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Ramanujans Function Armin Straub 06-May 2007 Armin Straub - - PowerPoint PPT Presentation

Introduction Modular Forms More Congruences for Fun Stuff Ramanujans Function Armin Straub 06-May 2007 Armin Straub Ramanujans Function Introduction Modular Forms More Congruences for Fun Stuff Outline Introduction The


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Introduction Modular Forms More Congruences for τ Fun Stuff

Ramanujan’s τ Function

Armin Straub 06-May 2007

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff

Outline

Introduction The τ Function Simple Congruences for τ Modular Forms Eisenstein Series Differentiating Modular Forms More Congruences for τ Differentiating ∆ An Exact Formula Modulus 691 Modulus 7 Fun Stuff Ramanujan Can Err Open Problems

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

The τ Function (I)

Definition

∆ q

  • n1

(1 − qn)24 =

  • n1

τ(n)qn.

Armin Straub Ramanujan’s τ Function

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SLIDE 4

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

The τ Function (I)

Definition

∆ q

  • n1

(1 − qn)24 =

  • n1

τ(n)qn.

Example

The first values are τ(1) = 1 τ(2) = −24 τ(3) = 252 τ(4) = −1472 τ(5) = 4830 τ(6) = −6048 τ(7) = −16744 τ(8) = 84480

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

The τ Function (II)

◮ A plot of log(τ(n)).

−25 25 100

Armin Straub Ramanujan’s τ Function

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SLIDE 6

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

The τ Function (II)

◮ A plot of log(τ(n)).

−25 25 100

◮ Ramanujan conjectured and Deligne proved

|τ(p)| 2p11/2.

Armin Straub Ramanujan’s τ Function

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SLIDE 7

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

The τ Function (II)

◮ A plot of log(τ(n)).

−25 25 100

◮ Ramanujan conjectured and Deligne proved

|τ(p)| 2p11/2.

. . .

Armin Straub Ramanujan’s τ Function

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SLIDE 8

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

Armin Straub Ramanujan’s τ Function

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SLIDE 9

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

◮ Hence,

q

  • n1

(1 − qn)24 ≡ q

  • n1

(1 − qn)3

n1

(1 − q7n)3 (mod 7)

Armin Straub Ramanujan’s τ Function

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SLIDE 10

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

◮ Hence, using Jacobi’s identity

q

  • n1

(1 − qn)24 ≡ q

  • n1

(1 − qn)3

n1

(1 − q7n)3 (mod 7) = q

  • n0

(−1)n(2n + 1)qn(n+1)/2

n1

(1 − q7n)3

Armin Straub Ramanujan’s τ Function

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SLIDE 11

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

◮ Hence, using Jacobi’s identity

q

  • n1

(1 − qn)24 ≡ q

  • n1

(1 − qn)3

n1

(1 − q7n)3 (mod 7) = q

  • n0

(−1)n(2n + 1)qn(n+1)/2

n1

(1 − q7n)3

◮ n(n+1) 2

+ 1 ≡ 0, 1, 2, 4 (mod 7)

Armin Straub Ramanujan’s τ Function

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SLIDE 12

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

◮ Hence, using Jacobi’s identity

q

  • n1

(1 − qn)24 ≡ q

  • n1

(1 − qn)3

n1

(1 − q7n)3 (mod 7) = q

  • n0

(−1)n(2n + 1)qn(n+1)/2

n1

(1 − q7n)3

◮ n(n+1) 2

+ 1 ≡ 0, 1, 2, 4 (mod 7)

◮ We conclude

τ(7n + 3), τ(7n + 5), τ(7n + 6) ≡ 0 (mod 7).

Armin Straub Ramanujan’s τ Function

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SLIDE 13

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (I)

◮ (a + b)p ≡ ap + bp

(mod p)

◮ Hence, using Jacobi’s identity

q

  • n1

(1 − qn)24 ≡ q

  • n1

(1 − qn)3

n1

(1 − q7n)3 (mod 7) = q

  • n0

(−1)n(2n + 1)qn(n+1)/2

n1

(1 − q7n)3

◮ n(n+1) 2

+ 1 ≡ 0, 1, 2, 4 (mod 7)

◮ We conclude

τ(7n), τ(7n + 3), τ(7n + 5), τ(7n + 6) ≡ 0 (mod 7).

. . .

Armin Straub Ramanujan’s τ Function

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SLIDE 14

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (II)

Theorem

τ(mn) = τ(m)τ(n) if gcd(m, n) = 1, τ(pn+1) = τ(p)τ(pn) − p11τ(pn−1) if p prime.

Armin Straub Ramanujan’s τ Function

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SLIDE 15

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (II)

Theorem

τ(mn) = τ(m)τ(n) if gcd(m, n) = 1, τ(pn+1) = τ(p)τ(pn) − p11τ(pn−1) if p prime.

◮ τ(p) ≡ 0 (mod p)

= ⇒ τ(np) ≡ 0 (mod p)

Armin Straub Ramanujan’s τ Function

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SLIDE 16

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (II)

Theorem

τ(mn) = τ(m)τ(n) if gcd(m, n) = 1, τ(pn+1) = τ(p)τ(pn) − p11τ(pn−1) if p prime.

◮ τ(p) ≡ 0 (mod p)

= ⇒ τ(np) ≡ 0 (mod p)

◮ τ(7) = −16744

= ⇒ τ(7n) ≡ 0 (mod 7)

Armin Straub Ramanujan’s τ Function

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SLIDE 17

Introduction Modular Forms More Congruences for τ Fun Stuff The τ Function Simple Congruences for τ

Simple Congruences (II)

Theorem

τ(mn) = τ(m)τ(n) if gcd(m, n) = 1, τ(pn+1) = τ(p)τ(pn) − p11τ(pn−1) if p prime.

◮ τ(p) ≡ 0 (mod p)

= ⇒ τ(np) ≡ 0 (mod p)

◮ τ(7) = −16744

= ⇒ τ(7n) ≡ 0 (mod 7)

◮ Only a few such primes are known:

2, 3, 5, 7, 2411 .

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Eisenstein Series

◮ Eisenstein Series En are an example of modular forms of

weight 2n. En = 1 − 4k B2k

  • n1

σ2k−1(n)qn

Armin Straub Ramanujan’s τ Function

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SLIDE 19

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Eisenstein Series

◮ Eisenstein Series En are an example of modular forms of

weight 2n. En = 1 − 4k B2k

  • n1

σ2k−1(n)qn

◮ Any modular form f can be obtained as a polynomial in E2

and E3.

Armin Straub Ramanujan’s τ Function

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SLIDE 20

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Eisenstein Series

◮ Eisenstein Series En are an example of modular forms of

weight 2n. En = 1 − 4k B2k

  • n1

σ2k−1(n)qn

◮ Any modular form f can be obtained as a polynomial in E2

and E3.

◮ What about E1?

Armin Straub Ramanujan’s τ Function

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SLIDE 21

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Eisenstein Series

◮ Eisenstein Series En are an example of modular forms of

weight 2n. En = 1 − 4k B2k

  • n1

σ2k−1(n)qn

◮ Any modular form f can be obtained as a polynomial in E2

and E3.

◮ What about E1? ◮ E1 is not modular but

E1(−1/z) = z2E1(z) + 12 2πiz.

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Differentiating Modular Forms

◮ If f is a modular form. What about

df dz ?

Armin Straub Ramanujan’s τ Function

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SLIDE 23

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Differentiating Modular Forms

◮ If f is a modular form. What about

qdf dq = 1 2πi df dz ?

Armin Straub Ramanujan’s τ Function

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SLIDE 24

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Differentiating Modular Forms

◮ If f is a modular form. What about

θf qdf dq = 1 2πi df dz ?

Armin Straub Ramanujan’s τ Function

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SLIDE 25

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Differentiating Modular Forms

◮ If f is a modular form. What about

θf qdf dq = 1 2πi df dz ?

◮ Not modular, but this is where E1 comes in.

Armin Straub Ramanujan’s τ Function

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SLIDE 26

Introduction Modular Forms More Congruences for τ Fun Stuff Eisenstein Series Differentiating Modular Forms

Differentiating Modular Forms

◮ If f is a modular form. What about

θf qdf dq = 1 2πi df dz ?

◮ Not modular, but this is where E1 comes in.

Lemma

If f is a modular form of weight k then θf − k 12E1f is a modular form of weight k + 2.

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Differentiating ∆

θf − k 12 E1f ◮ Applied to ∆ which is of weight 12,

θ∆ − E1∆ = 0.

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Differentiating ∆

E1 = 1 − 24 X

n1

σ1(n)qn ◮ Applied to ∆ which is of weight 12,

θ∆ − E1∆ = 0.

◮ This gives the recursion

(n − 1)τ(n) = −24

n−1

  • m=1

τ(m)σ1(n − m).

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Differentiating ∆

E1 = 1 − 24 X

n1

σ1(n)qn ◮ Applied to ∆ which is of weight 12,

θ∆ − E1∆ = 0.

◮ This gives the recursion

(n − 1)τ(n) = −24

n−1

  • m=1

τ(m)σ1(n − m).

◮ And the nice congruences

n ≡ 0, 2 (mod 6) = ⇒ τ(n) ≡ 0 (mod 24).

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

An Exact Formula

E3 = 1 − 504 X

n1

σ5(n)qn E6 = 1 + 65520 691 X

n1

σ11(n)qn ◮ ∆ = αE2 3 + βE6.

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

An Exact Formula

E3 = 1 − 504 X

n1

σ5(n)qn E6 = 1 + 65520 691 X

n1

σ11(n)qn ◮ ∆ = αE2 3 + βE6. ◮ This requires

= α + β, 1 = −2 · 504α + 65520 691 β.

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

An Exact Formula

E3 = 1 − 504 X

n1

σ5(n)qn E6 = 1 + 65520 691 X

n1

σ11(n)qn ◮ ∆ = αE2 3 + βE6. ◮ This requires

= α + β, 1 = −2 · 504α + 65520 691 β.

◮ 762048∆ = −691E2 3 + 691E6.

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

An Exact Formula

E3 = 1 − 504 X

n1

σ5(n)qn E6 = 1 + 65520 691 X

n1

σ11(n)qn ◮ ∆ = αE2 3 + βE6. ◮ This requires

= α + β, 1 = −2 · 504α + 65520 691 β.

◮ 762048∆ = −691E2 3 + 691E6. ◮ In other words,

τ(n) = 691 756σ5(n) − 691 3 σ5 ∗ σ5(n) + 65 756σ11(n).

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 691

◮ Consider the previous

τ(n) = 691 756σ5(n) − 691 3 σ5 ∗ σ5(n) + 65 756σ11(n).

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 691

◮ Consider the previous

τ(n) = 691 756σ5(n) − 691 3 σ5 ∗ σ5(n) + 65 756σ11(n).

◮ As desired,

τ(n) ≡ σ11(n) (mod 691).

Armin Straub Ramanujan’s τ Function

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SLIDE 36

Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 691

◮ Consider the previous

τ(n) = 691 756σ5(n) − 691 3 σ5 ∗ σ5(n) + 65 756σ11(n).

◮ As desired,

τ(n) ≡ σ11(n) (mod 691).

◮ For primes

τ(p) ≡ 1 + p11 (mod 691).

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 7

E2 = 1 + 240 X

n1

σ3(n)qn E3 = 1 − 504 X

n1

σ5(n)qn E4 = 1 + 480 X

n1

σ7(n)qn θf − k 12 E1f ◮ Let’s start with

E3 ≡ 1, E2

2 = E4

≡ E1 (mod 7).

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 7

E2 = 1 + 240 X

n1

σ3(n)qn E3 = 1 − 504 X

n1

σ5(n)qn E4 = 1 + 480 X

n1

σ7(n)qn θf − k 12 E1f ◮ Let’s start with

E3 ≡ 1, E2

2 = E4

≡ E1 (mod 7).

◮ Then

1728∆ = E3

2 − E2 3

≡ E1E2 − E3 = 3θE2 (mod 7).

Armin Straub Ramanujan’s τ Function

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SLIDE 39

Introduction Modular Forms More Congruences for τ Fun Stuff Differentiating ∆ An Exact Formula Modulus 691 Modulus 7

Modulus 7

E2 = 1 + 240 X

n1

σ3(n)qn E3 = 1 − 504 X

n1

σ5(n)qn E4 = 1 + 480 X

n1

σ7(n)qn θf − k 12 E1f ◮ Let’s start with

E3 ≡ 1, E2

2 = E4

≡ E1 (mod 7).

◮ Then

1728∆ = E3

2 − E2 3

≡ E1E2 − E3 = 3θE2 (mod 7).

◮ So that

τ(n) ≡ nσ3(n) (mod 7).

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Ramanujan Can Err

◮ We have

τ(n) ≡ nσ1(n) (mod 5) τ(n) ≡ nσ9(n) (mod 25) τ(n) ≡ n41σ29(n) (mod 125).

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Ramanujan Can Err

◮ We have

τ(n) ≡ nσ1(n) (mod 5) τ(n) ≡ nσ9(n) (mod 25) τ(n) ≡ n41σ29(n) (mod 125).

◮ Ramanujan conjectured that

τ(n) ≡ naσb(n) (mod 5k).

Armin Straub Ramanujan’s τ Function

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SLIDE 42

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Ramanujan Can Err

◮ We have

τ(n) ≡ nσ1(n) (mod 5) τ(n) ≡ nσ9(n) (mod 25) τ(n) ≡ n41σ29(n) (mod 125).

◮ Ramanujan conjectured that

τ(n) ≡ naσb(n) (mod 5k).

◮ This is false for k 4.

Armin Straub Ramanujan’s τ Function

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SLIDE 43

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Ramanujan Can Err

◮ We have

τ(n) ≡ nσ1(n) (mod 5) τ(n) ≡ nσ9(n) (mod 25) τ(n) ≡ n41σ29(n) (mod 125).

◮ Ramanujan conjectured that

τ(n) ≡ naσb(n) (mod 5k).

◮ This is false for k 4. ◮ Take n = 443. Since 4432 ≡ −1 (mod 54).

τ(443) = 328369848718692 ≡ 567 ≡ 443a(1+443b) (mod 54)

. . .

Armin Straub Ramanujan’s τ Function

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Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Open Problems (I)

◮ Of course, Lehmer’s conjecture

τ(n) = 0.

Armin Straub Ramanujan’s τ Function

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SLIDE 45

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Open Problems (I)

◮ Of course, Lehmer’s conjecture

τ(n) = 0.

◮ We found

τ(p) ≡ p(1 + p9) (mod 25) τ(p) ≡ p(1 + p3) (mod 7) τ(p) ≡ 1 + p11 (mod 691).

Armin Straub Ramanujan’s τ Function

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SLIDE 46

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Open Problems (I)

◮ Of course, Lehmer’s conjecture

τ(n) = 0.

◮ We found

τ(p) ≡ p(1 + p9) (mod 25) τ(p) ≡ p(1 + p3) (mod 7) τ(p) ≡ 1 + p11 (mod 691).

◮ Which implies

τ(p) = 0 = ⇒ p ≡ −1 (mod 52 · 691) p ≡ −1, 3, 5 (mod 7)

  • =

⇒ p = 863749, 1381999, 1589299, . . . .

. . .

Armin Straub Ramanujan’s τ Function

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SLIDE 47

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Open Problems (II)

◮ Is τ(n) ever a prime? Indeed,

τ(63001) = τ(2512) = τ(251)2 − 25111 = −80561663527802406257321747.

Armin Straub Ramanujan’s τ Function

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SLIDE 48

Introduction Modular Forms More Congruences for τ Fun Stuff Ramanujan Can Err Open Problems

Open Problems (II)

◮ Is τ(n) ever a prime? Indeed,

τ(63001) = τ(2512) = τ(251)2 − 25111 = −80561663527802406257321747.

◮ More of the following sort?

τ(p) ≡ 0 (mod p) = ⇒ p = 2, 3, 5, 7, 2411, . . . τ(p) ≡ 1 (mod p) = ⇒ p = 11, 23, 691, . . . τ(p) ≡ −1 (mod p) = ⇒ p = 5807, . . .

Armin Straub Ramanujan’s τ Function