Daniel Hug, Rolf Schneider, Ralph Schuster Firenze May...
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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
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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
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
1. Minkowski Functionals
Minkowski functionals (intrinsic volumes, quermassintegrals) are
basic functionals of convex bodies in Rn:
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
1. Minkowski Functionals
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
• Axiomatic characterization. The functionals
V0, V1, . . . , Vn−1, Vn
are real-valued additive (valuations), continuous and motion
invariant; Vk is homogeneous of degree k.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
• 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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
• 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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
• 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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
3. Tensor Valuations
• Schneider, Schneider & Hadwiger (’71, ’72)
• McMullen (’97)
• Alesker (’99)
• Schneider, Schneider & Schuster (’00, ’02, ’04)
• Beisbart, Mecke ... (’00 - ?)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
- 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!.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
- 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!.
- Rotation covariance: for all K ∈ Kn and U ∈ O(n),
ϕ(UK) = Uϕ(K).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
- 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!.
- Rotation covariance: for all K ∈ Kn and U ∈ O(n),
ϕ(UK) = Uϕ(K).
- Isometry covariance
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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}
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
A local parallel set:
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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〉.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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 ...
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Two main tasks to be discussed here:
1. Provide a complete system of integral geometric formulae for
tensor valuations. Problem: linear dependences
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Daniel Hug Valuations, Integral Geometry and 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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
5. A General Approach. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.
∫
Enk
Φj,r,s(K ∩ E)µnk (dE)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
The first step. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.
∫
Enk
Φj,r,s(K ∩ E)µnk (dE)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
The first step. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.
∫
Enk
Φj,r,s(K ∩ E)µnk (dE)
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
The first step. Let r, s ∈ N0, 0 6 j 6 k 6 n− 1.
∫
Enk
Φj,r,s(K ∩ E)µnk (dE)
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Case 1: j = k
Φ(Lt)k,r,s−2m(K ∩ Lt) = 0 if s− 2m 6= 0.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)
=1
(4π)s2 ( s
2 )!Φn,r,0(K)
∫
Lnk
Q(L⊥)s2 νn
k (dL)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)
=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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)),
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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 ).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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)
.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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 ),
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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 ).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
... 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 ).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
6. Linear Dependences
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.
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
6. Linear Dependences
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”
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Proof. Induction wrt r. Main case: r > 2, k > 1.
Assume∑
l,s
αl,sQlΦk−r+s+2l,r−s−2l,s = 0. (+)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Proof. Induction wrt r. Main case: r > 2, k > 1.
Assume∑
l,s
αl,sQlΦk−r+s+2l,r−s−2l,s = 0. (+)
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)
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Proof. Induction wrt r. Main case: r > 2, k > 1.
Assume∑
l,s
αl,sQlΦk−r+s+2l,r−s−2l,s = 0. (+)
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
RTN Workshop, Firenze, May 2005 Slide 57/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
RTN Workshop, Firenze, May 2005 Slide 58/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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.
RTN Workshop, Firenze, May 2005 Slide 59/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
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).
RTN Workshop, Firenze, May 2005 Slide 60/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Happy
RTN Workshop, Firenze, May 2005 Slide 61/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Birthday
RTN Workshop, Firenze, May 2005 Slide 62/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Dear
RTN Workshop, Firenze, May 2005 Slide 63/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Rolf
RTN Workshop, Firenze, May 2005 Slide 64/ 65
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Schneider
RTN Workshop, Firenze, May 2005 Slide 65/ 65