I consider the action of $GL(2, \mathbb{C})$ on function, inherited from the action of $GL(2, \mathbb{C})$ on $\mathbb{C}^2$, given by
$(g \cdot f)(\vec{v})=f(g^{-1}\cdot \vec{v})$
I want to study the subrepresentation of $GL(2, \mathbb{C})$ on $\mathbb{C}[x,y]$. For definition, $x(\vec{v})$ is the first coordinate of $\vec{v}$ and $g\cdot \vec{v}$ is the first coordinate of $g^{-1}\cdot \vec{v}$. For $y$ is analogous.
If $g = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ then $g^{-1}=\frac{1}{\Delta} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}$.
Therefore $(g\cdot x)( \vec{v}) = x( g^{-1} \vec{v})= x\Bigl(\frac{1}{\Delta} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}\Bigr)=\frac{1}{\Delta}(dv_1-bv_2)$ and $(g\cdot y)( \vec{v}) = y( g^{-1} \vec{v})= y\Bigl(\frac{1}{\Delta} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}\Bigr)=-\frac{1}{\Delta}(cv_1+av_2)$
If now I consider x and y as the dual base of $\mathbb{C}^2$, the actions is the controgradient of the standard action on $\mathbb{C}^2$ and the matrix is the transposed of the inverse $\frac{1}{\Delta}\begin{bmatrix}d&-c\\-b&a\end{bmatrix}$ and the action on $x \in $ the dual space of $\mathbb{C}^2$ is $g \cdot x = \frac{1}{\Delta}(xd -cy)$ and when applied to a vector $\begin{bmatrix}v_1\\v_2\end{bmatrix}$ gives $\frac{1}{\Delta}(dv_1-cv_2)$ that's different from what obtained before. I can't see where is the error.