In coordinate basis or holonomic basis (Wikipedia, Physics.SE, Math.SE and prior question) the primal basis vectors (covariant) can be interpreted as the derivatives with respect to the positional vector with respect to the different coordinates, i.e. $\left\{{\bf e}_1=\frac{\partial}{\partial u^1},\dots,{\bf e}_n=\frac{\partial}{\partial u^n}\right\}.$ The dual (contra-variant) basis are the exterior derivatives $\left\{{\bf e}^1={\rm d} u^1,\dots,{\bf e}^n={\rm d}u^n\right\},$ and act on the primal vector basis as follows:
$${\bf e}^i({\bf e}_j)=\delta^i_j$$
In this presentation online the "other set of basis vectors" (I presume the dual vectors) is plausibly defined as the gradient of the coordinate vectors normalized, i.e. $\frac{\vec \nabla (u^i)}{\vert\vec\nabla (u^i)\vert}$.
So for polar,
$$\begin{align} {\bf e}_1&=\frac{\partial R}{\partial r} = \cos\theta \hat i + \sin \theta \hat j\\ {\bf e}_2&=\frac{\partial R}{\partial \theta} = -r\sin\theta \hat i + r\cos \theta \hat j\\ {\bf e}^1&=\vec\nabla\left(\sqrt{r^2 \cos^2\theta+ r^2 \sin^2\theta}\right) = \frac{\cos\theta}r \hat i + \frac{\sin \theta }{r}\hat j\\ {\bf e}^2&=\vec\nabla\left(\arctan2 \left(\frac{\sin\theta}{\cos\theta}\right) \right)= -\frac{\sin\theta}{r^2} \hat i + \frac{\cos \theta}{r^2} \hat j\\ \end{align}$$
which indeed fulfill ${\bf e}^i({\bf e}_j)=\delta^i_j$ once these vectors and convectors are normalized.
The question is twofold:
- With vector basis and their dual defined this way, will this relationship always be preserved ${\bf e}^i({\bf e}_j)=\delta^i_j$ beyond this example of polar coordinates? Or only if the basis vectors happen to be orthogonal to each other?
- Is this a way of introducing the concept of vectors /covectors consistent with the understanding of different spaces? Is it a good / generalizable definition?
Based on what I can put together:
- In the example above the dual basis as vectors runs into the problem that an element of the dual can never be equal to an element of the original vector space (here and here).
- The gradient necessitates an scalar product defined (metric), whereas a one-form can be defined without a metric, yet containing all the information found in the gradient here. The primal/dual basis should ideally look more like in here:
$$\begin{align} \frac{\partial}{\partial x} &= \cos(\theta) \frac{\partial}{\partial r} - \sin(\theta) \frac{1}{r} \frac{\partial}{\partial \theta}\\[2ex] \frac{\partial}{\partial y} &= \sin(\theta) \frac{\partial}{\partial r} + \cos(\theta) \frac{1}{r} \frac{\partial}{\partial \theta}\\[2ex] dx &= \cos \theta dr - r\sin\theta d\theta\\[2ex] dy &= \sin \theta dr + r \cos \theta d\theta\\ \end{align} $$
- Interpreting the dual basis as the exterior derivative should be akin to the gradient in the presentation quoted above with the "gradient" not really the actual gradient (a vector), but the corresponding 1-form. This is touched upon here and here.
- Finally, considering how a one-form acts on a vector (in here):
$$\omega \left( V \right) = {\omega}_{i} d{x}^{i} \left( V \right) = {\omega}_{i}{V}^{i}$$
the orthogonality of the primal basis should not be a factor in the Kronecker relation of primal / dual basis.
However, an answer correcting misunderstandings in this attempt would be great.
