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Comptes Rendus
Mathématique
Rudy Rosas
Transversely product singularities of foliations in projective spaces
Volume 361 (2023), p. 1785-1787
Published online: 21 December 2023
https://doi.org/10.5802/crmath.528
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
Les Comptes Rendus. Mathématique sont membres du
Centre Mersenne pour l’édition scientifique ouverte
www.centre-mersenne.org
e-ISSN : 1778-3569
Comptes Rendus
Mathématique
2023, Vol. 361, p. 1785-1787
https://doi.org/10.5802/crmath.528
Functional analysis / Analyse fonctionnelle
Transversely product singularities of foliations
in projective spaces
Rudy Rosas a
a Departamento de Ciencias, Pontificia Universidad Católica del Perú, Av. Universitaria 1801, Lima, Perú
E-mail: rudy.rosas@pucp.pe
Abstract. We prove that a transversely product component of the singular set of a holomorphic foliation on
Pn is necessarily a Kupka component.
Funding. The author was supported by Vicerrectorado de Investigación de la Pontificia Universidad Católica
del Perú.
Manuscript received 4 May 2023, revised and accepted 23 June 2023.
1. Introduction
Let U be an open set of a complex manifold M and let k ∈N. Let η be a holomorphic k-form on
U and let Singη := {p ∈ U : η(p) = 0} denote the singular set of η. We say that η is integrable if
each point p ∈ U \Singη has a neighborhood V supporting holomorphic 1-forms ξ1, . . . ,ξk with
η|V = ξ1 ∧·· ·∧ξk , such that dξ j ∧η= 0 for each j = 1, . . . ,k. In this case the distribution
Dη : Dη(p) = {v ∈ Tp M : ivη(p) = 0}, p ∈U \Singη
defines a holomorphic foliation of codimension k on U \Singη. A singular holomorphic foliation
F of codimension k on M can be defined by an open covering (U j ) j∈J of M and a collection
of integrable k-forms η j ∈ Ωk (U j ) such that ηi = gi jη j for some gi j ∈ O∗(Ui ∩U j ) whenever
Ui ∩U j ̸= ;. The singular set SingF is the proper analytic subset of M given by the union of the
sets Singη j . From now on we only consider foliations F such that SingF has no component of
codimension one.
Given a singular holomorphic foliation F of codimension k on M as above, the Kupka singular
set of F , denoted by K (F ), is the union of the sets
K (η j ) = {p ∈U j : η j (p) = 0,dη j (p) ̸= 0}.
This set does not depend on the collection (η j ) of k-forms used to define F . It is well known
(see [7, 9]) that, given p ∈ K (F ), the germ of F at p is holomorphically equivalent to the product
of a one-dimensional foliation with an isolated singularity by a regular foliation of dimension
(dimF − 1). More precisely, if dim M = k + m + 1, there exist a holomorphic vector field X =
X1∂x1 + ·· · + Xk+1∂xk+1 on Dk+1 with a unique singularity at the origin, a neighborhood V of p
in M and a biholomorphism ψ : V →Dk+1 ×Dm , ψ(p) = 0, which conjugates F with the foliation
ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
1786 Rudy Rosas
FX of Dk+1 ×Dm generated by the commuting vector fields X ,∂y1 , . . . ,∂ym , where y = (y1, . . . , ym)
are the coordinates in Dm . If µ= dx1 ∧·· ·∧dxk+1, the foliation FX is also defined by the k-form
ω= iXµ and the Kupka condition dω(0) ̸= 0 is equivalent to the inequality div X (0) ̸= 0.
Following [8], we say that F is a transversely product at p ∈ SingF if as above there exist
a holomorphic vector field X and a biholomorphism ψ : V → Dk+1 ×Dm conjugating F with
FX , except that it is not assumed that div X (0) ̸= 0. We say that Γ is a local transversely product
component of SingF if Γ is a compact irreducible component of SingF and F is a transversely
product at each p ∈ Γ. In particular, if Γ⊂ K (F ) we say that Γ is a Kupka component — for more
information about Kupka singularities and Kupka components we refer the reader to [1–7]. If Γ is
a transversely product component of SingF , we can cover Γ by finitely many normal coordinates
like ψ, with the same vector field X : that is, there exist a holomorphic vector field X on Dk+1 with
a unique singularity at the origin and a covering of Γ by open sets (Vα)α∈A such that each Vα
supports a biholomorphism ψα : Vα→Dk+1 ×Dm that maps Γ∩Vα onto {0}×Dm and conjugates
F with the foliation FX . The sets (Vα) can be chosen arbitrarily close to Γ.
In [8], the author proves that a local transversely product component of a codimension one
foliation on Pn is necessarily a Kupka component. The goal of the present paper is to generalize
this theorem to foliations of any codimension.
Theorem 1. Let F a holomorphic foliation of dimension ≥ 2 and codimension ≥ 1 on Pn . Let Γ be
a transversely product component of SingF . Then Γ is a Kupka component.
This theorem is a corollary of the following result.
Theorem 2. Let F a holomorphic foliation of dimension ≥ 2 and codimension k ≥ 1 on a complex
manifold M. Suppose that F is defined by an open covering (U j ) j∈J of M and a collection of k-
forms η j ∈ Ωk (U j ). Let L be the line bundle defined by the cocycle (gi j ) such that ηi = gi jη j ,
gi j ∈ O∗(Ui ∩U j ). Let Γ be a transversely product component of SingF that is not a Kupka
component. Then c1(L|Γ) = 0.
2. Proof of the results
Proof of Theorem 2. Let dim M = k + m + 1. As explained in the introduction, there exist a
holomorphic vector field X on Dk+1 with a unique singularity at the origin and a covering of
Γ by open sets (Vα)α∈A such that each Vα is contained in V and supports a biholomorphism
ψα : Vα → Dk+1 ×Dm that maps Γ∩Vα onto {0}×Dm and conjugates F with the foliation FX
generated by the commuting vector fields X ,∂y1 , . . . ,∂ym . Notice that div(X )(0) = 0, because Γ is
not a Kupka component. Since FX is defined by the k-form ω= iXµ, where µ= dx1 ∧·· ·∧dxk+1,
we have that F |Vα is defined by the k-form ψ∗α(ω). If Vα∩Vβ ̸= ;, there exists θαβ ∈ O∗(Vα∩Vβ)
such that
ψ∗α(ω) = θαβψ∗β(ω). (1)
Therefore the cocycle (θαβ) define the line bundle L restricted to some neighborhood of Γ. Thus,
in order to prove that c1(L|Γ) = 0 it is enough to show that each θαβ|Γ is locally constant. Fix
some α,β ∈ A such that Vα∩Vβ ̸= ;. If we set ψ =ψα ◦ψ−1β and θ = θαβ ◦ψ−1β , from (1) we have
that ψ∗(ω) = θω, which means that ψ preserves the foliation FX . It suffices to prove that the
derivatives θy1 (p), . . . ,θym (p) vanish if p ∈ {0}×Dm . Since ∂y1 is tangent to FX , then the vector
field ψ∗(∂y1 ) is tangent to FX and so we can express
ψ∗(∂y1 ) =λX +λ1∂y1 +·· ·+λm∂ym ,
where λ,λ1, . . . ,λm are holomorphic. Then
Lψ∗(∂y1 )ω=LλXω=λLXω+dλ∧ iXω=λLXω=λdiv(X )ω,
Rudy Rosas 1787
where the last equality follows from the identity ω= iXµ. Thus, since
ψ∗
(
Lψ∗(∂y1 )ω
)
=L∂y1ψ
∗ω=L∂y1 (θω) = θy1ω,
we obtain that
θy1ω=ψ∗ (λdiv(X )ω) =λ(ψ)div(X )(ψ)θω
and therefore θy1 (p) = 0 if p ∈ {0}×Dm , because div(X ) vanishes along {0}×Dm . In the same way
we prove that θy2 (p) = ·· · = θym (p) = 0 if p ∈ {0}×Dm , which finishes the proof. □
Proof of Theorem 1. Suppose that Γ is not a Kupka component. Let L be the line bundle associ-
ated to F as in the statement of Theorem 2. We notice that c1(L) ̸= 0, otherwise F will be defined
by a global k-form on Pn , which is impossible. Then, if we take an algebraic curve C ⊂ Γ, we have
c1(L) ·C ̸= 0. Therefore, ifΩ is a 2-form on Pn in the class c1(L) and V is a tubular neighborhood
of Γ,
c1(L|Γ) ·C =
∫
C
Ω|Γ =
∫
C
Ω= c1(L) ·C ̸= 0,
which contradicts Theorem 2. □
References
[1] M. Brunella, “Sur les feuilletages de l’espace projective ayant une composante de Kupka”, Enseign. Math. 55 (2009),
no. 3-4, p. 227-234.
[2] O. Calvo-Andrade, “Foliations with a Kupka component on Algebraic Manifolds”, Bol. Soc. Bras. Mat. 30 (1999), no. 2,
p. 183-197.
[3] ——— , “Foliations of codimension greater than one with a Kupka component”, Qual. Theory Dyn. Syst. 8 (2009), no. 2,
p. 241-253.
[4] ——— , “Foliations with a radial Kupka set on projective spaces”, Bull. Braz. Math. Soc. (N.S.) 47 (2016), no. 4, p. 1071-
1083.
[5] O. Calvo-Andrade, M. Corrêa, A. Fernández-Pérez, “Higher codimensional foliations with Kupka singularities”, Int. J.
Math. 28 (2017), no. 3, article no. 1750019 (29 pages).
[6] D. Cerveau, A. Lins-Neto, “Codimension-one foliations in CP n , n ≥ 3, with Kupka components”, in Complex analytic
methods in dynamical systems, Astérisque, vol. 222, Société Mathématique de France, 1994, p. 93-132.
[7] I. Kupka, “The singularities of integrable structurally stable pfaffian forms”, Proc. Natl. Acad. Sci. USA 52 (1964),
p. 1431-1432.
[8] A. Lins-Neto, “Local transversely product singularities”, Ann. Henri Lebesgue 4 (2021), p. 485-502.
[9] A. S. de Medeiros, “Structural stability of integrable differential forms”, in Geometry and Topology, Lecture Notes in
Mathematics, Springer, 1977, p. 395-428.


Transversely product singularities of foliations in projective spaces

Timothée L. Labinet

Comptes Rendus
Mathématique
Timothée L. Labinet
Projective varieties have countably many real forms
Volume 361 (2023), p. 863-867
Published online: 18 July 2023
https://doi.org/10.5802/crmath.441
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
Les Comptes Rendus. Mathématique sont membres du
Centre Mersenne pour l’édition scientifique ouverte
www.centre-mersenne.org
e-ISSN : 1778-3569
Comptes Rendus
Mathématique
2023, Vol. 361, p. 863-867
https://doi.org/10.5802/crmath.441
Number theory / Théorie des nombres
Projective varieties have countably many real
forms
Timothée L. Labinet
Abstract. In this note, we check that a complex projective algebraic variety has (at most) countably many real
forms. We more generally prove it when the field of reals is replaced with a field that has only countably
many finite extensions up to isomorphism. The verification consists in gathering known results about
automorphism groups and Galois cohomology. This contrasts with the recent discovery by A. Bot of an affine
real variety with uncountably many real forms.
Mathematical subject classification (2010). 12G05, 12F10, 11S25, 11E72, 14J50.
Manuscript received 11 September 2022, revised 13 October 2022, accepted 28 October 2022.
1. Introduction
By countable we mean “at most countable”, possibly finite or even empty.
We say that a field k, with separable closure k, is tame if it has countably many finite Galois
extensions up to k-isomorphism, or equivalently if its absolute Galois group Gal(k/k) is metriz-
able, or equivalently if k has countable degree over k (see Proposition 11). For instance, every
non-discrete locally compact field of characteristic zero is tame. Also, every pseudofinite field is
tame (indeed having at most n finite extensions of degree m can be characterized by a first-order
formula), and so is every real-closed or algebraically closed field. The above examples actually
have finitely many extensions, up to isomorphism, in each given degree. Furthermore, countable
fields are tame, and if k is a tame field of characteristic zero, then so is the field k((t )) of Laurent
series. In contrast, the field C(t ) is not tame.
If k is a perfect field with an algebraic closure k ⊂ k, by k-group we mean a Gal(k/k)-group
(see Section 2). We say that a k-group is countable-by-algebraic if it has a normal k-subgroup of
countable index that is isomorphic to an algebraic k-group.
Theorem 1. Let k be a perfect tame field. Let G be a k-group that is countable-by-algebraic. Then
the Galois cohomology set H1(k,G) is countable.
The main example of a k-group we have in mind is the group of automorphisms of a k-variety
(see Section 4). Here, a k-variety is meant to be a separated noetherian scheme, locally of finite
type over k (no reducedness or irreducibility assumption). A classical result (see Theorem 12) is
that the group of automorphisms of every complete k-variety is countable-by-algebraic.
ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
864 Timothée L. Labinet
Corollary 2. Let k be a perfect tame field and k ⊂ k an algebraic closure. Then every complete
k-variety has countably many k-forms (up to isomorphism).
When k is the real field (which was our starting point), this number can be infinite, even in the
smooth projective case: the first examples are due to Lesieutre [8] and further ones were provided
satisfying additional hypotheses [4–6].
Note that the corollary holds for other objects provided the automorphism group has a similar
structure. For instance, it applies to k-forms of pairs (X ,Y ) consisting of a complete variety X and
a closed subvariety Y .
However, it does not apply to arbitrary varieties, and actually fails for affine varieties. Indeed,
Bot has constructed a smooth affine complex surface with continuum many real forms [2].
An anecdotic remark is that a perfect field k is tame if and only if the product k-algebra kn has
countably many k-forms for every n. This being observed, one can naturally wonder whether a
stronger converse of Corollary 2 is true, namely:
Question 3. Is it true that for every non-tame perfect field k, there exists an absolutely irreducible
(smooth?) projective k-variety with uncountably many k-forms?
If one drops the irreducibility requirement, this is therefore true: just considering the finite
k-variety Spec(kn) for an appropriate n. Question 3 has a positive answer in many cases: for
instance when k∗/k∗n is uncountable for some n ≥ 2. Indeed, the set of k-forms of the projective
space Pn−1 is in bijection with H 1(k,PGLn), which is in bijection with k∗/k∗n (this example was
suggested by the referee).
Another related question is the following:
Question 4. Does there exist a complex projective variety X with infinitely many real forms, for
which the group Aut(X )/Aut(X )0 is finitely generated (it has to be infinite anyway)?
Indeed, it is not finitely generated in the examples found so far [4–6, 8].
For completeness, let us formulate a more general result that follows from the proof. For α
an infinite cardinal, say that a field k is α-tame if it has at most α Galois extensions up to k-
isomorphism, or equivalently if Gal(k/k) has ≤α open subgroups, or equivalently if the degree of
k over k is ≤ α. We say that a k-group is (cardinal ≤ α)-by-algebraic if it has a normal algebraic
k-subgroup of index ≤α.
Theorem 5. Letα be an infinite cardinal, k anα-tame field and let G be a k-group that is (cardinal
≤α)-by-algebraic. Then the Galois cohomology set H1(k,G) has cardinal ≤α.
Let k be a perfect field with at most α extensions up to k-isomorphism. Then every complete
k-variety has at most α k-forms.
2. Group-theoretic background
We use standard conventions in Galois cohomology (as in [13]). If P is a profinite group, by P-
group G we mean a group G endowed with an action of P by group automorphisms, such that
the stabilizer Pg of each g ∈G is open in P . The action is denoted (s, g ) 7→ s ·g . By P-subgroup we
just mean a P-invariant subgroup. By GP we mean the subgroup of points fixed by P .
For reference we recall:
Definition 6. Let F be a finite group and G an F -group. By definition, H1(F,G) is the set of maps
b : F → G satisfying b(st ) = b(s)(s ·b(t )) for all s, t ∈ F , modulo identifying b and b′ if there exists
g ∈G such that b′(s) = g−1b(s)(s · g ) for all s ∈ F .
If P is a profinite group and G is a P-group, H1(P,G) is the inductive limit of H1(P/U ,GU ) where
U ranges over open normal subgroups of P.
Timothée L. Labinet 865
Lemma 7. Let P be a metrizable profinite group and G a countable P-group. Then H1(P,G) is
countable.
Proof. In this case the number of open subgroups U of G is countable, so it is enough to check
that each H1(P/U ,GU ) is countable. This is clear since the set of maps P/U → GU is then
countable. □
To prove Theorem 1 we need a dévissage result; namely there is one in [13, I.§5.5], and we need
its corollary:
Lemma 8 ([13, I.§5.5, Cor. 3]). Let P be a profinite group, let H be a P-group, N a normal P-
subgroup, Q = H/N . Then H1(P, H) is finite (resp. countable) if and only if its image in H1(P,Q)
is finite (resp. countable), and that for every cocycle b ∈ Z 1(P, H) the quotient H1(P, b N )/(bQ)P is
finite (resp. countable).
Here b N means N with the P-action twisted by the cocycle b (see [13, I.§5.3]).
Corollary 9. Under the same assumption, if H1(P,Q) is finite (resp. countable) as well as H1(P, b N )
for every b ∈ Z 1(P, H) then so is H1(P, H).
Recall that an abelian group is divisible if g 7→ g n is surjective for every positive integer n (or
equivalently every prime n).
Lemma 10. Let P be a metrizable profinite group. Let A be a divisible abelian P-group, whose
torsion subgroup T is countable. Then H1(P, A) is countable.
Proof. First we claim that H1(P, A) = {0} if A is torsion-free. In this case, A is naturally a vector
space over Q. Let U be an open subgroup of P . Since the action of P on A is Q-linear, we see that
AU is also divisible. Hence by the usual averaging argument, H1(P/U , AU ) = {0}. By Definition 6
we deduce H1(P, A) = {0}.
Under the more general assumption on A, we deduce H1(P, A/T ) = {0}, so by dévissage
(Corollary 9) we are reduced to checking that H1(P,T ) is countable, which holds by Lemma 7. □
3. Tameness
We write the following proposition for the record.
Proposition 11. Let k be a field with separable closure k. The following are equivalent:
(1) the absolute Galois group G = Gal(k/k) is metrizable;
(2) k has countable degree over k;
(3) k has countably many k-subfields that are finite-dimensional over k;
(4) k has countably many Galois k-subfields that are finite-dimensional over k;
(5) k has countably many finite separable extensions up to k-isomorphism (i.e., k is tame);
(6) k has countably many finite Galois extensions up to k-isomorphism.
Proof. A totally disconnected compact Hausdorff topological space is metrizable if and only if
it has countably many clopen subsets. Therefore, a profinite group is metrizable if and only if it
has countably many clopen subsets, if and only if it has countably many normal open subgroups.
This already implies the equivalence of (1) with each of (3) and (4).
Suppose (2). Let (kn) be an increasing sequence of subfields of k containing k, with union
k. These correspond to a decreasing sequence of open normal subgroups of G , with trivial
intersection. Hence G embeds into
∏
n Gn , and hence is metrizable. Conversely if G is metrizable,
it has such a sequence of open normal subgroups (enumerate open normal subgroups as (Hn)
and define Gn =⋂m≤n Hm) and we deduce the equivalence with (1).
866 Timothée L. Labinet
(5) clearly implies (6) and the converse is easy since each finite Galois extension contains only
finitely many subextensions and each finite separable extension is contained in a finite Galois
one.
The set of Galois k-subfields of k is in natural bijection with the set of k-isomorphism classes
of finite Galois extensions, whence the equivalence between (6) and (4). □
4. Automorphism groups
If k is a perfect field with an algebraic closure k ⊂ k, by k-group we mean a Gal(k/k)-group. If
H is an algebraic k-group, then H(k) is naturally a k-group in this sense. Say that a k-group A
is countable-by-algebraic if it has a normal k-subgroup N such that H/N is countable and such
that N is isomorphic as k-group to the group of k-points associated to some algebraic k-group.
We need the following classical result.
Theorem 12 (Matsusaka, Grothendieck, Matsumura–Oort). Let k be a field. For every complete
k-variety X , the group of k-automorphisms of X is a countable-by-algebraic k-group.
This is mostly due to Matsusaka [10, §11], who essentially proved it assuming X smooth
projective. The general case was formulated by Matsumura and Oort [9, Thm. 3.7], relying on
Grothendieck’s work in [7]. Note that “complete” means geometrically complete, i.e., without
reference to the field of definition.
Deduction of Corollary 2. This is basic Galois cohomology (see [13, III.§1]). For a k-variety with
no k-form, there is nothing to prove. Hence, let X be a complete k-variety. Then mapping
a k-form to its cohomology class induces an injective map from the set of k-isomorphism
types of k-forms of X into H1(k,Autk (X )). Since the k-group Autk (X ) is countable-by-algebraic
(Theorem 12), Theorem 1 applies. □
5. Proof of the theorem
The following is well-known and holds regardless of the characteristic.
Lemma 13. Let G be the group of k-points of a torus or abelian variety over an algebraically closed
field k. Then G is divisible with countable torsion subgroup.
Proof. The case of tori is straightforward (as G is then isomorphic to (k∗)dimG ). For abelian
varieties, this follows from [12, §II.6, Appl. 2], which says that G is divisible with finite n-torsion
for every n. The countability of the torsion subgroup immediately follows. □
We also use the following result about dévissage of algebraic group themselves.
Theorem 14 (Chevalley, see [3]). Let k be a perfect field. Every smooth connected algebraic k-
group G has a smallest normal algebraic subgroup L such that G/L is complete. Moreover, G/L is
an abelian variety, L is connected affine and smooth, and this formation commutes with taking
field extensions.
Proof of Theorem 5. We assume that k is a perfect tame field and G is a countable-by-algebraic
k-group, and we have to prove that H1(k,G) is countable. Let us start with three special cases:
• Suppose that G is countable. Then Lemma 7 does the job.
• Suppose that G is a torus or an abelian variety. Then Lemma 10 yields the conclusion (we
can apply it thanks to Lemma 13).
• Suppose that G is unipotent. Then H1(k,G) = {0} (see [15] for references and for coun-
terexamples when k is not perfect).
Timothée L. Labinet 867
By the dévissage Corollary 9, we can suppose that the k-group G is either countable (this case
is settled), or connected algebraic. So we now assume that G is a connected algebraic k-group.
Using Theorem 14 and again using dévissage, we can suppose that the connected algebraic
k-group is either an abelian variety (this case has been settled), or affine. So we now assume that
G is a connected affine algebraic k-group.
Let H be a Cartan subgroup of G that is defined over k. Then H1(k, H) → H1(k,G) is surjec-
tive [1, Cor. 2.14]. This reduces to the case when G0 is affine solvable, and hence, by dévissage
again, to the case when G is affine, solvable and connected. In turn, the latter reduces, once more
by dévissage, to when G is either unipotent or a torus, and these cases have been settled. □
Remark 15. If k = R one easily checks that H1(k,G) is finite for every abelian k-variety G (and
hence for every algebraic k-group G). For k = Qp this is not true: H1(k,G) is infinite for every
positive-dimensional abelian k-variety G . (Indeed, by [14], H1(k,G) is then isomorphic to the
Pontryagin dual of the compact p-adic Lie group Ĝ(Qp ), where Ĝ is the dual abelian variety; in
addition Ĝ(Qp ) has dimension equal to dim(Ĝ) = dim(G) [11] and in particular is infinite, so the
Pontryagin dual is infinite.)
On the proof of Theorem 5. The main difference is Lemma 7, in which case we get the bound α
instead of ℵ0. Then we need the obvious analogue of Corollary 9, which is not stated in [13] but
can be derived in the same fashion. Note that the countability result in Lemma 13, which is used
at several steps in the dévissage, does not require any tameness assumption on k. □
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08044, 2021.
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Projective varieties have countably many real forms

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