Let $V$ be the space of holomorphic polynomial functions in two complex variables $\xi,\eta$ and let $V^\ast$ be its dual space with subspace $W$ of linear functionals of the form $Df(1,0)$ where $D$ is a constant coefficient differential operator. Let the Lie algebra $sl(2,\mathbb C)$ act on $V^\ast$ by $(X \cdot \varphi)(f) = -\varphi(X \cdot f)$ where ($X \in sl(2,\mathbb C), \varphi \in V^\ast, f \in V$).
Why does this action of $sl(2,\mathbb C)$ on $W$ does not exponentiate to an action of the group $SL(2,\mathbb C)$ on $W$?
Let $H = \begin{pmatrix} 1&0\\0&-1\end {pmatrix}, E= \begin{pmatrix} 0&1\\0&0\end {pmatrix}, F= \begin{pmatrix} 0&0\\1&0\end {pmatrix}$ be the common basis of $sl(2,\mathbb C)$. The space $W$ has a basis consisting of $H$-eigenvectors: the operators $\varphi_k := \frac{\partial^k}{\partial \xi^k}|_{1,0}$ with $H \cdot \varphi_k = k \varphi_k$.
The above is the contragredient representation of $sl(2,\mathbb C)$ on $V$ which one gets by differentiating the left regular representation of $SL(2,\mathbb C)$ on functions.