Why is the identity element of the sandpile group self-similar?

I've been reading about the Abelian Sandpile Model and noticed the identity element of the sandpile group on the square has self-similar components.

The sandpile group of the 198x198 square of height 4 above is a finite abelian group. However, the sandpile corresponding to elements of this group can be fractal. Where is the complexity coming from?

Also, how does can you find the identity element (say, with a computer)?

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According to the paper you link to, this work arxiv.org/abs/0809.3416 studies the question of the fractal structure. –  j.c. Jul 8 '10 at 16:36

Since the time this question was asked, there have been some important breakthroughs on the convergence of the scaling limit of the sandpile model (and its identity element), which may explain some of the self-similarity you note.

See especially the papers: http://arxiv.org/abs/1105.0111 and http://arxiv.org/abs/1208.4839 and http://arxiv.org/abs/1309.3267

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Could you please fix the first link to go to the abstract rather than directly to the PDF? Thank you! –  Harry Altman Nov 25 '13 at 23:37
@HarryAltman: fixed. –  Sam Hopkins Nov 25 '13 at 23:39

As per your second question, the following algorithm allows one to compute the identity element.

Let $c$ denote the maximal stable configuration; i.e. $c = \sum_{v\in V}(d(v)-1) v$ This is always recurrent. Let $a^{\circ}$ denote the stabilization of a configuration $a$. Then this will give you the identity $e$:

$e =(2c - (2c)^\circ)^\circ$

If you are interested, check out this applet for doing a lot of this stuff (and it's pretty, too): http://people.reed.edu/~davidp/sand/program/program.html

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Sage (sagemath.org) has good support for computing with sandpiles: sagemath.org/doc/thematic_tutorials/sandpile.html –  Dima Pasechnik Nov 18 '11 at 8:01