Genus Characters, Gauss’s Duplication Theorem

Question: Which primes are represented by a given quadratic form $aX^2+bXY+cY^2$ of discriminant $D$ ?

If $p= ax_0^2+bx_0y_0+cy_0^2$ , then we can see that the form is equivalent to a form of the form $pX^2+BXY+CY^2$ . To observe this, choose a matrix $\left(\begin{array}{cc} x_0 & x_1 \\ y_0 & x_1\end{array}\right)$ of determinant $1$ and apply the coordinate transformation $(X, Y) \to (x_0X+x_1Y, y_0X+y_1Y)$ to get a form with coefficient $ax_0^2+bx_0y_0+cy_0^2 =p$ . We can always do that because $x_0, y_0$ are coprime (why?) –we just need $x_0y_1-y_0x_1=1 .$ The $X$ coefficient is always represented because $a=f(1, 0)$ , the above argument says that if we have a number $m$ that is represented primitively with $(x_0, y_0)=1$ , then we can transform the form to have to look like $mX^2+ ()XY+()Y^2$ . That is the coefficient of $X$ are precisely the numbers primitively represented by the form.

Now if $p$ is represented, the form is equivalent to $pX^2+BXY+CY^2$ and hence we have the discriminant equation $D=B^2-4pC$ .

This means that

$\displaystyle D \equiv B^2 \bmod (4p)$

On the other hand, if we have a prime satisfying this equation, we can construct $B, C$ and hence a form of discriminant $D$ that represents $p$ .

In general, any number $m$ is properly representable by “some” form of discriminant $D$ if and only if $D$ is a square modulo $4m.$ If $m$ is odd it’s enough to ask for solution to $D=b^2 \mod m$ .

But what about a given fixed form? How do we decide if a given number is represented by the form? Let’s assume we are given that $D$ is a square modulo $4m$ . (Note this already cuts down the possibilities to half). By using the solutions to $\displaystyle D \equiv B^2 \bmod (4m)$ , we can construct one form which represent $4m$ . If this form is equivalent to the given form, we are done. In general, for an odd prime $p$ we have

$(-n / p)=1$ iff $p$ is represented by one of the $h(-4 n)$ forms (up to equivalence) of discriminant $-4 n$ .

For instance, if the class group is trivial, that is if every two forms of discriminant $D$ are equivalent, numbers satisfying these congruence conditions $(m, D)=1, (D/4m)=1$ are represented by every form of discriminant $D$ . Take the example of $D=-4$ . Here the class group is trivial and any form is equivalent to $X^2+Y^2$ . If we have prime such that $-4 =b^2 \mod 4p$ , which is true for any prime $p =1\bmod 4$ , we have a form $pX^2+bXY+\frac{b^2+4}{4p}Y^2$ of discriminant $-4$ . Now because the class group is trivial, this form has to be equivalent to $X^2+Y^2$ and hence we have a solution to $X^2+y^2=p.$ Thus we have the following.

$\displaystyle p=x^{2}+y^{2} \Longleftrightarrow p=2$ or $\displaystyle p \equiv 1 \quad \bmod 4$
$\displaystyle p=x^{2}+4 y^{2} \Longleftrightarrow p \equiv 1 \quad \bmod 4$

Similarly we can get

$\displaystyle p=x^{2}+2 y^{2} \Longleftrightarrow p=2 \quad$ or $\displaystyle \quad p \equiv 1,3 \quad \bmod 8$
$\displaystyle p=x^{2}+3 y^{2} \quad \Longleftrightarrow p=3 \quad$ or $\displaystyle \quad p \equiv 1 \quad \bmod 3 .$
$\displaystyle p=x^{2}+7 y^{2} \Longleftrightarrow p \equiv 1,9,11,15,23,25 \bmod 28$

because the class groups are trivial. The residues are determined by the property $\left( \frac{-4n}{p}\right)=1$ , which define a subgroup of index 2 inside $(\mathbb Z/4n)^{*}.$

But the class numbers become large as $n$ becomes large. In fact we have

$h(-4 n)=1 \Longleftrightarrow n=1,2,3,4, 7$

For instance a non-trivial factorization $n=AC, (A, C)=1$ produces a form $AX^2+CY^2$ which is reduced if $A<C. .$ (This covers most cases $n$ except prime powers. The forms $4X^2+4XY+ (2^{r-2}+1)Y^2), n=2^r$ and $aX^2+2XY+cY^2, n+1=ac$ cover the rest of the cases. Also note that the case of odd negative discriminant is much more difficult $D=-3,-4, -7,-8,-11,-19,-43,-67,-163$ is the complete list with class number 1)

What about representation by $X^2+5Y^2$ ? Here is the class group has two reduced forms $X^{2}+5 Y^{2}, 2 X^{2}+2 XY+3 Y^{2}$ . Hence an odd prime which satisfies latex $\left(\frac{-20}{p}\right)=1$ is represented by one of these two forms and we have

$\displaystyle p=x^{2}+5 y^{2} \text { or } p =2x^2+2xy+3y^2 \Longleftrightarrow \left(\frac{-5}{p}\right)=1 \Longleftrightarrow p \equiv 1,3,7,9 \bmod 20$

The residue classes $1,3,7,9 \bmod 20$ are represented by one of the two reduced forms. But we can see that only half of these residues are represented by the principal form $X^2+5Y^2$ . And we have

$\displaystyle p=x^{2}+5 y^{2} \Longleftrightarrow p \equiv 1,9 \bmod 20$
$\displaystyle p=2 x^{2}+2 x y+3 y^{2} \Longleftrightarrow p \equiv 3,7 \bmod 20$

We said that the number coprime to $-4n$ represented by forms of discriminant $D=-4n$ have $D$ square modulo $4m$ , that is $\left(\dfrac{-4n}{4m} \right)=1$ . This forms a subgroup of index 2 inside $(\mathbb Z/4n)^{*}$ . Using the identity

$\displaystyle \left(X^{2}+n Y^{2}\right)\left(Z^{2}+n W^{2}\right)=(XZ - n Y W)^{2}+n(XW - Y Z)^{2}$

the numbers represented by the principal form $X^2+nY^2$ are a subgroup. In fact, each of the forms represent a coset of this particular subgroup. Like in the above case $\{1, 9\}$ is subgroup represented by the principal form and the coset $3\{1, 9\} \bmod 20$ is represented by the other reduced form.

Let us the take case $n=14$ . We now have 4 reduced form

$\displaystyle x^{2}+14 y^{2}, 2 x^{2}+7 y^{2}, 3 x^{2} \pm 2 x y+5 y^{2}$

So we have

$\displaystyle p=x^{2}+14 y^{2} \text{ or } 2 x^{2}+7 y^{2} \text{ or }3 x^{2} \pm 2 x y+5 y^{2} \Longleftrightarrow \left(\frac{-56}{p}\right)=1$

$\displaystyle \left(\frac{-56}{p}\right)=1 \Longleftrightarrow 1, 3, 5, 9, 13, 15, 19, 23, 25, 27, 39, 45 \mod 56$

But the classes represented by $X^2+14Y^2$ are only $1,9,15.23,25,39$ which form a subgroup of $(\mathbb{Z} / 56 \mathbb{Z})^{*}$

Let’s say we take a prime from one of these classes what can we say about the forms representing it. For $n=5$ , we had only one reduced form representing those classes, but in general the residues can be represented by the principal form and a set of other reduced forms. This collection of forms is a subgroup of the class group called the “genus”. For $n=14$ , we have

$\displaystyle p=x^{2}+14 y^{2} \text { or } 2 x^{2}+7 y^{2} \Longleftrightarrow p \equiv 1,9,15,23,25,39 \bmod 56$
$\displaystyle p=3 x^{2} \pm 2 x y+5 y^{2} \Longleftrightarrow p \equiv 3,5,13,19,27,45 \bmod 56$

So the forms $\displaystyle X^2+14Y^2, 2X^2+ 7Y^2$ form a subgroup of class group, and form a genus. And the coset of this subgroup contains $\displaystyle 3 X^{2} \pm 2 XY+5 Y^{2}$ which form another genus.

Thus in addition to the condition $\left( \frac{-4n}{p}\right)=1$ , if we know the coset of the subgroup of numbers represented by principal form to which $p$ belongs, we can deduce that $p$ is represented by one of the reduced forms of the coset of a subgroup of the class group called genus.

For instance $p$ is represented by the genus of the principal form $X^2+nY^2$ iff $p = k^2 \text{ or } k^2+n \mod 4n$ .
So using these congruence conditions, we can determine a genus of forms which could represent $p$ .

And It’s not possible to distinguish between the forms of a particular genus by congruence conditions alone! Every form in the class represents all the residue classes modulo $4n$ . In general there are no other congruence (abelian conditions) that can distinguish primes represented by forms in a genus.

Question: How many genera exist?

Because the squares are represented by principal form, all the squares inside $\left(\mathbb Z/4n\right)^{*}$ are represented. If $H$ is the subgroup represented by the principal genus, we see that every element of $\text{Ker} \chi$ , where $\chi:(\mathbb{Z} / 4n \mathbb{Z})^{*} \longrightarrow{\pm 1}$ is the character satisfying $\chi([p])=\left(\frac{-4n }{ p}\right)$ is of order 2 modulo $H$ . Thus number of genera have to be of size $2^s$ for some $s$

Theorem: If $r$ is the number of odd primes dividing $D=-4n$ , then the number genera is given by $2^{\mu-1}$ where

$\displaystyle \mu = r {\text { if }} n \equiv 3 \mod 4$
$\displaystyle \mu = r +1 {\text { if }} n \equiv 1, 2 \mod 4$
$\displaystyle \mu = r+2 {\text { if }} n \equiv 4 \mod 8$
$\displaystyle \mu = r+4 {\text { if }} n \equiv 0 \mod 8$

In general if we also take forms of discriminant $D =1 \mod 4$ , we have $\mu =r.$

Ambiguous classes: To explain this result, we first need to know the 2-torsion of the class group- that is forms of order at most $2$ called the ambiguous classes. A class is ambiguous precisely when it is equivalent to it’s inverse. Using the fact that positive definite forms are reduced only if $|b|\le a \le c$ and that $[a, -b, c]$ is the inverse of $[a, b, c]$ , we see that ambiguous forms precisely corresponds to reduced forms with

$b=0$ and $a=b$ and $a=c$ .

These conditions are enough to see that there are $2^{\mu-1}$ ambiguous classes. For instance in the $n =3\mod 4$ , $b=0 \implies n=ac$ and there are exactly $2^r$ factorisation with $(a,c)=1$ (each prime can either belong to $a$ or $c$ -out of which exactly half of them satisfy $a<c$ , so in total $2^{r-1}=2^{\mu-1}$ forms. Other cases the computations are similar.

Because the 2-torsion is of size $2^{mu-1}$ , the number of squared-classes if exactly equal to $h(D)/2^{\mu-1}$ . Look at the square map $\phi: Cl(D) \to Cl(D)$ , whose kernel is the ambiguous classes. Hence the image which is the set of squares has size $|Cl(D)|/|\text{Ker} \phi| = h(D)/2^{\mu -1}.$

Genus characters:

For any prime $p$ dividing the discriminant the character on the class group (primitive forms) by

$f= aX^2+bXY+cY^2 \to \left(\frac{aX_0^2+bX_0Y_0+cY_0^2}{p} \right)$

is well defined. In particular we can assume

$aX^2+bXY+cY^2 \to \left( \frac{a}{p}\right)$

To see this note that

$\displaystyle \left(aX_0^2+bX_0Y_0+cY_0^2\right)(\left(aX_1^2+bX_1Y_1+cY_1^2\right)=\left[a X_0X_1+\frac{b}{2}\left(X_0Y_1+X_1Y_0\right)+c Y_0 Y_1\right]^{2}-D\left(X_0 Y_1-X_1 Y_0\right)^{2} = \square \mod p$

So the values (co-prime to discriminant) represented by the forms are all quadratic residues or non-residues, and the character is well defined.

Thus we have a list of characters for each odd prime dividing the discriminant.

For $n =3 \bmod 4$ these is the complete list of “genus characters”. There are more characters for depending on other classes $n \mod 4$ or $n \mod 8$ . In total there are $\mu$ complete characters.

They are called genus characters precisely because all of the forms in a fixed genus have the same values for the complete list of genus characters. In fact, the principal genus corresponds to the forms that evaluate to $1$ on all these characters. But the number of possible value of this complete set of characters is $2^{mu-1}$ (precisely half of all the possible values since the values modulo $4n$ only belong to half the residue classes). Hence the number of genera is equal to $2^{\mu-1}.$

It’s easy to see that the classes of squares in the class group have all the genus characters equal to $1$ and hence belong to the principal genus. By the previous discussion on ambiguous classes, if the principal genus had more classes than these then the number of genera would be $h(D)/|\text{Principal Genus}|$ which would be smaller than $h(D)/{\text{Squares}} = h/(h/2^{\mu-1}) =2^{\mu-1}$ , which is not true. Thus the principal genus is exactly the set of squares (Duplicated forms). This is the beautiful Gauss’s duplication Theorem!

Question: Can you describe the discriminants for which each genus contains just a single reduced form?

Such discriminants are nicer because we can precisely pin point one reduced form with the congruence conditions.

For example in each of the following cases, there is only class per genus.

$\displaystyle p =x^{2}+6 y^{2} \Longleftrightarrow p \equiv 1,7 \bmod 24$
$\displaystyle p =x^{2}+10 y^{2} \Longleftrightarrow p \equiv 1,9,11,19 \bmod 40$
$\displaystyle p =x^{2}+13 y^{2} \Longleftrightarrow p \equiv 1,9,17,25,29,49 \bmod 52$
$\displaystyle p = x^{2}+15 y^{2} \Longleftrightarrow p \equiv 1,19,31,49 \bmod 60$
$\displaystyle p =x^{2}+21 y^{2} \Longleftrightarrow p \equiv 1,25,37 \bmod 84$

From the previous discussion, this can happen only if there are no non-trivial square classes. That is if square of every form is the principal class, equivalently a class is equivalent to it’s inverse. In this case the class number if exactly equal to $2^{\mu-1}.$

Gauss listed 65 discriminants with this property:

$\displaystyle 1,2,3,4,5,6,7,8,9,10,12,13,15,16,18,21,22,24,25,28,30,33,37,40,42,45,48,57,58,60$
$\displaystyle 70,72,78,85,88,93,102,105,112,120,130,133,165,168,177,190,210,232,240,253,273,280$
$\displaystyle 312,330,345,357,385,408,462,520,760,840,1320,1365, 1848$

(These numbers are first found by Euler- they are called convenient/idoneal numbers- any composite number represented by form has at least 2 solutions)

The list is known to be finite and conjectured to be be just this 65 numbers.

Other ways to define genus: We saw two ways to think of a genus–

As the set of forms representing the same values in $\mathbb Z/D\mathbbZ)^{*}$
As the set of forms which have the same complete characters–same values on the genus characters.

In fact the genus can be seen as the

3. set of forms equivalent modulo any integer $m$
4. set of form equivalent under $ $GL_2(\mathbb Q_S)$ , where $\mathbb Q_S$ corresponds to rationals with denominators coprime to a given integer $M$ .
5. set of form equivalent under $ $GL_2(\mathbb Q_S)$ , where $\mathbb Q_S$ corresponds to rationals with denominators coprime to a the integer $2D$ .
6. set of form which represent the same values modulo any integer $m$ .