Eisenstein Series
4.4 Gamma, zeta, and L -functions
4.3.3.(a) Show thatMk(N,1) =Mk(Γ0(N)).
(b) Show thatMk(N, χ) ={0} unlessχ(−1) = (−1)k. 4.3.4.(a) Show that Mk(Γ1(N)) =
χMk(N, χ). (A hint for this exercise is at the end of the book.)
(b) Show the same result for the cusp form spacesSk(N, χ).
(c) Show the same result for the Eisenstein spacesEk(N, χ).
in the next section.
TheRiemann zeta functionof the complex variable sis ζ(s) =
∞ n=1
1
ns, Re(s)>1.
The sum converges absolutely. An elegant restatement due to Euler of the Fundamental Theorem of Arithmetic (positive integers factor uniquely into products of primes) is
ζ(s) =
p∈P
(1−p−s)−1, Re(s)>1, whereP is the set of primes. Define
ξ(s) =π−s/2Γ(s/2)ζ(s), Re(s)>1.
Section 4.9 will prove that the functionξhas a meromorphic continuation to the entires-plane satisfying the functional equation
ξ(s) =ξ(1−s) for alls∈C
and having simple poles ats= 0 ands= 1 with respective residues−1 and 1.
By the properties of the gamma function, this shows that the Riemann zeta function has a meromorphic continuation to the entires-plane, with one simple pole ats= 1 having residue 1 and with simple zeros at s=−2,−4,−6, . . . It also shows thatζ(0) =−1/2 and gives the values ofζ(−1),ζ(−3),ζ(−5), . . . in terms ofζ(2),ζ(4),ζ(6), . . . , which were computed in Exercise 1.1.7(b).
See Exercise 4.4.3 for an appealing heuristic argument due to Euler in support of these valuesζ(1−k) for evenk≥2.
Every Dirichlet character χ modulo N has an associated Dirichlet L- functionsimilar to the Riemann zeta function,
L(s, χ) = ∞ n=1
χ(n) ns =
p∈P
(1−χ(p)p−s)−1, Re(s)>1. (4.17) Again this extends meromorphically to thes-plane and the extension is entire unlessχ=1N, in which case the L-function is essentially the Riemann zeta function (Exercise 4.4.4) and again has a simple pole ats= 1. Whenχ(−1) = 1, the functional equation satisfied byL(s, χ) is
π−s/2Γ(s2)NsL(s, χ) =π−(1−s)/2Γ(1−2s)g(χ)L(1−s,χ),¯ (4.18) and whenχ(−1) =−1 it is
π−(s+1)/2Γ(s+12 )NsL(s, χ) =−iπ−(2−s)/2Γ(2−2s)g(χ)L(1−s,χ).¯ (4.19) We will not prove these, but see the comment at the end of Section 4.7.
The modified zeta functions from Section 4.2, ζ+n(k) =
∞ m=1 m≡n(N)
1
mk, ζ+n(k, µ) = ∞ m=1 m≡n(N)
µ(m)
mk , n∈(Z/NZ)∗, are expressible in terms of L-functions. For any n ∈ (Z/NZ)∗ the second orthogonality relation shows that summing over Dirichlet characters picks out the desired terms of the full zeta function,
1 φ(N)
χ∈GN
χ(n−1)L(s, χ) = ∞ m=1
1 φ(N)
χ∈GN
χ(mn−1 (modN)) 1 ms
= ∞ m=1 m≡n(N)
1
ms =ζ+n(s), Re(s)>1.
Since the sum of L-functions extends meromorphically to the full s-plane with a simple pole ats= 1, so doesζ+n(s). Substituting the sum forζ+n(s) in relation (4.5) gives
Gvk(τ) = 1 Nφ(N)
χ,n
χ(n)L(k, χ)Eknv(τ), (4.20)
with the sum taken overχ∈(Z/NZ)∗ andn∈(Z/NZ)∗. To invert this rela- tion, take an arbitrary linear combination of seriesGmk−1voverm∈(Z/NZ)∗,
m
amGmk−1v= 1 Nφ(N)
m,χ,n
amχ(n)L(k, χ)Eknm−1v
= 1
Nφ(N)
m,χ,n
amχ(mn)L(k, χ)Eknv
= 1
Nφ(N)
m,χ,n
amχ(m)χ(n)L(k, χ)Eknv.
The left side is the inner product (without complex conjugation)a, Gwhere ais the vector with entriesam andG is the vector with entriesGmk−1v. The right side is the inner product 1/NaABk, E where A andBs (s ∈C) are the matrices
A=
χ(m) (φ(N))1/2
(m,χ)∈GN×GN
Bs=
χ(n)L(s, χ) (φ(N))1/2
(χ,n)∈GN×GN
andE is the vector with entries Eknv. By the second orthogonality relation, A∗A=IwhereA∗is the adjoint (transpose-conjugate), and soAis invertible.
By the first orthogonality relation,BsB∗s (where the adjoint operator conju- gates complex scalars but not the variable s) is the diagonal matrix with diagonal entriesL(s, χ)L(s,χ), and so¯ Bs is invertible as a matrix of mero- morphic functions onC. The product formula (4.17) shows thatL(s, χ)= 0 when Re(s)>1, and so the meromorphic functions in the inverse matrix have no poles when Re(s)>1, in particular whensis an integerk≥3. Lettinge1
denote the first standard basis vector and choosinga(k) =e1(ABk)−1 gives Ekv=N
m
am(k)Gmk−1v. (4.21) Comparing this to (4.6) shows thatζ+n(k, µ) =an−1(k) for integersk≥3, and in fact the argument establishes the same relation replacingkby any complex swith Re(s)>2. Thusζ+n(s, µ) is also a meromorphic function that continues to the fulls-plane.
The meromorphic continuation ofζ+n(s) quickly gives an analytic continu- ation of the functionζn(s) =
m≡n(N)1/msoccurring in the constant term of Gvk(τ), cf. Theorem 4.2.3 (Exercise 4.4.5(c)). We will need the formula (Exercise 4.4.5(d))
ζn(1) = πi N + π
N cot5πn N
6, gcd(n, N) = 1. (4.22)
in Section 4.8.
Exercises
4.4.1.(a) Show thatΓ(1) = 1.
(b) Show thatΓ(1/2) =√
π. (Hints for this exercise are at the end of the book.)
(c) Show thatΓ(s+ 1) =sΓ(s) when Re(s)>1.
(d) For any positive integernconsider the integral In(s) =
n t=0
1− t
n n
tsdt
t , s∈C, Re(s)>0.
Thus limn→∞In(s) =Γ(s). Change variables to get In(s) =ns
1 t=0
(1−t)nts−1dt,
integrate by parts to show thatIn(s) = (1/s)(n/(n−1))s+1In−1(s+ 1), and evaluateI1(s+n−1) to conclude that
In(s) = ns sn
m=1(1 +s/m).
Therefore
In(s)In(−s) = 1
−s2n
m=1(1−s2/m2). Lettingn→ ∞, use the formula sin(πs) =πs∞
m=1(1−s2/m2) and the iden- tity−sΓ(−s) =Γ(1−s) to prove (4.15). If you have the relevant background, make this heuristic argument rigorous by citing appropriate convergence the- orems for integrals and infinite products.
4.4.2.Prove formula (4.16). (A hint for this exercise is at the end of the book.) 4.4.3.This exercise presents an argument due to Euler for the functional equation ofζ(s).
(a) Lettbe a formal variable. Starting from the identity
t+t2+t3+t4+· · ·= (t−t2+t3−t4+· · ·) + 2(t2+t4+t6+t8+· · ·), show that applying the operatortdtd (i.e., differentiation and then multiplica- tion byt)ntimes gives
1nt+ 2nt2+ 3nt3+ 4nt4+· · ·
=
td dt
n t 1 +t
+ 2n+1(1nt2+ 2nt4+ 3nt6+ 4nt8+· · ·).
Formally, whent= 1 this is ζ(−n) =5 tdtd6n
t 1+t
t=1
+ 2n+1ζ(−n), giving a heuristic value forζ(−n). Thus for example, according to Euler, 1 + 1 + 1 + 1 +· · ·=−1/2 and 1 + 2 + 3 + 4 +· · ·=−1/12.
(b) Lett=eX and note thattdtd =dXd . Now we have (1−2n+1)ζ(−n) =
dn dXn
eX eX+ 1
X=0
forn∈N, giving the Taylor series
eX eX+ 1 =
∞ n=0
(1−2n+1)ζ(−n) n! Xn. Thus the function of a complex variable
F(z) = e2πiz e2πiz+ 1 =
∞ n=0
(1−2n+1)ζ(−n)(2πi)n
n! zn
generates (in the sense of generating function, cf. Section 1.2) the valuesζ(−n) for all natural numbersn.
LetG(z) =πcotπz. Recall from equations (1.1) that
G(z) =πie2πiz+ 1 e2πiz−1 = 1
z −2 ∞ k=1
ζ(2k)z2k−1.
The rational forms of F and G look similar, suggesting a relation between ζ(1−k) andζ(k) for evenk≥2. Use the rational forms ofF andGto show that in fact
1
πi(G(z)−2G(2z)) =−F(z) +F(−z), and then equate coefficients to obtain
ζ(1−k) = 2Γ(k)
(2πi)kζ(k) for evenk≥2.
The value of ζ(k) computed in Exercise 1.1.7 shows that ζ(1−k) = −Bk/k for evenk≥2 whereBk is the Bernoulli number.
(c) Use formula (4.16) to show that
π−k/2Γ(k2)ζ(k) =π−(1−k)/2Γ(1−2k)ζ(1−k) for evenk≥2, giving a partial version of the functional equation.
The ideas here can be turned into a rigorous proof of the meromorphic continuation and the functional equation ofζ, cf. [Hid93].
4.4.4.Show thatL(s,1N) =ζ(s)
p|N(1−p−s).
4.4.5.Take the nonpositive imaginary axis as a branch cut in thez-plane, so that arg(z) ∈ (−π/2,3π/2) on the remaining set and the function log(z) = ln|z|+iarg(z) is single-valued and analytic there. For any z off the branch cut define
zs=eslog(z), s∈C.
(a) Show that (−r)s = (−1)srs for any positive real number r and any s∈C, even though the rule (zw)s=zswsdoes not hold in general.
(b) Show that for anyn∈GN,
ζn(s) =ζ+n(s) + (−1)−sζ+−n(s), Re(s)>1.
(c) It follows that ζn(s) = 1
φ(N)
χ∈GN
(χ(n−1) + (−1)−sχ((−n)−1))L(s, χ), Re(s)>1.
This continues meromorphically to the fulls-plane with the only possible pole coming from the term whereχ=1N. Show that
slim→1
1
φ(N)(1N(n−1) + (−1)−s1N((−n)−1))L(s,1N) =πi N.
Thus the continuation ofζn(s) is entire. (Hints for this exercise are at the end of the book.)
(d) Forχ=1N, the analytic continuation of L(s, χ) is bounded ats= 1.
Show that forN >1, ζn(1) = πi
N + 1 φ(N)
χ=1N
(χ(n−1)−χ((−n)−1))L(1, χ)
= πi N + π
N cot5πn N
6.
4.4.6.Substitute u2=tin the definition ofΓ to get Γ(s) = 2
∞
u=0
e−u2u2sdu
u, s∈C, Re(s)>0.
Show that therefore fora, b >0, Γ(a)Γ(b) =Γ(a+b)
1 x=0
xa−1(1−x)b−1dx.
This last integral is aBeta integral. Show also that for any positive integerm, 1
t=0
√ dt
1−tm = 1 m
1 x=0
xm1−1(1−x)−12dx.
Use these results to explain the last equalities in the formulas for4 and3
at the end of Section 1.1. (A hint for this exercise is at the end of the book.)