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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.

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Hint : $G$ is connected, therefore generated by $\exp (\mathfrak{g})$. Moreover, if $\phi$ commutes with $X$, it commutes with $X^2,X^3,...$

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  • $\begingroup$ $\phi(\Pi_1(e^{tX}v))=\phi(e^{t{\pi_1(X)}}v)$ expanding the exponential series and using the hint $\phi(\pi_1^k(X)v)=\pi_2^k(X)\phi(v)$, I will be done right? $\endgroup$ Commented Apr 15, 2019 at 6:38
  • $\begingroup$ It's rather $ \pi_1(X)^k $ than $ \pi_1^k(X) $. But to be done you need to go from "it works for $ \exp (X) $" to "it works for all $ g $ ", with the hint : do you know how that works ? $\endgroup$ Commented Apr 15, 2019 at 8:26
  • $\begingroup$ yeah I meant $\pi_1^k(X)$ only. By connectivity of $G$, we know that every matrix in $G$ is a product of exponentials of matrices in $\mathfrak{g}$, right? $\endgroup$ Commented Apr 15, 2019 at 10:20
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    $\begingroup$ No you mean $\pi_1(X)^k$ :p yes, we do know that, and indeed with that you are done $\endgroup$ Commented Apr 15, 2019 at 10:31
  • $\begingroup$ @Max Dear Sir , I think $\pi $ commutes with representation $\Pi_i$ I do not understand how it commutes with X please can you elaborate $\endgroup$ Commented Apr 19, 2019 at 17:01

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