Reduced Forms

Definition 1.1 (Reduced)   A positive definite quadratic form $(a,b,c)$ is reduced if $\vert b\vert\leq a \leq c$ and if, in addition, when one of the two inequalities is an equality (i.e., either $\vert b\vert=a$ or $a=c$), then $b\geq 0$.

There is a geometric interpretation of reduced, which we will not use this later. Let $D=\disc(a,b,c) = b^2 - 4ac$ and set $\tau = \frac{-b+\sqrt{D}}{2a}$, so $\tau$ is the root of $ax^2 + bx + c$ with positive imaginary part. The right action of $\SL_2(\mathbb{Z})$ on positive definite binary quadratic forms corresponds to the left action of $\SL_2(\mathbb{Z})$ by linear fractional transformations on the complex upper half plane $\mathfrak{h}= \{z \in \mathbb{C}:$   Im$(z)>0\}$. The standard fundamental domain for the action of $\SL_2(\mathbb{Z})$ on  $\mathfrak{h}$ is

$$\mathcal{F} = \left\{\tau \in \mathfrak{h}: \mbox{\rm Re}(\tau)\i... ...rt>1 \text{ or } \vert\tau\vert=1\text{ and }\mbox{\rm Re}(\tau)\leq 0\right\}.$$

Then $(a,b,c)$ is reduced if and only if the corresponding complex number $\tau$ lies in $\mathcal{F}$. For example, if $(a,b,c)$ is reduced then Re$(\tau) = -b/2a\in[-1/2,1/2)$ since $\vert b\vert\leq a$ and if $\vert b\vert=a$ then $b\geq 0$. Also

$$\vert\tau\vert = \sqrt{\frac{b^2 + 4ac - b^2}{4a^2}} = \sqrt{\frac{c}{a}} \geq 1$$

and if $\vert\tau\vert=1$ then $b\geq 0$ so Re$(\tau) \leq 0$.

The following theorem (which is not proved in Davenport) highlights the importance of reduced forms.

Theorem 1.2   There is exactly one reduced form in each equivalence class of positive definite binary quadratic forms.

Proof. We have to prove two things. First, that every class contains at least one reduced form, and second that this reduced form is the only one in the class.

We first prove that there is a reduced form in every class. Let $\mathcal{C}$ be an equivalence class of positive definite quadratic forms of discriminant $D$. Let $(a,b,c)$ be an element of $\mathcal{C}$ such that $a$ is minimal (amongst elements of $\mathcal{C}$). Note that for any such form we have $c\geq a$, since $(a,b,c)$ is equivalent to $(c,-b,a)$ (use the matrix $\left( \begin{smallmatrix}0&-1\\ 1&\hfill 0\end{smallmatrix}\right)$). Applying the element $\left( \begin{smallmatrix}1&k\\ 0&1\end{smallmatrix}\right)\in\SL_2(\mathbb{Z})$ to $(a,b,c)$ for a suitably chosen integer $k$ (precisely, $k=\lfloor(a-b)/2a\rfloor$) results in a form $(a',b',c')$ with $a'=a$ and $b'\in (-a',a']$. Since $a'=a$ is minimal, we have just as above that $a'\leq c'$, hence $(a',b',c')$ is ``just about'' reduced. The only possible remaining problem would occur if $a'=c'$ and $b'<0$. In that case, changing $(a',b',c')$ to $(c'',b'',a'')=(c',-b',a')$ results in an equivalent form with $b''>0$, so that $(c'',b'',a'')$ is reduced.

Next suppose $(a,b,c)$ is a reduced form. We will now establish that $(a,b,c)$ is the only reduced form in its equivalence class. First, we check that $a$ is minimal amongst all forms equivalent to $(a,b,c)$. Indeed, every other $a'$ has the form $a'=ap^2 + bpr + cr^2$ with $p,r$ coprime integers (see this by hitting $(a,b,c)$ by $\left( \begin{smallmatrix}p&q\\ r&s\end{smallmatrix}\right)$). The identities

$$a p^2 + bpr + c r^2 = a p^2 \left(1+\frac{b}{a}\frac{r}{p}\right) + cr^2 = a p^2 + c r^2\left( 1 + \frac{b}{c}\frac{p}{r}\right)$$

then imply our claim since $\vert b\vert\leq a \leq c$ (use the first identity if $r/p<1$ and the second otherwise). Thus any other reduced form $(a',b',c')$ equivalent to $(a,b,c)$ has $a'=a$. But the same identity implies that the only forms equivalent to $(a,b,c)$ with $a'=a$ are obtained by applying a transformation of the form $\left( \begin{matrix}1&k\\ 0&1 \end{matrix}\right)$ (corresponding to $p=1$, $r=0$). Thus $b' = b+2ak$ for some $k$. Since $a=a'$ we have $b,b'\in(-a,a]$, so $k=0$. Finally

$$c'=\frac{(b')^2-D}{4a'}=\frac{b^2-D}{4a} = c,$$

so $(a',b',c')=(a,b,c)$. $\qedsymbol$