Does a representation of the universal cover of a Lie group induce a projective representation of the group itself?

Question

Suppose that $G$ is a connected Lie group, $\tilde{G}$ its universal cover, $p:\tilde{G}\to G$ the covering map. Does a representation $\rho$ of $\tilde{G}$ on a finite-dimensional vector space $V$ induce a projective representation of $G$? That is, given $\rho:\tilde{G}\to\mathrm{GL}(V)$, must there exist a Lie group homomorphism $\sigma:G\to\mathrm{PGL}(V)$ such that $\sigma\circ p=\pi\circ\rho$ (where $\pi:\mathrm{GL}(V)\to\mathrm{PGL}(V)$ is the quotient map)?

I know that something like this is true for irreducible representations, by Schur's Lemma (see the Wikipedia page on "Projective Representations"). However, I am mostly interested in the case of the adjoint representation of $\tilde{G}$ on its Lie algebra $\mathfrak{g}$, or more generally, when $\rho$ is almost faithful (i.e., has discrete kernel). If the answer to this question is "yes" in this case, then that should help resolve the previous question I asked on MO here: Which Lie groups are covers of matrix groups?.

Answer 1

This is not true for representations that are not irreducible, because the kernel of $p\colon \tilde G\to G$, which is central, doesn't have to act as a scalar. For example, take $G=\operatorname{\rm SO}(3)$, $\tilde G=\operatorname{SU}(2)$, and consider a direct sum of a two dimensional and a three dimensional irreducible representation of $\tilde G$.

On the other hand, the adjoint representation of $\tilde G$ on its Lie algebra factors through $G$, because the centre acts trivially.