5
$\begingroup$

In Local Statistics of Lattice Dimers we study a nice familiar object, domino tilings in the plane extending out to infinity.

His paper is going to discuss the frequency of various "motifs" in tiligns of infinite dimer regions and I am hoping to verify all such frequences lie in $\mathbb{Q}[\tfrac{1}{\pi}]$.

However, I am thrown off by the following phrase:

It is known that the translation-action of $\mathbb{Z}^2$ on dominos or lozenges is topologically mixing

I didn't even know where was a translation action on domino tilings. Kenyon proceeds to quantify this "mixing" as

$$ \mu(U_{T_1}\cap U_{v+T_2}) = \mu(U_{T_1})\,\mu(U_{v+T_2}) + O(\tfrac{1}{|v^2|}) $$

Apparently the translation $T_v: \mathbb{Z}^2 \to \mathbb{Z}^2$ induce a translation on the set of infinite dimer tilings on the plane $T_v: X \to X$.

This mixing result says any two motifs $T_1, T_2$ are relatively independent of each other. Then a tiling is expected to have a motif $T_1$ in one area with probability $\mu(U_{T_1})$ and $T_2$ with probability $\mu(U_{T_2})$. The odds of having both is close to $\mu(U_{T_1})\mu(U_{T_2}) $


Intuitively, I would have just embedded both regions $T_1$ and $T_2$ in a tilings of the square and computed the frequences of both motifs $T_1$ and $v + T_2$ occurring in random tilings as the size of the square got large.

However, the more they try to get precise, the more confused I get.

  • Translation is not mixing in the plane, and yet it is mixing on the space of domino tilings. How does that make sense?
  • What was the important of identifying the unique measure $\mu$ of maximal entropy? and showing $\mu$ was the same as that of uniform random tiling for large squares with entropy ? $$ H = \frac{1}{k^2} \log |\#\{ \text{ tilings of }\square_{k \times k}\}|$$
$\endgroup$
  • $\begingroup$ What do you perceive to be the connection between translation in the plane and translation on the space of domino tilings? The former is an action of $\mathbb{R}^2$ on itself, the latter is an action of $\mathbb{Z}^2$ on a closed subspace of $\{1,\ldots,k\}^{\mathbb{Z}^2}$. Or do you mean something else by "translation is not mixing in the plane"? $\endgroup$ – Ian Morris Jun 2 '14 at 12:01
  • $\begingroup$ @IanMorris The dominoes themselves can be shifted by a vector $v \in \mathbb{Z}^2$ to get another tiling. The translation action of $\mathbb{Z}^2$ on itself is definintely not mixing with respect to any measure. However, the action on the space of tilings doesn't appear to mix very much either. $\endgroup$ – john mangual Jun 2 '14 at 12:14
  • 5
    $\begingroup$ I'm not sure that that answers my intended question: why should the non-mixing of the action of $\mathbb{Z}^2$ on itself imply the non-mixing of an action of $\mathbb{Z}^2$ on the much richer space $X^{\mathbb{Z}^2}$? (Incidentally it seems to me that the number "2" is not particularly significant here. For example, $\mathbb{Z}$ induces a mixing action on $\{0,1\}^{\mathbb{Z}}$ equipped with Bernoulli measure $(\frac{1}{2}(\delta_0+\delta_1))^{\mathbb{Z}}$.) $\endgroup$ – Ian Morris Jun 2 '14 at 12:18
  • 1
    $\begingroup$ @IanMorris Truthfully that doesn't look very mixing either. All you did was take the same infinite word and move 5 spaces left. $\endgroup$ – john mangual Jun 2 '14 at 12:22
  • 3
    $\begingroup$ It seems to me then that you are confusing the effect of the action on open sets of configurations (which is what occurs in the definition of topological mixing) with the effect of that action on an individual configuration (an unilluminating isometry on $\mathbb{Z}^2$: but the action is typically not an isometry on the configuration space). The meaning of "mixing" in this context is that open sets of configurations will tend to be brought, by the action, into intersection with one another. $\endgroup$ – Ian Morris Jun 2 '14 at 12:27
3
$\begingroup$

Let's start with Ian's example of $\{ 0 , 1 \}^{\mathbb{Z}}$. I'll write a point of $\{ 0,1 \}^{\mathbb{Z}}$ as a doubly infinite sequence $(x_i)$. The topology on $\{ 0,1 \}^{\mathbb{Z}}$ has a basis of open sets of the form $\Omega(L,R, (a_L, \ldots, a_R)) := \{ (x_i) : (x_L, x_{L+1}, \ldots, x_R) = (a_L, a_{L+1}, \ldots, a_R) \}$ for some integers $L \leq R$ and some fixed bit sequence $(a_L, \ldots, a_R)$. The measure of this open set is $1/2^{R-L+1}$. Intuitively, we are talking about independent random coin flips.

Saying that translation $T$ is strongly mixing means that, for any $(L,R, (a_L, \ldots, a_R))$ and $(L', R', (a_{L'}, \ldots, a_{R'}))$, the probability that $x \in \Omega(L,R, (a_L, \ldots, a_R)) \cap T^N \Omega(L',R', (a_L', \ldots, a_R'))$ approaches $2^{-(R-L+1)} 2^{-(R'-L'+1)}$ as $N \to \infty$. Sure enough, as soon as $N$ is large enough that the intervals $[L,R]$ and $[L'+N, R'+N]$ don't overlap, the bits in those intervals become independent.

Note that it is exactly the fact that translation is NOT mixing on $\mathbb{Z}$ that allows it to be mixing on $\{ 0,1 \}^{\mathbb{Z}}$. If translation were mixing on $\mathbb{Z}$, then arbitrary translates of $[L',R']$ would continue to intersect $[L',R']$, so the bit strings would be correlated on these intervals and the probabilities wouldn't become independent.

Dimers are more complicated, because the condition that the region outside two planar patches is tileable imposes constraints even when the regions don't overlap. But it is intuitively plausible that this correlation decays as the regions become further separated, and the equation you quote makes this precise.


Your stated goal is true, and is pointed out by Kenyon immediately after Theorem 1. He expresses the frequency of motif with $k$ dimers as a $k \times k$ determinant, each entry of which is in $\mathbb{Q} + \mathbb{Q} (1/\pi)$.

$\endgroup$
  • $\begingroup$ "Ergodicity" is not a word I usually expect from a discussion of dimers and here the shift map is throwing me for a loop.- two of them since its $\mathbb{Z}^2$. For now the project is draw a large square domino tiling, count the motifs numerically and verify the determinant. $\endgroup$ – john mangual Jun 2 '14 at 19:01
  • $\begingroup$ I know that the uniform distribution is the "maximum entropy" distribution on a discrete set, unless you put a constraint like the expectation or the variance. The maximum entropy translation invariant measure $\nu$ they find for infinite lozenge tilings is a conjecture! Ergodicity of the shift map is a big them in Jockusch-Shor-Propp in that case the bits are coming from domino shuffling. $\endgroup$ – john mangual Jun 2 '14 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.