המחלקה למתמטיקה, בן-גוריון אשנב למתמטיקה ביום שלישי, 3 בינואר, 2017 בשעה 18:30 20:00 באולם 101- ההרצאה פונקציות הרמוניות וספירת מסלולים פשוטים תינתן ע

תמליל

1 המחלקה למתמטיקה, בן-גוריון אשנב למתמטיקה ביום שלישי, 3 בינואר, 2017 בשעה 18:30 20:00 באולם 101- ההרצאה פונקציות הרמוניות וספירת מסלולים פשוטים תינתן על-ידי אריאל ידין תקציר: המושג של פונקציה הרמונית חוזר לעבודות של לפלס ופוריה (במאה ה- 18!) ויש לו חשיבות עצומה בפיסיקה, הנדסה, ומדעים בכללי. בעזרת ההרמוניות השונות ניתן לתאר את כל הגלים האפשריים. זה, למשל, מהווה את הבסיס לקידוד mp3. במאות ה- 19 וה- 20 הכלילו את המושג גם לאוביקטים מתמטיים מודרניים יותר, והחשיבות הגיאומטרית שלו הובנה יותר. אנחנו נשוחח על הגדרה גיאומטרית של פונקציה הרמונית, שהיא כללית למדי. נסביר איך ניתן להשתמש בפונקציות כאלה כדי לספור מסלולים פשוטים - שזו בעיה קשה בפני עצמה. אני אשתדל להסביר את כל המושגים המופיעים בתקציר במהלך ההרצאה.

2 Harmonic Functions Ariel Yadin Ben Gurion University Eshnav, Jan 2017

3 classical notions random walks gambler s ruin counting simple paths

4 heat equation u(x, t) = heat at time t, point x [0, L] Fourier s law + conservation of energy: 2 u = α 2u t x Jean-Baptiste Joseph Fourier ( )

5 heat equation u(x, t) = heat at time t, point x [0, L] Fourier s law + conservation of energy: 2 u = α 2u t x Jean-Baptiste Joseph Fourier ( )

6 heat equation t u = u = 2 x 2 f(x + ε) = f(x) + f (x)ε f (x)ε 2 + O(ε 3 ) f(x ε) = f(x) f (x)ε f (x)ε 2 + O(ε 3 ) f(x + ε) + f(x ε) = 2f(x) + f (x)ε 2 + O(ε 3 ). u(x, t + δ) u(x, t) = δ 2 ( ) ε 2 u(x+ε,t)+u(x ε,t) 2 u(x, t)

7 heat equation t u = u = 2 x 2 f(x + ε) = f(x) + f (x)ε f (x)ε 2 + O(ε 3 ) f(x ε) = f(x) f (x)ε f (x)ε 2 + O(ε 3 ) f(x + ε) + f(x ε) = 2f(x) + f (x)ε 2 + O(ε 3 ). u(x, t + δ) u(x, t) = δ 2 ( ) ε 2 u(x+ε,t)+u(x ε,t) 2 u(x, t)

8 heat equation t u = u = 2 x 2 f(x + ε) = f(x) + f (x)ε f (x)ε 2 + O(ε 3 ) f(x ε) = f(x) f (x)ε f (x)ε 2 + O(ε 3 ) f(x + ε) + f(x ε) = 2f(x) + f (x)ε 2 + O(ε 3 ). u(x, t + δ) u(x, t) = δ 2 ( ) ε 2 u(x+ε,t)+u(x ε,t) 2 u(x, t)

9 discrete geometry - graphs A graph is a collection of vertices (nodes) and edges (connections).

10 graphs: cycle

11 graphs: finite line

12 graphs: Z

13 graphs: Z2

14 graphs: regular tree

15 graphs: hexagonal lattice

16 graphs: notation x y x, y are neigbors deg(x) = the number of neighbors of x paths give a metric (geometry) boundaries D = {x D : y x, y D} x x y y

17 graphs: notation x y x, y are neigbors deg(x) = the number of neighbors of x paths give a metric (geometry) boundaries x D = {x D : y x, y D}

18 graphs: notation x y x, y are neigbors deg(x) = the number of neighbors of x paths give a metric (geometry) boundaries D = {x D : y x, y D}

19 graphs: notation x y x, y are neigbors deg(x) = the number of neighbors of x paths give a metric (geometry) boundaries D = {x D : y x, y D}

20 harmonic function Definition A function is harmonic at x if 4 f(x) = 1 deg(x) f(y) y x

21 harmonic function Definition A function is harmonic at x if 4 f(x) = 1 deg(x) f(y) y x

22 harmonic function Definition A function is harmonic at x if f(x) = 1 deg(x) f(y) y x

23 harmonic function Definition A function is harmonic at x if f(x) := 1 deg(x) (f(y) f(x)) = 0 y x

24 harmonic functions maximum (minimum) principle boundary conditions uniquely define harmonic functions Liouville s Theorem: A bounded function harmonic on all of Z d is constant.

25 harmonic functions maximum (minimum) principle boundary conditions uniquely define harmonic functions Liouville s Theorem: A bounded function harmonic on all of Z d is constant.

26 harmonic functions maximum (minimum) principle boundary conditions uniquely define harmonic functions Liouville s Theorem: A bounded function harmonic on all of Z d is constant. Peter Gustav Lejeune Dirichlet ( ) Siméon Denis Poisson ( )

27 harmonic functions maximum (minimum) principle boundary conditions uniquely define harmonic functions Liouville s Theorem: A bounded function harmonic on all of Z d is constant. Joseph Liouville ( )

28 heat evolution u(x, t + δ) u(x, t) = δ 2 ( ) ε 2 u(x+ε,t)+u(x ε,t) 2 u(x, t) u(x, t + 1) u(x, t) = 1 deg(x) u(y, t) u(x, t) y x harmonic functions are stable under heat evolution

29 heat evolution u(x, t + δ) u(x, t) = ( ) u(x+ε,t)+u(x ε,t) 2 u(x, t) δ = 1 2 ε2 u(x, t + 1) u(x, t) = 1 deg(x) u(y, t) u(x, t) y x harmonic functions are stable under heat evolution

30 heat evolution u(x, t + δ) u(x, t) = ( ) u(x+ε,t)+u(x ε,t) 2 u(x, t) δ = 1 2 ε2 u(x, t + 1) u(x, t) = 1 deg(x) u(y, t) u(x, t) y x harmonic functions are stable under heat evolution

31 heat evolution u(x, t + δ) u(x, t) = ( ) u(x+ε,t)+u(x ε,t) 2 u(x, t) δ = 1 2 ε2 u(x, t + 1) u(x, t) = 1 deg(x) u(y, t) u(x, t) y x harmonic functions are stable under heat evolution

32 classical notions random walks gambler s ruin counting simple paths

33 random walk Given a graph G, walk randomly on a graph: at every time step, independently of the past, choose a uniform neighbor and move to it

34 Dirichlet problem solution in a graph G let D be some finite domain with a boundary D = {x D : y x, y D} f : D R boundary conditions the function u(x) = E[f(X T ) X 0 = x] is harmonic in D and coincides with the boundary conditions f. f

35 Dirichlet problem solution in a graph G let D be some finite domain with a boundary D = {x D : y x, y D} f : D R boundary conditions the function u(x) = E[f(X T ) X 0 = x] is harmonic in D and coincides with the boundary conditions f. f

36 Dirichlet problem solution in a graph G let D be some finite domain with a boundary D = {x D : y x, y D} f : D R boundary conditions the function u(x) = E[f(X T ) X 0 = x] is harmonic in D and coincides with the boundary conditions f. f

37 heat equation solution In a graph G let D be some finite domain with a boundary D = {x D : y x, y D}. Let f : D D R be some initial / boundary conditions. The function u(x, t) = E[f(X T t ) X 0 = x] is harmonic in D and coincides with the initial conditions f, and solves the heat equation u(x, t + 1) u(x, t) = u(x, t).

38 classical notions random walks gambler s ruin counting simple paths

39 fair game play the following game: at each step, a coin is tossed you gain one coin, or lose one coin, each with probability 1 2 let X 0, X 1,..., be the number of coins at step t question: starting with x coins, what is the probability to win N coins without going bankrupt?

40 fair game play the following game: at each step, a coin is tossed you gain one coin, or lose one coin, each with probability 1 2 let X 0, X 1,..., be the number of coins at step t question: starting with x coins, what is the probability to win N coins without going bankrupt?

41 fair game play the following game: at each step, a coin is tossed you gain one coin, or lose one coin, each with probability 1 2 let X 0, X 1,..., be the number of coins at step t question: starting with x coins, what is the probability to win N coins without going bankrupt?

42 gambler s ruin f(x) = probability to reach N before 0 started at x f(0) = 0, f(n) = 1 0 < x < N f(x) = 1 2 f(x + 1) + 1 2f(x 1) f is harmonic in (0, N) solution: f(x) = x N (unique by maximum principle)

43 gambler s ruin f(x) = probability to reach N before 0 started at x f(0) = 0, f(n) = 1 0 < x < N f(x) = 1 2 f(x + 1) + 1 2f(x 1) f is harmonic in (0, N) solution: f(x) = x N (unique by maximum principle)

44 gambler s ruin f(x) = probability to reach N before 0 started at x f(0) = 0, f(n) = 1 0 < x < N f(x) = 1 2 f(x + 1) + 1 2f(x 1) f is harmonic in (0, N) solution: f(x) = x N (unique by maximum principle)

45 gambler s ruin f(x) = probability to reach N before 0 started at x f(0) = 0, f(n) = 1 0 < x < N f(x) = 1 2 f(x + 1) + 1 2f(x 1) f is harmonic in (0, N) solution: f(x) = x N (unique by maximum principle)

46 gambler s ruin f(x) = probability to reach N before 0 started at x f(0) = 0, f(n) = 1 0 < x < N f(x) = 1 2 f(x + 1) + 1 2f(x 1) f is harmonic in (0, N) solution: f(x) = x N (unique by maximum principle)

47 classical notions random walks gambler s ruin counting simple paths

48 SAW in a graph G fix some root vertex o let SAW n be the set of all simple paths of length n started at o Problem: count SAW n =?

49 SAW easy using Fekete s Lemma: the limit exists: µ = µ(g) := lim n SAW n 1/n µ is called the connective constant

50 connective constant Example: µ(z) =

51 connective constant Example: µ(z) =

52 connective constant Example: regular tree, µ(t d ) =

53 connective constant Example: regular tree, µ(t d ) = d 1

54 connective constant Example: the ladder µ(l) =

55 connective constant Example: the ladder µ(l) =

56 connective constant Theorem (Duminil-Copin & Smirnov 2010) µ(h) =

57 connective constant Theorem (Duminil-Copin & Smirnov 2010) µ(h) = Hugo Duminil-Copin Stas Smirnov

58 calculating µ(h) 1 µ is the radius of convergence for the generating function P (z) = SAW n z n = v H n=0 ω:o v z ω

59 calculating µ(h) 1 µ is the radius of convergence for the generating function P (z) = SAW n z n = v H n=0 ω:o v z ω

60 reductions v P (z) = v H ω:o v z ω o

61 reductions v P (z) = v H ω:o v z ω o

62 reductions T v P (z) = v T ω:o v z ω o W H

63 define a function on mid-edges in T : F (p) = z ω e iαθ(ω) ω:o p where θ(ω) is the winding of ω extend F to the vertices by averaging F on the mid edges around each vertex: F (v) = (p v)f (p) + (q v)f (q) + (r v)f (r) where p, q, r are the mid-edges adjacent to v Question: can we find z, α so that F is holomorphic? that is, so that F is 0 on each vertex? q v r p

64 define a function on mid-edges in T : F (p) = z ω e iαθ(ω) ω:o p where θ(ω) is the winding of ω extend F to the vertices by averaging F on the mid edges around each vertex: F (v) = (p v)f (p) + (q v)f (q) + (r v)f (r) where p, q, r are the mid-edges adjacent to v Question: can we find z, α so that F is holomorphic? that is, so that F is 0 on each vertex? q v r p

65 define a function on mid-edges in T : F (p) = z ω e iαθ(ω) ω:o p where θ(ω) is the winding of ω extend F to the vertices by averaging F on the mid edges around each vertex: F (v) = (p v)f (p) + (q v)f (q) + (r v)f (r) where p, q, r are the mid-edges adjacent to v Question: can we find z, α so that F is holomorphic? that is, so that F is 0 on each vertex? q v r p

66 γ γ 0 = e i 4 3 π z k e iα 4 3 π + e i 4 3 π z k e iα 4 3 π + e i 4 3 π ze iα 1 3 π + e i 4 3 π ze iα 1 3 π z k exp(iα 4 3 π) zk exp( iα 4 3 π) end-mid= 4 3 π end-mid= 4 3 π γ γ γ z exp( iα 1 3 π) z exp(iα1 3 π) 1 end-mid= 4 3 π end-mid= 4 3 π end-mid= 0

67 Equations: 0 = cos ( (α + 1) 4 3 π) 1 = 2z cos ( (α + 4) 1 3 π)

68 Equations: 0 = cos ( (α + 1) 4 3 π) 1 = 2z cos ( (α + 4) 1 3 π) Solution: (α + 1) 4 3 π = (k )π

69 Equations: 0 = cos ( (α + 1) 4 3 π) 1 = 2z cos ( (α + 4) 1 3 π) Solution: α = 6k 5 8 z 1 = 2 cos( 2k+1 8 π)

70 Equations: 0 = cos ( (α + 1) 4 3 π) 1 = 2z cos ( (α + 4) 1 3 π) Solution: e.g. k = 0 and α = 5 8 and α = 6k 5 8 z 1 = 2 cos( 2k+1 8 π) z 1 = 2 cos π 8 = 2 + 2

71 summary so far E v P (z) = v T ω:o v z ω A o B Ē

72 summary so far E v P (z) = v T ω:o v z ω A o B Ē

73 summary so far E P (x) = ω:o x F (x) = ω:o x z ω z ω e iαθ(ω) A o v B Ē 0 = v T F (v) = F (A) + F (B) + λf (E) + λf (Ē) (where λ = e i 2 3 π ) the winding is constant on A, B, E and Ē! 0 = 1 e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē)

74 summary so far E P (x) = ω:o x F (x) = ω:o x z ω z ω e iαθ(ω) A o v B Ē 0 = v T F (v) = F (A) + F (B) + λf (E) + λf (Ē) (where λ = e i 2 3 π ) the winding is constant on A, B, E and Ē! 0 = 1 e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē)

75 summary so far E P (x) = ω:o x F (x) = ω:o x z ω z ω e iαθ(ω) A o v B Ē 0 = v T F (v) = F (A) + F (B) + λf (E) + λf (Ē) (where λ = e i 2 3 π ) the winding is constant on A, B, E and Ē! 0 = 1 e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē)

76 summary so far E P (x) = ω:o x F (x) = ω:o x z ω z ω e iαθ(ω) A o v B Ē 0 = v T F (v) = F (A) + F (B) + λf (E) + λf (Ē) (where λ = e i 2 3 π ) the winding is constant on A, B, E and Ē! 0 = 1 e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē)

77 1 = e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē) taking H (height) 1 = cos(απ) ω:o A z ω + ω:o B z ω cos(απ) = cos( 5 8 π) = cos( 3 8 π) =

78 1 = e iαπ P (A + ) e iαπ P (A ) + P (B) + λ 2 P (E) + λ 2 P (Ē) taking H (height) 1 = cos(απ) ω:o A z ω + ω:o B z ω cos(απ) = cos( 5 8 π) = cos( 3 8 π) =