Math 202B (Symmetric functions and groups)

55 Pages·2020·0.561 MB·English

by  Steven V. Sam

NOTES FOR MATH 202B
STEVEN V SAM
The goal of this course is to discuss some generalities on complex linear representation
theory of finite groups, the theory of symmetric functions, and how this connects with the
symmetric groups. At the end we’ll discuss a bit about explicit constructions of irreducible
representations of symmetric groups and polynomial functors. Some references are [Se] for
representation theory, [J] and [FH] for symmetric groups in particular, [Sta] and [Mac] for
symmetric functions.
Contents
1. Linear representations of finite groups 2
1.1. Definitions 2
1.2. Basic operations 3
1.3. Irreducible representations 3
1.4. Characters 5
1.5. Classification of representations 8
1.6. Examples 9
1.7. The group algebra 11
1.8. Restriction and induction 12
2. Constructing symmetric group representations 13
2.1. Partitions 13
2.2. Tabloids 15
2.3. Specht modules 15
2.4. Garnir relations and standard tableaux 18
3. Symmetric functions 20
3.1. Definitions 20
3.2. Monomial symmetric functions 21
3.3. Elementary symmetric functions 22
3.4. The involution ω 23
3.5. Complete homogeneous symmetric functions 23
3.6. Power sum symmetric functions 24
3.7. A scalar product 26
4. Schur functions and the RSK algorithm 28
4.1. Semistandard Young tableaux 28
4.2. RSK algorithm 30
4.3. Dual RSK algorithm 33
4.4. Determinantal formula 34
4.5. Multiplying Schur functions, Pieri rule 36
4.6. Jacobi–Trudi identity 38
Date: February 28, 2020.
1
2 STEVEN V SAM
5. Representation theory of the symmetric groups 39
5.1. Symmetric groups 39
5.2. The characteristic map 40
5.3. Murnaghan–Nakayama rule 42
6. Combinatorial formulas 44
6.1. Standard Young tableaux 44
6.2. Semistandard Young tableaux 47
6.3. Littlewood–Richardson coefficients 48
7. Polynomial functors 49
7.1. Basic multilinear algebra 50
7.2. Schur functors 51
7.3. Polynomial representations and characters 52
7.4. Re-interpreting symmetric function identities 54
References 55
1. Linear representations of finite groups
1.1. Definitions. LetGbeafinitegroup. Theidentityelementwillbecalled1 . A(linear)
G
representation of G over a field k is a homomorphism
ρ : G → GL(V)
V
for some k-vector space V, where GL(V) is the group of invertible linear operators on V.
Equivalently, giving a representation is the same as giving a linear action of G on V, i.e., a
function G×V → V which we think of as a multiplication g ·v for g ∈ G and v ∈ V such
that:
• g ·(v +v(cid:48)) = g ·v +g ·v(cid:48),
• (gg(cid:48))·v = g ·(g(cid:48) ·v),
• 1 ·v = v, and
G
• g ·(λv) = λ(g ·v) for any λ ∈ k.
The multiplication is obtained by setting g·v = ρ (g)(v). We will always assume that V is
V
finite-dimensional.
We will generally take the perspective that V “is” the representation, and the information
ρ is implicit but not always mentioned. So properties of a representation such as dimension,
V
being nonzero, etc. come from the vector space V.
Let V and V(cid:48) be two representations of G. A linear map f: V → V(cid:48) is G-equivariant if
for all g ∈ G, we have
f ◦ρ (g) = ρ (g)◦f,
V V(cid:48)
or more compactly: f(g · v) = g · f(v) for all v ∈ V. An isomorphism is a G-equivariant
map which is invertible; if an isomorphism exists we write V ∼= V(cid:48).
Example 1.1.1. (1) For any vector space V we can define ρ (g) to be the identity on
V
V. This is clearly a representation. When dimV = 1, this is called the trivial
representation.
(2) Let X be a finite set with a G-action. Recall this means that we have a function
G × X → X denoted (g,x) (cid:55)→ g · x such that 1 · x = x for all x ∈ X, and
G
g·(g(cid:48)·x) = (gg(cid:48))·x for all g,g(cid:48) ∈ G and x ∈ X. Let V = k[X] be the k-vector space
NOTES FOR MATH 202B 3
with basis {e | x ∈ X} and define ρ by ρ (g)e = e . This is the permutation
x V V x g·x
representation of X.
A special case of a group action is when X = G and g · x = gx is given by the
group operation. In that case, k[G] is called the regular representation. (cid:3)
Lemma 1.1.2. All of the eigenvalues of ρ(g) are roots of unity.
If k is algebraically closed and has characteristic 0, then ρ(g) is diagonalizable for all
g ∈ G.
Proof. Letλbeaneigenvalueofρ(g)witheigenvectorv. Thenρ(g)|G| = 1butalsoρ(g)|G|v =
λ|G|v, so λ|G| = 1.
Consider the Jordan normal form of ρ(g), which recall is an upper-triangular matrix whose
diagonal entries are the eigenvalues of ρ(g) and whose superdiagonal (the entries in positions
(i,i + 1)) are either 0 or 1. Then ρ(g) is diagonalizable if and only if the superdiagonal is
0. Furthermore, if any of those entries are 1, then no positive power of ρ(g) is equal to the
identity, but we know that ρ(g)|G| = 1. (cid:3)
Remark 1.1.3. The second part can fail in positive characteristic. For instance, let G =
(cid:20) (cid:21)
1 1
Z/2 = {1,z}andkbeafieldofcharacteristic2. Thenρ(z) = definesarepresentation
0 1
(since ρ(z)2 is the identity) but ρ(z) is not diagonalizable. (cid:3)
1.2. Basic operations. Let V and W be representations of G. There are a few basic
operations we will make use of:
• (Direct sum) The direct sum V ⊕W is a representation with multiplication given by
g ·(v,w) = (g ·v,g ·w).
• (Dual) Recall that the dual space V∗ is the vector space of linear functionals V → k.
It is a representation with multiplication given as follows: if f ∈ V∗, then g·f is the
functional defined by (g ·f)(v) = f(g−1 ·v).
• (Tensor product) Recall that the tensor product V ⊗ W is a vector space which is
spanned by symbols of the form v⊗w with v ∈ V and w ∈ W subject to the relations
– (v +v(cid:48))⊗w = v ⊗w+v(cid:48) ⊗w,
– v ⊗(w+w(cid:48)) = v ⊗w+v ⊗w(cid:48),
– λ(v ⊗w) = (λv)⊗w = v ⊗(λw) for any λ ∈ k.
(cid:80) (cid:80)
Then V ⊗W is a representation of G via g · (v ⊗w ) = (g ·v )⊗(g ·w ).
i i i i i i
• (Hom spaces) We let Hom(V,W) denote the vector space of linear maps V → W.
This is a representation of G via (g ·f)(v) = g ·f(g−1 ·v). Here we are using both
actions: g−1 · v is the multiplication for V and the other · is the multiplication for
W.
• (Invariants) VG = {v ∈ V | g ·v = v for all g ∈ G} is the space of G-invariants and
is clearly a subrepresentation of V.
When V and W are finite-dimensional, we have a natural isomorphism V∗ ⊗ W →
(cid:80) (cid:80)
Hom(V,W) given by f ⊗ w (cid:55)→ F where F(v) = f (v)w . Furthermore, this is a
i i i i i i
G-equivariant isomorphism (exercise).
It follows from the definitions that Hom(V,W)G is the space of G-equivariant maps V →
W. We will usually denote this by Hom (V,W).
G
1.3. Irreducible representations. Recallthefollowingfromlinearalgebra: ifV isavector
space with a subspace W, then a complement of W is another subspace W(cid:48) of V such that
4 STEVEN V SAM
W ∩W(cid:48) = 0 and W +W(cid:48) = V. In that case, we write V = W ⊕W(cid:48). Complements always
exist but are not unique if W (cid:54)= V and W (cid:54)= 0. The data of a complement is equivalent to
a projection π: V → V (i.e., a linear map satisfying π2 = π) whose image is W: given a
projection, we define W(cid:48) = kerπ; on the other hand, given W(cid:48), every vector v can be written
uniquely as v +v where v ∈ W and v ∈ W(cid:48), and we define π(v) = v .
1 2 1 2 1
A subrepresentation of a representation V is a subspace W ⊆ V such that ρ (g)w ∈ W
V
whenever g ∈ G and w ∈ W. We can then define a homomorphism ρ : G → GL(W) by
W
ρ (g) = ρ (g)| .
W V W
If f: V → V(cid:48) is G-equivariant, then kerf is a subrepresentation of V and imagef is a
subrepresentation of V(cid:48).
Lemma 1.3.1. Suppose that the characteristic of k is either 0 or is p > 0 and that p does
not divide |G|. Then given a subrepresentation W ⊆ V, there exists a subrepresentation
U ⊆ V which is a complement of W, i.e., V = W ⊕U.
Proof. First pick an arbitrary complement W(cid:48) of W. This gives a projection π: V → V.
Define a new linear map ψ: V → V by
1 (cid:88)
ψ = ρ (g)◦π ◦ρ (g)−1.
V V
|G|
g∈G
Note that ρ (h)◦ψ ◦ρ (h)−1 = ψ for any h ∈ G: by the above formula we have
V V
1 (cid:88)
ρ (h)◦ψ ◦ρ (h)−1 = ρ (h)◦ρ (g)◦π ◦ρ (g)−1ρ (h)−1
V V V V V V
|G|
g∈G
1 (cid:88)
= ρ (hg)◦π ◦ρ (hg)−1
V V
|G|
g∈G
The last sum is just the sum for ψ except indexed differently: do the change of variables
g (cid:55)→ h−1g. In particular, this means that ψ is G-equivariant.
Next, we claim that ψ2 = ψ and has image W. Pick v ∈ V. Then π(ρ (g)−1v) ∈ W since
V
π has image W. Since W is a subrepresentation, ρ (g)(π(ρ (g)−1v)) ∈ W as well, so ψ(v)
V V
is a sum of elements in W, and hence belongs to W. If v ∈ W, then ρ (g)−1v ∈ W, and so
V
π(ρ (g)−1v) = ρ (g)−1v since π2 is the identity on W. This gives
V V
1 (cid:88) 1
ψ(v) = ρ (g)(ρ (g)−1(v)) = |G|v = v.
V V
|G| |G|
g∈G
So we take U = kerψ and conclude that U is another complement W. Since ψ is G-
equivariant, U is a subrepresentation of V. (cid:3)
A nonzero representation V is irreducible if it has no nonzero subrepresentations other
than itself. A representation which is not irreducible is called reducible.
Example 1.3.2. Any 1-dimensional representation is automatically irreducible. An exam-
ple of a reducible representation would be to take V = k2 and have all g act by the identity.
In that case, any 1-dimensional subspace of V is a nonzero subrepresentation. This ex-
ample also shows that the complement found above is not unique: for any 1-dimensional
subrepresentation of V, any other 1-dimensional subspace gives a complement which is also
a subrepresentation. (cid:3)
NOTES FOR MATH 202B 5
Theorem 1.3.3 (Schur’s lemma). Let V and W be nonzero irreducible representations of G
and let ϕ: V → W be G-equivariant.
(1) Either ϕ is an isomorphism, or ϕ = 0.
(2) Suppose k is algebraically closed. If V = W, then ϕ is a scalar multiple of the identity.
Proof. (1) kerϕ is a subrepresentation of V. Since V is irreducible, this means that kerϕ = 0
or kerϕ = V. In the second case, ϕ = 0. In the first case, ϕ is injective and so the image
of ϕ is nonzero. Since W is irreducible and imageϕ is a subrepresentation, this means that
imageϕ = W. So ϕ is also surjective, so we conclude that ϕ is an isomorphism.
(2) Let λ be an eigenvalue of ϕ. Then ϕ−λ·id is G-equivariant, and has nonzero kernel.
V
In particular, it must be the 0 map by (a). (cid:3)
Example 1.3.4. Take k = R the field of real numbers and G = Z/4 = {1,z,z2,z3} and V =
(cid:20) (cid:21)
0 −1
R2. Then ρ(z) = uniquely determines a representation (since ρ(z)4 is the identity).
1 0
Furthermore, thisisanirreduciblerepresentation(any1-dimensionalsubrepresentationmust
be an eigenspace for ρ(z) but they are not realizable over R). Since G is abelian, ρ(z)
commutes with all ρ(g) and so in particular, ρ(z): V → V is a G-equivariant map which is
not a scalar multiple of the identity. The problem is that if we extend our scalars to C, then
V is no longer irreducible. (cid:3)
The previous example prompts a definition: we say that V is absolutely irreducible if it
remains irreducible upon enlarging the field of coefficients. Actually it suffices to know that
it remains irreducible upon enlarging to the algebraic closure k of k. We won’t go much into
the details of subtleties of non-algebraically closed fields, though.
Theorem 1.3.5 (Maschke). Suppose that the characteristic of k is either 0 or is p > 0 and
that p does not divide |G|. Every finite-dimensional representation of G is a direct sum of
irreducible subrepresentations.
Furthermore, this decomposition is unique in the following sense: if V = V ⊕ ··· ⊕ V
1 r
and V = V(cid:48)⊕···⊕V(cid:48) are two decompositions of V into irreducible subrepresentations, then
1 s
r = s and there exists a permutation σ of 1,...,r so that V ∼= V(cid:48) for all i.
i σ(i)
The empty direct sum is the 0 vector space, so in the proof, we assume that dimV > 0.
Proof. Induction on dimension of V. If dimV = 1, then V must be irreducible. Otherwise,
if dimV > 1 and V is not irreducible, pick a nonzero subrepresentation W ⊂ V with
W (cid:54)= V. By the previous result, we can find a subrepresentation U ⊂ V so that V = W ⊕U.
But dimW < dimV and dimU < dimV, so by induction, both W,U are direct sums of
irreducible subrepresentations.
For the second statement, consider the identity map on V. By Schur’s lemma, the com-
ponent maps V → V(cid:48) of the identity are either 0 or isomorphisms. Let W ,...,W be all of
i j 1 a
the V which are isomorphic to V and let W(cid:48),...,W(cid:48) be all of the V(cid:48) which are isomorphic
i 1 1 b j
to V . We conclude that the restriction of the identity to W ⊕···⊕W → W(cid:48) ⊕···⊕W(cid:48)
1 1 a 1 b
must be an isomorphism since the W have 0 maps to all other V(cid:48). This means that a = b;
i j
remove these summands from V and repeat to get the desired conclusion. (cid:3)
1.4. Characters. Now we assume that k = C.
6 STEVEN V SAM
The character of ρ is the function χ : G → C defined by χ (g) = Tr(ρ(g)) where Tr
ρ ρ
denotes the trace of a linear operator. This is constant on conjugacy classes of G:
χ (hgh−1) = Tr(ρ(h)ρ(g)ρ(h)−1) = Tr(ρ(g)) = χ (g).
ρ ρ
We’ll also write χ if V is the vector space of the representation. Since ρ(1 ) is the identity,
V G
we have χ (1 ) = dimV.
ρ G
We’ll use the following fact many times: the trace of a linear operator is the sum of its
eigenvalues.
Recall that an algebraic integer is a complex number α which is the solution to a
monic polynomial with integer coefficients, i.e., there exist integers c ,...,c so that
0 n−1
αn + (cid:80)n−1c αi = 0. The following results are standard results in algebra, and we will
i=0 i
not repeat their proofs:
Proposition 1.4.1. (1) The set of algebraic integers is a subring of the complex numbers:
they are closed under addition, subtraction, and multiplication.
(2) Every rational number which is an algebraic integer is in fact an integer.
An important example of algebraic integers are roots of unity: these are the solutions to
the equations zn −1 = 0 for some n.
√
Note that algebraic integers could be imaginary, for example, −1 is an algebraic integer.
Proposition 1.4.2. For any g ∈ G and representation V, χ (g) is an algebraic integer.
V
Proof. All of the eigenvalues of ρ(g) are roots of unity, which are algebraic integers. Since
χ (g) is the sum of these eigenvalues, it is also an algebraic integer. (cid:3)
V
Lemma 1.4.3. Let V be a representation with character χ . Then χ (g−1) = χ (g) and the
V V V
character of its dual is given by χ (g) = χ (g) where the bar denotes complex conjugation.
V∗ V
Proof. Let λ ,...,λ be the eigenvalues of ρ (g) with eigenvectors v ,...,v . Let f ,...,f
1 n V 1 n 1 n
be the dual basis for V∗. Then f is an eigenvector with eigenvalue λ−1 for ρ (g) (recall
i i V∗
the action is through g−1). Since the λ are roots of unity, we have λ−1 = λ (recall that
i i i
λλ = |λ| for any complex number and roots of unity have absolute value 1). In particular,
χ (g) = χ (g−1) = λ−1 +···+λ−1 = λ +···+λ = ρ (g). (cid:3)
V∗ V 1 n 1 n V
Lemma 1.4.4. Let V,W be representations. The character of V ⊕W is given by χ (g) =
V⊕W
χ (g)+χ (g).
V W
Proof. Letv ,...,v beaneigenbasisforρ (g)witheigenvaluesλ ,...,λ andletw ,...,w
1 n V 1 n 1 m
beaneigenbasisforρ (g)witheigenvaluesµ ,...,µ . Then{(v ,0),...,(v ,0),(0,w ),...,(0,w )}
W 1 m 1 n 1 m
is an eigenbasis for V ⊕W with eigenvalues λ ,...,λ ,µ ,...,µ . (cid:3)
1 n 1 m
Lemma 1.4.5. Let V,W be representations. The character of V ⊗W is given by χ (g) =
V⊗W
χ (g)χ (g).
V W
Proof. Letv ,...,v beaneigenbasisforρ (g)witheigenvaluesλ ,...,λ andletw ,...,w
1 n V 1 n 1 m
be an eigenbasis for ρ (g) with eigenvalues µ ,...,µ . Then {v ⊗ w } is an eigenba-
W 1 m i j i,j
(cid:80) (cid:80) (cid:80)
sis for ρ (g) with eigenvalues {λ µ }. So χ (g) = λ µ = ( λ )( µ ) =
V⊗W i j V⊗W i,j i j i i j j
χ (g)χ (g). (cid:3)
V W
Lemma 1.4.6. Let V = C[X] be a permutation representation on the set X. Then χ (g) =
V
|{x ∈ X | g ·x = x}|.
NOTES FOR MATH 202B 7
Proof. The elements of X give a basis for C[X]. In matrix form ρ(g) becomes a permutation
matrix and the number of 1’s on the diagonal is just the number of elements fixed by g. (cid:3)
Lemma 1.4.7 (Projection formula). Define ϕ: V → V by
1 (cid:88)
ϕ(v) = g ·v.
|G|
g∈G
ϕ is a projection and its image is VG. In particular, dimVG = Tr(ϕ).
Proof. First we prove the image is contained in VG: for any h ∈ G, we have
1 (cid:88) 1 (cid:88)
h·ϕ(v) = hg ·v = g ·v = ϕ(v).
|G| |G|
g∈G g∈G
where in the second equality, we reindexed the sum since {hg | g ∈ G} = {g | g ∈ G}.
On the other hand, given w ∈ VG, we have ϕ(w) = w, and so the image is all of VG.
These two facts imply that ϕ is a projection: ϕ2(v) = ϕ(ϕ(v)) and ϕ(v) ∈ VG which
implies that ϕ2(v) = ϕ(v).
For the last statement, note that the eigenvalues of a projection are either 0 or 1 (since
it’s a root of the polynomial t2 −t) and the multiplicity of 1 is its rank. (cid:3)
A function G → C which is constant on conjugacy classes is called a class function and
CF(G) is the set of class functions G → C. This is a C-vector space. Define a pairing on
CF(G) by:
1 (cid:88)
(ϕ,ψ) = ϕ(g)ψ(g)
G
|G|
g∈G
where the overline means complex conjugation. If we don’t need to specify G, we’ll just
write (,).
Lemma 1.4.8. (,) is an inner product on CF(G).
Proof. Let c be the number of conjugacy classes of G and order the conjugacy classes as
γ ,...,γ . Then CF(G) ∼= Cc by sending a class function f to ((cid:112)|γ |f(γ ),...,(cid:112)|γ |f(γ )).
1 c 1 1 c c
Under this isomorphism, (,) is 1/|G| times the standard inner product on Cc, and hence is
itself an inner product. (cid:3)
Lemma 1.4.9. Given two representations V,W, we have dimHom (V,W) = (χ ,χ ).
G V W
Proof. We have Hom (V,W) = Hom(V,W)G ∼= (V∗ ⊗ W)G. The character of V∗ ⊗ W is
G
given by χ (g) = χ (g)χ (g). From the projection formula, for any representation U,
V∗⊗W V W
we have dimUG = 1 (cid:80) χ (g). Applying this to U = V∗ ⊗W, we get
|G| g∈G U
1 (cid:88)
dim(V∗ ⊗W)G = χ (g)χ (g) = (χ ,χ ) = (χ ,χ ).
V W W V V W
|G|
g∈G
But note that dim(V∗ ⊗W)G is an integer, so in particular, (χ ,χ ) = (χ ,χ ). (cid:3)
V W V W
A priori, characters are complex-valued. For some special groups, the values never take
imaginary values.
Proposition 1.4.10. Suppose that g is conjugate to g−1 for all g ∈ G. Then χ (g) is a real
V
number for all g ∈ G and all representations V.
8 STEVEN V SAM
Proof. χ (g−1) = χ (g), but if they are conjugate, we also have χ (g) = χ (g−1), so
V V V V
χ (g) ∈ R. (cid:3)
V
Proposition 1.4.11. Suppose that for each integer m that is relatively prime to |G|, and
for all g ∈ G, we have that g is conjugate to gm. Then χ (g) is an integer for all g ∈ G and
V
all representations V.
Proof. Since χ (g) is an algebraic integer, it will suffice to show that it is a rational number
V
by Proposition 1.4.1. The proof is similar to the previous result, but uses a little bit of Galois
theory: let L be the field generated by Q and a primitive |G|th root of unity ω. For every m
that is relatively prime to |G|, there is an automorphism σ of L determined by replacing ω
m
by ωm, and an element of L is in Q if and only if it is fixed by all of these automorphisms.
From the previous results, we know that χ (g) ∈ L for all g ∈ G and all representations
V
V. Furthermore, σ (χ (g)) = χ (gm) since the trace of an mth power of a linear operator
m V V
is the sum of the mth powers of its eigenvalues, so our assumption together with the Galois
theory above implies that χ (g) ∈ Q for all g ∈ G and representations V. (cid:3)
V
1.5. Classification of representations.
Lemma 1.5.1. Let ρ: G → GL(V) be a representation and f ∈ CF(G) a class function.
Define
(cid:88)
ρ = f(g)ρ(g).
f
g∈G
If V is irreducible, then ρ = λ·id where
f V
|G|
λ = (f,χ ) .
V G
dimV
Proof. Pick h ∈ G. Then
(cid:88) (cid:88)
ρ(h)ρ ρ(h)−1 = f(g)ρ(hgh−1) = f(hgh−1)ρ(hgh−1) = ρ
f f
g∈G g∈G
where the second equality follows from f ∈ CF(G) and the third equality follows since
conjugation of h is a permutation of the elements of G. This implies that ρ commutes with
f
all h ∈ G and hence is a scalar by Schur’s lemma.
To determine λ, we have
(cid:88) (cid:88)
dim(V)λ = Tr(ρ ) = f(g)Tr(ρ(g)) = f(g)χ (g) = |G|(f,χ ) . (cid:3)
f V V G
g∈G g∈G
Theorem 1.5.2. • The characters of the irreducible representations form an orthonor-
mal basis for CF(G). In particular, the number of isomorphism classes of irreducible
representations of G is equal to the number of conjugacy classes of G.
• If two representations have the same character, then they are isomorphic.
∼
Proof. Let V,W be irreducible. By Schur’s lemma, dimHom (V,W) is 0 if V (cid:54)= W and is
G
∼ ∼ ∼
1 if V = W. In particular, (χ ,χ ) = 0 if V (cid:54)= W and is 1 if V = W. So if V ,V ,...
V W 1 2
are pairwise non-isomorphic irreducible representations, then χ ,χ ... are orthonormal.
V1 V2
In particular, they are linearly independent. Next, we need to show that they span CF(G).
Let f be a function in the orthogonal complement of the span of the χ . By Lemma 1.5.1,
Vi
ρ (notation used there) is 0 for all irreducible ρ, and hence for all representations ρ since
f
NOTES FOR MATH 202B 9
Maschke’s theorem implies every representation is a direct sum of irreducibles. Now consider
the regular representation C[G] of G. In that case we have
(cid:88)
0 = ρ (e ) = f(g)e .
f 1G g
g∈G
Since the e are a basis, this implies that f(g) = 0 for all g ∈ G, i.e., that f = 0. So CF(G)
g
is spanned by the χ .
Vi
Since the dimension of the space of class functions is the number of conjugacy classes of
G, we see that this is also the number of irreducible representations of G. Let V ,...,V be
1 c
a complete list of irreducible representations of G up to isomorphism.
Next, given a representation V, we have V ∼= V⊕m1 ⊕···⊕V⊕mc for some non-negative
1 c
integers m ,...,m (the multiplicities). Again, by Schur’s lemma and the previous result,
1 c
m = (χ ,χ ). If W is another representation, it is isomorphic to V if and only if the
coirresponVdiingV multiplicities agree with m ,...,m . Hence if χ = χ , then V ∼= W. (cid:3)
1 c V W
Corollary 1.5.3. The multiplicity of an irreducible representation V in the regular repre-
sentation C[G] is dimV.
Proof. The multiplicity is given by (χ ,χ ). Note that χ (1 ) = |G| and χ (g) = 0
V C[G] C[G] G C[G]
for g (cid:54)= 1 . In particular,
G
1
(χ ,χ ) = χ (1 )|G| = χ (1 ) = dimV. (cid:3)
V C[G] V G V G
|G|
Corollary 1.5.4. Let V ,...,V be the irreducible representations of G with dimensions
1 c
d ,...,d . Then d2 +···+d2 = |G|.
1 c 1 c
Proof. The dimension of C[G] is |G| and it contains V with multiplicity d , so |G| = (cid:80)c d2.
i i i=1 i
(cid:3)
Corollary 1.5.5. If G is abelian, Then all irreducible representations are 1-dimensional.
Proof. The conjugacy classes of abelian groups are singletons, so there are c = |G| many
conjugacy classes. The only solution to c = d2 +···+d2 where the d are positive integers
1 c i
are d = ··· = d = 1. (cid:3)
1 c
1.6. Examples.
1.6.1. Direct products. Suppose we are given two groups G ,G and representations ρ ,ρ
1 2 1 2
on vector spaces V ,V . Then G ×G has a linear action on V ⊗V by
1 2 1 2 1 2
(cid:88) (cid:88)
(g ,g )· v(1) ⊗v(2) = g ·v(1) ⊗g ·v(2) .
1 2 i i 1 i 2 i
i i
Hence we get a representation ρ ⊗ρ of G ×G . Its character is given by
1 2 1 2
χ (g ,g ) = χ (g )χ (g ).
ρ1⊗ρ2 1 2 ρ1 1 ρ2 2
This is the external tensor product of representations. To emphasize that it is a represen-
tation of the direct product, we will write V (cid:2) V . Here are some facts which are left as
1 2
exercises:
(1) If V,W are irreducible, then so is V (cid:2)W.
(2) Let V ,...,V and W ,...,W be complete lists of irreducible representations (up
1 n 1 m
to isomorphism) of G and G , respectively. Then {V (cid:2) W } is a complete list of
1 2 i j
irreducible representations (up to isomorphism) of G ×G .
1 2
10 STEVEN V SAM
1.6.2. Abeliangroups. WesawinCorollary1.5.5thatallirreduciblerepresentationsofabelian
groups are 1-dimensional. We will completely describe them starting with cyclic groups
G = Z/m. Let ω be a primitive mth root of unity, i.e., ωm = 1 but ωn (cid:54)= 1 for all n < m.
For a concrete example, we can take exp(2πi/m). For 0 ≤ i ≤ m−1, define a homomorphism
ρ : Z/m → GL (C) by ρ (g) = ωig. These are all irreducible representations (since they are
i 1 i
1-dimensional), they are not isomorphic to each other (their characters are all different), and
there are m of them, which is the number of conjugacy classes of Z/m, so we have described
all of them.
A general finite abelian group is isomorphic to a direct product of Z/m for various m, so
we can construct all of the irreducible representations using the previous example.
1.6.3. Dihedral groups. For n ≥ 3, let D be the symmetries of a regular n-gon. Then
n
|D | = 2n and we will use without proof that when n is odd, there are (n+3)/2 conjugacy
n
classes, and when n is even, there are (n+6)/2 conjugacy classes.
Actually, if we center the regular n-gon at the origin in the plane, then each element of
D (rotation or reflection) is a linear operator, so we get a representation ρ: D → GL (R),
n n 2
which is called the reflection representation. We can then extend the coefficients to C; I’ll
leave it as an exercise to show that the result is an irreducible representation. There is also
a 1-dimensional representation which sends g to the determinant of ρ(g) under the reflection
representation. We call this the sign representation.
For concreteness, consider D , which has 4 conjugacy classes and size 10. So it has 4
5
irreducible representations, let d ,...,d be their dimensions. We know that d2 + d2 +
1 4 1 2
d2 + d2 = 10, and so we must have d ≤ 2 for all i. The only solution is {1,1,2,2}. We
3 4 i
know three examples: the trivial representation, the sign representation, and the reflection
representation. Since the reflection representation is real-valued, it is isomorphic to its dual.
So that doesn’t create a new representation. Furthermore, we can create a 2-dimensional
representation by tensoring the reflection representation with the sign representation, is it
new? Turns out it is not – but we will stop with any further calculations.
1.6.4. Symmetric groups. A large portion of this course will be devoted to working out the
characters of the symmetric groups S . We will content ourselves now with some basic facts
n
and small examples. First, |S | = n!. The conjugacy classes are easy to describe:
n
Lemma 1.6.1. Two permutations are conjugate if and only if they have the same cycle type
decomposition. Hence, the conjugacy classes of S are naturally indexed by partitions of n,
n
and the number of irreducible representations of S is the partition number p(n)
n
Proof. if (i ,i ,...,i ) denotes the cycle i (cid:55)→ i (cid:55)→ ··· (cid:55)→ i (cid:55)→ i , then we can use the
1 2 k 1 2 k 1
identity
τ(i ,i ,...,i )τ−1 = (τ(i ),τ(i ),...,τ(i )). (cid:3)
1 2 k 1 2 k
Every character is integer-valued: this amounts to showing that for every permutation σ,
and every m coprime to n!, then σm and σ have the same cycle type. Note that m being
coprime to n! means it is coprime to every i = 1,...,n. If c ···c = σ is the decomposition
1 r
into disjoint cycles, then σm = cm···cm. Furthermore, if c has length i and m is coprime to
1 r i
i, then cm is again a cycle of length i. So σ and σm have the same cycle type and are hence
i
conjugate to one another.
Every S has a representation on Cn by sending each σ to the corresponding permutation
n
operator: ρ(σ)(e ) = e . This is the permutation representation of S . This is not
i σ(i) n