GAUSS-BONNET THEOREM AND CROFTON TYPE FORMULAS IN ...

GAUSS-BONNET THEOREM AND CROFTON TYPE
FORMULAS IN COMPLEX SPACE FORMS
JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Abstract. We give an expression, in terms of the so-called Hermitian
intrinsic volumes, for the measure of the set of complex
r-planes intersecting a regular domain in any complex space form.
Moreover, we obtain two different expressions for the Gauss-Bonnet-
Chern formula in complex space forms. One of them expresses the
Gauss curvature integral in terms of the Euler characteristic and
some Hermitian intrinsic volumes. The other one, which is shorter,
involves the measure of complex hyperplanes meeting the domain.
As a tool, we obtain new variation formulas in integral geometry
of complex space forms. Finally, we express the average over the
complex r-planes of the total Gauss curvature of the intersection
domain.
Prepublicació Núm 01, gener 2009.
Departament de Matemàtiques.
http://www.uab.cat/matematiques
1. Introduction
In the space of constant sectional curvature k, M n (k), Santaló [San04]
gave the expression of the measure of the set of r-planes meeting a regular
domain in terms of the mean curvature integrals. Suppose that L r
denotes the space of r-dimensional planes in M n (k) and dL r a density
of L r invariant under the isometry group of M n (k). If Ω ⊂ M n (k) is a
regular domain and r = 2l, then
(1)
∫
L 2l
χ(Ω ∩ L 2l )dL 2l = c 0 vol(Ω) +
l∑
c i k l−i M 2i−1 (∂Ω),
where c i are known and depend only on n, r and i, while M j (∂Ω)
denotes the mean curvature integral defined by
( ) −1 ∫ n − 1
(2) M j (∂Ω) =
σ j (II)dx
j
∂Ω
1991 Mathematics Subject Classification. Primary 53C65; Secondary 52A22,
53C55.
Key words and phrases. Complex space forms, Gauss-Bonnet formula, Crofton
formulas, Valuation.
Work partially supported by FEDER/MEC grant # MTM2006-04353 and by
the Departament d’Innovació Universitari i de Recerca from the Generalitat of
Catalonia and the ESF.
1
i=1

2 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
where σ j (II) is the j-th symmetric elementary function of the eigenvalues
of the second fundamental form. An analogous formula holds in
the case of odd-dimensional planes.
In the proof of formula (1), Santaló used the Gauss-Bonnet-Chern
theorem in M n (k), that for n even states
(3)
M n−1 (∂Ω)= O n
2 χ(Ω)−kc n−3M n−3 (∂Ω)−...−k n−2
2 c1 M 1 (∂Ω)−k n 2 vol(Ω),
and the reproductive property of the mean curvature integrals in M n (k),
∫
(4)
(∂Ω ∩ L r )dL r = cM i (∂Ω),
M (r)
i
L r
where M (r)
i (∂Ω∩L r ) denotes the i-th mean curvature integral of ∂Ω∩L r
as a hypersurface in L r and c is known and depends only on n, r and
i.
In this paper we generalize formula (1) and (3) to the space of constant
holomorphic curvature 4ɛ, CK n (ɛ). The role of L r in (1) will be
played by L C r , the space of complex r-planes (totally geodesic complex
submanifolds). For simplicity, we shall restrict to compact domains
Ω ⊂ CK n (ɛ) with smooth boundary, and call them regular domains.
The method used by Santaló cannot be applied in this situation.
Mean curvature integrals are defined in a complex space form as in (2),
but in the standard Hermitian space C n , it was not certain whether the
reproductive property remains true for the mean curvature integrals
when we integrate it over L C r . Moreover, in the complex projective
space and in the complex hyperbolic space, an explicit Gauss-Bonnet
formula is not known. Instead, we use variational arguments, as well
as some facts about the theory of valuations, which we briefly describe
next.
Definition 1.1. Let K(V ) denote the family of non-empty compact
convex subsets of a finite dimensional real vector space V of dimension
n. A scalar valued functional φ : K(V ) → C is called a valuation if
whenever A, B, A ∪ B ∈ K(V ).
φ(A ∪ B) = φ(A) + φ(B) − φ(A ∩ B)
The space of invariant valuations under the full group of isometries of
R n was studied by Hadwiger [Had57], who proved that the dimension
of this space is n + 1. Mean curvature integrals, volume and Euler
characteristic form a basis of this space.
On the standard Hermitian space C n with its isometry group IU(n) =
C n ⋊ U(n), Alesker [Ale03] proved that there are more linearly independent
valuations than on R n , the dimension of this space is ( )
n+2
2 .
Both the Euler characteristic, and the measure of complex planes intersecting
the domain belong to this space.

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 3
Bernig and Fu [BF08], consider several valuation bases on C n . Here
we will use the basis whose elements {µ k,q } {k,q} are called Hermitian
intrinsic volumes. These valuations where first introduced by Park
in complex space forms (cf. [Par02]). In section 2 we recall their
definition.
The main results of this paper can be stated as follows.
Theorem 1.2. Let Ω be a regular domain in CK n (ɛ). Then, for r =
1, ..., n − 1
∫
( ) −1 n − 1
χ(Ω ∩ L r )dL r = vol(G C n−1,r) (ɛ r (r + 1)vol(Ω)+
r
(5)
L C r
+
∑n−1
k=n−r
+
ɛ k−(n−r) ω 2n−2k
( n
k) −1
· ((k + r − n + 1)µ 2k,k +
∑k−1
q=max{0,2k−n}
( )
1 2k − 2q
µ
4 k−q 2k,q ))
k − q
where dL r denotes an invariant measure in the space of complex r-
planes L C r . Moreover,
(6)
O 2n−1 χ(Ω) = 2n(n + 1)ɛ n vol(Ω) +
⎛
⎞
∑n−1
ɛ c O 2n−2c−1
∑c−1
( )
+ ) ⎝
1 2c − 2q
µ
4 c−q 2c,q +(c + 1)µ 2c,c
⎠
c − q
c=0
( n−1
c
q=max{0,2c−n}
where ω i denotes the volume of the euclidean unit ball and O i the volume
of the euclidean unit sphere.
Formula (5) is a generalization of (1) in the sense that the measure
of the complex r-planes meeting a regular domain is expressed as a
linear combination of the volume and some other valuations related to
mean curvature integrals. This answers a question posed by Naveira
in [Nav05]. In case r = 1, formula (5) was already proved by different
methods in [Aba09]. For 2r ≥ n, and ɛ = 0, formula (5) was proved in
[BF08].
Formula (6) generalizes to complex space forms the Gauss-Bonnet
theorem (3). In complex dimensions n = 2, 3, formula (6) was obtained
in [Par02].
Combining expressions (5) and (6) we obtain
∫
M 2n−1 (∂Ω) = O 2n−1 χ(Ω) − 2nɛ χ(Ω ∩ L n−1 )dL n−1 −
k=1
L C n−1
∑n−1
( ) −1 n − 1
− ɛ k O 2n−2k−1 µ 2k,k − 2nɛ n vol(Ω).
k
This expression is similar to the following one for real space forms.

4 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Theorem ([Sol06]). Let S be a hypersurface bounding a compact domain
Q in a real space form with sectional curvature k and dimension
n. Then
∫
2(n − 1)
M n−1 (S) = O n−1 χ(Q) − k χ(Q ∩ L n−2 )dL n−2 .
O n−2 L n−2
The main idea for the proof of Theorem 1.2 is to take variation along
a vector field in CK n (ɛ), in both sides of equalities (5) and (6), and to
compare them.
In order to obtain a first expression of the variation of ∫ χ(∂Ω ∩
L C r
L r )dL r along a vector field in CK n (ɛ), we proceed as in [Sol06] (see
Section 3.1). In C n , the variation of the Hermitian intrinsic volumes
was obtained by Bernig and Fu [BF08]. Here we use the same method
to find the generalization for ɛ ≠ 0 (see Section 3.2).
Using formula (5), we prove in Section 6 that the total Gauss curvature
does not satisfy the reproductive property and we get in C n the
following expression:
∫
( ) −1 ( ) n − 1 n
M 2r−1 (∂Ω ∩ L r )dL r = 2rω2rvol(G 2 C n−1,r)
−1·
r r
L C r
(7)
⎛
· ⎝
∑n−r
q=max{0,n−2r}
( )
1 2n − 2r − 2q
4 n−r−q n − r − q
⎞
µ 2n−2r,q
⎠ .
In [Aba09], it is proved that the reproductive property (4) is not
satisfied by the mean curvature integral either.
Acknowledgments
We wish to thank Andreas Bernig for illuminating discussions during
the preparation of this work.
2. Hermitian intrinsic volumes on CK n (ɛ)
Let CK n (ɛ) be a (simply connected) complex space form with constant
holomorphic curvature 4ɛ. We denote by T ′ CK n (ɛ) the unit tangent
bundle of CK n (ɛ).
Definition 2.1. Let Ω be a regular domain in CK n (ɛ). The unit (inner)
normal bundle of ∂Ω is defined as
N(Ω) ={(p, v) ∈ T ′ CK n (ɛ) : p ∈ ∂Ω, v inner normal to T ′ p∂Ω}.
Given a 4n − 1 form ω in T ′ CK n (ɛ) we may consider, for every
regular domain Ω the integral over N(Ω) of the form ω. The resulting
functional is called a smooth valuation.
Let z ∈ CK n (ɛ), e 1 ∈ T ′ CK n (ɛ) and let {z; e 1 , ..., e n } be a moving
frame defined on an open subset U ⊂ T ′ CK n (ɛ). We denote by

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 5
{ω 1 , ω 2 , ..., ω n } the 1-forms on T ′ CK n (ɛ) defined as the dual basis of
{e 1 , ..., e n }, and by {ω ij } the connection forms of CK n (ɛ). That is, if
( , ) denotes the Hermitian product on CK n (ɛ) and ∇ the Levi-Civita
connection, then
ω j = (dz, e j ) and ω jk = (∇e j , e k ) where j, k ∈ {1, ..., n}.
Thus, these forms are C-valued. We denote
(8)
ω j = α j + iβ j ,
ω jk = α jk + iβ jk .
Remark 2.2. Forms α 1 , β 1 and β 11 are global forms in T ′ CK n (ɛ). We
denote by α, β, γ the forms α 1 , β 1 , β 11 , respectively. Note that α
coincides with the contact form of the unit tangent bundle.
Lemma 2.3. Let Ω ⊂ CK n (ɛ) be a regular domain. Forms α and dα
vanish at N(Ω) ⊂ T ′ CK n (ɛ).
Proof. Let V ∈ T (p,v) N(Ω). Then, α(V ) (p,v) = 〈dπ(V ), v〉 = 0 where
π : T ′ CK n (ɛ) → CK n (ɛ) is the canonical projection.
To prove that dα vanishes at the unit normal bundle, we consider
the inclusion of the unit normal bundle to the unit tangent bundle
i : N(Ω) → T ′ (Ω) and we use that exterior differential commutes with
the inclusion map to obtain
(9)
dα| N(Ω) = (i ∗ ◦ d)(α) = (d ◦ i ∗ )(α) = d(i ∗ α) = d(0) = 0.
Consider the following 2-forms in T ′ CK n (ɛ)
n∑
θ 0 = −Im((∇e 1 , ∇e 1 )) = −Im( ω 1i ⊗ ω 1i )
i=1
θ 1 = −Im((dz, ∇e 1 ) − (∇e 1 , dz))
n∑
n∑
= −Im( ω i ⊗ ω 1i − ω 1i ⊗ ω i )
i=1
i=1
θ 2 = −Im((dz, dz)) = −Im(
n∑
ω i ⊗ ω i )
i=1
θ s = Re((dz, ∇e 1 ) − (∇e 1 , dz))
n∑
n∑
= Re( ω i ⊗ ω 1i − ω 1i ⊗ ω i ).
i=1
This forms coincide with the invariant 2-forms, θ 0 , θ 1 , θ 2 and θ s
defined in T ′ C n by Bernig and Fu [BF08, p.14]. Note that, θ s is the
symplectic form of T CK n (ɛ).
Remark 2.4. Park [Par02] defined invariant 2-forms in T ′ CK n (ɛ) similar
to the ones in (9).
i=1
□

6 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
The forms β k,q and γ k,q defined in T ′ C n in [BF08], can be extended
to T ′ CK n (ɛ) from (9).
Definition 2.5. For positive integers k, q ∈ N with max{0, k − n} ≤
q ≤ k 2 < n, it is defined in Ω2n−1 (T ′ CK n (ɛ))
where
β k,q := c n,k,q β ∧ θ n−k+q
0 ∧ θ k−2q−1
1 ∧ θ q 2, k ≠ 2q
γ k,q := c n,k,q
2 γ ∧ θn−k+q−1 0 ∧ θ k−2q
1 ∧ θ2, q n ≠ k − q
1
c n,k,q :=
q!(n − k + q)!(k − 2q)!ω 2n−k
and ω 2n−k denotes the volume of the (2n − k)-dimensional euclidean
ball.
Given a regular domain Ω ⊂ CK n (ɛ), we define (for max{0, k − n} ≤
q ≤ k < n) 2
∫
∫
Bk,q(Ω) Ω := β k,q (k ≠ 2q), Γ Ω k,q(Ω) :=
(n ≠ k−q).
N(Ω)
γ k,q
N(Ω)
In C n , it is satisfied B Ω k,q (Ω) = ΓΩ k,q (Ω) since dβ k,q = dγ k,q . Next we
recall the exterior derivative of θ 0 , θ 1 and θ 2 , which can be found in
[BF08] when ɛ = 0, or in [Par02] for general ɛ.
Lemma 2.6 ([Par02]). In T ′ CK n (ɛ) it is satisfied
dα = −θ s , dθ 0 = −ɛ(α ∧ θ 1 + β ∧ θ s ),
dβ = θ 1 , dθ 1 = 0,
dγ = 2θ 0 − 2ɛθ 2 − 2ɛα ∧ β, dθ 2 = 0.
Next, we give the relation among {B Ω k,q (Ω)} and {ΓΩ k,q (Ω)} in CKn (ɛ)
which generalizes the relation in C n .
Proposition 2.7. In CK n (ɛ), for any positive integers k, q such that
max{0, k − n} < q < k/2 < n it is satisfied
c n,k,q
Γ Ω k,q(Ω) = Bk,q(Ω) Ω − ɛ B
c
k+2,q+1(Ω).
Ω
n,k+2,q+1
Proof. We denote by I the ideal generated by α, dα and the exact forms
in N(Ω). If λ, ρ are (2n − 1)-forms in N(Ω) equal modulo I, then by
Lemma 2.3
∫ ∫
λ = ρ.
Thus, it is enough to prove
N(Ω)
c n,k,q
N(Ω)
(10) γ k,q ≡ β k,q − ɛ β k+2,q+1 mod I.
c n,k+2,q+1

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 7
Consider the form η = (θ s − β ∧ γ) ∧ θ n−k+q−1
0 θ k−2q−1
1 θ q 2. As dη is
exact, we have dη ≡ 0 mod I. On the other hand, by Lemma 2.6 it
follows that modulo I
dη ≡ −γθ n−k+q−1
0 θ k−2q
1 θ q 2+2βθ n−k+q
0 θ k−2q−1
1 θ q 2−2ɛβθ n−k+q−1
0 θ k−2q−1
1 θ q+1
2 .
Using the definition of γ k,q and β k,q we obtain the relation in (10).
Remark 2.8. In complex dimensions n = 2, 3, the previous relations
were found in [Par02].
In view of the previous equalities, we define (for max{0, k − n} ≤
q ≤ k < n) 2
{ B
Ω
(11) µ k,q (Ω) := k,q (Ω) if k ≠ 2q
Γ Ω 2q,q(Ω) if k = 2q.
Remark 2.9. In [BF08], it is proved that valuations {µ k,q , vol} where
k ∈ {0, ..., n − 1} and q ∈ {max{0, k − n}, ..., ⌊k/2⌋} form a basis of
invariant continuous valuations on C n .
3. Variation formulas
3.1. Variation of Hermitian intrinsic volumes. In order to study
the variation on CK n (ɛ) of the Hermitian intrinsic volumes, we follow
the method used by Bernig and Fu [BF08] in Corollary 2.6. First, we
recall the definition of the Rumin operator, introduced in [Rum94], and
the definition of the Reeb vector field in a contact manifold.
Definition 3.1. Given µ ∈ Ω 2n−1 (T ′ CK n (ɛ)), let α be the contact form
of T ′ CK n (ɛ), and let α ∧ ξ ∈ Ω 2n−1 (T ′ CK n (ɛ)) be the unique form such
that d(µ + α ∧ ξ) is multiple of α (cf. [Rum94]). Then the Rumin
operator D is defined as
Dµ := d(µ + α ∧ ξ).
Definition 3.2. Let M be a contact manifold and let α be the contact
form. The Reeb vector field T is the unique vector field over M such
that
{
iT α = 1,
(12)
L T α = 0.
If the contact manifold is the unit tangent bundle of a Riemannian
manifold, then the Reeb vector field is the geodesic flow (cf. [Bla76, p.
17]).
Lemma 3.3. In T ′ CK n (ɛ), it is satisfied
i T α = 1, i T θ 1 = γ,
i T θ 2 = β, i T β = i T γ = i T θ 0 = 0.
□

8 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Proof. The first equality comes directly from the definition (cf. (12)).
As T is the geodesic flow, we have α i (T ) = β i (T ) = 0 and α 1i = β 1i = 0,
i ∈ {2, ..., n}. By Definition in (9), we obtain the result.
□
Given a smooth valuation µ, and a vector field X with flow ϕ t , we
are interested in computing
δ X µ(Ω) := d dt∣ µ(ϕ t (Ω)).
t=0
This can be done by means of the following result stated in [BF08].
Lemma 3.4 (Lemma 2.5 [BF08]). Suppose Ω ⊂ R n is a regular domain,
N is the inner normal field to ∂Ω, X is a smooth vector field on
C n and µ is a smooth valuation given by a (2n − 1)-form ρ. Then
∫
δ X µ(Ω) = 〈X, N〉 i T (Dρ)
N(Ω)
where T is the Reeb vector field on T ′ R n and Dρ is the Rumin operator
of ρ.
Although this result is stated and proved in C n , the given proof is
also valid in an arbitrary Riemannian manifold.
From Lemma 2.6, we obtain the exterior differential of the forms β k,q
and γ k,q .
Lemma 3.5. In CK n (ɛ)
dβ k,q = c n,k,q (θ n−k+q
0 ∧θ k−2q
1 ∧θ q 2−ɛ(n−k+q)α∧β∧θ n−k+q−1
0 ∧θ k−2q
1 ∧θ q 2)
and
dγ k,q =c n,k,q (θ n−k+q
0 ∧ θ k−2q
1 ∧ θ q 2 − ɛθ n−k+q−1
0 ∧ θ k−2q
1 ∧ θ q+1
2
− ɛα ∧ β ∧ θ n−k+q−1
0 ∧ θ k−2q
1 ∧ θ q 2
(n − k + q − 1)
− ɛ α ∧ γ ∧ θ n−k+q−2
0 ∧ θ k−2q+1
1 ∧ θ q 2
2
(n − k + q − 1)
− ɛ β ∧ γ ∧ θ s ∧ θ n−k+q−2
0 ∧ θ k−2q
1 ∧ θ q
2
2).
Notation 3.6. Let Ω be a regular domain in CK n (ɛ) and let N be a
normal field to ∂Ω. Let X be a smooth vector field on CK n (ɛ). We
denote
∫
∫
˜B k,q := ˜B k,q (Ω) = 〈X, N〉β k,q , ˜Γk,q := ˜Γ k,q (Ω) = 〈X, N〉γ k,q .
∂Ω
The variation of the valuations {µ k,q } on C n is given in [BF08, Proposition
4.6]. The next proposition gives this variation in the previous
notation.
∂Ω

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 9
Proposition 3.7 (Proposition 4.6 [BF08]).
δ X µ k,q = 2c n,k,q (c −1
n,k−1,q (k − 2q)2˜Γk−1,q − c −1
n,k−1,q−1 (n + q − k)q˜Γ k−1,q−1
+c −1
n,k−1,q−1 (n + q − k + 1 2 )q ˜B k−1,q−1 −c −1
n,k−1,q (k − 2q)(k − 2q − 1)˜B k−1,q ).
We extend this result to CK n (ɛ) as follows.
Proposition 3.8. Let X ∈ X(CK n (ɛ)) and let Ω ⊂ CK n (ɛ) be a regular
domain. Then
δ X B Ω k,q (Ω)= 2c n,k,q(c −1
n,k−1,q (k − 2q)2˜Γk−1,q −c −1
n,k−1,q−1 (n + q − k)q˜Γ k−1,q−1
+ c −1
n,k−1,q−1 (n + q − k + 1 2 )q ˜B k−1,q−1 − c −1
n,k−1,q (k − 2q)(k − 2q − 1) ˜B k−1,q
+ɛ(c −1
n,k+1,q+1 (k − 2q)(k − 2q − 1) ˜B k+1,q+1 −c −1
n,k+1,q (n − k + q)(q + 1 2 ) ˜B k+1,q ))
and
δ X Γ Ω 2q,q(Ω) =2c n,2q,q
(
− c −1
n,2q−1,q−1 (n − q)q˜Γ 2q−1,q−1
+ c −1
n,2q−1,q−1 (n − q + 1 2 )q ˜B 2q−1,q−1
+ ɛ(−c −1
n,2q+1,q ((n − q)(2q + 3 2 ) − 1 2 (q + 1)) ˜B 2q+1,q
+ c −1
n,2q+1,q (n − q − 1)(q + 1)˜Γ 2q+1,q
− ɛc −1
n,2q+3,q+1 (n − q − 1)(q + 3 2 ) ˜B 2q+3,q+1
) ) .
Proof. The proof is based on that of Proposition 4.6 in [BF08].
Consider first δ X Bk,q Ω (Ω). Lemma 3.4 provides an expression for the
variation of a valuation given by a smooth form. By this lemma and
Lemma 2.3, it is enough to find the Rumin operator of the form β k,q
modulo α, dα since both forms vanish over the unit normal fiber bundle.
In the case ɛ = 0 (cf. Proposition 4.6 in [BF08]) the Rumin differential
of β k,q is given by Dβ k,q = d(β k,q + α ∧ ξ) where ξ is an invariant form
fulfilling
(13)
ξ ≡ c n,k,q βγθ n+q−k−1
0 θ k−2q−2
1 θ q−1
2
∧ ( )
(n + q − k)qθ1 2 − (k − 2q)(k − 2q − 1)θ 0 θ 2
mod (α, dα).
For general ɛ we take a form ξ ɛ such that ξ ɛ (p,v) ≡ ξ (p ′ ,v ′ ) when we identify
T (p,v) T ′ CK n (ɛ) and T (p ′ ,v ′ )T ′ C n , for every (p, v) ∈ T ′ CK n (ɛ), (p ′ , v ′ ) ∈
T ′ C n . Then, it is clear from Lemma 3.5 that d(β k,q +α∧ξ ɛ ) ≡ 0 modulo
α, since the term multiple of ɛ in the expression of dβ k,q is also multiple
of α.
By Lemma 2.6, the exterior differential ξ for any ɛ is
dξ ≡ c n,k,q θ n+q−k−1
0 θ k−2q−2
1 θ q−1
2 ((n − k + q)qθ1 2 − (k − 2q)(k − 2q − 1)θ 0 θ 2 )
∧ (γθ 1 − 2βθ 0 + 2ɛβθ 2 ) mod α

10 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
and the contraction of dβ k,q with respect to the field T , by Lemma 3.3,
is
i T dβ k,q ≡ c n,k,q θ n+q−k−1
0 θ k−2q−1
1 θ q−1
2
∧ ((k − 2q)γθ 0 θ 2 + qβθ 0 θ 1 − ɛ(n − k + q)βθ 1 θ 2 ) mod α.
By substituting the last expressions in i T Dβ k,q ≡ i T dβ k,q − dξ (mod
α, dα), we get the result.
To compute δ X Γ 2q,q , note that dγ 2q,q has 3 terms which are not multiple
of α (cf. Lemma 3.5). Then the Rumin differential is given by
Dγ 2q,q = d(γ 2q,q + α ∧ (ξ + ξ 2 + ξ 3 )) where ξ corresponds, as above, to
the form in (13),
and
ξ 2 ≡ c n,2q+2,q+1 (n − q − 1)(q + 1)βγθ n−q−2
0 θ q 2 mod (α, dα)
n − q − 1
ξ 3 ≡ c n,2q,q βγθ n−q−2
0 θ q 2 mod (α, dα).
2
Indeed, as in Proposition 4.6 in [BF08], ξ cancels the first term of dγ 2q,q
modulo α and ξ 2 the second one. The third term is canceled by ξ 3 since
it is the same multiple of dα = −θ s .
□
3.2. Variation of the measure of complex r-planes intersecting
a regular domain. We denote by L C r , r ∈ {1, ..., n − 1} the space
of complex r-planes in CK n (ɛ). Complex r-planes are totally geodesic
submanifolds of complex dimension r isometric to CK r (ɛ) (cf. [Gol99,
Lemma 2.2.4]). Moreover, Santaló [San52] proved the following properties
of this space (as usual, J denotes the complex structure).
Lemma 3.9 ([San52]). L C r , is a homogeneous space and
where
L C r ∼ = U ɛ (n)/U ɛ (r) × U(n − r)
⎧
⎨ C n ⋉ U(n), if ɛ = 0,
U ɛ = U(1 + n), if ɛ > 0,
⎩
U(1, n), if ɛ < 0.
Let {g; g 1 , Jg 1 , ..., g n , Jg n } be a local orthonormal frame such that
{g 1 , Jg 1 , ..., g r , Jg r } generate the tangent space of a complex r-plane at
g. The density of L C r is given by
∣ ∣∣∣∣∣∣∣
n∧ ∧
(14) dL r = ∣ ω i ∧ ω i ω ij ∧ ω ij
∣
i=r+1
i=1,...,r
j=r+1,...,n
where {ω i , ω ij } {i,j} are defined as in (8).

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 11
On ∂Ω there is a canonical vector field given by JN, and a distribution
D = 〈N, JN〉 ⊥ . At every point x ∈ ∂Ω, D x is the maximal
complex linear subspace of T x CK(ɛ) contained in T x ∂Ω. We shall consider
the bundle G C n,r(T ∂Ω) whose fiber at every point x ∈ ∂Ω is the
Grassmanian G C n,r(T x ∂Ω) of r-dimensional complex subspaces of D x ;
i.e., G C n,r(T ∂Ω) = {(x, l)|x ∈ ∂Ω, l is a J-invariant r-dimensional linear
subspace of T x ∂Ω}.
Proposition 3.10. Suppose Ω ⊂ CK n (ɛ) is a regular domain, X is
a smooth vector field on CK n (ɛ), φ t is the flow associated to X and
Ω t = φ t (Ω), then
∫
∫
( ∫ )
d
dt∣ χ(Ω t ∩L r )dL r = 〈∂φ/∂t, N〉
σ 2r (II| V )dV dx
t=0 ∂Ω
G C n,r(T x∂Ω)
L C r
where N is the inner normal field and σ 2r (II| V ) denotes the 2r-th symmetric
elementary function of II restricted to V ∈ G C n,r(T x ∂Ω).
Proof. This proof is based on the one in [Sol06, Theorem 4]. For every
l ∈ G C n,r(T x ∂Ω), we make the parallel translation l t of l along φ t (x).
Recall that parallel translation preserves the complex structure (cf.
[O’N83, p. 326]). Then we project l t onto D φt(x), obtaining a complex
r-plane l ′ t (at least for small values of t). We define
As
γ : G C n,r(T ∂Ω) × (−ɛ, ɛ) −→ L C r
((x, l), t) ↦→ exp φt(x) l ′ t.
γ ∗ (dL r ) = ι ∂t (γ ∗ (dL r ))dt = γ ∗ t (ι dγ∂t dL r )dt
where γ t = γ(·, t), in the same way as in [Sol06], using the co-area
formula and taking derivatives with respect to t, we get
∫
∫
d
dt∣ χ(Ω t ∩ L r )dL r = signφ signK(L r )γ0(ι ∗ dφ∂t dL r ).
t=0 G C n,r (T ∂Ω)
L C r
Suppose that {g; g 1 , Jg 1 , ..., g n , Jg n } is a local orthonormal frame
defined on G C n,r(T ∂Ω 0 ) × (−ɛ, ɛ) such that g((x, l), t) = φ(x, t), γ =
〈g, g 1 , Jg 1 , ..., g r , Jg r 〉 ∩ CK n (ɛ) and Jg n ((x, l), t) = N t (N t denotes the
normal vector to ∂Ω t at φ t (x)). Consider the curve L r (t) given by the
parallel translation of L r along the geodesic given by N, the inner normal
vector to ∂Ω 0 . If P denotes the tangent vector to L r (t), then for
t = 0
ω i (P ) = 〈dg(P ), g i 〉 = 〈 d dt g(L r(t)), g i 〉 = 0,
ω n (P ) = 〈dg(P ), N〉 = 1,
ω kj (P ) = 〈dg k (P ), g j 〉 = 〈 d dt g k(L r (t)), g j 〉 = 0

12 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
where i ∈ {r + 1, r + 1, ..., n − 1, n − 1}, j ∈ {r + 1, r + 1, ..., n, n}, and
k ∈ {1, 1, ..., r, r}. By (14) and last equations we get
dL r = |ω n |ι P dL r
since ι P dL r = |ω n (P )| · | ∧ n
h=r+1 ω h ∧ ω h
∧
ωij |. Thus,
and
So,
ι dγ∂t dL r = |ω n (dγ∂t)|ι P dL r + |ω n |ι dγ∂t ι P dL r
ω n (dγ∂t) = 〈dg(dγ∂t), N〉 = 〈 ∂φ
∂t , N〉,
γ ∗ 0(ω n )(v) = 〈dg(dγ 0 (v)), N〉 = 0 ∀v ∈ T (p,l) G C n,r(T ∂Ω 0 ).
γ0(ι ∗ dγ∂t dL r ) = |〈 ∂φ
∂t , N〉|γ∗ 0(ι P dL r ).
Finally, using that γ0(ι ∗ P dL r ) = |K(l)|dL C r[x]
dx, we get the result. □
Remark 3.11. In real space forms it is known
∫
G n−1,r
σ r (II| V )dV = vol(G(n − 1, r))σ r (II),
but in complex space forms we do not know how to evaluate directly
the analogous integral over the complex Grassmanian.
If r = n − 1, then this integral can be easily evaluated in CK n (ɛ).
Corollary 3.12. Let Ω ⊂ CK n (ɛ) be a regular domain, let X be a
smooth vector field on CK n (ɛ), let φ t be the flow associated to X and
let Ω t = φ t (Ω). Then
∫
d
dt∣ χ(Ω t ∩ L n−1 )dL n−1 = ω 2n−1 ˜B1,0 (Ω).
t=0
L C n−1
Proof. Since there is only one complex hyperplane contained in the
tangent space of a hypersurface, using the following relations, we obtain
the result directly from Proposition 3.10 .
∫
∫
∫
d
dt∣ χ(Ω t ∩ L n−1 )dL n−1 = 〈∂φ/∂t, N〉 σ 2n−2 (II| V )dV dx
t=0 L C n−1
∂Q 0 G C n−1,n−1
∫
= 〈∂φ/∂t, N〉σ 2n−2 (II| D )dx
∂Q 0
=
∫∂Q 0
〈∂φ/∂t, N〉 β ∧ θn−1 0
(n − 1)!
= c−1 n,1,0
(n − 1)! ˜B 1,0 = ω 2n−1 ˜B1,0 .
□

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 13
4. Crofton type formulas
4.1. In the standard Hermitian space.
Theorem 4.1. Let Ω ⊂ C n , let X be a smooth vector field over C n ,
let φ t be the flow associated to X and let Ω t = φ t (Ω). Then
∫
( )
d
−1 ( ) n − 1 n
dt∣ χ(Ω t ∩ L r )dL r = vol(G C n−1,r)ω 2r+1 (r + 1)
−1·
t=0 L r r
C r
⎛
⎞
n−r−1
∑
( )
· ⎝
2q − 1 1
(15)
q 4 ˜B q−1 2n−2r−1,n−r−q (Ω) ⎠
and
∫
L C r
(16)
q=max{0,n−2r}
( ) −1 ( n − 1 n
χ(Ω ∩ L r )dL r = vol(G C n−1,r)ω 2r
r r
⎛
· ⎝
∑n−r
q=max{0,n−2r}
) −1·
1
4 n−r−q ( 2n − 2r − 2q
n − r − q
⎞
)
µ 2n−2r,q (Ω) ⎠ .
Proof. In order to simplify the following computations, we consider
(17) B ′ Ω
k,q (Ω) = c −1
n,k,q BΩ k,q(Ω),
and
(18) ˜B′ k,q = c −1
n,k,q ˜B k,q ,
Γ ′ Ω
k,q (Ω) = 2c −1
n,k,q ΓΩ k,q(Ω)
˜Γ′ k,q = 2c −1
n,k,q ˜Γ k,q .
The expression ∫ χ(L
L C r ∩ Ω)dL r is a valuation on C n with degree
r
of homogeneity 2n − 2r. Thus, it can be expressed as a linear combination
of the elements of a basis of valuations with the same degree of
homogeneity. Then, by Remark 2.9 and (11), we have
∫
n−r−1
∑
(19) χ(Ω ∩ L r )dL r =
C q B 2n−2r,q ′ + DΓ ′ 2n−2r,n−r
L C r
q=max{0,n−2r}
for certain constants C q , D which we wish to determine. This will be
done by comparing the variation of both sides of this equality. From
here on we assume 2r < n. The case 2r ≥ n can be treated in the same
way (cf. Remark 4.2).
By Proposition 3.8, the variation of the right hand side of (19) is a
linear combination of the following type
(20)
n−r−1
∑
q=n−2r−1
c q ˜B′ 2n−2r−1,q +
n−r−1
∑
q=n−2r
d q ˜Γ′ 2n−2r−1,q
where the coefficients c q and d q can be expressed in terms of a linear
combination with known coefficients of the variables C q and D, that
still remain unknown.

14 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
The variation of the left hand side of (19), by Proposition 3.10 is
∫
∫
∫
d
(21)
dt∣ χ(Ω t ∩ L r )dL r = 〈∂φ/∂t, N〉 σ 2r (II| V )dV dx.
t=0
L C r
∂Ω
G C n−1,r
After fixing a frame, the integral ∫ σ
G C 2r (II| V )dV is a polynomial
n−1,r
of the second fundamental form II restricted to the distribution D =
〈N, JN〉 ⊥ . When pulling-back the form γ k,q from N(Ω) to ∂Ω, one
gets a polynomial expression P k,q of degree k in the coefficients of II.
Each of the monomials of P k,q is a minor k × k of II containing the
entry II(JN, JN). For q ≠ q ′ , all the monomials of P k,q are different
from those of P k,q ′. Therefore, every linear combination of {P k,q } q
must contain the variable II(JN, JN). Since JN /∈ D, comparing
the expressions of (20) and (21), it follows that d q = 0 for all q ∈
{n − 2r, . . . , n − r − 1}.
As c q and d q depend on C q and D, we will obtain the value of c q
once we know the value of C q and D. We will get their value from the
equalities {d q = 0}. Note that this gives r equations, since q runs from
n − 2r to n − r − 1 in (20). As for the unknowns, we need to find r
constants C q plus the constant D in (19).
We will get an extra equation by taking II| D = Id and equating (21)
to (20). Then, for any pair (n, r) we have a compatible linear system
since constants in (19) exist. Next we find the solution, which turns
out to be unique.
Let us relate explicitly the coefficients {c q } and {d q } in (20) with
C q and D in (19). To simplify the range of the subscripts, we denote
d n−r−a with a = 1, . . . , r and c n−r−a with a = 1, . . . , r + 1.
Coefficient d n−r−1 .From the variation of B ′ k,q in Cn (Proposition 3.7),
the coefficient of ˜Γ ′ 2n−2r−1,n−r−1 comes from the variation of B ′ 2n−2r,n−r−1
and Γ ′ 2n−2r,n−r. Then,
(22)
d n−r−1 = −2r(n − r)D + (2n − 2r − 2(n − r − 1)) 2 C n−r−1
= 4C n−r−1 − 2r(n − r)D.
Coefficient d n−r−a , a = 2, ..., r. The coefficient of ˜Γ ′ 2n−2r−1,n−r−a
comes from the variation of B 2n−2r,n−r−a ′ and B 2n−2r,n−r−a+1. ′ Then,
(23)
d n−r−a = (2n − 2r − 2(n − r − a)) 2 C n−r−a
− (2r + n − r − a + 1 − n)(n − r − a + 1)C n−r−a+1
= 4a 2 C n−r−a − (r − a + 1)(n − r − a + 1)C n−r−a+1 .
Coefficient c n−r−1 . The coefficient of ˜B ′ 2n−2r−1,n−r−1 comes from the
variation of B ′ 2n−2r,n−r−1 and Γ ′ 2n−2r,n−r. Then,
(24)
c n−r−1 = 4(r + 1/2)(n − r)D − 4C n−r−1
= 2(2r + 1)(n − r)D − 4C n−r−1 .

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 15
Coefficient c n−r−a , a = 2, ..., r − 2. The coefficient of ˜B ′ 2n−2r−1,n−r−a
comes from the variation of B ′ 2n−2r,n−r−a and B ′ 2n−2r,n−r−a+1. Then,
c n−r−a = −2(2a)(2a − 1)C n−r−a + 2(r − a + 3/2)(n − r − a + 1)C n−r−a+1
(25)
= −4a(2a − 1)C n−r−a + (2r − 2a + 3)(n − r − a + 1)C n−r−a+1 .
Coefficient c n−2r−1 . The coefficient of ˜B′ 2n−2r−1,n−2r−1 comes from
the variation of B 2n−2r,n−2r. ′ Then,
(26)
c n−2r−1 = (2r − 2(r + 1) + 3)(n − r − (r + 1) + 1)C n−2r
= (n − 2r)C n−2r .
Now, we solve the linear system given by {d n−r−a = 0} where a ∈
{1, ..., r}. From equations (22) and (23) we have that the system is
given by:
{
r(n − r)D = 2Cn−r−1
4a 2 C n−r−a = (n − r − a + 1)(r − a + 1)C n−r−a+1 .
Thus,
(27)
(n − r − a + 1) · ... · (n − r)(r − a + 1)... · r
C n−r−a = D
2 · 4 a−1 a 2 (a − 1) 2 · ... · 1 2
(n − r)!r!
=
2 2a−1 (n − r − a)!(r − a)!a!a! D
= D ( )( n − r r
.
2 2a−1 a a)
To obtain the value of D, we calculate ∫ σ
G C 2r (p)dV and β 2n−2r−1,n−r−a
′ n−1,r
with II| D (p) = Id. On the one hand, we have
∫
σ 2r (p)(Id| V )dV = vol(G C n−1,r).
G C n−1,r
On the other hand, if II| D = Id, then the connection forms satisfy
α 1i = ω i and β 1i = ω i . Thus, θ 1 = 2θ 2 and θ 0 = θ 2 and we obtain
So, the equation
must be satisfied.
β ′ 2n−2r−1,n−r−a(p) = (β ∧ θ r−a+1
0 ∧ θ 2a−2
1 ∧ θ n−r−a
2 )(p)
= (β ∧ θ n−1
2 )(p) = 2 2a−2 (n − 1)!.
∑r+1
vol(G C n−1,r) = c n−r−a 2 2a−2 (n − 1)!
a=1

16 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Equations in (24), (25) and (26) give us the relation among c n−r−a ,
D and C n−r−a . Using these relations we have
vol(G C n−1,r)
= (2(2r + 1)(n − r)D − 4C n−r−1 )
(n − 1)!
r∑
+ 2 2a−2 ((2r − 2a + 3)(n − r + a + 1)C n−r−a+1 − 4a(2a − 1)C n−r−a )
a=2
+ 2 2r (n − 2r)C n−2r
= 2(2r + 1)(n − r)D + 4C n−r−1 ((2r − 1)(n − r − 1) − 1)
∑
(−2 2a−2 4a(2a − 1) + 2 2a (2r − 2a + 1)(n − r − a))C n−r−a
r−1
+
a=2
+ C n−2r (2 2r (n − 2r) − 2 2r−2 4r(2r − 1))
r∑
= 2(2r + 1)(n − r)D + 2 2a C n−r−a ((2r − 2a + 1)(n − r − a) − a(2a − 1))
a=1
(
)
(27)
r∑ (2r − 2a + 1)(n − r − a) − a(2a − 1)
= D 2(2r + 1)(n − r) + 2(n − r)!r!
a!a!
(n − r − a)!(r − a)!
a=1
(
)
n! − r!(n − r)!(2r + 1)
= D 2(2r + 1)(n − r) + 2(n − r)!r!
r!r!(n − r)!(n − r − 1)!
2 n!
= D
r!(n − r − 1)! .
Thus,
D = vol(GC n−1,r)
2 n!
C n−r−a = vol(GC n−1,r)
4 a n!
( ) −1 n − 1
,
r
( ) −1 ( )( n − 1 n − r r
r a a)
and, for 2r < n, we have
∫
r∑
χ(Ω ∩ L r )dL r = C n−r−a B 2n−2r,n−r−a ′ + DΓ ′ 2n−2r,n−r
L C r
a=1
= vol(GC n−1,r ) ( ) (
n − 1
−1 r∑ ( )( )
n − r r
2
2 n! r
a a)
−2a+1 B 2n−2r,n−r−a+ ′ Γ ′ 2n−2r,n−r
a=1
and
∫
d
dt∣ χ(Ω t ∩ L r )dL r = (2(2r + 1)(n − r)D − 4C n−r−1 )B 2n−2r−1,n−r−1
′
t=0 L C r
r∑
+ ((2r − 2a + 3)(n − r + a + 1)C n−r−a+1 −4a(2a − 1)C n−r−a )B 2n−2r−1,n−r−a
′
a=2
+(n − 2r)C n−2r B 2n−2r−1,n−2r−1
′
= vol(GC n−1,r ) ( ) (
n − 1
−1 ∑r+1
( n − r
n! r
a
a=1
)( r + 1
a
) )
a
4 ˜B ′ a−1 2n−2r−1,n−r−a .

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 17
Finally, we use the relation in (17) and (11) to obtain the result.
Remark 4.2. If 2r ≥ n, then formula (15) follows directly from the
relations among the different bases of valuations on C n given in [BF08]
and the following relation in [Ale03]
∫
χ(Ω ∩ L r )dL r = 1 M 2r−3 (∂Ω ∩ L r )dL r = cU 2(n−r),n−r
L O C r
2r−1
∫L C r
for a certain constant c coming from the different normalizations in
dL r .
4.2. In CK n (ɛ).
Corollary 4.3. Let Ω ⊂ CK n (ɛ), let X be a smooth vector field over
CK n (ɛ), let φ t be the flow associated to X and let Ω t = φ t (Ω). Then
∫
( )
d
−1 ( ) n − 1 n
dt∣ χ(Ω t ∩L r )dL r = vol(G C n−1,r)ω 2r+1 (r + 1)
−1·
t=0 L r r
C r
⎛
⎞
n−r−1
∑
( )
· ⎝
2q − 1 1
(28)
q 4 ˜B q−1 2n−2r−1,n−r−q (Ω) ⎠ .
q=max{0,n−2r−1}
Proof. Comparing equation (15) and Proposition 3.10 in case ɛ = 0
shows that
∫ ( ∫ )
〈X, N〉 σ 2r (II|V )dV dx
∂Ω
G C n,r
equals the right hand side of equation above. By taking a field X that
vanishes outside an arbitrarily small neigborhood of any x ∈ ∂Ω, we
deduce the following equality between forms
( ∫ )
σ 2r (II|V )dV dx = ω 2r+1
)( n
G C n,r
r)·
·
n−r−1
∑
q=max{0,n−2r−1}
( 2q − 1
q
( n−1
r
)
cn,2n−2r−1,n−r−q
β ∧ θ r−q+1
4 q−1 0 ∧ θ 2q−2
1 ∧ θ n−r−q
2
This equation extends obviously to CK n (ɛ) without change.
using Proposition 3.10 gives the result.
Theorem 4.4. Let Ω be a regular domain in CK n (ɛ). Then
∫
( ) −1 n − 1
χ(Ω ∩ L r )dL r = vol(G C n−1,r) (ɛ r (r + 1)vol(Ω)+
r
L C r
+
∑n−1
k=n−r
+
ɛ k−(n−r) ω 2n−2k
( n
k) −1
· ((k + r − n + 1)µ 2k,k +
∑k−1
q=max{0,2k−n}
( )
1 2k − 2q
µ
4 k−q 2k,q )).
k − q
□
Then,
□

18 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Proof. First, we write the formula to be proved in terms of {B k,q ′ } and
{Γ ′ k,q } defined in (17):
∫
χ(Ω ∩ L r )dL r = vol(GC n−1,r ) ( ) n − 1 −1
(ɛ r (r + 1)n!vol(Ω)+
n! r
+
L C r
(29)
n−1
∑
k=n−r
ɛ k−(n−r) ⎛
⎝
∑k−1
q=max{0,2k−n}
( )
1 n − k k
4 k−q B 2k,q
k − q)( ′ q
+ k + r − n + 1
2
Γ ′ 2k,k
We calculate the variation of both sides of the equation (29) to prove
that they coincide. The variation of the left hand side is given in
Corollary 4.3. To calculate the variation of the right hand side we use
Proposition 3.8 which we recall in the following table. We denote by
(δ X B k,q ′ , ˜B r,s) ′ the coefficient of ˜B r,s in the expression of δ X B k,q ′ .
⎧
2q(n + q − k + 1/2), if r = k − 1, s = q − 1
⎪⎨
(δ X B
k,q ′ , ˜B
−2(k − 2q)(k − 2q − 1), if r = k − 1, s = q
r,s)=
′ 2ɛ(k − 2q)(k − 2q − 1), if r = k + 1, s = q + 1
−2ɛ(n − k + q)(q + 1/2), if r = k + 1, s = q
⎪⎩
0, otherwise.
⎧
(δ X B
k,q ′ , ˜Γ
⎨ (k − 2q) 2 , if r = k − 1, s = q
′
r,s)= −(n + q − k)q, if r = k − 1, s = q − 1
⎩
0, otherwise.
⎧
4q(n − q + 1/2), if r = 2q − 1, s = q − 1
⎪⎨
(δ X Γ ′ 2q,q , ˜B r,s)=
′ 4ɛ((q + 1)/2 − (n − q)(2q + 3/2)), if r = 2q + 1, s = q
4ɛ ⎪⎩
2 (n − q − 1)(q + 3/2), if r = 2q + 3, s = q + 1
0, otherwise.
⎧
(δ X Γ ′ 2q,q , ˜Γ
⎨ −2(n − q)q, if r = 2q − 1, s = q − 1
′
r,s)= 2ɛ(n − q − 1)(q + 1), if r = 2q + 1, s = q
⎩
0, otherwise.
With these expressions, one can check that the variation of the right
hand side of (29) coincides with the right hand side of (28).
Finally, we take a deformation Ω t of Ω such that Ω t converges to a
point. Since both sides of (29) vanish in the limit, the result follows. □
5. Gauss-Bonnet theorem
Theorem 5.1. Let Ω be a regular domain in CK n (ɛ). Then
O 2n−1 χ(Ω) = 2n(n + 1)ɛ n vol(Ω) +
⎛
⎞
∑n−1
O 2n−2c−1 ɛ c ∑c−1
( )
+ ) ⎝
1 2c − 2q
(30)
µ
4 c−q 2c,q +(c + 1)µ 2c,c
⎠.
c − q
c=0
( n−1
c
q=max{0,2c−n}
⎞
⎠).

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 19
Proof. We proceed analogously to the proof of Theorem 4.4. In fact,
the same computations of the previous proof show (in case r = n) that
the right hand side of (30) has null variation.
If ɛ = 0, then we get the known Gauss-Bonnet formula in C n . Taking
some smooth deformation of Ω to get a domain contained in a ball
of arbitrarily small radius, we have that the equality is true for all
ɛ ∈ R.
□
Theorem 5.2. Let Ω be a regular domain in CK n (ɛ). Then
∫
M 2n−1 (∂Ω) = O 2n−1 χ(Ω) − 2nɛ χ(∂Ω ∩ L n−1 )dL n−1 −
k=1
L C n−1
∑n−1
( ) −1 n − 1
− ɛ k O 2n−2k−1 µ 2k,k − 2nɛ n vol(Ω).
k
Proof. First, we use the following relation between M 2n−1 (∂Ω) and
µ 0,0 (Ω)
∫
∫
γ ∧ θ0
n−1
M 2n−1 (∂Ω) = σ 2n−1 (II x )dx =
∂Ω
∂Ω (n − 1)!
∫
= 2c−1 n,0,0 c n,0,0
γ ∧ θ0 n−1 = 2c−1 n,0,0
(n − 1)! 2 ∂Ω
(n − 1)! µ 0,0(Ω)
(31)
= 2nω 2n µ 0,0 (Ω) = O 2n−1 µ 0,0 (Ω).
Now, from Theorems 4.4 and 5.1, it follows that
⎛
⎞
n−1
∑ ɛ c n−1
c! ∑
( )
χ(Ω) = ⎝
1 2c − 2q
π c
4 c−q µ 2c,q + (c + 1)µ 2k,k
⎠
c − q
c=0
+ ɛn (n + 1)!
π n
n−1
∑
= µ 0,0 +
c=1
q=max{0,2c−n}
vol(Ω)
⎛
ɛ c c!
π c
⎝
+ ɛn (n + 1)!
π n vol(Ω)
= µ 0,0 + ɛ n! n−1
∑
π n
+ ɛ n! n−1
∑
π n
c=1
c=1
= µ 0,0 + ɛ n! ∫
π n
( ɛ n (n + 1)!
+
π n
n−1
∑
q=max{0,2c−n}
⎛
ɛ c−1 c!π n−c
⎝
n!
ɛ c−1 c!π n−c
n!
⎞
( )
1 2c − 2q
4 c−q µ 2c,q + (c + 1)µ 2c,c
⎠
c − q
n−1
∑
q=max{0,2c−n}
µ 2c,c + ɛn (n + 1)!
π n vol(Ω)
n−1
∑
χ(∂Ω ∩ L n−1 )dL n−1 +
L n−1
− ɛn n!n
π n )
vol(Ω).
⎞
( )
1 2c − 2q
4 c−q µ 2c,q + cµ 2c,c
⎠
c − q
c=1
ɛ c c!
π c µ 2c,c

20 JUDIT ABARDIA, EDUARDO GALLEGO, GIL SOLANES
Remark 5.3. For n = 2 and n = 3, the Gauss-Bonnet-Chern formula
in CK n (ɛ) given in Theorem 5.1 was already stated in [Par02].
6. Total Gauss curvature integral
Theorem 6.1. Let Ω be a regular domain in C n . Then
∫
( ) −1 ( ) n − 1 n
M 2r−1 (∂Ω∩L r )dL r = 2rω2rvol(G 2 C n−1,r)
−1·
L r r
C r
⎛
⎞
∑n−r
( )
· ⎝
1 2n − 2r − 2q
µ
4 n−r−q 2n−2r,q
⎠ .
n − r − q
q=max{0,n−2r}
Proof. On the one hand, by Gauss-Bonnet formula in C n , we have
∫
∫
χ(Ω ∩ L r )dL r = µ 0,0 (Ω ∩ L r )dL r .
L C r
On the other hand, by Theorem 4.4 in C n , we have
∫
( ) −1 ( ) −1
n − 1 n
χ(∂Ω ∩ L r )dL r = vol(G C n−1,r) ω 2r
L r n − r
C r
⎛
⎞
∑n−r
( )
· ⎝
1 2n − 2r − 2q
µ
4 n−r−q 2n−2r,q
⎠ .
n − r − q
q=max{0,n−2r}
If we equate both expressions and we use the relation (31) between
the total Gauss curvature and the valuation µ 0,0 , we obtain the result.
□
Remark 6.2. The previous Theorem is not necessarily true in CK n (ε)
for ɛ ≠ 0. Indeed, Howard’s transfer principle can not be used here since
Theorem 6.1 is not local: it does not apply to general hypersurfaces,
but only to the closed embedded ones.
L C r
□
[Aba09]
[Ale03]
[BF08]
[Bla76]
[Gol99]
References
J. Abardia. Average of mean curvature integral in complex space forms.
Preprints de la Universitat Autònoma de Barcelona, 2009.
S. Alesker. Hard Lefschetz theorem for valuations, complex integral geometry,
and unitarily invariant valuations. J. Differential Geom., 63(1):63–
95, 2003.
A. Bernig and J. H. Fu. Hermitian integral geometry. arXiv:0801.0711v5,
2008.
David E. Blair. Contact manifolds in Riemannian geometry. Springer-
Verlag, Berlin, 1976. Lecture Notes in Mathematics, Vol. 509.
W. M. Goldman. Complex hyperbolic geometry. Oxford Mathematical
Monographs. The Clarendon Press Oxford University Press, New York,
1999. Oxford Science Publications.

GAUSS-BONNET THEOREM IN COMPLEX SPACE FORMS 21
[Had57] H. Hadwiger. Vorlesungen über Inhalt, Oberfläche und Isoperimetrie.
Springer-Verlag, Berlin, 1957.
[Nav05] A. M. Naveira. Two problems in real and complex integral geometry. In
Complex, contact and symmetric manifolds, volume 234 of Progr. Math.,
pages 209–220. Birkhäuser Boston, Boston, MA, 2005.
[O’N83] B. O’Neill. Semi-Riemannian geometry, volume 103 of Pure and Applied
Mathematics. Academic Press Inc. [Harcourt Brace Jovanovich Publishers],
New York, 1983. With applications to relativity.
[Par02] Heunggi Park. Kinematic formulas for the real subspaces of complex space
forms of dimension 2 and 3. PhD-thesis. University of Georgia, 2002.
[Rum94] M. Rumin. Formes différentielles sur les variétés de contact. J. Differential
Geom., 39(2):281–330, 1994.
[San52] L. A. Santaló. Integral geometry in Hermitian spaces. Amer. J. Math.,
74:423–434, 1952.
[San04] L. A. Santaló. Integral geometry and geometric probability. Cambridge
Mathematical Library. Cambridge University Press, Cambridge, second
edition, 2004. With a foreword by Mark Kac.
[Sol06] G. Solanes. Integral geometry and the Gauss-Bonnet theorem in constant
curvature spaces. Trans. Amer. Math. Soc., 358(3):1105–1115 (electronic),
2006.