1
$\begingroup$

For sake of concreteness, we can consider bond percolation on an infinite $\mathbb{Z}^d$ lattice. Let $p_b^{(d)}$ be the percolation threshold in $d$ dimension. It is known that $p^{(2)}_b = \frac{1}{2}$ exactly and $p_b^{(3)} \approx 0.245$ from simulations. This seems to suggest that $p^{(2)}_b > p^{(3)}_b$.

My question is the following: is it rigorously known that when the dimension is increased, the bond percolation threshold must strictly decrease? In other words, given any pair $d_2 > d_1$, $p^{(d_2)}_b < p^{(d_1)}_b$ must hold?

$\endgroup$

1 Answer 1

1
$\begingroup$

There's a short argument that $p_b^{(2d)} < p_b^{(d)}$.

To prove this, look at the $d$-dimensional sublattice of $\mathbb Z^{2d}$ formed by the points where the last $d$ coordinates are $0$. Between two adjacent points $$(x_1, \dots, x_i, \dots, x_d, 0, \dots, 0) \qquad (x_1, \dots, x_i+1, \dots, x_d, 0, \dots, 0)$$ we can find two edge-disjoint paths: one is just the edge between them, and one is the path which takes a step in dimension $d+i$, then dimension $i$, then dimension $d+i$ again. By alternating the sign of the $(d+i)$-dimension step depending on the parity of $x_i$, we can ensure that none of these paths share any edges.

With probability-$p$ bond percolation in $\mathbb Z^{2d}$, the probability that both paths are closed is $(1-p)(1 - p^3) = 1 - p - p^3 + p^4$, so bond percolation in this $d$-dimensional sublattice is dominated by probability-$(p+p^3-p^4)$ bond percolation in $\mathbb Z^d$. We can always choose a value of $p$ such that $p < p_b^{(d)} < p + p^3 - p^4$. For such a value of $p$, bond percolation on $\mathbb Z^{2d}$ will have an infinite component with probability $1$, so $p_b^{(2d)} < p$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .