Kindly refer to this paper: <https://arxiv.org/abs/1502.03050>


In this paper, Hugo Duminil-Copin and Vincent Tassion have given an alternative proof of the well known results. I was reading this paper and could not figure out an argument they have used in the proof of item 1 in theorem 1.1, page 6.


I am mentioning the relevant details in short.



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> Consider a transitive grah $G=(V,E)$ on which we want to study bond percolation. Each edge $e\_i$ is open with probability $p\_i$ and closed with probability $1-p\_i$ independent of all other edges.
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> For an event $A$, an edge $e$ is said to be pivotal if changing the status (open or closed) of the edge $e$ causes $E$ to occur and not occur alternatively.
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> Define for a finite set $S$ of vertices, an edge on the "external boundary" of $S$ to be an edge of the form $\{x,y\}$ where $x\in S,y\notin S$.
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> Suppose we have a finite set of vertices $\Lambda$ with $0\in\Lambda$. Then define the random subset $\mathcal S$ of $\Lambda$ as: $\mathcal S=\{v\in \Lambda:v$ is **not connected** to $\Lambda^c$ by an open path$\}$.
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> 


Consider now the increasing event $A:=\{0$ is connected to $\Lambda^c$ by an open path$\}$.


They have argued briefly that the event $[xy$ is pivotal for $A$ and $\mathcal S=S]$ is equivalent to $xy$ edge on the outer boundary of $S$ (with $x\in S,y\notin S$) and $0$ is connected to $x$ through vertices in $S$ by an open path, and $\mathcal S=S$.


I do not understand this argument. Why should my pivotal edges be only on the outer boundary of $S$?


If such a pivotal edge exists in $S^c$, then I think changing the status of that edge will not affect the connectivity of $0$ by an open path to $\Lambda^c$. But is it because there are other edges available that can make a path from $0$ to $\Lambda^c$? Why also can the pivotal edge not be in $S$?



Not understanding a part of Hugo and Vincent's proof on Percolation
I think the detail you missed was that they assumed that $0 \in S$


If $\{x,y\}$ is a pivotal edge, we need that at least one of then connects to $0$,
and one of then needs to connect to $\Lambda^c$. Otherwise, opening this edge wouldn't make $0 \longleftrightarrow \Lambda^c$. Suppose then that $0\longleftrightarrow x$ and $ y \longleftrightarrow \Lambda^c$. If $x \in S^c$, then we would have that $0 \in S^c$ as $0 \longleftrightarrow x \longleftrightarrow \Lambda^c$, which is a contradiction and therefore $x \in S$. On the other hand, by definition of $S$, we have that $y \in S^c$. Thus $\{x,y\}$ is in the external boundary of $S$.


