Lecture 4: Hermitian integral geometrymath.uga.edu/~fu/diablerets/diableret4.pdfAs valuations (i.e. as “complete integrals"), the second is 2× the ﬁrst. (J.H.G. Fu) The two faces

Lecture 4: Hermitian integral geometryjoint work with A. Bernig

The integral geometry of the complex space forms CPn,CHn, and of the corresponding flat caseof Cn under the group U(n), is much more complicated. Here we describe a pretty good pictureof the latter, worked out recently in collaboration with A. Bernig.The main obstruction to transferring to the curved cases is the fact that, unlike the situation withthe real space forms, the surjection CurvU(n)(Cn) → ValU(n)(Cn) has a nontrivial kernel.To illustrate, there are two invariant elements of CurvU(2)(C2) of degree 1. Note that the secondfundamental form of a hypersurface M = ∂A ⊂ C2 may be canonically decomposed as0@a b c

b A11 A12c A12 A22

1Acorresponding to the decomposition Tx M ' R⊕ C into its maximal complex subspace and thecomplementary line. The two invariant curvature measures correspond to the integrals of

det A, a · trace A + b2 + c2

As valuations (i.e. as “complete integrals"), the second is 2× the first.

(J.H.G. Fu) The two faces of Blaschkean integral geometry 1 / 21

Nonetheless, R. Howard (1986) showed that the transfer principle remains valid at the level ofCrofton formulas.This is illustrated by the 2003 work of Tasaki, based on his classification of the U(n) orbits of thereal Grassmannian Grk (Cn): for k ≤ n they are parametrized by the multiple Kähler angle ofE ∈ Grk (Cn),

Θ = (0 ≤ θ1 ≤ · · · ≤ θb k2 c≤π

2)

where the restriction of the Kähler form to E may be expressed

b k2 cX

i=1

cos θi α2i−1 ∧ α2i

with respect to some orthonormal basis αj of E∗. Thus Θ(E) ≡ 0 iff E contains a complexjk2

k-plane, and Θ(E) ≡ π

2 iff E is isotropic.


Theorem (Tasaki-Crofton formula 2003)

Given k ≤ n, there exists a symmetric (p + 1)× (p + 1) matrix T nk , p :=

jk2

k, such that for every

pair of compact C1 submanifolds Mk ,N2n−k ⊂ Cn,CPn or CHn, of dimension and codimensionk, respectively,Z

U(n)#(Mk ∩ gNn−k ) dg =

pXi,j=0

`T n

k´

ij

ZMσi (cos2 Θ(Tx M)) dx

ZNσj (cos2 Θ(Ty N⊥)) dy .

Furthermore

T n2 =

14n(n − 1)

„2n − 1 −1−1 2n − 1

«, T 3

3 =2

9π

„3 −1−1 5

3

«

As we will see, the array of the T nk for fixed n is equivalent to kCn,U(n)(χ). We can compute as

many as you like:


T n3 =

2n−2(n − 3)!

nπ(2n − 3)!!

„2n − 3 −1−1 2n−1

3

«,

T n4 =

(n − 4)!

16n!

0@3(2n − 5)(2n − 3) −3(2n − 3) 9−3(2n − 3) 2n2 − 4n + 3 −3(2n − 3)

9 −3(2n − 3) 3(2n − 5)(2n − 3)

1AT n

5 = 22n−1π(n−2)(2n−5)

×

0BBBBB@

“8 n3−60 n2+142 n−105

”((n−2)!)2

16(n−4)(n−3)n(n−2)(2 n−2)!− 1

128((n−4)!)2

(n−1) (n−4)n(2 n−7)!3

32(n−3)! (2 n−9)! (2 n−7)(n−2)! (n−4)!

(2 n−5)! n! (2 n−7)!

− 1128

((n−4)!)2

(n−1)(n−4)n(2 n−7)!1

80

“2 n2−6 n+7

”(2 n−3)((n−2)!)2

(n−4)(n−3)n(n−2)(2 n−2)!− 3

640(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!

332

(n−3)! (2 n−9)! (2 n−7)(n−4)! (n−2)!(2 n−5)! n! (2 n−7)!

− 3640

(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!1

640(2 n−3)((n−4)!)2

(n−4)n(n−1)(2 n−7)!

1CCCCCA

Although we don’t have nice closed forms, we do know

Theorem (Bernig-Fu 2007)

The T nk are positive definite. If k = 2p then there is an additional antidiagonal symmetry

(T n2p)p−i,p−j = (T n

2p)ij


T n3 =

2n−2(n − 3)!

nπ(2n − 3)!!

„2n − 3 −1−1 2n−1

3

«,

T n4 =

(n − 4)!

16n!

0@3(2n − 5)(2n − 3) −3(2n − 3) 9−3(2n − 3) 2n2 − 4n + 3 −3(2n − 3)

9 −3(2n − 3) 3(2n − 5)(2n − 3)

1AT n

5 = 22n−1π(n−2)(2n−5)

×

0BBBBB@

“8 n3−60 n2+142 n−105

”((n−2)!)2

16(n−4)(n−3)n(n−2)(2 n−2)!− 1

128((n−4)!)2

(n−1) (n−4)n(2 n−7)!3

32(n−3)! (2 n−9)! (2 n−7)(n−2)! (n−4)!

(2 n−5)! n! (2 n−7)!

− 1128

((n−4)!)2

(n−1)(n−4)n(2 n−7)!1

80

“2 n2−6 n+7

”(2 n−3)((n−2)!)2

(n−4)(n−3)n(n−2)(2 n−2)!− 3

640(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!

332

(n−3)! (2 n−9)! (2 n−7)(n−4)! (n−2)!(2 n−5)! n! (2 n−7)!

− 3640

(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!1

640(2 n−3)((n−4)!)2

(n−4)n(n−1)(2 n−7)!

1CCCCCA





2p)ij


T n3 =

2n−2(n − 3)!

nπ(2n − 3)!!

„2n − 3 −1−1 2n−1

3

«,

T n4 =

(n − 4)!

16n!

0@3(2n − 5)(2n − 3) −3(2n − 3) 9−3(2n − 3) 2n2 − 4n + 3 −3(2n − 3)

9 −3(2n − 3) 3(2n − 5)(2n − 3)

1AT n

5 = 22n−1π(n−2)(2n−5)

×

0BBBBB@

“8 n3−60 n2+142 n−105

”((n−2)!)2

16(n−4)(n−3)n(n−2)(2 n−2)!− 1

128((n−4)!)2

(n−1) (n−4)n(2 n−7)!3

32(n−3)! (2 n−9)! (2 n−7)(n−2)! (n−4)!

(2 n−5)! n! (2 n−7)!

− 1128

((n−4)!)2

(n−1)(n−4)n(2 n−7)!1

80

“2 n2−6 n+7

”(2 n−3)((n−2)!)2

(n−4)(n−3)n(n−2)(2 n−2)!− 3

640(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!

332

(n−3)! (2 n−9)! (2 n−7)(n−4)! (n−2)!(2 n−5)! n! (2 n−7)!

− 3640

(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!1

640(2 n−3)((n−4)!)2

(n−4)n(n−1)(2 n−7)!

1CCCCCA





2p)ij


T n3 =

2n−2(n − 3)!

nπ(2n − 3)!!

„2n − 3 −1−1 2n−1

3

«,

T n4 =

(n − 4)!

16n!

0@3(2n − 5)(2n − 3) −3(2n − 3) 9−3(2n − 3) 2n2 − 4n + 3 −3(2n − 3)

9 −3(2n − 3) 3(2n − 5)(2n − 3)

1AT n

5 = 22n−1π(n−2)(2n−5)

×

0BBBBB@

“8 n3−60 n2+142 n−105

”((n−2)!)2

16(n−4)(n−3)n(n−2)(2 n−2)!− 1

128((n−4)!)2

(n−1) (n−4)n(2 n−7)!3

32(n−3)! (2 n−9)! (2 n−7)(n−2)! (n−4)!

(2 n−5)! n! (2 n−7)!

− 1128

((n−4)!)2

(n−1)(n−4)n(2 n−7)!1

80

“2 n2−6 n+7

”(2 n−3)((n−2)!)2

(n−4)(n−3)n(n−2)(2 n−2)!− 3

640(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!

332

(n−3)! (2 n−9)! (2 n−7)(n−4)! (n−2)!(2 n−5)! n! (2 n−7)!

− 3640

(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!1

640(2 n−3)((n−4)!)2

(n−4)n(n−1)(2 n−7)!

1CCCCCA





2p)ij


T n3 =

2n−2(n − 3)!

nπ(2n − 3)!!

„2n − 3 −1−1 2n−1

3

«,

T n4 =

(n − 4)!

16n!

0@3(2n − 5)(2n − 3) −3(2n − 3) 9−3(2n − 3) 2n2 − 4n + 3 −3(2n − 3)

9 −3(2n − 3) 3(2n − 5)(2n − 3)

1AT n

5 = 22n−1π(n−2)(2n−5)

×

0BBBBB@

“8 n3−60 n2+142 n−105

”((n−2)!)2

16(n−4)(n−3)n(n−2)(2 n−2)!− 1

128((n−4)!)2

(n−1) (n−4)n(2 n−7)!3

32(n−3)! (2 n−9)! (2 n−7)(n−2)! (n−4)!

(2 n−5)! n! (2 n−7)!

− 1128

((n−4)!)2

(n−1)(n−4)n(2 n−7)!1

80

“2 n2−6 n+7

”(2 n−3)((n−2)!)2

(n−4)(n−3)n(n−2)(2 n−2)!− 3

640(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!

332

(n−3)! (2 n−9)! (2 n−7)(n−4)! (n−2)!(2 n−5)! n! (2 n−7)!

− 3640

(2 n−3)((n−4)!)2

(n−4)(2 n−5)(n−1)n(2 n−7)!1

640(2 n−3)((n−4)!)2

(n−4)n(n−1)(2 n−7)!

1CCCCCA





2p)ij


For example:

Theorem

Let M4,N5 ⊂ CP4 be real C1 submanifolds of dimension 4, 5 respectively. Let θ1, θ2 be theKähler angles of the tangent plane to M at a general point x and ψ the Kähler angle of theorthogonal complement to the tangent plane to N at y. Then

5π4

|U(5)|

ZU(5)

length(M ∩ gN) dg = 30 vol4(M) vol5(N)− 6 vol4(M)

ZN

cos2 ψ dy

− 3Z

M(cos2 θ1 + cos2 θ2) dx · vol5(N) + 7

ZM

(cos2 θ1 + cos2 θ2) dx ·Z

Ncos2 ψ dy .

Theorem

Let E ∈ Gr4(C4),F ∈ Gr3(C4); let θ1, θ2 be the Kähler angles of E and ψ the Kähler angle of F .If A ∈ K(E),B ∈ K(F ) then

1|U(4)|

ZU(4)

vol7(A + gB) dg =1

120vol4(A) vol3(B)×h

30− 6 cos2 ψ − 3(cos2 θ1 + cos2 θ2) + 7 cos2 ψ(cos2 θ1 + cos2 θ2)i.


To prove these things, begin with

Theorem

ValU(n)(Cn) ' R[s, t]/(fn+1, fn+2), where deg s = 2, deg t = 1 and fi (s, t) is the polynomial ofweighted degree i in the expansion

log(1 + s + t) =∞Xi=1

fi (s, t).

For the proof, for 2p ≤ k , let Grk,p(Cn) denote the space of real k -dimensional subspaces of Cn

that may be expressed as the orthogonal direct sum of a complex subspace of dimension p andan isotropic subspace. Define s ∈ ValU(n)

2 (Cn) by

s :=

ZGr2n−1,n−1(Cn)

χ(· ∩ H) dH =

ZGr2,1(Cn)

area(πP(·)) dP, (1)

normalized so that s(D1C) = 1. Alesker showed that the valuations

sp tk−2p, 0 ≤ k ≤ n, 0 ≤ p ≤ min(bk2c, b

2n − k2

c)

constitute a basis for ValU(n)(Cn).


We claim first thatt2k sn−k (Dn

C) =“2k

k

”, k = 0, . . . n. (2)

To see this we use the transfer principle from CPn. Define s ∈ V U(n+1)(CPn) by

s :=

ZGrCn−1(CPn)

χ(· ∩ P) dP

where dP is the invariant probability measure. Then s corresponds under the transfer principle tos ∈ ValU(n)

2 (Cn) (proof: s(CP1) = 1 = s(D1C) and area(CP1) = π

2 = area(D1C)). Since

vol(DnC) = vol(CPn) = πn

n!, the relation (2) follows from the identity

t2k sn−k (CPn) = t2k (CPk ) =(2k)!ω2k

π2kvol(CPk ) =

(2k)!

π2k

πk

k!

!2

=“2k

k

”.

Next we recall (thanks to I. Gessel) the Pfaff-Saalschütz identity

b n+12 cX

i=0

(−1)i

n + 1− i

“n + 1− ii

”“2n − 2i − 2kn − i − k

”=

(−2k)n

22n+1( 12 − k)n

(3)

where (a)n := a(a + 1) . . . (a + n − 1) .


In particularb n+1

2 cXi=0

(−1)i

n + 1− i

“n + 1− ii

”“2n − 2i − 2kn − i − k

”= 0, 0 ≤ 2k < n, (4)

which with (2) yields the following identities in ValU(n)(Cn):

tn−2k−1sk ·

0B@bn+1

2 cXi=0

(−1)i

n + 1− i

“n + 1− ii

”tn−2i+1si

1CA = 0, 0 ≤ 2k < n. (5)

Since the tn−2k−1sk in this range constitute a basis of ValU(n)n−1 (Cn), Alesker-Poincaré duality

implies that the sum in the second factor is zero. This sum is (−1)nfn+1(s, t).

Since fn+2(s, t) = 0 in ValU(n+1)(Cn+1), and the natural restriction mapValU(n+1)(Cn+1) → ValU(n)(Cn) respects the definitions of s, t , it follows that fn+2(s, t) = 0 inValU(n)(Cn) as well. Since fn+2(s, t) 6= t · fn+1(s, t), some algebra implies that these relationsgenerate the ideal of relations.


Exercise

Recall that H2∗(Gr2(CPn+2)) ' R[s, t]/(σn+1, σn+2), where

11 + s + t

=X

i

σi (s, t).

Thus these Betti numbers agree with those of ValU(n). Show that the two graded algebras areisomorphic if n = 2, but not if n ≥ 3.

Thus for each degree/codegree pair the matrix giving the Poincaré-Alesker pairings with respectto the basis of monomials si t j is a Hankel matrix of the form0BBB@

`00

´ `21

´ `42

´. . .

`2kk

´`21

´ `42

´. . . . . .

`2k+2k+1

´. . . . . . . . . . . . . . .`2k

k

´ `2k+2k+1

´. . . . . .

`4k2k

´1CCCA or

0BBB@`2

1

´ `42

´ `63

´. . .

`2k+2k+1

´`42

´ `63

´. . . . . .

`2k+4k+2

´. . . . . . . . . . . . . . .`2k+2k+1

´ `2k+4k+2

´. . . . . .

`4k+22k+1

´1CCCA

Question

What are the inverses of these matrices? Heck, what are their determinants? (Experimentallythey appear to be powers of 2.)

These inverses give kU(n)(χ) in terms of the si t j . Even so, it would be good to know more aboutthe valuations si t j and their relations with curvature integrals, Tasaki functions, etc.


Generalities about Val+(V )

Let V be an m-dimensional real vector space. It turns out that ValU(n) := ValU(n)(Cn) ⊂ Val+, sofor simplicity we restrict attention to even valuations on V .

(Klain embedding) The mapping Kl : Val+i → C∞(Gri ) given by

Klφ(E) := ω−1i φ(E ∩ B)

is injective.ψ ∈ Val+i admits a (generally not unique) signed Crofton measure mψ on Gri :

ψ(K ) =

ZGri|πE (K )| dmψ(E)

(Alesker Fourier transform) There is a linear isomorphismˆ: Val+i → Val+m−i characterizedby

Klψ(E) = Klψ(E⊥)

The pairing 〈φ, ψ〉 := φ · ψ on Val+i is symmetric, with

〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)

(Alesker hard Lefschetz theorem) For each k ≤ m2 , the map φ 7→ tm−2k · φ is an

isomorphism Valk → Valm−k .





Klφ(E) := ω−1i φ(E ∩ B)


ψ(K ) =





〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)







Klφ(E) := ω−1i φ(E ∩ B)


ψ(K ) =





〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)







Klφ(E) := ω−1i φ(E ∩ B)


ψ(K ) =





〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)







Klφ(E) := ω−1i φ(E ∩ B)


ψ(K ) =





〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)







Klφ(E) := ω−1i φ(E ∩ B)


ψ(K ) =





〈φ, ψ〉 =

ZGri

Klφ(E) dmψ(E)




The Fourier transform gives rise to another product on Val(V ), the convolution

φ ∗ ψ := φ · ψ.

For A ∈ Ksm(V ) define µA ∈ Val(V ) by µA(B) := |A + B|.

Theorem

For A,B ∈ Ksm(V ),µA ∗ µB = µA+B .

Suppose G ⊂ SO(V ) acts transitively on the unit sphere of V . Define the cocommutative,coassociative coproduct aG : ValG → ValG ⊗ValG by

aG(φ)(A,B) :=1|G|

ZGφ(A + gB) dg. (6)

ThenaG = (b⊗b) kG ˆ. (7)

It is clear that aG(vol) = kG(χ), even without knowing (7).


It can be useful to express valuations in terms of integrals over “ball bundles"

N1(A) = p∗(N(A)× [[0, 1]]) + [[A]] ∈ In(TV ).

where p(x , v , t) := (x , tv). Note that N1(A) is Lagrangian.

Since ∂N1(A) = N(A), if ϕ ∈ Ωn−1(UV ) is extended to all of TV then by Stokes’ theorem

νϕ(A) =

ZN1(A)

dϕ =: λdϕ(A)

This is particularly useful when ψ = dϕ has constant coefficients, i.e. is invariant not only undertranslations in the base space but also in the fiber. In this case the Fourier-Alesker transform isgiven by cλψ = λj∗ψ

where j : TV ' V × V → TV interchanges the base and fiber coordinates.


Application to ValU(n)

It turns out that all the elements of ValU(n) are constant-coefficient valuations. The algebraΛ∗(Cn ⊕ Cn)U(n) of invariants under the diagonal action is generated by four 2-forms: thesymplectic form, the Kähler forms θ2 =

Pdxi ∧ dyi , θ0 =

Pdξi ∧ dηi of the base and the fiber,

andθ1 :=

X(dxi ∧ dηi − dyi ∧ dξi ).

Put for 2p ≤ kθk,p := cn,k,pθ

n+p−k0 ∧ θk−2p

1 ∧ θp2 ,

and define the hermitian intrinsic volumes by

µk,p := λθk,p ∈ ValU(n)k .

and the Tasaki valuations by

τk,q =

bk/2cXi=q

“ iq

”µk,i , µk,q =

bk/2cXi=q

(−1)i+q“ i

q

”τk,i


The following observations are key. For k ≤ 2n, 2k − 2n ≤ 2p ≤ k , define the(k , p)-Grassmannian Grk,p ⊂ Grk (Cn) as the U(n) orbit of Cp ⊕ Rk−2p .

Proposition

1 For k ≤ n, each of the families µk,pp, τk,pp constitutes a basis of ValU(n)k

2 Klµk,p |Grl,q≡ δ

k,pl,q

3 dµk,p = µ2n−k,n−k+p

4 Klτk,p = σp(cos2 Θ), where σp is the pth elementary symmetric function

5 fk (s, t) = ckµk,0 for some constant ck

6 t · µk,p =ωk+1πωk

`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.

5: Both sides span ker“

ValU(k)k (Ck ) → ValU(k−1)

k (Ck−1)”

under the restriction map.

6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0

µ2n−k,n−k+p(K + rB) =

CR

N(K ) LT θ2n−k,n−k+p (Figure 4a), and the Lie derivative LT θl,q is easy to compute.

Here T (x , v) := (v , 0) is the Reeb vector field.



Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR





Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR





Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR





Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR





Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR





Proposition



k,pl,q





`2(p + 1)µk+1,p+1 + (k − 2p + 1)µk+1,p

´

Proof.



k (Ck−1)”


6: (t · µk,p)(K ) = C`t2n−1 ∗ µ2n−k,n−k+p

´(K ) = C d

dr

˛r=0


CR




The relations 1, 4 imply the following equivalent form of the Tasaki-Crofton formula:

Theorem

There are symmetric nondegenerate matrices T nk such that

kU(n)(χ) =Xk≤n

Xp,q

`T n

k´

pq τk,p ⊗dτk,q

The (diagonal) symmetry of T nk is a consequence of the symmetry of the pairing φ · ψ

above.It turns out that if we put u := 4s − t2 ∈ ValU(n)

2 then

τk,p =πk

ωk (k − 2p)!(2p)!up tk−2p.

Thus the rest of the array of kinematic formulas

kU(n)(τl,q) = (τl,q ⊗ χ) · kU(n)(χ), aU(n)(cτl,q) = (χ⊗ cτl,q) ∗ aU(n)(vol)

may also be computed in these terms once T nk is known.

The additional antidiagonal symmetry for k = 2p follows from the relation

τ2p,q · τ2p,r = τ2p,p−q · τ2p,p−r

which in turn hinges on the following algebraic fact:


Proposition

Put Σp := 〈σp,q : 0 ≤ q ≤ p〉 for the vector space spanned by the elementary symmetricfunctions in p variables. Define the maps ι : Σp → Σp , r : Σm → Σp,m ≥ p, by

ι(σp,q) := σp,p−q

r(f ) := f (x1, . . . , xp, 1, . . . , 1)

ThenΣm

ι−−−−−→ Σm??yr??yr

Σpι−−−−−→ Σp

commutes.

Proof.

It is enough to prove this for m = p + 1, in which case r(σp+1,i ) = σp,i + σp,i−1. Hence fori = 0, . . . , p + 1,

ι r(σp+1,i ) = ι(σp,i + σp,i−1) = σp,p−i + σp,p−i+1 = r(σp+1,p−i+1) = r ι(σp+1,i ).


The final ingredient in the calculation of the Tasaki matrices T nk is

Proposition

There are (degree-dependent) constants Ck ,C′k such that the degree and codegree 1 operators

Lφ := Ck t · φ, Λφ := C′k t2n−1 ∗ φ

generates an action of sl(2) on ValU(n). The primitive elements (i.e. the generators of ValU(n) asa sl(2)-module) are

π2r,r := (−1)r (2n − 4r + 1)!!rX

i=0

(−1)i (2r − 2i − 1)!!

(2n − 2r − 2i + 1)!!τ2r,i , 2r ≤ n.

(Here (−1)!! := 1.) Putting πk,r := Lk−2rπ2r,r , k ≥ 2r ,

dπk,r =(k − 2r)!

(2n − 2r − k)!π2n−k,r

The πk,r diagonalize the Poincaré pairing, with

πk,r · π2n−k,s = δrs

πn

ωkω2n−k

16r (2n − 4r)!(2n − 4r + 1)!n!

(2r)!(2n − 2r + 1)!(n − 2r)!2


Thus the relation between Alesker-Poincaré duality and k(χ) yields

Theorem

kU(n)(χ) =

π−nX

r

(2r)!(2n − 2r + 1)!(n − 2r)!2

16r (2n − 4r)!(2n − 4r + 1)!n!

Xk

ωkω2n−kπk,r ⊗ π2n−k,r

Corollary

The Tasaki matrices T nk are positive definite.


Say that φ ∈ Val(V ) is

positive if φ(A) ≥ 0 for A ∈ K(V )

monotone if φ(A) ≥ φ(B) whenever A,B ∈ K(V ) and A ⊃ B

Crofton positive if it admits a nonnegative Crofton measure

These three cones clearly satisfy CP ⊂ M ⊂ P.

Say that Ψ ∈ CurvV (V ) is positive if ΨA is a nonnegative measure whenever A ∈ K(V ).

There is a natural first variation map δ : Val(V ) → Curv(V ) given by

ddt

˛t=0

µ(Ft (A)) =

Z∂A〈ξ, nA〉 d(δµA)

whenever A ∈ Ksm(V ), where Ft is the flow of the vector field ξ. A classical identity states that

δµi = cn,iΦi−1, i = 0, . . . , n.

Proposition

A valuation φ is monotone iff its first variation δφ is a positive curvature measure.


Clearly CP ⊂ M ⊂ P, and

CP ∩ ValSO(V ) = M ∩ ValSO(V ) = P ∩ ValSO(V ) = 〈µ0, . . . , µn〉+

On the other hand, neither of the corresponding equalities holds for the intersections with ValU(n).

Theorem

Let νk,p be the dual basis with respect to the symmetric pairing 〈, 〉. Then

P ∩ ValU(n) = 〈µk,p〉+, CP ∩ ValU(n) = 〈νk,p〉+,

M ∩ ValU(n)k =

nXcqµk,q : (k − 2q)cq ≥ (k − 2q − 1)cq+1, max0, k − n ≤ q ≤

—k − 1

2

,

(n + q − k + 1)cq ≤ (n + q − k + 3/2)cq+1,max0, k − n − 1 ≤ q ≤—

k − 22

ff


Further developments, and some questions

Is there a better way to do all these calculations? maybe some variation on Schubertcalculus? Is there a reason that the T n

k ≥ 0?Bernig has worked out the integral geometry of the isotropic pairs(Cn,SU(n)), (C4,Spin7), (R7,G2). It turns out that there are only a handful of invariantvaluations that we have not met before. In the first case there are either four (n even) ortwo (n odd), all in the middle degree n. In this case, if n is odd then the analogue of theTasaki matrix T n

n is not positive definite— in fact its index is 2.The case (Hn,Sp(n)× Sp(1)) seems particularly interesting, but not much is known.Alesker has constructed an invariant valuation of degree n analogous to the “Kazarnovskiipseudovolume" µn,0 ∈ ValU(n), i.e. whose Klain function vanishes at every E ∈ Grn(Hn)that contains a complex line with respect to any of the usual complex structures on Hn.Although the Crofton formulas of the complex space forms are all identical, the same is nottrue for the full kinematic formulas. As a first step toward working them out, it will benecessary to understand the structure of the algebra of invariant valuations on CPn andCHn. We can calculate

t2i sj (CPn) =“2i

i

”“n − j + 1i + 1

”, t2i+1sj (CPn) = 0

Analogous to our approach to ValU(n)(Cn), it seems feasible to generate the relationsamong the t i sj by reverse engineering some identities among these quantities, then topass to CHn by analytic continuation in the curvature. This seems to work for n = 2, 3, 4.






t2i sj (CPn) =“2i

i

”“n − j + 1i + 1

”, t2i+1sj (CPn) = 0







t2i sj (CPn) =“2i

i

”“n − j + 1i + 1

”, t2i+1sj (CPn) = 0







t2i sj (CPn) =“2i

i

”“n − j + 1i + 1

”, t2i+1sj (CPn) = 0







t2i sj (CPn) =“2i

i

”“n − j + 1i + 1

”, t2i+1sj (CPn) = 0



Documents

Lecture 4: Hermitian integral geometrymath.uga.edu/~fu/diablerets/diableret4.pdfAs valuations (i.e. as “complete integrals"), the second is 2× the ﬁrst. (J.H.G. Fu) The two faces