Math 639: Lecture 24 the Gaussian Free Field and Liouville Quantum Gravity

Math 639: Lecture 24 The Gaussian Free Field and Liouville Quantum Gravity

Bob Hough

May 11, 2017

Bob Hough Math 639: Lecture 24 May 11, 2017 1 / 65 The Gaussian free ﬁeld

This lecture loosely follows Berestycki’s ‘Introduction to the Gaussian Free Field and Liouville Quantum Gravity’.

Bob Hough Math 639: Lecture 24 May 11, 2017 2 / 65 Itˆo’sformula

The multidimensional Itˆo’sformula is as follows. Theorem (Multidimensional Itˆo’sformula) Let tBptq : t ě 0u be a d-dimensional Brownian motion and suppose tζpsq : s ě 0u is a continuous, adapted stochastic process with values in m d m R and increasing components. Let f : R ` Ñ R satisfy Bi f and Bjk f , all 1 ď j, k ď d, d ` 1 ď i ď d ` m are continuous t 2 E 0 |∇x f pBpsq, ζpsqq| ds ă 8 then a.s.ş for all 0 ď s ď t s f pBpsq, ζpsqq ´ f pBp0q, ζp0qq “ ∇x f pBpuq, ζpuqq ¨ dBpuq ż0 s 1 s ` ∇ f pBpuq, ζpuqq ¨ dζpuq ` ∆ f pBpuq, ζpuqqdu. y 2 x ż0 ż0

Bob Hough Math 639: Lecture 24 May 11, 2017 3 / 65 Conformal maps

Deﬁnition 2 Let U and V be domains in R . A mapping f : U Ñ V is conformal if it is a bijection and preserves angles.

Viewed as a map between domains in C, this is equivalent to f is an analytic bijection.

Bob Hough Math 639: Lecture 24 May 11, 2017 4 / 65 Conformal invariance of Brownian motion

The following conformal invariance of Brownian motion may be established with Itˆo’sformula. Theorem Let U be a domain in the complex plane, x P U, and let f : U Ñ V be analytic. Let tBptq : t ě 0u be a planar Brownian motion started in x and

τU “ inftt ě 0 : Bptq R Uu

its ﬁrst exit time from the domain U. There exists a planar Brownian motion tB˜ptq : t ě 0u such that, for any t P r0, τU q,

t f pBptqq “ B˜pζptqq, ζptq “ |f 1pBpsqq|2ds. ż0 If f is conformal, then ζpτU q is the ﬁrst exit time from V by tB˜ptq : t ě 0u.

Bob Hough Math 639: Lecture 24 May 11, 2017 5 / 65 Conformal invariance of Brownian motion

Proof. We assume that the domains are bounded.

Recall that if f “ f1 ` if2 then the Cauchy-Riemann equations give 1 ∇f1 and ∇f2 are orthogonal, and |∇f1| “ |∇f2| “ |f |. Let σptq “ infts ě 0 : ζpsq ě tu Let tB˜ptq : t ě 0u be an independent Brownian motion, and deﬁne

W ptq “ f pBpσptq ^ τU qq ` B˜ptq ´ B˜pt ^ ζpτU qq, t ě 0.

It suﬃces to check that W ptq is a Brownian motion. Since it is almost surely continuous, it remains to check the f.d.d.

Bob Hough Math 639: Lecture 24 May 11, 2017 6 / 65 Conformal invariance of Brownian motion

Bob Hough Math 639: Lecture 24 May 11, 2017 7 / 65 Conformal invariance of Brownian motion

Proof. Calculate

E exλ,W ptqy W p0q “ f pxq ” ˇ ı 1 “ E exp xˇλ, f pBpσptq ^ τ qqy ` |λ|2pt ´ ζpσptq ^ τ qq . x ˇ U 2 U ˆ ˙ We use Itˆowith 1 F px, uq “ exp xλ, f pxqy ` |λ|2pt ´ sq . 2 ˆ ˙ Note ∆exλ,f pxqy “ |λ|2|f 1pxq|2exλ,f pxqy.

Bob Hough Math 639: Lecture 24 May 11, 2017 8 / 65 Conformal invariance of Brownian motion

Proof. 1 2 Recall F px, uq “ exp xλ, f pxqy ` 2 |λ| pt ´ uq . Set T “ σptq ^ τUn , with U “ tx P U : dpx, BUq ě 1{nu. n ` ˘ Hence

T F pBpT q, ζpT qq “ F pBp0q, ζp0qq ` ∇x F pBpsq, ζpsqq ¨ dBpsq ż0 T 1 T ` B F pBpsq, ζpsqqdζpsq ` ∆ F pBpsq, ζpsqqds. u 2 x ż0 ż0 Use dζpuq “ |f 1pBpuqq|2du to cancel the two terms on the bottom line. Also, the stochastic integral has mean 0.

Bob Hough Math 639: Lecture 24 May 11, 2017 9 / 65 Conformal invariance of Brownian motion

Proof. We thus calculate

xλ,W ptqy E e W p0q “ f pxq “ Ex rF pBpσptq ^ τU q, ζpσptq ^ τU qqs ˇ ” ı 1 2 “ lim Ex rFˇ pBpT q, ζpT qqs “ F px, 0q “ exp |λ| t ` xλ, f pxqy . nÑ8 ˇ 2 ˆ ˙

Bob Hough Math 639: Lecture 24 May 11, 2017 10 / 65 Discrete case

Let G “ pV , Eq be an undirected graph, and let B be a distinguished set of vertices, called the boundary. Let Vˆ “ V zB be the internal vertices. For x, y P V , write x „ y if x and y are neighbors.

Let tXnu be a random walk on G, in which at each step the walker chooses uniformly a random neighbor. Let P be the transition matrix, dpxq “ degpxq, which is an invariant measure for the walk, and let τ be the ﬁrst hitting time to the boundary.

Bob Hough Math 639: Lecture 24 May 11, 2017 11 / 65 Discrete case

Deﬁnition (Green function) The Green function Gpx, yq is deﬁned for x, y P V by putting

1 8 Gpx, yq “ Ex 1 . dpyq pXn“y;τąnq ˜n“0 ¸ ÿ Deﬁnition (Discrete Laplacian) The discrete Laplacian acts on functions on V by 1 ∆f pxq “ pf pyq ´ f pxqq. d x y„x p q ÿ

Bob Hough Math 639: Lecture 24 May 11, 2017 12 / 65 Discrete case

Proposition The following hold: 1 Let Pˆ denote the restriction of P to Vˆ . Then pI ´ Pˆq´1px, yq “ Gpx, yqdpyq for all x, y P V.ˆ 2 G is a symmetric nonnegative semideﬁnite function. That is, one has Gpx, yq “ Gpy, xq and if pλx qxPV is a vector then x,yPV λx λy Gpx, yq ě 0. Equivalently, all eigenvalues are non-negative. ř 3 Gpx, ¨q is discrete harmonic in Vˆ ztxu, and ∆Gpx, ¨q “ ´δx p¨q.

Bob Hough Math 639: Lecture 24 May 11, 2017 13 / 65 Discrete case

Deﬁnition

The discrete Gaussian free ﬁeld is the centered Gaussian vector phpxqqxPV with covariance given by the Green function G.

Note that if x PB, then Gpx, yq “ 0 for all y P V and hence hpxq “ 0 a.s.

Bob Hough Math 639: Lecture 24 May 11, 2017 14 / 65 Discrete case

Theorem (Law of the GFF)

The law of phpxqqxPV is absolutely continuous with respect to dx “ uPV dxu, and the joint pdf is given by

ś 1 1 Probphpxq P Aq “ exp ´ py ´ y q2 dy . Z 4 u v u A ˜ u„vPV ¸ uPV ż ÿ ź Z is a normalizing constant, called the partition function.

The quadratic form appearing in the exponential is called the Dirichlet energy.

Bob Hough Math 639: Lecture 24 May 11, 2017 15 / 65 Discrete case

Proof.

For a centered Gaussian vector pY1, ..., Ynq with covariance matrix V , the joint density is given by

1 1 exp ´ y T V ´1y Z 2 ˆ ˙

Bob Hough Math 639: Lecture 24 May 11, 2017 16 / 65 Discrete case

Proof. Restrict to only variables from Vˆ and calculate

hpxˆqT G ´1hpxˆq “ G ´1px, yqhpxqhpyq ˆ x,ÿyPV “ ´dpxqPˆpx, yqhpxqhpyq ` dpxqhpxq2 ˆ ˆ x,yPÿV :x„y xÿPV 1 “ ´ hpxqhpyq ` phpxq2 ` hpyq2q 2 x,yPV :x„y x,yPV :x„y ÿ ÿ 1 “ phpxq ´ hpyqq2. 2 x,yPV :x„y ÿ

Bob Hough Math 639: Lecture 24 May 11, 2017 17 / 65 Discrete case

Theorem (Markov property) Fix U Ă V . The discrete GFF hpxq can be decomposed as follows:

h “ h0 ` φ

where h0 is a Dirichlet boundary Gaussian free ﬁeld on U and φ is harmonic on U. Moreover, h0 and φ are independent.

We prove a continuum version of this theorem in the next section.

Bob Hough Math 639: Lecture 24 May 11, 2017 18 / 65 Continuum case

d D For D Ă R let pt px, yq be the transition kernel of Brownian motion killed on leaving D. Thus

D D pt px, yq “ pt px, yqπt px, yq

where |x´y|2 exp ´ 2t p px, yq “ t ´ d ¯ p2πtq 2 D and πt px, yq is the probability that a Brownian bridge of duration t remains in D.

Bob Hough Math 639: Lecture 24 May 11, 2017 19 / 65 Continuum case

Deﬁnition

The Green function Gpx, yq “ GD px, yq is given by

8 D Gpx, yq “ π pt px, yqdt. ż0 D Gpx, xq “ 8 for all x P D, since πt px, xq Ñ 1 as t Ñ 0. If D is bounded then Gpx, yq ă 8 for all x ‰ y.

Bob Hough Math 639: Lecture 24 May 11, 2017 20 / 65 Continuum case

Example Suppose D “ H is the upper half plane. Then H pt px, yq “ pt px, yq ´ pt px, yq and x ´ y G px, yq “ log . H x ´ y ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ

Bob Hough Math 639: Lecture 24 May 11, 2017 21 / 65 Continuum case

We now restrict attention to dimension 2. Proposition If T : D Ñ D1 is a conformal map (holomorphic and one-to-one), then GT pDqpT pxq, T pyqq “ GD px, yq.

Bob Hough Math 639: Lecture 24 May 11, 2017 22 / 65 Continuum case

Proof. Let φ be a test function and let x1, y 1 “ T pxq, T pyq. Then τ 1 1 1 1 1 1 1 GD1 px , y qφpy qdy “ Ex1 φpBt1 qdt 1 żD «ż0 ﬀ where B1 is Brownian motion and τ 1 is the ﬁrst exit time from D1. Since dy 1 “ |T 1pyq|2dy

1 1 1 1 1 2 GD1 px , y qφpy qdy “ GD1 pT pxq, T pyqqφpT pyqq|T pyq| dy. 1 żD żD

Bob Hough Math 639: Lecture 24 May 11, 2017 23 / 65 Continuum case

Proof. 1 Apply Itˆo’sformula to the RHS, writing Bt1 “ T pBF ´1pt1qq where t 1 2 F ptq “ 0 |T pBs q| ds for t ď τ. Calculateş τ 1 τ 1 1 1 Ex1 φpBt1 qdt “ Ex φpT pBs qqF psqds «ż0 ﬀ „ż0  τ 1 2 “ Ex φpT pBs qq|T pBs q| ds „ż0  1 2 “ GD px, yqφpT pyqq|T pyq| dy. żD This proves that GD1 pT pxq, T p¨qq “ GD px, ¨q as distributions.

Bob Hough Math 639: Lecture 24 May 11, 2017 24 / 65 Continuum case

Proposition The following properties hold

1 Gpx, ¨q is harmonic in Dztxu and ∆Gpx, ¨q “ ´2πδx p¨q. 2 Gpx, yq “ ´ logp|x ´ y|q ` log Rpx; Dq ` op1q where Rpx; Dq is the conformal radius of x P D, equal to |f 1p0q| where f is any conformal map from the disc D “ tz P C : |z| ă 1u to D satisfying f p0q “ x.

Bob Hough Math 639: Lecture 24 May 11, 2017 25 / 65 Continuum case

Proof. 1 Conformal maps preserve harmonicity, which proves the ﬁrst property. 2 A conformal map from D Ñ D which maps 0 to 0 is of the form f pzq “ eiθz. To check this, let f be such a map, which necessarily 8 n maps the boundary to itself, and write f pzq “ n“1 anz ,

1 ř 8 1 f pzq 1 1 2 1 “ dz “ f pzqf pzqdz “ n|an| 2πi f pzq 2πi |z|“1 |z|“1 n“1 ż ż ÿ 2 Since |an| “ 1 we get |a1| “ 1 and ai “ 0 for i ą 1. This proves that the conformal radius is well-deﬁned. ř 3 To calculate the formula for the Green’s function, observe that

GDp0, zq “ log |z|, as may be checked by using conformal invariance i´z and the map φpzq “ i`z from the upper half plane to the disc.

Bob Hough Math 639: Lecture 24 May 11, 2017 26 / 65 Continuum case

In the continuum case, one thinks of the GFF as a “random function” on a domain, with mean 0 and covariance given by the Green function. However, since the Green function is inﬁnite on the diagonal, the GFF is not deﬁned pointwise, and is instead in a negative Sobolev space.

Bob Hough Math 639: Lecture 24 May 11, 2017 27 / 65 Continuum case

Definition Let D be a domain on which the Green function is finite off the diagonal. Such a domain is called Greenian. Let M` be the space of positive measures with compact support on D satisfying

ρpdxqρpdyqGpx, yq ă 8. ż Let M be the collection of signed measures ρ “ ρ` ´ ρ´ with ρ˘ P M`. For ρ1, ρ2 P M , deﬁne bilinear form

Γpρ1, ρ2q :“ GD px, yqρ1pdxqρ2pdyq. 2 żD Deﬁne Γpρq “ Γpρ, ρq.

Bob Hough Math 639: Lecture 24 May 11, 2017 28 / 65 Continuum case

Theorem (Zero boundary GFF)

There exists a unique stochastic process phρqρPM , indexed by M , such

that, for every choice ρ1, ..., ρn, phρ1 , ..., hρn q is a centered Gaussian vector

with covariance matrix Covphρi , hρj q “ Γpρi , ρj q.

Deﬁnition

The process phρqρPM is called the Gaussian free ﬁeld in D with Dirichlet boundary conditions.

Bob Hough Math 639: Lecture 24 May 11, 2017 29 / 65 Continuum case

Proof. We need to check that the finite-dimensional distributions exist, are uniquely specified, and are consistent. The consistency is immediate, since the f.d.d. are Gaussian vectors. To check existence and uniqueness of the f.d.d. we need to show that Γpρi , ρj q is symmetric and positive semi-definite. Symmetry follows from symmetry of the Greens function, which D D follows from pt px, yq “ pt py, xq. To prove positivity, we need to show,

λi λj Γpρi , ρj q ě 0 i,j ÿ or, equivalently, Γpρq ě 0 for all ρ P M .

Bob Hough Math 639: Lecture 24 May 11, 2017 30 / 65 Continuum case

Proof. Recall that Green’s formula gives, for f , g in C 8pDq,

Bg ∇f ¨ ∇g “ ´ f ∆g ` f . Bn żD żD żBD 8 Let ρ P Cc pDq and

f pxq “ ´ GD px, yqρpyqdy ż so that ∆f “ 2πρ.

Bob Hough Math 639: Lecture 24 May 11, 2017 31 / 65 Continuum case

Proof. Hence 1 Γpρq “ ρpxq Gpx, yq∆ f pyqdydx 2π y żx ży 1 “ ρpxq ∆ Gpx, yqf pyqdydx 2π y żx ży “ ´ f pxq∆f pxqdx “ |∇f |2. żx żD 8 Thus Γpρ, ρq ě 0 for ρ P Cc pDq. The claim holds in general by density.

Bob Hough Math 639: Lecture 24 May 11, 2017 32 / 65 Continuum case

Going forward we write ph, ρq for hρ. The pairing is linear in ρ as may be checked by noting that the mean and variance of the diﬀerence ph, αρ ` βρ1q ´ αph, ρq ´ βph, ρ1q are both 0. The above description gives the GFF with Dirichlet boundary condition. If f (possibly random) is continuous on the conformal boundary of the domain D, then the GFF with boundary condition f is h “ h0 ` φ where φ is the harmonic extension of f to D, and where h0 is an independent solution of the Dirichlet GFF.

Bob Hough Math 639: Lecture 24 May 11, 2017 33 / 65 Distributions

8 Write DpDq “ Cc pDq for the space of ‘test functions’ on D. The topology is deﬁned by fn Ñ 0 in DpDq if there is a compact set K Ă D such that fn is supported in K, and fn and all derivatives converge to 0 uniformly. A continuous linear map u : DpDq Ñ R is a distribution on D. This space is written D1pDq. It is given the weak-* topology, so that 1 un Ñ u P D pDq if and only if unpρq Ñ upρq for all ρ P DpDq.

Bob Hough Math 639: Lecture 24 May 11, 2017 34 / 65 Dirichlet energy, Sobolev space

Deﬁnition For f , g P DpDq, their Dirichlet inner product is 1 pf , gq :“ ∇f ¨ ∇g. ∇ 2π żD 1 The Sobolev space H0 pDq is the completion of DpDq with respect to the Dirichlet inner product. This consists of L2pDq functions whose gradient is also in L2pDq.

Bob Hough Math 639: Lecture 24 May 11, 2017 35 / 65 The GFF as a random distribution

We now construct the GFF as a random distribution. Suppose h P D1pDq and f P DpDq. Set ρ “ ´∆f . By the deﬁnition of distributional derivatives 1 1 ph, f q “ ´ ph, ∆f q “ ph, ρq. ∇ 2π 2π

This makes h, f a centered Gaussian with variance Γpρq , p q∇ p2πq2

2 Varph, f q∇ “}f }∇.

By polarization, Covpph, f q∇, ph, gq∇q “ pf , gq∇.

Bob Hough Math 639: Lecture 24 May 11, 2017 36 / 65 The GFF as a random distribution

Suppose D is a bounded domain, let tfnu be an orthonormal basis of 1 H0 pDq which are eigenfunctions of ´∆ with increasing eigenvalue λn and let tXnu be a sequence of i.i.d. standard normals. Set

h “ Xnfn. n ÿ Weyl’s law gives that λn — n as n Ñ 8. Let en be fn scaled to be 2 ´1 λn orthonormal in L pDq. Since pfn, fnq∇ “ 2π pfn, ∆fnq2 “ 2π pfn, fnq2 we have 2π e “ f . λn n n s For s bP R, the Sobolev space H pDq is s 1 2 s H pDq “ tf P D pDq : pf , enq λn ă 8u n ÿ with inner product s pf , gqs “ pf , enqpg, enqλn. n ÿ Bob Hough Math 639: Lecture 24 May 11, 2017 37 / 65 The GFF as a random distribution

´ It follows that h “ n Xnfn converges a.s. in H pDq for every ą 0. For all f H1 D , P 0 p q ř

ph, f q∇ :“ Xnpfn, f q∇ n ÿ converges in L2pProbq and a.s. by the martingale convergence theorem. It’s limit is a Gaussian with variance

2 2 pfn, f q∇ “}f }∇. n ÿ

Bob Hough Math 639: Lecture 24 May 11, 2017 38 / 65 Markov property

Theorem (Markov property) Fix U Ă D open and take h a GFF with zero boundary condition on D. Then we may write h “ h0 ` φ where

1 h0 is a zero boundary condition GFF on U and is zero outside of U. 2 φ is harmonic in U.

3 h0 and φ are independent.

Bob Hough Math 639: Lecture 24 May 11, 2017 39 / 65 Markov property

Proof. 1 We ﬁrst check that H0 pDq “ SupppUq ‘ HarmpUq where SupppUq is the closure of smooth functions of compact support in U and HarmpUq is functions harmonic in U. The orthogonality of SupppUq and HarmpUq follows by the Gauss-Green formula. 1 Given f P H0 pDq, let f0 be the orthogonal projection onto SupppUq, φ “ f ´ f0.

For any test function ψ P DpUq, pφ, ψq∇ “ 0, so that

p∆φqψ “ p∆φqψ “ 0. żD żU so that ∆φ “ 0 as a distribution in U. By elliptic regularity, φ is a smooth function, and harmonic.

Bob Hough Math 639: Lecture 24 May 11, 2017 40 / 65 Markov property

Proof. 2 0 With the L decomposition, let tfn u be an o.n. basis of SupppUq, and tφnu an o.n. basis of HarmpUq.

Let pXn, Ynq be an i.i.d. sequence of standard Gaussians, and let 0 h0 “ n Xnfn and φ “ n Ynφn. We have h converges a.s. in 1 D since it is GFF on U. ř 0 ř D p q 1 Since h0 ` φ “ h, converges a.s. in D pDq, the series deﬁning φ converges a.s., hence is a C 8 harmonic on U.

Bob Hough Math 639: Lecture 24 May 11, 2017 41 / 65 Markov property

Proof. 2 0 With the L decomposition, let tfn u be an o.n. basis of SupppUq, and tφnu an o.n. basis of HarmpUq.

Bob Hough Math 639: Lecture 24 May 11, 2017 42 / 65 Conformal invariance

The Dirichlet inner product is conformally invariant. If ϕ : D Ñ D1 is conformally invariant then

∇pf ˝ φ´1q ¨ ∇pg ˝ φ´1q “ ∇f ¨ ∇g. 1 żD żD 1 ´1 Thus if tfnu is an o.n. basis of H0 pDq then tfn ˝ φ u is an o.n. basis of 1 1 H0 pD q. Thus we obtain the following theorem. Theorem If h is a random distribution on D1pDq with the law of the Gaussian free ﬁeld on D, then h ˝ φ´1 is a GFF on D1.

Bob Hough Math 639: Lecture 24 May 11, 2017 43 / 65 Circle average

Let D be a bounded domain, let z P D, and let 0 ă ă dpz, BDq. Let ρz, be uniform measure on the circle of radius centered at z. Notice that

ρz,pdxqρz,pdyqGpx, yq ă 8 2 żD 1 since 0 x log x ă 8, so that ρz, P M . Set ş hpzq “ ph, ρz,q.

Bob Hough Math 639: Lecture 24 May 11, 2017 44 / 65 Circle average

Theorem

Let h be a GFF on D. Fix z P D and let 0 ă 0 ă dpz, BDq. For t ě t0 “ logp1{0q, set Bt “ he´t pzq.

Then pBt , t ě t0q has the law of a Brownian motion started from Bt0 .

Bob Hough Math 639: Lecture 24 May 11, 2017 45 / 65 Circle average

Proof.

Suppose 1 ą 2 and we condition on h outside Bpz, 1q. Thus we 0 0 can write h “ h ` φ where φ is harmonic on U “ Bpz, 1q and h is GFF in U. 0 Then h2 pzq “ h2 pzq ` ψ where ψ is the circle average of φ on the

boundary of Bpz, 2q. By harmonicity of φ, ψ “ h1 pzq. Thus 0 h2 pzq “ h1 pzq ` h2 pzq which proves the independence of the increments. By applying the change of scale, w ÞÑ w´z , which conformally maps 1 0 the outer circle to the unit circle, we see that the increment h2 pzq depends only on r “ 2 . This provides the stationarity. 1

Bob Hough Math 639: Lecture 24 May 11, 2017 46 / 65 Circle average

Proof. To determine the rate of the Brownian motion, it suﬃces to check that the GFF on D satisﬁes Var hr p0q “ ´ log r for 0 ď r ă 1. We have

Varphr p0qq “ GDpx, yqρr pdxqρr pdyq. 2 żD By the mean value property of GDpx, ¨q, this is

Varphr p0qq “ GDpx, 0qρr pdxq “ ´ log r. żD

Bob Hough Math 639: Lecture 24 May 11, 2017 47 / 65 Circle average

Theorem There exists a modification of h such that phpzq, z P D, 0 ă ă dpz, BDqq is jointly Höldercontinuous of order 1 γ ă 2 on all compacts of pz P D, 0 ă ă dpz, BDqq. Note: A proof is given in Duplantier and Sheffield, 2011.

Bob Hough Math 639: Lecture 24 May 11, 2017 48 / 65 Thick points

Deﬁnition Let h be a GFF in D and let α ą 0. We say a point z P D is α-thick if

hpzq lim inf “ α. Ñ0 logp1{q

Bob Hough Math 639: Lecture 24 May 11, 2017 49 / 65 Thick points

Theorem

Let Tα denote the set of α-thick points. Almost surely

α2 dim T “ 2 ´ α 2 ˆ ˙`

and Tα is empty if α ě 2.

Bob Hough Math 639: Lecture 24 May 11, 2017 50 / 65 Thick points

Proof sketch. We sketch only the upper bound. Given ą 0,

Probphpzq ě α logp1{qq “ Probpηp0, logp1{q ` Op1qq ě α logp1{qq “ Probpηp0, 1q ě α logp1{q ` Op1qq 2 ď α {2. a

In the square D “ p0, 1q2, the number of sub-squares of side length with ´2`α2{2 center z and hpzq satisfying the bound is, on average, . A more α2 elaborate argument bounds the Minkowski dimension by 2 ´ 2 a.s.

Bob Hough Math 639: Lecture 24 May 11, 2017 51 / 65 Liouville measure

2 γhpzq γ {2 Given ą 0 deﬁne µpdzq :“ e dz. Theorem

Suppose γ ă 2. Then the random measure µ converges almost surely weakly to a random measure µ, the (bulk) Liouville measure, along the sequence “ 2´k . µ has a.s. no atoms, and for any A Ă D open, we have µpAq ą 0 a.s. In fact,

2 E µpAq “ Rpz, Dqγ {2dz P p0, 8q. żA

Bob Hough Math 639: Lecture 24 May 11, 2017 52 / 65 Liouville measure

Lemma

We have Var hpxq “ logp1{q ` log Rpx, Dq. In particular,

γ2{2 E µpAq “ Rpz, Dq dz. żA

Bob Hough Math 639: Lecture 24 May 11, 2017 53 / 65 Liouville measure

Proof. Calculate

Var hpxq “ Γpρx,q “ ρx,pdzqρx,pdwqGpz, wq ż “ ρx,pdzqGpz, xq. ż Gpx, ¨q “ ´ log |x ´ ¨| ` ξp¨q where ξ is the harmonic extension of ´ log |x ´ ¨| from the boundary. Thus

Var hpxq “ Gpxq “ Gpx, yqρx,pdyq “ logp1{q ` ξpxq. ż The conclusion follows since Gpx, yq “ logp1{|x ´ y|q ` ξpxq ` op1q as y Ñ x.

Bob Hough Math 639: Lecture 24 May 11, 2017 54 / 65 Liouville measure

Let S be bounded and open and set I “ µ pSq. Let ą 0 and δ “ . We ? 2 ﬁrst consider the easier case γ ă 2. Theorem 2 2´γ2 We have the estimate EppI ´ Iδq q ď C . In particular, I is a Cauchy sequence in L2pProbq and so converges to a limit in probability. Along “ 2´k this convergence occurs a.s.

Bob Hough Math 639: Lecture 24 May 11, 2017 55 / 65 Liouville measure

Proof. 2 γ2{2 Let hpzq “ γhpzq ´ pγ {2q Varphpzqq and let σpdzq “ Rpz, Dq . By Fubini

2 hpxq hδpxq hpyq hδpyq EppI ´Iδq q “ E e ´ e e ´ e σpdxqσpdyq. S2 ż ”´ ¯ ´ ¯ı Write

ehpxq ´ ehδpxq ehpyq ´ ehδpyq ´ ¯ ´ ¯ “ ehpxq`hpyq 1 ´ ehδpxq´hpxq 1 ´ ehδpyq´hpyq . ´ ¯ ´ ¯ The three terms now are independent of each other by the Markov property if |x ´ y| ą 2. In this case the expectation vanishes.

Bob Hough Math 639: Lecture 24 May 11, 2017 56 / 65 Liouville measure

Proof. When |x ´ y| ď 2,

2 EppI ´ Iδq q

ď Eppehpxq ´ ehδpxqq2q Eppehpyq ´ ehδpyqq2qσpdxqσpdyq ż|xý|ď2 b ď C Epe2hpxqq Epe2hpyqqσpdxqσpdyq ż|xý|ď2 b γ2 1 p2γq2 logp1{q ď C e 2 σpdxqσpdyq ż|xý|ď2 2 ď C2´γ .

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Lemma

Let X “ pX1, ..., Xnq be a Gaussian vector with law Prob, with mean µ and n covariance matrix V . Let α P R and deﬁne a new probability measure by d Prob1 exα,X y “ d Prob Erexα,X ys

Under Prob1, X is a Gaussian vector with covariance matrix V and mean µ ` V α.

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Proof. It suﬃces to let µ “ 0. The Laplace transform is given by

xλ`α,X y xλ,X y E e E 1 e “ Prob E exα,X y ” ı “ ‰ 1 xα`λ,V pα`λqy e 2 “ ‰ “ 1 xα,V αy e 2 1 xλ,V λy`xλ,V αy “ e 2 .

The term xλ, V λy corresponds to a Gaussian of variance V . The linear term xλ, V αy indicates the mean is V α.

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? Now consider γ P r 2, 2q. Let α ą 0 be fixed. Define ‘good’ event α G pxq “ thpxq ď α logp1{qu. Lemma (Liouville points are no more than γ-thick) For α ą γ we have hpxq E e 1 α , 1 p p G pxqq ě ´ p q where the function p may depend on α and for a fixed α ą γ, ppq Ñ 0 as Ñ 0 polynomially fast. The same estimate holds if hpxq is replaced by h{2pxq.

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Proof. Note

1 hpxq 1 α d Prob hpxq E e 1 α “ Prob pG pxqq, pxq “ e . G pxq d Prob ” ı 1 By the tilting lemma, under Prob the process Xs “ he´s pxq has the same covariance and its mean is γ CovpXs , Xt q. 1 Thus, under Prob , Xs is Brownian motion with drift γ. It follows that the probability that Xt ě αt is exponentially small in t for t large, or polynomially small in . Changing to {2 shifts t by log 2. The conclusion is the same.

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Fix α ą γ and introduce

hpxq 1 h{2pxq J “ e 1Gpxqσpdxq, J{2 “ e 1Gpxqσpdxq żS żS α with Gpxq “ G pxq. By the previous lemma, Ep|I ´ J|q ď ppq|S| and 1 Ep|I{2 ´ J{2|q ď ppq|S| also tends to zero.

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Lemma 1 2 r We have EppJ ´ J{2q q ď for some r ą 0. In particular, I is a Cauchy sequence in L1 and so converges to a limit in probability. Along “ 2´k , this convergence occurs almost surely.

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Proof.

Observe that if |x ´ y| ě 2 then the increments hpxq ´ h{2pxq and hpyq ´ h{2pyq are independent of each other and also the σ-algebra generated by h outside the balls of radius about x and y, in particular of Gpxq and Gpyq. By Cauchy-Schwarz,

1 2 E pJ ´ J{2q ´ ¯ 2hpxq 2hpyq ď C Epe 1Gpxqq Epe 1Gpyqqσpdxqσpdyq. |x´y|ď2 ż b

Bob Hough Math 639: Lecture 24 May 11, 2017 64 / 65 Liouville measure

Proof. Observe

2 2hpxq ´γ Epe 1Gpxqq ď Op1q Qphpxq ď α log 1{q

2h x where has density e p q . Q Ere2hpxqs By the tilting lemma hpxq is a normal random variable with mean 2γ logp1{q ` Op1q and variance log 1{ ` Op1q. Thus

p2γ ´ αq2 ph pxq ď α log 1{q ď Op1q exp ´ log 1{ . Q 2 ˆ ˙

2 1 2 1 2 2´γ 2 p2γ´αq Thus E pJ ´ J{2q ď Op1q . Choosing α suﬀ. close r to γ makes´ this ă ¯for some r ą 0.

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