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Now, the only formula for the volume element that we have at this point
(see Lemma 8.12.10 and its proof) is that, for a given tangent space, it is the wedge
product of the dual basis of an orthonormal basis for that tangent space. It is
convenient to have a formula for it in terms of any basis, not just an orthonormal
one.
8.12.18. Lemma.
Let
v
1
,
v
2
,...,
v
n
be an orthonormal basis for an n-dimensional
vector space
V
and
v
1
*,
v
2
*,...,
v
n
* its dual basis. Let
w
1
,
w
2
,...,
w
n
be any basis
for
V
and
w
1
*,
w
2
*,...,
w
n
* its dual basis. If
=
()
g
=∑
ww
and g
det
g
,
(8.42)
ij
i
j
ij
then
±
g
ww
*
Ÿ
*
Ÿ
....
Ÿ
wvv
*
=
*
Ÿ
*
Ÿ
...
Ÿ
v
*.
(8.43)
1
2
n
1
2
n
Proof.
Suppose that
n
Â
1
w
=
a
v
.
i
ij
j
j
=
It is easy to check that (g
ij
) = (a
ij
) (a
ij
)
T
and taking determinants of both sides we get
that g = (det (a
ij
))
2
. In particular, g ≥ 0. Since the vector space L
n
(
V
) has dimension 1,
to prove equation (8.43) we only need to show that both sides evaluate to the same
value on some nonzero element. It follows from equation (4.25d) that
g
(
ww
*
ŸŸŸ
(
*
...
www
*)
,
,...,
w
)
=
g
1
2
n
1
2
n
and
ŸŸ Ÿ
(
)
=
()
=
( *
vv
*
...
vww
*)
,
,...,
w
det
a
g
.
1
2
n
1
2
n
ij
The lemma is proved.
The significance of Lemma 8.12.18 is its application to finding volume elements
for manifolds.
8.12.19. Theorem.
Let
M
n
be an oriented Riemannian manifold. Let (
U
,j) be a
coordinate neighborhood for
M
and define functions
∂
∂
()
g
=
uu
and
∑
g
=
det
g
.
ij
ij
∂
∂
i
j
Then
dV
=±
g du
Ÿ
du
Ÿ
...
Ÿ
du
n
.
1
2
