# Completing compact surfaces with boundary to closed surfaces in $\mathbb R^3$

My question is whether any compact smooth surface in $\mathbb R^3$ (with smooth boundary) can be completed to a closed smooth surface in $\mathbb R^3$ without boundary? It is easy to complete it to an abstract smooth closed $2$-manifold by gluing disks to boundary circles. It seems, however, that in some cases this procedure can't be applied in $\mathbb R^3$ but in all examples that I can imagine we can glue together two (for example) different boundary circles by using a cylinder to obtain a closed surface. Is it a general case or there is an example of a smooth compact surface with smooth boundary in $\mathbb R^3$ which can't be completed to a closed surface in $\mathbb R^3$?

Edit:Lee Mosher in his answer showed that if the surface is not orientable, then it is easy to find a counterexample. What about orientable surfaces?

• @BalarkaSen I mean that $X$ is a part of some closed $\widetilde X$. Of course, this is not unique, so maybe the terminology is bad Apr 25, 2015 at 14:15
• yes, I realized that. Thanks for clarifying. Apr 25, 2015 at 14:16

## 1 Answer

The Mobius band cannot be extended to a closed surface.

Added: To answer the additional question, suppose that $\Sigma \subset \mathbb{R}^3$ is the given orientable surface. Then there is an embedding $f : \Sigma \times [0,1] \to \mathbb{R}^3$ such that $\Sigma = f(\Sigma \times 0)$. So then $\Sigma$ extends to a closed surface $$f\biggl(\bigl(\Sigma \times \{0,1\}\bigr) \cup \bigl(\partial \Sigma \times [0,1]\bigr)\biggr)$$ This is not yet smooth, but it can be easily smoothed.

• Thanks you, yes, but what if we require orientability? Apr 25, 2015 at 14:13
• Every orientable 2-manifold can be embedded in $\Bbb R^3$ by classifcation theorem for 2-manifolds, @Nimza. Apr 25, 2015 at 14:24
• @BalarkaSen yes, If I understand you correctly, you propose to complete my surface $X$ to an abstract closed $2$-manifold gluing balls to boundary circles and then embedding it into $\mathbb R^3$. The problem is that in my setting I'm already given the restriction of this embedding to $X$ so that in some cases there are no embedding to $\mathbb R^3$ with given restriction to $X$. Apr 25, 2015 at 14:28
• @Nimza: I added an answer to your orientable surface question. Apr 25, 2015 at 14:35
• @Nimza: It's just a standard theorem of differential topology, embedding the normal bundle of a submanifold as a neighborhood of the submanitold, although tailored slightly to cover the case of a submanifold-with-boundary. Apr 25, 2015 at 14:45