3
votes
1answer
50 views

Polynomial with given group of symmetries

Let $f$ be a polynomial in $n$ variables and $G$ - its group of symmetries (group of permutations of variables wich left $f$ in place). I'm trying to such $f$ for given group $G$. I have troubles when ...
1
vote
0answers
29 views

From linear invariants of group to general ones

There is a lot of information about classical/linear invariants of finite groups. But does it lead to general invariants of group (for example, when we consider some action of our group on finite ...
2
votes
2answers
90 views

Is the ring of polynomial invariants of a finite perfect group an UFD?

Let $G$ be a finite group. $G$ acts on $\mathbb K[x_1,...,x_n]$ by automorphisms fixing $K$. $\mathbb K[x_1,...,x_n]^G=\{ T\in \mathbb K[x_1,...,x_n],\forall \sigma \in G, T^{\sigma}=T\}$ is the ring ...
7
votes
1answer
121 views

Invariants of binary forms under a $\begin{pmatrix} 1& 1 \\ 0& 1 \end{pmatrix}$ action

The special linear group $\text{SL}_2(\mathbb{Z})$ of $2\times 2$ invertible matrices in $\mathbb{Z}$ acts on binary cubic forms $\{ax^3 + bx^2y + cxy^2 + dy^3\}$ by acting on the vector $(x,y)^T$. ...
1
vote
0answers
27 views

Degrees of parabolic subgroups

Suppose a finite reflection group $G$ has the degrees $d_1,\ldots,d_n$. Let $G^*$ be a parabolic subgroup of $G$. What are the degrees of $G^*$. Since $|G^*|$ divides $|G|$ it is clear that the ...
0
votes
1answer
26 views

Is it possible to have a point $P_1$ not $\chi$-semistable but $P_2$ $\chi$-semistable with these two points in the same orbit?

Let $G$ be a group acting on an affine variety $X\subseteq \mathbb{A}_{\mathbb{C}}^n$. Suppose $P_1$ and $P_2$ are two points in $X$ such that $g\circ P_1=P_2$ for some $g\in G$. This means that ...
2
votes
0answers
58 views

Computing $\mathbb{C}[x,y]^G$ or $\mathbb{C}[x,y,z]^G$ where $G$ is a finite subgroup of $GL_n(\mathbb{C})$

My question is related to this link: Ring of Invariant $\mathbf{Question \;1}$. Let $$ A = \left( \begin{array}{cc} 0 & -1 \\ 1& 0 \\ \end{array} \right). $$ Then $C= \langle A\rangle$ ...
8
votes
1answer
168 views

Any affine algebraic group is linear.

It is a well-known result that any affine algebraic group is a closed subgroup of some $\mathrm{Gl}_n(\Bbbk)$. However, I would like to see a proof for that, so I looked it up in various books, more ...
1
vote
0answers
62 views

$S_k$ action on $A/I$

Let $S_2$ be a finite group of order $2$ and let $S_2$ act on $k[x,y]$ by interchanging $x$ and $y$, where $k=\overline{k}$. Then since $$ R = \left( \dfrac{k[x,y]}{(x+y)} \right)^{S_2} = ...
3
votes
1answer
110 views

Invariant rings $\mathbb{C}[X,Y]^{GL_2}$ and $\mathbb{C}[X,Y]^{SL_2}$

I feel like I might have made a mistake on this question, and would appreciate some feedback from someone more experienced than me. If $G$ acts on $S$, we write $S^G = \{s \in S: gs = s \, \forall ...