I am a bit rusty about differential forms on manifolds, but I need to consider a pull-back of a 1-form defined on a curve. This curve $\gamma$ is diffeomorphic to, say, the interval $(0,1)$ through a map $f: (0,1) \mapsto \gamma$ and embedded in $\mathbb{R}^n$. The same question applies to $k$-dimensional manifolds diffeomorphic to open sets of $\mathbb{R}^k$, again with $k < n$. Is there a formula for writing down differential forms on the submanifolds in terms of the diffeomorphisms? For instance, given a 1-form $\omega$ in $\Lambda ^1 (\gamma)$ , can I write $\omega(p)$ somehow explicitly through the diffeomorphism $f$ between $(0,1)$ and $\gamma$?

What I need to, actually, is the pull-back of a 1-form supported on a curve $\gamma$ in $\mathbb{R}^n$, where the function I use for the pull-back is some smooth function $g: \mathbb{R}^n \to \gamma$. So considering such $g$ and such $\gamma$, I want an explicit formula for $g^* \omega$.

As an example: consider $\gamma=S^1 \subset \mathbb{R}^2$ and $f:[0, 2 \pi) \mapsto S^1$ defined as $f(\theta)= (\sin(\theta), \cos(\theta))$(ok this is not a diffeomorphism, but locally it is). Then consider some differential 1-form $\omega$ in $S^1$ and a smooth map $g: \mathbb{R}^2 \to S^1$, say, $g(x,y)= (\sin(x^2 + y^2), \cos(x^2 + y^2))$. How do I write $g^* \omega$ explicitly?

  • $\begingroup$ Your actual question is confusing. Yes, you can define $j$-forms on a submanifold $S$ (for any $j\le\dim S$) by pulling back by the inverse of a global parametrization (or by gluing together such for various parametrizations with a partition of unity). If $S\subset M$, you can also restrict any $j$-form on $M$ to $S$ (if you like, pull back by the inclusion map). If you happen to have global coordinates on $M$, then you can use them. We do this all the time with line integrals and surface integrals for submanifolds of $\Bbb R^3$. $\endgroup$ Apr 3, 2023 at 17:36
  • $\begingroup$ Let me rewrite everything more clearly, wait a second $\endgroup$
    – tommy1996q
    Apr 3, 2023 at 17:42
  • $\begingroup$ Made it shorter to go straight to the point, hope this helps $\endgroup$
    – tommy1996q
    Apr 3, 2023 at 17:46
  • $\begingroup$ No, that made it worse. :) What is $g$? We already had a diffeomorphism $f$. If $\gamma$ is a curve sitting in $\Bbb R^n$, how do you get a function $g\colon\Bbb R^n\to\gamma$? $\endgroup$ Apr 3, 2023 at 17:54
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    $\begingroup$ At some point, you might find my YouTube lectures on forms and integration on submanifolds helpful, if you can figure out which particular parts are relevant to refreshing your memory. :) The link to the 112 lectures is in my profile. The titles should lead you to relevant ones. $\endgroup$ Apr 3, 2023 at 18:03

1 Answer 1


Your map $g$ to the circle has nothing to do with the circle's embedding in $\Bbb R^2$. It could, in fact, be a map $g\colon\Bbb R^N\to S^1$ for an arbitrary $N$. The easiest way to do your example is to note that any $1$-form on the circle is of the form $\lambda(\theta)\,d\theta$ (with $\lambda(0)=\lambda(2\pi)$). I can naturally write $d\theta = -y\,dx + x\,dy$ in terms of the ambient coordinates on $\Bbb R^2$, with the circle given by $x^2+y^2=1$.

You have a mapping $g\colon\Bbb R^2\to S^1$ given by specifying that $x=g_1(u,v)$, $y=g_2(u,v)$ (I'm switching letters for obvious reasons). Then I just pull back in the usual way: $g^*x = g_1$, $g^*y = g_2$, $g^*dx = dg_1$, $g^*dy = dg_2$. So $g^*d\theta = g^*(-y\,dx+x\,dy) = -g_2\,dg_1 + g_1\,dg_2$, and you can put in whatever function $g$ you're fond of. If you have the coefficient function $\lambda$ as a function on the circle rather than as a function of the ambient coordinates $(x,y)$, it's a bit messier to work that out. Indeed, the moral of the story is that you'd much rather work with the embedded submanifold as it sits in the ambient space than with an (abstract) parametrization of it.


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