Let $G$ be a locally compact, Hausdorff and $2^{nd}$-countable group and let $G_{disc}$ be the same group with the discrete topology. We have a continuous (and bijective) homomorphism given by $$ id:G_{disc} \to G. $$ Now, let $\mathcal{L}(G) \subset \mathcal{B}(L^2(G))$ and $\mathcal{L}(G_{disc}) \subset \mathcal{B}(\ell^2(G))$ be the von Neumann algebras generated by the left regualar representations of each group.

Question:Does the identity map lift to a normal $\ast$-homomorphism $$ \pi: \mathcal{L}(G_{disc}) \to \mathcal{L}(G). $$

**Background:** If $G$ is Abelian, then $id: G_{disc} \to G$ induces a map $j: \widehat{G} \to \widehat{G}_{disc} = b(\widehat{G})$, where $b(\cdot)$ denotes the Bohr compactification. The map $j$ is an inclusion with dense range. If $\pi$ exists, in this case, it is given by $f \mapsto f{|}_{\mathrm{Im}(j)}$. That suggest that the map indeed exists.

On the other hand the image of $\pi$ will contain every $\lambda_G(g)$ and so it will be the whole of $\mathcal{L}(G)$. But every element of $\mathcal{L}(G_{disc})$ can be written unambigously as a sum $$ x = \sum_{g \in G} a_g \, \lambda_{G_{disc}}(g), $$ where $a_g = \langle \delta_e, x \, \delta_{g^{-1}} \rangle$. But using that $\mathcal{L}(G_{disc}) \subset \ell^2(G)$ will give that $(a_g)_g$ has numerable support. The kernel of $\pi$ will be given by those sequences $a = (a_g)_g$ in $\ell^2$ such that $\lambda_G(a) = 0$. If the kernel is $\{0\}$, that will force $\pi$ to be an isomorphism (a somewhat counterintuitive result).

Furthermore, we can understood both $\lambda_G$ and $\lambda_{G_{disc}}$ are representations of $G_{disc}$ in $L^2(G)$ and $\ell^2(G)$ respectively. The problem is equivalent to ask whether $$ id: \lambda_{G_{disc}}[G_{disc}]^{''} \to \lambda_{G}[G_{disc}]^{''} $$ is a well-defined normal $\ast$-homomorphism. That will follow when $\lambda_{G_{disc}} \leq \lambda_{G}$. Something that seems unlikely.