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.

  • $\begingroup$ When $G$ is abelian, it amounts to define a map $\ell^\infty(G)\to L^\infty(G)$. When $f\in\ell^\infty(G)$ is not measurable, how would you define the image of $f$? $\endgroup$
    – YCor
    Nov 28, 2017 at 14:48
  • 1
    $\begingroup$ @YCor: I don't think that's right. If $G$ is abelian, then as the OP says, $\mathcal L(G_{disc}) = L^\infty(b(\hat G))$ while $\mathcal L(G) = L^\infty(\hat G)$. $\endgroup$ Nov 28, 2017 at 14:58

1 Answer 1


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.


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