Homotopy groups of smooth manifolds? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:36:28Z http://mathoverflow.net/feeds/question/5745 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5745/homotopy-groups-of-smooth-manifolds Homotopy groups of smooth manifolds? Ilya Nikokoshev 2009-11-16T22:37:15Z 2010-01-09T21:43:18Z <blockquote> <p>For a fixed $d$, is there a relationship between the homotopy groups of smooth $d$-manifolds?</p> </blockquote> <p>The $d=1$ case is trivial, but I already don't know how to approach $d=3$ (I should have said that the case of $d=2$ is simple as well, since there is only a sphere to consider, but I don't know how to formulate the property of "having the same homotopy groups as $S^2$" in a simpler way).</p> <p>Note about the discussion on the comments: it's unreasonable to expect an easy <em>complete</em> characterization of homotopy groups of $S^2$, even less for other manifolds. But I think one could try some <em>partial</em> relations. An interesting relationship would be: for some $d$, the groups <code>$\pi_n$</code> can be determined from groups <code>$\pi_m$</code> for <code>$m&lt;N&lt;n$</code> (this is unlikely to be true though).</p> http://mathoverflow.net/questions/5745/homotopy-groups-of-smooth-manifolds/5749#5749 Answer by Ryan Budney for Homotopy groups of smooth manifolds? Ryan Budney 2009-11-16T23:07:55Z 2010-01-09T21:43:18Z <p>For $d=3$ the homotopy groups can be pretty elaborate. Consider the connect-sum of some lens spaces. The universal cover embeds in $S^3$ as the complement of a cantor set (except for a few degenerate cases where you have $\mathbb RP^3$ summands). So the homotopy-groups are pretty complicated ($\pi_2$ is finitely generated over $\pi_1$). You could probably make an argument that this is about the worst thing that can happen for the homotopy-groups of 3-manifolds. </p> <p>You might want to phrase your question as a question about the Postnikov towers of manifolds. Eilenberg-Maclane spaces are rarely compact boundaryless manifolds. </p> <p>edit: I guess another spin on your question could go like this. We know the fundamental groups of compact manifolds are all possible finitely presentable groups provided $n \geq 4$. So is there a sense in which the homotopy-algebras of manifolds can be anything finitely presentable? Say, for example, $\pi_2$. As a module over the group-ring of $\pi_1$, are there any restrictions beyond being finitely generated? I suppose you could construct a compact $6$-manifold with $\pi_2$ (almost) any finitely-presented thing over any finitely-presented $\pi_1$ pretty much the exact same way $4$-manifolds with any finitely presented $\pi_1$ are constructed. I think if $H_2(\pi_1)$ is non-trivial you might run into problems following the analogous construction, in that $\pi_2$ might strictly contain the $\pi_2$ you're trying to create.</p> <p>2nd edit: So regarding 3-manifolds I think your question has something of an answer now, right? $\pi_n M$ is $\pi_n$ of the universal cover provided $n > 1$. The universal cover of a geometric 3-manfold is homeomorphic to $\mathbb R^3$ or $S^3$. So by climbing up the JSJ and connect sum decomposition of a 3-manifold, the universal cover is diffeomorphic to a punctured $S^3$ -- the number of punctures is either $0$, $1$, $2$ or a Cantor set's worth of punctures. In the Cantor set case we're giving this complement the compactly generated topology induced from the Cantor set complement's subspace topology. So among other things, $\pi_2 M$ is a direct sum of copies of $\mathbb Z$, similarly $\pi_3 M$, torsion first appears in $\pi_4 M$. The complement you think of as a directed system of wedges of $S^2$'s so the Hilton-Milnor theorem tells you the homotopy groups. </p> http://mathoverflow.net/questions/5745/homotopy-groups-of-smooth-manifolds/6362#6362 Answer by Somnath Basu for Homotopy groups of smooth manifolds? Somnath Basu 2009-11-21T09:32:14Z 2009-11-21T09:32:14Z <p>The case for $d=4$ also runs into complications, pretty much like it happens for $d=3$. One can show that the homotopy groups of a simply connected smooth closed oriented four manifold $M$ is completely determined by its second Betti number $b_2$. More precisely, if $b_2=k$ then the homotopy groups of $M$ are given by the homotopy groups of $\sharp^{k-1}S^2\times S^3$. Again, these groups are very hard to compute.</p> http://mathoverflow.net/questions/5745/homotopy-groups-of-smooth-manifolds/6363#6363 Answer by Thomas Riepe for Homotopy groups of smooth manifolds? Thomas Riepe 2009-11-21T10:07:44Z 2009-11-21T10:07:44Z <p>Arapura wrote <a href="http://www.msri.org/publications/books/Book28/files/arapura.ps.gz" rel="nofollow" title="pdf">"a brief guide to some recent work on fundamental groups of varieties"</a>. <a href="http://de.arxiv.org/abs/math/0510245" rel="nofollow" title="arxiv">Pridham</a> gave new new restrictions on etale fundamental groups of smooth varieties. Perhaps Sullivan's work relates to the question. </p>