Let $\{M_i\}$ be a sequence of 2-dimensional orientable closed surfaces of genus $g$ with smooth Riemannian metrics with the Gauss curvature at least $-1$ and diameter at most $D$. By the Gromov compactness theorem, one can choose a subsequence converging in the Gromov-Hausdorff sense to a compact Alexandrov space with curvature at least $-1$ and Hausdorff dimension 0,1,or 2. One can show (see below) that if $g\geq 2$ then the limit space cannot be a point, thus the dimension of the limit space is at least 1 (while for $g=0,1$ it can be 0).

Let us assume that the limit space has dimension 1. Then it is either circle or segment.

Whether these both possibilities (circle and segment) can be obtained in the limit?

ADDED: It is not hard to see that one can get segment for $g=0$ and circle for $g=1$. I suspect (but cannot prove) that for $g\geq 2$ and $g=0$ one cannot get circle in the limit. In fact I do not even know whether in the case $g\geq 2$ a 2-dimensional limit is the only possibility.

UPDATE: Based on the answer by Igor Belegradek, let me summarize the situation. Let $\{M_i\}$ be a sequence of genus $g$ orientable closed surfaces with Riemannian metrics with Gauss curvature at least -1 which converges in the Gromov-Hausdorff sense to an Alexandrov space $X$.

1) If $g=0$ then $X$ is either a point, or a segment, or $X$ is homeomorphic to $S^2$ (by Perelman stability theorem), and all the three cases are possible.

2) If $g=1$ then $X$ is either a point, or a circle $S^1$, or homeomorphic to the 2-torus, and all the three cases are possible.

3) If $g\geq 2$ then $\dim X=2$ and hence $X$ is homeomorphic to an orientable genus $g$ closed surface.

ADDED: Let me add a proof that if $g\geq 2$ then a point cannot be the limiting space. Indeed otherwise we would have $d_i:=diam(M_i)\to 0$. Let us divide the metric of $M_i$ by $d_i$ and denote the new metric space by $N_i$. Then the sectional curvature of $N_i$ is at least $-d_i^2$ and diameter 1.

By the Gauss-Bonnet $$4\pi(1-g)=\int_{N_i}K\geq -d_i^2vol(N_i).$$ By the Bishop inequality $vol(N_i)$ is bounded from above. Hence the right hand side in the above inequality tends to 0. Hence $1-g\geq 0$ which is a contradiction.

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    $\begingroup$ If the limit is smooth (e.g. a circle), then by the Yamaguchi's fibration theorem $M_i$ is a fiber bundle over the limit (for large $i$). The only orientable closed surface that fibers over the circle is the torus (because the fiber is also a circle for dimension reasons). Reading Yamaguchi's more recent papers on low-dimensional collapse will help you understand the case when the limit is an interval. $\endgroup$ Commented Apr 12, 2016 at 16:46
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    $\begingroup$ Are your surfaces compact? You did not say this. If con-compact are permitted, take cylinders $I\times S$ where $I$ is an interval and $S$ is a small circle, and equip with the standard metric of curvature $0$. You obtain a segment in the limit. $\endgroup$ Commented Apr 12, 2016 at 20:17
  • $\begingroup$ I wish to add that if you wish to black box the collapsing theory (and you shouldn't!) you could just multiply your collapsing sequence by a circle, so that you now have a sequence of $3$-manifolds collapsing to a cylinder. Then Shioya-Yamaguchi's paper formally applies, see projecteuclid.org/euclid.jdg/1090347524. Of course this is an overkill. $\endgroup$ Commented Apr 12, 2016 at 21:46
  • $\begingroup$ @AlexandreEremenko: Thanks, corrected. $\endgroup$
    – asv
    Commented Apr 13, 2016 at 1:23
  • $\begingroup$ @IgorBelegradek: I am not sure, what is the conclusion of the use of the collapsing theory? $\endgroup$
    – asv
    Commented Apr 13, 2016 at 1:24

3 Answers 3


As mentioned in the comments, if the limit is a circle, then by the Yamaguchi fibration theorem, $M_i$ fibers over the circle, and hence it is a torus (or Klein bottle in the non-orientable case).

If the limit is a segment, then $M_i$ is $S^2$ for all large $i$. One (somewhat heavy handed) way to see this is to apply Corollary 0.4 of Shioya-Yamaguchi's paper. Indeed, the product of $M_i$ and a unit circle, collapses to the cylinder. Hence in Corollary 0.4 we have $g=0$ and $k=2$. Hence for large $i$ the fundamental group of $M_i\times S^1$ is a free product of $\mathbb Z$ and finitely many finite cyclic groups. Such a group cannot have a center unless all the finite cyclic groups are trivial. The circle factor of $M_i\times S^1$ is central in the fundamental group, so $\pi_1(M_i\times S^1)=\mathbb Z$. Hence $M_i$ is a sphere (for large $i$).

This argument uses orientability of $M_i$ because Shioya-Yamaguchi only deal with orientable manifolds.

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    $\begingroup$ Very nice but I wonder if there might be a straightforward argument using Gauss-Bonnet in the 2-dimensional case. $\endgroup$ Commented Apr 13, 2016 at 13:41
  • $\begingroup$ Great!! Thanks a lot. Though I agree with the comment by Mikhail Katz... $\endgroup$
    – asv
    Commented Apr 13, 2016 at 14:18
  • $\begingroup$ @MikhailKatz: On Gauss-Bonnet, what one has to rule out is small handles converging to points. One hopes that in the process curvature must tend to $-\infty$. If the handle is attached along a flat cylinder, then this seems doable: simply double along the boundary of the cylinder and argue that a closed hyperbolic surface cannot collapse to a point (which still requires more than Gauss-Bonnet, as far as I can see). Now in general the handle need not be attached along a flat cylinder and the curvature may go to $+\infty$ on the attaching region. Then I am not sure how to isolate that handle.. $\endgroup$ Commented Apr 13, 2016 at 14:48
  • $\begingroup$ @MikhailKatz: May I add also a simpler comment, though it might be obvious for experts. In order to exclude the case of the limiting space to be a point when $g\geq 2$, one has to use Gauss-Bonnet combined with the Bishop inequality. But the case of segment seems to be less elementary. $\endgroup$
    – asv
    Commented Apr 13, 2016 at 16:13
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    $\begingroup$ @user89334: still the set is not closed: rescaling a flat torus exactly as above gives a point in the limit while the curvature is zero. More importantly one gets many singular limits spaces. $\endgroup$ Commented Apr 14, 2016 at 22:39

Consider a shortest closed geodesic $\gamma$ on the surface of length sys, and the normal exponential map of $\gamma$. Using the lower curvature bound, we obtain an upper bound on the total area as $\text{sys}\cdot \sinh(D)$ where $D$ is the diameter. This follows just by applying Rauch bounds on Jacobi fields (this is an ingredient in the proof of Toponogov). Therefore the systole is bounded below by $ \frac{\text{area}}{\sinh D}$ and the area is bounded below by Gauss-Bonnet. Furthermore the filling radius is bounded below by the 1/6 of the systole by Gromov's inequality. The least Gromov-Hausdorff distance to a graph is bounded below by the filling radius. We therefore get a quantitative lower bound and not merely nonexistence of Yamaguchi-type collapse.

This proves that hyperbolic surfaces of curvature bounded below by $-1$ with diameter bounded above by $D$ cannot collapse so that a Gromov-Hausdorff limit is necessarily 2-dimensional.

For more details see https://arxiv.org/abs/1604.06782

  • $\begingroup$ Unfortunately I am unfamiliar with some of the background you use. What statement exactly you proved? $\endgroup$
    – asv
    Commented Apr 19, 2016 at 15:24
  • $\begingroup$ @sva, which concepts are you not familiar with? $\endgroup$ Commented Apr 19, 2016 at 15:26
  • $\begingroup$ Systole and filling radius. Also apparently you use their relation to the Gromov-Hausdorff distance (to a graph?). BTW in your statement should one add the condition that the genus of the surfaces in at least 2? $\endgroup$
    – asv
    Commented Apr 19, 2016 at 15:31
  • $\begingroup$ The systole is the least length of a noncontractible loop on the surface $M$. The filling radius of $M$ is defined as the least epsilon such that the inclusion of $M$ in its epsilon-neighborhood sends the fundamental homology class of $M$ to zero. Here the embedding is in the space of bounded functions on $M$ which sends a point to the distance function from that point. Actually this stage can be bypassed and one can show directly an inequality between the systole and the $(n-1)$-diameter. $\endgroup$ Commented Apr 19, 2016 at 15:37
  • $\begingroup$ @sva, the argument only works for hyperbolic surfaces (hence genus at least 2) since otherwise Gauss-Bonnet does not give a lower bound for the area. $\endgroup$ Commented Apr 19, 2016 at 17:07

I remembered another reason why closed surfaces of negative Euler characteristic cannot collapse under a lower bound on sectional curvature.

Much more is true: if a sequence of $n$-dimensional closed manifolds $M_i$ of Ricci curvature $\ge -k^2$ Gromov-Hausdorff converges to a compact space of (Hausdorff) dimension $<n$, then the simplical volume of $M_i$ is zero for large $i$.

Indeed, on the bottom of p.244 of of Gromov's Volume and bounded cohomology he shows that a lower Ricci curvature bound implies that the simplicial volume is bounded above in terms of volume and the dimension. If a sequence of manifolds converges to a space of dimension $<n$ under the lower Ricci bound, then their volume goes to zero; this is due to Colding, as mentioned e.g. on p.91 of Aspects of Ricci Curvature.

The simplicial volume of a closed surface of negative Euler charactersistic $\chi$ is $2|\chi|$, see p.217 in Gromov's paper, so it cannot collapse under a lower curvature bound. There are of course many high-dimensional manifolds of nonzero simplicial volume.


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