1 construction of haar measure - ucsd mathematicsmath.ucsd.edu/~nwallach/haarmeasure.pdf · 1...
TRANSCRIPT
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, GU2 = U2. Define
∞ > δ := inf{a : a > 0, aU2 ⊇ K1 −K1} ≥ 1
and U := δU2. For each 0 < ε < 1,
(1 + ε)U ⊇ K1 −K1 6⊆ (1− ε)U .
5
The family of relatively open sets {2−1U + k}, k ∈ K1, is a covering of
K1. Let {2−1U + k1, . . . , 2−1U + kn} be a finite sub-covering and let
p = (k1 + . . . kn)/n. If k is any point in K1, then ki − k ∈ 2−1U for
some 1 ≤ i ≤ n. Since ki − k ∈ (1 + ε)U for all i and all ε > 0, we have
p ∈ 1n
(2−1U + (n− 1) · (1 + ε)U
)+ k.
For ε = 14(n−1) , we have p ∈ (1− 1
4n )U + k for each k ∈ K1. Let
K2 = K1 ∩⋂
k∈K1
((1− 1
4n)U + k
)6= ∅.
6
Because (1− 14n )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 + Tk 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
sf(g) = f(s−1g) 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:
g(x) =∑
finite
aif(six) ai > 0,∑
finite
ai = 1
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 ε > 0, ∃ a neighborhood V = Vε of the identity element e ∈ Gsuch that:
y−1x ∈ V ⇒ |f(x)− f(y)| < ε
9
Since (s−1y)−1s−1x = y−1x, we also have
|sf(y)− sf(x)| < ε whenever y−1x ∈ V
Since the functions g are convex combinations of functions of the form
sf ,
|g(y)− g(x)| < ε 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 = Cf 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
sg = g (∀s ∈ G) ⇒ g(s−1) = sg(e) = c (∀s ∈ G)
for some constant c.
10
By the definition of the set Kf , given any ε > 0 there exists a finite set
{s1, . . . sn} in G and ai > 0 such that
n∑1
ai = 1 and
∣∣∣∣∣c−n∑1
aif(six)
∣∣∣∣∣ < ε (∀x ∈ G) (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 G′ of G ,
or the function f ′ = f(x−1)), obtaining a relatively compact set C′f with
compact convex closure K ′f containing a constant function c′. It will be
enough to show c = c′.(all constants c in Kf must be equal to one
chosen constant c′ of K ′f and conversely.)
11
There is certainly a finite combination of right translates which is close
to c′ namely
|c′ −∑
bjf(xtj)| < ε ( for some tj ∈ G, bj > 0 with∑
bj = 1)
Let us multiply this inequality by ai and put x = si to get
|c′ai −∑
aibjf(sitj)| < εai (1.2)
Summing over i, we obtain
|c′∑
ai −∑i,j
aibjf(sitj)| < ε∑
ai = ε (1.3)
12
Operating symmetrically on Equation (1.1) (multiplying by bj , putting
x = tj and summing over j), we find:
|c−∑i,j
aibjf(sitj)| < ε (1.4)
Subtracting (or adding) Equation (1.3) from (1.4) we get |c− c′| < 2ε.Since ε 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
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
|m(h)−∑
bjh(tjx)| < ε
for finitely many suitable tj ∈ G and bj > 0 with∑bj = 1. Using the
definition of h and m(h) = m(g), this implies
|m(g)−∑i,j
aibjg(sitjx)| < ε (1.5)
However multiplying Equation (1.1) by bj and replacing x by tjx and
summing over j we find
|m(f)−∑i,j
aibjf(sitjx)| < ε (1.6)
15
Adding Equation (1.5) and (1.6), this implies
|m(f) +m(g)−∑i,j
aibj(f + g)(sitjx)| < 2ε
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 ∫G
f(x)dx =∫
G
f(x−1)dx
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:
(x1h1)(x2h2) = x1h1x2h−11 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 ◦ τh′ = τhh′
for h, h′ ∈ H. Let GsH denote the cartesian product of G and H. For
(x, h) and (x′, h′) in GsH, define:
(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
respectively. The inverse of (x, h) is (τh−1(x−1), h−1). 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
(e1, h) · (x, e2) · (e1, h)−1 = (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 GsH in 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 π′ 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 → E′ of their
respective spaces with
A(π(s)v) = π′(s)Av (s ∈ G, v ∈ E)
More generally, continuous linear operators A : E → E′ satisfying all
commutation relations A(π(s)) = π′(s)A for all s ∈ G are called
intertwining operators or G -morphisms (from π to π′) and their set is
a vector space denoted either by
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 = H⊥1 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,
〈x,π(s)v〉 = 〈π(s)∗x, v〉 = 〈π(s−1)x, v〉 = 0
since by assumption π(s−1)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 −→ 〈π(s)v,π(s)w〉 (v, w ∈ H)
are also continuous (by continuity of scalar product in H ×H). We can
thus define
ϕ(v, w) =∫
G
〈π(s)v,π(s)w〉ds
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|| ≤M2||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) ≤M2||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.
ϕ(π(t)v,π(t)w) =∫
G
〈π(st)v,π(st)w〉ds =∫
G
f(st)ds
=∫
G
ft(s)ds =∫
G
f(s)ds = ϕ(v, w)
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
[r(s)f ](x) = f(xs) (f ∈ Lp(G))
With this definition, one has r(s) ◦ r(t) = r(st).
One can also consider the biregular representations of l × r of G×Gin Lp(G) defined as
[l × r(s, t)f ](x) = f(s−1xt) (f ∈ Lp(G))
and its restriction to the diagonal G→ G×G, s→ (s, s) which is the
adjoint representation of G . It is defined as
[Ad(s)f ](x) = f(s−1xs) (f ∈ Lp(G))
The regular representations are faithful, i.e π(s) = 11⇔ s = e
31
Let π : G→ Gl(E) and π′ : G′ → Gl(E′) be two representations. We
can define the external direct sum representation of G×G′ in E ⊕ E′
by
π ⊕ π′(s, s′) = π(s)⊕ π′(s′) (s ∈ G, s′ ∈ G′)
When G = G′, we can restrict this external direct sum to the diagonal
G of G×G, obtaining the usual direct sum of πand π′
π ⊕ π′ : G → Gl(E ⊕ E′)
s → π(s)⊕ π′(s)
The external tensor product π ⊗ π′ as a representation of G×G′ in
E × E′is defined as
π ⊗ π′(s, s′) = π(s)⊗ π′(s′) (s ∈ G, s′ ∈ G′)
32
We assume the two spaces E ,E′ 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 = G′) and is given by
π ⊗ π′(s) = π(s)⊗ π′(s) (s ∈ G)
33
For a given finite dimensional representation π : G→ Gl(E), define the
contragredient representation π. This representation acts in the dual
E′ of E (namely the space of linear forms on E) and
π(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 π(st) = π(s)π(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 〈·, ·〉− of H is defined as
〈v, w〉− = 〈v, w〉 = 〈w, v〉
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 〈v, w〉− = 〈v, w〉
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
basis of H , but a decomposition v =∑viei in H gives rise to the
decomposition
v =∑
viei (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 π 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 π′ are two representations of the same group
G (acting in respective Hilbert spaces H and H ′), show that the matrix
coefficients of π ⊗ π′ (with respect to bases (ei) in H and (e′j) in H ′
and ei ⊗ e′j 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
and 11n(s) = idCn 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:
(Kf)(x) =∫
G
k(x, y)f(y)dy
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) = f(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 ∗ fdefined as
ϕ(x) =∫
G
f(y)f(y−1x)dy
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
f = f0 +∑
fi , fi ∈ Ker(K − λi) = Hi (λi ∈ Spec(K))
implies that
ϕ = Kf =∑
Kfi =∑
λifi
where
fi =1λiKfi ∈ Im(K) ⊆ C(G)
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
cij : g −→ cij(g) = 〈ei,π(g)ej〉
More generally if u and v are elements of H, we can define the
(function) coefficient cuv of π on G by
g −→ cuv (g) = 〈u,π(g)v〉
44
These functions are obviously finite linear combinations of the previously
defined matrix coefficients cij . Introduce the subspace V (π) of C(G)spanned by the cij , or equivalently by all cuv for v, u ∈ Hπ . Observe that
the subspaces of C(G) attached in this way to two equivalent
representations π and π′ coincide namely, V (π) = V (π′). Thus we can
form the algebraic sum (a priori this algebraic sum is not a direct sum)
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
cuv of π and γst 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
particular the product of cvu and γst ) 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 .
(a) Let Apq denote the space of linear combinations of coefficients of the
representation
πpq = π⊗p ⊗ π⊗q in (V ∗)⊗p ⊗ V ⊗q = T p
q (V )
Prove that the sum of the subspaces Apq of C(G)is an algebra A (show
that ApqA
rs ⊆ Apr
qs), stable under conjugation (show that Apq = Aq
p),
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
representation of G appears as a subrepresentation of some πpq , 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
Un(C) −→ C , x = (xij) 7−→ xi
j
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
Proof. Take an orthonormal basis (χi) of V and recall the coefficients cijof π defined as
π(x)χi =∑
j
cji (x)χj , x ∈ G
For f =∑
i aiχi, we have
π(x)f =∑
i
aiπ(x)χi =∑i,j
aicji (x)χj
Hence
f(x) = [r(x)f ](e) =∑i,j
cji (x)aij (with ai
j = aiχj(e))
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) = Tr(A · π(x)). The right
translates of these coefficients are easily identified as
[r(s)cA](x) = cA(xs) = Tr(Aπ(x)π(s))
= Tr(π(s) ·Aπ(x)) = Tr(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
c : End(V ) −→ C(G) ⊆ L2(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 ′ is a r-invariant subspace of L2(G) such that the restriction of r to
V ′ is equivalent to π, then V ′ 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
Recall the G-morphism c : End(V)→ L2(G) , A 7→ cA := Tr(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
V ⊗ V −→ End(V ) (V = dual of V )
defined as follows: If u ∈ V and v ∈ V , the operator corresponding to
u⊗ v is
u⊗ v : x→ u(x)v = 〈u, x〉v
56
The image of u⊗ v consists of multiples of v, and u⊗ v has rank 1when 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
cuv = 〈u,π(x)v〉 = 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
A\ =∫
G
σ(g)Aπ(g)−1dg
is a G-morphism from Vπ to Vσ , namely A\ ∈ HomG(Vπ , Vσ).
Proof. We have
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)
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:
A\ =∫
G
π(g)Aπ(g)−1dg =Tr(A)dimV
idV
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
λATr(idV ) = Tr
(∫G
π(g)Aπ(g)−1dg
)=∫
G
Tr(π(g)Aπ(g)−1
)dg
=∫
G
Tr(A)dg = Tr(A)
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
〈cuv , cxy〉 =∫
G
〈u,π(g)v〉〈x,π(g)y〉dg = 〈u, x〉〈v, y〉/dimV
(c) More generally in the case π = σ, the inner product of general
coefficients is given by
〈cA, cB〉 =∫
G
Tr(Aπ(g))Tr(Bπ(g))dg = Tr(A∗B)/dimV
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 ∈ Vπ , y ∈ Vπ)
and apply the result of the proposition∫G
π(g)(v ⊗ y)π(g)−1dg =Tr(v ⊗ y)
dim VidV =
〈v, y〉dimV
idV
Let us apply this operator to the vector u and take the inner product
with the vector x
〈x,∫
G
π(g)(v ⊗ y)π(g)−1u dg〉 =〈v, y〉dimV
〈x, u〉 =〈u, x〉〈v, y〉
dimV
62
But we have
π(g)(v ⊗ y)π(g)−1u = π(g)〈v,π(g−1)u〉y= 〈π(g)v, u〉π(g)y = 〈u,π(g)v〉π(g)y
hence
〈x,∫
G
π(g)(v ⊗ y)π(g)−1u dg〉 =∫
G
〈u,π(g)v〉〈x,π(g)y〉 dg = 〈cuv , cxy〉
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 ),
||cA||2 = Tr(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
L2(G) =⊕
L2(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
AG =⊕
L2(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 E′ denote
the (topological) dual of E, namely E′ 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 x′ ∈ E′ with x′(x) 6= 0. For
u ∈ E′ and v ∈ E we can consider the corresponding coefficient of σ
cuv ∈ C(G) : g → cuv (g) = 〈u,σ(g)v〉
66
Letting v vary in E, we get a linear mapping
Q : E → C(G) ⊆ L2(G) , v → cuv
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
supg∈G||σ(g)|| = M <∞
Hence
||Qv|| = supg∈G|cuv (g)| = sup
g∈G|〈u,σ(g)v〉|
≤ ||u||E′ sup
g∈G||σ(g)v||
E≤M ||u||
E′ ||v||E
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
cuv (e) = 〈u, v〉 = 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).∑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
EQ−→ L2(G)
Pπ−→ L2(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)
cuσ(x)v(g) = 〈u,σ(g)σ(x)v〉 = cuv (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):
dim E ≤ dim L2(G,π) = (dim V )2
69
Definition 4.8. The dual G 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 ($ ∈ G). 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
L2(G) =⊕
$∈GL2(G,$)
Instead of π ∈ $ ∈ G we shall write more simply π ∈ G.
70
Proposition 4.9. Let G be a compact group. Then the following
properties are equivalent
(i) The dual G 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 G. Moreover G can always be embedded in a product∏
G
U(π) with U(π) ∼= Udim (π)(C) (metrizable group)
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 V be its
dual and for u ∈ V , u ∈ V denote by u⊗ v the operator x→ u(x)v as
defined in the beginning of this Section. Show
(a) Tr(u⊗ v) = 〈u, v〉 = u(v) (intrinsic definition of Tr),
(b) (u⊗ v) · (x⊗ y) = 〈u, y〉x⊗ v,(c) tAu⊗Bv = B · (u⊗ v) ·A for A,B ∈ End(V ).Moreover, identifying the dual of V ⊗ V with the dual of V ⊗ V in the
obvious canonical way show
(d) t(u⊗ v) = v ⊗ uAssume now that V is now a representation space for a group G. Using
(a) and (c) prove
(e) cu⊗v = cuv ( i.e. Tr(π(x) · u⊗ v) = 〈u,π(x)v〉).
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}
(a) Check that the operator P =∫
Gπ(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
Hom(Vπ , Vσ)G = HomG(Vπ , Vσ)
and deduce a proof of the Lemma 4.3 from this observation.
74
(c) Using the G-isomorphism
Vπ ⊗ Vσ −→ Hom(Vπ , Vσ)
with Vπ ⊗ Vσ being equipped with the representation π ⊗ σ, conclude
that the projector \ : Vπ ⊗ Vσ → (Vπ ⊗ Vσ)G is given by∫G
(π ⊗ 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
f ∗ g(x) =∫
G
f(y)g(y−1x)dy
Defining f∗(x) = f(x−1), we can also write
f ∗ g(x) =∫
G
f(xy)g(y−1)dy =∫
G
f(xy−1)g(y)dy
=∫
G
f∗(yx−1)g(y)dy = 〈r(x−1)f∗, g〉
76
The Cauchy-Schwartz inequality gives:
|f ∗ g(x)| ≤ ||f∗||2||g||2 = ||f ||2||g||2
whence
||f ∗ g||∞ = supx∈G|f ∗ g(x)| ≤ ||f ||2||g||2
Consequently the convolution product can be extended by continuity
from C(G) to L2(G) and by definition
∗ : L2(G)× L2(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:
〈f, g〉 = f∗ ∗ g(e) , ||f ||22 = 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,∫|f ∗ g(x)|dx ≤
∫dx
∫dy|f(y)g(y−1x)|
=∫dy|f(y)|
∫dx|g(y−1x)|
= ||g||1∫|f(y)|dy = ||f ||1||g||1 <∞
78
These inequalities prove that∫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
π1(f) =∫
G
f(x)π(x)dx (f ∈ L1(G))
These integrals converge absolutely in norm: ||π(x)|| = 1 implies∫||f(x)π(x)||dx =
∫|f(x)|dx = ||f ||1
Thus,
||π1(f)|| ≤ ||f ||1 (f ∈ L1(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
π : G→ Gl(H) ; π1 : L1(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
basis (ei) of V and assume that A is represented by the matrix (aij) in
the basis (ei). Obviously
||A||2HS =∑i,j
|aij |2
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
If B is another endomorphism, represented by the matrix (bij) (in the
same basis), a small computation shows that
Tr(A∗B) =∑ij
aijb
ij
This shows that the Hilbert-Schmidt norm is derived from the
Hilbert-Schmidt inner product
〈A,B〉HS =∑i,j
aijb
ij = Tr(A∗B)
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 ∈ G in some finite dimensional Hilbert space V = Vπ . The spaces
V ⊗ V , End(V ) , L2(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
cuv (x) = 〈u,π(x)v〉 (u ∈ V , v ∈ V )
and the more general coefficients (linear combinations of the preceding
ones)
cA(x) = Tr(Aπ(x)) (A ∈ End(V ))
with the relation
cA = cuv for A = u⊗ v
85
For fixed u ∈ V , the linear mapping
cu : V → L2(G,π) , v → cuv
is a G-morphism form π to r, the right regular representation.
86
Similarly if v ∈ V is fixed,
l(s)cuv (x) = cuv (s−1x) = 〈u,π(s−1)π(x)v〉
= 〈tπ(s−1)u,π(x)v〉 = cπ(s)uv (x)
shows that the linear mapping
cv : V → L2(G,π) , u→ cuv
is a G-morphism from π to l (the left regular representation). Summing
up,
c : V ⊗ V → L2(G,π) , u⊗ v → cuv
is a π ⊗ π to l × r (biregular representation) G×G-morphism.
87
Note that
[l × r(s, t)]cA(x) = cA(s−1xt) = Tr(Aπ(s−1)π(x)π(t))
= Tr(π(t)Aπ(s)−1π(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, V ⊗ V is equipped with the
inner product
〈u⊗ v, x⊗ y〉 = 〈u, x〉〈v, y〉
(We use the Riesz isomorphism between V and V ). This inner product
corresponds to the Hilbert-Schmidt norm
〈u⊗ v, x⊗ y〉 = Tr((u⊗ v)∗x⊗ y)
89
90
Schur’s orthogonality relations (Theorem 4.5) say
〈cuv , cxy〉 =1
dim π〈u, x〉〈v, y〉 (dim π = dim V ),
and hence
c : u⊗ v → cuv is√
dim π−1× an isometry map
91
The inverse of c is nearly the extension π1 of π. Let us compute
π1(cuv ). For this purpose, we apply this operator to a vector y and
compute the inner product of the result with a vector x
〈x,π1(cuv )y〉 =∫cuv (s)〈x,π(s)y〉ds = 〈cuv , cxy〉
This quantity will vanish when π is not equivalent to π.
92
However, Schur’s relations give
〈x,π1(cuv )y〉 = 〈cuv , cxy〉 = (dim π)−1〈u, x〉〈v, y〉
= (dim π)−1〈x, 〈v, y〉u〉 = 〈x, (dim π)−1v ⊗ u(y)〉
This implies
π1(cuv ) = (dim π)−1v ⊗ u
and by linearity
π1(cA) = (dim π)−1A∗
93
Since the cA are the coefficients of π, we see that on L2(G, π),f → π1(f) is of the form
π1∣∣L2(G,π)
=√
dim π−1× an isometry map
The composite:
End(V ) → L2(G,π)conj−→ L2(G, π) → End(V )
A → cA cA = f → π1(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 ∈ G 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
that the series∑
G fπ converges to f in L2(G). Then
(a) fπ(x) = dim π · Tr(π(f)π(x)). Here f(x) = f(x−1)(b) ||fπ ||22 = dim π · ||π(f)||2HS (Hilbert-Schmidt norm on RHS).
(c) ||f ||22
=∑
π∈G dim π · ||π(f)||22
(Parseval equality)
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
〈cB , f〉 = 〈cB , fπ〉 = 〈cB , cA〉 = (dim π)−1Tr(B∗A)
But
〈cB , f〉 =∫cB(x)f(x)dx =
∫Tr(Bπ(x))f(x)dx
=∫Tr(π(x−1)B∗)f(x)dx = Tr(B∗
∫π(x−1)f(x)dx)
= Tr(B∗π(f))
Comparing the two results obtained for all B, we indeed find
A = (dim π) · π(f)
96
This gives
fπ(x) = cA(x) = Tr(Aπ(x)) = (dim π) · Tr(π(f)π(x))
as asserted in part (a). Moreover Schur’s relations show that
||fπ ||22 = ||cA||22 = (dim π)−1||A||2HS = (dim π) · ||π(f)||2HS
Thus (b) is proved and (c) follows from the observation that f and f
have the same L2(G): we can interchange f and f . Also observe that
the dimensions of π and π 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 ∗ g
where f, g ∈ L1(G) and ϕ ∈ L2(G) (recall that g(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
L1inv = L1
inv(G) the closure of L1(G) of the subspace of continuous
functions f satisfying f(xy) = f(yx) (or equivalently f(y−1xy) = f(x))for all x, y ∈ G. Show that L1
inv is contained in the center of L1(G) (as
convolution algebra: prove that f ∗ g = g ∗ f for f, g ∈ L1inv). For any
irreducible representation π : G→ Gl(V ) prove that
π(f) = (dim π)−1〈χ, f〉1V (f ∈ L1inv)
where χ(g) = Tr(π(g)) and 〈χ, f〉 =∫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
L1(G); more generally 1 ∗ f = f ∗ 1 is the constant function∫f(x)dx.)
Exercise 5.7. Show that the “extended” left regular representation
l = l1 : L1(G)→ End(L2(G))
has trivial kernel {0}. (Hint: Let 0 6= f ∈ L1(G) and construct a
sequence (gn) ⊆ C(G) with gn ≥ 0,∫gn(x)dx = 1 and
l(f)(gn) = f ∗ gn → f 6= 0). Conclude that if 0 6= f ∈ L1(G), there
exists a π ∈ G such that π(f) 6= 0. (continued on the next page)
100
Finally prove that
L1(G) commutative⇐⇒ G commutative
Exercise 5.8. Let G be a compact group, π ∈ G 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×Gs → (s, e) s → (e, s))
Conclude that
l ∼= ⊕π ⊗ 1 ∼= ⊕dim π · π = dim π · πr ∼= ⊕1⊗ π ∼= ⊕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) = Tr(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 Tr(AB) = Tr(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
χ ∈ Cinv = {f ∈ C(G) : f(y−1xy) = f(x), x and y ∈ G}
Observe that
L1inv = closure of C(G) in L1(G)
L2inv = closure of C(G) in L2(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 π of π is given
by
χπ(x) = Tr(π(x)) = Tr(tπ(x−1)) = Tr(π(x−1)) = χ(x−1)
hence χπ = χ
When π is unitary, π(x−1) = π(x)∗ (π is equivalent to π) and χ 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
that χ(z) = λzdim (V ). Thus the restriction of (dim V )−1χ 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 ε > 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)− Tr(Aσ(x))| < ε (x ∈ G)
In this representation, replace x by one of its conjugates yxy−1:
|f(x)− Tr(Aσ(yxy−1))| = |f(x)− Tr(σ(y−1)Aσ(y)σ(x))| < ε
Integrating over y, we have
|f(x)− Tr(Bσ(x))| < ε 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
σ ∼=⊕πnππ ∼=
⊕π
π ⊗ 1nπ
the operator B will have the form
B =⊕
iddim π ⊗Bπ
and
Bσ(x) = σ(x)B ∼=⊕
π ⊗Bπ
Tr(Bσ(x)) =∑
aπχπ(x) (aπ = Tr(Bπ))
This shows that
|f(x)−∑finite
aπχπ(x)| < ε
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
〈χπ , χσ〉 = dim HomG(Vπ , Vσ)
111
Proof. By Lemma 4.3, we know that the integral∫π(x)⊗ σ(x)dx
is an expression for the projector
\ : Vπ ⊗ Vσ −→ (Vπ ⊗ Vσ)G
Hom(Vπ , Vσ) −→ HomG(Vπ , Vσ)
The dimension of the image space is the trace of this projector. Thus
〈χπ , χσ〉 =∫χπ(x)χσ(x)dx = dim HomG(Vπ , Vσ)
112
Corollary 6.3. Let π be a finite dimensional representation of G. Then
π is irreducible ⇐⇒ ||χπ ||2 =√〈χπ , χπ〉 = 1
Corollary 6.4. Let π,σ ∈ G. Then
〈χπ , χσ〉 = δπσ (= 1 if π is equivalent to σ, 0 otherwise)
113
Corollary 6.5. Let σ be a finite dimensional representation of G and
σ =⊕
I nππ (summation over a finite subset I ⊆ G) be a
decomposition into irreducible components. Then
(a) nπ = 〈χπ , χσ〉 is well determined
(b) ||χσ ||2 =∑
I n2π
Proof. Let Vσ =⊕
(Vτ ⊗ Cnτ ). Since each G-morphism Vσ → Vπ
must vanish on all isotypical components Vτ ⊗ Cnτ where τ is not
equivalent to π, we have
HomG(Vσ , Vπ) = HomG(Vπ ⊗ Cnπ , Vπ)
= Cnπ ⊗ HomG(Vπ , Vπ) = Cnπ
This proves assertion (a). Moreover (b) follows from Pythagoras
theorem and Corollary 6.3
114
Corollary 6.6. The set of characters (χπ)π∈G is an orthonormal basis
of the Hilbert space L2(G)inv. Every function f ∈ L2(G)inv can be
expanded in the series
f =∑G
〈χπ , f〉χπ (convergence in L2(G))
115
Theorem 6.7. Let G be a compact group. For π ∈ G, 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 · Tr(π(f)π(x))
Thus
fπ(x) = dim π · Tr(∫
f(y−1)π(y)π(x)dy)
= dim π ·∫f(y)Tr(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 ⊗ 1T ∈ 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
A(ej ⊗ x) =∑
i
ei ⊗Aijx (Ai
j ∈ End(H2)).
Using the commutations (Pj ⊗ 1)A = A(Pj ⊗ 1) where Pj is the
orthogonal projector on Cej , conclude that Aij = 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,
conclude that Aii = B ∈ End(H2) is independent of i.)
117
Exercise 6.9. Check that the formula
f =∑〈χπ , f〉χπ
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}that sends i to the unique j with Kgis
−1 = 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
The canonical projection Gp→ K\G pushes the Haar measure ds on G
forward to a measure dx on K\G characterized by∫K\G
f(x)dx =∫
G
f(p(s))ds (f ∈ C(K\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)) ∫K\G
[∫K
f(kx) dk]dx =
∫G
f(x)dx
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
||f ||2 =∫
G
||f(x)||2dx
The norm under the integral sign is computed in Vσ .
Note that L2(G,Vσ) is a Hilbert space with inner product
〈f, g〉 =∫
G
〈f(x), g(x)〉dx
The inner product under the integral sign is computed in Vσ .
124
The elements
f ∈ L2(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
||f ||2H =∫
K\G||f(x)||2 dx
〈f, g〉H =∫
K\G〈f(x), g(x)〉 dx
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.
The induced representation ρ = IndGK(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).
As before, write V Kσ for the K-fixed elements of Vσ (with respect to the
action given by σ).
Proposition 7.2. (1) The linear map HG → V Kσ 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
π ⊗ IndGK(σ) ∼→ IndG
K(π|K ⊗ σ) : Hπ⊗H → H
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
Thus, f(e) = σ(k)f(e) for all k ∈ K and so f(e) ∈ V Kσ , 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
IndGK(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
ρ = IndGK(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⊗H can be written
uniquely as∑ei ⊗ fi with
∑||fi||2 <∞, and such an element has its
image the function ϕ =∑ϕi given by
x 7→∑
π(x)ei ⊗ fi(x)
The norm of ϕ is
||ϕ||2H
=∫
K\G||ϕ(x)||2dx =
∫K\G||∑
π(x)ei ⊗ fi(x)||2dx
=∫
K\G
∑||fi(x)||2dx =
∑∫K\G||fi(x)||2dx
=∑||fi||2H = ||
∑ei ⊗ fi||2
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 :
Φ(x) =∑
π(x)ei ⊗ fi(x) (fi(x) ∈ V )
||Φ(x)||2 =∑||fi(x)||2
Therefore∑π(kx)ei ⊗ fi(kx) = Φ(kx) = [π(k)⊗ σ(k)]Φ(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
HomG(C,H) ∼= HG
and
HomK(C, Vσ) ∼= V Kσ
(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
representation of one of its closed subgroups K. Put ρ = IndGK(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
Hπ ⊗Hρ∼→ Hom(Hπ ,Hρ)
the representation π ⊗ ρ is transformed into the representation
A→ ρ(s)Aπ(s)−1 (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.
Hπ⊗Hρ∼→ MorHS(Hπ ,Hρ)
135
Thus G-morphisms Hπ → Hρ correspond to G-invariants in Hπ⊗Hρ.
In other words,
MorG(Hπ ,Hρ) ∼= (Hπ⊗Hρ)G
Since Hπ⊗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
(Hπ⊗Hρ)G ∼= HG ∼= (Hπ⊗Vσ)K
The conclusion follows from the identity
(Hπ⊗Vσ)K ∼= MorK(Hπ , Vσ)
136
Corollary 7.4. Consider (π,Hπ) ∈ G and (σ, Vσ) ∈ K. Then the
multiplicity of π in IndGK(σ) is the same as the multiplicity of σ in π|K
Proof. Denote the (infinite dimensional in general) space of
ρ = IndGK(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
⊕IHπ whence
MorG(Hπ ,⊕IHπ) =⊕
I
MorG(Hπ ,Hπ) =⊕
I
C
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
MorK(Hπ , Vσ) = MorK(Vσ ⊗ Cm, Vσ) ∼→ Mor(Cm,MorK(Vσ , Vσ))
138
To see the isomorphism
MorK(Vσ ⊗ Cm, Vσ) ∼→ Mor(Cm,MorK(Vσ , Vσ)),
note that if (ei) is a basis for Cm we can write any element of Vσ ⊗Cm
as∑
i vi ⊗ ei. Any map A ∈ MorK(Vσ ⊗Cm, Vσ), is specified by the m
maps v 7→ A(v ⊗ ei) = Ai belonging to MorK(Vσ , Vσ), and we can
think of these m maps as a single map from Cm to MorK(Vσ , Vσ)defined by ei 7→ Ai.
139
Since MorK(Vσ , Vσ) = CidV (Schur’s lemma), we have
MorK(Hπ , Vσ) ∼= dual of Cm
The isomorphism of the theorem implies equality of the dimension of the
spaces. They are respectively
Card(I) = multiplicity of π in ρ = IndGK(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
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
Sλ), by which we mean the
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
V λ = 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
Mλ = C{{t1}, . . . , {tk}}
where {t1}, . . . , {tk}, is a complete list of λ-tabloids. Then Mλ 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 Mλ is cyclic, generated by any given
λ-tabloid. In addition, dim Mλ = n!/λ!, the number of λ-tabloids.
Theorem 8.8. Consider λ ` n with the Young subgroup Sλ and tabloid
{tλ}, as before. Then V λ = CSnSλ and Mλ = CSn{tλ} are isomorphic
as Sn-modules.
Proof. Let π1, π2, . . . , πk be a transversal for Sλ. Define a map:
θ : V λ →Mλ
by θ(πiSλ) = {πitλ} 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
λ1 + λ2 + · · ·+ λi = number of elements in the first i rows of tλ
≥ 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,∑i−1j=1 λj =
∑i−1j=1 µj and
∑ij=1 λj >
∑ij=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
H+ =∑π∈H
π
and
H− =∑π∈H
sgn(π)π
For a tableau t, the element R+t 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
κtdef= C−t =
∑π∈Ct
sgn(π)π
Note that if t has columns C1, C2, . . . , Ck, then κt factors as
κt = κC1κC2 . . . 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. Rπt = πRtπ−1
2. Cπt = πCtπ−1
3. κπt = πκtπ−1
4. eπt = πet
Proof. 1. We have the following equivalent statements:
σ ∈ Rπt ←→ σ{πt} = {πt}
←→ π−1σπ{t} = {t}
←→ π−1σπ ∈ Rt
←→ σ ∈ πRtπ−1
The proofs of 2 and 3 are similar to that of part 1.
151
4. We have
eπt = κπt{πt} = πκtπ−1{πt} = πκt{t} = πet
Definition 8.16. For any partition λ, the corresponding Specht module,
Sλ is the submodule of Mλ 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
Recall that H− =∑
π∈H(sgnπ)π for any subset H ⊆ Sn. If H = {π},then we write π− for H−. We need the unique inner product on Mλ for
which
〈{t}, {s}〉 = 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 = (sgnπ)H−
Equivalently, π−H− = H−.
2. For any u,v ∈Mλ,
〈H−u,v〉 = 〈u,H−,v〉
3. If the transposition (b, c) ∈ H, then we can factor
H− = k(ε− (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
they cannot be in the same column of tλ, for if so then κt = k(ε− (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} = (sgnπ)κt{t} = ±et
157
Corollary 8.20. If u ∈Mµ and shape t = µ, then κtu is multiple of et.
Proof. We can write u =∑
i ci{si}, where the si are µ-tableaux. By
the previous corollary, κtu =∑
i±ciet.
158
Theorem 8.21 (Submodule Theorem). Let U be a submodule of
Mµ. 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
〈u, et〉 = 〈u, κt{t}〉 = 〈κtu, {t}〉 = 〈0, {t}〉 = 0
Since the et span Sµ, we have u ∈ Sµ⊥, as claimed.
160
Proposition 8.22. Suppose θ ∈ HomSn(Sλ,Mµ) 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 〈·, ·〉 is an inner product with complex scalars, Mλ = Sλ ⊕ Sλ⊥.
Thus we can extend θ to an element of HomSn(Mλ,Mµ) by setting
θ(Sλ⊥) = 0. So
0 6= θ(et) = θ(κt{t}) = κtθ({t}) = κt(∑
i
ci{si})
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
nonzero homomorphism θ ∈ HomSn(Sλ,Mµ), since Sµ ⊆Mµ. 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
fλ =∑λ−
fλ−
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
1. Sλ ↓Sn−1∼=⊕
λ− Sλ− , and
2. Sλ ↑Sn+1∼=⊕
λ+ Sλ+
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
(respectively, in row ri of tabloid {ti}), then ti (respectively, {ti} ) 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
such that V (i)/V (i−1) ∼= Sλi
as Sn−1 modules for 1 ≤ i ≤ k. Let V (i)
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.
Define maps θi : Mλ →Mλi
by linearly extending
{t} θi→
{ti} if n in row ri of {t},0 otherwise.
Verify that θi is an Sn−1-homomorphism. Furthermore, for standard t
we have
etθi→
eti if n is in row ri of t,
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,
θiV(i) = Sλi
and
V (i−1) ⊆ kerθi.
We can construct the chain
{0} = V (0) ⊆ V (1)∩kerθ1 ⊆ V (1) ⊆ V (2)∩kerθ2 ⊆ V (2) ⊆ · · · ⊆ V (k) = Sλ
But
dimV (i)
V (i) ∩ kerθi= dim θiV
(i) = fλi
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,
V (i)
V (i−1)∼=
V (i)
V (i) ∩ kerθi
∼= Sλi
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.
Let χλ be the character of Sλ. If Sλ ↑Sn+1∼= ⊕µ`n+1mµSµ, then by
taking characters, χλ ↑Sn+1∼=∑
µ`n+1mµχµ.
169
The multiplicities are given by
mµ = 〈χλ ↑Sn+1 , χµ〉
= 〈χλ, χµ ↓Sn〉
= 〈χλ,∑µ−
χµ−〉
=
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
f(x) =∑α
cαxα
(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
(c) xα stands for the monomial xα11 xα2
2 . . .
(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
Given λ = (λ1, λ2, . . .) ` n, define a symmetric function mλ(x) ∈ Λn by
mλ =∑α
xα
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
f =∑
α cαxα ∈ Λn 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
en = m1n =∑
i1<...<in
xi1 · · ·xin, n ≥ 1 (with e0 = m∅ = 1)
eλ = eλ1eλ2 · · · , 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
ri =∑
j
aij
cj =∑
i
aij
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
composition of n. Then the coefficient Mλα of xα 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,
eλ =∑µ`n
Mλµmµ (9.1)
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∏i,j
(1 + xiyj) =∑λ,µ
Mλµmλ(x)mµ(y) (9.2)
=∑
λ
mλ(x)eλ(y) (9.3)
Here λ and µ range over Par. (It suffices to take |λ| = |µ|, since
otherwise Mλµ = 0.)
178
Proof. A monomial xα11 xα2
2 · · · yβ11 yβ2
2 · · · = xαyβ appearing in the
expansion of∏
(1 + xiyj) is obtained by choosing a (0, 1)-matrix
A = (aij) with finitely many 1’s, satisfying∏i,j
(xiyj)aij = xαyβ
But ∏i,j
(xiyj)aij = xrow(A)ycol(A)
so the coefficient of xαyβ 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
Theorem 9.4. Let λ, µ ` n. Then Mλµ = 0 unless λ′ � µ, while
Mλλ′ = 1. Hence the set {eλ : λ ` n} is a basis for Λn (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 A′ be the matrix with
row(A′) = λ and with its 1′s left justified, i.e. A′ij = 1 precisely for
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 1′s in the first i columns of A, so
by definition of dominance order we have col(A′) � col(A) = µ.
But col(A′) = λ′, so λ′ � µ as desired. Moreover it is easy to see that
A′ is the only (0, 1)-matrix with row(A′) = λ and col(A′) = λ′, so
Mλ,λ′ = 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))′, . . . , (λ2)′, (λ1)′ is also
compatible with the dominance order. (Exercise: give an example of
such an order). Then the matrix (Mλµ), with the row order λ1, λ2, . . .
and column order (λ1)′, (λ2)′, . . . is upper triangular with 1′s on the
main diagonal. Hence it is invertible, so {eλ : λ ` n} is a basis for Λn.
(In fact it is a basis for ΛnZ since the diagonal entries are actually 1′s,
and not merely nonzero.)
182
The set {eλ : λ ∈ Par} consists of all monomials ea11 e
a22 . . . (where
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
hn =∑λ`n
mλ =∑
i1≤...≤in
xi1 · · ·xin(with h0 = m∅ = 1)
(9.4)
hλ = hλ1hλ2 · · · if λ = (λ1, λ2, . . .)
184
Proposition 9.5. Let λ ` n, and let α = (α1, α2, . . .) be a weak
composition of n. Then the coefficient Nλα of xα in hλ is equal to the
number of N-matrices A = (aij) satisfying row(A) = λ and col(A) = α.
That is,
hλ =∑µ`n
Nλµmµ, (9.5)
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∏i,j
(1− xiyj)−1 =∑λ,µ
Nλµmλ(x)mµ(y) (9.6)
=∑
λ
mλ(x)hλ(y) (9.7)
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
H(t) :=∑n≥0
hntn ∈ Λ [[t]]
E(t) :=∑n≥0
entn ∈ Λ [[t]]
Check the identities
H(t) =∏n
(1− xnt)−1 (9.8)
E(t) =∏n
(1 + xnt) (9.9)
Hence H(t)E(−t) = 1. Equating the coefficients of tn on both sides
yields
0 =n∑
i=0
(−1)ieihn−i, n ≥ 1 (9.10)
189
Conversely, if∑n
i=0(−1)iuihn−i = 0 for all n ≥ 1, for certain ui ∈ Λwith u0 = 1, then ui = ei. Now apply ω to Equation (9.10) to obtain
0 =n∑
i=0
(−1)ihiω(hn−i) = (−1)nn∑
i=0
(−1)iω(hi)hn−i,
whence ω(hi) = ei as desired.
190
Corollary 9.9. The set {hλ : λ ` n} is a basis for Λn (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
by ω(en) = hn is invertible (since ω = ω−1), 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:
pn = mn =∑
i
xni , n ≥ 1 (with p0 = m∅ = 1)
pλ = pλ1pλ2 · · · if λ = (λ1, λ2, . . .)
192
Proposition 9.10. Let λ = (λ1, . . . , λl) ` n, where l = l(λ), and set
pλ =∑µ`n
Rλµmµ (9.11)
Let k = l(µ). Then Rλµ is equal to the number of ordered partitions
π = (B1, B2, . . . Bk) of {1, . . . , l} such that
µj =∑i∈Bj
λi, 1 ≤ j ≤ k (9.12)
193
Proof. Rλµ is the coefficient of xµ = xµ11 xµ2
2 . . . in
pλ = (∑
xλ1i )(
∑xλ2
i ) · · ·
To obtain the monomial xµ in the expansion of the product, we choose
the term xλj
ijfrom each factor
∑x
λj
i so that∏
j xλj
ij= xµ. Define
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 = 0unless µ� λ, while
Rλλ =∏
i
mi(λ)! (9.13)
where λ = 〈1m1(λ)2m2(λ) · · · 〉. Hence {pλ : λ ` n} is a basis for Λn (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 = 〈1m1(λ)2m2(λ) · · · 〉 define
zλ = 1m1m1!2m2m2! · · · (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
permutations w ∈ Sn of a fixed cycle type ρ = 〈1m12m2 · · · 〉 is given by
#{w ∈ Sn : ρ(w) = ρ} =n!
1m1m1!2m2m2! · · ·= n!z−1
ρ (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 = 〈1m12m2 · · · 〉, define
ελ = (−1)m2+m4+··· = (−1)n−l(λ) (9.16)
Thus for any w ∈ Sn, εp(w) is +1 if w is an even permutation and −1otherwise, so the map Sn → {±1} defined by w 7→ εp(w) is the usual
“sign homomorphism”
198
Proposition 9.13. We have∏i,j
(1− xiyj)−1 = exp∑n≥1
1npn(x)pn(y)
=∑
λ
z−1λ pλ(x)pλ(y) (9.17)
∏i,j
(1 + xiyj) = exp∑n≥1
1n
(−1)n−1pn(x)pn(y)
=∑
λ
z−1λ ελpλ(x)pλ(y) (9.18)
199
Proof. We have
log∏i,j
(1− xiyj)−1 =∑i,j
log(1− xiyj)−1
=∑i,j
∑n≥1
1nxn
i ynj
=∑n≥1
1n
(∑i
xni
)∑j
ynj
=∑n≥1
1npn(x)pn(y)
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
ω∑
λ
z−1λ pλ(x)pλ(y) = ω
∏i,j
(1− xiyj)−1
=∑
v
mv(x)ωhv(y) (by (9.6))
=∑
v
mv(x)ev(y) ( by Theorem 9.8)
=∏i,j
(1 + xiyj) (by (9.2))
=∑
λ
z−1λ ελpλ(x)pλ(y) (by 9.18)
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
hn =∑λ`n
z−1λ pλ (9.19)
en =∑λ`n
ελz−1λ pλ (9.20)
(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.
〈mλ, hµ〉 = δλµ (9.22)
for all λ, µ ∈ Par. Notice that 〈·, ·〉 respects the grading of Λ, in the
sense that if f and g are homogeneous then 〈f, g〉 = 0 unless
deg f = deg g.
204
Proposition 9.16. The scalar product 〈·, ·〉 is symmetric, i.e.
〈f, g〉 = 〈g, f〉 for all f, g ∈ Λ.
Proof. The result is equivalent to Corollary 9.6. More specifically, it
suffices by linearity to prove 〈f, g〉 = 〈g, f〉 for some bases {f} and {g}of Λ. Take {f} = {g} = {hλ}. Then
〈hλ, hµ〉 =
⟨∑ν
Nλνmν , hµ
⟩= Nλµ (9.23)
Since Nλµ = Nµλ by Corollary 9.6, we have 〈hλ, hµ〉 = 〈hµ, hλ〉 as
desired.
205
Lemma 9.17. Let {uλ} and {vλ} be bases of Λ such that for all λ ` nwe have uλ, vλ ∈ Λn. Then {uλ} and {vλ} are dual bases if and only if∑
λ
uλ(x)vλ(y) =∏i,j
(1− xiyj)−1
206
Proof. Write mλ =∑
ρ ζλρuρ and hµ =∑
ν ηµνvν . Thus
δλµ = 〈mλ, hµ〉 =∑ρ,ν
ζλρηµν〈uρ, vν〉 (9.24)
For each fixed n ≥ 0, regard ζ and η as matrices indexed by Par(n), and
let A be the matrix defined by Aρν = 〈uρ, vν〉. 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
⇐⇒ δρν =∑
λ
ζλρηλν (9.25)
207
Now by Proposition 9.7 we have∏i,j
(1− xiyj)−1 =∑
λ
mλ(x)hλ(y)
=∑
λ
(∑ρ
ζλρuρ(x)
)(∑ν
ηλνvν(y)
)
=∑ρ,ν
(∑λ
ζλρηλν
)uρ(x)vν(y)
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
〈pλ, pµ〉 = zλδλµ (9.26)
Hence the pλ’s form an orthogonal basis of Λ. (They don’t form an
orthonormal basis, since 〈pλ, pλ〉 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 〈·, ·〉 is positive definite i.e.
〈f, f〉 ≥ 0 for all f ∈ Λ, with equality if and only if f = 0.
Proof. Write (uniquely) f =∑
λ cλpλ. Then
〈f, f〉 =∑
c2λzλ
The proof follows since each zλ > 0.
210
Proposition 9.20. The involution ω is an isometry, i.e.
〈ωf, ωg〉 = 〈f, g〉 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
xT = xα1(T )1 x
α2(T )2 · · ·
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
sλ/µ(x) =∑T
xT
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︸︷︷︸r=2
i+ 1 i+ 1 i+ 1 i+ 1︸ ︷︷ ︸s=4
i+ 1
i+ 1 i+ 1
219
In each such row convert the r i′s and s i+ 1’s to s i’s and r i+ 1’s
i
i i i i i i︸ ︷︷ ︸s=4
i+ 1 i+ 1︸ ︷︷ ︸r=2
i+ 1
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
sλ =∑α
Kλαxα
summed over all weak compositions α of n, so by Theorem 9.22 we have
sλ =∑µ`n
Kλµmµ (9.27)
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
sλ/ν =∑µ`n
Kλ/ν,µmµ (9.28)
222
Consider the number Kλ,1n , also denoted by fλ. By definition, fλ 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 fλ has several alternative
combinatorial interpretations as given by the following proposition.
223
Proposition 9.23. Let λ ∈ Par. Then the number fλ 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
just the finite order ideals of N× N. Then fλ = 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
(e) Lattice paths. Lattice 0 = v0, v1, . . . , vn in Rl (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.
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
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 Kλα = 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
for Λn, 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
follows that {sλ : λ ∈ Par(n)} is a Q-basis for Λn.
229
Corollary 9.27. The Schur functions sλ with λ ∈ Par(n) form a basis
for Λn, 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
follows that {sλ : λ ∈ Par(n)} is a Q-basis for Λn.
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
P1r by k. The element then “bumps” Pir := k′ into the second row, i.e.
insert k′ 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
wA =
i1 i2 i3 . . . im
j1 j2 j3 . . . jm
(9.29)
where (a) i1 ≤ i2 ≤ · · · ≤ im
(b) if ir = is, and r ≤ s, then jr ≤ js,
(c) for each pair (i, j), there are exactly aij values of r for which
(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 AA→ (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,
j occurs in P exactly∑
i
aij times (9.30)
i occurs in Q exactly∑
j
aij times (9.31)
(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
wA =
i1 i2 i3 . . . im
j1 j2 j3 . . . jm
(9.32)
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 ARSK→ (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 fλ denotes the number of SYTs of shape
λ, then we have the fundamental identity∑λ`n
(fλ)2 = n! (9.33)
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 wA.
247
Lemma 9.31. Let
wA =
i1 i2 i3 . . . in
j1 j2 j3 . . . jn
be a two-line array, and let
wA =
1 2 3 . . . n
j1 j2 j3 . . . jn
Suppose that wA
RSK→ (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
wARSK→ (P,Q). In other words, the operation wA 7→ wA “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 + ε, smaller than any
element of the row which is greater than j, then j + ε 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 wA, P with P , etc is called
standardization
249
9.10 Some consequences of the RSK algorithm
Theorem 9.32 (Cauchy identity). We have∏i,j
(1− xiyj)−1 =∑
λ
sλ(x)sλ(y) (9.34)
Proof. Write
∏i,j
(1− xiyj)−1 =∏i,j
∑aij≥0
(xiyj)aij
(9.35)
A term xαyβ in this expansion is obtained by choosing an N-matrix
At = (aij)t (the transpose of A) of finite support with row(A) = α and
col(A) = β.
250
Hence the coefficient of xαyβ 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∑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. 〈sλ, sµ〉 = δλµ
Proof. Combine Corollary 9.27 and Lemma 9.17.
252
Corollary 9.34. Fix partitions µ, ν ` n. Then∑λ`n
KλµKλν = Nµν = 〈hµ, hν〉
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
hµ =∑
λ
Kλµsλ (9.36)
In other words, if M(u, v) denotes the transition matrix from the basis
{vλ} to the basis {uλ} of Λ (so that uλ =∑
µM(u, v)λµvµ), then
M(h, s) = M(s,m)t
We give three proofs of this corollary, all essentially equivalent
First proof. Let hµ =∑
λ aλµsλ. By Corollary 9.33, we have
aλµ = 〈hµ, sλ〉. Since 〈hµ,mν〉 = δµν by definition (9.22) of the scalar
product 〈, 〉, we have from Equation (9.27) that 〈hµ, hsλ〉 = Kλµ.
254
Second Proof. Fix µ. Then
hµ =∑A
xcol(A)
=∑
(P,Q)
xQ by the RSK algorithm
=∑
λ
Kλµ
∑Q
xQ
=∑
λ
Kλµsλ
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∑λ
mλ(x)hλ(y) =∑
λ
sλ(x)sλ(y)
The two sides are equal by (9.7) and (9.34)
256
Corollary 9.36. We have
hn1 =
∑λ`n
fλsλ (9.37)
Proof. Take the coefficients of x1x2 . . . xn on both sides of (9.34). To
obtain a bijective proof, consider the RSK algorithm ARSK→ (P,Q) when
col(A) = 〈1n〉.
257
9.11 Symmetry of the RSK algorithm
Theorem 9.37. Let A be an N-matrix of finite support, and suppose
that ARSK→ (P,Q). Then At RSK→ (Q,P ), where t denotes the
transpose.
Proof. Let wA =(uv
)be the two-line array associated to A. Hence
w′A =(
vu
)sorted i.e., sort the columns of
(vu
)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
wA =
u1 . . . un
v1 . . . vn
=
u
v
Suppose the ui’s and vj ’s are distinct, define the inversion poset
I = I(A) = I((uv
)) as follows. The vertices of I are the columns of
(uv
).
For notational convenience, we denote a columna
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 < . . . < uini
vi1 > vi2 > . . . > vini
(9.39)
261
Lemma 9.39. Let I1, . . . , Id be the (nonempty) antichains defined
above, labeled as in (9.39). Let ARSK→ (P,Q). Then the first row of P
is v1n1v2n2 · · · 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
v
=
u1 u2 · · · un
v1 v2 · · · vn
,
u
v
=
u1 u2 · · · un−1
v1 v2 · · · vn−1
262
Let P (n− 1), Q(n− 1) be the tableaux obtained after inserting
v1, . . . , vn−1, and let the antichains I ′i := Ii((uv
)), 1 ≤ i ≤ e (where
e = d− 1 or e = d), be given by (ui1, vi1), . . . , (uimi , vimi) where
ui1 < · · · < uimi and vi1 > · · · vimi . By the induction hypothesis, the
first row of P (n− 1) is v1m1 v2m2 · · · veme , while the first row of Q is
u11u21 · · · ue1. Now we insert vn into P (n− 1). If vimi > vn, then
I ′i ∪ (un, vn) is an antichain of I((uv
)). Hence (un, vn) ∈ Ii(
(uv
)) if i is
the least index for which vimi > vn. If there is no such i, then (un, vn)is the unique element of the antichain Id(
(uv
)) of I(
(uv
)). 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
Proof. of Theorem 9.37 If the antichain Ii((uv
)) is given by 9.38 such
that (264) is satisfied, then by Lemma 9.38 the antichain Ii((
vu
)) is just
(vimi, uimi
), . . . (vi2, ui2), (vi1, ui1, )
where
vimi< . . . < vi2 < vi1
uimi> . . . ui2 > > ui1
Hence by Lemma 9.39, if At RSK→ (P ′, Q′), then the first row of P ′ is
u11u21 · · ·ud1, and the first row of Q′ is vim1v2m2 · · · vdmd. Thus by
Lemma 9.39, the first rows of P ′ and Q′ agree with the first rows of P
and Q, respectively.
264
When the RSK algorithm is applied to(uv
), 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
b
:=
u12 · · · u1m1 u22 · · · u2m2 · · ·ud2 · · ·udmd
v11 · · · v1m1−1 v21 · · · v2m2−1 · · · vd1 · · · vdmd−1
sorted
RSK→ (P , Q)
265
Similarly let (P ′, Q′) denote P ′ and Q′ with their first rows removed.
Applying the same argument to(
vu
)rather than
(uv
)yields
a′
b′
:=
v1m1−1 · · · v11 v2m2−1 · · · v21 · · · vdmd−1 · · · vd1
u1m1 · · · u12 u2m2 · · · u22 · · ·udmd · · ·ud2
sorted
RSK→ (P ′, Q′)
But(ab
)=(
b′
a′
)sorted, so by induction on n (or on the number of rows)
we have (P ′, Q′) = (Q, P ) and the proof follows.
266
Corollary 9.40. Let A be a N-matrix of finite support, and let
ARSK→ (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 ARSK→ (P, P ), and let
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∏i(1− xi) ·
∏i<j(1− xixj)
=∑
λ
sλ(x) (9.40)
summed over all λ ∈ Par.
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∑λ`n
fλ = #{w ∈ Sn : w2 = 1}
the number of involutions in Sn.
Proof. Let w ∈ Sn and wRSK→ (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∏(1 + xiyj) in the same way that the RSK algorithm itself is related to∏(1− xiyj)−1. We call this variation the dual RSK algorithm and
denote it by ARSK∗
→ (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∏i,j
(1 + xiyj) =∑
λ
sλ′(x)sλ(y)
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
ωy
∏(1− xiyj)−1 =
∏(1 + xiyj)
Proof. We have
ωy
∏(1− xiyj)−1 = ωy
∑λ
mλ(x)hλ(y) ( by Proposition 9.7)
=∑
λ
mλ(x)eλ(y) ( by Theorem 9.8 )
=∏
(1 + xiyj) ( by Proposition 9.3 )
274
Theorem 9.47. For every λ ∈ Par we have
ωsλ = sλ′
Proof. We have∑λ
sλ(x)sλ′(y) =∏
(1 + xiyj) ( by Theorem 9.45)
= ωy
∏(1− xiyj)−1 ( by Lemma 9.46)
= ωy
∑λ
sλ(x)sλ(y) ( by Theorem 9.32)
=∑
λ
sλ(x)ωy(sλ(y))
Take the coefficient of sλ(x) on both sides. Since the sλ(x)’s are linearly
independent, we obtain sλ′(y) = ωy(sλ(y)), or just sλ′ = ωsλ.
275
9.13 The Classical definition of the Schur functions
Let α = (α1, α2, . . . αn) ∈ Nn and w ∈ Sn. As usual write
xα = xα11 · · ·xαn
n and define
w(xα) = xαw(1)1 · · ·xαw(n)
n
Now define
aα = aα(x1, . . . , xn) =∑
w∈Sn
εww(xα) (9.41)
where
εw =
1 if w is an even permutation
−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
aα = det(xαj
i )ni,j=1
Note also that aα is skew-symmetric, i.e. w(aα) = εwaα, so aα = 0unless 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
aα = aλ+δ = det(x
λj+n−ji
)n
i,j=1(9.42)
277
For instance,
a421 = a211+210 =
∣∣∣∣∣∣∣∣x4
1 x21 x1
1
x42 x2
2 x12
x43 x2
3 x13
∣∣∣∣∣∣∣∣Note in particular that
aδ = det(xn−ji ) =
∏1≤i<j≤n
(xi − xj) (9.43)
the Vandermonde determinant. If for some i 6= j we put xi = xj in aα,
then because aα is skew-symmetric (or because the i-th row and j-th
row of the determinant (9.42) become equal), we obtain 0.
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
homogeneous of degree |α| − |δ| = |λ|. In other words, aα/aδ ∈ Λ|λ|n .
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 λ′ yields
eµ =∑
λ
Kλ′µsλ
Since the matrix (Kλ′µ) is invertible, it suffices to show that
eµ(x1, . . . , xn) =∑
Kλ′µaλ+δ
aδ
or equivalently (always working with n variables),
aδeµ =∑
λ
Kλ′µaλ+δ (9.44)
281
Since both sides of (9.44) are skew-symmetric, it is enough to show that
the coefficient of xλ+δ in aδeµ is Kλ′µ. We multiply aδ by eµ by
successively multiplying eµ1 , eµ2 , . . .. Each partial product aδeµ1 · · · eµk
is skew-symmetric, so any term xi11 · · ·xin
n appearing in aδeµ1 · · · eµkhas
all exponents ij distinct. When we multiply such a term xi11 · · ·xin
n by a
term xm1 · · ·xmjfrom 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 .
Hence to get the term xλ+δ, we must start with the term xδ in aδ and
successively multiply by a term xα1of eµ1 , then xα2
of eµ2 etc., keeping
the exponents strictly decreasing. The number of ways to do this is the
coefficient of xλ+δ in aδeµ.
282
Given the terms xα1, xα2
, . . . as above, define an SSYT
T = T (α1, α2, . . .) as follows: Column j of T contains an i if the
variable xj occurs in xαi
(i.e. the j-th coordinate of αi is equal to 1).
For example, suppose n = 4, λ = 5332, λ′ = 44311, λ+ δ = 8542, µ =3222211, xα1
= x1x2x3, xα2
= x1x2, xα3
= x3x4, xα4
= x1x2, xα5
=x1x4, x
α6= x1, x
α7= 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 λ′ 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 〈, 〉n by requiring that {sλ(x1, . . . , xn)} is an orthonormal
basis. If f, g ∈ Λ, then we write 〈f, g〉n as short for
〈f(x1, . . . .xn), g(x1, . . . , xn)〉n. Thus
〈f, g〉 = 〈f, g〉n
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
〈f, sλ〉n = [xλ+δ]aδf
the coefficient of xλ+δ in aδf .
Proof. All functions will be in the variables x1, . . . , xn. Let
f =∑
l(λ)≤n cλsλ. Then by Theorem 9.48 we have
aδf =∑
l(λ)≤n
cλaλ+δ,
so
〈f, sλ〉n = cλ = [xλ+δ]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
sνeµ =∑
λ
Lλ′
ν′µsλ,
or equivalently (multiplying by aδ),
aν+δeµ =∑
λ
Lλ′
ν′µaλ+δ (9.45)
Arguing as in the proof of Theorem 9.48 shows that Lλ′
ν′µ is equal to the
number of ways to write
λ+ δ = ν + δ + α1 + α2 + · · ·+ αk,
287
Here l(µ) = k, each αi is a (0, 1)-vector with µi 1’s, and each a partial
sum ν + δ + α1 + · · ·+ αi has strictly decreasing coordinates. Define a
skew SSYT T = Tλ′/ν′(α1, . . . , αk) 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
a 1. This establishes a bijection which shows that Lλ′
ν′µ is equal to the
skew Kostka number Kλ′/ν′,µ, the number of skew SSYTs of shape
λ′/ν′ and type µ (see Equation (9.28)). (If ν′ * λ′ then this number is
0.)
Corollary 9.50. We have
sνeµ =∑
λ
Kλ′/ν′,µsλ. (9.46)
Proof. Divide (9.45) by aδ and let n→∞.
288
Theorem 9.51. For any f ∈ Λ, we have
〈fsν , sλ〉 = 〈f, sλ/ν〉.
In other words, the two linear transformations Mν : Λ→ Λ and
Dν : Λ→ Λ defined by Mνf = sνf and Dνsλ = sλ/ν are adjoint with
respect to the scalar product 〈·, ·〉. In particular
〈sµsν , sλ〉 = 〈sµ, sλ/ν〉. (9.47)
289
Proof. Apply ω to (9.46) and replace ν by ν′ and λ by λ′. We obtain
sνhµ =∑
λ
Kλ/ν,µsλ
Hence
〈sνhµ, sλ〉 = Kλ/ν,µ = 〈hµ, sλ/ν〉, (9.48)
by (9.28) and the fact that 〈hµ,mρ〉 = δµρ by definition of 〈·, ·〉. 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 = sλ′/ν′ .
Proof. By equation (9.47) and the fact that ω is an isometry, we have
〈ω(sµsν), ωsλ〉 = 〈ωsµ, ωsλ/ν〉.
Hence by Theorem 9.47 we get
〈sµ′sν′ , sλ′〉 = 〈sµ′ , ωsλ/ν〉 (9.49)
On the other hand, substituting λ′, µ′, ν′ for λ, µ, ν respectively in (9.47)
yields
〈sµ′sν′ , sλ′〉 = 〈sµ′ , sλ′/ν′〉 (9.50)
From Equations (9.49) and (9.50) there follows ωsλ/ν = sλ′/ν′
291
9.14 The Jacobi-Trudi Identity
Theorem 9.53. Let λ = (λ1, . . . , λn) and µ = (µ1, . . . , µn) ⊆ λ. Then
sλ/µ = det(hλi−µj−i+j)ni,j=1 (9.51)
where we set h0 = 1 and hk = 0 for k < 0.
292
Proof. Let cλµν = 〈sλ, sµsν〉, so
sµsν =∑
λ
cλµνsλ sλ/µ =∑
ν
cλµνsν
Then ∑λ
sλ/µ(x)sλ(y) =∑λ,ν
cλµνsν(x)sλ(y)
=∑
ν
sν(x)sµ(y)sν(y)
= sµ(y)∑
ν
hν(x)mν(y)
293
Let y = (y1, . . . , yn). Multiplying by aδ(y) gives∑λ
sλ/µ(x)aλ+δ(y) =
(∑ν
hν(x)mν(y)
)aµ+δ(y)
=
(∑α∈Nn
hα(x)yα
)( ∑w∈Sn
εwyw(µ+δ)
)=∑
w∈Sn
∑α
εwhα(x)yα+w(µ+δ)
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
sλ/µ(x) =∑
w∈Sn
εwhλ+δ−w(µ+δ)(x) (9.52)
= det(hλi−µj−i+j(x))ni,j=1
294
Corollary 9.54 (Dual Jacobi-Trudi identity). Let µ ⊆ λ with λ1 ≤ n.
Then
sλ/µ = det(eλ′−µ′j−i+j)ni,j=1 (9.53)
295
Let fλ/µ be the number of SYT of shape λ/µ.
Corollary 9.55. Let |λ/µ| = N and l(λ) ≤ n. Then
fλ/µ = N !det
(1
(λi − µj − i+ j)!
)n
i,j=1
(9.54)
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× 2square.
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
sµpr =∑
λ
(−1)ht(λ/µ)sλ, (9.55)
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
aµ+δpr =n∑
j=1
aµ+δ+rεj, (9.56)
where εj is the sequence with a 1 in the j-th place and 0 elsewhere.
Arrange the sequence µ+ δ + rεj 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,
in which case aµ+δ+rεj= (−1)q−paλ+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
aµ+δpr =∑
λ
(−1)ht(λ/µ)aλ+δ
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
λi/λi+1 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
sµpα =∑
λ
χλ/µ(α)sλ, (9.57)
where
χλ/µ(α) =∑T
(−1)ht(T) (9.58)
summed over all border-strip tableaux of shape λ/µ and type α
303
Corollary 9.58. We have
pα =∑
λ
χλ(α)sλ (9.59)
where χλ(α) is given by (9.58) with µ = ∅.
304
Corollary 9.59. We have
sλ/µ =∑
ν
z−1ν χλ/µ(ν)pν , (9.60)
where χλ/µ(ν) is given by (9.58).
Proof. We have from (9.57) that
χλ/µ(ν) = 〈sµpν , sλ〉
= 〈pν , sλ/µ〉,
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∑λ
χλ(µ)χλ(ν) = zµδµν
(b) Fix λ, µ. Then ∑ν
z−1ν χλ(ν)χµ(ν) = δλµ
Proof. (a) Expand pµ and pν by (9.59) and take 〈pµ, pν〉.
(b) Expand sλ and sµ by (9.60) and take 〈sλ, sµ〉.
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
conjugacy classes) f : Sn → Q. Recall that CFn has a natural scalar
product defined by
〈f, g〉 =1n!
∑w∈Sn
f(w)g(w)
Sometimes by abuse of notation we write 〈φ, γ〉 instead of 〈f, g〉 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
ch f =1n!
∑w∈Sn
f(w)pρ(w)
=∑
µ
z−1µ f(µ)pµ
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 = 〈f,Ψ〉 (10.1)
309
Note that if fµ is the class function defined by
fµ(w) =
1, if ρ(w) = µ,
0, otherwise.
then chfµ = z−1µ 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.,
〈f, g〉CFn = 〈ch f, ch g〉Λn .
Proof. We have (using Proposition 9.18)
〈ch f, ch g〉 = 〈∑
µ
z−1λ f(λ)pλ,
∑µ
z−1µ g(µ)pµ〉
=∑
λ
z−1λ f(λ)g(λ)
= 〈f, g〉
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, . . . ,mwhile Sn permutes m+ 1,m+ 2, . . . ,m+ n. in symbols
f ◦ g = indSm+n
Sm×Sn(f × g).
312
Let CF = CF0 ⊕CF1 ⊕ · · · , and extend the scalar product on CFn to
all of CF by setting 〈f, g〉 = 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
associative commutative graded Q-algebra with the identity 1 ∈ CF0.
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
Proof. Let resGHf denote the restriction of the class function f on G to
the subgroup H. We then have
ch (f ◦ g) = ch (indSm+n
Sm×Sn(f × g))
= 〈indSm+n
Sm×Sn(f × g),Ψ〉 by (10.1)
= 〈f × g, resSm+n
Sm×SnΨ〉Sm×Sn (by Frobenius reciprocity)
=1
m!n!
∑u∈Sm
∑v∈Sn
f(u)g(v)Ψ(uv)
=1
m!n!
∑u∈Sm
∑v∈Sn
f(u)g(v)Ψ(u)Ψ(v)
= 〈f,Ψ〉Sm〈g,Ψ〉Sn
= (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
ch 1Sn=∑λ`n
z−1λ pλ = hn (10.2)
Corollary 10.3. We have ch 1Sn
Sα= hα
Proof. Since 1Sn
Sα= 1Sα1
◦ 1Sα2◦ · · · ◦ 1Sαl
, the proof follows from
Proposition 10.2 and Equation (10.2).
317
Now let Rn 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
Sαthat are irreducible characters of Sn. The
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
ch (ψλ) = sλ (10.3)
Since ch is an isometry (Proposition 10.1) we get 〈ψλ, ψµ〉 = δλµ. As
pointed out above, this means that the class functions ψλ are, up to
sign, the irreducible characters of Sn. Hence the ψλ for λ ` n form a Zbasis for Rn, and the image of Rn is the Z-span of the sλ’s which is
just ΛnZ as claimed.
320
Theorem 10.5. Regard the functions χλ (where λ ` n) of Section 9.15
as functions on Sn given by χλ(w) = χλ(µ), where w has cycle type µ.
Then the χλ are the irreducible characters of the symmetry group Sn.
321
By the Murnaghan-Nakayama rule (Corollary 9.59 ), we have
ch (χλ) =∑
µ
z−1µ χλ(µ)pµ = sλ
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) = fλ, so χλ is an irreducible
character.
322
By definition, ηλ is the character of the module Mλ. 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
〈ηα, χλ〉 = Kλα.
Proof. By Corollary 10.3 we have ch ηα = 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
Trϕ(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
char ϕλ = 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
char ϕ = (x1 · · ·xn)m = 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)
and the symmetric group Sk acts on V ⊗k 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
Sk ×GL(V ) on V ⊗k. A crucial fact is that the actions of GL(V ) and
Sk centralize each other. i.e the (invertible) linear transformations
V ⊗k → V ⊗k 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
V ⊗k =⊕
(Mλ ⊗ Fλ) (10.6)
where⊕
denotes the direct sum (the “double commutant theorem”).
330
Here the Mλ’s are nonisomorphic irreducible Sk modules, the Fλ’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
Mλ is the irreducible Sk module corresponding to λ ` k via Theorem
10.5. Thus we have constructed irreducible (or possibly 0)
GL(V )-modules Fλ. 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
tr(w ×A) =∑
λ
χλ(w) · tr(ϕλ(A)).
Let A have eigenvalues θ = (θ1, . . . , θn). A straightforward computation
shows that tr(w ×A) = pρ(w)(θ), so
pρ(w)(θ) =∑
λ
χλ(w)(char ϕλ)(θ)
But we know (Corollary 9.59) that
pρ(w) =∑
λ
χλ(w)sλ
332
Since the χλ’s are linearly independent, we conclude char ϕλ = 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
Ξn(f) :=∫
Tf dΞn =
∑j
f(νnj)
335
Note that if f : T→ C has Fourier expansion f(eiθ) =∑
j∈Z fjeijθ,
then
Ξn(f) = nf0 +∞∑
j=1
fj Tr (M jn) +
∞∑j=1
f−j Tr (M jn),
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
the asymptotic behaviour of∑∞
j=1(anj Tr (M jn) + bnj Tr (M j
n)) 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
(real) normal random variables with common variance 12 .
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
n ≥ (∑k
j=1 jaj) ∨ (∑k
j=1 jbj),
E
k∏j=1
(Tr (M j
n))aj(
Tr (M jn))bj
= δab
k∏j=1
jajaj !
= E
k∏j=1
(√jZj
)aj(√
jZj
)bj
.b) For any j, k,
E[Tr (M j
n) Tr (Mkn)]
= δjk(j ∧ n).
338
Proof. (a) Define the simple power sum symmetric function pj to be the
symmetric function pj(x1, . . . , xn) = xj1 + · · ·+ xj
n. Let µ be the
partition (1a1 , 2a2 , . . . , kak) of the integer K = 1a1 + 2a2 + · · ·+ kak
and set pµ =∏
j paj
j 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
pµ =∑λ`K
χλµsλ,
339
Here the sum is over all partitions of K, the coefficient χλµ 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
E[sλ(Mn)sπ(Mn)
]= δλπ1(`(λ) ≤ n),
Set ν = (1b1 , 2b2 , . . . , kbk) and L = 1b1 + 2b2 + · · ·+ kbk .
341
We have
E
k∏j=1
(Tr (M j
n))aj(
Tr (M jn))bj
= E
[pµ(Mn)pν(Mn)
]= E
(∑λ`K
χλµsλ(Mn)
)(∑π`L
χπν sπ(Mn)
)= δKL
∑λ`K
χλµχ
λν1(`(λ) ≤ n).
(11.1)
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
δKLδµν
k∏j=1
jajaj ! = δab
k∏j=1
jajaj !,
which coincides with the claimed mixed moment of√jZj , 1 ≤ j ≤ k.
343
(b) We have from (11.1) that
E[Tr (M j
n) Tr (Mkn)]
= δjk
∑λ`j
∣∣∣χλ(j)
∣∣∣2 1(`(λ) ≤ n),
where (j) is the partition of j consisting of a single part of size j. Now
χλ(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
χλ(j) = (−1)`(λ)−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
{anj : n ∈ N, j ∈ N}. Suppose there exists σ2 such that
limn→∞
∞∑j=1
|anj |2(j ∧ n) = σ2.
Suppose also that there exists a sequence of positive integers
{mn : n ∈ N} such that
limn→∞
mn/n = 0
and
limn→∞
∞∑j=mn+1
|anj |2(j ∧ n) = 0.
345
Then∑∞
j=1 anj Tr (M jn) converges in distribution as n→∞ to σZ,
where Z is a complex standard normal random variable.
346
Proof. Recall from Theorem 11.2 that E[ Tr (M jn)] = 0 and
E[ Tr (M jn) Tr (Mk
n)] = δjk(j ∧ n). Consequently, the series∑∞j=1 anj Tr (M j
n) converges in L2 for each n and
limn→∞ E[|∑∞
j=mn+1 anj Tr (M jn)|2] = 0.
It therefore suffices to show that σ−1∑mn
j=1 anj Tr (M jn) 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
E
mn∑j=1
anj Tr (M jn)
α
mn∑j=1
anj Tr (M jn)
β
= E
mn∑j=1
anj
√jZj
α
mn∑j=1
anj
√jZj
β
= E
mn∑
j=1
|anj |2j
1/2
Z0
α
mn∑j=1
|anj |2j
1/2
Z0
β
,
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
{anj : n ∈ N, j ∈ N} and {bnj : n ∈ N, j ∈ N}. Suppose there exist σ2,
τ2, and γ such that
limn→∞
∞∑j=1
|anj |2(j ∧ n) = σ2,
limn→∞
∞∑j=1
|bnj |2(j ∧ n) = τ2,
and
limn→∞
∞∑j=1
anjbnj(j ∧ n) = γ.
Suppose also that there exists a sequence of positive integers
{mn : n ∈ N} such that
limn→∞
mn/n = 0
349
and
limn→∞
∞∑j=mn+1
(|anj |2 + |bnj |2)(j ∧ n) = 0.
Then∑∞
j=1(anj Tr (M jn) + bnj Tr (M j
n)) converges in distribution as
n→∞ to X + iY , where (X,Y ) is a pair of centred jointly normal real
random variables with
E[X2] =12(σ2 + τ2 + 2<γ),
E[Y 2] =12(σ2 + τ2 − 2<γ),
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
fj :=12π
∫e−ijθf(θ) dθ, j ∈ Z,
for the Fourier coefficients of f .
Recall that a positive sequence {ck}k∈N is said to be slowly varying if
limk→∞
cbλkc
ck= 1, λ > 0,
351
Theorem 11.5. Suppose that f ∈ L2(T) is such that the sequence
{∑k
j=−k |fj |2j}k∈N is slowly varying. Then
Ξn(f)− E[Ξn(f)]√∑nj=−n |fj |2|j|
converges in distribution to a standard normal random variable as
n→∞.
352
Let H122 denote the space of functions f ∈ L2(T) such that
‖f‖212
:=∑j∈Z|fj |2|j| <∞,
and define an inner product on H122 by
〈f, g〉 12
:=∑j∈Z
fj¯jg|j|.
353
Alternatively, H122 is the space of functions f ∈ L2(T) such that
116π2
∫∫(f(φ)− f(θ))2
sin2(
φ−θ2
) dθ dφ <∞, (11.2)
Moreover,
〈f, g〉 12
=1
16π2
∫∫(f(φ)− f(θ)) (g(φ)− g(θ))
sin2(
φ−θ2
) dθ dφ
354
Theorem 11.6. If f1, . . . , fk ∈ H122 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 E[Ξ(fh)Ξ(f`)] = 〈fh, f`〉 12.
355
For 0 ≤ α < β < 2π write Nn(α, β) for the number of eigenvalues of
Mn of the form eiθ 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)]1π
√log n
, 0 ≤ α < β < 2π,
converge to those of a centred Gaussian process
{Z(α, β) : 0 ≤ α < β < 2π} with the covariance structure
E[Z(α, β)Z(α′, β′)] =
1, if α = α′ and β = β′,
12 , if α = α′ and β 6= β′,
12 , if α 6= α′ and β = β′,
− 12 , if β = α′,
0, otherwise.
357