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.