Modular Forms, PS 6

Bryden Cais October 21, 2002

Exercise 1. Suppose that l ∈ ℕ. Recall that

J_{l}(x) = ∑_{n=0}^{∞}(−1)^{n} \frac{(x/2)^{2n+l}}{n!(n + l)!}.

We have

∫_{−1/2}^{1/2} e^{2πilt}e^{−ix sin(2πt)} dt = ∑_{n=0}^{∞} \frac{(−ix)^{n}}{n!} ∫_{−1/2}^{1/2} e^{2πilt} sin^{n}(2πt) dt

= ∑_{n=0}^{∞} ∑_{j=0}^{n}(−1)^{j} \frac{(x/2)^{n}}{n!} \binom{n}{j} ∫_{−1/2}^{1/2} e^{2πilt}e^{2πi(2j−n)} dt

= ∑_{n=0}^{∞} ∑_{j=0}^{n}(−1)^{j} \frac{(x/2)^{n}}{n!} \binom{n}{j} ∫_{−1/2}^{1/2} e^{2πi(l+2j−n)} dt.

Now observe that the integral is zero unless n = l + 2j (in which case it is 1), and that for any j there is exactly one value of n for which this holds. Thus, we may reindex the sum over j to obtain

∫_{−1/2}^{1/2} e^{2πilt}e^{−ix sin(2πt)} dt = ∑_{j=0}^{∞}(−1)^{j} \frac{(x/2)^{2j+l}}{(2j + l)!} \binom{2j + l}{j}

= ∑_{j=0}^{∞}(−1)^{j} \frac{(x/2)^{2j+l}}{j!(j + l)!}

= J_{l}(x).

We thus have

|J_{l}(x)| ≤ ∫_{−1/2}^{1/2} |e^{2πilt}e^{−ix sin(2πt)}| dt

≤ sup_{t∈[−1/2,1/2]} |e^{2πilt}e^{−ix sin(2πt)}|

≤ 1 since both exponentials have purely imaginary argument.

Exercise 2. Let c = qr with (q, r) = 1. Then there exist integers u, v such that qu + rv = 1. Observe that

qu ≡ 1 mod r
rv ≡ 1 mod q,

so that u = q̄, v = r̄. We have

a = qua + rva
ā = quā + rvā.

Notice also that since aā ≡ 1 mod qr we have aā ≡ 1 mod r and aā ≡ 1 mod q. Finally, since (r, q) = 1 we see by the chinese remainder theorem that as a ranges over all values mod c, (a mod r, a mod q) ranges over all pairs (s mod r, t mod q). Thus, we have

S(m, n, c) = ∑_{aā≡1 mod c} e (\frac{am + ān}{c})

= ∑_{aā≡1 mod c} e (\frac{quam + rvam + quān + rvān}{rq})

= ∑_{aā≡1 mod c} e (\frac{uam + uān}{r} + \frac{vam + vān}{q})

= ∑_{aā≡1 mod c} e (\frac{amq̄ + ānq̄}{r}) e (\frac{amr̄ + vānr̄}{q}) .

Now since the value of

\frac{amq̄ + ānq̄}{r} mod 1

depends only on a mod r and similarly the value of

\frac{amr̄ + vānr̄}{q} mod 1

depends only on a mod q, by our observations above we have

S(m, n, c) = ∑_{ss̄≡1 mod r} e (\frac{smq̄ + s̄nq̄}{r}) ∑_{tt̄≡1 mod q} e (\frac{tmr̄ + vt̄nr̄}{q})
= S(mq̄, nq̄, r)S(mr̄, nr̄, q).

Exercise 3. We first prove some preliminary formulae. Suppose that (a, c) = 1. Then

S(am, n, c) = S(m, an, c).  (1)

For we have

S(am, n, c) = ∑_{xx̄≡1 mod c} e (\frac{amx + nx̄}{c})

= ∑_{xx̄≡1 mod c} e (\frac{mx + n\overline{xā}}{c})

= ∑_{xx̄≡1 mod c} e (\frac{mx + nax̄}{c})

= S(m, an, c),

where we have used the fact that as x runs over all units modulo c so does ax since (a, c) = 1. Now if p ∦(m, n) it follows from (1) that

S(m, n, p^{α}) = S(mn, 1, p^{α})

for any α ≥ 1. If, on the other hand, p|(m, n) then we have

S(m, n, p^{α}) = ∑_{xx̄≡1 mod p^{α}} e (\frac{mx + nx̄}{p^{α}})

= ∑_{\begin{matrix}xx̄≡1 mod p^{α−1} \\ k_{1},k_{2} mod p \end{matrix}} e (\frac{m(x + k_{1}p^{α−1}) + n(x̄ + k_{2}p^{α−1})}{p^{α}}) ,

where k_{1} runs over all integers mod p and k_{2} is uniquely determined by k_{1} and the condition

(x + p^{α−1}k_{1})(x̄ + p^{α−1}k_{2}) ≡ 1 mod p^{α}

where xx̄ ≡ 1 mod p. We are implicitly using Hensel’s lemma to rewrite the sum in this way. We then have

S(m, n, p^{α}) = ∑_{xx̄≡1 mod p^{α−1}} e (\frac{mx + nx̄}{p^{α}}) ∑_{k_{1},k_{2} mod p} e (\frac{mk_{1} + nk_{2}}{p} )

= ∑_{xx̄≡1 mod p^{α−1}} e (\frac{(m/p)x + (n/p)x̄}{p^{α−1}}) ∑_{k_{1},k_{2} mod p} 1

since p|m, n, so that the above is equal to

pS(m/p, n/p, p^{α−1})  (2)

since k_{2} is uniquely determined by k_{1} and k_{1} ranges over all values modulo p. Finally, observe that if p|(m, n) then

S(mn, 1, p^{α}) = ∑_{xx̄≡1 mod p^{α}} e (\frac{mnx + x̄}{p^{α}})

= ∑_{\begin{matrix}xx̄≡1 mod p^{α−1} \\ k_{1},k_{2} mod p \end{matrix}} e (\frac{mn(x + k_{1}p^{α−1}) + (x̄ + k_{2}p^{α−1})}{p^{α}})

= ∑_{xx̄≡1 mod p^{α−1}} e (\frac{mnx + x̄}{p^{α}}) ∑_{k_{1},k_{2} mod p} e (\frac{mnk_{1} + k_{2}}{p} ) .

But since p|m, n we have

∑_{k_{1},k_{2} mod p} e (\frac{mnk_{1} + k_{2}}{p} ) = ∑_{k_{1},k_{2} mod p} e (\frac{k_{2}}{p} )

= 0

since k_{2} ranges over all values modulo p. Hence if p|(m, n) then

S(mn, 1, p^{α}) = 0.  (3)

We are now prepared to prove the main statement. We first show by induction that

S(m, n, p^{α}) = ∑_{d|(m,n,p^{α})} dS(mn/d^{2}, 1, p^{α}/d).

For α = 0 the result follows by (1). Further, if p |̸(m, n) the result follows from (1), so we may assume that p|(m, n). If α = 1, we have

S(mn, 1, p) + pS(mn/p^{2}, 1, 1) = p ∑_{xx̄≡1 mod 1} e ((mn/p^{2})x + x̄) + ∑_{xx̄≡1 mod p} e (\frac{mnx + x̄}{p})

= p + ∑_{xx̄≡1 mod p} e (x̄/p) ,

= p − 1 while

S(m, n, p) = ∑_{xx̄≡1 mod p} e (\frac{mx + nx̄}{p})
= p − 1

since p|m, n.

If α > 1 then we have

∑_{d|(m,n,p^{α})} dS(mn/d^{2}, 1, p^{a}/d) = S(mn, 1, p^{α}) + ∑_{d|(m,n,p^{α−1})} pdS(mn/p^{2}d^{2}, 1, p^{α}/pd)

= p ∑_{d|(m,n,p^{α−1})} dS((m/p)(n/p)/d^{2}, 1, p^{α−1}/d)

by (3) and our assumption that p|(m, n)

= pS(m/p, n/p, p^{α−1})

by our induction hypothesis

= S(m, n, p^{α})

by(2). Therefore, for any p, α we have

S(m, n, p^{α}) = ∑_{d|(m,n,p^{α})} dS(mn/d^{2}, 1, p^{α}/d).  (4)

Now let c_{1}, c_{2} be arbitrary with (c_{1}, c_{2}) = 1. Then by Exercise 2 we have

S(m, n, c_{1}c_{2}) = S(mc̄_{1}, nc̄_{1}, c_{2})S(mc̄_{2}, nc̄_{2}, c_{1}).

Suppose that we have

S(mc̄_{1}, nc̄_{1}, c_{2}) = ∑_{d_{1}|(mc̄_{1},nc̄_{1},c_{2})} d_{1}S(mnc̄_{1}^{2}/d_{1}^{2}, 1, c_{2}/d_{1})
S(mc̄_{2}, nc̄_{2}, c_{1}) = ∑_{d_{2}|(mc̄_{2},nc̄_{2},c_{1})} d_{2}S(mnc̄_{1}^{2}/d_{2}^{2}, 1, c_{2}/d_{2}).

Observe that since (c_{1}, c_{2}) = 1 we may rewrite both sums as

∑_{d_{1}|(m,n,c_{2})} d_{1}S(mnc̄_{1}^{2}/d_{1}^{2}, 1, c_{2}/d_{1})

∑_{d_{2}|(m,n,c_{1})} d_{2}S(mnc̄_{2}^{2}/d_{2}^{2}, 1, c_{1}/d_{2}).

It follows by Exercise 2 that

S(m, n, c_{1}c_{2}) = ∑_{\begin{matrix}d_{1}|(m,n,c_{2}) \\ d_{2}|(m,n,c_{1}) \end{matrix}} d_{1}d_{2}S(mnc̄_{1}^{2}/d_{1}^{2}, 1, c_{2}/d_{1})S(mnc̄_{2}^{2}/d_{2}^{2}, 1, c_{1}/d_{2})

= ∑_{\begin{matrix}d_{1}|(m,n,c_{2}) \\ d_{2}|(m,n,c_{1}) \end{matrix}} d_{1}d_{2}S(mnc̄_{1}^{2}/d_{1}^{2}, 1, c_{2}/d_{1})S(mnc̄_{2}^{2}/d_{2}^{2}, 1, c_{1}/d_{2})

= ∑_{\begin{matrix}d_{1}|(m,n,c_{2}) \\ d_{2}|(m,n,c_{1}) \end{matrix}} d_{1}d_{2}S(mn/(d_{1}d_{2})^{2}(c̄_{1}^{2}/d̄_{2}^{2}), 1, c_{2}/d_{1})S(mn/(d_{1}d_{2})^{2}(c̄_{2}^{2}/d̄_{1}^{2}), 1, c_{1}/d_{2})

= ∑_{\begin{matrix}d_{1}|(m,n,c_{2}) \\ d_{2}|(m,n,c_{1}) \end{matrix}} d_{1}d_{2}S(mn/(d_{1}d_{2})^{2}(c̄_{1}/d̄_{2}), (c̄_{1}/d̄_{2}), c_{2}/d_{1})S(mn/(d_{1}d_{2})^{2}(c̄_{2}/d̄_{1}), (c̄_{2}/d̄_{1}), c_{1}/d_{2})

by Exercise 1

= ∑_{\begin{matrix}d_{1}|(m,n,c_{2}) \\ d_{2}|(m,n,c_{1}) \end{matrix}} d_{1}d_{2}S(mn/(d_{1}d_{2})^{2}, 1, c_{1}c_{2}/d_{1}d_{2}).

Again, since (c_{1}, c_{2}) = 1, as d_{1}, d_{2} run over all divisors of (m, n, c_{1}), (m, n, c_{2}) respectively, d_{1}d_{2} runs over all divisors of (m, n, c_{1}c_{2}). It follows that the above sum is

S(m, n, c_{1}c_{2}) = ∑_{d|(m,n,c_{1}c_{2})} dS(mn/d^{2}, 1, c_{1}c_{2}/d)  (5)

Therefore, by (4) and (5) we have

S(m, n, c) = ∑_{d|(m,n,c)} dS(mn/d^{2}, 1, c)

for all positive integers c and any m, n.

Exercise 4. Let us first show that for any prime p and any ϵ > 0 there exists a constant C_{p}(ϵ) independent of n such that

d(p^{n}) ≤ C_{p}(ϵ)p^{nϵ}.

We know that d(p^{n}) = n + 1. Since p ≥ 2, we obviously have

lim_{n→∞} \frac{n + 1}{p^{nϵ}} = 0

(by, say, l’Hopital’s rule). Since

\frac{n + 1}{p^{nϵ}}

is a continuous function of n, it follows that

sup_{n≥0} \frac{n + 1}{p^{nϵ}} = y < ∞.

Clearly y depends only on p, ϵ, so setting C_{p}(ϵ) = y + 1 gives he desired constant. Now observe that for p > 2^{1/ϵ} we have

\frac{n + 1}{p^{nϵ}} < \frac{n + 1}{2^{n}} ≤ 1

for all n ≥ 0. Therefore, for any p > 2^{1/ϵ} we may take C_{p}(ϵ) = 1. In any case, we always have C_{p}(ϵ) ≥ 1. Therefore, we have

d (∏_{i=1}^{r} p_{i}^{a_{i}}) = ∏_{i=1}^{r} d(p_{i}^{)}a_{i}

< ∏_{i=1}^{r} C_{p_{i}}(ϵ)p_{i}^{a_{i}ϵ}

< ∏_{p} C_{p}(ϵ) (∏_{i=1}^{r} p_{i}^{a_{i}})^{ϵ}

= ∏_{p<2^{1/ϵ}} C_{p}(ϵ) (∏_{i=1}^{r} p_{i}^{a_{i}})^{ϵ} .

Setting

C(ϵ) = ∏_{p<2^{1/ϵ}} C_{p}(ϵ)

(and noting that C(ϵ) depends only on ϵ) we see that for any ϵ > 0

d(m) < C(ϵ)m^{ϵ}.

Exercise 5. Let f ∈ S_{k}(Γ) and write f in terms of an orthogonal basis for S_{k}(Γ), that is,

f = ∑_{j=1}^{r} c_{j}f_{j}.

Recall that we have

\frac{Γ(k − 1)}{(4πn)^{k−1}} ∑_{j=1}^{r} \frac{|a_{f_{j}}(n)|^{2}}{∥f_{j}∥^{2}} = 1 + 2πi^{k} ∑_{c=1}^{∞} \frac{S(n, n, c)}{c}J_{k−1} (\frac{4πn}{c}) .

Also recall the bounds

J_{k−1} (\frac{4πn}{c}) ≪_{k} \begin{cases}(\frac{n}{c})k−1 & if c > n \\ \sqrt{\frac{c}{n}} & if c < n \end{cases}

and

|S(n, n, c)| ≤ d(c)(c, n)^{1/2}c^{1/2}.

It follows that

1 + 2πi^{k} ∑_{c=1}^{∞} \frac{S(n, n, c)}{c}J_{k−1} (\frac{4πn}{c}) ≪ 1 + ∑_{c<n} \frac{d(c)(c, n)^{1/2}}{\sqrt{n}} + ∑_{c>n} \frac{d(c)(c, n)^{1/2}}{\sqrt{c}} (\frac{n}{c})^{k−1} .

This suggests that we estimate the sum

S_{j}(n) := ∑_{jn<c<(j+1)n} d(c)(c, n)^{1/2}.

By Exercise 4 we have, for any ϵ > 0

S_{j}(n) ≤ C(ϵ) ∑_{jn<c<(j+1)n} c^{ϵ}(c, n)^{1/2}

≪ ((j + 1)n)^{ϵ} ∑_{jn<c<(j+1)n}(c, n)^{1/2}.

Now there are about x/a integers less than x that are divisible by a. Hence, there are about

\frac{(j + 1)n}{a} − \frac{jn}{a} = \frac{n}{a}

integers c between jn and (j + 1)n that are divisible by a. Therefore

S_{j}(n) ≪ ((j + 1)n)^{ϵ} ∑_{a|n} \frac{n}{a}a^{1/2}

≪ (j + 1)^{ϵ}n^{1/2+ϵ} ∑_{d|n} d^{1/2}

≪ (j + 1)^{ϵ}n^{1+ϵ}d(n)

≪ (j + 1)^{ϵ}n^{1+2ϵ}.

Replacing 2ϵ by ϵ gives

∑_{c<n} \frac{d(c)(c, n)^{1/2}}{\sqrt{n}} = \frac{S_{0}(n)}{\sqrt{n}} ≪ \frac{n^{1+ϵ}}{\sqrt{n}} ≪ n^{1/2+ϵ}.

Similarly, we have

∑_{c>n} \frac{d(c)(c, n)^{1/2}}{\sqrt{c}} (\frac{n}{c})^{k−1} = ∑_{j=1}^{∞} ∑_{jn≤c<(j+1)n} \frac{d(c)(c, n)^{1/2}}{\sqrt{c}} (\frac{n}{c})^{k−1}

≪ ∑_{j=1}^{∞} ∑_{jn≤c<(j+1)n} \frac{d(c)(c, n)^{1/2}}{\sqrt{jn}} (\frac{1}{j})^{k−1}

≪ n^{1/2+ϵ} ∑_{j=1}^{∞} \frac{(j + 1)^{ϵ}}{\sqrt{j}} (\frac{1}{j})^{k−1}

≪ n^{1/2+ϵ}

since the sum in j is clearly absolutely convergent and independent of n. Therefore, we have

\frac{Γ(k − 1)}{(4πn)^{k−1}} ∑_{j=1}^{r} \frac{|a_{f_{j}}(n)|^{2}}{∥f_{j}∥^{2}} ≪ n^{1/2+ϵ},

so that

∑_{j=1}^{r} \frac{|a_{f_{j}}(n)|^{2}}{∥f_{j}∥^{2}} ≪ n^{k−1+1/2+ϵ}.

It follows that

sup_{1≤j≤r} |a_{f_{j}}(n)|^{2} ≪ n^{k−1+1/2+ϵ}

so that

sup_{1≤j≤r} |a_{f_{j}}(n)| ≪ n^{\frac{k−1}{2} +1/4+ϵ}

for any ϵ > 0. Since f = c_{1}f_{1} + ⋯ + c_{r}f_{r} we have

|a_{f}(n)| ≤ |c_{1}||a_{f_{1}}(n)| + ⋯ + |c_{r}||a_{f_{r}}(n)| ≪ n^{\frac{k−1}{2} +1/4+ϵ}.