Page 119 - Vector Analysis

§4.3 The Surface Integrals 115

Proof. Using the permutation symbol and Kronecker’s delta, we have

}ψ,1 ˆψ,2 }R2 3 = 3 ( 3 εijkψj,1 ψk,2 )( 3 εirs ψ r ,1 ψs,2 )

ÿ ÿ ÿ

i=1 j,k=1 r,s=1

3 [( 3 εirs)ψj ψ k ,2 ψ r ,1 ψs,2 ]

= ÿ ÿ εijk ,1

j,k,r,s=1 i=1

= 3 ( δjs δkr )ψj ,1 ψ k ,2 ψ r ,1 ψs,2 ,
δj r δks
ÿ

´

j,k,r,s=1

where we use the identity (4.9)

3

ÿεijkεirs = δjrδks ´ δjsδkr

i=1

to conclude the last equality. Therefore,

}ψ,1 ˆψ,2 }R2 3 = 3 (ψj ,1 ψ k ,2 ψj ,1 ψ k ,2 ´ψj ,1 ψ k ,2 ψj ,2 ψ k ,1 )

ÿ

j,k=1

= g11g22 ´ g12g21 = det(g) = g .

Finally, (4.8) is concluded from the fact that g is positive definite. ˝

Remark 4.45. Let L P B(R2; TpΣ) be given by

L(ae1 + be2) = aψ,1 +bψ,2 ,

where B2 = te1, e2u is the standard basis of R2. Let B1 = te1, e2u be an orthonormal basis

of TpΣ, and B3 = te1, e2, e3u be the standard basis of R3. Then

[ ][ [e1]TB3 ] [ ...[ψ,2 ]
= [e2]BT3 [ψ,1
[L]B2,B1 = ψ,1 ¨e1 ψ,2 ¨e1 ]B3 ]B3 .
ψ,1 ¨e2 ψ,2 ¨e2

By the fact that te1, e2u is an orthonormal basis,

[ [ψ,1 ]BT3 ] [ ...[e2 ] [ [e1]BT3 ] [ ]B3 ...[ψ ]
= [ψ,2 ]BT3 ] [e1 [ψ,1
[L]TB2,B1 [L]B2,B1 ]B3 ]B3 [e,2 ]TB3 ] ,2 ]B3
[ [ψ,1 ]BT3 g12 ,
= [ψ,2 ]TB3 [ ...[ψ ] [ g11 g22
[ψ g21
,1 ]B3 ,2 ]B3 =

where [gαβ]2ˆ2 is the metric tensor associated with the parametrization tV, ψu. Therefore,
det([L]B2,B1) = ?g as long as B1 is an orthonormal basis of TpΣ.