Compactification theorem for differentiable manifolds ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T02:21:24Z http://mathoverflow.net/feeds/question/22441 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds Compactification theorem for differentiable manifolds ? Qfwfq 2010-04-24T17:20:45Z 2010-04-26T15:52:15Z <p>Just parallelling <a href="http://mathoverflow.net/questions/22437/a-variety-always-have-a-compactification-is-there-an-easy-proof" rel="nofollow">this</a> question, that seemed not to admit an easy answer at all, let's "soft down" the category and ask the same thing in the case of $\mathcal{C}^{\infty}$-differentiable manifolds [Edit: we consider only manifolds without boundary].</p> <p>Well, so:</p> <blockquote> <p>Is every differentiable manifold diffeomorphic to an open submanifold of a compact one?</p> </blockquote> <p><strong>Edit:</strong> As some comments have pointed out, there are manifolds for which the compactification theorem fails, so someone has suggested to change the question to the more meaningful:</p> <blockquote> <p>Which differentiable manifolds are diffeomorphic to an open submanifold of a compact one?</p> </blockquote> http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds/22444#22444 Answer by Richard Kent for Compactification theorem for differentiable manifolds ? Richard Kent 2010-04-24T17:26:10Z 2010-04-24T17:26:10Z <p>No. A surface of infinite genus is not a submanifold of a compact surface.</p> http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds/22445#22445 Answer by Paul for Compactification theorem for differentiable manifolds ? Paul 2010-04-24T17:38:27Z 2010-04-24T17:38:27Z <p>There is a long history on this problem, starting, in dimensions>4 with Browder-Levine-Livesay:</p> <p><a href="http://www.jstor.org/stable/2373259?origin=crossref" rel="nofollow">http://www.jstor.org/stable/2373259?origin=crossref</a></p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=189046" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=189046</a> </p> <p>Follow MR to get to results in dimensions 3,4, etc. You have to first eliminate issues like Richard mentions using finiteness obstructions.</p> http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds/22449#22449 Answer by Andy Putman for Compactification theorem for differentiable manifolds ? Andy Putman 2010-04-24T19:26:25Z 2010-04-24T19:26:25Z <p>As everyone has said, the answer is "no". You have to make assumptions to ensure that the "ends" of your manifold are sufficiently simple. It appears hard to find using the refs Paul posted, but the key result about this is Larry Siebenmann's thesis. I don't think this was ever published, but it is available on Andrew Ranicki's webpage <a href="http://www.maths.ed.ac.uk/~aar/surgery/thesis.pdf" rel="nofollow">here</a>. Another source (also on Ranicki's webpage) for this is some lecture notes of Kervaire, available <a href="http://www.maths.ed.ac.uk/~aar/papers/kervbrsi.pdf" rel="nofollow">here</a>.</p> <p>By the way, one obvious necessary condition is for your manifold to have a finitely presentable fundamental group (this is one of the problems with Richard's example). A classic example to show that this is still not enough (even in dimension 3) is the <a href="http://en.wikipedia.org/wiki/Whitehead_manifold" rel="nofollow">Whitehead manifold</a>.</p> <p>EDIT : I should also point out one beautiful recent about this. Marden's Tameness Conjecture (recently proved independently by Agol and Calegari-Gabai) says that if M is a hyperbolic 3-manifold with finitely generated fundamental group, then M is homeomorphic to the interior of a compact 3-manifold. The Whitehead manifold mentioned above shows that the assumption that M is hyperbolic is necessary.</p> http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds/22504#22504 Answer by Igor Belegradek for Compactification theorem for differentiable manifolds ? Igor Belegradek 2010-04-25T13:27:58Z 2010-04-25T14:31:56Z <p>I think none of the above posts answer the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?". Rather they answer "is every differentiable manifold diffeomorphic to the interior of a compact one?" The reason for the confusion could be the latter question is fundamental in geometric topology, while the former one has little significance. Anyway, </p> <p>The connected sum $V$ of infinitely many copies of $CP^3$'s is not diffeomorphic to an open subset of a compact manifold.</p> <p><i>EDIT: Hats off to Torsten Ekedahl who pointed out in comments that my argument below is incorrect (thus I don't know whether the above statement about $V$ is true). I decided not to delete it because it illuminates some subtleties of the original question.</i></p> <p>The point is that any diffeomorphism onto an open subset pulls back the tangent bundle, and in particular, pulls back the first Pontryagin class $p_1$. Thus if $V$ is an open subset of a compact manifold $M$, then its first Pontryagin class $p_1(V)$ lies in the image of $H^4(M)\to H^4(V)$, which is a <b>finitely generated</b> subgroup of $H^4(V)$, which is the infinite product of $\mathbb Z$'s corresponding to generators of $H^4(CP^3)$. The first Pontryagin class of $CP^3$ is a multiple of a generator of $H^4(CP^3)\cong\mathbb Z$, and removing a finite set of points from $CP^3$ does not affect the $4$th skeleton, so $p_1(V)$ does not lie in a finitely generated subgroup of $H^4(V)$.</p> <p>I am curious to see low-dimensional answers to the question "is every differentiable manifold diffeomorphic to an open submanifold of a compact one?"</p> http://mathoverflow.net/questions/22441/compactification-theorem-for-differentiable-manifolds/22547#22547 Answer by Agol for Compactification theorem for differentiable manifolds ? Agol 2010-04-25T23:34:22Z 2010-04-26T15:52:15Z <p>There are contractible 3-manifolds which cannot be embedded in any compact 3-manifold. <a href="http://www.ams.org/mathscinet-getitem?mr=144322" rel="nofollow">Kister and McMillan</a> constructed a variant of the Whitehead manifold $M'$ which is contractible but which cannot embed into $S^3$. From the Geometrization theorem, the universal cover of any compact 3-manifold embeds into $S^3$. So if $M'$ embedded into a compact 3-manifold $M'\subset M$, its lift $M'\subset \widetilde{M}\subset S^3$ to the universal cover would give a contradiction. </p>