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ABSORBING ANGLES, STEINER MINIMAL TREES, AND

ANTIPODALITY

HORST MARTINI, KONRAD J. SWANEPOEL, AND P. OLOFF DE WET

Abstract. We give a new proof that a star {opi : i = 1,...,k} in a normed plane is a Steiner

minimal tree of its vertices {o,p1,...,pk} if and only if all angles formed by the edges at o

are absorbing [Swanepoel, Networks 36 (2000), 104–113]. The proof is more conceptual and

simpler than the original one.

We also find a new sufficient condition for higher-dimensional normed spaces to share this

characterization. In particular, a star {opi : i = 1,...,k} in any CL-space is a Steiner minimal

tree of its vertices {o,p1,...,pk} if and only if all angles are absorbing, which in turn holds

if and only if all distances between the normalizations

mixed ?1 and ?∞ sum of finitely many copies of R.

Keywords: Steiner minimal tree, absorbing angle, antipodality, face antipodality, Minkowski

geometry.

1

?pi?pi equal 2. CL-spaces include the

1. Introduction

1.1. Minkowski geometry. Let Mddenote a d-dimensional normed space (or Minkowski

space) with origin o, i.e., Rdequipped with a norm ?·?. We call an M2a Minkowski plane.

Denote the unit ball by B = {x ∈ Rd: ?x? ≤ 1}. The dual Md

the dual norm

?x?∗:= max

where ?·,·? denotes the inner product on Rd. The dual unit ball

B∗= {x ∈ Rd: ?x,y? ≤ 1 ∀y ∈ B}

is also known as the polar body of B.

For example, the d-dimensional Minkowski spaces ?d

has the norm ?(x1,...,xd)?1:=?d

A vector x∗∈ Md

dual unit vector that attains its norm at x. In this case the hyperplane {x ∈ Rd: ?x∗,x? = 1}

supports the unit ball at

?x?x. Any hyperplane supporting the unit ball at

this way by some x∗dual to x. A unit vector v ∈ Mdis a regular direction if there is only one

hyperplane that supports B at v. Note that the norm function f(x) := ?x? is differentiable

at p ?= o if and only if

vector in Md

The exposed face of the unit ball B defined by a unit vector a∗∈ Md

[a∗] := {a ∈ B : ?a,a∗? = 1}.

Similarly, a unit vector a ∈ Mddefines an exposed face [a]∗of B∗. If B is a polytope then all

faces are exposed, and each face F of B corresponds to a face F∗of B∗as follows:

F∗:= {a∗∈ B∗: ?a,a∗? = 1 for all a ∈ F}.

∗of Mdis Rdequipped with

?y?≤1?x,y?,

1and ?d

∞are duals of each other, where ?d

∞has the norm ?(x1,...,xd)?∞:= max{|xi| :

1

i=1|xi| and ?d

i = 1,...,d}.

∗is dual to x ∈ Md, x ?= o, if ?x∗?∗= 1 and ?x∗,x? = ?x?, i.e., x∗is a

11

?x?x is given in

1

?p?p is a regular direction, and then the gradient ∇f(p) is the unique

∗dual to p.

∗is

Research supported by a grant from an agreement between the Deutsche Forschungsgemeinschaft in Germany

and the National Research Foundation in South Africa.

1

arXiv:1108.5046v1 [math.MG] 25 Aug 2011

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY2

1.2. Trees. Let S ⊂ Mdbe a finite, non-empty set of points. A spanning tree T of S is an

acyclic connected graph with vertex set S. Denote its edge set by E(T). A Steiner tree T of

S is a spanning tree of some finite V ⊂ Mdsuch that S ⊆ V and such that the degree of each

vertex in V \S is at least 3. The vertices in S are the terminals of the Steiner tree T, and the

vertices in V \ S the Steiner points of T. The length of a tree T in Mdis

?(T) :=

?

xy∈E(T)

?x − y?.

A Steiner minimal tree (SMT) of S is a Steiner tree of S of smallest length. The requirement

that Steiner points have degree at least 3 is for technical convenience, since Steiner points of

degree at most 2 can easily be eliminated using the triangle inequality without making the tree

longer. It is easily seen that the number of Steiner points is at most #S − 2. It then follows

by a simple compactness argument that any non-empty finite S has a SMT.

A star with center s is a tree in which the vertex s is joined to all other vertices. If s ∈ S has

neighbors s1,...,sk∈ V in some SMT, then clearly the star joining s to each si, i = 1,...,k,

is a SMT of {s,s1,...,sk}. Thus, to characterize the neighborhoods of terminals in SMTs, it

is sufficient to characterize SMTs which are stars with the center a terminal. This is the intent

of Theorem 1 below.

1.3. Angles. An angle ?x1x0x2in Mdis absorbing if the function

x ?→ ?x − x0? + ?x − x1? + ?x − x2?

attains its minimum at x0. Thus ?x1x0x2is absorbing if and only if the star {x0x1,x0x2} is

a SMT of {x0,x1,x2}.

Lemma 1. Let a and b be unit vectors in Md. Then the following are equivalent:

?aob is absorbing.(1)

There exist unit vectors a∗and b∗in Md

?a∗,a? = ?b∗,b? = 1 and ?a∗+ b∗?∗≤ 1.

The exposed faces [a]∗and [−b]∗= −[b]∗of the dual unit ball

are at distance ≤ 1.

Proof. (1)⇔(2) is part of Lemma 5.4 in [5]. (2)⇔(3) is trivial.

In particular, this is a property of the angle alone:

Corollary 1. If yi is a point on the ray− − →

xoxi, i = 1,2, then {x0x1,x0x2} is a SMT of

{x0,x1,x2} if and only if {x0y1,x0y2} is a SMT of {x0,y1,y2}. [5, Proposition 3.3].

Furthermore, an angle containing an absorbing angle is itself absorbing.

∗such that(2)

(3)

?

See also Proposition 3.3 and Lemma 5.4 of [5].

1.4. The planar case. We are now able to formulate the first result. In any Minkowski space,

all angles made by two incident edges of a SMT are clearly absorbing. (For angles in a minimal

spanning tree even more is true [4].) Remarkably, as shown in [8], for a Minkowski plane the

condition that all angles are absorbing is also sufficient for a star to be a SMT of its vertices.

Theorem 1. Let p1,...,pk?= o be points in a Minkowski plane M2. Then the star joining

each pito o is a SMT of {o,p1,...,pk} if (and only if) all angles ?piopj, i ?= j, are absorbing.

This result is used in [8] to show that the maximum degree of a vertex in a SMT in a

Minkowski plane is 6, with equality only if the unit ball is an affine regular hexagon; for all

other planes the maximum is 4 if there exist supplementary absorbing angles, and 3 otherwise.

The proof given in [8] employs a long case analysis. The new proof presented in Section 2 is

more conceptual.

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY3

1.5. Antipodality and higher dimensions. Theorem 1 does not hold anymore in Minkowski

spaces of dimension at least 3. For example, let the unit ball be the projection of a (d+1)-cube

along a diagonal. (When d = 3, this is the rhombic dodecahedron.) In this Minkowski space,

the star joining o to all 2d+1−2 vertices of the unit ball is not a SMT of these vertices if d ≥ 3,

despite all the angles being absorbing [9]. However, Theorem 1 extends to both ?d

Our second result is a generalization of this fact.

We first introduce some more notions, involving antipodality. Two boundary points of the

unit ball B are antipodal if there exist distinct parallel hyperplanes supporting the two points.

Equivalently, unit vectors a and b are antipodal if and only if ?a − b? = 2.

Lemma 2. If a and b are antipodal unit vectors in a Minkowski space, then ?aob is an

absorbing angle.

1and ?d

∞.

Proof. Since the union of the segments oa and ob form a shortest path from a to b, these two

segments form a SMT of {o,a,b}, hence ?aob is absorbing.

The converse of the above lemma is not necessarily true, as the Euclidean norm shows. We

call the unit ball of a Minkowski space Steiner antipodal if two points a and b on the boundary

of the unit ball are antipodal whenever ?aob is absorbing.

?

Theorem 2. Consider the following properties of a set {p1,...,pk} of unit vectors in a

Minkowski space Md.

All angles ?piopjare absorbing. (4)

All distances ?pi− pj? = 2.

k?

k?

Then the implications (5)⇒(6)⇒(7)⇒(4) hold. Furthermore, (4) to (7) are equivalent if, and

only if, the norm is Steiner antipodal.

(5)

The star

i=1

[o,pi] is a SMT of {p1,...,pk}.(6)

The star

i=1

[o,pi] is a SMT of {o,p1,...,pk}.(7)

Proof. (6)⇒(7) is true, since all Steiner trees of {o,p1,...,pk} are also Steiner trees of {p1,...,pk},

by considering o to be a Steiner point.

(7)⇒(4) holds, since if any star [o,pi] ∪ [o,pj] can be shortened, it would also shorten the

star?k

This leaves (5)⇒(6). Note that the given star is a Steiner tree of length k. It is sufficient to

show that all Steiner trees have length ≥ k. However, note that the open unit balls centered

at the piare pairwise disjoint, since ?pi− pj? = 2 for distinct i ?= j. Any Steiner tree will have

to join each pito the boundary of the unit ball with centre pi. The part of the Steiner tree

inside this ball must therefore have length at least 1. It follows that the length of any Steiner

tree must be at least k.

i=1[o,pi].

The implication (4)⇒(5) is equivalent to the definition of Steiner antipodality.

?

In order to apply this result, we need a characterization of Steiner antipodal norms in terms

of duality.

Proposition 1. The following are equivalent in any Minkowski space Md:

The norm is Steiner antipodal. (8)

The unit ball is a polytope and any two disjoint faces of

the dual unit ball are at distance > 1.

(9)

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY4

p1

p2

pk

q1

q2

qk

m

?

s

Figure 1.

Proof. (8)⇐(9) is immediate from the definition of Steiner antipodality and Lemma 1. (8)⇒(9)

follows upon noting that if a convex body is not a polytope, then there are disjoint exposed

faces that are arbitrarily close to each other.

?

A Minkowski space is a CL-space if for every maximal proper face F of the unit ball B we

have B = conv(F ∪ (−F)). It is easily seen from finite dimensionality that the unit ball of a

CL-space is a polytope. CL-spaces were introduced by R. E. Fullerton (see [7]), although the

notion has been studied before by Hanner [1], who proved that the unit balls of CL-spaces are

{0,1}-polytopes. McGregor [6] showed that CL-spaces are exactly those spaces with numerical

index 1. What is important for our purposes is that CL-spaces turn out to be Steiner antipodal.

Hanner [1] identified an important subclass of CL-spaces, namely those that can be built up

from the one-dimensional space R using ?1-sums and ?∞-sums. For two Minkowski spaces M

and N of dimension d and e we define their ?1-sum M ⊕1N and ?∞-sum M ⊕∞N to be the

Minkowski spaces on Rd+ewith norms ?(x,y)?1= ?x?+?y? and ?(x,y)?∞= max{?x?,?y?}.

Note that the unit ball of M ⊕1N is the convex hull of the unit ball of M when embedded

as M ⊕ {o} and the unit ball of N when embedded as {o} ⊕ N. The unit ball of M ⊕∞N

is the Cartesian product of the unit balls of M and N. The unit balls of these spaces are

called Hanner polytopes. We thus introduce the name Hanner space for these spaces. For more

information see [1, 3, 2, 7].

We summarize the above discussion as follows.

Proposition 2. All Hanner spaces are CL-spaces. All CL-spaces are Steiner antipodal.

Proof. It is clear and well-known that Hanner spaces are CL-spaces (see, e.g., [7]).

It is also well-known that the dual of a CL-space is a CL-space as well [6]. To prove the

second part of the proposition, it is by Proposition 1 sufficient to show that any two disjoint

faces F and G of the unit ball B are at distance > 1. Suppose that F is contained in the facet

F?. Then all vertices of B disjoint from F must lie in the opposite facet −F?. It follows that

G ⊆ −F?, and F and G are therefore at distance 2.

2. Proof of Theorem 1

?

Lemma 3. Let ? be a line passing through a Steiner point s of a SMT T in a Minkowski plane

M2. Assume that ? is parallel to a regular direction. Then T has edges incident to s in both

open half planes bounded by ?.

Proof. Without any assumption on ?, the edges incident to s cannot all lie in the same open

half plane bounded by ?. Indeed, such a tree can be shortened as follows (Fig. 1). Let some

line m intersect the interior of each edge spi, i = 1,...,k, in qi, say. Remove edges sp1, spk,

and sqi, i = 2,...,k − 1, and add edges p1q2, qiqi+1, 2 ≤ i ≤ k − 2, and qk−1pk, to obtain a

new Steiner tree T?, without the Steiner point s, but with new Steiner points qi, 2 ≤ i ≤ k−1.

By the triangle inequality, ?(T) − ?(T?) ≥?k−1

spkcannot be opposite edges both on ?, with all other spi, 2 ≤ i ≤ k − 1, on the same side of

? (Fig. 2). Let s2be a variable point on sp2with ?s2− s? small. Denote the intersection of

i=2?sqi? > 0, contradicting the minimality of T.

We now assume that ? is parallel to a regular direction. It is sufficient to show that sp1and

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY5

p1

pk

p2

p3

pk−1

s2

s3

sk−1

s

Figure 2.

s2pkand spiby si, i = 3,...,k − 1. Change the Steiner tree T as follows. Remove edges p1s,

pks and sis, i = 2,...,k − 1, and add edges p1s2and s2pk. This removes the Steiner point s

and introduces new Steiner points s2,...,sk−1. Denoting the new tree by T?, it follows that

the length changes by

?(T?) − ?(T)

=?s2− p1? + ?s2− pk? − ?s − p1? − ?s − pk? −

k−1

?

i=2

?s − si?

≤?(s − p1) + (s2− s)? − ?s − p1?

+ ?(s − pk) + (s2− s)? − ?s − pk? − ?s2− s?.

Since s − p1and s − pkare parallel to a regular direction, the norm is differentiable at both

points, i.e.,

?(s − p1) + (s2− s)? − ?s − p1?

?s2− s?

and

?(s − pk) + (s2− s)? − ?s − pk?

?s2− s?

(Since s−p1and s−pkare in opposite directions, their duals are opposite in sign.) It follows

that

?(T?) − ?(T) ≤ ?u∗,s2− s? + ?−u∗,s2− s? + o(?s2− s?) − ?s2− s?

= o(?s2− s?) − ?s2− s?,

which is negative if ?s2− s? is sufficiently small. Then ?(T?) < ?(T), a contradiction.

Proof of Theorem 1. Without loss of generality, the segments op1,...,opkare ordered around

o. Assume that all angles ?piopj are absorbing. We start off with an arbitrary SMT of

{o,p1,...,pk} and modify it in two steps without increasing the length. In Step 1 we eliminate

all Steiner points in the interiors of the angles ?piopi+1. In Step 2 we eliminate all edges

between vertices on different rays− →

opi. The edges of the final SMT are then all contained in

the the union of the segments opi, i = 1,...,k. This tree cannot have Steiner points, and so

has to be the star with centre o. This concludes the proof.

Step 1: For each angle ?piopi+1(where we let k +1 ≡ 0), choose a regular direction rinot

contained in the (closed) angle. Choose ? pi∈− →

moreover, α,β > 0) and define the measure of s to be

|s| := α + β.

Define the measure |T| of any Steiner tree T of {o,p1,...,pk} to be the sum of the measures

of all Steiner points of T not on any ray− →

opi. Let

µ = inf{|T| : T is a SMT of {o,p1,...,pk}}.

lim

s2→s

= ?u∗,s2− s?

lim

s2→s

= ?−u∗,s2− s?.

?

opiand ? qi∈− − − →

opi+1such that ? pi? qiis parallel to

ri(Fig. 3). For each point s in the interior of ?piopj, write s = αp + βq (uniquely, and then,

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY6

pi

pi+1

? pi

? qi

ri

o

Figure 3.

pi

pi+1

ri

o

s

?

x1

x2

x3

(a)

pi

pi+1

o

x1

x2

x3

s

s?

(b)

Figure 4.

Let Tnbe a sequence of SMTs of {o,p1,...,pk} with limn→∞|Tn| = µ. Since there are only

finitely many combinatorial types of Steiner trees on a set of k+1 points, we may, by passing to

a subsequence, assume without loss of generality that all Tnhave the same combinatorial type

with Steiner points s(n)

m , say. By taking further subsequences, we may assume that

each sequence of Steiner points converge, say s(n)

i

Steiner tree T0with ?(T0) = limn→∞?(Tn), hence T0is a SMT. Also, |T0| ≤ limn→∞|Tn|, since

the measure of a Steiner point is continuous in the interior of an angle, hence limn→∞|s(n)

if siis still in the interior of the same angle, otherwise limn→∞|s(n)

of the rays− →

opj. Therefore, |T0| = µ. It remains to show that µ = 0, since this will imply that

T0does not have any Steiner point in the interior of an angle.

Suppose that µ > 0. We obtain a contradiction by constructing a SMT T?with |T?| < µ.

Let s be a Steiner point of T0 in the interior of ?piopi+1, say (Fig. 4(a)). Without loss of

generality there is no point of T0in the translated angle s + ?piopi+1, since such a point is

necessarily another Steiner point s?and we may then repeatedly choose a new Steiner point s??

in s?+ ?piopi+1, until this procedure halts.

Let ? be the line through s parallel to ri. The points on ? in the interior of ?piopi+1all have

the same measure, and the points on the same side of ? as o have smaller measure. By Lemma 3

there is an edge sx1incident to s on the same side of ? as o. There are at least two more edges

sx2and sx3. Since not all edges are in an open half plane bounded by a line through s, we

may choose x2and x3such that the angle ?x2sx3contains the translated angle s+?p1op2in

its interior (with s excluded). It follows that there is a point s?on sx1sufficiently close to s

such that ?x2s?x3contains the translate s?+ ?p1op2, and so is still absorbing (Figure 4(b)).

We may therefore replace the edges sx2, sx3and ss?by s?x2and s?x3without lengthening T0,

to obtain a new SMT T?. However, |s?| < |s|, hence |T?| < |T| = µ, which gives the required

contradiction.

1,...,s(n)

→ si, i = 1,...,m. In the limit we obtain a

i

| = |si|

i

| ≥ |si| = 0 if siis on one

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ABSORBING ANGLES, STEINER MINIMAL TREES, AND ANTIPODALITY7

Step 2: Note that for any absorbing angle ?piopj,

?pi− o? + ?pj− o? + ?o − o? ≤ ?pi− pj? + ?pj− pj? + ?o − pj?,

i.e., ?pi− pj? ≥ ?pi?.

Suppose that the SMT T has an edge between two points on different segments, say between

qion opiand qjon opj. Without loss of generality, the unique path in T from o to qipasses

through qj(otherwise interchange qiand qj). Since ?qioqjis absorbing, ?qi− qj? ≥ ?qi?. We

can then replace the edge qiqj by oqj, without losing connectivity and without lengthening

T. This process may be repeated until all edges are on the segments opi, which finishes Step

2.

?

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225–241.

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(2006), 122–125.

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of Optimization Theory and Applications 115 (2002), 283–314.

6. C. McGregor, Finite-dimensional normed linear spaces with numerical index 1, J. London Math. Soc. (2) 3

(1971), 717–721.

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London Math. Soc. (2) 43 (1991), 137–148.

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Geometry 37 (2007), 419–442.

Fakult¨ at f¨ ur Mathematik, Technische Universit¨ at Chemnitz, D-09107 Chemnitz, Germany

E-mail address: martini@mathematik.tu-chemnitz.de

Fakult¨ at f¨ ur Mathematik, Technische Universit¨ at Chemnitz, D-09107 Chemnitz, Germany

E-mail address: konrad.swanepoel@gmail.com

Department of Decision Sciences, University of South Africa, PO Box 392, UNISA 0003, South

Africa

E-mail address: dwetpo@unisa.ac.za