Imaginary Hypercubes - Discrete and Computational Geometry and Graphs - page 177

infinitely many squares. In the three-dimensional case, there are many convex
double imaginary cubes, and H is the only 0/0.5/1 MCI among them as we
mentioned in Sect. 2 .For n

4, we show that 16-cell is the only convex double
imaginary n -cube shape. We prepare two lemmas, whose proofs are omitted.

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Lemma 7. For n ≥ 3 , the dimension of the ane hull of an imaginary n -cube

is n .

Note that this lemma does not hold for n = 2 because a line segment is an
imaginary 2-cube.
For an n -dimensional hyperplane G , we denote by r ( G ) he distance of G from
the origin.

n ) and G be an n -dimensional hyperplane.

Lemma 8. Let C =conv( {− 1 , 1 }

(1) If n> 4 and one of the open half spaces defined by G contains only one
vertex of C , then r ( G ) > 1 .
(2) If n =4 and one of the open half spaces defined by G contains only one
vertex v of C , then r ( G ) 1 .If r ( G )=1 , in addition, then the four

adjacent vertices of v are on G .

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Theorem 9. 16-cell is the only convex double imaginary 4-cube shape. For n>
4 , there is no double imaginary n -cube.

Proof. Let n 4. Suppose that B is a double imaginary cube of two n -cubes

C 1 and C 2 . One can see that A = C 1 ∩

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C 2 is a convex double imaginary cube

because we have B A . We consider the double imaginary cube A .

We can assume without loss of generality that C 1 =conv(

ↂ

n ) and that
the edge length of C 2 is less than or equal to the edge length of C 1 ,thatis,2.
Let P be a facet of C 2 and G be the hyperplane containing P . All the edges of
P must intersect with C 1 because A is an imaginary cube of C 2 . Hence P

{−

1 , 1

}

∩ C 1 is

an imaginary cube of an ( n − 1)-cube P . Since n ≥ 4, the dimension of its ane

hull is n 1 by Lemma 7 . On the other hand, it is immediate to show that each

facet of C 1 is not on G . Therefore, there exists a vertex v of C 1 in the open half-
space defined as the opposite side of C 2 with respect to G . Such a vertex of C 1
is unique because every edge of C 1 must intersect with C 2 . Therefore, if n> 4,
then r ( G ) > 1 by Lemma 8 . Since it also holds for the facet P which is parallel
to P , the edge length of C 2 is greater than 2, contradicting the assumption.
Therefore, we have n = 4. By Lemma 8 , the two 4-cubes have the same size
and P contains all the four adjacent vertices of v . Therefore, P

−

C 1 is a regular
tetrahedron. Since C 1 and C 2 have the same size, it holds for all the facets of C 1
and C 2 . Therefore, A is a 16-cell. Since a 16-cell is a minimal convex imaginary
4-cube, it is the only convex double imaginary 4-cube.

∩

3.3 Higher Dimensional Extensions of H and T

In this subsection, we make n -dimensional extensions of the four 0/0.5/1 MCIs
in Fig. 1 in each n 2. We regard the 0/0.5/1 n -MCI which has no v-vertices

as an imaginary n -cube corresponding to a cuboctahedron.

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