Given the set $P$ of polynomials of degree $3$ in $2$ variables.
I revised the definition of group action on $P$, but I do not see any reason in the definition that makes us say that $g.(pq) = (g.p)(g.q),$ could anyone explains this to me please?
EDIT:
I am asking about the answer I received in this question:
Why are the linear factors of $g.p$ are given by $(a_i, b_i)g^{-1}$?
so my question now (after the discussion in the comments) is, is the action of G on $Sym^3(\mathbb C^2)$ distributive over addition?
In the linked paper, they define the group action in question immediately before Remark 2.5.1: $$(g\cdot p)(x,y)=p(g^{-1}(x,y))$$ That is, compute $g^{-1}\cdot(x,y)$ via the usual action (see below) of $\mathrm{GL}(\mathbb{R}^2)$ on $\mathbb{R}^2$; then evaluate $p$ at the new point.
In comments to the answer to the linked question, you ask what it means to multiply a matrix by an ordered pair. Simple: treat it like a vector; that is, $$\mathcal{A}\cdot(x,y)=\mathcal{A}\begin{bmatrix}x\\y\end{bmatrix}$$ I call this the "usual action."
For example, suppose $p(x,y)=x^2+2xy$ and $g=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. Then $g^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}$, so that $$g^{-1}\cdot(x,y)=(x-y,y)$$ Thus \begin{align*} (g\cdot p)(x,y)&=p(x-y,y) \\ &=(x-y)^2+2(x-y)y \\ &=x^2-y^2 \end{align*}
This operation preserves addition and multiplication, because evaluation of polynomials is a homomorphism of rings. For example, if $g$ is as above, then \begin{align*} (g\cdot pq)(x,y)&=(pq)(x-y,y) \\ &=p(x-y,y)q(x-y,y) \\ &=(g\cdot p)(x,y)(g\cdot q)(x,y) \end{align*}