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Let
A
1
be a 16-cell given by vert(
A
1
)=
V
1
=
{
(
±
1
,
0
,
0
,
0)
,
(0
,
±
1
,
0
,
0)
,
4
) be a 4-cube. Let
V
2
and
V
3
be the subsets of vert(
C
1
) with even and odd numbers of 1
/
2-coordinates,
respectively. One can see that
C
2
=conv(
V
1
∪
(0
,
0
,
±
1
,
0)
,
(0
,
0
,
0
,
±
1)
}
,
and let
C
1
=conv(
{−
1
/
2
,
1
/
2
}
V
2
)are
also 4-cubes. Since
V
1
does not contain any star of
C
2
(resp.
C
3
), and every edge
of
C
2
(resp.
C
3
) contains a point in
V
1
, we can find that
A
1
is an imaginary cube
of
C
2
(resp.
C
3
) that has no e-vertices. Thus,
A
1
is a double imaginary 4-cube.
Note that
A
2
=conv(
V
2
)and
A
3
=conv(
V
3
) are also 16-cells.
As we described in Sect.
2
, T is a weak cross-polytope imaginary cube shape
and H is a double imaginary cube shape in the three-dimensional case. We show
that 16-cell is the only weak cross-polytope imaginary cube shape as well as the
only double imaginary cube shape in four and higher-dimensional cases. First,
we study weak cross-polytope imaginary cubes.
V
3
)and
C
3
=conv(
V
1
∪
Lemma 5.
A convex imaginary
n
-cube polytope has at least
2
n−
1
vertices.
Proof.
An
n
-cube has
n
2
n−
1
edges and a convex imaginary
n
-cube polytope
contains a vertex on each of these edges. Since a vertex is on at most
n
edges of
the cube, we have the result.
Proposition 6.
For
n
≥
3
,
T
and 16-cell are the only weak cross-polytope imag-
inary hypercube shapes.
Proof.
Since an
n
-dimensional weak cross-polytope has 2
n
vertices,
n
must sat-
isfy 2
n
≥
2
n−
1
by Lemma
5
, and hence
n
≤
4.
For
n
= 4, any weak cross-polytope has eight vertices. If a weak cross-
polytope
A
is an imaginary cube polytope of a 4-cube
C
, then
A
is a MCI
with no e-vertices, and each edge of
C
contains one vertex of
A
from the proof
of Lemma
5
.Thus
A
is a 16-cell.
For
n
= 3, assume that a weak cross-polytope
A
is an imaginary cube of a 3-
cube
C
. Note that
A
may not be an MCI of
C
.Weset
V
(
A
):=vert(
C
)
∩
A
. Since
a 3-cube has 12 edges and
A
has 6 vertices, we have 3#
V
(
A
)+(6
−
#
V
(
A
))
≥
12,
and get #
V
(
A
)
3.
If #
V
(
A
)=3,
A
has three e-vertices and they must be on the edges of
C
both of whose endpoints are not in
V
(
A
). Thus,
A
is an MCI of
C
, and we can
find that T is the only such polytope.
If #
V
(
A
)
≥
≥
4, there is a pair
{
v
1
,
v
2
}ↂ
V
(
A
) such that vert(
A
)
\{
v
1
,
v
2
}
is on a plane that is orthogonal to the line segment [
v
1
,
v
2
]. Suppose that
C
=
3
. Since [
conv
{
0
,
1
}
v
1
,
v
2
] contains an interior point of
C
, we can put
v
1
=
(0
,
0
,
0) and
v
2
=(1
,
1
,
1) without loss of generality. Suppose that the other four
vertices of
A
are on a plane defined as
). Since
A
has four or more v-vertices, we get
a
=1
,
2. If
a
=1,vert(
A
) must contain
(1
,
0
,
0)
,
(0
,
1
,
0) and (0
,
0
,
1). However, no line passes through two of them and
the origin (1
/
3
,
1
/
3
,
1
/
3) at the same time. Therefore, we have no weak cross-
polytope in this case. The case
a
= 2 is similar to the case
a
=1.
{
(
x, y, z
)
|
x
+
y
+
z
=
a
}
(
a
∈
R
Next, we study double imaginary
n
-cubes. A convex object can be an imaginary
2-cube of two or more squares. For example, a square is an imaginary cube of
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