Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 209 (Lattice Models)

The Harris-FKG inequality givesHere is an induction proof of the product-measure version, also called Harris' inequality. The result is immediate for one coordinate. For $n$ coordinates, defineThe sections of an increasing event are increasing events in the first $n-1$ coordinates, so the induction hypothesis givesMonotonicity gives $a_1\ge a_0$ and $b_1\ge b_0$. The difference between the right side andisThis closes the induction and proves positive association.

Suppose for contradiction that the critical probability for site percolation on the triangular lattice satisfied $p_c>1/2$. Then $p=1/2$ would be subcritical, so exponential decay of subcritical percolation would give constants $C,c>0$ such that the probability that a fixed site has an open path to distance $N$ is at most $C{e}^{-cN}$.

Every left-to-right open crossing of an $N$ by $N$ rhombus contains a site on its left side joined to distance at least $N$. There are $O(N)$ possible starting sites, so the union bound would implyPlanar self-duality and symmetry instead make this crossing probability exactly $1/2$ at every $N$. This contradiction proves $p_c\le 1/2$.

The Russo-Seymour-Welsh theorem and self-duality give a scale-independent $c>0$ such that every annuluscontains a closed circuit surrounding the origin with probability at least $c$. Choose every second annulus so that the corresponding site sets are disjoint. Their circuit events are then independent, and the Borel-Cantelli lemmas say that infinitely many of them occur almost surely.

Every such closed circuit separates the origin from infinity, so the origin cannot belong to an infinite open cluster. By translation invariance, the same holds for every site. Since the triangular lattice is countable, a countable union shows that almost surely no infinite open cluster exists at $p=1/2$.

Set $\tilde{\Gamma }(x)=\Gamma (x)-\Gamma (b)$. Then $\tilde{\Gamma }(b)=0$, and subtraction of a constant leaves every edge difference unchanged:The linear maphas determinant $\pm 1$: its inverse is $\gamma (x)=\tilde{\gamma }(x)-\tilde{\gamma }(a)$, because $\gamma (a)=0$. The change of variables formula therefore shows that the new density is proportional toon functions vanishing at $b$. This is precisely the discrete Gaussian free field with Dirichlet boundary condition at $b$.

Only the $2d$ edges incident to $x_0$ depend on $\Gamma (x_0)$. Conditional on the remaining field, writeCompleting the square givesThe conditional density is therefore proportional to $e^{-(u-m{)}^2}$, and henceThis is the Gibbs-Markov property of the discrete Gaussian free field in the normalization used by the paper.

The covariance isIndeed, the killed-walk Green function satisfiesOn the other hand, the conditional mean from part 2 makes the covariance harmonic in $x\ne y,a$. At $x=y$, the conditional variance $1/2$ gives the same equation with source $\frac{1}{2}{1}_{\{x=y\}}$. Uniqueness of the Dirichlet problem therefore identifies the covariance with $G_{\{a\}}/2$. The factor $1/2$ comes from the coefficient $1/(2d)$ in this paper's energy density.

Reveal the field in the order $x_1,\dots ,x_K$ and evaluate its joint density at zero. By the Spatial Markov property of the Gaussian free field, after the values onhave been fixed to zero, the remaining field is the GFF killed on $R_i$. Part 3 therefore givesFactoring the joint density into conditional probability densities now yieldsThe left side is intrinsic and does not depend on the order used to factor the density. Thereforeis independent of the ordering.

Run Wilson algorithm with root $a$, beginning with a simple random walk from $x_1$. Its chronological loop erasure is added as the first branch, so the resulting tree contains a path from $x_1$ to $a$ with exactly the law of the stated loop-erased random walk.

For completeness, the algorithm produces a uniform spanning tree. For any prescribed ordered collection of loop-erased branches, the random-walk decomposition gives a product of local transition factors and diagonal killed Green functions. The transition factors depend only on the resulting oriented tree, while the product of Green functions is independent of the vertex order by Question 3. Thus every spanning tree receives the same probability. Since a tree has a unique simple path between two vertices, its $x_1$-to-$a$ path has the claimed law.

Sample a uniform spanning tree $T$. Rooting Wilson's algorithm at $a$ shows that the path in $T$ oriented from $x_1$ to $a$ has the law of $\mathrm{LE}(x_1\to a)$. Rooting the same uniform law at $x_1$ shows that the same undirected path with its orientation reversed has the law of $\mathrm{LE}(a\to x_1)$. ConsequentlyThis is reversibility of loop-erased random walk.

Yes. Simple random walk in two dimensions is recurrent, so a walk started at $x_1$ hits $a$ almost surely and its loop erasure is a finite path. Exhaust ${\mathbb{Z}}^2$ by finite boxes, or use increasingly large tori with the marked vertices kept fixed. The probability that either walk reaches the boundary before hitting its target tends to zero by recurrence. The finite-graph reversal identity from part 2 therefore passes to the limit:

A site $v$ is pivotal for an increasing event $E_N$ in a configuration $\omega$ when changing only the state of $v$ changes whether $E_N$ occurs. Away from the boundary, pivotality for a left-to-right crossing has the geometric four-arm description: from neighbors of $v$ there are two disjoint open arms reaching the left and right sides and two disjoint closed arms reaching the top and bottom sides, in alternating cyclic order. The boundary version truncates the corresponding arms at the sides already adjacent to $v$.

The Margulis–Russo formula gives

Set $N=3^j$. At $p=1/2$, first use the Russo-Seymour-Welsh theorem to produce, with probability bounded below independently of $N$, two separated open crossings from left to right and a closed crossing between them. Explore the interface separating the lower open cluster from the adjacent closed cluster until it reaches the opposite macroscopic boundary. The explored interface supplies two alternating arms, while fresh RSW crossings in the unexplored regions supply the other two.

Repeat this construction in the geometrically separated scale bandsThe domain Markov property of a percolation exploration leaves unrevealed sites with their original independent critical law. RSW and the Harris-FKG inequality therefore give a uniform conditional probability $c>0$ that the required open and closed connections occur in each band. On that event the exploration identifies a site having four alternating arms to the four sides of $W_N$, hence a pivotal site for $E_N$.

Choose the candidates in disjoint scale bands, so successful bands give distinct pivotal sites. If $Y_k$ indicates success at scale $k$, thenThusfor $j\ge 2$, after adjusting the positive constant. This is the required logarithmic-in-$N$ lower bound.