7
$\begingroup$

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?

$\endgroup$
  • 6
    $\begingroup$ 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 $G$. $\endgroup$ – Ben McKay Nov 23 '18 at 15:48
  • 3
    $\begingroup$ 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. $\endgroup$ – Ben McKay Nov 23 '18 at 15:51
  • $\begingroup$ @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? $\endgroup$ – Rick Sternbach Nov 23 '18 at 17:41
  • 1
    $\begingroup$ 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. $\endgroup$ – Ben McKay Nov 23 '18 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.