Modular Forms, PS 9

Bryden Cais

November 11, 2002

Exercise 1. Suppose that χ is a primitive Dirichlet character modulo q and that (n, q) = d > 1. Then we have

∑_{a mod q} χ(a)e (\frac{an}{q}) = ∑_{a mod d} ∑_{b mod q/d} χ(a(q/d) + b)e (\frac{(a(q/d) + b)n}{q})

= ∑_{a mod d} ∑_{b mod q/d} χ(a(q/d) + b)e (\frac{bn}{q})

= ∑_{b mod q/d} e (\frac{bn}{q}) ∑_{\begin{matrix}j mod q \\ j≡b mod q/d \end{matrix}} χ(j).

We claim that there exists some c ≡ 1 mod q/d with (c, q) = 1 such that χ(c) ≠ 1. For suppose that χ(c) = 1 for all (c, q) = 1 that satisfy c ≡ 1 mod q/d. Then for any b we have χ(cb) = χ(c)χ(b) = χ(b) for all c ≡ 1 mod q/d, or equivalently, χ(a) = χ(b) for all a ≡ b mod q/d so that χ is then induced from a character modulo q/d. Since d > 1, this is a contradiction to our assumption that χ is primitive. So let c ≡ 1 mod q/d be such that (c, q) = 1 and χ(c) ≠ 1. Then we have

χ(c) ∑_{\begin{matrix}j mod q \\ j≡b mod q/d \end{matrix}} χ(j) = ∑_{\begin{matrix}j mod q \\ j≡b mod q/d \end{matrix}} χ(cj)

= ∑_{\begin{matrix}j mod q \\ cj≡b mod q/d \end{matrix}} χ(cj)

= ∑_{\begin{matrix}j mod q \\ j≡b mod q/d \end{matrix}} χ(j)

since c ≡ 1 mod q/d. Since χ(c) ≠ 1 it follows that

∑_{\begin{matrix}j mod q \\ j≡b mod q/d \end{matrix}} χ(j) = 0

for any b and hence that

∑_{a mod q} χ(a)e (\frac{an}{q}) = 0

whenever (n, q) > 1. This in conjunction with the result proved in lecture shows that

\overline{χ}(n) = \frac{1}{G(χ)} ∑_{a mod q} χ(a)e (\frac{an}{q})

for all n. Exercise 2. Let f ∈ S_{k}(1) be a Hecke eigenform, and let χ mod q be a primitive Dirichlet character. Recall from lecture that if

f = ∑_{n=1}^{∞} a(n)e(nz)

then

f_{χ} = ∑_{n=1}^{∞} a(n)χ(n)e(nz) ∈ S_{k}(q^{2}, χ^{2}).

Moreover, we showed that

T_{n,χ^{2}}(f_{χ}) = ∑_{m=1}^{∞} a_{n,χ}(m)e(mz)

where

a_{n,χ}(m) = ∑_{d|(m,n)} χ^{2}(d)d^{k−1}χ(mn/d^{2})a(mn/d^{2})

= χ(mn) ∑_{d|(m,n)} d^{k−1}a(mn/d^{2})

= χ(mn)a(m)a(n)

since f is a Hecke eigenform for T_{n} (normalized so as to have eigenvalue a(n)). But then we have T_{n,χ^{2}}(f_{χ}) = a(n)χ(n)f_{χ} so that f_{χ} is a normalized eigenform for the operators T_{n,χ^{2}} (since the n^{th} coefficient in the fourier expansion of f_{χ} is a(n)χ(n)). Recall that

E_{k}(z) = 1 + γ_{k} ∑_{n=1}^{∞} σ_{k−1}(n)e(nz)

(for some constant γ_{k}) is an eigenform for the T_{n}. We claim that

E_{k,χ}(z) = χ(0) + γ_{k} ∑_{n=1}^{∞} χ(n)σ_{k−1}(n)e(nz)

is a Hecke eigenform for the T_{n,χ^{2}}. The proof is precisely the same as above, since we have not used that f is a cusp form. Let us show that in general, E_{k,χ} is not a cusp form. Suppose that q is prime and that χ is primitive. By exercise 1, we can write

E_{k,χ}(z) = C ∑_{a mod q} \overline{χ}(a) ∑_{n=1}^{∞} σ_{k−1}(n)e (nz + n\frac{a}{q})

= C′ ∑_{a mod q} \overline{χ}(a)(E_{k}(γ_{a}z) − 1)

= C′ ∑_{a mod q} \overline{χ}(a)E_{k}(γ_{a}z)

where C, C′ are nonzero constants and

γ_{a} = (\begin{matrix}1 & a/q \\ 0 & 1 \end{matrix}) .

Let

γ = (\begin{matrix}1 & x \\ q & y \end{matrix})

where x, y are integers chosen so that γ ∈ SL_{2}(ℤ). We have

E_{k,χ}(z)|_{γ} = C′ ∑_{a mod q} \overline{χ}(a)(qz + y)^{−k}E_{k}(γ_{a}γz)

Now observe that if a ≢ −1 mod q then there exists v_{a} such that (a + 1)v_{a} ≡ ay mod q (since q is prime). It follows that in these cases, we have a v_{a} such that

(\begin{matrix}1 & a/q \\ 0 & 1 \end{matrix}) (\begin{matrix}1 & x \\ q & y \end{matrix}) = (\begin{matrix}a + 1 & (ay − (a + 1)v_{a})/q + x \\ q & y − v_{a} \end{matrix}) (\begin{matrix}1 & v_{a}/q \\ 0 & 1 \end{matrix})

and the matrix

(\begin{matrix}a + 1 & (ay − (a + 1)v_{a})/q + x \\ q & y − v_{a} \end{matrix}) ∈ SL_{2}(ℤ).

We can therefore write

E_{k,χ}(z)|_{γ} = C′ ∑_{\begin{matrix}a mod q \\ a≢−1 mod q \end{matrix}} \overline{χ}(a) (\frac{qz + y − v_{a}}{qz + y} )^{k} E_{k}(z + v_{a}/q)

+ \overline{χ}(−1)(qz + y)^{−k}E_{k} (\frac{qz + x + (q − 1)y/q}{qz + y} ) .

Since the value of E_{k,χ} at the cusp 1/q is lim_{z→∞} E_{k,χ}(z)|_{γ}, we see that this value is

lim_{z→∞} C′ ∑_{\begin{matrix}a mod q \\ a≢−1 mod q \end{matrix}} \overline{χ}(a) (\frac{qz + y − v_{a}}{qz + y} )^{k} E_{k}(z + v_{a}/q)

+ lim_{z→∞} \overline{χ}(−1)(qz + y)^{−k}E_{k} (\frac{qz + x + (q − 1)y/q}{qz + y} )

= ∑_{\begin{matrix}a mod q \\ a≢−1 mod q \end{matrix}} \overline{χ}(a) + lim_{z→∞} \overline{χ}(−1)(qz + y)^{−k}E_{k} (\frac{qz + x + (q − 1)y/q}{qz + y} )

since the value of E_{k}(z) at ∞ is 1. Recalling that

E_{k}(z) = ∑_{\begin{matrix}(c,d)∈ℤ^{2}\\{0\} \\ (c,d)=1 \end{matrix}}(cz + d)^{−k}

we see that

lim_{z→∞} \overline{χ}(−1)(qz + y)^{−k}E_{k} (\frac{qz + x + (q − 1)y/q}{qz + y} ) =

lim_{z→∞} \overline{χ}(−1) ∑_{\begin{matrix}(c,d)∈ℤ^{2}\\{0\} \\ (c,d)=1 \end{matrix}} (c(qz + x + (q − 1)y/q) + d(qz + y))^{−k} =

lim_{z→∞} 2\overline{χ}(−1) ((qz + x + (q − 1)y/q) − (qz + y))^{−k} =

2\overline{χ}(−1) (x − y/q))^{−k} =

2\overline{χ}(−1)\frac{−1}{q},

since every term in the sum vanishes in the limit, except for the two terms corresponding to (c, d) = (1, −1), (c, d) = (−1, 1), and since y − xq = 1. At long last, it follows that the value of E_{k,χ}(z) at the cusp 1/q is

∑_{\begin{matrix}a mod q \\ a≢−1 mod q \end{matrix}} \overline{χ}(a) − \frac{2χ(−1)}{q} = −χ(−1) (1 + \frac{2}{q}) ≠ 0,

and hence that E_{k,χ}(z) is not a cusp form.