3

In some recent computation I came across certain cubic forms and was wondering about analogue of following result for quadratic forms.

If $k^*/(k^*)^2$ is finite then there are only finitely many quadratic forms up to equivalence. This can be easily seen because of diagonalisation.

I have been wondering if we can say following: given $k^*/(k^*)^3$ is finite is it true that there are only finitely many cubic forms up to equivalence. Or can I say something about being isotropic.

Anupam Singh
  • 661
  • 1
    This sees obviously wrong: cubic curves have moduli, even over $\mathbb{C}$, – Alex Degtyarev May 15 '16 at 11:56
  • An argument is that the dimension of symmetric trilinear forms is $n(n+1)(n+2)/6$, which is $>n^2=\dim(\mathrm{GL}_n)$ as soon as $n\ge 3$. This only leaves some hope for $n=2$; I haven't checked but certainly somebody here knows. – YCor May 15 '16 at 12:09
  • In addition, finiteness for cubic forms implies the same for quadratic forms: consider the forms $QL$ where $Q$ is quadratic and $L$ is linear. – Laurent Moret-Bailly May 15 '16 at 14:19
  • Not all cubic forms can be diagonalized. For example, in $3$ variables a (nondegenerate) diagonalized cubic form over $\mathbf C$ is equivalent to $x^3 + y^3 + z^3$, which can be transformed by an invertible change of variables into the Weierstrass form $y^2z - x^3 + 432z^3$, whose zero set is an elliptic curve over $\mathbf C$ with $j$-invariant $0$. So the Weierstrass equation of any elliptic curve with nonzero $j$-invariant provides an example of a cubic form in $3$ variables that can't be diagonalized. A specific example is $y^2z - x^3 + xz^2$. – KConrad May 15 '16 at 14:56
  • For the case $n=2$ over $\mathbf C$ see the comment by Frank Thorne at http://mathoverflow.net/questions/77702/transformation-of-a-cubic-form. – KConrad May 15 '16 at 14:58
  • Continuing my first comment (ran out of time to edit it), for each $j \in \mathbf C$ there is an ell. curve over $\mathbf C$ with $j$-invariant $j$, so there are infinitely many inequivalent cubic forms over $\mathbf C$ in $3$ variables. – KConrad May 15 '16 at 15:03
  • I assumed the question was about binary cubic forms, which have no moduli. – Noam D. Elkies May 15 '16 at 16:02
  • @LaurentMoret-Bailly, is it obvious that $Q L$ is equivalent to $Q'L'$ only if $Q$ is equivalent to $Q'$? – LSpice May 15 '16 at 17:11
  • @KConrad thanks for the link. The argument indeed shows, if I understand correctly, that $GL_2(\mathbf{C})$ has exactly 4 orbits of representatives of homogeneous polynomials of degree 3, with representatives given by $x^3+y^3$, $x^3+x^2y$, $x^3$, and 0. The question of describing for which fields the classification yields finitely many cases is probably doable. – YCor May 16 '16 at 11:57

1 Answers1

3

To expand a bit on the comments. For simplicity, let's assume $k$ is characteristic $0$ (or at least, not characteristic 2 or 3). Let $f(X,Y)=aX^3+bX^2Y+cXY^2+dY^3$ be a cubic form. The case where $f$ is reducible comes down to classifying quadratic forms and linear forms, so let's assume that $f$ is irreducible. More generally, assume that $f(X,1)$ has distinct roots $e_1,e_2,e_3\in\overline k$. Let $\lambda=(e_1-e_2)/(e_1-e_3)$. Then the quantity $j(f)=(\lambda^2-\lambda+1)^3/\lambda^2(\lambda-1)^2$ is an invariant of $f$. And conversely, if $j(f)=j(f')$, then there is a transformation defined over $\overline k$ taking $f$ to $f'$. So what you need to do is fix a value $j_0\in k$ and then look at all (separable) $f\in k[X,Y]$ satisfying $j(f)=j_0$. Fixing one such $f_0$, you should then be able to classify the $\overline k/k$-twists of $f_0$ using Galois cohomology in the usual way.

Joe Silverman
  • 45,660
  • 1
    $\lambda$ and $j$ are not well-defined unless you fix the point at infinity. Under ${\rm PGL}_2({\bar k})$ all binary cubic forms with distinct roots are equivalent. – Noam D. Elkies May 15 '16 at 16:04
  • @NoamD.Elkies Right you are. I was thinking of cubic forms up to equivalence by affine transformations. I think that what I said is okay for affine equivalence. But you are, of course, right that for full PGL equivalence, all separable binary cubic forms are equivalent. – Joe Silverman May 15 '16 at 17:43