The Hunt for Totally Real Number
Fields
Vandita Patel
University of Warwick
YRM 2015
August 17, 2015
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

The Problem ...
Question
Given integers n and ∆ , can we find all number fields of degree n and discriminant ∆ ?
Restrictions:
1
2
Primitive number fields ( K primitive if there does not exist
L such that
Q (
L
(
K ).
Totally Real Number Fields.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

The Problem ...
Question
Given integers n and ∆ , can we find all number fields of degree n and discriminant ∆ ?
Restrictions:
1
2
Primitive number fields ( K primitive if there does not exist
L such that
Q (
L
(
K ).
Totally Real Number Fields.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

The Problem ...
Question
Given integers n and ∆ , can we find all number fields of degree n and discriminant ∆ ?
Restrictions:
1
2
Primitive number fields ( K primitive if there does not exist
L such that
Q (
L
(
K ).
Totally Real Number Fields.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Current Method ...
Theorem (Hunter’s Theorem)
Let K be a primitive totally real number field of degree n and discriminant ∆ . Then there exists α ∈ O
K
\
Z such that:
1
0 ≤ T r ( α ) ≤ n
,
2
2
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Current Method ...
Theorem (Hunter’s Theorem)
Let K be a primitive totally real number field of degree n and discriminant ∆ . Then there exists α ∈ O
K
\
Z such that:
1
0 ≤ T r ( α ) ≤ n
,
2
2
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Current Method ...
Theorem (Hunter’s Theorem)
Let K be a primitive totally real number field of degree n and discriminant ∆ . Then there exists α ∈ O
K
\
Z such that:
1
0 ≤ T r ( α ) ≤ n
,
2
2
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Let n = 2 and ∆ ≤ 5. Find all totally real number fields of degree 2 and discriminant less than or equal to 5.
Hunter’s Theorem gives us some bounds.
0 ≤ T r ( α ) ≤ n
2
= 1 ,
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
≤
( T r ( α ))
2
2
+ γ n − 1
+ γ
1
5
2
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Let n = 2 and ∆ ≤ 5. Find all totally real number fields of degree 2 and discriminant less than or equal to 5.
Hunter’s Theorem gives us some bounds.
0 ≤ T r ( α ) ≤ n
2
= 1 ,
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
≤
( T r ( α ))
2
2
+ γ n − 1
+ γ
1
5
2
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Let n = 2 and ∆ ≤ 5. Find all totally real number fields of degree 2 and discriminant less than or equal to 5.
Hunter’s Theorem gives us some bounds.
0 ≤ T r ( α ) ≤ n
2
= 1 ,
0 ≤ n
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
≤
( T r ( α ))
2
2
+ γ n − 1
+ γ
1
5
2
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example if T r ( α ) = 0 −→ T r ( α
2
) ≤
5 if T r ( α ) = 1 −→ T r ( α
2
) ≤
2
1
2
+
5
2
= 3 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
We are looking for a polynomial f ( x ) = x
2
+ ax + b . We let α and β be the roots of f . Then f ( x ) = ( x − α )( x − β ) = x
2 − ( α + β ) x + αβ .
1
2
3
4
T r ( α ) = α + β .
αβ =
( α + β )
2 − α
2 − β
2
2
− 1 ≤ αβ ≤ 0.
=
1
2
T r ( α )
2 − T r ( α
2
If T r ( α ) = 0, then T r ( α
2
) ≤ 2 and αβ ≥
) .
0 − 2
2
If T r ( α ) = 1, then
− 1 ≤ αβ ≤ 0.
T r ( α
2
) ≤ 3 and αβ ≥
1 − 3
2 and so and so
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
We are looking for a polynomial f ( x ) = x
2
+ ax + b . We let α and β be the roots of f . Then f ( x ) = ( x − α )( x − β ) = x
2 − ( α + β ) x + αβ .
1
2
3
4
T r ( α ) = α + β .
αβ =
( α + β )
2 − α
2 − β
2
2
− 1 ≤ αβ ≤ 0.
=
1
2
T r ( α )
2 − T r ( α
2
If T r ( α ) = 0, then T r ( α
2
) ≤ 2 and αβ ≥
) .
0 − 2
2
If T r ( α ) = 1, then
− 1 ≤ αβ ≤ 0.
T r ( α
2
) ≤ 3 and αβ ≥
1 − 3
2 and so and so
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
We are looking for a polynomial f ( x ) = x
2
+ ax + b . We let α and β be the roots of f . Then f ( x ) = ( x − α )( x − β ) = x
2 − ( α + β ) x + αβ .
1
2
3
4
T r ( α ) = α + β .
αβ =
( α + β )
2 − α
2 − β
2
2
− 1 ≤ αβ ≤ 0.
=
1
2
T r ( α )
2 − T r ( α
2
If T r ( α ) = 0, then T r ( α
2
) ≤ 2 and αβ ≥
) .
0 − 2
2
If T r ( α ) = 1, then
− 1 ≤ αβ ≤ 0.
T r ( α
2
) ≤ 3 and αβ ≥
1 − 3
2 and so and so
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
We are looking for a polynomial f ( x ) = x
2
+ ax + b . We let α and β be the roots of f . Then f ( x ) = ( x − α )( x − β ) = x
2 − ( α + β ) x + αβ .
1
2
3
4
T r ( α ) = α + β .
αβ =
( α + β )
2 − α
2 − β
2
2
− 1 ≤ αβ ≤ 0.
=
1
2
T r ( α )
2 − T r ( α
2
If T r ( α ) = 0, then T r ( α
2
) ≤ 2 and αβ ≥
) .
0 − 2
2
If T r ( α ) = 1, then
− 1 ≤ αβ ≤ 0.
T r ( α
2
) ≤ 3 and αβ ≥
1 − 3
2 and so and so
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
We are looking for a polynomial f ( x ) = x
2
+ ax + b . We let α and β be the roots of f . Then f ( x ) = ( x − α )( x − β ) = x
2 − ( α + β ) x + αβ .
1
2
3
4
T r ( α ) = α + β .
αβ =
( α + β )
2 − α
2 − β
2
2
− 1 ≤ αβ ≤ 0.
=
1
2
T r ( α )
2 − T r ( α
2
If T r ( α ) = 0, then T r ( α
2
) ≤ 2 and αβ ≥
) .
0 − 2
2
If T r ( α ) = 1, then
− 1 ≤ αβ ≤ 0.
T r ( α
2
) ≤ 3 and αβ ≥
1 − 3
2 and so and so
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Possible polynomials:
1
2
3
4 x
2 − 0 x + 0 −→ reducible x
2 − 0 x − 1 −→ reducible x
2 − x −→ reducible x
2 − x − 1 X
T r ( α
2
) = d
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Possible polynomials:
1
2
3
4 x
2 − 0 x + 0 −→ reducible x
2 − 0 x − 1 −→ reducible x
2 − x −→ reducible x
2 − x − 1 X
T r ( α
2
) = d
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Possible polynomials:
1
2
3
4 x
2 − 0 x + 0 −→ reducible x
2 − 0 x − 1 −→ reducible x
2 − x −→ reducible x
2 − x − 1 X
T r ( α
2
) = d
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Possible polynomials:
1
2
3
4 x
2 − 0 x + 0 −→ reducible x
2 − 0 x − 1 −→ reducible x
2 − x −→ reducible x
2 − x − 1 X
T r ( α
2
) = d
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Example
Possible polynomials:
1
2
3
4 x
2 − 0 x + 0 −→ reducible x
2 − 0 x − 1 −→ reducible x
2 − x −→ reducible x
2 − x − 1 X
T r ( α
2
) = d
X
σ i
( α
2
) ≤ i =1
( T r ( α ))
2 n
+ γ n − 1
| ∆ | n
1 n − 1
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Motivation ...
Primitive degree 8 Number Fields.
1 x
8 − 28 x
6 − 28 x
5
+ 196 x
4
+ 308 x
3 − 224 x
2 − 528 x − 182
2
3
4 x
8 − 42 x
6 − 56 x
5
+ 420 x
4
+ 1092 x
3
+ 406 x
2 − 852 x − 609 x
8 − 4 x
7 − 28 x
6
+ 60 x
5
+ 102 x
4 − 20 x
3 − 36 x
2 − 4 x + 1 x
8 − 2 x
7 − 34 x
6
+ 28 x
5
+ 280 x
4 − 210 x
3 − 686 x
2
+ 588 x + 49
∆
1.
2 18 7 10
2.
2 8 3 10 7 8
3.
2 16 7 4 11 6
4.
2 12 7 6 11 6
HB min(Tr( α
2
)) Dim( M
4
( N )) Dim( M
7 / 2
( N ))
130
133
157
184
56
84
72
72
184
536
150
150
152
864
124
124
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Approach via Modular Forms ...
Theorem (Hecke,Schoenberg)
1
2
3
4
Let K be a primitive, totally real number field with degree n and discriminant ∆ . Let O
K basis { b
1
, . . . , b n
} .
be its ring of integers, with
Pick any α ∈ O
K i.e.
α = x
1 b
1
Let
R
Q
(
Q m
( x ) =
) = #
P n i =1
{ x ∈
σ i
( α
2
) .
Z n | Q ( x ) = m } .
+ · · · + x n b n
.
Then f =
∞
X
R
Q
( m ) q m m =0
∈ M n/ 2
( N, χ ) .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Approach via Modular Forms ...
Theorem (Hecke,Schoenberg)
1
2
3
4
Let K be a primitive, totally real number field with degree n and discriminant ∆ . Let O
K basis { b
1
, . . . , b n
} .
be its ring of integers, with
Pick any α ∈ O
K i.e.
α = x
1 b
1
Let
R
Q
(
Q m
( x ) =
) = #
P n i =1
{ x ∈
σ i
( α
2
) .
Z n | Q ( x ) = m } .
+ · · · + x n b n
.
Then f =
∞
X
R
Q
( m ) q m m =0
∈ M n/ 2
( N, χ ) .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Approach via Modular Forms ...
Theorem (Hecke,Schoenberg)
1
2
3
4
Let K be a primitive, totally real number field with degree n and discriminant ∆ . Let O
K basis { b
1
, . . . , b n
} .
be its ring of integers, with
Pick any α ∈ O
K i.e.
α = x
1 b
1
Let
R
Q
(
Q m
( x ) =
) = #
P n i =1
{ x ∈
σ i
( α
2
) .
Z n | Q ( x ) = m } .
+ · · · + x n b n
.
Then f =
∞
X
R
Q
( m ) q m m =0
∈ M n/ 2
( N, χ ) .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Approach via Modular Forms ...
Theorem (Hecke,Schoenberg)
1
2
3
4
Let K be a primitive, totally real number field with degree n and discriminant ∆ . Let O
K basis { b
1
, . . . , b n
} .
be its ring of integers, with
Pick any α ∈ O
K i.e.
α = x
1 b
1
Let
R
Q
(
Q m
( x ) =
) = #
P n i =1
{ x ∈
σ i
( α
2
) .
Z n | Q ( x ) = m } .
+ · · · + x n b n
.
Then f =
∞
X
R
Q
( m ) q m m =0
∈ M n/ 2
( N, χ ) .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Approach via Modular Forms ...
Theorem (Hecke,Schoenberg)
1
2
3
4
Let K be a primitive, totally real number field with degree n and discriminant ∆ . Let O
K basis { b
1
, . . . , b n
} .
be its ring of integers, with
Pick any α ∈ O
K i.e.
α = x
1 b
1
Let
R
Q
(
Q m
( x ) =
) = #
P n i =1
{ x ∈
σ i
( α
2
) .
Z n | Q ( x ) = m } .
+ · · · + x n b n
.
Then f =
∞
X
R
Q
( m ) q m m =0
∈ M n/ 2
( N, χ ) .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Let
K =
Q
(
√
5) .
α = a + b
1 +
2
√
5
!
∈ O
K
, a, b, ∈
Z
.
Q = σ
1
( α
2
) + σ
2
( α
2
) = 2 a
2
+ 2 ab + 3 b
2
.
Then f = 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · ∈ M
1
(Γ
1
(20) , χ ) where
χ (11) = χ (17) = − 1 , character modulo 20, conductor 20 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Let
K =
Q
(
√
5) .
α = a + b
1 +
2
√
5
!
∈ O
K
, a, b, ∈
Z
.
Q = σ
1
( α
2
) + σ
2
( α
2
) = 2 a
2
+ 2 ab + 3 b
2
.
Then f = 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · ∈ M
1
(Γ
1
(20) , χ ) where
χ (11) = χ (17) = − 1 , character modulo 20, conductor 20 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Let
K =
Q
(
√
5) .
α = a + b
1 +
2
√
5
!
∈ O
K
, a, b, ∈
Z
.
Q = σ
1
( α
2
) + σ
2
( α
2
) = 2 a
2
+ 2 ab + 3 b
2
.
Then f = 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · ∈ M
1
(Γ
1
(20) , χ ) where
χ (11) = χ (17) = − 1 , character modulo 20, conductor 20 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Let
K =
Q
(
√
5) .
α = a + b
1 +
2
√
5
!
∈ O
K
, a, b, ∈
Z
.
Q = σ
1
( α
2
) + σ
2
( α
2
) = 2 a
2
+ 2 ab + 3 b
2
.
Then f = 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · ∈ M
1
(Γ
1
(20) , χ ) where
χ (11) = χ (17) = − 1 , character modulo 20, conductor 20 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
Let
K =
Q
(
√
5) .
α = a + b
1 +
2
√
5
!
∈ O
K
, a, b, ∈
Z
.
Q = σ
1
( α
2
) + σ
2
( α
2
) = 2 a
2
+ 2 ab + 3 b
2
.
Then f = 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · ∈ M
1
(Γ
1
(20) , χ ) where
χ (11) = χ (17) = − 1 , character modulo 20, conductor 20 .
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Backwards Compatibility ...
Question
Given M k
( N, χ ) , can we find all number fields of degree 2 k and discriminant ∆ where N | ∆ | N
2 k
?
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2
− 2 q
3
+ q
4
+ q
5
+ 2 q
6
− 2 q
7
− q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

An Example ...
M
1
(20 , χ ) has dimension 2, where χ (11) = χ (17) = − 1, character modulo 20, conductor 20.
Question
Is f := x
1 f
1
+ x
2 f
2 a modular form coming from a number field?
f
1
:= 1 + 2 q
2
+ 4 q
3
+ 4 q
7
+ 2 q
8
+ · · · f
2
:= q − q
2 − 2 q
3
+ q
4
+ q
5
+ 2 q
6 − 2 q
7 − q
8
+ · · · x
1
= 1.
Equate coefficients: q : x
2
≥ 0 q
2
: 2 − x
2 and 2 − x
2
≥ 0 −→ x
2
≥ 2 −→ x
2
≤ 2
≤ 0.
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Further Work ...
1
2
3
This method gives us a very high dimensional linear programming problem.
As the dimension increases, our solution space looks like an unbounded cone.
A bound from above!
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Further Work ...
1
2
3
This method gives us a very high dimensional linear programming problem.
As the dimension increases, our solution space looks like an unbounded cone.
A bound from above!
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Further Work ...
1
2
3
This method gives us a very high dimensional linear programming problem.
As the dimension increases, our solution space looks like an unbounded cone.
A bound from above!
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Bound From Above ...
1
2
3
4
5
Let n ≥ 2 be a natural number.
Let Q ( x ) = P n i,j =1 a ij x i x j be a positive definite quadratic form with symmetric matrix.
Let A ( X ) be the number of lattice points in the region
Q ( x ) ≤ X
Then, A ( X ) = V ( X ) + P ( X ).
Application: The sum of the coefficients of the modular form must be less than or equal to A ( X ).
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Bound From Above ...
1
2
3
4
5
Let n ≥ 2 be a natural number.
Let Q ( x ) = P n i,j =1 a ij x i x j be a positive definite quadratic form with symmetric matrix.
Let A ( X ) be the number of lattice points in the region
Q ( x ) ≤ X
Then, A ( X ) = V ( X ) + P ( X ).
Application: The sum of the coefficients of the modular form must be less than or equal to A ( X ).
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Bound From Above ...
1
2
3
4
5
Let n ≥ 2 be a natural number.
Let Q ( x ) = P n i,j =1 a ij x i x j be a positive definite quadratic form with symmetric matrix.
Let A ( X ) be the number of lattice points in the region
Q ( x ) ≤ X
Then, A ( X ) = V ( X ) + P ( X ).
Application: The sum of the coefficients of the modular form must be less than or equal to A ( X ).
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Bound From Above ...
1
2
3
4
5
Let n ≥ 2 be a natural number.
Let Q ( x ) = P n i,j =1 a ij x i x j be a positive definite quadratic form with symmetric matrix.
Let A ( X ) be the number of lattice points in the region
Q ( x ) ≤ X
Then, A ( X ) = V ( X ) + P ( X ).
Application: The sum of the coefficients of the modular form must be less than or equal to A ( X ).
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Bound From Above ...
1
2
3
4
5
Let n ≥ 2 be a natural number.
Let Q ( x ) = P n i,j =1 a ij x i x j be a positive definite quadratic form with symmetric matrix.
Let A ( X ) be the number of lattice points in the region
Q ( x ) ≤ X
Then, A ( X ) = V ( X ) + P ( X ).
Application: The sum of the coefficients of the modular form must be less than or equal to A ( X ).
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick

Thank you for Listening ...
Vandita Patel
The Hunt for Totally Real Number Fields
University of Warwick