Wheeler, Mechanics

Now, defining
the inner product becomes
( ) ∂
g ij = g
∂x i , ∂
∂x j
g (A, B) = A i B j g ij
When the two vectors are the same, we recover eq.(20).
4.3.2 Duality and linear maps on vectors
We may always define a duality relationship between vectors and forms. Starting with the coordinate basis,
∂
∂x
, for vectors and the corresponding coordinate basis, dx i , for forms, we define a linear bracket relation
i
by
〈 〉 ∂
∂x j , dxi = δj
i
Now suppose the space of vectors, V ∗ is given an arbitrary set of basis vectors, e i . We define the basis dual
to e i by
〈
ej , e i〉 = δ i j
Since each basis set can be expanded in terms of a coordinate basis,
〈
〉
m ∂
ej
∂x m , e i
n dx n = δ j i
we may use linearity to find
δ i j = e
= e
= e
j
j
j
m
e
n
i
m i
en
δm
n
m i
em
〈 ∂
∂x m , dxn 〉
i
It follows that the matrix en
giving the form basis in terms of the coordinate differentials is inverse to the
matrix giving the dual basis for vectors.
Now consider an arbitrary vector, v ∈ V ∗ and form, ω ∈ V ∗ . The duality relation becomes a map,
〈v, ω〉 = 〈 v j e j , ω i e i〉
= v j ω i
〈
ej , e i〉
= v j ω i δ i j
= v i ω i
Using this map together with the metric, we define a unique 1-1 relationship between vectors and forms. Let
w be an arbitrary vector. The 1 -form ω corresponding to w is defined by demanding
for all vectors v ∈ V ∗ . In components this relationship becomes
g (v, w) = 〈v, ω〉 (21)
g ij v i w j = v i ω i
In order for this to hold for all vectors v i , the components of the form ω must be related to those of w by
ω i = g ij w j
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