the configuration space of Branched polymers R. Kenyon, P. Winkler

Transcription

1 the configuration space of Branched polymers R. Kenyon, P. Winkler

2 artificial blood catalyst recovery Branched polymers (dendrimers) in modern science artificial photosynthesis

3 A branched polymer is a connected collection of unit disks with non-overlapping interiors (generically, a tree of disks) X n = space of branched polymers with n labelled disks.

4 2D examples: 1 2 Volume(X 2 ) = 2π π 4π Volume(X 3 ) = 2(2π) 2 2π 4π π 4π 3

5 1 (2π) 3 5 (2π) (2π)4 27 (2π)4 27 (2π)4

6 Questions 1. What does a random BP looks like? ( diameter? number of leaves? scaling limit? ) 2. Does the space X n have a nice geometric structure? Theorem [Brydges/Imbrie 2002] 2D: The volume of X n, measured in terms of angles, is (2π) n 1 (n 1)!. 3D: The volume of X n, measured in terms of angles, is (2π) n 1 n n 1. Generally, they related BPs in R D+2 with the hard sphere model in R D.

7 Questions 1. What does a random BP looks like? ( diameter? number of leaves? scaling limit? ) 2. Does the space X n have a nice geometric structure? Theorem [Brydges/Imbrie 2002] 2D: The volume of X n, measured in terms of angles, is (2π) n 1 (n 1)!. 3D: The volume of X n, measured in terms of angles, is (2π) n 1 n n 1. Generally, they related BPs in R D+2 with the hard sphere model in R D.

8 Key observation: Invariance Lemma in 2D: The volume of X n doesn t change when you change the radii. ([BI] proved this using localization/ equivariant cohomology) 3 2 1

9 Key observation: Invariance Lemma in 2D: The volume of X n doesn t change when you change the radii. ([BI] proved this using localization/ equivariant cohomology)

10 Key observation: Invariance Lemma in 2D: The volume of X n doesn t change when you change the radii. ([BI] proved this using localization/ equivariant cohomology) 3 2 θ 3 θ 2 1 θ 1 Vol=2π(2π 2θ 1 )

11 Key observation: Invariance Lemma in 2D: The volume of X n doesn t change when you change the radii. ([BI] proved this using localization/ equivariant cohomology) 3 2 θ 3 θ 2 1 θ 1 Vol=2π(2π 2θ 1 ) 2π(2π 2θ 2 ) + 2π(2π 2θ 3 ) 2π(6π 2π)

12 Assuming this Lemma, we can compute Vol(X n ) as follows: Take disks of radii 1, ɛ, ɛ 2,..., ɛ n 1. In this case, we can ignore BPs like: since they have small volume ( 0 as ɛ 0). So we need consider only configurations where balls are attached in order of decreasing radius.

13 Attach i to one of 1, 2,..., i 1 2π (i 1) + O(ɛ) choices Total volume n i=2 2π(i 1) = (2π)n 1 (n 1)! in limit ɛ 0.

14 Proof of Invariance Lemma The proof is based on the following (trivial) fact: Lemma: If v 1,..., v n R n 1 ( ) A = v 1 v 2... v n and the n n matrix is singular, then n ( 1) j v 1 v j v n = 0. j=1 Now, change the radii of the disks continuously. The volumes of the components of X n (one for each tree) change because their boundaries in the space of angles move.

15 Codimension-1 boundaries are where the BP has a cycle: A BP with a k-cycle is in the boundary of k components (break one of the k edges of the cycle). Show using Lemma that the net volume lost to each component sums to zero:

16 Codimension-1 boundaries are where the BP has a cycle: θ 2 θ 1 P θ 3 θ 4 θ 5 Let θ i be the angle of edge i of P.

17 θ 2 P θ 1 θ 3 θ 4 R When increasing R, this component loses volume near P.

18 θ 2 θ 1 R P θ 5 θ 4 When increasing R, this componenet gains volume near P.

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20 Computing volume changes: Let t 1,..., t k 2 preserving edge lengths. be coordinates for perturbations of a k-gon P Let R be a perturbation of P increasing lengths at the same rate. θ 1 t 1... B =. θ 1 t k 2... θ 1 R... two adjacent edge θ k t 1. θ k t k 2 θ k R. Note that (1, 1..., 1) is in the span of the t i (rotate P ). So by the lemma, k j=1 ( 1)j dθ 1 dθ j dθ k = 0.

21 θ 2 θ 1 φ φ θ 5 θ 4 The volume changes are proportional to the projection lengths along the angle bisector. (Calculation omitted.)

22 Perfect simulation of a 2D branched polymer.

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25 1000 balls balls

26 What if you weaken the hard core interaction? that is, allow certain pairs of disks to overlap each other. Define an interaction graph G: v v if v and v are not allowed to overlap. E.g. Two species of balls: blues & reds (G is complete bipartite graph) Bipartite polymers Theorem The volume of the configuration space is (2π) n 1 T G (1, 0). T G (1, 0) = spanning subgraphs ( 1) edges T G (x, y) is the Tutte polynomial of G.

27 Branched polymers in 3D Project the ball centers to the x-axis: x 1 x 2 x 3 x 4 x 5 Let G be the graph with vertices x i and edges when x i x j < 1.

28 Lemma: The n-dimensional volume of the set of 3DBP s whose centers project to x 1 < x 2 < < x n is (2π) n 1 n 2 i=1 n i where n i is the number of x j with j > i, with x j x i < 1. Proof: Compute T G (1, 0). Use Archimedes theorem. Theorem: [Archimedes] The Lebesgue measure on S 2 projects to uniform Lebesgue measure on a segment in the x-direction.

29 Question: when we project the points of a 3DBP, what is their distribution? Answer: (1) (2) (3) Construct a random labelled tree on n vertices. Assign a uniform [0,1] length to each edge, independently. Dangle the tree from the root. (4) Let x i be the distance of the ith vertex from the root. (and x 1 = 0) Theorem: These points are equidistributed with the projected points of a BP.

30 x 1... x n Corollary. The 3DBP with n balls has diameter of order n.

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32 Proof of Theorem Let x 1,..., x n R be the projections of the centers. Build a tree by gluing each x k to some x j to its left with x k x j < 1 x 1 x 2 x n The number of ways to do this is n 1 i=1 n i Integrate over positions of the x i : Each tree contributes [0,1] n 1 dx So get n n 1 (total number of rooted trees).

33 Open questions 1. Dynamics on polymers 2. 4D polymers ( 2D hard disk model). 3. Diameter of a 2DBP 4. X n [0, 1] M T n 1 [0, 1] M

34 A puzzle: [Spitzer] Starting at origin in R 2, take a random walk, each step being a uniformly chosen unit vector. Show: the probability that you are within 1 of the origin after n steps is 1 n+1. Hint: Use G=n-cycle.

35 Computing Z 4D. Take points x 1,..., x n R 2, with x 1 = 0. Define a graph G by: x i x j if x i x j < 1. Z 4D = T G (1, 0)dx 2... dx n.