I was wondering how I can see if a manifold has a boundary just by looking at the surface?
The thing is that I want to understand how to apply the Gauß Bonnet theorem to surfaces and there I need to integrate over the boundary, too.
If you are just dealing with sets in $\mathbb{R}^n$ it is pretty simple to find out if a point is on the boundary, as soon as you have a visual picture of the set, cause you can just ask yourself if this point would be in the closure of the complement or if there is an open set containig this point which is still part of the set.
This seems to be no longer possible for manifolds.
Despite, it seems to be still possible to say that no point on the sphere for example is not in the interior which is different from the triangle in the plane, where the edges are visually not part of the interior.
So somehow this boils down to: Think you are standing on a $n-$ dimensional manifold and you are standing at a point $p$. If you can walk in any of the $n-$ directions, this point is not part of the boundary, otherwise it is. Still, I don't like this characterization, because a cube for example does not have a boundary, although standing on any of the edges might give you the idea, that this should be part of a boundary.
So my question is: Does anybody have a good way to find out if a point belongs to the boundary of a surface or not?
First, and perhaps foremost, manifolds don't have boundaries. Suppose, by way of contradiction, that $M$ is a manifold with nonempty boundary and that $p$ is a point on that boundary. Then, no neighborhood of $p$ is homeomorphic to $\mathbb{R}^m$. Contradiction.
What you actually mean is manifold with boundary. A manifold with boundary has a boundary if and only if it is not (strictly speaking) a manifold. Like I stated above, $p$ is on the boundary if and only if none of its neighborhoods is homeomorphic to $\mathbb{R}^m$.
To determine if an oriented smooth manifold with boundary has a boundary, we can use Stoke's theorem: if there exists an $m$-form $\alpha$ with compact support such that $$\int_M\mathrm{d}\alpha=\int_{\partial M}\alpha\neq 0,$$ then the boundary is nonempty.