NOTES
Edited by Sergei Tabachnikov
Hyperplane Sections of the
n-Dimensional Cube
Rolfdieter Frank and Harald Riede
Abstract. We deduce an elementary formula for the volume of arbitrary hyperplane sections
of the n-dimensional cube and show its application in various dimensions.
1. INTRODUCTION. Intersecting a cube with a plane leads to quite different intersection polygons, depending on the position of the plane. In each single case, these
polygons
can be found easily by calculation, and their possible areas range from 0 to
p
2 times the area of a face of the cube.
However, the situation changes dramatically if we intersect the four-dimensional
cube with a three-dimensional hyperplane. Evers showed in [5] that at least 30 different combinatorial types of intersection polyhedrons occur! Therefore, it is quite
remarkable that in each case, regardless of which dimension, there is a simple general
formula for the volume of the intersection polytope. It was found by Ball [1]; Berger
[4] and Zong [7] report on it.
Theorem 1 (Ball 1985). Let Cn = [ 1, 1]n be an n-dimensional cube with n
and let
2
H = {x 2 Rn | a • x = b} with a = (a1 , . . . , an ) 2 (R \ {0})n , b 2 R,
be a hyperplane. Then
|a| n
Vol(Cn \ H ) =
·2
⇡
1
Z
1
1
n
Y
sin(ak t)
k=1
ak t
!
· cos(bt) dt.
(1)
Here and subsequently, Vol denotes the (n
1)-dimensional volume.
p
Ball uses formula (1) to prove Vol(Cn \ H )  2n 1 2. This bound is best possible for each n and was conjectured by Hensley in 1979. For details see [1], [4],
or [7].
But trying to apply this amazing formula to any concrete case quickly dampens the
initial enthusiasm. Berger writes in [4], p. 468: “It seems that there is no direct geometric formula; any such formula would have to be different according to the various
ways in which H meets the edges.” We succeeded in calculating the integral in (1) (see
[6]) and obtained formula (2).
http://dx.doi.org/10.4169/amer.math.monthly.119.10.868
MSC: Primary 51M20, Secondary 51M25
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[Monthly 119
Theorem 2. Let the same assumptions hold as in Theorem 1. Then
! 1
n
Y
|a|
Vol(Cn \ H ) =
·
ak
2(n 1)!
k=1
·
X
w2{ 1,1}n
(a • w + b)
n 1
sgn(a • w + b) ·
(2)
n
Y
wk .
k=1
Here and subsequently, sgn denotes the sign of a real number.
We will prove Theorem 2 directly, without making use of (1). This is easier than
calculating the integral and works without the methods of probability theory, which
are used in Ball’s proof of Theorem 1.
Remark 1. The expression (1) is a slight modification of that found in [1], [4], and
[7]. There, |a| = 1 is assumed and Cn is the unit cube [ 12 , 12 ]n .
Remark 2. Our assumption ak 6 = 0 for all k is no restriction. Namely, if a1 , . . . , am 6 =
0 and am+1 , . . . , an = 0, then Cn \ H = (Cm \ H 0 ) ⇥ Cn m with H 0 = {x 2 Rm |
a1 x1 + · · · + am xm = b} and therefore
Vol(Cn \ H ) = Volm 1 (Cm \ H 0 ) · Voln
m (C n m ).
Now, Volm 1 (Cm \ H 0 ) can be calculated by means of (2).
Remark 3. Theorem 2 can be generalized to hyperplane sections of parallelotopes. Let A be a real invertible n ⇥ n-matrix. Then A(Cn ) is a parallelotope. Since
A 1 (H ) = {x 2 Rn | A T a • x = b}, we obtain Vol(Cn \ A 1 (H )) if we replace a by
A T a in formula (2). If subsets of Rn are multiplied by A, then their n-dimensional
volumes are multiplied by |det A|. With this we deduce that
Vol(A(Cn ) \ H ) = Vol A Cn \ A 1 (H )
= |a| · |A T a|
1
· |det A| · Vol Cn \ A 1 (H ) .
Remark 4. Each facet of Cn \ H is a section of H with a facet of Cn , which is an
(n 1)-dimensional cube. Hence, for n 3, Theorem 2 yields a formula for the sum
of the (n 2)-dimensional volumes of all facets of Cn \ H . For n = 3, this sum is
just the circumference of the intersection
q polygon. For n = 4, it is the surface of the
|a|2
intersection polyhedron. With |a|i =
n
n
X
Y
|a|i ·ai
·
ak
2(n
2)!
i=1
k=1
!
1
·
X
ai2 , this sum is
(a • w+b)n 2 sgn(a • w+b) · wi ·
w2{ 1,1}n
n
Y
wk . (3)
k=1
Remark 5. In H = {x 2 Rn | a • x = b}, we may assume |a| = 1. Then |b| is the
distance of the hyperplane H from the origin, and integration of equation (2) over b
from b = 1 to b = b0 yields a formula for the volume of the intersection of Cn with
the half space {x 2 Rn | a • x  b0 }. Such a formula has already been deduced in [3]
by Barrow and Smith. Vice versa, we could obtain our equation (2) by differentiating
their formula with respect to b ( in the Barrow and Smith formula).
December 2012]
NOTES
869
2. PROOF OF THEOREM 2. As above, let Cn = [ 1, 1]n be an n-dimensional
cube with n 2 and let H = {x 2 Rn | a • x = b} with a = (a1 , . . . , an ) 2 (R \
{0})n , b 2 R, be a hyperplane. For k = 1, 2, . . . , n, we define the functions Ak by
Ak (t) = 1 if |ak |  t  |ak |, Ak (t) = 0 otherwise. By
R 1Fn we denote the convolution A1 ? · · · ? An ; thus we have F1 = A1 and Fn (b) = 1 Fn 1 (b t) · An (t) dt for
n > 1 and b 2 R.
We prove the following lemma by induction on n and include the case n = 1 in
order to facilitate the basis of induction.
Lemma 1. Using the notation defined above, it follows that
Vol(Cn \ H ) = |a| ·
for all n
n
Y
k=1
ak 1 · Fn (b)
1.
Proof. Since the reflection of Rn in the hyperplane xk = 0 is a symmetry of Cn , we
may assume ak > 0 for all k = 1, 2, . . . , n.
The intersection of the interval [ 1,q
1] with H = {x 2 R | a1 x = b} has the 0-
dimensional volume A1 (b) = F1 (b) = a12 · a1 1 F1 (b). Hence, Lemma 1 is correct
for n = 1. Now let n > 1 and assume Lemma 1 is true for n 1. For every t 2 R, let
Ht denote the hyperplane xn = t. Furthermore, we introduce a 0 = (a1 , . . . , an 1 , 0),
Q
Q
= nk=1 ak , 0 = nk=11 ak , and the hyperplane H 0 = {y 2 Rn | a 0 • y = 0}.
R1
It follows that 1 Voln 2 (Cn \ H \ Ht ) dt is the volume of the orthogonal projection of Cn \ H onto the hyperplane H 0 . Hence, we obtain Vol(Cn \ H ) if we
divide this integral by the cosine of the angle between H and H 0 . This cosine is
a•a 0
1)-dimensional cube, we have by hypothequal to |a|·|a
0 | . Since C n \ Ht is an (n
esis Voln 2 (Cn \ H \ Ht ) =
|a 0 |
· Fn 1 (b
0
|a|
Vol(Cn \ H ) = 0 ·
|a |
Z
an t). Therefore, we get
1
1
Voln 2 (Cn \ H \ Ht ) dt
Z 1 0
|a |
|a|
· Fn 1 (b an t) dt
= 0 ·
0
|a |
1
Z an
|a|
1
= 0 ·
Fn 1 (b t) ·
dt
an
an
Z 1
|a|
|a|
·
· Fn (b).
Fn 1 (b t) · An (t) dt =
=
1
Lemma 2. Using our previous notation, it follows that
Fn (b) =
for all n
1
2(n
1)!
·
X
w2{ 1,1}n
(a • w + b)
n 1
sgn(a • w + b) ·
2 and b 2 R.
Proof. We proceed by induction on n and define µ0 (w) =
Qn 1
2 we have that
k=1 wk . First, for all n
870
c
Pn
1
k=1
n
Y
wk
k=1
ak wk and ↵ 0 (w) =
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[Monthly 119
Fn (b) =
Z
1
1
t) · An (t) dt =
Fn 1 (b
Z
an
Fn 1 (b
t) dt.
an
For the next transformation, we use in the case n = 2 that
F1 (t) = A1 (t) =
1
(sgn(a1 + t)
2
t 2 R \ {a1 , a1 };
sgn( a1 + t)) ,
in the case n > 2, we use the induction hypothesis and get in both cases
Z an
Fn 1 (b t) dt
an
=
Z
=

=
an
an
1
2(n
2)!
1
2(n
1
2(n
1)!
·
X
w2{ 1,1}n 1
1)!
·
·
X
X
w2{ 1,1}n 1
w2{ 1,1}n
(µ0 (w) + b
t)n
2
sgn(µ0 (w) + b
t) · ↵ 0 (w) dt
an
0
(µ (w) + b
(a • w + b)
n 1
t)
n 1
0
sgn(µ (w) + b
sgn(a • w + b) ·
n
Y
0
t) · ↵ (w)
an
wk .
k=1
Now we obtain formula (2), if in Lemma 1 we replace Fn (b) by the expression
calculated in Lemma 2.
3. APPLICATIONS OF THEOREM 2. We can check formula (2) geometrically
by finding the vertices of Cn \ H as intersection points of H with the edges of Cn , and
calculating the volume of the convex hull of these vertices. We give two examples in
dimensions 4 and 5 and invite the reader to produce their own in dimensions 2 and 3.
For H = {x 2 Rn | x1 + · · · + xn = b}, formula (2) simplifies to
p
✓ ◆
n
X
n
i n
·
(n
( 1)
Vol(Cn \ H ) =
2(n 1)! i=0
i
2i + b)n
1
sgn(n
2i + b) .
(4)
Example 1. Let the hyperplane H in R4 be given by x1 + x2 + x3 + x4 = 2. Then (4)
yields
p
4
(4 + 2)3 4(2 + 2)3 + 6(0 + 2)3 + 4( 2 + 2)3 ( 4 + 2)3 )
Vol(C4 \ H ) =
12
8
= .
3
Now, C4 \ H has as vertices thep4 permutations of (1, 1, 1, 1), and is thus a regular tetrahedron with edge length 2 2.
Example 2. Let the hyperplane H in R5 be given by x1 + x2 + x3 + x4 + x5 = 0.
Then formula (4) yields
p
5 4
115 p
Vol(C5 \ H ) =
5
5 · 34 + 10 · 14 + 10 · 14 5 · 34 + 54 =
5.
48
12
December 2012]
NOTES
871
Now, C5 \ H has as vertices the 30 permutations of (1, 1, 0, 1, 1). Here the
geometric calculation of Vol(C5 \ H ) is more difficult. In fact, C5 \ H has 10 congruent facets in the 10 hyperplanes xi = ±1 (i = 1, . . . , 5). The facet with x1 = 1
has as vertices the 12 permutations of (1, 1, 0, 1) (with the x1 -component omitted)
and the center (of gravity of the vertices) ( 1, 14 , 14 , 14 , 14 ). It is bounded by 4 regular sixgons with equations xi = 1 (i = 2, . . . , 5)
p and 4 regular triangles with equations xi = 1 (i = 2, . . . , 5) with
p side length 2. Hence, each facet is a truncated
. So we can dissect C5 \ H into
tetrahedron with edge length 2 and volume 23
3
p
and height 12 5 to get
10 congruent 4-dimensional pyramids with base volume 23
3
p
p
1
115
·
5
=
5.
Vol(C5 \ H ) = 10 · 14 · 23
3
2
12
4. OPEN QUESTIONS.
Question 1. Ball’s proof of Hensley’s conjecture Vol(Cn \ H )  2n
complicated. Is there a simple proof using Theorem 2?
1
p
2 is quite
Question 2. Ball generalized Theorem 1 to subspaces Hk of arbitrary dimension k <
p n k
n and used this generalization to prove Volk (Cn \ Hk )  2k 2 . This bound is the
best possible if 2k n, but, for example, the maximal area of C5 \ H2 is unknown
(see [2] or [7]). Can we generalize Theorem 2 and obtain an elementary formula for
Volk (Cn \ Hk )?
REFERENCES
1. K. Ball, Cube slicing in Rn , Proc. of the AMS 97 no. 3 (1986) 465–473.
2.
, Volumes of sections of cubes and related problems, Lecture Notes in Math. 1376 (1989) 251–260,
available at http://dx.doi.org/10.1007/BFb0090058.
3. D.L. Barrow, P.W. Smith, Spline notation applied to a volume problem, Amer. Math. Monthly 86 (1979)
50–51, available at http://dx.doi.org/10.2307/2320304.
4. M. Berger, Geometry Revealed, Springer-Verlag, Berlin, 2010.
5. D. Evers, Hyperebenenschnitte des vierdimensionalen Würfels, Wissenschaftl. Prüfungsarbeit, Universität
Koblenz-Landau, Koblenz, 2010.
R 1 ⇣Qn sin(ak x) ⌘
· cos(bx) d x (to appear), available
6. R. Frank, H. Riede, Die Berechnung des Integrals 1
k=1 ak x
at http://www.uni-koblenz-landau.de/koblenz/fb3/mathe/Forschung/wuerfelintegral.
pdf.
7. C. Zong, The Cube: A Window to Convex and Discrete Geometry, Cambridge Univ. Press, Cambridge,
2006.
Mathematisches Institut der Universität Koblenz-Landau, Campus Koblenz, D 56070 Koblenz, Germany
[email protected]
[email protected]
Tiling Hamiltonian Cycles on the 24-Cell
Jacob A. Siehler
Abstract. We present a construction for tiling the 24-cell with congruent copies of a single
Hamiltonian cycle, using the algebra of quaternions.
http://dx.doi.org/10.4169/amer.math.monthly.119.10.872
MSC: Primary 00A08, Secondary 05C45; 51M20
872
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[Monthly 119