# A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Manifolds with Boundary

Authors: Jian Wang; Yong Wang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We give a brute-force proof of the Kastler-Kalau-Walze type theorem for 7-dimensional manifolds with boundary.


---

## Body

## 1. Introduction

The noncommutative residue found in [1, 2] plays a prominent role in noncommutative geometry. For one-dimensional manifolds, the noncommutative residue was discovered by Adler [3] in connection with geometric aspects of nonlinear partial differential equations. For arbitrary closed compact n-dimensional manifolds, the noncommutative residue was introduced by Wodzicki in [2] using the theory of zeta functions of elliptic pseudodifferential operators. In [4], Connes used the noncommutative residue to derive a conformal 4-dimensional Polyakov action analogy. Furthermore, Connes made a challenging observation that the noncommutative residue of the square of the inverse of the Dirac operator was proportional to the Einstein-Hilbert action in [5]. Let s be the scalar curvature and let Wres denote the noncommutative residue. Then, the Kastler-Kalau-Walze theorem gives an operator-theoretic explanation of the gravitational action and says that, for a 4-dimensional closed spin manifold, there exists a constant c
0, such that
(1)
Wres
(
D
-
2
)
=
c
0
∫
M
s
d
vo
l
M
.
In [6], Kastler gave a brute-force proof of this theorem. In [7], Kalau and Walze proved this theorem in the normal coordinates system simultaneously. And then, Ackermann proved that the Wodzicki residue Wres
(
D
-
2
) in turn is essentially the second coefficient of the heat kernel expansion of D
2 in [8].On the other hand, Fedosov et al. defined a noncommutative residue on Boutet de Monvel’s algebra and proved that it was a unique continuous trace in [9]. In [10], Schrohe gave the relation between the Dixmier trace and the noncommutative residue for manifolds with boundary. For an oriented spin manifold M with boundary ∂
M, by the composition formula in Boutet de Monvel’s algebra and the definition of   Wres
~ [11], Wres
~
[
(
π
+
D
-
1
)
2
] should be the sum of two terms from interior and boundary of M, where π
+
D
-
1 is an element in Boutet de Monvel’s algebra  [11]. It is well known that the gravitational action for manifolds with boundary is also the sum of two terms from interior and boundary of M [12]. Considering the Kastler-Kalau-Walze theorem for manifolds without boundary, then the term from interior is proportional to gravitational action from interior, so it is natural to hope to get the gravitational action for manifolds with boundary by computing Wres
~
[
(
π
+
D
-
1
)
2
]. Based on the motivation, Wang [13] proved a Kastler-Kalau-Walze type theorem for 4-dimensional spin manifolds with boundary
(2)
Wres
~
[
(
π
+
D
-
1
)
2
]
=
-
Ω
3
3
∫
M
s
d
vo
l
M
,
where Ω
3 is the canonical volume of S
3. Furthermore, Wang [14] found a Kastler-Kalau-Walze type theorem for higher dimensional manifolds with boundary and generalized the definition of lower-dimensional volumes in [15] to manifolds with boundary. For 5-dimensional spin manifolds with boundary [14], Wang got
(3)
Wres
~
[
(
π
+
D
-
2
)
2
]
=
π
i
2
Ω
2
vo
l
∂
M
,
and for 6-dimensional spin manifolds with boundary,
(4)
Wres
~
[
(
π
+
D
-
2
)
2
]
=
-
5
Ω
5
3
∫
M
‍
s
d
vo
l
M
.In order to get the boundary term, we computed the lower-dimensional volumeVo
l
6
(
1,3
) for 6-dimensional spin manifolds with boundary associated with D
-
1 and D
-
3 in [16] and obtained the volume with the boundary term
(5)
Wres
~
[
π
+
D
-
1
·
π
+
D
-
3
]
=
-
5
Ω
4
3
∫
M
‍
s
d
vo
l
M
+
π
Ω
3
∫
∂
M
‍
K
d
vo
l
∂
M
,
where K is the extrinsic curvature.In [17], Wang proved a Kastler-Kalau-Walze type theorem for general form perturbations and the conformal perturbations of  Dirac operators for compact manifolds with or without boundary. Let M be 4-dimensional compact manifolds with the boundary ∂
M and let Ψ be a general differential form on M, from Theorem 10 in [17]; then
(6)
Wres
~
[
(
π
+
(
D
~
Ψ
)
-
1
)
2
]
=
4
π
2
∫
M
‍
T
r
[
-
s
12
-
1
2
c
(
Ψ
)
c
(
e
i
)
c
(
Ψ
)
c
(
e
i
)
k
k
k
k
k
k
k
k
k
k
k
k
k
+
(
c
(
Ψ
)
)
2
-
s
12
]
d
vo
l
M
.Recently, we computedWres
~
[
π
+
D
-
p
1
·
π
+
D
-
p
2
] for n-dimensional spin manifolds with boundary in case of n
-
p
1
-
p
2
≤
2. In the present paper, we will restrict our attention to the case of n
-
p
1
-
p
2
=
3. We compute Wres
~
[
(
π
+
D
-
2
)
2
] for 7-dimensional manifolds with boundary. Our main result is as follows.Main Theorem. The following identity for 7-dimensional manifolds with boundary holds:
(7)
Wres
~
[
(
π
+
D
-
2
)
2
]
=
π
4
48
∫
∂
M
‍
(
-
47
2
K
2
-
49
s
M
|
∂
M
-
77
4
s
∂
M
)
d
vo
l
∂
M
,
where s
M and s
∂
M are, respectively, scalar curvatures on M and ∂
M. Compared with the previous results, up to the extrinsic curvature, the scalar curvature on ∂
M and the scalar curvature on M appear in the boundary term. This case essentially makes the whole calculations more difficult, and the boundary term is the sum of fifteen terms. As in computations of the boundary term, we will consider some new traces of multiplication of Clifford elements. And the inverse 4-order symbol of the Dirac operator and higher derivatives of -1-order and -3-order symbols of the Dirac operators will be extensively used.This paper is organized as follows. In Section2, we define lower-dimensional volumes of compact Riemannian manifolds with boundary. In Section 3, for 7-dimensional spin manifolds with boundary and the associated Dirac operators, we compute Wres
~
[
(
π
+
D
-
2
)
2
] and get a Kastler-Kalau-Walze type theorem in this case.


## 2. Lower-Dimensional Volumes of Spin Manifolds with Boundary

In this section, we consider ann-dimensional oriented Riemannian manifold (
M
,
g
M
) with boundary   ∂
M equipped with a fixed spin structure. We assume that the metric g
M on M has the following form near the boundary:
(8)
g
M
=
1
h
(
x
n
)
g
∂
M
+
d
x
n
2
,
where g
∂
M is the metric on ∂
M. Let U
⊂
M be a collar neighborhood of  ∂
M which is diffeomorphic ∂
M
×
[
0,1
). By the definition of h
(
x
n
)
∈
C
∞
(
[
0,1
)
) and h
(
x
n
)
>
0, there exists h
~
∈
C
∞
(
(
-
ɛ
,
1
)
) such that h
~
|
[
0,1
)
=
h and h
~
>
0 for some sufficiently small ɛ
>
0. Then, there exists a metric g
^ on M
^
=
M
⋃
∂
M
‍
∂
M
×
(
-
ɛ
,
0
] which has the form on U
⋃
∂
M
‍
∂
M
×
(
-
ɛ
,
0
]
(9)
g
^
=
1
h
~
(
x
n
)
g
∂
M
+
d
x
n
2
,
such that g
^
|
M
=
g. We fix a metric g
^ on the M
^ such that g
^
|
M
=
g.Let us give the expression of Dirac operators near the boundary. SetE
~
n
=
∂
/
∂
x
n and E
~
j
=
h
(
x
n
)
E
j
(
1
≤
j
≤
n
-
1
), where {
E
1
,
…
,
E
n
-
1
} are orthonormal basis of T
∂
M. Let ∇
L denote the Levi-Civita connection about g
M. In the local coordinates {
x
i
;
1
≤
i
≤
n
} and the fixed orthonormal frame {
E
~
1
,
…
,
E
~
n
}, the connection matrix (
ω
s
,
t
) is defined by
(10)
∇
L
(
E
~
1
,
…
,
E
~
n
)
t
=
(
ω
s
,
t
)
(
E
~
1
,
…
,
E
~
n
)
t
.
The Dirac operator is defined by
(11)
D
=
∑
j
=
1
n
‍
c
(
E
j
~
)
[
E
j
~
+
1
4
∑
s
,
t
‍
ω
s
,
t
(
E
j
~
)
c
(
E
s
~
)
c
(
E
t
~
)
]
.
By Lemma 6.1 in [18] and Propositions 2.2 and 2.4 in [19], we have the following lemma.Lemma 1.
Letf
=
(
1
/
h
) and M
~
=
I
×
f
M be a Riemannian manifold with the metric g
f
=
d
x
n
2
+
f
2
(
x
n
)
g. For vector fields X and Y in L
(
M
), then
(12)
(
1
)
∇
~
∂
x
n
∂
x
n
=
0
;
(
2
)
∇
~
∂
x
n
X
=
∇
~
X
∂
x
n
=
(
lnf
)
′
X
;
(
3
)
∇
X
Y
=
∇
X
M
Y
-
g
(
X
,
Y
)
f
grad
(
f
)
.DenoteA
j
s
t
=
2
〈
∇
E
j
L
,
∂
M
E
s
,
E
t
〉; then we obtain the following lemma.Lemma 2.
The following identity holds:(13)
(
1
)
〈
∇
E
~
i
L
∂
x
n
,
E
~
j
〉
=
-
h
′
2
h
;
(
2
)
〈
∇
E
~
i
L
E
~
j
,
∂
x
n
〉
=
h
′
2
h
;
(
3
)
〈
∇
E
~
j
L
E
~
s
,
E
~
t
〉
=
h
2
A
j
s
t
.
Others are zeros.By Lemma2, we have the following definition.Definition 3.
The following identity holds in the coordinates near the boundary:(14)
D
=
∑
β
=
1
n
‍
c
(
E
β
~
)
E
β
~
-
h
′
h
c
(
d
x
n
)
+
h
8
∑
s
,
α
,
β
<
n
‍
A
β
s
α
c
(
E
β
~
)
c
(
E
s
~
)
c
(
E
α
~
)
.To define the lower-dimensional volume, some basic facts and formulae about Boutet de Monvel’s calculus which can be found in Section 2 in [11] are needed.Denote by(15)
F
:
L
2
(
R
t
)
⟶
L
2
(
R
v
)
;
F
(
u
)
(
v
)
=
∫
e
-
i
v
t
u
(
t
)
d
t
the Fourier transformation and Φ
(
R
+
¯
)
=
r
+
Φ
(
R
) (similarly, define Φ
(
R
-
¯
)
), where Φ
(
R
) denotes the Schwartz space and
(16)
r
+
:
C
∞
(
R
)
⟶
C
∞
(
R
+
¯
)
;
f
⟶
f
∣
R
+
¯
;
R
+
¯
=
{
x
≥
0
;
x
∈
R
}
.
We define H
+
=
F
(
Φ
(
R
+
¯
)
) and H
0
-
=
F
(
Φ
(
R
-
¯
)
) which are orthogonal to each other. We have the following property: h
∈
H
+
(
H
0
-
) iff h
∈
C
∞
(
R
) which has an analytic extension to the lower (upper) complex half-plane {
Im
⁡
ξ
<
0
}
(
{
Im
⁡
ξ
>
0
}
) such that, for all nonnegative integers l,
(17)
d
l
h
d
ξ
l
(
ξ
)
~
∑
k
=
1
∞
‍
d
l
d
ξ
l
(
c
k
ξ
k
)
as |
ξ
|
→
+
∞, Im
⁡
ξ
≤
0
(
Im
⁡
ξ
≥
0
).LetH
′ be the space of all polynomials and let H
-
=
H
0
-
⊕
H
′; H
=
H
+
⊕
H
-. Denote by π
+
(
π
-
), respectively, the projection on H
+
(
H
-
). For calculations, we take H
=
H
~
=
{rational functions having no poles on the real axis} (H
~ is a dense set in the topology of H). Then, on H
~,
(18)
π
+
h
(
ξ
0
)
=
1
2
π
i
lim
⁡
u
→
0
-
∫
Γ
+
h
(
ξ
)
ξ
0
+
i
u
-
ξ
d
ξ
,
where Γ
+ is a Jordan close curve which included Im
⁡
ξ
>
0 surrounding all the singularities of h in the upper half-plane and ξ
0
∈
R. Similarly, define π
′ on H
~:
(19)
π
′
h
=
1
2
π
∫
Γ
+
h
(
ξ
)
d
ξ
.
So, π
′
(
H
-
)
=
0. For h
∈
H
⋂
‍
L
1
(
R
), π
′
h
=
(
1
/
2
π
)
∫
R
‍
h
(
v
)
d
v and for h
∈
H
+
⋂
‍
L
1
(
R
), π
′
h
=
0.LetM be an n-dimensional compact oriented manifold with boundary ∂
M. Denote by B Boutet de Monvel’s algebra; we recall the main theorem in [9].Theorem 4 (Fedosov-Golse-Leichtnam-Schrohe).
LetX and ∂
X be connected, let dim
⁡
X
=
n
≥
3, and let A
=
(
π
+
P
+
G
K
T
S
)
∈
B, and denote by p, b, and s the local symbols of   P
,
G, and S, respectively. Define
(20)
Wres
~
(
A
)
=
∫
X
∫
S
tr
E
[
p
-
n
(
x
,
ξ
)
]
σ
(
ξ
)
d
x
+
2
π
∫
∂
X
∫
S
′
{
tr
E
[
(
tr
⁡
b
-
n
)
(
x
′
,
ξ
′
)
]
k
k
k
k
k
k
k
k
k
k
k
k
+
tr
F
[
s
1
-
n
(
x
′
,
ξ
′
)
]
}
σ
(
ξ
′
)
d
x
′
.
Then, (a) Wres
~
(
[
A
,
B
]
)
=
0, for any A
,
B
∈
B; (b) it is a unique continuous trace on B
/
B
-
∞.Letp
1 and p
2 be nonnegative integers and let p
1
+
p
2
≤
n. Then, by Section 2.1 of   [13], we have the following definition.Definition 5.
Lower-dimensional volumes of spin manifolds with boundary are defined by(21)
Vo
l
n
(
p
1
,
p
2
)
M
:
=
Wres
~
[
π
+
D
-
p
1
·
π
+
D
-
p
2
]
.Denote byσ
l
(
A
) the l-order symbol of an operator A. An application of (2.1.4) in [11] shows that
(22)
Wres
~
[
π
+
D
-
p
1
·
π
+
D
-
p
2
]
=
∫
M
‍
∫
|
ξ
|
=
1
‍
trace
S
(
TM
)
[
σ
-
n
(
D
-
p
1
-
p
2
)
]
σ
(
ξ
)
d
x
+
∫
∂
M
‍
Φ
,
where
(23)
Φ
=
∫
|
ξ
′
|
=
1
∫
-
∞
+
∞
∑
j
,
k
=
0
∞
∑
(
-
i
)
|
α
|
+
j
+
k
+
1
α
!
(
j
+
k
+
1
)
!
trace
S
(
TM
)
kkkkkkkkkkkkkkkkk
×
[
∂
x
n
j
∂
ξ
′
α
∂
ξ
n
k
σ
r
+
(
D
-
p
1
)
(
x
′
,
0
,
ξ
′
,
ξ
n
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
∂
x
′
α
∂
ξ
n
j
+
1
∂
x
n
k
σ
l
(
D
-
p
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
′
,
0
,
ξ
′
,
ξ
n
)
∂
x
n
j
∂
ξ
′
α
∂
ξ
n
k
σ
r
+
(
D
-
p
1
)
]
d
ξ
n
σ
(
ξ
′
)
d
x
′
,
and the sum is taken over r
-
k
+
|
α
|
+
l
-
j
-
1
=
-
n, r
≤
-
p
1, l
≤
-
p
2.


## 3. A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Spin Manifolds with Boundary

In this section, we compute the lower-dimensional volume for 7-dimensional compact manifolds with boundary and get a Kastler-Kalau-Walze type formula in this case. From now on, we always assume thatM carries a spin structure so that the spinor bundle and the Dirac operator are defined on M.The following proposition is the key of  the computation of lower-dimensional volumes of  spin manifolds with boundary.Proposition 6 (see [14]).
The following identity holds:(24)
(
1
)
w
h
e
n
p
1
+
p
2
=
n
,
t
h
e
n
,
Vol
n
(
p
1
,
p
2
)
M
=
c
0
Vo
l
M
;
(
2
)
when
p
1
+
p
2
≡
n
mod
⁡
1
,
Vol
n
(
p
1
,
p
2
)
M
=
∫
∂
M
‍
Φ
.Nextly, for 7-dimensional spin manifolds with boundary, we computeVo
l
7
(
2,2
). By Proposition 6, for 7-dimensional compact manifolds with boundary, we have
(25)
Wres
~
[
(
π
+
D
-
2
)
2
]
=
∫
∂
M
‍
Φ
.Recall the Dirac operatorD of Definition 3. Write
(26)
k
k
l
D
x
α
=
(
-
-
1
)
|
α
|
∂
x
α
;
σ
(
D
2
)
=
p
2
+
p
1
+
p
0
;
σ
(
D
-
2
)
=
∑
j
=
1
∞
‍
q
-
j
.
By the composition formula of pseudodifferential operators, then we have
(27)
1
=
σ
(
D
2
·
D
-
2
)
=
∑
α
‍
1
α
!
∂
ξ
α
[
σ
(
D
2
)
]
D
x
α
[
σ
(
D
-
2
)
]
=
(
p
2
+
p
1
+
p
0
)
(
q
-
2
+
q
-
3
+
q
-
4
+
⋯
)
+
∑
j
‍
(
∂
ξ
j
p
2
+
∂
ξ
j
p
1
+
∂
ξ
j
p
0
)
kkkk
×
(
D
x
j
q
-
2
+
D
x
j
q
-
3
+
D
x
j
q
-
4
+
⋯
)
+
∑
i
,
j
‍
∂
ξ
i
∂
ξ
j
(
p
2
+
p
1
+
p
0
)
D
x
i
D
x
j
kkkkk
×
(
q
-
2
+
q
-
3
+
q
-
4
+
⋯
)
=
p
2
q
-
2
+
(
p
2
q
-
3
+
p
1
q
-
2
+
∑
j
‍
∂
ξ
j
p
2
D
x
j
q
-
2
)
+
(
p
2
q
-
4
+
p
1
q
-
3
+
p
0
q
-
2
+
∑
j
‍
∂
ξ
j
p
2
D
x
j
q
-
3
kkkk
k
+
∑
j
‍
∂
ξ
j
p
1
D
x
j
q
-
2
+
∑
i
,
j
=
1
n
‍
∂
ξ
i
∂
ξ
j
p
2
D
x
i
D
x
j
q
-
2
)
+
⋯
.
Thus, we get
(28)
q
-
2
=
p
2
-
2
;
q
-
3
=
-
p
2
-
1
[
p
1
q
-
2
-
i
∑
j
=
1
n
‍
∂
ξ
j
p
2
∂
x
j
q
-
2
]
;
q
-
4
=
-
p
2
-
1
[
p
0
q
-
2
+
p
1
q
-
3
-
i
∑
j
=
1
n
‍
∂
ξ
j
p
2
∂
x
j
q
-
3
hhhhhhhhhhh
-
i
∑
j
=
1
n
‍
∂
ξ
j
p
1
∂
x
j
q
-
2
hhhhhhhhhhh
-
1
2
∑
i
,
j
=
1
n
‍
∂
ξ
i
∂
ξ
j
p
2
∂
x
i
∂
x
j
q
-
2
]
.DefineΓ
k
=
∑
i
,
j
<
n
‍
∑
l
<
n
‍
g
i
j
g
l
k
〈
∇
∂
i
L
∂
j
,
∂
l
〉
+
∑
l
<
n
‍
g
l
k
〈
∇
∂
n
L
∂
n
,
∂
l
〉  and  σ
k
(
x
0
=
(
1
/
4
)
∑
s
,
t
‍
ω
s
,
t
(
∂
i
)
c
(
e
s
~
)
c
(
e
t
~
). By Theorem 1 in [6] and Lemma 2.1 in [13], we have the following.Lemma 7.
Consider the symbol of the Dirac operator(29)
σ
-
2
(
D
-
2
)
=
|
ξ
|
-
2
;
σ
-
3
(
D
-
2
)
=
-
-
1
|
ξ
|
-
4
ξ
k
(
Γ
k
-
2
σ
k
)
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
j
g
α
β
;
σ
-
4
(
D
-
2
)
=
-
|
ξ
|
-
6
ξ
μ
ξ
ν
(
Γ
μ
-
2
σ
μ
)
(
Γ
ν
-
2
σ
ν
)
+
2
|
ξ
|
-
8
ξ
μ
ξ
ν
ξ
α
ξ
β
(
Γ
ν
-
2
σ
ν
)
∂
μ
x
g
α
β
+
|
ξ
|
-
4
(
∂
x
μ
σ
μ
+
σ
μ
σ
μ
-
Γ
μ
σ
μ
)
-
1
4
|
ξ
|
-
4
s
(
x
)
-
2
|
ξ
|
-
6
ξ
μ
ξ
ν
∂
μ
x
(
Γ
ν
-
2
σ
ν
)
+
12
|
ξ
|
-
10
ξ
μ
ξ
ν
ξ
α
ξ
β
ξ
γ
ξ
δ
∂
μ
x
g
α
β
∂
ν
x
g
γ
δ
-
4
|
ξ
|
-
8
ξ
μ
ξ
α
ξ
γ
ξ
δ
∂
μ
x
g
ν
α
∂
ν
x
g
γ
δ
-
4
|
ξ
|
-
8
ξ
μ
ξ
ν
ξ
γ
ξ
δ
∂
μ
ν
x
g
γ
δ
+
|
ξ
|
-
6
ξ
α
ξ
β
(
Γ
μ
-
2
σ
μ
)
∂
μ
x
g
α
β
-
|
ξ
|
-
6
ξ
α
ξ
β
g
μ
ν
∂
μ
ν
x
g
α
β
+
2
|
ξ
|
-
8
ξ
α
ξ
β
ξ
γ
ξ
δ
g
μ
ν
∂
μ
x
g
α
β
∂
ν
x
g
γ
δ
,
where
(30)
Γ
k
(
x
0
)
=
{
0
,
i
f
k
<
n
;
3
h
′
(
0
)
ξ
n
,
i
f
k
=
n
,
σ
k
(
x
0
)
=
{
1
4
h
′
(
0
)
∑
k
<
n
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
,
i
f
k
<
n
;
0
,
i
f
k
=
n
.SinceΦ is a global form on ∂
M, so for any  fixed point x
0
∈
∂
M, we can choose the normal coordinates U of x
0 in ∂
M    (not in M) and compute Φ
(
x
0
) in the coordinates U
~
=
U
×
[
0,1
) and the metric (
1
/
h
(
x
n
)
)
g
∂
M
+
d
x
n
2. The dual metric of g
M on U
~ is h
(
x
n
)
g
∂
M
+
d
x
n
2. Write g
i
j
M
=
g
M
(
∂
/
∂
x
i
,
∂
/
∂
x
j
) and g
M
i
j
=
g
M
(
d
x
i
,
d
x
j
); then,
(31)
[
g
i
,
j
M
]
=
[
1
h
(
x
n
)
[
g
i
,
j
∂
M
]
0
0
1
]
;
[
g
M
i
,
j
]
=
[
h
(
x
n
)
[
g
∂
M
i
,
j
]
0
0
1
]
,
∂
x
s
g
i
j
∂
M
(
x
0
)
=
0
,
1
≤
i
,
j
≤
n
-
1
;
g
i
,
j
M
(
x
0
)
=
δ
i
j
.Let{
E
1
,
…
,
E
n
-
1
} be an orthonormal frame field in U about g
∂
M which is parallel along geodesics and E
i
=
(
∂
/
∂
x
i
)
(
x
0
); then, {
E
1
~
=
h
(
x
n
)
E
1
,
…
,
E
n
-
1
~
=
h
(
x
n
)
E
n
-
1
,
E
n
~
=
d
x
n
} is the orthonormal frame field in U
~ about g
M. Locally, S
(
T
M
)
∣
U
~
≅
U
~
×
∧
C
*
(
n
/
2
). Let {
f
1
,
…
,
f
n
} be the orthonormal basis of ∧
C
*
(
n
/
2
). Take a spin frame field σ
:
U
~
→
Spin
(
M
) such that π
σ
=
{
E
1
~
,
…
,
E
n
~
}, where π
:
Spin
(
M
)
→
O
(
M
) is a double covering; then, {
[
σ
,
f
i
]
,
1
≤
i
≤
6
} is an orthonormal frame of S
(
T
M
)
∣
U
~. In the following, since the global form Φ is independent of the choice of the local frame, we can compute t
r
S
(
T
M
) in the frame {
[
σ
,
f
i
]
,
1
≤
i
≤
6
}. Let {
E
^
1
,
…
,
E
^
n
} be the canonical basis of R
n and let c
(
E
^
i
)
∈
cl
⁡
C
(
n
)
≅
Hom
(
∧
C
*
(
n
/
2
)
,
∧
C
*
(
n
/
2
)
) be the Clifford action. By [13], then
(32)
c
(
E
i
~
)
=
[
(
σ
,
c
(
E
^
i
)
)
]
;
c
(
E
i
~
)
[
(
σ
,
f
i
)
]
=
[
σ
,
(
c
(
E
^
i
)
)
f
i
]
;
∂
∂
x
i
=
[
(
σ
,
∂
∂
x
i
)
]
;
then, we have (
∂
/
∂
x
i
)
c
(
E
i
~
)
=
0 in the above frame. By Lemma 2.2 in [13], we have the following.Lemma 8.
With the metricg
M on M near the boundary,
(33)
∂
x
j
(
|
ξ
|
g
M
2
)
(
x
0
)
=
{
0
,
i
f
j
<
n
;
h
′
(
0
)
|
ξ
′
|
g
∂
M
2
,
i
f
j
=
n
,
∂
x
j
[
c
(
ξ
)
]
(
x
0
)
=
{
0
,
i
f
j
<
n
;
∂
x
n
(
c
(
ξ
′
)
)
(
x
0
)
,
i
f
j
=
n
,
where ξ
=
ξ
′
+
ξ
n
d
x
n.Then, the following lemma is introduced.Lemma 9.
The following identity holds:(34)
∂
x
i
[
A
β
s
α
]
(
x
0
)
=
{
R
β
i
s
α
∂
M
(
x
0
)
,
i
f
i
<
n
;
0
,
i
f
i
=
n
.Proof.
From Lemma 5.7 in [20], we have
(35)
A
β
s
α
=
R
β
l
s
α
x
l
+
O
(
|
x
|
2
)
.
Then, we obtain ∂
x
i
[
A
β
s
α
]
(
x
0
)
=
R
β
i
s
α
∂
M
(
x
0
).Lemma 10.
Letg
M be the metric on 7-dimensional spin manifolds M near the boundary; then,
(36)
∂
x
γ
Γ
k
(
x
0
)
|
|
ξ
′
|
=
1
=
{
5
6
∑
i
<
n
‍
R
i
γ
i
k
∂
M
(
x
0
)
,
i
f
γ
<
n
,
k
<
n
;
0
,
i
f
γ
<
n
,
k
=
n
;
0
,
i
f
γ
=
n
,
k
<
n
;
3
h
′′
(
0
)
-
9
2
(
h
′
(
0
)
)
2
,
i
f
γ
=
n
,
k
=
n
,
∂
x
γ
σ
k
(
x
0
)
|
|
ξ
′
|
=
1
=
{
1
8
∑
s
≠
t
<
n
‍
R
k
γ
s
t
∂
M
(
x
0
)
c
(
E
s
~
)
c
(
E
t
~
)
,
i
f
γ
<
n
,
k
<
n
;
0
,
i
f
γ
<
n
,
k
=
n
;
∑
t
<
n
‍
(
3
8
(
h
′
(
0
)
)
2
-
1
4
h
′′
(
0
)
)
×
c
(
E
n
~
)
c
(
E
t
~
)
,
i
f
γ
=
n
,
k
<
n
;
1
8
∑
t
<
n
‍
(
(
h
′
(
0
)
)
2
-
h
′′
(
0
)
)
×
c
(
E
s
~
)
c
(
E
t
~
)
.
i
f
γ
=
n
,
k
=
n
.Proof.
From Lemma 2.3 in [13], we have
(37)
Γ
k
=
∑
i
,
j
<
n
‍
∑
l
<
n
‍
g
i
j
g
l
k
〈
∇
∂
i
L
∂
j
,
∂
l
〉
+
∑
l
<
n
‍
g
l
k
〈
∇
∂
n
L
∂
n
,
∂
l
〉
.
Then,
(38)
∂
x
γ
Γ
k
=
∑
i
,
j
<
n
‍
∑
l
<
n
‍
∂
x
γ
(
g
i
j
)
g
l
k
〈
∇
∂
i
L
∂
j
,
∂
l
〉
+
∑
i
,
j
<
n
‍
∑
l
<
n
‍
g
i
j
∂
x
γ
(
g
l
k
)
〈
∇
∂
i
L
∂
j
,
∂
l
〉
+
∑
i
,
j
<
n
‍
∑
l
<
n
‍
g
i
j
g
l
k
∂
x
γ
(
〈
∇
∂
i
L
∂
j
,
∂
l
〉
)
+
∑
l
<
n
‍
∂
x
γ
(
g
l
k
)
〈
∇
∂
n
L
∂
n
,
∂
l
〉
+
∑
l
<
n
‍
g
l
k
∂
x
γ
(
〈
∇
∂
n
L
∂
n
,
∂
l
〉
)
.
Let H
n
n
=
〈
∂
n
,
E
n
~
〉
=
1, H
n
j
=
〈
∂
n
,
E
j
~
〉
=
0, and H
i
j
=
〈
∂
j
,
E
i
~
〉
M
=
(
1
/
h
(
x
n
)
)
〈
∂
j
,
E
i
〉
∂
M. Then,
(39)
〈
∇
∂
i
L
∂
j
,
∂
l
〉
=
〈
∇
∂
i
L
(
〈
∂
j
,
E
j
1
~
〉
E
j
1
~
)
,
〈
∂
l
,
E
l
1
~
〉
E
l
1
~
〉
=
∂
i
(
〈
∂
j
,
E
j
1
~
〉
)
〈
∂
l
,
E
l
1
~
〉
〈
E
j
1
~
,
E
l
1
~
〉
+
〈
∂
l
,
E
l
1
~
〉
〈
∂
j
,
E
j
1
~
〉
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
=
∂
i
(
H
j
j
1
)
〈
∂
l
,
E
l
1
~
〉
δ
j
1
l
1
+
〈
∂
l
,
E
l
1
~
〉
〈
∂
j
,
E
j
1
~
〉
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
.
When i
<
n, j
<
n,
(40)
〈
∇
∂
i
L
∂
j
,
∂
l
〉
(
x
0
)
=
∂
i
(
H
j
j
1
)
〈
∂
l
,
E
l
1
~
〉
δ
j
1
l
1
(
x
0
)
+
〈
∂
l
,
E
l
1
~
〉
〈
∂
j
,
E
j
1
~
〉
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
(
x
0
)
=
δ
l
1
l
δ
j
1
j
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
(
x
0
)
=
h
(
x
0
)
2
A
i
j
l
(
x
0
)
=
0
,
〈
∇
∂
n
L
∂
n
,
∂
l
〉
(
x
0
)
=
0
.
Then
(41)
∂
x
γ
Γ
k
(
x
0
)
=
∑
i
,
l
<
n
‍
∂
x
γ
(
〈
∇
∂
i
L
∂
j
,
∂
l
〉
)
(
x
0
)
+
∑
l
<
n
‍
∂
x
γ
(
〈
∇
∂
n
L
∂
n
,
∂
l
〉
)
(
x
0
)
=
∂
γ
∂
i
(
H
j
j
1
)
〈
∂
l
,
E
l
1
~
〉
+
∂
i
(
H
j
j
1
)
∂
γ
(
〈
∂
l
,
E
l
1
~
〉
)
(
x
0
)
+
(
∂
γ
(
〈
∂
l
,
E
l
1
~
〉
)
〈
∂
j
,
E
j
1
~
〉
+
〈
∂
l
,
E
l
1
~
〉
∂
γ
(
〈
∂
j
,
E
j
1
~
〉
)
)
×
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
(
x
0
)
+
〈
∂
l
,
E
l
1
~
〉
〈
∂
j
,
E
j
1
~
〉
∂
γ
(
〈
∇
∂
i
L
E
j
1
~
,
E
l
1
~
〉
)
(
x
0
)
+
∂
γ
∂
i
(
H
n
n
1
)
〈
∂
l
,
E
l
1
~
〉
+
∂
i
(
H
n
n
1
)
∂
γ
(
〈
∂
l
,
E
l
1
~
〉
)
(
x
0
)
+
(
∂
γ
(
〈
∂
l
,
E
l
1
~
〉
)
〈
∂
n
,
E
n
1
~
〉
+
〈
∂
l
,
E
l
1
~
〉
∂
γ
(
〈
∂
n
,
E
n
1
~
〉
)
)
×
〈
∇
∂
n
L
E
n
1
~
,
E
l
1
~
〉
(
x
0
)
+
〈
∂
l
,
E
l
1
~
〉
〈
∂
n
,
E
n
1
~
〉
∂
γ
(
〈
∇
∂
n
L
E
n
1
~
,
E
l
1
~
〉
)
(
x
0
)
=
∂
γ
∂
i
(
H
j
l
)
(
x
0
)
+
∂
γ
(
〈
∇
∂
i
L
E
j
~
,
E
l
~
〉
)
(
x
0
)
=
-
1
3
∑
i
l
<
n
‍
R
i
γ
l
i
∂
M
(
x
0
)
+
∂
γ
(
h
(
x
n
)
2
A
i
j
l
)
(
x
0
)
=
5
6
∑
i
<
n
‍
R
i
γ
i
k
∂
M
(
x
0
)
.
Similarly, when γ
<
n, k
=
n, or γ
=
n, k
<
n, ∂
x
γ
Γ
k
(
x
0
)
=
0. When γ
=
k
=
n, ∂
x
γ
Γ
k
(
x
0
)
=
3
h
′′
(
0
)
-
(
9
/
2
)
(
h
′
(
0
)
)
2.
On the other hand, from definitions (10) and (11), then
(42)
σ
k
(
x
0
)
=
1
4
∑
s
,
t
‍
ω
s
,
t
(
∂
i
)
c
(
E
s
~
)
c
(
E
t
~
)
.
When γ
<
n, k
<
n,
(43)
∂
x
γ
σ
k
(
x
0
)
=
∂
x
γ
(
1
4
∑
s
,
t
‍
ω
s
,
t
(
∂
k
)
c
(
E
s
~
)
c
(
E
t
~
)
)
(
x
0
)
=
1
4
∂
x
γ
(
∑
s
,
t
‍
ω
s
,
t
(
∂
k
)
)
(
x
0
)
c
(
E
s
~
)
c
(
E
t
~
)
=
1
4
(
∑
s
,
t
<
n
‍
∂
x
γ
ω
s
,
t
(
∂
k
)
+
∑
s
=
n
,
t
<
n
‍
∂
x
γ
ω
s
,
t
(
∂
k
)
h
h
h
h
h
h
h
+
∑
s
<
n
,
t
=
n
‍
∂
x
γ
ω
s
,
t
(
∂
k
)
)
(
x
0
)
c
(
E
s
~
)
c
(
E
t
~
)
=
1
4
(
∑
s
,
t
<
n
‍
∂
x
γ
〈
∇
∂
x
k
L
E
s
~
,
E
t
~
〉
c
(
E
s
~
)
c
(
E
t
~
)
h
h
h
h
h
h
h
+
∑
s
=
n
,
t
<
n
‍
∂
x
γ
〈
∇
∂
x
k
L
E
s
~
,
E
t
~
〉
c
(
E
s
~
)
c
(
E
t
~
)
h
h
h
h
h
h
h
+
∑
s
<
n
,
t
=
n
‍
∂
x
γ
〈
∇
∂
x
k
L
E
s
~
,
E
t
~
〉
c
(
E
s
~
)
c
(
E
t
~
)
)
(
x
0
)
=
1
4
∑
s
,
t
<
n
‍
∂
γ
(
h
(
x
n
)
2
A
k
s
t
)
(
x
0
)
=
1
8
∑
s
≠
t
<
n
‍
R
k
γ
s
t
∂
M
(
x
0
)
c
(
E
s
~
)
c
(
E
t
~
)
.
Similarly, when γ
=
n, k
<
n, ∂
x
γ
σ
k
(
x
0
)
=
∑
t
<
n
(
(
3
/
8
)
(
h
′
(
0
)
)
2
-
(
1
/
4
)
h
′′
(
0
)
)
c
(
E
n
~
)
c
(
E
t
~
). When γ
<
n, k
=
n, ∂
x
γ
σ
k
(
x
0
)
=
0. When γ
=
k
=
n, ∂
x
γ
σ
k
(
x
0
)
=
(
1
/
8
)
∑
t
<
n
(
(
h
′
(
0
)
)
2
-
h
′′
(
0
)
)
c
(
E
s
~
)
c
(
E
t
~
)
.Lemma 11.
Whenγ
<
n,
(44)
∂
x
γ
(
σ
-
3
(
D
-
2
)
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
5
i
6
ξ
k
(
1
+
ξ
n
2
)
2
∑
i
<
n
‍
R
i
γ
i
k
∂
M
(
x
0
)
+
i
4
ξ
k
(
1
+
ξ
n
2
)
2
∑
s
≠
t
<
n
‍
R
k
γ
s
t
∂
M
(
x
0
)
c
(
E
s
~
)
c
(
E
t
~
)
+
2
i
3
1
(
1
+
ξ
n
2
)
3
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
j
ξ
α
ξ
β
.
When γ
=
n,
(45)
∂
x
n
(
σ
-
3
(
D
-
2
)
)
(
x
0
)
|
|
ξ
′
|
=
1
=
2
i
h
′
(
0
)
(
1
+
ξ
n
2
)
3
(
-
1
2
h
′
(
0
)
∑
k
<
n
‍
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
+
3
h
′
(
0
)
ξ
n
)
-
i
(
1
+
ξ
n
2
)
2
×
(
ξ
n
(
3
h
′′
(
0
)
-
9
2
(
h
′
(
0
)
)
2
)
∑
s
≠
t
<
n
‍
kkk
-
2
ξ
k
(
3
8
(
h
′
(
0
)
)
2
-
1
4
h
′′
(
0
)
)
∑
t
<
n
‍
c
(
E
n
~
)
c
(
E
t
~
)
k
k
k
-
1
4
ξ
n
(
(
h
′
(
0
)
)
2
-
h
′′
(
0
)
)
∑
s
≠
t
<
n
‍
c
(
E
s
~
)
c
(
E
t
~
)
)
.Proof.
Whenγ
<
n, from Lemmas 7 and 8 and ∂
x
γ
(
c
(
d
x
j
)
)
(
x
0
)
=
0, we get
(46)
∂
x
γ
(
σ
-
3
(
D
-
2
)
)
=
∂
x
γ
(
-
-
1
|
ξ
|
-
4
ξ
k
(
Γ
k
-
2
σ
k
)
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
j
g
α
β
)
=
-
-
1
|
ξ
|
-
4
ξ
k
∂
x
γ
(
Γ
k
-
2
σ
k
)
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
x
γ
∂
j
g
α
β
=
-
-
1
|
ξ
|
-
4
(
ξ
k
∂
x
γ
(
Γ
k
-
2
σ
k
)
+
ξ
n
∂
x
γ
(
Γ
n
-
2
σ
n
)
)
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
x
γ
∂
j
g
α
β
.
Substituting Lemma 10 into (47), conclusion (45) then follows easily. Similarly, we can obtain (45).Next, we can computeΦ (see formula (23) for definition of Φ). Since the sum is taken over -
r
-
l
+
1
+
k
+
j
+
|
α
|
=
7, r
,
l
≤
-
2, then we have that ∫
∂
M
‍
Φ is the sum of the following fifteen cases.Case 1.
Considerr
=
-
2, l
=
-
2, k
=
0, j
=
1
, and |
α
|
=
1.
From (23), we have
(47)
Case
1
=
i
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
x
n
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
l
×
∂
x
′
α
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 8, for i
<
n, we have
(48)
∂
x
i
σ
-
2
(
D
-
2
)
(
x
0
)
=
∂
x
i
(
|
ξ
|
-
2
)
(
x
0
)
=
0
.
So Case 1 vanishes.Case 2.
Considerr
=
-
2, l
=
-
2, k
=
0, j
=
2, and |
α
|
=
0.
From (23), we have
(49)
Case
2
=
i
6
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
j
=
2
‍
trace
[
∂
x
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
∂
ξ
n
3
σ
-
2
(
D
-
2
)
]
kkkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 7, a simple computation shows
(50)
∂
ξ
n
3
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
∂
ξ
n
3
(
1
1
+
ξ
n
2
)
=
24
ξ
n
-
24
ξ
n
3
(
1
+
ξ
n
2
)
4
,
(51)
∂
x
n
2
σ
-
2
(
D
-
2
)
(
x
0
)
=
2
|
ξ
|
6
∂
x
n
(
|
ξ
|
2
)
∂
x
n
(
|
ξ
|
2
)
(
x
0
)
-
∂
x
n
2
(
|
ξ
|
2
)
|
ξ
|
6
(
x
0
)
=
2
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
3
-
h
′′
(
0
)
(
1
+
ξ
n
2
)
2
.
By (18) and the Cauchy integral formula, then
(52)
π
ξ
n
+
[
c
(
ξ
)
(
1
+
ξ
n
2
)
2
]
=
π
ξ
n
+
[
c
(
ξ
′
)
+
ξ
n
c
(
d
x
n
)
(
1
+
ξ
n
2
)
2
]
=
1
2
π
i
lim
⁡
u
→
0
-
∫
Γ
+
‍
(
(
c
(
ξ
′
)
+
η
n
c
(
d
x
n
)
(
η
n
+
i
)
2
(
ξ
n
+
i
u
-
η
n
)
)
hhhhhhhhhhhhhhh
×
(
η
n
-
i
)
-
2
c
(
ξ
′
)
+
η
n
c
(
d
x
n
)
(
η
n
+
i
)
2
(
ξ
n
+
i
u
-
η
n
)
)
d
η
n
=
[
c
(
ξ
′
)
+
η
n
c
(
d
x
n
)
(
η
n
+
i
)
2
(
ξ
n
-
η
n
)
]
(
1
)
|
η
n
=
i
=
-
i
c
(
ξ
′
)
4
(
ξ
n
-
i
)
-
c
(
ξ
′
)
+
i
c
(
d
x
n
)
4
(
ξ
n
-
i
)
2
.
Similarly, we obtain
(53)
π
ξ
n
+
[
1
(
1
+
ξ
n
2
)
2
]
=
-
2
-
i
ξ
n
4
(
ξ
n
-
i
)
2
,
π
ξ
n
+
[
1
(
1
+
ξ
n
2
)
3
]
=
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
16
(
ξ
n
-
i
)
3
.
From (51) and (53), we get
(54)
∂
x
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
8
(
ξ
n
-
i
)
3
(
h
′
(
0
)
)
2
+
2
+
i
ξ
n
4
(
ξ
n
-
i
)
2
h
′′
(
0
)
.
Note that tr
⁡
[
id
]
=
8; then, from (50), (54), and direct computations, we obtain
(55)
trace
[
∂
x
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
ξ
n
3
σ
-
2
(
D
-
2
)
]
(
x
0
)
|
|
ξ
′
|
=
1
=
(
h
′
(
0
)
)
2
(
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
+
h
′′
(
0
)
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
4
.
Therefore,
(56)
Case
2
=
i
6
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
(
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
4
(
(
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
)
(
24
ξ
n
-
24
ξ
n
3
)
)
K
K
K
K
K
K
K
K
K
K
×
(
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
)
-
1
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
4
×
(
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
)
-
1
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
+
i
6
h
′′
(
0
)
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
4
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
6
(
h
′
(
0
)
)
2
Ω
5
×
∫
Γ
+
‍
(
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
7
(
ξ
n
+
i
)
4
d
ξ
n
d
x
′
+
i
6
h
′′
(
0
)
Ω
5
∫
Γ
+
‍
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
-
i
)
6
(
ξ
n
2
+
i
)
4
d
ξ
n
d
x
′
=
i
6
(
h
′
(
0
)
)
2
2
π
i
6
!
×
[
(
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
+
i
)
4
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
+
i
6
h
′′
(
0
)
2
π
i
5
!
×
[
(
4
+
2
i
ξ
n
)
(
24
ξ
n
-
24
ξ
n
3
)
(
ξ
n
2
+
i
)
4
]
(
5
)
|
ξ
n
=
i
Ω
5
d
x
′
=
(
7
8
(
h
′
(
0
)
)
2
-
3
8
h
′′
(
0
)
)
π
Ω
5
d
x
′
,
where Ω
5 is the canonical volume of S
5.Case 3.
Considerr
=
-
2, l
=
-
2, k
=
0, j
=
0, and |
α
|
=
2.
From (23), we have
(57)
Case
3
=
i
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
2
‍
trace
[
∂
ξ
′
α
π
ξ
n
+
σ
-
2
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
D
-
2
)
∂
x
′
α
∂
ξ
n
σ
-
2
(
D
-
2
)
]
kkkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 7, a simple computation shows
(58)
∂
ξ
′
α
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
∂
ξ
j
∂
ξ
i
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
1
(
1
+
ξ
n
2
)
2
∂
ξ
i
∂
ξ
j
(
|
ξ
|
2
)
(
x
0
)
+
2
(
1
+
ξ
n
2
)
3
∂
ξ
j
(
|
ξ
|
2
)
∂
ξ
i
(
|
ξ
|
2
)
(
x
0
)
=
-
2
δ
i
j
(
1
+
ξ
n
2
)
2
+
8
(
1
+
ξ
n
2
)
3
ξ
i
ξ
j
.
By (53) and (58), we obtain
(59)
π
ξ
n
+
∂
ξ
′
α
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
(
2
+
i
ξ
n
)
δ
i
j
2
(
ξ
n
-
i
)
2
+
-
3
i
ξ
n
2
-
9
ξ
n
+
8
i
2
(
ξ
n
-
i
)
3
ξ
i
ξ
j
.
On the other hand, by Lemmas 7 and 8, we obtain
(60)
∂
x
′
α
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
1
(
1
+
ξ
n
2
)
2
∂
x
i
∂
x
j
(
|
ξ
|
2
)
(
x
0
)
+
2
(
1
+
ξ
n
2
)
3
∂
x
j
(
|
ξ
|
2
)
∂
x
i
(
|
ξ
|
2
)
(
x
0
)
=
1
3
(
1
+
ξ
n
2
)
2
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
α
ξ
β
+
2
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
3
.
Hence, in this case,
(61)
∂
x
′
α
∂
ξ
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
4
ξ
n
3
(
1
+
ξ
n
2
)
3
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
α
ξ
β
+
-
12
ξ
n
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
4
.
From (59), (61), and direct computations, we obtain
(62)
trace
[
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
x
′
α
∂
ξ
n
σ
-
2
(
D
-
2
)
]
(
x
0
)
=
-
4
ξ
n
-
2
i
ξ
n
2
3
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
3
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
α
ξ
β
+
-
16
i
ξ
n
+
18
ξ
n
2
+
6
i
ξ
n
3
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
×
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
i
ξ
j
ξ
α
ξ
β
+
(
h
′
(
0
)
)
2
-
12
ξ
n
-
6
i
ξ
n
2
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
4
+
(
h
′
(
0
)
)
2
-
48
i
ξ
n
+
54
ξ
n
2
+
18
i
ξ
n
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
.
Similar to (16) in [6], we have
(63)
∫
‍
ξ
μ
ξ
ν
=
1
6
[
μ
ν
]
,
∫
‍
ξ
μ
ξ
ν
ξ
α
ξ
β
=
c
0
[
μ
ν
α
β
]
,
where [
μ
ν
α
β
] stands for the sum of products of g
α
β determined by all “pairings” of μ
ν
α
β and c
0 is a constant. Using the integration over S
5 and the shorthand ∫
‍
=
(
1
/
π
3
)
∫
S
5
‍
d
5
ν, we obtain Ω
5
=
π
3. Let s
∂
M be the scalar curvature ∂
M; then,
(64)
∑
i
,
α
,
j
,
β
<
n
‍
R
i
α
j
β
∂
M
(
x
0
)
∫
|
ξ
′
|
=
1
‍
ξ
α
ξ
β
ξ
i
ξ
j
σ
(
ξ
′
)
=
c
π
3
∑
i
,
α
,
j
,
β
<
n
‍
R
i
α
j
β
∂
M
(
x
0
)
(
δ
α
β
δ
i
j
+
δ
α
i
δ
β
j
+
δ
α
j
δ
β
i
)
=
0
,
where c is a constant. Therefore,
(65)
Case
3
=
i
2
Ω
5
(
s
∂
M
∫
-
∞
+
∞
‍
-
4
ξ
n
-
2
i
ξ
n
2
9
(
ξ
n
-
i
)
2
(
1
+
ξ
n
2
)
3
d
ξ
n
kkkkk
kk
+
(
h
′
(
0
)
)
2
∫
-
∞
+
∞
‍
4
i
ξ
n
-
9
ξ
n
2
-
3
i
ξ
n
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
d
ξ
n
)
d
x
′
=
i
2
Ω
5
(
s
∂
M
2
π
i
4
!
[
-
4
ξ
n
-
2
i
ξ
n
2
9
(
ξ
n
+
i
)
3
]
(
4
)
|
ξ
n
=
i
kkkkkk
+
(
h
′
(
0
)
)
2
2
π
i
6
!
[
4
i
ξ
n
-
9
ξ
n
2
-
3
i
ξ
n
3
(
ξ
n
+
i
)
4
]
(
6
)
|
ξ
n
=
i
)
d
x
′
=
π
6
s
∂
M
Ω
5
d
x
′
+
11
π
128
(
h
′
(
0
)
)
2
Ω
5
d
x
′
,
where ∑
t
,
l
<
n
R
t
l
t
l
∂
M
(
x
0
) is the scalar curvature s
∂
M.Case 4.
Considerr
=
-
2, l
=
-
2, k
=
1, j
=
1, and |
α
|
=
0.
From (23) and the Leibniz rule, we obtain
(66)
Case
4
=
i
6
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
x
n
∂
ξ
n
π
ξ
n
+
σ
-
2
kkkkkkkkkkkkkkkkkkk
×
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By (54), we obtain
(67)
∂
x
n
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
h
′
(
0
)
-
3
-
i
ξ
n
4
(
ξ
n
-
i
)
3
.
From (50) and (51), we obtain
(68)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
h
′
(
0
)
4
-
20
ξ
n
2
(
1
+
ξ
n
2
)
4
.
Therefore,
(69)
Case
4
=
i
6
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
-
24
-
8
i
ξ
n
+
120
ξ
n
2
+
40
i
ξ
n
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
6
(
h
′
(
0
)
)
2
2
π
i
6
!
×
[
-
24
-
8
i
ξ
n
+
120
ξ
n
2
+
40
i
ξ
n
3
(
ξ
n
+
i
)
4
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
=
-
5
8
(
h
′
(
0
)
)
2
π
Ω
5
d
x
′
.Case 5.
Considerr
=
-
2, l
=
-
2, k
=
1, j
=
0, and |
α
|
=
1.
From (23), we have
(70)
Case
5
=
i
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
′
α
∂
ξ
n
π
ξ
n
+
σ
-
2
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
∂
ξ
′
α
∂
ξ
n
π
ξ
n
+
σ
-
2
×
(
D
-
2
)
∂
x
′
α
∂
ξ
n
∂
x
n
σ
-
2
(
D
-
2
)
]
kkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
From Lemmas 7 and 8, for i
<
n, we obtain
(71)
∂
x
′
∂
x
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
1
(
1
+
ξ
n
2
)
2
∂
x
i
∂
x
n
(
|
ξ
|
2
)
(
x
0
)
+
2
(
1
+
ξ
n
2
)
3
∂
x
n
(
|
ξ
|
2
)
∂
x
i
(
|
ξ
|
2
)
(
x
0
)
=
0
.
Therefore, Case 5 vanishes.Case 6.
Considerr
=
-
2, l
=
-
2, k
=
2, j
=
0, and |
α
|
=
0.
From (23), we have
(72)
Case
6
=
i
6
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
k
=
2
‍
trace
[
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
ξ
n
kkkkkkkkkkkkkkkkkkk
×
∂
x
n
2
σ
-
2
(
D
-
2
)
]
kkkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
From (50)–(53), we have
(73)
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
i
(
ξ
n
-
i
)
3
,
(74)
∂
ξ
n
∂
x
n
2
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
4
ξ
n
h
′′
(
x
0
)
(
1
+
ξ
n
2
)
3
+
-
12
ξ
n
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
4
.
Therefore,
(75)
Case
6
=
i
6
h
′′
(
0
)
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
-
32
i
ξ
n
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
d
ξ
n
σ
(
ξ
′
)
d
x
′
+
i
6
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
96
i
ξ
n
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
4
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
6
h
′′
(
0
)
2
π
i
5
!
[
-
32
i
ξ
n
(
ξ
n
+
i
)
3
]
(
5
)
|
ξ
n
=
i
Ω
5
d
x
′
+
i
6
(
h
′
(
0
)
)
2
2
π
i
6
!
[
96
i
ξ
n
(
ξ
n
+
i
)
4
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
=
(
-
3
8
h
′′
(
0
)
+
7
8
(
h
′
(
0
)
)
2
)
π
Ω
5
d
x
′
.Case 7.
Considerr
=
-
2, l
=
-
3, k
=
0, j
=
1, and |
α
|
=
0.
From (23) and the Leibniz rule, we obtain
(76)
Case
7
=
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
ξ
n
∂
x
n
π
ξ
n
+
σ
-
2
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
kk
×
(
D
-
2
)
∂
ξ
n
σ
-
3
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
-
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
ξ
n
2
∂
x
n
π
ξ
n
+
σ
-
2
kkkkkkkkkkkkkkkkkkkkkk
×
(
D
-
2
)
σ
-
3
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 8, we have
(77)
π
ξ
n
+
∂
x
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
h
′
(
0
)
2
+
i
ξ
n
4
(
ξ
n
-
i
)
2
.
Then,
(78)
∂
ξ
n
2
π
ξ
n
+
∂
x
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
h
′
(
0
)
4
+
i
ξ
n
2
(
ξ
n
-
i
)
4
.
In the normal coordinate,g
i
j
(
x
0
)
=
δ
i
j and ∂
x
j
(
g
α
β
)
(
x
0
)
=
0, if j
<
n
;
=
h
′
(
0
)
δ
β
α, if j
=
n. By Lemma A.2 in [13] and Lemma 7, we obtain
(79)
σ
-
3
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
-
1
|
ξ
|
-
4
ξ
k
(
Γ
k
-
2
δ
k
)
(
x
0
)
|
|
ξ
′
|
=
1
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
j
g
α
β
(
x
0
)
|
|
ξ
′
|
=
1
=
-
i
(
1
+
ξ
n
2
)
2
(
-
1
2
h
′
(
0
)
∑
k
<
n
‍
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
+
3
h
′
(
0
)
ξ
n
)
-
2
i
h
′
(
0
)
ξ
n
(
1
+
ξ
n
2
)
3
.
We note that∫
|
ξ
′
|
=
1
‍
ξ
1
⋯
ξ
2
q
+
1
σ
(
ξ
′
)
=
0, so the first term in (79) has no contribution for computing Case 7. Combining (78), (79), and direct computations, we obtain
(80)
trace
[
∂
ξ
n
2
∂
x
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
σ
-
3
(
D
-
2
)
]
(
x
0
)
=
(
h
′
(
0
)
)
2
-
80
i
ξ
n
+
20
ξ
n
2
-
48
i
ξ
n
3
+
12
ξ
n
4
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
.
Therefore,
(81)
Case
7
=
-
1
2
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
(
-
80
i
ξ
n
+
20
ξ
n
2
-
48
i
ξ
n
3
+
12
ξ
n
4
(
ξ
n
+
i
)
3
(
-
80
i
ξ
n
+
20
ξ
n
2
-
48
i
ξ
n
3
+
12
ξ
n
4
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
)
-
1
-
80
i
ξ
n
+
20
ξ
n
2
-
48
i
ξ
n
3
+
12
ξ
n
4
(
ξ
n
+
i
)
3
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
-
1
2
(
h
′
(
0
)
)
2
2
π
i
6
!
×
[
-
80
i
ξ
n
+
20
ξ
n
2
-
48
i
ξ
n
3
+
12
ξ
n
4
(
ξ
n
+
i
)
3
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
=
21
8
(
h
′
(
0
)
)
2
π
Ω
5
d
x
′
.Case 8.
Considerr
=
-
2, l
=
-
3, k
=
0, j
=
0, and |
α
|
=
1.
From (23) and the Leibniz rule, we obtain
(82)
Case
8
=
-
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
l
k
×
∂
x
′
α
∂
ξ
n
σ
-
3
(
D
-
2
)
]
kkkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
n
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
∂
x
′
α
σ
-
3
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 7, a simple computation shows
(83)
∂
ξ
′
α
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
∂
ξ
i
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
2
ξ
i
(
1
+
ξ
n
2
)
2
.
From (53) and (83), we obtain
(84)
∂
ξ
n
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
∂
ξ
i
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
3
-
i
ξ
n
2
(
ξ
n
-
i
)
3
ξ
i
.
By (44), (84), and direct computations, we obtain
(85)
trace
[
∂
ξ
n
∂
ξ
′
α
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
x
′
α
σ
-
3
(
D
-
2
)
]
(
x
0
)
=
8
ξ
n
-
24
i
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
×
∑
α
,
β
<
n
‍
(
R
i
α
j
β
∂
M
(
x
0
)
+
R
i
β
j
α
∂
M
(
x
0
)
)
ξ
i
ξ
j
ξ
α
ξ
β
+
30
i
-
10
ξ
n
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
∑
i
<
n
‍
R
i
γ
i
k
∂
M
(
x
0
)
ξ
k
ξ
γ
.
From (63), (64), and (85), we obtain
(86)
Case
8
=
1
9
s
∂
M
Ω
5
∫
-
∞
+
∞
‍
15
i
-
5
ξ
n
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
d
ξ
n
d
x
′
=
1
9
s
∂
M
2
π
i
4
!
[
15
i
-
5
ξ
n
(
ξ
n
+
i
)
2
]
(
4
)
|
ξ
n
=
i
Ω
5
d
x
′
=
5
16
s
∂
M
π
Ω
5
d
x
′
.Case 9.
Considerr
=
-
2, l
=
-
3, k
=
1, j
=
0, and |
α
|
=
0.
From (23) and the Leibniz rule, we obtain
(87)
Case
9
=
-
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
kkkkkkkkkkkkkkkkkkkkkkkkk
×
∂
ξ
n
∂
x
n
σ
-
3
(
D
-
2
)
]
kkkkkkkkkkkkkkkkk
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
x
n
σ
-
3
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
From (73), we have
(88)
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
i
(
ξ
n
-
i
)
3
.
Combining (45) and (88), we obtain
(89)
trace
[
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
x
n
σ
-
3
(
D
-
2
)
]
(
x
0
)
|
|
ξ
′
|
=
1
=
(
h
′
(
0
)
)
2
(
84
ξ
n
+
36
ξ
n
3
)
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
+
-
24
ξ
n
h
′′
(
0
)
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
.
Therefore,
(90)
Case
9
=
1
2
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
(
84
ξ
n
+
36
ξ
n
3
)
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
d
ξ
n
σ
(
ξ
′
)
d
x
′
+
1
2
h
′′
(
0
)
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
-
24
ξ
n
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
1
2
(
h
′
(
0
)
)
2
Ω
5
∫
Γ
+
‍
84
ξ
n
+
36
ξ
n
3
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
3
d
ξ
n
d
x
′
+
1
2
h
′′
(
0
)
Ω
5
∫
Γ
+
‍
-
24
ξ
n
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
d
ξ
n
d
x
′
=
1
2
(
h
′
(
0
)
)
2
2
π
i
5
!
[
84
ξ
n
+
36
ξ
n
3
(
ξ
n
+
i
)
3
]
(
5
)
|
ξ
n
=
i
Ω
5
d
x
′
+
1
2
h
′′
(
0
)
2
π
i
4
!
[
-
24
ξ
n
(
ξ
n
+
i
)
2
]
(
4
)
|
ξ
n
=
i
Ω
5
d
x
′
=
(
9
8
h
′′
(
0
)
-
27
8
(
h
′
(
0
)
)
2
)
π
Ω
5
d
x
′
.Case 10.
Considerr
=
-
3, l
=
-
2, k
=
0, j
=
1, and |
α
|
=
0.
From (23), we have
(91)
Case
10
=
-
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
x
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By the Leibniz rule, trace property, and “+
+” and “-
-” vanishing after the integration over ξ
n in [9], then
(92)
∫
-
∞
+
∞
‍
trace
[
∂
x
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
d
ξ
n
=
∫
-
∞
+
∞
‍
trace
[
∂
x
n
σ
-
3
(
D
-
2
)
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
d
ξ
n
-
∫
-
∞
+
∞
‍
trace
[
∂
x
n
σ
-
3
(
D
-
2
)
∂
ξ
n
2
π
ξ
n
+
σ
-
2
(
D
-
2
)
]
d
ξ
n
.
Combining these assertions, we obtain
(93)
Case
10
=
Case
9
-
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
x
n
σ
-
3
(
D
-
2
)
k
k
h
h
h
h
h
h
h
h
h
h
h
h
k
k
k
k
k
×
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 7, a simple computation shows
(94)
∂
ξ
n
2
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
6
ξ
n
2
-
2
(
1
+
ξ
n
2
)
3
.
Combining (45) and (94), we obtain
(95)
trace
[
∂
x
n
σ
-
3
(
D
-
2
)
∂
ξ
n
2
σ
-
2
(
D
-
2
)
]
(
x
0
)
=
(
h
′
(
0
)
)
2
(
84
i
ξ
n
+
36
i
ξ
n
3
)
(
-
2
+
6
ξ
n
2
)
(
1
+
ξ
n
2
)
6
+
h
′′
(
0
)
24
i
ξ
n
(
2
-
6
ξ
n
2
)
(
1
+
ξ
n
2
)
5
.
We note that
(96)
∫
-
∞
+
∞
‍
(
84
i
ξ
n
+
36
i
ξ
n
3
)
(
-
2
+
6
ξ
n
2
)
(
1
+
ξ
n
2
)
6
d
ξ
n
=
2
π
i
5
!
[
(
84
i
ξ
n
+
36
i
ξ
n
3
)
(
-
2
+
6
ξ
n
2
)
(
ξ
n
+
i
)
6
]
(
5
)
|
ξ
n
=
i
=
0
,
∫
-
∞
+
∞
‍
24
i
ξ
n
(
2
-
6
ξ
n
2
)
(
1
+
ξ
n
2
)
5
d
ξ
n
=
0
.
Therefore,
(97)
Case
10
=
(
9
8
h
′′
(
0
)
-
27
8
(
h
′
(
0
)
)
2
)
π
Ω
5
d
x
′
.Case 11.
Considerr
=
-
3, l
=
-
2, k
=
0, j
=
0, and |
α
|
=
1.
From (23), we have
(98)
Case
11
=
-
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
∑
|
α
|
=
1
‍
trace
[
∂
ξ
′
α
π
ξ
n
+
σ
-
3
(
D
-
2
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
K
×
∂
x
′
α
∂
ξ
n
σ
-
2
(
D
-
2
)
]
K
K
K
K
K
K
K
K
K
K
K
K
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 8, for i
<
n, we have
(99)
∂
x
i
σ
-
2
(
D
-
2
)
(
x
0
)
=
∂
x
i
(
|
ξ
|
-
2
)
(
x
0
)
=
0
.
So Case 11 vanishes.Case 12.
Considerr
=
-
3, l
=
-
2, k
=
1, j
=
0, and |
α
|
=
0.
From (23) and the Leibniz rule, we have
(100)
Case
12
=
-
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
x
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
∂
x
n
σ
-
2
(
D
-
2
)
]
KKKKKKKKKKK
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
KKKKKKKKKKK
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By the Leibniz rule, trace property, and “+
+” and “-
-” vanishing after the integration over ξ
n in [9], then
(101)
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
d
ξ
n
=
∫
-
∞
+
∞
‍
trace
[
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
d
ξ
n
-
∫
-
∞
+
∞
‍
trace
[
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
]
d
ξ
n
.
Combining these assertions, we see
(102)
Case
12
=
Case
7
+
1
2
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
KKKKKKKKKKK
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
From (68) and direct computations, we obtain
(103)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
4
-
20
ξ
n
2
(
1
+
ξ
n
2
)
4
h
′
(
0
)
.
Combining (79) and (103), we obtain
(104)
trace
[
σ
-
3
(
D
-
2
)
∂
ξ
n
2
∂
x
n
σ
-
2
(
D
-
2
)
]
(
x
0
)
=
(
h
′
(
0
)
)
2
-
20
i
ξ
n
+
88
i
ξ
n
3
+
60
i
ξ
n
5
(
1
+
ξ
n
2
)
7
.
We note that
(105)
∫
-
∞
+
∞
‍
-
20
i
ξ
n
+
88
i
ξ
n
3
+
60
i
ξ
n
5
(
1
+
ξ
n
2
)
7
d
ξ
n
=
2
π
i
6
!
[
-
20
i
ξ
n
+
88
i
ξ
n
3
+
60
i
ξ
n
5
(
ξ
n
+
i
)
7
]
(
6
)
|
ξ
n
=
i
=
0
.
Therefore,
(106)
Case
12
=
21
8
(
h
′
(
0
)
)
2
π
Ω
5
d
x
′
.Case 13.
Considerr
=
-
3, l
=
-
3, k
=
0, j
=
0, and |
α
|
=
0.
From (23) and the Leibniz rule, we have
(107)
Case
13
=
-
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
3
(
D
-
2
)
∂
ξ
n
σ
-
3
(
D
-
2
)
]
KKKKKKKKKKKK
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
ξ
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
σ
-
3
(
D
-
2
)
]
KKKKKKKKKKK
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By (79), we obtain
(108)
σ
-
3
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
-
1
|
ξ
|
-
4
ξ
k
(
Γ
k
-
2
δ
k
)
(
x
0
)
|
|
ξ
′
|
=
1
-
-
1
|
ξ
|
-
6
2
ξ
j
ξ
α
ξ
β
∂
j
g
α
β
(
x
0
)
|
|
ξ
′
|
=
1
=
-
i
(
1
+
ξ
n
2
)
2
(
-
1
2
h
′
(
0
)
∑
k
<
n
‍
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
+
3
h
′
(
0
)
ξ
n
)
-
2
i
h
′
(
0
)
ξ
n
(
1
+
ξ
n
2
)
3
=
i
2
(
1
+
ξ
n
2
)
2
h
′
(
0
)
∑
k
<
n
‍
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
+
h
′
(
0
)
-
5
i
ξ
n
-
3
i
ξ
n
3
(
1
+
ξ
n
2
)
3
.
By (18) and the Cauchy integral formula, then
(109)
π
ξ
n
+
[
-
5
i
ξ
n
-
3
i
ξ
n
3
(
1
+
ξ
n
2
)
3
]
=
π
ξ
n
+
[
-
5
i
ξ
n
-
3
i
ξ
n
3
(
ξ
n
+
i
)
3
(
ξ
n
-
i
)
3
]
=
1
2
π
i
lim
⁡
u
→
0
-

∫
Γ
+
‍

(
-
5
i
η
n
-
3
i
η
n
3
)
/
(
(
η
n
+
i
)
3
(
ξ
n
+
i
u
-
η
n
)
)
(
η
n
-
i
)
3
d
η
n
=
1
2
[
-
5
i
η
n
-
3
i
η
n
3
(
η
n
+
i
)
3
(
ξ
n
-
η
n
)
]
(
1
)
|
η
n
=
i
=
9
i
-
7
ξ
n
8
(
ξ
n
-
i
)
3
.
Then, we obtain
(110)
∂
ξ
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
3
i
-
ξ
n
4
(
ξ
n
-
i
)
3
h
′
(
0
)
∑
k
<
n
‍
ξ
k
c
(
e
k
~
)
c
(
e
n
~
)
+
h
′
(
0
)
7
ξ
n
-
10
i
4
(
ξ
n
-
i
)
4
.
By the relation of the Clifford action and tr
⁡
A
B
=
tr
⁡
B
A, then we have the equalities
(111)
tr
⁡
[
c
(
ξ
′
)
c
(
d
x
n
)
]
=
0
;
tr
⁡
[
c
(
d
x
n
)
2
]
=
-
8
;
tr
⁡
[
c
(
ξ
′
)
2
]
(
x
0
)
|
|
ξ
′
|
=
1
=
-
8
;
tr
⁡
[
c
(
E
j
~
)
c
(
E
n
~
)
c
(
E
k
~
)
c
(
E
n
~
)
]
=
-
tr
⁡
[
id
]
δ
j
k
=
-
8
δ
j
k
.
Then,
(112)
trace
[
∂
ξ
n
π
ξ
n
+
σ
-
3
(
D
-
2
)
σ
-
3
(
D
-
2
)
]
(
x
0
)
=
(
-
3
-
i
ξ
n
)
h
′
(
0
)
8
(
ξ
n
-
i
)
3
(
1
+
ξ
n
2
)
2
×
∑
k
<
n
‍
ξ
j
ξ
k
trace
[
c
(
E
j
~
)
c
(
E
n
~
)
c
(
E
k
~
)
c
(
E
n
~
)
]
+
(
h
′
(
0
)
)
2
2
(
10
i
-
7
ξ
n
)
(
5
i
ξ
n
+
3
i
ξ
n
3
)
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
=
(
h
′
(
0
)
)
2
-
3
i
-
96
ξ
n
-
72
i
ξ
n
2
-
56
ξ
n
3
-
41
i
ξ
n
4
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
.
Therefore,
(113)
Case
13
=
i
(
h
′
(
0
)
)
2
×
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
(
(
-
3
i
-
96
ξ
n
-
72
i
ξ
n
2
-
56
ξ
n
3
-
41
i
ξ
n
4
)
(
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
)
-
1
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
)
-
1
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
(
h
′
(
0
)
)
2
Ω
5
×
∫
Γ
+
‍
-
3
i
-
96
ξ
n
-
72
i
ξ
n
2
-
56
ξ
n
3
-
41
i
ξ
n
4
(
ξ
n
-
i
)
4
(
1
+
ξ
n
2
)
3
d
ξ
n
d
x
′
=
i
(
h
′
(
0
)
)
2
2
π
i
6
!
×
[
-
3
i
-
96
ξ
n
-
72
i
ξ
n
2
-
56
ξ
n
3
-
41
i
ξ
n
4
(
ξ
n
+
i
)
3
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
=
-
57
8
(
h
′
(
0
)
)
2
π
Ω
5
d
x
′
.Case 14.
Considerr
=
-
2, l
=
-
4, k
=
0, j
=
0, and |
α
|
=
0.
From (23) and the Leibniz rule, we have
(114)
Case
14
=
-
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
2
(
D
-
2
)
∂
ξ
n
σ
-
4
(
D
-
2
)
]
K
K
K
K
K
K
K
K
K
K
K
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
σ
-
4
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
From (73), we have
(115)
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
i
2
(
ξ
n
-
i
)
2
.
From Lemmas 7 and 10, we obtain
(116)
σ
-
4
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
3
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
-
9
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
3
ξ
n
3
ξ
μ
ξ
ν
+
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
2
ξ
μ
ξ
ν
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
-
1
4
(
1
+
ξ
n
2
)
2
s
(
x
0
)
-
5
3
(
1
+
ξ
n
2
)
3
ξ
μ
ξ
ν
∑
i
<
n
‍
R
i
μ
i
ν
∂
M
(
x
0
)
-
6
(
1
+
ξ
n
2
)
3
h
′′
(
0
)
ξ
n
2
-
4
3
(
1
+
ξ
n
2
)
4
ξ
μ
ξ
ν
ξ
γ
ξ
δ
∑
γ
,
δ
<
n
‍
(
R
μ
γ
ν
δ
∂
M
(
x
0
)
+
R
ν
γ
μ
δ
∂
M
(
x
0
)
)
+
4
h
′′
(
0
)
(
1
+
ξ
n
2
)
4
ξ
n
2
-
1
3
(
1
+
ξ
n
2
)
3
ξ
α
ξ
β
∑
α
,
β
<
n
‍
(
R
μ
α
ν
β
∂
M
(
x
0
)
+
R
ν
β
μ
α
∂
M
(
x
0
)
)
+
h
′′
(
0
)
(
1
+
ξ
n
2
)
3
+
2
+
3
ξ
n
+
10
ξ
n
2
+
12
ξ
n
3
-
4
ξ
n
4
+
9
ξ
n
5
(
1
+
ξ
n
2
)
5
(
h
′
(
0
)
)
2
.
From (115), (116), and direct computations, we obtain
(117)
trace
[
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
σ
-
4
(
D
-
2
)
]
(
x
0
)
|
|
ξ
′
|
=
1
=
i
2
(
ξ
n
-
i
)
2
×
-
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
3
tr
⁡
[
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
]
-
i
2
(
ξ
n
-
i
)
2
×
9
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
3
ξ
n
3
ξ
μ
ξ
ν
tr
⁡
[
id
]
+
i
2
(
ξ
n
-
i
)
2
×
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
2
ξ
μ
ξ
ν
tr
⁡
[
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
]
-
i
2
(
ξ
n
-
i
)
2
×
1
4
(
1
+
ξ
n
2
)
2
s
(
x
0
)
tr
⁡
[
id
]
-
i
2
(
ξ
n
-
i
)
2
×
5
3
(
1
+
ξ
n
2
)
3
ξ
μ
ξ
ν
∑
i
<
n
‍
R
i
μ
i
ν
∂
M
(
x
0
)
tr
⁡
[
id
]
-
i
2
(
ξ
n
-
i
)
2
×
6
(
1
+
ξ
n
2
)
3
h
′′
(
0
)
ξ
n
2
tr
⁡
[
id
]
-
i
2
(
ξ
n
-
i
)
2
×
4
3
(
1
+
ξ
n
2
)
4
ξ
μ
ξ
ν
ξ
γ
ξ
δ
×
∑
γ
,
δ
<
n
‍
(
R
μ
γ
ν
δ
∂
M
(
x
0
)
+
R
ν
γ
μ
δ
∂
M
(
x
0
)
)
tr
⁡
[
id
]
-
i
2
(
ξ
n
-
i
)
2
×
1
3
(
1
+
ξ
n
2
)
3
ξ
α
ξ
β
×
∑
α
,
β
<
n
‍
(
R
μ
α
ν
β
∂
M
(
x
0
)
+
R
ν
β
μ
α
∂
M
(
x
0
)
)
tr
⁡
[
id
]
+
i
2
(
ξ
n
-
i
)
2
×
4
h
′′
(
0
)
(
1
+
ξ
n
2
)
4
ξ
n
2
tr
⁡
[
id
]
+
i
2
(
ξ
n
-
i
)
2
×
h
′′
(
0
)
(
1
+
ξ
n
2
)
3
tr
⁡
[
id
]
+
i
2
(
ξ
n
-
i
)
2
×
2
+
3
ξ
n
+
10
ξ
n
2
+
12
ξ
n
3
-
4
ξ
n
4
+
9
ξ
n
5
(
1
+
ξ
n
2
)
5
×
(
h
′
(
0
)
)
2
tr
⁡
[
id
]
.
Combining (64), (113), and (117), we obtain
(118)
Case
14
=
s
(
x
0
)
Ω
5
∫
-
∞
+
∞
‍
1
(
ξ
n
-
i
)
4
(
ξ
n
+
i
)
2
d
ξ
n
d
x
′
+
s
∂
M
(
x
0
)
Ω
5
∫
-
∞
+
∞
‍
14
9
(
ξ
n
-
i
)
5
(
ξ
n
+
i
)
3
d
ξ
n
d
x
′
+
(
h
′
(
0
)
)
2
Ω
5
×
∫
-
∞
+
∞
(
14
9
(
ξ
n
-
i
)
5
(
ξ
n
+
i
)
3
-
4
(
1
-
ξ
n
2
-
6
ξ
n
4
)
(
ξ
n
-
i
)
6
(
ξ
n
+
i
)
4
(
-
53
-
72
ξ
n
-
33
ξ
n
2
-
288
ξ
n
3
+
525
ξ
n
4
-
216
ξ
n
5
‍
m
m
m
h
m
m
+
217
ξ
n
6
)
×
(
6
(
ξ
n
-
i
)
7
(
ξ
n
+
i
)
5
)
-
1
-
4
(
1
-
ξ
n
2
-
6
ξ
n
4
)
(
ξ
n
-
i
)
6
(
ξ
n
+
i
)
4
)
d
ξ
n
d
x
′
+
h
′′
(
0
)
Ω
5
∫
-
∞
+
∞
‍
-
4
(
1
-
ξ
n
2
-
6
ξ
n
4
)
(
ξ
n
-
i
)
6
(
ξ
n
+
i
)
4
d
ξ
n
d
x
′
=
s
(
x
0
)
2
π
i
3
!
[
1
(
ξ
n
+
i
)
2
]
(
3
)
|
ξ
n
=
i
Ω
5
d
x
′
+
s
∂
M
(
x
0
)
2
π
i
4
!
[
14
9
(
ξ
n
+
i
)
3
]
(
4
)
|
ξ
n
=
i
Ω
5
d
x
′
+
(
h
′
(
0
)
)
2
2
π
i
6
!
×
[
(
6
(
ξ
n
+
i
)
5
)
-
1
(
-
53
-
72
ξ
n
-
33
ξ
n
2
-
288
ξ
n
3
+
525
ξ
n
4
m
h
-
216
ξ
n
5
+
217
ξ
n
6
)
×
(
6
(
ξ
n
+
i
)
5
)
-
1
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
+
h
′′
(
0
)
2
π
i
5
!
[
-
4
(
1
-
ξ
n
2
-
6
ξ
n
4
)
(
ξ
n
+
i
)
4
]
(
5
)
|
ξ
n
=
i
Ω
5
d
x
′
=
(
(
343
192
-
3
i
2
)
+
(
343
192
-
3
i
2
)
(
h
′
(
0
)
)
2
+
13
16
h
′′
(
0
)
-
1
4
s
(
x
0
)
-
35
96
s
∂
M
(
x
0
)
m
m
+
(
343
192
-
3
i
2
)
(
h
′
(
0
)
)
2
+
13
16
h
′′
(
0
)
)
π
Ω
5
d
x
′
.Case 15.
Considerr
=
-
4, l
=
-
2, k
=
0, j
=
0, and |
α
|
=
0.
From (23), we have
(119)
Case
15
=
-
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By the Leibniz rule, trace property, and “+
+” and “-
-” vanishing after the integration over ξ
n in [9], then
(120)
∫
-
∞
+
∞
‍
trace
[
π
ξ
n
+
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
d
ξ
n
=
∫
-
∞
+
∞
‍
trace
[
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
d
ξ
n
-
∫
-
∞
+
∞
‍
trace
[
σ
-
4
(
D
-
2
)
∂
ξ
n
π
ξ
n
+
σ
-
2
(
D
-
2
)
]
d
ξ
n
.
Combining these assertions, we see
(121)
Case
15
=
Case
14
-
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
.
By Lemma 7, a simple computation shows
(122)
∂
ξ
n
σ
-
2
(
D
-
2
)
(
x
0
)
|
|
ξ
′
|
=
1
=
-
2
ξ
n
(
1
+
ξ
n
2
)
2
.
From (116), (122), and direct computations, we obtain
(123)
trace
[
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
(
x
0
)
|
|
ξ
′
|
=
1
=
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
-
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
3
tr
⁡
[
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
9
(
h
′
(
0
)
)
2
(
1
+
ξ
n
2
)
3
ξ
n
3
ξ
μ
ξ
ν
tr
⁡
[
id
]
+
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
(
h
′
(
0
)
)
2
4
(
1
+
ξ
n
2
)
2
ξ
μ
ξ
ν
×
tr
⁡
[
c
(
E
μ
~
)
c
(
E
n
~
)
c
(
E
ν
~
)
c
(
E
n
~
)
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
1
4
(
1
+
ξ
n
2
)
2
s
(
x
0
)
tr
⁡
[
id
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
5
3
(
1
+
ξ
n
2
)
3
ξ
μ
ξ
ν
∑
i
<
n
‍
R
i
μ
i
ν
∂
M
(
x
0
)
tr
⁡
[
id
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
6
(
1
+
ξ
n
2
)
3
h
′′
(
0
)
ξ
n
2
tr
⁡
[
id
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
4
3
(
1
+
ξ
n
2
)
4
ξ
μ
ξ
ν
ξ
γ
ξ
δ
×
∑
γ
,
δ
<
n
‍
(
R
μ
γ
ν
δ
∂
M
(
x
0
)
+
R
ν
γ
μ
δ
∂
M
(
x
0
)
)
tr
⁡
[
id
]
-
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
1
3
(
1
+
ξ
n
2
)
3
ξ
α
ξ
β
×
∑
α
,
β
<
n
‍
(
R
μ
α
ν
β
∂
M
(
x
0
)
+
R
ν
β
μ
α
∂
M
(
x
0
)
)
tr
⁡
[
id
]
+
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
4
h
′′
(
0
)
(
1
+
ξ
n
2
)
4
ξ
n
2
tr
⁡
[
id
]
+
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
h
′′
(
0
)
(
1
+
ξ
n
2
)
3
tr
⁡
[
id
]
+
-
2
ξ
n
(
1
+
ξ
n
2
)
2
×
2
+
3
ξ
n
+
10
ξ
n
2
+
12
ξ
n
3
-
4
ξ
n
4
+
9
ξ
n
5
(
1
+
ξ
n
2
)
5
(
h
′
(
0
)
)
2
tr
⁡
[
id
]
.
Combining (64), (111), and (123), we obtain
(124)
-
i
∫
|
ξ
′
|
=
1
‍
∫
-
∞
+
∞
‍
trace
[
σ
-
4
(
D
-
2
)
∂
ξ
n
σ
-
2
(
D
-
2
)
]
×
(
x
0
)
d
ξ
n
σ
(
ξ
′
)
d
x
′
=
-
i
s
(
x
0
)
Ω
5
∫
-
∞
+
∞
‍
4
ξ
n
(
1
+
ξ
n
2
)
4
d
ξ
n
d
x
′
-
i
s
∂
M
(
x
0
)
Ω
5
∫
-
∞
+
∞
‍
56
ξ
n
9
(
1
+
ξ
n
2
)
5
d
ξ
n
d
x
′
-
i
h
′′
(
0
)
Ω
5
∫
-
∞
+
∞
‍
-
16
ξ
n
+
16
ξ
n
3
+
96
ξ
n
5
(
1
+
ξ
n
2
)
6
d
ξ
n
d
x
′
-
i
(
h
′
(
0
)
)
2
Ω
5
∫
-
∞
+
∞
‍
(
(
-
100
ξ
n
-
144
ξ
n
2
-
66
ξ
n
3
-
576
ξ
n
4
(
3
(
1
+
ξ
n
2
)
7
)
-
1
hhhhhhhhhhhhhhhhhlhhh
+
1050
ξ
n
5
-
432
ξ
n
6
+
434
ξ
n
7
)
hhhhhhhhhhhhhhhhhhlh
×
(
3
(
1
+
ξ
n
2
)
7
)
-
1
)
d
ξ
n
d
x
′
=
-
i
s
(
x
0
)
2
π
i
3
!
[
4
ξ
n
(
ξ
n
+
i
)
4
]
(
3
)
|
ξ
n
=
i
Ω
5
d
x
′
-
i
s
∂
M
(
x
0
)
2
π
i
4
!
[
56
ξ
n
9
(
ξ
n
+
i
)
5
]
(
4
)
|
ξ
n
=
i
Ω
5
d
x
′
-
i
h
′′
(
0
)
2
π
i
5
!
[
-
16
ξ
n
+
16
ξ
n
3
+
96
ξ
n
5
(
ξ
n
+
i
)
6
]
(
5
)
|
ξ
n
=
i
Ω
5
d
x
′
-
i
(
h
′
(
0
)
)
2
2
π
i
6
!
[
(
-
100
ξ
n
-
144
ξ
n
2
-
66
ξ
n
3
-
576
ξ
n
4
×
(
3
(
ξ
n
+
i
)
7
)
-
1
hhhhhhhhhhhhhhhhh
+
1050
ξ
n
5
-
432
ξ
n
6
+
434
ξ
n
7
)
k
k
k
k
k
k
k
k
k
k
k
k
k
k
×
(
3
(
ξ
n
+
i
)
7
)
-
1
]
(
6
)
|
ξ
n
=
i
Ω
5
d
x
′
=
3
i
(
h
′
(
0
)
)
2
π
Ω
5
d
x
′
.
Therefore,
(125)
Case
15
=
(
-
1
4
s
(
x
0
)
-
35
96
s
∂
M
(
x
0
)
+
(
343
192
+
3
i
2
)
k
k
k
k
×
(
h
′
(
0
)
)
2
+
13
16
h
′′
(
0
)
)
π
Ω
5
d
x
′
.Now,Φ is the sum of the case (1,2
,
…
,
15), so
(126)
Φ
=
∑
I
=
1
15
‍
case
I
=
(
-
1475
384
(
h
′
(
0
)
)
2
+
25
8
h
′′
(
0
)
kkkkkkkkkkkkkk
-
1
2
s
-
77
192
s
∂
M
)
π
Ω
5
d
x
′
.
Hence, we conclude that, for 7-dimensional compact manifold M with the boundary ∂
M,
(127)
Vo
l
7
(
2,2
)
=
1
2
∫
∂
M
‍
(
-
1475
192
(
h
′
(
0
)
)
2
+
25
4
h
′′
(
0
)
k
k
k
k
k
k
k
-
s
-
77
96
s
∂
M
)
π
Ω
5
d
vol
∂
M
.Next, we recall the Einstein-Hilbert action for manifolds with boundary (see [13] or [14]):
(128)
I
Gr
=
1
16
π
∫
M
‍
s
d
vo
l
M
+
2
∫
∂
M
‍
K
d
vo
l
∂
M
:
=
I
Gr
,
i
+
I
Gr
,
b
,
where
(129)
K
=
∑
1
≤
i
,
j
≤
n
-
1
‍
K
i
,
j
g
∂
M
i
,
j
;
K
i
,
j
=
-
Γ
i
,
j
n
,
and K
i
,
j is the second fundamental form or extrinsic curvature. Take the metric in Section 2; then, by Lemma A
.
2 in [13], we have K
i
,
j
(
x
0
)
=
-
Γ
i
,
j
n
(
x
0
)
=
-
(
1
/
2
)
h
′
(
0
) for i
=
j
<
n; Γ
s
,
t
i
(
x
0
)
=
0, if i
<
n. For n
=
7, then
(130)
K
(
x
0
)
=
∑
i
,
j
‍
K
i
.
j
(
x
0
)
g
∂
M
i
,
j
(
x
0
)
=
∑
i
=
1
6
‍
K
i
,
i
(
x
0
)
=
-
5
2
h
′
(
0
)
.
So
(131)
I
Gr
,
b
=
-
5
h
′
(
0
)
Vo
l
∂
M
.On the other hand, by Proposition 2.10 in [21], we have the following lemma.Lemma 12.
LetM be a 7-dimensional compact manifold with the boundary ∂
M; then,
(132)
s
M
(
x
0
)
=
3
2
(
h
′
(
0
)
)
2
-
6
h
′′
(
0
)
+
s
∂
M
(
x
0
)
.Proof.
From Proposition 2.10 in [21], let B
=
[
0,1
), b
2
=
(
1
/
h
(
x
n
)
), and F
=
∂
M; we obtain s
B
=
0, |
gra
d
B
b
|
2
=
(
b
′
)
2, and
(133)
s
M
(
x
0
)
=
12
b
′′
(
x
0
)
-
30
(
b
′
(
x
0
)
)
2
+
s
∂
M
(
x
0
)
.
By a simple computation, the lemma as follows.Hence, from (127) and (133), we obtain the following.Theorem 13.
LetM be a 7-dimensional compact manifold with the boundary ∂
M; then,
(134)
Wres
~
[
(
π
+
D
-
2
)
2
]
=
π
4
48
∫
∂
M
‍
(
-
47
2
K
2
-
49
s
M
|
∂
M
-
77
4
s
∂
M
)
d
vo
l
∂
M
.



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*Source: 465782-2014-10-14.xml*