Exercise 3.5.1 (Reduce $7x^2 + 25 xy + 23y^2$)

Question

Find a reduced form that is equivalent to the form $7x^2 + 25 xy + 23y^2$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

Notations: We write $f = (a,b,c)$ for the quadratic form $f(x,y)= ax^2 + bxy +cy^2$. Moreover , if $M = \begin{pmatrix} \alpha & \beta\ \gamma & \delta\end{pmatrix} \in \mathrm{GL}_2(\Z)$, let us denote
$g =f\cdot M$ if $g(x,y) = f(\alpha x + \beta y, \gamma x + \delta y)$. Then $(f \cdot M) \cdot N = f\cdot (MN)$. If $M  \in \Gamma = \mathrm{SL}_2(\Z)$, we write $f \overset{+}{\sim} g$. 
Put $S =  \begin{pmatrix} 0 & 1\ -1 & 0\end{pmatrix}, \ T =  \begin{pmatrix} 1 & 1\ 0 & 1\end{pmatrix}$. Then
\begin{align*}
(a,b,c) \cdot S &= (c, - b , a),\
(a,b,c) \cdot T^m &= (a, 2am + b, am^2 + bm +c).
\end{align*}
\begin{proof}

Put $f(x,y) = 7x^2 + 25 xy + 23y^2$, so $f = (7, 25, 23)$.

We apply $T^{-2} = \begin{pmatrix} 1 & -2\ 0 & 1\end{pmatrix} \in \Gamma = \mathrm{SL}_2(\Z)$ to obtain
$$f_1 = f\cdot T^{-2} = (7, -3, 1).$$
Since $7 > 1$, we apply $S$:
$$f_2 = f_1 \cdot S = (1,3,7).$$
Then
$$f_3 = f_2 \cdot T^{-1} = (1,1, 5).$$

So $x^2 + xy + 5 y^2$ is a reduced form, properly equivalent to  $7x^2 + 25 xy + 23y^2$.
\end{proof}

{\bf Check:}
$$M = T^{-2}ST^{-1} = \begin{pmatrix} -2 & 1\ 1 & -1\end{pmatrix} \in \Gamma,$$
and
\begin{align*}
(f\cdot M) (x, y) &= f(-2x + y, x- y)\
 &= 7(-2x + y)^2 + 25 (-2x+y)(x-y) + 23 (x-y)^2\
 &= x^2 + xy + 5y^2\
 &= f_3(x,y).
\end{align*}
All these forms have discriminant $-19$.

Exercise 3.5.1 (Reduce $7x^2 + 25 xy + 23y^2$)

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Exercise 3.5.1 (Reduce $7x^2 + 25 xy + 23y^2$)

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