618 CHAPTER 31. CURVILINEAR COORDINATES

I want {ek}nk=1 to be a basis. Thus, from Proposition 31.1.5,

det(

∂Mi

∂xk

)≡ det(Dy (x))≡ det(D(M)(x)) ̸= 0. (31.15)

Letyi = Mi (x) i = 1, · · · ,n (31.16)

so that the yi are the usual rectangular coordinates with respect to the usual basis vectors{ik}n

k=1 of the point y =M (x) . Letting x ≡(x1, · · · ,xn

), it follows from the inverse

function theorem (See Chapter 24) that M (D) is open, and that 31.15, 31.13, and 31.14imply the equations 31.16 define each xi as a C2 function of y≡

(y1, · · · ,yn

). Thus, abusing

notation slightly, the equations 31.16 are equivalent to

xi = xi (y1, ...,yn) , i = 1, · · · ,n

where xi is a C2 function of the rectangular coordinates of a point y. It follows from thematerial on the gradient described earlier,

∇xk (y) =∂xk (y)

∂y j i j.

Then

∇xk (y) ·e j =∂xk

∂ys is · ∂yr

∂x j ir =∂xk

∂ys∂ys

∂x j = δkj

by the chain rule. Therefore, the dual basis is given by

ek (x) = ∇xk (y (x)) . (31.17)

Notice that it might be hard or even impossible to solve algebraically for xi in termsof the y j. Thus the straight forward approach to finding ek by 31.17 might be impossible.Also, this approach leads to an expression in terms of the y coordinates rather than thedesired x coordinates. Therefore, it is expedient to use another method to obtain thesevectors in terms of x. Indeed, this is the main idea in this chapter, doing everything interms of x rather than y. The vectors, ek (x) may always be found by using formula 31.9and the result is in terms of the curvilinear coordinates x. Here is a familiar example.

Example 31.3.1 D ≡ (0,∞)× (0,π)× (0,2π) and y1

y2

y3

=

 x1 sin(x2)

cos(x3)

x1 sin(x2)

sin(x3)

x1 cos(x2)

(We usually write this as  x

yz

=

 ρ sin(φ)cos(θ)ρ sin(φ)sin(θ)

ρ cos(φ)

where (ρ,φ ,θ) are the spherical coordinates. We are calling them x1,x2, and x3 to preservethe notation just discussed.) Thus

e1 (x) = sin(x2)cos

(x3)i1 + sin

(x2)sin

(x3)i2 + cos

(x2)i3,