Let $G$ be a group and $CG$ the complex group algebra over the field $C$ of complex number. The group von Neumann algebra $NG$ is the completion of $CG$ wrt weak operator norm in $B(l^2(G))$, the set of all bounded linear operators on Hilbert space $l^2(G)$. Let $f:G \to H$ be any homomorphism of groups. My question is: is there a homomorphism of the group von Neumann algebra $NG \to NH$ induced from $f$?.
If $NG$ is replaced with $CG$, it's obvious true. If $NG$ is replaced with $C^\ast_r(G)$, the reduced group $C^\ast$ algebra, it's not necessary true.
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By the way, here's the "correct" functorial property. If G and H are abelian, and $f:G\rightarrow H$ is a continuous group homomorphism, then we get a continuous group homomorphism $\hat f:\hat H\rightarrow \hat G$ between the dual groups. By the pull-back, we get a *-homomorphism $\hat f_*:C_0(\hat G) \rightarrow C^b(\hat H)$. We should think of $C^b(\hat H)$ as the multiplier algebra of $C_0(\hat G)$. Then $C_0(\hat G) \cong C^*_r(G)$, and so we do get a *-homomorphism $C^*_r(G) \rightarrow M(C^*_r(H))$; the strict-continuity extension of this is a *-homomorphism $M(C^*_r(G)) \rightarrow M(C^*_r(H))$ which does indeed send $\lambda(s)$ to $\lambda(f(s))$.
For non-abelian group (in fact, non-amenable groups) it's necessary to work with $C^*(G)$ instead.
We cannot ensure a map to $C^*_r(H)$ itself, as we cannot ensure a map from $C_0(\hat G)$ to $C_0(\hat H)$; indeed, this would only happen when $\hat f$ were a proper map.
Similarly, we don't get maps at the von Neumann algebra level, as we don't get a map $L^\infty(\hat G) \rightarrow L^\infty(\hat H)$: we would need that $\hat f$ pulled-back null sets in $\hat G$ to null sets in $\hat H$.