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I've been trying to understand differential forms but still have some parts of confusion. In particular, it is not clear to me when to use charts to restrict a differential form and when not.

For example, consider

$$\omega = \sum_{k=1}^{n+1} x_k dy_k - y_k dx_k$$

This is a contact form on $S^{2n +1}\subseteq \mathbb R^{2n + 2}$. It has $2n + 2$ coordinates in the expression.

On the other hand, if one picks charts for the sphere and restricts $\omega$ to the sphere using the charts then the resulting local expression will only have $2n+1$ coordinates since the sphere is $2n+1$ dimensional.

Now what I don't understand is when I use one or the other. It's clear to me that some forms don't have a global expression.

But given a global expression when in general is it advisable to write it in local coordinates (restrict with charts)?

self-learner
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1 Answers1

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First of all, $\omega$ is a form on $\mathbb R^{2n+2}$ not on the sphere. It has a "global expression" because $\mathbb R^{2n+2}$ admits for a global chart (which is a nice thing). If you truly want a form on the sphere, you have to pull it back to it and express it in local coordinates.

Writing such a form in e.g. spherical coordinates can be very useful. Consider e.g. geodesics on the sphere; they are most conveniently expressed in spherical coordinates. The same holds for vector fields on the sphere and you immediately know the action of a form on them.

However, embeddings are also very useful and it is often easier to work with objects which are defined on the embedding space, since the latter usually admits for a global chart. Furthermore, if you consider maps between manifolds, writing them as maps between the embedding spaces (often this is done "on the fly") makes it possible to write out easily their derivatives etc. At the end, of course, everything has to be projected to the manifolds of interest.

Markus Heinrich
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  • Thank you so much. Your answer is so helpful. Your first paragraph resolves about half of all my confusion and also answers a previous question of mine. – self-learner Jul 04 '16 at 07:42
  • But the pullback would be done using the inclusion (=identity) map. The differential of that is the identity, too. So then the pullback has no effect because $i^\ast \omega = \omega$. No? – self-learner Jul 04 '16 at 07:44