Determine the ring of invariants $\mathbb C [x,y,z]^\Gamma$ for:
$$\Gamma :=\{ \begin{pmatrix} \pm1 & 0 & 0 \\ 0 & \pm1 & 0 \\ 0 & 0 & \pm1 \\ \end{pmatrix} \} \subset GL_3(\mathbb C).$$
So I read this as: what 3x3 invertible matrices (with columns x,y,z) multiplied by $\pm$ Identity stays the same? The only thing that stays the same when multiplied by both +1 and -1 is $0$. That's a very small ring. Am I misunderstanding this? How should I be tackling this problem?
On
Let $R$ be Reinolds average operator for the group $\Gamma$ $$ R=\frac{1}{8} \sum_{g \in \Gamma} g. $$ For example, we have
$R(x)=R(y)=R(z)=0,$
$R(xy)=R(yz)=R(xz)=R(xyz)=0,$
$R(x^2)=x^2, R(y^2)=y^2,R(z^2)=z^2.$
It is a proector from algebra $C[x,y,z]$ into the algebra $C[x,y,z]^{\Gamma}.$ Also, $C[x,y,z]^{\Gamma}$ is generated by invariants polynomials up to degree $8$ (the Noether's bound). So, the generating set of algebra of invariants $C[x,y,z]^{\Gamma}$ consists of all the elements $R(x^\alpha y^\beta z^\gamma)$, $\alpha+\beta+\gamma\leq 8.$
Also, for the operator $R$ we have that $R(f \cdot a)=a \cdot R(f),$ for any invariants $a \in C[x,y,z]^{\Gamma}.$
Now, by direct calculation you may get that all invariants of to degree $8$ has the form $x^{2i} y^{2j} z^{2k}, i+j+k \leq 4.$ Thus $C[x,y,z]^{\Gamma}=C[x^2,y^2,z^2].$
For the group of order $8$, the invariants are $\mathbb C[x^2,y^2,z^2]$.
For the group $\pm I$ of order $2$, the invariants are $\mathbb C[\text{ even degree polynomials }]=\mathbb C[x^2,y^2,z^2,xy,yz,zx]$.