MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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 '11 at 6:17
When the degree is greater than 2, it does. – algori May 30 '11 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 '11 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 '11 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 '11 at 23:41

You should check the case of binary sectics since that is the best understood. The $GL_2(\mathbb C)$-invariants are called absolute invariants and denoted by $i_1, i_2, i_3$. Two binary sectics are projectively equivalent if they have the same absolute invariants; see the following for details.

  1. Invariants of Binary Forms, V. Krishnamoorthy, T. Shaska, H. Voelklein
  2. Jun-ichi Igusa, Arithmetic variety of moduli for genus two. Ann. of Math. (2) 72 1960 612–649.

For the case of binary octavics such absolute invariants are also explicitly known. There are $t_1, \dots , t_6$ invariants which satisfy some algebraic relation among them. Their definitions can be found in

  1. T. Shaska, Some remarks on the hyperelliptic moduli of genus 3, Communications in Algebra, Volume 42, Issue 9, September 2014, pages 4110-4130.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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