Computer Algebra, Ramanujan Congruences, and Polynomials

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WS 3: Computer Algebra and Polynomials/ 1

RICAM Special Semester on

Applications of Algebra and Number Theory Johann Radon Institute (RICAM), Nov 28, 2013

Computer Algebra, Ramanujan Congruences, and Polynomials

Peter Paule

(joint work with Cristian-Silviu Radu)

Johannes Kepler University Linz

Research Institute for Symbolic Computation (RISC)

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WS 3: Computer Algebra and Polynomials/Introduction 2

Introduction

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1 1 2 3 5

7 11 15 22 30

42 56 77 101 135

176 231 297 385 490

627 792 1002 1255 1575

1958 2436 3010 3718 4565

5604 6842 8349 10143 12310

14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 204226 239943 281589 329931 386155 451276 526823 614154 715220 831820 966467 1121505 1300156 1505499 1741630 2012558 2323520 2679689 3087735 3554345 4087968 4697205 5392783 6185689 7089500 8118264 9289091 10619863 12132164 13848650

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Theorem

LetS be a compact Riemann surface and z a meromorphic function onS having npoles (incl. multiplicity). Let f be any other meromorphic function onS. Then there exist rational functionsck(x)∈C(x) such that

fⁿ+c1(z)fⁿ⁻¹+· · ·+cn−1(z)f+cn(z) = 0.

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Theorem

LetS be a compact Riemann surface and z a meromorphic function onS having npoles (incl. multiplicity). Let f be any other meromorphic function onS. Then there exist rational functionsck(x)∈C(x) such that

fⁿ+c₁(z)fⁿ⁻¹+· · ·+cn−1(z)f+c_n(z) = 0.

Moreover, if

PoleSet(z)⊆PoleSet(f), then thec_k(x)are polynomials; i.e., c_k(x)∈C[x].

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WS 3: Computer Algebra and Polynomials/Partition Numbers 6

Partition Numbers

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Example: p(4) = 5: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.

Theorem

(p(n))_n≥0 is not holonomic.

Proof. The generating function is

∞

X

n=0

p(n)qⁿ =

∞

Y

n=1

(1−qⁿ)⁻¹

= (1 +q¹+q¹⁺¹+q¹⁺¹⁺¹+. . .)

× (1 +q²+q²⁺²+q²⁺²⁺²+. . .)

× etc.

= . . .+q¹⁺¹⁺¹q²⁺²· · ·+. . .

This implies non-holonomicity, because holonomic functions have only finitely many singularities.

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Example: p(4) = 5: 4, 3 + 1, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.

Theorem

(p(n))_n≥0 is not holonomic.

Proof. The generating function is

∞

X

n=0

p(n)qⁿ =

∞

Y

n=1

(1−qⁿ)⁻¹

= (1 +q¹+q¹⁺¹+q¹⁺¹⁺¹+. . .)

× (1 +q²+q²⁺²+q²⁺²⁺²+. . .)

× etc.

= . . .+q¹⁺¹⁺¹q²⁺²· · ·+. . .

This implies non-holonomicity, because holonomic functions have only finitely many singularities.

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Growth ofp(n)

Example [MacMahon 1916]:

p(200) = 3,972,999,029,388

Theorem

p(n)∼ 1 4n√

3exp π r2n

3

!

as n→ ∞.

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Growth ofp(n)

Example [MacMahon 1916]:

p(200) = 3,972,999,029,388

Theorem

p(n)∼ 1 4n√

3exp π r2n

3

!

as n→ ∞.

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Back to our table

I 34 of the 80 entries are even; 46 are odd.

I 4,996 of the first 10,000 entries are even; 5,004 are odd. Does this pattern continue?

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Back to our table

I 4,996 of the first 10,000 entries are even; 5,004 are odd.

Does this pattern continue?

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Back to our table

I 4,996 of the first 10,000 entries are even; 5,004 are odd.

Does this pattern continue?

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DEEPER INSIGHT: The Subbarao Conjecture (1966)

I Ono [1996]: Given integers 0≤B < A:

p(An+B)≡0 (mod 2) for infinitely manyn.

I Radu [2012]: Given integers 0≤B < A:

p(An+B)6≡0 (mod 2) for infinitely manyn.

Remark. Radu’s algorithmic insight enabled him to invoke a result by Deligne and Rapoport [1979]. The same applies to his proof [2012] of a conjecture of Ahlgren and Ono [2002].

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I Radu [2012]: Given integers0≤B < A:

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I Radu [2012]: Given integers0≤B < A:

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Ahlgren and Ono determined:

I 3,313, 3,325, and 3,362 of the first 10,000 entries are congruent respectively to 0, 1, and 2 modulo 3.

BUT:

I 3,611 (many more than the expected one-fifth) of the first 10,000 values of p(n) are divisible by 5.

WHY? Let’s look again at our table:

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BUT:

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BUT:

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1 1 2 3 5

7 11 15 22 30

42 56 77 101 135

176 231 297 385 490

627 792 1002 1255 1575

1958 2436 3010 3718 4565

5604 6842 8349 10143 12310

14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 204226 239943 281589 329931 386155 451276 526823 614154 715220 831820 966467 1121505 1300156 1505499 1741630 2012558 2323520 2679689 3087735 3554345 4087968 4697205 5392783 6185689 7089500 8118264 9289091 10619863 12132164 13848650

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WS 3: Computer Algebra and Polynomials/Partition Congruences and Combinatorics 13

Partition Congruences and Combinatorics

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I There is a combinatorial explanation for the divisibility of p(5n+ 4) by 5 (Dyson’s rank statistics). But concerning further connections between combinatorics and modular forms, not too much is known so far.

As an example, we present a recent combinatorial construction of modular forms which involves Dedekind’s eta function:

η(τ) :=q²⁴¹

∞

Y

n=1

(1−qⁿ) (q =e^2πiτ).

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Partition Diamonds

c a1

AAA U c

a2

- ac3

AA AA

AA - U

ac4

- ac5

AA AA

AA - U

ac6

. . . . ac7

. . . .a2kc−1

AA AA

AA - U

a2kc−2

-

a2k+1c AUAA a2kc

ca2k+2

Ak-elongated partition diamond of length 1

c a1

AAA U c

a2

. . . . ac3

. . . . c a2k

AAA a2k+1c

U c a2k+2 c

AAA U c a2k+3

. . . . a2k+4c

. . . . c a4k+1

AAA a4k+2c

U c

a4k+3 . . . . c AAA U c

. . . . c

. . . . c a_(2k+1)n−1

AAA a(2k+1)nc

U ca(2k+1)n+1

Ak-elongated partition diamond of lengthn

1

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Generating function:

hn,k(q) = Qn−1

j=0(1 +q^(2k+1)j+2)(1 +q^(2k+1)j+4)· · ·(1 +q^(2k+1)j+2k) Q(2k+1)n+1

j=1 (1−q^j)

This generating function is beautiful indeed, but George Andrews wanted more!

From: George E Andrews<[email protected]> Date: Jan 2005

To: [email protected], [email protected] Subject: I had a brainstorm

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Generating function:

hn,k(q) = Qn−1

j=0(1 +q^(2k+1)j+2)(1 +q^(2k+1)j+4)· · ·(1 +q^(2k+1)j+2k) Q(2k+1)n+1

j=1 (1−q^j)

This generating function is beautiful indeed, but George Andrews wanted more!

From: George E Andrews<[email protected]>

Date: Jan 2005

To: [email protected], [email protected] Subject: I had a brainstorm

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Dear Axel and Peter,

I have had a BRAINSTORM! (I can almost see the smile on Peter’s face ...) To begin with let me draw your attention to Thm.2 in PAXI and Cor.2.1 in PAVIII ... However, THIS IS JUST THE BEGINNING.

Namely, I always had a slight disappointment in both the earlier results because the generating function ... was not a modular form. However, I now know how to produce the appropriate directed graph to provide ... a generating function that is not only a modular form but, in fact, a product of instances of Dedekind’s eta-function.

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Recall the generating function:

h_n,k(q) = Qn−1

j=0(1 +q^(2k+1)j+2)(1 +q^(2k+1)j+4)· · ·(1 +q^(2k+1)j+2k) Q(2k+1)n+1

j=1 (1−q^j) The result of Andrews’ brainstorm: delete the source:

h^∗_n,k(q) = Qn−1

j=0(1 +q^(2k+1)j+1)(1 +q^(2k+1)j+3)· · ·(1 +q(2k+1)j+2k−1) Q(2k+1)n

j=1 (1−q^j) andglue the diamonds together:

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Recall the generating function:

h_n,k(q) = Qn−1

j=0(1 +q^(2k+1)j+2)(1 +q^(2k+1)j+4)· · ·(1 +q^(2k+1)j+2k) Q(2k+1)n+1

j=1 (1−q^j) The result of Andrews’ brainstorm: delete the source:

h^∗_n,k(q) = Qn−1

j=0(1 +q^(2k+1)j+1)(1 +q^(2k+1)j+3)· · ·(1 +q(2k+1)j+2k−1) Q(2k+1)n

j=1 (1−q^j) andglue the diamonds together:

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∞

X

n=0

∆_k(n)qⁿ:=h_n,k(q)h^∗_n,k(q)

=

∞

Y

n=1

(1−q²ⁿ)(1−q^(2k+1)n) (1−qⁿ)³(1−q^(4k+2)n)

=q^(k+1)/12η(2τ)η((2k+ 1)τ) η(τ)³η((4k+ 2)τ)

c b_(2k+1)n+1

AAAc b_(2k+1)n−1

. . . . b(2k+1)nc

. . . .c AAA c

c

K

. . . c AAA

c. . . .

c. . . . K

bc7

AA AA

AA

bc6

K

bc5

K bc4

AA AA

AA

bc3

bc2

b2k+2

b2k+1

b2k

c a1

AAA U c

a2

. . . . ac3

. . . .c a2k

AAA a2k+1c

U c

a2k+2 . . . c

AAA U c

. . . . c

. . . .c a_(2k+1)n−1

AAA a(2k+1)nc

U c a_(2k+1)n+1

A brokenk-diamond of length 2n

1

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∞

X

n=0

∆_k(n)qⁿ:=h_n,k(q)h^∗_n,k(q)

=

∞

Y

n=1

(1−q²ⁿ)(1−q^(2k+1)n) (1−qⁿ)³(1−q^(4k+2)n)

=q^(k+1)/12η(2τ)η((2k+ 1)τ) η(τ)³η((4k+ 2)τ)

c b_(2k+1)n+1

AAAc b_(2k+1)n−1

. . . . b(2k+1)nc

. . . .c AAA c

c

K

. . . c AAA

c. . . .

c. . . . K

bc7

AA AA

AA

bc6

K

b5c

K b4c

AA AA

AA

bc3

bc2

b2k+2

b2k+1

b2k

c a1

AAA U c

a2

. . . . ac3

. . . .c a2k

AAA a2k+1c

U c

a2k+2 . . . c

AAA U c

. . . . c

. . . .c a_(2k+1)n−1

AAA a(2k+1)nc

U c a_(2k+1)n+1

A brokenk-diamond of length 2n

1

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Numerous Congruences for∆_k(n) Example [Sellers & Radu, 2012]

∞

X

n=0

∆₂(3n+ 1)qⁿ≡2q

∞

Y

n=1

(1−q¹⁰ⁿ)⁴

(1−q⁵ⁿ)² (mod 3)

This implies for alln≥0:

∆2(15n+ 1) ≡ 0 (mod 3),

∆₂(27n+ 16) ≡ 0 (mod 3),

∆₂(147n+ 58) ≡ 0 (mod 3), etc.

Note. More people became interested in such congruences, e.g., S.H. Chan, W.Y.C. Chen, S. Cui, A.R.B. Fan, S.S. Fu, N. Gu, M.D. Hirschhorn, M. Jameson, E. Mortenson, X. Xiong, R.T. Yu.

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∞

X

n=0

∆₂(3n+ 1)qⁿ≡2q

∞

Y

n=1

(1−q¹⁰ⁿ)⁴

(1−q⁵ⁿ)² (mod 3)

∆2(15n+ 1) ≡ 0 (mod 3),

∆₂(27n+ 16) ≡ 0 (mod 3),

∆₂(147n+ 58) ≡ 0 (mod 3), etc.

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∞

X

n=0

∆₂(3n+ 1)qⁿ≡2q

∞

Y

n=1

(1−q¹⁰ⁿ)⁴

(1−q⁵ⁿ)² (mod 3)

∆2(15n+ 1) ≡ 0 (mod 3),

∆₂(27n+ 16) ≡ 0 (mod 3),

∆₂(147n+ 58) ≡ 0 (mod 3), etc.

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WS 3: Computer Algebra and Polynomials/Ramanujan’s Congruences 21

Ramanujan’s Congruences

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Ramanujan’s Congruences

p(5n+ 4) ≡ 0 (mod 5), p(7n+ 5) ≡ 0 (mod 7), p(11n+ 6) ≡ 0 (mod 11)

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Ramanujan’s Congruences

p(5n+ 4) ≡ 0 (mod 5), p(5²n+ 24) ≡ 0 (mod 5²), p(5³n+ 99) ≡ 0 (mod 5³),

etc.

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Ramanujan’s Conjecture (1919)

For`∈ {5,7,11} andα∈ {1,2,3, . . .}:

p(`^αn+µ_α,`)≡0 (mod `^α), whereµα,`∈ {0, . . . , `^α−1} is uniquely defined by

24µα,`≡1 (mod`^α).

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A Bit of History

I Ramanujan [1917-1919]: proof sketches for `= 5 and `= 7;

see Berndt and Ono [1999].

I Watson [1938]: `= 5 and`= 7 (corrected version)

I Atkin [1967]: `= 11

I Gordon [1983]:

p(11^αn+µ_α,11)≡0 (mod 11^α+), where is a fixed non-positive integer to be determined. Following Gordon’s proof one can easily determine = 0.

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A Bit of History

I Atkin [1967]: `= 11

I Gordon [1983]:

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A Bit of History

I Atkin [1967]: `= 11

I Gordon [1983]:

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A Bit of History

I Atkin [1967]: `= 11

I Gordon [1983]:

p(11^αn+µα,11)≡0 (mod 11^α+), where is a fixed non-positive integer to be determined.

Following Gordon’s proof one can easily determine = 0.

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Our proof for ` = 11 and ` = 5, 7:

I In his generalization of Watson’s method, Gordon needs to compute thestructure constants of a certain algebra.

Although this is not needed in the case`= 5,7.

I For `= 11we succeeded to do without structure constants, (i.e., we stay with the module structure), so the ingredients here are the same as in the original proof of Watson.

I This way we obtain a unified frameworkfor all three cases

`= 5,7,11.

I The price to pay: our approach is more expensive computationally.

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Our proof for ` = 11 and ` = 5, 7:

`= 5,7,11.

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Our proof for ` = 11 and ` = 5, 7:

`= 5,7,11.

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Our proof for ` = 11 and ` = 5, 7:

`= 5,7,11.

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WS 3: Computer Algebra and Polynomials/The Modular Function Setting 26

The Modular Function Setting

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Basic Notions

The level`congruence subgroup Γ0(`):=

a c b d

∈SL2(Z):c≡0 (mod `)

acts onH^∗ :=H∪Q∪ {i∞}(extended upper half plane) via a

c b d

(τ)7→ aτ+b cτ+d.

This induces an action on functionsf :H^∗→C∪ {i∞}by (f|γ)(τ) :=faτ+b

cτ +d

where γ = a

c b d

.

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Basic Notions

The level`congruence subgroup Γ0(`):=

a c b d

∈SL2(Z):c≡0 (mod `)

acts onH^∗ :=H∪Q∪ {i∞}(extended upper half plane) via a

c b d

(τ)7→ aτ+b cτ+d.

This induces an action on functionsf :H^∗→C∪ {i∞}by (f|γ)(τ) :=faτ+b

cτ +d

where γ = a

c b d

.

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Compact Riemann Surfaces X₀(`)

X0(`) := orbit space of the action ofΓ0(`) onH^∗

= {[τ] :τ ∈H^∗}

Note.

aτ+b cτ +d −→ a

c when τ −→i∞.

and a

0 =i∞. Set of cusps: For anyprime `,

{[x]∈X₀(`):x∈Q∪ {i∞}}={[0],[i∞]}.

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= {[τ] :τ ∈H^∗}

Note.

and a

0 =i∞.

Set of cusps: For anyprime `,

{[x]∈X₀(`):x∈Q∪ {i∞}}={[0],[i∞]}.

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= {[τ] :τ ∈H^∗}

Note.

and a

0 =i∞.

Set of cusps: For anyprime `,

{[x]∈X₀(`):x∈Q∪ {i∞}}={[0],[i∞]}.

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Modular Functions (MF)on Compact Riemann Surfaces X₀(`) f :H^∗→C∪ {i∞}induces a MFf^∗:X0(`)→C∪ {i∞} if

I f is holomorphic onH;

I f is constant on the orbits[τ],τ ∈H^∗;

I f is meromorphic at points inQ∪ {i∞}.

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Modular Functions (MF)on Compact Riemann Surfaces X₀(`) f :H^∗→C∪ {i∞}induces a MFf^∗:X0(`)→C∪ {i∞} if

I f is holomorphic onH;

I f is constant on the orbits[τ],τ ∈H^∗;

I f is meromorphic at points inQ∪ {i∞}.

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Fourier expansions off^∗ at cusps [a/c]∈X₀(`)

Forτ ∈neighborhood(i∞):

f^∗

aτ +b cτ+d

) = (f|γ)(τ) = X

n≥ord_[a/c](f^∗)

a_[a/c](n)q_[a/c]ⁿ .

whereγ =

ab cd

∈SL2(Z),q_[a/c]= e^2πiτ^gcd(c

2,`)

` .

Note. If[a/c] = [a/0] = [i∞]thenq_[a/c]=q.

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Fourier expansions off^∗ at cusps [a/c]∈X₀(`)

Forτ ∈neighborhood(i∞):

f^∗

aτ +b cτ+d

) = (f|γ)(τ) = X

n≥ord_[a/c](f^∗)

a_[a/c](n)q_[a/c]ⁿ .

whereγ =

ab cd

∈SL2(Z),q_[a/c]= e^2πiτ^gcd(c

2,`)

` .

Note. If[a/c] = [a/0] = [i∞]thenq_[a/c]=q.

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Back to Ramanujan’s congruences L2β−1,` := q

∞

Y

n=1

(1−q^`n)

∞

X

n=0

p(`^2β−1n+µ2β−1,`)qⁿ, and

L_2β,` :=q

∞

Y

n=1

(1−qⁿ)

∞

X

n=0

p(`^2βn+µ_2β,`)qⁿ.

We consider these series asFourier expansions at[i∞]of modular functionsf^∗:X0(`)→C∪ {i∞} from a suitablesubring R(`).

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The RingsR(5), R(7), andR(11) Recall η(τ) =q²⁴¹

∞

Y

n=1

(1−qⁿ) (q=e^2πiτ).

From the literature it is well known that we can take

R(5)=Z[z5], R(7)=Z[z7], and

R(11)=hJ₀, J₁, . . . , J₄i_Z[z₁₁_] where

z5 := η⁶(5τ)

η⁶(τ) , z7 := η⁴(7τ)

η⁴(τ) , z11:= η¹²(11τ) η¹²(τ) ,

and with theJk such thatord[i∞](J_k^∗) =k as in Atkin (1967).

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The RingsR(5), R(7), andR(11) Recall η(τ) =q²⁴¹

∞

Y

n=1

(1−qⁿ) (q=e^2πiτ).

From the literature it is well known that we can take

R(5)=Z[z5], R(7)=Z[z7], and

R(11)=hJ₀, J₁, . . . , J₄i_Z[z₁₁_] where

z5 := η⁶(5τ)

η⁶(τ) , z7:= η⁴(7τ)

η⁴(τ) , z11:= η¹²(11τ) η¹²(τ) ,

and with theJk such thatord[i∞](J_k^∗) =k as in Atkin (1967).

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TheU-operator

U`(f^∗):=X

n

ằn)qⁿ

wheref^∗([τ]) =P

năn)qⁿ is the Fourier expansion off^∗ at[i∞].

Definition. For prime `, U_`⁽¹⁾(f^∗):=U`

η(`²τ) η(τ) f^∗

and U_`⁽²⁾(f^∗):=U`(f^∗).

Lemmạ

f^∗ ∈R(`) ⇒ U_`⁽ⁱ⁾(f^∗)∈R(`).

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TheU-operator

U`(f^∗):=X

n

ằn)qⁿ

wheref^∗([τ]) =P

η(`²τ) η(τ) f^∗

and U_`⁽²⁾(f^∗):=U`(f^∗).

Lemmạ

f^∗ ∈R(`) ⇒ U_`⁽ⁱ⁾(f^∗)∈R(`).

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TheU-operator

U`(f^∗):=X

n

ằn)qⁿ

wheref^∗([τ]) =P

η(`²τ) η(τ) f^∗

and U_`⁽²⁾(f^∗):=U`(f^∗).

Lemmạ

f^∗ ∈R(`) ⇒ U_`⁽ⁱ⁾(f^∗)∈R(`).

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Recursive representation of L_α,` in R(`)

Lemma. For `∈ {5,7,11},

L_α,`∈R(`);

theL_α,` satisfy a recursion in R(`) as follows:

L0,`= 1,

L2β−1,`=U_`⁽¹⁾(L2β−2,`) andL_2β,` =U_`⁽²⁾(L2β−1,`).

This recursion is used in the induction proof of our “Main Theorem”:

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Main Theorem

For suitably definedsubsetsXs,` ofR(`) we have:

Theorem. For`∈ {5,7,11}and all positive integersβ there existf_β,` ∈X_1,` ifβ is odd, andf_β,` ∈X_2,` if β is even, such that

Lβ,`=`^βfβ,` for `= 5,11, and

L_β,` =`^d^β+1² ^ef_β,` for `= 7.

Note. Ramanujan’s congruences are obtained as an immediate corollary.

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TheR(`)-subsets X_s,` (`= 11: Atkin) Set

A_`:= 12

gcd(`−1,12)

`

`+ 1 and z_`:=η_` η

gcd(`−1,12)²⁴

.

Definition. For `∈ {5,7,11} ands= 1,2:

Xs,` :=

(_n

`−1

X

i=0

Ji,`

∞

X

j=0

ai(j)`^d^A`^` ^(`j+ξⁱ^(s,`)^)ez_`^j : a0(0) = 0 and

with ai(j)∈Znonzero for only finitely many j )

The integer exponentsξ_i^(s,`) are defined as follows:

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The Exponentsξ_i^(s,`)

Definition. For `∈ {5,7,11} ands= 1,2define

ξ^(s,`) :{0, . . . , n_`−1} →Z, i7→ξ_i^(s,`)

by

(ξ^(1,11)₀ , . . . , ξ₄^(1,11)) := (−5,−1,1,2,6), (ξ^(2,11)₀ , . . . , ξ₄^(2,11)) := (−4,0,2,3,7);

and

ξ₀^(1,7):=−7and ξ₀^(2,7) :=−10;

and

ξ₀^(1,5):=−6and ξ₀^(2,5) :=−5.

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Main Tool: The Fundamental Lemma

Fundamental Lemma. Let `∈ {5,7,11}. For any analytic w:H→Candj∈Z:

U_`(wz^j_`) =−

d`−1

X

i=0

a^(`)_i (z_`)U_`(wz^j+i−d_` ^`).

Settingwto the generators of the moduleR(`), we obtain the U_`⁽ⁱ⁾ actions on all the module elements — provided

we know the2·d_`·n_` (2·5·1,2·7·1,2·55·5) initial actions.

This action then serves to expressLβ,` in terms of the generators which reveal the divisibilty properties of these elements.

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WS 3: Computer Algebra and Polynomials/Proof of the Main Theorem 39

Proof of the Main Theorem

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The Fundamental Polynomial for any prime ` ≥ 5

The main ingredient in our induction proof is

Theorem. For any prime`≥5there exists a monic irreducible polynomialF`(X, Y)∈Q[Y][X]of degreed` := gcd(`−1,12)^`(`−1) in X (andY) such that

F_`(z_`(τ), z_`(`τ)) = 0.

Recall that

z`(τ) =η(`τ) η(τ)

gcd(`−1,12)²⁴

.

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Proving the Existence of the Fundamental Polynomial

Proof.

The proof uses a variant of a classical fact from Riemann surfaces;

namely, that on a compact Riemann surface two meromorphic functions are connected by an algebraic relation.

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The Fundamental Polynomial for ` ∈ {5, 7, 11}

Theorem. For`∈ {5,7,11}the fundamental polynomial F`(X, Y) =X^d^`+

d`−1

X

i=0

a^(`)_i (Y)Xⁱ

has coefficient polynomialsa^(`)_i (Y)∈Q[Y]of the form

a^(`)_i (Y)=

d`

X

j=d^d`_`⁻ⁱe

s_`(i, j)`^d^A`^` ^(`j+i−d^`^)eY^j

withs`(i, j)∈Z (determined using computer algebra).

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Corollary: The Fundamental Lemma

Fundamental Lemma. Let `∈ {5,7,11}. For any w:H→C andj∈Z:

U_`(wz^j_`) =−

d`−1

X

i=0

a^(`)_i (z_`)U_`(wz^j+i−d_` ^`).

Proof.

ApplyingU` to wz^j−d_` ^`F`(z`(τ), z`(`τ)) = 0gives the desired result.

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Corollary: The Fundamental Lemma

Fundamental Lemma. Let `∈ {5,7,11}. For any w:H→C andj∈Z:

U_`(wz^j_`) =−

d`−1

X

i=0

a^(`)_i (z_`)U_`(wz^j+i−d_` ^`).

Proof.

ApplyingU_` to wz^j−d_` ^`F_`(z_`(τ), z_`(`τ)) = 0gives the desired result.