Let $G=SL_2(\mathbb{R})$ and $U=M_2(\mathbb{C})$. Then for all $g \in G$ consider the representation $\pi(g): U \to U$, sending $X \mapsto Xg$.
In a textbook they wanted to write $U$ as a direct sum of irreducible subspaces as an exercise. They gave us the hint to understand the map applied to the basis elements of the Lie algebra of $G$.
The Lie algebra of $G$ is $\langle X, Y, Z \rangle$, where $$ X:= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\, Y:= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\, Z:= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. $$
If I write $g = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \in G$, i.e. $ad-bc=1$, then $$ \pi(g)(X) = \begin{pmatrix} a & -c \\ b & -d \end{pmatrix},\, \pi(g)(Y) = \begin{pmatrix} c & 0 \\ d & 0 \end{pmatrix},\, \pi(g)(Z) = \begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix}. $$
But somehow I don't see how this could be useful. Can someone help me?
$$M_2(\mathbb C)=\bigg\{\begin{pmatrix}a&b\\0&0\end{pmatrix}:a,b\in\mathbb C\bigg\}\oplus\bigg\{\begin{pmatrix}0&0\\a&b\end{pmatrix}:a,b\in\mathbb C\bigg\}$$ is a decomposition into $G$-stable subspaces. These are both isomorphic to the standard representation $\mathbb C^2$ of $\mathrm{SL}_2(\mathbb C)$. The standard representation is irreducible, since if $0\ne V\subseteq \mathbb C^2$ then for $v\ne0\in V$, we know there exists a matrix $g$ such that $gv$ is linearly independent from $v$, hence $V=\mathbb C^2$.