# Lie Algebra homomorphism from $\mathfrak{sl}(2,\mathbb{R})$ to $\mathfrak{gl}(m,\mathbb{R})$ is the derivtive of a unique Lie group homomorphism

I want to show that every lie algebra homomorphism $$\phi$$: $$\mathfrak{sl}(2,\mathbb{R}) \rightarrow \mathfrak{gl}(m,\mathbb{R})$$ is the derivtive of a unique Lie group homomorphism $$\Phi: Sl(2,\mathbb{R}) \rightarrow GL(m,\mathbb{R})$$ at the identity ie $$\phi = T_e\Phi$$ I know that the statement would be true if $$SL(2,\mathbb{R})$$ was simply connected.

I think the uniqueness should work like that:

Since $$\Phi(exp(X)) = exp(\phi(X))$$ for all $$X \in \mathfrak{sl}(2,\mathbb{R}), \Phi$$ is uniquely determined on an open neighborhood of e. Because $$GL(2,\mathbb{R})$$ is path-connected, any open neighborhood generates G, so $$\Phi$$ is unique if it exists.

Existence seems a lot harder...

I think there has to be a shortcut that I am missing.

The hint says: "consider the relation between $$\mathfrak{sl}(2,\mathbb{R})$$ and $$\mathfrak{sl}(2,\mathbb{C})$$

Any help would be much appreciated :)

• A priori you only get a map from the universal cover of $SL_2$ to the universal cover of $GL_m$ (and neither is simply connected). I don't know what the intended solution is here but you can do this by just classifying all finite-dimensional representations of $\mathfrak{sl}_2(\mathbb{R})$ and just checking directly that they all exponentiate to representations of $SL_2(\mathbb{R})$. Jun 30, 2022 at 20:12

Complexifying $$\phi$$ gives a Lie algebra homomorphism $$\mathfrak{sl}(2, \mathbb C) \to \mathfrak{gl}(m, \mathbb C)$$. Because $$\operatorname{SL}(2, \mathbb C)$$ is simply connected, this one does lift to the groups and restricting gives a Lie group homomorphism $$\operatorname{SL}(2, \mathbb R) \to \operatorname{GL}(m, \mathbb C)$$ with derivative $$\phi$$. The image is generated by $$\exp$$ of a neighborhood of $$0$$ in $$\mathfrak{gl}(m, \mathbb R)$$, so it lands in $$\operatorname{GL}(m, \mathbb R)$$.
The alternative proof, as in the comment, is to use that all irreducible representations of $$\mathfrak{sl}(2, \mathbb R)$$ are symmetric powers of the standard representation, and all representations are sums of irreducibles, so they all lift to the group because the standard representation does.