1. A NOTE ON CHARACTER THEORY
FAN HUANG
In this note, we explore two examples. We try to glimpse both the analogy between
character theory of finite groups and representation theory of the infinite compact group,
and the power of character theory for nonabelian groups.
1. Basic Facts of Character Theory of Finite Groups
Let G be a finite group and ρ : G → GL(V ) be a linear representation of G in the complex
finite dimensional vector space V . For each g ∈ G, put χρ(g) = Tr(ρ(g)). The complex
valued function χρ on G is called the character of the representation ρ and it is a class
function– those are any functions from G into F which are constant on the conjugacy classes
of G, i.e., f : G → F such that f(g−1
xg) = f(x) for all g, x ∈ G. It characterizes the
representation ρ in the sense that representations are determined by their characters, up to
isomorphisms. Let
ρ : G → GL(V )
τ : G → GL(W)
be two representations of a group G on vector spaces V and W. An isomorphism from V
and W is a G-linear map f : V → W such that
τ(g) ◦ f(v) = f ◦ ρ(g)(v), ∀g ∈ G, ∀v ∈ V
i.e., for all g ∈ G, the below square commutes.
V V
W W
ρ(g)
f
τ(g)
f
In that case, we say that two representations ρ and τ are isomorphic. In terms of matrices,
two matrix representations are isomorphic if and only if they describe the same representation
in different bases. Specifically, choose a basis A = {a1, . . . , an} for V and a basis B =
{b1, . . . , bn} for W, and let
ρA : G → GLn(C)
τB : G → GLn(C)
be the two matrix representations obtained by writing ρ and τ in these bases. Then ρ and
τ are isomorphic if and only if ρA and τB are isomorphic.
Proof. It can be illustrated via the commutative graphs:
Date: August 2013.
1

2. A NOTE ON CHARACTER THEORY 2
Cn
Cn
W W
V V
Cn
Cn
τB(g)
B B
f f
A A
τ(g)
ρ(g)
ρA(g)
S S−1
There is an inner product on the vector space of class functions given by
f, h =
1
|G| g∈G
f(g)h(g),
where |G| is the order of the group G. The characters of irreducible representations form an
orthonormal basis for the space of class functions on G, denoted as Cl(G). Every represen-
tation is a direct sum of irreducible representations. Let ρ be any representation of G and ρi
be an isomorphism class of irreducible representations of G. Let χ and χi be the characters
of ρ and ρi respectively. Then the multiplicity of ρi in ρ is the inner product χi, χ .
2. The representation theory of the circle group S1
(the simplest infinite compact group)
Consider the unit circle S1
= {z ∈ C×
| |z|= 1} = {(x, y) ∈ R2
| x2
+ y2
= 1} in the
complex plane. If we view the circle as the set of points {eiθ
: θ ∈ R}, the natural group
operation is multiplication of complex numbers. We treat S1
as an abelian topological group
under multiplication with its topology as a subset of C. (A topological group is a group G
equipped with a topology such that the product map G×G → G and the inverse map G → G
are both continuous. More generally, a representation of a topological group on a vector
space V is a continuous group homomorphism ρ : G → GL(V ) with the topology of GL(V )
inherited from the space End(V ) of continuous linear operators). However, if we identify
points on S1
with their angle θ, then S1
becomes R modulo 2π, denote it R/2πZ, where the
operation is addition.The group isomorphism R/2πZ → S1
is given by t → eit
= cos t+i sin t.
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3. FAN HUANG 3
From [JM74], since it is a continuous bijection between compact topological spaces, it is a
homeomorphism.
In this note we will focus on unitary representations of S1
, i.e., representations on a
Hilbert space which preserve the inner product. Unitary representations are those for which
the operators ρg are unitary, i.e. preserve the inner product. On the complex plane, this
means ρgv, ρgw = v, w = wv, ∀v, w ∈ V, g ∈ G. which means equivalently that ρg
t
ρg = I.
We denote the unitary representation of S1
as ρn : S1
→ U(V ).
It is not hard to see that for each integer n, the formula ρn(z) = zn
gives a 1-dimensional
irreducible representation of S1
. We shall show that
(1) Every irreducible unitary representation of S1
has dimension 1.
(2) Every continuous irreducible representation of S1
is of this form.
(3) For n = k, ρn and ρk are not isomorphic.
That is, Z is isomorphic to irreducible representations of S1
with bijection given by n → ρn.
First, we note that if W is an irreducible unitary representation of S1
, then dim W = 1.
To see this, we take as a fact that ρ is determined by ρ(g) where g ∈ S1
is a topological
generator. Then by density, we know where every other element goes. The spectral theorem
from linear algebra asserts that any unitary transformation is diagonalizable. Put another
way, ρ (g) = Sρ(g)S−1
, where S ∈ U(V ) and ρ (g) is a diagonal matrix. The eigenvalues of
unitary matrices all have absolute value 1. It is obvious that for each i the diagonal term
ρi(g) is a 1-dimensional unitary irreducible representation. This follows from ρ = n
i=1 ρi,
ρ = n
i=1 SρiS−1
and thus dim W = 1 as desired.
Proposition 2.1. Every irreducible unitary representations of S1
are the continuous homo-
morphisms of the form z → zn
for some n ∈ Z.
To prove this proposition, we need two Lemmas.
R R
S1
S1
φ
ψ
ρ
ψ
Lemma 2.1. Consider (R, +). If φ : R → R is a continuous homomorphism, then φ is a
multiplication by a scalar.
Proof. Set c = φ(1). Then φ(n) = nc, n ∈ Z. Also mφ( 1
m
) = c and then φ( 1
m
) = c
m
, m ∈ Z.
Thus φ( n
m
) = c n
m
and so φ(x) = cx, for x ∈ Q. Since Q is dense in R and φ is continuous,
we have φ(x) = cx for all x ∈ R.
Lemma 2.2. If ψ : R → S1
is a continuous homomorphism, then there exists c ∈ R such
that ψ(x) = eicx
for all x ∈ R.
Proof. We claim that there is a unique continuous homomorphism l : R → R such that
ψ(x) = ei·l(x)
. The exponential map ε : R → S1
given by ε(x) = eix
maps the real line around
the unit circle with period 2π and it is a universal cover. For any continuous ψ : R → S1
such that ψ(0) = 1, there exists a unique continuous lift l of this function to the real line
such that l(0) = 0, In other words, there exists a unique continuous function l : R → R such
that l(0) = 0 and ψ(x) = ε(l(x)) for all x, so the following diagram commutes:
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4. A NOTE ON CHARACTER THEORY 4
R
S1R
ε
ψ
l
We also claim that if ψ is a homomorphism, then its lift l is also a homomorphism and by
Lemma 2.1 l(x) = cx for some c.
Note that ψ(s + t) = ψ(s)ψ(t), thus ε(l(s + t) − l(s) − l(t)) = 1 = e0
. It follows that
l(s + t) − l(s) − l(t) = 2πn for some n ∈ Z which depends only on s, t.
Since s and t varies continuously, we find n is a constant. We set s = t = 0 to conclude
that n = 0. Thus l is a homomorphism as promised, and so l(x) = cx for some c ∈ R by
Lemma 2.2.
Proof of proposition 2.1. Going back to the irreducible unitary representations of S1
, given
a representation ρ : S1
→ C∗
, it has a compact and thus bounded image. It follows that the
image lies on S1
. Thus ρ : S1
→ S1
is a continuous homomorphism. Precompose ρ with the
exponential map ε, then we have the following diagram commutes:
S1
S1
R
ρ
ε
ρε
By Lemma 2.2 there exists c ∈ R with ρε(x) = eicx
for all x. Thus we have ρ(eix
) = eicx
.
Note that 1 = ρ(1) = ρ(e2πi
) = ρε(2π) = e2πic
and thus c ∈ Z. So we have ρ(z) = ρn(z) =
zn
.
Finally, to show for n = k, that ρn and ρk are not isomorphic, because let f : C → C
be the G-linear map. Note that zn
f(v) = f(zn
v) = zk
f(v) for all z ∈ S1
, then we have
zn−k
= 1, i.e., n = k.
Because of the isomorphism from S1
to R/2πZ, another way to view the characters on S1
is as a map R/2πZ → C. Then the characters are given by en(x) = einx
with n ∈ Z. We
shall show all characters have this form by following an exercise from [ST03].
First we denote by ˆG the class of all characters of G, which is called the dual group of G.
We observe that the trivial character e(g) = 1 plays the role of the unit, so ˆG inherits the
structure of an abelian group under the multiplication
χ1χ2(g) = χ1(g)χ2(g)
for all g ∈ G and characters χ1, χ2.
For x ∈ G the mapping χ → χ(x) is an irreducible character of ˆG and so an element of
the dual
ˆˆG of the dual
ˆˆG of ˆG. Let φx be the element be the map that maps x ∈ ˆG to
χx. Then consider the map Φ : G →
ˆˆG given by x → φx. We see it is a homomorphism
since Φ(xy) = φxy(χ) = φx(χ)φy(χ) = Φ(x)Φ(y). Also if for x, x ∈ G, Φ(x) = Φ(x ), then
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5. FAN HUANG 5
χ(x) = χ(x ) and thus χ(xx −1
) = 1 for χ ∈ ˆG, which implies xx −1
= 1. So Φ is injective.
Since G and
ˆˆG have same orders, we conclude that it is an isomophism.
Lemma 2.3. If F : R → R is continuous and F(x+y) = F(x)F(y), then F is differentiable
and F is of the form eAx
for some constant A.
Proof. If F(x) = 0 is trivial, then we have F(0) = F(0)2
. It follows that F(0) = 1 and it is
clearly differentiable and is of the form e0x
for constant 0.
Consider F(0) = 0, then for an appropriate δ, c =
δ
0
F(y) dy = 0. cF(x) =
δ
0
F(y)F(x) dy =
δ
0
F(x + y) =
δ+x
x
F(y)dy. Then
c lim
t→0
F(x + t) − F(x)
t
= lim
t→0
δ+x+t
x+t
F(y) dy −
δ+x
x
F(y) dy
t
= lim
t→0
δ+x+t
δ+x
F(y) dy
t
− lim
t→0
x+t
x
F(y) dy
t
= F(δ + x) − F(x)
Thus F is differentiable and we have
cF (x) = F(δ + x) − F(x)
= F(δ)F(x) − F(x)
F (x) =
F(δ) − 1
c
F(x)
So F is of the form eAx
for some constant A.
By applying Lemma 2.3 and noting that the characters on S1
satisfy the assumptions, we
see that all the characters are of the form eAx
for constant A . The dual group of the circle
S1
is ˆS1 = {en}n∈Z, where en(x) = einx
. Moreover, en → n gives an isomorphism between
ˆS1 and the integers Z.
Now we have a complete list of irreducible unitary representations of S1
. From [Serre77],
we already know the orthogonality relations of irreducible characters of finite groups.
(a) If χ is the character of an irreducible representation, χ, χ = 1 (i.e., χ is “of norm 1”).
(b) If χ and χ are the characters of two nonisomorphic irreducible representations, χ, χ =
0 (i.e., χ and χ are orthogonal).
(c) Let V be a linear representation of G, with character φ, and suppose V decomposes into
a direct sum of irreducible representations: V = k
i=1 Wi. Then, if W is an irreducible
representation with character χ, the number of Wi isomorphic to W is equal to the scalar
product φ, χ . Moreover, for f ∈ Cl(G), f = k f, χ χ.
As we shall see, the basic theorems from Fourier analysis give the analogy of representation
theory listed above for G = S1
and V = L2
(S1
). The space L2
(S1
) is the Hilbert space of
complex measurable functions f on S1
with
f = f 2=
1
0
|f|2
1
2
≤ ∞.
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6. A NOTE ON CHARACTER THEORY 6
Also L2
(S1
) is endowed with an inner product f, g = 1
2π t∈G
f(t)g(t) dt for f, g ∈ L2
(S1
).
Note that the linear combinations of these characters f(x) = n cneinx
(x ∈ R) are trigono-
metric polynomials. With complex coefficients an trigonometric polynomial is actually a
Fourier series. Then naturally, we can find analogy with fourier analysis.
Attention is now focused on the space L2
(S1
). By Stone-Weierstrass approximation theo-
rem, the space spanned by the positive and negative powers of e2πix
is dense in the space of
continuous functions and the space C∞
(S1
) of infinitely differentiable functions is dense in
L2
(S1
). Note S1
acts by left multiplication on f ∈ L2
(S1
), i.e., (ρ(t) · f)(θ) = f(t + θ) for
t, θ ∈ S1
. This is well-defined on L2
since Lebesgue measure is invariant under translation.
Put another way, this action is indeed the regular representation and also the representation
induced from the identity element, denoted as IndG
{1}. We will talk more of the induced
representation in the next example.
From [DM72],the basic theorems of Fourier series on S1
are:
Theorem 2.1. The orthogonal family of unit length ( en = 1)
en(x) = e2πinx
, n ∈ Z
is a basis for L2
(S1
), that is, any function f ∈ L2
(S1
) can be expanded into a Fourier Series
f =
∞
−∞
ˆf(n)en
with coefficients
ˆf(n) = (f, en) = fen dx = f(x)e−2πinx
dx,
the sum being understood in the sense of distance in L2
(S1
). The map f → ˆf is therefore
an isomorphism of L2
(S1
) onto L2
(Z) and there is a Plancherel identity:
f 2
2=
1
0
|f|2
= f 2
=
∞
−∞
| ˆf(n)|2
.
Theorem 2.2. For any 1 ≤ p < ∞ and any f ∈ Cp
(S1
), the partial sums
Sn = Sn(f) =
|k|≤n
ˆf(k)ek
converge to f, uniformly as n ↑ ∞; in fact, Sn − f ∞ is bounded by a constant multiple of
n−p+1
2 .
Theorem 2.3. For functions f of class C(S1
), the arithmetic means n−1
(S0 + · · · + Sn−1)
of the partial sums Sn = |k|≤n
ˆf(k)ek converge uniformly to f.
In our example, we replace the form “ 1
|G| t∈G f(t)” with “ 1
2π
2π
0
f(t) dt ”. So the ana-
logues for the above in S1
are:
(a) The characters is of unit lenth en, en = 1
2π
2π
0
eint
e−int
dt = 1
2π
2π
0
1 dt = 1.
(b) For m = k, en, ek = 1
2π
2π
0
eint
e−ikt
dt = 1
2π
2π
0
ei(n−k)t
dt = 0.
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7. FAN HUANG 7
(c) For f ∈ L2
(S1
), f = ∞
−∞ f, en en, converges for fixed n, where f, en = 1
2π
2π
0
e−int
f(t) dt,
leading to the Fourier inversion formula.
Among others, we see that the characters ρn = einx
form a complete orthonomal basis of
L2
(S1
).
3. An example: the symmetric group Σ3
(the simplest nonabelian group)
The symmetric group Σ3 is isomorphic to the dihedral group, D3, and has 6 elements (the
identity 1, three transpositions of order two: (12), (13) and (23) and two permutations of
order three: (123) and (132). It can be presented as x, y | x3
, y2
, xyxy , where x = (123)
and y = (12). Its regular representation has dimension 3! = 6 and since it is a nonabelian
finite group, its subrepresentation cannot be all 1-dimensional. In Σn, conjugacy classes
are determined by cycle decomposition sizes: two permutations are conjugate if and only
if they have the same number of cycles of each length. For Σ3, there are 3 conjugacy
classes, so there are 3 different irreducible representations over C. In the decomposition
into irreducible representations with respective dimensions d1, d2, d3, the order of the group
|Σ3|= 6 = d2
1 + d2
2 + d2
3. Thus there must be two 1-dimensional and one 2-dimensional
representation . Explicitly, they are:
(a) The trivial representation with character χT , that take the value 1 identically.
(b) The sign representation with character χΣ given by πsgn(g)v = sgn(g)v, where sgn(g) =
±1 is the sign of the permutation.
(c) The standard representation with character χA, that leaves invariant the orthogonal
subspace V = {(z1, z2, z3) ∈ C3
| z1 + z2 + z3 = 0}. This representation is 2-dimensional.
The irreducible characters of a group can be assembled into a character table, where the
first row in a character table lists representatives of conjugacy classes, the second row the
numbers of elements in the conjugacy classes, and the other rows list the values of the
characters on the conjugacy classes–which are obtained by explicitly computing traces in the
irreducible representations. The characters are constant on conjugacy classes.
The character table of Σ3 is:
Σ3 (1) (123) (12)
# 1 2 3
χT 1 1 1
χΣ 1 1 -1
χA 2 -1 0
Let H be a subgroup of G. Let ρH be the restriction of ρ, the representation of G, to H.
Let W be a sub representation of ρH, that is, a vector space of V -stable under the ρt, t ∈ H.
Denote by θ : H → GL(W) the representation of H in W thus defined. We call ρ of G in V
is induced by the representation θ of H in W and write
V ∼=
σ∈G/H
σWσ
where σW = gσW and gσ is any representative of the coset σ. Denote V =IndG
H(Wσ). Sup-
pose (V, ρ) is induced by (W, θ) and let χρ and χθ be the corresponding characters of G and
Math 597 Fan Huang

8. A NOTE ON CHARACTER THEORY 8
of H. (W, θ) determines (V, ρ) up to isomorphism.
Example 1. Let H be the trivial subgroup {1}. Let W = C be the trivial representation.
Then G/H ∼= G and thus V = g∈G gC. G acts by permuting these copies of W and we
can see that V is the regular representation of G.
As is shown in [Serre77]:
Theorem 3.1. Let h be the order of H and let R be a system of representatives of G/H.
For each u ∈ G, we have
χρ(u) =
r∈R
r−1ur∈H
χθ(r−1
ur) =
1
h s∈G
s−1us∈H
χθ(s−1
us).
In particular, χρ(u) is a linear combination of the values of χθ on the intersection of H with
the conjugacy class of u in G.
We shall apply above theorem to calculate the IndΣ3
H for the subgroups of Σ3. We note
that characters do play an important role in the algorithm. Let {ρi}r
i=1 be the irreducible
representations of G and {χi}r
i=1 be the corresponding characters. To find how the induced
representation decomposes in terms of irreducible representations of the group Σ3, that
is, IndG
Hχρ(u) = i=r
i=1 aiχi, Since the characters follow the same relation and thus ai =
χρ, χi = 1
|G| t∈G χρ(t)χi(t).
Example 2. Recall that the cyclic group Cn, a finite abelian group, has n irreducible rep-
resentations of degree 1 whoses characters are primitive roots of unity. Let H be the cyclic
subgroup C3 = {1, (123), (132)}, which is a normal subgroup of Σ3. Thus there is short exact
sequence of Σ3:
1 H = C3 Σ3
ϕ
G/H = Z2
ψ
1
The character table of C3 with 3 1-dimensional irreducible characters is
C3 (1) (123) (132)
# 1 1 1
χ1 1 1 1
χ2 1 ω ω2
χ3 1 ω2
ω
where ω = e2πi/3
.
In this example we choose a system representatives of G/H: R = {(1), (12)}. Since
H G, we have r−1
ur ∈ H for all r ∈ R and u ∈ G.
Let W1 be the 1-dimensional representation of H given by (123) → 1, i.e, the trivial
representation of C3. Then
χρ(1) = χθ(1) + χθ(1) = 2
χρ(12) = χρ(13) = χρ(23) = 0
χρ(123) = χθ(123) + χθ(132) = 2
χρ(132) = χθ(132) + χθ(123) = 2
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9. FAN HUANG 9
Since IndΣ3
C3
χ1 is of the form aχT + bχΣ + cχA and we can calculate
a = χρ1 , χT =
1
6
· [2 · 1 + 2 · 1 + 2 · 1 + 0 + 0 + 0] = 1
b = χρ1 , χΣ =
1
6
· [2 · 1 + 2 · 1 + 2 · 1 + 0 + 0 + 0] = 1
c = χρ1 , χA =
1
6
· [2 · 2 + 2 · (−1) + 2 · (−1) + 0 + 0 + 0] = 0
So IndΣ3
C3
ρ1
∼= ρT ⊕ ρΣ.
Let W2 be the 1-dimensional representation of H given by (123) → ω. Then
χρ(1) = χθ(1) + χθ(1) = 2
χρ(12) = χρ(13) = χρ(23) = 0
χρ(123) = χθ(123) + χθ(132) = ω + ω2
= −1
χρ(132) = χθ(132) + χθ(123) = ω + ω2
= −1
Since IndΣ3
C3
χ2 is of the form aχT + bχΣ + cχA and we can calculate
a = χρ2 , χT =
1
6
· [2 · 1 + (−1) · 1 + (−1) · 1 + 0 + 0 + 0] = 0
b = χρ2 , χΣ =
1
6
· [2 · 1 + (−1) · 1 + (−1) · 1 + 0 + 0 + 0] = 0
c = χρ2 , χA =
1
6
· [2 · 2 + (−1) · (−1) + (−1) · (−1) + 0 + 0 + 0] = 1
So IndΣ3
C3
ρ2
∼= ρA, and same for IndΣ3
C3
ρ3 by symmetry.
Example 3. Let H be the cyclic subgroup C2 = {1, (12)} and its character table is:
C2 (1) (12)
# 1 1
χ1 1 1
χ2 1 -1
The three cosets are {(1), (12)}, {(13), (123)}, {(23), (132)}. We choose a system represen-
tatives of G/H: R = {(1), (123), (132)}. A bit different from previous example,H is not a
normal subgroup in Σ3. We need to check if r−1
ur ∈ H for all r ∈ R and u ∈ G.
For example, if u = (12), then
r ∈ R r−1
ur in H?
1 (12) Yes
(123) (13) No
(132) (23) No
if u = (123), then
r ∈ R r−1
ur in H?
1 (123) No
(123) (123) No
(132) (123) No
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10. A NOTE ON CHARACTER THEORY 10
Let W1 be the 1-dimensional representation of H given by (12) → 1, i.e, the trivial repre-
sentation of C2. Then
χρ(1) = χθ(1) + χθ(1) + χθ(1) = 3
χρ(12) = χρ(13) = χρ(23) = χθ(12) = 1
χρ(123) = χρ(132) = 0
Since IndΣ3
C2
χ1 is of the form aχT + bχΣ + cχA and we can calculate
a = χρ1 , χT =
1
6
· [3 · 1 + 0 + 0 + 1 · 1 + 1 · 1 + 1 · 1] = 1
b = χρ1 , χΣ =
1
6
· [3 · 1 + 0 + 0 + 1 · (−1) + 1 · (−1) + 1 · (−1)] = 0
c = χρ1 , χA =
1
6
· [3 · 2 + 0 + 0 + 0 + 0 + 0] = 1
So IndΣ3
C2
ρ1
∼= ρT ⊕ ρA.
Let W2 be the 1-dimensional representation of H given by (12) → −1. Then
χρ(1) = χθ(1) + χθ(1) + χθ(1) = 3
χρ(12) = χρ(13) = χρ(23) = χθ(12) = −1
χρ(123) = χρ(132) = 0
Since IndΣ3
C2
χ1 is of the form aχT + bχΣ + cχA and we can calculate
a = χρ2 , χT =
1
6
· [3 · 1 + 0 + 0 + (−1) · 1 + (−1) · 1 + (−1) · 1] = 0
b = χρ2 , χΣ =
1
6
· [3 · 1 + 0 + 0 + (−1) · (−1) + (−1) · (−1) + (−1) · (−1)] = 1
c = χρ2 , χA =
1
6
· [3 · 2 + 0 + 0 + 0 + 0 + 0] = 1
So IndΣ3
C2
ρ2
∼= ρΣ ⊕ ρA.
Recall that Σ3 acts on the cosets of G/H by ggσH = gτ H and we can actually compute the
explicit induced representation. For each g ∈ G, first note if it flips cosets to decide whether
the matrix is diagonal or flipped. Then note which element of H appears and decides the
actual entries in the matrix. Specifically, the matrix representation induced from W2 in
Example 2 are as follows:
1 →
1 0
0 1
, (12) →
0 1
1 0
, (13) →
0 w
w2
0
,
(23) →
0 w2
w 0
, (123) →
w 0
0 w2 , (132) →
w2
0
0 w
Again, it is clear that this induced representation is ρA by directly calculating the trace for
each matrix .
We see from the above examples that normal subgroups give an easier way to compute
induced characters.
Another way to look at induced representation is by seeing that W is a left C[H]-module
and C[G] is a right C[H]-module. We define V := C[G] ⊗C[H] W. In fact C[G] is a free
Math 597 Fan Huang

11. FAN HUANG 11
C[H]-module via multiplication on the right with basis of gσ, representatives of left cosets
of H. Here, for example,
(123) · ((1) ⊗ e) = (123) ⊗ e = (1)((123) ⊗ e) = (1) ⊗ (123) · e.
(123) · ((12) ⊗ e) = (123)(12) ⊗ e = (13) ⊗ e = (12)(132) ⊗ e
These results lead to further questions. For example, why do characters decide the rep-
resentation up to isomorphism? Another question is how to recover the symmetric function
from the trace?
Acknowledgement
The author would like to thank Professor Matthew Ando for precious and generous guid-
ance, also appreciate undergraduate student Yan Zhou for helpful suggestions and graduate
student Sarah Loeb for careful peer review.
References
[Serre77] Jean-Pierre Serre, Linear Representations of Finite Groups, Springer-Verlag, NY,1977.
[DM72] H.Dym and H.P.Mckean, Fourier Series and Integrals, Academic Press, NY, 1972.
[ST03] Elias M. Stein and Rami Shakarchi, Princeton Lectures in Analysis Fourier Analysis: An Introduc-
tion, Princeton University Press, NJ, 2003, 231–237.
[MA10] Michael Artin, Algebra, Pearson,2 edition, 2010.
[JM74] James R.Munkres, Topology; A First Course, Prentice Hall College Div, 1974.
[CT05] Constantin Teleman, Lecture Notes of Representation Theory, CT, Lent 2005, available at math.
berkeley.edu/~teleman/math/RepThry.pdf.
Department of Mathematics, University of Illinois, Urbana-Champaign, IL 61801, USA
E-mail address: fhuang5@illinois.edu
Math 597 Fan Huang