Download Intro to Complex Analysis in Number Theory: Riemann Zeta & Analytic Continuation and more Study notes Number Theory in PDF only on Docsity!
Math 259: Introduction to Analytic Number Theory The Riemann zeta function and its functional equation (and a review of the Gamma function and Poisson summation)
Recall Euler’s identity:
[ζ(s) :=]
∑^ ∞
n=
n−s^ =
p prime
∑^ ∞
cp=
p−cps
p prime
1 − p−s^
We showed that this holds as an identity between absolutely convergent sums and products for real s > 1. Riemann’s insight was to consider (1) as an identity between functions of a complex variable s. We follow the curious but nearly universal convention of writing the real and imaginary parts of s as σ and t, so s = σ + it.
We already observed that for all real n > 0 we have |n−s| = n−σ^ , because
n−s^ = exp(−s log n) = n−σ^ eit^ log^ n
and eit^ log^ n^ has absolute value 1; and that both sides of (1) converge absolutely in the half-plane σ > 1, and are equal there either by analytic continuation from the real ray t = 0 or by the same proof we used for the real case. Riemann showed that the function ζ(s) extends from that half-plane to a meromorphic function on all of C (the “Riemann zeta function”), analytic except for a simple pole at s = 1. The continuation to σ > 0 is readily obtained from our formula
ζ(s) −
s − 1
∑^ ∞
n=
[
n−s^ −
∫ (^) n+
n
x−s^ dx
]
∑^ ∞
n=
∫ (^) n+
n
(n−s^ − x−s) dx,
since for x ∈ [n, n + 1] (n ≥ 1) and σ > 0 we have
|n−s^ − x−s| =
∣s
∫ (^) x
n
y−^1 −s^ dy
∣ ≤ |s|n
− 1 −σ
so the formula for ζ(s) − (1/(s − 1)) is a sum of analytic functions converging absolutely in compact subsets of {σ + it : σ > 0 } and thus gives an analytic function there. (See also the first Exercise below.) Using the Euler-Maclaurin summation formula with remainder, we could proceed in this fashion, extending ζ to σ > −1, σ > −2, etc. However, once we have defined ζ(s) on σ > 0 we can obtain the entire analytic continuation at once from Riemann’s functional equation relating ζ(s) with ζ(1 − s). This equation is most nicely stated by introducing the meromorphic function ξ(s) defined by^1
ξ(s) := π−s/^2 Γ(s/2)ζ(s) (^1) Warning: occasionally one still sees ξ(s) defined as what we would call (s (^2) − s)ξ(s) or (s^2 − s)ξ(s)/2, as in [GR 1980, 9.561]. The factor of (s^2 − s) makes the function entire, and does not affect the functional equation since it is symmetric under s ↔ 1 − s. However, for most uses it turns out to be better to leave this factor out and tolerate the poles at s = 0, 1.
for σ > 0. Then we have:
Theorem (Riemann). The function ξ extends to a meromorphic function on C, regular except for simple poles at s = 0, 1 , which satisfies the functional equation
ξ(s) = ξ(1 − s). (2)
It follows that ζ also extends to a meromorphic function on C, which is regular except for a simple pole at s = 1, and that this analytic continuation of ζ has simple zeros at the negative even integers − 2 , − 4 , − 6 ,.. ., and no other zeros outside the closed critical strip 0 ≤ σ ≤ 1.
[The zeros − 2 , − 4 , − 6 ,... of ζ outside the critical strip are called the trivial zeros of the Riemann zeta function.]
The proof has two ingredients: properties of Γ(s) as a meromorphic function of s ∈ C, and the Poisson summation formula. We next review these two topics.
The Gamma function was defined for real s > 0 by Euler^2 as the integral
Γ(s) :=
0
xse−x^
dx x
We have Γ(1) =
0 e
−x (^) dx = 1 and, integrating by parts,
sΓ(s) =
0
e−xd(xs) = −
0
xsd(e−x) = Γ(s + 1) (s > 0),
so by induction Γ(n) = (n − 1)! for positive integers n. Since |xs| = xσ^ , the integral (3) defines an analytic function on σ > 0, which still satisfies the re- cursion sΓ(s) = Γ(s + 1) (proved either by repeating the integration by parts or by analytic continuation from the positive real axis). That recursion then extends Γ to a meromorphic function on C, analytic except for simple poles at 0 , − 1 , − 2 , − 3 ,.. .. (What are the residues at those poles?) For s, s′^ in the right half-plane σ > 0 the Beta function^3 B(s, s′), defined by the integral
B(s, s′) :=
0
xs−^1 (1 − x)s
′− 1 dx,
is related with Γ by Γ(s + s′)B(s, s′) = Γ(s)Γ(s′). (4)
(This is proved by Euler’s trick of calculating
0
0 x
s− (^1) ys′− (^1) e−(x+y) (^) dx dy
in two different ways.) Since Γ(s) > 0 for real positive s, it readily follows that Γ has no zeros in σ > 0, and therefore none in the complex plane.
This is enough to derive the poles and trivial zeros of ζ from the functional equation (2). [Don’t take my word for it — do it!] But where does the func- tional equation come from? There are several known ways to prove it; we give (^2) Actually Euler used Π(s − 1) for what we call Γ(s); thus Π(n) = n! for n = 0, 1 , 2 ,.. .. (^3) a.k.a. “Euler’s first integral”, with (3) being “Euler’s second integral”.
Theorem. Let f : R→C be a C^2 function such that (|x|r^ + 1)(|f (x)| + |f ′′(x)|) is bounded for some r > 1 , and let fˆ be its Fourier transform
f^ ˆ (y) =
−∞
e^2 πixy^ f (x) dx.
Then (^) ∞ ∑
m=−∞
f (m) =
∑^ ∞
n=−∞
f^ ˆ (n), (6)
the sums converging absolutely.
[The hypotheses on f can be weakened, but this formulation of Poisson sum- mation is more than enough for our purposes.]
Proof : Define F : R/Z→C by
F (x) :=
∑^ ∞
m=−∞
f (x + m),
the sum converging absolutely to a C^2 function by the assumption on f. Thus the Fourier series of F converges absolutely to F , so in particular
F (0) =
∑^ ∞
n=−∞
0
e^2 πinxF (x) dx.
But F (0) is just the left-hand side of (6), and the integral is
∑
m∈Z
0
e^2 πinxf (x + m) dx =
m∈Z
∫ (^) m+
m
e^2 πinxf (x) dx =
−∞
e^2 πinxf (x) dx
which is just fˆ (n), so its sum over n ∈ Z yields the right-hand side of (6).
Now let f (x) = e−πux
2
. The hypotheses are handily satisfied for any r, so (6) holds. The left-hand side is just θ(u). To evaluate the right-hand side, we need the Fourier transform of f , which is u−^1 /^2 e−πu − (^1) y 2 . [Contour integration reduces this claim to
−∞ e
−πux^2 dx = u− 1 / (^2) , which is the well-known Gauss
integral — see the Exercises.] Thus the right-hand side is u−^1 /^2 θ(1/u). Multi- plying both sides by u^1 /^2 we then obtain (5), and finally complete the proof of the analytic continuation and functional equation for ξ(s).
Remarks. We noted already that to each number field K there corresponds a zeta function
ζK (s) :=
I
|I|−s^ =
℘
(1 − |℘|−s)−^1 (σ > 1),
in which |I| is the norm of an ideal I, the sum and product extend respectively over ideals I and prime ideals ℘ of the ring of integers OK , and their equality
expresses unique factorization. As in our case of K = Q, this zeta function extends to a meromorphic function on C, regular except for a simple pole at s = 1. Moreover it satisfies a functional equation ξK (s) = ξK (1 − s), where
ξK (s) := Γ(s/2)r^1 Γ(s)r^2 (4−r^2 π−n|d|)s/^2 ζK (s),
in which n = r 1 + 2r 2 = [K : Q], the exponents r 1 , r 2 are the numbers of real and complex embeddings of K, and d is the discriminant of K/Q. The factors Γ(s/2)r^1 , Γ(s)r^2 may be regarded as factors corresponding to the “archimedean places” of K, as the factor (1 − |℘|−s)−^1 corresponds to the finite place ℘. The functional equation can be obtained from generalized Poisson summation as in [Tate 1950]. Most of our results for ζ = ζQ carry over to these ζK , and yield a Prime Number Theorem for primes of K; L-series generalize too, though the proper generalization requires some thought when the class and unit groups need no longer be trivial and finite as they are for Q. See for instance H.Heilbronn’s “Zeta-Functions and L-Functions”, Chapter VIII of [CF 1967].
Exercises
Concerning the analytic continuation of ζ(s):
- Show that if α : Z→C is a function such that
∑n m=1 α(m) =^ O(1) (for instance, if α is a nontrivial Dirichlet character) then
n=1 α(n)n
−s (^) converges
uniformly, albeit not absolutely, in compact subsets of {σ + it : σ > 0 }, and thus defines an analytic function on that half-plane. Apply this to
(1 − 21 −s)ζ(s) = 1 −
2 s^
3 s^
4 s^
(with α(n) = (−1)n−^1 ) and to (1 − 31 −s)ζ(s) to obtain a different proof of the analytic continuation of ζ to σ > 0.
- Prove that the Bernoulli polynomials Bn (n > 0) have the Fourier expansion
Bn(x) = −n!
k
′ (^) e^2 kπix (2kπi)n^
for 0 < x < 1, in which
k is the sum over nonzero integers^ k. Deduce that
ζ(n) =
(2π)n^
|Bn| n!
(n = 2, 4 , 6 , 8 ,.. .),
and thus that ζ(1 − n) = −Bn/n for all integers n > 1. For example, ζ(−1) = − 1 /12. What is ζ(0)?
It is known that in general ζK (−m) ∈ Q (m = 0, 1 , 2 ,.. .) for any number field K. In fact the functional equation for ζK indicates that once [K : Q] > 1 all the ζK (−m) vanish unless K is totally real and m is odd, in which case the rationality of ζK (−m) was obtained in [Siegel 1969].
A further application of (7):
- Now let χ 4 be the character mod 4 defined by χ 4 (±1) = ±1. Show that, again under the Poisson hypotheses,
∑^ ∞
m=−∞
χ 4 (m)f (m) =
∑^ ∞
n=−∞
χ 4 (n) fˆ (n/4).
This time, taking f (x) = e−πux 2 does not accomplish much! Use f (x) = xe−πux 2 instead to find a functional equation for L(s, χ 4 ).
We shall see that the L-function associated to any primitive Dirichlet character χ satisfies a similar functional equation, with the Gamma factor depending on whether χ(−1) = +1 or χ(−1) = −1.
- For light relief after all this hard work, differentiate the identity (5) with respect to u, set u = 1, and conclude that eπ^ > 8 π −2. What is the approximate size of the difference?
Further applications of Poisson summation:
- Use Poisson summation to evaluate
n=1 1 /(n
(^2) +c (^2) ) for c > 0. [The Fourier
transform of 1/(x^2 + c^2 ) is a standard exercise in contour integration.] Verify that your answer approaches ζ(2) = π^2 /6 as c→0.
- [Higher-dimensional Poisson, and more on zeta functions of quadratic forms] Let A be a real positive-definite symmetric matrix of order r, and Q : Rr^ →R the associated quadratic form Q(x) = (x, Ax). The theta function of Q is
θQ(u) :=
n∈Zr
exp(−πQ(n)u).
For instance, if r = 1 and A = 1 then θQ(u) is just θ(u). More generally, show that if A is the identity matrix Ir (so Q(x) =
∑r j=1 x
2 j ) then^ θQ(u) =^ θ(u)
r (^).
Prove an r-dimensional generalization of the Poisson summation formula, and use it to obtain a generalization of (5) that relates θQ(u) with θQ∗^ (1/u), where Q∗^ is the quadratic form associated to A−^1. Using this formula, and a Mellin integral formula for
ζQ(s) =
nn∈ 6 =0Zr
Q(n)s^
conclude that ζQ extends to a meromorphic function on C that satisfies a func-
tional equation relating ζQ with ζQ∗. Verify that when r = 2 and A = I 2 your functional equation is consistent with the identity ζQ(s) = 4ζ(s)L(s, χ 4 ) and the functional equations for ζ(s) and L(s, χ 4 ).
References
[CF 1967] Cassels, J.W.S., Fr¨ohlich, A., eds.: Algebraic Number Theory. Lon- don: Academic Press 1967. [AB 9.67.2 / QA 241.A42]
[GR 1980] Gradshteyn, I.S., Ryzhik, I.M.: Table of Integrals, Series, and Prod- ucts. New York: Academic Press 1980. [D 9.80.1 / basement reference QA55.G6613]
[Siegel 1969] Siegel, C.L.: Berechnung von Zetafunktionen an ganzzahligen Stellen, G¨ott. Nach. 10 (1969), 87–102.
[Tate 1950] Tate, J.T.: Fourier Analysis in Number Fields and Hecke’s Zeta- Functions. Thesis, 1950; Chapter XV of [CF 1967].
