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First passage percolation on rotationally invariant fields
Allan Sly Princeton University September 2016 Joint work with Riddhipratim Basu (Stanford) and Vladas Sidoravicius (NYU Shanghai)
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First Passage Percolation
Model: π π,π an IID random field of numbers π π₯,π¦ minimum sum along paths from x to y. 1 3 5 6 9 7 8 4 2 By Subadditive Ergodic Theorem: lim 1 π π 0,π π₯ = π π₯ π.π .
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Variance Central question: What is the variance?
By Poincare inequality [Kesten β91] πππ π π π₯ =π(π) Using hypercontractivity for Boolean case πππ π π π₯ =π(π/ log π ) [Benjamini, Kalai, Schramm β03] Extended to a wider range of distributions [Damron, Hanson, Sosoe β15] For oriented last passage percolation with exponential or geometric entries πππ π π π₯ ~ πΆ π 2/3 [Johansson β00]
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Rotationally Invariant models
Our model: Take Ξ¦: β 2 β β 2 rotationally invariant, smooth and compactly supported. Let Ξ:ββ(π,π) be continuous and strictly increasing. Set π π₯,π¦ =Ξ Ξ¦ π’βπ₯,π£βπ¦ ππ΅(π’,π£) Define the distance as π π₯,π¦ = min πΎ πΎ π
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Main Result Main result (Basu, Sidoravicius, S. β16) For some π>0,
πππ π π =π( π 1βπ ) The specifics of the model are not that important, should hold for models with Rotational invariance FKG Property Short range of dependence. E.G. Graph distances for supercritical random geometric graphs.
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Basic Approach Mutli-scale: V n βπππ π π . Set π π = π log π π = π 1βπ . We show that π π π β€ π π π = π 2 π . Enough to show that for all π, π π β€ π π β π ππ β€ π ππ = π 2 π π . Block version of Kestenβs bounds π βπ β€πΆ β π π Chaos estimate β path is highly sensitive to noise.
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Kestenβs martingale argument
Reveal the sites one by one: π π =πΌ π π β± π ] Then πππ π π = Ξ£ π πππ π π β π πβ1 β€ Ξ£ π πΌπππ T n | β± π c The value of block i will only matter if it is on the optimal path so Ξ£ π πΌ πππ T n | β± π c β πΆπΌ #{π :πβπΎ}β πΆπ With some extra tricks one can also get concentration bounds.
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Multiscale version of Kesten argument
Split grid into blocks length π, height W n = π 1/2 π π 1/4 Revealing blocks - analyze Doob martigale of π βπ What we need Relate point to point with side to side Variance: π βπ β€ πΆ β π π Concentration: β π βπ βπΌ π βπ β₯π₯ β π π β€ πΆ π βπ π₯ 2/3 πΌ π βπ βπ β π β€πΆ β π π Transversal Fluctuations of order β 3/4 π π
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Side to side Diagonal Length
π 2 + π π 2 = π 2 +π π π 1/2 βπ π π 1/2 And π π 1/2 is the bound on the standard deviation.
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Transversal fluctuations
To move up π blocks, extra length is 2 π π π . For midpoint π πβπππππ πππ’ππ‘π’ππ‘πππ β€ πΆ π βπ π 4/3 For other dyadic points use chaining. At least on segment must deviate from its mean by at least π π π π π π
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Side to side To compare the maximum side to side length π π + with point to point π π . Use chaining πΌ π π + βπΌ π π β€ πΆ π π 1/2
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Side to side To compare the minimum side to side length π π β with point to point π π + . Split up path πΌ π π + β€πΌ π 4π 5 β +2πΌ max ππ π π 10 ,π,π πΆ π π 1/2 Max Min Max Max
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Relating mean to π By subadditivity πΌ π π >π π.
By enumerating over long paths we show that for C large if πΌ π π β β₯ππ+πΆ π π then lim 1 βπ π βπ > π.
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Concentration π= Ξ£ π # ππππππ ππ πΆπππ’ππ π 2
π= Ξ£ π # ππππππ ππ πΆπππ’ππ π 2 Similarly to transversal fluctuations β π> πΆ+π₯ β β€ π β π₯β 2/3 Apply Doob martingale and Kestenβs concentration argument revealing columns one at a time. Can not take union bound over all paths because of sub-exponential tails
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Proof by contradiction
Case 1: Either for some 1β€ββ€π we have π βπ β€ πΏ β π π in which case we show that π ππ β€ πΆ πΏ π π π β€ π ππ . Case 2: For all ββ€π π βπ β₯ πΏ β π π Use chaos argument. This case never actually happens as we believe π π β π 2/3 .
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Super-concentration β chaos
In the context of FPP: Super-concentration: Better than Poincare inequality i.e. π π =π(π) Chaos: with πΎ the optimal path and πΎβ² the optimal path after resampling π fraction of the field then πΎβ© πΎ β² β€π(π) Super-concentration β Chaos [e.g. Chatterjee β14 ] Works well for block version.
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Proving Chaos Aim: Resample π fraction of the blocks and find good alternatives to the original path. Need to understand the field conditioned on the path before and after resampling Similar to [BSS β14]
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Percolation type estimates
We have control of the transversal fluctuations of the path. A percolation estimate says that all paths with reasonable fluctuations spend most of their time in βtypicalβ regions. Atypical
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FKG type estimates Conditioning on the location and value of the path is a positive event for the rest of the field. We can use FKG to sample create regions that are very positive which the optimal path must avoid (before and after resampling.
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Planting a configuration
For a region A, suppose that if π π΄ , π π΄ π =( π³ π΄ , π³ π΄ π ) such that πΎ does not intersect π΄ then β π π΄ = π³ π΄ π π΄ π = π³ π΄ π ,πΎ] β₯β[ π π΄ = π³ π΄ ] So we can plant configurations provided they avoid A. A
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Big changes Using our assumption that πππ π βπ β₯ πΏ π βπ we show that we can find regions with very long and very short geodesics. By interpolation between them in 1/π steps we can find regions with a large change with positive probability after π resampling. We look for regions which become much shorter.
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Pulling the paths apart
We design a collection of events which together separate the old and new paths. Positive probability at each location. Concentration estimates separate path at πΏ fraction of location w.h.p.
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Multi-scale Improvements
We look for improvements on a range of scales. Show that πΎβ© πΎ β² β€πΏ|πΎ|. Conclude π ππ β€ π 2 π π = π ππ
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Lattice models Can rotational invariance be relaxed?
Should be sufficient that the limiting shape is smooth and has positive curvature in a neighbourhood of the direction.
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