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First, let me formulate the question in the language of elliptic curves. Almost every elliptic curve has a Weierstrass normal form $y^2=x^3+px+1$, but this form is bad from the point of view of invariant theory, and I have heard that there is another normal form written in terms of symmetric polynomials.

Second, from the point of view of invariant theory, the group $\operatorname{SL}_3$ acts standardly on $S^3\mathbb C^3$, there is an open invariant subset $W \subset S^3\mathbb C^3$ and some subset $$U=\{ y^2z-x^3-pxz^2-qz^3 \} \subset S^3\mathbb C^3,$$ such that any $\operatorname{SL}_3$-orbit in $W$ intersects $U \cap W$ exactly once. Namely, $W$ consists of polynomials that define smooth irreducible curve in $\mathbb{CP}^3$, the intersection $U \cap W$ is defined by non-vanishing of the discriminant of $x^3+px+q$, and its elements are called Weierstrass normal forms.

Then the invariants $p$ and $q$ can be defined via some complicated contractions of tensors involving many indices, but it gives little insight. Could you help me with an anologous normal form, but involving symmetric polynomials, and its invariants? I am in particular interesting in the case of $\operatorname{SL}_3$, not just $\operatorname{GL}_3$. It is probably very classical subject, but I can not find the answer, so any reference is welcome!

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    $\begingroup$ I think the invariants you seek are the same ones denoted $I(f)$ and $J(f)$ here, maybe up to some scaling factor - do you find this definition more conceptual? arxiv.org/pdf/1007.0052v2.pdf $\endgroup$
    – Will Sawin
    Dec 24, 2016 at 14:13
  • $\begingroup$ @Evgeny: 1) why so much hate for contractions of tensors? 2) what do you mean by symmetric polynomials? or rather symmetric in which variables? For binary forms one can write invariants as symmetric polynomials in the roots, but here you are dealing with a ternary form. Unless you already have a reduction to the form $y^2=$ quartic in $x$, I don't see how to bring symmetric polynomials into the picture. In any case it is hard to understand from the phrasing of the question what you really want. $\endgroup$ Dec 26, 2016 at 17:16
  • $\begingroup$ @AbdelmalekAbdesselam, in all, like $a(x^3+y^3+z^3)+bxyz$ (the simplest candidate for a representative of a general $SL_3$-orbit). $\endgroup$
    – evgeny
    Dec 26, 2016 at 20:16

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I think what you're looking for are the standard invariants (and covariants) of the space of non-singular cubic polynomials $$ f(x,y,z) := ax^3+by^3+cz^3+dx^2y+exy^2+fx^2z+gxz^2+hy^2z+iyz^2+jxyz $$ relative to the natural action of $\text{SL}_3$. If you're looking for formulas, they're in (among other places) Lessons introductory to the modern higher algebra, George Salmon, 1876 (3rd ed). As Will Sawin noted, these invariants are often denoted by $I(f)$ and $J(f)$. There are also covariants which can be used to define a finite map from the curve $f(x,y,z)=0$ to the elliptic curve $$Y^2Z=X^3+I(f)XZ^2+J(f)Z^3.$$

Final quibble: It is not true that "Every elliptic curve has a Weierstrass normal form $y^2=x^3+px+1$," although that is almost true over algebraically closed fields of characteristic 0. But even in that case, I think you're missing the isomorphism class of the curve $y^2=x^3+x$, i.e., the curve with endomorphism ring $\mathbb Z[i]$.

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