# Riemann surfaces with bounded curvature

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?

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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 '11 at 12:29
What do you mean by "circles"? –  Igor Rivin Nov 16 '11 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 '11 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 '11 at 15:37
Why do you ask? This is much too broad a question. –  Igor Rivin Nov 16 '11 at 15:54

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.

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The circle also cannot happen is the area is fixed. –  Sergei Ivanov Nov 16 '11 at 20:36
yes, that's a good point. –  Vitali Kapovitch Nov 16 '11 at 21:51
So if the metrics has higher order curvature bound then the smoothness of the limit can be improved too? About the torus: can it collapse with bounded volume and diameter? –  zalver Nov 17 '11 at 13:46
yes, in the noncollapsed case if you have bounds on higher order derivatives of the metric then you get similar bounds for the limit metric by Cheeger compactness. Torus can certainly collapse with bounded diameter e.g. by rescaling a flat metric. –  Vitali Kapovitch Nov 17 '11 at 15:21

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,

1. The pointed Gromov-Hausdorff limit is a line.

2. 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.

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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 '11 at 15:30
@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 '11 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 '11 at 15:54
lol..I just thought it's same as the positive curved case. thanks. –  zalver Nov 17 '11 at 13:36