colloqium talk

Hadwiger’s Characterization Theorem

Brian Whetter

Western Washington University

May 5, 2015

Introduction

I Hadwiger’s Characterization Theorem was originally proved in1957.

I The original proof by Hadwiger was very long and arduous.

I Daniel Klain found a shorter proof which I studied for myproject.

Introduction

Statement

Theorem (Hadwiger’s Characterization Theorem)

A continuous rigid-motion-invariant valuation µ on Kn can bewritten as

µ =n∑

where ci ∈ R and µi are the intrinsic volumes.

I Functions of interest to geometers can be written as linearcombinations of functions that are well understood.

Statement

µ =n∑

I Functions of interest to geometers can be written as linearcombinations of functions that are well understood.

Questions

I What is a valuation?

I What does it mean for a valuation to be continuous andrigid-motion invariant?

I What is the set Kn?

I What are the intrinsic volumes?

Questions

Overview

I First, I will explain what the statement of the theorem means.

I I will then give an outline of Klain’s proof.

Overview

I First, I will explain what the statement of the theorem means.

I I will then give an outline of Klain’s proof.

Valuations

DefinitionLet S be a set and let G be a family of subsets of S closed underfinite intersections. A valuation µ is a function µ : G → R suchthat for all A,B ∈ G , with A ∪ B ∈ G ,

µ(A ∪ B) = µ(A) + µ(B)− µ(A ∩ B), and µ(∅) = 0.

I Every finitely additive measure is a valuation.

Valuations

DefinitionLet S be a set and let G be a family of subsets of S closed underfinite intersections. A valuation µ is a function µ : G → R suchthat for all A,B ∈ G , with A ∪ B ∈ G ,

µ(A ∪ B) = µ(A) + µ(B)− µ(A ∩ B), and µ(∅) = 0.

I Every finitely additive measure is a valuation.

Euler Characteristic

I Let G be the set of coordinate boxes in Rn (i.e, parallelotopeswith facets parallel to the coordinate hyperplanes).

I Note that the intersection of two coordinate boxes is again acoordinate box.

I Given B ∈ G , µ0(B) = 1 if B 6= ∅ and µ0(B) = 0 if B = ∅.I This is a special valuation called the Euler Characteristic.

I Given B ∈ G , µ0(B) = 1 if B 6= ∅ and µ0(B) = 0 if B = ∅.

I This is a special valuation called the Euler Characteristic.

The Other Intrinsic Volumes

I For a coordinate box B ∈ Rn with edges of lengtha1, a2, . . . , an, define µnk(B) = ek(a1, a2, . . . , an), where

ek(x1, x2, . . . , xn) =n∑

1≤i1≤···≤ik≤nxi1xi2 · · · xik , 1 ≤ k ≤ n,

denotes the kth symmetric function.

I These are the other intrinsic volumes. They independent ofthe dimension n. We then write µmk (B) = µnk(B) = µk(B),and call it the kth intrinsic volume.

I We get that µn is the same as n-volume, µn−1 is half of thesurface area, and µ1 is a multiple of mean width.

ek(x1, x2, . . . , xn) =n∑

1≤i1≤···≤ik≤nxi1xi2 · · · xik , 1 ≤ k ≤ n,

ek(x1, x2, . . . , xn) =n∑

1≤i1≤···≤ik≤nxi1xi2 · · · xik , 1 ≤ k ≤ n,

The set Kn

I Geometers care about the set Kn, the set of all compactconvex sets in Rn.

I A set K ∈ Rn is said to be convex if for all x , y ∈ K and allt ∈ [0, 1],

(1− t)x + ty ∈ K .

I The intersection of two convex sets is convex.

The set Kn

(1− t)x + ty ∈ K .

The set Kn

(1− t)x + ty ∈ K .

Example

Figure 1: A convex set

Non-example

Figure 2: Not a convex set

The set Kn

I A set K ∈ Rn is said to be compact if it is closed andbounded.

I The intersection of two compact sets is compact, so the setKn is a proper set on which to define a valuation.

The set Kn

I A set K ∈ Rn is said to be compact if it is closed andbounded.

I The intersection of two compact sets is compact, so the setKn is a proper set on which to define a valuation.

Hausdorff Distance

I What does it mean for a valuation to be continuous? Weneed to introduce a notion of distance between two sets.

I First, given a set K ⊂ Kn and x ∈ Rn, the distance from thepoint x to the set K is given by

d(x ,K ) = mink∈K||x − k ||.

I For K , L ⊂ Kn, the Hausdorff distance δ(K , L) is defined by

δ(K , L) = max

(maxa∈K

d(a, L),maxb∈L

d(b,K )

I Note that maxa∈K d(a, L) and maxb∈L d(b,K ) eachcorrespond to a unique point since K and L are compactconvex sets.

Hausdorff Distance

d(x ,K ) = mink∈K||x − k ||.

δ(K , L) = max

(maxa∈K

d(a, L),maxb∈L

d(b,K )

Hausdorff Distance

d(x ,K ) = mink∈K||x − k ||.

δ(K , L) = max

(maxa∈K

d(a, L),maxb∈L

d(b,K )

Hausdorff Distance

d(x ,K ) = mink∈K||x − k ||.

δ(K , L) = max

(maxa∈K

d(a, L),maxb∈L

d(b,K )

Continuous Valuations

I Hausdorff distance is a metric on Kn.

I A sequence of sets Kj ∈ Kn converges to a set K , orKj → K , if δ(Kj ,K )→ 0 as j →∞.

I A valuation µ is continuous on Kn if µ(Kj)→ µ(K ) asKj → K .

Rigid-Motion Invariance

Next I will talk about what it means for a valuation to berigid-motion invariant. First, let us think about different types ofrigid motions in Rn. We have...

Types of Rigid Motions

Figure 3: Reflection

Types of Rigid Motion

Figure 4: Translation

Types of Rigid Motions

Figure 5: Rotation

I In general, a rigid motion is an isometry from Rn onto itself(an isometry is a distance preserving map).

I Every rigid motion can be written as a composition of arotation (either proper or improper) and a translation.

I The set of all rigid motions of Rn forms the Euclidean groupdenoted En.

I A valuation µ on Kn is said to be rigid-motion invariant ifgiven g ∈ En, and K ∈ Kn, µ(K ) = µ(gK ).

Intrinsic Volumes of Coordinate Boxes

I The intrinsic volumes for coordinate boxes are...

I Valuations

I Continuous

I Rigid-motion invariant

I Valuations

I Continuous

I Valuations

I Continuous

I Valuations

I Continuous

Hadwiger’s Theorem Again

µ =n∑

Everything should be clear now except what µi are on Kn.

µ =n∑

Everything should be clear now except what µi are on Kn.

The Need to Extend

I Recall that for a coordinate box B ∈ Rn with edges of lengtha1, a2, . . . , an, we defined µnk(B) = ek(a1, a2, . . . , an).

I This definition clearly won’t work for more general sets in Kn.

The Need to Extend

I Recall that for a coordinate box B ∈ Rn with edges of lengtha1, a2, . . . , an, we defined µnk(B) = ek(a1, a2, . . . , an).

I This definition clearly won’t work for more general sets in Kn.

Haar Measure

I Let Graff(n, k) denote the set of k-dimensional planes in Rn.

I Given K ∈ Kn, Graff(K ; k) is the set of k-dimensional planesthat meet K .

I There exists an invariant measure λnk on Graff(n, k).

I This measure is called the Haar measure, and it actually existson any locally compact topological group.

Haar Measure

Definition for µk on Kn

TheoremThere exist constants Cn

k such that

µn−k(B) = Cnk λ

nk(Graff(B; k))

for any coordinate box B.

We now define for K ∈ Kn,

µn−k(K ) = Cnk λ

nk(Graff(K ; k)).

Definition for µk on Kn

TheoremThere exist constants Cn

k such that

µn−k(B) = Cnk λ

nk(Graff(B; k))

for any coordinate box B.

We now define for K ∈ Kn,

µn−k(K ) = Cnk λ

nk(Graff(K ; k)).

Properties of the Intrinsic Volumes

I µk are valuations.

I Continuous (you have to look more closely at the definition ofλnk).

I Invariant (this follows from the invariance of λnk).

I µn is the same as n-volume, µn−1 is half of the surface area,and µ1 is a multiple of mean width.

µ =n∑

Now I will give a sketch of how you prove the theorem.

µ =n∑

Now I will give a sketch of how you prove the theorem.

Outline of the Proof

Both Hadwiger and Klain make use of the following theorem.

Theorem (The Volume Theorem)

Suppose µ is a continuous rigid-motion-invariant simple valuationon Kn. Then there exists c ∈ R such that µ(K ) = cµn(K ), for allK ∈ Kn.

I A valuation µ is simple if µ(K ) = 0 whenever K hasdimension less than n.

I Once you have the Volume Theorem, Hadwiger’s Theorem isquite simple to prove.

Klain shortened the road to the Volume Theorem by first provingthe following...

Theorem (Klain’s Lemma)

Suppose that µ is a continuous translation-invariant simplevaluation on Kn. Suppose also that µ([0, 1]n) = 0 and thatµ(K ) = µ(−K ), for all K ∈ Kn. Then µ(K ) = 0, for all K ∈ Kn.

Klain shortened the road to the Volume Theorem by first provingthe following...

Theorem (Klain’s Lemma)

Suppose that µ is a continuous translation-invariant simplevaluation on Kn. Suppose also that µ([0, 1]n) = 0 and thatµ(K ) = µ(−K ), for all K ∈ Kn. Then µ(K ) = 0, for all K ∈ Kn.

I The proof makes use of µ being translation-invariant andsimple to make “cut and paste” arguments. You proveµ(K ) = 0 for increasingly complicated shapes.

1. Do the base case n = 1.

2. Right cylinders.

3. Slanted cylinders, oblique cylinders, or prisms.

Figure 6: Right vs Oblique Cylinder

2. Right cylinders.

Centrally Symmetric Sets

4. Next we go to centrally symmetric sets. This step is morecomplicated and creative, so I will go into more detail anddevelop a couple more tools.

I For K ∈ Kn if K = −K , we say K is origin symmetric. A setK is centrally symmetric if some translate of K is originsymmetric. We denote by Kn

c the set of all centrallysymmetric members of Kn.

Figure 7: Centrally Symmetric Sets

Zonotopes and Zonoids

I Zonotopes and zonoids are particular types of centrallysymmetric objects.

I A zonotope is a finite Minkowski sum of line segments wherethe Minkowski sum of two sets A and B is

A + B = {a + b : a ∈ A, b ∈ B}.

I Every zonotope is a convex polytope (the intersection of afinite number of half spaces).

I A compact convex set is a zonoid if it is the limit in theHausdorff metric of a sequence of zonotopes.

A + B = {a + b : a ∈ A, b ∈ B}.

Figure 8: A zonotope from four vectors

Figure 9: Other Zonotopes

Not every centrally symmetric set is a zonoid!

Figure 10: Left: Zonoid Right: Not a zonoid

Not every centrally symmetric set is a zonoid!

Figure 10: Left: Zonoid Right: Not a zonoid

Support Function

I If K ∈ Kn, its support function hK : Sn−1 → R is defined byhK (u) = maxx∈K{x · u}.

Figure 11: The Support Function

I A non-empty compact convex set is uniquely determined byits support function.

Support Function

I If K ∈ Kn, its support function hK : Sn−1 → R is defined byhK (u) = maxx∈K{x · u}.

Figure 11: The Support Function

I A non-empty compact convex set is uniquely determined byits support function.

Power of the Support Function

For K ∈ Kn, if hK ∈ C∞(Sn−1), then we say K is smooth.

TheoremLet K ∈ Kn

c be smooth. There exist zonoids Y1 and Y2 such thatK + Y2 = Y1.

The proof makes heavy use of the support function by representingK , Y1, and Y2 as functions!

TheoremLet K ∈ Kn

Back to step 4

I Take a convex polytope P. Given a vector v , let v be the linesegment connecting 0 and v .

I Let P1,P2, . . . ,Pm be the facets of P with correspondingoutward unit normal vectors u1, u2, . . . , um.

I We can assume without loss of generality that P1,P2, . . . ,Pj

are the facets of P with ui · v > 0 for 1 ≤ i ≤ j .

I Now P + v = P ∪

(Pi + v)

)and using the fact that µ is

simple and Pi + v is a prism we get

µ(P + v) = µ(P) +

j∑i=1

µ(Pi + v)

= µ(P).

Back to step 4

I Now P + v = P ∪

(Pi + v)

µ(P + v) = µ(P) +

j∑i=1

µ(Pi + v)

= µ(P).

Back to step 4

I Now P + v = P ∪

(Pi + v)

µ(P + v) = µ(P) +

j∑i=1

µ(Pi + v)

= µ(P).

Back to step 4

I Now P + v = P ∪

(Pi + v)

µ(P + v) = µ(P) +

j∑i=1

µ(Pi + v)

= µ(P).

Onward

I We could continue this process over and over. From inductionwe get for any convex polytope P and any zonotope Z ,

µ(Z ) = 0 and µ(P + Z ) = µ(P).

I From continuity and the fact that K ∈ Kn can beapproximated by polytopes, we also get for K ∈ Kn and anyzonoid Y ,

µ(Y ) = 0 and µ(K + Y ) = µ(K ).

I Now for K ∈ Knc there exist zonoids Y1 and Y2 such that

K + Y2 = Y1 and

µ(K ) = µ(K + Y2) = µ(Y1) = 0.

Onward

µ(Z ) = 0 and µ(P + Z ) = µ(P).

µ(Y ) = 0 and µ(K + Y ) = µ(K ).

K + Y2 = Y1 and

µ(K ) = µ(K + Y2) = µ(Y1) = 0.

Onward

µ(Z ) = 0 and µ(P + Z ) = µ(P).

µ(Y ) = 0 and µ(K + Y ) = µ(K ).

K + Y2 = Y1 and

µ(K ) = µ(K + Y2) = µ(Y1) = 0.

Onward

Finally, since any centrally symmetric compact convex set can beapproximated by smooth ones, we are done by again applyingcontinuity!

Onward

5. Simplices (an n-simplex is the n-dimensional convex hull ofn + 1 points). This step should not be obvious, but it can bedone!

6. Polytopes, by using triangulation.

Figure 12: Triangulation

7. Kn, from using continuity again.

Onward

Inching Forward

The theorem we just proved is equivalent to...

TheoremSuppose that µ is a continuous translation-invariant simplevaluation on Kn. Then there exists c ∈ R such thatµ(K ) + µ(−K ) = cµn(K ), for all K ∈ Kn.

This theorem seems much closer to the volume theorem. In factwe are done if K is origin symmetric.

Inching Forward

The theorem we just proved is equivalent to...

TheoremSuppose that µ is a continuous translation-invariant simplevaluation on Kn. Then there exists c ∈ R such thatµ(K ) + µ(−K ) = cµn(K ), for all K ∈ Kn.

This theorem seems much closer to the volume theorem. In factwe are done if K is origin symmetric.

Proof of the Forward Direction

I Suppose µ is a continuous translation-invariant simplevaluation on Kn.

I For K ∈ Kn, define

ν(K ) = µ(K ) + µ(−K )− 2µ([0, 1]n)µn(K ).

I ν satisfies the conditions of Klain’s lemma, so for allK ∈ (K )n, ν(K ) = 0 and

µ(K ) + µ(−K ) = cµn(K ),

where c = 2µ([0, 1]n).

ν(K ) = µ(K ) + µ(−K )− 2µ([0, 1]n)µn(K ).

µ(K ) + µ(−K ) = cµn(K ),

where c = 2µ([0, 1]n).

ν(K ) = µ(K ) + µ(−K )− 2µ([0, 1]n)µn(K ).

µ(K ) + µ(−K ) = cµn(K ),

where c = 2µ([0, 1]n).

Review

µ =n∑

Review

µ =n∑

Proof of Hadwiger’s Theorem

I Proof by induction. Let µ be a continuous,rigid-motion-invariant valuation on Kn. Now consider µ|Hwhere H is a hyperplane. We have that µ|H is a valuation onKn−1.

I Now using the induction hypothesis, for all A ⊂ H, we have

µ|H(A) =n−1∑i=0

ciµi (A).

I Since µ is rigid-motion invariant, the coefficients ci will be thesame for any choice of hyperplane H.

µ|H(A) =n−1∑i=0

ciµi (A).

µ|H(A) =n−1∑i=0

ciµi (A).

I We now have that µ−∑n−1

i=0 ciµi is a continuous,rigid-motion-invariant, simple valuation and using the theVolume Theorem we get

µ−n−1∑i=0

ciµi = cnµn

for some cn ∈ R.

I We finish the proof upon rearrangement!

I We now have that µ−∑n−1

i=0 ciµi is a continuous,rigid-motion-invariant, simple valuation and using the theVolume Theorem we get

µ−n−1∑i=0

ciµi = cnµn

for some cn ∈ R.

I We finish the proof upon rearrangement!

Thank You

I Professor Gardner.

I Daniel Klain for responding to questions during the project.

I You all for coming!

I Mom, Dad, and sister.

Thank You

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