Nakahara M. Geometry, Topology and Physics
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These vectors generate the Lie algebra (3). This reflects the fact that S
2
is the homogeneous space SO(3)/SO(2) and the metric on S
2
retains this SO(3)
symmetry (see example 5.18(a)). In general S
n
= SO(n + 1)/SO(n) with the
usual metric has dim SO(n + 1) = n(n + 1)/2 Killing vectors and they form
the Lie algebra
(n + 1). The sphere S
n
with the usual metric is a maximally
symmetric space. We may squash S
n
so that it has fewer symmetries. For
example, if S
2
considered here is squashed along the z-axis it has a rotational
symmetry around the z-axis only and there exists one Killing vector field L
z
=
∂/∂φ.
7.7.2 Conformal Killing vector fields
Let (M, g) be a Riemannian manifold and let X ∈
(M). If an infinitesimal
displacement given by ε X generates a conformal transformation, the vector field
X is called a conformal Killing vector field (CKV). Under the displacement
x
µ
→ x
µ
+ ε X
µ
, this condition is written as
∂(x
κ
+ εX
κ
)
∂x
µ
∂(x
λ
+ ε X
λ
)
∂x
ν
g
κλ
(x + εX) = e
2σ
g
µν
(x ).
Noting that σ ∝ ε,wesetσ = εψ/2, where ψ ∈
(M). Then we find that g
µν
and X
µ
satisfy
X
g
µν
= X
ξ
∂
ξ
g
µν
+ ∂
µ
X
κ
g
κν
+ ∂
ν
X
λ
g
µλ
= ψg
µν
. (7.129a)
Equation (7.129a) is easily solved for ψ to yield
ψ =
X
ξ
g
µν
∂
ξ
g
µν
+ 2∂
µ
X
µ
m
(7.129b)
where m = dim M.Weverifythat
(i) a linear combination of CKVs is a CKV: (
aX+bY
g)
µν
= (aϕ + bψ)g
µν
where a, b ∈ ,
X
g
µν
= ϕg
µν
and
Y
g
µν
= ψg
µν
;
(ii) the Lie bracket [X, Y ] of a CKV is again a CKV:
[X,Y ]
g
µν
= (X[ψ]−
Y [ϕ])g
µν
.
Example 7.12. Let x
µ
be the coordinates of (
m
,δ). The vector
D ≡ x
µ
∂
∂x
µ
(7.130)
(dilatation vector) is a CKV. In fact,
D
δ
µν
= ∂
µ
x
κ
δ
κν
+ ∂
ν
x
λ
δ
µλ
= 2δ
µν
.
7.8 Non-coordinate bases
7.8.1 Definitions
In the coordinate basis, T
p
M is spanned by {e
µ
}={∂/∂x
µ
} and T
∗
p
M by {dx
µ
}.
If, moreover, M is endowed with a metric g, there may be an alternative choice.
Let us consider the linear combination,
ˆe
α
= e
α
µ
∂
∂x
µ
{e
α
µ
}∈GL(m, ) (7.131)
where det e
α
µ
> 0. In other words, {ˆe
α
} is the frame of basis vectors which is
obtained by a GL(m,
)-rotation of the basis {e
µ
} preserving the orientation. We
require that {ˆe
α
} be orthonormal with respect to g,
g( ˆe
α
, ˆe
β
) = e
α
µ
e
β
ν
g
µν
= δ
αβ
. (7.132a)
If the manifold is Lorentzian, δ
αβ
should be replaced by η
αβ
. We easily reverse
(7.132a),
g
µν
= e
α
µ
e
β
ν
δ
αβ
(7.132b)
where e
α
µ
is the inverse of e
α
µ
; e
α
µ
e
α
ν
= δ
µ
ν
, e
α
µ
e
β
µ
= δ
α
β
. [We have used
the same symbols for a matrix and its inverse. So long as the indices are written
explicitly it does not cause confusion.] Since a vector V is independent of the
basis chosen, we have V = V
µ
e
µ
= V
α
ˆe
α
= V
α
e
α
µ
e
µ
. It follows that
V
µ
= V
α
e
α
µ
V
α
= e
α
µ
V
µ
. (7.133)
Let us introduce the dual basis {
ˆ
θ
α
} defined by
ˆ
θ
α
, ˆe
β
=δ
α
β
.
ˆ
θ
α
is given
by
ˆ
θ
α
= e
α
µ
dx
µ
. (7.134)
In terms of {
ˆ
θ
α
}, the metric is
g = g
µν
dx
µ
⊗ dx
ν
= δ
αβ
ˆ
θ
α
⊗
ˆ
θ
β
. (7.135)
The bases {ˆe
α
} and {
ˆ
θ
α
} are called the non-coordinate bases.Weuseκ, λ,
µ, ν, ... (α, β, γ, δ,...) to denote the coordinate (non-coordinate) basis. The
coefficients e
α
µ
are called the vierbeins if the space is four dimensional and
vielbeins if it is many dimensional. The non-coordinate basis has a non-vanishing
Lie bracket. If the {ˆe
α
} are given by (7.131), they satisfy
[ˆe
α
, ˆe
β
]|
p
= c
αβ
γ
( p) ˆe
γ
|
p
(7.136a)
where
c
αβ
γ
( p) = e
γ
ν
[e
α
µ
∂
µ
e
β
ν
− e
β
µ
∂
µ
e
α
ν
]( p). (7.136b)
Example 7.13. The standard metric on S
2
is
g = dθ ⊗ dθ + sin
2
θ dφ ⊗ dφ =
ˆ
θ
1
⊗
ˆ
θ
1
+
ˆ
θ
2
⊗
ˆ
θ
2
(7.137)
where
ˆ
θ
1
= dθ and
ˆ
θ
2
= sin θ dφ. The ‘zweibeins’ are
e
1
θ
= 1 e
1
φ
= 0
e
2
θ
= 0 e
2
φ
= sin θ.
(7.138)
The non-vanishing components of c
αβ
γ
are c
12
2
=−c
21
2
=−cot θ .
Exercise 7.19. (a) Verify the identities,
δ
αβ
= g
µν
e
α
µ
e
β
ν
g
µν
= δ
αβ
e
α
µ
e
β
ν
. (7.139)
(b) Let γ
α
be the Dirac matrices in Minkowski spacetime, which satisfy
{γ
α
,γ
β
}=2η
αβ
. Define the curved spacetime counterparts of the Dirac matrices
by γ
µ
≡ e
α
µ
γ
α
. Show that
{γ
µ
,γ
ν
}=2g
µν
. (7.140)
7.8.2 Cartan’s structure equations
In section 7.3 the curvature tensor R and the torsion tensor T have been defined
by
R(X, Y )Z =∇
X
∇
Y
Z −∇
Y
∇
X
Z −∇
[X,Y ]
Z
T (X, Y ) =∇
X
Y −∇
Y
X −[X, Y ].
Let {ˆe
α
} be the non-coordinate basis and {
ˆ
θ
α
} the dual basis. The vector fields
{ˆe
α
} satisfy [ˆe
α
, ˆe
β
]=c
αβ
γ
ˆe
γ
. Define the connection coefficients with respect to
the basis {ˆe
α
} by
∇
α
ˆe
β
≡∇
ˆe
α
ˆe
β
=
γ
αβ
ˆe
γ
. (7.141)
Let ˆe
α
= e
α
µ
e
µ
. Then (7.141) becomes e
α
µ
(∂
µ
e
β
ν
+e
β
λ
ν
µλ
)e
ν
=
γ
αβ
e
γ
ν
e
ν
,
from which we find that
γ
αβ
= e
γ
ν
e
α
µ
(∂
µ
e
β
ν
+ e
β
λ
ν
µλ
) = e
γ
ν
e
α
µ
∇
µ
e
β
ν
. (7.142)
The components of T and R in this basis are given by
T
α
βγ
=
ˆ
θ
α
, T ( ˆe
β
, ˆe
γ
)=
ˆ
θ
α
, ∇
β
ˆe
γ
−∇
γ
ˆe
β
−[ˆe
β
, ˆe
γ
]
=
α
βγ
−
α
γβ
− c
βγ
α
. (7.143)
R
α
βγδ
=
ˆ
θ
α
, ∇
γ
∇
δ
ˆe
β
−∇
δ
∇
γ
ˆe
β
−∇
[ˆe
γ
, ˆe
δ
]
ˆe
β
=
ˆ
θ
α
, ∇
γ
(
ε
δβ
ˆe
ε
) −∇
δ
ε
γβ
ˆe
ε
) − c
γδ
ε
∇
ε
ˆe
β
=ˆe
γ
[
α
δβ
]−ˆe
δ
[
α
γβ
]+
ε
δβ
α
γε
−
ε
γβ
α
δε
− c
γδ
ε
α
εβ
.
(7.144)
We define a matrix-valued one-form {ω
α
β
} called the connection one-form by
ω
α
β
≡
α
γβ
ˆ
θ
γ
. (7.145)
Theorem 7.3. The connection one-form ω
α
β
satisfies Cartan’s structure
equations,
d
ˆ
θ
α
+ ω
α
β
∧
ˆ
θ
β
= T
α
(7.146a)
dω
α
β
+ ω
α
γ
∧ ω
γ
β
= R
α
β
(7.146b)
where we have introduced the torsion two-form T
α
≡
1
2
T
α
βγ
ˆ
θ
β
∧
ˆ
θ
γ
and the
curvature two-form R
α
β
≡
1
2
R
α
βγδ
ˆ
θ
γ
∧
ˆ
θ
δ
.
Proof. Let the LHS of (7.146a) act on the basis vectors ˆe
γ
and ˆe
δ
,
d
ˆ
θ
α
(ˆe
γ
, ˆe
δ
) +[ω
α
β
, ˆe
γ
ˆ
θ
β
, ˆe
δ
−
ˆ
θ
β
, ˆe
γ
ω
α
β
, ˆe
δ
]
={ˆe
γ
[
ˆ
θ
α
, ˆe
δ
] − ˆe
δ
[
ˆ
θ
α
, ˆe
γ
] −
ˆ
θ
α
, [ˆe
γ
, ˆe
δ
]}+ {ω
α
δ
, ˆe
γ
−ω
α
γ
, ˆe
δ
}
=−c
γδ
α
+
α
γδ
−
α
δγ
= T
α
γδ
where use has been made of (5.70). The RHS acting on ˆe
γ
and ˆe
δ
yields
1
2
T
α
βε
[
ˆ
θ
β
, ˆe
γ
ˆ
θ
ε
, ˆe
δ
−
ˆ
θ
ε
, ˆe
γ
ˆ
θ
β
, ˆe
δ
] = T
α
γδ
which verifies (7.146a).
Equation (7.146b) may be proved similarly (exercise).
Taking the exterior derivative of (7.146a) and (7.146b), we have the Bianchi
identities
dT
α
+ ω
α
β
∧ T
β
= R
α
β
∧
ˆ
θ
β
(7.147a)
dR
α
β
+ ω
α
γ
∧ R
γ
β
− R
α
γ
∧ ω
γ
β
= 0. (7.147b)
These are the non-coordinate basis versions of (7.81a) and (7.81b).
7.8.3 The local frame
In an m-dimensional Riemannian manifold, the metric tensor g
µν
has m(m +1)/2
degrees of freedom while the vielbein e
α
µ
has m
2
degrees of freedom. There are
many non-coordinate bases which yield the same metric, g, each of which is
related to the other by the local orthogonal rotation,
ˆ
θ
α
−→
ˆ
θ
α
( p) =
α
β
( p)
ˆ
θ
β
( p) (7.148)
at each point p. The vielbein transforms as
e
α
µ
( p) −→ e
α
µ
( p) =
α
β
( p)e
β
µ
( p). (7.149)
Unlike κ,λ,µ,ν,... which transform under coordinate changes, the indices
α,β,γ,... transform under the local orthogonal rotation and are inert under
coordinate changes. Since the metric tensor is invariant under the rotation,
α
β
satisfies
α
β
δ
αδ
δ
γ
= δ
βγ
if M is Riemannian (7.150a)
α
β
η
αδ
δ
γ
= η
βγ
if M is Lorentzian. (7.150b)
This implies that {
α
β
( p)}∈SO(m) if M is Riemannian with dim M = m and
{
α
β
( p)}∈SO(m −1, 1) if M is Lorentzian. The dimension of these Lie groups
is m(m − 1)/2 = m
2
− m(m + 1)/2, that is the difference between the degrees
of freedom of e
α
µ
and g
µν
. Under the local frame rotation
α
β
( p), the indices
α,β,γ,δ,... are rotated while κ,λ,µ,ν,... (world indices) are not affected.
Under the rotation (7.148), the basis vector transforms as
ˆe
α
−→ ˆe
α
=ˆe
β
(
−1
)
β
α
. (7.151)
Let t = t
µ
ν
e
µ
⊗ dx
ν
be a tensor field of type (1, 1). In the bases {ˆe
α
}
and {
ˆ
θ
α
},wehavet = t
α
β
ˆe
α
⊗
ˆ
θ
β
,wheret
α
β
= e
α
µ
e
β
ν
t
µ
ν
. If the new
frames {ˆe
α
}={ˆe
β
(
−1
)
β
α
} and {
ˆ
θ
α
}={
α
β
ˆ
θ
β
} are employed, the tensor t
is expressed as
t = t
α
β
ˆe
α
⊗
ˆ
θ
β
= t
α
β
ˆe
γ
(
−1
)
γ
α
⊗
β
δ
ˆ
θ
δ
from which we find the transformation rule,
t
α
β
−→ t
α
β
=
α
γ
t
γ
δ
(
−1
)
δ
β
.
To summarize, the upper (lower) non-coordinate indices are rotated by (
−1
).
The change from the coordinate basis to the non-coordinate basis is carried out
by multiplications of vielbeins.
From these facts we find the transformation rule of the connection one-form
ω
α
β
. The torsion two-form transforms as
T
α
−→ T
α
= d
ˆ
θ
α
+ ω
α
β
∧
ˆ
θ
β
=
α
β
[d
ˆ
θ
β
+ ω
β
γ
∧
ˆ
θ
γ
].
Substituting
ˆ
θ
α
=
α
β
ˆ
θ
β
into this equation, we find that
ω
α
β
β
γ
=
α
δ
ω
δ
γ
− d
α
γ
.
Multiplying both sides by
−1
from the right, we have
ω
α
β
=
α
γ
ω
γ
δ
(
−1
)
δ
β
+
α
γ
(d
−1
)
γ
β
(7.152)
where use has been made of the identity d
−1
+ d
−1
= 0, which is derived
from
−1
= I
m
.
The curvature two-form transforms homogeneously as
R
α
β
−→ R
α
β
=
α
γ
R
γ
δ
(
−1
)
δ
β
(7.153)
under a local frame rotation .
7.8.4 The Levi-Civita connection in a non-coordinate basis
Let ∇ be a Levi-Civita connection on (M, g), which is characterized by the metric
compatibility ∇
X
g = 0, and the vanishing torsion
λ
µν
−
λ
νµ
= 0. It is
interesting to see how these conditions are expressed in the present approach.
The components
λ
µν
and
α
βγ
are related to each other by (7.142). Let (M, g)
be a Riemannian manifold (if (M, g) is Lorentzian, we simply replace δ
αβ
all
below by η
αβ
). If we define the Ricci rotation coefficient
αβγ
by δ
αδ
δ
βγ
the
metric compatibility is expressed as
αβγ
= δ
αδ
e
δ
λ
e
β
µ
∇
µ
e
γ
λ
=−δ
αδ
e
γ
λ
e
β
µ
∇
µ
e
δ
λ
=−δ
γδ
e
δ
λ
e
β
µ
∇
µ
e
α
λ
=−
γβα
(7.154)
where ∇
µ
g = 0 has been used. In terms of the connection one-form ω
αβ
≡
δ
αγ
ω
γ
β
, this becomes
ω
αβ
=−ω
βα
. (7.155)
The torsion-free condition is
d
ˆ
θ
α
+ ω
α
β
∧
ˆ
θ
β
= 0. (7.156)
The reader should verify that (7.156) implies the symmetry of the connection
coefficient
λ
µν
=
λ
νµ
in the coordinate basis. The condition (7.156) enables
us to compute the c
αβ
γ
of the basis {ˆe
α
}. Let us look at the commutation relation
c
αβ
γ
ˆe
γ
=[ˆe
α
, ˆe
β
]=∇
α
ˆe
β
−∇
β
ˆe
α
(7.157)
where the final equality follows from the torsion-free condition. From (7.141),
we find that
c
αβ
γ
=
γ
αβ
−
γ
βα
. (7.158)
Substituting (7.158) into (7.144) we may express the Riemaun curvature tensor in
terms of only,
R
α
βγδ
=ˆe
γ
[
α
δβ
]−ˆe
δ
[
α
γβ
]+
ε
δβ
α
γε
−
ε
γβ
α
δε
− (
ε
γδ
−
ε
δγ
)
α
εβ
. (7.159)
Example 7.14. Let us take the sphere S
2
of example 7.13. The components of
e
α
µ
are
e
1
θ
= 1 e
1
φ
= 0 e
2
θ
= 0 e
2
φ
= sin θ. (7.160)
We first note that the metric condition implies ω
11
= ω
22
= 0, hence ω
1
1
=
ω
2
2
= 0. Other connection one-forms are obtained from the torsion-free
conditions,
d(dθ) + ω
1
2
∧ (sin θ dφ) = 0 (7.161a)
d(sin θ dφ) + ω
2
1
∧ dθ = 0. (7.161b)
From the second equation of (7.161), we easily see that ω
2
1
= cos θdφ and the
metric condition ω
12
=−ω
21
implies ω
1
2
=−cos θ dφ. The Riemann tensor is
also found from Cartan’s structure equation,
ω
1
2
∧ ω
2
1
=
1
2
R
1
1αβ
ˆ
θ
α
∧
ˆ
θ
β
(7.162a)
dω
1
2
=
1
2
R
1
2αβ
ˆ
θ
α
∧
ˆ
θ
β
(7.162b)
dω
2
1
=
1
2
R
2
1αβ
ˆ
θ
α
∧
ˆ
θ
β
(7.162c)
ω
2
1
∧ ω
1
2
=
1
2
R
2
2αβ
ˆ
θ
α
∧
ˆ
θ
β
. (7.162d)
The non-vanishing components are R
1
212
=−R
1
221
= sin θ , R
2
112
=
−R
2
121
=−sin θ . The transition to the coordinate basis expression is carried
out with the help of e
α
µ
and e
α
µ
. For example,
R
θ
φθφ
= e
α
θ
e
β
φ
e
γ
θ
e
δ
φ
R
α
βγδ
=
1
sin
2
θ
R
1
212
=
1
sin θ
.
Example 7.15. The Schwarzschild metric is given by
ds
2
=−
1 −
2M
r
dt
2
+
1
1 −
2M
r
dr
2
+r
2
(dθ
2
+ sin
2
θ dφ
2
)
=−
ˆ
θ
0
⊗
ˆ
θ
0
+
ˆ
θ
1
⊗
ˆ
θ
1
+
ˆ
θ
2
⊗
ˆ
θ
2
+
ˆ
θ
3
⊗
ˆ
θ
3
(7.163)
where
ˆ
θ
0
=
1 −
2M
r
1/2
dt
ˆ
θ
1
=
1 −
2M
r
−1/2
dr
ˆ
θ
2
= r dθ
ˆ
θ
3
= r sin θ dφ.
(7.164)
The parameters run over the range 0 < 2M < r ,0≤ θ ≤ π and 0 ≤ φ<2π .
The metric condition yields ω
0
0
= ω
1
1
= ω
2
2
= ω
3
3
= 0 and the torsion-free
conditions are:
d[(1 − 2M/r )
1/2
dt]+ω
0
β
∧
ˆ
θ
β
= 0 (7.165a)
d[(1 −2M/r )
−1/2
dr]+ω
1
β
∧
ˆ
θ
β
= 0 (7.165b)
d(r dθ) + ω
2
β
∧
ˆ
θ
β
= 0 (7.165c)
d(r sin θ dφ) + ω
3
β
∧
ˆ
θ
β
= 0. (7.165d)
The non-vanishing components of the connection one-forms are
ω
0
1
= ω
1
0
=
M
r
2
dt ω
2
1
=−ω
1
2
=
1 −
2M
r
1/2
dθ
ω
3
1
=−ω
1
3
=
1 −
2M
r
1/2
sin θ dφω
3
2
=−ω
2
3
= cos θ dφ.
(7.166)
The curvature two-forms are found from the structure equations to be
R
0
1
= R
1
0
=
2M
r
3
ˆ
θ
0
∧
ˆ
θ
1
R
0
2
= R
2
0
=−
2M
r
3
ˆ
θ
0
∧
ˆ
θ
2
R
0
3
= R
3
0
=−
M
r
3
ˆ
θ
0
∧
ˆ
θ
3
R
1
2
=−R
2
1
=−
M
r
3
ˆ
θ
1
∧
ˆ
θ
2
R
1
3
=−R
3
1
=−
M
r
3
ˆ
θ
1
∧
ˆ
θ
3
R
2
3
=−R
3
2
=
2M
r
3
ˆ
θ
2
∧
ˆ
θ
3
.
(7.167)
7.9 Differential forms and Hodge theory
7.9.1 Invariant volume elements
We have defined the volume element as a non-vanishing m-form on an m-
dimensional orientable manifold M in section 5.5. If M is endowed with a
metric g, there exists a natural volume element which is invariant under coordinate
transformation. Let us define the invariant volume element by
M
≡
|g|dx
1
∧ dx
2
∧ ...∧ dx
m
(7.168)
where g = det g
µν
and x
µ
are the coordinates of the chart (U,ϕ).Them-form
M
is, indeed, invariant under a coordinate change. Let y
λ
be the coordinates of
another chart (V ,ψ)with U ∩ V =∅. The invariant volume element is
9
det
∂x
µ
∂y
κ
∂x
ν
∂y
λ
g
µν
dy
1
∧ ...∧ dy
m
in terms of the y-coordinates. Noting that dy
λ
= (∂y
λ
/∂x
µ
) dx
µ
, this becomes
det
∂x
µ
∂y
κ
|g|det
∂y
λ
∂x
ν
dx
1
∧ dx
2
∧ ...∧ dx
m
=±
|g|dx
1
∧ dx
2
∧ ...∧ dx
m
.
If x
µ
and y
κ
define the same orientation, det(∂ x
µ
/∂y
κ
) is strictly positive on
U ∩ V and
M
is invariant under the coordinate change.
Exercise 7.20. Let {
ˆ
θ
α
}={e
α
µ
dx
µ
} be the non-coordinate basis. Show that the
invariant volume element is written as
M
=|e|dx
1
∧ dx
2
∧ ...∧ dx
m
=
ˆ
θ
1
∧
ˆ
θ
2
∧ ...∧
ˆ
θ
m
(7.169)
where e = det e
α
µ
.
Now that we have defined the invariant volume element, it is natural to define
an integration of f ∈
(M) over M by
M
f
M
≡
M
f
|g|dx
1
dx
2
...dx
m
. (7.170)
Obviously (7.170) is invariant under a change of coordinates. In physics, there
are many objects which are expressed as volume integrals of this type, see
section 7.10.
7.9.2 Duality transformations (Hodge star)
As noted in section 5.4,
r
(M) is isomorphic to
m−r
(M) on an m-dimensional
manifold M.IfM is endowed with a metric g, we can define a natural
isomorphism between them called the Hodge ∗ operation. Define the totally
anti-symmetric tensor ε by
ε
µ
1
µ
2
...µ
m
=
+1if(µ
1
µ
2
...µ
m
) is an even permutation of (12 ...m)
−1if(µ
1
µ
2
...µ
m
) is an odd permutation of (12 ...m)
0otherwise.
(7.171a)
Note that
ε
µ
1
µ
2
...µ
m
= g
µ
1
ν
1
g
µ
2
ν
2
...g
µ
m
ν
m
ε
ν
1
ν
2
...ν
m
= g
−1
ε
µ
1
µ
2
...µ
m
. (7.171b)
The Hodge ∗ is a linear map ∗:
r
(M) →
m−r
(M) whose action on a basis
vector of
r
(M) is defined by
∗ (dx
µ
1
∧ dx
µ
2
∧ ...∧ dx
µ
r
)
=
√
|g|
(m −r)!
ε
µ
1
µ
2
...µ
r
ν
r+1
...ν
m
dx
ν
r+1
∧ ...∧ dx
ν
m
. (7.172)
It should be noted that ∗1 is the invariant volume element:
∗1 =
√
|g|
m!
ε
µ
1
µ
2
...µ
m
dx
µ
1
∧ ...∧ dx
µ
m
=
|g|dx
1
∧ ...∧ dx
m
.
For
ω =
1
r!
ω
µ
1
µ
2
...µ
r
dx
µ
1
∧ dx
µ
2
∧ ...∧ dx
µ
r
∈
r
(M)
we have
∗ω =
√
|g|
r!(m −r)!
ω
µ
1
µ
2
...µ
r
ε
µ
1
µ
2
...µ
r
ν
r+1
...ν
m
dx
ν
r+1
∧ ...∧ dx
ν
m
. (7.173)
If we take the non-coordinate basis {θ
α
}={e
α
µ
dx
µ
},the∗ operation
becomes
∗(
ˆ
θ
α
1
∧ ...∧
ˆ
θ
α
r
) =
1
(m −r )!
ε
α
1
...α
r
β
r+1
...β
m
ˆ
θ
β
r+1
∧ ...∧
ˆ
θ
β
m
(7.174)
where
ε
α
1
...α
m
=
+1if(α
1
...α
m
) is an even permutation of (12 ...m)
−1if(α
1
...α
m
) is an odd permutation of (12 ...m)
0otherwise
(7.175)
and the indices are raised by δ
αβ
or η
αβ
.
Theorem 7.4.
∗∗ω = (−1)
r(m−r)
ω. (7.176a)
if (M, g) is Riemannian and
∗∗ω = (−1)
1+r(m−r)
ω (7.176b)
if Lorentzian.
Proof. It is simpler to prove (7.176a) with a non-coordinate basis. Let
ω =
1
r!
ω
α
1
...α
r
ˆ
θ
α
1
∧ ...∧
ˆ
θ
α
r
.
Repeated applications of ∗ on ω yield
∗∗ω =
1
r!
ω
α
1
...α
r
1
(m −r )!
ε
α
1
...α
r
β
r+1
...β
m
×
1
r!
ε
β
r+1
...β
m
γ
1
...γ
r
ˆ
θ
γ
1
∧ ...∧
ˆ
θ
γ
r
=
(−1)
r(m−r)
r!r!(m −r )!
αβγ
ω
α
1
...α
r
ε
α
1
...α
r
β
r+1
...β
m
ε
γ
1
...γ
r
β
r+1
...β
m
×
ˆ
θ
γ
1
∧ ...∧
ˆ
θ
γ
r
=
(−1)
r(m−r)
r!
ω
α
1
...α
r
ˆ
θ
α
1
∧ ...∧
ˆ
θ
α
r
= (−1)
r(m−r)
ω
where use has been made of the identity
βγ
ε
α
1
...α
r
β
r+1
...β
m
ε
γ
1
...γ
r
β
r+1
...β
m
ˆ
θ
γ
1
∧ ...∧
ˆ
θ
γ
r
= r !(m −r )!
ˆ
θ
α
1
∧ ...∧
ˆ
θ
α
r
.
The proof of (7.176b) is left as an exercise to the reader (use det η =−1).
Thus, we find that (−1)
r(m−r)
∗∗(or (−1)
1+r(m−r)
∗∗) is an identity map
on
r
(M). We define the inverse of ∗ by
∗
−1
= (−1)
r(m−r)
∗ (M, g) is Riemannian (7.177a)
∗
−1
= (−1)
1+r(m−r)
∗ (M, g) is Lorentzian. (7.177b)
7.9.3 Inner products of r-forms
Take
ω =
1
r!
ω
µ
1
...µ
r
dx
µ
1
∧ ...∧ dx
µ
r
η =
1
r!
η
µ
1
...µ
r
dx
µ
1
∧ ...∧ dx
µ
r
.