Can a general binary sextic form be put into the following form?

Question

Let $F(x,y) = a_6 x^6 + a_5 x^5 y + \cdots + a_0 y^6$ be a binary sextic form with real coefficients and non-zero discriminant. Can one always find an element $U = \begin{pmatrix} u_1 & u_2 \\ u_3 & u_4 \end{pmatrix}$ in $\operatorname{GL}_2(\mathbb{R})$ such that the form

$$\displaystyle F_U(x,y) = F(u_1 x + u_2 y, u_3 x + u_4 y)$$

is of the shape

$$\displaystyle F_U(x,y) = A_6 x^6 + A_4 x^4 y^2 + A_2 x^2 y^4 + A_0 y^6?$$

If the answer is yes, can one give a criterion to decide whether $U$ can be chosen to have coefficients in $\mathbb{F}$, which is the smallest subfield of $\mathbb{R}$ containing all coefficients of $F$?

Answer 1

You may know this already, but, if not, the following may be helpful to you:

The ring of invariants for binary sextics is generated in degrees 2, 4, 6, 10, and 15. (See this reference on binary sextics.) Thus, in your notation (at least when the field $\mathbb{F}$ has characteristic $0$) for $k=2,4,6,10, 15$ there exists a polynomial $Q_k:S^6(\mathbb{F}^2)\to \mathbb{F}$ that is homogeneous of degree $k$ that satisfies $$ Q_k\bigl(F_U(x,y)\bigr) = \det(U)^{3k}\,Q_k\bigl(F(x,y)\bigr)\tag1 $$ for all $U\in\mathrm{GL}(2,\mathbb{F})$, and, moreover, every polynomial $Q: S^6(\mathbb{F}^2)\to \mathbb{F}$ that satisfies $$ Q\bigl(F_U(x,y)\bigr) = Q\bigl(F(x,y)\bigr) $$ for all $U\in\mathrm{SL}(2,\mathbb{F})$ is, in fact, a polynomial in $\{1,Q_2,Q_4,Q_6,Q_{10},Q_{15}\}$. It is known that ${Q_{15}}^2$ is a polynomial in $\{Q_2,Q_4,Q_6,Q_{10}\}$ and that this is the only relation among these polynomials. Consequently, the polynomials $Q_{k}$ are irreducible.

Taking $U = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$, one sees that when $F(x,y)$ is in the desired form, one has $F_U(x,y) = F(x,y)$. Conversely, $F(x,y)$ can be put into the desired form over the algebraic closure $\mathbb{F}^+$ of $\mathbb{F}$ if and only if there exists a $U\in\mathrm{GL}(2,\mathbb{F}^+)$ satisfying $U^2 = I$ and $U\not=I$ such that $F_U(x,y) = F(x,y)$. Using (1) with $k=15$, one finds that the existence of such a $U$ for a given $F(x,y)$ implies $$ Q_{15}\bigl(F(x,y)\bigr) = 0. $$ Thus, the hypersurface $Q_{15}=0$ in $S^6(\mathbb{F})$ must be an irreducible component of the hypersurface $L\subset S^6(\mathbb{F})$ consisting of sextics that can be written in the desired form after a linear change of variables.

Meanwhile, it seems likely that $L$ is irreducible, since, over $\mathbb{F}^+$ anyway, we know, by Noam's observation/comment that it can be 'parametrized' by a connected algebraic variety. Assuming this detail, it would follow that the necessary and sufficient condition for $F(x,y)$ to be written in the above form over $\mathbb{F}^+$ is that it lie in the zero locus of $Q_{15}$. (I don't know how to tell when it can be done over $\mathbb{F}$, but perhaps this can be deduced by knowing something about the values $Q_k\bigl(F(x,y)\bigr)$ for $k=2,4,6,10$.)

Answer 2

This can be done only if the corresponding genus 2 curve has an extra involution. That is equivalent with $\chi_{35}=0$ or $J_{30}=0$, both $SL_2(\mathbb C)$ invariants given in terms of the coefficients of the sextic. In such case, the curve is defined over its field of moduli.

Answer 3

There is an old method of Clebsch called typical representation which shows how to write a sextic in canonical form with coefficients $A$ given by invariants. I did not look at it in detail but my colleague Llyod West did in his recent work "The Moduli Space of Cubic Rational Maps". See in particular the reference therein to the paper by Mestre.