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If we choose a coordinate neighborhood (
U
,j) so that j is orientation preserving, then
dV
=
g du
Ÿ
du
Ÿ
...
Ÿ
du
n
.
1
2
Proof.
One simply has to use Lemma 8.12.18 on each tangent space T
p
(
M
) for
p
Œ
U
. Recall that the vectors ∂/∂u
i
form a basis for the tangent spaces and the differen-
tial forms du
i
are the dual basis. If j is orientation preserving, the determinant of the
matrix (a
ij
) in the proof of Lemma 8.12.18 will be positive.
Theorem 8.12.19 shows that the volume element for an oriented Riemannian man-
ifold is readily determined from any orientation-preserving coordinate neighborhood.
To define the volume we had to integrate and we only showed how to do this over
oriented manifolds. What if a manifold is not orientable? Well, we can still integrate
something similar to the volume element. Intuitively, this makes sense since volume
is gotten by integration and it is natural to think of every manifold as having a
“volume” (with respect to a given Riemannian metric, of course, since area, volume,
etc. is meaningless without a metric).
Given a manifold
M
n
, a function w :
M
Æ
R
of the form
Definition.
()
=
L
(
()
)
wh
p
for some
h
Œ
T
M
,
p
p
p
that is,
( (
)
=
(
)
≥
()
w
pv v
,
,...,
v
h
v v
,
,...,
v
0
,
for
v
Œ
T
M
,
12
n
p
12
n
i
p
is called a
volume element
of
M
.
Note that we are not saying that a volume element is necessarily a differential
form. Furthermore, we are setting it up so that it will lead to an unsigned volume.
With respect to a coordinate neighborhood (
U
,j) it will look like
w=
f du
Ÿ
du
Ÿ
...
Ÿ
du
,
where f
≥
0
.
1
2
n
By Theorem 8.12.19,
w=
gdu
Ÿ
du
Ÿ
...
Ÿ
du
n
1
2
is one such volume element. One can show that it is possible to define an integral
Ú
w
over any manifold
M
for any volume element w on
M
that has compact support. Sec-
tions 9.2 and 9.8 will have more to say about volume.
Example 8.12.20.
Consider a surface
S
with a given Riemannian metric and
define a volume element by
( (
)
=
()
w
pv w
,
area of paralle
log
ram spanned by
v w
,
Œ
T
S
.
(8.44)
pp
pp p
