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THESIS

C\D‘

  
      

  

LIBRARY

Michigan State
University

[,M‘K

 

 

ABSTRACT

PTOLBMAIC METRIC SPACES AND THE CHARACTERIZATION
OF GBODESICS BY VANISHING METRIC CURVATURE

by David CliffOrd Kay

In a metric space, the characterization of a geodesic among all

rectifiable arcs by the identical vanishing of its metric curvature--from

now on referred to as the fundamental theorems-has curiously required the

space to be locally ptolemaic. Such a metric space is one whose metric
xy satisfies the inequality

pcrrs + pr'qs > ps°qr
in.some neighborhood of each point of the space. Haantjes did obtain
the theorem for his concept of curvature in a class of metric Spaces
that includes, for example, spherical space, but his proof involves
certain variations of the ptolemaic inequality. Counterexamples in
Ll Spaces show that some condition is necessary. Part I of this
thesis is devoted to discovering how restrictive the ptolemaic in-
, equality is, with emphasis on those metric Spaces introduced and
studied previously by H. Busemann called G-sEaces. On the other hand,
it is possible that other inequalities may imply the fundamental theorem.
This is explored in.Part II, and finally, a new concept of curvature
is prOposed for which the fundamental theorem can be proved in a wide
class of metric Spaces which includes locally ptolemaic G-Spaces, two-
dimensional Riemannian spaces, and Banach spaces with strictly convex

unit sphere.

 

David Clifford Kay

More specificalLy, in Part I a concept of Space curvature
called £59153 curvature is introduced, analogous to that studied by
Busemann. This concept is compared with that of Busemann and the
following theorems are proved:

A G-space having non—positive median space curvature is locally
ptolemaic.

In a Riemannian space the conditions non~positive median curva-
ture, nonrpositive Busemann curvature, non-positive classical curvature
at every point, and being locally ptolemaic, are equivalent.

Other theorems of Part I are:

A straight G-space with convex differentiable spheres is
euclidean iff it is ptolemaic and satisfies the parallel axiom.

A Finsler space is Riemannian with non-positive Busemann curva-
ture iff it is locally ptolemaic.

A symmetric Hilbert geometry (where the absolute is a symmetric
convex surface) is hyperbolic iff it is locally ptolemaic.

In Part II, two weaker forms of "non-positive median curvature"
_are intnnduced: If for each point p in a G-Space there exists a
neighborhood V in which the Space has unique joins and a constant
7p with O<yp< 1 such that if a, b, and c be any three points
of Vp and m the midpoint of the segment joining a and b, then
the inequality

7p-mc<\/%a:2+%bc2-§iab2

 

holds, the space is said to have weakly non-positive median space

 

curvature, while if the inequality

 

David Clifford Kay

mcgfi (ac + bc)

holds, the space is said to have feebly non:positive median space

 

curvature. (Non-positive median curvature is defined by the former
inequality with 7p = l for all p.) The theorem obtained is:

In any G-Space which has both weakly and feebly non-positive
median SpaCe curvature, the fundamental theorem is valid for both
the Menger and the Haantjes definitions of the curvature of an arc.

The class of Spaces for which this theorem is valid includes
all Riemannian surfaces and certain (non-euclidean) Minkowski spaces.
Also in Part II is introduced the following concept of curvature for
arcs in G-spaces: If {qg} , irgg , and {If} are any three
sequences of points on an arc such that for each i, qi< ri<si,
qiri = risi, and mi is the unique midpoint of the segment joining

qi and s then that arc is said to have transverse curvature

i)

 

KTQO) at p iff

. 8 m.r.
11m 1 1 = KT\p).
1—»90 2

qisi

 

This curvature is compared with classical concepts in euclidean space
and a formula for it is derived in certain Minkowski spaces. Then
the fundamental theorem is proved for this curvature in G-Spaces

having feebly non-positive median Space curvature.

PTOLEMAIC METRIC SPACES AND THE CHARACTERIZATION

OF GEODESICS BY VANISHING METRIC CURVATURE

By

David Clifford Kay

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1963

23m

The author is deeply indebted to his major professor, Dr. L. M.
Kelly, for his inSpiration during the preparation of this thesis.
Appreciation is due him especially for his patience during the actual
writing and preparation of the final draft. The author also wishes
to extend his appreciation to his parents, Mr. and Mrs. Clifford I.

Kay for providing for the typing and printing of the manuscript.

 

To Helen

 

TABLE 933 CONTENTS

 

SECTION

Introduction . . . .

Part I . . . . . . . . . . .

1. Preliminaries . .

2. The Ptolemaic and Median Inequalities Related
3. The Ptolemaic Inequality and Riemannian Spaces
h. Metric Curvatures in Riemannian Space

5. The Ptolemaic Inequality and G-Spaces

Part II . . . . . .

6. Preliminaries . . . . .

 

OOOOOO

O O O O

.....

7. Menger Curvature and Haantjes' Proof of the

Fundamental Theorem

0

O

8. Haantjes' Proof in NOn-Ptolemaic Metric Spaces .

9. Transverse Curvature . . . . . . . . . . . . . .

lO. Transverse Curvature and the Fundamental Theorem

Bibliography . . . . . . . .

iv

PAGE

19
25

51
51

Sb
65
7h
85

89

INTRODUCTION

Various definitions of curvature for arcs in general metric
Spaces have been advanced. Menger [1h] was perhaps the first to
propose an acceptable definition, and he made immediate efforts to
show that an arc in a metric space is a segment if and only if his
curvature identically vanishes along the arc. He concocted a variety
of examples showing that this is not the case, but eventually did
Show with the help of Alt and Beer that the property is character-
istic of segments in E“. Schoenberg [16] extended this theorem to
arcs in metric ptolemaic spaces (see p. 5 for the definition) while
Haantjes [ll] , using a definition of curvature suggested by Finsler,
established a similar theorem in a class of metric Spaces somewhat
more general than ptolemaic spaces. Simple examples in L1 spaces
of dimension two Show that the chamcterimtion theorem is not ex-
tendible to general linear spaces using any of the definitions tlms
far considered. It has been conjectured by L. Kelly and others that
an arc with identically vanishing Menger or Haantjes-Finsler curvature
in a strictly convex Banach space must be a segment, and’one of the
principal objectives of this thesis iS the study of this conjecture.

Somewlat more generally we attempt here to extend the funda-

mental characteriaation theorem to as wide a class of metric Spaces

 

as possible, while at the same time we seek to gain a deeper insight
into the scope of the class of ptolemaic metric spaces. Our efforts

to verify the conjecture are unsuccessful, but they lead to the intro-
duction of a new curvature Kur (defined in Section 9) which coincides
with the variously studied curvatures in classical spaces and for which
we can prove the following satisfying theorem:

Theorem: An arc .A in a strictly convex Banach space is a
segment if and only if for each point peAyKT(p) = 0.

Actually the class of Spaces for which this theorem is valid is
much wider than the strictly convex Banach Spaces. It includes all
spaces each two points of which can be joined by a unique segment
(straight Spaces in the language of Busemann [6D and in which the
length of the median of a triangle is less than or equal to one-half
the sum of the adjacent sides. The theorem can be "localized" and in
the local form such Spaces include spherical, euclidean, hyperbolic,
.elliptic and in fact all Riemannian spaces. In Section 8 we obtain
the precise relationship between KT in a Minkowski Space and -
classical curvature in the associated euclidean space thus making con-
tact with Busemann's work [5] in this direction.

Freese [Q] in his Missouri dissertation proves that two-
dimensional Riemannian space with everywhere non-positive Gauss
curvature is ptolemaic, a fact which is actually implicit in the work
of wald on the metrization of Gauss curvature [20] . We introduce
a new concept of space curvature somewhat analogous to that of
Busemann and prove that any space with non-positive curvature in

this sense is ptolemaic. This implies the Freese result in any

 

 

dimension.

We had hoped to establish that any G-space (defined in Section 5)
having negative curvature in the sense of Busemann is ptolemaic. This
proved to be rather difficult and in our efforts to construct examples
we discovered the interesting fact that any symmetric ptolemaic
Hilbert geometry (defined in Section 5) is hyperbolic. The concept
of curvature we introduce also provides the analogue of the P. Kelly
and Strauss theorem, fl3] to the effect that any Hilbert geometry with
unisigned curvature is hyperbolic.

We should note in conclusion two other possibly significant
failures. 'We have been unable to find an example of a "reSpectable"
Space (e.g., a G-Space) in which.the fundamental curvature characteriza-
tion theorem.for geodesics is not valid. We have also been unable to
construct an example of a Space of negative curvature in the sense of

Busemann.which is not Riemannian.

PART I

1. Preliminaries. If M is any set, a real-valued function

 

d defined over MXM is a distance function on M if for any pie M,
(a) d(P1:PQ) >/ O-
(b) d(p1,p2) = 0 iff (if and only if) p1: p2.
(C) d(p1,p2) = d(p2,p1).
(d) d(p1,p3) é d(p1,p2) + d(p2,p3), referred to as the

triangle inequality. A set M with a distance function is a metric

 

 

sage and (’(Pi’Pj) is the distance from pi to pj. The distance
will be represented here more simply by the symbol pipj and the
number pipj itself by a Greek character, the Latin alphabet being
reserved primarily for points in M.

The t0pologica1 conventions of Busemann [6] will be adopted.

For the convenience of the reader a list of the more frequently used
concepts appears below.

The sequence {pi} , i = l, 2, 3, °°°, where pie M, converges
£9 pEM iff ili’rgopip: O.

A function f:M1—->M2 mapping M1 into M2 is continuous iff
{f(pi)} converges to f(p) whenever {pi}, i= 1, 2, 3, -°-, converges to
p. A function full—>142 is an isometzl (or a congruence) iff f
is an onto mapping and pipj = f<pi) f(pj) for all pkg M1. Such a

mapping is clearly continuous and in fact a homeomorphism.

A metric space is compact iff every sequence contains a convergent
subsequence or alternatively iff every infinite subset has an accumulation

element. It is said to be finitely compact iff every bounded subset is

 

compact.

A neighborhood about peaM, to be denoted usually by Vp, is the
3233322353 V(p,8) for some 6‘>O, that is, the sets of points xeM
such that px < 6.

A compact subset S of M has the property that every cpen
covering by neighborhoods .{Vé}, pEES, can be refined to a finite sub-
covering, Vbl, sz, -°', Vb“.

.A definition of particular importance to this thesis is the
fbllowing:

Definition 1.1: A quadruple of points p1, p2, p3, and ph
in a metric Space is ptolemaic iff

pipypkpm + pipk-pjpm >pipm'pJ-pk
where (i, j, k, m) is any permutation of (l, 2, 3, b). A Space M is
said to be ptolemaic iff each of its quadruples are ptolemaic. It is lg-
‘gglly ptolemaic iff each point peaM has a neighborhood Vb which is

ptolemaic.

Euclidean space of dimension n, denoted by 3“, consists of all

ordered n-tuples (x1,x2,--°,xn) of real numbers xi metrized by

defining

 

2
XY = \//(X1 - Y1)? + (X2 ‘ Y2) + "' + (xn - y )2

n
where x = (x1,x2,'°',xfi) and y = (y1,y2,°°°,yn).

.An isometric image in M of the euclidean segment is a £32332
segment in M; The images of the endpoints of the euclidean segment
are the endpoints of the metric segment and the metric segment is said

to be a segment joining those endpoints. .S(p,q) will denote the

segment joining p and q whenever it is unique in M. If S(p,q)
exists for each pair of points (p,q) in M it is said to have
gn_i_q_u_e segment .

The point q is metrically between p and r iff pq + qr = pr
and p t q # r, and the symbol (pqr) will denote this situation. If
p, q, and r are arv three distinct points of a segment then one is
always metrically between the other two.

A metric space is m; iff each pair of points has a point
between them.

Theorem (Menger): Each two points of a complete convex metric
space are joined by a metric segment. (The proof of this may be found
in Busemann Ufl .)

A 32333 in a metric space M is a continuous map of E1 into
M; an m is a homeomorphism of a euclidean segment. A. geodesic is
a curve which is locally a metric segment, that is, each point of
die curve has a neighborhood in which each of its subarcs is a metric
segment.

If (pmq) holds and pm = mq then m is called a midpoint
of the pair p, q. If S(p,q) exists m must be in S(p,q) and
it is then unique and called thg midpoint of the segment S(p,q),
denoted by mpq.

If the points p, q, and r in M determine unique segments
then the set union of those segments is the triangle with vertices
p, q, and r, and is denoted by T(p,q,r). The E22332 of the triangle
T(p,q,r) £32m p is the segment S(p,mqr), if it exists.

This section will be concluded with a definition which proves

to be relevant to ptolemaic spaces.

Definition 1.2: The median inequality for a complete convex metric
space M is for any point triple (p,q,r) and any midpoint m of
q and r the inequality

2 2
pmzéipq2+ipr -2‘rqr.

395515: The significance of this inequality may be seen from
the fact that if (p,q,r) '—>(p',r',q') is an isometry of M into
32 with m' the midpoint of S(q',r'), and if the median inequality
holds in M, the-n from the formula for the length of the median of a
triangle in terms of its sides in 52 it follows tint pm g p'm';
this applies even if p', q', and r' are collinear. Moreover,
if. 212.2. radian umamy Raids is a 99393.2? 9,9112% 226.9519 15251.92 M,
the. mags ELL! 1131.3 9.9.1393 segments For, Stippose m is a midpoint
of q and r, and p is any other midpoint. Then in the isometry
(p,q,r)—-—~>-(p',q',r') into 52, p' coincides with m' and the
above observation produces pm S 0. That is, pm = O, and p = m.
Thus, midpoints are unique in M, and it easily follows tint segments

are unique al so .

2. The Ptolemaic and Median Inequalities Related. A stronger
form of the median inequality is obtained, from which it will be
proved tint in any complete convex metric space the median inequality
implies the ptolemaic inequality.

Lemma 2.12 Let M be any complete convex metric Space in
which the median inequality holds, and therefore in which unique seg-
ments exist. Then if p, q, r, and s are any four points with

s e S(q,r), the inequality

 

pszg :1: .pg2 + :5- vprz- qs-sr
qr qr

holds. Moreover, if M is such tint the strict median inequality
holds for any three non-linear points (that is, not on a segment),
then the above inequality is also strict for p,q,r non-linear points,

qistr.

p
Proof: Since S(q,r) is iso-
metric to the interval 0 S E g qr
by the mapping x ———>Z, x 6 S(q,r), q s r

put’°‘c' = qx/qr = E/qr, which defines

a linear map from S(q,r) to the unit

interval 0 g T Q 1; then denote x by x.,. The desired inequality
for x1, = s then reads

(2.1) pX§é(1-’E)pq2+’5'pr2- 7«'(1-7) qr?

This will first be proved for 'L’ a diadic rational m/2n, by
induction on n. Let R(n) = {1/2“, 3/2“, 5/2“, W, (2n - Men}.
For n= 1, if ’L’ is in R(n) then ’L’=% and x1, is the midpoint
of S(q,r), so the proposition follows by the median inequality itself.
Suppose it has been proved for all integers k less than n and
tel-Kit). Let T be in R(n) and suppose ’b'= (2m- l)/2n where

1g m ran-1. Put ‘0: (m .- 1)/2”"‘1 and cr= m/2n"1 so that F

and 0’ are each in some R(k) for k < n. Then the induction
hypothesis gives p
(2.2)

pxg é (1-?)qu + ("pr2 -= #1?)qu

(2.3)

pxgé (14’)qu + cr-pr2 - <7(l~c)qrg q x? x.r x0. r

Note that x1 is the midpoint of S(x€,x¢), since

XrXr= qu — qxf = (r-f)qr
while

szu= (IX. - cps, = (Cr-”chm
and (’C-f)qr = 2"“oqr = (o’-1)qr. Observe also that 17= gt”).
Applying the median inequality to T(p,x(,,x¢) it follows that

PX: <:% pxg + i pr ~ % Xexg,
or, using xrx, = (d’-p)qr,
QM) pééipfi+ipfiwi(np%g-
From (2.2) and (2.3),

mi S %(1-f)pq2 + if'prg - i ((0 - 92hr? +

a (14qu + Igor-pr? - i (cr- «2)qu - %(0'-f’)2q1‘2

(2.5) = (1-if-iotn2 + %(f+6)pr2 - (%f+%°'-iPZ-%e°‘%”%qr2
= (1- '6)pq2 + 1-pr2 - (1 - 12hr?
which is the desired inequality for ’b’ in R(n). Hence the inequal-
ity (2.1) holds by induction for all n and 'L’eR(n), that is, if 17
is a diadic rational. Since the diadic rationals are dense on the
interval (0,1) a sequence {1n} of diadic rationals can be found
which converge to 1’, given s = x1, so that the corresponding sequence
{ x‘n‘J converges to x,” and by the continuity of the metric,
gimme¢n = pX1. But since (2.1) holds for each ’L‘n we have for each n,
(2.6) pmfi g (1 - 1,1)qu + 1n~pr2 - 1,,(1 ., Tn)qr2.
Taking the limit as n—'-00 , the desired result (2.1) is obtained.
Now suppose the strict median inequality holds for non-linear

points in M, and let p, q, and r be non-linear, with q =k s :1: r.

Put 5 -= x, and locate x and x, any two points on S(q,r) such

6

 

 

10

that xC is the midpoint of s(x(,,x,). Again it follows that Inga”)
except that here, P! er, and '1: are not necessarily diadic rationals.
Nevertheless the inequalities (2.2) and (2.3) are now established

for arbitrary lo and a‘, and the strict inequality holds in (2.1;).

It then follows that in the subsequent steps the inequalities are
strict in (2.5) which gives the strict inequality in (2.1), completing
tie proof.

m: It follows from the lemma that if the median in-
equality holds in M and (p,q,r)—*(p',q',r*) is an isometry
into 152 with s' that point on S(q',r') such tint q's' = qs,
then ps g p's'. For it is clear that

ps g E}: 'pq + 33 ’pr - qs‘sr
qr qr

= S'r' .pqu + Q's, .plr! ... q.s',s|r'
qlrl qul

 

= p'S',
making use of an elementary formula valid in 52.

Theorem 2.1: If the median inequality holds in a coznplete
convex metric space M, then M is ptolemaic.

111323: It is trivial tint if four points are not distinct, or
if they are linear, then they are ptolemaic. It may then be assumed
that the four points are distinct and non-linear. Let the points be
represented by p, q, r, and s, so ordered that among pq.rs, proqs,
and ps-qr, the value ps~qr is maximal. It then suffices to prove
the single inequality
(2.7) pq-rs + mm 2 ps-qr

for all cases. Suppose (p,q,r)—-—>(p',q',r') is an isometry

‘I!

11

into E2. Chooseasecond isometry (q,r,s)—+(q',r',s') into 52

so that if both triples (p,q,r) and (q,r,s) are non-linear, p'
ql

M q s ffl‘Illlllliilllll.-" s'
r I“

and 5' fall on Opposite sides of the lines through q' and r'.
The points p’ and s' are then distinct, for otherwise both triples
(p,q,r) and (q,r,s) would be linear, and since M has unique seg-
ments from the median inequality this would imply that all four points
were linear, contrary to assumption. Again, the line through p' and
s' in 132 will always have exactly one point x' in common with the
line through q' and r' or else these lines would coincide implying
tint p, q, r, and s are linear. There are then two cases: (a) x|
falls on S(q',r') or, (b) x' is exterior to S(q',r').

(a) x' falls on S(q',r'). Let x be that point on S(q,r)
such that qx = q'x'. By the lemma and the remark following it,
(2.8) px£p'x' and sxgs'x'
so that px + xs ép'x' + x'sll = p's' and the triangle inequality
ps gpx + xs imply
(2.9) ps ép's'.
From the fact that 32 is ptolemaic, (2.9) implies
(2.10) pqors + proqs = p'ql-r'sI + p'r'oq's' 2p's'-q'r' gps-qr.

(b) x' is exterior to S(q',r3). Consider the sum of the

2 and the sum of angle

angles (p',q',r') and (r',q',s') in E
(p',r',q') and angle (q',r',s'). One of these sums must be greater

than a straight angle. Since the following argument my be applied

 

 

12

to either case it may be assumed that the

latter sum is greater than a straight angle.

If both angles in this sum are > 11 then
/ ’2

p'q'>q‘r' and s'q'>q'r' so that

 

pq>qr and qs>qr and it follows that
pq-rs + pr-qs > qr-rs + pr-qr = (rs + pr)or>/ ps-qr.

The remaining cases are when one of the angles of this sum is acute.

Again it may be assumed without loss of generality that angle (q*,r',s')‘

is acute and angle (p',r',q') is obtuse or straight. Finally, it may be

assumed that ps >p’s', for otherwise (2.9) holds and (2.10) follows as
2

it did in (a). Introduce a coordinate system (Eng) in E with the ori-

gin at r', p' on the negative E-axis, and s' in the fourth quad-
rant or on the negative 71- axis. Since 11 (angle (p',r',q')<1'r, q‘
will either lie in the first quadrant or on“the positive Z-axis, hence its
coordinates (05,8) will be such that 06>O, {320. Let the coordinates
of s' be (7,8) with ?’>O’ 3<O. Consider the minor arc A of the
circle with center at r’ and radius ’0 equal to r's‘, whose endpoints
are 5’ = (71,3) and t' = ((0,0). The real-valued function {(2‘) = p‘z'
for 2’61). is continuous, and f(s‘) = p’s'< ps while f(t') = p't’ =
p'r' + r’s‘ = pr + rsgps. Hence there exists a point 5" + s' on A

such that f(s”) =pS. If s” has coordinates (7256") then 7-97) and

8'>8 and it follows that
cps-'2 - ens"? = (a m2 + ((5 - n2 - (a - 7')? - (,9 - .902
. 2w :7) + 25(3- - s) + (72 + 32) - (7'2 + 3:2)
>(O 2 _’02 = o
01‘ q‘s' >q‘s" and therefore

(2.11) qs > (1:13".

 

 

13

Applying the ptolemaic inequality to

the quadruple p',q',r', and s" q'

in E2,
p' r' '
plqi.r's" + ptrI-qlsn 2 plsn-qirl "\T::::: s" E;
5'
or, since p's" = ps,

pq'rs + pr°CI'S" Zps'qr

 

which, together with (2.11), implies
ptrrs * pr‘qs > pq-rs + pr.qu" Z ps-qr,

completing the proof. I

539555: In 82 the ptolemaic inequality is strict for distinct
points in all but two cases, so the applications made of it in the
above proof result in the strict inequality in all but those two cases.
The two cases when equality holds for points in E2 are when the
points are linear or cyclic. An examination of the proof of Case (a)
above reveals that since the points p', q', r', and 3' cannot be
linear, the only possible way for the ptolemaic equality to hold for
those points is for them to be the vertices of a cyclic quadrilateral
with diagonals S(p',s') and S(q',r'). Then x' is an interior
point of S(q',r') which implies from the lemma that the strict
inequality for (2.9), (2.10), and therefore for (2.7). In Case (b)
the argument either reduced to Case (a) or fine strict inequality was
obtained for (2.7). This proves

Corollary 2.1: If the strict median inequality holds for non-
linear triples in M and p, q, r, and s are any four points in M,
the only occurrence of equality for (2.7) is the case when either

the four points are not distinct or when they are linear.

 

 

1h

3. The Ptolemaic Inequality and Riemannian Spaces. A brief

 

description of a Riemannian nnmanifold M will be given for the
convenience of the reader. many terms will be encountered which
involve basic concepts of topology and are not defined here. The
reader is referred to Synge and Schild D8] for details concerning
tensor calculus.

(a) M is an nemanifbld with a differentiable structure of

class 02 at least. That is, M is a connected separable topological

 

32323 such that for each point p in M there is an open neighbor-
hood Vfi homeomorphic to an open neighborhood of E“, n fixed for all
g>€lL such that if this homeomorphism be denoted by hp and its inverse
by hp"1 then whenver two neighborhoods Vb and Vq have non-void
intersection the transformation hth'1 from En(x1,x2,°°°,xn) into
En(y1,y2,"‘,yn) is of the form ‘
(3_1) yi = yi(x1,x2,'°°,xn), i = 1, 2’ ..., n
where the yi have continuous second order partial derivatives at
least.

(b) There exists, with respect to the coordinate systems of
M corresponding to these homeomorphisms into E“ a positive definite
fundamental or metric covariant tensor gij(x1,x2,‘°',xn) of the second
order defined at each point for each coordinate system. That is,
there exists an n)(n positive definite matrix [gij] which trans-
fbrnw by means of (3.1) from one coordinate system to another as
follows (summation indicated by repeated indices):

9 3X? 3X§
rs gyi §§J

g'iJ

where 9' corresponds to the components of the matrix in the

ij

IS

coordinate system (yl,y2,~~',yn) . One usually makes the assumption
that the gij are each of class 03 at least.
(c) In terms of this tensor the arc length of curves of

class C1 at least, xi: xi(’v), i=1, 2, °°°, n, is defined as

Af/‘jdq: dxide d’L’

by means of which M is metrized by taking xy to be the number

Inff/gij dxide d1
MW

taken over all curves xi = xi(L’) such that x corresponds to T = 06

 

 

and y to 7= ,5.
Observe that in the case n = 3 and with

gij(x1,x2,x3) E 1 when i = j,
EEO otherwise,

the above formula reduces to the usual one for differentiable curves

in 83,
A f/(eé) (ems)? ..

and thus is E3 in thi)s case.

 

A few well-known results from Riemannian geometry will be needed.
One of these is that every point has a neighborhood in which each pair
of points may be joined by a unique metric segment, and that there
exists a family of geodesics of class C2, passing through each point,
a unique one for each direction.

The angle between two geodesics may be defined as follows:
Suppose two geodesics represented by xi - xift) and x; - XECF),
i s l, 2, o-o, n, emanate from a certain point a 6 M? with the respec-

tive directions (u§,u3,-H,u3),j=l,2, where u‘dei/d’cm; = dxé/dt’,

 

 
 

 

16

i = l, 2, '~', n. The measure A of the angle between those geodesics

i .
A = Cos'1 gijulué

 

gijuiu‘i gijuéug

There is a satisfactory concept of curvature in M, referred
to as Gauss curvature in the case n = 2 which is a generalization
of Gauss curvature of surfaces in B3. The curvature K at a point
p in M for n>2 is defined corresponding to each two-dimensional
surface through p as the Gauss curvature of that surface. It is
possible to characterize locally euclidean spaces (those for which
each point has a neighborhood which is isometric to a neighborhood
in 13“) among all Riemannian manifolds by the condition K = O at
every point (see Synge and Schi 1d [18] ). The following results con-
cerning curvature are from Cartan [7] , p. 261, for non-positive curva-
ture, and from Alexandrov [l] for positive curvature.

The Cosine Inequality: If K é O in a Riemannian space M
there exists a neighborhood V of each point in which the sides of

every triangle T(a,b,c) with vertices in V satisfy

(3.2) aha? ac2 + bc2 - 2ac~bc cos C

where C is the measure of the angle at c. If c

K > O at some point p in M, then there exists

a surface which contains two geodesics through b
a

p such that on that surface there is a
neighborhood V in which the sides of every triangle T(a,b,c) with
vertices in V satisfy (writing 5?; as the metric of the surface)

(3.3) a32< a-c2 + 552 - 252m? cos C.

 

17

The following theorem was obtained for the 2-dimensional case
by Freese [9] , using methods which cannot be extended to higher
dimensions. The following theorem, therefore, is somewhat stronger,
and its proof does not involve dimension in any way.

Theorem 3.1: If M is any Riemannian Space of arbitrary
dimension and with K g 0 at every point, then M is locally
ptolemaic.

£3293: Since K Q 0 at any point pEM there exists a
neighborhood Vp which ins unique segments and in which (3.2) holds
for every triangle with vertices in VP. It will be shown that VP
is a convex metric Space in which the median inequality holds. Vp
is alreaw a convex metric space by assumption. Let a, b, and c
be any three points of Vp and m the midpoint of a and b.
Since T(a,c,m) and T(c,m,b) exist (3.2) may be applied. Let D
be the measure of angle (a,m,c) and put 06= bc, f3= ac, 7’= ab,
and )\= cm. Then

(32):”,2 + X2 - 7!) cos D
42>“)? + >\2 + 9’)\ cos D

 

which implies ,
“2*p2>%72+2>\2v a %7 mi? b
That is, Vs i062 + #32 - i712, which is the median inequality.
By Theorem 1.2 VP is ptolemaic which proves M is locally ptolemaic.
It might be conjectured that in certain cases the local con-

dition of non-positive curvature could guarantee the ptolemaic inequal-
ity in the large. But that this is not the case even for K strictly
negative everywhere may be seen in the example of the surface of

2

revolution in E3(x,y,z) obtained by rotating the curve y = x + l,

18

z = 0 about the x-axis. This surface has negative curvature every-
where and by the above theorem is locally ptolemaic. But observe

the points a, b, d, and 0 equally spaced on the circle generated
by the point (0,1) on the generating curve. With xy denoting the
metric on this surface, ab = bd = cd = ac = '1 while ad = bc = 1T ,

but abecd + ac~bd = while adobc = 1T2 so that

Ti 2

T
abocd + ac'bd < ad-bc

violating the ptolemaic inequality for arbitrary points.

The converse of Theorem 3.1 also holds.

Theorem3.2: If M is a Riemannian space and M is locally
ptolemaic then Ké O everywhere.

_P_ro_o_f_: Suppose K > O for some point p of M. This means
that there is a two-dimensional neighborhood about p such that every
triangle T(a,b,c) in it satisfies (3.3) Let L1 and L2 be two
geodesics in that neighborhood meeting at p orthogonally. Consider
points a and d on L1 on opposite sides of p with 513 = _p = 0‘
and points b and c on L2 on opposite sides of p so that
13-p- = E = 06. Then from (3.3), E, 133, Ed, and are are each less
than \/-2_ot and

ah-cd+ac-bd $a—bc—d + a_c'b_d < 203 + 20:? = 20¢~2oc= 57$ch = adrbc
for all sufficiently snail at, so
regardless of the n—dimensional neigh- L1
borhood taken about p there always
exist a quadruple of points violating /a
the ptolemaic inequality. But this b w a. c
contradicts the-fact tlat M is locally \
d

ptolemaic, which then proves KS 0.

 

1;. Metric Curvatures in Riemannian Space. In this section we
stuchv two metric concepts of curvature in the context of metric spaces
with locally unique segments. In particular we investigate the
relationships of these curvatures to the ptolemaic inequality in
Riemannian spaces. After Busemann [6] we state

Definition 11.1: A metric space M which locally has unique
segments is said to have nonapositive Busemann curvature denoted by
KB<O iff each point peM has a neighborhood vp such that if a, b,
and c are any three points in Vp then mac and mbc exist and
04'” macmbcé % ab.

The space is said to have zero Busemann mac

C

mbc
curvature denoted by KB = 0 iff

 

(14.2) macmbc = fi- ab a ‘3
always holds, and non=negative Busemann curvature denoted by KB ) 0 iff
(’4‘3) macmbc > % ab

always holds. If the strict inequality holds in (h.l) for non-linear
triples a, b, and c the Space is said to have negative Busemann £3133-
t_u£e denoted by KB< O, and if strict inequality holds in (14.3) for non-
linear triples it is said to have positive Busemann curvature denoted by

KB>O.

20

Definition 11_._2: A metric space M which locally has unique

segments is said to have non-positive median curvature denoted by

 

KM< 0 iff each point p6 M has a neighborhood Vp such that if
a, b, and c are any three points in Vp then "’ab exists and
(Lt-h) rnabc2 g fi- ac2 + 5 bc2 - ii ab2. C

The space is said to have 5.519. m

curvature denoted by KM = 0 iff

2 2 a mah b

(h.5) mabc2 =4§~ac2 + the - 2E ab

always holds, and non-negative median curvature denoted by KM 2 0 iff
(11.6) mabc2 > 15 ac2 + I! be2 - é ab2

always holds. If the strict inequalities (11.14) or (14.6) hold for non-

 

linear triples, as in Definition 14.1 the Space is said to have respec-

tively negtive and positive median curvature denoted in the obvious way.

 

It should be noticed that the inequality (1.2) referred to
earlier as the median inequality is precisely the condition (L14)
affirming that the Space has non-positive median curvature. It then
follows by a previous theorem that a Space which has non-positive
median curvature is locally ptolemaic. It is clear that this is not
the case with non-positive Busemann curvature, for this class of spaces
includes Minkowskian spaces which are not in general ptolemaic.

It is important to observe that for both space curvatures KB

and K the condition l'non-spositive curvature" does not necessarily

M
include the conditions "negative curvature" and "zero curvature" as
mutually exclusive cases. That is, non-zero curvature for a space

does not imply that the space has either positive or negative curvature.

21

Theorem h.l: In any metric space which has unique segments
locally, the conditions negative, non-positive, zero, non-negative,
and positive median curvature reSpectively imply negative, non-
positive, zero, non-negative, and positive Busemann curvature.

23223: To handle the above cases at once, let the following
notation be used. Ifametric space S has KMéO write KM(S)<O,
and similarly for the other cases. Corresponding to each metric space
S for which KM(S) is defined, define a binary relation Rs(d,P) on
the real numbers as follows:

RS(°‘:13) iff d<F if lip/1(5) < o
RS(OL,P) iff océfi if KM(s) s o
RS(06,P) iff 0L=’3 if KM(S) = o
Rs(oc,p) iff océfi if KM(S) 2 o
Rs(oc,p) iff oc>fi if KM(S)> 0
Let p be any point in the space. In each of the five cases there
exists a neighborhood Vp in which for any three non-linear points
x, y, and z in Vp and w the midpoint of x and z,
Rs(yw2, % xy2 + fi yz2 - % x22) holds. Let Mp be the
neighborhood about p with radius one~half that of Vb and suppose
a, b, and c are any three points of WP. If f is the radius of
Vfi and mbc is the midpoint of b and c, then
Pmbc$ PC + cmho
= pc + % bc
épc + i001) + p6)
<ie+i (ie+i(’) =7

so that a, c, and ”be lie in Vp. Let mac be the midpoint of a

22

and c and put 06 = be, fi= ac, 7= ab, 7‘ = macmbc, and 72" = mbca.
If a, b, and c are linear, it is clear that 7‘ = is)’. Otherwise,

a, b, and c are non-linear and therefore a, mbc’ and c are non-

 

 

linear and lie in VP. Therefore
2 1 2 2 2
RS(’)/' , 3'01 + £7!" - 21:19 1 c
holds, or
2 2
(14.7) RS(2 7| ‘ ’1ka 4. lip 0/112)
Also,
2 2
(h.8) R59." , if? + 2,72 _ 21(042).
. . . . . a B
The tranSlthlty of R then lmplles 7/

S
from (11.7) and (11.8)

RS(27’I2 _ $052 41%162’ $92 + 5:72 _ 2£062)

or RS( 7’ '2, £721
Since 7' and 7/ are positive then RS(7" $7) holds, which com-
pletes the proof.

Busemann proves in [6] pp. 269-70 that in a Riemannian Space
with classical curvature x defined at each point, KB g o and K g o
everywhere are equivalent conditions in the Space, that also KB = O
and K = O at each point are equivalent, and that K < O at every
point implies KB < 0. It is now possible to give a similar compari-
son between K and KM“ This is done in the next two theorems.

Theorem 11.2: In a Riemannian Space M, the following conditions
are equivalent;

a. K< 0 at every point.

b. K .
M<O

c. The space is locally ptolemaic.

 

23

2102:: (a) implies (b) since (a) implies the cosine inequality
(3.2) and, as in the proof of Theorem 3.1, the cosine inequality implies
ng O. (b) implies (c) by Theorem (2.1), and (c) implies (a) by
Theorem 3.2.

53111313: It follows that in a Riemannian space, non-positive
Busemann curvature and non-positive median curvature are equivalent.
Indeed, KB g 0 and the three conditions of Theorem h.2 are equivalent
for a Riemannian Space.

Theorem h.3: In a Riemannian Space M, K = O at every point
iff KM = O, and if K< O at every point then KM< O and M is
locally strictly ptolemaic.

M: The first statement follows easily from the fact that
K = O at every point rakes M locally euclidean (isometric to B“
locally) so that KM = 0 holds, and conversely it will be seen later
that KM = 0 implies M is locally euclidean which means K = 0 at
every point. For the second assertion, let K < O at every point.

Then KB< 0 from Busemann's theorem, and the cosine inequality (3.2)
holds locally. Let V

p
p .6 M such that for any non-linear triple x, y, and z in VP’ mxz

be a neighborhood about an arbitrary point

and myz exist and both inequalities

(he) may. < i xy
(14.10) xyz) yz2 + x22 - 2 yz-xz-cos Z

are valid, where Z - angle (x,z,y). Let WP be the neighborhood
about p whose radius is one—fourth that of Vp and suppose a, b,

and c are any three non-linear points in WP. The midpoint m of

21L

a and b lies in a neighborhood about p of radius one-half that of
VP’ and therefore the midpoints m1 and m2 of m and a and

of m and c respectively lie in V . c

The non-linearity of a, b, and c

implies that of a, c, and m. Hence “1?

(14.9) implies

(11.11) m1m2< i. ac. a m'l ' b

But from (1i.lO), with C = angle (m1,m,m2),

 

 

a!

m1m22 > ”111112 + 11121112 ._ 2.mlmom2m-COS C

= l . ab2 + l-mc2 - l -ab~mc-cos C
I6 “I: T

and from (1i.ll),
(14.12) ac2> % ab2 + no2 - abcmc~cos C

Also from (ll. 10) ,

(b.13) bc2 2 i ab2 + me2 + ab-chcos C
Summing,
ac2 + be2 > % ab2 + 2 me2
or
(14.11;) mc2< % ac2 + <5 bc2 - 21; abz.

This proves that for any non-linear triple a, b, c in "P the
strict inequality (int) holds, hence KM< 0. That the strict ptole-
maic inequality holds for distances in Wp (for quadruples of
distinct non-linear points). now follows from Corollary 2.1.

The above theorems show that any one of the concepts we lave been
oonsideri ng serves to distinguish the non-positive Sign of curvature in

Riemannian geometry. This suggests the use of either KM or the local

25

ptolemaic condition--both metric concepts-—as a means of defining more
generally non-positive curvature in G-Spaces, analogous to Busemann's

condition KBsg O. This will be treated in the next section.

5. The Ptolemaic Inequality and G-Spaces. Metric Spaces which

 

satisfy the following list of axioms have been termed G—spaces, or
geodesic spaces, by Busemann 33:

a. The space is finitely compact

b. The space is convex.

c. The Space is localky prolongable, that is, for p any point
of the space there exists a neighborhood Vp of p such that given q
and r any two distinct points in Vfi there exists a point S with
(qrs) satisfied.

d. The space is uniquely prolongable, that is, if p and q are
distinct points then (pqu) and (pqr2) imply r1 = r2 whenever
qu = qr2.

The immediate important consequences of these axioms are:

a. Each point has a neighborhood in which each pair of points
can be joined by a unique segment.

b. Each segment can be prolonged to a unique geodesic of the
space which contains that segment as a sub-arc. Accordingly, we Shall
denote the unique geodesic obtained in this manner by L(p,q) where p
and q are the end-points of the segment.

The stock of "standard" examples of G-spaces consists of
Riemannian space, along with the special cases of hyperbolic, euclidean,

and spherical Space, Finsler space, along with the special case of

 

 

26

Minkowski space, Hilbert geometry, and Cartesian products of any of the
above metrized by

d< (x1,><2),(ypy2) ) = my,“ + xoyghv“
with aj>l, where xi and yi are in Mi’ a G-Space with metric

Xiyi> i = l, 2.

In this section we extend the study of G-Spaces to include our con-
cept of median curvature of space, with particular emphasis on G-spaces
which are locally ptolemaic. Our original objective at this point was to
establish or disprove the conjecture that a G—Space with KBl< O is local-
ly ptolemaic. In this we were unsuccessful but the investigation led to
many interesting results which we include here. The argument presented in
the proof of Theorem h.l is equally applicable to a G-Space. It then fol-
lows that in any G~Space KMé 0 implies KB< 0, while in a Riemannian
space, these conditions are equivalent. This points up how desirable it
would be to know the precise relationship between PiM and KB hia G-Space.

In [61 BUSemann shows that a G-space with KB = O is a locally
Minkowshian Space, the term "locally" being used in the strong sense
indicating that each point has a neighborhood isometric to one in a
Minkowskian G-space. Since flue condition KM = 0 implies both that
KB = O and that the Space is locally ptolemaic, then it implies in a
G-Space Ulat the Space is a locally ptolemaic Minkowskian space. This,
together with the fact (Schoenberg D7] ) that a ptolemaic Banach

space is an inner product Space proves

Theorem 5.1: A G-space is localty euclidean iff KM = 0.
It should be remarked here that this theorem is implicit in the
work of Blumenthal [2,3] since the condition KM = o is locally

what he terms the feeble euclidean four-point property. Our

 

 

 

I ill [.1 .l.l ml :iv

 

 

27

proof of the above theorem then provides an independent proof of some
of Blumenthal's characterization theorems.

The proof of Theorem 5.1 given above employs Busemann's theorem
which characterizes locally Minkowskian G-Spaces as those for which
KB = 0. Since Busemann's argument establishing this fact is scattered
throughout a number of other important results, it seems desirable to
present a new and somewhat more direct proof of this fact here. Then, a
more direct proof of Theorem 5.1 will be indicated, after which, as stated
earlier, we shall explore the implications of the locally ptolemaic con-
dition in certain G-spaces.

Preliminary to the proof of the Busemann result we sketch a few of
the relevant facts concerning normed linear spaces which will be useful
throughout the remainder of the thesis.

Definition 5.1: A vector spaceMover the real field F is said
to be M, or to Egg a flu, iff there is a mapping of M into the
set of non-negative reals, denoted by llxll for xeM, satisfying the
axioms

[XH= 0 iff x = o, the zero of M as a vector space.

 

a.
b. “XXII-‘- IMHXH for xeM, )xEF.
c. “X + yll < “X" + “Y“ for KEM, YEN.
M is said to have a semi-norm ||x|| iff just (a) and (b) hold.
It is easily seen that M becomes a metric space by defining the
metric as xy=||x - y” . The topological terms connected with norms are

with this und erstandi ng.

Definition 5.2: A space M is called a (real) Banach sEce iff

it is a vector space with a complete mrm, tint is, a norm for which

 

 

28

every Cauchy sequence has a limit. The dimension of M is understood
to mean its dimension as a vector space.
Definition 5.3: The unit sphere U of a semi-normed vector space

is the set of all points x such tlat Hxll= l. A point x is called an

interior or an exterior point of U according as either Hxll < l or
”X“ > 1»

Definition 5.1;: The unit Sphere U is said to be convex iff for

 

each XEU, yeU then
(5.1) “Xx + (l - My“ é l, O<>\<l,
and it is called strictlyw iff the strict inequality holds in (5.1)
Whenever x:|: y.
It follows automatically trat a harmed vector Space, and therefore
also a Banach space, l'as a convex unit sphere, for if O<>\<1 then since
“xll = Hy” = 1.
”ix + (1 -X)yll<llkxll + “(1 - My” =1>Hlxll +l(1-A)Hl‘y|l . 1.
The set ny, sometimes called a "one-flat," defined as the set of
all x in a Banach space M such that
z = Z) = Xx + (l -)\)y, )‘eF,
is isometric to El. For, if f(z>‘) = )vxy where xy is the metric
of M defined as above, it follows that
lez )‘2 =I|Z>1‘ Z )2“
= Ha, - >\2)x~(k1— kpyll
H1’ >\2l "X ' Y“
ml. in .xy

 

29

while
r(z.1)r(zi2> = |f(z)\1) - dz”)!

= ”new - X2094

= U1 - )2‘ 'xy
or z )‘12y2 = f(z>‘1)f(z>\2). The subset SXy of ny defined by
restricting X to ogh<l is therefore isometric to the euclidean seg-
ment O<T<xy, 7631 and is then a segment joining x and y. Segments
are unique in M iff U is strictly convex (see Busemann [6] , p. 96).
The only axiom for a G-space not satisfied by a Banach space with strictly
convex unit sphere is therefore that of finite compactness.

Definition 5.5: A G-espace is called a straight space iff it has

the preperty that each two of its points can be joined by a unique seg-
ment.

Definition 5.6: A Minkowski sgce is a finite-dimensional Banach

 

space.

It then follows that a Minkowski m M with strictly w
3235. E22513 is a straight G=SEaC8, with the unique segment joining a
pair of points x and y in M and the unique geodesic through x
and y being respectively the sets Sxy and ny defined above, and we
may now employ our previous notation S(x,y) and L(x,y) to represent
these sets. Now consider any three points x, y, and z in M and let
mxz and "‘yz be the midpoints of S(x,z) and S(y,z) respectively.
Then

mxz=ix+(1-i) z=i<x+ 2)
and similarly myz = i (y + 2) so that
mxzmyz= Hi (x+ 2) -i(y+ 2)” =lli hwy)” =i‘x7

 

 

30

and therefore KB = 0, showing that this condition is necessary for a
G-space to be locally Minkowskian.
It is useful to express the Minkowski metric differently and to

View the space in a more familiar setting. Suppose a basis for an
n-dimensional Banach space M consists of the points b1, b2, on,bn
in M. By the linear independence of the bi each point xeM has the
9.233.? representation

x = OLLb1 + oc2b2 + ... + ocnbn
where the 061 are real numbers. Thus the map f of M-—>l£n given by

foo = (0.1, «pm, an)
is a one-to-one onto correspondence. It moreover follows that

f(x + y) = f(X) + f(y)
and

f()\x) = >\f(x)
for each x and y in M and all real X, so that M and En in
this way prove to be isomorphic vector Spaces and thus their topologies
are equivalent. From now on the distinction will not be made between
the points of M and En although, naturally, their metrics differ in
general. Now let x and y be two given points in M, x1= y, and let

the metric of B“ be denoted by 3. Put 2 = x — y and let L(o,z)

cut the unit sphere U in u and -u (since Hull = H-uH =1).
Then
(5.2) xy= IIx-yn = Ilzll

and since the points 0, x, y, and 2. are vertices of a parallelogram,
(5.3) 'x32’ = '65.

There exists a real >\ such that z = )\u and therefore

 

 

 

31

(5.1;) 02= ”le = "M“ = Nllull = N .

It has been observed that scalar multiplication in M and s“

Hence, by definition of scalar multiplication in E“,

(5.5) 6-2 = his:

The result is

(5.6) xy = i,
E

and so it follows that the metric xy of M is given by dividing the
euclidean distance from x to y by the euclidean “radius" of the
unit sphere that is parallel to L(x,y).

It is clear from the preceding remarks that if U is any
euclidean sphere, M is isometric to E“. A much stronger result is:
Two n-dimensional Minkowski spaces are isometric iff there exists an

‘affine transformation which maps the unit sphere of one onto that of

theoUuer, from which it follows that a Minkowski metric 13 euclidean iff

 

its 2233 sphege is an ellipsoid. A proof of this may be found in
Bus emann [6] .

It is now possible to prove the theorem mentioned earlier:

Theorem 5.2: A straight G-Space is Minkowskian iff KB = O
in the large.

2322;: The necessity of the condition was shown in the re-
marks following Definition 5.6. The proof of the sufficiency will be
given in a sequence of lemmas. In each of these the assumption will be
that M is a straight GaSpace in which the distance between the mid-

points of two sides of any triangle equals one-half the third side.

coincide.

 

 

32

lfgmm i: If a, a', b', and b are any four points of M with
m and m' the respective midpoints of S(a,h) and S(a',b'), then
mm'< i (aa’ + bb'). b

PrOof: Let c be the midpoint

n aw»,-

    

,_>

I}

l

  

of S(a,a') and m1, m2 the midpoints
of S(b,c) and 5(b',c) respectively. m
Then
mn'sgnmh + mlm2 + m2m'
= % aa' + % bb' + i a3'
= % (aa' + bb‘).
page“? Suppose in triangle T(a,h,b‘) that m€S(a,b) and

ln'E'S(a,b‘) are two points such that

a
am = am'
ab ab'
holds. Then triangles T(a,m,m') and
T(a,b,h') are similar, that is, m m'
b_ 4‘bl

am am' mm‘

3522:: Define m1 and mi1 reSpectively on S(a,b) and S(a,b')
and S(a,b') with (Elma/ab = (amt')/ab' = 77, ogrgi. It must be
shown that for all real 1' in this range

mmmz' = T-bb'.
This will be slnwn first for the diadic rationals by induction on n
where R(n) = _l_:._2_""’ __££%fi%_£{} for n a positive integer.

2n 2n
For n = l, m%m%' = % bb' follows by hypothesis. Assume it has been

 

 

 

 

33

proved for rL'6R(k), k<n. Let ’7.“ be in R(n). Then it may be assumed
that

1 = 2 2" l ,
2“
where p is some integer, 1<p< 2n—l. If p =1 so tiat ’6= _l_,
2n
put cr= 2.5 = 11 so that 0‘6R(k) for some k<n, which implies
2n-

 

by the induction hypothesis,
(5.7) mfmp' = U‘bb'.
Also, m, and mf' are the respective midpoints of S(a,m,) and
S(a,m,‘) so it follows that
Imam,t = % mm,‘ = %o’-'bb‘ = Tobb'
and the assertion holds for p = 1.

It may then be assumed that 2 g p 4 2“"1. Write

a =22_h=E-2
Zn 211-].

 

= 2p - 3
(9 2..

F‘ZE' = p-l
2n. 2n-1

(1:2p-l)

2“
a‘ =32= 2
2n 2““1

so that a, (0, and 0' are each in some R(k) for k<n and the
induction hypothesis applies to mgmg', mfmt1 V, and mdmcfli
(5.8) mama,I = av'bb', umP' = ‘O-bb', m,m,,.' = (T—bb'

where 0H0 and 0'1=l. But if m0 and mo‘ are defined as a,

 

 

3b

and m1 and ml‘ as b and b' respectively, then (5.8) holds even
for oc= o and 0'= 1, so that for all p, 2<p< 2’“, (5.8) holds.
It is clear that m,, m,, and mt, divide S(mu,m,) into equal segments,

and similarly for mg', me', and mt'. Thus Lemma 1 implies

 

mcmr\ <£ (meme' + msmcr') a
= 2 ({o+c’) bb‘
= % (%é++ BET) bbl
= 32;; .be
2n
or,
(5-9) “Limz' é T-bb'.

 

 

13mm lamoimnks

(5.10) (cobb’ = mfmf'g 43- (mfimfi' + mfmt')
and .
(5.11) f-bblg %[%(mwmxl + mfnif’) + meTI] .

Using (5.8) this becomes
seesaw 1.10)be i mm,
and therefore

man. > (ip- ion) bb'
;
2‘

= - 2 bl
( 2n_ 'En: ) b

= .11....2 '1.bb'.
2!)
That is,

m%121.bbi,

 

 

35

and in view of (5.9) this proves that mfimw' = T'bb‘. Induction
then carries, and the proposition has been proved for ’t a diadic
rational. The continuity of the metric then implies mtmf' = Tobb'
for arbitrary Oé’Eél.

Lemma 3: The medians of a triangle are concurrent in a point
which is two-thirds the distance on each median from the vertex to the
midpoint of the opposite side.

3329f: Let T(a,b,c) be any triangle and consider
the median S(m,c) from 0. Let
x be the point on S(m,c) such
that x =—§-mc, and let d and d‘

be those points on S(a,b) and

 

 

S(a,c) reSpectively such that b C
ad = g-ab and ad' = -§-ac. The lemma will be proved by first show-

ing that x is the midpoint of S(d,d'). By Lemma 2,
(5.12) dd' = gm...
But bd = %°2 bm = —§-bm and similarly cdl = gcm', so again Lemma 2
implies dx = %bc and xd‘ = %-m' = -§-% bc = é-bc, or dx = xd' = -13-bc.
From (5.12) it follows dx + xd‘ = dd'. Therefore x is the midpoint
of s(d,d').

Now if y is that point on the median S(b,m') from b such
that by = 23‘ bm', it follows in exactly the same manner that y is the
midpoint of S(d,d'). Therefore, x = y thus proving that any two
medians of a triangle intersect in a point x which is two-thirds the

distance on each median from the vertex to the midpoint of the Opposite

side. This establishes the lemma.

   

 

 

36

m: For the proper interpretation of the next lemma, let
the following conventions be established. Whenever x = y the midpoint
of S(x,y) will be taken as x, and L(x,y) will consist of the single
point x. Also, (xyz) is interpreted loosely as meaning simply
xy + yz = xz without any demands on distinctness of x, y, and 2.

Lemma ii: Let 0 be any fixed point and x and y arbitrar-
ily given points in M. Define the point z = x + y by locating the
midpoint m of S(x,y) and taking 2
as that point of L(o,m) such that X
(omz) holds and cm = mz. The opera- m x + y
tion x + y thus defined makes M an
additive abelian group. y

£33932: It is clear from the definition that:

a. The Operation is well-defined and closed by the fact that
M is a straight Space.

b. x+y=y+x for xeM, yéM.

c. o + x=x for all xeM.

d. -x is that point z on L(o,x) such that (zox) holds
and 20 = ox.

All that remains to prove is the associative law. Let x, y,
and 2 be three given distinct points, each distinct from o. This
involves no loss of generality, for by the continuity of x + y in M
this case implies those cases where at least one pair of points coin-
cide. In the definition of (x + y) + 2 let m1 be the midpoint of
S(x,y) thus locating x + y, and let me be the midpoint of S(x + y,z)

which then defines (x + y) + z. Next, let ml' be the midpoint of

 

 

 

37

S(y,z) which defines y + z and let m2' be the midpoint of S(x,y + 2)

thus locating x + (y + z). By Lemma 3, in triangle T(o,x, y + z) med-

   

. ' l
lans S(o,m2 ) and S(x,ml ) X + (y + 2)

intersect in a point r such

 

that (X + y) + z
(5.13) xr =33-xm1'

and

(5.110 or =33-orr12'. y+ 2

In triangle T(o, x+y,z) medians S(o,m2) and S(z,m1) intersect

in a point 5 such that

(5-15) ZS = 332m1
and
(5.16) OS = é-omg.

Finally, in triangle T(x,y,z) medians S(x,ml') and S(z,m1) inter-

sect in a point t such that

(5.17) xt = éxmlr
and
(5.18) zt = 352ml.

Then it follows that r and t are both on S(x,m1') with xr = —§-me'
from (5.13) and xt = g—xml' from (5.17), hence xr = xt and therefore
r = t. Also, 5 and t are both on S(z,m1) with 25 = %2m1 from
(5.15) and zt = gtzml from (5.18), hence 25 = zt and therefore 5 = t.
Then r = s and therefore m2 and m2' are such that (ormz) and
,(orm2') hold. But also om2 = om2' from (5.1)4) and (5.16) which im-
plies r1112 = rm2'. Therefore m2 = m2' from which it follows that

(x+y)+z=x+(y+z).

 

38

Lemma 5: Let x be given and >\ any real non-negative number.
Define >\x as that point y on L(o,x) such that one of (oxy) or
(oyx) holds and oy = >w0x. Extend the definition to negative >\ by
defining )(x = (->\) (-x) . With this as scalar multiplication, M iS
a real vector space.

23331:: Directly from the definition it follows that 1.x = x
and it is an elementary exercise to verify that (>\,U-)x = )\(,u. x) and
()\ +/I.)x = >\x +’.LX for real >\, [1. To prove that Mx + y) = >\x +>\y,
for positive X, let m be the midpoint of S(x,y) which defines
x + y and >\(x + y). Let m' be the midpoint of S()\x, )iy) which
defines )\x + )y. Put m" = )\m,

x' =>\x, and y' =)\y. It will be
proved that m" is the midpoint of
S(X',y') . Now (ox')/ox = (oy')/oy =)\
so that (x'y')/xy = X. Similarly,

(x'm")/xm = )\ and (y'm")/ym = X, or

xlm" = :C'm" =xI;:! =x'::' = xizcl
xm ym xy 2 xm 2 ym

and therefore x'm" = y‘m" = % x‘y' so tlat x'm" + y'm" = x'y'.
Hence m" is the midpoint of S(x',y') or S( )\X, Xy) and it
follows that m" coincides with m'. Writing w1= >\(x+ y) and
w2 = Xx + Xy, both “1 and w2 lie on the single geodesic L(o,m).
But if z = x + y then 02 = 2 om so that w1=>iz means

owl = X-oz = 2>\-om. Also, ow2 = 2 om' = 2 cm" = 2).~om since m"
was defined as hm. Hence ow1 = ow2 and the betweenness relations

then imply w1 = w2 or that

 

 

39

>\(x+ y) =>\x+)\y.
For h==0, the axiom in question is trivial; for X <.O, the above
proof implies C-X) (~X-Y) = (-X) (-X) + (- X) (-Y), 50 by the
definition of th for this case,

MK + y) = (->\) [-(X + y)]

= (->\) (- X - y)
= (->\) (-X) + (->\) (-y)
=)\x+>\y

which verifies the axiom for all real X.
Lemma 6: Define “x“ = ox for XEIW. Then “x“ is a norm
for M and under this norm, M is a Banach Space.
3332;: That "x" is a norm follows from:
a. “x” = ox = 0 iff x = o by the fact that the metric xy
behaves this way.
b. If y = Xx, Ilkxn =|lyu = oy = lhiox = [)\uxH, for all real i.
c. Let x + y = 2. Then if m is the midpoint of S(x,y)
and m' that of S(o,x),
m‘m = % xz, m'm = % oy
so that xz = oy. Hence from the

triangle inequality of the metric

 

it follows that
IIXWII = "2" = 02<ox+xz=ox+oy= ”XII + HY”-
That the norm is complete is clear from the finite compactness of M.

Therefore, M is a Banach space.
Lemma 7: M is finite dimensional, and therefore M is a

Minkowski space.

 

 

to

3329;: This also follows by the finite compactness of M by
standard arguments (see, for example, Taylor [19] p. 97).

These seven lemmas complete the proof of Theorem 5.2. The pre-
ceding arguments may all be localized to give:

Corollafl 5. : A G-Space is locally Minkowskian iff KB = 0.

It was stated earlier that a direct proof of Theorem 5.1 would be
indicated. That proof proceeds as follows: The local condition KM = 0
implies KB = 0, so the proofs of Lemmas 1-7 follow locally, and it then
follows that some neighborhood V of an arbitrary point 06M is Min-
kowskian. The following theorem of Schoenberg (see [17] for the proof)
may now be employed, stated here as a lemma Since it will be found useful
later:

Lemma 5.1: A linear space M is an inner product Space (and
therefore euclidean) iff for each two points x and y on the unit
Sphere,

(5.19) "X + yu 2 + ux - in 22 u.

Since the neighborhood V either contains the unit Sphere or is
isometric to one which does, there is no loss in generality in assuming
that V contains the unit Sphere U. Now let x and y be two given
points in V. Applying the condition KM =0 to the median S(o,i-(x+y))
of triangle T(o’,x,y) it follows that

I|i<x + on 2 = i “xi 2 + inn 2 - iux - in2
or

ux + yn 2 + In - yu 2 = 2 lell 2 + 2th 2

(the familiar parallelogram law) which, in particular, implies (5.19)

 

lil

if llx“ = “y“ = 1, that is, if x and y are on U. Therefore,
the metric of V is euclidean and M is locally euclidean. Conversety,
if M is locally euclidean, it is clear that KM = O.

The characterization theorems mentioned earlier wdll now be pre-
sented. Certain preliminary definitions and remarks are necessary for
each theorem.

Definition 5.7: If the sphere S in a straight G-Space is con-
vex, then it is called differentiable (Busemann Efl ) at x on S iff
no preper subset of U18 set‘ W, formed by the points on the supporting
lines of S at x, decomposes the Space.'

The Parallel £532§.£2£ 233 Dimensions: The Parallel Axiom holds
in a straight two-dimensional G-Space iff for a given line (geodesic) L
and a given point x not on L, exactly one line thrOugh x exists
which does not intersect L.

The parallel axiom may be formulated in straight G-Spaces of
higher dimension whenever the Space is Desarguesian or hyperplanes
exist. In this connection Busemann E3 , p. 1&7, proves that a
straight G-Space of dimension greater than two in which the Spheres
are convex and differentiable and the parallel axiom holds, is a .
Minkowski space with differentiable spheres. He proves another theorem.
(p. 157) which handles the case for two dimensions: A two-dimensional
straight G-Space is Minkowskian iff it satisfies the parallel axiom and
an 0L)l exists such that py“é i-(pxd' + p2“) whenever y is the
midpoint of S(x,z). It is easy to combine these two theorems into the

following characterization of E“ by means of the ptolemaic inequality.

 

h2

Theorem 5.3: A straight G-space M in which the Spheres are con-
vex and differentiable is euclidean iff it is ptolemaic and satisfies the
parallel axiom.

3323;: The necessity of the conditions are obvious; it suffices
to prove for the sufficiency that M is Minkowskian, for then the
ptolemaic inequality implies (5.19) from which it follows by Lemma 5.1
that M is euclidean. To obtain (5.19) from the ptolemaic inequality,
let x and y be any two points on the unit Sphere, and consider the
quadruple x, y, ~y, and -x. The ptolemaic inequality gives

Xy-(-y)(-X) + X(-y)‘y(-X) >X(-X)-y(-y)
or

Hx - yu 2 + ix + in 22 2.0.2.... = l.
Now if the dimension of M is higher than two the theorem of Busemann
mentioned above already implies M is Minkowskian. For dimension two,
it need only be shown that if y is the midpoint of S(x,z) and p
is any other point, there exists an «)1 with
(5.20) py‘Zi (px°“ + p2“).
But the ptolemaic inequality applied to p, x, y, 2 gives

pZ‘Xy + pX'yz 9 pyixz
or, since xy = yz = % xz,

% (P2 + PX) >133”

which is (5.20) with 06 = 1, thus completing the proof of the theorem.

The next theorem concerns itself with Fins ler Spaces, the essen-
tial axioms of which are now presented.

a. The Space (denoted from now on by M) is an n-manifold with a

differentiable structure of class Ch. Neighborhoods of M are

h3

coordinatized by means of the homeomorphisms into En (just as they are
in Riemannian Space). Thus, overlapping neighborhoods have more than
one coordinate system, so the differentiability structure sets up a
tensor calculus there.

b. Let a curve of class C1 be represented by xi = xi (I),
ué’USF. Then $2.1 is a contravariant tensor. Let a function
F(x1,x2,"',xn,E},E?,-'-,En) be defined over the 2n variables
x1, x2, ..., xn,El;g?, -.-,EP and let it be of class C3 where the
domain of the x1 is the neighborhood where this coordinate system is
valid, and that of 'Ei all of 8“. Further, let F satisfy

(1) F(X,E) > O for E+O, where x = (x1,x2, ---,xn) and
a= (£1,22,-'°,E“).

(2) Poem = Nuts).

(3) The surface F(x, E) = l in 'E- space for each fixed

x has everywhere positive Gauss curvature. Then the definition

*9 l 2 n dxl dx2 dxn d?
= F 1 1 '°' 7 ~___’___, ’__.
A, J; <x<),x<), “be. d1 d1)
will be a suitable arc-length for curves locally (a classical result in

the calculus of variations).

c. The metric of M is taken as

. Inf f F(x,dX) d’t’
(xlonec‘ at 3?
XY
where @xy is the class of C1. curves in M with
x = (x1<ac>,x2<oc),-~.xn(oc)), y = (x1(p),><2((6),“'.xn(f9)).
Theorem 5.11: A Finsler space is Riemannian with KB< 0 iff it

is locally ptolemaic.

 

lib

3133‘: The necessity is clear from Theorem li.2 and the remark
following it. A result of Busemann and Mayer [ll] will be used to prove
the sufficiency. Define for peM, 563“,

rpm) = Fp(al,az,~~,a“) = F<p1.p2.---.p‘“;21,22,m,t“).
Then for p = (pl,p2,"',Pn), a fixed point in M, the points in En
satisfying 1713(3) = 1 define a strictly convex surface U. If x and
y now be given in M and if E= x - y then since FPOCE) = IMFPG),
it follows that
Yy = mp(X.y) = Fp(x - y)
defines a Minkowski metric at p (since a Minkowski metric is deter-
mined uniquely up to an isometry by Specifying its unit sphere). The

result of Busemann and Mayer we shall use is:

lim 4— = l.
x—_>p Xy
y—"P

Suppose U‘ is not an ellipsoid. Then mp(x,y) is not euclidean,
and in view of Lemma 5.1 this means there exist a pair of points a and
b in M lying on U satisfying

Na + in2 + Na - bui<lu
If c = —b and d = -a, this is equivalent to
33-35 + EE-Bd< BEBE,
which becomes, with 06 = (Etta/(53R) and (8: (atom/(355’) ,
at + f9< 1. '
Define the points x =>\a, y = Xb, z = )(c, and w = )\d, O< )x <1,
in M. Then since 35y = )\-55, E2 = X-a'c', and so on, we have

L3” = 06 if”: =fi.

xwyz xwoyz

115

where 06 and F are constant with respect to >\. Hence

lim (LVN + w” >

 

 

X—>O xw~yz xw~yz
= n... xy-zw . ass... . xzyw . wrap)
>‘_>O $5 xw-yz 7y—w xw-yz

=ot+p

< 1
which means there is a sequence of points Xi’ yi, zi, and wi converg—
ing to p for which
XiYi‘ZiWi * Xizl'yiwi< Xiwi”yizi
holds, i = l, 2, This contradicts the fact that M is locally
ptolemaic.
Therefore, U must be an ellipsoid with the equation (using
summation convention)
gij Ei‘gj = 1
where [gij] is a symmetric positive definite an matrix defined at
p. Therefore F:(E)E gij(p) Eigj and in general
F2(x1,x2,°",xn;dxl,dx2,---,dxn) = gij(xl,x2,~--,xn)dxidxj
thus proving that M is a Riemannian Space, with RES 0.
The final theorem of this section requires the definition of

Hilbert geometry, which will now be given.

Definition 5.8: A Hilbert geometgy M correSponding to any convex

 

body U with boundary U in 8“ consists of the set of points 1.] - U
metrized as follows. If x and y be any two points of M, then the

affine line through x and y cuts U in two points u and v; define

lié \

xy= log XV'yu

 

 

fi-y’v

where 55 denotes the euclidean metric. A symmetric Hilbert geometry
is one for which the convex surface U is symmetric, that is, a surface
for which there exists a point 0 called its 3933:: such that for all
x6 U, the affine line through 0 and x intersecting U also in x',
3)? = 072' .

It can be shown (see Busemann [6] , p. 96) that M is a straight
G-space whenever U is strictly convex, with the lines and segments coin-
ciding with the intersection of M with the affine lines and segments.
The reader may not ice that Hilbert geometry is a generalization of hyper-
bolic geometry when U is an ellipsoid. The convex surface U is
often called the absolute.

We The fact that a Hilbert geometry remains unaltered
under affine transformations will be used later. This can be observed
from the fact that the expression in the logarithm which defines the
metric is a cross ratio, an invariant under affine transformations, so
that although the points themselves have been transformed, the new space
is isometric to the old.

142%: In any Hilbert geometry M with absolute U, if {xi}
and {yi} are two sequences of points in M converging to beM in
such a way that the intersections u. and Vi of L(Xi’yi) and U

1

for each i converge to aGU and c€U reSpectively, then

lim Xiyi ac .
i—'°° Xiyi 35-35

 

h?

Proof: For i sufficiently large, xiyi< Xiui and it may be

assumed without loss of generality that for all such i x1 is between

ui and yi. For any such 1, put Xiyi =’6’, Xiui = da, and Xivi =fi.

Then ’U/ab<1 and T/p< 1, so by using a Taylor expansion,

 

 

 

l = 72' log iii'iiiii
Xiy]: Xiui'yivi

= __1__,.1 é(oo+’b’;
’5 09 ac (8-1

= %[Iog(1+3’f.) ~1og (lug—)1

= _L[.§_-_Z:+_:-_H+_Z_+_’§:+_E:+...]
"L’ 06 20¢“ 01.3 fl 2’61 33

= 1+——1— 1 ‘1... 1 +_1__.’rz+ no
06 ,3 (29‘ 20c“) (3P3 3o?)

Letting i-—>°°, then ’17—’0, 06—'>EB, and ,8—*bc, so the above limit
is (1/ EB) + (1/ FE), proving the lemma.

Theorem 5.5: A symmetric Hilbert geometry is hyperbolic iff it
is locally pto lemaic.

1:32;: Again the necessity is clear; assume for the sufficiency
that M is locally ptolemaic symmetric Hilbert geometry, but not hyper-
bolic. Then the symmetric surface U with center 0 is not an ellipsoid,
and there accordingly exists a plane section C of U through 0 which
is not an ellipse. The assertion is made: There exists an ellipse E
inside C with center at o, and with four points in common with C.

To prove this, first observe that there exists among all ellipses in-

terior to C and with center at 0, an ellipse E with maximal area.*

 

*This argument is due to Schoenberg [l7] .

li8

This follows from the fact that the set consisting of C and all its in-

terior points is closed. It is clear that E must have at least one

point a in common with C, or it would not have maximal area. By the

fact that C is symmetric about 0, then also —a is a common point.

Now perform an affine transformation taking 3 into the unit circle

E' (the maximal prOperty of E is preserved under such a mapping),

C into some convex curve C', a into a', -a into -a', and assume

a coordinate system (5,71) such that o is (0,0), a' is (1,0) and

-a' is (-l,0). Suppose B' and C' have no other points in common

besides a' and -a', and consider the one-parameter family of ellipses

Be" passing through (I ___l__ , i ___l__ given by
VT \/’2’

m£2+(2-¢)722=1, l<a$<2
with conjugate radii of respective lengths 1/ Jo? and l/ m,

and hence with area

Ax = ____1_T____
V Zen-m2
Thus, since “>1, 206-062< 1 so that A¢>1T for each ellipse in the
family. But taking 06 close enough to unity, there exists an ellipse of
this family completely interior to C', and with area greater than that
of E". This contradicts the naximal prOperty of B' and proves that
E' and C' must have at least a third point b', and therefore a fourth,
-b' in common. If b and -b are the correSponding points under the
mapping, then B has the four points a, -a, b, and -b in common with

C.

 

h9

It is convenient to argue in terms of the above transformed
ellipse E' and curve C'. The assumption made above now implies
ti'at E' is interior to, and distinct from, C', with the four points
of contact a', b', -a", and -b'. Since the Hilbert geometry M'
defined by C' is isometric to the original one defined by C, M' is
also locally ptolemaic. Let f = f(6) be the polar representation
of C'. It follows that f(9) 2 l, o g 9g 11. The distinctness
of E' and Cl implies the existence of at least one 9 for which
r(9) > 1, and hence by continuity of 1(9) an interval ac< 6<fl
for which f(9) > 1. him the existence of the four points of con-
tact, it follows that 06 and p may be taken so that Mac) = l,
11p) C: l, o<a<m o<p< 11, and o<|at-/3\<Tr. Let c and d be
the points of C' corresponding to 6=06 and 938 , respectively,
and let d.’ = flawfi) and p' = %(¢+’8): .721, choosing the sign so
that Ogfi' < '11. For some neighborhood of o, the ptolemaic inequality
is valid; let x, y, z, and w be in that neighborhood and such that
they are the vertices of a rectangle
with diagonals along L(o,c) and
L(o,d), with x65(o,c) and
y€S(o,d). Hence

xy-zw + xzoyw )/ )CW‘yZ

and, dividing by H.353, this

may be written in the form

 

 

 

xyozw 35.72:; xzuyw 35-23%
(5.21) :"':'........+ .......‘:":::
xyozw xw'yz x2oyw xw-

50

But if 7/ is the angle between S(x,z) and S(z,y) it follows that

 

 

 

 

 

 

xy _ zw xz
......_"..._. = siny, = = cosy
3'6"»: 3?? W 373
and (5.21) becomes
xz ‘xw z
.1. sin27/ + . W coszy>_=__. y
is; '2‘»? 32% Tu W yz

which, in the limit as x, y, z, and w tend to 0, becomes by the

lemma,

 

 

 

(“my n.2,. . (“We“)? .0052, >2f<«>'2r<p>

r2< p .) r2( aw) ’ firzcooorzge)
or,
2
(5.22) 31— + 319321. >, 1.

f2(at') f2(fi')‘

But f(ot') > 1 and f(!9')>1. provides the contradiction

 

2 . 2
(5.23) 1< (:8 7 + 352.2:— < 0032?) + sin27 = l.
f (06') f ('3')

Therefore U must be an ellipsoid which means that M is a hyperbolic

space.

PART II

6. Preliminaries. We turn now from the study of ptolemaic
Spaces to a consideration of the class of metric spaces in which
geodesics are characterized by the identical vanishing of one of the
metric curvatures thus far introduced. We have had a measure of suc-
cess in extending the fundamental theorem to spaces which are not
localLy ptolemaic. In particular, we extend the theorem to include a
restricted class of strictly convex Banach Spaces--Banach spaces whose
unit spheres are strictly convex--but we are still unable to settle
the question with that restriction removed. However, we introduce
another quite natural definition of curvature,ocT, and for this
definition geodesics in strictly convex Banach Spaces are character-

ized by the vanishing of is at every point.

T
In considering Menger curvature, KM? an alternative definition

its suggested itself which seemed equivalent to K. In Section 7 we

M'
prove this equivalence and utilize this result to effect a new proof
of the fundamental theorem in ptolemaic Spaces, based on Haantjeso
original argument. Then in the final sections we introduce the curva-
ture KT, study the relationship between KT in a "smooth" Minkowski

Space and the classical curvature relative to the associated euclidean

metric, and prove the fundamental theorem for it in a broad class

51

of metric spaces, which includes strictly convex Banach spaces.

We introduce the notation to be used in connection.with curves,
arcs, and arc-length. A curve in a metric Space M was defined pre-
viously as a continuous image of E1, while an arc is a homeomorphism
of an interval. Rectifiability is defined here in terms of arcs and
not curves.

Definition 6.1: .A lattice P of an arc A with (distinct)
endpoints a and b, henceforth denoted A2, is any finite set of
points pienfi, i = l, 2, ..., n, ordered by < by means of the natural
ordering of the interval of which A: is a homeomorphism, such that

a =p0<p1<p2<w (pn=b.
The harm of P is defined as
max [pi-lpil i=1, 2, “'°, n]

and the length of P‘ is

pcpl + p1‘92 + p2p3 + + pn-lpn

where pq is the metric of M.

Definition;6;2: If BA is any lattice of arc A: with norm

 

A , L(PA) is its length, and R is the set of positive reals, con-

 

sider the number .EEER L(BA). The arc is said to be rectifiable with
b . . . . .
arc-length.dfixa' lff this number 18 finite.
.An important theorem used hater is

Theorem 6.1: If an are .A: is rectifiable of length [Xi then

b
lim. L P =
A om Aa
(A proof may be found in Busemann [6] , pp. 19-20.)

It is clear that if an arc .A: is rectifiable then each of

its subarcs A; must also be rectifiable. If A3): is defined as

 

53

-1X; for ysgx, it can be shown that for arbitrary points x, y, and
z on a rectifiable arc .A:,

A1 + A; = A: .
From.this it follows that if th) is the homeomorphism of the in-
terval “gag/g into M defining AZ, then A251) is a continuous,
oneutouone function of 1’, and therefore of xéAE. A parametrization
y(c'), ogchg, of a: by arc-length a’ can now be given. Define
y(0) = a, and for each a, o<a<A:, define y(0’) as that point y
on .A: such that

A geodesic was previously defined as a curve which was "localLy"
a metric segment. For greater ease of eXposition we shall confine our
attention to "geodesic arcs" which are now given a precise definition:

Definition 6.3: A.geodesic‘a£g is an arc which has the preperty
that each point on it has a neighborhood in.which that are is a metric
segment.

It is clear that an arc A: parametrized by arc- length y(a') ,
ogogAg, is a metric segment iff A: = ab with y(0’) as the
isometry of the interval into M which defines A2. It follows that
a rectifiable arc .A is a geodesic arc iff Ztfi = xy holds locally

on it.

.Another theorem which we shall use is the "n-lattice theorem“

of Schoenberg:

Theorem_6.2: For any arc A: and each positive integer n

 

there exists an n-lattice, that is, a lattice

{3 = p0) Pl: P2; "‘i Pn=b}

Sh

of A: such that

popl = p1p2 = p2p3 = non = pn_lpn°
(Schoenberg‘s proof of this is found in Blumenthal [3] , pp. 73-h.)

7. Menger Curvaturemand Haanties‘ Proof of the Fundamental

 

Theorem. Following Menger [114] , Menger curvature for curves is defined,
then a simplified version of it is given which is later proved to be
equivalent to it. '

Definition Z.1: Let p be any point on an arc A and let

q, r, and s be any triple of distinct points on A. Define

 

JTqr + rs + sq)qu + rs - sq)(qr - rs + qu-qr + rs + sq)
K'M(Cbr:5) = w v w
qr-rs-sq

 

Then A is said to have at p Menger curvature KM(p) iff to each
€>0 there corresponds a 8>0 such tint for pq, pr, and ps each
less than 3,
IKM(Q:r:S) ' K‘M(p)| < 6"
Note that there are no difficulties arising due to permutations
of q, r, and 5, since the eXpression KM(q,r,s) is symmetric in
q, r, and 5.

Definition 7.2: Let p be a given point of an arc A with

 

q, r, and 3 any triple of distinct points of A, with sq maximal

among the values qr, rs, and sq. Define

 

S(qr + is - sq?
K'S(Q)r:5) = e
qr-rSosq

 

 

Then A is said to have at p simple Menger curvature Ks(p) iffto each €>O
there corresponds a 8>O such that for pq, pr, and ps each less than 5,
‘ |‘S(q,r,5) - Ks(p)| < 6-
‘Since the restriction qr = rs in the definition of KM(q,r,s)
above effects a close resemblance between Menger curvature and another
curvature considered by Haantjes and Finsler, from now on referred to
simply as Haantjes curvature, its definition is stated here.

Definition 7.3: Let A be a rectifiable arc and p any point

 

on it, with q and r a pair of points of A such that q<r. Writ—

ing arc-length Ai on A as A(x,y), define

KH(q.r) = 2b / A(q,r) ~ qr '
A3(q,r)

Th th arc's aidt hv Ha t'scrvatr K. ‘fftech
ene 15 oae anleu ue H(p)l oa

 

€>O there correSponds a $>O such that for pq and pr each less
than S,
\KH(q,r) - KH(p)\ < 6.

Pauc [15] has proved that the existence of Menger curvature at
a point of a curve C in an arbitrary metric space implies some sub-
arc of C containing that point is a rectifiable arc, so the two
curvatures KM and ”H are comparable, that is, they apply to the
same class of curves--those which possess a sub-arc at one of their
points which is a rectifiable arc. Haantjes [10] proved that in
M implies that of ”H at

a point on an are but not conversely, and that in E2 the two are

abstract metric spaces the existence of K.

equivalent. It is well known that for curves in B” of differentia-

bility class Cu or higher if the classical curvature K(p) + O at

56

a point p of the curve, the curvatures 'CM and KH each exist there
and agree with the classical curvature. Note that in En the eXpres-

sion 1/ KM(q,r,s) is the radius of the circle C(q,r,s) passing

through q, r, and 3, provided q, r, and s are nonslinear. For

curves of differentiability class Cu or higher with. K(p) >'0, then

if q, r, and, 3 each approach p along the curve, C(q,r,s) approaches
the (free) osculating circle at p with l/ K(p) as radius. We now prove

Theorem 3:}: In any metric space and fbr an arbitrary arc .A,

 

KM(p) exists for pe A iff Ks(p) exists, and the two are equal.
Proof: (1) Suppose Ks(p) exists. Let {q,} , {ri} , and

.{si} be three sequences of points on A which converge to p, with

qi, ri, and si distinct for i = l, 2, °°', and consider i¢M(qi,ri,s i).

It may be assumed without loss of generality that the ordering is such

that for each~ i, Siqi is maximal in qiri, risi, Siqi° Then

¥SS(qi,ri,si) converges to Ks(p) and therefore risi°siqi' K H(q*, i,s. )

2
q. iri’ s. 1iq 'l( &(q ,r i’si)’ and qiri”risi' ts(qi,ri,si) each converge to

zero, hence

 

 

 

 

 

 

Q.r +r s.+s.q. qi r +r.s i-s.1q
1
lim 1 l111=1m1|: q11+2]=2
Siqi
q r -ris.+siqi qi r H+r S SiQ.
(7.1) lim ii 1 =liml:-: s;__:+2:l=2
qii
-qiri+risi+siqi qi r i+r is i-s. 1qi
lim _ = lim __ + 2 = 2
risi

But

S7

KM(qi’ri,Si) =

'ra+r.SQ+SD o 0 r0 Sv .r.-roSI+So p _ .rg+r.s.+S. .
\/q1 1 1 1 1q1.K'S(q1’ 1’ 1) . \/q1 1 1 1 1q1 \/q1 1 1 1 1q1
. . ' W ' .r. ° w . .
Slq1 \/—8- ql 1 F151

which converges to

K‘s‘P)
VFTT’ v,1,.

proving that KM(p) exists and equals Ks(p).~

 

 

 

 

 

 

 

' ”fl— : K3(9)

(2) Let KM(p) exist, Consider any three sequences {qi} ,
{r1} , and {Si} of points on A which converge to p with Siqi
maximal in qiri, risi, siqi, and with qi, ri, qi distinct, for
i= 1, 2, Since KMmi’ri’si) converges to KM(p) then
(Siqi)2'Kfq(qi’ri’si) converges to zero. That is,
q1r1*risi*siqi qiri*r151'siqi qiri'ri5i+siqi ’qiri* r151*siqi

Orl r. O or. r O
q1 1 151 q1 1 isi

 

 

 

 

converges to zero. In the four above expressions, the first, third,
and fourth are bounded away from zero, therefore the second must con-
verge to zero. For, in the first expression risi+siqi > qiri implies
that

qiri+risi+siqi qiri+qiri

/m=2,
qi’i qiri

 

in the third eXpression, siqi g risi implies

qiri‘r‘s°*siqi qiri + 0 =

 

1 —- v mull- 1’
qir‘ " girl
and in the last, Siqi > qiri implies
'qiri*r131*siqi 0 + r151 g 1_

 

 

. w / O C
rlsi r151

Therefore

. ¢+ . . - a ..
qlrl r151 Slql

(7.2) lim r—S. = 0.

Now, since Siqi/risi )1, then from

Siqi ° qiri+ri5i'5iqi =1im qiri+riSi-siqi = o
risi Siqi risi

lim

 

 

it follows that

0+ o 0— c c
qirl r131 Slql

(7.3) lim siqi = O.

 

Next it will be shown that

qiri*risi'siqi
(7.u) 11m qiriv—~—— — o.

 

Suppose (7.b) is false. Then there exists a subsequence {j} of {i}
and some €3>O (since the expression in (7.h) is non-negative by the

triangle inequality for every i), such that for all j in the subse-

 

 

 

quence,
(7.5) qffrjsj'sjqj >e.
quj
Consider
. . . .+ . .- . . o .
quJ . quJ rJSJ SJqJ = quJ
~— ~ 1 W )
quJ+rjsJ'SJqJ sjqj sjqj

for each j. From (7.5) the first eXpression on the left is bounded
while the second converges to zero by (7.3). Hence
q r'
(7.6) lim 3 J = O,
3 quj

which implies, also from (7.3),

 

 

59

r.s.
(7.7) lim. .2.2 = 1.
J quJ

Hence there exists a jo such that for all j) jo
r,s

(7.8) .41 > i
quj

. . . 2 2
Next con31der, for J > JO, (rjsj) . KM(qj,rj,sj), or

fjrj*rj3j*sjqj quj+rjsj'sjqj quj'rjsj*3jqj ‘quj*r38j*5jqj

i?“— I —' ‘— 0 O

 

 

 

 

sjqj quj quj quj
By the triangle inequality,

quj*rjsj*sjqj sjqj*sjqj . 2
‘ /

 

 

 

SJqJ sjqj
by (7-5) )
q.r .+r .s .-s .q.
J J J J J J
r"‘>€’
qj j
from quj ) rjsj,

 

 

quj'rjsj*sjqj quj * 0 . 1,
I‘ V/ -I‘
qjj qJJ

and by (7.8), for j) jo, and using 5

'QJrj*erj*SJQJ 0 * erj

 

 

Sfii %%
Therefore,

2
(rjsj) 'K§(qj,rj,sj) > 24-12% = e

 

60

for all j) jo. But this contradicts the fact that KM(qj,rj,sj) con-
verges, hence (7.h) holds.
Making use of the limits (7.2), (7.3), and (7.11,) it follows

that

KS(qi’rl’Si) =

 

0 0+ I o- 0 I
qlrl 1‘15 1 Slql

 

II
a
GI
H4
h:
H
HI
”’1
Poi
(I)
H
.9
H

V 8 ' KM(qi’ri’si)

O .+ o l+ o o o O- o c+ o - o a o o u
fqlrl r131 Slql qlrl 1T151 slqi C111‘14'I‘is’1'r5‘:iqi

 

 

 

 

Siqi qiri 1‘15:

\/ 8 ° KM(qi,ri,si)

 

i i i i- i i
Siqi .

 

 

2+-

 

 

converges to

x/T-KM(p)
V5.7? w—

proving that Ks(p) exists and equals IcM(p). This conpletes the

_. = K'M(p)

 

proof of Theorem 7.1.
Remark: Suppose that KM(p) exists for peC. Then let
{qi} , {r1} , and {Si} be three sequences of points on C which

converge to p in such a way that for each i = l, 2,

Hence

 

61

2 qiri

+5171 2 ‘YiIi'Siqi ,

 

 

2 2 _
(risi) 'KM(qi’ri’Si) qiri qiri
which converges to zero. Since (2 qiri+siqi)/(q3ri)j).2 for each

i, it follows that

 

 

 

 

o 1' 2 qiri'siqi 2 1, Siqi ,
= 1m ‘ = - 1m
qiri qiri
and therefore
q.r. r.s.
(7.9) lim 1 1 = lim 1 1 = %.
Siqi Siqi

It follows that there exists an iO such that for all i‘> lo
qiri = risi‘<_siqi, and hence by Theorem 7.1,
Km(p) = lim KS(qi’ri’Si)'

Mq- 8-)
If X(qi,si) is defined as 2 qiri, then by (7.9) lim 1’ 1 = l

 

 

 

 

 

qisi
and therefore from a
< ) 8 ( 2 qiri-siqi) _\/32 ( >((qi,si)-siqi) . )\(qi,si)
K q. r. s. = 2 - 3 “"""
S ’ ’ .
1 1 1 (qiri) Siqi X‘(qi’si) Siqi

the result is

 

)\(Qiasij - qisi '

lim (/3§ °' '

 

(7.10) KM(p)

This eXpression bears a striking resemblance to the one for Haantjes
curvature and suggests the possibility of a proof of the fundamental
theorem for Menger curvature that closely parallels the proof Haantjes
gave.

For the purpose of simplicity, throughout the remainder of
this thesis we shall confine our attention to the characterization
of geodesic arcs by identically vanishing curvature among all resti-

fiable arcs.

 

 

 

62

Theorem 7.2: In any locally ptolemaic metric Space M a
rectifiable are A is a geodesic arc iff KM(p) = O for each point
pGA.

25293: The neCessity of the condition being obvious, suppose A
is a rectifiable arc with KM(p) = O for p61). Restricting the arc
to a neighborhood Vp of p in which M is ptolemaic, there
exists an arc A' with endpoints a and b, a<b, such that A'CA
and A'CVp. It suffices to show that A(a,b) = ab proving that
A' is a metric segment. Let e>O be given and consider xeA’.
Since KM(x) = O, by (7.10) there exists a 3x>0 such that for every
triple of points q, r, and s in ll' with qr = rs and xr<8x,
xs<3x,

X(q,s) - C15 6
)\B(q,s) 16 A3
holds, where )\(q,s) = 2 qr and A = A(a,b) > 0. But if

 

 

(7.11) o 4

WX = {w | wx < 8x} then {wx} x e A' is an Open covering of the

.0. w .
X2, ) xm
It is at once clear that there exists a 8>0 such that if now q, r,

compact set A', whence there is a finite subcovering le,

and ‘ s are any three points of A' with qr == rs and qr<S then
q, r, and s all three will lie in at least one of the wx and hence

I
will satisfy (7.11). By Theorem 6.2 there is an n-lattice of A'

a=po<pl<p2< <Pn=b
where popl = plp2 = ... = pn-lpn = )‘n° By the rectifiability of A'
and Theorem 6.1 the limit of the lengths of these n— lattices, n X“, as
n—voo is A(a,b), hence there is an n1 such that for all n > hi,

O< A(a,b) - an<§ , and since >‘n< (l/n) A(a,b) there is an

 

63

n2 such that for all n > n2, )(n < 3. Let n be any integer larger
than either nl and n2 and consider the correSponding n-lattice

{pk} , k = o, 1, 2, m, n. It then follows that
(7.12) 04 mam - nxn<g

and for k = l, 2, -°-, n-l,

 

 

 

o < >\(pk_1,pk+1) - pk_lpk+l e ’
\~ 1“.
3 3
>\ (pk-l’pkH) 16 A
which reduces to
)‘3 e
(7.13) O<2Xn-pk_1pk+l< n , k= 1, 2, ”°, n-l

2A3
from >\(pk-1’pk+l) = 2)‘n‘

The points p0, pk_1, pk, and pk+1 for each h, k=1, 2, °°°,
n-l, are ptolemaic, hence

popk-1’pkpk+1 + popk+1'pk-1pk > polD k’pk- 1pk+ 1

which reduces by (7.13) and pkpk+l = pk_1pk = kn to

 

2
Xne
popk-1+ popk+1 > pOpk (2 ' 2A3)

or, 2
in e
2 pOpk < popk-1 + popmi + popk° 2 A3
and using popk é; [\(a,pk) << I\(a,b) = I\ this becomes
2
( 1b) 2 ME
7. . popk < popk_1 + popk+1 + “-1- °

2A

Define

ak=kxn-p0pk’ k=031929 "'sn:

 

621

and substitute into (7.1h) The result 18 the set of n-l inequalities

X 2
(7.15) 06 -0c (on -a. + “6
k+l k K k-l 2 A2

 

, k = l, 2, --o, n-l.

For each integer j = l, 2, -'~, n-l sum (7.15) over k = l, 2, °--, j

to obtain the n-l inequalities

 

2
j X
“n+1 - d-‘< d, - (x + _~_£LE.
J l J O 2
2 A
But “1 = 040 = o by definition and j xn< nkngA so
>\ e
(7.16) 06.+1< a. + n , j= 1, 2, m, n-l.
,J J 2A

II

Summing (7.16) over j l, 2, °--, n-l,

 

-1
062+oc3+ooo+ocn<wl+ 062+"°+O(, 1+(n)_&\in_1€
n...
2A
and again since (n-l) )\n< n)\ éA this reduces to
n
6
ocn< 2

From (7.12) it follows that
O< [\(a,b) - ab
A(a,b) - min + n kn - popIn
(A(a)b) " nxn) + “n

e E
< '2"?

and therefore
0 g A(a,b) .- ab < 6.
Since [\(a,b) and ab are constants and 62>O was arbitrary then
A(a,b) - ab = 0 must hold which proves that A' is a metric

65
segment and that the given arc A is a geodesic arc.

8. Haantjes'Eroof_in Non-Ptolemaic Metric Spaces. The attempt
was made to adapt the Haantjes proof of the fundamental theorem (essen-
tialty the proof just presented in the preceding section) to Banach
spaces with strictly convex unit spheres. It is clear that if strict
convexity is not required, then the two-dimensional space (x,y) with
unit sphere as [X] + \yl = l affords a counterexample: the curve
y = x is not a geodesic but it has zero 1cm and KH at all its
points. The resulting partial solution to this problem applies to cer-

tain non-linear metric spaces.

 

Definition 8.1: .A metric Space M ‘will be said to haV€.HEE§$X
non~positive median (Space) curvature iff there exists a positive
constant 71p<l and a neighborhood Up for each point pEM such
that if a, b, and c are any three points in Vb and m is any

midpoint of a and b, then

 

(8-1) yp-czn < J95: 2102 +12 bc2 — £21132 .

_Qefinition 8.2: .A metric Space M is said to have feebly

 

nonrpositive median (Space) curvature iff there exists a neighborhood
Vé of each point 13614 such that if’ a, b, and c are any three
points in VP with m any midpoint of a and b, then
(8.2) cm<%(ac + be).

Unfortunately, the condition for "feeble" does not seem to be
a special case of the condition for "weakly" nonppositive median curva-
ture. Both conditions are needed in the proof of the fundamental

theorem given in this section, as is the existence of midpoints, so

66

there M will be assumed to be (metrically) convex and complete. If
this is not assumed, then (8.1) and (8.2) could be satisfied trivially
without imposing any condition on the space.

It is interesting that in Banach Spaces (8.1) is a form of what
will be called the "weak" parallelogram law.

Definition_8.3: The norm ”x" of a Banach space M for xesM

 

is said to satisfy the weak parallelogram law iff there exists a posi-

 

tive constant 7W<l such that for every x and y in M,
Ma) nx+wfi+7%x-wfigznmfi+2nm2.

To see that (8.3) implies (8.1) in any strictly convex Banach
Space, given a, b, and c, set x = a - c and y = -(b - c). Since
m = %(a + b) is the only midpoint of a and b in this case, it
follows from (8.3) that

h-bF+17%aa+m-cV<2Ha-d2+2h-cfi
or,

ab2 + byzacng 2 ac2 + 2 ch

which is (8.1). In euclidean Space, trivially (8.1), (8 2), and
(8.3) all three are satisfied in the large with 7’= 1, while for
arbitrary metric spaces, M has weakly nonrpositive median curva-
ture if it has nonrpositive median curvature, has feebly non-positive
median curvature if it is locally ptolemaic (see the proof to Theorem
5.3), and in a Banach space, the weak parallelogram law holds if the
parallelogram law holds. Following the proof of the next theorem,

the question of what non-ptolemaic metric spaces have weakly and

feebly nonppositive curvature will be discussed.

67

Theorem 8.1: In any complete, convex metric space M having

 

both weak and feeble non-positive median space curvature, a rectifiable
arc .A is a geodesic arc iff it has vanishing Menger curvature at

each of its points, and iff it has vanishing Haantjes curvature at each
of its points.

5223:: The necessity of the condition is clear. For the suffi-
ciency, let KN! and KH vanish identically along a rectifiable arc A.
There exists a sub-arc A' of A containing an arbitrary point peaA
which is contained in a neighborhood V? of p in which (8.1) and
(8.2) are valid. As in the proof of Theorem 7.2, let a and b, a‘( b,
be the endpoints of A'; it suffices to prove A(a,b) = ab. Let O<E<A(a,b)
be given. In order to handle both KM’ and KH’ simultaneousty, the
following notation will be used. Let
(8.11) a=po<pl<p2<°~<pn=b

be an n-lattice of A‘ in the case of '(M with kn for

z Pk- 18h
k = 1, 2, ..., n, where n is so large that both

(8.5) _2w}n.+ Pk-lpk+l <1 6' , k = 1, 2’ ..., n-l,
(2 kn)3

the choice of e' to be made later, and

(8.6) o g A(a,b) - n)\n<§ .

In the case of KH, let (8.b) be a lattice of A' such that
A<po:p]_) = A(P1,P2) = '°' = A(pn_1>pn)

with 'Xn = [Hpk 1,pk) for k = l, 2, °-', n where n is so large

that

(807) KH(pk-1’pk+1)< e" k = 1) 2) ...9 “‘1,

 

68

and, in this case for any choice of n,
- = E .

(8.8) A(a,b) nkn O<§
But (8.7) is the same inequality as (8.5) and (8.8) and (8.6) are the
same, so from this point on the proofs for KM and NH will be iden-
tical. Moreover, it is sufficient to Show how to obtain the set of n-l
inequalities of (7.15) for, from that point on the proof will be identi-
cal to that of Theorem 7.2.

For each k = l, 2, °°°, n-l let
mk be any midpomt of pk—l and

pk”, the existence of at least one

 

being assured by the fact that M is
convex and complete. From popké pomk + mkpk and by (8.1) and (8.2),

it fol lows that

 

pOpk Q %(p0pk_1 + pOpk+1)+( 1/7) \/ 324131“ lpkfgpki); 1)2-%(Pk_ 113k... 1) 2

while pk-lpkgxn for k = l, 2, °~-, n reduces this to

 

2 9091: < PoPh-1 * Popku " (1/7’) \/ (2 >\rr1"h--1Ph+1’(2 >‘n+pk-lpk+l)

But by (8.5)

3
2 >‘n - 1Dk-lpkfl <8 )‘n 6'

and

2 >\n * pk-lpk+l< 2 >‘n + pk-lpk + pkpk+1<2kn + )‘n + \‘n = M‘n

so therefore,

 

2 pOpk< pOpk-l “ pOpk+l * (1/7’) \/ 32 >‘nu 6' '

Now 5' is chosen as (7)2 €2)/(123 Ab) and “k = k >‘n " Pop}; 3
= 000 ° . 8 - 3 ' M-
for k 0, l, 2, , n are defined, with 042 2km p0p2<8>\n e < 16A}.

By substitution,

69

 

X26
n
- - -d -
2k).n 2“k<(k1)>‘n k_1+(k+1) >‘n dk+l+ 2
2A
which simplifies to
X26
8. a - w d - -——-2“
( 9) k+l k< k ak_1+2A , k=1, 2, ..., n-l,

identical to (7.15). The remainder of the proof proceeds as in Theorem
7.2, hence A is a geodesic arc.

The following discussion will serve as a sketch of the proof
that any Riemannian surface has both feebly and weakly non-positive
median curvature. It is clear that in view of Theorem 11.2 a
Riemannian me with non-positive classical curvature has non-
positive median curvature and is locally ptolemaic, and therefore
will have both feebly and weakly non-positive curvature.

Consider any 2-Sphere Sf. with radius ‘0 in 153. Letting
{ai} , {bi} , and {Ci} be any three sequences of points on 5‘.
converging to a point p 6 Sf” for all sufficiently large i the
following is true:

(a) mi = maibi exists.

(b) The formula from Spherical trigonometry

 

C03

is valid.
b.c. a.c. . .
Let “1 = c-IF—l , pi = .1: ’ and 71 = Cit-i . Again, for 1

large enough, the following manipulation with Taylor series is permis-

sible:

7O
2<1- W = 2 - E°S “131151.
(2 cos hii
2 2 222212 ' 22:2 + 0(di3) + 2(212)
1 .. .5.in + 0(7),?)
2 2[2222222221222(212)")(121) 2][122222+°(712)]
22 222212 27212132222222 22?*0(d3)+o(p?)+o(73)

 

 

2-

 

Therefo re,

01112 cm.2

i i i 1
_ ._""2.fi = 2137 {13227 12#12 ”(22225 222712 1
2 1 2 C2212— Baciz

 

 

fimi 222i - 3% 2(l-cos .175J) 8&2 - it”?
Now
lim 1° - 1

 

i—+°° 2(l-cos 2.1.3.)

and it is easily verified that
2 2 2 2 3 3 3

1im°‘171*p1')’i +0<ai)+o(pi)+o(2yi) =0

. w” 7 _ 7 2

1—roo -

8% * 821 h7i

regardless of the way in which the values wi’fii’Yi converge to zero.

Hence there cannot exist three sequences of points converging to p for

 

 

which .
2°imi2
we no: + in: +°<£>1+°<7131> .
2 (l-cosigii.) 2 82212 2 8&2 2 27212 J

so there evidently exists a 2? depending only on (0 such that if a,
b, and c are any three points of St with ap<¢$,, bp<6, cp<8r

and m the midpoint of a and b writing as before 22221222 (93752"

and 7’ = —, then the above emression in at, p, 7;, c, and m is not greater

71

than 14, that is,

 

 

C1712
“'2-
‘° — — < 11
1:062 + £182 - 9:72 ‘
OI‘
cm? )4

fibcz :fac2 - fiab2
which is (8.1) with 7p = g. It is well known from Spherical geometry
that the median of every triangle in a small enough neighborhood is
less than half the sum of the adjacent sides, which is (8.2). This,
together with the preceding paragraph shows that non-ptolemaic spheri-
cal geometry has both weakly and feebly non-pos itive median curvature.
The result of Wald [20] for Riemannian surfaces is now

employed. Define Sk for each real number k to be the 2-sphere in

BB 2

of radius l/k if k > o, to be the euclidean plane E if k = o,
and to be the hyperbolic plane of curvature k if k < O. For a
Riemannian surface M, let V be any neighborhood of an arbitrary
point p having unique segments. The theorem of Wald states that

each four points of V can be isometrically imbedded in some 5k,
where it lies between the upper and lower bounds of the Gauss curva-
ture K in V. Then already (8.2) is clear. For (8.15, if 11-40
there is no problem; if k > 0, using the above remarks, if the im-

bedded points lie outside the neighborhood of radius 8 then the

l/k
neighborhood V is chosen smaller, but It is changed (perhaps).
However, 81/}: is evidently a decreasing function of it so if k'

is the least upper bound of it, it suffices to take the radius of V

72

to be 81/1“, for then the imbedded points will lie in a neighborhood

of radius 8 g Sl/k; That is, those points lie in the desired

l/k‘
neighborhood on the sphere 51/}: and therefore (8.1) holds. It is
natural to conjecture from the locally euclidean character of any
Riemannian space, that a Riemannian n-manifold has both feebly and
weakly non-positive median curvature.

There are some strictly convex Banach Spaces which satisfy (8.3)
and therefore (8,1)“ The condition (8.2) is of course always satisfied
by a simple application of the triangle inequality for the norm“ The

following theorem gives a sufficient condition for (8035

Theorem 8.22 Let M be a Minkowski space with strictly convex

 

unit sphere U with center or If there exists an upper bound (0 for
the diameters of all ellipses with center 0 and passing through three
distinct points of U, then the norm of M satisfies the weak parallel-
ogram law.

£3922: Let '55 denote an associated euclidean metric of M.

Then put 0’ = Min EU and take
uEU

7= Min(i ,V/f),
hence, O< 7(1. The inequality
ux + yu2 + 72~ ux - yu2< 2 M2 + 2 M2
is trivially true for the cases x = o, y = o, and y =)\x, )k real.
Assume y + Xx for all real X and that x + o and y 1: oc If
x', y', and z' are the points where the rays from 0 through x, y,
and z = x + y respectively intersect U, then these will be three

distinct points of U. By the strict convexity of U, there is a unique

73

ellipse B through xi, yi, and 2‘ with center oi The argument

may now be restricted to the plane of E, Put w = x - y and let

w' be the point of intersection of the ray from 0 through w and U,

and w" the point where that ray intersects E. Letting “X“B repre-
sent the norm with unit Sphere E, since E is an ellipse it follows
that "x"E defines an inner-product
Space and hence satisfied the parallel-

ogram law
h+yM2+h-y%2=dd§+2hhi

But as the unit spheres U and E of

the two spaces coincide at x', y‘, and

 

2', it follows that

2 2 2 2
(8:10) "X + Y" + “X " YHE = 2 "X" + 2 ”3’" -
Now a" < Diameter FAQ? so that l/{o < l/ .675", while ago}?! im-

plies that O’/(o<( 317v / BEE" ). That is,

m

 

(8‘11) 7 < if: e
ow"
Then from (8.10), __ 2 __ 2
ix + y“? + flux - yn 2< “X + yu2 + (332) (it)
67¢" ml
2

"X + yll2 + (g)
ow"

2 2
nx+yu+ux-ynB

2 2
, 2 llxll + My“ .
We: The two-dimensional Minkowski spaces {(x,y)} given

parametrically by taking vi)? as the curve

7h

2
{x + Xxy + y2 l, for (x,y) in first and third quadrants

x2 - luxy + y2 l, for (x,y) in second and fourth quadrants
where o < >\<2 and o </‘<2’ clearly satisfy the hypothesis of the
above theorem, and so is a class of non-ptolemaic Banach spaces in
which the fundamental theorem ho lds°

I'tzfollows in a similar manner that a two-dimensional lp Space,
1 < pg 2, satisfies the condition of Theorem 8.2 since its unit Sphere
is given by

IXIP + IYIP = 1

which has everywhere positive curvature, infinite curvature at (:1, O),
(0,.tl) if p + 2. Also by curvature considerations it can be shown--
although it will not be undertaken here--that finite dimensional lp
Spaces, l< p g 2 in general satisfy that condition.

A Banach space which do es _n9_t satisfy the weak parallelogram
law is the set {(x,y)} metrized by taking as unit Sphere

’4 = l, [)4 4 1. Consider the points on the unit Sphere

 

 

lyl +x
a=(‘l’,l-Th) b=(-’C’,l-’6'h) O<1’<l.
a + b = (0,2-2’6L‘) and a - b = (217,0) imply that
|Ia+b|| =2-21’L‘, Ila-bl] =2’6.
But then '
enan 2+__2_ non 2- na+ b|12 = h-jz - ”if = 2.12 _ 16
[la—on? M2

which converges to zero as ’C——>O.

9. Transverse Curvature. Attention now turns to a different

concept of curvature for arcs from those considered earlier. The

 

7S

definition will be given in as general a metric space as possible, and
for as general a curve as is possible, then emphasis will be on apply-
ing the definition to arcs in CE-Spaceso

Definition 9.1: Let p be any point of a curve C

 

in a metric space having unique local segments, and suppose Up is a
neighborhood of p in which M has unique segments. Given q, r,
and S three points on C and in Vp such that q<r<s and qr = rs,

define

8mr

K'T(CI:r:S) = --—3;:- °
qs

Then the curve C is said to have transverse curvature KT(p) at
the point p iff to each €>O there corresponds a 8>O such that
whenever qp, rp, and Sp are each less than 8 then

lKT(q,r,s) - KT(p)| < E.

It will be apparent that this curvature has some advantages
over the others, although it is not known precisely how it compares
with them. We do not exp lore that problem here. However, we do make
a few comparisons of this curvature with so-called classical curvature
for euclidean Space. We take as the definition of classical curvature
the following:

Definition 9.2: For any arc in En the classical curvature

 

K(p) exists at a point p of that arc iff for any sequence of circles
{Ci} which have three points qi, ri, and si in common with the
arc for i = l, 2, such that lim qi = lim ri = lim si = p, then

the sequence {l/oci} of inverses of the radii “i of C., for
1

76

each i, converges to K,(p). (A straight line in En is defined to
be a circle with zero as the inverse of its radius.)

This strictly geometric definition of curvature actually coin-
cides with that of Menger curvature when applied to the metric of En.

Now suppose C is a circle in 32 with radius at. To find
the transverse curvature of C at a point peC, let {qi} , {r1} ,
and {Si} be three sequences of points on C with qiri = risi
and each sequence converging to p. Then if 0 is the center of C,
writing mi for mqisi for each i, it follows that

1 1

K. ( 8 m.r.
T q12r135i) ------2
(qisi)

8(OL - omi)
.w 2 __
Mmisi)

 

2((1 - omi)
2

 

0L2 - om,
1

l

 

a - f miri
Since lim m r. = 0 then
{-900 1
limit (q,r,s)=.]:.="(p)= K'(ID)
1*” T I, 1’ i m T

n
More generally, let A be any arc in E whose classical curva-

ture exists at a point p 6 A. Let {qi} , {ri} , and {Si} be
three sequences of points on A each converging to p such that for
each 1 = l, 2, on, qi’ ri, si are distinct pOlnts and qiri= risi.

Put mi mq S . Let {j} be that subsequence of {i} obtained
i i

77

by considering those triples qi, ri, si which are linear, and let {k}
be the sequence remaining. Consider the case when both subsequences

{j} and {k} are infinite. By the definition of {j} , mjr' = 0
J
for every j and Q5.) 0 50 K' (Ci.:1‘.,3.) = O, and therefore
3 J T J J J

lim IC (q.,r.,s.) = 0.
j T J J J

For each k, the radius 06 of the circle through q

k {, rk, and s

1 k

exists, and the calculations above Show that
l
06k - Q mkrk

The linearity of an infinite subsequence of triples qi, ri, Si implies

 

5

K =
(9 l) T(qk,rk,sk)

that K.(p) = O, and its existence implies for the subsequence {k}
that oak-+00 , hence

limK( rs)=O
k qu’k’h

proving that KT(p) exists and equals K.(p). The case in which

{k} is a finite subsequence leads to the same result. Finally, when
{j} is a finite sequence, it may be assumed without loss of general-
ity that {i} = {k} and if (Mk—+00 then again (T(p) exists

and equals K.(p) = 0. Otherwise, lim “k exists and since

mkrkg rksk + mksk = rksk + £ qksk then lim mkrk = 0 so (9.1)
again shows trat KT(p) exists and equals Mp). This proves the
theorem:

0 n -I
Theorem 9.1: For arcs in E the transverse curvature eXlsts

 

 

at a point of the arc whenever the classical curvature exists, and the

two are equal.

78

In order for Definition 9.1 to have any meaning in a Banach
Space, it must have unique segments and therefore strictly convex unit
sphere. The fOIIOwing property is needed for the next theorem.

Definition 9.}: ,A Minkowski Space is said to be 322222 iff its
unit Sphere has at each of its points a unique hyperplane of support.

In.a smooth Minkowski Space, if p is any point in M let Up
be the unit sphere with center p. The sphere U? will have the same
"smoothness" prOperties as U since they are congruent bodies in 3“.
Suppose plane ‘nb passes through p and cuts Ub in Cp. Then Cp
is a strictly convex curve in ’fl? with unique support lines at each
point up of Op. The diameter S(up,-ub) of Cp is uniqueky deter-

. . J. .L
mined u as is the con u ate diameter 5 u

,-u§L ) where u
and -uéL are defined as the points of intersection of the line

through p parallel to the unique line of support at up of Cp.
If L is any line through p cutting Cp at the points 111 the

P P

line L(u -'-,-u -'~) will be denoted L‘L .
P P P
The euclidean area of a parallelogram with adjacent sides of

unit Minkowski length along two distinct lines L and Li
will be denoted A(L,L'), while an.) will denote the euclidean
area of a square with side of Minkowski unit length along line L. It is
clear that n(L,Ll) and A(L) are respectively equal to A(K,Ki)
and A(K) whenever K is a line parallel to L and K' is one
parallel to L'.

Finally,'we need definitions for the existence of tangents and

osculating planes. For greater simplicity these concepts will be

79

stated in algebraic terms, the first definition after Kelly and
ewald [l2] .

Definition 9.b: If for any two sequences -{q€} and .{rg}
of points on an arc A converging to p e A and qi<( ri holds for
each i the limit

1‘. — q ]
lim p + _£ 1 = t
i—voo [ r,q, p

1 1

 

exists, the line L(p,tp) is called the free tangent to A at p.

 

 

Definition 9.5: If for any three sequences -{q{} , {rg} , and
{:si} of points on an arc A. converging to peEA, and qi<: ri<: Si holds
-qi

s
for each i with x any limit point of the sequence {:p + and

m, - r
yp any limit point of the sequence {:p + g r i}. , where
i i

the plane 1fp = 11(p,xp,y ) determined by p, XB’ y is unique, then
P p

 

that plane is called the free osculating plane to A. at p.

 

It is easily shown that the free tangent and osculating planes
of an arc are geametric objects of En, independent of the particular
Minkowski metric defining them. That is, the free tangent of an arc
exists in En iff it exists in M, and the two are equal. A similar
statement holds for the free osculating plane. Standard examples Show
that the existence of the free tangent does not guarantee that of the
free osculating plane, nor does the existence of the free osculating
plane ensure the existence of the free tangent, as a plane curve with-
out a tangent shows.

BEES? A result of L. Kelly and o. Bwald [12] which will be
implicitly assumed in the statement of the following theorem states

that (as applied to our case) if the classical curvature of an arc

80

exists at a point of that arc, then it has a free tangent there.

Theorem 9.2: If M is any smooth Minkowski space with

 

strictly convex unit sphere and A is an arc in M for which both
the classical curvature and the free osculating plane exist at some

point p€A regarded as an arc in B“, then the transverse curvature
K-.I.(p) exists atpMoreover, if the free tangent to A at p is

Tp = L(p,tp) , then KT(p) is given by the formula
2
2

MP) A (T)

if
ATT
(10,10)

 

“T(p) =

Proof: Without loss of generality it may be assumed that p = 0.
Let {qi} , {r1} , {Si} be any three sequences of pomts on A
converging to p such that for each i, qi< r1 < Si and qiri= risi.

From the existence of the free tangent at p (and since p = o), if

 

 

I‘ "" q S - I‘. So - q.
ai = 1 1 , bi = _i_____l , Ci = 1 1 ’
r. 0 Or. I D
lql s1 1 slql

then the sequences {ai} , {bi} , and {Ci} each converge to tp,

some point on U. It is now shown that if tp'L and -tp are the

endpoints of the diameter of U which is conjugate to S(tp,-t ), the
p

sequence of elements

a = 1 1 i=1, 2’ 000

 

either converges to t:, converges to -t:‘ , or can be divided into

two subsequences, one converging to tgL, the other to --tp . Since

'n'iie U for each i, ii, is bounded and thus has at least one limit

point. Consider in any such limit point and let {j} be any

81

subsequence of {i} such that

lim 5. = m.

j J
The line L through tp parallel to
L(p,m) is the set of points

{tp+ Xml >\ real},

Suppose L is not a line of support

 

at tp. Then there exists a second
point q‘+ tp in common between U

and L. Then for some X,

 

q = tp + )\m
which means that
"kn" = Ill = Na - tpll = qtp
or )x = :tqtp.

Since the argument for both cases is similar, it may be assumed that
)\ = qtp and that therefore
= t + t 0
q p (q p) m
Put 1‘ = fi- (tp + q) = tp + § (qtp) m. By the strict convexity of U,
r is an interior point of U. Therefore “r“ <1. Consider the

sequence of points

r.=b.+ (qt)r'fi..
J i p J

 

J
This converges to r since lim b. = t and limil'l, = m. But since
J J P J
.. r - it (q. + S)
m a J
J r.
JmJ
r q 1‘ q s r s r
= % ‘—-L—1 o .J—I—Iil - "£0 .J—ai o uni-.3
r m r.q, rm 5
J J J J J J J J

 

 

q.r,
=%s—J.—J (a-b)
m.r. J
J J
a. - b.
ajbj
then
a. - b.
Y‘.=b,+ t M: .a.+(l- .)b.
J J %(qp)a.b. ’uJJ "LJ J
J J
where = t /2 a.b.. Since lim a.b. = t t = 0 for all ‘
(‘3' (qp) JJ jJJ pp ’ J

sufficiently large ij will be greater than unity and the corres-

ponding ?. will be exterior to the segment S(aj’bj) and therefore
J

to U, hence ”Ej" )>l which results in the contradiction
1>l|r|l =11... it,“ >1.
J

Therefore, L is a line of support.

Next, L is in fi%, the free osculating plant of .A at p, for

‘3

by definition ‘fl' is the plane determined by p, lim a,, and lim'mJ
p o o

J
or p, tp, and m. Hence tp + m is also in that plane. But tp and

tp + m are then two distinct points of L lying on ‘flL, so that L
is in 11p. L must accordingly be the unique support line of U'r)r$

at t . Since L(p,m¥d.coincides with TEL where Tp is the tangent
P

F

to .A at p, m is either tJ' or -t . Thus the only limit points of

pro

the sequence {ii} are +tp'l' and -t , proving the assertion. It

*0

follows that since pt = p(-t:), then iffy is a euclidean metric for M

P

-.L
lim pm, = pt .
Also, it follows from lim ci'= tp that
i—voo
(9.3) .lim pci = p p .

1—->oo

83

Now if t, is a point on A r ti
1
for each i such that ‘E-i—S-i = Eiqi’

let

 

and suppose that Obi = m(ri’ti’mi)

where m(x,y,z) denotes the

 

(euclidean) measure of angle (x,y,z),
and, fli = m(ti’ri’mi) .. a1 equals either m(xi,p,yi) or 1T- m(xi,p,yi)
so that sin “i = sin m(xi,p,yi); also pi equals either m(yi,p,mi). or
“If - m(yi,p,'n'ii) and therefore sin fii = sin m(yi,p,7ni). Again by the
existence of the free tangent to A at p, the only limit points of {yi}
are itp~ Also, since L(p,xi) is always orthogonal to L(p,ci) then
{Xi} has at most two limit points, x and -x such that L(x,-x) is
orthogonal to Tb. Since
sin m(x,p,tp) = sin m(tx,p,:tp)
this proves the only limit point for {sin m(xi,p,yi)} is
sin m(x,p,tp) , that is

(9.)4) lim sin at.i = sin m(x,p,tp) = Sin {Tr = l

i—v°°
and similarly,

. . _ .L
(9.5) 11m Sin Pi sin m(tp,p,tp ).

i~>°°

Therefore,

 

8h

 

 

 

 

sin “i 35

 

qisi sin fl, pini

with the limit, since K(P) = lim (8 3121,) / (32392,

i—voo
1 F3132
Mp) ' . ..

 

. .L
Sin m(tp,p,tp ) 35:
Mp) (3383

5'5 555* - sin m(t ,p,t ‘1‘)
P p P p

 

proving that KT(p) exists at p and that
3
MP) AUTP)

.L
TT
A(p.p)

 

blip) =

The resemblance of the above formula to one obtained by
Busemann [5] is close. His formula involved a different definition
of curvature x(S) where S is Minkowski arc length and the same
curvature SKS) under the associated euclidean metric. The formula is

x<s> =§<§> crap 63(5)

where 0" (t2) is the area of a certain two-dimensional region and

85

a’(t1) is the length of a certain line segment. The corresponding

factors in our formula are therefore inverted.

lO. Transverse Curvature and the fundamental Theorem. By a
theorem of Busemann proved in L6] a G-space having non-positive
Busemann curvature also has feebly nonppositive median curvature; that
is, each point of the space has a neighborhood in which every triangle
has the property that any median is less than or equal to one-half the
sum of the adjacent sides. Fer, Busemann proves that in a G-space with
IkBé; 0, each point has a neighborhood V’ with unique segments and in
which, if y(t) is the arc-length parametrization.of a geodesic arc in

V and x is any fixed point in V not on that geodesic, the function

xyCK) is convex. Specifically, this means that for any two values

11 and 172 of 1’,

lad—i734 4 i xylrp + i amp.

Applying this to a triangle T(a,b,c) with vertices and sides in V
with the midpoint m of S(a,b) proves c
(10.1) cm<fiac+§bc.

The following theorem applies more
generally to all metric spaces with locally a m b
unique segments and having feebly non-positive median curvature. This
class of metric spaces is indeed large, for it includes

(1) G-spaces with non-positive Busemann curvature.

(2) Banach Spaces with strictly convex unit spheres.

(3) Locally ptolemaic metric spaces with locally unique seg-

~ ments.

 

86

()4) Riemannian surfaces, and, apparently, Riemnnian Space.

(S) The three classical geometries, hyperbolic, euclidean, and
spherical spaces.

Theorem 10.1: If M is a metric space with locally unique seg—
ments and having feebly non-positive median Space curvature, then any
rectifiable arc of M is a geodesic iff its transverse curvature van-
ishes identically along the arc.

M: For the necessity, let A be a geodesic arc and p
any point on A. There exists a neighborhood VI) of p in which M
has unique segments and in which A is a metric segment. Let {qi} ,
{ri} , and {Si} , be three sequences of points on A converging to
p, with qi< ri< si and qiri = risi for each i. For all suf-
ficiently large i qi, ri, and si lie in Vp and hence, as distinct
points on a metric segment, one of (qirisi)’ (riqisi), or (risiqi)
must hold. But qi < ri< 51 implies that (qirisi) is the only
possibility and hence ri is the midpoint of qi and Si. There-
fore rimqisi = riri = O and KT(qi,ri,si) = 0 for all sufficiently
large i, proving that KT(p) exists and equals zero, for p e A.

Now let A be a rectifiable arc with KT(p) = 0 for p e A.
Let Vp be a neighborhood of p in which M has unique segments and
(10.1) holds for every triangle with vertices in V . Let A' be any

P

sub-arc of A contained in V with endpoints a and b, with
P

A(x,y), x<y, as arc-length along A. Suppose e>0 be given.

Following the method of proof of Theorem 7.2, choose an n-lattice of

A?

a = p0<p1<p2< (pr, = b

 

87

with n so large that

.. E .-..-
(10.2) o g A(a,b) nkn< s, , kn pkulpk
and
6 _ .0-
(1003) KT(pk-l’pk’pk+ 1) < m 2 k " l: 2: J “‘1:

where A = A(a,b). Again, it will be sufficient to obtain the set
of n-l inequalities of (7.15). With mk as the midpoint of pk”1

and pk”, (10.3) becomes

2 2 2
(pk:1pk+l) e (Pk-1pk "’ pkpk-l- 1) e = h>‘n 5

 

 

8 “hp < ~ <
k 2 A2 \ 2 A2 2 A2
or, 2
(10.11) mkpk < 3.1;, a = 1, 2, '“n-l.

11A

substituting (10.1) and (10.h) in the

triangle inequality popké porn.k + mkpk’

 

it follows that

 

(10.5) 2 popk< popk_1+ popk+1+ £315.;- , k = l, 2, on, n-l.
Put
06k=k)\n-p0pk, k=O, 1,2, '°',n,
and substitute into (10.5): 2
2 k kn - 2“k<(k'1)>‘n - dk_1+ (k+l))\n - on”, + 'x“ 62
2 A
._ which simplifies to 2
(10.6) (“n+1 - at (a - a +>‘“ e , k= 1, 2, m, n-l,
k k k-l 2‘17

the desired set of inequalities (7.15). The remainder of the proof

follows exactly as before, proving that A(a,b) = ab. Therefore A'

 

88

is a metric segment and .A is a geodesic arc.

The result of this theorem is that transverse curvature not
only affords the same characterization of geodesics in practically
the same class of metric spaces, namely'ptolemaic spaces, as the
Menger and Haantjes curvatures originally afforded, but widens the
class to include such general spaces as Banach spaces with strictly
convex unit spheres--spaces which were not covered even by Haantjes'
theorems covering certain non-ptolemaic spaces such as Spherical
space. The fact that the transverse curvature agrees with classical
curvature when the space is euclidean makes it a "reasorable" concept
of curvature. Moreover, the existence of a readily applied formula
in the case of Minkowski Spaces makes it a useful generalization of
euclidean curvature as far as those Spaces are concerned. (The
Menger and Haantjes curvatures are not so readily calculated in
Minkowski spaces.) In conclusion, many interesting questions re-

garding this concept of curvature await their answers.

 

10.

ll.

12.

BIB LIOGRAPH!

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Berlin, 1935.

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Blumenthal, L.M , Theo dA lications of Lwistance Geome me.t£y
London: Oxfor Press,

Busemann, H., and W. Mayer. "0n the Foundations of Calculus of
Variations." Transactions of the American lhthematical

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Busemann, H. "The Foundations of Minkowskian Geometry."

Commentarii Mathematica Helvetica, vol. 21;, pp. 156-187,
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Busemann, H. The Geometfl_ of Geodesics. New York: Academic
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Cartan, E. Lecons sur _l_a Geometrie des BSpaces g3 Riemann.
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Day, M. M. "Some Characterizations of Inner—Product Spaces."
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Ml EAT-m

Freese, R. W. Ptolemaic MetricArm 8 aces. University of Missouri
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1961.

Haantjes, J. aDistance Geometry. Curvature in Abstract Metric
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Haantjes, J. "A Characteristic Local PrOperty of Geodesics in
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Kelly, L. M., and G. Ewald. "’l‘angents in Real Banach Spaces."

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90

Kelly, P., and B. Strauss. "Curvature in Hilbert Geometries."
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‘Menger, K. "Zur Metrik der Kurven." Mathematische.Annalen, vol.
103, pp. h66-501, 1930.

Pauc, C. "Coubure dans les ESpaceS Metriques." Rendiconti della
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Schoenberg, I. J. "On Metric Arcs of Vanishing Menger Curvature."
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Inner-Product Spaces and a Conjecture of L. M. Blumenthal."
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pp 0 96 1-;6E2—i9gz

Synge, J. L., and.A. Schild. Tensor Calculus. Toronto, Canada:
University of Toronto Press, 19 .

 

Taylor, A. B. Introduction to Functional.Analysis. New York:
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Wald,.A. "Begrundung einer Koordinatenlosen Differentialgeometrie
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(Wien), vol. 7, pp. 2hifi6, I935.

 

 

   

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