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I have some problem to understand the following:

Let $X=\left\{0,1,2\right\}$ and consider $X^{\mathbb{Z}^d}, d\geq 2$ as being the set of all function from $Z^d$ to $X$. So for $\eta\in X^{\mathbb{Z}^d}$ it is either $\eta(x_i)=0$, $\eta(x_i)=1$ or $\eta(x_i)=2$.

Each $x\in Z^d$ is supposed to have 2d neighbours (for $d=2$ this means the point above, the point below, the point to the left and the point to the right).

Now the following definition is given.

We call a set of distinct points $x_0,x_1,\ldots,x_n=x_0$ a defect for $\eta$, if $x_{i+1}$ is a neighbor of $x_i$ for each $i$ and $$ \frac{1}{3}\sum_{k=0}^{n-1}\left\{\eta(x_{k+1})-\eta(x_k)\right\}\neq 0, $$ where the summands are all chosen mod 3 to be -1, 0 or 1 and the sum is ordinary addition, not mod 3.

My problem is: I do not see why the summands, when mod 3, are 0, 1 or -1.

What, if $\eta(x_{k+1})=2, \eta(x_k)=0$? Then the summand is 2 mod 3 =2....

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    Usually it doesn't matter what convention you use for the representatives of the classes in modular arithmetic. Here it does because you use the representatives in ordinary addition. Consequently it matters that here you are choosing the representatives ${ -1,0,1 }$ and not ${ 0,1,2 }$. – Ian Dec 29 '14 at 17:52

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It's telling you that for this purpose $2 - 0 = 2$ is replaced by $-1$ (since $2 \equiv -1 \mod 3$), and similarly $0 - 2 = -2$ is replaced by $1$. As to "why", I can't really say, because you haven't shown us what's going to be done with this definition.

Robert Israel
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  • I think it has something to do with the index? The authot writes: "Intuitively, if we view 0,1,2 as sitting on a circle, then as we go along the points $x_0,x_1,\ldots,x_{n-1}$ and return to $x_0$, the corresponding path on the circle has nonzero index. - - So this is maybe the reason to use 1,0,-1? Or what do you think? –  Dec 29 '14 at 18:01
  • To be honest, I do not really understand what this has to do with the index of a path?! –  Dec 29 '14 at 18:21