## Classical invariant theory: absolute rational invariants and $GL(2)$-orbits

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I have a question concerning classical invariant theory. Consider binary $n$-forms (i.e. all homogeneous polynomials of degree $n$ of two variables) over the field of complex numbers. Clearly, the group $GL(2,C)$ acts on the space of all such forms by changes of the variables. A classical relative invariant is a polynomial function $I$ in the coefficients of the form such that under the $GL(2,C)$ action the value of $I$ changes only by multiplication by $\det C$ to some power $k$ ($k$ is called the weight of $I$). One can now form rational absolute invariants by taking ratios of relative invariants of equal weights.

My question is: $GL(2,C)$-orbits of what forms can be distinguished by such rational absolute invariants? How about forms with non-zero discriminant, for example? I have found some classical results by Clebsch of the 19th century and a result by Olver of 1990, but they do not quite give the result that I want. Also, Geometric Invariant Theory seems to deal only with $SL(2,C)$-actions. For $SL(2,C)$-actions the orbits can be distinguished just by polynomial invariants, but this is a completely different situation.

In some cases (e.g. for quintics) I can prove what I need, but I am wondering if there is perhaps a general result.

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## 1 Answer

The space of non-degenerate binary forms is an affine variety, since it is the complement of a hypersurface in an affine space. The reductive group $GL_2(\mathbb{C})$ acts on it with finite stabilizers, so the quotient is again affine, and its elements are distinguished by regular functions, which lift to functions of the form $\frac{f}{\triangle^k}$ on the space of binary forms. Here $f$ is a polynomial, $\triangle$ is the discriminant and $k$ is a non-negative integer.

If, on the other hand, we allow two roots to coincide, the quotient will be projective, and there will be no functions on it at all.

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Thank you for that, but does $GL_2({\Bbb C})$ really act with finite stabilizers? For example, for the quadratic form $xy$ all the maps $x\mapsto c x$, $y\mapsto 1/c y$ are in the stabilizer. – Alexander Isaev May 30 2011 at 6:17
When the degree is greater than 2, it does. – algori May 30 2011 at 6:36
The easiest way to see this is this: if we projectivize the space of non-degenerate forms, we get an unordered configuration space of $\mathbb{P}^1$. Now, a subgroup of $PGL_2(\mathbb{C})$ of positive dimension can not preserve a subset with more than 2 elements. – algori May 30 2011 at 6:44
I have thought about what you said, and I now agree. I am wondering if you let me mention this fact in one of my papers. Can I refer to "personal communications" with you? If you do not mind my doing that, what is your name? Thank you again for your answer. – Alexander Isaev May 30 2011 at 23:00
Dear Alexander -- of course I don't mind if you mention this conversation in a paper. Re my name: I will shortly send you an email. – algori May 30 2011 at 23:41
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