Gauss Duplication Theorem

One way to define the genus of quadratic forms is by the values represented by the form or by the complete characters( the values of genus characters). In terms of the class group, forms in the principal genus are exactly the square classes, form in a particular genus form a coset of the square classes. This fact that every class in the principal genus (seen in terms of the value represented) is a square is called the duplication theorem.

One way to prove this by by showing that the number of possible complete characters is $2^{\mu-1}$ , so that the number of genera is $2^{\mu-1}$ , and then also show that the ambiguous classes (2-torsion) are also of the same number– so that the number of cosets of the square classes is $2^{\mu-1}$ which should be at most the number of genera. Because the square-classes are in the principal genus, we see that this equality implies that the group of square classes has to coincide with the principal genus.

Gauss proves that every class in the principal genus is a square directly by using reduction of ternary quadratic forms!

Gauss’s Proof:

a) Any form in the principal genus represents $1 \bmod D$ , because the values represented by the principal genus is a subgroup containing $1 \bmod D$ . For instance we have $F_0(X, 0)=1$ for the principal form.

b) Using the representation $F(m, n) =1 + kD$ , we can construction a ternary form of determinant $\frac{-1}{4}$ given by the matrix

$\displaystyle \mathbf{A}=\left(\begin{array}{ccc} a & b / 2 & -n / 2 \\ b / 2 & c & m / 2 \\ -n / 2 & m / 2 & -k \end{array}\right)$

where $\displaystyle F(x, y) = ax^2+bxy+cy^2$

because the determinant of $\mathbf A$ can be computed to be

$\displaystyle \det (\mathbf A)=a \left|\begin{array}{cc}c & m / 2 \\m / 2 & -k \end{array}\right|-\frac{b}{2} \left|\begin{array}{cc} b / 2 & m / 2 \\ -n / 2 & -k \end{array}\right|+ \frac{-n}{2} \left|\begin{array}{cc}b / 2 & c \\-n / 2 & m / 2\end{array}\right|$

$\displaystyle = -\frac{1}{4}\left(a m^{2}+b mn+cn^{2}-k\left(b^{2}-4 a c\right)\right)=-\frac{1}{4}(F(m,n)-kD)=\frac{-1}{4}$

c) By reduction of ternary forms, every ternary of form of discriminant $\frac{-1}{4}$ is equivalent to the form $X^2-YZ$ under some change of variables in $GL_3(\mathbb Z)$

d) We have $\displaystyle F(x, y) = ax^2+bxy+cy^2= \left(\begin{array}{lll} x & y & 0 \end{array}\right) \mathbf{A}\left(\begin{array}{l} x \\y \\ 0 \end{array}\right)$

By the above reduction (change of variables we have),

$\displaystyle F(x,y) = Y^2-XZ = \left(p_2x+q_2y+r_2z\right)^2-\left(p_1x+q_1y+r_1z\right)\left(p_3x+q_3y+r_3z\right)=\left(p_2x+q_2y\right)^2-\left(p_1x+q_1y\right)\left(p_3x+q_3y\right)$

We can in addition assume that $(p_1q_2-p_2q_1, D)=1$ by the action of $SL_2(\mathbb Z)$ on the coordinates $x, y$ .

e) Every class in the principal genus can be represented by by a form that looks like

$\displaystyle m^2x^2+bxy+cy^2, (m, D)=1$

This is because setting $(x, y) =(-q_1, p_1)$ , we get that $F(x, y) =(p_1q_2-p_2q_1)^2$ represents a square.

Choosing $m_1=p_1q_2-p_2q_1$ we see that $F(x, y)$ represents $m_1$ . If it’s primitively represented then we can change coordinates to transform it to look like $m_1x^2+bxy+cy^2$ , and we are done. If it’s not primitive, divide the common factors, to get $m_1^2 =m^2e^2$ , and the form primitively represents $m^2$ .

f) Consider form $[m, b, mc]$ , whose square under the composition turns out to be $[m^2, b, c]$ . (The form is “concordant”/ united with itself, so we can apply Dirichlet composition)

So we showed that every form in the principal class is equivalent to a form in the square class and we are done!

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