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I have a $d$-dimensional hypercube in a $d$-dimensional space. I want to find out the number of "faces" of the hypercube that a 2D plane can intersect simultaneously.

The 2D plane can be written as the intersection of $d-2$ hyperplanes:

$$\langle \pmb{p} - \pmb{v_1}, \pmb{n_1} \rangle = 0, ..., \langle \pmb{p} - \pmb{v_{d-2}}, \pmb{n_{d-2}} \rangle = 0$$

So the intersection with a hyperplane, requires adding the hyperplane's equation to the system, and in the general case results in a 1D intersection set (in special cases it may be a 2D set, or a 0D set). The issue is that faces are bounded along each dimension (let's assume to $(0,1)$). So some of the resulting planar intersections need not be counted.

My conjecture is that at most 4 faces may be intersected (as in the 3d case), but I am unsure how to prove it/disprove it. If the intersection is of the plane coinciding with a face the intersection set is 2D, if the intersection is with a corner the set is 0D, in all other cases it is 2D.

My conjecture for a $k$-D plane is that the maximum number of intersected faces is $2^k$, but I am unsure how I can formalize that idea.

lightxbulb
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2 Answers2

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Let the cube be $0\le x,y,z\le1$. The plane $x+y+z=3/2$ cuts all six faces of the cube.

Empy2
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  • Then is there any upper bound in general? – lightxbulb Oct 19 '19 at 08:36
  • Try to adapt it to 4d, and can you find 1 or 2 points from the plane on each face. If not, why won't it work? – Empy2 Oct 19 '19 at 08:41
  • Because if it had an intersection with every possible face in 4D, the plane will not be a 2D set anymore? Does this generalize for hyperplanes of higher dimensions. I think it that case the number of intersections would increase greatly even for a 3D plane. – lightxbulb Oct 19 '19 at 08:46
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The main idea here is that of the Petrie polygon of the respective polytope.

It is defined to be a skeletal polygon contained within a regular polytope as pseudo face, which outlines in d-dimensional subelements a sequence of d+1 consecutive edges and then continues accordingly with the neighbouring d-dimensional subelement, where it already had outlined d consecutive edges. As induction start one clearly can take a flat regular polygon. Here the polygon itself and its Petrie polygon coincides. With respect to polyhedra e.g. the Petrie polygons thus are nothing but the equatorial antiprismatic zig-zags. In fact, these Petrie polygons quite generally are eqatorial non-flat zig-zags, allowing for some projection onto a 2-dimensional subspace. Thereby the polytope then will be contained completely within the projection of that zig-zag, which by itself then becomes a flat convex regular polygon.

Thus you would just have to use that 2D plane through the center of your nD hypercube and will readily derive the maximal possible section each.

Nij
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Dr. Richard Klitzing
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