Let $G$ be a weakly amenable group, in the sense that it has a net of finitely supported functions $\varphi:G\to \mathbb{C}$ which converge point wise to 1 and their cb norm is bounded uniformly by some constant $C$.

It is known that this is equivalent to completely bounded approximation property (CBAP) for the reduced group $C^*$-algebra of $G$.

Question 1: If $G$ is weakly amenable, does the full group $C^*$-algebra have CBAP?

Question 2: Do the functions $\varphi$ as multipliers extend to completely bounded maps on the full group $C^*$-algebra, with the same cb norm?

Edit: Yemon Choi pointed out that the answer to Question 1 is negative. In that case, are there other $C^*$-completions of the group ring $\mathbb{C}G$ for which the answer(s) would be affirmative?

  • 2
    $\begingroup$ CBAP implies exactness (this is somewhere in Effros-Ruan's book, I think, but I don't have my copy at hand). It is well known that the full group Cstar algebra of a nonabelian free group is non-exact, so it can't have CBAP. $\endgroup$
    – Yemon Choi
    Mar 7 '17 at 3:35
  • 1
    $\begingroup$ As for other Cstar completions, there are other negative results in recent work of Ruan and Wiersman arxiv.org/abs/1505.00805 $\endgroup$
    – Yemon Choi
    Mar 7 '17 at 3:36
  • $\begingroup$ Finally: I don't quite understand the part of your question that starts "Namely as multipliers..." Why would CBAP for the full group Cstar algebra be equivalent to the stated property of cb-multipliers? (Perhaps I am overlooking something obvious) $\endgroup$
    – Yemon Choi
    Mar 7 '17 at 3:38
  • $\begingroup$ @YemonChoi Thanks! That answers the question. With the second part -thanks for pointing this out - I'll edit the question. $\endgroup$ Mar 7 '17 at 6:30
  • $\begingroup$ Moving comments here so as not to clog up Mateusz's fine answer: if $\varepsilon$ IS the trivial repn then ${\rm C}^*_\varepsilon(G)={\bf C}$. Your comments/questions about "does $\pi$ containing the trivial representation acts an obstruction?" seem to be jumping to conclusions without much thought -- think of the case of a finite group $\endgroup$
    – Yemon Choi
    Mar 8 '17 at 13:31

The answer to the second question is also negative. Let $(u_g)_{g\in G}$ be the generating unitaries of $C^{\ast}(G)$. Suppose that $(\varphi_i)_{i\in I}$ is a net a functions whose associated multipliers $m_{\varphi_{i}}: C^{\ast}_{r}(G) \to C^{\ast}_r(G)$ give CBAP of $C^{\ast}_{r}(G)$. Suppose now that the assignments $u_g \mapsto \varphi_i(g) u_g$ extend to uniformly completely bounded multipliers on the full $C^{\ast}$-algebra $C^{\ast}(G)$. It would follow that $C^{\ast}(G)$ has the CBAP because, by uniform boundedness, it would suffice to check convergence on finite sums $\sum a_g u_g$, which are dense in $C^{\ast}(G)$, and this is easy.

Let me now discuss, where the subtlety lies. The multipliers giving CBAP for non-amenable groups are not uniformly bounded in $B(G)$, the Fourier-Stieltjes algebra, and therefore do not give nice multipliers on $C^{\ast}(G)$. Indeed, for a function $\varphi: G \to \mathbb{C}$ with $\|\varphi\|_{B(G)}\leqslant C$ we get an associated multiplier $m_{\varphi}:C^{\ast}(G) \to C^{\ast}(G)$ via composition $C^{\ast}(G) \stackrel{\Delta}{\rightarrow} C^{\ast}(G) \otimes C^{\ast}(G) \stackrel{id \otimes \varphi}{\rightarrow} C^{\ast}(G)$, where $\Delta$ is a $\ast$-homomorphism such that $\Delta(u_g)=u_g\otimes u_g$, given by the universal property, and $\varphi:C^{\ast}(G)\to \mathbb{C}$ is the functional given by $\varphi$. Since $\ast$-homomorphisms are completely contractive, we get $\|m_{\varphi}\|_{cb}\leqslant C$.

The problematic behaviour of multipliers manifests itself in the fact that they are not decomposable as maps from $C^{\ast}_r(G)$ to itself. What I mean by this is that if we decompose them as combinations of positive multipliers, then the norms of the terms involved are not controlled by the cb norm of the multiplier itself. A positive multiplier on $C^{\ast}_r(G)$ necessarily comes from a positive definite function on $G$, hence automatically extends to a multiplier on $C^{\ast}(G)$, so I think that decomposability is the real issue here.

  • $\begingroup$ Great answer. Do you know if there are other completions for which the answer would be positive? For instance consider the completion $C^*_{\pi}$ under a particular, fixed unitary representation $\pi$. Would the cb functions on $G$ extend to cb maps on $C^*_{\pi}$? Is it possible that $\pi$ contains the trivial representation? It sounds like this might be the problem here. $\endgroup$ Mar 7 '17 at 20:32
  • $\begingroup$ @user10439561 Can you clarify if by "completion" you are assuming $\pi$ is faithful? Also, at this rate your question is in danger of becoming a bit of a fishing expedition. Why do you want other completions for which this works, and why do you think there might even be any? $\endgroup$
    – Yemon Choi
    Mar 8 '17 at 3:24
  • $\begingroup$ @YemonChoi What do you mean by asking questions ``at this rate''? I thought this is what this site is for. Also, what is wrong with a fishing expedition? This is often how research is done. I'd like to understand what is the obstruction to extending the cb maps on $G$ to cb multipliers on the completion and there is not much on this in the literature. I would guess it is related to whether $\pi$ contains the trivial representation but I'd like to hear from someone who has an understanding of the issue. $\endgroup$ Mar 8 '17 at 6:36
  • $\begingroup$ To clarify what I mean by the completion: take a unitary representation $\pi$ of $G$ on a Hilbert space $H$ and the norm given by $\Vert f\Vert_\pi=\Vert\pi(f)\Vert_{B(H)}$. Denote the closure of $\mathbb{C} G$ in this norm by $C^*_{\pi}(G)$. For $\pi$ universal we get the maximal group $C^*$-algebra, for $\pi$ regular we get the reduced. $\endgroup$ Mar 8 '17 at 7:38
  • 1
    $\begingroup$ @user1043956: I don't really have much to say about these issues. It is true that if the trivial representation is weakly contained in $\pi$ then automatically all multipliers must come from $B(G)$. It seems to me that the case of the regular representation is really special: we have a canonical trace and the Fourier decomposition, and completely bounded multipliers always extend to Schur multipliers on $B(\ell^2(G))$. I am not sure how one could replace these essential tools in a more general context. One more issue: it is not clear if positive definite functions give rise to multipliers. $\endgroup$ Mar 8 '17 at 9:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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