Stats 359: Topics in Mathematical Physics Spring 2017 Lecture 1:...

Stats 359: Topics in Mathematical Physics Spring 2017

Lecture 1: April 3Lecturer: Sourav Chatterjee Scriber: Horia Margarit

Disclaimer: These notes may contain factual and/or typographic errors.

Ising Model

Comes from statistical mechanics, and was originally applied in 1924 to study ferromagnets. Let Λ ⊆ Zd.Let σi ∈ −1, 1 denote the ”spin” at vertex i ∈ Λ. Define a probability measure µΛ on the space ofconfigurations Σ ≡ −1, 1Λ with density proportional to exp (β

∑i,j∈ηΛ

σiσj) where ηΛ is the set of allpairs of neighboring vertices in Λ. Setting β = 0 yields the i.i.d measure while setting β > 0 tries to getneighboring spins aligned in the same direction.

Phase Transitions of the Ising Model

Expectations of the Ising Model

EµΛ(σiσj) ≡ Z−1

∑σ∈−1,1Λ

σiσj exp(βH(σ)

) H(σ) ≡∑i,j∈ηΛ

σiσj

Z ≡∑

σ∈−1,1Λexp

(βH(σ)

)Ising proved that there is never any magnetization when d = 1. More precisely, |EµΛ(σiσj)| ≤ c exp

(− c(β)|i− j|

)when d = 1 and Λ = [−N,N ] for large enough N , and for some constants c and c(β) which do not dependon Λ.

P(σ) ∝ exp(β∑k

σkσk+1

)by symmetry on the line

P(σi|σjj 6=i) ∝ exp (βσi(σi−1 + σi+1))

EµΛ(σi|σjj 6=i) = tanh (β(σi−1 + σi+1))

= tanh (2β)σi−1 + σi+1

2

= δσi−1 + σi+1

2

EµΛ(σiσj) = EµΛ

(EµΛ

(σi∣∣σjj 6=i)σj) by the Tower property

=δ

2

(EµΛ

(σi−1σj

)+ EµΛ

(σi+1σj

))Iterating on the last expression to eliminate the expectation yields the exponentially decaying result.

Peierls (1936) proved that the Ising Model has a phase transition in d ≥ 2. That is to say there exists

1-1

1-2 Lecture 1: April 3

some β below which there is never any magnetization, and (perhaps another) β above which there is mag-netization. Formally, Pierls demonstrated that ∃β0 such that, if β ≥ β0 then |EµΛ(σiσj)| ≥ c(β) > 0 ∀i, jin any infinite volume limit (free boundary), but ∃β1 > 0 such that, if β < β1 then |EµΛ(σiσj)| ≤c exp (−c(β) |i− j|) ∀i, j and where |i−j| is the Euclidean distance between i and j. Prior to illuminatingPeierls’ proof, we need some definitions and lemmas.

Infinite Volume Limit

Consider the Ising Model on Λ ⊆ Zd and extend it to a probability measure on −1, 1Zd by defining arbi-

trarily outside Λ. Call this measure µΛ and note that −1, 1Zd is a compact metric space in the producttopology. Taking Λ1,Λ2, . . . as subsets which increase to Zd, µΛn∞n=1 will have a subsequetial limit. Anysuch subsequential limit is called an infinite volume Ising model with free boundary conditions.

If f is a function of finitely many spins, then∫fdµΛn →

∫fdµΛ as n ↑ ∞ where µΛ is our subsequence

limit.

Lemma 1

Choose any two edge sets η(1)Λ ⊆ η

(2)Λ . Define η

(∗)Λ ≡ η

(2)Λ \ η(1)

Λ . Then define the expectation under the i.i.dmeasure Zη = Ei.i.d(exp

(β∑i,j∈η σiσj

)). We claim that

2−|η(1)Λ | ≤

Zη

(∗)Λ

Zη

(2)Λ

≤ 2|η(1)Λ |.

This is proved by induction on lemma 2.

Lemma 2

For any edge set η of Λ and any edge q = (i, j) ∈ η and small enough β, we have that

1

2≤ Zη\qZη

≤ 2

Proof:

Zη = Ei.i.d

(exp

(β∑i,j∈η

σiσj))

= Ei.i.d

( ∏i,j∈η

exp(βσiσj

))= Ei.i.d

(∏e∈η

(1 + ρe

))= Ei.i.d

((1 + ρq

) ∏e∈η\q

(1 + ρe

))

Zη −Zη\q = Ei.i.d

((1 + ρq

) ∏e∈η\q

(1 + ρe

)−

∏e∈η\q

(1 + ρe

))= Ei.i.d

(ρq

∏e∈η\q

(1 + ρe

))

ρq ≡ exp(βσiσj

)− 1

Lecture 1: April 3 1-3

Say that two edges are connected if they share a vertex. For ν ⊆ η with q = (i, j) ∈ ν, decompose intoconnected components ν ≡ ν1∪ν2 with q ∈ ν1 and with ν2 ≡ ν \ν1. Observe that ν 7→ (ν1, ν2) is a bijection.Therefore

Zη −Zη\q =∑

(ν1,ν2)

Ei.i.d

(∏e∈ν

ρe

)=∑

(ν1,ν2)

Ei.i.d

( ∏e∈ν1

ρe

)Ei.i.d

( ∏e∈ν2

ρe

)=∑ν1

Ei.i.d

( ∏e∈ν1

ρe

) ∑ν2⊆η\ν1

Ei.i.d

( ∏e∈ν2

ρe

)=∑ν1

Ei.i.d

( ∏e∈ν1

ρe

)Zη\ν1

where we defined ν1 ≡ ν1 ∪ all edges which share a vertex with some member of ν1. (To be continued.)


Lecture 2: April 5Lecturer: Sourav Chatterjee Scriber: Horia Margarit


Phase Transitions of the Ising Model

(Continued from previous lecture.) Recall that we showed

Zη −Zη\q =∑ν1

Ei.i.d

( ∏e∈ν1

ρe

)Zη\ν1

where we defined ν1 ≡ ν1 ∪ all edges in η which are adjacent to some edge in ν1. Since |ρe| ≤ c(β) wherec(β) is a constant that depends only on β, we get∣∣∣∣ ZηZη\q

− 1

∣∣∣∣ ≤∑ν1

(c(β)

)|ν1| Zη\ν1

Zη\q.

Using this inequality, it is now easy to prove by induction on the size of η that if β is sufficiently small, then

1

2≤ Zη\qZη

≤ 2, as was to be shown.

This is because c(β)→ 0 as β → 0, and the number of connected sets of edges ν1 of size k that contains q isat most Ck, where C depends only on the dimension.

Now we may prove that for sufficiently small β the expectation of any two spins being aligned is exponentiallydecaying in distance. Recall that if f is a function of finitely many spins, then

∫fdµΛn →

∫fdµΛ as n ↑ ∞

where µΛ is the (free boundary) infinite volume subsequence limit of our probability density. Taking σ assome arbitrary configuration of spins for an Ising Model Λn ⊆ Zd, and by definition of expectation, we havethat

EµΛn

(f(σ)

)=

Ei.i.d

(f(σ) exp

(βH(σ)

))Ei.i.d

(exp

(βH(σ)

))=

1

ZηEi.i.d

(f(σ)

∏i,j∈η

exp(βσiσj

))=

1

ZηEi.i.d

(f(σ)

∏e∈η

(1 + ρe

))=

1

Zη∑ν⊆η

Ei.i.d

(f(σ)

∏e∈ν

ρe

)Let F be the set of vertices which determines f . Let ψ be the set of edges incident to some vertex in F . Letν1 be the union of all connected components of ν that contain some element of ψ. Defining ν2 ≡ ν \ ν1 yields

2-1


the bijection ν 7→ (ν1, ν2). This enables the expansion of the expectation as was done in the previous lecture

EµΛn

(f(σ)

)=

1

Zη∑

(ν1,ν2)

Ei.i.d

(f(σ)

( ∏e∈ν1

ρe

)( ∏e∈ν2

ρe

))=

1

Zη∑ν1

Ei.i.d

(f(σ)

∏e∈ν1

ρe

) ∑ν2⊆η\ν1

Ei.i.d

( ∏e∈ν2

ρe

)=

1

Zη∑ν1

Ei.i.d

(f(σ)

∏e∈ν1

ρe

)Zη\ν1

=

∞∑k=0

∑ν1 : |ν1|=k

Ei.i.d

(f(σ)

∏e∈ν1

ρe

)Zη\ν1

Zη

where the number of ν1 such that |ν1| = k is bounded by Ck. Suppose f(σ) = σiσj and suppose k < |i− j|then no connected component of ν1 of size k can contain edges incident to both i and j. Such ν1 decomposesas ν1 = ν′1 ∪ ν′′1 where ν′1 is the union of all connected components of edges incident on i and ν′′1 is the unionof all connected components of edges incident on j. Therefore ν′1 ∩ ν′′1 = ∅. By independence,

Ei.i.d

(f(σ)

∏e∈ν1

ρe

)= Ei.i.d

(f(σ)

∏e∈ν′1

ρe

)Ei.i.d

(f(σ)

∏e∈ν′′1

ρe

)= 0.

Therefore,

EµΛn(f(σ)) ≤

∑k≥|i−j|

c(β)kCk

≤ c1 exp (−c2 |i− j|) as was to be shown.

Next we prove that for sufficiently large β, Ising model is magnetized in d ≥ 2. For any A ⊆ Λ ⊆ Zd let DAbe all edges with one vertex in A and one vertex in Λ \ A. Any set of edges ϕ will be called a contour ifϕ = DA for some A such that A and Λ \A are both connected. Given some arbitrary configuration of spinsσ ∈ −1, 1Λ, a contour ϕ will be called an interface if σu 6= σv ∀u, v ∈ ϕ.

Unlikely Interfaces (Lemma 3)

Given a contour ϕ, the probability that ϕ is an interface is upper bounded by exp (−2β|ϕ|)

Proof: Let Xϕ be the set of all σ such that ϕ is an interface with respect to σ. Then

P(ϕ is an interface

)=

∑σ∈Xϕ exp (βH(σ))∑

σ exp (βH(σ))

Consider ϕ fixed and given any σ ∈ Xϕ let T (σ) be the spin configuration obtained by flipping all the spinson one side of the contour ϕ, where the choice of the side is predetermined. Finally letting Yϕ = range(T )yields

P(ϕ is an interface

)=

∑σ∈Xϕ exp (βH(σ))∑σ∈Yϕ exp (βH(σ).)

(To be continued.)


Lecture 3: April 10Lecturer: Sourav Chatterjee Scriber: Morris (Jie Jun) Ang


3.1 Nondecay of Correlations at High β for d ≥ 2

Theorem 1 Let d ≥ 2. Take any inifinite volume limit of the Ising model on Zd at inverse temperature β.If β is sufficiently large, then there exists C(β) > 0 such that for all i, j, we have

〈σiσj〉 ≥ C(β).

To prove the theorem, it suffices to consider the model in a finite connected set Λ which will eventually growto Zd. Recall the following definitions:

Definition 2 For any A ⊂ Λ let ∂A be the set of all edges with one endpoint in A and one endpoint in Λ\A.

Definition 3 A set of edges γ is called a contour if γ = ∂A for some A such that A and Λ\A are eachconnected.

Definition 4 Given a configuration σ, a contour γ is called an interface if σu 6= σv for all 〈u v〉 ∈ γ.

From the previous lecture, we have the following lemma:

Lemma 5 Given a contour γ,P (γ is an interface) ≤ e−2β|γ|.

We will also need the following two facts:

Fact 6 A contour is a connected set under an appropriate graph structure on the edges of Λ.

A reference for this is Lemma 2 from Adam Timar, ”Boundary-connectivity via Graph Theory” – Proc.AMS 2013. Here’s the proof from the paper:Proof: Let C be the set of 4-cycles surrounding some 2-face in a unit cube in Zd. In what follows we performaddition modulo 2 (so technically we’re in the space (F2)E(Λ)). This allows us to sum elements of C. Weremark that C generates all finite cycles in Zd.

Consider the graph structure G′ on the edges E(Λ) defined as follows: It has vertex set V (G′) = E(Λ), andtwo edges e = (u1, u2), e′ = (v1, v2) ∈ E(Λ) are adjacent in G′ if d(ui, vj) ≤ 1 for some i, j. We will showthat if γ = ∂A with A,Λ\A each connected, then γ is a connected set in G′.

Consider any partition γ = S1 ∪ S2 with S1, S2 6= ∅ and S1 ∩ S2 = ∅; it suffices to show that in G′ we haved(S1, S2) = 1.

3-1


Arbitrarily pick x ∈ A and y ∈ Λ\A. Pick paths P1, P2 from x to y such that P1 ∩ S2 = ∅ and P2 ∩ S1 = ∅.Since P1 + P2 is a cycle, we may represent it as a sum

P1 + P2 =∑C∈A

C for some finite subset A ⊂ C.

Let A1 ⊂ A be the elements intersecting S1, and A2 = A\A1, so

P1 +∑C∈A1

C = P2 +∑C∈A2

C.

Observe that P2 contains an odd number of terms in S2 (since it enters and leaves A an odd number of times,and is disjoint from S1), and each C ∈ C contains an even number of terms in S2. Hence RHS contains anodd number of terms in S2, and so LHS contains some element of S2; since P1 ∩ S2 = ∅, there exists someO ∈ A1 which intersects S2. By definition this O also intersects S1. Since O is a 4-cycle we conclude thatsome edge of S1 is within distance 1 of some edge of S2, and so d(S1, S2) = 1, as needed.

Fact 7 If G is a graph with maximal degree ∆, then the number of connected sets of vertices of size kcontaining a given vertex is at most eC(∆)k.

Proof: It suffices to prove that when G is a ∆-ary tree, the number of connected k-subsets of G containingthe root r is at most eC(∆)k. Consider the following procedure for choosing such a subset S. Start withS = ∅ and a queue initially containing only r. At each step, we dequeue the front element v from the queue.If we choose to include v in S we enqueue its ∆ children in some fixed order, and otherwise we do nothing.We stop when |S| = k. Clearly this procedure can give us any k-subset containing r, and moreover, any runof this procedure takes at most k∆ steps. Each step is either ”keep” or ”reject”, and since we wish to keepprecisely k vertices, there is an injection from S ⊂ G | |S| = k and r ∈ S into the set of all binary stringsof length k∆ which contain exactly k 1’s. Thus using a Stirling approximation,

#k − subsets of G containing r ≤(k∆

k

)≤ (k∆)k

k!≤ kk∆k

√2πk (k/e)

k≤ eC(∆)k.

Combining these facts, we conclude that for any given edge e,

#Contours of size k containing e ≤ eC(d)k.

The last ingredient we need is the following lemma:

Lemma 8 If σi 6= σj, then there must be an interface separating i and j.

Proof: Consider the set of all sites having the spin σi, and let B be the connected component of this setcontaining i. Let A be the connected component of Λ\B that contains j. It’s easy to check that both A andΛ\A are connected. Then γ = ∂A is the desired interface separating i and j.

Now we are ready to prove the theorem.

Proof: Consider i 6= j. We first bound the number of length-k contours that separate i and j. Observethat any such contour must pass within distance k from either i or j. There are at most Ckd edges withindistance k from either i or j, and each such edge has at most eC(d)k contours passing through it. Thus,defining

Sk = length-k contours separating i and j,


we have#Sk ≤ CkdeC(d)k.

Now we bound the probability that σi 6= σj . We have

P (σi 6= σj) ≤ P (there exists an interface separating i, j)

≤∞∑k=1

P (there exists a length-k interface separating i, j)

≤∞∑k=1

∑γ∈Sk

P (γ is an interface)

≤∞∑k=1

CkdeC(d)ke−2βk

<1

2

if β is sufficiently large (size depending only on d, not on Λ or i, j).

Finally, we observe that given the above bound P (σi 6= σj) <12 , we have that 〈σiσj〉 = 1 − 2P (σi 6= σj) is

uniformly bounded away from zero. Thus correlations do not decay in the d ≥ 2, β large regime.

We remark that this proof further shows that as β →∞, the correlations between spins goes to 1.

This theorem was important because it revealed to physicists that phase transitions can arise without largechanges in dynamics.

3.2 Summary

For the Ising model, we can have shown that:

• d = 1: Exponential decay of correlation at all β.

• d ≥ 2: Exponential decay of correlation at small β, no decay at large β.

A generalization of the Ising model is the O(n) model. The sites are still vertices of Zd, but instead of havingspins in −1, 1, we have spins in Sn−1. The probability measure on a configuration σ is then given by thedensity

1

Zexp

∑〈i j〉

σi · σj

where σi · σj is the usual inner product on Sn−1.

• O(1) is the Ising model,

• O(2) is called the XY model or rotator model,

• O(3) is called the XYZ model or Heisenberg model.

What is known for O(n)?


• Exponential decay of correlation at small β for all n, d.

• The O(1) model has no decay at large β if d ≥ 2, and has exponential decay at all β if d = 1.

• O(n) models have no decay if d ≥ 3 and β large. See Frohlich, Spencer, etc’s work using the methodof infrared bounds.

• For d = 2, O(2) model has decay in correlation for all β. Remarkably we have exponential decay atsmall β, but polynomial decay at large β. This is called the Kosterlitz-Thouless transition.

• For d = 2 and n ≥ 3, it is known that polynomial decay occurs at all β. However, the PolyakovConjecture states that exponential decay should happen at all β! This is hard to prove because wecurrently have no good techniques to prove exponential decay for all β.

3.3 Exponential Decay of Correlation at Low β by Coupling

Although we’ve already seen a proof of the exponential decay of correlation for low β, coupling is an importanttechnique that frequently works in proving such bounds for Markov random fields.

The Ising model is an example of a Markov Random Field. That is, the conditional distribution of (σi)i∈Agiven (σj)j∈Λ\A depends only on those j ∈ Λ\A which are neighbors of some i ∈ A.

Theorem 9 Consider the Ising model for any d, in a box with center i and sidelength k. Fix arbitraryboundary conditions. If β ≤ β0(d) then |〈σi〉| ≤ C1e

−C2k.

Note that this statement is much stronger than the exponential decay of correlation, and implies it as follows:Writing τ to be the conditional expectation of σi given all spins outside a box with center i and side lengthk = |i− j| (so j lies outside the box), we have

〈σiσj〉 = 〈σjτ〉,

but the theorem together with the Markov property shows that |τ | ≤ C1e−C2k, and hence 〈σiσj〉 decays

exponentially in k = |i− j|.Now we prove the theorem.

Proof: Consider two different boundary conditions on the box. Let σ, η be configurations from the two Isingmodels with different boundary conditions. We will show that

|〈σi〉 − 〈ηi〉| ≤ C1e−C2k if β is small. (3.1)

This proves the original claim, since we can average this relation over all boundary conditions for η, and bysymmetry the average of 〈ηi〉 over all boundary conditions is zero.

In order to prove 3.1, we use the technique known as coupling. Coupling originated in the mid 1990s; inphysics by Salas-Sokal and in concentration inequalities by Martin.

We start with independent samples σ(0), η(0) from the two boundary conditions. We keep updating them toget (σ(1), η(1)), (σ(2), η(2)), · · · such that for all n,

σ(n) d= σ(0),

η(n) d= η(0),


where Xd= Y means that X,Y have the same distribution. We will prove that

lim supn→∞

P (σ(n)i 6= η

(n)i ) ≤ C1e

−C2k, (3.2)

which would imply 3.1 if we take the lim sup of

|〈σi〉 − 〈ηi〉| = |〈σ(n)i 〉 − 〈η

(n)i 〉| ≤ 2P (σ

(n)i 6= η

(n)i ).

Updating (σ(n), η(n)):Choose a point j ∈ Λ uniformly at random. Define

µ = law of σ(n)j given (σ

(n)k )k 6=j ,

ν = law of η(n)j given (η

(n)k )k 6=j .

Generate (σ(n+1), η(n+1)) such that σ(n+1)j ∼ µ, η

(n+1)j ∼ ν, and P (σ

(n+1)j 6= η

(n+1)j ) = dTV (µ, ν) (see below

for definition). Also update σ(n+1)k = σ

(n)k and η

(n+1)k = η

(n)k for k 6= j.

(Proof to be completed in next lecture)

Definition 10 If µ, ν are probability measures on the same σ-algebra, then the total variation distancebetween µ, ν is

dTV (µ, ν) = supA|µ(A)− ν(A)|,

where the supremum is taken over A in the σ-algebra.

The alternative (equivalent) version of this is the following: Consider any pair of random variables (X,Y )with X ∼ µ and Y ∼ ν. Then the minimum possible value of P (X 6= Y ) is dTV (µ, ν).


Lecture 4: April 12Lecturer: Spring 2017 Scriber: Sourav Chatterjee

Disclaimer: These notes may contain factual and/or typographic errors. Erik Bates

4.0 Recap

Recall that we are considering the Ising model on Λ ⊂ Zd, which for simplicity we assume is a box withcenter i ∈ Λ and side length k. Our goal was to prove exponential decay of spin-spin correlations at hightemperatures (small β) via a coupling argument, a technique that arises frequently in probabilistic analysis.In particular, we seek to show that

|〈σi〉| ≤ C1e−C2k, (4.3)

where σ ∈ ±1Λ is a random spin configuration chosen according to an arbitrary boundary condition, andthe constants C1 and C2 are positive and depend only on β and d. We discussed in the previous lecturethat (4.3) implies exponential decay of correlations. Moreover, by an averaging principle we reasoned thatto show (4.3), it suffices to prove

|〈σi〉 − 〈ηi〉| ≤ C1e−C2k, (4.4)

where σ and η are random spin configurations associated to two arbitrary boundary conditions.

4.1 Proof of (4.4) at high temperatures

Let µ and ν be the probability measures on ±1Λ associated to two arbitrary (but fixed) boundary condi-tions. We will define a sequence of couplings of these two laws such that the marginals of the spin at site ibecome exponentially close in k. Here “closeness” is measured by the total variation distance between twoprobability measures:

dTV(P1, P2) = inf(X,Y )∼(P1,P2)

P (X 6= Y ), (4.5)

where the infimum is over random variables X (distributed as P1) and Y (distributed as P2) defined on thesame probability space (Ω, P ).

The coupling procedure is prescribed inductively as follows. First generate spin configurations σ(0) ∼ µ andη(0) ∼ ν independently at random. Then given σ(n) and η(n), pick a coordinate j ∈ Λ uniformly at random,

and let µ(n)j be the conditional distribution of σ

(n)j given (σ

(n)k )k 6=j . Similarly, let ν

(n)j denote the conditional

distribution of η(n)j given (η

(n)k )k 6=j . We will update the spins at site j to σ

(n+1)j and η

(n+1)j by sampling

from µ(n)j and ν

(n)j , respectively, but not independently. Spins at all other sites k 6= j are unchanged. Having

described the update procedure, let us pause to make some observations:

Observation 11 The conditional distributions under consideration are probabilities on ±1, so there surely

exists a coupling of µ(n)j and ν

(n)j on some (Ω, P ) such that P (σ

(n+1)j 6= η

(n+1)j ) achieves the infimum in the

4-1


definition (4.5). That is, we may generate the pair (σ(n+1)j , η

(n+1)j ) such that the new spins are equal with

the highest possible probability:

P (σ(n+1)j 6= η

(n+1)j ) = dTV(µ

(n)j , ν

(n)j ). (4.6)

For example, if the marginals must satisfy P (σ(n+1)j = +1) = 0.3 and P (η

(n+1)j = +1) = 0.5, then we simply

choose a coupling so that

P (σ(n+1)j = a, η

(n+1)j = b) =

0.3 (a, b) = (+1,+1)

0.5 (a, b) = (−1,−1)

0.2 (a, b) = (−1,+1)

0 (a, b) = (+1,−1).

Observation 12 The conditional distributions actually depend only on the spins at neighbors of j, a set we

will write as N (j) ⊂ Λ. In particular, if σ(n)k = η

(n)k for all k ∈ N (j), then

dTV(µ(n)j , ν

(n)j ) = 0. (4.7)

Observation 13 When β = 0, regardless of neighboring spins, the conditional distribution of the spin atsite j is uniform over ±1, provided j /∈ ∂Λ. So (4.7) always holds in this case. For β > 0, there is still anupper bound C depending only on β and d:

dTV(µ(n)j , ν

(n)j ) ≤ C, (4.8)

again provided that j /∈ ∂Λ. Moreover, C → 0 as β → 0. (Exercise: Check that C can be taken equal totanh(2dβ).)

Observation 14 Since σ(n+1)j has the same distribution as σ

(n)j , and all other spins are unchanged, the law

of σ(n+1) is the same as that of σ(n). Therefore, σ(n) ∼ µ for every n. By the same logic, η(n) ∼ ν for everyn. Therefore,

|〈σi〉 − 〈ηi〉| = |2P(σ(n)i = 1, η

(n)i = −1)− 2P(σ

(n)i = −1, η

(n)i = 1)|

≤ 2P(σ(n)i 6= η

(n)i ) for all n

⇒ |〈σi〉 − 〈ηi〉| ≤ 2 lim supn→∞

P(σ(n)i 6= η

(n)i ). (4.9)

Now let N = |Λ|. At the n-th update, there is a 1/N probability that j is picked as the update site. If j isnot picked, then trivially we have

P(σ(n+1)j 6= η

(n+1)j | σ(n), η(n), j not picked) = 1σ(n)

j 6=η(n)j

.

On the other hand, if j is picked and j /∈ ∂Λ, then

P(σ(n+1)j 6= η

(n+1)j | σ(n), η(n), j picked)

(4.6)= dTV(µ

(n)j , ν

(n)j )

(4.7)= dTV(µ

(n)j , ν

(n)j )1σ(n)

k 6=η(n)k for some k∈N (j)

(4.8)

≤ C1σ(n)k 6=η

(n)k for some k∈N (j)

≤ C∑

k∈N (j)

1σ(n)k 6=η

(n)k

.


Finally, if j is picked but j ∈ ∂Λ, then no updates can be made because of the boundary conditions.Consequently, the probability of equality is 0 or 1, depending on if the two boundary conditions agree at sitej:

P(σ(n+1)j 6= η

(n+1)j | σ(n), η(n), j picked) = 1σ(n)

j 6=η(n)j

.

The three cases we just considered can be expressed in a single inequality:

P(σ(n+1)j 6= η

(n+1)j | σ(n), η(n)) ≤

(1− 1

N

)1σ(n)

j 6=η(n)j

+C

N

∑k∈N (j)

1σ(n)k 6=η

(n)k

+hjN,

where

hj :=

1 j ∈ ∂Λ

0 else.

Averaging over all possible realizations of σ(n) and η(n), we find

P(σ(n+1)j 6= η

(n+1)j ) ≤

(1− 1

N

)P(σ

(n)j 6= η

(n)j ) +

C

N

∑k∈N (j)

P(σ(n)k 6= η

(n)k ) +

hjN. (4.10)

We have derived a recursive inequality (4.10). Given the interaction between neighboring sites, it is usefulto express this inequality in terms of matrix-vector products. Upon defining the vectors

x(n) := (x(n)j )j∈Λ, x

(n)j := P(σ

(n)j 6= η

(n)j ),

h := (hj)j∈Λ,

and the matrix

Q := (qjk)j,k∈Λ, qjk :=

C k ∈ N (j)

0 else,

we can write the coordinate-wise inequality

x(n+1) ≤(

1− 1

N

)x(n) +

1

NQx(n) +

h

N.

As the above display holds for every n, we in fact have

x ≤(

1− 1

N

)x+

1

NQx+

h

N,

where

xj := lim supn→∞

x(n)j , j ∈ Λ.

Rearranging terms gives

(I −Q)x ≤ h. (4.11)

Recall the definition of the matrix norm

‖Q‖∞→∞ := sup‖Qz‖∞ : ‖z‖∞ ≤ 1.


Since each row of Q has at most 2d nonzero entries,

‖Q‖∞→∞ ≤ 2dC.

Here, for the first and only time, we make an assumption on β. Namely, we may assume by Observation 13that β is sufficiently small so that

2dC < 1. (4.12)

It follows that if R :=∑∞m=0Q

m, then

‖R‖∞→∞ ≤∞∑m=0

‖Qm‖∞→∞ ≤∞∑m=0

‖Q‖m∞→∞ <∞.

Furthermore,

R(I −Q) = R−RQ = R−∞∑m=0

Qm+1 = R− (R− I) = I.

Since all entries of R are nonnegative, (4.11) implies

x ≤ Rh =

∞∑m=0

Qmh,

where the j-th entry of Qmh is

(Qmh)j =∑

j1,j2,...,jm

qjj1qj1j2 · · · qjm−1jmhjm .

Notice that the product qjj1qj1j2 · · · qjm−1jmhjm is zero if either jm ∈ ∂Λ or (j, j1, . . . , jm) is not a latticepath. In particular, when m < k/2 and j = i is at the center of Λ (which has side length k), at least one ofthese two statements must be true. Hence

|〈σi〉 − 〈ηi〉|(4.9)

≤ 2xi ≤ 2∑

m≥k/2

(Qmh)i ≤ 2∑

m≥k/2

‖Qmh‖∞ ≤ 2∑

m≥k/2

‖Q‖m∞→∞‖h‖∞ ≤ 2(2dC)k/2

1− 2dC.

Setting our constants to be

C1 :=2

1− 2dC, C2 := −1

2log(2dC),

we have proved (4.4) for β such that (4.12) holds.

4.2 O(2) model: polynomial decay for low temperatures in d = 2

Recall the O(n) model in which spins assume values in the sphere §n−1, and the Hamiltonian is the sum ofinner products between neighboring spins:

H(σ) :=∑〈j k〉

σj · σk


In this section, we begin a proof (due to McBryan and Spencer [2]) that spin-spin correlations in two-dimensional O(n) models decay faster than a polynomial for sufficiently large β. More precisely, for anyn ≥ 2, for all sufficiently large β,

〈σj · σk〉 ≤ C1|j − k|−C2 ,

where C1 and C2 are positive constants depending only on β and n. We will consider just the case n = 2,also known as the XY model.

Remark 15 In the previous lecture we mentioned that correlations decay exponentially for small β. Itis interesting, then, that the proof we present works only for large β. Although, the Kosterlitz-Thoulesstransition [1] says that for n = 2, exponential decay does not hold for large β. Meanwhile, for n ≥ 3,Polyakov [3] conjectured that exponential decay holds for all temperatures.

Remark 16 The result under discussion is a special case of the Mermin-Wagner theorem in physics, whichroughly says, ”Systems with continuous symmetries have no ordered phases in dimension two.” Here theterm ordered means having long-range interactions. Indeed, since we are considering n ≥ 2, the spins in§n−1 have continuous symmetry.

The setting we consider is Λ ⊂ Z2. A spin configuration will be thought of as an element of [−π, π)Λ, andfor simplicity we assume all boundary spins are 0. With this parameterization, the Hamiltonian is written

H(θ) =∑〈j k〉

cos(θj − θk).

The correlation we wish to study is

〈cos(θu − θv)〉,

for u, v ∈ Λ \ ∂Λ.

Claim 17 If β is large enough, then

|〈cos(θu − θv)〉| ≤ C1|u− v|−C2(β).

We now begin the proof. To simplify notation, denote Λ := Λ \ ∂Λ. By symmetry, we must have 〈sin(θu −θv)〉 = 0, and so

〈cos(θu − θv)〉 = 〈ei(θu−θv)〉 =1

Z

∫[−π,π]Λ

ei(θu−θv)eβH(θ) dθ,

where dθ denotes the usual Lebesgue integration, and the normalizing constant

Z =

∫[−π,π)Λ

eβH(θ) dθ

makes the integral a valid expectation.

Let a = (aj)j∈Λ be any vector of real numbers such that aj = 0 for all j ∈ ∂Λ (later we will choose the other

values in an optimal way). We can order the elements of Λ as j1, j2, . . . , jN so that∫[−π,π)Λ

ei(θu−θv)eβH(θ) dθ =

∫[−π,π)

· · ·∫

[−π,π)

ei(θu−θv)eβH(θ) dθ1 · · · dθN .


+ iaj + iaj

Figure 4.1: Contour of integration

For each j (starting at j1, then j2, and so on), we can rewrite the integral over [−π, π) as the same integralover the line segment [−π + iaj , π + iaj). To see this, one can consider the same integrand, but integratedover the closed contour shown in Figure 4.1. Notice that the integrals along the vertical segments will cancelone another, since

ei(π+iau−θv) = e2πiei(−π+iau−θv) = ei(−π+iau−θv),

and similarly

cos(π + iaj − θk) = cos(−π + iaj − θk),

for any θv, θk (real or complex). Therefore, by Cauchy’s integral formula, the integral over [−π, π) is equalto the integral over the line segment [−π + iaj , π + iaj). As claimed, then, we can write

〈cos(θu − θv)〉 =1

Z

∫[−π,π)Λ

ei(θu+iau−θv−iav)eβH(θ+ia) dθ.

Upon recalling the identity

cos(x+ iy) = cos(x) cos(iy)− sin(x) sin(iy)

= cos(x) cosh(y)− i sin(x) sinh(y),

we can expand the Hamiltonian as

H(θ + ia) =∑〈j k〉

cos(θj + iaj − θk − iak)

=∑〈j k〉

cos(θj − θk) cosh(aj − ak)− i∑〈j k〉

sin(θj − θk) sinh(aj − ak).

Hence

|eβH(θ+ia)| = exp

(β∑〈j k〉

cos(θj − θk) cosh(aj − ak)

)

= exp

(β∑〈j k〉

cos(θj − θk)(cosh(aj − ak)− 1)

)exp

(β∑〈j k〉

cos(θj − θk)

)

≤ exp

(β∑〈j k〉

(cosh(aj − ak)− 1)

)eβH(θ),


where the inequality is justified by the fact that cosh(x) ≥ 1 for all x ∈ R. Using these observations, we find

|〈cos(θu − θv)〉| ≤1

Z

∫[−π,π)Λ

|ei(θu+iau−θv−iav)eβH(θ+ia)| dθ

≤ 1

Zexp

(− (au − av) + β

∑〈j k〉


)∫[−π,π)Λ

eβH(θ) dθ

= exp

(− (au − av) + β

∑〈j k〉


).

In the next lecture we will choose the aj to minimize this upper bound.

References

[1] John Michael Kosterlitz and David James Thouless. Ordering, metastability and phase transitions intwo-dimensional systems. Journal of Physics C: Solid State Physics, 6(7):1181, 1973.

[2] Oliver A McBryan and Thomas Spencer. On the decay of correlations inso (n)-symmetric ferromagnets.Communications in Mathematical Physics, 53(3):299–302, 1977.

[3] Alexander M Polyakov. Interaction of goldstone particles in two dimensions. applications to ferromagnetsand massive yang-mills fields. Physics Letters B, 59(1):79–81, 1975.


Lecture 5: April 17Lecturer: Sourav Chatterjee Scriber: Morris (Jie Jun) Ang


5.3 Continuation: Polynomial decay of correlation at low temper-ature in 2-dimensional O(2) model

Let’s recall some notation. We have the collection of sites Λ ⊂ Z2, with spins θ = (θu)u∈Λ ∈ [−π, π)Λ. Weimposed the boundary condition θu = 0 for u ∈ ∂Λ.

Recall from the last lecture that, given any u, v ∈ Λ and any choice of real numbers (aj)j∈Λ with aj = 0 on∂Λ, we have

〈cos(θu − θv)〉 ≤ exp

−(au − av) +β

2

∑〈j k〉


.

We want to choose the (aj)j∈Λ so that the bound is strong. We reparametrize, writing bj = βaj . We’llchoose (bj)j∈Λ ∈ RΛ to be the solution to

minimizeb

− (bu − bv) +1

2

∑〈j k〉

(bj − bk)2

subject to bj = 0, j ∈ ∂Λ.

Recall the standard definition of the discrete Laplacian operator

(∆b)j =∑

k∈N(j)

(bk − bj), where N(j) is the set of vertices adjacent to j.

Taking finite differences, the minimizer of the above system satisfies

(∆b)j =

0 if j 6= u, v1 if j = u,−1 if j = v.

We can invert this using the discrete Green’s function, to get

bj =∑k

G(j, k)lk,

wherelu = 1, lv = −1, lk = 0 for all other k.

Substituting this into the given expression, we see that the minimizer of the above system is

bj = G(j, u)−G(j, v).

5-1


We will see later in the course that

bu − bv = G(u, u) +G(v, v)− 2G(u, v) ≈ C log |u− v|.

Moreover, it’s well known that at this optimal (bj)j∈Λ we have∑〈j k〉

(bj − bk)2 = bu − bv.

Given a uniform bound |bj − bk| ≤ C for neighboring j, k, we get a corresponding bound |aj − ak| ≤ Cβ , and

so a Taylor expansion of cosh yields

cosh(aj − ak)− 1 ≤ 1 + ε

2(aj − ak)2,

where ε = ε(β)→ 0 as β →∞. Choose β sufficiently large so that ε < 1, then

−(au − av) +β

2

∑〈j k〉

(cosh(aj − ak)− 1) ≤ −(au − av) +β

2

∑〈j k〉

(aj − ak)2

= −(au − av) +1

2β

∑〈j k〉

(bj − bk)2

= −(au − av) +1

2β(bu − bv)

= −(au − av) +1

2(au − av)

= −1

2(au − av)

≈ − C2β

log |u− v|.

Returning to our original inequality, we finally get

〈cos(θu − θv)〉 ≤ exp

−(au − av) +β

2

∑〈j k〉


≤ |u− v|− C2β .

5.4 Existence of ordered phase in O(n) models in d ≥ 3 (Frohlich,Simon, Spencer)

For simplicity, we work in the discrete torus TdL = −L+ 1,−L+ 2, . . . , L− 1, Ld. Our spin configuration

σ lies in (Sn−1)TdL , and H(σ) =

∑〈uv〉 σu · σv. For convenience, write V = V (TdL).

Theorem 18

1

|V |2∑x,y∈V

〈σx · σy〉 ≥ Cn,d,β > 0 if d ≥ 3 and β sufficiently large.

Crucially, this lower bound Cn,d,β is independent of L.


This proof will depend on three important concepts: Gaussian domination, the infrared bound, and reflectionpositivity.

Let E be the expectation with respect to the iid measure on (Sn−1)Λ (that is, when we take β = 0). For anyfunction τ : V → Rn, let

Z(τ) = E exp

−β2

∑〈u v〉

‖σu + τu − σv − τv‖2 ,

where we understand the spin σu ∈ Sn−1 to be a unit vector in Rn.

Lemma 19 (Gaussian domination)

Z(τ) ≤ Z(0) for all τ : V → Rn.

Lemma 20 (Infrared bound)

1

|V |∑u,v∈V

eik·(v−u)〈σu · σv〉 ≤n

2β∑nj=1(1− cos kj)

where k = (kj)nj=1 ∈ V ∗\0 and V ∗ = π

LV .

We note that by translation invariance of the torus, letting 0 denote the origin,

1

|V |∑u,v∈V

eik·(v−u)〈σu · σv〉 =1

|V |∑v∈V

eik·v〈σ0 · σv〉 = f(k)

is a Fourier transform.

The proof of the theorem uses the infrared bound, and the proof of the infrared bound depends on Gaussiandomination. Gaussian domination is proved via reflection positivity.

We can write V = V0 ∪ V1, where

V0 = (x1, . . . , xd) ∈ V : x1 ≤ 0,

V1 = (x1, . . . , xd) ∈ V : x1 ≥ 1.Let θ(x1, . . . , xd) = (−x1 + 1, x2, . . . , xd) be the reflection across the plane x1 = 1

2 . We can write anyτ : V → Rn as τ = (τ0, τ1), with τi : Vi → Rn for i = 0, 1.

For τ0 : V0 → Rn, let θ(τ0) : V1 → Rn be defined by

θ(τ0)(x) = τ0(θ(x)).

Likewise, for τ1 : V1 → Rn, define θ(τ1) : V0 → Rn.

Lemma 21 (Reflection positivity) For all τ0, τ1,

Z(τ0, τ1)2 ≤ Z(τ0, θ(τ0))Z(θ(τ1), τ1).

Equipped with this, we can prove Gaussian domination:

Proof:[Gaussian domination] We first prove that there exists some τ maximizing Z(τ). Clearly Z is boundedabove by 1, so supτ Z(τ) exists. Let τ1, τ2, · · · be a sequence that approaches supτ Z(τ). Since Z(τ) =


Z(τ + c) for any constant c, we may assume that τn0 = 0 for all n. We claim that there exists some C suchthat τnv < C for all n ∈ N, v ∈ V . Indeed, if not, then there exists a sequence τnivi which grows to infinity,then since τni0 = 0 and ‖σvi‖ = ‖σ0‖ = 1, we have ‖σvi + τnivi − σ0 − τni0 ‖ → ∞ also. Then Z(τni) → 0, acontradiction. Thus τnv < C for all n ∈ N, v ∈ V , so by compactness there exists some increasing sequencen1, n2, . . . such that τni converges pointwise to some τ∞. Since Z is continuous in τ , we conclude thatZ(τ∞) = supτ Z(τ). Hence some function τ attains the maximal value of Z.

For any τ , letk(τ) = #〈u v〉 : τu 6= τv

be the number of pairs of adjacent sites whose τ -values disagree. Among all τ maximizing Z, pick τminimizing k(τ). We claim that k(τ) = 0. Suppose otherwise; let 〈u v〉 be one edge for which τu 6= τv.Without loss of generality (by translation invariance), assume u ∈ V0 and v ∈ V1.

Let τ = (τ0, τ1), and define τ ′ = (τ0, θ(τ0)) and τ ′′ = (θ(τ1), τ1). Observe that performing the reflection θremoves the disagreement at the edge 〈u v〉, so either k(τ) > k(τ ′) or k(τ) > k(τ ′′); WLOG it’s the former.

By reflection positivity, we haveZ(τ)2 ≤ Z(τ ′)Z(τ ′′),

and since Z(τ) maximizes Z, we conclude that Z(τ ′) = Z(τ ′′) = Z(τ). Recall however that k(τ ′) < k(τ),contradicting how τ was chosen.

Thus, the constant function is a maximizer of Z. Since Z(τ) = Z(τ + c) for any constant c, we concludethat Z(τ) ≤ Z(0) for all τ .


Lecture 6: April 19Lecturer: Sourav Chatterjee Scriber: Erik Bates


6.0 Recap

We are considering the O(n) model in dimension d ≥ 3, where n ≥ 1 is arbitrary. To simplify analysis, wedefined on the model on the torus with side length 2L:

TdL = −L+ 1, . . . , Ld,

with edge set E = E(TdL) and vertex set V = V (TdL). Spin configurations σ ∈ (§n−1)TdL are subject to the

Hamiltonian

H(σ) =∑〈u v〉∈E

σu · σv.

In the last lecture, we stated some preliminary lemmas with the goal of proving the following result.

Theorem 22 For any d ≥ 3, and all β and L sufficiently large (depending on d), there exists a constantC = C(n, d, β) > 0 such that

1

|V |2∑x,y∈V

〈σx · σy〉 ≥ C.

Let us recall some notation. We considered a partition of V = V0 ∪ V1, where

V0 := (x1, . . . , xd) ∈ V : x1 ≤ 0V1 := (x1, . . . , xd) ∈ V : x1 ≥ 1,

along with a reflection map θ : V → V sending vectors between the two sets:

θ(x1, x2, . . . , xd) = (1− x1, x2, . . . , xd).

Given a map τ : V → Rn, we denoted its restrictions τ0 = τ∣∣V0

and τ1 = τ∣∣V1

, and made the identification

τ = (τ0, τ1). Moreover, given τ0 : V0 → Rn, we defined the function θ(τ0) : V1 → Rn by

θ(τ0)(x) := τ0(θ(x)), x ∈ V1.

Analogously, for τ1 : V1 → Rn, we have

θ(τ1)(x) := τ1(θ(x)), x ∈ V0.

The key quantity we started to examine is

Z(τ) := E exp

(− β

2

∑〈u v〉∈E

‖σu + τu − σv − τv‖2),

where E(·) denotes expectation with respect to the i.i.d. measure on spins. We stated but did not prove thefollowing lemma, which resembles a Cauchy-Schwarz inequality.

6-1


Lemma 23 (Reflection positivity) For any τ = (τ0, τ1),

Z(τ)2 ≤ Z(τ0, θ(τ0)) · Z(θ(τ1), τ1).

Assuming this result, we did prove the next lemma. (The name comes from the fact that a Gaussian systemwould give equality.)

Lemma 1 (Gaussian domination) For any τ : V → Rn,

Z(τ) ≤ Z(0).

Finally, we stated without proof the so-called “infrared bound”, which gives a uniform bound on the nonzeroFourier coefficients of the spin correlation function v 7→ 〈σ0 · σv〉.

Lemma 2 (Infrared bound) Let V ∗ = (π/L)V . For any k ∈ V ∗ \ 0,

1

|V |∑u,v∈V

eik·(v−u)〈σu · σv〉 ≤n

2β∑dj=1(1− cos kj)

.

6.1 Proof of Lemma 1a

Fix τ : V → Rn. Let σ : V → §n−1 ⊂ Rn be a spin configuration such that (σu)u∈V are i.i.d. uniformrandom variables on §n−1. Consider the edges bridging V0 and V1:

E′ := 〈u v〉 : u ∈ V0, v ∈ V1.

In this notation,

Z(τ) = Z(τ0, τ1) = E exp

(f0(σ0, τ0) + f1(σ1, τ1)− β

2

∑〈u v〉∈E′

‖σu + τu − σv − τv‖2),

where we have separated the contributions from just V0 or just V1:

f0(σ0, τ0) = −β2

∑〈u v〉∈E(V0)

‖σu + τu − σv − τv‖2

f1(σ1, τ1) = −β2

∑〈u v〉∈E(V1)

‖σu + τu − σv − τv‖2.

This will be a convenient representation after we apply a trick sometimes called “Gaussian disintegration”to linearize the exponent. Recall that if X ∼ N (0, I) is a standard Gaussian random vector in Rn, then itsFourier transform is

EX(eiθ·X) = e−‖θ‖2/2, θ ∈ Rn. (6.13)


Hence

E exp

(− β

2

∑〈u v〉∈E′

‖σu + τu − σv − τv‖2)

= E

[ ∏〈u v〉∈E′

exp(− β

2‖σu + τu − σv − τv‖2

)](6.13)

= E

[ ∏〈u v〉∈E′

EXuv exp(i√β(σu + τu − σv − τv) ·Xuv)

]

= E

[EX

( ∏〈u v〉∈E′

exp(i√β(σu + τu − σv − τv) ·Xuv)

)]

= E

[EX exp

( ∑〈u v〉∈E′

i√β(σu + τu − σv − τv) ·Xuv

)],

where X = (Xuv)〈u v〉∈E′ is a collection of i.i.d. standard Gaussian random vectors in Rn. Everything beingintegrable, we are allowed to reverse the order of E and EX (by Fubini’s theorem). Upon doing so, we obtain

Z(τ0, τ1) = EX

[E(eY0(σ0,τ0,X)eY1(σ1,τ1,X))

],

where

Y0(σ0, τ0, X) := f0(σ0, τ0) + i√β

∑〈u v〉∈E′

(σu + τu) ·Xuv

Y1(σ1, τ1, X) := f1(σ1, τ1)− i√β

∑〈u v〉∈E′

(σv + τv) ·Xuv.

Notice that given X, the random variables Y0 and Y1 are independent. It follows that

Z(τ0, τ1) = EX

[E(eY0(σ0,τ0,X))E(eY1(σ1,τ1,X))

](6.14)

≤√

EX

[|E eY0(σ0,τ0,X)|2

]EX

[|E eY1(σ1,τ1,X)|2

], (6.15)

where the inequality is an application Cauchy-Schwarz. We now calculate

|E eY0(σ0,τ0,X)|2 = E(eY0(σ0,τ0,X))E(eY0(σ0,τ0,X)) = E(eY0(σ0,τ0,X))E(eY0(σ0,τ0,X))

= E(eY0(σ0,τ0,X))E(eY0(θ(σ1),τ0,X)),

where the last equality follows from the fact that σ0 has the same law as θ(σ1). Upon noticing

f0(θ(σ1), τ0) = f1(σ1, θ(τ0)), τu = (θ(τ0))v for 〈u v〉 ∈ E′,we can write

E(eY0(θ(σ1),τ0,X)) = E exp

(f0(θ(σ1), τ0)− i

√β

∑〈u v〉∈E′

(σu + τu) ·Xuv

)

= E exp

(f1(σ1, θ(τ0))− i

√β

∑〈u v〉∈E′

(σv + (θ(τ0))v) ·Xuv

)= E(eY1(σ1,θ(τ0),X)).

Therefore,

EX

[|E(eY0(σ0,τ0,X))|2

]= EX

[E(eY0(σ0,τ0,X))E(eY1(σ1,θ(τ0),X))

] (6.14)= Z(τ0, θ(τ0)).

Similarly, we can obtain

EX

[|E eY1(σ1,τ1,X)|2

]= Z(θ(τ1), τ1),

and so (6.15) gives the desired inequality.


6.2 Proof of Lemma 2

Take any τ : V → Rn. For α ∈ R, define the function

f(α) := Z(ατ) = E(eA(α)), A(α) = −β2

∑〈u v〉∈E

‖σu + ατu − σv − ατv‖2.

By Lemma 1, f is maximum at the zero function. In particular,

f ′′(0) ≤ 0. (6.16)

To make use of this condition, we make the computation

f ′′(α) = E[(A′′(α) + [A′(α)]2)eA(α)

],

where

A′(α) = −β∑〈u v〉∈E

(σu + ατu − σv − ατv) · (τu − τv)

= −β∑〈u v〉∈E

[(σu − σv) · (τu − τv) + α‖τu − τv‖2

]⇒ A′′(α) = −β

∑〈u v〉∈E

‖τu − τv‖2.

We now have

f ′′(0) = E

[− β

∑〈u v〉∈E

‖τu − τv‖2 +

(β∑〈u v〉∈E

(σu − σv) · (τu − τv))2

eA(0)

].

Observe that

eA(0) = exp

(− β

2

∑〈u v〉∈E

‖σu − σv‖2)

= exp

(− β

∑〈u v〉∈E

(1− σu · σv))

= e−β|E|eβH(σ),

which means we can express f ′′(0) in terms of an expectation with respect to the Gibbs measure:

f ′′(0) = e−β|E|E(eβH(σ))

⟨− β

∑〈u v〉∈E

‖τu − τv‖2 +

(β∑〈u v〉∈E

(σu − σv) · (τu − τv))2⟩

.

Now (6.16) shows ⟨(β∑〈u v〉∈E

(σu − σv) · (τu − τv))2⟩

≤ β∑〈u v〉∈E

‖τu − τv‖2. (6.17)

More generally, if τ : V → Cn, we can apply (6.17) to Re τ and Im τ separately and then obtain⟨∣∣∣∣β ∑〈u v〉∈E

(σu − σv) · (τu − τv)∣∣∣∣2⟩ ≤ β ∑

〈u v〉∈E

‖τu − τv‖2, (6.18)

where now x · y denotes the standard inner product in Cn:

x · y =

n∑i=1

xiyi.


Next we make some simplifying calculations. First,∑〈u v〉∈E

(σu − σv) · (τu − τv) =∑〈u v〉∈E

σu · (τu − τv)−∑〈u v〉∈E

σv · (τu − τv)

=1

2

∑u∈V

∑v∈N (u)

σu · (τu − τv)−1

2

∑v∈V

∑u∈N (v)

σv · (τu − τv)

= −1

2

∑u∈V

σu · (∆τ)u −1

2

∑v∈V

σv · (∆τ)v

= −∑u∈V

σu · (∆τ)u,

(6.19)

where

(∆τ)u :=∑

v∈N (u)

τv − τu.

Replacing σu − σv in the above computation by τu − τv, we similarly obtain∑〈u v〉∈E

‖τu − τv‖2 = −∑u∈V

τu · (∆τ)u.

Let 1 ∈ Rn be the vector of all 1’s, and make the specific choice of τ given by

τu = eik·u1, k ∈ V ∗ \ 0.

In this case,

(∆τ)u =∑

v∈N (u)

(eik·v − eik·u)1 =

d∑j=1

[eik·(u+ej) + eik·(u−ej) − 2eik·u

]1

=

( d∑j=1

2(cos(kj)− 1)eik·u)

1.

Therefore,

∑〈u v〉∈E

‖τu − τv‖2 = −∑u∈V

τu · (∆τ)u = −∑u∈V

2n

d∑j=1

(cos(kj)− 1) = n|V |d∑j=1

2(1− cos kj). (6.20)

Also, ⟨∣∣∣∣ ∑u∈V

σu · (∆τ)u

∣∣∣∣2⟩ =

( d∑j=1

2(1− cos kj)

)2⟨∣∣∣∣ ∑u∈V

(σu · 1)eik·u∣∣∣∣2⟩, (6.21)

where ⟨∣∣∣∣ ∑u∈V

(σu · 1)eik·u∣∣∣∣2⟩ =

⟨(∑u∈V

(σu · 1)eik·u)(∑

v∈V(σv · 1)eik·v

)⟩=∑u,v∈V

eik·(u−v)〈(σu · 1)(σv · 1)〉.


The proof will be complete once we show

〈(σu · 1)(σv · 1)〉 = 〈σu · σv〉, (6.22)

since then

1

|V |∑u,v∈V

eik·(v−u)〈σu · σv〉(6.22)

=1

|V |∑u,v∈V

eik·(v−u)〈(σu · 1)(σv · 1)〉

(6.21)=

1

|V |

⟨∣∣∣∣ ∑u∈V

σu · (∆τ)u

∣∣∣∣2⟩( d∑j=1

2(1− cos kj)

)−2

(6.19)=

1

|V |

⟨∣∣∣∣ ∑〈u v〉∈E

(σu − σv) · (τu − τv)∣∣∣∣2⟩( d∑

j=1

2(1− cos kj)

)−2

(6.18)

≤ 1

|V |

(1

β

∑〈u v〉∈E

‖τu − τv‖2)( d∑

j=1

2(1− cos kj)

)−2

(6.20)=

n

β

( d∑j=1

2(1− cos kj)

)−1

,

which is exactly the claim.

To verify (6.22), consider the following fact. For any u 6= v in V and indices i 6= j, the pair of coordinates((σu)i, (σv)j) has the same joint distribution as ((σu)i,−(σv)j) under the Gibbs measure. This is because

H(σ) = H(σ′),

where σ′ is the spin configuration obtained from σ by flipping the j-th coordinate at every site:

(σ′w)k =

(σw)k k 6= j

−(σw)j k = j.

This symmetry implies 〈(σu)i(σv)j〉 = 0 whenever i 6= j. Therefore,

〈(σu · 1)(σv · 1)〉 =

⟨( n∑i=1

(σu)i

)( n∑j=1

(σv)j

)⟩=

n∑i,j=1

〈(σu)i(σv)j〉 =

n∑i=1

〈(σu)i(σv)i〉 = 〈σu · σv〉.

6.3 Proof of Theorem 1

Take any a ∈ Zn, and consider the quantity

∑k∈V ∗

eik·a =

L∑t1,...,td=−L+1

d∏j=1

eiπtjaj/L =

d∏j=1

L∑t=−L+1

eiπtaj/L,

which one may recognize as the d-dimensional discrete Fourier transform of the constant 1 function, evaluatedat the frequency vector a. It is not surprising, then, that when aj 6= 0, the product is 0. Indeed, by a geometricseries computation, aj ∈ Z \ 0 implies

L∑t=−L+1

eiπtaj/L =eiπaj(−L+1)/L(1− ei2πaj )

1− eiπaj/L = 0.


Of course, if aj = 0 for every j, then ∑k∈V ∗

eik·a = |V ∗| = |V |.

This discussion shows

∑k∈V ∗

eik·(v−u)〈σu · σv〉 =

|V | u = v

0 u 6= v.

Using Lemma 2, we thus have

|V | = 1

|V |∑u,v∈V

∑k∈V ∗

eik·(v−u)〈σu − σv〉

=1

|V |∑u,v∈V

〈σu − σv〉+1

|V |∑

k∈V ∗\0

∑u,v∈V

eik·(v−u)〈σu · σv〉

≤ 1

|V |∑u,v∈V

〈σu − σv〉+n

2β

∑k∈V ∗\0

1∑dj=1(1− cos kj)

,

which upon rearrangement reads

1

|V |2∑u,v∈V

〈σu · σv〉 ≥ 1− n

2β|V |∑

k∈V ∗\0


. (6.23)

It is now a standard observation of Riemann integration that

limL→∞

1

|V |∑

k∈V ∗\0


=1

(2π)d

∫(−π,π]d

1∑dj=1(1− cosxj)

dx,

which is finite if d ≥ 3. To why this is the case, recall that 1− cosx ≈ x2/2 for x close to 0. Of course,

d∑j=1

x2j = |x|2,

and so we are really observing that∫B(0,1)

1

|x|2 =

∫ 2π

0

∫ π

0

· · ·∫ π

0

∫ 1

0

1

r2rd−1 sind−2(φ1) sind−3(φ2) · · · sin(φd−2) dr dφ1 · · · dφd−2 dφd−1

=∞ d = 1, 2

<∞ d ≥ 3.

The result now follows from (6.23) by taking β and L sufficiently large.


Lecture 7/8: April 24 & 26Lecturer: Sourav Chatterjee Scriber: Yue Hui


0.4 Correlation inequalities

0.4.1 FKG inequality

First, we discuss the FKG inequality.Let S ∈ R be a finite or countable subset of R, and let ρ be a probability mass function that is strictlypositive.First, we introduce some definitions:

Definition 3 For x, y ∈ Sn, x ≤ y if and only if xi ≤ yi, ∀i.

Definition 4 The probability mass is said to satisfy the FKG lattice condition if

∀x, y ∈ Sn, ρ(x)ρ(y) ≤ ρ(x ∧ y)ρ(x ∨ y)

Note that a probability measure will always satisfies the lattice condition.

Definition 5 A function f : Sn → R is monotone increasing if

x ≤ y ⇒ f(x) ≤ f(y)

Theorem 6 Consider ρ that is strictly positive and satisfies lattice condition. Let X be distributed as ρ. Letf, g be two monotone increasing functions s.t.

Ef(X)2 <∞, Eg(X)2 <∞.

Then,

Cov (f(X), g(X)) ≥ 0

Proof: By induction on nFor n=1, let X,Y be i.i.d. ρ distributed.Then,

E[(f(X)− f(Y )) (g(X)− g(Y ))] = 2Cov (f(X), g(X))

Since n=1, X ≥ Y , or X ≤ Y , and in either case, as f, g are monotone increasing,

(f(X)− f(Y )) (g(X)− g(Y )) ≥ 0.

0-1

0-2 Lecture 7/8: April 24 & 26

Therefore,

E[(f(X)− f(Y )) (g(X)− g(Y ))] ≥ 0

Now, suppose that n > 1 and the inequality holds in all small dimensions.Define

f1(a) =E (f(X)|X1 = a)

We have an easy identity

Cov(A,B) =E[Cov(A,B|C)] + Cov (E(A|C), E(B|C)) ,

Hence, we have

Cov (f(X), g(X)) =E [f(X)g(X)− f1(x1)g1(x1)] + E [f1(x1)g1(x1)]− E[f(X)]E[g(X)]

Let X ′ = (X2, · · · , Xn), then the conditional probability mass for X ′ given X1 = a is propotional toρ(a, x2, ·, xn). Clearly, this also satisfies the lattive condition, and by the induction hypothesis, we have

Cov(f(X), g(X)|X1 = a) ≥ 0, ∀a

Next, we want to show that f1, g1 are monotone increasing functions, and this will complete the proof bythe n = 1 case

f1(b)− f1(a) =

∑x′∈§n−1

f(b, x′)ρ(b, x′)

∑x′∈§n−1

ρ(b, x′)− f1(a)

Let τ(x′) = ρ(b,x′)ρ(a,x′)

f1(b)− f1(a) =

∑x′∈Sn−1

f(b, x′)τ(x′)ρ(a, x′)∑x′∈Sn−1

τ(x′)ρ(a, x′)− f1(a)

≥

∑x∈Sn−1

ρ(a, x′)τ(x′)f(a, x′)∑x∈Sn−1

τ(x′)ρ(a, x′)− f1(a)

=

∑x∈Sn−1

f(a, x)τ(x)f(a, x)

∑ρ(a,x)∑

x∈Sn−1

τ(x)f(a, x)

∑rho(a,x)

− f1(a)

=E(f(X)τ(X ′)|X1 = a)

E(τ(X ′)|X1 = a)− E(f(X)|x1 = a)

=Cov(f(a,X ′), τ(X ′)|X1 = a)

E(τ(X ′)|X1 = a)

Lecture 7/8: April 24 & 26 0-3

We have that the distribution of X ′ given X1 = a satisfies the lattice condition and f(a, ·) is monotoneincreasing function(m.i.f.).Thus,to show that, we only need to show that τ is a m.i.f.This is true as for x′, y′ ∈ Sn−1, x′ ≤ y′, ρ satisfies the lattice condition.

ρ(b, y′)ρ(a, x′)− ρ(a, y′)ρ(b, x′)

0.4.1.1 Application to Ising model

Let σ, σ′ be two configuration, τ = σ ∧ σ′, η = σ ∨ σ′.We want to show that ∑

<i,j>

σiσj +∑<i,j>

σ′iσ′j ≤

∑<i,j>

τiτj +∑<i,j>

ηiηj

This is true as σiσj + σ′iσ′j ≤ τiτj + ηiηj , for each i, j , because of a simple fact that given a1, · · · , an and

b1, · · · , bn,∑i aibπ(1) maximized when bπ(1), · · · , bπ(n) in the same order as a1, · · · , an

0.4.2 Griffin’s inequality

Let µ1, ·, µn be probability measure on R, which are symmetric around 0.For each subset A ⊂ 1, · · ·n, let JA be a non-neg real numberDefine a prob measure γ on Rn

dγ(η) =1

Zexp

(∑A

JAηA

)Πdµi(ηi)

where ηA = Πi∈Aηi.

Theorem 7 Let ¡¿ denote expectation w.r.t. γ.Then,∀A, < ηA >≥ 0,∀A,B,< ηAηB >≥< ηA >< ηB >

Now, we begin setting up notations and lemmas for proving the inequality.For any i and any non-neg integer m ∫

ξmi dµi(ξi) ≥ 0 (0.24)

Then

Given ξ, χ ∈ Rn, let

q =ξ − χt =ξ + χ


Lemma 8 ∀i, ∀a, b non-neg integer ∫qai t

bidµ(ξi)dµ(χi) ≥ 0

Proof: By the symmetry of µi, the integral remains the same if we replace ξi → −χi, χi → −ξi Then, theintegral becomes ∫

(−qi)atbidµ(ξi)dµ(χi)

Thus, if a odd, then the integral becomes 0. Similarly, if b is odd, then the integral is also 0. If a, b botheven, then clearly the integral is nonnegative.

Now, we state another lemma.

Lemma 9 For any A, ξA + χA and ξA − χA are both polynomials in q, t with non-neg coefficients.

Proof:

ξA ± χA = 2−|A|((q + t)A ± (q − t)A

)To show that this polynomial in (q, t) has non-neg coefficients, observe that a typical term in (q+ t)A is hasa corresponding term in (q − t)AThe two terms either equal(when |A − B| is even) and contribute to a 2 or cancel out in adding ξA + χA,and the reverse happens when doing ξA − χA.Hence, all the coefficients are non-neg. Now, let ξ, χ be independent copies from γ. Then ∀A,B

< ξAξB > − < ξA >< ξB >=< ξA(ξB − χB) >=< a polynomial in (q, t) with non-neg coefficients >

Now, if we can prove the following lemma, we will finish the proof for the inequality.

Lemma 10 Let P be the set of polynomials in (q, t) with non-neg coefficients.Then , for all f ∈ P , < f >≥ 0

Proof:

< f >=1

Z2

∫f exp(

∑JA)(ξA + χA)dµ(ξ)dµ(χ)

=1

Z2

∞∑k=0

1

k!

∫f(∑

JA(ξA + χA))k

Πdµi(ξi)Πdµi(χi)

≥0 , by lemma 8

0.4.3 Examples

We’ll give some examples of the application of these two inequalities.

Lecture 7/8: April 24 & 26 0-5

0.4.3.1 FKG

Clearly, σ → σi is a monotone increasing function for all iThen we have that < σiσj >≥< σi >< σj > in the Ising model. In particular, for free or periodic boundary,< σiσj >≥ 0

0.4.3.2 Application of Griffin’s inequality

Consider the Ising model on Λ ⊂ Zdwith free bdry.Suppose that we add one vertex v to Λ to get Λ′ with prob density proportional to exp(

∑σiσj)

Take any A ⊂ Λ s.t. no member of A is a neighbour of v.

∂

∂J< σA >= < σA

∑j∈N(v)

σiσj

> − < σA ><

∑j∈N(v)

σiσj

>

≥0

When J = 0, < σA > is the expectation of σA in the Ising model on ΛJ = β, < σA > is the expectation of σA in the Ising model on Λ′.This means that the new expectation ≥ old expectation wherever you added the new spin.But 0 ≤< σA >≤ 1, so if we have a sequence Λn increase to Zd, < σA > must increase to a limit, and isthis limit independent of the choice of the sequence.

Lastly, observe that expectation like < σA > completely determines the finite dim distribution of any prob

measure on −1, 1Zd with the product σ−algebra

0.4.3.3 Ising model with an external field

By Griffin’s inequality, this model has an infinite volume limit.

Definition 11 Expected magnetizition

E[

1

|Λ|

]= m(β, h,Λ)

We want to know What is m(β, h) = limΛ→Zd m(β, h,Λ). If h = 0, then this is 0.

We will approach Zd through Λn = [−n, n]d ∩ ZdLet mn = m(β, h,Λn)

Theorem 12 limmn exists for almost every h and is non-decreasing in h

Theorem 13 If β is large enough, then m(β, h) has a jump discontinuous at 0.

Theorem 14 Lee-yang theorem for magnetizationOn (0,∞) and (−∞, 0), m(β, h) is analytic of h, for any β, and hence no other phase transition.

Let Zn(h) = Z(β, h,Λn), with β fixed, and Fn(h) = 1|Λn| log zn(h) be the free energy.


Lemma 15 ∀h ∈ R,F (h) = limFn(h) exists

Simple calculation gives

F ′n(h) = E(magnetization)Fn”(h) = |Λn|V ar(magnetization)

Hence Fn is convex and F is also convex of h

Fact a convex function is a.e. differentiable

Then the limit exists and equals F ′(h).


Lecture 9: May 1Lecturer: Sourav Chatterjee Scriber: Jun Yan


Today we’ll prove Claim 1 and Theorem 2 below. First let’s recall where we were last time. Write Λn =[−n, n]d ∩Zd. Consider

Zn =∑

σ∈−1,1nexp

(β∑

σjσk + h∑

σj

).

Fix β, then we can write Zn as a function of h : Zn(h). Let Fn = 1|Λn| logZn. We showed that ∀h,

limn→∞ Fn(h) exists.

Call the limit as F (h). Then <magnetization>= F ′(h). Fn is convex=⇒ F ′n is increasing in h, =⇒ limF ′n(h)exists for a.e. h (where F is differentiable). We’ll prove the following claim.

Claim 16 F ′ has a jump discontinuity at 0 if β is large.

Proof: Take any h > 0. Let hn = α|Λn| . By the monontonicity of F ′n, F ′n(h) ≥ F ′n(hn) if n is large. Write

m = m(σ) =∑σj

|Λn| . Then

F ′n(hn) = 〈m〉hn =〈meαm〉0〈eαm〉0

=〈m〉0 + α

⟨m2⟩

0+O(α2)

1 + α 〈m〉0 +O(α2),

where O(α2) is a quantity bounded by Cα2 for a universal constant C. Notice that 〈m〉0 = 0, thus if β islarge

〈m〉hn =α⟨m2⟩

0+O(α2)

1 +O(α2)=

1

|Λn|2∑〈σj , σk〉 ≥ C(β) > 0

by taking a small enough α. Thus F ′(h) = limF ′n(h) ≥ C(β) > 0. Similarly, F ′(−h) ≤ −C(β) < 0, whichimplies the jump discontinuity.

Next we prove the following Lee-Yang Theorem.

Theorem 17 (Lee-Yang) For any fixed β, F is analytic on (0,∞) and (−∞, 0). (meaning that F hasconvergent power series on the neighborhood of every points.)

Proof: Fix β, and take any complex h. Define Zn(h) as before. Let Ω = z ∈ C :Re (z) ≥ |Im(z)|. Leeand Yang proved that all zeros of Zn lie on the imaginary axis. We’ll prove that Zn has no zeros in Ω, whichis enough for our purpose.

Note the following fact: If f is an analytic function on a simply connected domain Ω and f has no zeros inΩ, then ∃ analytic g on Ω s.t. eg = f .

Thus, on our Ω, Fn(h) = 1|Λn| logZn is defined. Moreover, since Zn is positive on (0,∞), logZn must have

imaginary part = integer times 2π on (0,∞), =⇒ the integer must be same throughout. Thus we can choosea version of logZn such that it agrees with our usual logZn on (0,∞).

9-1

9-2 Lecture 9: May 1

Consequence: Fn has an analytic extension to Ω. Let Gn := |Zn|1|Λn| , then because

|Zn| =∣∣∣∑ exp(β

∑σjσk + h

∑σj)∣∣∣

≤∑

exp(β∑

σjσk + Re(h)∑

σj) = Zn(Re h),

we see that |Gn(h)| ≤ exp(Fn(Re h)). Note that:

(1) Gn converges pointwisely on (0,∞).

(2) Gn is uniformly bounded on compact subsets of Ω (with respect to n).

By (1), (2) and the Vitali’s convergence theorem for analytic function, we see that G = limGn exists on Ωand is analytic. We’ll prove that:

(3) ∀h ∈ Ω, |Zn| ≥ Zn(Re h− |Im h|).If (3) is true, ∀h ∈ Ω we have |Gn(h)| ≥ G(Re h− |Im h|) > 0, and thus G has no zeroes in Ω, =⇒ it has ananalytic logarithm. Thus logG can be made to agree with F on (0,∞), =⇒ F has a continuous extensionto Ω.

Proof of (3): Fix n. Note that

|Zn(h)|2 =∑σ,τ

exp(β∑

(σjσk + τjτk) + h∑

σj + h∑

τj

).

Let zj = (σj + τj)/2 + (σj − τj)/2, then |zj | = 1. After some algebra we have Re(zjzk) = (σjσk + τjτk) /2,and thus

σjσk + τjτk = zjzk + zkzj .

Similarly we can verify that

hσj + hτj = (Re h+ Im h) zj + (Re h− Im h) zj .

Hence

|Zn(h)|2 =∑σ,τ

exp(β∑

(zjzk + zkzj) +∑

(Re h+ Im h) zj +∑

(Re h− Im h) zj

).

If h ∈ Ω, then the expression in the bracket is a polynomial in (zj , zj) with non-negative coeffcients. =⇒exp(...) is a convergent power series in (zj , zj) with non-negative coefficients. Moreover, the coefficients aremonotone increasing function of (Re h+Im h,Re h−Im h). It is easy to see that for any n ∈ Z,

∑τj ,σj∈−1,1

znj =

0, if n is odd0, if 2 | n, 4 - n4, if 4 | n

.

Note that |zj | = 1 implies that zj = z−1j , =⇒∀n,m ∈ Z,

∑τj ,σj∈−1,1 z

mj z

nj ≥ 0. Thus,

∑σ,τ (any monomial in (zj , zj)j∈Λn) ≥

0. This implies that |Zn(h)|2 is a sum of non-negative terms, each of which is a non-negative monotone in-creasing function of (Re h+Im h,Re h−Im h). Let h′ =Re h− |Im h|. Then

(Re h′ + Im h′,Re h′ − Im h′) ≤ (Re h+ Im h,Re h− Im h),

and it implies that |Zn(h)|2 ≥ |Zn(h′)|2 = Zn(Re h− |Im h|)2.


Lecture 10: May 3Lecturer: Sourav Chatterjee Scriber: Jun Yan


10.5 Lattice Gauge Theory

In this lecture we introduce the lattice gauge theory.

Motivation: First we do some comparisons between classical particle and field. We denote a path as q(t),0 ≤ t ≤ T , q(0) = a, q′(0) = v. Let S be an action on the path q, such that

S(q) =

∫ T

0

(1

2q′(t)2 − V (q(t)))dt.

The minimizer of S(q) gives Newton’s equation: q′′(t) = −V ′(q(t)). This is the classical particle case. Forfield, we can do the same thing, that is, we can calculate the minimizer of some function acting on fields.Quantization in the particle case: consider the measure exp(−S(q))D(q), where D(q) =”Lebesgue measure

on path space”. This is actually Wiener measure weighted by exp(−∫ T

0V (q(t))dt). We want to put measure

on fields, too. However, the building of the measure is not developed well, so people took a look at thediscrete one, whih is the lattice gauge theory.

Yang Mills Theory: Components of the standard model of Quantum Mechanics, not mathematicallywell-defined.

Lattice Gauge Theory: Discrete version of the Yang Mills Theory, mathematically well defined.

Definition: Write d-dimension, G-compact Liegroup (SU(N), SO(n), U(n), O(n)). In physics, SO(3) corre-sponds to strong force, and SO(2) corresponds to weak force (no one knows why). Denote ρ as a represen-tation of G, which means that ρ is a group homomorphism from G to GL(V ), where V is a complex vectorspace, and GL(V ) is the inversible linear maps on V . We further require ρ to be irreducible, meaning thatV doesn’t have non-trivial invariant subspace for ρ. Recall the following fact:

Fact: G is a compact Lie group =⇒ V must be finite dimensional.

Write χρ : χρ(g) =Tr(ρ(g)) for g ∈ G, which is the chatacter of ρ. Take Λ ∈ Zd, denote by E(Λ) =edge setof Λ (undirected). Denote the space of configurations as GE(Λ). Each edge has a positively and a negativelyoriented version. If g = (g(e))e∈E(Λ) is a configuration, then we follow the convention that gxy = g−1

yx = g(e),if e is the edge < x, y > and xy is the positive orientation. We define the Plaquatte p as a unit square, withits four edges e1 → e2 → e3 → e4. It is always a two-dim object, even though the space might be in higherdimension. Write gp = ge1ge2ge3ge4 . The lattice gauge theory defines a probability measure on the space ofconfigurations, with density proportional to exp(β

∑p Re(χ(gp))), with respect to the Haar measure on G.

The Yang Mills continuity problem is that: this measure has a continuous version after some scaling.

Write β = 1g20, where g0 =coupling strenth. Then β small corresponds to strong coupling regime, β large

corresponds to weak coupling regime.

YM mass gap problem (discrete version): Prove exp decay of correlation at all large β in d = 4, G =SU(3), ρ =standard representation.

10-1


Wilson loop: Take any loop l = e1e2...en in Zd, define the wilson loop variable wl = Re(χ(g(e1)...g(en))).Here is the Wilson conjecture: In d = 4, G-non-abelian, we should have |〈wl〉| ≤ exp(−c(β)area(l)), wherearea(l) =minimum surface area enclosed by l.

This is supposed to explian confinement of quarks. Since suppose that two quarks are at distance R of eachother, let l be a rectangle with edge length R, T,R, T . Then Wilson gave a phisics argument to show thatthe potential V (R) between the quarks can be obtained as limT→∞− 1

T log < wl >. So if the area law holds,then V (R) ≥ cR, which confines quarks. The area law for small β for any dimensions and G = SU(n) orU(1), l =retangles is proved. We’ll do a probability proof of this one.

Remark: However this result is not useful for the general area law, since this method works for the abeliangroup, but the general one shouldn’t hold for the abelian group.


Lecture 11/12: May 8 & 10Lecturer: Sourav Chatterjee Scriber: Mark Perlman


0.6 Lattice Gauge Theory

We’re on Zd for d ≥ 2. G is one of the following groups: SU(n), U(n), O(n) or SO(n) for n ≥ 1, and ρ isthe standard representation. We let Λ ⊂ Zd, and denote by E(Λ) the set of positively oriented edges. Thenthe set of configurations is GE(Λ).

For a parameter β, p.m. has density with respect to Haar measure given by:

1

2exp

(β∑p

Re(Tr(gp))

),

where gp = ge1ge2ge3ge4 for a square with edges e1, e2, e3 and e4:

Denote by l a loop in Zd, that is, l is a sequence of edged e1e2 · · · em forming a closed path. Definewl = Re(Tr(ge1 · · · gem)).

Theorem 18 If β is small enough, depending on d, then

|〈wl〉| ≤ e−c(β,d)area(l).

We will only prove this when l is a rectangle. Note that the bound is independent of Λ as long as Λ is largeenough. This theorem tells us that the infinite volume limit exists by similar compactness arguments as inthe Ising model.

It is conjectured that the infinite volume limit is unique for all β; this has only been proved for small β.

Exercise 1 Establish exponential decay of correlations at small β. Using this, prove the uniqueness of theinfinite volume limit.

Definition 19 Take any infinite volume limit. The quark-antiquark tension at separation R is defined as

V (R) = limT→∞

− 1

Tlog〈wl〉

0-1

0-2 Lecture 11/12: May 8 & 10

assuming the limit exists, where l is an R× T rectangle:

A model is said to exhibit quark confinement if limR→∞ V (R) =∞.

Proposition 20 If the area law holds, then so does quark confinement.

Proof: d = 2, layers are independent after conditioning on the horizontal edges:

If d = 3, pick one of the 3 directions of the edges and condition on edges in the other two directions, toobtain 2d layers after conditioning. In general d-dimension, pick one cardinal direction, and condition onedges not in that direction. That is, condition on edges that are not of the form

((x1, ..., xi, ..., xd), (x1, ..., xi + 1, ..., xd))

for a single fixed i. After conditioning, the system consists of independent (d − 1)-dimensional layers, andeach layer is like a spin system in Zd−1 in terms of neighbor interaction.

Fix any layer, and consider the edges as vertices in Zd−1. The conditional probability density of a configu-ration within a layer is proportional to

exp

β∑〈ef〉

Re(Tr(geAefgfBef ))

,

where the matrices Aef and Bef are deterministic and defined through the conditioning. This density hasg → −g symmetry, and therefore the conditional expectation of ge is zero for every edge.

Moreover, by a coupling argument as before, we get exponential decay of correlations at small β in thisconditional law. The exponent does not depend on the conditioning, because for small β, the law is uniformlyclose to Haar measure. A and B are elements of a compact group and are therefore bounded, so if e, f areedges at distance 2m, then for any two elements gije and gklf ,

|〈gije gklf 〉cond| ≤ e−C(β−d)m,

where here 〈·〉cond denotes the conditional expectation. Denoting the edges of the rectangle l as

Lecture 11/12: May 8 & 10 0-3

we have

wl = Re(Tr(UAge1 . . . geRUBg∗fR . . . g

∗f1

)).

We will show that, for any conditioning:

|〈wl〉cond| ≤ e−CTR,again, the above denoting conditional expectation. This implies the area law. We the trace inside wl as

Tr(. . . ) =∑

i1,...,i2R+2

U i1i2A gi2i3e1 . . . giR+1iR+2eR U

iR+2iR+3

B ¯gfRiR+4iR+3 . . . gg1

i1i2R+2 .

Then taking conditional expectation,

〈Tr(. . . )〉cond =∑

U i1i2A UiR+2iR+3

B 〈gi2i3e1 gf1

i1i2R+2〉cond · · · 〈. . . 〉cond.

And therefore each summand is bounded by (e−C(β)T )R in absolute value. The number of summands isbounded by eCR, so if we choose β small enough that C(β) > C, we obtain the desired inequality.

0.7 Gaussian Free Field

Definition 21 Let G = (V,E) be a finite undirected graph with no self groups, which we’ll assume isconnected. The Gaussian free field on G is a Gaussian field (φv)v∈V with density proportional to

ρ = exp

−1

2

∑(u,v)∈E

(φu − φv)2

This definition immediately runs into the problem that

∫ρ = ∞ because of translation invariance. There

are two ways to solve this problem:

1. Designate a subset ∂V ⊂ V as the “boundary” of V and fix the value of φ on ∂V (e.g. zero). If∂V = φ, the density is integrable.

2. Put a “mass” term in the Hamiltonian:

exp

−1

2

∑〈u,v〉∈E

(φu − φv)2 − m2

2

∑u

φ2u

.

This is referred to as a massive free field.

0-4 Lecture 11/12: May 8 & 10

Example 1 Let V = 0, . . . , n−1d with the natural boundary ∂V = x ∈ V : some coordinate of x is 0 or n−1.

Example 2 Let V = 0, . . . , n − 1 with ∂V = 0, φ0 = 0, then the Gaussian free field is a random walkwith Gaussian increments.

Example 3 Let V = 0, . . . , n−1 with ∂V = 0, n−1 and a zero boundary condition. Then the Gaussianfree field is a BB evaluated at 0, . . . , n− 1.

Remark 22 From here on we’ll assume a zero boundary condition on ∂V .

Definition 23 Given a graph G, the discrete Laplacian of a function f : G→ R is given by

∆f(u) =∑

v∈N(u)

f(v)− f(u),

where N(u) is the neighborhood of u.

Proposition 24 ∆ is given by a strictly negative definite matrix.

Proof: We calculate∑〈u,v〉

(f(u)− f(v))2 =∑〈u,v〉

f(u)(f(u)− f(v))−∑〈u,v〉

f(v)(f(u)− f(v))

=1

2

∑u

∑w∈N(u)

f(u)(f(u)− f(w))− 1

2

∑v

∑w∈N(v)

f(v)(f(w)− f(v))

= −1

2

∑u

f(u)∆f(u)− 1

2

∑v

f(v)∆f(v)

= −∑u

f(u)∆f(u).

Corollary 25 The covariance matrix of the Gaussian free field is (−∆)−1.

0.7.1 Random Walk Representation of Covariance

Discrete Green’s functions are typically complicated objects, but become more manageable when viewed inthe following random walk representation:

Attach to each edge of G a Poisson clock that rings at rate 1; that is, every edge has a clock that rings aftera random time distributed as an exponential distribution with rate 1, and then starts another exponentialclock. The clocks on each edge are assumed independent of each other. Pick a vertex u ∈ int(V ), whereint(V ) = V \ ∂V . Start a random walk at u which remains at u until an edge attached to u rings, at whichpoint cross that edge. Continue doing this until the random walk hits ∂V , at which point the walk stops.

Recall that a stopping time τ is a random time such that the event τ ≤ t depends only on what hashappened up to time t. Recall the memoryless property of Poisson processes: for any stopping time τ , theprocess on [τ,∞) is the same as another Poisson process with the same rate. Thus in the above randomwalk, when the process jumps across an edge (i.e. we pick τ to be the first time the process jumps across anedge), all the clocks “reset.” That is to say, this random walk is a Markov process.

Lecture 11/12: May 8 & 10 0-5

Theorem 26 Denote G(u, v) = Cov(φu, φv) = E(φuφv). For each v ∈ V ,

G(u, v) = Eu(time spent at v),

where Eu denotes expectation conditional on the random walk starting at u.

Proof: Fix v. Define f(u) = Eu(time spent at v), and let g(u) = E(φuφv). We want to prove that f = g.We already know that f = g = 0 on ∂V .

Let u ∈ int(V ), and assume first that u 6= v. We condition on the first jump. The fact that the walk isequally likely to jump to any neighbor of u, combined with the Markov property, tell us that

f(u) =1

|N(u)|∑

w∈N(u)

f(w),

and therefore ∆f(u) = 0.

Now assume u = v. Then we again condition on the first jump, but now once we jump away from v, we’vespent some time at v. Thus,

f(v) =1

|N(u)|∑

w∈N(v)

f(w) + E(time until first jump from v).

The time until the first jump is the minimum of |N(v)| i.i.d. exponential random variables (the clocks onevery edge touching v), which is equal in distribution to an exponential with parameter |N(v)|. Thereforethe expected time until the first jump is 1

|N(v)| :

f(v) =1

|N(u)|∑

w∈N(v)

f(w) +1

|N(v)| .

Therefore ∆f(v) = −1.

Now we prove that g satisfies the same Laplacian equations. Assume first that u 6= v. From the density ofthe Gaussian free field, we see that given (φw)w 6=u, the density of φu is proportional to

exp

−1

2

∑w∈N(u)

(φu − φw)2

= exp

−1

2|N(u)|φ2

u +∑

w∈N(u)

φuφw

= exp

−|N(u)|2

φu − 1

|N(u)|∑

w∈N(u)

φw

2 .

Therefore, the conditional law of φu is normal with parameters

N

1

|N(u)|∑

w∈N(w)

φw,1

|N(u)|

,

0-6 Lecture 11/12: May 8 & 10

and thus we calculate

g(u) = E(φuφv)

= E[E(φu | (φw)w 6=u)φv]

= E

1

|N(u)|∑

w∈N(u)

φwφv

=

1

|N(u)|∑

w∈N(u)

g(w).

Therefore ∆g(u) = 0. On the other hand, when u = v,

g(v) = E(φ2v)

= E[E(φ2v | (φw)w 6=v)]

= E[Var(φv | (φw)w 6=v)] + E[E(φv | (φw)w 6=v)2]

=1

|N(v)| + E[φvE(φv | (φw)w 6=v)]

=1

|N(v)| + E

φv 1

|N(v)|∑

w∈N(v)

φw

=

1

|N(v)| +1

|N(v)|∑

w∈N(v)

g(w).

Therefore ∆g(v) = −1.

If we let h = f − g, then ∆h = 0 on int(V ), and h = 0 on ∂V . Since ∆ is strictly negative definite, thisimplies that h = 0, i.e. f = g.

Corollary 27 Since G(u, v) is symmetric in u and v, we have Eu(time spent at v) = Ev(time spent at u).

Example 4 In d = 2, let V = 0, . . . , n − 12 with the natural boundary and zero boundary condition.Consider a typical interior vertex u. Then

Var(φu) = G(u, u) = Eu(time spent at u before hitting the boundary).

Up to a constant, the random walk described above moves approximately as fast as a simple random walk. IfXt is a simple random walk on Z2, then Pu(Xt = u) ∼ constant

t+1 . So if T is some fixed large time, then

Eu(time spent at u before T ) =

∫ T

0

Pu(Xt = u)dt ∼ C log T.

Let τ be the time to hit the boundary, so τ is approximately the time for a simple random walk to attain amaximum or minimum of n/2, which takes about n2 steps.

Exercise 2 Show that G(u, v) ∼ C log n.

Exercise 3 We have E(φu − φv)2 = G(u, u) + G(v, v) − 2G(u, v). Using the random walk representation,show that this quantity is ∼ C log(|u− v|+ 1) when |u− v| n.

Lecture 11/12: May 8 & 10 0-7

0.7.2 Continuum Limit of Gaussian Free Field

Take V = 0, . . . , n − 12, scaled by 1n so V ⊂ [0, 1]2. Define a Gaussian free field as before on the scaled

grid, where we don’t scale the field (as compared to the d = 1 case, where we scaled the field along with thegrid). Extend this to a field on [0, 1]2 in some sensible way. Call this field φ, still depending on n. Let f bea smooth function with support in (0, 1)2. Then∫

fφ ≈ 1

n2

∑u∈grid

f(u)φu

is a normal random variable with mean 0 and variance

∼ 1

n4

∑u,v

f(u)f(v)Gn(u, v),

where Gn is the Green’s function of the Gaussian free field on the scaled grid (i.e. the covariance matrix).

Fact 28 If un, vn are two sequences of grid points approaching two distinct points x, y ∈ (0, 1)2, then

Gn(un, vn)→ G(x, y),

where G(x, y) is the inverse of the Dirichlet Laplacian on [0, 1]2, and is ∼ C log |x− y| for |x− y| 1.

Remark 29 The continuum Gaussian free field is a random distribution, not a function. However, we canstill see the covariance matrix; the only problem is that it’s infinite on the diagonal.


Lecture 13: May 15Lecturer: Sourav Chatterjee Scriber: Sicong (Scott) Zhang


Gaussian Free Field with zero boundary condition

Let G = (V,E), ∂V ⊆ V be connected undirected graph. Probability density is given by

exp

−1

2

∑〈u,v〉

(φu − φv)2

,

where φu = 0 for u ∈ ∂V . Connected graph implies that it is integrable and gives a probability measure.

Random walk representation is given by:

〈φu, φv〉 = Eu(time spent at v before hitting ∂V )

Massive Free Field

Another way to get a probability measure is a massive free field:

exp

−1

2

∑〈u,v〉

(φu − φv)2 − m2

2

∑u

φ2u

.

Its random walk representation is:

〈φu, φv〉 = Eu(time spent at v)

but it dies after exp(m2) random time.

Exercise: prove this.

Question from class: why is the mass “m2”? Answer: This comes from the potential of simple harmonicoscilator in QFT.

13-1


Continue limits in 2D

Free field is the simplest Euclidean QFT, rigorously defined. Formally, it gives a probability measure on(generalized) functions on R2 with probability density proportional to

exp

(−1

2

∫R2

|∇φ|2 − m2

2

∫R2

φ2u

)with respect to a “Lebesgue measure” on function space.

Start with GFF on εZ2, which is a correct discretization:

1

2

∫R2

|∇φ|2 − m2

2

∫R2

φ2u →

1

2ε∑〈u,v〉

(φu − φv

ε)2 +

m2ε2

2

∑u

φ2u

=1

2ε

∑〈u,v〉inεZ2

(φu − φv)2 +m2ε2

2

∑u

φ2u.

Define this for finite boxes in εZ2, and then take the infinite volume limit.

Why do we have uniqueness of the limit? You can see this by the random walk representation: random walkstarting at point u will (with high probability) not “see” the boundary of the box before it dies, if the boxis very big.

Therefore, 〈φu, φv〉 tends to a limit as boxes in εZ2.

Since Gaussian distributions are determined by covariances, the infinite volume limit is well-defined.

Now take u, v ∈ R2, u 6= v, and let uε, vε be the closest points to u, v in εZ2. Let Cε(uε, vε) = 〈φuε , φvε〉 inεZ2. Then because transition kernel

pεt(x, y) = px(random walk at y at time t),

Cε(uε, vε) =

∫ ∞0

pεt(uε, vε)P (t < τε)dt

=

∫ ∞0

pεt(uε, vε)e−m2ε2tdt

=

∫ ∞0

1

ε2pεs/ε2(uε, vε)e

−m2sds (change of variable: s = ε2t)

As ε→ 0, pt(u, v) = 14πte

− |u−v|2

4t , so

Cε(uε, vε)→ C(u, v) =

∫ ∞0

1

4πse|u−v|

2/4se−m2sds.

Let φε be a field on εZ2, extended to R2 in some reasonable way. Let f be a smooth function with exponentialdecay at infinity in R2.∫

fφε ∼ ε2∑u∈εZ2

f(u)φεu ∼ N(

0, ε4∑u∈εZ2

f(u)f(v)Cε(u, v)

)−−−→ε→0

∫R

∫Rf(x)f(y)C(x, y)dxdy

Lecture 13: May 15 13-3

The integral is finite because

C(x, y) ∼ − 1

2πlog(m|x− y|)

when |x− y| → 0.

Minlos’s theorem

Let S(R2) be the space of Schwartz functions, i.e. smooth functions f such that f and any of its higherderivatives decay faster then any polynomial at infinity. Namely, let Z+ be non-negative integers, α =

(α1, α2) ∈ Z2+, define multi-index |α| = α1 + α2, with Dα = ∂|α|

∂xα11 ∂x

α22

, x = (x1, x2) ∈ R2, xα = xα11 xα2

2 ,

pα,β(f) = supx∈R2 |xαDβf(x)|.Define a metric on C∞(R2),

d(f, g) =∑α,β

2−|α|−|β|pα,β(f − g)

1− pα,β(f − g),

then S(R2) is the complete metric space under this metric. It is a topological vector space.

S ′(R2)= the continuous dual of S(R2), the space of tempered distributions.

S(R2) → S ′(R2) is given by φ 7→ 〈φ, f〉 on S ′(R2), where f ∈ S(R2).

The σ-algebra generated by these maps is called the cylinder σ-algebra on S ′(R2).

Given any probability measure on the cylinder σ-algebra, define its “character function”

S(f) =

∫ei〈φ,f〉dv(φ).

Theorem 30 (Minlos’s theorem) : function S : S(R2) → C is a character for a probability measure onthe σ-algebra of S ′(R2) iff S(0) = 1, positive semi-definite, and S is continuous.

Proof: [Forward direction only] Let Sε be a character function of φε. ∀f, 〈φ, f〉 converges in distribution

⇒ E(ei〈φ,f〉) convergers to a limit

⇒ S(f) = limε→0 Sε(f) exists for all f .

then S(0) = 1, S is positive semi-definite, because each Sε is.

|Sε(f)− Sε(g)| = |E(ei〈φε,f〉 − ei〈φε,g〉)|

≤ E(|〈φε, f − g〉|)≤√

Var(〈φε, f − g〉)

−−−→ε→0

√∫∫(f(x)− g(x)) (f(y)− g(y))C(x, y)dxdy

⇒ Sεε>0 is uniformly equicontinuous, so by Arzela–Ascoli theorem, S is continuous.


Physics Motivation

Quantum mechanics for single particle in 1D

let q be location, t be time.

A state is ψ(q, t) ∈ L2(R) with ||ψ||L2 = 1. |ψ(q, t)|2 is the probability density at q at time t.

Schrodinger’s equation:

i∂ψ

∂t= −1

2

∂2ψ

∂q2+ V (q)ψ(q, t) = Hψ

where V (q) is the potential function, and H = − 12∂2

∂q2 + V is the energy (Hamiltonion) function.

Solution to Schrodinger’s equation:ψ(·, t) = e−itHψ(·, 0)

where e−itH is unitary because H is symmetric (Hermetian?)

Feynman’s path integral solution: “integrate over function space”.

ψ(q, t) =

∫Kt(q, q

′)ψ(q′, 0)dq′

where the Schrodinger kernel

Kt(q0, q1) =

∫exp

[i

(1

2

∫ T

0

q(t)2dt−∫ T

0

V (q(t))dt

)]Dq

where Dq is the “Lebesgue measure” on paths (might not exist rigorously.) The integral is over all pathssuch that q(0) = q0, q(T ) = q1.

Mark Kac developed the real version of path integral solution, for the heat equation

∂ψ

∂t=

1

2

∂2ψ

∂q2− V ψ = −Hψ

Solutions areψ(·, t) = e−tHψ(·, 0)

Kac proved rigorously that indeed

ψ(q, t) =

∫Kt(q, q

′)ψ(q′, 0)dq′

where

Kt(q0, q1) =

∫exp

(−1

2

∫ T

0

q(t)2dt−∫ T

0

V (q(t))dt

)Dq.

Although Dq is still undefined, exp(− 12

∫ T0

(q)2dt)Dq is well-defined: it is the probability law of Brownianmotion.

Take any path q(t), 0 ≤ t ≤ τ , imagine q has analytic continuation to the imaginary axis. Wick rotationtrick: q1(t) := q(it).

i

∫ T

0

[1

2q(t)2 − V (q(t))]dt =

∫ T

0

[1

2q(it)2 − V (q(it))]dt


2 particle in 1D

Let ψ(q1, q2, t) be the quantum state of 2 particles in 1D. Schrodinger equation becomes

i∂ψ

∂t= −1

2∂2q1ψ −

1

2∂2q2ψ + V (q1, q2)ψ(q1, q2, t)

For example, V (q1, q2) can be q21 + q2

2 + (q1 − q2)2.

Feynman representation of the Schrodinger kernel is similarly∫exp

[i

(1

2

∫ T

0

q1(t)2dt+1

2

∫ T

0

q2(t)2dt−∫ T

0

V (q(t))dt

)]Dq

Consider n particles in 1D, each by itself will behave like a simple harmonic oscilator, but there is some weakinteraction between any pair, i.e. V (q1, q2, · · · , qn) =

∑q2i + ε

∑(qi − qi+1)2.

QFT deals with infinitely many particles, one at each x ∈ R. (We are considering 1 space dimension and 1time dimension.) Then q1(t), q2(t), · · · becomes a function q(x, t).

q = qt : R→ R, V (q) =∫q(x)2dx+

∫(∂xq)

2dx or V (q) =∫U(q(x))dx+

∫(∂xq)

2dx

It is impossible to write down the Schrodinger PDE. However, Feynman integral can still be written down:∫exp

[i

(1

2

∫R

∫ T

0

(∂tq)2dtdx−

∫R

∫ T

0

U(q(x, t))dtdx− 1

2

∫R

∫ T

0

(∂xq)2dtdx

)]Dq

integral is over the space of surfaces. (NOTE: where did the last 1/2 come from?)

Wick rotation, t→ it gives∫exp

(−1

2

∫R

∫ T

0

(∂tq)2dtdx−

∫R

∫ T

0

U(q(x, t))dtdx− 1

2

∫R

∫ T

0

(∂xq)2dtdx

)Dq

which is exactly the 2D Gaussian free field.

Constructible QFT

Arthur Wightman gave a list of criteria that allow you to do the analytic continuation rigorously, and builtan actual QFT. Simplification was given by Osterwalder and Schrader.

Yang-Mill theory is the limit of lattice gauge theory. But QFT is hard to construct. Many of them areGaussian, i.e. “trivial”. It has connection with random surface theory, and Liouville quantum gravity.


Lecture 14: May 17Lecturer: Sourav Chatterjee Scriber: Sicong (Scott) Zhang


Physics Motivation

Quantum mechanics for single particle in 1D

let q be location, t be time.

A state is ψ(q, t) ∈ L2(R) with ||ψ||L2 = 1. |ψ(q, t)|2 is the probability density at q at time t.

Schrodinger’s equation:

i∂ψ

∂t= −1

2

∂2ψ

∂q2+ V (q)ψ(q, t) = Hψ

where V (q) is the potential function, and H = − 12∂2

∂q2 + V is the energy (Hamiltonion) function.

Solution to Schrodinger’s equation:ψ(·, t) = e−itHψ(·, 0)

where e−itH is unitary because H is symmetric (Hermetian?)

Feynman’s path integral solution: “integrate over function space”.

ψ(q, t) =

∫Kt(q, q

′)ψ(q′, 0)dq′

where the Schrodinger kernel

Kt(q0, q1) =

∫exp

[i

(1

2

∫ T

0

q(t)2dt−∫ T

0

V (q(t))dt

)]Dq

where Dq is the “Lebesgue measure” on paths (might not exist rigorously.) The integral is over all pathssuch that q(0) = q0, q(T ) = q1.

Mark Kac developed the real version of path integral solution, for the heat equation

∂ψ

∂t=

1

2

∂2ψ

∂q2− V ψ = −Hψ

Solutions areψ(·, t) = e−tHψ(·, 0)

Kac proved rigorously that indeed

ψ(q, t) =

∫Kt(q, q

′)ψ(q′, 0)dq′

14-1


where

Kt(q0, q1) =

∫exp

(−1

2

∫ T

0

q(t)2dt−∫ T

0

V (q(t))dt

)Dq.

Although Dq is still undefined, exp(− 12

∫ T0

(q)2dt)Dq is well-defined: it is the probability law of Brownianmotion.

Take any path q(t), 0 ≤ t ≤ τ , imagine q has analytic continuation to the imaginary axis. Wick rotationtrick: q1(t) := q(it).

i

∫ T

0

[1

2q(t)2 − V (q(t))]dt =

∫ T

0

[1

2q(it)2 − V (q(it))]dt

Finitely many particle in 1D

Let ψ(q1, q2, t) be the quantum state of 2 particles in 1D. Schrodinger equation becomes

i∂ψ

∂t= −1

2∂2q1ψ −

1

2∂2q2ψ + V (q1, q2)ψ(q1, q2, t)

For example, V (q1, q2) can be q21 + q2

2 + (q1 − q2)2.

Feynman representation of the Schrodinger kernel is similarly∫exp

[i

(1

2

∫ T

0

q1(t)2dt+1

2

∫ T

0

q2(t)2dt−∫ T

0

V (q(t))dt

)]Dq

Consider n particles in 1D, each by itself will behave like a simple harmonic oscilator, but there is some weakinteraction between any pair, i.e. V (q1, q2, · · · , qn) =

∑q2i + ε

∑(qi − qi+1)2.

infinitely many particles

QFT deals with infinitely many particles, one at each x ∈ R. (We are considering 1 space dimension and 1time dimension.) Then q1(t), q2(t), · · · becomes a function q(x, t).

q = qt : R→ R, V (q) =∫q(x)2dx+

∫(∂xq)

2dx or V (q) =∫U(q(x))dx+

∫(∂xq)

2dx

It is impossible to write down the Schrodinger PDE. However, Feynman integral can still be written down:∫exp

[i

(1

2

∫R

∫ T

0

(∂tq)2dtdx−

∫R

∫ T

0

U(q(x, t))dtdx− 1

2

∫R

∫ T

0

(∂xq)2dtdx

)]Dq

integral is over the space of surfaces. (NOTE: where did the last 1/2 come from?)

Wick rotation, t→ it gives∫exp

(−1

2

∫R

∫ T

0

(∂tq)2dtdx−

∫R

∫ T

0

U(q(x, t))dtdx− 1

2

∫R

∫ T

0

(∂xq)2dtdx

)Dq

which is exactly the 2D Gaussian free field.


Constructible QFT

Arthur Wightman gave a list of criteria that allow you to do the analytic continuation rigorously, and builtan actual QFT. Simplification was given by Osterwalder and Schrader.

Yang-Mill theory is the limit of lattice gauge theory. But QFT is hard to construct. Many of them areGaussian, i.e. “trivial”. It has connection with random surface theory, and Liouville quantum gravity.


Lecture 15: May 22Lecturer: Sourav Chatterjee Scriber: Leila Sloman


15.8 Hermite Polynomials & the Wick-ordered Power

First we’ll define the Hermite polynomials Hk(x).

Definition 31 For k ≥ 1, the k-th Hermite polynomial Hk is given by

Hk(x) = (−1)kεx2/2 d

k

dxk(ε−x

2/2)

Hermite polynomials can also be defined without the factor of 12 in the the exponential.

Example 5

H1(x) = x, H2(x) = x2 − 1, H3(x) = x3 − 3x, H4(x) = x4 − 6x2 + 3

Let γ be the standard Gaussian measure on R, i.e. with density 1√2πε−x

2/2 with respect to Lebesgue measure.

For f, g ∈ L2(γ), define

(f, g) =

∫fgdγ

Then (f,Hk) = (f ′, Hk−1) for f ∈ L2(γ):

(f,Hk) =

∫fHkdγ

=

∫ ∞−∞

f(x)(−1)kεx2/2 d

k

dxk(ε−x

2/2) · 1√2πε−x

2/2dx

=1√2π

∫ ∞−∞

f(x)(−1)kdk

dxk(ε−x

2/2)dx

=

∫ ∞−∞

f ′(x)(−1)k−1εx2/2 d

k−1

dk−1(ε−x

2/2) · 1√2πε−x

2/2dx

= (f ′, Hk−1)

Since Hk is a polynomial with leading term xk, if j < k, the above shows that

(Hj , Hk) = (H ′j , Hk−1) = · · · = (j!, Hk−j) = 0

Thus Hk is a system of orthogonal polynomials for the inner product defined by γ; moreover, it is acomplete basis of L2(γ) (because of density of polynomials in L2(γ)).

Now, suppose (X,Y ) is a jointly normal random vector with mean 0, variance 1, and variance ρ.

15-1


Claim 32 Y can be written as ρX +√

1− ρ2Z for X,Z i.i.d ∼ N (0, 1).

This claim will not be proven. Suppose X,Y is jointly Gaussian with mean 0, Var(X) = σ2X , Var(Y ) = σ2

Y ,and Cov(X,Y ) = σXY .

Definition 33 The Wick-ordered k-th power of X is denoted : Xk :, and given by

: Xk :=σkX√k!Hk

(X

σX

)

Example 6

: X4 : =σ4X√24

((X

σX

)4

− 6

(X

σX

)2

+ 3

)

=1√24

(X4 − 6σ2XX

2 + 3σ4X)

For any k, E[: Xk :] = 0, and

Cov(: Xk :, : Y k :) =σkXσ

kY

(√k!)2

E[Hk

(X

σX

)Hk

(Y

σY

)]= σkXσ

KY ρ

kXY

where

ρXY = Cov

(X

σX,Y

σY

)=

σXYσXσY

Thus Cov(: Xk :, : Y k :) = σkXY .

15.9 φ4 theory on the 2D torus

We will now see a sketch of the construction of φ4 theory on the 2D torus. Consider an ε-discretization ofthe torus [0, 1]2 ∩ εZ2.

Let φ be a Gaussian free field on this grid with mass mε, i.e. with density proportional to

exp

−1

2

∑〈uv〉

(φu − φv)2 − m2ε2

2

∑u

φ2u

which converges to the GFF. Let σε2 = Var(φu) ∼ C log

(1ε

). Let V (φ) = λε2

∑u : φ4

u :. Then

V (φ) = λε2∑u

(φ4u − 6σ2

εφ2u + 4σ4

ε )

Consider the probably measure µε with density proportional to

exp

−1

2

∑〈uv〉

(φu − φv)2 − m2ε2

2

∑u

φ2u − V (φ)


which is in turn proportional to

exp

−1

2

∑〈uv〉

(φu − φv)2 + (6λσ2ε −

m2

2)ε2∑u

φ2u − λε2

∑u

φ4u

We will show that there is a convergent subsequence of these measures. To do this, we’ll prove “ultravioletstability:, which says that for all α > 0, there exist c1, c2 > 0 dependent only on m,λ, α such that

c1 ≤ E[ε−V (φ)] ≤ c2 (15.25)

for φ ∼ GFF at scale ε.

For any nonnegative function F on the space of configurations,∫Fdµε =

E[F (φ)ε−V (φ)]

E[ε−V (φ)]≤√E[F (φ)2]E[ε−2V (φ)]

E[ε−V (φ)]≤ const. ·

√E[F (φ)2]

So if F is a function such that E[F (φ)2] is uniformly bounded, irrespective of ε, so is∫Fdµ. This can be

used to prove the tightness of linear functionals under µ.

The same inequality can be used to prove uniform equicontinuity of characteristic functions under µ; then,one can get a subsequential limit by Arzela-Ascoli.

By the definition of V (φ) as λε2∑u : φ4

u :, E[V (φ)] = 0, and

Var(V (φ)) = λ2ε4∑u,v

Cov(: φ4u :, : φ4

v :) := λ2ε4∑u,v

Cε(u, v)4

where Cε(u, v) = Cov(φu, φv). This converges to λ2∫ ∫

C(x, y)4dxdy where C(x, y) is the covariance kernelof the GFF with mass on the torus.

Remark 34 If V (φ) were defined as λε2∑u φ

4u, rather than using the Wick-ordered power, it would blow

up as ε→ 0.

Note that by Jensen’s inequality,E[ε−V (φ)] ≥ ε−E[V (φ)] = 1

Now suppose X is a multivariate normal random variable with mean 0 and covariance matrix Σ, and supposeΣ can be decomposed as a sum of positive definite matrices Σ1 and Σ2. Then, if Y ∼ N (0,Σ1), Z ∼ N (0,Σ2),with Y and Z independent,

Y + ZD= X

One common way to do this kind of decomposition is to write

Σ =

n∑i=1

λiuiuTi (spectral decomposition)

and take

Σ1 =

k∑i=1

λiuiuTi , Σ2 =

n∑i=k+1

λiuiuTi


The same can be done for the discrete GFF: take φ = φk + φ′k, φk and φ′k independent, where the level kdepends on where you truncate. For this decomposition,

E

(ε2∑u

: φ4u : −ε2

∑u

: φ4k,u :

)2 = ε4

∑u,v

E[(

: φ4u : − : φ4

k,u :) (

: φ4v : − : φ4

k,v :)]

= ε4∑u,v

[Cov(φu, φv)

4 + Cov(φk,u, φk,v)4]

− ε4∑u,v

[Cov(φu, φk,v)

4 + Cov(φv, φk,u)4]

Since Cov(φ′k,u, φk,v) = 0 by independence, this simplifies to

E

(ε2∑u

: φ4u : −ε2

∑u

: φ4k,u :

)2 = ε4

∑u,v

[Cov(φu, φv)

4 − Cov(φk,u, φk,v)4]

For each k, we choose the decomposition such that as ε→ 0,

Cov(φk,u, φk,v) −−−−−−→u→x,v→y

Ck(x, y)

where |C − Ck| = O(k−δ). Then, for all j, there will be an integrable bound such as:

E

(ε2∑u

: φ4u : −ε2

∑u

: φ4k,u :

)2j ≤ (C1j)

C2jk−δj


Lecture 16: May 24Lecturer: Sourav Chatterjee Scriber: Leila Sloman


16.10 Bounding E[ε−V (φ)]

Let φ a Gaussian free field, φk a smoothed version of φ. We want to show that

E[ε−V (φ)] ≤ C

where C is independent of the scaling ε, and V (φ) = λε2∑u : φ4

u :. We can choose k so that

−V (φk) ≤ C(log k)α

and

E[|V (φ)− V (φj)|j ] ≤ (C1j)C2jk−δj ∀j

where C1, C2, δ are constants, by what we did last time.

Using the moment inequality and Chebyshev’s inequality, we get

P(|V (φ)− V (φk)| ≥ 1) ≤ E[|V (φ)− V (φk)|j ] ≤ (C1j)C2jk−δj

Choosing j = kε for some small enough ε, we see that this probability is ≤ C1ε−C2k

ε

. Thus

P(ε−V (φ) ≥ a) = P(ε−(V (φ)−V (φk))−V (φk) ≥ a)

≤ P(ε−(V (φ)−V (φk))εC(log k)α ≥ a)

Choose k such that ε(C log k)α ∼ a, which means that k ∼ εC′(log a)γ . Then

P(ε−V (φ) ≥ a) ≤ P(|V (φ)− V (φk)| ≥ const.) ≤ C1ε−C2k

ε

= C1ε−C2ε

C3(log a)γ

16.11 Decay of Correlations

Recall the O(2) model on the 2D torus T2n = 0, 1, . . . , n− 12. The O(2) model has spins σx ∈ S1 on each

x ∈ V (T2n), σx = (cos θx, sin θx). The probability density is proportional to exp(β

∑〈xy〉 cos(θx − θy)).

Claim 35 ρxy = 〈σx · σy〉 = 〈cos(θx − θy)〉 decays like |x− y|−α for some α.

16-1


Intuition. Suppose we approximate cos(θx − θy) by its first order Taylor expansion:

cos(θx − θy) ≈ 1− 1

2(θx − θy)2

Then we get back the Gaussian free field. If θ is a configuration from the Gaussian free field on T2n, then

cos(θx − θy) is a normal random variable, so

〈cos(θx − θy)〉 = ε−12E[(θx−θy)2] ∼ |x− y|−α

since E[(θx − θy)2] ∼ log |x− y|.For small β, there is exponential decay of correlations, so this is not the correct answer. For large β,cos(θx − θy) tends to 1, so that θx, θy tend to be close for x, y neighbors.

Patrascioiu-Seiler proposed O(2) model restricted to configurations where cos(θx − θy) ≥ 1√2

for all 〈xy〉.The idea is that θx, θy “close enough” is sufficient for a power law decay. Aizenman proved power law decaylower bound for this model.

Theorem 36 For this model, for any `,

max‖x−y‖1≥`

ρxy ≥1

2`2

Proof: Write σx = (σ1x, σ

2x). Then, since e.g. σ1

x = |σ1x|εx where εx = sgn(σ1

x),

ρxy = E[σ1xσ

1y + σ2

xσ2y]

= 2E[σ1xσ

1y]

= 2E[(|σ1x||σ1

y|εxεy)

= 2E[|σ1x||σ1

y|E[εxεy|σ2]]

for σ2 = (σ2x)x∈T2

n. Let U : [−1, 1]→ (−∞,∞],

U(x) =

∞ if x < 1√

2

−βx if x ≥ 1√2

Then the density of σ is proportional to exp(−∑〈uv〉 U(σu · σv)). This can be written as

exp(−∑〈uv〉

U(|σ1u||σ1

v |εuεv + σ2uσ

2v))

So if

fuv(ε) = exp

−∑〈uv〉

U(|σ1u||σ1

v |εuεv + σ2uσ

2v) +

∑〈uv〉

U(−|σ1u||σ1

v |+ σ2uσ

2v)

− 1,

then the conditional density of ε given σ2 is proportional to∏〈uv〉(fuv(ε) + 1).

Note that U is a nonincreasing function, which implies that fuv(ε) ≥ 0. Moreover, if εu 6= εv, then fuv(ε) = 0;if εu = εv, then fuv(ε) depends only on σ2.


The conditional density of ε given σ2 can be written as

1

Z

∑E⊂E[T2

n]

∏〈uv〉∈E

fuv(ε)×1

Z(E)× Z(E)

whereZ(E) =

∑ε∈±1V (T2

n)

∏〈uv〉∈E

fuv(ε)

This model can be reinterpreted as follows: pick a set of edges E with probability Z(E)/Z. Given E, pick εwith density proportional to

∏〈uv〉∈E fuv(ε).

Note that given E, we get a division of V (T2n) into connected components. Any configuration ε where εu is not

the same everywhere within a connected component has probability 0. Moreover, any allowed configurationhas the same probability. Thus, given E, ε is uniformly distributed on the set of all configurations where εis constant in each connected component. Therefore

E[εxεy|σ2, E] = 1x↔y∈E

This shows in particular that ρxy is nonnegative, even in the original O(2) model—we have not used yetthat cos(θx − θy) ≥ 1√

2.

Let V0 = v ∈ V (T2n) : |σ1

v | ≥ 1√2. Define a new adjacency structure on T2

n by adding edges along the

diagonals of the lattice. Then if u, v ∈ V0 and u, v are neighbors in the new graph, then εu = εv, since theangle between neighbors in the new graph is at most π/2.

Let Exy = x↔ y ∈ V0 with new adjacency structure.Exy depends only on σ2. Exy true implies εx = εy, which in turn implies that x must be connected to y inthe graph determined by E. Thus, E[εxεy|σ2, E] ≥ 1Exy . So

ρxy = E[σx · σy] = 2E[|σ1x||σ1

y|E[εxεy|σ2]] ≥ 2E[|σ1x||σ1

y|1Exy ] ≥ P(Exy)

Let R = 1, . . . , `2 ⊂ V (T2n), and

A = ∃ top-bottom crossing of R w. vertices in V0 in new adj. structure,

B = ∃ left-right crossing of R w. vertices not in V0 in old adj. structure.

Fact: either A or B must happen. Thus 1 ≤ P(A) + P(B). But by σ1 → σ2 symmetry,

P(B) = P(B)

whereB = ∃ left-right crossing of R in old adj. through V0

P(B) ≤ P(A), ince the new adj. has more edges. Thus

1 ≤ 2P(A) ≤ 2∑

x∈top of R,y∈bottom of R

P(Exy)


Lecture 17: June 5Lecturer: Sourav Chatterjee Scriber: Song Mei


In this lecture, we are talking about Liouville Quantum Gravity.

17.12 Gaussian Free Field

Let D be a bounded domain in R2. We would like to consider a Gaussian free field (GFF) on D with zeroboundary condition. One way to define this GFF is to approximate D by a grid and then take the gridspacing to zero. The limiting object is a random distribution h on D.

Definition 37 (Gaussian Free Field) We say h is a Gaussian Free Field on D with zero boundary con-dition, if for any smooth compactly supported f on D, we have∫

Dfh ∼ N

(0,

∫D

∫Df(x)f(y)GD(x, y)δxδy

), (17.26)

where GD(x, y) is Green’s function on D with Dirichlet boundary condition. This means, we have

GD(x, y) = E[

∫ ∞0

pD(t, x, y)δt], (17.27)

where pD(t, x, y) is the transition kernel of a Brownian motion killed when hitting ∂D.

An alternative representation of GD gives: Consider the Laplacian ∆ acting on smooth functions on Dthat vanishes at the boundary. Under mild condition on D, there exists an L2(D) orthonormal sequencesunn∈N+

on D, where each un is smooth and vanishes at the boundary such that un is an eigenfunction of

∆, and unn∈N+ is a basis of L2(D). Then the Green’s function can be formally represented as

GD(x, y) =∞∑n=1

1

λnun(x)un(y), x 6= y, (17.28)

where

∆un = −λnun. (17.29)

Also, the GFF can be formally represented as

h(x) =

∞∑n=1

Xn√λnun(x), (17.30)

where Xii.i.d.∼ N (0, 1).

As d ≥ 2, this sum diverges. For example, D = (0, 1)d, then the eigenfunctions are∏dj=1 sin(πkjxj) indexed

by (k1, . . . , kd). The eigenvalues are λ(k1, . . . , kd) = π2(k21 + · · · + k2

d). Alternatively using n as indices, wehave λn ≈ n2/d.

But for Equation (17.30), the convergence is still valid on various spaces of distributions.

17-1

17-2 Lecture 17: June 5

17.13 Domain Markov Properties

Proposition 38 Let U ⊂ D. Let h be a GFF on D with zero boundary condition. Then h can be written ash = h0 + ϕ, with

1. h0 is independent of ϕ.

2. h0 is a GFF on U with zero boundary condition and h0 = 0 on D \ U .

3. ϕ is harmonic on U .

For the discrete GFF, Markovian property is obvious from the definition. Moreover, given the value on theboundary of a sub-graph U , the field inside has the same covariance structure as a GFF with zero boundarycondition, but the mean is a discrete harmonic function with the prescribed boundary values.

Thus, inside U , the conditioned distribution of the GFF given the values on ∂U is the law of a zero boundedGFF on U plus a harmonic function with given boundary values.

One can get the continuum decomposition by passing to the continuum limit from the discrete. But one canalso get it by a direct continuum argument.

17.14 Circle Averages

Let h be a GFF on a domain D ⊂ R2 with zero boundary condition. For any z ∈ D, and ε < dist(z, ∂D), let

hε(z) = 1/(2πε) ·∫∂B(z,ε)

h(x)δx. (17.31)

Take any z ∈ D and ε0 < dist(z, ∂D). For t ≥ t0 = log(1/ε0), let Bt = he−t(z), then (Bt)t≥t0 is a standardBrownian motion starting at Bt0 .

The reason is that, we take two radius ε1 > ε2. On B(z, ε1), write h = h0 + φ, with h0 a GFF withzero boundary in B(z, ε1), and ϕ a harmonic function in B(z, ε1) and agrees with h outside. h0 and ϕ areindependent.

Then

hε2(z) =h0,ε2(z) + ϕε2(z), (17.32)

hε1(z) =h0,ε1(z) + ϕε1(z). (17.33)

But h0,ε1(z) = 0, since h0 is GFF on B(z, ε1) with zero boundary. Therefore, hε1(z) = ϕε1(z). Since ϕ isharmonic in B(z, ε1), ϕε1(z) = φε2(z). Therefore hε2(z) = h0,ε2(z) + hε1(z). Therefore hε2(z) − hε1(z) =h0,ε2(z) which is independent of ϕε1(z) = hε1(z).

Therefore, we proved that he−t(z) as a random function of t has independent increment. The parameteriza-tion using t makes the variance exactly to be t.

Definition 39 (Thick points) A point z is called α-thick for a GFF h if

lim infε→0

hε(z)/ log(1/ε) = α. (17.34)

It is known that a 2D GFF has α-thick points for ∀α ∈ [0, 2). Moreover, dim(Set of α thick points) =2− α2/2.

Lecture 17: June 5 17-3

17.15 Liouville Quantum Gravity

Definition 40 LQG on a bounded domain D ⊂ R2 is a Riemannian metric on D, which is proportional to[exp(γh(z)) 0

0 exp(γh(z))

](17.35)

at z.

The volume form for this metric is proportional to exp(γh(z))δz. This made rigorous by Duplantier andSheffield.

Let µε(δz) = exp(γhε(z)− γ2/2 · log(1/ε))δz. We have the following theorem

Theorem 41 µε has a weak limit µ as ε→ 0 though 2−k. As Leb(A) > 0, we have µ(A) > 0. We also haveµ(D) <∞.

We will prove this theorem in the next lecture.


Lecture 18: June 7Lecturer: Sourav Chatterjee Scriber: Song Mei


In this lecture, we are talking about Liouville Quantum Gravity.

18.16 Liouville Quantum Gravity

Definition 42 LQG on a bounded domain D ⊂ R2 is a Riemannian metric on D, which is proportional to[exp(γh(z)) 0

0 exp(γh(z))

](18.36)

at z.

The volume form for this metric is proportional to exp(γh(z))δz. This made rigorous by Duplantier andSheffield.

Let µε(δz) = exp(γhε(z)− γ2/2 · log(1/ε))δz. We have the following theorem

Theorem 43 µε has a weak limit µ as ε→ 0 though 2−k. As Leb(A) > 0, we have µ(A) > 0. We also haveµ(D) <∞.

18.17 Proof

Let h be a 2D Gaussian Free Field on a bounded domain D ⊂ R2. Let hε(z) be the average of h in acircle of radius ε around z. Let Bt = he−t(z). Then Bt is a standard Brownian motion. Let µε(δz) =expγhε(z) − γ2/2 · log(1/ε)δz. Claim: µε converges to a random limit measure µ, if γ ∈ [0, 2). This µ isthe volume form of LQG(γ).

We will show that for any Borel set A ⊂ D, µε(A) converges to a finite limit almost surely as ε→ 0 through2−k.

The proof is easier when γ <√

2. We have

µε(A) =

∫A

expγhε(z)− γ2/2 · log(1/ε)δz. (18.37)

A direct calculation shows that

E[(µε(A)− µε/2(A))2] ≤ Cε2−γ2

. (18.38)

So if γ <√

2, then µ2−k(A)k≥1 is almost surely Cauchy.

18-1

18-2 Lecture 18: June 7

The proof for all γ ∈ [0, 2) is more complicated. For some ε0 > 0, some α > γ, let

Gε(z) = hγ(z) ≤ α log(1/γ),∀γ ∈ [ε, ε0]. (18.39)

Let Iε = µε(A), and Jε =∫A

1Gε(z)δµε(z). We claim that

1. supε E[J2ε ] <∞,

2. E[ehε(z)(1−1Gε(z))] ≤ C(ε0),∀z, where C(ε0)→ 0 as ε0 → 0, and hε(z) = γhε(z)− γ2/2 · log(1/ε).

Observation: (1) and (2) imply that Iεε>0 is uniformly integrable. Actually, take any δ > 0. Find ε0 sosmall that C(ε0) ≤ δ/100. Then (2) implies that EλIε − Jελ ≤ δ/100. (1) implies that Jεε>0 is uniformlyintegrable. Find k so that E[λJελ;λJελ > k] ≤ δ/100.

Using these, we get a uniform bound on E[λJελ;λJελ > k]. Uniformly integrability and a martingaleargument gives the convergence of Iε.

Therefore

E[ehε(z)(1− 1Gε(z))] =1− P (Gε(z)), (18.40)

δP

δP(z) =ehε(z). (18.41)

We have

E[eγBT−γ2/2·T f((Bt)0≤t≤T )] = E[f((Bt)0≤t≤T )], (18.42)

where Bt = Bt + γt.

Proof of (1) sketch:

E[J2ε ] =

∫A×A

E[ehε(z)+hε(y)1Gε(z) ∩Gε(y)]δxδy (18.43)

=

∫A×A

E[ehε(z)+hε(y)]P(Gε(x) ∩Gε(y))δxδy (18.44)

≤(ε′)−γ2

P(hε(x) ≤ α log(1/ε)). (18.45)

where ε′ = maxλx− yλ, ε. Under P, hε(z) ∼ N (2r log(1/ε), log(1/ε)). Therefore

E[J2ε ] ≤ C

∫A×A

λx− yλ−γ2+(2γ−α)2/2δxδy. (18.46)

If γ < 2, then we can find α > r such that (2γ − α)2/2− γ2 > −2.

18.18 KPZ relation

Definition 44 Let A ⊂ D. The Euclidean scaling exponent of A gives

x = limε→0

P(B(z, ε) ∩A 6= ∅)log(ε2)

. (18.47)

where z ∼ Unif(D).

Lecture 18: June 7 18-3

Under some condition on A, we have

dim(A) = 2(1− x). (18.48)

Definition 45 Let A ⊂ D. The Quantum scaling exponent of A gives

∆ = limδ→0

P(Bδ(z) ∩A 6= ∅)log(δ)

. (18.49)

where z ∼ Unif(D) and Bδ(z) is the ball centered at z that has LQG(z) mass δ.

The KPZ relation gives: for any A, we have x = γ2/4 ·∆2 + (1− γ2/4)∆.

Stats 359: Topics in Mathematical Physics Spring 2017 Lecture 1:...

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