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\begin{document}

\title{Topological problems of the theory of  \\
asymptotic curves}
\author{V.I. Arnold$^1$}
\date{}

\footnotetext[1]{Partially supported by the Russian Basic Research Foundation,
project {\bf 96-01-01104} and Institut Universitaire de France}

\maketitle

\vspace{2mm}
\centerline{to S.P. Novikov admiringly}

\vspace{6mm}
\noindent \underline{\bf {I. Introduction.}}

\vspace{4mm}
An asymptotic curve on a surface in the euclidean or projective $3$-space is an
integral curve of the field of asymptotic directions (directions along which the
second quadratic form vanishes). It is proved below that (typical)
asymptotic curves
on hyperbolic surfaces are exactly the (typical) space curves having no
flattening
points. These curves might also be defined as the smooth curves whose dual
curves are
also smooth. I call them rotating curves. A rotating curve may have inflection
points (where the curve curvature vanishes).

\vspace{2mm}
The rotating curves in the three-dimensional projective space have remarkable
topological properties. These properties imply, for instance, that an asymptotic
curve of a surface, which is a graph of a smooth function, can't be a star-like
curve (this result find by D. Panov was the starting point of the present
work). In
spite of this, there exist closed asymptotic curves on some hyperbolic surfaces
which are graphs of functions of two variables. Some examples are presented
below
(to construct them is an unexpectedly complicated task). I do not know what
is the
minimal number of the inflections points on such a curve. (Using the method
of the
present paper, F.\nolinebreak Aicardi has recently constructed curves with
only two
inflections).

\vspace{2mm}
The dynamical systems, defined by the asymptotic direction fields on the
(doubled)
hyperbolic domain, have not been studied yet, as far as I know. On a cubic
surface
there are $27$ asymptotic straight lines (some of which may be real). But
even the
integrability of the asymptotic lines equations in the victinity of these
lines has
not been studied.

\vspace{2mm}
All the selfdual smooth curves in the three-dimensional space are rotating
curves
(the simplest example being the curve $x = t$, $y = t^a$, $z = t^{a+1}$). An
immersion of  of a circle into a projective space is called selfdual, if
the dual
immersion sends the circle onto a curve in the dual projective space, which is
projectively equivalent to the image of the original immersion. Thus a
selfduality
defines a diffeomorphism of the immersed circle onto itself. The case where this
diffeomorphism (or its power) is the identity mapping deserves a special study.

\vspace{2mm}
The projective topology, that is the study of the discrete invariants of smooth
objects (curves, surfaces, ...) in projective spaces, has been founded
already by
M\"{o}bius and Jacobi, but its development was slow and for a long time it was
considered as a dull and unfashionable domain of differential geometry.
However it
has been understood recently that all this domain belongs rather to
symplectic and
contact topology. The projective topology provides a way to see the simplest
manifestations of the complicated symplectic and contact topology
phenomena, whose
general theory is yet to be constructed.

\vspace{4mm}
\noindent \underline{\bf {II. Finite type curves and rotating curves}}

\vspace{4mm}
Consider a curve, smoothly immersed into the euclidean space. The curve is
{\it {of
finite type}} at a point, if some derivatives of the immersion at this
point generate
all the tangent space at it. The parametric equation of the curve can then be
written (near this point) in suitable affine coordinates in $\BbR^n$ in the form
\begin{equation}
x_k = t^{^{a_k}} + o(t^{^{a_k}}), \quad 1 = a_1 < a_2 < ... <a_n \ .
\end{equation}
The set of the degrees $a_k$ is called the {\it symbol} of the point.
Formula (1)
implies that a finite type curve has a continuously rotating osculating flag of
osculating spaces of all dimensions.

The property to have a finite type (and also the symbol itself) are invariant under
the projective transformations of the space (and under the diffeomorpisms
of the axis
of the parametre $t$). Hence the finite type curves, as well as the symbols
of their
points, are well defined for the curves smoothly immersed into the
projective spaces
(real or complex).

All the curves, having a point of infinite type, form a codimension
infinity set in
the space of curves. For example, an analytic curve has a finite type at
every point
if it does not lie in a hyperplane.

We shall study below the properties of the asymptotic curves on hyperbolic
surfaces
(surfaces of negative curvature). All such surfaces, having an asymptotic curve
carrying an infinite type point, form a codimension infinity set in the space of
surfaces.

Hence we shall suppose everywhere below (sometimes implicitly) that all our
curves
are of finite type.

The {\it dual curve} of a finite type curve immersed into the projective
space lives
in the dual projective space and is formed by the osculating hyperplanes of the
original curve. The dual curve is not always immersed. Its symbol (at the point
corresponding to the point (1)) is dual to the original one :
$$a^\vee \ = \ (a_n - a_{n-1} < ... < a_n - a_1 < a_n ) \ . $$

For instance, a typical curve, immersed into the $3$-space, has symbol
($1<2<3$) at
the generic points and symbol $(1 <2<4$) at some isolated {\it flattening
points}
(where the curvature remains nonzero, but the torsion vanishes).

The dual curve has the usual symbol ($1 < 2 < 3)$ at the ordinary points,
but the
flattening points of the original curve correspond to the semicubical {\it
cusps}
(with symbol $(2 < 3<4$)) of the dual curve.

\vspace{4mm}
\noindent \underline{\bf Definition}. A finite type curve in three-dimensional
projective space is called {\it a rotating curve} if this curve as well as
the dual
curve are smoothly immersed (that is, if their symbols at every point have the
selfdual form ($1<a<a+1)$).

The simplest degeneration of the symbol of a rotating curve is not the
flattening
(which is impossible for a rotating curve) but the inflection (whose symbol
$(1<3<4$) is selfdual) which is the zero of the curvature and not of the
torsion.
The set of the curves having an inflection point is a hypersurface (a
codimension
one set) in the space of all curves. It means that the (simplest)
inflection points
occur for isolated parametre values in typical one-parametre families of
curves. But
in the space of the rotating curves the inflection is a common event.

The theorems proved below show that the rotating curves (called this way for the
following reason : the helical rotation orientation does not change along such a
curve, the accompanying Frenet frame might be continued smoothly across the
inflection point, provided that the curvature changes the sign at every generic
inflection point) have interesting topological properties and do deserve a
special
study.

\vspace{4mm}
\noindent \underline{\bf {III. Asymptotic curves on hyperbolic surfaces.}}

\vspace{4mm}
We shall show below that the asymptotic curves on the hyperbolic surfaces
form the
same class of curves as the rotating curves in the ambient projective $3$-space.

A {\it hyperbolic surface} in the real projective three-dimensional space is a
surface whose second quadratic form is hyperbolic. At the neighbourhood of
any of
its point the hyperbolic surface may be defined (in suitable affine coordinates)
by the equation $z = xy+...$, where the dots denote the terms of higher
order of the
Taylor series.

{\it The asymptotic directions} of a hyperbolic surface are the directions along
which the second quadratic form of the surface vanish. In the above coordinate
system the asymptotic directions (at point $0$) are the directions of the
coordinate
axis $x$ and $y$. The asymptotic directions depend on the projective
structure of
the space rather than on its metric : they are the directions of the
branches of the
intersection curve of the surface with the projective plane tangent to it
(and are
hence invariant under the projective transformations of the space).

An {\it asymptotic curve} on a hyperbolic surface is a curve whose direction at
every point is an asymptotic direction.

We shall see that the asymptotic curves on a hyperbolic surface are essentially
those curve, whose osculating planes are tangent to the surface. More
precisely, the
following simple theorem holds :

\vspace{4mm}
\noindent \underline{\bf Theorem 1}. {\it (i) If the osculating plane of a curve
(lying on a hyperbolic surface) is tangent to the surface at some point of the
curve, then the direction of the curve at this point is an asymptotic direction.

(ii) if a nonzero curvature curve, lying on a hyperbolic surface, has an
asymptotic
direction at a point, then the osculating plane of the surface at this point is
tangent to the surface.

(iii) The osculating planes of a finite type asymptotic curve on a hyperbolic
surface are everywhere tangent to the surface.}

\vspace{4mm}
\noindent \underline{Proof}. Choose affine coordinates, in which the surface is
defined in a neighbourhood of the point of study by the equation of the
form $z = xy
+ f_3 +...$ ($f_3$ is the cubical part of the Taylor series, the higher
order terms
being denoted by the dots).

(i). A curve of a nonasymptotical direction at the origin has the equation
of the
form $y = Ax + ...$, $A \not = 0$. Along this curve $z = Ax^2 + ...$, hence the
tangent plane of the surface $(z = 0$) is not the osculating plane of the curve
(which has with its osculating plane at least three infinitesimaly-close common
points).

(ii). A curve of asymptotic direction at the origin, whose space curvature
is not
vanishing, has the equation of the form $y = Ax^2 + ...$, $A \not = 0$.
Along this
curve $z = Ax^3 + f_3 (x,0)+ ...$. Hence the tangent plane $(z = 0$) of the
surface
is the osculating plane of the curve.

\vspace{4mm}
\underline{Remark}. An osculating plane of an asymptotic curve of zero curvature
may be transversal to the surface.

\vspace{4mm}
(iii). The curvature of a finite type curve is different from zero almost
everywhere. The osculating plane of a finite type asymptotic curve is therefore
tangent to the plane almost everywhere (see (ii)). Both planes --- the one
tangent to
the surface ans the osculating plane of the asymptotic line\nolinebreak ---
depend
continuously on the point of the finite type curve. Hence they coincide
everywhere and the Theorem is proved.

\vspace{4mm}
\noindent \underline{\bf Theorem 2}. {\it An asymptotic curve of finite
type on a
hyperbolic surface is a rotating curve.}

\vspace{4mm}
\noindent \underline{Proof}. Use the affine coordinates, such that the surface
equation takes the form $z = cxy + f(x,y)$ (where $c \not = 0$ and $f$
contains the
cubic and higher order terms of the Taylor series). According to Theorem 1,
(iii),
the asymptotic curve with symbol ($1 < a<b$) at the origin has the equations
$$y = Ax^a + ..., \quad z = Bx^b + ..., \quad A \not = 0 \ , \quad B \not =
0 . $$
Theorem 2 states that $b = a+1$. We shall now prove it. Consider the derivative
$h(x) = dy / dx = A ax^{a-1} + ...$. The asymptotic direction definition
$$F_{xx} (dx)^2 + 2 F_{xy} \  dx dy + F_{yy} (dy)^2 \ = \ 0$$
takes for $F = cxy + f$, $dy = h \ dx$, the form
$$f_{xx} + 2(c + f_{xy} ) h + (f_{yy}) h^2 \ = \ 0. $$
The terms of order smaller than $a$ in $x$ (along the curve $y = Ax^a +
...$) are
only provided by the terms $f_{xx} (x, 0) + 2ch$. We deduce that $f(x,0) =
-(2cA /
(a+1)$) $x^{a+1} + ...$. Hence, along the  asymptotic line the variable $z$
has the
Taylor series
$$z = (c Ax^{a+1} + ...) + f(x, Ax^a + ...) = \bigl( (a-1) / (a+1) \bigr) Ac x
^{a+1} + ... \ , $$
whence $b = a+1 \Bigl( B = \bigl( (a-1) / (a+1) \bigr) \ Ac \not = 0
\Bigr)$, which
proves the Theorem.

\vspace{4mm}
\noindent \underline{Remark}. Theorem 2 shows, that rotating curves form an
interesting space\nolinebreak : one should study the topological invariants
of this
space, the natural strati- fication of its boundary, the corresponding
Vassiliev invariants and so on. Indeed, the following simple complement to Theorem 2
holds :

\vspace{4mm}
\noindent \underline{\bf Theorem 3}. {\it Every rotating space curve of
finite type
is an asymptotic curve on a suitable hyperbolic surface}.

\vspace{4mm}
\noindent \underline{Proof}. Choose a straight line, transversal to the
tangent line
of the curve, in the osculating plane of the curve at each point (it should
depend
smoothly of the point, it suffice to take the normal line in some metric).
The union
of these straight lines form a smooth surface in a neighbourhood of the
curve. The
osculating planes of the curve are tangent to the surface by the
con\nolinebreak struction. We shall prove that the surface thus constructed is
hyperbolic. Theorem 1(i) would then imply that the original rotating curve
is its
asymptotic line.

To prove the hyperbolicity of the surface at a point of the curve choose an
affine
coordinate system centred at the point, whose $x$ axis is tangent to the
curve and
whose $y$ axis belongs to the surface, whose equation takes the form
$$z = H(x,y), \quad H = c xy + f(x,y), \quad f = O \Bigl( |x|^3 + |y|^3
\Bigr) \ .
$$
We shall prove that $c \not = 0$ (which means hyperbolicity). Write the
equation of
the original curve (lying on the surface) in the parametrical form
$$ x = t, \quad y = g(t), \quad z = h(t), \quad {\rm where \ } \ h (t) =
H(t, g(t))
\ . $$
The tangency condition (of the osculating plane of the curve to the
surface) claims
that the vector of the second derivative $(0, g^{^{..}}, h^{^{..}})$
(generating the
osculating plane together with the first derivative vector at a generic
point of a
finite type curve) should verify the tangent plane equation, $dz = H_x dx +
H_y dy$.
We get thus the relation
\begin{equation} h^{^{..}} \ = \ H_y (t, g(t)) \ g^{..} (t) \end{equation}
identically along the curve.

\vspace{2mm}
Suppose that point $t = 0$ is of type $(1 < a < a+1)$, whence one has
$$g(t) = At^a + O
(t^{a+1}), \  h(t) = B t^b + O (t^{b+1}), \ b = a+1, \ AB \not = 0. $$
We shall see that
these formulae imply $c \not =  0$. From the definitions of $g$ and of $H$
we get
the relations
$$g^{..} = pt^{a-2} + O (t^{a-1}), \quad H_y = cx + O (|x|^2 + |y|^2) = ct
+ O(t^2)$$
where $p = Aa (a-1)$. For $h$ we get the expression $h(t) = cAt^{a+1} =
f(t, 0) + O
(t^{a+2})$, implying  that $h^{^{..}} = q c t^{a-1} + f^{^{..}} (t, 0) +
O(t^a)$,
where $q = Aa (a+1)$. Finally, we have reduced equation (2) to the form
$$qct^{a-1} + f^{^{..}} (t, 0) + O (t^a) = (ct +O (t^2)) (pt ^{a-1} +
O(t^{a-1})) \
.$$
 If $c$ were equal to zero, we would have $f^{^{..}} (t, 0) = O(t^a)$.
Since $f = O
(t^3)$, this relation would imply that $f(t, 0) = O (t^{a+2})$. But then
one would
have $h(t) = O (t^{a+2})$, and the curve would not be a rotating curve
(since for a
rotating curve $B \not = 0$).

Thus $c \not = 0$ and hence the surface is hyperbolic. The curve is then
asymptotic
(by Theorem 1(i)). Theorem 3 is proved.

\vspace{2mm}
\noindent \underline{Remark}. If a finite type curve is defined in some affine
coordinates by the equations $y = g(x)$, $z = h(x)$, the rotating
conditions takes
the form $(h^{''} / g^{''} ) ^{^{'}} \not = 0$ (the ratio of the second
derivatives
is continued to the point where $g^{''} = 0$ as a smooth function).

\vspace{2mm}
Indeed, choose affine coordinates for which the point we are studying is
the origin,
the $x$ axing being directed along the curve and the osculating plane at
the origin
being the plane $ z = 0$. This change of variables will transform functions
$g$ and
$h$ into functions
\begin{eqnarray*}
\tilde g (x) & = & g(x_0 + x) - y_0 - v_0 x \ = \ Ax^a + ... , \\
\tilde h (x) & = & h(x_0 + x) - z_0 - w_0 x - u_0 \tilde g (x) \ = \ B
x^{a+1} + ...
\ . \\
\end{eqnarray*}
Hence the derivatives of the ratios of the second derivaties coincide in the
corresponding points :
$$(\tilde h^{''} / \tilde g ^{''})^{^{'}} \ = \ (h^{''} / g^{''} )^{^{'}} \ . $$
But in the new coordinates we have
$$\tilde h^{''} / \tilde g^{''} \ = \ (B (a+1) / (A(a-1))) x + ... \ . $$
Hence function $h^{''} / g^{''}$ has a smooth continuation to the points where
$g^{''}$ vanishes. The rotation condition, $B \not = 0$ takes hence the form
$(h^{''} / g^{''})' \not = 0$ both at the ordinary points, where $g^{''}
\not = 0$,
and at the inflection points, where $g^{''} = 0$.

\vspace{4mm}
\noindent \underline{\bf Corollary}. {\it An asymptotic line on a periodic
surface
$$z \ = \ F (x, y), \ F (x+1, y) \equiv F (x,y) $$
cannot be defined globally by the periodic equation
$$y = g(x), \ g (x+1) \equiv g(x) \ . $$ }

\vspace{4mm}
\noindent \underline{ Proof}. The ratio $h^{''} / g^{''} $ is a smooth periodic
function along such a curve (it is periodic even at the zeros of the
denominator).
If its derivative were sign-preserving, it would violate the Morse inequality.

\vspace{4mm}
\noindent \underline{Remark}. Strictly speaking, this is only proved for
the finite
type asymptotic curves. But the type finiteness condition is probably
irrelevant,
since any neighbourhood  of a closed asymptotic curve on a periodic surface
contains
a finite type periodic asymptotic curve on a neighbouring periodic surface.

\vspace{4mm} \newpage  % added by SVD
\noindent \underline{\bf {IV. The unavoidable bends of asymptotic and rotating
curves}}

\vspace{4mm}
The union of the tangent straight lines of a space curve is called its {\it
tangent
developable } surface.

\vspace{4mm}
\noindent \underline{\bf Theorem 4}. {\it Suppose that the osculating
planes of a
closed rotating curve in $\BbR P^3$ do not contain some point $O$. Then the
tangent
developable surface of this curve intersects every straight line containg
point $O$.}

\vspace{4mm}
\noindent \underline{Proof}. If a straight line $l$, containing $O$, were not
intersecting the tangent developable surface, then every osculating plane of the
curve would intersects $l \backslash O$. The intersection point would move along
$l$ (when the osculation point moves positively along the curve) always in
the same
direction.

\noindent Indeed, the momentary rotation axis of the osculating plane is
the tangent
line of the original curve. This line does not intersect $l$. The infinitesimal
rotation direction is constant along the rotating curve (by the very
definition of
the rotating curve). This implies the constancy of the direction of the
motion of
the intersection point of the osculating plane with $l$.

But a circle can't be mapped monotonically on the interval $l \backslash
O$. This
proves the Theorem.

\vspace{4mm}
\noindent \underline{Example}. Call point $O$ the {\it vertical direction
infinite
point} and call the straight lines containing point $O$ {\it vertical
lines}. The
vertical lines project the complement to point $O$ onto the complementary {\it
horizontal plane} $\BbR P^2$. The Theorem implies the following

\vspace{4mm}
\noindent \underline{\bf Corollary}. {\it Suppose that a closed rotating
curve has no
vertical osculating planes. Then the tangent lines of the projection of
this curve
to the horizontal plane cover this projective plane completly.
}

\vspace{2mm}
Indeed, the tangent lines of the original curve are projected onto the
tangent lines
of the projected curve. If a point of the horizontal plane were uncovered by the
projected  tangent lines, the vertical line, connecting this point to $O$,
would not
intersect the tangent developable surface, in contradiction with Theorem 4.

\vspace{4mm}
\noindent \underline{\bf Corollary}. {\it A closed asymptotic finite type
curve on
a hyperbolic surface, immersed into $\BbR P^3$ and transversal to the straight
lines containing point $O$, is projected from $O$ to $\BbR P^2$ as a curve,
whose
projections are covering $\BbR P^2$. }

\vspace{4mm}
\noindent \underline{Proof}. Theorem 2 implies that this curve is rotating and
that its osculating planes are tangent to the surface. Since the surface is
nowhere
vertical, these osculating planes do not contain point $O$. According to
Theorem 4,
every vertical straight line intersects one of the tangent lines of the
asymptotic
curve. The projection of this tangent line contains the point of $\BbR P^2$
corresponding to the vertical line and is tangent to the projection of the
asymptotic
curve to the horizontal plane.

\vspace{4mm}
\noindent \underline{\bf Corollary} (D. Panov). {\it  The projection of a closed
asymptotic line of a surface $z = f(x,y)$ to the plane $\lbrace (x,y)
\rbrace$ can
not be a star-line curve (in particular, this projection can't be a convex
curve). }

\vspace{4mm}
\noindent \underline{Remark}. The type finiteness condition is perhaps
irrelevant for
these corollaries, since one might approximate the given closed asymptotic
curve by
a finite type asymptotic curve on a neighbouring surface.

\vspace{4mm}
\noindent \underline{\bf {V. Examples of closed asymptotic curves on the
surfaces}}
z = f(x,y).

\vspace{4mm}
To construct such a curve it suffices (according to Theorem 3) to construct
a closed
rotating curve whose osculating plane is nowhere vertical and whose
projection into
the horizontal plane is an embedding.

I shall construct first a periodic curve with these properties (it is
closed rather in
the solid torus $ \lbrace (x$ mod $1,y, z) \rbrace$ than in the Euclidean
space).

To transform it into a rotating curve in the Euclidean space, we bend
slightly a long
segment of the periodic curve (say, of length $N$) to obtain a rotating
curve, whose
projection to the $(x, y)$-plane, seen from far, would look like a circular
arc of
length $N$ with a small central angle $\theta = 2 \pi /M$. The union of $M$ such
bended arcs forms a closed curve embedded into the plane (looking from far as a
circle). This large plane curve is the projection of the required rotating space
curve (having no vertical osculating planes) to the horizontal plane.

The periodic rotating curve will be constructed from the elements of two
types : the
regular helical curves (whose projection to the  horizontal plane will be almost
circular arcs) and from the small inflection segments, attached to the helical
curves at the curvature maxima points on the segments. The smallness of the
inflection segments will be controlled by a parametre $\epsilon$. Thus, the
projection of the rotating curve to the horizontal plane will consist of the
alternating long arcs of approximate circles of radius $1$ and short arcs of
approximate sinusoids of length of order $\epsilon$. The rotating curve will be
constructed as a space perturbation of this horizontal curve (of the form
$z = \mu
h$, where $\mu$ is a small parametre whose value will be choosen after the
choice of
the small value of $\epsilon$).
The space perturbation will shift slightly the horizontal projection of the
curve,
but this shift will be of order $\mu^2$ (and hence we may neglect it in the
calculations of the first order terms in $\mu$ of perturbation theory).

I am starting from the general formulae of the space perturbation theory of any
plane curve
$$ \Bigl(x, \  y = g(x), \quad z = \mu h (x) \Bigr), \quad \mu \rightarrow
0 \ . $$
Direct calculations lead to the following

\vspace{4mm}
\noindent \underline{\bf Lemma 1}. {\it The curvature and the torsion of an
almost
plane curve are equal, in the first approximation, to
$$ k = | g^{''} | / v^3 + O (\mu^2), \ \kappa = \mu (h^{'''} g^{''} -
h^{''} g^{'''} )
(g^{''})^{-2} + O (\mu^3) \ , $$
where $v^2 = 1 + (g')^{^2}$.

The Frenet accompagnying frame of an almost plane curve is given by the formulae
$$\tau = (A, B, \mu C), \quad n = s(P, Q, \mu R), \quad b= s (\mu U, \mu V,
W) \ , $$
where the coefficients are given, with an error at most $O (\mu^2)$, by the
formulae
$s = $ sign $ (g^{''})$,
$$\begin{array}{lll}
 A  =  1 / v, &  B =  g' / v, & C =  h' / v \ ; \\
	P  =  -g'/v, &  Q =  1/v, & R = (h^{''} v^2 - h'g' g^{''} / (|g^{''} | v) \\
 U  =  (h^{''} g' - h' g^{''} ) / (g^{''}), & V = -h^{''} / (g^{''}), & W =  1 \
. \\
\end{array}$$ }

\vspace{4mm}
\noindent \underline{Remark}.  In these formulae $g^{''} \not = 0$, and the
frame
$(\tau, n , b)$ orients the space as the coordinate system $(x, y, z)$, the
normal
direction being choosen by the condition that the curvature $k$ should be
positive.
The Frenet formulae have the usual form
$$ \dot \tau = kn, \quad \dot n = - k\tau + \kappa b, \quad \dot b =
-\kappa n \ . $$
At the crossing of a simplest inflection point (of type ($1 < 3< 4$)) the frame
jumps, turning at an angle $\pi$ around the tangent direction $\tau$ (this
jump is
reflected by the appearance of the sign $s$ in the formulae for the unit
vectors $n$
and $b$).

To everse  the rotating curve I shall use the following special curve
having a point
of the ordinary inflection $X = 0$ :
\begin{equation} (X, \quad  Y = \sin X, \quad Z = \xi^2 - \sin^2 \xi) \ ,
\end{equation} where $\xi = \lambda X,  0 < \lambda < 1,  0 \leq X < \pi/2 \ . $

\vspace{4mm}
\noindent \underline{{\bf Lemma 2}} . { \it This curve is  a rotating
curve}.
 At the point $X = 0$ the curve has an ordinary inflection point (described by
the symbol $(1 < 3 < 4)$). At the points where $X \not = 0$ the derivative
of the
ratio of the second derivatives of the components with respect to $X$ does not
change the sign :
$$g^{''} / g^{'''} \ \not =  \ H^{''} / H^{'''} \ , $$
where $g = $sin $ X$, $H(X) = \xi^2 - \sin^2 \xi$. Indeed,
$$g^{''} / g^{'''} \ = \ {\rm tg} \ X, \quad H^{''} = 4 \lambda^2 \ \sin^2
\xi, \quad
H^{'''} = 4 \lambda^3 \sin 2 \xi \ , $$
whence $H^{''} / H^{'''} = ({\rm tg}  \ \lambda X) / (2 \lambda)$. But for
the values
of $X$ belonging to the interval $(0, \pi/2$) and for the values of $\lambda$ ,
belonging to the interval $0 < \lambda < 1$, we have tg $\lambda X <
\lambda $ tg $
X$. Hence on our interval tg $X > $ tg $(\lambda X) / (2 \lambda)$ and therefore
curve (3) is a rotating curve.

Now we shall translate curve (3) and shall shrink it to get a curve whose
projection
on the $(x,y)$ plane is tangent to the circle $x = \sin \varphi$, $y = \cos
\varphi$
at the point $\varphi = 0$ having there the same curvature as the circle :
\begin{equation} x = \epsilon (X - \pi/2), \quad y = \epsilon^2 Y +1-\epsilon^2,
\quad z = \mu Z \ . \end{equation}
Lemma 1 implies that this translated curve (4) has a curvature maximum
point (whose
critical value differs at most by $O (\mu^2$) from $1$) in a small
neighbourhood of
the point $X = \pi/2$ (that is $x = 0$, $y= 1$, $z = 0$), provided that $\mu$ is
small.

Choose a helical line having at this point the same $3$-jet (that is, the same
acompagnying frame, curvature and torsion) as the curve (4). To choose such a
helical curve is possible just because we have choosen the curvature
maximum point : the $3$-jet takes also into account the curvature
derivative, which is
vanishing for the helical curve as well as for curve ($4$) at the curvature
maximum
point.

Lemma 1 provides the formulae for the parameters of this helical line in first
approximation with respect to $\mu$. For $\mu = 0$ curve $(4)$ is a plane
curve. Its
curvature centre (at point $(0,1)$ on the curve) is $(0, 0)$. Therefore the
required
helical line lies on a cylinder which is close to the cylinder $x^2 + y^2 =
1$ whose
axis forms an angle of order $\mu$ with the $z$ axis direction and whose
radius is
close to $1$ (its difference from 1 being at most $O(\mu^2)$ according to
Lemma 1).
The torsion of curve (4) at the connection point is of order $\mu$. Hence the
helical line, $3$-connected to curve (4) at the curvature maximum point, may be
obtained (up to the perturbations of order $\mu^2$) from the circle ($x^2 +
y^2 =
1$, $z = 0$) by the following three perturbations of order $\mu$ : 1) a vertical
shift along the $z$-axis ; 2) an inclination of the cylinder axis at an angle of
order $\mu$ to the $z$-axis ; 3) the replacement of the circle $z = 0$ by the
helical curve $z = \kappa \varphi$, the constant $\kappa$ being
proportional to $\mu$
and being the torsion of curve (4) at point $x = 0$.

These three operations may be performed independently, and the
result\nolinebreak ing
perturbations of order $\mu$ should be simply added to each other (since we
neglect
the nonlinear effects of order $\mu^2$). The inclination of the cylinder axis
transforms the circle into an ellipse, but the shortening of the smaller
axis is of
the order of $\mu^2$ only. Therefore we may neglect this inclination
calculating the
corrections of order $\mu$ to the coordinates $x$ and $y$.

It results that the helical line having the third order contact with curve
(4) at
the curvature maximum point has the equations of the form (neglecting
$O(\mu^2)$)
$$x \ = \ \sin \varphi, \quad y \ = \ \cos \varphi, \quad z = \mu (u_1 +
u_2 + u_3) \
, $$
where $\mu u_1$ --- the influence of the shift of the contact point along
the $z$ axis
\nolinebreak --- is a constant, independent of $\varphi$, $\mu u_2$ denotes
the effect
of the cylinder axis inclination and $\mu u_3 = \kappa \varphi$ being the
effect of
the torsion of curve (4) at the contact point.

To calculate $u_2$, write the equation of the plane, containing the origin and
inclinated at an angle of order $\mu$ to the horizontal plane $z = 0$, in
the form $z
= \mu (ax + by)$. We obtain, substituing $x = \sin \varphi$ and $y = \cos
\varphi$,
the perturbation of the $z$ coordinate due to the cylinder axis
inclination,  to be
$$ u_2 = a \sin \varphi + b \cos \varphi$$
 (up to the corrections of order $\mu^2$).
Taking all the three perturbations into account, we get finally the third order
contact conditions in the following form :

\vspace{4mm}
\noindent \underline{\bf Lemma 3}. {\it The helical curve, which has a
third order
contact with curve (4) at its curvature maximum point, close to $(x = 0$,
$y = 1$,
$z = 0)$, is defined (up to the corrections of order $\mu^2$) by the parametric
equations
$$x = \sin \varphi, \quad y = \cos \varphi, \quad z = \mu F (\varphi) \ , $$
where $F (\varphi) = c + a \sin \varphi + b \cos \varphi + d \varphi$, the
coefficients being related to the values of function $Z = h(x)$ in (4) and
of its
derivatives at point $x = 0$ by the formulae
$$c = h_0 + h^{''}_0, \quad a = h'_0 - h^{'''}_0 , \quad b = -h^{''}_0,
\quad d =
h^{'''}_0 \ . $$ }

\noindent \underline{Proof}. The values of the coefficients result from the
comparision of the Taylor series of $h(x)$ and of $F$ (arc $\sin x$) at point
$0$ taking into account that $\varphi = x + x^3 / 6 + ...)$.

The normal to the perturbed helical curve is generated by the vector of the
second
derivatives with respect  to $\varphi$ (up to correction of order $\mu^2$),
$(- \sin
\varphi$, $-\cos \varphi$, $\mu F^{^{..}})$. We deduce that on the helical
curve there exists a point, where this normal is horizontal (perpendicular
to the
$z$-axis). This point is defined (in first approximation) by the equation $
F^{^{..}} = 0$, that is $a \sin \varphi + b \cos \varphi = 0$. Substituting the
values of the coefficients $a$ and $b$ from Lemma 3 and returning from function
$h(x)$ to function $H(X) = h(\epsilon (X - \pi/2))$, we get the normal
horizontality
condition in the form
\begin{equation}
{\rm tg} \ \varphi_0 \ = \ - \epsilon h^{''}_\ast \ \Bigl( H^{'''}_\ast -
\epsilon^2
H'_\ast \Bigr) \ , \end{equation}
where the derivatives are derivatives with respect to the variable $X$ and
the stars
mean the evaluation at $X = \pi/2$. (The author is greatfull to F. Aicardi
who has
pointed an error in the original version of formula (5)).

\vspace{4mm}
\noindent \underline{\bf Lemma 4}. {\it Equation $H^{'''}_\ast = \epsilon^2
H'_\ast$
has a smooth solution $\lambda = \lambda_\ast (\epsilon)$, $\lambda_\ast
(0) = 1$,
$\lambda_\ast (\epsilon) < 1$ for small $\epsilon > 0$. }

Indeed, $H^{'''}_\ast = 4 \lambda^3$ $\sin (\pi \lambda) > 0$ and $H'_\ast =
\lambda^2 \pi - \lambda \sin (\pi \lambda) > 0$, while $(d / d\lambda)
(H^{'''}_\ast)
< 0 $ at $\lambda = 1$.

Thus choosing $\lambda (\epsilon)$ we may get $\varphi_0 = \pi /2$.

The fact we really need is the inequality $\varphi_0 < 5 \pi /6$ for
sufficiently
small values of $\epsilon$. We fix $\varphi_0$ and $\lambda (\epsilon)$.

Now we are ready to construct the periodic space curve. We start from a segment
$\Gamma$ of this curve, consisting of curve (4) and of the helical curve
third order
tangent to it at the curvature maximum point, close to the point $x = 0$.
We include
only the arc of this helical curve bounded by the point on the horizontal
normal $N$
(close to $\varphi = \varphi_0)$. The value of the parametre $\epsilon$ is
choosen
(after the choice of $\lambda$) to be so small, that the angle $\varphi_0$
is close
to $\pi/2$. Then the value of parametre $\mu$ is choosen so small (even with
respect to $\epsilon^3$, in fact) that the error of our first approximation
in $\mu$
will be small.

The helical curve is strictly symmetric with respect to any of its normals, for
instance with respect to $N$ (the reflection $A$ in this normal $N$ maps
the helical
curve onto itself, reversing its orientation). Consider the binormal $B$ to
curve
(4) at the inflection  point $X = 0$ (this binormal is the vertical line $x = -
\epsilon \pi /2$, $y = 1-\epsilon^2)$. Curve (4) ($|X| < \pi/2)$ is sent exactly
onto itself by the reflection $C$ in the binormal $B$ (reversing the curve
orientation). Lines $N$ and $B$ are strictly orthogonal.

Add to the segment $\Gamma$ the symmetric segments $A \Gamma$ and $C \Gamma$
reflecting $\Gamma$ in the bounding normal $N$ and binormal $B$. The segment $A
\Gamma$ is bounded by an inflection point with a vertical binormal $AB$. The
reflection $C'$ in this binormal sends the segment $A \Gamma$ onto the
segment $C' A
\Gamma$, bounded by a point on the normal $C' N$. This normal is horizontal
and is a
symmetry axis of the helical curve contained in the segment $C'A\Gamma$. The
reflection $A'$ in the axis $C'N$ sends the segment $C'A \Gamma$ onto the
segment
$A' C' A \Gamma$, bounded by a point on the vertical binormal $A' AB$.

The four segments $(\Gamma,, \ A \Gamma, \ C'A \Gamma, A' C' A \Gamma)$ form
together one period of the periodic curve that we are constructing. The minimal
period is the translation $A' A$ along the horizontal $(x, y)$-plane,
sending $B$
onto $A' AB$.

The projection of the resulting curve to the horizontal plane is formed by a
sequence of arcs, which are close to the arcs of the circles of radius $1$,
supporting the angles of equal magnitude $2 \varphi_0$ and almost tangent
to each
other at a point, close to the place where the binormal $(B, AB, ...)$ is
vertical.
More precisely, the almost circular arcs are connected by small inflection
segments
(of lengths of order $\epsilon$). Our rotating periodic curve is a chain of such
tangent arcs, whose projections are convex alternatively toward the
directions of
positive and of negative $x$. The minimal period is close to the pair of tangent
circular arcs supporting the angles $2 \varphi_0$.

If $\varphi_0 < 5 \pi / 6$, all the circular arcs in such a chain are not
intersecting each other. Hence for the sufficiently small values of parametre
$\epsilon$ the periodic curve that we have constructed has a smoothly embedded
projection to the horizontal plane.

Our construction provides a curve of class $C^3$ (we were controlling only the
continuity of the derivatives up to order $3$). But a standard smoothening
allow us
to make our rotating curve as smooth as we wish (even analytic). We conclude
therefore that the following results holds :

\vspace{4mm}
\noindent \underline{\bf Theorem 5}. {\it There exists a smooth closed
nonselfintersecting curve in the plane $(x, y)$, over which on a suitable
hyperbolic
surface $z = f(x,y)$ lives a closed asymptotic curve (a limit cycle of the
asymptotic directions field of the surface).}

\vspace{4mm}
\noindent \underline{\bf Theorem 6}. {\it The set of the embedded (immersed)
rotating curves, equipped with their osculating planes fields, is
$C^0$-dense in the
space of the embedded (immersed) smooth curves in $\BbR P^3$, equipped with
continuous
fields of tangent planes. }


\vspace{8mm}
\hfill CEREMADE, UMR CNRS ${\rm N}^o \ 7534$

\hfill Universit\'e Paris-Dauphine

\hfill Place du Mar\'echal de Lattre de Tassigny

\hfill 75775 PARIS CEDEX 16 (FRANCE)

\smallskip
\hfill and

\smallskip

\hfill Steklov Mathematical Institute

\hfill 8, Gubkina Street

\hfill 117966 MOSCOW,  GSP-1 , (RUSSIA)


























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