Compute the associated induced Lie algebra action $\text{d}\pi$ Let $G=\mathrm{SL}_2(\mathbb{C})$ and consider the action of $G$ on the space of smooth functions on column vectors $\mathbb{C^2}$ given by $\big(\pi(g)\phi\big)(v)=\phi\left({g^\top}\,v\right)$ for all smooth functions $\phi:\mathbb{C}^2\to\mathbb{C}$ and $v\in\mathbb{C}^2$.

Compute the associated derived Lie algebra action $\text{d}\pi$ for $Y$=$\begin{bmatrix} 0 & 0 \\1  &  0\end{bmatrix} \in \mathfrak{sl}_2(\mathbb{C})$.

I know that $\pi$ defines a representation.
How do we use the matrix to compute its associated derived Lie algebra action?
 A: I am assuming that "smooth" means holomorphic (otherwise, you can do a similar computation by treating $\mathbb{C}$ as $\mathbb{R}^2$).  Let $V$ denote the space of holomorphic functions from $\mathbb{C}^2$ to $\mathbb{C}$. Then, $\pi:\text{SL}_2(\mathbb{C})\to\text{GL}(V)$ is indeed a representation.  We want to determine $\text{d}\pi:\mathfrak{sl}_2(\mathbb{C})\to\mathfrak{gl}(V)$.  Since $\pi\circ\exp^{\mathfrak{sl}_2(\mathbb{C})}=\exp^{\mathfrak{gl}(\mathbb{C})}\circ\text{d}\pi$, we have that, for each $x\in \mathfrak{sl}_2(\mathbb{C})$, $\phi\in V$, and $v\in\mathbb{C}^2$,  $$\begin{align}
\big(\text{d}\pi(x)\,\phi\big)(v)&=\left.\frac{\text{d}}{\text{d}\lambda}\right|_{\lambda=0}\,\Bigl(\pi\big(\exp(\lambda\,x)\big)\,\phi\Bigr)(v)
\\&=\left.\frac{\text{d}}{\text{d}\lambda}\right|_{\lambda=0}\,\Biggl(\pi\left(\text{id}_V+\lambda\,x+\frac{\lambda^2}{2}\,x^2+\ldots\right)\,\phi\Biggr)(v)
\\
&=\left.\frac{\text{d}}{\text{d}\lambda}\right|_{\lambda=0}\,\phi\Biggl(\left(\text{id}_V+\lambda\,x^\top+\frac{\lambda^2}{2}\,\left(x^\top\right)^2+\ldots\right)v\Biggr)
\\
&=\text{d}\phi(v)\,\left(x^\top\,v\right)\,.
\end{align}$$
For example, $\big(\text{d}\pi(Y)\,\phi\big)(v)=\text{d}\phi(v)\,\left(Y^\top\,v\right)=\begin{bmatrix}\partial_1\phi(v)&\partial_2\phi(v)\end{bmatrix}\,\begin{bmatrix}v_2\\0\end{bmatrix}=\partial_1\phi(v)\,v_2$, where $\partial_i$ is the partial derivative with respect to the $i$-th coordinate and $v=\begin{bmatrix}v_1\\v_2\end{bmatrix}$.  If "smooth" has the usual meaning, then $\big(\text{d}\pi(Y)\,\phi\big)(v)=\partial_{1,\text{Re}}\phi(v)\,\text{Re}\left(v_2\right)+\partial_{1,\text{Im}}\,\text{Im}\left(v_2\right)$, where $\partial_{i,\text{Re}}$ and $\partial_{i,\text{Im}}$ are the partial derivatives with respect to the real part and the imaginary part of the $i$-th coordinate, respectively.
Observe that $\text{d}\pi(x)\,\text{d}\pi(y)-\text{d}\pi(y)\,\text{d}\pi(x)=\text{d}\pi\big([x,y]\big)$ for all $x,y\in\mathfrak{sl}_2(\mathbb{C})$.  Hence, $\text{d}\pi$ is indeed a representation of $\mathfrak{sl}_2(\mathbb{C})$ in $\mathfrak{gl}(V)$.
