The Möbius strip (without boundary) $ S $ can be realized as a regular surface of $ R^3 $ (regular surface is meant in the sense of do Carmo's book). Using a 'manifold' language it can be proved that $ S $ is an embedded submanifold of $ R^3 $. Therefore it is for itself a smooth manifold (that we realize as embedded into $ R^3 $).

My trouble is that we cannot obtain the Möbius strip with boundary as an embedded subanifold of $ R^3 $. If we want to realize the Möbius band with boundary it seems natural to consider the topological closure of $ S $ as a subset of $ R^3 $ and then proving that this closed subset is an embedded submanifold with boundary into $ R^3 $. However this argument should fail since we know that no closed subset of $ R^3 $ cannot be homeomorphic to a non orientable $ 2 $-manifold.

My question is : why does this argument should fail? I haven't checked the detail but i'm looking for an intuitive answer for it.

N.B: an embedded submanifold $ M \subset R^3 $ is defined as a subset such that for every $ p \in M$ there exists a diffeomorphism $ \phi:U \rightarrow V $, where $ p \in U $, $ \phi(p) =0 $ and $ U,V $ are open subsets of $ R^3 $ such that $ \phi(U \cap M)=V \cap R^2 $

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    $\begingroup$ You could probably find this answer interesting. $\endgroup$ – A.P. May 6 '13 at 16:46

Let $M\subset \Bbb R^3$ be a $2$ dimensional submanifold with boundary (say the Möbius strip) and let $p\in M$ be a boundary point. Let $U$ be an open ball of $\Bbb R^3$ centered on $p$, like in your definition of embedded submanifold. Then $U\cap M$ will be an open neighbourhood of $p$ in $M$, diffeomorphic to an open half-disk $D$ in $\Bbb R^2$ ("diameter" included). Finally, observe that there is no open subset of $\Bbb R^2$ diffeomorphic to $D$, so there is no $V\subseteq \Bbb R^3$ open such that $\phi(U\cap M)=V\cap \Bbb R^2$.

Note that the difference between a manifold with an one without boundary lies exactly in the fact that every point of the former has a neighbourhood diffeomorphic to $\Bbb R^n$, while this fails for some points of the latter, which admit open neighbourhoods diffeomorphic to an open half-ball in $\Bbb R^n$.

  • $\begingroup$ Thanks for your answer. I'm not sure about what you want to say... Have you proved that there are no embedded submanifolds in $ R^n $? $\endgroup$ – user55449 May 6 '13 at 19:09
  • $\begingroup$ No, I showed you why you cannot embed any manifold with boundary (orientable or not) in $\Bbb R^3$, at least with your definition of embedded submanifold. $\endgroup$ – A.P. May 6 '13 at 19:12
  • $\begingroup$ Clearly the definition of embedded submanifold that i've given in my question is more precisely the definition of 'embedded submanifolds without boundary'. An embedded submanifold with boundary could be defined as the image of a manifold with boundary trhought an embedding. Note that this definition can be also used in the case without boundary $\endgroup$ – user55449 May 6 '13 at 19:13
  • $\begingroup$ Yes, but that's the scope of the result you cite, i.e. that every compact 2-dimensional submanifold of $\Bbb R^3$ is orientable. That is exactly the point made in the answer I linked in my comment to your question. Sorry if I wasn't clear about this. $\endgroup$ – A.P. May 6 '13 at 19:18
  • $\begingroup$ I haven't seen the linked commment! In any case my comments are referred to the answer. Now i've red your link. Thanks $\endgroup$ – user55449 May 6 '13 at 19:25

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