How to calculate the pullback of a $k$-form explicitly

I’m having trouble doing actual computations of the pullback of a $k$-form. I know that a given differentiable map $\alpha: \mathbb{R}^{m} \rightarrow \mathbb{R}^{n}$ induces a map $\alpha^{*}: \Omega^{k}(\mathbb{R}^{n}) \rightarrow \Omega^{k}(\mathbb{R}^{m})$ between $k$-forms. I can recite how this map is defined and understand why it is well defined, but when I’m given a particular $\alpha$ and a particular $\omega \in \Omega^{k}(\mathbb{R}^{n})$, I cannot compute $\alpha^{*}\omega$.

For example I found an exercise (Analysis on Manifolds, by Munkres) where $\omega = xy \, dx + 2z \, dy – y \, dz\in \Omega^{k}(\mathbb{R}^{3})$ and $\alpha: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}$ is defined as $\alpha(u, v) = (uv, u^{2}, 3u + v)$, but I got lost wile expanding the definition of $\alpha^{*} \omega$. How can I calculate this?

Note: This exercise is not a homework, so feel free to illustrate the process with any $\alpha$ and $\omega$ you wish.

Answer

Instead of thinking of $\alpha$ as a map, think of it as a substitution of variables:
$$
x = uv,\qquad y=u^2,\qquad z =3u+v.
$$
Then
$$
dx \;=\; \frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv \;=\; v\,du+u\,dv
$$
and similarly
$$
dy \;=\; 2u\,du\qquad\text{and}\qquad dz\;=\;3\,du+dv.
$$
Therefore,
$$
\begin{align*}
xy\,dx + 2z\,dy – y\,dz \;&=\; (uv)(u^2)(v\,du+u\,dv)+2(3u+v)(2u\,du)-(u^2)(3\,du+dv)\\[1ex]
&=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
\end{align*}
$$
We conclude that
$$
\alpha^*(xy\,dx + 2z\,dy – y\,dz) \;=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv.
$$

Attribution
Source : Link , Question Author : Tony Burbano , Answer Author : Jim Belk

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