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$\ldots$
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. 🙂
Best Answer
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 $\operatorname{GL}(m, \mathbb R)$.
This is related: Why does the universal cover of $SL_{2}(\mathbb{R})$ have no finite-dimensional representations?
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.