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Consider the infinite cylinder defined with its axis being the z-axis and radius 1.

This is a manifold. Now, what does it mean exactly to have a differential form $\omega$ on this manifold, $M$?

A differential form is a map from vector fields to differentiable functions. So what would be the vector field here? A vector field maps points in $M$ to the tangent bundle $TM$, which itself is a manifold.

So then is it true that $\omega$ maps tangent vectors of $M$ to differential functions?

Then can you explain to me what $\omega = dx\wedge dy$ means, intuitively?

If you have a point $p$ on the cylinder $M$, the tangent plane at $p$ is the plane passing through $p$ that is orthogonal to $p$.

So then what would $\omega = dx\wedge dy$ mean in that case?

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closed as unclear what you're asking by Aloizio Macedo, Parcly Taxel, Xam, Yiorgos S. Smyrlis, user223391 Feb 4 '18 at 2:34

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You're starting with a $2$-form (not a $1$-form), so you need to apply it to a pair of vector fields.

If you integrate your $2$-form $\omega = dx\wedge dy$ over a little piece $\Sigma$ of surface, you'll find the (oriented) area of the projection of $\Sigma$ into the $xy$-plane. In the case of your cylinder, you'll get $0$. (Why?) On the other hand, if you take $\omega = dx\wedge dz$, then you'll get the (oriented) area of the projection on the $xz$-plane. Draw some pictures and play with it. If you take $\omega = d\theta\wedge dz$, you'll get precisely the area $2$-form for your cylinder (so you'll get the surface area of $\Sigma$).

[It might help you to watch some of my lectures on YouTube, starting with Day 24 of MATH 3510. (That particular lecture has a complete version at the very bottom of the list.) There are lots of concrete examples, as well as theoretical material.]

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