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

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?.

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.