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Let $G$ and $H$ be discrete groups and $f:G \rightarrow H$ be any homomorphism of these groups. I have three questions about it:

1) How to prove the functoriality of the construction of universal $C^*$-algebra of discrete group (the existence of induced homomorphism $C^*(G) \to C^*(H)$)?

2) How to prove that the construction of reduced $C^*$-algebra of discrete group is not functorial (I am especially interested in counterexamples) and in which case (I mean conditions for group homomorphism) it'll be functorial?

3) Let us consider the case when $G = \mathbb{Z}$ (integers) and $H = \mathbb{Z}/ n\mathbb{Z}$. How to describe the kernel and the image of the induced homomorphism of group $C^*$-algebras?

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    $\begingroup$ Maybe you could give some brief detail about why you want to know, what level of study/research you are at, etc? These are natural questions but they seem like they could come from a course on $C^*$-algebras. $\endgroup$
    – Yemon Choi
    Commented Jun 17, 2010 at 13:56
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    $\begingroup$ By the way, here's a hint for the first part of Q2: $C_r^*({\mathbb F}_2)$ is simple, hence has no non-trivial closed ideals, even though there is an obvious homomorphism from ${\mathbb F}_2$ onto ${\mathbb Z}^2$. Here ${\mathbb F}_2$ and ${\mathbb Z}_2$ are, respectively, the free group and the free abelian group on two generators. $\endgroup$
    – Yemon Choi
    Commented Jun 17, 2010 at 14:00
  • $\begingroup$ Q3 should also not be that hard if you understand the definitions and have been told/have learned/can see what $C^*({\mathbb Z})$ is. $\endgroup$
    – Yemon Choi
    Commented Jun 17, 2010 at 14:02
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    $\begingroup$ Thank you very much for your response! As far as my (and your) questions: Q1. I am interested in this question because I think that it is very natural but I can't find an explanation of this facts in the literature (I've read Davidson and Pedersen, for instance) Q2. Thank you very much for your hint, I understand it well now Q3. I know that C*(Z) = C(T), where T is unit circle, and C*(Z/nZ) = C^n.. As far as the nature of my interest - I am a low-dimensional topologist and I started to learn C* - algebras in connection with K-theory that could be useful in my science. $\endgroup$
    – skripka
    Commented Jun 18, 2010 at 8:23
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    $\begingroup$ This is almost the same question as mathoverflow.net/questions/14995/… $\endgroup$
    – Andreas Thom
    Commented Apr 20, 2011 at 5:28

2 Answers 2

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Regarding Q2: the reduced $C^\star$-algebra is functorial wrt homomorphisms with amenable kernels. Indeed, let $N$ be a normal, amenable subgroup of $G$; since the trivial representation of $N$ is weakly contained in the regular representation of $N$ (by amenability), by continuity of induction the regular representation of $G/N$ is weakly contained in the regular representation of $G$, which means that the reduced $C^\star$-algebra of $G$ maps onto the one of $G/N$.

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Actually, I'm not quite sure where this is explained completely basically in the literature. So maybe here's a hint, at least for Q1. The definition of $C^*(G)$ is that it's the completion of L^1(G) with respect to the biggest C*-norm. So if $\phi:G\rightarrow H$ is a continuous group homo, we immediately get a *-homomorphism $\ell^1(G) \rightarrow \ell^1(H)$, and so by inclusion, a *-homo $\ell^1(G) \rightarrow C^*(H)$. But this defines some C*-norm on $\ell^1(G)$, so the norm on $C^*(G)$ must dominate this, and hence we get a continuous extension to $C^*(G) \rightarrow C^*(H)$.

For Q2, find a proof in the literature (I think this goes back to Godemont?) that $C^*_r(G) = C^*(G)$ if and only if the left-regular representation contains the trivial representation, if and only if G is amenable. Put another way, if G is not amenable, then the trivial homomorphism $G\rightarrow\{1\}$ doesn't induce a map $C^*_r(G)\rightarrow C^*(\{1\}) = \mathbb C$.

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  • $\begingroup$ Matt, thank you very much for your response! Your explanation of Q1 is great, I hope I understand all things maybe except one: could you explain how our * - homo l^1 (G)-> C*(H) defines a C*-norm on l^1(G)? As far as Q2, thank you, it is a good advice and we can provide a counterexample with a classical non - amenable group (free group of 2 generators) and free abelian group with 2 generators. Thank you very much one more time! $\endgroup$
    – skripka
    Commented Jun 18, 2010 at 18:06
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    $\begingroup$ Matt, I think some of your "norms" should perhaps be "seminorms"? Also, "Godement". $\endgroup$
    – Yemon Choi
    Commented Jun 18, 2010 at 19:07
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    $\begingroup$ Yeah, okay, I was being vague. If $\ell^1(G)\rightarrow C^*(H)$ is injective then you're done: pull-back the norm from $C^*(H)$. Otherwise, you can only pull-back a seminorm, but this doesn't matter, as it's still dominated by the $C^*(G)$ norm. $\endgroup$
    – Matthew Daws
    Commented Jun 18, 2010 at 19:11
  • $\begingroup$ @Yemon: Yes, and, ahem, yes! $\endgroup$
    – Matthew Daws
    Commented Jun 18, 2010 at 19:11
  • $\begingroup$ Matt, Yemon, thank you very much, I've got it now. As far as reduced C* -algebra: is it functorial only for proper homomorphisms or condition of injectivity is sufficient? Thanks a lot. $\endgroup$
    – skripka
    Commented Jun 18, 2010 at 20:06

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