Law of the SLE tip

We analyze the law of the SLE tip at a fixed time in capacity parametrization. We describe it as the stationary law of a suitable diffusion process, and show that it has a density which is a unique solution of a certain PDE. Moreover, we identify the phases in which the even negative moments of the imaginary value are finite. For the negative second and negative fourth moments we provide closed-form expressions.


Introduction
The Schramm-Loewner evolution (SLE κ ) is a family of random planar fractal curves indexed by the real parameter κ ≥ 0, introduced by Schramm in [Sch00].These random fractal curves are proved to describe scaling limits of a number of discrete models that are of great interest in planar statistical physics.For instance, it was proved in [LSW04] that the scaling limit of loop-erased random walk (with the loops erased in a chronological order) converges in the scaling limit to SLE κ with κ = 2 .Moreover, other two-dimensional discrete models from statistical mechanics including Ising model cluster boundaries, Gaussian free field interfaces, percolation on the triangular lattice at critical probability, and uniform spanning tree Peano curves were proved to converge in the scaling limit to SLE κ for values of κ = 3, κ = 4, κ = 6 and κ = 8 respectively in the series of works [Smi10], [SS09], [Smi01] and [LSW04].There are also other models of statistical physics in 2D that are conjectured to have SLE κ , for some value of κ, as a scaling limit, among which is the two-dimensional self-avoiding walk which is conjectured to converge in the scaling limit to SLE 8/3 .For a detailed exposure and pedagogical introduction to SLE theory, we refer the reader to [RS05], [Law05], and [Kem17].
Questions concerning the behaviour of the SLE trace at the tip can be found in the existing body of SLE literature, for example in [JVL12] where the almost sure multi-fractal spectrum of the SLE trace near its tip is computed, and in [Zha16] in which the ergodic properties of the harmonic measure near the tip of the SLE trace are studied.

Main results
First, let us introduce the basic notation.For a domain D ⊂ R k , k ⩾ 1, let C ∞ (D, R) be a set of functions D → R which have derivatives of all orders.The set of functions from C ∞ (D, R) which are bounded and have bounded derivatives of all orders will be denoted by C ∞ b (D, R).As usual, for a function f : D → R d , d ⩾ 1, we will denote its supremum norm by ∥f ∥ ∞ := sup x∈D |f (x)|.Let H be the open complex upper half-plane {Im(z) > 0}.
Until the end of the paper we fix κ ∈ (0, ∞).Let g t : H t → H, t ⩾ 0, be the forward SLE flow, that is the solution to the Loewner ODE (2.1) Throughout the paper we use the notations γ(t) and γ t interchangeably.
Our main result is the following statement.
We have attached in Figure 1 numerical simulations of γ(1) with various values of κ.There we have chosen the coordinates (α, y) where α = arg γ(1) and y = Im γ(1) so that they fit well in the plot.
As an application of Theorem 2.1, we show that the following quantities can be explicitly calculated.
Theorem 2.2.The following holds: (i) For any measurable set Λ ⊂ H one has dx dy.
(2.4) for κ < 8/3. (2.6) Remark 2.3.Note that the left-hand side of (2.3) is an average amount of time SLE spends in a set Λ. A version of this identity has previously appeared in [Zha19, Corollary 5.3].However, in that paper the constant in front of the integral has been implicitly specified as 1/C κ,1 with . In particular, our result implies κ ) As we will point out in Section 3.2, our Theorem 2.2 may seem like a simple consequence of Theorem 2.1 that can be heuristically deduced from integration by parts arguments.However, it is surprisingly tricky to control the boundary behaviour of ψ and its derivatives.Therefore it requires more work to rigorously prove Theorem 2.2.One of our initial motivations was to know more about the marginal law of α = arg γ(1).We believe that the marginal density should behave like α 8/κ as α ↘ 0. We did not succeed in proving this; instead, we prove the following in Section 3.2.Denote (α, y) = (arg γ(1), Im γ(1)) and let q(α, y) = ψ(y cot α, y) y sin 2 α the density in these coordinates.Then for n ⩾ 1 we have 2 0 y −2n q(α, y) dy ≈ α 8/κ−2n as α ↘ 0. Remark 2.4.The support of the density is quite natural since the half-plane capacity of γ[0, t] is always at least 1 2 Im γ(t) 2 , and hence we always have Im √ t which is driven by the constant driving function.
To obtain these results we establish the following lemma which links the law of SLE κ with invariant measure of a certain diffusion process.Introduce the reverse SLE flow where B is the time-reversed Brownian motion, that is, where B ′ is a Brownian motion independent of B. It is obvious that B is a Brownian motion.

Proofs
3.1 Proofs of Lemma 2.5 and Theorem 2.1 We begin with the proof of Lemma 2.5.
Proof of Lemma 2.5.
Recalling the definition of f and taking t = 1, we obtain (3.1).
Next, we note that the following scaling property holds: for any c > 0 Indeed, using again the definition of h in (2.7), we see that for any t ⩾ 0 Since the process ( 1 c √ κ B c 2 t ) t⩾0 has the same law as ( √ κ B t ) t≥0 and the solution of the Loewner differential equation is a deterministic function of the driver, we see that (3.2) holds.
where we have also used the fact that converges in law to γ(1) as u ↘ 0. This implies the statement of the lemma.
Recall the definition of the reverse SLE flow h in (2.7) and the reversed Brownian Motion B in (2.8).Lemma 2.5 implies the following result.
Corollary 3.1.Let ( X t , Z t ) t⩾0 be the stochastic process that satisfies the following equation with the initial data ; here B t := − t 0 e −s/2 d B e s and the filtration F t := σ( B r , r ∈ [0, e t ]).Then ( X t , Z t ) → (Re(γ 1 ), log(Im(γ 1 ))) in law as t → ∞. (3.5) Note that the initial value of the process ( X, Z) is random but measurable with respect to F 0 .
Then, it follows from (2.7) that X 0 = 0, Y 0 = 1.For t ⩾ 0, let X t := e −t/2 X e t and Y t := e −t/2 Y e t .We apply Itô's formula to derive Clearly, B is a standard Brownian motion with respect to the filtration F t .By definition, we also have X 0 = X 1 , Y 0 = Y 1 .The change of variables Z t := log Y t and another application of Itô's formula implies that the process ( X, Z) t⩾0 satisfies SDE (3.3)-(3.4)with the initial conditions Furthermore, It follows from Corollary 3.1 that to prove Theorem 2.1 one needs to study invariant measures of (3.3)-(3.4).PDE (2.2) is then the Fokker-Planck-Kolmogorov equation for this process.However, since the coefficients have a singularity at 0, a bit of care is needed to make the statements rigorous.
First, we show that this SDE is well-posed and is a Markov process.We will need the following notation.For a vector field U : R 2 → R 2 denote its derivative matrix by (DU ) i,j := ∂ x j U i .The Lie bracket between two vector fields U, V : R 2 → R 2 is given by We begin with the following technical statement.
Lemma 3.3.For any random vector ( x 0 , z 0 ) independent of B the stochastic differential equation (3.3)-(3.4)has a unique strong solution with ( X 0 , Z 0 ) = ( x 0 , z 0 ).This solution is a Markov process in the state space R 2 and its transition kernel P t is strong Feller for any t > 0.
Proof.First, we consider the case when the initial data ( x 0 , z 0 ) is deterministic.Then it is immediate to see that for any T > 0 a solution to (3.3)-(3.4)satisfies Strong existence for the case of arbitrary initial data follows now from [Kal96, Theorem 1], and strong uniqueness from [IW89, Remark IV.1.4].Moreover, [KS91,Theorem 5.4.20]shows that ( X t , Z t ) t⩾0 is a Markov process with the state space R 2 equipped with the Borel topology.Now let us show that (P t ) t⩾0 is strong Feller.Let f be an arbitrary bounded measurable function Without loss of generality we can assume that z n 0 ⩾ −2|z 0 | for all n ∈ Z + .Fix t > 0.Then, denoting by (P ε t ) t⩾0 the transition kernel associated with SDE (3.12)-(3.13),we derive where ε := exp(−4|z 0 | − t) and we used here (3.16) and (3.17).By Lemma 3.2, we have here we used once again (3.16) and (3.17).Combining (3.18) and (3.19), we see that P t is strong Feller.
To show uniqueness of the invariant measure of (P t ), we will need the following support theorem.For δ > 0, v ∈ R 2 let B δ,v be the ball of radius δ centred at v. Lemma 3.4.For any (x 0 , z 0 ) ∈ R 2 , δ > 0, there exists T > 0 such that where x(0) = x 0 , z(0) = z 0 and U ∈ C 1 ([0, T ]; R) is a non-random function with U 0 = 0. We claim that we can find T > 0 and U such that x T = 0 and |z T − log 2| < δ/2.First, we take a , be a solution to (3.21) with the initial condition z 0 (for x constructed above).
Consider now the equation with the initial condition z 1 constructed above.It is easy to see that there exists Finally, let U t , t ∈ [0, T ], be a C 1 path such that (3.20) holds for x, z constructed above and U 0 = 0.The desired control U has been constructed.Now for arbitrary ε > 0, consider the event It is well-known (see, e.g., [Fre71,Theorem 38]) that P(A ε ) > 0. Let ( X t , Z t ) t∈[0,T ] be the solution of (3.3)-(3.4)with the initial condition (x 0 , z 0 ).Then where we used (3.22) and the fact that the Lipschitz constant of the drift of SDE . By the Gronwall inequality and (3.23), we have on Choose now ε small enough, such that the right-hand side of the above inequality is less than δ/2.Then recalling that x T = 0 and |z T − log 2| < δ/2, we finally deduce Lemma 3.5.The measure π := Law Re(γ 1 ), log(Im(γ 1 )) is the unique invariant measure for the process (3.3)-(3.4).
Proof.The fact that the measure π is invariant follows by a standard argument.Denote, as usual, for a measurable bounded function f : R 2 → R and a measure ν on R 2 Consider the measure µ := Law Re(h ) .Rewriting (3.5), we see that P t µ → π weakly as t → ∞. (3.24) Fix any s ⩾ 0. Let us show that P s π = π.Indeed, let f : R 2 → R be an arbitrary continuous bounded function.Then where the second identity follows from (3.24) and the fact that P s f is a bounded continuous function (this is guaranteed by the Feller property of P ).Since f was arbitrary bounded continuous function, we see that P s π = π for any s ⩾ 0. Thus, the measure π is invariant for SDE (3.(3.25) We claim now that the point (0, log 2) belongs to the support of both of these measures.Indeed, fix arbitrary δ > 0. Take any (x 0 , z 0 ) ∈ supp(ν).Then, by Lemma 3.4, there exists T > 0, ε > 0 such that P T ((x 0 , z 0 ), B δ,(0,log 2) ) > ε.By the strong Feller property of P T , the function (x, z) → P T ((x, z), B δ,(0,log 2) ) is continuous.Therefore, there exists δ ′ > 0 such that where the last inequality follows from the fact that (x 0 , z 0 ) ∈ supp(ν).Since δ was arbitrary, we see that (0, log 2) ∈ supp(ν).Similarly, (0, log 2) ∈ supp( ν), which contradicts (3.25).Therefore, SDE (3.3)-(3.4)has a unique invariant measure.
Let L be the generator of the semigroup P As usual, the adjoint of L will be denoted by L * .
Lemma 3.6.The measure π := Law Re(γ 1 ), log(Im(γ 1 )) has a smooth density p with respect to the Lebesgue measure.Further, p is the unique solution in the class of densities of the Fokker-Planck-Kolmogorov equation (3.26) Finally, p(x, z) = 0 for x ∈ R, z ⩾ log 2, and p(x, z) > 0 for x ∈ R, z < log 2.
Proof.Since the measure π is invariant for P , we have (in the weak sense) Let us now check that L * satisfies the (standard) Hörmander condition.
Denote by b the drift of Thus, we assume the contrary and suppose that p ′ is another probability density which solves (3.26).Let π ′ be the measure with density p ′ .We claim that π ′ is another invariant measure for (P t ).
Consider a Lyapunov function V (the suggestion to take this specific function is due to Stas Shaposhnikov) .
Combining this with (3.29), we see that for any T > 0 Therefore, by the generalized Ambrosio-Figalli-Trevisan superposition principle [BRS21, Theorem 1.1] and the standard equivalence between weak solutions of SDE and the martingale problems, see, e.g., [KS91, Proposition 5.4.11],there exists a weak solution to SDE (3.3)-(3.4) on the interval [0, T ] such that for any t ⩾ 0 we have Law( X t , Z t ) = π ′ .Thus, the measure π ′ is also invariant for the semigroup (P t ).However, this contradicts Lemma 3.5.Therefore, (3.26) has a unique solution in the class of probability densities.Finally, let us prove the results concerning the support of p.Note that if Z 0 (ω) > log 2, then Z 0 (ω) > Z 1 (ω).Let f : R → [0, ∞) be an increasing function such that f (x) = 0 for x ⩽ log 2 and f (x) > 0 for x > log 2. Then f ( Z 0 ) − f ( Z 1 ) ⩾ 0. On the other hand, by invariance This implies that P π a.s.we have f ( Z 0 ) = f ( Z 1 ).By the definition of f this implies that P π ( Z 0 > log 2) = 0 and thus π(R × (log 2, ∞)) = 0. Since the density p is continuous we have p(x, z) = 0, x ∈ R, z ⩾ log 2. (3.30) Figure 2: Support of the density p (yellow and red regions).The process Z t is increasing when ( X t , Z t ) is in the red region, and decreasing whenever ( X t , Z t ) is in the yellow region.The dashed line, which touches the red region, is z = log 2. Now let us show that p(x, z) > 0 for any z < log 2. The idea of this part of the proof is due to Stas Shaposhnikov.Suppose the contrary that for some x 0 ∈ R, z 0 < log 2 we have p(x 0 , z 0 ) = 0. We claim that this implies that p ≡ 0. Note that the set {z = z 0 } is the set of elliptic connectivity for operator L * in the sense of [OR73, Chapter III.1] (see also [Hil70, Section 2]).Therefore, the maximum principle for degenerate elliptic equations [OR73, Theorem 3.1.2](see also [Hil70, Theorem 1], [Ale58, Theorem 4]) implies that p(x, z 0 ) = 0 for any x ∈ R.
Note that in the domain PDE (3.26) becomes a parabolic equation in (z, x) and on its complement (3.26) is a backward parabolic equation.This corresponds to the fact that the process Z t is increasing on D and decreasing on R 2 \ D, see Figure 2.
Fix now small δ such that δ 2 + exp(2z 0 ) < 4 (this is possible since z 0 < log 2).Consider now the domain D for certain smooth functions a, b, c and Therefore, by the Harnack inequality for parabolic equations (see, e.g., [Eva98, Section 7.1, Theorem 10]), we get for arbitrary z 1 ⩽ z 0 , and C > 0 sup Using again the maximum principle for degenerate elliptic equations, we deduce from this that p(x, z 1 ) = 0 for any x ∈ R. Since z 1 ⩽ z 0 was arbitrary we have that We use a similar argument to treat the case z ⩾ z 0 .Consider now the domain p(x, z 0 ) = 0. and thus, as above, the maximum principle implies that p ≡ 0 on R × [z 0 , ∞).
Therefore the function p is identically 0 which is not possible since p is a density.This contradiction shows that p(x, z) > 0 for any x ∈ R, z < log 2. Together with (3.30) this concludes the proof of the theorem.
It follows from Theorem 2.1 that the density φ is the unique solution to the corresponding Fokker-Planck-Kolmogorov equation, which in the new coordinates is given by Recall that we can consider this equation on a larger domain (0, π) × (0, ∞), but since ψ(x, y) = 0 for y ⩾ 2, we have φ(α, u) = 0 for u ⩾ 4. Note that this PDE can be rewritten as The crucial statement on the way to prove Theorem 2.2 is the following lemma.

This yields that
Lemma 3.8.Let S, T ∈ R, S ⩽ T .Suppose x k : [S, T ] → R d , k ∈ N, are continuous functions that solve the integral equation where Moreover, suppose that there exist Then there exists a continuous function x : [S, T ] → R d such that along some subsequence (k j ) j∈Z + we have x k j → x uniformly as j → ∞ and (3.36) Proof.First, we show that x k are uniformly bounded.Indeed, by our assumptions we have for any t ∈ [S, T ] and an application of Grönwall's inequality implies x k are uniformly bounded.Consequently, we can assume F to be bounded.It follows that the family Hence, by the Arzelà-Ascoli theorem, we have x k j → x uniformly along some subsequence.Equation (3.36) follows now from (3.35) by taking limits.
We will later also frequently use integation by parts arguments.In order to control the boundary terms that appear, the following lemma will be useful.Lemma 3.9.Let T > 0, and f : (0, T ] → R be a differentiable function such that T ε f (s) ds neither diverges to +∞ nor −∞ as ε ↘ 0. Let h : (0, T ] → (0, ∞) be a non-increasing differentiable function such that T 0 h(s) ds = +∞.Then there exists a sequence t k ↘ 0 such that Proof.First we note that there must exist a sequence s k ↘ 0 such that |f (s k )| < h(s k ) for all k ∈ Z + , otherwise we would have To control f ′ , we distinguish two cases.
Case 1: We have |f (t)| ⩽ h(t) for all small t.In that case, consider g(t) := h(t) − f (t).The function g cannot be always increasing for small t, otherwise we would have T 0 f (s) ds = ∞.Consequently there must be a sequence t k ↘ 0 such that g ′ (t k ) ⩽ 0.
Case 2: We have |f (r k )| > h(r k ) along a sequence r k ↘ 0. We can pick the sequence such that either f (r k ) > h(r k ) for all k or f (r k ) < −h(r k ) for all k.In the former case ) is well-defined and tends to 0 due to the existence of (s k ) as above).Then Corollary 3.10.Consider the same setup as Lemma 3.9, and suppose additionally that f ⩾ 0. Then there exists a sequence t k ↘ 0 such that (3.37) Proof.Let (t k ) k∈Z + be a sequence as in Lemma 3.9.Fix now k |} (this set is non-empty, otherwise we would have f (t) → −∞ as t ↘ 0).By definition, we have f ′ (t) > |h ′ (t)| for t ∈ ]s k , t k ], and hence also f (s k ) < f (t k ).Moreover, we find some

Since the derivative of any differentiable function satisfies the intermediate value theorem, we find some
We now proceed to the main part of our proof.In the following, we denote for n ∈ Z + , α ∈ (0, π), and ε > 0 From the equation (3.32), we will deduce a recursive system of ODEs that are satisfied for the functions I n .In fact, the relation is satisfied for general n ∈ R but we will use it only with n ∈ Z + .
Thus we see that all the conditions of Lemma 3.8 are satisfied.By passing to the limit as ε → 0 in (3.40) and using continuity of g, we get (3.39).
To make this heuristic precise, we find a suitable subsequence α k ↘ 0 where we can apply a similar argument.
Let us now proceed to the rigorous induction on n.Base case.n = 1.In this case (3.51), (3.52) and continuity of I 1 was already proven in (3.34).The fact that π 0 I 1 (α) dα is finite if and only if κ < 8 is immediate.Inductive step.Suppose that the statement of the lemma is valid for n ∈ Z + .Let us prove it for n + 1.
If κ ⩾ 8/(2n − 1), then π 0 I n (α) dα = ∞, and this obviously implies that π 0 I n+1 (α) dα = ∞.Therefore it is sufficient to consider the case κ < 8/(2n − 1).By the inductive step, for these values of κ we have π 0 I n (α) dα < ∞.Hence, Lemma 3.12 implies that condition (3.43) holds.This, together with continuity of I n , shows that all the conditions of Lemma 3.11 are met.Note that we cannot have I n+1 = ∞ for all α ∈ (0, π).Indeed, in this case the left-hand side of identity (3.42) would be infinite but the right-hand side of this identity is finite (because it is equal to C π 0 I n (α) dα).Thus, Lemma 3.11 implies that I n+1 is twice differentiable and satisfies (3.39).
Therefore inequality (3.58) holds for a certain r ∈ [−3, s) which is a contradiction as before.
To show (2.6), fix κ < 8/3.Note that in this regime by Lemma 3.14, we have Unfortunately, we do not have an explicit formula for π 0 In(α) sin 2 α dα for n ⩾ 2. This prevents us from getting explicit formulas of negative moments of Im(γ 1 ) of higher order.
Remark 3.16.Another possible approach to find explicit formulas for I n would be through its Fourier coefficients Formally expanding (3.62), we obtain a (countable) system of linear equations for (a j ) j⩾0 in terms of the Fourier coefficients (b j ) j⩾0 of the function I n−1 / sin 2 α.However, it seems difficult to solve the system of equations explicitly.Only for a 0 , a 1 we get a system of two equations in terms of b 0 , b 1 which correspond exactly to what we obtain from the proof above.
2(X e t + iY e t ) = X t + i Y t .
Theorem 7.4.3],L * is hypoelliptic. 1Therefore, (3.27) implies that the Schwarz distribution π ∈ C ∞ (R 2 ).Thus, the measure π has a C ∞ density p with respect to the Lebesgue measure and (3.26) holds.Now let us show that (3.26) does not have any other solutions.We have already seen that semigroup (P t ) has a unique invariant measure (this has been established in Lemma 3.5).In general, without extra conditions, this does not immediately imply uniqueness of solutions to (3.26) in the class of probability measures, see [BKRS15, hint to exercise 9.8.48].This is because not every probability solution to the Fokker-Planck-Kolmogorov equation corresponds to a solution of the martingale problem; we refer to [BRS21, p. 719] for further discussion.