Timeline for Lacunary hyperbolic groups and weak amenability
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2012 at 4:17 | comment | added | Denis Poulin | @Mark: Yes my goal is to construct a HAagerup Traski monster and then to see if it is possible to make it weakly amenable. | |
| Mar 10, 2012 at 5:35 | comment | added | user6976 | I do not know what $A(G)$ is, unfortunately. A group containing expander does not act non-trivially on Hilbert spaces (by isometry). I know what Haagerup property (a-T-menability) is. Your goal is to construct a Haagerup Tarski monster? | |
| Mar 8, 2012 at 21:43 | comment | added | Denis Poulin | @Mark: I am not familiar at all with the notion of expander. However, any information or possible link between a Tarski's monster group $G$ its Fourier algebra $A(G)$ is welcome. I would like to mentioned that even if can only prove that $A(G)$ has an approximate identity or the Haagerup property would be a really nice result. | |
| Mar 8, 2012 at 18:57 | comment | added | user6976 | @Denis: I see. There are many different Tarski monsters. One can most probably make a Tarski monster that contains an expander, has property (T) and so on. Would that help? | |
| Mar 8, 2012 at 16:09 | comment | added | Denis Poulin | Dear Professor Mark Sapir, the Arens regularity is a property which passes to closed subalgebra. Hence, if $A(G)$ is Arens regular where $G$ is discrete, then for any subgroup $H$, $A(H)$ is also Arens regular. Now, by our result, it means that $G$ cannot contains any infinite weakly amenable subgroup. This explain why Tarski's monster group are the only class of groups which are possible counter-example as they have no proper infinite subgroups. | |
| Mar 8, 2012 at 10:57 | comment | added | user6976 | I still do not quite understand the question. Why Tarski monsters are the only possible counterexamples? How about Gromov's groups containing expanders, satisfying property (T), etc? These are also lacunary hyperbolic. | |
| Mar 7, 2012 at 16:08 | comment | added | Denis Poulin | Dear all, the main reason of this question is a long standing conjecture which states that the Arens regularity of the Fourier algebra implies that the group is finite. M. Neufang and I proved this for all weakly amenable groups. Now, the only possible counter examples of this conjecture are Tarski's monster groups described with the characterization lacunary hyperbolic groups. However, the general question can be formulated as if $G$ is the direct limits of weakly amenable groups $G_i$ with $\Lambda_{G_i} < M$ for all $i$, is $G$ weakly amenable or at least have the Haagerup property. | |
| Mar 7, 2012 at 6:52 | comment | added | Mikael de la Salle | Lattices in $Sp(n,1)$ are hyperbolic and have Cowling Haagerup constant $2n-1$. It is precisely from this result that the term "Cowling-Haagerup constant" comes from. | |
| Mar 7, 2012 at 3:36 | comment | added | YCor | Is there any reason to focus the question on lacunary hyperbolic groups rather than arbitrary direct limits of surjections of hyperbolic groups? | |
| Mar 7, 2012 at 3:16 | comment | added | Denis Poulin | Let say that an approximate identity $e_{\alpha}$ in $A(G)$ is $C$-completely bounded the completely bounded norm of the net $e_{\alpha}$ is bounded by $C$. The constant $\Lambda_{G}$ for a locally compact group $G$ is the infimum of such $C$ over all the $C$-completely bounded approximate identity of $A(G)$. For the free group over 2 generators $F_2$, U. Haagerup proved that $\Lambda_{F_2} = 1$. | |
| Mar 7, 2012 at 3:14 | comment | added | Denis Poulin | Let say that an approximate identity $e_{\alpha}$ is $C$-completely bounded the completely bounded norm of the net $E_{\alpha}$ is bounded by $C$. The constant $\Lambda_{G}$ for a locally compact group $G$ is defined as $$ \Lambda_{G} = \\{C~|~\exists \textrm{$C$-completely bounded approximate identity in A(G)}\\}.$$ For the free group over 2 generators $F_2$, U. Haagerup proved that $\Lambda_{F_2} = 1$. | |
| Mar 7, 2012 at 2:11 | comment | added | user6976 | Perhaps it would be better if you explain what $\Lambda$ is. What is that constant for the free group or a virtually free group? | |
| Mar 6, 2012 at 22:42 | history | asked | Denis Poulin | CC BY-SA 3.0 |