General properties

As above we setψa(x) = ( ¯ψa(x), ta(x)) whereψ_a(x) = (ψ_a¹(x),· · · , ψⁿ⁻¹_a (x)).

Lemma 10. For a0 ∈ ∂M choose conformal Fermi-coordinates around a0. In these coordinates we obtain

(a) _∂a^∂_i|a0 ψ_a(x)

2=−2ψ_a0(x)ⁱ+O(dga(x, a)³), (b) _∂aj∂a^∂2_i|a0

ψ_a(x)

2=−2δ_ij+O(d_ga(x, a)²), (c) _∂a^∂_i|a0t_a(x) =O(d_ga(x, a)²),

(d) _∂aj∂a^∂2_i|a0t_a(x) =O(d_ga(x, a)).

Proof. To prove this Lemma we need the FunctionF in (A.3) and some of its derivatives.

F(y, x, t, z) =c(y, x, t)−z, where c(y, x, t) is the solution of the initial value problem in the proof of Proposition 22. First we compute the rst derivatives at zero.

Claim:

1. D₁F(0,0,0,0)[h] = (h,0)for h∈Rⁿ⁻¹, 2. D2F(0,0,0,0)[h] = (h,0)for h∈Rⁿ⁻¹, 3. _dt|t=0^d F(0,0, t,0) =e_n,

4. D₄F(0,0,0,0)[h] =−hfor h∈Rⁿ.

Proof of the claim. F(y,0,0,0) = (y,0) which proves 1. Furthermore F(0, th,0,0) = (γ(0, th,1),0) = (γ(0, h, t),0) which proves 2. The equations in 3. and 4. are obvious from the denition ofF.

We also need the second derivatives of F. Since F(y,0,0,0) = y, it follows that D₁²F(0,0,0,0) = 0. Observe that F(0, se_i+re_j,0,0) is γ(s, r,1), where γ(s, r,·) is the solution of (

¨

γ^k(s, r, t) + Γ^k_ij(γ(s, r, t)) ˙γⁱ(s, r, t) ˙γ^j(s, r, t) = 0 γ(s, r,0) =y, γ˙(s, r,0) =sei+rej

. Setβ(t) = _∂r∂s^∂2 γ(0,0, t),thenβ solves the initial value problem

β(t) = 0;¨ β(0) = 0, β(0) = 0˙

since γ(0,0, t) = 0 and Γ^k_ij(0) = 0 in Fermi-coordinates. Hence β = 0 which proves D₂²F(0,0,0,0) = 0.

Now we compute the second derivative w.r.t. t. We have F(0,0, t,0) =c(0,0, t). From the initial value problem, which is solved by c, we infer c¨^k(0) = −Γ^k_nn(0) = 0. This implies _dt2^d|t=0² F(0,0, t,0) = 0.

In the following we show that also the mixed second derivatives are zero. We begin with D1D2F(0,0,0,0).

D₂F(ty,0,0,0)[x] = d dt|t= 0

n−1

X

i=1

e_i(ty)x_i,

where (e1(y),· · · , en−1(y)) is a local g-orthonormal frame. Since Fermi coordinates are normal coordinates at the boundary, the metric vanishes up to second order and therefore D₁D₂F(0,0,0,0) = 0.NowF(y,0, t,0) =c(y,0, t), wherec(y,0,·)solves the ODE

(

¨

c^k(y,0, t) + Γ^k_ij(y, c(y,0, t)) ˙cⁱ(y,0, t) ˙c^j(y,0, t) = 0 c(y,0,0) =y, c(y,˙ 0,0) =en.

ThusD₁D₃F(0,0,0,0) = 0.With the same arguments as above we get

D3F(0, x,0,0) = en. Hence D2D3F(0,0,0,0) = 0. Altogether we have proved that the second derivative of F vanishes at (0,0,0,0).

Taking the derivative of0 =F(y, x(y, z), t(y, t), z) at (0,0,0,0)yields

D_yx(0,0)[e_i] =−e_i, D_yt(0,0) = 0, D_zx(0,0)[e_i] =e_i, D_zt(0,0)[e_i] = 0if i < n;

Dzx(0,0)[en] = 0, Dzt(0,0)[en] = 1. (A.4) Furthermore, taking the second derivatives at(0,0,0,0)gives

D²_yx(0,0) = 0, D²_zx(0,0) = 0, D²_yt(0,0) = 0, D_z²t(0,0) = 0,

D_yD_zx(0,0) = 0, D_yD_zt(0,0). (A.5) We use (A.4), (A.5) and Taylor expansion to derive

∂

∂ai|a₀

ψ_a(ψ_a0(z))

2 = ∂

∂yi|0

ψ_Φ(y,0) ψ_a⁻¹₀ (z)

2 = ∂

∂yi|0|x(y, z)|²,

=−2z_i+O(|z|³)

∂²

∂a_j∂a_i|a₀

ψ_a(ψ_a0(z))

2=−2δ_ij+O(|z|²),

∂

∂a_i|a₀t_a ψ_a⁻¹₀ (z)

= ∂

∂y_i|0t(y, z) =O(|z|²) as well as

∂²

∂a_j∂a_i|a₀t_a ψ_a⁻¹₀ (z)

=O(|z|).

Therefore the Lemma is proved.

In this chapter we expand the scalar product of two dierent bubbles under the assump-tion

(A) εij :=





1

λi

λj +^λ_λ^j

i +λ_iλ_jd_g(a_i, a_j)²



→0 and λi, λj → ∞.

Proposition 24. Under (A) it holds Z

∂M

u_aiδˆ_ai,λi2

u_ajδˆ_aj,λjdσ_g=I₁χ_ρ(|ψ_ai(a_j)|)u_aj(a_i) 1

λi

λj +λiλjdg_aj(ai, aj)²

!

+o(ε_ij) provided λ_i≥λ_j.Here

I₁ = Z

R³

1 1 +|x|²

2

dx.

Proof. From now on we assumeλ_i≥λ_j. We would like to expand the integral Z

∂M

u_aiδˆ_ai,λi2

u_ajδˆ_aj,λjdσ_g = Z

∂M

u_aj uai

δˆ_a²

i,λi

δˆ_aj,λjdσ_gai. (B.1) Therefore, we need an expansion of ˆδaj,λj in conformal Fermi-coordinates at ai. Since (a, x)7→ |ψ_a(x)|²is smooth ford(a, x)<8ρ0,we get the following expansion if|z| ≤4ρ0:

|ψ_aj(ψ_a⁻¹_i (z))|²=dg_aj(aj, ai)²+

3

X

i=1

∂

∂zi|0|ψ_aj(ψ_a⁻¹_i (z))|²zi+O(|z|²), (B.2) in which

|∇|ψ_aj(ψ_a⁻¹_i (0))||² =|∇_ai|ψ_aj(ai)|²| ≤Cdg_aj(aj, ai).

Under the assumption |x| ≤ ε^λ_λⁱ

j or |x| ≤ ε q

λiλjdaj(ai, aj) the following expansion is

due to (B.2):

1

λ_iˆδ_aj,λj(ψ_a⁻¹_i (x

λ_i)) =χ_ρ

|ψ_aj(ψ⁻¹_ai (x λ_i))|

1

λi

λj +λ_jλ_i|ψ_aj(ψ⁻¹_ai (_λ^x

i))|²

!

=χ_ρ(d_gaj(a_j, a_i)) 1

λi

λj +λjλidg_aj(aj, ai)²

!

1−λj∇|ψ_aj(ψ_a⁻¹_i (0))|²·x

λi

λj +λjλidg_aj(aj, ai)²

!

+

λj

λiO(|x|²) λi

λj +λjλidg_aj(aj, ai)²

2. (B.3)

We now expand (B.1) under the assumption ε_ij →0.Sinceλ_i ≥λ_j either λiλjd(ai, aj)²≤ λi

λj

→ ∞ or λi

λj

≤λiλjd(ai, aj)² → ∞.

We rst assume λiλjd(ai, aj)² ≤ ^λ_λⁱ

j.In this case a rst expansion yields Z

Bε λj(ai)

uaj

u_aiδˆ²_ai,λiˆδ_aj,λjdσ_gai

= Z

|x|≤ε^λi

λj

1 1 +|x|²

2

u_aj(a_i) +O(|x|

λ_i) 1

λ_iˆδ_aj,λj(ψ_a⁻¹_i (x

λ_i))(1 +O(|x|¹⁰ λ¹⁰_i ))dx

=I1uaj(ai)χρ(dg_aj(aj, ai)) 1

λi

λj +λ_jλ_id_gaj(a_j, a_i)²

!

+o(εij). (B.4)

Since ^λ_λjⁱ+λ_iλ_jd_gaj(x, a_j)²≥ ¹₂

λi

λj +λ_iλ_jd_gaj(a_i, a_j)²

in this case, we have the following estimate:

Z

B2ρ(ai)\Bε λj(ai)

u_aj

u_aiδˆ²_ai,λiˆδ_aj,λjdσg_ai ≤Cεij

Z

|x|≥ε^λi

λj

1 1 +|x|²

2

dx=o(εij). (B.5) Hence form (B.4) and (B.5) we derive the expansion

Z

∂M

u_aiδˆ_ai,λi2

u_ajδˆ_aj,λjdσ_g=I₁u_aj(a_i)χ_ρ(d_gaj(a_j, a_i)) 1

λi

λj +λ_jλ_id_gaj(a_j, a_i)²

!

+o(εij).

From now on we assume _λ^λi_j ≤λ_iλ_jd(a_i, a_j)² → ∞. We rst expand Z

B

ε

√λj

√λjd(ai,aj)

(ai)

u_aj u_ai

ˆδ_a²_i,λiδˆ_aj,λjdσg_ai

= Z

|x|≤ε√

λiλjd(ai,aj)

1 1 +|x|²

2

u_aj(a_i) +O(|x|

λi

) 1

λi

δ_aj,λj(ψ_a⁻¹_i (x λi

))dx+o(ε_ij)

=I₁u_aj(a_i)χ_ρ(d_gaj(a_j, a_i)) 1

λi

λj +λ_jλ_id_gaj(a_j, a_i)²

!

+o(ε_ij), (B.6)

where we used (B.3) again. SetB :=B¹

10(aj). Then it holdsd(x, ai)≥ ₁₀⁹ d(ai, aj)onB, which implies

Z

B

u_aj u_ai

δˆ_a²_i,λiδˆ_aj,λjdσg_ai ≤C

1 1 +λ²_id(a_i, a_j)²

2 λj

λ_i Z

|x|≤λ_id(ai,aj)





 1 1 +^λ

2 j

λ²_i|x|²





dx

≤C ε_ij

λ_id(a_i, a_j) =o(ε_ij).

Finally onC= (B∪B

ε

√λj

√λjd(ai,aj)(a_i))^c we have the estimate Z

C

u_aj uai

δˆ_a²_i,λiδˆaj,λjdσg_ai

≤C 1

λi

λj +λ_jλ_id_gaj(a_j, a_i)²

!Z

|x|≥ε√

λiλjd(ai,aj)

1 1 +|x|²

2

dx=o(εij).

Hence the proof is completed.

With the help of Proposition 24 we now proof an expansion of the interaction between two dierent bubbles.

Proposition 25. If2≤λ_iρ, λ_jρ and λ_j ≤λ_i, then

< ϕai,λi, ϕaj,λj >=I1uaj(ai)χρ(|ψ_aj(ai)|)





1

λi

λj +^λ_λ^j

i +λjλidg_aj(aj, ai)²



+o(εij) +I1

(1−χρ(|ψ_aj(ai)|))G(aj, ai) λjλi

+O

1 λ²_iλ_jρ³

+O(ρεij).

Proof.

< ϕ_ai,λi, ϕ_aj,λj >= Z

M

Lgϕ_ai,λiϕ_aj,λj+ Z

∂M

Bgϕ_ai,λiϕ_aj,λjdσg

Z

∂M

Bgϕai,λiϕaj,λjdσg = Z

∂M

u_aj uai

Bg_aiϕˆai,λiϕˆaj,λjdσg_ai

= Z

∂M

u_aj u_ai

2χρδˆ_a²_i,λi+hg_aiχρδˆ_ai,λi

χρδˆ_aj,λj+ (1−χρ)G_aj(a_j,·) λ_j

dσg_ai. (B.7) Due to Proposition 24 we have

Z

∂M

uaj

u_aiχ_ρˆδ²_ai,λi

χ_ρδˆ_aj,λj+ (1−χ_ρ)Gaj(aj,·) λ_j

dσ_gai

=I1uaj(ai)χρ(|ψ_aj(ai)|)





1

λi

λj +^λ_λ^j

i +λjλidg_aj(aj, ai)²



+o(εij) +

Z

∂M

1 uai

χ_ρδˆ_a²

i,λi (1−χ_ρ)G(a_j,·) λj

dσ_gai. Furthermore a Taylor expansion yields

1

u_ai(ψ_a⁻¹_i (z))

(1−χ_ρ)G(aj, ψ⁻¹_ai (z)) λ_j

=

(1−χρ(|ψ_aj(ai)|))G(aj, ai) λj

+IN+O |z|²

ρ⁴λj

, where IN is a linear term (in z). The O-term does neither depend on ρ nor on a_j. Therefore

Z

∂M

1

u_aiχρδˆ²_ai,λi (1−χρ)G(aj,·) λ_j

dσg_ai

= Z

Bρ(ai)

1 uai

ˆδ²_ai,λi (1−χρ)G(a_j,·) λj

dx+O 1

λ²_iλ_jρ³

= Z

B_λiρ

1 1 +|x|²

2

1

u_ai(ψ⁻¹_ai (x λ_i))

(1−χ_ρ)G(a_j, ψ_a⁻¹_i (_λ^x

i)) λ_jλ_i

dx+O

1 λ²_iλjρ³

=I1

(1−χρ(|ψ_aj(ai)|))G(aj, ai) λjλi

+O

1 λ²_iλ_jρ³

. (B.8)

Hence

Z

∂M

uaj

u_aiχ_ρˆδ²_ai,λi

χ_ρδˆ_aj,λj+ (1−χ_ρ)Gaj(aj,·) λ_j

dσ_gai

=I1uaj(ai)χρ(|ψ_aj(ai)|)





1

λi

λj +^λ_λ^j

i +λjλidg_aj(aj, ai)²



+o(εij) +I1

(1−χρ(|ψ_aj(ai)|))G(aj, ai) λ_jλ_i

+O

1 λ²_iλ_jρ³

.

Moreover observe that

|ϕˆ_aj,λj(x)| ≤C λj

1 +λ²_jd_gaj(a_j, x)²

!

if 2≤ρλj. Since

cdga(x, y)≤dg(x, y)≤Cdga(x, y) for a∈∂M, x, y ∈M and two universal constants c, C we have

|ϕˆaj,λj(x)| ≤C λj

1 +λ²_jd_gai(a_j, x)²

! . This estimate implies

Z

∂M

uaj

u_aih_gaiχ_ρδˆ_ai,λiϕˆ_aj,λjdσ_gai

≤C Z

B2ρ(ai)

λ_i

1 +λ²_idg_ai(aj, x)²

! λ_j

1 +λ²_jdg_ai(aj, x)²

!

dσ_gai ≤Cρε_ij, provided 2≤ρλ_j. Here we integrated over the setsB_2ρ∩Aand B_2ρ∩A^c, where

A:=

x∈∂M |2dai(aj, x)≤ 1 λi

+dg_ai(ai, aj)

.

Finally we have to estimate the integral inM. From Proposition 2 we derive the estimate

|L_gaiϕˆ_ai,λi| ≤C λ²_i

(1 +λ_id_gai(a_i, x))³ +1 ρ

λi

1 +λ²_idg_ai(ai, x)²

!!

1_d

gai(ai,x)≤2ρ. Set

A:=

x∈M |2dai(aj, x)≤ 1

λ_i +dg_ai(ai, aj)

. As above, integration overB2ρ∩A and B2ρ∩A^c yields

Z

M

L_gϕ_ai,λiϕ_aj,λjdV_g

≤Cρε_ij.

We add the previous expansions and estimates which proves the Proposition.

Remark 4. Since cd_g(a, x) ≤d_ga(a, x) ≤Cd_g(a, x) uniformly for a, x ∈∂M, we easily derive the estimate

|< ϕ_ai,λi, ϕ_aj,λj >| ≤Cεij

from the previous Proposition.

C.1. Selnteractions

In this section we expand the scalar product of a bubble with its derivatives w.r.t. λand a.

Proposition 26. It holds (a)

< ϕ_a,λ, λ ∂

∂λϕ_a,λ>= 2|S₊³|H_a(a) λ² +O

|Π(a)|²log(λρ)

λ² + ρ

λ² + 1 (λρ)³

, (b)

< ϕ_a,λ,1 λ

∂

∂a^mϕ_a,λ>

≤C 1 λ². Proof. We rst prove (a):

Since

w_a,λ:=λ ∂

∂λϕˆ_a,λ+ ˆϕ_a,λ= 2χ_ρδ_a,λ

(1 +λt) (1 +λt)²+λ²|x|²

(C.1) the scalar product is given by

< ϕ_a,λ, λ ∂

∂λϕ_a,λ>=−||ϕ_a,λ||²+ Z

M

∇ϕˆ_a,λ· ∇w_a,λ+1

6R_gaϕˆ_a,λw_a,λdV_ga +

Z

∂M

h_gaϕˆ_a,λw_a,λdσ_ga. (C.2)

Moreover Z

Bρ⁺

∇ϕˆa,λ· ∇w_a,λdVga = Z

B⁺ρ

∇ϕˆa,λ· ∇w_a,λdxdt+O 1

λ³

= Z

Bρ⁺

∇δ_λ· ∇w_λdxdt+ Z

B⁺ρ

(g^ij−δij)∂iϕˆ_a,λ∂jw_λdxdt+O 1

λ³

= Z

Bρ⁺

∇δ_λ· ∇w_λdxdt+O

|Π(a)|²

λ² +log(λρ) λ³

= 2 Z

Bρ

δ_λ²w_λdx+ Z

S_ρ,+³

∂_νδ_λw_λdS+O

|Π(a)|²

λ² +log(λρ) λ³

= 4 Z

Bρ

λ³

(1 +λ²|x|²)⁴dx+O

|Π(a)|²

λ² +log(λρ) λ³ + 1

(λρ)³

= 2I0+O

|Π(a)|²

λ² +log(λρ) λ³ + 1

(λρ)³

and

1 6

Z

B⁺ρ

RgaδλwλdVga+ Z

Bρ

hgaδλwλdσga =O

|Π(a)|²

λ² +log(λρ) λ³

. Hence

Z

B⁺ρ

∇ϕˆa,λ· ∇w_a,λ+1 6

Z

B⁺ρ

Rgaϕˆa,λwλdVga+ Z

Bρ

hgaϕˆa,λwλdσga

= 2I0+O

|Π(a)|²

λ² +log(λρ) λ³ + 1

(λρ)³

. (C.3)

Furthermore, partial integration yields Z

M\B⁺ρ

∇ϕˆ_a,λ· ∇w_a,λ+1 6

Z

B⁺ρ

R_gaϕˆ_a,λw_λdV_ga+ Z

∂M\Bρ

h_gaϕˆ_a,λw_λdσ_ga

= Z

M\B_ρ⁺

L_gaϕˆ_a,λw_a,λdV_ga+ Z

∂M\B_ρ

B_gaϕˆ_a,λw_λdσ_ga− Z

Sρ,+

∂_νϕˆ_a,λw_a,λdS_ga

=O 1

(λρ)³

(C.4) due to Proposition 2. Therefore (a) is proved by adding (C.2), (C.3), (C.4) and the expansion of||ϕˆ_a,λ||².

Proof of (b): We take the derivative at a with respect to ψa Fermi-coordinates. In Lemma 10 we proved the expansions

∂

∂a^m|ψ¯_a(x)|² =−2ψ_a(x)^m+O(d_ga(a, x)³), ∂

∂a^mt_a(x) =O(d_ga(a, x)²).

In general 1 λ

∂

∂a^mϕ_a,λ= 1 λ

∂

∂a^mu_aχ_ρδ_a,λ+u_a1 λ

∂

∂a^m

δ_a,λ+ (1−χ_ρ)G(a,·) λ

, (C.5) form which we derive the inequality

1 λ

∂

∂a^mϕ_a,λ

≤Cϕ_a,λ

provided 2≤λρ.Furthermore, the previous expansions yield onB⁺_ρ 1

λ

∂

∂a^mϕ_a,λ ψ⁻¹_a (x, t)

= 1 λ

∂

∂a^mu_aχ_ρδ_a,λ+u_a4λ²x^m+λ²O(|x, t|³) +O 1 +λt)|(x, t)|²λ (1 +λt)²+λ²|x|²2 . We compute

< ϕ_a,λ,1 λ

∂

∂a^mϕ_a,λ>= Z

M

Lgϕ_a,λ1 λ

∂

∂a^mϕ_a,λdVg+ Z

∂M

Bgϕ_a,λ1 λ

∂

∂a^mϕ_a,λdσg

= Z

M

L_gaϕˆ_a,λ 1 u_a

1 λ

∂

∂a^mϕ_a,λdV_ga+ Z

∂M

B_gaϕˆ_a,λ 1 u_a

1 λ

∂

∂a^mϕ_a,λdσ_ga. First observe

Z

M\B⁺ρ

Lgaϕˆa,λ

1 ua

1 λ

∂

∂a^mϕa,λdVga+ Z

∂M\Bρ

Bgaϕˆa,λ

1 ua

1 λ

∂

∂a^mϕa,λdσga

=O

|Π(a)|

λ²ρ + ρ

λ² + 1 (λρ)³

. Moreover

Z

B⁺ρ

Lgaϕˆ_a,λ 1 u_a

1 λ

∂

∂a^mϕ_a,λdVga + Z

Bρ

Bgaϕˆ_a,λ 1 u_a

1 λ

∂

∂a^mϕ_a,λdσga

= Z

B⁺ρ

Lgaϕˆ_a,λ1 λ

∂

∂a^mϕˆ_a,λdVga+ Z

Bρ

Bgaϕˆ_a,λ1 λ

∂

∂a^mϕˆ_a,λdσga

+O

|Π(a)|²

λ² +log(λρ) λ³

=O 1

λ²

, which nally proves the Proposition.

C.2. Interaction with the derivatives

For the expansion of the gradient we also need the interaction between a bubble and the derivative of a dierent bubble.

Proposition 27. It holds (a)

< ϕ_ai,λi, λ_j ∂

∂λ_jϕ_aj,λj >= 2I₁λ_j ∂

∂λ_jI(ε_ij) +O(ρε_ij) +o(ε_ij)

(b)

< ϕ_ai,λi, 1 λj

∇_ajϕ_aj,λj· ∇K(a_j)>

≤C|∇K(a_j)|ε_ij.

Proof. Proof of (a). We rst assume that λj ≤λi in which case we compute

< ϕ_ai,λi, λ_j ∂

∂λ_jϕ_aj,λj >= Z

M

L_gϕ_ai,λiλ_j ∂

∂λ_jϕ_aj,λjdV_g+ Z

∂M

B_gϕ_ai,λiλ_j ∂

∂λ_jϕ_aj,λjdσ_g

= Z

∂M

B_gϕ_ai,λiλ_j ∂

∂λj

ϕ_aj,λjdσ_g+O(ρε_ij),

where we estimate the rst integral like in Proposition 25 using the estimate

λ_j∂λ^∂

jϕ_aj,λj

≤Cϕ_aj,λj.Furthermore, the derivative on∂M is given by:

λ_j ∂

∂λ_jϕ_aj,λj(x) =u_ajχ_ρˆδ_aj,λj 1−λ²_j|ψ_aj(x)|² 1 +λ²_j|ψ_aj(x)|²

!

−(1−χ_ρ)G(a_j,·) λ_j . Integration as in proof of Proposition 24 and 25 yields

Z

∂M

B_gϕ_ai,λiλ_j ∂

∂λj

ϕ_aj,λjdσ_g = 2I₁λ_j ∂

∂λj

I(ε_ij) +O(ρε_ij) +o(ε_ij).

Therefore (a) is proved under the assumptionλj ≤λi.From now on we assumeλi ≤λj. In this case

< ϕ_ai,λi, λ_j ∂

∂λ_jϕ_aj,λj >

= Z

M

Lg

λj

∂

∂λ_jϕ_aj,λj

ϕ_ai,λidVg+ Z

∂M

Bg

λj

∂

∂λ_jϕ_aj,λj

ϕ_ai,λidσg

= Z

M

λ_j ∂

∂λj

L_g ϕ_aj,λj

ϕ_ai,λidV_g+ Z

∂M

λ_j ∂

∂λj

B_g ϕ_aj,λj

ϕ_ai,λidσ_g

Due to Proposition 2 and 3, λ_j∂λ^∂

jL_g ϕ_aj,λj

satises the same estimate as L_gϕ_aj,λj. Hence the following inequality holds true:

Z

M

λj

∂

∂λ_jLg ϕ_aj,λj

ϕ_ai,λidVg

≤Cρεij. Furthermore,

∂

∂λj

Bg ϕaj,λj

=u²_ajχρ

2 ∂

∂λj

δ_a²_j,λj+hg_aj

∂

∂λj

δaj,λj

, which implies

Z

∂M

λ_j ∂

∂λj

B_g ϕ_aj,λj

ϕ_ai,λidσ_g = Z

∂M

u²_ajχ_ρ

2λ_j ∂

∂λj

δ²_aj,λj

ϕ_ai,λidσ_g+O(ρε_ij)

= 2I1λj

∂

∂λj

I(εij) +O(ρεij) +o(εij).

Here we integrated like in Proposition 24 and 25. Therefore part (a) is proved.

Proof of (b): Since

1 λj

∇_ajϕaj,λj · ∇K(a_j)

≤C|∇K(a_j)|ϕ_aj,λj, we deduce

< ϕ_ai,λi, 1

λ_j∇_ajϕ_aj,λj · ∇K(a_j)>

≤C|∇K(a_j)|

Z

M

|L_gϕai,λi|ϕ_aj,λjdVg+ Z

∂M

|B_gϕai,λi|ϕ_aj,λjdσg

≤C|∇K(a_j)|

Z

B2ρ(ai)

δˆ_a²_i,λi λj

1 +λ²_jd_gai(a_i, a_j)²

!

dσg_ai+Cρεij

!

≤C|∇K(a_j)|ε_ij,

where we integrate over the setA∩B_2ρ(a_i)andA^c∩B_2ρ(a_i)to estimate the last integral.

Here

A:=

x∈∂M |2dg_ai(aj, x)≤ 1 λi

+dg_ai(ai, aj)

. Therefore the proof is completed.

C.3. Further estimates

We need more expansions and estimates which we mainly use in chapter 7.

Lemma 11. It holds

(a) < λ_∂λ^∂ ϕ_a,λ, λ_∂λ^∂ϕ_a,λ>=I₂+O _λ¹ , (b) < λ_∂λ^∂ ϕ_a,λ,_λ¹_∂a^∂

iϕ_a,λ>=O_log(λ)

λ

, (c) < ¹_λ∂a^∂

jϕa,λ,¹_λ∂a^∂

iϕa,λ>=I3δij+O ¹_λ , (d)

< _λ¹

j

∂

∂a^m_j ϕ_aj,λj,_λ¹

i

∂

∂a^k_iϕ_ai,λi >

+

< _λ¹

j

∂

∂a^m_j ϕ_aj,λj, λ_i∂λ^∂

iϕ_ai,λi >

=O(ε_ij), (e)

< λ_j∂λ^∂

jϕ_aj,λj, λ_i∂λ^∂

iϕ_ai,λi >

=O(ε_ij) where

I₂ = 4 Z

R³

1 1 +|x|²

3

1− |x|² 1 +|x|²

2

dx and

I3= 16 3

Z

R³

|x|² (1 +|x|²)⁵dx.

Proof. (a)

< λ ∂

∂λϕ_a,λ, λ ∂

∂λϕ_a,λ>

= Z

M

λ ∂

∂λL_gaϕˆ_a,λ

λ ∂

∂λϕˆ_a,λdV_ga+ Z

∂M

λ ∂

∂λB_gaϕˆ_a,λ

λ ∂

∂λϕˆ_a,λdσ_ga. Due to Proposition 3

Z

M

λ ∂

∂λL_gaϕˆ_a,λ

λ ∂

∂λϕˆ_a,λdV_ga

≤C1 λ. Furthermore

Z

∂M

λ ∂

∂λBgaϕˆ_a,λ

λ ∂

∂λϕˆ_a,λdσga = 4 Z

Bρ

δ³_λ

1−λ²|x|² 1 +λ²|x|²

2

dx+O 1

λ³

=I2+O 1

λ³

. Hence (a) is proved.

(b) In this case we expand as follows

< λ ∂

∂λϕ_a,λ,1 λ

∂

∂ai

ϕ_a,λ>

= Z

M

λ ∂

∂λL_gaϕˆ_a,λ1 λ

∂

∂ai

ˆ

ϕ_a,λdV_ga+ Z

∂M

λ ∂

∂λB_gaϕˆ_a,λ1 λ

∂

∂ai

ˆ

ϕ_a,λdσ_ga.

Like in the previous case the interior integral is bounded by C

λ.On the boundary we get Z

∂M

λ ∂

∂λBgaϕˆa,λ

1 λ

∂

∂a_iϕˆa,λdσga

= 8 Z

Bρ

δ³_λ

1−λ²|x|² 1 +λ²|x|²

2λx_i+O(λ|x|³) 1 +λ²|x|²

dx+O

log(λ) λ

=O

log(λ) λ

, which proves (b).

(c) Due to Proposition 4 the same estimates as in (a) yield

Z

M

1 λ

∂

∂a_jLgϕ_a,λ1 λ

∂

∂a_iϕ_a,λdVg

≤C1 λ. We are left to expand the boundary integral

Z

∂M

1 λ

∂

∂aj

Bgϕa,λ

1 λ

∂

∂ai

ϕa,λdσg

= 2 Z

Bρ

1 λ

∂

∂aj

u²_aδ_a,λ² 1 λ

∂

∂ai

(u_aδ_a,λ)dσ_g+O

log(λ) λ

= 2 Z

Bρ

1 λ

∂

∂a_jδ_a,λ² 1 λ

∂

∂a_iδ_a,λdx+O

log(λ) λ

= 16 Z

Bρ

δ³_λ

λxj+O(λ|x|³) 1 +λ²|x|²

λxi+O(λ|x|³) 1 +λ²|x|²

dx+O

log(λ) λ

=I3δij +O

log(λ) λ

. Hence (c) is proved.

(d)+(e) follow easily by using the same arguments as in the proof of Proposition 27 (b).

Lemma 12. There exists a constantC >0, independent of a, such that (a) |ϕ_a,λ|,|λ_∂λ^∂ ϕa,λ|,|_λ¹∇_aϕa,λ| ≤C

λ 1+λ²dg(a,x)²

, (b) |λ²_∂λ^∂2₂ϕ_a,λ|,|_λ¹₂∇²_aϕ_a,λ|,|_∂λ^∂∇_aϕ_a,λ| ≤C

λ 1+λ²dg(a,x)²

. Proof. Ifdg(a, x)≤2δ and k∈Na computation yields

∂

∂ai

(1 +λta(x))²+λ²|ψ_a(x)|²−^k₂

=−k

2 (1 +λta(x))²+λ²|ψ_a(x)|²−^k

2−1

2(1 +λta(x))λ ∂

∂a_ita(x) +λ² ∂

∂a_i|ψ_a(x)|²

, (D.1) hence Lemma 10 implies

∂

∂ai

(1 +λt_a(x))²+λ²|ψ_a(x)|²−^k₂

≤Cλ (1 +λt_a(x))²+λ²|ψ_a(x)|²−^k₂

. (D.2) Furthermore, with the use of (D.1) and Lemma 10 we estimate

∂²

∂a_i∂a_j (1 +λt_a(x))²+λ²|ψ_a(x)|²−^k

2

≤Cλ² (1 +λt_a(x))²+λ²|ψ_a(x)|²−^k

2 (D.3) as well as

∂

∂λ

∂

∂a_i (1 +λta(x))²+λ²|ψ_a(x)|²−^k

2

≤C (1 +λta(x))²+λ²|ψ_a(x)|²−^k

2. (D.4)

In addition, there holds

|∇_aG(a, x)| ≤C 1

d_g(a, x)³, |∇²_aG(a, x)| ≤C 1 d_g(a, x)⁴. Since(1 +λta(x))²+λ²|ψ_a(x)|²≥(1 +λ²|ψ_a(x)|²)≥c 1 +λ²dg(a, x)²

, ifdg(a, x)≤2ρ, the Lemma follows from the previous estimates.

Lemma 13. Let a, b∈∂M such that d_ga(a, b)≤ρ₀ andv∈H¹(M).If ¹₂ ≤ ^λ_µ ≤2 then (a) |< v, ϕ_b,µ−ϕ_a,λ>| ≤C||v|| λd_g(a, b) +

1− ^µ_λ

, (b) |< ϕ_a,λ, ϕ_b,µ−ϕ_a,λ>| ≤C

λ²d_g(a, b)²+ 1−^µ_λ

2 , (c) |< v, µ_∂µ^∂ ϕb,µ−λ_∂λ^∂ϕa,λ>| ≤C||v|| λdg(a, b) +

1−^µ_λ

, (d) |< ϕ_b,µ−ϕ_a,λ, λ_∂λ^∂ ϕ_a,λ>| ≤C

λ²d_g(a, b)²+ 1−^µ_λ

2 , (e) |< v,¹_µ∂a^∂

iϕ_b,µ− ¹_λ∂a^∂

iϕ_a,λ>| ≤C||v|| λd_g(a, b) + 1−^µ_λ

, (f) |< ϕ_b,µ−ϕ_a,λ,_λ¹_∂a^∂

iϕ_a,λ>| ≤C

λ²d_g(a, b)²+ 1−^µ_λ

2 .

Proof. Since Lgϕa,λ =u³_aLgaϕˆa,λ and Bgϕa,λ =u²_aBgaϕˆa,λ, the following estimates are an immediate consequence of Lemma 10 and the proof of Lemma 12.

1

λ∇_aLgϕa,λ

L⁴³(M)

+

1

λ²∇²_aLgϕa,λ

L⁴³(M)

≤C, (D.5)

1

λ∇_aBgϕa,λ

L³²(∂M)

+

1

λ²∇²_aBgϕa,λ

L³²(∂M)

≤C, (D.6)

λ ∂

∂λLgϕ_a,λ L⁴³(M)

+

λ² ∂²

∂λ²Lgϕ_a,λ L⁴³(M)

≤C, (D.7)

λ ∂

∂λB_gϕ_{a,λ L}³_2(∂M)

+

λ² ∂²

∂λ²B_gϕ_{a,λ L}³_2(∂M)

≤C, (D.8)

∇_a ∂

∂λL_gϕ_a,λ L⁴³(M)

+

∇_a ∂

∂λB_gϕ_a,λ L³²(∂M)

≤C. (D.9)

With the help of these estimates we can now proof the assertions.

(a)

< v, ϕb,µ−ϕa,λ>=

Z

M

Lg(ϕb,µ−ϕa,λ)vdVg+ Z

∂M

Bg(ϕb,µ−ϕa,λ)vdσg. In conformal Fermi coordinates atalet x=ψ⁻¹_a (y). We compute

L_g(ϕ_b,µ−ϕ_a,λ) (x) = Z 1

0

∂

∂yi|L_gϕ_y(t),λ(t)(x)ψ_a(b)ⁱ+ ∂

∂λL_gϕ_y(t),λ(t)(x)(µ−λ)

dt,

where(y(t), λ(t)) = (tψ_a(b), λ+t(µ−λ)).We use Hoelder inequality forp= ⁴₃ to obtain

|L_g(ϕ_b,µ−ϕ_a,λ) (x)|^p ≤C_p Z 1

0

|∇_yL_gϕ_y(t),λ(t)(x)|^pdt |ψ_a(b)ⁱ|^p +C_p

Z 1 0

| ∂

∂λL_gϕ_y(t),λ(t)(x)|^pdt |µ−λ|^p.

Hence integration over the manifold and switching the integration, combined with (D.5) and (D.7), yields

||L_g(ϕ_b,µ−ϕ_a,λ)||

L⁴³(M)≤C

λ|ψ_a(b)|+|µ λ−1|

≤C

λd_g(a, b) +|µ λ −1|

. If we use the same arguments as above we can also prove that

||B_g(ϕb,µ−ϕa,λ)||

L³²(∂M) ≤C

λdg(a, b) +|µ λ−1|

. Claim (a) follows from the last two estimates.

(b) We again choose conformal Fermi-coordinates around a. In this coordinates b = ψ⁻¹_a (z).By Taylor expansion we obtain

ϕ_b,µ(x)−ϕ_a,λ(x) = ∂

∂λϕ_a,λ(x)(µ−λ) + ∂

∂a_iϕ_a,λ(x)zⁱ+ Z ₁

0

(1−t) ∂²

∂λ²ϕ_a(t),λ(t)(µ−λ)²+ ∂²

∂λ∂a_iϕ_a(t),λ(t)(µ−λ)zⁱ+ ∂²

∂a_i∂a_jϕ_a(t),λ(t)zⁱz^j

dt (D.10) where

(a(t), λ(t)) = (tz, λ+t(µ−λ)) in coordinates. We expand

< ϕ_a,λ, ϕ_b,µ−ϕ_a,λ>=

Z

M

Lgϕ_a,λ(ϕ_b,µ−ϕ_a,λ)dVg+ Z

∂M

Bgϕ_a,λ(ϕ_b,µ−ϕ_a,λ)dσg. As an example we estimate:

Z

M

L_gϕ_a,λ Z 1

0

(1−t)λ(t)² ∂²

∂λ²ϕ_a(t),λ(t)dtdV_g

=

Z ₁

0

(1−t) Z

M

L_gϕ_a,λλ(t)² ∂²

∂λ²ϕ_a(t),λ(t)dV_gdt

≤C Z 1

0

||L_gϕa,λ||

L⁴³(M)||λ(t)² ∂²

∂λ²ϕ_a(t),λ(t)||_L4(M)dt≤C

due to Lemma 12. The same estimate also holds for the other integrals in (D.10). Hence

< ϕ_a,λ, ϕ_b,µ−ϕ_a,λ>=< ϕ_a,λ, λ ∂

∂λϕ_a,λ>(µ

λ−1)+< ϕ_a,λ,1 λ

∂

∂a_iϕ_a,λ> λzⁱ +O

µ λ −1

2

+λ²|z|²

=< ϕ_a,λ, λ ∂

∂λϕ_a,λ>(µ

λ−1)+< ϕ_a,λ,1 λ

∂

∂a_iϕ_a,λ> λzⁱ +O

µ λ −1

2

+λ²dg(a, b)²

=O 1

λ²

µ λ−1

+λdg(a, b)

+O

µ λ−1

2

+λ²dg(a, b)²

where we used Proposition 26. Therefore (b) is proved.

(c) We write

< v, µ ∂

∂µϕ_b,µ−λ ∂

∂λϕ_a,λ>

= Z

M

L_g

µ ∂

∂µϕ_b,µ−λ ∂

∂λϕ_a,λ

vdV_g+ Z

∂M

B_g

µ ∂

∂µϕ_b,µ−λ ∂

∂λϕ_a,λ

vdσ_g

= Z

M

Lg

µ ∂

∂µϕb,µ−µ ∂

∂λϕa,λ

vdVg+ Z

∂M

Bg

µ ∂

∂µϕb,µ−µ ∂

∂λϕa,λ

vdσg

+ (µ λ−1)

Z

M

λ ∂

∂λLgϕa,λvdVg+ (µ λ−1)

Z

∂M

λ ∂

∂λBgϕa,λvdσg

= Z

M

L_g

µ ∂

∂µϕ_b,µ−µ ∂

∂λϕ_a,λ

vdV_g+ Z

∂M

B_g

µ ∂

∂µϕ_b,µ−µ ∂

∂λϕ_a,λ

vdσ_g +O

µ λ−1

||v||

.

With the notations of above we expand µ ∂

∂µϕ_b,µ−µ ∂

∂λϕ_a,λ=µ Z 1

0

∂²

∂λ²ϕ_a(t),λ(t)(λ−µ) + ∂²

∂λ∂ai

ϕ_a(t),λ(t)zⁱdt.

Hence the same method as in (a) implies Z

M

L_g

µ ∂

∂µϕ_b,µ−µ ∂

∂λϕ_a,λ

vdV_g+ Z

∂M

B_g

µ ∂

∂µϕ_b,µ−µ ∂

∂λϕ_a,λ

vdσ_g

=O

µ λ−1

+λd_g(a, b)

||v||, which proves (c).

(d) From the same Taylor expansion of ϕ_b,µ−ϕ_a,λ as in (b) we derive the following

estimate

< ϕb,µ−ϕa,λ, λ ∂

∂λϕa,λ>

=< λ ∂

∂λϕ_a,λ, λ ∂

∂λϕ_a,λ>µ λ −1

+< 1 λ

∂

∂ai

ϕ_a,λ, λ ∂

∂λϕ_a,λ> λzⁱ +O

µ λ −1

2

+λ²dg(a, b)²

=I₂µ λ−1

+O

log(λ) λ

µ λ−1

+λd_g(a, b) +O

µ λ−1

2

+λ²d_g(a, b)²

. The last expansion follows by Lemma 11.

(e) Similar to (c) we get

< v,1 µ

∂

∂ai

ϕb,µ− 1 λ

∂

∂ai

ϕa,λ>

= Z

M

1 µ

∂

∂ai

Lgϕb,µ− ∂

∂ai

Lgϕa,λ

vdVg+ Z

∂M

1 µ

∂

∂ai

Bgϕb,µ− ∂

∂ai

Bgϕa,λ

vdσg

+ λ

µ−1 Z

M

1 λ

∂

∂a_iϕ_a,λvdVg+ λ

µ −1 Z

∂M

1 λ

∂

∂a_iϕ_a,λvdσg

=O

µ λ−1

+λd_g(a, b)

||v||, which proves (e).

(f) The same ideas as in (b) and (d) yield

< ϕ_b,µ−ϕ_a,λ,1 λ

∂

∂ai

ϕ_a,λ>

=< λ ∂

∂λϕa,λ,1 λ

∂

∂ai

ϕa,λ>

µ λ−1

+< 1

λ

∂

∂aj

ϕa,λ,1 λ

∂

∂ai

ϕa,λ> λz^j +O

µ λ−1

2

+λ²d_g(a, b)²

=I₃λψ_a(b)ⁱ+O

log(λ) λ

µ λ−1

+λd_g(a, b) +O

µ λ−1

2

+λ²d_g(a, b)²

, which proves (f).

Lemma 14. If ˜λiλidg(˜aj, ai)→0,^λ_λ^˜i

i →1 then (a) < ϕ_˜a

i,˜λi−ϕai,λi, ϕaj,λj >=O(εij)

λidg(˜aj, ai) +

λ˜i

λi −1

(b) < ϕ_˜a

i,˜λi−ϕ_ai,λi, λ_j∂λ^∂

jϕ_aj,λj >=O(ε_ij)

λ_id_g(˜a_j, a_i) +

˜λi

λi −1

(c) < ϕ_˜a

i,˜λi−ϕai,λi,_λ¹

j

∂

∂a^m_j ϕaj,λj >=O(εij)

λidg(˜aj, ai) +

λ˜i

λi −1

Proof. (a) Since dg(˜aj, ai) →0, we chooseψai Fermi coordinates and writeψai(˜ai) =z.

Furthermore we set

(z(t), λi(t)) = (tz, λi+t(˜λi−λi)); ai(t) =ψ⁻¹_ai (tz) and obtain

ϕ_˜a

i,˜λi(x)−ϕai,λi(x) = Z 1

0

∂

∂λi

ϕ_ai(t),λi(t)(x)

λ˜i−λi

+ ∂

∂a^m_i ϕ_ai(t),λi(t)(x)z^mdt.

Hence

< ϕ_˜a

i,˜λi−ϕai,λi, ϕaj,λj >

= Z

M

Lgϕ_aj,λj

ϕ_˜a

i,˜λi−ϕ_ai,λi

dVg+ Z

∂M

Bgϕ_aj,λj

ϕ_˜a

i,λ˜i −ϕ_ai,λi

dσg

= Z 1

0

< ϕaj,λj, λi(t) ∂

∂λi

ϕ_ai(t),λi(t) >

˜λ_i−λ_i λ(t) dt +

Z ₁

0

< ϕ_aj,λj, 1 λ_i(t)

∂

∂a^m_i ϕ_ai(t),λi(t) > λ_i(t)z^mdt

= Z 1

0

O(εij(t))dt λidg(˜aj, ai) +

λ˜i

λ_i −1

!

≤Cεij λidg(˜aj, ai) +

λ˜i

λ_i −1

! , which proves the assertion. (b)-(c) follow similarly.

Letψabe conformal Fermi-coordinates arounda∈∂M. In what follows(M, g)is a four-dimensional compact Riemannian manifold with boundary and positive Sobolev-quotient Q(M, ∂M,[g]). In this chapter we prove an appropriate expansion of the Green's function G_a(·) at awith respect to the conformal operator (L_ga, B_ga). We will use that

detg_a(x, t) = 1 +O(|(x, t)|¹⁰) (E.1) in these coordinates, which implies

hga(x) =O(|x|⁹). (E.2)

We would like to expand Ga as follows

G_a(x) = Γ(ψ_a(x)) +H_a(x), where

Γ(x, t) = 1

|(x, t)|² 1 +ψ(x, t)

with some appropriate functionψ ∈C^∞(R⁴\ {0}) andH_a ∈C^2,α(M).In the following we always write ginstead of ga. Locally around0∈R⁴₊ we dene the operators

Ku:=∂_i (g^ij −δ^ij)∂_ju

, Lu=−|(x, t)|²∆u+ 4<∇u,(x, t)> . Then, due to (E.1)

|(x, t)|⁴LgΓ =Lψ− |(x, t)|⁴K ψ

|(x, t)|²

− |(x, t)|⁴K 1

|(x, t)|²

+1

6Rg|(x, t)|² 1 +ψ +O |(x, t)|¹⁰|D²ψ|

. (E.3)

Here and in the following g(y) =O(f(y))means

|∇^kg(y)| ≤C(k)|∇^kf(y)|, k= 0,1,2.

Furthermore

BgΓ =− ∂_tψ

|(x, t)|² +hgΓ. (E.4)

We need to nd ψ such that L_gΓ ∈ C^0,α and ∂_tψ = 0. Then B_gΓ ∈ C^1,α, which is crucial for our argument. We will successively remove the singularities in (E.3) by using homogeneous function of increasing degree.

Let H_k be the space of (smooth) homogeneous functions on R⁴+ and C^l the space of functions u ∈ C^∞(R⁴₊ \ {0}) such that u(x, t) = O |(x, t)|^l

. Now we state a rst Lemma.

Lemma 15. If ψ∈H_k, k≤2, then (a) ∂^αψ∈Hk−|α| for all multiindizesα (b) −|(x, t)|⁴K

1

|(x,t)|²

+¹₆Rg|(x, t)|² 1 +ψ

∈P4

l=k+1Hl+C⁵.

Since we work with homogeneous functions onR⁴+, boundary value problems on R⁴+ can be reduced to boundary value problems on the upper half sphereS₊⁴. Therefore we need the following Propositions.

Proposition 28. Let f¯∈C^∞(S₊ⁿ⁻¹) and ψ¯∈C^∞(S₊ⁿ⁻¹) be a solution of the boundary value problem

(−∆_Sⁿ⁻¹

+

ψ¯+k(n−2−k) ¯ψ= ¯f S₊ⁿ⁻¹

∂_νψ¯= 0 ∂S₊ⁿ⁻¹, (E.5)

then ψ(x, t) :=|(x, t)|^kψ¯ _(x,t)

|(x,t)|

is a homogeneous function of degree k, which solves the boundary value problem

(Lnψ=f Rⁿ+\ {0}

∂_tψ= 0 ∂Rⁿ+\ {0}. (E.6)

Here f(x, t) :=|(x, t)|^kf¯_(x,t)

|(x,t)|

andL_nu=−|(x, t)|²∆u+ 2(n−2)<∇u,(x, t)> .

Proof. Withn-dimensional polar-coordinates the Laplacian is given by

∆u= 1 rⁿ⁻¹

d

dr rⁿ⁻¹ d dru

+ 1 r²∆_Sⁿ⁻¹

+ u.

If we write (x, t) =rω withω∈S₊ⁿ⁻¹, then

∆ψ(x, t) =k(k+n−2)r^k−2ψ(ω) +¯ r^k−2∆_Sⁿ⁻¹

+

ψ(ω).¯

Furthermore, sinceψ is homogeneous of degree k we have the equality<∇ψ,(x, t) >=

kψ. The previous identities yield L_nψ(x, t) =−r^k∆_Sⁿ⁻¹

+

ψ(ω)¯ −k(k+n−2)r^kψ(ω) + 2(n¯ −2)kr^kψ(ω)¯

=r^kf(ω) =¯ f(x, t).

The boundary equation follows easily. Therefore the proof is completed.

If we want to solve (E.5) we need to have knowledge about the sectrum σ(−∆_Sn−1 + ) of the Laplacian onS₊ⁿ⁻¹ with respect to Neumann boundary conditions.

Proposition 29. For f¯∈C^∞(S₊ⁿ⁻¹) the boundary value problem (−∆_Sn−1

+

ψ¯−λψ¯= ¯f S₊ⁿ⁻¹

∂νψ¯= 0 ∂S₊ⁿ⁻¹ (E.7)

has a solution if¯∈ker −∆_Sⁿ⁻¹

+

−λ⊥

. Moreover σ(−∆_Sn−1

+ ) ={l(l+n−2) : l∈N0}.

Furthermore ifλ /∈σ(−∆_Sn−1 + ), then

||ψ||¯ _C2,α ≤C||f¯||_C0,α.

Proof. The statement follows from standard elliptic theory (see [26, 33]) and the identity σ(−∆_Sⁿ⁻¹

+ ) =σ(−∆_Sⁿ⁻¹).

Now we are prepared to begin the expansion of the Green's function. For ψ = 0, (E.3) yields

−|(x, t)|⁴K 1

|(x, t)|²

+1

6R_g|(x, t)|² =

4

X

k=1

g_k+C⁵, wheregk are homogeneous function of degree k onR⁴₊\ {0}.More precisely

g₁(x, t) =−16th_ijx_ix_j

|(x, t)|² ,

where h_ij are the coecients of the umibilicity tensor Π at a ∈ ∂M in ψ_a Fermi-coordinates. We setg¯1 =g1|_S3

+ and observe that the boundary value problem (−∆_S3

+

ψ¯+ ¯ψ=−¯g1 S₊³

∂_νψ¯= 0 ∂S₊³,

has a unique solutionψ¯₁. Proposition 28 implies that the function ψ₁(x, t) :=|(x, t)|ψ¯₁

(x, t)

|(x, t)|

solvesLψ1 =−g₁ and ∂tψ1 = 0.Furthermore Proposition 29 implies

|∇^kψ1| ≤C|Π(a)||(x, t)|^1−k for k= 0,1,2.In addition

Z

S³₊

ψ₁dS= 0, becausehg(a) = 0.

We setΓ₁(x, t) = _|(x,t)|¹ 2 1 +ψ₁(x, t)

and observe by (E.3) and Lemma 15 that

|(x, t)|⁴L_gΓ₁=

4

X

l=2

b¹_l +C⁵,

where b¹_l ∈ H_l. Now we want to remove the function b¹₂. This can be done with the ansatz

Γ2(x, t) = 1

|(x, t)|² 1 +ψ1(x, t) +ψ2(x, t)

if the unknown function ψ2 solves the equation Lψ2 = −b¹₂ and ∂tψ2(x,0) = 0.This is equivalent to

(−∆_S3 +

ψ¯=−¯b²₁ S₊³

∂_νψ¯= 0 ∂S₊³, which is not solvable unless Z

S₊³

¯b²₁dS= 0,

Since this identity is wrong in general we need a further Proposition to continue the expansion.

Proposition 30. Ifk∈ {2,3,4}, m∈N0andp∈H_k, then there existp₀,· · · , p_m+1∈H_k such that ∂tpi = 0 on ∂R⁴₊ and

L

m+1

X

i=0

p_ilog(|(x, t)|)ⁱ

!

=plog(|(x, t)|)^m.

Proof. We prove this statement by induction on m. Therefore we frequently use the formula

L(qlog(r)^m) =Lqlog(r)^m+m(2−2k)qlog(r)^m−1−m(m−1)qlog(r)^m−2 (E.8) for m≥0,q ∈H_k andr =|(x, t)|.

m= 0 :Let<e¯1,· · · ,e¯_l>=kern −∆_S³

+ +k(2−k)id

.Furthermore we write

¯ p=

l

X

j=1

hp,¯ e¯_ji_L2(S₊³)e¯_j+



p¯−

l

X

j=1

h¯p,¯e_ji_L2(S³₊)e¯_j



=: ¯p⁰+ ¯p⁰⁰. Then there existsq¯⁰⁰ such that

(−∆_S3

+q¯⁰⁰+k(2−k)¯q⁰⁰= ¯p⁰⁰ S₊³

∂νq¯⁰⁰= 0 ∂S₊³.

Letp, p⁰, p⁰⁰, q⁰⁰ ∈H_k be the homogeneous extensions of the previous functions then p=p⁰+p⁰⁰, L(p⁰) = 0, L(q⁰⁰) =p⁰⁰

and (E.8) implies

L

q⁰⁰+ 1

2−2kp⁰log(r)

=p⁰⁰+p⁰=p.

Hence the case m= 0 is proved.

m−1 7→ m. As above we write plog(r)^m = p⁰log(r)^m+p⁰⁰log(r)^m, where L(p⁰) = 0, L(q⁰⁰) =p⁰⁰ for some q⁰⁰∈H_k. Again (E.8) yields

L

q⁰⁰log(r) + p⁰

(m+ 1)(2−2k)log(r)^m+1

−p⁰log(r)^m−p⁰⁰log(r)^m

=m(4−2k)q⁰⁰log(r)^m−1−m(m+ 1)p⁰log(r)^m−1−m(m−1)q⁰⁰log(r)^m−2. By induction we can solve the last equation, which proves the Proposition.

Now we return to the expansion of the Green's function. Since ker(−∆_S3

+) = R, L(|(x, t)|²) = 0. The proof of the previous Proposition in case m = 0 implies the existence of p∈H2 and c∈Rsuch that

L p+c|(x, t)|²log(r)

=−b¹₂.

Moreover Z

S₊³

pdS= 0 and c=− 1 2|S₊³|

Z

S₊³

b²₁dS.

Furthermore, due to (A.1),

c= 1 2|S₊³|

Z

S₊³

b²₁dS=β|Π(a)|²

for some constantβ >0.We set ψ₂ :=p+c|(x, t)|²log(r)and observe

|(x, t)|⁴L_gΓ₂(x, t) =

4

X

l=3

b²_l +c₁|(x, t)|⁴log(r) +C⁵(1 + log(r)), c₁ ∈R

withb²_l ∈H_l. Finally, due to Proposition 30 we nd p₁, p₂ ∈H₃ andq₁, q₂, q₃∈H₄ such that

|(x, t)|⁴LgΓ4(x, t) =C⁵(1 + log(r)²), if we set

Γ₄(x, t) = 1

|(x, t)|² 1 +

4

X

i=1

ψ_i(x, t)

!

and

ψ₃ =p₁+p₂log(r), ψ₄ =q₁+q₂log(r) +q₃log(r)².

Furthermore, due to our construction ∂_tΓ₄(x,0) = 0. Finally on M, we dene Γ(x) = χρ(|ψ_a(x)|)Γ₄(ψa(x)) which is in C^∞(M \ {a}). More precisely, our construction and assumption (1.2) yield

(L_gΓ∈C^0,12(M),

BgΓ∈C^1,12(∂M) (E.9)

as well as Γ,|∇Γ| ∈ L¹(M). Using the Green formulas, we easily deduce the following Proposition.

Proposition 31. For ϕ∈C²(M) the following identity holds true:

2|S₊³|ϕ(a)

= Z

M

Γ(x)L_gϕdV_g+ Z

∂M

ΓB_gϕdσ_g− Z

M

L_gΓϕdV_g− Z

∂M

B_gΓϕdσ_g.

SinceQ(M, ∂M,[g])>0, the operator

(L_g, B_g) :C^2,α(M)→ C^0,α(M), C^1,α(∂M)

, u7→(L_gu, B_gu)

is an isomorphism. Due to (E.9), there exits exactly one function H_a ∈ C^2,12(M) such that

LgHa=−L_gΓ, BgHa=−B_gΓ.

We set

Ga(x) := Γ(x) +Ha(x) and observe

2|S₊³|ϕ(a) = Z

M

Ga(x)LgϕdVg+ Z

∂M

Ga(x)Bgϕdσg ∀ϕ∈C²(M), henceGa is the normalized Green function ata.

Summing up we have proved the following expansion of the normalized Green's function Ga in conformal Fermi-coordinates.

Proposition 32. Let (M, g) be a four-dimensional compact Riemannian manifold with boundary and positive Sobolev-quotient. Ifg is a metric such that (E.1) holds, then

G_a(x) = Γ(ψ_a(x)) +H_a(x), where Γ is singular at 0 andH_a∈C^2,12(M). More precisely:

Ga(ψ_a⁻¹(x, t)) = 1

|(x, t)|² 1 +ψ1(x, t) +ψ2(x, t)

+c|Π(a)|²log(r) +Ha(a) +O(rlog(r)), where ψi are homogeneous functions of degree i such that

Z

S₊³

ψidS= 0 andΠ(a) is the umbilicity tensor at a.

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