Higher Dimensional Conundra
Steven G. Krantz1
Abstract: In recent years, especially in the subject of harmonic
analysis, there has been interest in geometric phenomena of RN
as N → +∞. In the present paper we examine several spe-
cific geometric phenomena in Euclidean space and calculate the
asymptotics as the dimension gets large.
0 Introduction
Typically when we do geometry we concentrate on a specific venue in a par-
ticular space. Often the context is Euclidean space, and often the work is
done in R2 or R3. But in modern work there are many aspects of analysis
that are linked to concrete aspects of geometry. And there is often interest
in rendering the ideas in Hilbert space or some other infinite dimensional set-
ting. Thus one wants to see how the finite-dimensional result in RN changes
as N → +∞.
In the present paper we study some particular aspects of the geometry of
N and their asymptotic behavior as N → ∞. We choose these particular
examples because the results are surprising or especially interesting. One
may hope that they will lead to further studies.
1 Volume in RN
Let us begin by calculating the volume of the unit ball in RN and the surface
area of its bounding unit sphere. We let ΩN denote the former and ωN−1
denote the latter. In addition, we let Γ(x) be the celebrated Gamma function
of L. Euler. It is a helpful intuition (which is literally true when x is an
integer) that Γ(x) ≈ (x− 1)!. We shall also use Stirling’s formula which says
k! ≈ kk · e−k ·
1We are happy to thank the American Institute of Mathematics for its hospitality and
support during this work.
http://arxiv.org/abs/0704.1041v1
or, more generally,
Γ(x) ≈ (x− 1)x−1e−(x−1)
2π(x− 1)
for x ∈ R, x > 0.
Lemma 1 We have that
e−π‖x‖
dx = 1.
Proof: The case N = 1 is familiar from calculus. We write
hence
e−π‖x‖
(polar coordinates)
‖x‖=1
r ds(x)dr
hence S = 1.
For the N−dimensional case, write
e−π|x|
1dx1 · · ·
and apply the one-dimensional result.
Let σ be the unique rotationally invariant area measure on SN−1 = ∂BN .
Lemma 2 We have
ωN−1 =
2πN/2
Γ(N/2)
Proof: Introducing polar coordinates we have
e−π|x|
rN−1dr
Letting s = r2 in this last integral and doing some obvious manipulations
yields the result.
Corollary 3 The volume of the unit ball in RN is
2πN/2
Γ(N/2) ·N
Proof: We calculate that
1 dV (x)
(polar coordinates)
‖x‖=1
1·rN−1 dσ(x)dr = ωN−1·
That completes the proof.
Now the first nontrivial fact that we wish to observe about the volume of
the Euclidean unit ball in N -space is that that volume tends to 0 at N → ∞.
More formally,
Proposition 4 We have the limit
Ω(N) = 0 .
Proof: We calculate that
(Volume of Unit Ball) =
2πN/2
Γ(N/2) ·N
2πN/2
((N − 2)/2)(N−2)/2e−(N−2)/2
2π[(N − 2)/2] ·N
(2πe)N/2 · 2
N (N−1)/2 ·
(2πe)N/2 · 2
N (N+1)/2 ·
This expression clearly tends to 0 as N → +∞.
In fact we can actually say something about the rate at which the volume
of the ball tends to zero. We have
Proposition 5 We have the estimate
0 ≤ ΩN ≤ 2 ·
20N/2
N (N+1)/2
Proof: Follows by inspection of the last line of the proof of Proposition 4.
In fact something more is true about the volumes of balls in high-dimensional
Euclidean space.
Proposition 6 Let R > 0 be fixed. Then
Vol(B(0, R)) = 0 .
In other words, the volume of the ball of radius R tends to 0.
Proof: From the formula for the volume of the unit ball we have that
Vol(B(0, R)) = lim
2πeR2
This expression clearly tends to 0 as N → +∞.
We leave the proof of the next result as an exercise for the reader; simply
examine the formula for ωN−1:
Proposition 7 Let R > 0. Then the surface area of the sphere of radius R
in RN tends to 0 as N → +∞.
The following very simple but remarkable fact comes up in considerations
of spherical summation of Fourier series.
Proposition 8 As N → +∞, the volume of the unit ball in RN is con-
centrated more and more out near the boundary sphere. More precisely, let
δ > 0. Then
volume(B(0, 1) \B(0, 1− δ))
volume(B(0, 1)
= 1 .
Proof: We have
volume(B(0, 1) \B(0, 1− δ))
volume(B(0, 1)
= lim
[1− (1− δ)N ] · [2πN/2]/[Γ(N/2) ·N ]
[2πN/2]/[Γ(N/2) ·N ]
= lim
1− (1− δ)N
= 1 .
That is the desired conclusion.
2 A Case of Leakage
The title of this section gives away the punchline of the example. Or so it
may seem to some.
Consider at first a square box of side two with sides parallel to the coor-
dinate axes in the Euclidean plane. We may inscribe in this box four discs of
diameter 1, as shown in Figure 1. These discs will be called primary discs.
Once those four discs are inscribed, we may inscribe a small, shaded disc in
the middle as shown in Figure 2. We set
area of shaded disc
area of large box
The same construction may be performed in Euclidean dimension 3. Ex-
amine Figure 3. It suggests a rectangular parallelepiped with all sides equal
Figure 1: The configuration in dimension 2.
Figure 2: The shaded disc in dimension 2.
Figure 3: The configuration in dimension 3.
to 2, and 8 unit balls inscribed inside in a canonical fashion. These eight
primary balls determine a unique inscribed shaded ball in the center. We set
volume of shaded ball
volume of large box
A similar construction may be performed in any dimension N ≥ 2, with
2N balls inscribed in a rectangular box of side 2. The ratio RN is then
calculated in just the same way. The question is then
What is the limit limN→∞RN as N → +∞?
It is natural to suppose, and most people do suppose, and that this limit
(assuming it exists) is between 0 and 1. All other things being equal, it is
likely equal to either 0 or 1. Thus it comes as something of a surprise that
this limit is in fact equal to +∞. Let us now enunciate this result and prove
Proposition 9 The limit
RN = +∞ .
Figure 4: The disc trapped in dimension 2.
Of course this result is counter-intuitive, because we all instinctively be-
lieve that the shaded ball, in any dimension, is contained inside the big box.
Such is not the case. We are being fooled by the 2-dimensional situation de-
picted in Figure 1. In that special situation, any of the two adjacent primary
discs actually touch in such a way as to trap the shaded disc in a particular
convex subregion of the big box (see Figure 4). So certainly it must be that
R2 < 1. But such is not the case in higher dimensions. There is actually a
gap on each side of the box through which the shaded ball can leak. And
indeed it does.
This is what we shall now show. First we shall perform the calculation of
RN for each N and confirm that the expression tends to +∞ as N → +∞.
Then we shall calculate the first dimension in which the shaded ball actually
leaks out of the box.
Proof of the Proposition: Notice that the center of one of the primary
balls is at the point (1, 1, . . . , 1). It is a simple matter to calculate that a
boundary point of this ball that is nearest to the center of the box is located
at P ∗ ≡ (1 − 1/
N, 1 − 1/
N, . . . , 1 − 1/
N . Since the shaded ball will
osculate the primary ball at that point, we see that the shaded ball has center
the origin and radius equal to
dist(0, P ∗) =
(1− 1/
N)2, 1− 1/
N)2, . . . , 1− 1/
N)2 =
N + 1− 2
Thus we see that the volume of the shaded ball is
[N + 1− 2
N ]N/2 · ΩN .
The ratio RN is then
[N + 1− 2
N ]N/2 · ΩN
Now we may simplify this last expression to
2 · πN/2
Γ(N/2) ·N
· [N + 1− 2
N ]N/2
After some simplification we find that
2(π/4)N/2 · [N + 1− 2
N ]N/2
Γ(N/2) ·N
By Stirling’s formula, this last expression is approximately equal to
2 · (π/4)N/2(N + 1− 2
N)N/2
N − 2
)(2−N)/2
· e(N−2)/2 · 1√
π(N − 2)
(N + 1− 2
N) · 2
N − 2
· N − 2
π(N − 2)
After some manipulation, we finally find that
RN = lim
N + 1
N − 2
· N − 2
π(N − 2)
= lim
)N/2 (
N − 2
· N − 2
π(N − 2)
Now, in the limit, we may replace expressions like N − 2 by N . And we may
reparametrize N as 3N . The result is
N − 2
π(N − 2)
= lim
)3N/2
N − 2
π(N − 2)
= lim
N − 2
π(N − 2)
What we see now is that this last equals
Plainly, because πe/2 > 4, this limit is +∞. That proves the result.
And now we turn to the question of when the shaded ball starts to leak
out of the big box. This is in fact easy to analyze. We need only determine
when the radius of the shaded ball exceeds 1. First notice that the radius of
the shaded ball is monotone increasing in N . Now we need to solve
N + 1− 2
N > 1 .
This is a simple algebra problem, and the solution is N > 4. Thus, beginning
in dimension 5, the shaded ball will “leak out of” the large box.
It may be noted that RichardW. Cottle has made a study of mathematical
phenomena that change (in the manner of a catastrophe—see [ZEE]) between
dimensions 4 and dimensions 5. The results may be found in [COT].
3 Centroids
This final section of the paper will be more like an invitation to further
exploration. We cannot include all the details of the calculations, as they
are too recondite and complex. Yet the topic is very much in the spirit of
the theme of this paper, and we cannot resist including a few pointers to this
new and interesting work (for which see [KRA1] and [KRMP]).
Figure 5: Centroids for a triangle.
The inspiration for this work is the following somewhat surprising obser-
vation. Let T be a triangle in the plane (see Figure 5). There are three ways
to calculate the centroid of this figure: (i) average the vertices, (ii) average
the edges, or (iii) average the 2-dimensional solid figure. And the question
is: are these three versions of the centroid the same? The answer is that
(i) and (iii) are always the same. Generically (ii) is different. In fact the
three versions of the centroid coincide if and only if the triangle is equilateral
[KRMP].
We used this fact as a springing-off point to investigate analogous ques-
tions in higher dimensions. Consider the simplex S in RN that is the con-
vex hull of the points 0 = (0, 0, . . . , 0), (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . ,
(0, 0, . . . , 0, 1). Refer to Figure 6. Such an N -dimensional geometric figure
comes equipped with (N+1) notions of centroid: one can average the vertices
(or 1-dimensional skeleton) S0, or one can average over the 1-dimensional
skeleton S1, or one can average over the two-dimensional skeleton S2A, or
. . . one can average over the (N − 1)-dimensional skeleton SN−1, or one can
average over the N -dimensional solid SN . There results the centroids C0,N ,
C1,N , . . . , CN,N . And the question is: Are these different notions of centroid
all the same? And here is the somewhat surprising answer:
In dimensions 2 through 12 (for the ambient space), the skeletons
S0 and SN for the simplex S have the same centroid. In those
Figure 6: A simplex in RN .
same dimensions, the skeletons S1, S2, . . . , SN−1 all have different
centroids, and the centroids all differ from the common centroid
for S0 and SN . But in dimension 13 things are different. In fact
in that dimension the skeletons S3 and S8 have the same centroid.
Let us say a word about why these facts are true. Let ej denote the j
coordinate vector in RN (i.e., the vector with a 1 in the jth position and
0s in all other slots). Then a sophisticated computation with elementary
calculus yields that the centroid of the k-skeleton Sk of the simplex which is
the convex hull of 0, e1, e2, . . . , eN is
Ck,N =
k + (N − k)
k + 1
(k + 1) + (N − k)
k + 1
(e1 + e2 + · · ·+ eN) .
From this formula it can immediately be verified that
S0 = SN =
N + 1
(e1 + e2 + · · ·+ eN ) .
It can also be checked that, in dimensions 2 through 12, all the intermediate
skeletons have distinct centroids. But, in dimension N = 13, we observe that
C3,13 = C8,13 =
13 · 24
(e1 + e2 + · · ·+ eN ) .
One may well ask whether dimension N = 13 is the only dimension in
which there are two intermediate skeletons with the same centroid. The
answer is “no”; there are in fact infinitely many such dimensions (although
they are quite sparse—sparser than the prime integers). One may verify this
assertion by using the following Diophantine formula.
Theorem 10 Fix a dimension N ≥ 2. Consider the simplex S as described
above. There are skeletons of dimension k1 and k2, 1 ≤ k1 < k2 ≤ N − 1,
of the simplex S which have the same centroid if and only if k1 = a
2 − 1,
k2 = b
2 − 1 (for positive integers a and b) and, in addition,
N = (b2 + ab+ a2)− (b+ a)− 1 . (⋆)
Obviously this theorem gives us a tool for finding dimensions in which
the simplex S has two intermediate skeletons with the same centroid. The
following table gives some values of the dimension, and of the intermediate
dimensions of skeletons which have the same centroid. Of course this data
may be confirmed by direct calculation with the formula (⋆). We stress
that there are in fact infinitely many dimensions in which this phenomenon
occurs. The proof of this statement is a nontrivial exercise in elementary
number theory (see [KRMP]).
value of N value of k1 value of k2 approx. coord. of centroid
13 3 8 0.0737179487
21 3 15 0.0464285714
29 8 15 0.0340038314
31 3 24 0.0317204301
40 8 24 0.0247619047
43 3 35 0.0229789590
51 15 24 0.0194852941
53 8 35 0.0187368973
57 3 48 0.0173872180
65 15 35 0.0153133903
We conclude this discussion by recording the fact that it is impossible in
any dimension for there to be three intermediate skeletons with the same
centroid.
Proposition 11 For no dimension N can there exists 3 distinct number
1 ≤ k1 < k2 < k3 ≤ N − 1 such that the centroids Ck1,N , Ck2,N , Ck3,N for the
simplex S coincide.
Proof: We let
Q(a, b) = (b2 + ab+ a2)− (b+ a)− 1 .
It suffices for us to show that there do not exist natural numbers a < b < c
such that Q(a, b) = Q(a, c). Seeking a contradiction, we suppose that such a
triple does indeed exist.
b2 + ab− b = c2 + ac− c
b2 + (a− 1)b = c2 + (a− 1)c .
Since a ≥ 1, the function b 7→ b2+(a−1)b is strictly increasing, which yields
a contradiction.
The exploration of centroids for simplices of high dimension is a new venue
of exploration. There are many new phenomena, and more to be discovered.
See [KRMP] for more results along these lines. The reference [ZON] is also
of interest.
References
[COT] R. W. Cottle, Quartic barriers, Computational Optimization and Ap-
plications 12(1999), 81–105.
[KRA1] S. G. Krantz, A Matter of gravity, Amer. Math. Monthly 110(2003),
465–481.
[KRMP] S. G. Krantz, J. E. McCarthy, and H. R. Parks, Geometric characteri-
zations of centroids of simplices, Journal of Mathematical Analysis and
Applications 316(2006), 87–109.
[ZEE] E. C. Zeeman, Catastrophe Theory. Selected Papers, 1972–1977, Addison-
Wesley, Reading, MA, 1977.
[ZON] C. Zong, Strange Phenomena in Convex and Discrete Geometry, Springer-
Verlag, New York, 1996.
STEVEN G. KRANTZ received his B.A. degree from the University of
California at Santa Cruz in 1971. He earned the Ph.D. from Princeton Uni-
versity in 1974. He has taught at UCLA, Princeton University, Penn State,
and Washington University in St. Louis. Krantz is the holder of the UCLA
Alumni Foundation Distinguished Teaching Award, the Chauvenet Prize,
and the Beckenbach Book Prize. He is the author of 150 papers and 50
books. His research interests include complex analysis, real analysis, har-
monic analysis, and partial differential equations. Krantz is currently the
Deputy Director of the American Institute of Mathematics.
American Institute of Mathematics, 360 Portage Avenue,
Palo Alto, CA 94306
skrantz@aimath.org
	Introduction
	Volume in RN
	A Case of Leakage
	Centroids

  We study asymptotics of various Euclidean geometric phenomena as the
dimension tend to infinity.

Introduction
Typically when we do geometry we concentrate on a specific venue in a par-
ticular space. Often the context is Euclidean space, and often the work is
done in R2 or R3. But in modern work there are many aspects of analysis
that are linked to concrete aspects of geometry. And there is often interest
in rendering the ideas in Hilbert space or some other infinite dimensional set-
ting. Thus one wants to see how the finite-dimensional result in RN changes
as N → +∞.
In the present paper we study some particular aspects of the geometry of
N and their asymptotic behavior as N → ∞. We choose these particular
examples because the results are surprising or especially interesting. One
may hope that they will lead to further studies.
1 Volume in RN
Let us begin by calculating the volume of the unit ball in RN and the surface
area of its bounding unit sphere. We let ΩN denote the former and ωN−1
denote the latter. In addition, we let Γ(x) be the celebrated Gamma function
of L. Euler. It is a helpful intuition (which is literally true when x is an
integer) that Γ(x) ≈ (x− 1)!. We shall also use Stirling’s formula which says
k! ≈ kk · e−k ·
1We are happy to thank the American Institute of Mathematics for its hospitality and
support during this work.
http://arxiv.org/abs/0704.1041v1
or, more generally,
Γ(x) ≈ (x− 1)x−1e−(x−1)
2π(x− 1)
for x ∈ R, x > 0.
Lemma 1 We have that
e−π‖x‖
dx = 1.
Proof: The case N = 1 is familiar from calculus. We write
hence
e−π‖x‖
(polar coordinates)
‖x‖=1
r ds(x)dr
hence S = 1.
For the N−dimensional case, write
e−π|x|
1dx1 · · ·
and apply the one-dimensional result.
Let σ be the unique rotationally invariant area measure on SN−1 = ∂BN .
Lemma 2 We have
ωN−1 =
2πN/2
Γ(N/2)
Proof: Introducing polar coordinates we have
e−π|x|
rN−1dr
Letting s = r2 in this last integral and doing some obvious manipulations
yields the result.
Corollary 3 The volume of the unit ball in RN is
2πN/2
Γ(N/2) ·N
Proof: We calculate that
1 dV (x)
(polar coordinates)
‖x‖=1
1·rN−1 dσ(x)dr = ωN−1·
That completes the proof.
Now the first nontrivial fact that we wish to observe about the volume of
the Euclidean unit ball in N -space is that that volume tends to 0 at N → ∞.
More formally,
Proposition 4 We have the limit
Ω(N) = 0 .
Proof: We calculate that
(Volume of Unit Ball) =
2πN/2
Γ(N/2) ·N
2πN/2
((N − 2)/2)(N−2)/2e−(N−2)/2
2π[(N − 2)/2] ·N
(2πe)N/2 · 2
N (N−1)/2 ·
(2πe)N/2 · 2
N (N+1)/2 ·
This expression clearly tends to 0 as N → +∞.
In fact we can actually say something about the rate at which the volume
of the ball tends to zero. We have
Proposition 5 We have the estimate
0 ≤ ΩN ≤ 2 ·
20N/2
N (N+1)/2
Proof: Follows by inspection of the last line of the proof of Proposition 4.
In fact something more is true about the volumes of balls in high-dimensional
Euclidean space.
Proposition 6 Let R > 0 be fixed. Then
Vol(B(0, R)) = 0 .
In other words, the volume of the ball of radius R tends to 0.
Proof: From the formula for the volume of the unit ball we have that
Vol(B(0, R)) = lim
2πeR2
This expression clearly tends to 0 as N → +∞.
We leave the proof of the next result as an exercise for the reader; simply
examine the formula for ωN−1:
Proposition 7 Let R > 0. Then the surface area of the sphere of radius R
in RN tends to 0 as N → +∞.
The following very simple but remarkable fact comes up in considerations
of spherical summation of Fourier series.
Proposition 8 As N → +∞, the volume of the unit ball in RN is con-
centrated more and more out near the boundary sphere. More precisely, let
δ > 0. Then
volume(B(0, 1) \B(0, 1− δ))
volume(B(0, 1)
= 1 .
Proof: We have
volume(B(0, 1) \B(0, 1− δ))
volume(B(0, 1)
= lim
[1− (1− δ)N ] · [2πN/2]/[Γ(N/2) ·N ]
[2πN/2]/[Γ(N/2) ·N ]
= lim
1− (1− δ)N
= 1 .
That is the desired conclusion.
2 A Case of Leakage
The title of this section gives away the punchline of the example. Or so it
may seem to some.
Consider at first a square box of side two with sides parallel to the coor-
dinate axes in the Euclidean plane. We may inscribe in this box four discs of
diameter 1, as shown in Figure 1. These discs will be called primary discs.
Once those four discs are inscribed, we may inscribe a small, shaded disc in
the middle as shown in Figure 2. We set
area of shaded disc
area of large box
The same construction may be performed in Euclidean dimension 3. Ex-
amine Figure 3. It suggests a rectangular parallelepiped with all sides equal
Figure 1: The configuration in dimension 2.
Figure 2: The shaded disc in dimension 2.
Figure 3: The configuration in dimension 3.
to 2, and 8 unit balls inscribed inside in a canonical fashion. These eight
primary balls determine a unique inscribed shaded ball in the center. We set
volume of shaded ball
volume of large box
A similar construction may be performed in any dimension N ≥ 2, with
2N balls inscribed in a rectangular box of side 2. The ratio RN is then
calculated in just the same way. The question is then
What is the limit limN→∞RN as N → +∞?
It is natural to suppose, and most people do suppose, and that this limit
(assuming it exists) is between 0 and 1. All other things being equal, it is
likely equal to either 0 or 1. Thus it comes as something of a surprise that
this limit is in fact equal to +∞. Let us now enunciate this result and prove
Proposition 9 The limit
RN = +∞ .
Figure 4: The disc trapped in dimension 2.
Of course this result is counter-intuitive, because we all instinctively be-
lieve that the shaded ball, in any dimension, is contained inside the big box.
Such is not the case. We are being fooled by the 2-dimensional situation de-
picted in Figure 1. In that special situation, any of the two adjacent primary
discs actually touch in such a way as to trap the shaded disc in a particular
convex subregion of the big box (see Figure 4). So certainly it must be that
R2 < 1. But such is not the case in higher dimensions. There is actually a
gap on each side of the box through which the shaded ball can leak. And
indeed it does.
This is what we shall now show. First we shall perform the calculation of
RN for each N and confirm that the expression tends to +∞ as N → +∞.
Then we shall calculate the first dimension in which the shaded ball actually
leaks out of the box.
Proof of the Proposition: Notice that the center of one of the primary
balls is at the point (1, 1, . . . , 1). It is a simple matter to calculate that a
boundary point of this ball that is nearest to the center of the box is located
at P ∗ ≡ (1 − 1/
N, 1 − 1/
N, . . . , 1 − 1/
N . Since the shaded ball will
osculate the primary ball at that point, we see that the shaded ball has center
the origin and radius equal to
dist(0, P ∗) =
(1− 1/
N)2, 1− 1/
N)2, . . . , 1− 1/
N)2 =
N + 1− 2
Thus we see that the volume of the shaded ball is
[N + 1− 2
N ]N/2 · ΩN .
The ratio RN is then
[N + 1− 2
N ]N/2 · ΩN
Now we may simplify this last expression to
2 · πN/2
Γ(N/2) ·N
· [N + 1− 2
N ]N/2
After some simplification we find that
2(π/4)N/2 · [N + 1− 2
N ]N/2
Γ(N/2) ·N
By Stirling’s formula, this last expression is approximately equal to
2 · (π/4)N/2(N + 1− 2
N)N/2
N − 2
)(2−N)/2
· e(N−2)/2 · 1√
π(N − 2)
(N + 1− 2
N) · 2
N − 2
· N − 2
π(N − 2)
After some manipulation, we finally find that
RN = lim
N + 1
N − 2
· N − 2
π(N − 2)
= lim
)N/2 (
N − 2
· N − 2
π(N − 2)
Now, in the limit, we may replace expressions like N − 2 by N . And we may
reparametrize N as 3N . The result is
N − 2
π(N − 2)
= lim
)3N/2
N − 2
π(N − 2)
= lim
N − 2
π(N − 2)
What we see now is that this last equals
Plainly, because πe/2 > 4, this limit is +∞. That proves the result.
And now we turn to the question of when the shaded ball starts to leak
out of the big box. This is in fact easy to analyze. We need only determine
when the radius of the shaded ball exceeds 1. First notice that the radius of
the shaded ball is monotone increasing in N . Now we need to solve
N + 1− 2
N > 1 .
This is a simple algebra problem, and the solution is N > 4. Thus, beginning
in dimension 5, the shaded ball will “leak out of” the large box.
It may be noted that RichardW. Cottle has made a study of mathematical
phenomena that change (in the manner of a catastrophe—see [ZEE]) between
dimensions 4 and dimensions 5. The results may be found in [COT].
3 Centroids
This final section of the paper will be more like an invitation to further
exploration. We cannot include all the details of the calculations, as they
are too recondite and complex. Yet the topic is very much in the spirit of
the theme of this paper, and we cannot resist including a few pointers to this
new and interesting work (for which see [KRA1] and [KRMP]).
Figure 5: Centroids for a triangle.
The inspiration for this work is the following somewhat surprising obser-
vation. Let T be a triangle in the plane (see Figure 5). There are three ways
to calculate the centroid of this figure: (i) average the vertices, (ii) average
the edges, or (iii) average the 2-dimensional solid figure. And the question
is: are these three versions of the centroid the same? The answer is that
(i) and (iii) are always the same. Generically (ii) is different. In fact the
three versions of the centroid coincide if and only if the triangle is equilateral
[KRMP].
We used this fact as a springing-off point to investigate analogous ques-
tions in higher dimensions. Consider the simplex S in RN that is the con-
vex hull of the points 0 = (0, 0, . . . , 0), (1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . ,
(0, 0, . . . , 0, 1). Refer to Figure 6. Such an N -dimensional geometric figure
comes equipped with (N+1) notions of centroid: one can average the vertices
(or 1-dimensional skeleton) S0, or one can average over the 1-dimensional
skeleton S1, or one can average over the two-dimensional skeleton S2A, or
. . . one can average over the (N − 1)-dimensional skeleton SN−1, or one can
average over the N -dimensional solid SN . There results the centroids C0,N ,
C1,N , . . . , CN,N . And the question is: Are these different notions of centroid
all the same? And here is the somewhat surprising answer:
In dimensions 2 through 12 (for the ambient space), the skeletons
S0 and SN for the simplex S have the same centroid. In those
Figure 6: A simplex in RN .
same dimensions, the skeletons S1, S2, . . . , SN−1 all have different
centroids, and the centroids all differ from the common centroid
for S0 and SN . But in dimension 13 things are different. In fact
in that dimension the skeletons S3 and S8 have the same centroid.
Let us say a word about why these facts are true. Let ej denote the j
coordinate vector in RN (i.e., the vector with a 1 in the jth position and
0s in all other slots). Then a sophisticated computation with elementary
calculus yields that the centroid of the k-skeleton Sk of the simplex which is
the convex hull of 0, e1, e2, . . . , eN is
Ck,N =
k + (N − k)
k + 1
(k + 1) + (N − k)
k + 1
(e1 + e2 + · · ·+ eN) .
From this formula it can immediately be verified that
S0 = SN =
N + 1
(e1 + e2 + · · ·+ eN ) .
It can also be checked that, in dimensions 2 through 12, all the intermediate
skeletons have distinct centroids. But, in dimension N = 13, we observe that
C3,13 = C8,13 =
13 · 24
(e1 + e2 + · · ·+ eN ) .
One may well ask whether dimension N = 13 is the only dimension in
which there are two intermediate skeletons with the same centroid. The
answer is “no”; there are in fact infinitely many such dimensions (although
they are quite sparse—sparser than the prime integers). One may verify this
assertion by using the following Diophantine formula.
Theorem 10 Fix a dimension N ≥ 2. Consider the simplex S as described
above. There are skeletons of dimension k1 and k2, 1 ≤ k1 < k2 ≤ N − 1,
of the simplex S which have the same centroid if and only if k1 = a
2 − 1,
k2 = b
2 − 1 (for positive integers a and b) and, in addition,
N = (b2 + ab+ a2)− (b+ a)− 1 . (⋆)
Obviously this theorem gives us a tool for finding dimensions in which
the simplex S has two intermediate skeletons with the same centroid. The
following table gives some values of the dimension, and of the intermediate
dimensions of skeletons which have the same centroid. Of course this data
may be confirmed by direct calculation with the formula (⋆). We stress
that there are in fact infinitely many dimensions in which this phenomenon
occurs. The proof of this statement is a nontrivial exercise in elementary
number theory (see [KRMP]).
value of N value of k1 value of k2 approx. coord. of centroid
13 3 8 0.0737179487
21 3 15 0.0464285714
29 8 15 0.0340038314
31 3 24 0.0317204301
40 8 24 0.0247619047
43 3 35 0.0229789590
51 15 24 0.0194852941
53 8 35 0.0187368973
57 3 48 0.0173872180
65 15 35 0.0153133903
We conclude this discussion by recording the fact that it is impossible in
any dimension for there to be three intermediate skeletons with the same
centroid.
Proposition 11 For no dimension N can there exists 3 distinct number
1 ≤ k1 < k2 < k3 ≤ N − 1 such that the centroids Ck1,N , Ck2,N , Ck3,N for the
simplex S coincide.
Proof: We let
Q(a, b) = (b2 + ab+ a2)− (b+ a)− 1 .
It suffices for us to show that there do not exist natural numbers a < b < c
such that Q(a, b) = Q(a, c). Seeking a contradiction, we suppose that such a
triple does indeed exist.
b2 + ab− b = c2 + ac− c
b2 + (a− 1)b = c2 + (a− 1)c .
Since a ≥ 1, the function b 7→ b2+(a−1)b is strictly increasing, which yields
a contradiction.
The exploration of centroids for simplices of high dimension is a new venue
of exploration. There are many new phenomena, and more to be discovered.
See [KRMP] for more results along these lines. The reference [ZON] is also
of interest.
References
[COT] R. W. Cottle, Quartic barriers, Computational Optimization and Ap-
plications 12(1999), 81–105.
[KRA1] S. G. Krantz, A Matter of gravity, Amer. Math. Monthly 110(2003),
465–481.
[KRMP] S. G. Krantz, J. E. McCarthy, and H. R. Parks, Geometric characteri-
zations of centroids of simplices, Journal of Mathematical Analysis and
Applications 316(2006), 87–109.
[ZEE] E. C. Zeeman, Catastrophe Theory. Selected Papers, 1972–1977, Addison-
Wesley, Reading, MA, 1977.
[ZON] C. Zong, Strange Phenomena in Convex and Discrete Geometry, Springer-
Verlag, New York, 1996.
STEVEN G. KRANTZ received his B.A. degree from the University of
California at Santa Cruz in 1971. He earned the Ph.D. from Princeton Uni-
versity in 1974. He has taught at UCLA, Princeton University, Penn State,
and Washington University in St. Louis. Krantz is the holder of the UCLA
Alumni Foundation Distinguished Teaching Award, the Chauvenet Prize,
and the Beckenbach Book Prize. He is the author of 150 papers and 50
books. His research interests include complex analysis, real analysis, har-
monic analysis, and partial differential equations. Krantz is currently the
Deputy Director of the American Institute of Mathematics.
American Institute of Mathematics, 360 Portage Avenue,
Palo Alto, CA 94306
skrantz@aimath.org
	Introduction
	Volume in RN
	A Case of Leakage
	Centroids
