SLE Seminar

7/26/2019 SLE Seminar

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Schramm Lwner Evolutions

Manjunath Krishnapur

Indian Institute of Science


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ABSTRACT. Lecture notes from SLE seminar in Summer and Fall of 2010 at IISc,Bangalore. Mainly, we follow (some chapters of) Wendelin Werners St. Flour lecturenotes on SLE, but filling in most proofs. Much of this latter is in Greg Lawlers bookfrom which we borrow a lot of things. Some proofs are a little different. However, wewant to keep it streamlined to a minimal introduction to SLE. In particular, we stickto chordal versions of all theorems.


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Contents

Chapter 1. Complex analysis background 11.1. Boundary behaviour of conformal maps 11.2. The standard setting 21.3. Brownian motion 31.4. Harmonic measure 41.5. Beurlings theorem 5

1.6. Schwarzs lemma and its many variants 71.7. Half-plane capacity 81.8. A heuristic derivation of Lwners equation 91.9. Closeness of conformal maps 111.10. Chordal version of Lwners differential equation 131.11. Continuously growing hulls 141.12. Lwner evolution in reverse 15

v


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CHAPTER 1

Complex analysis background

1.1. Boundary behaviour of conformal maps

When does a conformal map on a region extends continuously to the boundarypoint? We present two results - Schwarzs reflection principle which applies to gen-eral holomorphic function, and Theorem1.1.3which applies only to conformal maps.

Theorem 1.1.1(Schwarzs reflection principle). Let f: +

be analytic and sup-

pose thatIm f(z) 0 as Imz 0. Then, f extends as an analytic function toandthe extension satisfies f(z) =f(z).

Needless to say, one can also perform reflection along other curves, for example ifI is a circular arc and fis an analytic function on such that Imf(z) 0 (or if|f(z)| 1)asz I, then fextends as an analytic function acrossIby reflection. Thiscan be deduced from Theorem1.1.3by mapping Ito a line segment by a conformalmap. For example, ifI , then the map w i 1+w1w works.Definition 1.1.2. Let be a region. We say that zn converge to along acurveif there exists a curve : [0,1] such that (t) if t < 1 and (1) = and(tn) =znfor somet 1 < t2 < ....

Declare (zn)(wn) if the interlaced sequence z1, w1,z2, w2,... converges to aboundary pointalong a curve. We call the equivalence classes as prime ends. (The

usual definition is more generally applicable, but we need only this much. This isbecause boundary points that we consider will have at least one prime end corre-sponding to them).

Theorem 1.1.3(Rudin 14.18). Let be a bounded s.c. region. Let f : be anyRiemann map (one-one onto conformal map).

(1) If zn along a curve, then f(zn) has a limit that belongs to =. Thelimit is the same for all sequences in the same prime end.

(2) The limits thus obtained are different for distinct prime ends.

Remark 1.1.4.The proof of Theorem1.1.3given in Rudins book depends on a resultthat a bounded holomorphic function f on has radial limits at a.e point of thecircle. This is a result of interest in itself and is proved using maximal functions (seechapter 11 of Rudin). For the purpose of proving Theorem 1.1.3,it clearly suffices toshow thatfor a.e. point on the circle, there is some curve converging to it from insidethe disk, along which the limit of f exists. This fact will fall out of other probabilisticconsiderations later.

Riemann mapping theorem asserts that given a s.c. , there does exist aone-one onto conformal map f : . The space of all such conformal maps isthree-dimensional, since the group of conformal automorphisms of the disk is. Forexample, to get uniqueness, specify interior points z0 and 0 and ask f(z0) = 0

1


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2 1. COMPLEX ANALYSIS BACKGROUND

and f(z0) > 0. Any three constraints should do as well, and as a consequence ofTheorem1.1.3, these constraints can be on the boundary.

Corollary 1.1.5. Let be a bounded s.c. region. Let f : be any Riemann map(one-one onto conformal map). Then the conclusions of Theorem1.1.3are valid, with

=R{}in place of.PROOF. (z) := zi

z+i maps onto conformally and is a continuous bijectionfrom onto . Apply Theorem1.1.3to fand tranfer results back by 1.

1.2. The standard setting

Definition 1.2.1. A bounded set K is said to be a hullifK=K and \K iss.c. Let r(K) = sup{|z| :z K}.Example 1.2.2. +, (0, i] and (1,1+ i]

(1, 1+2i] etc, are hulls while{1+ i y : y > 0}

or 2 i + or{z :x 1, x y 1}etc., are not.

Theorem 1.2.3(Riemann mapping theorem). Let K be a hull. Then, there exists aunique conformal map : \K such that (z) = z + a

z+ O(z2) as z . The

number a is non-negative.

PROOF. By Riemann mapping theorem, there is a conformal map :\K.Corollary 1.1.5 asserts that t:= () is a well-defined number in R{}. Post-compose by z z

t1z1 which maps conformally onto itself to get a conformal :\K such that () = .

Since K is bounded, for all real x with|x| >x0=r(K), as z x from above,we get f(z) R from above. Schwarzs reflection principle applies and extendsanalytically to \(K

K) and the extension maps (,x0]

[x0,) into R. We want

to take the power series expansion ofat. More precisely, set F(w) = 1/(1/w)for|w| < 1/r(K) and write F(w) = b1w + b2w2 + ... as F(0) = 0. Since Fmaps reals toreals,b iare all real numbers.

Thus, for |z| > r(K), we can write (z) =F(1/z)1 = a1z + a0 + a1z1 + a2z2 + . . .where a i are real, which is what we mean by the power series of at. Moreovera1 > 0 or else would map parts of upper half plane into the lower half plane. Asa1 > 0 anda0is real,w (wa0)/a1maps conformally onto itself. Compose with to get the conformal map (z) = z + a

z+O(z2) wherea = a1/a1. The uniqueness

is also clear.It remains to show that a 0. This is really a boundary version of Schwarzs

lemma. For z0 , let z0 (z) := (z z0)/(z z0) is a Riemann map from onto theunit disk taking z0to 0. Let w0:=(z0). Then ifg :=w0 1z0, then g1 maps into with g(0) = 0. Hence by Schwarzs lemma,|(g1)(0)| 1 with equality if andonly ifg(z) =z. Thus,

1 |g(0)| = |w0 (w0)||(z0)||z0 (z0)| =(Imz0) |

(z0)|Im(z0) =1

+a y

2

0 +O(y

3

0 )

1 a y20 +O(y30 )where in the end we chose z0=i y0. Letting y0 , we see that a must be non-negative. Further, a = 0 if and only ifKis empty. Remark 1.2.4. Schwarzs lemma (or Schwarz-Pick lemma) has the following mean-ing. Consider the disk with the hyperbolic metric (1|z|2)2(dx2 + d y2). If f: is analytic, the push-forward of the metric is |f

(z)|2(1|f(z)|2)2 (dx

2 + d y2). Schwarz-Pick


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1.3. BROWNIAN MOTION 3

lemma is the statement that any analytic function f : is a contraction in thehyperbolic metric, and if it is not a strict contraction at some z, then fis a Mbiustransformation of the disk onto itself, which is an isometry of the hyperbolic metric.

It is easy to check that with the metric dx2+d y2y2

is isomorphic to the the disk

with its hyperbolic metric, and hence Schwarz-Pick lemma gives that for any holo-

morphic g :, we have |g(z)|Img(z) 1Imz which is what we used above.

Example 1.2.5. Suppose K=[x,x + it] where x is real and t>0. Then K(z)=(z x)2 + t2 where the principal brach of square root

rei = rei/2 for (0,2).

We can write the expansion K(z) = z + t2/2z

+ O(z2). For K= r+:= {z : |z| r}we can see that K(z) = z + r

2

z. A not so trivial exercise is to find Kfor the oblique

slit [0, te i] for some 0 < 0.

1.3. Brownian motion

What is Brownian motion? We give a quick definition and a convenient way tovisualize it. It can be shown that Brownian motion does exist.

Definition 1.3.1. Brownian motion in the plane is W=(Wt)t0 is a collection ofcomplex-valued random variables on a probability space such that (a)W0 = 0 w.p.1.(b) For any t1< t2< . . . < tk the random variables Wti Wt i1 are independent, andfor any s 0, imagine a particlewhich picks a random direction and moves a distance

2hin a time duration ofh,

then picks a random direction again and moves a distance

2hetc. As h 0, thisrandom trajectory converges to Brownian motion in a precise sense.

What we really use in the sequel are the following basic properties of Brownianmotion.

(1) Symmetries: e iW() d= W() (rotation invariance). W(r) d= rW() (scale in-variance). W

d= W(reflection invariance).(2) Strong Markov property: Let be a stopping time. Then, conditional on

(Wt)t0, the future path W(+)W() is a standard Brownian motion.Our interest in Brownian motion for now is because of its usefulness as a tool toprove things about harmonic functions and related things! Here is the basic resultthat connects Brownian motion to harmonic functions and the Dirichlet problem.

Theorem 1.3.2. Let be a region such that c has a connected componentcontaining more than one point1. Let := inf{t : Wt }. Let f : Rbe a boundedmeasurable function.

(1) Pz( < ) = 1 for any z and u(z) := E [f(W)] is well defined and har-monic in .

(2) Assume that f is continuous at a point and that there is a connectedset K c that contains and has more than one point. Then u(z) f()as z .

1Here and elsewhere, we think of as a subset of the sphere

{}, so is included in c .


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(3) Assume that for every , there is a set K as in (2). Then, for anyf C(), the function u is the unique solution to the

Dirichlet problem: Find v C() such that v = 0in , and v =f on.PROOF. (1) We leave the part about Pz( < ) = 1 to Exercise1.3.3. Fix zand

r such that z + r . Define the stopping time T:= inf{t: |Wt z| = r}. By strongMarkov property ofWwe getE

u(z)

WT= u(WT). Take expectations again to getu(z) = E [u(WT)] =

20 u(z + rei) d2 by rotation invariance ofW. Thus uhas mean-

value property. Therefore u = 0 in .(2) Fix > 0 and pick< dia(K) so that |f()f()| < for any B(,). If

z , then Pz

WexitsB(,) before hitting K 0 [Why?]. On the complement of

this eventW B(,) and hence |f(W)f()| < . Taking expectations, we getE [|f(W)f()|] +2fPz

WexitsB(,) before hittingK

0as z . That is u(z) f(), henceu C() and u =f on.

(3) u is a solution by part (2). Uniqueness is by maximum principle.

Exercise 1.3.3. Let be a region such that c has a connected componentcontaining more than one point and let := inf{t:Wt}. Then, Pz( < ) = 1 forany z . [Hint: By symmetry, there is a positive probability that Wstarted at apoint inside (z, r) exits the disk through a given arc I (z, r). Try to constructa sequence of disks and arcs so that ifWexits those disks in those arcs, then it hits

c].

Remark 1.3.4.Martingales are another way to link harmonic functions to Brownianmotion! Suppose uis a harmonic function on . Then, using the mean value prop-erty ofu, one can check thatu(Wt) is a (bounded and continuous) martingale. By themartingale convergence theorem there is a random variable Xsuch that u(Wt)

a.s.X.This means that for a.e Brownian trajectory starting at 0, the limit ofu along the

trajectory exists. In particular, for a.e point on

, there is at least one trajectoryconverging to it, along which the limit ofu exists. This (applied to Ref and Imf)is precisely what we needed to complete the proof of Theorem1.1.3(see the remarkfollowing that theorem)!

1.4. Harmonic measure

Definition 1.4.1. Let be a region such that c has a connected componentcontaining more than one point. Then := inf{t: B t} is finite w.p.1. Define thez(A) = Pz(W A) for Borel set A. We call z theharmonic measure on in as

seen from z. It is a probability measure supported on .

An alternative definition is to define for each z the linear functional Tz onCb() by Tzf= u(z) where f Cb() and u is the unique solution to the Dirichlet

problem with boundary value f(the solution is unique for every bounded continuousf, by part (3) of Theorem1.3.2). ThenTzis a positive linear functional with Tz1 = 1,whence by Riesz representation theorem,2 there is a unique Borel probability mea-surez on such that Tzf=

f dz for all f Cb(). That this definition agrees

with the earlier one follows immediately from Theorem1.3.2.

2By the footnote on the previous page, is a compact subset of the sphere

{}and hence anycontinuous function f on is also bounded. Of course, then fmust be continuous at also to startwith.


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1.5. BEURLINGS THEOREM 5

Example 1.4.2. On ,z(dt) =P(z,eit )dtwhere P (z,eit ) = 121|z|2

|zeit |2 is the Poisson

kernel for the unit disk. On ,z(dt) = y(x

t)2

+y2

dt where z =x + i y. On +can you

findz([1,1])?These examples are atypical. Usually we cannot find z exactly but only esti-

mate it. The next section will introduce one such estimate which will be of great useto us.

1.5. Beurlings theorem

IfKis a compact subset of the plane we let K:= {|z| :z K}be itsradial projec-tion.

Theorem 1.5.1(Beurlings projection theorem and estimate). Let K be compactand let V=\K and let V =\K.

(1) For any z we haveVz(K) V

|z|(K).

(2) If K is connected,0 K and K, then

V

z(

) c0 |z|.PROOF. (1) By rotating we may assume that z = |z|. Let L = {xe i :x

R}. Apply Lemma1.5.2with the lines L/2,L/4,L/8,. .. successively toget setsK1 =R/2(K),K2 =R/4R/4(K1),K3 =R/8R/8(K2) etc. with

Vjz (Kj) increasing in j. Then, Kjis confined to the sector {2j argz 2j}.

Vjz (Kj) being the probability for a BM started at zto hit Kj before

exiting , it can be proved that it converges to V

z (K).

(2) Clearly K= [0,1]. Although part (a) applies to compact subsets of thedisk, in the form Vz() V

z () it applies equally toK . ThusVz(K)

[0,1]r ([0,1]) and one can explicitly compute that

[0,1]r ()= 4arctan

r

c0

r.

Lemma 1.5.2. Let z and let K be a compact subset of. Let L be a line passingthrough0 and let RL denote reflection across L. Let K+ be the part of K on the same

side as z and let K be the part of K on the other side. Let K:= KRL(K+). Set

V=\K and V=\K. Then,Vz(K) Vz( K).

PROOF. [Bernt ksendal] Without loss of generality, we assume Imz > 0 so thatK+=K+ and K=K. We just write A for RL(A). IfH:= K+

(K\K+),

then H K while H= K. Thus, proving the lemma for H implies the lemmafor K. Thus we may assume that K+ K= . In addition, one can show fromproperties of Brownian motion that ifKn K, then Pz(Whits Knbefore ) convergestoPz(Whits Kbefore ) and similarly for decreasing limits. Thus we may assumethat K+ and K are compact and that the mutual distances between K+,K and[

1,1] are strictly positive.

Let A = K =K+

K and B = K =K+

K and let I= [1,1]. Start a BM at zand run it till it hits

Kand successively record which of the sets A,I,Kare hit.

For example, A I A K denotes the set of all paths that hit A first, then I, then AandthenK. Then,

(1.5.1) Vz(K) = Pz(K)+ Pz(AK)+ Pz(IK)+ Pz(A IK)+Pz(I AK)+ Pz(A I AK)+ . . .The sum is over all strings in which A and Ialternate a number of times and is

terminated by K. Similarly, we can writeVz( K) by recording which of the sets B,I


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and Kare hit. But now, B can only be followed by I, so we get

(1.5.2) V

z( K)=

Pz(I K)+

Pz(BI K)+

Pz(IB I K)+

Pz(BI BI K)+

. . .

We show that each term here is equal to one of the terms in(1.5.1). More precisely,

Claim : Pz(I K) = Pz(IK), Pz(BI K) = Pz(A IK), Pz(IB I K) = Pz(I A IK), etc.

For clarity of explanation, we first show that Pz(I K) = Pz(IK). Indeed, write I K=IK

IK+ and I K=IK

IK+ with the obvious meaning for the new symbols. For

any path IK+, let =inf{t :t I} and define a new path (t) := (t)1t +R((t))1t>. Then, IK+. Thus, Pz(IK+) = Pz(IK+), by the strong Markov prop-erty. ThatPz(IK) = Pz(I K) follows.

More generally, say to show that Pz(IB I K) = Pz(I A IK), let V=\K andU=V\Iand write

Pz(AI A IK)=A IA I

Uz(dw1)

Uw

1

(dt1)Vt

1

(dw2)Uw

2

(dt 2)Vt

2

(K)

and similarly

Pz(BI BI K) =

B

I

B

I

Uz(dw1)

Uw1

(dt1)Vt1

(dw2)Uw2

(dt2)Vt2

(K).

By considering reflection as before, Vt2

(K) = Vt2 (K) for any t2I. Further, as B =R(A), the integrals are also equal (make a change of variables from w ito R(wi)).

Beurlings estimate is very useful to us later. Here is one example that we uselater.

Corollary 1.5.3. Let be a s.c. domain (other than the whole plane). Then, for any

z , if d = dist(z,c), then for any x > 0we have

Pz (W travels at least distance xd away from z before exiting ) c0x

.

PROOF. Let with|z | = d. Consider B(,xd) with K= . By assump-tion that is s.c., it follows that Kis connected and connects to B(,xd) (ifxd issmall enough, which is all we need to consider). Apply Beurlings estimate. A standard application of harmonic measure in complex analysis is to Phragmen-Lindelf type theorems. This may be omitted and is included only for general inter-est.

Corollary 1.5.4. Let f be an analytic function on a neighbourhood of= {0 Rez 1} and suppose that|f(z)| MR:= exp{exp{CR 1}} for some > 0. If|f(z)| M on{Rez

=0or 1}and Then,

|f(z)

| M for all z

.

PROOF. For R > 0, let SR:= [0,1] [R,R] and let u (z) be a harmonic functionon SR whose boundary values are equal to log |f|. Since log |f| is subharmonic, weget log |f(z)| u(z) for any zSR . Fix z and let z(R) denote the harmonic mea-sure of the top and bottom edges of SR , as seen from z. Thus, log |f(z)| u(z)z(R)logMR + logM, by the Brownian motion solution to the Dirichlet problem. Weneed an estimate for z(R). Divide SR into2R squares of side 1 and argue thatz(R) ecR for some c > 0. LettingR we get |f(z)| M.


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1.6. SCHWARZS LEMMA AND ITS MANY VARIANTS 7

1.6. Schwarzs lemma and its many variants

Schwarzs lemma is the well-known statement that a holomorphic f : that fixes 0 satisfies|f(0)| 1 and|f(z)| |z| for all z and strict inequalitieshold everywhere unless fis one-one and onto the disk. In which case it is a linearfractional transformation f(z) = (az +b)/(cz+d) for somea, b, c, d with |a|2|b|2 =1. Pick generalized this statement to say that for any holomorphic f : , wehave

(1.6.1) |f(z)f(w)|

|1f(z)f(w)| |z w||1zw| and

|f(z)|1|f(z)|2

1

1|z|2 for any z , w

with strict inequalities everywhere unless f is one-one and onto the disk. Theseinequalities follow from Schwarzs statement by pre-composing (and post-composing)fwith linear fractional transformation that takes 0 to z (respectively, f(z) to 0).

In the form given by Pick, Schwarzs lemma attains a deep geometric meaning.On the disk, define a Riemannian metric (1

|z

|2)2(dx2

+d y2) called the hyperbolic

metric. This means that an curve :[0,1] has length() =

10(1|t|2)1|t|dt .The length minimizing curves are calledgeodesicsand they may be seen to be exactlyarcs of circles that intersect at right angles (this includes diameters ofwhichare arcs of circles of infinite radius).

If we call each z apointand each geodesic aline, then this system of pointsand lines satisfy the axioms of the hyperbolic geometrydiscovered by Bolyai, Gaussand Lobachevsky. The axioms are the same as that of Euclidean geometry, except forthe parallel postulate, which is replaced by a statement that given a line and a pointnot on it, there exist at least two lines through the point that do not intersect the given

line. We say that the disk with the hyperbolic metric forms a model for hyperbolicgeometry.

Exercise 1.6.1. Show that with the metric y2(dx2 + d y2) is isometric to thehyperbolic metric on . What are the geodesics?

Returning to Schwarz-Pick lemma, the inequalities1.6.1state precisely that anyholomorphic map from the disk to itself is a contraction of the hyperbolic metric. Andif it is not strictly contracting even at one point, then it must be an isometry of thedisk, in other words a linear fractional transformation that maps the disk injectivelyonto itself. Here are some variants of Schwarzs lemma that we shall need.

Lemma 1.6.2. (1) Let f: be holomorphic. Then,|f(z)|

Imf(z) 1

Imz and

f(z)f(w)f(z)f(w) |z w||z w|

with strict inequalities unless f(z) = (az + b)/(cz + d) for some a, b, c, d Rwith ad

bc

=1.

(2) Let f : be any holomorphic function that extends analytically to, andfixes two distinct points a, b . Then, f(a)f(b) 1 with equality if andonly if f is a hyperbolic isometry of.

(3) Let f: be holomorphic function that extends analytically to, and fixes. Suppose f(z) =z+O(1)as z . If f extends analytically to some x Rand f(x) R, then |f(x)| .

PROOF. (1) Let(z) = ziz+i and apply Schwarz-Pick to f1.


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(2) Derive this one from part (3) using a linear fractional transformation thatmapsa and b to 0 and . We have not omitted absolute values by mistake.Both f

(a) and f

(b) must be positive as f maps

(a,) and

(b,)

into .(3) must be positive as f maps into . Without loss of generality take

x = 0 = f(x) and write f(z) = z + O(z2) as z 0 so that f(0) = which isalso positive as f maps into near 0. Now take z = i yand w = iv andapply part (1) to get iy iv +O(1) iy+ iv +O(1)

|i y iv + O(1)||i y+ iv + O(1)| .

The left hand side is 12yv+O(y2/v2) while the right hand side is 1 2y

v+

O(y2/v2) from which it follows that as claimed.

1.7. Half-plane capacity

Definition 1.7.1. For a hull K, Theorem1.2.3yields a unique conformal map K:H\Ksuch that K(z) = z + a(K)z + O(z2) as z . The number a(K) is calledthehalf-plane capacityofKas seen from infinity.

Example 1.7.2. If K= r+, then K(z) =z + r2

z+ O(z2), hence a(K) = r2. IfK=

[0, it], then K(z) =

z2 + t2 =z + t22z + O(z2) whencea(K) = t2/2.Lemma 1.7.3. We have the following properties of half-plane capacity.

(1) If K1 K2are hulls, then a(K1) a(K2)with equality if and only if K1 =K2.(2) a(rK+ t) = r2a(K)if r 0and t R.(3) LetK:= inf{t : Wt K

R}. Then, a(K) = lim

yyEi yImWK

.

PROOF. (1) We saw that a(K) 0 with equality if and only ifK=. Now,

let K3:=

K1 (K2\K1). Then,

K2= K3 K1 from which it follows thata(K2) = a(K1) + a(K3) a(K1) with equality if and only ifK3= which isequivalent toK1 =K2.

(2) True, since rK+t(z) =rK

ztr

+ t = (z t) + r2 a(K)zt + O((z t)2) + t = z +

r2 a(K)z

+ O(z2).(3) Im{K(z) z}is a bounded harmonic function on \Kand equal toImz

for z KR. By Theorem1.3.2we haveEz ImWK= Imz ImK(z). Setz = i y, multiply by yto get yEi y

ImWK

= y2 yy a(K)y

+ O(y2) a(K)

as y .

Exercise 1.7.4. IfKis a hull with r(K) < 1, thena(K) = 20

Eeit [ImB]sin t d t.

With the probabilistic interpretation, we can estimate a(K) using Brownian motion

hitting probabilities which in turn are estimated by Beurlings estimate. Thus wehave the following fact.

Theorem 1.7.5. Let K1, K2 be two hulls.

(1) If K1 K2 and every point of(\K2) (which may be smaller than K2R)

is within distance of K1R. Then a(K1) a(K2) a(K1)+ c0

r(K2)3/2.

(2) K1,K2 general with r(Ki) r. If every point of(\K2)is within distanceof K1

Randvice versa, then |a(K1) a(K2)| c0

r 3/2.


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1.8. A HEURISTIC DERIVATION OF LWNERS EQUATION 9

PROOF. (1) We know that a(K1) a(K2). To show the second inequality,start Brownian motion Wat i yand let 2 1 be the hitting times of(2r

2)R, ofK

2Rand ofK

1R, respectively.

Let A:= {|W| = 2r2}. If A c occurs, then W R and 1= 2= andhence ImW2 ImW1= 0. On the event A:= {|W| = 2r2}, we may deducefrom corollary1.5.3that P

ImW2 ImW1>x|A

c0 /xfor x < r2 andequals 0 for x > r2. This is because W2 is within distance of some pointinK1

R.

Further, P i y(A) r2y as y , hence using the probabilistic interpre-tation of half-plane capacity,a(K2) a(K1) is the limit as y of

yEi yImW2

yEi y ImW1 = yPi y(A)E i y ImW2 ImW1 A

c0r2r2

0

x

dx = c01/2r3/22 .

(2) Immediate upon applying the first part to the pairsK1

K1K2and K2 K1K2.

Example 1.7.6. Let Kt= {ei : 0 < t} for t < 1 and K1=+:= {z : |z| 1}.Then, a(Kt) is continuous in t, including t=1. Had our condition in part (1) ofTheorem1.7.5said K2

R instead of (\K2), continuity at t= 1 would not have

followed!Suppose is a simple curve in with 0 R. Then, \[0, t] has a unique un-

bounded component whose complement we denote by Kt. Then Kt is an increasingfamily of hulls. Again, a(Kt) is continuous in t. This too follows easily from Theo-rem1.7.5because(\Kt) [0, t] andis continuous. A special case is when is asimple curve witht Rfor any t > 0 and0 R. In this case, Kt = [0, t].Remark 1.7.7. There is a well-known quantity called logarithmic capacity. Thereare many ways to define it, but for a connected compact set K, its capacity c(K)turns out to be equal to |()| where is a Riemann map from \Konto :=\with () = . Here() is the reciprocal of the coefficient a in the expansion(z) = az + b + cz1 + .. . near .

IfKis a hull in the upper half-plane, is a(K) related to c(K) or perhaps c(K

K)?Ifmaps \Konto \+, then it extends to a map from \(K

K) onto . The

expansion at may be written as K(z) = az +b+cz1+.... Then K(z) = a1((z)+(z)1)b/a. This has the expansionK(z) =z+a1(c+a1)z1+O(z2) which showsthata(K) = a1 c+a2. Of coursec(KK) = a1. There does not seem to be any simplerelationship betweena(K) and c(K

K).

1.8. A heuristic derivation of Lwners equation

Let t be a simple curve with (0)

Rand (t)ift

>0. Let gt be conformal

maps from \[0, t] . We want to understand how the maps gtvary with t , theultimate goal being to derive Lwners differential equation

gt(z) =at

gt(z)Utwhere at= a([0, t]), Ut=g t(t) and at,gt denote time (t) derivatives. Among sev-eral issues raised by this statement, we have addressed a few. For exampleUt iswell-defined because tcorresponds to only one prime end in \[0, t]. What about


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differentiability ofa t? Theorem1.7.3shows that a t is strictly increasing and Theo-rem1.7.5shows that a t is continuous. Therefore, can always be reparameterizedso that a

tis smooth. In particular, if we set

s =

a1

(2s), defined for s

0we have gt+h = Gh gt. Fromthis, it is clear that is also a simple curve in its natural parametrization (see theproof of the first part of Lemma1.7.3). If we heuristically replace [0, h] by a verticalslit [Ut,Ut + i2

h] having the same half-plane capacity 2h, then by Example1.2.5,

we get g t+h(z)

(gt(z)Ut)2 +4h. Differentiate by h and seth = 0 to get

gt(z) = 2

gt(z)Utwhich is exactly Lwners equation! If you have been able to calculate the conformalmaps for oblique slits asked for in Example 1.2.5,then check that replacing [0, h]by any oblique slit with the same half-plane capacity 2hwill also result in the same

differential equation.The gaps in this heuristic are that we only calculated the right derivative of gt

above and that we did not justify how to replace [0, h] by a slit. Both these issuesare addressed by proving several estimates in the next section about how conformalmaps Kdiffer if the hulls Kdiffer only a little. Then we put together the proof insection1.10.


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1.9. CLOSENESS OF CONFORMAL MAPS 11

1.9. Closeness of conformal maps

We develop some required estimates in this section. The first theorem makes

precise in what sense the approximations K(z) zor K(z) z+ a(K)z1 hold. Thelast one gets an estimate for how much a set near Kcan be blown up by K.

Theorem 1.9.1. Let K be a hull and Kas before. Then,|K(z) z| 3r(K) for allz \K.

PROOF. By scaling it suffices to assume thatr(K) < 1 and prove that |K(z)z| 3 for all z \K. For any fixed x 1 (or x 1) we claim that K(x) is increasing(respectively, decreasing) in K, as long as r(K) < 1. Assuming this claim, apply it tothe hulls K+to get

x K(x) x +1

x for x 1 and x K(x) x +

1

x for x 1.

In particular,

|K(x)

x

| 3 for

|x

| 1. For z

+\K, we must have K(z)

[K(1),K(1)] [2,2]. Thus |zK(z)| 3. Thus we have shown that limsup|K(z)z| 3 whenever z (\K) (note that the limit value is 0 as z ). By maximummodulus principle, |K(z)z| 3 for any z \K.

It remains to prove the claim that K(x) L(x) for x 1 ifKLwith r(L) < 1.If J= K(L\K), then L= JK. Hence it suffices to show that K(x) x fora single hull K. The map K(z) K(x) +x fixes x and and has derivative 1 at. Applying part (3) of Lemma1.6.2to 1

K we get

K(x) 1. For large y, K(y) =

y+ a(K)y1 + O(y2) > y. Therefore K(x) >xfor all x > 1. Theorem 1.9.2. If K is a hull, then

K(z) z a(K)z c 0a(K)r(K)|z|2 whenever|z| >

16r(K).

The proof we give is exactly as in Lawler. Later we describe an attempt at a

slightly different reasoning that did not quite work. Do let me know if you see howto fix it.

PROOF. Scale and assume that r(K) = 1. Let g(z) = K(z) z aKz and let v =Img. By harmonicity ofv and Theorem1.3.2,we write

v(z) = Ez [Im{K(W)W}] a(K)Im(1/z) = Ez [ImW] a(K)Im(1/z)whereis the hitting time ofK

R by the Brownian motionW. Ifdenote the hitting

time ofR, then by the strong Markov property, we can write

v(z) = Ez

Eeit [ImW] a(K)Im(1/z)

=

0

Eeit [ImW]

p(z,eit )2

sin t

dt

where p(z,eit ) is the density ofWon and by writinga(K) as in Exercise1.7.4. InLemma lem:hittingofcircle we calculate p(z,eit ) explicitly and show that p(z,eit ) =Im(1/z) 2sin t (1+O(1/|z|)). Hence, we get

|v(z)| c0 Im(1/z)1

|z|

20

Eeit [ImW]2

sin td t c0a(K)

Im(z)

|z|3 .

.


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Now, consider|w| 2. Then the disk (w, |w|/2) is contained in {|z| >1} andhence, we can use Poisson integral formula to write for |w w| 1 and := inf{t : Wt R}. Let p(z,eit )be the density of W

conditional on W . Then, p(z,eit ) = Im(1/z) 2sin t (1+ O(1/|z|))for all t [0,].PROOF. Let |z| > 1. For any arcI, the harmonic measure z(I) isthe value at

zof the harmonic function which has boundary values 1 on Iand 0 on R

{}(\I).By conformal invariance of harmonic functions, we can map it forward by z z+1/zwhich takes to [2,2]. By knowledge of the Poisson kernel on, the hitting densityofW, started from w = z + 1/zis given by p(s) = Im w

|ws|2 for s R. Pulling it back tothe original problem, we find that

p(z,eit

) = Im(z

+z1)z +z1 2cos t 2

2sin t

which can easily be seen to be equal to Im(1/z) 2

sin t (1+ O(1/|z|)). Remark 1.9.4. Taking z = i yin Lemma1.9.3and letting y , in conjunction withpart (3) of Lemma1.7.3solves Exercise1.7.4

Theorem 1.9.5. Let K be a hull and let :[0,1] be a curve such that (0) KRand (t)\K for t(0,1]. Let d be the diameter of and let :=supt Imt.

Assume that d10. Then, there exists a constant c 0 such that diaK()

c0

d .

The main idea in the proof is to replace diameter by a conformally invariantquantity that is comparable to the diameter of a set. The conformally invariantquantity we use will be the harmonic measure. For instance, the length of an interval

on the line can be estimated by the hitting probability by a Brownian motion startedfar off, that is, if= inf{t : Wt R}, then lim

yyPi y(W [a, b]) =1 (ba). The following

exercise provides the needed generalization of this.

Exercise 1.9.6. Then there exists c 1, c2such that for any connected hull Kwe have

c1dia(K) limyyPi y(W [a, b]) c2dia(K)

where is the hitting time ofKRby W.


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1.10. CHORDAL VERSION OF LWNERS DIFFERENTIAL EQUATION 13

PROOF. [Proof of Theorem1.9.5] The idea is as follows. LetWbe a Brownianmotion started at i yfor a large y and letbe the hitting time ofK

R. We use an ar-

gument (analogous to what is needed for the upper bound in Exercise1.9.6)to boundthe probability thatWhitsbefore time. This probability is the harmonic measureofas seen from i y in \K. By conformal invariance of harmonic functions, it isthe same as the harmonic measure ofK() as seen from K(i y) in . Then usingthe lower bound in Exercise1.9.6, we get the desired bound for dia(K()).

Now for details. Let L=K(0,10) and let be the hitting time of LRby W. It s easy to prove that Pi y(W 0 + )c0y1 (left as exercise). IfW 0 +, then there is no hope forWto hit before time. On the other hand,ifW = 0 +then we claim that P(Whitsbefore time) c0

d/. This

follows by Beurlings estimate but with a small twist. See Exercise1.9.7.Puting everything together, we get yP iy(Whitsbefore time) c0

d. We

get the same bound for yPK(i y)(Whits K() before time) by conformal invarinceof harmonic functions. Since K(i y) = y + O(y1), using the lower bound in Exer-cise1.9.6we get dia(K()) c0

d.

Exercise 1.9.7. Let Kbe a connected set in the plane connecting 0 to . Then forany , we have P(Whits before hitting K) c0

.

1.10. Chordal version of Lwners differential equation

Let be a simple curve in with0 = 0 andt for t > 0. Let gt:\[0, t] be the unique conformal map such that gt(z) = z + atz+ O(z2) as z . Hereat= a([0, t]) is the half-plane capacity. Then, by Theorem 1.7.5 at continuous.Lemma1.7.3asserts that it is also strictly increasing, as is simple. Thus, thereis a unique way to reparameterize so that at= 2t for all t. We call this thestan-dard parameterization.

Theorem 1.10.1 (Lwner). Let be a simple curve in standard parameterization andgt the associated conformal maps normalized hydrodynamically. Then, Ut:= g t(t)is continuous in t and tgt(z) = 2gt(z)Ut for z \[0, t].

PROOF. Let Kt= [0, t]. Fix T> 0 and consider t 2r(Kt,s).If we fix and z \Ks and let ts, then the last condition holds for t close

enough tos. Divide (1.10.1)by t sand let t sto get gs(z) = 2gs(z)Us . Here we used


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the continuity ofgt(z) andUtin t. Similarly, we can fix z \Kt, divide by s tandlets tin (1.10.1). The condition |gt(z)Ut| > 2r(Kt,s) holds again for s close enoughtot and we get g

t+(z)

= 2

gt(z)Ut. Thus we have proved Lwners differential equation

gt(z) = 2gt(z)Ut . Remark 1.10.2. In the course of the proof we claimed that the continuity ofUt intfollows fromgt gs C,T

dia([t,s]). As a matter of fact, this inequality can

be used to prove the existence ofUt:= limgt(z) as z t in \Kt without havingto invoke the theorem on boundary values of conformal maps. We leave this as anexercise or to look up the proof of Lemma 4.2 in Lawlers book.

Remark 1.10.3. Suppose Kt is a one-parameter family of hulls such that Kt Ksfor t 0Kt,t+ = {Ut}for every t. We then say that the

family of hulls is right continuous. It is clear that a right continuous family of hullscan be reparameterized so that at:= a(Kt) = 2t. Then the proof of Theorem1.10.1goes through exactly as stated above. In particular, Utis continuous in t and gt(z) =2gt(z)Ut for any z \Ktand any t.

Uis called the driving function.

1.11. Continuously growing hulls

SupposeKtis a one-parameter family of hulls such that Kt Ksfor t 0Kt,t+ = {Ut}. We then say that the family of hulls is right continuous with

driving function U.Froma(Ks) = a(Kt)+ a(Kt,s) and Theorem1.7.5we see that a(Kt) is continuous.

Hence a right continuous family of hulls can be reparameterized so that a t:= a(Kt) =2t.

In this case, one half of the proof of Theorem 1.10.1goes through exactly asstated above.

Proposition 1.11.1. Let K be right-continuous with driving function U . Assumethat the K is in its natural parameterization. Then Ut isright continuous in t andgt+(z) = 2gt(z)Ut for any z \Kt and any t.

PROOF. As usual it suffices to check this att = 0 withU0 = 0. For h > 0 and z K2hwe have |gh(z)| |gh(z)z|+|z| which is bounded by 3r(Kh)+ r(K2h) which goesto zero as h 0. SinceU0 = 0 andUh Kh,2h =g h(K2h), it follows thatUh U0 0as h 0. This shows right continuity ofU. The right derivative of gt can be foundexactly as before. Exercise 1.11.2. Let be a curve in and let Kt be the hull generated by [0, t],

that is \Ktis the unique unbounded component of\[0, t]. Assume that Kt Ksfor any t


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1.12. LWNER EVOLUTION IN REVERSE 15

taket = 2it for t 1 andt = iei(t1) for 1 t 1+2. Att0 = 1+, the curve crossesitself. There are two prime ends corresponding to the boundary point 1 of\K1+andU

t0andU

t0are the limits ofg

t0(z) as z

1 along these two prime ends.

Definition 1.11.3. IfK is right continuous with driving functionU, and t Ut iscontinuous, then we say that K is a continuously growing family of hulls.

Since this definition puts in exactly what is needed to make the proof of Theo-rem1.10.1, we get

Proposition 1.11.4. Let K be a continuously growing family of hulls with drivingfunction U. Assume that the K is in its natural parameterization. Then gt(z)=

2gt(z)Ut for any z \Kt and any t.

1.12. Lwner evolution in reverse

We start with a converse to Proposition1.11.4.Theorem 1.12.1. Let Ut be a continuous real valued function with U0= 0. Thenthere exists a unique continuously increasing family of hulls K that is naturally

parameterized and whose driving function is U .

PROOF. Consider the differential equation xt = 2/(xt Ut) in the complex plane.For each z, let Tz be the maximal time for which the solution exists, starting withthe initial condition x0 = z. This unique solution we denote by gt(z). Thus g0(z) = zandtgt(z) = 2gt(z)Ut fort [0, Tz). Clearly, Tz = inf{t :g t(z) = Ut}.

Let Kt:= {z : Tz t}. By continuity of solutions in the initial conditions, Ktis closed in . IfUt = max{Us:s t}, then the bound on the speed| gt(z)| = 2/Utimplies that Kt is bounded. Further, if z Kt, then gs(z) Kt for any s


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Definition 1.12.2. Let B be a standard one-dimensional Brownian motion. For 0, letKtbe the increasing family of hulls with driving function Ut:=

Btas assured

by Theorem1.12.1. The family of random hulls Kis called SLE().

It is a non-trivial theorem of Rohde and Schramm that SLE()is generated bya random curve as in Exercise 1.11.2. Usually that curve is called SLE(). Formany results, it suffices to think of SLE()as a family of increasing hulls, but weshall anyway assume the result of Rohde and Schramm without proof and simplytalk of SLE()curves. Sometimes, the end result, K (or ) is what we care about.It is a hull that connects 0 to(if K, then K would be bounded and henceat = 2twould be bounded above!).

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