Pushing forward vectorfields to compute flows. So, I talked to my professor today, asking him for a way to compute flows of vectorfields. An example was given by $ v = x\partial_x + y\partial_y$ living in $\mathbb{R}^2$.
Now I know that the way to go would be to solve the system of ODE's given by $x'(t) = x(t), y'(t)=y(t)$ given varying initial values.
However, drawing the vectorfield one sees how it acts on rays that turn radially, so it looks like an easy one-dimensional problem when only looking at the rays. And since it is all radial we went on to look at the problem in polar coordinates. Form here on I get a little lost:
Let $\Phi(r,\varphi) := (r\cos\varphi, r\sin\varphi)$. We next want to get a description of $v$ in these coordinates. As I see it we cannot compute $\Phi_*(v)$ simply because at first $v$ is given in the wrong coordinates. I thought what we want is somewhat $\Phi^{-1}_*(v)$ and then replacing $x=r\cos\varphi$ a.s.o.
What we did however was something along these lines:
$$\begin{align}x\partial_x + y\partial_y &= r\cos(\varphi)\partial(r\cos\varphi) + r\sin\varphi \partial(r\cos\varphi)\\
&= r\cos(\varphi)\left(\frac{\partial\Phi_1(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_1(r,\varphi)}{\partial \varphi}\partial_\varphi\right) + r\sin\varphi \left(\frac{\partial\Phi_2(r,\varphi)}{\partial r}\partial_r + \frac{\partial\Phi_2(r,\varphi)}{\partial \varphi}\partial_\varphi\right)\\
&= r\partial_r
\end{align}$$
And then we have found the "pushforward" of $v$ somewhat magically. Now I do not get why this works? Why is this exactly what we want? These seem like magical calculations that randomly give the right result to me. Maybe I remember his steps wrong, but in the end we never computed $\Phi^{-1}$ or the inverse of the jacobian and yet came to the right result.
(The next steps would be to get the flow in the coordinates for $\Phi$ and write it back in the coordinates of $(x,y)$ I guess).
I would be happy for clarifications on why this should work, where it comes from and also maybe a more formal proof on how to compute the flow of the given vector field when first changing coordinates and then getting back.
 A: It looks like your professor is exploiting the standard metric and the musical isomorphisms to avoid calculating the inverse matrix, because pulling back differential forms $(x,y)$ to $(r,\theta)$ works the same direction as we want.  We just need to be able to freely change vector fields to forms, and that is provided by the metric (= inner product, or if want indefinite signatures, = nondegenerate symmetric bilinear form) inducing a natural isomorphism between tangent and cotangent vectors $v\leftrightarrow v^\flat:=\langle v,-\rangle$: if the metric in $x^i$ coordinates is $\sum_i h^i(\mathrm{d}x^i)^2$ then $(\partial_{x_i})^\flat = h^i\,\mathrm{d}x^i$ and the other way is $(\mathrm{d}x^i)^\sharp=\frac1{h^i}\partial_{x_i}$.
Now what you have given looks like the usual calculation for $x\,\mathrm{d}x+y\,\mathrm{d}y=r\,\mathrm{d}r$.  Since the standard metric $\mathrm{d}x^2+\mathrm{d}y^2=\mathrm{d}r^2+r^2\,\mathrm{d}\theta^2$, both $(x,y)$ and $(r,\theta)$ are orthogonal coordinates making the calculation easy.  So "sharp"ing it gives $x\partial_x+y\partial_y=r\partial_r$ indeed.
If we try a similar approach on $x\partial_y-y\partial_x$, we have
$$x\,\mathrm{d}y-y\,\mathrm{d}x=r^2\,\mathrm{d}\theta$$
so when you sharp it, you need to remember the coefficient $r^2$ in front of $\mathrm{d}\theta^2$ in $\mathrm{d}r^2+r^2\mathrm{d}\theta^2$ means $(r^2\,\mathrm{d}\theta)^\sharp=\partial_\theta$ so the full calculation goes
$$
x\partial_y-y\partial_x=(x\,\mathrm{d}y-y\,\mathrm{d}x)^\sharp=(r^2\mathrm{d}\theta)^\sharp=\partial_\theta.
$$
