The isomorphism of the theorem implies equality of the dimension of the spaces. They are respectively

G Card(I) = multiplicity of π in ρ = IndK (σ)

m = multiplicity of σ in π|K

140 8 Representations of the symmetric group

8.1 Young subgroups, tableaux and tabloids

If λ = (λ1, λ2, . . . , λl) is a partition of n, then write λ ` n. We also use P the notation |λ| = i λi, so that a partition of n satisfies |λ| = n Definition 8.1. Suppose λ = (λ1, λ2, . . . , λl) ` n. The Ferrers diagram or shape, of λ is an array of n dots having l left-justified rows with row i containing λi dots for 1 ≤ i ≤ l.

Deﬁnition 8.2. For any set T , let ST be the set of permutations of T .

Let λ = (λ1, λ2, . . . , λn) ` n. Then the corresponding Young subgroup of Sn is

Sλ = S{1,2,...λ1} × S{λ1+1,λ1+2,...,λ1+λ2} × S{n−λl+1,n−λl+2,...,n}

141 Now consider the representation (1 ↑Sn ), by which we mean the Sλ representation of Sn induced by the trivial representation of the subgroup Sλ. If π1, π2, . . . , πk is a transversal for Sλ, then the vector space

λ V = C{π1Sλ, π2Sλ, . . . , πkSλ} is a module for our induced representation . Deﬁnition 8.3. Suppose λ ` n.A Young tableau of shape λ, is an array t obtained by replacing the dots of the Ferrers diagram with the numbers 1, 2, . . . , n bijectively.

Deﬁnition 8.4. Two λ-tableaux t1 and t2 are row equivalent, t1 ∼ t2, if the corresponding rows of the two tableaux contain the same elements. A tabloid of shape λ or λ-tabloid is then

{t} = {t1|t1 ∼ t} where shape(t) = λ

142 Now π ∈ Sn acts on a tableau t = (ti,j) of shape λ ` n as follows:

πt = (π(ti,j))

This induces an action on tabloids by letting

π{t} = {πt}

Exercise: Check that this is well deﬁned, namely independent of the choice of t. Deﬁnition 8.5. Suppose λ ` n. Let

λ M = C{{t1},..., {tk}}

λ where {t1},..., {tk}, is a complete list of λ-tabloids. Then M is called the permutation module corresponding to λ.

143 Deﬁnition 8.6. Any G-module M is cyclic if there is a v ∈ M such that

M = CGv where Gv = {gv|g ∈ G}. In this case we say that M is generated by v. Proposition 8.7. If λ ` n, then M λ is cyclic, generated by any given λ-tabloid. In addition, dim M λ = n!/λ!, the number of λ-tabloids.

Theorem 8.8. Consider λ ` n with the Young subgroup Sλ and tabloid λ λ λ λ {t }, as before. Then V = CSnSλ and M = CSn{t } are isomorphic as Sn-modules.

Proof. Let π1, π2, . . . , πk be a transversal for Sλ. Deﬁne a map:

θ : V λ → M λ

λ by θ(πiSλ) = {πit } for i = 1, 2, . . . , k and linear extension. It is not hard to verify that θ is the desired Sn-isomorphism of modules.

144 8.2 Dominance and Lexicographic ordering

Deﬁnition 8.9. Suppose λ = (λ1, . . . , λl) and µ = (µ1, µ2, . . . , µm) are partitions of n. Then λ dominates µ, written as λ ¤ µ if

λ1 + λ2 + ... + λi ≥ µ1 + µ2 + . . . µi for all i ≥ 1. If i > l (respectively, i > m), then we take λi (respectively,

µi) to be zero.

145 Lemma 8.10 (Dominance lemma for partitions). Let tλ and sµ be tableaux of shape λ and µ respectively. If for each index i, the elements of row i of sµ are all in diﬀerent columns in tλ, then λ ¤ µ.

Proof. By hypothesis, we can sort the entries in each column of tλ so that the elements of rows 1, 2, . . . , i of sµ all occur in the ﬁrst i rows of tλ. Now note that

λ λ1 + λ2 + ··· + λi = number of elements in the ﬁrst i rows of t ≥ number of elements of sµ in the ﬁrst i rows of tλ

= µ1 + µ2 + ··· + µi

146 Deﬁnition 8.11. Let λ = (λ1, λ2, . . . , λl) and µ = (µ1, µ2, . . . , µm) be partitions of n. Then λ < µ in lexicographic order if for some index i,

λj = µj for j < i and λi < µi

Proposition 8.12. If λ, µ ` n with λ ¤ µ then λ ≥ µ

Proof. If λ 6= µ, then find the first index i where they differ. Thus, Pi−1 Pi−1 Pi Pi j=1 λj = j=1 µj and j=1 λj > j=1 µj (since λ £ µ). So λi > µi.

147 8.3 Specht Modules

Deﬁnition 8.13. Suppose now that the tableau t has rows

R1,R2,...Rl and columns C1,C2,...,Ck. Then

Rt = SR1 × SR2 × ... × SRl and

Ct = SC1 × SC2 × ... × SCk are the row-stabilizer and column-stabilizer of t respectively.

Note that our equivalence classes can be expressed as {t} = Rtt.

148 In general, given a subset H ⊆ Sn, we can form the group algebra elements X H+ = π π∈H and X H− = sgn(π)π π∈H + For a tableau t, the element Rt is already implicit in the corresponding tabloid by the remark at the end of the previous paragraph. However we will also need to make use of

def − X κt = Ct = sgn(π)π

π∈Ct

Note that if t has columns C1,C2,...,Ck, then κt factors as

κt = κC1 κC2 . . . κCk

149 Deﬁnition 8.14. If t is a tableau, then the associated polytabloid is

et = κt{t}

150 Lemma 8.15. Let t be a tableau and π be a permutation. Then −1 1. Rπt = πRtπ −1 2. Cπt = πCtπ −1 3. κπt = πκtπ

4. eπt = πet

Proof. 1. We have the following equivalent statements:

σ ∈ Rπt ←→ σ{πt} = {πt} ←→ π−1σπ{t} = {t} −1 ←→ π σπ ∈ Rt −1 ←→ σ ∈ πRtπ

The proofs of 2 and 3 are similar to that of part 1.

151 4. We have

−1 eπt = κπt{πt} = πκtπ {πt} = πκt{t} = πet

Deﬁnition 8.16. For any partition λ, the corresponding Specht module, λ λ S is the submodule of M spanned by the polytabloids et, where t is of shape λ

152 Proposition 8.17. The Sλ are cyclic modules generated by any given polytabloid.

Proof. This follows from part 4 of Lemma 8.15

153 8.4 The Submodule theorem

− P Recall that H = π∈H (sgn π)π for any subset H ⊆ Sn. If H = {π}, then we write π− for H−. We need the unique inner product on M λ for which

h{t}, {s}i = δ{t},{s} (8.1)

154 Lemma 8.18 (Sign Lemma). Let H ≤ Sn be a subgroup. Then 1. If π ∈ H, then πH− = H−π = (sgn π)H− Equivalently, π−H− = H−. 2. For any u, v ∈ M λ,

hH−u, vi = hu,H−, vi

3. If the transposition (b, c) ∈ H, then we can factor

H− = k( − (b, c)) where k ∈ C[Sn] 4. If t is a tableau with b, c in the same row of t and (b, c) ∈ H, then

H−{t} = 0

155 Proof. Exercise

156 Corollary 8.19. Let t = tλ be a λ-tableau and s = sµ be a µ-tableau, where λ, µ ` n. If κt{s}= 6 0, then λ ¤ µ. And if λ = µ, then

κt{s} = ±et

Proof. Suppose b and c are two elements in the same row of sµ. Then λ they cannot be in the same column of t , for if so then κt = k( − (b, c)) and κt{s} = 0 by parts 3 and 4 in the preceding lemma. Thus the dominance lemma 8.10 yields λ ¤ µ.

If λ = µ, then we must have {s} = π{t} for some π ∈ Ct by the same argument that established the dominance lemma. Using part 1 of the preceeding lemma yields

κt{s} = κtπ{t} = (sgn π)κt{t} = ±et

157 µ Corollary 8.20. If u ∈ M and shape t = µ, then κtu is multiple of et.

P Proof. We can write u = i ci{si}, where the si are µ-tableaux. By P the previous corollary, κtu = i ±ciet.

158 Theorem 8.21 (Submodule Theorem). Let U be a submodule of M µ. Then U ⊇ Sµ or U ⊆ Sµ⊥ In particular the Sµ are irreducible.

Proof. Consider u ∈ U and a µ-tableau t. By the preceding corollary, we know that κtu = fet for some ﬁeld element f. There are two cases, depending on which multiples can arise. Suppose that there exists a u and a t with f 6= 0. Then since u is in the submodule U, we have fet = κtu ∈ U. Thus et ∈ U (since f is nonzero) and Sµ ⊆ U (since Sµ is cyclic).

159 On the other hand, suppose we always have κtu = 0. We claim that this forces U ⊆ Sµ⊥. Consider any u ∈ U. Given an arbitrary µ- tableau t, we can apply part 2 of the sign lemma to obtain

hu, eti = hu, κt{t}i = hκtu, {t}i = h0, {t}i = 0

µ µ⊥ Since the et span S , we have u ∈ S , as claimed.

160 λ µ Proposition 8.22. Suppose θ ∈ HomSn (S ,M ) is nonzero. Then λ ¤ µ. Moreover, if λ = µ, then θ is multiplication by a scalar.

Proof. Since θ 6= 0, there is some basis vector et such that θ(et) 6= 0. Because h·, ·i is an inner product with complex scalars, M λ = Sλ ⊕ Sλ⊥. λ µ Thus we can extend θ to an element of HomSn (M ,M ) by setting θ(Sλ⊥) = 0. So X 0 6= θ(et) = θ(κt{t}) = κtθ({t}) = κt( ci{si}) i where the si are µ-tableaux. By Corollary 8.19, we have λ ¤ µ.

In the case λ = µ, Corollary 8.20 yields θ(et) = cet for some constant c. So for any permutation π,

θ(eπt) = θ(πet) = πθ(et) = π(cet) = ceπt Thus θ is multiplication by c.

161 Theorem 8.23. The Sλ for λ ` n form a complete list of irreducible

Sn-modules.

Proof. The Sλ are irreducible by the submodule theorem. Since we have the right number of modules for a full set, it suﬃces to show that they are pairwise inequivalent. But if Sλ =∼ Sµ, then there is a λ µ µ µ nonzero homomorphism θ ∈ HomSn (S ,M ), since S ⊆ M . Thus λ ¤ µ (Proposition 8.22). Similarly µ ¤ λ, so λ = µ.

162 8.5 Standard Tableaux and a Basis for Sλ

Deﬁnition 8.24. A tableau t is standard if the rows and columns of t are increasing sequences. In this case we also say that the corresponding tabloid and polytabloid are standard. Theorem 8.25. The set

{et : t is a standard λ-tableau} is a basis for Sλ.

Proof. Omitted

163 8.6 The Branching Rule

Deﬁnition 8.26. If λ is a diagram, then an inner corner of λ is a node (i, j) ∈ λ whose removal leaves the Ferrers diagram of a partition. Any partition obtained by such a removal is denoted by λ−. An outer corner of λ is a node (i, j) ∈/ λ whose addition produces the Ferrers diagram of a partition. Any partition obtained by such an addition is denoted by λ+

164 Lemma 8.27. We have X − f λ = f λ λ−

Proof. Every standard tableau of shape λ ` n consists of n in some inner corner together with a standard tableau of shape λ− ` n − 1. The result follows.

165 Theorem 8.28 (Branching Rule). If λ ` n, then λ ∼ L λ− 1. S ↓Sn−1 = λ− S , and + λ Sn+1 ∼ L λ 2. S ↑ = λ+ S

Proof. 1. Let the inner corners of λ appear in rows r1 < r2 < ··· < rk. For each i, let λi denote the partition of λ− obtained by removing the corner cell in row ri. In addition, if n is at the end of row ri of tableau t i i (respectively, in row ri of tabloid {ti}), then t (respectively, {t } ) will be the array obtained by removing the n. Now given any group G with module V and submodule W , it is easy to see that V =∼ W ⊕ (V/W ), where V/W is the quotient space. Thus it suﬃces to ﬁnd a chain of subspaces {0} = V (0) ⊂ V (1) ⊂ V (2) ⊂ · · · ⊂ V (k) = Sλ

166 (i) (i−1) ∼ λi (i) such that V /V = S as Sn−1 modules for 1 ≤ i ≤ k. Let V be the vector space spanned by the standard polytabloids et where n appears in t at the end of one of rows r1 through ri. We show that the V (i) are our desired modules as follows.

λ λi Deﬁne maps θi : M → M by linearly extending   {ti} if n in row ri of {t}, {t} →θi  0 otherwise.

Verify that θi is an Sn−1-homomorphism. Furthermore, for standard t we have  i θi  et if n is in row ri of t, et →  0 if n is in row rj of t, where j < i.

167 This is because any tabloid appearing in et, t standard, has n in the same row or higher than in t. Since the standard polytabloids form a basis for the corresponding Specht module, (i) λi θiV = S and (i−1) V ⊆ kerθi. We can construct the chain

(0) (1) (1) (2) (2) (k) λ {0} = V ⊆ V ∩kerθ1 ⊆ V ⊆ V ∩kerθ2 ⊆ V ⊆ · · · ⊆ V = S

But (i) V i dim = dim θ V (i) = f λ (i) i V ∩ kerθi

168 By the preceding lemma, the dimensions of these quotients add up to dim Sλ. Since this leaves no space to insert extra modules, the chain must have equality for the ﬁrst, third, etc. containments. Furthermore,

(i) (i) V V i ∼ ∼ Sλ (i−1) = (i) = V V ∩ kerθi as desired. 2. We will show that this part follows from the ﬁrst by Frobenius reciprocity. In fact, parts 1 and 2 can be shown to be equivalent by the same method.

λ λ λ Sn+1 ∼ µ Let χ be the character of S . If S ↑ = ⊕µ`n+1mµS , then by λ Sn+1 ∼ P µ taking characters, χ ↑ = µ`n+1 mµχ .

169 The multiplicities are given by

λ Sn+1 µ mµ = hχ ↑ , χ i λ µ = hχ , χ ↓Sn i X − = hχλ, χµ i µ−   1 if λ = µ− =  0 otherwise   1 if µ = λ+ =  0 otherwise

170 9 Symmetric Functions

9.1 Symmetric functions in general

Let x = (x1, x2,...) be a set of indeterminates. A homogeneous symmetric function of degree n over Q is a formal power series X α f(x) = cαx α

(a) α = (α1, α2,...) ranges over all sequence of non-negative integers that sum to n (i.e. weak compositions of n)

(b) cα ∈ Q

α α1 α2 (c) x stands for the monomial x1 x2 ...

(d) f(xw(1), xw(2),...) = f(x1, x2,...) for every permutation w of the positive integers.

171 The set of all homogeneous symmetric functions of degree n over Q is denoted as Λn. If f ∈ Λm and g ∈ Λn, then it is clear that fg ∈ Λm+n (where fg is a product of the formal power series). Hence if we deﬁne

Λ = Λ0 ⊕ Λ1 ⊕ · · · (vector space direct sum)

Then Λ has the structure of a Q-algebra

172 9.2 Monomial Symmetric Functions

n Given λ = (λ1, λ2,...) ` n, deﬁne a symmetric function mλ(x) ∈ Λ by

X α mλ = x α where the sum ranges over all distinct permutations α = (α1, α2,...) of the entries of the vector λ = (λ1, λ2,...).

We call mλ a monomial symmetric function. Clearly if P α n P f = α cαx ∈ Λ then f = λ`n cλmλ. The set {mλ : λ ` n} is a (vector space) basis for Λn and hence that

dim Λn = p(n) the number of partitions of n. Moreover the set {mλ : λ ∈ Par} is a basis for Λ.

173 9.3 Elementary Symmetric Functions

We deﬁne the elementary symmetric functions eλ for λ ∈ Par by the formula X n en = m1 = xi1 ··· xin , n ≥ 1 (with e0 = m∅ = 1)

i1<...

eλ = eλ1 eλ2 ··· , if λ = (λ1, λ2,...)

174 Suppose that A = (aij)i,j≥1 is an integer matrix with ﬁnitely many nonzero entries with row and column sums X ri = aij j X cj = aij i

Deﬁne the row-sum vector row(A) and column-sum vector col(A) by

row(A) = (r1, r2,...)

column(A) = (c1, c2,...)

A (0, 1)-matrix is a matrix all of whose entries are either 0 or 1.

175 Proposition 9.1. Let λ ` n, and let α = (α1, α2,...) be a weak α composition of n. Then the coeﬃcient Mλα of x in eλ is equal to the number of (0, 1)-matrices A = (aij)i,j≥1 satisfying row(A) = λ and col(A) = α. That is, X eλ = Mλµmµ (9.1) µ`n

176 Corollary 9.2. Let Mλµ be given by Equation (9.1). Then Mλµ = Mµλ.

That is, the transition matrix between the bases {mλ : λ ` n} and

{eλ : λ ` n} is a symmetric matrix.

177 Proposition 9.3. We have Y X (1 + xiyj) = Mλµmλ(x)mµ(y) (9.2) i,j λ,µ X = mλ(x)eλ(y) (9.3) λ Here λ and µ range over Par. (It suﬃces to take |λ| = |µ|, since otherwise Mλµ = 0.)

178 α1 α2 β1 β2 α β Proof. A monomial x1 x2 ··· y1 y2 ··· = x y appearing in the Q expansion of (1 + xiyj) is obtained by choosing a (0, 1)-matrix

A = (aij) with ﬁnitely many 1’s, satisfying

Y aij α β (xiyj) = x y i,j But

Y aij row(A) col(A) (xiyj) = x y i,j α β Q so the coeﬃcient of x y in the product (1 + xiyj) is the number of (0, 1)-matrices satisfying row(A) = α and col(A) = β. Hence equation (9.2) follows. Equation (9.3) is then a consequence of (9.1)

179 0 Theorem 9.4. Let λ, µ ` n. Then Mλµ = 0 unless λ ¤ µ, while n Mλλ0 = 1. Hence the set {eλ : λ ` n} is a basis for Λ (so

{eλ : λ ∈ Par} is a basis for Λ). Equivalently, e1, e2,... are algebraically independent and generate Λ as a Q-algebra, which we write as

Λ = Q[e1, e2,...]

180 Proof. Suppose Mλµ 6= 0 so by Proposition 9.1 there is a (0, 1)-matrix A with row(A) = λ and col(A) = µ. Let A0 be the matrix with 0 0 0 row(A ) = λ and with its 1 s left justified, i.e. Aij = 1 precisely for 0 0 1 ≤ j ≤ λi. For any i the number of 1 s in the first i columns of A clearly is not less than the number of 10s in the first i columns of A, so by definition of dominance order we have col(A0) ¤ col(A) = µ. But col(A0) = λ0, so λ0 ¤ µ as desired. Moreover it is easy to see that A0 is the only (0, 1)-matrix with row(A0) = λ and col(A0) = λ0, so

Mλ,λ0 = 1.

181 The previous argument shows the following. Let λ1, λ2, . . . , λp(n) be an ordering of Par(n) that is compatible with the dominance order and such that the “reverse conjugate” order (λp(n))0,..., (λ2)0, (λ1)0 is also compatible with the dominance order. (Exercise: give an example of 1 2 such an order). Then the matrix (Mλµ), with the row order λ , λ ,... and column order (λ1)0, (λ2)0,... is upper triangular with 10s on the n main diagonal. Hence it is invertible, so {eλ : λ ` n} is a basis for Λ . (In fact it is a basis for Λn since the diagonal entries are actually 10s, Z and not merely nonzero.)

182 a1 a2 The set {eλ : λ ∈ Par} consists of all monomials e1 e2 ... (where P ai ∈ N, ai < ∞). Hence the linear independence of {eλ : λ ∈ Par} is equivalent to the algebraic independence of e1, e2,... as desired

183 9.4 Complete Homogeneous Symmetric functions

Deﬁne the complete homogeneous symmetric functions (or just complete symmetric functions) hλ for λ ∈ Par by the formulas X X hn = mλ = xi1 ··· xin (with h0 = m∅ = 1)

λ`n i1≤...≤in (9.4)

hλ = hλ1 hλ2 ··· if λ = (λ1, λ2,...)

184 Proposition 9.5. Let λ ` n, and let α = (α1, α2,...) be a weak α composition of n. Then the coeﬃcient Nλα of x in hλ is equal to the number of N-matrices A = (aij) satisfying row(A) = λ and col(A) = α. That is, X hλ = Nλµmµ, (9.5) µ`n

185 Corollary 9.6. Let Nλµ be given by Equation (9.5). Then Nλµ = Nµλ , i.e the transition matrix between the bases {mλ : λ ` n} and

{hλ : λ ` n} is a symmetric matrix. Note that by a Corollary in the next section (Corollary 9.9), it follows that the set {hλ : λ ` n} is indeed a basis.

186 Proposition 9.7. We have

Y −1 X (1 − xiyj) = Nλµmλ(x)mµ(y) (9.6) i,j λ,µ X = mλ(x)hλ(y) (9.7) λ where λ and µ range over Par (and where it suﬃces to take |λ| = |µ|).

187 9.5 An Involution

Since Λ = Q[e1, e2,...], an algebra endomorphism f :Λ → Λ is determined uniquely by its values f(en), n ≥ 1; and conversely any choice of (f(en)) ∈ Λ determines an endomorphism f. Deﬁne an endomorphism ω :Λ → Λ by ω(en) = hn, n ≥ 1. Thus (since ω preserves multiplication), ω(eλ) = hλ for all partitions λ. Theorem 9.8. The endomorphism ω is an involution i.e. ω2 = 1 (the identity automorphism), or equivalently ω(hn) = en. (Thus, ω(hλ) = eλ for all partitions λ.)

188 Proof. Consider the formal power series

X n H(t) := hnt ∈ Λ [[t]] n≥0 X n E(t) := ent ∈ Λ [[t]] n≥0 Check the identities

Y −1 H(t) = (1 − xnt) (9.8) n Y E(t) = (1 + xnt) (9.9) n Hence H(t)E(−t) = 1. Equating the coeﬃcients of tn on both sides yields n X i 0 = (−1) eihn−i, n ≥ 1 (9.10) i=0

189 Pn i Conversely, if i=0(−1) uihn−i = 0 for all n ≥ 1, for certain ui ∈ Λ with u0 = 1, then ui = ei. Now apply ω to Equation (9.10) to obtain

n n X i n X i 0 = (−1) hiω(hn−i) = (−1) (−1) ω(hi)hn−i, i=0 i=0 whence ω(hi) = ei as desired.

190 n Corollary 9.9. The set {hλ : λ ` n} is a basis for Λ (so {hλ : λ ∈ Par} is a basis for Λ). Equivalently, h1, h2,... are algebraically independent and generate Λ as a Q-algebra, which we write as

Λ = Q[h1, h2,...]

Proof. Theorem 9.8 shows that the endomorphism ω :Λ → Λ deﬁned −1 by ω(en) = hn is invertible (since ω = ω ), and hence is an automorphism of Λ. The proof now follows from Theorem 9.4.

191 9.6 Power Sum Symmetric Functions

We deﬁne a fourth set pλ of symmetric functions indexed by λ ∈ Par and called the power sum symmetric functions as follows:

X n pn = mn = xi , n ≥ 1 (with p0 = m∅ = 1) i

pλ = pλ1 pλ2 ··· if λ = (λ1, λ2,...)

192 Proposition 9.10. Let λ = (λ1, . . . , λl) ` n, where l = l(λ), and set X pλ = Rλµmµ (9.11) µ`n

Let k = l(µ). Then Rλµ is equal to the number of ordered partitions

π = (B1,B2,...Bk) of {1, . . . , l} such that X µj = λi, 1 ≤ j ≤ k (9.12)

i∈Bj

193 µ µ1 µ2 Proof. Rλµ is the coeﬃcient of x = x1 x2 ... in

X λ1 X λ2 pλ = ( xi )( xi ) ··· To obtain the monomial xµ in the expansion of the product, we choose the term xλj from each factor P xλj so that Q xλj = xµ. Deﬁne ij i j ij Br = {j : ij = r}. Then (B1,...Bk) will be an ordered partition of {1, . . . , l} satisfying Equation (9.12), and conversely every such ordered partition gives rise to a term xµ.

194 Corollary 9.11. Let Rλµ be as in Equation (9.11). Then Rλµ = 0 unless µ ¤ λ, while Y Rλλ = mi(λ)! (9.13) i

m1(λ) m2(λ) n where λ = h1 2 · · · i. Hence {pλ : λ ` n} is a basis for Λ (so

{pλ : λ ∈ Par} is a basis for Λ). Equivalently, p1, p2,... are algebraically independent and generate Λ as a Q-algebra, i.e.

Λ = Q[p1, p2,...]

195 We now consider the eﬀect of the involution ω on pλ. For any partition λ = h1m1(λ)2m2(λ) · · · i deﬁne

m1 m2 zλ = 1 m1!2 m2! ··· (9.14)

If w ∈ Sn, then the cycle type ρ(w) of w is the partition

ρ(w) = (ρ1, ρ2,...) ` n such that the cycle lengths of w (in its factorization into disjoint cycles) are ρ1, ρ2,.... The number of m1 m2 permutations w ∈ Sn of a ﬁxed cycle type ρ = h1 2 · · · i is given by n! #{w ∈ Sn : ρ(w) = ρ} = m m 1 1 m1!2 2 m2! ··· −1 = n!zρ (9.15)

The set {v ∈ Sn : ρ(v) = ρ} is just the conjugacy class in Sn containing w. For any ﬁnite group G, the order #Kw of the conjugacy class Kw is equal to the index [G : C(w)] of the centralizer of w.

196 Hence:

Proposition 9.12. Let λ ` n. Then zλ is equal to the number of permutations v ∈ Sn that commute with a ﬁxed wλ of cycle type λ.

197 For a partition λ = h1m1 2m2 · · · i, deﬁne

m2+m4+··· n−l(λ) ελ = (−1) = (−1) (9.16)

Thus for any w ∈ Sn, εp(w) is +1 if w is an even permutation and −1 otherwise, so the map Sn → {±1} deﬁned by w 7→ εp(w) is the usual “sign homomorphism”

198 Proposition 9.13. We have Y X 1 (1 − x y )−1 = exp p (x)p (y) i j n n n i,j n≥1 X −1 = zλ pλ(x)pλ(y) (9.17) λ Y X 1 (1 + x y ) = exp (−1)n−1p (x)p (y) i j n n n i,j n≥1 X −1 = zλ ελpλ(x)pλ(y) (9.18) λ

199 Proof. We have

Y −1 X −1 log (1 − xiyj) = log(1 − xiyj) i,j i,j X X 1 = xnyn n i j i,j n≥1 !   X 1 X X = xn yn n i  j  n≥1 i j X 1 = p (x)p (y) n n n n≥1

200 Proposition 9.14. Let λ ` n. Then

ωpλ = ελpλ

In other words, pλ is an eigenvector for ω corresponding to the eigenvalue ελ.

Proof. Regard ω as acting on symmetric functions in the variables y = (y1, y2,...). Those in the variables x are regarded as scalars.

201 Apply ω to Equation (9.17). We obtain

X −1 Y −1 ω zλ pλ(x)pλ(y) = ω (1 − xiyj) λ i,j X = mv(x)ωhv(y)(by (9.6)) v X = mv(x)ev(y)( by Theorem 9.8) v Y = (1 + xiyj)(by (9.2)) i,j X −1 = zλ ελpλ(x)pλ(y)(by 9.18) λ

Since the pλ(x)’s are linearly independent, their coeﬃcients in the ﬁrst and last sums of the above chain of equalities must be the same. In other words, ωpλ(y) = ελpλ(y), as desired.

202 Proposition 9.15. We have

X −1 hn = zλ pλ (9.19) λ`n X −1 en = ελzλ pλ (9.20) λ`n (9.21)

Proof. Substituting y = (t, 0, 0,...) in Equation (9.17) immediately yields Equation (9.19). Equation (9.20) is similiarly obtained from (9.18), or by applying ω to (9.19).

203 9.7 A Scalar product

Deﬁne a scalar product on Λ by requiring that {mλ} and {hµ} be dual bases i.e.

hmλ, hµi = δλµ (9.22) for all λ, µ ∈ Par. Notice that h·, ·i respects the grading of Λ, in the sense that if f and g are homogeneous then hf, gi = 0 unless deg f = deg g.

204 Proposition 9.16. The scalar product h·, ·i is symmetric, i.e. hf, gi = hg, fi for all f, g ∈ Λ.

Proof. The result is equivalent to Corollary 9.6. More speciﬁcally, it suﬃces by linearity to prove hf, gi = hg, fi for some bases {f} and {g} of Λ. Take {f} = {g} = {hλ}. Then * + X hhλ, hµi = Nλν mν , hµ = Nλµ (9.23) ν

Since Nλµ = Nµλ by Corollary 9.6, we have hhλ, hµi = hhµ, hλi as desired.

205 Lemma 9.17. Let {uλ} and {vλ} be bases of Λ such that for all λ ` n n we have uλ, vλ ∈ Λ . Then {uλ} and {vλ} are dual bases if and only if

X Y −1 uλ(x)vλ(y) = (1 − xiyj) λ i,j

206 P P Proof. Write mλ = ρ ζλρuρ and hµ = ν ηµν vν . Thus X δλµ = hmλ, hµi = ζλρηµν huρ, vν i (9.24) ρ,ν For each ﬁxed n ≥ 0, regard ζ and η as matrices indexed by Par(n), and let A be the matrix deﬁned by Aρν = huρ, vν i. Then (9.24) is equivalent to I = ζAηt, where t denotes the transpose and I is the identity matrix. Therefore

{uλ} and {vλ} are dual bases ⇐⇒ A = I ⇐⇒ I = ζηt ⇐⇒ I = ζtη X ⇐⇒ δρν = ζλρηλν (9.25) λ

207 Now by Proposition 9.7 we have

Y −1 X (1 − xiyj) = mλ(x)hλ(y) i,j λ ! ! X X X = ζλρuρ(x) ηλν vν (y) λ ρ ν ! X X = ζλρηλν uρ(x)vν (y) ρ,ν λ

Since the power series uρ(x)vν (y) are linearly independent over Q, the proof follows from (9.25).

208 Proposition 9.18. We have

hpλ, pµi = zλδλµ (9.26)

Hence the pλ’s form an orthogonal basis of Λ. (They don’t form an orthonormal basis, since hpλ, pλi= 6 1)

Proof. By Proposition 9.13 and Lemma 9.17 we see that {pλ} and

{pµ/zµ} are dual bases, which is equivalent to (9.26).

209 Corollary 9.19. The scalar product h·, ·i is positive deﬁnite i.e. hf, fi ≥ 0 for all f ∈ Λ, with equality if and only if f = 0.

P Proof. Write (uniquely) f = λ cλpλ. Then

X 2 hf, fi = cλzλ

The proof follows since each zλ > 0.

210 Proposition 9.20. The involution ω is an isometry, i.e. hωf, ωgi = hf, gi for all f, g ∈ Λ.

Proof. By the bilinearity of the scalar product, it suﬃces to take f = pλ and g = pµ. The result then follows from Propositions 9.14 and 9.18.

211 9.8 The Combinatorial Deﬁnition of Schur Functions

The fundamental combinatorial objects associated with Schur functions are semistandard tableaux. Let λ be a partition. A semistandard

(Young) tableaux (SSYT) of shape λ is an array T = (Tij) of positive integers of shape λ (i.e., 1 ≤ i ≤ l(λ), 1 ≤ j ≤ λi) that is weakly increasing in every row and strictly increasing in every column. The size of an SSYT is its number of entries.

212 If T is an SSYT of shape λ then we write λ = sh(T ). Hence the size of T is just |sh(T )|. We may also think of an SSYT of shape λ as the Ferrers diagram of λ whose boxes have been ﬁlled with positive integers (satisfying certain conditions).

213 We say that T has type α = (α1, α2,...), denoted α = type(T ), if T has αi = αi(T ) parts equal to i. For any SSYT T of type α (or indeed for any multiset on P with possible additional structure), write

T α1(T ) α2(T ) x = x1 x2 ···

214 There is a generalization of SSYTs of shape λ that ﬁts naturally into the theory of symmetric functions. If λ and µ are partitions with µ ⊆ λ (i.e.

µi ≤ λi for all i), then deﬁne a semistandard tableau of (skew) shape

λ/µ to be an array T = (Tij) of positive integers of shape λ/µ (i.e.

1 ≤ i ≤ l(λ), µi < j ≤ λi) that is weakly increasing in every row and strictly increasing in every column.

215 We can similarly extend the deﬁnition of a Ferrers diagram of shape λ to one of shape λ/µ. Thus an SSYT of shape λ/µ may be regarded as a Ferrers diagram of shape λ/µ whose boxes have been ﬁlled with positive integers (satisfying certain conditions), just for “ordinary shapes” λ.

216 The deﬁnitions of type(T ) and xT carry over directly from SSYTs T of ordinary shape to those of skew shape. Deﬁnition 9.21. Let λ/µ be a skew shape. The skew Schur function sλ/µ = sλ/µ(x) of shape λ/µ in the variables x = (x1, x2,...) is the formal power series X T sλ/µ(x) = x T summer over all SSYTs T of shape λ/µ. If µ = ∅, then we call sλ(x) the Schur function of shape λ.

217 Theorem 9.22. For any skew shape λ/µ, the skew Schur function sλ/µ is a symmetric function.

Proof. It suﬃces to show that sλ/µ is invariant under interchanging xi and xi+1. Suppose that |λ/µ| = n and that α = (α1, α2,...) is a weak composition of n. Let

α˜ = (α1, α2, . . . , αi−1, αi+1, αi, αi+2,...).

If Tλ/µ,α denotes the set of all SSYTs of shape λ/µ and type α, then we seek the bijection ϕ : Tλ/µ,α → Tλ/µ,α˜.

218 Let T ∈ Tλ/µ,α. Consider the parts of T equal to i or i + 1. Some columns of T will contain no such parts, while some others will contain two such parts, viz., one i and i + 1. These columns we ignore. The remaining parts equal to i or i + 1 occur once in each column, and consist of rows with a certain number r of i’s followed by a certain number s of i + 1’s. (Of course r and s depend on the row in question.) For example, a portion of T could loook as follows: i i i i i i + 1 i + 1 i + 1 i + 1 i + 1 |{z} | {z } r=2 s=4 i + 1 i + 1

219 In each such row convert the r i0s and s i + 1’s to s i’s and r i + 1’s i i i i i i i i + 1 i + 1 i + 1 | {z } | {z } s=4 r=2 i + 1 i + 1

It’s easy to see that the resulting array ϕ(T ) belongs to Tλ/µ,α˜, and that ϕ establishes the desired bijection.

220 If λ ` n and α is a weak composition of n, then let Kλα denote the number of SSYTs of shape λ and type α. Kλα is called a Kostka number. By Deﬁnition 9.21 we have

X α sλ = Kλαx α summed over all weak compositions α of n, so by Theorem 9.22 we have X sλ = Kλµmµ (9.27) µ`n

221 More generally, we can deﬁne the skew Kostka number Kλ/ν,α as the number of SSYTs of shape λ/ν and type α, so that if |λ/ν| = n then X sλ/ν = Kλ/ν,µmµ (9.28) µ`n

222 λ λ Consider the number Kλ,1n , also denoted by f . By deﬁnition, f is the number of ways to insert the numbers 1, 2, . . . , n into the shape λ ` n, each number appearing once, so that every row and column is increasing. Such an array is called a standard Young tableau(SYT) (or just standard tableau) of shape λ. The number f λ has several alternative combinatorial interpretations as given by the following proposition.

223 Proposition 9.23. Let λ ∈ Par. Then the number f λ counts the objects in items (a)-(e) below. We illustrate these objects with the case λ = (3, 2). (a) Chains of partitions. Saturated chains in the interval [∅, λ] of Young’s lattice Y , or equivalently, sequences ∅ = λ0, λ1, . . . , λn = λ of partitions (which we identify with their diagrams) such that λi is obtained from λi−1 by adding a single square. ∅ ⊂ 1 ⊂ 2 ⊂ 3 ⊂ 31 ⊂ 32 ∅ ⊂ 1 ⊂ 2 ⊂ 21 ⊂ 31 ⊂ 32 ∅ ⊂ 1 ⊂ 2 ⊂ 21 ⊂ 22 ⊂ 32 ∅ ⊂ 1 ⊂ 11 ⊂ 21 ⊂ 31 ⊂ 32 ∅ ⊂ 1 ⊂ 11 ⊂ 21 ⊂ 22 ⊂ 32

(b) Linear extensions. Let Pλ be the poset whose elements are the squares of the diagram of λ, with t covering s if t lies directly to the

224 right or directly below s(with no squares in between). Such posets are λ just the ﬁnite order ideals of N × N. Then f = e(Pλ), the number of linear extensions of Pλ (c) Ballot sequences. Ways in which n voters can vote sequentially in an election for candidates A1,A2,..., so that for all i, Ai receives λi votes, and so that Ai never trails Ai+1 in the voting. (We denote such a voting sequence as a1a2 ··· an, where the k-th voter votes for Aak .) 11122 11212 11221 12112 12121

(d) Lattice permutations. Sequences a1a2 ··· an in which i occurs λi times, and such that in any left factor a1a2 ··· aj, the number of i’s is at least as great as the number of i + 1’s (for all i). Such a sequence is called a lattice permutation(or Yamanouchi word or ballot sequence) of type λ. 11122 11212 11221 12112 12121

225 l (e) Lattice paths. Lattice 0 = v0, v1, . . . , vn in R (where l = l(λ)) from the orign v0 to vn = (λ1, λ2, . . . , λl), with each step a unit coordinate vector, and staying within the region (or cone) x1 ≥ x2 ≥ · · · ≥ xl ≥ 0.

226 Define a reverse SSYT or column-strict plane partition of (skew) shape λ/µ to be an array of positive integers of shape λ/µ that is weakly decreasing in rows and strictly decreasing in columns. Define the type α of a reverse SSYT exactly as for ordinary SSYT. ˆ Define Kλ/µ,α to be the number of reverse SSYTs of shape λ/µ and type α. Proposition 9.24. Let λ/µ be a skew partition of n, and let α be a ˆ weak composition of n. Then Kλ/µ,α = Kλ/µ,α.

Proof. Suppose that T is a reverse SSYT of shape λ and type

α = (α1, α2,...). Let k denote the largest part of T . The transformation Tij 7→ k + 1 − Tij shows that Kbλα = Kλα¯, where α¯ = (αk, αk−1, . . . , α1, 0, 0,...). But by Theorem 9.22 we have

Kλα¯ = Kλα, and the proof is complete.

227 Proposition 9.25. Suppose that µ and λ are partitions with |µ| = |λ| and Kλµ 6= 0. Then λ ¤ µ. Moreover Kλλ = 1.

Proof. Suppose Kλµ 6= 0. By deﬁnition, there exists an SSYT T of shape λ and type µ. Suppose that a part Tij = k appears below the k-th row (i.e. i > k). Then we have 1 ≤ T1k < T2k < ··· < Tik = k for i > k, which is impossible. Hence the parts 1, 2, . . . , k all apppear in the

ﬁrst k rows, so that µ1 + µ2 + ··· + µk ≤ λ1 + λ2 + ··· + λk, as desired.

Moreover, if µ = λ then we must have Tij = i for all (i, j), so

Kλλ = 1.

228 Corollary 9.26. The Schur functions sλ with λ ∈ Par(n) form a basis n for Λ , so {sλ : λ ∈ Par} is a basis for Λ. In fact, the transition matrix

Kλµ which expresses the sλ’s in terms of the mµ’s, with respect to any linear ordering of Par(n) that extends dominance order, is lower triangular with 1’s on the main diagonal.

Proof. Proposition 9.25 is equivalent to the assertion about Kλµ. Since a lower triangular matrix with 1’s on the main diagonal is invertible, it n follows that {sλ : λ ∈ Par(n)} is a Q-basis for Λ .

229 Corollary 9.27. The Schur functions sλ with λ ∈ Par(n) form a basis n for Λ , so {sλ : λ ∈ Par} is a basis for Λ. In fact, the transition matrix

Kλµ which expresses the sλ’s in terms of the mµ’s, with respect to any linear ordering of Par(n) that extends dominance order, is lower triangular with 1’s on the main diagonal.

230 9.9 The RSK Algorithm

The basic operation of the RSK algorithm consists of the row insertion

P ← k of a positive integer k into a nonskew SSYT P = (Pij). The operation P ← k is defined as follows: Let r be the largest integer such that P1,r−1 ≤ k. (If P11 > k then let r = 1.) If P1r doesn’t exist (i.e. P has r − 1 columns), then simply place k at the end of the first row. The insertion process stops, and the resulting SSYT is P ← k. If on the other hand, P has at least r columns, so that P1r exists, then replace 0 P1r by k. The element then “bumps” Pir := k into the second row, i.e. insert k0 into the second row of P by the insertion rule just described. Continue until an element is inserted at the end of a row (possibly as the first element of the next row). The resulting array is P ← k.

231 Lemma 9.28. (a) When we insert k into an SSYT P , then the insertion path moves to the left. More precisely, if (r, s), (r + 1, t) ∈ I(P ← k) then t ≤ s. (b) Let P be an SSYT, and let j ≤ k. Then I(P ← j) lies strictly to the left of I((P ← j) ← k). More precisely, if (r, s) ∈ I(P ← j) and (r, t) ∈ I((P ← j) ← k), then s < t. Moreover, I((P ← j) ← k) does not extend below the bottom of I(P ← j). Equivalently

#I((P ← j) ← k) ≤ #I(P ← j)

232 Proof. (a) Suppose that (r, s) ∈ I(P ← k). Now either Pr+1,s > Prs (since P is strictly increasing in columns) or else there is no (r + 1, s) entry of P . In the first case, Prs cannot get bumped to the right of column s without violating the fact that the rows of P ← k are weakly increasing, since Prs would be to the right of Pr+1,s on the same row. The second case is clearly impossible, since we would otherwise have a gap in row r + 1. Hence (a) is proved. (b) Since a number can only bump a strictly larger number, it follows that k is inserted in the first row of P ← j strictly to the right of j. Since the first row of P is weakly increasing, j bumps an element no larger than the element k bumps. Hence by induction I(P ← j) lies strictly to the left of I((P ← j) ← k).

233 The bottom element b of I(P ← j) was inserted at the end of its row. By what was just proved, if I((P ← j) ← k) has an element c in this row, then it lies to the right of b. Hence c was inserted at the end of the row, so the insertion procedure terminates. It follows that I((P ← j) ← k) can never go below the bottom of I(P ← j).

234 Corollary 9.29. If P is an SSYT and k ≥ 1, then P ← k is also an SSYT.

Proof. It is clear that the rows of P ← k are weakly increasing. Now a number a can only bump a larger number b. By Lemma 9.28(a), b does not move to the right when when it is bumped. Hence b is inserted below a number that is strictly smaller than b, so P ← k remains an SSYT.

235 Now let A = (aij) be a N-matrix with finitely many nonzero entries. We will say that A is an N-matrix of finite support. We can think of A as either an infinite matrix or as an m × n matrix when aij = 0 for i > m and j > n.

236 Associate with A a generalized permutation of two-line array wA deﬁned by   i1 i2 i3 . . . im wA =   (9.29) j1 j2 j3 . . . jm where (a) i1 ≤ i2 ≤ · · · ≤ im

(b) if ir = is, and r ≤ s, then jr ≤ js,

(ir, jr) = (i, j)

237 It is easily seen that A determines a unique two line array wA satisfying (a) − (c), and conversely any such array corresponds to a unique A.

238 We now associate with A (or wA) a pair (P,Q) of SSYTs of the same shape, as follows. Let wA be given by (9.29). Begin with (P (0),Q(0)) = (∅, ∅) (where ∅ denotes the empty SSYT). If t < m and (P (t),Q(t)) are deﬁned, then let

(a) P (t + 1) = P (t) ← jt+1;

(b) Q(t + 1) be obtained from Q(t) by inserting it+1 (leaving all parts of Q(t) unchanged) so that P (t + 1) and Q(t + 1) have the same shape. The process ends at (P (m),Q(m)), and we deﬁne (P,Q) = (P (m),Q(m)). We denote this correspondence by A →A (P,Q) and call it the RSK algorithm. We call P the insertion tableau and Q the recording tableau of A or of wA

239 Theorem 9.30. The RSK algorithm is a bijection between N-matrices A = (aij)i,j≥1 of ﬁnite support and ordered pairs (P,Q) of SSYT of the same shape. In this correspondence, X j occurs in P exactly aij times (9.30) i X i occurs in Q exactly aij times (9.31) j (These last two conditions are equivalent to type(P ) = col(A), type(Q) = row(A)).

240 Proof. By Corollary 9.29, P is an SSYT. Clearly, by definition of the RSK algorithm P and Q have the same shape, and also (9.30) and (9.31) hold. Thus we must show the following: (a) Q is an SSYT , and (b) the RSK algorithm is a bijection, i.e., given (P,Q) , one can uniquely recover A. To prove (a), first note that since the elements of Q are inserted in weakly increasing order, it follows that the rows and columns of Q are weakly increasing. Thus we must show that the columns of Q are strictly increasing, i.e. no two equal elements of the top row of wA can end up in the same column of Q. But if ik = ik+1 in the top row, then we must jk ≤ jk+1. Hence by Lemma 9.28(b), the insertion path of jk+1 will always lie strictly to the right of the path for jk, and will never extend below the bottom of jk’s insertion path. It follows that the bottom elements of the two insertion paths lie in different columns, so the columns of the Q are strictly increasing as desired.

241 The above argument establishes an important property of the RSK algorithm: Equal elements of Q are inserted strictly left to right. It remains to show that the RSK algorithm is a bijection. Thus given

(P,Q) = (P (m),Q(m)), let Qrs be the rightmost occurrence of the largest entry of Q (where Qrs is the element of Q in row r and column s). Since equal elements of Q are inserted left to right, it follows that

Qrs = im, Q(m − 1) = Q(m) \ Qrs (i.e., Q(m) with the element Qrs deleted), and that Prs was the last element of P to be bumped into place after inserting jm into P (m − 1). But it is then easy to reverse the insertion procedure P (m − 1) ← jm.

242 Prs must have been bumped by the rightmost element Pr−1,t of row r − 1 of P that is smaller than Prs. Hence remove Prs from P , replace

Pr−1,t with Prs, and continue by replacing the rightmost element of row r − 2 of P that is smaller than Pr−1,t with Pr−1,t,etc. Eventually some element jm is removed from the ﬁrst row of P . We have thus uniquely recovered (im, jm) and (P (m − 1),Q(m − 1)). By iterating this procedure we recover the entire two-line array wA. Hence the RSK algorithm is injective.

243 To show surjectivity, we need to show that applying the procedure of the previous paragraph to an arbitrary pair (P,Q) of SSYTs of the same shape always yields a valid two-line array

  i1 i2 i3 . . . im wA =   (9.32) j1 j2 j3 . . . jm

Clearly, i1 ≤ i2 ≤ · · · ≤ im, so we need to show that if ik = ik+1 then jk ≤ jk+1. Let ik = Qrs and ik+1 = Quv, so r ≥ u and s < v. When we begin to apply inverse bumping to Puv, it occupies the end of its row (row u).

244 Hence when we apply inverse bumping to Prs, its “inverse insertion path” intersects row u strictly to the left of the column v. Thus at row u the inverse insertion path of Prs lies strictly to the left of that of Puv. By a simple induction argument (essentially the “inverse” of Lemma

9.28(b)), the entire inverse insertion path of Prs lies strictly to the left of that of Puv. In particular, before removing ik+1 the two elements jk and jk+1 appear in the ﬁrst row with jk to the left jk+1. Hence jk ≤ jk+1 as desired, completing the proof.

245 When the RSK algorithm is applied to a permutation matrix A (or a permutation w ∈ Sn), the resulting tableaux P,Q are just standard Young tableaux (of the same shape). Conversely, if P and Q are SYTs of the same shape, then the matrix A satisfying A RSK→ (P,Q) is a permutation matrix. Hence the RSK algorithm sets up a bijection between the symmetric group Sn and pairs (P,Q) of SYTs of the same shape λ ` n. In particular, if f λ denotes the number of SYTs of shape λ, then we have the fundamental identity X (f λ)2 = n! (9.33) λ`n

246 Although permutation matrices are very special cases N-matrices of finite support, in fact the RSK algorithm for arbitrary N-matrices A can be reduced to the case of permutation matrices. Namely, given the two line array wA, say of length n, replace the first row by 1, 2, . . . , n. Suppose the second row of wA has ci i’s. Then replace the 1’s in the second row from left-to-right with 1, 2, . . . , c1, next the 2’s from left to-right with c1 + 1, c2 + 1, . . . , c1 + c2 etc. until the second row becomes a permutation of 1, 2, . . . , n. Denote the resulting two-line array by wÃ.

247 Lemma 9.31. Let   i1 i2 i3 . . . in wA =   j1 j2 j3 . . . jn be a two-line array, and let   1 2 3 . . . n w˜A =   ˜j1 ˜j2 ˜j3 ... ˜jn

RSK Suppose that w˜A → (P,˜ Q˜). Let (P,Q) be the tableaux obtained from

P˜ and Q˜ by replacing k in P˜ by ik, and ˜jk in Q˜ by jk. Then RSK wA → (P,Q). In other words, the operation wA 7→ w˜A “commutes” with the RSK algorithm.

248 Proof. Suppose that when the number j is inserted into a row at some stage of the RSK algorithm, it occupies the k-th position in the row. If this number j were replaced by a larger number j + , smaller than any element of the row which is greater than j, then j + would also be inserted in at the k-th position. From this we see that the insertion procedure for elements j1, j2, . . . , jn exactly mimics that for

˜j1, ˜j2,... ˜jn, and the proof follows.

The process of replacing wA with w˜A, P with P˜, etc is called standardization

249 9.10 Some consequences of the RSK algorithm

Theorem 9.32 (Cauchy identity). We have

Y −1 X (1 − xiyj) = sλ(x)sλ(y) (9.34) i,j λ

Proof. Write  

Y −1 Y X aij (1 − xiyj) =  (xiyj)  (9.35)

i,j i,j aij ≥0

A term xαyβ in this expansion is obtained by choosing an N-matrix t t A = (aij) (the transpose of A) of ﬁnite support with row(A) = α and col(A) = β.

250 α β Hence the coeﬃcient of x y in (9.35) is the number of Nαβ of N-matrices A with row(A) = α and col(A) = β. This statement is also equivalent to (9.6). On the other hand the coeﬃcient of xαyβ in P λ sλ(x)sλ(y) is the number of pairs (P,Q) of SSYT of the shape λ such that type(P ) = α and type(Q) = β. The RSK algorithm sets up a bijection between the matrices A and the tableau pairs (P,Q), so the proof follows.

251 Corollary 9.33. The Schur functions form an orthonormal basis for Λ, i.e. hsλ, sµi = δλµ

Proof. Combine Corollary 9.27 and Lemma 9.17.

252 Corollary 9.34. Fix partitions µ, ν ` n. Then X KλµKλν = Nµν = hhµ, hν i λ`n where Kλµ and Kλν denote Kostka numbers, and Nµν is the number N-matrices A with row(A) = µ and col(A) = ν.

Proof. Take the coeﬃcient of xµyν on both sides of (9.34).

253 Corollary 9.35. We have X hµ = Kλµsλ (9.36) λ In other words, if M(u, v) denotes the transition matrix from the basis P {vλ} to the basis {uλ} of Λ (so that uλ = µ M(u, v)λµvµ), then M(h, s) = M(s, m)t

We give three proofs of this corollary, all essentially equivalent P First proof. Let hµ = λ aλµsλ. By Corollary 9.33, we have aλµ = hhµ, sλi. Since hhµ, mν i = δµν by deﬁnition (9.22) of the scalar product h, i, we have from Equation (9.27) that hhµ, hsλi = Kλµ.

254 Second Proof. Fix µ. Then

X col(A) hµ = x A X = xQ by the RSK algorithm (P,Q) X X Q = Kλµ x λ Q X = Kλµsλ λ where (i) A ranges over all N-matrices with row(A) = µ (ii) (P,Q) ranges over all pairs of SSYT of the same shape with type(P ) = µ and (iii) Q ranges over all SSYT of shape λ.

255 Third proof Take the coeﬃcient of mµ(x) on both sides of the identity X X mλ(x)hλ(y) = sλ(x)sλ(y) λ λ The two sides are equal by (9.7) and (9.34)

256 Corollary 9.36. We have

n X λ h1 = f sλ (9.37) λ`n

Proof. Take the coeﬃcients of x1x2 . . . xn on both sides of (9.34). To obtain a bijective proof, consider the RSK algorithm A RSK→ (P,Q) when col(A) = h1ni.

257 9.11 Symmetry of the RSK algorithm

Theorem 9.37. Let A be an N-matrix of ﬁnite support, and suppose that A RSK→ (P,Q). Then At RSK→ (Q, P ), where t denotes the transpose.

u Proof. Let wA = v be the two-line array associated to A. Hence 0 v v wA = u sorted i.e., sort the columns of u so that the columns are weakly increasing in lexicographic order. It follows from Lemma 9.31 that we may assume u and v have no repeated elements.

258 Consider     u1 . . . un u wA =   =   v1 . . . vn v

Suppose the ui’s and vj’s are distinct, deﬁne the inversion poset u u I = I(A) = I( v ) as follows. The vertices of I are the columns of v . a For notational convenience, we denote a column b as ab. Deﬁne ab < cd in I if a < c and b < d.

259 Lemma 9.38. The map ϕ : I(A) → I(At) deﬁned by ϕ(ab) = ϕ(ba) is an isomorphism of posets.

260 Now given the inversion poset I = I(A), deﬁne I1 to be the set of minimal elements of I, then I2 to be the set of minimal elements of

I − I1, then I3 to be the set of minimal elements of I − I1 − I2 etc.

Note that since Ii is an antichain of I, its elements can be labeled

(ui1, vi1), (ui2, vi2),..., (uini , vini ) (9.38) where ni = #Ii such that

ui1 < ui2 < . . . < uin i (9.39)

vi1 > vi2 > . . . > vini

261 Lemma 9.39. Let I1,...,Id be the (nonempty) antichains defined above, labeled as in (9.39). Let A RSK→ (P,Q). Then the first row of P is v1n1 v2n2 ··· vdnd , while the first row of Q is u11u21 ··· ud1. Moreover, if (uk, vk) ∈ Ii, then vk is inserted into the i-th column of the first row of the tableau P (k − 1) in the RSK algorithm.

Proof. Induction on n, the case n = 1 being trivial. Assume the assertion for n − 1, and let         u u1 u2 ··· un u˜ u1 u2 ··· un−1   =   ,   =   v v1 v2 ··· vn v˜ v1 v2 ··· vn−1

262 Let P (n − 1),Q(n − 1) be the tableaux obtained after inserting 0 u˜ v1, . . . , vn−1, and let the antichains Ii := Ii( v˜ ), 1 ≤ i ≤ e (where e = d − 1 or e = d), be given by (˜ui1, v˜i1),..., (˜uimi , v˜imi ) where u˜i1 < ··· < u˜imi and v˜i1 > ··· v˜imi . By the induction hypothesis, the

first row of P (n − 1) is v˜1m1 v˜2m2 ··· v˜eme , while the first row of Q is u˜11u˜21 ··· u˜e1. Now we insert vn into P (n − 1). If v˜imi > vn, then 0 u u Ii ∪ (un, vn) is an antichain of I( v ). Hence (un, vn) ∈ Ii( v ) if i is the least index for which v˜imi > vn. If there is no such i, then (un, vn) u u is the unique element of the antichain Id( v ) of I( v ). These conditions mean that vn is inserted into the i-th column of P (n − 1), as claimed. We start a new i-th column exactly when vn = vd1, in which case un = ud1, so un is inserted into the i-th column of the first row of Q(n − 1), as desired.

263 u Proof. of Theorem 9.37 If the antichain Ii( v ) is given by 9.38 such v that (264) is satisﬁed, then by Lemma 9.38 the antichain Ii( u ) is just

(vimi , uimi ),... (vi2, ui2), (vi1, ui1, ) where

vimi < . . . < vi2 < vi1

uimi > . . . ui2 > > ui1

Hence by Lemma 9.39, if At RSK→ (P 0,Q0), then the first row of P 0 is 0 u11u21 ··· ud1, and the first row of Q is vim1 v2m2 ··· vdmd . Thus by Lemma 9.39, the first rows of P 0 and Q0 agree with the first rows of P and Q, respectively.

264 u When the RSK algorithm is applied to v , the element vij, 1 ≤ j < mi, gets bumped into the second row of P before the element 1 ≤ s < mr, if and only if ui,j+1 < ur,s+1. Let P¯ and Q¯ denote P and Q with their ﬁrst rows removed. It follows that     a u12 ··· u1m1 u22 ··· u2m2 ··· ud2 ··· udmd   :=   b v ··· v v ··· v ··· v ··· v 11 1m1−1 21 2m2−1 d1 dmd−1 sorted RSK→ (P,¯ Q¯)

265 Similarly let (P¯0, Q¯0) denote P 0 and Q0 with their ﬁrst rows removed. v u Applying the same argument to u rather than v yields

    0 a v1m1−1 ··· v11 v2m2−1 ··· v21 ··· vdmd−1 ··· vd1   :=   b0 u ··· u u ··· u ··· u ··· u 1m1 12 2m2 22 dmd d2 sorted RSK→ (P¯0, Q¯0)

a b0 But b = a0 sorted, so by induction on n (or on the number of rows) we have (P¯0, Q¯0) = (Q,¯ P¯) and the proof follows.

266 Corollary 9.40. Let A be a N-matrix of ﬁnite support, and let A RSK→ (P,Q). Then A RSK→ (P,Q). Then A is symmetric i.e. (A = At) if and only if P = Q.

Proof. Immediete from the fact that At RSK→ (Q, P ).

267 Corollary 9.41. Let A = At and A RSK→ (P,P ), and let P α = (α1, α2,...) where αi ∈ N and αi < ∞. Then the map A 7→ P establishes a bijection between symmetric N-matrices with row(A) = α and SSYTs of type α.

Proof. Follows from Corollary 9.40 and Theorem 9.30.

268 Corollary 9.42. We have 1 X Q Q = sλ(x) (9.40) (1 − xi) · (1 − xixj) i i

Proof. The coeﬃcient of xα on the left side is the number of symmetric N-matrices A with row(A) = α while the coeﬃcient of xα on the right hand side is the number of SSYTs of type α. Now apply Corollary 9.41.

269 Corollary 9.43. We have

X λ 2 f = #{w ∈ Sn : w = 1} λ`n the number of involutions in Sn.

RSK Proof. Let w ∈ Sn and w → (P,Q) where P and Q are SYT of the same shape λ ` n. The permutation matrix corresponding to w is symmetric if and only if w2 = 1. By Theorem 9.37 this is the case if and only if P = Q, and the proof follows.

Alternatively, take the coeﬃcient of x1, ··· xn on both sides of (9.40)

270 9.12 The dual RSK Algorithm

There is a variation of the RSK algorithm that is related to the product Q (1 + xiyj) in the same way that the RSK algorithm itself is related to Q −1 (1 − xiyj) . We call this variation the dual RSK algorithm and ∗ denote it by A RSK→ (P,Q). The matrix A will now be a (0, 1) matrix of ∗ ﬁnite support. Form the two line array wA just as before. The RSK algorithm proceeds exactly like the RSK algorithm, except that an element i bumps the leftmost element ≥ i, rather than the leftmost element > i. (In particular, RSK and RSK∗ agree for permutation matrices.) It follows that each row of P is strictly increasing.

271 Theorem 9.44. The RSK∗ algorithm is a bijection between (0, 1)-matrices A of ﬁnite support and pairs (P,Q) such that P t (the transpose of P ) and Q are SSYTs with sh(P ) = sh(Q). Moreover, col(A) = type(P ) and row(A) = type(Q).

272 Theorem 9.45. We have Y X (1 + xiyj) = sλ0 (x)sλ(y) i,j λ

273 Lemma 9.46. Let ωy denote ω acting on the y variables only (so we regard the xi’s as constants commuting with ω). Then

Y −1 Y ωy (1 − xiyj) = (1 + xiyj)

Proof. We have

Y −1 X ωy (1 − xiyj) = ωy mλ(x)hλ(y)( by Proposition 9.7) λ X = mλ(x)eλ(y)( by Theorem 9.8 ) λ Y = (1 + xiyj)( by Proposition 9.3 )

274 Theorem 9.47. For every λ ∈ Par we have

ωsλ = sλ0

Proof. We have X Y sλ(x)sλ0 (y) = (1 + xiyj)( by Theorem 9.45) λ Y −1 = ωy (1 − xiyj) ( by Lemma 9.46) X = ωy sλ(x)sλ(y)( by Theorem 9.32) λ X = sλ(x)ωy(sλ(y)) λ

Take the coeﬃcient of sλ(x) on both sides. Since the sλ(x)’s are linearly independent, we obtain sλ0 (y) = ωy(sλ(y)), or just sλ0 = ωsλ.

275 9.13 The Classical deﬁnition of the Schur functions

n Let α = (α1, α2, . . . αn) ∈ N and w ∈ Sn. As usual write α α1 αn x = x1 ··· xn and deﬁne

α αw(1) αw(n) w(x ) = x1 ··· xn Now deﬁne X α aα = aα(x1, . . . , xn) = εww(x ) (9.41)

w∈Sn where   1 if w is an even permutation εw =  −1 if w is an odd permutation

276 Note that the right-hand side of equation (9.41) is just the expansion of a determinant, namely

αj n aα = det(xi )i,j=1

Note also that aα is skew-symmetric, i.e. w(aα) = εwaα, so aα = 0 unless all the αi’s are distinct. Hence assume that

α1 > α2 > ··· > αn ≥ 0, so α = λ + δ, where λ ∈ Par, l(λ) ≤ n, and

δ = δn = (n − 1, n − 2,..., 0). Since αj = λj + n − j, we get n λj +n−j aα = aλ+δ = det xi (9.42) i,j=1

277 For instance,

x4 x2 x1 1 1 1 4 2 1 a421 = a211+210 = x x x 2 2 2 4 2 1 x3 x3 x3 Note in particular that

n−j Y aδ = det(xi ) = (xi − xj) (9.43) 1≤i

278 Hence aα is divisible by xi − xj and thus by aδ (in the ring Z[x1, . . . xn]). Thus aα/aδ ∈ Z[x1, . . . , xn]. Moreover, since aα and aδ are skew-symmetric, the quotient is symmetric, and is clearly |λ| homogeneous of degree |α| − |δ| = |λ|. In other words, aα/aδ ∈ Λn .

279 Theorem 9.48. We have

aλ+δ/aδ = sλ(x1, . . . , xn)

280 Proof. There are many proofs of this result. We give one that can be extended to give an important result on skew Schur functions(Theorem 9.51). Applying ω to 9.36 and replacing λ by λ0 yields X eµ = Kλ0µsλ λ

Since the matrix (Kλ0µ) is invertible, it suﬃces to show that

X aλ+δ eµ(x1, . . . , xn) = Kλ0µ aδ or equivalently (always working with n variables), X aδeµ = Kλ0µaλ+δ (9.44) λ

281 Since both sides of (9.44) are skew-symmetric, it is enough to show that λ+δ the coeﬃcient of x in aδeµ is Kλ0µ. We multiply aδ by eµ by successively multiplying eµ1 , eµ2 ,.... Each partial product aδeµ1 ··· eµk i1 in is skew-symmetric, so any term x1 ··· xn appearing in aδeµ1 ··· eµk has i1 in all exponents ij distinct. When we multiply such a term x1 ··· xn by a term xm1 ··· xmj from eµk+1 (so j = µk+1), either two exponents become equal, or the exponents maintain their relative order. If two exponents become equal then that term disappears from aδeµ1 ··· eµk+1 . λ+δ δ Hence to get the term x , we must start with the term x in aδ and α1 α2 successively multiply by a term x of eµ1 , then x of eµ2 etc., keeping the exponents strictly decreasing. The number of ways to do this is the λ+δ coeﬃcient of x in aδeµ.

282 1 2 Given the terms xα , xα ,... as above, deﬁne an SSYT T = T (α1, α2,...) as follows: Column j of T contains an i if the αi i variable xj occurs in x (i.e. the j-th coordinate of α is equal to 1). For example, suppose n = 4, λ = 5332, λ0 = 44311, λ + δ = 8542, µ = α1 α2 α3 α4 α5 3222211, x = x1x2x3, x = x1x2, x = x3x4, x = x1x2, x = α6 α7 x1x4, x = x1, x = x3. Then T is given by

1113 2235 447 5 6

283 It is easy to see that the map (α1, α2,...) 7→ T (α1, α2,...) gives a bijection between ways of building up the term xλ+δ from xδ (according to the rules above) and SSYT of shape λ0 and type µ, so the proof follows.

284 From the combinatorial deﬁnition of Schur functions it is clear that sλ(x1, . . . , xn) = 0 if l(λ) > n. It is not hard to check thatt dim (Λn) = #{λ ∈ Par : l(λ) ≤ n}. It follows that the set

{sλ(x1, . . . , xn): l(λ) ≤ n} is a basis for Λn. (This also follows frm a simple extension of the proof of Corollary 9.27). We deﬁne on Λn a scalar product h, in by requiring that {sλ(x1, . . . , xn)} is an orthonormal basis. If f, g ∈ Λ, then we write hf, gin as short for hf(x1, . . . .xn), g(x1, . . . , xn)in. Thus

hf, gi = hf, gin provided that every monomial appearing in f involves at most n distinct variables e.g., if deg f ≤ n

285 Corollary 9.49. If f ∈ Λn, l(λ) ≤ n, and δ = (n − 1, n − 2,..., 1, 0), then λ+δ hf, sλin = [x ]aδf λ+δ the coeﬃcient of x in aδf.

Proof. All functions will be in the variables x1, . . . , xn. Let P f = l(λ)≤n cλsλ. Then by Theorem 9.48 we have X aδf = cλaλ+δ, l(λ)≤n so λ+δ hf, sλin = cλ = [x ]aδf.

286 Let us now consider a “skew generalization” of Theorem 9.48. We continue to work in n variables x1, . . . , xn. For any λν ∈ Par, l(λ) ≤ n, l(ν) ≤ n, consider the expansion

X λ0 sν eµ = Lν0µsλ, λ or equivalently (multiplying by aδ),

X λ0 aν+δeµ = Lν0µaλ+δ (9.45) λ

λ0 Arguing as in the proof of Theorem 9.48 shows that Lν0µ is equal to the number of ways to write

λ + δ = ν + δ + α1 + α2 + ··· + αk,

287 i Here l(µ) = k, each α is a (0, 1)-vector with µi 1’s, and each a partial sum ν + δ + α1 + ··· + αi has strictly decreasing coordinates. Deﬁne a 1 k 0 0 skew SSYT T = Tλ0/ν0 (α , . . . , α ) of shape λ /ν and type µ by the condition that i appears in column j of T if the j-th coordinate of αi is λ0 a 1. This establishes a bijection which shows that Lν0µ is equal to the skew Kostka number Kλ0/ν0,µ, the number of skew SSYTs of shape λ0/ν0 and type µ (see Equation (9.28)). (If ν0 * λ0 then this number is 0.) Corollary 9.50. We have X sν eµ = Kλ0/ν0,µsλ. (9.46) λ

Proof. Divide (9.45) by aδ and let n → ∞.

288 Theorem 9.51. For any f ∈ Λ, we have

hfsν , sλi = hf, sλ/ν i.

In other words, the two linear transformations Mν :Λ → Λ and

Dν :Λ → Λ deﬁned by Mν f = sν f and Dν sλ = sλ/ν are adjoint with respect to the scalar product h·, ·i. In particular

hsµsν , sλi = hsµ, sλ/ν i. (9.47)

289 Proof. Apply ω to (9.46) and replace ν by ν0 and λ by λ0. We obtain X sν hµ = Kλ/ν,µsλ λ Hence

hsν hµ, sλi = Kλ/ν,µ = hhµ, sλ/ν i, (9.48) by (9.28) and the fact that hhµ, mρi = δµρ by deﬁnition of h·, ·i. But equation (9.48) is linear in hµ, so since {hµ} is a basis for Λ, the proof follows.

290 Theorem 9.52. For any λ, ν ∈ Par we have ωsλ/ν = sλ0/ν0 .

Proof. By equation (9.47) and the fact that ω is an isometry, we have

hω(sµsν ), ωsλi = hωsµ, ωsλ/ν i.

Hence by Theorem 9.47 we get

hsµ0 sν0 , sλ0 i = hsµ0 , ωsλ/ν i (9.49)

On the other hand, substituting λ0, µ0, ν0 for λ, µ, ν respectively in (9.47) yields

hsµ0 sν0 , sλ0 i = hsµ0 , sλ0/ν0 i (9.50)

From Equations (9.49) and (9.50) there follows ωsλ/ν = sλ0/ν0

291 9.14 The Jacobi-Trudi Identity

Theorem 9.53. Let λ = (λ1, . . . , λn) and µ = (µ1, . . . , µn) ⊆ λ. Then

n sλ/µ = det(hλi−µj −i+j)i,j=1 (9.51) where we set h0 = 1 and hk = 0 for k < 0.

292 λ Proof. Let cµν = hsλ, sµsν i, so

X λ X λ sµsν = cµν sλ sλ/µ = cµν sν λ ν Then

X X λ sλ/µ(x)sλ(y) = cµν sν (x)sλ(y) λ λ,ν X = sν (x)sµ(y)sν (y) ν X = sµ(y) hν (x)mν (y) ν

293 Let y = (y1, . . . , yn). Multiplying by aδ(y) gives ! X X sλ/µ(x)aλ+δ(y) = hν (x)mν (y) aµ+δ(y) λ ν ! ! X α X w(µ+δ) = hα(x)y εwy n α∈N w∈Sn X X α+w(µ+δ) = εwhα(x)y

w∈Sn α

Now take the coeﬃcient of yλ+δ on both sides (so we are looking for terms where λ + δ = α + w(µ + δ)). We get X sλ/µ(x) = εwhλ+δ−w(µ+δ)(x) (9.52)

w∈Sn n = det(hλi−µj −i+j(x))i,j=1

294 Corollary 9.54 (Dual Jacobi-Trudi identity). Let µ ⊆ λ with λ1 ≤ n. Then n s = det(e 0 0 ) (9.53) λ/µ λ −µj −i+j i,j=1

295 Let f λ/µ be the number of SYT of shape λ/µ. Corollary 9.55. Let |λ/µ| = N and l(λ) ≤ n. Then

1 n f λ/µ = N!det (9.54) (λi − µj − i + j)! i,j=1

296 9.15 The Murnaghan-Nakayama Rule

A skew shape λ/µ is connected if the interior of the diagram of λ/µ, regarded as a union of solid squares, is a connected (open) set. A border strip (or rim hook or ribbon) is a connected skew shape with no 2 × 2 square.

Given positive integers a1, . . . , ak, there is a unique border strip λ/µ (up to translation) with ai squares in row i (i.e. ai = λi − µi). It follows that the number of border strips of size n (up to translation) is 2n−1. Deﬁne the height ht(B) of a border strip B to be one less than its number of rows.

297 Theorem 9.56. For any µ ∈ Par and r ∈ N we have

X ht(λ/µ) sµpr = (−1) sλ, (9.55) λ summed over all partitions λ ⊇ µ for which λ/µ is a border strip of size r.

298 Proof. Let δ = (n − 1, n − 2,..., 0), and let all functions be in the variables x1, . . . , xn. In equation 9.41 let α = µ + δ and multiply by pr. We get n X aµ+δpr = aµ+δ+rj , (9.56) j=1 where j is the sequence with a 1 in the j-th place and 0 elsewhere.

Arrange the sequence µ + δ + rj in descending order. If it has two terms equal, then it will contribute nothing to (9.56). Otherwise there is some p ≤ q for which

µp−1 + n − p + 1 > µq + n − q + r > µp + n − p,

q−p in which case aµ+δ+rj = (−1) aλ+δ, where λ is the partition

λ = (µ1, . . . , µp−1, µq + p − q + r, µp + 1, . . . , µq−1 + 1, µq+1, . . . , µn)

299 Such partitions are precisely those for which λ/µ is a border strip B of size r, and q − p is just ht(B). Hence

X ht(λ/µ) aµ+δpr = (−1) aλ+δ λ

Divide by aδ and let n → ∞ to obtain 9.55

300 Let α = (α1, α2,...) be a weak composition of n. Deﬁne a border-strip tableaux (or rim-hook tableaux) of shape λ/µ (where |λ/µ| = n) and type α to be an assignment of positive integers to the squares of λ/µ such that (a) every row and column is weakly increasing

(b) the integer i appears αi times, and (c) the set of squares occupied by i forms a border strip. Equivalently, one may think of a border-strip tableau as a sequence µ = λ0 ⊆ λ1 ⊆ · · · ⊆ λr ⊆ λ of partitions such that each skew shape i i+1 λ /λ is a border strip of size αi (including the empty border-strip ∅ when αi = 0).

301 Deﬁne the height ht(T ) of a border-strip tableau T to be

ht(T ) = ht(B1) + ht(B2) ··· + ht(Bk) where B1,...,Bk are the (nonempty) border strips appearing in T .

302 Theorem 9.57. We have

X λ/µ sµpα = χ (α)sλ, (9.57) λ where X χλ/µ(α) = (−1)ht(T) (9.58) T summed over all border-strip tableaux of shape λ/µ and type α

303 Corollary 9.58. We have

X λ pα = χ (α)sλ (9.59) λ where χλ(α) is given by (9.58) with µ = ∅.

304 Corollary 9.59. We have

X −1 λ/µ sλ/µ = zν χ (ν)pν , (9.60) ν where χλ/µ(ν) is given by (9.58).

Proof. We have from (9.57) that

λ/µ χ (ν) = hsµpν , sλi

= hpν , sλ/µi, and the proof follows from Proposition 9.18.

305 The orthogonality properties of the bases {sλ} and {pλ} translate into orthogonality relations satisﬁed by the coeﬃcients χλ(µ). Proposition 9.60. (a) Fix µ, ν. Then

X λ λ χ (µ)χ (ν) = zµδµν λ (b) Fix λ, µ. Then X −1 λ µ zν χ (ν)χ (ν) = δλµ ν

Proof. (a) Expand pµ and pν by (9.59) and take hpµ, pν i.

(b) Expand sλ and sµ by (9.60) and take hsλ, sµi.

306 10 Characters of the Symmetric and Unitary Groups

10.1 Characters of the Symmetric Group

Let CFn denote the set of all class functions (i.e. functions constant on n conjugacy classes) f : Sn → Q. Recall that CF has a natural scalar product deﬁned by 1 X hf, gi = f(w)g(w) n! w∈Sn Sometimes by abuse of notation we write hφ, γi instead of hf, gi when φ and γ are representations of Sn with characters f and g.

307 If α = (α1, . . . , αl) is a vector of positive integers and

|α| := α1 + ··· + αl = n, then recall the Young subgroup Sα ⊆ Sn given by

Sα = Sα1 × Sα2 × · · · × Sαl where Sα1 permutes 1, 2, . . . , α1; Sα2 permutes α1 + 1, α1 + 2, . . . α1 + α2 etc.

308 Consider the following linear transformations ch : CFn → Λn called the Frobenius characteristic maps. If f ∈ CFn, then 1 X ch f = f(w)p n! ρ(w) w∈Sn X −1 = zµ f(µ)pµ µ where f(µ) denotes f(w) for any type ρ(w) = µ. Equivalently, extending the ground ﬁeld Q to the algebra Λ and deﬁning Ψ(w) = pρ(w), we have ch f = hf, Ψi (10.1)

309 Note that if fµ is the class function deﬁned by   1, if ρ(w) = µ, fµ(w) =  0, otherwise.

−1 then chfµ = zµ pµ.

NOTE. Let ϕ : Sn → GL(V ) be a representation of Sn with character χ. Some times by abuse of notation we will write ch ϕ or ch V instead of ch χ.

310 Proposition 10.1. The linear transformation ch is an isometry i.e.,

hf, giCFn = hch f, ch giΛn .

Proof. We have (using Proposition 9.18)

X −1 X −1 hch f, ch gi = h zλ f(λ)pλ, zµ g(µ)pµi µ µ X −1 = zλ f(λ)g(λ) λ = hf, gi

311 We now want to deﬁne a product on class functions that will correspond to the ordinary product of symmetric functions under the characteristic map ch . Let f ∈ CFm and g ∈ CFn. Deﬁne the pointwise product f × g ∈ CF(Sm × Sn) by

(f × g)(u, v) = f(u)g(v).

If f and g are characters of representations of ϕ and ψ, then f × g is just the character of the tensor product representation ϕ ⊗ ψ of

Sm × Sn. Now deﬁne the induction product f ◦ g of f and g to be the induction of f × g to Sm+n where as before Sm permutes 1, 2, . . . , m while Sn permutes m + 1, m + 2, . . . , m + n. in symbols

f ◦ g = indSm+n (f × g). Sm×Sn

312 Let CF = CF0 ⊕ CF1 ⊕ · · · , and extend the scalar product on CFn to all of CF by setting hf, gi = 0 if f ∈ CFm , g ∈ CFn, and m 6= n. The induction product on characters extends to all of CF by (bi)linearity. It is not hard to check that this takes CF into an 0 associative commutative graded Q-algebra with the identity 1 ∈ CF . Similarly we can extend the characteristic map ch to a linear transformation ch : CF → Λ.

313 Proposition 10.2. The characteristic map ch : CF → Λ is a bijective algebra homomorphism, i.e. ch is one-to-one and onto, and satisﬁes

ch (f ◦ g) = (ch f)(ch g)

314 G Proof. Let resH f denote the restriction of the class function f on G to the subgroup H. We then have

ch (f ◦ g) = ch (indSm+n (f × g)) Sm×Sn = hindSm+n (f × g), Ψi by (10.1) Sm×Sn = hf × g, resSm+n Ψi (by Frobenius reciprocity) Sm×Sn Sm×Sn 1 X X = f(u)g(v)Ψ(uv) m!n! u∈Sm v∈Sn 1 X X = f(u)g(v)Ψ(u)Ψ(v) m!n! u∈Sm v∈Sn

= hf, ΨiSm hg, ΨiSn = (ch f)(ch g)

315 Moreover, from the deﬁnition of ch and the fact that the power sums pµ form a Q-basis for Λ it follows that ch is bijective.

316 Note that by Equation (9.19) and the deﬁnition of ch we have

X −1 ch 1Sn = zλ pλ = hn (10.2) λ`n

Corollary 10.3. We have ch 1Sn = h Sα α

Proof. Since 1Sn = 1 ◦ 1 ◦ · · · ◦ 1 , the proof follows from Sα Sα1 Sα2 Sαl Proposition 10.2 and Equation (10.2).

317 n Now let R denote the set of all virtual characters of Sn, i.e. functions of Sn that are the diﬀerence of two characters (= integer linear combinations of irreducible characters). Thus Rn is a lattice (discrete subgroup of maximum rank) in the vector space CFn. The rank of Rn is p(n), the number of partitions of n, and a basis consists of the irreducible characters of Sn. This basis is the unique orthonormal basis of Rn up to sign and order, since the transition matrix between two such bases must be a integral orthogonal matrix and hence a signed permutation. Deﬁne R = R0 ⊕ R1 ⊕ · · · .

318 Proposition 10.4. The image of R under the characteristic map ch is

ΛZ. Hence ch : R → ΛZ is a ring isomorphism.

319 Proof. It will suffice to find integer linear combinations of the characters ηα of the representations 1Sn that are irreducible characters of S . The Sα n Jacobi-Trudi identity (Theorem 9.53) suggests we define the (possibly virtual) characters ψλ = det(ηλi−i+j), where the product used in evaluating the determinant is the induction product. Then by the Jacobi-Trudi identity and Proposition 10.1 we have

λ ch (ψ ) = sλ (10.3)

λ µ Since ch is an isometry (Proposition 10.1) we get hψ , ψ i = δλµ. As pointed out above, this means that the class functions ψλ are, up to λ sign, the irreducible characters of Sn. Hence the ψ for λ ` n form a Z n n basis for R , and the image of R is the Z-span of the sλ’s which is just Λn as claimed. Z

320 Theorem 10.5. Regard the functions χλ (where λ ` n) of Section 9.15 λ λ as functions on Sn given by χ (w) = χ (µ), where w has cycle type µ. λ Then the χ are the irreducible characters of the symmetry group Sn.

321 By the Murnaghan-Nakayama rule (Corollary 9.59 ), we have

λ X −1 λ ch (χ ) = zµ χ (µ)pµ = sλ µ

Hence by Equation (10.3), we get χλ = ψλ. Since the ψλ, up to sign, are the irreducible characters of Sn, it remains only to determine wether χλ or −χλ is a character. But χλ(1n) = f λ, so χλ is an irreducible character.

322 By deﬁnition, ηλ is the character of the module M λ. It can be shown that χλ is the character of the Specht module Sλ. Proposition 10.6. Let α be a composition of n and λ ` n. Then the multiplicity of the irreducible character χλ in the character ηα is just the

Kostka number Kλα. In symbols

α λ hη , χ i = Kλα.

α Proof. By Corollary 10.3 we have ch η = hα. Then the proof follows from Corollary 9.35 and Theorem 10.5.

323 10.2 The characters of GL(n, C)

A linear representation of GL(V ) is a homomorphism ϕ : GL(V ) → GL(W ), where W is a complex vector space. From now on we assume that all representations are ﬁnite dimensional i.e dim (W ) < ∞. We call dim (W ) the dimension of the representation ϕ, denoted dim (ϕ).

324 The representation ϕ is a polynomial representation if, after choosing ordered bases for V and W , the entries of ϕ(A) are polynomials in the entries of A ∈ GL(n, C). It is clear that the notion of polynomial representations is independent of the choice of ordered bases of V and W , since linear combinations of polynomials remain polynomials.

325 Fact: If ϕ is a polynomial representation of GL(V ), then there is a symmetric polynomial char ϕ in dim V variables such that

T rϕ(A) = char ϕ(θ1, . . . , θn) for all A ∈ GL(V ), where θ1, . . . , θn are the eigenvalues of A.

326 Theorem 10.7. The irreducible polynomial representations ϕλ of GL(V) can be indexed by partitions λ of length at most n so that

λ char ϕ = sλ(x1, . . . , xn)

327 Examples:

• If ϕ(A) = 1 (the trivial representation), then char ϕ = s∅ = 1. • If ϕ(A) = A (the deﬁning representation), then

char ϕ = x1 + ··· + xn = s1. • If ϕ(A) = (detA)m for an positive integer m, then m char ϕ = (x1 ··· xn) = smn .

328 Proof. (Sketch) Let V be an n-dimensional complex vector space. Then GL(V ) acts diagonally on the k-th tensor power V ⊗k i.e

A · (v1 ⊗ · · · vk) = (A · v1) ⊗ · · · ⊗ (A · vk), (10.4)

⊗k and the symmetric group Sk acts on V by permuting tensor coordinates, i.e.

w · (v1 ⊗ · · · vk) = vw−1(1) ⊗ · · · vw−1(k) (10.5)

329 The actions of GL(V ) and Sk commute, so we have an action of ⊗k Sk × GL(V ) on V . A crucial fact is that the actions of GL(V ) and

Sk centralize each other. i.e the (invertible) linear transformations ⊗k ⊗k V → V that commute with the Sk action are just those given by Eqn. (10.4), while conversely the linear transformations that commute with the GL(V ) actions are those generated (as a C algebra) by Eqn (10.5). From this it can be shown that V ⊗k decomposes into irreducible

Sk × GL(V )-modules as follows M V ⊗k = (M λ ⊗ F λ) (10.6) where L denotes the direct sum (the “double commutant theorem”).

330 λ Here the Mλ’s are nonisomorphic irreducible Sk modules, the F ’s are nonisomorphic irreducible GL(V ) modules, and λ ranges over some index set. We know (Theorem 10.5) that the irreducible representations of Sk are indexed by partitions λ of k, so we choose the indexing so that λ M is the irreducible Sk module corresponding to λ ` k via Theorem 10.5. Thus we have constructed irreducible (or possibly 0) GL(V )-modules F λ. These modules aﬀord polynomial representations ϕλ, and the nonzero ones are inequivalent.

331 Next we compute the character of ϕλ. Let w × A be an element of

Sk × GL(V ), and let tr(w × A) denote the trace of w × A acting on V ⊗k. Then by Equation (10.6) we have X tr(w × A) = χλ(w) · tr(ϕλ(A)). λ

Let A have eigenvalues θ = (θ1, . . . , θn). A straightforward computation shows that tr(w × A) = pρ(w)(θ), so

X λ λ pρ(w)(θ) = χ (w)(char ϕ )(θ) λ But we know (Corollary 9.59) that

X λ pρ(w) = χ (w)sλ λ

332 λ λ Since the χ ’s are linearly independent, we conclude char ϕ = sλ. A separate argument shows that there are no other irreducible polynomial characters.

333 Fact: The ϕλ remain irreducible when restricted to U(V ) because the (dim V )2 entries of a general unitary matrix are algebraically independent, and so every irreducible polynomial representation of GL(V ) is still irreducible when restricted to U(V ).

334 11 Eigenvalues of random matrices

For n ∈ N, let Mn be a random n × n unitary matrix with distribution given by Haar measure on the unitary group. The eigenvalues of Mn lie on the unit circle T of the complex plane C. Write Ξn for the random measure on T that places a unit mass at each of the eigenvalues of Mn. That is, if the eigenvalues are {νn1, . . . , νnn}, then Z X Ξn(f) := f dΞn = f(νnj) T j

335 iθ P ˆ ijθ Note that if f : → has Fourier expansion f(e ) = fje , T C j∈Z then ∞ ∞ ˆ X ˆ j X ˆ j Ξn(f) = nf0 + fj Tr (Mn) + f−j Tr (Mn), j=1 j=1 where Tr denotes the trace.

Questions about the asymptotic behaviour of cn(Ξn(fn) − E[Ξn(fn)]) for a sequence of test functions {fn} and sequence of norming constants

{cn} may therefore be placed in the larger framework of questions about P∞ j j the asymptotic behaviour of j=1(anj Tr (Mn) + bnj Tr (Mn)) for arrays of complex constants {anj : n ∈ N, j ∈ N} and {bnj : n ∈ N, j ∈ N}.

336 Deﬁnition 11.1. A complex random variable is said to be standard complex normal if the real and imaginary parts are independent centred 1 (real) normal random variables with common variance 2 .

337 11.1 Moments of Traces

Theorem 11.2. a) Consider a = (a1, . . . , ak) and b = (b1, . . . , bk) with aj, bj ∈ {0, 1,...}. Let Z1,Z2,...Zk be independent standard complex normal random variables. Then for Pk Pk n ≥ ( j=1 jaj) ∨ ( j=1 jbj),

 k  k a bj Y j j j Y aj E  Tr (Mn) Tr (Mn)  = δab j aj! j=1 j=1

 k  Y p aj p bj = E  jZj jZj  . j=1

b) For any j, k, h i j k E Tr (Mn) Tr (Mn ) = δjk(j ∧ n).

338 Proof. (a) Deﬁne the simple power sum symmetric function pj to be the j j symmetric function pj(x1, . . . , xn) = x1 + ··· + xn. Let µ be the a1 a2 a partition (1 , 2 , . . . , k k ) of the integer K = 1a1 + 2a2 + ··· + kak Q aj and set pµ = j pj to be the corresponding compound power sum symmetric function. Associate µ with the conjugacy class of the symmetric group on K letters that consists of permutations with aj j–cycles for 1 ≤ j ≤ k. We have the expansion

X λ pµ = χµsλ, λ`K

339 λ Here the sum is over all partitions of K, the coeﬃcient χµ is the character of the irreducible representation of the symmetric group associated with the partition λ evaluated on the conjugacy class associated with the partition µ, and sλ is the Schur function corresponding to the partition λ

340 Given an n × n unitary matrix U, write sλ(U) (resp. pµ(U)) for the function sλ (resp. pµ) applied to the eigenvalues of U. Writing `(λ) for the number of parts of the partition λ (that is, the length of λ), the functions U 7→ sλ(U) are irreducible characters of the unitary group when `(λ) ≤ n and sλ(U) = 0 otherwise. Thus h i E sλ(Mn)sπ(Mn) = δλπ1(`(λ) ≤ n),

b1 b2 b Set ν = (1 , 2 , . . . , k k ) and L = 1b1 + 2b2 + ··· + kbk .

341 We have

 k  bj Y j aj j E  Tr (Mn) Tr (Mn)  j=1 h i = E pµ(Mn)pν (Mn)  ! ! (11.1) X λ X π = E  χµsλ(Mn) χν sπ(Mn)  λ`K π`L

X λ λ = δKL χµχν 1(`(λ) ≤ n). λ`K

342 When K ≤ n, all partitions of K are necessarily of length at most n, and so, by the second orthogonality relation for characters of the symmetric group, the rightmost term of (11.1) becomes

k k Y aj Y aj δKLδµν j aj! = δab j aj!, j=1 j=1 √ which coincides with the claimed mixed moment of jZj, 1 ≤ j ≤ k.

343 (b) We have from (11.1) that h i 2 j k X λ E Tr (Mn) Tr (Mn ) = δjk χ(j) 1(`(λ) ≤ n), λ`j where (j) is the partition of j consisting of a single part of size j. Now λ χ(j) = 0 unless λ is a hook partition (that is, a partition with at most one part of size greater than 1), in which case

λ `(λ)−1 χ(j) = (−1) Since there are j ∧ n hook partitions of j of length at most n, part (b) follows.

344 11.2 Linear combination of Traces

Theorem 11.3. Consider an array of complex constants 2 {anj : n ∈ N, j ∈ N}. Suppose there exists σ such that ∞ X 2 2 lim |anj| (j ∧ n) = σ . n→∞ j=1 Suppose also that there exists a sequence of positive integers {mn : n ∈ N} such that

lim mn/n = 0 n→∞ and ∞ X 2 lim |anj| (j ∧ n) = 0. n→∞ j=mn+1

345 P∞ j Then j=1 anj Tr (Mn) converges in distribution as n → ∞ to σZ, where Z is a complex standard normal random variable.

346 j Proof. Recall from Theorem 11.2 that E[ Tr (Mn)] = 0 and j k E[ Tr (Mn) Tr (Mn )] = δjk(j ∧ n). Consequently, the series P∞ j 2 j=1 anj Tr (Mn) converges in L for each n and lim [| P∞ a Tr (M j )|2] = 0. n→∞ E j=mn+1 nj n −1 Pmn j It therefore suﬃces to show that σ j=1 anj Tr (Mn) converges in distribution as n → ∞ to a complex standard normal random variable.

Let Z0,Z1,Z2,... be a sequence of independent complex standard normals.

347 From Theorem 11.2 we know that  α  β  mn   mn   X j X j  E  anj Tr (Mn) anj Tr (Mn)  j=1  j=1 

 α  β  mn   mn   X p X p  = E  anj jZj anj jZj  j=1  j=1     1/2 α  1/2 β  mn   mn   X 2 X 2  = E   |anj| j Z0  |anj| j Z0 ,    j=1   j=1  provided that αmn ≤ n and βmn ≤ n. The result now follows by convergence of moments for complex normal distributions and the assumption that mn/n → 0.

348 Theorem 11.4. Consider arrays of complex constants 2 {anj : n ∈ N, j ∈ N} and {bnj : n ∈ N, j ∈ N}. Suppose there exist σ , τ 2, and γ such that ∞ X 2 2 lim |anj| (j ∧ n) = σ , n→∞ j=1 ∞ X 2 2 lim |bnj| (j ∧ n) = τ , n→∞ j=1 and ∞ X lim anjbnj(j ∧ n) = γ. n→∞ j=1 Suppose also that there exists a sequence of positive integers {mn : n ∈ N} such that

lim mn/n = 0 n→∞

349 and ∞ X 2 2 lim (|anj| + |bnj| )(j ∧ n) = 0. n→∞ j=mn+1

P∞ j j Then j=1(anj Tr (Mn) + bnj Tr (Mn)) converges in distribution as n → ∞ to X + iY , where (X,Y ) is a pair of centred jointly normal real random variables with 1 [X2] = (σ2 + τ 2 + 2<γ), E 2 1 [Y 2] = (σ2 + τ 2 − 2<γ), E 2 and E[XY ] = =γ.

350 Given f ∈ L2(T) (where we deﬁne L2(T) to be the space of real–valued square–integrable functions), write 1 Z fˆ := e−ijθf(θ) dθ, j ∈ , j 2π Z for the Fourier coeﬃcients of f.

Recall that a positive sequence {ck}k∈N is said to be slowly varying if c lim bλkc = 1, λ > 0, k→∞ ck

351 Theorem 11.5. Suppose that f ∈ L2(T) is such that the sequence Pk ˆ 2 { j=−k |fj| j}k∈N is slowly varying. Then

Ξn(f) − E[Ξn(f)] q Pn ˆ 2 j=−n |fj| |j| converges in distribution to a standard normal random variable as n → ∞.

352 1 2 2 Let H2 denote the space of functions f ∈ L (T) such that 2 X ˆ 2 kfk 1 := |fj| |j| < ∞, 2 j∈Z

1 2 and deﬁne an inner product on H2 by X ˆ ¯ hf, gi 1 := fj ˆg|j|. 2 j j∈Z

353 1 2 2 Alternatively, H2 is the space of functions f ∈ L (T) such that 1 ZZ (f(φ) − f(θ))2 dθ dφ < ∞, (11.2) 16π2 2 φ−θ sin 2

Moreover, 1 ZZ (f(φ) − f(θ)) (g(φ) − g(θ)) hf, gi 1 = dθ dφ 2 16π2 2 φ−θ sin 2

354 1 2 R Theorem 11.6. If f1, . . . , fk ∈ H2 with E[Ξn(fh)] = n fj(θ) dθ = 0 for 1 ≤ h ≤ k, then the random vector (Ξn(f1),..., Ξn(fk)) converges in distribution to a jointly normal, centred random vector

(Ξ(f1),..., Ξ(fk)) with [Ξ(fh)Ξ(f`)] = hfh, f`i 1 . E 2

355 For 0 ≤ α < β < 2π write Nn(α, β) for the number of eigenvalues of iθ Mn of the form e with θ ∈ [α, β]. That is, Nn(α, β) = Ξn(f) where f is the indicator function of the arc {eiθ : θ ∈ [α, β]}. Note that E[Nn(α, β)] = n(β − α)/2π.

356 Theorem 11.7. As n → ∞, the ﬁnite–dimensional distributions of the processes

Nn(α, β) − E[Nn(α, β)] 1 √ , 0 ≤ α < β < 2π, π log n converge to those of a centred Gaussian process {Z(α, β) : 0 ≤ α < β < 2π} with the covariance structure  1, if α = α0 and β = β0,   1 0 0  , if α = α and β 6= β ,  2 [Z(α, β)Z(α0, β0)] = 1 0 0 E 2 , if α 6= α and β = β ,  − 1 , if β = α0,  2  0, otherwise.

357