Timeline for Abelian sandpile models
Current License: CC BY-SA 3.0
11 events
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| Nov 21, 2011 at 7:31 | comment | added | Dima Pasechnik | @Mark, at least, if your infinite graph is a directed tree, with each vertex of finite degree, (I guess a DAG would work, I didn't check the details though-for finite DAGs it's OK) then the monoid of stable configurations is a group (which must coincide with the sandpile group), with the all-0 configuration as neutral element. Let $S$ be a stable configuration. If $x$ is a minimal vertex on which $S$ has a chip, i.e. $S(x)>0$, then $S^{-1}(x)=deg(x)-S(x)+1$, and $S^{-1}(y)=deg(y)-S(y)$ for each $y$ in the subtree rooted at $x$. In $S+S^{-1}$ all such $x$ can be fired, getting 0 in the limit. | |
| Nov 18, 2011 at 15:56 | comment | added | user6976 | @Dima: I do not know if sandpile groups of infinite graphs can be defined. Even "recurrent configurations" are not obviously define. So the inverse limit of the sandpile groups of finite subgraphs should be considered a definition of the sandpile group of the infinite graph. Then the question of uniqueness becomes even more important. | |
| Nov 18, 2011 at 14:37 | comment | added | Dima Pasechnik | @Mark: meant: "the sandpile groups of these de Bruijn digraphs..." | |
| Nov 18, 2011 at 14:36 | comment | added | Dima Pasechnik | @Mark: these de Bruijn digraphs, they do form an inverse system. In the general setting of L.Levine's paper, probably, too, it is true... I guess I know what the inverse limit in the de Bruijn digraphs case looks like, but have no clue how to confirm this. Neither I know if this limit can be realized as the sandpile group; I should look at the article of your student. | |
| Nov 18, 2011 at 11:52 | comment | added | user6976 | @Dima: Do these groups always form an inverse system? In that case a weaker question would indeed be whether the inverse limit does not depend on the sequence of subgraphs. Anyway, the question was related to my student's paper front.math.ucdavis.edu/1110.6263. I just asked him to define the sandpile model not for an infinite graph but for a sequence of finite graphs converging to an infinite graph. This way there is not ambiguity. | |
| Nov 18, 2011 at 8:18 | comment | added | Dima Pasechnik | There are situations when the sandpile groups of $\Gamma_k$'s form an inverse system (e.g. when $\Gamma_k$ are binary de Bruijn digraphs, for which Lionel Levine actually computed these groups (math.cornell.edu/~levine/directed-line.pdf), the homomorphisms are just taking $2^{i-j}$ powers ). IMHO the corresponding profinite group ought to be connected with $\Gamma$ – perhaps it's the right way to define its sandpile group? Perhaps this might help to understand the question asked :-) | |
| Oct 1, 2011 at 20:10 | history | edited | user6976 | CC BY-SA 3.0 |
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| Sep 27, 2011 at 1:43 | history | edited | user6976 | CC BY-SA 3.0 |
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| Sep 27, 2011 at 1:42 | comment | added | user6976 | @Noam: Of course you are right. I'll fix the question. | |
| Sep 27, 2011 at 1:26 | comment | added | Noam D. Elkies | How can the answer to "Why does the limit exist?" be "(obviously) yes"? Did you mean to ask "Does the limit exist?"? | |
| Sep 27, 2011 at 1:12 | history | asked | user6976 | CC BY-SA 3.0 |