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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 $C= \langle A\rangle$ is a cyclic, finite group of order $4$.

Suppose $A$ acts on $\mathbb{C}[x,y]$ linearly.

Then what is the subring $\mathbb{C}[x,y]^C$ of invariant functions in $\mathbb{C}[x,y]$? What is the basic strategy?

Note that $$ C = \left\{ \left( \begin{array}{cc} 0 & -1 \\ 1& 0 \\ \end{array} \right), \left( \begin{array}{cc} -1 & 0 \\ 0& -1 \\ \end{array} \right), \left( \begin{array}{cc} 0 & 1 \\ -1& 0 \\ \end{array} \right), \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \right\}. $$


$\mathbf{Question \;2}$. Now, suppose the dihedral group $D_6 = \langle \rho, \psi : \rho^6 = \psi^2 =e,\psi \rho\psi^{-1}=\rho^{-1} \rangle$ acts on $\mathbb{C}[x,y,z]$, with the action defined by the matrices $$ \rho = \left( \begin{array}{ccc} 1/2 & -\sqrt{3}/2 & 0 \\ -\sqrt{3}/2 & 1/2 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \mbox{ and } \psi = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & -1 \\ \end{array}\right). $$

Then what is $\mathbb{C}[x,y,z]^{D_6}$?


$\mathbf{Question \;3}$. What is the general strategy, if we have something like the subgroup generated by $B$ and $-B$ in $GL_3(\mathbb{C})$ acting on a polynomial ring $\mathbb{C}[x,y]$ of only two variables, where $$ B = \left( \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & -1 \\ 0 & 0& 1 \\ \end{array} \right)? $$

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    The basic strategy I'm aware of is to write down some invariant polynomials and then use Molien's theorem to check that you've found a set of generators.2012-07-14
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    I agree with Qiaochu. There is a very nice book which treats this material: Mara Neusel's *Invariant Theory*. See e.g. http://www.amazon.com/Invariant-Theory-Student-Mathematical-Library/dp/08218413272012-07-14
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    @QiaochuYuan: Thanks for the reply!2012-07-14
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    @PeteL.Clark Thanks for the reply! I studied group theory but I just realized that I've never worked with finite groups acting on a polynomial ring before. That's why I'm asking these questions. And I just learned how to "ping" people, and I think you're supposed to ping one person at a time.2012-07-14

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