2. THE VARIATION FORMULA 9
Now if u = 0 on ∂D(0), du = ∂u + ∂u = 0 along ∂D(0) and we obtain
∗du =
−2in
n
a,b=1
Ggab
∂u
∂zb
dza dz1 dz1 · · · dza dza · · · dzn dzn.
Applying this to u = g(0,z) on ∂D(0), we get
∂g
∂za
∗dg(0,z) =
−2in
n
i,j=1
Ggij
∂g
∂za
∂g
∂zj
dzi dz1 dz1 · · · dzi dzi · · · dzn dzn.
Again we use the fact that g(0,z) = 0 on ∂D(0) so that dg(0,z) = 0 along ∂D(0).
Then, if i = a,
∂g
∂za
dza in the above formula can be replaced by
∂g
∂zi
dzi. It follows
that
∂g
∂za
dg(0,z) =
−2in
(
G
n
i,j=1
gij
∂g
∂zj
∂g
∂zi
)
dza dz1 dz1
· · · dza dza · · · dzn dzn,
where the term in parenthesis is independent of a = 1, ..., n. We plug this into (II )
where
(II) =
−cn
2n
∂D(0)
F
and
F :=
2
∑n
a,b
(gab
∂2g
∂zb∂t
∂g
∂za
)
∂g
∂t ∑n=1
a,b=1
gab
∂g
∂za
∂g
∂zb
dg(0,z)
= −4
in
(
n
a,b=1
Ggab
∂2g
∂zb∂t
∂g
∂t
)
dza dz1 dz1 · · · dza dza · · · dzn dzn .
Note that the denominator cancels.
Next we use the relation
∗∂(
∂g
∂t
) =
−in
n
a,b=1
Ggab
∂2g
∂zb∂t
dza dz1 dz1 · · · dza dza · · · dzn dzn
to obtain
F = 4
∂g
∂t
(∗∂(
∂g
∂t
))
on ∂D(0).
Thus we obtain
∂2λ
∂t∂t
(0) = cn
∂D(0)
k2(0,z)[
n
a,b=1
gab
∂g
∂za
∂g
∂zb
]dσz

cn
2n−2
∂D(0)
∂g
∂t
(∗∂(
∂g
∂t
)) =: −I J
where
J :=
cn
2n−2
∂D(0)
f
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