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Logarithmic correction to resistance

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We study the trace of the incipient infinite oriented branching random walk in Zd×Z+\mathbb{Z}^d \times \mathbb{Z}_+ when the dimension is d=6d = 6. Under suitable moment assumptions, we show that the electrical resistance between the root and level n is O(nlogξn)O(n \log^{-\xi}n ) for a ξ>0\xi > 0 that does not depend on details of the model.
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arXiv:2006.03988v1 [math.PR] 6 Jun 2020
Logarithmic correction to resistance
Antal A. J´arai1Dante Mata L´opez2
June 9, 2020
Abstract
We study the trace of the incipient infinite oriented branching random walk in Zd×Z+
when the dimension is d= 6. Under suitable moment assumptions, we show that the
electrical resistance between the root and level nis O(nlogξn) for a ξ > 0 that does not
depend on details of the model.
1 Introduction
Consider a critical branching random walk in Zdconditioned to survive forever, starting with
a single individual at the origin o. The space-time points visited by this process can be turned
into a multi-graph, by placing an edge between (x, n) and (y, n + 1), whenever a particle located
at xat time nproduces an offspring located at yat time n+ 1. We call this multi-graph the
trace. We regard the trace as an electrical network, where each edge has unit conductance. Let
R(n) denote the expected resistance between (o, 0) and level nof the trace.
Barlow et al. [4, Example 1.8(iii)] showed that when d > 6, one has R(n)nAnswering a
question of [4], J´arai and Nachmias [14] showed that for d5 one has R(n) = O(n1α) for a
universal constant α > 0, under suitable moment assumptions. In the present paper we show,
under the same moment assumptions as in [14], that in d= 6 dimensions R(n) is sub-linear by
at least a logarithmic factor.
Theorem 1.1. Consider the trace of a branching random walk in dimension d= 6 with progeny
distribution that is critical, has positive variance and finite third moment, conditioned to survive
forever. Assume that the random walk steps are symmetric, non-degenerate and have exponential
tails. Then there exists a universal ξ > 0such that
R(n) = Onlogξn.
1Department of Mathematical Sciences, University of Bath, Claverton Down, Bath, BA2 7AY, United King-
dom. E-mail: A.Jarai@bath.ac.uk
2Department of Probability and Statistics, Centro de Investigaci´on en Matem´aticas A.C. Calle Jalisco S/N,
C.P. 36240, Guanajuato, Mexico. E-mail: dante.mata@cimat.mx
1
Let us explain some background and our motivation. Given an infinite graph G, suitable
bounds on the volume growth of Gand electrical resistances in Gprovide quantitative informa-
tion on the behaviour of the simple random walk on G, such as bounds on exit times from balls,
and the heat kernel; see e.g. [4], [17]. One can distinguish two regimes.
On the one hand, several examples are known where the behaviour of the walk is charac-
terised by the so-called Alexander-Orbach (AO) exponents [1]. See [15] for the case of a critical
branching tree conditioned to survive, and [5] for more detailed estimates in the case of perco-
lation on regular trees. The same exponents were also shown for: the incipient infinite cluster
of oriented (spread-out) percolation in dimensions d+ 1 >6 + 1 [4]; for the oriented critical
branching random in dimension d+ 1 >6 + 1 [4, Example 1.8(iii)] ; for unoriented percolation
under the triangle condition [16], [12]; and for the uniform spanning forest in d > 4 and on non-
amenable graphs [13]. In all of these examples, the underlying graph is similar, in a quantifiable
sense, to a critical branching process, and the scaling limit of the walk is conjectured, and in
some cases rigorously known, to be Brownian motion on the continuum random tree, see e.g. [7],
[6].
On the other hand, there are ‘low-dimensional’ examples where the random walk exponents
are different (either conjecturally, or rigorously). See [15], [9] for the incipient infinite percolation
cluster in 2D; see [3] for the uniform spanning tree in 2D and [2] for the uniform spanning
tree in 3D. Despite this progress, it is a major challenge, for example, to establish the scaling
of resistances for 2D critical percolation. This is due to lack of a clear connection between
resistance and conformally invariant quantities.
A consequence of [14] is that for the trace of oriented critical branching random walk, the
AO exponents cannot hold in any dimension d+ 1 5 + 1. This results from the fact that the
resistance does not scale linearly: R(n) = O(n1α) for some α > 0. Although branching random
walk is one of the simplest statistical physics models, determining the exact behaviour of the
resistance for d6 is already very challenging in this case. To the best of our knowledge, no
polynomial lower bound on R(n) is known in 3 d5, where such is expected. In this paper
we consider the trace of oriented branching random walk when d= 6, which is the conjectured
critical dimension. By analogy with other statistical physics models, one expects a logarithmic
correction: R(n)nlogξnfor some ξ>0, and Theorem 1.1 provides an upper bound of this
form. We expect that with additional work one can prove a lower bound of the same form (with
an exponent ξ′′ <), and we outline a possible strategy for this at the end of Section 1.3. We
also explain there why this is easier than proving meaningful lower bounds in 3 d5.
We follow a very similar setup to that of [14], in that we establish our upper bound by showing
that sufficiently many intersections are present in the trace to reduce the resistance from O(n)
to O(nlogξn). The main difference is that in d= 6 the intersections are more sparse than
in d5. In particular, on each scale, there is only a logarithmically small probability to
find intersecting paths on that scale. Establishing this intersection estimate is more delicate
compared to its analogue in [14], and some of the other estimates also need improvement.
In order to facilitate the import of the setup from [14], we use the following convention: all
2
notation that has the same meaning as in [14] is identical in the present paper, and notation that
has closely related meaning is denoted by a prime. For clarity of the proofs, we found it necessary
to spell out even smaller changes compared to [14]. However, we do take some arguments without
change from [14], and hence familiarity with that paper is essential to understand our arguments.
1.1 Assumptions
Let p(k), k0 be a progeny distribution that satisfies:
(i) Pkkp(k) = 1;
(ii) Pkk(k1)p(k) = σ2(0,);
(iii) Pkk3p(k)C3<.
The incipient infinite branching process is obtained by conditioning on survival up to time n
and taking the weak limit as n→ ∞. The limiting object admits the following alternative con-
struction [15], [19]. Consider an infinite path (V0, V1,...), and attach to each Vi, independently,
a branching tree that in its first generation follows the size-biased distribution:
ep(k) = (k+ 1)p(k+ 1), k 0,
and follows pafterwards.
Let p1(x, y) be a one-step random walk transition probability in Zdthat satisfies:
(i) PxZdeb|x|p1(o, x)<for some b > 0;
(ii) {xZd:p1(o, x)>0}generates Zdas a group;
(iii) p1(x, y) = p1(y, x).
We will denote by pn(x, y) the n-step transition probabilities. The incipient infinite branching
random walk is obtained by first drawing a sample Tof the incipient infinite branching process,
and then applying a random walk map Φ : T Zd×Z+defined as follows. We initialize Φ
by requiring that the root ρof Tis mapped to (o, 0). Then, recursively, if {U, V }is an edge
of Tbetween generations nand n+ 1, such that Φ(U) = (x, n) has already been defined, we
set Φ(V) = (y, n + 1) with the displacement yxchosen according to p1(x, y), independently
between different edges. By the trace of the branching random walk we mean the multi-graph
with vertex set Φ(T), and edge set consisting of {Φ(U),Φ(V)}for every edge {U, V }of T.
1.2 Electrical resistance
For background on electrical resistance, see [20]. We denote by Reff (xy) the effective resis-
tance between vertices xand y. We will frequently use the triangle inequality:
Reff (xz)Reff (xy) + Reff (yz).(1.1)
We will also use the following parallel law : if G1= (V, E1) and G2= (V, E2) are graphs on the
same vertex set but with disjoint edge sets, and G= (V , E1E2), and if R1=Reff (xG1
y),
3
R2=Reff (xG2
y), then
Reff (xG
y)1
R1
+1
R21
1
4(R1+R2),(1.2)
where the second inequality uses that the harmonic mean is at most the arithmetic mean.
1.3 A finite approximation
Given n1 and m2n, we define the random tree Tn,m as in [14]:
(i) consider a backbone V0,...,Vnwith marked root ρ=V0;
(ii) attach to each Via critical tree that has distribution epin the first step, and pafterwards,
and is conditioned to die out by time mi.
We define γ(n, x) as in [14]:
γ(n, x) = sup
m2n
ETn,m hReff ((o, 0) Φ(Vn)) Φ(Vn) = (x, n)i, x Zd.
The following theorem is our main technical result, and is an analogue of [14, Theorem 1.2]. It
is expressed in terms of the norm:
kxk:= v
u
u
t1
d
d
X
i,j=1
xiQ1
ij xj,
where Qij =PxZdxixjp1(o, x) is the covariance matrix of the random walk step distribution.
Theorem 1.2. Assume d= 6. There exists a universal constant ξ(0,1/2) and A=
A(σ2, C3,p1)<such that for all n2we have
γ(n, x)
An(log n)ξwhen kxk ≤ n;
An(log n)ξ1log(kxk2/n)
log nξ
when n < kxk ≤ n/2;
An when kxk> n/2.
Proof of Theorem 1.1 assuming Theorem 1.2. As in [14], we have
R(n)lim
m→∞ ETn,m hReff ((o, 0) Φ(Vn)) i
sup
m2n
ETn,m hReff ((o, 0) Φ(Vn)) i
=X
xZd
pn(o, x)γ(n, x).
4
From the bound in Theorem 1.2 we have
R(n)An(log n)ξ"X
x:kxk≤n
pn(o, x) + X
x:n<kxk≤n3/4
pn(o, x)C
+X
x:n3/4<kxk≤n/2
pn(o, x)C(log n)ξ+X
x:n/2<kxk
pn(o, x) (log n)ξ#.
The first two sums are bounded by 1 and C, respectively. In the third and fourth sums we use
that
(log n)ξC n1/2Ckxk2
n,when kxk> n3/4.
This gives the upper bound
CX
x
pn(o, x)kxk2
n=C.
1.4 Strategy for a lower bound
Let us now present a possible strategy for a lower bound on R(n) in d= 6. Consider independent
copies T1and T2of Tn,2n, and random walk mappings Φ1and Φ2initialized by Φ1(ρ1) = (o, 0) and
Φ2(ρ2) = (x, 0), where kxk ≍ n. The first and second moments of the number of intersections
between Φ1(T1) and Φ2(T2) are 1 and log n, respectively, in d= 6, as we will see. A key
estimate we prove in this paper is the following (presented somewhat informally at this stage):
P(#intersections of Φ1(T1) and Φ2(T2)clog n)c
log n.(1.3)
Suppose for what follows that one could complement this with the bound:
P1(T1)Φ2(T2) = )1C
log n.(1.4)
Consider the event that in Φ(T2n,m), the edge {Φ(Vn),Φ(Vn+1)}is pivotal for connecting (o, 0)
to level 2nof the trace. Let Trkdenote the Φ-image of all critical trees attached to the path
{Vn2k,...,Vn2k1},k= 1,2,...,log2n, and let f
Tr denote the Φ-image of all critical trees
attached to the path {Vn+1,...,V2n}. By the FKG inequality, and a heuristic based on (1.4),
we get
P({Φ(Vn),Φ(Vn+1)}is pivotal)
log2n
Y
k=1
P(Trkf
Tr = )
log2n
Y
k=1 1C
log 2kc(log n)O(1).
This would imply that there are at least nlogξ′′ npivotals along the backbone of T2n,m, and
hence a lower bound on the resistance. For the same ideas to be fruitful in 3 d5, one would
also need the exponent to be <1, not only O(1).
5
1.5 Random walk estimates
The following proposition collects some random walk estimates we take without change from
[14]. Let S(n), n0 denote a random walk with S(0) = o, and step distribution p1.
Proposition 1.3 ([14, Proposition 1.3]).There exists k1=k1(p1), C > 0and δ1=δ1(d)>0
such that the following hold.
(i) Whenever k1kn,kxk ≤ 4n/k, we have
EhkS(k)k2S(n) = xiCk.
(ii) Whenever k1kδ1n,kxk ≤ 4n/k, we have
EhkS(k)k2S(n) = x, kS(k)k>kiCk.
(iii) Whenever k1kδ1n,k1knkand kxk ≤ min{4n/k, 4n/k}, we have
EhkS(k+k)S(k)k2S(n) = x, kS(k)k>kiCk.
Let us write
D:= det(Q)1/2d.
The following lemma is a special case of [14, Lemma 1.4] (take β= 0 there).
Lemma 1.4. There exists C=C(d)such that the following hold.
(i) There exists n1=n1(p1)such that for all yZdwe have
P(S(n) = y)2C
Ddnd/2,(1.5)
when nn1.
(ii) For any 0< ε < 1and 0< L < there exists n2=n2(p1, ε, L)such that for all yZd
such that kyk ≤ Lnwe have
P(S(n) = y)C(1 + ε)
Ddnd/2edkyk2/(2n),
P(S(n) = y)C(1 ε)
Ddnd/2edkyk2/(2n).
(1.6)
when nn2.
6
Above we assumed that the random walk has period 1. Trivial modifications can be made
when the period is 2, and we will not make this explicit in our arguments.
Below we collect a few more frequently used facts. These are all standard (see [14, Section
1.5] for more information). There exists a constant c > 0 such that we have
X
x:kxk≤L
1cDdLd, L 1.(1.7)
When d3, the Green function G(x) := P
n=0 pn(o, x) satisfies
G(x)C(d)
Ddkxk2d,when kxk ≥ L1=L1(p1). (1.8)
2 Induction Scheme
We set up the induction scheme as in [14], apart from the definition of the event B(i, c0). For
convenience of the reader, we provide Definitions 2.1–2.6 below, that are from [14]. Given an
instance of Tn,m, consider a small δ > 0, such that δn is an integer. We write
Xi=Viδn, i = 0,1,...,δ1,
and write xiZd,i= 0,1,...,δ1for the spatial location of Xi, so that Φ(Xi) = (xi, iδn).
Write Tn,m() for the subtree of Tn,m emanating from Voff the backbone (including V).
Fix an integer Kand write n=NKδn +Kδn + ˜n, with 0 K< K an integer and
δn ˜n < 2δn. The definitions to follow are illustrated in Figure 1.
Definition 2.1. For satisfying iδn ℓ < (i+ 1)δn we say that a backbone vertex Vhas the
unique descendant property (UDP) if among its descendants at level (i+ 1)δn in Tn,m() there
is a unique one that reaches level (i+ 2)δn. For any other vertex Vof Tn,m at level iδn we say
that Vhas UDP if among its descendants at level (i+ 1)δn there is a unique one that reaches
level (i+ 2)δn.
Definition 2.2. Given an integer K1, a number δ > 0 such that Kδ (1/2) and an instance
of Tn,m we say that a sequence (i, i + 1,...,i+K) of length K+ 1 is K-tree-good if the following
holds:
(1) There exists a unique iδn 1<(i+ 1)δn such that Tn,m(1) reaches height (i+ 2)δn.
Moreover, this unique 1satisfies (i+ 1/4)δn 1(i+ 3/4)δn.
(2) V1has UDP. We call the unique descendant Yi+1. For all isatisfying i+ 2 ii+K
we inductively define the vertices Yiof Tn,m(1) as follows. We require that Yi1has
UDP and call the unique descendant Yi.
7
(3) There exists a unique (i+K1)δn 2<(i+K)δn such that Tn,m(2) reaches height
(i+K+ 1)δn. Moreover, this unique 2satisfies (i+K3/4)δn 2(i+K1/4)δn.
(4) V2has UDP, and we call the unique descendant X
i+K. The vertex X
i+Khas UDP, and
we call the unique descendant X
i+K+1. Similarly, Yi+Khas UDP, and we call the unique
descendant Yi+K+1.
(a)
PSfrag replacements
Xi
V1V+
1
Xi+1
Xi+2
Xi+K1
V2V+
2
Xi+KX
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1
v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
(b)
PSfrag replacements
Xi
V1
V+
1
Xi+1
Xi+2
Xi+K1
V2
V+
2
Xi+K
X
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
Figure 1: (a) Illustration of K-tree-good. (b) Illustration of K-spatially-good. All spatial
distances between consecutive vertices are at most time difference and the spatial distance
between x
i+Kand yi+Kis at most δn. [Figure reused by permission of Springer, from Com-
mun. Math. Phys. 331 (2014), 67–109 (Electrical resistance of the low-dimensional critical
branching random walk. A.A. J´arai and A. Nachmias) c(2014).]
Given a K-tree-good sequence (i, . . . , i +K) we denote by V+
1(respectively V+
2) the child of
V1(respectively V2) leading to Yi+1 (respectively X
i+K). We further define the spatial locations
yiby Φ(Yi) = (yi, iδn) for i+ 1 ii+K+ 1, and we similarly define x
i+K,x
i+K+1,v1,
v+
1,v2,v+
2.
8
We will write UWto denote that Wis a descendant of U, and write h(U), h(W) for their
respective heights in the tree (in particular, h(W)> h(U)).
Definition 2.3. Let UWbe two tree vertices and let u, w Zdbe defined by Φ(U) =
(u, h(U)) and Φ(W) = (w, h(W)). We say that Uand Ware typically-spaced if kwuk ≤
ph(W)h(U). Denote this event by T S(U, W ).
Definition 2.4. We say that a K-tree-good sequence (i, . . . , i +K) is K-spatially-good if the
following holds.
(5) T S(Xi, V1),
T S(V1+1, Xi+1),
For each i+ 1 ji+K2 we have T S(Xj, Xj+1),
T S(Xi+K1, V2),
T S(V2+1, Xi+K),
(6) T S(V+
1,Yi+1),
For each i+ 1 ji+K1 we have T S(Yj,Yj+1),
T S(V+
2,X
i+K),
• kx
i+Kyi+Kk ≤ δn.
Definition 2.5. When a sequence (i, . . . , i +K) is both K-tree-good and K-spatially-good we
say that it is K-good. Let A(i) be the event that (i, . . . , i +K) is K-good.
Next, let (i, . . . , i +K) be a K-good sequence and let U1, U2be two vertices at the same
height such that U1≻ X
i+Kand U2≻ Yi+K. Given these, we write Z1for the highest common
ancestor of U1and X
i+K+1 and Z2for the highest common ancestor of U2and Yi+K+1 (see
Figure 2). Further, we denote by Z+
1(respectively Z+
2) the child of Z1(respectively Z2) leading
to U1(respectively U2).
Definition 2.6. We say that U1, U2intersect-well if the following conditions hold:
1. U1≻ X
i+K,U2≻ Yi+K,
2. (i+K+ (5/6))δn h(U1) = h(U2)(i+K+ 1)δn;
3. (i+K+ (1/2))δn h(Z1), h(Z2)(i+K+ (4/6))δn;
4. T S(X
i+K, Z1), T S(Z+
1, U1), T S(Yi+K, Z2), T S(Z+
2, U2);
5. Φ(U1) = Φ(U2).
9
PSfrag replacements
Xi
V1
V+
1
Xi+1
Xi+2
Xi+K1
V2
V+
2
Xi+K
X
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1
v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
Φ(X
i+K) Φ(Yi+K)
Φ(X
i+K+1) Φ(Yi+K+1 )
Φ(U1)=Φ(U2)
Φ(Z1)
Φ(Z2)
Φ(Z+
1)
Φ(Z+
2)
Figure 2: The labelling of vertices in the two (potentially) intersecting trees emanating from
X
i+Kand Yi+K. [Figure reused by permission of Springer, from Commun. Math. Phys. 331
(2014), 67–109 (Electrical resistance of the low-dimensional critical branching random walk.
A.A. J´arai and A. Nachmias) c(2014).]
And define the random set Iby
I={(U1, U2) : U1and U2intersect-well},(2.1)
Now we define the event B(i, c
0) where c
0>0 is a constant to be chosen later:
B(i, c
0) = A(i)|I| ≥ c
0σ4
Ddlog(δn).
The following theorem replaces [14, Theorem 2.1], and will be proved in Section 3.
Theorem 2.7 (Intersections exist).Assume d= 6. There exist constants c
0, c
1>0and for
any K2there exists c2=c2(K)>0, and n
3=n
3(σ2, C3,p1, K)such that for any 0< δ <
(K+ 4)1, whenever δn n
3and xsatisfies kxk ≤ p2n/δ, we have
P(A(i)|Φ(Vn) = (x, n)) c2,
and
P(B(i, c
0)|A(i),Φ(Vn) = (x, n)) c
1
log(δn),
for i= 0, K, 2K, . . . , (N1)K.
10
As in [14], we define
γ(n) = sup
x:kxk≤n
γ(n, x).
The proof of the following theorem will be completed in Section 6.
Theorem 2.8 (Analysis of good blocks).There exists K
0<and n
4=n
4(σ2, C3,p1)such
that if KK
0and δn n
4, we have
EhReff (Φ(Xi)Φ(Xi+K))A(i),B(i, c
0),Φ(Vn) = (x, n)i3K
4max
1kδn γ(k),
for i= 0, K, 2K, . . . , (N1)K.
3 Existence of Intersections
In this section we prove Theorem 2.7. The statement about the probability of A(i) is unchanged
compared to [14, Theorem 2.1], and hence requires no proof. On the other hand, there are
substantial changes to the proof of the estimate on the probability of B(i, c
0), that we now
detail. Some of the required estimates appeared in the MSc thesis [18], in a slightly different
form (with less technical restrictions on the intersections). Here we adapt and complete the
analysis of [18] in a form that suits the requirements of the present paper.
3.1 Sufficient intersections
We now proceed to prove the second statement of Theorem 2.7. We will need to assume that
the progeny distribution is bounded by M, and approximate the original progeny distribution
with bounded distributions pM(k), in such a way that
1 = X
0kM
pM(k) = X
0kM
kpM(k)
σ2= lim
M→∞ X
0kM
k(k1)pM(k)
C3X
0kM
k3pM(k).
Given any nand msuch that m2nwe regard the random tree Tn,m as a subtree of an infinite
M-ary tree TMwith root ρas follows: the root of Tn,m is mapped to ρand if Wis a vertex of
Tn,m with kchildren we map the kedges randomly amongst the M
kpossible choices in TM.
Denote by VnTMthe random vertex where the last backbone vertex of Tn,m was mapped to.
The triple (Tn,m, ρ, Vn) is a doubly rooted tree.
11
Let (T1, ρ1,V1) and (T2, ρ2,V2) be two independent copies of (Tδn,2δn , ρ, Vδn ), randomly
imbedded into TM. Let Φ1and Φ2be two independent random walk mappings of TMsuch
that Φ1(ρ1) = Φ(X
i+K) and Φ2(ρ2) = Φ(Yi+K). However, for notational convenience, and with-
out loss of generality, we assume that Φ1(ρ1) = (o, 0) and Φ2(ρ2) = (x, 0), with kxk ≤ δn.
Recall that the random variable |I| introduced in (2.1) has the same distribution as the random
variable (also denote |I| here):
|I| =X
U1∈T1, U2∈T2
1(U1, U2) intersect well.
Our goal in this section is to show that when d= 6, we have
P|I| ≥ c
0σ4Ddlog(δn)c
1
log(δn)(3.1)
for suitable c
0, c
1>0. Note that once we prove this estimate for offspring distribution pM(k)
(uniformly in M), we can let M→ ∞, and obtain the second statement of Theorem 2.7 in full
generality.
For technical reasons, we will also need a slight modification of I. The difference is that we
make a stronger restriction on the spatial displacement between Φ(Z+
1) and Φ(U1) and between
Φ(Z+
2) and Φ(U2), as well as a stronger restriction on the height of the intersection:
I=
(U1, U2)∈ I :
h(U1) = h(U2)(11/12)δn,
kz+
1u1k ≤ 1
2qh(U1)h(Z+
1),
kz+
2u2k ≤ 1
2qh(U2)h(Z+
2)
.
The starting point for proving (3.1) is to look at the moments of |I|. The following theorem
is an analogue of [14, Theorem 3.8].
Theorem 3.1. Assume that d= 6 and kxk ≤ δn. There exist constants C<and c>0
and n
9=n
9(σ2, C3,p1)<such that for δn n9we have
E|I| ≥ E|I| ≥ cσ4
D6,(3.2)
and
E|I|2Cσ8
D12 log(δn).(3.3)
Since the proof of this theorem requires only minor adaptations compared to the proof of
[14, Theorem 3.8], we omit it.
Applying the Paley-Zygmund inequality to |I| shows that P(|I| >0) c/log(δn), but
comes short of proving (3.1). In order to prove this, we will consider the number of intersections
12
conditional on (U1, U2)∈ Ifor fixed U1, U1TM, and show that this is at least of order log(δn)
with conditional probability bounded away from 0.
For the statement of the next lemma we fix
0k1δn 1k1+ 1 huδn.
Given VTMat level δn and UTMat level hulet ZTMbe the highest common ancestor
of Vand Uand let Z+be the unique child of Zleading towards U. We assume that Zis at
height k1. Given a tree tTMsuch that V, U tand Vdoes not have any children in t,
we have a unique decomposition of tinto edge disjoint trees (tA, ρ, Z),(tB, Z +, U), tCand tD,
see Figure 3. The doubly rooted tree (tA, ρ, Z) contains all the descendants of ρthat are not
descendants of Z. The doubly rooted tree (tB, Z +, U) contains all the descendants of Z+that
are not descendants of U. The tree tCcontains all the descendants of Uand finally the tree tD
contains all other edges, namely, all the descendants of Zthat are not descendants of Z+(in
particular, the edge Z, Z+is in tD).
For WTMlet ΘWdenote the tree isomorphism that takes Wto ρand the descendants
subtree of Wonto TM.
The following lemma is taken without change from [14].
Lemma 3.2 ([14, Lemma 6.4]).Let V, U TMbe at heights δn and hu, respectively, and let
(T, ρ, V)be distributed as (Tδn,2δn, ρ, Vδn ). Conditionally on the event {V =V, U T } we have
that
(TA, ρ, Z)d
= (Tk1,2δn, ρ, Vk1)|Vk1=Z,
and
ΘZ+((TB, Z+, U)) d
= (Thuk11,2δnk11, ρ, Vhuk11)|Vhuk11= ΘZ+(U).
Let us fix vertices V1, V2TMat height δn, and vertices U1, U2TM, both at height hu,
where (5/6)δn hu(11/12)δn, such that (1/2)δn h(Z1), h(Z2)(4/6)δn. Let us denote
the vertices on the path in TMbetween Z+
1and U1as
Z+
1=W(k1+1)
1, W (k1+2)
1,...,W(hu1)
1, W (hu)
1=U1,
and vertices on the path in TMbetween Z+
2and U2as
Z+
2=W(k2+1)
2, W (k2+2)
2,...,W(hu1)
2, W (hu)
2=U2.
For any choice of (V1, V2, U1, U2) as above, let us define the event
IW=IW(V1, V2, U1, U2) = {V1=V1,V2=V2} ∩ {(U1, U2)∈ I}.
We will also use the shorthand
1IW=1IW(V1,V2,U1,U2).
13
PSfrag replacements
Xi
V1
V+
1
Xi+1
Xi+2
Xi+K1
V2
V+
2
Xi+K
X
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1
v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
Φ(X
i+K)
Φ(Yi+K)
Φ(X
i+K+1)
Φ(Yi+K+1)
Φ(U1)=Φ(U2)
Φ(Z1)
Φ(Z2)
Φ(Z+
1)
Φ(Z+
2)
δn
2δn
k1
hu
0
V
U
Z
Z+
ρ
tA
tB
tC
tD
Figure 3: Illustration of the decomposition into edge-disjoint trees tA, tB, tC, tDappearing in
Lemma 3.2 (2δn and δn are not to scale). [Figure reused by permission of Springer, from
Commun. Math. Phys. 331 (2014), 67–109 (Electrical resistance of the low-dimensional critical
branching random walk. A.A. J´arai and A. Nachmias) c(2014).]
14
For the next definition, recall the constants n1and n2of Lemma 1.4, and put
n= max{n1, n2(p1, ε = 1/2, L = 1)}.
When the event IWoccurs, we say that a pair of vertices (Y1, Y2) is a good extra intersection,
if the following events all occur:
Y1∈ T B
1and Y2∈ T B
2;
Y1is a descendant of W(r1)
1for some (9/12)δn r1hun;
Y2is a descendant of W(r2)
2for some (9/12)δn r2hun;
(Y1, Y2) intersect-well;
huh(Y1) = h(Y2).
Observe that the last two conditions require in particular that (5/6)δn huh(Y1) = h(Y2)
δn, and Φ1(Y1) = Φ2(Y2). We define the random set:
b
IV,U =({(Y1, Y2) : (Y1, Y2) is a good extra intersection}when IWoccurs;
when IWdoes not occur,
and the random variable:
J=J(c
0) = X
V1,V2TM
U1,U2TM
1|b
IV,U |c
0σ4
Ddlog(δn).
A crucial property is that we have the following inclusion of events:
{J > 0} ⊂ V1, V2, U1, U2:b
IV,U c
0
σ4
Ddlog(δn)
|I| ≥ c
0
σ4
Ddlog(δn)=B(c
0).
Hence in order to conclude, we need to show that P(J > 0) c
1/log(δn) for suitable c
0, c
1>0
This follows immediately from the theorem and proposition stated below, that are the main
results of this section.
Theorem 3.3. Assume d= 6 and kxk δn. There exists n
9=n
9(σ2, C3,p1)<such that
for δn n
9and all V1, V2, U1, U2TMwe have
Ehb
IV,U IW(V1, V2, U1, U2)icσ4
D6log(δn),
and
Ehb
IV,U 2IW(V1, V2, U1, U2)iCσ8
D12 log2(δn).
In particular, taking c
0=1
2c, we have
Pb
IV,U c
0
σ4
D6log(δn)IW(V1, V2, U1, U2)4(c)2
C>0.(3.4)
15
Proposition 3.4. Assume d= 6 and kxk ≤ δn. For δn n
9and with the choice of c
0in the
conclusion (3.4) of Theorem 3.3, we have
EJ(c
0)cσ4
D6,
and
EJ(c
0)2Cσ8
D12 log(δn).
In particular, we have
P(B(c
0)) P(J(c
0)>0) (c)2
C
1
log(δn).
Proof of Proposition 3.4 assuming Theorem 3.3. Due to (3.4) of Theorem 3.3, we have
EJ(c
0) = X
V1,V2
U1,U2
P(IW(V1, V2, U1, U2)) Pb
IV,U c
0
σ4
D6log(δn)IW(V1, V2, U1, U2)
cX
V1,V2
U1,U2
P(IW(V1, V2, U1, U2)) = cE|I|.
Due to Theorem 3.1, this is at least cσ4D6, proving the first statement.
For the second statement we use that J(c
0)≤ |I|, and hence the upper bound on EJ(c
0)2
follows immediately from Theorem 3.1.
The last statement follows from the Paley-Zygmund inequality:
P(J(c
0)>0) [EJ(c
0)]2
EJ(c
0)2.
Hence it remains to prove Theorem 3.3, that we do in the next two sections.
3.2 Lower bound on the first moment
We start with the arguments for the lower bound on the first moment. For the next lemma,
assume that the event IW(V1, V2, U1, U2) occurs. Recall that this defines vertices Z1, Z +
1∈ T1
and Z2, Z+
2∈ T2at levels k1, k1+ 1 and k2, k2+ 1, respectively, as well as the vertices W(r1)
1at
levels k1+1 r1huand W(r2)
2at levels k2+1 r2hu. We will write Φ(W(r1)
1) = (w(r1)
1, r1)
and Φ(W(rr)
r) = (w(r2)
2, r2).
16
Lemma 3.5. There exists an absolute constant c>0with the following property. Assume
that the event IW(V1, V2, U1, U2)occurs. Then for δn 12n2(p1, ε = 1/2, L = 1) and any
(9/12)δn r1, r2hun2we have
Pkw(rj)
jz+
jk ≤ 1
2prjkj1,kujw(rj)
jk ≤ 1
8phurj, j = 1,2IWc.(3.5)
Proof. Condition on the event IW, and let us further condition on the spatial locations z+
1,
z+
2and u1=u2=u. Due to Lemma 3.2, the conditional distribution of the path between
(z+
j, kj+ 1) and (u, hu), is a random walk started at z+
jand conditioned to arrive at uat time
hukj1, j= 1,2. Let us write
j=wZd:kwz+
jk ≤ 1
2prjkj1,kuwk ≤ 1
8phurj.
Therefore, the probability in the statement of the lemma equals
Y
j∈{1,2}Pwjprjkj1(z+
j, w)phurj(w, u)
phukj1(z+
j, u).
Since hukj1(5/6)δn (4/6)δn 1(1/12)δn n2, and hurjn2, the local limit
theorem (Lemma 1.4) implies that there exists C=C(d) and c=c(d)>0, such that for all
wjwe have
phukj1(z+
j, u)C
Dd(hukj1)d/2C
Dd(δn)d/2
prjkj1(z+
j, w)c
Dd(rjkj1)d/2c
Dd(δn)d/2
phurj(w, u)c
Dd(hurj)d/2.
(3.6)
Since we are conditioning on IW, the restriction kuz+
jk ≤ 1
2phukj1 holds. We claim
that this implies that
|j| ≥ cDd(hurj)d/2.(3.7)
Indeed, writing t=hukj1, αt =hurj, (1 α)t=rjkj1, we have α1/2, and this
implies the inequalities 1
8αt 1
2p(1 α)tand 1
2t1
2p(1 α)t. These in turn imply
that at least half of the ball of radius αt centred at u(namely the half lying in the direction of
z+
j) is included in Ωj. Putting together the bounds (3.6) and (3.7) gives the statement of the
lemma.
For the next lemma, assume the event IW(V1, V2, U1, U2), and consider vertices Y1 T B
1
and Y2 T B
2at common height huh(Y1) = hy=h(Y2)δn. Let Y1be a descendant of
W(r1)
1, and Y2be a descendant of W(r2)
2.
17
Lemma 3.6. Fix V1, V2, U1, U2, and assume the event IW(V1, V2, U1, U2). Let Y1 T B
1and
Y2∈ T B
2be vertices at height huh(Y1) = hy=h(Y2)δn, and assume they are descendants
of W(r1)
1and W(r2)
2, respectively. Assume also that hyr1, hyr2n2(p1, ε = 1/2, L = 1).
We have
P(Y1, Y2)b
IV,U T1,T2,IW(V1, V2, U1, U2)c
Dd
1
(2hyr1r2)d/2.(3.8)
Proof. Without loss of generality, we assume that hyr1hyr2(the opposite case is handled
analogously). Condition on the event in the statement, and let us further condition on the spatial
locations of W(r1)
1and W(r2)
2, that we denote by w1and w2, for short. Due to Lemma 3.5, we
may assume, at the cost of a constant factor, that the event in (3.5) hold. Assuming that this
is the case, let
Ω = {yZd:kw1yk ≤ 1
2phyr1,kw2yk ≤ 1
2phyr2}.
We show that || ≥ cDd(hyr1)d/2. For this, it is enough to show that all points ysatisfying
the condition on kw1yk ≤ 1
8phyr1automatically satisfy the condition on kw2ykin the
definition of Ω.
Write a=hur1,b=hur2,c=hyhu, so that 0 aband c0. Then we have
kw2w1k ≤ 1
8a+1
8b1
4a+b.
Hence we are left to show that
1
8a+c+1
4a+b1
2b+c.
It is easy to see that this follows from a+cb+cand a+b2(b+c).
With the estimate on the size of Ω at hand, and using that hyr1, hyr2n2, we can
apply the local CLT (Lemma 1.4) to get that the conditional probability in (3.8) is at least:
X
y
phyr1(w1, y)phyr2(w2, y)≥ ||c
Dd(hyr1)d/2c
Dd(hyr2)d/2
c
Dd(hyr2)d/2c
Dd(hyr2+hyr1)d/2,
as claimed.
Lemma 3.7. Assume d= 6. Then for δn (24n)2(where n= max{n1, n2(p1, ε = 1/2, L =
1)}), we have
E|b
IV,U |IWcσ4
Ddlog(δn).
18
Proof. Due to Lemma 3.6, we have
E|b
IV,U |IWc
Dd
hun2
X
r1=(9/12)δn
hun
X
r2=(9/12)δn
δn
X
hy=hu
1
(2hyr1r2)3EL(hy, r1)EL(hy, r2),(3.9)
where, analogously to [14, Lemma 3.10], L(hy, r1) denotes the number of vertices Y1of TB
1at
level hythat are descendants of W(r1)
1. As in [14, Lemma 3.10], we have EL(hy, r1)2and
EL(hy, r2)2. This gives that the right hand side of (3.9) is at least
cσ4
Dd
hun
X
r1=(9/12)δn
hun
X
r2=(9/12)δn
δn
X
hy=hu
1
(2hyr1r2)3.
Let us write s1=hur1,s2=hur2and h=hyhu, so that the last expression satisfies
cσ4
Dd
(1/24)δn
X
s1=n
(1/24)δn
X
s2=n
(1/12)δn
X
h=0
1
(2h+s1+s2)3
cσ4
Dd
(1/24)δn
X
s1=n
(1/24)δn
X
s2=n2
1
(s1+s2)2
cσ4
Dd
(1/24)δn
X
s1=n
1
s1
cσ4
Ddlog(δn),
using in the last step that log(n) + log(24) 1
2log(δn).
3.3 Upper bound on the second moment
We fix V1, V2, U1, U2such that (5/6)δn h(U1) = hu=h(U2)(11/12)δn, and recall that we
condition on the event IW. Fix heights huhy, heyδn, and consider a pair of vertices (Y1,e
Y1)
that are both in TB
1such that h(Y1) = hyand h(e
Y1) = hey. There then exist unique heights
(9/12)δn r1,er1hun1such that Y1W(r1)
1and e
Y1W(er1)
1. Let e
Z1denote the highest
common ancestor of Y1and e
Y1, and let k
1=h(e
Z1). Note that we can have r1=er1, in which
case k
1r1=er1, while if r16=er1, we have k
1=r1er1. Let us write L(hy, hey, k
1, r1,er1) for the
number of pairs (Y1,e
Y1) that satisfy the above height restrictions with given hy, hey, k
1, r1,er1.
The following is an analogue of [14, Lemma 3.11].
19
Lemma 3.8. We have
EhL(hy, hey, k
1, r1,er1)IWi
σ4when r1=er1< k
1< hy, hey;
σ2when r1=er1< k
1=hyhey;
C3when r1=er1=k
1< hy, hey;
σ4when r16=er1,k
1=r1er1< hy, hey;
(Observe that there is no case r1=er1=k
1=hyhey, since r1=er1< huhy, hey.)
Proof. Conditional on IWthe distribution of TB
1is the same as that of Thuk11,2δnk11
(cf. Lemma 3.2). Hence the proof boils down to the same (straightforward) branching process
calculations as the proof of [14, Lemma 3.11].
The following ‘diagrammatic estimate’ is taken without change from [14]. Recall the constant
n1=n1(p1) from Lemma 1.4(i), and the constant L1=L1(p1) from (1.8). See Figure 4(a).
Lemma 3.9 ([14, Lemma 3.12]).Suppose d3. There are constants C=C(d)>0and
C2=C2(p1)such that for all ez1,ez2Zdwe have
X
h:k
1k
2hδn
p2hk
1k
2(ez1,ez2)C
Ddf(k
1, k
2,ez1,ez2),
where
f(k
1, k
2,ez1,ez2) :=
|k
1k
2|(2d)/2if kez1ez2k ≤ |k
1k
2|1/2
and |k
1k
2| ≥ n1;
C2if kez1ez2k ≤ |k
1k
2|1/2<n1;
kez1ez2k2dif kez1ez2k>|k
1k
2|1/2
and kez1ez2k ≥ L1;
C2if |k
1k
2|1/2<kez1ez2k< L1.
Proposition 3.10. Assume d= 6. We have
E|b
IV,U |2IWCσ8
D2dlog2(δn).
Proof. The proof broadly follows the outline of the proof of [14, Theorem 3.8]. In addition to
the event IW, let us further condition on the spatial location uof the intersection (U1, U2)∈ I,
as well as the spatial locations z+
1and z+
2. Let Y1,e
Y1 T B
1and Y2,e
Y2 T B
2be pairs of tree
20
PSfrag replacements
Xi
V1
V+
1
Xi+1
Xi+2
Xi+K1
V2
V+
2
Xi+K
X
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1
v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
Φ(X
i+K)
Φ(Yi+K)
Φ(X
i+K+1)
Φ(Yi+K+1)
Φ(U1)=Φ(U2)
Φ(Z1)
Φ(Z2)
Φ(Z+
1)
Φ(Z+
2)
δn
2δn
k1
hu
0
V
U
Z
Z+
ρ
tA
tB
tC
tD
0
k
1
k
2
h
hy
hey
ez1ez2
y
ey
δn
PSfrag replacements
Xi
V1
V+
1
Xi+1
Xi+2
Xi+K1
V2
V+
2
Xi+K
X
i+K
X
i+K+1
Yi+1
Yi+2
Yi+K1
Yi+K
Yi+K+1
δn
xi
v1
v+
1
xi+1
xi+2
xi+K1
v2
v+
2
xi+K
x
i+K
x
i+K+1
yi+1
yi+2
yi+K1
yi+K
yi+K+1
δn
Φ(X
i+K)
Φ(Yi+K)
Φ(X
i+K+1)
Φ(Yi+K+1)
Φ(U1)=Φ(U2)
Φ(Z1)
Φ(Z2)
Φ(Z+
1)
Φ(Z+
2)
δn
2δn
k1
hu
0
V
U
Z
Z+
ρ
tA
tB
tC
tD
0
k
1
k
2
h
hy
hey
ez1ez2
y
ey
δn
Figure 4: (a) Illustration of the quantity bounded in Lemma 3.9. The curves represent random
walk transition probabilities between the indicated space-time points. The expression is summed
over yto obtain p2hk
1k
2(ez1,ez2) and then summed over h. When kez1ez2k>|k
1k
2|1/2,
we get the Green function decay from the spatial separation kez1ez2k. When kez1ez2k ≤
|k
1k
2|1/2we get decay from the time separation |k
1k
2|. (b) Illustration of the quantity
appearing in qa,1that contains “two copies” of f. [Figure adapted by permission of Springer,
from Commun. Math. Phys. 331 (2014), 67–109 (Electrical resistance of the low-dimensional
critical branching random walk. A.A. J´arai and A. Nachmias) c(2014).]
21
vertices, such that h(Y1) = hy=h(Y2) and h(e
Y1) = hey=h(e
Y2). Recall the notation introduced
at the beginning of this section, and extend it to the tree TB
2; e.g. e
Z2is the highest common
ancestor of Y2,e
Y2at height k
2, etc.
We give separate bounds in the following four cases.
Case (a). r1=er1< k
1and r2=er2< k
2;
Case (b1). r1=er1< k
1and r2er2=k
2;
Case (b2). r1er1=k
1and r2=er2< k
2;
Case (c). r1er1=k
1and r2er2=k
2.
We then have
E|b
IV,U |2IW,Φ(U1) = (u, hu) = Φ(U2),Φ(Z+
1) = (z+
1, k1+ 1),Φ(Z+
2) = (z+
2, k2+ 1)
=Sa+Sb1+Sb2+Sc,
(3.10)
where the four terms represent contributions from intersecting pairs satisfying the criteria of the
respective cases. Then the proposition follows from the three lemmas below.
Lemma 3.11. We have
SaCσ8
D2dlog2(δn).
Lemma 3.12. We have
Sb1+Sb2Cσ8
D2dlog2(δn).
Lemma 3.13. We have
ScCσ8
D2dlog2(δn).
Proof of Lemma 3.11. By symmetry, we can restrict to r1r2. We then have
Sa2X
(9/12)δnr2r1hun1X
r1<k
1δn
r2<k
2δn
δn
X
hy,hey=k
1k
2hu
EL(hy, hey, r1, r1, k
1)EL(hy, hey, r2, r2, k
2)
×pa(hy, hey, k
1, k
2, r1, r2),
where pais the probability, given the conditioning in (3.10) that Φ(Y1) = Φ(Y2) and Φ( e
Y1) =
Φ(e
Y2). This indeed only depends on the heights in the argument of pa, and can be written
as follows. Write e
p(w1),e
p(w2) for the conditional distributions of the spatial locations w1, w2,
respectively, we have:
e
p(w1) = pr1k11(z+
1, w1)phur1(w1, u)
phuk11(z+
1, u)w1Zd;
e
p(w2) = pr2k21(z+
2, w2)phur2(w2, u)
phuk21(z+
2, u)w2Zd.
22
Then we have
pa(hy, hey, k
1, k
2, r1, r2) = X
w1,w2ZdX
ez1,ez2ZdX
y,eyZde
p(w1)e
p(w2)pk
1r1(w1,ez1)pk
2r2(w2,ez2)
×phyk
1(ez1, y)phyk
2(ez2, y)pheyk
1(ez1,ey)pheyk
2(ez2,ey)
=X
w1,w2ZdX
ez1,ez2Zde
p(w1)e
p(w2)pk
1r1(w1,ez1)pk
2r2(w2,ez2)
×p2hyk
1k
2(ez1,ez2)p2heyk
1k
2(ez1,ez2).
We perform the summation over hy, heyusing Lemmas 3.8 and 3.9. Restricting the sum to
hy, hey> k
1k
2we get an upper bound of the form:
Sa,1=C σ8
D2dX
(9/12)δnr2r1hun1X
r1<k
1δn
r2<k
2δn
qa,1(k
1, k
2, r1, r2),
where
qa,1(k
1, k
2, r1, r2) = X
w1,w2ZdX
ez1,ez2Zde
p(w1)e
p(w2)pk
1r1(w1,ez1)pk
2r2(w2,ez2)f(k
1, k
2,ez1,ez2)2.
Similarly, summing over hy> k
1k
2and hey=k
1k
2, and when the roles of yand eyare
interchanged, yields the upper bound:
Sa,2=C σ6
DdX
(9/12)δnr2r1hun1X
r1<k
1δn
r2<k
2δn
qa,2(k
1, k
2, r1, r2),
where
qa,2(k
1, k
2, r1, r2) = X
w1,w2ZdX
ez1,ez2Zde
p(w1)e
p(w2)pk
1r1(w1,ez1)pk
2r2(w2,ez2)
×p|k
1k
2|(ez1,ez2)f(k
1, k
2,ez1,ez2).
And finally, our bound when hy=hey=k
1k
2is
Sa,3=C σ4X
(9/12)δnr2r1hun1X
r1<k
1δn
r2<k
2δn
qa,3(k
1, k
2, r1, r2),
where
qa,3(k
1, k
2, r1, r2) = X
w1,w2ZdX
ez1,ez2Zd
pk
1r1(w1,ez1)pk
2r2(w2,ez2)p|k
1k
2|(ez1,ez2)p|k
1k
2|(ez1,ez2).
23
This altogether gives an upper bound of the form:
SaSa,1+Sa,2+Sa,3,
and hence we bound each of the three terms separately.
We start with bounding Sa,1, and we split the summation over ez1,ez2into:
(I)+ kez2ez1k ≤ |k
1k
2|1/2and k
2hu;
(I)– kez2ez1k ≤ |k
1k
2|1/2and r2< k
2< hu;
(II)+ kez2ez1k>|k
1k
2|1/2and k
2hu;
(II)– kez2ez1k>|k
1k
2|1/2and r2< k
2< hu.
We start with (I)+, and initially restrict to |k
1k
2| ≥ n1. Then using Lemma 3.9 and the local
limit theorem (Lemma 1.4) to bound pk
2r2(w2,ez2), we have
qa,1,(I)+(k
1, k
2, r1, r2)C
Dd|k
1k
2|2d(k
2r2)d/2X
w1,w2ZdX
ez1ZdX
ez2:kez2ez1k≤|k
1k
2|1/2
e
p(w1)e
p(w2)pk
1r1(w1,ez1)
C|k
1k
2|2d/2(k
2r2)d/2X
w1ZdX
ez1Zde
p(w1)pk
1r1(w1,ez1)
C|k
1k
2|2d/2(k
2r2)d/2
=C|k
1k
2|1(k
2r2)3.
Performing the sum over k
1yields a factor log(δn), while performing the sums over k
2,r2and
r1yield: X
(9/12)δnr1hun1X
r2r1X
k
2hu
(k
2r2)3Clog(δn).
The contribution to (I)+ from |k
1k
2|< n1is of lower order, since:
qa,1,(I)+ C C2
2
Dd(k
2