2
$\begingroup$

Let $V_d$ be the complex vector space of binary forms of degree $d$ endowed with the natural action of the special orthogonal group $SO(2).$ Consider the corresponding action of the group $SO(2)$ on the coordinate ring $\mathbb{C}[V_d]$ and let $\mathbb{C}[V_d]^{SO(2)}$ the algebra of $SO(2)$-invariant polynomial functions.

What we know about the algebra $\mathbb{C}[V_d]^{SO(2)}$: minimal generating set, Poincare series, ... ?

In the case of the group $SL_2$ the situation is classical and algebra of invariants of binary form is well-studied but what about binary form invariants of the group $SO(2)?$

I hope the problem was already solved in 19th century. Can anyone tell me a good reference?

Improve this question
$\endgroup$
5
  • $\begingroup$ $SO(2)$ is Abelian so all its irreducible representations are abelian. You now need to be more specific about how you define $V_d$: are they polynomial maps $\mathbb{R}^2\to\mathbb{C}$ or are they polynomial maps $\mathbb{C}^2\to\mathbb{C}$? In the first case the only invariants are functions of $r^2=x^2+y^2$. In the second case there are no invariant polynomials in the complex variables $z_1,z_2$. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 11, 2018 at 19:57
  • $\begingroup$ @LiviuNicolaescu, all its irreducible representations are Abelian = all its irreducible representations are 1-dimensional, of course. $\endgroup$
    – LSpice
    Commented Dec 11, 2018 at 20:10
  • $\begingroup$ @LSpice Yes of course. But the span of the monomial $z_1^mz_2^n$ is an irreducible representation with weight $m+n$. On the other hand, if we allow complex conjugate variable then the span of $x=z^m\;\bar{z}^n$ has weight $m-n$ so we get invariant monomials when $m=n$. Thus the need to precisely define $V_d$. $\endgroup$
    – Liviu Nicolaescu
    Commented Dec 11, 2018 at 20:18
  • 1
    $\begingroup$ @LiviuNicolaescu, right; I was just commenting on an obvious typo (for the poster's benefit), not arguing about mathematics. $\endgroup$
    – LSpice
    Commented Dec 11, 2018 at 20:38
  • 1
    $\begingroup$ @Liviu Nicolaescu. I am talking about polynomial function invariant, so any such invariant is a polynomial of coefficients of the binary form. So $x^2+y^2$ is not invariant but the discriminant $b^2-ac$ of the binary form $ax^2+2bxy+cy^2$ is a sample of such invariants $\endgroup$
    – Leox
    Commented Dec 11, 2018 at 21:44

0

Reset to default

You must log in to answer this question.