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(This question is a special case of this old question, in the hopes of making progress on a more tractable piece of the puzzle.)

Let $H = [0,1]^4\subseteq \mathbb R^4$ be the unit 4-hypercube, and let $P$ be some plane, specified e.g. by the set of solutions to two simultaneous linear equations. What is the maximum area of the set $H\cap P$?

By taking $P = \{(x,y,z,w)\ |\ x=y, z=w\}$, we obtain a $\sqrt{2}\times\sqrt{2}$ square of area $2$. Is this maximal?

One approach that might be promising is to consider $P$ fixed as the $xy$-plane, and ask instead about the choice of orthogonal unit vectors $v_1,v_2,v_3,v_4$ such that the resulting cube intersects $P$ well; this amounts to intersecting some bands that are the projection of the inequalities $0\le (x,y,z,w)\cdot v\le 1$.

It's possible that the question is more tractable if we require that $P$ contain the center of the hypercube; I have a strong intuition that this ought to be the case, but can't immediately prove it.

RavenclawPrefect
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  • You have intuition that $P$ contains the origin, or the center of the hypercube? The former doesn't seem obvious to me, unless you simply mean that it just has to contain a vertex. – Brian Tung Oct 09 '23 at 20:33
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    A comment here suggests that Ball proved that an area of $2$ is indeed maximal. Haven't verified that yet, though. – Brian Tung Oct 09 '23 at 20:37
  • @BrianTung: I meant the center of the hypercube - I've edited the post, sorry for the typo. Thanks for the pointer to the comment, I'd missed that the last time I read the linked post - would love to find a citation for that! – RavenclawPrefect Oct 09 '23 at 21:47

1 Answers1

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This 1989 paper “Volumes of sections of cubes and related problems” of Keith Ball shows that $2$ is indeed maximal for $P$ passing through the center of the hypercube; in general it shows that the maximal central $k$-dimensional section of an $n$-cube is at most $2^{(n-k)/2}$, a bound which is tight when $k\ge n/2$. Thanks to Brian Tung for the pointer to this paper's existence!

I'm still curious about whether we can relax the condition that $P$ pass through the center, but this mostly suffices to answer my question.

MJD
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RavenclawPrefect
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  • A general approach to showing that one can assume $P$ passes through the center is to suppose that it passes through $(x, 1/2, 1/2, \ldots, 1/2)$ wlog. Whatever the intersection is, show that it will have no less measure if you translate it by $1/2-x$. – Brian Tung Oct 09 '23 at 23:36