In M. Barlow's paper: arxiv.org/pdf/math/0302004.pdf, P17- (2.7) formula. 



> 
> Let $k\geq 10$, and consider a tiling of $\mathbb{Z}^2$ by disjoint squares
>  $$T(x):=\{y\in \mathbb{Z}^2: x\_i\leq y\_i< x\_i+k, i=1, 2\}$$
>  with side $k-1$. 
>  Let $\widehat{Q}$ be a macroscopic square of side $m$, and associate with $\widehat{Q}$ the microscopic square $$Q=\cup\{T(x), x\in \widehat{Q}\}$$
> 
> 
> 


How to understand $T(x)$ and $Q$? I am confused about why $T(x)$ is a disjoint tiling of $\mathbb{Z}^2$ and what is $Q$? 



A tiling of $\mathbb{Z}^2$ from M. Barlow's paper
You have made a few copying errors ($T$ for $T^+$, for instance). 


For the tiling, my guess is that the claim is not that $\{ T(x) \, : \, x \in \mathbb{Z}^d \}$ is a tiling (that would be false), but rather one just takes enough $x$'s to get a tiling, for instance $\{ T(x) \, : \, x \in k\mathbb{Z}^d \}$.


For the $Q$ vs. $\hat{Q}$ (Barlow uses $\tilde{Q}$, and I'm not sure why you changed this), note from Definition 2.1 that $T^+(x) = \mathbb{Z}^d \cap A\_1$ where $A\_1$ is the cube with the same center as $T(x)$ but side length $3k/2$. The centers of the cubes $T^+(x)$, for $x \in \tilde{Q}$, then form a cube inside $\tilde{Q}$ with $2m/3k$ cube centers per edge. In particular, this is smaller than $m$, which is why this is called microscopic. My guess is that this has something to do with the overall renormalization argument.


If this doesn't clear things up, I recommend emailing Barlow about it. I'm definitely not an expert in this area.


