Is every linear representation of a group $G$ on $k[x_1,\dots,x_n]$ a dual representation?

Question

Let $\rho\colon G\to GL(V)$ be a linear representation of $G$ on a $k$-vector space $V$. The dual representation is

$$G\to GL(V^*),\quad g\mapsto(\varphi\mapsto\varphi\circ\rho(g^{-1})).$$

By the same rule, we get a linear representation on the coordinate ring $k[V]$ of $V$, which is the $k$-algebra generated by $V^*$, with pointwise multiplication. Now if $V$ is finite-dimensional, $e_1,\dots,e_n$ the standard basis, and $x_1,\dots,x_n$ the dual basis, then $k[V]\cong k[x_1,\dots,x_n]$. Hence there is a linear representation $\bar\rho$ of $G$ on $k[x_1,\dots,x_n]$.

My question is: Is there a way to go "back"? That is, given a linear action of a group $G$ on $k[x_1,\dots,x_n]$, does it come from a linear representation of $G$ on an $n$-dimensional vector space $V$?

More generally, given a linear representation $\bar\rho:G\to GL(k[V])$, does it come from a representation $G\to GL(V)$?

Edit: As an example, consider $G\subseteq GL_n(k)$ acting on $k[x_1,\dots,x_n]$ via $(A,f)\mapsto f(A\cdot x)$. Does this action come from a linear representation of $G$ on $k^n$? I could consider $G$ acting on $k^n$ by multiplication, but the action on the coordinate ring corresponding to this should be $(A,f)\mapsto f(A^{-1}\cdot x)$, or am I mistaken here?

This is motivated by the fact that in invariant theory, one studied actions of subgroups $G\subseteq GL_n(k)$ on the polynomial ring $k[x_1,\dots,x_n]$ as above. But many theorems or facts in invariant theory are formulated in the language of representations, and the invariant ring there is $k[V]^G$. Now if I know that under certain circumstances $k[V]^G$ is finitely generated, I wanted to "come back" to the initial problem of linear actions on $k[x_1,\dots,x_n]$. But I can't really connect these two yet. As written above, any representation of $G$ on some f.d. vector space $V$ induces an action on the coordinate ring $k[V]$, but why do these include the above actions on the polynomial ring? And if they do, how does the initial action on $k^n$ look, given $\rho:G\to GL(k[V])$?

Edit5: I finally found a way to hopefully make clearer what I'd like. Let $G\subseteq GL_n(k)$. Can every linear $G$-action on $k[x_1,\dots,x_n]$ be written as $\bar\rho:G\to GL(k[V])$, coming from some $\rho:G\to GL(V)$? So that the $k[V]^G$ really are a "generalization" of the invariant ring $k[x_1,\dots,x_n]^G$.

Answer 1

If I understand your question correctly, the answer is no, not every representation of a group $G$ on the polynomial algebra $k[V]$ comes from an action of $G$ on $V$.

One way to construct examples to see this is by using the fact that the polynomial ring $k[V]$ is graded: $$ k[V] = \bigoplus_{m=0}^\infty k[V]_m, $$ where $k[V]_m$ consists of the polynomials of degree $m$.

Then a group $G$ can act on just one component of this direct sum without acting on the others. For instance, let $G = \mathbb{Z}/2\mathbb{Z}$, written multiplicatively with generator $\alpha$ (so $\alpha^2 = e$), and for $f \in k[V]_m$, define $$ \alpha \cdot f = \begin{cases} -f & \text{ if } m=0\\ f & \text{ if } m>0 \end{cases} $$ Essentially, $G$ acts on a polynomial by changing the sign of the constant term and leaving everything else alone. This clearly doesn't come from an action of $G$ on $V$.

Answer 2

Yes. This is related to Peter Weyl Theorem that every representation has a dual representation.

\[ L^2(G) \simeq \bigoplus_{\mathrm{Irr}(G)} R \otimes \overline{R} \]

Physicists have a shorthand for this using "bra-ket" notation: the "ket" or left-action $ g \,|v \rangle $ is dual to "bra" right-action $ \langle v | g$.

In these notes on representation theory, in section 1.3, they talk about left- and right-modules.