Modular Forms, PS 11

Bryden Cais November 23, 2002

Exercise 1. Let {f_{j}} be a basis of normalized Hecke eigenforms for the space S_{k}^{new}(N, χ) and write

f_{j}(z) = ∑_{n=1}^{∞} λ_{j}(n)e(nz)

so that

T_{n,χ}(f_{j}) = λ_{j}(n)f_{j}

for all n. Recall that we have the formula

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

a special case of which is

λ_{j}(p^{r−1})λ_{j}(p) = \begin{cases}λ_{j}(p^{r}) + χ(p)p^{k−1}λ_{j}(p^{r−2}) & if (p, N) = 1 \\ λ_{j}(p^{r}) & if p|N \end{cases}  (1)

since χ is a Dirichlet character modulo N. Write

λ_{j}(p) = α_{j}(p) + β_{j}(p)

with

α_{j}(p)β_{j}(p) = χ(p)p^{k−1}.

We now show by induction that for (p, N) = 1 we have

λ_{j}(p^{n}) = ∑_{i=0}^{n} α_{j}(p)^{n−i}β_{j}(p)^{i}.  (2)

The cases n = 1, 0 follow by definition. Suppose that (2) holds for n < r. Applying formula (1) yields

λ_{j}(p^{r}) = λ_{j}(p^{r−1})λ_{j}(p) − χ(p)p^{k−1}λ_{j}(p^{r−2})

= λ_{j}(p) ∑_{i=0}^{r−1} α_{j}(p)^{r−1−i}β_{j}(p)^{i} − χ(p)p^{k−1} ∑_{i=0}^{r−2} α_{j}(p)^{r−2−i}β_{j}(p)^{i}

= ∑_{i=0}^{r−1} (α_{j}(p)^{r−i}β_{j}(p)^{i} + α_{j}(p)^{r−1−i}β_{j}(p)^{i+1}) − ∑_{i=0}^{r−2} α_{j}(p)^{r−1−i}β_{j}(p)^{i+1}

= (α_{j}(p)^{r} + α_{j}(p)^{r−1}β_{j}(p)) + ∑_{i=1}^{r−1} α_{j}(p)^{r−1−i}β_{j}(p)^{i+1}

= (α_{j}(p)^{r} + α_{j}(p)^{r−1}β_{j}(p)) + ∑_{i=2}^{r} α_{j}(p)^{r−i}β_{j}(p)^{i}

= ∑_{i=0}^{r} α_{j}(p)^{r−i}β_{j}(p)^{i}

by our induction hypothesis. Therefore, if (p, N) = 1 we have

(1 − \frac{λ_{j}(p)}{p^{s}} + \frac{χ(p)p^{k−1}}{p^{2s}} )^{−1} = (1 − \frac{α_{j}(p) + β_{j}(p)}{p^{s}} + \frac{α_{j}(p)β_{j}(p)}{p^{2s}})^{−1}

= (1 − \frac{α_{j}(p)}{p^{s}})^{−1} (1 − \frac{β_{j}(p)}{p^{s}})^{−1}

= (1 + \frac{α_{j}(p)}{p^{s}} + \frac{α_{j}(p)^{2}}{p^{2s}} + ⋯) (1 + \frac{β_{j}(p)}{p^{s}} + \frac{β_{j}(p)^{2}}{p^{2s}} + ⋯)

= ∑_{n=0}^{∞} ( ∑_{u+v=n} α_{j}(p)^{u}β_{j}(p)^{v}) p^{−ns}

= ∑_{n=0}^{∞} \frac{λ_{j}(p^{n})}{p^{ns}},

while if p|N then (1) tells us that

λ_{j}(p^{n}) = λ_{j}(p)^{n}

so that in these cases we have

∑_{n=0}^{∞} \frac{λ_{j}(p^{n})}{p^{ns}} = ∑_{n=0}^{∞} (\frac{λ_{j}(p)}{p^{s}})^{n}

= (1 − \frac{λ_{j}(p)}{p^{s}})^{−1} .

Exercise 2. Suppose that f ∈ S_{k}(N, χ) and let K be defined by

K(f(z)) = \overline{f(−z̄)}.

Then we claim that if g = Kf then g ∈ S_{k}(N, χ̄). For let

γ = (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ Γ_{0}(N).

Then we have

g|_{γ} = (cz + d)^{−k}g (\frac{az + b}{cz + d})

= (cz + d)^{−k}\overline{f (−\frac{az̄ + b}{cz̄ + d})}

= (cz + d)^{−k}\overline{f (\frac{a(−z̄) − b}{−c(−z̄) + d})}

= (cz + d)^{−k}\overline{χ(a)(−c(−z̄) + d)^{k}f (−z̄)}

= (cz + d)^{−k}(−c(−z) + d)^{k}\overline{χ(a)}\overline{f (−z̄)}

= \overline{χ(a)}g(z),

where we have used the fact that ( \begin{matrix}a & −b \\ −c & d \end{matrix} ) ∈ Γ_{0}(N) (obviously). Clearly, then, g ∈ S_{k}(N, χ̄), as claimed.