Timeline for Quasi-isometry classes of elementary amenable groups
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2011 at 20:05 | comment | added | user6976 | @Denis: If you still do not know how to do it by January 3 when I return to Nashville, we can discuss it then. I do not think we should take more space here. | |
| Dec 26, 2011 at 19:30 | comment | added | Simon Thomas | @Denis: In fact, I would not fix the sequences in advance. I would construct a complete binary tree of sequences at the same time that I was inductively constructing the groups. | |
| Dec 26, 2011 at 18:26 | comment | added | user6976 | @Denis: If we know $c_i$ (the parameters of the construction), we know the lengths of relations of the virtually free groups approximating the amenable group. It is easy to see which of these sequences of relations tend to a non-trivial loop in the asymptotic cone. One does not need small cancelation for that. In fact a sequence gives a non-trivial loop if the relations of that sequence do not follow from smaller relations of the group. That is the Gromov "divisibility" property. It remains to choose $c_i$ in a proper manner which is, I repeat, easy. | |
| Dec 26, 2011 at 17:43 | comment | added | Denis Osin | @Simon: Of course this kind of "strengthening" is an easy exercise for your construction since you have small cancellations there and you can effectively prove that a given cone is not an R-tree. This is not so in our construction: we know that some cones are not R-trees, but it is not obvious which ones. So fixing sequences in advance does not work without some additional arguments. See my comment to Mark. | |
| Dec 26, 2011 at 17:40 | comment | added | Denis Osin | @Mark: OK, here is a very particular question. We are given a sequence $(d_n)$. How do we create an elementary amenable group as in our paper with asymptotic cones corresponding to $(d_n)$ not R-trees? (And how do we prove that they are not R-trees?). This is not similar to Thomas-Velicovic, since they do know how to answer this question, while in our construction this is not at all obvious. | |
| Dec 26, 2011 at 15:59 | comment | added | Simon Thomas | @Mark: In my graduate course on geometric group theory, I usually show that there are T-V groups $G_{\alpha}$ for $\alpha < 2^{\omega}$ such that for every subset $S \subseteq 2^{\omega}$, there is an ultrafilter such that the corresponding cone of $G_{\alpha}$ isn't simply connected if $\alpha \in S$ and is an $\mathbb{R}$-tree otherwise. As you have been pointing out, this kind of "strengthening" is an easy exercise. | |
| Dec 26, 2011 at 7:20 | comment | added | user6976 | @Denis: Create enough sequences in advance, then create groups. It is similar to the fact that there are uncountably many q.i. classes of Thomas-Velickovic groups (I think it also has not been mentioned in their paper or elsewhere). | |
| Dec 26, 2011 at 1:51 | comment | added | Denis Osin | @Mark: Yes, I understand this. So we have 2 non-q.i. groups. In a similar way we can construct infinitely many. But how do we pass to continuously many? | |
| Dec 25, 2011 at 23:23 | comment | added | user6976 | @Denis: If a group is infinitely presented one of the cones has non-trivial fundamental group. The non-trivial loop comes from a sequence of defining relations. In our group, we know the lengths of defining relations. That gives scaling constants for a non-tree cone. In another group the same scaling sequence will get a tree. | |
| Dec 25, 2011 at 22:44 | comment | added | Denis Osin | @Mark: Right, but as I said, it is not clear to me how to prove that a particular cone is not an R-tree. | |
| Dec 25, 2011 at 22:43 | comment | added | Denis Osin | @Yves: Yes, this is the way to go. But the problem is that we know how to prove that some ultrafilters belong to N(G), but we do not know how to prove that an ultrafilter is not in N(G). At least this is not immediate from our paper. | |
| Dec 25, 2011 at 22:43 | comment | added | user6976 | @Denis: I think you are make the problem too complicated. The problem whether all cones have cut points is much more difficult. One does not need to consider all cones. Only those that are guaranteed to be trees and those that are guaranteed to be non-trees. @Yves: What about solvable groups? | |
| Dec 25, 2011 at 22:33 | comment | added | YCor | @Denis: to any f.g. group $G$, you can associate the set $N(G)$ of ultrafilters $\omega$ such that $Cone(G,\omega,(1/n))$ is a real tree. So $N(G)$ is a QI-invariant of $G$. I think the idea is to show that elementary amenable groups achieve continuum many sets $N(G)$. It's not obvious (and I haven't checked) but I guess you can do this. | |
| Dec 25, 2011 at 22:30 | comment | added | Denis Osin | @Mark: Of course, each our group has a sequence of relations of certain fast growing lengths. But we do not know anything about the geometric shape of these relations in the Cayley graph, so it is not clear how to show that the corresponding asymptotic cone is not an R-tree. If we could do this, we could probably also decide whether or not all cones of our groups have cut points (which is an open problem in our paper). | |
| Dec 25, 2011 at 22:25 | comment | added | user6976 | @Denis: It does not seem to be difficult: each group should have a sequence of relations of certain fast growing lengths while in all other groups this sequence misses all relations. I think it is easy. A solvable example is more difficult. One would have to deal with Abels' group or Kharlampovich's group and its factor-groups over central subgroups. Perhaps again asymptotic cones can help. But it is not clear. | |
| Dec 25, 2011 at 22:10 | comment | added | Denis Osin | @Mark: I think it does not follow directly, although one can probably prove it this way. One problem is that there exist 2 elementary amenable groups G, H as in our paper (corresponding to different parameters) such that for any scaling sequence and any ultrafilter, the corresponding cone of at least one of G, H is an R-tree. So any third group cannot be distinguished from both G and H by means of asymptotic cones. It is not obvious to me how to choose those "sufficiently different sequences of parameters" to avoid this problem. | |
| Dec 25, 2011 at 20:41 | comment | added | Bill Johnson | @Denis: You are too young to be forgetting your own theorems. :) | |
| Dec 25, 2011 at 20:35 | history | answered | user6976 | CC BY-SA 3.0 |