Given the one form $\nu(\vec y)=y^2dy^1-y^1dy^2$ in $(y^1,y^2,y^2)$ coordinates, how do you get the pullback $\mu(\vec x)$ in the $(x^1,x^2,x^3)$ coordinates?
We have $F(x^1,x^2,x^3)=(-x^2,x^1,x^3), \text{ } \vec y = F(\vec x)$ and
$$DF= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ where $\frac{\partial F^i}{\partial x^j}=DF_{ij}$.
I know $\mu (\vec x) = \nu_1dx^1-\nu_1dx^2+\nu_3dx^3$
So $\mu (\vec x)=-y^1dx^1-y^2dx^2$
But doesn't this mean $\mu (\vec x)=x^2dx^1-x^1dx^2$?
Finally I changed the $y$ coordinates into $x$ by saying $y_i=F_i(\vec x)$
I am confused about how to change the coordinates.
Update:
Yes, your result is right. But actually, you should not simply thinks $y_i = F_i(x)$ Because in general case, when you talk about two different base space, $y_i$ is a function with domain in the later space. While $F$ is a map from the first to the later space. And in that case, $F_i$ actually means the composition $y_i$$F$.(I wrote this paragraph because I could not understand how you works in the last 3 and 4 lines
I know.....So......)Do not Consider $y_i$s simply as a number but a function which assign each point a number (called coordinates.) And then to change the coordinates is simply to composition the function with the given map F.
You should never mix the coordinate of the point in the base space with the coordinate of vectors in the tangent space associated with that point.