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Zagiers Class Number Definition of Binary Quadratic Forms
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If I understood the definition of class numbers of discriminant $D$ for binary quadratic forms (b.q.f.) correctly, the class number $h(D)$ for a given discriminant $D$ is the number of equivalence classes $(*)$ of b.q.f. with discriminant $D$.
$(*)$ Two binary quadratic forms $f_1, f_2$ are equivalent if there exists $M in textSL(2 times 2, mathbbZ)$ such that $f_1(x, y) = f_2(Mcdot(x, y)^t)$.
I'm currently reading Don Zagiers "Zetafunktionen and quadratische Körper". He writes that there are two more very imported invariants of binary quadratic forms and is going to use them to further refine the classification. Namely:
- the gcd of the coefficients of the b.q.F.
- the sign of the first coefficient
In the case $D<0$ the first coefficients of two equivalent b.q.f. have the same sign. He writes: In the case $D < 0 $ we therefore only need to look at b.q.f with positive first coefficient (positive definite). He also writes, that we only need to consider b.q.f. where the gcd of the coefficients equals $1$ (a primitive b.q.f), since a b.q.f. of Discriminant $D$, where the gcd equals $r$ is $r$ times a primitive b.q.f of Discriminant $fracDr$. He than defines the class number for discriminant $D$ as:
$$h(D) :=
begincases
textnumber of equivalence classes of primitive b.q.f. \ textwith discriminant $D$ if $D>0$ \[10pt]
textnumber of equivalence classes of primitive, positive definite b.q.f. \ textwith discriminant $D$ if $D < 0$
endcases$$
Somehow this isn't clear to me. Is this definition equivalent to the original definition I gave in the first paragraph. Is the statement here that we can find a representative of a specific form? Or do we truly refine the number of equivalence classes?
number-theory algebraic-number-theory quadratic-forms greatest-common-divisor discriminant
asked Jul 16 at 12:09
Quadrat
485
add a comment |Â
up vote
3
down vote
favorite
If I understood the definition of class numbers of discriminant $D$ for binary quadratic forms (b.q.f.) correctly, the class number $h(D)$ for a given discriminant $D$ is the number of equivalence classes $(*)$ of b.q.f. with discriminant $D$.
$(*)$ Two binary quadratic forms $f_1, f_2$ are equivalent if there exists $M in textSL(2 times 2, mathbbZ)$ such that $f_1(x, y) = f_2(Mcdot(x, y)^t)$.
I'm currently reading Don Zagiers "Zetafunktionen and quadratische Körper". He writes that there are two more very imported invariants of binary quadratic forms and is going to use them to further refine the classification. Namely:
- the gcd of the coefficients of the b.q.F.
- the sign of the first coefficient
In the case $D<0$ the first coefficients of two equivalent b.q.f. have the same sign. He writes: In the case $D < 0 $ we therefore only need to look at b.q.f with positive first coefficient (positive definite). He also writes, that we only need to consider b.q.f. where the gcd of the coefficients equals $1$ (a primitive b.q.f), since a b.q.f. of Discriminant $D$, where the gcd equals $r$ is $r$ times a primitive b.q.f of Discriminant $fracDr$. He than defines the class number for discriminant $D$ as:
$$h(D) :=
begincases
textnumber of equivalence classes of primitive b.q.f. \ textwith discriminant $D$ if $D>0$ \[10pt]
textnumber of equivalence classes of primitive, positive definite b.q.f. \ textwith discriminant $D$ if $D < 0$
endcases$$
Somehow this isn't clear to me. Is this definition equivalent to the original definition I gave in the first paragraph. Is the statement here that we can find a representative of a specific form? Or do we truly refine the number of equivalence classes?
number-theory algebraic-number-theory quadratic-forms greatest-common-divisor discriminant
asked Jul 16 at 12:09
Quadrat
485
1
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
If I understood the definition of class numbers of discriminant $D$ for binary quadratic forms (b.q.f.) correctly, the class number $h(D)$ for a given discriminant $D$ is the number of equivalence classes $(*)$ of b.q.f. with discriminant $D$.
$(*)$ Two binary quadratic forms $f_1, f_2$ are equivalent if there exists $M in textSL(2 times 2, mathbbZ)$ such that $f_1(x, y) = f_2(Mcdot(x, y)^t)$.
I'm currently reading Don Zagiers "Zetafunktionen and quadratische Körper". He writes that there are two more very imported invariants of binary quadratic forms and is going to use them to further refine the classification. Namely:
- the gcd of the coefficients of the b.q.F.
- the sign of the first coefficient
In the case $D<0$ the first coefficients of two equivalent b.q.f. have the same sign. He writes: In the case $D < 0 $ we therefore only need to look at b.q.f with positive first coefficient (positive definite). He also writes, that we only need to consider b.q.f. where the gcd of the coefficients equals $1$ (a primitive b.q.f), since a b.q.f. of Discriminant $D$, where the gcd equals $r$ is $r$ times a primitive b.q.f of Discriminant $fracDr$. He than defines the class number for discriminant $D$ as:
$$h(D) :=
begincases
textnumber of equivalence classes of primitive b.q.f. \ textwith discriminant $D$ if $D>0$ \[10pt]
textnumber of equivalence classes of primitive, positive definite b.q.f. \ textwith discriminant $D$ if $D < 0$
endcases$$
Somehow this isn't clear to me. Is this definition equivalent to the original definition I gave in the first paragraph. Is the statement here that we can find a representative of a specific form? Or do we truly refine the number of equivalence classes?
number-theory algebraic-number-theory quadratic-forms greatest-common-divisor discriminant
asked Jul 16 at 12:09
Quadrat
485
If I understood the definition of class numbers of discriminant $D$ for binary quadratic forms (b.q.f.) correctly, the class number $h(D)$ for a given discriminant $D$ is the number of equivalence classes $(*)$ of b.q.f. with discriminant $D$.
$(*)$ Two binary quadratic forms $f_1, f_2$ are equivalent if there exists $M in textSL(2 times 2, mathbbZ)$ such that $f_1(x, y) = f_2(Mcdot(x, y)^t)$.
I'm currently reading Don Zagiers "Zetafunktionen and quadratische Körper". He writes that there are two more very imported invariants of binary quadratic forms and is going to use them to further refine the classification. Namely:
- the gcd of the coefficients of the b.q.F.
- the sign of the first coefficient
In the case $D<0$ the first coefficients of two equivalent b.q.f. have the same sign. He writes: In the case $D < 0 $ we therefore only need to look at b.q.f with positive first coefficient (positive definite). He also writes, that we only need to consider b.q.f. where the gcd of the coefficients equals $1$ (a primitive b.q.f), since a b.q.f. of Discriminant $D$, where the gcd equals $r$ is $r$ times a primitive b.q.f of Discriminant $fracDr$. He than defines the class number for discriminant $D$ as:
$$h(D) :=
begincases
textnumber of equivalence classes of primitive b.q.f. \ textwith discriminant $D$ if $D>0$ \[10pt]
textnumber of equivalence classes of primitive, positive definite b.q.f. \ textwith discriminant $D$ if $D < 0$
endcases$$
Somehow this isn't clear to me. Is this definition equivalent to the original definition I gave in the first paragraph. Is the statement here that we can find a representative of a specific form? Or do we truly refine the number of equivalence classes?
number-theory algebraic-number-theory quadratic-forms greatest-common-divisor discriminant
asked Jul 16 at 12:09
Quadrat
485
edited Jul 16 at 12:15
asked Jul 16 at 12:09
Quadrat
485
asked Jul 16 at 12:09
Quadrat
485
asked Jul 16 at 12:09
Quadrat
485
485
1
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12
add a comment |Â
1
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12
1
1
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12
add a comment |Â
1 Answer
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votes
up vote
1
down vote
the class number is always for primitive forms. Among other things, this allows Gauss composition to make the set of classes into a group.
For definite forms, the negative definite forms are just a copy of the positive definite forms, each negated. Furthermore, the composition of two positive forms is positive. So, we always take the positive definite forms.
He does automorphism group page 63. Finally he does reduction page 120, for indefinite forms 122 (6) (this is his version, different from Gauss/Lagrange).
I suggest you fill in with other sources for the quadratic form parts. From about 1929, Introduction to the Theory of Numbers by Leonard Eugene Dickson. For composition, genera, Primes of the Form $x^2 + n y^2$ by David A. Cox. This has a very clean presentation of Dirichlet's method of computing composition, which has the advantage of actually resembling a mulitplication
answered Jul 16 at 17:40
Will Jagy
97.2k594196
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
the class number is always for primitive forms. Among other things, this allows Gauss composition to make the set of classes into a group.
For definite forms, the negative definite forms are just a copy of the positive definite forms, each negated. Furthermore, the composition of two positive forms is positive. So, we always take the positive definite forms.
He does automorphism group page 63. Finally he does reduction page 120, for indefinite forms 122 (6) (this is his version, different from Gauss/Lagrange).
I suggest you fill in with other sources for the quadratic form parts. From about 1929, Introduction to the Theory of Numbers by Leonard Eugene Dickson. For composition, genera, Primes of the Form $x^2 + n y^2$ by David A. Cox. This has a very clean presentation of Dirichlet's method of computing composition, which has the advantage of actually resembling a mulitplication
answered Jul 16 at 17:40
Will Jagy
97.2k594196
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
add a comment |Â
up vote
1
down vote
the class number is always for primitive forms. Among other things, this allows Gauss composition to make the set of classes into a group.
For definite forms, the negative definite forms are just a copy of the positive definite forms, each negated. Furthermore, the composition of two positive forms is positive. So, we always take the positive definite forms.
He does automorphism group page 63. Finally he does reduction page 120, for indefinite forms 122 (6) (this is his version, different from Gauss/Lagrange).
I suggest you fill in with other sources for the quadratic form parts. From about 1929, Introduction to the Theory of Numbers by Leonard Eugene Dickson. For composition, genera, Primes of the Form $x^2 + n y^2$ by David A. Cox. This has a very clean presentation of Dirichlet's method of computing composition, which has the advantage of actually resembling a mulitplication
answered Jul 16 at 17:40
Will Jagy
97.2k594196
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
add a comment |Â
up vote
1
down vote
up vote
1
down vote
the class number is always for primitive forms. Among other things, this allows Gauss composition to make the set of classes into a group.
For definite forms, the negative definite forms are just a copy of the positive definite forms, each negated. Furthermore, the composition of two positive forms is positive. So, we always take the positive definite forms.
He does automorphism group page 63. Finally he does reduction page 120, for indefinite forms 122 (6) (this is his version, different from Gauss/Lagrange).
I suggest you fill in with other sources for the quadratic form parts. From about 1929, Introduction to the Theory of Numbers by Leonard Eugene Dickson. For composition, genera, Primes of the Form $x^2 + n y^2$ by David A. Cox. This has a very clean presentation of Dirichlet's method of computing composition, which has the advantage of actually resembling a mulitplication
answered Jul 16 at 17:40
Will Jagy
97.2k594196
the class number is always for primitive forms. Among other things, this allows Gauss composition to make the set of classes into a group.
For definite forms, the negative definite forms are just a copy of the positive definite forms, each negated. Furthermore, the composition of two positive forms is positive. So, we always take the positive definite forms.
He does automorphism group page 63. Finally he does reduction page 120, for indefinite forms 122 (6) (this is his version, different from Gauss/Lagrange).
I suggest you fill in with other sources for the quadratic form parts. From about 1929, Introduction to the Theory of Numbers by Leonard Eugene Dickson. For composition, genera, Primes of the Form $x^2 + n y^2$ by David A. Cox. This has a very clean presentation of Dirichlet's method of computing composition, which has the advantage of actually resembling a mulitplication
answered Jul 16 at 17:40
Will Jagy
97.2k594196
answered Jul 16 at 17:40
Will Jagy
97.2k594196
answered Jul 16 at 17:40
Will Jagy
97.2k594196
answered Jul 16 at 17:40
Will Jagy
97.2k594196
97.2k594196
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
add a comment |Â
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
What do you mean is always for primitive forms? Is there always a primitive representative?
– Quadrat
Jul 16 at 19:31
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
@Quadrat the (form) class group is primitive forms only, up to equivalence over $SL_2 mathbb Z.$ The class number is the count of those classes.
– Will Jagy
Jul 16 at 19:33
add a comment |Â
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1
Why do some people feel the need to downvote without giving a reason? I'll happily edit my question to include further information if this is the problem.
– Quadrat
Jul 16 at 12:12