468 CHAPTER 25. CURVILINEAR COORDINATES

The proof of the remaining formula in 25.10 is similar.To verify 25.11,

gi jg jk = ei ·e je j ·ek =((ei ·e j)e j

)·ek = ei ·ek = δ

ik.

This shows the two determinants in 25.12 are non zero because the two matrices are in-verses of each other. It only remains to verify that one of these is greater than zero. Lettingei = a j

i i j = biji

j, we see that since i j = i j,a ji = bi

j. Therefore,

ei ·e j = ari ir ·b j

kik = ar

i bjkδ

kr = ak

i b jk = ak

i akj.

It follows that for G the matrix whose i jth entry is ei ·e j, G = AAT where the ikth entry ofA is ak

i . Therefore, det(G) = det(A)det(AT)= det(A)2 > 0. It follows from 25.11 that if

H is the matrix whose i jth entry is gi j, then GH = I and so H = G−1 and

det(G)det(G−1)= det

(gi j)det(G) = 1.

Therefore, det(G−1

)> 0 also. ■

Note that det(AAT

)≥ 0 always, because the eigenvalues are nonnegative.

As noted above, we have the following definition.

Definition 25.1.7 The matrix (gi j) = G is called the metric tensor.

25.2 Exercises1. Let e1 = i+j,e2 = i−j,e3 = j+k. Find e1,e2,e3, (gi j) ,

(gi j). If

v = i+2j+k, find vi and v j, the contravariant and covariant components of thevector.

2. Let e1 = 2i+j,e2 = i−2j,e3 = k. Find e1,e2,e3, (gi j) ,(gi j). If

v = 2 i− 2j+k, find vi and v j, the contravariant and covariant components of thevector.

3. Suppose e1,e2,e3 have the property that ei ·e j = 0 whenever i ̸= j. Show the sameis true of the dual basis and that in fact, ei is a multiple of ei.

4. Let e1,· · · ,e3 be a basis for Rn and let v = viei = viei,w= w je j = w je

j be twovectors. Show

v ·w= gi jviw j = gi jviw j.

5. Show if {ei}3i=1 is a basis in R3

e1 =e2×e3

e2×e3 ·e1, e2 =

e1×e2

e1×e3 ·e2, e3 =

e1×e2

e1×e2 ·e3.

6. Let {ei}ni=1 be a basis and define

e∗i ≡ei

|ei|, e∗i ≡ ei |ei| .

Show e∗i ·e∗j = δij.