Now that I’ve got your attention….

Sadhu and Dhar say in the paper I linked in the last post that it remains open whether the sandpile ring A is isomorphic to R^{2^N} when N is not congruent to 3 mod 4. I think there’s an easy argument that shows the answer is yes. Let’s give it a go. We will freely use the notation from this afternoon’s post.

It’s not hard to see that the only way two different length-N words in {T,T’} can yield identical homomorphisms g and g’ from A to R is if g(x_i) = g'(x_i) = 0 for some x_i. So this amounts to showing that when N is not 3 mod 4, no x_i goes to 0.

If x_i = 0, then we must have

x_{i-1} = -x, x_i = 0, x_{i+1} = x

for some real value x, and

x_0 = x_{N+1} = 1.

The choice of x has the effect of scaling everything, so this is the same as saying

[x_{i-1},x_i,x_{i+1}] = [-1,0,1]

and then we want to get x_0 = x_{N+1}.

Here’s an example of this with N=3:

[2,-1,0,1,2]

Note that we can think of this as running the random walk in SL_2(Z) starting from [0,1] (as we move rightward from i) or starting from [0,-1] (as we move leftward.) So what we’re trying to prove amounts to the following:

Start with [0,1], and for each word w in {T,T’}, we let f(w) be the second coordinate of the pair of integers obtained by applying w to [0,1]. Write |w| for the length of w. It suffices to show:

**Claim:** If f(w) = -f(w), then |w| + |w’| = 2 mod 4.

(For instance, in the example above, f(T) = 2 and f(T’) = -2, and the summed length is 2.)

But now just look at what happens mod 8. Why mod 8? Because I tried 2 and 4 and they didn’t work. Starting with [0,1], and applying either T or T’ at each stage, you get

0,1,±2,3, 4 or 0, 5, ±2, 7, 4 or 0, 1, …..

and it cycles thereafter, with period 8. And now the claim follows by inspection: for instance, if f(w) = 2 or -2 mod 8, then |w| is 1 or 5 mod 8. If f(w) = 3 mod 8, then |w| is 2 mod 8, while if f(w) = -3 mod 8, then |w| is 4 mod 8. And so on.

Sound good?

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