NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Contents 1. Introduction 1 1.1. Goals of the subject and the course 1 1.2. Buffon needle problem 2 1.3. Integral geometry of the euclidean plane 2 1.4. Relaxation of smoothness conditions. 6 2. Classical integral geometry 6 2.1. Steiner’s formula 6 2.2. Statement of Hadwiger’s theorem, and its integral geometric consequences 10 2.3. Support functions, the cosine transform and zonoids 14 2.4. Proof of Hadwiger’s theorem 17 2.5. Even valuations: Klain functions, Crofton measures and the Alesker Fourier transform 19 2.6. The McMullen decomposition 20 2.7. Nijenhuis’s theorem 20 2.8. The Alesker product, part I 23 2.9. Review of differential forms 24 2.10. Integration of differential forms and the normal cycle 28 2.11. Curvature measures and valuations. First variation and the kernel theorem 31 2.12. Kinematic formulas for invariant curvature measures 34 2.13. The transfer principle 37 2.14. Integral geometry of real space forms 40 3. Alesker theory for affine spaces 46 3.1. Alesker irreducibility and its consequences 46 3.2. Convolution 49 3.3. Constant coefficient valuations 52 n 4. Integral geometry of (C ,U(n)) 54 4.1. Hermitian intrinsic volumes and Tasaki valuations 54 4.2. On kU(n)(χ) 56 4.3. Algebraic structure of ValU(n) 57 4.4. Dictionaries 61 n 4.5. Global kinematic formulas in C 63 4.6. Two dualities 64 References 66

1. Introduction 1.1. Goals of the subject and the course. Geometric probability. Rota: “continuous combi- natorics” Geometry of singular spaces: language of valuations and curvature measures is natural.

Date: December 5, 2011. 1 2 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

To give an account of the basic vocabulary, which (as usual in geometry) includes a variety of complementary approaches. To describe and (partially) prove recent results about the integral geometry of the unitary group. 1.2. Buffon needle problem. Reference: [21] Imagine the euclidean plane ruled by a system of parallel lines one inch apart. Drop a needle of length ` ≤ one inch in a random position on the plane. What is the probability that the needle crosses one of the lines? First we reframe the question as: what is the expectation E(`) of the number of lines crossed by a needle of unrestricted length? If ` ≤ 1 then this is the same as the first question. Observe that E(`1 +`2) = E(`1)+E(`2): for we may think of the long needle as the concatenation of the two shorter ones. Although the two crossing events are not independent, dropping one of the needles at random also gives a random position for the other. In fact it is not even necessary that the two needles be colinear, and furthermore one may concatenate any number of needles and the same additivity will result. Finally, (by continuity and perhaps a leap of faith) the needle may even be curved. So the function E(`) is linear, and gives the expected number of crossings of any rectifiable curve of length `. Thus E(`) = c` for some constant c. To evaluate the constant c, take the curve to be a circle C of diameter 1. In this case C crosses exactly two lines regardless of its position. Thus 2 cπ = E(π) = 2 =⇒ c = . π This argument is an instance of the “template method”: once we know that an integral-geometric formula of a certain kind exists, in order to specify the constants one finds one or more convenient geometric objects for which everything is directly computable. Then the constants can be computed by working out the resulting linear equations. Ideally the templates can be chosen so that the system is diagonal. 1.3. Integral geometry of the euclidean plane.

2 2 1.3.1. Haar measures on Gr1(R ) and SO(2). It is convenient to identify R ' C. The euclidean 1 iθ group SO(2) is diffeomorphic to S × C by identifying (e , z0) to the map iθ z 7→ z0 + e z. iθ Thus z0 is the image of the origin and e is the (constant) derivative of the map. 3 Consider the the measure on SO(2) given by identifying it locally with R via the coordinates θ, x, y and taking Lebesgue measure there. Since iθ iψ i(θ+ψ) iθ (e , z0) · (e , w0) = (e , e w0 + z0) Fubini’s theorem shows that this measure is invariant under the action of the group on itself: if E ⊂ SO(2) 3 g then m(gE) = m(E). The affine Grassmannian Gr1(C) is a 2-dimensional manifold. It can be viewed as a quotient of SO(2) as follows. Clearly SO(2) acts transitively on Gr1(C). The subgroup H := {±1}×R ⊂ SO(2) is the stabilizer of R ⊂ C. Hence the map g 7→ gR gives an identification of the left coset space SO(2) /H ' Gr1(C). iθ −iθ The locally defined functions (e , z0) 7→ θ, =(e z0) on SO(2) are invariant, up to sign, under the right action of H. Therefore they are lifts of locally defined functions θ, p on Gr1. Geometrically, θ is the angle that the line makes with the x-axis. Exercise 1.1. Show that p = ± the distance from the line to the origin.

Exercise 1.2. θ, p give local coordinates on Gr1. The resulting measure dpdθ is invariant under the action of SO(2). NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 3

1.3.2. Crofton’s formula for the length of a rectifiable plane curve. Recall that a continuous curve 2 γ : [0, 1] → R is rectifiable if N X |γ| := sup |γ(ti) − γ(ti−1| < ∞. P={t0<···

Lemma 1.3. Let f : R → R. Then the set of all local extreme values of f is countable.

Proof. A point x ∈ R is a local max if there is an open interval (p, q) 3 x such that f(x) = maxp,q f. Clearly p, q may be taken to be rational. Thus the set of all local maximum values is contained in the countable set {sup f : p, q ∈ Q}. (p,q) The set of local minimum values is likewise countable.  Lemma 1.4. Let f : R → R be continuous. Suppose that c is not a local extreme value of f, and that f(x0) = c. Let  > 0 be given. Then for any sufficiently fine partition P of R there are points a1, a2 ∈ P, with |x − ai| < , such that f(a1) < c < f(a2). 2 Theorem 1.5. Let γ ⊂ R be a rectifiable curve. Then Z |γ| = 4 #(γ ∩ `(θ, p)) dθdp. Gr1

Proof. Given a partition P, let γP denote the corresponding piecewise linear path, with vertices at the points γ(t), t ∈ P. On the other hand, one easily calculates that for a line segments σ Z #(σ ∩ `(θ, p)) dθdp = 4|σ| Gr1 and therefore Z #(γP ∩ `(θ, p)) dθdp = 4|γP |. Gr1 Thus Z sup #(γP ∩ `(θ, p)) dθdp = 4|γ|. (1) P Gr1 and it remains to show that the left hand side equals R #(γ ∩ `(θ, p)) dθdp. Gr1 Let P1, P2,... be a sequence of successive refinements with mesh → 0. By the intermediate value theorem, given any line `

#(` ∩ γPi ) ≤ #(` ∩ γPi+1 ) ≤ #(` ∩ γ). By (1) and the dominated convergence theorem,

#(` ∩ γP ) ≤ N(`) := lim #(` ∩ γP ) < ∞ i i→∞ i for all i. Let u be a unit vector and consider the continuous function f; = u · γ. The level sets f −1(c) correspond to the intersections between γ and the lines ⊥ to u. By Lemma 1.3, among these lines ` only countably many correspond to local extreme values c. By Lemma 1.4, if ` is one of the generic lines that do not, then for sufficiently large i

#(` ∩ γPi ) = #(` ∩ γ).

Thus the functions ` 7→ #(γPi ∩ `) increase to ` 7→ #(γ ∩ `) at a.e. line `, which with (1) and dominated convergence concludes the proof.  4 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

1.3.3. Poincar´eformula. Theorem 1.6. Let β, γ be C1 curves in the plane. Then Z #(β ∩ gγ) dg = 4|β||γ| (2) SO(2) Proof. Consider the map F : S1 × β × γ → SO(2), (eiθ, p, q) 7→ (eiθ, β(s) − eiθγ(t)). Clearly F (p, q, eiθ)q = p, and the cardinality of F −1(g) is #(β ∩ gγ). Thus the change of variables formula gives Z Z #(β ∩ gγ) dg = |det DF | (3) SO(2) S1×β×γ — though strictly speaking we should replace F in the right hand integral by its expression in local coordinates (θ ◦ F, x ◦ F, y ◦ F ). For simplicity of notation we abbreviate this as (θ, x, y). Let s, t be the respective arclength coordinates on β, γ and ψ the angular parameter on the S1 factor in the domain. Then ∂θ = 1 ∂ψ ∂θ ∂θ = = 0 ∂x ∂y so the determinant in question is  ∂x ∂x  ∂s ∂t 0 iθ 0 0 iθ 0 0 iθ 0 det ∂y ∂y = <β =(e γ ) − =β <(e γ ) = hβ , ie γ i. ∂s ∂t The integral with respect to θ of the absolute value of this expression is clearly independent of s, t, so by (3) the formula (2) holds for some value of the constant. To evaluate the constant we use the template method, taking β, γ to be circles of radius 1. In this case #(gγ ∩ β) = 2 whenever the image of the center of γ under g lies within distance 2 of the center of β, and the circles are disjoint if the distance is greater (if the distance is exactly 2 then the circles coincide, but the set of such g has measure zero). Hence the left hand side of (2) 2 is 2 × 2π × 4π = 16π , which agrees with the right hand side.  Remark: Poincar´efor pairs of rectifiable curves.

2 2 1.3.4. Kinematic formulas. Let D ⊂ R be a compact domain with boundary consisting of the C pairwise disjoint curves γ1, . . . , γn. Parametrize these curves by arclength and consistently with the orientation of D. Let T be the unit tangent field and N the unit outward-pointing normal field. The curvature of γi at p = γ(s0) is 0 κ(p) = κ(s0) = T (s0),N(s0) . Theorem 1.7 (Umlaufsatz). n X Z κ ds = 2πχ(D). i=1 γi Proof. The Euler characteristic of D is given by the number of components minus the number of bounded components of the complement. This is equal to the number of boundary curves γi that NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 5 are oriented counterclockwise minus the number oriented clockwise. Thus we only need to show that if γ is oriented counterclockwise then Z κ ds = 2π. (4) γ To see this, observe that there is a smooth isotopy taking γ to a round circle, again oriented counterclockwise. It is easy to check that R κ remains unchanged during this process (the integral is continuous in the variation parameter, and takes values in 2πZ). Meanwhile, it is clear that (4) holds for the circle.  This fact extends also to certain cases where the boundary is less smooth. The simplest is the 2 one in which the γi are only piecewise C . Let θij denote the exterior angles at the junctures 2 between C pieces; thus −π ≤ θij ≤ π, as the sign is + if the angle juts out of the domain and − if it juts in. Corollary 1.8. n X Z X κ ds + θij = 2πχ(D). i=1 γi ij

Now consider two such piecewise smooth domains A, B with boundary curves αi, βj. Lemma 1.9. For a.e. g ∈ SO(2), the domain A ∩ gB is again a piecewise smooth domain.

Proof. In fact by the Poincar´eformula the number of points of αi ∩ gβj is finite for all i, j and a.e. g.  2 2 Theorem 1.10 (Principal kinematic formula in R ). If A, B ⊂ R are compact piecewise smooth domains then Z χ(A ∩ gB) dg = 2πχ(A) area(B) + length(∂A) length(∂B) + 2π area(A)χ(B). (5) SO(2) Proof. We use the Umlaufsatz. For generic g the boundary of A ∩ gB consists of three parts: (1) the part of ∂A lying in the interior of gB (2) ∂A ∩ ∂B (3) the part of g∂B lying in the interior of A. These three pieces each account for one term in (5). To deal with the first one, let α be one of the components of the boundary of A. Put D := {(s, g) ∈ α × SO(2) : s ∈ gB}. We compute using Fubini’s theorem Z Z Z κ(s) ds dg = κ(s) ds dg SO(2) α∩gB D Z = κ(s) dg({g : s ∈ gB}) ds α The set {g : s ∈ gB} is the image of S1 × B under the map (eiθ, q) 7→ (eiθ, s − eiθq). Since the Jacobian determinant of this map is 1, the measure of this set is 2π area(B). The contributions of the angles of α can be dealt with similarly, so by the Umlaufsatz this part of the integral gives the first term of (5). By the same token the third part of the integral gives the last term. To evaluate the contribution of the second part, let β be a component of ∂B with arclength parameter t. Consider the map g : S1 × α × β → SO(2), g(eiθ, s, t) := (eiθ, s − eiθt). 6 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 and the function iθ 0 iθ 0 A(e , s, t) := ∠(α (s), e β (t)). This function is defined at all smooth points (s, t), which is clearly a set of full measure. The Jacobian determinant of g is easily computed to be | sin A|. Hence by the change of variables formula and Fubini Z X Z ∠(α, gβ) dg = A| sin A|dθ ds dt SO(2) S1×α×β Z π = |α||β| |ψ sin ψ| dψ. −π Of course the integral can be evaluated easily to get the constant, but it is also possible to read it off using the template method. We take A, B to be thin tubes around curves α, β, so that the length of the boundaries are approximately 2 length α, 2 length β. As the thickness of the tubes tends to zero, the integrand in (5) tends to that of the Poincar´eformula (2).  1.4. Relaxation of smoothness conditions. We may relax the conditions on A, B and still have the principal kinematic formula. It is more or less obvious that the boundary curves may be taken to be only piecewise C2, and in fact the same proof works. More generally still, the boundary curves γ may be taken to have finite total curvature, i.e. we need assume only that the unit dγ tangent field T = ds have bounded variation. In this case the curvature becomes a signed Radon measure, so the analysis is slightly more complicated, but there is no substantial change. We observe, however, that this is not the largest possible class of domains for which this all works. There are many such domains which are not finite disjoint unions of simple curves, e.g. the region of the disk bounded by the graphs of the functions ( ± exp(− 1 )(1 + sin x), x > 0 f ± (x) := x2 0 x ≤ 0

2 Say that a compact set A ⊂ R is geometrically good if there exists a sequence of smooth S domains A1 ⊃ A2 ⊃ ... with A = i Ai and such that Z |κ| ds ≤ C ∂Ai for some constant C. Conjecture 1.11. The principal kinematic formula holds for pairs of geometrically good sets.

2. Classical integral geometry 2.1. Steiner’s formula. Theorem 2.1 (Steiner’s formula). If A ∈ K then n X n−i vol(Ar) = ωn−i µi(A) r , r ≥ 0. (6) i=0

The functionals µi (thus defined) are called the intrinsic volumes, and are equal up to scale to the Quermassintegrale introduced by Minkowski. It is easy to prove (6) if A ∈ Ksm. In this case

Ar = A ∪ expA(bdry A × [0, r]) (7) where expA(x, t) := x + tnA(x) (8) NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 7

n−1 and nA : bdry A → S is the Gauss map. It is clear that expA gives a diffeomorphism bdry A × n (0, ∞) → R − A: in fact the inverse map may be written in terms of

x = πA(expA(x, t)), t = δA(expA(x, t)) n where πA : R → A is the nearest point projection and δA is the distance from A. Clearly

Dx,t0 expA(v + c∂t) = v + t0Lx(v) + cnA(x) (9) n−1 n for v ∈ Tx bdry A, where Lx : Tx bdry A → Tx bdry A = TnA(x)S ⊂ R is the Weingarten map. Thus the area formula gjves Z Z r

vol(Ar) = vol(A) + d voln−1 x dt det(IdTx bdry A +tLx) bdry A 0 n−1 Z Z r Y = vol(A) + d voln−1 x dt (1 + tkj) bdry A 0 j=1 n−1 X rj+1 Z = vol(A) + σ (k , . . . , k ) d vol x j + 1 j 1 n−1 n−1 j=0 bdry A where the kj are the principal curvatures. In particular 1 Z µi(A) = σn−i−1(k1, . . . , kn−1) (10) (n − i)ωn−i bdry A in this case. To extend Steiner’s formula to general A ∈ K we first express the coefficients in a more robust 1 form. Note first that µn−1 is 2 the perimeter. Observe also that the codimension 1 Grassmannian Grn−1 admits a natural SO(n)-invariant measure: this may be constructed, for example, as the n−1 n−1 image ⊥∗ m of the invariant measure m on the sphere S under the map ⊥: S → Grn−1. Here m may be defined by x m(E) := vol ({x ∈ n : 0 < |x| ≤ 1, ∈ E}) n R |x| and the pushforward measure ⊥∗ m by −1 (⊥∗ m)(F ) := m(⊥ (F )) for F ⊂ Grn−1. (Note that the pushforward of any Borel measure under any continuous map is a well-defined Borel measure.) For the moment at least we will renormalize the measure to be a probability measure (mass 1). Exercise 2.2.   n n ωn µi(B ) = . i ωn−i sm n Proposition 2.3 (Crofton’s formula). Let A ∈ K (R ). Then Z nωn µn−1(A) = |πE(A)| dE (11) 2ωn−1 Grn−1 where dE is the appropriately normalized Haar measure on Grn−1.

Proof. Fixing E ∈ Grn−1 with unit normal vector v = vE, and x ∈ ∂A, the Jacobian determinant of the restriction of det DπE to Tx∂A is J det Dπ | = |v · n (x)|. E Tx∂A A 8 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Hence

Z Z Z |πE(A)| dE = dE |v · nA(x)| dx (12) Grn−1 Grn−1 ∂A Z Z = dx |v · u| dv (13) ∂A Sn−1 = µn−1(A) (14) where u ∈ Sn−1 is an arbitrarily chosen unit vector, provided the Haar measures dE, dv are nor- malized appropriately. 

Next we want to express every µk as an integral over Grk, where the invariant measure on Grk may be constructed as follows. Consider first the Stiefel manifold Σn−k := {(v1, . . . , vn−k) ∈ n−1 n−k (S ) : vi ⊥ vj for i 6= j}. In other words, for (v1, . . . , vn−k) ∈ Σn−k we may think of v2 as an ⊥ ⊥ element of the sphere of v1 , v3 as an element of the sphere of hv1, v2i , etc. Each of these spheres has a probability measure that is invariant under the orthogonal group of the associated plane. Then for E ⊂ Σn−k we put

Z Z Z −1 −1
m(E) := dv1 dv2 ... dvn−k E ∩ π1 (v1) ∩ · · · ∩ πn−k−1(vn−k−1) n−1 ⊥ ⊥ S S(v1 ) S(hv1,...,vn−k−1i )

Now the invariant measure dG on Grk may be constructed as the pushforward of this measure under the ⊥ map Σn−k → Grk.

sm n Corollary 2.4. Let A ∈ K (R ). If the Haar measure dG is appropriately normalized then

  Z n ωn µk(A) = volk(πG(A)) dG, k = 0, . . . , n − 1. (15) k ωkωn−k Grk

sm Proof. Observe first that, by the definition of K , the Gauss map nA has nonsingular derivative at every point of bdry A. Thus if we fix E ∈ Grn−1 and take v to be a unit vector ⊥ E then the locus {p ∈ bdry A : nE(p) · v = 0} is a smooth submanifold ME ⊂ bdry A, and since DnA = IIA is positive definite it follows that every tangent space TpME 63 v. Therefore the projection πE(ME) = sm bdry πE(A) is smooth, and in fact πE(A) ∈ K (E). Thus we may apply Steiner’s formula to each πE(A), noting that

πE(Ar) = [πE(A)]r.

Polarizing Steiner’s formula we compute easily that

d µk(A) = c µk+1(Ar) dr r=0 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 9 hence we may apply Crofton’s formula to obtain

d µn−2(A) = c µn−1(Ar) dr r=0 Z d = c |[πE(A)]r| dE dr r=0 Grn−1 Z d = c |[πE(A)]r| dE Grn−1 dr r=0 Z = c µn−2(πE(A)) dE Grn−1 Z Z = c dE voln−2(πF (A)) dF Grn−1 Grn−2(E) Z = c voln−2(πF (A)) dF n Grn−2(R ) since πF ◦ πE = πF when F is a subspace of E. Continuing in this way we arrive at the desired n conclusion, where the constant may be evaluated by taking A = B .  Now we may give the proof of Steiner’s formula for general A ∈ K. Observe first that K is naturally a metric space under the Hausdorff metric

d(A, B) := min (min{r : B ⊂ Ar}, min{s : A ⊂ Bs}) . (16) n Exercise 2.5. a) In fact (16) defines a metric on the set of all compact subsets A, B ⊂ R . Call the resulting metric space C. b) K is a closed subset of C. c) For R > 0, the set of all S ∈ C such that S ⊂ clos B(0,R) is compact. d) (Blaschke selection theorem) The set of all A ∈ K such that A ⊂ clos B(0,R) is compact.

Lemma 2.6. The function voln is continuous on the metric space K. For r > 0 the map A 7→ Ar is an isometry K → K. The Minkowski sum map (A, B) 7→ A + B is continuous K × K → K.

Proof. Given Ai → A, the characteristic functions 1Ai → 1A off of the boundary of A, which has measure zero. Therefore dominated convergence implies that |[Ai]| → |A|. The second assertion follows from the relation (Ar)s = Ar+s.  sm Note that there exist A1,A2, · · · ∈ K such that Ai → A in the Hausdorff metric (we will come back to this point shortly). By Lemma 2.6, for each fixed r > 0 the volumes voln((Ai)r) → voln(Ar). On the other hand, Corollary 2.4 implies that the coefficients of the Steiner polynomials for the Ai are uniformly bounded. Thus there exists a subsequence of the Ai such that the corresponding Steiner polynomials converge, so the limiting polynomial must give the volume of the tube Ar for each r > 0. (In fact, since this conclusion is independent of the chosen subsequence, the entire sequence converges.) Similarly, Corollary 2.4 holds for general A ∈ K as well. This can also be written in the following n equivalent form. Let Grn−k denote the space of affine planes of dimension n − k in R . Then Z µk(A) = c χ(A ∩ H¯ ) dH,¯ k = 0, . . . , n − 1 (17) Grn−k where χ is the Euler characteristic, and dH¯ is the Haar measure obtained by viewing Grn−k as the total space of the tautological bundle over Grk and taking the product of dG from Prop. 2.4 with 10 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 the Lebesgue measure on the fibers. In particular, the formula (17) serves to give a definition of the µk for the polyconvex sets. Exercise 2.7. Deduce that Z m µk(A) = ck µk(πE(A)) dE, m ≥ k Grm Z j ¯ ¯ = dk µk−j(A ∩ P ) dP , n ≥ k ≥ j Grn−j m j Is it possible to choose the normalizations of the Haar measures so that ck , dk ≡ 1? n N Note that the µi are independent of the ambient dimension, i.e. if j : R → R is a linear isometry then µi(j(A)) = µi(A). In fact, the normalization is chosen so that µi(A) = voli(A) whenever dim A = i.

Proposition 2.8. The µk are continuous on K and SO(n)-invariant, and are valuations on the class P K of polyconvex sets, i.e.

µk(A ∪ B) = µk(A) + µk(B) − µK (A ∩ B). If t ∈ R and A ∈ P K then k µk(tA) = |t| µk(A). n n We denote the vector space of continuous, translation-invariant valuations on R by Val(R ), SO(n) SO(n) n and the SO(n)-invariant subspace by Val = Val (R ). 2.2. Statement of Hadwiger’s theorem, and its integral geometric consequences. Theorem 2.9 (Hadwiger’s characterization theorem). SO(n) Val = hµ0, . . . , µni. This theorem has the following important consequences. Note first that the group SO(n) carries a natural invariant measure. One way to see this is to identify SO(n) with the component of the n Stiefel manifold Σn consisting of positively oriented bases for R , and then impose the invariant n probability measure defined above. Furthermore the oriented euclidean group SO(n) := SO(n)nR n also has an invariant measure: identifying SO(n) with the product SO(n) × R by letting (g, x) · y := gy + x we observe that the product of the measure defined above with Lebesgue measure is invariant. Fact 2.10. The map g 7→ g−1 is measure-preserving in both cases. Note that n dim SO(n) = (n − 1) + (n − 2) + ··· + 1 = 2 n dim SO(n) = + n. 2

l l Theorem 2.11. [Blaschke kinematic formulas] There are constants cij = cji, 0 ≤ i, j, l ≤ n (depending also on n), such that for A, B ∈ P K Z X l µl(A ∩ gB¯ ) dg¯ = cijµi(A)µj(B). (18) SO(n) i+j=n+l NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 11

It is convenient to denote the left-hand side by

kSO(n)(µl)(A, B).

The theorem may then be rephrased as the statement: kSO(n) is a operator of degree n from ValSO(n) → ValSO(n) ValSO(n) (symmetric product).

Proof. Fix B, and consider the functional νB(A) defined by the left hand side of (18). Clearly ν is an SO(n)-invariant valuation; we wish to show that ν is continuous. Consider the set Γ ⊂ SO(n) consisting of allg ¯ ∈ SO(n) such that A∩gB¯ 6= ∅ but A, gB¯ are subsets of complementary halfspaces. Using Lemma 2.21 below, the continuity of µk, and the dominated convergence theorem, it is enough to show that dg¯(Γ) = 0. n−1
In fact, Γ is a rectifiable set (a notion we will discuss later) of dimension 2(n − 1) + 2 = n
n−1+ 2 < dim SO(n). To see this we note that the sets N(A),N(B) are rectifiiable of dimension n − 1, where n−1 N(C) := {(x, v) ∈ ∂C × S : A ⊂ Hv,x := {y : y · v ≤ x · v}}. Nowg ¯ ∈ Γ iff there are (x, v) ∈ N(A), (y, w) ∈ N(B) such that gy¯ = x, gw = v. For fixed (x, v), (y, w), the set of suchg ¯ is a diffeomorphic copy of SO(n − 1): given any v ∈ Sn-1, there is a neighborhood V ⊂ Sn-1 of v and a smooth map γ : V → SO(n) such that

γ(w) · e1 = w where e1 is the first standard basis vector. To see this, choose vectors v2, . . . , vn such that n v, v2, . . . , vn is a positively oriented basis for R . Then w, v2, . . . , vn is a positively oriented ba- sis for any w ∈ Sn-1 sufficiently close to v. Applying the Gram-Schmidt process to this basis, and assembling the resulting basis as column vectors of a matrix, yields the desired γ(w). Note that if −1 −1 g ∈ SO(n) is any other element with ge1 = w then g γ(w) · e1 = e1, so g γ(w) belongs to the copy of SO(n − 1) fixing e1. Such a map γ is called a local section of the projection (fiber bundle) SO(n) → Sn-1 given by g 7→ g · e1. n-1 Now if v, w ∈ S have neighborhoods V,W and corresponding local sections γV , γW , then for v0 ∈ V, w0 ∈ W the set of all elements of SO(n) taking w0 7→ v0 is

γ(v) · SO(n − 1) · γ(w0)−1. Thus Γ can be locally realized as the image of N(A) × N(B) × SO(n − 1) under a differentiable map. But this set has the claimed dimension. SO(n) Therefore νB ∈ Val , so by Hadwiger’s theorem there are functionals φ0(B), . . . , φn(B) such that n X kSO(n)(µl)(A, B) = φi(B)µi(A). i=0 Now if we fix A and let B vary, as before we see that this sum defines an element of ValSO(n). It is clear that we may find A0,...,An ∈ K such that the matrix [µi(Aj)]0≤i,j≤n is nonsingular (e.g. sm SO(n) take Aj := jA for some A ∈ K ), so we conclude that each φi ∈ Val . Hadwiger’s theorem P then implies that φi = j cijµj. To see that the indices are restricted as stated follows from a scaling argument. Given t > 0 −1 n consider the automorphism ρt of SO(n) given by ρt(g, x) := (g, t x). Clearly ρt∗(dg¯) = t dg¯. 12 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Therefore Z kSO(n)(µl)(tA, tB) = µl(tA ∩ g¯(tB)) dg¯ SO(n) Z = µl(t[A ∩ (ρtg¯)(B)]) dg¯ SO(n) Z l = t µl(A ∩ (ρtg¯)(B)) dg¯ SO(n) Z l+n ¯ ¯ = t µl(A ∩ h(B)) dh SO(n) l+n = t kSO(n)(µl)(A, B) Finally, the symmetry of the coefficients follows from Fact 2.10.  Definition 2.12. A subset S of a metric space X is a rectifiable of dimension k if there k k exist countable families {Ei} of subsets of R and of Lipschitz maps fi : R → X such that S∞ S = i=0 fi(Ei). Definition 2.13. An oriented affine hyperplane P¯ is a support hyperplane at x for A ∈ K if

x ∈ A ∩ P,A¯ ⊂ P¯− where P− is the halfspace having P¯ as its boundary. Proposition 2.14. Let A ∈ K.

(1) The distance function δA(x) := minp∈A |p − x| is 1-Lipschitz. n (2) For each x ∈ R there is a unique point πA(x) ∈ A such that |x − πA(x)| = δA(x). (3) πA is 1-Lipschitz. Proof. (1): This is true even if A is not convex. p+q (2): If there were distinct points p, q ∈ A with this property then 2 ∈ A, and one computes easily that it is closer  Proposition 2.15. Let A ∈ K and suppose x∈ / A. Then the oriented hyperplane with normal x − πA(x) is a support plane for A through πA(x). Every support hyperplane can be obtained in this way.

Proof. Denote this hyperplane by P . If there is y ∈ A − P− then the line segment [y, πA(x)] ⊂ A. But it is clear that the points of this segment near πA(x) are closer to x than πA(x) is. To prove the second assertion, we note that

(x, v) ∈ N(A) =⇒ πA(p + tv) = p, t ≥ 0. (19)  n Definition 2.16. Let S ⊂ R be bounded. The closed convex hull of S is \ conv S := {A : A ∈ K,A ⊃ S}. Corollary 2.17. Every A ∈ K is the intersection of the closed halfspaces that contain it. In fact, n given any bounded set S ⊂ R , its closed convex hull conv S is the intersection of the halfspaces that contain it. Proof. For the first assertion: clearly A ⊂ this intersection. Conversely, if x∈ / A then the oriented hyperplane with normal x−πA(x) through πA(x) bounds one of these halfspaces that does not 3 x. The second assertion is an immediate consequence.  NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 13

Definition 2.18. For A ∈ K we put N(A) := {(x, v): x ∈ ∂A, v⊥ + x is a support hyperplane for A through x} n n-1 ⊂ R × S . sm Observe that if A ∈ K then N(A) = {(x, nA(x)) : x ∈ ∂A}. Proposition 2.19. N(A) is rectifiable of dimension n − 1. Proof. Consider the map   n x − πA(x) ΠA : R − A → A, ΠA(x) := πA(x), . |x − πA(x)| n By Proposition 2.14, this is clearly Lipschitz when restricted to any R − Ar, r > 0. Let R > 0 be large enough that A ⊂ B(0,R) and put SR := ∂B(0,R). That ΠA(SR) = N(A). This follows from (19), since for each (x, v) ∈ N(A) there is clearly some t > 0 such that x + tv ∈ SR.  Actually the additivity condition in Hadwiger’s theorem can be relaxed: an apparently weaker but actually equivalent definition of a valuation φ insists only that A, B, A ∪ B ∈ K =⇒ φ(A ∪ B) = φ(A) + φ(B) − φ(A ∩ B). (20) Using this characterization we may establish the following. Put for A, B ∈ K Z aSO(n)(µl)(A, B) = µl(A + gB) dg. (21) SO(n) Theorem 2.20 (Additive kinematic formulas). SO(n) SO(n) SO(n) aSO(n) : Val → Val Val and has degree 0.

Proof. It is clear that aSO(n)(µl)(·, ·)) is a continuous function on K × K → R. Thus if we can show that is a valuation in each variable separately then we may argue as above. But this follows from the fact that if A1,A2,A1 ∪ A2,B ∈ K then

(A1 ∪ A2) + B = (A1 + B) ∪ (A2 + B),

(A1 ∩ A2) + B = (A1 + B) ∩ (A2 + B). The first assertion is obvious, as is ⊂ in the second. To prove ⊃, suppose

a1 + b1 = a2 + b2 = p ∈ (A1 + B) ∩ (A2 + B) where ai ∈ Ai and b1, b2 ∈ B. Since A1 ∪ A2 ∈ K, there is t ∈ [0, 1] such that a := ta1 + (1 − t)a2 ∈ A1 ∩ A2. Therefore, putting b := tb1 + (1 − t)b2 ∈ B,

(A1 ∩ A2) + B 3 a + b = (a1 + (1 − t)a2) + (b1 + (1 − t)b2)

= t(a1 + b1) + (1 − t)(a2 + b2) = p.  Lemma 2.21. Suppose A, B ∈ K, and either A ∩ B = ∅ or are not subsets of complementary halfspaces. Then the map ∩ : K × K → K is continuous at (A, B).

Proof.  Lemma 2.22. vol is a continuous function K → R. 14 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

2.3. Support functions, the cosine transform and zonoids. The support function hA : n R → R of A ∈ K is hA(v) := sup x · v. x∈A Obviously hA is positively homogeneous of degree 1:

t ≥ 0 =⇒ hA(tv) = thA(v). Furthermore

hA(v + w) = sup x · (v + w) x∈A ≤ sup x · v + sup x · w x∈A x∈A = hA(v) + hA(w).

Assuming homogeneity this is equivalent to the assertion that hA is a convex function:

hA(tv + (1 − t)w) ≤ thA(v) + (1 − t)hA(w). Such a function is said to be sublinear. sm Note that if A ∈ K and y = tnA(x), x ∈ ∂A, t > 0, then

hA(y0) = x0 · y0. (22) Proposition 2.23. Let A, B ∈ K.

(1) A ⊂ B iff hA ≤ hB (2) hA+B = hA + hB (3) max(hA, hB) = hconv(A∪B) (4) If additionally A ∪ B ∈ K then

hA∩B = min(hA, hB). Proof. The first assertion is obvious, and the second follows from the second assertion of Corollary 2.17. 

Proposition 2.24. The map A 7→ hA|Sn-1 is an isometric imbedding, with respect to the uniform norm on C(Sn-1). In particular this map is injective. Every sublinear function is the support function of some A ∈ K. Proof. The Hausdorff distance between A, C ∈ K may be expressed as the smallest r such that

A ⊂ C + Br,C ⊂ A + Br. Since hBr (v) = r|v| it follows that this distance may also be expressed as the smallest r such that

hA|Sn-1 ≤ hC |Sn-1 + r, hC |Sn-1 ≤ hA|Sn-1 + r, which is precisely khA|Sn-1 − hC |Sn-1 k∞. This proves the first assertion. n To prove the second assertion, let f : R → R be a sublinear function. It is not hard to show that f is continuous (in fact any finite-valued convex function is continuous). Put n n A := {x ∈ R : x · v ≤ f(v) for all v ∈ R }.

Clearly A ∈ K, with hA ≤ f. We must show that A 6= ∅ and hA ≥ f. Consider the epigraph of f, n+1 epi f := {(v, t) ∈ R : t ≥ f(v)}. Since f is a convex function, epi f is a closed convex set; since f is homogeneous, epi f is a cone with vertex 0 (i.e. invariant under multiplication by any positive number). It follows that epi f has n a support plane at every boundary point, i.e. every point (v, f(v)), v ∈ R , and that every such plane must pass through the origin. Furthermore such a support plane cannot be vertical. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 15

Let v 6= 0 be given, and let P be a support plane for epi f at (v, f(v)). By the remarks above, P has an oriented normal vector of the form (x, −1) (i.e. the last coordinate is negative: P is not vertical, and the normal must point downward). Since P passes through 0, every point (w, f(w)) of the epigraph lies above P , i.e. satisfies w · x − f(w) ≤ 0 ⇐⇒ w · x ≤ f(w). n This is true for all w ∈ R , hence x ∈ A, so A 6= ∅. On the other hand we have equality here if w = v, so hA = f.  Definition 2.25. Z ∈ K is a zonotope if it is a Minkowski sum of a finite collection of line segments. Z is a zonoid if is the limit of a sequence of zonoids. A cube is a zonotope. A ball is a zonoid. Proposition 2.26. Every zonoid Z has a center of symmetry, i.e. there is a point z ∈ Z such that Z = 2z − Z. A symmetric convex body Z with center at 0 is a zonoid iff there is an even measure m on Sn-1 such that Z n hZ (x) = |x · v| dm(v), x ∈ R . (23) Sn-1 Proof. Applied to zonotopes, the first assertion is obvious: the center of symmetry is just the sum of the midpoints of the constituent line segments. The general statement then follows by continuity. To prove the second statement, assume first that Z is a zonoid, with approximating zonotopes Pi Zi := j=1 σij, where the σj = [−vij, vij] are line segments with midpoints at 0. Then i X X Z hZi (w) = hσij (w) = |vij · w| = |w · v| dmi(v) n-1 j=1 S ! i P v n-1 where the measure mi = j=1 |vij| δ ij + δ −vij is a sum of Dirac masses on S . In particular |v | ij |vij | the total mass of mi is Z

C hZi (w) dw ≤ C max hZi = C max |x|. Sn-1 Sn-1 x∈Zi Since the Zi → Z, they are all ⊂ B(0,R) for some R > 0, so the total masses of the mi are all ≤ CR. It follows that there is a weakly convergent subsequence mi0 * m, and we easily check that (23) holds. Conversely, if the support function of Z is given by (23) then we may approximate the even measure m weakly by a sequence of even atomic measures. As above, the corresponding convex bodies are zonotopes, which converge to Z because their support functions converge uniformly to that of Z.  Theorem 2.27. The cosine transform Z Cf(v) := |w · v|f(w) dw Sn-1 ∞ n-1 ∞ n-1 yields a linear isomorphism C+ (S ) → C+ (S ). ∞ n-1 Here C+ (S ) denotes the space of smooth even functions on the sphere. Unfortunately we don’t have time to go into the proof of this fact.

sm Corollary 2.28. If A ∈ Ko then there exists r > 0 and a zonotope Z such that

A + Br = Z. 16 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

∞ n-1 Proof. By Theorem 2.27 and Proposition 2.29 below, there exists g ∈ C+ (S ) such that Z Cg = |() · w| g(w) dw = hA. Sn-1 Put −M := min(g, 0) and take r to be the constant value of the cosine transform of the constant function M. Then g + M ≥ 0 and

C(g + M) = Cg + CM = hA + r = hA+Br . So the result follows from Proposition 2.26. 

Proposition 2.29. If hA is smooth away from 0 then for v 6= 0 v = n (∇h (v)). (24) |v| A A sm 2 Furthermore A ∈ K iff hA is smooth away from 0 with D hA(v) v⊥ everywhere nonsingular.

Proof. Suppose first that hA is smooth away from 0. Given v 6= 0, the point x ∈ A for which v x · v = hA(v) clearly lies on the boundary of A, with |v| = nA(x). To prove (24), we must show that the directional derivatives of hA at v are given by

d DwhA := hA(v + tw) = x · w. (25) dt t=0 Clearly hA(v + tw) ≤ (v + tw) · x = hA(v) + tw · x. It is a general fact that one-sided directional derivatives of convex functions exist (this follows from same assertion for convex functions of one variable, which are characterized by the condition that the derivative exists off a countable set and is increasing), so by the last relation so the one-sided directional derivative

+ d −1 Dw hA := hA(v + tw) := lim t (hA(v + tw) − hA(v)) ≤ w · x dt+ t=0 t↓0 and similarly − Dw hA ≥ w · x. Since the directional derivative itself exists by hypothesis, (25) follows. The “if” part of the last assertion now follows at once. Suppose conversely that A ∈ Ksm. Consider the n-dimensional smooth submanifold n n M := {(x, tnA(x)) : x ∈ ∂A, t > 0} ⊂ R × R , the diffeomorphic image of ∂A × (0, ∞) under the map

f(x, t) := x + tnA(x). Thus the tangent spaces of M are given by

T(x0,t0nA(x0))M = graph(t0DnA(x0)) ⊕ h(0, nA(x))i

⊂ Tx0 ∂A ⊕ Tx0 ∂A ⊕ h(0, nA(x))i (26) n n ⊂ R ⊕ R . sm Since A ∈ K , the image of DnA(x) is the whole of Tx∂A, and therefore the projection of the n n−1 tangent space to the second R factor is surjective. Additionally, the map nA : ∂A → S is a bijection, so the restriction to M of the projection π2 to the second factor is a diffeomorphism to n n n R − {0}. To put it another way, if ι : R × R is the interchange map ι(p, q) := (q, p), then ι(M) n is the graph of a smooth vector field V on R − {0} which is positively homogeneous of degree 0, NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 17

n i.e. V (ty) = V (y) for t > 0. Likewise, at a general point y0 = t0nA(x0) ∈ R − {0} the derivative DV (y0) is given by

graph DV (y0) = ι(T(x0,t0nA(x0))M). n n In other words, decomposing Ty0 R = R as the orthogonal direct sum Tx0 ∂A ⊕ hnA(x0)i, this derivative may be expressed −1 DV (y0)(v, t) = (DnA(x0) v, 0). ⊥ Since DnA(x0) is self-adjoint and nondegenerate when restricted to y0 , it follows that DV (y0) n is also. In other words, V = ∇f for some smooth function f : R − {0} → R, and by the 0- homogeneity of V this f may be taken to be positively homogeneous of degree 1, and extended to n all of R by taking f(0) = 0. By path-independence we may evaluate f(y0) as Z |y0| y0 f(y0) = ∇f(ty0) · dt 0 |y0| Z |y0| y0 = V (y0) · dt 0 |y0| Z |y0| y0 = x0 · dt 0 |y0| = x0 · y0

= hA(y0) by (22). Thus hA is smooth, as claimed.  2.4. Proof of Hadwiger’s theorem. Definition 2.30. A valuation µ is • even (resp. odd) if µ(−A) = µ(A) (resp. µ(−A) = −µ(A)) for all A ∈ K. • is simple if µ(A) = 0 whenever dim A < n. Clearly every valuation is the sum of an even valuation and an odd valuation. To prove Hadwiger’s theorem we use induction on the dimension n, the case n = 0 begin trivial. Assuming the statement true through dimension n − 1, suppose µ ∈ ValSO(n). By induction, there are constants c0, . . . , cn−1 such that n−1 X µ(A) = ciµi(A) i=0 n−1 0 Pn−1 SO(n) whenever A ⊂ K(R ). By invariance it follows that ν := µ − ciµi ∈ Val is simple. 0 We must prove that ν is a multiple of µn. Lemma 2.31. ν0 is even. n Proof. We show that any simplex ∆ ⊂ R may be decomposed as

∆ = P1 ∪ · · · ∪ PN , where each Pi is a polytope which is congruent to −Pi, and the intersection of any two distinct Pi,Pj has dimension < n. Let ∆0,..., ∆n be the faces of ∆, and let z ∈ ∆ be the center of the sphere inscribed in ∆. This sphere touches each face ∆i at a point zi. Put Pij for the convex hull of z, zi, zj, ∆i ∩ ∆j. Clearly each Pij is taken to itself by the reflection in the hyperplane through z, ∆i ∩ ∆j. Thus by additivity and the simplicity assumption X X ν(∆) = ν(Pi) = ν(−Pi) = ν(−∆). 18 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Since every polytope Q may be decomposed into finitely many essentially disjoint simplices, it follows that ν(Q) = ν(−Q). Now any A ∈ K is the limit of a sequence of polytopes (e.g. the sequence obtained by taking the convex hulls of the first few points in a dense sequence), so by continuity ν(A) = ν(−A) and ν is even.  The proof is now completed with the following result, which is of independent interest. Theorem 2.32. Any continuous, translation-invariant, even, simple valuation ν0 is a multiple of the volume.

0 0 n Put ν := ν − ν ([0, 1] )µn. We must show that ν = 0. Note that by subdividing into and then reassembling a cube into small essentially disjoint subcubes we may see (using continuity) that ν vanishes on any rectanglular solid. Put K+ := {A ∈ K : A = −A}.

n−1 Lemma 2.33. If A = σ + C, where σ is a line segment and C ∈ K(R ), then ν(A) = 0. n−1 Proof. Suppose first that σ ⊥ R . Fixing σ, the inductive hypothesis implies that the valuation n−1 on R given by C 7→ ν(σ + C) is zero. In the general case, note that if C is not too wide in comparison with the length of σ then we may cut off one end of σ +C by a hyperplane ⊥ σ, then translate it and reattach it to the other end to obtain a new convex body which is the sum of a translate of C with a perpendicular segment. By the last paragraph the value of ν on this new body is zero, and by additivity, simplicity and invariance this value is just ν(σ + C). If C is too wide we may subdivide it by hyperplanes and apply the argument to the pieces.  Lemma 2.34. If P ∈ K is a polytope and σ is a line segment then ν(P ) = ν(P + σ).

Proof. Orienting σ, let F1,...,FN be the codimension 1 faces of P whose outward normals make an acute angle with the direction of σ. Then P + σ is congruent to the essentially disjoint union of P together with copies of the Fi + σ. By the last lemma each ν(Fi + σ) = 0, so the Lemma follows by additivity and simplicity. 

Lemma 2.35. If A ∈ K+ then ν(A) = 0. Proof. By induction, continuity and Lemma 2.34, if P is a polytope and Z is a zonoid then ν(P + Z) = ν(P ). By continuity this holds for general A ∈ K as well. Applying this result to a A = a point, it follows that ν(Z) = 0 for any zonoid Z. Now suppose sm A ∈ K+ . By Corollary 2.28, ν(A) = ν(A + Br) = ν(Z) = 0 for some r > 0 and some zonoid Z. sm Finally, any A ∈ K+ may be approximated by a sequence in K+ , so the Lemma follows by continuity.  Now we may complete the proof of Hadwiger’s theorem. As above, it is enough to show that ν(∆) = 0 for simplices ∆. We may assume that one vertex of ∆ lies at the origin, and denote the Pn ∗ others by v1, . . . , vn. Consider the parallelotope P := i=0[0, vi]. Put ∆ to be the convex hull P P P ∗ of V := i vi, i6=1 vi, ··· i6=n vi; thus ∆ = V − ∆, so by translation-invariance ane evenness ν(∆∗) = ν(∆). Since P is centrally symmetric, we know that ν(P ) = 0, and the same is true of Q := clos(P − ∆ − ∆∗). Thus 0 = ν(P ) = ν(∆) + ν(∆∗) + ν(Q) = 2ν(∆).

So ν(∆) = 0, as claimed.  NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 19

2.5. Even valuations: Klain functions, Crofton measures and the Alesker Fourier trans- + + n form. Theorem 2.32 has the following additional important consequence. Put Valk = Valk (R ) for the vector space of continuous, translation-invariant, even valuations of degree k. This is a normed space under the norm kφk := sup |φ(A)|. A∈K,A⊂B(0,1)

+ n Theorem 2.36. If φ ∈ Valk and E ∈ Grk(R ) then the restriction φ|E = c volk for some constant + c =: Klφ(E). The function Klφ : Grk → R is continuous, and the map Kl : Valk → C(Grk) is a continuous injection. Proof. We may suppose that φ 6= 0. Let j ≥ 0 be the smallest nonnegative integer for which n there exists a j-dimensional A ∈ K such that φ(A) 6= 0. Thus if E ∈ Grj(R ) then the restriction of φ to K(E) is simple, so φ|E = c volj for some constant c = cE, by Theorem 2.32. Since volj is homogeneous of degree j, we must have j = k. Since φ is continuous it follows that cE is a + continuous function of E ∈ Grk, and the continuity of the resulting function Kl : Valk → C(Grk) is clear from the definition of the norm above. 

Klφ is called the Klain function of φ. This argument also shows: + Lemma 2.37. If φ ∈ Valk and dim A < k then φ(A) = 0. Putting Z νm(A) := volk(πE(A)) dm(E) Grk + it is clear that νm ∈ Valk . If φ = νm then m is a Crofton measure for φ. For example, a suitable multiple of the Haar measure is a Crofton measure for the intrinsic volume µk. Unless k = 1, n − 1 then the Crofton measure is not unique. Note that the Klain function of νm is the cosine transform of m, Z

Klνm (E) = (Cos m)(E) := cos(E,F ) dm(F ) Grk where cos(E,F ) is the Jacobian determinant of the restriction of the orthogonal projection πE to F . + Suppose φ ∈ Valk admits a Crofton measure m. We define the Alesker Fourier transform of φ to be ˆ + φ := ν(⊥∗m) ∈ Valn−k, where ⊥∗ m is the image of the signed measure m under the orthogonal complement map ⊥: Grk → Grn−k. Alternatively, this valuation may be characterized by the property ∗ Klφˆ =⊥ Klφ . This follows from the fact that cos(E⊥,F ⊥) = cos(E,F ). (27) By Theorem 2.36, this characterization implies that the Alesker Fourier transform is well-defined, even though the Crofton measure of φ may not be unique. k n Exercise 2.38. The vector space Λ (R ) admits a natural euclidean structure by taking the ele- k n−k ments {dxI : I ⊂ {1, . . . , n}, #I = k} to be an orthonormal basis. The linear map p :Λ → Λ defined by dxI 7→ dxI¯, where the orderings are determined by dxI ∧ dxI¯ = +dx1 ∧ · · · ∧ dxn, is an isometry. Given E,F ∈ Grk, let v1, . . . , vk; w1, . . . , wk be orthonormal bases for E,F respectively. Then

cos(E,F ) = |(v1 ∧ · · · ∧ vk) · (w1 ∧ · · · ∧ wk)| 20 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 where the dot product is with respect to this euclidean structure. Conclude that (27) holds.

− n n 2.6. The McMullen decomposition. Put Valk (R ) ⊂ Val(R ) for the subspace of odd valua- tions of degree k. Theorem 2.39. n n M M  n Val(R ) = Valk(R ). k=0 =± Furthermore, Val0 and Valn are 1-dimensional, spanned respectively by the Euler characteristic χ and the volume voln. 2.7. Nijenhuis’s theorem. Nijenhuis [25] proved that by altering the basis of ValSO(n) to be certain multiples of the intrinsic volumes, the structure constants in the Blaschke kinematic formulas may all be taken equal to unity. He asked whether this might be explained by some algebraic structure. Here we give a description of such a structure and prove Nijenhuis’s theorem. It is interesting to note that this proof does not rely on any technology developed in the intervening years, so one might say the failure to find this proof was due to a psychological mental block. Consider the vector space n n Val = Val(R ) := {continuous, translation invariant valuations : K(R ) → R} and define the linear operator L : Val → Val to be Z Lφ(A) := φ(A ∩ P¯) dP¯ Grn−1 where mn−1 := dP¯ is an invariant measure on Grn−1. Observe that by Crofton’s formula (Corollary 2.4) Lχ = cµ1 (28) k Note that there is a rational map Jk : Grn−1 99K Grn−k, given by intersections in general position. This map is almost proper, in the sense that the pre-image of any compact subset of Grn−k has compact closure. Moreover, the “diagonal” consisting of those k-tuples (P¯1,..., P¯k) that are not in general position clearly has measure zero. It follows that the pushforward measure k mn−k := Jk∗((mn−1) ) is a well-defined Radon measure in Grn−k. Since Jk is SO(n)-equivariant, this measure is again invariant, and clearly (R φ(· ∩ Q¯) dmn−kQ,¯ k ≤ n Lkφ = Grn−k (29) 0, k > n. n (Note that L φ(A) = c voln(A)φ(point), and clearly L voln = 0.) In particular, ( k cklµl+k, l + k ≤ n L µl = (30) 0 l + k > n. for some constants ckl > 0 (cf. Exercise 2.7). Since the intrinsic volumes are linearly independent, we deduce at once that L0 = id, L, . . . , Ln are linearly independent elements of End(Val), and in fact the linear map determined Li 7→ Liχ is a linear isomorphism hL0,...,Lni → ValSO(n). Identifying the two spaces, this induces a com- mutative product on ValSO(n). Putting t := Lχ, (31) NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 21

SO(n) Val becomes a commutative algebra over R generated by t. In fact SO(n) n+1 Val ' R[t]/(t ). (32) SO(n) SO(n) SO(n) This algebra is related to the kinematic coproduct kSO(n) : Val → Val ⊗ Val in the following way. Given µ ∈ ValSO(n) we compute Z k k kSO(n)(t · µ)(A, B) = (t · µ)(A ∩ gB¯ ) dg¯ SO(n) Z Z = µ(P¯ ∩ A ∩ gB¯ ) dP¯ dg¯ SO(n) Grn−k Z Z = µ(P¯ ∩ A ∩ gB¯ ) dg¯ dP¯ Grn−k SO(n) Z ¯ ¯ = kSO(n)(µ)(P ∩ A, B) dP Grn−k h k i = (t ⊗ 1) · kSO(n)(µ) (A, B).

Thus (using the cocommutativity of kSO(n)) k k k kSO(n)(t · µ) = (t ⊗ 1) · kSO(n)(µ) = (1 ⊗ t ) · kSO(n)(µ). (33) Introduce a symmetric bilinear pairing on ValSO(n) by (µ, ν) = degree n component of µ · ν. ∗ This pairing is clearly perfect, i.e. induces a linear isomorphism p : ValSO(n) → ValSO(n) . Theorem 2.40 (Ftaig for SO(n)). Let m : ValSO(n) ⊗ ValSO(n) → ValSO(n) denote the multiplica- ∗ ∗ ∗ tion map above and m∗ : ValSO(n) → ValSO(n) ⊗ValSO(n) its adjoint. Then the following diagram commutes: k ValSO(n) −−−−→SO(n) ValSO(n) ⊗ ValSO(n)   p p⊗p (34) y y ∗ ∗ ∗ ∗ ValSO(n) −−−−→m ValSO(n) ⊗ ValSO(n)

Proof. The fact that k = kSO(n) is a cocommutative, coassociative coproduct implies that its adjoint ∗ k∗ is a commutative, associative product on ValSO(n) , and therefore induces a product on ValSO(n) via the map p. In other words µˆ· ν := p−1 ◦ k∗(pµ ⊗ pν) is a commutative, associative product on ValSO(n). We must show that this is the same as the multiplication defined above. We claim that in these terms the relation (33) is equivalent to (µ · ν)ˆ· φ = µ · (νˆ· φ). (35) To prove this claim we show that for any ψ ∈ ValSO(n) hψ, k∗(p(µ · ν) ⊗ pφ))i = hψ, p(µ · (νˆ· φ))i. (36) By definition the lefthand side may be written

hk(ψ), p(µ · ν) ⊗ pφ)i = [k(ψ) · ((µ · ν) ⊗ φ)]n,n

= [k(ψ · µ · ν · φ)]n,n 22 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 by (33). Meanwhile the righthand side is

∗ [ψ · µ · (νˆ· φ)]n = hψ · µ, k (pν ⊗ pφ)i = hk(ψ · µ), pν ⊗ pφi

= [k(ψ · µ) · (ν ⊗ φ)]n,n

= [k(ψ · µ · ν · φ)]n,n. Thus if we can show that χ acts as the identity element also for the product ˆ· , then, setting ν = χ in (35), we will be done. Unpacking the definition of ˆ· , the relation (35), with ν = χ, is equivalent to the assertion that ∗ k (pµ ⊗ τn) = pµ (37) for all µ ∈ ValSO(n). ∗ Denote by hφ, α∗i the pairing between φ ∈ ValSO(n) and α∗ ∈ ValSO(n) , and also for the pairing SO(n) SO(n) SO(n)∗ SO(n)∗ SO(n)∗ between Val ⊗ Val and Val ⊗Val . Denote by τn ∈ Val the dual element i i ht , τni = δn.

Observe that τn = pχ. Take the Haar measure dg¯ on SO(n) so that dg¯({g¯ :go ¯ ∈ S} = tn(S)

n n for all S ⊂ R (recall that t is some constant multiple of Lebesgue measure). Then it is clear that for any φ ∈ ValSO(n), if we expand X j k(φ) = βj ⊗ t j

∗ SO(n)∗ then βn = φ. Thus if α ∈ Val then ∗ ∗ ∗ hkSO(n)(φ), α ⊗ τni = hφ, α i. (38)

Therefore for µ, φ ∈ ValSO(n)

∗ hk (pµ ⊗ τn), φi := hpµ ⊗ τn, k(φ)i = hpµ, φi.

Since this holds for all φ this is sufficient to establish (37).  Exercise 2.41. Given the multiplicative structure of ValSO(n) and the definition of the Poincar´e duality map p, show that the ftaig is just a fancy way of saying that X k(tl) = ti ⊗ tj. i+j=n+l

The valuation t ∈ ValSO(n) and its powers may also be viewed as independent of the dimension n, SO(∞) SO(n+1) n+1 as follows. We define Val to be the inverse limit of the system of algebras Val (R ) → SO(n) n Val (R ) under restriction. Normalizing t so that 2 t = µ (39) π 1 SO(∞) we may identify Val ' R[[t]], the field of formal power series in t. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 23

2.8. The Alesker product, part I. The product on ValSO(n) is essentially a special case of the following more general product. n Let m be a finite signed Borel measure on K(R ) with compact support. Then Z Z µm(A) := χ(A ∩ (B + x)) dx dm(B) (40) K Rn n defines a valuation ∈ Val(R ) (continuity follows by dominated convergence). We denote the vector n space of all such valuations by M(R ). If m1, m2 are two such measures we put Z Z

(µm1 · µm2 )(A) := χ(A ∩ (B + x) ∩ (C + y)) dx dy dm1(B)dm2(C). K×K Rn×Rn Observe that this valuation again ∈ M. By Fubini’s theorem, Z Z

(µm1 · µm2 )(A) := µm2 (A ∩ (B + x)) dy dm1(B). (41) K Rn Using this relation we may show that this is actually a product of valuations, i.e. is independent of the representations (40)— from what we already know about the integral geometry of SO(n), for 0 0 instance, there are many such ways to represent the intrinsic volumes. For if m1, m2 are different measures on K, with µ 0 = µ , then by (41) mi mi µ · µ = µ · µ 0 = µ 0 · µ = µ 0 · µ 0 . m1 m2 m1 m2 m2 m1 m2 m1 Clearly the Euler characteristic χ acts as the multiplicative identity.  0 Proposition 2.42. The Alesker product is bigraded, i.e. if µm ∈ Vali , µm0 ∈ Valj then µm · µm0 ∈ +0 Vali+j . Proof. 

Define the function sin : Grk × Grl → R by

volk+l(BE + BF ) = sin(E,F )ωkωl. Proposition 2.43. The space of (even) valuations that admit a Crofton is a subspace of M +. If m1, m2 are Crofton measures for µ1 ∈ Mk, µ2 ∈ Ml respectively, then +∗(sin m1 × m2) is a Crofton measure for µ1 · µ2, where + : Grk × Grl 99K Grk+l is the sum map. Proof. Given a signed measure m on Grk, the associated valuation is the degree k part of the + 0 valuation in Mk , with measure m on K given by 0 m ({BE : E ∈ S ⊂ Grk}) = m(S). + So Crofton valuations ⊂ M .  n n Proposition 2.44. Let V ⊂ R be a vector subspace. Then the restriction map rV maps M(R ) to M(V ), and rV is a homomorphism of algebras. n n−k Proof. We show first that if A ∈ K(R ) then rV (µA) ∈ M(V ). Put k := dim V and W for the orthogonal complement of V , and for w ∈ W put Aw := (A ∩ (V + w)) − w ∈ K(V ) for the slices of A by planes parallel to V . Then for B ∈ K(V ) Z rV (µA)(B) = µA(B) = χ(B ∩ (A + x)) dx Rn Z Z = χ(B ∩ (A − w + v)) dv dw W V Z Z = χ(B ∩ (Aw + v)) dv dw W V 24 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Thus rV (µA) = µm, where m is the measure on K(V ) given by

m({Aw : w ∈ E}) = voln−k(E), m(S) = 0 if S ∩ {Aw}w = ∅. n Now the extension to general µ ∈ M(R ) is straightforward. To show that rV is multiplicative it is again sufficient to show that for C ∈ K(V ) Z Z

rV (µA) · rV (µB) = χ(C ∩ (Aw1 + v1) ∩ (Bw2 + v2)) dv1 dv2 dw1 dw2 W ×W V ×V Z = χ(C ∩ (A + x1) ∩ (B + x2)) dx1 dx2 Rn×Rn = (µA · µB)(C).  2.9. Review of differential forms.

2.9.1. exterior algebra.

∗ n n n n 2.9.2. An action of sl2 on Λ (R ⊕R ). Reference: [18] Let Q = R ⊕R be a symplectic real vector P 2 ∗ ∗−1 space of dimension 2n with symplectic form ω = dxα ∧ dyα ∈ Λ Q. Put iα, jα :Λ Q → Λ Q for the interior products with ∂ , ∂ respectively. Clearly ∂xα ∂yα

iαdyβ = −dyβiα, jαdxβ = −dxβjα,

iαdxβ = −dxβiα, jαdyβ = −dyβjα, α 6= β

iαiβ = −iβiα, jαjβ = −jβjα, iαjβ = −jβiα

dxαiα = πα, iαdxα = Id − πα

dyαjα = λα, jαdyα = Id − λα where πα (resp. λα) is the projection onto the subspace of multiples of dxα (resp. dyα). Define the operators L, Λ:Λ∗Q → Λ∗Q, of degree 2, −2 respectively, by X Lφ := ω ∧ φ, Λ := iαjα. α Then !     ! X X X X [L, Λ] = dxα ∧ dyα  iβjβ −  iβjβ dxα ∧ dyα α β β α X = (dxα ∧ dyα)(iαjα) − (iαjα)(dxα ∧ dyα) α X + (dxα ∧ dyα)(iβjβ) − (iβjβ)(dxα ∧ dyα) α6=β In view of the relations above, the second sum in the last expression vanishes, while the first is X X (Id − πα)(Id − λα) − παλα = Id − (πα + λα) α α X = nId − (πα + λα). α Now if φ ∈ ΛkQ then X (πα + λα)φ = kφ, α NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 25 hence the restriction of H := [L, Λ] to ΛkQ is (n − k) times the identity. Clearly [L, H] = −2L, [Λ,H] = 2Λ. (42)

This means that we now have an action of the Lie algebra sl2(R) of traceless 2 × 2 matrices on Λ∗Q. For if we put 1 0  0 1 0 0 Z := ,X := ,Y := 0 −1 0 0 1 0 then [X,Y ] = Z, [X,Z] = −2X, [Y,Z] = 2Y. We can now apply the general theory of such representations to prove Proposition 2.45. The restriction of L to Λn−1Q is an isomorphism. In other words, every element of Λn+1Q may be written as ω ∧ ψ for a unique ψ ∈ Λn−1Q. The Proposition is a consequence of the following more structural statement. Say that 0 6= φ ∈ ΛkQ is primitive if Λφ = 0. Theorem 2.46. (1) Every primitive element has degree ≤ n. (2) φ ∈ ΛkQ is primitive iff Ln−kφ 6= 0,Ln−k+1φ = 0. (43) i (3) In this case, the vector space spanned by the L φ is an sl2 submodule. (4) Λ∗Q is the direct sum of such submodules. Proof. Suppose φ is primitive with deg φ = k. We claim that ΛLiφ = [i(i − 1) − i(n − k)]Li−1φ (44) This is obvious if i = 0, and for i > 0 we have inductively ΛLi+1φ = (−H + LΛ)Liφ = −(n − k − 2i)Liφ + L(ΛLiφ) = [−(n − k) + 2i + i(i − 1) − i(n − k)]Liφ = [(i + 1)i − (i + 1)(n − k)]Liφ. This establishes that (43) holds, and also conclusions (1) and (3). Next we prove conclusion (4). It is enough to show the following. Let M ⊂ Λ∗Q be a direct sum of modules of the type above, and suppose that M 6= Λ∗Q. Then there exists a primitive element ∗ φ ∈ Λ Q − M, and the sl2 module it generates meets M only at 0. To see this, φ0 be an element of minimal degree in Λ∗Q − M. Then Λφ0 ∈ M, so from the structure of M we know that there is ψ ∈ M such that Λψ = Λφ0. Thus φ := φ0 − ψ is primitive. Since φ = cΛiLiφ whenever Liφ 6= 0, it follows that all Liφ∈ / M. Finally, conclusion (2) is an immediate consequence of conclusion (4) and the (proven) equation (43) for primitive φ. 

∗ n n 2.9.3. Another action of sl2 on Λ (R ⊕ R ). Consider adapted coordinates

x1, . . . , xn, y1, . . . , yn n n P ∗ n n ∗ n n for R ⊕R , so that ω = dxi∧dyi is the usual symplectic form. Put j :Λ (R ⊕R ) → Λ (R ⊕R ) for the algebra isomorphism that interchanges the coordinates:

j(dxi) = dyi, j(dyi) = dxi, i = 1, . . . , n. 26 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

∗ n n ∗ n n There are derivations `, λ :Λ (R ⊕ R ) → Λ (R ⊕ R ), of degrees ±1 respectively, determined by

`(dyi) := dxi, `(dxi) := 0

λ(dxi) := dyi, λ(dyi) := 0, j ◦ ` = λ ◦ j. (45)

In fact, putting mt(x, y) := (x + ty, y)

d ∗ λϕ = mt ϕ (46) dt t=0 ∗ n n and a similar formula holds for `. It is easy to see that for a monomial ϕ ∈ Λ (R ⊕ R ) [`, λ] ϕ = (r − s)ϕ (47) where r, s respectively denote the number of dxi and dyi factors in ϕ. Putting h for the operator ϕ 7→ (r − s) ϕ, clearly [h, `] = 2`, [h, λ] = −2λ. (48)

The discussion of the last section goes through for this sl2 representation as well.

2.9.4. Differential forms. d and f ∗. Let M be a smooth manifold. A differential form on M is a smooth section of the bundle Λ∗TM. In other words, a differential form φ of degree k assigns to each k point p ∈ M an element φp ∈ Λ (TpM), with the property that if v1, . . . , vk are smooth vector fields on M then p 7→ φp(v1(p), . . . , vk(p)) is a smooth function on M. The space of differential forms of degree k on M is denoted Ω∗(M), and subspace of k-forms Ωk(M). If F : M → N is a smooth map to another smooth manifold, then the pullback under F is the map F ∗ : Ω(N) → Ω(M) given by ∗ (F ψ)p(v1, . . . , vk) := ψF (p)(F∗v1,...,F∗vk) where F∗ : TM → TN is the derivative. Since this operation is defined pointwise it follows that F ∗(ψ ∧ θ) = (F ∗ψ) ∧ (F ∗θ). and (F ◦ G)∗ = G∗ ◦ F ∗. n k If M = U is an open subset of R then Ω (U) is spanned by the φ = fdxi1 ∧ . . . dxik , where f ∈ C∞(U). In this case we define the exterior derivative by X dφ := ∂jfdxj ∧ dxi1 ∧ · · · ∧ dxik , extending to all of Ω∗(U) by linearity. It is clear that d(φ ∧ ψ) = dφ ∧ ψ + (−1)deg φφ ∧ dψ. (49) m ∞ Suppose F = (F1,...,Fm): U → V ⊂ R is a smooth map, and g ∈ C (V ). Then n ∗ X F dyj = ∂iFjdxi i=1 and so by the chain rule m n ! m ∗ ∗ X X X ∗ ∗ d(F g) = d (g ◦ F ) = [(∂jg) ◦ F ] ∂iFjdxi = [(∂jg) ◦ F ]F dyj = F dg j=1 i=1 j=1 It now follows from (49) that F ∗ ◦ d = d ◦ F ∗. (50) NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 27

The most important consequence of this relation is that d is a well-defined operator on Ω(M) for smooth manifolds M. To see this we let X = (x1, . . . , xn); Y = (y1, . . . , yn) be two sets of overlapping local coordinates on M. Then any φ ∈ Ω∗(M) may be written locally as φ = X∗φ0 = Y ∗φ00 0 00 n for some locally defined forms φ , φ on R . Then dφ0 = d[(Y ◦ X−1)∗φ00] = (Y ◦ X−1)∗dφ00 = X−1∗ ◦ Y ∗dφ00 and thus we may define unambiguously dφ := Y ∗dφ00 = (X−1 ◦ X)∗Y ∗dφ00 = X∗X−1 ∗ Y ∗dφ00 = X∗dφ0.

Another important derivation on Ω(M) is the Lie derivative Lv with respect to a smooth vector field v on M, defined as follows. Let Ft be the flow of v, i.e. the family of diffeomorphisms of M such that

d Ft = v. dt t=0 Then

d ∗ Lvφ := Ft φ. dt t=0 ∗ This operation may be expressed in terms of the exterior derivative as follows. Let iv :Ω (M) → Ω∗−1(M) be the interior product, i.e.

ivφ(w1, . . . , wk) := φ(v, w1, . . . , wk). Lemma 2.47.

Lv = d ◦ iv + iv ◦ d. (51) Proof. It is straightforward to check that each side defines a derivation, which in this case means

Lv(φ ∧ ψ) = (Lvφ) ∧ ψ) + φ ∧ (Lvψ)

(d ◦ iv + iv ◦ d)(φ ∧ ψ) = ((d ◦ iv + iv ◦ d)φ) ∧ ψ) + φ ∧ ((d ◦ iv + iv ◦ d)ψ) Thus it is enough to check that (51) holds when applied to functions f ∈ Ω0(M) and exact 1-forms df ∈ Ω1(M), i.e. to check that

Lvf = ivdf, Lvdf = divf The first of these relations is obvious— it states that the derivative of f in the direction v is given in the usual way by the total derivative. To prove the second relation we compute

d ∗ Lvdf = Ft df dt t=0

d ∗ = d(Ft f) dt t=0 −1 ∗ = lim t (d(Ft f) − df) t→0 −1 ∗ = lim d(t (Ft f − f)) t→0

−1 ∗ Since the function (t, p) 7→ Ft ∗ (p) is smooth, it follows that t (Ft f − f), together with all its first derivatives, converges uniformly to Lvf = ivdf. Thus the limit in the last expression commutes with d to yield divdf, as claimed.  28 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

2.9.5. Contact manifolds and the Rumin operator. Recall that a smooth manifold N 2n+1 equipped with a totally non-integrable smooth field of hyperplanes Qx ⊂ TxN is called a contact manifold. An equivalent statement of the defining condition is that locally there exists a 1-form α on N such that α ∧ dαn 6= 0. In this case Q = α⊥. The sphere bundle N = SM of a smooth Riemannian manifold is a contact manifold, with global n+1 contact form αξ · v := hξ, π∗vi. This is particularly clear in M = R . Letting the coordinates be P 2 x1, . . . , xn+1, v1, . . . vn+1, vi = 1, the contact form becomes n+1 n+1 X X α = hv, dxi = vidxi, dα = dvi ∧ dxi. i=1 i=1 Each hyperplane Qx of the contact distribution of a contact manifold is a symplectic vector space of dimension 2n, with symplectic form dα. In general α is only defined up to multiplication by a nonvanishing function f, so since d(fα) = df ∧ α + fdα the symplectic form is in general only defined up to scale. Let N be a contact manifold with global contact form α. There is a unique vector field T on N such that α · T ≡ 1 and LT α = 0. By Lemma 2.47, the second condition may be replaced by iT dα = 0. This field is called the is Reeb field. The flow generated by a Reeb field is a family of contactomorphisms. n n n−1 If N = SR then Tx,v := (v, 0) ∈ TxR ⊕ TvS defines a Reeb field. More generally, if M is a Riemannian manifold then the vector field on SM generating the geodesic flow is a Reeb field. Proposition 2.48 ([28]). Let N 2n+1 be a smooth contact manifold equipped with a global contact form α. Given ψ ∈ Ωn(N), there is a unique differential form Dψ ∈ Ωn+1(N) such that (1) Dψ is a multiple of α, i.e. annihilates the contact distribution Q (2) Dψ − dψ = d(α ∧ θ) for some θ ∈ Ωn−1(N). In other words, we may perturb ψ by a multiple of α so that the exterior derivative of the perturbed form is again a multiple of α, and this exterior derivative is uniquely determined. Proof. Since for each x ∈ N the restricted exterior derivative dα | is nondegenerate, by Propo- x Qx sition 2.45 there is a unique θ0 ∈ Λn−1Q such that dα ∧ θ0 = − dψ| . If T is the Reeb field, this x Qx 0 n−1 θ may be extended to give θ ∈ Ω (N) by stipulating that iT θ = 0. Then Dψ := d(ψ + α ∧ θ) = dψ + dα ∧ θ − α ∧ dθ. The sum of the first two terms annihilates Q by construction, and the third obviously does too since Q = α⊥. ¯ ¯ 0 To prove uniqueness, suppose θ is another form satisfying the requirements. Then θ Q = θ = ¯ θ|Q, so θ − θ = α ∧ β for some β, and d(ψ + α ∧ θ¯) = d(ψ + α ∧ (θ + α ∧ β)) = Dψ.  2.10. Integration of differential forms and the normal cycle. Let φ be a measurable k-form k on R , i.e. φ = g(x)dx1 ∧ · · · ∧ dxk where g is a Lebesgue measurable function. Given a measurable set E ⊂ k we put R Z Z φ := g. E E k k If M is a smooth manifold and f : R → M is Lipschitz we put for φ ∈ Ω (M) Z Z ∗ f∗[[R]](φ) = φ := f φ. f∗[[E]] E This is well-defined because of NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 29

n Theorem 2.49 (Rademacher’s theorem). A Lipschitz function f : R → R is differentiable almost everywhere. A Lipschitz k-chain is a finite sum N X T := fi∗[[Ei]] (52) i=1 k k where the Ei ⊂ R are compact and the fi : R → M are Lipschitz with respect to some Riemannian metric on M. Observe that by compactness any map that is Lipschitz with respect to one such metric is k Lipschitz with respect to any other. Also, it is irrelevant whether the fi are defined over all of R or only over Ei, since it is easy to extend Lipschitz maps. Lemma 2.50. Let X be a metric space, S ⊂ X, and f : S → R. Then there is f¯ : X → R with Lip(f¯) ≤ Lip(f). Proof. Put L := Lip(f) and put f¯(y) := inf f(x) + Ld(x, y). x∈S  ∗ ∗ ∗ Unfortunately, the basic relation fi d = dfi is not immediately available since f φ is in general not even continuous, hence its exterior derivative cannot be defined directly. Nevertheless everything k n goes through because of the following. Define the comass of φ ∈ Ω (R ) by

kφk := sup |φx(v1, . . . , vk)| n x∈R ,|vi|=1

Observe that if n = k then the supremum is attained whenever the vi are orthonormal. k n Lemma 2.51. If f : R → R has Lipschitz constant L, or more generally if kDfk∞ ≤ L, then kf ∗φk ≤ Lk kφk . k n Lemma 2.52. Let E ⊂ R be a compact domain with piecewise smooth boundary, f, g : E → R 1 k n be C maps, and φ ∈ Ω (R ). Then Z ∗ ∗ k f φ − g φ ≤ kf − gk∞ (kDfk∞ + kDgk∞) kdφk volk(E) E k−1 + kf − gk∞ (kDfk∞ + kDgk∞) kφk volk−1(∂E). k+1 n Proof. Define the map F : R → R by F (t, x) := g(x) + t(f(x) − g(x)). Then by Stokes’ theorem

Z Z Z ∗ ∗ ∗ ∗ f φ − g φ = F dφ − F φ E E×[0,1] ∂E×[0,1] ∗ ∗ ≤ volk(E) kF dφk + volk−1(∂E) kF φk k Letting v1, . . . , vk ∈ R be orthonormal, ∗ ∗ kF dφk = sup |F dφ(∂t, v1, . . . , vk)|

= sup |dφ(F∗∂t,F∗v1,...,F∗vk)|

≤ kdφk |F∗∂t||F∗v1| ... |F∗vk| k ≤ kdφk |f(x) − g(x)|(kDfk∞ + kDgk∞) . 30 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

The bound for the other term is similar.  n ∞ k Lemma 2.53. Let E be as above and let f : E → R be Lipschitz. Let h ∈ C (R ) be a com- R −k −1 pactly supported function with k h = 1, let h(x) :=  h( x) be the corresponding approximate R k n identity, and put f := f ∗ h. Then for any φ ∈ Ω (R ) Z Z ∗ lim f φ = φ ↓0 E E ∗ ∗ Proof. In fact f φ(x) → f φ(x) at every Lebesgue point x of f, so the result follows by dominated convergence.  k Theorem 2.54. Let E ⊂ R be a compact domain with piecewise smooth boundary, and f1, f2, ··· : n E → R be a sequence of Lipschitz maps with Lip(fi) ≤ L, and converging uniformly to f0 (which k n therefore has Lip(f0) ≤ L). Let φ ∈ Ω (R ). Then Z Z ∗ ∗ fi φ → f0 φ. (53) E E 1 Proof. By Lemma 2.53, for each i there are C maps gi, hi such that Z ∗ ∗ kDgik∞ ≤ L, kgi − fik∞ → 0, fi φ − gi φ → 0. E and Z ∗ ∗ kDhik∞ ≤ L, khi − f0k∞ → 0, hi φ − f0 φ → 0. E Then

kgi − hik∞ → 0 so by Lemma 2.52 Z Z Z Z ∗ ∗ ∗ ∗ lim fi φ = lim gi φ = lim hi φ = f0 φ. E E E E  n Theorem 2.55. Suppose A1,A2, · · · ∈ K(R ), with Ai → A0 in the Hausdorff metric. Then

N(Ai) → N(A0) weakly, i.e. R φ → R φ for every φ ∈ Ωn−1(S n). N(Ai) N(A0) R Proof. The normal cycle of any A ∈ K may be constructed as follows. For x∈ / A put   x − πA(x) ΠA(x) := πA(x), |x − πA(x)| where πA is the nearest point projection. Let R be large enough that A ⊂ B(0,R). Then

N(A) = ΠA∗[∂B(0,R)]. (54)

Now let R be large enough that all Ai ⊂ B(0,R). Then the restrictions of the πAi to ∂B(0,R) are uniformly Lipschitz, with πAi → πA0 uniformly. It follows that the same is true of the ΠAi , so the conclusion follows from (54) and Theorem 2.54.  Corollary 2.56. The normal cycle of any A ∈ K annihilates multiples of the contact form α and of its exterior derivative dα. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 31

Proof. This is clearly true if A ∈ Ksm. A general A may be approximated in the Hausdorff metric sm by elements of A1, A2, · · · ∈ K , and by Theorem 2.55 the normal cycles N(Ai) → N(A) weakly. n−2 n In particular, if φ ∈ Ω (SR ) then Z Z α ∧ φ = lim α ∧ φ = 0 i→∞ N(A) N(Ai) and a similar relation holds for multiples of dα. 

2.11. Curvature measures and valuations. First variation and the kernel theorem. Let M n be a connected smooth oriented manifold and µ ∈ V(M). Given a vector field V on M, denote by Ft : M → M the flow generated by V . We consider the first variation of µ with respect to V given by

d δV µ(A) := µ(Ft(A)) (55) dt t=0 where A ⊂ M is nice. Clearly µ = 0 iff δµ = 0 and µ({p}) = 0 for some point p ∈ M. n n−1 ∗ If µ = Ψ(θ,ϕ) ∈ Ω (M) × Ω (S M) then the first variation may be represented as follows. Put ˜ ∗ ∗ ∗ Ft := F−t : S M → S M for the corresponding flow of contact transformations of the cosphere bundle. Then F˜ is the flow of a vector field V˜ on S∗M. Then

"Z Z # d δV µ(A) = θ + ϕ ∗ dt t=0 Ft(A) N (Ft(A)) "Z Z # d = θ + ϕ ∗ dt t=0 Ft(A) F˜t∗N (A)) "Z Z # d ∗ ˜∗ = Ft θ + Ft ϕ dt t=0 A N ∗(A) Z Z d ∗ d ˜∗ = Ft θ + Ft ϕ (56) A dt t=0 N ∗(A) dt t=0 Z Z = LV θ + LV˜ ϕ A N ∗(A) Z Z = diV θ + (diV˜ + iV˜ d)ϕ A N ∗(A) Z ∗ = iV˜ (π θ + dϕ) N ∗(A) (57) since π∗N(A) = ∂[[A]] and ∂N(A) = 0. In particular this last expression is independent of the choice of differential forms θ, ϕ representing µ.

Lemma 2.57. ω ∧ iT Dϕ = 0. Proof. This is equivalent to the relation dα ∧ Dϕ = 0. But

dα ∧ Dϕ = d(α ∧ Dϕ) = 0 since α ∧ Dϕ = 0 by construction.  32 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Now we can rewrite (56) as Z ∗ δV µ(A) = iV˜ (π θ + Dϕ) N ∗(A) Z ∗ = π (iV θ) + iV˜ (α ∧ iT Dϕ) N ∗(A) Z ∗ = π (iV θ) + iV˜ (α) ∧ iT Dϕ (58) N ∗(A) Z ∗ = π (iV θ) + hα, V i ∧ iT Dϕ N ∗(A) since N ∗(A) is Legendrian. Since this expression does not involve derivatives of V , this shows that the first variation operator takes values among covector-valued curvature measures. ∗ R Theorem 2.58 (Bernig-Br¨ocker [7]). The valuation Ψθ,ϕ = 0 iff π θ + Dϕ = 0 and ∗ ϕ = 0 Sp M for some point p ∈ M.

Proof. To prove “if”, by finite additivity it is enough to prove that Ψθ,ϕ(A) = 0 for contractible sets A. In this case, if Ft is the flow of a smooth vector field V contracting A to a point p, then ∗ N(Ft(A)) → N({p}) = [[Sp M]]. Hence by (58) Z Ψθ,ϕ(A) = Ψθ,ϕ({p}) = ϕ = 0. ∗ Sp M ∗ Conversely, it is clear that if iT π θ+iT Dϕ annnihilates every Lagrangian subspace of every Qx,v, and if R ϕ = 0, then Ψ = 0. By Lemma 2.57 i π∗θ + i Dϕ determines a primitive element SpM θ,ϕ T T n of each Λ Qx,v. Thus the proof will thus be concluded by Lemma 2.62 below.  2.11.1. A detour into the linear algebra of euclidean symplectic vector spaces. Our main aim in n n this section is to prove the following. As before, we let Q := R ⊕ R , equipped with its standard euclidean and symplectic structures. Lemma 2.59. If 0 6= φ ∈ ΛnQ is primitive then there exists a Lagrangian subspace V ⊂ Q such that φ|V 6= 0. Before we begin it will be necessary to understand more about the geometry of Q, and the operators L, Λ on Λ∗Q in particular. Consider the group G of linear transformations of Q generated by the diagonal action of O(n) g · (v, w) := (gv, gw) and the action of the torus T n

iθ1 iθn (e , . . . , e ) · (v1, . . . , vn, w1, . . . , wn) :=

(v1 cos θ1, . . . , vn cos θn, w1 sin θ1, . . . , wn sin θn).

Clearly the torus action takes each term dxi ∧ dyi to itself, hence leaves the symplectic form ω invariant. Since ω may be expressed as

ω((v1, w1), (v2, w2)) = v1 · w2 − w1 · v2 it follows that the O(n) action does the same. In particular, if g ∈ G then g∗ ◦ L = L ◦ g∗. In fact the analogous relation is true also of the operator Λ. Again, it is clear that each term i i commutes with the torus action. To check that Λ commutes with the O(n) action, consider ∂xj ∂yj NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 33

first the case of φ ∈ Λ2Q. Since Q is euclidean, there is a natural identification of Q with its dual ∗ 2 ∗ ∗ space Q , hence any φ ∈ Λ Q is a sum of terms of the form (v1, w1) ∧ (v2, w2) . Clearly ∗ ∗ Λ[(v1, w1) ∧ (v2, w2) ] = v1 · w2 − w1 · v2 as before, from which it follows that Λ ◦ g = g ◦ Λ when restricted to Λ2Q. To extend this to all 1 degrees, we observe that if θ1, . . . , θk ∈ Λ Q and k ≥ 2 then X ˆ ˆ Λ(θ1 ∧ · · · ∧ θk) = ±Λ(θi ∧ θj) ∧ θ1 ∧ · · · ∧ θi ∧ · · · ∧ θj ∧ · · · ∧ θk. i

Let us say that v1, . . . , vk ∈ Q are orthoisotropic if

ω(vi, vj) ≡ 0, vi · vj ≡ 0 for i 6= j, vi · vi ≡ 1. Lemma 2.60. Any two systems of k orthoisotropic vectors are equivalent under the action of G. Proof. We use induction on k. If k = 1 the statement is simply that G acts transitively on the sphere of unit vectors. To see this note first that the action of the torus, any unit vector v is 0 n 0 equivalent under the action of G to some unit vector v ∈ R ⊕ 0. Now the action of O(n) takes v to any other unit vector in this subspace. Assuming that the statement is true for k, to prove it for k + 1, by the construction of the last paragraph it is enough to show that if both v1, . . . , vk, ∂xn and w1, . . . , wk, ∂xn are orthoisotropic then there is g ∈ G such that gvi = wi and g(en, 0) = (en, 0). Note that the subspace {u ∈ n−1 n−1 Q : u · ∂xn = ω(u, ∂xn ) = 0} is precisely R := R ⊕ R . So we may apply the inductive hypothesis, with the group H ⊂ G of all transformations that fix ∂xn and ∂yn , to arrive at the desired conclusion. 

A subspace V ∈ Grk(Q) is said to be isotropic if ω|V = 0. If k = n then V is Lagrangian. Corollary 2.61. Any isotropic subspace of Q has dimension ≤ n, and is contained in an isotropic subspace of maximal dimension n. In fact we use induction to prove a more general statement. It is a fact that k ≤ n in this case, and if k < n then there exists an isotropic W ∈ Grk+1(Q) with V ⊂ W .Thus a Lagrangian subspace is the same thing as a maximal isotropic subspace. k Lemma 2.62. If 0 6= φ ∈ Λ Q is primitive then there exists an isotropic subspace V ∈ Grk(Q) such that φ|V 6= 0. n n Proof of Lemma 2.62. Recalling that we may take Q = R ⊕ R , we use induction on n, the cases n = 0, 1 being trivial. n−1 n−1 n−1 n ∗ ∗−2 Put R := R ⊕R under the usual embedding R → R , and let λ :Λ R → Λ R denote the Lefschetz operator of R. We decompose a given φ as above as

φ = φ1 + φ2 ∧ dxn + φ3 ∧ dyn + φ4 ∧ dxn ∧ dyn where k k−1 k−2 φ1 ∈ Λ R, φ2, φ3 ∈ Λ R, φ4 ∈ Λ R. Then 0 = Λφ = λφ1 + (λφ2) ∧ dxn + (λφ3) ∧ dyn + (λφ4) ∧ dxn ∧ dyn + φ4 so

λφ2 = λφ3 = λφ4 = 0 (59)

λφ1 + φ4 = 0. (60) 0 If W ∈ Grk−1(R) is isotropic as a subspace of R, then W := W ⊕h∂xn i is isotropic as a subspace of Q. Thus 0 = φ| 0 = φ ∧ dx | , which implies that φ | = 0. Therefore φ = 0 by induction, W 2 n W1 2 W 2 and we conclude similarly that φ3 = 0. 34 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Now suppose U ∈ Grk−2(R) is isotropic in R. By Lemma 2.60, we may assume that U = h∂x1 , . . . , ∂xk−2 i. Recalling that k ≤ n by Corollary 2.61, the subspace

U ⊕ h∂xn−1 + ∂xn , ∂yn−1 − ∂yn i ∈ Grk(Q). is isotropic in Q. Therefore

0 = φ(∂x1 , . . . , ∂xk−2 , ∂xn−1 + ∂xn , ∂yn−1 − ∂yn )

= (λφ1 − φ4)(∂x1 , . . . , ∂xk−2 ).

Comparing with (60), we conclude that φ4 annihilates U. Thus φ4 annihilates every isotropic subspace of R, and since φ4 is primitive by (59), the inductive hypothesis implies that φ4 = 0. So φ = φ1, and the inductive hypothesis implies that φ1 = 0.  The discussion above works equally well in euclidean space and in a smooth connected Riemann- ian manifold M. In this case the contact form α ∈ Ω1(SM) is defined as before, and the Reeb field T is the vector field on SM that generates the geodesic flow, i.e. Tξ ∈ TξSM is the unique tangent vector that is horizontal with respect to the Levi-Civita connection and has π∗Tξ = ξ ∈ TπξM. We can now characterize completely the kernels of the maps Φ:Ωn(M) × Ωn−1(SM) → Curv(M) Ψ:Ωn(M) × Ωn−1(SM) → V(M) Theorem 2.63. ker Φ = 0 × {α ∧ β + dα ∧ γ} (61) Z ∗ ker Ψ = {(θ, φ): iT (π θ + Dφ) = 0, φ = 0} (62) SpM (63) where p ∈ M is an arbitrarily chosen point. 2.12. Kinematic formulas for invariant curvature measures. Let M be a connected Rie- mannian manifold of dimension n and G a Lie group acting effectively and isotropically on M, i.e. acting by isometries and such that the induced action on the tangent sphere bundle SM is transitive. The main examples are the (real) space forms M with their groups G of orientation- n n preserving isometries; the complex space forms CP and CH with their groups of holomorphic n isometries, and M = C with G = U(n); and the quaternionic space forms. Leaving aside the volume, every G-invariant curvature measure Φ ∈ CurvG is the image of some G-invariant element φ ∈ Ωn−1(SM)G. Let L ⊂ H ⊂ G be the subgroups fixing distinguished points o¯ ∈ SM, o = π(¯o) ∈ M. Then it is clear that n−1 G n−1 L Ω (SM) ' Λ (To¯SM) . (64) Thus G n−1 L Curv (M) ' hvolM i ⊕ Λ (To¯SM) /(α, dα). Theorem 2.64. There is a linear map G G G K = KG : Curv (M) → Curv (M) ⊗ Curv (M) such that for any open sets U, V ⊂ M and any sufficiently nice compact sets A, B ⊂ M Z K(Φ)(A, U; B,V ) = Φ(A ∩ gB, U ∩ gV ) dg. (65) G This map is cocommutative, coassociative coproduct. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 35

Corollary 2.65. There is a linear map G G G k = kG : V (M) → V (M) ⊗ V (M) such that for any sufficiently nice compact sets A, B ⊂ M Z k(φ)(A, B) = φ(A ∩ gB) dg. (66) G This map is cocommutative, coassociative coproduct.

n The first example is that of M = R under the action of SO(n). In this case it is not hard ∗ n SO(n) SO(n) n to see that Ω (SR ) is generated by α, dα, κ0, . . . , κn−1. In particular Curv (R ) ' SO(n) n Val (R ), and the operator K is essentially the same as the kinematic operator defined in section 2.2 above. Associated with this are the real space forms Sn ,Hn of constant curvature λ+ λ− λ+ > 0, λ− < 0. n The second example is M = C . Geometrically this is of course simply a euclidean space of n even dimension, but we take the group G to be U(n) := U(n) n C . In this case the kernel of the projection CurvU(n) → ValU(n) is not zero for n ≥ 2. Again there are curved analogues, namely n n CP and CH . Across each of these two families we will see that at the level of curvature measures, the kinematic operators K are in a definite sense identical. This is not so at the level of valuations. The case of the real space forms is essentially classical, but some interesting wrinkles appear once the Alesker product of valuations is introduced below. The case of the complex space forms has only recently been worked out.

k m 2.12.1. Fiber integration. Suppose V ⊂ R is a relatively compact oriented submanifold with finite ∗ m n ∗−k n volume. There is then an operation πV ∗ :Ω (R × R ) → Ω (R ) called fiber integration, m n defined as follows. Denote the coordinates on R by x1, . . . , xm and on R by y1, . . . , yn. If #I = k n we put for y0 ∈ R Z 

(πV ∗[f(x, y)dxI dyJ ])y0 := f(x, y0) dxI dyJ . M If #I 6= k then the left hand side is defined to be 0. Now extend by linearity. This operation is well-defined, i.e. independent of choice of coordinates. In fact it even works in situations more general than products, viz. smooth fiber bundles. We recall the definition. Let M,F be smooth manifolds and G a Lie group acting by diffeomorphisms on F . Suppose E is a smooth manifold, equipped with a surjective submersive map π : E → M, such that M is covered by a family of open sets U for which π−1U ' U × F. (67) This equivalence is required to intertwine the projections to U ⊂ M. In addition, given two open sets U, V of this type the resulting transition map

tUV : U × F → V × F is required to have the form (p, x) 7→ (p, g(p) · x) where g : U ∩ V → G is smooth. This structure is called a smooth fiber bundle with base M, fiber F and group G. The space E is the total space of the bundle. Now suppose that V ⊂ F is as above, and furthermore that g|V is a diffeomorphism of V for each g ∈ G. Then the fiber integral over V , defined locally via the local product structure (67), is well defined. 36 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

2.12.2. Construction of the kinematic operator KG. Let G be a connected Lie group that acts transitively on the sphere bundle SM. Let Go ⊂ G be the stabilizer of o ∈ M and Go¯ ⊂ Go be the stabilizer ofo ¯ ∈ SoM. We construct a smooth fiber bundle B over SM × SM with fiber n−1 F := Go × S . For ξ ∈ SM, denote by [ξ] the projection of ξ to M. The total space of our bundle is then E := {(g, ξ, η, ζ) ∈ G × (SM)3 :[ξ] = g[η] = g[ζ]} The projection takes such a point to (ξ, η) ∈ (SM)2. In the literal sense, the fiber is obviously Fξ,η,ζ := {(g, ζ): g[η] = g[ζ] = [ξ]}, which is diffeomorphic to F via maps of the following type. Let g0, g1 ∈ G such that g0o = [ξ], g1o = [η], and take F → Fξ,η,ζ by

−1 (h, v) 7→ (g0hg1 , g1v).

The group of this bundle is Go¯, with the diagonal action k · (h, v) := (kh, k · v). (68)

To see this, observe that for given ξ, η there is a distinguished family of diffeomorphisms as above 0 0 −1 such that g0o¯ = ξ ∈ SM. Notice that g0o¯ = ξ iff g0g0 ∈ L. Thus if we have two local trivializations of B, where the fiber maps are of this restricted type, then over each point the transition map is given by map of the type (68), i.e. the group may be taken to be L, as claimed. It follows that any suitable Go¯-invariant submanifold C ⊂ F gives rise to a well-defined family of submanifolds living in the fibers of E → (SM)2, and thereby to a fiber integration operator ∗ ∗−dim C πC∗ :Ω (E) → Ω (SM × SM). The one that works is

C := {(h, ζ) ∈ Go × SoM : ho¯ 6= −o,¯ ζ ∈ o,¯ ho¯} (69) where o,¯ ho¯ denotes the minimizing geodesic on SoM joining the two indicated points. This manifold has finite volume and compact closure, and inherits a natural orientation from those of Go and SoM, by orienting these minimizing geodesics to run fromo ¯ to ho¯. The first main point is that, given nice subsets A, B ⊂ M, we can use C to build a formal picture of the family of normal cycles N(A ∩ gB) as g varies over the group G. Put p : E → G × SM, p(g, ξ, η, ζ) := (g, ζ). Let us say that A, B ⊂ M meet transversely if

ξ ∈ N(A), η ∈ N(B), πξ = πη =⇒ ξ 6= −η.

Proposition 2.66. Let A, B ⊂ M be nice subsets. Then A and gB meet transversely for a.e. g ∈ G, with

−1 −1 N(A ∩ gB) = (N(A) ∩ π gB) ∪ (gN(B) ∩ π A) ∪ (p(N(A) × N(B) ×B C)). (70) Proof. The transversality statement is just a dimension count: if dim M = n, then

dim SM = dim G − dim Go¯ = 2n − 1.

The set of g ∈ G taking a given η ∈ SM to another point −ξ ∈ SM is a coset of Go¯, hence the set S of all g such that A, gB do not meet transversely is the image under a smooth map of a bundle over N(A) × N(B) with fibers diffeomorphic to Go¯. Thus dim S ≤ 2n − 2 + dim Go¯ < dim G. Tracing through the maps, the given portrayal of N(A ∩ gB) is just our standard picture of the normal cycle of the intersection of two sets X,Y in general position: there is the part of N(X) that sits over the interior of Y , and vice versa, plus the union of the arcs interpolating between normals ξ ∈ N(X), η ∈ N(Y ) at which [ξ] = [η] ∈ ∂X ∩ ∂Y .  NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 37

Conclusion of the proof of Theorem 2.64. We now have a diagram p E −−−−→ G × SM     (71) y y SM × SM −−−−→ M × M (The bottom right corner is not important for us at this point, but will be useful soon. Here the projection G×SM → M ×M is (q, ζ) 7→ (g[ζ], [ζ]). Thus E is precisely the fiber product of G×SM and SM × SM with respect to the projections to M × M.) The second and final main point is that that G × G acts on E by (g0, g1) · (g, ξ, η, ζ) := −1 (g0gg1 , g0ξ, g1η, g1ζ). Clearly there are G × G actions on G × SM and SM × SM that are intertwined by these maps, i.e. the diagram is G × G equivariant. Furthermore G × G stabilizes the family of submanifolds C of the fibers. Let dg denote the (biinvariant) volume form of G. Given φ ∈ Ωn−1(SM)G, we thus obtain dg ∧ φ ∈ Ω∗(G × SM)G×G. Pulling this back to E via p and ∗ G×G pushing down to SM × SM via the fiber integral πC∗, we obtain an element of Ω (SM × SM) . Projecting to the subspace Ωn−1(SM)G ⊗ Ωn−1(SM)G, Proposition 2.66 implies that the resulting map descends to give a map CurvG → CurvG ⊗ CurvG. The kinematic operator K is then the sum of this map with Φ 7→ Φ ⊗ vol + vol ⊗Φ.  2.13. The transfer principle. Let the isotropic space (M,G) be given. Then the stabilizer Go ⊂ G of a point o ∈ M acts transitively on the sphere SoM, so the affine space ToM, equipped with the euclidean structure coming from the Riemannian metric on M, is isotropic under the action of Go := Go n ToM. Thus the space of Go-invariant curvature measures on this euclidean space is again a coalgebra as above. We show here that this coalgebra is canonically isomorphic to the original coalgebra CurvG(M). We first show that the two are naturally isomorphic as vector spaces. The Riemannian connection induces a Go¯-invariant decomposition ⊥ To¯SM = H ⊕ V ' ToM ⊕ o¯ (72) into horizontal and vertical subspaces, where πM∗ induces an isomorphism H → ToM and V is ⊥ naturally isomorphic too ¯ ⊂ ToM. Since ⊥ To¯S(ToM) = ToM ⊕ To¯SoM = ToM ⊕ o¯ (73) we get a natural identification To¯SM ' To¯S(ToM). Note also that the natural actions on these spaces of the stabilizer Go¯ ⊂ Go ofo ¯ are intertwined by this identification. We thus obtain an isomorphism ∗ Go¯ ∗ Go¯ (Λ To¯SM) ' (Λ To¯S(ToM)) (74) between the respective subalgebras of Go¯-invariant elements. This in turn induces an isomorphism

∗ G ∗ Go Ω (SM) ' Ω (SToM) (75) between the spaces of invariant differential forms on the sphere bundles. Proposition 2.67. This isomorphism induces an isomorphism

G Go τ : Curv (M) ' Curv (ToM) (76) between the corresponding spaces of invariant curvature measures.

Proof. Puttingα ¯ for the canonical 1-form of S(ToM), by Theorem 2.63, p it is enough to show that α, α¯ and dα, dα¯ correspond to each other under the identification (74). That this is true of α andα ¯ is trivial. Note that the second correspondence is not a formal consequence, since the isomorphism (75) only respects the linear structure but does not intertwine 38 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 the exterior derivative. Representing elements of To¯S(ToM) as indicated by the decomposition (73), we have dα¯((v1, w1), (v2, w2)) = w1 · v2 − v1 · w2. (77) In order to compare this with dα¯, let ξ, η be smooth vector fields defined in a neighborhood of o¯ ∈ SM. Assume that π∗ξ, π∗η are everywhere linearly independent. Then there exists a smooth unit vector field Z defined in a neighborhood of o such that ξo¯, ηo¯ are tangent to the graph of Z at o¯. Define vector fields X,Y on a neighborhood of o ∈ M by X(p) := π∗ξ(Z(p)),Y (p) := π∗η(Z(p)). With respect to the decomposition (72), we may write

ξZ = (X, ∇X Z), ηZ = (Y, ∇Y Z). Since αZ (ξ) = Z · π∗ξ. we may compute

(dα)Z (ξ, η) = Dξ(Z · π∗η) − Dη(Z · π∗ξ) − Z · π∗[ξ, η]

= DX (Z · Y ) − DY (Z · X) − Z · [X,Y ]

= (∇X Z) · Y − (∇Y Z) · X. Comparing with (77), this concludes the proof under the linear independence assumption above. The general case follows by continuity.  If we normalize the Haar measure dg on the group G so that dg({g : go ∈ S}) = vol(S) (78) for measurable S ⊂ M, then Theorem 2.68. The correspondence (76) is an isomorphism of coalgebras.

Proof. Since (To¯SToM, Go) is again an isotropic pair, we have another diagram similar to (71) in this case. Note that the proof of Theorem 2.64 actually constructs a map H :Ω∗(SM)G → Ω∗(SM)G ⊗ Ω∗(SM)G In view of (64), this is equivalent to a map

∗ Go¯ ∗ Go¯ ∗ Go¯ H˜ :Λ (To¯SM) → Λ (To¯SM) ⊗ Λ (To¯SM) . (79)

∗ Go¯ In the affine case we get similar map involving Λ (To¯SToM) , the corresponding space for our affine model. Since the two are canonically isomorphic, and by Proposition 2.67 this isomorphism respects the projection to curvature measures, we need to check that the H˜ maps correspond. To this end we consider the derived diagram TE| −−−−→ T (G × SM)| = TG| × TSM| F F Go SoM    p (80) y y π To¯SM ⊕ To¯SM −−−−→ ToM ⊕ ToM where F = Go×SoM is the fiber over (¯o, o¯), which includes the submanifold C. The map (79) is then constructed as follows. An element of Λ∗(T SM)Go¯ induces a G -invariant section of Λ∗ TSM| . o¯ o SoM We take wedge product with the invariant volume form of G, pull back via the top map, and finally push down using the fiber integral over C. Likewise we have the analogous diagram for the euclidean model:

˜
T E −−−−→ T Go × SM = T Go × TSToM|S M . F F Go o     (81) y p˜y π˜ To¯SToM ⊕ To¯SToM −−−−→ ToM ⊕ ToM NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 39

The analogue of (79) for the isotropic euclidean space (ToM, Go) is constructed in the same way. The identification To¯SM ' To¯SToM that arises from (72) and (73) induces an isomorphism between the lower left spaces, intertwining the projections π, π˜. In the same way we may identify the second factors of the top right spaces, intertwining the restrictions of p, p˜. Finally, we wish to identify the first factors of the top right spaces. The identification we give will not be canonical, but it is enough for our purposes. To this end we realize Go the cartesian product Go × ToM, acting on ToM by

(g, v) · w := g∗(w + v).

Thus T Go ' TGo ⊕ ToM. Recalling thatp ˜ is the derivative of the map (g, ζ) 7→ (g[ζ], [ζ]) G0 from map on the right of the euclidean analogue of (71), we see that for (g, ζ) ∈ Go × SoM, and (γ, v) ∈ TGo × ToM, w ∈ Tζ SToM,

p˜(γ, v; w) = (g∗π∗w + g∗v, π∗w) (82) where π : SToM → ToM is the projection. We may represent TG| in a similar fashion. Let m ⊂ g be a linear complement to g in the Go 0 Lie algebra g = TeG. Then π∗ induces an isomorphism m → ToM where π : G → M = G/Go is the quotient map. Pushing m forward by the derivative of left multiplication by g ∈ Go, we obtain an injection −1 π∗ Lg∗ ToM −−−−→ m −−−−→ g∗m ⊂ TgG. which gives an identification TG| ' TG ⊕ T M. With respect to this decomposition, the map Go o o p acts formally exactly as in (82). Thus the diagrams (84), (81), except for the top left corners, may be identified, intertwining all of the maps that appear. It follows that the top left corners may also be identified: Lemma 2.69. Let M,E0 be smooth fiber bundles over N, and E their fiber product: E −−−−→ E0    p y y (83) f M −−−−→ N Let m ∈ M and F the fiber of E0 over f(m), which is the same as the fiber of E over m. Then the 0 restricted tangent bundle TE|F is the fiber product over Tf(m)N of TmM and TE |F : 0 TE|F −−−−→ TE |F     y y (84)

TmM −−−−→ Tf(m)N Proof. Since f, p are submersions, the implicit function theorem implies that the original fiber product E = {(m, e) ∈ M × E0 : f(m) = p(e)} is a smooth submanifold, with 0 T(m,e)E = {(v, w) ∈ TmM ⊕ TeE : f∗v = p∗w}. Thus 0 TE|F = {(m, v; e, w) ∈ TmM ⊕ TE F : f∗v = p∗w} as claimed.  40 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Thus the maps (79) likewise correspond, provided the selected volume forms dg correspond when restricted to Go. But this is ensured by the convention (78): each corresponding volume form is then the product of the probability measure on Go with the pullback of the volume form of M under the projection G → M. (This shows, incidentally, that the choice of the complement m above was immaterial.)  2.14. Integral geometry of real space forms. Let us see what all this tells us about the classical n case of the real space forms. Put Mλ for the space form of curvature λ; this is a sphere if λ > 0 and a hyperbolic space is λ < 0. The main point is that in either of these cases the stabilizer Go is n isomorphic to SO(n), acting on T0M ' R in the usual way. Thus all of the kinematic coalgebras over all choices of curvature λ are isomorphic to the colgebra arising in curvature λ = 0, and hence to each other. ∗ n SO(n) Proposition 2.70. Ω (SR ) is generated by α, dα and n forms given at the point (0, en) as

κ0 = dv1 ∧ · · · ∧ dvn−1,

κ1 = dx1 ∧ dv2 ∧ · · · ∧ dvn−1 + dv1 ∧ dx2 ∧ dv3 · · · ∧ dvn−1 + ··· + dv1 ∧ . . . dvn−2 ∧ dxn−1, ...

κn−1 = dx1 ∧ · · · ∧ dxn−1,

n ∗ SO(n) Proof. Since To¯SR ' R ⊕ Q as above, it is enough to show that Λ (Q) is generated by the symplectic form ω and the κi. We use induction on n, the case n = 1 being trivial. Let φ ∈ Λk(Q)SO(n). Since the Lefschetz operator L :Λk−2Q → ΛkQ is surjective if k > n, we 0 00 may assume that k ≤ n. Writing φ in terms of the basis dxI dyJ , we split it as φ = φ + φ where φ0 is the sum of all terms omitting some index i entirely and φ00 is the sum of the others. Since 00 k ≤ n, by invariance it is clear that φ is a linear combination of the κi. Consider the sum σ of all terms of φ0 that omit the index n. By induction, σ is expressible in 0 0 ∗ n−1 n−1 SO(n−1) terms of the corresponding forms ω , κi ∈ Λ (R ⊕ R ) . The latter cannot occur, since 0 0 0 0 the maps ei 7→ −ei, en 7→ −en, i < n, belong to SO(n), while taking κi 7→ −κi, ω 7→ ω . Thus σ is a multiple of some power of ω0. The same is true for any other omitted index, and by invariance under diagonal elements of SO(n) we have the same multiple of the same power in each case. It 0 follows that φ is a multiple of a power of ω.  n SO(n) Denote the space of SO(n)-invariant curvature measures on R by Curv . By the transfer SO(n) n principle, we may think of Curv as living simultaneously in each of the Mλ , and equipped with a single comultiplication SO(n) SO(n) SO(n) Kn : Curv → Curv ⊗ Curv n n independent of λ. Denote the space of invariant valuations on Mλ by Vλ . As previously noted, SO(n) n the projections Curv → Vλ are all isomorphisms. We may also take the inverse limit of the CurvSO(n) under restriction to obtain the infinite-dimensional vector space CurvSO(∞). Taking λ = 0, let t be the generator of ValSO(∞) defined in Section 2.7 and normalized as in (39). SO(∞) i Denote by Ti ∈ Curv the curvature measure that corresponds to t under the isomorphism SO(∞) SO(∞) Curv → Val . Then Ti induced by some multiple of κi, although the constant will in general depend on n. Since the transfer principle is based on the decomposition of the tangent n space of a sphere bundle into its horizontal and vertical parts, if A ⊂ Mλ is a compact smooth domain then Z Ti(A, E) = ci σn−i−1(k1, . . . , kn−1) E∩∂A for some constant ci, independent of λ. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 41

SO(n) n For Ψ ∈ Curv , put [Ψ]λ ∈ Vλ for the corresponding valuation in curvature λ. Put λ τi := [Ti]λ. 0 i Thus τi = t . We can use the transfer principle to compute the relations between the ti and the intrinsic SO(n) volumes µi ∈ Val . Put Φi := ciTi, where ci is chosen so that 1 i i ciτi (S1) = [Φi]1(S1) = 2. From the original normalization of t, 1 1 τ1 (S1 ) = 4, so c1 = 2. Note also that c0 = 1. Proposition 2.71. i+1 i i!ωi 2 t = i µi = µi. (85) π αi Proof. Using the totally geodesic spheres as templates, we compute Z n i j i j i+j k1 ([Φl]1)(S ,S ) = [Φl]1(S ∩ gS ) dg = 2αnδn+l SO(n+1) under our usual convention for the Haar measure on the group. Thus

αn X K (Φ ) = Φ ⊗ Φ . n l 2 i j i+j=n+l n Transferring back to R , αn X kn(χ) = kn([Φ ] ) = [Φ ] ⊗ [Φ ] . 0 0 0 0 2 i 0 j 0 i+j=n n Since k0 is multiplicative, αn X αn X ([Φ ] ⊗ χ) · [Φ ] ⊗ [Φ ] = ([Φ ] ⊗ χ) · kn(χ) = kn([Φ ] ) = [Φ ] ⊗ [Φ ] . 2 l 0 i 0 j 0 l 0 0 0 l 0 2 i 0 j 0 i+j=n i+j=n+l

It follows that [Ψi]0 · [Ψj]0 ≡ [Ψi+j]0, so i i i t = 2 [ψ1]0 i = 2 [Ψi]0 2i+1 = µi αi whence the second relation of (85) follows. The first then follows from the identity 2n+1πn ω ω = . n n+1 (n + 1)!

 Corollary 2.72.   i + j ωi+j µi · µj = µi+j. i ωiωj 42 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Remark. The relations (85) may also be expressed

X 1 X µi exp(πt) = ωiµi, = . (86) 2 − t αi n For λ > 0 there are two other natural bases for V (Sλ ). The first we have seen before: since n n+1 SO(n+1) n+1 Sλ ⊂ R , we may restrict the elements of Val (R ) to the sphere to obtain elements 2 n n i i i j t, t , . . . , t ∈ V (Sλ ). Since t (Sλ) 6= 0 and t (Sλ) = 0 for i > j, it follows that together with χ they constitute a basis. n For the second, note first that just as in the case of R there is a natural way to multiply elements n n of V (Sλ ). Given a nice set A ⊂ Sλ , put Z µA := χ(gA ∩ · ) dg. SO(n+1) n Clearly µA ∈ V (Sλ ). Now define, for any valuation ν on the sphere, Z µA · ν := ν(gA ∩ · ) dg. SO(n+1) Just as in the euclidean case it is easy to see that this gives a well-defined product on valuations, i.e. does not depend on the representation as above. Note that the ti are naturally expressed as integrals of measured families of valuations µA: they are given by intersections with generic planes n+1 n of codimension i in R , or alternatively for subsets of Sλ by measured intersections with all subspheres (not only the totally geodesic ones) of codimension i. In particular, the restriction map SO(n+1) n+1 n Val (R ) → V (Sλ ) is a homomorphism of algebras. The Euler characteristic is again the multiplicative identity, and by Fubini’s theorem the kine- matic operator is again multiplicative: n Proposition 2.73. If ν, µ ∈ Vλ then n kλ (ν · µ) = (ν ⊗ χ) · kn(µ). λ Now consider the space Grn−1 of geodesic hyperspheres H, equipped with its invariant Haar measure, normalized so that their total measure is √2 , and put λ Z φ := χ( · ∩ H) dH. λ Grn−1 Then Z φi := χ( · ∩ J) dJ λ Grn−i i − i where the Haar measure dJ now has total measure 2 λ 2 .

λ i i 2.14.1. Relations between the bases τi , φ , t . We have shown αn X K (T ) = T ⊗ T , l = 0, . . . , n. (87) n l 2n+1 i j i+j=n+l λ j n n Taking λ > 0, the values of the τi on the totally geodesic spheres S (λ) ⊂ S (λ) = Mλ are  2 i τ λ(Sj(λ)) = δi · 2 √ . i j λ Therefore n b 2 c i X λ χ = τ (88) 4 2i i=0 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 43

n in Mλ , λ > 0, and by analytic continuation this formula holds for λ ≤ 0 as well. n Therefore the principal kinematic formula in Mλ may be written b n c   2  i ∞  j X λ αn X X λ kn(χ) = kn(τ λ ) = τ λ ⊗ τ λ . (89) λ 4 λ 2i 2n+1 i  4 n−i+2j i=0 i j=0

Returning to the case λ > 0, let φ denote the invariant valuation given by √2 times the average λ Euler characteristic of the intersection with a totally geodesic hypersphere. Then k n  2  λ 2 Z φk := √ χ(· ∩ gSn−k(λ)) dg λ αn SO(n+1) k n−k 2 λ 2 n−k = kSn(λ)(χ)(·,S (λ)) (90) αn ∞ j X λ = τ λ , k = 0, . . . , n. 4 2j+k j=0 In particular, if N k is a compact piece of a totally geodesic submanifold of dimension k, then k+1 k λ 2 φ (N) = τk (N) = |N|. αk Comparing (88) and (90) we find that λ χ = τ λ + φ2 (91) 0 4 or Z λ λ τ0 = χ − χ(· ∩ H) dH. (92) 4 Grλ,n−2 Now the principal kinematic formula (89) may be written αn X kn(χ) = τ λ ⊗ φj. (93) λ 2n+1 i i+j=n By the multiplicative property, αn X   kn(τ ) = τ λ ⊗ φj · τ λ . (94) λ k 2n+1 i k i+j=n Comparing with (87) we find that j λ λ φ · τi = τi+j (95) This is the reproductive property of the τi. At this point we may write   X i j kn(ψ) = (ψ ⊗ τ0) ·  φ ⊗ φ  (96) i+j=n n whenever ψ is an invariant valuation on Mλ . By analytic continuation these formulas are valid for λ ≤ 0 as well, where φ is the corresponding integral over the space of totally geodesic hyperplanes, 2 normalized so that φ(γ) = π |γ| for curves γ. λ By (90), the valuations φ, τ1 and the generator t of the Lipschitz-Killing algebra all coincide if the curvature λ = 0. For general λ, the relations (91) and (95) yield λ τ λ = φi − φi+2. (97) i 4 44 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

n There is also a simple general relation between φ and t. To put it in context, denote by Vλ the n n n+1 algebra of invariant valuations on Mλ . Since each Mλ embeds essentially uniquely into Mλ , and n n+1 every isometry of Mλ extends to an isometry on Mλ , there is a natural surjective restriction n+1 n ∞ ∞ homomorphism Vλ → Vλ . Put Vλ for the inverse limit of this system. Thus Vλ is isomorphic to the ring of formal power series in one variable, which may be taken to be either φ or t. The ∞ valuations τi may also be regarded as living in Vλ since they behave well under the restriction ∞ maps. The relations among valuations given above are also valid in Vλ except for those that depend explicitly on the kinematic operators. These depend on the dimension n and definitely do ∞ not lift to Vλ . Proposition 2.74. φ t t = , φ = (98) q λφ2 q λt2 1 − 4 1 + 4 Proof. We use the template method to prove that ! φ2 λ λ2 t2 = = φ2 1 + φ2 + φ4 + ... (99) λφ2 4 4 1 − 4 for λ > 0, so t must be given by the square root that assigns positive values (lengths) to curves. The generalization to all λ ∈ R follows by analytic continuation. Since everything scales correctly it is enough to check this for λ = 1. In fact it is enough to check that that the values of the right- and left-hand sides of (99) agree on spheres S2l of even dimension. Clearly ( 2 · 4k, k ≤ l φ2k(S2l) = 0, k > l so the right-hand side yields 8l. To evaluate the left-hand at S2l we recall that 4ω 4 t2(S2l) = 2t2(B2l+1) = 2 µ (B2l+1) = µ (B2l+1). π2 2 π 2 2l+1 Meanwhile, Theorem 2.1 for the volume of the r-tube about B 1 yields λ− 2 2l+1 X 2l+1 2l+1−i ω2l+1(1 + r) = ω2l−i+1µi(B )r . Equating the coefficients of r2l−1 we obtain   2l+1 ω2l+1 2l + 1 µ2(B ) = = 2πl ω2l−1 2 in view of the identity ω 2π n = . ωn−2 n  2.14.2. Analytic continuation. These formulas also hold if λ < 0, although there is a subtle difficulty in dealing with Crofton formulas in hyperbolic spaces. These are essentially kinematic formulas for certain noncompact objects. In the euclidean case it is easy to derive them from the standard kinematic formula by replacing affine planes by large balls within them, renormalizing the integrals, and passing to the limit, but this device does not work in hyperbolic space: the problem is that the perimeter of a large hyperbolic ball grows at the same (exponential) rate as its volume. The way around this is to use analytic continuation in the curvature λ, as follows. The main point n is that we may find local coordinates on Mλ that work for all λ simultaneously. Furthermore these NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 45 coordinates have the remarkable property that the hyperplanes (i.e. totally geodesic hypersurfaces) appear as ordinary euclidean hyperplanes. − 1 For the sphere of curvature λ (and radius R = λ 2 ) the coordinates only cover the north- ern hemisphere, and are given by the radial projection onto the tangent plane at the north pole (0,..., 0,R). It is clear that this map takes totally geodesic spheres to affine planes. One easily writes down the formula     n R R R R n Sλ 3 (p1, . . . , pn, pn+1) 7→ p1,..., pn,R 7→ p1,..., pn ∈ R pn+1 pn+1 pn+1 pn+1 for this map, and (x, R) x 7→ R p|x|2 + R2 n−1 n for its inverse. In terms of polar coordinates (r, v) 7→ rv, R+ × S → R , the induced metric on n R is dr2 r2dv2 dr2 r2dv2 ds2 = + = + r2 2 r2 (1 + λr2)2 1 + λr2 (1 + R2 ) 1 + R2 where dv2 is the usual metric on the unit sphere. The induced volume form is n−1 n−1 dr r d volSn−1 r d volλ = 2 n−1 = n+1 dr d volSn−1 (100) 1 + λr (1 + λr2) 2 (1 + λr2) 2 with total volume Z ∞ n−1 n αn r − 2 λ = αn−1 n+1 dr (101) 2 0 (1 + λr2) 2 From the geometric description of these coordinates it is clear that the image of any totally n n geodesic subsphere of dimension k of Sλ is an affine plane of the same dimension in R : the n subsphere is the intersection of Sλ with a (k + 1)-plane P through the origin, and its projection n to the tangent plane T at the north pole is P ∩ T . Therefore the invariant measure on Grn−1(Sλ ) n induces a measure on the affine Grassmannian Grn−1(R ). n Let us compute this induced measure in terms of the usual coordinates ϕ : Grn−1(R ) → [0, ∞)× Sn−1: recall that if P¯ = {x : v · x = t} for |v| = 1 and t > 0 then ϕ(P¯) = (v, t). Up to scale, the n n measure on Grn−1(Sλ ) may be obtained by pulling back the measure from Sλ via the ⊥ map. A simple picture shows that the composition n−1 n n n n n−1 [0, ∞) × S → Grn−1(R ) → Grn−1(Sλ ) → Sλ → R → (0, ∞) × S R2 1 is simply the map (t, v) 7→ ( t , v) = ( λt , v). Pulling back the volume form (100) via this map we obtain dt d volSn−1 n−1 n+1 . λ 2 (1 + λt2) 2 In view of (101) and our standing convention that the total measure of Gr (Sn) be √2 , we n−1 λ λ renormalize to obtain the measure 4 dt d volSn−1 n+1 . (102) αn (1 + λt2) 2 Now all of these formulas may be extended verbatim to λ ≤ 0. Setting λ = 0 gives the usual − 1 euclidean metric. If λ < 0 then this same expression gives a metric on the ball of radius (−λ) 2 ; n this is the Klein model for Hλ . By analytic continuation, the curvature of this metric is λ. In these coordinates all of the various totally geodesic hypersurfaces appear as hyperplanes (or their n intersections with the disk) of R , and the expression (102) gives a measure on the corresponding Grassmannian. 46 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

It remains to show that the measure (102) is invariant under the isometry group for λ < 0. Again this can be accomplished using analytic continuation at the level of Lie algebras, starting in positive curvature λ > 0. Recall that the Lie algebra so(n+1) = hEij −Ejii : 1 ≤ i, j ≤ n+1. Each n element induces a vector field on each Sλ , which integrates to an isometry, and these isometries generate the whole isometry group. Via the coordinate map we also get corresponding vector fields n on R , although they generally do not generate global flows since the induced metric is incomplete. Nevertheless, the isometry property is reflected in the fact that the Lie derivative of the induced metric under one of these flows vanishes. n Clearly so(n + 1) ' so(n) ⊕ R , where so(n) corresponds to the subgroup S)(n) ⊂ SO(n + 1) n fixing the north pole (0,..., 0,R) of the sphere, and R ' hEi,n+1 − En+1,i : i = 1, . . . , ni. The so(n) part is not a problem, since these elements do integrate to rotations, and act in the same n way regardless of the curvature λ. So let us consider the n-parameter family of vector fields on R induced by the second factor. By symmetry it is enough to consider the 1-parameter family induced n by the 1-dimensional subspace hE1,n+1 − En+1,1i. The flow on Sλ induced by E1,n+1 − En+1,1 is Ft(p1, . . . , pn+1) = (p1 cos t + pn+1 sin t, p2, . . . , pn, −p1 sin t + pn+1 cos t). n so the induced flow on R is R Gt(x1, . . . , xn) = (x1 cos t + R sin t, x2, . . . , xn) (−x1 sin t + R cos t) generated by the vector field

d x1 Gt(x) = (x1, . . . , xn) + (R, 0,..., 0) dt t=0 R 1 − 1 = x1λ 2 (x1, . . . , xn) + (λ 2 , 0,..., 0) − 1 = λ 2 [(1, 0,..., 0) + λx1(x1, . . . , xn)].

Reparametrizing in time, the Lie derivative of the metric with respect to the vector field Xλ := (1, 0,..., 0) + x1λ(x1, . . . , xn) vanishes for λ > 0. Now analytic continuation applies, and we conclude that the same is true for λ ≤ 0. Finally, since we know a priori that the flows for λ > 0 induce measure-preserving flows on the ˜ n spherical Grassmannians, each Xλ induces a vector field Xλ on Grn−1(R ), and the Lie derivatives of the corresponding volume forms (102) also vanish. It now follows by analytic continuation that this is still true for λ ≤ 0. Strictly speaking, we should check that the X˜λ also depend analytically on λ. To see this, let ¯ n Xλ denote the induced flow on the cosphere bundle S ∗ R , which clearly depends analytically on λ ∗ n n ˜ ¯ by the chain rule. But S R projects canonically to Grn−1(R ), and Xλ is the image of Xλ under the derivative of this projection.

3. Alesker theory for affine spaces n 3.1. Alesker irreducibility and its consequences. It is clear that GL(n, R) acts on Val(R ) by (gµ)(A) := µ(g−1A)  and in fact this action stabilizes each McMullen summand Valk, k = 0, . . . , n,  = ±. The McMullen decomposition also implies that Val is a Banach space under the norm kµk := sup |µ(A)| . A⊂B(0,1) Theorem 3.1 (Alesker irreducibility). Each of these summands is irreducible as a GL(n) repre- sentation.  In other words, there are no nontrivial closed subspaces of Valk that are stabilized by GL(n). NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 47

Corollary 3.2. The subspace of valuations

Z n n−1 n R νϕ := ϕ, ϕ ∈ Ω (SR ) N(·) L is dense in k

Recall that in section 2.8 we constructed valuations µm for measures m on K, and defined a natural product on their span, and that the product respects the grading. Denote by M ⊂ Val the space of all such valuations. From the asymmetric form (41) for the product, it follows that for fixed µ ∈ M the map ν 7→ µ · ν is continuous. In fact

kµ · νk ≤ Cµ kνk (103) for some constant Cµ depending only on µ. Furthermore Lemma 3.5. For µ, ν ∈ M and g ∈ GL(n) g(µ · ν) = gµ · gν.

Proof. It is enough to check this for µ = µA, ν = µB, in which case −1 2 gµA · gµB = | det g | µgA · µgB Z Z = | det g−1|2 χ( · ∩ (x + gA) ∩ (y + gB)) dx dy R R Z Z = | det g−1|2 χ( g−1(·) ∩ (g−1x + A) ∩ (g−1y + B)) dx dy R R Z Z = | det g−1|2 χ( g−1(·) ∩ (ξ + A) ∩ (η + B)) d(gξ) d(gη) R R Z Z = χ( g−1(·) ∩ (ξ + A) ∩ (η + B)) dξ dη R R = g(µA · µB).  48 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Since Valn = hvoli ' R, we may define a symmetric pairing on M by

(µ, ν) := (µ · ν)n. As above, for fixed µ this pairing is a continuous function of ν. By Lemma 3.5, it is equivariant with respect to the GL(n) action: (gµ, gν) = | det g−1|(µ, ν). (104)

Note that the components of each µ = µm with respect to the McMullen decomposition have the same form. In fact n X i µtA = t µA,n−i; i=0 Pn i since, for any polynomial p(t) = i=0 cit of degree n n n−1  X X i ∆p(t) := p(t + 1) − p(t) = c tj i  j  i=1 j=0 it follows that the µA,n−i may be expressed as a linear combination of µ0·A = vol, µA, . . . , µnA with coefficients independent of A. To separate into the even and odd parts we use the relations 1 1 µ+ = (µ + µ ), µ− = (µ − µ ). A 2 A −A A 2 A −A

Corollary 3.6 (Alesker Poincar´eduality). This pairing is perfect, i.e. for any µ = µm 6= 0 there exists µm0 such that (µm, µm0 ) 6= 0.  ¯ Proof. Given k,  let N ⊂ Mk be the subspace annihilated by the pairing. By (103), N ∩ M = N.  Furthermore, the GL(n) action stabilizes Mk, and by equivariance it also stabilizes N. Therefore ¯  by Alesker irreducibility N is either 0 or all of Valk. So the proof may be concluded by showing k + that µ ∈ M 6= N. Since the intrinsic volume µk ∈ Mk − N, this concludes the proof in the case  = +.  For the rest of this section we assume that G ⊂ SO(n) is a closed subgroup acting transitively on Sn−1. Because of Corollary 3.3, the proof of Theorem 2.11 yields Theorem 3.7. There exists a cocommutative, coassociative coproduct G G G kG : Val → Val ⊗ Val such that for µ ∈ ValG Z µ(A ∩ gB¯ ) dg¯ = kG(µ)(A, B) G for any A, B ∈ K. Corollaries 3.4 and 3.3 implies that ValG is spanned by valuations of the form Z G µA := χ( · ∩ gA¯ ) dg¯ ∈ M,A ∈ K. G Corollary 3.6 implies that for every µ ∈ ValG there is ν ∈ M such that (µ, ν) 6= 0. In view of (104), (µ, gν) = (µ, ν) for all g ∈ G, and by linearity and continuity Z (µ, gν dg) = (µ, ν). G R G G G∗ Since G gν ∈ Val , it follows that there is a Poincar´eduality isomorphism p : Val → Val such that (µ, ν) = hpµ, νi. Now the proof of Theorem 2.40 yields the following. NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 49

Theorem 3.8 (The ftaig). Let m : ValG ⊗ ValG → ValG denote the multiplication map above and ∗ m∗ : ValG → ValG∗ ⊗ ValG∗ its adjoint. Then the following diagram commutes: k ValG −−−−→G ValG ⊗ ValG   p p⊗p (105) y y ∗ ValG∗ −−−−→m ValG∗ ⊗ ValG∗ Finally, Corollary 3.2 implies Theorem 3.9. The projection CurvG → ValG is surjective, and the diagram K CurvG −−−−→G CurvG ⊗ CurvG     (106) y y k ValG −−−−→G ValG ⊗ ValG commutes. 3.2. Convolution. We define the convolution of µ, ν ∈ M + by µ ∗ ν := µˆd· ν.ˆ Clearly deg(µ ∗ ν) = deg µ + deg ν − n (i.e. the convolution product is graded by the codegree), and vol acts as the identity for this product. The product of valuations is a kind of extension of the intersection of two sets. It turns out that the convolution arises in similar fashion from the Minkowski sum. Theorem 3.10. If A, B ∈ K+ then µA ∗ µB = µA+B. + + Lemma 3.11. Given ν ∈ Val and 0 ≤ k ≤ n, let νk ∈ Valk denote the degree k component of ν. n Then for E ∈ Grk(R ) ν(rBE) Klνk (E) = lim r→∞ volk(rBE) where BE is the unit ball in E.

Proof. By Lemma 2.37, if j > k then νj(BE) = 0. Thus P ν(rB ) νj(rBE) lim E = lim j≤k r→∞ volk(rBE) r→∞ volk(rBE) P j j≤k νj(BE)r = lim k r→∞ r volk(BE)

= Klνk (E) since νk(BE) = Klνk (E) volk(BE).  + Lemma 3.12. Let A ∈ K and E ∈ Grk. Then

KlµA,k (E) = voln−k(πE⊥ A), (107)

Kl(ˆµA)k (E) = volk(πEA). (108) In particular,

rE(ˆµA) = µ[πE A (109) where the expression on the right is the Fourier transform in Val(E). 50 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Proof. By Lemma 3.11,

µA(rBE) Klνk (E) = lim r→∞ volk(rBE) vol(A + rB ) = lim E r→∞ volk(rBE) k−1 vol (rB ) vol (π ⊥ A) + O(r ) = lim k E n−k E r→∞ volk(rBE)

= voln−k(πE⊥ A), which is (107). To prove (108), ⊥ Kl(ˆµA)k (E) = Kl(µA)n−k (E ) = volk(πEA) by (107). Since πF (πEA) = πF A for linear subspaces F ⊂ E, it follows that the Klain functions of the homogeneous components of the valuations on the right- and left-hand sides of (109) agree, hence the valuations themselves agree. 

Lemma 3.13. Suppose µ, ν ∈ Mk. Let m be a Crofton measure for ν. Then Z (µ, νˆ) = (ˆµ, ν) = Klµ dm. Grk Proof. Let m0 be a Crofton measure for µ. Proposition 2.43 implies that Z Z (µ, νˆ) = sin(E,F ⊥) dm0(E) dm(F ) Grk Grk Z Z = cos(E,F ) dm0(E) dm(F ) Grk Grk Z = Klµ(F ) dm(F ). Grk  Proof of Theorem 3.10. We show that for k = 0, . . . , n Kl = Kl . (110) (ˆµA·µˆB )k (µ\A+B )k For k = n this relation may be restated as

(ˆµA, µˆB) voln = (µ\A+B)n. (111) The left hand side is n n X X (ˆµA)k · (ˆµB)n−k = voln ((ˆµA)k, µdB,k) k=0 k=0 n X Z = voln Kl(ˆµA)k dmB,k k=0 Grk n X Z = voln volk(πEA) dmB,k(E) k=0 Grk = µB(A) voln

= voln(A + B) voln NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 51 by Lemmas 3.12 and 3.13, where mB,k is a Crofton measure for µB,k. Since µA+B,0 = vol(A + B)χ andχ ˆ = voln, this establishes (111). To prove (110) for general k, let E ∈ Grk. Let ((·, ·)) denote the Poincar´epairing for Val(E). By Proposition 2.44, we must show that

((rEµˆA, rEµˆB)) volk = (rE(µ\A+B))k. (112) By (109), and applying (111) with n replaced by k, the left-hand side is (taking Fourier transforms in Val(E))

(µ[πE A · µ[πE B)k = volk(πEA + πEB) volk = volk(πE(A + B)) volk

= (rE(µ\A+B))k.  Corollary 3.14. If ν ∈ M then

(µA ∗ ν) = ν( · + A)

Proof. It is enough to prove this for ν = µB,B ∈ K, in which case by Theorem 3.10

µA ∗ µB(C) = µA+B(C) = vol(A + B + C) = µB(C + A).  In terms of differential forms, the convolution is easier to describe than the product. We illustrate this with the important special case of convolution with the codegree 1 intrinsic volume µn−1. n−1 n n Proposition 3.15. Let ϕ ∈ Ω (SR )R . Then 1 ν ∗ µ = ν . (113) ϕ n−1 2 LT ϕ

Proof. Given A ∈ K, the normal cycle of A+Br is N(A+Br) = fr∗N(A), where fr(x, v) = (x+rv, v) n for (x, v) ∈ SR . In other words, the fr constitute the flow of the Reeb vector field T . By Corollary 3.14 it follows that

d d (µBr ∗ νϕ)(A) = νϕ(A + Br) dr r=0 dr r=0 Z d = ϕ dr r=0 N(A+Br) Z d = ϕ dr r=0 fr∗N(A) Z d ∗ = fr ϕ dr r=0 N(A) Z d ∗ = fr ϕ N(A) dr r=0 Z = LT ϕ. N(A) On the other hand, Steiner’s formula states that n X i µBr = ωir µn−i i=0 52 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 and therefore   d d (µBr ∗ ν) = µBr ∗ ν = 2µn−1 ∗ ν dr r=0 dr r=0  SO(n) Note that (Val , ∗) is an algebra, with identity element voln = µn, and generated by µn−1 (or tn−1). Using Proposition 2.71, one computes   ω2n−i−j 2n − i − j µi ∗ µj = µi+j−n. (114) ωn−iωn−j n − i n 3.3. Constant coefficient valuations. Given A ∈ K(R ) we define the n-dimensional Lipschitz n n n submanifold of T R ' R × R

N1(A) := {(x, tv):(x, v) ∈ N(A), 0 ≤ t ≤ 1} ∪ (A × {0}), with boundary ∂N1(A) = N(A). n−1 n n−1 n Thus if ϕ ∈ Ω (SR ) is the restriction of someϕ ¯ ∈ Ω (T R ) then by Stokes’ theorem Z Z νϕ = ϕ = dϕ.¯ N(·) N1(·) Vn n n For ϕ ∈ (C ⊕ C ) we recycle our notation by putting Z νϕ := ϕ. N1(·) n Extend the Reeb field T to T R by T (x, y) := (y, 0). Then the flow of T again takes N1(A) to the N1(A + Br), so the relation (113) holds verbatim with respect to the recycled notation. n n n If dϕ ∈ Ω (R × R ) is invariant under translations in both the horizontal and the vertical n n n directions— i.e., if in fact dϕ ∈ Λ (R ⊕ R )— then νϕ is said to be a constant coefficient valuation. Every valuation in ValSO(n) is constant coefficient, since X dκi = I dxI dyI¯ #I=i where I = ±1 is determined by the relation I dxI dxI¯ = dx1 ∧ · · · ∧ dxn. n n Denote by CCV(R ) the vector space of constant coefficient valuations on R . ∗ n n Proposition 3.16. Let `, λ be the operators on Λ (R ⊕R ) defined in section 2.9.3. Let i(x, y) := n n Vn n n (y, x) be the interchange map on R ⊕ R . Suppose ϕ ∈ k (R ⊕ R ). (1) The Klain function of νϕ is obtained as follows. Given E ∈ Grk, let e¯1,..., e¯n be a positively n oriented basis for R , such that e¯1,..., e¯k span E, and let

ei := (¯ei, 0), i := (0, e¯i), i = 1, . . . , n. Then

Klνϕ = ωn−k ϕ(e1, . . . , ek, k+1, . . . , n). n (2) CCV(R ) is closed under the Alesker Fourier transform, with k(n−k) ωn−kωk+1 νˆϕ = (−1) νi∗ϕ. ωn−k−1ωk

for νϕ ∈ CCVk. (3) CCV is a module over ValSO(n), under the Alesker product, with n−1 (−1) ωn−k t · νϕ = ν`ϕ. π ωk NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 53

(4) CCV is a module over (ValSO(n), ∗), under convolution, with 1 µ ∗ ν = ν . n−1 ϕ 2 λϕ

Proof. (2): Let ϕ := dxI dyJ , #I = k, #J = n − k. If E ∈ Grk then N1(BE) = BE × BE⊥ + boundary stuff. Thus 1 Z Z Klνϕ (E) = dxI dxJ . ωk BE BE⊥ k(n−k) Since i∗(BE × BE⊥ ) = (−1) (BE⊥ × BE) k(n−k) Z Z k(n−k) (−1) (−1) ωk Klνi∗ϕ (E) = dxI dxJ = Klνϕ (E). ωn−k ωn−k BE BE⊥

(4): This is (??), since LT¯ = λ by (46). (3): Suppose νϕ ∈ CCVk. Then

t[· νϕ = tˆ∗ νˆϕ

k(n−k) 2 ωn−k = (−1) µˆ1 ∗ νi∗ϕ π ωk k(n−k) 2 ωn−k = (−1) µn−1 ∗ νi∗ϕ π ωk k(n−k) (−1) ωn−k = νλi∗ϕ π ωk k(n−k) (−1) ωn−k = νi∗`ϕ π ωk by (45). Since ν`ϕ ∈ CCVk+1, the result follows from item (2). 

Corollary 3.17. For ν ∈ CCVk, put

2ωk Lν := µ1 · ν (115) ωk+1 2ωn−k Λν := µn−1 ∗ ν (116) ωn−k+1 Hν := (2k − 2n)ν. (117) Then the map H 7→ 2k − 2n X 7→ L Y 7→ Λ defines a representation of sl(2, R) on CCV . Furthermore Λϕ = ([Lϕˆ) and L, Λ act on the intrinsic volumes by

Lµk = (k + 1)µk+1 (118)

Λµk = (n − k + 1)µk−1. (119) 54 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

n 4. Integral geometry of (C ,U(n)) U(n) n 4.1. Hermitian intrinsic volumes and Tasaki valuations. Every element of Val (C ) is a constant coefficient valuation, as√ follows. Let (z1, . . .√ , zn, ζ1, . . . , ζn) be canonical coordinates on n n n n T C ' C × C , where zi = xi + −1yi and ζi = ξi + −1ηi. The natural action of U(n) on T C n n corresponds to the diagonal action on C × C . Following Park [26], we consider the elements n X θ0 := dξi ∧ dηi i=1 n X θ1 := (dxi ∧ dηi − dyi ∧ dξi) i=1 n X θ2 := dxi ∧ dyi i=1 2 n n ∗ n n in Λ (C ⊕ C ) . Thus θ2 is the pullback via the projection map T C → C of the K¨ahlerform of n C , and θ0 +θ1 +θ2 is the pullback of the K¨ahlerform under the exponential map exp(z, ζ) := z +ζ. Pn Together with the symplectic form ω = i=1(dxi ∧ dξi + dyi ∧ dηi), the θi generate the algebra of ∗ n n all U(n)-invariant elements in Λ (C × C ). n It turns out that the algebra of invariant forms on the sphere bundle SC is generated by the restrictions of the θi, together with the restrictions of the 1-forms n X α = ξidxi + ηidyi i=1 n X β := ξidyi − ηidxi i=1 n X γ := ξidηi − ηidxi i=1 Clearly dα = −ω

dβ = θ1

dγ = 2θ0 from which it follows that

U(n) n Lemma 4.1. Val ⊂ CCV(C ). k For positive integers k, q with max{0, k − n} ≤ q ≤ 2 ≤ n, we now set n+q−k k−2q q 2n n n θk,q := cn,k,qθ0 ∧ θ1 ∧ θ2 ∈ Λ (C × C ) for cn,k,q to be specified below, and put

µk,q := νθk,q . (120) It will be useful to understand the Klain functions associated to the elements of ValU(n). Clearly n they are invariant under the action of U(n) on the (real) Grassmannian of C . Tasaki classified n n the U(n) orbits of the Grk(C ) by defining for E ∈ Grk(C ) the multiple K¨ahlerangle Θ(E) := π k (0 ≤ θ1(E) ≤ · · · ≤ θp(E) ≤ 2 ), p := b 2 c, via the condition that there exist an orthonormal basis NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 55

∗ n α1, . . . , αk of the dual space E such that the restriction of the K¨ahlerform of C to E is b k c X2 cos θi α2i−1 ∧ α2i. i=1 This is a complete invariant of the orbit. If k > n then   Θ(E) = 0,..., 0, Θ(E⊥) . | {z } k−n k,q We put Grk,q for the orbit of all E = E that may be expressed as the orthogonal direct sum of a q-dimensional complex subspace and a complementary (k − 2q)-dimensional subspace isotropic with respect to the K¨ahlerform, i.e.  π π  Θ(Ek,q) = 0,..., 0, ,..., . | {z } 2 2 q | {z } p−q n In particular, Gr2p,p is the Grassmannian of p-dimensional complex subspaces and Grn,0(C ) is the Lagrangian Grassmannian. Let k ≤ n, and assume for definiteness that k = 2p is even. Given E ∈ Grk and Θ(E) = (θ1 ≤ · · · ≤ θp,) we may find a basis e1, . . . , en, orthonormal with respect to the Hermitian form, such that √ √ E = hcos θ1e1 + sin θ1ep+1, −1e1,..., cos θpep + sin θpe2p, −1epi

=: hv1, . . . , v2pi √ √ ⊥ E = h− sin θ1e1 + cos θ1ep+1, −1ep+1,..., − sin θpep + cos θpe2p, −1e2p, √ √ e2p+1, −1e2p+1,..., −1eni

=: hw1, . . . , w2p, u1, . . . , u2n−2ki. We parametrize E × E⊥ by X X X  n n (x1, . . . , x2p, y1, . . . , y2p, t1, . . . , t2n−2k) 7→ xivi, yjwj + tlul ∈ C ⊕ C under which θ0, θ1, θ2 pull back respectively to p n−k X X cos θj dy2j−1 ∧ dy2j + dt2l−1 ∧ dt2l j=1 l=1 p X sin θi(dx2i−1 ∧ dy2i + dy2i−1 ∧ dx2i) i=1 p X cos θi dx2j−1 ∧ dx2j i=1 Now it is easy to deduce U(n) Lemma 4.2. The Klain function map gives a linear isomorphism between Valk and the vector 2 2 2 space spanned by the elementary symmetric functions in cos Θ(E) := (cos θ1(E),..., cos θp(E)) for k ≤ n, and in cos2 Θ(E⊥) if k > n. Corollary 4.3. ( k  U(n) 2 + 1, k ≤ n dim Valk =  k  U(n) n − 2 + 1 = dim Val2n−k, k > n. 56 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

U(n) Another way of stating this last result is that the Poincar´eseries for Val∗ is n+1 n+2 X U(n) (1 − x )(1 − x ) xk dim Val = . k (1 − x)(1 − x2) k To see this, note first that there is agreement in degrees k ≤ n. Then observe that both series are palindromic with center in degree n.

Definition 4.4. We choose the constants cnkq so that k,p p Klµkq (E ) = δq .

The resulting valuations µkq are the Hermitian intrinsic volumes. The Tasaki valuations τkq are determined by the relations k 2 Klτkq (E ) = σq(cos Θ(E)) where σq is the qth elementary symmetric function. The Hermitian intrinsic volumes and the Tasaki valuations behave correctly under the restriction maps ValU(n) → ValU(n−1), and therefore give a basis for the inverse limit ValU(∞). Proposition 4.5.

Lµk,q = 2(q + 1)µk+1,q+1 + (k − 2q + 1)µk+1,q

Λµk,q = 2(n − k + q + 1)µk−1,q + (k − 2q + 1)µk−1,q−1,

Lτk,p = (k − 2p + 1) τk+1,p,

Λτk,p = (2n − 2p − k + 1) τk−1,p + (k − 2p + 1) τk−1,p−1,

Proof. This is a direct calculation, using the fact that the operators `, λ are derivations, together with the relations

`θ0 = θ1, λθ0 = 0

`θ1 = 2θ2, λθ1 = 2θ0

`θ2 = 0, λθ2 = θ1 

4.2. On kU(n)(χ). What we will do in the last part of the course is to write down the principal kinematic formula, kU(n)(χ), in as clear terms as possible, and and give a method for computing any given value of the kinematic operator kU(n). An understanding of the sl2 action via Proposition 4.5 already yields a lot of information about the first goal. n Examining the dimensions given in Corollary 4.3, we see that there must be exactly b 2 c + 1 n primitive elements, living in degrees 0, 2,..., b 2 c. We denote these as π0,0, π2,1, π4,2,... . The U(n) k−2q entire algebra Val is spanned by the πk,q := L π2q,q. Lemma 4.6. If p 6= q then πk,p · π2n−k,q = 0. Proof. We compute k−2p 2n−k−2q πk,p · π2n−k,q = L π2p,p · L π2q,q 2n−2p−2q = cπ2p,p · L π2q,q for some constant c (recall that L is some dimension-dependent multiple of multiplication by t). 2n−4q+1 We may suppose that p < q. Then 2n − 2p − 2q ≥ 2n − 4q + 1, while L π2q,q = 0 since π2q,q is primitive of degree 2q.  NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 57

Theorem 4.7. Fix the constants dkp by the relation πkp · π2n−k,p = dk,p vol2n. Then

n−1 X X −1 X −1 kU(n)(χ) = 2 dk,p πk,p π2n−k,p + dn,p πn,p ⊗ πn,p (121) k=0 2p≤k 2p≤n ∗ Proof. By the ftaig, k(χ) = (p−1 ⊗ p−1) ◦ m∗ ◦ p(χ) where p : ValU(n) → ValU(n) is the Poincar´e ∗ duality map hpφ, ψi := (φ, ψ). In particular, pχ = vol2n, where this last dual element annihilates all valuations in degrees < 2n. We compute

∗ ∗ ∗ k,p hm vol , πkp ⊗ π2n−j,qi := hvol , πkp · π2n−j,qi = dkpδj,q . Thus −1 −1 ∗ ∗ X −1 ∗ −1 ∗ k(χ) = (p ⊗ p ) ◦ m vol = dkpp (πkp) ⊗ p (π2n−k,p). kp −1 But p(πkp) = dkp π2n−k,p.  Moreover, using Proposition 4.5 one computes easily r X (2r − 2i − 1)!! π := (−1)r(2n − 4r + 1)!! (−1)i τ . (122) 2r,r (2n − 2r − 2i + 1)!! 2r,i i=0

Furthermore it is essentially straightforward to express the other πkqs in terms of the τkqs. Therefore we may regard our first goal as attained by working out the constants ckq above. Since the kinematic operator is multiplicative, to stand in the same position with regard to our second goal it would then be enough to work out the multiplication table for ValU(n) in terms of some natural (i.e. geometrically comprehensible) basis. It turns out that the Tasaki valuations provide such a basis, and enjoy a number of interesting properties.

4.3. Algebraic structure of ValU(n). We have already defined the valuation t ∈ ValSO(n); iden- n 2n U(n) tifying C with R , this valuation also belongs to Val . C Definition 4.8. Put Grk = Gr2k,k for the Grassmannian of complex k-dimensional vector subspaces n U(n) of C . Define s ∈ Val2 by Z s(A) := area(π`A) d` C Gr1 Z = χ(A ∩ P¯) dP¯ C Grn−1 where the Haar measures are normalized so that 1 s(DC) = 1. Here D1 denotes the unit disk in a complex line. C Lemma 4.9. 2n − 2k skt2n−2k(Dn) = . C n − k In other words, n! 2n − 2k skt2n−2k = vol . πn n − k 2n 58 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

Proof. Since skt2n−2k is a translation-invariant valuation of degree 2n on n, it is a multiple of n C vol . Since vol (Dn) = ω = π , the second relation is equivalent to the first. 2n 2n C 2n n! To prove this we will use the transfer principle Theorem 2.68. The first point is that this theorem n applies to G = SU(n + 1),M = CP , and that the corresponding affine model (the tangent space n n to CP at a representative point) is precisely C under the action of U(n). We wish to carry out n our calculation with curvature measures on CP , then use the transfer principle to come to the n parallel conclusion in C . n The idea is that the valuation s corresponds to a valuations ¯ on CP , given literally as the average Euler characteristic with a projective complex hyperplane. Since such a hyperplane meets a generic P 1 in a point, we find thats ¯( P 1) = 1 = s(D1 ). Since area P 1 = π = area D1 , this C C C C C shows that s, s¯ indeed correspond. Clearlys ¯k now gives the average Euler characteristic of the intersection of a complex projective subspace of codimension k. Therefore Z k 2n−2k n 2n−2k s¯ t (CP ) = t (P ) dP C Grn−k 2n−2k n−k = t (CP ) (2n − 2k)!ω = 2n−2k vol ( P n−k) π2n−2k 2n−2k C 2n − 2k = n − k

πn−k n−k k 2n−2k since ω2n−2k = (n−k)! = vol2n−2k(CP ). Buts ¯ t is a multiple of vol2n, and since s, s¯ k 2n−2k correspond it follows that s t is the same multiple of vol2n. Of course the context of the discussion of the last paragraph is valuations, and not curvature measures, which is the setting in which the transfer principle holds. To make formal sense of it, U(n) consider the basis for Curv corresponding to the monomials in β, γ, θ0, θ1, θ2. Denote by U(n) n−1 • Γ2n−2 ∈ Curv2n−2 the curvature measure corresponding to cγθ2 , with c to be determined, U(n) 0 n−2 0 • Γ2 ∈ Curv2 the curvature measure corresponding to c γθ2θ0 , with c determined by the condition 1 1 glob0(Γ2)(DC) = s(DC) = 1 U(n) U(n) where glob0 : Curv → Val is the usual projection, −1 2n • and by Φ0 the curvature measure corresponding to α2n−1γθ0 , with glob0(Φ0) = χ ∈ ValU(n).

It is clear that Γ2n−2 is the only basis element that yields a nonzero measure when applied to a totally geodesic complex hypersurface, and in fact when applied to such a submanifold it yields a constant multiple of vol2n−2. The same holds for Γ2 with respect to totally geodesic complex curves. U(n) U(n) U(n) Recalling the kinematic operator K = KU(n) : Curv → Curv ⊗ Curv , it is clear that we may choose the constant c above so that the expansion of K(Φ0) in the basis described above includes the term Γ2 ⊗ Γ2n−2. Under this choice of normalization, ∗ s = glob0(K(Φ0)(·;Γ2n−2)) ∗ where Γ2n−2 denotes the corresponding dual element. By the same token, ∗ s¯ = glob1(K(Φ0)(·;Γ2n−2)) U(n) n where now glob1 is the projection Curv → V (CP ). U(n) 2n−2k U(n) Fix a curvature measure T2n−2k ∈ Curv that projects to t ∈ Val , or for that matter (2n−2k)!ω2n−2k any valuation whose value on a (2n−2k)-dimensional submanifold M is π2n−2k vol2n−2k(M). NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 59

Putting K¯ : CurvU(n) → (CurvU(n))⊗(k+1) for the iterated kinematic operator Z K¯ (Φ)(A0,S0; ... ; Ak,Sk) = Φ(A0 ∩ g¯1A ∩ · · · ∩ g¯kAk; S0 ∩ g¯1S ∩ g¯kSk) dg¯1 . . . dg¯k (U(n))k for Φ ∈ CurvU(n), we obtain ¯ ∗ ∗ K(T2n−2k)(·, ·;Γ2n−2; ... ;Γ2n−2) = c vol2n . n n k 2n−2k k 2n−2k On the other hand, globalizing the left hand side in C and in CP yields s t ands ¯ t respectively, and we conclude that these two valuations represent the same multiple of vol2n, as claimed above.  Lemma 4.10 (Aigner [1], Radoux [27]).  0
2
4
6
2k
 0 1 2 3 ... k 2
4
6
2k
2k+2
 ...  k det  1 2 3 k k+1  = 2  ......  2k
2k+2
4k
k k+1 ...... 2k Proof.  Exercise 4.11. Give your own proof of Lemma 4.10. Easier (?) version: show that the given matrix is nonsingular.

Proposition 4.12. s, t generate ValU(n).

U(n) U(n) Proof. Since multiplication by t gives an isomorphism Val2k → Val2k+1 if 2k + 1 ≤ n, and 2n−2k U(n) U(n) multiplication by t gives an isomorphism Valk → Val2n−k if k ≤ n, it is enough to show U(n) k 2 k−1 2k U(n) that Val2k is spanned by s , t s , . . . , t for 2k ≤ n. Since we know the dimension of Valk , it is in fact enough to show that these elements are independent. Consider the matrix of Poincar´epairings

k   h k−i 2i n−j−k 2j i  n−i−j 2i+2j n  2i + 2j (s t , s t ) = (s t )(DC) ij = . i,j=0 i + j ij This is the Hankel matrix above, and hence nonsingular, which implies the conclusion. 

n−k n−k−1 2 n−2k 2k U(n) Remark: this also shows that s , s t , . . . , s t span Val2n−2k. Lemma 4.13.

b n+1 c X2 (−1)i n + 1 − i2n − 2k − 2i (−1)n−k  k  = . (123) n + 1 − i i n − k − i n + 1 n − k i=0 In particular, the left hand side is zero if 2k < n. Proof. In terms of generating functions the identity (123) that we desire may be written

X (−1)i(n + 1)n + 1 − i2n − 2k − 2i 1 xnyk = . (124) n + 1 − i i n − k − i 1 − xy(1 − x) n,k,i 60 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

But using well-known generating functions [?] the left hand side may be expressed as the sum of X n − i2n − 2k − 2i X 2mm + k (−1)i xnyk = (−1)i xm+k+iyk i − 1 n − k − i m i − 1 n,k,i m,k,i X 2m = − (1 − x)m+kxm+k+1yk m m,k X 2m X = −x (1 − x)mxm (1 − x)kxkyy m m k −x = p1 − 4x(1 − x)(1 − xy(1 − x)) and X n + 1 − i2n − 2k − 2i X 2mm + k (−1)i xnyk = (−1)i xm+k+iyk i n − k − i m i − 1 n,k,i m,k,i X 2mm + k + 1 = (−1)i xm+k+iyk m i m,k,i 1 − x = , p1 − 4x(1 − x)(1 − xy(1 − x)) hence the sum is 1 − 2x 1 = p1 − 4x(1 − x)(1 − xy(1 − x)) 1 − xy(1 − x) as claimed.  U(n) n Theorem 4.14. Val (C ) ' R[s, t]/(fn+1, fn+2), where 2 X k log(1 + sx + tx) = fk(s, t)x k

The fk satisfy the relations

f1 = t t2 f = s − 2 2 ksfk + (k + 1)tfk+1 + (k + 2)fk+2 = 0, k ≥ 1. (125) Proof. For general algebraic reasons (viz. (??) and Alesker-Poincar´eduality), the ideal of relations is generated by independent homogeneous generators in degrees n + 1, n + 2. Since the natural U(n+1) n+1 U(n) n restriction map Val (C ) → Val (C ) is an algebra homomorphism, and maps s to s and U(n) t to t, it is enough to show that fn+1(s, t) = 0 in Val . C n To prove this we use the template method and the transfer principle. Let Grn−1(CP ) denote n the Grassmannian of C-hyperplanes P ⊂ CP , equipped with the Haar probability measure dP , and define Z U(n+1) n s¯ := χ(· ∩ P ) dP ∈ V2 (CP ). (126) C n Grn−1(CP ) U(n) n It is not hard to see that the valuations s, t ∈ Val (C ) correspond via the transfer principle to U(n+1) n s,¯ t ∈ V (CP ) modulo filtration 3 and 2 respectively. Since πn 2n vol( P n) = vol(Dn) = =⇒ t2n( P n) = t2n(DC) = C C n! C n n NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 61 by (2.71), it follows that 2n − 2k skt2n−2k(Dn) =s ¯kt2n−2k( P n) = t2n−2k( P n−k) = , k = 0, . . . , n. (127) C C C n − k

U(n) n With (127), the lemma yields the following identities in Val (C ):

b n+1 c  X2 (−1)i n + 1 − i tn−2k−1sk · tn−2i+1si = 0, 0 ≤ 2k < n. (128)  n + 1 − i i  i=0

n−2k−1 k U(n) n Since the t s in this range constitute a basis of Valn−1 (C ), Alesker-Poincar´eduality implies n that the sum in the second factor is zero. This sum is (−1) fn+1(s, t). 

4.4. Dictionaries. Let u := 4s − t2. Lemma 4.15. 1  1  s = µ + µ π 2,1 2 2,0 2 u = µ π 21 Proof. Since s(D1 ) = 1, it follows that s = π−1(µ + aµ ) for some constant a. On the other C 21 20 hand, in ValU(2) there is the relation 1 1 0 = f (s, t) = −st + t3 = cL(−s + t2) 3 3 3 From (118) and Proposition 4.5 we compute that 1 0 = πL(−s + t2) 3 1 = −L(µ + aµ ) + L(2µ ) 21 20 3 2 = (1 − 2a)µ3

U(2) 1 since µ30 = µ32 = 0 in Val . So a = 2 .  Lemma 4.16. For k ≥ 1, 2ω (k − 1)! f (s, t) = (−1)k+1 k µ k (2π)k k0

Proof. Each side of this relation spans the kernel of the restriction map ValU(k−1) → ValU(k−2), so fk(s, t) = γkµk0 for some constant γk. We use induction to evaluate the constants; from our 1 calculations above, γ1 = 1 and γ2 = − 2π . In view of (125), for k ≥ 3

(k + 2)fk+2(s, t) = −t(k + 1)fk+1(s, t) − skfk(s, t) t2 k − 2 = −(k + 1)γ tµ − kγ µ − γ uµ k+1 k+1,0 k 4 k0 4 k k0

We know that this expression = (k + 2)γk+2µk+2,0. We claim that if we expand the last term on the right in the hermitian intrinsic volumes then the coefficient of µk+2,0 is zero: in fact, this is true 62 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

U(n) k+2,0 of any multiple of u. For if m is a Crofton measure for some φ ∈ Valk and E is an isotropic subspace, then for any A ⊂ Ek+2,0 Z u · φ(A) = u(A ∩ P¯) dm(P¯) = 0 Gr2n−k k+2,0 2,0 since A ∩ P¯ ⊂ E ∩ P¯, which is necessarily an E , on which u ∼ µ2,1 vanishes. Now it is a trivial matter to evaluate the coefficient of µk+2,0 above using Proposition 4.5.  Lemma 4.17. bk/2c 1 X  k  f = (−1)q tk−2quq. (129) k k(−2)k−1 2q q=0 √ Proof. We introduce the formal complex variable z := t + −1v, where v is formally real and v2 = u. Then X  t2 v2  f = log(1 + s + t) = log 1 + t + + k 4 4 k  z 2 = log 1 + 2   z  = 2 Re log 1 + 2 X 1 √ k = Re t + −u
. k(−2)k−1 k  Proposition 4.18. k   π k q k−2q τkq = u t k!ωk 2q Proof. We use induction on q. From the definitions of the hermitian intrinsic volumes and the P Tasaki valuations, it is clear that τk0 = µk = q µkq, which yields the case q = 0. By Lemma 4.16, Lemma 4.17 and induction, q ! (2q − 1)! X (2q − 1)! − (−1)kτ = − µ 22q−1πqq! 2q,k 22q−1πqq! 2q,0 k=0 2ω (2q − 1)! = − 2q µ (2π)2q 2q,0 = f2q q 1 X 2q = (−1)p t2q−2pup 2q(−2)2q−1 2p p=0   q−1    −1 1 X 2q (2q)!ω2q 2q = (−1)quq + (−1)p τ 2q(−2)2q−1  2p π2q 2p kp p=0

 q−1  1 (2q)!ω2q X = (−1)quq + (−1)pτ . 2q(−2)2q−1  q!πq kp p=0 Solving for uq yields the identity for k = 2q. For k > 2q the result now follows from Proposition 4.5.  NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 63

Corollary 4.19. 2(2q + 1)(2q + 2) u · τ = τ kq π(k + 2) k+2,q+1 n 4.5. Global kinematic formulas in C . We calculate the constants dkp from Theorem 4.7. Lemma 4.20. bk/2c X i τ = µ k,q q k,i i=q bk/2c X i µ = (−1)i+q τ . k,q q k,i i=q Lemma 4.21. (k − 2r)! π = π dk,r (2n − k − 2r)! 2n−k,r

n−k U(n) U(n) Proof. By symmetry we may suppose that k ≤ n. Since Λ : Val2n−k → Valn is injective, n−k (k−2r)! n−k n−k n\−k it is enough to show that Λ πdk,r = (2n−2r−k)! Λ π2n−k,r. Since Λ πdk,r = L πk,r and the U(n) Fourier transform acts trivially on Valn , the left hand side is just πn,r. To evaluate the right hand side, we recall the basic relations for sl2 [H,Li] = 2iLi [Li, Λ] = iLi−1 ◦ H + i(i − 1)Li−1 which are easily proved by induction, giving l−2r Λπl,r = ΛL π2r,r l−2r l−2r = −[Λ,L ]π2r,r + L Λπ2r,r l−2r = −[Λ,L ]π2r,r

= (l − 2r)(2n − 2r − l + 1)πl−1,r. After iterating n − k times this gives (2n − 2r − k)! Λn−kπ = π , 2n−k,r (k − 2r)! n,r as claimed.  Proposition 4.22. 8pπn  n (2n − 4p)! (2n − 4p + 1)!! dkp = (130) ωkω2n−k 2p (n − p)! (2n − 2p + 1)!! Proof. By Lemma 4.21 it is equivalent to show that 8pπn  n  (k − 2p)!(2n − 4p)! (2n − 4p + 1)!! πkp · πckp = vol2n (131) ωkω2n−k 2p (2n − 2p − k)!(n − p)! (2n − 2p + 1)!! We show first that 8(2n − 2r + 3)(n − 2r + 1)(n − 2r + 2) u · π = π , (132) d2r,r π(2n − 4r + 3)(2n − 4r + 5) 2\r−2,r−1 64 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011

U(n) U(n) for 2r ≤ n. Both sides of (132) lie in the kernel of the map L : Val2n−2r+2 → Val2n−2r+3, which is one-dimensional. By the definition (122) and Lemma 4.20, (2n − 4r + 1)!!(2r − 1)!!  2(2r − n − 1)  π ≡ (−1)r µ + µ (133) 2r,r (2n − 2r + 1)!! 2r,0 2r − 1 2r,1

mod hµ2r,i : i > 1i, 2r ≤ n. and therefore (2n − 4r + 1)!!(2r − 1)!!  2(2r − n − 1)  π ≡ (−1)r µ + µ (134) d2r,r (2n − 2r + 1)!! 2n−2r,n−2r 2r − 1 2n−2r,n−2r+1

mod hµ2n−2r,n−2r+i : i > 1i. Meanwhile, if 2p ≤ k then using Corollary 4.19 and Lemma 4.20 4(p + 1) u · µ ≡ ((2p + 1)µ − 2(p + 2)µ ) (135) k,p π(k + 2) k+2,p+1 k+2,p+2

mod hµk+2,i : i > p + 2i.

Using these relations, the coefficients of µ2n−2r+2,n−2r+2 on the two sides of (132) agree (note that U(n) µ2n−2r+2,n−2r+1 = 0 in Val ). Now by induction  8 r n! (2n − 4r + 1)!! (ur, π ) = (136) d2r,r π (n − 2r)! (2n − 2r + 1)!! since πd0,0 = µc0 = µ2n. To prove (131) observe that by 2r r π r π r! r π2r,r ≡ τ2r,r = u = u mod t. ω2r(2r)! (2r)! Since t · πd2r,r = const. t · π2n−2r,r = 0, the case k = 2r follows from (136). If k > 2r then (k − 2r)! π · π = π · π k,r dk,r (2n − 2r − k)! k,r 2n−k,r (k − 2r)! = (Lk−2rπ ) · (L2n−k−2rπ ) (2n − 2r − k)! 2r,r 2r,r

(k − 2r)! ω2rω2n−2r 2n−4r = π2r,r · L π2r,r (2n − 2r − k)! ωkω2n−k (k − 2r)!(2n − 4r)! ω2rω2n−2r = π2r,r · πd2r,r. (2n − 2r − k)! ωkω2n−k  4.6. Two dualities. The ftaig and Lemma 3.13 immediately imply the following U(n) Proposition 4.23. For k ≤ n, let φ0, . . . , φ k be a basis for Val . Then b 2 c k X X X kU(n)(χ) = (Mn)pq φnp ⊗ φdnq + (Mk)pq φkp φckq p,q k

Proposition 4.24. ω ω (T n) = (−1)i+j k 2n−k k ij πn k b 2 c " −1 X  n  (2n − 2r − k)!(n − r)!(k − 2i)!(k − 2j)! 2r 8r(k − 2r)!(2n − 4r)!(2r − 2i)!(2r − 2j)! r=max(i,j) (2n − 2r + 1)!!(2n − 4r + 1)!!(2r − 2i − 1)!!(2r − 2j − 1)!! × . (2n − 2r − 2i + 1)!!(2n − 2r − 2j + 1)!!

n The Tk are the Tasaki matrices. Since the Tasaki valuations are essentially monomials, and since kU(n) is multiplicative, this form of the principal kinematic formula is convenient for computing other values of the kinematic operator. Beyond this, for k = 2p the Tasaki matrices enjoy the following mysterious additional symmetry.

Theorem 4.25.

n n (T2p)i,j = (T2p)p−i,p−j, 0 ≤ i, j ≤ l. (137)

By the ftaig, this is equivalent to the relation

τ2p,i · τd2p,j = τ2p,p−i · τ\2p,p−j

U(∞) U(∞) To prove this we introduce the linear involution ι : Val2∗ → Val2∗ on the subspace of valuations of even degree, determined by its action on Tasaki valuations:

ι(τ2l,q) := τ2l,l−q.

Lemma 4.26. (1) ι is an algebra automorphism. U(n) (2) ι covers an algebra automorphism of every Val2∗ . U(n) (3) The action of ι on the top degree component Val2n is trivial. (4) ι commutes with the Fourier transform.

U(∞) Proof of Lemma 4.26. (1): Any element of Val2∗ may be expressed as polynomial in t and v, involving only even powers of each√ variable. We may regard this as a (real) polynomial function√ p(z) in the complex variable z = t+ −1v. By Proposition 4.18, in these terms ι(p(z)) = p( −1 z), which is of course an algebra isomorphism. (2): To prove that ι descends to an automorphism of ValU(n) it is enough to show that ι stabilizes U(∞) U(n) the kernel of the map Val2∗ → Val2∗ . This kernel consists of the even degree elements of the ideal (fn+1, fn+2). By (1), it is enough to show that ι(f2k) ∈ (f2k) and that ι(tf2k−1) ∈ (tf2k−1, f2k). But by the proof of Lemma 4.17,

1 ι(f ) = − ι(Re z2k) 2k k 22k 1 √ = − Re( −1z)2k k 22k (−1)k+1 = Re z2k k 22k k = (−1) f2k 66 NOTES ON INTEGRAL GEOMETRY UGA, FALL 2011 and 1 ι(tf ) = ι(Re[tz2k−1]) 2k−1 (2k − 1) 22k−2 1 √ = Re[−v( −1 z)2k−1] (2k − 1) 22k−2 1 h √ √ √ i = Re ( −1 z)2k − −1 t( −1 z)2k−1 (2k − 1) 22k−2 4k (−1)k+1 = (−1)k+1 f + Re(tz2k−1) 2k − 1 2k (2k − 1)22k−2  4k  = (−1)k+1 f + tf . 2k − 1 2k 2k−1 n
(3): Since locally µ2n,k = 0 for k < n, Lemma 4.20 shows that τ2n,k = τ2n,n−k = k µ2n,n. (4): Put Σp for the vector space spanned by the elementary symmetric polynomials σp,0 := 1, σp,1 := x1 + ··· + xp, . . . , σp,p := x1x2 . . . xp in the p variables x1, . . . , xp. As noted above, Σp is U(∞) canonically isomorphic to Val2p via σp,q 7→ τ2p,q, where the map ι corresponds to σp,q 7→ σp,p−q, which we again denote by ι. U(n) U(n) Fixing n ≥ 2p, the Fourier transform b: Val2n−2p → Val2p corresponds to the linear surjection r :Σn−p → Σp given by r(f) = f(x1, . . . , xp, 1,..., 1). The assertion thus reduces to the claim that for m = n − p ≥ p the diagram ι Σm −−−−→ Σm   r r y y ι Σp −−−−→ Σp commutes. It is enough to prove this for m = p + 1, in which case r(σp+1,i) = σp,i + σp,i−1. Hence for i = 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).  Proof of Thm. 4.25. By Lemma 4.26,

τ2p,i · τd2p,j = τ2p,i · (ιτ\2p,p−j) = τ2p,i · ι(τ\2p,p−j)
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