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Let $V$ be a body in $\mathbb{R}^3$ bounded by a smooth sphere with principal curvatures at most 1 (by absolute value). Is it true that $$\mathop{\rm vol} V\ge \mathop{\rm vol} B,$$ where $B$ denotes the unit ball?

Comments

  • I heard this question from Dima Burago.

  • In $\mathbb{R}^2$ an analogous statement follows from the Pestov–Ionin theorem.

  • One might think that $V$ contains a copy of $B$, but this is not true; an example was constructed by Lagunov (Sibirsk. Mat. Zh. 1961)

Postscript. There is a closed surface with normal curvatures at most 1 that bounds volume arbitrarily close to $\tfrac{22 }3\cdot\pi-2\cdot\pi^2\approx 3.3$. This volume is swapped by rotating the hatched figure in $2\times2$ square around its lower side.

enter image description here

This volume is smaller than the volume of unit ball which is $\tfrac43\cdot\pi\approx 4.2$.

However, the surface is homeomorphic to the sphere with two handles. (It is an optimized version of one of Lagunov's examples.) I tried to optimize the spherical example and was not able to get below $\tfrac43\cdot\pi$. Maybe one can prove that it is impossible, but most likely it is going to be tedious.

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    $\begingroup$ Any chance you could describe Lagunov’s example? $\endgroup$
    – Deane Yang
    Jun 26, 2022 at 15:26
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    $\begingroup$ @DeaneYang It can be obtained by thinkening Bing's house. A slightly simpler example is discussed in our book; see pp 107--108 arxiv.org/pdf/2012.11814.pdf $\endgroup$ Jun 26, 2022 at 20:23
  • $\begingroup$ This remind me of a question I posed in my course of "differential geometry" in midterm as follows: Let $\gamma$ be closed curve included in unit circle prove that there is a point whose curvature is $\geq 1$ $\endgroup$ Feb 10, 2023 at 15:20
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    $\begingroup$ @OtisChodosh yes $\endgroup$ Feb 13, 2023 at 14:48
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    $\begingroup$ @L.F.Cavenaghi I could not imagine that it is working if $V$ is nonconvex (which is the only interesting case). $\endgroup$ Feb 15, 2023 at 20:33

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