Last time we outlined the four steps of the proof of Harris-Kesten and worked through the first. This time, we’ll do the second: showing that for , there is an infinite open cluster with probability 1, allowing us to conclude that . We begin with two lemmas.

**1. Some Lemmas **

Recalling our discussion of the combinatorial hypercube, we say that a bond is pivotal for an event in a configuration if switching the state of takes into or out of . The influence of on is

For increasing, this simplifies to

where are configurations that differ from only at .

**Lemma 1.** * Let be an rectangle in . For a bond in ,
*

for .

*Proof:* For this proof, we ignore all bonds outside . The argument is a little weird: we’ll prove that

and

Then, noting that is an increasing function of , the first allows us to conclude for and the second for .

Since and , any horizontal crossing of uses . Thus, in , each endpoint of is joined by an open path to either end of , at least one of which must have length , so the first bound given above follows.

Since , a previous lemma tells us that . Similarly, , so contains an open dual path crossing and using the dual edge of . So that edge has an endpoint which is the start of an open dual path of length , and so we get the second bound stated above.

**Lemma 2.** * Fix and an integer . There exist constants (that depends on ) and (that depends on and ) such that
*

for all .

*Proof:* Recall that in a remark last time, we mentioned that

Setting gives

and it follows from this that

for . (We’re using some bounds on from last post here.) Since , there must be a constant such that . The consequence of this we need here is that

Now, in the proof of Harris’s theorem, we showed that

and combining this with the previous lemma gives

for all bonds and . Let . Using a lemma proved in the second post in this series, we get the bound

But another result from that second post tells us that the sum on the LHS is the derivative of , so we let

and write

Our observation above that says that is bounded below by a constant depending on . So for , we get , and the result follows.

**2. Kesten’s Theorem **

This is the last piece in the proof that .

**Theorem 3 (Kesten’s Theorem).** * Let be the event that there is an infinite open cluster. Then for bond percolation in , if , we have *

*Proof:* Take and choose (depending on ) and be as in the second lemma with . Choose some large , and let be a family of rectangles whose bottom left corner is the origin and which have sides of length and . The longer side is the vertical for even and the horizontal for odd .

Suppose is the event that is crossed in the longer direction by an open path. Any two such paths must meet, so if every holds, so does (since the open cluster containing the origin has unbounded radius). Now we write

But by the final lemma of the previous section, and the fact that is arbitrarily large,

So , and thus 1.

The result proven at the end of the last post was that while this result shows that , so we conclude that in fact . In the next and final post, we’ll turn our attention to .