Let $V_\mathbb{R}$ denote the $\mathbb{R}$-vector space of binary quartic forms. The group $\operatorname{GL}_2(\mathbb{R})$ acts on $V_\mathbb{R}$ via the standard substitution action. That is, if $F(x,y) \in V_\mathbb{R}$ and $T = \begin{pmatrix} t_1 & t_2 \\ t_3 & t_4 \end{pmatrix} \in \operatorname{GL}_2(\mathbb{R})$, then $T$ sends $F$ to $F_T(x,y) = F(t_1 x + t_2 y, t_3 x + t_4 y)$. The action induced by the subgroup $\operatorname{GL}_2(\mathbb{Z})$ of $\operatorname{GL}_2(\mathbb{R})$ has two invariants, $I(F)$ and $J(F)$, which are algebraically independent and generate the ring of invariants under the action of $\operatorname{GL}_2(\mathbb{Z})$.

We will denote by the height of $F$, as in Bhargava and Shankar's paper (see references below), as $H(F) = \max\{|I(F)|^3, J(F)^2/4\}$.

The action of $\operatorname{GL}_2(\mathbb{R})$ on $F \in V_\mathbb{R}$ has a stabilizer, which we will denote by $\operatorname{Aut}_\mathbb{R} (F)$.

We denote by $V_\mathbb{Z}$ the subring of $V_\mathbb{R}$ consisting of binary quartic forms with integer coefficients. We will say an element $U \in \operatorname{Aut}_\mathbb{R}(F)$ almost rational if it is of the form

$$\displaystyle U = U(\alpha, \beta, \gamma) = \frac{1}{\sqrt{D}} \begin{pmatrix} \beta & 2 \gamma \\ -2 \alpha & -\beta \end{pmatrix},$$

where $\alpha, \beta, \gamma$ are co-prime integers and $D = |\beta^2 - 4 \alpha \gamma|$. We will say that $U$ is reduced if $f(x,y) = \alpha x^2 + \beta xy + \gamma y^2$ is a reduced binary quadratic form (in the sense of Gauss).

We will say that a binary quartic form $F$ of height less than $Z$ has a "large" stabilizer if $\operatorname{Aut}_\mathbb{R} (F)$ contains a reduced almost rational element $U(\alpha, \beta, \gamma)$ such that $D \gg Z^{1/6}$. How does one count the number of irreducible forms with "large" stabilizer of height up to $Z$?

There is a reason for the exponent of $1/6$. Consider the following family of forms fixed by the matrix

$$\displaystyle U(1, 0, D) = \frac{1}{\sqrt{D}} \begin{pmatrix} 0 & D \\ -1 & 0 \end{pmatrix}.$$

The family of forms $F$ fixed by $U(1, 0, D)$ are the forms of the shape

$$\displaystyle F(x,y) = a_4 x^4 + a_3 x^3 y + a_2 x^2 y^2 - a_3 Dxy^3 + a_4 D^2 y^4,$$

and the $I$-invariant of the generic form in this family is given by

$$\displaystyle I(F) = 12 a_4^2 D^2 + 3 a_3^2 D + a_2^2.$$

Since for every form $F$ in the family the height $H(F)$ is given by $H(F) = I(F)^3$ (one checks that every form in this family has non-negative discriminant), it follows that the height condition is equivalent to

$$\displaystyle 12a_4^2 D^2 + 3a_3^2 D + a_2^2 \leq Z^{1/3}.$$

Thus, if $D \gg Z^{1/6}$, then $a_4 = 0$, which means that all such forms are reducible, which we do not want to count. However, this argument does not seem to work for an arbitrary family, so I am wondering if there is a more subtle principle at work.


M. Bhargava, A. Shankar, Binary quartic forms having bounded invariants, and the boundedness of the average rank of elliptic curves, Annals of Mathematics 181 (2015), 191-242.

  • $\begingroup$ Partial answer: any such form must have discriminant divisible by $D^2$, so no such forms exist if $D \gg Z^{1/2}$. This falls short of the expectation that $D \gg Z^{1/6}$ suffices. $\endgroup$ – Stanley Yao Xiao Mar 20 '16 at 13:45

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