Smoothability of compact Alexandrov surfaces with curvature bounded from below. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:59:20Z http://mathoverflow.net/feeds/question/86807 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86807/smoothability-of-compact-alexandrov-surfaces-with-curvature-bounded-from-below Smoothability of compact Alexandrov surfaces with curvature bounded from below. Thomas Richard 2012-01-27T10:01:55Z 2012-01-29T03:16:01Z <p>Let $(X,d)$ be compact metric space of curvature greater than $-1$ (in the sense of comparison triangles), assume that its Hausdorff dimension is $2$. Then a result of Perelman says that $X$ is a 2-dimensional manifold.</p> <h2>Claim</h2> <p>$(X,d)$ can be Gromov-Hausdorff approximated by a sequence of Riemannian surfaces $(M_i,g_i)$ such that $\int_{M_i}|K_{g_i}|dv_{g_i}$ is bounded.</p> <p>To see this :</p> <ol> <li><p>A version of the Gauss-Bonnet Theorem holds (<a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jmsj/1225113599%20%22Machigashira%22" rel="nofollow">Machigashira, The Gaussian curvature of an Alexandrov surface</a>), which implies that $(X,d)$ is an Alexandrov surface with bounded <em>integral</em> curvature.</p></li> <li><p>Any such surface can be approximated by smooth Riemannian surfaces with bounded integral curvature. See Reshetnyak, Geometry IV, Encyclopaedia of Mathematical Sciences. </p></li> </ol> <h2>Question</h2> <p>Is any compact Alexandrov surface of curvature greater than $-1$ approximated by a sequence of smooth compact Riemannian surfaces with curvature bounded from below (by -1, or something else if this helps) ?</p> <p>Maybe this is classic but I didn't found explicit results of this kind.</p> <p>Thanks.</p> http://mathoverflow.net/questions/86807/smoothability-of-compact-alexandrov-surfaces-with-curvature-bounded-from-below/86868#86868 Answer by Vitali Kapovitch for Smoothability of compact Alexandrov surfaces with curvature bounded from below. Vitali Kapovitch 2012-01-27T23:31:45Z 2012-01-29T03:16:01Z <p><em>Edit: Addressing Igor's comment I'd like to correct the references I gave. The correct reference for the exact argument I sketch should be the original book by Alexandrov <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=29518" rel="nofollow">"Intrinsic Geometry of Convex Surfaces"</a>(Chapter 7, section 6, Lemmas 1-3). In the later book by Alexandrov and Zalgaller a similar (but necessarily more complicated) argument is given for the case of surfaces with bounded integral curvature (Theorem 10, page 84). One can reconstruct the original proof in the easier case of curvature bounded below from that one but it requires some work.</em></p> <p>This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=216434" rel="nofollow">"Intrinsic geometry of surfaces"</a>).</p> <p>The general structure the proof is as follows. You take a very fine triangulation of $X$ and substitute the curved triangles by triangles with the same sides in the space form of constant curvature $-1$. This will give you a polyhedral surface of curvature $\ge -1$ (you only need to check that the cone angles at vertices are $\le 2\pi$ which is immediate from the definition of an Alexandrov space). Away from the cone points the metric will be smooth of constant curvature $-1$. The cone points can then be easily smoothed to get a smooth metric of $\sec\ge -1$. The hard part is to show that the resulting polyhedral surface is close to the original space $X$.</p> <p>A more interesting result is another theorem of Alexandrov ( <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=29518" rel="nofollow">"Intrinsic Geometry of Convex Surfaces"</a>) that locally any 2-dimensional Alexandrov space of $curv\ge -1$ is isometric to a level set of a convex function in $\mathbb H^3$. When the lower bound is 0 he proved an even sharper result that any 2-sphere of curvature $\ge 0$ is isometric to the boundary of a convex body in $\mathbb R^3$.</p> <p>BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too. </p>