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},