Timeline for Aperiodic graphs
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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Mar 10, 2015 at 9:18 | answer | added | Jesús Álvarez | timeline score: 0 | |
Mar 8, 2015 at 17:25 | answer | added | Jamie Walton | timeline score: 4 | |
Mar 7, 2015 at 21:42 | answer | added | domotorp | timeline score: 0 | |
Mar 7, 2015 at 21:00 | comment | added | Jesús Álvarez | You're right. Sorry. Perhaps one has to use more numbers to get an aperiodic decoration of $\mathbb Z$. Of course it's very easy if one uses a decoration with infinite values; for instance, assigning $n$ to each $n$. But, to justify my question, one should be able to construct an aperiodic decoration of $\mathbb Z$ using a finite number of values. I'll think more about it. | |
Mar 7, 2015 at 20:25 | comment | added | domotorp | Won't this have $\dots−0−0−1−1-0−0−1−1-\dots$ in its hull? | |
Mar 7, 2015 at 20:19 | comment | added | Jesús Álvarez | I think so. Let us begin with the decoration $\alpha_0$ given by $\cdots - 0 - 1 -0 - 1 - \cdots$. By induction on $n=1,2,\dots$, define the decoration $\alpha_n$ by changing the decoration $\alpha_{n-1}$ in each interval $[10^{10n}m+1,10^{10n}m+2n]$ ($m\in2\mathbb Z+1$) by a decoration of the form $0 - 0 - 1 - 1 - 0 - 0 - 1 - 1 - \cdots - 0 - 0 - 1 - 1$. The sequence $\alpha_n$ has a limit decoration $\alpha$ in the obvious sense, and $\alpha$ is aperiodic. | |
Mar 7, 2015 at 17:17 | comment | added | domotorp | Btw, do you have a simple aperiodic decoration for the Cayley graph of $\mathbb Z$? | |
Mar 6, 2015 at 23:11 | comment | added | domotorp | I see, so a graph G' is in the hull of G, if when you "look around" in G' for a finite number of steps, you might as well be in G. | |
Mar 6, 2015 at 22:17 | comment | added | Jesús Álvarez | Consider for instance the Cayley graph $G$ of the $\mathbb Z$, with the decoration $\alpha$ of the form: $\cdots-1-0-0-0-1-0-0-1-0-1-0-0-1-0-0-0-1-\cdots$. Then the decorated graph $\cdots-0-0-0-\cdots$ is in the hull of $(G,\alpha)$ because its a union of decorated balls with the same center and increasing radius that are isometric to decorated balls in $(G,\alpha)$. | |
Mar 6, 2015 at 21:05 | comment | added | domotorp | Could you maybe provide a link for these notions or give an example for the hull of an infinite graph? | |
Mar 6, 2015 at 15:58 | comment | added | Jesús Álvarez | The terms non-periodic and aperiodic are used with different meanings here. The considered class consists of connected countable graphs with finite degree at each vertex, but, in the question, one takes a graph with uniformly bounded degree for all vertices. If the graph is infinite, its hull also consists of infinite graphs. If a graph is finite, its hull only has that graph. | |
Mar 6, 2015 at 15:05 | comment | added | domotorp | I don't understand the question. Aperiodic and non-periodic are two different things? What is "this class"? Countable, finite degree graphs? The hull is a family of finite or infinite graphs? | |
Mar 6, 2015 at 8:10 | history | asked | Jesús Álvarez | CC BY-SA 3.0 |