I want to prove the following: Given two representations of a connected matrix Lie group are equivalent if and only if the associated Lie algebra representations are equivalent.
Definition: Let $G$ be a matrix Lie group, let $\Pi$ be a representation of
$G$ acting on the space $V$, and let $\Sigma$ be a representation of $G$ acting on the space $W$. A linear map $\phi : V \rightarrow W$ is called an intertwining map of representations if $\phi(\Pi(A) v)=\Sigma(A) \phi(v)$ for all $A \in G$ and all $v \in V$.
The intertwining maps of representations of a Lie algebra are defined analogously. If $\phi$ is an intertwining map of representations and, in addition, $\phi$ is invertible, then $\phi$ is said to be an equivalence of representations. If there exists an isomorphism between $V$ and $W,$ then the representations are said to be equivalent.
My attempt:
Assume that $\pi_1, \pi_2$ are associated Lie algebra representations that are equivalent. Then $\exists$ $\phi$ such that $\phi(\pi_1(X) v)=\pi_2(X) \phi(v)$ for all $X \in \mathfrak{g}$ and all $v \in V $. Also, $\pi_i(X)=\frac{d}{d t} \Pi_i\left.\left(e^{t X}\right)\right|_{t=0}$, where $i=1,2$.
Now I need to use both of this to show that $\Pi_1,\Pi_2$ are equivalent. Any hint to proceed will be helpful. Thanks.
Best Answer
Hint : $G$ is connected, therefore generated by $\exp (\mathfrak{g})$. Moreover, if $\phi$ commutes with $X$, it commutes with $X^2,X^3,...$