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If we consider an orientable and regular surface $\Sigma \subset \Bbb R^3$, as we would consider in Stokes' theorem, do we have any criteria that would guarantee its boundary to be empty ?

For example, I know that the sphere $\Sigma_1 = \{(x, y, z) :x^2 + y^2 + z^2 = 1\}$ has an empty boundary, i.e. $\partial \Sigma_1 = \emptyset$, but the cylinder $\Sigma_2 = \{(x, y, z) : x^2 + y^2 = 1, 0 \leq z \leq 1\}$ has not, $\partial \Sigma_2 = \{(\cos\theta, \sin\theta, 0) \mid \theta : 0 \rightarrow 2\pi \} \cup \{(\cos\theta, \sin\theta, 1) \mid \theta : 2\pi \rightarrow 0 \}$.

Desura
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  • It is hard to see what sort of criterion you are looking for... – Mariano Suárez-Álvarez Dec 26 '16 at 17:47
  • Anything. In fact, I don't really understand why the sphere has an empty boundary. So I thought that any standard convex surface would have an empty boundary as well but that's not the case. I'm looking for "classes" of surface with empty boundary or just a lot of examples where this happens. – Desura Dec 26 '16 at 18:02
  • @Desura: what meaning of "boundary" are you thinking about? The boundary of a subset in a toplogical space, or the boundary of a surface/submanifold with boundary in $\mathbb{R}^3$? – Catalin Zara Dec 26 '16 at 18:08
  • @Catalin Zara The boundary of a surface in $\Bbb R^3$, as in (Kelvin-)Stokes' theorem. It is indeed not the boundary of a subset of a topological space, even if it has the same notation.. – Desura Dec 26 '16 at 18:15
  • Well, a boundaryless compact orientable surface must have nontrivial second homology. – symplectomorphic Dec 26 '16 at 18:26
  • What convex body has boundary with a non-empty boundary? – Mariano Suárez-Álvarez Dec 26 '16 at 20:51
  • Well, a cylinder am I wrong ? It is weird because I read that "the boundary (in surface terms) of a boundary (in topological terms) is always the empty set", But isn't the considered cylinder the boundary of a cylinder (as a 3D volume) ? – Desura Dec 27 '16 at 10:22

1 Answers1

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Based on your comments, it sounds like you don't understand what the boundary is. Why does the sphere have empty boundary and cylinder does not? If you understand what is meant by the boundary (not the definition, but the idea), then these are obvious.

A boundary point on a manifold is a point where you can travel in some directions tangent to the manifold, but you cannot move in others. A simple example is the half-plane $H = \{(x,y)\mid x, y \in \Bbb R, x \ge 0\}$. From points with $x > 0$, you can travel in all directions on the plane without leaving $H$, even if only for a small distance. But for points where $x = 0$, you can only travel in half of the directions. The other half would take you immediately out of $H$.

The sphere is symmetric in all of its points. Each has a neighborhood that looks like a disk. No matter which direction along the sphere you stay within the sphere. (Perhaps a more accurate way of saying this is that given any vector tangent to the sphere at the point, there are curves within the sphere passing through the point and having that as their tangent vector. I.e., there is nothing about the direction itself that forces one to leave the sphere.)

For the cylinder, this is not the case. When $0 < z < 1$, you can travel in all directions for the point. But for points where $z = 0$, you cannot travel in any direction that has a negative $z$ component, as any curve traveling in such a direction would have to move through points with $z < 0$, which is not in the surface. And similarly for the $z = 1$ boundary: in that case if the direction vector has positive $z$ component, you cannot travel in that direction without leaving the surface.

The boundary is the edge of the surface, the place you cannot sail past without falling off.

It is not a matter of convexity either. Note that the open cylinder $\{(x, y, z) : x^2 + y^2 = 1, 0 < z < 1\}$ has no boundary, though is almost identical to your $\Sigma_2$. The difference is that the boundary points have been removed.

Paul Sinclair
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