Timeline for specific qi on free groups
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2016 at 9:30 | comment | added | HJRW | The question as stated is certainly not correct: a better question would be to ask whether $\phi$ is at most finite distance from some $h\mapsto g\psi(h)$. (When one defines the quasi-isometry group of a metric space, this is the notion of equivalence one uses.) As @YCor's example shows, however, the answer is still 'no'. | |
| Dec 1, 2016 at 17:07 | comment | added | YCor | The group $G=PSL_2(\mathbf{Q}_p)$ is uncountable and QI to $F_n$. Its left translations are self-QIs and are pairwise non-equivalent (=not at finite distance). Conjugating by a QI $G\to F_n$ you get the same uncountable number of pairwise non-equivalent self-QIs on $F_n$. | |
| Dec 1, 2016 at 16:44 | comment | added | Ashot Minasyan | Now, $T_1$ is a connected subset of the Cayley tree, which has vertices arbitrarily far from $v$. Therefore $T_1$ must contain a vertex corresponding to an element $h \in H$ (because $H$ has finite index, so every element of $F_n$ is within a bounded distance from an element of $H$). Hence $\phi(h) \neq h$, as $\phi(h)$ is a vertex of $T_2$, which is disjoint from $T_1$. Thus $\phi$ cannot induce the identity on $H$, and the above claim follows. | |
| Dec 1, 2016 at 16:36 | comment | added | Ashot Minasyan | First, let's restrict ourselves only to isometries fixing the identity (there are still $2^\omega$ of them). But this means that $g=1$ for any such isometry. Next, let's show that any pair $(H,\psi)$, using your notation, has at most one preimage in the group of isometries of $F_n$ (fixing $1$). Indeed, if $\phi:F_n \to F_n$ is a non-trivial isometry fixing $1$, then it must interchange at least two branches $T_1$ and $T_2$, of the Cayley graph, which originate at some fixed vertex $v$. | |
| Dec 1, 2016 at 14:50 | comment | added | Jiang | @AshotMinasyan, if I understand your comment correctly, you are trying to use cardinality argument, but I am not sure why the map $\{QI~~ on ~~F_n\}\to \{finite ~~index~~ subgroups~~ of ~~F_n\}\times F_n $ should be at most countable to one. | |
| Dec 1, 2016 at 14:40 | comment | added | Ashot Minasyan | As a metric space $F_n$ has a continuum of isometries, as they correspond to automorphisms of its Cayley graph, which is an infinite $2n$-regular tree. Therefore I do not think that any such description of isometries is possible. | |
| Dec 1, 2016 at 14:19 | history | asked | Jiang | CC BY-SA 3.0 |