2. THE VARIATION FORMULA 11
Now since g is a function, δg = δg = 0; thus we rewrite Δg + cg = 0 as
(δ∂ + δ∂)g + cg = −(∗∂ + ∗∂ ∂)g + cg = 0.
Applying to this relation and using K = K, we have
−(∂ + ∂)g + ∗cg = 0.
This last equation can be written using D1 and D2 as
−(D2g + D1g + D2g + D1g) + ∗cg = 0.
Hence
D2g =
−1
2
[D1g + D1g ∗cg]. (2.8)
Equation (2.8) is valid in D(0) \ {p0} indeed, in each D(t) \ {p0} as an equality
of (n, n) forms. Differentiating (2.8) with respect to t, we thus obtain
D2(
∂g
∂t
) =
−1
2
[D1(
∂g
∂t
) + D1(
∂g
∂t
) ∗c
∂g
∂t
] (2.9)
in D(t).
We use (2.9) in (ii):
∂(∗∂(
∂g
∂t
)) = D2(
∂g
∂t
) + D1(
∂g
∂t
) =
−1
2
[D1(
∂g
∂t
) D1(
∂g
∂t
) ∗c
∂g
∂t
]
=
−1
2
−1
i
ω
∂g
∂t

1
i
ω
∂g
∂t
∗c
∂g
∂t
.
Inserting parts (i) and (ii) of ∂f back in to J =
cn
2n−2 D(0)
∂f, we obtain
J =
cn
2n−2
D(0)

∂g
∂t
∗∂
∂g
∂t
+
1
2
∂g
∂t
1
i
ω
∂g
∂t
+
1
i
ω
∂g
∂t
+ ∗c
∂g
∂t
.
Hence,
J =
cn
2n−2
D(0)

∂g
∂t
∧∗∂
∂g
∂t
+ (
1
2
∂g
∂t
1
i
∂∗ω∧∂
∂g
∂t
+
1
i
∂∗ω∧∂
∂g
∂t
)+
1
2
c|
∂g
∂t
|2
ωn
n!
.
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
||∂
∂g
∂t
||D(0)2
+
1
2
||

c
∂g
∂t
||D(0)
2
+
1
2
D(0)
∂g
∂t
1
i
ω
∂g
∂t
+
1
i
ω
∂g
∂t
which is (2.3) for t = 0. The second order variation formula (2.3) is thus proved.
Remark 2.1. Note that (2.3) reduces to (1.1) if M =
Cn, ds2
=
|dz|2
is the
Euclidean metric, and c 0.
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