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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?

flag	asked Nov 16 2011 at 11:47 zalver 51●4

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I don't understand. If $\Sigma$ is compact, then certainly the curvature and the area will be bounded. Do you mean that there is a uniform bound for all $n$? – Spiro Karigiannis Nov 16 2011 at 12:29

What do you mean by "circles"? – Igor Rivin Nov 16 2011 at 13:23

Yeah. I mean unifom bounded. The circle may appear in a scaling way: E.g. shrink the torus $n S^1* (1/n)S^1$ to $S^1*(1/n^2)S^1$.. – zalver Nov 16 2011 at 13:48

For $C^{1,\alpha}$ convergence, it is enough to have a uniform upper bound on diameter, a uniform lower bound on volume, and a uniform bound on sectional curvature. This is due to Cheeger and Gromov. The proof works for Riemannian manifolds for any dimension. For the case of collapse, I think your question is subsumed by the papers "Collapsing Riemannian Manifolds while keeping their Curvature Bounded" parts I and II, by Cheeger and Gromov, but you are probably more interested in simplifications of this general result in your case. – Ken Knox Nov 16 2011 at 15:37

Why do you ask? This is much too broad a question. – Igor Rivin Nov 16 2011 at 15:54

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Vitali Kapovitch · Accepted Answer · 2011-11-16 22:09:59Z UTC

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).

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 $i$.

Lastly note that collapsing with bounded diameter can only happen for a torus and a Klein bottle. This is immediate from Gauss-Bonnet.

macbeth · Answer 2 · 2011-11-16 14:01:06Z UTC

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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,

The pointed Gromov-Hausdorff limit is a line.
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.

answered Nov 16 2011 at 14:01

macbeth
1,194●6●14

Yeah it depends on what 'limit' means. There must be some materials discussing this concept. But for simplicity let's add some other condition, say the curvature $K_n$ is unfiomly bounded by $a<K_n<b<0$. Then the above example (the diameter tends to infinity) seems not to happen. Then is it easier to talk about the 'limit'? – zalver Nov 16 2011 at 15:30

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@zalver The curvature can be constantly $-1$ and the Gromov-Hausdorff limit could be a line. Just pick a sequence of hyperbolic surfaces whose injectivity radii tend to zero, and pick basepoints on the shortest geodesic. – Richard Kent Nov 16 2011 at 15:54

If your metrics are pinched negatively curved, diameter can certainly go to infinity (not for the torus), but then the limits correspond to curves pinching off. – Igor Rivin Nov 16 2011 at 15:54

lol..I just thought it's same as the positive curved case. thanks. – zalver Nov 17 2011 at 13:36

Riemann surfaces with bounded curvature

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