Modular Forms, PS 3

Bryden Cais September 23, 2002

Exercise 1. Recall that

E_{2k}(z) = 1 − \frac{2k}{B_{k}} ∑_{n=1}^{∞} σ_{k−1}(n)q^{n},

where B_{k} is defined by the generating function

\frac{t}{e^{t} − 1} = ∑_{k=0}^{∞} B_{k} \frac{t^{k}}{k!}.

It is not difficult to determine the first few B_{k}. We have

1 = ∑_{m=0}^{∞} \frac{t^{m}}{(m + 1)!} ∑_{k=0}^{∞} B_{k} \frac{t^{k}}{k!}

= ∑_{n=0}^{∞} ∑_{m+k=n} \frac{B_{k}}{k!(m + 1)!}t^{n},

which gives that B_{0} = 1 and

∑_{k=0}^{n} \binom{n + 1}{k}B_{k} = 0.

From this, we can readily calculate the B_{k} and find, in particular, that

B_{2} = \frac{1}{6}  B_{4} = \frac{1}{30}  B_{6} = \frac{1}{42}
B_{8} = \frac{1}{30}  B_{10} = \frac{5}{66}  B_{12} = \frac{691}{2730}.

Now consider the function

f(z) := E_{12}(z) − E_{6}^{2}(z)

Clearly f(z) is a modular form of weight 12. We readily determine the first few terms of the q expansion of f:

f(z) = 1 + \frac{65520}{691} ∑_{n=1}^{∞} σ_{11}(n)q^{n} − (1 − 504 ∑_{n=1}^{∞} σ_{5}(n)q^{n})^{2}

= \frac{762048}{691}q + · · · .

Observe that f is thus a cusp form of weight 12. Since S_{12}(Γ) is one dimensional, with Δ(z) = q+⋯ as a basis, we must have

E_{12}(z) − E_{6}^{2}(z) = \frac{2^{6}3^{5}7^{2}}{691} Δ(z).

Now observe that the coefficients of E_{6} are integers. Therefore, the coefficients of 691E_{6}^{2}(z) are all divisible by 691. By comparing coefficients modulo 691, we therefore have

65520σ_{11}(n) ≡ 762048τ(n) mod 691.

Since 65520 ≡ 762048 ≡ 566 mod (691), we see that τ(n) ≡ σ_{11}(n) mod 691.

Exercise 2. Let f ∈ M_{k}(Γ). Then f(z + 1) = f(z) and f(−1/z) = z^{k}f(z). Differentiating these identities gives f′(z + 1) = f′(z) and f′(−1/z) = z^{k+2}f′(z) + kz^{k+1}f(z). Recall that E_{2}(z) satisfies E_{2}(z + 1) = E_{2}(z) and

E_{2}(−1/z) = z^{2}E_{2}(z) + \frac{12z}{2πi} .

Now define

g(z) = \frac{f′(z)}{2πi} − \frac{k}{12}f(z)E_{2}(z).

We clearly have g(z + 1) = g(z). Using our formulae above, we find that

g(−1/z) = \frac{z^{k+2}f′(z) + kz^{k+1}f(z)}{2πi} − \frac{k}{12}z^{k}f(z) (z^{2}E_{2}(z) + \frac{12z}{2πi} )

= z^{k+2} \frac{f′(z)}{2πi} + \frac{k}{2πi}z^{k+1}f(z) − \frac{k}{12} \frac{12z}{2πi} z^{k}f(z) − \frac{k}{12}z^{k+2}f(z)E_{2}(z)

= z^{k+2} \frac{f′(z)}{2πi} − \frac{k}{12}z^{k+2}f(z)E_{2}(z)

= z^{k+2}g(z).

Therefore, since z ↦ z + 1 and z ↦ −1/z generate Γ, we see that g(z) ∈ M_{k+2}(Γ).

Exercise 3. Let us now use Exercise 2 to derive “Ramanujan type” relations between σ_{5}(n), σ_{3}(n) and σ_{1}(n), and between σ_{7}(n), σ_{5}(n), and σ_{1}(n). We know that for k < 6 the space M_{2k}(Γ) is one dimensional, a basis being given by E_{2k}. Therefore, since, by Exercise 2, the functions

f_{6}(z) := \frac{E4^{′}(z)}{2πi} − \frac{1}{3}E_{4}(z)E_{2}(z),

and

f_{8}(z) := \frac{E6^{′}(z)}{2πi} − \frac{1}{2}E_{6}(z)E_{2}(z)

are modular forms of weights 6 and 8 respectively, we have

f_{6}(z) = c_{6}E_{6}(z),  (1)

and

f_{8}(z) = c_{8}E_{8}(z)  (2)

for some constants c_{6}, c_{8}. To determine these constants, we write down the q series expansions of both sides of equations (1) and (2) and equate constant terms. We therefore obtain

1 − 504 ∑_{k=1}^{∞} σ_{5}(k)q^{k} = (1 + 240 ∑_{n=1}^{∞} σ_{3}(n)q^{n}) (1 − 24 ∑_{m=1}^{∞} σ_{1}(m)q^{m}) − 720 ∑_{n=1}^{∞} kσ_{3}(k)q^{k}

and

1 + 480 ∑_{k=1}^{∞} σ_{7}(k)q^{k} = (1 − 504 ∑_{n=1}^{∞} σ_{5}(n)q^{n}) (1 − 24 ∑_{m=1}^{∞} σ_{1}(m)q^{m}) + 1008 ∑_{k=1}^{∞} kσ_{5}(k)q^{k}.

Now equate coefficients of q^{k} on both sides of the above two formulae to obtain

7σ_{5}(k) = 80 ∑_{\begin{matrix}m+n=k \\ m,n≥0 \end{matrix}} σ_{3}(m)σ_{1}(n) + 10kσ_{3}(k)

10σ_{7}(k) = 252 ∑_{\begin{matrix}m+n=k \\ m,n≥0 \end{matrix}} σ_{5}(m)σ_{1}(n) + 21kσ_{5}(k),

where we define

σ_{1}(0) = −\frac{1}{24},
σ_{3}(0) = \frac{1}{240},

and

σ_{5}(0) = −\frac{1}{504}.

Exercise 4. Given that there exist α_{p}, β_{p} such that

τ(p) = p^{11/2}(α_{p} + β_{p}),

we show that

∑_{n=0}^{∞} \frac{τ(p^{n})}{p^{n(s+11/2)}} = (1 − \frac{α_{p}}{p^{s}} )^{−1} (1 − \frac{β_{p}}{p^{s}} )^{−1} .  (3)

First let us prove by induction that

τ(p^{n}) = p^{11n/2} ∑_{\begin{matrix}m+k=n \\ m,k≥0 \end{matrix}} α_{p}^{m}β_{p}^{k}.

We are already given the identity for n = 0, 1 so suppose that the identity holds for n = r. Using the multiplicativity of the τ function, we have

τ(p^{r+1}) = τ(p)τ(p^{r}) − p^{11}τ(p^{r−1})

= p^{11(r+1)/2}(α_{p} + β_{p}) ∑_{k=0}^{r} α_{p}^{k}β_{p}^{r−k} − p^{11(r+1)/2} ∑_{k=0}^{r−1} α_{p}^{k}β_{p}^{r−k−1}

= p^{11(r+1)/2} (∑_{k=0}^{r} α_{p}^{k+1}β_{p}^{r−k} + ∑_{k=0}^{r} α_{p}^{k}β_{p}^{r+1−k} − ∑_{k=0}^{r−1} α_{p}^{k}β_{p}^{r−1−k})

= p^{11(r+1)/2} ∑_{k=0}^{r+1} α_{p}^{k}β_{p}^{r+1−k},

which is the required result. Now, expanding the right hand side of (3)as a geometric series in p^{−s}, we have

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

= 1 + ∑_{n=1}^{∞} (p^{11n/2} ∑_{\begin{matrix}m+k=n \\ m,k≥0 \end{matrix}} α_{p}^{m}β_{p}^{k}) p^{−n(s+11/2)}

= ∑_{n=0}^{∞} \frac{τ(p^{n})}{p^{n(s+11/2)}} ,

and we are done.

Exercise 5. Suppose that for all integers n and primes p we have

τ(p)τ(n) = ∑_{d|(n,p)} d^{11}τ(np/d^{2}).

We show by induction that

τ(p^{r})τ(n) = ∑_{d|(n,p^{r})} d^{11}τ(np^{r}/d^{2})

for all r > 0. We are given the result for r = 0, 1. We have

τ(p^{r})τ(n) = τ(p)τ(p^{r−1})τ(n) − p^{11}τ(p^{r−2})τ(n)  (4)
= ∑_{d|(n,p^{r−1})} d^{11}τ(p)τ(np^{r−1}/d^{2}) − ∑_{d|(n,p^{r−2})}(dp)^{11}τ(np^{r}/(dp)^{2})  (5)
= ∑_{d|(n,p^{r−1})} ∑_{e|(np^{r−1}/d^{2},p)}(ed)^{11}τ(np^{r}/(ed)^{2}) − ∑_{dp|(np,p^{r−1})}(dp)^{11}τ(np^{r}/(dp)^{2})  (6)
= ∑_{d|(n,p^{r−1})} ∑_{e|(np^{r−1}/d^{2},p)}(ed)^{11}τ(np^{r}/(ed)^{2}) − ∑_{\begin{matrix}d|(np,p^{r−1}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}).  (7)

Now observe that every divisor of (n, p^{r}), with the exception of 1 and p^{r} if it divides n can be represented as ed where d|(n, p^{r−1}) and e|(np^{r−1}/d^{2}, p) in exactly two ways. To see this, let s be the highest power of p dividing n.

1. If s < r − 1, then (n, p^{r−1}) = p^{s}. Then for t ≤ s, we can write p^{t} = ed where d = p^{t} and e = 1 or where d = p^{t−1} and e = p (This is possible since p|(p^{r+s−1−2(t−1)}, p) since r − 2 > s > t). Since e must divide p, these are the only two such representations. Observe,however, that we obtain one more term that does not correspond to a divisor of (n, p^{r}) = p^{s}. This is the term d = p^{s} = (n, p^{r−1}) and e = p|(np^{r−1}/p^{2s}, p) since r − s − 1 ≥ 1.

2. If s > r − 1, then (n, p^{r−1}) = p^{r−1}. For t < r we have p^{t} = ed with d = p^{t}|(n, p^{r−1}) and e = 1 or with d = p^{t−1} and e = p|(np^{r−1}/d^{2}, p) as before. This gives precisely two representations for t < r. If t = r, there is a unique such representation: d = p^{r−1}, e = p.

3. If s = r − 1, then (n, p^{r−1}) = (n, p^{r}) = p^{r−1}. For t < r there are two representations, as before, while if d = p^{r−1} then (np^{r−1}/d^{2}, p) = 1 so we have a unique representation of the divisor p^{r−1} of (n, p^{r}).

We therefore can write equality (7) as

1. If s ≥ r − 1:

τ(p^{r})τ(n) = τ(np^{r}) + (n, p^{r})^{11}τ(n/(n, p^{r})) + 2 ∑_{\begin{matrix}d|(n,p^{r}) \\ 1<d<(n,p^{r}) \end{matrix}} d^{11}τ(np^{r}/d^{2}) − ∑_{\begin{matrix}d|(np,p^{r−1}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) (8)

2. If s < r − 1:

τ(p^{r})τ(n) = τ(np^{r}) + 2 ∑_{\begin{matrix}d|(n,p^{r}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) + p^{11(s+1)}τ(np^{r−2s−2}) − ∑_{\begin{matrix}d|(np,p^{r−1}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) (9)

We now claim that the divisors of (np, p^{r−1}) are essentially in one to one correspondence with the nontrivial divisors of (n, p^{r}). As before, let s be the highest power of p dividing n.

1. If s < r − 1 then (np, p^{r−1}) = p^{s+1} while (n, p^{r}) = p^{s}. Therefore,

∑_{\begin{matrix}d|(np,p^{r−1}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) = ∑_{\begin{matrix}d|(n,p^{r}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) + p^{11(s+1)}τ(np^{r}/p^{2(s+1)}).

2. If s > r − 1 then (np, p^{r−1}) = p^{r−1} and (n, p^{r}) = p^{r}. In this case,

∑_{\begin{matrix}d|(np,p^{r−1}) \\ d>1 \end{matrix}} d^{11}τ(np^{r}/d^{2}) = ∑_{\begin{matrix}d|(n,p^{r}) \\ 1<d<(n,p^{r}) \end{matrix}} d^{11}τ(np^{r}/d^{2}).

3. If s = r−1 then (np, p^{r−1}) = p^{r−1} and (n, p^{r}) = p^{r−1}. In this case we have an actual bijection.

Putting all three cases together and combining with the two cases above, we find that in any case,

τ(p^{r})τ(n) = τ(np^{r}) + (n, p^{r})^{11}τ(n/(n, p^{r})) + ∑_{\begin{matrix}d|(n,p^{r}) \\ 1<d<(n,p^{r}) \end{matrix}} d^{11}τ(np^{r}/d^{2})

= ∑_{d|(n,p^{r})} d^{11}τ(np^{r}/d^{2}),

which is what we wanted to prove. Now observe that for p, q relatively prime we have

τ(p^{r})τ(q^{s})τ(n) = ∑_{d|(n,q^{s})} d^{11}τ(p^{r})τ(nq^{s}/d^{2})

= ∑_{d|(n,q^{s})} ∑_{e|(nq^{s}/d^{2},p^{r})}(ed)^{11}τ(nq^{s}p^{r}/(ed)^{2})

= ∑_{d|(n,q^{s}p^{r})} d^{11}τ(np^{r}q^{s}/d^{2}),

since the divisors of (n, p^{r}q^{s}) are in one to one correspondence with pairs of divisors d, e of (n, p^{r}), (n, q^{s}) respectively. It therefore follows that for any m, n one has

τ(m)τ(n) = ∑_{d|(m,n)} d^{11}τ(mn/d^{2}).