4 Conformal Maps

4 Conformal maps

4.1 Deﬁnition, Riemann mapping theorem We start by restating a deﬁnition we made before.

Definition 13 Let D and D0 be open subsets of R2. A map f : D D0 is said to → preserve angles if for every two differentiable curves γ1 and γ2 in D defined on the time interval ( , ) which intersect at t = 0, the angle formed by their tangents at γi(0) is − 0 equal to the angle formed by the tangents to f γ and f γ at f(γi(0)). A conformal map from D to D0 is a one to one, onto, differentiable◦ function◦ that preserves angles.

Let f(x, y) = (u(x, y), v(x, y)) be a differentiable map. We leave it to the reader to show that it preserves angles at a point if and only if its derivative (which is a 2 by 2 matrix) is equal to a positive constant times a rotation matrix. So there is an a > 0 and a θ such that ∂u ∂u cos(θ) sin(θ) ∂x ∂y = a − (67) ∂v ∂v sin(θ) cos(θ) ∂x ∂y This implies that the Cauchy-Riemann equations are satisfied. Conversely, if the Cauchy- Riemann equations are satisfied and the derivative at the point is not zero, then one can show there is an a > 0 and θ such that the above is true. So the map preserves angles. Thus a map is a conformal map if and only if it is a one to one, onto analytic function of D to D0. Note if f is a conformal map of D onto D0, then f −1 is a conformal map of D0 onto D. There is a special family of conformal maps - the linear fractional transformations. They are of the form az + b f(z) = (68) cz + d where a, b, c, d are complex numbers with ad bc = 0. Linear fractional transformations − 6 map circles onto circles if we think of lines in the plane as circles. We will use D to denote the open unit disc with center at the origin and H to denote the upper half plane. The Riemann sphere Cˆ = C is the complex plane with the point at added and the usual topology. Linear fractional∪ ∞ transformations are conformal maps of C∞ˆ onto itself. The linear fractional transformations that map D onto D are of the form z w f(z) = eia − , w D, a R (69) 1 wz ∈ ∈ − The extended half plane Hˆ is H with added. The linear fractional transformations that ∞ map Hˆ onto Hˆ are of the form az + b f(z) = , a, b, c, d, R, ad bc > 0 (70) cz + d ∈ −

38 a conformal map of the unit disc D to the upper half plane H is z i f(z) = − (71) z + i All the conformal maps of D onto H are obtained by following this map with a conformal map of H onto itself.

Theorem 14 (Schwarz lemma) If f : D D is analytic with f(0) = 0 then f(z) z for z D. If equality holds for a nonzero→z, then f(z) = eiθz for some real θ.| | ≤ | | ∈ Proof: Apply the maximum principle to f(z)/z. We will use “domain” to mean an open subset of the complex plane. A domain is simply connected if it does not have any holes:

Deﬁnition 14 A domain D is simply connected if the region bounded by every simple closed curve in D is contained in D, i.e., every simple closed curve in D may be continuously contracted to a point without leaving D. Equivalently, D is simply connected if Cˆ D is connected. \ Recall that Cauchy’s theorem says that if D is simply connected and f is analytic on D and γ is a diﬀerentiable closed curve in D, then

f(z)dz = 0 (72) Zγ We can take the logarithm of a non-zero analytic function on a simply connected domain.

Proposition 5 Let D be a simply connected domain, f an analytic function on D which never vanishes. Then there exists an analytic function g on D such that f = eg.

To see that the simply connected hypothesis is needed, consider f(z) = z on the plane with the origin removed. There is no way to deﬁne log(z) on this region. You must have a branch cut somewhere. We will use the following theorem on a daily basis.

Theorem 15 (Riemann mapping theorem) Let D be a simply connected region which is not all of C and let w D. Then there is a unique conformal transformation f of D onto the unit disc D such that∈ f(w) = 0 and f 0(w) = 0.

39 Corollary: Any two simply connected domains have a conformal map between them. A proof can be found in Lawler’s book or any ﬁrst year graduate complex variables book. We do not give it in full, but there is a consequence of the proof we will need later so we give the main idea. The proof considers the collection of conformal maps h of D to D with h(w) = 0 and h0(w) > 0 which are not necessarily onto. The proof shows that there is an f in this collection which maximizes h0(w). One then shows that this f is onto. Thus for any conformal map h from D into D with h(w) = 0 and h0(w) > 0 we have h0(w) f 0(w). One of the≤ amazing aspects of this theorem is that it does not require any smoothness of the boundary. The boundary need not even be a curve. If we want to extend the conformal map so that it maps the boundary of D onto the boundary of D then we need some condition on the boundary of D

Deﬁnition 15 A closed set K C is locally connected if for every > 0 there is a δ > 0 ⊂ such that if z, w K with z w < δ then there exists a connected set K1 K with z, w K and diam∈ (K ) < |. − | ⊂ ∈ 1 1 Theorem 16 Let D be a bounded simply connected domain. Let f be a conformal map of the unit disc D onto D. Then f has a continuous extension to D if and only if C D is locally connected. \

We will not give a proof. A reference for a proof may be found in Lawler’s book. The boundary of a simply connected domain need not be the image of a curve. If it is the image of a continuous curve, then ∂D is locally connected. With a little more work one can then show that C D is locally connected. Conversely, if C D is locally connected, \ \ then by the above theorem and the Riemann mapping theorem there is a conformal map of D onto D which extends continuously to the boundary of D. Consider the curve f(eit) where 0 t 2π. It will traverse the boundary of D. Thus the boundary is a curve if ≤ ≤ and only if C D is locally connected. Consider the\ slit disc D = D [0, 1) where [0, 1) denotes the subset of the complex \ plane consisting of real numbers in the interval. This is a simply connected domain and its boundary is a curve. However, the points along the slit are traversed twice when we traverse the curve that gives the boundary. A closed curve γ : [a, b] C is a Jordan → curve if γ is one to one on [a, b). A bounded domain is a Jordan domain if its boundary is a simple curve. For a Jordan domain D, if f is a conformal map of D onto D, then the continuous extension of f to D will be one to one.

Proposition 6 Let D,D0 be Jordan domains. 0 0 0 0 (i) Let z1, z2, z3 ∂D and z1, z2, z3 ∂D . Both sets of points are oriented clockwise. ∈ ∈ 0 0 Then there is a unique conformal map f of D onto D such that f(zi) = zi, i = 1, 2, 3. (ii) Let z ∂D, w D and z0 ∂D0, w0 D0. Then there is a unique conformal map f ∈ ∈ ∈ ∈ of D onto D0 such that f(z) = z0 and f(w) = w0.

40 Proof: Let f be a conformal map of D onto D and g a conformal map of D0 onto D. 0 D D So f(zi) and g(zi) belong to ∂ . If h is a conformal map of onto itself such that 0 −1 0 h(f(zi)) = g(zi) for i = 1, 2, 3, then g h f is the desired map of D to D . The conformal maps of D onto itself are linear fractional◦ ◦ transformations of the form (69). We leave it as an exercise to show that a suitable choice of a and θ gives the needed h. Roughly speaking, the family of conformal maps from one simply connected domain to another has three real degrees of freedom. In (i) they are determined by three real constraints. In (ii) the constraint f(w) = w0 is a complex constraint and so uses two real degrees of freedom. Another common way to impose a real constraint is to require that the derivative at some interior point be positive as we did in the Riemann mapping theorem. We end this section with a useful result to extending analytic functions. If f(z) is analytic, then it is easy to check that f(z) is too (deﬁned on the obvious domain). We would like to use this fact to take an analytic function deﬁned on a subset of the upper half plane and extend it to the refected domain in the lower half plane. Clearly f(z) will need to be real on the part of the real axis in the original domain if this is to work.

Theorem 17 (Schwarz reﬂection principle) Let D be a domain which is symmetric about the real axis. Let D = D H. Let f be a function which is continuous on D , analytic + ∩ + in D+ and real valued on the set of reals in D. Then f may be extended to an analytic function on all of D which satisﬁes f(z) = f(z).

4.2 Univalent functions A function is called univalent if it is analytic and one to one. Let be the set of simply A connected domains not equal to the entire plane which contain the origin. Let ∗ denote the set of univalent functions on D with f(0) = 0 and f 0(0) > 0. The RiemannSmapping theorem says that f f(D) gives a one to one correspondence between ∗ and . We let denote the set of→f ∗ with f 0(0) = 1. S A SA compact hull K is a∈connectedS compact subset of C which contains more than one point and such that C K is connected. Let K0 be the image of K under the map z 1/z. 0 \ → 0 Then C K is a simply connected set. So there is a conformal map fK of D onto C K \ 0 \ with fK (0) = 0 and fK (0) > 0. Deﬁne 1 FK(z) = (73) fK (1/z)

C D C 0 Then F is a conformal map of onto K with limz→∞ FK(z)/z = fK (0) > 0. The (logarithmic) capacity of K is \ \

0 FK (z) cap(K) = log fK (0) = log lim (74) − z→∞ z

41 Exercise: Show that if K and K are compact hulls with K K , then cap(K ) 1 2 1 ⊂ 2 1 ≤ cap(K2). Show that if the two capacities are equal then K1 = K2. Hint: see the remarks on the proof of the Riemann mapping theorem.

The power series expanion of fK about the origin shows that the Laurent expansion of FK is of the form ∞ b F (z) = bz + b + n (75) K 0 zn n=1 X where b = ecap(K).

Proposition 7 (Area theorem) Let K be a compact hull containing the origin with cap(K) = 0. Then the coeﬀecients in the Laurent expansion above satisfy

∞ area(K) = π 1 n b 2 (76) − | n| " n=1 # X In particular, ∞ n b 2 1. n=1 | n| ≤ P Proof: If γ is a simple, closed smooth curve, oriented clockwise and A is the region enclosed by γ, then by Green’s theorem we have

z dz = [(x dx + y dy) + i(x dy y dx)] = 2i dx dy = 2i area(A) (77) − Zγ Zγ ZA

Let r > 1 and let γ be the image of the circle of radius r under the map FK . Then γ encloses a region which contains K and in the limit r 1+, the area of the region → iθ enclosed by γ converges to the area of K. The curve γ is parameterized by FK(re ) with 0 θ 2π. So ≤ ≤ 2π ∞ zdz = F (reiθ) ireiθ F 0 (reiθ) dθ = 2πi[r2 n b 2r−2n] (78) K K − | n| γ 0 n=1 Z Z X Letting r 1+ gives the proposition. → Proposition 8 If f , then there is an odd function h such that h(z)2 = f(z2) for z D. ∈ S ∈ S ∈ Proof: Consider f(z)/z. Since f(0) = 0, the singularity at 0 is removable and so f(z)/z is analytic. Since f(z) is not zero for z = 0 and f 0(0) = 0, f(z)/z is not zero. Thus f(z)/z has a square root, i.e., there is an analytic6 function g6 (z) on D such that g(z)2 = f(z)/z. Deﬁne h(z) = zg(z2). Then h(z) is odd. And h(z)2 = z2g(z2)2 = f(z2). Clearly, h(0) = 0 0 and h (0) = 1. It remains to be shown that h is one to one. If h(z1) = h(z2) then

42 2 2 2 2 f(z1 ) = f(z2 ). Since f is one to one, this implies z1 = z2 . So either z1 = z2 or z1 = z2. The oddness of h rules out the second case. − Let f . Then the power series expansion of f about the origin is of the form ∈ S n f(z) = z + anz (79) n=2 X The study of bounds on the an is a famous problem. Proposition 9 (Bierbach) If f then a 2. ∈ S | 2| ≤ Proof: Let h be the odd function in given by the above proposition so that h(z)2 = S f(z2) for z D. Let g(z) = 1/h(1/z). Then the Laurent expanion of g(z) begins with ∈ a g(z) = z 2 + (80) − 2z · · · So by the area theorem a 2. | 2| ≤ Bierbach conjectured that an n for all n 2. This was proved by de Branges in 1985. (Note that a = 1 by the| deﬁnition| ≤ of .) ≥ 1 S Let f . Let r < 1. Then f(rD) is an open set containing the origin. So it contains a disc about∈ Sthe origin. Amazingly, there is a lower bound on the radius of this disc which is uniform over f . We will use B(a, r) to denote the disc of radius r centered at a. ∈ S Theorem 18 (Koebe 1/4 theorem) If f and 0 < r 1, then B(0, r/4) f(rD). ∈ S ≤ ⊂ Proof: We prove it for r = 1. The other values of r follow by considering the function f(rz)/r. Let z / f(D). We must show z 1/4. Deﬁne 0 ∈ | 0| ≥ z f(z) fˆ(z) = 0 (81) z f(z) 0 − This is a composition of f(z) and a linear fractional transformation, so it is one to one. Let f(z) = z + a z2 . By the Bierbach theorem, a 2. The expansion of fˆ(z) is 2 · · · | 2| ≤ 1 2 fˆ(z) = z + (a2 + )z + (82) z0 · · · Applying the Bierbach theorem to fˆ we have a +1/z 2. Combining this with a 2 | 2 0| ≤ | 2| ≤ leads to z 1/4. | 0| ≥ Theorem 19 (Growth and distortion theorems) Let f and z D. Then ∈ S ∈ z z | | f(z) | | (83) (1 + z )2 ≤ | | ≤ (1 z )2 | | − | | 1 z 1 + z − | | f 0(z) | | (84) (1 + z )3 ≤ | | ≤ (1 z )3 | | − | |

43 4.3 Half plane capacity We continue to use H to denote the upper half plane. We do not include the real axis, so this is an open set.

Deﬁnition 16 A bounded subset A of H is a “compact H-hull” if A = H A and H A ∩ \ is simply connected.

Proposition 10 If A is a compact H-hull, then there is a unique conformal map gA : H A H such that \ → lim [gA(z) z] = 0 (85) z→∞ − Proof: The Riemann mapping theorem says there exist a conformal map g of H A onto H which maps to itself. Since A is bounded it is contained in a ball B(0, r) of\ radius r about 0 for some∞ r. Consider U = z : z > r . On H U, g is an analytic function { | | } ∩ that can be continued to the boundary. It must map the boundary to the boundary of H, i.e, the real axis. So by the Schwarz reflection principle, g(z) = g(z) defines an analytic continuation of g to all of U. So if we let f(z) = 1/g(1/z), then f is analytic on z : z < r . g( ) = implies f(0) = 0. So the power series expansion of f(z) about {the origin| | is}of the∞ form∞ ∞ n f(z) = anz (86) n=1 X which implies g has an expansion about of the form ∞ ∞ −n g(z) = b−1z + b0 + bnz (87) n=1 X A little thought shows that since f maps parts of the real axis onto the real axis, all the b must be real. If we let g (z) = (g(z) b )/b− , then g satisfies (85). We leave it to i A − 0 1 A the reader to prove uniqueness.

Deﬁnition 17 Let A be a compact H-hull. Let gA be the unique conformal map given by the proposition. So the Laurent expansion is of the form

∞ b g (z) = z + n (88) A zn n=1 X The half plane capacity, hcap(A), is b1.

Proposition 11 Let A1 A2 be compact H-hulls. Then hcap(A1) hcap(A2) with equality only if the sets are⊂equal. If A is a compact H-hull and r > 0, x≤ R then ∈ hcap(rA) = r2 hcap(A), hcap(A + x) = hcap(A) (89)

44 Exercise: Prove the above theorem. Example: Let g(z) = √z2 + 1. Then g is a conformal map of H (0, i] onto H which satisfies the normalization in the definition of half plane capacity\. From g(z) = z + 1/(2z) + we see that hcap((0, i]) = 1/2. · · · Exercise: The goal of this exercise is to see what happens to the capacity of the previous example when we rotate the segment (0, i]. Let 0 < α < 1 and define

f(z) = [z + 1 α]α[z α]1−α (90) − − Show that f(z) conformally maps H onto H A where A is a line segment from 0 to reiαπ \ for some r. Let g(z) = f −1(z), so g conformally maps H A onto H. Show that g(z) satisﬁes (85). Find r as a function of α and from this compute\ the capacity of the segment of length 1 from 0 to eiαπ. You can ﬁnd the answers in example 3.37 in Lawler’s book. The next proposition gives an expression for the half plane capacity involving complex Brownian motion. We use Im to denote imaginary part.

Proposition 12 Let A be a compact H-hull. Let Bt be a complex Brownian motion starting at z. (z will be in the upper half plane). Let τ be the hitting time for R A. Let g be the conformal map used to deﬁne the half plane capacity of A. Then for z ∪ H A, A ∈ \ z Im(z) = Im(gA(z)) + E [Im(Bτ )] (91)

The half plane capacity is given by

iy hcap(A) = lim yE [Im(Bτ )] (92) y→∞

If rad(A) < 1, then π 1 iθ hcap(A) = Ee [Im(B )] sin(θ) dθ (93) π τ Z0

Proof: In chapter 6 we will show that if φ is a bounded harmonic function, then φ(Bt) is a martingale. The real and imaginary parts of an analytic function are harmonic. Thus φ(z) = Im[z gA(z)] is a harmonic function, and so φ(Bt) is a martingale. It is also bounded. So b−y the optional sampling theorem,

Ezφ(B ) = Ezφ(B ) = φ(z) = Im(z) Im(g (z)) (94) τ 0 − A Consider φ(B ) = Im(B ) Im(g (B )). By the deﬁnition of τ, B is on the boundary τ τ − A τ τ of H A. So gA(Bτ ) is on the bounday of H, i.e., it is real and so Im(gA(Bτ )) = 0. This proves\ the ﬁrst part of the theorem. Taking z = iy in the above we have

EiyImB = y Im(g (iy)) (95) τ − A 45 As y , g (iy) = iy + hcap(A)/iy + O(1/y2). So → ∞ A y[y Im(g (iy))] = hcap(A) + O(1/y) (96) − A The second statement in the proposition follows. Now suppose rad(A) < 1. For y > 1, we start a Brownian motion at iy and let σ be the hitting time for the boundary of H D. Let p(iy, z) be the probability density for Bσ where z ∂(H D). Then the strong Mark\ ov property implies ∈ \ π iy eiθ iθ E [Im(Bτ )] = E [Im(Bτ )] p(iy, e ) dθ (97) Z0 It can be shown (lot of work) that on the boundary of the circle, 2 p(iy, eiθ) = sin(θ)[1 + O(y−1)], 0 < θ < π (98) yπ which leads to the last equation in the proposition. The deﬁnition of half-plane capacity says that

hcap(A) g (z) = z + + O(1/z2) (99) A z The next proposition will give some uniformity to this statement over A.

Proposition 13 There is a constant c such that for all compact H hulls A, for z | | ≥ 2rad(A) we have hcap(A) rad(A)hcap(A) g (z) z c (100) | A − − z | ≤ z 2 | |

Proof: We prove it for rad(A) = 1 and leave the general case as an exercise. Exercise: Use a scaling argument and the theorem for rad(A) = 1 to prove the theorem for general A.

46 This page intentionally left blank for the proof of the previous thm.

47 4.4 Loewner differential equation We recall one of the key ideas of SLE. Given a simple curve in the upper half plane starting at the origin, the half plane with the image of the curve up to time t removed is a simply connected domain. So there is a conformal map gt from this “slit” half plane onto the half plane. If the curve is random, then this one parameter family of conformal maps gt is random. SLE studies the random curve by studying the random conformal maps gt. The following theorem says that gt satisfies a differential equation known as Loewner’s equation. Note that the theorem is deterministic; there is no randomness yet.

Theorem 20 Let γ be a simple curve with γ(0) = 0 and γ(0, ) H. Let b(t) = ∞ ⊂ hcap(γ[0, t]), and suppose that b(t) is C 1 and b(t) as t . Let H = H γ[0, t]. → ∞ → ∞ t \ This is a compact H-hull, so we can let gt be the unique conformal map of Ht onto H such that as z , → ∞ hcap(γ[0, t]) 1 g (z) = z + + O( ) (101) t z z 2 | | Then for z H, g (z) is the solution of ∈ t b˙(t) g˙ (z) = , g (z) = z (102) t g (z) U 0 t − t where U = g (γ(t)). If z H γ(0, ), then the diﬀerential equation holds for all t 0. t t ∈ \ ∞ ≥ It z = γ(t0), then the diﬀerential equation holds for t < t0 and

lim− gt(z) = Ut0 (103) t→t0

Partial Proof: The following proof has a couple of gaps. We will only consider a right sided derivative in (102) and we will need to make a continuity type assumption that while plausible is non-trivial to prove. The gaps may be ﬁlled in by consulting Lawler’s book. It can be shown that C γ[0, t] is locally connected. (This is a non-trivial baby analysis \ −1 exercise.) So if we let f = g , then f has a continuous extension to H. Let U R be t t t t ∈ such that ft(Ut) = γ(t). (How do you know Ut is unique?) Fix s > 0. We want to compute

˙ gs+(z) gs(z) gt(z) = lim − (104) →0 We will only compute the right-sided derivative, i.e., we restrict to positive in the above limit. We look at the small portion of the curve from time s to s + and consider its image under gs: A = gs(γ[s, s + ]) (105)

48 This is a curve starting at the point U on the real axis. We have g = g g . Letting s s+ A ◦ s w = gs(z), we have gs+(z) gs(z) = gA(w) w. We are going to apply (13) to gA. The − − 0 set A is very small, but it is not close to the origin. So we define A = A Us. It is trivial 0 − to check that gA0 (z) = gA(z + Us) Us. We also note that hcap(A ) = hcap(A). The bound (13) says that for w 2rad−(A0) we have | | ≥ hcap(A) rad(A0)hcap(A) g 0 (w) w c (106) | A − − w | ≤ w 2 | | and so hcap(A) rad(A0)hcap(A) g (w + U ) w U c (107) | A s − − s − w | ≤ w 2 | | By a trivial change of variables this gives hcap(A) rad(A0)hcap(A) g (w) w c (108) | A − − w U | ≤ w U 2 − s | − s| for w U 2rad(A0). And so | − s| ≥ hcap(A) rad(A0)hcap(A) g (z) g (z) c (109) | s+ − s − g (z) U | ≤ g (z) U 2 s − s | s − s| for g (z) U 2rad(A0). We have | s − s| ≥ hcap(γ[0, s + ]) = hcap(γ[0, s]) + hcap(A) (110) So hcap(A) lim = b˙(s) (111) →0+ 0 + We assume that rad(A ) 0 as 0 . For z H γ[0, s], gs(z) will be in H and so → →0 ∈ \ we will have gs(z) Us 2rad(A ) for small enough . Thus we have proved that the Loewner equation| (102)− |holds≥ for the right sided derviative. What if b(t) = hcap(γ[0, t]) is not C1 ? b(t) is strictly increasing and it can be shown to be continuous (not trivial). So if we assume it goes to as t , then it has a ∞ → ∞ continuous inverse defined on [0, ). Define ∞ γ(˜t) = γ(b−1(2t)) (112)

Then hcap(γ(˜t)) = 2t. With this parameterization the Loewner equation is just 2 g˙ (z) = , g (z) = z (113) t g (z) U 0 t − t We will call gt arising from this equation a Loewner chain with driving function Ut. A gt arising from the diﬀerential equation in the theorem will be called a generalized Loewner chain.

49 The real valued function Ut in the above theorem is known as the “driving function.” The previous theorem starts with a simple curve and produces a driving function Ut. Now we go the other way. We assume we have some continuous real valued function Ut and see what we can get out of the Loewner equation. If Ut is only continuous, then we need not get a simple curve. The following theorem is a special case of theorem 4.5 in Lawler’s book.

1 Theorem 21 Let Ut be a continuous real-valued function on [0, ). Let b(t) be a C real-valued function on [0, ) which is increasing. For z H, let g∞(z) be the solution of ∞ ∈ t b˙(t) g˙ (z) = , g (z) = z (114) t g (z) U 0 t − t The denominator can go to zero, causing the solution to fail to exist after some finite time. Define T = sup t : g (z) exists (115) z { t } Define H = z H : T > t , K = z H : T t = H H (116) t { ∈ z } t { ∈ z ≤ } \ t Then g (z) is the unique conformal map of H onto H such that g (z) z 0 as z . t t t − → → ∞ Furthermore , b(t) 1 g (z) = z + + O( ), z (117) t z z 2 → ∞ | |

Proof: By standard ODE theorems, gt(z) is analytic in z and continuous in t when the solution exists. We show next that for each t, gt(z) is one to one. Fix z = w and let ∆ = g (z) g (w). From the dif. eq. we have 6 t t − t b˙(t)∆ ∆˙ = − t (118) t (g (z) U )(g (z) U ) t − t s − s So d ln(∆ ) b˙(t) t = − (119) dt (g (z) U )(g (w) U ) t − t t − t We solve this dif. eq. using the initial condition ∆ = z w. This gives 0 − t b˙(s) ∆ = (z w) exp − ds (120) t − − (g (z) U )(g (w) U ) " Z0 s − s s − s # This shows ∆ = 0, and so g is one to one. t 6 t

50 Next we need to show the range of gt is H. We split gt into its real and imaginary parts, gt(z) = rt(z) + ivt(z) where rt and vt are real valued. Then we have

b˙(t)(r (z) U ) r˙ (z) = t − t (121) t (r (z) U )2 + v (z)2 t − t t and b˙(t)v (z) v˙ (z) = − t (122) t (r (z) U )2 + v (z)2 t − t t ˙ Note that vt(z) is always decreasing and on the real axis vt(z) = 0. We leave it to the reader to show that gt(z) cannot leave the upper half plane. So gt(Ht) is a subset of H. To show gt(Ht) is all of H we consider the dif. eq. run backwards in time. Fix t > 0 and define ht(z) by b˙(t s) h˙s(w) = − − , h0(w) = w (123) h (w) U − s − t s Now the imaginary part of hs(z) will always increase, so the solution to this differential equation will exist for all time, in particular for 0 s t. Now G = h − satisfies ≤ ≤ s t s the original Loewner equation and has Gt(w) = h0(w) = w and G0(z) = ht(z). So gt(ht(z)) = w. Finally, consider the dif. eq. for large z,

b˙(t) b˙(t) b˙(t) g˙ (z) = = + O( g −2) = + O( z −2) (124) t g (z) U g (z) | t| z | | t − t t

Since g0(z) = z, this implies

t t b˙(t) b(t) g (z) z = g˙ (z) ds = [ + O( z −2)] ds = + O( z −2) (125) t − s z | | z | | Z0 Z0 which verifies (117). Suppose Kt is a one parameter family of compact H-hulls that is increasing in the sense that K K for t < s. Can we find a real valued driving function such that the t ⊂ s above proposition produces the given Kt. The following example shows the answer is no. Exercise: Define H = z H : z > 1 (126) 1 { ∈ | | } The conformal map of H1 onto H is 1 g (z) = z + (127) 1 z Now define K = z H : z t (128) t { ∈ | | ≤ } 51 Find gt and show that there is no real valued Ut such that the Loewner equation produces this Kt.

In the above example, Kt grows at all points along its boundary. A natural question is what condition(s) must the Kt satisfy to come from a Loewner chain. One might guess that K must always grow locally, that is, the diameter of K K must converge to zero t t+δ \ t as δ 0. This is not true as the following example shows. → Example: Arc pinching oﬀ half disc.

Lemma 1 Let gt be the Loewner chain with driving function Ut. Deﬁne

Rt = max √t, sup Us (129) { 0≤s≤t | |}

Then K B(0, 4R ). Moreover, if z 4R , then t ⊂ t | | ≥ t g (z) z R , for 0 s t (130) | s − | ≤ t ≤ ≤ Proof: Fix t. If z 4R , then the bound (130) implies | | ≥ t g (z) U = (g (z) z) + z U z g (z) z U 4R R R = 2R (131) | s − s| | s − − s| ≥ | | − | s − | − | s| ≥ t − t − t t This shows that g (z) is defined for s t. In fact, by continuity it shows it is defined for s ≤ s slightly greater than t. So z Ht, i.e., z / Kt. Thus (130) implies Kt B(0, 4Rt). So it suffices to prove (130). ∈ ∈ ⊂ Suppose there is a time s < t with g (z) z R . Let σ be the first such time. So | s − | ≥ t gs(z) z < Rt for s σ. By the above this implies g˙s(z) 1/Rt. Since g0(z) = z, this |implies−that| ≤ | | ≤ σ t gσ(z) z < Rt (132) | − | ≤ Rt Rt ≤ where we have used t R2. But this contradicts g (z) z R . ≤ t | σ − | ≥ t The lemma says that initially Kt grows at the origin. Now we use the lemma to study the growth at other times. Fix s > 0. We saw that Ks+t Ks need need not be small for small t. Consider instead g (K K ). We will use the\lemma to show its diameter s s+t \ s goes to zero as t 0. Let U˜ = U , g˜ = g g−1 Then g˜ satisfies the Loewner → t s+t t s+t ◦ s t equation with driving function Ut and the initial condition g˜0(z) = z. So the lemma says K˜t B(0, R˜t), where R˜t is defined using U˜t. In particular diam(K˜t) 0 as t 0. We claim⊂ that K˜ is g (K K ). By definition K˜ is the set of z H for→which the→solution t s s+t \ s t ∈ g˜t(z) does not exist at time t. This is equivalent to saying that the solution gt+s does not exist if we use the initial condition g−1(z). So

K˜ = z H : g−1(z) K (133) t { ∈ s ∈ t+s}

52 −1 Note that g (z) is always in Ht, i.e., not in Kt. So the above set equals

z H : g−1(z) K K = g (K K ) (134) { ∈ s ∈ t+s \ t} s t+s \ t

In fact there is a rather succint characterization of those Kt that can arise from a Loewner chain. A sketch of the proof of the following can be found in Werner’s St. Flour notes, p. 23.

Theorem 22 An increasing family Kt of compact H-hulls comes from a Loewner chain if and only if the following are true. hcap(K ) = 2t for t 0. And for all T > 0 and t ≥ > 0 there is a δ > 0 such that for t T there exists a bounded connected set S Ht with diam(S) such that S disconne≤cts K K from inﬁnity in H . ⊂ ≤ t+δ \ t t The above has all been the chordal case. We state but do not prove the analog of theorem 21 for the radial case.

Theorem 23 Let Ut be a continuous function on [0, ) which takes values on the unit circle. Let b(t) be a C1 real-valued function on [0, ) ∞which is increasing. For z H, let ∞ ∈ gt(z) be the solution of U + g (z) g˙ (z) = g (z)b˙(t) t t , g (z) = z (135) t t U g (z) 0 t − t Deﬁne T = sup t : g (z) exists (136) z { t } Deﬁne D = z D : T > t (137) t { ∈ z } D 0 Then gt(z) is the unique conformal map of Dt onto such that gt(0) = 0 and gt(0) > 0. Furthermore , 0 ln(gt(0)) = b(t) (138)

Theorems 21 and 23 produce increasing families of sets, Kt. If the driving function is nice enough, Kt will be a simple curve. In the chordal case Ut will be √κBt where Bt is a standard Brownian motion. Interestingly, it turns out the smoothness of Brownian motion is right at the borderline of “nice enough.” So the deterministic theorems about the Loewner equation will not say anything about SLE. As we will see, whether Kt is a simple curve depends on the value of κ. Even when Kt is not a simple curve, it will turn out that is generated by a non-simple curve in the following sense.

Deﬁnition 18 We will say that the Loewner chain gt is generated by the curve γ : [0, ) H if for each t, the domain H of g is the unbounded component of H γ[0, t]. ∞ → t t \ 53 For example, when κ = 6, Kt is not a simple curve. The percolation exploration process corresponds to the curve that generates Kt. We end this chapter with a couple of examples of Ut for which the Loewner equation may be solved explicitly. 2 If gt(z) = √z + 4t, then gt solves the Loewner equation with Ut = 0. The hull Kt is just the vertical line segment from 0 to 2ti. Let 0 < α < 1. Deﬁne

φ(z) = [z + (1 α)]α[z α]1−α (139) − − Then φ is a conformal map of H onto H minus a line segment starting at 0 and forming an angle of (1 α)π with the positive real axis. The length of the segment depends on − α. By computing this length one ﬁnds that if we let

4t α(1 α) f (z) = φ(z − ) (140) t α(1 α) 4t s − r −1 and let gt = ft then gt satisﬁes the Loewner equation and Kt is a line segment starting at 0 and forming an angle of (1 α)π with the positive horizontal axis whose length grows − with t. The driving function works out to be Ut = cαt where cα = 2(2α 1) α(1 α). Note that c is positive for α < 1/2 while it is negative for α > 1/2. − − α p