Contemporary Mathematics
Some Applications of the Fourier Transform
in Algebraic Coding Theory
Jay A. Wood
In memory of Shoshichi Kobayashi, –
Abstract. This expository article describes two uses of the Fourier transform
that are of interest in algebraic coding theory: the MacWilliams identities and
the decomposition of a semi-simple group algebra of a finite group into a direct
sum of matrix rings.
Introduction
Because much of the material covered in my lectures at the ASReCoM CIMPA
research school has already appeared in print ([18], an earlier research school sponsored by CIMPA), the organizers requested that I discuss more broadly the role of
the Fourier transform in coding theory. I have chosen to discuss two aspects of the
Fourier transform of particular interest to me: the well-known use of the complex
Fourier transform in proving the MacWilliams identities, and the less-well-known
use of the Fourier transform as a way of understanding the decomposition of a
semi-simple group algebra into a sum of matrix rings. In the latter, the Fourier
transform essentially acts as a change of basis matrix.
I make no claims of originality; the content is discussed in various forms in such
sources as [1, 4, 15, 16], to whose authors I express my thanks. The material is
organized into three parts: the theory over the complex numbers needed to prove
the MacWilliams identities, the theory over general fields relevant to decomposing
group algebras, and examples over finite fields. Each part is further divided into
numbered sections.
Personal Remark. This article is dedicated to the memory of Professor Shoshichi
Kobayashi, my doctoral advisor, who died 29 August 2012. I admired Professor
2010 Mathematics Subject Classification. Primary 94B05, Secondary 16P10, 20C15, 43A40.
I thank the organizers of the CIMPA school, Edgar Martı́nez Moro and Mustapha Lahyane,
for their support and the invitation to write this article. I thank Steve Szabo for suggesting the
topic of the Fourier transform as a change of basis in a group algebra.
c
2013
American Mathematical Society
1
2
JAY A. WOOD
Kobayashi’s ability to write such wonderful research monographs, and I cherish the
memory of his patience and kindness towards me.
I. Fourier transform over C for finite abelian groups
1. Linear codes and weight enumerators
We begin with a short summary of terminology from coding theory.
Let R be a finite associative ring with 1; we do not assume that R is commutative. Any module over R will be assumed to be unital, i.e., 1 ∈ R acts as the
identity. Fix a finite left R-module A; A will be the alphabet for linear codes. A
(left) R-linear code of length n over A is a left R-submodule C ⊂ An . An important
special case is when A = R, the ring itself. (One could also study right linear codes,
of course.)
Of interest in coding theory are the weights of codewords. The weights are
often studied by means of weight enumerators. There are a number of different
(but related) weight enumerators, of two main types. Our primary interest is in
the first type.
Composition-type. We begin with the most general form of weight enumerator,
the complete weight enumerator.1 For any a ∈ A and x = (x1 , . . . , xn ) ∈ An , define
the counting function
ca (x) = |{i : xi = a}|,
i.e., the
x ∈ An that equal a particular a ∈ A. Note
P number of entries in the vector
n
that a∈A ca (x) = n for all x ∈ A . For a fixed x ∈ An , the numbers ca (x), a ∈ A,
form the “composition” of x.
For every a ∈ A, let Za be an indeterminate. Form the polynomial ring C[Za :
a ∈ A] over the complex numbers. Given an R-linear code C ⊂ An , the complete
weight enumerator of C is
cweC (Z) =
n
XY
Z xi =
x∈C i=1
XY
Zaca (x) .
x∈C a∈A
The complete weight enumerator counts the number of codewords in C of any given
composition.
Experience has shown it useful to cluster certain compositions together; i.e.,
certain compositions are to be regarded as equivalent. To that end, let ∼ be an
equivalence relation on the alphabet A. Denote the equivalence class of a ∈ A
by [a]. For x ∈ An and equivalence class [a], a ∈ A, define a symmetrized weight
composition swc by
X
swc[a] (x) = |{i : xi ∼ a}| =
cb (x).
b∈[a]
Then define the symmetrized weight enumerator of a linear code C ⊂ An by
sweC (S) =
n
XY
x∈C i=1
S[xi ] =
X
Y
swc[a] (x)
S[a]
,
x∈C [a]∈A/∼
1There is the even-more-general full (or exact) weight enumerator, which is a copy of the
code itself. See [14, 18].
FOURIER TRANSFORM
3
where A/∼ denotes the set of all ∼-equivalence classes and the S[a] , [a] ∈ A/ ∼,
are indeterminates. Note that the specialization of variables Za 7→ S[a] induces a
homomorphism of polynomial rings C[Za : a ∈ A] → C[S[a] : [a] ∈ A/ ∼] that maps
cweC (Z) to sweC (S).
Suppose there are two equivalence relations on A and one is finer than the
other. That is, a ∼1 b implies a ∼2 b; or every equivalence class of ∼1 is contained
in an equivalence class of ∼2 . Then the specialization of variables S[a]1 7→ T[a]2 is
well-defined and induces a ring homomorphism C[S[a]1 : [a]1 ∈ A/ ∼1 ] → C[T[a]2 :
(1)
(2)
[a]2 ∈ A/ ∼2 ] taking sweC (S) to sweC (T ).
There are two examples of equivalence relations of special note. The Hamming
equivalence relation has just two equivalence classes: {0} and {a ∈ A : a 6= 0}. It is
traditional to write wt(x) = |{i : xi 6= 0}|, the Hamming weight of x ∈ An . Define
the Hamming weight enumerator by
X
(1.1)
hweC (X, Y ) =
X n−wt(x) Y wt(x) .
x∈C
The specialization of variables Z0 7→ X and, for a 6= 0, Za 7→ Y , induces a homomorphism C[Za : a ∈ A] → C[X, Y ] carrying cweC (Z) to hweC (X, Y ). It is very
common to see hweC (X, Y ) denoted WC (X, Y ).
The other special case is an equivalence relation arising from a group action.
Let Aut(A) be the group of all invertible R-linear transformations from A to itself.
Because A is a left R-module, we will write mappings on the right, so that Rlinearity is expressed as (ra)φ = r(aφ), for r ∈ R, a ∈ A, and φ ∈ Aut(A). Fix a
subgroup G ⊂ Aut(A) and define an equivalence relation via the action of G on A:
a ∼G b if there exists φ ∈ G with a = bφ. We will see more of this example later,
in Remark 1.3.
When A = R, the ring itself, Aut(A) is just the group of units of R. Fixing
a subgroup G of the group of units, the equivalence relation is then a ∼G b when
a = bu for some u ∈ G. As a concrete example, let R = Z/4Z and let G = {±1},
the full group of units. Then 1 ∼G 3 in Z/4Z, and there are three equivalence
classes: {0}, {1, 3}, and {2}. This case arises when studying the Lee weight on
Z/4Z.
Example 1.1. Let R = A = Z/4Z, and let C ⊂ R3 be the submodule generated
by 123 and 022. Then
C = {000, 123, 202, 321, 022, 101, 220, 303}.
Let G = {±1} determine an equivalence relation. The three weight enumerators
described above are:
cweC (Z) = Z03 + Z0 Z12 + 3Z0 Z22 + Z0 Z32 + 2Z1 Z2 Z3
3
2
2
2
sweC (S) = S[0]
+ 2S[0] S[1]
+ 3S[0] S[2]
+ 2S[1]
S[2]
hweC (X, Y ) = X 3 + 5XY 2 + 2Y 3 .
Weight-type. Although we have defined weight enumerators, we have not defined weights, except for the Hamming weight. A very general definition is: a weight
on an alphabet A is any function w : A → C with w(0) = 0. (One may wish to
impose additional hypotheses on w, such as the triangle inequality, P
but we will not
n
do so here.) The weight is then extended to An by setting w(x) = i=1 w(xi ), for
n
x = (x1 , . . . , xn ) ∈ A . The Hamming weight satisfies w(a) = 1 for nonzero a ∈ A.
4
JAY A. WOOD
Suppose a weight w on an alphabet A has values in the nonnegative integers,
and suppose that M is the largest value of w on A. Define the w-weight enumerator
of a linear code C ⊂ An by
n
XY
X
wweC (X, Y ) =
X M −w(xi ) Y w(xi ) =
X nM −w(x) Y w(x) .
x∈C i=1
x∈C
This weight enumerator counts the number of codewords of any given weight. For
the Hamming weight, this definition agrees with that of hweC given in (1.1).
Example 1.2. The Lee weight wL on R = A = Z/4Z is defined by wL (0) = 0,
wL (1) = wL (3) = 1, and wL (2) = 2. Note that M = 2. For the code given in
Example 1.1, we have
wweC (X, Y ) = X 6 + 2X 4 Y 2 + 5X 2 Y 4 .
Remark 1.3. Given a weight w on A, there is a symmetrized weight enumerator
naturally determined by w. Define the right symmetry group G of w by G =
{φ ∈ Aut(A) : w(aφ) = w(a), for all a ∈ A}. As above, the group G determines an
equivalence relation on A and hence a symmetrized weight enumerator sweC for
any linear code C ⊂ An . In fact, the symmetrized weight numerator determines
the w-weight enumerator. The specialization of variables S[a] 7→ X M −w(a) Y w(a)
is well-defined (by the definition of G) and induces a ring homomorphism C[S[a] :
[a] ∈ A/ ∼] → C[X, Y ] that takes sweC (S) to wweC (X, Y ).
Example 1.4. Note that wweC does not determine sweC , as there exist codes with
different swe’s but the same wwe’s. Let R = A = Z/4Z with the Lee weight wL .
Let C1 ⊂ R3 be generated by 123, and let C2 ⊂ R3 be generated by 220 and 022.
Thus,
C1 = {000, 123, 202, 321},
C2 = {000, 220, 022, 202}.
Then,
3
2
2
sweC1 (S) = S[0]
+ S[0] S[2]
+ 2S[1]
S[2] ,
3
2
sweC2 (S) = S[0]
+ 3S[0] S[2]
,
wweC1 (X, Y ) = wweC2 (X, Y ) = X 6 + 3X 2 Y 4 .
2. Characters and the Fourier transform over C
In this section we develop the theory of characters of finite abelian groups, the
Fourier transform, and the Poisson summation formula. These are the tools we will
use later to prove the MacWilliams identities on weight enumerators.
Characters. Let A be a finite abelian group, written additively. A character
of A is a group homomorphism π : A → C× from A to the multiplicative group
of nonzero complex numbers. I.e., a character satisfies π(a1 + a2 ) = π(a1 )π(a2 ),
for all a1 , a2 ∈ A. Because every element a ∈ A has finite order, every character
value π(a) ∈ C× is a root of unity. In particular, |π(a)| = 1. It follows that
π(−a) = (π(a))−1 = π(a), the complex conjugate of π(a).
b the set of all characters of A; A
b is itself an abelian group, the charDenote by A
acter group, under pointwise multiplication of functions: (π1 π2 )(a) = π1 (a)π2 (a).
FOURIER TRANSFORM
5
The identity of the character group is the principal character whose value is 1 at
every a ∈ A; the principal character will sometimes be denoted by 1. Defining
b and π −1 = π.
π(a) = π(a), we see that π ∈ A
Example 2.1. Let A = Z/nZ, an additive cyclic group of order n; let ζ be a
primitive nth root of 1 in C, e.g., ζ = exp(2πi/n). For any b ∈ Z/nZ, define a
character πb of A by πb (a) = ζ ab , a ∈ A. Every character of A is of this form, and
b∼
A
= A.
We state without proof the following well-known properties of character groups.
Proofs may be found in such sources as [15, 16].
Theorem 2.2. Let A be a finite abelian group. Then
b = |A|;
(1) |A|
b
(2) A ∼
= A, but not naturally so;
(3) the mapping a 7→ (π 7→ π(a)) is a natural isomorphism from A to the
b b;
double-character group (A)
b1 × A
b2 , for finite abelian groups A1 , A2 .
(4) (A1 × A2 ) b ∼
=A
Several fundamental properties of characters are summarized in the following
lemmas.
Lemma 2.3. For any finite abelian group A, the following formulas hold.
(
(
X
X
|A|, a = 0,
|A|, π = 1,
π(a) =
π(a) =
0,
a 6= 0.
0,
π 6= 1;
a∈A
b
π∈A
Proof. Obvious if π = 1. If π 6= 1, then there exists a0 ∈ A with π(a0 ) 6=
1.
P Reindex the
P sum with a = a0 +Pb and use the homomorphism property:
π(a)
=
a∈A
b∈A π(a0 + b) = π(a0 )
b∈A π(b). Since π(a0 ) 6= 1, the sum must
vanish.
Let F (A, C) = {f : A → C}, the set of all functions from A to C; F (A, C)
is a vector space over C under pointwise addition and scalar multiplication. The
dimension of F (A, C) is |A|.
Define an inner product on F (A, C) by
1 X
(2.1)
hf, gi =
f (a)g(a),
f, g ∈ V,
|A|
a∈A
where the bar denotes complex conjugation. This inner product is complex linear
in the left entry and conjugate linear in the right entry. It is hermitian-symmetric:
hg, f i = hf, gi, as well as positive definite: hf, f i ≥ 0, with hf, f i = 0 if and only if
f = 0.
b form an orthonormal basis of F (A, C) under
Lemma 2.4. The characters π ∈ A
h·, ·i.
b Then
Proof. Suppose ψ, π ∈ A.
1 X
1 X
hψ, πi =
ψ(a)π(a) =
(ψπ)(a) =
|A|
|A|
a∈A
a∈A
(
1, ψ = π,
0, ψ =
6 π,
using Lemma 2.3. This shows that the characters form an orthonormal set under
b = dim F (A, C). h·, ·i. That they form a basis now follows from the fact that |A|
6
JAY A. WOOD
Fourier transform. The Fourier transform is a complex-linear transformab C) defined by
tion ˆ : F (A, C) → F (A,
X
fˆ(π) =
π(a)f (a),
f ∈ F (A, C).
a∈A
Note that fˆ(π) = |A|hf, πi. The Fourier tranform is invertible.
Lemma 2.5 (Fourier inversion). For any a ∈ A and f ∈ F (A, C),
1 X
π(−a)fˆ(π).
f (a) =
|A|
b
π∈A
Proof. Expand the sum by using the definition of fˆ, interchange the order of
summation, and use Lemma 2.3.
P Because π(a) = π(−a), a ∈ A, this relation can also be viewed as f =
bhf, πi π, as one expects for an orthonormal basis. Thus, the Fourier transform
π∈A
can be interpreted as associating to any f ∈ F (A, C) its coefficients with respect
to the basis of characters.
In order to present the Poisson summation formula, we need to define annihilators of subgroups. Let B be a subgroup of the finite abelian group A. Define the
b by
annihilator of B in A
b : B) = {π ∈ A
b : π(B) = 1}.
(A
b : B) ∼
b : B)| = |A|/|B|, and that (A : (A
b:
It follows that (A
= (A/B) b , that |(A
B)) = B.
Lemma 2.6. Suppose B is a subgroup of a finite abelian group A. Then for any
a ∈ A,
(
X
b : B)|, a ∈ B,
|(A
π(a) =
0,
a 6∈ B.
b
π∈(A:B)
Proof. The case a ∈ B is obvious. In the other case, use an argument dual
to that in Lemma 2.3.
Lemma 2.7 (Poisson summation formula). Let A be a finite abelian group with
subgroup B. Then, for any a ∈ A,
X
X
1
f (a + b) =
fˆ(π)π(−a).
b : B)|
|(
A
b∈B
b
π∈(A:B)
In particular, if a = 0, then
X
(2.2)
f (b) =
b∈B
1
b : B)|
|(A
X
fˆ(π).
b
π∈(A:B)
Proof. Expand fˆ, interchange order of summation, and use Lemma 2.6.
In Section 4, we will prove the MacWilliams identities by identifying the expressions in (2.2) with corresponding coding-theoretic expressions. Some of those
identitfications will depend upon additional hypotheses placed on the ring R, which
is the topic of Section 3.
FOURIER TRANSFORM
7
3. Frobenius rings
In this section we quickly describe some features of finite Frobenius rings. References include [12, 17, 18].
Let R be a finite associative ring with 1. The Jacobson radical rad(R) of R is
the intersection of all maximal left ideals of R; rad(R) is itself a two-sided ideal. A
left R-module M is simple (or irreducible) if M has no nonzero proper submodules.
Since the annihilator of a simple module is a maximal ideal, the radical rad(R)
annihilates every simple R-module.
For any left R-module M , the socle soc(M ) is the sum of all the simple submodules of M ; there is a corresponding notion for right modules. The ring R can
be considered as both a left and a right R-module (denoted R R and RR ). The left
socle soc(R R) and the right socle soc(RR ) are both two-sided ideals of R, but they
need not be equal. (They are necessarily equal if the ring is semiprime; for finite
rings, this is equivalent to being semisimple.)
Let R be the quotient ring R/ rad(R); R is both a left and a right R-module.
A ring is Frobenius if R R ∼
= soc(R R) and RR ∼
= soc(RR ). When R is finite, it is a
theorem of Honold [10] that one of the isomorphisms suffices.
b the character group of A, becomes a
Let A be a finite left R-module, then A,
b
right R-module, as follows. For π ∈ A and r ∈ R, the right scalar multiplication is
b
defined by π r (a) = π(ra), a ∈ A. Similarly, if B is a finite right R-module, then B
b and r ∈ R.
is a left R-module via r ψ(b) = ψ(br), for b ∈ B, ψ ∈ B,
b is also
Considering R as a left and as a right R-module, the character group R
a left and a right R-module. Finite Frobenius rings are characterized as those finite
b∼
rings satisfying R
= R, either as left R-modules or as right R-modules ([9, 17]).
b as a left R-module if and only
For a finite Frobenius ring, a character generates R
b as a right R-module ([17, Theorem 4.3]); any such generator is
if it generates R
called a generating character of R.
4. The MacWilliams identities
The MacWilliams identities give a relationship between the weight enumerator
of a linear code and that of its dual code. In this section, we will define dual
codes and prove the MacWilliams identities for the complete and Hamming weight
enumerators by use of the Poisson summation formula.
We begin with a very general version of the MacWilliams identities. Let R
be a finite ring and A a finite left R-module. For any R-linear code C ⊂ An , the
bn : C) will serve as a type of dual code. Observe
character-theoretic annihilator (A
n
b : C) is a right R-submodule of A
bn ; i.e., (A
bn : C) is a right R-linear code in
that (A
bn . As such, (A
bn : C) has the various weight enumerators defined in Section 1. The
A
bn : C) are related by a linear change of variables.
weight enumerators of C and (A
Theorem 4.1. Let C be a left R-linear code in An . Then
1
cweC (Z) =
cwe(Abn :C) (ζ) P
,
bn : C)|
ζπ = a∈A π(a)Za
|(A
1
hweC (X, Y ) =
hwe(Abn :C) (X + (|A| − 1)Y, X − Y ).
bn : C)|
|(A
8
JAY A. WOOD
bn : C) ⊂ A
bn is a polynomial in variThe complete weight enumerator of (A
b The notation in the theorem is to replace (or evaluate) ζπ by
ables ζπ , π ∈ A.
P
π(a)Z
;
i.e.,
the ζ-variables are replaced by the Fourier transforms of the
a
a∈A
Z-variables.
Proof. Apply the Poisson summationQformula (2.2) with C ⊂ An playing
n
the roles of B ⊂ A and f (x1 , . . . , xn ) = i=1 Zxi playing the role of f (b). A
Qn P
computation then shows that fˆ(π1 , . . . , πn ) = i=1 a∈A πi (a)Za .
As for the Hamming weight enumerator, one can again apply (2.2) directly, or
(doing essentially the same work) P
verify that the specialization of variables Z0 7→ X,
Za 7→ Y (for a 6= 0) carries ζπ = a∈A π(a)Za to X + (|A| − 1)Y , if π = 1, and to
X − Y , if π 6= 1. Indeed, by using Lemma 2.3,
X
X
ζπ =
π(a)Za 7→ π(0)X +
π(a)Y
a∈A
a6=0
=X+
X
π(a)Y
a6=0
(
=
X + (|A| − 1)Y, π = 1,
X − Y,
π=
6 1.
Theorem 4.1 is not the typical version of the MacWilliams identities that one
sees in textbooks. We now turn to the more customary format, which involves some
form of inner product on An in order to define dual codes. The reader interested
in a general alphabet A is referred to [14] (also, [19]). We will now assume that
the alphabet A is the ring R itself.
The standard dot product on Rn is
x·y =
n
X
xi yi ,
i=1
for x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn . Notice that the dot product is left Rlinear in the left variable and right R-linear in the right variable. The dot product
is nondegenerate; i.e., if x · y = 0 for all y ∈ Rn , then x = 0 (and vice versa). Unless
R is commutative, the dot product is not necessarily symmetric.
For a left R-linear code C ⊂ Rn , define its right annihilator by
r(C) = {y ∈ Rn : x · y = 0, for all x ∈ C}.
There is a parallel notion of left annihilators of right R-linear codes. Over a finite
Frobenius ring R, there is a group isomorphism between r(C) and the characterbn : C), as follows.
theoretic annihilator (R
Lemma 4.2. Let R be a finite Frobenius ring with generating character χ. Then the
bn , x 7→ βx , with βx (y) = χ(y · x), y ∈ Rn , is an isomorphism
mapping β : Rn → R
bn : C).
of left R-modules taking r(C) to (R
bn : C) is only a right submodule of R
bn , β restricts to
Remark 4.3. Because (R
n
b
just a group isomorphism between r(C) and (R : C).
Proof. To show that β is left R-linear we compute that
βrx (y) = χ(y · rx) = χ(yr · x) = βx (yr) = (r βx )(y).
FOURIER TRANSFORM
9
To show that β is an isomorphism, we must exploit a property of generating
characters. Namely, a character χ is a generating character of R if and only if
ker χ contains no nonzero left ideal of R. Indeed, any character χ determines a left
b by left scalar multiplication: r 7→ r χ. Because
R-linear homomorphism R → R
b = |R|, χ will be a generating character if and only if this homomorphism is
|R|
injective. But r is in the kernel of the homomorphism if and only if r χ = 1; i.e.,
1 = (r χ)(s) = χ(sr), for all s ∈ R. The latter occurs if and only if Rr ⊂ ker χ. The
property now follows.
We show that β is injective. If x ∈ ker β, then βx = 1. Thus, 1 = βx (y) =
χ(y · x), for all y ∈ Rn . This means that Rn · x ⊂ ker χ. Since Rn · x is a left ideal of
R and χ is a generating character, we have Rn · x = 0, which means x = 0. Because
bn | = |Rn |, the injectivity of β implies that β is an isomorphism.
|R
bn : C). Conversely, if βx ∈ (R
bn : C),
It is clear that x ∈ r(C) implies βx ∈ (R
then 1 = βx (C) = χ(C · x). Thus the left ideal C · x ⊂ ker χ, and C · x = 0 because
χ is a generating character. Thus, x ∈ r(C).
Theorem 4.4 (MacWilliams identities). Let R be a finite Frobenius ring with
generating character χ. Let C be a left R-linear code in Rn , with right annihilator
r(C). Then
1
cwer(C) (ζ)ζ =P
,
cweC (Z) =
r
s∈R χ(sr)Zs
|r(C)|
1
hweC (X, Y ) =
hwer(C) (X + (|R| − 1)Y, X − Y ).
|r(C)|
As in Theorem 4.1, cwer(C) is P
a polynomial in variables ζr , r ∈ R. On the right
side, cwer(C) is evaluated at ζr = s∈R χ(sr)Zs .
Proof. Use the isomorphism β of Lemma 4.2 to modify the right sides of the
equations in
The new ζr corresponds to the old ζβr . Then notice
PTheorem 4.1. P
that ζβr = s∈R βr (s)Zs = s∈R χ(sr)Zs .
5. Additional MacWilliams identities
In Section 4, the MacWilliams identities for the complete and Hamming weight
enumerators were proved using the Poisson summation formula. What about symmetrized and w-weight enumerators? For those weight enumerators the situation is
more complicated, and we will discuss only a portion of what is known. Additional
references are [2, 6, 7, 11, 20, 21].
Because a symmetrized weight enumerator is a specialization of the complete
weight enumerator (as is a w-weight enumerator), the ideal situation would be
where the specialization of variables commutes with the Fourier transform (as for
the Hamming weight enumerator). In Section 4, the MacWilliams identities for the
complete weight enumerator were presented in two contexts: a general charactertheoretic setting (Theorem 4.1) and the setting over finite Frobenius rings (Theorem 4.4). In the latter setting, one would like for there to be only one specialization
of variables, one that works on both sides of the identities. In the general charactertheoretic setting, perhaps there is one specialization of variables for the alphabet
b This is indeed the case.
A and another for the character module A.
Given an equivalence relation ∼p on A (‘p’ for primal ), define an equivalence
P
b by π1 ∼d π2 when P
relation ∼d (‘d’ for dual ) on A
b∈[a] π1 (b) =
b∈[a] π2 (b), for
10
JAY A. WOOD
b by
all equivalence classes [a] ∈ A/ ∼p . Denote the ∼d -equivalence
class of π ∈ A
P
[π]. In Theorem 4.1, the variable ζπ is evaluated at a∈A π(a)Za . If we specialize
ζπ 7→ σ[π] and Za 7→ S[a] , then notice that
X
X
X
X
π(a)Za 7→
π(a)S[a] =
π(b) S[a] .
a∈A
a∈A
[a]∈A/∼p
b∈[a]
P
By the definition of the equivalence relation ∼d , the coefficient b∈[a] π(b) of S[a]
depends only on the equivalence class [π]. Thus, the specializations are compatible
with the change of variables, and we have proved the following.
b as
Theorem 5.1. Let ∼p and ∼d be related equivalence relations on A and A,
n
above. Let C ⊂ A be a left R-linear code. Then the symmetrized weight enumerators associated to ∼p and ∼d satisfy the MacWilliams identities
1
(d)
(p)
.
swe(Abn :C) (σ)
sweC (S) =
P
P
bn : C)|
σ[π] = [a]∈A/∼p ( b∈[a] π(b))S[a]
|(A
b will in turn define an equivalence
Remark 5.2. An equivalence relation ∼d on A
P
P
relation ∼dd (double dual ) on A; a ∼dd b when ψ∈[π] ψ(a) = ψ∈[π] ψ(b), for all
b came from ∼p on A, there is no guarantee
equivalence classes [π]. But if ∼d on A
that the new equivalence relation ∼dd on A agrees with the original ∼p . See [7] for
details.
However, when the equivalence relation ∼p on A is given by the orbits of a
right action of a subgroup G ⊂ Aut(A), then ∼d is given by orbits of the induced
b In this case, forming the new equivalence relation ∼dd on A
left action of G on A.
does indeed yield the original ∼p .
Remark 5.3. In the setting of Theorem 4.4 over finite Frobenius rings, the situabn is a left
tion is complicated. As noted in Remark 4.3, the mapping β : Rn → R
n
b
isomorphism, while r(C) and (R : C) are right R-modules. This leads to a mixingup of left and right when trying to understand the equivalence relations ∼p and ∼d ,
even in the situation where the equivalence relations are given by group actions.
If the group G is central, then everything works out fine ([17, Theorem 8.4]). In
particular, there are no problems for the case of group actions over commutative
Frobenius rings.
Remark 5.4. The situation for w-weight enumerators is also complicated. As
noted in Remark 1.3, every weight w on A determines a right symmetry group G,
which in turn defines an equivalence relation ∼G on G. But the weight itself defines
an equivalence relation: a ∼w b when w(a) = w(b). By the definition of G, ∼G is
finer than ∼w ; i.e., if a ∼G b, then a ∼w b.
When the two equivalence relations are the same, then the validity of the
MacWilliams identities for the symmetrized weight enumerator of ∼G (for example,
over finite commutative Frobenius rings) will imply the MacWilliams identities for
the w-weight enumerator. When ∼G is strictly finer than ∼w , then the MacWilliams
identities for the w-weight enumerator may fail.
Example 5.5. Let R = A = Z/N Z, equipped with the Lee weight. If we view
elements of R as being uniquely represented by integers in the range −N/2 <
r ≤ N/2, then the Lee weight wL (r) = |r|, the ordinary real absolute value of
FOURIER TRANSFORM
11
the representative r. The symmetry group G of wL is G = {±1}. Note that in
this example ∼G is equal to ∼wL . Thus the MacWilliams identities hold for the
wL -weight enumerator ([8]).
Example 5.6. Let R = A = Z/8Z, equipped with the following weight w (an
example of a homogeneous weight due to Constantinescu and Heise [3]):
0, r ≡ 0 mod 8,
w(r) = 2, r ≡ 4 mod 8,
1, otherwise.
The symmetry group G is the full group of units of R; namely, G = {±1, ±3}.
Because w(1) = w(2) = 1, ∼G is strictly finer than ∼w . The MacWilliams identities
hold for the symmetrized weight enumerator determined by ∼G , but they fail for
the w-weight enumerator. (A short computation shows that no linear substitution
yields the correct identities for all linear codes of length 1.)
6. Convolutions
In this section we begin to discuss the decomposition of group algebras. The
Fourier transform has the property that the transform of the convolution product
of two functions on a finite abelian group is equal to the pointwise product of the
transforms. This shows that the complex group algebra of a finite abelian group is
a direct sum of copies of the complex number field.
Convolution products. As in Section 2, let A be a finite abelian group,
written additively, and let F (A, C) be the vector space of all C-valued functions on
A. While F (A, C) has a multiplication given by the pointwise product of functions,
such a multiplication makes no use of the group structure on A. We will instead
be interested in the convolution product:
X
(f ∗ g)(a) =
f (a − b)g(b),
a ∈ A, f, g ∈ F (A, C).
b∈A
P
By making the change of variable c = a−b, we see that (f ∗g)(a) = c∈A f (c)g(a−
c), as well. The convolution product makes F (A, C) a commutative algebra over C.
Lemma 6.1. For f, g ∈ F (A, C), we have
(f ∗ g) ˆ (π) = fˆ(π)ĝ(π),
b
π ∈ A.
Proof. Straight-forward exercise. Use the homomorphism property of characters to write π(a) = π(a − b + b) = π(a − b)π(b).
Lemma 6.1 says that the Fourier transform is an algebra homomorphism from
b C) with the pointwise product. As
F (A, C) with the convolution product to F (A,
the Fourier transform is invertible (Lemma 2.5), this is actually an algebra isomorphism.
12
JAY A. WOOD
Group algebras. An equivalent way of looking at F (A, C) with the convolution product is as the complex group algebra C[A]. More broadly, let k be any
(commutative) field and let G be any finite group (abelian or not) written multiplicatively. The group algebra k[G] is the set of all formal k-linear combinations of
elements of G:
X
k[G] =
cg g : cg ∈ k .
g∈G
P
P
Addition is defined by adding corresponding coefficients: g∈G cg g + g∈G dg g =
P
g + dg )g, and scalar multiplication by c ∈ k multiplies every coefficient cg
g∈G (cP
P
by c: c g∈G cg g = g∈G (ccg ) g. The multiplication in k[G] is induced by the
multiplication in the group G:
!
X
X
X
cg g
dh h =
cg dh gh
g∈G
h∈G
g,h∈G
=
X
X
cg dh s
s∈G
gh=s
X
X
s∈G
h∈G
!
=
csh−1 dh
s.
With these operations, k[G] is a k-algebra of dimension |G|; k[G] is commutative if
and only if G is abelian.
The group algebra k[G] is isomorphic to F (G, k) with the (now, noncommutative) convolution product
X
(f1 ∗ f2 )(g) =
f1 (gh−1 )f2 (h),
g ∈ G, f1 , f2 ∈ F (G, k).
h∈G
The element
g ∈ G.
P
g∈G cg
g ∈ k[G] corresponds to the function c : G → k, c(g) = cg ,
Idempotents. Let us examine Lemma 6.1 more carefully in the language of
C[A], where A is a finite abelian group. The group elements a ∈ A form a basis
for the group algebra C[A], as is obvious from the definition of C[A] as a set of
formal linear combinations. Treated as elements of F (A, C), the elements a ∈ A
correspond to the indicator functions δa : A → C, δa (b) = 1 when b = a and
δa (b) = 0 otherwise.
b we need to remember
While we have been treating characters as elements of A,
b
that characters are functions from A to C, so that A ⊂ F (A, C). Viewed as an
b is π = P
b
element of C[A], a character π ∈ A
a∈A π(a) a. Recall that |A| = |A|, so
there are as many characters as the dimension of C[A]. The characters will turn
b
out to form a very nice basis for
PC[A]. More specifically, for each character π ∈ A,
define eπ = (1/|A|)π = (1/|A|) a∈A π(a) a ∈ C[A].
b form a set of primitive orthogonal idempotents
Proposition 6.2. The eπ , π ∈ A,
in C[A]. These idempotents sum to 1 ∈ A
That is, e2π = eπ ; eπ1 eπ2 = 0 for π1 6= π2 ; and each eπ cannot be expressed as
the sum of two nonzero orthogonal idempotents.
FOURIER TRANSFORM
13
Proof. This is Lemma 2.4 in disguise. We compute eπ1 eπ2 = 0:
!
!
X
X
2
|A| eπ1 eπ2 =
π1 (a) a
π2 (b) b
a∈A
b∈A
!
=
X
X
c∈A
b∈A
π1 (c − b)π2 (b) c
!
=
X
X
π1 (c)
c∈A
π1 (−b)π2 (b) c
b∈A
= |A|hπ2 , π1 i
X
π1 (c) c
(which, by Lemma 2.4)
c∈A
( P
|A| c∈A π1 (c) c,
=
0,
π1 = π2 ,
π1 6= π2 .
Dividing both sides by |A|2 yields the first two claims. Notice that the eπ being
orthogonal idempotents reflects the characters π being orthonormal with respect to
h·, ·i.
For the last claim, we prove more; namely, that the ideal generated by any eπ is
of dimension one. Since the sum of two orthogonal idempotents would generate an
ideal of dimension at least two, we conclude that each eπ is primitive. To examine
the ideal generated by eπ we calculate c eπ for c ∈ C[A]:
!
!
X
X
|A|c eπ =
ca a
π(b) b
a∈A
b∈A
!
=
X
X
d∈A
a∈A
ca π(d − a) d
!
=
X
ca π(−a)
a∈A
!
X
π(d) d .
d∈A
Thus, c eπ =P
|A|hc, πieπ , and the ideal generated by eπ has dimension one.
Finally, π∈Ab eπ = 1 follows from Lemma 2.3.
Change of basis in C[A]. Left multiplication by an element c ∈ C[A] defines
a linear transformation Lc of C[A] to itself, called the left regular representation of
C[A]. We examine the matrices representing this transformation with respect to
the two bases of C[A] discussed above, the group elements and the idempotents.
For a given basis {v1 , . . . , vn }, n = |A|, of C[A], the (i, j)-entry ai,j of the
matrix representing Lc with respect to this basis is obtained from
Lc (vj ) =
n
X
ai,j vi .
i=1
Let c =
P
d∈A cd
d ∈ C[A]. In terms of the basis of group elements, we calculate
!
X
X
Lc (b) =
cd d b =
ca−b a.
d∈A
a∈A
14
JAY A. WOOD
Thus, the matrix representing Lc with respect to the basis of group elements has
(a, b)-entry equal to ca−b . Notice that multiplying this matrix times a column
vector (pb ) yields the expression for the convolution product.
In terms of the basis of idempotents, we obtain Lc (eψ ) = c eψ = |A|hc, ψieψ ,
as in the proof of Proposition 6.2. Thus, the (π, ψ)-entry of the matrix is 0 when
π 6= ψ, and |A|hc, ψi when π = ψ. The basis of idempotents diagonalizes the
transformations Lc simultaneously.
When we write elements of C[A] in the basis of idempotents, the multiplication
has the expression
ab =
X
aπ eπ
b
π∈A
X
bψ e ψ =
b
ψ∈A
X
(aπ bπ )eπ .
b
π∈A
That is, multiplication is componentwise multiplication of coefficients. This proves
the next result.
Theorem 6.3. Let A be a finite abelian group. The mapping C[A] → C ⊕ · · · ⊕ C,
P
b is an
(with |A| summands) sending a = π∈Ab aπ eπ to its coefficients aπ , π ∈ A,
isomorphism of C-algebras.
Again, this result is just another form of Lemma 6.1. Notice that the algebra
structure of C[A] depends just on |A|, so that C[C4 ] ∼
= C[C2 × C2 ], where C4 is a
cyclic 4-group and C2 × C2 is the Klein 4-group, even though the groups themselves
are not isomorphic.
Finally, what is the change of basis matrix P that intertwines the two matrices
representing Lc ? The (a, π)-entry of P expresses the new
Pbasis element eπ in terms
of the old basis of group elements. But eπ = (1/|A|) a∈A π(a) a, so the (a, π)entry of P is exactly π(a)/|A|. By using Lemma 2.3, one sees that the (π, a)-entry
of the inverse matrix P −1 is π(−a) = π(a). Remember that the coefficients of
c ∈ C[A] expressed in the idempotent basis eπ are of the form cπ = |A|hc, πi. Thus,
up to the factor |A|, the matrices P −1 and P represent the Fourier transform and
Fourier inversion, respectively.
Example 6.4. Let A be a cyclic group C3 of order 3; let a ∈ A be a generator,
so that A = {e, a, a2 }, with a3√= e. Let ω be a primitive third root of 1 in C,
say ω = exp(2πi/3) = (−1 + i 3)/2. The three characters of A are defined by
πj (ak ) = ω jk , for j, k = 0, 1, 2. Thus, the idempotent basis for C[A] is
1
e + a + a2 ,
3
1
=
e + ωa + ω 2 a2 ,
3
1
e + ω 2 a + ωa2 .
=
3
eπ0 =
eπ1
eπ2
One verifies, for c = c0 e + c1 a + c2 a2 ∈ C[A], that
c = (c0 + c1 + c2 )eπ0 + (c0 + ω 2 c1 + ωc2 )eπ1 + (c0 + ωc1 + ω 2 c2 )eπ2 .
FOURIER TRANSFORM
15
II. Fourier transform over a field for finite groups
Many of the techniques and results in Section 6 carry over to group algebras
k[A] of a finite abelian group A with coefficients in a field whose characteristic does
not divide |A|. Moreover, there are generalizations for k[G], where G is any finite
group, abelian or not, still assuming that the characteristic of k does not divide |G|.
The main tools come from the representation theory of finite groups. References
include [1, 4, 15, 16].
7. Group algebras and representation theory
In this section we review some of the basic terminology from the representation
theory of finite groups, especially the equivalence of representations and modules
over the group algebra.
Terminology. Let G be a finite group (abelian or not), and let k be a field.
A (finite-dimensional) representation of G over k consists of a finite-dimensional
k-vector space V and a group homomorphism ρV : G → GL(V ), where GL(V ) is
the group of invertible k-linear transformations of V to itself. Two representations
(V, ρV ), (W, ρW ) are equivalent if there exists a k-linear isomorphism T : V → W
that intertwines the representations; i.e., ρW (g) ◦ T = T ◦ ρV (g), for every g ∈ G.
The character of a representation is the function from G to k defined by χρV =
tr ρV , where tr : GL(V ) → k is the matrix trace function. (See Remark 7.7 for a
comparison between this usage of character and that from Section 2.)
If a k-linear subspace W ⊂ V satisfies (ρV (g))(W ) ⊂ W , for all g ∈ G, we say
that W is invariant under the representation ρV . The restriction of each ρV (g) to
an invariant subspace W defines a subrepresentation (W, ρV |W ). A representation
(V, ρV ) is irreducible if it is nonzero and has no nontrivial subrepresentations (no
nontrivial invariant subspaces); the ever-present trivial cases are {0} and V itself.
A representation is decomposable if there exist nontrivial invariant subspaces
W1 , W2 ⊂ V , with V = W1 ⊕W2 . A representation is indecomposable if it is nonzero
and not decomposable. Note that every irreducible representation is indecomposable. The converse will be true when the characteristic of k does not divide |G|
(Section 8), but the converse is false when char k divides |G| (the ‘modular’ case).
Example 7.1. Let G = C2 be a cyclic group of order 2, say G = {e, a} with a2 = e.
Let k = F2 . Taking V = k 2 , define ρV by
1 0
0 1
ρV (e) =
,
ρV (a) =
.
0 1
1 0
The subspace W = {00, 11} is invariant, so ρV is reducible. However, there are no
other invariant subspaces of V , so ρV is indecomposable.
If instead char k 6= 2, let W1 = {(x, y) : y = x} ⊂ k 2 . Again, W1 is invariant
under ρV . But now we can define W2 = {(x, y) : y = −x} ⊂ k 2 . Then W2 is also
invariant, and W1 ⊕ W2 = k 2 . Indeed, any pair (x, y) can be written uniquely as
(x, y) = ((x + y)/2, (x + y)/2) + ((x − y)/2, −(x − y)/2). Of course, in characteristic
2, W2 = W1 .
16
JAY A. WOOD
Intertwining. Consider two k-representations of G, (V, ρV ) and (W, ρW ), and
the set of linear transformations that intertwine them:
I(ρV , ρW ) = {f : V → W : f is k-linear and
f ◦ ρV (g) = ρW (g) ◦ f, for all g ∈ G}.
The set I(ρV , ρW ) is a k-vector space.
Lemma 7.2. Suppose f ∈ I(ρV , ρW ). Then the kernel ker f is an invariant subspace of V , and the image Im f is an invariant subspace of W .
Proof. Suppose v ∈ ker f and g ∈ G. Then f ρV (g)v = ρW (g)f v = 0, so
that ρV (g)v ∈ ker f . Likewise, if w = f v, then ρW (g)w = ρW (g)f v = f ρV (g)v ∈
Im f .
Now assume that both representations are irreducible.
Lemma 7.3. Suppose the representations (V, ρV ) and (W, ρW ) are irreducible.
Then every nonzero element f ∈ I(ρV , ρW ) is an isomorphism.
Proof. By Lemma 7.2, ker f is an invariant subspace of V . If f is nonzero,
then ker f 6= V , so the irreducibility of ρV implies ker f = 0. Likewise, Im f is a
nonzero invariant subspace of W . By the irreducibility of ρW , Im f = W .
We now consider self-intertwining maps of a single irreducible representation.
Proposition 7.4. Let (V, ρV ) be an irreducible k-representation of a finite group
G. Then I(ρV , ρV ) is a division algebra over k.
Proof. The k-vector space I(ρV , ρV ) is a k-algebra under composition. It
contains the identity transformation 1V of V and all scalar multiples α1V , α ∈ k.
By Lemma 7.3, any nonzero f ∈ I(ρV , ρV ) is an isomorphism, and one checks that
f −1 is also in I(ρV , ρV ).
Corollary 7.5 (Schur’s Lemma). Let (V, ρV ) be an irreducible k-representation
of a finite group G, and assume that k is algebraically closed. Then I(ρV , ρV ) =
{α1V : α ∈ k} ∼
= k.
Proof. We already know that {α1V : α ∈ k} ⊂ I(ρV , ρV ). Suppose f ∈
I(ρV , ρV ). Let α be an eigenvalue of f ; α ∈ k because k is algebraically closed.
Then f − α1V ∈ I(ρV , ρV ). But f − α1V is not an isomorphism because α is an
eigenvalue, so that f − α1V = 0 by Lemma 7.3. Thus f = α1V , and I(ρV , ρV ) =
{α1V : α ∈ k}.
An irreducible k-representation (V, ρ) is absolutely irreducible if its intertwining
algebra I(ρV , ρV ) = {α1V : α ∈ k} ∼
= k. (This is as small as I(ρV , ρV ) can be,
as we always have {α1V : α ∈ k} ⊂ I(ρV , ρV ).) A field k is a splitting field for G
if every irreducible k-representation of G is absolutely irreducible. Thus, Schur’s
Lemma says that an algebraically closed field is a splitting field.
Corollary 7.6. Let (V, ρV ) be a k-representation of a finite abelian group A that
is absolutely irreducible. Then dimk V = 1.
FOURIER TRANSFORM
17
Proof. Because the group is abelian, we claim ρV (a) ∈ I(ρV , ρV ), for all
a ∈ A. Indeed, ρV (a) ◦ ρV (b) = ρV (ab) = ρV (ba) = ρV (b) ◦ ρV (a), for all a, b ∈ A.
By absolute irreducibility, every ρV (a) has the form αa 1V , for some αa ∈ k. But
any subspace of V is invariant under such transformations, so the irreducibility of
ρV implies dimk V = 1.
Remark 7.7. Corollary 7.6 allows us to reconcile the two meanings of the term
‘character’. In general, a character is the trace of a representation. When the group
is abelian and k is algebraically closed (such as k = C), any irreducible representation has dimension one, so that the character equals the representation itself. In
particular, characters of one-dimensional representations are homomorphisms into
GL1 (k) = k × .
Modules over group algebras. We now describe how group representations
are essentially the same as modules over the group algebra. We will assume that
linear transformations of a vector space V have their inputs written on the right:
v 7→ T (v). With this convention, group representations will correspond to left
modules over the group algebra. (If inputs are on the left, then one uses right
modules.)
As above, let G be a finite group and k a field. Let V be a left k[G]-module.
Then V is a k-vector space, using the scalar multiplication by the elements αe ∈
k[G], α ∈ k, where e ∈ G is the identity element. We define a representation ρV
on V by (ρV (g))(v) = gv, g ∈ G, v ∈ V , where the right side is the k[G]-scalar
multiplication viewing g ∈ k[G]. Conversely, given a k-representation ρV of G on
V , we define a k[G]-module structure on V by
X
X
X
αg g v =
αg (ρV (g)) (v),
αg g ∈ k[G], v ∈ V.
g∈G
g∈G
g∈G
Irreducible representations correspond to simple k[G]-modules, i.e., nonzero
modules having no nonzero proper submodules. Indecomposable representations
correspond to indecomposable k[G]-modules, i.e., those nonzero modules which cannot be written as a direct sum of two nonzero proper submodules. An intertwining
map f : (V, ρ) → (W,P
φ) corresponds to a homomorphism
of k[G]-modules.
Indeed,
P
P
for v ∈ V and α = g αg g ∈ k[G]: f (αv) = f ( g αg ρ(g)v) = g αg f (ρ(g)v) =
P
g αg φ(g)(f v) = αf (v).
8. Maschke’s theorem
Maschke’s Theorem is one of the first major results in the representation theory
of finite groups, dating from 1898, [13]. We now assume that char k does not divide
|G|. The theorem says a number of equivalent things: k[G] is a semi-simple algebra;
every invariant subspace has an invariant complement; every indecomposable k[G]module is simple. The key technique is to average over the group (sum and then
divide by |G|); because char k does not divide |G|, |G| is invertible in k.
Theorem 8.1 (Maschke). Let (V, ρV ) be a k-representation of a finite group G.
Assume that char k does not divide |G|. If W ⊂ V is an invariant subspace under
ρV , then there exists an invariant subspace W 0 ⊂ V with V = W ⊕ W 0 .
Proof. Let U be any vector space complement of W in V ; i.e., U is any
subspace satisfying W ⊕ U = V . The subspace U determines a projection to W ; we
18
JAY A. WOOD
will average this projection over the group G and obtain an invariant complement
for W as the kernel of the invariant projection.
The direct sum decomposition V = W ⊕ U defines a k-linear transformation
p = pU : V → V by p(v) = w, where v = w + u, w ∈ W , u ∈ U , in the direct sum
decomposition. The transformation p is a projection, i.e., p2 = p. In particular,
p(w) = w for any w ∈ W .
The projection p is an element of End(V ), the k-algebra of all k-linear transformations of V to itself. The group GL(V ) is the group of units of End(V ). Define
a new element of End(V ) by averaging conjugates of p over the group G:
1 X
ρV (g −1 ) ◦ p ◦ ρV (g).
(8.1)
p0 =
|G|
g∈G
A re-indexing argument will show that p0 intertwines ρV , i.e., ρV (g)◦p0 = p0 ◦ρV (g),
or equivalently p0 = ρV (g −1 ) ◦ p0 ◦ ρV (g), for all g ∈ G. Indeed, for any h ∈ G (and
writing ρ in place of ρV ):
X
1
ρ(g −1 ) ◦ p ◦ ρ(g) ◦ ρ(h)
ρ(h−1 ) ◦ p0 ◦ ρ(h) = ρ(h−1 ) ◦
|G|
g∈G
1 X
ρ((gh)−1 ) ◦ p ◦ ρ(gh) = p0 .
=
|G|
g∈G
Before we proceed further, we observe that p ◦ ρ(g) ◦ p = ρ(g) ◦ p, for all g ∈ G.
Indeed, for any v ∈ V , p(v) ∈ W . Because W is invariant, (ρ(g) ◦ p)(v) ∈ W as
well. Since p(w) = w for all w ∈ W , we conclude that p ◦ ρ(g) ◦ p = ρ(g) ◦ p, for all
g ∈ G.
We claim that p0 (w) = w for all w ∈ W . By invariance, ρ(g)w ∈ W , so that
pρ(g)w = ρ(g)w. Thus,
1 X
1 X
ρ(g −1 ) p ρ(g)w =
ρ(g −1 )ρ(g)w = w.
p0 w =
|G|
|G|
g∈G
g∈G
0
We next show that p is a projection. We will again use the observation that
p ◦ ρ(g) ◦ p = ρ(g) ◦ p, for all g ∈ G. We have that
!
X
1
1 X
0
0
−1
−1
ρ(g ) ◦ p ◦ ρ(g) ◦
ρ(h ) ◦ p ◦ ρ(h)
p ◦p =
|G|
|G|
g∈G
h∈G
1 X
=
ρ(g −1 ) ◦ p ◦ ρ(gh−1 ) ◦ p ◦ ρ(h)
|G|2
g,h∈G
1 X
=
ρ(g −1 ) ◦ ρ(gh−1 ) ◦ p ◦ ρ(h)
|G|2
g,h∈G
1 X
=
ρ(h−1 ) ◦ p ◦ ρ(h) = p0 .
|G|2
g,h∈G
0
Thus p is a projection that intertwines ρV , with p0 (V ) = W .
Let 1V ∈ End(V ) denote the identity automorphism of V . Set q 0 = 1V −p0 ; q 0 is
another projection that intertwines ρV . (Said another way, p0 and q 0 are orthogonal
idempotents in the k-algebra End(V ), and p0 +q 0 = 1V .) Define W 0 = Im q 0 = ker p0 .
FOURIER TRANSFORM
19
Because p0 , q 0 are orthogonal idempotents that sum to 1V , we have V = W ⊕ W 0 .
Because p0 , q 0 intertwine ρV , W 0 is invariant.
Example 8.2. We continue Example 7.1, but assume char k 6= 2. The subspace
W1 = {(x, y) : y = x} ⊂ k 2 is invariant. We choose the subspace U = {(0, y) : y ∈
k} ⊂ k 2 as a vector space complement. The projection p to W1 with respect to the
splitting W1 ⊕ U = k 2 has matrix
1 0
p=
.
1 0
As in (8.1), we calculate p0 :
1
1 0
1 0
1 0
0 1
1
0
p =
+
0 1
1 0
0 1
1 0
1
2
1 1 1
1
1 0
0 1
+
=
.
=
1 0
0 1
2
2 1 1
0
0
0
1
1
0
Note that ker p0 = W2 = {(x, y) : y = −x} ⊂ k 2 , which is indeed an invariant
complement to W1 .
Corollary 8.3. Let (V, ρV ) be a k-representation of a finite group G. Assume that
char k does not divide |G|. Then (V, ρV ) splits into a direct sum of irreducible representations V ∼
= ⊕ti=1 Vi , where (Vi , ρVi ) is irreducible. The splitting is unique: if
t
V ∼
= ⊕i=1 Vi ∼
= ⊕sj=1 Vj0 , with Vi and Vj0 irreducible, then s = t and, after reindexing,
0
Vi ∼
= Vi , for all i.
Proof. For existence, use induction on the (finite) dimension of the representation V . Every representation of dimension 1 is irreducible. Assuming the result
for all dimensions < d, consider a representation (V, ρV ) of dimension d. If V is itself irreducible, there is nothing more to prove. If V is reducible, there is a nonzero
proper subrepresentation W ⊂ V . By Maschke’s Theorem, the representation V
splits as V = W ⊕ W 0 . Now apply the induction hypothesis to W and W 0 .
For uniqueness, suppose V ∼
= ⊕ti=1 Vi ∼
= ⊕sj=1 Vj0 , with Vi and Vj0 irreducible. For
j = 1, . . . , s, consider the intertwining mappings fj : V1 ,→ ⊕ti=1 Vi ∼
= ⊕sj=1 Vj0 Vj0 .
0
Because the Vi and Vj are irreducible, Lemma 7.3 implies that each fj is either zero
or an isomorphism. If all the fj equal zero, we have a contradiction to V1 injecting
into the direct sum. Thus some fj is an isomorphism. By reindexing, we may
assume j = 1. Then
⊕ti=2 Vi ∼
= V /V1 ∼
= V /V10 ∼
= ⊕sj=2 Vj0 ,
and we proceed by induction.
9. Structure of group algebras
In this section we study the structure of group algebras in the situation where
char k does not divide |G|. We mimic the approach of Terras [16], but for arbitrary
fields. Certain results will hold over any field k, while additional, stronger results
will hold over splitting fields.
As usual, let G be a finite group and k a field such that char k does not divide
|G|. All representations discussed are assumed to be finite dimensional. Denote by
b the set of isomorphism classes of irreducible k-representations of G.
G
20
JAY A. WOOD
Remark 9.1. When G = A is abelian and k is algebraically closed (as when k = C),
then every irreducible representation is of dimension one (Corollary 7.6) and hence
b given in Section 2.
equal to its character. This matches the definition of A
Let F (G, k) be the k-vector space of all k-valued functions on G; when equipped
with the convolution product, F (G, k) is isomorphic to the group algebra k[G].
Define a k-valued inner product h·, ·i on F (G, k) by
1 X
f1 (g)f2 (g −1 ).
hf1 , f2 i =
|G|
g∈G
Remark 9.2. When k = C and G = A is abelian, this inner product differs from
that in (2.1). However, the two inner products agree when the second entry f2 is
b
equal to a character π ∈ A.
Matrix entries. Let (V, ρ) be a k-representation of G of dimension dρ . If we
fix a basis of V , the various ρ(g), g ∈ G, can be viewed as dρ × dρ matrices. Denote
the (i, j)-entry of this matrix by ρi,j (g); ρi,j is a function from G to k called a
matrix entry of the representation ρ.
We want to give some properties of the matrix entries and characters of irreducible representations. For some of the properties, we will need the stronger
property of being absolutely irreducible.
Proposition 9.3. Suppose (V, ρ) and (W, φ) are inequivalent irreducible k-representations of G. Then:
(1) there are no intertwining maps: I(ρ, φ) = 0;
(2) matrix entries are orthogonal: hρi,j , φm,n i = 0, for all i, j, m, n;
(3) characters are orthogonal: hχρ , χφ i = 0.
Proof. Suppose f ∈ I(ρ, φ). If f 6= 0, then Lemma 7.3 implies that f is an
isomorphism. This would provide an equivalence between ρ and φ, contrary to
hypothesis.
Suppose dimk V = dρ and dimk W = dφ . Fix bases in V and W so that
matrix entries are defined. Let M be any dφ × dρ matrix over k. Define a k-linear
transformation M 0 : V → W by
X
M0 =
φ(g −1 )M ρ(g).
g∈G
0
We claim that M ∈ I(ρ, φ). This follows from a re-indexing argument (with
` = gh); for any h ∈ G,
X
X
M 0 ρ(h) =
φ(g −1 )M ρ(gh) =
φ(h`−1 )M ρ(`) = φ(h)M 0 .
g∈G
`∈G
Since I(ρ, φ) = 0, we have M 0 = 0, no matter which M is chosen.
Pick M so that its only nonzero entry is a 1 in position (n, i). Then a matrix
calculation shows that the (m, j)-entry of M 0 equals |G|hρi,j , φm,n i. Since M 0 = 0
(and |G| =
6 0 in k), we have the orthogonality of matrix entries.
Expand the characters in terms of the diagonal matrix entries; e.g., χρ (g) =
Pdρ
P
i=1 ρi,i (g). Then a calculation shows that hχρ , χφ i =
i,m hρi,i , φm,m i, which
vanishes by the previous result.
FOURIER TRANSFORM
21
Now suppose that the representations are the same and are absolutely irreducible.
Proposition 9.4. Suppose (V, ρ) is an absolutely irreducible k-representation of G
of dimension dρ . Then:
(1) the characteristic char k does not divide dρ ;
(2) the matrix entries satisfy:
(
1
, i = n, j = m,
hρi,j , ρm,n i = dρ
0,
otherwise;
(3) hχρ , χρ i = 1.
Proof. We apply the same ideas as in the proof of Proposition 9.3. For any
dρ × dρ matrix M , define
X
(9.1)
M0 =
ρ(g −1 )M ρ(g).
g∈G
Then M 0 ∈ I(ρ, ρ) = k, by the hypothesis of absolute irreducibility. Thus M 0 =
αM 1V for some αM ∈ k (the value of αM depends upon the matrix M ). In particular, the off-diagonal entries of M 0 vanish.
Since the matrix trace is conjugation invariant, taking the trace of (9.1) yields
tr M 0 = |G| tr M . But M 0 = αM 1V , so that tr M 0 = αM dρ . Thus αM dρ = |G| tr M .
Pick any matrix M0 with tr M0 = 1, say the matrix whose only nonzero entry is a
1 in position (1, 1). Then αM0 dρ = |G|. Since char k does not divide |G|, it does
not divide dρ either. Note that, in general, αM = |G| tr M/dρ .
As in the proof of Proposition 9.3, now pick M to be the matrix whose only
nonzero entry is a 1 in position (n, i). The same computation shows that the
(m, j)-entry of M 0 equals |G|hρi,j , ρm,n i. But M 0 = αM 1V , so that
(
αM = |G|
dρ tr M, j = m,
|G|hρi,j , ρm,n i =
0,
j 6= m.
However, tr M = 0 when i 6= n, and tr M = 1 when i = n. The stated formula now
follows.
For thePcharacter formula, the same calculation as in Proposition 9.3 yields
hχρ , χρ i = i,m hρi,i , ρm,m i, which simplifies to 1.
Finite number of irreducible representations. We will apply the orthogonality results to prove that there are only a finite number of irreducible krepresentations (up to isomorphism). We will have more precise results when k is a
splitting field for G, i.e., when every irreducible k-representation of G is absolutely
irreducible. Recall that Schur’s Lemma implies that any algebraically closed field
is a splitting field.
The left regular representation ρ of a finite group G is defined by having V =
k[G] and ρ(g) be left multiplication by g in k[G]. With respect to the basis of group
elements, the matrix representing ρ(g) has entries equal to 1 at positions (gb, b),
b ∈ G, and zeros elsewhere. In particular, all the diagonal entries are zero, except
in the case g = e where ρ(e) = 1V is the identity. The next result is now clear.
22
JAY A. WOOD
Lemma 9.5. The character χρ of the left regular representation is
(
|G|, g = e,
χρ (g) =
0,
g 6= e.
Lemma 9.6. Let ρ be the left regular representation of G, and let (W, φ) be any
other representation of G. Then ρ intertwines φ; i.e., I(ρ, φ) 6= 0.
Proof. This is essentially the same as φ being
of G. Pick
P a representation
P
any nonzero w ∈ W . Define f : k[G] → W by f ( g αg g) = g αg φ(g)w; because
P
w 6= 0, f is not the zero map. We will show that f ∈ I(ρ, φ). Let α = h αh h ∈
k[G] and g ∈ G. Then
X
X
f (ρ(g)α) = f (gα) = f (
αh gh) =
αh φ(gh)w
h
=
X
h
αh φ(g)φ(h)w = φ(g)f (α).
h
By Corollary 8.3, the left regular representation ρ splits uniquely into a sum of
irreducible k-representations
(9.2)
ρ=
t
X
n i ρi ,
i=1
where each ρi is irreducible and ρi ∼
6 ρj if i 6= j.
=
Proposition 9.7. Every irreducible k-representation of G is isomorphic to one of
the irreducible representations ρi appearing in the splitting (9.2) of the left regular
representation ρ. The number of inequivalent irreducible k-representations of G is
finite.
Proof. Let (W, φ) be an irreducible k-representation of G. Let f be a nonzero
element of I(ρ, φ), by Lemma 9.6. By (9.2), k[G] splits as a representation space
into k[G] = ⊕ti=1 ⊕njii=1 Vi,ji .
For any irreducible Vi,ji , consider the composite mapping of representations
fi,ji : Vi,ji ,→ k[G] → W , where the map k[G] → W is f . This composite map fi,ji
intertwines (Vi,ji , ρi ) and (W, φ). If φ is not equivalent to any of the ρi , then each
fi,ji is zero, by Proposition 9.3. But this implies f itself is zero, contrary to the
choice of f . Thus, φ must be isomorphic to one of the ρi .
Suppose the irreducible representations ρi in (9.2) satisfy dimk Vρi = dρi . Then
Pt
by counting the dimensions of the representations, (9.2) implies |G| = i=1 ni dρi .
Since ni , dρi ≥ 1, the number of inequivalent irreducible representations of G is
bounded above by |G|. We will have stronger results when k is a splitting field for
G.
Lemma 9.8 (Frobenius and Schur, 1906). Suppose k is a splitting field for G. Let
ρ1 , . . . , ρt be all the inequivalent irreducible k-representations of G, with dimensions
dρ1 , . . . , dρt . Then the matrix entries of ρ1 , . . . , ρt are linearly independent, and
Pt
2
i=1 dρi ≤ |G|.
Proof.
Pt Consider all the matrix entries of all the representations ρ1 , . . . , ρt ;
there are i=1 d2ρP
such matrix entries. To show that they are linearly independent
i
over k, suppose i,m,n ci;m,n ρi;m,n = 0, with ci;m,n ∈ k. Fix an arbitrary choice
FOURIER TRANSFORM
23
i; m, n of indices, and apply h·, ρi;n,m i to the equation. By Propositions 9.3 and 9.4
(since k is a splitting field), the inner product simplifies to ci;m,n = 0. Thus, the
matrix entries are linearly independent.
Each matrix entry is an element of the vector space F (G, k). Since the matrix
Pt
2
entries are linearly independent, we conclude that
i=1 dρi ≤ dimk F (G, k) =
|G|.
The group G acts on itself by conjugation: g ∈ G defines h 7→ ghg −1 . The
orbits {ghg −1 : g ∈ G} are called conjugacy classes of G. A function f ∈ F (G, k)
is a class function if f is constant on conjugacy classes; i.e., if f (ghg −1 ) = f (h),
for all g, h ∈ G. Let Cl(G, k) ⊂ F (G, k) be the k-linear subspace consisting of all
class functions; dimk Cl(G, k) equals the number of conjugacy classes of G.
The next result is often called the Peter-Weyl Theorem.
Theorem 9.9 (Peter-Weyl Theorem). Suppose k is a splitting field for G. Let
ρ1 , . . . , ρt be all the inequivalent irreducible k-representations of G, of dimensions
dρ1 , . . . , dρt . Then,
(1) the left regular representation is equivalent to ⊕ti=1 dρi ρi ;
Pt
2
(2)
i=1 dρi = |G|;
(3) the matrix entries ρi;mi ,ni , i = 1, . . . , t, mi , ni = 1, . . . , dρi , are a basis for
F (G, k);
(4) the characters χρ1 , . . . , χρt are a basis for Cl(G, k);
(5) the number t of inequivalent irreducible representations equals the number
of conjugacy classes of G.
Proof. As in (9.2), suppose the left regular representation ρ splits as ρ =
⊕ti=1 ni ρi , with the ni being nonnegative integers. By taking traces, we see that
(9.3)
χρ =
t
X
ni χρi .
i=1
Fixing an index i and applying h·, χρi i to (9.3) yields hχρ , χρi i = ni (this step
uses that k is a splitting field via Proposition 9.4). But a direct calculation yields
hχρ , χρi i = dρi . Thus ni = dρi and ρ = ⊕ti=1 dρi ρi
Pt
On the other hand, evaluating (9.3) at g = e yields |G| =
i=1 ni dρi . As
Pt
2
ni = dρi , we conclude that i=1 dρi = |G|. Since we know the matrix entries are
linearly independent (Lemma 9.8) and we have the right number of them, they form
a basis of F (G, k).
Because the matrix trace is conjugation invariant, every character is a class
function. Thus χρ1 , . . . , χρt ∈ Cl(G, k). These characters are also linearly indepenPt
dent. Indeed, suppose i=1 ci χρi = 0, for some ci ∈ k. Fix an arbitrary index i
and apply h·, χρi i to the equation. This simplifies to ci = 0 (again, k splitting), and
the characters are linearly independent.
To show that the characters χρ1 , . . . , χρt span Cl(G, k), take any f ∈ Cl(G, k).
For any index i, define a dρi × dρi matrix Mi whose (m, n)-entry is hf, ρi;m,n i. We
24
JAY A. WOOD
claim that Mi ∈ I(ρi , ρi ). For any g ∈ G, the (j, l)-entry of ρi (g)Mi ρi (g −1 ) is:
X
ρi;j,m (g)hf, ρi;m,n iρi;n,l (g −1 )
ρi (g)Mi ρi (g −1 ) j,l =
m,n
1 XX
ρi;j,m (g)f (h)ρi;m,n (h−1 )ρi;n,l (g −1 )
|G|
h∈G m,n
1 X
=
f (h)ρi;j,l (gh−1 g −1 )
|G|
h∈G
1 X
f (g −1 rg)ρi;j,l (r−1 )
=
(for r = ghg −1 )
|G|
r∈G
1 X
f (r)ρi;j,l (r−1 )
(f class function)
=
|G|
=
r∈G
= hf, ρi;j,l i = Mi;j,l
Since ρi is absolutely irreducible (k splitting), Mi = αi Idρi , for some αi ∈ k.
P
The matrix entries are a basis for F (G, k), so f = i;mi ,ni ci;mi ,ni ρi;mi ,ni for
some coefficients ci;mi ,ni ∈ k. By the inner product formulas (k splitting), we see
that hf, ρi;mi ,ni i = ci;ni ,mi /dρi . From Mi = αi Idρi , we deduce that ci;mi ,mi = αi dρi ,
P
and ci;mi ,ni = 0 if mi 6= ni . Then the sum f = i;mi ,ni ci;mi ,ni ρi;mi ,ni simplifies
P
P
to f = i;mi αi dρi ρi;mi ,mi = i αi dρi χρi . Thus, the characters span Cl(G, k) and
consequently form a basis. The number t of characters then equals the dimension
of Cl(G, k), which is the number of conjugacy classes in G.
Remark 9.10. A word of caution. As usual, let G be a finite group and k be
a field whose characteristic does not divide |G|. Let K be an extension field of
k. One can then consider k-representations of G as well as K-representations of
G. Because of the inclusion k ⊂ K, every k-representation ρ can be viewed as
a K-representation. (In terms of modules over group algebras, this is tensoring
over k with K.) If ρ is irreducible when viewed as a K-representation, then ρ is
also irreducible as a k-representation. The converse fails in general (unless ρ is
absolutlely irreducible); this is why there are stronger results over splitting fields.
The number of inequivalent irreducible k-representations of G depends on the field
k; the number can increase by going to an extension field K of k.
Example 9.11. Let G = C4 be a cyclic group of order 4;
a4 = e. Define ρ with e, a, a2 , a3 sent respectively to
1 0
0 −1
−1
0
0
,
,
,
0 1
1
0
0 −1
−1
G = {e, a, a2 , a3 }, with
1
.
0
If char k = 2, this is essentially Example 7.1 composed with the natural quotient
map C4 C2 ; ρ is indecomposable, but not irreducible. When char k 6= 2, ρ is
irreducible if and only if −1 is not a square in k. Indeed, an invariant subspace
for ρ would be spanned by an eigenvector of ρ(a), and the eigenvalues of ρ(a) are
zeroes of the characteristic polynomial det(xI − ρ(a)) = x2 + 1. For example, ρ is
irreducible over the real numbers R but not over the complex numbers C. Over C,
ρ splits into two 1-dimensional irreducibles.
FOURIER TRANSFORM
25
Fourier transform. Over a finite abelian group A, the Fourier transform of
b C); F (A,
b C) was viewed as a direct sum of copies
Section 2 mapped F (A, C) → F (A,
b Over a general finite group G and a field k whose
of C, one for each character in A.
characteristic does not divide |G|, the Fourier transform will map F (G, k) into a
b
direct sum of matrix rings over k, one for each irreducible representation in G.
When the field k is a splitting field, the Fourier transform will be an isomorphism
of F (G, k) to a direct sum of matrix rings over k.
Assume G is a finite group and k is a field whose characteristic does not divide
b is the set of (equivalence classes of) irreducible k-representaions
|G|. Recall that G
of G. As in (9.2), let ρ1 , . . . , ρt be all the inequivalent irreducible k-representations
of G, of dimensions dρ1 , . . . , dρt , and with n1 , . . . , nt being the multiplicities of ρi
in the left regular representation of G. Let
∆=
t
a
Mdρi ×dρi (k),
i=1
the disjoint union of the k-algebras of all dρi ×dρi matrices over k. Abusing notation,
define
b k) = {τ : G
b → ∆ : τ (ρi ) ∈ Md ×d (k)}.
F (G,
ρi
ρi
b k), is defined by
The Fourier transform ˆ : F (G, k) → F (G,
X
fˆ(ρi ) =
f (g)ρi (g),
f ∈ F (G, k).
g∈G
b k) → F (G, k) by
In the other direction, define ˇ : F (G,
t
1 X
ni tr ρi (g −1 )τ (ρi ) .
τ̌ (g) =
|G| i=1
Proposition 9.12. The Fourier transform is injective: if τ = fˆ, then f = τ̌ . If
k is a splitting field for G, then the Fourier transform is an isomorphism, with
inverse given by ˇ.
Proof. We start with f ∈ F (G, k), set τ = fˆ, and show that τ̌ = f . We will
expand fˆ(ρi ), change the order of summations, reindex the sum (r = g −1 h), and
appeal to the left regular representation ρ and its character χρ , Lemma 9.5.
!
t
t
X
X
X
−1 ˆ
−1
|G|τ̌ (g) =
ni tr ρi (g )f (ρi ) =
ni tr ρi (g )
f (h)ρi (h)
i=1
=
X
h∈G
=
X
r∈G
=
X
i=1
f (h)
t
X
ni tr ρi (g −1 h) =
i=1
f (gr) tr
X
r∈G
t
X
h∈G
f (gr)
t
X
ni tr ρi (r)
i=1
!
ni ρi (r)
i=1
=
X
f (gr) tr ρ(r)
r∈G
f (gr)χρ (r) = f (g)|G|.
r∈G
b k), set f = τ̌ , and show
Now assume k is a splitting field. Start with τ ∈ F (G,
ˆ
ˆ
that f = τ . We find the (m, n)-entry of |G|f (ρi ) by expanding τ̌ (g), expanding the
26
JAY A. WOOD
matrix trace, changing the order of summations, using the orthogonality properties
of matrix entries (k splitting), and using the fact that ni = dρi over a splitting field:
|G|fˆ(ρi )m,n =
t
XX
dρj tr ρj (g −1 )τ (ρj ) ρi;m,n (g)
g∈G j=1
=
t
XX
X
dρj
ρj;λ,µ (g −1 )τµ,λ (ρj ) ρi;m,n (g)
g∈G j=1
=
t
X
dρj
j=1
=
t
X
dρj
j=1
= dρi
λ,µ
X
τµ,λ (ρj )
X
ρj;λ,µ (g −1 )ρi;m,n (g)
λ,µ
g∈G
X
τµ,λ (ρj )hρi;m,n , ρj;λ,µ i|G|
λ,µ
X
τµ,λ (ρi )hρi;m,n , ρi;λ,µ i|G|
λ,µ
= dρi τm,n (ρi )
1
|G| = |G|τm,n (ρi ).
dρi
Recall that F (G, k) is a ring under the convolution product. We show that the
Fourier transform converts the convolution product into matrix products, generalizing Lemma 6.1.
b we have
Proposition 9.13. For f1 , f2 ∈ F (G, k) and ρi ∈ G,
(f1 ∗ f2 )ˆ(ρi ) = fˆ1 (ρi )fˆ2 (ρi ),
where the right side is matrix multiplication.
Proof. As with Lemma 6.1, one expands the convolution and uses the homomorphism property of representations: ρi (g) = ρi (gh−1 h) = ρi (gh−1 )ρi (h). A
re-indexing argument finishes the proof.
Theorem 9.14. Assume G is a finite group and k is a field whose characteristic
does not divide |G|. The Fourier transform induces an injective homomorphism of
k-algebras
(9.4)
k[G] ∼
= F (G, k) →
t
M
i=1
Mdρi ×dρi (k),
f 7→
t
M
fˆ(ρi ).
i=1
When k is a splitting field for G, this injective homomorphism is an isomorphism
of k-algebras.
Proof. Recall that the group algebra k[G] is isomorphic to F (G, k) equipped
b k) is isomorphic to ⊕t Md ×d (k)
with the convolution product. The space F (G,
ρi
ρi
i=1
t
b
by sending τ ∈ F (G, k) to ⊕i=1 τ (ρi ). Proposition 9.12 shows that the Fourier
transform is an injective homomorphism (resp., isomorphism, for k splitting) of
k-vector spaces, while Proposition 9.13 shows that the Fourier transform is a ring
homomorphism.
FOURIER TRANSFORM
27
Remark 9.15. For k splitting, let us re-examine the left regular representation in
light of the isomorphism (9.4). By virtue of the splitting, it is enough to understand
left multiplication in the matrix ring R = Md×d (k). Left multiplication by a fixed
element c of R defines a k-linear transformation Lc : R → R. Fixing a basis of R
as a k-vector space (of dimension d2 ) allows us to represent Lc as a d2 × d2 matrix.
Note that c itself is a d×d matrix. Depending on the exact choice of basis for R, the
matrix for Lc will essentially be d copies of c along the diagonal.
This phenomenon
Pt
explains why the left regular representation splits as ρ = i=1 dρi ρi for k splitting.
As an example, consider the case where d = 2. Choose as basis for R the
following matrices:
1 0
0 0
0 1
0 0
,
,
,
.
0 0
1 0
0 0
0 1
Then below are c and the matrix for Lc with respect to
α β 0
γ δ 0
α β
c=
, Lc =
0 0 α
γ δ
0 0 γ
the given basis:
0
0
.
β
δ
Remark 9.16. There is a famous theorem due to Wedderburn and Artin ([1,
Theorem 1.3.5]) that implies that k[G], with char k not dividing |G|, is isomorphic as
k-algebras to a direct sum of matrix rings over division algebras defined over k. Here
is a sketch of the proof. The ring k[G] is isomorphic to its ring of endomorphisms,
viewed as acting on k[G] on the right. Split the left regular representation as in (9.2).
Each intertwining algebra I(ρi , ρi ) is a division algebra Di over k (Proposition 7.4).
Write k[G] = ⊕ti=1 Vi , where Vi is the sum of the ni representation spaces isomorphic
to ρi . Endomorphisms of k[G] must preserve this splitting (Proposition 9.3). The
endomorphism rings of the Vi are then matrix rings (of size ni ×ni ) over the division
algebra Di . When k is a splitting field, each Di = k.
What are we seeing in Theorem 9.14? When k is a splitting field, we are seeing
exactly the Wedderburn-Artin splitting as a direct sum of matrix rings over k. In
the nonsplitting case, the homomorphism of (9.4) is not surjective. Its image inside
Mdρi ×dρi (k) is a copy of Mni ×ni (Di ).
Remark 9.17. What happens to the Wedderburn-Artin splitting under field extensions? Let k ⊂ K, and suppose
k[G] ∼
= ⊕t Mn ×n (Di ),
i=1
i
i
where the Di are division algebras over k. Tensoring with K over k then yields
K[G] ∼
= k[G] ⊗k K ∼
= ⊕t Mn ×n (Di ⊗k K).
i=1
i
i
The difficulty is that Di ⊗k K need not be a division algebra over K; it can be a
sum of such. For example, C ⊗R C ∼
= C ⊕ C as C-algebras. We will re-visit this
issue in Section 10.
Example 9.18. Continue with Example 9.11. The representation ρ is the only 2dimensional irreducible representation of G = C4 over R. The intertwining algebra
is I(ρ, ρ) ∼
= C (which is why one sees matrices of the form ρ(a) as models for
complex structures). There are also two 1-dimensional irreducible representations
over R: the trivial representation taking all elements of G to 1 ∈ R, and the
‘square’ of ρ taking e, a2 to 1 ∈ R and a, a3 to −1 ∈ R. Then R[C4 ] ∼
= R ⊕ R ⊕ C ,→
28
JAY A. WOOD
R ⊕ R ⊕ M2×2 (R), with the image in the last summand being all matrices of the
form
α −β
.
β
α
When we tensor with C, we get
C[C2 ] ∼
= R[C2 ] ⊗R C ∼
= (R ⊕ R ⊕ C) ⊗R C ∼
= C ⊕ C ⊕ C ⊕ C,
which reflects the splitting of the R-irreducible ρ into two C-irreducibles. The
inclusion C ,→ M2×2 (R) after tensoring becomes C ⊕ C ,→ M2×2 (C), with image
the diagonal matrices.
By using Theorem 9.14, we can prove the existence of a splitting field that is a
finite extension of its prime field.
Proposition 9.19. Assume G is a finite group and k is a field whose characteristic
does not divide |G|. Then there is a finite extension K of k that is a splitting field
for G.
Proof. By Schur’s Lemma, we know that an algebraic closure k̄ of k is a
splitting field for G. By Theorem 9.14, k̄[G] ∼
= ⊕ti=1 Mdρi ×dρi (k̄). This isomorphism
says that we have two bases for k̄[G] as a vector space over k̄: the group elements
g ∈ G and the matrix units ei;m,n (a dρi × dρi matrix whose only nonzero entry is
a 1 in position (m, n)). Write down the transitions between the two bases:
X
X
g=
αi;m,n (g)ei;m,n , ei;m,n =
βi;m,n (g)g,
g
i;m,n
where the coefficients belong to k̄.
Let K be the extension field of k generated by all the coefficients αi;m,n (g),
βi;m,n (g); there being only finitely many such α, β, the field K is a finite extension
of k. We see that each matrix unit ei;m,n ∈ K[G], so that K[G] ∼
= ⊕ti=1 Mdρi ×dρi (K).
This in turn implies that K is a splitting field for G.
III. Examples
As coding theorists pursue the study of linear codes defined over finite rings
or, more generally, finite modules, there is increasing interest in understanding the
structure of finite rings, especially finite Frobenius rings. Group algebras of finite
groups over finite fields are some of the most fundamental finite Frobenius rings,
and they may serve as a starting point for further investigations.
10. Representations over finite fields
We collect a few results about finite fields and discuss representations over finite
fields.
A finite field is a field with a finite number of elements. The number of elements
is necessarily a prime power q = pf , p prime, f a positive integer. For every such
q = pf there exists a finite field with q elements, and it is unique up to isomorphism.
Typical notations are Fq or GF (q) (for Galois field, in honor of Galois).
FOURIER TRANSFORM
29
Lemma 10.1. Let Fq be a finite field with q = pf , p prime. Suppose n is an integer
not divisible by p. Then there exists a finite extension field K over Fq that contains
n distinct nth roots of 1.
Proof. The integers n and q are relatively prime, so q is a unit in the ring
Z/nZ. Let m be the order of q (mod n), i.e., the smallest positive integer so that
q m ≡ 1 (mod n). Let K = Fqm , a degree m extension of Fq . The multiplicative
group of K is cyclic of order q m − 1. By construction, n divides q m − 1, so the
multiplicative group of K contains a cyclic subgroup of order n. The elements of
this subgroup are nth roots of 1, and there are n of them.
Lemma 10.2. Let Fq be a finite field, with extension fields Fqa and Fqb . Let
d = gcd(a, b) be the greatest common divisor of a and b, and let m = lcm(a, b) be
their least common multiple. Then, as Fq -algebras,
Fqa ⊗Fq Fqb ∼
= Fqm ⊕ · · · ⊕ Fqm ,
where there are d summands on the right side.
Proof. This is a special case of Exercise I.2 in [5].
In light of Lemma 10.2, let us re-visit Remark 9.17. As usual, let G be a finite
group whose order |G| is prime to p, and let k = Fq be a finite field of order q = pf .
Split the left regular representation of G as in (9.2). Then the Wedderburn-Artin
theorem says that k[G] splits into a direct sum of matrix rings over the (finitedimensional) division algebras Di = I(ρi , ρi ) over k. In particular, the division
algebras I(ρi , ρi ) are finite. Another theorem of Wedderburn says that all finite
division algebras are commutative, so the intertwining algebras I(ρi , ρi ) are finite
extensions Fqδi of k, where δi = dimk I(ρi , ρi ). Thus, the splitting of k[G] is
(10.1)
Fq [G] ∼
=
t
M
Mni ×ni (Fqδi ).
i=1
The irreducible k-representation ρi of dimension dρi is obtained by restriction of
scalars to k from an irreducible Fqδi = I(ρi , ρi )-representation of dimension ni ,
with dρi = ni δi .
Now let L = Fqb be any finite extension field of k = Fq . The group algebra
L[G] can be obtained from (10.1) by tensoring with L over k:
(10.2)
Fqb [G] ∼
=
t
M
Mni ×ni (Fqδi ⊗Fq Fqb ).
i=1
Some of the summands Mni ×ni (Fqδi ⊗Fq Fqb ) split further, by Lemma 10.2, when
δi and b are not relatively prime. This reflects that fact that an irreducible krepresentation ρi may fail to be irreducible as an L-representation.
We can also look at this splitting from another perspective. By Proposition 9.19, there is a finite extension K = Fqm of k that is a splitting field for
G. Over K, the left regular representation ρ of G splits into a sum of absolutely
0
irreducible K-representations: ρ0 ∼
= ⊕tj=1 n0j ρ0j (using primes (0 ) to distinguish the
irreducible K-representations from the irreducible k-representations). The representation ρ0j has dimension dρ0j = n0j over K, by the Peter-Weyl Theorem. The
30
JAY A. WOOD
splitting of group algebras is then
0
(10.3)
K[G] ∼
=
t
M
Mn0j ×n0j (K).
j=1
Let us now reconcile (10.1), (10.2) and (10.3), using the splitting field K = Fqm
in the role of L = Fqb . By taking a larger K if necessary, we may assume that all
the intertwining algebras Fqδi = I(ρi , ρi ) ⊂ K = Fqm . Then δi divides m, and
Fqδi ⊗k K = δi K, i.e., the direct sum of δi copies of K.
Applying these relations to (10.2), we see that K[G] ∼
= ⊕ti=1 δi Mni ×ni (K). That
is, each irreducible k-representation ρi (which has I(ρi , ρi ) = Fqδi and dimk ρi =
ni δi ) has ρi ⊗k K split into δi irreducible K-representations of dimension ni over
K.
Since this splitting must match the splitting in (10.3), we conclude that t0 =
Pt
0
i=1 δi and that the nj equal the ni from which they arise.
Let us look more carefully at the relationship between k-representations of
G and K-representations of G. As above, we assume k = Fq ⊂ K = Fqm is a
finite extension with dimk K = m. In a K-representation (V, ρ) of G, V is a finitedimensional K-vector space of dimension dimK V = dρ . Because k ⊂ K, restricting
scalars to k also allows us to view V as a k-vector space. Then dimk V = mdρ .
The K-linear transformations ρ(g), g ∈ G, are also k-linear, so (V, ρ) is also a
k-representation; we will refer to it as ρ↓k .
We can go in the other direction by tensoring. If (V, ρ) is a k-representation of
dimension dρ , then (V ⊗k K, ρ⊗k K) is a K-representation with dimK (V ⊗k K) = dρ ,
so that dimk (V ⊗k K) = mdρ .
We know from Galois theory that any finite extension of a finite field k = Fq ⊂
K = Fqm is a Galois extension with a cyclic Galois group. Let Gal(K/k) be the
Galois group of this extension, i.e., the group of field automorphisms of K that fix
elements of k. The Galois group Gal(K/k) is cyclic of order m, generated by the
qth power map α 7→ αq .
Let τ be a field automorphism of K. Given a vector space V over K, we
define another K-vector space V τ . Let V τ equal V as abelian groups (i.e., same
elements and same addition), but define a different scalar multiplication (denoted
α · v), which is twisted by the automorphism τ : α · v = τ (α)v, α ∈ K, v ∈ V ,
and using the original scalar multiplication to perform τ (α)v. One observes that
any K-linear transformation of V is also a K-linear transformation of V τ . Thus, a
K-representation (V, ρ) gives rise to another K-representation (V τ , ρ); we will refer
to the latter as ρτ . If ρ is K-irreducible, so is ρτ . In terms of a basis for V , the
matrix entries are related by ρτm,n (g) = τ (ρm,n (g)).
Example 10.3. Let us continue Example 9.11, with G = C4 a cyclic group of
order 4; write C4 = {e, a, a2 , a3 }, with a4 = e. Let k = F3 and K = F9 . Note
that primitive 4th roots of unity exist in K but not in k. The third-power map
τ : K → K, τ (α) = α3 , is a field automorphism of order 2; τ maps a primitive 4th
root of unity to its inverse.
Fix a primitive 4th root of unity ω ∈ K. The four irreducible K-representations
of C4 are given by ρi (aj ) = ω ij , for i, j = 0, 1, 2, 3. The reader is invited to verify
that ρτ0 ∼
= ρ0 , ρτ2 ∼
= ρ2 , ρτ1 ∼
= ρ3 , and ρτ3 ∼
= ρ1 . Also, ρ1 ↓k ∼
= ρ3 ↓k . (Using the same
FOURIER TRANSFORM
31
matrices over R ⊂ C, this says that rotation through 90◦ is equivalent to rotation
through −90◦ , by interchanging the order of the basis vectors.)
b of isomorphism classes of irThe Galois group Gal(K/k) acts on the set G
reducible K-representations by τ ∈ Gal(K/k) : ρ 7→ ρτ . Because τ ∈ Gal(K/k)
fixes elements of k, the k-vector space structures of V and V τ are equal; thus
ρτ ↓k ∼
= ρ↓k .
What happens to irreducibles between k and K? Let ρ be an irreducible Krepresentation. Define the stabilizer subgroup S ⊂ Gal(K/k) of ρ under the action
b by S = {τ ∈ Gal(K/k) : ρτ ∼
of Gal(K/k) on G
= ρ}; i.e., S is the group of all
automorphisms τ for which ρτ is equivalent to ρ. This group action of Gal(K/k)
b partitions G
b into orbits; the orbit of ρ has order |Gal(K/k)|/|S|.
on G
By the fundamental theorem of Galois theory, the subgroup S ⊂ Gal(K/k)
determines a fixed field L = K S ⊂ K by L = {α ∈ K : τ (α) = α, for all τ ∈ S}.
We also know that L ⊂ K is a Galois extension, with Gal(K/L) = S. Then,
dimL K = |S| and dimk L = |Gal(K/k)|/|S|.
Continue to consider the irreducible K-representation ρ. It is stabilized by any
τ ∈ S = Gal(K/L). By restricting scalars to k, we get a k-representation ρ↓k . This
restriction will split as ρ↓k ∼
= (dimL K)φ, where φ is an irreducible k-representation
with I(φ, φ) = L. In particular, ρ↓k is irreducible when L = K, i.e., when S = {1}.
Going the other direction: tensoring this φ with K, we get
M
ρτ ,
φ ⊗k K ∼
=
τ ∈Gal(K/k)/S
the sum of the dimk L distinct K-representations in the Gal(K/k)-orbit of ρ. (These
are the δ = dimk L irreducible K-representations described after (10.3).) Because
ρτ ↓k ∼
= ρ↓k , any other representation in the same orbit gives the same picture.
We will illustrate this phenomenon is subsequent sections.
11. Finite abelian groups
In this section we discuss the structure of group algebras Fq [A], where A is a
finite abelian group.
Every finite abelian group A splits as a product of cyclic groups of prime power
order. For example, there are two isomorphism classes of abelian groups of order
12: the cyclic group C12 ∼
= C4 × C3 , as well as C2 × C2 × C3 ∼
= C2 × C6 .
Lemma 11.1. For any field k and finite groups G1 and G2 , we have
k[G1 × G2 ] ∼
= k[G1 ] ⊗k k[G2 ].
Proof. This is Exercise 12.8 of [4].
Lemma 11.1 reduces the study of k[A] for a general finite abelian group to
understanding k[Cpn ] for p prime. Although most of this article has dealt with
fields k whose characteristic does not divide the order of the group G, we make an
exception here.
Proposition 11.2. Let p be a prime, and let q = pf . Then
n
Fq [Cpn ] ∼
= Fq [u]/(up ),
which is a chain ring with ideals
n
Fq [u]/(up ) = (1) ⊃ (u) ⊃ (u2 ) ⊃ · · · ⊃ (up
n
−1
) ⊃ (0).
32
JAY A. WOOD
n
ap
n
xp
n
Proof. Clearly Fq [Cpn ] ∼
= Fq [x]/(xp − 1) since a generator a of Cpn satisfies
n
n
= e. Let u = x − 1. Because we are in characteristic p, up = (x − 1)p =
− 1 = 0.
Now let us turn to the case where k = Fq , q = pf , and A = Cn , where p
and n are relatively prime. (We could further factor n into primes, but it is just
as easy this way.) By Lemma 10.1, there exists a finite extension K of k, say
K = Fqm , in which there are n nth roots of unity. This field K is a splitting
field for Cn . Indeed, we can write down n 1-dimensional (necessarily absolutely
irreducible) representations over K. Let ω ∈ K be a primitive nth root of unity,
and let a ∈ Cn be a generator of the cyclic group. Define representations ρi by
ρi (aj ) = ω ij , i, j = 0, 1, . . . , n − 1. Then K[Cn ] ∼
= K ⊕ · · · ⊕ K (n summands).
To understand k[Cn ], and indeed L[Cn ] for any intermediate field k ⊂ L ⊂ K,
we can use the Galois theory arguments outlined in Section 10. We will work an
example that illustrates this general theory.
Example 11.3. Let p = 3 and n = 16, so that k = F3 and A = C16 . The first
time that 16 divides the order of the multiplicative group of F3f is for f = 4, where
16 divides 34 − 1 = 80. Thus, K = F81 will be a splitting field for A. There are 16
1-dimensional representations over K, as above; call them ρi , i = 0, 1, . . . , 15. The
Galois group Gal(K/k) is cyclic of order 4, generated by τ : K → K, τ (α) = α3 .
b we have ρτ ∼
In the action of Gal(K/k) on A,
i = ρ3i , where the index is computed
(mod 16).
b together with the stabilizer subgroup Si and
The orbits of Gal(K/k) on A,
fixed field Li = K Si are:
orbit
Si
{ρ0 }
Gal(K/k)
{ρ1 , ρ3 , ρ9 , ρ11 } {1}
{ρ2 , ρ6 }
{1, τ 2 }
{ρ4 , ρ12 }
{1, τ 2 }
{ρ5 , ρ15 , ρ13 , ρ7 } {1}
{ρ8 }
Gal(K/k)
{ρ10 , ρ14 }
{1, τ 2 }
Li
k = F3
K = F81
F9
F9
K = F81
k = F3
F9
From this we can write down k[A] and, by tensoring with L = F9 , L[A] as well. We
keep the ordering of the orbits from the table above.
F3 [C16 ] ∼
= F3 ⊕ F81 ⊕ F9 ⊕ F9 ⊕ F81 ⊕ F3 ⊕ F9
F9 [C16 ] ∼
= F9 ⊕ (F81 ⊕ F81 ) ⊕ (F9 ⊕ F9 ) ⊕ (F9 ⊕ F9 )
⊕ (F81 ⊕ F81 ) ⊕ F9 ⊕ (F9 ⊕ F9 )
∼
F81 [C16 ] = F81 ⊕ · · · ⊕ F81 (16 summands)
Remark 11.4. The analysis in Example 11.3 should be familiar from the study of
cyclic codes over finite fields, because cyclic codes are exactly ideals in the group
algebra k[Cn ].
Let us examine these representations more carefully. The values of the representations ρi are powers of ω i ∈ K. Because ω 16 = 1, the even powers of ω are 8th
roots of unity and are invariant under τ 2 : τ 2 (ω 2i ) = ω 18i = ω 2i . Thus the even
powers of ω are elements of the fixed field of {1, τ 2 }; i.e., ω 2i ∈ F9 ⊂ F81 . Similarly,
ω 8 = −1 ∈ F3 ⊂ F9 ⊂ F81 .
FOURIER TRANSFORM
33
The linear transformation ρi (aj ), aj ∈ C16 , is left-multiplication by ω ij on
K, considered as a 1-dimensional vector space over K. The linear transformation
ρi ↓L (aj ) is also left-multiplication by ω ij on K, but now K is considered as a
2-dimensional vector space over L. Similarly, the linear transformation ρi ↓k (aj ) is
left-multiplication by ω ij on K, but with K considered as a 4-dimensional vector
space over k. While ρi1 is not equivalent to ρi2 , i1 6= i2 , as K-representations,
those representations
in the same τ 2 -orbit are equivalent when restricted to L; i.e.,
2
ρi ↓L ∼
= ρτi y . Similarly, those representations in the same τ -orbit are equivalent
L
when restricted to k.
To be very explicit, consider the polynomial p(x) = 1 + x + x2 + x3 + x4 over
F3 . One notes that p(x) is irreducible over F3 . (Because (x − 1)p(x) = x5 − 1, any
zero of p(x) is a 5th root of unity. But F3 and F9 have no 5th roots of unity (other
than 1).) We will use F81 = F3 [x]/(p(x)). Every element of F81 can be written
uniquely in the form α = α0 + α1 x + α2 x2 + α3 x3 , with αi ∈ F3 . Multiplication is
carried out mod(p(x)), so that x4 = −1 − x − x2 − x3 . The reader will verify the
following facts about F81 :
(1) F9 = {a + b(x2 + x3 ) : a, b ∈ F3 } ⊂ F81 ;
(2) ζ := x2 + x3 ∈ F9 is a primitive 8th root of unity;
(3) ω := 1 − x − x2 ∈ F81 satisfies ω 2 = ζ and is a primitive 16th root of unity;
(4) 1 ∈ F81 is a basis for F81 over F81 ;
(5) 1, x ∈ F81 is a basis for F81 over F9 ;
(6) 1, x, x2 + x3 , −1 − x − x2 (= x(x2 + x3 )) ∈ F81 is a basis for F81 over F3 .
The linear transformation ρ1 (a) = ω is (left) multiplication by ω on elements of
F81 . This mapping can be viewed as a linear transformation over F81 , F9 , and F3 ;
these are ρ1 (a), ρ1 ↓F9 (a), and ρ1 ↓F3 (a), resp. In terms of the bases listed above,
the respective matrices representing ρ1 (a) are:
−1
0
0 −1
0
1
1
0
−1 −ζ
.
(ω), A =
, B=
0 −1 −1
1
ζ
1
1
0 −1
1
To verify, for example, that ρ1 ↓ ∼
= ρ9 ↓ , here is an intertwining over F9 .
F9
F9
The form is AM = M A9 = M (−A) (using ω 8 = −1):
−1 −ζ
0 1
0 1
1
ζ
=
.
ζ
1
1 0
1 0
−ζ −1
Similarly, one verifies that ρ1 ↓F3 ∼
= ρ3 ↓F3 , of form BP = P B 3 :
−1
0
0 −1
0
1 −1
0
0
1
1
0
0 −1
0
0
0 −1 −1
1
1 −1
0 −1
1
0 −1
1
1
0
0
0
0
1 −1
0
0 −1 −1
1
0
1
0
−1
0
0
−1
1
=
1 −1
0 −1 −1
1
1
1
1
0
0
0
−1
1 −1 −1
.
Going in the other direction, consider ρ1 ↓F3 ⊗F3 F81 . Evaluated at a ∈ C16 , this
involves looking at the matrix B above, but as a matrix over F81 . The characteristic
34
JAY A. WOOD
polynomial of B is λ4 + λ2 − 1. Over F81 , this factors as
λ4 + λ2 − 1 = (λ − ω)(λ − ω 3 )(λ − ω 9 )(λ − ω 11 ),
as the reader is invited to verify. This means that the matrix B diagonalizes over
F81 , with diagonal entries ω, ω 3 , ω 9 , ω 11 . Said another way, we have
∼ ρ1 ⊕ ρ3 ⊕ ρ9 ⊕ ρ11 ,
ρ1 ↓ ⊗F F81 =
F3
3
which is a sum over a Gal(F81 /F3 )-orbit.
The linear transformation ρ2 (a) = ω 2 = ζ is (left) multiplication by ζ ∈ F9 on
elements of F81 . When viewed as a linear transformation over F9 , i.e., ρ2 ↓F9 (a),
the transformation diagonalizes. Indeed, the matrix representing ρ2 ↓F9 (a) is just
A2 from above, and
ζ 0
2
A =
.
0 ζ
This reflects the general theory regarding splittings of ρ↓L .
12. Some non-abelian groups of small order
Example 12.1. Let G = Σ3 , the symmetric group on three letters. The group Σ3
has 6 elements;
Σ3 = {e, σ, σ 2 , τ, τ σ, τ σ 2 },
3
2
where σ = e, τ = e, and στ = τ σ 2 .
Let k be any finite field with char k 6= 2, 3. Then there are three irreducible
k-representations whose values on σ and τ are as displayed:
Representation ρ
ρ1
ρ2
ρ(σ)
1
1
0 −1
1 −1
ρ(τ )
1
−1
−1 1
ρ3
0 1
∼ k ⊕ k ⊕ M2×2 (k).
All the representations are absolutely irreducible, and k[Σ3 ] =
Example 12.2. Let G = Q8 , the quaternionic 8-group. The elements are
Q8 = {±e, ±i, ±j, ±k},
where i, j, k multiply like unit quaternions: i2 = j 2 = k 2 = −e, ij = k, jk = i,
ki = j, and distinct i, j, k anticommute.
Let K be any finite field with char K 6= 2. The subgroup {±e} ⊂ Q8 is central,
with quotient group Q8 /{±e} ∼
= C2 × C2 . The abelian group C2 × C2 has four
irreducible K-representations (of dimension 1), which, via the quotient map, yield
four irreducible K-representations of Q8 :
ρ
ρ1
ρ2
ρ3
ρ4
ρ(−1) ρ(i) ρ(j) ρ(k)
1
1
1
1
1
1 −1
−1
1 −1
1
−1
1 −1 −1
1
This already accounts for 4 dimensions in the 8-dimensional K[Q8 ], so that K[Q8 ] ∼
=
K ⊕ K ⊕ K ⊕ K ⊕ M , where dimK M = 4. By the Wedderburn-Artin Theorem,
M must be a sum of matrix rings over extension fields of K. Because Q8 is a
nonabelian group, the group algebra K[Q8 ] is noncommutative. Thus, there must
FOURIER TRANSFORM
35
be at least one matrix ring of size n × n with n ≥ 2. If K ⊂ K 0 is a finite extension,
then dimK Mn×n (K 0 ) = n2 dimK K 0 . Since dimK M = 4, the only possibility is
M = M2×2 (K). Thus, K[Q8 ] ∼
= K ⊕ K ⊕ K ⊕ K ⊕ M2×2 (K). (For some features
of K[Q8 ] in characteristic zero, see [15, Exercise 12.3].)
Example 12.3. Let G = Dn , the dihedral group with 2n elements:
Dn = {σ i , τ σ i : i = 0, 1, . . . , n − 1},
with σ n = e, τ 2 = e, and στ = τ σ −1 = τ σ n−1 .
Let k be a finite field Fq with characteristic not dividing 2n, and let K be a
finite extension of k that has n nth roots of unity (Lemma 10.1). Following [15,
Section 5.3], we can write down irreducible K-representations of Dn , as follows.
Let ω ∈ K be a primitive nth root of unity.
Case of n even. There are four 1-dimensional representations and several 2dimensional representations:
ρ
ρ01
ρ02
ρ03
ρ04
ρi
ρ(σ j )
1
1
(−1)j
j
ij(−1) ω
0
0 ω −ij
ρ(τ σ j )
1
−1
(−1)j
j+1
(−1) −ij 0 ω
ω ij
0
The 1-dimensional representations are well-defined: ρ0 (σ n ) = (±1)n = 1, because
n is even. As for the 2-dimensional representations, ρn−i ∼
= ρi , and ρ0 and ρn/2 are
reducible. The representations ρi for 0 < i < n/2 are irreducible. The sums of the
squares of their dimensions is 4 · 12 + (n/2 − 1) · 22 = 2n = |Dn |, so there are no
other irreducibles.
Case of n odd. The only difference when n is odd is that there are just two
1-dimensional representations: ρ01 and ρ02 . The dimension count is now 2 · 12 + (n −
1)/2 · 22 = 2n = |Dn |.
Galois action. The Galois group Gal(K/k) is generated by the qth power map:
α ∈ K 7→ αq . Because the representations ρ01 , ρ02 , (and ρ03 , ρ04 , when n is even) have
coefficients ±1 ∈ k, these representations are invariant under Gal(K/k). They have
intertwining algebras I(ρ0i , ρ0i ) = k for all i.
Because the qth power map sends ω i to ω qi , the representation ρi is sent to
ρqi . Also recall that ρn−i ∼
= ρi . With these two facts, one can work out the orbits
of the Galois action and the intertwining algebras for any particular example.
Example 12.4. Let us examine in more detail the case of G = D5 . Let k = Fp
be a finite field of prime order, with p 6= 2, 5. In order to guarantee that a finite
extension K = Fpm of k has 5th roots of unity, m should be the order of p (mod 5),
by Lemma 10.1. It is easy to check that the order m of p (mod 5) is
1, p ≡ 1 (mod 5),
m = 2, p ≡ 4 (mod 5),
4, p ≡ 2, 3 (mod 5).
When p ≡ 1 (mod 5), k = Fp is itself a splitting field for D5 . As described
above, there are two 1-dimensional irreducible k-representations (ρ01 and ρ02 ) and
36
JAY A. WOOD
two 2-dimensional irreducible k-representations (ρ1 and ρ2 ). Then k[D5 ] ∼
= k⊕k⊕
M2×2 (k) ⊕ M2×2 (k). (Also, the pth power map fixes each of the representations.)
When p ≡ 4 (mod 5), the pth power maps sends ρ1 to ρ4 ∼
= ρ1 . Similarly, ρ2
is sent to ρ8 = ρ3 ∼
= ρ2 . Thus, both representations are Galois-invariant, and their
intertwining algebras are k itself. Then k[D5 ] ∼
= k ⊕ k ⊕ M2×2 (k) ⊕ M2×2 (k).
When p ≡ 2, 3 (mod 5), the pth power map interchanges ρ1 ∼
= ρ4 and ρ2 ∼
= ρ3 .
∼
∼
Thus ρ1 ↓k = ρ2 ↓k , with intertwining algebra Fp2 . Then k[D5 ] = k ⊕k ⊕M2×2 (Fp2 ).
Case where p = 3. We will use the notations for F9 and F81 from Example 11.3.
One can verify that the following matrices provide representations of D5 over F9 :
ρ1 (σ) =
ρ2 (σ) =
0
−1
0
−1
1
ζ
1
0
1
1
0
0
1
1
0
ρ1 (τ ) =
ρ2 (τ ) =
ζ3
While these representations are inequivalent over F9 , their restrictions over F3 are
equivalent:
0
0
ρ1 (σ) =
−1
0
0
0
ρ2 (σ) =
−1
0
0 1
0
0 0
1
0 0
1
−1 1 −1
0
1
0
0
0
1
0 −1 −1
−1 −1
0
0
0
ρ1 (τ ) =
1
0
0
0
ρ2 (τ ) =
1
0
0
0
0
1
1
0
0
0
0
0
0
1
1
0
0
0
0
1
0
0
0
1
0
0
These are intertwined by any matrix of the form
0
1
a
0
0
−1 0
0
−1 1
0 1
0 0
0
+ b
0 0
0 0 −1
0 1
0
0 0
0 0
0 0
,
−1 1
0 1
a, b ∈ F3 .
Case when p = 11. Because F11 contains 5th roots of unity (the integers 3, 4, 5, 9
(mod 11)), F11 is itself a splitting field for D5 . Then F11 [D5 ] ∼
= F11 ⊕ F11 ⊕
M2×2 (F11 ) ⊕ M2×2 (F11 ). By using the inverse Fourier transform, Proposition 9.12,
the minimal ideals in the direct sum decomposition can be viewed as linear codes
of length 10 over F11 with respect to the basis e, σ, σ 2 , σ 3 , σ 4 , τ, τ σ, τ σ 2 , τ σ 3 , τ σ 4
FOURIER TRANSFORM
37
(we will use a to denote 10, as in hexadecimal notation):
C1 :1111111111
C2 :11111aaaaa
C3 :9541300000
0000093145
0000095413
9314500000
C4 :9435100000
0000091534
0000094351
9153400000
Each of these codes exhibits D5 symmetry, and Ci is orthogonal to Cj , i 6= j, under
the standard dot product over F11 .
Case when p = 19. We saw above that k[D5 ] ∼
= k ⊕ k ⊕ M2×2 (k) ⊕ M2×2 (k).
One can verify that the following matrices provide irreducible representations of
D5 over F19 :
ρ1 (σ) =
ρ2 (σ) =
2
4
7
3
15
2
16
7
0
1
1
0
0
1
1
0
ρ1 (τ ) =
ρ2 (τ ) =
By using the inverse Fourier transform, Proposition 9.12, the minimal ideals in
the direct sum decomposition can be viewed as linear codes of length 10 over F19
with respect to the basis e, σ, σ 2 , σ 3 , σ 4 , τ, τ σ, τ σ 2 , τ σ 3 , τ σ 4 (we use c, g, and i to
38
JAY A. WOOD
denote 12, 16, and 18, à la hexadecimals):
C1 :1111111111
C2 :11111iiiii
C3 :489980g7c3
03c7g48998
0g7c348998
4899803c7g
C4 :498890cg37
073gc49889
0cg3749889
49889073gc
Each of these codes exhibits D5 symmetry, and Ci is orthogonal to Cj , i 6= j, under
the standard dot product over F19 .
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