Nakahara M. Geometry, Topology and Physics
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(c) A complex q-form ω is uniquely written as
ω =
r+s=q
ω
(r,s)
(8.36a)
where ω
(r,s)
∈
r,s
(M). Thus, we have the decomposition
q
(M) =
6
r+s=q
r,s
(M). (8.36b)
The proof is easy and is left to the reader. Now any q-form ω is decomposed
as
ω =
r+s=q
ω
(r,s)
=
r+s=q
1
r!s!
ω
µ
1
...µ
r
ν
1
...ν
s
dz
µ
1
∧ ...∧ dz
µ
r
∧ dz
ν
1
∧ ...∧ dz
ν
s
(8.37)
where
ω
µ
1
...µ
r
ν
1
...ν
s
= ω
∂
∂z
µ
1
,...,
∂
∂z
µ
r
,
∂
∂z
ν
1
,...,
∂
∂z
ν
s
. (8.38)
Exercise 8.6. Let dim
M = m.Verifythat
dim
r,s
p
(M) =
m
r
m
s
if 0 ≤ r, s ≤ m
0otherwise.
Show also that dim
q
p
(M) =
r+s=q
dim
r,s
p
(M) =
2m
q
.
8.3.3 Dolbeault operators
Let us compute the exterior derivative of an (r , s)-form ω. From (8.35), we find
dω =
1
r!s!
∂
∂z
λ
ω
µ
1
...µ
r
ν
1
...ν
s
dz
λ
+
∂
∂z
λ
ω
µ
1
...µ
r
ν
1
...ν
s
dz
λ
× dz
µ
1
∧ ...∧ dz
µ
r
∧ dz
ν
1
∧ ...∧ dz
ν
s
. (8.39)
dω is a mixture of an (r + 1, s)-form and an (r, s + 1)-form. We separate the
action of d according to its destinations,
d = ∂ +
∂ (8.40)
where ∂ :
r,s
(M) →
r+1,s
(M) and ∂ :
r,s
(M) →
r,s+1
(M). For example,
if ω = ω
µν
dz
µ
∧ dz
ν
, its exterior derivatives are
∂ω =
∂ω
µ
ν
∂z
λ
dz
λ
∧ dz
µ
∧ dz
ν
∂ω =
∂ω
µ
ν
∂z
λ
dz
λ
∧ dz
µ
∧ dz
ν
=−
∂ω
µ
ν
∂z
λ
dz
µ
∧ dz
λ
∧ dz
ν
.
The operators ∂ and
∂ are called the Dolbeault operators.
If ω is a general q-form given by (8.37), the actions of ∂ and
∂ on ω are
defined by
∂ω =
r+s=q
∂ω
(r,s)
∂ω =
r+s=q
∂ω
(r,s)
. (8.41)
Theorem 8.1. Let M be a complex manifold and let ω ∈
q
(M) and ξ ∈
p
(M) .Then
∂∂ω = (∂
∂ + ∂∂)ω = ∂∂ω = 0 (8.42a)
∂
ω = ∂ω, ∂ω = ∂ω (8.42b)
∂(ω ∧ ξ) = ∂ω ∧ ξ + (−1)
q
ω ∧ ∂ξ (8.42c)
∂(ω ∧ ξ) = ∂ω ∧ ξ + (−1)
q
ω ∧ ∂ξ. (8.42d)
Proof. It is sufficient to prove them when ω is of bidegree (r, s).
(a) Since d = ∂ +
∂,wehave
0 = d
2
ω = (∂ + ∂)(∂ + ∂)ω = ∂∂ω + (∂∂ + ∂∂)ω + ∂ ∂ω.
The three terms of the RHS are of bidegrees (r + 2, s), (r + 1, s + 1)
and (r, s + 2) respectively. From proposition 8.1(c), each term must vanish
separately.
(b) Since d
ω = dω,wehave
∂
ω + ∂ω = dω = (∂ + ∂)ω = ∂ω + ∂ω.
Noting that ∂ω and
∂ω are of bidegree (s + 1, r ) and ∂ω and ∂ω are of
(s, r + 1), we conclude that ∂
ω = ∂ω and ∂ω = ∂ω.
(c) We assume ω is of bidegree (r, s) and ξ of (r
, s
). Equation (8.42c) is
proved by separating d (ω ∧ ξ) = dω ∧ ξ + (−1)
q
ω ∧ dξ , into forms of
bidegrees (r + r
+ 1, s + s
) and (r +r
, s + s
+ 1).
Definition 8.3. Let M be a complex manifold. If ω ∈
r,0
(M) satisifies ∂ω = 0,
the r-form ω is called a holomorphic r-form.
Let us look at a holomorphic 0-form f ∈ (U) on a chart (U,ϕ).The
condition
∂ f = 0 becomes
∂ f
∂z
λ
= 01≤ λ ≤ m = dim M. (8.43)
A holomorphic 0-form is just a holomorphic function, f ∈ (U ) .Letω ∈
r,0
(M),where1≤ r ≤ m = dim M. On a chart (U,ϕ),wehave
ω =
1
r!
ω
µ
1
...µ
r
dz
µ
1
∧ ...∧ dz
µ
r
. (8.44)
Then
∂ω = 0 if and only if
∂
∂z
λ
ω
µ
1
...µ
r
= 0
namely if ω
µ
1
...µ
r
are holomorphic functions on U .
Let dim
M = m. The sequence of -linear maps
r,0
(M)
∂
−→
r,1
(M)
∂
−→···
···
∂
−→
r,m−1
(M)
∂
−→
r,m
(M) (8.45)
is called the Dolbeault complex. Note that
∂
2
= 0. The set of ∂-closed (r, s)-
forms (those ω ∈
r,s
(M) such that ∂ω = 0) is called the (r, s)-cocycle and is
denoted by Z
r,s
∂
(M).Thesetof∂-exact (r, s)-forms (those ω ∈
r,s
(M) such
that ω =
∂η for some η ∈
r,s−1
(M)) is called the (r, s)-coboundary and is
denoted by B
r,s
∂
(M). The complex vector space
H
r,s
∂
(M) ≡ Z
r,s
∂
(M)/B
r,s
∂
(M) (8.46)
is called the (r, s)th
∂-cohomology group, see section 8.6.
8.4 Hermitian manifolds and Hermitian differential geometry
Let M be a complex manifold with dim
M = m and let g be a Riemannian
metric of M as a differentiable manifold. Take Z = X +iY, W = U +iV ∈ T
p
M
and extend g so that
g
p
(Z , W ) = g
p
(X, U ) − g
p
(Y, V ) + i[g
p
(X, V ) + g
p
(Y, U )]. (8.47)
The components of g with respect to the bases (8.13) are
g
µν
( p) = g
p
(∂/∂ z
µ
,∂/∂z
ν
) (8.48a)
g
µν
( p) = g
p
(∂/∂ z
µ
,∂/∂z
ν
) (8.48b)
g
µ
ν
( p) = g
p
(∂/∂ z
µ
,∂/∂z
ν
) (8.48c)
g
µν
( p) = g
p
(∂/∂ z
µ
,∂/∂z
ν
). (8.48d)
We easily verify that
g
µν
= g
νµ
, g
µν
= g
νµ
, g
µν
= g
νµ
, g
µν
= g
µν
, g
µν
= g
µν
.
(8.49)
8.4.1 The Hermitian metric
If a Riemannian metric g of a complex manifold M satisfies
g
p
(J
p
X, J
p
Y ) = g
p
(X, Y ) (8.50)
at each point p ∈ M and for any X, Y ∈ T
p
M, g is said to be a Hermitian metric.
The pair (M, g) is called a Hermitian manifold. The vector J
p
X is orthogonal
to X with respect to a Hermitian metric,
g
p
(J
p
X, X ) = g
p
(J
2
p
X, J
p
X) =−g
p
(J
p
X, X ) = 0. (8.51)
Theorem 8.2. A complex manifold always admits a Hermitian metric.
Proof.Letg be any Riemannian metric of a complex manifold M.Defineanew
metric ˆg by
ˆg
p
(X, Y ) ≡
1
2
[g
p
(X, Y ) + g
p
(J
p
X, J
p
Y )]. (8.52)
Clearly ˆg
p
(J
p
X, J
p
Y ) =ˆg
p
(X, Y ). Moreover, ˆg is positive definite provided
that g is. Hence, ˆg is a Hermitian metric on M.
Let g be a Hermitian metric on a complex manifold M. From (8.50), we find
that
g
µν
= g
∂
∂z
µ
,
∂
∂z
ν
= g
J
∂
∂z
µ
, J
∂
∂z
ν
=−g
∂
∂z
µ
,
∂
∂z
ν
=−g
µν
hence g
µν
= 0. We also find that g
µν
= 0. Thus, the Hermitian metric g takes
the form
g = g
µν
dz
µ
⊗ dz
ν
+ g
µν
dz
µ
⊗ dz
ν
. (8.53)
[Remark: Take X, Y ∈ T
p
M
+
. Define an inner product h
p
in T
p
M
+
by
h
p
(X, Y ) ≡ g
p
(X, Y ). (8.54)
It is easy to see that h
p
is a positive-definite Hermitian form in T
p
M
+
. In fact,
h(X, Y ) = g(X, Y ) = g(X, Y ) = h(Y, X)
and h(X, X) = g(X,
X) = g(X
1
, X
1
) + g(X
2
, X
2
) ≥ 0forX = X
1
+iX
2
.This
is why a metric g satisfying (8.50) is called Hermitian.]
8.4.2 K
¨
ahler form
Let (M, g) be a Hermitian manifold. Define a tensor field whose action on
X, Y ∈ T
p
M is
p
(X, Y ) = g
p
(J
p
X, Y ) X, Y ∈ T
p
M. (8.55)
Note that is anti-symmetric, (X, Y ) = g(JX, Y ) = g(J
2
X, JY) =
−g( JY, X) =−(Y, X ). Hence, defines a two-form called the K
¨
ahler form
of a Hermitian metric g. Observe that is invariant under the action of J ,
(JX, JY) = g( J
2
X, JY) = g(J
3
X, J
2
Y ) = (X, Y ). (8.56)
If the domain is extended from T
p
M to T
p
M , is a two-form of bidegree
(1, 1). Indeed, for the metric (8.53), it is found that
∂
∂z
µ
,
∂
∂z
ν
= g
J
∂
∂z
µ
,
∂
∂z
ν
= ig
µν
= 0.
We also have
∂
∂z
µ
,
∂
∂z
ν
= 0,
∂
∂z
µ
,
∂
∂z
ν
= ig
µν
=−
∂
∂z
ν
,
∂
∂z
µ
.
Thus, the components of are
µν
=
µν
= 0
µ
ν
=−
ν
µ
= ig
µ
ν
. (8.57)
We may write
= ig
µν
dz
µ
⊗ dz
ν
− ig
νµ
dz
ν
⊗ dz
µ
= ig
µν
dz
µ
∧ dz
ν
. (8.58)
is also written as
=−J
µν
dz
µ
∧ dz
ν
(8.59)
where J
µν
= g
µλ
J
λ
ν
=−ig
µν
. is a real form;
=−ig
µ
ν
dz
µ
∧ dz
ν
= ig
ν
µ
dz
ν
∧ dz
µ
= . (8.60)
Making use of the K¨ahler form, we show that any Hermitian manifold, and
hence any complex manifold, is orientable. We first note that we may choose an
orthonormal basis {ˆe
1
, J ˆe
1
,...,ˆe
m
, J ˆe
m
}. In fact, if g( ˆe
1
, ˆe
1
) = 1, it follows
that g(J ˆe
1
, J ˆe
1
) = g( ˆe
1
, ˆe
1
) = 1andg(ˆe
1
, J ˆe
1
) =−g( J ˆe
1
, ˆe
1
) = 0. Thus ˆe
1
and J ˆe
1
form an orthonormal basis of a two-dimensional subspace. Now take ˆe
2
which is orthonormal to ˆe
1
and J ˆe
1
and form the subspace {ˆe
2
, J ˆe
2
}. Repeating
this procedure we obtain an orthonormal basis {ˆe
1
, J ˆe
1
,...,ˆe
m
, J ˆe
m
}.
Lemma 8.1. Let be the K¨ahler form of a Hermitian manifold with dim
M =
m.Then
∧ ...∧
,
-. /
m
is a nowhere vanishing 2m-form.
Proof. For the previous orthonormal basis, we have
(ˆe
i
, J ˆe
j
) = g( J ˆe
i
, J ˆe
j
) = δ
ij
(ˆe
i
, ˆe
j
) = (J ˆe
i
, J ˆe
j
) = 0.
Then it follows that
∧ ...∧
,
-. /
m
(ˆe
1
, J ˆe
1
,...,ˆe
m
, J ˆe
m
)
=
P
(ˆe
P(1)
, J ˆe
P(1)
)...(ˆe
P(m)
, J ˆe
P(m)
)
= m!(ˆe
1
, J ˆe
1
)...(ˆe
m
, J ˆe
m
) = m!
where P is an element of the permutation group of m objects. This shows that
∧ ...∧ cannot vanish at any point.
Since the real 2m-form ∧...∧ vanishes nowhere, it serves as a volume
element. Thus, we obtain the following theorem.
Theorem 8.3. A complex manifold is orientable.
8.4.3 Covariant derivatives
Let (M, g) be a Hermitian manifold. We define a connection which is compatible
with the complex structure. It is natural to assume that a holomorphic vector
V ∈ T
p
M
+
parallel transported to another point q is, again, a holomorphic vector
˜
V (q) ∈ T
q
M
+
. We show later that the almost complex structure is covariantly
conserved under this requirement. Let {z
µ
} and {z
µ
+z
µ
} be the coordinates of
p and q, respectively, and let V = V
µ
∂/∂z
µ
|
p
and
˜
V (q) =
˜
V
µ
(z +z)∂/∂z
µ
|
q
.
We assume that (cf (7.9))
˜
V
µ
(z + z) = V
µ
(z) − V
λ
(z)
µ
νλ
(z)z
ν
. (8.61)
Then the basis vectors satisfy (cf (7.14))
∇
µ
∂
∂z
ν
=
λ
µν
(z)
∂
∂z
λ
. (8.62a)
Since ∂/∂
z
µ
is a conjugate vector field of ∂/∂z
µ
,wehave
∇
µ
∂
∂z
ν
=
λ
µ ν
∂
∂z
λ
(8.62b)
where
λ
µ ν
=
λ
µν
.
λ
µν
and
λ
µ ν
are the only non-vanishing components of
the connection coefficients. Note that ∇
µ
∂/∂z
ν
=∇
µ
∂/∂z
ν
= 0. For the dual
basis, non-vanishing covariant derivatives are
∇
µ
dz
ν
=−
ν
µλ
dz
λ
∇
µ
dz
ν
=−
ν
µλ
z
λ
. (8.63)
The covariant derivative of X
+
= X
µ
∂/∂z
µ
∈ (M)
+
is
∇
µ
X
+
= (∂
µ
X
λ
+ X
ν
λ
µν
)
∂
∂z
λ
(8.64)
where ∂
µ
≡ ∂/∂z
µ
.ForX
−
= X
µ
∂/∂z
µ
∈ (M)
−
,wehave
∇
µ
X
−
= ∂
µ
X
λ
∂
∂z
λ
(8.65)
since
λ
µν
=
λ
µ
ν
= 0. As far as anti-holomorphic vectors are concerned, ∇
µ
works as the ordinary derivative ∂
µ
. Similarly, we have
∇
µ
X
+
= ∂
µ
X
λ
∂
∂z
λ
(8.66)
∇
µ
X
−
= (∂
µ
X
λ
+ X
ν
λ
µ ν
)
∂
∂z
λ
. (8.67)
It is easy to generalize this to an arbitrary tensor field. For example, if t =
t
µν
λ
dz
µ
⊗ dx
ν
⊗ ∂/∂z
λ
,wehave
(∇
κ
t)
µν
λ
= ∂
κ
t
µν
λ
− t
ξν
λ
ξ
κµ
− t
µξ
λ
ξ
κν
(∇
κ
t)
µν
λ
= ∂
κ
t
µν
λ
+ t
µν
ξ
λ
κξ
.
We require the metric compatibility as in section 7.2. We demand that
∇
κ
g
µν
=∇
κ
g
µν
= 0. In components, we have
∂
κ
g
µν
− g
λν
λ
κµ
= 0 ∂
κ
g
µν
− g
µλ
λ
κ µ
= 0. (8.68)
The connection coefficients are easily read off:
λ
κµ
= g
νλ
∂
κ
g
µν
λ
κ ν
= g
λµ
∂
κ
g
µν
(8.69)
where {g
νλ
}is the inverse matrix of g
µν
; g
µλ
g
λν
= δ
µ
ν
, g
νλ
g
λµ
= δ
ν
µ
. A metric-
compatible connection for which (mixed indices) = 0iscalledtheHermitian
connection. By construction, this is unique and given by (8.69).
Theorem 8.4. The almost complex structure J is covariantly constant with respect
to the Hermitian connection,
(∇
κ
J )
ν
µ
= (∇
κ
J )
ν
µ
= (∇
κ
J )
ν
µ
= (∇
κ
J )
ν
µ
= 0. (8.70)
Proof. We prove the first equality. From (8.22), we find
(∇
κ
J )
ν
µ
= ∂
κ
iδ
ν
µ
− iδ
ξ
µ
ξ
κν
+ iδ
ν
ξ
µ
κξ
= 0.
Other equalities follow from similar calculations.
8.4.4 Torsion and curvature
The torsion tensor T and the Riemann curvature tensor R are defined by
T (X, Y ) =∇
X
Y −∇
Y
X −[X, Y ] (8.71)
R(X, Y )Z =∇
X
∇
Y
Z −∇
Y
∇
X
Z −∇
[X,Y ]
Z. (8.72)
We find that
T
∂
∂z
µ
,
∂
∂z
ν
= (
λ
µν
−
λ
νµ
)
∂
∂z
λ
T
∂
∂z
µ
,
∂
∂z
ν
= T
∂
∂z
µ
,
∂
∂z
ν
= 0
T
∂
∂z
µ
,
∂
∂z
ν
= (
λ
µν
−
λ
νµ
)
∂
∂z
λ
.
The non-vanishing components are
T
λ
µν
=
λ
µν
−
λ
νµ
= g
ξλ
(∂
µ
g
νξ
− ∂
ν
g
µξ
) (8.73a)
T
λ
µν
=
λ
µν
−
λ
νµ
= g
λξ
∂
µ
g
νξ
− ∂
ν
g
µξ
. (8.73b)
As for the Riemann tensor, we find, for example, that
R
κ
λµν
= ∂
µ
κ
νλ
− ∂
ν
κ
µλ
+
η
νλ
κ
µη
−
η
µλ
κ
νη
.
If (8.69) is substituted, we find that
R
κ
λµν
= ∂
µ
g
ξκ
∂
ν
g
λξ
+ g
ξκ
∂
µ
∂
ν
g
λξ
− ∂
ν
g
ξκ
∂
µ
g
λξ
− g
ξκ
∂
µ
∂
ν
g
λξ
+ g
ξη
∂
ν
g
λξ
g
ζκ
∂
µ
g
ηζ
− g
ξη
∂
µ
g
λξ
g
ζκ
∂
ν
g
ηζ
= 0
where use has been made of the identity g
ζκ
∂
µ
g
ηζ
=−g
ηζ
∂
µ
g
ζκ
etc. In general,
we find that
R
κ
λAB
= R
κ
λAB
= R
A
Bκλ
= R
A
B
κλ
= 0 (8.74)
where A and B are any (holomorphic or anti-holomorphic) indices. As a result,
we are left only with the components R
κ
λ
µ
ν
, R
κ
λµ
ν
, R
κ
λµν
and R
κ
λµν
. Note that
we have a trivial symmetry R
κ
λ
µν
=−R
κ
λν
µ
. So the independent components
are reduced to R
κ
λ
µ
ν
and R
κ
λµν
= R
κ
λ
µ
ν
.Wefindthat
R
κ
λ
µν
= ∂
µ
κ
νλ
= ∂
µ
(g
ξκ
∂
ν
g
λξ
) (8.75a)
R
κ
λµν
= ∂
µ
κ
νλ
= ∂
µ
(g
κξ
∂
ν
g
ξ λ
). (8.75b)
Exercise 8.7. Show that
R
κ
λ
µ
ν
≡g
κ
ξ
R
ξ
λ
µ
ν
= ∂
µ
∂
ν
g
λ
κ
− g
ηξ
∂
µ
g
κ
ξ
∂
ν
g
λ
η
(8.76a)
R
κλµν
≡g
κξ
R
ξ
λµν
= ∂
µ
∂
ν
g
λκ
− g
ηξ
∂
µ
g
κξ
∂
ν
g
λη
(8.76b)
R
κλµν
≡g
κξ
R
ξ
λµ
ν
=−R
κλνµ
(8.76c)
R
κλµν
≡g
κξ
R
ξ
λµν
=−R
κλνµ
. (8.76d)
Verify the symmetries
R
κλµν
=−R
λκµν
R
κλµν
=−R
λκµν
. (8.77)
Let us contract the indices of the Riemann tensor as
µ
ν
≡ R
κ
κµ
ν
=−∂
ν
(g
κξ
∂
µ
g
κξ
) =−∂
ν
∂
µ
log G (8.78)
where G ≡ det(g
µ
ν
) =
√
g. To obtain the last equality, we used an identity
δG = Gg
µν
δg
µν
; see (7.204). We define the Ricci form by
≡ i
µν
dz
µ
∧ dz
ν
= i∂∂ log G. (8.79)
is a real form; =−i∂∂ log G =−i∂∂ log G = . From the identity
∂
∂ =−
1
2
d (∂ − ∂),wefind is closed; d ∝ d
2
(∂ − ∂)log G = 0. However,
this does not imply that
is exact. In fact, G is not a scalar and (∂−∂)log G is not
defined globally.
defines a non-trivial element c
1
(M) ≡[ /2π]∈H
2
(M; )
called the first Chern class. We discuss this further in section 11.2.
Proposition 8.2. The first Chern class c
1
(M) is invariant under a smooth change
of the metric g → g + δg.
Proof. It follows from (7.204) that δ log G = g
µν
δg
µ
ν
.Then
δ
= δi∂∂ log G = i∂∂g
µν
δg
µ
ν
=−
1
2
d (∂ − ∂)ig
µν
δg
µ
ν
.
Since g
µν
δg
µν
is a scalar, ω ≡−
1
2
(∂ −∂)g
µν
δg
µν
is a well-defined one-form on
M. Thus, δ
= dω is an exact two-form and [ ]=[ + δ ], namely c
1
(M) is
left invariant under g → g + δg.
8.5 K
¨
ahler manifolds and K
¨
ahler differential geometry
8.5.1 Definitions
Definition 8.4. A K
¨
ahler manifold is a Hermitian manifold (M, g) whose K¨ahler
form is closed: d = 0. The metric g is called the K
¨
ahler metric of M.
[Warning: Not all complex manifolds admit K¨ahler metrics.]
Theorem 8.5. A Hermitian manifold (M, g) is a K¨ahler manifold if and only if
the almost complex structure J satisfies
∇
µ
J = 0 (8.80)
where ∇
µ
is the Levi-Civita connection associated with g.
Proof. We first note that for any r-form ω,dω is written as
dω =∇ω ≡
1
r!
∇
µ
ω
ν
1
...ν
r
dx
µ
∧ dx
ν
1
∧ ...∧ dx
ν
r
. (8.81)
[For example,
∇ =
1
2
∇
λ
µν
dx
λ
∧ dx
µ
∧ dx
ν
=
1
2
(∂
λ
µν
−
κ
λµ
κν
−
κ
λν
µκ
) dx
λ
∧ dx
µ
∧ dx
ν
=
1
2
∂
λ
µν
dx
λ
∧ dx
µ
∧ dx
ν
= d
since is symmetric.] Now we prove that ∇
µ
J = 0 if and only if ∇
µ
= 0. We
verify the following equalities:
(∇
Z
)(X, Y ) =∇
Z
[(X, Y )]−(∇
Z
X, Y ) − (X, ∇
Z
Y )
=∇
Z
[g( JX, Y )]−g( J ∇
Z
X, Y ) − g(JX, ∇
Z
Y )
= (∇
Z
g)( JX, Y ) + g(∇
Z
JX, Y ) − g( J ∇
Z
X, Y )
= g(∇
Z
JX − J ∇
Z
X, Y ) = g((∇
Z
J )X, Y )
where ∇
Z
g = 0 has been used. Since this is true for any X, Y, Z, it follows that
∇
Z
= 0 if and only if ∇
Z
J = 0.
Theorems 8.4 and 8.5 show that the Riemann structure is compatible with
the Hermitian structure in the K¨ahler manifold.
Let g be a K¨ahler metric. Since d = 0, we have
(∂ +
∂)ig
µ
ν
dz
µ
∧ dz
ν
= i∂
λ
g
µν
dz
λ
∧ dz
µ
∧ dz
ν
+ i∂
λ
g
µν
dz
λ
∧ dz
µ
∧ dz
ν
=
1
2
i(∂
λ
g
µ
ν
− ∂
µ
g
λ
ν
) dz
λ
∧ dz
µ
∧ dz
ν
+
1
2
i(∂
λ
g
µ
ν
− ∂
ν
g
µλ
) dz
λ
∧ dz
µ
∧ dz
ν
= 0
from which we find
∂g
µ
ν
∂z
λ
=
∂g
λν
∂z
µ
∂g
µ
ν
∂z
λ
=
∂g
µλ
∂z
ν
. (8.82)