I'm trying to do my first pull-back of a differential form.
I know that $\omega=(2xy+x^{2}+1)dx+(x^{2}-y)dy$ is a differential form on $\mathbb{R}^{2}$.
I have $f : \mathbb{R}^{3} \to \mathbb{R}^{2}$ which is
$$f(u,v,w)=(u-v,v^{2}-w)$$
and I have to calculate the pullback. I was told that by definition
$$(f^{*}\omega)(X) = \omega(f_{*}(X)),$$
and so I calculated
$$f_{*}=\begin{pmatrix}
1 & -1 & 0\\
0 & 2v & 1
\end{pmatrix}$$
But then I don't really know how to proceed. Should I take a general vector and calculate the form, should I substitute $x,y$ with $u,v,w$? Do you have a general recipe to proceed?
Best Answer
$\newcommand{\Blank}{\underline{\qquad}}$(Good-natured note: This isn't the first pullback you've computed. You've been computing pullbacks since you learned the chain rule and method of substitution.)
It's easiest to start by turning the clock back to 1850 or so. You have \begin{align*} x &= u - v, \\ y &= v^{2} - w, \end{align*} so the chain rule gives \begin{align*} dx &= \Blank\, du + \Blank\, dv + \Blank\, dw, \\ dy &= \Blank\, du + \Blank\, dv + \Blank\, dw. \end{align*} Now, to express this in modern terms, replace the $1$-forms on the left by pullbacks: $dx \to f^{*}dx$, etc.