Introduction to SLE Lecture Notes
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1 Inroducion o SLE Lecure Noe May 13, 16 - The goal of hi ecion i o find a ufficien condiion of λ for he hull K o be generaed by a imple cure. I urn ou if λ 1 < 4 hen K i generaed by a imple curve. We will prove hi fac in hi ecion. Propoiion. If here exi a coninuou funcion v : R + R + uch ha v() = and for all we have hen ˆ ɛ (g 1 ) (λ + iy) dy v(ɛ), β() := lim g 1 ɛ (λ + iɛ) exi and i coninuou, and he hull are generaed by a curve, namely β(). Proof. lim y g 1 (λ + iy) exi becaue he lengh of γ := g 1 (i[, y]) i finie. Preciely, we can prove hi by conradicion; if lim y g 1 (λ + iy) doe no exi, hen γ ha a lea diinc limi poin. γ ravel back and forh beween hee wo poin, o i ha infinie lengh. Moreover, β() g 1 (λ + iɛ) v(ɛ),. Since he map g 1 (λ + iɛ) i coninuou (Exercie), δ ɛ >.. g 1 (λ + iɛ) g 1 (λ + iɛ) v(ɛ) 1
2 if δ ɛ. Then β() β() β() g 1 (λ +iɛ) + g 1 (λ +iɛ) g 1 (λ +iɛ) + β() g 1 (λ +iɛ) 3v(ɛ). If v(ɛ) a ɛ, hen β i coninuou. Remark 1. A remark here i ha, if g 1 (λ + iy) Cy α for ome α < 1, all and y < 1, hen he aumpion in he propoiion above i fulfilled becaue ˆ ɛ (g 1 ) (λ + iy) dy C ˆ ɛ Then β i coninuou and generae he hull. y α dy = C ɛ 1 α. Theorem (By Marhall-R, Lind, R-Tran, Zinmeier). If λ 1 < 4 for all and, hen he L.E. generae imple curve. Moreover, he curve i a quaiconformal arc. Noe. 1. λ 1 < 4 λ() λ() < 4 1 for all and.. If φ : D D analyic, hen φ (z) 1 z. 3. g aifie he ODE ġ = g λ bu g 1 doe no aify any ODE. Inead we can wrie i in erm of a PDE, namely g(, g 1 z (z)) = z. 4. B T B T i a Brownian moion.
3 Proof. (Skech) The idea here i o how he Holder coninuiy: α = α( λ ) < 1.. (g 1 ) (λ + iy) C y α. Then by remark 1 and propoiion above we have he deire concluion. Fir we ry o e up he formula for he backward flow of g. Fix ime T >, le λ = λ T (The ilde migh be dropped laer). Conider he backward flow, g T (z). By he chain rule, g T (z) = Thu if aifie he ODE f (z) = g T (z) λ T. wih he iniial condiion f f (z) λ (z) = z, hen f T = g 1 T. Noe ha i i no rue in general ha f = g 1. Now we apply a ranlaion o ha λ ay fixed, namely, Z = X +iy = f (z) λ T. Then we have f (z) = (Z + λ ) = = (X iy ). Z X + Y Taking he real and imaginary par we have (X + λ ) = X X + Y (1) Y = Y X + Y >, () ince Y >. Noe ha () implie we are flowing upward a we hoped. 3
4 Now conider log f (z). log f (z) = Re log f (z) = Re log f (z) = Re f (z) f (z) I follow ha, = Re z f (z) f (z) = Re z f (z) λ f (z) = Re (X iy ) (X + Y ) = X Y (X + Y ) f (z) = exp log f (z) = exp = exp ˆ ˆ log f (z) d X Y X + Y Le u = log Y, hen u = Y Y = d X +Y we have ˆ u f (z) = exp u ( Xu Y u ) 1 ( Xu Y u ) + 1 d ( ) X + Y = Re (Quoien rule) Z by (). Applying he ubiuion, ˆ u Wu 1 du = exp u Wu + 1 du, where W u := Xu Y u. If we aume W < k for ome k independen on, hen W u 1 W u+1 x 1 < α for ome α < 1 (Conider he graph of funcion ). Then he x+1 heorem follow from Remark 1 ince ˆ u f (z) = exp Wu 1 u Wu + 1 ˆ u du < exp u αdu = exp(α(u u )) = e α Y log( Y ) = ( Y ) α. Y Now i remain o how our aumpion ha k independen on wih W < k for all i rue. To achieve hi we inroduce lemma a he following. Lemma 1. If λ 1 4 and X =, hen C = C( λ 1 ),.. X CY. 4
5 Lemma. For every λ 1 x y c 1, [, T ], hen x y c for all. < 4 here exi c 1 < c.. if x y = c and Noe. 1. Lemma 1 implie our aumpion above wih α = C +1 C 1.. Lemma implie lemma 1. To ee hi, we ue he reeing clock argumen. Take he la ime, namely, when Z ecape he cone {y > c 1 x}. Then Z aifie he condiion for lemma. By lemma, he concluion of lemma 1 i rue wih C = c. So i uffie o how lemma. 3. Boh lemma hold a long a λ 1 < 4. In cla we gave a proof of lemma for λ 1 = 1. Here we will give a direc proof of lemma 1 for λ 1 <. The proof become raher echnical for λ 1 < 4. Reader who are inereed may go hrough heorem in hi paper for more deail. Proof. (of lemma 1) Le σ = λ 1 < 4. Fir we claim ha if X < σ for ome σ < and all CY, hen Y L for all, where L = min ( 1 C, 4 σ ) >. Fir auming he claim. By equaliy (1), if X >= hen X + λ decreae. By ymmery and conidering he la ime ha X i of he ame ign a X, we have X X + up{ λ λ : [, ]} + σ. 5
6 Therefore he aumpion of he claim i aified wih arbirarily mall C. I flow from he claim ha X Y < σ 4 σ, for greaer han arbirarily mall C. So i i rue for all, which conclude he proof. Nex we how he claim by conradicion. Le = CY. Since L 1/C and Y i increaing, we have L /C Y Y. If he claim were no rue, hen ince Y > = L, here exi a minimal >.. Y and Y L on [, ]. Then i follow from () ha = L Y = 4Y X + Y L M + L L for all (, ), which implie Y Y L( ). Thi conradic o our aumpion ha Y Y = L Y < L( ). 6
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