# The narrow class group of an order of a quadratic field and the genera of binary quadratic forms

Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$. Let $F = ax^2 + bxy + cy^2 \in \mathfrak{F}$. Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $SL_2(\mathbb{Z})$. We write $f^\alpha(x, y) = f(px + qy, rx + sy)$. Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $SL_2(\mathbb{Z})$ acts on $\mathfrak{F}$.

We say $D = b^2 - 4ac$ is the discriminant of $F$. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive. If $D < 0$ and $a > 0$, we say $ax^2 + bxy + cy^2$ is positive definite.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). We denote the set of primitive binary quadratic forms of discriminant $D$ by $\mathfrak{F}_0(D)$. We denote the set of primitive positive definite binary quadratic forms of discriminant $D$ by $\mathfrak{F}^+_0(D)$. It is easy to see that $\mathfrak{F}_0(D)$, $\mathfrak{F}^+_0(D)$ are $SL_2(\mathbb{Z})$-invariant.

If $D < 0$, we denote the set of $SL_2(\mathbb{Z})$-orbits on $\mathfrak{F}^+_0(D)$ by $C(D)$.

If $D > 0$, we denote the set of $SL_2(\mathbb{Z})$-orbits on $\mathfrak{F}_0(D)$ by $C(D)$.

Let $F \in \mathfrak{F}^+_0(D)$ if $D < 0$, $F \in \mathfrak{F}_0(D)$ if $D > 0$. We denote by $[F]$ the $SL_2(\mathbb{Z})$-orbit represented by $F$.

Let $\Phi_1,\dots,\Phi_{\mu}$ be the system of genus characters of discriminant $D$[see this question].

We defined a homomorphism $\Phi\colon(\mathbb{Z}/D\mathbb{Z})^\times \rightarrow (\mathbb{Z}^\times)^\mu$ by $\Phi([n]) = (\Phi_1(n),\dots,\Phi_{\mu}(n))$ in this question.

Let $F \in \mathfrak{F}^+_0(D)$ if $D < 0$, $F \in \mathfrak{F}_0(D)$ if $D > 0$. Let $m$ be an integer which is represented by $F$ and gcd($m, D) = 1$. Then, by the proposition of this question, $\Phi_1([m]),\dots,\Phi_{\mu}([m])$ do not depend on the choice of $m$. Since the set of integers represented by $F$ is determined by the class $[F]$, each $\Phi_i$ defines a map $C(D) \rightarrow \mathbb{Z}^\times$. By abuse of notation, we denote this map by $\Phi_i$. We write, by abuse of notation, $\Phi([F]) = (\Phi_1([F]),\dots,\Phi_{\mu}([F]))$. Hence we get a map $\Phi\colon C(D) \rightarrow (\mathbb{Z}^\times)^\mu$. We can define an equivalence relation on $C(D)$ by $\Phi([F]) = \Phi([G])$. Each equivalence class of this relation is called a genus(Gauss: Disquisitiones Arithmeticae, art.231). The genus which is represented by the prinicipal form is called the prinicipal genus(for the definition of the principal form, please see this question) of discriminant $D$. We would like to investigate the set of genera.

Let $S = \{ [m] \in (\mathbb{Z}/D\mathbb{Z})^\times; m$ is represented by $F\}$. $S$ is determined by $[F]$.

Let $H = \{ [m] \in (\mathbb{Z}/D\mathbb{Z})^\times$; $m$ is represented by the principal form of discriminant $D\}$.

By the proposition of this question, $S$ is a residue class of Ker$(\chi)/H$, where $\chi\colon (\mathbb{Z}/D\mathbb{Z})^\times\rightarrow \mathbb{Z}^\times = \{-1, 1\}$ is the homomorphism defined in the proposition 2 of this question.

Hence we get a map $\Psi\colon C(D) \rightarrow$ Ker$(\chi)/H$. Since Ker$(\Phi) = H$ by this question, $\Psi([F]) = \Psi([G])$ if and only if $[F]$ and $[G]$ belong to the same genus.

On the other hand, by the proposition of this question, there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$. And there exists a bijection $\psi\colon Cl^+(R) \rightarrow C(D)$ by the proposition of this question. If we identify $C(D)$ with $Cl^+(R)$ by $\psi$, we get a map $\Psi\colon Cl^+(R) \rightarrow$ Ker$(\chi)/H$.

My question Is the following proposition true? If yes, how do we prove it?

Proposition The map $\Psi\colon Cl^+(R) \rightarrow$ Ker$(\chi)/H$ is a homomorphism.