Modular Forms, PS 10

Bryden Cais

November 18, 2002

Exercise 1. Suppose that f ∈ S_{k}(1) is a Hecke eigenform for all T_{n}. We show that if p|N then f considered as an element of S_{k}(N, χ_{0}) is not an eigenform for T_{p,χ}. Recall from class that we showed that for p|n one has

T_{p,χ_{0}} = \frac{1}{p} ∑_{b mod p} \binom{1 b}{p}

while

T_{p} = \frac{1}{p} ∑_{b mod p} \binom{1 b}{p} + p^{k−1} (\begin{matrix}p \\ & 1 \end{matrix}) .

Suppose that f is an eigenform for T_{pχ_{0}}. Then we have

\frac{1}{p} ∑_{b mod p} f (\frac{z + b}{p}) = μ_{p}f(z)

and

\frac{1}{p} ∑_{b mod p} f (\frac{z + b}{p}) + p^{k−1}f(pz) = λ_{p}f(z).

We may assume that λ_{p} ≠ μ_{p} for otherwise it is clear that f = 0 identically. However, subtracting these two equations yields

f(z) = \frac{p^{k−1}}{λ_{p} − μ_{p}} f(pz).

Applying this formula twice gives

f(z) = (\frac{p^{k−1}}{λ_{p} − μ_{p}} )^{2} f(p^{2}z),

from which it follows that if we write

f(z) = ∑_{n=1}^{∞} a(n)e(nz)

then a(n) = 0 unless p^{2}|n. In particular, a(l) = 0 for all primes l. Now since f was supposed a normalized eigenform for T_{n} for all n, it follows that we have the relation

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

and thus that a(n) = 0 for all n. This gives f = 0 identically, a contradiction.

Exercise 2. Suppose that f ∈ S_{k}(1) is a normalized eigenform for all T_{n} and let l|n. We show that f(lz) as considered as an element of S_{k}(N, χ_{0}) is an eigenform for all T_{n,χ_{0}} with (n, N) = 1 and eigenvalue a(n). Recall that

T_{n,χ_{0}}(f(z)) = \frac{1}{n} ∑_{ad=n} χ_{0}(a)a^{k} ∑_{b mod d} \binom{a b}{d} f(z).

It follows that

T_{n,χ_{0}}(f(lz)) = \frac{1}{n} ∑_{ad=n} χ_{0}(a)a^{k} ∑_{b mod d} (\begin{matrix}l \\ & 1 \end{matrix}) \binom{a b}{d} f(z)

= \frac{1}{n} ∑_{ad=n} χ_{0}(a)a^{k} ∑_{b mod d} \binom{a bl}{d} (\begin{matrix}l \\ & 1 \end{matrix}) f(z).

Now since (n, N) = 1 and l|N, we have (l, N) = 1. Thus, as b runs over all classes modulo d so does lb (since d|n). Moreover, since a|n and (n, N) = 1, we see that χ_{0}(a) = 1 for all a|n since χ_{0} is the trivial character modulo N. We then have

T_{n,χ_{0}}(f(lz)) = \frac{1}{n} ∑_{ad=n} a^{k} ∑_{b mod d} \binom{a b}{d} (\begin{matrix}l \\ & 1 \end{matrix}) f(z)
= a(n)f(lz),

since f was assumed to be an eigenform for all T_{n}, normalized to have eigenvalue a(n).