Daniel Hug, Rolf Schneider, Ralph Schuster Firenze May...
Transcript of Daniel Hug, Rolf Schneider, Ralph Schuster Firenze May...
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
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1. Minkowski Functionals
Minkowski functionals (intrinsic volumes, quermassintegrals) are
basic functionals of convex bodies in Rn:
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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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• 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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• 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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• 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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• 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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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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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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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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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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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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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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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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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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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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A local parallel set:
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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Case 1: j = k
Φ(Lt)k,r,s−2m(K ∩ Lt) = 0 if s− 2m 6= 0.
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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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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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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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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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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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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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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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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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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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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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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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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
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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.
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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.
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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).
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Happy
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Birthday
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Dear
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Rolf
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Daniel Hug Valuations, Integral Geometry and Linear Dependences
Schneider
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