Riemann surfaces with bounded curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:21:27Z http://mathoverflow.net/feeds/question/81058 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81058/riemann-surfaces-with-bounded-curvature Riemann surfaces with bounded curvature zalver 2011-11-16T11:47:08Z 2011-11-16T22:09:59Z <p>Say there are metrics $g_n$ on a compact Riemann surface $\Sigma$ with bounded curvature and bounded area, or even with the same area element . What can we say about the 'limit' of $(\Sigma, g_n)$? Maybe collapsing to Riemann surfaces with lower genus+circles? </p> http://mathoverflow.net/questions/81058/riemann-surfaces-with-bounded-curvature/81070#81070 Answer by macbeth for Riemann surfaces with bounded curvature macbeth 2011-11-16T14:01:06Z 2011-11-16T14:01:06Z <p>Whatever notion of limit you're using, you need a few more things in your "limit set." Consider the sequence of flat tori $\mathbb{R}^2/\Lambda_n$, where $\Lambda_n$ is the lattice generated by $(0,n)$ and $(1/n,0)$. We have uniform bounds 0 on curvature and 1 on area. However, </p> <ol> <li><p>The pointed Gromov-Hausdorff limit is a line. </p></li> <li><p>Pulling back by the diffeomorphisms $(x,y)\mapsto (nx, 1/n y)$, we get a sequence of metrics $n^2dx^2+1/n^2dy^2$ all on the "same" torus $\mathbb{R}^2/\Lambda_1$, satisfying the above and also your extra assumption of having the same area element. These metric tensors have no ($\mathcal{W}^{k,p}$, say) limit.</p></li> </ol> http://mathoverflow.net/questions/81058/riemann-surfaces-with-bounded-curvature/81091#81091 Answer by Vitali Kapovitch for Riemann surfaces with bounded curvature Vitali Kapovitch 2011-11-16T17:11:42Z 2011-11-16T22:09:59Z <p>You need to specify what limit you are talking about as the question makes no sense otherwise. The weakest natural topology to consider in this setting is pointed Gromov-Hausdorff topology. Gromov-Hausdorff convergence with two sided curvature bounds is very well understood by the theory developed by Cheeger, Fukaya and Gromov and is particularly easy in dimension 2. If collapsing occurs then the limit is either a point (can not happen if you fix volume), or a 1-dimensional manifold without boundary (so a line or a circle). The elements of the sequence in this case locally fiber over the limit with circle fibers (globally fiber over the limit if the limit is a circle).</p> <p>If the limit is 2-dimensional then it's an Alexandrov space with 2-sided curvature bounds. It's a $C^{1,\alpha}$ Riemannian manifold (again without boundary). Moreover, in this case you have topological stability on larger and larger balls as $i\to\infty$. In particular if you fix a bound on diameter then you have diffeomorphism stability and the limit has the same genus as the elements of the sequence for large $i$.</p> <p>Lastly note that collapsing with bounded diameter can only happen for a torus and a Klein bottle. This is immediate from Gauss-Bonnet.</p>