Exercise 2.9.9

Question

Verify Example 2.5.9.

Show that the automorphism group of the integral primitive forms of discriminant $-3$ is the cyclic group generated by 
$$\begin{pmatrix} 0 & -1\ 1 & 1 \end{pmatrix}.$$
It is of order $6$.

Richard Ganaye · Accepted Answer

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\begin{proof}
Since the automorphism groups of all integral primitive forms of discriminant $-3$ are isomorphic, we can use the form $f = (1,1,1)$, of discriminant $-3$.

By theorem 2.5.5, the automorphisms are in bijection with the set of solutions of $x^2 + 3y^2 = 4$, and the bijection is given by
$$(x,y) \mapsto U(f,x,y) = \begin{pmatrix} \frac{x - yb}{2} & -cy \ay & \frac{x+yb}{2} \end{pmatrix} = \begin{pmatrix} \frac{x - y}{2} & -y \y & \frac{x+y}{2} \end{pmatrix}.$$

Since $x^2 + 3 y^2 = 4$, $y^2 \leq \left \lfloor \frac{4}{3} ight floor = 1$, thus $y \in \{-1,0,1\}$. The set of solutions is
$$G = \{(-2,0),(2,0), (1,1),(1,-1),(-1,1),(-1,-1)\}.$$
We know that $G$ is a group for the law $*$ , with neutral element $e = (2,0)$, given by
$$(x,y) * (x',y') = \left(\frac{xx'- 3yy'}{2}, \frac{xy'+yx'}{2}ight)$$
(see Ex 2.9.7).

Since $(1,1)^2 = (-1,1)
e e, (1,1)^3 = (-2,0) 
e e, (1,1)^6 =(2,0) = e$, $(1,1)$ is of order $6$. The corresponding automorphism is 
$$A = \begin{pmatrix} 0 & -1\ 1 & 1 \end{pmatrix}.$$
Therefore the group $\mathrm{Aut}(f)$ is cyclic of order $6$, generated by $A$
$$\mathrm{Aut}(f) = \left\{\begin{pmatrix} 1 & 0\ 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & -1\ 1 & 1 \end{pmatrix}, \begin{pmatrix} -1&-1\ 1 & 0 \end{pmatrix},
                                         \begin{pmatrix} -1 & 0\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0&1,\ -1 & -1 \end{pmatrix},\begin{pmatrix} 1 & 1\ -1 & 0 \end{pmatrix}
 ight\}.$$
 As a verification, $$(f\cdot A)(x,y) = f(-y,x+y) = y^2 - y(x+y) + (x+y)^2 = x^2 + xy +y^2 = f(x,y).$$
 \end{proof}

Exercise 2.9.9

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Exercise 2.9.9

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