Bijection between the set of classes of positive definite quadratic forms and the set of classes of quadratic numbers in the upper half plane

Solution 1:

Proof of (1) We define a map $\psi\colon \mathcal{H}(D) \rightarrow \mathfrak{F}^+_0(D)$ as follows. Let $\theta \in \mathcal{H}(D)$. $\theta$ is a root of the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$. $D = b^2 - 4ac$. We define $\psi(\theta) = ax^2 + bxy + cy^2$. Clearly $\psi$ is the inverse map of $\phi$.

Proof of (2) Let $f = ax^2 + bxy + cy^2$. Let $\sigma = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right) \in \Gamma$. Let $f^\sigma = kx^2 + lxy + my^2$. Let $\theta = \phi(f)$. Let $\gamma = \sigma^{-1}\theta$. Then $\theta = \sigma \gamma$. $a\theta^2 + b\theta + c = 0$. Hence $a(p\gamma + q)^2 + b(p\gamma + q)(r\gamma + s) + c(r\gamma + s)^2 = 0$. The left hand side of this equation is $f(p\gamma + q, r\gamma + s) = k\gamma^2 + l\gamma + m$. Since $f^\sigma$ is positive definite by this question, $k \gt 0$. Since gcd$(k, l, m)$ = 1 by the same question, $\psi(\gamma) = f^\sigma$. Hence $\phi(f^\sigma) = \gamma = \sigma^{-1}\theta$. This proves (2).