I have very few ideas to how solve this problem.

Let $ D = \{ z^2 : 0 \leq \Re(z) , \Im(z) \leq 1 \}$. Discretize $D$ with $D_{\delta}$ by a honeycomb lattice of mesh size $\delta$ and color the hexagons of $D_{\delta}$ in black or white with probability 1/2 independently. Let $P_{\delta} $ denote the probability that the points $1/4$ and $-1/4$ are separated by a black path from the points $3/4 + i $ and $-3/4 + i $ in $D_{\delta}$. Show that $$ \lim_{\delta \to 0 } P_{\delta} = 1/2 $$

Problem 1: I don't get very well what shape has the domain $D$.

Idea: Carleson's formulation of Cardy's formula says what follows: If we have a Jordan domain $( \Omega, a_1,a_2,a_3,a_4)$, with $a_1,a_2,a_3,a_4 \in \partial \Omega $ in counterclockwise order, then it is known that there exist a unique conformal from $ \Omega$ to the equilateral triangle $ \Delta$ with vertices $1, \pm \frac{\sqrt{3}}{3} i $. Then if we discretize $\Omega$ by $\Omega_{\delta}$ $$ \lim_{\delta \to 0} \mathbb{P} \{ [a_1a_2] \leftrightarrow_w [a_3a_4] \} = \Re( \varphi(a_4) )$$ where $\leftrightarrow_w$ means that there exists a white path connecting $[a_1a_2]$ to $[a_3a_4]$.

But I truly believe improbable that I have to found explicitly a conformal map between $D$ and $\Delta$, unless it is easier than what I expect, so maybe there is some percolation argumentation that make this problem more easy than one would expect. Some ideas?

  • 1
    $\begingroup$ The points are the images under $z\mapsto z^2$ of the edge midpoints of the unit square. If you performed the discretization in the unit sphere, the result would follow by standard percolation duality. The hard part is showing that performing the discretization in the image $D$ doesn't break the symmetry. BTW, here's what $D$ looks like: desmos.com/calculator/teob3d32qv. $\endgroup$
    – joriki
    Jan 12, 2023 at 7:03
  • $\begingroup$ Mmh... I don't think D looks like the red region there since there is no $(10+i/2)^2$ inside, there is even not $x^2$ for alla real $x>1$ which are in $D$. I'dont know of this change something in your argumentation $\endgroup$
    – 3m0o
    Jan 12, 2023 at 10:15
  • $\begingroup$ We have seen a thm saying that if two Jordan domains $ ( \Omega',a_1',a_2',a_3',a_4') $ and $ ( \Omega,a_1,a_2,a_3,a_4)$ are conformally equivalent then $ \lim_{\delta \to 0} \mathbb{P}_{\Omega_{\delta}} \{ [a_1 a_2] \leftrightarrow [a_3a_4] \} = \lim_{\delta \to 0} \mathbb{P}_{\Omega_{\delta}'} \{ [a_1' a_2'] \leftrightarrow [a_3'a_4'] \}$. The unit square in $ \{ z : \Re(z) \geq 0 , \Im(z) < 0 \} $ and its image in $D$, via $z^2$ are conformal. But why we can consider only the unit square in the original domain? Why other region doesn't change probabilty in the square (via FKG inequality) ? $\endgroup$
    – 3m0o
    Jan 12, 2023 at 10:43
  • 1
    $\begingroup$ I took your notation $0 \leq \Re(z) , \Im(z) \leq 1$ to mean $0 \leq \Re(z)\leq 1 \land0 \leq \Im(z) \leq 1$, i.e., the unit square. If you mean $0 \leq \Re(z)\land \Im(z) \leq 1$, my comments are probably irrelevant. $\endgroup$
    – joriki
    Jan 12, 2023 at 11:06
  • $\begingroup$ I think you are right... it is at least two weeks that I think about this problem and I never considered the eventuality that by $ 0 \leq \Re(z), \Im(z) \leq 1$ means the unit square... and if otherwise I think that the probability is not $1/2$. $\endgroup$
    – 3m0o
    Jan 12, 2023 at 11:21


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