Edgewise Subdivision of a Simplex
Discrete Comput Geom
Geometr Y
Edgewise Subdivision of a Simplex 0
H. Edelsbrunner! 0
D. R. Grayson? 0
0 2Department of Mathematics, University of Illinois at UrbanaChampaign , Urbana, IL 61801 , USA
1 Department of Computer Science, Duke University , Durham, NC 27708, USA and Raindrop Geomagic , Research , Triangle Park, NC 27709 , USA
In this paper we introduce the abacus model of a simplex and use it to subdivide a dsimplex into kd dsimplices all of the same volume and shape characteristics. The construction is an extension of the subdivision method of Freudenthal [3] and has been used by Goodman and Peters [4] to design smooth manifolds. * Research by both authors was partially supported by NSF under Grant DMS 9873945. The research of the first author was also supported by NSF under Grant CCR9619542, and by ARO under Grant DAAG559810177.

1. Introduction
It is easy to subdivide a triangle into four similar triangles all of the same area: cut
each edge into halves and connect the three dividing points. Since the triangles are
similar we can repeat the operation and refine while preserving the triangle shape.
Such a construction does not exist in general for tetrahedra and higherdimensional
simplices.
Result. A subdivision of a dsimplex is a decomposition into dsimplices such that each
pair is either disjoint or meets along a common face. In other words, the dsimplices
and their faces form a simplicial complex. Such complexes are used in engineering and
science to represent geometric shapes and domains for the purpose of design, analysis,
simulation, and visualization. The main result of this paper is an elementary framework
for generalizing the above subdivision method beyond triangles. We give a detailed
analysis of the generalized subdivision and show that it preserves most of the symmetry
we find in the triangle case.
Main Theorem For every integer k 2: 1 and every dsimplex a there is a subdivision
into kd dsimplices a, with the following properties:
(
1
) All a, have the same ddimensional volume.
(
2
) The a, fall into at most d!/2 congruence classes.
(
3
) Thefaces ofa are subdivided the same way.
(
4
) Repeated subdivision has the same effect as increasing k.
(
5
) Except for boundary effects the neighborhoods of vertices are translates of each
other.
Figures 47 show subdivisions satisfying the Main Theorem. Properties (
1
)(
5
) are
statements of symmetry. Observe that the number of congruence classes claimed in
Property (
2
) is independent of k. Property (
3
) means that the effect of the subdivision
on a face of a is exactly the subdivision of that face with the same k and the same
method. Property (
4
) says that instead of subdividing each a, into ld dsimplices we can
subdivide a into (kl)d dsimplices and reach the same result. Property (
5
) implies that
we can think of the subdivision as the intersection of a with a periodic tiling of IRd that
looks the same from every vertex. The subdivision described by the Main Theorem is
not unique but rather one among d!/2.
Previous and Related Work. For k = 2, the subdivisions of the Main Theorem have
been described by Freudenthal in 1942 [3]. In contrast to our algebraic approach, his
construction is geometric, as are the related triangulations of the ddimensional cube
by Coxeter [2] and Kuhn [6]. The twodimensional case of Freudenthal's subdivision
is the 4division of the triangle illustrated in Fig. 4. It is frequently used for generating
twodimensional grids as well as for modeling surfaces in space. The threedimensional
case is the 8division of the tetrahedron; see Fig. 5. It was studied by Bey [1] and by
Liu and Joe [7] in the context of adaptive mesh refinement for finite element analysis.
Freudenthal's subdivision for general dimensions was also considered by Moore [8] who
studied hierarchies for adaptive meshing. Goodman and Peters [4] extend Freudenthal's
method to the same generality reached in this paper and use it to design smooth manifolds.
Their description is less elementary than ours and because of their different focus they
do not study the subdivisions themselves in any amount of detail. The work in this paper
takes its inspiration from the algebraic constructions of Grayson [5]. He subdivides
simplices into Cartesian products of simplices in the context of Ktheory. This approach
leads to a simple and purely algebraic description of simplices and subdivisions, which
is amenable to computer implementations.
Outline. Section 2 describes an algorithm that maps a point to a simplex containing
the point. Section 3 proves the Main Theorem by collecting the simplices generated by
the algorithm. Section 4 illustrates and discusses sample subdivisions in dimensions 2,
3, and 4. Section 5 counts subdivisions and simplices.
8
o
o
4
Abacus Model of a Simplex
We think of a dsimplex primarily as a sequence of d + 1 numbers in the unit interval.
With the introduction of this idea we prepare an algebraic interpretation of a simplex
that frees us from depending on our geometric intuition.
Colored Rectangle. Let Po, PI, ... , Pd be affinely independent points that span a sim
plex a in jRd. The ordering of the points is important, as it affects the following con
struction. If bo, bv, ... , bd are nonnegative real numbers that sum to 1, then
is a point in a. The bx are the barycentric coordinates of X. We may portray them
graphically by drawing the unit interval as a rectangle with regions colored from 0
through d, making sure to arrange the colors from left to right, so that bx is the fraction
of points with color x. Fig. 1 illustrates this for d = 7 and the point with barycen
tric coordinates (0.26, 0.11, 0.07, 0.11, 0.19, 0.08, 0.04, 0.14). The coordinates of the
dividing lines are displayed above the rectangle. They are provided by the partial sums
o = Bo, B 1, ••• , Bd, Bd+l = 1 with Bx = bo + bl + ... + bx l . BI through Bd can be
any nondecreasing sequence in the unit interval.
Stack of Rectangles. Suppose we chop the rectangle in Fig. 1 into k pieces of equal
width, stack them on top of each other, and expand the scale by a factor of k in the
horizontal direction for clarity; see Fig. 2. The coordinates of the dividing lines are
obtained by multiplying the coordinates of the earlier dividing lines by k and discarding
the integer part, keeping only the fractional part. We discard any duplicates, letting j +2 be
the number of distinct values remaining. Call these numbers 0 = Co, C1, .•. , Cj, Cj +l =
1, making sure to sort them in increasing order. We extend the dividing lines vertically
until they span the entire stack and label each region by its color; see Fig. 3. The widths
of the regions in the stack are c, = Ci +l  C), for 0 ::s i ::s j.
8
o
2
I
5
o
3
I 6
00
In
o
7
4
Fig. 2. The rectangle in Fig. 1 is chopped into three pieces, which are stacked and expanded.
a
o
N
N
o
Color Scheme. The number of regions in each row of the stack is j + 1. Forgetting the
positions of the vertical dividing lines we get a matrix
M=
XI,O
X2,0
XI,1
X2,1
Xk,O
Xk, I
XI,j
X2,j
Xk,j
of k(j + 1) colors. We call M a color scheme because it determines the combinatorics but
not the geometry of the coloring. The matrices that may occur are those whose entries
are drawn from the set to, 1, ... ,d}, whose entries are in nondecreasing order when read
like text:
XI,O ~ XI,1 ~ .•. ~ XI,j ~ X2,0 ~ ... ~ Xk,j,
and whose columns are pairwise different.
Combination ofPoints. The numbers c, sum to 1 and can therefore be used as barycen
tric coordinates to express our original point X in terms of other points. For knot
necessarily distinct colors Xl through Xk we introduce the notation
PXIX2"'Xk =
(PX1 + P X2 + · · . + P Xk)
k
.
This way we obtain exactly all points in our dsimplex (J' whose barycentric coordinates
are integer multiples of 1/ k. Each column in a color scheme corresponds to such a point.
We now show that the c, are the barycentric coordinates of X with respect to the points
that correspond to the columns of the color scheme M = M(X, k).
Combination Lemma
X =
L'/=oC;' P X1,;X2,;"'Xk,;'
Proof The product k . bx is the total width of regions with color X. It follows that the
sum, over all entries of the color scheme, of Px/ k times the width of the corresponding
region is the sum of bx . Px' By definition this is our original point X. Now consider
Y =
j
L c, . P XI,iX2,;"'Xk,;
;=0
1 j
= k' L
;=0
c, · (PXI,i + PX2,i + ··· + PXt,i)'
which is a sum over all entries of the same color scheme. Each term corresponds to a
column, and the contribution of an entry with color X is again Px/ k times the width of
the corresponding region. Hence X = Y. 0
Independence of Points. We prove that the points in the Combination Lemma are
affinely independent. This implies that a color scheme with j + 1 columns specifies a
j simplex. The color scheme depends on the sequence in which the vertices of a are
presented. This sequence defines a directed path obtained by concatenating the shape
vectors Vx = Px  Px 1 for 1 ~ X ~ d. Since a is nondegenerate, the d shape vectors
are linearly independent.
Independence Lemma The points PXl,; X2,i •••Xk.i defined by the columns ofa color scheme
Mare affinely independent.
Proof. Assume first that M has d + 1 columns and consider the vectors connecting
points of adjacent columns:
Because the color scheme is full, the transition from column i  I to column i changes
only a single color, and that color increases by 1. Suppose that the color increases in
row i. We define a permutation n of {I, 2, ... , d} by setting 7r(i) = Xl,i = Xl,il + 1. n
is indeed a permutation because the transition to 7r(i) occurs exactly once in the matrix X.
We find that
1
V;, = k· (P]f(i)  P]f(i)l)
1
= k· V]f(i).
The linear independence of the Vi thus implies the linear independence of the Vi', which
implies the affine independence of the points PXl,; X2,i'" xu A color scheme with j + 1
columns is obtained by removing d  j of the d + 1 columns of a full color scheme.
The corresponding operation for the vectors is addition. Specifically, the d linearly
independent vectors are grouped into j sums, which are again linearly independent
vectors and thus define a j simplex. 0
3. Subdivision of a Simplex
Section 2 described an algorithm which takes a point of a and produces a smaller simplex
within a that contains the point. We use that algorithm to construct a subdivision of a .
Edgewise Subdivision. Each point X E adefinesauniquecolorschemeM = M(X,k)
with k rows. This color scheme defines a unique simplex au, namely the one spanned by
the points PX l,i X2,i " ' Xk,i that correspond to the columns of M. The edgewise subdivision
of a consists of all simplices defined this way by points of a:
Esdj,o' = {aM 1M = M(X, k), X E a}.
If we delete one or more columns from M we get another color scheme M(Y, k), where
Y is obtained from X by setting one or more of the barycentric coordinates to zero
and increasing the others accordingly. It follows that for each simplex aM the edgewise
subdivision contains all faces of that simplex. A set of simplices with that property is a
subdivision of a iff the interiors of the simplices are pairwise disjoint and the union of
interiors is a. We establish the latter two properties for Esdka .
Subdivision Lemma
For each k 2:: 1, Esdjo' is a subdivision ofa.
Proof. The algorithm in Section 2 is deterministic and maps a point X E a to a set of
barycentric coordinates that express X in terms of the vertices of a j simplex aM 5; a.
These coordinates are positive, which implies that X lies in the interior of au It follows
that the union of interiors of simplices in Esdka is a .
What happens if we start at the other end? Suppose we take a j simplex aM E Esdso ,
choose nonzero barycentric coordinates Co + CI + ... + Cj = 1, and define
j
Y = LCi .PX 1,i X2,i " ' Xk,i .
i=O
Applying our algorithm to this point Y will lead to the same c, and aM because each
step in the algorithm is uniquely reversible. It follows that the interiors of the simplices
in Esdj,o are pairwise disjoint. 0
Volume. For a simplex a with vertices Po, PI, ... , Pd in ~d it is known that the
ddimensional volume is
1
vola = d! . Idet(PI  Po, P2  Po, ... , Pd  Po)l·
We prove Property (1) of the Main Theorem, which claims that all dsimplices in the edgewise subdivision have the same volume.
Volume Lemma
vol aM = vola/kd.
Proof. By definition of the shape vectors we have Px  Po = Vx + Vx I + ... + VI.
Using elementary column operations we find that volume can also be expressed using shape vectors:
The ddimensional volume of every dsimplex aM
E
Esdro is
Permuting the vectors does not affect the absolute value of the determinant, and shrinking
them to (1/ k )th their original lengths decreases the determinant by a factor of 1/ kd • 0
The Volume Lemma implies that the number of ddimensional simplices in the edge
wise subdivision of a dsimplex is kd , as claimed in the Main Theorem. Counting lower
dimensional simplices is somewhat more involved.
Simplex Types. We have seen in the proof of the Independence Lemma that the shape
vectors of a dsimplex in the edgewise subdivision are obtained by permuting and shrink
ing the shape vectors of G. We use this to prove Property (
2
) of the Main Theorem.
Congruence Lemma If the dimension ofG is d, then the number ofpairwise noncon
gruent dsimplices in Esdj,o' is at most d!/2.
Proof Esdro contains at most d! types of dsimplices, each defined by a permutation
of the d shape vectors of G. Reversing the sequence (without reversing the directions of
the vectors) effectively reflects the d simplex through the origin. Reflected simplices are
congruent, which implies the claimed upper bound. 0
In general the upper bound is tight. For sufficiently large k all permutations occur,
and for generic shape vectors two permutations define congruent dsimplices only if
the permutations are equal or the reverse of each other. For special dsimplices G the
congruence classes may further collapse. For example, if every permutation of the d
shape vectors can be geometrically realized by an orthogonal transformation then we
have only one congruence class. Such shape vectors are determined by their common
length and the same angle between every pair of vectors. The angle can be chosen
anywhere in the open interval between 0 and adI, where adl is the angle formed by
the vectors connecting the barycenter with two vertices of a regular (d  1)simplex.
For each angle in this interval we get a d simplex that tiles jRd with congruent copies of
itself. The oneparameter family of such simplices was studied already by Moore [8].
According to Senechal [9] the problem of characterizing all d simplices that tile jRd is
open even for d = 3.
Symmetry of Dimension. Each face r of G is a simplex which is subdivided as a
consequence of the subdivision of G. We prove Property (
3
) of the Main Theorem.
Face Lemma Let r be a face ofG and assume the vertices ofr are ordered as specified
by the ordering of the vertices ofG. Then Esd, r ~ Esdjo .
Proof Let E = {O, 1, ... , d} and T C :E such that Px is a vertex of r iff X E T. A
point X = L bx . Px E G belongs to r iff bx = 0 for all X E :E  T. As a consequence,
the color scheme M defined by X contains only colors in T. Every color scheme with
colors drawn from T can be obtained from a color scheme with colors drawn from ~ by
dropping all columns that contain colors from :E  T. 0
Symmetry ofScale. It is plausible that a finer edgewise subdivision of G is a subdivision
of a coarser edgewise subdivision provided this is true for the edges of G. We prove
Property (
4
) of the Main Theorem.
Refinement Lemma
EsdEsdsc' = Esdkia.
Proof. We show that every simplex in Esdkia belongs to Esdi of a simplex in Esdso ,
Since Esdkia and Esd/Esdj,« both subdivide a this implies that the two subdivisions are
the same.
Let X = L bx . Px be a point in a, and let Bx be the sum of the first X barycentric
coordinates, as usual. Let M = M(X, k) and M' = M(X, kl) be the corresponding
color schemes. The coordinates of the dividing lines in the corresponding stacks are
k Bx mod 1 and kt: Bx mod 1. Equivalently, if C is a dividing line for M then i· C mod 1
is a dividing line for M'. In other words, the stack of M' is obtained from that of M by
chopping each rectangle into l pieces, stacking the pieces, and expanding by l in the
horizontal direction for clarity. It follows that aM' E EsdiaM . 0
Symmetry of Location. Ignoring boundary effects, the neighborhoods of any two
vertices in an edgewise subdivision are the same. We prove Property (
5
) of the Main
Theorem.
Translation Lemma Let P, P' be vertices and let a' be a simplex in Esdro . Then
a' + (P  P') either belongs to Esdka or its interior lies outside a.
Proof. Consider a point X in the interior of a' and assume Y = X + (P  P') lies
within a. The coordinates of P  P' are integer multiples of 1/ k. It follows that the
vertical dividing lines of the stacks defined by X and Y are the same, but the rows defining
the dividing lines may be different. Specifically, the row indices increase ~r decrease
depending on the coordinates of P  P'. The color scheme M = M(Y, k) differs from
M' = M(X, k) by increasing or decreasing colors accordingly. Since all points X in
the interior of a' define the same color scheme, all Y = X + (P  P') define the same
transformed color scheme and therefore belong to the interior of the same simplex in the
edgewise subdivision. This simplex is a' + (P  P'). 0
4. Examples
This section presents examples of the edgewise subdivision for simplices of dimension 2,
3, and 4.
Triangle. Figure 4 shows the edgewise subdivisions Esd2 and Esd, of a triangle. The
three vertices are labeled kOO, OkO, OOk indicating the sequence used in the construction
of the subdivisions. Not that the sequence matters, however, since all 6 = 3! permutations
lead to the same subdivision consisting of k2 similar triangles. The vertex labels encode
barycentric coordinates:
x y z
xyz =  ·kOO+  ·OkO+  ·OOk.
k k k
8Division of Tetrahedron. The smallest nontrivial edgewise subdivision of a tetrahe
dron a is the 8division illustrated in Fig. 5. The tetrahedra adjacent to the four vertices
UU2
2(X) tI¥.. 020
JJ()
)(X) .+... 030
210 120
are similar to a. The four tetrahedra subdividing the center octahedron are generally
not similar to a because their shape vectors come in a different sequence. Observe that
exactly one diagonal of the octahedron belongs to the 8division. Every proper subset
T C {a, 1, 2, 3} defines a nontrivial partition into two sets and a halfway plane of points
with barycentric coordinates LiET b i = LjfjT b j. There are seven halfway planes,
and the 8division can be defined by cutting a with six of them. The only halfway
plane not used is defined by T = {O,2} and crosses the diagonal that belongs to the
subdivision.
27Division of Tetrahedron. The 27division of a shares many of the features of the
8division; see Fig. 6. It embeds four 8divisions, one adjacent to every vertex of a.
Any two such 8divisions overlap in a single tetrahedron adjacent to the middle third of
the edge joining the two vertices. We have 4 . 8  (~) = 26, which shows that the four
8divisions cover all tetrahedra except the one in the center of the 27division.
16Division of 4Simplex. We are not able to draw the entire 16division, but we can
draw the 4simplex a and argue about its 16division. A 3simplex is usually drawn as
a convex quadrilateral in }R2 simultaneously subdivided into two triangles in the two
ways possible. Two of the triangles are in the back and the other two are in the front. By
analogy we think of a 4simplex as a convex double tetrahedron in }R3 simultaneously
subdivided into two tetrahedra and into three tetrahedra. Maybe the three tetrahedra are
in the front and the two are in the back. Figure 7 shows two copies of the 4simplex
Fig. 5. The octahedron in the center of the 8division is decomposed into four tetrahedra surrounding the
space diagonal.
with front tetrahedra 1203,2403,4103 and back tetrahedra 0124, 3124. For k = 2 the
color scheme of an edge is a 2by2 matrix with entries from {O, 1,2,3, 4}. At least
one of the colors is not used, which implies that all edges in the 16division belong to
the subdivisions of proper faces. We can therefore draw all edges by superimposing the
edges of the five 8divisions in the boundary of a.
The simplices that decompose the interior of a correspond to color schemes that use
all five colors. There are only five triangles because there are only that many 2by3
matrices with five colors:
( 02 03 14) '(20 31 41) '(20 31 42) ' 3(0 31 42)' 3(0 41 42)·
As illustrated in Fig. 7, the triangles form a Mobius strip one time around the center of
a. The five solid edges form the boundary of the strip and the five dotted edges cut it
into five triangles. Similarly we find 20 tetrahedra that do not lie in the boundary and
decompose a into 24 = 164simplices.
Variation. The abacus model can also be used to decompose a dsimplex into Cartesian
products of simplices. Such decompositions are byproducts of edgewise subdivisions.
We return to the stack of rectangles as illustrated in Fig. 2. Each rectangle contains
dividers which are allowed to move freely between the left and the right end, except they
cannot switch places. Each rectangle represents a simplex, and the stack represents the
3 0 =~~~\ ~
Fig. 7. Two views of a 4simplex. To the left the dotted lines show the 8division of one tetrahedral face. To
the right the dotted lines decompose a Mobius strip in the interior of the 4simplex into five triangles.
Cartesian product of these simplices. In ~3 we have three types of threedimensional
Cartesian products: tetrahedra, triangular prisms, and parallelepipeds. Figure 8 illustrates
the decomposition for k = 3, which is the smallest integer for which all three types arise.
Compare this with the 27division in Fig. 6.
Counting
This section counts the edgewise subdivisions of a simplex and the simplices in an
edgewise subdivision.
Number ofSubdivisions. A triangle has six orderings of its vertices each leading to the
same edgewise subdivision. A tetrahedron has twentyfour orderings and three different
edgewise subdivisions. For k = 2 these are characterized by the pair of vertex disjoint
edges whose midpoints span an edge in the subdivision. We extend the two data points
to a function of d.
Subdivision Counting Lemma For every dsimplex a and every k ~ 2 the number
ofpairwise different edgewise subdivisions is d!/2.
Proof Call two orderings of the d + 1 vertices equivalent if they differ at most by
a cyclic rotation possibly followed by a reversal. Goodman and Peters [4] showed that
two equivalent orderings define the same edgewise subdivision. For example, for d = 3
we have three classes of eight orderings each. We also have three different edgewise
subdivisions, one per pair of vertex disjoint edges of a.
We now return to d ~ 3 dimensions and consider two nonequivalent orderings. Let
1C be the permutation that takes one ordering to the other. Since 1C is not a cyclic rotation
(of the identity or its reverse) there is a restriction of 1C to four vertices that is also not
a cyclic rotation. In other words, there is a threedimensional face or tetrahedron such
that the induced orderings of its four vertices are nonequivalent. The Face Lemma and
the above discussion imply that the two subdivisions of the tetrahedron are different. It
follows that the two subdivisions of a are different. Each equivalence class consists of
2(d + 1) orderings, which implies there are (d + 1)!/2(d + 1) = d!/2 classes and the
same number of different edgewise subdivisions, as claimed. 0
Number ofSimplices. We derive a closed form expression for the number of j simplices
in the edgewise subdivision of adsimplex a . To simplify notation we fix d and k and
we let Sj be the number of j simplices in Esdso , As mentioned earlier, for j = d we
have Sj = kd.A color scheme M is defined by setting d dividing lines, each on one of k
levels. These dividing lines can merge with each other or with the right ending line, and
the result defines a j simplex iff exactly j different dividing lines remain.
Let Q = Q(j+l)k+l be the set of all limiting and dividing positions: the start position, k
times j interior positions, and k ending positions. These are the limits and dividers if we
arrange the k rectangles back into one piece. We always choose the starting and ending
positions. The number of jsimplices is equal to the number of dmultisets chosen from
Q with at least one token at every dividing line. Recall that
is the number of ways we can choose d tokens from (j + l)k+ 1 with repetition, and note
that Cj + 1 is larger than Sj because it also counts multisets where some of the dividing
lines receive no token. Specifically,
C}.+1 _ (d + (jd + l)k)
Sj = Cj+1
i=1
i: (~)Sji
l
(
1
)
because there are ({) different ways to miss i of the j dividing lines. We use (
1
) to derive
a formula for the number of j simplices.
Simplex Counting Lemma
The number of jsimplices in Esd, of a dsimplex is
Sj = t <  l ) i ·
i=O
(~)
l
. Cj +1 i •
Proof We use induction. To prove the equation for Sj we collect the coefficients of
Cj +1 i , for 0 ~ i ~ j. For i = 0 there is only one occurrence of Cj +1 i in (
1
), and its
coefficient agrees with the coefficient in the claimed equation. For i > 0 we have an
occurrence each for Sjl, Sj2, ••. , Sji in (
1
). The coefficient of Cj +1 i in the expression
forsj_m is (_I)im.({=:). The coefficient of Cj.q.., forsj is the sum of these coefficients,
each multiplied by 1 and the binomial coefficient as in (
1
):
We need to prove
x = (_l)i ({)
and to do that we use
We have
x =
c
~ i)' j;(_I)iml (~)
= G). [;(_I)iml(~)  (_I)il(~)]
= G)· [(1  I)i + (_I)i)
= (_I)i.
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