# Is the group von Neumann algebra construction functorial?

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.

• Didn't this use to have a fa.functional-analysis tag? Is that somehow less relevant than the NCG tag? (If anything, this should have a ct.category-theory tag, IMHO) – Yemon Choi Feb 12 '10 at 8:14

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 $\hat H$.

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.

• Nice to know, but how to use amenability or finitness of \kerf? Can you give some hints or references? – m07kl Oct 23 '10 at 14:35
• If $\ker(f)$ is finite, then the normalized sum over all elements in $\ker(f)$ is a central projection $p \in N(G)$ and the extension is given by the cut-down composed with the inclusion of $G/\ker(f) \subset H$, i.e. $N(G) \to pN(G)p = N(G/\ker(f)) \subset N(H)$. – Andreas Thom Oct 23 '10 at 16:03
• If $\ker(f)$ is amenable, then take a sequence of Foelner sets $F_n \subset \ker(f)$ and define vectors $$\xi_n = |F_n|^{-1} \cdot \sum_{g \in F_n} \delta_g.$$ Then, the states $\phi_n(a) = \langle a \xi_n, \xi_n \rangle$ on $C^{\ast}_{red}(G)$ will have a weak limit and the associated GNS-representation gives a $\ast$-homomorphism $C^{\ast}_{red}(G) \to C^{\ast}_{red}(G/\ker(f))$. Again, this can be composed with the inclusion $C^{\ast}_{red}(G/\ker(f)) \subset C^{\ast}_{red}(H). – Andreas Thom Oct 23 '10 at 16:09 • Compare with mathoverflow.net/questions/28502/… – Alain Valette May 1 '11 at 18:19 Well, for$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... • Is this somehow connected to the fact that, since the inverse image of an element of$H$may be infinite, the map $G\to H$ does not induce a map $\ell^2(H)\to\ell^2(G)$? – Harald Hanche-Olsen Feb 11 '10 at 16:39 • Matt: what was wrong with the statement you struck through? If G is a closed subgroup of H then the sub-vN-algebra of N(H) that is generated by "left translation by elements of G" is isomorphic to N(G), is it not? I thought this was in Takesaki & Tatsuuma, for instance. – Yemon Choi Feb 11 '10 at 18:41 • Takesaki & Tatsuuma certainly study how one can "see" closed subgroups of G as certain subalgebras of NG, but they seem to studiously avoid claiming that the subalgebras are isomorphic to NH, at least on my reading. I struck through the sentence, as the proof I had was (obviously) wrong. – Matthew Daws Feb 11 '10 at 21:11 • Fixed my proof (modulo not getting the LaTeX to render correctly). – Matthew Daws Feb 11 '10 at 22:17 • Just to note (mostly for amusement) that in Matt's counterexample, one could take$n=1$. (Thus: the augmentation character is well-defined as a functional on the$\ell^1$-group algebra, here being evaluation of the Fourier transform at$1$; but it doesn't extend in the right way to VN(Z), because one can't evaluate an element of$L^\infty(T)$at the point$1\$. – Yemon Choi Feb 12 '10 at 6:26