# Which representations of the Lie algebra of a Lie group come from representations of the group itself?

I think this is very classic mathematics, but I can't find a complete answer in the literature.

Let $$G$$ be a Lie group, $$\mathfrak{g}$$ the Lie algebra of $$\mathfrak{g}$$. Suppose $$\rho : \mathfrak{g} \to \mathfrak{gl}(V)$$ is an arbitrary finite-dimensional representation of the Lie algebra $$\mathfrak{g}$$ on the real vector space $$V$$. I want to ask when there exists a representation $$\rho_G : G \to GL(V)$$ of the Lie group $$G$$, such that the tangent map of $$\rho_G$$ at $$e$$ (the identity for $$G$$) is exactly $$\rho$$.

I think when $$G$$ is simply connected, then every (finite-dimensional) representation of $$\mathfrak{g}$$ is induced by a representation of $$G$$ as a Lie group. But what if $$G$$ is merely supposed to be connected, or even more generally, what if $$G$$ is an arbitrary Lie group?

• If $G$ is connected then the universal covering group of $G$ maps to $G$ by a Lie group homomorphism, and you need to see if the representation is trivial on the kernel of that homomorphism. That kernel is identified with the fundamental group of $G$. – Ben McKay Nov 23 '18 at 15:48
• If $G$ is not connected, then it is often a semidirect product with a small finite group, and then the extension problem is not so difficult. But in general, if $G$ is not connected, the problem seems unlikely to have any straightforward resolution, as I don't think we know the representations of all discrete groups. – Ben McKay Nov 23 '18 at 15:51
• @BenMcKay In general, it is not very intuitive to find an explicit description of the universal cover group $\widetilde{G}$ of $G$ to test whether the kernel acts trivially or not (e.g. when $G=SL_2(\mathbb{R})$). However, loops based on $e$ in $G$ are sent to loops in ${\rm Aut}(\mathfrak{g})$ based on ${\rm Id}_{\mathfrak{g}}$ via the adjoint representation of $G$. Is it possible to reformulate the triviality of the action on the kernel for the universal covering as some invariant conditions for the repsentation $\rho$ with respect to the automorphisms in the image of these loops? – Rick Sternbach Nov 23 '18 at 17:41
• Typically, simple Lie groups have finite fundamental group; indeed $SL_2(\mathbb{R})$ is one of 2 exceptions. Using $KAN$ decomposition, you can write out the universal covering group and covering map for $SL_2(\mathbb{R})$, and see which representations drop. Loops in the adjoint representation are no good: they drop to loops in the adjoint form, i.e. they are the same for all semisimple Lie groups with the same Lie algebra. – Ben McKay Nov 23 '18 at 18:43