112 CHAPTER 3. MATHEMATICAL FOUNDATIONS
be the embedding function. Consider the vector product ofnvectors inRn:
[
∂~r
∂x^1
,···,
∂~r
∂xn
]
=
∣
∣
∣
∣
∣∣
∣
∣
∣
~e 1 ··· ~en+ 1
∂ 1 r 1 ··· ∂ 1 rn+ 1
..
.
..
.
∂nr 1 ··· ∂nrn+ 1
∣
∣
∣
∣
∣∣
∣
∣
∣
,
where{~e 1 ,···,~en+ 1 }is an orthogonal basis ofRn+^1. Then the volume elementΩdxis
Ωdx=
∣
∣
∣
∣
[
∂~r
∂x^1
,···,
∂~r
∂xn
]∣∣
∣
∣dx.
Bygij=∂∂~xri·∂∂x~rj, the norm of the vector[∂ 1 ~r,···,∂n~r]is
|[∂ 1 ~r,···,∂n~r]|=
√
−g, g=det(gij).
Thus (3.1.14) follows.
We now verify the invariance of the volume element. Under thetransformation (3.1.9),
(3.1.15) dx ̃=dx ̃^1 ∧ ··· ∧d ̃xn
=
(
∂ φ^1
∂x^1
dx^1 +···+
∂ φ^1
∂xn
dxn
)
∧ ··· ∧
(
∂ φn
∂x^1
dx^1 +···+
∂ φn
∂xn
dxn
)
=det
(
∂ φi
∂xj
)
dx^1 ∧ ··· ∧dxn.
On the other hand,
(3.1.16) ( ̃gij) =
(
∂ ψi
∂yj
)
(gij)
(
∂ ψi
∂yj
)T
, ψ=φ−^1.
Hence
(3.1.17)
√
det(g ̃ij) =det
(
∂ φi
∂xj
)− (^1) √
det(gij).
We deduce from (3.1.15) and (3.1.17) that
√
det( ̃gij)d ̃x=
√
det(gij)dx.
Namely, both the volume and the volume element in (3.1.13)-(3.1.14) are invariant.
- The metric{gij}gives rise to an inner product structure on the tangent spaceof a
Riemann manifoldM, and defines the angle between two tangent vectors.
Letp∈Mbe a given point, andTpMbe the tangent space atp∈M. For two vectors
X,Y∈TpM,
X={X^1 ,···,Xn}, Y={Y^1 ,···,Yn},