Exercise 2.9.12

Question

Prove that for any discriminant there is exactly one $\Gamma$-orbit $(1,b,c) \Gamma$ of forms of discriminant $\Delta$. Construct this $\Gamma$-orbit.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

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

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
If $b = 2k$ is even, then $(1,b,c)\cdot T^{-k} = (1,0,\cdot)$, and if $b= 2k + 1$ is odd, then $(1,b,c) \cdot T^{-k} = (1,1,\cdot)$. Therefore a $\Gamma$-orbit of a form $(1,b,c) \Gamma$ contains a form $(1,B,C)$, with $B \in \{0,1\}$. Moreover the discriminant is $\Delta = B^2 - 4C  \equiv B^2 \pmod 4$, thus $B$ and $\Delta$ are of same parity.

\begin{enumerate}
\item[$\bullet$] If $\Delta \equiv 0 \pmod 4$, $B = 0$, and $C = -\Delta/4$, thus there is only one $\Gamma$-orbit, the orbit of the principal form $f_0 = (1,0, -\Delta/4)$ with  discriminant $\Delta$..

\item[$\bullet$] If $\Delta \equiv 1 \pmod 4$, $B = 1$, and $C = (1 - \Delta /4)/4$, thus there is only one $\Gamma$-orbit, the orbit of the principal form $f _1= (1, 1, (1 - \Delta)/4)$ with discriminant $\Delta$.
\end{enumerate}

Since $(a,b,c)\cdot T^k = (a,b+2ka, a k^2+bk+c)$, the unique $\Gamma$-orbit of forms representing 1 of discriminant $\Delta$ is

$$\Gamma f_0 = \left \{ \left(1, 2k, k^2 - \frac{\Delta}{4} ight) \mid k \in \Zight \} 	ext { if } \Delta \equiv 0 \pmod 4,$$
and 
$$\Gamma f_1 = \left \{ \left(1, 1+2k, k^2 + k + \frac{1- \Delta}{4} ight) \mid k \in \Zight \} 	ext { if } \Delta \equiv 0 \pmod 4.$$
\end{proof}

Exercise 2.9.12

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Exercise 2.9.12

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