If an infinite percolation cluster exists almost surely, the probability that a large square box intersects one tends to one. The four rotationally symmetric exterior-side arm events are increasing. By the Harris-FKG inequality, their common probability is at least , where is their union. Thus a pair of specified opposite sides both has an infinite exterior arm with probability at least , tending to one. The rays start at a boundary graph vertex and immediately leave the graph vertex box.
Connection decay rate in a percolation strip 2026-10-06
For nearest-neighbour bond percolation with on the strip , put . The Harris-FKG inequality and translation invariance imply . Thus is a subadditive sequence, and the Fekete lemma givesThe direct horizontal graph path gives . Enlarging the strip increases every percolation two-point connection probability, so is nonincreasing in and has a nonnegative limit of a sequence.
One-arm extension estimate Created 2026-09-24 Updated 2026-10-06
The Russo-Seymour-Welsh theorem and the Harris-FKG inequality imply for a scale-independent constant .
Past exam of the mathematics course of the University of Cambridge 2002 iii Paper 28 1 Solution Created 2026-10-03 Updated 2026-10-14
In bond percolation on the square lattice, every nearest-neighbour edge independently has state one, meaning open, with probability , and state zero, meaning closed, with probability . Thus the configuration law is the product measure . An open path in bond percolation uses only open edges. An increasing event satisfies : opening further edges cannot destroy the event.
The Harris-FKG inequality for two increasing events isFor disjoint occurrence of increasing events, write when disjoint sets of open edges separately witness and , so each event is forced by its own witness regardless of the other edges. The Van den Berg-Kesten inequality statesFor events given by cylinder sets these are finite-coordinate assertions. Connection events have finite open-path witnesses; truncating to finite boxes and taking increasing limits gives the same disjoint-occurrence bound for them. Sharing a vertex between the two witness paths is permitted; sharing an edge is not.
For , take a simple open path from the origin to the radius- boundary and let be its first vertex on the radius- boundary. Its initial segment witnesses . If is its final vertex, then . Consequently the remaining segment reaches , and its initial portion until that visit witnesses . The two segments use disjoint edges. HenceThe union bound, Van den Berg-Kesten inequality and translation invariance of the product measure now give the weighted BK boundary-splitting estimateFor the same bound follows directly from the union bound, with . With the conventional , gives equality.
Write for the percolation cluster of the origin. By summing its membership indicators, the percolation susceptibility isIf it is finite, then . For , choose with . Iterating the boundary-splitting estimate with this fixed givesFor , , so . To handle all smaller positive radii without an unspecified prefactor, observe that . ThereforeFor , the second chosen rate gives , and for larger the first rate applies. At both sides equal one. The endpoint has for every , so any positive rate works; is excluded by finite susceptibility. This proves exponential one-arm decay from finite susceptibility with the required unit prefactor.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 30 2 a Solution Created 2026-10-03 Updated 2026-10-07
Write . The complements are decreasing events. The Harris-FKG inequality for a product measure givesThe same correlation inequality for two decreasing events follows from that for their increasing complements, since their covariance is unchanged. ThusBoth and the square root are nonnegative, so rearranging gives the square-root bound for increasing events:No independence between the events themselves is asserted; independence is the coordinate property of the underlying product measure.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 30 2 b ii Solution Created 2026-10-03 Updated 2026-10-07
Let be the increasing events that an infinite simple open ray in a graph starts at the indicated side and immediately leaves the box, with all later graph vertices outside it. A corner exit is assigned according to the direction of its outgoing edge. The four events have the same probability by rotational symmetry. Their union is from the preceding step.
The Harris-FKG inequality applies to these exterior-ray events by approximation with finite connections to larger boxes and passage to decreasing limits. Positive association of the four decreasing events , by iterating the Harris-FKG inequality, givesEquivalently, one may apply the preceding square-root bound for increasing events first to the two opposite-side unions and then to the two individual sides of one pair. ThereforeA union bound now showsChoose such that . ThenThe infinite exterior clusters touched by these two sides need not be the same. The same argument works for dual boxes and any sufficiently large size; this permits the matching primal/dual contours used in the full proof.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 26 2 Solution Created 2026-10-03 Updated 2026-10-07
Order configurations coordinatewise. The FKG lattice condition on the masses of a finite Boolean lattice isThe FKG inequality states that under this condition, increasing real-valued functions satisfy . In particular increasing events satisfy . For a product measure with coordinate weights , the pair has the same two entries as . Multiplying over coordinates gives equality in the lattice condition, including degenerate Bernoulli parameters without division by zero.
For independent bond percolation on the nearest-neighbor cubic lattice, write . Define the percolation critical probability and the connective constant bywhere counts rooted -step self-avoiding walks. Translation invariance makes the root irrelevant, and part (i) proves existence of the latter limit.
First work on the square lattice, whose dual is another translated square lattice. Put and assume . If the open cluster at the origin is finite, its exterior edge boundary contains a closed dual simple circuit surrounding the origin. To see this planar fact, surround its finitely many vertices by their unit square cells and follow the exterior boundary: its crossing primal edges are closed. Resolving repeated boundary vertices into simple circuits leaves a circuit separating the origin from infinity.
Let count such dual circuits of length . Every one crosses the positive horizontal ray at distance at most , because it surrounds the origin and its horizontal span is at most its length. Choosing a ray-crossing edge, an orientation and all but the last edge encodes it by one of at most rooted length- self-avoiding walks. Thus . Choose with . The root-count limit gives , soThis summable tail alone need not make the probability of every enclosing circuit less than one. Choose a large and let be the absence of enclosing closed dual circuits of length at least , with . Let require every primal edge inside to be open. It has positive probability. Both events are increasing. The finite-measure Harris-FKG inequality extends to by decreasing limits over finitely many circuit exclusions, so .
On the origin is connected to every graph vertex of that box. Any closed dual circuit enclosing the origin must then enclose the whole box, hence have length at least . On no such circuit exists, so the origin cluster is infinite. This proves the connective-constant Peierls bound, . Finally the lattice in dimension contains a coordinate copy of the square lattice, whose edge law is unchanged. Percolation in that subgraph implies percolation in the full graph, and hence
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 28 2 Solution Created 2026-10-03 Updated 2026-10-06
Use the coordinatewise partial order: means for every . An increasing event is upward closed: and imply .
Under the independent Bernoulli distribution product measure on these coordinates, the Harris-FKG inequality statesfor increasing events . For , let be the cylinder of configurations agreeing with on . The disjoint occurrence of increasing events consists of configurations for which there exist disjoint with and . For increasing events, witnesses can be taken to prescribe only open coordinates. The BK inequality isPositive association rewards simultaneous occurrence; the disjoint-witness requirement in the BK inequality gives a bound in the opposite direction. The inequalities also apply to different independent Bernoulli parameters in different coordinates.
Regard the boxes as sets of lattice graph vertices. Write , and put . On the event defining , choose an open self-avoiding walk from the origin to its first visit to . Let be its first visit to . Its initial segment witnesses . Its remaining segment ends at a point withFor , truncate this remaining segment on its first visit to . The two segments use disjoint edges, so they are disjoint witnesses. For , the second event is the sure event with empty witness. Thus the union bound, the BK inequality and translation invariance giveAn unrestricted connection event can depend on infinitely many edges. Apply the finite-coordinate BK inequality first to connections confined to growing boxes and take increasing limits; each occurrence has a finite path witness. This justifies its use here. The estimate is the weighted BK boundary-splitting estimate.
By summing cluster indicators and using Tonelli theorem, the percolation susceptibility isIf , then . At all vanish, so any positive exponential rate works. Otherwise choose with . Iterating the weighted BK boundary-splitting estimate gives, for with ,For , , so . To include the finitely many smaller indices without a prefactor, note that under finite susceptibility andTherefore one may takeFor this follows from , and for it follows from the iterated estimate. This proves exponential one-arm decay from finite susceptibility with exactly the requested unit prefactor.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 28 3 Solution Created 2026-10-03 Updated 2026-10-06
The purpose of the RSW lemma is to turn local crossing information into control at every fixed shape and scale. Work with independent bond percolation on the square lattice. Let be the event of an open left-to-right graph path in , and write .
A useful precise uniform version of the RSW lemma is the following uniform RSW crossing estimate. If for a fixed parameter and some ,then for every fixed aspect ratio there is , independent of , such thatIn particular this bounds length in terms of the uniform square bound at length . The constants need not be sharp. The hypotheses used in the proof are planarity, translation and reflection symmetries, the Harris-FKG inequality and independence of unexplored edges. At a self-dual parameter the same argument for dual crossings gives an upper bound strictly below one as well. Neither an exact value of nor the existence of an infinite cluster is an assumption of this estimate.
Here is the gluing picture behind the RSW lemma. For increasing crossing or attachment events, the Harris-FKG inequality provides a lower bound on their joint probability. If an event of probability at least is a union of two reflection-related increasing alternatives , the square-root trick for positively associated events givesIndeed their decreasing complements are positively associated too, so . This prevents a square crossing from concentrating all its useful attachment locations on just one side.
The nontrivial first gluing step enlarges a square to aspect ratio . Explore an extremal square crossing, revealing the edges on the explored side but leaving its other side unexamined. In that unexplored region the conditional law is still independent bond percolation. Compare possible attachments to the crossing with their reflected alternatives. Reflection symmetry and the preceding square-root estimate give a positive bound, depending only on , for the required attachment after averaging over the explored crossing. Carry out the reflected construction at the other end and use the Harris-FKG inequality to combine the increasing attachment events with the original crossing. Planarity ensures that the relevant transverse paths actually meet. This RSW reflection extension lemma yields a lower bound for the -by- crossing. The exploration is important: a reflection compares the laws of fresh configurations; reflecting the picture of an open path does not make the reflected edges open.
From that first extension, longer rectangles are obtained by a genuinely transverse gluing. Two -by- rectangles shifted by overlap in an -by- square. Require a horizontal crossing in each long rectangle and a vertical crossing of the overlap square. Each horizontal crossing crosses that overlap from left to right, and therefore meets its vertical crossing. Their union crosses the -by- rectangle. The Harris-FKG inequality gives a lower bound . Repeat a bounded number of times for any fixed . Integer rounding uses neighboring lattice rectangles and bounded additional gluing steps; finitely many smallest scales can be absorbed into the constant. The constants deteriorate with but not with . Merely requiring horizontal crossings in adjacent squares would not suffice, since their endpoints need not coincide; the overlap crossing solves that problem.
I apply the RSW lemma to the exact threshold for bond percolation on the square lattice. The planar duality for rectangle crossings says that an open horizontal crossing and a closed dual vertical crossing are complementary. At , the dual edge states have the same law as the primal ones. For the balanced lattice rectangle with side lengths , the rotated dual crossing rectangle has those same side lengths: its transverse lengths before rotation are . Thus the exact self-dual rectangle crossing probability is . Restricting a crossing of this rectangle to its first visit to the shorter vertical side givesThe one-unit balance avoids assuming an exact one-half probability for every finite vertex-square convention. The RSW lemma now gives scale-independent positive bounds for all fixed-aspect-ratio primal and closed dual rectangle crossings.
Arrange four appropriately overlapping long rectangles around a square ring. Closed dual crossings along the four sides, with transverse overlap crossings if needed, join to a closed dual graph cycle surrounding the inner square. The Harris-FKG inequality, applied to the closed dual states, and the RSW lemma give a constant for this circuit event, uniformly over ring size. Choose disjoint rings with radii increasing, for example, by a factor of four. Their circuit events depend on disjoint edge sets, so they are independent. An open graph path from the origin to infinity would have to avoid every one of these dual barriers. Its probability is at most after rings and therefore zero. HenceThis is the independent annular barriers for percolation argument. It is worth separating it from the other inequality: the absence of an infinite cluster at one parameter alone does not prove that every larger parameter percolates.
For the reverse inequality use sharpness of the percolation transition: below , independent bond percolation on the cubic lattice has exponentially decaying connection probabilities. One can see why this is the relevant general ingredient through the finite-set criterion for percolation sharpness. For a finite set containing the origin putIf , split a long open self-avoiding walk at its first exit from . Its internal connection, exit edge and subsequent connection have disjoint witnesses. The BK inequality gives a contraction by each time distance decreases by the diameter of plus a fixed step. Iteration proves exponential decay of subcritical percolation at such .
To identify this finite-set threshold with , the Margulis–Russo formula expresses the derivative of as the sum of pivotal-edge probabilities. Explore the cluster attached to the box boundary, and let be its complement. On failure of the origin-to-boundary event, ; all edges from to the boundary cluster are closed, while the internal edges of remain fresh. A boundary edge is pivotal precisely when its endpoint in is connected to the origin inside . Removing the factor for a closed pivotal edge yields the general differential inequalityFor every finite-set quantity on the right is at least one. Integrating from any gives a positive lower bound for independent of , and taking gives . Combined with the contraction below , this proves and the stated sharpness conclusion. This outline supplies the extra threshold argument rather than assuming the desired critical value.
If , sharpness would give . But a square crossing starting somewhere on the left side entails an open connection from that starting graph vertex to distance . A union bound over the possible starting graph vertices givescontradicting exponential decay. Therefore , and the two directions establishThis is the Harris-Kesten theorem. The overall mechanism is that self-duality supplies a square crossing, the RSW lemma transports it between shapes and creates barriers at every scale, and sharpness converts the finite-scale crossing information into the exact threshold.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 204 1 b Solution Created 2026-10-03 Updated 2026-10-14
Order configurations coordinatewise: means for every edge. An increasing event is a measurable set such that and imply . Opening additional edges cannot destroy an increasing event. The Harris-FKG inequality, the product measure version of the FKG inequality, statesfor two increasing events. Equivalently, bounded coordinatewise increasing random variables satisfy . The event inequality also holds for two decreasing events, by taking complements.
For disjoint occurrence of increasing events, a finite open edge set witnesses in if every configuration agreeing with on belongs to . Since is an increasing event and all edges of are open, this says that the configuration with exactly open already forces . Define to consist of configurations admitting disjoint finite witnesses for . For events depending on finitely many edges, this is the usual disjoint-occurrence definition; it also applies to finite-connection events on the infinite graph. The Van den Berg-Kesten inequality statesOn a countable graph, its finite-witness version follows by taking increasing unions over finite edge sets. The Harris-FKG inequality concerns simultaneous occurrence, whereas the Van den Berg-Kesten inequality requires separate certificates using disjoint coordinates.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 204 1 d Solution Created 2026-10-03 Updated 2026-10-06
The union bound gives . Hence some boundary graph vertex has percolation two-point connection probability at least . The rotations and reflections of the square lattice let us choose such a graph vertex as , . Reflection in the vertical line through sends to and fixes . ThereforeThe Harris-FKG inequality applied to these two increasing events proves the reflection lower bound for two-point percolation:Every graph path from to reaches , so . Taking roots givesBoth outside expressions have limit of a sequence , proving the even case. For , the Harris-FKG inequality with the last horizontal edge gives , and also . The lower bound has rootThe upper bound has the same limit of a sequence. At all positive-distance connection probabilities vanish. Thus for every .
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 204 4 a Solution Created 2026-10-03 Updated 2026-10-06
Write and for the horizontal and vertical open-crossing increasing events of a rectangle in the rotated square lattice. At parameter , planar duality for rectangle crossings says that failure of is exactly a closed dual vertical crossing. Quarter-turn symmetry and the boundary convention of the rotated rectangles identify the two alternatives on a square. Thus its crossing probability is ; a rectangle obtained by a harmless one-step boundary adjustment has crossing probability at least . This is the finite-domain exact self-dual rectangle crossing probability, not an assumption about infinite percolation clusters.
We first prove a RSW reflection extension lemma. In coordinate units adapted to the rotated drawing, take , , and . Reveal the rightmost open vertical crossing of , when one exists, using only its edges and the region to its right. Reflect across the horizontal axis. The part of to the left of these two paths remains unexposed and has the independent parameter- law. A horizontal crossing of must reach their union. Its probability is at least . Reflection symmetry and the union bound give a probability at least of reaching the upper path through . Conditional opening of the edges on can only help. Since has probability at least , the increasing event that a vertical crossing of is connected inside to its left side has probability at least .
Reflect this construction in the vertical midline of to obtain an increasing event of the same probability, connecting a vertical crossing of to the right side of . On , the horizontal crossing of meets both vertical crossings, so all three graph paths join and give . The Harris-FKG inequality now yieldsThe exploration only conditions the exposed side of the extremal graph path; reflecting a drawn path does not assert that its reflected edges are open. This distinction is what makes the extension argument valid.
To pass from aspect ratio to , put four rectangles of width and height next to each other, with successive left boundaries separated by . Adjacent rectangles overlap in a square of side . Require horizontal crossings of the four rectangles and vertical crossings of the three overlap squares. Every consecutive pair of horizontal crossings meets the vertical crossing in its overlap; together they form a horizontal crossing of width . The Harris-FKG inequality gives a strictly positive bound, for these compatible scales, ofThe construction works on the rotated square lattice with the boundary sites used in the source drawing. Integer roundings do not create a scale restriction: for an odd height use the largest smaller compatible even height, and use eight horizontal rectangles and seven overlap squares, whose union has auxiliary aspect ratio , so that its width exceeds the desired width for all sufficiently large . Truncate its horizontal crossing on first reaching the desired right side. The number of extra rectangles and overlap-square crossings is bounded independently of , and every factor is bounded below by the same positive constants. The finitely many smallest rectangles each have positive crossing probability, since a specified finite open graph path suffices. Therefore a constant works at every integer scale. Taking ensures the requested strict inequality:This proves the required instance of the Russo-Seymour-Welsh theorem through reflection, exploration and gluing. The parameter in this calculation is .