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 $b (v_{1}, v_{2}) = ϕ (v_{1}) (v_{2}) .$
The alternating case of this correspondence is the symplectic vector space.

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.

Claudio

The transpose as the pullback on the dual. Given a linear map $F : V \to W$ with matrix $A$ in bases ${e^{i}}$ (of $V$ ) and ${f^{j}}$ (of $W$ ), the pullback $F^{*} : W^{*} \to V^{*}$ , $F^{*} ω = ω \circ F$ , has matrix $A^{t}$ w.r.t. the dual bases ${e_{i}}$ , ${f_{j}}$ : indeed

(F^{*} f_{j}) (e^{i}) = f_{j} (F e^{i}) = f_{j} (\sum_{k} A_{k i} f^{k}) = A_{j i} .

So the pullback acts on coordinates with the transpose: $ω = \sum_{j} η_{j} f_{j}$ is sent to $\sum_{i} (\sum_{j} A_{j i} η_{j}) e_{i}$ , i.e. the new coordinate column is $A^{t} η$ .

Warning: transpose is not inverse. The pullback dualizes $V$ into $V^{*}$ ; it does not reverse the map. $A^{t} = A^{- 1}$ holds only when $A$ is orthogonal w.r.t. the pairing (rotations!). For a shear $P = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ , $P^{t} = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) \neq P^{- 1} = (\begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix})$ .

Passive reading. For a change of basis $b_{2} = T \circ b_{1}$ with matrix $P$ , the pullback of the coordinate change reads as the coordinate change itself: $(T^{*} ω) (u_{i}) = ω (T u_{i}) = ω (v_{i})$ , so $T^{*} ω$ expressed in the old dual basis has exactly the coordinates $ω$ gets in the new one — covectors are covariant, see the Claudio annotation in covariance and contravariance in linear algebra.

On manifolds this is the Jacobian story of covariance and contravariance on manifolds: vector coordinates convert by $J = \partial y / \partial x$ , covector coordinates by $η_{y} = η_{x} \cdot \partial x / \partial y$ (row), whose row-to-column switch is $(\partial x / \partial y)^{t}$ — the transpose of line 19.