Some Properties of Annulus SLE

An annulus SLE$_\kappa$ trace tends to a single point on the target circle, and the density function of the end point satisfies some differential equation. Some martingales or local martingales are found for annulus SLE$_4$, SLE$_8$ and SLE$_{8/3}$. From the local martingale for annulus SLE$_4$ we find a candidate of discrete lattice model that may have annulus SLE$_4$ as its scaling limit. The local martingale for annulus SLE$_{8/3}$ is similar to those for chordal and radial SLE$_{8/3}$. But it seems that annulus SLE$_{8/3}$ does not satisfy the restriction property.


Introduction
Schramm-Loewner evolution (SLE) is a family of random growth processes invented by O. Schramm in [12] by connecting Loewner differential equation with a one-dimentional Browinian motion. SLE depend on a single parameter κ ≥ 0, and behaves differently for different value of κ. Schramm conjectured that SLE(2) is the scaling limit of some loop-erased random walks (LERW) and proved his conjuecture with some additional assumptions. He also suggested that SLE (6) and SLE (8) should be the scaling limits of certain discrete lattice models. After Schramm's paper, there were many papers working on SLE. In the series of papers [4][5] [6], the locality property of SLE (6) was used to compute the intersection exponent of plane Brwonian motion. In [14], SLE (6) was proved to be the scaling limit of the cite percolation explorer on the triangle lattice. It was proved in [7] that SLE(2) is the scaling limit of the corresponding loop-erased random walk (LERW), and SLE(8) is the scaling limit of some uniform spanning tree (UST) Peano curve. SLE(4) was proved to be the scaling limit of the harmonic exploer in [13]. SLE(8/3) satisfies restriction property, and was conjectured in [8] to be the scaling limit of some self avoiding walk (SAW). Chordal SLE(κ, ρ) processes were also invented in [8], and they satisfy one-sided restriction property. For basic properties of SLE, see [11], [3], [16], [15].
The SLE invented by O. Schramm has a chordal and a radial version. They are all defined in simply connected domains. In [17], a new version of SLE, called annulus SLE, was defined in doubly connected domains as follows.
Then A p is bounded by C p and C 0 . Let ξ(t), 0 ≤ t < p, be a real valued continuous function. For z ∈ A p , solve the annulus Loewner differential equation ∂ t ϕ t (z) = ϕ t (z)S p−t (ϕ t (z)/ exp(iξ(t))), 0 ≤ t < p, ϕ 0 (z) = z, where for r > 0, For 0 ≤ t < p, let K t be the set of z ∈ A p such that the solution ϕ s (z) blows up before or at time t. Then for each 0 ≤ t < p, ϕ t maps A p \ K t conformally onto A p−t , and maps C p onto C p−t . We call K t and ϕ t , respectively, 0 ≤ t < p, the annulus LE hulls and maps, respectively, of modulus p, driven by ξ(t), 0 ≤ t < p. If (ξ(t)) = √ κB(t), 0 ≤ t < p, where κ ≥ 0 and B(t) is a standard linear Brownian motion, then K t and ϕ t , 0 ≤ t < p, are called standard annulus SLE κ hulls and maps, respectively, of modulus p. Suppose D is a doubly connected domain with finite modulus p, a is a boundary point and C is a boundary component of D that does not contain a. Then there is f that maps A p conformally onto D such that f (1) = a and f (C p ) = C. Let K t , 0 ≤ t < p, be standard annulus SLE κ hulls. Then (f (K t ), 0 ≤ t < p) is called an annulus SLE κ (D; a → C) chain. It is known in [17] that annulus SLE κ is weakly equivalent to radial SLE κ . so from the existence of radial SLE κ trace, we know the existence of a standard annulus SLE κ trace, which is β(t) = ϕ −1 t (exp(iξ(t))), 0 ≤ t < p. Almost surely β is a continuous curve in A p , and for each t ∈ [0, p), K t is the hull generated by β((0, t]), i.e., the complement of the component of A p \ β((0, t]) whose boundary contains C p . It is known that when κ = 2 or κ = 6, lim t→p β(t) exists and lies on C p almost surely. In this paper, we prove that this is true for any κ > 0. And we discuss the density function of the distribution of the limit point. The density function should satisfy some differential equation.
When κ = 2, 8/3, 4, 6, or 8, radial and chordal SLE κ satisfy some special properties. Radial SLE 6 satisfies locality property. Since annulus SLE 6 is (strongly) equivalent to radial SLE 6 , so annulus SLE 6 also satisfies the locality property. Annulus SLE 2 is the scaling limit of the corresponding loop-erased random walk. In this paper, we discuss the cases κ = 4, 8, and 8/3. We find martingales or local martingales for annulus SLE κ in each of these cases. From the local martingale for annulus SLE 4 , we may construct a harmonic explorer whose scaling limit is annulus SLE 4 . The martingales for annulus SLE 8/3 are similar to the martingales for radial and chordal SLE 8/3 , which are used to show that radial and chordal SLE 8/3 satisfy the restriction property. However, the martingales for annulus SLE 8/3 does not help us to prove that annulus SLE 8/3 satisfies the restriction property. On the contrary, it seems that annulus SLE 8/3 does not satisfy the restriction property.

Annulus Loewner Evolution in the Covering Space
We often lift the annulus Loewner evolution to the covering space. Let e i denote the map z → e iz . For p > 0, let S p = {z ∈ C : 0 < Im z < p}, R p = ip + R, and For 0 ≤ t < p, let K t be the set of z ∈ S p such that ϕ t (z) blows up before or at time t. Then for each 0 ≤ t < p, ϕ t maps S p \ K t conformally onto S p−t , and maps R p onto R p−t . And for any k ∈ Z, ϕ t (z + 2kπ) = ϕ t (z) + 2kπ. We call K t and ϕ t , 0 ≤ t < p, the annulus LE hulls and maps, respectively, of modulus p in the covering space, driven by ξ(t), 0 ≤ t < p. Then we have K t = (e i ) −1 (K t ) and e i • ϕ t = ϕ t • e i . If (ξ(t)) 0≤t<p has the law of ( √ κB(t)) 0≤t<p , then K t and ϕ t , 0 ≤ t < p, are called standard annulus SLE κ hulls and maps, respectively, of modulus p in the covering space. It is clear that H r is an odd function. It is analytic in C except at the set of simple poles {2kπ + i2mr : k, m ∈ Z}. And at each pole z 0 , the principle part is 2 z−z 0 . For each z ∈ C, H r (z + 2π) = H r (z), and H r (z + i2r) = H r (z) − 2i. Let Then f r is an odd function. It is analytic in C except at the set of simple poles {z ∈ C : i π r z = 2kπ + i2mπ 2 /r, for some k, m ∈ Z} = {2mπ − i2kr : m, k ∈ Z}. And at each pole z 0 , the principle part is 2 z−z 0 . We then compute Let g r (z) = f r (z) − H r (z). Then g r is an odd entire function, and satisfies g r (z + 2π) = g r (z) + 2π/r, g r (z + i2r) = g r (z) + 2i, for any z ∈ C. Thus g r (z) = z/r. So we have

Long Term Behaviors of Annulus SLE Trace
In this section we fix κ > 0 and p > 0. Let ϕ t and K t , 0 ≤ t < p, be the annulus LE maps and hulls, respectively, of modulus p driven by ξ(t) = √ κB(t), 0 ≤ t < p. Let ϕ t and K t be the corresponding annulus LE maps and hulls in the covering space. Let β(t) be the corresponding annulus SLE κ trace.
. From Ito's formula and equation (3) we have which implies that Proof. Suppose the lemma is not true. Then there is z 0 ∈ R p and a > 0 such that the probability that |X s (z 0 )| < a for all s ∈ [π 2 /p, ∞) is positive. Let X s denote X s (z 0 ). Then we have Let T a be the first time that |X s | = a. If such time does not exist, then let T a = ∞. Let Then 2 sinh(X s ) cosh(2ks) + cosh(X s ) ds. Let where B 2 (r) is another standard one dimensional Brownian motion. And T a is a stopping time w.r.t. B 2 (r). Let ; Thus the Nivikov's condition is satisfied. Let P denote the original measure for B 2 (r). Define Q on F Ta such that dQ (ω) = M Ta (ω)dP (ω). Then ( U r , 0 ≤ r < T a ) is a one dimensional Brownian motion started from 0 and stopped at time T a w.r.t. the probability law Q . For 0 ≤ s < T a , is a one dimensional Brownian motion w.r.t. Q , so on the event that T a = ∞, Q {lim sup r→∞ | U r | < ∞} = 0. Thus Q {lim sup r→∞ | U r | = ∞} > 0. Since P and Q are equivalent probability measures, so P {lim sup r→ Ta | U r | = ∞} > 0. Thus P {lim sup s→Ta |U s | = ∞} > 0. This contradicts the fact that for all s ∈ [π 2 /p, ∞), U s ∈ (0, C(κ)) and C(κ) < ∞. Thus the hypothesis is wrong, and the proof is completed. ✷ From this lemma and the definition of X t , we know that for any z ∈ R p , (Re ϕ t (z) − √ κB(t))/(p − t) is not bounded on t ∈ [0, p) a.s.. Since for any k ∈ Z and z ∈ R p ,

Lemma 3.2
For every z ∈ R p , almost surely lim s→∞ X s (z)/s exists and the limit is an odd integer.
Proof. Fix ε 0 ∈ (0, 1/2) and z 0 ∈ R p . Let X s denote X s (z 0 ). There is b > 0 such that the probability that | √ κB(t)| ≤ b + ε 0 t for any t ≥ 0 is greater than 1 − ε 0 . Since coth(x/2) → ±1 as x ∈ R and x → ±∞, so there is R > 0 such that when ±x ≥ R, ± coth(x/2) ≥ 1 − ε 0 . Let T = R + b + 1. If for any s ≥ 0, |X s − 2ks| < T for some k = k(s) ∈ Z, then there is k 0 ∈ Z such that |X s − 2k 0 s| < T for all s ≥ T . From the argument after Lemma 3.1, the probability of this event is 0. Let s 0 be the first time that |X s − 2ks| ≥ T for all k ∈ Z. Then s 0 is finite almost surely. There is k 0 ∈ Z such that 2k 0 s 0 + T ≤ X s 0 ≤ 2(k 0 + 1)s 0 − T . Let s 1 be the first time after s 0 such that From equation (4), we have that for s ∈ [s 0 , s 1 ), from which follows that has probability greater than 1 − ε 0 . Since we may choose ε 0 > 0 arbitrarily small, so a.s. lim s→∞ Re ϕ s (z 0 )/s exists and the limit is 2k 0 + 1 for some k 0 ∈ Z. The proof is now finished by the facts that If the event that m − < m + has a positive probability, then there is a ∈ R such that the event that m − < a < m + has a positive probability. From the definitions, m − < a < m + implies that lim s→∞ X s (a + ip)/s ∈ (−1, 1), which is an event with probability 0 by Lemma 3.2. This contradiction shows that Therefore lim s→∞ X s (x + ip)/s = 2k + 1.
Let K p = ∪ 0≤t<p K t and K p = ∪ 0≤t<p K t . Then K p = e i ( K p ), and so K p = e i ( K p ).
, so by Koebe's 1/4 theorem, the distance between c(s) + ip and K p−π 2 /s tends to 0 as s → ∞. This is a contradiction.
From equation (5), for all s ≥ s 0 , , and L t , 0 ≤ t < q, are standard annulus SLE κ hulls of modulus q, then the probability that ∪ 0≤t<q L t ⊂ {e iz : |Re z| ≤ C 0 q} is greater than 1 − ε.
Proof. Let q 0 = 2π 2 ln (2) . Suppose q ∈ (0, q 0 ]. Let L t and ψ t , 0 ≤ t < q, be the annulus LE hulls and maps of modulus q driven by √ κB(t), 0 ≤ t < q. Let L t and ψ t , 0 ≤ t < q, be the corresponding annulus LE hulls and maps in the covering space.
Let R = ln(64) and where we use the fact that e −R ≤ 1 64 and e −2s ≤ e −2s 0 = e −2π 2 /q ≤ e −2π 2 /q 0 ≤ 1 2 . Thus Then we have This contradicts that Re Then the claim is justified, and the proof is finished. ✷

For two nonempty sets
Form conformal invariance and comparison principle of extremal distance, we have that for any d > 0, there is h(d) > 0 such that for any p > 0, if for j = 1, 2, A j is a union of connected subsets of A p , each of which touches both C p and C 0 , and the extremal distance between A 1 and A 2 in A p is greater than h(d), then d a (A 1 , A 2 ) > dp. Proof. From Lemma 3.3, the distance from e −p+im to K t tends to 0 as t → p a.s.. Since K t is the hull generated by β((0, t]), so the distance from e −p+im to β((0, t]) tends to 0 as t → p a.s.. Suppose the theorem does not hold. Then there is a, δ > 0 such that the event that lim sup t→p |e −p+im − β(t)| > a has probability greater than δ. Let E 1 denote this event. Let ε = δ/4. Let C 0 depending on ε be as in Lemma 3.4. Let R = min{a, e −p } and r = min{1 − e −p , R exp(−2πh(2C 0 + 1))}, where h is the function in the argument before this theorem. Since K t is generated by β((0, t]), and e −p+im ∈ K p a.s., so the distance between e −p+im and β((0, t]) tends to 0 a.s. as t → p. So there is t 0 ∈ (0, p) such that the event that the distance between e −p+im and β((0, t 0 ]) is less than r has probability greater than 1 − ε. Let E 2 denote this event. Let q 0 = 2π 2 ln(2) , T = max{t 0 , p − q 0 , − ln(r + e −p )}, T (exp(iξ(T ))z)/ exp(iξ(T )) for 0 ≤ t < p T . Then one may check that K T,t and ϕ T,t , 0 ≤ t < p T , are the annulus LE hulls and maps of modulus p T driven by ξ T . Since ξ T (t) has the same law as √ κB(t) and p T = p − T ≤ q 0 , so from Lemma 3.4, the event that diam a (∪ 0≤t<p T K T,t ) ≤ 2C 0 p T has probability greater than 1 − ε. Let E 3 denote this event. Since P (E c This means that the events E 1 , E 2 and E 3 can happen at the same time. We will prove that this is a contradiction. Then the theorem is proved. , resp.) that touch C p . From the properties of β in the event E 1 and E 2 , we see that A r and A R both intersect K p \ K T . Since the distance between e −p+im and K T is less than r, and r < R, so both A r and A R are unions of two curves which touch both C p and C 0 ∪ K T . Let B r = e −iξ(T ) ϕ T (A r ) and B R = e −iξ(T ) ϕ T (A R ). Then both B r and B R are unions of two curves in A p T that touch both C p T and C 0 .
The extremal distance between A r and A R in A p \ K T is at least ln(R/r)/(2π) ≥ h(2C 0 + 1). Thus the extremal distance between B r and B R in A p T is at least h(2C 0 + 1). So the angular distance between B r and B R is at least . Then β is a simple curve in S p started from 0, and β(t) = e i ( β(t)). From Theorem 3.1, lim t→p β(t) exists and lies on R p . We call β an annulus SLE κ trace in the covering space. Let m p + ip denote the limit point, where m p is a real valued random variable.
Suppose the distribution of m p is absolutely continuous w.r.t. the Lebesgue measure, and the density function λ(p, x) is C 1,2 continuous. This hypothesis is very likely to be true, but the proof is still missing now. We then have R λ(p, x)dx = 1 for any p > 0. Since the distribution of β is symmetric w.r.t. the imaginary axis, so is the distribution of lim t→p β(t). Thus λ(p, −x) = λ(p, x). Moreover, we expect that when p → 0 the distribution of (m p + ip) * π p tends to the distribution of the limit point of a strip SLE κ trace introduced in [18], whose density is cosh(x/2) −4/κ /C(κ) for some C(κ) > 0. If this is true, then the distribution of m p tends to the point mass at 0 as p → 0.
For 0 ≤ t < p, let F t be the σ−algebra generated by ξ(s), Then ∂ t ϕ T,t (z) = H p T −t ( ϕ T, (z) − ξ T (t)), and ϕ T,0 (z) = z. Thus ϕ T,t (z), 0 ≤ t < p T , are annulus LE maps of modulus p T in the covering space driven by ξ T (t), 0 ≤ t < p T , and so are independent of F T . Let for 0 ≤ t < p T . Then β T (t), 0 ≤ t < p T , is a standard annulus SLE κ trace of modulus p T in the covering space, and is independent of F T . Thus lim t→p T β T (t) exists and lies on R p T a.s.. Let m p T + ip T denote the limit point. Then m p T is independent of F T , and the density of m p T w.r.t. the Lebesgue measure is λ(p T , ·). From equation (6), we see Since m p T has density λ(p T , ·) and is independent of F T , and X T is F T measurable, so where ∂ 1 and ∂ 2 are partial derivatives w.r.t. the first and second variable, respectively. Let Λ(p, x) = x 0 λ(p, s)ds for p > 0 and x ∈ R. Then for any p > 0, Λ(p, ·) is an odd and increasing function, lim x→±∞ Λ(p, x) = ± 1 2 , and λ(p, x) = ∂ 2 Λ(p, x). Thus for any r > 0 and x ∈ R, Since Λ(r, ·) is an odd function and H r (0) = 0, so Thus for any r > 0 and x ∈ R, we have And we expect that for any x ∈ R \ {0}, lim r→0 Λ(r, x) → sign 1 2 . On the other hand, if Λ(r, x) satisfies (8), then λ(r, x) := ∂ 2 Λ(r, x) satisfies (7).
Let λ(r, x) = k∈Z λ(r, x + 2kπ). Then λ(r, ·) has a period 2π, and is the density function of the distribution of the argument of lim t→r β(t), where β is a standard annulus SLE κ trace of modulus r. So it satisfies π −π λ(r, x)dx = 1. And λ(r, ·) is an even function for any r > 0. Since H r has a period 2π, so λ(r, x) also satisfies equation (7). Let Λ(r, x) = x 0 λ(r, s)ds. Then Λ(r, x) satisfies (8). But Λ(r, x) does not satisfies lim x→±∞ Λ(r, x) = ±1. Instead, we have Λ(r, x + 2π) = Λ(r, x) + 1. In the case that κ = 2, we have some nontrivial solutions to (8). From Lemma 3.1 in [17], we see −∂ r H r + H r H ′ r + H ′′ r = 0, where the function S r in [17] is the function H r here. From the definition of H r , we may compute that −∂ r H r + H r H ′ r + H ′′ r = 0. Thus Λ 1 (r, x) = H r (x) and Λ 2 (r, x) = rH r (x) + x satisfy equation (8). So λ 1 (r, x) = H ′ r (x) and λ 2 (r, x) = rH ′ r (x) + 1 are solutions to (7). In fact, λ 2 (r, x)/(2π) is the distribution of the argument of the end point of a Brownian Excursion in A r started from 1 conditioned to hit C r . From Corollary 3.1 in [17], this is also the distribution of the argument of the limit point of a standard annulus SLE 2 trace of modulus r. So we justified equation (7) in the case κ = 2.
From the equation for H r and the definition of G s , we have −∂ s G s + G s G ′ s + G ′′ s = 0. Thus P 1 (s, y) = G s (y) and P 2 (s, y) = sG s (y) + y are solutions to (9). In fact, P 1 (s, y) corresponds to −Λ 2 (r, x)/π, and P 2 (s, y) corresponds to −πΛ 1 (r, x).
For each r > 0, suppose J r is the conformal map from A r/2 onto {z ∈ C : |Im z| < 1} \ [−a r , a r ] for some a r > 0 such that ±1 is mapped to ±∞. This J r is symmetric w.r.t. both x-axis and y-axis, i.e., J r (z) = J r (z), and J r (−z) = −J r (z). And Im J r is the unique bounded harmonic function in A r/2 that satisfies (i) Im J r ≡ ±1 on the open arc of C 0 from ±1 to ∓1 in the ccw direction; and (ii) Im J r ≡ 0 on C r/2 . Let J r = J r • e i .
Proof. Since Im J r ≡ 0 on R r/2 , by reflection principle, J r can be extended analytically across R r/2 . And we have Im J ′ r = ∂ x Im J r ≡ 0 and Im J ′′ r = ∂ 2 x Im J r ≡ 0 on R r/2 . From the equality Im J r (x + ir/2) = 0, we have ∂ r Im J r + ∂ y Im J r /2 ≡ 0 on R r/2 . On R r/2 , note that Im T r + 1 2 J ′′ r . Then Im F r ≡ 0 on R r/2 . For any k ∈ Z, we see that J r (z) is equal to (−1) k+1 2 π ln(z − kπ) plus some analytic function for z ∈ A r/2 near kπ. So we may extend Re J r (z) harmonically across R \ {kπ : k ∈ Z}. Since Im J r takes constant value (−1) k on each interval (kπ, (k + 1)π), k ∈ Z, we have Re J r (z) = Re J r (z). Moreover, the following properties hold: ∂ r J r is analytic in a neighborhood of R, J ′ r and J ′′ r are analytic in a neighborhood of R \ {kπ : k ∈ Z}. The fact that Im J r takes constant value (−1) k on each (kπ, (k + 1)π), k ∈ Z, implies that Im ∂ r J r , Im J ′ r and Im J ′′ r vanishes on R \ {kπ : k ∈ Z}. Since Im T (2) r also vanishes on R \ {kπ : k ∈ Z}, so we compute Im F r ≡ 0 on R \ {kπ : k ∈ Z}.
From J r (z) = J r (z), we find that J r (−z) = J r (z). So Re J r (z) = Re J r (−z) = Re J r (−z). This means that Re J r is an even function, so is ∂ r J r and J ′′ r . And J ′ r is an odd function. Note that T (2) r is an odd function, so F r is an even function. Since T (2) r (z) is equal to 1/(2z) plus some analytic function for z near 0, so the pole of F r at 0 has order at most 2. However, the coefficient of 1/z 2 is equal to 2/π * 1/2 − 1/2 * 2/π = 0. And 0 is not a simple pole of F r because F r is even. So 0 is a removable pole of F r . Similarly, π is also a removable pole of F r . Since F r has period 2π, so every kπ, k ∈ Z, is a removable pole of F r . So F r can be extended analytically across R, and Im F r ≡ 0 on R. Thus Im F r ≡ 0 in S r/2 , which implies that F r ≡ C for some constant C ∈ R.
Finally, J r (−z) = −J r (z) implies that J r (z + π) = − J r (z). Since π is a period of r , we compute F r (z + π) = −F r (z). So C has to be 0. ✷ Note that ξ(t)/2 = B(t). From Ito's formula and the last lemma, we have Thus J p−t (Z t ), 0 ≤ t < p, is a local martingale. For any z ∈ A p/2 , there is z 0 ∈ S p/2 such that z = e i (z 0 ). Then So J p−t (ψ t (z)/e iξ(t)/2 ), 0 ≤ t < p, is a local martingale. Since |Im J r (z)| ≤ 1 for any r > 0 and z ∈ A r/2 , so Im J p−t (ψ t (z)/e iξ(t)/2 ), 0 ≤ t < p, is a bounded martingale. ✷ Let h t (z) = J p−t (ψ t (z)/e iξ(t)/2 ). Then h t maps A (p−t)/2 \ L t conformally onto {z ∈ C : |Im z| < 1} \ [−a p−t , a p−t ] so that α ± (t) is mapped to ±∞. So Im h t is the unique bounded harmonic function in A p/2 \ L t that vanishes on C p/2 , equals to 1 on the arc of C 0 from 1 to −1 in the ccw direction and the north side of α + (0, t) and α − (0, t), and equals to −1 on the arc of C 0 from −1 to 1 in the ccw direction and the south side of α + (0, t) and α − (0, t).
We have another choice of J r . Let J r be the conformal map of A r/2 onto the strip {z ∈ C : |Im z| < 1} \ [−ib r , ib r ] for some b r > 0 so that ±1 is mapped to ±∞. Then Im J r is the bounded harmonic function in A r/2 determined by the following properties: (i) Im J r ≡ ±1 on the open arc of C 0 from ±1 to ∓1 in the ccw direction; and (ii) the normal derivative of Im J r vanishes on C r/2 . Let J r = J r • e i . The proposition and lemma in this subsection still hold for J r and J r defined here. The proofs are almost the same. The only difference is at the step when we prove Im F r ≡ 0 on R r/2 . Here we have Re J ′ r = ∂ y Im J r vanishes on R r/2 . Use an argument similar to the proof of the lemma, we can show that Re F ′ r vanishes on R r/2 . So ∂ y Im F r vanishes on R r/2 . Since Im F r vanishes on R, Im F r has to vanish in S r/2 . Use J r (−z) = −J r (z), we then conclude that F r ≡ 0. If we let h t (z) = J p−t (ψ t (z)/e iξ(t)/2 ), then Im h t is the unique bounded harmonic function in A p/2 \ L t that satisfies the following properties: equals to 1 on the arc of C 0 from 1 to −1 in the ccw direction, and the north side of α + (0, t) and α − (0, t), equals to −1 on the arc of C 0 from −1 to 1 in the ccw direction, and the south side of α + (0, t) and α − (0, t), and the normal derivatives vanish on C p/2 .
Suppose E is a doubly connected domain such that 0 lies in the bounded component is the union of two disjoint simple curve started from v + and v − , respectively. Let α ± denote the curve started from v ± . Then D is a symmetric (−D = D) doubly connected domain, and α − (t) = −α + (t) for 0 ≤ t < p.
. Let γ ± t denote the boundary arc of ∂ o D t from α ± (t) to α ∓ (t) in the ccw direction. Then γ ± t contains a boundary arc of ∂ o D, one side of α + ([0, t]) and one side of α − ([0, t]). Let H t be the bounded harmonic function in D t which has continuations at ∂ i D and γ ± t such that H t ≡ 0 on ∂ i D and H t ≡ ±1 on γ ± t . By the definition of SLE 4 (E; v → ∂ i E) and conformal invariance of harmonic functions, for any fixed z 0 ∈ D, H t (z 0 ), 0 ≤ t < p, is a bounded martingale. This H t corresponds to Im h t defined right after Proposition 4.1. We may replace the condition H t ≡ 0 on ∂ i D by ∂ n H t ≡ 0 on ∂ i D. Then this H t corresponds to the Im h t defined in the last paragraph. So for any fixed z 0 ∈ D, it is still true that H t (z 0 ), 0 ≤ t < p, is a bounded martingale. Now we construct two curves α + and α − as follows. Let α ± (0) = v ± . Let α ± (1) be a neighbor vertex of α ± (0) such that [α ± (0), α ± (1)] is shared by a white hexagon and a black hexagon. At time n ∈ N, if α ± (n) ∈ ∂ i D, then α ± (n) is a vertex shared by a black hexagon, a white hexagon, and an uncolored hexagon, denoted by f n ± . Let H n be the function defined on faces, which takes value 1 on the black faces, −1 on the white faces, 0 on faces that touch ∂ i D, and is discrete harmonic at other faces of D. Then H n (f n − ) = −H n (f n + ). We then color f ± black with probability equal to (1+H n (f n ± ))/2 and white with probability equal to (1−H n (f n ± ))/2 such that f n + and f n − are colored differently. Let α ± (n + 1) be the unique neighbor vertex of α ± (n) such that [α ± (n), α ± (n + 1)] is shared by a white hexagon and a black hexagon. Increase n by 1, and iterate the above process until α + and α − hit ∂ i D at the same time. We always have α − (n) = −α + (n), f n − = −f n + , and H n (−g) = −H n (g). From the construction, conditioned on α ± (k), k = 0, 1, . . . , n, the expected value of

Harmonic Explorers for Annulus SLE
And if a face f is colored before time n, then its color will not be changed after time n, so H n+1 (f ) = H n (f ). Since H n+1 and H n both vanish on the faces near ∂ i D, and are discrete harmonic at all other uncolored faces at time n + 1 and n, resp., so for any face f of D, the conditional value of H n+1 (f ) w.r.t. α ± (k), k = 0, 1, . . . , n is equal to H n (f ). Thus for any fixed face f 0 of D, H n (f 0 ) is a martingale.
Note that if the side length of the hexagons is very small compared with the size of D, then for any face f of D, H n (f ) is close to the value of H n at the center of f , where H n is the bounded harmonic function defined on D n , which has a continuation to ∂D \ {v + , v − } and the two sides of α ± ([0, t)) such that H n ≡ 0 on ∂ i D, and H n ≡ ±1 on the curve on ∂ o D n from α ± (n) to α ∓ (n) in the ccw direction. From the last section, we may guess that the distribution of α ± tends to that of the square root of an annulus SLE 4 (P 2 (D); P 2 (v ± ) → ∂ i P 2 (D)) trace when the mesh tends to 0. If at each step of the construction of α ± , we let H n be the function which is is equal to 1 on the black faces, −1 on the white faces, and is discrete harmonic at all other faces of D including the faces that touch ∂ i D, then we get a different pair of curves α ± . If the mesh is very small compared with the size of D, then for any face f of D, H n (f ) is close to the value of H n at the center of f , where H n is the bounded harmonic function defined on D n , which has a continuation to ∂D \ {v + , v − } and the two sides of α ± ([0, t)) such that ∂ n H n ≡ 0 on ∂ i D, and H n ≡ ±1 on the curve on ∂ o D n from α ± (n) to α ∓ (n) in the ccw direction. So we also expect the law of α ± constructed in this way tends to that of the square root of an annulus SLE 4 (P 2 (D); P 2 (v ± ) → ∂ i P 2 (D)) trace when the mesh tends to 0.

Proposition 4.2 For any
Proof. Let G r := G r • e i . For any z ∈ A p/4 , there is w ∈ S p/4 such that z = e i (w). Then To prove this proposition, it suffices to show that for any w ∈ S p/4 , G p−t ( ψ t (w) − ξ(t)/4) is a local martingale. Let Z t = ψ t (w) − ξ(t)/4, then Thus by Ito's formula, So it suffices to prove the following lemma.

Annulus SLE 8/3 and the Restriction Property
In this section, we fix κ = 8/3 and α = 5/8. Let ϕ t and K t , 0 ≤ t < p, be the annulus LE maps and hulls of modulus p, driven by ξ(t) = √ κB(t), 0 ≤ t < p. Let ϕ t and K t , 0 ≤ t < p, be the corresponding annulus LE maps and hulls in the covering space. Let A = ∅ be a hull in A p w.r.t. C p (i.e., A p \ A is a doubly connected domain whose one boundary component is Then h is a continuous increasing function with h(0) = 0. So h maps [0, T A ) onto [0, S A ) for some S A ∈ (0, p 0 ]. From Proposition 2.1 in [17], L s = K h −1 (s) , 0 ≤ s < S A , are the annulus LE hulls of modulus p 0 , driven by some real continuous function, say η(s). Let ψ s , 0 ≤ s < S A , be the corresponding annulus LE maps. Let ψ s and L s , 0 ≤ s < S A , be the annulus LE maps and hulls, respectively, in the covering space. Thus positive measure µ t supported by C 0 of total mass (p − t) − (p 0 − h(t)) such that for any z ∈ A p 0 −h(t) , − ln(g t (z)/z) = C 0 S p 0 −h(t) (z/θ)dµ t (θ) + iC.
Then M t → 1 as t → T A on the event that T A = p. From the Markov property, we have Recall that p 0 is the modulus of A p \ A. Let K p = ∪ 0≤t<p K t . Then If A is not a smooth hull, we may find a sequence of smooth hulls A n that approaches A. Then ϕ ′ An (0) → ϕ ′ A (0) and the modulus of A p \ A n tends to the modulus of A p \ A, so equation (19) still holds. Now suppose B is a hull in A p 0 w.r.t. C p 0 . Let D = A ∪ ϕ −1 A (B). Then D is a hull in A p w.r.t. C p . Let p 1 be the modulus of A p \ D, which is also the modulus of A p 0 \ B. dr .
If L s , 0 ≤ s < p 0 , are standard annulus SLE 8/3 hulls of modulus p 0 , and L p 0 = ∪ 0≤s<p 0 L s , then Thus conditioned on the event that K p ∩ A = ∅, ϕ A (K p ) has the same distribution as L p 0 . Then we proved the restriction property of annulus SLE 8/3 under the assumption (18). Unfortunately, the assumption (18) is actually always false. From equation (3) one may compute that S r is of order Θ(1/r 2 ) as r → 0. From (17), (p − t) − (p 0 − h(t)) = |µ t | is of order O((p − t) 2 ). In fact, one could prove that it is of order Θ((p − t) 2 ). So p−t p 0 −h(t) S r dr is uniformly bounded away from 0. Thus it does not tend to 0 as t → p. Therefore we guess that annulus SLE 8/3 does not satisfy the restriction property.
Recently, Robert O. Bauer studied in [2] a process defined in a doubly connected domain obtained by conditioning a chordal SLE 8/3 in a simply connected domain to avoid an interior contractible compact subset. The process describes a random simple curve connecting two prime ends of a doubly connected domain that lie on the same side, so it is different from the process we study here. That process automatically satisfies the restriction property from the restriction property of chordal SLE 8/3 . And it satisfies conformal invariance because the set of boundary hulls generates the same σ-algebra as the Hausdorff metric on the space of simple curves.