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?