8 KANG-TAE KIM, NORMAN LEVENBERG AND HIROSHI YAMAGUCHI
We note that, in Cn, from the definition of the ∗-operator, we have
∗dg(t0,z) =
−2n∇zg(t0,z)
dSz =
2n−1
∂g
∂nz
(t0,z)dSz 0
for z ∂D(t0), where dSz and nz are the Euclidean area element and the unit
outer normal vector for ∂D(t0) at z.
We may assume t0 = 0. Now using the fact that −g(t, z) is a defining function
for D, we can write, from (2.4),
k2(0,z) · (
n
a,b=1
gab
∂g
∂za
∂g
∂zb
)3/2
=
∂2g
∂t∂t
(
n
a,b=1
gab
∂g
∂za
∂g
∂zb
)
+ 2 [(
n
a,b=1
gab
∂2g
∂zb∂t
∂g
∂za
)
∂g
∂t
] +
1
2
|
∂g
∂t
|2Δg.
(2.6)
Since Δg+cg = 0 on ∂D(0) (here we use the fact that g is of class
C∞
on D(0)\{p0})
and g = 0 on ∂D(0), we have from (2.5) and (2.6)
∂2λ
∂t∂t
(0) =
−cn
2n
∂D(0)
−k2(0,z)
n
a,b=1
(gab
∂g
∂za
∂g
∂zb
)1/2
+
2
∑n
a,b=1
(gab
∂2g
∂zb∂t
∂g
∂za
)
∂g
∂t
∑n
a,b=1
gab
∂g
∂za
∂g
∂zb
dg(0,z)
(I) + (II).
In general, for any
C∞
defining function v on D(0) (v = 0 on ∂D(0) and v 0
on D(0)),
∗dv =
2n(
n
a,b=1
gab
∂v
∂za
∂v
∂zb
)1/2dσz
0
for z ∂D(0), where dσz is the area element on ∂D(0) at z with respect to the
metric
ds2.
We apply this to v = −g(0,z), plug this into (I ), and obtain the formula
(I) = −cn
∂D(0)
k2(0,z)
n
a,b=1
(gab
∂g
∂za
∂g
∂zb
)dσz.
Now we work with (II ). We need to calculate
∂g
∂za
dg(0,z)
on ∂D(0). To this end, we first note that for a function u, we have the following
formulas for ∗∂u and ∗∂u, where A means “omit A”:
∗∂u =
−in
n
a,b=1
Ggab
∂u
∂zb
dza dz1 dz1 · · · dza dza · · · dzn dzn;
∗∂u =
in
n
a,b=1
Ggba
∂u
∂zb
dza dz1 dz1 · · · dza dza · · · dzn dzn.
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