Ramanujan’s τ -function Michael Bennett Values of the Ramanujan τ -function Michael Bennett (joint with Adela Gherga, Vandita Patel and Samir Siksek) University of British Columbia (and University of Manchester, Warwick University) Fields Institute : November 2020
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Ramanujan’sτ -function
MichaelBennett
Values of the Ramanujan τ -function
Michael Bennett (joint with Adela Gherga, Vandita Pateland Samir Siksek)
University of British Columbia (and University of Manchester, WarwickUniversity)
Fields Institute : November 2020
Ramanujan’sτ -function
MichaelBennett The Ramanujan τ -function
The Ramanujan τ -function τ(n) is defined via the expansion
It was conjectured by Ramanujan and proved by Mordell thatτ(n) is a multiplicative function, i.e. that
τ(n1n2) = τ(n1)τ(n2),
for all coprime pairs of positive integers n1 and n2.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : parity
Further, we have
∞∑n=1
τ(n)qn ≡ q∞∏n=1
(1 + q8n)3 ≡ q∞∏n=1
(1− q8n)(1 + q8n)2
and so∞∑n=1
τ(n)qn ≡∞∑m=0
q(2m+1)2 mod 2,
via Jacobi’s triple product formula, whence τ(n) is oddprecisely when n is an odd square and, in particular, τ(p) iseven for every prime p.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : arithmetic
Amongst the many open questions about the possible values ofτ(n), the most notorious is an old conjecture of Lehmer to theeffect that τ(n) never vanishes.
There are many papers in the literature proving results aboutthe arithmetic nature of τ(n) under various hypotheses : onaverage, for most n, subject to GRH, etc. See e.g. work ofMurty-Murty, Garavev, Shparlinski, Konyagin, etc.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : upper bounds
In terms of the size of values of τ , one has the upper bound ofDeligne (originally conjectured by Ramanujan) :
|τ(p)| ≤ 2 · p11/2,
valid for prime p.
Ramanujan’sτ -function
MichaelBennett The Ramanujan τ -function : lower bounds
In the other direction, Atkin and Serre conjectured (as astrengthening of Lehmer’s conjecture) that, for ε > 0,
|τ(p)| �ε p9/2−ε,
so that, in particular, given a fixed integer a, there are at mostfinitely many primes p for which
τ(p) = a.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : odd values
While this problem remains open, in the special case where theinteger a is odd, Murty, Murty and Shorey (1987) proved thatthe equation
τ(n) = a,
has at most finitely many solutions in integers n (note that, inthis case, n is necessarily an odd square). More precisely, theydemonstrated the existence of an effectively computablepositive constant c such that if τ(n) is odd, then
|τ(n)| > (log(n))c.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : odd values continued
A number of recent papers have treated the problem ofexplicitly demonstrating that the equation
τ(n) = a
has, in fact, no solutions, for various odd values of a, including
a = ±1 (Lygeros and Rozier)
|a| < 100 an odd prime (Balakrishnan, Craig and Ono,Balakrishnan, Craig, Ono and Tsai, Dembner and Jain)
|a| < 100 an odd integer (Hanada and Madhukara)
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : odd values continued
A number of recent papers have treated the problem ofexplicitly demonstrating that the equation
τ(n) = a
has, in fact, no solutions, for various odd values of a, including
a = ±1 (Lygeros and Rozier)
|a| < 100 an odd prime (Balakrishnan, Craig and Ono,Balakrishnan, Craig, Ono and Tsai, Dembner and Jain)
|a| < 100 an odd integer (Hanada and Madhukara)
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : odd values continued
A number of recent papers have treated the problem ofexplicitly demonstrating that the equation
τ(n) = a
has, in fact, no solutions, for various odd values of a, including
a = ±1 (Lygeros and Rozier)
|a| < 100 an odd prime (Balakrishnan, Craig and Ono,Balakrishnan, Craig, Ono and Tsai, Dembner and Jain)
|a| < 100 an odd integer (Hanada and Madhukara)
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : odd values continued
A number of recent papers have treated the problem ofexplicitly demonstrating that the equation
τ(n) = a
has, in fact, no solutions, for various odd values of a, including
a = ±1 (Lygeros and Rozier)
|a| < 100 an odd prime (Balakrishnan, Craig and Ono,Balakrishnan, Craig, Ono and Tsai, Dembner and Jain)
|a| < 100 an odd integer (Hanada and Madhukara)
Ramanujan’sτ -function
MichaelBennett
A non-Archimidean analogue of Murty-Murty-Shorey
Theorem (B., Gherga, Patel, Siksek 2020)
There exists an effectively computable constant κ > 0 suchthat if τ(n) is odd, with n ≥ 25, then either
P (τ(n)) > κlog log log n
log log log log n,
or there exists a prime p | n for which τ(p) = 0.
Ramanujan’sτ -function
MichaelBennett
A non-Archimidean analogue of Murty-Murty-Shorey
Recall that a powerful number (also known as squarefull or2-full) is defined to be an integer n with the property that if aprime p | n, then necessarily p2 | n.
Equivalently, we can write such an integer as n = a2b3, wherea and b are integers. Our techniques actually show thefollowing (from which the preceding Theorem is an immediateconsequence).
Ramanujan’sτ -function
MichaelBennett
A non-Archimidean analogue of Murty-Murty-Shorey
Theorem (B., Gherga, Patel, Siksek 2020)
We havelimn→∞
P (τ(n)) =∞,
where the limit is taken over powerful numbers n for whichτ(p) 6= 0 for each p | n. More precisely, there exists aneffectively computable constant κ > 0 such that if n ≥ 25 ispowerful, either
P (τ(n)) > κlog log log n
log log log log n,
or there exists a prime p | n for which τ(p) = 0.
Ramanujan’sτ -function
MichaelBennett
If Lehmer’s Conjecture is false.....
The restriction that n have no prime divisors p for whichτ(p) = 0 is, in fact, necessary if one wishes to obtain a lowerbound upon P (τ(n)) that tends to ∞ with n. Indeed, one mayobserve that, if τ(p) = 0, then
P(τ(p2k)
)= P
((−1)kp11k
)= p
is bounded, independent of k. While Lehmer’s conjectureremains unproven, we do know that if there is a prime p forwhich τ(p) = 0, then
p > 816212624008487344127999,
by work of Derickx, van Hoeij and Zeng.
Ramanujan’sτ -function
MichaelBennett
What we really prove....
We will actually show that, given m ≥ 2 and prime p for whichτ(p) 6= 0,
P (τ(pm))� log log(pm), for m ∈ {2, 3, 4, 5}, say,
and
P (τ(pm))� log log(pm)
log log log(pm),
more generally, where the implied constants are absolute.
Ramanujan’sτ -function
MichaelBennett
Explicit results
To demonstrate that these results and the techniquesunderlying them are somewhat practical, we prove the followingcomputational “coda” :
Theorem (B., Gherga, Patel, Siksek 2020)
If n is a powerful positive integer, then either n = 8, where wehave
τ(8) = 29 · 3 · 5 · 11,
orP (τ(n)) ≥ 13.
Ramanujan’sτ -function
MichaelBennett
Explicit results
Corollary (B., Gherga, Patel, Siksek 2020)
If n is a positive integer for which τ(n) is odd, then
P (τ(n)) ≥ 13.
In other words, the equation
τ(n) = ±3α5β7γ11δ
has no solutions in integers n ≥ 2 and α, β, γ, δ ≥ 0.
Ramanujan’sτ -function
MichaelBennett
The Ramanujan τ -function : prime values
It is conjectured that |τ(n)| takes on infinitely many primevalues, the smallest of which corresponds to
τ(2512) = −80561663527802406257321747.
Our arguments enable us to eliminate the possibility of powersof small primes arising as values of τ . By way of example, wehave the following.
Theorem (B., Gherga, Patel, Siksek 2020)
The equationτ(n) = ±qα
has no solutions in prime q with 3 ≤ q < 100 and α, n ∈ Z.
Ramanujan’sτ -function
MichaelBennett
Our techniques
Our proofs rely upon two quite different approaches. The first,to derive inequalities of the shape
P (τ(n)) > κlog log log n
log log log log n,
proceeds in the same way as the original work ofMurty-Murty-Shorey, appealing to bounds for linear forms inlogarithms (though we necessarily must use both bounds forcomplex and p-adic logarithms). The second (for our explicitresults), uses Frey curves and the modularity of associatedGalois representations.
Ramanujan’sτ -function
MichaelBennett
Generalities: coefficients of cuspidal newforms
It is worth observing that the techniques we employ are readilyextended to treat more generally coefficients λf (n) of cuspidalnewforms of (even) weight k ≥ 4 for Γ0(N), with trivialcharacter and λf (p) even for suitably large prime p; our resultscorrespond to the case of ∆(z), where k = 12 and N = 1. Werestrict our attention to this latter case for simplicity.
Ramanujan’sτ -function
MichaelBennett
Lucas sequences
A Lucas pair is a pair (α, β) of algebraic numbers such thatα+ β and αβ are non-zero coprime rational integers, and α/βis not a root of unity. In particular, associated to the Lucas pair(α, β) is a characteristic polynomial
X2 − (α+ β)X + αβ ∈ Z[X].
This polynomial has discriminant D = (α− β)2 ∈ Z \ {0}.Given a Lucas pair (α, β), the corresponding Lucas sequenceis given by
un =αn − βn
α− β, n = 0, 1, 2, . . . .
Ramanujan’sτ -function
MichaelBennett
Lucas sequences : primitive divisors
Let (α, β) be a Lucas pair. A prime ` is a primitive divisor ofthe n-th term of the corresponding Lucas sequence if ` dividesun but ` fails to divide (α− β)2 · u1u2 . . . un−1.
We shall make essential use of the celebrated Primitive DivisorTheorem of Bilu, Hanrot and Voutier.
Theorem (Bilu, Hanrot and Voutier)
Let (α, β) be a Lucas pair. If n ≥ 5 and n 6= 6 then un has aprimitive divisor.
Ramanujan’sτ -function
MichaelBennett
Lucas sequences : rank of apparition
Let ` be a prime. The smallest positive integer m such that` | um is called the rank of apparition of `; we denote this bym`.
The important result is that, if m` is finite, then either m` = `,or m` | `± 1.
Ramanujan’sτ -function
MichaelBennett From Ramanujan τ to Lucas sequences
Let us fix a prime p and consider the sequence{1, τ(p), τ(p2), τ(p3), . . .
}.
We will associate to this a Lucas pair and a correspondingLucas sequence. Our starting point is the identity
τ(pm) = τ(p)τ(pm−1)− p11τ(pm−2),
valid for all integer m ≥ 2. Once again, this was conjectured byRamanujan and proved by Mordell.
Ramanujan’sτ -function
MichaelBennett
The degenerate case
If Lehmer’s conjecture fails and there exists a prime p for whichτ(p) = 0, then our recursion implies that
τ(pn) = 0 for all odd n, and
τ(pn) = (−1)n/2p11n/2 for even n.
In particular, this does not correspond to a Lucas sequence.
Ramanujan’sτ -function
MichaelBennett
The degenerate case
If Lehmer’s conjecture fails and there exists a prime p for whichτ(p) = 0, then our recursion implies that
τ(pn) = 0 for all odd n, and
τ(pn) = (−1)n/2p11n/2 for even n.
In particular, this does not correspond to a Lucas sequence.
Ramanujan’sτ -function
MichaelBennett
The degenerate case
If Lehmer’s conjecture fails and there exists a prime p for whichτ(p) = 0, then our recursion implies that
τ(pn) = 0 for all odd n, and
τ(pn) = (−1)n/2p11n/2 for even n.
In particular, this does not correspond to a Lucas sequence.
Ramanujan’sτ -function
MichaelBennett
Otherwise
If we have that pk | τ(p), we may conclude that k ≤ 5, viaDeligne’s bounds. Supposing that pr ‖ τ(p), we write
un =τ(pn−1)
pr(n−1), n ≥ 1.
We may show that {un} is a Lucas sequence with
α =α′
pr, β =
β′
pr,
for α′ and β′ the roots of the quadratic equation
X2 − τ(p)X + p11 = 0.
Ramanujan’sτ -function
MichaelBennett
An aside : p | τ(p)
We note that p | τ(p) for
p = 2, 3, 5, 7, 2411, 7758337633, . . .
We expect that p | τ(p) for infinitely many primes p; see workof Lygeros and Rozier for a discussion of this problem andrelated computations.
Ramanujan’sτ -function
MichaelBennett Towards the proofs
We haveτ(p2) = τ2(p)− p11,
τ(p3) = τ(p)(τ2(p)− 2p11)
andτ(p4) = p22 − 3p11τ2(p) + τ4(p).
We can rewrite this last equation as
4τ(p4) =(2τ2(p)− 3p11
)2 − 5p22.
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ∈ {2, 3, 4, 5}
In each case, we have that
P (τ(pm)) ≥ P (Y 2 −DX11),
for D ≤ 5, X and Y integers.
Appealing to bounds for linear forms in complex and q-adiclogarithms (say due to Bugeaud and Gyory), we find that
P (τ(pm)) ≥ c1 log log(pm), for m ∈ {2, 3, 4, 5},
for some positive constant c1.
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ∈ {2, 3, 4, 5}
In each case, we have that
P (τ(pm)) ≥ P (Y 2 −DX11),
for D ≤ 5, X and Y integers.
Appealing to bounds for linear forms in complex and q-adiclogarithms (say due to Bugeaud and Gyory), we find that
P (τ(pm)) ≥ c1 log log(pm), for m ∈ {2, 3, 4, 5},
for some positive constant c1.
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ≥ 6
More generally, for m ≥ 6, we can write
τ(pm) = τ δ(p)Fm(p11, τ2(p)),
where δ ∈ {0, 1}, δ ≡ m mod 2 and Fm(X,Y ) is a binaryform with integer coefficients and degree [m/2].Explicitly, expanding
1
1−√Y T +XT 2
=
∞∑m=0
Y δ/2 Fm(X,Y )Tm,
we find, for example, that
F6(X,Y ) = −X3 + 6X2Y − 5XY 2 + Y 3.
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ≥ 6
Note that from the fact that the Lucas sequences {un} aredivisibility sequences, we have
Fm(X,Y ) | Fn(X,Y ) if n ≡ −1 mod m+ 1,
whence many of these forms are, in fact, reducible. Explicitly,we can write
Fm(X,Y ) =
[m/2]∑j=1
(−1)j(m− jj
)Y [m/2]−jXj
and so
Fm(X,Y ) =
[m/2]∏j=1
(Y − 4X cos2
(πj
m+ 1
)).
Ramanujan’sτ -function
MichaelBennett τ(pm),m ≥ 6
We observe that
Fm(X,Y ) = Gm(X,Y − 2X), (1)
where
Gm(X,Y ) =
[m/2]∏j=1
(Y − 2X cos
(2πj
m+ 1
)). (2)
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ≥ 6 : Theorem 1
We write
τ(pm) = prmum+1 = prm · um+1
uqα· uq
α
uqα−1
· uqα−1 ,
where qα is the largest prime power divisor of m+ 1.
After ruling out some small cases, we have that the termuqα
uqα−1
is an irreducible binary form F (X,Y ) in X = p11−2r andY = p−2rτ2(p), of degree φ(qα)/2 ≥ 3, where φ denotes theEuler phi-function. Explicitly, we have
F (X,Y ) =
φ(qα)/2∏j=1
gcd(j,q)=1
(Y − 4X cos2
(πj
qα
)).
Ramanujan’sτ -function
MichaelBennett
τ(pm),m ≥ 6 : Theorem 1
We write
τ(pm) = prmum+1 = prm · um+1
uqα· uq
α
uqα−1
· uqα−1 ,
where qα is the largest prime power divisor of m+ 1.
After ruling out some small cases, we have that the termuqα
uqα−1
is an irreducible binary form F (X,Y ) in X = p11−2r andY = p−2rτ2(p), of degree φ(qα)/2 ≥ 3, where φ denotes theEuler phi-function. Explicitly, we have
F (X,Y ) =
φ(qα)/2∏j=1
gcd(j,q)=1
(Y − 4X cos2
(πj
qα
)).
Ramanujan’sτ -function
MichaelBennett
Thue-Mahler equations
We haveP (F (X,Y )) ≤ P (τ(pm)).
Putting this all together (using bounds for solutions ofThue-Mahler equations due to Bugeaud and Gyory, we find that
P (τ(pm))� log log(pm)
log log log(pm).
If we choose pm to be the largest prime power divisor of ourpowerful number n, then pm � log n, whereby
P (τ(n))� log log log n
log log log log n,
as claimed.
Ramanujan’sτ -function
MichaelBennett
P (τ(pm)) : explicit results
To deal with, for example, the equation
τ(pm) = ±2α3β5γ7δ11ω,
our approach is as follows
Use Ramanujan’s congruences for τ(n) modulo powers of2, 3, 5, 7 and 23 to reduce the possibilities for theexponents
Appeal to the Primitive Divisor Theorem to restrict m
Solve the Thue-Mahler equations associated with m ≥ 6
Treat the equations corresponding to τ(pm), m ≤ 5.
Ramanujan’sτ -function
MichaelBennett
P (τ(pm)) : explicit results
To deal with, for example, the equation
τ(pm) = ±2α3β5γ7δ11ω,
our approach is as follows
Use Ramanujan’s congruences for τ(n) modulo powers of2, 3, 5, 7 and 23 to reduce the possibilities for theexponents
Appeal to the Primitive Divisor Theorem to restrict m
Solve the Thue-Mahler equations associated with m ≥ 6
Treat the equations corresponding to τ(pm), m ≤ 5.
Ramanujan’sτ -function
MichaelBennett
P (τ(pm)) : explicit results
To deal with, for example, the equation
τ(pm) = ±2α3β5γ7δ11ω,
our approach is as follows
Use Ramanujan’s congruences for τ(n) modulo powers of2, 3, 5, 7 and 23 to reduce the possibilities for theexponents
Appeal to the Primitive Divisor Theorem to restrict m
Solve the Thue-Mahler equations associated with m ≥ 6
Treat the equations corresponding to τ(pm), m ≤ 5.
Ramanujan’sτ -function
MichaelBennett
P (τ(pm)) : explicit results
To deal with, for example, the equation
τ(pm) = ±2α3β5γ7δ11ω,
our approach is as follows
Use Ramanujan’s congruences for τ(n) modulo powers of2, 3, 5, 7 and 23 to reduce the possibilities for theexponents
Appeal to the Primitive Divisor Theorem to restrict m
Solve the Thue-Mahler equations associated with m ≥ 6
Treat the equations corresponding to τ(pm), m ≤ 5.
Ramanujan’sτ -function
MichaelBennett P (τ(p2)) :Frey curves
If we suppose that
τ(p2) = ±2α3β5γ7δ11ω,
for p odd, then we necessarily have α = γ = δ = 0 andβ ∈ {0, 1}.
We consider the Frey curves{Ep : Y 2 = X(X2 + 2τ(p)X + τ(p)2 − p11) if p ≡ 1 mod 4,
Ep : Y 2 = X(X2 + 2τ(p)X + p11) if p ≡ 3 mod 4.
Ramanujan’sτ -function
MichaelBennett P (τ(p2)) :Frey curves
If we suppose that
τ(p2) = ±2α3β5γ7δ11ω,
for p odd, then we necessarily have α = γ = δ = 0 andβ ∈ {0, 1}. We consider the Frey curves{Ep : Y 2 = X(X2 + 2τ(p)X + τ(p)2 − p11) if p ≡ 1 mod 4,
Ep : Y 2 = X(X2 + 2τ(p)X + p11) if p ≡ 3 mod 4.
Ramanujan’sτ -function
MichaelBennett
P (τ(p2)) :Frey curves
It follows from the modularity theorem and Ribet’s levellowering theorem that there is a normalized newform
f = q +
∞∑n=1
cnqn
of weight 2 and level 25 · 3δ111δ2 and a prime $ | 11 in theintegers of K = Q(c1, c2, . . . ) so that
ρEp,11 ∼ ρf,$.
In particular, we have
NormK/Q(a`(Ep)− c`) ≡ 0 mod 11.
Ramanujan’sτ -function
MichaelBennett
P (τ(p2)) :Frey curves
Recall that we had
τ2(p)− p11 = ±3β11ω, β ∈ {0, 1}.
Sieving using
NormK/Q(a`(Ep)− c`) ≡ 0 mod 11,
for ` - 66, ` < 200, we find that either
β = 1, ω = 0 or β = 0, 11 | ω.
Ramanujan’sτ -function
MichaelBennett P (τ(p2)) :Frey curves, final steps
We eliminate the first of these cases by reducing it to the Thueequation