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.



Does the inclusion of the discretized group into itself lift to the group von Neumann algebras?
I think this is only true if $G$ is discrete. As $\pi$ is assumed normal, it's pre-adjoint would give a map
$$ \pi\_\*: A(G) \rightarrow A(G\_{disc}) $$
between the associated Fourier algebras. As $\pi(\lambda\_{G\_{disc}}(g)) = \lambda\_G(g)$, the map $\pi\_\*$, considered as a map of function spaces, would be the formal identity on functions.


But for any group $H$, there is an inclusion $A(H) \rightarrow C\_0(H)$ which has dense range. So for $\pi\_\*$ to exist, we need that every $f\in A(G)$ is a member of $c\_0(G)$. That's only going to be true if $G$ is discrete.


Or another way: If $\pi$ existed then composing $\pi$ with a normal functional would show that every coefficient of $\lambda\_G$, when viewed as a function on $G\_{disc}$ would vanish at infinity, which holds only when $G$ is discrete.


