Consider a vector $\vec{A} = A_x \hat{x} + A_y \hat{y} = A^i e_i$.
Intuitively, it seems as though the coordinates ($A_x,A_y$) must transform in the opposite way as the bases ($\hat{x},\hat{y}$). For example if you rotate the bases by some angle, then to get the components in the different bases, you must rotate in the opposite direction to compensate. i.e. if you can write $e'_i = M e_i$, where $M$ is some rotation (or other linear transformation), then $(A')^i = M^{-1} A^i$. And this also seems familiar from my vague memory of covariance vs contravariance in general relativity.
But considering a particular example: $\vec{A} = A_x \hat{x} + A_y \hat{y} = A_r \hat{r} + A_\theta \hat{\theta}$, and the transformation between bases is given by:
$$\pmatrix{\hat{r} \\ \hat{\theta}} = \pmatrix{\cos\theta & \sin\theta \\ -\sin\theta & \cos\theta} \pmatrix{\hat{x} \\ \hat{y}}$$
This gives,
$$\vec{A} = A_r(\hat{x}\cos\theta + \hat{y}\sin\theta) + A_\theta(-\hat{x}\sin\theta + \hat{y}\cos\theta) \\ = \hat{x}(A_r \cos\theta – A_\theta \sin\theta) + \hat{y}(A_r\sin\theta + A_\theta\cos\theta)$$
Doing the same for $\hat{x}$ and $\hat{y}$ shows that the components transform in the same way as the bases, i.e. $A_r = A_x \cos\theta + A_y \sin \theta, \,\,\, A_\theta = – A_x \sin \theta + A_y \cos\theta$, instead of transforming in the inverse manner.
Am I doing something wrong here? Or are the components and bases supposed to transform in the same way?
Best Answer
I think using charts helps to keep the ideas clear, because then it's just a change of basis calculation. More precisely, let $\varphi:\mathbb R^2\to \mathbb R^2$ by $\varphi(r,\theta)=(r\cos\theta,r\sin\theta)$ and $\psi:\mathbb R^2\to \mathbb R^2$ by $\psi=id$, the identity on $\mathbb R^2.$
For any $p\in \mathbb R^2$, the tangent space $T_p\mathbb R^2$ has basis $\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\}$ in the coordinates given by $\varphi$ and $\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\}$ in the coordinates of $\psi$, so that $x$ and $y$ are the coordinate projections on the first and second coordinate, respectively.
To see how the components of vectors transform, write
$\frac{\partial}{\partial r}=a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}.$ Applying the projections $x$ and $y$, we get $a=\cos\theta$ and $b=\sin \theta.$
Similarly, $\frac{\partial}{\partial \theta}=c\frac{\partial}{\partial x}+d\frac{\partial}{\partial y}$ and $c=-r\sin\theta$ and $d=r\cos\theta.$
Therefore, if the vector in the $\varphi$ system has coordinates $(u,v)$ then the coordinates in the $\psi$ system are
$$\begin{pmatrix} \cos\theta &-r\sin\theta \\ \sin\theta&r\cos\theta \end{pmatrix}\begin{pmatrix} u\\v \end{pmatrix}$$
and the matrix of the transformation is displayed explicitly.