6
$\begingroup$

Suppose I am looking at $GL(4,K)$ acting on a cubic form in say four variables $x,y,z,w$ over $K$ via the usual induced action on a polynomial. Does anyone know what is/where I can find how to compute the ring of invariants? The case of personal interest is when $K$ is a finite field but the the answer over $\mathbb{C}$ would of course be more than useful.

$\endgroup$

3 Answers 3

5
$\begingroup$

According to Dolgachev in http://arxiv.org/abs/math/0408283 the ring of invariants is generated by 6 invariants of degrees 6, 16, 24, 32, 40 and 100 the last one being a polynomial in the other ones. In addition to the references to Salmon and Clebsch indicated by Dolgachev, I would also look up the book by Salmon "A treatise on the analytic geometry of three dimensions" pp 392-400. It is downloadable from google books: http://books.google.com/books?id=LTgDAAAAQAAJ&printsec=frontcover#v=onepage&q&f=false The article http://arxiv.org/abs/1006.0721 also contains some references which might help.

$\endgroup$
2
  • $\begingroup$ These sources are really useful, thanks. In the last paper it gives explicit formulas for the fundamental invariants in terms of elementary symmetric polynomials in the coefficients of the surface in Sylvester normal form (which is what is proved in the various sources I suppose). Can something similar be done over finite fields? (i.e. Are there similar normal forms which one can then write down the ring of invariants in terms of) $\endgroup$ Jul 4, 2011 at 20:02
  • $\begingroup$ I am not aware of an analogous theorem in char p. I don't see other approach except carefully following the Salmon/Clebsch proof and seeing if it adapts to char p. $\endgroup$ Jul 4, 2011 at 21:56
4
$\begingroup$

Don't be misled into thinking that the answer over $\Bbb C$ tells you very much about the answer over your finite field $K$. The space of cubic forms in 4 variables is 20 dimensional. The group $GL(4,K)$ is a finite group and so the ring of invariants you seek has Krull dimension 20. In particular, it has at least 20 generators. Probably, however, it has a few thousand generators if not many more.

There are computer algebra packages (e.g. Magma) that include routines that can compute generators for your ring of invariants in principle, but in practice you will find that they run out of memory long before getting very far on a problem of this size. They will be able to compute vector space bases for the invariants in low degrees. On the other hand, it is doubtful that having a list of thousands of generators (or even just their degrees) for the ring of invariants will help you very much. On the other hand if you want to know something about the properties of the ring of invariants as a ring, then it's possible that may be known.

$\endgroup$
1
$\begingroup$

There is also a nice book by Sturmfels called Algorithms in Invariant Theory that discusses the problem of finding invariants. It may be worth a look.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.