Tube formulas for valuations in complex space forms

Abstract

Given an isometry invariant valuation on a complex space form we compute its value on the tubes of sufficiently small radii around a set of positive reach. This generalizes classical formulas of Weyl, Gray and others about the volume of tubes. We also develop a general framework on tube formulas for valuations in Riemannian manifolds.

1 Introduction

For a compact convex set $A\subset {\mathbb {R}}^m$, the Steiner formula computes the volume of the set $A_t$ consisting of points at distance smaller than t from A as follows

$$\begin{aligned} \textrm{vol}(A_t) = \sum _{i = 0}^m \omega _{m-i}\mu _i(A)t^{m-i}. \end{aligned}$$

(1)

Here the functionals $\mu _i$ are the so-called intrinsic volumes, and the normalizing constant $\omega _k$ is the volume of the k-dimensional unit ball. By Hadwiger’s characterization theorem, the intrinsic volumes span the space of valuations (finitely additive functionals on convex bodies) that are continuous and invariant under rigid motions.

The famous tube formula of H. Weyl [28] is the assertion that (1) holds true for $A\subset {\mathbb {R}}^m$ a smooth compact submanifold and $t\ge 0$ small enough, with the additional insight that the coefficients $\mu _i(A)$ depend only on the induced Riemannian structure of A. Even more generally, Federer extended the validity of (1) to the class of compact sets of positive reach. Later on, the same formula has been proven to hold for bigger classes of sets (see e.g. [14, 16]). As for the coefficients $\mu _i$, the current perspective is to view them as smooth valuations in the sense of Alesker’s theory of valuations on manifolds (see [4]).

Already in Weyl’s original work, the tube formula was extended to the sphere and to hyperbolic space. In that case, instead of a polynomial on the radius t one has a polynomial in certain functions $\sin _\lambda (t),\cos _\lambda (t)$ whose definition we recall in (52). Later, Gray and Vanhecke computed the volume of tubes around submanifolds of rank one symmetric spaces (cf. [19, 20]).

All these classical tube formulas are most naturally expressed in the language of valuations on manifolds. Furthermore, this theory has allowed for the determination of kinematic formulas (a far-reaching generalization of tube formulas) in isotropic spaces. These spaces are Riemannian manifolds under the action of a group of isometries that is transitive on the sphere bundle. For instance, in [10] and [11] the kinematic formulas of complex complex space forms (i.e. complex euclidean, projective and hyperbolic spaces) were obtained, and Gray’s tube formulas on such spaces were recovered.

Tube formulas, however, exist also for other valuations than the volume, and these do not follow from the kinematic formulas. For instance, differentiating the Steiner formula one easily obtains

$$\begin{aligned} \mu _k(A_t) = \sum _{j=0}^{k}\left( {\begin{array}{c}m-j\\ m-k\end{array}}\right) \dfrac{\omega _{m-j}}{\omega _{m-k}}\mu _{j}(A) t^{k-j},\qquad A\subset {\mathbb {R}}^m. \end{aligned}$$

(2)

In real space forms (i.e. the sphere and hyperbolic space), Santaló obtained similar tube formulas for all isometry invariant valuations (see [26]). For rank one symmetric spaces, the tube formulas of a certain class of valuations (integrated mean curvatures) were found in [19], still with a differential-geometric viewpoint. There are however many invariant valuations on these spaces that were not considered.

In this paper we prove the existence of tube formulas for any smooth valuation in a Riemannian manifold. Then we develop a method to determine these formulas for the invariant valuations of an isotropic space. Using this method we compute all tube formulas explicitly in the case of complex space forms. In fact, our approach also reveals some interesting aspects in the case of real space forms.

Let us briefly describe our results. First, given a Riemannian manifold M we construct a family ${\textbf{T}}_t$ of tubular operators on the space ${\mathcal {V}}(M)$ of smooth valuations of M such that for any $\mu \in {\mathcal {V}}(M)$ and every compact set of positive reach $A\subset M$ one has

$$\begin{aligned} \mu (A_t)={\textbf{T}}_t\mu (A), \end{aligned}$$

for $t\ge 0$ small enough (see Definition 4.1 and Corollary 4.7). Differentiating ${\textbf{T}}_t$ at $t=0$ yields an operator $\partial :{\mathcal {V}}(M)\rightarrow {\mathcal {V}}(M)$. If G is a group of isometries of M acting transitively on the sphere bundle SM, the subspace ${\mathcal {V}}(M)^G$ of G-invariant valuations is finite dimensional, and the determination of the tube operators ${\textbf{T}}_t$ reduces to the computation of the flow generated by $\partial$.

Once this general framework is established we concentrate on the complex space forms ${\mathbb {C}}P^{n}_{\lambda }$. For $\lambda =0$ this refers to complex euclidean space ${\mathbb {C}}^n$ under the group of complex isometries, and for $\lambda \ne 0$ this is the n-dimensional complex projective or hyperbolic space of constant holomorphic curvature $4\lambda$, under the full group of isometries G. We simply denote $\mathcal V_{\lambda ,{\mathbb {C}}}^n:= {\mathcal {V}}({\mathbb {C}}P^n_\lambda )^G$.

For $\lambda =0$, we will readily obtain the tube formulas ${\textbf{T}}_t\mu$ of all translation-invariant and U(n)-invariant continuous valuations $\mu$ thanks to the existence of an $\mathfrak {sl}_2$-module structure on the space ${\text {Val}}^{U(n)}$ of such valuations. This structure, discovered by Bernig and Fu in [10], is induced by two natural operators $\Lambda ,L$, the first of which is a normalization of $\partial$.

Remarkably, it turns out that also for $\lambda \ne 0$ the derivation operator $\partial$ is closely related to the operators $\Lambda ,L$ of the flat space. Indeed, in Theorem 4.11 we find an isomorphism $\Phi _\lambda :{\text {Val}}^{U(n)}\rightarrow {\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ such that

$$\begin{aligned} \left. \partial \right| _{{\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n}=\Phi _\lambda \circ (\Lambda -\lambda L)\circ \Phi _\lambda ^{-1}. \end{aligned}$$

(3)

Using the decomposition of ${\text {Val}}^{U(n)}$ into $\mathfrak {sl}_2$-irreducible components, the computation of the tubular operator boils down to the solution of a Cauchy problem in some abstract model spaces, yielding our main result.

Theorem

There exists a basis $\{\sigma _{k,r}^\lambda \}$ of the space ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ of invariant valuations of ${\mathbb {C}}P^{n}_{\lambda }$ such that

$$\begin{aligned} \textbf{T}_t\sigma _{k,r}^{\lambda }= \sum _{j = 0}^{2n-4r}\phi _{2n-4r,k-2r,j}^\lambda (t)\sigma _{j+2r,r}^{\lambda }, \end{aligned}$$

(4)

where

$$\begin{aligned} \phi _{m,k,j}^\lambda (t) = \sum _{h \ge 0}(-\lambda )^{j-h}\left( {\begin{array}{c}m-j\\ k-h\end{array}}\right) \left( {\begin{array}{c}j\\ h\end{array}}\right) \sin _\lambda ^{k+j-2h}(t)\cos _\lambda ^{m-k-j+2h}(t). \end{aligned}$$

We describe the basis $\sigma _{k,r}^\lambda$ explicitly in terms of the previously known valuations $\tau _{k,p}^\lambda$ of [11]. The tube formulas for the $\tau _{k,p}^\lambda$ can be easily obtained from the previous ones, as we also provide the expression of these valuations in terms of the $\sigma _{k,r}^\lambda$.

Curiously, the expressions (4) are extremely similar to those obtained by Santaló in the real space form ${\mathbb {S}}^{m}_{\lambda }$ of constant curvature $\lambda$. Indeed, for a certain basis $\{\sigma _i\}_{i=0}^m$ of the space $\mathcal V_{\lambda ,{\mathbb {R}}}^m$ of isometry invariant valuations of ${\mathbb {S}}^{m}_{\lambda }$ one has

$$\begin{aligned} {\textbf{T}}_t\sigma _i^{\lambda }= \sum _{j = 0}^{m-1}\phi ^\lambda _{m-1,i,j}(t)\sigma ^\lambda _{j}, \quad 0 \le i \le m-1. \end{aligned}$$

The tube formula for $\sigma _m={\text {vol}}$ is however quite different. As an explanation for these similarities, we show in Theorem 4.12 the existence of a phenomenon similar (but not completely analogous) to (3).

The paper concludes with a detailed study of the spectrum and the eigenspaces of the derivative operator $\partial$ in $\mathcal V_{\lambda ,{\mathbb {C}}}^n$ and ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$. In particular, we compute the kernel of $\partial$ in ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$; i.e. we determine the invariant valuations of ${\mathbb {C}}P_\lambda ^n$ for which the tube formulas are constant. We also identify the images $\partial ({\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n)$ and $\partial (\mathcal V_{\lambda ,{\mathbb {R}}}^m)$, and we compute the preimage by $\partial$ of any element belonging to these subspaces.

2 Background

2.1 Valuations

Let V be a finite-dimensional real vector space, and let ${\mathcal {K}}(V)$ be the space of convex compact subsets of V, endowed with the Hausdorff metric. A valuation on V is a map $\varphi :{\mathcal {K}} (V) \rightarrow {\mathbb {C}}$ such that

$$\begin{aligned} \varphi (A \cup B)=\varphi (A)+\varphi (B)-\varphi (A \cap B), \end{aligned}$$

for $A,B,A\cup B\in {\mathcal {K}}(V)$. The space of translation-invariant, continuous valuations on V is denoted by ${\text {Val}}(V)$.

The notion of valuation was extended to smooth manifolds by Alesker (cf. [2,3,4, 6]). For simplicity we will focus on the case of a Riemannian manifold $M^n$. It is also natural to consider here the class of compact sets of positive reach in M, which we denote ${\mathcal {R}}(M)$. The definition and some basic properties of such sets are recalled in Sect. 4.2.

Let SM be the sphere bundle of M consisting of unit tangent vectors, and let $\pi :SM\rightarrow M$ be the canonical projection.

Definition 2.1

(Smooth valuation) A smooth valuation on M is a functional $\varphi :{\mathcal {R}}(M) \rightarrow {\mathbb {C}}$ of the form

$$\begin{aligned} \varphi (A)= \int _{N(A)}\omega + \int _{A}\eta , \end{aligned}$$

where $\omega \in \Omega ^{n-1}(SM)$ and $\eta \in \Omega ^{n}(M)$, are complex-valued differential forms, and N(A) is the normal cycle of A (cf. e.g. [14]). We will denote $\varphi =[\![\omega ,\eta ]\!]$ in this case. For any subgroup $G \le {\text {Diff}}(M)$, we will denote by ${\mathcal {V}}^{G}(M)$ the space of G-invariant valuations; i.e. $\mu \in {\mathcal {V}}(M)$ such that $\mu (gA) = \mu (A)$ for all $A \in {\mathcal {R}}(M)$ and $g \in G$.

The kernel of the map $(\omega ,\eta )\mapsto [\![\omega ,\eta ]\!]$ was determined by Bernig and Bröcker in [8] as follows. Given $\omega \in \Omega ^{n-1}(SM)$, there exists $\xi \in \Omega ^{n-2}(SM)$ such that

$$\begin{aligned} D\omega := d(\omega + \alpha \wedge \xi ), \end{aligned}$$

is a multiple of $\alpha$, the canonical contact form on SM. The unique n-form $D\omega$ satisfying this condition is called the Rumin differential of $\omega$ (see [24]). Then $[\![\omega ,\eta ]\!]= 0$ if and only if

$$\begin{aligned} D\omega + \pi ^* \eta = 0, \qquad \text{ and }\qquad \int _{S_xM}\omega = 0,\quad \forall x \in M. \end{aligned}$$

(5)

One of the most striking aspects of Alesker’s theory of valuations on manifolds is the existence of a natural product on ${\mathcal {V}}(M)$, which turns this space into an algebra with $\chi$ as the unit element. The realization by Fu that this product is closely tied to kinematic formulas opened the door to the recent development of integral geometry in several spaces, including the complex space forms [1, 10, 11].

Another important algebraic structure is the convolution of valuations found by Bernig and Fu in linear spaces (cf. [9], but also [5]). This is a product on the dense subspace ${\text {Val}}^\infty (V):={\text {Val}}(V)\cap {\mathcal {V}}(V)$ characterized as follows. Given $A \in {\mathcal {K}}(V)$, with smooth and positively curved boundary, we have $\mu _A(\cdot ):= {\text {vol}}(\cdot + A)\in {\text {Val}}^\infty (V)$. The convolution is determined by

$$\begin{aligned} \mu _A * \varphi := \varphi (\cdot + A),\qquad \varphi \in {\text {Val}}^{\infty }(V), \end{aligned}$$

(6)

where $+$ refers to the Minkowski sum. In particular, ${\text {vol}}$ is the unit element of this operation.

2.2 Real space forms

The fundamental examples of valuations in Euclidean space ${\mathbb {R}}^m$ are the intrinsic volumes $\mu _k$. These are implicitly defined by the Steiner formula

$$\begin{aligned} \textrm{vol}_{{\mathbb {R}}^m}(A + t{\mathbb {B}}^m) = \sum _{k=0}^m t^{m-k} \omega _{m-k} \mu _k(A), \quad A \in {\mathcal {K}}({\mathbb {R}}^m), \end{aligned}$$

(7)

where ${\mathbb {B}}^m$ is the unit ball and $\omega _i$ is the volume of the i-dimensional unit ball. In particular $\mu _0 = \chi$, $\mu _{m-1} = \frac{1}{2}\,\textrm{perimeter}$, and $\mu _n = \textrm{vol}_{m}$ are intrinsic volumes.

We will denote by ${\mathbb {S}}^{m}_{\lambda }$ the m-dimensional complete and simply connected Riemannian manifold of constant curvature $\lambda$. That is, the sphere $S^m(\sqrt{\lambda })$ for $\lambda > 0$, Euclidean space ${\mathbb {R}}^n$ for $\lambda = 0$, and hyperbolic space $H^m(\sqrt{-\lambda })$ for $\lambda < 0$. Let $G_{\lambda ,{\mathbb {R}}}$ be the group of orientation preserving isometries of ${\mathbb {S}}^{m}_{\lambda }$; i.e. $G_{\lambda ,{\mathbb {R}}}\cong SO(m+1)$ for $\lambda >0$, and $G_{\lambda ,{\mathbb {R}}} \cong SO(m) \ltimes {\mathbb {R}}^m$ for $\lambda = 0$, while $G_{\lambda ,{\mathbb {R}}}\cong PSO(m,1)$ for $\lambda < 0$. We will denote by ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ the space of $G_{\lambda ,{\mathbb {R}}}$-invariant valuations of ${\mathbb {S}}_\lambda ^m$.

Let $\kappa _0,\dots ,\kappa _{m-1}\in \Omega ^{m-1}(S{\mathbb {S}}^{m}_{\lambda })^{{{G}}_{\lambda ,{\mathbb {R}}}}$ be the differential forms defined in [13, §0.4.4]. In the same paper it was shown that the ${\mathbb {R}}$-algebra of ${G}_{\lambda ,{\mathbb {R}}}$-invariant differential forms is generated by $\kappa _0,\dots ,\kappa _{m-1},\alpha ,d\alpha$. It follows by [11, Prop. 2.6] that the following valuations constitute a basis of ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^{m}$

$$\begin{aligned} \sigma _i^{\lambda }&:= [\![\kappa _i,0]\!],\qquad 0 \le i \le m-1\\ \sigma _m^{\lambda }&:= {\text {vol}}_{{\mathbb {S}}^{m}_{\lambda }}. \end{aligned}$$

In euclidean space ${\mathbb {R}}^m$ these valuations are proportional to the intrinsic volumes:

$$\begin{aligned} \sigma _i^\lambda = {(m-i)\omega _{m-i}}\mu _i,\qquad \lambda =0. \end{aligned}$$

For general $\lambda$, the $\sigma _i^\lambda$ are proportional to the valuations $\tau _i^\lambda$ appearing in [7, 17]

$$\begin{aligned} \sigma _i^\lambda&=\dfrac{\pi ^i(m-i)\omega _{m-i}}{i! \omega _i}\tau _i^{\lambda } , \quad 0 \le i \le m-1, \end{aligned}$$

(8)

$$\begin{aligned} \sigma _m^{\lambda }&=\frac{\pi ^m}{m!\omega _m}\tau _m^\lambda . \end{aligned}$$

(9)

As we will see, the normalization taken for the $\sigma _i^\lambda$ makes the tube formulas in ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ specially simple. A stronger reason in favor of this normalization is Theorem 4.12.

2.3 Complex space forms

We denote by ${\mathbb {C}}P^{n}_{\lambda }$ the complete, simply connected n-dimensional Kähler manifold of constant holomorphic curvature $4\lambda$; i.e. the complex projective space (with the suitably normalized Fubini-Study metric) for $\lambda > 0$, the complex euclidean space ${\mathbb {C}}^n$ for $\lambda = 0$, and the complex hyperbolic space for $\lambda < 0$. For $\lambda \ne 0$ we let $G_{\lambda ,{\mathbb {C}}}$ be the full isometry group of ${\mathbb {C}}P^{n}_{\lambda }$. For $\lambda =0$ we put $G_{\lambda ,{\mathbb {C}}} = U(n) \ltimes {\mathbb {C}}^n$. We denote by ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ the space of $G_{\lambda ,{\mathbb {C}}}$-invariant valuations on ${\mathbb {C}}P^{n}_{\lambda }$.

Let $\{\beta _{k,q},\gamma _{k,q}\} \subset \Omega ^{2n-1}(S{\mathbb {C}}P^{n}_{\lambda })^{G_{\lambda ,{\mathbb {C}}}}$ be the differential forms introduced in [10] for $\lambda =0$, and extended to the curved case $\lambda \ne 0$ in [11]. Let also

$$\begin{aligned}&\mu _{k,q}^\lambda := [\![\beta _{k,q},0]\!], \quad k>2 q, \end{aligned}$$

(10)

$$\begin{aligned}&\mu _{2q,q}^\lambda := \sum _{i = 0}^{n-q-1}\left( \dfrac{\lambda }{\pi }\right) ^i \dfrac{(q+i)!}{q!}[\![\gamma _{2q+2i,q+i},0]\!]+ \left( \dfrac{\lambda }{\pi }\right) ^{n-q} \dfrac{n!}{q!}[\![0,d\textrm{vol}]\!] \end{aligned}$$

(11)

where $d\textrm{vol}$ is the Riemannian volume element. It was shown in [10, 11] that these valuations $\mu _{k,q}^\lambda$ with $\max \{0, k-n\} \le q \le \frac{k}{2} \le n$ constitute a basis of ${\mathcal {V}}^{n}_{\lambda ,{\mathbb {C}}}$. It is convenient to emphasize that the $\mu _{k,q}^\lambda$ do not coincide with the hermitian intrinsic volumes $\mu _{k,q}^M$ for $M={\mathbb {C}}P^{n}_{\lambda }$ introduced in [12].

For $\lambda =0$ we simply write $\mu _{k,q}$ instead of $\mu _{k,q}^0$. We will also use the so-called Tasaki valuations

$$\begin{aligned} \tau _{k,q}:=\sum _{i=q}^{\lfloor k / 2\rfloor }\left( {\begin{array}{c}i\\ q\end{array}}\right) \mu _{k,i}, \qquad 0, k-n \le q \le \frac{k}{2} \le n. \end{aligned}$$

It will be useful to consider the following linear isomorphisms:

$$\begin{aligned} {\mathcal {F}}_{\lambda ,{\mathbb {C}}}:{\text {Val}}^{U(n)}\longrightarrow \mathcal V_{\lambda ,{\mathbb {C}}}^n,&\quad \mathcal F_{\lambda ,{\mathbb {C}}}(\mu _{k,q})=\mu _{k,q}^\lambda . \end{aligned}$$

More generally, whenever we have a valuation $\nu$ in ${\text {Val}}^{U(n)}$ we will denote $\nu ^\lambda := {\mathcal {F}}_{\lambda ,{\mathbb {C}}}(\nu )$. For instance $\tau _{k,q}^\lambda =\mathcal {F_{\lambda ,{\mathbb {C}}}}(\tau _{k,q})$.

3 Tube formulas in linear spaces

Let V be an m-dimensional euclidean vector space. Given $t\ge 0$, let ${\textbf{T}}_t:{\text {Val}}(V)\rightarrow {\text {Val}}(V)$ be given by

$$\begin{aligned} ({\textbf{T}}_t\mu )(A)=\mu (A + t{\mathbb {B}}^m)=(\mu _{t{\mathbb {B}}^m}*\mu )(A)\qquad A\in {\mathcal {K}}(V), \end{aligned}$$

(12)

where ${\mathbb {B}}^m$ is the unit ball. We will call ${\textbf{T}}_t$ the tubular operator. Let also $\partial :{\text {Val}}(V) \rightarrow {\text {Val}}(V)$ be the operator given by

$$\begin{aligned} \partial \mu = \left. \dfrac{d}{dt}\right| _{t = 0}{\textbf{T}}_t\mu . \end{aligned}$$

(13)

This operator has sometimes been denoted by $\Lambda$ in the literature, but following [10] we reserve the symbol $\Lambda$ for a certain normalization of $\partial$ (see (18)).

The properties of the Minkowski sum ensure that ${\textbf{T}}_{t+s} = {\textbf{T}}_t \circ {\textbf{T}}_s={\textbf{T}}_s \circ {\textbf{T}}_t$. Differentiating with respect to s at zero yields

$$\begin{aligned} \dfrac{d}{dt}{\textbf{T}}_t\mu = {\textbf{T}}_t\partial \mu = \partial {\textbf{T}}_t\mu . \end{aligned}$$

(14)

It follows that

$$\begin{aligned} \partial ^i\mu = \left. \dfrac{d^i}{dt^i}\right| _{t = 0}{\textbf{T}}_t\mu . \end{aligned}$$

(15)

For each $\mu \in {\text {Val}}(V)$, the map $t \mapsto {\textbf{T}}_t\mu$ is a polynomial in t of degree m by (12) and the Steiner formula (7) (or by [23]). Hence

$$\begin{aligned} {\textbf{T}}_t\mu&= \sum _{i = 0}^m \dfrac{t^i}{i!}\left. \dfrac{d^i}{dt^i}\right| _{t = 0}{\textbf{T}}_t\mu \end{aligned}$$

(16)

$$\begin{aligned}&= \sum _{i = 0}^m \dfrac{t^i}{i!}\partial ^i \mu . \end{aligned}$$

(17)

Note also that, by (15) and (16), the derivative operator $\partial$ is $(m+1)$-nilpotent; i.e. $\partial ^{m+1} = 0$.

Let us compute the tube formula for the intrinsic volume $\mu _i$ for each $0 \le i \le m$ using (17). For that purpose we first compute $\partial$. Since ${\textbf{T}}_{t+s} = {\textbf{T}}_s\circ {\textbf{T}}_t$ we have

$$\begin{aligned} {\textbf{T}}_{t+s}{\text {vol}} = \sum _{j = 0}^m \omega _{m-j}t^{m-j}{\textbf{T}}_s\mu _j. \end{aligned}$$

On the other hand

$$\begin{aligned} {\textbf{T}}_{t+s}{\text {vol}} = \sum _{j = 0}^m \omega _{m-j}(t+s)^{m-j}\mu _j, \end{aligned}$$

Differentiating at $s=0$ and comparing coefficients yields

$$\begin{aligned} \partial \mu _j = \dfrac{\omega _{m-j+1}}{\omega _{m-j}}(m-j+1)\mu _{j-1}. \end{aligned}$$

Finally, using (17), we get

$$\begin{aligned} {\textbf{T}}_t\mu _k&= \sum _{i = 0}^m \frac{t^i}{i!}\partial ^i\mu _k =\sum _{i = 0}^k \frac{t^i}{i!}\frac{\omega _{m-k+i}}{\omega _{m-k}}\frac{(m-k+i)!}{(m-k)!}\mu _{k-i}\\&= \sum _{j=0}^k \left( {\begin{array}{c}m-j\\ k-j\end{array}}\right) \frac{\omega _{m-j}}{\omega _{m-k}} t^{k-j}\mu _j, \end{aligned}$$

which is (2).

In order to compute the tube formulas for invariant valuations in ${\mathbb {C}}^n$ (i.e. to determine ${\textbf{T}}_t$ on ${\text {Val}}^{U(n)}$), it will be useful to recall the $\mathfrak {sl}_2$-module structure of ${\text {Val}}^{U(n)}$ found in [10]. Consider the linear maps $\Lambda , L, H:{\text {Val}}^\infty (V)\rightarrow {\text {Val}}^\infty (V)$, defined as follows

$$\begin{aligned} \Lambda \nu := \dfrac{\omega _{m-k}}{\omega _{m-k+1}}\partial _{}\nu ,\qquad L \nu := \dfrac{2\omega _k}{\omega _{k+1}} \mu _1 \cdot \nu ,\qquad H\nu =(2k-m)\nu ,\qquad \nu \in {\text {Val}}_k^{\infty }(V), \end{aligned}$$

(18)

where $\cdot$ refers to the Alesker product.

Proposition 3.1

On ${\text {Val}}^{O(m)}$ the operators $\Lambda ,L$ are given by

$$\begin{aligned}&L \mu _k = (k+1)\mu _{k+1}, \end{aligned}$$

(19)

$$\begin{aligned}&\Lambda \mu _k = (m-k+1)\mu _{k-1}, \end{aligned}$$

(20)

while on ${\text {Val}}^{U(n)}$ one has

$$\begin{aligned}&L \mu _{k, p} =(k-2 q+1) \mu _{k+1, q}+2(q+1) \mu _{k+1, q+1} \end{aligned}$$

(21)

$$\begin{aligned}&\Lambda _{} \mu _{k, p}= (k-2 q+1) \mu _{k-1, q-1}+2(n-k+q+1) \mu _{k-1, q}, \end{aligned}$$

(22)

which implies

$$\begin{aligned}&L\tau _{k,q} = (k-2q+1)\tau _{k+1,q} \end{aligned}$$

(23)

$$\begin{aligned}&\Lambda \tau _{k,q} = (k-2 q+1) \tau _{k-1, q-1}+(2 n-2 q-k+1) \tau _{k-1, q} \end{aligned}$$

(24)

Proof

The first two equalities are [7, eqs. (2.3.12) and (2.3.13)]. The rest is [10, Lemma 5.2]. $\square$

Proposition 3.2

([7, Prop. 2.3.10 (3)]) The operators $\Lambda , L,H$ define an $\mathfrak {sl}_2$-module structure on both ${\text {Val}}^{O(m)}$ and ${\text {Val}}^{U(n)}$; i.e. $[L,\Lambda ]=H$, $[H,L]=2L, [H,\Lambda ]=-2\Lambda$.

The decomposition into irreducible components is as follows

$$\begin{aligned} {{\text {Val}}^{O(m)}\cong V^{(m)}},\qquad {\text {Val}}^{U(n)}\cong \bigoplus _{0 \le 2r\le n}V^{(2n-4r)} \end{aligned}$$

(25)

where $V^{(m)}$ is the $(m+1)-$dimensional irreducible $\mathfrak {sl}_2$-representation. In particular, for $0 \le 2r \le n$, there exists a unique, up to a multiplicative constant, primitive element (i.e. anihilated by $\Lambda$) in each irreducible component of ${\text {Val}}^{U(n)}$. By the so-called Lefschetz decomposition, the L-orbits of these primitive elements consitute a basis of ${\text {Val}}^{U(n)}$. This basis was explicitly computed in [10] as follows.

Proposition 3.3

([10, eq. (76)]) The following valuations

$$\begin{aligned} \pi _{2r,r}&:= (-1)^{r}(2 n-4 r+1) ! ! \sum _{i=0}^{r}(-1)^{i} \frac{(2 r-2 i-1) ! !}{(2 n-2 r-2 i+1) ! !} \tau _{2 r, i},&0 \le 2r \le n, \end{aligned}$$

(26)

are $\Lambda$-primitive; i.e. $\Lambda \pi _{2r,r}=0$. The family

$$\begin{aligned} \pi _{k, r}&:=L^{k-2 r} \pi _{2 r, r} \end{aligned}$$

(27)

$$\begin{aligned}&=(-1)^{r}(2 n-4 r+1) ! ! \sum _{i=0}^{r}(-1)^{i} \frac{(k-2 i) !}{(2 r-2 i) !} \frac{(2 r-2 i-1) ! !}{(2 n-2 r-2 i+1) ! !} \tau _{k, i},\nonumber \\ {}&\quad 2r \le k \le 2n-2r \end{aligned}$$

(28)

forms a basis of ${\text {Val}}^{U(n)}$.

In particular the irreducible components of ${\text {Val}}^{U(n)}$ are the following subspaces

$$\begin{aligned} {\mathcal {I}}_{0}^{n,r} := \left\{ \pi _{k,r} : 2r \le k \le 2n-2r \right\} , \quad 0 \le 2r \le n. \end{aligned}$$

(29)

We are now able to compute the tube formulas in the complex case using (17).

Theorem 3.4

 

$$\begin{aligned} {\textbf{T}}_t\pi _{k,r}&= \dfrac{(k-2r)!}{\omega _{2n-k}}\sum _{j = 0}^{k-2r} \left( {\begin{array}{c}2n-4r-j\\ k-2r-j\end{array}}\right) t^{k-2r-j}\dfrac{\omega _{2n-2r-j}}{j!}\pi _{j+2r,r}. \end{aligned}$$

(30)

Proof

By [10, Lemma 5.6],

$$\begin{aligned} \Lambda \pi _{k,r} = (k-2r)(2n-k-2r+1)\pi _{k-1,r}, \end{aligned}$$

(31)

and then

$$\begin{aligned} \Lambda ^{i} \pi _{k,r}&= \dfrac{(k-2r)!(2n-k-2r+i)!}{(k-2r-i)!(2n-k-2r)!}\pi _{k-i,r}. \end{aligned}$$

(32)

Using (17), we obtain the tube formula

$$\begin{aligned} {\textbf{T}}_t\pi _{k,r}&= \sum _{i = 0}^{2n} \dfrac{t^i}{i!} \frac{\omega _{2n-k+i}}{\omega _{2n-k}}\Lambda ^i\pi _{k,r}\\&=\dfrac{(k-2r)!}{\omega _{2n-k}}\sum _{i = 0}^{k-2r} \dfrac{t^i}{i!}\omega _{2n-k+i}\dfrac{(2n-k-2r+i)!}{(k-2r-i)!(2n-k-2r)!}\pi _{k-i,r} \\&=\dfrac{(k-2r)!}{\omega _{2n-k}}\sum _{j = 0}^{k-2r} \left( {\begin{array}{c}2n-4r-j\\ k-2r-j\end{array}}\right) t^{k-2r-j}\dfrac{\omega _{2n-2r-j}}{j!}\pi _{j+2r,r}. \end{aligned}$$

$\square$

These tube formulas can also be given in terms of the valuations $\tau _{k,q}$. To this end, we next compute their Lefschetz decomposition.

Proposition 3.5

The Lefschetz decomposition of $\tau _{k,r}$ is given by

$$\begin{aligned} \tau _{k,r}&= \frac{1}{(k-2r)!} \sum _{i = 0}^r \left( {\begin{array}{c}n-2i\\ r-i\end{array}}\right) \frac{(2n-2i-2r) !}{(2 n-4 i) !} \pi _{k,i}. \end{aligned}$$

(33)

Proof

Consider the linear map $\psi : {\text {Val}}^{U(n)} \rightarrow {\text {Val}}^{U(n)}$ mapping $\tau _{k,r}$ to the left hand side of (33). We need to show that $\psi ={\text {id}}$. Let us check that this endomorphism commutes with both $\Lambda$ and L. To check commutation with $\Lambda$, we only need to verify the following

$$\begin{aligned}&(k-2r)!\psi (\Lambda (\tau _{k,r})) \\&\quad =\sum _{i=0}^{r-1} \frac{(n-2i) !(2n-2i-2r+2)!}{(r-i-1) !(n-r-i+1)!(2n-4i) !}\pi _{k-1, i}\\&\quad \quad +(k-2r)(2n-k-2r+1)\sum _{i=0}^r \frac{(n-2 i) !(2n-2i-2r)!}{(r-i)!(n-i-r)!(2n-4i)!}\pi _{k-1, i}\\&\quad =\sum _{i = 0}^r \dfrac{(n-2i)!(2n-2i-2r)!}{(r-i)!(n-i-r)!(2n-4i)!}(k-2i)(2n-k-2i+1)\pi _{k-1,i} \\&\quad = (k-2r)!\Lambda \psi (\tau _{k,r}). \end{aligned}$$

Comparing term by term, the previous identities boil down to

$$\begin{aligned} 2 (r - i) (2 n - 2 i - 2 r + 1) + (k - 2 r) (2 n - k - 2 r + 1) = (k - 2 i) (2 n - k - 2 i + 1) \end{aligned}$$

which is trivial.

Commutation with L is straightforward using $L\pi _{k,i} = \pi _{k+1,i}$.

Given that $\psi$ commutes with the operators $\Lambda$ and L and ${\text {Val}}^{U(n)}$ is multiplicity-free, Schur’s lemma implies that for each $0 \le 2r \le n$, there exists a constant $c_r$ such that $\left. \psi \right| _{{\mathcal {I}}_{0}^{n,r}} = c_r {\text {id}}$.

Let $a_{2r,j}$ and $b_{2r,i}$ be the coefficients of $\pi _{2r,j}$ and $\tau _{2r,i}$ in (33) and (26) respectively, so that $\psi (\tau _{2r,i}) = \sum _{j= 0}^i a_{2r,j} \pi _{2r,j}$ and $\pi _{2r,r} = \sum _{i = 0}^r b_{2r,i} \tau _{2r,i}$. Then

$$\begin{aligned} c_r \pi _{2r,r} = \psi (\pi _{2r,r})=\sum _{i = 0}^rb_{2r,i}\left( \sum _{j = 0}^{i} a_{2r,j} \pi _{2r,j}\right) =\sum _{j = 0}^{r}\sum _{i = j}^r b_{2r,i}a_{2r,j} \pi _{2r,j}. \end{aligned}$$

Comparing the coefficient of $\pi _{2r,r}$ on both sides we get $c_r=b_{2r,r}a_{2r,r}=1$ for each $0 \le 2r \le n$. Hence $\psi = {\text {id}}$, which proves (33). $\square$

By plugging (28) and (33) in (30) one gets the tube formulas ${\textbf{T}}_t\tau _{k,p}$ in terms of the $\tau _{i,j}$.

4 Tube formulas in Riemannian manifolds

4.1 Tubular and derivative operators

Next we extend to any complete Riemannian manifold M the tubular operator ${\textbf{T}}_t$ introduced in the previous section on linear spaces. Let T be the Reeb vector field on SM, which is characterized by $i_T \alpha = 1$ and ${\mathcal {L}}_T\alpha = 0$, where ${\mathcal {L}}$ is the Lie derivative. The Reeb flow $\phi : SM \times {\mathbb {R}}\rightarrow SM$, defined as the flow of T, is a family of contactomorphisms and coincides with the geodesic flow on SM (see e.g. [18, Theorem 1.5.2]).

Definition 4.1

(Tubular and derivative operators) Given $t\ge 0$, we define the tubular operator ${\textbf{T}}_t$ by

$$\begin{aligned} {\textbf{T}}_t :{\mathcal {V}}(M) \longrightarrow {\mathcal {V}}(M), \quad [\![\omega ,\eta ]\!]\longmapsto [\![\phi _t^{*}\omega + (p_t)_*(\pi \circ \phi )^*\eta , \eta ]\!], \end{aligned}$$

where $p_t :SM \times [0,t] \rightarrow SM$ is the projection on the first factor, and $\phi _t=\phi (\cdot ,t)$. We define the derivative operator $\partial =\partial _M$ by

$$\begin{aligned} \partial _{M} :{\mathcal {V}}(M) \longrightarrow {\mathcal {V}}(M), \quad \mu \longmapsto \left. \dfrac{d}{dt}\right| _{t = 0} {\textbf{T}}_t\mu . \end{aligned}$$

To show that these definitions are consistent, suppose $\mu = [\![\omega ,\eta ]\!]= 0$, and let us check that ${\textbf{T}}_t\mu = 0$ for all $t \ge 0$, i.e.

$$\begin{aligned} \int _{N(A)}\phi _t^{*}\omega + \int _{N(A)} (p_t)_*(\pi \circ \phi )^*\eta + \int _A \eta = 0, \quad \forall A \in {\mathcal {R}}(M). \end{aligned}$$

By (5) we have $\pi ^*\eta = -D\omega =-d(\omega +\xi \wedge \alpha )$. Hence

$$\begin{aligned} \begin{aligned} \int _{N(A)} (p_t)_*(\pi \circ \phi )^*\eta&= - \int _{N(A)}(p_t)_*\circ \phi ^*D\omega = - \int _{N(A)\times [0,t]}\phi ^* d(\omega + \xi \wedge \alpha ) \\&=- \int _{N(A) \times [0,t]}d\phi ^{*}(\omega +\xi \wedge \alpha ) = -\int _{N(A)\times \{0,t\}}\phi ^* \omega +\phi ^*\xi \wedge \alpha \\&= \int _{N(A)}\phi _{0}^*\omega - \int _{N(A)}\phi _t^{*}\omega = \int _{N(A)}\omega - \int _{N(A)}\phi _t^{*}\omega , \end{aligned} \end{aligned}$$

as $\alpha$ vanishes on N(A). Since $[\![\omega ,\eta ]\!]= 0$, we have $\int _{N(A)}\omega = -\int _A\eta$. Therefore ${\textbf{T}}_t \mu = 0$.

Let us next establish some basic properties of these operators.

Lemma 4.1

 

$$\begin{aligned} \frac{d}{dt}(p_t)_*\phi ^*\rho =i_T \phi _t^*\rho ,\qquad \rho \in \Omega ^*(SM) \end{aligned}$$

Proof

Given a compact smooth submanifold $N\subset SM$,

$$\begin{aligned} \int _N (p_t)_*\phi ^*\rho&=\int _{N\times [0,t]} \phi ^{*}\rho =\int _{N\times [0,t]} i_{\frac{\partial }{\partial t}}\phi ^{*}\rho \wedge dt= \int _0^t\left( \int _{N} \phi _t^{*}i_{\frac{\partial \phi }{\partial t}}\rho \right) dt, \\ \end{aligned}$$

Since $i_T$ and $\phi _t^*$ commute, the result follows. $\square$

Proposition 4.2

For $\mu = [\![\omega ,\eta ]\!]$,

• (i) $\partial \mu = [\![i_T \left( d\omega + \pi ^*\eta \right) ,0]\!]$

• (ii) ${\textbf{T}}_{t+s}\mu =({\textbf{T}}_{t}\circ {\textbf{T}}_{s})\mu$

Proof

Modulo exact forms we have

$$\begin{aligned} \frac{d}{dt} \phi _{t}^*\omega =\left. \frac{d}{ds}\right| _{s=0} \phi _{t+s}^*\omega ={\mathcal {L}}_T \phi _t^*\omega \equiv i_T \phi _t^*d\omega . \end{aligned}$$

(34)

Together with Lemma 4.1, this yields

$$\begin{aligned} \frac{d}{dt}{\textbf{T}}_t \mu =[\![i_T(d\phi _t^*\omega +\phi _t^*\pi ^*\eta ),0]\!]. \end{aligned}$$

(35)

Evaluating at $t=0$, this gives (i).

In order to prove (ii), it is enough to check that both sides have the same derivative with respect to s, as they clearly agree for $s=0$. By (35), we have

$$\begin{aligned} \frac{d}{ds}{\textbf{T}}_t\circ {\textbf{T}}_s (\mu )&= {\textbf{T}}_t\circ \frac{d}{ds} {\textbf{T}}_s(\mu ) \\&={\textbf{T}}_t [\![i_T(d\phi _s^*\omega +\phi _s^*\pi ^*\eta ),0]\!]\\&= [\![\phi _t^* i_T(d\phi _s^*\omega +\phi _s^*\pi ^*\eta ),0]\!]. \end{aligned}$$

Since $\phi _t^*$ and $i_T$ commute, it follows from (35) that $\frac{d}{ds}{\textbf{T}}_{t+s}=\frac{d}{ds}{\textbf{T}}_{t}\circ {\textbf{T}}_{s}$. $\square$

Fix $\mu \in {\mathcal {V}}(M)$. It follows from Proposition 4.2 (ii) that

$$\begin{aligned} \frac{d}{dt} {\textbf{T}}_t\mu =\partial _{} {\textbf{T}}_t\mu . \end{aligned}$$

(36)

If $\mu \in {\mathcal {V}}(M)^G$ for a group G acting on M by isometries, then also ${\textbf{T}}_t\mu \in {\mathcal {V}}(M)^G$. Hence, in case ${\mathcal {V}}^G(M)$ is finite-dimensional, computing ${\textbf{T}}_t\mu$ boils down to solving the Cauchy problem (36) with initial condition ${\textbf{T}}_0\mu =\mu$; i.e.

$$\begin{aligned} {\textbf{T}}_t\mu = {\text {exp}}(t\partial )\mu = \sum _{i \ge 0}\dfrac{t^i}{i!}\partial ^i\mu . \end{aligned}$$

(37)

This is the approach we will follow to obtain the tube formulas for invariant valuations in complex space forms. Note that (37) coincides with (16) except that $\partial$ does not need to be nilpotent for general M.

4.2 Tubes in Riemannian manifolds

Let M be a complete Riemannian manifold and let $d:M \times M \rightarrow [0,\infty )$ be the Riemannian distance on M. For $t\ge 0$, the tube of radius t around a subset $A\subset M$ is defined as

$$\begin{aligned} A_t:= \left\{ p \in M: d_A(p) \le t \right\} , \end{aligned}$$

where

$$\begin{aligned} d_A(p):= \inf \left\{ d(p,q): q \in A \right\} . \end{aligned}$$

Next we review some basic facts about tubes around sets of positive reach (introduced by Federer in euclidean spaces and by Kleinjohann in Riemannian manifolds). For such sets A we will prove that ${\textbf{T}}_t \mu (A)=\mu (A_t)$ for any $\mu \in {\mathcal {V}}(M)$ and sufficiently small t.

Definition 4.2

(Sets of positive reach) A set of positive reach in M is a closed subset $A\subset M$ for which there exists an open neighborhood $U_A\supset A$ such that for every $p\in U_A{\setminus } A$ there exists a unique point $f_A(p)\in A$ such that $d(p,f_A(p))=d_A(p)$, and a unique minimizing geodesic joining p with $f_A(p)$. We denote by ${\mathcal {R}}(M)$ the class of compact sets of positive reach in M.

By the previous definition, there is a well-defined map

$$\begin{aligned} F_A : U_A {\setminus } A \longrightarrow SM,\qquad F_A(p) = \left( \gamma (0),\gamma '(0)\right) \end{aligned}$$

(38)

where $\gamma$ is the unique minimizing geodesic such that $\gamma (0) = f_A(p)$ and $\gamma (d_A(p)) = p$.

It was shown by Kleinjohann [22, Satz 3.3] that $N(A):= F_A(U_A{\setminus } A)$ is a naturally oriented compact Lipschitz submanifold of SM. The corresponding current, also denoted by N(A), is called the normal cycle of A. It follows from Proposition 4.6 below that N(A) is legendrian (i.e. it vanishes on multiples of $\alpha$).

Proposition 4.3

([22, Satz 3.3, Korollar 2.7]) Given a set of positive reach A in M there exists $r=r_A>0$ such that $A_r\subset U_A$ and

• (i) for $0<t<r$ the restriction $\left. F_A\right| _{\partial A_t}$ gives a bilipschitz homeomorphism between $\partial A_t$ and N(A), preserving the natural orientations,

• (ii) the distance function $d_A$ is of class $C^1$ in $A_{r}{\setminus } A$ and

  $$\begin{aligned} \phi _{d_A(p)}(F_A(p))=(p,\nabla d_A(p)),\qquad \partial A_t=d_A^{-1}(\{t\}) \end{aligned}$$

  for $0<t<r$. In particular, each level set $\partial A_t$ with $0<t<r$ is a $C^1$-regular hypersurface with unit normal vector field $\nabla d_A$.

The following propositions are certainly well-known.

Proposition 4.4

For $0<s<r=r_A$ the set $A_s$ has positive reach and on $A_r{\setminus } A_s$ we have

$$\begin{aligned} d_{A_s}=d_A-s,\qquad F_{A_s}=\phi _s\circ F_A. \end{aligned}$$

(39)

In particular $(A_s)_t=A_{t+s}$ for $t+s<r$.

Proof

Let $p \in A_r{\setminus } A_s$, and put $d=d_A(p)$. Let $\gamma :[0,d]\rightarrow A_r$ be the unique minimizing geodesic with $\gamma (0)=f_A(p)$ and $\gamma (d)=p$. In particular $|\gamma '|=1$ and thus $\gamma (s)\in A_s$.

Assume that $\left. \gamma \right| _{[s,d]}$ does not minimize the distance between p and $A_s$, i.e., there exists a smooth curve $\alpha : [0,1] \rightarrow M$ with $q:=\alpha (0)\in A_s$, $\alpha (1)=p$ and length $\ell (\alpha ) < d-s$. It follows that

$$\begin{aligned} d_A(p)\le \ell (\alpha )+d_A(q)\le \ell (\alpha )+s<d_A(p), \end{aligned}$$

a contradiction. We conclude that $\gamma |_{[s,d]}$ realizes the distance $d_{A_s}(p)$. Hence $d_{A_s}(p)=d_A(p)-s$ and

$$\begin{aligned} F_{A_s}(p)=(\gamma (s),\gamma '(s))=\phi _s(\gamma (0),\gamma '(0))=\phi _s(F_A(p)). \end{aligned}$$

$\square$

Proposition 4.5

For $0<s<r_A$, the restriction $\left. \phi _s\right| _{N(A)}$ is a bilipschitz homeomorphism between N(A) and $N(A_s)$.

Proof

Take t with $s<t<\min (r_A,s+r_{A_s})$. By Proposition 4.3, both $\left. F_A\right| _{\partial A_t}:\partial A_t \rightarrow N(A)$ and $\left. F_{A_s}\right| _{\partial A_t} :\partial A_t \rightarrow N(A_s)$ are bilipschitz homeomorphisms. By (39) we have

$$\begin{aligned} \left. \phi _s\right| _{N(A)}=\left. F_{A_s}\right| _{\partial A_t}\circ (\left. F_A\right| _{\partial A_t})^{-1}. \end{aligned}$$

The statement follows. $\square$

Proposition 4.6

For $0<t<r_A$ the composition $\pi \circ \phi$ gives a bijective Lipschitz map between $N(A) \times (0,t]$ and $A_t {\setminus } A$.

Proof

Since $\pi ,\phi$ are smooth, the restriction of $\pi \circ \phi$ to the Lipschitz manifold $N(A)\times (0,t]$ is clearly Lipschitz.

Given $(\xi ,s)\in N(A)\times (0,t]$, we know by the previous proposition that $\phi (\xi ,s)\in N(A_s)$ and thus $\pi \circ \phi (x,s)\in \partial A_s\subset A_t{\setminus } A$.

To check surjectivity, given $p\in A_t{\setminus } A$ take $\xi =F_A(p), s=d_A(p)$ and note that $\pi \circ \phi (\xi ,s)=p$.

As for injectivity, suppose $\pi \circ \phi (\xi _1,t_1)=\pi \circ \phi (\xi _2,t_2)=:p$ for some $(\xi _1,t_1),(\xi _2,t_2)\in N(A)\times (0,t]$. By the previous proposition p belongs to both $\partial A_{t_1},\partial A_{t_2}$, so $t_1=t_2$. For $s\in [0,t_1]$, the geodesics $\gamma _1(s)=\pi \circ \phi (\xi _1,s),\gamma _2(s)=\pi \circ \phi (\xi _2,s)$ realize the distance between p and A. Since $A_s\subset A_{r_A}\subset U_A$, we have $\gamma _1=\gamma _2$ and thus $\xi _1=\xi _2$. $\square$

Corollary 4.7

For every $A\in {\mathcal {R}}(M)$ and $\mu \in {\mathcal {V}}(M)$ we have $\mu (A_t) = {\textbf{T}}_t\mu (A)$ for $0\le t\le r_A$.

Proof

Let $\mu = [\![\omega ,\eta ]\!]$. By Propositions 4.5 and 4.6 and the coarea formula,

$$\begin{aligned} \mu (A_t)&= \int _{N(A_t)}\omega + \int _{A_t} \eta \int _{\phi _t(N(A))} \omega \\&\quad + \int _{(\pi \circ \phi )(N(A)\times (0,t])} \eta + \int _A \eta = {\textbf{T}}_t\mu (A). \end{aligned}$$

$\square$

Remark

In the subclass ${\mathcal {P}}(M)\subset {\mathcal {R}}(M)$ of compact submanifolds with corners, the normal cycle is more naturally defined as follows. For $A \in {\mathcal {P}}(M)$ and $p \in A$, let

$$\begin{aligned} T_p A&=\left\{ \gamma '(0) \in T_p M : \gamma \in C^1([0,1),A), \gamma (0) = p\right\} \\ N'(A)&= \{(p,v) \in SM : p \in A, \langle v, w\rangle \le 0\ \forall w \in T_p A\}. \end{aligned}$$

Let us check that indeed $N'(A)$ equals $N(A)=F_A(U_A)$. Covering A by local charts (locally modelled on ${\mathbb {R}}^k\times [0,\infty )^l\subset {\mathbb {R}}^m$), and considering the copy of $N'(A)$ in the cosphere bundle of M, one sees that $N'(A)$ is a compact topological manifold.

It is also easy to show that $N(A)\subset N'(A)$. It then follows by the invariance of domain theorem that N(A) is an open subset of $N'(A)$. Since $N'(A)$ is a Hausdorff space and N(A) is compact, we also have that N(A) is a closed subset of $N'(A)$. Since the number of connected components of both $N(A),N'(A)$ clearly equals the number of connected components of A, we necessarily have $N(A)=N'(A)$.

4.3 Derivative operators in ${\mathbb {S}}^{m}_{\lambda }$ and ${\mathbb {C}}P^{n}_{\lambda }$

Given $\lambda \in {\mathbb {R}}$ let $\partial _{\lambda ,{\mathbb {R}}}:{\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m\rightarrow {\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ be the restriction of $\partial _{{\mathbb {S}}^{m}_{\lambda }}$ to ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$, and let $\partial _{\lambda ,{\mathbb {C}}}$ be the restriction of $\partial _{{\mathbb {C}}P^{m}_{\lambda }}$ to ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$.

Proposition 4.8

 

$$\begin{aligned}&\partial _{\lambda ,{\mathbb {R}}} \sigma _i^{\lambda } = (m-i)\sigma _{i-1}^{\lambda }-\lambda (i+1)\sigma _{i+1}^{\lambda },&0 \le i \le m-2, \end{aligned}$$

(40)

$$\begin{aligned}&\partial _{\lambda ,{\mathbb {R}}} \sigma _{m-1}^{\lambda } = \sigma _{m-2}^{\lambda }, \end{aligned}$$

(41)

$$\begin{aligned}&\partial _{\lambda ,{\mathbb {R}}} \sigma _{m}^{\lambda } = \sigma _{m-1}^{\lambda }, \end{aligned}$$

(42)

where it is understood that $\sigma _{-1}^\lambda =0$.

Let us emphasize that (40) would make formal sense but does not hold for $i=m-1$.

Proof

By [15, Lemma 3.1], putting $\kappa _m=0$, we have

$$\begin{aligned} d\kappa _{i} = \alpha \wedge \left( (m-i) \kappa _{i-1}-\lambda (i+1) \kappa _{i+1}\right) , \quad 0 \le i \le m-1. \end{aligned}$$

Contracting with T yields

$$\begin{aligned} i_T d\kappa _i = (m-i)\kappa _{i-1} - \lambda (i+1)\kappa _{i+1}, \quad \mod (\alpha ,d\alpha ). \end{aligned}$$

By Proposition 4.2 the result follows. $\square$

Lemma 4.9

The following equalities hold modulo $\alpha ,d\alpha$

• (i) For $k > 2q$

  $$\begin{aligned} \begin{aligned} \dfrac{\omega _{2n-k}}{\omega _{2n-k+1}}i_{T} d\beta _{k,q}&\equiv 2 (n-k+q+1)\gamma _{k-1,q}\\ {}&\quad +{(k-2q+1)}\beta _{k-1,q-1} \\&\quad -\frac{\lambda }{2\pi } {(k-2q+1)(2n-k+1)}\beta _{k+1,q}. \end{aligned} \end{aligned}$$

  (43)

• (ii) For $n > k-q$

  $$\begin{aligned} \begin{aligned} \dfrac{\omega _{2n-2q}}{\omega _{2n-2q+1}}i_T d\gamma _{2q,q}&\equiv \beta _{2q-1,q-1} \\&\quad -\frac{\lambda }{2\pi } \frac{(q+2)(2n-2q+1)}{n-q}\beta _{2q+1,q} \\&\quad -\frac{\lambda }{2\pi } \frac{(n-q-1)(2n-2q+1)}{n-q}\gamma _{2q+1,q}. \end{aligned} \end{aligned}$$

  (44)

Proof

This is a straightforward computation using [1, Lemma 3.3, Lemma 3.6]. $\square$

Proposition 4.10

For $k > 2q$

$$\begin{aligned} \dfrac{\omega _{2n-k}}{\omega _{2n-k+1}}\partial _{\lambda ,{\mathbb {C}}}\mu _{k, q}^\lambda&=(k-2q+1) \mu _{k-1,q-1}^\lambda + 2 (n-k+q+1)\mu _{k-1,q}^\lambda \nonumber \\&\quad - \dfrac{\lambda }{2 \pi } (2 n-k+1)\left( (k-2 q+1) \mu _{k+1,q}^\lambda +2 (q+1) \mu _{k+1,q+1}^\lambda \right) \end{aligned}$$

(45)

and

$$\begin{aligned} \dfrac{\omega _{2n-2q}}{\omega _{2n-2q+1}} \partial _{\lambda ,{\mathbb {C}}} \mu _{2q,q}^\lambda = \mu _{2q-1,q-1} - (2n-2q+1) \dfrac{\lambda }{2 \pi }\mu _{2q+1,q}. \end{aligned}$$

(46)

Proof

Equality (45) follows from Proposition 4.2, using (43) and the following (see [1, Proposition 2.7])

$$\begin{aligned}{}[\![\gamma _{k,q},0 ]\!]= \mu _{k,q}^\lambda - \lambda \dfrac{(2n-k)(q+1)}{2\pi (n-k+q)}\mu _{k+2,q+1}^\lambda , \quad n-k+q > 0. \end{aligned}$$

(47)

Let us now prove (46). Note first that from (44) and (47) we get

$$\begin{aligned} \begin{aligned}{}[\![i_T d\gamma _{2j,j},0]\!]&=\dfrac{\omega _{2n-2j+1}}{\omega _{2n-2j}} \mu _{2j-1,j-1}^\lambda \\&\quad -\dfrac{\omega _{2n-2j+1}}{\omega _{2n-2j}} \dfrac{(2n-2j+1)(n+1)}{n-j}\dfrac{\lambda }{2\pi }\mu _{2j+1,j}^\lambda \\&\quad + \dfrac{\omega _{2n-2j+1}}{\omega _{2n-2j}}\dfrac{(2n-2j+1)(2n-2j-1)(j+1)}{n-j}\dfrac{\lambda ^2}{4\pi ^2}\mu _{2j+3,j+1}^\lambda \\&=: a_j \mu _{2j-1,j-1}^\lambda +b_j\frac{\lambda }{\pi }\mu _{2j+1,j}^\lambda +c_j\frac{\lambda ^2}{\pi ^2}\mu _{2j+3,j+1}^\lambda \end{aligned} \end{aligned}$$

(48)

Then, by Proposition 4.2 and observing that $a_{n} = 2$

$$\begin{aligned} \partial _{\lambda ,{\mathbb {C}}}\mu _{2q,q}^\lambda&= \sum _{i = 0}^{n-q-1}\left( \dfrac{\lambda }{\pi }\right) ^i \dfrac{(q+i)!}{q!}[\![i_T d\gamma _{2q+2i,q+i},0]\!]+ 2 \left( \dfrac{\lambda }{\pi }\right) ^{n-q}\frac{n!}{q!}\mu _{2n-1,n-1}^\lambda \\&= \sum _{i = 0}^{n-q}\left( \dfrac{\lambda }{\pi }\right) ^i \dfrac{(q+i)!}{q!}a_{q+i}\mu _{2q+2i-1,q+i-1}^\lambda \\&\quad +\sum _{i = 0}^{n-q-1}\left( \dfrac{\lambda }{\pi }\right) ^{i+1} \dfrac{(q+i)!}{q!}b_{q+i}\mu _{2q+2i+1,q+i}^\lambda \\&\quad +\sum _{i = 0}^{n-q-2}\left( \dfrac{\lambda }{\pi }\right) ^{i+2} \dfrac{(q+i)!}{q!}c_{q+i}\mu _{2q+2i+3,q+i+1}^\lambda \\&= a_q\mu _{2q-1,q-1} + \frac{\lambda }{\pi }((q+1)a_{q+1} + b_q)\mu _{2q+1,q}^\lambda \\&\quad + \sum _{j = 2}^{n-q} \left( \dfrac{\lambda }{\pi }\right) ^j\left( \dfrac{(q+j)!}{q!}a_{q+j} + \dfrac{(q+j-1)!}{q!}b_{q+j-1} \right. \\&\left. \quad +\dfrac{(q+j-2)!}{q!}c_{q+j-2}\right) \mu _{2q+2j-1,q+j-1}^\lambda \end{aligned}$$

A straightforward computation using $k\omega _k=2\pi \omega _{k-2}$ shows

$$\begin{aligned} j(j-1)a_j+(j-1)b_{j-1}+c_{j-2}=0 \end{aligned}$$

and the result follows. $\square$

Note that by (18) the linear map $\Phi _0:{\text {Val}}^{U(n)}\rightarrow {\text {Val}}^{U(n)}$ given by $\left. \Phi _0\right| _{{\text {Val}}_k^{U(n)}}=\omega _{2n-k}{\text {id}}$ satisfies

$$\begin{aligned} \partial _{0,{\mathbb {C}}}=\Phi _0\circ \Lambda \circ \Phi _0^{-1}. \end{aligned}$$

Remarkably, a similar identity holds for all $\lambda$, which will be crucial for our determination of tube formulas in ${\mathbb {C}}P^{n}_{\lambda }$.

Theorem 4.11

The linear isomorphism

$$\begin{aligned} \Phi _\lambda ={\mathcal {F}}_{\lambda ,{\mathbb {C}}}\circ \Phi _0:{\text {Val}}^{U(n)} \longrightarrow {\mathcal {V}}^n_{\lambda ,{\mathbb {C}}}, \qquad \mu _{k,q}\longmapsto \omega _{2n-k}\mu _{k,q}^\lambda . \end{aligned}$$

fulfills

$$\begin{aligned} \partial _{\lambda ,{\mathbb {C}}} = \Phi _\lambda \circ \left( \Lambda - \lambda L\right) \circ \Phi _\lambda ^{-1}. \end{aligned}$$

Proof

By combining Proposition 4.10, Proposition 3.1 and the fact $\frac{\omega _n}{\omega _{n-2}} = \frac{2\pi }{n}$, this is straightforward to check:

$$\begin{aligned}&\Phi _\lambda \circ (\Lambda - \lambda L)(\mu _{k,q}) \\&\quad = (k-2q+1)\omega _{2n-k+1}\mu _{k-1,q-1}^\lambda +2(n-k+q+1)\omega _{2n-k+1}\mu _{k-1,q}^\lambda \\&\quad \quad - \lambda (k-2q+1)\omega _{2n-k-1}\mu _{k+1,q}^\lambda - 2\lambda (q+1)\omega _{2n-k-1}\mu _{k+1,q+1}^\lambda \\&\quad = \omega _{2n-k+1}\left( (k-2q+1) \mu _{k-1,q-1}^\lambda + 2 (n-k+q+1)\mu _{k-1,q}^\lambda \right. \\&\quad \quad - \dfrac{\lambda }{2 \pi } (2 n-k+1)\left( (k-2 q+1) \mu _{k+1,q}^\lambda +2 (q+1) \mu _{k+1,q+1}^\lambda \right) \\&\quad = \omega _{2n-k}\partial _{\lambda ,{\mathbb {C}}}\mu _{k,q}^\lambda = \partial _{\lambda ,{\mathbb {C}}}\circ \Phi _\lambda (\mu _{k,q}). \end{aligned}$$

$\square$

A similar phenomenon holds in real space forms, but restricted to a hyperplane of ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$.

Theorem 4.12

The linear monomorphism

$$\begin{aligned} \Psi _\lambda : {\text {Val}}^{O(m)} \longrightarrow {\mathcal {V}}^{m+1}_{\lambda ,{\mathbb {R}}}, \qquad \mu _k \longmapsto \sigma _k^{\lambda } \end{aligned}$$

fulfills

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}} \circ \Psi _\lambda = \Psi _\lambda \circ \left( \Lambda - \lambda L\right) . \end{aligned}$$

Proof

By Proposition 4.8 and Theorem 3.4

$$\begin{aligned} \begin{aligned} \partial _{\lambda ,{\mathbb {R}}} \circ \Psi _\lambda (\mu _k)&= \partial _{\lambda ,{\mathbb {R}}}\sigma _k^{\lambda } = (m-k+1)\sigma _{k-1}^\lambda - \lambda (k+1)\sigma _{k+1}^\lambda \\&= \Psi _\lambda ((m-k+1)\mu _{k-1} - \lambda (k+1)\mu _{k+1}) \\&= \Psi _\lambda (\Lambda \mu _k - \lambda L\mu _{k}). \end{aligned} \end{aligned}$$

$\square$

Note the difference of dimensions between the source and the target of $\Psi _\lambda$. We will show that there is no isomorphism between ${\text {Val}}^{O(m)}$ and ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ intertwining $\partial$ and $\Lambda -\lambda L$. This is essentially due to the fact that (41) and (42) differ from (40).

5 A model space for tube formulas

We next perform some abstract computations that will easily lead to the tube formulas in both complex and real space forms via (62) and (64). The same approach will allow us to determine the kernel, the image, and the spectrum of the derivative operator $\partial$ on these spaces.

5.1 A system of differential equations

It is well-known that the operators $X = x\frac{\partial }{\partial y}$, $Y = y \frac{\partial }{\partial x}$ and $H=[X,Y]$ induce an $\mathfrak {sl}_2$-structure on ${\mathbb {C}}[x,y]$. The decomposition into irreducible components is ${\mathbb {C}}[x,y]=\bigoplus _{m\ge 0} V^{(m)}$ where $V^{(m)}$ is the subspace of m-homogeneous polynomials:

$$\begin{aligned} V^{(m)}:= {\text {span}}_{\mathbb {C}}\displaystyle \{x^ky^{m-k}\}_{k = 0}^m. \end{aligned}$$

One has $H(x^ky^{m-k})=(m-2k) x^k y^{m-k}$.

Motivated by Theorem 4.11, we consider $Y_\lambda = Y - \lambda X$, which is a derivation on ${\mathbb {C}}[x,y]$. It will be sometimes convenient to consider the monomials ${m\atopwithdelims ()k}x^ky^{m-k}$. In these terms

$$\begin{aligned} {m\atopwithdelims ()k}Y_\lambda (x^ky^{m-k}) = (m-k+1){m\atopwithdelims ()k-1}x^{k-1}y^{m-k+1} - \lambda (k+1){m\atopwithdelims ()k+1} x^{k+1}y^{m-k-1}. \end{aligned}$$

(49)

Our goal here is to solve the following Cauchy problem: find $p_k:{\mathbb {R}}\rightarrow V^{(m)}$ such that

$$\begin{aligned} \partial _t p_k(t)=Y_\lambda p_k(t) , \quad p_k(0) = \left( {\begin{array}{c}m\\ k\end{array}}\right) x^{k}y^{m-k}, \quad 0 \le k \le m, \end{aligned}$$

(50)

i.e. to compute

$$\begin{aligned} p_k(t) = \left( {\begin{array}{c}m\\ k\end{array}}\right) \exp (tY_\lambda )(x^{k}y^{m-k}), \quad 0 \le k \le m. \end{aligned}$$

(51)

We will use the standard notation

$$\begin{aligned} {\text {sin}}_{\lambda }(t):=\left\{ \begin{array}{cc} \dfrac{\sin (\sqrt{\lambda } t)}{\sqrt{\lambda }} &{}\quad \lambda >0, \\ \\ t &{}\quad \lambda =0, \\ \\ \dfrac{\sinh (\sqrt{|\lambda |} t)}{\sqrt{|\lambda |}} &{}\quad \lambda <0, \end{array}\right. \end{aligned}$$

(52)

which is an analytic function in both $\lambda$ and t, and $\cos _\lambda (t):= \frac{d}{dt}\sin _\lambda (t)$.

Proposition 5.1

For any $\lambda ,t \in {\mathbb {R}}$, we have

$$\begin{aligned} \exp (tY_\lambda )x = x \cos _\lambda (t) + y \sin _\lambda (t){=:u}, \quad \exp (tY_\lambda )y = y \cos _\lambda (t)-\lambda x \sin _\lambda (t){=:v}. \end{aligned}$$

Proof

Since clearly

$$\begin{aligned} Y_\lambda ^{2k}x = (-\lambda )^k x, \quad Y_\lambda ^{2k+1}x = (-\lambda )^k y, \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} \exp (tY_\lambda )x&=\sum _{k \ge 0}\dfrac{t^k}{k!}Y_{\lambda }^{k}x\\&=\sum _{k \ge 0} \dfrac{t^{2k}}{(2k)!}(-\lambda )^k x+ \sum _{k \ge 0}\dfrac{t^{2k+1}}{(2k+1)!}(-\lambda )^k y \\&= x\cos _\lambda (t) + y \sin _\lambda (t). \end{aligned} \end{aligned}$$

In the same way we can compute $\exp (tY_\lambda )y$. $\square$

The following standard and elementary fact will be useful.

Lemma 5.2

Let ${\textbf{A}}$ be a finite-dimensional algebra. A vector field on ${\textbf{A}}$ is a derivation iff its flow $\phi _t$ satisfies

$$\begin{aligned} \phi _t(pq) = \phi _t(p)\phi _t(q), \quad \forall p,q \in {\textbf{A}},\forall t\in {\mathbb {R}}. \end{aligned}$$

In other words, each $\phi _t$ is an ${\textbf{A}}$-morphism.

Theorem 5.3

The solution of the Cauchy problem (50) is

$$\begin{aligned} p_k(t)&= \left( {\begin{array}{c}m\\ k\end{array}}\right) u^k v^{m-k} \end{aligned}$$

(53)

$$\begin{aligned}&=\left( {\begin{array}{c}m\\ k\end{array}}\right) (x\cos _\lambda (t) + y \sin _\lambda (t))^k (y \cos _\lambda (t) - \lambda x \sin _\lambda (t))^{m-k} \end{aligned}$$

(54)

$$\begin{aligned}&= {\sum _{j = 0}^{m}\phi _{m,k,j}^{\lambda }(t)\left( {\begin{array}{c}m\\ j\end{array}}\right) x^j y^{m-j}}, \end{aligned}$$

(55)

where

$$\begin{aligned} \phi _{m,k,j}^\lambda (t) = \sum _{h \ge 0}(-\lambda )^{j-h}\left( {\begin{array}{c}m-j\\ k-h\end{array}}\right) \left( {\begin{array}{c}j\\ h\end{array}}\right) \sin _\lambda ^{k+j-2h}(t)\cos _\lambda ^{m-k-j+2h}(t). \end{aligned}$$

(56)

Proof

Since $Y_\lambda$ is a derivation, $\exp (tY_\lambda )$ is a ${\mathbb {C}}[x,y]$-morphism by the previous lemma. Hence

$$\begin{aligned} \exp (tY_\lambda )x^{k}y^{m-k} = (\exp (tY_\lambda ) x)^k (\exp (tY_\lambda ) y)^{m-k} = u^k v^{m-k}. \end{aligned}$$

Comparing with (51) yields (54).

It remains to prove (55). Putting $s=\sin _\lambda (t),c=\cos _\lambda (t)$ we have

$$\begin{aligned}&{m\atopwithdelims ()k}(xc + y s)^k(yc - \lambda x s)^{m-k}\\&\qquad = {m\atopwithdelims ()k} \sum _{a,b}{k\atopwithdelims ()a}(ys)^a (xc)^{k-a}{m-k\atopwithdelims ()b}(-\lambda xs)^b(yc)^{m-k-b}\\&\qquad = {m\atopwithdelims ()k} \sum _{a,b}{k\atopwithdelims ()a}{m-k\atopwithdelims ()b}(-\lambda )^b s^{a+b} c^{m-a-b} x^{k-a+b} y^{m-k+a-b}\\&\qquad ={m\atopwithdelims ()k}\sum _{j,h} {k\atopwithdelims ()h} {m-k\atopwithdelims ()j-h} (-\lambda )^{j-h} s^{j+k-2h}c^{m-j-k+2h} x^j y^{m-j} \end{aligned}$$

where we changed $a=k-h,b=j-h$. Using

$$\begin{aligned} {m\atopwithdelims ()k}{k\atopwithdelims ()h}{m-k\atopwithdelims ()j-h}={m-j\atopwithdelims ()k-h}{j\atopwithdelims ()h}{m\atopwithdelims ()j} \end{aligned}$$

(57)

yields (55). $\square$

5.2 Eigenvalues and eigenvectors of $Y_\lambda$

Given $f:V \rightarrow V$ an endomorphism of ${\mathbb {C}}$-vector spaces, we denote by ${\text {spec}}(f)$ the set of eigenvalues of f and by $E_\alpha (f)$ the eigenspace associated to each $\alpha \in {\text {spec}}(f)$.

Lemma 5.4

The endomorphism $\left. Y_\lambda \right| _{V^{(m)}}$ is diagonalizable with simple multiplicities and

$$\begin{aligned} {\text {spec}}(\left. Y_\lambda \right| _{V^{(m)}})= & {} \left\{ (2k-m)\sqrt{-\lambda }: 0 \le k \le m \right\} , \\{} & {} E_{(2k-m)\sqrt{-\lambda }}(\left. Y_\lambda \right| _{V^{(m)}}) = {\text {span}} \{e_1^ke_2^{m-k}\}, \end{aligned}$$

where $e_1:= \sqrt{-\lambda }x + y$ and $e_2:= -\sqrt{-\lambda }x + y$.

Proof

The result is trivial to check for $m=1$ as

$$\begin{aligned} Y_\lambda (e_1)=\sqrt{-\lambda }y-\lambda x=\sqrt{-\lambda }e_1,\qquad Y_\lambda (e_2)=-\sqrt{-\lambda }y-\lambda x=-\sqrt{-\lambda }e_2. \end{aligned}$$

Since $Y_\lambda$ is a derivation

$$\begin{aligned} \begin{aligned}&Y_\lambda e_1^{k} = k e_{1}^{k-1}Y_\lambda e_1 = k \sqrt{-\lambda } e_1^{k},\\&Y_\lambda e_2^{m-k} = (m-k) e_2^{m-k-1}Y_\lambda e_2 = -\sqrt{-\lambda }(m-k) e_2^{m-k}. \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} Y_\lambda (e_1^{k}e_2^{m-k}) = (2k-m) \sqrt{-\lambda } e_1^{k}e_2^{m-k}, \end{aligned}$$

as stated. $\square$

Remark

It is interesting to notice that the spectra of $Y_\lambda$ and $\sqrt{-\lambda } H$, when restricted to each $V^{(m)}$, are identical. These two operators are thus intertwined e.g. by the linear isomorphism $x^ky^{m-k}\mapsto e_1^{k}e_2^{m-k}$.

5.3 Image of $Y_\lambda$

Using Lemma 5.4, we can conclude that $\left. Y_\lambda \right| _{V^{(m)}}$ is bijective if and only if m is odd. If m is even, then the kernel is one-dimensional. An explicit description is the following.

Proposition 5.5

If m is even, then

$$\begin{aligned} {\text {im}}(\left. Y_\lambda \right| _{V^{(m)}}) = \ker Z_{m,\lambda }, \end{aligned}$$

(58)

where

$$\begin{aligned} Z_{m,\lambda }:= \left( \dfrac{\partial ^{2}}{\partial x^{2}} + \lambda \dfrac{\partial ^{2}}{\partial y^{2}}\right) ^{m/2}. \end{aligned}$$

Proof

By the binomial formula

$$\begin{aligned} Z_{m,\lambda }(x^k y^{m-k})&= \sum _{i = 0}^{m/2} \lambda ^{m/2- i}\left( {\begin{array}{c}m/2\\ i\end{array}}\right) \dfrac{\partial ^{m}}{\partial x^{2i}\partial y^{m-2i}}x^k y^{m-k}\delta _{k,2i}\nonumber \\&=\lambda ^{\frac{m-k}{2}} \left( {\begin{array}{c}m/2\\ k/2\end{array}}\right) k!(m-k)! \end{aligned}$$

(59)

if k is even, and $Z_{m,\lambda }(x^k y^{m-k}) = 0$ if k is odd. Therefore

$$\begin{aligned} Z_{m,\lambda }\circ Y_\lambda (x^{2l+1}y^{m-2l-1})&= Z_{m,\lambda }((2l+1) x^{2l}y^{m-2l} - \lambda (m-2l-1)x^{2l+2}y^{m-2l-2}) \\&= \lambda ^{\frac{m}{2}-l} \left( {\begin{array}{c}m/2\\ l\end{array}}\right) (2 l+1)! (m-2l)!\\&\quad -\lambda ^{\frac{m}{2}-l} \left( {\begin{array}{c}m/2\\ l+1\end{array}}\right) (2l+2)! (m-2l-1)! = 0\\&Z_{m,\lambda }\circ Y_\lambda (x^{2l}y^{m-2l}) =\, 0. \end{aligned}$$

This shows that $\text {im}(Y_\lambda )$ is a subspace of $\ker Z_{m,\lambda }$. Given that $Z_{m,\lambda }$ is not zero, we have $\text {dim} \ker Z_{m,\lambda } = m$, and by Lemma 5.4, we know that the image of $\left. Y_\lambda \right| _{V^{(m)}}$ has the same dimension. This yields (58). $\square$

Next we compute, for even m and given $\varphi$ in the image of $\left. Y_\lambda \right| _{V^{(m)}}$, the preimage $Y_\lambda ^{-1}(\{\varphi \})$.

Consider

$$\begin{aligned} P_{m,k} := \sum _{j \ge 0} \lambda ^j \dfrac{(k+2j-1)!! (m-k-2j-1)!!}{(k-1)!!(m-k+1)!!}\left( {\begin{array}{c}m\\ k+2j\end{array}}\right) x^{k+2j}y^{m-k-2j} \in V^{(m)}. \end{aligned}$$

(60)

A simple computation using (49) shows

$$\begin{aligned} Y_\lambda P_{m,k}=\left( {\begin{array}{c}m\\ k-1\end{array}}\right) x^{k-1}y^{m-k+1}-c_{m,k} x^m \end{aligned}$$

where $c_{m,k}=0$ if $m-k$ is even, and otherwise

$$\begin{aligned} c_{m,k}=\lambda ^{\frac{m-k+1}{2}}\frac{m!!}{(k-1)!!(m-k+1)!!}. \end{aligned}$$

With these ingredients at hand, for even m, we can now compute a preimage by $Y_\lambda$ of any element in ${\text {im}} Y_\lambda$ as follows.

Proposition 5.6

Let $\Pi :V^{(m)} \rightarrow V^{(m)}$ be given by $\left( {\begin{array}{c}m\\ k\end{array}}\right) x^k y^{m-k} \mapsto P_{m,k+1}$. If m is even then

$$\begin{aligned} {Y_\lambda \circ \Pi }(\varphi ) =\varphi ,\qquad \forall \varphi \in {\text {im}}\left. Y_\lambda \right| _{V^{(m)}} \end{aligned}$$

(61)

Proof

Let $0< k < m$. Since $(m-k+1)c_{m,k} - \lambda (k+1)c_{m,k+2} = 0$, using (49) we get

$$\begin{aligned}&Y_\lambda \circ \Pi \circ Y_\lambda \left( {\begin{array}{c}m\\ k\end{array}}\right) x^k y^{m-k} \\&\quad = Y_\lambda \circ \Pi \left( (m-k+1)\left( {\begin{array}{c}m\\ k-1\end{array}}\right) x^{k-1}y^{m-k+1} - \lambda (k+1)\left( {\begin{array}{c}m\\ k+1\end{array}}\right) x^{k+1}y^{m-k-1}\right) \\&\quad =(m-k+1)Y_\lambda P_{m,k} - \lambda (k+1)Y_\lambda P_{m,k+2} \\&\quad = (m-k+1)\left( {\begin{array}{c}m\\ k-1\end{array}}\right) x^{k-1}y^{m-k+1} - \lambda (k+1)\left( {\begin{array}{c}m\\ k+1\end{array}}\right) x^{k+1}y^{m-k-1}\\&\quad \quad -((m-k+1)c_{m,k}-\lambda (k+1)c_{m,k+2})x^m \\&\quad = Y_\lambda \left( {\begin{array}{c}m\\ k\end{array}}\right) x^k y^{m-k}. \end{aligned}$$

For $k = 0$ and $k = m$,

$$\begin{aligned} Y_\lambda \circ \Pi \circ Y_\lambda (y^m)&= -\lambda Y_\lambda (P_{m,2})=-\lambda (mx y^{m-1} - c_{m,2}x^m)=Y_\lambda (y^m) +\lambda c_{m,2}x^m, \\ Y_\lambda \circ \Pi \circ Y_\lambda (x^m)&= Y_\lambda (P_{m,m})=m x^{m-1}y - c_{m,m}x^m =Y_\lambda (x^m) + c_{m,m}x^m. \end{aligned}$$

Since $c_{m,m} = 0$, and $c_{m,2} = 0$ if m is even, the result follows. $\square$

6 Tube formulas in ${\mathbb {S}}^{m}_{\lambda }$ and ${\mathbb {C}}P^{n}_{\lambda }$

Here we will obtain our main result: the tube formulas for invariant valuations of ${\mathbb {C}}P^{n}_{\lambda }$ (i.e. the tubular operator ${\textbf{T}}_t$ on ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$). We will also recover Santaló’s tube formulas for ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ (cf. [25]) in a way that explains the similarities between the real and the complex space forms.

6.1 Tube formulas in complex space forms

Recalling (25) and Proposition 3.3, we get an isomorphism $I:W_n\rightarrow {\text {Val}}^{U(n)}$ of $\mathfrak {sl}_2$-modules from

$$\begin{aligned} W_{n}:=\bigoplus _{0 \le 2r \le n} V^{(2n-4r)} \end{aligned}$$

to ${\text {Val}}^{U(n)}$ by putting $I(y^{2n-4r})=\pi _{2r,r}$ (i.e. mapping Y-primitive elements to $\Lambda$-primitive elements) and

$$\begin{aligned} \begin{aligned} {2n-4r\atopwithdelims ()k-2r}I(x^{k-2r} y^{2n-k-2r})&=\frac{1}{(k-2r)! }I(X^{k-2r}(y^{2n-4r}))\\&= \frac{1}{(k-2r)! }L^{k-2r}I(y^{2n-4r})=\frac{1}{(k-2r)! }\pi _{k,r}. \end{aligned} \end{aligned}$$

By Theorem 4.11, the map $J_{\lambda ,{\mathbb {C}}}:=\Phi _\lambda \circ I:W_{n}\rightarrow \mathcal V_{\lambda ,{\mathbb {C}}}^n$ fulfills

$$\begin{aligned} \partial _{\lambda ,{\mathbb {C}}} \circ J_{\lambda ,{\mathbb {C}}} =J_{\lambda ,{\mathbb {C}}} \circ Y_\lambda . \end{aligned}$$

(62)

We define

$$\begin{aligned} \sigma _{k,r}^\lambda := {2n-4r\atopwithdelims ()k-2r}J_{\lambda }(x^{k-2r} y^{2n-k-2r})= \dfrac{\omega _{2n-k}}{(k-2r)!}\pi _{k,r}^\lambda \end{aligned}$$

(63)

and arrive at our main theorem.

Theorem 6.1

The tubular operator ${\textbf{T}}_t$ in ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ is given by

$$\begin{aligned} {\textbf{T}}_t(\sigma _{k,r}^{\lambda }) = \sum _{j = 0}^{2n-4r}\phi _{2n-4r,k-2r,j}^\lambda (t)\sigma _{j+2r,r}^{\lambda }, \end{aligned}$$

where

$$\begin{aligned} \phi _{m,i,j}^\lambda (t) = \sum _{h \ge 0}(-\lambda )^{j-h}\left( {\begin{array}{c}m-j\\ i-h\end{array}}\right) \left( {\begin{array}{c}j\\ h\end{array}}\right) \sin _\lambda ^{i+j-2h}(t)\cos _\lambda ^{m-i-j+2h}(t). \end{aligned}$$

Proof

By (37), using (62) and (63), and putting $m=2n-4r$, we get

$$\begin{aligned} {\textbf{T}}_t \sigma _{k,q}^\lambda&=\textrm{exp}(t \partial _{\lambda ,{\mathbb {C}}})(\sigma _{k,q}^\lambda )\\&={m\atopwithdelims ()k-2r}\textrm{exp}(t \partial _{\lambda ,{\mathbb {C}}})\circ J_{\lambda ,{\mathbb {C}}}(x^{k-2r} y^{m-k+2r})\\&={m\atopwithdelims ()k-2r}J_{\lambda ,{\mathbb {C}}}\circ \textrm{exp} (t Y_\lambda )(x^{k-2r} y^{m-k+2r})\\&= J_{\lambda ,{\mathbb {C}}} (p_{k-2r}(t)). \end{aligned}$$

Using (55) the result follows. $\square$

The tube formulas in terms of the $\tau _{k,i}^\lambda$ can be obtained from Theorem 6.1 using (28) and (33) which hold verbatim replacing $\pi ^\lambda _{k,r},\tau _{k,r}^\lambda$ for $\pi _{k,r},\tau _{k,r}$.

Remark

The tube formula for the volume $\sigma _{2n,0}^{\lambda } = {\text {vol}}_{{\mathbb {C}}P^{n}_{\lambda }}$ is given by the following simple expression

$$\begin{aligned} {\text {vol}}_{{\mathbb {C}}P^{n}_{\lambda }}(A_t)= \sum _{j = 0}^{2n}\sin _{\lambda }^{2n-j}(t)\cos _\lambda ^{j}(t)\sigma _{j,0}^{\lambda }(A), \end{aligned}$$

which is Theorem 4.3 of [11], since $\sigma _{j,0}^\lambda = \omega _{2n-j}\tau _{j,0}^{\lambda } = \Phi _\lambda (\mu _j)$. The tube formulas ${\textbf{T}}_t \sigma _{2n-2r,r}$ are equally simple.

Remark

An interesting feature of the previous tube formulas is the following self-similarity property, which is explained by (62). Let

$$\begin{aligned} \textbf{G}^{n,j}:{\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n\longrightarrow {\mathcal {V}}^{n+2j}_{\lambda ,{\mathbb {C}}},\qquad \textbf{G}^{n,j}(\sigma _{k,r}^{\lambda })=\sigma _{k+2j,r+j}^\lambda . \end{aligned}$$

Then one has ${\textbf{T}}_t\circ {\textbf{G}}^{n,j}={\textbf{G}}^{n,j}\circ {\textbf{T}}_t$.

Remark

It is also worth noting that ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n=\bigoplus _{0 \le 2r \le n}{\mathcal {I}}_{\lambda }^{n,r}$ where

$$\begin{aligned} {\mathcal {I}}_{\lambda }^{n,r}:= J_{\lambda ,{\mathbb {C}}}(V^{(2n-4r)}) = \left\{ \sigma _{k,r}^\lambda : 2r \le k \le 2n-2r \right\} , \end{aligned}$$

and that these subspaces are $\partial _{\lambda ,{\mathbb {C}}}$-invariant. In particular, given $\varphi \in {\mathcal {I}}_{\lambda }^{n,r}$ one has ${\textbf{T}}_t(\varphi )\in {\mathcal {I}}_{\lambda }^{n,r}$.

6.2 Tube formulas in real space forms

Let $I:V^{(m)}\rightarrow {\text {Val}}^{O(m)}$ be the isomorphism of irreducible $\mathfrak {sl}_2$-representations determined by $I(y^{m})=\chi$; i.e.

$$\begin{aligned} {m \atopwithdelims ()i }I(x^i y^{m-i})&= \frac{1}{i!} I(X^i(y^m))=\frac{1}{i!}L^i(I(y^m))\\&=\frac{1}{i!}L^i(\mu _0)=\mu _i \end{aligned}$$

where we used (19). By Theorem 4.12, the map $J_{\lambda ,{\mathbb {R}}}=\Psi _\lambda \circ I$ satisfies

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}} \circ J_{\lambda ,{\mathbb {R}}} = J_{\lambda ,{\mathbb {R}}} \circ Y_\lambda . \end{aligned}$$

(64)

The map $J_{\lambda ,{\mathbb {R}}}$ is explicitly given by

$$\begin{aligned} J_{\lambda ,{\mathbb {R}}} :V^{(m)} \longrightarrow {\mathcal {V}}_{\lambda ,{\mathbb {R}}}^{m+1}, \qquad \left( {\begin{array}{c}m\\ i\end{array}}\right) x^{i}y^{m-i} \longmapsto \sigma _i^{\lambda }. \end{aligned}$$

(65)

The image of $J_{\lambda ,{\mathbb {R}}}$ is the hyperplane $\mathcal H_\lambda ^{m+1}:={\text {im}} J_{\lambda ,{\mathbb {R}}}= {\text {span}}\left\{ \sigma _{0}^\lambda ,\dots ,\sigma _{m}^\lambda \right\}$.

Theorem 6.2

The tubular operator on ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^{m+1}$ is given as follows. For $i = 0,\dots ,m$,

$$\begin{aligned} {\textbf{T}}_t\sigma _i^{\lambda } = \sum _{j = 0}^{m}\phi ^\lambda _{m,i,j}(t)\sigma ^\lambda _{j}. \end{aligned}$$

(66)

In particular

$$\begin{aligned} {\textbf{T}}_t\sigma _{m}^\lambda = \sum _{j = 0}^{m}\sin _\lambda ^{m-j}(t)\cos _\lambda ^{j}(t)\sigma ^\lambda _{j}, \end{aligned}$$

(67)

and thus

$$\begin{aligned} {\textbf{T}}_t\sigma _{m+1}^\lambda = \sum _{j = 0}^{m}\left( \int _{0}^{t}\sin _\lambda ^{m-j}(s)\cos _\lambda ^{j}(s)ds\right) \sigma ^\lambda _{j} + \sigma _{m+1}^\lambda . \end{aligned}$$

(68)

These formulas where first obtained by Santaló [25].

Proof

By (37), (64) and (65), we have for $0 \le i \le m$,

$$\begin{aligned} {\textbf{T}}_t \sigma _i^{\lambda }&=\textrm{exp}(t\partial _{\lambda ,{\mathbb {C}}})(\sigma _i^{\lambda })\\&={m\atopwithdelims ()i}\textrm{exp}(t \partial _{\lambda ,{\mathbb {C}}})\circ J_{\lambda ,{\mathbb {C}}}(x^{i} y^{m-i})\\&={m\atopwithdelims ()i}J_{\lambda ,{\mathbb {C}}}\circ \textrm{exp} (t Y_\lambda )(x^{i} y^{m-i})\\&=J_{\lambda ,{\mathbb {C}}} (p_{i}(t)). \end{aligned}$$

This proves (66) of which (67) is a particular case. Integrating with respect to t yields (68). $\square$

Remark

It is worth pointing out the similarity between tube formulas in real and complex space forms. More precisely, note that the isomorphism

$$\begin{aligned} \textbf{F}_{n,r}: \mathcal H_{\lambda }^{2n-4r+1} \longrightarrow {\mathcal {I}}_{\lambda }^{n,r}, \qquad \sigma _{j}^\lambda \longmapsto \sigma _{j+2r,r+j}^\lambda \end{aligned}$$

(69)

between the subspaces $\mathcal H_{\lambda }^{2n-4r+1}\subset {\mathcal {V}}^{2n-4r+1}_{\lambda ,{\mathbb {R}}}$ and ${\mathcal {I}}_{\lambda }^{n,r}\subset {\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ commutes with the tubular operator ${\textbf{T}}_t$. This is explained by (62) and (64).

Remark

Recently, Hofstätter and Wannerer [21] have found a map $\mathcal {V}^{2n+1}_{\lambda ,{\mathbb {R}}}\rightarrow {\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ which also commutes with $\textbf{T}_t$. Next we describe their results and how they relate to $\textbf{F}_{n,0}^\lambda$. For $\lambda >0$ let $\pi _\lambda : {\mathbb {S}}^{2n+1}_{\lambda } \rightarrow {\mathbb {C}}P^{n}_{\lambda }$ be the Hopf fibration. For a proper submersion $p:M\rightarrow N$, Alesker proved the existence of a push-forward of valuations $p_*:\mathcal V(M)\rightarrow \mathcal V(N)$ characterized by

$$\begin{aligned} p_*\varphi (A)=\varphi (p^{-1}(A)),\qquad A\in \mathcal R(N). \end{aligned}$$

Hofstätter and Wannerer have computed the push-forward of invariant valuations through the Hopf fibration. More presicely they have shown that $(\pi _\lambda )_*$ commutes with $\textbf{T}_t$ and deduced from this fact that

$$\begin{aligned} (\pi _\lambda )_*\sigma _{k}^\lambda = 2\pi \left( (2n-k+1)\sigma _{k-1,0}^\lambda - \lambda (k+1)\sigma _{k+1,0}^\lambda \right) . \end{aligned}$$

(70)

It follows from (40) that

$$\begin{aligned} (\pi _\lambda )_* = 2 \pi \textbf{F}^{\lambda }_{n,0}\circ \partial _{\lambda ,\mathbb {R}}. \end{aligned}$$

(71)

6.3 Spectral analysis of the derivative map

Here we compute the eigenvalues and eigenvectors of $\partial _{\lambda ,{\mathbb {R}}}$ and $\partial _{\lambda ,{\mathbb {C}}}$. Note that the tube formulas for such valuations are extremely simple: if $\partial _{} \mu =a\mu$ with $a\in {\mathbb {C}}$, then ${\textbf{T}}_t\mu =e^{at}\mu$.

Proposition 6.3

For $0 \le 2r \le n$, the restriction of $\partial _{\lambda ,{\mathbb {C}}}$ to ${{\mathcal {I}}_{\lambda }^{n,r}}$ has the following (simple) eigenvalues and eigenspaces:

$$\begin{aligned} {\text {spec}}\left( \left. \partial _{\lambda ,{\mathbb {C}}}\right| _{{\mathcal {I}}_{\lambda }^{n,r}}\right)&= \left\{ 0,\pm 2\sqrt{-\lambda },\pm 4\sqrt{-\lambda },\dots ,\pm 2(n-2r)\sqrt{-\lambda }, \right\} ,\\ {E_{(2k-2n+4r)\sqrt{-\lambda }}}&={\text {span}}_{{\mathbb {C}}}\left\{ J_{\lambda ,{\mathbb {C}}}(e_1^{k}e_{2}^{2n-4r-k}) \right\} ,\qquad 0 \le k \le 2n-4r. \end{aligned}$$

Hence $\partial _{\lambda ,{\mathbb {C}}}$ diagonalizes on $\mathcal V_{\lambda ,{\mathbb {C}}}^n$ with the following eigenspaces:

$$\begin{aligned} E_{2j\sqrt{-\lambda }}(\partial _{\lambda ,{\mathbb {C}}})&= {\text {span}}_{\mathbb {C}}\left\{ J_{\lambda ,{\mathbb {C}}} (e_1^{j+n-2r}e_2^{n-2r-j}) : 0 \le 2r \le {\min \{n-j,n+j\}} \right\} , \end{aligned}$$

for $-n\le j\le n$.

Proof

Everything follows from Lemma 5.4 and (62). $\square$

Proposition 6.4

 

• (i) In ${\mathbb {S}}^{2n}_{\lambda }$ the derivative operator is diagonalizable with

  $$\begin{aligned} {\text {spec}}(\partial _{\lambda ,{\mathbb {R}}})&= \left\{ 0,\pm \sqrt{-\lambda },{\pm 3\sqrt{-\lambda }},\dots , \pm (2n-1)\sqrt{-\lambda } \right\} , \end{aligned}$$

  (72)
  $$\begin{aligned} E_0(\partial _{\lambda ,{\mathbb {R}}})&= {\text {span}}_{\mathbb {C}}\{\chi \} \end{aligned}$$

  (73)
  $$\begin{aligned} E_{(2k-2n+1)\sqrt{-\lambda }}(\partial _{\lambda ,{\mathbb {R}}})&= {\text {span}}_{\mathbb {C}}\{J_{\lambda ,{\mathbb {R}}}(e_1^{k}e_2^{2n-k-1})\}, \quad 0 \le k \le 2n-1 \end{aligned}$$

  (74)

• (ii) In ${\mathbb {S}}^{2n+1}_{\lambda }$ the derivative operator is not diagonalizable since

  $$\begin{aligned} {\text {spec}}(\partial _{\lambda ,{\mathbb {R}}})&= \left\{ 0,0,\pm 2\sqrt{-\lambda },{\pm 4\sqrt{-\lambda }},\dots ,\pm 2n\sqrt{-\lambda } \right\} , \end{aligned}$$

  (75)
  $$\begin{aligned} E_0(\partial _{\lambda ,{\mathbb {R}}})&= {\text {span}}_{\mathbb {C}}\{\chi \} \end{aligned}$$

  (76)
  $$\begin{aligned} E_{(2k-2n)\sqrt{-\lambda }}(\partial _{\lambda ,{\mathbb {R}}})&= {\text {span}}_{\mathbb {C}}\{J_{\lambda ,{\mathbb {R}}}(e_1^{k}e_2^{2n-k})\}, \quad 0 \le k \le 2n. \end{aligned}$$

  (77)

Proof

 

(i):

By Lemma 5.4 and (64) we have that $(2k-2n+1)\sqrt{-\lambda }$, $0 \le k \le 2n-1$, is an eigenvalue of ${\partial _{\lambda ,{\mathbb {R}}}}$ with eigenspace given by (74). The Euler characteristic is clearly an eigenvector with zero eigenvalue. We thus have at least $2n+1$ eigenvalues. Since this is precisely the dimension of ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^{2n}$, the statement follows.

(ii):

In light of Lemma 5.4 and (64), we ascertain that $(2k-2n)\sqrt{-\lambda }$, $0 \le k \le 2n$, is an eigenvalue of ${\partial _{\lambda ,{\mathbb {R}}}}$ and the corresponding eigenspace is described by (77).

Our next objective is to prove that while the algebraic multiplicity of the zero eigenvalue is two, its geometric multiplicity is only one. This will entail finding a valuation $\mu$ that satisfies $\partial _{\lambda ,{\mathbb {R}}}^2 \mu = 0$, while also ensuring that $\partial _{\lambda ,{\mathbb {R}}} \mu \ne 0$. Consider $\sigma _{2n}^\lambda =J_{\lambda ,{\mathbb {R}}}(x^{2n}) \in {\mathcal {V}}^{2n+1}_{\lambda ,{\mathbb {R}}}$. In the notation of Lemma 5.4,

$$\begin{aligned} x = \frac{1}{2\sqrt{-\lambda }}(e_1 - e_2), \qquad x^{2n} = (-4\lambda )^{-n}\sum _{i = 0}^{2n}(-1)^i \left( {\begin{array}{c}2n\\ i\end{array}}\right) e_1^{i}e_{2}^{2n-i}. \end{aligned}$$

Hence

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}}\sigma _{2n+1}^\lambda = \sigma _{2n}^\lambda = (-4\lambda )^{-n}\sum _{i = 0}^{2n}(-1)^i \left( {\begin{array}{c}2n\\ i\end{array}}\right) J_{\lambda ,{\mathbb {R}}}(e_1^{i}e_{2}^{2n-i}). \end{aligned}$$

Consider

$$\begin{aligned} \nu := (-4\lambda )^{-n} \sum ^{2n}_{\begin{array}{c} i = 0\\ i\ne n \end{array}}\left( {\begin{array}{c}2n\\ i\end{array}}\right) \dfrac{(-1)^i}{(2i-2n)\sqrt{-\lambda }} J_{\lambda ,{\mathbb {R}}}(e_1^{i}e_{2}^{2n-i}), \end{aligned}$$

and note that, by Lemma 5.4,

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}}\nu := (-4\lambda )^{-n} \sum ^{2n}_{\begin{array}{c} i = 0\\ i\ne n \end{array}}\left( {\begin{array}{c}2n\\ i\end{array}}\right) (-1)^{i} J_{\lambda ,{\mathbb {R}}}(e_1^{i}e_{2}^{2n-i}), \end{aligned}$$

since $e_1^{n}e_2^n \in \ker Y_\lambda$. Finally, we define $\mu = \sigma _{2n+1}^\lambda -\nu$. Then

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}} \mu = (-4\lambda )^{-n}\left( {\begin{array}{c}2n\\ n\end{array}}\right) (-1)^n J_{\lambda ,{\mathbb {R}}}(e_1^{n}e_2^{n})\ne 0, \end{aligned}$$

while

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}}^2 \mu&= (-4\lambda )^{-n}\left( {\begin{array}{c}2n\\ n\end{array}}\right) (-1)^n \partial _{\lambda ,{\mathbb {R}}} J_{\lambda ,{\mathbb {R}}}(e_1^{n}e_2^{n}) \\&= (-4\lambda )^{-n}\left( {\begin{array}{c}2n\\ n\end{array}}\right) (-1)^n J_{\lambda ,{\mathbb {R}}}(Y_{\lambda }(e_1^{n}e_2^{n})) = 0. \end{aligned}$$

It follows that $\dim \ker \partial _{\lambda ,{\mathbb {R}}}<\dim \ker \partial _{\lambda ,{\mathbb {R}}}^2$. Noting that $\chi \in \ker \partial _{\lambda ,{\mathbb {R}}}$ this implies the statement.

$\square$

Remark

We conclude from Prosposition 6.4 and Lemma 5.4 that there is no isomorphism between ${\text {Val}}^{O(m)}$ and ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ intertwining $\Lambda - \lambda L$ and $\partial _{\lambda ,{\mathbb {R}}}$. Indeed, these two operators have different spectra no matter the parity of m.

6.4 Stable valuations in complex space forms

We say that a valuation $\varphi \in {\mathcal {V}}(M)$ on a Riemannian manifold M is stable if $\partial \mu =0$, or equivalently, if ${\textbf{T}}_t\mu =\mu$ for all t. By Propositions 6.4 and 6.3, up to multiplicative constants, the Euler characteristic is the unique isometry-invariant stable valuation in ${\mathbb {S}}^{m}_{\lambda }$. The complex case is more interesting.

Proposition 6.5

The unique (up to multiplicative constants) stable valuation on ${\mathcal {I}}_\lambda ^{n,r}$ is given by

$$\begin{aligned} \psi _{2r} = \sum _{i = r}^{n-r}\left( {\begin{array}{c}n-2r\\ i-r\end{array}}\right) \left( {\begin{array}{c}2n-4r\\ 2i-2r\end{array}}\right) ^{-1}\lambda ^{i-r} \sigma _{2i,r}^\lambda . \end{aligned}$$

Proof

By Lemma 5.4 the kernel of $Y_\lambda$ on the space $V^{(m)}$ of homogeneous polynomials of degree $m=2n-4r$ is spanned by

$$\begin{aligned} \begin{aligned} e_1^{n-2r}e_2^{n-2r}&= (y + \sqrt{-\lambda } x)^{n-2r}(y - \sqrt{-\lambda } x)^{n-2r} \\&= (y^2 + \lambda x^2)^{n-2r} = \sum _{j = 0}^{n-2r}\left( {\begin{array}{c}n-2r\\ j\end{array}}\right) \lambda ^j x^{2j} y^{m-2j} \\&= \sum _{i = r}^{n-r}\left( {\begin{array}{c}n-2r\\ i-r\end{array}}\right) \left( {\begin{array}{c}2n-4r\\ 2i-2r\end{array}}\right) ^{-1}\lambda ^{i-r} \left( {\begin{array}{c}2n-4r\\ 2i-2r\end{array}}\right) x^{2i-2r} y^{2n-2i-2r} \end{aligned} \end{aligned}$$

Therefore the kernel of $\partial _{\lambda ,{\mathbb {C}}}$ in ${\mathcal {I}}_{\lambda }^{n,r}$ is spanned by $\psi _{2r}=J_\lambda (e_1^{n-2r}e_2^{n-2r})$, for each $0 \le 2r \le n$. $\square$

Next we express the Euler characteristic as a combination of the stable valuations $\psi _{2r}$. Note in particular that $\chi$ is not confined to any $\partial$-invariant subspace ${\mathcal {I}}_{\lambda }^{n,r}$.

Proposition 6.6

 

$$\begin{aligned} \begin{aligned} \chi&= \sum _{0 \le 2r \le n} \left( \frac{\lambda }{4\pi }\right) ^r\left( {\begin{array}{c}2r\\ r\end{array}}\right) \dfrac{r!}{\omega _{2n-2r}}\psi _{2r}. \end{aligned} \end{aligned}$$

Proof

Since $\chi$ is stable, it can be expressed as $\chi =\sum _j a_j \psi _{2j}$. By [11, Theorem 3.11]

$$\begin{aligned} \begin{aligned} \chi&=\left. \sum _{k,p \ge 0}\left( \frac{\lambda }{\pi }\right) ^{k+p} \frac{\partial ^{k+p}}{\partial \xi ^{k} \partial \eta ^{p}}\frac{1}{\sqrt{1-\xi }\sqrt{1-\eta }}\right| _{(0,0)} \tau _{2 k+2 p, p}^{\lambda }. \end{aligned} \end{aligned}$$

The coefficient of $\tau _{2r,r}^{\lambda }$ in this expansion is

$$\begin{aligned} {[}\tau _{2r,r}^{\lambda }](\chi )&=\left( \frac{\lambda }{\pi }\right) ^r\left. \frac{\partial ^{r}}{\partial \eta ^r}\frac{1}{\sqrt{1-\eta }}\right| _{0}\\&={\left( \frac{\lambda }{\pi }\right) ^r} \left( {\begin{array}{c}2r\\ r\end{array}}\right) r!4^{-r}. \end{aligned}$$

By Proposition 3.3, we have

$$\begin{aligned} {[}\tau _{2r,r}^{\lambda }](\sigma _{k,r}^\lambda )=\frac{\omega _{2n-k}}{(k-2r)!}[\tau _{2r,r}^{\lambda }](\pi _{k,r}^\lambda ) = \frac{\omega _{2n-k}}{(k-2r)!}\delta _{k,2r}, \end{aligned}$$

whence

$$\begin{aligned} \begin{aligned} {[}\tau _{2r,r}^\lambda ]\left( \sum _j a_j \psi _{2j}\right)&= a_r[\tau _{2r,r}^\lambda ](\sigma _{2r,r}^\lambda ) \\&= a_r\omega _{2n-2r}. \end{aligned} \end{aligned}$$

Hence

$$\begin{aligned} a_r = {\left( \frac{\lambda }{\pi }\right) ^r}\left( {\begin{array}{c}2r\\ r\end{array}}\right) \dfrac{r!}{4^r \omega _{2n-2r}} \end{aligned}$$

and the result follows. $\square$

6.5 Image of $\partial _{\lambda ,{\mathbb {C}}}$ and $\partial _{\lambda ,{\mathbb {R}}}$

Next we describe the image of the operators $\partial _{\lambda ,{\mathbb {C}}}$ and $\partial _{\lambda ,{\mathbb {R}}}$, and we compute the preimage of any element belonging to them.

Proposition 6.7

Given any $\varphi = \sum _{k,r} a_{k,r}\sigma _{k,r}^\lambda \in {\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$, we have $\varphi \in {\text {im}}\partial _{\lambda ,{\mathbb {C}}}$ if and only if

$$\begin{aligned} \sum _{l = r}^{n-2r} a_{2l,r} \left( {\begin{array}{c}n-2r\\ l-r\end{array}}\right) \lambda ^{n-l-r} = 0,\qquad \text{ for } \quad 0 \le 2r \le n. \end{aligned}$$

(78)

Proof

Note that $\varphi =\sum _{r}\varphi _r$ with $\varphi _r=\sum _{k} a_{k,r}\sigma _{k,r}^\lambda$ is the decomposition of $\varphi$ corresponding to ${\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n=\bigoplus _{r=0}^{\left\lfloor n/2\right\rfloor } {\mathcal {I}}_{\lambda }^{n,r}$. By (62) and Proposition 5.5 we have $\varphi \in {\text {im}}\partial _{\lambda ,{\mathbb {C}}}$ if and only if for every r

$$\begin{aligned} 0= Z_{2n-4r,\lambda }(\varphi _r)&= \sum _{k=2r}^{2n-4r} a_{k,r} \left( {\begin{array}{c}2n-4r\\ k-2r\end{array}}\right) Z_{2n-4r,\lambda }(x^{k-2r}y^{2n-k-2r}) \end{aligned}$$

(79)

$$\begin{aligned}&= \sum _{l=r}^{n-2r}a_{2l,r}\left( {\begin{array}{c}2n-4r\\ 2l-2r\end{array}}\right) \left( {\begin{array}{c}n-2r\\ l-r\end{array}}\right) \lambda ^{n-l-r}(2l-2r)!(2n-2l-2r)! \end{aligned}$$

(80)

$$\begin{aligned}&=(2n-4r)!\sum _{l = r}^{n-2r} a_{2l,r} \left( {\begin{array}{c}n-2r\\ l-r\end{array}}\right) \lambda ^{n-l-r} \end{aligned}$$

(81)

where we used (59). $\square$

Proposition 6.8

Given $\varphi = \sum _{k,r} a_{k,r}\sigma _{k,r}^\lambda \in {\mathcal {V}}_{\lambda ,{\mathbb {C}}}^n$ satisfying (78) we have

$$\begin{aligned} \partial _{\lambda ,{\mathbb {C}}}^{-1}(\{\varphi \})=\sum _{k,r} a_{k,r} J_{\lambda ,{\mathbb {C}}}(P_{2n-4r,k-2r+1})+{\text {span}}\{\psi _{2r}:0\le 2r \le n\} \end{aligned}$$

where $P_{m,l}$ is given by (60).

Proof

This follows at once from Proposition 5.6 after decomposing $\varphi =\sum _r\varphi _r$ as in the previous proof. $\square$

Proposition 6.9

The image of $\partial _{\lambda ,{\mathbb {R}}}$ in ${\mathcal {V}}_{\lambda ,{\mathbb {R}}}^m$ is the hyperplane ${\mathcal {H}}_\lambda ^m$ generated by $\sigma _0^{\lambda },\dots ,\sigma ^{\lambda }_{m-1}$. Moreover

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}}\phi ^{k} =\dfrac{k! \omega _k}{\pi ^{k}\omega _{m-k}}\sigma _{k-1}^\lambda , \qquad 1 \le k \le m, \end{aligned}$$

(82)

where $\phi ^k=\sum _{j \ge 0}\left( \frac{\lambda }{4}\right) ^j \tau _{k+2j}^\lambda$. In particular

$$\begin{aligned} \partial _{\lambda ,{\mathbb {R}}}^{-1}(\{\sigma ^{\lambda }_{k-1}\})=\dfrac{\pi ^{k}\omega _{m-k}}{k! \omega _k}\phi ^k+{\mathbb {C}}\cdot \chi . \end{aligned}$$

Recall from [7, eq. (118)] that $\phi ^k= \int _{G_{\lambda ,{\mathbb {C}}}}\chi (\cdot \cap g {\mathbb {S}}^{m-k}_{\lambda })dg$ where dg is a properly normalized Haar measure on $G_{\lambda ,{\mathbb {C}}}$, and ${\mathbb {S}}^{m-k}_{\lambda }$ is an $(m-k)$-dimensional totally geodesic submanifold in ${\mathbb {S}}^{m}_{\lambda }$.

Proof

By (8) and (9)

$$\begin{aligned} \phi ^k = \sum _{j=0}^{\left\lfloor \frac{m-k-1}{2}\right\rfloor }\left( \dfrac{\lambda }{4\pi ^2}\right) ^j\dfrac{(k+2j)!\omega _{k+2j}}{\pi ^{k}(m-k-2j)\omega _{m-k-2j}}\sigma _{k+2j}^\lambda +\left( \left( \frac{\lambda }{4}\right) ^{\frac{m-k}{2}}\frac{m!\omega _m}{\pi ^m}\sigma _m^\lambda \right) , \end{aligned}$$

where the term between brackets appears only if $m-k$ is even. Using Proposition 4.8, this yields (82). The rest of the statement follows. $\square$

Remark

Equation (82) also follows from Theorem 4 in [27].