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Let $R=k[u,v,w]$ and $p\in R$ be a cubic form. Let $G$ be the group of graded automorphisms of $R$ which preserve $p$, i.e., $G$ is the subgroup of $GL_3(k)$ consisting of elements $g$ such that $g(p) \in k p$. My question: is $G$ some well known algebraic group?

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    $\begingroup$ I assume that your form $p$ is irreducible. Your group $G$ contains the central subgroup $\mathbb{G}_m\subset \mathrm{GL}_3$ of homotheties; the quotient is the group of projective automorphisms of the curve $E$ defined by $p=0$, and this is well-known to be finite. $\endgroup$
    – abx
    May 12, 2014 at 12:29
  • $\begingroup$ @abx I think the group of automorphisms of a smooth elliptic curve $E$ as an algebraic variety is infinite (at least when $k$ is infinite), it has the group automorphisms as well as translations. $\endgroup$
    – Kenneth
    May 12, 2014 at 18:18
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    $\begingroup$ I was talking about the group of projective automorphisms of a plane cubic. $\endgroup$
    – abx
    May 12, 2014 at 18:31
  • $\begingroup$ @abx Thanks for the clarification. So by Steven's comment below this quotient is actually $\mathbb{Z}/3 \times \mathbb{Z}/3$ (in the case $E$ is smooth and $k=\bar{k}$). $\endgroup$
    – Kenneth
    May 12, 2014 at 22:29
  • $\begingroup$ Except that Steven forgot that there are other automorphisms than translations... You have to add a term $\mathbb{Z}/2$ in any case (so that the quotient is $(\mathbb{Z}/3)^2\rtimes \mathbb{Z}/2$), and in some particular cases a slightly larger cyclic group. $\endgroup$
    – abx
    May 13, 2014 at 4:45

4 Answers 4

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To elaborate on abx's comment: modding out by scalars, i.e., working in $PGL_3$ instead of $GL_3$, by definition the stabilizer of $p$ is the group of projective automorphisms of the curve $p=0$ which preserves the embedding of the curve.

If we assume that the curve is smooth and that there is a rational point, for example, if the field is algebraically closed, then this is an elliptic curve embedded by a degree 3 line bundle. For simplicity, assume $3 \ne 0$ in the field.

EDIT: (taking into account abx's comments) This stabilizer is the subgroup of automorphisms of the elliptic curve which preserves its 3-torsion. This includes translations by 3-torsion elements (which is a finite group abstractly isomorphic to $Z/3 \times Z/3$) but also automorphisms of the curve which preserve the identity This is generically $Z/2$ given by the inversion map but can be bigger for special elliptic curves. So we get a semidirect product of this automorphism group with the 3-torsion.

If we do not assume there is a rational point (but still assume the curve $C$ is smooth), then we can say that $C$ is a torsor for an elliptic curve $E$ and identify the stabilizer with the semidirect product of the automorphism group of $E$ preserving the identity with the 3-torsion $E[3]$ of $E$. The latter is a group scheme which is isomorphic to $Z/3 \times Z/3$ after passing to an algebraically closed field.

The extension of $E[3]$ by the scalars $G_m$ is an example of a Heisenberg group scheme, and one can use the Weil pairing to reconstruct it.

I'm not exactly sure what to say when $p$ is not smooth -- in the case of mild singularities probably one gets a similar description.

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I think so. This can be deduced from the classification of cubic forms in 3 varibles; see Tadayuki Abiko, Classification of cubic forms with three variables, Hokkaido Math. J. Vol. 10, 1981, 239-248. Google is the quickest way to find this paper which is freely available via Project Euclid.

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The best known situation of this type involving an algebraic group would occur in type $E_6$, where there is a long history and quite a bit of literature. Is this the "well known algebraic group" you have in mind? In case you have access to MathSciNet, you can find a list of literature cited by a fairly recent article:

MR2381940 (2008m:20077).
Vavilov, N. A.(RS-STPTMM); Luzgarev, A. Yu.(RS-STPTMM), The normalizer of Chevalley groups of type $E_6$. Algebra i Analiz 19 (2007), no. 5, 37-64; translation in St. Petersburg Math. J. 19 (2008), no. 5, 699–718.

As Premet notes, there is also more general literature about the classification of cubic forms and related groups, going back to the early work in invariant theory.

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If the field is $\mathbb{C}$ (or algebraically closed of characteristic $\not=2,3$), then you can put any smooth cubic into the Hessian form: $$X^3+Y^3+Z^3+\lambda XYZ=0$$ for some $\lambda\in \mathbb{C}$. This corresponds to put the nine inflection onto the intersection of the curve with $XYZ=0$.

Then, the pencil generated by the cubic and $XYZ$ is the Hessian pencil (which corresponds to let $\lambda$ vary in the equation) and is preserved by the classical Hessian group $G\subset \mathrm{PGL}(3,\mathbb{C})$, of order $216$. The groups acts on the set of parameters, paramertised by $\mathbb{P}^1$, and the kernel of this action contains the group $(\mathbb{Z}/3\mathbb{Z})^3\rtimes \mathbb{Z}/2\mathbb{Z}$ generated by the diagonal action $$[X:Y:Z]\mapsto [X:\theta Y:\theta^2Z]$$ $\theta^3=1$ and the group of permutations. This shows that you have generically this group as a group of automorphisms of $\mathbb{P}^2$ preserving the curve. For some values of $\lambda$, you have other automorphisms. Indeed, each of the elements of the Hessian group has two fixed points on $\mathbb{P}^1$.

I suggest to read the nice article "The Hesse pencil of plane cubic curves" of M. Artebani and I. Dolgachev on this subject http://arxiv.org/abs/math/0611590 .

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