Dual Vector Space

Given a vector space $V$ , we define $V^{*}$ as the space of linear mappings from $V$ to $R$ (linear forms). Given a basis ${e^{i}}$ in $V$ , we call the set ${e_{i}}$ the dual basis, where the $e_{i}$ are forms satisfying:

e_{i} (e^{j}) = δ_{i}^{j} (Kronecker delta).

No natural isomorphism

At a first glance, there is no natural isomorphism between $V$ and $V^{*}$ . But if we have in $V$ an inner product $g$ , we automatically do have a natural isomorphism from $V$ into $V^{*}$ :

T (v) = g (\cdot, v)

And reciprocally: given an isomorphism we recover a bilinear form (not necessarily an inner product), which is non degenerated. Given $ϕ : V \to V^{*}$ we can define

b : V \times V \to R

by means of $b (v_{1}, v_{2}) = ϕ (v_{1}) (v_{2}) .$

But you can argue that we have a natural isomorphism: the one which sends $e^{i}$ to $e_{i}$ . But it is needed to fix a basis. In this case, the bilinear for associated is the one with matrix $(\begin{array}{ccc} 1 & 0 & \dots \\ 0 & 1 & \dots \\ \dots & \dots & \dots \end{array})$ in the basis ${e^{i}}$ .
So in a vector space is equivalent:

Fixing a basis
Fixing an non degenerated bilinear form.
Fixing an isomorphism with its dual, with conditions.

Moreover, the original inner product induces another on $V^{*}$ . This can be seen "by hand" or by considering that, just as there is a correspondence between inner products on $V$ and isomorphisms from $V$ to $V^{*}$ , there is also a correspondence between inner products on $V^{*}$ and isomorphisms from $V^{*}$ to $V^{* *}$ . But since $V^{* *} = V$ , the sequence is as follows: the inner product $g$ induces the isomorphism $T$ , and because $T^{- 1}$ is also an isomorphism, it induces an inner product $g^{*}$ on $V^{*}$ .

Moreover, the inner products $g$ and $g^{*}$ are inverse, in the sense that for a vector $v$

g^{*} (g (v,_{)}, -) = v

In Penrose abstract index notation we would have:

g_{a b} g^{b c} = δ_{a}^{c}

Covariant components of a vector

Let $v = \sum_{i} v^{i} e^{i}$ . It is clear that it has a correspondent $v^{*} \in V^{*}$ such that $v^{*} = \sum_{i} g (v, e^{i}) e_{i}$ . We write $v_{i} = g (v, e^{i})$ and call it the covariant components of $v$ , while $v^{i}$ are the contravariant ones. Since $g$ has an inverse, we can recover the contravariant components from the covariant ones: $v^{i} = g^{*} (v^{*}, e_{i})$ . Using the matrix form of $g = (g_{i j})$ and $g^{*} = (g^{i j})$ respect to the chosen basis we would write

\sum_{i} v^{i} e^{i} = \sum_{i} \sum_{j} g^{i j} v_{j} e^{i} = \sum_{i} \sum_{j} g^{i j} g (v, e^{j}) e^{i}

and so

v^{i} = \sum_{j} g^{i j} g (v, e^{j})

Compare with the usual orthogonal case $v^{i} = g (v, e^{i})$ .

Here you can compare the $v^{i}$ s with the $v_{i}$ s:

This ideas, I think, can be generalized to general frames in homogeneous spaces. See general covariance and contravariance.