Co-ordinate tangent vectors and one-forms.

Let $(U, \psi)$ be a set of co-ordinates on $M$ where $\psi = (\psi^1,\dots, \psi^n)$. Then each of the component functions $\psi^i$ is a real function so we can define $n$ one-forms $d\psi^i(x) \in T_x^{*}M$ called the co-ordinate one-forms.

We have seen that

$$d\psi^{-1} \colon T_xM \to \mathbb{R}^n$$

is a linear isomorphism. We denote by

$$\frac{\partial\phantom{\psi^i}}{\psi^i}(x) = d\psi^{-1}(\psi(x))(e^i)$$

the image under this map of the standard basis vector $e^i$ in $\mathbb{R}^n$. We call the set of these the basis of co-ordinate tangent vectors. Consider what happens when we apply $d\psi^i(x)$ to $\partial /\partial \psi^j(x)$. We have

$$d\psi^i(x)(\frac{\partial\phantom{\psi^j}}{\psi^j}(x)) = d\psi^i(x) (d\psi^{-1}(x)(e^j)) = d(\psi^i\circ \psi^{-1})(\psi(x))(e^j)).$$

Notice that if $y=(y^1,\dots, y^n)$ is a point in $\mathbb{R}^n$ then $\psi^i\circ \psi^{-1}(y) = y^i$ is a linear map so equal to its own derivative. Hence $d(\psi^i\circ \psi^{-1})(\psi(x))(e^j)$ is the $i$th component of the vector $e^j$ or just $\Delta_{ij}$. It follows from linear algebra that $d\psi^1(x),\dots, d\psi^n(x)$ is a basis of $T_xM$ and, in fact, the dual basis to the basis $\partial /\partial \psi^i$.