Eisenstein Series
4.9 The Fourier transform and the Mellin transform
Similarly, use Theorem 4.5.2 to find formulas forr(n,6) andr(n,8).
(d) The proof of Proposition 5.9.1 to follow will establish that the Fourier coefficients of any cusp formf ∈ Sk(Γ) satisfy |an(f)| ≤Cnk/2 for someC.
Lets≥10 be even. Argue that the Fourier coefficients of the Eisenstein series in Ms/2(Γ1(4)) dominate those of the cusp forms inSs/2(Γ1(4)) as n→ ∞, and so the methods of this exercise give an asymptotic solution to thessquares problem by ignoring the cusp forms.
0.2 0.4 0.6 0.8 1 2
4 6 8
Figure 4.1.The first few terms ofϑ(it,1) and their sum
ˆ ϕ(x) =
∞
y=−∞
e−π(y2+2iyx−x2)e−πx2dy=e−πx2 ∞
y=−∞
e−π(y+ix)2dy.
Complex contour integration shows that the integral is just the Gaussian integral ∞
−∞e−πy2dy (Exercise 4.9.3(a)), and this is 1 (Exercise 4.9.3(b)).
Thus ˆf = f when l = 1. To prove the same result for l > 1, note that f takes the product formf(x) =ϕ(x1)· · ·ϕ(xl) for any vectorx= (x1, . . . , xl), and consequently the Fourier transform is ˆf(x) = ˆϕ(x1)· · ·ϕ(xˆ l), i.e., ˆf =f as desired (Exercise 4.9.3(c)). Also, for any function h ∈ L1(Rl) and any positive number r, the Fourier transform of the function h(xr) isr−lh(x/r)ˆ (Exercise 4.9.3(d)), so in particular the Fourier transform of f(xt1/2) is t−l/2f(xt−1/2).
For any functionh∈ L1(Rl) such that the sum
d∈Zlh(x+d) converges absolutely and uniformly on compact sets and is infinitely differentiable as a function ofx, thePoisson summation formulais
d∈Zl
h(x+d) =
m∈Zl
ˆh(m)e2πim,x.
For example, formula (1.2), which we have used to compute the Fourier expansions of Eisenstein series, is a special case of one-dimensional Pois- son summation known as the Lipschitz formula (Exercise 4.9.4). Letting h(x) be the Gaussian f(xt1/2), Poisson summation with x = 0 shows that f(nt1/2) =t−l/2
f(nt−1/2), i.e.,
ϑ(i/t, l) =tl/2ϑ(it, l), t >0. (4.36) By the Uniqueness Theorem from complex analysis this relation extends to
ϑ(−1/τ, l) = (−iτ)l/2ϑ(τ, l), τ∈ H.
Here −iτ lies in the right half plane, and the square root is defined for all complex numbers except the negative reals, extending the real positive square root function of positive real numbers. This transformation law should look
familiar from Section 1.2 since the function that arose there from the four squares problem isθ(τ, l) =ϑ(2τ, l).
The properties of the Riemann zeta function are established by examining theMellin transformof (essentially) the theta function. In general, the Mellin transform of a functionf :R+−→Cis the integral
g(s) = ∞
t=0
f(t)tsdt t
for s-values such that the integral converges absolutely. (See Exercise 4.9.5 for the inverse Mellin transform.) For example, the Mellin transform ofe−t is Γ(s). Setting l = 1 and writing ϑ(τ) rather than ϑ(τ,1) for the duration of this section, consider the Mellin transform of the function∞
n=1e−πn2t= 1/2(ϑ(it)−1) fort >0,
g(s) = ∞
t=0
∞ n=1
e−πn2ttsdt t =1
2 ∞
t=0
(ϑ(it)−1)tsdt
t . (4.37) Sinceϑ(it) converges to 1 ast→ ∞, the transformation law (4.36) shows that as t → 0, ϑ(it) grows at the same rate as t−1/2, and therefore the integral g(s) converges at its left endpoint if Re(s)>1/2. And since the convergence ofϑ(it) to 1 as t→ ∞is rapid, the integral converges at its right end for all values ofs. Rapid convergence lets the sum pass through the integral in (4.37) to yield, after a change of variable,
g(s) = ∞ n=1
(πn2)−s ∞
t=0
e−ttsdt
t =π−sΓ(s)ζ(2s), Re(s)>1/2.
Thus the integral g(s/2) is the function ξ(s) = π−s/2Γ(s/2)ζ(s) from Sec- tion 4.4 when Re(s)>1. That is, the Mellin transform of the 1-dimensional theta function is the Riemann zeta function multiplied by a well understood factor. Thinking in these terms, the factorπ−s/2Γ(s/2) is intrinsically asso- ciated toζ(s), makingξ(s) the natural function to consider.
The second integral in (4.37) provides the meromorphic continuation and functional equation of ξ. Computing part of g(s/2) by splitting off a term, replacingt by 1/t, using the transformation law (4.36) for ϑ(it) withl = 1, and splitting off another term gives (Exercise 4.9.6)
1 2
1 t=0
(ϑ(it)−1)ts/2dt t =1
2 1
t=0
ϑ(it)ts/2dt t −1
s
=1 2
∞
t=1
ϑ(i/t)t−s/2dt t −1
s
=1 2
∞
t=1
ϑ(it)t(1−s)/2dt t −1
s
=1 2
∞
t=1
(ϑ(it)−1)t(1−s)/2dt t −1
s − 1 1−s.
And combining this with the remainder of the integralg(s/2), ξ(s) = 1
2 ∞
t=1
(ϑ(it)−1)(ts/2+t(1−s)/2)dt t −1
s− 1
1−s, Re(s)>1. (4.38) But the integral in (4.38) is entire ins, making the right side holomorphic everywhere in thes-plane except for simple poles at s = 0 ands = 1. And the right side is clearly invariant unders→1−s. Thus the functionξ(s) has a meromorphic continuation to the full s-plane and satisfies the functional equation ξ(1−s) = ξ(s). The properties of the Riemann zeta function as stated in Section 4.4 now follow.
Exercises
4.9.1.Show that the Fourier transform is continuous. (A hint for this exercise is at the end of the book.)
4.9.2.This exercise shows that the seriesϑ(τ, l) =
n∈Zleπi|n|2τ is holomor- phic.
(a) Let Sm ={n ∈Zl : max{|n1|, . . . ,|nl|}=m} for each natural num- berm. Note that|n|2≥m2for alln∈Sm. Show that|Sm| ≤l(2m+ 1)l−1.
(b) Let K ⊂ H be compact. Thus there exists some y0 > 0 such that Im(τ)≥y0for allτ∈K. Show that for anyτ ∈Kandn∈Sm,
|eπi|n|2τ| ≤e−πm2y0.
(c) There exists a natural number M0 such that if m≥M0 thenl(2m+ 1)l−1 < em and πm2y0 >2m. Show that for any τ ∈ K and any M ≥M0, the corresponding tail of the seriesϑ(τ, l) satisfies
n∈Zl
|n|2≥M
|eπi|n|2τ| ≤ ∞ m=M
e−m=e−M(1−e−1)−1.
This is less than arbitrary ε > 0 for large enough M independently of τ, showing that the seriesϑ(τ, l) converges absolutely and uniformly on compact subsets ofH. Since each summand is holomorphic, so is the sum.
4.9.3.(a) Use complex contour integration to show that ∞
t=−∞e−π(t+ix)2dt equals∞
−∞e−πt2dt.
(b) Show that this Gaussian integral is 1. (A hint for this exercise is at the end of the book.)
(c) Show that if f ∈ L1(Rl) takes the form of a product f(x) = f1(x1)· · ·fl(xl) then its Fourier transform is the corresponding product fˆ(x) = ˆf1(x1)· · ·fˆl(xl).
(d) Show that for any functionh∈ L1(Rl) and any positive numberr, the Fourier transform of the functionh(xr) isr−lh(x/r).ˆ
4.9.4.Formula (1.2) is a special case of the Lipschitz formulaforτ∈ H and q=e2πiτ,
d∈Z
1
(τ+d)s = (−2πi)s Γ(s)
∞ m=1
ms−1qm, Re(s)>1.
Prove the Lipschitz formula by applying Poisson summation to the function f ∈ L1(R) given by
f(x) =
xs−1e2πixτ ifx >0,
0 ifx≤0.
(A hint for this exercise is at the end of the book.)
4.9.5.Ifg is a holomorphic function of the complex variablesin some right half plane then itsinverse Mellin transformis
f(t) = 1 2πi
σ+i∞ s=σ−i∞
g(s)t−sds
for positivet-values such that the integral converges absolutely. Complex con- tour integration shows that this is independent ofσ.
(a) Let t=ex,s=σ+ 2πiy,f(t) = e−σxfσ(x), andg(s) = gσ(y). Show that
g(s) = ∞
t=0
f(t)tsdt
t ⇐⇒ gσ(y) = ∞
x=−∞
fσ(x)e2πixydx and that
f(t) = 1 2πi
σ+i∞ s=σ−i∞
g(s)t−sds ⇐⇒ fσ(x) = ∞
y=−∞
gσ(y)e−2πiyxdy.
The right side conditions are equivalent by Fourier inversion, and so the left side conditions are equivalent as well. This is theMellin inversion formula.
(b) Evaluate the integral σ+i∞
s=σ−i∞Γ(s)x−sds for any σ > 0. (A hint for this exercise is at the end of the book.)
4.9.6.Verify the calculations leading up to formula (4.38).