Daniel Hug, Rolf Schneider, Ralph Schuster Firenze May 2005web.math.unifi.it/users/salani/workshop/talks/Hug.pdfDaniel Hug Valuations, Integral Geometry and Linear Dependences •

Daniel Hug, Rolf Schneider, Ralph Schuster

Mathematisches Institut

Universitat Freiburg

Valuations, Integral Geometry and Linear Dependences

Daniel Hug, Rolf Schneider, Ralph Schuster

Firenze

May 2005

Daniel Hug Valuations, Integral Geometry and Linear Dependences

Contents

1. Minkowski Functionals

2. Valuations and Integral Geometry

3. Tensor Valuations

4. Special Results

5. A General Approach

6. Linear Dependences

RTN Workshop, Firenze, May 2005 Slide 2/ 65



Minkowski functionals (intrinsic volumes, quermassintegrals) are

basic functionals of convex bodies in Rn:




Minkowski functionals (intrinsic volumes, quermassintegrals) are

basic functionals of convex bodies in Rn:

• Coefficients of a Steiner formula. Let K ∈ Kn and ε > 0:

Hn(K + εBn) =n

∑

i=0

εiκn−iVi(K),

where Bn denotes the unit ball in Rn, κn its volume, and Hn is

the Hausdorff measure (volume).



• Axiomatic characterization. The functionals

V0, V1, . . . , Vn−1, Vn

are real-valued additive (valuations), continuous and motion

invariant; Vk is homogeneous of degree k.



• Axiomatic characterization. The functionals

V0, V1, . . . , Vn−1, Vn

are real-valued additive (valuations), continuous and motion

invariant; Vk is homogeneous of degree k.

Theorem.[Hadwiger] Let ψ : Kn → R be additive, continuous and

motion invariant. Then there exist constants c0, . . . , cn such that

ψ =n

∑

i=0

ciVi.



• Objects of integral geometry. Let Enk be the Grassmannian of

k-flats in Rn, let µnk be a Haar measure on En

k . Then, for

K ∈ Kn and 0 6 j 6 k 6 n, we have the Crofton formula∫

Enk

Vj(K ∩ E)µnk(dE) = αnjkVn+j−k(K).



• Objects of integral geometry. Let Enk be the Grassmannian of

k-flats in Rn, let µnk be a Haar measure on En

k . Then, for

K ∈ Kn and 0 6 j 6 k 6 n, we have the Crofton formula∫

Enk

Vj(K ∩ E)µnk(dE) = αnjkVn+j−k(K).

Let G(n) be the motion group, and µ a Haar measure on G(n).

For K,L ∈ Kn and j ∈ {0, . . . , n}, the kinematic formula states

∫

G(n)

Vj(K ∩ gL)µ(dg) =n

∑

k=j

αnjkVk(K)Vn+j−k(L).



2. Valuations and integral geometry

Results about valuations have been used for the proofs of integral

geometric results.

Hadwiger’s abstract integral geometric formula

Let ϕ : Kn → R be an additive, continuous function. Put

ϕn−q(K) :=

∫

Enq

ϕ(K ∩ E)µnq (dE), K ∈ Kn.



2. Valuations and integral geometry

Results about valuations have been used for the proofs of integral

geometric results.

Hadwiger’s abstract integral geometric formula

Let ϕ : Kn → R be an additive, continuous function. Put

ϕn−q(K) :=

∫

Enq

ϕ(K ∩ E)µnq (dE), K ∈ Kn.

Theorem.[Hadwiger] For any K,L ∈ Kn and ϕ as above,

∫

G(n)

ϕ(K ∩ gL)µ(dg) =n

∑

q=0

ϕn−q(K)Vq(L).

Hints to the literature



3. Tensor Valuations

• Schneider, Schneider & Hadwiger (’71, ’72)

• McMullen (’97)

• Alesker (’99)

• Schneider, Schneider & Schuster (’00, ’02, ’04)

• Beisbart, Mecke ... (’00 - ?)



Background

- Space of symmetric tensors of rank r: Tr; T

symmetric tensor product of a, b ∈ T: ab

- Tensor valuation: ϕ : Kn → ⋃r

s=0 Ts



Background




s=0 Ts

- Translation covariance: there are ϕ0, . . . , ϕr : Kn → T such

that, for all K ∈ Kn and t ∈ Rn,

ϕ(K + t) =r

∑

j=0

ϕr−j(K)tj

j!.



Background




s=0 Ts



ϕ(K + t) =r

∑

j=0

ϕr−j(K)tj

j!.

- Rotation covariance: for all K ∈ Kn and U ∈ O(n),

ϕ(UK) = Uϕ(K).



Background




s=0 Ts



ϕ(K + t) =r

∑

j=0

ϕr−j(K)tj

j!.

- Rotation covariance: for all K ∈ Kn and U ∈ O(n),

ϕ(UK) = Uϕ(K).

- Isometry covariance



A detour: Support measures

Fix K ∈ Kn, ε > 0, η ∈ Σ := Rn × S

n−1, x ∈ Rn \K.

• Metric projection: p(K,x)

• Direction vector: u(K,x) := (x− p(K,x))/‖x− p(K,x)‖

• Local parallel set:

Mε(K, η) := {x ∈ (K + εBn) \K : (p(K,x), u(K,x)) ∈ η}



A detour: Support measures

Fix K ∈ Kn, ε > 0, η ∈ Σ := Rn × S

n−1, x ∈ Rn \K.

• Metric projection: p(K,x)

• Direction vector: u(K,x) := (x− p(K,x))/‖x− p(K,x)‖

• Local parallel set:

Mε(K, η) := {x ∈ (K + εBn) \K : (p(K,x), u(K,x)) ∈ η}

• Local Steiner formula:

Hn(Mε(K, η)) =n−1∑

i=0

εn−iκn−iΛi(K, η);

Λn(K, ·) := λnxK on Rn



A local parallel set:



Basic Examples. K ∈ Kn, r, s ∈ N0, 0 6 k 6 n− 1:

Φk,r,s(K) :=1

r!s!

ωn−k

ωn−k+s

∫

Σ

xrusΛk(K, d(x, u))

and

Φn,r,0(K) :=1

r!

∫

Rn

xrΛn(K, dx);

in all other cases, Φk,r,s := 0. Further, Q ∈ T2 is defined by

Q(x, y) := 〈x, y〉.



Basic Examples. K ∈ Kn, r, s ∈ N0, 0 6 k 6 n− 1:

Φk,r,s(K) :=1

r!s!

ωn−k

ωn−k+s

∫

Σ

xrusΛk(K, d(x, u))

and

Φn,r,0(K) :=1

r!

∫

Rn

xrΛn(K, dx);

in all other cases, Φk,r,s := 0. Further, Q ∈ T2 is defined by

Q(x, y) := 〈x, y〉.

Theorem. [Alesker] Let p ∈ N0, and let ϕ : Kn → Tp be a con-

tinuous, isometry covariant valuation. Then ϕ is a linear com-

bination, with constant real coefficients, of the basic valuations

QmΦk,r,s, where m, k, r, s ∈ N0 are such that 2m+ r + s = p.



Example. For p = 2 one can show that the tensor valuations

• QVj , j = 0, . . . , n,

• Φj,2,0, j = 0, . . . , n, and

• Φj,0,2, j = 1, . . . , n− 1.

form a basis ...



Example. For p = 2 one can show that the tensor valuations

• QVj , j = 0, . . . , n,

• Φj,2,0, j = 0, . . . , n, and

• Φj,0,2, j = 1, . . . , n− 1.

form a basis ...

The special linear relationships

2π∑

s

sΦk−r+s,r−s,s = Q∑

s

Φk−r+s,r−s,s−2,

for k, r ∈ N0, have been found by McMullen.



Two main tasks to be discussed here:

1. Provide a complete system of integral geometric formulae for

tensor valuations. Problem: linear dependences



Two main problems to be discussed here:

1. Provide a complete system of integral geometric formulae for

tensor valuations.

2. Find all linear relationships between the basic tensor valuations,

determine the dimension of the corresponding vector space.



4. Special Results

The case s = 0:∫

Enk

Φj,r,0(K ∩ E)µnk (dE) = αnjkΦn+j−k,r,0(K),

∫

G(n)

Φj,r,0(K ∩ gL)µ(dg) =n

∑

k=j

αnjkΦk,r,0(K)Vn+j−k(L).



4. Special Results

The case s = 0:∫

Enk

Φj,r,0(K ∩ E)µnk (dE) = αnjkΦn+j−k,r,0(K),

∫

G(n)

Φj,r,0(K ∩ gL)µ(dg) =n

∑

k=j

αnjkΦk,r,0(K)Vn+j−k(L).

The case j = n− 1:∫

Enn−1

Φn−1,r,s(K ∩ E)µnn−1(dE) = δ(n, s)Q

s2 Φn,r,0(K),

∫

G(n)

Φn−1,r,s(K∩gL)µ(dg) = Φn−1,r,s(K)Vn(L)+δ(n, s)Qs2 Φn,r,0(K)Vn−1(L).



The case j = n− 2:

∫

Enn−1

Φn−2,r,s(K∩E)µnn−1(dE) =

b s2 c

∑

m=0

α(n, s,m)QmΦn−1,r,s−2m(K)

and

∫

Enn−2

Φn−2,r,s(K ∩ E)µnn−2(dE) = β(n, s)Q

s2 Φn,r,0(K).

Further results follow from McMullen’s relationships.



5. A General Approach. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.

∫

Enk

Φj,r,s(K ∩ E)µnk (dE)



Theorem. For K ∈ Kn, r, s ∈ N0 and 0 6 k 6 n− 1,

∫

Enk

Φk,r,s(K ∩ E)µnk (dE) =

0, if s is odd,

αQs2 Φn,r,0(K), if s is even.



Theorem. For K ∈ Kn and k, j, r, s ∈ N0 with 0 6 j < k 6 n− 1,∫

Enk

Φj,r,s(K ∩ E) dµnk(E)

=

b s2 c

∑

z=0

χ(1)n,j,k,s,zQ

zΦn+j−k,r,s−2z(K) +

with explicitly known constants χ(1)n,j,k,s,z and χ

(2)n,j,k,s,z independent

of r.



Theorem. For K ∈ Kn and k, j, r, s ∈ N0 with 0 6 j < k 6 n− 1,∫

Enk

Φj,r,s(K ∩ E) dµnk(E)

=

b s2 c

∑

z=0

χ(1)n,j,k,s,zQ

zΦn+j−k,r,s−2z(K) +

b s2 c−1∑

z=0

χ(2)n,j,k,s,zQ

z

×s−2z−1

∑

l=0

(2πlΦn+j−k−s+2z+l,r+s−2z−l,l(K)

−QΦn+j−k−s+2z+l,r+s−2z−l,l−2(K))

with explicitly known constants χ(1)n,j,k,s,z and χ

(2)n,j,k,s,z independent

of r.



The first step. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.

∫

Enk





∫

Enk


For L ∈ Lnk , t ∈ L⊥ and Lt := L+ t (McMullen):

Φj,r,s(K ∩ Lt) =∑

m>0

Q(L⊥)m

(4π)mm!Φ

(Lt)j,r,s−2m(K ∩ Lt).




∫

Enk


For L ∈ Lnk , t ∈ L⊥ and Lt := L+ t:

Φj,r,s(K ∩ Lt) =∑

m>0

Q(L⊥)m

(4π)mm!Φ

(Lt)j,r,s−2m(K ∩ Lt).

Thus∫

Enk

Φj,r,s(K ∩ E)µnk(dE) =

∫

Lnk

∫

L⊥

Φj,r,s(K ∩ Lt)Hn−k(dt)νnk (dL)

=∑

m>0

Q(L⊥)m

(4π)mm!

∫

Lnk

∫

L⊥

Φ(Lt)j,r,s−2m(K ∩ Lt)Hn−k(dt)νn

k (dL).



Case 1: j = k

Φ(Lt)k,r,s−2m(K ∩ Lt) = 0 if s− 2m 6= 0.



Case 1: j = k

Φ(Lt)k,r,s−2m(K ∩ Lt) = 0 if s− 2m 6= 0.

If s is odd,

Φk,r,s(K ∩ Lt) = 0,

and thus∫

Enk

Φk,r,s(K ∩ E)µnk(dE) = 0.



Case 1: j = k

Φ(Lt)k,r,s−2m(K ∩ Lt) = 0 if s− 2m 6= 0.

If s is odd,

Φk,r,s(K ∩ Lt) = 0,

and thus∫

Enk

Φk,r,s(K ∩ E)µnk(dE) = 0.

If s is even,

Φk,r,s(K ∩ Lt) =Q(L⊥)

s2

(4π)s2 ( s

2 )!r!

∫

K∩Lt

xrHk(dx).



Hence, if s is even,

∫

Enk

Φk,r,s(K ∩ E)µnk (dE)

=1

(4π)s2 ( s

2 )!r!

∫

Lnk

Q(L⊥)s2

∫

L⊥

∫

K∩Lt

xrHk(dx)Hn−k(dt)νnk (dL)




∫

Enk


=1

(4π)s2 ( s

2 )!r!

∫

Lnk

Q(L⊥)s2

∫

L⊥

∫

K∩Lt


=1

(4π)s2 ( s

2 )!Φn,r,0(K)

∫

Lnk

Q(L⊥)s2 νn

k (dL)




∫

Enk


=1

(4π)s2 ( s

2 )!r!

∫

Lnk

Q(L⊥)s2

∫

L⊥

∫

K∩Lt


=1

(4π)s2 ( s

2 )!Φn,r,0(K)

∫

Lnk

Q(L⊥)s2 νn

k (dL)

= αQs2 Φn,r,0(K),

where α is explicitly known.



Lemma. For m ∈ N0 and k ∈ {1, . . . , n},∫

Lnk

Q(L)mνnk (dL) =

Γ(k2 +m)Γ(n

2 )

Γ(n2 +m)Γ(k

2 )Qm.



Case 2: 0 6 j 6 k − 1, P ∈ Pn.

Φj,r,s(P ∩ Lt) =∑

m>0

Q(L⊥)m

(4π)mm!Φ

(Lt)j,r,s−2m(P ∩ Lt)

=∑

m>0

Q(L⊥)mαj,k,s,mωk−j

r!

∫

Lt×(Sn−1∩L)

xrus−2mΛ(Lt)j (P ∩ Lt, d(x, u)),



Case 2: 0 6 j 6 k − 1, P ∈ Pn.

Φj,r,s(P ∩ Lt) =∑

m>0

Q(L⊥)m

(4π)mm!Φ

(Lt)j,r,s−2m(P ∩ Lt)

=∑

m>0

Q(L⊥)mαj,k,s,mωk−j

r!

∫

Lt×(Sn−1∩L)

xrus−2mΛ(Lt)j (P ∩ Lt, d(x, u)),

We need the translative integral (Crofton) formula (Rataj ’99):∫

L⊥

∫

Lt×(Sn−1∩L)

g(x, v)Λ(Lt)j (P ∩ Lt, d(x, v))Hn−k(dt)

=1

ωk−j

∑

F∈Fn+j−k(P )

∫

F×(N(P,F )∩Sn−1)

g(x, πL(u))‖pL(u)‖j−k

× [F,L]2 (Hn+j−k ⊗Hk−j−1)(d(x, u)).



By Fubini’s theorem,

∫

Lnk

∫

L⊥

Φj,r,s(P ∩ Lt)Hn−k(dt)νnk (dL)

=∑

m>0

∑

F∈Fn+j−k(P )

αj,k,s,m

1

r!

∫

F

xrHn+j−k(dx)

∫

N(P,F )∩Sn−1

×∫

Lnk

Q(L⊥)mπL(u)s−2m‖pL(u)‖j−k[F,L]2νnk (dL).



Proposition.∫

Lnk

Q(L⊥)mπL(u)s−2m‖pL(u)‖j−k[F,L]2νnk (dL)

= βn,j,k

b s2 c

∑

z=0

ζ(1)n,j,k,s,z,mQ

zus−2z

+ βn,j,k

k − n

n+ j − k

b s2 c−1∑

z=0

ζ(2)n,j,k,s,z,mQ

zus−2z−2Q(F ).



Some constants:

βn,j,k :=(k − 1)!(n+ j − k)!√

πj!(n− 1)!

Γ(n2 )Γ(n+1

2 )

Γ(k2 )Γ(k+1

2 ),

γ(1)n,k,l,p,q :=

q∑

y=0

(−1)l+y

(

q

y

)

Γ(k−12 + l − p+ y)

Γ(n+12 + l − p+ y)

(

k−12 + l − p+ y

)

,

γ(2)n,k,l,p,q :=

q∑

y=0

(−1)l+y

(

q

y

)

Γ(k−12 + l − p+ y)

Γ(n+12 + l − p+ y)

(l − p+ y)

with γ(2)n,k,l,p,q = 0 if l − p+ q = 0.



More constants:

ζ(1)n,j,k,s,z,m :=

m∑

l=max{0,m−z}

l∑

p=0

b s2 c−m+p∑

q=max{0,z−m+p}

(−1)m−p+q−zγ(1)n,k,l,p,q

(

m

l

)(

l

p

)

×(

s− 2m+ 2p

2q

)(

l − p+ q

z −m+ l

)

Γ( s+j2 −m+ p− q + 1)Γ(q + 1

2 )

Γ( s+n−k+j2 −m+ p+ 1)

and

ζ(2)n,j,k,s,z,m :=

m∑

l=max{0,m−z}

l∑

p=0

b s2 c−m+p∑

q=max{0,z−m+p+1}

(−1)m−p+q−z−1γ(2)n,k,l,p,q

(

m

l

)(

l

p

)

×(

s− 2m+ 2p

2q

)(

l − p+ q − 1

z −m+ l

)

Γ( s+j

2 −m+ p− q + 1)Γ(q + 12 )

Γ( s+n−k+j

2 −m+ p+ 1).



Some combinatorial identities of the form:

Lemma. For m ∈ N0, n ∈ N, k ∈ {1, . . . , n− 1} and a > 0,

m∑

i=0

(

m

i

)

Γ(

k+a2 +m− i

) Γ(n−k2 + i)Γ(a

2 + 1)Γ(n2 )

Γ(n−k2 )Γ(a

2 + 1 − i)Γ(n2 + i)

(−1)i

=Γ(k

2 +m)Γ(n+a2 +m)Γ(n

2 )Γ(k+a2 )

Γ(k2 )Γ(n+a

2 )Γ(n2 +m)

.



Putting things together and by some additional calculations, we get

∫

Lnk

∫

L⊥

Φj,r,s(P ∩ Lt)Hn−k(dt)νnk (dL)

= βn,j,k

b s2 c

∑

z=0

ξ(1)n,j,k,s,z(s− 2z)!ωs−2z−j+k Q

zΦn+j−k,r,s−2z(P )

+ βn,j,k

k − n

n+ j − k

b s2 c−1∑

z=0

ξ(2)n,j,k,s,z(s− 2z − 2)!ωs−2z−2−j+k

× Qz∑

F∈Fn+j−k(P )

Q(F )Υr(F )Θs−2z−2(P, F ),



where, for a k-face F of P ,

Υr(F ) :=1

r!

∫

F

xrHk(dx)

and

Θs(P, F ) :=1

s!

∫

N(P,F )

xse−π‖x‖2Hn−k(dx).

Hence

Φk,r,s(P ) =∑

F∈Fk(P )

Υr(F )Θs(P, F ).



The final step: a lemma due to McMullen ...

∑

F∈Fn+j−k(P )

Q(F )Υr(F )Θs−2z−2(P, F )

= QΦn+j−k,r,s−2z−2(P ) − 2π(s− 2z)Φn+j−k,r,s−2z(P )

+∑

G∈Fn+j−k+1(P )

Q(G)Υr−1(G)Θs−2z−1(P,G).



... can be iterated to give

∑

F∈Fn+j−k(P )

Q(F )Υr(F )Θs−2z−2(P, F )

=∑

l>s−2z

QΦn+j−k−s+2z+l,r+s−2z−l,l−2(P )

− 2π∑

l>s−2z

lΦn+j−k−s+2z+l,r+s−2z−l,l(P ).




Theorem. For k, r ∈ N0,

2π∑

s

sΦk−r+s,r−s,s −Q∑

s

Φk−r+s,r−s,s−2 = 0. (∗)

Any linear dependence among those tensor valuations QlΦk,r,s,

which do not vanish trivially, can be obtained by multiplying by

Q relations of the form (∗) and by taking linear combinations of

relations obtained in this way.




Theorem. For k, r ∈ N0,

2π∑

s

sΦk−r+s,r−s,s −Q∑

s

Φk−r+s,r−s,s−2 = 0. (∗)

Any linear dependence among those tensor valuations QlΦk,r,s,

which do not vanish trivially, can be obtained by multiplying by

Q relations of the form (∗) and by taking linear combinations of

relations obtained in this way.

Proof. Consider, for r ∈ N0 and k = 0, . . . , n+ r,

Gk,r := lin{Ql′Φk′,r′,s′ : k′ + r′ = k, r′ + s′ + 2l′ = r}

“Tensors of rank r, homogeneous of degree k”



Proof. Induction wrt r. Main case: r > 2, k > 1.

Assume∑

l,s

αl,sQlΦk−r+s+2l,r−s−2l,s = 0. (+)




Assume∑

l,s


Substitute K + t, for t ∈ Rn, and use translation covariance:

∑

l,s

αl,sQlΦk−r+s+2l,r−s−2l−1,s = 0. (in Gk−1,r−1)




Assume∑

l,s


Substitute K + t, for t ∈ Rn, and use translation covariance:

∑

l,s

αl,sQlΦk−r+s+2l,r−s−2l−1,s = 0. (in Gk−1,r−1)

Via the induction hypothesis, (+) is equivalent to

α2πr−1∑

s=1

sΦk−r+s,r−s,s+α0rΦk,0,r+∑

l>1,s>0

αl,sQlΦk−r+s+2l,r−s−2l,s = 0



or to

(α0,r − 2πrα)Φk,0,r +∑

l>1,s>0

αl,sQlΦk−r+s+2l,r−s−2l,s = 0.

This is of the form

(α0,r − 2πrα)Φk,0,r = Qv, v ∈ Tr−2.



or to

(α0,r − 2πrα)Φk,0,r +∑

l>1,s>0

αl,sQlΦk−r+s+2l,r−s−2l,s = 0.

This is of the form

(α0,r − 2πrα)Φk,0,r = Qv, v ∈ Tr−2.

Lemma. For all s, k ∈ N with 1 6 k 6 n − 1 and s > 2, there

exists a convex body K ∈ Kn such that

Φk,0,s(K) 6= Qv

for all v ∈ Ts−2.



Theorem. Let r ∈ N0, n > 1 and 0 6 k 6 n+ r. Put

j0 := min

{⌊

n+ r − k

2

⌋

,⌊r

2

⌋

}

, j1 := max

{

−1,

⌊

r − k

2

⌋}

.

Then

dim(Gk,r) = j0(min{1, n− k}+ r− j0) + 1− (j1 + 1)(r − k − j1).



Happy



Birthday



Dear



Rolf



Schneider


Daniel Hug, Rolf Schneider, Ralph Schuster Firenze May 2005web.math.unifi.it/users/salani/workshop/talks/Hug.pdfDaniel Hug Valuations, Integral Geometry and Linear Dependences •

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