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

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 $k[x_1,\dots,x_n]^G$.

• If $V$ is finite dimensional then $V^{**} \cong V$ as $kG$-modules, so you can "go back" that way. Jul 14, 2013 at 11:30
• The "example" in your edit is not an example. $k\left[x_1,x_2,...,x_n\right]$ is not a "dual" of $k^n$, but rather the symmetric algebra, or the symmetric algebra of a dual. Jul 14, 2013 at 12:08
• @mt_ Oh yes, forgot about that, gotta do that next. Jul 14, 2013 at 12:53
• @darijgrinberg I didn't mean that $k[x]$ was the dual space of $V=k^n$, but rather that it's (isomorphic to) the coordinate ring $k[V]$. What's confusing me is: given any action as above on $k[x]$, then I can consider the space $V=k^n$. Then how do I connect the action of $G$ on $k[V]$ to the construction of the action on the coordinate ring from an action on $V$? The title is very badly chosen, as I'm not interested in the "dual" representation itself, but the representation on the coordinate ring. But maybe I just didn't understand your comment. Jul 14, 2013 at 13:00
• OK, let me try to sidestep the confusion around here and just answer Edit5: No, it cannot. There are many more linear $G$-actions on $k\left[x_1,...,x_n\right]$ than linear $G$-actions on $V$. A former comes from a latter if and only if every $g\in G$ acts on $k\left[x_1,...,x_n\right]$ by an algebra automorphisms. Jul 14, 2013 at 13:19

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 $V$.

• Thanks for your comment and answer so far :) To make everything clear for me: what is the answer in the special case $G\subseteq GL_n(k)$, with action $Af=f(Ax)$ on $k[x]$? Since this is the "traditional" setting in invariant theory, and one nowadays makes statements about $k$-algebras of the form $k[V]^G$, I assume the traditional setting is covered by this? Jul 14, 2013 at 13:30
• As per my comment on the original question, you need $Af(x) = f(A^{-1}x)$, and that is exactly the representation coming from the usual action of $G$ on $k^n$. As darij pointed out in his comment on your question, the linear actions of $G$ on $k[V]$ that come from an action of $G$ on $V$ itself are precisely those actions that are compatible with the algebra structure of $k[V]$, i.e. where the representation is actually a group homomorphism $\sigma : G \to \mathrm{Aut}_{alg}(k[V])$.
– MTS
Jul 14, 2013 at 13:44
• Thank you very much! I also just now found this primer, which, too, was helping me in understanding this. My confusion is finally gone :) Jul 14, 2013 at 13:49

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.

• John, I don't think that the Peter Weyl Theorem implies that every representation of a group on a polynomial algebra comes from an action on the degree one part.
– MTS
Jul 14, 2013 at 13:26