Trace Formulae on Finite Abelian Groups

A First Course in Number Theory

4.5 Trace Formulae on Finite Abelian Groups

7. Letf ∈L²(G×G). Use Poisson summation to prove that

x∈G

f(x, x) = 1

|G|

χ∈G

(x,y)∈G×G

f(x, y)χ(x)χ(y).

Note that this identity is also an immediate consequence of the or- thogonality relations.

8. This is another example that shows that the lower bound in the uncer- tainty principle (Theorem 4.10) is best possible. LetH be a subgroup ofG, and define δH ∈L²(G) by

δH(x) =

1 ifx∈H 0 ifx∈H. (a) Prove that

supp(δH) =H.

(b) Prove that ifχ∈G, then δ!H(χ) =

* |H| ifχ∈G^H 0 ifχ∈G^H. (c) Prove that

supp(δH)supp δ!H

=|G|.

4.5 Trace Formulae on Finite Abelian Groups 145 then then×nmatrixA= (aij) = [T]_Bis called the matrix of the operator T with respect to the basisB.

LetB={v₁, . . . , v_n} be another basis forV, and let T(v_j) =

n i=1

a_ijv_i. (4.11)

ThenA = (a_ij) = [T]_B is the matrix ofT with respect to the basisB.

Each vectorv_j ∈ B is a linear combination of the vectors in the basisB, v_j=

n i=1

rijvi, (4.12)

and each vector vj ∈ B is a linear combination of the vectors in the basis B,

vj= n i=1

sijv_i. (4.13)

Consider the n×n matrices R = (rij) and S = (sij). Then S = R⁻¹ (Exercise 2). We have

T(v_j) = T _n

=1

r_jv

= n

=1

r_jT(v)

= n

=1

rj

n k=1

akvk

= n

=1

rj

n k=1

ak

n i=1

sikv_i

= n i=1

_n

k=1

n

=1

s_ika_kr_j

v_i.

Comparing this with (4.11), we obtain a_ij =

n k=1

n

=1

s_ika_kr_j

for alli, j= 1, . . . , n, and so

A =SAR=R⁻¹AR.

Identity (4.10) implies that

tr(A) =tr(R⁻¹AR) =tr(ARR⁻¹) = tr(A).

It follows that we can define the trace of a linear operatorT on a vector spaceV as the trace of the matrix ofT with respect to some basis forV, and that this definition does not depend on the choice of basis.

The vectorv ∈V is called aneigenvectorfor the operatorT witheigen- valueλifv= 0 andT(v) =λv. The operatorT isdiagonalizableif there exists a basis for V consisting of eigenvectors, that is, there exist nonzero vectorsv₁, . . . , v_n ∈V and numbersλ₁, . . . , λ_n such thatB={v₁, . . . , v_n} is a basis forV andT(v_i) =λ_iv_i fori= 1, . . . , n. In this case, the matrix forT with respect to the basisB is the diagonal matrix

D=









λ1 0 0 · · · 0 0 0 λ2 0 · · · 0 0 0 0 λ3 · · · 0 0

... ...

0 0 0 0 0 λn







 ,

and so

n i=1

aii= tr(A) = tr(D) = n i=1

λi.

We restate this important identity as a theorem.

Theorem 4.12 (Elementary trace formula) Let T be a linear opera- tor on an n-dimensional vector space V, let B be a basis for V, and let A = (aij) be the matrix of T with respect to B. If T is diagonalizable, then V has a basisB={v₁, . . . , v_n} of eigenvectors with T(v_i) =λ_iv_i for i= 1, . . . , n, and the trace ofA is equal to the sum of the eigenvalues ofT, that is,

n i=1

a_ii= n i=1

λ_i.

We shall show that both the Fourier inversion theorem and the Poisson summation formula are consequences of this elementary trace formula.

Let G be a finite abelian group of order n, and let L²(G) be the n- dimensional vector space of complex-valued functions onG. For everya∈G there is a linear operatorT_a onL²(G) defined byT_a(f)(x) =f(x−a). The operatorTa is called translation bya.

Another class of operators on L²(G) are integral operators. A function K ∈ L²(G×G) induces a linear operator ΦK on the vector space L²(G) as follows: For f ∈L²(G), let

Φ_K(f)(x) =

#

G

K(x, y)f(y)dy=

y∈G

K(x, y)f(y).

The map ΦK is called an integral operatoronL²(G) withkernelK(x, y).

4.5 Trace Formulae on Finite Abelian Groups 147 LetG={x1, . . . , xn}. Associated to the kernelKis a matrixA= (aij)∈ Mn(C) defined by

aij=K(xi, xj). (4.14) Conversely, to every matrixA= (a_ij)∈M_n(C) there is a functionK(x, y)∈ L²(G×G) defined by (4.14), and an associated integral operator Φ_K. Theorem 4.13 Let G={x1, . . . , xn} be an abelian group of order n. Let K ∈L²(G×G) and let ΦK be the associated integral operator on L²(G).

The matrix ofΦ_K with respect to the orthonormal basis{δ_xi :i= 1, . . . , n} is(K(x_i, x_j)), and the trace of Φ_K is

tr(ΦK) = n i=1

K(xi, xi). (4.15)

Proof. The matrix of the operator ΦK is (cij), where cij is defined by Φ_K(δ_xj) =

n i=1

c_ijδ_xi.

Then

cij = ΦK(δxj)(xi) =

y∈G

K(xi, y)δxj(y) =K(xi, xj).

This completes the proof.2

Theorem 4.14 Let G be a finite abelian group. LetK∈L²(G×G)with ΦK the associated integral operator onL²(G). The operator ΦK commutes with all translationsTa, that is,

TaΦK(f) = ΦKTa(f)

for alla∈Gandf ∈L²(G), if and only if there exists a functionh∈L²(G) such thatK(x, y) =h(x−y)for allx, y∈G. In this case,ΦK is convolution by h, that is,

ΦK(f)(x) =h∗f(x) =

#

G

h(x−y)f(y)dy, and the trace ofΦ_K is

tr(ΦK) =nh(0).

Proof. Let f, h ∈ L²(G). We define the convolution operator C_h on L²(G) by

Ch(f)(x) =h∗f(x) =

#

G

h(x−y)f(y)dy=

y∈G

h(x−y)f(y).

(See Exercise 10 in Section 4.3.) DefineK(x, y)∈L²(G×G) byK(x, y) = h(x−y). Then

ΦK(f)(x) =

#

G

K(x, y)f(y)dy=

#

G

h(x−y)f(y)dy=Ch(f)(x), and ΦK is convolution byh. Fora, x∈G, we have

T_aC_h(f)(x) = C_h(f)(x−a)

=

y∈G

h(x−a−y)f(y)

=

y∈G

h(x−y)f(y−a)

=

y∈G

h(x−y)Ta(f)(y)

= ChTa(f)(x),

and so TaCh=ChTa, that is, convolution commutes with translations.

Conversely, let K(x, y)∈L²(G×G). For a, x∈ Gand f ∈L²(G), we have

TaΦK(f)(x) = ΦK(f)(x−a) =

y∈G

K(x−a, y)f(y) and

Φ_KT_a(f)(x) =

y∈G

K(x, y)T_a(f)(y)

=

y∈G

K(x, y)f(y−a)

=

y∈G

K(x, a+y)f(y).

If Φ_K commutes with translations, thenT_aΦ_K = Φ_KT_a, and

y∈G

K(x−a, y)f(y) =

y∈G

K(x, a+y)f(y).

Applying this identity to the function f(x) =δ0(x) =

1 ifx= 0 0 ifx= 0.

we obtain K(x−a,0) = K(x, a) for all a, x ∈ G. Define the function h∈L²(G) by

h(x) =K(x,0).

4.5 Trace Formulae on Finite Abelian Groups 149 Then

K(x, y) =K(x−y,0) =h(x−y)

for all x, y ∈ G, and the operator Φ_K is convolution by h(x). Moreover, tr(ΦK) =nh(0) by (4.15). This completes the proof. 2

Theorem 4.15 (Trace formula) Forh∈L²(G), let C_h be the convolu- tion operator on L²(G), that is, C_h(f) = h∗f for f ∈ L²(G). The dual groupGis a basis of eigenvectors forC_h. Ifχ is a character inG, thenχ has eigenvalueh(χ), that is,

Ch(χ) =h(χ)χ, and

nh(0) =

χ∈G

h(χ).

Proof. This is a straightforward calculation. Forx∈G, we have C_h(χ)(x) = h∗χ(x) =χ∗h(x)

=

y∈G

χ(x−y)h(y)

=





y∈G

h(y)χ(y)



χ(x)

= h(χ)χ(x),

and so χis an eigenvector of the convolutionC_h with eigenvalueh(χ). By Theorem 4.12, since Gis a basis for L²(G), the trace ofC_h is the sum of the eigenvalues, that is,

tr(Ch) =

χ∈G

h(χ).

By Theorem 4.14, we also have

tr(C_h) =nh(0).

This completes the proof.2

We can immediately deduce the Fourier inversion formula (Theorem 4.8) from Theorem 4.15. Iff ∈L²(G), then

f(0) = 1 n

χ∈G

f(χ). (4.16)

This trace formula can also be obtained by computing the Fourier series forf atx= 0. On the other hand, if we simply apply (4.16) to the function T_−a(f) and use Exercise 9 in Section 4.3, then we obtain

f(a) = T_−a(f)(0)

= 1

n

χ∈G

"

T₋a(f)(χ)

= 1

n

χ∈G

f(χ)χ(a).

This is the Fourier inversion formula.

Next, we derive the Poisson summation formula (Theorem 4.11) from the elementary trace formula.

LetHbe a subgroup ofG, and letπ:G→G/Hbe the natural map. For x∈G, definex =π(x) =x+H ∈G/H. There is an orthonormal basis for the vector space L²(G/H) that consists of the functions δ_x, where

δ_x(y) =

1 ifx=y 0 ifx=y. Forf ∈L²(G), define the functionf∈L²(G/H) by

f(x+H) =

y∈H

f(x+y).

Let C_f be convolution by f onL²(G/H). The operatorC_f has matrix f(x−y)

, with respect to the basis{δx}. By Theorem 4.14, the trace ofC_f is

tr(C_f) =|G/H|f(0) = |G|

|H|

y∈H

f(y).

By Theorem 4.15, the character group G/H" is a basis of eigenvectors for the convolution operatorC_y. Ifχ∈G/H" andχ=π(χ)∈G^H, then

C_f(χ) =f(χ)χ, with eigenvalue

f(χ) =

x∈G/H

f(x)χ(x)

=

x∈G/H

y∈H

f(x+y)χ(x)

=

x∈G/H

y∈H

f(x+y)χ(x+y)

=

x∈G

f(x)χ(x).