What would be counterexamples to the following statement:

It is not true that any $n$-manifold with boundary is a $n$-manifold with finitely many embedded disjoint open disks removed, since that would mean that its boundary is a disjoint union of spheres. E.g. the solid torus $B^2\times S^1$ has boundary the torus $S^1\times S^1$, which is not a disjoint union pf spheres.

Fuzzy question: If one had a classification of manifolds (up to diffeomorphism/homeomorphism/h-equivalence), could one automatically obtain a clasification of manifolds with boundary?

I know that to each manifold $M$ with boundary, there is associated a double manifold $DM$, which is the disjoint union of two copies of $M$, glued along their boundary. But if knew precisely what $DM$ was (in the classification scheme), could we classify $M$ itself?

Is any manifold a double of some other manifold, i.e. can every manifold be split into equal halves?

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    $\begingroup$ I found your question very confusing initially, because under the standard definitions, "manifold" means "manifold without boundary" (see this nLab entry), and hence "compact" means "closed". May I suggest saying "compact $n$-manifolds with boundary" to clarify? $\endgroup$ – Zev Chonoles Jul 7 '13 at 0:57
  • $\begingroup$ In my faculty, manifolds can have boundary by default. But I fixed the formulation. Thanks for the advice. $\endgroup$ – Leo Jul 7 '13 at 1:00
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    $\begingroup$ Doesn't your condition simply imply that the boundary components of $X$ are homemorphic/diffeomorphic to spheres? This is easily shown to be false - $S^1\times S^1$ is a not a sphere and is a boundary of $D^2\times S^1$. What am I missing? $\endgroup$ – Jason DeVito Jul 7 '13 at 1:04
  • $\begingroup$ @JasonDeVito: Eh, you're right. I guess that answers my question. But may I change it a little to get a more meaningful thread? $\endgroup$ – Leo Jul 7 '13 at 1:06
  • $\begingroup$ I wouldn't mind if you did, and I'd guess no one else would really, but of course, I can't really speak for everyone. $\endgroup$ – Jason DeVito Jul 7 '13 at 1:10

Simpler, $\mathbb RP^2$, the real projective plane, is not the double of a manifold with boundary.

If a manifold $M$ has boundary, notice that $\partial (M \times [0,1])$ is the double of $M$. So doubles have the property that they are boundaries of higher-dimensional manifolds. In dimension 2, a manifold is a double if and only if it is null-cobordant, which is if and only if it has even Euler characteristic.

You could construct some much sharper obstructions to a manifold being a double. Doubles have an effective action of $\mathbb Z_2$ with co-dimension $1$ fixed point set, the action switching the orientation of the normal bundle. With a little more work you could refine this to an if and only if statement for when a manifold is a double.

  • $\begingroup$ Thank you! What about the classification of $\partial$-manifolds out of a classification of manifolds? $\endgroup$ – Leo Jul 7 '13 at 13:52
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    $\begingroup$ Well, if you had a classification of closed manifolds with $\mathbb Z_2$-actions that have a co-dimension 1 fixed point set, then you would have a classification of manifold with boundary via the doubling operation. $\endgroup$ – Ryan Budney Jul 7 '13 at 21:31

In 3-manifold theory, any hyperbolic manifold that splits as a double must contain an embedded, $\pi_1$-injective surface (namely, the old boundary itself), and so must be Haken. There are manifolds that are not Haken, so not every manifold is a doubled manifold.

  • $\begingroup$ Thank you! How about my first question? $\endgroup$ – Leo Jul 7 '13 at 13:59

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