# 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.

• Can you post your calculation so we can see where you get lost? Nov 22, 2013 at 0:57

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

An answer that uses the definitions. I tried to write down everything with the hope to clarify the ideas. I do not know if this is helpful or simply too verbose...

If $$\omega\in\Omega^1(N)$$ and $$\alpha:M\rightarrow N$$. The aim of the pullback is to define a form $$\alpha^*\omega\in\Omega^1(M)$$ from a form $$\omega\in\Omega^1(N)$$.

A 1-form $$\omega$$ evaluated at $$n=(x,y,z)\in N$$ is $$\omega[n]=\omega_x(n)dx+\omega_y(n)dy+\omega_z(n)dz$$ where

• $$(\omega_x(n),\omega_y(n),\omega_z(n))\in\mathbb{R}^3$$
• $$dx, dy, dz$$ are also 1-forms, the dual basis of $$\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}$$.

$$\omega[n]\in T_nN^*$$ eats one vector $$b\in T_nN$$ (a tangent vector at point $$n$$):

\begin{align*} \omega[n](b^x\frac{\partial}{\partial x}+b^y\frac{\partial}{\partial y}+b^z\frac{\partial}{\partial z}) &=\omega_x(n)b^x+\omega_y(n)b^y+\omega_z(n)b^z\in\mathbb{R} \end{align*}

Now the objective is to define $$\alpha^*\omega\in\Omega^1(M)$$ from $$\omega\in\Omega^1(N)$$. Given a point $$m=(u,v)\in M$$ and a vector $$a\in T_m M$$, a natural idea is to use the point $$n=\alpha(m)\in N$$ and to pushforward a vector $$a\in T_mM$$ to get a vector $$\alpha_*(a)\in T_{\alpha(m)}N$$. Actually this is how $$\alpha^*\omega$$ is defined: $$(\alpha^*\omega)[m](a)=\omega[\alpha(m)](\alpha_*(a))$$

Now we must compute $$\alpha_*(a)\in T_{\alpha(m)}N$$, see at the end for details, we get:

$$\alpha_*(a)=d\alpha^x[\alpha(m)](a)\frac{\partial}{\partial x}+d\alpha^y[\alpha(m)](a)\frac{\partial}{\partial y}+d\alpha^z[\alpha(m))](a)\frac{\partial}{\partial z}$$ Thus: \begin{align*} (\alpha^*\omega)[m](a)&=\overbrace{\omega_x[\alpha(m)]d\alpha^x[\alpha(m)](a)}^{\text{term }dx(\frac{\partial}{\partial x})=1}+ \overbrace{\omega_y[\alpha(m)]d\alpha^y[\alpha(m)](a)}^{\text{term }dy(\frac{\partial}{\partial y})=1}+ \overbrace{\omega_z[\alpha(m)]d\alpha^z[\alpha(m)](a)}^{\text{term }dz(\frac{\partial}{\partial z})=1} \in \mathbb{R} \end{align*} We can drop the argument, vector $$a=a^u\frac{\partial}{\partial u}+a^v\frac{\partial}{\partial v}$$, it remains: $$(\alpha^*\omega)[m]=\omega_x[\alpha(m)]d\alpha^x[\alpha(m)]+ \omega_y[\alpha(m)]d\alpha^y[\alpha(m)]+ \omega_z[\alpha(m)]d\alpha^z[\alpha(m)]\in T_mM^*$$ In the example: $$(\omega_x,\omega_y,\omega_z)=(xy,2z,-y)$$ and $$\alpha: (u,v)\mapsto (uv,u^2,3u+v)$$, therefore:

• $$(\omega_x[\alpha(m)],\omega_y[\alpha(m)],\omega_z[\alpha(m)])=(u^3v,6u+2v,-u^2)$$
• $$d\alpha^x[\alpha(m)]=vdu+udv$$
• $$d\alpha^y[\alpha(m)]=2udu$$
• $$d\alpha^z[\alpha(m)]=3du+dv$$

By substitution we get the expected result: $$(\alpha^*\omega)[m]=(u^3v)(vdu+udv)+(6u+2v)(2udu)+(-u^2)(3du+dv)$$

$$\alpha^*\omega=(u^3v^2+9u^2+4uv)du+(u^4v-u^2)dv\in\Omega^1(M)$$

It remains to explain how to compute $$\alpha_\star(a)\in T_{\alpha(m)}N$$ from a tangent vector $$a\in T_mM$$

One can interpret a vector $$a\in T_mM$$ as a first order differential operator acting on functions $$f:M\to\mathbb{R}$$.

More explicitly we have: $$a[f]=(a^u\frac{\partial}{\partial u}+a^v\frac{\partial}{\partial v})f=df[m](a) \in\mathbb{R}$$ The pushforward $$\alpha_*$$ transform a vector $$a\in T_mM$$ into a vector $$\alpha_\star(a)\in T_{\alpha(m)}N$$, thus it must act on functions $$g:N\to\mathbb{R}$$. A natural definition is: $$\alpha_\star(a)[g]=d(g\circ\alpha)[m](a)\in\mathbb{R}$$ Expanding this formula we get: \begin{align*} d(g\circ\alpha)[m](a)&=\left((\frac{\partial g}{\partial x}\frac{\partial \alpha^x}{\partial u}+\frac{\partial g}{\partial y}\frac{\partial \alpha^y}{\partial u}+\frac{\partial g}{\partial z}\frac{\partial \alpha^z}{\partial u})du+(\frac{\partial g}{\partial x}\frac{\partial \alpha^x}{\partial v}+\frac{\partial g}{\partial y}\frac{\partial \alpha^y}{\partial v}+\frac{\partial g}{\partial z}\frac{\partial \alpha^z}{\partial v})dv\right)(a) \\ &= \left((\frac{\partial \alpha^x}{\partial u}du+\frac{\partial \alpha^x}{\partial v}dv)\frac{\partial g}{\partial x}+(\frac{\partial \alpha^y}{\partial u}du+\frac{\partial \alpha^y}{\partial v}dv)\frac{\partial g}{\partial y}+(\frac{\partial \alpha^z}{\partial u}du+\frac{\partial \alpha^z}{\partial v}dv)\frac{\partial g}{\partial z}\right)(a) \\ &= (\frac{\partial \alpha^x}{\partial u}du+\frac{\partial \alpha^x}{\partial v}dv)(a)\frac{\partial g}{\partial x}+(\frac{\partial \alpha^y}{\partial u}du+\frac{\partial \alpha^y}{\partial v}dv)(a)\frac{\partial g}{\partial y}+(\frac{\partial \alpha^z}{\partial u}du+\frac{\partial \alpha^z}{\partial v}dv)(a)\frac{\partial g}{\partial z} \\ &= \left( d\alpha^x[m](a)\frac{\partial}{\partial x}+d\alpha^y[m](a)\frac{\partial}{\partial y}+d\alpha^z[m](a)\frac{\partial}{\partial z} \right) g \end{align*} we get the expected result:

## $$\alpha_\star(a)=d\alpha^x[m](a)\frac{\partial}{\partial x}+d\alpha^y[m](a)\frac{\partial}{\partial y}+d\alpha^z[m](a)\frac{\partial}{\partial z}\in T_{\alpha(m)}N$$

CAVEAT: we can always pullback differential forms, but only pushforward vectors (and not vector fields, unless $$\alpha$$ is a diffeomorphism (which is obviously not the case here)). See wikipedia, pushforward for further details.

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 $\omega$!