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 :)

  • $\begingroup$ 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 $SL_2(\mathbb{R})$. $\endgroup$ Jun 30, 2022 at 20:12

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .