Eisenstein Series
4.3 Dirichlet characters, Gauss sums, and eigenspaces
Computing the Eisenstein series for the groups Γ1(N) and Γ0(N) requires some machinery, to be given over the next two sections. For any positive integer N, letGN denote the multiplicative group (Z/NZ)∗, of order φ(N) where φ is the Euler totient, when this doesn’t conflict with the notation for Eisenstein series. ADirichlet character moduloN is a homomorphism of multiplicative groups,
χ:GN −→C∗.
For any two Dirichlet charactersχ andψ modulo N, the product character defined by the rule (χψ)(n) =χ(n)ψ(n) forn∈GN is again a Dirichlet char- acter moduloN. In fact, the set of Dirichlet characters moduloN is again a multiplicative group, called thedual groupofGN, denoted GN, whose iden- tity element is thetrivial character modulo N mapping every element ofGN
to 1, denoted 1N or just1 when N is clear. Since GN is a finite group the values taken by any Dirichlet character are complex roots of unity, and so the inverse of a Dirichlet character is its complex conjugate, defined by the rule ¯χ(n) = χ(n) for all n ∈ GN. (So here the overline denotes complex conjugation, not reduction moduloN.) As explained in the previous section, G1={0}, and so the only Dirichlet character modulo 1 is the trivial charac- ter11.
For any primepthe groupGpis cyclic of orderp−1. Letgbe a generator and letµp−1 be a primitive (p−1)st complex root of unity. Then the group of Dirichlet characters modulopis again cyclic of order p−1, generated by the character takinggto µp−1. In general (see, for example, [Ser73]),
Proposition 4.3.1.LetGN be the dual group ofGN. ThenGN is isomorphic toGN. In particular, the number of Dirichlet characters moduloN isφ(N).
The two groups are noncanonically isomorphic, meaning that construct- ing an actual isomorphism fromGN toGN involves arbitrary choices of which elements map to which characters. The groupsGN andGN satisfy theorthog- onality relations(Exercise 4.3.1),
n∈GN
χ(n) =
φ(N) ifχ=1, 0 ifχ=1,
χ∈GN
χ(n) =
φ(N) ifn= 1, 0 ifn= 1.
Let N be a positive integer and let d be a positive divisor of N. Every Dirichlet characterχ modulo d lifts to a Dirichlet characterχN moduloN, defined by the ruleχN(n(modN)) =χ(n(modd)) for alln∈Z relatively prime to N. That is, χN = χ◦πN,d where πN,d : GN −→ Gd is natural projection. For example, the Dirichlet character modulo 4 taking 1 to 1 and 3 to−1 lifts to the Dirichlet character modulo 12 taking 1 and 5 to 1 and 7 and 11 to−1. Going in the other direction, from modulus N to modulus d, isn’t always possible. Every Dirichlet characterχmoduloN has aconductor, the smallest positive divisordofN such thatχ=χd◦πN,dfor some character χd modulod, or, equivalently, such thatχis trivial on the normal subgroup
KN,d= ker(πN,d) ={n∈GN :n≡1 (modd)}.
For example, the Dirichlet character modulo 12 taking 1 and 7 to 1 and 5 and 11 to−1 has conductor 3. A Dirichlet character moduloN is primitive if its conductor isN. The only character moduloN with conductor 1 is the trivial character1N, and the trivial character1N moduloN is primitive only forN= 1.
Every Dirichlet characterχmoduloNextends to a functionχ:Z/NZ−→
C whereχ(n) = 0 for noninvertible elementsn of the ringZ/NZ, and then extends further to a functionχ: Z−→ C where (abusing notation)χ(n) = χ(n(modN)) for alln∈Z. Thusχ(n) = 0 for allnsuch that gcd(n, N)>1.
The extended function is no longer a homomorphism, but it still satisfies χ(nm) =χ(n)χ(m) for alln, m. In particular, the trivial character moduloN extends to the function
1N(n) =
1 if gcd(n, N) = 1, 0 if gcd(n, N)>1.
Thus the extended trivial character is no longer identically 1 unlessN = 1.
The extension of any Dirichlet characterχmoduloN satisfies χ(0) =
1 ifN = 1, 0 ifN >1.
Summing overn= 0 to N−1 in the first orthogonality relation and taking n∈Zin the second gives modified versions,
N−1 n=0
χ(n) =
φ(N) ifχ=1, 0 ifχ=1,
χ∈GN
χ(n) =
φ(N) ifn≡1 (N), 0 ifn≡1 (N).
TheGauss sumof a Dirichlet characterχmoduloNis the complex number g(χ) =
N−1 n=0
χ(n)µnN, µN =e2πi/N.
Ifχis primitive moduloN then (Exercise 4.3.2) for any integerm,
N−1 n=0
χ(n)µnmN = ¯χ(m)g(χ). (4.11) It follows that the Gauss sum of a primitive character is nonzero. Indeed, the square of its absolute value is
g(χ)g(χ) =
N−1 m=0
¯
χ(m)g(χ)µ−Nm=
N−1 m=0
N−1 n=0
χ(n)µnmN µ−Nm by (4.11) (4.12)
=
N−1 n=0
χ(n)
N−1 m=0
µ(nN−1)m=N, (4.13)
the last equality holding because the inner sum is N when n = 1 and 0 otherwise. Formula (4.11) and the following lemma will be used in the next section.
Lemma 4.3.2.Let N be a positive integer. If N = 1 or N = 2 then every Dirichlet characterχmoduloN satisfiesχ(−1) = 1. IfN >2then the number of Dirichlet characters modulo N is even, half of them satisfying χ(−1) = 1 and the other half satisfyingχ(−1) =−1.
Proof. The result forN = 1 andN = 2 is clear. IfN >2 then 4|N orp|N for some odd prime p. The nontrivial character modulo 4 takes −1 (mod 4) to−1, and for every odd primepthe character moduloptaking a generatorg ofGp to a primitive (p−1)st complex root of unity takes−1 (modp) to −1 since−1≡g(p−1)/2 (modp). In either case the character lifts to a character moduloN taking−1 (modN) to−1.
Let GN denote the group of Dirichlet characters modulo N. The map GN −→ {±1}taking each characterχtoχ(−1) is a homomorphism. We have just seen that the homomorphism surjects ifN >2, and so the result follows from the First Isomorphism Theorem of group theory.
We are interested in Dirichlet characters because they decompose the vec- tor spaceMk(Γ1(N)) into a direct sum of subspaces that we can analyze in- dependently. For each Dirichlet characterχmoduloN define theχ-eigenspace ofMk(Γ1(N)),
Mk(N, χ) ={f ∈ Mk(Γ1(N)) :f[γ]k =χ(dγ)f for allγ∈Γ0(N)}. (Here dγ denotes the lower right entry of γ.) In particular the eigenspace Mk(N,1) isMk(Γ0(N)) (Exercise 4.3.3(a)). Also note thatMk(N, χ) is just {0}unless χ(−1) = (−1)k (Exercise 4.3.3(b)). The vector space Mk(Γ1(N)) decomposes as the direct sum of the eigenspaces (Exercise 4.3.4(a)),
Mk(Γ1(N)) =
χ
Mk(N, χ),
and the same result holds for cusp forms (Exercise 4.3.4(b)), so it holds for the quotients as well (Exercise 4.3.4(c)),
Ek(Γ1(N)) =
χ
Ek(N, χ).
Exercises
4.3.1.Prove the orthogonality relations. (A hint for this exercise is at the end of the book.)
4.3.2.Let χ be a primitive Dirichlet character modulo N and let m be an integer. Prove formula (4.11) as follows.
(a) Form= 0, confirm the formula forχ=11 and for nontrivialχ.
(b) Now assumem= 0. Let g= gcd(m, N),N =Ng, andm=mghfor the minimalhsuch that gcd(m, N) = 1. Show that since ¯χ(m)χ(m) = 1,
N−1 n=0
χ(n)µnmN = ¯χ(m)
N−1 n=0
⎛
⎜⎜
⎝
N−1 n=0 n≡n(N)
χ(n)
⎞
⎟⎟
⎠µnNh. (4.14)
Let K = KN,N = {n ∈ (Z/NZ)∗ : n ≡ 1 (modN)}, the subgroup of (Z/NZ)∗ introduced in the section. Use the fact that χ is primitive to show that
n∈K
χ(n) =
1 ifg= 1, 0 ifg >1.
Since (Z/NZ)∗=
nnKwhere the coset representativesntaken moduloN run through (Z/NZ)∗, show that the inner sum in (4.14) is
n∈nKχ(n), and that this isχ(n) ifg= 1 and 0 ifg >1. Show that formula (4.11) follows in both cases.
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).