The τ of Ramanujan

V. Kumar MurtyUniversity of Toronto

Indian Institute of Science Education and Research, BhopalBhopal, India

October 10, 2011

What is the τ function?

It is defined by the equality

q∞∏n=1

(1− qn)24 =∞∑n=1

τ(n)qn.

Thus, τ(1) = 1, τ(2) = −24, τ(3) = 252.

Why is it interesting?

I The main aim of this talk is to explain why Ramanujan’sstudy of the arithmetic properties of this function is not onlyinteresting, but one of the epoch-making insights of ourcentury.

I As with much of Ramanujan’s work, many different strandsare tied together here and we have to separate them tounderstand what is going on.

Why is it interesting?

I The main aim of this talk is to explain why Ramanujan’sstudy of the arithmetic properties of this function is not onlyinteresting, but one of the epoch-making insights of ourcentury.

I As with much of Ramanujan’s work, many different strandsare tied together here and we have to separate them tounderstand what is going on.

A table of the first few values

n τ(n) n τ(n) n τ(n)

1 1 11 534,612 21 -4,219,488

2 -24 12 -370,944 22 -12,830,688

3 252 13 -577,738 23 18,643,272

4 -1472 14 401,856 24 21,288,960

5 4830 15 1,217,160 25 -25,499,225

6 -6048 16 987,136 26 13,865,712

7 -16744 17 -6,905,934 27 -73,279,080

8 84480 18 2,727,432 28 24,647,168

9 -113,643 19 10,661,420 29 128,406,630

10 -115,920 20 -7,109,760 30 -29,211,840

Growth Estimate

I The table suggests the values are growing quite rapidly.

I Ramanujan proved that τ(n) < n6 so it is in fact at mostpolynomial growth.

I He conjectured a stronger estimate

|τ(n)| ≤ d(n)n11/2

where d(n) denotes the number of positive divisors of n.

Growth Estimate

I The table suggests the values are growing quite rapidly.

I Ramanujan proved that τ(n) < n6 so it is in fact at mostpolynomial growth.

I He conjectured a stronger estimate

|τ(n)| ≤ d(n)n11/2

where d(n) denotes the number of positive divisors of n.

Growth Estimate

I The table suggests the values are growing quite rapidly.

I Ramanujan proved that τ(n) < n6 so it is in fact at mostpolynomial growth.

I He conjectured a stronger estimate

|τ(n)| ≤ d(n)n11/2

where d(n) denotes the number of positive divisors of n.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

The Ramanujan Hypothesis

I The question attracted the attention and efforts of manyauthors who proved significant but incremental improvementsof Ramanujan’s estimate.

I Hardy gave it the name “The Ramanujan Hypothesis”.

I It was finally proved by Deligne by reducing it to the Weilconjectures.

I The Weil conjectures are a set of conjectures about zetafunctions associated to algebraic varieties over finite fields. Inparticular, they give an analogue of the Riemann Hypothesisfor such zeta functions.

I It is not at all evident that there is any connection to theRamanujan Hypothesis.

I We don’t have an alternate proof.

Convergence and HolomorphyI Thinking of q as a complex number with |q| < 1 and taking

the logarithm of∞∏n=1

(1− qn)24

we get

−24∞∑n=1

∞∑m=1

qnm

m= −24

∞∑m=1

1

m

qm

1− qm.

This converges for |q| < 1.

I For z ∈ C with positive imaginary part, q = e2πiz satisfies|q| < 1. Thus, the function

∆(z) = q∞∏n=1

(1− qn)24

is a holomorphic function on the upper half plane. Moreover,it is non-vanishing there.

Convergence and HolomorphyI Thinking of q as a complex number with |q| < 1 and taking

the logarithm of∞∏n=1

(1− qn)24

we get

−24∞∑n=1

∞∑m=1

qnm

m= −24

∞∑m=1

1

m

qm

1− qm.

This converges for |q| < 1.I For z ∈ C with positive imaginary part, q = e2πiz satisfies|q| < 1. Thus, the function

∆(z) = q∞∏n=1

(1− qn)24

is a holomorphic function on the upper half plane. Moreover,it is non-vanishing there.

The Discriminant Function

I Replacing z by z + 1 doesn’t change the value of q = e2πiz so

∆(z + 1) = ∆(z).

I Less evident is that replacing z by −1/z also gives rise to asymmetry

∆(−1/z) = z12∆(z).

I These properties, together with the holomorphy and the“vanishing at ∞” makes ∆ a modular (cusp) form of weight12 for the group SL2(Z).

The Discriminant Function

I Replacing z by z + 1 doesn’t change the value of q = e2πiz so

∆(z + 1) = ∆(z).

I Less evident is that replacing z by −1/z also gives rise to asymmetry

∆(−1/z) = z12∆(z).

I These properties, together with the holomorphy and the“vanishing at ∞” makes ∆ a modular (cusp) form of weight12 for the group SL2(Z).

The Discriminant Function

I Replacing z by z + 1 doesn’t change the value of q = e2πiz so

∆(z + 1) = ∆(z).

I Less evident is that replacing z by −1/z also gives rise to asymmetry

∆(−1/z) = z12∆(z).

I These properties, together with the holomorphy and the“vanishing at ∞” makes ∆ a modular (cusp) form of weight12 for the group SL2(Z).

The Discriminant Function

I ∆(z) had been studied by other authors before Ramanujan,notably Dedekind.

I The transformation properties were known and used by theseauthors.

I ∆(z) occurs in many places such as the Kronecker LimitFormula, construction of Ramachandra units, transformationproperties of Dedekind sums, ...

The Discriminant Function

I ∆(z) had been studied by other authors before Ramanujan,notably Dedekind.

I The transformation properties were known and used by theseauthors.

I ∆(z) occurs in many places such as the Kronecker LimitFormula, construction of Ramachandra units, transformationproperties of Dedekind sums, ...

The Discriminant Function

I ∆(z) had been studied by other authors before Ramanujan,notably Dedekind.

I The transformation properties were known and used by theseauthors.

I ∆(z) occurs in many places such as the Kronecker LimitFormula, construction of Ramachandra units, transformationproperties of Dedekind sums, ...

∆(z) and sums of squares

I Ramanujan might have noticed it in the context of formulasfor the number of ways of writing an integer as a sum ofsquares.

I Clasically, Jacobi and others had explicit formulae for writinga number as a sum of 2 squares and 4 squares in terms offunctions like the function σ(n), the sum of the positivedivisors of n.

I The ∆ function arises when you ask about writing a numberas a sum of 24 squares. Here one gets an explicit formula interms of elementary functions like σk(n), the sum of the k-thpowers of the positive divisors of n (for various values of k)and one “mysterious” function, namely the τ function.

I In other words, the τ function appears as an “error” term.

∆(z) and sums of squares

I Ramanujan might have noticed it in the context of formulasfor the number of ways of writing an integer as a sum ofsquares.

I Clasically, Jacobi and others had explicit formulae for writinga number as a sum of 2 squares and 4 squares in terms offunctions like the function σ(n), the sum of the positivedivisors of n.

I The ∆ function arises when you ask about writing a numberas a sum of 24 squares. Here one gets an explicit formula interms of elementary functions like σk(n), the sum of the k-thpowers of the positive divisors of n (for various values of k)and one “mysterious” function, namely the τ function.

I In other words, the τ function appears as an “error” term.

∆(z) and sums of squares

I Ramanujan might have noticed it in the context of formulasfor the number of ways of writing an integer as a sum ofsquares.

I Clasically, Jacobi and others had explicit formulae for writinga number as a sum of 2 squares and 4 squares in terms offunctions like the function σ(n), the sum of the positivedivisors of n.

I The ∆ function arises when you ask about writing a numberas a sum of 24 squares. Here one gets an explicit formula interms of elementary functions like σk(n), the sum of the k-thpowers of the positive divisors of n (for various values of k)and one “mysterious” function, namely the τ function.

I In other words, the τ function appears as an “error” term.

∆(z) and sums of squares

I Ramanujan might have noticed it in the context of formulasfor the number of ways of writing an integer as a sum ofsquares.

I Clasically, Jacobi and others had explicit formulae for writinga number as a sum of 2 squares and 4 squares in terms offunctions like the function σ(n), the sum of the positivedivisors of n.

I The ∆ function arises when you ask about writing a numberas a sum of 24 squares. Here one gets an explicit formula interms of elementary functions like σk(n), the sum of the k-thpowers of the positive divisors of n (for various values of k)and one “mysterious” function, namely the τ function.

I In other words, the τ function appears as an “error” term.

Ramanujan’s great insight

I Ramanujan’s original contribution and great insight was torealize that the coefficients τ(n) might be interesting in andof themselves and not just as an “error” term.

I It is the beginning of the modern arithmetic theory ofautomorphic forms.

I Work of Hardy before coming into contact with Ramanujanmight be termed as all related to GL1 theory.

I Ramanujan’s insight opened the door to the study of GL2theory by number theorists, including Hardy.

Ramanujan’s great insight

I Ramanujan’s original contribution and great insight was torealize that the coefficients τ(n) might be interesting in andof themselves and not just as an “error” term.

I It is the beginning of the modern arithmetic theory ofautomorphic forms.

I Work of Hardy before coming into contact with Ramanujanmight be termed as all related to GL1 theory.

I Ramanujan’s insight opened the door to the study of GL2theory by number theorists, including Hardy.

Ramanujan’s great insight

I Ramanujan’s original contribution and great insight was torealize that the coefficients τ(n) might be interesting in andof themselves and not just as an “error” term.

I It is the beginning of the modern arithmetic theory ofautomorphic forms.

I Work of Hardy before coming into contact with Ramanujanmight be termed as all related to GL1 theory.

I Ramanujan’s insight opened the door to the study of GL2theory by number theorists, including Hardy.

Ramanujan’s great insight

I Ramanujan’s original contribution and great insight was torealize that the coefficients τ(n) might be interesting in andof themselves and not just as an “error” term.

I It is the beginning of the modern arithmetic theory ofautomorphic forms.

I Work of Hardy before coming into contact with Ramanujanmight be termed as all related to GL1 theory.

I Ramanujan’s insight opened the door to the study of GL2theory by number theorists, including Hardy.

Related Products

I ∆(z) has a product structure. We can look at other productsthat have interesting coefficients when expanded.

I Euler studied the product

∞∏n=1

(1− qn)

and proved that it is equal to∑n∈Z

(−1)nq(3n2+3n)/2.

I Note that this product does not have a nice transformationproperty like ∆ (in particular, it is not a modular form).

Related Products

I ∆(z) has a product structure. We can look at other productsthat have interesting coefficients when expanded.

I Euler studied the product

∞∏n=1

(1− qn)

and proved that it is equal to∑n∈Z

(−1)nq(3n2+3n)/2.

I Note that this product does not have a nice transformationproperty like ∆ (in particular, it is not a modular form).

Related Products

I ∆(z) has a product structure. We can look at other productsthat have interesting coefficients when expanded.

I Euler studied the product

∞∏n=1

(1− qn)

and proved that it is equal to∑n∈Z

(−1)nq(3n2+3n)/2.

I Note that this product does not have a nice transformationproperty like ∆ (in particular, it is not a modular form).

Euler and Jacobi

I Jacobi studied the cube

∞∏n=1

(1− qn)3

which is also not a modular form.

I He proved that it is equal to

∞∑n=0

(2n + 1)qn(n+1)/2.

I By raising Euler’s formula to the 24-th power, or Jacobi’sformula to the 8-th power, we get an expression for the τfunction.

Euler and Jacobi

I Jacobi studied the cube

∞∏n=1

(1− qn)3

which is also not a modular form.

I He proved that it is equal to

∞∑n=0

(2n + 1)qn(n+1)/2.

I By raising Euler’s formula to the 24-th power, or Jacobi’sformula to the 8-th power, we get an expression for the τfunction.

Euler and Jacobi

I Jacobi studied the cube

∞∏n=1

(1− qn)3

which is also not a modular form.

I He proved that it is equal to

∞∑n=0

(2n + 1)qn(n+1)/2.

I By raising Euler’s formula to the 24-th power, or Jacobi’sformula to the 8-th power, we get an expression for the τfunction.

A formula for the τ function

Using the identities of Euler and Jacobi, we get

(n−1)τ(n) =∑

1≤|m|≤an

(n − 1− 25m

2(3m + 1)

)(n − m

2(3m + 1)

)where

an =1

6

(1 + (1 + 24n)

12

).

This is due to Lehmer. There is another formula due toRamanujan that expresses τ(n) as an alternating weighted sum ofτ(m) with m�

√n.

A computational question

I The formula above shows that the value of τ(n) can becomputed in O(

√n) steps.

I Is there a faster way? In particular, can τ(n) be computed inpolynomial time?

A computational question

I The formula above shows that the value of τ(n) can becomputed in O(

√n) steps.

I Is there a faster way? In particular, can τ(n) be computed inpolynomial time?

Multiplicative Property

I Something that is not at all evident is that τ is multiplicative:

τ(mn) = τ(m)τ(n) if (m, n) = 1.

I Moreover, for powers of a fixed prime p, there is the degreetwo recursion

τ(pa+1) = τ(p)τ(pa)− p11τ(pa−1).

I Ramanujan conjectured both of these properties in his 1916paper. They were later proved by Mordell and then generalizedby Hecke in his theory of Hecke operators, a fundamentalcomponent of the modern theory of automorphic forms.

Multiplicative Property

I Something that is not at all evident is that τ is multiplicative:

τ(mn) = τ(m)τ(n) if (m, n) = 1.

I Moreover, for powers of a fixed prime p, there is the degreetwo recursion

τ(pa+1) = τ(p)τ(pa)− p11τ(pa−1).

I Ramanujan conjectured both of these properties in his 1916paper. They were later proved by Mordell and then generalizedby Hecke in his theory of Hecke operators, a fundamentalcomponent of the modern theory of automorphic forms.

Multiplicative Property

I Something that is not at all evident is that τ is multiplicative:

τ(mn) = τ(m)τ(n) if (m, n) = 1.

I Moreover, for powers of a fixed prime p, there is the degreetwo recursion

τ(pa+1) = τ(p)τ(pa)− p11τ(pa−1).

I Ramanujan conjectured both of these properties in his 1916paper. They were later proved by Mordell and then generalizedby Hecke in his theory of Hecke operators, a fundamentalcomponent of the modern theory of automorphic forms.

Back to Computation

I A result of Edixhoven (generalizing the famous algorithm ofSchoof for elliptic curves) shows that τ(p) can be computedin polynomial time.

I This means that τ(n) can be computed with the sameefficiency as factoring. With current knowledge, this meanssub-exponential time

O(exp((log n)12+ε).

Back to Computation

I A result of Edixhoven (generalizing the famous algorithm ofSchoof for elliptic curves) shows that τ(p) can be computedin polynomial time.

I This means that τ(n) can be computed with the sameefficiency as factoring. With current knowledge, this meanssub-exponential time

O(exp((log n)12+ε).

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Is the converse true? That is, given a method to computeτ(n) can n be factored?

I For an RSA modulus, n = pq with p and q large primes.

I Thenτ(n2) = (τ(p)2 − p11)(τ(q)2 − q11)

I Simplifying,

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I This means that we know τ(p)2q11 + τ(q)2p11.

I We also know the product τ(p)2q11τ(q)2p11 = τ(n)2n11.

I Hence, we know the individual terms τ(p)2q11 and τ(q)2p11.

τ(n) and factoring

I Dividing by n11, we get τ(p)2/p11 and τ(q)2/q11.

I From this, we can determine p and q.

I Summarizing, if n is an RSA modulus and we know τ(n) andτ(n2), we can factor n.

I Is there a general algorithm?

τ(n) and factoring

I Dividing by n11, we get τ(p)2/p11 and τ(q)2/q11.

I From this, we can determine p and q.

I Summarizing, if n is an RSA modulus and we know τ(n) andτ(n2), we can factor n.

I Is there a general algorithm?

τ(n) and factoring

I Dividing by n11, we get τ(p)2/p11 and τ(q)2/q11.

I From this, we can determine p and q.

I Summarizing, if n is an RSA modulus and we know τ(n) andτ(n2), we can factor n.

I Is there a general algorithm?

τ(n) and factoring

I Dividing by n11, we get τ(p)2/p11 and τ(q)2/q11.

I From this, we can determine p and q.

I Summarizing, if n is an RSA modulus and we know τ(n) andτ(n2), we can factor n.

I Is there a general algorithm?

τ(n) and primality

I If p is prime, we have

τ(p2) = τ(p)2 − p11.

I If for some n, we have

τ(n2) = τ(n)2 − n11

is n prime?

τ(n) and primality

I If p is prime, we have

τ(p2) = τ(p)2 − p11.

I If for some n, we have

τ(n2) = τ(n)2 − n11

is n prime?

τ(n) and primality

I We can certainly see that n cannot be of the form pq.

I Indeed, if n = pq,

τ(n2) = (τ(p)2 − p11)(τ(q)2 − q11).

I Expanding and rearranging, this is

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I The conditionτ(n2) = τ(n)2 − n11

would force p6|τ(p) and q6|τ(q) and this in turn forcesτ(p) = τ(q) = 0.

I But then τ(n) = 0 and so τ(n2) = τ(p2)τ(q2) = n11.

I In particular, τ(n2) 6= τ(n)2 − n11.

τ(n) and primality

I We can certainly see that n cannot be of the form pq.

I Indeed, if n = pq,

τ(n2) = (τ(p)2 − p11)(τ(q)2 − q11).

I Expanding and rearranging, this is

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I The conditionτ(n2) = τ(n)2 − n11

would force p6|τ(p) and q6|τ(q) and this in turn forcesτ(p) = τ(q) = 0.

I But then τ(n) = 0 and so τ(n2) = τ(p2)τ(q2) = n11.

I In particular, τ(n2) 6= τ(n)2 − n11.

τ(n) and primality

I We can certainly see that n cannot be of the form pq.

I Indeed, if n = pq,

τ(n2) = (τ(p)2 − p11)(τ(q)2 − q11).

I Expanding and rearranging, this is

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I The conditionτ(n2) = τ(n)2 − n11

would force p6|τ(p) and q6|τ(q) and this in turn forcesτ(p) = τ(q) = 0.

I But then τ(n) = 0 and so τ(n2) = τ(p2)τ(q2) = n11.

I In particular, τ(n2) 6= τ(n)2 − n11.

τ(n) and primality

I We can certainly see that n cannot be of the form pq.

I Indeed, if n = pq,

τ(n2) = (τ(p)2 − p11)(τ(q)2 − q11).

I Expanding and rearranging, this is

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I The conditionτ(n2) = τ(n)2 − n11

would force p6|τ(p) and q6|τ(q) and this in turn forcesτ(p) = τ(q) = 0.

I But then τ(n) = 0 and so τ(n2) = τ(p2)τ(q2) = n11.

I In particular, τ(n2) 6= τ(n)2 − n11.

τ(n) and primality

I We can certainly see that n cannot be of the form pq.

I Indeed, if n = pq,

τ(n2) = (τ(p)2 − p11)(τ(q)2 − q11).

I Expanding and rearranging, this is

τ(n2) = τ(n)2 + n11 − (τ(p)2q11 + τ(q)2p11).

I The conditionτ(n2) = τ(n)2 − n11

would force p6|τ(p) and q6|τ(q) and this in turn forcesτ(p) = τ(q) = 0.

I But then τ(n) = 0 and so τ(n2) = τ(p2)τ(q2) = n11.

I In particular, τ(n2) 6= τ(n)2 − n11.

Parity of the τ -function

The congruence

(1− qn)24 ≡ (1 + q8n)3 (mod 2)

and Jacobi’s identity

∞∏n=1

(1 + q8n)3 =∞∑

m=0

q4m(m+1)

gives the congruence

∞∑n=1

τ(n)qn ≡ q∞∑

m=0

q4m(m+1) ≡∞∑

m=0

q(2m+1)2 (mod 2).

In particular, τ(n) is odd if and only if n is an odd square. Inparticular, τ(p) is even for every prime p.

Relations discovered by Ramanujan

Ramanujan proved other congruences

I τ(p) ≡ 1 + p (mod 3)

I τ(p) ≡ p + p10 (mod 52)

I τ(p) ≡ p + p4 (mod 7)

I τ(p) ≡ 1 + p11 (mod 691)

Relations discovered by Ramanujan

Ramanujan proved other congruences

I τ(p) ≡ 1 + p (mod 3)

I τ(p) ≡ p + p10 (mod 52)

I τ(p) ≡ p + p4 (mod 7)

I τ(p) ≡ 1 + p11 (mod 691)

Relations discovered by Ramanujan

Ramanujan proved other congruences

I τ(p) ≡ 1 + p (mod 3)

I τ(p) ≡ p + p10 (mod 52)

I τ(p) ≡ p + p4 (mod 7)

I τ(p) ≡ 1 + p11 (mod 691)

Relations discovered by Ramanujan

Ramanujan proved other congruences

I τ(p) ≡ 1 + p (mod 3)

I τ(p) ≡ p + p10 (mod 52)

I τ(p) ≡ p + p4 (mod 7)

I τ(p) ≡ 1 + p11 (mod 691)

Galois Representations

I In the theory of Serre and Swinnerton-Dyer, these arise fromreducibility properties of certain families of Galoisrepresentations

ρ` : Gal(Q/Q) −→ GL2(Z`).

Here ` runs over rational primes.

I Deligne’s proof of the Ramanujan Hypothesis also constructsa family of such representations with the property that ρ` isunramified at all primes other than ` and for p 6= `, aFrobenius automorphism at p has the characteristicpolynomial

T 2 − τ(p)T + p11.

Galois Representations

I In the theory of Serre and Swinnerton-Dyer, these arise fromreducibility properties of certain families of Galoisrepresentations

ρ` : Gal(Q/Q) −→ GL2(Z`).

Here ` runs over rational primes.

I Deligne’s proof of the Ramanujan Hypothesis also constructsa family of such representations with the property that ρ` isunramified at all primes other than ` and for p 6= `, aFrobenius automorphism at p has the characteristicpolynomial

T 2 − τ(p)T + p11.

Reciprocity Laws

I To understand the significance of this, we have to recognizethat a fundamental theme in number theory is a reciprocitylaw: a law that determines the factorization of primes.

I If we consider a field K ⊇ Q of finite degree over Q, there isan analogue for K of the ordinary integers Z denoted OK .

I A prime ideal pZ in Z can be lifted to an ideal pOK which ingeneral will fail to be prime. For example if K = Q(

√−1),

then OK = Z[√−1] and

2OK = ((1 +√−1)OK )2

is the square of a prime ideal.

Reciprocity Laws

I To understand the significance of this, we have to recognizethat a fundamental theme in number theory is a reciprocitylaw: a law that determines the factorization of primes.

I If we consider a field K ⊇ Q of finite degree over Q, there isan analogue for K of the ordinary integers Z denoted OK .

I A prime ideal pZ in Z can be lifted to an ideal pOK which ingeneral will fail to be prime. For example if K = Q(

√−1),

then OK = Z[√−1] and

2OK = ((1 +√−1)OK )2

is the square of a prime ideal.

Reciprocity Laws

I To understand the significance of this, we have to recognizethat a fundamental theme in number theory is a reciprocitylaw: a law that determines the factorization of primes.

I If we consider a field K ⊇ Q of finite degree over Q, there isan analogue for K of the ordinary integers Z denoted OK .

I A prime ideal pZ in Z can be lifted to an ideal pOK which ingeneral will fail to be prime. For example if K = Q(

√−1),

then OK = Z[√−1] and

2OK = ((1 +√−1)OK )2

is the square of a prime ideal.

Reciprocity Laws

I A law that specifies how “primes factorize” is called areciprocity law.

I The most famous of these is the quadratic reciprocity law. Inthe special case of Q(

√−1) it says that pOK remains a prime

ideal if and only if p ≡ 3 (mod 4).

I In general, if K/Q is a Galois extension with Abelian group,the reciprocity law can be stated in terms of congruences: psplits in a certain way depending on its congruence classmodulo the “conductor” of K .

Reciprocity Laws

I A law that specifies how “primes factorize” is called areciprocity law.

I The most famous of these is the quadratic reciprocity law. Inthe special case of Q(

√−1) it says that pOK remains a prime

ideal if and only if p ≡ 3 (mod 4).

I In general, if K/Q is a Galois extension with Abelian group,the reciprocity law can be stated in terms of congruences: psplits in a certain way depending on its congruence classmodulo the “conductor” of K .

Reciprocity Laws

I A law that specifies how “primes factorize” is called areciprocity law.

I The most famous of these is the quadratic reciprocity law. Inthe special case of Q(

√−1) it says that pOK remains a prime

ideal if and only if p ≡ 3 (mod 4).

I In general, if K/Q is a Galois extension with Abelian group,the reciprocity law can be stated in terms of congruences: psplits in a certain way depending on its congruence classmodulo the “conductor” of K .

Reciprocity Laws

I In the general (non-Abelian) case, a reciprocity law is specifiednot by a congruence satisfied by p, but rather the Frobeniusconjugacy class in the Galois group associated to the prime.

I In particular, if the Galois group of K/Q is GL2(Z/`Z), givingthe characteristic polynomial (almost) specifies the conjugacyclass.

Reciprocity Laws

I In the general (non-Abelian) case, a reciprocity law is specifiednot by a congruence satisfied by p, but rather the Frobeniusconjugacy class in the Galois group associated to the prime.

I In particular, if the Galois group of K/Q is GL2(Z/`Z), givingthe characteristic polynomial (almost) specifies the conjugacyclass.

Reciprocity Laws

I In particular, reducing the representations of Deligne modulo `gives

ρ` : Gal(Q/Q) −→ GL2(Z/`Z)

and so the image is finite.

I The fixed field of the kernel of this representation is a finiteextension K`/Q.

I Moreover, Frobp belongs to a conjugacy class withcharacteristic polynomial

T 2 − (τ(p) (mod `))T + (p11 (mod `)).

Reciprocity Laws

I In particular, reducing the representations of Deligne modulo `gives

ρ` : Gal(Q/Q) −→ GL2(Z/`Z)

and so the image is finite.

I The fixed field of the kernel of this representation is a finiteextension K`/Q.

I Moreover, Frobp belongs to a conjugacy class withcharacteristic polynomial

T 2 − (τ(p) (mod `))T + (p11 (mod `)).

Reciprocity Laws

I In particular, reducing the representations of Deligne modulo `gives

ρ` : Gal(Q/Q) −→ GL2(Z/`Z)

and so the image is finite.

I The fixed field of the kernel of this representation is a finiteextension K`/Q.

I Moreover, Frobp belongs to a conjugacy class withcharacteristic polynomial

T 2 − (τ(p) (mod `))T + (p11 (mod `)).

Reciprocity Laws

This is not a special case. In fact by Serre’s Conjecture (nowproved by Khare and Wintenberger), an odd two dimensionalrepresentation

ρ` : Gal(Q/Q) −→ GL2(Z/`Z)

has to come by reducing modulo ` the Galois representationassociated to a modular form.

Ramanujan’s Congruences

I In this context, if for a given `, it happens that

ρ` = 1 ⊕ χ11

is a reducible representation given by the two charactersabove, then

τ(p) ≡ 1 + p11 (mod `).

I In particular, this happens for ` = 3, 691. Other congruencesare explained similarly.

I Thus, the question of congruences has been transformed intoa question of studying the reducibility of the Galoisrepresentations associated to ∆.

Ramanujan’s Congruences

I In this context, if for a given `, it happens that

ρ` = 1 ⊕ χ11

is a reducible representation given by the two charactersabove, then

τ(p) ≡ 1 + p11 (mod `).

I In particular, this happens for ` = 3, 691. Other congruencesare explained similarly.

I Thus, the question of congruences has been transformed intoa question of studying the reducibility of the Galoisrepresentations associated to ∆.

Ramanujan’s Congruences

I In this context, if for a given `, it happens that

ρ` = 1 ⊕ χ11

is a reducible representation given by the two charactersabove, then

τ(p) ≡ 1 + p11 (mod `).

I In particular, this happens for ` = 3, 691. Other congruencesare explained similarly.

I Thus, the question of congruences has been transformed intoa question of studying the reducibility of the Galoisrepresentations associated to ∆.

Nonlinear relations discovered by Ramanujan

The 1916 paper in which Ramanujan formulated his famousconjectures about the τ function, actually begins with thefollowing quadratic relation:

σ1(1)σ3(n)+σ1(3)σ3(n−1)+· · ·+σ1(2n+1)σ3(0) =1

240σ5(n+1).

Here,σk(n) =

∑d |n,d>0

dk

and

σk(0) =1

2ζ(−k).

Other non-linear relations

I The expression above can be seen as a quadratic relationamongst Eisenstein series.

I There are also relations involving cusp forms, in particular ∆.Lahiri was the first to give a relation of this kind.

I Many quadratic identities have been produced by S. Gun, B.Ramakrishnan and B. Sahu, as well as other authors.

I One such identity is

τ(n) = n2σ7(n)− 540n−1∑m=1

m(n −m)σ3(m)σ3(n −m).

I What do they mean?

Other non-linear relations

I The expression above can be seen as a quadratic relationamongst Eisenstein series.

I There are also relations involving cusp forms, in particular ∆.Lahiri was the first to give a relation of this kind.

I Many quadratic identities have been produced by S. Gun, B.Ramakrishnan and B. Sahu, as well as other authors.

I One such identity is

τ(n) = n2σ7(n)− 540n−1∑m=1

m(n −m)σ3(m)σ3(n −m).

I What do they mean?

Other non-linear relations

I The expression above can be seen as a quadratic relationamongst Eisenstein series.

I There are also relations involving cusp forms, in particular ∆.Lahiri was the first to give a relation of this kind.

I Many quadratic identities have been produced by S. Gun, B.Ramakrishnan and B. Sahu, as well as other authors.

I One such identity is

τ(n) = n2σ7(n)− 540n−1∑m=1

m(n −m)σ3(m)σ3(n −m).

I What do they mean?

Other non-linear relations

I The expression above can be seen as a quadratic relationamongst Eisenstein series.

I There are also relations involving cusp forms, in particular ∆.Lahiri was the first to give a relation of this kind.

I Many quadratic identities have been produced by S. Gun, B.Ramakrishnan and B. Sahu, as well as other authors.

I One such identity is

τ(n) = n2σ7(n)− 540n−1∑m=1

m(n −m)σ3(m)σ3(n −m).

I What do they mean?

Other non-linear relations

I The expression above can be seen as a quadratic relationamongst Eisenstein series.

I There are also relations involving cusp forms, in particular ∆.Lahiri was the first to give a relation of this kind.

I Many quadratic identities have been produced by S. Gun, B.Ramakrishnan and B. Sahu, as well as other authors.

I One such identity is

τ(n) = n2σ7(n)− 540n−1∑m=1

m(n −m)σ3(m)σ3(n −m).

I What do they mean?

Quadratic Relations in the Representation Ring

I If we consider the ring of mod ` representations of modularforms (under ⊗ product), the above identities can be seen asquadratic relations in this ring.

I Can we classify all such relations?

I In understanding quadratic relations, we seem to be wherecongruences were before the Serre/Swinnerton-Dyer theory.

I Much work to be done!

Quadratic Relations in the Representation Ring

I If we consider the ring of mod ` representations of modularforms (under ⊗ product), the above identities can be seen asquadratic relations in this ring.

I Can we classify all such relations?

I In understanding quadratic relations, we seem to be wherecongruences were before the Serre/Swinnerton-Dyer theory.

I Much work to be done!

Quadratic Relations in the Representation Ring

I If we consider the ring of mod ` representations of modularforms (under ⊗ product), the above identities can be seen asquadratic relations in this ring.

I Can we classify all such relations?

I In understanding quadratic relations, we seem to be wherecongruences were before the Serre/Swinnerton-Dyer theory.

I Much work to be done!

Quadratic Relations in the Representation Ring

I If we consider the ring of mod ` representations of modularforms (under ⊗ product), the above identities can be seen asquadratic relations in this ring.

I Can we classify all such relations?

I In understanding quadratic relations, we seem to be wherecongruences were before the Serre/Swinnerton-Dyer theory.

I Much work to be done!

Thank You!