TITLE: What is the explicit ${\rm GL}_n$-action on the Weyl algebra?
QUESTION [5 upvotes]: This is a typical classical invariant theory setup.  Let $V = \mathbb C^n$, and let $\mathbb C[V] \cong \mathbb C[x_1,\ldots,x_n]$ be the space of polynomial functions on $V$.  The group $G={\rm GL}(n,\mathbb C)$ acts on $V$ in the obvious way, inducing the usual action on $\mathbb C[V]$ via $(g \cdot f)(v) = f(g^{-1}v)$ for $g \in G$, $f \in \mathbb C[V]$, and $v \in V$.  Hence $G \subset {\rm End}(\mathbb C[V])$.
Now consider the Weyl algebra $\mathbb D(V)$ consisting of polynomial-coefficient differential operators on $\mathbb C[V]$.  Clearly $\mathbb D(V) \subset {\rm End}(\mathbb C[V])$ as well, and in fact $G$ normalizes $\mathbb D(V)$; that is, we have a group action of $G$ on $\mathbb D(V)$ given by
$$
g \cdot D = g \circ D \circ g^{-1} \in \mathbb D(V),
$$ for $g \in G$ and $D \in \mathbb D(V)$.  Given the action above, I have seen several sources state without proof that the $G$-action on the span of the differential operators $\partial_i = \frac{\partial}{\partial x_i} \in \mathbb D(V)$ is equivalent to the defining representation $V$, whereas the $G$-action on the span of the multiplication operators $x_i \in \mathbb D(V)$ is equivalent to the dual $V^*$.
This has been driving me insane; below is my attempt to convince myself of this fact, using a concrete example.  Is this the correct way to think about the actions here?  Or, is there a more obvious way to see it?
Example. Let $n=2$, so that $\mathbb C[V] = \mathbb C[x,y]$, and $\mathbb D(V)$ is generated by the set $\{x,y,\partial_x, \partial_y\}$.  Consider $g=\left[ \begin{smallmatrix} 3 & 5\\1 & 2 \end{smallmatrix}\right] \in G$, so that $g^{-1} = \left[ \begin{smallmatrix} \phantom{-}2 & -5\\-1 & \phantom{-}3 \end{smallmatrix}\right]$.  If my interpretation above is correct, then we should have the following:
\begin{align*}
g \cdot \partial_x &= 3 \partial_x + \partial_y\\
g \cdot \partial_y &= 5 \partial_x + 2 \partial_y\\
g \cdot x &= 2x - 5y\\
g \cdot y &= -x + 3y.
\end{align*}
(Note that I have $G$ acting via inverse transpose on $x$ and $y$, since those are suppposed to play the role of the basis vectors of $V^*$.)  Here is my attempted justification for the first equation above, using $u$ and $v$ to substitute in the third line:
\begin{align*}
(g\cdot \partial_x) \cdot f(x,y) &= (g \circ \partial_x \circ g^{-1})\cdot f(x,y)\\
&= (g \circ \partial_x) \cdot f(3x+5y, \: x+2y)\\
&= (g \circ \partial_x) \cdot f(u,v)\\
&= g \cdot f_x(u,v)\\
&= g \cdot (f_u u_x + f_v v_x) & \text{(chain rule)}\\
&= g \cdot (3 f_u(u,v) + f_v(u,v))\\
&= 3 g \cdot f_u(u,v) + g \cdot f_v(u,v)\\
&= 3 f_x(x,y) + f_y(x,y)\\
&= (3 \partial_x + \partial_y)\cdot f(x,y).
\end{align*}
This is the "correct" anticipated result, but am I applying all the transformations in the correct order?
Likewise, for the third equation above (the action of $g$ on $x$), I have this so far:
\begin{align*}
(g \cdot x) \cdot f(x,y) &= (g \circ x \circ g^{-1}) \cdot f(x,y)\\
&= (g \circ x) \cdot f (u,v) \\
&= g \cdot x f(u,v).
\end{align*}
This is where I'm not sure how to finish rigorously; one one hand, $g$ should transform $f(u,v)$ back into $f(x,y)$, and on the other hand, since $g^{-1} \cdot \left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right] = \left[\begin{smallmatrix} 2x - 5y \\ -x + 3y \end{smallmatrix}\right]$, I suppose that the $g$-action here transforms the "extra" $x$ into $2x-5y$?  So the final result is to multiply the original $f(x,y)$ by $2x-5y$, just as anticipated?
Is there a better way to understand the $G$-action here?

REPLY [2 votes]: The Weyl algebra can be seen as $T(V^{\star} \oplus V)$ factored by the ideal generated by the elements $\phi \otimes \psi - \psi \otimes \phi$, $v \otimes w - w\otimes v$ and $ v\otimes \phi - \phi \otimes v - \phi(v)$, for all $v \in V$, and $\phi\in V^{\star}$.
Remember that $v$ is a (constant) vector field on $V$ so a differential operator, and $\phi$ is a linear function on $V$
($V = \mathbb{C}^n$, but can be any vector space over a field $k$).
Now, $G = GL(V)$ acts on $V$ ( standard) and on $V^{\star}$ (dual action) , and we have $(g\phi) (g v) = \phi(v)$, so $G$ invariates the ideal. We get therefore an action  of $G$ on the Weyl algebra of $V$.