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On the backbone exponent Christophe Garban Université Lyon 1 joint work with Jean-Christophe Mourrat (ENS Lyon) Cargèse, September 2016 C. Garban (univ. Lyon 1) On the backbone exponent 1 / 30

Critical percolation C. Garban (univ. Lyon 1) On the backbone exponent 2 / 30

SLE κ processes (Schramm) H γ t+s γ t g t g s Domain Markov Property g t+s (law) = g s g t Definition (Schramm, 1999) { t = t1 +... + t k g t (law) = g (k) t k... g (2) t 2 g (1) t 1 Let κ R +. SLE κ is the Loewner chain driven by a Brownian motion κb t, i.e. 2 t g t (z) = g t (z) z H, t < T (z) κb t C. Garban (univ. Lyon 1) On the backbone exponent 3 / 30

Different phases of SLE κ C. Garban (univ. Lyon 1) On the backbone exponent 4 / 30

Zoology SLE 3 SLE 6

Critical exponents R Theorem (Lawler, Schramm & Werner 2002) α 1 (R) = R 5/48+o(1) 0 Corollary As η 0, large clusters converge to random compact sets in the plane of dimension d H = 91 48 C. Garban (univ. Lyon 1) On the backbone exponent 6 / 30

Critical exponents One-arm event Polychromatic two-arm event Monochromatic two-arm event R R R 0 Expect R α1 α 1 = 5 48 LSW 2002 Expect R α2 α 2 = 1/4 LSW and Smirnov-Werner Expect R α 2 α 2 =?? (Backbone exponent) C. Garban (univ. Lyon 1) On the backbone exponent 7 / 30

Picturing the backbone In a finite box R R, this looks as follows: C. Garban (univ. Lyon 1) On the backbone exponent 8 / 30

How does one compute a critical exponent? Two main steps: I. Find an exponent for the scaling limit using SLE 6 P D ɛ ɛ α II. Connect this with the actual discrete model (using quasi-multiplicativity). P [ ε ω D ] ε α P [ 0 B(0, R) ] = R α+o(1) C. Garban (univ. Lyon 1) On the backbone exponent 9 / 30

Requires a radial version of SLE κ θ 0 g t θ t g t is the conformal map D \ γ([0, t]) D such that g t(0) = e t Radial version of Loewner s Theorem { t g t (z) = g t (z) gt(z)+eiλ t g 0 (z) = z g t(z) e iλ t z D, t < T (z) Consequence Radial SLE 6 corresponds to λ t 6B t. It follows that t θ t satisfies the SDE: dθ t = 6dB t + cot( θ t 2 )dt

How did LSW find α 1 = 5/48? e iθ Introduce h(θ, t) := P [ r(θ) e t]

How did LSW find α 1 = 5/48? e iθ Introduce h(θ, t) := P [ r(θ) e t] Q(θ)

How did LSW find α 1 = 5/48? e iθ Introduce h(θ, t) := P [ r(θ) e t] Q(θ) Claim The function h(θ, t) solves the following PDE: t h = 3 θh 2 + cot( θ 2 ) θh h(θ, t = 0) := 1 where: Dirichlet b.c. on {θ = 0, t > 0} Neumann b.c. on {θ = 2π, t > 0}

Short sketch of proof e iθ e iθ g u e iθ u Q(θ) h(θ, t) := P [ r(θ) e t]

Short sketch of proof e iθ e iθ g u e iθ u Q(θ) h(θ, t) := P [ r(θ) e t] = E [ P [ r(θ) e t] Fu ]

Short sketch of proof e iθ e iθ g u e iθ u Q(θ) h(θ, t) := P [ r(θ) e t] = E [ P [ r(θ) e t] ] Fu E [ P [ r(θ u ) e u t]] = E [ h(θ u, t u) ]

Short sketch of proof e iθ e iθ g u e iθ u Q(θ) h(θ, t) := P [ r(θ) e t] = E [ P [ r(θ) e t] Fu ] E [ P [ r(θ u ) e u t]] = E [ h(θ u, t u) ] 1 From the above argument, u h(θ u, t u) is a martingale. 2 u θ u evolves according to the SDE dθ u = 6dB u + cot( θu 2 )du. 3 From Itô s formula, get t h = 3 2 θ h + cot( θ 2 ) θh

Now that h(θ, t) := P [ r(θ) e t] is known to solve the PDE t h = 3 2 θ h + cot( θ 2 ) θh, it remains to prove: Theorem (Lawler, Schramm, Werner 2002) The one-arm exponent α 1 is the leading eigenvalue of L where L is the following differential operator on [0, 2π]: L = 3 2 θ + cot( θ 2 ) θ with Neumann b.c. at 0 and Dirichlet b.c. at 2π. This gives α 1 = 5 48, i.e. P [ 0 B R ] = R 5/48+o(1) By a quite elaborate use of the maximum principle, they prove that 1 C e α1t (sin(θ/4)) 1/3 h(θ, t) Ce α1t (sin(θ/4)) 1/3

The one-arm exponent from a direct exploration procedure e iθ θ t 2π θ t In the bulk, dθ t = 6dB t + cot( θt 2 )dt From this exploration path perspective, want to consider: where T := inf{u > 0, θ u = 0}. h(θ, t) := P θ [ T > t ], If one carefully characterizes the reflexion on 2π as being Neumann, then indeed h satisfies the same PDE as h, with same initialisation h(t = 0) = h(t = 0) 1 so that h = h

Polychromatic two-arm exponent α 2 = 1/4 Theorem (LSW and Smirnov, Werner 2002) The (polychromatic) two-arm exponent α 2 is the leading eigenvalue of the same! differential operator on [0, 2π]: L = 3 2 θ + cot( θ 2 ) θ with different b.c: Dirichlet at 0 and 2π. This gives α 2 = 1 4. e iθ θ t 2π θ t Here, the situation is somewhat easier as the diffusion does not touch the Bdy.

Backbone exponent α 2 Warm-up remark: it is a priori not clear at all that SLE 6 will help here! C. Garban (univ. Lyon 1) On the backbone exponent 16 / 30

Backbone exponent α 2 Warm-up remark: it is a priori not clear at all that SLE 6 will help here! Yet, the following very nice approach is sketched in the appendix of LSW02: Keep track of TWO angles: α, β (and call γ := 2π α β). eiβ e iγ - Let Q(α, β) be the set of points which have two dispoint open paths to the γ-grey arc. N.B. these path may use the β-black arc! e iα C. Garban (univ. Lyon 1) On the backbone exponent 16 / 30

Backbone exponent α 2 Warm-up remark: it is a priori not clear at all that SLE 6 will help here! Yet, the following very nice approach is sketched in the appendix of LSW02: Keep track of TWO angles: α, β (and call γ := 2π α β). eiβ e iγ - Let Q(α, β) be the set of points which have two dispoint open paths to the γ-grey arc. N.B. these path may use the β-black arc! e iα - Introduce the quantity G(α, β, t) := P [ r(α, β) e t] C. Garban (univ. Lyon 1) On the backbone exponent 16 / 30

eiβ e iγ G(α, β, t) := P [ r(α, β) e t] Claim: the function G(α, β, t) solves the following parabolic PDE on {α + β 2π}. 2π e iα t G = ( 3 α β 2 + cot( α 2 ) α + cot( β 2 ) β) G := L[G] With boundary conditions: 1 Neumann on bottom 2 Constant on left, i.e. G t (0, β) C t β 3 Dirichlet on {(0, 2π)} {(2π, 0)} diagonal. 0 Neumann α 2π C. Garban (univ. Lyon 1) On the backbone exponent 17 / 30

In the appendix of LSW02, it is stated that the Monochromatic two-arms exponent α 2 is the leading eigenvalue of the differential operator on the triangle: L = 3 2 α β + cot( α 2 ) α + cot( β 2 ) β with the above b.c.s (Neumann+Dirichlet+"Constant"). A meta-statement All examples of known critical exponents in percolation are obtained as leading eigenvalues of some differential operator. Examples 1 α 1, α 2, α 2. 2 Plane exponents α j, j 3 (through derivative exponents which also require a leading eigenvalue analysis). 3 Brownian intersection exponents such as ξ(3, 3) etc...

Stochastic process associated to the backbone As in the one-arm case, it is natural to consider (if well-defined) the stochastic process X t associated to the generator 2π L = 3 2 α β + cot( α 2 ) α + cot( β 2 ) β β With previous boundary conditions: 1 Neumann on bottom 2 Constant on left. 3 Dirichlet on {(0, 2π)} {(2π, 0)} diagonal. 0 Neumann α 2π This stochastic process is reminiscent of a striking algorithm to detect a monochromatic two-arm event! A closer inspection of this stochastic process leads to a few mathematical issues on the initial PDE approach. C. Garban (univ. Lyon 1) On the backbone exponent 19 / 30

Mathematical issues raised L = 3 2 α β + cot( α 2 ) α + cot( β 2 ) β 1 A.s., the process X t never reaches the diagonal {γ = 2π} (except both corners). What does it mean to require Dirichlet boundary conditions there? 2 The process is highly non-reversible. There are no measure µ and no domain for which the differential operator L is self-adjoint. How can one justify the existence of a (leading) positive eigenfunction? 3 (facultative) is the process X t well-defined? (Rather unusual b.c.s). 4 Finally, assuming the process X t is well-defined, let Ĝ(α, β, t) := P α,β [ Tsurvival > t ] then Ĝ satisfies the same PDE also starting from Ĝ(t = 0) 1 (so that Ĝ G). t Ĝ = ( 3 2 α β + cot( α 2 ) α + cot( β 2 ) β)ĝ = L Ĝ C. Garban (univ. Lyon 1) On the backbone exponent 20 / 30

[ Ĝ(α, β, t) := P α,β Tsurvival > t ] = G(α, β, t) Two possible strategies: 2π 2π β OR β 0 α 2π 0 α 2π

[ Ĝ(α, β, t) := P α,β Tsurvival > t ] = G(α, β, t) Two possible strategies: 2π 2π β OR β 0 α 2π 0 α 2π The second strategy corresponds to the polychromatic two-arm event. As such, we would have e α 2 t G( π 2, π, t) e α2t 2

Where does issue 4 come from? eiβ e iα e iγ - Recall G(α, β, t) := P [ r(α, β) e t] - Let T be the stopping time of touching Triangle. - It is tempting to write as in the one-arm case G(α, β, t) = E α,β [ G(XT, t T ) ] It thus appears that the previous heuristics is not true here: stopping time T DOES have a non-trivial effect! C. Garban (univ. Lyon 1) On the backbone exponent 22 / 30

Where does issue 4 come from? eiβ e iα e iγ - Recall G(α, β, t) := P [ r(α, β) e t] - Let T be the stopping time of touching Triangle. - It is tempting to write as in the one-arm case G(α, β, t) = E α,β [ G(XT, t T ) ] It thus appears that the previous heuristics is not true here: stopping time T DOES have a non-trivial effect! t G [ 3 2 α β + cot( α 2 ) α + cot( β 2 ) β] G C. Garban (univ. Lyon 1) On the backbone exponent 22 / 30

How to proceed to deal with these difficulties? A) Ignore them! and rely on Camia/Newman or Schramm/Smirnov + Beffara/Nolin. C. Garban (univ. Lyon 1) On the backbone exponent 23 / 30

How to proceed to deal with these difficulties? A) Ignore them! and rely on Camia/Newman or Schramm/Smirnov + Beffara/Nolin. Still, having a direct SLE 6 proof is desirable C. Garban (univ. Lyon 1) On the backbone exponent 23 / 30

How to proceed to deal with these difficulties? A) Ignore them! and rely on Camia/Newman or Schramm/Smirnov + Beffara/Nolin. Still, having a direct SLE 6 proof is desirable B) Look for a direct PDE proof C) Look for a proof based on the suggested exploration procedure (process X t ) C. Garban (univ. Lyon 1) On the backbone exponent 23 / 30

Let s investigate the PDE approach Issues: The operator L is not self-adjoint Existence of a positive eigenfunction? No Hilbertian structure... Which PDE? t G LG. Boundary conditions are rather singular Which domain for this operator? Viscosity solutions? Maximum Principle and generalized principal eigenvalue for degenerate elliptic operators. Berestycki, Capuzzo Dolcetta, Porretta, Rossi, 2013. Due to the possible loss of regularity, as well as of boundary conditions, which is caused by degeneracy of ellipticity, the appropriate framework to deal with this problem is, even in the linear case, that of viscosity solutions. C. Garban (univ. Lyon 1) On the backbone exponent 24 / 30

Instead, look for a proof based on the suggested exploration procedure This goes into three steps: 1 Prove the convergence of the exploration procedure towards a well-defined stochastic process X t 2 Obtain a candidate (λ) for the exponent from a sub-additivity property. 3 Identify λ α 2 C. Garban (univ. Lyon 1) On the backbone exponent 25 / 30

Step 1: convergence of the exploration procedure 2π L = 3 2 α β + cot( α 2 ) α + cot( β 2 ) β β 0 α 2π dα t dβ t = κdb t + cot( αt 2 )dt = κdb t + cot( βt 2 )dt Plus Neumann on β t Plus Constant bdy condition C. Garban (univ. Lyon 1) On the backbone exponent 26 / 30

Step 2: a sub-additivity property. For f in the domain of L, we expect P t+s f (x) P t f (x)p s f (x) Note that if needed, one may rely here on Hörmander s theorem C. Garban (univ. Lyon 1) On the backbone exponent 27 / 30

Step 2: a sub-additivity property. For f in the domain of L, we expect P t+s f (x) P t f (x)p s f (x) Note that if needed, one may rely here on Hörmander s theorem There are more directs ways to extract an exponent. For example, introduce p δ t := P (0,0) [ u [t, t + δ], so that Xu = (0, 0) ] Easy observation: p 2δ t+s p δ t p δ s lemma There is a constant C > 0 so that for any u 0, p 2δ u Cp δ u+δ This gives for any t, s > 0, C p δ t+s+δ pδ t p δ s which implies the existence of λ := lim t log pδ t t C. Garban (univ. Lyon 1) On the backbone exponent 27 / 30

Step 3: identifying λ α 2 Let A t be the backbone event from e t to D. Our goal is to show that P [ A t ] = e λt+o(t) The lower bound is clear: P [ A t ] e λ(t 10)+o(t)

Step 3: identifying λ α 2 Let A t be the backbone event from e t to D. Our goal is to show that P [ A t ] = e λt+o(t) The lower bound is clear: P [ A t ] e λ(t 10)+o(t) For the upper bound: tempting to decompose as follows: P [ ] A t = P [ A t, Grey angle of X t [2 k, 2 k+1 ] ] k=1 But, this is not at all suitable to analysis! Instead, decompose as follows: P [ A t ] = P [ A t, T [t (k + 1), t k] ] k=1 k k p δ=1 t k P (0,0) [ { T > k} At ] e λ(t k)+o(t k)

Step 3: identifying λ α 2 Let A t be the backbone event from e t to D. Our goal is to show that P [ A t ] = e λt+o(t) The lower bound is clear: P [ A t ] e λ(t 10)+o(t) For the upper bound: tempting to decompose as follows: P [ ] A t = P [ A t, Grey angle of X t [2 k, 2 k+1 ] ] k=1 But, this is not at all suitable to analysis! Instead, decompose as follows: P [ A t ] = P [ A t, T [t (k + 1), t k] ] k=1 k p δ=1 t k P (0,0) [ { T > k} At ] Case study: 1 What if λ > α 3? 2 λ = α 3? k e λ(t k)+o(t k) e α3 k

Backbone exponent for κ 6? { dα t dβ t = κdb t + κ 4 2 cot( αt = κdb t + 1 cot( βt 2 2 )dt 6 κ )dt + 2 cot( αt 2 )dt Which gives the following PDE: t G = L κ [G] with L κ := κ 2 2 α β + κ 4 2 One can check L 6 = L. cot( α ( 2 ) α + cot( β 2 ) + 6 κ cot( α ) 2 2 ) β The vector field is very different when κ 6! C. Garban (univ. Lyon 1) On the backbone exponent 29 / 30

Conclusion SLE have provided a way to derive critical exponents rigorously (as opposed to Conformal Field Theory CFT.) The earlier meta-statement that most exponents are obtained as leading eigenvalues... may still hold if one wants to include the backbone exponent, but then what we mean by leading eigenvalue needs to be generalized accordingly (Viscosity solutions?... ) NUMERICS: viscosity solutions? Use this strengthened understanding of α 2 to compute a very good approximate value of α 2. This sheds some light on why the backbone exponent α 2 from the others and why it looks harder to compute singles out C. Garban (univ. Lyon 1) On the backbone exponent 30 / 30

Conclusion SLE have provided a way to derive critical exponents rigorously (as opposed to Conformal Field Theory CFT.) The earlier meta-statement that most exponents are obtained as leading eigenvalues... may still hold if one wants to include the backbone exponent, but then what we mean by leading eigenvalue needs to be generalized accordingly (Viscosity solutions?... ) NUMERICS: viscosity solutions? Use this strengthened understanding of α 2 to compute a very good approximate value of α 2. This sheds some light on why the backbone exponent α 2 from the others and why it looks harder to compute singles out Conjecture (Beffara, Nolin) α 2 = α 1 + α 2 C. Garban (univ. Lyon 1) On the backbone exponent 30 / 30