Let $G$ be a semi-simple lie group (simply connected for simplicity), $\mathfrak{g}$ its lie algebra. Write $\overline{G}=Inn(\mathfrak{g})$ for the adjoint form of $G$ which we identify here with the group of inner automorphisms of $\mathfrak{g}$.
For a representation $M$ of $\mathfrak{g}$ and an automorphism $\varphi$ of $\mathfrak{g}$, define the twisting of $M$ by $\varphi$ as the $\mathfrak{g}$ module $M^{\varphi}$, which as a vector space is $M$ and the $\mathfrak{g}$-action is given by $X\cdot m=\varphi^{-1}(X)m$ (the inverse is to simplify things later). Note that if $M$ is infinite-dimensional this twist may not be isomorphic to $M$.
Now suppose that $M$ is a simple representation of $\mathfrak{g}$ such that for every inner automorphism $\varphi$ of $\mathfrak{g}$, we have an isomorphism of $\mathfrak{g}$-modules $M^{\varphi}\cong M$. By Schur's lemma, the isomorphism $M^{\varphi}\to M$ is unique up to a nonzero scalar, hence we obtain a map $\overline{G}\to PGL(M)$. One can check that this is in fact an homomorphism of groups.
My question is: how does this projective representation of $G$ relate to the originally defined $\mathfrak{g}$-representation? I would really like the following to be true: in the given situation, we in fact must have that $M$ is integrable (i.e. a direct sum of finite-dimensional representations), and hence the projective representation $\overline{G}\to PGL(M)$ is coming from a representation of $G$ on $M$.
However some obvious difficulties are that we know nothing about the map $\overline{G}\to PGL(M)$ besides that it is a group homomorphism, and further with $M$ infinite dimensional, I don't know of a natural topology to put on $PGL(M)$.
