Here $h_1, h_2, u_1, u_1, v_1, v_2$ are binary quadratic forms with integer coefficients, and $\Delta(h_1) = \Delta(h_2)$ (here $\Delta$ is the discriminant of a binary form). Put

$$\displaystyle F_1 = h_1(u_1(x,y), v_1(x,y)), F_2 = h_2(u_2(x,y), v_2(x,y)),$$

so that $F_1, F_2$ are binary quartic forms. Suppose that $F_1, F_2$ are $\operatorname{GL}_2(\mathbb{Z})$-equivalent; that is, there exists $T = \left(\begin{smallmatrix} t_1 & t_2 \\ t_3 & t_4 \end{smallmatrix}\right) \in \operatorname{GL}_2(\mathbb{Z})$ such that

$$\displaystyle F_1(t_1 x + t_2 y, t_3 x + t_4 y) = F_2(x,y),$$

and that $u_1, v_1, u_2, v_2$ are pairwise non-proportional. Does it follow that $h_1, h_2$ are $\operatorname{GL}_2(\mathbb{Z})$-equivalent?

  • $\begingroup$ What kind of nondegeneracy are you assuming for $u_i, v_i$? Because the answer could trivially be no: $u_1 = a xy, u_2 = b xy, v_1 = c xy, v_2 = d xy$, and some suitable choice of $h_1, h_2$. $\endgroup$ – user44191 Mar 13 '17 at 23:24
  • $\begingroup$ I think the only necessary condition to assume is that $u_1, v_1, u_2, v_2$ are pairwise not proportional... I will change the question to reflect this. $\endgroup$ – Stanley Yao Xiao Mar 14 '17 at 1:30
  • $\begingroup$ $u_1 = xy, v_1 = (x+y)^2, v_2 = (x-y)^2, h_1 = 4a^2, h_2 = (a-b)^2$. Then $F_1 = F_2$, but the minimal divisors for $h_1, h_2$ are different, so they are not equivalent. $\endgroup$ – user44191 Mar 14 '17 at 20:02
  • $\begingroup$ Your requirement that the discriminants be equal feels out of place here. In some sense, it's a requirement that depends on the "size" of the inputs, and the "size" appears nowhere else. I found the previous answer by trying to "scale" the $u,v$ in different ways. $\endgroup$ – user44191 Mar 14 '17 at 20:09

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