A curious identity involving a covariant of binary cubic forms

Let $F(x,y) = a_3 x^3 + a_2x^2 y + a_1 xy^2 + a_0 y^3$ be a binary cubic form, say with real coefficients. Put $H(x,y) = H_F(x,y)$ for the Hessian covariant of $F$, defined by

$$\displaystyle H_F(x,y) = \frac{1}{4} \begin{vmatrix} F_{xx} & F_{xy} \\ F_{xy} & F_{yy} \end{vmatrix},$$

and put

$$\displaystyle G(x,y) = G_F(x,y) = \begin{vmatrix} F_x & F_y \\ H_x & H_y \end{vmatrix}.$$

$H,G$ are both covariants of $F$ under the substitution action of $\operatorname{GL}_2$, and in particular, $H, G, F$ and the invariant $\Delta(F)$, the discriminant of $F$, generate the ring of polynomial covariants. They are connected by a single syzygy, given by

$$\displaystyle 4H(x,y)^3 + G(x,y)^2 = -27 \Delta(F) F(x,y)^2.$$

As can be verified by immediate calculation, we have

$$\displaystyle \Delta(G) = 729 \Delta(F)^3,$$

which is a perfect cube.

My question is, suppose that $G$ is a binary cubic form with integer coefficients satisfying $\Delta(G) = 729 n^3$ for some non-zero integer $n$. Does it follow that $G$ is the $G_F$-covariant for some binary cubic form $F$ with integer (rational) coefficients? If not, what is a counter-example, and what would be a stricter condition that guarantees $G$ is such a covariant?

On a related note, if $G'$ is the non-trivial cubic covariant of $G_F$, then $G' = -729 \Delta(F)^2 F(x,y)$.

• Shouldn't the involutive property you noted at the end answer your surjectivity question? – Abdelmalek Abdesselam Dec 28 '17 at 22:02

As Abdelmalek notes, if $G$ is to be the cubic covariant $G_F$ of another form $F$, and has a nonzero discriminant of the expected shape $\Delta(G)=3^6n^3$, then $F$ must divide $G'=G_G$ - indeed the only candidate (comparing discriminants and exploiting homogeneity) is $F=-G_G/(3^6n^2)$, which will have rational coefficients when $n$ is rational, and integral coefficients when (and only when) those of $G_G$ are integral multiples of $3^6n^2$.
And when this is the case, this $F$ does indeed work: $$G_F = -G_{G_G}/(3^6n^2)^3 = 3^6\Delta(G)^2G/(3^{18}n^6) = 3^6(3^6n^3)^2G/(3^{18}n^6) = G\,.$$
Not exactly an answer but some elaboration on the remark at the end of the OP's question. It seems to me that if an $F$ exists such that $G=G_F$ then taking the cubicovariant of $G$ should allow us to recover the candidate $F$ modulo a little bit of algebra/arithmetic.
The operation $F\rightarrow$ the cubicovariant $G_F$ essentially is an involution. Namely, the cubicovariant of the cubicovariant should be a numerical multiple of $F\times \Delta(F)^2$. See the Maple calculation below (using a routine for calculating transvectants courtesy of Jaydeep Chipalkatti):