My question might be an easy or could be a bit complicate and classic. Actually I am trying to understand why the discriminant of a binary quadratic form is a "the fundamental invariant" under $GL(2,\mathbb{Z})$-action i.e any other invariant is a polynomial of the discriminant. Also I am interested to know about general binary forms. More precisely I would like to know the number of fundamental invariant of binary form of degree n under $GL(2,\mathbb{Z})$-action.
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2$\begingroup$ @Ryan: Well, I guessed I have explained it. I want to know how many fundamental invariants exists for the $GL(2,\mathbb{Z})$-action on binary form of degree n. For example for binary quadratic and cubic form, there exists only one invariant which is the discriminant. I hope I have made it clear. $\endgroup$– M.BDec 13, 2010 at 5:49
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4$\begingroup$ Dear Ryan, I don't think the question is vague, although I don't have a good answer off the top of my head. $\endgroup$– EmertonDec 13, 2010 at 6:13
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2$\begingroup$ Dear Qiaochu, A fundamental invariant is a member of a generating set for the full ring of invariants. $\endgroup$– EmertonDec 13, 2010 at 6:14
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2$\begingroup$ @Denis Serre: if I may, I wonder whether your comment is predicated on a conflation of two different notions of "fundamental system of invariants". You seem to be intending it in the sense "a collection of properties of an object which collectively determine it up to isomorphism", whereas the OP means something very different: "the subring of elements fixed under the action of a group on a [polynomial, classically] ring". (Along with Emerton, I don't find the question to be vague. In fact, I would be interested in reading answers to it.) $\endgroup$– Pete L. ClarkDec 13, 2010 at 8:28
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4$\begingroup$ You should have a look at "Classical Invariant Theory" by Olver. $\endgroup$– Franz LemmermeyerDec 13, 2010 at 8:47
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1 Answer
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Finding the number of generators for the invariants of binary forms is a classic and very hard problem in invariant theory. For forms of small degree one can find a description in Hilbert's book on invariant theory (ISBN 978-0521449038). In the 19th century it was solved for forms of degree at most 8, and it has recently been pushed to degree 10 in http://dx.doi.org/10.1016/j.jsc.2010.03.002 using computer calculations (where one needs 106 generators). The latter link gives the history of the problem in more detail.
answered Dec 13, 2010 at 15:36
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4$\begingroup$ Perhaps one should add to this that the polynomial invariants for $\mathrm{GL}(2,\mathbb Z)$ are the same as for its Zariski closure which consists of the matrices whose determinant has a square equal to $1$. This connects up the question to classical invarant theory. $\endgroup$ Dec 13, 2010 at 19:26