10 KANG-TAE KIM, NORMAN LEVENBERG AND HIROSHI YAMAGUCHI
and
f :=
∂g
∂t
(∗∂(
∂g
∂t
))
is a
C∞
(n, n 1) form on D(0). We want to convert J into a volume integral. To
do so we must compute ∂f since f is an (n, n 1) form. We obtain
J =
cn
2n−2
D(0)
df =
cn
2n−2
D(0)
∂f.
We get two terms whose sum comprise ∂f:
(i) = (∂
∂g
∂t
) (∗∂(
∂g
∂t
));
(ii) =
∂g
∂t
∂(∗∂(
∂g
∂t
)).
Now
D(0)
(i) =
D(0)
(∂
∂g
∂t
) (∗∂(
∂g
∂t
)) = ||∂
∂g
∂t
||D(0).2
We note that in
Cn,
we have ||∂
∂g
∂t
||D(0)
2
=
2n
D(0)
(
∑n
i=1
|
∂2g
∂t∂zi
|2)dVz,
where dVz
is the Euclidean volume element of
Cn,
so that the first term
cn
2n−2 D(0)
(i) of J
coincides with the last term of formula (1.1).
Next, for (ii),
∂(∗∂(
∂g
∂t
)) =
in
n
a,b=1
∂(Ggab)
∂za
∂2g
∂zb∂t
+
Ggab
∂3g
∂za∂zb∂t
dz1 dz1 · · · dzn dzn.
Using the relation
ω =
−in+1
n
a,b=1
∂(Ggab)
∂zb
dz1 dz1 · · · dza dza · · · dzn dzn, (2.7)
with an analogous formula for ω, if u is a complex-valued function, we obtain
(∂ ω) ∂u =
in+1
n
a,b=1
∂(Ggab)
∂za
∂u
∂zb
dz1 dz1 · · · dzn dzn.
Also
∂u =
in
n
a,b=1
[Ggba
∂2u
∂za∂zb
+
∂(Ggba)
∂za
∂u
∂zb
] dz1 dz1 · · · dzn dzn
=: D2u + D1u.
Similarly,
∂u = D2u + D1u.
Finally, it is straightforward to verify that
(∂ ω) ∂u = i D1u.
Using these relations, we obtain
(∂ ω)
∂g
∂t
= i D1(
∂g
∂t
)
and
(∂ ω)
∂g
∂t
= −iD1(
∂g
∂t
).
Previous Page Next Page