THE ANALYTIC AND FORMAL NORMAL FORM FOR THE
NILPOTENT SINGULARITY
·
· A̧DEK
EWA STRÓ ZYNA
AND HENRYK ZO×
Abstract. We study orbital normal forms for analytic planar vector …elds
with nilpotent singularity. We show that the Takens normal form is analytic.
In the case of generalized cusp we present the complete formal orbital normal
form; it contains functional moduli. We interprete the coe¢ cients of these
moduli in terms of the hidden holonomy group.
1. Introduction
1.1. Around the Takens prenormal form. Our subjects of investigation are
germs at 0 2 C2 of analytic vector …elds of the form
(1.1)
x_ = 2y + : : : ; y_ = : : : ;
i.e. with nilpotent linear part. (We will also write x_ = y + : : :.)
We will pay particular attention to the case when this system is close to the
Hamiltonian system with the Hamilton function
H = y2
(1.2)
xs ; s
3:
It means that
x_ = 2y + : : : ; y_ = sxs
(1.3)
1
+ :::
where the dots mean the terms of ”higher order”. Later we shall specify this notion.
A special case of the system (1.3) forms the Bogdanov–Takens singularity
x_ = y; y_ = x2 +axy+: : :, where a 6= 0 (see [B1]); here we skip the latter assumption.
In the paper of R. Moussu [M2] such singularity (i.e. the system (1.3) with s = 3)
is called dégénérée transverse. In the work of P. Elizarov, Yu. Il’yashenko, A.
Shcherbakov and S. Voronin [EISV] this class is denoted by J .
There was done some work on the formal orbital normal forms for the systems
with nilpotent linear part. F. Takens [T] in 1974 proved that the system (1.1) can
be formally reduced to
(1.4)
x_ = y + a(x); y_ = b(x)
r
where a(x) = ar x + ar+1 x
+ : : : ; b(x) = bs 1 xs 1 + bs xs + : : : are some formal
power series. Some authors use the following, equivalent to (1.4), prenormal form
(1.5)
r+1
x_ = y; y = b(x) + yc(x):
Date : February 14, 2001.
1991 Mathematics Subject Classi…cation. Primary 34A20, 58F36; Secondary 34C20.
Key words and phrases. Singular point of vector …eld, normal form, monodromy group.
Supported by Polish KBN Grant No 2 P03A 022 08 and No 2 P03A 041 15.
1
2
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
Since the forms (1.4) and (1.5) are not the complete normal forms, we shall call
them the Takens prenormal forms; (similarly, the systems (1.5) and (1.6) below are
called the Bogdanov–Takens prenormal forms).
In order to obtain the form (1.4) Takens uses only changes of variables x; y but
not the time. A. Baider and J. Sanders [BS] have continued investigations in this
direction and obtained some new results about non-orbital normal form. They still
are not complete, the …nal (non-orbital) formal normal form is not known.
We have two possibilities: either s = 1, (i.e. b(x) 0) or s < 1. If s = 1, then
the system (1) has non-isolated critical point and its phase portrait is the same as
the ‡ow-box foliation. Later we assume that s < 1.
It is easy to reduce the form (1.4) to
(1.6)
x_ = y + a(x); y_ = sxs
1
(see Lemma 3 in Section 3 below and [B2]). In the case s < 2r (i.e. the system
(1.3)) an equivalent to (1.6) normal form is
(1.7)
x_ = 2y + 2xc(x); y = sxs
1
+ syc(x)
(see [L3] and Lemma 3 in Section 3 below). The latter form has the property that
the “cusp” curve y 2 = xs is invariant.
In the present paper we shall show that the prenormal forms (1.4) (1.5) (1.6)
and (1.7) can be chosen analytic (see Theorem 6 in the next section).
The form (1.6) is not the …nal orbital normal form. Some (in…nitely many)
coe¢ cients in the Taylor expansion of c(x) can be reduced to zero. However, this
reduction cannot be realized by means of analytic change of variables and time (see
Remark 18 in Section 5.1). The analyticity property reaches its limit in the formula
(1.6).
We shall reduce all the terms from c(x) which can be reduced. Therefore we
shall obtain the complete formal orbital normal form in the case ar bs 1 6= 0 and
(1.8)
s < 2r;
i.e. for the system (1.3) (see Theorem 7 in the next section). We will call this case
the generalized cusp case.
(The case with s > 2r will be called the generalized saddle–node case and
the case with s = 2r we call the generalized saddle case.)
There appeared the work of R. Bogdanov [B2] with formal orbital normal forms
for a large class of singularities including the nilpotent ones. His method uses some
homological machinery (spectral sequences etc.) and probably was not understood
by other specialists. Moreover, our results contradict one of his theorems.
Similar to [B2] results, with much simpler proofs, were obtained by A. Sadovski
[S2]. However, in the case (1.8), he repeats the theorem of R. Bogdanov (with the
same mistake and without proof).
Recently F. Loray [L3] has obtained the same formal orbital normal form as the
our form from Theorem 7. Our proof is di¤erent (more algebraic) than the Loray’s;
so we present it here.
1.2. Resolution of singularity and hidden holonomy. During the last 20 years
a lot of work have been done in the theory of analytic orbital normal forms (Ecalle–
Voronin moduli, Martinet–Ramis moduli, Stokes operators etc.). In particular, this
theory was applied to the nilpotent cusp singularity, …rstly by R. Moussu and D.
NORM AL FORM FOR NILPOTENT SINGULARITY
3
Cerveau [CM], [M2], next by the group of Yu. Il’yashenko [EISV] and by F. Loray
and R. Meziani [L1], [L2], [LM].
The geometrical picture is following. It is well known that a singular point of
a vector …eld V from the class J is resolved in three elementary blowing-ups. In
the general case the resolution of the vector …eld (1.1) satisfying (1.8) depends on
the parity of the exponent s. This resolution is the same as the resolution of the
Hamilton function y 2 xs (see Figure 1).
If s = 2k + 1, then we need k + 2 elementary blowing–ups. The last blowing–up
gives the divisor Ek+2 with the three singular points of the induced foliation near
it (see Figure 1):
p0 with 1 : (4k + 2) resonance and with a separatrix
representing an
invariant analytic curve y 2 x2k+1 + : : : = 0,
p1 = Ek+2 \ Ek with 1 : 2 resonance and
p2 = Ek+2 \ Ek+1 with k : (2k + 1) resonance.
If s = 2k, then we need k blowing–ups. The last blowing–up gives the divisor
Ek with the three singular points:
p0 = Ek \ Ek 1 with (k 1) : k resonance and
p1;2 with 1 : 2k resonance and with separatrices 1;2 representing two invariant analytic curves y xk + : : : = 0.
There is a hidden holonomy group G associated with the germ of vector …eld
V . It is the monodromy group (or holonomy group) G associated with the
punctured divisor E = E n fp0 ; p1 ; p2 g, where E is the last divisor appearing in the
resolution. It is a subgroup of the group of germs of holomorphic di¤eomorphisms
of a holomorphic disc D transverse to E and is de…ned by lifts to the leaves of the
corresponding foliation of loops in E .
This holonomy is called (by R. Moussu) hidden, in order to distinguish it from the
holonomy maps associated with the separatrix (or separatrices). It may occur that
two germs of vector …elds have the same holonomies associated with separatrices,
but are not analytically equivalent. This holds in the case when G is solvable and
s is odd.
The group G is also called the projective monodromy group.
In [CM] and [M2] series of results about the group G were proved. It is generated
by two maps f1 and f2 corresponding to the simple loops around p1 and p2 .
In the case s = 2k + 1 we have
f1 (z) = ei z + : : :
f2 (z) = e2 ik=(2k+1) z + : : :
f0 (z) = f1 f2 (z) = e i =(2k+1) z + : : :
(1.9)
and there are two relations
[2]
(1.10)
[2k+1]
f1 = f2
= id
[j]
where f = f f : : : f (j times).
In the case s = 2k we have
f1;2 (z) = ei =k z + : : :
(1.11)
f0 (z) = f1 f2 (z) = e2
and there is one relation
(1.12)
[k]
f0 = id:
i=k
z + :::
4
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
The relations (1.10) and (1.12) are consequences of the analytic linearizability of the
corresponding singular points; that linearizability is obtained from contractibilities
of some loops in certain divisors at Figure 1: E1 and Ek+1 .
In what follows we shall mean by a monodromy group, a subgroup of the group
of germs of analytic di¤eomorphisms of (C; 0) with a distinguished system of its
generators. Therefore we have G = hf1 ; f2 i, where f1;2 satisfy (1.9)–(1.10) or
(1.11)–(1.12).
The next theorem, which says that the pair (f1 ; f2 ) of germs of di¤eomorphisms
(modulo equivalence) constitutes a complete invariant of orbital classi…cation of
generalized cusp singularities, was proved by R. Moussu and D. Cerveau [M2],
[CM] (see also [EISV]). Before its formulation we give two de…nitions.
De…nition 1. Two germs V; V 0 of analytic vector …elds in (C2 ; 0) are formally
(respectively analytically) orbitally equivalent i¤ there is a formal (respectively analytic) di¤eomorphism G of (C2 ; 0) transforming the phase curves of V to
the phase curves of V 0 . This means that there is a formal (respectively analytic)
function ; (0) 6= 0 such that
V 0 = G 1 V G.
De…nition 2. Two groups G = hf1 ; : : : ; fl i ; G0 = hf10 ; : : : ; fl0 i of germs of
conformal di¤eomorphisms of (C; 0) are analytically equivalent i¤ there is a
germ h of analytic di¤eomorphism of (C; 0) conjugating the corresponding maps
from G and G0 . Thus
fi0 = h fi h 1 :
G and G0 are formally equivalent if the above holds at the level of formal
power series.
Theorem 1. (a) The monodromy group, modulo analytic equivalence, associated
with the generalized cusp singularity constitutes an invariant of the orbital analytic
classi…cation for such germs of vector …elds.
(b) Two germs with generalized cusp singularity having analytically equivalent
monodromy groups are orbitally analytically equivalent.
(c) Each groups G=hf1 ; f2 i satisfying (1.9)–(1.10) or (1.11)–(1.12) is realized
as a hidden monodromy group of a generalized cusp singularity.
(d) The same statements hold when we replace the orbital analytical equivalence
of vector …elds and the analytic equivalence of monodromy groups by the corresponding formal equivalences.
1.3. The theory of groups of germs of 1–dimensional di¤eomorphisms and
its application to generalized cusps. Here we present results about classi…cation of …nitely generated groups germs of analytic di¤eomorphism of a complex line.
They are standard for specialists. We present them in full generality, (although we
need them in a restricted form), in order to show their panorama for unexperienced
readers. For details we refer the reader to the survey article by P. Elizarov, Yu.
Il’yashenko, A. Shcherbakov and S. Voronin [EISV]).
Theorem 2. (Formal classi…cation of groups). ([CM]) (a) Any …nitely
generated abelian group G of germs of conformal di¤ eomorphisms of (C; 0) is
formally equivalent either to a group consisting of linear maps
z ! z;
2C ;
NORM AL FORM FOR NILPOTENT SINGULARITY
5
or to a group consisting of maps of the for m
t
gw
t
z ! gw
;
p
= 1; t 2 C
where
is the ‡ow map generated by the vector …eld w = wp;
=
z p+1 =(1 + z p ) @z . Here the …eld wp; is …xed for the whole group and at least
one t 6= 0.
(b) Any …nitely generated solvable nonabelian group G of germs of conformal
di¤ eomorphisms of (C; 0) is formally equivalent to a group consisting of maps of
the form
z ! gzt p+1 (z); 2 C ; t 2 C
where gzt p+1 (z) = z(1 ptz p ) 1=p is the ‡ow map generated by the vector …eld
wp;0 = z p+1 @z . Here the integer p is …xed for the whole group and at least one
t 6= 0.
From this Dit follows
E that any abelian group de…ned by (1.9)–(1.10) is formally
1
b
b
equivalent to f1 ; f2 = h z; zi ; = e2 ik=(2k+1) or to z; gw
; w = wp; ; p =
(2k+1)m. Any such solvable nonabelian group is formally equivalent to z; gz1p+1 .
Analogous groups de…ned by (1.11)–(1.12) are formally equivalent to:
1
; p = 2km (abelian); or
e2 i=k z; e i=k z (abelian); or e2 i=k z; e i=k gw
e2 i=k z; e i=k gz1p+1 (solvable nonabelian).
We associate with a solvable group G two other groups: the additive group
t
b
TG = ft : gw
2 Gg;
w = wp; or w = wp;0
and the multiplicative group (of multipliers if G)
G
= ff 0 (0) : f 2 Gg:
De…nition 3. (a) An abelian group G of conformal di¤eomorphisms of (C; 0) is
called formally linear i¤ it is formally equivalent to a group consisting of linear
transformations.
If G is abelian but not formally linear, then we call it either: exceptional, if
the additive group TG is cyclic, or typical otherwise.
If G is formally linear and G is …nite, then we say that the group is …nite; (it
is …nite in fact).
(b) A solvable nonabelian group of conformal di¤eomorphisms of (C; 0) is called
exceptional i¤ the additive group TG is cyclic. A non-exceptional solvable group
is called typical.
Remark 1. If G is formally linear, then it can be …nite or in…nite. If G is …nite,
then G consists of roots of unity and G is cyclic and analytically equivalent to G .
t
b is cyclic
If G is abelian exceptional, then the additive group TG = ft : gw
2 Gg
and the multiplicative group G consists of roots of unity of order p, pG = f1g.
If G is solvable nonabelian exceptional, then G is a subgroup of the group of
roots of unity of degree 2p, pG = f 1g.
In the solvable case with p = 1 the maps z ! gzt 2 = z=(1 tz) are conjugated
with the a¢ ne automorphisms ! a + b of the complex line. In the exceptional
case the group TG is identi…ed with a subgroup b0 Z C, which must be preserved
by the group of multipliers G = (group of a’s). Thus, either G = f1g and G is
abelian or G = f 1g and G is nonabelian.
6
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
If p > 1, then one has only semi-conjugation with a subgroup of the group of
a¢ ne maps.
Theorem 3. (Rigidity of groups). ([R], [CM]) If two groups G and G0 are
either:
(i) …nite, or
(ii) abelian typical, or
(iii) solvable typical, or
(iv) non-solvable
and are formally equivalent, then they are also analytically equivalent.
Remark 2. In [CM] and [EISV] an analytic classi…cation of exceptional abelian
and solvable exceptional groups is given (Theorems 1.3 and 2.4 in [EISV]). In contrary to the typical solvable case they contain functional modulus. It is the Ecalle–
Voronin modulus of one of the generators of the group. In the cases of typical
(abelian and solvable) groups the Ecalle–Voronin modulus turns out trivial; this is
the reason for their rigidity.
We do not discuss here the problem of rigidity and analytic classi…cation of
groups which are formally linear, but in…nite. This problem is related to the problem of small divisors (theorems of A. Briuno and J.-C. Yoccoz).
Application of the above results to the monodromy group of a generalized cusp
singularity of vector …elds has lead the authors to the following result.
Theorem 4. (Rigidity of generalized cusps). ([CM], [EISV], [L1], [LM])
(a) Let s = 2k + 1. Any group de…ned by (1.9)–(1.10) is either …nite or solvable
typical or non-solvable (i.e. it is rigid).
(b) Let s = 2k. Any group de…ned by (1.11)–(1.12) is either …nite or abelian
exceptional or solvable typical or non-solvable.
(c) This implies the following (rigidity property):
if two germs of analytic vector …elds with generalized cusp singularity with
non-exceptional monodromy groups are formally equivalent, then they are analytically equivalent; (they are rigid).
In particular, any generalized cusp with s = 2k + 1 is rigid.
From Theorem 4 it follows that, when the group is non-solvable or solvable
typical or …nite and the formal orbital normal form of V is polynomial, then the
conjugation transformation is always analytic.
Remark 3. F. Loray and R. Meziani [L1], [LM] obtained analytical classi…cation
of germs of vector …elds with solvable groups in the case of odd s. The corresponding
orbital normal forms are polynomial and are similar to our formal normal form given
in Theorem 7 below.
Remark 4. Some authors study also the topological rigidity of the groups
de…ned by (1.9)–(1.10) (or (1.11)–(1.12)) and of generalized cusp singularities. We
do not need them here and we omit formulation of the corresponding theorems.
We complete this point by presenting a criterion of solvability of the monodromy
in the cases of generalized cusps of small multiplicity. We thank the referee for
detecting a mistake in our previous formulation of this result (we claimed that an
analogous criterion holds for any s).
NORM AL FORM FOR NILPOTENT SINGULARITY
7
Theorem 4. (Solvability criterion). (a) If s = 3, then a group de…ned by
[6]
(1.9) and (1.10) is solvable i¤ f0 = id. For vector …eld this means the following:
its monodromy group is solvable i¤ the saddle p0 of the resolved …eld is linearizable.
[4]
(b) If s = 4 and G is de…ned by (1.11) and (1.12), then f1;2 = id i¤ either G is
solvable nonabelian or G is …nite. In terms of the vector …eld, this property of its
monodromy means that the saddles p1;2 of the resolved vector …eld are linearizable.
1.4. About the results of this paper. At …rst approach we hoped to obtain a
polynomial formal normal forms and then we planned to use the above analytic
machinery to obtain results about analytical classi…cation. However it turned out
conversely; the normal form is not polynomial.
In the present paper we give the complete formal orbital classi…cation in the case
(1.8) (Theorem 7). The proof of this formal orbital form relies on solution of the
homological equations with respect to the monomial components of the changes of
coordinates and time. The problem is reduced to the solution of series of linear
algebraic equations.
The analyticity of the conjugating di¤eomorphism is obtained only for the preliminary normal form, i.e. the Bogdanov–Takens form (1.6) (Theorem 6). This is
the main result of the paper. It is also the most technical part of the work.
We show analyticity directly by estimating the coe¢ cients of the Taylor series
of the changes of variables. The key point of the proof is that the essential part
of the linear operator acting on vector …elds in the homological equation can be
decomposed into a sum of a diagonal operator and a quasi–nilpotent operator. Next
we use a variant of the Newton’s method and techniques from the KAM theory.
Knowing the formal normal form we are able to recognize the type of its projective holonomy (Theorem 8). In the cases of …nite and solvable nonabelian monodromy group the formal normal forms are polynomial, what means that these
forms are also analytic orbital normal forms.
In the case s = 3, we interpret the …rst coe¢ cient of the Taylor expansion of
the formal functional modulus in terms of the saddle quantities of the saddle p0 of
the resolved …eld (Theorem 9(a)). In the even s case we interpret the …rst terms of
the formal normal form in terms of the saddle quantities of the saddles p1;2 of the
blown-up …eld (Theorem 9(b)).
2. Results
The …rst of our theorems is the main result of the work. It solves the old problem
of analyticity of the prenormal form obtained by F. Takens.
Theorem 6. The Bogdanov–Takens prenormal form x = y + a(x); y = sxs 1
of the singularity x = y + : : : ; y = sxs 1 : : : (with the only restriction 2 s < 1)
is analytic.
Remark 5. Let s = 2. Then the singular point is a saddle with the 1 : 1
resonance. Theorem 6 says that the Bogdanov–Takens prenormal form (y@x +
x@y ) + c(x)(x@x + y@y ) holds and is analytic. The …rst nonzero coe¢ cient in the
Taylor expansion of the function c(x) is the saddle quantity (the analogue of the
focus quantity).
8
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
The further results concern only the generalized cusp singularities, i.e. the vector
…elds in the form (1.7). Let
XH = 2y@x + sxs
1
@y
be the Hamiltonian vector …eld and
EH = 2x@x + sy@y
be the quasi-homogenous Euler vector …eld. We shall deal with the vector …elds
XH + c(x)EH , where c(x) is are germs of analytic function vanishing at x = 0. We
can assume that c(x) = xr 1 + : : : (we …x the …rst coe¢ cient by rescaling) or that
c(x) 0 (i.e. r = 1).
We introduce also the notation
r 1
(2.1)
n0 =
:
s 2
Theorem 7. Any analytic germ XH + c(x)EH is formally orbitally equivalent
to one of the normal forms Jr;s , indexed by exponents
and by formal power series
r 6= 0 (mod s) or r = 1
= (x) =
X
0
cj xj
(where the summation runs over some subset of integers), de…ned below:
s
(i) J1;0
: XH ;
(ii)
(2.2)
Jr;s : XH + xr
1
(1 + (x))EH
NORM AL FORM FOR NILPOTENT SINGULARITY
with
if r < 1 and n0 62 Z;
(iii) the …eld (2.2) with
X
0
9
X
=
j6=0; r (m o d s)
= cn0 s xn0 s
if r < 1 and n0 2 Z;
(iv) the …eld (2.2) with
X
0
=
X
j2fn0 s;j0 g
+
X
j > j0
j 6= 0; r (mod s)
j 6= j0 + n0 s
if r < 1, n0 2 Z and there exists a nonzero coe¢ cient cj0 with j0 6= 0; r (mod s)
(we mean that cj0 6= 0 is the …rst such coe¢ cient).
0
If two vector …elds with the normal forms Jr;s and Jrs0 ; 0 are formally orbitally
equivalent, then r = r0 , s = s0 and 0 (x)
( x) for some constant
which
satis…es 2r s = 1 (when r 6= 1).
Remark 6. The normal forms from Theorem 7 need explanations. We explain
them on the examples with s = 3 and s = 4.
If s = 3, then we have two series of normal forms (with one functional modulus)
XH + x3m (1 + x (x3 ))EH ;
XH + x3m+1 (1 + x2 (x3 ))EH ;
and one exceptional (of in…nite codimension) singularity XH .
If s = 4, then r can take the values 1; 2; 3 (mod 4). If r = 1 (mod 4) or
r = 3 (mod 4), then n0 62 Z and the normal form has two functional moduli. This
gives two series of normal forms (the case (ii))
XH + x4m (1 + x 1 (x4 ) + x2 2 (x4 ))EH ;
XH + x4m+2 (1 + x2 1 (x4 ) + x3 2 (x4 ))EH :
If r = 2 (mod 4), then n0 2 Z, r = 4n0 + 2 and the term c4n0 x4n0 cannot be
eliminated from (x). We have two possibilities: either the whole (x) equals to
c4n0 x4n0 (the case (iii)) or it contains some nonzero term cj_ 0 xj0 ; j0 6= 0; r (mod 4)
(the case (iv)). We assume that j0 is the minimal index with this property.
In the case (iii) we get one series (without functional moduli)
XH + x4n0 +1 (1 + x4n0 )EH :
In the case (iv) we have two series, (indexed by r = 4n0 + 2 and by j0 = 4m + 1,
or by j0 = 4m + 3 and containing two functional moduli):
XH + x4n0 +1 [1 + c4n0 x4n0 + x4m+1
XH + x4n0 +1 [1 + c4n0 x4n0 + x4m+3
(n )
4
1 (x )
4
1 (x )
+ x4m+3
+ x4m+5
4
2 (x )]EH ;
4
2 (x )]EH ;
where 1 (0) 6= 0 = 1 0 (0). This means that the monomial x4n0 +j0 is absent in
(x).
There remains the exceptional form (of in…nite codimension) XH .
Note also that the change (x; y) ! ( 2 x; s y) induces the changes XH !
s 2
XH and xr 1 EH ! 2r 2 xr 1 EH . Therefore when 2r s = 1, this change
10
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
does not in‡uence the orbital type of the r–th jet of the …eld Jr;s . This means that
the normal forms are unique modulo action of the …nite group Z=(2r s)Z. We
thank the referee for insisting on this point.
The reason for the elimination of xn0 s+j0 is following. All eliminations (proved
in Theorem 7) are made using linear homological equation with respect to change
of variables. It means that, in each step of the reduction, we solve a system of
linear algebraic equations. If n0 is not integer, then the corresponding system has
always unique solution.
If n0 is integer, then, in certain step of the reduction, the determinant of the
corresponding linear matrix vanishes. There is a term which cannot be eliminated
(it is xn0 s in (x)) and there remains one term in the transformation, which is not
used in this step. We use the latter term to eliminate the additional term xj0 +n0 s
in (x).
Remark 7. The normal form from Theorem 7 can be written in the following
equivalent way
Jr;s : [y + xr (1 + (x))]@x + xs 1 @y :
In the papers [B2] and [S2] R. Bogdanov and A. Sadovski present their normal form
as
X
x_ = y + xr (1 +
gj xj ); y_ = xs 1
where gj = 0 for j = sm r; j = sm; j > 3 (without restrictions onto the …rst
term xr @x ). In our form the case with r = sm is absent and in the case of integer
n0 the normal form is more complicated.
Remark 8. In the case (ii) of Theorem 7 an equivalent form is XH + xr 1 (1 +
Ps 1
i
i=1;i6=s r x i (H))EH .
In the case (iii) of Theorem 7 the more natural normal form, equivalent to
Jr;s ; = cn0 s xn0 s , is
(2.3)
XH + xs=2
1
H n0 (1 + H n0 )
1
EH :
Analogously one can rewrite the normal form in the case (iv). Such kind of
normal form was obtained by F. Loray in [L3].
Remark 9. The result of Theorems 7 can be extended to the non-nilpotent case
s = 2, i.e. 1 : 1 resonant saddle. It is known that the formal orbital normal form
of such saddle can be written as XH + (H m + H 2m )EH . It coincides with the case
(iii) of Theorem 7 with r = 2m + 1; n0 = m.
Because the complete analytic normal form of resonant vector …eld contains functional moduli and is di¤erent from the formal normal form, we get an explanation
why the form from Theorem 7 is not analytic.
In a previous version of this work we obtained a formal normal form with the
set of exponents only like in the case (ii). Consideration of the case s = 2 allowed
to reveal the mistake. We also strived to show that the formal normal form is
analytic, what cannot be true in the case s = 2. A close examination of the proofs
has shown that the proof of analyticity of the Bogdanov -Takens normal form (1.6)
was correct but the reduction of additional terms from c(x)EH can be divergent.
In order to obtain a complete analytic normal form one must construct a sort
of Martinet–Ramis moduli: divide a neighborhood of (0; 0) into conic-like domains
NORM AL FORM FOR NILPOTENT SINGULARITY
11
and perform analytic reductions in the cones to some standard systems, (e.g. with
solvable monodromy group). The di¤erences between the reducing transformations
in intersections of the cones could give functional moduli. Probably these moduli
will not be ‡at (contrary to the Martinet–Ramis moduli) and the coe¢ cients of their
Taylor expansion should correspond to the coe¢ cients cj in the formal functional
modulus .
Something like this is already done with the hidden holonomy group (see Theorem 1). But we think that it is not end of the story.
In the next theorem we classify the holonomy groups accordingly to the formal
classi…cation from Theorem 7.
Theorem 8. The hidden holonomy group associated with the germ XH +c(x)EH
is:
–non-solvable in the following cases from Theorem 7: (ii) with 6 0 and (iv);
– typical solvable (nonabelian) in the case (ii) with
0;
– abelian in the cases: (i) (then it is …nite) and (iii) (then it is exceptional).
In the cases of solvable and …nite monodromy group the formal orbital normal
form is the same as the analytic orbital normal form.
P0
This theorem says that the coe¢ cients in the formal function =
cj xj from
Theorem 7 are the obstacles to solvability of the monodromy group. We see that in
a generic case the holonomy group is non-solvable, the solvability is the phenomenon
of in…nite codimension (very rare). More precisely, the set of germs of type (1.7)
with solvable monodromy forms a pro-algebraic subset (i.e. is given by algebraic
equations in …nite jets) of in…nite codimension.
We do not know the full interpretation of the coe¢ cients cj in terms of known
objects. We have only partial results in this direction.
Theorem 9. (a) If s = 3, then the …rst coe¢ cient of the Taylor expansion of
the function plays the role of the saddle quantity of the saddle p0 ; it is obstacle
to its linearization.
(b) If n0 2 Z, then the term xr 1 EH is the …rst obstacle to the linearization of
the resolved vector …eld near any of the saddles p1 ; p2 2 Ek . In the case (iii) the
coe¢ cient cn0 s plays the role of the modulus of formal orbital classi…cation of the
resolved vector …eld near p1 as well as near p2 .
Remark 10. E. Stróz·yna [St] has recently obtained analogous results in the
generalized saddle–node, (i.e. when the inequality (1.8) is reversed).
After resolution of the singularity, (x; y) ! (x; u = y=xr ), one obtains a 1 : r
resonant saddle p0 : u = x = 0 and a saddle–node p1 : x = 0; u = 1 with nonzero
eigenvalue in the direction of x = 0. Two germs of such vector …elds are orbitally
analytically (formally) equivalent i¤ their monodromy groups are analytically (formally) equivalent.
A natural analogue of the Bogdanov-Takens prenormal form is (y xr )@x +
yc(x)@y with analytic c(x), c(x) = xs r 1 + : : :. Denote EH = x@x + ry@y and
n0 = s=r 2.
The …nal orbital formal normal form is equal either:
(i) Jr1;0 : (y xr )@x , or
(ii)
Jrs; : (y xr )@x + xs r 1 (1 + (x))EH
12
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
P0
P0 P
where (x) =
cj xj and
= j6=0 (m o d r) , if n0 62 Z, or
s;
n0 r
(iii) Jr with P= cnP
x , if n0 2
0r
PZ, or
0
(iv) Jrs; with
= j2fn0 r;j0 g + j6=0 (m o d r);j6=t0 +n0 r , if n0 2 Z and there
exists a (…rst) nonzero coe¢ cient cj0 with j0 6= 0 (mod r).
The monodromy group associated with the singularity is: non-solvable in the
case (ii) with 6 0 and (iv); solvable nonabelian in the case (ii) with = 0 (it can
be typical or exceptional) and abelian in the cases: (i) (it is …nite) and (iii) (it is
exceptional).
The term xs r 1 EH responds for the …rst term in the normal form near the
saddle–node p1 . In the case (iii) the coe¢ cient cn0 r is a formal invariant of p1 .
Remark 11. Theorem 6 has applications in the problem of nilpotent centers.
Recall that the (real) system in the Bogdanov-Takens prenormal form
x_ = y + a(x); y_ = xs
1
;
a(x) = ar xr + : : : ; = 1, has center i¤: (a) = 1, (b) s = 2k is even, (c)
r > k or r = k and rar < 4 and (d) a(x)
a( x) . This result was obtained
independently by A. Sadovski [S1] and by R. Moussu [M1].
M. Berthier and R. Moussu [BM] have proven the following results about such
centers:
1) It is analytically reversible. It means that there is an analytic vector …eld
V 0 in (R2 ; 0) and a local holomorphic map
: (R2 ; 0) ! (R2 ; 0) of the fold type,
2
1 0
(x; y) ! (x + : : : ; y + : : :), such that V = F
V
for some holomorphic
function F .
2) Two vector …elds V1 ; V2 with nilpotent centers are analytically orbitally equivalent i¤ the corresponding vector …elds V10 ; V20 are analytically orbitally equivalent.
The …rst result follows from Theorem 6, because the function a(x) is analytic
and the existence of center means that a(x) = A(x2 ). Thus we can apply the map
(x; y) = (x2 ; y). The proof of 1) in [BM] is di¤erent. It relies on a subtle analysis
of the monodromy maps.
A. Sadovski in [S3] investigated the problem of existence of analytic …rst integral
for nilpotent centers. He has shown that there is such integral i¤ the system is
s
. Using Theorem 7 and Theorem 8 we can
formally equivalent to the case J1;0
assert the following.
Corollary 1. The system x = y + : : : ; y = : : : with center has local analytic …rst
integral i¤ it is analytically orbitally equivalent to the Hamiltonian system
x_ = y; y_ =
x2k
1
:
A. Sadovski informed the authors that the problem of analytic …rst integral for
nilpotent centers was stated by A. Lyapunov.
The plan of the further parts of the paper is following. In the next section we
recall the proofs of the Takens and Bogdanov–Takens prenormal forms. In Section
4 we prove Theorem 6. In Section 5 we prove Theorem 7. Section 6 contains the
proof of Theorem 8 and Theorem 9 is proved in Section 7.
NORM AL FORM FOR NILPOTENT SINGULARITY
13
3. The Takens homological equation
Here we prove the formulas (1.4)–(1.7) at the level of formal power series. In
Section 4 we shall prove that they are analytic. We also present an expansion
formula for dependence of a transformed vector …eld on the change of variables;
this formula will be next used in Section 4.
Lemma 1. ([T]). There is a formal change of the variables (x; y) which reduces
the vector …eld y@x + : : : to the Takens prenormal form
(3.1)
[y + a(x)]@x + b(x)@y :
Proof. We use a composition of in…nite series of transformations of the form
id + Z (k) , i.e.
(x; y) ! (x1 ; y1 ) = (x; y) + Z (k) (x; y);
x1 = x + z1 (x; y); y1 = y + z2 (x; y), where z1 (x; y) and z2 (x; y) are homogeneous
polynomials of degree k. We shall treat also Z (k) as a vector …eld, Z (k) = z1 @x +
z2 @y .
P
Let V = y@x + j>1 V (j) (X), where V (j) are homogeneous summands of the
vector …eld V . Comparing the terms of degree k in the transformed vector …eld
(id + Z (k) ) V = (id + Z (k) )
1
V
(id + Z (k) )
we obtain the homological equation
(3.2)
[Z (k) ; y@x ] + V (k) = (V (k) )T akens
where (V (k) )T akens is a term from the Takens prenormal form, (V (k) )T akens =
ak xk @x + bk xk @y .
The operator [ ; y@x ], written in the components, acts on z1 @x + z2 @y as follows
(3.3)
z1 @x + z2 @y !
z2
@z1
y @x
@x
@z2
y@y :
@x
The equation (3.2) is the Takens homological equation and the operator (3.3) is the
Takens homological operator.
Applying the formula (3.3) to the equation (3.2), we see that we can cancel all
the terms in the y–component of V (k) divisible by y and, when z2 is …xed, we can
also cancel the terms divisible by y in the x–component. In this way we obtain the
formal Takens prenormal form (3.1).
From the proof of Lemma 1 it follows that the components z1 (x; y) and z2 (x; y)
can be chosen as divisible by x, and with this choice they are determined uniquely.
Let us formulate this property in a separate lemma.
Lemma 2. For any homogeneous vector …eld V (k) , k
2, the homological
equation (3.2) has unique solution Z (k) in the class of (homogeneous) vector …elds
of the form
e(k 1) :
Z (k) = xZ
This means that the homological operator (3.3) de…nes an isomorphism between
e and the space
the space (xC[[x; y]])2 of formal vector …elds of the form Z = xZ
2
f
(yC[[x; y]]) of formal vector …elds of the form W = y W (complementary to the
space of Takens prenormal forms).
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
14
Remark 12. The equivalent to (3.1) preliminary normal forms are
x_ = y; y_ = b(x) + yc(x)
and, if the condition s < 2r holds,
(3.4)
x_ = y + xc(x); y_ = b1 (x) + yc(x);
> 0:
Proof. In the …rst case we make the substitution y1 = y + a(x). (Note that here
we do not need any restrictions).
In the second case we make the substitution y1 = y + d(x) with d(x) satisfying
the equation xd 0 = d + a(x). We have
Z x
1
s1= 1 a(s)ds
d(x) =
x 1=
0
We see that, if > 0, then d(x) is well de…ned; as a formal power series if a(x)
is formal, and as an analytic function if a(x) is analytic.
We see also that d(x) =const xr + : : : if a(x) =const xr + : : :.
Lemma 3. ([B2]) (a) Let s < 1. There exists a change of the variables (x; y)
and of time which reduces the vector …eld (3.1) to the Bogdanov–Takens prenormal
form
[y + a(x)]@x + sxs 1 @y :
(b) Let s < 2r < 1, i.e. the generalized cusp case. There exists a change of the
variables (x; y) and of time which reduces the vector …eld (3.1) to the Bogdanov–
Takens prenormal form
(2y@x + sxs
(3.5)
1
@y ) + c(x)(2x@x + sy@y );
which has the property that the “cusp” curve y 2 = xs is invariant.
(c) The changes from the points (a) and (b) are analytic when the …eld (3.1) is
analytic and are formal when (3.1) is formal.
Proof. (a) Assume that b(x) = sxs 1 b1 (x); b1 (0) 6= 0. Our aim is to reduce b1 (x)
to 1.
We apply the change x1 = x (x); dt=dt1 = (x), which should satisfy the
equations dx1 =dt1 = y + a1 (x1 ); dy=dt1 = sx1s 1 . Comparing the terms sxs1 1 @y
and y@x1 we get the conditions
(x) (x (x))0 = 1;
or the Bernoulli equation x
0
+
(x) = x
=
1
(x)b1 (x) = s
1 s
1
.
s b1 (x)
Z
s 1
(x);
It has the solution
1=s
x
s 1
b1 ( )d
:
0
This solution is an analytic function when b1 (x) is analytic.
(b) We apply the change x1 = x (x), y1 = y+d(x1 ), dt=dt1 = (x) = (dx1 =dx) 1
(from the point (a) and Remark 12) to the vector …eld [2y + a(x)]@x + xs 1 b1 (x)@y .
We write a(x) = e
a(x1 ), (x) = e(x1 ).
We obtain the system
dx1
dt1
dy1
dt1
= 2y1 + [e
a(x1 ) 2d(x1 )]
=
(x)[b(x) + d0 (x1 )(e
a(x1 )
2d(x1 ))] + y1 2e(x1 )d0 (x1 ):
Following the proof of the formula (3.4), we choose d(x1 ) such that e
a(x1 )
2d(x1 ) = 2x1 c(x1 ) and y1 2e(x1 )d0 (x1 ) = sy1 c(x1 ).
NORM AL FORM FOR NILPOTENT SINGULARITY
15
By the assumption s < 2r we have d0 (x1 )(e
a(x1 ) 2d(x1 )) = O(x2r 1 ) = o(xs 1 )
0
and we can write b(x) + d (x1 )(e
a(x1 ) 2d(x1 )) as xs 1 [b1 (0) + xb2 (x; (x))]. We
obtain the following di¤erential equation for (x): x 0 + = 1s 1 s [b1 (0)+xb2 (x; )].
Introduce the variable = s . Then the latter equation changes to
x 0 + s = b1 (0) + xeb2 (x; ):
(3.6)
The initial condition (i.e. the limit as x ! 0) is (0) = b1 (0)=s.
The graphic = f(x; (x))g of the solution (x) to the latter problem forms a
phase curve of the analytic vector …eld x = x; = b1 (u) s + xeb2 . We see that
forms a separatrix of the singular point (0; (0)) of this vector …eld. This singular
point is a saddle, what means that is unique analytic invariant curve through
(0; (0)). Therefore the function (x) (as well as (x)) as analytic, if the function
eb2 is analytic, i.e. if the system (3.1) is analytic.
If (3.1) is only formal, then the equation (3.6) admits a solution in the form of
a power series, which is unique.
We …nish this section by deriving the expansion of the changed vector …eld
(id + Z) V = ((I + DZ) 1 V ) (id + Z) with respect to Z.
Lemma 4. We have the following formula
(id + Z) V
(3.7)
= V
+[Z; V ]
+ 21 (D2 V (id + Z)) Z; Z
(DZ DV ) (id + 0 Z) Z
D2 Z (id + 0 Z) V (id + 0 Z); Z
+((DZ)2 (I + DZ) 1 V ) (id + Z):
Here DU and D2 U mean the …rst and second derivatives of a vector …eld U =
(u1 ; u2 ) with respect to x; y. The expressions like U (id + Z), = ( 1 ; 2 ) 2
[0; 1] [0; 1], mean that the both components of the …eld U take the form u1 (x +
as in u).
1 z1 ; y + 1 z2 ), u2 (x + 2 z1 ; y + 2 z2 ) (the same index in
Proof. The formula (3.7) is a consequence of Mean Value Theorem, which states
that:
if f (x), x 2 Ck is an analytic scalar function and m is natural, then f (x+h) =
1
f (x) + Df (x) h+ 12 D2 f (x)h; h + : : : + m!
hDm f (x + h); (h; : : : ; h)i for some
2 [0; 1].
The …rst row in the right–hand side of (3.7) contains the term constant with
respect to Z. The second row
Pcontainsn the linear terms. The last row arises from
1
the expansion
(I
+
DZ)
=
( DZ) and takes into account only the summands
P
with n 2, n 2 ( DZ)n = (DZ)2 (I + DZ) 1 .
Mean Value Theorem is applied to the expressions V (id + Z) = V + DV Z +
1
hDV
(id + Z) Z; Zi and (DZ V ) (id + Z) = DZ V D(DZ V ) (id +
2
0
Z) Z; (in fact, it is applied independently to each component).
There reader should note that in (3.7) the argument is shifted in V as well as in
Z (and in their derivatives).
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
16
4. The analytic Takens prenormal form
4.1. Preliminaries. In this section we prove Theorem 6. In order to show analyticity of the Takens prenormal form, we must prove convergence of the series of
transformations described in Section 3.
There are two general ways of the proof of analyticity: estimation of the coe¢ cients of the …elds Z, or reduction the problem to a …xed point equation for some
contractible operator in some space of analytic vector–valued functions. The …rst
method is used in the linearization theorems in the Siegel domain, when there are
no resonances (see [A]), the second method was used by N. Brushlinskaya [Br] in
the Poincaré domain. Our method is a combination of both.
The main ingredients of the proof are: the Newton’s method of solution of
nonlinear equation and certain modi…cation of the norm of nonlinear terms.
Let us make the problem more precise. Lemma 2 (from Section 3) says that the
f (k 1) ,
formal reduction of terms outside the Takens form, i.e. the terms W (k) = y W
(k)
is obtained by means of compositions of transformations id + Z , where Z (k) =
e(k) are homogeneous of degree k. Moreover the Takens homological operator
xZ
Z ! [Z; y@x ] is an isomorphism between the spaces (xC[[x; y]])2 and (yC[[x; y]])2
of formal vector …elds. Therefore we shall apply formal changes in the form id + Z,
e and try show that the power series for Z
e is convergent.
Z = xZ
Denote
(4.1)
U (x; y) = U (x; y)
U (x; 0)
the projection onto the space of formal vector …elds divisible by y (i.e.
! (yC[[x; y]])2 ) and by
: (C[[x; y]])2
1
y
(4.2)
the operator of division by y.
The condition that the nonlinear terms in the …eld (id+Z) V = y@x +: : : depend
only on x reads as [(id + Z) V y@x ] = 0. We rewrite it in the (homogeneous)
form
1
e
(4.3)
f(I + Z) V y@x g = 0; Z = xZ:
y
e
It is an equation for Z.
The equation (4.3) contains the linear part
1
y
e V ];
[xZ;
a constant part (which has to be reduced) and a nonlinear part (see Lemma 4).
The linear part strongly depends on the vector …eld V . This means that during
iterations we shall deal with varying problems. In order to impose some stability, we
distinguish the part of the latter expression associated with the Takens prenormal
form part of V T akens .
We have
V = V T akens + W;
T akens
V
= [y + a(x)]@x + b(x)@y ;
(4.4)
f = (V y@x ):
W = yW
NORM AL FORM FOR NILPOTENT SINGULARITY
17
De…ne the linear operator
e!
L:Z
(4.5)
1
y
e V T akens ]:
[xZ;
The operator L is invertible in the space (C[[x; y]])2 of formal vector …elds (by
Lemma 2). It will vary during iterations, but it stabilizes with the number of iterations going to in…nity. Moreover, the operator L is simpler than the commutator
operator with the whole V and we will be able to estimate its inverse in a suitable
Banach space.
By the formula (3.7) from Lemma 4, the equation (4.3) can be rewritten in the
form
e = 0:
e+W
f+ 1
[Z; W ] + (nonlinear terms in Z)
(4.6)
LZ
y
We shall solve the equation (4.6) successively, in each step getting better approximation to the normal form. In each step we solve it approximately. We put
e=
Z
(4.7)
L
1f
W:
f is of the same order as Z.
e
Therefore W
The transformed vector …eld V 0 = (id+Z) V again has the form V 0 = V 0T akens +
f 0 . From the formula (3.7) we have
W 0 , W 0 = (V 0 y@x ) = y W
(4.8)
V 0T akens
W0
= V T akens + [Z; V ]T akens + (nonlin: terms)T akens ;
=
[Z; W ] + (nonlin: terms);
where is the projection operator de…ned in (4.1) and (nonlin: terms) denote the
terms nonlinear in Z in (3.7). We see that W 0 is of second order with respect to
W and V 0T akens V T akens is of …rst order with respect to W .
Moreover, the stabilization of the Taylor expansions of V and V T akens takes
place. Namely, if W begins with terms of order m, then W 0 begins with terms of
order m + 1. It follows from Lemma 1.
In the further stages of the proof of Theorem 1 we shall assume that
(4.9)
s = 3;
i.e. b(x) = x2 + b3 x3 + : : :. The general case (s = 2 or s > 3) is proved in the same
way. Our restriction serves to simpli…cation of the exposition.
e ! LZ
e (see (4.5)), written in the basis
4.2. The operator L. The operator Z
i j
i j
(x y @x ; x y @y ) is divided into two parts, diagonal and quasi–nilpotent. The latter
means that for each element of the above basis some power of the operator nulli…es
this element. (In functional analysis by a quasi-nilpotent operator people mean an
operator in a Banach space whose whole spectrum is concentrated at 0).
e = z~1 @x + z~2 @y = (~
Let Z
z1 ; z~2 ). We have
h
i
e= 1
e 0 + x~
e 0 + x~
e0 :
LZ
y(xZ)
z2 @x + x~
z1 a0 @x a(xZ)
z1 b0 @y xbZ
x
x
y
y
The diagonal part is
e=
L0 Z
1
e 0x ] =
[y(Z)
y
e 0x ;
(xZ)
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
18
or
L0 xi y j @x;y =
(4.10)
The operator L
We have
(i + 1)xi y j @x;y :
L0 is divided into three parts
e=
L1 Z
e=
L2 Z
e
L3 Z =
1
y
1
y
1
y
[x~
z2 @x ];
[xa0 z~1 @x
ey0 ]:
[ xbZ
e 0 + z~1 b0 @y ];
a(xZ)
x
L = L0 (I + L0 1 L1 + L0 1 L2 + L0 1 L2 ):
Lemma 5. The expressions of the operators L0 1 Lj in the monomial basis are
the following
(4.11)
L0 1 L1 xi y j @y
L0 1 L1 xi y j @y
L0 1 L2 xi y j @x
L0 1 L2 xi y j @y
L0 1 L3 xi y j @x;y
1 i+1 j 1
x y @x
= i+2
= 0
i
= ii+31 a2 xi+2 y j 1 @x + i+4
a3 xi+3 y j 1 @x + : : :
2
3
i+2 j 1
y @y i+4 b3 xi+3 y j 1 @y + : : :
i+3 b2 x
i+1
i+3 j 1
= i+3 a2 xi+2 y j 1 @y + i+2
y @y + : : :
i+4 a3 x
j
j
i+3 j 2
i+4 j 2
= i+4 b2 x y @x;y + i+5 b3 x y @x;y + : : :
Here we put zero whenever we get a negative power of y in the above formulas and
b2 = 1.
Figure 2 shows how the operators L0 1 Lj act.
4.3. The weighted norm. From Lemma 5 we see that the operators L1;2 are
bounded relatively to L0 . L3 is not relatively bounded. However the fact that it
is quasi–nilpotent allows us to treat it as bounded with respect to L0 . As in the
…nite–dimensional case, an operator of the form semi–simple plus quasi–nilpotent
is invertible. In the …nite–dimensional case it is used to introduce a new norm such
that the nilpotent part is small relatively to the semi–simple part.
The natural norm in the space of vector–valued power series, giving
P the Banach space of series representing analytic vector …elds is jjU jj =
ij (ju1;ij j +
i+j
ju2;ijP
j)
for some > 0 (representing the radius of convergence of the series
U = i;j (u1;ij @x + u2;ij @y ) xi y j ); so jjxi y j @x;y jj = i+j .
We modify this norm in the following way. We put jjxi y j @x;y jjw; = wi;j i+j ,
where wi;j are weights de…ned below.
De…nition 4. De…ne rational numbers w
~i;j ; (i; j) 2 Z2+ as follows:
(4.12)
w
~i;0
w
~i;1
=
=
w
~i;j
=
The weight wi;j is equal to
1;
1;
o
n
~i+3;j
max 1; ji+32 w
j
p
2;
for j
(4.13)
wi;j =
j+1 w
~i;j
where
> 1 is some constant which will be …xed later.
2:
NORM AL FORM FOR NILPOTENT SINGULARITY
19
P
We denote by Xw; the Banach space of series U = i;j (u1;ij @x + u2;ij @y ) xi y j
with the norm
X
jjU jjw; =
(ju1;ij j + ju2;ij j)wij i+j :
ij
Let us comment this de…nition. The choice of the numbers w
~i;j is motivated
1
by the requirement that the operator L0 L3 be bounded. The leading matrix
j
elements of this operator (corresponding to xi y j @x;y ! xi+3 y j 2 @x;y ) equal i+4
. It
w
~
i;j
. However, the ratio ji+32 has the geometrical
is almost the same as ji+32 = w~i+3;j
2
interpretation as the slope coe¢ cient of the radius vector of the endpoint of the
segment S, joining the point (i; j) with (i + 3; j 2) and presented at Figure 2.
Note also that the numbers w
~i;j are chosen such that they are equal 1 whenever
j
i + 5, i.e. below the shifted diagonal; (but above this shifted diagonal they
are strictly increasing as we move along the lines with the slope 2=3 in the left
direction). The reason for this is that the matrix elements of the operators L0 1 L1;2
are of order O(1) and we cannot use the ratios ji+32 , which are smaller than 1.
The factor j allows us to control the norm of the operator L0 1 L1 ; the greater
is, the smaller the norm jjL0 1 L1 jjw; is. The constant will be determined in
Lemma 6 below.
p In fact, we will …x = 2.
The factor j + 1 is responsible for the estimate from Lemma 8(e) below (which
is needed for estimation of norms of products of series); so it is introduced for
technical reasons.
If the integer s from b(x) = sxs 1 + : : : is di¤erent than 3 (i.e. (4.9) does
not hold), then
~i;j is modi…ed as follows:
n theoinductive de…nition of the numbers w
w
~i;j = max 1; ji+s2 w
~i+s;j
2.
Remark 13. The idea of changing the norm was used in [Br]. There the
Lyapunov metric was introduced.
4.4. The weighted norm at work. In this point we demonstrate how useful
our weighted norms are. Namely, we prove that the operator L is invertible in the
Banach space Xw; with bounded inverse and that the elements from Xw; represent
Taylor series of analytic vector …elds.
We will use the notation
X
U m=
(u1;ij @x + u2;ij @y ) xi y j
i+j m
for a vector …eld U =
P
i;j
(u1;ij @x + u2;ij @y ) xi y j .
Lemma 6. (Bound for L 1 ). Assume that V T akens = [y + a(x)]@x + b(x)@y =
y@x + V 2 is an analytic vector …eld such that
X
b 2
(4.14)
jjV 2 jjw; =
(jan j + jbn j) n C
n 2
b does not depend on :
for small ’s, where the constant C
If the constant
(from De…nition 4) is su¢ ciently large and is su¢ ciently
small, then the operator L 1 is bounded in Xw; with the norm 4.
Proof. We use the formula L
1
= (I + L0 1 L1 + L0 1 L2 + L0 1 L3 )
1
L0 1 .
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
20
Because L0 is diagonal with the eigenvalues
1.
(i+1) (see (4.10)) then jjL0 1 jjw; =
We use the following properties of weights:
(i) wi+1;j wi;j ,
1
(ii) wi+1;j 1
wi;j .
The inequality (i) is equivalent to the inequality w
~i+1;j w
~i;j and follows from
the formula (4.12) (see also the point (a) of Lemma 8 below). The inequality (ii)
constitutes the point (b) of Lemma 8 below and will be proved later.
Note that (i) implies wi+k;j wi;j for any k > 0 and the both inequalities imply
l
wi+k;j l
wi;j for k > l 0.
Let e = ei;j = xi y j @x;y . By Lemma 5, the inequality (ii) and De…nition 4 of
weights, we have
jjL0 1 L1 ejjw;
jjejjw;
1
i+2
1
i+2
1
i+2
=
jjei+1;j 1 jj
jjei;j jj
wi+1;j 1
wi;j
1
:
1
We choose the constant such that jjL0 L1 jjw;
1=4, i.e. we put = 2.
Next, by Lemma 5, the inequalities (i) and (ii), De…nition 4 and formula (4.14),
we have
jjL0 1 L2 ejj
w
w
1
1
+ ja3 j 2 i+3;j
+ :::
ja2 j i+2;j
jjejj
wi;j
wi;j
wi+2;j 1
2 wi+3;j 1
+jb2 j
+ :::
wi;j P+ jb3 j
wi;j
1 wi+2;j 1
n
(ja
j
+
jb
j)
n
n
wi;j
1
jjV 2 jjw;
b = :
C
For small we get jjL0 1 L2 jj
Finally, we have
jjL0 1 L3 ejj
jjejj
1=4.
wi+3;j 2
w
j
j
2
+ i+5
jb3 j 2 i+4;j
+
i+4 jb2 j
wi;j
wi;j
P
~i+3;j 2
j
1
n
2 w
jbn j
i+4
w
~o
i;j
n
j
i+3
1
jjV 2 jjw;
2
i+4 min 1; j 2
1
2
:::
3 2 jjV jj
b = 2:
3C
n
o
j
min 1; ji+32
3). We get jjL0 1 L3 jj 1=4 for small .
(we have used i+4
The reader can see that the same estimates hold in the case when s = 3 is
replaced by arbitrary integer 2. The estimates for L0 , L0 1 L1 and L0 1 L2 are the
same. There are only minor di¤erences in the estimate for L0 1 L3 .
Lemma 7. (Analyticity). We have
i+j
jjxi y j @x;y jjw;
(C1 )i+j
for some constant C1 which does not depend on i; j; .
This implies that any element U of Xw; represents an analytic vector …eld in
the ball B with center at 0 and radius and that
(4.15)
sup
(x;y)2B
Moreover:
jU (x; y)j
jjU jjw; :
NORM AL FORM FOR NILPOTENT SINGULARITY
21
(a) If a series U (x; y) is such that jjU k jjw; < const ( = 0 )k ; k = 1; 2; : : :,
then U represents an analytic function in the ball B 0 .
(b) Conversely, if a series U (x; y) represents an analytic function convergent
in an open set containing the ball B 0 ; 0 > C1 , then U 2 Xw; and jjU k jjw;
K k , jjDU k jjw; < K (2 )k , where
(4.16)
= C1 =
0
and K is a constant depending on f .
Proof. This lemma follows directly from the inequality
1
wi;j
C1i+j
which is proved in the point (c) of Lemma 8 from the next subsection.
4.5. Estimates for the weights and the weighted norms. This subsection is
devoted to formulation additional technical lemmas, necessary to obtain recursive
estimates in the proof of Theorem 1. These lemmas are formulated in the same
way for s = 3 and for other s. In fact only in the proof of Lemma 8 there are minor
di¤erences for di¤erent s. The proofs are put o¤ to the last part of this section.
Lemma 8. (Properties of weights). The weights wi;j satisfy the following
properties:
(a) wi+k;j wi;j for k > 0;
k
(b) wi+k;j k
wi;j wi;j for k > 0,
i+j
(c) 1 wi;j C1 ,
(d) wi;j C1 (i + j + 1)k+l wi+k;j+l for k; l 0, 0 k + l 2,
(e) wi+k;j+l C1 wi;j wk;l ,
where C1 is a constant not depending on the indices.
The inequalities (a), (b) and (c)from Lemma 8 have been used in the previous
subsection. The estimates (b), (d) and (e) are successively used in the proofs of the
lemmas 9–11 below.
The further lemmas will deal with functions rather than with vector …elds.
Namely
we introduce the Banach space Fw; of scalar power series f (x; y) =
P
i j
i;j fij x y , fij 2 C with the norm
X
jjf jjw; =
jfij jwi;j i+j :
The estimates will be formulated in terms of elements from Fw; , but they are also
true for vector–valued (and for matrix–valued) series. This approach was suggested
by the referee and its aim is to simplify the exposition.
Lemma 9. (Estimate of derivatives). If f 2 Fw; ,
and k + l 2, then
C3
jjf jjw;
@xk @yl f w; 0
k+l
( 2)
where C3 is a constant not depending on f; k; l; ; .
0
= =1
2 ( 12 ; 1)
The proof, which uses the estimate (d) of Lemma 8, will be given later.
Lemma 10. (Estimate of products). Let f be a series in Fw; .
(a) If g is a function from Fw; , then f g is in Fw; and
jjf gjjw;
C2 jjf jjw;
jjgjjw; :
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
22
(b) The series
1
y
(f ) 2 Fw; (see (4.1) and (4.2)) and
1
y
C2
(f )
w;
jjf jjw; :
Here the constant C2 does not depend on f; g; .
The proof uses the estimate (e) of Lemma 8 and is put o¤ to the last subsection.
Lemma 11. (Estimate of compositions). Let Z 2 Xw; be a vector …eld of
e and let 0 = = 1
e w;
the form xZ
2 ( 21 ; 1) be such that jjZjj
2C2 , where C2 is
the constant from Lemma 8. If a function f 2 Fw; , then f (id + Z) 2 Fw; 0 and
!
e w;
C4 jjZjj
jjf jjw;
jjf (id + Z)jjw; 0
1+
3
where C4 is some constant not depending on f; Z; ; .
3
e w;
In particular, if we assume that jjZjj
=C4 (what will take place in applications), then
(4.17)
jjf
(id + Z)jjw;
0
2jjf jjw; :
The proof uses the point (b) of Lemma 8 and will be given in the end of this
section. We remark only here that the thesis of Lemma 11 ceases to be true when
e The reason is that
we remove the restriction onto the form of Z (i.e. for Z 6= xZ).
when we expand f (id+Z) into series, then we obtain the terms Dm f Z m where the
e then Z m = (xm ) Z
em and
norms jjDm f jj behave badly for large m. When Z = xZ,
the large power of x gives small norm of Z m (see De…nition 4). This compensates
the growth of jjDm f jj.
4.6. Description of the iterative procedure. We begin with application of the
Lemmas 9–11 to the transformed vector …eld (id + Z) V . There we must replace
scalar series by vector–valued or by matrix–valued series; this is not a problem.
Corollary 2. There exists a constant C5 not depending on V; Z; ; such that
for 0 < 0 < << 1 satisfying 0 = = 1
2 ( 12 ; 1), any V 2 Xw; and any
3
e such that jjZjj
e w;
Z = xZ
=C4 (see Lemma 11) the following inequalities hold
jj(id + Z) V
jj[Z; V ]jjw;
V [Z; V ]jjw;
0
0
C5
jjV jjw;
6 jjV jjw;
2
C5
e w; ;
jjZjj
e w; )2 :
(jjZjj
Proof. The commutator [Z; V ] equals DZ V
DV Z. The expression (id +
Z) V
V
[Z; V ] consists of the nonlinear terms from (3.7), i.e.
= (1=2) D2 V (id + Z) Z; Z
(DZ DV ) (id + 0 Z) Z
D2 Z (id + 0 Z) V (id + 0 Z); Z + ((DZ)2 (I + DZ) 1 V ) (id + Z). We shall
use also the notion
(4.18)
(nonlin: terms) = (id + Z) V
V
[Z; V ]:
We introduce the intermediate norm jj jjw; 00 , where 00 = 21 ( + 0 ). Firstly we
estimate the derivatives with non-shifted argument in this new norm using Lemma
2
00
9. Each derivative gives the factor (1
= ) 2 < const
. Next, we apply
NORM AL FORM FOR NILPOTENT SINGULARITY
23
Lemma 11 (the formula 4.17) to estimate the corresponding factors in summands
with shifted argument (like D2 V (id + Z)) using the norm jj jjw; 0 . Finally, we
estimate the products in the norm jj jjw; 0 using Lemma 10(a).
Recall that one summand takes the form [(DZ)2 (I + DZ) 1 V ] (id + Z). Due
e w; = O( 3 ) the 2 2–matrix valued function
to the assumptions << 1 and jjZjj
DZ is small and the term (I + DZ) 1 is bounded.
Since the nonlinear terms contain expressions where the derivative appears at
most three times in each summand, we get the factor 1= 6 . The linear terms acquire
only the factor 1= 2 .
We pass to the description of iterations. Assume that we begin with the vector
…eld V = V1 which is analytic in a ball B 0 . We have the splitting V1 = V1T akens +
f1 . We can assume that W1 has order 4 at the origin.
W1 , W1 = y W
Choose a radius 1 in such a way that the parameter
= C1
1= 0
from Lemma 7 (the formula (4.16)) is small. We obtain the starting estimates (see
Lemma 7)
(4.19)
V1T akens
jjV1 jjw;
y@x w;
jjW1 jjw;
where the constant K does not depend on
Introduce the radii
1
(4.20)
k = 1
k2
1
K ;
K 2;
K 4;
1
1
1
(only on V ).
k 1;
k
2:
We have lim k = 1 =const 1 > 0. 1 will be the radius of the ball B 1 of
analyticity of the Takens prenormal form.
fk and di¤eomorDe…ne inductively the vector …elds Zk , Vk , VkT akens , Wk = y W
phisms Gk :
ek ;
Zk = xZ
fk ;
e
Zk =
L 1W
T akens
Vk+1 = (id + Zk ) Vk = Vk+1
+ Wk+1 ;
Gk = (id + Z1 ) : : : (id + Zk ):
e ! 1 [xZ;
e V T akens ]. Therefore L also depends
Recall that the operator L equals Z
k
y
on k, L = Lk .
The Taylor expansion of Zk starts from terms of degree 3+k. Hence the terms
of degrees k + 3 in Gk are …xed. In this sense Gk tend to some formal map. The
same holds for Vk . We have to show that Gk and Vk converge in the ball B 1 .
4.7. Proof of Theorem 6. This proof relies on the following proposition.
Main Proposition. For any k
(4.21)
jjVkT akens
1 and
VkT akens
jjw;
1
jjWk jjw;
jjZk jjw;
small enough one has
k
k
k
K(2 )k ;
K k+3 ;
K k+2 ;
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
24
so that
(4.22)
jjVkT akens
jjVk jjw;
y@x jjw;
k
4K
1 2
k
6K
8K ;
2
:
where K is the constant from (4.19):
If the estimates (4.21) for Zk hold, then by the bound (4.15) from Lemma 7
jZk (x; y)j K k+2 as (x; y) 2 B k . Moreover Zk has small derivative in a slightly
smaller ball B( k + k+1 )=2 , jDZk (x; y)j jjDZk jjw;( k + k+1 )=2 const k 4 k+2 for
(x; y) 2 B( k + k+1 )=2 .
Therefore the map id + Zk represents a holomorphic di¤eomorphism of the ball
k
B( k + k+1 )=2 to a domain containing the ball B k+1 . Indeed, jG(k) j
is much
2
smaller than the di¤erence of the radii ( k + k+1 )=2
=
O(k
)
and
then
k+1
(id + Zk )(B( k + k+1 )=2 ) B k+1 .
The in…nite composition of the maps id + Zk is convergent in the ball B 1 and
represents an analytic di¤eomorphism G1 = lim Gk : B( k + k+1 )=2 !
G1 (B( k + k+1 )=2 ).
The limit of Vk ’s de…nes a vector …eld V1 holomorphic in the ball B 1 . It is in
the Takens prenormal form and the di¤eomorphism G1 conjugates V with V1 .
This completes the proof of Theorem 6.
Remark 14. The convergence of the successive approximations is very rapid. If
n
the perturbation is of order ", then after n steps we would obtain error of order "2 .
Such convergence is characteristic for the Newton’s method of …nding approximate
zero of a di¤erentiable functions. It was used in KAM theory (see also [A]). In our
proof we generally have followed the Newton’s method but we did not use as good
estimates as in [A] (Section 3, §12). We have used also some ideas from [P].
4.8. Proof of Main Proposition. The inequalities (4.22) are rather obvious.
The estimate for Zk in (4.21) follows from the estimate for Wk . Indeed, we
fk . Using Lemma 10(b) we estimate
fk , Zk = xZ
ek and Z
ek = L 1 W
have Wk = y W
f
(W ) by C2 k 1 jjWk jjw; k . The assumption of Lemma 7, i.e.
W
= y1
w;
k
that the nonlinear part of VkT akens is small enough, holds:
6K
2
6K 2
2
1
o
f
4 W
obtain
For small
w;
k
b
C
2
k,
b =const K= 2 . Therefore L
where C
0
VkT akens
y@x
w;
ek
4 and Z
1
k
w;
k
ek . We
. Next, we use Lemma 10(a) to estimate the product x Z
jjZk jjw;
k
we have 4C22 K
4C22 jjWk jjw;
k
(4C22 K )
k+2
:
K and the third estimate from (4.21) holds.
The …rst two estimates in (4.21) for k = 1 hold by (4.19): we have
V1T akens V0T akens = V1T akens w;
K
K(2 )1 and jjW1 jjw;
1
1
K
1+3
.
T akens
The inductive estimates for Vk+1
VkT akens and Wk+1 follow from the formuT akens
las (4.8) and Corollary 2. Indeed, we have Vk+1 = (id + Zk ) Vk = Vk+1
+ Wk+1 ,
NORM AL FORM FOR NILPOTENT SINGULARITY
25
where by (4.8)
T akens
Vk+1
Wk+1
= VkT akens + [Zk ; Vk ]T akens + (nonlin: terms)T akens ;
=
[Zk ; Wk ] + (nonlin: terms):
Here is the projection operator (4.1) and the notion (nonlin: terms) is de…ned in
(418). We use Corollary 2 with V = Vk , Z = Zk , = k , 0 = k+1 , = (k + 1) 2
and with the inductive assumptions jjVk jjw; k 8K , jjZk jjw; k K k+2 . We get
T akens
jjVk+1
for
VkT akens jjw;
jjWk+1 jjw;
k+1
k+1
C5 (k + 1)4 8K K k+2
C5 (k + 1)12 8K (K k+2 )2
K(2 )k+1 ;
K (k+1)+3
small.
The proof of Main Proposition is complete.
4.9. Proof of the technical lemmas. (A) Proof of Lemma 8. Thorough the
proof we shall use the notion of asymptotic equivalence of functions:
if f (t) and g(t) are positive functions of t
0, then f (t)
g(t) i¤ there
exists a constant C > 0 such that C1 g(t) f (t) Cg(t) and f (t) = O(g(t)) when
f (t) Cg(t) for some constant C.
p
For example, the Stirling formula says that (t + 1) tt e t t + 1 and implies
the formula (t + a + 1) (t + 1)a (t + 1) for a …xed or bounded.
Recall the de…nition of weights:
p
wi;j = j j + 1w
~i;j ;
where
> 1 and
w
~i;0 = w
~i;1 = 1; w
~i;j = max 1;
j 2
i+3
w
~i+3;j
2:
We obtain that
(4.23)
w
~i;j =
N
Y
j 2n
;
i + 3n
n=1
where N is the integer de…ned by N < j 5 i N + 1 if j i > 5 and the product is
void, i.e. w
~i;j = 1, if j i 5. Hence N = j 5 i
, where 0 <
1.
j 2n
The factors i+3n have the interpretation of slopes w(S
~ n ) of the radius vectors
from the origin to the right endpoints of the segments S = Sn joining the points
(i
Q+ 3n 3; j 2n + 2) and (i + 3n; j 2n). The product in (4.23) is the product
w(S
~ n ) along the chain = (S1 ; : : : ; SN ) joining the point (i; j) with the “target”
set = fj i = 5g (See Figure 3).
The point (a) of Lemma 8 states that wi+k;j
wi;j for k > 0. This inequality
is equivalent to the inequality w
~i+k;j
w
~i;j , which easily follows from the above
recurrent formula.
k
wi;j for k > 0. Since
q The point (b) of Lemma 8 states that wi+k;j k
j+1
j k+1
> 1, this inequality follows from the inequality
w
~i+k;j
k
w
~i;j :
The latter inequality becomes evident when we use the slope interpretation of
the quantities w
~i;j . This situation is presented at Figure 3(a). We associate with
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
26
each segments S from the chain starting at (i; j) the segments S 0 = S + (k; k)
from the chain starting at (i + k; j k). We see that w(S 0 ) < w(S).
The point (c) of Lemma 8, i.e. the estimate 1
wij
C1i+j . Of course, it is
e i+j for some constant C.
e This is a consequence
enough to show that 1 w
~i;j C
of the following representation of the quantities w
~i;j .
Lemma 12. Let
=
i
i+j
1
2;
w
~i;j
=1
s
=
j
i+j .
We have
i+1
( )i+j
j+1
where ( ) 1 is uniformly
from above continuous function.
p bounded
i + 1 j ( )i+j .
This means that wi;j
Proof. We use the formula (4.23) for j > i + 5, which can be rewritten as
w
~i;j
j=2 N
2 N j=2 1
3
i=3+1 : : : i=3+N
(i=3+1)
(j=2)
2 N
3
(j=2 N )
(i=3+N +1)
(i=3+1) (j=2+1)
2 (j i)=5
3
((2i+3j)=10+1) ((2i+3j)=15+1)
=
(4.24)
=
We use the Stirling formula.
Putting i = (i + j); j =
q
i+1
( )i+j where
j+1
i+j+1
j+1 :
(i + j), after some
calculation, we get w
~ij
(4.25)
( )
=
=3
=2
(2 +3 )=6
5
2 +3
;
1
2:
0
The function ( ); 2 [0; 12 ] is continuous and bounded: 1 = ( 12 )
( )
q
5
(0) = 3 . It is Lipschitz continuous away from the point = 0; at = 0 the
derivative is equal to 1.
In the case of general nilpotent singularity, not just s = 3, Lemma 12 is also
true, but the function
=s
( ) takes the form
=2
s+2
2 +s
(2 +s )=2s
:
The point (e) of Lemma 8 says that the ratios
wi;j
wi+k;j+l
(4.26)
w
w
~
i;j
i;j
are bounded by const (i + j + 1)k+l for 0 k + l 2. We have wi+k;j+l
w
~i+k;j+l .
When j i
5 and/or j + l i k
5 the numerator and the denominator of
the latter ratio are uniformly bounded and the estimate is true. Otherwise we use
the formulas (4.24) and (m + a + 1) (m + 1)a (m). We obtain that (4.26) is
const
i+j+1
i+1
k=3
k=3
const (i + j + 1)
i+j+1
j+1
l=2
const
i+j+1
i+1
k+l
const (i + j + 1)
.
The point (d) of Lemma 8 states that the ratio
Rijkl =
wi+k;j+l
wi;j wk;l
k=3
(since i
O(j)). This is
NORM AL FORM FOR NILPOTENT SINGULARITY
q
j+l+1
is uniformly bounded. Due to the inequality (j+1)(l+1)
to the problem of estimating the ratio
w
~i+k;j+l
w
~i;j w
~k;l
27
1, the problem reduces
uniformly with respect to the indices.
We distinguish three cases:
(i) j i + 5 and j + l i + k + 5;
(ii) j i + 5 and j + l > i + k + 5;
l
(iii) j > i + 5; l > k + 5; here we assume ji
k.
In the case (i) we have w
~i;j = w
~i+k;j+l = 1; w
~k;l
1 and the inequality from
Lemma 8(e) is obvious.
In the case (ii) we have w
~i;j = 1 and the problem is to estimate the ratio
n 2
e = w~i+k;j+l . Recall that w
~
R
i+k;j+l is the product of slopes w(S) = m+3 over
w
~k;l
segments S = [(m; n); (m + 3; n 2)] from the chain joining the point (i + k; j + l)
with the diagonal.
The situation is presented at Figure 3(b). With each segment S = [(i + m; j +
n); (i + m + 3; j + n 2)] from the chain we associate the unique segment S 0 =
[(m; n); (m+3; n 2)] = S (i; j) from the chain 0 starting at (k; l). Because w(S) <
w(S 0 ) and w(S 00 ) 1 for other segments from the chain 0 , we get w
~i+k;j+l w
~k;l .
The situation from the case (iii) is presented at Figure 3(c). With the segments
S from the chain , which lie above the line y = (j=i)x, we associate the segment
S 0 = S (i; j) from the chain 0 . We have w(S) < w(S 0 ).
This shows that it is enough to prove the estimate frompLemma 8(e) in the case
i
k
i+k
0
00
when = i+j
= k+l
= i+j+k+l
. We get wi;j
i + 1 j ( )i+j , wk;l
p
p
l
k+l
j+l
i+j+k+l
( ) , wi+k;j+l
i+k+1
( )
and hence Rijkl
qk + 1
i+k+1
(i+1)(k+1)
= O(1).
Unfortunately this is not end of the proof of this point. It is correct when the
equalities = 0 = 00 are exact. However, they can be not exact and the function
( ) from Lemma 12 is not Lipschitz continuous at 0. So we have to perform more
estimates.
The di¤erences between ’s are of the form
a
a
0
=
; 00
=
k+l
i+j+k+l
where jaj 3. This can be seen from Figure 3(c): a is the di¤erence between the
x–coordinates of the points: (i + k; j + l) and the point of intersection of the line
y = ji x with the line y + x = i + j + k + l. We can assume that i + j k + l. We
1
have either: i > 0 (and then
= 0, 0 6= k = a and
i+j ), or i = 0 ( and then
0
00
j+l
l ).
By (4.25) we have ( ) = =3 e ( ), where the function e is di¤erentiable and
hence Lipschitz continuous. Thus we have to estimate the two expressions:
"
# 13
00
e ( 00 )i+j+k+l
( 00 ) (i+j+k+l)
and
(4.27)
:
(i+j) ( 0 ) 0 (k+l)
e ( )i+j e ( 0 )k+l
The logarithm of the …rst equals
(4.28)
(i + j + k + l) [
00
1
3
log
times
00
log ]
(k + l)[
0
log
0
log ]:
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
28
For
00
00
> 0 we represent
log
00
00
+(
Similarly we have
) log
0
00
log
i+j
k+l
(4.28) is bounded by O
as
a
(1 + log ) + O
i+j+k+l
=
0
log
log
log
=
a
k+l (1
1
(k+l)2
+ log ) + O
= O(1), as i + j
00
1
(i + j + k + l)2
:
. Therefore
k + l.
j+l
l
If = 0, then (4.28) equals a log 0
log
O(1), since j k + l = O(l).
e
Using the function ( ) = log ( ) we represent the second ratio in (4.27) as
the di¤erence of two terms (i + j + k + l)[ ( 00 )
( )] and (k + l) [ ( 0 )
( )].
Both are bounded by La, where L is the Lipschitz constant of .
(Another proof of the bound for Rijkl uses the asymptotic formula wij = O(1)
j
( 3i + 1) j 1=2 i=3 ( (0))j for i << j. Under the assumption ki
l we get
3=2
1=2
Rijkl O(1) (i + 1)
j
.)
Remark 15. The reader can notice here that the choice
Q of de…nition of weights
is not very rigid. For example, instead of the product w(S) in (4.23) one could
use the formula (4.24), or its modi…cation with N = j 5 i . Nevertheless, the requirements of …niteness of the norm L0 1 L3 says that they should take form of
j
product of terms
i . Another restriction is imposed by the just proven property
(e) of Lemma 8.
P
(B) Proof of Lemma 9. Let f =
fij xi y j . We have to estimate the derivak l
tives @x @y f for k + l 2.
We have
P k l
jvij j wi k;j l ( 0 )i+j k l
k@xk @yl f kw; 0
i;j i j
jjV jjw;
By the point (e) of Lemma 8 wi
ik j l
wi
k;j
wij
l
(1
)i+j
( 0)
k l
k;j l =wi;j
k;j
l
wij
( 0 = )i+j :
<const (i + j + 1)k+l . Thus
const (i + j + 1)2(k+l) (1
const
N
wi
maxij ik j l
t
2(k+l)
)i+j+1
:
N
(We have used the inequality t (1
)
(N=e ) for small ).
Now the estimate of the norm of the derivative by const jjV jj=(
2 k+l
)
follows.
P
P
of Lemma
10. (a) Let f =
fijP
xi y j , g =
gij xi y j , f g =
P (C) i Proof
P
hij x y j , jjf jjw; =
jfij jwij i+j < 1, jjgjjw; =
jgij jwij i+j < 1. Then we
have
XX
wkl
jjf gjjw;
(jfij jwij i+j ) (jgk i;l j jwk i;l j k i+l j ):
w
w
ij
k
i;l
j
i;j
k;l
By Lemma 8(d) wkl C1 wij wk i;l j for 0 i k; 0 j l. From this the
…rst estimate from Lemma 10 follows.
(b) The second estimate is proved in the same way, using the inequality wi;j 1 <
wi;j . The latter is a consequence of the inequalities wi;j 1 wi+1;j 1 and wi+1;j 1
1
wi;j < wi;j given in Lemma 8 (the points (a) and (b)).
P
(D) Proof of Lemma 11. Let f =
fij xi y j . We estimate f (id + Z) f
e
for Z = xZ = (x~
z1 ; x~
z2 ).
NORM AL FORM FOR NILPOTENT SINGULARITY
Expanding this expression we get
P
P
jjf (id + Z) f jjw; 0
jfij j k+l
i;j
P
P
i;j jfij j
k+l
i
1 k
i
1 k
j
l
j
l
jj(x~
z1 )k (x~
z2 )l xi k y j l jjw;
jj(xi+l y j l )~
z1k z~2l jj 0 :
29
0
Because f 2 Fw; , we have jfij j
jjf jjw; =(wij i+j ). By de…nition of the
i+l j l
weighted norm, jjx y jjw; 0 wi+l;j l ( 0 )i+j , where wi+l;j l < wi;j by Lemma
8(b). Using the estimate for products from Lemma 10(a), we get
k+l
P
P
e
jjf (id + Z) f jjw; 0
jjf jj i;j ( 0 = )i+j k+l 1 ki jl C2 jjZjj
i+j
P
e
jjf jj i;j (1
)i+j 1 + C2 jjZjj
1
P
e
e i+j 1
const jjf jj jjZjj
)(1 + C2 jjZjj]
i;j (i + j)[(1
P
i+j
1
e
const jjf jj jjZjj
(i + j)[1
=2]
Pi;j
e
const jjf jj jjZjj n (n + 1)n[1
=2]n 1
3
e (2= )
const jjf jj jjZjj
C4
e
jjf jj jjZjj:
3
Here we have used the inequalities (1 + t)n 1
e
1
=2 (following from the assumption jjZjj
3
2(1 t) .
e
nt(1 + t)n 1 , (1 )(1 + C2 jjZjj)
P
n 1
=(2C2 )) and n>0 (n + 1)nt
=
5. The formal normal form
5.1. Reduction of additional terms using the Hamiltonian part and solvable monodromy. In Section 3 we obtained the Bogdanov–Takens prenormal
form [2y + a(x)]@x + sxs 1 @y ; (where, according to Theorem 6, the function a(x)
is analytic). In this section we assume the following restrictions.
We have a(x) = ar xr + : : : ; ar 6= 0; b(x) = sxs 1 with the condition
s < 2r:
Lemma 13. The system
x_ = 2y + asm xsm + O(xsm+1 ); y_ = sxs
1
can be transformed to
x_ = 2y + O(xsm+1 ); y_ = sxs
1
:
Proof. The sm–th order jet of this system can be transformed, via the change
(x; y) ! (h = y 2 xs ; y), to
h_ =
asm (y 2
h)m ; y_ = 1:
This is a nonsingular system with an analytic …rst integral H = h+: : : = y 2 xs +: : :.
The function H(x; y) has the As 1 type singularity and is equivalent (by an analytic
change of coordinates) to Y 2 X s (see [AVG]).
Lemma 14. If r 6= 0 (mod s), then the terms of order r of the system
x_ = 2y + 2ar xr ; y_ = sxs
1
+ sar xr
1
y
cannot be cancelled by means of any orbital transformation preserving the form
XH + c(x)EH .
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
30
Proof. We can assume that ar = 1 and we can consider only the r–th jet of
the vector …eld. We shall …nd a rational change of variables reducing it to some
integrable vector …eld. This change is the same as the composition of the blowing–
up maps described in the introduction and the transformed system describes the
resolved …eld of directions in some a¢ ne chart near the …nal divisor E of the
resolution (see Introduction). The cases with odd and even s are treated separately.
(a) The case s = 2k + 1. Here the change is following
(5.1)
u=
y
x2k+1
; v= k
2
y
x
(then x = uv 2 ; y = uk v 2k+1 ). We obtain the system
(5.2)
u_ = 2(2k + 1)u(1
u); v_ = [(2k + 1)u
2k]v + ur
k 2(r k)
v
:
The line v = 0 represents the a¢ ne line E n p1 (punctured divisor). The other
singular points are p2 = (0; 0); p0 = (1; 0) and the separatrix is given by u = 1;
(1; 0) represents the point p1 .
The system (5.2) can be integrated. The equation for the phase curves is the
Bernoulli equation with the …rst integral of the Darboux–Schwartz–Christo¤el type
Z u
2k 2r+1
+r k 1
(1
) 1d
(5.3)
F =v
u (1 u)
0
where
=
k(2k 2r+1)
(2k+1) ;
=
(2k 2r+1)
(4k+2) .
Remark 16. Loray in his thesis calls the function (5.3) a hypergeometric function. We think that our notion is more suitable. The hypergeometric functions are
given in certain integral form with the argument playing role of the parameter of
the subintegral function. Their monodromies are usually non-solvable. The argument in the integral in (5.3) stays in the limit and the function (5.3) can serve as
a model example of a function with solvable monodromy.
By the assumption of Lemma 14, r 6= 0 (mod s) and, as we shall see, the integral
(5.3) is not of the Darboux type. This means that its monodromy group is solvable
and non-abelian.
(By the monodromy group of a multivalued holomorphic function F on Cn ,
with rami…cations along some algebraic hypersurface S, we mean certain group of
1
permutations of a typical …ber Ma =
(a) of the covering M n S ! Cn n S, where
M is the Riemann surface of F with the natural projection . These permutations
are induced by analytic continuation of analytic elements (above a neighborhood of
a) along loops in Cn n S which start at a. In the case of solvable monodromy this
group coincides with the group of deck automorphisms of the covering M n S !
Cn n S.)
The rami…cation curves of F are u = 0; u = 1 and u = 1. The monodromy
maps associated to loops around u = 0 and u = 1 are
h!
0 h;
h!
1h
+c
R1
+r k 1
where 0 = e2 i ; 1 = e2 i ; c = (1
(1
) 1 d . The
1) 0
latter integral diverges at = 1 (because < 0). Therefore its value is given by a
NORM AL FORM FOR NILPOTENT SINGULARITY
31
suitable regularization.
R 1 Here we buse1 the analytic continuation of the Euler Beta–
function B(a; b) = 0 a 1 (1
) d = (a) (b)= (a + b) as a function of (a; b).
If r 6= 0 (mod s), then c 6= 0.
Of course, the monodromy group of the …rst integral is associated with the monodromy group of the divisor E n fp0 ; p1 ; p2 g; one needs to parametrize the disc D
transversal to E (at a point q) by F . Here the transition from the natural parametrization of the disc D by v = vjD to the parametrization h = F jD = Av 2k 2r+1 +B
is not one-to-one; it is semi-conjugation of the monodromy group with a subgroup of
the group of a¢ ne maps (see Remark 1 in Section1). Nevertheless the fundamental
properties of the groups (like solvability) are preserved under such semi-conjugation.
The map f1 (i.e. holonomy around p1 ) corresponds to a loop around u =
[2]
[2k+1]
1; f1 =id, the map f2 corresponds to a loop around u = 0; f2
=id and
f0 = f1 f2 corresponds to a loop around u = 1. Applying the above semiconjugation we get that fj (v) = j v(1 Cj v p ) 1=p ; p = 2r 2k 1.
[4k+2]
We have f0
=id, because the map f0 in the chart h is a rotation around its
…xed point. Thus by Theorem 5(a) (Section 1) the monodromy group is solvable.
On the other hand it is non-abelian.
If there were a transformation reducing the r–th order terms, then the monodromy group of this jet should be abelian. This shows the thesis of our lemma.
(b) The case s = 2k. The change is of the form
y
u = k; v = x
x
and we obtain the system
(5.4)
u2 ); v_ = uv + v r
u_ = k(1
k+1
:
The line v = 0 represents the a¢ ne line E n f1g, p0 = (1; 0); p1;2 = ( 1; 0)
and the separatrices 1;2 are u = 1. The corresponding …rst integral equals
Z
r k u
2
F = [v(1 u2 )1=2k ]k r +
(1
) (k+r)=2k d
k
As in the previous case, we check that, for r 6= 0 (mod k), the monodromy group
of F is solvable non-abelian. (This case corresponds to the situation with n0 62 Z
in Theorem 7). Indeed, we get the situation exactly as in the case of odd s: F has
rami…cations at u = 1; u = 1 and the monodromy maps are of the form
h!e
i(k r)=k
h + c1;2
where c1;2 6= 0.
If r = (2m + 1)k, then
F =v
2mk
(1
2
u )
m
+ 2m
Z
u
(1
2
)
m 1
d :
contains logarithmic singularities d1;2 ln(1 u). The monodromy maps of F are
h ! h + c1;2 and generate an abelian group. The initial parametrization of the
transversal D to the divisor E is given by v with the relation h = F jD = Av 2mk +B
and, after applying this semi-conjugation, we obtain again abelian group:
v!e
i=k
v(1
C1;2 v p )
1=p
; p = 2mk
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
32
We see that this group consists of maps gvt p+1 ; p = 1, like the exceptional
abelian group from Theorem 2(a) and De…nition 3(a) from Introduction (with the
vector …eld w = wp;0 ). Theorem 2(a) says that this group cannot be formally
reduced to a …nite abelian group. This is the case with n0 2 Z from Theorem 7;
we have m = n0 .
Lemma 14 is complete.
s
Corollary 3. The monodromy group of the system (5.2) or (5.4) (i.e. Jr;0
) is
solvable nonabelian for n0 62 Z and exceptional abelian for n0 2 Z. The generators
take the forms
f0 (v) =
t0
0 gv p+1 ;
f2 (v) =
2 v;
p = 2(r
k)
1
for s = 2k + 1 and
f1;2 (v) =
t1;2
1;2 gv p+1 ;
p=r
k
for s = 2k.
In particular, the monodromy group associated with any vector …eld Jr;s cannot
be formally linear.
Moreover, if n0 2 Z, then the monodromy group of the initial germ x = 2y +
: : : ; y = sxs 1 : : : cannot be solvable nonabelian.
Remark 17. Consider the cases, which are excluded from the assumption of
Lemma 14.
Let s = 2k + 1. If r is of the form (2k + 1)m, then the exponent in the integral
(5.3)
is integer, = k(2k 2r + 1)=(2k + 1) = k(1 m); = 1=2 m. The integral
R u (k+1)m
1
(1
) 1 d in (5.3) can be expressed as (1 u)
(polynomial).
Because half-integer, the square of the …rst integral F is a rational …rst integral
for the system (5.2).
Let s = 2k. If r = 2mk, then the Schwartz–Christo¤el integral in the expression
for F takes the form
Z u
2
(1
) m 1=2 d :
We claim that it is also of the form (1 u2 ) m+1=2 polynomial.
R1
2
Indeed, calculation of the regularized integral P 1 (1
) m 1=2 (with the
R1
1
2
=
substitution = 1
and using the Beta–function) gives:
0 m+1=2 (1 )1=2
(1=2) (1=2 m)= ( m) = 0. Thus the regularized ”values” of the Schwarz–
Christo¤el integral at the branching points are equal. Putting this value equal zero
we extract from the Schwarz–Christo¤el integral the factor (1 u2 )1=2 m . The
other factor must be a polynomial P (u).
In fact we can …nd this polynomial explicitly as P (u) = u(a0 + a2 u2 + : : : +
am 1 u2(m 1) ), where the coe¢
the recurrent relations
aj =
R u a0 =2 1; 5=2
R u cients 2satisfy
m j
) 3=2 = u(1 u2 ) 1=2 ;
(1
)
=
aj 1 . For example,
(1
2 2j+1
Ru
3
2
3=2
2
1=2
1
(u 2u =3)(1 u )
. However
(1
)
= sin (u).
This gives another explanation why the monodromy is abelian in the case r =
0 (mod s).
Notice that repeating the proof of Lemma 13 we can reduce all the terms
asm xsm @x . To be precise, we should take into account not the homogeneous
NORM AL FORM FOR NILPOTENT SINGULARITY
33
…ltration of the space of germs generated by the degree, but rather the quasihomogeneous …ltration given by the exponents
d(x) = 2; d(y) = s; d(@x ) =
2; d(@y ) =
s:
Then the part of the vector …eld of the lowest quasi-homogeneous degree (equal to
s 2) is XH = 2y@x + sxs 1 @y . Because we can reduce xsm @x using XH , the same
holds when we replace XH by XH +(higher order terms).
We see that the terms xsm @x are the only terms which can be reduced when one
uses the …rst part of the vector …eld, XH .
Conclusion: Using the Hamiltonian part, we can reduce the system either to
x_ = 2y; y_ = sxs
s
(the case J1;0
), or to
x_ = 2y + xr +
X
1
aj xj ; y_ = sxs
j6=sm
1
; r 6= 0 (mod s):
In the second case the r–th order part of the system, i.e. x_ = 2y + xr ; y_ =
sx , does not determine the coordinate system uniquely. Namely, the changes
(x; y; t) ! ( 2 x; s y; s 2 t) give the system x_ = 2y + 2r s xr ; y_ = sxs 1 . So the
linear coordinates are …xed modulo action of the group Z=(2r s)Z. This group
acts on the remaining terms aj xj , also after the further reductions of some of them.
Further we assume the second possibility.
s 1
Remark 18. Here is place to explain why the method of the proof of analyticity
of the Bogdanov–Takens prenormal form cannot be adapted in some way to get
analyticity of the whole formal normal form.
Consider the reduction of the terms xms @x by means of LXH , the commutator
with XH . One can introduce the spaces: of orbital changes (id + Z; ) (where
f + (xs )@x . It is possible to
is a function) and of cancelled vector …elds y W
introduce certain weighted norm in these spaces such that the operator LX1H would
be bounded and the series bounded in that norm would represent analytic functions.
Unfortunately, the operators Lxj @x are neither bounded relatively to LXH nor
are quasi-nilpotent in this setting.
5.2. The …nal formal reduction. (A) The general scheme of the reduction.
We use a formal orbital change
b
(X; dt) ! (exp Z(X);
(1 + ^ )dt)
b = zb1 @x + zb2 @y is a formal vector …eld, exp Z
b = g 1 is the phase
where X = (x; y); Z
b
Z
‡ow di¤eomorphism and ^ (X) is a formal nonzero function. Applying it to a vector
…eld V we obtain the …eld
b ^ ) = Ad b
PV (Z;
exp Z
(1 + ^ )V:
b ^ ) ! PV (Z;
b ^ ) is clearly non-injective. For example, if
The non-linear map (Z;
b
b preserves the phase portrait
Z = ^ V (^ –a formal function), then the map exp Z
of V and the …eld PV (^ V; 0) is parallel to V , equals to ^V . In order to avoid this
ambiguity, one uses the notion of bi-vector …eld introduced by Bogdanov in [B2].
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
34
b = zb1 @x + zb2 @y and V = P @x + Q@y , then we de…ne the bi-vector …eld
If Z
b ^ V = b dx ^ dy
Z
(5.5)
b = Qb
where b = V (Z)
z1 P zb2 . One can say that b measures the component of
b transversal to V . If V (Z)
b = 0 and V has isolated singularity, then Z
b = ^ V for
Z
some series ^ .
This shows that the problem of formal reduction of V is equivalent to the problem
of formal reduction of terms transversal to V by means of b . One has the (nonlinear) map from C[[x; y]] to C[[x; y]]
(5.6)
b ! PV (Z;
b 0) ^ V = dx ^ dy = LV b + : : :
b satis…es
where LV denotes the linear part (the bi-vector homological operator ) and Z
b
b
b
b
the identity V (Z) = . LV is linear in V (for …xed ).
Next, one solves the corresponding (non-linear) equation for b . Using the quasihomogeneous …ltration one reduces the non-linear problem to the linear one, i.e. to
description of the action of LV .
We begin with the operator LXH . One describes a linear space NXH (of preliminary bi-vector normal forms) which is complementary to Im LXH . This analysis
repeats in a sense the calculations from the previous subsections.
It turns out that LXH has non-trivial kernel (in…nite dimensional). This indicates that NXH may be not a good candidate for the space of bi-vector normal
forms. Some b ’s from Ker LXH may give contribution to the image of the nonlinear operator (5.6) This contribution can be linear (from higher terms of V ) or
nonlinear.
As the next approximation of the homological operator we use LXH +xr @=@x . We
de…ne a linear space NXH +xr @=@x , complementary to Im LXH +xr @=@x . It turns out
that in one case, when the number n0 from Theorem 7 is not integer, Ker LXH +xr @=@x
= 0. The standard arguments, involving the quasi-homogeneous …ltration, show
that NXH +xr @=@x is good space for bi-vector normal forms. It is ”transversal” to
the image of the operator (5.6).
If the number n0 is integer, then Ker LXH +xr @=@x is 1–dimensional. In that case
one uses the operator LXH +xr @=@x+at xr+t @=@x for some index t.
(B) The formula for LV and the bi-vector homological equation in
b 0); V = P @x + Q@y ,
the Bogdanov–Takens case. The linear part of PV (Z;
b
Z = zb1 @x + zb2 @y is
(zb_ 1
Px0 zb1
Py0 zb2 )@x + (zb_ 2
Q0x zb1
Q0y zb2 )@y
where the dot denotes the derivative in the direction of V . From this, (5.5) and
(5.6) one easily obtains that
(5.7)
_
LV b = b
divV
b:
If W = w1 @x + w2 @y is the part of the vector …eld V devoted to killing, then we
have the bi-vector homological equation
(5.8)
LV b + h = 0;
h = W ^ V = dx ^ dy:
NORM AL FORM FOR NILPOTENT SINGULARITY
35
In the Bogdanov-Takens case we have
V = [2y + a(x)]@x + sxs 1 @y ;
b = sxs 1 zb1 [2y + a(x)]b
z2 ;
h = sxs 1 w1 [2y + a(x)]w2 ;
and the bi-vector homological equation takes the form
b_ a0 (x) b + h = 0:
(5.9)
What we are looking for is a bi-vector normal form space NV
C[[x; y]] com_
0
b
b
b
plementary to the image of the linear map
!
a (x) . Because we expect
P
the normal form for V as (2y + xr (1 + j2J aj xj ))@x + sxs 1 @y , the space NV is
P
identi…ed with the space fxr+s 1 j2J aj xj g of series. Here the summation runs
over restricted set J of indices. Our problem is to describe the set J. We formulate
this statement in a separate lemma.
Lemma 15. Assume that V = [2y + a(x)]@x + sxs 1 @y ; a(x) = xr + : : : has the
property ker LV = 0.
P
Then the space NV can be chosen in the form fxr+s 1 j2J aj xj g. In this case
P
the formal orbital normal form for V is (2y + xr (1 + j2J aj xj ))@x + sxs 1 @y .
(C) The image and the kernel of LXH . We have
_
(5.10)
LXH b = b = 2y b 0x + sxs 1 b 0y
and the equations (5.8) is replaced by
(5.11)
2y b 0x + sxs
1 b0
y
+ h = 0:
Because LXH acts quasi-homogeneously, we investigate its restrictions to the
…nite dimensional spaces of b = with …xed quasi-homogeneous degree.
Let this degree be 2(ms + i); i = 0; 1; : : : s 1, i.e.
= ! 0 xsm+i + ! 1 xs(m
1)+i 2
y + : : : + ! m xi y 2m :
The function h in (5.11) is of degree 2(ms + i) + s
ms+i 1
h = h0 x
if i 6= 0 and
h = h0 xs(m
2, i.e.
i 1 2m+1
y + : : : + hm x
1)+s 1
y + : : : + hm
if i = 0.
The matrix of LXH expressed in these bases
2
ms + i
s
0
6
0
ms
s
+
i
2s
6
6
0
0
ms s + i
6
6 :::
:
:
:
:::
(5.12)
6
6
0
0
0
6
4
0
0
0
0
0
0
y
s 1 2m 1
y
1x
is equal to 2 times
:::
0
:::
0
:::
0
:::
:::
: : : 2s + i
:::
0
:::
0
0
0
0
:::
ms s
s+i
0
0
0
0
:::
0
ms
i
3
7
7
7
7
7
7
7
7
5
where there is no last row in the above matrix for i = 0.
We see that if i 6= 0, then it is invertible matrix. If i = 0, then we get m rows
and m+1 columns; so, it has 1-dimensional kernel. This kernel equals to (xs y 2 )C
(because then _ = 0).
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
36
If the quasi-homogeneous degree of
h
h
is 2(ms + i) + s with
= ! 0 xms+i y + : : : + ! m xi y 2m+1
= h0 x(m+1)s+i 1 + : : : ;
= h0 xms+s 1 + : : : ;
then the corresponding matrix is
2
s
0
0
6 2(ms + i)
3s
0
6
6
0
2(ms
s
+
i)
5s
6
6
:::
:::
:::
(5.13)
6
6
0
0
0
6
4
0
0
0
0
0
0
i 6= 0
i=0
:::
0
:::
0
:::
0
:::
:::
: : : (2m 1)s
:::
2(s + i)
:::
0
0
0
0
:::
0
(2m + 1)s
2i
3
7
7
7
7
7
7
7
7
5
where again the last row is absent for i = 0.
We see that if i 6= 0, then we have (m + 1) m matrix and the image of the
corresponding operator is m-dimensional (and of codimension one). We project it
to the space generated by the monomials x y +1 .
If i = 0, then the matrix (5.13) (without the last row) is invertible.
Remark 19. Sadovski in [S2] considered the case r = 2s and had written down
an equation analogous to (5.11).
We have proven the following lemma, whose …rst part repeats the result of the
previous subsection.
Lemma 16. (i) The equation (5.11) has solution
for any h of the form
P
s 1~
s
~
~
y h(x; y) + x h(x ). This means that the space NXH = fxs 1 j6=o (m od s) aj xj g
is complementary to Im LXH .
(ii) ker LXH = C[[y 2 xs ]]. It means that for any two solutions 1 ; 2 of (5.11)
the di¤ erence 1
2 is a function of H.
(D) Reduction by means of LXH +xr @=@x : As in the previous point the problem reduces itself to solution of the equation
e
e
( e + e ) _ divV ( e + e ) + h = 0
e
where e ; e ; h are quasi-homogeneous, @ e =@XH 0; V = XH + xr @x .
e
Thus
= e + e = ! 1 (xs y 2 )n + ! 0 xsm+i y + : : : + ! m xi y 2m+1 and the
quasi-homogeneous part of LXH +xr @=@x equals
(5.14)
[xr e 0x
rxr
The quasi-homogeneity implies
sn + r
and h = h0 xs(m+1)+i
(5.15)
1
1e
e
] + [2y e 0x + sxs
1 = s(m + 1) + i
1e
e 0 ]:
y
1
+ : : : The corresponding matrix takes the form
A=
A B
0 D
NORM AL FORM FOR NILPOTENT SINGULARITY
where A and D are equal respectively
2
(ns r) n0
s
6 (ns s r) n 2(ms + i)
1
6
6 (ns 2s r) n
0
2((m
2
6
6
:
:
:
:
:
:
6
4
:::
:::
r nn
0
2
6
6
6
6
4
2((m
n)s + i)
0
:::
0
0
0
3s
1)s + i)
:::
:::
0
(2n + 3)s
2((m n 1)s + i)
:::
0
0
:::
0
:::
0
:::
0
:::
:::
: : : 2(r + s)
:::
0
0
(2n + 5)s
:::
0
0
:::
0
:::
0
:::
:::
: : : 2(s + i)
:::
0
37
0
0
0
:::
(2n 1)s
2r
3
7
7
7
7
7
7
5
0
0
:::
(2m + 1)s
2i
Lemma 17. The determinant of the matrix A equals to
(5.16)
( 2s)m n+1 ( si )1m n+1
s( 2s)n ( r 1)( rs 32 )n1
where (a)01 = 1; (a)k1 = a(a
1) : : : (a
1
r
s
[n
3
7
7
7
7
5
1
2]
k + 1).
Proof. The …rst factor in the above formula is det D. The second factor, i.e.
det A, is calculated using the expansion with respect to the …rst column. (Like
in the calculation of determinant of the characteristic polynomial for higher order
di¤erential linear equation). We have used the formulas
n
X
n
(a)k1 (b)n1
k
k
= (a + b)n1 ;
k=0
n
X
k
k=0
following from the identity
Pn
n
k
(a)k1 (b)n1
k=0 k
k
j
=1
=
=
=
n
(a)k1 (b)n1
k
k
= na(a + b
1)n1
1
@
@ n a b
j = = =1
@ n+ @
@
a
(
+
)
(
)b j =1; =0
@
n
(a + b)1
and its derivative with respect to , equal to n
applied them with a = (r + 1)=s; b = 1=2.
@
@
+
@
@
n 1 @
@
a b
. We have
(E) The …nal step. Lemma 17 shows that det A = 0 i¤ n = n0 + 1, where
n0 =
r
s
1
2
is the number de…ned in Theorem 7.
If n0 is not integer, then ker LXH +xr @=@x = 0 and all the terms
xr+s 1 x(n 1)s are in Im LXH +xr @=@x ; these terms correspond to xr+(n 1)s @x in
P
V . Thus NXH +xr @=@x = fxr+s 1 j6=0; r (m o d s) aj xj g. By Lemma 15 we have the
case (ii) of Theorem 7.
If n0 is integer, then the term cn0 s xr+n0 s @x remains in the normal form for V .
Here we have two possibilities: either all the further terms vanish, or there is a
term cj0 xr+j0 @x 6= 0; j0 6= 0; r (mod s).
In the …rst possibility we have
(2y + xr (1 + xn0 s ))@x + sxs
1
@y ;
38
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
i.e. the case (iii) of Theorem 7.
In the second case we use the bi-vector …eld e = !
xr+j0 @x , to reduce the term
xr+j0 +n0 s @x :
s
1 (x
y 2 )n0 +1 , applied to
This gives the case (iv) from Theorem 7.
Because there are no more possibilities of reductions, Theorem 7 is complete.
Remark 20. From the Loray’s proof in [L3] it follows that the reductions performed in the points (D) and (E) can be made analytic (provided that we start from
an analytic form from Lemma 16). Therefore the only place where the analyticity
fails is the reduction by means of LXH .
6. Proof of Theorem 8
(A) Theorem 8 says how to recognize the type of monodromy group from the
formal orbital normal form. On the basis of this classi…cation the cases when
the formal normal form is the same as the analytic one are extracted. Below we
successively associate to given classes of groups the corresponding formal normal
forms and at the end of this section we discuss the problem of analyticity of these
forms.
The projective monodromy group G associated with a germ of nilpotent analytic
vector …eld is either …nite or exceptional abelian or solvable nonabelian or nonsolvable.
(B) From Corollary 3 (in Section 5.1) it follows that, if G is …nite, then we have
the case (i) of Theorem 7. Here G is analytically isomorphic to a …nite subgroup
of the group of rotations.
(C) If G is exceptional abelian, then Corollary 3 implies that n0 2 Z (and hence
s = 2k). By Theorem 2(a) (Section 1) the generators of G are simultaneously
formally equivalent to
t1;2
f1;2 = gw
where = e i=k ; w = wp; . One can see also from the proof of Lemma 14 that
p = 2n0 k and p = 1 (see Corollary 3).
Lemma 18. The exceptional abelian monodromy group corresponds to the case
(iii) from Theorem 7.
Proof. We have to calculate the monodromy group associated with the normal
form from the case (iii) of Theorem 7. However, for this purpose that normal form
is not the best choice. We shall choose the form from Remark 8 (Section 2), i.e.
XH + xk 1 H n0 (1 + H n0 ) 1 EH .
Here we can integrate the analogue of the system (5.4), describing the phase
portrait after resolution of the singularity. In the variables u = y=xk ; H = y 2
x2k = (u2 1)v 2k we have
(6.1)
dH
2H n0 +1
=
du
(1 + H n0 )(1
u2 )
:
Now it is clear that, when we parametrize the disc transversal to the exceptional
divisor E by H, then the monodromy maps are the ‡ow maps of the vector …eld
H_ = H n0 +1 =(1 + H n0 ).
NORM AL FORM FOR NILPOTENT SINGULARITY
39
Remark 21. The equation (6.1) can be integrated. Its …rst integral takes the
so-called generalized Darboux form
F =
H
n0
+ n0 ln H
2n0 ln[(1 + u)=(1
u)]:
(D) If G is solvable non-abelian, then it is formally conjugated to
p
(6.2)
f1 (z) = ei ; f2 (z) = e2 ik=(2k+1) z= p 1 pz p ;
s = 2k + 1, or to
(6.3)
f0 (z) = e2
i=k
z; f1 (z) = e
i=k
p
z= p 1
pz p ;
s = 2k (see Theorem 2(b) in Section 1 and [CM]). It is also known that in the case
s = 2k + 1 this group is typical (see Theorem 3 in Section 1).
Let s = 2k + 1. The typicality of G means that p 6= 0 (mod s): G = fe il=2s :
l = 0; 1; : : :g and pG 6= f 1g (see Remark 1 in Section 1).
2k+1
This group is the holonomy group of the singularity Jr;0
, i.e. the case (ii) of
Theorem 7 with the formal functional invariant
0. We have p = 2(r k) 1
(see Corollary 3 in Section 5.1) and the condition r 6= 0 (mod s) from Theorem 7
is equivalent to the condition p 6= 0 (mod s) (i.e. typicality of G).
Let s = 2k. The case p = 0 (mod s) is excluded because then G is abelian:
il=k
: l = 0; 1; : : :g and pG = f1g (see Remark 1). Moreover, because G is
G = fe
typical, we have p 6= 0 (mod k).
2k
This group is the holonomy group of the singularity Jr;0
. We have p = r k (see
1
Corollary 3) and n0 = r=s 1=2 62 2 Z (for p 6= 0 (mod k)). Therefore, we have the
case (ii) of Theorem 7 with the formal functional modulus
0. The conditions
r 6= 0 (mod s); n0 62 Z from Theorem 7 are equivalent to the typicality of solvable
G.
(E) Because all solvable groups de…ned by (1.9)–(1.10) and by (1.11)–(1.12)
are formally equivalent to the above ones, the other normal forms Jr;s ; 6 0 or
6= cn0 s xn0 s have non-solvable holonomy group.
(F) In this point we prove analyticity of the normal form from Theorem 7 in the
cases of solvable non-abelian and …nite holonomy group.
Take a germ of analytic vector …eld V with solvable non-abelian monodromy
group G. Because G is typical, it is analytically conjugated with its standard
form (6.2) or (6.3) (see Theorem 3(iii) in Introduction). On the other hand the
s
polynomial …eld Jr;0
has the monodromy equal to (6.2) or (6.3) and is analytic.
s
By Theorem 1(b) the analytic conjugation of the holonomies of V and Jr;0
implies
analytic orbital conjugation of these vector …elds.
s
The analyticity of the formal normal form J1;0
(i.e. the Hamiltonian …eld) can
be proven in three ways.
One way uses the rigidity of …nite group (Theorem 3(i)) and Theorem 1(b)).
This proof is the same as in the solvable case.
One can also use the following theorem of Mattei and Moussu [MM]:
if a singular point of a planar holomorphic foliation has only …nite number of
separatrices (i.e. analytic invariant curves through the singularity) and any other
leaf does not accumulate at the singular point, then the foliation has a local analytic
…rst integral.
In the generalized cusp case we have only …nite number of separatrices (one or
two) and the …niteness of the holonomy group implies that the leaves are separated
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
40
from the singular point. So, the assumptions of the Mattei–Moussu’s theorem hold
and there is an analytic …rst integral. It has the form y 2 = xs + : : :. This function
is equal to Y 2 X s with analytic X = x + : : : ; Y = y + : : : (see [AVG]).
Below we propose another proof based on the following lemma.
Lemma 19. The case with …nite monodromy group occurs onlyPwhen the Bogdanov–Takens normal form contains only powers of xs , i.e. XH + ( bm xsm 1 ) EH .
Proof. Indeed, repeating the arguments from Section 5.1, we can show that any
vector …eld of the form
!
X
sm 1
r 1
XH +
csm 1 x
+ cr 1 x
EH ; M s < r < (M + 1)s
m<M
with cr 1 6= 0 has either non-abelian or exceptional abelian monodromy group.
If s = 2k + 1 is odd, then in the coordinates (5.1) we get
X
dv
= A1 (u)v +
csm 1 Asm 1 (u)v 2(sm k) + cr 1 Ar 1 (u)v 2(r k) :
du
The fact, that the terms xsm 1 EH can
Pbe removed, means that, after applying
certain change of the form v ! v1 = v + m<M Bm (u)v 2(sm k) , the 2(sM k)–th
part of the latter equation disappears. There remains the term cr 1 Ar 1 (u)v 2(r k)
and other terms which depend on the coe¢ cients csm 1 . However, the latter terms
arise only from the action of the change v ! v1 on the term with v 2(r k) and are
of higher degree.
2(r k)
Therefore, dv1 =du = A1 (u)v1 + cr 1 Ar 1 (u)v1
+ : : : and the 2(r k)–th jet
of the monodromy group contains the map v ! 1 v + const cr v 2(r k) + : : :, where
const 6= 0. This shows that the monodromy group is non-abelian.
The case with even s is analogous.
If the (analytic) Bogdanov–Takens normal form is equal to XH + xf (xs )EH ,
then repeating the proof of Lemma 13 (form Section 5.1) we see that it has analytic
…rst integral y 2 xs + : : :, which is analytically equivalent to Y 2 X s . Therefore
s
the form J1;0
is analytic.
This completes the proof of Theorem 8.
7. Proof of Theorem 9
In Theorem 8 we have given interpretation of the …rst part of the formal orbital
normal form in terms of the holonomy group. Theorem 9 says about interpretation
of certain higher order terms from the formal orbital normal form in terms of
singularities of the resolved vector …eld and in terms of the holonomy. Firstly we
interpret the …rst coe¢ cient of the formal functional modulus (x) for s = 3 and
then we interpret the coe¢ cient cn0 s in the exceptional abelian case.
(A) In the case s = 3 we can prove the non-solvability in another way than
in Section 6. Note that solvability of G means that the resolved vector …eld near
the singular point p0 : u = 1; v = 0 is formally (and analytically) linearizable (see
Theorem 5(a) in introduction). Here we show that we can express the obstacles to
the linearization in the terms of the coe¢ cients of the expansion of the functional
modulus from Theorem 7.
NORM AL FORM FOR NILPOTENT SINGULARITY
41
Recall that by Theorem 7 the formal normal form is of one of the two types:
either (r = 3m + 1)
x_ = 2[y + x3m+1 (1 + c1 x + c4 x4 + c7 x7 + : : :)];
y_ = 3[x2 + x3m y(1 + c1 x + c4 x4 + c7 x7 + : : :)];
or (r = 3m + 2)
x_ = 2[y + x3m+2 (1 + c2 x2 + c5 x5 + c8 x8 + : : :)];
y_ = 3[x2 + x3m+1 y(1 + c2 x2 + c5 x5 + c8 x8 + : : :)]:
In the coordinates (5.1), when we additionally perform the change u = 1 + z, we
get the equation
dv
= A1 (z)v + Ar (z)v 2r
dz
r
2
+ dt At (z)v 2t
2
t
2
+ :::
2
1+3z
; Ar = (1+z)6z ; At = (1+z)
and dt ; t 6= 0; r (mod 6) is
where A1 = 6z(1+z)
6z
the …rst non-zero coe¢ cient of the Taylor expansion of (dt is the …rst of cj ’s).
Recall also that the formal orbital normal form of a 1 : 6 resonant saddle is
following
d~
v
v~
=
(1 + e1 (~
v 6 z~) + e2 (~
v 6 z~)2 + : : :):
d~
z
6~
z
Here z~ = z(1 + : : :); v~ = v(1 + : : :).
Remark 22. The formal orbital normal form is simpler. There remain only two
coe¢ cients ej , namely ek = 1 and e2k , where the latter plays a role of the formal
invariant. In our situation the range of changes is smaller. So, we get the richer
normal form near p0 .
We know that if dt = dt+3 = : : : = 0, then all the coe¢ cients ej vanish. So,
…rstly we should apply a change which cancels the term Ar v 2r 2 .
Denote q = 2r 2. The change is following
v = w 1 + B(z)wq
1
1=(q 1)
where the function B satis…es the equation B 0 + (q 1)A1 B + (q 1)Ar = 0 with
the solution
Z z
q 1 (q 1)=6
(q+5)=6
B=
z
(1 + z)(q 1)=3
(1 + )(q 4)=6 d = 1 + O(z):
6
The transformed equation becomes equal to
dw
2(r t)=(2r 3)
= A1 w + dt At w2t 2 1 + Bw2r 3
+ ::::
dz
(Note that if q = 6p + 1, then the expansion of B(z) contains the logarithmic
term
M z log z:
Namely the coe¢ cients M before such logarithms will play a role of the saddle
quantities. Above we have q 6= 6p + 1, but we will use the coe¢ cient M below.)
Expanding the last expression in the equation for w, we get the term
dt At w2t 2 (which we reduce in the same way as Ar v 2r 2 ) and the next term of
the form
At Bw2r+2t 5 :
·
·
EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
42
It turns out that, in our situation, always 2r + 2t 5 = 1 (mod 6): either
3
3m + 1; t = 3n + 2 in the case J3m+1;
, or r = 3m + 2; t = 3n + 1 in the
3
J3m+2; .
Repeating the proof of cancelling the Ar , with q = 2r + 2t 5, we get a
B(z), i.e.
Z z
(3 r t)=3
(1 + )2(3 r t)=3 At ( )B(
B1 = const z (r+t 3)=3 (1 + z)2(r+t 3)=3
1
We are interested in the coe¢ cient before
function.
r =
case
new
)d :
in the expansion of the subintegral
Lemma 20. This coe¢ cient is non-zero.
Proof. Let us write this subintegral function explicitly (up to a constant)
Z
(2t+3)=6
(2r+3)=6
(1 + )(t 3)=3
(1 + )(r 3)=3 d :
3
In the case J3m+1;
; t = 3n + 2, the expansion of (1 + ) and the integration give
"1
# "1
#
l
X n 1=3
X m 2=3
m n 1
k
k
l
l m + 1=6
k=0
and its residuum at
l=0
= 0 is
m+n
X
n 1=3
m+n l
m
2=3
l
l=0
It can be rewritten in the form
hP
m+n
n 1=3
l
1
:
m + 1=6
i
( 1)l
m+n
l=0
l m+1=6
m+n
l
R 1 m 5=6
n 1=3
(1 t)m+n dt
m+n P 0 t
n 1=3
m+n B( m + 1=6; m + n + 1)
R
=
=
6=
0:
The latter integral is divergent and P means its regularization. One uses the
analytic continuation of the Beta function B(a; b).
3
In the case J3m+2;
, the calculations are quite similar and one obtains the coef…cient
n 2=3
B( m 1=6; m + n + 1) 6= 0:
m+n+1
Remark 23. In the proof of Proposition 1 we have applied two local (near p0 )
changes and detected the resonant term. Unfortunately, this works only in the case
s = 3. If s > 3, then we need much more local changes and the calculations become
very complicated. We were not able to complete them.
Remark 24. Note that if the formal functional modulus 6 0, then only the
…rst non-vanishing coe¢ cient al of the expansion of is important in non-solvability
[6]
f0 (z) = z + al z 6q+1 + : : : :
Maybe the other coe¢ cients of the expansion of are responsible for new relations
in group G. Something like this: f1 f2 f1 f2 : : : f2 1 (z) = z + an z r(n) + : : :
xk
(B) Let n0 2 Z; s = 2k. Then we can take the vector …eld in the form XH +
1 n0
H EH + : : :. Its (2n0 + 1)s–jet is integrable: dH=du = 2H n0 +1 =(1 u2 ) in
NORM AL FORM FOR NILPOTENT SINGULARITY
43
the separating variables from the proof of Lemma 18. It is clear that the saddle
u = 1; v = 0 in the divisor of resolution is not linearizable.
In the case (iii) we get dH=du = 2H n0 +1 =[(1 + H n0 )(1 u2 )]. Here
=
const cn0 s . It is clear that cn0 s plays a role of the formal invariant of the saddle
u = 1; v = 0.
Theorem 9 is complete.
Acknowledgments. The main problem for us during the work on this subject
was our isolation. In this context very fruitful were discussions with Frank Loray
and Ra…k Meziani during their visits in Warsaw. We thank Anton Sadovski for
his remarks, Yuri Il’yashenko for critical comments about previous versions of the
text. We thank the referee, whose clever suggestions helped to simplify the proof
of Theorem 6 and avoid some mistakes.
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44
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EWA STRÓ ZYNA
AND HENRYK ZO×
A̧DEK
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E-mail address : [email protected]
Current address : Institute of Mathematics, University of Warsaw,, ul. Banacha 2, 02-097
Warsaw, Poland
E-mail address : [email protected]