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The question is in the title:

Question: Which inscribed $n$-dimensional polytope (inscribed in the unit sphere) with $2^n$ vertices has the largest possible volume?

Is it the $n$-dimensional cube? If not, how much larger can its volume be?

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    $\begingroup$ This seems to be false in dimension 3 : the convex hull of the north pole, the south pole, and a regular hexagon inscribed in the equator has volume $\sqrt{3}$, larger than the volume of the inscribed cube. $\endgroup$ Commented Feb 15, 2022 at 13:29
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    $\begingroup$ I guess a natural follow-up question would be. does the cube maximise volume among inscribed polytopes combinatorically equivelant to the cube? $\endgroup$
    – Nick L
    Commented Feb 16, 2022 at 13:08
  • $\begingroup$ @NickL more generally, we can ask if the cube maximizes volume among inscribed $n$-polytopes with $2n$ facets. This looks plausible indeed. We can also ask whether the regular cross-polytope maximizes volume among inscribed polytopes with $2n$ vertices. The answers to both questions are positive if we restrict a priori to centrally symmetric polytopes, but it is not obvious why maximizers should be symmetric. $\endgroup$ Commented Feb 16, 2022 at 13:39
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    $\begingroup$ @Nick L Did you ever find an answer to your question about polytopes combinatorially equivalent to the hypercube? I am also interested in this question, not just for volume, but for surface area as well. $\endgroup$
    – user3816
    Commented Jul 20, 2022 at 22:50

1 Answer 1

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For $n=3$, the maximal volume polytope with 8 vertices is described in that paper.

Berman, J. D.; Hanes, K., Volumes of polyhedra inscribed in the unit sphere in (E^3), Math. Ann. 188, 78-84 (1970). ZBL0187.19604.The link is paywalled, but a summary with the optimal shape drawn is available in this answer.

It is combinatorially very different from a cube: it is simplicial, and each vertex has degree 4 or 5. The correct optimal polyhedron also has only $D_2$ symmetry.

Edit. An inscribed polytope which maximizes the volume given the number of vertices is always simplicial (see Lemma 1 in Horváth, Ákos G.; Lángi, Zsolt, Maximum volume polytopes inscribed in the unit sphere, Monatsh. Math. 181, No. 2, 341-354 (2016). ZBL1354.52016.), so the answer is "no" as well in any dimension $>2$.

The gap is even quantitative. Let $\mu_n$ be the largest volume of an $n$-dimensional polytope with $2^n$ vertices inscribed in the ball of radius $\sqrt{n}$ (this normalization is better since it counterbalances the fact that the ball of unit radius has a tiny volume for large $n$). The cube gives the lower bound $\mu_n \geq 2^n$. By taking direct products, we have $\mu_{m+n} \geq \mu_m\mu_n$, and therefore $\mu_{3n} \geq \mu_3^n \geq (2+\epsilon)^{3n}$, exponentially better that the cube. We cannot do much better since the upper bound $\mu_n \leq C^n$ for some constant $C$ holds trivially by comparing with the ball itself.

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  • $\begingroup$ Greate answer, thank you! The question that remains for me: can we put an upper bound on $2+\epsilon$ better than the obvious upper bound $2\pi e$ coming from the volume of the ball? $\endgroup$
    – M. Rumpy
    Commented Feb 18, 2022 at 16:06
  • $\begingroup$ Is it rather $\sqrt{2 \pi e}$ ? This is a very good question. At the moment I cannot find an argument giving an estimate $\mu_n \leq (\sqrt{2 \pi e} - \delta)^n$ for some $\delta > 0$. $\endgroup$ Commented Feb 21, 2022 at 9:15
  • $\begingroup$ What might actually work is the dual proposition using cross polytopes (similar to the octahedron in three dimensions): $2n$ vertices at the surface of a ball of $n$ dimensions. The regular cross polytope clearly has maximal volume in both dimensions $2$ and $3$, and to the degree that I can visualize things in higher dimensions the generalization "looks right". $\endgroup$ Commented Sep 1, 2022 at 13:23

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