466 CHAPTER 25. CURVILINEAR COORDINATES

Definition 25.1.2 Let {ei}pi=1 form a basis for Rp. Then

{ei}p

i=1 is called the dual basis if

ei ·e j = δij ≡

{1 if i = j0 if i ̸= j

. (25.1)

Theorem 25.1.3 If {ei}pi=1 is a basis then

{ei}p

i=1 is also a basis provided 25.1 holds.

Proof: Supposev = vie

i. (25.2)

Then taking the dot product of both sides of 25.2 with e j,yields

v j = v ·e j. (25.3)

Thus there is at most one choice of scalars v j such that v = v jej and it is given by 25.3.(

v−v ·e jej) ·ek = 0

and so, since {ei}pi=1 is a basis, (

v−v ·e jej) ·w = 0

for all vectors w. It follows v−v ·e jej = 0 and this shows

{ei}p

i=1 is a basis. ■In the above argument are obtained formulas for the components of a vector v, vi,

with respect to the dual basis, found to be v j = v ·e j. In the same way, one can find thecomponents of a vector with respect to the basis {ei}p

i=1 . Let v be any vector and let

v = v je j. (25.4)

Then taking the dot product of both sides of 25.4 with ei we see vi = ei ·v.Does there exist a dual basis and is it uniquely determined?

Theorem 25.1.4 If {ei}pi=1 is a basis for Rp, then there exists a unique dual basis,

{e j}p

j=1satisfying

e j ·ei = δji .

Proof: First I show the dual basis is unique. Suppose{f j}p

j=1 is another set of vectors

which satisfies f j ·ei = δji . Then

f j = f j ·eiei = δ

ji e

i = e j.

Note that from the definition, the dual basis to{i j}p

j=1 is just i j = i j. It remains to verifythe existence of the dual basis. Consider the matrix gi j ≡ ei ·e j. This is called the metrictensor. If the resulting matrix is denoted as G, does it follow that G−1 exists? Suppose youhave ei ·e jx j = 0. Then, since i is arbitrary, this implies e jx j = 0 and since

{e j}

is a basis,this requires each x j to be zero. Thus G is invertible. Denote by gi j the i jth entry of thisinverse matrix. Consider e j ≡ g jkek. Is this the dual basis as the notation implies?

e j ·ei = g jkek ·ei = g jkgki = δji

so yes, it is indeed the dual basis. This has shown both existence and uniqueness of thedual basis. ■

From this is a useful observation.

466 CHAPTER 25. CURVILINEAR COORDINATESDefinition 25.1.2 Let {ei}? form a basis for R?. Then {e! P is called the dual basis ifi=!F . lifi=jele; =.= i= J (25.1)OifiFjTheorem 25.1.3 /f {e; a , is a basis then fe! yey is also a basis provided 25.1 holds.Proof: Supposev=vie'. (25.2)Then taking the dot product of both sides of 25.2 with e;,yieldsVj =v-e;j. (25.3)Thus there is at most one choice of scalars v; such that v = v;e/ and it is given by 25.3.(v —v-eje/) -e, =0and so, since {e;}?_, is a basis,(v—v-eje/)-w=0for all vectors w. It follows v — v- eje/ = 0 and this shows {e'}?_, is a basis. llIn the above argument are obtained formulas for the components of a vector v, vj,with respect to the dual basis, found to be v; = v-e;. In the same way, one can find thecomponents of a vector with respect to the basis {e;}/_, . Let v be any vector and letv =we;. (25.4)Then taking the dot product of both sides of 25.4 with e! we see v' = e!-v.Does there exist a dual basis and is it uniquely determined?Theorem 25.1.4 /f {ei}?, is a basis for R? , then there exists a unique dual basis, {e/ asatisfyinge!-e; = 6).Proof: First I show the dual basis is unique. Suppose { fi ye _, is another set of vectors; j=lwhich satisfies f/ -e; = 6/. Thenfi =f! -ee' = dle =e’.Note that from the definition, the dual basis to {4 iti is just 4/ = 4;. It remains to verifythe existence of the dual basis. Consider the matrix g;; = e;-e;. This is called the metrictensor. If the resulting matrix is denoted as G, does it follow that G~! exists? Suppose youhave e;-e;x/ = 0. Then, since i is arbitrary, this implies e ;x/ = 0 and since {e;} is a basis,this requires each x/ to be zero. Thus G is invertible. Denote by g’/ the ij” entry of thisinverse matrix. Consider e/ = g/*e,. Is this the dual basis as the notation implies?- ok gie! -e; = gl exe; = B!" gui = 8;So yes, it is indeed the dual basis. This has shown both existence and uniqueness of thedual basis.From this is a useful observation.