# Lie group-algebra representations

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.

Hint : $$G$$ is connected, therefore generated by $$\exp (\mathfrak{g})$$. Moreover, if $$\phi$$ commutes with $$X$$, it commutes with $$X^2,X^3,...$$
• $\phi(\Pi_1(e^{tX}v))=\phi(e^{t{\pi_1(X)}}v)$ 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? – toric_actions Apr 15 '19 at 6:38
• 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 ? – Maxime Ramzi Apr 15 '19 at 8:26
• 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? – toric_actions Apr 15 '19 at 10:20
• No you mean $\pi_1(X)^k$ :p yes, we do know that, and indeed with that you are done – Maxime Ramzi Apr 15 '19 at 10:31
• @Max Dear Sir , I think $\pi$ commutes with representation $\Pi_i$ I do not understand how it commutes with X please can you elaborate – idon'tknow Apr 19 '19 at 17:01