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=
⇒
D
1
=
(
κ
g
+
κ
n
)
D
1
+(
κ
g
−
κ
n
τ
g
)
D
2
+(
κ
g
τ
g
+
κ
n
)
D
3
,
D
1
=
G
K
U
K
,
D
1
=
G
K,L
U
K
U
L
+
G
K
U
K
,
−
(E.9)
D
1
=
G
K,LM
U
K
U
L
U
M
+
G
K,L
(
U
K
U
L
+
U
K
U
L
)+
G
K,L
U
K
U
L
+
G
K
U
K
,
G
K,LM
=
=
N
G
N
+
N
G
N,M
+
H
KL,M
G
3
+
H
KL
G
3
,M
,
KL
KL
,M
G
K,LM
=
(E.10)
H
KL
H
MM
2
G
M
2
K
2
G
K
2
+
=
−
+
N
1
M
2
N
1
M
G
M
2
+
N
1
H
N
1
,M
G
3
+
N
,M
G
N
+
H
KL,M
G
3
,
KL
KL
KL
D
1
=
=
U
M
+(2
U
K
U
L
+
U
K
U
L
)
M
+
U
K
U
L
U
M
2
KL
M
KL
,M
2
+
M
1
M
M
1
M
2
−
(E.11)
KL
H
KL
H
M
1
M
2
G
M
1
M
)]
G
M
+[(2
U
K
U
L
+
U
K
U
L
)
H
KL
+
−
+
U
K
U
L
U
M
2
M
1
H
M
1
M
2
+
H
KL,M
2
G
3
.
KL
M
KL
=
M
,
“symmetry” of Christoffel symbols leads to
LK
D
1
=
=
U
M
+3
U
K
U
L
M
+
U
K
U
L
U
N
KL
M
KL
,N
+
Q
M
NQ
H
KL
H
NQ
G
QM
G
M
+
−
(E.12)
KL
+
3
U
K
U
L
H
KL
+
U
K
U
L
U
N
Q
H
NQ
+
H
KL,N
D
3
.
KL
τ
g
then is obtained as
κ
g
τ
g
+
κ
n
=3
U
K
U
L
H
KL
+
U
K
U
L
U
N
(
Q
H
NQ
+
H
KL,N
)
,
KL
κ
n
=
H
KL,N
U
K
U
L
U
N
+2
H
KL
U
K
U
L
⇒
κ
g
τ
g
=
H
KL
U
K
U
L
+
Q
H
NQ
U
K
U
L
U
N
(E.13)
KL
⇒
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