Diametric completions

Diametric completions

Rolf Schneider

Universit ¨at Freiburg

Szeged Workshop in Convex and Discrete Geometry May 21–23, 2012

LetMbe a nonempty bounded set in a metric space(X, ρ).

ThediameterofMis

diamM:=sup{ρ(x,y) :x,y ∈M}.

Misdiametrically completeif

diam(M∪ {x})>diamM ∀x ∈X \M.

A(diametric) completionofMis any diametrically complete set containingMand with the same diameter.

Every nonempty bounded set has a completion (many, in general); for example, by Zorn’s lemma.

E. Akin: Maximalr-diameter sets and solids of constant width.

arXiv:1003.5824v2

InEuclidean spaces,Meissner (1911)proved:

K is diametrically complete ⇐⇒ K is of constant width.

Recall that a convex bodyK in Euclidean spaceRⁿisof

constant widthd if any two parallel supporting planes ofK have distanced.

Equivalent:

The support function ofK satisfies

h(K,u) +h(K,−u) =d ∀u.

Equivalent:

K + (−K) =B(o,d), ball of radiusd.

The space of bodies of constant widthd inRⁿis an infinite-dimensional closedconvexset inKⁿ, the space of convex bodies:

IfK₁,K₂are bodies of constant widthd, then (1−λ)K₁+λK₂ (0≤λ≤1) is a body of constant widthd.

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What about diametrically complete sets in Minkowski spaces?

Some answers are given in:

The space of bodies of constant widthd inRⁿis an infinite-dimensional closedconvexset inKⁿ, the space of convex bodies:

IfK₁,K₂are bodies of constant widthd, then (1−λ)K₁+λK₂ (0≤λ≤1) is a body of constant widthd.

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What about diametrically complete sets in Minkowski spaces?

Some answers are given in:

Joint work withJos ´e Pedro Moreno:

•Local Lipschitz continuity of the diametric completion mapping.

Houston J. Math.

•Diametrically complete sets in Minkowski spaces.

Israel J. Math.

•The structure of the space of diametrically complete sets in a Minkowski space.

Discrete Comput. Geom.

•Canonical diametric completions in Minkowski spaces.

(work in progress)

Minkowski spaces

X = (Rⁿ,k · k) Minkowski space:

a finite-dimensional real normed space

The normk · kdefinesdistanceρ(x,y) :=kx−yk,width, diameter, unit ballB :={x ∈Rⁿ:kxk ≤1},ballsλB+z.

Bodies ofconstant widthand(diametrically) complete setsare defined as before. Facts:

•Every set of diameterd is contained in a complete set of diameterd.

•K is of constant width =⇒ K is diametrically complete

Meissner stated the converse, and this was believed for more than 50 years (and ‘reproved’), untilEggleston (1965)gave counterexamples. The situation is worse:

Minkowski spaces

X = (Rⁿ,k · k) Minkowski space:

a finite-dimensional real normed space

The normk · kdefinesdistanceρ(x,y) :=kx−yk,width, diameter, unit ballB :={x ∈Rⁿ:kxk ≤1},ballsλB+z.

Bodies ofconstant widthand(diametrically) complete setsare defined as before. Facts:

•Every set of diameterd is contained in a complete set of diameterd.

•K is of constant width =⇒ K is diametrically complete Meissner stated the converse, and this was believed for more than 50 years (and ‘reproved’), untilEggleston (1965)gave counterexamples. The situation is worse:

Theorem 1.For a Minkowski space X , the following are equivalent:

•Every complete set is of constant width.

•The set of complete sets is convex.

•The set of completions of any given set is convex.

Two-dimensional spaces have these properties.

Theorem 2.(Yost 1991, M–S 2010)

Let n≥3. In the space of all n-dimensional Minkowski spaces, a dense open set of Minkowski spaces has the following properties:

•The only bodies of constant width are balls.

•The sum of a complete body and a ball need not be complete.

•The set of completions of a given set need not be convex.

Theorem 1.For a Minkowski space X , the following are equivalent:

•Every complete set is of constant width.

•The set of complete sets is convex.

•The set of completions of any given set is convex.

Two-dimensional spaces have these properties.

Theorem 2.(Yost 1991, M–S 2010)

Let n≥3. In the space of all n-dimensional Minkowski spaces, a dense open set of Minkowski spaces has the following properties:

•The only bodies of constant width are balls.

•The sum of a complete body and a ball need not be complete.

•The set of completions of a given set need not be convex.

Description of complete bodies

K,M ∈ Kⁿ, dimK =n,d >0 Supporting slabofK:

set bounded by two parallel supporting hyperplanes ofK A supporting slab ofK isM-regularif at least one of the bounding hyperplanes of the parallel supporting slab ofM contains a smooth boundary point ofM.

Theorem 3.K is a diametrically complete body of diameter d if and only if(a)and(b)hold:

(a)Every B-regular supporting slab of K has width≤d . (b)Every K -regular supporting slab of K has width=d .

G

On the space of diametrically complete sets

LetD_X be the space of translation classes of diametrically complete sets of diameter 2 inX.

Theorem 4.If X = (Rⁿ,k · k)is polyhedral (i.e., the unit ball B is a polytope), thenD_X is the union set of a finite polytopal

complex.

The proof uses representations of polyhedral sets introduced byMcMullen (1973)(a variant of the Gale diagram technique).

Corollary.If X is polyhedral, thenD_X has only finitely many extreme points.

Open problem:Does this characterize polyhedral norms?

(Yes, ifn=2)

LetD₂be the space of diametrically complete sets of diameter 2 inX.

In Euclidean space,D₂is convex.

In a typical Minkowski space (in the sense of Baire category), D₂is not even starshaped.

A positive result:

Theorem 5.The space D₂is contractible.

LetD₂be the space of diametrically complete sets of diameter 2 inX.

In Euclidean space,D₂is convex.

In a typical Minkowski space (in the sense of Baire category), D₂is not even starshaped.

A positive result:

Theorem 5.The space D₂is contractible.

γ(K) :=set of completions ofK vmax(K) :=max{V(M) :M∈γ(K)}

γmv(K) :={M ∈γ(K) :V(M) =vmax(K)}

Groemer (1986):γmv(K)consists of translates of one body τ(K) :=M−s(M)for anyM∈γmv(K),sSteiner point loc. Lip. continuity ofγ (M–S 2010)⇒ v_maxis continuous

⇒ τ is continuous

ForK ∈D₂andλ∈[0,1], letK_λ := (1−λ)K +λBand F(K, λ) :=τ

2 diamK_λK_λ

+ (1−λ)s(K) +λs(B).

ThenF :D₂×[0,1]→D₂is continuous andF(K,0) =K, F(K,1) =B. Hence,D₂is contractible.

Canonical completions

Is ‘completion’ continuous?

Letγ(K)be the set of all completions ofK ∈ Kⁿ.

Theorem 6.The mappingγ :Kⁿ→ C(Kⁿ)is locally Lipschitz continuous, with respect to the Hausdorff metricδ induced by the norm onKⁿ and the Hausdorff metric∆induced byδ on C(Kⁿ), the space of nonempty compact subsets ofKⁿ. In general,γ is many-valued.

For example, ifK is a segment of lengthd in Euclidean space, then a suitable translate of any body of constant widthd is a completion ofK.

Doesγ have a continuous selection?

The usual constructions of completions involve many arbitrary choices and hence cannot yield continuous completions.

This raises the question for ‘canonical’ completions.

A construction byMaehara (1984)can be slightly generalized.

Definition.ForK ∈ Kⁿof diameterd, let η(K) := \

x∈K

B(x,d), θ(K) := \

x∈η(K)

B(x,d).

Then

µ(K) := 1

2[η(K) +θ(K)] is theMaehara setofK.

Doesγ have a continuous selection?

The usual constructions of completions involve many arbitrary choices and hence cannot yield continuous completions.

This raises the question for ‘canonical’ completions.

A construction byMaehara (1984)can be slightly generalized.

Definition.ForK ∈ Kⁿof diameterd, let η(K) := \

x∈K

B(x,d), θ(K) := \

x∈η(K)

B(x,d).

Then

µ(K) := 1

2[η(K) +θ(K)]

is theMaehara setofK.

Always true:µ(K)is atight coverofK (i.e., containsK and has diameterd).

In Euclidean spaces:µ(K)is of constant width, and hence a completionofK.

Where does this work?

Definition.The normk · kwith unit ballB has thes-propertyif B∩(B+x)is a summand ofB, for eachx withkxk ≤1.

Maehara (1984), Sallee (1987), Balashov & Polovinkin (2000), Karas ¨ev (2001):

Theorem.The Maehara setµ(K), for any K ∈ Kⁿ, is of constant width and hence a completion of K , if and only if the norm of X has the s-property.

Theorem 7.Suppose that the norm of X has the s-property, and let2C denote the Jung constant of X .

Let K,L∈ Kⁿbe convex bodies with

δ(K,L)≤≤ 1

3(1−C)min{d_K,d_L}.

Then

δ(µ(K), µ(L))≤ 7−C

1−C≤ 13

2 (n+1). In particular, if X is a Euclidean space, then

δ(µ(K), µ(L))≤20,

and if X is a two-dimensional Minkowski space, then

δ(µ(K), µ(L))≤ 15 2 .

Two new properties of the Maehara completion in Euclidean spaces:

Recall thatγ(K)denotes the set of all completions ofK. Theorem 8.The Maehara completion of K is a metric centre of γ(K), that is, it minimizes the maximal Hausdorff distance from the elements ofγ(K).

The Maehara completion of a convex body is at least as smooth as the body itself:

Theorem 9.Every normal cone of the Maehara completion µ(K)is contained in some normal cone of K .

Back to Minkowski spaces:

The only known examples of Minkowski spaces with the s-property are Euclidean spaces, two-dimensional Minkowski spaces, and⊕_∞sums of such spaces.

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Therefore, forgeneral Minkowski spaces, a different canonical completion procedure is needed.

For this, we extend a Euclidean method ofReinhardt (1922) (n=2) andB ¨uckner (1936)(n=3).

In a first step, we replace eachK by 1

2[η(K) +K],

which is a tight cover ofK and has inradius at least (2(n+1))⁻¹(diamK).

The only known examples of Minkowski spaces with the s-property are Euclidean spaces, two-dimensional Minkowski spaces, and⊕_∞sums of such spaces.

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Therefore, forgeneral Minkowski spaces, a different canonical completion procedure is needed.

For this, we extend a Euclidean method ofReinhardt (1922) (n=2) andB ¨uckner (1936)(n=3).

In a first step, we replace eachK by 1

2[η(K) +K],

which is a tight cover ofK and has inradius at least (2(n+1))⁻¹(diamK).

The generalized B ¨uckner completion

A convex bodyK of diameterd is complete if and only if K = \

x∈K

B(x,d)

(thespherical intersection property).

Aiming at completing a convex bodyK of diameterd, one is therefore tempted to consider

η(K) := \

x∈K

B(x,d),

thewide spherical hullofK.

However, in general,diamη(K)>d.

B ¨ucknerhad the idea to consider a ‘one-sided’ version of the wide spherical hull, namely

Cu(K) :=η(K)∩Z⁺(K,u) for givenu6=o, with

Z⁺(K,u) :={x+λu :x ∈K, λ≥0}.

Cu(K)is a tight cover ofK!

ButC_u(K)is generally not complete.

However, it is‘partially complete’:

Each ‘upper’ boundary point (w.r.t. u) is the endpoint of a diameter segment ofC_u(K).

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9{U

Finitely many iterations, formfixed directionsu₁, . . . ,u_m (where mdepends only onn) yield a completionC(K)ofK.

We callC thegeneralized B ¨uckner completion.

Theorem 10.The generalized B ¨uckner completion is locally Lipschitz continuous.

Perfect norms

Recall:

Theorem 1.For a Minkowski space X , the following are equivalent:

•Every complete set is of constant width.

•The set of complete sets is convex.

•The set of completions of any given set is convex.

Two-dimensional spaces have these properties.

Definition.(Karas ¨ev) A Minkowski space with these properties and its norm are calledperfect.

Eggleston (1965)andChakerian & Groemer (1983)have asked for a determination of all perfect Minkowski spaces.

This is still open.

We know fromMaeharaandKaras ¨ev:

Theorem.If the norm of X has the s-property, then X is perfect.

Conjectures.LetK be a convex body of dimension≥3 with the s-property. IfK is either smooth (R. S. 1974) or strictly convex (Karas ¨ev 2001), then it is an ellipsoid.

Theorem(Karas ¨ev 2001).A Minkowski space with a strictly convex norm is perfect if and only if its norm has the s-property.

For general norms, this is not true.

Example:

A new necessary condition:

Theorem 11.If B is the unit ball of a perfect norm, then 1

2 B∩(B+x) is a summand of B for all x withkxk ≤1.

D ¨urer’s (1514)octahedron shows that the constant ¹₂ is best possible.

The proof of Theorem 11:

Lemma.Let K,L∈ Kⁿ. If to each supporting hyperplane H of K there are a point x∈H∩K and a vector t ∈Rⁿsuch that x ⊂L+t ⊂K , then L is a summand of K .

The steps to prove Theorem 11:

kuk ≤1,Harbitrary support plane toB∩(B+u) y ∈H∩B∩(B+u)

S:= ¹₂((B∩(B+u)−y) +y ⇒ diamS∪ {o,u} ≤1 C:=completion ofS∪ {o,u} ⇒ C ⊂B∩(B+u)

⇒ HsupportsC aty

H⁰:=support plane ofCparallel toH k · kis perfect ⇒ dist(H,H⁰) =1 Lety⁰∈C∩H⁰ ⇒C⊂B+y⁰

HsupportsB+y⁰, sincedist(H,y⁰) =1

SinceH was arbitrary, the lemma shows that ¹₂(B∩(B+u))is a summand ofB.

Thank you for your attention!

And the organizers for the Workshop!

Thank you for your attention!

And the organizers for the Workshop!