Timeline for A group acting acylindrically on a fine hyperbolic graph with infinite edge stabilizers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2015 at 14:46 | comment | added | HJRW | @SvenjaKnopf, it's easy to modify Lee's example to satisfy condition (d). For instance, if the infinite vertex stabilizer $G_V$ has a $\mathbb{Z}/2$ quotient, then Lee's extra loop can be replaced by a pair of loops, swapped by $\mathbb{Z}/2$. | |
| Oct 26, 2015 at 8:48 | comment | added | Svenja Knopf | @Seirios If $T$ has a vertex $v$ of infinite valency (which it has to have to satisfy c) and d)), then the 1-skeleton of $T\times [0,1]$ is not a fine graph: the edge between $(v,0)$ and $(v,1)$ is contained in any of the infinitely many circuits of length 4 starting at $(v,0)$ running through $(w,0), (w,1), (v,1)$ back to $(v,0)$ where $w$ is any vertex in $T$ adjacent to $v$. | |
| Oct 24, 2015 at 9:02 | comment | added | Seirios | What about the following construction: Let $G \curvearrowright T$ be an action on a tree satisfying the conditions a), c) and d). Now, $G \times \mathbb{Z}_2$ acts on (the 1-squeleton of) $T \times [0,1]$, where $\mathbb{Z}_2$ permutes the two copies $T \times \{ 0 \}$ and $T \times \{ 1 \}$ of $T$. | |
| Oct 24, 2015 at 6:15 | history | edited | Svenja Knopf | CC BY-SA 3.0 |
fixed a typo
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| Oct 24, 2015 at 6:14 | comment | added | Svenja Knopf | @Misha Replacing one edge in a tree with infinitely many edges between its two endpoints results in a graph that is not fine. So this is excluded by a). | |
| Oct 24, 2015 at 6:08 | history | edited | Svenja Knopf | CC BY-SA 3.0 |
added "connected" in Condition d) to clarify the condition
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| Oct 23, 2015 at 18:16 | comment | added | Misha | @SvenjaKnopf: You should add "connected" subgraph to your condition (d), otherwise, you can, say, omit all the edges. But even this is not enough: You can start with a group action on a tree and replace an edge $e$ (with infinite stabilizer $G_e$) with infinitely many edges (connecting the same pair of vertices) and let $G_e$ permute these new edges transitively (but maybe not simply transitively). | |
| Oct 23, 2015 at 8:34 | history | edited | Svenja Knopf | CC BY-SA 3.0 |
added Condition d) to exclude artificial examples
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| Oct 23, 2015 at 8:24 | comment | added | Svenja Knopf | @LeeMosher thanks for the example! When asking the question I had a minimal action of $G$ on $\Gamma$ in mind though. I edited the question accordingly. | |
| Oct 22, 2015 at 14:42 | comment | added | Lee Mosher | "Fine" is defined by Bowditch, meaning that for each edge $E$ and integer $L \ge 1$ the graph has only finitely many circuits of length $L$ passing through $E$. homepages.warwick.ac.uk/~masgak/papers/bhb-relhyp.pdf | |
| Oct 22, 2015 at 14:41 | comment | added | Lee Mosher | One could easily concoct artificial examples from known examples. For instance, take a relatively hyperbolic example, so c) fails, let $V$ be a vertex with infinite stabilizer, and for each vertex in the orbit $G \cdot V$, attach an edge with both endpoints at that vertex. | |
| Oct 22, 2015 at 14:36 | comment | added | YCor | Could you define what a fine graph is? | |
| Oct 22, 2015 at 13:01 | review | First posts | |||
| Oct 22, 2015 at 13:33 | |||||
| Oct 22, 2015 at 12:53 | history | asked | Svenja Knopf | CC BY-SA 3.0 |