Colloqium Talk

Hadwiger’s Characterization Theorem
Brian Whetter
Western Washington University
May 5, 2015

Introduction
Hadwiger’s Characterization Theorem was originally proved in
1957.

Introduction
1957.
The original proof by Hadwiger was very long and arduous.

Introduction
1957.
The original proof by Hadwiger was very long and arduous.
Daniel Klain found a shorter proof which I studied for my
project.

Statement
Theorem (Hadwiger’s Characterization Theorem)
A continuous rigid-motion-invariant valuation µ on Kn can be
written as
µ =
n
i=0
ci µi
where ci ∈ R and µi are the intrinsic volumes.

Statement
written as
µ =
n
i=0
ci µi
Functions of interest to geometers can be written as linear
combinations of functions that are well understood.

Questions
What is a valuation?

Questions
What does it mean for a valuation to be continuous and
rigid-motion invariant?

Questions
What is the set Kn?

Questions
What is the set Kn?
What are the intrinsic volumes?

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

Overview
First, I will explain what the statement of the theorem means.
I will then give an outline of Klain’s proof.

Valuations
Deﬁnition
Let S be a set and let G be a family of subsets of S closed under
ﬁnite intersections. A valuation µ is a function µ : G → R such
that for all A, B ∈ G, with A ∪ B ∈ G,
µ(A ∪ B) = µ(A) + µ(B) − µ(A ∩ B), and µ(∅) = 0.

Valuations
Definition
Let S be a set and let G be a family of subsets of S closed under
finite intersections. A valuation µ is a function µ : G → R such
that for all A, B ∈ G, with A ∪ B ∈ G,
µ(A ∪ B) = µ(A) + µ(B) − µ(A ∩ B), and µ(∅) = 0.
Every finitely additive measure is a valuation.

Euler Characteristic
Let G be the set of coordinate boxes in Rn (i.e, parallelotopes
with facets parallel to the coordinate hyperplanes).

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

coordinate box.
Given B ∈ G, µ0(B) = 1 if B = ∅ and µ0(B) = 0 if B = ∅.

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

The Other Intrinsic Volumes
For a coordinate box B ∈ Rn with edges of length
a1, a2, . . . , an, deﬁne µn
k(B) = ek(a1, a2, . . . , an), where
ek(x1, x2, . . . , xn) =
n
1≤i1≤···≤ik ≤n
xi1 xi2 · · · xik
, 1 ≤ k ≤ n,
denotes the kth symmetric function.

a1, a2, . . . , an, deﬁne µn
k(B) = ek(a1, a2, . . . , an), where
ek(x1, x2, . . . , xn) =
n
1≤i1≤···≤ik ≤n
, 1 ≤ k ≤ n,
These are the other intrinsic volumes. They independent of
the dimension n. We then write µm
k (B) = µn
k(B) = µk(B),
and call it the kth intrinsic volume.

a1, a2, . . . , an, deﬁne µn
k(B) = ek(a1, a2, . . . , an), where
ek(x1, x2, . . . , xn) =
n
1≤i1≤···≤ik ≤n
, 1 ≤ k ≤ n,
These are the other intrinsic volumes. They independent of
the dimension n. We then write µm
k (B) = µn
k(B) = µk(B),
and call it the kth intrinsic volume.
We get that µn is the same as n-volume, µn−1 is half of the
surface area, and µ1 is a multiple of mean width.

The set Kn
Geometers care about the set Kn, the set of all compact
convex sets in Rn.

The set Kn
convex sets in Rn.
A set K ∈ Rn is said to be convex if for all x, y ∈ K and all
t ∈ [0, 1],
(1 − t)x + ty ∈ K.

The set Kn
convex sets in Rn.
A set K ∈ Rn is said to be convex if for all x, y ∈ K and all
t ∈ [0, 1],
(1 − t)x + ty ∈ K.
The intersection of two convex sets is convex.

Example
Figure 1: A convex set

Non-example
Figure 2: Not a convex set

The set Kn
A set K ∈ Rn is said to be compact if it is closed and
bounded.

The set Kn
A set K ∈ Rn is said to be compact if it is closed and
bounded.
The intersection of two compact sets is compact, so the set
Kn is a proper set on which to deﬁne a valuation.

Hausdorﬀ Distance
What does it mean for a valuation to be continuous? We
need to introduce a notion of distance between two sets.

Hausdorﬀ Distance
First, given a set K ⊂ Kn and x ∈ Rn, the distance from the
point x to the set K is given by
d(x, K) = min
k∈K
||x − k||.

Hausdorff Distance
d(x, K) = min
k∈K
||x − k||.
For K, L ⊂ Kn, the Hausdorff distance δ(K, L) is defined by
δ(K, L) = max max
a∈K
d(a, L), max
b∈L
d(b, K) .

Hausdorff Distance
d(x, K) = min
k∈K
||x − k||.
For K, L ⊂ Kn, the Hausdorff distance δ(K, L) is defined by
δ(K, L) = max max
a∈K
d(a, L), max
b∈L
d(b, K) .
Note that maxa∈K d(a, L) and maxb∈L d(b, K) each
correspond to a unique point since K and L are compact
convex sets.

Continuous Valuations
Hausdorﬀ distance is a metric on Kn.

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

A sequence of sets Kj ∈ Kn converges to a set K, or
Kj → K, if δ(Kj , K) → 0 as j → ∞.
A valuation µ is continuous on Kn if µ(Kj ) → µ(K) as
Kj → K.

Rigid-Motion Invariance
Next I will talk about what it means for a valuation to be
rigid-motion invariant. First, let us think about diﬀerent types of
rigid 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

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

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

The set of all rigid motions of Rn forms the Euclidean group
denoted En.

The set of all rigid motions of Rn forms the Euclidean group
denoted En.
A valuation µ on Kn is said to be rigid-motion invariant if
given g ∈ En, and K ∈ Kn, µ(K) = µ(gK).

Intrinsic Volumes of Coordinate Boxes
The intrinsic volumes for coordinate boxes are...

Valuations

Valuations
Continuous

Valuations
Continuous
Rigid-motion invariant

Hadwiger’s Theorem Again
written as
µ =
n
i=0
ci µi

written as
µ =
n
i=0
ci µi
Everything should be clear now except what µi are on Kn.

The Need to Extend
Recall that for a coordinate box B ∈ Rn with edges of length
a1, a2, . . . , an, we deﬁned µn
k(B) = ek(a1, a2, . . . , an).

The Need to Extend
Recall that for a coordinate box B ∈ Rn with edges of length
a1, a2, . . . , an, we deﬁned µn
k(B) = ek(a1, a2, . . . , an).
This deﬁnition clearly won’t work for more general sets in Kn.

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

Haar Measure
Given K ∈ Kn, Graﬀ(K; k) is the set of k-dimensional planes
that meet K.

Haar Measure
that meet K.
There exists an invariant measure λn
k on Graﬀ(n, k).

Haar Measure
that meet K.
There exists an invariant measure λn
k on Graﬀ(n, k).
This measure is called the Haar measure, and it actually exists
on any locally compact topological group.

Deﬁnition for µk on Kn
Theorem
There exist constants Cn
k such that
µn−k(B) = Cn
k λn
k(Graﬀ(B; k))
for any coordinate box B.

Definition for µk on Kn
Theorem
There exist constants Cn
k such that
µn−k(B) = Cn
k λn
k(Graff(B; k))
for any coordinate box B.
We now define for K ∈ Kn,
µn−k(K) = Cn
k λn
k(Graff(K; k)).

Properties of the Intrinsic Volumes
µk are valuations.

µk are valuations.
Continuous (you have to look more closely at the deﬁnition of
λn
k).

µk are valuations.
λn
k).
Invariant (this follows from the invariance of λn
k).

µk are valuations.
λn
k).
Invariant (this follows from the invariance of λn
k).
µn is the same as n-volume, µn−1 is half of the surface area,
and µ1 is a multiple of mean width.

written as
µ =
n
i=0
ci µi
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 valuation
on Kn. Then there exists c ∈ R such that µ(K) = cµn(K), for all
K ∈ Kn.

K ∈ Kn.
A valuation µ is simple if µ(K) = 0 whenever K has
dimension less than n.

K ∈ Kn.
A valuation µ is simple if µ(K) = 0 whenever K has
dimension less than n.
Once you have the Volume Theorem, Hadwiger’s Theorem is
quite simple to prove.

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

Klain shortened the road to the Volume Theorem by ﬁrst proving
the following...
Theorem (Klain’s Lemma)
Suppose that µ is a continuous translation-invariant simple
valuation on Kn. Suppose also that µ([0, 1]n) = 0 and that
µ(K) = µ(−K), for all K ∈ Kn. Then µ(K) = 0, for all K ∈ Kn.

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

1. Do the base case n = 1.

2. Right cylinders.

2. Right cylinders.
3. Slanted cylinders, oblique cylinders, or prisms.
Figure 6: Right vs Oblique Cylinder

Centrally Symmetric Sets
4. Next we go to centrally symmetric sets. This step is more
complicated and creative, so I will go into more detail and
develop a couple more tools.

For K ∈ Kn if K = −K, we say K is origin symmetric. A set
K is centrally symmetric if some translate of K is origin
symmetric. We denote by Kn
c the set of all centrally
symmetric members of Kn.

For K ∈ Kn if K = −K, we say K is origin symmetric. A set
K is centrally symmetric if some translate of K is origin
symmetric. We denote by Kn
c the set of all centrally
symmetric members of Kn.
Figure 7: Centrally Symmetric Sets

Zonotopes and Zonoids
Zonotopes and zonoids are particular types of centrally
symmetric objects.

symmetric objects.
A zonotope is a ﬁnite Minkowski sum of line segments where
the Minkowski sum of two sets A and B is
A + B = {a + b : a ∈ A, b ∈ B}.

symmetric objects.
A + B = {a + b : a ∈ A, b ∈ B}.
Every zonotope is a convex polytope (the intersection of a
ﬁnite number of half spaces).

symmetric objects.
A + B = {a + b : a ∈ A, b ∈ B}.
Every zonotope is a convex polytope (the intersection of a
ﬁnite number of half spaces).
A compact convex set is a zonoid if it is the limit in the
Hausdorﬀ metric of a sequence of zonotopes.

Figure 8: A zonotope from four vectors

Figure 9: Other Zonotopes

Not every centrally symmetric set is a zonoid!

Not every centrally symmetric set is a zonoid!
Figure 10: Left: Zonoid Right: Not a zonoid

Support Function
If K ∈ Kn, its support function hK : Sn−1 → R is deﬁned by
hK (u) = maxx∈K {x · u}.
Figure 11: The Support Function

Support Function
If K ∈ Kn, its support function hK : Sn−1 → R is deﬁned by
hK (u) = maxx∈K {x · u}.
Figure 11: The Support Function
A non-empty compact convex set is uniquely determined by
its support function.

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

Theorem
Let K ∈ Kn
c be smooth. There exist zonoids Y1 and Y2 such that
K + Y2 = Y1.

Theorem
Let K ∈ Kn
c be smooth. There exist zonoids Y1 and Y2 such that
K + Y2 = Y1.
The proof makes heavy use of the support function by representing
K, Y1, and Y2 as functions!

Back to step 4
Take a convex polytope P. Given a vector v, let v be the line
segment connecting 0 and v.

Back to step 4
Let P1, P2, . . . , Pm be the facets of P with corresponding
outward unit normal vectors u1, u2, . . . , um.

Back to step 4
We can assume without loss of generality that P1, P2, . . . , Pj
are the facets of P with ui · v > 0 for 1 ≤ i ≤ j.

Back to step 4
We can assume without loss of generality that P1, P2, . . . , Pj
are the facets of P with ui · v > 0 for 1 ≤ i ≤ j.
Now P + v = P ∪
j
i=1
(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).

Onward
We could continue this process over and over. From induction
we get for any convex polytope P and any zonotope Z,
µ(Z) = 0 and µ(P + Z) = µ(P).

Onward
µ(Z) = 0 and µ(P + Z) = µ(P).
From continuity and the fact that K ∈ Kn can be
approximated by polytopes, we also get for K ∈ Kn and any
zonoid Y ,
µ(Y ) = 0 and µ(K + Y ) = µ(K).

Onward
µ(Z) = 0 and µ(P + Z) = µ(P).
From continuity and the fact that K ∈ Kn can be
approximated by polytopes, we also get for K ∈ Kn and any
zonoid Y ,
µ(Y ) = 0 and µ(K + Y ) = µ(K).
Now for K ∈ Kn
c there exist zonoids Y1 and Y2 such that
K + Y2 = Y1 and
µ(K) = µ(K + Y2) = µ(Y1) = 0.

Onward
Finally, since any centrally symmetric compact convex set can be
approximated by smooth ones, we are done by again applying
continuity!

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

Onward
done!
6. Polytopes, by using triangulation.
Figure 12: Triangulation

Onward
done!
6. Polytopes, by using triangulation.
Figure 12: Triangulation
7. Kn, from using continuity again.

Inching Forward
The theorem we just proved is equivalent to...
Theorem
valuation on Kn. Then there exists c ∈ R such that
µ(K) + µ(−K) = cµn(K), for all K ∈ Kn.

Inching Forward
The theorem we just proved is equivalent to...
Theorem
valuation on Kn. Then there exists c ∈ R such that
µ(K) + µ(−K) = cµn(K), for all K ∈ Kn.
This theorem seems much closer to the volume theorem. In fact
we are done if K is origin symmetric.

Proof of the Forward Direction
Suppose µ is a continuous translation-invariant simple
valuation on Kn.

valuation on Kn.
For K ∈ Kn, deﬁne
ν(K) = µ(K) + µ(−K) − 2µ([0, 1]n
)µn(K).

valuation on Kn.
For K ∈ Kn, deﬁne
ν(K) = µ(K) + µ(−K) − 2µ([0, 1]n
)µn(K).
ν satisﬁes the conditions of Klain’s lemma, so for all
K ∈ (K)n, ν(K) = 0 and
µ(K) + µ(−K) = cµn(K),
where c = 2µ([0, 1]n).

Review
K ∈ Kn.

Review
K ∈ Kn.
written as
µ =
n
i=0
ci µi

Proof of Hadwiger’s Theorem
Proof by induction. Let µ be a continuous,
rigid-motion-invariant valuation on Kn. Now consider µ|H
where H is a hyperplane. We have that µ|H is a valuation on
Kn−1.

Kn−1.
Now using the induction hypothesis, for all A ⊂ H, we have
µ|H(A) =
n−1
i=0
ci µi (A).

Kn−1.
Now using the induction hypothesis, for all A ⊂ H, we have
µ|H(A) =
n−1
i=0
ci µi (A).
Since µ is rigid-motion invariant, the coeﬃcients ci will be the
same for any choice of hyperplane H.

We now have that µ − n−1
i=0 ci µi is a continuous,
rigid-motion-invariant, simple valuation and using the the
Volume Theorem we get
µ −
n−1
i=0
ci µi = cnµn
for some cn ∈ R.

We now have that µ − n−1
i=0 ci µi is a continuous,
rigid-motion-invariant, simple valuation and using the the
Volume Theorem we get
µ −
n−1
i=0
ci µi = cnµn
for some cn ∈ R.
We ﬁnish the proof upon rearrangement!

Thank You
Professor Gardner.
Daniel Klain for responding to questions during the project.

Thank You
Professor Gardner.
You all for coming!

Thank You
Professor Gardner.
You all for coming!
Mom, Dad, and sister.

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