Dirichlet Composition of Binary Quadratic Forms

If we multiply two expressions of the form $X^2+DY^2$ , we get an expression of the same form. That is

$\displaystyle \left(X_1^2+DY_1^2\right) \left(X_2^2+DY_2^2\right) = \left(X_3^2+DY_3^2\right)$

where $\displaystyle X_3 =X_1X_2-DY_1Y_2$ and $\displaystyle Y_3 = X_1Y_2+X_2Y_1$

A way to see this is by writing $X^2+DY^2 = \left(X+\sqrt{-D}Y\right)\left(X-\sqrt{-D}Y\right)$ , as a product of conjugates elements (norm) in $\mathbb Q(\sqrt{-D})$ . So we have

$\displaystyle \left(X_1^2+DY_1^2\right) \left(X_2^2+DY_2^2\right)$
$\displaystyle =\left(X_1+\sqrt{-D}Y_1\right)\left(X_1-\sqrt{-D}Y_1\right)\left(X_2+\sqrt{-D}Y_2\right)\left(X_2-\sqrt{-D}Y_2\right)$
$\displaystyle =\left[\left(X_1+\sqrt{-D}Y_1\right)\left(X_2+\sqrt{-D}Y_2\right)\right] \left[\left(X_1-\sqrt{-D}Y_1\right)\left(X_2-\sqrt{-D}Y_2\right)\right]$
$\displaystyle= \left(X_3+\sqrt{-D}Y_3\right)\left(X_3-\sqrt{-D}Y_3\right)=\left(X_3^2+DY_3^2\right)$

Note the forms we are multiplying have the same discriminant ( $-4D$ in our case). What happens if we take another two forms of the same discriminant and multiply them together?

Let say $\displaystyle \left(a_1X_1^2+b_1X_1Y_1+c_1Y_1^2\right) \left(a_2X_2^2+b_2X_2Y_2+c_2Y_2^2\right)$

We can again write this as $\displaystyle a_1\left(X_1 + \frac{-b_1+ \sqrt{D} }{2a_1}\right)\left(X_1 + \frac{-b_1- \sqrt{D} }{2a_1}\right)$ times
$\displaystyle a_2\left(X_2 + \frac{-b_2+ \sqrt{D} }{2a_2}\right)\left(X_2 + \frac{-b_2- \sqrt{D} }{2a_2}\right)$

and group the first and third terms together, and their conjugate second and fourth terms together, to get another quadratic form of the same discriminant $D=b_1^2-4a_1c_1=b_2^2-4a_2c_2$ .

Equivalences: A form of discriminant $D$ obtained by change of coordinates $aX+bY, cX+dY$ also represents the same form– that is they represent the same set of integers for instance. For $aX+bY, cX+dY$ to be a change of coordinates we need $M= \left[\begin{array}{cc} a & b \\c & d\end{array}\right]$ be invertible. Hence if we are concerned with integral quadratic forms, we need the determinant to be $\pm 1$ . Two form $f, g$ such that $g(X, Y) = f(M(X,Y)) = f(aX+bY, cX+dY) 0$ are called equivalent and properly equivalent if the matrix $M$ has determinant $1$ , improperly equivalent if the determinant is $1 .$

Example: Using the change of coordinates $\displaystyle T=\left(\begin{array}{ll}1 & 1 \\ 0 & 1 \end{array}\right)$ we see that the form $\displaystyle [a,b,c] =aX^2+bXY+cY^2$

is equivalent to

$\displaystyle a(X+Y)^2+b(X+Y)Y+cY^2 =aX^2+(b+2a)XY+(c+b+a)Y^2$

$\displaystyle =[a, b+2a, c+b+a]$

Similary using $\displaystyle S=\left(\begin{array}{cc}0 & -1 \\ -1 & 0 \end{array}\right)$ we see that the form $\displaystyle [a,b,c] =aX^2+bXY+cY^2$

is equivalent to

$\displaystyle a(-Y)^2+b(-Y)(-X)+c(-X)^2 =cX^2+bXY+aY^2=[c, b, a]$

Under equivalence, the discriminant of the forms is preserved. To see that note that

$\displaystyle f(X,Y) = aX^2+bXY+cY^2= (X, Y)\left(\begin{array}{cc} a & b/2 \\ b/2 & c\end{array}\right)\left(\begin{array}{l} X \\ Y \end{array}\right)$

and hence the action

$\displaystyle f(M(X,Y)) =(X, Y)M^{T}\left(\begin{array}{cc} a & b/2 \\ b/2 & c\end{array}\right)M\left(\begin{array}{l} X \\ Y \end{array}\right)$

so the action on the coefficients is given by

$\displaystyle \left(\begin{array}{cc} a & b/2 \\ b/2 & c\end{array}\right) \to M^{T}\left(\begin{array}{cc} a & b/2 \\ b/2 & c\end{array}\right)M$

which preserves the determinant which is equal to

$\displaystyle \det \left(\begin{array}{cc} a & b/2 \\ b/2 & c\end{array}\right) = ac-b^2/4 = -D/4$

and hence the discriminant.

OK, so we have these forms and equivalences under the action of these matrices. How many such equivalence classes exist? Rather can we get all the forms starting from a single form by these equivalence (changing the coordinates). That answer is no! It depends on the discriminant and the set of proper equivalences classes is called the class group $Cl(D) .$

Reduction Theory: Is there some way to describe the equivalence classes? (Maybe by listing the representatives?) There is a way to reduce a given form under equivalence so that all of the forms are equivalent to one of the reduced forms. The theory now crucially depends on the sign of the discriminant.

Negative Discriminant (positive definite forms): $\displaystyle f=[a, b, c]$ is reduced if

$\displaystyle |b| \leq a \leq c$

We can apply the transformations $S, T$ to the form to make sure that $\tau =\frac{-b + \sqrt{-D}}{2a}$ has the highest imaginary part, one we finish the process we get a reduced form. The conditions will give $|b| \leq \sqrt{D / 3}$ and hence the classes are finite. Also except for $[a, b, a] \sim [a, -b, a]$ and $[a, a, c] \sim [a, -a, c]$ , reduced forms represent distinct equivalence classes. The reduction is related to the fundamental domain of the action of the matrices of determinant one $SL_2(\mathbb Z)$ on the upper-half of the complex plane: $F=\{z \in \mathbb{H}:|z|>1,-1 / 2\le \Re(z) \le 1 / 2\}$ . Note that the root $\tau \frac{-b + \sqrt{-D}}{2a} \in \mathbb H$ only for negative discriminants.

Positive Discriminant: The reduction theory is slightly more complicated here. The roots are real numbers. A form is called reduced if

$\displaystyle 0<b<\sqrt{D}, \sqrt{D}-b<2|a|<\sqrt{D}+b$

Again it’s easy to see that there are only finitely many reduced forms, and that every form is equivalent to a reduced form. But the new feature is that the reduced forms can be partioned into cycles of forms which are all equivalent. Each cycle of the form corresponds to the continued fraction of the root $\tau$ which is periodic for quadratic irrationals.

And we described a way to compose form in the beginning- We can in fact check that the composition respects the equivalences and we have a composition law on the class group (these equivalences classes). In fact, a beautiful observation of Gauss is that the composition is a “group” law – that is we have natural properties like associativity and invertibility. The class of inverse of $[a, b, c]$ is represented by $[a, -b, c] .$ (Note that this proper equivalence, if we allow improper equivalence, these forms are in fact equivalent under $X\to -X, Y\to Y .$ The identity element is the form $X^2+dY^2$ or $X^2+XY + \frac{1-d}{4}Y^2$ (depends on the discriminant $D=-4d$ or $D=d$ , that is $D\equiv 0, 1 \mod 4$ )

We already saw that composition is related to arithmetic in $\mathbb Q({\sqrt{-D}})$ , in fact in modern terms the composition, group law, invertibility etc can be seen phrased in terms of invertible ideals of the order of the given discriminant. The map is

$aX^2+bXY+cY^2 \to \Big\langle a, \frac{-b+\sqrt{D}}{2}\Big\rangle$

The equivalence in the forms will correspond to change of basis for the ideals and the equivalence of ideals up to principal ideals, that is $I \sim J$ if $I=(\alpha) J$ . The composition of form is just multiplication of ideals.

Explicit descriptions of the composition is helpful to understand the arithmetic of the class group better.

Dirichlet’s “United” Forms: Two forms $f= \left[a_{1}, b_{1}, c_{1}\right]$ and $g=\left[a_{2}, b_{2}, c_{2}\right]$ of the same discriminant $D$ are called united if the gcd of $a_{1}, a_{2}$ , and $\left(b_{1}+b_{2}\right) / 2$ is $1 .$

If we have united forms we have transform them into form that look like

$\displaystyle [a_1, b_1, c_1] \sim [a_1, B, a_2C]$
$\displaystyle [a_2, b_2, c_2] \sim [a_2, B, a_1C]$

Now the composition of the form is defined to be

$\displaystyle \left(a_{1} X_{1}^{2}+B X_{1} Y_{1}+a_{2}CY_{1}^{2}\right)\left(a_{2}X_{2}^{2}+BX_{2}Y_{2}+a_{1} C X_{2}^{2}\right)$
$\displaystyle =a_{1}a_{2}X^{2}+BXY+CY^{2}$

In fact, if given united form $f, g$ , we can choose $B \mod 2a_1a_2$ such that

$\displaystyle B \equiv b \bmod 2 a_1 , \quad B \equiv b^{\prime} \bmod 2 a_2, \quad B^{2} \equiv D \bmod 4 a_1 a_2$

and the composition is

$\displaystyle [a_1, b_1, c_1] \circ [a_2, b_2, c_2] = [a_1a_2, B, \frac{B^2-D}{4a_1a_2}]=a_1a_2X^2+BXY+ \frac{B^2-D}{4a_1a_2}Y^2$

If the forms are primitive that is if the coefficients are coprime $(a_1, b_1, c_1)=(a_2, b_2, c_2)=1$ , we can choose representatives such that $(a_1, a_2)=1$ , so that the forms are united. If they are not primitive we can always separate the common factor to get primitive forms.

Example: Let $D =0 \mod 4$ , then the identity form is $X^2-\frac{D}{4}Y^2$ . Because the coefficient is 1, any form is united with this form. Take $f=[a, b, c]$ . Then $B=b$ satisfies the above properties.

And the composition is

$\displaystyle [1, 0, \frac{-D}{4}] \circ [a, b, c] =[a, b, \frac{b^2-D}{4a}]=[a, b, c]$

Similarly if $D = 1 \mod 4$ , the identity is $X^2+XY +\frac{1-D}{4}Y^2$ and we have $B=b$ satisfying all the conditions. Therefore the composition is

$\displaystyle [1, 1, \frac{1-D}{4}] \circ [a, b, c] =[a, b, \frac{b^2-D}{4a}]=[a, b, c]$

Now we want to show that primitive forms $f=[a, b, c]$ and $g=[a, -b, c]$ represent inverse classes.

The forms as given are not necessarily united because $a_1=a_2=a$ . But use $f \sim f_1= [c, -b, a]$ and now $f_1, g$ are united. Again $B=b$ satisfies the required properties. So we have

$\displaystyle [c, -b, a] \circ [a, -b, c] = [ac, b, \frac{b^2-D}{4ac}] =[ac, b, 1]$

Now apply $\displaystyle (X, Y) \to (-Y, X), [ac, b, 1] \to [1, -b, ac]$ and then

$\displaystyle (X,Y) \to (X+\frac{b}{2}Y), [1, -b, ac] \to [1, 0, \frac{-D}{4}]$

for $D =0 \mod 4$ and

$\displaystyle (X,Y) \to (X+\frac{b+1}{2}Y), [1, -b, ac] \to [1, 0, \frac{1-D}{4}]$

for $D =1 \mod 4$

We saw invertibility and the identity operations for the composition law. The associativity can also be similarly seen in fact given three forms $f, g, h$ , we can choose representative that look like

$\displaystyle f \sim [a_1, B, c_1], g\sim [a_2, B, c_2], h \sim [a_3, B, c_3]$ and now we have

$\displaystyle f\circ g = [a_1a_2, B, \frac{B^2-D}{4a_1a_2}], g \circ h = [a_2a_3, B, \frac{B^2-D}{4a_2a_3}]$

$\displaystyle (f \circ g) \circ h =[a_1a_2a_3, B, \frac{B^2-D}{4a_1a2a_3}] = f\circ (g \circ h)$

Remarks: Proper vs Improper equivalence- the distinction is important – the group structure is only defined on the proper equivalence classes. If you add the improper equivalence we put the relation $f \sim x f^{-1}$ on the proper equivalence classes, and this is not consistent with the group operation. For instance, nothing happens to ambiguous classes, other classes are identified with their inverses. And we get that every squared class is improperly equivalent to the principal class. A direct way to see it is the following identity

$\displaystyle \left(a X^{2}+2 b X X+c Y^{2}\right)\left(a Z^{2}+2 b Z W+c W^{2}\right) =(a X Z+b X W+b Y Z+c Y W)^{2}+n(X W-Y Z)^{2}$

Gauss’s Direct Composition: Gauss “direct composition” defined below makes some choices- the choices are not-canonical but up to proper equivalence well defined.

For instance there are many choices to compose to have the following

$\displaystyle f(X, Y) g(Z, W)=F\left(B_{1}(X, Y ; Z, W), B_{2}(X, X ; X, W)\right) &fg=00000$ where

$B_1, B_2$ are integral bilinear forms

$\displaystyle B_{i}(X, Y ; Z, W)=a_{i} X Z+b_{i} X W+c_{i} Y Z+d_{i} YW$

The formula for the square of form given above is one such expression. But not all of them give group laws!

$\displaystyle f=[a, b, c], g=[a', b', c'] \implies a_1b_2-a_2b_1=\pm a, a_1c_2-a_2c_1 =\pm a'$

Only the choices of positive signs give the group law.

The composition up to proper equivalence loses some structure- for instance it’s possible to define classes with just weaker equivalence under action by the matrix $\displaystyle T=\left(\begin{array}{ll}1 & 1 \\ 0 & 1 \end{array}\right)$ and define a group structure.

We can also consider equivalence under “congruence” subgroup of $SL_2(\mathbb Z)$ , like $\Gamma_0(N)$ which will be give “ray” class groups– ideals coprime to $N$ modulo ideals principal $(\alpha), \alpha=1 \mod N$