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.

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

3 Answers 3


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 $\omega$!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.