Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the two-dimensional lattice $G=(\mathbb{Z}^2,\mathbb{E}^2)$, where edge set $\mathbb{E}^2$ is given by the pairs of nearest neighbors in the $\ell^1$ norm in $\mathbb{Z}^2$. Let $V\subset\mathbb{Z}^2$ an infinite subset and suppose that $G[V]$, the induced subgraph, is connected. Let $\Lambda_n=([-n,n]\times[-n,n])\cap \mathbb{Z}^2$ be a sequence of squares on the two-dimensional lattice. Suppose additionally that $$ \limsup_{n\to\infty} \frac{|V\cap \Lambda_n|}{|\Lambda_n|}=0. $$ Question: under the above conditions is it true that the independent bond percolation, with parameter $p$, on $G[V]$ is trivial, in the sense that for any $p\in [0,1)$ we do not have almost surely an infinite cluster ?

I suspect that the answer is affirmative and this is considered in the literature, but until now I only found trivial examples of such graphs $G[V]$, basically constructed from the one-dimensional lattice, where there is no percolation.

share|improve this question
add comment

1 Answer

up vote 5 down vote accepted

This is not true without additional assumptions on your subset. E.g. consider $$V = \{(x,y) \in \mathbb Z^2 : x>0, |y| < \sqrt x \}.$$ It clearly has zero asymptotic density, but on the other hand its critical point is equal to $1/2$ like that of $\mathbb Z^2$ itself. (You can prove that it percolates for every $p>1/2$ using exponential decay of dual clusters in the whole lattice, for instance.)

share|improve this answer
    
Hi Vincent, thanks for the answer. –  Leandro Apr 2 '13 at 15:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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