AN Ω-THEOREM FOR RAMANUJAN’S τ-FUNCTION

R. BALASUBRAMANIAN AND M. RAM MURTY

1. Introduction

Let τ(n) be defined by

Δ(z) = ∑_{n=1}^{∞} τ(n)q^{n} = q ∏_{n=1}^{∞}(1 − q^{n})^{24}, q = e^{2πiz}.

This function was first studied by Ramanujan [6]. He wrote, for every prime p,

τ(p) = 2p^{11/2} cos θ_{p}

and conjectured that θ_{p} is real. This was proved by Deligne [2]. It is known that

τ(p^{α}) = p^{11α/2} \frac{sin(α + 1)θ_{p}}{sin θ_{p}} .

If d(n) denotes the number of divisors of n, then it follows that

|τ(n)| ≤ n^{11/2}d(n),

as τ is a multiplicative function. Therefore, for some constant c_{1} > 0,

τ(n) = O (n^{11/2} exp (\frac{c_{1} log n}{log log n})) .

It is conjectured that

τ(n) = Ω (n^{11/2} exp (\frac{c_{2} log n}{log log n}))  (1)

for some constant c_{2} > 0.

A conjecture of Sato and Tate states that the angles θ_{p} are equidistributed in [0, 2π] with respect to the measure

\frac{2}{π} sin^{2} θ dθ.

It is easy to see that the conjecture of Sato-Tate implies (1). In fact, if

card(p ≤ x : 0 ≤ θ_{p} ≤ φ) ≫ x^{δ}

for some φ < \frac{π}{3} and some δ > 0. Then (1) follows easily. Both assertions about the distribution of the angles θ_{p} remain unproved.

With respect to unconditional results, Rankin [5] showed

lim sup_{n→∞} \frac{|τ(n)|}{n^{11/2}} = +∞,

and Joris [3] proved

τ(n) = Ω(n^{11/2} exp(c(log n)^{(1/22)−ε})).

We shall show below that

τ(n) = Ω(n^{11/2} exp(c(log n)^{(2/3)−ε})).

For an arbitrary normalized Hecke eigenform

f = ∑_{n=1}^{∞} a_{n}e^{2πinz}

of weight k, a similar result is true if we assume that

∑_{n=1}^{∞} \frac{|a_{n}|^{2}}{n^{s}}

has no real zeros in the critical strip k − 1 ≤ σ ≤ k.

Nevertheless, by an elementary method, one can show that

a_{n} = Ω (n^{\frac{k−1}{2}} exp (\frac{c(log n)^{\frac{1}{k}}}{log log n} )) .

This result remains true if f is a normalized eigenform of even weight for an arbitrary congruence subgroup of SL_{2}(ℤ).

Notation. For the sake of brevity, we write

τ_{n} = τ(n)/n^{11/2},

and

f(s) = ∑_{n=1}^{∞} \frac{τ_{n}^{2}}{n^{s}} .

2. Real Zeroes of f(s)

We show that f(s) has no real zeroes in the critical strip 0 ≤ σ ≤ 1.

Let z = x + iy and set

ϕ(z, s) = \frac{s(s − 1)}{2} (\frac{y}{π} )^{s} Γ (s)^{∑′} |mz + n|^{−2s} ,

where the dash on the summation indicates we sum over all pairs of integers (m, n) ≠ (0, 0). If we let

K(z, w) = ^{∑′} exp(−\frac{πw}{y} |mz + n|^{2}),

then it is easily seen that

ϕ(z, s) = \frac{1}{2} + \frac{s(s − 1)}{2} ∫_{1}^{∞}(w^{s−1} + w^{−s})K(z, w) dw  (2)

as

1 + K(z, w) = \frac{1}{w} \{1 + K (z, \frac{1}{w} )\} .

Letting

ψ(s) = (2π)^{−2(s+11)}Γ (s + 11)Γ (s)ζ(2s)f(s)s(s − 1)

we see that

ψ(s) = ∫∫_{D} y^{12} · |Δ(z)|^{2} · ϕ(z, s)\frac{dxdy}{y^{2}} ,  (3)

where D denotes the standard fundamental domain for the full modular group acting on the upper half-plane. Also, ψ satisfies the functional equation

ψ(s) = ψ(1 − s).

In view of this functional equation and the fact that f(s) has a simple pole at s = 1, it suffices to consider \frac{1}{2} ≤ s < 1 in our search for real zeroes.

Lemma 1.

∫_{1}^{∞}K(z, w) dw ≤ log (\frac{e^{γ+1}}{4π} ) − 2 log(y^{\frac{1}{2}} |η(z)|^{2}),

where γ is Euler’s constant and

η(n) = e^{πiz/12} ∏_{n=1}^{∞}(1 − e^{2πinz}).

Proof. From Kronecker’s limit formula (see e.g. Ramachandra [7]), it follows that

lim_{s→1} [(\frac{y}{π} )^{s} Γ (s)^{∑′} |mz + n|^{−2s} − \frac{1}{s − 1}] = log (\frac{e^{γ}}{4π} ) − 2 log(y^{\frac{1}{2}} |η(n)|^{2}).

But

\frac{2ϕ(z, s)}{s(s − 1)} = \frac{1}{s − 1} − 1 + ∫_{1}^{∞}(1 + \frac{1}{w} ) K(z, w) dw + higher powers of (s − 1)

so that

−1 + ∫_{1}^{∞}(1 + \frac{1}{w} ) K(z, w) dw = log (\frac{e^{γ}}{4π} ) − 2 log(y^{\frac{1}{2}} |η(n)|^{2})

from which the result follows.

Corollary 1.

(i) for \frac{\sqrt{3}}{2} ≤ y ≤ 2, ∫_{1}^{∞}K(z, w) dw ≤ \frac{1}{2},

(ii) for y ≥ 1, ∫_{1}^{∞}K(z, w) dw ≤ \frac{π}{3} y.

Proof. We have

log |η(n)| = −\frac{πy}{12} + ∑_{n=1}^{∞} log |1 − e^{2πinz}| .

So that

− log |η(n)| ≤ \frac{πy}{12} + \frac{e^{−2πy}}{(1 − e^{−2πy})^{2}} .  (4)

It follows that

∫_{1}^{∞}K(z, w) dw ≤ \frac{π}{3} y − log y − 0.92

as

log (\frac{e^{γ+1}}{4π} ) = −0.95 …

4 R. BALASUBRAMANIAN AND M. RAM MURTY

and

\frac{e^{−2πy}}{(1 − e^{−2πy})^{2}} ≤ 5.0 × 10^{−3}

for y ≥ \sqrt{3}/2. Both (i) and (ii) are now easily deduced. Theorem 1. ψ(s) ≠ 0 for \frac{1}{2} ≤ s ≤ 1. Proof. From (2) and (3), we observe that

ψ(s) = \frac{1}{2}(Δ, Δ) + \frac{s(s − 1)}{2} ∫∫_{D} y^{12} |Δ(n)|^{2} \{∫_{1}^{∞}(w^{s−1} + w^{−s})K(z, w) dw\} \frac{dxdy}{y^{2}}

where (·, ·) denotes the Petersson inner product. It is apparent that for \frac{1}{2} ≤ s ≤ 1,

|ψ(s) − \frac{1}{2}(Δ, Δ)| ≤ \frac{1}{4} ∫∫_{D} y^{12} |Δ(z)|^{2} (∫_{1}^{∞}K(z, w) dw) \frac{dxdy}{y^{2}} .

By the corollary to Lemma 1, this is

≤ \frac{1}{8}(Δ, Δ) + \frac{π}{12} ∫∫_{\begin{matrix}D \\ y≥2 \end{matrix}} y^{11} |Δ(z)|^{2} dxdy.

Noting that

log |Δ(z)| = 24 log |η(n)| ≤ −2πy + 24 · \frac{e^{−2πy}}{(1 − e^{−2πy})^{2}} ,

we deduce

|Δ(z)| ≤ (1 · 1)e^{−2πy}

for y ≥ \sqrt{3}/2.

This estimate implies

∫∫_{\begin{matrix}D \\ y≥2 \end{matrix}} y^{11} |Δ(z)|^{2} dxdy ≤ (1 · 21) ∫_{2}^{∞}y^{11}e^{−4πy} dy ≤ (1 · 21) (\frac{2}{e})^{4π} \frac{e^{−4π}}{4π}
≤ (0 · 04)\frac{e^{−4π}}{4π} .

It follows that

ψ(s) ≥ \frac{3}{8}(Δ, Δ) − (0.01)\frac{e^{−4π}}{4π} .

We note that, if F (x) = ∑ c_{ne}^{2πinz}, then for k ≥ 2,

∫∫_{\begin{matrix}|x|< \frac{1}{2} \\ y>1 \end{matrix}} y^{k} |F (z)|^{2} \frac{dxdy}{y^{2}} = ∑_{n=1}^{∞} |c_{n}|^{2} ∫_{1}^{∞}y^{k−2}e^{−4πny} dy
≥ |c_{1}|^{2} ∫_{1}^{∞}e^{−4πy} dy
= |c_{1}|^{2} \frac{e^{−4π}}{4π} .

Taking in particular, k = 12, c_{n} = τ(n), we have

(Δ, Δ) ≥ \frac{e^{−4π}}{4π} .

We finally obtain

ψ(s) ≥ \frac{e^{−4π}}{16π} > 0

for \frac{1}{2} ≤ s ≤ 1.

Remarks. 1. Lehmer [4] has computed (Δ, Δ) = 1.036 × 10^{−6}.

2. It is possible to estimate

∫_{1}^{∞}K(z, w) dw

without appealing to Kronecker’s limit formula. We split the sum

∑′ ∫_{1}^{∞}exp (−\frac{πw}{y} |mz + n|^{2}) dw

into four parts corresponding to n = 0, m = 0, |n| ≤ |m| y, and |n| ≥ |m| y, where in the latter two cases, we utilise the inequalities |mz + n|^{2} ≥ m^{2}y^{2} and |mz + n|^{2} ≥ \frac{3}{4} |n|^{2} in the respective cases. The resulting four sums are easily estimated and the main contribution arises from the term corresponding to n = 0.

We indicate another proof of Theorem 1 which can be based on the following idea. From Chowla-Selberg [1, p. 106] we know

\frac{2ϕ(z, s)}{s(s − 1)} = \frac{ξ(2s)y^{s}}{s(2s − 1)} + \frac{ξ(2s − 1)y^{1−s}}{(s − 1)(2s − 1)} + R(y, s)

where

ξ(s) = \frac{s(s − 1)}{2}π^{−s/2}Γ (\frac{s}{2}) ζ(s)

and

|R(y, s)| ≤ \frac{8}{π\sqrt{y}} · \frac{1}{e^{πy} − 1}

for \frac{1}{2} ≤ s ≤ 1.

A simple calculation reveals |R(y, s)| ≤ 0.01. Utilizing the fact that

ξ(s) = \frac{1}{2} + \frac{s(s − 1)}{2} ∫_{1}^{∞}ψ(x)(x^{s−1} + x^{−s}) dx,

where

ψ(x) = ∑_{1}^{∞} e^{−n^{2}πx},

it is straightforward to show that for y ≤ 2,

\frac{2ϕ(z, s)}{s(s − 1)} ≤ \frac{1}{2s − 1} (\frac{y^{s}}{s} − \frac{y^{1−s}}{1 − s}) + 0.15.

A simple application of Rolle’s theorem reveals that

\frac{2ϕ(z, s)}{s(s − 1)} ≤ −0.1

for y ≤ 2.

A similar argument shows that for y ≥ 2, and \frac{1}{2} < s ≤ 1,

\frac{2ϕ(z, s)}{s(s − 1)} ≤ y^{2}.

These two inequalities are enough for Theorem 2 to be deduced.

3. Zeroes of ψ(s) in the Critical Strip

Let N(T, ψ) be the number of zeroes of ψ(s) satisfying 0 < σ < 1 and 0 < t < T.

Lemma 2.

N(T, ψ) = \frac{2}{π} T log T + O(T).

Proof. Let R be the rectangle with vertices \frac{3}{2}, \frac{3}{2} + iT, −\frac{1}{2} + iT, −\frac{1}{2}. In view of the functional equation and the fact that ψ(s) has no real zeroes in −\frac{1}{2} ≤ σ ≤ \frac{3}{2}, we see that

πN(T, ψ) = Δ_{L} arg ψ(s),

where Δ_{L} denotes the variation in the argument as s traverses from \frac{3}{2} to \frac{3}{2} + iT and then to \frac{1}{2} + iT. Stirling’s formula easily gives

Δ_{L} arg((2π)^{−2s−22}s(s − 1)Γ (s)Γ (s + 11)) = 2T log T + O(T).

Moreover, as ψ(s) is of order 1, it is deduced, in a standard way, that

∑_{ρ} \frac{1}{1 + (T − γ)^{2}} = O(log T),

as ρ runs through the zeroes of ψ(s).

It follows that the number of zeroes σ + iγ with |T − γ| < 1 is O(log T) and

\frac{2ζ′}{ζ} (2s) + \frac{f′}{f} (s) = \begin{matrix}∑ & ′ \\ ρ \end{matrix} \frac{1}{(s − ρ)} + O(log T),

where the dash on the summation indicates the sum is over zeroes of ψ(s) for which |t − γ| < 1 ρ = σ + iγ. We have

Δ_{L} arg(ζ(2s)f(s)) = ∫_{L} Im (2\frac{ζ′}{ζ} (2s) + \frac{f′}{f} (s)) ds

= O(1) − ∫_{\frac{1}{2} +iT}^{\frac{3}{2} +iT} Im (2\frac{ζ′}{ζ} (2s) + \frac{f′}{f} (s)) ds,

the O(1) term coming from the variation along σ = \frac{3}{2}. As

∫_{\frac{1}{2} +iT}^{\frac{3}{2} +iT} Im(s − ρ)^{−1}ds = Δ arg(s − ρ) = O(1)

for those zeroes ρ satisfying |t − γ| < 1 we deduce

Δ_{L}(arg(ζ(2s)f(s))) − O(log T).

This completes the proof.

4. Other Lemmas

Lemma 3. Let τ_{p}^{2} > 1. For such a prime p, there is an m = m(p) and an absolute constant c such that τ_{p^{m}} ≥ c > 1 and

m(p) ≪ \frac{1}{τ_{p}^{2} − 1}.

Proof. If τ_{p}^{2} − 1 > 10^{−10}, then take m(p) = 1. Now suppose

0 < τ_{p}^{2} − 1 < 10^{−10}.

Then θ_{p} is close to \frac{π}{3} or \frac{2π}{3} . We consider the case θ_{p} close to \frac{π}{3} , the other case being similar. Also, 0 < θ_{p} < \frac{π}{3} . If θ_{p} < \frac{π}{6} , we may take m(p) = 1. So we may assume \frac{π}{6} < θ_{p} < \frac{π}{3} .

Choose m ≡ 0 (mod 6) such that

\frac{\frac{π}{10}}{\frac{π}{3} − θ_{p}} < m + 1 < \frac{\frac{π}{10}}{\frac{π}{3} − θ_{p}} + 20

so that sin(m + 1)θ_{p} = sin ((m + 1)\frac{π}{3} + (m + 1) (θ_{p} − \frac{π}{3} )) ≥ sin (\frac{π}{3} + \frac{π}{10}) , as

|θ_{p} − \frac{π}{3} | < 2 |sin θ_{p}| |θ_{p} − \frac{π}{3} | ≤ |2 cos θ_{p} − 1| ≤ τ_{p}^{2} − 1 ≤ 10^{−10}.

Therefore,

\frac{sin(m + 1)θ_{p}}{sin θ_{p}} ≥ \begin{matrix}sin & (\frac{π}{3} + \frac{π}{10}) \\ sin & \frac{π}{3} \end{matrix} > 1.

Moreover, m satisfies

m ≪ \frac{1}{\frac{π}{3} − θ_{p}} ≤ \frac{\sqrt{3}}{τ_{p} − 1} ≤ \frac{3\sqrt{3}}{τ_{p}^{2} − 1}.

This completes the proof.

Lemma 4.

∑_{\begin{matrix}p \\ τ_{p}^{2}>1 \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{β}} = +∞,

for β < \frac{1}{2}.

Proof. Set θ(s) = \frac{ζ(2s)}{ζ(s)}f(s). We know

log θ(s) = ∑_{p,n} \frac{2 cos nθ_{p} + 1}{np^{ns}} = ∑_{p} \frac{τ_{p}^{2} − 1}{p^{s}} (1 + \frac{τ_{p}^{2} − 3}{2p^{s}} + ⋯) .

Now write

∑_{p} \frac{τ_{p}^{2} − 1}{p^{s}} = f_{+}(s) − f_{−}(s)

where

f_{+}(s) = ∑_{τ_{p}^{2}>1} \frac{τ_{p}^{2} − 1}{p^{s}}

and

f_{−}(s) = − ∑_{τ_{p}^{2}<1} \frac{τ_{p}^{2} − 1}{p^{s}}.

Suppose that f_{+}(\frac{1}{2}) < ∞. Then, for σ > \frac{1}{2}, f_{+}(s) is analytic. By Lemma 2, log θ(s) has singularities with Re s ≥ \frac{1}{2} arising from the zeroes of ψ(s). The set of singularities of log θ(s) coincides with the set of singularities of f_{−}(s) for Re s > \frac{1}{2}. If this set is not empty, f_{−}(s) has a real singularity by Landau’s theorem. Therefore, ϕ(s) has a real zero which contradicts Theorem 1. Therefore, all the singularities of log θ(s) lie on the line σ = \frac{1}{2}. As log θ(s) is analytic at s = \frac{1}{2}, both f_{+}(s) and f_{−}(s) have a singularity at s = \frac{1}{2}. Therefore,

∑_{\begin{matrix}p \\ τ_{p}^{2}>1 \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{\frac{1}{2} −ε}} = +∞

and

∑_{\begin{matrix}p \\ τ_{p}^{2}<1 \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{\frac{1}{2} −ε}} = −∞.

This completes the proof of the lemma.

5. Main Theorem

Theorem 2. Suppose ∑_{τ_{p}^{2}>1} \frac{τ_{p}^{2}−1}{p^{β}} = +∞. Then

τ_{n} = Ω(exp(c(log n)^{\frac{1}{2−β} −ε})).

Proof. Clearly, the set

S = \{m : ∑_{\begin{matrix}e^{m}<p<e^{m+1} \\ τ_{p}>1 \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{β}} ≥ \frac{2}{m^{2}} \}

is infinite. Since

∑_{\begin{matrix}e^{m}<p<e^{m+1} \\ 0<τ_{p}^{2}−1< \frac{1}{p^{1−β} log p} \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{β}} ≤ ∫_{e^{m}}^{e^{m+1}} \frac{dt}{t log^{2} t} ≤ \frac{1}{m^{2}} ,

we see that

T = \{m : ∑_{\begin{matrix}e^{m}<p<e^{m+1} \\ 3>τ_{p}^{2}−1> \frac{1}{p^{1−β} log p} \end{matrix}} \frac{τ_{p}^{2} − 1}{p^{β}} ≥ \frac{1}{m^{2}} \}

is infinite.

AN Ω-THEOREM FOR RAMANUJAN’S τ-FUNCTION 9

For each m ∈ T, we know

∑_{\frac{1}{100} <τ_{p}^{2}−1<3} + ∑_{p^{−ε}<τ_{p}^{2}−1< \frac{1}{100}} + ⋯ + ∑_{\frac{p^{β−1}}{log p} <τ_{p}^{2}−1<p^{β−1+ε}} \frac{τ_{p}^{2} − 1}{p^{β}} ≥ \frac{1}{m^{2}} .

The number of sums is O_{ε}(1). Therefore, for some γ,

∑_{p^{−γ−ε}<τ_{p}^{2}−1<p^{−γ}} \frac{τ_{p}^{2} − 1}{p^{β}} ≥ \frac{c(ε)}{m^{2}}.

Hence if

W = {p : e^{m} < p < e^{m+1}, p^{−γ−ε} < τ_{p}^{2} − 1 < p^{−γ}},

then

|W | e^{−m(β+γ)} ≥ \frac{c(ε)}{m^{2}}.

Now, set B ⊆ W such that |B| = \frac{(e^{m})^{β+γ}}{m^{2}} , and define

n = ∏_{\begin{matrix}p≤x \\ p∈B \end{matrix}} p^{m(p)}

where m(p) is defined by Lemma 3. Then by Lemma 3,

τ_{n} ≥ exp (\frac{cx^{β+γ}}{(log x)^{2}} ) .

Since,

log n ≤ ∑_{\begin{matrix}p≤x \\ p∈B \end{matrix}} p^{γ+ε} log p ≤ \frac{x^{β+2γ+ε}}{(log x)}

we have

x^{β+2γ+ε} ≥ (log n)(log x).

As

log n ≥ ∑_{\begin{matrix}p≤x \\ p∈B \end{matrix}} log p ≥ \frac{x^{β+γ}}{(log x)^{2}} ,

we have

τ_{n} ≥ exp (\frac{c((log n)(log log n))^{\frac{β+γ}{β+2γ+ε}}}{(log log n)^{2}} ) .

Noting that,

0 ≤ γ + ε ≤ 1 − β,

we finally deduce

τ_{n} ≥ exp(c(log n)^{\frac{1}{2−β} −ε}),

as desired. Corollary 2. τ_{n} = Ω(exp(c(log n)^{\frac{2}{3} −ε})). Proof. By Lemma 4 any β < \frac{1}{2} satisfies the condition of the theorem. This gives the result. Remarks. 1. By utilizing the fact that

∑_{τ_{p}^{2}<1} \frac{τ_{p}^{2} − 1}{p^{\frac{1}{2} −ε}} = −∞

and repeating the above argument, one can deduce that

|τ_{n}| < exp(−c(log n)^{\frac{2}{3} −ε})

for an infinity of n.

2. The argument can be extended to any real valued multiplicative function c satisfying

(i) c(p)^{2} − 1 = c(p^{2})

(ii) c(n) = O(n^{ε})

(iii) the Dirichlet series \frac{ζ(2s)}{ζ(s)} ∑_{n=1}^{∞} \frac{c(n)^{2}}{n^{s}} = L(s) (say) has an analytic

continuation to Re s = 0.

Also, if L(s) has only non-real zeroes in Re s ≥ \frac{1}{2}, then

c(n) = Ω(exp(c(log n)^{\frac{1}{2} −ε})).

In fact, if Re s ≥ σ is the largest zero-free half-plane for L(s), then

c(n) = Ω(exp(c(log n)^{σ−ε})).

6. General Results

Let f(z) = ∑_{n=1}^{∞} a(n)e^{2πinz} be a normalized Hecke eigenform of weight k for the full modular group. Then k is even and as the a(n)’s are integers, we have for primes p,

|a(p)^{2} − p^{k−1}| ≥ 1.

if we let

a_{p} = \frac{a(p)}{p^{\frac{k−1}{2}}} ,

then

|a_{p}^{2} − 1| ≥ \frac{1}{p^{k} − 1}.

Lemma 5. There is an m = m(p) such that |a_{p^{m}}| ≥ c > 1, where c is a fixed constant and

m(p) ≪ \frac{1}{|a_{p}^{2} − 1|}.

The proof of this lemma proceeds exactly as in Lemma 3 and therefore, we suppress it.

By the preceding remarks, it is evident that m(p) in Lemma 5 satisfies

m(p) ≪ p^{k−1}.

Therefore, if we let

n = ∏_{p≤x} p^{m(p)},

then

a_{n} ≥ c^{π(x)}.

But

log n = ∑_{p≤x} m(p) log p ≤ x^{k}.

This proves the following theorem.

Theorem 3. If f(z) = ∑_{1}^{∞} a(n)e^{2πinz} is a normalized Hecke eigenform, weight k, then

a(n) = Ω (n^{\frac{k−1}{2}} exp (\frac{c(log n)^{\frac{1}{k}}}{log log n} )) .

Acknowledgements. We would like to thank Professor A. Selberg for many of the ideas contained in this paper and Professor S. Chowla for his kind encouragement.

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School of Mathematics, Institute for Advanced Study, Princeton, NJ 08540, USA