Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

If $G \leq GL_2(\mathbb{C})$ is generated by matrix $\begin{pmatrix} 1 & 2\\0 & -1 \end{pmatrix}$, acting on the polynomial ring $\mathbb{C}[X,Y]$, then how can we find the ring of invariants $\mathbb{C}[X,Y]^G$?

I've got it in the form $\{f \in \mathbb{C}[X,Y] : f(x+2y,-y)=f(x,y)\}$, but I think a nicer form is required. Maybe something to do with eigenvectors? Is the $G$-invariance of the zero set $Z(f)$ relevant?

Thanks for any help with this!

share|cite|improve this question
I presume what the question wants is a description via generators and relations. Are you familiar with doing this using Molien's theorem? – Qiaochu Yuan Mar 30 '12 at 18:59
@Qiaochu Yuan - no, but since the eigenvalue-eigenvector pairs are $\big(1,\begin{pmatrix} 1\\0 \end{pmatrix}\big)$ and $\big(-1,\begin{pmatrix} 1\\-1 \end{pmatrix}\big)$, does this mean the answer is $\mathbb{C}[X,(X-Y)^2]$? (This looks right but I can't see how to prove it!) – Harry Macpherson Mar 30 '12 at 19:09
Your finite group $G$ is of order 2 (square your upper triangular matrix to get the identity matrix) and all $G$ orbits look like lines (on the complex $2$-plane) of the form $Y = -X/2 + c$, where $c\in\mathbb{C}$. So isn't your ring of invariance just $\mathbb{C}[X]$? – math-visitor Jun 5 '12 at 22:46
@math-visitor given that the group is order 2, the $G$-orbits can contain at most two points each, so they are not lines. – Ben Blum-Smith Oct 5 '15 at 21:06

Your Answer


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

Browse other questions tagged or ask your own question.