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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?

Thomas Andrews
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weird
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    Presumably, $p$ and $q$ are element of the set, and $g$ is the element of a group. But then what would $pq$ mean, since the set is just a set? – Thomas Andrews Jul 22 '22 at 00:06
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    This is meaningless because there's not necessarily any operation defined on the members of the set being acted upon. – Robert Shore Jul 22 '22 at 00:06
  • @ThomasAndrews, yes for your first comment – weird Jul 22 '22 at 00:09
  • you are correct but the set is the set of polynomials of degree 3 in 2 variables @ThomasAndrews – weird Jul 22 '22 at 00:09
  • @RobertShore you are correct, but the set is the set of polynomials of degree 3 in 2 variables – weird Jul 22 '22 at 00:11
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    Don't you think that is something worth mentioning in the question? It still doesn't make sense, because of $p,q$ are polynomials of degree $3$ or less, then $pq$ is not. Seems like there are big holes in your question. – Thomas Andrews Jul 22 '22 at 00:12
  • I think in my case, the vector space of homogeneous polynomials of degree 3 in 2 variables – weird Jul 22 '22 at 00:12
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    Then why use $pq$ rather than $p+q?$ A key point in writing mathematics is to write as if you don't think the people reading what you have written can read your mind. Edit your question and include missing information. Answerers should not have to read the comments to understand your question. – Thomas Andrews Jul 22 '22 at 00:13
  • @ThomasAndrews I am asking this question because of the answer I got in this question https://math.stackexchange.com/questions/4496314/why-are-the-linear-factors-of-g-p-are-given-by-a-i-b-ig-1 – weird Jul 22 '22 at 00:14
  • @ThomasAndrews that is how it was written in the paper I mentioned in the question ... ok, so my question now is, is the action of G on $Sym^3(\mathbb C^2)$ distributive over addition? – weird Jul 22 '22 at 00:16
  • Stop telling me what you are saying, and write it in the question. – Thomas Andrews Jul 22 '22 at 00:17
  • "... but the set is the set of polynomials of degree 3 in 2 variables" Don't put that information in the comments, edit your question so it makes sense. – jjagmath Jul 22 '22 at 00:19
  • It is still unclear what you mean by $pq,$ since the set you've given is not closed under multiplication. Also, what do you mean by "I revised the definition...?" As normally written, that means changed, but you don't tell us what definition you saw previously and how you revised it. – Thomas Andrews Jul 22 '22 at 00:50
  • @ThomasAndrews I think I referred you to an answer that contains $pq$ in it when I provided the link above. so I do not know the intension of the author there – weird Jul 22 '22 at 04:56
  • @ThomasAndrews: "Revised" is British English for reviewed. – Jacob Manaker Jul 22 '22 at 07:56

1 Answers1

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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*}

Jacob Manaker
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