Dimension and Measure of SLE on the Boundary
Tom Alberts
joint work with Scott Sheffield
September 29, 2008
Explanation of the Title
• In the range 4 < κ < 8, the chordal SLE(κ) has a nontrivial intersection with the real line.
• The set γ∩R is very fractal and has a non-trivial Hausdorff
dimension.
SLE(6)
• We compute the Hausdorff dimension of γ ∩ R.
• We also put a measure on γ ∩ R that we call the covariant
measure.
Dimension and Measure of Random Sets
• Many stochastic processes create random sets that are
very fractal and that can be difficult to analyze.
• For example:
– The zeros of a one-dimensional Brownian motion,
– The graph of a d-dimensional Brownian motion
– The outer boundary of a 2-dimensional Brownian
motion run for one unit of time.
• Let C be a fractal subset of Rn. We commonly try to
describe the size of C via its Hausdorff dimension.
• In many cases even though C is random, dimH C is not.
Dimension and Measure of Random Sets
• There is a general procedure for computing these
Hausdorff dimensions.
• Two steps: show that a given number is both an upper
bound and a lower bound for dimH C.
• Upper bound is usually easy, lower bound is often difficult.
Dimension and Measure of Random Sets
• Create “ǫ-approximations” C ǫ of C such that
– C ǫ is an ǫ-thickening of C
ǫ′
– C ⊂ C ǫ for 0 < ǫ′ < ǫ
T
– C = ǫ>0 C ǫ
• Obtain a “one-point bound”
P (x ∈ C ǫ) ≤ c1ǫβ
for some constant c1 > 0 and some exponent 0 ≤ β ≤ 1.
Then by explicitly covering C ǫ with dyadic intervals such
that [k2−n, (k + 1)2−n] ∩ C ǫ 6= ∅, get
with probability one.
dimH C ≤ 1 − β
Dimension and Measure of Random Sets
• Obtain a “two-point bound”
2β
ǫ
P (x, y ∈ C ǫ) ≤ c2
|x − y|β
to get the existence of a constant ρ > 0 such that
P (dimH C ≥ 1 − β − δ) ≥ ρ for all δ > 0
• Use some property of C, usually scale invariance, and
some 0-1 Law to get that dimH C is constant with
probability one.
• Conclude that dimH C = 1 − β with probability one.
Dimension and Measure of Random Sets
• Example: C = {t ≥ 0 : Bt = 0}
• Set C ǫ = {t ≥ 0 : |Bt| ≤ ǫ}
• Get explicit bounds on
and
P (x ∈ C ǫ)
P (x, y ∈ C ǫ)
to conclude that dimH C = 1/2 almost surely.
Dimension and Measure of Random Sets
• Two parts to this talk:
– Use this method to compute dimH γ ∩ R,
– Show a way of strengthening the two-point bound
argument, via the Doob-Meyer decomposition, to come up
with a measure-valued function µ of the SLE curve γ that
is supported on γ ∩ R.
Overview of SLE
• SLE is a measure on paths in a simply connected subset
of C, where the paths have the property that they do not
self cross.
• We will consider paths in H from 0 to ∞.
• The SLE process is realized by inputting some
randomness into the classical Loewner equation from the
1920s.
Loewner Equation
• Consider a non self crossing curve γ : [0, ∞) → H such
that γ(0) = 0, γ(∞) = ∞.
• Then ∀t ≥ 0, H\γ[0, t] is a simply connected domain.
gt
γ([0, t])
0
Ut = gt (γ(t))
• Riemann Mapping Theorem says there exists a conformal
map gt : H\γ[0, t] → H.
• Loewner Equation describes how the maps gt evolve as
the curve γ[0, t] grows.
Loewner Equation
• Consider a non self crossing curve γ : [0, ∞) → H such
that γ(0) = 0, γ(∞) = ∞.
• Then ∀t ≥ 0, H\γ[0, t] is a simply connected domain.
gt
γ([0, t])
0
Ut = gt (γ(t))
2
∂tgt(z) =
,
gt(z) − Ut
g0(z) = z
Using the Loewner Equation to Produce Curves
• So given the curve γ, the maps gt must satisfy
2
∂tgt(z) =
,
gt(z) − Ut
g0(z) = z
where Ut = gt(γ(t)).
• Can turn the situation around.
Given the driving
function Ut : [0, ∞) → R, the ODE can be solved to
determine gt. The gt then determine γ[0, t].
• (Chordal) SLE(κ) refers to the random curves
produced
√
when the inputted driving function is Ut = κBt, with κ >
0 and Bt a standard Brownian motion with B0 = 0.
Domain Markov Property of SLE
• Domain Markov: γ ∗(s) has the law of a chordal SLE
curve in H from Ut to ∞ that is independent of γ[0, t].
gt
γ([0, t])
γ ∗ (s) = gt (γ(t + s))
0
Ut = gt (γ(t))
• It will be convenient to use the shifted maps
so that ht(γ(t)) = 0.
ht(z) = gt(z) − Ut
Scaling Property of SLE
√
• From the driving function Ut = κBt, the SLE curves
inherit the Brownian scaling relation
rγ[0, t] ≡ γ[0, r2t]
• Important Corollary: dimH γ ∩ R is constant almost
surely.
This is a consequence of the scaling and
Blumenthal 0-1 Law.
The Dependence on κ
• κ looks like an innocent parameter, but it actually plays a
very important role.
• κ = 0:
2
,
∂tgt(z) =
gt(z)
gt(z) =
√
√
2 t
0
g0(z) = z
z 2 + 4t
The Dependence on κ
• For κ ≤ 4, the curve is almost surely simple (i.e. doesn’t
touch itself).
• Moreover, it doesn’t intersect the real line, i.e. γ ∩ R = {0}
• For a fixed z ∈ H, almost surely the curve does not hit z.
SLE(2)
The Dependence on κ
• For 4 < κ < 8, the curve touches itself (but does not cross
itself).
• The curve intersects part of the real line, i.e. γ ∩ R ( R.
• For a fixed z ∈ H, almost surely the curve does not hit z,
but does make a loop around z.
• Let Tz be the random time at which point z is swallowed.
SLE(6)
The Dependence on κ
• For κ ≥ 8, the curve is space filling.
• Hits every point on the real line, i.e. γ ∩ R = R.
• For a fixed z ∈ H, almost surely the curve does hit z.
SLE(κ > 8)
The Range 4 < κ < 8
• In the rest of this talk we only work with 4 < κ < 8, and
the intersection of γ with R+.
• Random set to work with is C = γ ∩ R+.
• Approximate with C ǫ = {x ∈ R+ : γ ∩ [x, x + ǫ] 6= ∅}.
One-Point Bound
• [RS05] computes P (x ∈ C ǫ) explicitly.
Γ(4/κ)
P(curve hits in [y, x]) =
Γ(8/κ − 1)Γ(1 − 4/κ)
Z
x−y
x
0
du
u2−8/κ(1 − u)4/κ
• Up to constants the right hand side is
P(curve hits in [y, x]) ≍
x − y β
x
,
β = 8/κ − 1
• Here f (s) ≍ g(s) means there exists constants 0 < C1 <
C2 such C1f (s) ≤ g(s) ≤ C2f (s) that for all s
• Corollary: If x is not too big or too small
ǫ β
≍ ǫβ
P (x ∈ C ǫ) = P (γ ∩ [x, x + ǫ] 6= ∅) ≍
x+ǫ
Swallowing Probabilities
• [Dub03]: Given a point z ∈ H, can compute the
swallowing probabilities
P(Tz < T1), P(Tz = T1), P(Tz > T1)
z
0
1
Swallowing Probabilities
• [Dub03]: Given a point z ∈ H, can compute the
swallowing probabilities
P(Tz < T1), P(Tz = T1), P(Tz > T1)
• Apply the Schwarz-Christoffel map F that sends H to the
interior of an isosceles triangle, with 0, 1, ∞ going to the
vertices and angle βπ at F (1).
F (1)
z
F
F (z)
0
1
F (∞)
F (0)
Swallowing Probabilities
• [Dub03]: Given a point z ∈ H, can compute the
swallowing probabilities
P(Tz < T1), P(Tz = T1), P(Tz > T1)
• Apply the Schwarz-Christoffel map F that sends H to the
interior of an isosceles triangle, with 0, 1, ∞ going to the
vertices and angle βπ at F (1).
• The barycentric coordinates of F (z) in the triangle give the
F (1)
three hitting probabilities.
z
F
F (z)
0
1
F (∞)
F (0)
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
dimH
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
1
4
8
κ
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
dimH
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
1
4
8
κ
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
dimH
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
1
4
8
κ
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
dimH
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
1
4
8
κ
Upper Bound on the Hausdorff Dimension
• Hence we get β = 8/κ − 1 in the one-point bound, so
1 − β = 2 − 8/κ.
dimH
• Result: [AS08] The Hausdorff dimension of γ ∩ R is
almost surely d = 1 − β = 2 − κ8 .
1
4
• How do we prove the lower bound?
8
κ
Two-Interval Hitting Probability
• Goal of Next 10 Minutes: Want to show the two-point
bound
ǫ2β
P (x, y ∈ C ) = P (γ ∩ [x, x + ǫ] 6= ∅, γ ∩ [y, y + ǫ] 6= ∅) ≤ c2
|x − y|β
ǫ
Two-Interval Hitting Probability
• Already know P(y ∈ C ǫ) ≍ ǫβ , so only need to know the
conditional probability of hitting the second interval given
the curve has hit the first.
Two-Interval Hitting Probability
• Already know P(y ∈ C ǫ) ≍ ǫβ , so only need to know the
conditional probability of hitting the second interval given
the curve has hit the first.
• Use the Domain Markov property.
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• Already know P(y ∈ C ǫ) ≍ ǫβ , so only need to know the
conditional probability of hitting the second interval given
the curve has hit the first.
• Use the Domain Markov property.
0
y
x
y+ǫ
x+ǫ
hT y
0
hTy (x)
hTy (x + ǫ)
Two-Interval Hitting Probability
• Already know P(y ∈ C ǫ) ≍ ǫβ , so only need to know the
conditional probability of hitting the second interval given
the curve has hit the first.
• Use the Domain Markov property.
0
y
x
y+ǫ
x+ǫ
hT y
0
hTy (x)
hTy (x + ǫ)
Two-Interval Hitting Probability
• Given FTy , the probability of hitting in [x, x + ǫ] is therefore
0
hTy (x + ǫ) − hTy (x)
hTy (x + ǫ)
y
β
x
y+ǫ
x+ǫ
hT y
0
hTy (x)
hTy (x + ǫ)
Two-Interval Hitting Probability
• Given FTy , the probability of hitting in [x, x + ǫ] is therefore
hTy (x + ǫ) − hTy (x)
hTy (x + ǫ)
β
• Problem: hTy is random so we have very little idea how
big this quantity is!
0
y
x
y+ǫ
x+ǫ
hT y
0
hTy (x)
hTy (x + ǫ)
Two-Interval Hitting Probability
• Need to know something about hTy . Recall it’s the unique
conformal map that removes the hull from the half-plane
and is like the identity at ∞.
• So the geometry of the hull tells us what hTy is doing.
• But what part of the geometry is important?
Two-Interval Hitting Probability
• Need to know something about hTy . Recall it’s the unique
conformal map that removes the hull from the half-plane
and is like the identity at ∞.
• So the geometry of the hull tells us what hTy is doing.
• But what part of the geometry is important?
• Look at what we’re trying to estimate:
hTy (x + ǫ) − hTy (x)
hTy (x + ǫ)
β
≈
ǫh′Ty (x)
hTy (x)
!β
Two-Interval Hitting Probability
• Need to know something about hTy . Recall it’s the unique
conformal map that removes the hull from the half-plane
and is like the identity at ∞.
• So the geometry of the hull tells us what hTy is doing.
• But what part of the geometry is important?
• Look at what we’re trying to estimate:
hTy (x + ǫ) − hTy (x)
hTy (x + ǫ)
β
≈
ǫh′Ty (x)
hTy (x)
!β
• In fact using Growth Theorem can show that
hTy (x + ǫ) − hTy (x)
hTy (x + ǫ)
β
≤
4ǫh′Ty (x)
hTy (x)
!β
Two-Interval Hitting Probability
h′Ty (x)
hTy (x)
• Extend the map hTy across the real line.
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
h′Ty (x)
hTy (x)
• Extend the map hTy across the real line.
x
Two-Interval Hitting Probability
h′Ty (x)
hTy (x)
• Extend the map hTy across the real line.
x
hT y
0
hTy (x)
Two-Interval Hitting Probability
h′Ty (x)
hTy (x)
• Extend the map hTy across the real line.
D
x
hT y
0
D′
hTy (x)
Two-Interval Hitting Probability
h′Ty (x)
hTy (x)
• Extend the map hTy across the real line.
D
x
hT y
0
D′
hTy (x)
h′Ty (x) h′Ty (x)
1
Koebe 1/4 Theorem =⇒
=
≍
′
D
D
hTy (x)
Two-Interval Hitting Probability
• Conditional probability of hitting the second interval
satisfies
!β
β
′
ǫ β
4ǫhTy (x)
hTy (x + ǫ) − hTy (x)
≤
≍
hTy (x + ǫ)
hTy (x)
D
Two-Interval Hitting Probability
• Conditional probability of hitting the second interval
satisfies
!β
β
′
ǫ β
4ǫhTy (x)
hTy (x + ǫ) − hTy (x)
≤
≍
hTy (x + ǫ)
hTy (x)
D
• So the relevant geometrical information from the hull at
time Ty is the distance of the hull from the point x.
Two-Interval Hitting Probability
• Conditional probability of hitting the second interval
satisfies
!β
β
′
ǫ β
4ǫhTy (x)
hTy (x + ǫ) − hTy (x)
≤
≍
hTy (x + ǫ)
hTy (x)
D
• So the relevant geometrical information from the hull at
time Ty is the distance of the hull from the point x.
• The two-interval hitting probability can then be bounded
above by
ǫ β P(x, y ∈ C ǫ) ≤ E 1{y ∈ C ǫ}
D
Two-Interval Hitting Probability
ǫ β E 1{y ∈ C ǫ}
D
Two-Interval Hitting Probability
ǫ β E 1{y ∈ C ǫ}
D
• To evaluate this we need to estimate
P ({y ∈ C ǫ} ∩ {D ≤ r})
Two-Interval Hitting Probability
ǫ β E 1{y ∈ C ǫ}
D
• To evaluate this we need to estimate
P ({y ∈ C ǫ} ∩ {D ≤ r})
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
ǫ β E 1{y ∈ C ǫ}
D
• To evaluate this we need to estimate
P ({y ∈ C ǫ} ∩ {D ≤ r})
r
0
y
y+ǫ
x
x+ǫ
• This is the proportion of curves that hit the semi-circle of
radius r centered at x before hitting the real line in [y, y +ǫ]
Two-Interval Hitting Probability
ǫ β E 1{y ∈ C ǫ}
D
• To evaluate this we need to estimate
P ({y ∈ C ǫ} ∩ {D ≤ r})
r
0
y
y+ǫ
x
x+ǫ
• This is the proportion of curves that hit the semi-circle of
radius r centered at x before hitting the real line in [y, y +ǫ]
• Note r small is the important part. In this case the
probability should be extremely small, since a curve that
first hits the semi-circle will rarely hit in [y, y + ǫ].
Two-Interval Hitting Probability
• Let τr be the first time at which the curve hits the semicircle.
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• Let τr be the first time at which the curve hits the semicircle.
• From this time compute the conditional probability of
hitting in [y, y + ǫ] and show that it’s small.
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• Let τr be the first time at which the curve hits the semicircle.
• From this time compute the conditional probability of
hitting in [y, y + ǫ] and show that it’s small.
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• Let τr be the first time at which the curve hits the semicircle.
• From this time compute the conditional probability of
hitting in [y, y + ǫ] and show that it’s small.
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
• Same type of conditional probability as before!
β
hτr (y + ǫ) − hτr (y)
hτr (y + ǫ)
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
• Same type of conditional probability as before!
β
hτr (y + ǫ) − hτr (y)
hτr (y + ǫ)
• Again have to estimate using the geometry of the hull.
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
• Same type of conditional probability as before!
β
hτr (y + ǫ) − hτr (y)
hτr (y + ǫ)
• Again have to estimate using the geometry of the hull.
• But first an interlude about intervals on the line.
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
hm (z, H; [x2, x3])
x3 − x2
= lim
x3 − x1 z→∞ hm (z, H; [x1, x3])
z
x1
x2
x3
Two-Interval Hitting Probability
hm (z, H; Green)
hτr (y + ǫ) − hτr (y)
= lim
z→∞ hm (z, H; P urple)
hτr (y + ǫ)
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
hm (z, H; Green)
hτr (y + ǫ) − hτr (y)
= lim
z→∞ hm (z, H; P urple)
hτr (y + ǫ)
• Using conformal invariance, we can convert this into a
ratio of harmonic measures in the top domain H\Kτr
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
hm (z, H\Kτr ; Green)
hτr (y + ǫ) − hτr (y)
= lim
z→∞ hm (z, H\Kτr ; P urple)
hτr (y + ǫ)
• Using conformal invariance, we can convert this into a
ratio of harmonic measures in the top domain H\Kτr
r
0
y
x
y+ǫ
x+ǫ
hτr
0
hτr (y)
hτr (y + ǫ)
Two-Interval Hitting Probability
hm (z, H\Kτr ; Green)
hτr (y + ǫ) − hτr (y)
= lim
z→∞ hm (z, H\Kτr ; P urple)
hτr (y + ǫ)
• Using conformal invariance, we can convert this into a
ratio of harmonic measures in the top domain H\Kτr
• The Brownian motion from near ∞ must pass through the
left arc of the semi-circle to get to the green or purple
boundaries.
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
hm (z, H\Kτr ; Green)
hτr (y + ǫ) − hτr (y)
= lim
z→∞ hm (z, H\Kτr ; P urple)
hτr (y + ǫ)
• Using basic harmonic measure estimates on Green and
Purple, we prove there is a universal constant c > 0 such
that
hτr (y + ǫ) − hτr (y)
ǫr
≤c
hτr (y + ǫ)
(x − y)2
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
ǫ
P(D ≤ r)
P ({y ∈ C } ∩ {D ≤ r}) ≤
2
(x − y)
r
0
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
ǫ
P(D ≤ r)
P ({y ∈ C } ∩ {D ≤ r}) ≤
2
(x − y)
r
0
y
y+ǫ
x
x+ǫ
• Can be shown that P(D ≤ r) ≍ P(Tx+ir < Ty ), for small r.
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
ǫ
P(D ≤ r)
P ({y ∈ C } ∩ {D ≤ r}) ≤
2
(x − y)
r
0
y
y+ǫ
x
x+ǫ
• Can be shown that P(D ≤ r) ≍ P(Tx+ir < Ty ), for small r.
• Using Dubédat’s swallowing probabilities, can compute
P(Tx+ir < Ty ).
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
ǫ
P(D ≤ r)
P ({y ∈ C } ∩ {D ≤ r}) ≤
2
(x − y)
r
0
y
y+ǫ
x
x+ǫ
P(D ≤ R) ≤ CP(Tx+ir < Ty ) ≤ Cr(x − y)β−1
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
β−1
ǫ
r(x
−
y)
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)2
r
0
y
y+ǫ
x
x+ǫ
P(D ≤ R) ≤ CP(Tx+ir < Ty ) ≤ Cr(x − y)β−1
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
β−1
ǫ
r(x
−
y)
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)2
r
0
• Simplify:
y
y+ǫ
x
x+ǫ
Two-Interval Hitting Probability
• To compute P ({y ∈ C ǫ} ∩ {D ≤ r}), all that’s needed is
the proportion of curves that hit the semi-circle before y is
swallowed.
β
ǫr
β−1
ǫ
r(x
−
y)
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)2
r
y
0
y+ǫ
x
• Simplify:
ǫβ rβ+1
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)β+1
ǫ
x+ǫ
Two-Interval Hitting Probability
• Recall
ǫ β P(curve hitting in [y, y + ǫ] & [x, x + ǫ]) ≤ E 1{y ∈ C ǫ}
D
Two-Interval Hitting Probability
• Recall
ǫ β P(curve hitting in [y, y + ǫ] & [x, x + ǫ]) ≤ E 1{y ∈ C ǫ}
D
• Have the bound
β β+1
ǫ
r
ǫ
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)β+1
Two-Interval Hitting Probability
• Recall
ǫ β P(curve hitting in [y, y + ǫ] & [x, x + ǫ]) ≤ E 1{y ∈ C ǫ}
D
• Have the bound
β β+1
ǫ
r
ǫ
P ({y ∈ C } ∩ {D ≤ r}) ≤ C
(x − y)β+1
• Combine the two, using that 0 ≤ D ≤ x−y and integrating
to get the desired bound
2β
ǫ
P(y, x ∈ C ǫ) ≤ C
(x − y)β
The Schramm Zhou Method
• Our upper bound on the two-interval hitting probability
uses mostly conformal mapping and harmonic measure
techniques.
• Schramm & Zhou [SZ07] give an (almost) purely
probabilistic proof of the same bound.
• Consider again
ht(x + ǫ) − ht(x)
P (curve hits in [x, x + ǫ]| γ[0, t]) ∼
ht(x + ǫ)
′ β
ǫht(x)
∼
ht(x)
β
as ǫ ↓ 0. Define
′ β
P (curve hits in [x, x + ǫ]| γ[0, t])
ht(x)
= lim
Mt(x) :=
ǫ↓0
ht(x)
ǫβ
The Schramm Zhou Method
• In this sense, Mt(x) is the conditional probability that the
curve hits x, given γ[0, t].
• Important Fact: Mt(x) is a local martingale that is not a
martingale. But since it’s positive, it is a supermartingale.
• Not So Important Fact:
computed exactly!
M0(x)
E [Mt(x)] =
12−κ
Γ 2κ
[Yor01] E [Mt(x)] can be
Z
0
x2 /2t
u
12−κ
2κ −1
e−u du
The Schramm Zhou Method
• We’re interested in points x for which Mt(x) grows large.
Ctǫ := x > 0 : sup Ms(x) ≥ ǫ−β
0≤s≤t
Ct :=
\
Ctǫ
ǫ>0
• Not surprisingly, Mt(x) also describes the distance from x
to the curve. By the Koebe-1/4 Theorem argument,
x ∈ Ctǫ =⇒ dist(x, γ[0, t]) ≤ 4ǫ,
x ∈ Ct =⇒ x ∈ γ[0, t] ∩ R
The Schramm Zhou Method
• Schramm & Zhou devise a two-point bound
2β
ǫ
P (y ∈ Ctǫ, x ∈ Ctǫ) ≤ c β
y (x − y)β
• They do it by finding a two-point martingale
ht(y)
Mt(y)Mt(x)
u
ht(x)
where
u(z) = (1 − z)−β 2F1(4/κ, 1 − 8/κ, 8/κ; 1 − z).
• The two-point bound follows by a stopping time argument.
The Schramm Zhou Method
• Schramm & Zhou devise a two-point bound
2β
ǫ
P (y ∈ Ctǫ, x ∈ Ctǫ) ≤ c β
y (x − y)β
• They do it by finding a two-point martingale
ht(y)
Mt(y)Mt(x)
u
ht(x)
where
u(z) = (1 − z)−β 2F1(4/κ, 1 − 8/κ, 8/κ; 1 − z).
• The two-point bound follows by a stopping time argument.
• Conclusion: dimH Ct ≥ d, and since Ct ⊂ γ[0, t] ∩ R the
same lower bound holds for the curve.
The Schramm Zhou Method
• Same stopping time argument shows that for some c
depending only on κ
E [Mτ (x)Mτ (y)] ≤ cy −β (x − y)−β
where τ is an a.s. finite stopping time.
The Schramm Zhou Method
• Same stopping time argument shows that for some c
depending only on κ
E [Mτ (x)Mτ (y)] ≤ cy −β (x − y)−β
where τ is an a.s. finite stopping time.
Constructing a Measure on γ ∩ R
• dimH γ ∩ R = 2 − 8/κ is at best a qualitative description of
the size of γ ∩ R.
• Goal is to construct a measure-valued function of γ that
is supported on γ ∩ R.
• Measure will necessarily be singular with respect to
Lebesgue.
• Key idea behind the construction is an abstract appeal to
the Doob-Meyer decomposition.
A New Measure on γ ∩ R
• Let I ⊂ R+ be an interval. At any time t ≥ 0 can always
divide it into a left interval of swallowed points and a right
interval of unswallowed points.
I ∩ Kt
I\Kt
• For any Borel measure µ on R+,
µ(I) = µ(I ∩ Kt) + µ(I\Kt)
A New Measure on γ ∩ R
• Since I ∩ Kt consists of swallowed points that cannot be
hit again, it is reasonable to assume that µ(I ∩ Kt) is
completely determined at time t, i.e. it is Ft-measurable.
I ∩ Kt
I\Kt
• For any Borel measure µ on R+,
µ(I) = µ(I ∩ Kt) + µ(I\Kt)
A New Measure on γ ∩ R
• Since I ∩ Kt consists of swallowed points that cannot be
hit again, it is reasonable to assume that µ(I ∩ Kt) is
completely determined at time t, i.e. it is Ft-measurable.
I ∩ Kt
I\Kt
• For any Borel measure µ on R+, under this assumption,
E [µ(I)| Ft] = µ(I ∩ Kt) + E [µ(I\Kt)| Ft]
A New Measure on γ ∩ R
• Since I ∩ Kt consists of swallowed points that cannot be
hit again, it is reasonable to assume that µ(I ∩ Kt) is
completely determined at time t, i.e. it is Ft-measurable.
I ∩ Kt
I\Kt
• For any Borel measure µ on R+, under this assumption,
E [µ(I)| Ft] = µ(I ∩ Kt) + E [µ(I\Kt)| Ft]
• This is the Doob-Meyer decomposition of the positive
supermartingale E [µ(I\Kt)| Ft].
A New Measure on γ ∩ R
• Identify an explicit supermartingale E [µ(I\Kt)| Ft], and
then use the increasing part of its Doob-Meyer
decomposition to build the measure.
• Two assumptions
supermartingale:
on
µ
fix
the
choice
of
– Assume µ has a certain scaling rule
– Impose a specific Domain Markov property on µ.
the
Scaling Rule for µ
• Scaling properties of SLE suggest that µ should satisfy
µ(r·) ≡ rdµ(·)
• Consequently
E [µ(rI)] = rdE [µ(I)]
which implies that
E [dµ(x)] = xd−1 dx = x−β dx.
Domain Markov Property of µ
• The measure µ∗ corresponding to the future curve γ ∗ is
independent of γ[0, t] but has the same law as µ.
ht
γ([0, t])
γ ∗ (s) := ht (γ[t, t + s])
0
0
• Need a rule for how the measure transforms under
conformal maps.
dµ∗(ht(x)) = |h′t(x)|ddµ(x)
• Taking conditional expectations yields
E [dµ(x)| Ft] = h′t(x)−dht(x)−β d(ht(x)) =
h′t(x)
ht(x)
β
dx
Conditional Expected Density of µ
• Recall
Mt(x) :=
h′t(x)
ht(x)
β
is the conditional probability that x is hit by γ. Hence it is a
very natural conditional density.
Conditional Expected Density of µ
• This leads to the choice
E [µ(I\Kt)| Ft] =
Z
Mt(x) dx.
I\Kt
• Already know that
|{x ∈ I ∩ Kt : Mt(x) 6= 0}| = 0
so this is the same as
Z
Z
Mt(x) dx = Mt(x) dx =: Xt(I)
I\Kt
I
Doob-Meyer Decomposition of Xt(I)
• Questions:
– Is Xt(I) actually a supermartingale?
– Is Xt(I) a local martingale?
– Does it have a Doob-Meyer decomposition?
– Is the increasing part of the decomposition not
identically zero?
• Fact: A positive supermartingale can always be uniquely
decomposed as a local martingale minus a predictable,
increasing process.
• If Xt(I) is a local martingale, then its increasing part is
necessarily zero.
Doob-Meyer Decomposition of Xt(I)
• Want to show that Xt(I) is a supermartingale with a
Doob-Meyer decomposition as a martingale minus an
increasing process.
• The required ingredient is uniform integrability of the
family
{Xτ (I) : τ an almost surely finite stopping time }
This is called the class D property.
Doob-Meyer Decomposition of Xt(I)
• Thanks to Schramm & Zhou, this is pretty easy:
"Z
2 #
Mτ (x) dx
2
E Xτ (I) = E
I
Z Z
=
E [Mτ (x)Mτ (y)] dy dx
IZ IZ
≤ c
y −β (x − y)−β dy dx
< ∞
I
I
and the bound is independent of τ .
R
• Note that I Mt(x) dx has a Doob-Meyer decomposition
as a martingale minus an increasing part, but Mt(x) does
not.
• This is an example of the Doob-Meyer decomposition not
commuting with integration.
Doob-Meyer Decomposition of Xt(I)
Xt(I) = Nt(I) − At(I)
where Nt(I) is a martingale and At(I) is an increasing
process.
• Properties:
• At(I) is a continuous function of t.
continuity of Xt(I).
• At(I) is flat at times for which γ(t) 6∈ I.
Follows from the
Definition and Properties of the Measure
Xt(I) = Nt(I) − At(I)
where Nt(I) is a martingale and At(I) is an increasing
process.
• Define µ by
µ(I) = lim At(I)
t↑τI
where τI is the stopping time at which I is entirely
swallowed.
• µ is countably additive on intervals, hence it extends to a
measure on R+.
• µ is supported exactly on C∞.
• µ is free of atoms.
• µ assigns mass to an open interval if and only if the curve
hits that interval.
Definition and Properties of the Measure
• Side Benefit: Can compute E [µ(I ∩ Kt)] exactly by
E [µ(I ∩ Kt)] = E [Nt(I)]
Z − E [Xt(I)]
= X0 − E [Mt(x)] dx
I
Z
=
M0(x) − E [Mt(x)] dx
Z ∞
ZI
−β
12−κ
x
2κ −1 e−u du dx
u
=
12−κ
x2 /2t
I Γ
2κ
• For the measure of γ ∩ R
Z
Z
E [µ(I)] = M0(x) dx = x−β dx
I
I
Relation to Hausdorff and Minkowski Measures
• Open Questions: Is our measure the same as the
Hausdorff measure of γ ∩ R+ for some gauge function?
Or is it related to the Minkowski measure?
Relation to Hausdorff and Minkowski Measures
• Open Questions: Is our measure the same as the
Hausdorff measure of γ ∩ R+ for some gauge function?
Or is it related to the Minkowski measure?
• Conjecture: Yes to both.
Relation to Hausdorff and Minkowski Measures
• Open Questions: Is our measure the same as the
Hausdorff measure of γ ∩ R+ for some gauge function?
Or is it related to the Minkowski measure?
• Conjecture: Yes to both.
• For Hausdorff we have no idea what the right gauge
should be. Don’t even know if the power function works
or not.
Relation to Hausdorff and Minkowski Measures
• Open Questions: Is our measure the same as the
Hausdorff measure of γ ∩ R+ for some gauge function?
Or is it related to the Minkowski measure?
• Conjecture: Yes to both.
• For Hausdorff we have no idea what the right gauge
should be. Don’t even know if the power function works
or not.
• For Minkowski we believe they are the same. Minkowski
measure of I would be
lim ǫ−β |{x ∈ I : dist(x, γ) ≤ ǫ}| .
ǫ↓0
• This Minkowski measure is difficult to work with because
it’s hard to analyze distances from points to the curve.
• Instead work with a conformal distance that comes from
the Koebe-1/4 Theorem.
The Conformal Minkowski Measure
• Recall that x ∈ Ctǫ =⇒ dist(x, γ[0, t]) ≤ 4ǫ.
• Hence if we can show the existence of
lim ǫ−β |I ∩ Ctǫ|
ǫ↓0
then we have a natural measure of γ[0, t] ∩ I.
• Using the Doob-Meyer construction, we can show the
limit exists in L1.
The Conformal Minkowski Measure
Z
ǫ−β |I ∩ Ctǫ| = ǫ−β 1 {x ∈ Ctǫ} dx
I
• If the martingale Mt(x) reaches the threshold ǫ−β then
stop it. Call the stopped martingale Mtǫ(x).
Z
ǫ−β |I ∩ Ctǫ| = Mtǫ(x)1 {x ∈ Ctǫ} dx
I
and this is part of the trivial decomposition
Z
Z
Z
Mtǫ(x) dx = Mtǫ(x)1 {x 6∈ Ctǫ} dx + Mtǫ(x)1 {x ∈ Ctǫ} dx
I
I
I
The Conformal Minkowski Measure
Z
ǫ−β |I ∩ Ctǫ| = ǫ−β 1 {x ∈ Ctǫ} dx
I
• If the martingale Mt(x) reaches the threshold ǫ−β then
stop it. Call the stopped martingale Mtǫ(x).
Z
ǫ−β |I ∩ Ctǫ| = Mtǫ(x)1 {x ∈ Ctǫ} dx
I
and this is part of the trivial decomposition
Z
Z
Z
Mtǫ(x) dx = Mtǫ(x)1 {x 6∈ Ctǫ} dx + Mtǫ(x)1 {x ∈ Ctǫ} dx
I
I
I
• This is the Doob-Meyer decomposition
supermartingale
Z
Mtǫ(x)1 {x 6∈ Ctǫ} dx
I
for
the
The Conformal Minkowski Measure
• Can compute
lim
ǫ↓0
Z
I
Mtǫ(x)1 {6∈ Ctǫ} dx
The Conformal Minkowski Measure
• Can compute
lim
ǫ↓0
Z
I
Mtǫ(x)1 {6∈ Ctǫ} dx
• First note that Mtǫ(x)1 {x 6∈ Ctǫ} = Mt(x)1 {x 6∈ Ctǫ}.
The Conformal Minkowski Measure
• Can compute
lim
ǫ↓0
Z
I
Mtǫ(x)1 {6∈ Ctǫ} dx
• First note that Mtǫ(x)1 {x 6∈ Ctǫ} = Mt(x)1 {x 6∈ Ctǫ}.
• The sets Ctǫ decrease to Ct as ǫ ↓ 0, so
1 {x 6∈ Ctǫ} ↑ 1 {x 6∈ Ct}
The Conformal Minkowski Measure
• Can compute
lim
ǫ↓0
Z
I
Mtǫ(x)1 {6∈ Ctǫ} dx
• First note that Mtǫ(x)1 {x 6∈ Ctǫ} = Mt(x)1 {x 6∈ Ctǫ}.
• The sets Ctǫ decrease to Ct as ǫ ↓ 0, so
1 {x 6∈ Ctǫ} ↑ 1 {x 6∈ Ct}
• By monotone convergence and dimH Ct ≤ dimH C < 1
lim
ǫ↓0
Z
Z
Mtǫ(x)1 {x 6∈ Ctǫ} dx = lim Mt(x)1 {x 6∈ Ctǫ} dx
I
Zǫ↓0 I
=
Mt(x)1 {x 6∈ Ct} dx
ZI
=
Mt(x) dx
I
The Conformal Minkowski Measure
• So if
lim ǫ−β |I ∩ Ctǫ|
ǫ↓0
exists, it should be the increasing part of the Doob-Meyer
decomposition for
Z
Xt(I) = Mt(x) dx
I
The Conformal Minkowski Measure
• A theorem of Dellacherie and Meyer [DM82] says that
because the Doob-Meyer decomposition of Xt(I) exists
and
R
Mt(x) dx
=
Nt
−
At
↑
R
R
−β
ǫ
ǫ
ǫ
ǫ
|
(x)
dx
−
ǫ
|I
∩
C
M
}
dx
=
(x)1
{x
∈
6
C
M
t
t
t
t
I
I
I
then the corresponding increasing parts also converge, i.e.
for every t ≥ 0,
lim ǫ−β |I ∩ Ctǫ| = At(I)
ǫ↓0
where the limit is in L1.
Is Conformal Minkowski the Right Measure?
• SLE curves are the scaling limit of curves arising in
discrete models.
• The discrete model curves induce very natural measures
on the boundary.
• Open Question: Do the measures arising from the
discrete models, when properly scaled, converge to the
conformal Minkowski measure?
Is Conformal Minkowski the Right Measure?
• SLE curves are the scaling limit of curves arising in
discrete models.
• The discrete model curves induce very natural measures
on the boundary.
• Open Question: Do the measures arising from the
discrete models, when properly scaled, converge to the
conformal Minkowski measure?
• Our conjecture is yes.
Is Conformal Minkowski the Right Measure?
• SLE curves are the scaling limit of curves arising in
discrete models.
• The discrete model curves induce very natural measures
on the boundary.
• Open Question: Do the measures arising from the
discrete models, when properly scaled, converge to the
conformal Minkowski measure?
• Our conjecture is yes.
• We have (mild) numerical evidence to back this up.
Is Conformal Minkowski the Right Measure?
Percolation on a triangular domain.
Is Conformal Minkowski the Right Measure?
Percolation exploration path on an equilateral triangle.
Is Conformal Minkowski the Right Measure?
Measure induced by the exploration path.
Is Conformal Minkowski the Right Measure?
Measure induced by the exploration path.
Is Conformal Minkowski the Right Measure?
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
x
0.6
0.8
1
40 triangles per side, averaged over 100,000 paths
Is Conformal Minkowski the Right Measure?
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
x
0.6
0.8
1
100 triangles per side, averaged over 1,000,000 paths
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Empirical occupation measure, 40 triangles per side,
averaged over 100,000 paths
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Theoretical occupation measure, 40 triangles per side
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Ratio of empirical measure to theoretical one
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Empirical occupation measure, 100 triangles per side,
averaged over 1,000,000 paths
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Theoretical occupation measure, 100 triangles per side
Is Conformal Minkowski the Right Measure?
0
0.2
0.4
0.6
0.8
1
Ratio of empirical measure to theoretical one
Slides Produced With
Asymptote: The Vector Graphics Language
symptote
http://asymptote.sf.net
(freely available under the GNU public license)
References
[AS08] Tom Alberts and Scott Sheffield. Hausdorff dimension
of the sle curve intersected with the real line. Electron.
Jour. Probab., 13:1166–1188 (electronic), 2008.
[DM82] Claude
Dellacherie
and Paul-André
Meyer.
Probabilities and potential. B, volume 72 of NorthHolland Mathematics Studies.
North-Holland
Publishing Co., Amsterdam, 1982.
Theory of
martingales, Translated from the French by J. P.
Wilson.
[Dub03] Julien Dubédat. SLE and triangles. Electron. Comm.
Probab., 8:28–42 (electronic), 2003.
[RS05] Steffen Rohde and Oded Schramm. Basic properties
of SLE. Ann. of Math. (2), 161(2):883–924, 2005.
[SZ07] Oded Schramm and Wang Zhou. Boundary proximity
of SLE. arXiv:0711.3350v2 [math.PR], 2007.
[Yor01] Marc Yor. Exponential functionals of Brownian motion
and related processes. Springer Finance. SpringerVerlag, Berlin, 2001. With an introductory chapter by
Hélyette Geman, Chapters 1, 3, 4, 8 translated from
the French by Stephen S. Wilson.