Gradient Vector Fields Do Not Generate Twister Dynamics

Journal of Differential Equations 174, 91100 (2001) Gradient Vector Fields Do Not Generate Twister Dynamics P. Fortuny and F. Sanz 1 Departamento de A lgebra,Geometr@a y Topolog@a,Universidad de Valladolid, Prado de la Magdalena sn,47008 Valladolid,Spain E-mail: pfortunyagt.uva.es, fsanzagt.uva.es Received August 2, 1999 Thom's Gradient Conjecture states that a solution #of an analytic gradient vector field Xapproaching to a singularity Pof Xhas a tangent at P. A stronger version asserts that #does not meet an analytic hypersurface an infinite number of times (it is non-oscillating). We prove, in dimension 3, that if #is ``infinitely near'' an analytic curve 1not composed of singularities of X, then #is non-oscillating and, moreover, it does not spiral around 1in a precise sense. 2001 Academic Press Key Words:trajectories of vector fields; gradient conjecture; oscillation; spiraling. INTRODUCTION Let X={ g fbe the gradient of an analytic function in R n with respect to an analytic Riemannian metric g. In [3], S. 4ojasiewicz proved that if #is a solution of Xwhich remains bounded in a relatively compact set, then # accumulates at a single point P. Afterwards, R. Thom conjectured (cf. [4]) that # has a tangent at P. That is to say, the transformation of #after the blowing-up of R n at Paccumulates at a single point in the exceptional divisor. This result is proved in a recent manuscript of Kurdyka, Mostowski and Parusinski [2]. In a more general way, R. Moussu has proposed the ``strong gradient conjecture'': solutions of analytic gradient vector fields are non-oscillating. This means that #meets each analytic hypersurface not containing it only a finite number of times. In [1], Cano et al. they study the relation between nonoscillation and the existence of ``all iterated tangents,'' that is, the existence of a tangent for the transform of #after any number of point blow-ups. In an ambient space doi:10.1006jdeq.2000.3926, available online at http:www.idealibrary.com on 91 0022-039601 35.00 Copyright 2001 by Academic Press All rights of reproduction in any form reserved. 1 Both authors partially supported by The European Commission, TMR Network ``Singularidades de Ecuaciones Diferenciales y Foliaciones'' ERBF MRXCT 96-0040. The second author thanks the University of Bourgogne for several stays during which this work was developed. of dimension 3, nonoscillation for #is equivalent to the following conditions: (a) The solution #has all iterated tangents, and (b) There is no ``spiraling axis'' 1for #. The aim of this paper is to prove property (b) for analytic gradient vector fields when 1is not a union of stationary points. These axes are called ``twister axes,'' and they give rise to a twister dynamics locally around them, as shown in [1]. To be precise, our main result is the following Theorem 5.1. Let g be an analytic metric on a three-dimensional analytic manifold M.Given an analytic function f on M,the gradient vector field { g f has no non-degenerate twister axes. This paper is structured as follows: in Section 1 we present the notions of spiraling, oscillation, iterated tangents, and twister axis. In Section 2, we establish a necessary condition for a smooth invariant axis 1not to be a spiraling axis. This condition is given in terms of the linear term of the ``normal'' component of Xwith respect to 1. Then we prove (Section 3) that spiraling axes are preserved by ramifications. This leads in a natural way to the study, made in Section 4, of a generalization of gradient vector fields: those arising from a bilinear form obtained as the ramification of a Riemannian metric. We prove that if X={ g fis an ordinary gradient in M, 1is a semi-branch and \:M$ÄMis a ramification with \ &1 (1) smooth, then this new branch cannot be a spiraling axis for the generalized gradient X$={ \*g \*f. The essence of this proof is the appearance of the Hessian of \*fin the normal component of Xwith respect to 1. The symmetry inherent to the Hessian is the obstruction to the existence of spiraling. Theorem 5.1 is deduced in the last section as a consequence of all these results. 1. PRELIMINARIES Let Xbe an analytic vector field on a three-dimensional real analytic manifold M. Let #be a trajectory of Xwhose |-limit set |(#) is a single point P.Byananalytic semi-branch at P, we mean the image of (0, =)by a non-constant analytic map _:(&=,=)ÄMwith _(0)=P. The semibranch 1is smooth (or non-singular)atPif _(&=,=) is a non-singular analytic curve. Suppose 1is a semi-branch at Pand let ?:M ÄMbe a sequence of blow-ups of points such that ? &1 (1) is a smooth semi-branch at a point P #? &1 (P). Fix a local coordinate system (x,y,z)atP such that 92 FORTUNY AND SANZ ? &1 (1)=[x=y=0, z>0]. Let #~ =? &1 (#). The definition of spiraling axis for Xis introduced in [1]. We give here an equivalent property: Definition 1.1. The curve # spirals around 1 (or equivalently, 1is a spiraling axis for #)if#~ admits a parametrization #~ (t)=(\(t) cos(.(t)), \(t) sin(.(t)), z(t)), t#(0,), where \,.,zare analytic in t, lim(\(t)z(t) n )=0 for all n0, and lim tÄ (.(t))=. Any spiraling axis is invariant for X(see [1]) and the property is preserved by blowing up P. Moreover, if 1is non-singular, #spirals around it and ':M$ÄMis the blowing-up of Mat Pwith center 1; then ' &1 (#) is the whole projective line ' &1 (P). Definition 1.2. The curve #is said to have all the iterated tangents if for any sequence of point blow-ups M n wwÄ ? n&1 M n&1 wwÄ ? n&2 }}}wwÄ ? 1M 1 wwÄ ? 0M 0 =M with ? 0 centered at P, the |-limit set of ? &1 n&1 (#) is a single point. If this is the case, then one defines the sequence TI(#)=[P n ]recursively: P 0 =P, P i+1 =|(? &1 i (# i )), where ? i is the blowing-up of M i with center P i and # i is the pull-back of #by ? i&1 b}}}b? 0 . Notice that if 1is a spiraling axis for #, then #has all the iterated tangents and TI(#) is exactly the sequence of infinitely near points of 1.We remark also that #has all the iterated tangents and TI(#) is the sequence of infinitely near points of an analytic branch 1if and only if for any semianalytic open set V#1, one has #(t)#Vfor t>>0. Definition 1.3. The trajectory #is said to oscillate at Pif there is an analytic surface f=0 such that #/ 3(f=0) but the set ( f=0)&|#|is infinite. The main result from [1] which we are going to use is the following Proposition [1]. If the trajectory # has all the iterated tangents at P and oscillates,then there is an analytic semi-branch 1 which is a spiraling axis for #. Definition 1.4. The semi-branch 1is a twister axis for Xif there is a positively invariant neighbourhood Uof 1such that for every Q#U&1, the trajectory of Xpassing through Qspirals around 1. 93 GRADIENT VECTOR FIELDS When 1is non-degenerate (which means that it is not composed of singularities of X), then 1is a spiraling axis if and only if it is a twister axis [1]. Thus, in order to know if a non-degenerate semi-branch is a twister axis, one need only test the existence of one solution spiraling around it. Let ?:M ÄMbe the blowing-up of Mwith center P. Denote by 1 ,#~ , and X the transforms of 1,#, and Xby ?. We know (see [1]) that 1is a spiraling axis for #, if and only if 1 is a spiraling axis for #~ . 2. SUFFICIENT CONDITIONS FOR NON-SPIRALING Let X,Mbe as above. In this section, 1will denote a non-singular analytic semi-branch at a point P, invariant and non-degenerate for X. Let r0 be the order of X along 1, that is, the algebraic order at the origin of the restriction of Xto the analytic curve in Mcontaining 1. Consider a system of coordinates (x,y,z) in a neighbourhood of Psuch that 1# [x=y=0, z>0]. In these coordinates, r=& z (X(z)). Write Xas follows, X=:  i=0 L i z i +v(x,y,z) z+X , where the L i are linear vector fields in the variables (x,y) and X =a(x,y,z) x+b(x,y,z) y, with & x,y (a,b)2. Let k=min[i :L i is not a radial vector field]. This number kis independent of the chosen coordinate system. As 1is nondegenerate, then v(0, 0, z)=z r u(z), with :=u(0){0. Proposition 2.1. If X and 1 satisfy one of the following conditions: (0) :>0, (1) :<0 and kr, (2) :<0, kr&1, and L k has two different real eigenvalues, then 1 is not a twister axis for X. Proof. Each of those conditions is stable under blowing-up of P. After making 2rblow-ups we can write Xas follows, up to multiplication by an analytic unit X=: r&1 i=0 L i z i +z r \: z+Z+, 94 FORTUNY AND SANZ where L i are linear vector fields in (x,y), and Zis a vector field with Z(z)#0 and & x,y (Z(x), Z(y))1. If condition (0) holds, then 1is not positively invariant. Assume that either (1) or (2) holds. Let #be a trajectory of Xsuch that |(#)=Pand suppose, in order to obtain a contradiction, that #spirals around 1. By the expression of Xabove, there is a parametrization #(z)=(x(z), y(z), z) for z>0 (not necessarily analytic at z=0). Let 'be the blowing-up of Malong the z-axis, with local equations x=x$, y=x$y$, z=z$. The curve #$=' &1 (#) admits a parametrization #$(z$)=(x$(z$), y$(z$), z$). If condition (1) holds, then we have d dz$(y$(z$))=A(x$(z$), y$(z$), z$), where Ais an analytic function. As #$ accumulates along the whole ' &1 (0), there is a z$ 0 such that 1>z$ 0 >0 and |y$(z$ 0 )|<1. Then | d dz$ (y$(z$))| is bounded for 0<z$<z$ 0 , which implies that |y$(z$)| is also bounded near z$=0, contradicting the accumulation of #$ along ' &1 (0). If condition (2) holds, then the transform of Xby 'is X=*(z$) x$ x$++z$ k y$ y$+z$ k+1 Y, where k#Z + ,*is a polynomial in z$ of degree at most k,+is a non-zero constant, and Yis analytic. This gives d dz$(y$(z$))= 1 :z$ r&k (+y$(z$)+z$B(x$(z$), y$(z$), z$)), Bbeing an analytic function. From this equation we infer that near the points P 1 =(0, 1, 0), P 2 =(0, &1, 0), d dz$ (y$(z$)) has opposite signs for z$ positive. This contradicts again the accumulation of y$(z) along the whole real line. K 3. RAMIFICATIONS Let 1be a (not necessarily smooth) semi-branch at a point P#M. Take local analytic coordinates (x,y,z)atPsuch that 1/[z>0]. We consider the ramification-rectification morphism \:M$ÄM (x$, y$, t)[(x,y,z)=(x$+:(t), y$+;(t), t q ), 95 GRADIENT VECTOR FIELDS where (:(t), ;(t), t q ) is a Puiseux parametrization of 1for t>0. Denote 1$=\ &1 (1). We shall assume that \is an algebraic morphism, which can be accomplished after an analytic coordinate change in M(see [5]). Given an analytic vector field Xin M, let us consider the analytic vector field X$inM$ such that \ C (X$)=zX. Notice that 1is a spiraling axis for Xif and only if it is so for zX. Proposition 3.1. Suppose 1 is a spiraling axis for X.Then 1$is also a spiraling axis for X$. Proof. Let #be a trajectory of Xspiraling around 1and take #$=\ &1 (#), which is a trajectory of X$. We may suppose that |#|/[z>0] and |#$|/[t>0].As\is a homeomorphism between [t0]and [z0], the set |(#)=\ &1 (P) consists of a single point. Let us prove that #$ spirals around 1$. It is clearly oscillating, since #is so. Thus, we shall finish if we prove that #$ has all the iterated tangents and TI(#$) is the set of infinitely near points of 1$. With the system of coordinates we are using, it suffices to prove that for any k=1, 2, ..., #$(t) is in the open cone C$ k =[x$ 2 +y$ 2 <t 2k ] for t>>0. Since these cones are algebraic, they project by \into semianalytic sets C k containing 1in their interior. Thus, #(t)isinC k for t>>0 because TI(#) coincides with the sequence of infinitely near points of 1. So, for any k=1, 2, ..., #$(t)#C$ k for t>>0, which implies that TI(#$) exists and is the sequence of infinitely near points of 1$, which completes the proof. K Note that the converse is also true. 4. RAMIFIED GRADIENTS Let gbe an analytic symmetric bilinear form on M. Given a point P, denote by M P the field of germs of meromorphic functions at P, that is, the field of quotients of the ring O P of germs of analytic functions at P. Let 3 P be the O P -module of germs of analytic vector fields. The bilinear operator g P :3 P _3 P ÄO P induces a symmetric bilinear form g~ :3 P _3 P ÄM P , with 3 P =3 P M P .Ifg~ is non-degenerate, we shall call 9 g~ to the natural isomorphism 9 g~ :3 P Ä3 C P =Hom(3 P ,M P ) 96 FORTUNY AND SANZ induced by g~ . Taking into account that 3 C P is the M P -vector space of germs of meromorphic 1-forms at P, we give the following Definition 4.1. For f#M P , the generalized gradient of fwith respect to gis the meromorphic vector field { g f=9 &1 g~ (df ). Consider a non-singular divisor D/Mcontaining P. Let gbe as before and let q0 be a non-negative integer, Definition 4.2. The form gis a metric of q-ramified type relative to D if there is a local coordinate system (x,y,z)atPwith D=[z=0]such that 1. The restriction g N of g~ to the O P -module Ngenerated by the meromorphic vector fields [x,yand z &q z] is an O P -bilinear form, that is, g N (N_N)/O P . 2. The specialization g N (P)=g N Rdefines a positive definite bilinear form on the three-dimensional real vector space NR. In these conditions, we shall say that the coordinate system (x,y,z)is appropriate for g. Notice that a coordinate change of the form x$[.(x,y)+z q+1 . 1 ,y$[(x,y)+z q+1  1 ,z$[zu(x,y,z) respects the lattice N/A3 P and gives another appropriate system. If the matrix of g N (P) is the identity for the base of NRcorresponding to [x,y,z &q z], we say that the coordinate system (x,y,z)isnormal for g. Making coordinate changes as above, we can always get a normal system; we remark that the curve x=y=0 need not be preserved under these changes. Proposition 4.1. Let g be a metric of q-ramified type relative to a divisor D at P.Consider an appropriate and normal system of coordinates (x,y,z)and a germ of analytic function f #O P .The meromorphic vector field X=z 2q { g f is in fact analytic.If the curve Y=[x=y=0] is invariant and non-degenerate for X,then the branch 1=Y&[z>0] is not a twister axis for X. 97 GRADIENT VECTOR FIELDS Proof. Let Abe the matrix of g N in the basis [x,y,z &q z]. Notice that Ais invertible and, in fact, the coefficients of A &1 belong to O P , since A(0) is the identity matrix. Put z q z q G=(G ij )= \ z q + A &1 \ z q + 11 The vector field X=z 2q { g f=a  x +b  y +c  z is analytic, for (abc)=(f x f y f z )G t , where subindices indicate partial derivation. From now on, we shall write h to indicate the restriction of an analytic function hin Mto the curve Y. The following bounds follow from the fact that A &1 (0) is the identity matrix & z (G ii )=2q,& z (G ii x ,G ii y )2q & z (G 12 )2q+1, & z (G 12 x ,G 12 y )2q(1) & z (G i3 )q+1, & z (G i3 x ,G i3 y )q,& z (G 33 )=0, where i=1, 2. Let us show that Xand 1satisfy one of the conditions of Proposition 2.1. Let rbe the order of Xalong the curve Yand let kbe the least integer such that L k is not radial in the expression X=:  i=0 L i z i +v(x,y,z) z+X  used in Section 2. If kr, then condition (0) or (1) holds. Thus, assume k<r. First, we note that the invariant ris equal to & z (f z ). This follows from the fact that X| Y =z 2q { g| Yfand that the matrix of g| Y is exactly (G 33 ) in the basis [z &q z]. In the expression of Xabove, one can write the linear part normal to 1: :  i=0 L i z i =(xy)(N+H)\x y+, where Nis the 2_2 matrix \G 11 x f x +G 12 x f y +G 13 x f z +G 13 f xz G 11 y f x +G 12 y f y +G 13 y f z +G 13 f yz G 12 x f x +G 22 x f y +G 23 x f z +G 23 f xz G 12 y f x +G 22 y f y +G 23 y f z +G 23 f yz + 98 FORTUNY AND SANZ and His obtained from the Hessian of fas follows: H=\G 11 G 12 G 12 G 22 +\f xx f xy f xy f yy +. Notice that this Hessian Hcan be written H=z 2q (A i z i )(B i z i ) where A 0 is the identity and A i ,B i are symmetric matrices. Then, the first nonradial term appearing in the power series expansion of His symmetric and so it has two different real eigenvalues. Thus, we shall finish if we prove that & z (N)r, for we are assuming k<r, from which we infer that the first non-radial term in the linear part of Xnormal to 1comes, in fact, from H. To see that & z (N)r, take l=min[& z (f x ), & z (f y ), & z (f z )]. From the bound (1), one sees that & z (N)q+l. There are two possibilities: if l= & z (f z )=rthen & z (N)r.Ifl<rassume, by symmetry, that l=& z (f x ). As Yis an invariant curve for X, one must have a =G 11 f x +G 12 f y +G 13 f z =0. Since & z (G 12 f y )>& z (G 11 f x ) then & z (G 11 f x )=& z (G 13 f z ), whence 2q+l= & z (G 13 )+rq+1+r, from where & z (N)=q+l>rand we are done. K 5. GRADIENTS DO NOT GENERATE TWISTER AXES Theorem 5.1. Let g be an analytic metric on a three-dimensional analytic manifold M.Given an analytic function f on M,the gradient vector field { g f has no non-degenerate twister axes. Proof. Suppose, on the contrary, that 1is a twister axis for Xat P. Let (x,y,z) be a coordinate system appropriate for gand normal at P. Moreover, we can take (x,y,z) such that 1is tangent to x=y=0 and is contained in z>0. Let \:M$ÄMbe an algebraic ramification-rectification morphism as in Section 3 \(x$, y$, z$)=\x$+_(z$), y$+{(z$), 1 q+1 z$ q+1 + such that 1$=\ &1 (1)=[x$=y$=0, z$>0]. Let g$=\ C gbe the transformed bilinear form. A computation shows that g$ is a metric of q-ramified type relative to [z$=0]and (x$, y$, z$) is a coordinate system appropriate and normal for g$. Let X$={ g$ (\ C f). By Proposition 4.1, 1$ cannot be a twister axis for z 2q X$, in contradiction with Proposition 3.1, since \ C X$=X.K 99 GRADIENT VECTOR FIELDS