most recent 30 from http://mathoverflow.net 2013-05-24T11:51:35Z http://mathoverflow.net/feeds/question/106169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106169/closed-manifold-has-no-nontrivial-totally-convex-subset jiangsaiyin 2012-09-02T10:12:49Z 2012-09-02T13:39:58Z

<p>Peter Peterson's book "Riemannnian Geometry" p351 says: </p> <ol> <li><p>Closed manifold has no nontrivial totally convex subset. Using the energy functional if $A\subset M$ is totally convex, then $A\subset M$ is $k$-connected for any $k$. </p></li> <li><p>It is however not possible for a closed n-manifold to have $n$-connected nontrivial subsets as this will violate Poincare duality. </p></li> </ol> <p>Why are these statements true?</p>

http://mathoverflow.net/questions/106169/closed-manifold-has-no-nontrivial-totally-convex-subset/106172#106172 Igor Belegradek 2012-09-02T11:23:35Z 2012-09-02T11:30:11Z

<p>The inclusion from a closed totally convex subset to the ambient manifold is a homotopy equivalence. Details can be found in <a href="http://www.intlpress.com/JDG/archive/1981/16-2-333.pdf" rel="nofollow">"Totally convex sets in complete Riemannian manifolds"</a> by Bangert, JDG, 1981. </p> <p>For the second assertion, Cheeger-Gromoll prove in their paper on the soul theorem that any closed totally convex subset is a manifold with boundary, so if the boundary is non-empty, the manifold has zero top-dimensional homology, hence it cannot be homotopy equivalent to a closed manifold of that top dimension.</p>