Is the group von Neumann algebra construction functorial?

Question

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.

Answer 1

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 $\hat H$.

Answer 2

Let $f : G \to H$ be a homomorphism of discrete groups.

The homomorphism $f$ extends to a homomorphism of reduced group $C^{\ast}$-algebras if and only of $\ker(f)$ is amenable, and extends to a homomorphism of group von Neumann algebras if and only if $\ker(f)$ is finite.

Answer 3

Well, for $s\in G$ let $\lambda(s)$ be the left-translation operator by $s$; all such operators are in the group von Neumann algebra. I guess that the hoped for homomorphism $F:NG \rightarrow NH$ should satisfy $F(\lambda(s)) = \lambda(f(s))$ for $s\in G$, and we should have that $F$ is an (ultraweakly) continuous $*$-homomorphism. In particular, $F$ is contractive.

Then $F$ need not exist. Let $G=\mathbb Z$ and $H=\mathbb Z/n\mathbb Z$. Then $NH = CH$, so we have a trace on $NH$ which sends $\lambda(0)$ to $1$. So if we apply $F$, and then take the trace, we should get an ultraweakly continuous functional $\phi$ on $NG$ which satisfies $\phi(\lambda(ns)) = 1$ for all $s\in\mathbb Z$.

But this can't happen: maybe we can see this via the Fourier transform. Then $NG \cong L^\infty(\mathbb T)$ and $\phi$ induces $h\in L^1(\mathbb T)$ which satisfies $\int h(\theta) e^{ins\theta} d\theta = 1$ for all $s\in\mathbb Z$. This violated Reimann-Lebesgue.

On the other hand, if $G \subseteq H$ is an inclusion (of discrete groups, to avoid topology) then we do get an inclusion $NG \rightarrow NH$. Here's a construction. Find an index set $I$ and $(h_i)$ in $H$ such that $H$ is the disjoint union of $\{Gh_i\}$. Then define $V:l^2(H)\rightarrow l^2(G)\otimes l^2(I)$ by $V(\delta_h) = \delta_g\otimes\delta_i$ if $h=gh_i$ (so defined on point-masses, and extend by linearity). So $V$ is unitary, and $\theta:x\mapsto V^*(x\otimes 1)V$ is a normal $*$-homomorphism $NG\rightarrow B(l^2(H))$. Then, for $r\in G$, $V^*(\lambda(r)\otimes 1)V(\delta_h) = V^*(\delta_{rg}\otimes\delta_i) = \delta_{rh}$ as $rg\in G$. So $\theta$ maps into $NH$, and does what we want.

Surely there is some general result, but I'm not sure of it...