Diamond Aggregation Lionel Levine Toronto Probability Seminar - PowerPoint PPT Presentation

Diamond Aggregation Lionel Levine Toronto Probability Seminar March 30, 2009 Joint work with Wouter Kager 1

❘ Talk Outline Joint work with Wouter Kager: ◮ Internal DLA: from random walk to growth model ◮ Uniformly layered walks ◮ Limiting shape and fluctuations ◮ 2

Talk Outline Joint work with Wouter Kager: ◮ Internal DLA: from random walk to growth model ◮ Uniformly layered walks ◮ Limiting shape and fluctuations ◮ Joint work with Yuval Peres: ◮ Multiple point sources ◮ Smash sum of two domains in ❘ d ◮ 2

❩ From random walk to growth model Internal DLA Given a Markov chain on state space ❩ 2 . ◮ Start with n particles at the origin. ◮ Each particle walks until it finds an unoccupied site, stays there. ◮ 3 Internal DLA

From random walk to growth model Internal DLA Given a Markov chain on state space ❩ 2 . ◮ Start with n particles at the origin. ◮ Each particle walks until it finds an unoccupied site, stays there. ◮ A ( n ) : the resulting random set of n sites in ❩ 2 . ◮ Growth rule: Let A ( 1 ) = { o } , and ◮ A ( n + 1 ) = A ( n ) ∪ { X n ( τ n ) } 3 Internal DLA

From random walk to growth model Internal DLA Given a Markov chain on state space ❩ 2 . ◮ Start with n particles at the origin. ◮ Each particle walks until it finds an unoccupied site, stays there. ◮ A ( n ) : the resulting random set of n sites in ❩ 2 . ◮ Growth rule: Let A ( 1 ) = { o } , and ◮ A ( n + 1 ) = A ( n ) ∪ { X n ( τ n ) } where X 1 , X 2 , . . . are independent random walks, and τ n = min � t | X n ( t ) � A ( n ) � . 3 Internal DLA

The growth rule illustrated 4 Internal DLA

The growth rule illustrated 4 Internal DLA

The growth rule illustrated 4 Internal DLA

Example: simple random walk Main questions 1. Limiting shape? 5 Internal DLA

Example: simple random walk Main questions 1. Limiting shape? 2. Fluctuation size? 5 Internal DLA

Simple random walk Lawler-Bramson-Griffeath ’92 The limiting shape is a disk: ∀ ǫ > 0, with probability 1 B ( 1 − ǫ ) n ⊂ A ( π n 2 ) ⊂ B ( 1 + ǫ ) n eventually . 6 Internal DLA

Simple random walk Lawler-Bramson-Griffeath ’92 The limiting shape is a disk: ∀ ǫ > 0, with probability 1 B ( 1 − ǫ ) n ⊂ A ( π n 2 ) ⊂ B ( 1 + ǫ ) n eventually . Lawler ’95 Strengthened this to show B n − f ( n ) ⊂ A ( π n 2 ) ⊂ B n + f ( n ) eventually for f ( n ) = n 1 / 3 log 4 n . 6 Internal DLA

Simple random walk Lawler-Bramson-Griffeath ’92 The limiting shape is a disk: ∀ ǫ > 0, with probability 1 B ( 1 − ǫ ) n ⊂ A ( π n 2 ) ⊂ B ( 1 + ǫ ) n eventually . Lawler ’95 Strengthened this to show B n − f ( n ) ⊂ A ( π n 2 ) ⊂ B n + f ( n ) eventually for f ( n ) = n 1 / 3 log 4 n . Someone in the audience ’09 (?) The true order of fluctuations f ( n ) is only logarithmic in n . 6 Internal DLA

What about other walks? Modify transition probabilities on the axes: Steps toward the origin along the x - and y -axes are reflected ◮ away from the origin instead. So for x > 0, P (( x , 0 ) , ( x + 1 , 0 )) = 1 2 P (( x , 0 ) , ( x , ± 1 )) = 1 4 . 7 Diamond aggregation

What about other walks? Modify transition probabilities on the axes: Steps toward the origin along the x - and y -axes are reflected ◮ away from the origin instead. So for x > 0, P (( x , 0 ) , ( x + 1 , 0 )) = 1 2 P (( x , 0 ) , ( x , ± 1 )) = 1 4 . Off the axes, same as simple random walk. ◮ 7 Diamond aggregation

What about other walks? Modify transition probabilities on the axes: Steps toward the origin along the x - and y -axes are reflected ◮ away from the origin instead. So for x > 0, P (( x , 0 ) , ( x + 1 , 0 )) = 1 2 P (( x , 0 ) , ( x , ± 1 )) = 1 4 . Off the axes, same as simple random walk. ◮ Instead of a disk, limiting shape is now a diamond! ◮ 7 Diamond aggregation

Diamond Aggregation 8 Diamond aggregation

❩ Diamond Layers Notation � � � = | x | + | y | . � ( x , y ) ◮ � � L n = the diamond layer of radius n ◮ z ∈ ❩ 2 : � z � = n � � = 9 Diamond aggregation

Diamond Layers Notation � � � = | x | + | y | . � ( x , y ) ◮ � � L n = the diamond layer of radius n ◮ z ∈ ❩ 2 : � z � = n � � = D n = the diamond of radius n ◮ z ∈ ❩ 2 : � z � ≤ n � � = d n = # D n = 2 n ( n + 1 ) + 1 ◮ 9 Diamond aggregation

Diamond Layers Notation � � � = | x | + | y | . � ( x , y ) ◮ � � L n = the diamond layer of radius n ◮ z ∈ ❩ 2 : � z � = n � � = D n = the diamond of radius n ◮ z ∈ ❩ 2 : � z � ≤ n � � = d n = # D n = 2 n ( n + 1 ) + 1 ◮ Uniformly layered walk: Distribution of X ( t ) is a mixture of uniform distributions on layers L n . 9 Diamond aggregation

P P Uniformly Layered Walk Discrete time Markov chain X ( t ) on state space ❩ 2 satisfying � � � � � � � − � ≤ 1 (U1) � X ( t + 1 ) � X ( t ) � � � � � � � � (U2) For all n ≥ 1, P o ( X ( t ) ∈ L n for some t < ∞ ) = 1 . 10 Diamond aggregation

Uniformly Layered Walk Discrete time Markov chain X ( t ) on state space ❩ 2 satisfying � � � � � � � − � ≤ 1 (U1) � X ( t + 1 ) � X ( t ) � � � � � � � � (U2) For all n ≥ 1, P o ( X ( t ) ∈ L n for some t < ∞ ) = 1 . (U3) For all k ≥ 0, n ≥ 1 and all x ∈ L n = 1 � � � X ( t ) = x � X ( t ) ∈ L n P k � 4 n where P k is the law of the walk started from uniform on layer L k . 10 Diamond aggregation

Shape Theorem Theorem (Kager-L.) For any uniformly layered walk, with probability 1 D n − 4 √ n log n ⊂ A ( d n ) ⊂ D n + 20 √ eventually . n log n 11 Diamond aggregation

Shape Theorem Theorem (Kager-L.) For any uniformly layered walk, with probability 1 D n − 4 √ n log n ⊂ A ( d n ) ⊂ D n + 20 √ eventually . n log n So all uniformly layered walks have the diamond as their limiting ◮ shape. 11 Diamond aggregation

Shape Theorem Theorem (Kager-L.) For any uniformly layered walk, with probability 1 D n − 4 √ n log n ⊂ A ( d n ) ⊂ D n + 20 √ eventually . n log n So all uniformly layered walks have the diamond as their limiting ◮ shape. � Is n log n the right order of fluctuations? ◮ Or do the fluctuations depend on the particular u.l. walk? ◮ 11 Diamond aggregation

Proof sketch: Containing a large diamond Fix a site z ∈ L n − ρ . Want an upper bound on P ( z � A ( d n )) . D n z ρ 12 Diamond aggregation

Proof sketch: Containing a large diamond Fix a site z ∈ L n − ρ . Want an upper bound on P ( z � A ( d n )) . Among the first d n − 1 walks, let ◮ D n M = # that first hit L n − ρ at z . z L = # that first hit L n − ρ at z ρ after dropping their particle . 12 Diamond aggregation

Proof sketch: Containing a large diamond Fix a site z ∈ L n − ρ . Want an upper bound on P ( z � A ( d n )) . Among the first d n − 1 walks, let ◮ D n M = # that first hit L n − ρ at z . z L = # that first hit L n − ρ at z ρ after dropping their particle . � � Then z � A ( d n ) ⊂ { L = M } . ◮ 12 Diamond aggregation

Proof sketch: Containing a large diamond Fix a site z ∈ L n − ρ . Want an upper bound on P ( z � A ( d n )) . Among the first d n − 1 walks, let ◮ D n M = # that first hit L n − ρ at z . z L = # that first hit L n − ρ at z ρ after dropping their particle . � � Then z � A ( d n ) ⊂ { L = M } . ◮ Both L and M are sums of indicator RV’s. ◮ Main difficulty: The summands of L are dependent. ◮ 12 Diamond aggregation

❊ ❊ P Finding independence Estimating L Start one new walk from every site of D n − ρ − 1 , and let z L ′ = # of new walks that first hit L n − ρ at z . ρ Since at most one particle can attach to the cluster at a given site, L ≤ L ′ . 13 Diamond aggregation

Finding independence Estimating L Start one new walk from every site of D n − ρ − 1 , and let z L ′ = # of new walks that first hit L n − ρ at z . ρ Since at most one particle can attach to the cluster at a given site, L ≤ L ′ . Strategy Since both L ′ and M are sums of independent indicators, show ❊ L ′ < ❊ M and use large deviations to bound P ( L ′ ≥ M ) . 13 Diamond aggregation

❊ P P Separating ❊ M and ❊ L ′ Writing ℓ = n − ρ , we have ❊ M = ( d n − 1 ) P o ( X ( τ ℓ ) = x ) = 2 n ( n + 1 ) > n + ρ . 4 ℓ 2 14 Diamond aggregation

Separating ❊ M and ❊ L ′ Writing ℓ = n − ρ , we have ❊ M = ( d n − 1 ) P o ( X ( τ ℓ ) = x ) = 2 n ( n + 1 ) > n + ρ . 4 ℓ 2 ❊ L ′ = � P y ( X ( τ ℓ ) = x ) y ∈D ℓ − 1 −{ o } ℓ − 1 � = 4 k P k ( X ( τ ℓ ) = x ) k = 1 14 Diamond aggregation

Separating ❊ M and ❊ L ′ Writing ℓ = n − ρ , we have ❊ M = ( d n − 1 ) P o ( X ( τ ℓ ) = x ) = 2 n ( n + 1 ) > n + ρ . 4 ℓ 2 ❊ L ′ = � P y ( X ( τ ℓ ) = x ) y ∈D ℓ − 1 −{ o } ℓ − 1 � = 4 k P k ( X ( τ ℓ ) = x ) k = 1 ℓ − 1 < n − ρ 4 k 4 ℓ = ℓ − 1 � = . 2 2 k = 1 14 Diamond aggregation