No Percolation at Criticality: the θ(p_c) = 0 Proof, Visualized

Part 1 · The question. Bond percolation: each edge of the lattice is open independently with probability p. Open edges are drawn; the largest open cluster is highlighted. As p grows, clusters merge. Beyond a critical value p_c (here 1/2) an infinite cluster appears, and in a finite window the largest cluster spans it. θ(p) is the probability that the origin belongs to an infinite open cluster. It is zero below p_c and positive above. The question is what happens exactly at p_c: does θ rise continuously from zero, or does it jump, so that an infinite cluster already exists at p_c? The proof rules out the jump: there is no infinite cluster at p_c, that is, θ(p_c) = 0. Known before: d = 2 (Harris 1960, Kesten 1980) and d ≥ 11 (Hara–Slade 1990, Fitzner–van der Hofstad 2017). The cases 3 ≤ d ≤ 10 stayed open for decades. The new preprint claims every quasi-transitive graph with p_c < 1, which is Benjamini and Schramm's 1996 conjecture, and hence every ℤᵈ with d ≥ 2.

Part 2 · The proof at a glance. Suppose, for contradiction, that an infinite cluster exists at p_c. The argument splits by how fast balls grow in the graph. Exponential growth was settled by Hutchcroft (2016), and an intermediate case has its own two-cluster argument. ℤᵈ has polynomial growth: that is the case we follow. There the strategy is: find likely local moves at p_c, which by continuity stay likely slightly below p_c; then chain them through corridors into an infinite cluster at some p′ < p_c. That contradicts the definition of p_c. Let's watch the strategy in motion. If an infinite cluster exists at p_c, it is unique and reaches out in every direction, so a fixed small ball touches it with high probability. From that ball, open paths then reach a small box one step ahead and up, and one step ahead and down, each with probability at least 1 − η (after rescaling the coordinates). Each move is witnessed by finitely many edges, so its probability is a polynomial in p, hence continuous. It therefore stays above 1 − η at some p′ slightly below p_c. Chain the moves: sites of an oriented lattice are joined by corridors, regions of the original graph. Two corridors meet only inside the small rectangle at a shared endpoint. Test sites layer by layer. Testing a site reveals only its incoming corridor; its two outgoing corridors are still untouched. So, whatever has been revealed so far, a tested site fails with probability at most ε. Each time a site passes, the probability that its fresh corridors will carry the cluster onward is computed and stored; that stored prediction is what makes the bound hold at the next test. (Illustration: failures drawn at random.) If only finitely many sites were good, the boundary of the good set would be a closed trail of some length l passing at least l/4 failed sites, and there are at most 4ˡ such trails. For small ε the total is below 1. So with positive probability infinitely many sites are good, each reached by an actual open path from the seed: an infinite open cluster at p′ < p_c. Contradiction, so θ(p_c) = 0. Inside a corridor, the cluster is handed from one relay region to the next, all the way to the child site. Each hand-off must fail with probability at most κ, whatever happened before; summing over the hand-offs controls the whole corridor. The catch: the cluster arrives at a point of Bᵢ chosen by the very edges that must carry it onward. Making the hand-off bound uniform is the delicate part. That is the whole strategy. One ingredient is far more delicate than the rest: the hand-off bound, called joint gluing. Its proof is the most involved step. Let's zoom in on it. Joint gluing says: reaching the relays and the target is at least as likely as reaching the relays, times the worst relay's chance of reaching the target. We take it in three steps.

Joint gluing, step 1 · Why the hand-off is subtle. One hand-off, abstractly: the cluster of o has reached a set A of relays, and each relay, on its own, reaches the target T with probability at least 1 − λ, close to 1. Must o then reach T? The catch: which relay o arrives at is decided by the same random edges that must carry it onward. Conditioning on the arrival can spoil the continuation. The smallest example is a 4-cycle with relays a₁ and a₂ and target b. Each relay reaches b either directly or around the cycle, so this probability tends to 1 as p → 1. Split according to the first relay reached, taking a₁ before a₂. Suppose a₂ comes first: o reaches a₂ but not a₁. That forces o–a₁ closed and o–a₂ open. It also forbids a₂–b and b–a₁ from both being open, since then o would reach a₁ through b. Three cases remain. The conditional chance that a₂ reaches b is the green share: p(1 − p) / (1 − p²) = p/(1 + p). Now let p → 1. It tends to 1/2, although each relay alone reaches b with probability tending to 1. Being reached first rules out the likeliest configuration, both edges at b open, and leaves two equally likely cases, only one of which connects. So a hand-off cannot be argued relay by relay through 'the first relay reached': that conditioning can cost a factor 1/2. A different decomposition is needed.

Joint gluing, step 2 · The inequality and its relay weights. Joint gluing holds for independent edges with any probabilities, including 0 and 1, so it still applies after conditioning on edges already revealed. Subtracting it from ℙ(o ↔ A) gives the hand-off bound: if every relay reaches the target with probability at least 1 − λ, then reaching the relays but missing the target has probability at most λ, however many relays there are. Proof idea: find relay weights αₐ ≥ 0 that add up to ℙ(o ↔ A) and multiply the plain chances ℙ(a ↔ T), with no conditioning on which relay came first. Bounding the sum by its smallest term gives the theorem. List the relays x₁, x₂, … and put o last. Column k of the matrix H describes the cluster of xₖ, conditioned to avoid the earlier ones; entry (i, k) is the chance that it contains xᵢ. So H is unit lower triangular. Key lemma: H⁻¹ = I − Γ where no entry of Γ is negative. The relay weights are the last row of Γ, and they add up to ℙ(o ↔ A). Applying the same positivity to the vector of target chances gives the weighted bound. On this 4-cycle the weights are 0.516 and 0.474, nothing like the first-relay masses 0.973 and 0.017. They are not probabilities of any 'first' event, so the trap of step 1 does not apply. The bound checks out exactly: 0.963 ≤ 0.964. Everything rests on one fact: no entry of Γ is negative. Step 3 shows why.

Joint gluing, step 3 · Why the weights are non-negative. Ingredient 1, fresh territory. To know the cluster of S exactly, you only look at edges touching it: open ones inside, closed ones on its boundary (dashed). Every other edge is untouched: given the cluster, the rest of the graph is still independent percolation with the original probabilities. Ingredient 2, alternating resampling. Given the cluster U of s, regrow the cluster W of S with U removed; then, given W, regrow the cluster of s with W removed. Repeat. The law of the cluster of s, conditioned to avoid S, is stationary for this two-step chain (van den Berg–Häggström–Kahn). One round splits the column for an increasing function F into the column for its one-round average PF, plus leftover terms carried by later relays. The leftovers are non-negative: removing more vertices can only shrink the cluster of s. Ingredient 3, forgetting. With probability at least a, every edge at s is closed and s is isolated, wherever the chain started. So repeated averaging flattens F to its mean, geometrically fast. Iterating, the column for F minus its mean is a non-negative mix of later columns. With F = 'the cluster meets a later relay', this is exactly H⁻¹ = I − Γ with Γ ≥ 0.

Recap. If an infinite cluster existed at p_c, likely local moves would persist slightly below p_c, and corridors with an adaptive exploration would chain them into an infinite cluster below p_c. Every hand-off along the way is controlled by joint gluing. The result is a contradiction, so θ(p_c) = 0. The gluing inequality itself uses only the independence of the edges: no symmetry and nothing specific to the lattice. Its non-negative relay weights replace the first-relay decomposition that fails on a 4-cycle.