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arXiv:1005.4908v2 [gr-qc] 10 Jun 2010

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The BKL Conjectures for

Spatially Homogeneous Spacetimes

Michael Reiterer, Eugene Trubowitz

Department of Mathematics, ETH Zurich, Switzerland

Abstract:We rigorouslyconstructandcontrola genericclass ofspatiallyhomogeneous

(Bianchi VIII and Bianchi IX) vacuum spacetimes that exhibit the oscillatory BKL

phenomenology. We investigate the causal structure of these spacetimes and show that

there is a “particle horizon”.

1. Introduction

The goal of this paper is to rigorously construct and explicitly control a generic class of

solutions Φ = α ⊕ β : [0,∞) → R3⊕ R3, with independent variable τ ∈ [0,∞) and

with1(α1+ α2+ α3)|τ=0< 0, to the autonomous system of six ordinary differential

equations

0 = −d

0 = −d

def

= {(1,2,3),(2,3,1),(3,1,2)},subject to the quadratic constraint2

0 = α2α3+α3α1+α1α2−(β1)2−(β2)2−(β3)2+2β2β3+2β3β1+2β1β2 (1.1c)

Here, α = (α1,α2,α3), β = (β1,β2,β3). The system (1.1) are the vacuum Einstein

equations for spatially homogeneous (Bianchi) spacetimes, see Proposition 2.1.

ThepioneeringcalculationsandheuristicpictureofBelinskii, Khalatnikov,Lifshitz3

[BKL1] and Misner [Mis] suggest that a generic class of solutions to (1.1) are oscilla-

tory as τ → +∞ and that the dynamics of one degree of freedom is closely related

to the discrete dynamics of the Gauss map G(x) =

dταi− (βi)2+ (βj)2+ (βk)2− 2βjβk

dτβi+ βiαi

(1.1a)

(1.1b)

for all (i,j,k) ∈ C

1

x− ⌊1

x⌋, a non-invertible map

1If τ ?→ Φ(τ) is a solution to (1.1), so is τ ?→ −Φ(−τ). The condition (α1+α2+α3)|τ=0< 0 breaks

this symmetry. Solutions to (1.1) with (α1+ α2+ α3)|τ=0< 0 do not break down in finite positive time,

that is, they extend to [0,∞). A proof of this fact is given later in this introduction.

2As a quadratic form on R3⊕ R3, the right hand side of (1.1c) has signature (+,+,−,−,−,−).

3The work of Belinskii, Khalatnikov, Lifshitz concerns general (inhomogeneous) spacetimes, but relies

on intuition about the homogeneous case.

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from (0,1)\Q to itself. Every element of (0,1)\ Q admits a unique infinite continued

fraction expansion

?k1,k2,k3,...? =

1

k1+

1

k2+

1

k3+...

(1.2)

where (kn)n≥1are strictly positive integers. The Gauss map is the left-shift,

G??k1,k2,k3,...??= ?k2,k3,k4,...?

(1.3)

Rigorous results about spatially homogeneous spacetimes have been obtained by

Rendall [Ren] and Ringstr¨ om [Ri1], [Ri2]. See also Heinzle and Uggla [HU2]. We

refer to the very readable paper [HU1] for a detailed discussion.

The first rigorous proofs that there exist spatially homogeneous vacuum spacetimes

whose asymptotic behavior is related, in a precise sense, to iterates of the Gauss map,

have been obtained recently by B´ eguin [Be] and by Liebscher, H¨ arterich, Webster and

Georgi [LHWG]. These theorems apply to a dense subset of (0,1)\Q. A basic restric-

tion of both these works is that the sequence (kn)n≥1has to be bounded, a condition

fulfilled only by a Lebesgue measurezero subset of (0,1)\Q. The results of the present

paperapplyto anysequence(kn)n≥1that growsat most polynomially.Thecorrespond-

ing subset of (0,1) \ Q has full Lebesgue measure one.

We point out some properties of the system (1.1a), (1.1b), not assuming (1.1c):

(i) The right hand side of (1.1c) is a conserved quantity.

(ii) If τ ?→ Φ(τ) is a solution, so is τ ?→ pΦ(pτ + q), for all p,q ∈ R.

(iii) The signatures (sgnβ1,sgnβ2,sgnβ3) are constant.

(iv)

(v) We have4 d

d

dτ|β1β2β3|2= 2(α1+ α2+ α3)|β1β2β3|2.

dτ(α1+ α2+ α3) ≥ −3|β1β2β3|2/3.

If in addition we assume (1.1c), then:

(vi)

d

dτ(α1+ α2+ α3) = α2α3+ α3α1+ α1α2≤1

Let Φ = α⊕β be any solution to (1.1), that is (1.1a), (1.1b), (1.1c), on the half-open

interval [0,τ1) with 0 < τ1< ∞. Set α / = α1+ α2+ α3and suppose α /(0) < 0. Then

α /(τ) ≤ −|α /(0)|/?1 +1

by (vi). Consequently,|β1β2β3| is bounded,by (iv), and α / is boundedbelow, by (v), on

[0,τ1). The constraint(1.1c) implies that5(α1)2+(α2)2+(α3)2≤ 6|β1β2β3|2/3+α /2

is bounded. Now (1.1b) implies that (β1)2+ (β2)2+ (β3)2is bounded. Therefore,

solutions to (1.1) with α /(0) < 0 can be extended to [0,∞). The solutions considered

in this paper belong to this general class. We are especially interested in their τ → +∞

asymptotics.

3(α1+ α2+ α3)2.

3|α /(0)|τ?< 0

for all τ ∈ [0,τ1)

(1.4)

4(β1)2+(β2)2+(β3)2−2β2β3−2β3β1−2β1β2+3|β1β2β3|2/3≥ 0 holds for all β1,β2,β3∈ R,

see [HU1]. The only nontrivial cases are β1,β2,β3> 0 or β1,β2,β3< 0. In these cases, the inequality is

a direct consequence of the polynomial identity

x6+ y6+ z6− 2y3z3− 2z3x3− 2x3y3+ 3x2y2z2=

1

2

?x2+y2+z2+yz+zx+xy??

(y−z)2(y+z−x)2+(z−x)2(z+x−y)2+(x−y)2(x+y−z)2?

5Use 2(α2α3+ α3α1+ α1α2) = α /2− (α1)2− (α2)2− (α3)2.

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For every solution to (1.1) with α /(0) < 0, as in the last paragraph, the half-infinite

interval [0,∞) actually corresponds to a finite physical duration of the associated spa-

tially homogeneousvacuum spacetime (given in Proposition 2.1). In fact, an increasing

affine parameter along the timelike geodesics orthogonal to the level sets of τ is given

by τ ?→?τ

called Bianchi VIII or IX models). We now give an informal description of the solu-

tions that we construct, the phenomenologicalpicture of [BKL1]. The structure of each

of these solutions is described by three sequences of compact subintervals (Ij)j≥1,

(Bj)j≥1, (Sj)j≥1of [0,∞), for which:

(a.1) The left endpoint of I1is the origin, and the right endpoint of Ij, henceforth

denoted τj, coincides with the left endpoint of Ij+1, for all j ≥ 1. Set τ0= 0.

(a.2)?

(a.4) Sjis the closed interval of all points between Bjand Bj+1, for all j ≥ 1.

Here is a picture:

0exp(1

2

?s

0α /)ds, with uniform upper bound 6|α /(0)|−1, by (1.4).

In this paper, we consider only solutions to (1.1) for which β1,β2,β3 ?= 0 (also

j≥1Ij= [0,∞), that is, limj→+∞τj= +∞.

(a.3) Bjis contained in the interior of Ij, and 0 < |Bj| ≪ |Ij|, for all j ≥ 1.

Ij

Bj+1

Sj+1

Bj

Ij+1

Sj

τj

τj+1

Let S3be the set of all permutations (a,b,c) of the triple (1,2,3). The solution is

further described by a sequence (πj)j≥1in S3, with πj= (a(j),b(j),c(j)), so that:

(b.1) On Ij, the components βb(j), βc(j)are so small in absolute value that the local

dynamics of Φ = α ⊕ β is essentially unaffected if βb(j), βc(j)are set equal to

zero in the four equations (1.1a) and (1.1c).

(b.2) On Ij\ Bj, the component βa(j)is so small in absolute value that the local dy-

namics of Φ = α ⊕ β is essentially unaffected if βa(j)is set equal to zero in the

four equations (1.1a) and (1.1c). The component βa(j)is not small on Bj, but the

mixed products βa(j)βb(j)and βa(j)βc(j)are still small.

(b.3) Items (b.1) and (b.2) imply that mixed products of components of β are small on

all of [0,∞), and that all three components of β are small on?

(b.5) None of the properties listed so far distinguishes b(j) from c(j). By (b.4), this

ambiguity can be consistently eliminated by stipulating b(j) = a(j + 1).

j≥1Sj.

(b.4) a(j) ?= a(j + 1) for all j ≥ 1.

We can draw the following heuristic consequences from the eight heuristic properties

above. Separately on each interval Sj, j ≥ 1:

(c.1) The components of α are essentially constant, by (1.1a) and (b.3), and log|β1|,

log|β2|, log|β3| are essentially linear functions with slopes α1, α2, α3, by (1.1b).

(c.2) The constraint (1.1c) essentially reduces to α2α3+ α3α1+ α1α2 = 0. As be-

fore, we require α / = α1+ α2+ α3 < 0. Furthermore, we make the generic

assumption that all components of α are nonzero. These conditions imply that

two components of α are negative, one component of α is positive, and the sum

of any two is negative.

(c.3) The single positive component of α has to be αb(j)= αa(j+1). In fact, we know

that |βa(j+1)| is very small on Sjbut is not small on Bj+1. Therefore, the slope

of log|βa(j+1)|, which is αa(j+1)by (c.1), has to be positive on Sj.

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(c.4) By the last three items and (b.4), there is at most one point in Sjwhere |βa(j)| =

|βa(j+1)|. By (b.1), (b.2), there is such a point, because |βa(j)| is going from not

small to small on Sj, and |βa(j+1)| is going from small to not small on Sj. By

convention, this point is τj.

Separately on each interval Ij, j ≥ 1 (in particular on Bj⊂ Ij):

(d.1) αa(j)+ αb(j)and αa(j)+ αc(j)are essentially constant, by (1.1a), (b.1), and

they are both negative, by (c.1), (c.2). Also, log|βa(j)βb(j)|, log|βa(j)βc(j)| are

essentially linear functionswith slopes αa(j)+αb(j)and αa(j)+αc(j), by (1.1b).

(d.2) Essentially (αa(j)+αb(j))(αa(j)+αc(j)) = (αa(j))2+(βa(j))2, by (1.1c). Since

the left hand side is essentially constant by (d.1), so is the right hand side.

(d.3) By (d.1),it only remains to understandthe behaviorof αa(j), βa(j). By (1.1a), we

essentially have

d

dταa(j)= −(βa(j))2

d

dτβa(j)= αa(j)βa(j)

(1.5)

A special solution is αa(j)= −tanhτ and βa(j)= ±sechτ = ±(coshτ)−1.

The general solution is obtained from the special solution by applying the affine

symmetry transformation (ii) above, with p > 0. Since Bjis essentially the in-

terval on which |βa(j)| is not small, see (b.2), we must have p ∼ |Bj|−1(here ∼

means “same order of magnitude”). See [BKL1], Section 3, in particular pages

534 and 535.

(d.4) Recall (c.1). By (d.3), we have αa(j)|Sj−1= −αa(j)|Sj, since the hyperbolic

tangent just flips the sign. Therefore, by (d.1), the net change across Bjof the

components of α, from right to left, is given by

αa(j)|Sj−1= αa(j)|Sj− 2αa(j)|Sj

αb(j)|Sj−1= αb(j)|Sj+ 2αa(j)|Sj

αc(j)|Sj−1= αc(j)|Sj+ 2αa(j)|Sj

These equations make sense only for j ≥ 2, since S0has not been defined.

In this paper, we turn the heuristic picture of [BKL1], sketched above,into a mathemat-

ically rigorous one, globally on [0,∞), for a generic class of solutions. The first step is

to construct a discrete dynamical system, that maps the state Φ(τj) to the state Φ(τj−1)

at the earlier time τj−1 < τj, for all j ≥ 1. That is, the construction proceeds from

right-to-left, beginning at τ = +∞. We refer to the discrete dynamical system maps as

transfer maps.

For each j ≥ 0, two components of β(τj) have the same absolute value, see (c.4),

and Φ(τj) satisfies the constraint (1.1c). Therefore, the states of the discrete dynamical

system have 4 continuous degrees of freedom. By the symmetry (ii), the transfer maps

commute with rescalings. Taking the quotient, one obtains a 3-dimensional discrete

dynamical system. The three “dimensionless” quantities that we use to parametrize the

discrete states are denoted fj= (hj,wj,qj). Morally, they are interpreted as follows:

• hj∼ |Bj|/|Ij| > 0. In the billiard picture of [Mis], it is the dimensionless ratio of

the collision and free-motion times. By (a.3), one has 0 < hj≪ 1. In fact, hjis the

all-important small parameter in our construction. It goes to zero rapidly as j → ∞.

This is necessary for us to make a global construction on [0,∞). The precise rate

depends on the sequence (kn)n≥1. The rate is the same as in Proposition 4.4, up to

even smaller corrections.

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• The components of α are essentially constant on Sjand subject to the reduced con-

straint equation in (c.2). Thus, modulo the scaling symmetry (ii), only one degree of

freedomis requiredto parametrizeα|Sj. We use wj≈ −(αb(j)/(αa(j)+αb(j)))|Sj.

By (c.2)and(c.3),we havewj> 0. Theleft-to-rightdiscretedynamicsof wj(which

is opposite to the right-to-left direction of our transfer maps) is closely related to a

variant of the Gauss map, sometimes referred to as the BKL map or Kasner map.

• The meaning of qjwill be explained in a more indirect way. As pointed out above,

theleft-to-rightdynamicsofwjisrelatedtotheGauss map,whichis anon-invertible

left-shift, see (1.3). The non-invertibilityof the Gauss map seems to be at odds with

the invertible dynamics of the system of ordinary differential equations (1.1). The

parameterqjis introducedso that the joint left-to-rightdiscretedynamicsof(wj,qj)

is closely related to the left-shift on two-sided sequences (kn)n∈Zof strictly positive

integers, which is invertible. Accordingly, the right-to-left transfer maps are related

to the right shift on two-sided sequences (kn)n∈Z.

This concludes the informal discussion. We emphasize that the notation used above is

specific to the introduction. In particular, (Ij)j≥1, (Bj)j≥1, (Sj)j≥1do not appear in

the main text. Starting from Section 2, all the notation is introduced from scratch.

We now state simplified, self-contained versions of our results. References to their

stronger counterparts are given. Here is a short guide:

Definition 1.1 (equivalent to Definition 3.12). Introduces the state vectors Φ⋆(π,f,σ∗)

of the 3-dimensionaldiscrete dynamical system. The dynamics of the signature vec-

tor σ∗is trivial, by (iii), but it affects the dynamics of (π,f) in a non-trivial way.

Definition 1.2 (this is Definition 3.16). Introduces explicit maps PL, QL, λLthat turn

out to be verygoodapproximationsto the transfermaps.It is shown in Section4 that

iteratesofQLcanbeunderstoodintermsoftheGaussmap/continuedfractionsand,

by a change of variables, in terms of solutions to certain linear equations.

Definition 3.19 (only in the main text). The essential smallness condition on h > 0 is

quantitatively encoded in an open subset F ⊂ (0,1) × (0,∞) × ((0,∞) \ {1}). It

determines the domain of definition of the transfer maps.

Proposition 1.1 (slimmed-down version of Proposition 3.3). It asserts the existence of

transfer maps. The pair (PL,Π) and the triple (PL,Π,Λ) constitute the transfer

maps for the 3-dimensional and 4-dimensional systems, respectively, and they are

very close to (PL,QL) and (PL,QL,λL). Explicit error bounds and precise esti-

mates for the transfer solution appear only in the full version, Proposition 3.3.

Theorem 1.1 (simplified version of Theorems 6.2, 6.3). Gives a generic class of iterates

to (PL,Π) that are super-exponentially close to iterates of (PL,QL). That is, it

asserts the existence of solutions to the 3-dimensional discrete dynamical system.

The overview is as follows. Every solution to the 3-dimensional discrete dynamical

system as in Theorem 1.1 can be lifted to a unique solution to the 4-dimensional dis-

crete dynamical system, up to an overall scale, through the map Λ in Proposition 1.1.

This solution corresponds to the sequence of states (Φ(τj))j≥0in the informal discus-

sion. Proposition 1.1 gives solutions to (1.1) on compact intervals that connect next-

neighbor states. Symmetry (ii) is used to translate these compact intervals and place

them next to each other, beginning at τ = 0, just like the (Ij)j≥1in the informal dis-

cussion. As in (a.2) of the informal discussion, the union of these intervals is indeed

[0,∞), and a semi-global solution to (1.1) is obtained. To see this, denote the states

by λjΦ⋆(πj,gj,σ∗) with λj > 0 and πj ∈ S3and gj = (h′

j ≥ 0. One has λj = Λ[πj+1,σ∗](gj+1)λj+1 ≥ λj+1 by the definition of Λ and

j,w′

j,q′

j) ∈ F, where

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h′

bounded from above by λ0> 0. By Proposition 1.1, the length of each of the intervals

is bounded from below by (2λ0)−1> 0.

Definition 1.1 (State vectors). Let π = (a,b,c) ∈ S3and σ∗ ∈ {−1,+1}3and

f = (h,w,q) ∈ (0,∞)2×R. Let Φ⋆= Φ⋆(π,f,σ∗) = α⊕β ∈ R3⊕R3be the vector

given by (sgnβ1,sgnβ2,sgnβ3) = σ∗and by

j∈ (0,1) by the definition of F. In particular, the sequence of products (λjh′

j)j≥0is

αa= −1

αb=

hlog|1

hlog|1

hlog|1

2βa| = −1+w

2βb| = −1+w

2βc| = −(1 + w)q −w(1+w)

1+2w(1 + hlog2)

w

1+w 1+2w(1 + hlog2)

αc= −w − µ

where µ = µ(π,f,σ∗) ∈ R is uniquely determined by requiring that (1.1c) holds.

Definition 1.2 (Approximate transfer maps). Introduce three maps

1+2w−1+3w+w2

1+2w

hlog2

PL:

QL:

λL:

S3× (0,∞)3→ S3

(0,∞)3→ (0,∞)2× R

(0,∞)3→ (0,∞)

where f = (h,w,q) and qL= num1L/denLand hL= num2L/denL, and:

• if q ≤ 1:

(a′,b′,c′) = (c,a,b)num1L= (1 + w)(1 − q) − hlog2 + hwlog(2 + w)

wL=

1+w

num2L= h(2 + w)

λL= 2 + w denL= (1 + w)(q − hlog2) + h(3 + w)log(2 + w)

• if q > 1:

(a′,b′,c′) = (b,a,c) num1L= (1 + w)(q − 1 − hlog2) − hwlog2+w

wL= 1 + w num2L= h(2 + w)

λL=2+w

1+w

denL= (1 + w) − hlog2 + h(3 + 2w)log2+w

Observe that denL> 0.

((a,b,c),f) ?→ (a′,b′,c′)

f ?→ (hL,wL,qL)

f ?→ λL

1

1+w

1+w

Proposition 1.1 (Transfer maps). Fix σ∗∈ {−1,+1}3and π = (a,b,c) ∈ S3. There

exist maps6

Π[π,σ∗] : F → (0,∞)2× R

such that for every λ > 0 and f = (h,w,q) ∈ F, the solution to (1.1) starting at

λΦ⋆(π,f,σ∗) at time 0 passes through λ′Φ⋆(π′,f′,σ∗) at an earlier time τ′< 0, with

1

2≤ hλ|τ′| ≤ 3. Here f′= Π[π,σ∗](f) and λ′= λΛ[π,σ∗](f) and π′= PL(π,f).

Schematically, the transition is

?

Furthermore (informal): Π and Λ are approximated by the maps QLand λL, with

errors that go to zero exponentially as h ↓ 0 (for fixed w, q). See Proposition 3.3

6Caution: The maps Π cannot immediately be iterated /composed, because (0,∞)2× R ?⊂ F.

and

Λ[π,σ∗] : F → [1,∞)

λΛ[π,σ∗](f) Φ⋆

PL(π,f), Π[π,σ∗](f), σ∗

?

←− λΦ⋆(π,f,σ∗)

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Theorem 1.1. Fix σ∗∈ {−1,+1}3andπ0∈ S3.FixconstantsD ≥ 1, γ ≥ 0.Suppose

the vector f0= (h0,w0,q0) ∈ (0,∞)3satisfies

(i) w0∈ (0,1) \ Q and q0∈ (0,∞) \ Q.

(ii) kn ≤ D max{1,n}γfor all n ≥ −2, where the two-sided sequence of strictly

positive integers (kn)n∈Zis given by

(1 + q0)−1= ?k0,k−1,k−2,...?

(iii) 0 < h0< A♯where A♯= A♯(D,γ) = 2−56D−4(4(γ + 1))−4(γ+1).

Then f0and π0are the first elements of a unique sequence (fj)j≥0in F and a unique

sequence (πj)j≥0in S3, respectively, with πj = PL(πj+1,fj+1) and fj = QL(fj+1)

for all j ≥ 0. Furthermore,there exists a sequence(gj)j≥0in F such that for all j ≥ 0,

gj= Π[πj+1,σ∗](gj+1)

and

2(1 +√5),

?

If γ > 1 and D >

log 2

(iii) has positive Lebesgue measure.

w0= ?k1,k2,k3,...?

πj= PL(πj+1,gj+1)

and, with ρ+=1

?gj− fj?R3 ≤ exp−

1

h0A♯ρ((D−1j)1/(γ+1))

+

?

1

γ

γ−1, then the set of all vectors f0∈ (0,∞)3that satisfy (i), (ii),

The class of solutions that we construct is generic in the sense of the last sentence

of Theorem 1.1. It would be desirable to have a stronger genericity statement, namely

a genericity statement for “the g0rather than the f0”.

For the causal structure and particle horizons, see Proposition 2.2 and Section 7.

It is a pleasure to thank J. Fr¨ ohlich, G.M. Graf and T. Spencer for their support and

encouragement.

2. Spatially homogeneous vacuum spacetimes

Proposition 2.1. Let α⊕β : (τ0,τ1) → R3⊕R3be a solution to (1.1) and let Ω ⊂ R3

be open, with Cartesian coordinates x = (x1,x2,x3). Fix any τ∗∈ (τ0,τ1) and let

v1=?3

be three smooth vector fields on Ω that are a frame at each point and satisfy

µ=1v1µ(x)

∂

∂xµ

v2=?3

µ=1v2µ(x)

∂

∂xµ

v3=?3

µ=1v3µ(x)

∂

∂xµ

[vj,vk] = βi(τ∗)vi

on Ω

for all (i,j,k) ∈ C

e0= eζ(τ) ∂

def

= {(1,2,3),(2,3,1),(3,1,2)}.Introduce

∂τ

ei= eζi(τ)vi

ζi(τ) = −1

i = 1,2,3

ζ(τ) = ζ1(τ) + ζ2(τ) + ζ3(τ)

2

?τ

τ∗dsαi(s)

i = 1,2,3

on the domain (τ0,τ1) × Ω ⊂ R4. Then, the Lorentzian metric g with inverse

g−1= −e0⊗ e0+ e1⊗ e1+ e2⊗ e2+ e3⊗ e3

is a solution to the vacuum Einstein equations Ric(g) = 0 on (τ0,τ1) × Ω.

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Proof. In this proof, everywhere (i,j,k) ∈ C. It follows from

ζi(τ∗) = 0 and (1.1b) that βi(τ) = βi(τ∗)e−2ζi(τ). Now, by direct calculation,

d

dτe−2ζi= αie−2ζiand

[e0,ei] = −1

2eζαiei

[ej,ek] = eζβiei

Let ∇ be the Levi-Civita connection associated to g. Then, for all a,b,c = 0,1,2,3,

g?∇eaeb, ec

By direct calculation,

?=1

2

?

g?[ea,eb],ec

?− g?[eb,ec],ea

?+ g?[ec,ea],eb

??

∇e0e0= 0

∇e0ei= 0

∇eie0=1

∇eiei=1

∇ejek=1

∇ekej=1

2eζαie0

2eζ(+βi− βj+ βk)ei

2eζ(−βi− βj+ βk)ei

2eζαiei

and

Riem(ei,ej,ei,ej)

4e2ζ?

Riem(e0,ea,ei,ej)

=1

(+βi− βj− βk)(+βi− βj+ βk) + 2βk(+βi+ βj− βk) + αiαj

?

=1

4e2ζδak

?

(−βi+ βj− βk)αi+ (+βi− βj− βk)αj+ 2αkβk

Riem(e0,ea,e0,ei)

?

Furthermore, Riem(ea,eb,ec,ed) = 0 unless {a,b} = {c,d} with a ?= b. The Rie-

mann curvature tensor is completely specified by these identities and by its algebraic

symmetries. It follows that

?

= −1

4e2ζδai

2d

dταi− (αj+ αk)αi

?

Ric(e0,e0) = −1

Ric(e0,ei) = 0

Ric(ei,ei) = +1

2e2ζ d

dτ(α1+ α2+ α3) +1

2e2ζ(α2α3+ α3α1+ α1α2)

2e2ζ d

dταi+1

2e2ζ?+ (βi)2− (βj)2− (βk)2+ 2βjβk

?

Ric(ej,ek) = 0

The right hand sides of the first and third equation vanish by (1.1a) and (1.1c).

⊓ ⊔

Proposition 2.2. In the context of Proposition 2.1, let γ : (τ′

a smooth curve given by γ(τ) = (τ,γ♯(τ)), where γ♯is a curve on Ω. Let g♯be the

Riemannian metric on Ω defined by g♯(va,vb) = δabfor all a,b = 1,2,3. If γ is

non-spacelike with respect to g, then the length of γ♯with respect to g♯is bounded by

0,τ′

1) → (τ0,τ1) × Ω be

Lengthg♯(γ♯) ≤

?τ′

1

τ′

0

dτ max

(i,j,k)∈Ce−ζj−ζk

(2.1)

The integral on the right hand side may be divergent.

Page 9

9

Proof. Write the velocity

d

dτγ as

∂

∂τ+?3

i=1Xivi= e−ζe0+?3

i=1Xie−ζiei

with smooth coefficients Xi= Xi(τ). By assumption, γ is non-spacelike:

0 ≥ g(d

i=1(Xi)2≤ max(i,j,k)∈Ce−2ζj−2ζk. Now, the claim follows from

?τ′

dτγ,

d

dτγ) = −e−2ζ+?3

i=1(Xi)2e−2ζi

Consequently,?3

Lengthg♯(γ♯) =

1

τ′

0

dτ

?

g♯(d

dτγ♯,

d

dτγ♯) =

?τ′

1

τ′

0

dτ

??3

i=1(Xi)2

⊓ ⊔

3. Construction of the transfer maps

Let (τ0,τ1) ⊂ R be a finite or infinite open interval, parametrized by τ ∈ (τ0,τ1)

(“time”).In this paper,the unknownfield is a vectorvaluedmap Φ ∈ C∞((τ0,τ1),R6):

Φ = α[Φ] ⊕ β[Φ] : (τ0,τ1) → R3⊕ R3

If no confusion can arise, we just write Φ = α ⊕ β.

Definition 3.1. To every field Φ = α ⊕ β ∈ C∞((τ0,τ1),R6), every constant h > 0

and every n ∈ R3, associate a field

a[Φ,h,n] ⊕ b[Φ,h,n] ⊕ c[Φ,h,n] : (τ0,τ1) → R3⊕ R3⊕ R

by

(3.1)

ai[Φ,h,n] = −hd

bi[Φ,h,n] = −hd

c[Φ,h,n] =?

dταi− (niβi)2+ (njβj− nkβk)2

dτβi+ βiαi

?αjαk− (niβi)2+ 2njnkβjβk

(3.2a)

(3.2b)

(i,j,k)∈C

?

(3.2c)

for all (i,j,k) ∈ C. For later use, it is convenient to introduce, for all m,n ∈ R3,

ai[Φ,h,n,m] = ai[Φ,h,n] − ai[Φ,h,m]

= −(niβi)2+ (njβj− nkβk)2+ (miβi)2− (mjβj− mkβk)2

Definition 3.2.

(3.3)

B1= (1,0,0)B2= (0,1,0)B3= (0,0,1)Z = (1,1,1)

These vectors will play the role of the vector n ∈ R3that appears in Definition 3.1.

Proposition 3.1 (Global symmetries). Let χ : (τ0,τ1) → (τ′

morphism between finite or infinite intervals, χ(τ) = pτ +q with p > 0, and let A > 0

be a constant. Then

?

for all fields Φ = α ⊕ β ∈ C∞((τ′

0,τ′

1) be a linear diffeo-

(a,b,c)A(Φ ◦ χ),

1

pAh, n

?

= A2?

(a,b,c)[Φ,h,n] ◦ χ

?

0,τ′

1),R6), all constants h > 0 and all n ∈ R3.

Page 10

10

Corollary 3.1. In Proposition3.1, the field (a,b,c)[A(Φ◦χ),

tically on (τ0,τ1) if and only if (a,b,c)[Φ,h,n] vanishes identically on (τ′

1

pAh, n] vanishes iden-

0,τ′

1).

Remark 3.1. The equations (a,b,c)[Φ,1,Z] = 0 are identical to (1.1). The equations

(a,b,c)[Φ,h,Z] = 0 are equivalent to (1.1), for any h > 0, by Corollary 3.1.

Proposition 3.2. Recall Definition 3.1. For all Φ = α ⊕ β ∈ C∞((τ0,τ1),R6), all

h > 0 and all n ∈ R3, we have

?

0 = − hd

dτc +

(i,j,k)∈C

?

− αjak− αkaj+ 2(ni)2βibi− 2njnkβjbk− 2njnkβkbj

?

(3.4)

with (a,b,c) = (a,b,c)[Φ,h,n]. In particular, if (a,b) = 0 identically on (τ0,τ1),

then c vanishes identically on (τ0,τ1) if and only if c vanishes at one point of (τ0,τ1).

Proof. Replace all occurrences of a, b and c on the right hand side of (3.4) by the

respective right hand sides of (3.2). Then, verify that everything cancels.

⊓ ⊔

Definition 3.3. For all h > 0 andall vectors Φ = α⊕β ∈ R3⊕R3with β1,β2,β3?= 0,

define Am[Φ] ∈ (0,∞) and ϕm[Φ] ∈ R by

?

|βm|

Am[Φ] =

ϕm[Φ] = −arcsinhαm

|αm|2+ |βm|2> |αm| ≥ 0

for all m = 1,2,3. Equivalently,

αm=

βm= (sgnβm) Am[Φ] sech ϕm[Φ]

−Am[Φ] tanhϕm[Φ]

(3.5a)

(3.5b)

Furthermore, define ξm[Φ,h] ∈ R by

ξm[Φ,h] = h log??1

2βm

??

for all m = 1,2,3. Furthermore, for all m,n = 1,2,3, introduce the abbreviations

αm,n[Φ] = αm+ αn

ξm,n[Φ,h] = ξm[Φ,h] + ξn[Φ,h]

If no confusion can arise, we drop the explicit dependence [Φ] or [Φ,h]. For instance,

we write Am= Am[Φ]. If Φ is not an element of R3⊕ R3, but rather a function of the

real variable τ with values in R3⊕ R3, with β1,β2,β3?= 0 everywhere, then Am,ϕm,

ξm, ξm,n, αm,n, with m,n = 1,2,3, are functions of τ, too. In this case, we define the

additional functions θm[Φ,h], m = 1,2,3, through

ϕm[Φ](τ) =1

h

?τ − θm[Φ,h](τ)?Am[Φ](τ)

Remark 3.2. In the context of Definition 3.3, we have, for all m = 1,2,3:

h|ϕm| = −ξm+ h log

?

|1

2αm| +

?

|1

2αm|2+ exp(1

h2ξm)

?

Page 11

11

Lemma 3.1. Recall Definitions 3.1, 3.2, 3.3. For all h > 0 and all Φ = α ⊕ β ∈

C∞((τ0,τ1),R6) such that β1, β2, β3never vanish on (τ0,τ1), we have

d

dτ

?Ai

θi

?

=

1

(Ai)2

?1

h(Ai)2tanhϕi

ϕitanhϕi− 1

1

h(Ai)2sechϕi

sinhϕi+ ϕisechϕi

??

ai[Φ,h,Bi]

−σibi[Φ,h,Bi]

?

for i = 1,2,3 and σi = sgnβi ∈ {−1,+1}. The matrix on the right hand side has

determinant1

h(Ai)2coshϕi?= 0.

Proof. We have ai[Φ,h,Bi] = −hd

Replace all occurrences of αiand βiby the right hand sides of (3.5), respectively. Use

d

dτϕi=

dταi− (βi)2and bi[Φ,h,Bi] = −hd

dτβi+ αiβi.

1

Ai(d

dτAi)ϕi+1

hAi(1 −

d

dτθi). Now, solve for

d

dτAiand

d

dτθi.

⊓ ⊔

Remark 3.3. So far, we have stated all definitions and propositions for a C∞-field Φ =

α ⊕ β, defined on an open interval. This was just for convenience. We will, from now

on, use these definitions and propositions even when the C∞-requirement is not met,

or when the field is defined on, say, a closed interval rather than an open interval. It will

be clear in each case, that the respective definition or proposition still makes sense.

Definition 3.4. Set S3 = {(1,2,3),(2,3,1),(3,1,2),(3,2,1),(1,3,2),(2,1,3)}, the

set of all permutations of (1,2,3).

Definition 3.5. Forall σ∗∈ {−1,+1}3letD(σ∗) betheset ofallΦ = α⊕β ∈ R3⊕R3

with?sgnβ1,sgnβ2,sgnβ3

Definition 3.6. For all π = (a,b,c) ∈ S3and h > 0 and σ∗∈ {−1,+1}3define two

functions D(σ∗) × D(σ∗) → [0,∞) by

???Aa[Φ] − Aa[Ψ]??

??αc,a[Φ] − αc,a[Ψ]??,

and

d /D(σ∗),h(Φ,Ψ) = max

i=1,2,3

?= σ∗. For all τ0,τ1∈ R with τ0< τ1let E(σ∗;τ0,τ1)

be the set of all continuous maps Φ : [τ0,τ1] → D(σ∗).

dD(σ∗),(π,h)(Φ,Ψ) = max

,

??hϕa[Φ]

??ξc,a[Φ,h] − ξc,a[Ψ,h]??

??ξi[Φ,h] − ξi[Ψ,h]???

Aa[Φ]− hϕa[Ψ]

Aa[Ψ]

??

,

??αb,a[Φ] − αb,a[Ψ]??,??ξb,a[Φ,h] − ξb,a[Ψ,h]??,

?

???αi[Φ] − αi[Ψ]??,

Then (D(σ∗),dD(σ∗),(π,h)) and (D(σ∗),d /D(σ∗),h) are metric spaces.

Definition 3.7. For all π ∈ S3and h > 0 and σ∗∈ {−1,+1}3and τ0,τ1∈ R with

τ0< τ1define a function E(σ∗;τ0,τ1) × E(σ∗;τ0,τ1) → [0,∞) by

dE(σ∗;τ0,τ1),(π,h)(Φ,Ψ) = supτ∈[τ0,τ1]dD(σ∗),(π,h)(Φ(τ),Ψ(τ))

Then (E(σ∗;τ0,τ1),dE(σ∗;τ0,τ1),(π,h)) is a metric space.

Lemma 3.2. Let π = (a,b,c) ∈ S3and h > 0 and σ∗∈ {−1,+1}3. Suppose h ≤ 1.

Let C,D ≥ 1 be constants. Then, for all Φ,Ψ ∈ D(σ∗) such that

C−1≤ Aa[X] ≤ C

for both X = Φ and X = Ψ and such that sgnϕa[Φ] = sgnϕa[Ψ], we have:

D−1≤ h|ϕa[X]| ≤ D

Page 12

12

(a) d /D(Φ,Ψ) ≤ 23C2DdD(Φ,Ψ)

(b) If exp(−1

Here, dD= dD(σ∗),(π,h)and d /D= d /D(σ∗),h.

hC−2D−1) ≤ 2−6C−4D−2, then dD(Φ,Ψ) ≤ 25C3Dd /D(Φ,Ψ)

Proof. In this proof, A, B, α, ξ play the roles of Aa, hϕa/Aa, αa, ξa, respectively.

To show (a), let P : (0,∞) × R → R2, (A,B) ?→ (α(A,B),ξ(A,B)), where

α(A,B) = −Atanh(1

This is a diffeomorphism. The Jacobian J of P is given by

hAB)ξ(A,B) = hlog(1

2Asech(1

hAB))

J =

?∂α

∂A

∂A

∂ξ

∂α

∂B

∂ξ

∂B

?

=

?

−1

hAB sech2(1

h

A− B tanh(1

hAB) − tanh(1

hAB)

hAB)−1

hA2sech2(1

−Atanh(1

hAB)

hAB)

?

Let pi= (Ai,Bi) ∈ (0,∞) × R and set (αi,ξi) = P(pi), where i = 0,1. Set γ(t) =

(A(t),B(t)) = (1 − t)p0+ tp1where t ∈ [0,1]. We have

?α1−α0

Suppose C−1≤ Ai ≤ C and (CD)−1≤ |Bi| ≤ CD and sgnB0 = sgnB1. Then,

C−1≤ A(t) ≤ C and (CD)−1≤ |B(t)| ≤ CD for all t ∈ [0,1]. Observe that

|ϕsech2ϕ| ≤

implies (a).

We show that underthe assumptions of (b), we have |detM| ≥ 2−3C−1, and therefore

|(M−1)ij| ≤ 24C3D for all i,j ∈ {0,1}. This would imply (b). We have |detM| ≥

|M00M11|−|M01M10|. Set ϕ(t) =1

assumption of (b), we have e−|ϕ(t)|≤ 2−6C−4D−2, for all t ∈ [0,1]. We will also use

the general inequalities 0 ≤ 1 − tanh|ϕ| ≤ 2e−2|ϕ|and |ϕsech2ϕ| ≤ 4|ϕ|e−2|ϕ|≤

4e−|ϕ|. We have |−ϕsech2ϕ−tanhϕ| ≥ tanh|ϕ| = 1−(1−tanh|ϕ|) ≥ 2−1. The

last inequality holds for all t ∈ [0,1] and implies |M00| ≥ 2−1, because ϕ has constant

sign. We have |M11| ≥ 2−1C−1and |M10| ≤ 2CD and |M01| ≤ 2−4C−2D−1. This

implies |detM| ≥ 2−3C−1.

Definition 3.8. Let X = D(σ∗) or X = E(σ∗;τ0,τ1). For all δ ≥ 0 and Φ ∈ X and

π ∈ S3and h > 0, set BX,(π,h)[δ,Φ] = {Ψ ∈ X | dX,(π,h)(Φ,Ψ) ≤ δ}.

Definition 3.9 (The reference field Φ0). For all π = (a,b,c) ∈ S3, f = (h,w,q) ∈

(0,∞)3, σ∗∈ {−1,+1}3let Φ0= Φ0(π,f,σ∗) : R → D(σ∗) be given by

Aa[Φ0](τ) = 1

θa[Φ0,h](τ) = 0

αb,a[Φ0](τ) = −(1 + w)−1

αc,a[Φ0](τ) = −(1 + w)

ξb,a[Φ0,h](τ) = −1 − hlog2 − (1 + w)−1τ

ξc,a[Φ0,h](τ) = −(1 + w)q − hlog2 − (1 + w)τ

(see Definition 3.3) for all τ ∈ R.

Remark 3.4. The field Φ0is, up to renaming, given by equation (3.12) in [BKL1].

ξ1−ξ0

?= M?A1−A0

B1−B0

?

with

M =?M00 M01

M10 M11

?=?1

0dtJ(γ(t))

1

2for all ϕ ∈ R. We have |Mij| ≤ 2C2D for all i,j ∈ {0,1}. This

hA(t)B(t). We have |ϕ(t)| ≥1

hC−2D−1. By the

⊓ ⊔

(3.6a)

(3.6b)

(3.6c)

(3.6d)

(3.6e)

(3.6f)

Page 13

13

Lemma 3.3. Let Φ0be as in Definition 3.9. Then (a,b,c)[Φ0,h,Ba] = 0 on R.

Proof. Let α = α[Φ0], β = β[Φ0], ξ = ξ[Φ0,h]. We have (aa,ba)[Φ0,h,Ba] = 0 by

Lemma3.1. Forp ∈ {b,c},wehaveaa[Φ0,h,Ba]+ap[Φ0,h,Ba] = −hd

that is ap[Φ0,h,Ba] = 0. We also have β−1

−d

αa,bαa,c= −A2

Definition 3.10. For all f = (h,w,q) ∈ (0,∞)3set

τ−(f) = −?1 −

τ+(f) = 1 +1

w

dταa,p= 0,

aba[Φ0,h,Ba] + β−1

pbp[Φ0,h,Ba] =

dτξa,p+ αa,p= 0, that is bp[Φ0,h,Ba] = 0. Finally, c[Φ0,h,Ba] = −α2

a+ αa,bαa,c= 0. Here, Aa= Aa[Φ0].

a− β2

a+

⊓ ⊔

1

2+w

?min{1,q}< 0

> 0

Lemma 3.4 (Technical Lemma 1). Let π = (a,b,c) ∈ S3, f = (h,w,q) ∈ (0,∞)3,

σ∗∈ {−1,+1}3and fix δ > 0, ǫ−∈ (0,−τ−), ǫ+∈ (0,τ+) where τ±= τ±(f). Set

τ0−= τ−+ ǫ−< 0Φ0= Φ0(π,f,σ∗)??

[τ0−,τ0+]

(3.7a)

τ0+= τ+− ǫ+> 0E = E(σ∗;τ0−,τ0+)

(3.7b)

Then Φ0∈ E. Furthermore, if the inequality

δ ≤ 2−4min?1, w, ǫ−,

ǫ+

τ+τ0+

?

(3.8)

holds, then for all Φ = α ⊕ β ∈ BE,(π,h)[δ,Φ0] the estimates

max?|βb|2, |βc|2, |βbβa|, |βcβa|?

≤ 24exp?−

1

h2(1 + |τ|)

|βa| ≤ 2

1

4hmin{1, ǫ−,

ǫ+

τ+}?

|Aa[Φ] − 1| ≤ 2−1

|ϕa[Φ]| ≤

hold on [τ0−,τ0+].

Proof. The following estimates hold for the components of Φ, for all τ ∈ [τ0−,τ0+]:

|βbβa| = 4exp?1

≤ 2exp?−1

≤ 2exp?−1

4h

|βcβa| = 4exp?1

≤ 2exp?−1

≤ 2exp?−1

hξb,a

hξb,a[Φ0,h] +1

h−1

h−1

h+1

1

?

hξc,a[Φ0,h] +1

h(1 + w)q −1

h(1 + w)q −1

h(1 + w)q +1

1

2hǫ−

?

≤ 4exp?1

≤ 2exp?−1

≤ 2exp?−

≤ 4exp?1

≤ 2exp?−1

≤ 2exp?−

hδ?

h(1 + w)−1τ +1

h(1 + w)−1τ−+1

h(2 + w)−1+1

hδ?

hδ?

hδ?

hξc,a

?

hδ?

h(1 + w)(τ−+ ǫ−) +1

(1+w)2

2+wq −1

h(1 + w)τ +1

hδ?

hδ?

hhǫ−+1

hδ?

?

Page 14

14

|ϕa| =1

hAa|τ − θa|

≤1

≤1

≤1

|βa|−1= |Aa|−1coshϕa

≤ 2exp?|ϕa|?

|βb| = |βbβa| · |βa|−1

≤ 4exp?−1

≤ 4exp?1

≤ 4exp?−

The last step uses δ ≤ 2−3ǫ+

ǫ+≤1

h(1 + δ)?|τ| + δ?

h

h2(1 + |τ|)

?|τ| + δ|τ| + 2δ?

≤ 2exp?1

h|τ| +1

hδ|τ| +1

h2δ?

h−1

h(1 + w)−1τ +1

hmax?− 1 −2+w

≤ 4exp?−1

h|τ| +1

hδ|τ| +1

h3δ?

h3δ?

≤ 4exp?1

1+wτ0−− δτ0−, −1 +

τ++ δτ0+} +1

16min{ǫ−,ǫ+

2hmin{ǫ−,ǫ+

w

1+wτ0++ δτ0+

?+1

h3δ?

hmax{−ǫ−− δτ0−,−ǫ+

15

h

τ+} +1

τ+}?

h3δ?

1

τ+. In the case ǫ+≥

1

2τ+, this follows from δ ≤ 2−4. If

τ+τ0+, because τ0+= τ+−ǫ+≥1

2τ+, then this follows from δ ≤ 2−4

ǫ+

2τ+≥1

2.

|βc| = |βcβa| · |βa|−1

≤ 4exp?−1

h(1 + w)q −1

h(1 + w)q

h(1 + w)τ +1

h|τ| +1

hδ|τ| +1

h3δ?

≤ 4exp?−1

≤ 4exp?−1

≤ 4exp?−1

+1

hmax?− (2 + w + δ)τ0−,−(w − δ)τ0+

h4δ?

?+1

h3δ?

h(1 + w)q +1

h2ǫ−+1

?

⊓ ⊔

h(2 + w)|τ−| −1

h(2 + w + δ)ǫ−+1

h4δ?

≤ 4exp?−1

hǫ−

This concludes the proof.

Lemma 3.5. Recall Definitions 3.1 and 3.2. We have

aa[Φ,h,Z,Ba] = +β2

ab[Φ,h,Z,Ba] = −β2

ac[Φ,h,Z,Ba] = +β2

b+ β2

b+ β2

b− β2

c− 2βbβc

c− 2βaβc

c− 2βaβb

for all (a,b,c) ∈ S3.

Remark 3.5. Lemma 3.5 displays the differences between the equations a[Φ,h,Z] =

0 and a[Φ,h,Ba] = 0. Lemma 3.4 gives bounds for the terms that appear in these

differences. Informally,they tend exponentiallyto zero as h ↓ 0. This quantifies a basic

guiding intuition of [BKL1].

Page 15

15

Definition 3.11. For all vectors Φ = α ⊕ β ∈ R3⊕ R3with β1,β2,β3?= 0, all π =

(a,b,c) ∈ S3and all h > 0, define four real numbers by

I1[Φ,h,π] = −1

I2[Φ,h,π] =?Aa[Φ]?−2aa[Φ,h,Z,Ba]

I(3,p)[Φ,h,π] =1

haa[Φ,h,Z,Ba] tanhϕa[Φ]

?

1 − ϕa[Φ]tanhϕa[Φ]

haa[Φ,h,Z,Ba]

?

hap[Φ,h,Z,Ba] +1

where p ∈ {b,c}. If Φ is not an element of R3⊕ R3, but rather a function with values

in R3⊕R3, with β1,β2,β3?= 0 everywhere, then I1, I2, I(3,b), I(3,c)are functions, too.

Lemma 3.6 (Technical Lemma 2).In the contextof Lemma 3.4, if δ > 0 satisfies (3.8),

then, for all Φ,Ψ ∈ BE,(π,h)[δ,Φ0] and all S ∈ {1,2,(3,b),(3,c)}, the estimates

??IS[Φ]??≤ 211max{1,1

h,1

h|τ|} exp?−

1

4hmin{1,ǫ−,ǫ+

1

4hmin{1,ǫ−,ǫ+

τ+}?

(3.9a)

??IS[Φ] − IS[Ψ]??≤ 217?max{1,1

hold on [τ0−,τ0+]. Here, IS[Φ] = IS[Φ,h,π], IS[Ψ] = IS[Ψ,h,π] and dE= dE,(π,h).

h,1

h|τ|}?2exp?−

τ+}?dE(Φ,Ψ)

(3.9b)

Proof. In this proof, we simplify the notation by suppressing h > 0 and abbreviating

M = exp(−1

4hmin{1,ǫ−,ǫ+

τ+})M1= max{1,1

h,1

h|τ|}

Lemmas 3.4, 3.5 imply??ai[Φ,h,Z,Ba]??≤ 26M, i = 1,2,3, and |ϕa[Φ]| ≤ 22M1and

??ϕa[Φ] − ϕa[Ψ]??≤1

≤ 22M1dE(Φ,Ψ)

??ξa[Φ] − ξa[Ψ]??≤ h|logAa[Φ] − logAa[Ψ]|

≤ 23hM1dE(Φ,Ψ)

??βp[Φ]βa[Φ] − βp[Ψ]βa[Ψ]??≤1

≤ 24M1M dE(Φ,Ψ)

??βp[Φ] − βp[Ψ]??≤1

≤ 26M1M1/2dE(Φ,Ψ)

??βp[Φ]βq[Φ] − βp[Ψ]βq[Ψ]??≤ 29M1M dE(Φ,Ψ)

??ai[Φ,h,Z,Ba] − ai[Ψ,h,Z,Ba]??≤ 211M1M dE(Φ,Ψ)

with Lipschitz-constant L > 0 determined by LtanhL = 1, in particular L < 2.

(Aa[Φ])−2≤ 22. This implies (3.9a). To show (3.9b), observe that (here p,q ∈ {b,c})

h|Aa[Φ] − Aa[Ψ]||τ| +1

≤1

h|Aa[Φ]θa[Φ] − Aa[Ψ]θa[Ψ]|

h2|θa[Φ] − θa[Ψ]|

h(1 + |τ|)|Aa[Φ] − Aa[Ψ]| +1

+ h|logcoshϕa[Φ] − logcoshϕa[Ψ]|

hmax?|βp[Φ]βa[Φ]|,|βp[Ψ]βa[Ψ]|?

×??ξa,p[Φ] − ξa,p[Ψ]??

hmax?|βp[Φ]|,|βp[Ψ]|???ξp[Φ] − ξp[Ψ]??

≤1

h22M1/2???ξa,p[Φ] − ξa,p[Ψ]??+??ξa[Φ] − ξa[Ψ]???

Consequently, for i = 1,2,3,

With these estimates, (3.9b) follows. Observe that R → R,x ?→ xtanhx is Lipschitz

⊓ ⊔