Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

Claim

$(X,d)$ 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.

To see this :

  1. A version of the Gauss-Bonnet Theorem holds (Machigashira, The Gaussian curvature of an Alexandrov surface), which implies that $(X,d)$ is an Alexandrov surface with bounded integral curvature.

  2. Any such surface can be approximated by smooth Riemannian surfaces with bounded integral curvature. See Reshetnyak, Geometry IV, Encyclopaedia of Mathematical Sciences.

Question

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) ?

Maybe this is classic but I didn't found explicit results of this kind.

Thanks.

share|improve this question
add comment

1 Answer 1

up vote 10 down vote accepted

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 "Intrinsic Geometry of Convex Surfaces"(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.

This is indeed classical and is due to Alexandrov ( see the book by Alexandrov and Zalgaller "Intrinsic geometry of surfaces").

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 $X$.

A more interesting result is another theorem of Alexandrov ( "Intrinsic Geometry of Convex Surfaces") 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 $\mathbb R^3$.

BTW, the result you attribute to Perelman that a 2-dimensional Alexandrov space is a topological manifold is actually due to Alexandrov too.

share|improve this answer
    
Where exactly is this argument in Alexandrov-Zalgaller's book? –  Igor Belegradek Jan 28 '12 at 0:06
    
@Igor sorry, I don't have the book and I haven't looked at in in quite a while but my recollection is that I read it there. –  Vitali Kapovitch Jan 28 '12 at 0:22
    
Thanks, I didn't realized that Alexandrov already proved the result about the topology of 2-dimensional Alexandrov surfaces. Maybe my mistake come from "Geometry IV" of Reshetnyak, which defines surfaces of bounded integral curvature assuming at the beginning that the underlying space is a topological surface. By the way, Alexandrov-Zalgaller is on my desk these days, I'll have a look at it. Thanks again. –  Thomas Richard Jan 28 '12 at 13:32
    
After looking at Alexandrov-Zalgaller's argument, I must admit I would not be comfortable quoting it as an answer to the OP's question. Given that the question is of considerable interest it is unfortunate that there seem to be no proof in the literature. –  Igor Belegradek Jan 28 '12 at 14:15
    
@Vitali, on your edit: which chapter is this in? I have the new edition of Alexandrov's "Intrinsic Geometry of Convex Surfaces" published in his collected works and almost every chapter has section 6. –  Igor Belegradek Jan 28 '12 at 17:54
show 7 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.