Smooth and analytic structures on low dimensional euclidian spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:58:04Z http://mathoverflow.net/feeds/question/52620 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52620/smooth-and-analytic-structures-on-low-dimensional-euclidian-spaces Smooth and analytic structures on low dimensional euclidian spaces Hugo Chapdelaine 2011-01-20T13:56:49Z 2011-01-20T19:52:40Z <p>So it is relatively easy to show that there exists only one smooth structure on the real line \$\mathbb{R}\$. So here are 2 natural questions:</p> <p>Q1: Up to equivalence, is there only one real analytic structure on \$\mathbb{R}\$? If so, then do we have a simple proof of that? </p> <p>Q2: Where can I find the simplest proofs that there exists only one smooth structure on \$\mathbb{R}^2\$ and \$\mathbb{R}^3\$?</p> <p>So I've heard that on \$\mathbb{R}^4\$ there are infinitly (in fact uncountably) many non-equivalent smooth structures, so what about real analytic strucutres? Is there some kind of moduli space of smooth structures on \$\mathbb{R}^4\$. if so, in how many ways is it possible to deform a smooth structure into a real analytic one?</p> http://mathoverflow.net/questions/52620/smooth-and-analytic-structures-on-low-dimensional-euclidian-spaces/52682#52682 Answer by Ryan Budney for Smooth and analytic structures on low dimensional euclidian spaces Ryan Budney 2011-01-20T19:45:16Z 2011-01-20T19:45:16Z <p>Regarding Q1, put an analytic Riemann metric on your 1-manifold. Integrating a unit speed vector field gives an analytic diffeomorphism to \$\mathbb R\$. Another way to prove analytic structures are unique is to notice the same argument that one uses to prove that the group of \$C^k\$-diffeomorphisms of \$\mathbb R\$ has the homotopy type of \$\mathbb Z_2\$ works for analytic diffeomorphisms -- simply take the straight-line homotopy between your original diffeomorphism and either the identity or the negative identity, appropriately. </p> <p>Regarding Q2, I don't know much in the way of really simple proofs. But when \$n=2\$ you've got the Uniformization Theorem from complex analysis. That's relatively simple. </p>