"Affine communication" for topological manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T11:31:31Z http://mathoverflow.net/feeds/question/31568 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31568/affine-communication-for-topological-manifolds "Affine communication" for topological manifolds Tyler Lawson 2010-07-12T15:58:14Z 2010-07-12T18:27:27Z <p>There is a situation that comes up regularly in algebraic topology when giving proofs of facts about manifolds, like Poincare duality and the like. The typical sequence goes like this:</p> <ul> <li>Prove something for \$\mathbb{R}^n\$.</li> <li>Then it follows for open disks.</li> <li>Use a Mayer-Vietoris argument to prove it for finite unions of convex sets in \$\mathbb{R}^n\$.</li> <li>Use a colimit argument to prove it for arbitrary open subsets of \$\mathbb{R}^n\$.</li> <li>It then follows for open subsets of a manifold admitting a homeomorphism to an open subset of \$\mathbb{R}^n\$.</li> <li>Use a Mayer-Vietoris argument to prove it for finite unions of such subsets of a manifold.</li> <li>Use a colimit argument to prove it for arbitrary open subsets of a manifold.</li> </ul> <p>Obviously there is some redundancy here, and it makes the technical details in these proofs overwhelm the underlying ideas. In the smooth category one can do better, but usually only by appealing to machinery which is useful but takes extra time to prove.</p> <p>The answers/comments in the following question point towards a very useful way to compare coordinate charts in different affine covers of a given scheme:</p> <p><a href="http://mathoverflow.net/questions/28496/what-should-be-learned-in-a-first-serious-schemes-course" rel="nofollow">http://mathoverflow.net/questions/28496/what-should-be-learned-in-a-first-serious-schemes-course</a></p> <p>Namely, the intersection of any two affine opens has a cover by open sets which are simultaneously distinguished in both.</p> <p>I feel like I should know a reference to whether this is true in the topological category - and I suspect that it's not - but I shamefully don't know. One can phrase this in terms of continuous local homeomorphisms from \$\mathbb{R}^n\$ to itself, but I'll instead just ask:</p> <p><em>Given two coordinate charts on a topological manifold M and a point in their intersection, is there a neighborhood of this point which is simultaneously a convex open set in both charts? Is there a simple counterexample?</em></p> http://mathoverflow.net/questions/31568/affine-communication-for-topological-manifolds/31597#31597 Answer by Tom Goodwillie for "Affine communication" for topological manifolds Tom Goodwillie 2010-07-12T18:27:27Z 2010-07-12T18:27:27Z <p>There are piecewise linear counterexamples in dimension \$2\$. </p> <p>Arrange \$2m\$ evenly spaced rays \$R_i\$ around the origin, \$m\ge 3\$. If \$C\$ is a convex neighborhood of the origin, let \$r_i\$ be the reciprocal of the length of the portion of \$R_i\$ in \$C\$. For some number \$K>0\$ (depending on \$m\$), convexity implies \$r_{i+1}+r_{i-1}\ge 2Kr_i\$. Thus if \$e(C)\$ and \$o(C)\$ are the sums of \$r_i\$ over even and odd \$i\$ we have \$e(C)\ge Ko(C)\$ and \$o(C)\ge Ke(C)\$. Now apply a homeomorphism \$h\$ that linearly stretches even-numbered rays by \$A>0\$ and odd-numbered rays by \$B>0\$. If \$h(C)\$ is a convex set then you would have numbers \$e(h(C))=e(C)/A\$ and \$o(h(C))=o(C)/B\$, whence \$e(C)/A\ge Ko(C)/B\$, contradicting \$o(C)\ge Ke(C)\$ if \$A/B\$ is chosen big enough. </p>