# What is the correct notion of boundary of a submanifold?

This question has been bothering me for a while now and I fear is a result of a gap in my education.

Just so we are all on the same page I'll define submanifold... Suppose we have a subset $N\subset M$ where $M$ is a $n$-manifold. We say that $N$ is a $k$-dimensional submanifold if for each point $p\in N$ there is a neighborhood $U$ of $p$ in $M$ so that the component of $U\cap N$ containing $p$ may be parameterized by a $C^\infty$ map of an open subset of $\mathbb{R}^k$. For the sake of concreteness (as I think it might matter) let us suppose $M$ has a metric and is complete with respect to this metric.

My question is: Is there a correct way to talk about $\partial N$ the boundary of $N$ or does one need more structure? For instance if $N$ has the structure of an integer rectifiable current then the boundary is defined unambiguously.

To clarify I bit what I'm worried about I'll give some examples:

Suppose one starts with a closed submanifold $N$ and deletes a finite number of points what is the boundary of new submanifold? What if you instead delete a Cantor set?

$N$ is non-proper. For instance if the induced metric on $N$ is complete but $N$ lies entirely within a compact region of $M$ (so $N$ has points of accumulation in $M$). Does $N$ not have a boundary? (This would I believe, prevent $N$ from being a current).

Clarification

The actual situation where this arose for me was the following: I have an open ball $B$ in euclidean space and a submanifold $N$ of $B$ as discussed above. I want to be able to say that for (almost*) every closed ball $B'\subset B$ that $N\cap B'$ is a submanifold with boundary in the usual well understood sense AND $\partial (B'\cap N) \subset \partial B'$. To ensure this one would like to say that $\partial N\subset \partial B$. This of course raises legitimate questions as to what $\partial N$ means. It seems a reasonable definition for my purposes is $\partial N=\bar{N}\backslash N$ (here $\bar{N}$ is the closure in the ambient space). However, I worry there are subtle problems with such a definition and was wondering if there was a canonical definition (which perhaps there is not).

*One can't expect to have this for every closed ball due to a failure of transverality between $\partial B'$ and $N$.

• Not every open manifold can be realized as the interior of a manifold with boundary. This is the topic of Larry Siebenmann's dissertation. In particular your cantor set complement is such an example. So it seems like you really ought to be asking not "what is the boundary" but "which manifolds can be given a boundary" ? For that the answer is: L. C. Siebenmann, Finding a boundary for an open manifold, Notices Amer. Math. Soc 12 (1965), 337. – Ryan Budney Sep 12 '10 at 20:16
• Can you, please, clarify what you mean by the boundary of $N$? The general topological notion of boundary (=frontier) applies, but it needs not have any structure besides being a closed set (just consider removing an arbitrary closed subset from $\mathbb{R}^n$). On the other hand, if you want to consider a submanifold $\textit{with boundary}$ of $M,$ you need to modify your definition of submanifold. – Victor Protsak Sep 12 '10 at 20:24
• In general I don't see the utility of "what the correct notion of..." type questions, as it begs the question "what is the purpose of..." – Ryan Budney Sep 12 '10 at 20:37
• I have added a bit of explanation of what my motivation for asking the question is. Perhaps the orginal question should have asked whether there is a concept of boundary defined for all open submanifolds that agrees with the correct notion of boundary when the manifold arises as the interior of a submanifold with boundary. Or failing existence of such a notion what is the best one can do. – Rbega Sep 12 '10 at 21:21
• Consider the case of a disc embedded in $\mathbb R^3$ and it will be positioned so that with respect to a height function there is only one extrema on the disc, the absolute minimum. Imagine as you walk up level-sets the circle levels evolve from being a small round circle to the various stages of a space-filling curve in $[0,1]^2$. Taking the closure of this open, smooth embedded disc you get a disc whose boundary in the point-set topology sense is the unit square. Are you happy with this kind of situation -- boundary not lowering dimension? – Ryan Budney Sep 12 '10 at 21:38

For your stated goal it suffices to define the boundary as manifold's boundary (i.e., it is empty for a submanifold as defined in your first paragraph). Then you'll have $\partial(B'\cap N)\subset\partial B'$ automatically.
However I suspect that the notion you are seeking is "properly embedded submanifold": a submanifold $N$ of $M$ is properly embedded if the inclusion map $i:N\to M$ is proper (a map is called proper if the pre-image of every compact set is compact). In the case when $M$ is an open ball $B$ the intuitive meaning of this is that all ends of $M$ accumulate at the boundary of $B$.
This property also implies that the boundary of $N$ as a current (if it is one) is supported in the boundary of $B$.