Timeline for Can Thompson's group F be realized as a semigroup of continuous transformations of a tree?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2012 at 21:03 | answer | added | Matt Brin | timeline score: 3 | |
| Aug 24, 2011 at 22:52 | comment | added | Ian Agol | The semigroup action would have to be invertible on the image of the identity element (like in your example) which should be a subtree by continuity. So if you restrict to the subspace which is the image of the identity, you would get bijections of the subtree. It might be more natural to consider maps of trees up to homotopy, such as in the action of Out(F_n) on outer space. | |
| Aug 24, 2011 at 20:12 | answer | added | Benjamin Steinberg | timeline score: 3 | |
| Aug 24, 2011 at 19:59 | comment | added | dan | So, just to clarify, I'm not looking for a group action of F, it would have to be a semigroup action. | |
| Aug 24, 2011 at 19:54 | comment | added | dan | Consider, for example, two functions $a$ and $e$ from $\{1,2,3\}$ to $\{1,2,3\}$ defined by $a(1)=2, a(2)=1, a(3)=1$ and $e(1)=1,e(2)=2$, and $e(3)=2$. Then you can check that $a^2=e$, $e^2=e$, and $ea=ae=e$. So the semigroup generated by $a$ and $e$ is a group, and in fact it is the cyclic group of order 2. So you have a group defined by functions that are not injective or surjective. I'm just wondering if something like this can happen with F on a regular rooted tree, where the transformations are continuous and level-preserving. | |
| Aug 24, 2011 at 19:48 | answer | added | R W | timeline score: 2 | |
| Aug 24, 2011 at 18:07 | comment | added | André Henriques | Thomson's group $F$ is a group. How do you plan to define the inverse of a transformations that is not injective or surjective? | |
| Aug 24, 2011 at 17:44 | history | asked | dan | CC BY-SA 3.0 |