AN ABSTRACT OF THE THESIS OF
MARTIN J. STYNES for the degree of DOCTOR OF PHILOSOPHY
in
MATHEMATICS
presented on
January 13, 1977
Title: AN ALGORITHM FOR THE NUMERICAL CALCULATION OF
THE DEGREE OF A MAPPING
Abstract approved:
Let
Redacted for Privacy
Philip M. Anselone
el be a continuous function from a connected
n-dimensional polyhedron Pn
vanish on the boundary b(Pn),
1.n
to
Rn. Assume ei does not
so that the topological degree of
relative to the origin is defined. In this dissertation an algorithm
for the computation of the degree is obtained. It utilizes certain
subdivisions of
b(P) and depends on the modulus of continuity
restricted to b(Pn).
An Algorithm for the Numerical Calculation of
the Degree of a Mapping
by
Martin J. Stynes
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Doctor of Philosophy
June 1977
APPROVED:
Redacted for Privacy
Professor of Mathematics
in charge of major
Redacted for Privacy
Chair' Man of bepartment of Mathematics
Redacted for Privacy
Dean of Graduate School
Date thesis is presented
Typed by Clover Redfern for
January 13, 1977
Martin 3. Stvnes
ACKNOWLEDGMENTS
First thanks m, st go to my major professor, Dr. Philip M.
Anselone, for his patience and most able guidance. The 0.S. U.
mathematics department has always been ready and willing to discuss
my problems; in particular I wish to thank Professors D. H. Carlson,
W. J. Firey, and F. J. Flaherty for their time and effort.
The writing of a thesis is a long and painful road, and I am
proud of the unfailing encouragement I have received from my family
in Ireland. The path was also made smoother by many friends in
Corvallis; in particular Tom Kelley, Weldon A. Lodwick, Brian T.
McCay and Dave Borer stand out in my memory. Even above these
I remember Jeanne Ng with affection.
They were all instrumental
in making my time here a very happy one.
TABLE OF CONTENTS
Page
Chapter
I.
INTRODUCTION
1, Summary
2. Definitions of Topological Degree
II. BACKGROUND MATERIAL
III. SUFFICIENT REFINEMENTS AND COMPUTATION
OF THE DEGREE
IV. IMPARTIAL REFINEMENTS
Motivation
Impartial Refinements Definition and Properties
V. ALGORITHM FOR COMPUTATION OF THE DEGREE
USING IMPARTIAL REFINEMENTS
Introduction and Description of Algorithm
Proof of Convergence of Algorithm
VI.
1
1
2
8
18
35
35
41
47
47
51
SIMPLIFICATION OF THE COMPUTATIONS
60
BIBLIOGRAPHY
78
AN ALGORITHM FOR THE NUMERICAL CALCULATION
OF THE DEGREE OF A MAPPING
I. INTRODUCTION
1. Summary
The topological degree of a function
f
Rn
Rn
is an integer
which when non-zero guarantees the existence of solutions
x
to
equations of the type f(x) = p. The concept was first introduced by
Konecker [9] in 1869. Since then it has found many applications, most
notably in the fields of differential and integral equations (cf. [2,3,8]).
However these applications generally make some assumption regard-
ing the value of the degree because, as J. Cronin observes [2, p. 37]:
"the problem of computing the degree is, to a considerable extent,
unsolved except in the plane. " Thus F. Stenger's paper [11], which
gave an algorithm for computing the degree in Rn,
represents a
significant advance.
In Chapter III of this thesis a new proof is given of the basic
formula used to compute the degree in that paper. It is remarkable
that this formula, first given in [11], was anticipated in a simpler
context as long ago as 1904 by J. Hadamard [5, pp. 452-460]. The
algorithm of [11] has at least one serious drawback: it generates a
sequence of numbers which eventually equal the degree, but it is in
general impossible to decide when this equality has occurred. Our
2
main result, proven in Chapters IV and V, shows that in many cases
an alternative procedure is available which unambiguously computes
the degree. The calculations needed to compute the degree using the
basic formula of [11] are very time-consuming, involving the evaluan x n determinants. In Chapter VI of this thesis a
tion of many
simplification is given which replaces the determinant evaluations by
a "scanning" of the matrices associated with the determinants, i. e.
a search for certain entries in those matrices.
It would be of considerable interest to extend the foregoing
results to the infinite dimensional situation (cf. [2,3]).
Z. Definitions of Topological Degree
We begin with an intuitive "definition" of topological degree,
after which the two principal properties of the degree are listed.
Let
be a bounded region (open connected set) in Rn
D
its closure. Denote the boundary of
b(D).
P
Let F
D
Rn
D
be continuous.
with
D \D) by
(i. e.
Choose
p E R.11
with
Fn(b(D)).
If
(2;10
x
traces out
b(D)
once "ccu.nterclockwise" then the
ical de
is an
nof
integer measuring how many times
surrounds
Fn(x)
"counterclockwise" manner. In the case
n
2
familiar "winding number" of complex analysis.
p
in a
the degree is the
3
Fn(x)
d(Fn, D, p) = +1
G(x)
d(Gn, D, p) = -2
There are basically two important properties of d(Fn, D, P):
Homotopy property: if H(t, x)
[0, lj x
VtE [0, it
H(t,x)
p
H(0,x)
Fn(x)
H(1,x)
G(x) VxED
Vx
--'Rn
is continuous with
b(D)
VxED
then d(Fn, D, p) = d(Gn, D, p).
Existence of solutions roperty: if
XED
such that
d(Fn, D p) i 0 then
3
Fn (x)p.
Next we give several strict definitions of the degree which can
be shown to be equivalent. In this section we will take
to be the origin
On,
since any other point
q
p
in
Rn
may be dealt with
4
i. e. , by defining d(Fn, D, q)
by translating Fn,
to be
d( Fn-q, D, en).
Let
D
be a bounded region in R. Let Fn U
F (x)
continuous with
The case
where
n=1
On
Rn
V x E b(D).
is treated separately: suppose
D
(a, b),
q
are
-co < a < b < +00. Set
sgn Fl(a)},
{sgn Fl(b)
d(F1 ,D, U1)
where
+1
if
t>
0,
0
if
t
0,
-1
if
t < 0.
sgn t =
=
For the rest of the section take n > 1.
If
Bi
Oo
i 1,
bi2,
b.
.
iq
)
= 1,2, .
,Notation.
vectors, then
1
b
B)
A (B 1
21
b12
blq
b
b
22
B.
is the
ith
2q
=
ql b q2
In this array
be
row, for
1<
b
i < q.
qq
5
1st Definition (Kronecker).
(cf. [1, pp. 465-467]).
Ln(Fn,
d(Fn, D, On) =
n-1
n
X (U
X
n-1 )
dul...
a Fn
.7'11
b(D) 11F11 n
'57 )
1
dun
n/2
where
n-1
=
U
F
is the Euclidean norm in Rn,
(ul'u
,un-1)
F(ni2)' and
is a parametrization of b(D) oriented
n-1
in a certain way.
Here we assume that Fn is of class
C
1
on
If
b(D).
is merely continuous on b(D) we can approximate it by
CI
Fn
func-
tions and show that the value of the degree is the same for all such
functions close enough to
finally define
Fn;
d(Fn, D, On)
to be
this common value.
Note that the integral is a direct generalization of the winding
number formula of complex analysis. In general it is very difficult to
evaluate.
2nd Definition (E. Heinz
d(Fn, D, On
Here we assume that
(cf. [6]).
.
54
co
(11Fnli )j(F')dx
Fn E CI (D)
.
.
and that the Jacobian
6
aFn
j(F) =,
eFn
n
ax 1
<r
Fn,
and
The func-
Fn(x) = On.
is continuous with support of co C [r1,r2] where
co
0 < r1
axn
such that
is non-zero at every point x E D
tion
'
< 00,
and
are small, depending on
r
and
r1
with
o)dx1
Fn
If
dx = 1.
n...
is merely continuous we can again define
d(Fn, D, On)
by approximation, using Sardis theorem.
Note that the integral is again difficult to evaluate.
Fn
3rd Definition. (cf. [10]). Suppose
Vx
such that Fn(x)
D
-
Let N+ (N)
Fri(x)
N-
on
such that
on
we have
E
C1 (D).
j(F)(x)
denote the number of solutions
j(F)(x)
Assume that
0.
x
in
is positive (negative). Both
D
of
N+ and
are finite since these solutions must be isolated points.
Then
If
define
Fn
d(Fn, D, On) =N+ - N
is only continuous on
D
approximate once more to
d(Fn, D, On).
Once again this is a difficult definition to use for calculating the
degree, and in practice is not very helpful because to use it we need
7
to know the solutions of
Fn(x) -= en;
unfortunately we usually want
to use the topological degree to guarantee the existence of such
solutions.
Our 4th and final definition is the one which will be used for the
remainder of this thesis. It might be called the "simplicial homology"
definition (although no reference is ever made to homology theory!).
The next chapter is a careful development of the definition.
8
II. BACKGROUND MATERIAL
Almost all of the material presented in this chapter has been
condensed from [2, Chapter 11,
Notation. In general for the remainder of this thesis super -
scripts will indicate dimension while subscripts are used as indices.
A set of points
a}
q
{a0, a1,
Rn
in
is linearly
independent if the points are not contained in any subspace of dimen-
sion < q-l.
A ci -simplex
independent points
Sq = (a
of
01
q
).
Sq
is the closed convex hull of
a0, al, ...
The points
a, a l
l:)
,
.
.
linearly
and is denoted by
Rn,
in
,a
q+1
are called the vertices
,a
Sq.
0
- simplex
a0
2 - simplex
1
a0
3
- simplex
al
- simplex
a0
An r -dimensional face of Sq
convex hull of any
r+1
of the points
(0 < r < q) is the closed
a0,a1,
,a
.
Observation 2.1.
of
p
in Rn,a
1 < k < n,
then
Pk
If
the
.jk
and
is the
pk
coordinate of a.
kth
coordinate
kth
in
Rn,
J
p E Sq
=
p E Rn
X
ja jk
iff
with all X. > 0
and
X>i
= 1.
J
j=1
j=1
Given a q-simplex we say that two orderings on its vertices
are equivalent if one can be obtained from the other by an even permutation of symbols. This gives rise to two equivalence classes of
orderings on a given q-simplex, if q > 0: one we call "positive",
the other "negative". For
q
0
call the only possible ordering the
positive one.
We adopt the following convention: if
with
x. = (xi
,
12
X.
for each
i
x.
Rn, 0 < i <
_ n,
(Cartesian coordinates),
then
orientation (xoxi.. .x)
sgn zsn+1((1, xo), (1, xi),
sgn
(1, xn))
x01
x02 *** x On
1
xll
x12
xln
1
x
x
x
nl
n2
nn
10
This is not zero if the
are linearly independent and does not
x,
change value under an even permutation of vertices.
Note: Here, given a Cartesian basis for Rn,
essentially ordering
Rn
we are
itself this is what an "oriented Rn" will
mean in the sequel. It can be shown that given a basis and orientation
on
Rn,
the orientation with respect to another basis is the same iff
the determinant of the affine mapping taking one basis to the other is
positive. Consequently we see that an orientation on Rn induce s
an orientation on all its subspaces and their translations in a natural
way.
<x0x1... xn> is an oriented n-simplex (i. e.
simplex with associated orientation). We write
Notation:
,
a
<x1x0x2... xn> = -<x0x1x2... xn> etc.
A q -chain is a finite algebraic sum of oriented q-sirnplexes with
integer coefficients, e.g. ,
2<z0a1a2>
3<a3a1a4> is a 2-chain.
In any given q-chain we assume that all possible cancellations have
been made (i. e. , if
u.S.
cq
3
any
iZ'
IfSq =
and all the
u.
<a0 a1 ...a q>,
is a q-chain then
3
i=0
31
are non-zero).
define its 12.1anc
(-1)i<a 01
a ...a...
i
b(Sq) =
sc.'
a>
to be
± sq.
32
for
11
denotes omission. Thus
where
A
q>0
(we take
b(S q )
is a (q-1)-chain, if
b(S0) = 0).
For example, if
S2
= <a a a >
denotes orientation
S
2
then
b(S2 = <a1a2> - <a0a2> + <a0a1>
= <a1a2> + <a2a0> + <a0a1> .
b(S2):
where the
Given a q-chain
define
u.
are integers,
so the boundary of a q-chain is a
b(oq) =
(q-1)-chain if q>
Remark 2.2. It is easy to prove that b(b(Sq))
Sq,
hence
b(b(cq))
Notation:
Let
0
S.
for any chain
0
for any
cq.
be a collection of (possibly oriented)
q-simplexes indexed by j. Then
12
{p: p E
S.(
for some
3
S.q _- p: p E
for all
{(m
If
means
u.Sq
cq
pE
is a 4-chain and
then A \ cq
Intersection Numbers.
let
then
pE
cq
for some
Se:1
If A C Rn,
R n;
V j,
u, 4' 0
Let
Sn
be an oriented n-simplex in
We say that Sn
b(Sn).
p E Rn
{x E A:
<p>
and
are in gen-
eral position.
If
p
Sn
set
i(Sn,p) = 0.
If
p E Sn
set
i(Sn,p) = orientation of S.
We call
Let
i
Hn-I
the intersection number.
be a hyperplane in an oriented Rn;
a line in the same Rn. Suppose
<pa1...an-1>
and
<pb>
H
n-1
G
1
{p}.
let
G1
be
Choose
positively oriented in Hn-1
and
G1
respectively.
Then set i(Hn-1 ,G ) = orientation <pa1 2 .. a n-1 b>. It can be
checked that <pa1a2... an_ib> is an n-simplex and that i is
1
independent of the choice of points a1,a2, .
Suppose
Sn
-1
and
T1
.
are oriented simplexes in Rn. We
say that they are in general position if
S
n-1
T1
is a single point
13
or the null set. Suppose they are in general position.
plane
If
S
S1
T
If
Sn-1
T1
Hn-1
set
point. Suppose
GI
Observation 2.3.
n-1
\ b(S),
and
If
U.
determine a hyper-
T1
b1
-n
oriented so that
Rn
=-
respectively with Hn-1
orientations. Then set i(sn-1, T1)
p E S n-1
T 1)
then Sn-1
4' 4)
and line
i(Sn
Sn
-1
and
a single
G1
have positive
T1
_(Hnl,C1)
= <a1 a....
an>
2
and
then orientation <a1a2...an> = orientation
<pa2... an>. It is elementary to check this from the definition of
orientation, since by Observation 2. 1 we can write
with each
jk
j
3'1
and a multiple of one row of a
= 1,
and
X.. > 0
Pk
j=1
determinant can be subtracted from another row without altering the
value of the determinant. Consequently if in Rn Sn-1
point p with
i(S
n -1
T1)
p
b(S1)
1
b(T ),
where
T1
rm
is a
T1
then
= <cb>,
= orientation <pa2...anb>
Given two chains
cP
u.X13
dq
where
vkYkq
u.,
3
vk
are integers and the simplexes X..k ,
with either p = n-1
and
q=
1
or
p
that the chains are in genera/ position if
eral position for every pair
Y
k
n
X
lie in an o:i°iented
and
q
Yq
Rn,
we say
0
are in gen-
(j, k). In this case we define
14
jkjk
P
q)
u.v iCK,,Y
j,
Theorem 2.4 [2, Theorem 2. 2]. If the chains
cn, b(d1 ), also
are in general position then
b(cn), d1
1
i(cn,b(d1)) = (-1)ni(b(cn
This is elementary but tedious; one considers various
Proof.
cases.
Order of a Point Relative to a Boundary. Suppose zn1
some chain
cn
and
p
relative to
c
n
(we say that
zn-1
is an (n-1)-boundary). Suppose
are in general position. Then we define the order of
z
n-1
to be
,
vtz
n-1
It is not difficult to check that
of the particular
p
cn
=ic
v
satisfying
pi
.
is well-defined, i. e. , independent
b(cn) = zn-1
Remark 2.5.
v(zn-1, p) 3/ 0 with b(c) = zn--1, then p on
for all integers
v(azin-1 +bz2n-1 ,p) av(zin-1 , p) + bv(z2n-1 ,
If
a, b
E
(provided the right-hand side is defined),
15
by a con-
We now extend this definition by replacing
tinuous image of itself.
Definition.
An n-dimensional polyhedron
is the empty set of S. (Th Sn
simplexes either Sn
Kn
i=1
r-
\hese
Sr.1
Sn
and
S. We
depending on context.
Kn =
S
such
is a common face, i. e. , an r-simplex
(0 < r < n) whose vertices are vertices of both
write
is a union of a
S', i= 1, 2, ... ,m,
finite number of oriented n-simplexes
that for every pair Sri, Snj of t
Kn
i=1
Example.
The shaded area is a 2-dimensional polyhedron.
Definition. An n-region is a connected n-dimensional polyhedron.
Definition. Let
2
z
-n 1
Kn
Rn
Kn
be an n-region,
be continuous, and let
is an (n-1)-boundary such that
tinuously into
e(b(Kn))
define the degree of
d(e, Kn, p)
=
v(zn
1
nz1
Rn \ Vi(lb(le7,:.
Then if
can be deformed con-
without passing through the point
el on K
p).
p
Let
Kn
relative to
p
to be
1D,
we
16
It is shown in [2] that d(e,Kn,p)
for all zn-1
is well-defined, i. e. , that
satisfying the conditions of the definition
has the same value. The existence of such
breaking up each S'
'I'n
Kn
-n 1
is proven by
a finite sum of n-simplexes,
expressing each S'
function
z
v(zn-1, p)
e.,
i
an n-region, then proving existence of a
Rn which maps n-chains formed by sums of these
smaller simplexes into n-chains in Rn in such a way that
l'7,(b(Kn))
is an (n-1)-boundary
This decomposition of Kn
the function
'T'n
with respect to p.
n-1
just as in the definition.
is called a
licia1 subdivision, and
is called a chain approximation to
Kn
It is shown that
is arbitrarily close to
'T':,(b(Kn))
on
and
Kn,
on
b(t'1:41,(Kn)).
Note that there exist arbitrarily small perturbations of
which will not destroy any of the properties attributed above to
Observation 2. 6. For
n = 1,
Then from the definition we can take
z
0
= b(<1(x01(x
)>).
suppose
z
0
b(K1)
1
(x
)>
---- <x
>
1
<.Ht'
(x0)
<x0
>.
so
By examining the various cases we see that
17
1
,K1 ,p) =
1
1
{sgn('T.1(xm)-p)-sgn('t' (x0 )-p))
Thus, since the computation of d('T'll,Kn,p)
is trivial for
n
1,
this case is mentioned in the sequel only to illustrate definitions and
concepts and to supply the first step in inductive definitions and
proofs.
Remark 2.7.
If
p
del(Kn),
d(e,Kn,p) = 0 from the definition
then
and Remark 2. 5(i);
From the definition
Clearly
d(,Kn,p)
in Section I. 2.
d(
Kn, p)
is always an integer;
has the homotopy property described
18
III. SUFFICIENT REFINEMENTS AND COMPUTATION
OF THE DEGREE
In this chapter we will use the definition of topological degree
given in Chapter II to reprove the basic computational formula of [11].
This formula was originally proven using the 3rd definition of Section
1.2.
Let
Pn
Pn
be an n-region. We shall always assume that
b(Pn)
is written as a sum of oriented n-simplexes in such a way that
coincides with the topological boundary of
Pn when Pn is
regarded as lying in Rn. It is also assumed that all decompositions
of
b(Pn)
pre serve or ientations
Example for Case n = 1.
I-
x°
x2
x5
x4
4
1
<xixi+1>
i=0
SO
(<x i+1 >-<x.>) = x
b(<x x i+1 >)
b(P1 )
i=0
5
-x
0
i=0
In the future we shall often assume that for n > 1
b(Pn)
is
19
written in such a way that the coefficient of each (n-1)-simplex is
+1;
this can be achieved by changing the order of the vertices in an
oriented simplex where necessary.
Let
e(p)
'n
= (col, (P
sufficiently refined relative to
p1 (x0) q1(x)
sgn
be continuous with
R
V p E b(Pn).
,L
Inductive Definition. If
to
Pn
el
0.
if
n
b(P1) = <x > -
1,
1
sgn
If n > 1, b(Pn)
<xo>
(say) is
if
sgn (pi
is sufficiently refined relative
b(P) has been subdivided so that it may be written
as a union of a finite number of (n-1)-regions
n-1
p
p1
n-1
p
n-1
in such a way that
the (n-1)-dimensional interiors of the n-1
p1
pairwise
disjoint;
at least one of the functions
2°.°9CP
zero on each region Pin-1
n-1
on Pin- 1
(iii) if (pr
then b(o
p')
relative to
n-1
sgnr
where
D
-1
°
n = 2,
suppose
-
b(P
<
j 0
cp
is non-
sufficiently refined
signifies omission).
Example 3. 1. In the case
say
xjx j+1>
20
xx .
where
This is sufficiently refined relative to sgn
0
if at least one of
is non-zero on each line segment (x.x.,
j
cp1, y92
),
and neither is zero at any x.. Here we are taking
n-1
J.
=
0 < j < m-1.
<xxi+1>,
There are some apparent differences between this definition and
that given in [11]. We shall explain and justify them before proceed-
ing further.
n > 1. Fix
Let
Then a Qn-set
i
and fix
,n}
E {1,2,
is a connected set of points
= n. or
A
lying in b(Pn) such
q
that
=A.
)
is
two Q'-sets are associated with different
-1.
If
and/or different
A's
in this definition, they are said to be of different types
Note:
refinement,
(a) From part (iii) of our definition of a sufficient
if
n-I
p E b(13 i
for some
)
n-1
r.. Of course
p
cannot lie in any Q'-set. Since the
)
then
also implies that
j
p E b(Pi_
i
yo,(p) / 0
yo
(p) / 0. Thus
Q'-sets are connected, this
shows that each of them must lie in the interior of some
(b) Because
sgn
an-1
cpr
/0
on
n-1
Pi
is constant and non-zero on
for some
and
n-I
P.
pn-1i
131,1-1
is connected,
and it follows that no
can contain points from Qn-sets of different types.
Our conditions for a sufficient refinement are repeated in the
definition given in [11] together with the requirements that (A) each
21
Qn-set must lie in the interior of some
pri. -1
and (B) each
pni -1
must contain at most one Q -set. We have show in (a) above that (A)
follows from the other hypotheses, and it is evident that (B) can be
weakened to the conclusion of (b) in the proof given in [11] of the
computational formula for the degree. Since sufficient refinements
are introduced for the purpose of deriving this formula, our definition
would have sufficed in [11].
We shall prove a theorem which gives an inductive degree
Rn-1 n-1)
relation between d(e, p n on) and the d(e-1
r
i
1,2, ... , m. For the proof we need two lemmas.
Lemma 3. 2.
n > 1. Suppose
Let
a/ On lies on one of the
coordinate axes in Rn with its rth coordinate non-zero. Let
sn-1 be an (n-1)-simplex with a function Fn Sn-1
Rn,
Fn = (f1, f2,
, fn),
such that Fn(Sn-1)
is an (n-1)-simplex.
Suppose that
(i)f
0
on
sn - 1
with
sgn fr sn- I
sgn (rth coordinate of
a)
(ii) i(Fri(sn-1), <ona>)
(here
n-1
=
Fr
Then
i
and
i(rn-1(sn-1), 0n-1
is the intersection number of Chapter II and
A
(f1'
are defined
fn) )
22
(-1)r+ni(Fn-1(Sn-1), On-1) s-n fr
i(Fn(Sn-1), <Ono>)
(Fn(sn-1), <ena >)
Proof. If
then
0
n-1 n-1
) r <0 n-1 > =
also by (i)
Fr (5
n-1 n-1 ), 0 n-1 ) = 0; the conclusion is verified. We may
i(Fr (S
Fn(Sn-1)
and
<Ono> = 0,
so
therefore assume that pn(sn-1)
P
.
(0, 0,
. . .
, p, .
.
.
, 0)
(m
Rn, p t 0
in
pn(sn-1)
\ b(Fn(Sn-1))
say, where
--.
rth
lying in the
sgn fr
choice of a. Note that (i) sgn p
(ii)
<Ana> = {P}
position by
i(Fn1(sn-1), en1) is
because
defined.
Set
P°
= (0, 0,
, 2p,
,
0).
Fn(Sn-1) = <
Suppose
yn>
Then
i(Fn(S n-1 ), <0 a>) = orientation <Py
y
n
(follows from observation 2.3)
sgn
,
P), (1, y,
1
0
0
1
y21
An+
1
(
(1
(1,
yn),
(1,P '))
0
Y2n
sgn
1
Ynl
yn2 *
0
where ym (yml yma,
ymn)
in
0
R
Ynr
Ynn
...
2p
for
1 < m < n.
...
0
23
Subtract the first row from the last one and expand in terms of last
row;
0
1
0
0
A
n+r
i(Fn(Sn-1), <0na>
sgn p sgn
1
Y21 Y22
A
=
(-1)n+r sgn p sgn
1
Ynl YnZ
1
yll Y12
in
1
Y21 Y22
Y2n
ynl Yn2
nn
1
(by an argument like that of observation 2.3)
(_on+r
r
by definition; we know that On-1
E
ui(pn-1(sn-1), on-1)
r
n -1n-1
(S
Fr
)
from our initial
assumption.
n
Lemma 3.3. Let Pn be an n-region with
; Pn R
continuous and .t,n, on on b(Pn). Suppose that b(P) has been
subdivided into a finite number of (n-1) regions
f3n
-1
,
jE
J,
so that
it is sufficiently refined relative to sgn
Let .T.n be a chain
approximation to VI on pn with respect to On, with
{Sn
-1
:k E K}
the associated simplicial subdivision of
24
n-1
b(Pn) =
Let the point
lie on one of the coordinate axes
a
EJ
in
rth
with its
Rn
coordinate non-zero and
> max 11
'1'11(x) 11
x E Pn
(usual Euclidean norm). Then we may assume that
relative to sgn
(i) the sufficient refinement of b(Pn)
relative to
also a sufficient refinement of b(Pn)
with
yo
r, / 0
on
n-1
p.
is
sgn
implying that
n-1
on
Pi
also;
(ti) 114 > max
11 '1'11(x)II
X E Pn
S'
both i(e..,(Skn 1), <0na>) and
(iii) for every
i(In-1(5n-1),0n-1) are defined.
*r
k
l
Proof. (i) We know that we may take
Pn;
on
the proof is then easy using induction on
fact that t.,n a chain approximation to
on
implies
respect to
.I,
0
n-1
*r
Pn
on
'T'n
a chain approximation to
.i-,_].
r,
4
n-1 i
Since
Pn,
arbitrarily close to
and the
n
with respect to
on
pi:)._ 1
with
J
3
n
may be chosen arbitrarily close to
on
we may assume that (ii) holds.
If
i((S:-1), <011a>)
is not defined then
n n-1 rm (B a)is a (non-trivial) line segment, implying that
k
)
n-ln-1
(S
*r
if
k
is an (n-2)-simplex and
)
on-1(sn-1), on-1 )
*r
k
0
n-1
is not defined then
lies in this (n-2)-simplex.
en-1
E
b(e1-1(Sn-1)),
*r
k
25
i.e. ,
0
n-1
lies in an (n-2)-chain.
n n-1 ), <0n>) and
Thus to ensure that both i(t.,(Sk
n-1
n-1 n)
are defined for the given finite set {Sn-1: k E 10,
0
(S1
k
it is sufficient that
does not lie in a certain finite collection
On-1
of (n-2)-simplexes which are in (n-1)-dimensional space. Therefore
if necessary we can perturb
<Ona>
arbitrarily slightly in a direc-
tion perpendicular to itself so that the above intersection numbers are
defined for all
k.
<0na> as it was
This is equivalent to leaving
originally, then perturbing
slightly in the opposite direction.
This perturbation can be chosen so small that all the properties of
as a chain approximation are preserved.
Rn
be an n-region and let
Pn, n > 1,
Theorem 3. 4. Let
be continuous with
On
on
d(e, p n on) is defined. Suppose that b(Pn)
into (n-1)-regions
relative to sgn
on
n-1
R.
n -1
p.
n on)
j
J,
E
Then if
'T'n.
for each
,
j
'T.n
b(Pn)
so that
has been subdivided
so that it is sufficiently refined
(yo.
2'
. .
n
)
with
/0
ccr
E
r,+1
21
)
d(er
.
EJ
1
n-1
0.
j
,
0
n- 1
sgn r.
J
n-1
PJ
26
n- 1
r. pn.1 e ) is defined since by definition
is sufficiently refined relative to sgn n-1 , which implies
Proof. Each
b(p1.1-1)
that
n-1
r.
n-1
V
d(
n-1
9
n
b(p-1.
). Note that the sum above is well-
on
3
j E J both
defined insofar as if for some
zero on
_n-1
because
el-1 V en-1
.
then
-1
d(
1
and
n-1-
'T't
,
0
I0
n-1
and
cos
n-1
) = d(.t't
n-1
, p.
are non-
cpt
,e
n-1
)
=o
n-1
onand
so it is
n-1
133
immaterial whether we associate
Choose a point
(with say
(Ps
or
(Pt
with
lying on one of the coordinate axes in Rn
a
rth coordinate non-zero) such that 114 > max e(x)11.
X E Pn
Let
respect to
el,1/4
be a chain approximation to
On;
on
Pn
by Lemma 3.3 we may assume that
with
satisfies
certain conditions. Now
d(e, pn, en)
=
y(b(Pn)), n)
(by definition)
Oh( e,(Pn)), 011)
(property of chain approximation)
(pn
(1)
(by definition of v)
iie(pn),
i(e(pn), a)
(by Lemma. 3.3 we assume
n(n) )
i(,In(pn), -On
27
-ie(Pn),b(<011a>))
= (-1)
n
n+1
n ), <0na>)
(Theorem 2.4)
(_on+1e,pn, <ona>)
<ona>)
1)n+1,*F;
jE J
where by Lemma 3.3 we assume each term in the sum is defined.
Then using the definition of intersection number
d(e, pn, en)
n n-1 ), <0na>)
L1)n+1
(2)
J
j EJa
where
(Ona) I ci)}
Ja={jEJ: '1'7.(13r.1-1)
J
Fix
jE j. Then r.
r
with
n-1
sgn (rth
P.
coordinate of a).
Let
sgn co*r.1
sn-1
n-1
k
in the simplicial subdivision of Pn.
k E K.
J
Then
n n-1
i((),
<0na>)
<Ona,>)
J
kEK.
(the sum is defined by Lemma 3.3)
28
r .+n
(-1)3
on-1
*ri(e-1(sn-1\
k
) sgn 9,,
n-1
.
k K.
E
by Lemma 3.2 and (iii) of Lemma 3.3.
Thus from (2) above
dee, _n en)
r.+n i(e,-1(sn-1) on-1)
(-1)3
-1)n+I
jE
ja
E
K.
n_i
X sgn cp
PJ
r.+1
-1)
n-I (p.n-I ),e n-1
3
.
E
)
sgn cp
3
J
3
(3)
n-1
'1'1-.1
133°
Ja
Now there are
sets
2n
depending on the half of the
Ja,
coordinate axis on which a lies, and we get (3) for each
Adding these
.1a.
equations gives
2n
r.+1
)3 nr,-1(pn.-1 en-1)
2n d(e, Pn, On)
(4)
s
n-1
3
P.
E
using (i) of Lemma 3.3 to replace sgn
by
sgn co
.
Here the
3
summation over all
does not lie in any
all possible
a. -.
jE
Ja,
J
is justified as follows: if some j
this means that
,,
10
3
.t,n(p.7-1)
n-1
on
(31.1-1,
(Ona)
E
cp
giving
for
29
n-1 n-1 n-1 ) = 0;
*r, (p.
therefore j
can be included in the sum
without altering its value. On the other hand it is impossible for
some
j
to lie in two different sets
EJ
would imply that no cp, 1 < r < n,
a
J a'
because this
n-1
was non-zero on
and by
pi
(i) of Lemma 3.3 this is not so.
n-1J
Finally since
respect to
n
:.'
implies
On
_ri
a chain approximation to
-1
*r
W
a chain approximation to
,t.r1
.
P.
on
with respect to en-1
I-'
.4-1
Tr
J
J
to get the required result.
We are almost ready to prove our computational formula for
d(e, p n on,) using the inductive degree relation of the theorem.
2
Recall that if xi = (x.11' x.,
, xi)
in
Rn,
0 < i < n,
then
n( 1,x 2,
x11 x12
xln
x21 x22
x2n
,xn)
x
nl
x
n2
nn
and
x01 x02
A
n+1
((IL,
x0), (1,x1),
,
with
on
we can use (1) to rewrite (4)
J,
jE
.._,.n.
(1,x ))
xOn
lx 11 x12
1
xn1
n2
nn
30
Lemma 3.5. Let
(j) (j)
01x
Kn
(j) >
be an n-region,
° xn
°
j=1
with
(j)
b(Kn)
tj <y1 ... y
(j) >,
t
+ 1.
Suppose
1'
3=1
t..6.n(G(y(j)),...,G(0))).
1
n+1'((1, G(x(j)),...,(1,G(x(j))))
Then
0
j=1
G:
j
j=1
Proof.
An(( 1, G(4)),..
.
(1, G(xn(j))))
j=1
G(x.
)
A
,
G(x(0))
(1)
21)iLn(G(xo(j)),
by expanding
,6n+1
along its first column. Now
AO)
(-1) <xj) ...x.
`...xn(j)>
(
b(Kn)
j=1 i=0
by definition; also
t.<y (j)... yn(j)>
b(Kn)
J
1
j=1
by hypothesis, where the y's
are just a relabelling of certain x's.
Consequently from (1) we see that
31
n (G(yi(i) ),...
,6n((1,G(x0(i))),...,(1,G(x(i))))=
j=1
, G(yn(i)))
.
j=1
Theorem 3. 6 (com utational formula for the deree Let Pn
pn Rn is continuous
,c9n)
(col,
be an n-region. Suppose
e
with e On
b(Pn). Suppose that b(Pn)
on
into (n-1)-regions
w1-1
refined relative to
sgn el;
so that it is sufficiently
1 < k < m,
vk
has been subdivided
(j) ...y (j)
t.<y
n
j
1
let b(Pn)
t. = + L
j=1
Then
/
V t.,6 n (sgn 1.n(y1(j) ), ... , sgn
1
d(el,Pn, On)
2
n
n!
L...)
n
t.
(j)
(yn ))
,
J
j--,--1
where
sgn el(y) = (sgn cai(y),
, sgn (pn(y)).
Proof. This is the same proof as that in Section 4.3 of [li],
We use induction on n.
For
n=1
this clearly reduces to the formula given in
observation 2.6 for del.1, Pl, 01).
Fix
n>1
and assume that the theorem holds in the
n-1
case. We have
111.
dee, pn, on)
zln
-1)rk+ld(.T.n-1, pn-1,0n-1) sgn
k=1
rk
k
(1)
o
rk Pn-1
32
from Theorem 3.4.
b(p1)
Let
i
Z(i) >
n-1
1
1 < k < m T,= + 1.
By
1E1k
definition this is sufficiently refined relative to
so by the
rk
induction hypothesis we have
n-1
n-1
n-1 ,O1)
Pk.
-
rk
n-1
1
2n-1
(11_1)1
(T..6 sgn
n-1
i)
,
1
sgn_1(zo.)1))
rk
1E
1
't.Ln((1,
2n-1 (11_01
sgn
ei(i))), ..., (1, sgn e1-1(yn(j))))
rk
E
1
(j))
by Lemma 3.5, where pn-1
n
1
tk,Y(j)
(1) becomes
d(e,Pn
on)
rk+1
t.
(-1)
1
2n n!
k=1 j E Jk
X pn((1, sgn
rk
sgnn-1(j))))
rk
X sgn (Pr
-k
1
2n n! k=1 jE jk
t..6n(sgn
J
sgn '1'n(y(i)))
Thus
33
(moving first column of
ing it by
rk1
columns to the right then multiply-
sgn
1n
t.Ls(sgn n (yi(j) ),
2n n1
n
, sgn
(j)
(yn
)).
j=1
Example. P = unit square in
orientation. Let
We take
R
2
with counterclockwise
(x, y) = (x, x -2y) = x', y)
say.
b(P2) = <y1y2> + <y2y3> + <y3y4> + <y4 y1> .
Example 3. 1 this gives a sufficient refinement of b(P2)
sgn2.
Then by Theorem 3. 6
d('T.2, P2, 02)1
=
1
1
-1
-II +
= -8-{-2 -Z -2 -2}
-1.
1
I
-1
-I
II +
-1
1
1
11
By
relative to
34
Historical note. It is a remarkable fact that 3. Hadamard in
[5, pp. 452-460] essentially verified Equation (3) of Theorem 3.4 for
the special case
r.
1.
He used the Kronecker integral definition of
Chapter I, simplifying the integral by ingenious manipulation. It is
easy to see how to extend his work to the general case where
E
fr.
1, 2,
.
.
,
n}.
Since Equation (3) is the crucial step needed to
J
prove the computational formula, we can say with some truth that
Theorem 3. 6 might well have been proven 60 years before it was.
35
IV. IMPARTIAL REFINEMENTS
1. Motivation
Let
Pn
be an n-region with e
continuous and satisfying
/ On
(co
,n)
Pn
Rn
b(Pn). It is shown in [11]
on
that if we keep refining the simplexes in the original representation of
b(Pn)
in such a way that their mesh (i.e. , their largest diameter)
tends to zero, then after a finite number of steps we will obtain a
sufficient refinement provided that the zeroes of the
co.
satisfy cer-
taion conditions (these conditions involve the Qn-sets defined near the
beginning of Chapter III and are similar to insisting that no
'p.
can
have an infinite number of isolated zeroes on b(Pn) ). The proof of
this statement relies on a contradiction argument and consequently
gives no estimate regarding the number of steps needed to reach a
sufficient refinement in a given situation. Moreover, as the next
example illustrates, it is in general impossible to tell from the computations involved in this algorithm when a sufficient refinement has
been reached. However once we have a sufficient refinement it is
clear from the definition that any further refinement will not 'lose'
this property, so there is no question of over-refining.
Example.
orientation. Let
P2 = unit square in
2
R2
with counterclockwise
= (x',y')
(x,y) = ((y---x)(y-4x-1), y-1)
3
9
9
say.
36
y
=0
(-1, 1/3)
I
1
= (1,1)
xt = 0
Y -0
xt = 0
=0
I
Y6
, -2/9)
I
=0
y4
Y7
3
Suppose that we initially have subdivision points
Yr Y2' Y3' y4
<y4y1> . If this is a
so that b(P2) = <y1y2> + <y2y3> + <y3y4>
sufficient refinement then by Theorem 3.6
2
d(
P
2
2
I
1
1
-1
11
)= 8{1_11
,1
1
1
1
11
-1
+
1
1
-1
-11
1
Ii
-1
11}
1
which is impossible since the degree is always an integer. Therefore
we do not have a sufficient refinement and must refine further.
37
Introduce four more subdivision points y5, y6, Y7' Y8 as
indicated on the diagram so that
b(P2 )
=
<y1y5>
+
<y5y2>
+
+
If this is a sufficient refine-
<y8y1>.
ment then
sgn co1(y1) sgn
=1{1
8
sgn
(y
1
,5
sgn
)
c2(y1)d(2,P2,02)
sgn. yo1(y8
+
+
2(y5)
5gn2(y8)
sgn (pi(y1) sgn q'2(y1)
= 1.
Since we know the positions of the zeroes of
on
we see that this is not a sufficient refinement because neither
b(P2)
x'
and y°
x°
nor
y°
is non-zero on the 1-simplex
<y8y1>. If we introduce
further subdivision points and use Theorem 3. 6 to compute a possible
value for d('1.2, P2 02)
each time, we will still get the value
until we introduce a point such as
If b(P2) = <Y1Y5>
y9
<Y5Y2>
or
y10
<Y8Y9>
1
on the diagram.
<Y9Y10>
<non>
we now have a sufficient refinement by Example 3. 1 and consequently
29 P2, 02)
0
using Theorem 3.6.
The point here is that if the zeroes of
(P1
and
co, 2
on b(P2)
were unknown, the procedure of subdividing, computing the possible
degree from Theorem 3. 6, further subdividing etc. might yield a
sequence of possible degree values in which the term
1
occurs
38
many successive times. We might thus be misled into believing that
2
,P 2
0
2
In this case the calculations contained in the pro-
1.
)
cedure would not indicate that we do not have a sufficient refinement
because it is easy to find a function
sgn
4)2
= sgn 'T.
2
2,
11
kx, y)
1
(-9x-y+1, y3
b(P2) \ {(x, y) x = 1,
subdivision points
b(P2)
relative to
For
n>2
P2
R2
for which
at all previously sampled points and yet b(P2)
sufficiently refined relative to
,
qi2
2
<
sgn
.
Thensgn )4)2
y<
with
_4(5._
For example take
sgn
on
d(4)2, P2, 02) = 1,
and the
yield a sufficient refinement of
Yi, Y2,
sgn
2
is
4)2.
it is in general a complicated matter to check that
a decomposition of b(Pn)
yields a sufficient refinement, because of
condition (iii) of the definition. We will define what we call an
impartial refinement U; this is a stronger property than that of
"sufficient refinement" but its conditions are simpler to verify. Consider the following motivating example: P3 = <a0a1a2a3> (a tetrahedron) and
diagram.
T.3
P3
R3
with
sgn cpi
as indicated in the
39
a1
Then
sgn .T.3(a2) = sgn 'T'3(a4) = (1,1,1).
Let
sgn 3(a0)
= (1, 1, -1), sgn1) = (-1, 1, 1), sgn(a3) =
(1, -1, 1).
Assume that
sgn yo.(a.)
j
sgn yo.(ak ) = A / 0 => sgn co, =
on
<ajak>,
all i, j, k.
Now
b(P3) = <a0a1a4> + <a0a4a3> + <a1a2a3> -F <a0a2
> + <a3a2a0> .
If this decomposition into 5 2-regions yields a sufficient refinement of
b(P3)
relative to
sgn
(13,
then
40
3
3
0
3
1
-
23 3!
1
1
-1
-1
1
1
11
+
-1
+
11
11
1
1
1
-1
1
1
1
1
1
1
1
1
-1
1
1
1
11
1
1
1
1
-1
-1
1
-.1
1
1
12
which is of course impossible. Thus we do not have a sufficient
refinement. The problem is that the boundary of the region
<a1a2a3>
is not sufficiently refined relative to
b(<a1a2a3>) = <a1a2>
and both
sgn
(a3)
yol
and
cp2
-sgn yoi(a1),
sgn(cpi, yoz).
For
<a2a3> + <a3a1>
+
have zeroes on
<a3a1>
since
i= 1, 2. Thus
<a1a3>
cannot be used as
a 1-region in a sufficient refinement of b(<a1a2a3>).
If <a1a2a3> is divided into <a1a2a4> + <a4a2a3> this
problem is removed because now
b(<a1a2a3> + <a4a2a3>) = <a1a2> + <a2a3> + <a3a4>
+
<a4a1>
which is sufficiently refined by Example 3. 1 .
We can now apply
41
Theorem 3. 6 to get d(1.3, P3, 03) = 0.
Here we obtain a sufficient refinement by making the decomposition of b(P3) more impartial insofar as
is no longer
<a1a3>
regarded as a single 1-simplex in the region <a1a2a3> while being
regarded as <a1a4> + <a4a2> (a sum of two 1-simplexes) in the
region (<a0a1a4>
+
<a0a4a3>).
2. Impartial Refinements: Definition and Properties
Let
Pn be an n-region. Let
el(p) / On
be continuous with
1'1
b(Pn)
Rnn
V p E b(Pn).
Definition. If n = 1, b(P1
sgn
,(pn): P
'T.11 = ((pi, (p2,
is impartially refined relative to
)
if it is sufficiently refined relative to sgn
is impartially refined relative to
If
if
sgri 'In
b(Pn)
n > 1,
has been
subdivided so that it may be written as a union of a finite number of
fan - 1
-1
in such a way that
the (n-1)-dimensional interiors of the ,n-1
(n-1) -regions
P1
Pi
are pairwise
disjoint;
at least one of the functions
non-zero on each region
(iii) each region
e _1
n-1
13.
r,n-
c913 °
9
say
1
is maximal insofar as if
with
j 7 i,
then
r,
r,;
cor
is
42
(iv) if
Sn-1.
is an (n-1)-simplex in
n-1
S.1 -1
such that
Pit
n-1
has a face of dimension n-2 lying on b(pi, ),
then this
face is also a face of at least one (n-1)-simplex S-1 lying
n4
in some
where
1.
j
Before proving the fundamental theorem which states that all
impartial refinements are sufficient refinements we obtain two pre-
liminary results.
Lemma 4.1. The boundary of an n-dimensional polyhedron is an
(n-1)-dimensional polyhedron.
Proof.
and
T
-n 1
show that
Let Kn
be an n-dimensional polyhedron.
Let
S
n-1
be (n-1)-simplexes in the (n-1)-chain b(Kn). We must
Sn-1
is either the empty set or a common face
Tn-1
of the simplexes.
Suppose
Tn-1 fi
Sn-1
boundary of an n-simplex
comes from a
Tn
in
ep.
(say)
Sn
Let
Kn.
Sn-1
Now
in
Sn
Kn;
is part of the
likewise
Tn-1
<s s 1. . . sn>,
0
Tn = <t0t1... tn> . Without loss of generality take
5
If Sn
because
Kn
n-1 = <s s
12
Tr
sn>,
T
n-1 = <tot1...tn_i> .
it is easy to finish the proof, so assume not. Then
is a polyhedron
is a common r-dimensional
43
face, with 0 < r < n. Without loss of generality take
r-
SnT
P
X.s.,
Li X. > 0, 2 X. = 1
P
i=0
i=0
i=0
i=0
q:q=
=
using observation 2.1. By linear independence and the uniqueness of
the extreme points in Sn
{
:
Os.
s. = t.
i < r} = ft.
<
:
n
Tn
we must have
0 < i < r}. We may assume in fact that
0< i< r.
for
Thus
Sn-1
n
/
(Snn Tn)
Tn-1
=
p:p=
sn -1
X. s
,
X. > 0,
(sn-lr Tn. -1) c sn-1
(snr.\ Tn)
=> r > 1).
Likewise
T' n (S n
T
as
r < n-1.
Tn)
:q=
> 0,
1
44
Recall that
1
1
Tn-1
sn-1
0 < i < r°
for
s. = t.
sn
rmTn-1)
(sn
thus
T)n
(snrm Tn)]
= .,.
x: x=
X.isi, X.i >
and this is a common face of Sn-1
1 , 2, .
,
n-1
b(11)
n},
Without loss of generality take
(if m
m>1
is
is impartially
sgnn-1
refined relative to
Proof.
as required.
sgnn with notation as in the defi-
impartially refined relative to
E
Tn-1
and
and suppose that b(Pn)
Theorem 4. 2. Let n > 1
nition. Then for any
(snrm Tri)]
then
1
b(Pn
-1
)
1
r7= 1, it = 1,
and
satisfies the "impartially refined"
definition vacuously).
Let
{S1.1-1}.I
LE
be the (finite) set of (n-1)-simplexes in
having an (n -2)-dimensional face lying on b( p1
-1)
Let
).
p
be the corresponding simplexes given by part (iv) of the definition
(i.---' j(i)
sn - 1
i
S..
ij
is not in general a function).
as lying in
n-1
(j --' j'
say
For each
Consider
n-1
S.
1
(Th
E
J
we take
is a many-one map).
133
for the (n-2)-simplex
j
Sn-1
j(i)
.
Write
45
S2 = {S11-2
ij
i
:
E
I,
j
J, Sn -1 C 13n
j."
E
1
r
= 2}.
By Lemma 4.1 the intersection of any two distinct (n-2)n
S-2
ij
sim plexes
all
lie in
Sn..-2
is either a common face or the empty set since
n-1
b(131
We can therefore write
).
as a dis-
S2
joint union of (n-2)-regions in a unique way (by taking connected components).
Continue thus: consider, for
S
k
{Sn -2
and write Sk \ (S
regions.
:
I, j E 3,
iE
...
S
3 < k < n,
-1
k-1 )
Taking all these (n-2)-regions
a decomposition of
n-1
n-1
13(31
)
k},
as a disjoint union of (n-2)n-2
Pi
-
1
satisfy r., / 1. We claim
that in fact this gives us an impartial refinement of
tive to
(say) gives us
since by part (iv) of the definition all
/ 1) which intersect
PJ
r.
n-1
b(131
)
rela-
sgn
Clearly (i) and (iii) of the definition are satisfied.
-1
n-1
Recall that for any i S..n-2
and consequently
13
n-1
Sj(i)
,
By maximality of
r., / 1, and so a
r / 0 on sn-2.
1
ij
J
ji
n., -1
component of T 1
is non-zero on each 1P-2, i.e. , (ii) of the
q)
13
definition is satisfied (this argument also takes care of the exceptional
case n = 2).
46
For condition (iv) (when
n > 2)
,n-2 be any (n-2)'31
let
I6fl-2 such that an (n-3) -diensional
simplex in .1
rri
face of S r1-2 lies
n-2
n-1
in b(13
131
). Now
is a region and b(b(131n-1 )) = 0 (Remark
n-1
2.2) so any (n-3)-simplex in _nb(2) must, since n-2
I
C b(131
-2
also be an (n-3)-simplex in
b(f3nt°
),
sn -2
sequently a face of an (n-2)-simplex
t'
some
),
It is con-
1.
in
The principal result is the following.
Theorem 4.3. If b(Pn)
is impartially refined relative to
then it is also sufficiently refined relative to
sgn
Proof. Use induction on
For
n
1
n.
the assertion is trivial. Fix n > 1 and assume
pn-1
that the theorem holds for any
Suppose that
using the regions
sgnn.
is impartially refined relative to
b(Pn)
n-1
n -1
p1
,
and
Pin
.
sgn
We only have to show that (iii) of
.
the "sufficiently refined" definition holds, the rest being automatically
satisfied.
For any
4. 2
b(1311
1)
iE
{1, 2,
.
,
cpr.
/0
n-1
on
.
is impartially refined relative to
is sufficiently refined relative to
hypothesis, and we are done.
sgn n
1-
and by Theorem
nn-1)
sgn 1i
by the inductive
47
V. ALGORITHM FOR COMPUTATION OF THE DEGREE
USING IMPARTIAL REFINEMENTS
1. Introduction and Description of Algorithm
Let
with
Pn be an n-region. Let
on
.T,n
(ion)
Pn
Rn
b(Pn). We now describe an algorithm for corn-
on
d(e, P,,,n,), assuming that we can major ize the modulus
puting
ell
of continuity of
by a known function
which is
Q(5)
b(Pn)
ell
0(5). This happens for example if we know that
satisfies
b(Pn)
a Lipschitz condition of order
a > O.
The algorithm constructs an impartial refinement of b(Pn)
then terminates. By Theorem 4.3 this impartial refinement is also
p
d('1,n
on)
a sufficient refinement. We can thus compute
,
with
assurance using the formula of Theorem 3. 6.
Take
n>1
as the computation of d(t.1, Pl, 01)
Let w(.
) be the modulus of continuity of
i
i = 1, 2,
the
i
.
,n. Suppose that
LOS) <
1
are known functions and
2.(6)
1
<n
1 i<,
C°i
for
S2.(5)
is
is trivial.
0(6)
b(Pn
for each
where
i.
1
For any
i
and any
xE
b(Pn)
with
cp. (X)
0,
choose
L
5.>
0
such that
Q.(5.)
<
1
1
1 coi(x)1
Let
norm in Rn. Then for any y E b(P)
have
1.11
denote the Euclidean
such that
ll x-ylj <_
,
we
48
(x)
icoi(y)I
I co,(x).(y)1
6.
( x> )1
1
1
ca.(x)1
> 0.
Let
pE
b(Pn).
Choose
5(p) =
max
1<i<n I
are chosen as large as possible such that
pyo.( ) = 0
take
6(p) >
S-2.(5.)
<
1
1
cP.(p)I
1
6.
(if
6. = 0). For at least one
min 6:
<i<n
mme(x)1 >
m
IC9i(P)I
so
where the
6.,
xEb(P--)
where each
6! > 0
possible such that 0.(6!) < m. Thus
is chosen as large as
say, where
(5(p)
c
is
positive and independent of p.
From the preceding paragraph we see that if
satisfies
pendent of
II 13-0 < 6(P),
then
B
for some
b(Pn)
i
inde-
y.
Thus:, given any point
ball
co.(y)i > 0
yE
b(Pn),
p
of radius at least
such that some component of
Definition.
(a fixed positive constant)
c
VI
we can surround it by a
is non-zero on B rm b(Pn).
A simplex is acceptable if at least one component of
is known to be non-zero thereon.
49
Remark 5.1.
Any subdivision of an acceptable simplex yields
acceptable simplexes.
Where necessary our algorithm will subdivide simplexes in
b(Pr)
until all simplexes in b(Pn)
original representation of b(Pn)
are acceptable. Also, the
is a polyhedron by Lemma 4. 1 and
the algorithm will preserve this property.
An edge of a simplex is a one-dimensional face of
Definition.
that simplex.
Let
Sn1
n,S1
.
1
b(Pn)
with
b(Pn)
be a list of the (oriented) simplexes in
At first all these simplexes are unac-
=
j=1
ceptable. Let PI'
'13
be a list of the vertices of the S.
Algorithm.
Set
If
i=1
m,
i
and go to step 3.
terminate. Otherwise replace
i
by
i+1
and
j
by
j+1
and
continue.
Compute
Set
If
j
j=/
continue.
O(p.)
1
as described above.
and go to step 6.
go to step 2. Otherwise replace
6. If
n-1
is acceptable or if pi is not a vertex of
S.
50
SiS
to to step 5. Otherwise continue.
n-
7. Without loss of generality assume that
<PiY2*.' Yn>
and go to step 9.
If k = n list sn-1 as acceptable and to to step 5. OtherSet
k -= 2
wise replace
Compute
1
If
by
k
the length of the edge
k,
5(P.),
If lk > 8(p.),
Lk>
<PiY
and continue.
k+1
<PiYk>
go to step 8.
let pm+1
be the point lying on
min{6(pi), 12 id from
at a distance
In
step 2 replace m by m+1. Replace the oriented
simplex Sn - 1
by the two oriented simplexes
J
S.
"iY2'
Yk - 1Pm± lYk+ 1
°Y >
and
n-1
4.1
=
n
<prn+1 Y2
In step 5 replace
f
by
>
1 +1 .
In exactly the same way
replace every other oriented simplex having <PiYk> as
an edge by two new oriented simplexes whose sum" is the
original oriented simplex, increasing i
in step 5 to
each time a new simplex is created. Got to step 8.
f+1
51
Summary of Algorithm. Cons ider the vertices
turn. For each p.
having
pi
consider in turn those unacceptable simplexes
as a vertex. For each such simplex
edges emanating from
exceeds
p.
Ei(pi)
pi , p2,
consider the
Sn-1
Subdivide those edges whose length
so that the new length of each edge of
as an endpoint is at most
Then
.5(p.).
S'
S.
S'1
S
having
must be acceptable.
Here we also subdivide all other simplexes sharing the above edges
n-1
with
S.
in order to preserve the polyhedral property of b(Pn).
2. Proof of Convergence of Algorithm
Observation 5. 2.
By virtue of step 6, it is sufficient to prove
that after a finite number of iterations all simplexes are acceptable.
For this proof we use the following two lemmas.
Lemma 5.3.
Let
c
and
k
be positive constants. Write
AB' etc. Given a triangle
or equal to
k, AB > c,
ABC
and
D
AB < 2c,
for 'length of
as shown with diameter less than
lying on AB
constraints:
(a) if
'AB
then AD =
1
AB;
subject to the
52
(b) if
AB > 2c,
c < AD <
then
1
AB.
Then
c2
CD2 < k2 - ---4
Proof. We break up the proof into three cases:
AB < 2c
and AB > BC
AB < 2c
and
AB < BC
AB > 2c.
Case (i). Now BC2 = AC2
BC
2
- AB 2
=
AC2
+
AB2 - ZAC AB cos a
- 2AC AB cos a
so
and
BC < AB <=>AC2 - ZAC AB cos a < 0
- 2 AB cos a < 0
<=> cos a > AC
AB
Since AB < 2c,
CD
we have by hypothesis AD = -1-2 AB. Thus
22,
AC+ AD
= AC2 +
From above AB > BC--=> cos a
1
7
AB
- 2AC AD cos a
2
AC
> 2 AB
AC AB cos a
so from (1)
(1)
53
CD2 < AC2 +
1
AC2
+
=
Ca____Aeiiil.
If
2
AB > AC,
AB2
AC AB 2AC
AB
1AB2
4
then by symmetry we can obtain
the same result as in Case (i). Therefore take AB < AC. By
symmetry we can without loss of generality take
BC2
=
- 2AC AB cos a
AC2 + AB2
BC2 - AC2
AB2
=
BC > AC. Then
(2)
- 2AC AB cos a
SO
AB2
- 2AC AB cos a > 0
AB - 2AC cos a >
cos a <
AB
AC
(3)
Combining (2) with (1) gives
CD
' BC2 - 4AB2 + AC AB cos a
3
<
BC2
=
BC2 -
< k2 -
-
c
+ AC AB
4
1
AB2
AC
54
Case (iii). Now
c < AC <
CD2
For the given triangle
d(CD
d(AD)
=
-1
AB.
- 2AC AD cos a
AC2 f- AD
(4)
CD depends only on AD. Differentiating,
- 2 AD - 2 AC cos a
for
=0
this gives a minimum value for
cos a
AD
AC
CD.
Thus
absolute maximum value wither at AD = c
i, e. , CD J AB;
achieves its
CD
or at
AD =
1
AB. If
this maximum is at AD =1AB we are done by Cases (i) and (ii)
2
(note that in these cases the fact that AB < 2c
deduce that AD =
1
AB). We therefore suppose that the absolute
maximum occurs when AD = c,
AC2
(AD = c
+
c2
was used only to
- 2 AC c cos a > AC2
on left-hand side,
c
(from (4)) that
i. e.
1
-4-
- AC AB cos a
AB2
1
AD =2 AB on right-hand side) so
21
AB
4
2
> (AC cos a)(2c-AB)
(c+ 12AB)(c- AB)
> ("LAC cos a)(c-2AB)
2
1
=>
c+
1
AB < 2AC cos a,
1
since
c
1
AB < 0
55
1
c + 7AB
< cos a.
2 AC
Now with AD = c
CD2
=
AC2 + c
< AC2
2
- 2AC c cos a
2
+c - 2AC c
1
c
2
AB
2 AC
1
=AC 2 - 2c
AB
< k2
-
1
c2
Finally, combining the results of the three cases yields the
conclusion of the lemma (use the fact that k > AB > c).
Definition.
When dealing with a vertex
p.
in the algorithm
(i.e. running through steps 3 -9), a new edge is an edge formed by
subdivision which is not part of any edge previously present.
Example.
NNN.
Suppose that when dealing with
D,
then join
are not.
D
to
C.
Now CD
A
we divide the edge AB at
is a new edge;
AD and
DB
56
the algorithm
When dealing with a fixed pi
Observation 5.4.
does not increase the number of simplexes having
after considering all
P.
of which
n-1
SiConsequen-ly,
as a vertex.
is a vertex,
pi
step 6 will always return us to step 5 and so we eventually get
pi by acceptable
returned to step 2. We have now "surrounded"
simplexes, i.e. , every simplex of which pi is a vertex is now
acceptable, by virtue of steps 7, 8, and 9. This property will be
retained for the remainder of the algorithm (recall Remark 5. 1).
Notation. Let
simplexes
Sin
-1
,
S
denote the largest diameter of the original
0
,7
-1
.
Let
Sr, r
1,
denote the largest
diameter of the unacceptable simplexes present after we have surrounded
pi, p2,
'1r by acceptable simplexes (we take
Sr =
0
iff all simplexes are acceptable).
Lemma 5.5. When dealing with any vertex
pi
in the
algorithm, any new edge constructed in an unacceptable simplex lies
in a triangle such that the edge and triangle satisfy the properties
ascribed to CD and ABC
k = S1
1.
and
c
Proof. If
(P.j
respectively in Lemma 5.3, with
that given in Section V. L
(p.p.,)
is an edge which we divide at.
PJ
Pi ,P i° ) then the new edges resulting from this division are of
the form
kj)
where
k
(see step 9(b)). Any such edge
57
will lie in the triangle
(pipi,pk)
which is a face of an unacceptable
simplex by hypothesis. Here we have a correspondence
P.,i ---4" B,
-)k ' C,
A,
P.
to Lemma 5.3.
p.
3
We must have length
(PiPi0 > c
as otherwise this edge would
not have been divided (see Section V. 1 and step 9 of the algorithm).
The diameter of the triangle
of
(pipi ,pk)
by definition
is at most S.1
1-
S1.
i- Finally, the restrictions on the length of (p.p.) follow
from step 9(b) of the algorithm and the fact that
6,(pi) > c.
Theorem 5. 6. After a finite number of iterations of the
algorithm all simplexes are acceptable.
Proof.
Clearly
0 < S r+1 < S r
for r
0, 1, 2,
.
.
.
.
We
show that for some R, SR < c; from this point onwards step 9(a)
of the algorithm will always return us to step 8. As a result all
simplexes will eventually be acceptable (recall Observation 5.4).
Consider the original vertices Pi, P2,
pm. After we have
dealt with these vertices, any part of an original edge which now lies
in an unacceptable simplex has either resulted from a bisection or
from a subdivision which removed a length of at least
length is at most max{
ceptable simplexes (i. e.
2
,
0
,
S0
c.
Thus its
-c}. Regarding other edges of unac-
those new edges constructed in the
algorithm), by Lemma 5. 5 such edges arise precisely as CD does
in Lemma 5.3, and it follows that the length of any such edge is at
58
most
(S
21
2
1
c
4
0
2
Combining these results gives
< max{
m
S2
We now have a list
4
m
,
0 (S0
-c)2 , SO -
1
pm+i,
12
, pm,
2
2
,(Sm -c),s m
< max{ T s,m
C2}
.
, Pm, Pm+1,
1)1,
Apply the same argument to
S2
2
S2
Pmf
say of vertices.
to get
1
2
Continue thus. It is clear that the sequence
} .
S ,S ,S
,
'
must eventually be less than or equal to c. By the opening remarks
the proof is complete.
After a finite number of iterations the algorithm
Corollary 5. 7.
terminates.
Proof. Use Observation 5. 2.
Observation 5. 8.
sentation of b(Pn),
some
Sn-1
1
in
Since
for the original repre-
every (n-2)-s implex lying in
b(Pn)
other (n-1)-simplex
b(b(Pn)) = 0
2
must also lie in
in
b(Sn2 1)
b(Sni 1)
for
for at least one
b(Pn). If in the algorithm this
(n-2)-simplex is subdivided, it is subdivided in all (n-1)-simplexes
containing it and from step 9(b) we see that new (n-2)-simplexes
appear in exactly as many boundaries of (n-1)-simplexes as their
59
ancestors did, and with exactly the same orientation. Thus
b(b1(Pn)) = 0,
where
b 1'
(13n)
is the new representation of b(P)
obtained 13\ means of the algorithm.
Computation of the Degree.
Since
Pn
is an n-region, by
Lemma 4.1 the original representation of b(Pn) is a polyhedron.
The subdivisions of the algorithm are such that it preserves this
property of b(Pn). We will thus obtain a polyhedral decomposition
of
b(Pn)
into acceptable simplexes. This gives an impartial
refinement of
b(Pn)
relative to
sgnn: take maximal connected
collections of simplexes to be (n-1)-regions, avoiding overlapping,
and use the fact that
b(bi(Pn)) = 0
(iv) of the definition.
The degree
(Observation 5. 8) to check part
d(
puted using the formula of Theorem 3. 6.
Pn, On)
can then be corn-
60
SIMPLIFICATION OF THE COMPUTATIONS
VI.
In this chapter a proceduro is g:ven which replaces the computation of the
n x n determinan_s in Theorem 3. 6 by a "scanning?'
of the matrices associated with these determinants, that is a counting
of
in certain positions within the matrices.
+
As usual let
(col,
cot.)
Pn
pn
be an n-region with
Rn
The next lemma and its
continuous.
two corollaries will be used later when we need to choose simplexes
el have certain signs.
on which certain components of
is impartially refined
Lemma 6. 1. Suppose that b(Pn)
relative to
sgnn
in such a way that it is a polyhedron. Then if
any coordinate function
vertices of any simplex
sgn (p. = A
d(
has the same sign A (A = + 1)
(p'
i
_n -1
S
i
in
we may assume that
b(Pn),
without altering the value of
on all of Sn-1
Pn, On
Proof. Suppose that we do not have
n-1
S.
i
S7-1'
J
Now
S.
l_n-
lies in some region
sgn (pi = A
n -1
6 ,;
.
i
where of course
and so
on all of
(pr.,
i
0
on
i
i.
.r
3
By continuity
S1.
at all
p
0
on some open neighborhood M of
M so small that it does not contain any vertex (of any
simplex in the impartial refinement of b(Pn)) at which sgn (pi = 0
Or
-A
(this can be done because of the assumed polyhedral property
of the impartial refinement). Choose another open set
S' CNCNCM. Now homotopically deform
S.
nary so that inside
it has constant sign
N
(Pi
such that
N
where neces-
while outside M
A,
it is unchanged.
Since
yo
r.,
0
on
[z, Theorem 6. 4, p. 31]
on
M,
ionn,
,0n
d(
the homotopy and by
is unaffected. Note that the
)
same impartial refinement can be retained since the sign of
cp.
is
unchanged at all points considered in the definition.
Corollary 6. 2.
Under the same hypotheses as those of Lemma
6. 1, we may assume that
n-1
S.
impartial refinement such that
lies in a region
n-1
of the
r., =i.
Proof. Since the refinement bl(Pn) of b(Pn) is polyhedral
and refinements are always assumed to preserve orientations,
b(bi(Pn)) = 0
Form
(see Observation 5.8).
n-1
p. ,
as the maximal connected anion containing Sn-1
of those (n1)s implexes in bl(Pn) on which sgn cp. A. Then
after redefining some other (n-1)-regions to eliminate overlapping if
necessary,
b 1 (Pn)
is still an impartial refinement because it is a
polyhedron and because b(b (Pn)) = 0.
Notation. If
B. = (bi 1, biz, .
b.
= 1, 2,
,q
are
62
vectors, then
(q x q matrix).
Of course
,B
)) = Acl(B
det(Mq(B1'
in the notation of
)
Section 1. 2.
Let
(k)
tk<yi
...yn(k)>
be a sufficient refinement
where
Its associated matrices are
b(Pn) =
k= 1
relative to
sgn 'T.11,
tk = + 1.
n,y(k),,
`Icjvi
n-1
(
1
,n,
(k)
k
k" ))
gn
)
1
2
'.." f
(1)
be the set of regions13 .n-1 in the
J
n}
given sufficient refinement for which the associated coordinate funcLet
tion is
(Fir
{p.
:JE
with
sgn r
n-1
n 13.
=A
here
;
r
E
{1, 2,
...
andAn E {-1, 1}.
Take
pn -I.
=
(k)
/1
<Y1
b(p1)
(k),
for each
(k)
(-1) i+1<Y1
kEIK, i=1
^(k)
SO
jE
(k)
>
63
with associated matrices
i+1 n-1
(-1)M
n-1 (y,(k) ),
rn
a- 1
, sgn
(k)
(yi
, sgn
),
n-1
(k)
(yn ))
1(sgn
= 1,2,...,n,
K., jE
(2)
JAn
Corollary 6. 3. Suppose that the above sufficient refinement is
actually a polyhedral impartial refinement. Then the matrices
Mn-1
of (2) can be obtained as follows: choose from (1) those Mn
rnth column consists entirely of An's, then delete the
ith row and rnth column from these matrices, where i ranges
whose
over the values
Proof.
mn(sgn
,n.
1, 2,
Consider the matrices
(k)
(y1 )'
(k)..
"
sgn
(k)
(k)
pn-1
yn
(yn
(3)
JAn
Each matrix in (3) is a matrix in (1) whose
consists entirely of
An s ;
rnth column
by Corollary 6. 2 we may assume that
all such matrices in (1) are actually matrices in (3). Thus choosing
rnth column consists entirely of
is a means of listing the matrices in (3). Now each matrix in
from (1) those matrices whose
An's
(2) is obtained by deleting the
ith
row and
rnth
column from a
64
matrix in (3). This completes the proof.
Using the above notation, consider the matrices
tkMn(sgn 11((k))
.
, sgn 'Tn(y(k))), k = 1, 2,
(1)
,
Recall that their associated determinants An were used to compute
d(e, p n on,) in Theorem 3. 6. In the procedure below a signature is
assigned to certain of these matrices, then the signatures are added;
it is shown later that this yields dce,Pn, On).
Procedure.
V
i.
Let
{6,
:
I<i<
Let {r.
1<
C {1, 2,
.
with r, <
i
i < n} C {-1,1}.
Choose from (1) all tkMn whose rnth column consists
r +1
(-1)
n
entirely of 6n Ts. Assign a temporary signature tkAn
to
each such matrix; assign the signature zero to every other
Delete the
Mn.
r th column from each chosen matrix to form an
array. In the rn-1th column of this array pick all
combinations of n-1 rows having n-1 6 n-1 s as entries, if any
n x (n-1)
such combination exists (if not, discard the matrix, i. e. , assign it the
signature zero) (if the matrix is not discarded there will be either
1
or
n
such combinations).
If there is one such combination, suppose that the
is the unique row with entry
0
or
-6 n-1
.
q
n-1 th
row
Delete this row to give
65
an
tk
(n-1) x (n-1)
n n-1
n
(-1)r+1+rn-1+q
If there are
tains
in the
matrix and assign a temporary signature
nn-l's)
n-1
to the matrix.
such combinations (i. e. ,
n
rn- 1th
then deleting each row in turn gives
rn-lth column. Do this, obtaining
n
column conn-1
Ln-l's
(n-1) x (n -1)
matrices with associated temporary signatures
n(-1)rn+1+r
tk nA n- 1
deleted (so
qn -1
,+43
'n -1
where the
runs through the values
row was the one
1, 2, ... ,n).
matrix just as each original
(n-1) x (n-1)
Now deal with each
qn-1th
n x n matrix was dealt with after assigning the first temporary
throughout. Continue reducing
signature, replacing
n
until left with
matrices. At this stage the sum of the tem-
1x1
n-1
by
porary signatures of those
1x1
matrices whose ancestor was a
particular Mn is taken as the signature of that Mn. Finally add
all the signatures of the Mn.
We now recall some equations which will be used to prove that
the procedure computes d( pn, en).
Again suppose that
(k)
<1().
y1
.. yn
> is sufficiently
b(Pn
k=1
refined relative to
of Theorem 3.4:
sgn
.T.n,
where
tk
+ 1.
Recall Equation (3)
66
d(el,
On)
rj+1ite-1(Rn-1,i,on-1,) sgn
*r.
=
n-1
E Ja
P.
J
J
Using (1) of that theorem and Lemma 3.3 this can be written as
(-1)rj
d(e, Pn, On) =
r.
) sgn co
j
n-1
r
E Ja
PJ
Changing the notation a little we have
r +1
(-on
tn-1, en-1)
d(e-1,
n) =
rn
jEJ
An
r +1
= (-1) n
_n-1
E
(1)
rn
An
J,6
)
-1
where the summation is over all regions
associated coordinate function
here
rn
E
{1, 2,
For each
.
j
..
E
,
and
n}
is(pr n
{-1,
Ln E
with
1}
sgn
(k)
1
1
k E K.
(pr
b(Pn)
1
n
n-1
whose
n
.
PJ
are arbitrary but fixed.
take
Jzs
in
-
67
SO
(k)
b(pni
<yl
i
A(k)
°
yi
°
°
(k)
>.
yn
=kEK.
Consider the associated matrices
A
sgn n-1
(-1)i+1Mn-1(sgn n-1 (y'(k) ),
rn
n
n, kE
i= 1, 2,
pl
(k))),
y.(k)'),...,sgn T_n-1,
r (yn
1
n
(2)
K.,An
Observation 6.4. Consider the special case n 1. Let
m-1
with associated
<x.x.
> so that b(P1) = <x > - <x >
m
1 1+1
0
i=0
matrices1(x),1(x
).
m
0
1x1
Of necessity r1 =
1.
As usual
it is assigned the signature
6
,r1+1
1
(-1)
the signature zero. If.11(x 0 ) = 6
r141
(-1).61
-6 1
;
Ai E {-I, 1}.
=
;
If
'1(x)
=A
if not, it is assigned
it is assigned the signature
1
if not, it is assigned the signature zero.
(-1)
Finally the signatures are added.
Theorem 6.5.
continuous. Suppose
relative to
above.
sgn
Let Pn be an n- region with
b(Pn)
P
Rn
has a polyhedral impartial refinement
so that it is sufficiently refined with notation as
Then the procedure computes d(el,Pn
68
Proof. Use induction on
for the case n = 1
By inspection the theorem holds
n.
(see Observations 6. 4 and 2. 6). Now fix
and assume that the theorem is true for the n-1
n>1
case. Then the
procedure deals with the matrices
(k)
, sgn
tkMn(sgn ei( y(k)1),
yn
)),
1, 2,
k
. .
.
,1
(3)
By Theorem 4. 2 and Lemma 4.1 we are justified in applying
the inductive hypothesis to calculate
)
r
Thus,
E
letting j vary over
choose from (2) all those matrices
J.6n,
(-1) i+ln-1
whose
rn-1th column consists entirely of
and assign a temporary signature
(_1)i+1,6
i-nrn_1+1
n-1'
'
n -1
(-1)
irn-1+
to each. Apply the procedure from this point to compute the signature
of each
(-1)i+1Mn-1,
To now calculate
ture of each
then add these signatures.
d(, Pn, On),
(-1)i+1Mn-1
r
by (-1)n
use (1): multiply the signa-
+.1
n
and then add these values.
By inspection, using Corollary 6.3, this method for computing
d(e,pn, on) is seen to coincide with the procedural method
applied
to the matrices in (3).
69
Corollary 6. 6.
matrix
nxn
Any
(n > 1)
having two
or of
columns each of which consists entirely of +1's
-1's
will
be assigned the signature zero.
Proof.
sth
column of
Take
rth
Suppose that the
G
Pn
s
Is,
column consists of
.6.
and
r
where,r
{-1, 1}
}
S
the
r °s,
s.
to be a n-simplex with a continuous function
such that
on all of P
sgncos = As
and
rA
on
n
vertices of P
sgn cor =
-Ar
Then b(Pn)
sgn
on the remaining vertex.
is a polyhedron and is impartially refined relative to
taking all of b(P1)
as an (n-1)-region on which
We may therefore apply Theorem 6. 5 with rn
r
and
cos
/ 0.
An = Ar.
In the procedure only one matrix will be chosen because for only
one simplex in b(Pn)
does
sgn cor
matrix has an rth column
ofAr 's
at all vertices. This
and an
sth
column of
As
the other entries (if any) are arbitrary. Its signature must be
+ d(ea, pn, 0n)
by Theorem 6.5; however this is zero by Remark
fl/ 0 on pn.
2. 7(i) because q / 0 on pn
70
Theorem 6. 7.
Let Pn be an n-regio.n with
.t.n
:
Pn
Rn
is sufficiently refined relative to
Then the procedure computes d(e, p n en).
continuous.
sgn
Suppose
b(Pn)
Proof. The proof is almost identical to that of Theorem 6. 5.
No alteration is needed for the
n=1
case. For n > 1
the
same basic argument holds, but some changes are needed in the
results quoted from elsewhere.
Instead of quoting Theorem 4. 2 and Lemma 4.1 we appeal to
part (iii) of the definition of a sufficient refinement.
We cannot prove a version of Corollary 6. 3 for arbitrary
sufficient refinements. It guaranteed that any matrix Mn whose
rnth
n-1 ,
13.
corresponded to a simplex in some
column consisted of An s
j
(notation of Theorem 6.5), and so should be chosen
E
JAn
at the beginning of the procedure. However any such matrix corres
pond ing to a simplex in a
Rn -
1
where
columns each consisting of +1's
or of
k
J
-1's
will then have two
because of the suf-
ficient refinement, and by Corollary 6. 6 may be included among the
chosen matrices because it will be assigned the signature zero in any
case. This observation should be used in place of Corollary 6,3 in
the proof.
Corollary 6. 6 indicates that there is superfluous computation
involved in the original procedure.
We now give a modified
71
procedure designed to circumvent this. This modified procedure will
also explicitly discard matrices having a column consisting entirely
of zeroes since such matrices are clearly discarded at some stage.
Modified Procedure.
relative to sgn
with
b(P) is sufficiently refined
Suppose
b(P)
t <y (ik)...
(k)
yn >,
tk
+ 1.
Consider the associated matrices
mn(sgn
"1 )
,,n(_(k),
k = 1, 2, ...
))
Yn
(1)
A signature is assigned to certain of these matrices, then the signatures are added; by what has gone before this will give us
d(e, pn, on).
Let
{A.
{ri
1<i<
with r, <
} C {1, 2,
i
V i. Let
1 < i < n} C. {-1,1}.
Choose from (1) all matrices
Mn whose
rnth column
consists entirely of An's. Assign a temporary signature
rn+1
(-1)
to each such matrix.
tk An
tk
If any other column of the matrix is constant (i.e. , all its
entries have the same value) assign the signature zero to the matrix,
i.e. , discard Mn. Otherwise delete the
rnth column to form an
n x (n-1) array. In the rn- 1th column of this array pick the
combination of n-1 rows having n-1 An - 1 's as entries, if such a
72
combination exists (if not, discard Mn) (there will be at most one
such combination otherwise
Mn would have been discarded already).
Suppose the qth row is the unique row with entry
-An-1 .
Delete this row to give an
0
or
matrix and assign
(n-1) x (n-1)
- 1+qn - 1
to tk Mn.
a temporary signature tk znAn-1 (-1) +rn l+rn
Deal with this (n-1) x (n-1) matrix just as we dealt with the
nxn
situation after choosing the matrices Mn and assigning the
first temporary signature, replacing
n
tinue reducing until left with a
matrix. The temporary signa-
1x1
ture at this stage is the signature of
signature (tkMn)
by
n-1
thoughout. Con-
tkMn,i.e.
rn+1+rn-1+cin-1+.°.+TYcli
(-1)
tkLxn,An-1.. . A1
Finally add the signatures of the chosen Mn 's.
Example. We recompute d(3, P3, 03) for the tetrahedron
example of Section IV .1.
This is done for two different sets
{(ri, Ai), (r2,A2), (r3,G3)}.
In the notation of Section IV. 1, a sufficient refinement of
b(P3)
is given by
b(P3 )
<a0a1a4>
+
+
<a0a4a3> + <a1a2a4>
<a0a2a1 > + <a3a2a0> .
+
<a4 a 2 a 3>
73
The associated matrices are
-1 \
/1
1
1
1
(1
1
1
1
1
1/
\1
1
1
1
1
1
/1
1
-1
/1
-1
1
1
1
1
1
1
ci
1
1
-1
1
\-1
1
1
1
-1
(-1
(1
1
Case (i).
(-1
1
1
r3 = 1,
A3
-1
= -1, r2,r1' A2' Al
1
arbitrary. Begin
by choosing those matrices whose first column consists entirely of
-1's. Since there are none, all matrices are discarded and
3, P3, 03) = 0.
Case (ii)
r3
=
r2
=
r1 = 1°
A3 =
A2
=
1,
= -1. Choose all
matrices whose first column consists entirely of +1's;
temporary signature
(+1)( -1)1+1
=
assign a
to each, writing it in front of
1
the matrix:
1
(+1)
(1
1
-1
1
1
1
1
(+1
1\
1
1
-1
1
,
(+1)
1
-1
1
1
1
1
\1
1
Discard the second matrix because its third column is also constant.
Delete the first column from the remaining two matrices:
74
1
1/
-1
1
assign
Choose the arrays whose first column contains two +1's;
the temporary signature
(+1)(+1)(-1)1+i
row is the one containing the
delete the
-1;
(1
-1\
1
1)
(+1)
to each where the
ith
ith
row:
(+1)
,
Delete the first column from each matrix:
iN
(+1)
,
\k
(+1)
1/
Choose the arrays whose first column contains one
temporary signature
(+1)(-1)(-1)1+i
is the one containing the
1x1
assign the
to each where the ith row
delete the
+1;
-1;
ith
row. Since this gives
matrices the temporary signature becomes the signature.
Thus the signatures here are +1, -1.
Finally d( 3, P3, 03) = sum of signatures
=
1
= 0.
-
1
75
Remark 6. 8. In the modified procedure it is clear that the
-1, 0, or +1.
signature of any matrix is
Corollary 6. 9 (to Theorem 6. 7).
n > 1,
with
.1'n
Pn
:
Rn
sufficiently refined relative to
Then
b(Pn)
Let
Pn
ba an n-region,
continuous. Suppose that
b(Pn)
is
Let m
sgn
is subdivided into at least m n 2n-1
simplexes.
Proof. By Theorem 6. 7 the modified procedure gives
d(e, pn, on,) for any choice of {r. 1 < i < n} C {1, 2,
,
n}
1
satisfying
ri <
V i,
i
There are
2n
and
{A
1<
i < n}
{-1, 1}.
possibilities for the (ordered) pair
(rn, An).
Now any matrix chosen for two distinct pairs will hrve signature zero
by the modified procedure (two columns will be constant). Conse-
quently we can assume that there is no overlap in choice; count the
minimum number of matrices needed to give
for a fixed pair
Fix
(rn, An
(rn, An).
)
Id(el,Pn,°11)1
then multiply this number by
Consider the matrices
Mn
2n.
chosen for this
pair. Delete the
rnth column from each Mn to leave an
n x n-1 array. Choose rn--1 satisfying 1 < r1
n- < n-1. Then
the matrices chosen at this stage for the pair (rn-1, 1) must not
(rn-1, -1) if n > 2 because
overlapping implies the existence of an n x (n-1) array with a
overlap with those chosen for the pair
column containing
(n-1) +lis
and
1
-1 Is.
Thus the minimum
76
number of matrices needed for the pair
the minimum number needed for the pair
(rn,L,n)
(r
is at least twice
n-1 1).
We can apply the laEt argument repeatedly going from the
to
(i-1)th
stage for
n, n-1,
i
.
:4,3. Each stage yields a factor
of two, so with the original factor of
2n
this gives a factor
2n l.
(2n)(2)11-2 = n
When
ith
2
i
is reached all that can be said is that by Remark
6. 8 at least m matrices are needed. This gives the final lower
bound of m n 2n-1
Corollary 6. 10. Let Pn be an n-region, n > 1, with
pn Rn continuous. Suppose that b(P) is sufficiently
refined relative to
sgn2.
subdivided into at least
Proof.
If d(e, Pn, On)
n 212-1
0,
then b(P)
simplexes.
Immediate from Corollary 6. 9.
Example.
Consider once more the tetrahedron example of
Section IV. 1. Here
n
3
so
n 2n-1 = 12.
However we have a
sufficient refinement of b(P3) consisting of 6 2-simplexes. By
Corollary
6.103, P3,03)
Remark.
= 0.
The result of Corollary 6. 9 can be improved for
n >6 using the Theorem 3.6 formula
is
77
d(e, pn, en)
1
2n n! k=1
(k)),
tk An(sgn e(y1
)
sgn
(Yn(k
Hadamard's determinant theorem [4] tells us that for any determinant
in this sum,
A
nn/2. Consequently under the hypotheses
<
of Corollary 6. 9 the above sum must contain at least m 2n n! /nn12
terms, and for n > 6
it is easy to check that this is greater than
m n 2n-1 We conjecture that in fact the lower bound can be increased
2m n!. Then Corollary 6. 10 would have the lower bound
2 n!.
If so, this is certainly the best possible estimate: let Pn be the
cube of side
+ 1.
We give
in
2
Rn
the standard "counterclockwise" orientation. If
PR
ei(x) = x for all
with vertices all of whose coordinates are
x
in
Pn,
then it is easy to show that
d(el, Pn, On) = 1. Now [7] gives a simplicial decomposition of
pn
which, as can be checked, readily yields a sufficient refinement of
b(Pn)
2 n!
relative to
sgn el,
(n-1) -simplexes.
and this sufficient refinement contains
78
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