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

Question

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 1

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 $$ \alpha^*(xy\,dx + 2z\,dy - y\,dz) \;=\; (u^3v^2+9u^2+4uv)\,du\,+\,(u^4v-u^2)\,dv. $$

Answer 2

In some of my research I encountered $2$-forms that were given by $$\omega = dx \wedge dp + dy \wedge dq$$

and a map $$i : (u,v) \mapsto (u,v,f_u,-f_v)$$

for a general smooth map $f : (u,v) \mapsto f(u,v)$. I wanted to calculate the pullback of this map, i.e. $i^*\omega$. So, \begin{align} i^*\omega &= i^*(dx \wedge dp + dy \wedge dq) \\ &= d(x \circ i)\wedge d(p \circ i) + d(y \circ i)\wedge d(q \circ i). \end{align} Now, calculating each terms gives \begin{align} d(x \circ i) &= d(u) = du, \\ d(y \circ i) &= d(v) = dv, \\ d(p \circ i) &= d(f_u) = f_{uu}du + f_{uv}dv, \\ d(q \circ i) &= d(-f_v) = -f_{vu}du - f_{vv}dv. \end{align} Then, the pullback is given by \begin{align} i^*\omega &= du \wedge (f_{uu}du + f_{uv}dv) - dv \wedge (f_{vu}du + f_{vv}dv) \\ &= du \wedge (f_{uu}du) + du \wedge (f_{uv}dv) - dv \wedge (f_{vu}du) - dv \wedge (f_{vv}dv). \end{align} Now here, since the wedge product is $C^\infty(M)$-bilinear rather than just $\Bbb R$-bilinear, and $du \wedge dv = -dv\wedge du$. Using this property we have

$$i^*\omega =2f_{uv} du \wedge dv.$$

I realise that these are based solely on $2$-forms and you were asking about a general $k$-form but you did write pick any $\alpha$ and $\omega$!