Matrices of finite abelian groups, Finite Fourier Transform and codes

0.17 MB·English

by  Kanemitsu Shigeru

MATRICES OF FINITE ABELIAN GROUPS,
FINITE FOURIER TRANSFORM AND CODES
3 S. KANEMITSU AND M. WALDSCHMIDT
1
0
2 Abstract. Finite(orDiscrete)FourierTransforms(FFT)arees-

n sentialtoolsinengineeringdisciplinesbasedonsignaltransmission,
a which is the case in most of them. FFT are related with circulant
J
matrices, which can be viewed as group matrices of cyclic groups.

7 In this regard, we introduce a generalization of the previous in-

vestigations to the case of finite groups, abelian or not. We make
]
T clear the points whichwerenotrecognizedas underlyingalgebraic
N structures. Especially, all that appears in the FFT in engineering
. has been elucidated from the point of view of linear representa-
h
tions of finite groups. We include many worked-out examples for
t
a the readers in engineering disciplines.
m
[

1
1. The matrix of a finite abelian group
v
8
1.1. Matrix of a finite group. Let G be a finite group of order n,
4
2 and let F be a field of characteristic not dividing n. This setting will
1
be used throughout; we also assume that F contains a primitive n-th
.
1
root ζ = ζ of 1, but sometimes we will consider subfields of F which
0 n
3 do not satisfy this condition. We use the symbols j = ζ3 = e2πi/3 and
1 i = ζ = e2πi/4 to mean a primitive cube and fourth root of unity, the
4
:
v latter expressions valid in characteristic 0.
i
X Let X := (X ) bean n-tuple of variables indexed by the elements
σ σ∈G
r of G. The group matrix
a
AG := Xτ−1σ τ,σ∈G ∈ Matn×n F[X] ,
(depending on a labeli(cid:0)ng of e(cid:1)lements of G) h(cid:0)as be(cid:1)en introduced by
Dedekind in the course of his investigation on normal bases for Galois
extensions. In 1886, Frobenius gave a complete factorization of the
determinant of A into irreducible factors in F[X] – this was the start
G
of the theory of linear representations and characters of finite groups.
1.2. Matrix of a finite abelian group. We will consider the general
case of a finite group in §4; here we assume the group G to be abelian
and we take F = C. Let G be the dual of G, which is the group
Date: 13/11/2012.
b
1
Hom(G,C×) of characters of G. We will consider n–tuples of complex
numbers; when they are indexed by the elements of G, we say that
they are in CG; when they are indexed by the elements of G, we say
b
that they are in CG.
b
For each χ ∈ G, the vector
χ(σ) ∈ CG
b σ∈G
is an eigenvector of the ma(cid:0)trix A(cid:1)G belonging to the eigenvalue given
by the linear form
(1.1) Y := χ(σ)X .
χ σ
σ∈G
X
This follows from the relation, for χ ∈ G,
χ(σ)Xτ−1σ = χ(τ)b χ(σ)Xσ.
σ∈G σ∈G
X X
Therefore the n×n matrix
(1.2) P := χ(σ) ∈ Mat (C)
σ∈G,χ∈Gb n×n
is regular and (cid:0) (cid:1)
(1.3) A P = PD,
G
where D is the diagonal n×n matrix
D := Diag(Yχ)χ∈Gb := Yχδχ,ψ χ,ψ∈Gb.
We have used Kronecker’s symbol (cid:0) (cid:1)
1 if χ = ψ,
δ =
χ,ψ
0 if χ 6= ψ.
(
In particular the determinant of A , called determinant of the group
G
G (Gruppendeterminant in German – see the historical note of [B]), is
(1.4) detA = Y = χ(σ)X .
G χ σ
χY∈Gb χY∈GbσX∈G
This formula is used by Hasse [H] to give an explicit formula for the
number of ideal classes of an algebraic number field (see also [Yam]). It
is also useful for computing the p–adic rank of the units of an algebraic
number field [Ax].
The dual G of G, which is called the bidual of G, is canonically
isomorphic to G, the characters of G being given by χ 7→ χ(σ) for
bb b 2
b
σ ∈ G. Denoting by U the group of n–th roots of unity in C, namely
n
the set of complex roots of the polynomial Xn −1, the pairing
G×G −→ U
n
(σ,χ) 7−→ χ(σ)
b
is non–degenerate.
In parallel to the case of the dual of G, we introduce their counter-
parts. Correspondence can be seen in the table below.
Let T := (Tχ)χ∈Gb be an n-tuple of variables indexed by G. The
matrix Ab ∈ C[X] of the dual of G is:
G
b
AGb = Tψ−1χ ψ,χ∈G ∈ Matn×n C[X] .
For each σ ∈ G, the ve(cid:0)ctor (cid:1) (cid:0) (cid:1)
b
χ(σ) ∈ CG
b
χ∈G
is an eigenvector of the ma(cid:0)trix A(cid:1) b belonging to the eigenvalue given
G
by the linear form
(1.1’) U := χ(σ)T .
σ χ
b
χX∈G
This follows from the relation, for σ ∈ G,
χ(σ)Tψ−1χ = ψ(σ) χ(σ)Tχ.
b b
χX∈G χX∈G
Therefore the transpose tP of the matrix P given by (1.2), namely
(1.2’) tP := χ(σ) ∈ Mat (C)
χ∈Gb,σ∈G n×n
satisfies (cid:0) (cid:1)
(1.3’) AbtP =tPD,
G
where D is the diagonal n×n matrix
b
b D := Diag(Uσ)σ∈Gb.
b
Table. Correspondence between G and G
group n-tuples eigenvalues vectors
G X Y χ(σ) b
σ χ σ∈G
G Tχ Uσ χ(σ)χ∈Gb
3
b
1.3. Matrix of a cyclic group. We consider here the special case
where the group G is the cyclic group C of order n. Let σ be a gener-
n 1
ator of G and χ a generator of the cyclic group G. Then the number
1
ζ = χ (σ ) is a primitive n–th root of unity which we have assumed
1 1
to belong to F. We have G = {1,σ ,...,σn−1}, Gb= {1,χ ,...,χn−1}
1 1 1 1
and we set X = X and Y = Y . Then
i σ1i ℓ σ1ℓ
b
X X X ··· X
0 1 2 n−1
X X X ··· X
n−1 0 1 n−2
ACn =  ... ... ... ... ... 
 X X X ··· X 
 2 3 4 1 
 X X X ··· X 
 1 2 3 0 
 
is a circulant (see [Dav]) and may be expressed as n−1X Kℓ (cf.
ℓ=0 ℓ
§1.4), where K is the n×n matrix which is the specialization of A
P Cn
at
(X ,X ,X ,...,X ) = (0,1,0,...,0).
0 1 2 n−1
Since χ (σℓ) = ζiℓ, (1.1) reads
i 1
n−1
Y = ζiℓX .
ℓ i
i=0
X
The matrix P is
1 1 1 ··· 1
1 ζ ζ2 ··· ζn−1
P = ζij = 1 ζ2 ζ4 ··· ζ2(n−1) ,
0≤j,j≤n−1 ... ... ... ... ... 
(cid:0) (cid:1)  
1 ζn−1 ζ2(n−1) ··· ζ(n−1)(n−1)
 
 
where the exponent of ζ is given by the multiplication table of the ring
Z/nZ. The determinant ∆ of P is considered by Massey in [Mas]. It
n
is a Vandermonde determinant, with its value
n−1ℓ−1
∆ = ζi(ζℓ−i −1).
n
ℓ=1 i=0
YY
For instance
1 1
∆ = 1, ∆ = det = −2,
1 2 1 −1
(cid:18) (cid:19)
1 1 1
∆ = det 1 j j2 = 3j(j −1),
3
 
1 j2 j
4
 
1 1 1 1
1 i −1 −i
∆ = det  = 16i,
4 1 −1 1 −1
1 −i −1 i 
 
where j = ζ and i = ζ arethe primitive third root and fourth root of
3 4
1, respectively. In general, the sum of the n rows is
(n,0,0,...0),
andthe determinant ofP isntimes thedeterminant ∆′ ofthe(n−1)×
n
(n−1) matrix ζij . If n is prime, after a suitable permutation
1≤i,j≤n−1
of the rows, one can write ∆′ as a circulant determinant with first row
(cid:0) (cid:1) n
(ζ,ζ2,...,ζn−1).
There are various subfields of F over which one can decompose the
group determinant into a product of irreducible factors.
Firstly, over F itself (which contains the n-th roots of unity), the
decomposition is given by (1.4).
Secondly, in characteristic zero, over Q, the polynomial Xn−1 splits
as
(1.5) Xn −1 = Φ (X),
d
d|n
Y
whereΦ isthecyclotomicpolynomialofindexd,whichisanirreducible
d
polynomial in Z[X] of degree ϕ(d). Let ζ be a root of Φ (i.e. a
d d
primitive d–th root of unity). Then it generates the d-th cyclotomic
field over Q
(1.6) Γ := Q[X]/(Φ (X)) = Q(ζ ).
d d d
Accordingly, the group determinant (1.4) of a cyclic group of order
n splits into a product of irreducible polynomials in Q[X]
detA = ψ (X),
G d
d|n
Y
where the homogeneous polynomial ψ ∈ Q[X] is given by the norm
d
N of Γ
Γd/Q d
ψ (X) = N (X +ζ X +···+ζn−1Xn−1).
d Γd/Q 0 d 1 d
For instance, for n = 3, the determinant of the cyclic group C of order
3
3 is
X X X
0 1 2
detA = X X X = X3 +X3 +X3 −3X X X ,
C3 (cid:12) 2 0 1(cid:12) 0 1 2 0 1 2
(cid:12)X X X (cid:12)
1 2 0
(cid:12) (cid:12)
(cid:12) (cid:12) 5
(cid:12) (cid:12)
(cid:12) (cid:12)
over C the decomposition (1.4) is
(1.7) (X +X +X )(X +jX +j2X )(X +j2X +jX ),
0 1 2 0 1 2 0 1 2
while over Q the decomposition is
(1.8) (X +X +X )(X2 +X2 +X2 −X X −X X −X X ),
0 1 2 0 1 2 0 1 1 2 2 0
where the second factor is
N (X +jX +j2X ).
Q(j)/Q 0 1 2
Thirdly, in finite characteristic, over a finite field F with q elements
q
(and gcd(q,n) = 1), the decomposition of the determinant of the cyclic
group C is given by the decomposition of the cyclotomic polynomials
n
Φ , with d ranging over the set of divisors of n, over F . For such a
d q
d, let r be the order of q in the multiplicative group (Z/dZ)×. Then
Φ splits in F [X] into ϕ(d)/r polynomials, all of the same degree r. If
d q
H is the subgroup generated by the class q modulo d in (Z/dZ)×, the
choice of a primitive d-th root of unity ζ gives rise to an irreducible
d
factor
P (X) = (X −ζh),
H d
h∈H
Y
and all factors of Φ are obtained by taking the ϕ(d)/r classes of
d
(Z/dZ)× modulo H; for any m ∈ (Z/dZ)×, set
P (X) = (X −ζmh).
mH d
h∈H
Y
Then the decomposition of Φ into irreducible factors over F is
d q
Φ (X) = P (X).
d mH
mH∈(ZY/dZ)×/H
Here is another description of the decomposition of the polynomial
Xn−1intoirreducible factorsoverF . The2n factorsofthepolynomial
q
Xn −1 over a field containing a primitive n–th root of unity ζ are
Q (X) = (X −ζℓ), (L ⊆ {1,...,n})
L
ℓ∈L
Y
(with Q = 1, as usual), and such a polynomial belongs to F [X] if
∅ q
and only if Q (X)q = Q (Xq). This condition is satisfied if and only
L L
if the label set L of {1,...,n} is stable under multiplication by q in
Z/nZ. Hence the irreducible factors of Xn−1 over F are the Q with
q L
L stable under multiplication by q and minimal for this property.
Once we know the decomposition of Xn −1, we deduce the decom-
position of the determinant of the cyclic group G of order n.
6
Example. Consider the cyclic group C of order 3, assuming that the
3
characteristic isnot3. Forq ≡ 1 mod 3, thepolynomialX3−1hasthe
decomposition (1.7) with three homogeneous linear factors (because F
q
contains the primitive cubic roots of 1), while for q ≡ 2 mod 3, the
polynomial X3−1 has the decomposition (1.8) with one homogeneous
linear factor and one irreducible factor of degree 2 (because X2+X+1
is irreducible over F ).
q
1.4. The group ring of a cyclic group and the algebra of circu-
lants. Recall that F is a field whose characteristic does not divide n.
Let K denote the n×n circulant matrix with its first row (0,1,0,...,0)
(often referred to as the shift-forward matrix), where a circulant matrix
is one whose rows consists of the n cycles
(c ,...,c ), (c ,c ,...,c ), ...,
0 n−1 n−1 0 n−2
which therefore can be written as
c c c ··· c
0 1 2 n−1
c c c ··· c
n−1 0 1 n−2
 ... ... ... ... ...  = c0I +c1K +···+cn−1Kn−1,
 c c c ··· c 
 2 3 4 1 
 c c c ··· c 
 1 2 3 0 
 
so that the algebra of circulant n × n matrices is nothing other than
F[K], which we denote by R subsequently. Further, the minimal poly-
nomial of K is Tn −1. Hence F[K] is isomorphic to F[T]/(Tn −1).
If C denotes a cyclic group of order n, then the algebra F[C ],
n n
called the group ring of C over F, is also isomorphic to F[T]/(Tn−1).
n
Altogether,
(1.9) R = F[K] ≃ F[T]/(Tn −1) ≃ F[C ].
n
Assume F contains a primitive n–th root ζ of unity. We split the
polynomial Tn −1 into irreducible factors over F, say
n−1
Tn −1 = (T −ζℓ).
ℓ=0
Y
Then the algebra F[C ] splits accordingly into a product of n algebras,
n
all isomorphic to F:
n−1
(1.10) F[C ] ≃ F[T]/(T −ζℓ).
n
ℓ=0
Y 7
For 0 ≤ ℓ ≤ n − 1, denote by R the subset of R which is the image
ℓ
of the factor F[T]/(T −ζℓ) (0 ≤ ℓ ≤ n−1) on the right hand side of
(1.10). Then R is a simple F– algebra and R = R ×···×R . Let
ℓ 0 n−1
E be the unity element of R . Then we have the decomposition into
ℓ ℓ
orthogonal idempotents
R = RE , 1 = E +···+E and E E = δ (0 ≤ i,ℓ ≤ n−1).
ℓ ℓ 0 n−1 i ℓ iℓ
For the structure theorem of semi–simple rings, see for instance [La],
Th. 4.4 in Chap. XVIIor [B]. The special case ofthe algebra R = F[K]
of circulants of order n is worked out in [Wil]: the solution is
n−1
1
(1.11) E = ζ−hℓKℓ, (0 ≤ h ≤ n−1).
h
n
ℓ=0
X
In the other direction we have
n−1
Kh = ζhℓE , (0 ≤ h ≤ n−1).
ℓ
ℓ=0
X
These formulae are easy to check, but it is interesting to explain
where they come from. The isomorphism (1.10) from F[G] to the
product of the algebras F[T]/(T−ζℓ) maps the class modulo Xn−1 of
a polynomial P to the n–tuple (P(ζℓ)) . We want to explicitly
0≤ℓ≤n−1
write down the inverse isomorphism. Given an n–tuple (b ) , one
ℓ 0≤ℓ≤n−1
deduces from the Chinese Remainder Theorem that there is a unique
polynomial P of degree ≤ n−1 such that P(ζh) = b for 0 ≤ h ≤ n−1.
h
To write down the solution P amounts to solving the associated inter-
polation problem, which is done by classical interpolation formulae. In
this specific case, they lead us to introducing the polynomial
1 Xn −1 1
P (X) = · = (Xn−1 +···+X +1).
0
n X −1 n
It satisfies P (1) = 1 and P (η) = 0 for any n–th root of unity η
0 1
not equal to 1. Hence for 0 ≤ h ≤ n − 1, the polynomial P (X) :=
h
P (X/ζℓ), which is
0
n−1
1
P (X) = ζ−hℓXℓ,
h
n
ℓ=0
X
satisfies
P (ζℓ) = δ for 0 ≤ h,ℓ ≤ n−1.
h h,ℓ
8
This is how (1.11) arises: E = P (K). Also, the solution of the
h h
interpolation problem is therefore the following: the polynomial
n−1
P(X) = b P (X)
h h
h=0
X
satisfies P(ζ ) = b for 0 ≤ h ≤ n−1.
h h
In characteristic 0, there is another basis for the circulant algebra,
which is rational over Q. Let n be a positive integer. We consider
the decomposition, into a product of simple algebras over Q, of the
semi–simple algebra Q[X]/(Xn−1) associated with the decomposition
(1.5) of the polynomial Xn −1 into irreducible factors over Q:
Q[X]/(Xn −1) = Γ ,
d
d|n
Y
where Γ is the d–th cyclotomic field defined by (1.6). For each divisor
d
d of n, define
Xn −1
Ψ (X) = = Φ (X).
n,d d′
Φ (X)
d
Yd′|n
d′6=d
Since Φ and Ψ are relatively prime, there is a unique polynomial
d n,d
Ψ of degree ≤ ϕ(d)−1 which is the inverse of Ψ modulo Φ :
n,d n,d d
Ψ Ψ ≡ 1 mod Φ .
n,d n,d d
e
Then a basis of the Q–algebra Q[C ] is given by
n
e
(1.12) E | 0 ≤ j ≤ ϕ(d)−1, d | n ,
d,j
where
(cid:8) (cid:9)
E ≡ XjΨ (X)Ψ mod (Xn −1).
d,j n,d n,d
As an example, consider the case where n = p is a prime. We have
Xp −1 e Xp −1
Ψ = = Φ , Ψ = = X −1 = Φ ,
p,1 p p,p 1
X −1 Φ
p
hence Ψ = 1/p. To compute Ψ , we start by taking the derivative
p,1 p,p
of Xp −1 = (X −1)Φ :
p
e pXp−1 = Φ (Xe)+(X −1)Φ′(X).
p p
Hence the polynomial
1 Xp−1 −1
Ψ := Φ′ −
p,p p p X −1
satisfies
e
1
(X −1)Ψ = 1− Φ .
p,p p
p
9
e
Therefore a basis of the circulant algebra with n = p is given by
((1/p)Φ ,F ,F ,...,F ), with
p 0 1 p−2
F ≡ Xj(X −1)Ψ (X) mod (Xp −1) (0 ≤ j ≤ p−2).
j p,1
For instance, when p = 3, we have
e
1 1
Ψ (X) = − (X +2), (X −1)Ψ (X) = − (X +2)(X −1),
3,1 3,1
3 3
e 1 e 1
(X−1)XΨ (X) = − (X+2)(X−1)X ≡ − (X−1)2 mod (X3−1),
3,1
3 3
and the basis of Q[X]/(X3−1) which is associated to the basis (1,0),
e
(0,1), (0,X) of the product
Q[X]/(X −1)×Q[X]/(X2 +X +1)
under the natural isomorphism is given by the classes modulo X3 −1
of the polynomials
1 1 1
(X2 +X +1), − (X2 +X −2), − (X2 −2X +1).
3 3 3
Remark. There is no element J in the algebra Q[X]/(X3 − 1) which
satisfies 1+J +J2 = 0. This is analogous to the fact that the product
algebra Q×Q[i] does not contain a square root of −1. In a product
A ×A of two algebras, there are in general no subalgebras isomorphic
1 2
to the factors A and A .
1 2
1.5. The group ring F[G] of a finite abelian group G. We extend
theresults of theprevious section tothe algebra F[G]of a finiteabelian
group G. Here we assume that F× contains a subgroup of order n,
where n is the order of G. Hence the characteristic of F does not
divide n.
According to Maschke’s Theorem (see for instance [La] Chap. XVIII,
§1, Th. 1.2; see also [Se] Chap. 6, Prop. 9 for the characteristic zero
case), the algebra F[G] is semi–simple: it is a product of n algebras
isomorphic to F. Under such an isomorphism, the canonical basis of
FG is associated with a basis (eχ)χ∈Gb of F[G] satisfying
e e = δ e .
χ ψ χ,ψ χ
An explicit solution is given by
1
e = χ−1(σ)σ (χ ∈ G).
χ
n
σ∈G
X 10
b