Differential forms should be invariant under coordinate transformations I am wondering why, if we transform the following differential form, it does not seem to be invariant under the coordinate transformation. The $1$-form on $\mathbb R^2$ is
$$
 \omega = \sqrt{x^2 + y^2} dx + 0\cdot dy
$$
and as the coefficient function for $dx$ is radially symmetric, it should be easier to express this form in polar coordinates. Then with
$$
 dx = \cos\theta dr - r\sin\theta d\theta \quad
 dy = \sin\theta dr + r\sin\theta d\theta 
$$
and $f(x(r,\theta), y(r,\theta)) = r$ for $f(x,y) = \sqrt{x^2+y^2}$ we have
$$
 \omega = r(\cos\theta dr - r\sin\theta d\theta)
        = r\cos\theta dr - r^2\sin\theta d\theta.
$$
So, now it might be reasonable to expect that the value of this differential form is not changed by the coordinate transformation. So consider the point $(1,1)$ in cartesian coordinates, and the tangent vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ to that point, then evaluating $\omega$ in cartesian coordinates we have
$$
 \omega(1,1)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
  = \sqrt{2} \cdot dx\begin{pmatrix} 1 \\ 0 \end{pmatrix}
  = \sqrt{2}
$$
as $dx\begin{pmatrix} x \\ y \end{pmatrix} = x$. Now evaluating the form in polar coordinates, the same point is $(\sqrt 2, \pi/4)$ expressed in polar coordinates, so evaluating $\omega$ in polar coordinates we get
\begin{align*}
 \omega(\sqrt 2, \pi/4)\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right)
 & = \sqrt 2 \cos(\pi/4) dr\begin{pmatrix} 1 \\ 0 \end{pmatrix}
 - 2\sin(\pi/4) d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
 & = 1\cdot dr\begin{pmatrix} 1 \\ 0 \end{pmatrix} - \sqrt 2 \cdot d\theta\begin{pmatrix} 1 \\ 0 \end{pmatrix} \\
 & = 1 \cdot 0 - \sqrt 2 \cdot 0 \\
 & = 1
\end{align*}
which is different (again using that $dr, d\theta$ maps on the first and second component). So maybe the tangent vector has to be transformed too, so
$$
 \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\theta \\ \sin\theta\end{pmatrix}
 = \frac{1}{2}\begin{pmatrix}\sqrt 2 \\ \sqrt 2 \end{pmatrix}.
$$
So now doing the second computation again:
\begin{align*}
 \omega(\sqrt 2, \pi/4)\left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) 
 & = \sqrt 2 \cos(\pi/4) dr \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) - 2\sin(\pi/4) d\theta \left( \frac{1}{2} \begin{pmatrix} \sqrt 2 \\ \sqrt 2 \end{pmatrix} \right) \\
 & = \frac{1}{2}\sqrt 2 - \sqrt 2 \frac{1}{2}\sqrt 2 \\
 & = \frac{1}{2}\sqrt 2 - 1
\end{align*} 
which is again different.
So whats wrong here? Have I done any transformations wrong? The form should yield the same value for the same point and tangent vector, regardless of the coordinate system in which it is written, or not?
 A: The problem with your computation is that the polar coordinates of the basis tangent vector $\partial_x$ at $(1,1)$ are $(\sqrt 2 /2,-1/2)$:
$$v = \partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)} = \frac{\sqrt 2}{2}\partial_r\rvert_{(1,1)} - \frac{1}{2}\partial_\theta\rvert_{(1,1)}.$$
Therefore we find $\omega_{(1,1)}(v) = dr_{(1,1)}(v) - \sqrt 2 d\theta_{(1,1)}(v) = \frac{\sqrt2}{2} + \frac{\sqrt 2}{2} = \sqrt 2$.

Viewing $\mathbb{R}^2 - 0$ as a manifold, you have a polar coordinate chart $\phi = (r,\theta)$ which maps some neighborhood of $(1,1)$ onto a neighborhood of $(\sqrt 2,\pi/4)$. The basis vectors $\partial_r\rvert_{(1,1)}$ and $\partial_\theta\rvert_{(1,1)}$ of the tangent space at $(1,1)$ in terms of this coordinate chart are defined as pushforwards of the standard basis vectors of the tangent space at $(\sqrt 2,\pi/4)$:
$$\partial_r\rvert_{(1,1)} = (\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}), \quad \partial_\theta\rvert_{(1,1)} = (\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}).$$
By definition of the pushforward, we have $$\phi_*(\partial_x\rvert_{(1,1)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_x\rvert_{(\sqrt 2, \pi/4)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_y\rvert_{(\sqrt 2, \pi/4)}$$
and so
$$\partial_x\rvert_{(1,1)} = \frac{\partial r}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_x\rvert_{(\sqrt 2,\pi/4)}) + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}(\phi^{-1})_*(\partial_y\rvert_{(\sqrt 2,\pi/4)}) = \frac{\partial r}{\partial x}\rvert_{(1,1)}\partial_r\rvert_{(1,1)} + \frac{\partial \theta}{\partial x}\rvert_{(1,1)}\partial_\theta\rvert_{(1,1)}.$$
