For $\mathbb{C}^*$-action on $\mathbb{C}^2$ by $t\circ(x,y)=(t^{-1}x,ty)$, the ring of invariant polynomials is $\mathbb{C}[xy]$.
For $\mathbb{C}^*$-action on $\mathbb{C}^4$ by $t\circ(x,y,z,w)=(t^{-1}x,t^{-1}y,tz,tw)$, then the ring of invariants is $\mathbb{C}[xz,xw,yz,yw]$, which is a quadratic cone since there is a relation among the generators: $\mathbb{C}[X,Y,Z,W]/\langle XW-YZ\rangle$.
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For $G = GL_2(\mathbb{C})$ acting on $M_2(\mathbb{C})$ by conjugation ($g\circ m=gmg^{-1}$), the ring of invariants is $\mathbb{C}[\det(m),\textrm{tr}(m)]$.
For $G$ acting on $M_2(\mathbb{C})\times M_2(\mathbb{C})$ by $g\circ (m,n)=(gmg^{-1},gng^{-1})$, why is it that the ring of invariant polynomials under the group action is only $\mathbb{C}[\textrm{tr}(m),\det(m),\textrm{tr}(n),\det(n)]$? More precisely, why doesn't there exist other $G$-invariants using a combination of $m$ and $n$?
If we restrict $G$-action to $S= \{ (m,n): mn=nm\} \subset M_2\times M_2$, then is it true that $\mathbb{C}[S]^{G} = \mathbb{C}[\textrm{tr}(m),\det(m),\textrm{tr}(n),\det(n)]$?