modular forms modulo 2 - brandeis...
TRANSCRIPT
Modular forms modulo 2Mathematics Department Colloquium, Columbia University
Joel Bellaıche
Brandeis University, Yale University
November 30, 2012
Outline
I. Modular forms and Hecke algebra modulo 2 (published workof Nicolas and Serre)
II. Galois representation on the Hecke algebra modulo 2(published work of me)
III. Special modular forms modulo 2 (work in progress of Nicolas,Serre, and me)
IV. Density of modular forms modulo 2 (work in progress of me)
Outline
I. Modular forms and Hecke algebra modulo 2 (published workof Nicolas and Serre)
II. Galois representation on the Hecke algebra modulo 2(published work of me)
III. Special modular forms modulo 2 (work in progress of Nicolas,Serre, and me)
IV. Density of modular forms modulo 2 (work in progress of me)
Outline
I. Modular forms and Hecke algebra modulo 2 (published workof Nicolas and Serre)
II. Galois representation on the Hecke algebra modulo 2(published work of me)
III. Special modular forms modulo 2 (work in progress of Nicolas,Serre, and me)
IV. Density of modular forms modulo 2 (work in progress of me)
Outline
I. Modular forms and Hecke algebra modulo 2 (published workof Nicolas and Serre)
II. Galois representation on the Hecke algebra modulo 2(published work of me)
III. Special modular forms modulo 2 (work in progress of Nicolas,Serre, and me)
IV. Density of modular forms modulo 2 (work in progress of me)
Part I.
Modular forms and Hecke algebra modulo 2
(after Jean-Louis Nicolas, Jean-Pierre Serre)
Modular formsA modular form of weight k (and level 1) is an analytic function on{q ∈ C, |q − 1| < 1},
f (q) =∞∑n=0
anqn ∈ C[[q]]
such that, setting q = e2iπz , one has
f (−1/z) = zk f (z)
Examples:
∆(q) = q∏n≥1
(1− qn)24 =∞∑n=1
τ(n)qn.
∆k(q) =∞∑n=k
τk(n)qn.
Modular formsA modular form of weight k (and level 1) is an analytic function on{q ∈ C, |q − 1| < 1},
f (q) =∞∑n=0
anqn ∈ C[[q]]
such that, setting q = e2iπz , one has
f (−1/z) = zk f (z)
Examples:
∆(q) = q∏n≥1
(1− qn)24 =∞∑n=1
τ(n)qn.
∆k(q) =∞∑n=k
τk(n)qn.
Modular formsA modular form of weight k (and level 1) is an analytic function on{q ∈ C, |q − 1| < 1},
f (q) =∞∑n=0
anqn ∈ C[[q]]
such that, setting q = e2iπz , one has
f (−1/z) = zk f (z)
Examples:
∆(q) = q∏n≥1
(1− qn)24 =∞∑n=1
τ(n)qn.
∆k(q) =∞∑n=k
τk(n)qn.
The space M of modular forms modulo 2
If f (q) =∑∞
n=0 anqn ∈ Z[[q]] is a modular form, define
f =∑
anqn ∈ F2[[q]].
Example (Jacobi): ∆ =∑
n odd qn2 = 1 + q9 + q25 + q49 + . . .
Fact (Swinnerton-Dyer): the subspace of F2[[q]] generated by f forall modular forms f ∈ Z[[q]] is F2[∆].
Remark: if f =∑
anqn ∈ F2[[q]], then f 2 =∑
anq2n is not reallynew. Hence we only consider ∆k with k odd. We define, for everyodd integer k ,
Mk = F2∆⊕ F2∆3 ⊕ · · · ⊕ ∆k , M = ∪k≥1,k odd Mk
The space M will be the main object of this talk.
The space M of modular forms modulo 2
If f (q) =∑∞
n=0 anqn ∈ Z[[q]] is a modular form, define
f =∑
anqn ∈ F2[[q]].
Example (Jacobi): ∆ =∑
n odd qn2 = 1 + q9 + q25 + q49 + . . .
Fact (Swinnerton-Dyer): the subspace of F2[[q]] generated by f forall modular forms f ∈ Z[[q]] is F2[∆].
Remark: if f =∑
anqn ∈ F2[[q]], then f 2 =∑
anq2n is not reallynew. Hence we only consider ∆k with k odd. We define, for everyodd integer k ,
Mk = F2∆⊕ F2∆3 ⊕ · · · ⊕ ∆k , M = ∪k≥1,k odd Mk
The space M will be the main object of this talk.
The space M of modular forms modulo 2
If f (q) =∑∞
n=0 anqn ∈ Z[[q]] is a modular form, define
f =∑
anqn ∈ F2[[q]].
Example (Jacobi): ∆ =∑
n odd qn2 = 1 + q9 + q25 + q49 + . . .
Fact (Swinnerton-Dyer): the subspace of F2[[q]] generated by f forall modular forms f ∈ Z[[q]] is F2[∆].
Remark: if f =∑
anqn ∈ F2[[q]], then f 2 =∑
anq2n is not reallynew. Hence we only consider ∆k with k odd. We define, for everyodd integer k ,
Mk = F2∆⊕ F2∆3 ⊕ · · · ⊕ ∆k , M = ∪k≥1,k odd Mk
The space M will be the main object of this talk.
The space M of modular forms modulo 2
If f (q) =∑∞
n=0 anqn ∈ Z[[q]] is a modular form, define
f =∑
anqn ∈ F2[[q]].
Example (Jacobi): ∆ =∑
n odd qn2 = 1 + q9 + q25 + q49 + . . .
Fact (Swinnerton-Dyer): the subspace of F2[[q]] generated by f forall modular forms f ∈ Z[[q]] is F2[∆].
Remark: if f =∑
anqn ∈ F2[[q]], then f 2 =∑
anq2n is not reallynew. Hence we only consider ∆k with k odd. We define, for everyodd integer k ,
Mk = F2∆⊕ F2∆3 ⊕ · · · ⊕ ∆k , M = ∪k≥1,k odd Mk
The space M will be the main object of this talk.
The space M of modular forms modulo 2
If f (q) =∑∞
n=0 anqn ∈ Z[[q]] is a modular form, define
f =∑
anqn ∈ F2[[q]].
Example (Jacobi): ∆ =∑
n odd qn2 = 1 + q9 + q25 + q49 + . . .
Fact (Swinnerton-Dyer): the subspace of F2[[q]] generated by f forall modular forms f ∈ Z[[q]] is F2[∆].
Remark: if f =∑
anqn ∈ F2[[q]], then f 2 =∑
anq2n is not reallynew. Hence we only consider ∆k with k odd. We define, for everyodd integer k ,
Mk = F2∆⊕ F2∆3 ⊕ · · · ⊕ ∆k , M = ∪k≥1,k odd Mk
The space M will be the main object of this talk.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q; (for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q; (for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q;
(for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q; (for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q; (for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Motivation
The study of M will give informations on the parity of manyarithmetic functions, including:
I the generalized Ramanujan functions τk(n);
I the number rq(n) of representations of n by a a certain typeof quadratic form q; (for example,
∆9 = ∆8∆ =∑
a,b, odd
q8a2+b2
hence the coefficient ap of ∆9 is 1 if and only if p ≡ 1(mod 8) and n is not represented by a2 + 32b2.)
I More indirectly, and speculatively, the partition function p(n),which is conjectured to be odd or even half of the time inaverage.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M. Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M.
Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M. Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.
The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M. Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M. Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
Hecke operators and the Hecke algebra A
For every prime `, the Hecke operator T` on F2[[q]] is defined by
T`(∑
anqn) =∑
anqn` + `∑
an`qn,
The T`’s stabilize Mk and M. Let Ak be the subalgebra ofEndF2(Mk) generated by the Hecke operators T` for ` odd prime.The inclusion Mk ⊂ Mk+2 induces a surjective morphismAk+2 → Ak .
SetA = lim←−Ak .
A is called the Hecke algebra of modular forms modulo 2, and ourmain tool to study M, which is an A-module.
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8).
Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.
Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What is the structure of the Hecke algebra A?
Three easy facts known in the late 1960’s
1.– There is ony one semi-simple Galois representationGQ,2 → GL2(F2), namely 1⊕ 1 (trace: 0).
2.– Consequence (Deligne): T` is nilpotent on Mk .
3.– Consequence: the algebra A is a complete local ring, with theT` in its maximal ideal mA.
Theorem (Nicolas-Serre, 2010-2012)
A = F2[[x , y ]] where x = T3, y = T5. Hence A is a noetheriencomplete regular local ring of dimension 2.
We can also take x = T`, ` ≡ 3 (mod 8), y = T`′ , `′ ≡ 5
(mod 8). Their proof is hard, very long, completely elementary.Remark: same is true with 2 replaced by p > 2 (joint work withKhare).
What are the consequences for the structure of M?
The A-module M is the dualizing module of A.
Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
What are the consequences for the structure of M?
The A-module M is the dualizing module of A. Hence,
Corollary
There exists a unique basis m(a, b)a≥0,b≥0 of M such that:
I m(0, 0) = ∆
I T3m(a, b) = m(a− 1, b) if a > 0, and T3m(0, b) = 0.
I T5m(a, b) = m(a, b − 1) if b > 0, and T5m(a, 0) = 0.
I the first coefficient a1 of m(a, b) is 0 except if a = b = 0.
Such a basis is called an adapted basis (for X = T3 and Y = T5)
More structure: a simple grading by (Z/8Z)∗
Grading of M: for i ∈ (Z/8Z)∗ = {1, 3, 5, 7}, let M i ⊂ Mgenerated by the ∆k , k ≡ i (mod 8).
M i = {f =∑
anqn ∈ M, an = 1⇒ n ≡ i (mod 8)}.
M = M1 ⊕M3 ⊕M5 ⊕M7.
Also, let A1 = F2[[x2, y2]], A3 = xA1, A5 = yA1, A7 = xyA1. Then
A = A1 ⊕ A3 ⊕ A5 ⊕ A7,
A is a graded algebra and M a graded A-module.That is
AiAj ⊂ Aij , AiM j ⊂ M ij .
One has Tp ∈ A p(mod 8). The form m(a, b) is in M 3a5b(mod 8).
More structure: a simple grading by (Z/8Z)∗
Grading of M: for i ∈ (Z/8Z)∗ = {1, 3, 5, 7}, let M i ⊂ Mgenerated by the ∆k , k ≡ i (mod 8).
M i = {f =∑
anqn ∈ M, an = 1⇒ n ≡ i (mod 8)}.
M = M1 ⊕M3 ⊕M5 ⊕M7.
Also, let A1 = F2[[x2, y2]], A3 = xA1, A5 = yA1, A7 = xyA1. Then
A = A1 ⊕ A3 ⊕ A5 ⊕ A7,
A is a graded algebra and M a graded A-module.That is
AiAj ⊂ Aij , AiM j ⊂ M ij .
One has Tp ∈ A p(mod 8). The form m(a, b) is in M 3a5b(mod 8).
More structure: a simple grading by (Z/8Z)∗
Grading of M: for i ∈ (Z/8Z)∗ = {1, 3, 5, 7}, let M i ⊂ Mgenerated by the ∆k , k ≡ i (mod 8).
M i = {f =∑
anqn ∈ M, an = 1⇒ n ≡ i (mod 8)}.
M = M1 ⊕M3 ⊕M5 ⊕M7.
Also, let A1 = F2[[x2, y2]], A3 = xA1, A5 = yA1, A7 = xyA1. Then
A = A1 ⊕ A3 ⊕ A5 ⊕ A7,
A is a graded algebra and M a graded A-module.
That is
AiAj ⊂ Aij , AiM j ⊂ M ij .
One has Tp ∈ A p(mod 8). The form m(a, b) is in M 3a5b(mod 8).
More structure: a simple grading by (Z/8Z)∗
Grading of M: for i ∈ (Z/8Z)∗ = {1, 3, 5, 7}, let M i ⊂ Mgenerated by the ∆k , k ≡ i (mod 8).
M i = {f =∑
anqn ∈ M, an = 1⇒ n ≡ i (mod 8)}.
M = M1 ⊕M3 ⊕M5 ⊕M7.
Also, let A1 = F2[[x2, y2]], A3 = xA1, A5 = yA1, A7 = xyA1. Then
A = A1 ⊕ A3 ⊕ A5 ⊕ A7,
A is a graded algebra and M a graded A-module.That is
AiAj ⊂ Aij , AiM j ⊂ M ij .
One has Tp ∈ A p(mod 8). The form m(a, b) is in M 3a5b(mod 8).
More structure: a simple grading by (Z/8Z)∗
Grading of M: for i ∈ (Z/8Z)∗ = {1, 3, 5, 7}, let M i ⊂ Mgenerated by the ∆k , k ≡ i (mod 8).
M i = {f =∑
anqn ∈ M, an = 1⇒ n ≡ i (mod 8)}.
M = M1 ⊕M3 ⊕M5 ⊕M7.
Also, let A1 = F2[[x2, y2]], A3 = xA1, A5 = yA1, A7 = xyA1. Then
A = A1 ⊕ A3 ⊕ A5 ⊕ A7,
A is a graded algebra and M a graded A-module.That is
AiAj ⊂ Aij , AiM j ⊂ M ij .
One has Tp ∈ A p(mod 8). The form m(a, b) is in M 3a5b(mod 8).
Part II.
Galois representation on the Hecke algebra modulo 2
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
),
where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).
I If P is a prime of height 1, let k(P) = Frac(A/P),rP : G → GL2(k(P)). Then rP is
I irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP is
I irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;
I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
The universal Galois representation on A
Let G be the maximal pro-2-quotient on GQ,2.
Theorem
I There exists a unique continuous r : G → GL2(A) such thattr r(Frob `) = T` for ` 6= 2. It is absolutely irreducible, anddet r = 1.
I r (mod mA) '(
1 η0 1
), where η is the character attached to
Q(√
2).I If P is a prime of height 1, let k(P) = Frac(A/P),
rP : G → GL2(k(P)). Then rP isI irreducible for all P;I absolutely irreducible for all P but one:P0 = (x + y + x3 + x5 + x9 + x11 + x129 + . . . );
I strongly absolutely irreducible for all P excepted P0, (x), (y).
I’m going to explain the proof of the first point.
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension.
Chenevier’spseudorepresentation always work but are much more complicatedto define. I will only define them in dimension 2.
Let A be a commutative ring, G a group. A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension. Chenevier’spseudorepresentation always work
but are much more complicatedto define. I will only define them in dimension 2.
Let A be a commutative ring, G a group. A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension. Chenevier’spseudorepresentation always work but are much more complicatedto define.
I will only define them in dimension 2.
Let A be a commutative ring, G a group. A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension. Chenevier’spseudorepresentation always work but are much more complicatedto define. I will only define them in dimension 2.
Let A be a commutative ring, G a group. A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension. Chenevier’spseudorepresentation always work but are much more complicatedto define. I will only define them in dimension 2.
Let A be a commutative ring, G a group.
A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Chenevier’s pseudorepresentations
Traditional pseudorepresentations (Taylor, Rouquier) work only incharacteristic greater than the dimension. Chenevier’spseudorepresentation always work but are much more complicatedto define. I will only define them in dimension 2.
Let A be a commutative ring, G a group. A pseudorepresentationof G on A is a pair (t, d) of applications from G to A such that:
(i) d : G → A∗ is a group morphism.
(ii) t(1) = 2.
(iii) t(gh) = t(hg).
(iv) t(gh) + d(h)t(gh−1) = t(g)t(h).
Facts about Chenevier’s pseudorepresentations
I If r : G → GL2(A) is a representation, (tr r , det r) is apseudo-representation.
I If K is an algebraically closed field, any pseudorepresentation(t, d) of G to K is (tr r , det r) for a unique semi-simplerepresentation r : G → GL2(K ).
I If B is a subring of A, (t, d) pseudorepresentation of G to Asuch that t(G ) ⊂ B, d(G ) ⊂ B∗, then (t, g) is apseudorepresentation on G to B. (”pseudorepresentationsdescend and glue well”).
Facts about Chenevier’s pseudorepresentations
I If r : G → GL2(A) is a representation, (tr r , det r) is apseudo-representation.
I If K is an algebraically closed field, any pseudorepresentation(t, d) of G to K is (tr r , det r) for a unique semi-simplerepresentation r : G → GL2(K ).
I If B is a subring of A, (t, d) pseudorepresentation of G to Asuch that t(G ) ⊂ B, d(G ) ⊂ B∗, then (t, g) is apseudorepresentation on G to B. (”pseudorepresentationsdescend and glue well”).
Facts about Chenevier’s pseudorepresentations
I If r : G → GL2(A) is a representation, (tr r , det r) is apseudo-representation.
I If K is an algebraically closed field, any pseudorepresentation(t, d) of G to K is (tr r , det r) for a unique semi-simplerepresentation r : G → GL2(K ).
I If B is a subring of A, (t, d) pseudorepresentation of G to Asuch that t(G ) ⊂ B, d(G ) ⊂ B∗, then (t, g) is apseudorepresentation on G to B. (”pseudorepresentationsdescend and glue well”).
Facts about Chenevier’s pseudorepresentations
I If r : G → GL2(A) is a representation, (tr r , det r) is apseudo-representation.
I If K is an algebraically closed field, any pseudorepresentation(t, d) of G to K is (tr r , det r) for a unique semi-simplerepresentation r : G → GL2(K ).
I If B is a subring of A, (t, d) pseudorepresentation of G to Asuch that t(G ) ⊂ B, d(G ) ⊂ B∗, then (t, g) is apseudorepresentation on G to B. (”pseudorepresentationsdescend and glue well”).
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible.
If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab.
Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2.
But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9,
Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler).
Contradiction.
Proof of the existence of the representation r
Step 1 Construct a continuous pseudorepresentation(t, d) : G → A such that t(Frob p) = Tp, d = 1. One gluesDeligne’s representations attached to eigenforms of level 1 incharacteristic 0 and reduces modulo 2.
Step 2 Let K = Frac(A), K = alg. closure of K . By Chenevier’stheorem, there exists r : G → SL2(K ), such thattr r(Frob p) = Tp. (Remark: I don’t claim that r is continuous)
Step 3 Claim: r is (absolutely) irreducible. If not, r , hence t = tr rfactors through G ab. Therefore for any integer k, t(Frob p) = Tp
in Ak ⊂ End(Mk) depends only on Frob p in a finite abelianextension of Q unramified outside 2. But for k = 9, Tp∆9 = ∆ ifand only if p ≡ 1 (mod 8) and p is not represented by a2 + 32b2,and this condition does not depend on Frob p in an abelianextension of Q (because 32 is not idoneal – Euler). Contradiction.
Proof of the existence of the representation r
Step 4 The representation r is defined over K .
If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even.
So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian.
Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2.
The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Proof of the existence of the representation r
Step 4 The representation r is defined over K . If not, sincetr r(G ) ⊂ K and r is absolutely irreducible, there is a field ofquaternions H over K such that r factors through G → H∗. Thenfor c ∈ G the complex conjugation, r(c)2 = r(c2) = 1, so(r(c)− 1)2 = 0 and r(c) = 1 since H is a field, and r is even. So rfactors through the maximal totally real quotient of G , which isabelian. Contradiction.
Step 5 Since tr (r) ⊂ A and r :→ SL2(K ) is absolutely irreducible,r stabilizes an A-lattice M ⊂ K 2. The bidual M ′′ of M is areflexive module over A which is a regular local ring of dimension2, hence is free. Hence a representation r : G → SL2(A) of trace t.
Step 6 Since r is absolutely irreducible, and tr r = t is continuous,r is continuous.
Universality of A
TheoremThe algebra A, with the pseudorepresentation (t, 1) constructed instep 1 above, is the universal deformation ring R of thepseudorepresentation (0, 1) : G → F2, with the condition d = 1and t(c) = 0.
This is an ”R = T ” theorem, excepted that T is called A. This isalso the main tool to prove the other points of the thorem.
Universality of A
TheoremThe algebra A, with the pseudorepresentation (t, 1) constructed instep 1 above, is the universal deformation ring R of thepseudorepresentation (0, 1) : G → F2, with the condition d = 1and t(c) = 0.
This is an ”R = T ” theorem, excepted that T is called A. This isalso the main tool to prove the other points of the thorem.
Part III.
Special modular forms modulo 2
(joint with Jean-Louis Nicolas, Jean-Pierre Serre)
Definitions of special modular forms.
For every f , Tpf depends only on f in some finite extension L ofQ, unramified outside 2, of degree a power of 2. Let L(f ) thesmallest such extension.
One says that f ∈ M is abelian if L(f ) is abelian. Examples: ∆9 isnot abelian, but ∆3, ∆5, ∆7 are.
One says that f ∈ M is Q(i)-dihedral, resp. Q(√−2)-dihedral if
L(f ) is a dihedral extension of Q containing Q(i) (resp. Q(√−2)).
One says that f is special if it is a linear combination of abelianand dihedral forms.
Definitions of special modular forms.
For every f , Tpf depends only on f in some finite extension L ofQ, unramified outside 2, of degree a power of 2. Let L(f ) thesmallest such extension.
One says that f ∈ M is abelian if L(f ) is abelian. Examples: ∆9 isnot abelian, but ∆3, ∆5, ∆7 are.
One says that f ∈ M is Q(i)-dihedral, resp. Q(√−2)-dihedral if
L(f ) is a dihedral extension of Q containing Q(i) (resp. Q(√−2)).
One says that f is special if it is a linear combination of abelianand dihedral forms.
Definitions of special modular forms.
For every f , Tpf depends only on f in some finite extension L ofQ, unramified outside 2, of degree a power of 2. Let L(f ) thesmallest such extension.
One says that f ∈ M is abelian if L(f ) is abelian. Examples: ∆9 isnot abelian, but ∆3, ∆5, ∆7 are.
One says that f ∈ M is Q(i)-dihedral, resp. Q(√−2)-dihedral if
L(f ) is a dihedral extension of Q containing Q(i) (resp. Q(√−2)).
One says that f is special if it is a linear combination of abelianand dihedral forms.
Definitions of special modular forms.
For every f , Tpf depends only on f in some finite extension L ofQ, unramified outside 2, of degree a power of 2. Let L(f ) thesmallest such extension.
One says that f ∈ M is abelian if L(f ) is abelian. Examples: ∆9 isnot abelian, but ∆3, ∆5, ∆7 are.
One says that f ∈ M is Q(i)-dihedral, resp. Q(√−2)-dihedral if
L(f ) is a dihedral extension of Q containing Q(i) (resp. Q(√−2)).
One says that f is special if it is a linear combination of abelianand dihedral forms.
Definitions of special modular forms.
For every f , Tpf depends only on f in some finite extension L ofQ, unramified outside 2, of degree a power of 2. Let L(f ) thesmallest such extension.
One says that f ∈ M is abelian if L(f ) is abelian. Examples: ∆9 isnot abelian, but ∆3, ∆5, ∆7 are.
One says that f ∈ M is Q(i)-dihedral, resp. Q(√−2)-dihedral if
L(f ) is a dihedral extension of Q containing Q(i) (resp. Q(√−2)).
One says that f is special if it is a linear combination of abelianand dihedral forms.
Characterization of special modular forms
TheoremA form f ∈ M is Q(i)-dihedral iff xf = 0 or ∆. A form f ∈ M isdihedral of type Q(
√−2) iff yf = 0 or ∆. A form f ∈ M is abelian
iff f is killed by P20 .
Corollary
The space of Q(i)-dihedral forms is generated by the m(0, b),b ∈ N and m(1, 0) = ∆3. The space of Q(
√−2)-dihedral forms is
generated by the m(a, 0), a ∈ N and m(0, 1) = ∆5.
Corollary
Special forms are sparse: in the subspace of M generated by them(a, b) with a + b < n, of dimension n(n + 1)/2, the dimension ofthe subspace of abelian forms is 2n − 1, and dimension of thesubspace of dihedral forms of each type is n + 1.
Characterization of special modular forms
TheoremA form f ∈ M is Q(i)-dihedral iff xf = 0 or ∆. A form f ∈ M isdihedral of type Q(
√−2) iff yf = 0 or ∆. A form f ∈ M is abelian
iff f is killed by P20 .
Corollary
The space of Q(i)-dihedral forms is generated by the m(0, b),b ∈ N and m(1, 0) = ∆3. The space of Q(
√−2)-dihedral forms is
generated by the m(a, 0), a ∈ N and m(0, 1) = ∆5.
Corollary
Special forms are sparse: in the subspace of M generated by them(a, b) with a + b < n, of dimension n(n + 1)/2, the dimension ofthe subspace of abelian forms is 2n − 1, and dimension of thesubspace of dihedral forms of each type is n + 1.
Characterization of special modular forms
TheoremA form f ∈ M is Q(i)-dihedral iff xf = 0 or ∆. A form f ∈ M isdihedral of type Q(
√−2) iff yf = 0 or ∆. A form f ∈ M is abelian
iff f is killed by P20 .
Corollary
The space of Q(i)-dihedral forms is generated by the m(0, b),b ∈ N and m(1, 0) = ∆3. The space of Q(
√−2)-dihedral forms is
generated by the m(a, 0), a ∈ N and m(0, 1) = ∆5.
Corollary
Special forms are sparse: in the subspace of M generated by them(a, b) with a + b < n, of dimension n(n + 1)/2, the dimension ofthe subspace of abelian forms is 2n − 1, and dimension of thesubspace of dihedral forms of each type is n + 1.
Characterization of special modular forms
TheoremA form f ∈ M is Q(i)-dihedral iff xf = 0 or ∆. A form f ∈ M isdihedral of type Q(
√−2) iff yf = 0 or ∆. A form f ∈ M is abelian
iff f is killed by P20 .
Corollary
The space of Q(i)-dihedral forms is generated by the m(0, b),b ∈ N and m(1, 0) = ∆3. The space of Q(
√−2)-dihedral forms is
generated by the m(a, 0), a ∈ N and m(0, 1) = ∆5.
Corollary
Special forms are sparse: in the subspace of M generated by them(a, b) with a + b < n, of dimension n(n + 1)/2, the dimension ofthe subspace of abelian forms is 2n − 1, and dimension of thesubspace of dihedral forms of each type is n + 1.
Powers of ∆ that are special
Proposition
For any r , ∆2r+1 and for any odd r , ∆(2r+1)/3 are dihedral. Fork = 1, 3, 5, 7, 19, 21, ∆k is abelian.
Conjecture
The only forms ∆k that are special are the ones given in theproposition.
In particular, there are only 6 values of k (explicitly1, 3, 5, 7, 19, 21), conjecturally, such that ∆k is abelian.This isvaguely reminescent of the finiteness of idoneal numbers(conjectured by Euler, proved by Hecke and Landau under GRH,then unconditionally).
Our conjecture is proved for most k ’s.
Powers of ∆ that are special
Proposition
For any r , ∆2r+1 and for any odd r , ∆(2r+1)/3 are dihedral. Fork = 1, 3, 5, 7, 19, 21, ∆k is abelian.
Conjecture
The only forms ∆k that are special are the ones given in theproposition.
In particular, there are only 6 values of k (explicitly1, 3, 5, 7, 19, 21), conjecturally, such that ∆k is abelian.
This isvaguely reminescent of the finiteness of idoneal numbers(conjectured by Euler, proved by Hecke and Landau under GRH,then unconditionally).
Our conjecture is proved for most k ’s.
Powers of ∆ that are special
Proposition
For any r , ∆2r+1 and for any odd r , ∆(2r+1)/3 are dihedral. Fork = 1, 3, 5, 7, 19, 21, ∆k is abelian.
Conjecture
The only forms ∆k that are special are the ones given in theproposition.
In particular, there are only 6 values of k (explicitly1, 3, 5, 7, 19, 21), conjecturally, such that ∆k is abelian.This isvaguely reminescent of the finiteness of idoneal numbers(conjectured by Euler, proved by Hecke and Landau under GRH,then unconditionally).
Our conjecture is proved for most k ’s.
Powers of ∆ that are special
Proposition
For any r , ∆2r+1 and for any odd r , ∆(2r+1)/3 are dihedral. Fork = 1, 3, 5, 7, 19, 21, ∆k is abelian.
Conjecture
The only forms ∆k that are special are the ones given in theproposition.
In particular, there are only 6 values of k (explicitly1, 3, 5, 7, 19, 21), conjecturally, such that ∆k is abelian.This isvaguely reminescent of the finiteness of idoneal numbers(conjectured by Euler, proved by Hecke and Landau under GRH,then unconditionally).
Our conjecture is proved for most k ’s.
Part IV.
Density of modular forms modulo 2
Density: definitions and reductions
For f =∑
anqn ∈ M, the set {p, ap = 1} has a density, denotedδ(f ) (by Chebotarev, since ap depends only on Frob p inGal(L(f )/Q)).
One has 0 ≤ δ(f ) ≤ 1, δ(f ) ∈ Z[1/2].
Aim: compute δ(f ) for f ∈ M.
Remark: If f = f1 + f3 + f5 + f7 with fi ∈ M i ,δ(f ) = δ(f1) + δ(f3) + δ(f5) + δ(f7). This reduces the study ofdensity to the case of homegenous f , i.e. f ∈ M i for somei ∈ (Z/8Z)∗. For f homogeneous, δ(f ) ≤ 1/4.
Density: definitions and reductions
For f =∑
anqn ∈ M, the set {p, ap = 1} has a density, denotedδ(f ) (by Chebotarev, since ap depends only on Frob p inGal(L(f )/Q)). One has 0 ≤ δ(f ) ≤ 1, δ(f ) ∈ Z[1/2].
Aim: compute δ(f ) for f ∈ M.
Remark: If f = f1 + f3 + f5 + f7 with fi ∈ M i ,δ(f ) = δ(f1) + δ(f3) + δ(f5) + δ(f7). This reduces the study ofdensity to the case of homegenous f , i.e. f ∈ M i for somei ∈ (Z/8Z)∗. For f homogeneous, δ(f ) ≤ 1/4.
Density: definitions and reductions
For f =∑
anqn ∈ M, the set {p, ap = 1} has a density, denotedδ(f ) (by Chebotarev, since ap depends only on Frob p inGal(L(f )/Q)). One has 0 ≤ δ(f ) ≤ 1, δ(f ) ∈ Z[1/2].
Aim: compute δ(f ) for f ∈ M.
Remark: If f = f1 + f3 + f5 + f7 with fi ∈ M i ,δ(f ) = δ(f1) + δ(f3) + δ(f5) + δ(f7). This reduces the study ofdensity to the case of homegenous f , i.e. f ∈ M i for somei ∈ (Z/8Z)∗. For f homogeneous, δ(f ) ≤ 1/4.
Density: definitions and reductions
For f =∑
anqn ∈ M, the set {p, ap = 1} has a density, denotedδ(f ) (by Chebotarev, since ap depends only on Frob p inGal(L(f )/Q)). One has 0 ≤ δ(f ) ≤ 1, δ(f ) ∈ Z[1/2].
Aim: compute δ(f ) for f ∈ M.
Remark: If f = f1 + f3 + f5 + f7 with fi ∈ M i ,δ(f ) = δ(f1) + δ(f3) + δ(f5) + δ(f7). This reduces the study ofdensity to the case of homegenous f , i.e. f ∈ M i for somei ∈ (Z/8Z)∗. For f homogeneous, δ(f ) ≤ 1/4.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: qualitative results
TheoremIf f ∈ M, f 6= 0, ∆, then δ(f ) > 0.
Corollary
Let f =∑
anqn, and g =∑
bnqn. If ap = bp for every prime p(except possibly a set of density 0), then f = g or f = g + ∆(hence an = bn for all n except odd squares).
Corollary
For every odd k ≥ 3, there exists infinitely many p such that τk(p)is odd.
TheoremIf f is homogenous and f 6= ∆3, ∆5, δ(f ) < 1/4. For any f ∈ M,δ(f ) < 1
Corollary
For every odd k, there exists infinitely many p such that τk(p) iseven.
Density: quantitative results and a conjecture
For f in M, its nilpotence index is the smallest a such thatxnymf = 0 for every n,m such that n + m > a.
TheoremIf f is special and homogeneous, of nilpotence index a, thenδ(f ) = 2−v(a)−u(a)−1, where v(a) is the 2-valuation of a, u(a) isthe numbers of digits 1 in the expansion of a in basis 2.
Conjecture
If f is non-special and homogenous, then δ(f ) = 1/8.
Computed with sage the density of ∆k , k < 4000 and linearcombinations of ∆k , k < 80, for primes up to a 1 million. Veryclose to 0.125.
TheoremThe conjecture is true for all forms of nilpotence index at most 12,for all the forms m(a, 1) (or m(1, a)) and linear combinationsthereof, for all forms m(2r − 1, 2), etc.
Density: quantitative results and a conjecture
For f in M, its nilpotence index is the smallest a such thatxnymf = 0 for every n,m such that n + m > a.
TheoremIf f is special and homogeneous, of nilpotence index a, thenδ(f ) = 2−v(a)−u(a)−1, where v(a) is the 2-valuation of a, u(a) isthe numbers of digits 1 in the expansion of a in basis 2.
Conjecture
If f is non-special and homogenous, then δ(f ) = 1/8.
Computed with sage the density of ∆k , k < 4000 and linearcombinations of ∆k , k < 80, for primes up to a 1 million. Veryclose to 0.125.
TheoremThe conjecture is true for all forms of nilpotence index at most 12,for all the forms m(a, 1) (or m(1, a)) and linear combinationsthereof, for all forms m(2r − 1, 2), etc.
Density: quantitative results and a conjecture
For f in M, its nilpotence index is the smallest a such thatxnymf = 0 for every n,m such that n + m > a.
TheoremIf f is special and homogeneous, of nilpotence index a, thenδ(f ) = 2−v(a)−u(a)−1, where v(a) is the 2-valuation of a, u(a) isthe numbers of digits 1 in the expansion of a in basis 2.
Conjecture
If f is non-special and homogenous, then δ(f ) = 1/8.
Computed with sage the density of ∆k , k < 4000 and linearcombinations of ∆k , k < 80, for primes up to a 1 million. Veryclose to 0.125.
TheoremThe conjecture is true for all forms of nilpotence index at most 12,for all the forms m(a, 1) (or m(1, a)) and linear combinationsthereof, for all forms m(2r − 1, 2), etc.
Density: quantitative results and a conjecture
For f in M, its nilpotence index is the smallest a such thatxnymf = 0 for every n,m such that n + m > a.
TheoremIf f is special and homogeneous, of nilpotence index a, thenδ(f ) = 2−v(a)−u(a)−1, where v(a) is the 2-valuation of a, u(a) isthe numbers of digits 1 in the expansion of a in basis 2.
Conjecture
If f is non-special and homogenous, then δ(f ) = 1/8.
Computed with sage the density of ∆k , k < 4000 and linearcombinations of ∆k , k < 80, for primes up to a 1 million. Veryclose to 0.125.
TheoremThe conjecture is true for all forms of nilpotence index at most 12,for all the forms m(a, 1) (or m(1, a)) and linear combinationsthereof, for all forms m(2r − 1, 2), etc.
Density: quantitative results and a conjecture
For f in M, its nilpotence index is the smallest a such thatxnymf = 0 for every n,m such that n + m > a.
TheoremIf f is special and homogeneous, of nilpotence index a, thenδ(f ) = 2−v(a)−u(a)−1, where v(a) is the 2-valuation of a, u(a) isthe numbers of digits 1 in the expansion of a in basis 2.
Conjecture
If f is non-special and homogenous, then δ(f ) = 1/8.
Computed with sage the density of ∆k , k < 4000 and linearcombinations of ∆k , k < 80, for primes up to a 1 million. Veryclose to 0.125.
TheoremThe conjecture is true for all forms of nilpotence index at most 12,for all the forms m(a, 1) (or m(1, a)) and linear combinationsthereof, for all forms m(2r − 1, 2), etc.
Tools for the proofs: the structure of the pro-2-group G
Let F be the Frattini subgroup of G . Then
G/F = Gal(Q(µ8)/Q) = (Z/8Z)∗
We write G i for the preimage of i = 1, 3, 5, 7 in G/F . We haveFrob p ∈ Gp(mod 8) and t(G i ) ∈ Ai .Let c ∈ G be a complex conjugation. Hence c ∈ G 7.
Lemma (Serre)
Any element of order 2 of G is conjugate to c.
LemmaLet u be any element in G 3 or G 5. Then G has the presentation〈u, c |c2 = 1〉 in the category of pro-2-groups.
The proof uses a result of Shafarevich.Let us choose u and c as in the lemma, with u ∈ G 3 to fix ideas.Let us write t(u) = X ∈ A, t(cu) = Y ∈ A. Then X ,Y areanalogs of x , y .
Tools for the proofs: the structure of the pro-2-group G
Let F be the Frattini subgroup of G . Then
G/F = Gal(Q(µ8)/Q) = (Z/8Z)∗
We write G i for the preimage of i = 1, 3, 5, 7 in G/F . We haveFrob p ∈ Gp(mod 8) and t(G i ) ∈ Ai .Let c ∈ G be a complex conjugation. Hence c ∈ G 7.
Lemma (Serre)
Any element of order 2 of G is conjugate to c.
LemmaLet u be any element in G 3 or G 5. Then G has the presentation〈u, c |c2 = 1〉 in the category of pro-2-groups.
The proof uses a result of Shafarevich.Let us choose u and c as in the lemma, with u ∈ G 3 to fix ideas.Let us write t(u) = X ∈ A, t(cu) = Y ∈ A. Then X ,Y areanalogs of x , y .
Tools for the proofs: the structure of the pro-2-group G
Let F be the Frattini subgroup of G . Then
G/F = Gal(Q(µ8)/Q) = (Z/8Z)∗
We write G i for the preimage of i = 1, 3, 5, 7 in G/F . We haveFrob p ∈ Gp(mod 8) and t(G i ) ∈ Ai .Let c ∈ G be a complex conjugation. Hence c ∈ G 7.
Lemma (Serre)
Any element of order 2 of G is conjugate to c.
LemmaLet u be any element in G 3 or G 5. Then G has the presentation〈u, c |c2 = 1〉 in the category of pro-2-groups.
The proof uses a result of Shafarevich.
Let us choose u and c as in the lemma, with u ∈ G 3 to fix ideas.Let us write t(u) = X ∈ A, t(cu) = Y ∈ A. Then X ,Y areanalogs of x , y .
Tools for the proofs: the structure of the pro-2-group G
Let F be the Frattini subgroup of G . Then
G/F = Gal(Q(µ8)/Q) = (Z/8Z)∗
We write G i for the preimage of i = 1, 3, 5, 7 in G/F . We haveFrob p ∈ Gp(mod 8) and t(G i ) ∈ Ai .Let c ∈ G be a complex conjugation. Hence c ∈ G 7.
Lemma (Serre)
Any element of order 2 of G is conjugate to c.
LemmaLet u be any element in G 3 or G 5. Then G has the presentation〈u, c |c2 = 1〉 in the category of pro-2-groups.
The proof uses a result of Shafarevich.Let us choose u and c as in the lemma, with u ∈ G 3 to fix ideas.
Let us write t(u) = X ∈ A, t(cu) = Y ∈ A. Then X ,Y areanalogs of x , y .
Tools for the proofs: the structure of the pro-2-group G
Let F be the Frattini subgroup of G . Then
G/F = Gal(Q(µ8)/Q) = (Z/8Z)∗
We write G i for the preimage of i = 1, 3, 5, 7 in G/F . We haveFrob p ∈ Gp(mod 8) and t(G i ) ∈ Ai .Let c ∈ G be a complex conjugation. Hence c ∈ G 7.
Lemma (Serre)
Any element of order 2 of G is conjugate to c.
LemmaLet u be any element in G 3 or G 5. Then G has the presentation〈u, c |c2 = 1〉 in the category of pro-2-groups.
The proof uses a result of Shafarevich.Let us choose u and c as in the lemma, with u ∈ G 3 to fix ideas.Let us write t(u) = X ∈ A, t(cu) = Y ∈ A. Then X ,Y areanalogs of x , y .
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Tools for the proofs: the Fricke polynomialsLet Γ be the discrete subgroup of G generated by u and c .
Proposition
For γ ∈ Γ, t(γ) ∈ F2[X ,Y ].
Classical Fricke polynomials are polynomials in Z[X ,Y ,Z ] arisingin representation theory and quantum physics. The t(γ) are thereduction mod (Z , 2) of those polynomials. Example: t(un) is theChebychev polynomial of degree n in X modulo 2.
Corollary
One has A = F2[[X ,Y ]]. One has (x) = (X ), (y) = (Y ),P0 = (X + Y ).
Proof: t(G ) ⊂ F2[[X ,Y ]] and t(G ) generates A as an algebra byuniversality.
Proposition
The closed subspace of A generated by t(G ) is the maximal idealmA.
Proof: computations.
Proof of the theorem of positive density
TheoremIf f ∈ M, f 6= 0,∆, δ(f ) > 0
Proof: Define
H = Hf = {T ∈ A, a1(Tf ) = 0}.
Then H is an open hyperplan of A, and H 6= mA because f 6= ∆.Since ap = a1(Tpf ) is 1 if and only if Tp ∈ H, and Tp = t(Frob p),one has by Chebotarev
δ(f ) = µG (t−1(H)),
where µG is the Haar probablity measure on G .But t−1(H) is openand non-empty since t(G ) is dense in mA. QED.
Proof of the theorem of positive density
TheoremIf f ∈ M, f 6= 0,∆, δ(f ) > 0
Proof: Define
H = Hf = {T ∈ A, a1(Tf ) = 0}.
Then H is an open hyperplan of A, and H 6= mA because f 6= ∆.
Since ap = a1(Tpf ) is 1 if and only if Tp ∈ H, and Tp = t(Frob p),one has by Chebotarev
δ(f ) = µG (t−1(H)),
where µG is the Haar probablity measure on G .But t−1(H) is openand non-empty since t(G ) is dense in mA. QED.
Proof of the theorem of positive density
TheoremIf f ∈ M, f 6= 0,∆, δ(f ) > 0
Proof: Define
H = Hf = {T ∈ A, a1(Tf ) = 0}.
Then H is an open hyperplan of A, and H 6= mA because f 6= ∆.Since ap = a1(Tpf ) is 1 if and only if Tp ∈ H, and Tp = t(Frob p),one has by Chebotarev
δ(f ) = µG (t−1(H)),
where µG is the Haar probablity measure on G .
But t−1(H) is openand non-empty since t(G ) is dense in mA. QED.
Proof of the theorem of positive density
TheoremIf f ∈ M, f 6= 0,∆, δ(f ) > 0
Proof: Define
H = Hf = {T ∈ A, a1(Tf ) = 0}.
Then H is an open hyperplan of A, and H 6= mA because f 6= ∆.Since ap = a1(Tpf ) is 1 if and only if Tp ∈ H, and Tp = t(Frob p),one has by Chebotarev
δ(f ) = µG (t−1(H)),
where µG is the Haar probablity measure on G .But t−1(H) is openand non-empty since t(G ) is dense in mA. QED.
Idea of the proof of δ(f ) = 1/8 for some f ∈ M i
Proposition
There exists unique continuous action of Out(G ) on A such that ifψ ∈ Out(G ), g ∈ G , ψ · t(g) = t(ψ(g)).
Idea of proof: A is the universal deformation of thepseudo-deformation (0,1) with the conditions d = 1, t(c) = 0.Since t(ψ(c)) = 0 for ψ ∈ Aut (G ) because of Serre’s lemma, ψdefines an automorphism on A.
LemmaFor any v ∈ G 3, there exists a unique automorphism of G suchthat ψ(u) = v, ψ(c) = c. This automorphisms stabilizes G i fori ∈ (Z/8Z)∗.
Since f ∈ M i , H i = H ∩ Ai is an hyperplan of Ai . Using thelemma, one constructs for suitable f ’s a ψ ∈ Out(G ) such thatψ · H i = Ai − H i , ψ · (Ai − H i ) = H i . Hence
µG (t−1(H)) = µG (t−1(Hi )) = 1/2µG (G i ) = 1/2 · 1/4 = 1/8.
Idea of the proof of δ(f ) = 1/8 for some f ∈ M i
Proposition
There exists unique continuous action of Out(G ) on A such that ifψ ∈ Out(G ), g ∈ G , ψ · t(g) = t(ψ(g)).
Idea of proof: A is the universal deformation of thepseudo-deformation (0,1) with the conditions d = 1, t(c) = 0.Since t(ψ(c)) = 0 for ψ ∈ Aut (G ) because of Serre’s lemma, ψdefines an automorphism on A.
LemmaFor any v ∈ G 3, there exists a unique automorphism of G suchthat ψ(u) = v, ψ(c) = c. This automorphisms stabilizes G i fori ∈ (Z/8Z)∗.
Since f ∈ M i , H i = H ∩ Ai is an hyperplan of Ai . Using thelemma, one constructs for suitable f ’s a ψ ∈ Out(G ) such thatψ · H i = Ai − H i , ψ · (Ai − H i ) = H i . Hence
µG (t−1(H)) = µG (t−1(Hi )) = 1/2µG (G i ) = 1/2 · 1/4 = 1/8.
Idea of the proof of δ(f ) = 1/8 for some f ∈ M i
Proposition
There exists unique continuous action of Out(G ) on A such that ifψ ∈ Out(G ), g ∈ G , ψ · t(g) = t(ψ(g)).
Idea of proof: A is the universal deformation of thepseudo-deformation (0,1) with the conditions d = 1, t(c) = 0.Since t(ψ(c)) = 0 for ψ ∈ Aut (G ) because of Serre’s lemma, ψdefines an automorphism on A.
LemmaFor any v ∈ G 3, there exists a unique automorphism of G suchthat ψ(u) = v, ψ(c) = c. This automorphisms stabilizes G i fori ∈ (Z/8Z)∗.
Since f ∈ M i , H i = H ∩ Ai is an hyperplan of Ai . Using thelemma, one constructs for suitable f ’s a ψ ∈ Out(G ) such thatψ · H i = Ai − H i , ψ · (Ai − H i ) = H i .
Hence
µG (t−1(H)) = µG (t−1(Hi )) = 1/2µG (G i ) = 1/2 · 1/4 = 1/8.
Idea of the proof of δ(f ) = 1/8 for some f ∈ M i
Proposition
There exists unique continuous action of Out(G ) on A such that ifψ ∈ Out(G ), g ∈ G , ψ · t(g) = t(ψ(g)).
Idea of proof: A is the universal deformation of thepseudo-deformation (0,1) with the conditions d = 1, t(c) = 0.Since t(ψ(c)) = 0 for ψ ∈ Aut (G ) because of Serre’s lemma, ψdefines an automorphism on A.
LemmaFor any v ∈ G 3, there exists a unique automorphism of G suchthat ψ(u) = v, ψ(c) = c. This automorphisms stabilizes G i fori ∈ (Z/8Z)∗.
Since f ∈ M i , H i = H ∩ Ai is an hyperplan of Ai . Using thelemma, one constructs for suitable f ’s a ψ ∈ Out(G ) such thatψ · H i = Ai − H i , ψ · (Ai − H i ) = H i . Hence
µG (t−1(H)) = µG (t−1(Hi )) = 1/2µG (G i ) = 1/2 · 1/4 = 1/8.
Personal reminder
observeremainingtime T (inminutes)
is T > 5?is
T < −5?
apologizego to
restaurant
takequestions
has part Vbeen
covered?
coverpart V yes
no
yes
no
no yes
Part V.
Speculation on the partition function modulo 2
Partitions and ∆−1/3
Let
P(q) =∞∑n=0
p(n)qn =∏n≥1
(1− qn)−1
Thus,qP(q)−24 = ∆(q)
Orq−1/3P(q)8 = ∆(q)−1/3 ∈ Z((q1/3)).
Reducing mod 2,
q−1/3P(q8) = ∆(q)−1/3.
Hence to understand P, one needs to understand ∆−1/3
Analyzing ∆−1/3
In some sense,∆−1/3 = lim
r→∞∆(2r−1)/3
For odd r , ∆(2r−1)/3 is in M, non-special according to our firstconjecture, hence of density 1/8 according to our secondconjecture.
Hence, ∆−1/3 should also have density 1/8, and this would implythat for j = 0, 1
Aj(x) := #{n ≤ x , p(n) ≡ 1 (mod 2)} > 1
2(x/ log(x))+o(x/ log(x))
much closer to the conjectures Aj(x) ∼ 12x that anything known
today (best lower bounds are in x1/2 log(x).)To make this argument work, one needs, besides our twoconjectures, a strong form of effective Chebotarev, that one canperhaps obtain by Large Sieve methods. At least, the method issound numerically.
Analyzing ∆−1/3
In some sense,∆−1/3 = lim
r→∞∆(2r−1)/3
For odd r , ∆(2r−1)/3 is in M, non-special according to our firstconjecture, hence of density 1/8 according to our secondconjecture.Hence, ∆−1/3 should also have density 1/8, and this would implythat for j = 0, 1
Aj(x) := #{n ≤ x , p(n) ≡ 1 (mod 2)} > 1
2(x/ log(x))+o(x/ log(x))
much closer to the conjectures Aj(x) ∼ 12x that anything known
today (best lower bounds are in x1/2 log(x).)
To make this argument work, one needs, besides our twoconjectures, a strong form of effective Chebotarev, that one canperhaps obtain by Large Sieve methods. At least, the method issound numerically.
Analyzing ∆−1/3
In some sense,∆−1/3 = lim
r→∞∆(2r−1)/3
For odd r , ∆(2r−1)/3 is in M, non-special according to our firstconjecture, hence of density 1/8 according to our secondconjecture.Hence, ∆−1/3 should also have density 1/8, and this would implythat for j = 0, 1
Aj(x) := #{n ≤ x , p(n) ≡ 1 (mod 2)} > 1
2(x/ log(x))+o(x/ log(x))
much closer to the conjectures Aj(x) ∼ 12x that anything known
today (best lower bounds are in x1/2 log(x).)To make this argument work, one needs, besides our twoconjectures, a strong form of effective Chebotarev, that one canperhaps obtain by Large Sieve methods.
At least, the method issound numerically.
Analyzing ∆−1/3
In some sense,∆−1/3 = lim
r→∞∆(2r−1)/3
For odd r , ∆(2r−1)/3 is in M, non-special according to our firstconjecture, hence of density 1/8 according to our secondconjecture.Hence, ∆−1/3 should also have density 1/8, and this would implythat for j = 0, 1
Aj(x) := #{n ≤ x , p(n) ≡ 1 (mod 2)} > 1
2(x/ log(x))+o(x/ log(x))
much closer to the conjectures Aj(x) ∼ 12x that anything known
today (best lower bounds are in x1/2 log(x).)To make this argument work, one needs, besides our twoconjectures, a strong form of effective Chebotarev, that one canperhaps obtain by Large Sieve methods. At least, the method issound numerically.