Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

3 Answers

up vote 5 down vote accepted

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 $\hat H$.

share|improve this answer
add comment

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.

share|improve this answer
    
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
1  
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
1  
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
add comment

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...

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.