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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?

Najib Idrissi
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Picard
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    Refer to this http://math.stackexchange.com/questions/576638/how-to-calculate-the-pullback-of-a-k-form-explicitly – inquisitive Mar 02 '16 at 12:27

2 Answers2

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$\newcommand{\Blank}{\underline{\qquad}}$(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.

Andrew D. Hwang
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The nice thing about forms is that you are intuitively doing the right thing, which is: just plug in $x=u-v$ etc. as Andrew already wrote.

To see the connection with the formal definitions, note that the exterior derivative and pullbacks commute and hence $f^*dx=d(f^*x)=d(x\circ f)=d(u-v)$. Furthermore, pullbacks respect wedge products, so e.g. $$f^*(x^2dx)=f^*(x^2 \wedge dx)=(f^*x^2) \wedge (f^*dx)=(u-v)^2d(u-v)$$

Markus Heinrich
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