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Let $\{(\mathbb{T}^2,g_i)\}_{i=1}^\infty$ be a sequence of 2-dimensional tori with smooth Riemannian metrics with Gauss curvature at least $-1$. It was explained in the final answer to the post Gromov-Hausdorff limits of 2-dimensional Riemannian surfaces that such a sequence cannot converge to a segment in the Gromov-Hausdorff sense.

The argument there used a reduction to the case of collapse of 3-manifolds studied in the paper Shioya–Yamaguchi - Collpapsing three-manifolds under a lower curvature bound.

I would be interested to have a more direct proof of the above fact that 2-tori with Gauss curvature at least $-1$ cannot converge to a segment.

This fact seems to me to be more elementary than the case of 3-manifolds studied in the above paper. A reference would be very helpful.

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  • $\begingroup$ I edited in a link to the answer to mathoverflow.net/questions/236001/… that was 'final' in the chronological sense, but maybe you meant 'final' in some other sense (e.g., sort order, which can change). Please feel free to change the link to the appropriate answer if I got it wrong. (Also, it's 'tori', not 'torii'.) $\endgroup$
    – LSpice
    Commented Oct 16, 2019 at 19:39

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This is not quite an answer, bur certainly too long to be a comment! Here are two sketchy ideas.

By following the arguments of Shioya and Yamaguchi, you could make rigorous this argument that the only orientable 2-manifold collapsing to an interval is $S^2$. The result will not be so elementary as an application of the Gauss--Bonnet Theorem, but it is surely more straightforward than going up to dimension three.

The portion of $M^2$ which is collapsing to the interior of the interval fibers over it, so is homeomorphic to $S^1 \times I$. Now you need to cap this off with two pieces corresponding to the ends of the intervals. Let $p_i$ be a sequence of points converging to an end of the interval. Rescale the sequence $(M^2, p_i)$ so that you obtain a non-negatively curved 2-dimensional limit space, which can have only one end. This is either $D^2$ or the Möbius strip -- by orientability it must be $D^2$. Therefore $M^2$ is given by $S^2$.

Something a little more elementary might be as follows. Suppose $M^2$ were a torus. Cut the interval in two and lift this cut through the fibration to divide $M^2$ along a circle. This describes $M^2$ as a connected sum, and the only way to do that with a torus is as $T^2 \# S^2$. Carefully piece two copies of the $(T^2 \setminus D^2)$ portion together preserving the lower curvature bound -- fiddly, but almost certainly doable. In this way we obtain a genus two surface collapsing to an interval and so a contradiction.

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  • $\begingroup$ Thank you. This is a little bit too concise for me, particularly the last two sentences of the first approach. $\endgroup$
    – asv
    Commented Aug 19, 2019 at 9:55
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    $\begingroup$ Essentially it's Theorem 0.5 of Shioya-Yamaguchi, but in lower dimension. Working through that argument will show how it works in dim 2. To expand a little, a non-compact non-negatively curved 2-dimensional limit space will deformation retract onto a 'soul', which is a point or a circle. In the former case the space is a disk, while in the latter it is a line bundle over the circle. The trivial line bundle, though, has two ends, so it is the Möbius strip. However, that is not orientable. Therefore the space is $D^2 \cup (S^1 \times I) \cup D^2$, i.e. it is $S^2$. Hope that helps. $\endgroup$
    – John Harvey
    Commented Aug 21, 2019 at 20:14

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