%From: "Sergei Anisov" <anisov@mccme.ru>
%Date: Thu, 11 Feb 1999 09:00:53 GMT
%To: duzhin@botik.ru
%Subject: Paper in MIRAN Vol.221

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\topmatter
\title Convex Curves in $\RP^n$ \endtitle
\author S.~S.~Anisov \endauthor
\translator the author\endtranslator
\address Mathematical College of the Moscow Independent University \endaddress
\email anisov\@mccme.ru, anisov\@mech.math.msu.su \endemail
\thanks This research was supported by the ISSEP under grant~No.~a96-537 and by
the Russian Foundation for Basic Research under grant No.~95-01-00846.
\endthanks
\toc
\widestnumber\head{\S7}
\specialhead{} Introduction \endspecialhead
\head \S1. Preliminaries \endhead
\head \S2. Main results and methods \endhead
\head \S3. Proof of Theorem 7 \endhead
\head \S4. Topology of the space of convex curves \endhead
\head \S5. Generalizations of the concept of convexity \endhead
\head \S6. Structure of the sets $M_k(\G)$ \endhead
\head \S7. Convex projections of convex curves \endhead
\endtoc
\endtopmatter

\document
	\head Introduction. \endhead
Convex curves, that is, curves in $\RP^n$ intersecting any hyperplane at most
$n$ times, arise between singularity theory, projective geometry, differential
geometry, and theory of Chebyshev systems of functions. Author's research was
stimulated by V.~I.~Arnold's paper~\cite4, which is devoted to the following
theorem: if $\G$ is a curve in $\RP^{n+1}$ such that its projection $\g=p_A(\G
)$ (where $A\notin\G$) is a convex curve in $\RP^n$, then there are at least $n
+2$ different flattening points on~$\G$. (Recently A.~O.~Viro proved the
converse theorem~\cite9: for a nonconvex curve $\g\subset\RP^n$ there exists
its lifting with less than $n+2$ flattening points.)

Many questions and conjectures concerning convex curves are stated in
~\cite{4,~5}. This paper contains answers for most of them. In \S3 we prove
that any convex curve $\g\subset\RP^n$ is a projection $p_A(\G)$ of some convex
curve $\G\subset\RP^{n+1}$ passing through $A$; here $A$ is an arbitrary point
in $\RP^{n+1}\setminus\RP^n$. This enables us to find the homotopy type of the
space of convex curves in $\RP^n$ (which is a subset of $C^n(S^1,\RP^n)$): it
is homotopy equivalent to the orthogonal group $\SO(n+1,\R)$ or to the group
$\PSO(n+1,\R)$, depending on the parity of $n$ (see \S4; before M.~Shapiro has
proved~\cite{19} that this space is connected). In \S7 we prove that for a
convex curve in even-dimensional Euclidean space there exists a subspace of
codimension~2 such that the projection of the curve on this subspace also is
convex. To this end, it is necessary to study the structure of the sets $M_k(\G
)$ of the hyperplanes that intersect $\G$ exactly $k$ times (see \S6). As a
by-product, this implies that any Chebyshev system of functions on the circle
contains (up to linear transformation, normalization, and reparametrization)
the functions $\sin\f$ and $\cos\f$ simultaneously. In \S5 we consider various
generalizations of the concept of convexity. A more detailed review of
principal results can be found in \S2; the following sections contain the
proofs. Definitions are collected in \S1, which also contains two isolated
theorems that are not used below: Theorem~4 describes the convex hull of a
convex curve, and Theorem~5 shows that a convex curve cannot be longer than $n
\pi$.

We consider only closed convex curves. Nonclosed convex curves were studied in
M.~Shapiro's thesis~\cite{19}. Intersections are counted with multiplicities,
unless explicitly stated otherwise. Some of the results of this paper were
announced in~\cite1.

The author is grateful to V.~I.~Arnold, S.~M.~Gusein-Zade, M.~\'E.~Kazarian,
A.~G.~Khovanskii, M.~Z.~Shapiro, V.~A.~Vassiliev, and A.~O.~Viro for useful
discussions and for their sincere interest in my work, and to V.~D.~Sedykh for
a number of useful remarks.

	\head \S1. Preliminaries. \endhead
\definition\nofrills{Definition}\ \cite4. Let $\f\pmod{2\pi}$ be a parameter
on~$S^1$. Consider a smooth embedding $\G\:S^1\to\RP^n$. A curve $\G(\f)$ is
called {\it convex\/} if it intersects any hyperplane no more than $n$ times,
taking multiplicities into account.			\enddefinition

Here we consider intersections of order up to~$n$. So we assume that $\G\in C^n
(S^1,\RP^n)$. Further, a transversal intersection of a parameterized curve $\G(
\f)$ and a hyperplane can have multiplicity higher than one if $\frac d{d\f}\G(
\f)=0$. However, the last condition cannot hold for an embedded curve $\G(\f)$.

\example{Examples} 1.~$\G(\f)=(1:t:t^2:\ldots:t^n)$, where $t=\tg\frac\f2$, is
a convex curve in $\RP^n$;

2.~$\G(\f)=\big(\sin\psi:\cos\psi:\sin3\psi:\cos3\psi:\ldots:\sin(2n+1)\psi:
\cos(2n+1)\psi\big)$, where $\psi=\f/2$, is a convex curve in $\RP^{2n+1}$;

3.~$\G(\f)=\big(1:\sin\f:\cos\f:\sin2\f:\cos2\f:\ldots:\sin n\f:\cos n\f\big)$
is a convex curve in $\RP^{2n}$;

4.~The boundary of a strictly convex bounded domain in $\R^2$ is a convex curve
in $\RP^2=\R^2\cup l$, where $l$ is the infinite line.		\endexample

A complex analog of convex curves is furnished by curves of degree $d$ in $\CP^
d$ \cite{16, Chapter~5}, \cite{11, Chapter~2}.

Let $A_1,\dots,A_k$ be different points of $\G(\f)$, and let $m_1,\dots,m_k$ be
natural numbers such that $m_1+\ldots+m_k=n$. Then there exists a hyperplane
that crosses $\G$ at the points $A_i$ with multiplicities $m_i$. For a convex
curve $\G$, this hyperplane is unique and, by virtue of convexity, there are no
other intersection points (indeed, if there are two hyperplanes with this
property, then we can draw a hyperplane through their $(n-2)$-dimensional
intersection and an arbitrary point of $\G$ different from all $A_i$; this
hyperplane crosses $\G$ more than $n$ times). In particular, for $k=1$ and $m_1
=n$, we get the {\it osculating hyperplane\/} $\Pi(\f)$ of $\G$ at $A=\G(\f)$.

The following Theorems~1 and~2 are well known.

\proclaim{Theorem 1} A convex curve in odd-dimensional projective space
intersects any hyperplane.					 \endproclaim

Let us recall that a subset of projective space is {\it affine\/} if it does
not intersect a certain ``infinite'' hyperplane.

\proclaim{Theorem 2} A convex curve in even-dimensional projective space is
affine.								\endproclaim

Denote by $M_k(\G)\subset(\RP^n)^*$ the set of hyperplanes that intersect $\G$
exactly $k$ times, counting multiplicities.

\proclaim{Theorem 3} Let $n$ be an even number and $\G$ be a convex curve in
$\RP^n$. Then the set $M_0(\G)$ is contractible. Moreover, it is affine and
even convex.							\endproclaim

Theorem 1 follows from the fact that the intersection index of $\G$ and a
hyperplane in $\RP^{2n+1}$ is odd, and therefore nonzero. A plane required for
Theorem~2 is obtained by a small perturbation of an osculating hyperplane.

\demo{Proof of Theorem $3$} By Theorem~2, the set $M_0$ is not empty. Let $\Pi$
be a plane from $M_0$. Regard $\Pi$ as the infinite plane. Then $\G$ is
contained in the affine chart $\P\setminus\Pi$. We can assume that the origin
lies on~$\G$. Then a hyperplane $\Pi'\in M_0$ is determined by an equation
	$$\sum a_ix_i=d,\qquad\sum a_i^2=1,\quad d>0\tag1$$
(here $d\ne0$, because the origin lies on $\G$ and thus cannot belong to $\Pi'
\in M_0$). Let a hyperplane $\Pi'(t)$ be determined by the equation
	  $$\sum a_ix_i=d\(1+\tg\frac{\pi t}2\).\tag2$$
Formula \eq(2) describes a retraction of the set~$M_0$ to its point~$\Pi$. In
fact, for $t=0$ we have the equation of~$\Pi'$, while for $t=1$ relation~\eq(2)
becomes the ``equation'' $\sum a_ix_i=\infty$, which determines~$\Pi$.
Moreover, $\Pi'(t)$ depends on $t$ and $\Pi'$ continuously. Thus $M_0$ is a
contractible set. Further, hyperplanes passing through the origin (or through
any other point of $\G$) form a hyperplane in $\P^*$ that does not intersect~$M
_0$. Now it follows that $M_0$ is an affine set.

Let $\Pi_1$ and $\Pi_2$ be hyperplanes. By $T=\Pi_1\cap\Pi_2$ denote their
intersection, $\dim T=n-2$. The pencil of hyperplanes in $\P$ that pass through
$T$ forms a certain line $l$ in $\P^*$. This line passes through points $\Pi_1$
and $\Pi_2$ of $\P^*$. Let $O$ be a point in $\P\setminus T$ (the origin). Let
us regard the hyperplane in $\P^*$ dual to $O$ as the infinite hyperplane. Draw
the hyperplane $\pi$ through $T$ and $O$. Then $\pi\in l$ is the intersection
point of $l$ and the infinite plane in $\P^*$. Then the segment with the
endpoints $\Pi_1$ and $\Pi_2$ in $\P^*$ is the part of $l$ that does not
contain $\pi$, i.e., does not cross the infinite hyperplane.

It follows from eq.~\eq(2) that $M_0$ contains the segment with the ends $\Pi'$
and $\Pi$. In fact, all the $\Pi'(t)$ pass through $T=\Pi'\cap\Pi$, because
they are parallel to $\Pi'$ in the chosen affine chart and, moreover, $\Pi'(0)=
\Pi'$, $\Pi'(1)=\Pi$, and planes $\Pi'(t)$ do not pass through the origin. Note
that for $\Pi'$, see \eq(1), we can take any point of $M_0$; a point $\Pi$ also
is an arbitrary point of $M_0$. Thus equation \eq(2) implies that any segment
$\Pi\Pi'$ lies in $M_0$ whenever $\Pi$ and $\Pi'$ belong to~$M_0$. Therefore,
$M_0$ is a convex set. This completes the proof.		\qed\enddemo

Note that we use in this proof only the fact that $\G$ is a nonempty affine
set. The sets $M_k(\G)\subset\P^*$ with $k>0$ are studied in \S6 below.

Choose an affine chart so that $\G(\f_0)=0$. Consider vectors $v_j=\left.\frac{
d^j}{d\f^j}\G(\f)\right|_{\f=\f_0}$, where $j=1,\dots,n$. If they are linearly
independent, $\G(\f_0)$ (or simply $\f_0$) is called a {\it regular point\/} of
$\G$; otherwise it is called a {\it flattening point}.

\definition{Definition} A {\it regular\/} (or {\it nondegenerate\/}) {\it 
curve\/} is a curve having no flattening points.		\enddefinition

Any convex curve is regular. The converse statement is not true: if we pass a
convex curve twice, we get a regular curve that is not convex. The heating
fiber of an electric lamp also forms a regular but not convex curve.

Still an osculating hyperplane of a convex curve intersects it only at $\G(\f_0
)$. Curves with this property are called {\it locally convex curves}.
V.~I.~Arnold proposed a question~\cite4 whether any locally convex curve is
convex. It will be shown in \S5 that the answer is negative for the case of
dimension $n=3$. R.~Uribe constructed examples of locally convex but not convex
curves in Euclidean space of any even dimension $n>2$.

An osculating hyperplane is defined at an isolated flattening point $\G(\f_0)$
provided that the vectors $v_1,\dots,v_{n-1}$ are linearly independent (we say
that these points are {\it nondegenerate\/}). This hyperplane contains $v_n$
and coincides with the limit of hyperplanes $\Pi(\f)$ as $\f\to\f_0$. If $\Pi(
\f_0)$ contains $v_1,\dots,v_n,v_{n+1},\dots,v_{n+k-1}$ and does not contain $v
_{n+k}$, then $k$ is said to be the {\it order}, or the {\it multiplicity\/} of
flattening at $\G(\f_0)$. The multiplicity of intersection of $\G$ and $\Pi(\f_
0)$ at flattening point of order $k$ equals $n+k$. The order of a nonisolated
flattening point is infinite.

\subhead Convex curves and their connections with Chebyshev systems of
functions and disconjugate differential equations		\endsubhead

\definition\nofrills{Definition}\ \cite{15}. A system of functions $f_0,f_1,
\dots,f_n$ with the same domain is said to be a {\it Chebyshev system\/} if any
linear combination $\sum\lambda_if_i$ has at most $n$ zeroes according the
multiplicities whenever there is a nonzero~$\lambda_i$.		\enddefinition

Only the segment and, for even $n$, the circle can carry a Chebyshev system.
For example, $1,\allowmathbreak x,x^2,\dots,x^n$ is a Chebyshev system on the
segment, and $f_0=1$, $f_{2k-1}=\sin kx$, $f_{2k}=\cos kx$, where $k=1,2,\dots,
n/2$, is a Chebyshev system on the circle. The case of the segment was studied
by M.~Shapiro in~\cite{19}.

Consider the curve $t\mapsto\(f_0(t):f_1(t):\ldots:f_n(t)\)$ in $\RP^n$. Here
$t$ is the parameter on the segment or (for $n$ even) on the circle. This curve
is convex if and only if $f_0,f_1,\dots,f_n$ is a Chebyshev system of
functions. Indeed, zeroes of $\sum\lambda_if_i$ correspond to intersections of
the curve and the hyperplane $\sum\lambda_ix_i=0$. The following
transformations preserve the Chebyshev property of systems of functions as
well as the convexity of corresponding curves:
  \roster
  \item"{1.}" Changes of the parameter: $t\mapsto\tau(t)$, where $\tau$ is a
		diffeomorphism.
  \item"{2.}" Multiplication by nonvanishing functions: $f_i\mapsto gf_i$,
		where $g$ does not vanish.
  \item"{3.}" Linear transformations:
	     $$\(\matrix f_0\cr \vdots\cr f_n\cr\endmatrix\)\mapsto B
	       \(\matrix f_0\cr\vdots\cr f_n\cr\endmatrix\),
	       \qquad B\in \SL(n+1,\R). 			$$
  \endroster
Transformations 1 and 2 here do not change the curve $\G\subset\RP^n$ at all,
and transformation~3 can be considered as a projective transformation of~$\RP^
n$. Transformations 1--3 preserve the Young diagram of a system of
functions~\cite{14}; the Young diagram expresses the degeneracy of a flattening
point of a curve $(f_0:f_1:\ldots:f_n)$ \cite{7, Chapter.~3, \S4}.

In the case of odd-dimensional projective space, convex curves correspond to
systems of $2\pi$-periodic odd functions $f_0,f_1,\dots,f_n$ (a function $f_i$
is odd if $f_i(x+\pi)=-f_i(x)$) such that any nonzero linear combination of
them has at most $n$ zeroes at the semiperiod. These systems can be regarded as
systems of sections of the M\"obius bundle over~$S^1$, see~\cite6 and
\cite{4,~\S10}.

\definition\nofrills{Definition}\ (see~\cite{19} or~\cite{7, Chapter~3, \S5}).
A linear $n$\snug th order differential equation
       $$ L(t)x=x^{(n)}+a_1(t)x^{(n-1)}+\ldots+a_n(t)x=0\tag3$$
on the segment or on the circle is said to be {\it disconjugate\/} if any its
nonzero solution has no more than $n$ zeroes, taking multiplicities into
account.							\enddefinition

For the circle, we consider only equations with the identical (or scalar)
monodromy operator.

The curve
	$$\G(t)=(f_1(t):\ldots:f_n(t))\subset\RP^{n-1}$$
is assigned to an equation of this type, where $f_1(t),\dots,f_n(t)$ is the
fundamental system of solutions. The condition that equation \eq(3) is
disconjugate is equivalent to convexity of $\G(t)$, and also to the Chebyshev
property of the system $f_1,\dots,f_n$. The transformations~3 (see above) act
transitively on the set of fundamental systems of solutions of~\eq(3).

\subhead Convex hulls of convex curves \endsubhead

Here we prove Theorem~4 (see below), which shows that convex curves are very
specific objects. This theorem will not be used below; however, the idea of its
proof will be used in the proof of Lemma~5, see~\S7.

Recall that the {\it convex hull\/} $\conv M$ of a subset $M$ of $\R^n$ is the
smallest convex set that contains~$M$. It is easy to show that $\conv M$ is the
union of all $n$-dimensional simplices with vertices in~$M$. In some
situations, the dimension of involved simplices can be reduced. For example, if
$M$ is a connected set, then it suffices to take simplices of dimension $n-1$,
the author is grateful to O.~F.~Kryzhanovskii for this remark. The proof is
based on the following idea. Let $A$ be an interior point of a simplex with
vertices $A_0,A_1,\dots,A_n$ that belong to $M$. A simplex of this type exists
for any $A\in\conv M$. If $A$ lies on the boundary of this simplex, then our
statement holds. Consider all the $(n-1)$-dimensional simplices containing $A$
that have all vertices except for one in the set $\{A_0,A_1,\dots,A_n\}$. By
$N$ denote the locus of the remaining vertices of these simplices, see Fig.~1.
If the set $M\cap N$ is nonempty, then we get an $(n-1)$-dimensional simplex
with the vertices in~$M$ that contains~$A$. Otherwise $N$ separates the
connected components of points $A_i$ of the set~$M$.

Hence, for a connected set $M$ the dimension of the simplices that constitute
$\conv M$ can be reduced by~1. However, if $M$ is a convex curve in $\R^n$ with
even $n$, $n=2k$, then this dimension can be divided by~2\,!

\proclaim{Theorem 4} Let $\G$ be a closed convex curve in $\R^n$, and $n=2k$ be
an even number. Then $\conv\G$ is formed by the simplices of dimension~$k$ with
vertices on~$\G$.						\endproclaim

\demo{Proof} By $M$ denote the union of the interiors of the $k$-dimensional
simplices with the vertices on~$\G$, and by $N$ denote the union of the same
but closed simplices. Since the curve $\G$ is compact, we have $N=\ol M$. So
the theorem states that $\conv\G=\ol M$. It is clear that $\ol M\subset\conv
\G$, and it suffices to show that $M$ is a nonempty open set and $\partial M
\subset\partial\conv\G$.

Obviously, the set $M$ is not empty. Now consider a point $A$ of $M$ and a
simplex $\Delta=A_0A_1\ldots A_k$ that contains $A$; its vertices $A_0,\dots,A_
k$ are geometrically different. Then we can write $A=\sum\l_{i=0}^k\lambda_iA_
i$, where the $\lambda_i$ are the barycentric coordinates of $A$; therefore
$\lambda_i>0$ and $\sum\lambda_i=1$. A shift of $A_i$ by a vector $v$ induces
the shift of~$A$ by $\lambda_iv$. Consequently,
	$$T_AM\supset T_A\Delta\oplus\<v_0,v_1,\dots,v_k\>,\tag4$$
where $v_i$ is the tangent vector of $\G$ at $A_i$. If the dimension of the
space in the right-hand side of \eq(4) equals $n$, then $A$ is an interior
point of~$M$, and for local coordinates in the neighborhood of~$A$ we can take
some $k$ of the numbers $\lambda_0,\lambda_1,\dots,\lambda_k$; other $k$
coordinates correspond to $k$ vectors of the set $\{v_0,v_1,\dots,v_k\}$ that
are linearly independent modulo $T_A\Delta$. Now assume that the dimension of
this space is less than~$n$. Then the subspace containing $A_0,\dots,A_k$ and
the tangent vectors of $\G$ at these points is contained in a hyperplane, which
intersects $\G$ at least $(2k+2)$ times. This contradicts the convexity
of~$\G$. Therefore, any point $A$ of the set $M$ is its interior point, that
is, $M$ is an open set.

It is easy to prove that $\partial M=N\setminus M$ is the union of the
$(k-1)$-dimensional simplices with vertices on~$\G$. If $A\in\partial M$, then
$A$ belongs to some simplex $A_0\ldots A_{k-1}$ with vertices on $\G$
(possibly, $A$ lies in the boundary of a simplex of this type). Draw a
hyperplane $\Pi$ that contains $A$ and intersects $\G$ twice at any~$A_i$. Note
that $\G$ lies at one side from $\Pi$. Indeed, there are $2k$ intersections of
$\Pi$ and $\G$ at $A_0,\dots,A_{k-1}$ (counting the multiplicities). So $\Pi$
and $\G$ do not intersect one another anymore. On the other hand, the
intersections at $A_i$ are of multiplicity~$2$, and thus $\G$ does not pass to
the other side from $\Pi$ at these points. Therefore, $\conv\G$ also lies at
one side from $\Pi$. So if $A\in\partial M$, then $A\in\conv\G$ and $A$
belongs to a hyperplane of support of $\conv\G$. In this case $A$ is not an
interior point of $\conv\G$, that is, $A\in\partial\conv\G$. Theorem~4 is
proved.								\qed\enddemo

\remark{Remarks} 1.~For nonclosed curves, the statement similar to Theorem~4 is
contained in~\cite{15, Chapter~3, Theorem~5.1} (with more sophisticated proof).

2.~As a by-product, we have proved again the Schoenberg's result~\cite{21} on
the structure of the boundary of the convex hull of a convex curve lying in
even-dimensional Euclidean space.

3.~Our arguments work for the case of odd dimension as well. For $A\in\G$,
choose an affine chart so that $A$ is the unique infinite point of $\G$. For
example, we can take the osculating hyperplane $\Pi(A)$ as the infinite plane
(the transversality of an intersection at $A$ is extraneous, we only use that
there are no other intersection points). By $\conv_A\G$ denote the convex hull
of $\G$ with respect to this affine chart. Note that $\conv_A\G$ is independent
of the choice of an infinite hyperplane that passes through $A$ and does not
intersect $\G$ anymore. By definition, put $\conv\G=\bigcup\l_{A\in\G}\conv_A
\G$. By the way, this definition makes sense for even-dimensional case as well
(then the multiplicity of intersection of $\G$ and the infinite plane at $A$ is
even); however, in this case for any $A\in\G$ the set $\conv_A\G$ is nothing
but $\conv\G$. The convex curve $\conv\G$ (defined above for a convex curve $\G
\subset\RP^{2k+1}$) is the union of all $k$-dimensional simplices with vertices
on $\G$ (here the same set of vertices determines different simplices for
different affine charts, and it is necessary to consider all simplices that can
be obtained in all admissible charts, i.e., in charts that contain the whole
curve $\G$ except for one point). Further, the boundary $\partial\conv\G$
consists of the $k$-dimensional simplices having two infinite vertices in the
opposite directions (in some affine chart). The proofs are similar to that of
Theorem~4.							\endremark

\subhead Projective duality and wave fronts \endsubhead

Let $M^{n-1}$ be a smooth hypersurface in $\RP^n$. Since all the arguments of
this subsection are local, we can assume that the hypersurface is cooriented.
Its tangent hyperplanes are points of the dual space $\P^*$; they constitute
the {\it dual hypersurface\/} $M^*\subset\P^*$. It turns out that $M^{**}
\subset\P^{**}=\P$ is nothing but the original hypersurface~$M$. One of the
proofs of this statement is based on regarding $M$ as a wave front moving on
the sphere $S^n$ (which is the space of $2$-fold covering over $\RP^n$) with
unit velocity at $t=0$, while $M^*$ is regarded as the same wave front at $t=
\pi/2$, and $(M^*)^*$ is the same front at $t=\pi$. Then $M^{**}\subset S^n$ is
centrally symmetric to $M$, see~\cite2.

Now let us get back to curves.

\definition{Definition} Let $\G$ be a curve in $\P^n$ of finite type (i.e., a
curve with only finite number of flattenings of finite order and without
flattenings of infinite order, see~\cite{22}). Its {\it dual curve\/} $\G^*\!
\subset\P^*$ consists of osculating hyperplanes of~$\G$.	\enddefinition

This terminology suggests that $\G^{**}=\G$. If $n=2$, then $\G\subset\P^2$ is
a hypersurface, and the dual curve is nothing but the dual hypersurface,
because the osculating hyperplane is nothing but the tangent line. For $n>2$,
the coincidence $\G^{**}=\G$ is proved in~\cite{22}. Let us give another proof
of this fact, similar to the proof mentioned above of the fact that $M^{**}=M$.

Consider the lifting of $\G$ to $S^n$. At any regular point $x\in\G$, there is
a marked basis $\{x,e_1(x),\dots,e_n(x)\}$ of $T_x\R^{n+1}=\R\oplus T_xS^n$.
This basis is obtained by the Gram--Schmidt orthogonalization algorithm from
the basis $\left\{x=\G(0),\left.\frac{d^k}{d\f^k}\G(\f)\right|_{\f=0}\right\}$,
$k=1,2,\dots,n$. The osculating hyperplane at $x$ is the span of $e_1(x),\dots,
e_{n-1}(x)$. Draw a geodesic (that is, a great circle) through $x$ in the
direction $e_n(x)$. By $g_t(x)$ denote the point of this circle that lies at
the distance $t$ from $x$. Similarly, by $g_t(\G)$ denote the curve formed by
all points $g_t(x)$, where $x\in\G$. We have $g_{t+2\pi}(x)=g_t(x)$, $g_{2\pi}(
x)=x$, and $g_\pi(x)=-x$. It is easy to show that $\G^*=g_{\pi/2}(\G)$. This
suggests the conclusion that $\G^{**}=g_{\pi/2}(g_{\pi/2}(\G))=g_\pi(\G)=-\G$
in $S^n$ (and thus, $\G^{**}=\G$ in $\P^n$); we have only to check that $e_n(g_
{\pi/2}(x))=\pm g_{\pi/2}(e_n(x))$, where the vector $e_n(g_{\pi/2}(x))$ is
constructed from the curve $\G^*=g_{\pi/2}(\G)$, and $g_{\pi/2}(e_n(x))$
denotes the image of $e_n(x)$ under the parallel transport along $g_t(x)$, $0
\le t\le\pi/2$.

Unlike the previous case, the relation $e_n(g_t(x))=g_t(e_n(x))$ does not hold
for an arbitrary $t$. Indeed, $(e_n(g_t(x)))^\perp$ is a plane such that the
distance between it and $g_t(\G(\f))$ is $O(\f^n)$ as $\f\to0$; we suppose that
$x=\G(0)$. On the other hand, the Frenet formulas imply that the distance
between $(g_t(e_n(x)))^\perp$ and $g_t(\G(\f))$ equals $\pm\frac12\kappa_{n-1}(
x)t\f^2+\dots$ as $t\to0$, $\f\to0$, where $\kappa_{n-1}$ is the $(n-1)$\snug
th twisting coefficient at $x$; it does not vanish provided that $x$ is not a
flattening point of $\G$. Nevertheless, for $t=\pi/2$ the relation $e_n(g_t(x))
=g_t(e_n(x))$ holds. Moreover, there is the following statement.

\proclaim{Lemma} For $0\le i\le n$ we have $f_i=\pm e_{n-i}(x)$ \rom(these
objects are regarded here as vectors in $\R^{n+1}$\rom). Here $f_i=e_i(g_{\pi/2
}(x))$ for $i=1,\dots,n$\rom; further, $f_0=g_{\pi/2}(x)$ and $e_0=x$.
								\endproclaim

Since $g_{\pi/2}(e_n(x))=-x$ (as vectors in $\R^{n+1}$), Lemma implies that $e_
n(g_{\pi/2}(x))=\pm g_{\pi/2}(e_n(x))$, and thus $\G^{**}=\G$.

\demo{Proof} The basis $\{f_0,f_1,\dots,f_n\}$ of $T_y\R^{n+1}=\R\oplus T_yS^n$
(here $y=g_{\pi/2}(x)=f_0$) is obtained by the Gram--Schmidt orthogonalization
algorithm, followed by normalization, from the basis
	$$\left.\frac{d^k}{d\f^k}g_{\pi/2}(\G(\f))\right|_{\f=0},
						\qquad k=0,1,\dots,n.$$
Obviously, we have $g_{\pi/2}(\G(\f))=e_n(\G(\f))$; in particular, $f_0=e_n(\G(
0))=e_n(x)$. Applying the Frenet formulas, we get
	$$\frac d{d\f}g_{\pi/2}(\G(\f))=\frac d{d\f}e_n(\G(\f))=
						-\kappa_{n-1}e_{n-1}.$$
Therefore, $f_1=\pm e_{n-1}$, according to sign of $\kappa_{n-1}$. Further,
	$$\frac{d^2}{d\f^2}g_{\pi/2}(\G(\f))=-\frac d{d\f}\kappa_{n-1}
				e_{n-1}=\alpha e_{n-2}+\beta e_{n-1}+\g e_n,$$
where $\alpha$, $\beta$, and $\g$ are certain coefficients; note that $\alpha=
\kappa_{n-1}\kappa_{n-2}\ne0$. If we orthogonalize the basis $\{f_0=e_n,f_1=\pm
e_{n-1},\,\alpha e_{n-2}+\beta e_{n-1}+\g e_n\}$, we obtain $f_2=\pm e_{n-2}$.
Applying the Frenet formulas again, we get $f_3=\pm e_{n-3},\dots,f_n=\pm e_0$.
In fact, we obtain
	$$\align\frac{d^k}{d\f^k}g_{\pi/2}(\G(\f))=&(-1)^k\kappa_{n-1}\kappa
		_{n-2}\dots\kappa_{n-k}e_{n-k}\\
  &+\{\hbox{a linear combination of\ }e_{n-k+1},e_{n-k+2},\dots\};\endalign$$
the coefficient of $e_{n-k}$ is nonzero provided that $x$ is not a flattening
point of $\G(\f)$. For isolated flattening points of finite order these
arguments can be ``extended by continuity''. Lemma is proved.	\qed\enddemo

\remark{Remark on signs} It follows from the construction of the moving
$n$-hedron that all the $\kappa_i$ (possibly except for $\kappa_{n-1}$) are
positive, while the sign of $\kappa_{n-1}$ can changes only at flattening
points. Consequently, for a convex curve $\G$ we have $f_i=(-1)^ie_{n-i}$; and
in the general case either relations $f_i=(-1)^ie_{n-i}$ or relations $f_i=(-1)
^{i-1}e_{n-i}$ hold for $i=1,\dots,n$.	\endremark

We conclude this section with an example of making use of the projective
duality for estimation the length of a convex curve.

\proclaim{Theorem 5} Let $\G$ be a convex curve in $\RP^n$. Suppose that the
metric on $\RP^n$ is induced by the standard metric on the unit sphere $S^n
\subset\R^{n+1}$. Then we have\rom:
  \roster
  \item"{a)}" $l(\G)\le n\pi$\rom;
  \item"{b)}" $l(\G)\ge\pi$ for $n$ odd.
  \endroster
Here $l(\G)$ denotes the length of $\G$\rom; for $n=1$ we have $l(\G)=\pi$,
while for $n>1$ both inequalities are strict.			\endproclaim

\demo{Proof} Lift $\G$ to $S^n$. For $x\in\G$, set $\Pi(x)=S^n\cap(
\overrightarrow{Ox})^\perp$. Points of $\Pi(x)$ are dual to hyperplanes passing
through $x$. Let $M_k$, where $k=0,1,\dots$, be the set of points of $S^n$ that
belong to exactly $k$ of planes $\Pi(x)$. Since $\G$ is a convex curve, we have
	$$\Pi(x_1)\cap\Pi(x_2)\cap\ldots\cap\Pi(x_{n+1})=\pusto$$
for all $x_1,\dots,x_{n+1}\in\G$. Then $M_{n+1}=M_{n+2}=\dots=\pusto$, and $S^n
=M_0\cup M_1\cup\ldots\cup M_n$. Thus $\sigma_n=\Vol S^n=\sum_{k=0}^n\Vol M_k$
(the sets $M_k$ are measurable, because $M_k=N_k\setminus N_{k+1}$, where $N_i=
M_i\cup M_{i+1}\cup\ldots\cup M_n$, $i=0,1,\dots,n$, are closed subsets of $S^
n$). The Crofton formula~\cite{17, chapter~18} implies that
	$$\sum_{k=0}^\infty k\Vol M_k=\frac{\sigma_n}\pi l(\G).$$
Therefore, $l(\G)=\frac\pi{\sigma_n}\sum k\Vol M_k$, where $k\le n$ and $\sum
\Vol M_k=\sigma_n$. So inequality a) follows. By Theorem~1, we have $M_0=
\pusto$ for $n$ odd. This implies inequality b). Furthermore, it is clear that
$M_n\ne\pusto$. Consequently, $l(\G)>\pi$ for odd $n>1$. Finally, it follows
from Theorems~1 and~2 that $M_0\cup M_1\ne\pusto$. So $l(\G)<n\pi$ whenever $n>
1$. Theorem is proved.						\qed\enddemo

\remark{Remarks} 1.~Recall~\cite{16} that for a curve of degree $d$ in $\CP^d$
the equality $\Vol_1(\G)=d\Vol_1(L)$ holds, where $\Vol_1$ is the
one-dimensional volume and $L$ is the straightline. Thus in our situation a
complex manifold maximizes the volume.

2.~Both estimates of Theorem~5 are sharp. Indeed, since $M_n\ne\pusto$, there
is a linear transformation $A_\e\in\SL(n+1,\R)$ such that the image of $M_n$
occupies more than $1-\e$ of the volume of $S^n$, where $\e$ is an arbitrary
positive number. Then we see that $l(A^*_\e(\G))>(1-\e)n\pi$.	\endremark

All objects defined in this section are projectively invariant.

	\head \S2. Main results and methods. \endhead
Central projection is widely used in the theory of convex curves. Let $A$ be a
point of $\P^n$ and let $X$ be a subset of $\P^n$. Lines in $\P^n$ that pass
through $A$ form projective space $\P^{n-1}$ of dimension $n-1$. The lines that
intersect $X$ form a subset $p_A(X)$ of $\P^{n-1}$. This subset is a {\it
projection\/} of $X$ from~$A$.

\proclaim{Lemma} If $\G\subset\P^n$ is a convex curve and $A\in\G$, then the
projected curve $\g=p_A(\G)$ is convex in $\P^{n-1}$.		\endproclaim

\remark{Remark} If the original curve $\G$ is $C^k$-smooth, $k\ge n$ (see~\S1),
then the mapping $\g\:S^1\to\RP^{n+1}$ also is $C^k$-smooth. Since $\G$ is a
convex curve, its projection $\g$ also is $C^k$-smooth anywhere except for $p_A
(A)$, where $p_A(A)$ is a tangent line to $\G$ at $A$. The smoothness of $\g$
at $p_A(A)$ is $C^{k-1}$.					\endremark

\demo{Proof} If a hyperplane $\beta\subset\P^{n-1}$ and the curve $\g$
intersect one another $k$ times, then the preimage $B=p^{-1}_A(\beta)\subset\P^
n$ and $\G$ have $k+1$ points of intersection, because the multiplicity of
intersection of $B$ and $\G$ at $A$ is one greater than the multiplicity of
intersection of $\beta$ and $\g$ at $p_A(A)$, while all other multiplicities
do not change under projection. Consequently, if $\beta$ intersects $\g$ more
than $n-1$ times, then $B$ intersects $\G$ more than $n$ times, and thus $\G$
is not a convex curve.						\qed\enddemo

\proclaim{Corollary} Let $\G\subset\RP^{2n+1}$ be a convex curve, and $A\in\G$
be its point. Then there exists a hyperplane $B$ that intersects $\G$
transversally at $A$ and nowhere else.				\endproclaim

Recall that it was shown in~\S1 that a convex curve always is affine in
projective space of even dimension, and cannot be affine in odd-dimensional
space.

\demo{Proof} By Lemma, the curve $p_A(\G)=\g\subset\RP^{2n}$ is convex; by
Theorem~2, it is affine. Let $\beta\subset\RP^{2n}$ be a hyperplane that has no
intersections with $\g$. Then $B=p_A^{-1}(\beta)$ is a required hyperplane
in~$\RP^{2n+1}$.						\qed\enddemo

The following theorem describes the boundary of the space of convex curves. It
was used, for example, in~\cite{4,~20}. Nevertheless, it seems to be
unpublished.

\proclaim{Theorem 6} Consider a (continuous in $t$) family $\G_t(\f)$ of $C^{n+
1}$-smooth closed curves in~$\RP^n$. Suppose that all curves $\G_t$ are convex
whenever $t<0$, while $\G(\f)=\G_0(\f)$ is not a convex curve. Then $\G(\f)$
contains a flattening point of multiplicity at least two. Moreover, if a
hyperplane intersects $\G(\f)$ more than $n$ times, then it is the osculating
hyperplane at some point of multiple flattening.		\endproclaim

\remark{Remark} The curve $\G(\f)$ contains no ordinary (that is, non-multiple)
flattening points. Otherwise close curves $\G_t(\f)$ with $|t|$ small enough
also have flattenings, and thus are not convex.			\endremark

\demo{Proof} The proof is by induction on the dimension. The case of curves in
$\RP^2$ is simple; however, we take $n=1$ for the induction base. For this
case, the statement of the theorem becomes obvious: ``Let $C^2$-smooth mappings
$\G_t(\f)\:S^1\to\RP^1$ are diffeomorphisms for $t<0$, while a mapping $\G(\f)$
is not a diffeomorphism. Then there exists a $\f_0$ such that $\G'(\f_0)=\G''(
\f_0)=0$ (although $\G$ is injective). Moreover, if $\G'(0)=0$ for some $\f$,
then $\G''(0)=0$.''

Now suppose that the theorem is proved for curves in $\RP^1$, $\RP^2$, $\ldots,
\RP^{n-1}$. Let us prove it for curves in $\RP^n$. Assume the converse. Then
the curve $\G(\f)$ is not convex and there is a hyperplane $B$ that intersects
$\G(\f)$ more than $n$ times (according to multiplicities) and has at least two
different points of intersection with $\G$. There is a point $A$ of a multiple
intersection: otherwise all curves $\G_t$ close to $\G$ also intersect $B$ more
than $n$ times. We can assume that $A=\G(0)$; moreover, without loss of
generality it can be assumed that $\G_t(0)\equiv A$. Consider the central
projection $p_A\:\RP^n\to\RP^{n-1}$. Put $\g_t(\f)=p_A(\G_t(\f))$. By Lemma,
all curves $\g_t(\f)$ with $t<0$ are convex. The proof of Lemma also implies
that a hyperplane $\beta=p_A(B)$ (which really is a hyperplane, because $A\in
B$) intersects $\g(\f)=\g_0(\f)$ more than $(n-1)$ times. Further, there are at
least two different point of intersection of $\beta$ and $\g$, namely $\g(0)$
(indeed, an intersection of $B$ and $\G$ at $\G(0)$ is multiple) and the
projection of a certain other (i.e., different from $A$) point of intersection.
This contradicts the induction hypothesis.			\qed\enddemo

The following theorem of M.~Shapiro~\cite{20} is often used in this paper.

\proclaim{Theorem} The space of convex curves in $\RP^n$ is connected.
								\endproclaim

In other words, any two convex curves in~$\RP^n$ are homotopic to each other
in the class of convex curves. (Since the space of convex curves is an open
subset of~$C^n(S^1,\RP^n)$ (as well as the space of curves having no flattening
points), we see that connectedness of this subspace is equivalent to its
arcwise connectedness.) Probably this statement is well known. However, I have
not found a reference. Let us outline the proof, following the private
communication~\cite{20}. In~\cite{19}, it was shown that there are only a
little number (2 to 10, according to the parity of $n$ and to the presence of
orientations of a curve and $\RP^n$) of the connected components of the space
of nondegenerate curves in~$\RP^n$. By Theorem~6, a loss of convexity is
accompanied by a birth of (multiple) flattening point. Thus the boundary of the
space of convex curves is contained in the boundary of the space of
nondegenerate curves, and so the space of convex curves is formed by some
connected components of the space of nondegenerate curves. However, all
components of the latter space, except for one of them, contain curves that are
not convex. This completes the proof.					\qed

\remark{Remark} 1.~There is also the following version of the Shapiro theorem:
{\sl for any $k\ge0$ the space of convex $C^{n+k}$-smooth curves in $\RP^n$ is
a connected subspace of $C^{n+k}(S^1,\RP^n)$}. We use this theorem in \S4; its
proof is quite similar to that is described above.

2.~In the Shapiro theorem, we do not distinguish curves that differs only by
the orientation change in~$\RP^n$ (for $n$ odd).		\endremark

\proclaim{Corollary 1} For $n$ even, the space of convex curves in $\R^n$ is
connected \rom(and for $n$ odd, this space is empty by virtue of
Theorem~\rom{1)}.						\endproclaim

Indeed, let $\G_0$ and $\G_1$ be convex curves in $\R^n$. By the Shapiro
theorem, there is a homotopy $\G_t$ connecting $\G_0$ and $\G_1$ in the class
of convex curves in~$\RP^n$. Let $A_t\in\SL(n+1,\R)$ be a family of projective
transformations of $\RP^n$ continuous in $t$ such that $A_0=A_1=E$ and all
curves $A_t(\G_t)$ are affine. The sufficient condition here is that the
infinite plane $\Pi_t$ corresponding to $A_t$ belongs to the set $M_0(\G_t)$ of
planes that do not intersect~$\G_t$. By Theorem~3, the set $M_0$ is arcwise
connected. Thus we can choose $A_t$ so that the previous condition holds. Then
the family $A_t(\G_t)$ is a homotopy between $\G_0$ and $\G_1$ in the class of
convex curves in $\R^n$.						\qed

\remark{Remark} For $n$ even and $k$ nonnegative, the set of convex $C^{n+k
}$-smooth curves in $\R^n$ also is a connected subset of $C^{n+k}(S^1,\R^n)$.
								\endremark

\proclaim\nofrills{Corollary 2}~\cite{4, p.~19}. The curve $\G^*$ projectively
dual to a convex curve $\G$ is convex as well.			\endproclaim

Theorem 7, 8, and 9 stated before are the principal results of the paper. The
proofs are contained in following sections.

Theorem 7 shows that the projection of a convex curve from its point can be in
a sense inverted.

\proclaim{Theorem 7} Suppose that $\g\subset\P^n$ is a $C^{n+1}$-smooth convex
curve and $A\in\P^{n+1}\setminus\P^n$. Then there exists a convex curve $\G
\subset\P^{n+1}$ such that $A\in\G$ and $p_A(\G)=\g$.		\endproclaim

Theorem 7 is proved in \S3. It enables us to find the homotopy type of the
space of convex curves, see Theorem~9 at the end of this section.

The following two propositions concern problems similar to those of Theorem~7.
Recall that a convex curve in even-dimensional space is affine.

\proclaim{Theorem 7$'$} Any $C^{2k+2}$-smooth convex curve $\g\subset\R^{2k}$
is a projection of some convex curve $\G\subset\R^{2k+2}$.	\endproclaim

\demo{Proof of Theorem $7'$} Let $\g$ be a $C^{n+2}$-smooth convex curve in $\R
^n$, where $n$ is an even number. Applying Theorem~7 twice, we get a curve $\G
\subset\RP^{n+2}$ and a line $l$ such that $l$ intersects $\G$ two times and $p
_l(\G)=\g$. It can be assumed that $l$ is the tangent line of $\G$ at $A$ (that
is, two points of intersection of $l$ and $\G$ coincide). Choose an affine
chart so that the subspace $\R^n$ containing $\g$ has the equations $x_{n+1}=x_
{n+2}=0$, while $l$ is the infinite line in the coordinate plane $Ox_{n+1}x_{n+
2}$. By $p(\G)$ denote the projection of $\G$ on this plane. The curve $p(\G)$
tangents $l$ at one point and does not intersect it anymore. So there exists a
slight move that takes $l$ to the line having no common points with $p(\G)$; in
other words, there exists a projective transformation $x_{n+1}\mapsto f_1(x_{n+
1},x_{n+2})$, $x_{n+2}\mapsto f_2(x_{n+1},x_{n+2})$ (where $f_1$ and $f_2$ are
linear fractional functions with the same denominator) that transforms $p(\G)$
to an affine curve. Extend this transformation to $\RP^{n+2}$ by setting $x_i
\mapsto x_i$ for $i=1,2,\dots,n$. This transformation (which is not a
projective one) takes $\G$ to a curve $\G'\subset\R^{n+2}$ such that its
projection on $\R^n$ is $\g$. If the transformation $x_{n+1}\mapsto f_1$, $x_{n
+2}\mapsto f_2$ is sufficiently close to identity, then $\G'$ is a curve
sufficiently close (in $C^{n+2}(S^1,\RP^{n+2})$) to $\G$, and thus $\G'$ is a
convex curve. The proof is complete.				\qed\enddemo

\proclaim\nofrills{Theorem 8} \ \rom(V.~I.~Arnold's conjecture\cite5\rom). Any
convex curve $\G\subset\R^{2k}$ has a convex projection $\g\subset\R^{2k-2}$ on
$\R^{2k-2}$.							\endproclaim

For the proof, see \S7.

\proclaim{Corollary} Let $\{f_0,f_1,\dots,f_{2k}\}$ be a Chebyshev system of
functions on the circle. Using transformations \rom1--\rom3 defined in \S\rom1
\rom(i.e., changes of a parameter, multiplying by a nonzero function, and
linear transformation\rom), we can obtain a system containing
\roster
\item"a)" a constant function\rom;
\item"b)" the sine and cosine functions simultaneously.
\endroster							\endproclaim

\demo{Proof} Consider a convex curve $\G\subset\RP^{2k}$, where $\G(\f)=(f_0(\f
):f_1(\f):\ldots:f_{2k}(\f))$. By Theorem~2, the curve $\G$ is affine, that is,
it does not intersect some hyperplane $h=0$, where $h=\sum\l_{i=0}^{2k}a_ix_i$.
Then the function $f=\sum\l_{i=0}^{2k}a_ix_i$ has no zeroes on $\G$. Put $g=1/
f$. Then $\sum\l_{i=0}^{2k}a_i\,gf_i\equiv1$. The first statement is proved.

In what follows, $f_0\equiv1$ and $\G(\f)=(f_1,\dots,f_{2k})$ is a convex curve
in $\R^{2k}$. Applying Theorem~8 $k-1$ times, we conclude that the projection
of $\G$ on some two-dimensional plane $Oxy$ is a convex curve. Denote this
projection by~$\g$. Thus we have $\g(\f)=(g(\f),h(\f))$, where $g$ and $h$ are
linear combinations of $f_0,f_1,\dots,f_{2k}$. We can assume that the origin
$O$ lies inside $\g$, because we can add a constant to $g$ and $h$: $g(\f)+c
\equiv g(\f)+cf_0(\f)$. Write the equation of $\g$ in the polar coordinates:
$\g(\f)=(\psi(\f),r(\f))$, where $\psi$ is the polar angle. Since $O\in\Int\g$,
the coordinates $\psi$ and $\f$ are involved in some one-to-one correspondence.
Now a change of the parameter yields $\g(\psi)=(\psi,r(\psi))$, where $r(\psi)
>0$. Then $h(\psi)/r(\psi)=\sin\psi$ and $g(\psi)/r(\psi)=\cos\psi$.
								\qed\enddemo

\proclaim{Theorem 9} The space $S_n$ of parameterized convex curves in $\RP^n$
is homotopy equivalent to the special orthogonal group $\SO(n+1,\R)$ for $n$
even and to the group $\PSO(n+1,\R)=\SO(n+1,\R)/\pm\Id$ for $n$ odd \rom(in
this case we consider only curves with positively oriented moving
$n$-hedron\rom).						\endproclaim

For the proof, see \S4.

        \head \S3. Proof of Theorem 7. \endhead
In this section we prove that any convex curve $\g\subset\P^n$ is a projection
from a given point $A\in\P^{n+1}\setminus\P^n$ of a certain convex curve $\G
\subset\P^{n+1}$ that passes through $A$.

So we want to construct a curve $\G(\f)\subset\P^{n+1}$ such that $\G(0)=A$ and
$p_A(\G(\f))$ $=\g(\f)$. For $n$ even, choose an affine chart so that $\g(\f)$
is a bounded curve in a ``horizontal'' plane $x_{n+1}=0$, while the center $A$
of the projection is an infinite point in the direction of the ``vertical''
axis $x_{n+1}$. For $n$ odd, we suppose that $\g(\f)$ lies in the plane $x_{n+1
}=0$ and tends to infinity only as $\f\to0\pmod{2\pi}$ (see Corollary of Lemma
in \S2), while $A$ is the same as before. In this coordinate system, to
construct a curve $\G$ means to find a proper real function $f$ on $(0,2\pi)$.
Here $f\to-\infty$ as $\f\to0$ and $f\to+\infty$ as $\f\to2\pi$ for $n$ even,
while $f\to+\infty$ as $\f\to0$ and as $\f\to2\pi$ for $n$ odd.

Let us call a regular curve {\it positive\/} if near any point $\G(\f_0)$ it is
placed at the same side from the osculating hyperplane $\Pi(\f_0)$ as the graph
of $(\f-\f_0)^{n+1}$ with respect to the axis $\f$, i.e., below this hyperplane
for $n$ even and $\f-\f_0<0$, and above it otherwise. The definition of {\it
nonnegative\/} curve is obtained by replacing the words ``below'' and `above''
by ``not above'' and ``not below'', respectively. The definitions of {\it
negative\/} and {\it nonpositive\/} curves are similar to those given above. If
we change the direction of the axis $x_{n+1}$, then positive curves becomes
negative, and otherwise. Any regular curve is either positive or negative.
Conversely, all positive (as well as all negative) curves are regular. We
consider in Theorem~10 (see below) regular curves that are positive. In fact,
for a negative curve we have $f(\f)\to-\infty$ as $\f\to2\pi$.

\proclaim{Theorem 10} Let $\g(\f)$ be a regular $C^{n+1}$-smooth curve in the
plane $x_{n+1}=0$ \rom(here $\g(\f)$ is finite for $0<\f<2\pi$\rom), and let $F
(\g)$ is the set of functions $f$ on $(0,2\pi)$ such that $\G(\f)=(\g(\f),f(\f)
)\subset\P^{n+1}$ is a regular positive curve \rom(we put here $\G(0)=A$\rom).
Then $F$ is a nonempty open convex cone in the space $C^{n+1}\((0,2\pi),\R\)$.
								\endproclaim

Besides Theorem~7, we deduce from this statement an information about the
homotopy type of the space of convex curves. Namely, below we construct a
fibering with the space of reduced (see \S4) convex curves in $\P^{n+1}$ as the
total space, the space of convex curves in $\P^n$ as the base, and an affine
subspace $F_0\subset F$ as a fiber.

\demo{Proof of Theorem $10$} A curve $\G(\f)$ is regular if and only if its
last twisting coefficient does not vanish. The expression for this coefficient
involves functions $f(\f)$ and $x_i(\g(\f))$ (where $i=1,\dots,n$), and their
derivatives up to order $n+1$. Thus $F$ is an open set.

The osculating hyperplane at $\G(\f)$ is not vertical if $\f\ne0$, because $\g(
\f)$ is a regular curve. Indeed, the distance between $\G(\f)$ and its
osculating hyperplane $\Pi(\f_0)$ is $O(\f-\f_0)^{n+1}$. If $\Pi(\f_0)$ is a
vertical plane, then $A\in\Pi(\f_0)$ and the distance between $p_A(\Pi(\f_0))$
and $\g(\f)=p_A(\G(\f))$ also is $O(\f-\f_0)^{n+1}$, i.e., an $(n+1)$-multiple
intersection of $\g$ and a hyperplane occurs. This contradicts the regularity
of $\g$.

If $f$ determines a positive curve $\G$, then obviously $\a f$ (where $\a>0$)
also determines a convex curve, which is obtained from $\G$ by the
transformation $x_{n+1}\mapsto\a x_{n+1}$. Hence, $F$ is a cone.

The {\it sum\/} of curves $\G_1(\f)=(\g(\f),f_1(\f))$ and $\G_2(\f)=(\g(\f),f_2
(\f))$ is the curve $\G(\f)=(\g(\f),f_1(\f)+f_2(\f))$. Similarly, the {\it
sum\/} of nonvertical hyperplanes $\Pi_1:x_{n+1}=L_1(x_1,\dots,x_n)$ and $\Pi_2
:x_{n+1}=L_2(x_1,\dots,x_n)$ (where $L_1$ and $L_2$ are linear functions of the
form $\sum\l_ik_ix_i+b$) is the hyperplane $\Pi:x_{n+1}=L_1(x_1,\dots,x_n)+L_2(
x_1,\dots,x_n)$. It is easy to see that $\Pi(\f_0)=\Pi_1(\f_0)+\Pi_2(\f_0)$ for
any $\f_0\in(0,2\pi)$ whenever $\G=\G_1+\G_2$ and $\G_1$, $\G_2$ are regular
curves. Now it is clear that the sum of positive curves is positive. Therefore
$F$ is a convex open cone, and it remains to prove that $F$ is not an empty
set.

Let us draw a $C^{n+1}$-smooth curve $\G$ through $A$ so that $p_A(\G)=\g$. The
set of its flattening points is closed; $A$ is not a flattening point, because
otherwise $\g(0)$ is a flattening point of $\g$. Thus we have construct a
function $f_0$ on $(0,\e)\cup(2\pi-\e,2\pi)$ such that the curve $\G(\f)=(\g(\f
),f_0(\f))$ is regular outside of $[\e,2\pi-\e]$. If $\G(\f)$ is a negative
curve in this neighborhood of $A$, then we replace $f_0$ by $-f_0$. Put
   $$\G_0=\left.\(\g(\f),f_0(\f)\)\right|_{\f\in(0,\e)\cup(2\pi-\e,2\pi)}.$$
Then $\G_0$ is a segment of the required positive curve. Now let us construct a
function $f_1$ on $[\e/3,2\pi-\e/3]$ such that the curve $\G_1=\left.\(\g(\f),f
_1(\f)\)\right|_{\f\in[\e/3,2\pi-\e/3]}$ is positive. This condition is
equivalent to the inequality
$$Lf=\det\pmatrix\dot\g&\ddot\g&\hdots&\g^{(n)}&\g^{(n+1)}\\
	   \vphantom{}&       &      &	       &	  \\
		 \dot f&\ddot f&\hdots& f^{(n)}& f^{(n+1)}\endpmatrix>0,\tag5$$
where $\g^{(i)}$ is the $i$\snug th derivative of the vector-function $\g(\f)$
with respect to $\f$. So $L$ is a differential operator of order $n+1$. Instead
of inequality~\eq(5), consider the following equation:
				$$Lf=1.\tag6$$
The leading coefficient,
	$$\Delta=\det\pmatrix\dot\g&\ddot\g&\hdots&\g^{(n)}\endpmatrix,$$
does not vanish because of the regularity of $\g$. Other coefficients also can
be expressed in terms of the derivatives of $\g$, and depend on $\f$
continuously. Hence~\eq(6) is solvable on $[\e/3,2\pi-\e/3]$. Let $f_1$ be one
of its solutions, and $\G_1$ be the corresponding positive curve. Fig.~2 shows
the graphs of $f_0$ and $f_1$ on $[0,2\pi]$. Now let us try to make the graph
of $f_0$ like it is shown on Fig.~3. Namely, our goal is a function $f_0$
(extended to $[\e,2\pi-\e]$ by zero) that is $C^{n+1}$-smooth and defines a
nonnegative $C^{n+1}$-smooth curve $\G$ that is positive outside of $[\e,2\pi-
\e]$. To this end, we modify separately ``right'' and ``left'' ``branches'' of
$f_0$ so that the corresponding curve $\G_0$ remains positive. We do not change
$f_0$ in some neighborhood of $A\in\G_0$, and thus it is sufficient to check
that the obtained curve is positive (and $C^{n+1}$-smooth) only on the
intervals $(0,\e)$ and $(2\pi-\e,2\pi)$ but not at the point $A$. Below we
describe the necessary changes on $(0,\e)$ for even $n$. For the interval $(2
\pi-\e,2\pi)$, we do the same, and the case of $n$ odd requires (due to the
fact that the curve $\g\subset\R^n$ is not compact) only insignificant
modifications of the reasoning presented below.

So on the interval $(0,\e)$ we have $Lf>0$ (below the index is omitted in the
notation for $f_0$). Let $y=Lf$. Put
$$z(\f)=\cases
y(\f)				   	&\hbox{for }\f\le\e/2  	  \\
2y\(\dfrac\e2\)\cdot\dfrac{\e-\f}\e	&\hbox{for }\e/2\le\f\le\e\\
0					&\hbox{for }\f>\e.	\endcases$$
Consider the equation $Lf=z(\f)$. By $\tilde f$ denote its solution that equals
$f$ for $0<\f<\e/2$. Then the curve $(\g(\f),\tilde f(\f))$ is positive
whenever $\f<\e$ and is plane whenever $\f\ge\e$, that is, for $\f\ge\e$ we
have $\tilde f(\f)=l(\g(\f))$, where $l$ is a linear function $k_1x_1+\ldots+k_
nx_n+b$ in the coordinates $x_1,\dots,x_n$ of $\g(\f)$. Set $g=\tilde f-l$. The
function $g$ determines a positive curve if $\f<\e$; for $\f\ge\e$ we have $g=
0$. Now let us find a function $w$ such that $h=gw$ is a $C^{n+1}$-smooth
function that defines a positive (for $\f<\e$) curve (i.e., $Lh>0$), equals $f$
at some neighborhood of zero, and equals $g$ for $\f>\delta$, where $\delta\in(
0,\e)$.

\proclaim{Lemma} Let $1+c_1\f+c_2\f^2+\ldots$ be the germ at zero of a $C^{n+1
}$-smooth function $\tilde f/g=1+l/g$. Then there exists a $C^{n+1}$-smooth
function $w$ that has the same germ at zero and identically equals $1$ for $\f
\ge\delta$, and that satisfies inequalities $\|w^{(k)}\|\le M\delta^{-(k-1)}$
for $k=1,2,\dots,n+1$, where $M$ does not depend on $\delta$\rom; here $\|\,
\cdot\,\|$ denotes the maximum of the absolute value.		\endproclaim

\demo{Proof of Lemma} Let $u(x)$ be a $C^{n+1}$-smooth function on $[0,\infty)$
shown on Fig.~4. Also put $u_\delta(x)=u(x/\delta)$. Define a function $w(\f)$
by the formula
	$$w(\f)=1+u_\delta(\f)\cdot\(\frac{\tilde f(\f)}{g(\f)}-1\).\tag7$$
Let us determine the derivatives of $w$. We have:
$$w^{(k)}=\sum_{i+j=k}C^i_ku^{(i)}_\delta\(\frac{\tilde f}g-1\)^{(j)}.\tag8$$
Let us prove that the $i$\snug th summand in~\eq(8) is bounded from above by
$M_i\delta^{-(k-1)}$, where the constant $M_i$ is independent of $\delta$.
Obviously,
	$$u^{(i)}_\delta(\f)=\frac1{\delta^i}u^{(i)}\(\frac\f\delta\)
		\le M'_i\delta^{-i},\qquad\hbox{where}\quad M'_i
				=\max\l_{\f\in[0,1]}u^{(i)}(\f).$$
Further, for $j=0$ we have
$$\(\frac{\tilde f}g-1\)^{(j)}=\frac lg\le M''_0\delta^{+1},\qquad\hbox{where}
		\quad M''_0=\max\l_{\f\in(0,\e)}\frac{l(\f)}{\f g(\f)},$$
and thus $M''_0$ is independent of $\delta$. For $j>0$, the derivatives $(lg^{-
1})^{(j)}$ are bounded by constants $M''_j$ (instead of the functions $M''_j
\delta^{1-j}$), because $g^{-1}$ is one of coordinates of the curve $(\g(\f),
\tilde f(\f))$ (which is $C^{n+1}$-smooth in the neighborhood of $A$) in a
proper affine chart. More precisely, we consider here a chart with the point
$A$ as the origin and with the infinite hyperplane $x_{n+1}=l(x_1,\dots,x_n)$;
a coordinate $x'_{n+1}$ here equals $(x_{n+1}-l(x_1,\dots,x_n))^{-1}$.
Then~\eq(8) implies that
	$$|w^{(k)}|\le\sum\l_{i+j=k,\,j>0}C^i_kM'_iM''_j\delta^{-i}
				+M'_kM''_0\delta^{1-k}\le M\delta^{1-k},$$
where $M$ is independent of $\delta$. Thus~\eq(7) provides the required
function $w$. Lemma is proved.					\qed\enddemo

Now let us show that the function $h=gw$ determines a positive curve on $(0,\e
)$, that is, $L(gw)>0$ on $(0,\e)$. Substitute $h=gw$ for $f$ in~\eq(5) and
expand the determinant in the last raw:
	$$Lh=h^{(n+1)}\Delta_0+\sum_{k=1}^n(-1)^{k-1}h^{(k)}\Delta_k,\tag9$$
where
  $$\aligned\Delta_k=&\det\pmatrix\dot\g&\ddot\g&\dots&\widehat{\g^{(k)}}&\dots
		     &\g^{(n+1)}\endpmatrix\qquad\hbox{for}\quad k=1,\dots,n,\\
	    \Delta_0=&\det\pmatrix\dot\g&\ddot\g&\dots&\g^{(n)}
						\endpmatrix.\endaligned$$
Since $\g\in C^{n+1}(S^1,\RP^n)$, all the $\Delta_k$ are bounded on $[0,\e]$.
Further, $h^{(k)}=\sum\l_{i+j=k}C^i_kg^{(i)}w^{(j)}$, where $|w^{(j)}|\le M
\delta^{1-j}$ for $j>0$ by Lemma, and $C^i_kg^{(i)}(\f)\le N_i\f^{-(i+1)}$.
(The last inequality follows from the fact that the function $\f g(\f)$ is $C^
n$-smooth on $[0,\e]$. Indeed, $g^{-1}(\f)$ is one of the coordinates of a $C^{
n+1}$-smooth curve $(\g(\f),g(\f))$ in an affine chart with the infinite
hyperplane $x_{n+1}=l(x_1,\dots,x_n)$ and the origin $A=\G(0)$. Then $g^{-1}(\f
)=\f c(\f)$, where $c(\f)$ is a $C^n$-smooth function such that $c(0)\ne0$, and
thus the function $\f g(\f)=(c(\f))^{-1}$ also is $C^n$-smooth on $[0,\e]$.) So
we have
	$$|h^{(k)}(\f)|\le\sum_{i+j=k,\,j>0}N_i\f^{-1-i}M\delta^{1-j}
		+g^{(k)}w\le M\(\sum N_i\)\f^{-k}+g^{(k)}w,\tag10$$
i.e., $|h^{(k)}(\f)|<P_k\f^{-(k+1)}$ for $\f\in(0,\delta)$ with some constant
$P_k$. On the other hand,
	$$h^{(n+1)}=g^{(n+1)}w+\sum_{j=0}^nC^j_{n+1}g^{(j)}w^{(n+1-j)}.\tag11$$
The second summand in~\eq(11) admits the following estimate similar to~\eq(10):
	$$\left|\shave{\sum_{j=0}^n}C^j_{n+1}g^{(j)}w^{(n+1-j)}\right|
						\le N\f^{-(n+1)},\tag12$$
while it is easily shown that
	$$g^{(n+1)}w>C\f^{-(n+2)},\qquad C>0.\tag13$$
Relations \eq(9)--\eq(13) implies that
	$$Lh(\f)>C\Delta_0\f^{-(n+2)}-\(N\Delta_0+
		\sum\l_{k=1}^nP_k|\Delta_k|\)\f^{-(n+1)}\tag14$$
for $\f\in(0,\delta)$. Here $\Delta_0$ is positive and bounded away from zero,
because the curve $\g(\f)$ is positive in the neighborhood of $\g(0)$; the
constants $C$, $N$, $P_k$ are independent of $\delta$. Consequently, for
$\delta$ small enough we have
	$$Lh(\f)>0\qquad\hbox{for}\quad\f\in(0,\delta),$$
that is, the curve $(\g(\f),h(\f))$ is positive on the interval $(0,\delta)$.
Since at the neighborhood of the origin $h(\f)$ equals $f(\f)$, this curve is
positive also at $\f=0$. Since $h(\f)=g(\f)$ whenever $\f>\delta$, this curve
is positive on the whole interval $(0,\e)$. So for the case of $n$ even, we get
the required modification of $\G_0$; for $n$ odd the same arguments work but we
choose here an affine chart so that $\g(0)$ is not an infinite point: it is
necessary to write estimates~\eq(9) and~\eq(14) of the determinants $\Delta_i$,
where $i=0,1,\dots,n$.

So we obtain a nonnegative $C^{n+1}$-smooth curve $\G$ that is positive outside
of $[\e,2\pi-\e]$, see Fig.~3.

Now consider the function $f_1$, see Fig.~2. Let us multiply it by a $C^{n+1
}$-smooth function $u$ with graph as shown on Fig.~5. Then we get the function
$uf_1$ that determines a curve positive on $\(2\e/3,2\pi-2\e/3\)$. Consider a
curve determined by $h+\a uf_1$, where $\a>0$. This curve is positive on the
intervals $\(2\e/3,2\pi-2\e/3\)$ and $[0,\e/3)\cup(2\pi-\e/3,2\pi]$. In the
first case we have $L(h+\a uf_1)=0+L(\a uf_1)=\a Lf_1>0$, and in the second
case $L(h+\a uf_1)=Lh>0$. Finally, on the set $U=\[\e/3,2\e/3\]\cup\[2\pi-2\e/3
,2\pi-\e/3\]$, the relation $L(h+\a uf_1)=Lh+\a L(uf_1)$ holds. This sum is
positive provided that $\a<\min\l_{\f\in U}\left|\dfrac{L(h)(\f)}{L(uf_1)(\f)}
\right|$. Then the function $h+\a uf_1$ determines a positive curve. Now the
proof of Theorem~10 is complete.				\qed\enddemo

\demo{Proof of Theorem $7$} Let $\g(\f)$ be a convex curve in $\RP^n$, and let
$\G_0(\f)$ be some convex curve in $\RP^{n+1}$ such that $A=\G_0(0)\notin\RP^
n$. By $\g_0(\f)$ denote the projection $p_A(\G_0(\f))\subset\RP^n$ of $\G_0$
on $\RP^n$ from $A$; it follows from Lemma of \S2 that this curve also is
convex. By the Shapiro theorem, there exists a continuous (in $C^{n+1}(S^1,\RP^
n)$) family $\g_t(\f)\subset\RP^n$, $t\in[0,1]$, such that $\g_1(\f)=\g(\f)$
and all curves $\g_t(\f)$ are convex. Choose an affine chart as in the proof of
Theorem~10. Set $\F=\bigcup\l_{t\in[0,1]}(f,t)$, where $f\in F(\g_t)$ (hence,
$\F\subset C^{n+1}((0,2\pi),\R)\times[0,1]$). Consider the mapping $\p\:\F\to[0
,1]$ defined by $\p(f,t)\mapsto t$. This mapping has the covering homotopy
property. Indeed, by Theorem~10 all fibers $F(\g_t)$ are nonempty open convex
cones. If $f\in F(\g_t)$, then it is easy to see that $f\in F(\g_s)$ for any
$s$ from some neighborhood of $t$ (it follows from the fact that the
coefficients of the differential operator $L$ that corresponds to $\g(t)$,
see~\eq(50), are continuous functions of $t$, and from the compactness of these
$\g_t$, which are homeomorphic to the circle). The segment $[0,1]$ can be
covered by a finite set of these neighborhoods $U(0),U(t_1),\dots,U(t_k)$. For
each of them, there exists a function $f_i\in\bigcap\l_{t\in U(t_i)}F(\g_t)$;
here $i=1,\dots,k$; similarly, $f_0\in\bigcap\l_{t\in U(0)}F(\g_t)$, where $f_
0$ is the function that determines $\G_0$. Let $s_i\in(t_{i-1},t_i)$ and
$\delta_i$ be positive numbers such that $[s_i-\delta_i,s_i+\delta_i]\subset U(
t_{i-1})\cap U(t_i)$; also we set $t_0=0$. Put
$$f_t=\cases\dfrac{f_{i-1}+f_i}2+\dfrac{t-s_i}{\delta_i}\,\dfrac{f_i-f_{i-1}}2
	\qquad  &\hbox{for}\quad t\in[s_i-\delta_i,s_i+\delta_i],\\
	f_{i-1} &\hbox{for}\quad
	t\in(s_{i-1}+\delta_{i-1},s_i-\delta_i)\endcases$$
(we assume that the segments $[s_i-\delta_i,s_i+\delta_i]$ do not intersect one
another and that $s_0+\delta_0=0$). Then for all $t\in[0,1]$ we have $f_t\in F(
\g_t)$. Thus we proved the covering homotopy property for the mapping $\p$.
Consequently, there exists a continuous (in the space $C^{n+1}(S^1,\RP^{n+1})$)
family $\G_t$, $t\in[0,1]$, such that $\G_t(0)=A$ and $p_A(\G_t(\f))=\g_t(\f)$,
and all curves $\G_t$ are positive. Put $\G=\G_1$. The curve $\G$ and the
convex curve $\G_0$ belongs to the same connected component of the space of
regular curves. By the Shapiro theorem, the curve $\G$ is convex as well.
Theorem~7 is proved.						\qed\enddemo

	\head \S4. Topology of the space of convex curves. \endhead
Consider the space of the classes of projectively equivalent parameterized
convex curves in $\RP^n$. Using Theorem~10, we prove that this space is
contractible. Then we find the homotopy type of the space of all parameterized
convex curves in $\RP^n$.

If a curve $\G(\f)\subset\RP^n$ is convex, then $A=\G(0)$ is a regular point,
and the coordinates of $\G(\f)$ have the following canonical form in a proper
affine chart:
$$\aligned x_{n+1-i}(\G(\f))&=\f^i+O(\f^{i+1})\quad\hbox{ for }0<i<n,\\
	       x_1(\G(\f))&=\f^n\!+o(\f^n),		\endaligned\tag15$$
where $\f\to0$. Then the coordinates $x_1,\dots,x_n$ are said to be {\it
canonical\/} for $\G$. Their $\underline{\hbox{existence}}$ follows from the
regularity of $\G$ at $\G(0)$, see~\S1. Further, the oriented coordinate flag
$\{\<0\>,\<e_1\>_+,\<e_1,e_2\>_+,\dots,\R^n_+\}$ is defined by $\G$ in a unique
way. If we fix a scalar product (for example, by choosing a Euclidean metric on
the unit sphere $S^n\subset\R^{n+1}$, which is the space of the 2-fold covering
over $\RP^n$), then this flag contains a $\underline{\hbox{unique}}$
orthonormal basis, and there exists a unique linear transformation (in our
affine chart) that makes this basis canonical for $\G$. Note that the matrix of
this transformation is diagonal with respect to our basis.

\definition{Definition} A curve (possibly non convex) $\G(\f)$ is called {\it
reduced\/} if its germ at zero is of the form~\eq(15).		\enddefinition

Let us fix a scalar product in $\R^{n+1}$. Then a metric on $\RP^n$ is defined,
and $\RP^n$ is identified with $\(\RP^n\)^*$. Fix also some affine chart, as
well as a coordinate system in it. By $\tilde S_{n,k}$ denote the space of
reduced (with respect to the chosen coordinates) convex curves in $\RP^n$ that
are $C^{n+k}$-smooth everywhere except zero and $C^n$-smooth at the
neighborhood of zero. The topology in $\tilde S_{n,k}$ is induced from the
space $C^{n+k}(S^1\setminus0,\RP^n)\cap C^n(S^1,\RP^n)$.

\proclaim{Theorem 11} The space $\tilde S_{n,k}$ is contractible \rom(here $n
\ge1$ and $k\ge0$\rom).						\endproclaim

\demo{Proof} The proof is by induction on $n$ (here the sum $n+k$ is fixed).

{\it The base}, $n=1$. An element of $\tilde S_{1,k}$ is an orientation
preserving $C^{k+1}$-diffeomorphism of the pair $(S^1,x)$ (where $x$ is a base
point) such that its differential at $x$ is identical. In other words, it is a
diffeomorphism $f\:[0,2\pi]\to[0,2\pi]$ such that $f(0)=0$, $f(2\pi)=2\pi$, and
$f'(0)=f'(2\pi)=1$. The retraction of $\tilde S_{1,k}$ can be determined by the
formula
		$$F(f,t)(\f)=t\f+(1-t)f(\f).\tag16$$
For $t=1$, we get $F(f,1)(\f)\equiv\f$.

{\it The induction step}. Let $\G(\f)\in\tilde S_{n,k}$ be a reduced convex
curve in $\RP^n$. In the chart $x_{n+1}=1$ relations \eq(15) hold as $\f\to0$.
Let $\RP^{n-1}$ be embedded into $\RP^n$ as the hyperplane $x_{n+1}=0$. Put $\g
(\f)=p_A(\G(\f))$, where $A=\G(0)=(0,\dots,0,1)$. Then $\g(\f)$ is a convex
curve in $\RP^{n-1}$. It follows from \eq(15) that in the chart $x_n=1$ we have
$$\aligned x_{n-i}(\g(\f))&=\f^i+O(\f^{i+1})\qquad\hbox{for}\quad0<i<n-1,\\
	   x_1    (\g(\f))&=\f^{n-1}+o(\f^{n-1}),	\endaligned\tag17$$
as $\f\to0$, that is, $\g(\f)\in\tilde S_{n-1,k+1}$. Thus we have constructed
a mapping
		$$\p\:\tilde S_{n,k}\to\tilde S_{n-1,k+1}.\tag18$$
For $\g\in\tilde S_{n-1,k+1}$, by $F(\g)$ denote the set appeared in
Theorem~10. Obviously, we have $\p^{-1}(\g)\subset F_0(\g)$, where $F_0(\g)$ is
a subset of $F(\g)$ that consists of reduced convex curves with the required
smoothness. Here we identify functions from $F(\g)$ and curves defined by these
functions. On the other hand, $F_0(\g)\subset\p^{-1}(\g)$, because $F_0(\g)$
consists of regular curves and, by virtue of Theorem~7, contains convex curves.
The arguments used in the proof of Theorem~10 imply that $F_0(\g)$ is a convex
(and thus connected) subspace in the space of functions of required smoothness.
Further, if $f\in F(\g)$, then $cf\in F_0(\g)$ for a certain $c>0$, because
\eq(15) obviously hold for $i>1$, while
		$$x_n(\g(\f),f(\f))=c\f+O(\f^2);\tag19$$
This implies that $x_n(\g(\f),cf(\f))=\f+O(\f^2)$ and the curve $(\g(\f),cf(\f)
)$ is reduced. So, in this case the set $F_0(\g)$ is nonempty. Moreover, we
have show that $F_0(\g)=F(\g)\cap N(\g)$, where $N(\g)$ is an affine subspace
of $C^n(S^1,S^1=\R\cup\infty)\cap C^{n+k}(I=S^1\setminus0,\R)$ defined by the
equation $c=1$; here $c$ is the coefficient in~\eq(19) (in fact, the subspace
$N=N(\g)$ does not depend on $\g$, because $\g\subset\RP^{n-1}$ is a reduced
curve). Since $F_0(\g)$ is the intersection of a convex set $F(\g)$ with the
linear subspace $N$, we see that this space is connected and convex. Since the
space $F_0(\g)$ is connected, it consists of convex curves; relations~\eq(17)
imply that these curves are reduced. It is easy to see that these curves are
sufficiently smooth. Consequently, we have $\p^{-1}(\g)=F_0(\g)$.

Let us prove that the mapping~\eq(18) is a locally trivial fibering. Assume
that $\g\in\tilde S_{n-1,k+1}$ and $\G\in\p^{-1}(\g)$. There exist an open (in
$N$) $\e$-neighborhood $U(f_0)$ of $f_0$, where $\G=(\g_0,f_0)$, and an open
(in $\tilde S_{n-1,k+1}$) neighborhood $U(\g_0)$ of $\g_0$ such that the curve
$(\tilde\g(\f),\tilde f(\f))$ lies in $\tilde S_{n,k}$ whenever $\tilde f\in U(
f_0)$ and $\tilde\g\in U(\g_0)$. (It is ensured by inequality~\eq(5), which
includes derivatives of $\tilde\g$ and $\tilde f$ of order no greater than $n$;
at $\f=0$ the curve $\G$ is regular, because it is reduced.) For any $\g\in U(
\g_0)$ we construct a homeomorphism $\Phi_\g$ (that depends on $\g$
continuously) between $\p^{-1}(\g)$ and $U(f_0)=B(f_0,\e)$, where $B(f_0,\e)$
is the ball of radius $\e$ with center $f_0$ in the space $C^{n+k}(I,\R)$ with
norm $\|f\|=\max\l_{\f\in(0,2\pi)}\sum\l_{i=0}^{n+k}\f^{i+1}(2\pi-\f)^{i+1}|f^{
(i)}(\f)|$. A function $f\in\p^{-1}(\g)$ has a unique representation $f=f_0+tg$
with $t>0$ and $\|g\|=1$. Put $\tau=\arctg t$. Then there exists a number $\tau
(g,\g)\in[\arctg\e,\pi/2]$ that depends on $g$ and $\g$ continuously and such
that the function $f=f_0+tg$ belongs to $\p^{-1}(\g)$ if and only if $0\le t<
\tg(\tau(g,\g))$. Indeed, the fact that the set $\{(g,\g,t)\mid{}$the curve $(
\g(\f),f_0(\f)+tg(\f))$ is positive for any $\f\in(0,2\pi)\}$ is open implies
that the function $\tau(g,\g)$ is lower semicontinuous. Similarly, the fact
that this function is upper semicontinuous follows from the fact that the set
$\{(g,\g,t)\mid{}$the curve $(\g(\f),f_0(\f)+tg(\f))$ is negative for at least
one $\f\in(0,2\pi)\}$ is open. So the mapping
	$$\Phi_{\g}\:f=f_0+tg\mapsto f_0+\e\frac{\arctg t}{\tau(g,\g)}g$$
ia a required homeomorphism of $\p^{-1}(\g)$ on $U(f_0)$.

Thus mapping~\eq(18) is a fibering. Its fiber $\p^{-1}(\g)$ is contractible,
because it is a convex subset of a linear space, while the base $\tilde S_{n-1,
k+1}$ is contractible by the induction hypothesis. This implies that the total
space $\tilde S_{n,k}$ is contractible as well. The theorem is proved.
								\qed\enddemo

\demo\nofrills{}

Now we can study the topology of the space $S_n$ of all (reduced and
nonreduced) parameterized convex curves in $\RP^n$. For $n$ odd, we consider
only curves with positively oriented moving $n$-hedron. Let $\G\in S_n$ be a
convex curve in $\RP^n$, let $\tilde\G\subset S^n\subset\R^{n+1}$ be its
lifting to the sphere $S^n$, and let $G\in\SL(n+1,\R)$ be a projective
transformation that converts $\G$ into a reduced curve. Consider a polar
decomposition $G=TU$ of $G$, where $U\in\SO(n+1,\R)$ and $T$ is an upper
triangle matrix with positive diagonal entries. Further, $T=T_1D$, where $T_1$
is a unipotent upper triangular matrix and $D$ is a diagonal matrix with unit
determinant. The operator $D$ in the decomposition $G=T_1DU$ is determined
uniquely by the condition that $G(\G)$ is a reduced curve. The operator $U$ is
defined by the same condition in a unique way for $n$ even and up to sign for
$n$ odd. This ambiguity is due to the fact that the centrally symmetric
covering curve $\tilde\G\subset S^n$ is connected for $n$ odd but consists of
two components for $n$ even. In the latter case a moving $n$-hedron of exactly
one of the components has positive orientation, and this enables us to avoid
ambiguity. The above-diagonal entries of $T_1$ can be chosen arbitrarily. So
let $T_1=E$. Then the mapping
		$$S_n\to\tilde S_{n,0},\tag20$$
is defined, which takes a curve $\G\in S_n$ to a (uniquely defined) reduced
curve $DU(\G)\subset\tilde S_{n,0}$. It is easy to see that~\eq(20) is a
fibering. Its fiber $F$ (that is, the set of matrices of the form $DU$) is
homeomorphic to $\R^n\times\SO(n+1,\R)$ for $n$ even and to $\R^n\times\PSO(n+1
,\R)$ for $n$ odd. Indeed, the mapping $(a_1,\dots,a_n)\mapsto\diag(e^{a_1},
\dots,e^{a_n},e^{-(a_1+\ldots+a_n)})$ is a homeomorphism between $\R^n$ and the
set of diagonal matrices with positive entries and unit determinant, and the
fibering $F\to\SO(n+1,\R)$ (or $F\to\PSO(n+1,\R)$) with fiber $\R^n$ is trivial
(because the polar decomposition $DU$ is unique provided that the entries of
$D$ are positive. However, we do not use the triviality of the fibering: all we
need is the homotopy type of~$F$. Thus we have fibering~\eq(20). Its base is
contractible by virtue of Theorem~11. So its total space $S_n$ is homotopy
equivalent to the fiber $F$, which is homotopy equivalent to $\SO(n+1,\R)$ or
$\PSO(n+1,\R)$, according to the parity of~$n$. This proves Theorem~9.
\qed\enddemo

\remark{Remark} Put $T_n=\partial S_n$, that is, $T_n$ is the set of curves $\G
\subset\RP^n$ such that any hyperplane crossing $\G$ more than $n$ times is an
osculating hyperplane of $\G$ at some flattening point of multiplicity at
least~2, and does not intersect $\G$ anymore (see Theorem~6). Also put $U_n=
\ol S_n=S_n\cup T_n$. Similarly to the previous arguments, we can define the
spaces $\tilde T_{n,k}$ and $\tilde U_{n,k}$ of reduced ``boundary convex'' and
``not strongly convex'' curves (a curve $\G(\f)\in T_n$ is convex if $\G(0)$ is
a flattening point and in a proper affine chart we have $x_n(\G(\f))=O(\f)$ and
$x_{n-i}(\G(\f))=O(\f x_{n-(i-1)}(\G(\f)))$ as $\f\to0$, compare with~\eq(15)).
There are the fiberings $\tilde T_{n,k}\to\tilde T_{n-1,\,k+1}$ and $\tilde U_{
n,k}\to\tilde U_{n-1,\,k+1}$ defined by analogy with~\eq(18). Similarly to
Theorem~11, it can be proved that the spaces $\tilde T_{n,k}$ and $\tilde U_{n,
k}$ are contractible. However, there is no analog of fibering~\eq(20), and we
cannot conclude anything on the topology of $T_n$ and $U_n$. This is caused by
the fact that the coordinate flag corresponding to the canonical coordinates of
a reduced curve $\G\in T_n$ (or $\G\in U_n$) is not well-defined. Thus we have
no mapping analogous to~\eq(20).
\endremark

	\head \S5. Generalizations of the concept of convexity. \endhead
In this section we consider various generalization of the concept of convex
curves. More precisely, we study locally convex curves, curves convex in the
sense of Barner~\cite8, and convex surfaces. We show that convex surfaces do
exist only if either dimension or codimension is equal to~1. Further, the
class of locally convex curves differs from the class of convex curves;
however, some properties of locally convex curves are similar to those of
convex curves.

\subhead Locally convex curves \endsubhead

\definition{Definition} A regular curve $\G$ is said to be {\it locally
convex\/} if the osculating hyperplane $\Pi(\f)$ intersects it only at $\G(\f)$
for any $\f$.							\enddefinition

Obviously, any convex curve is a locally convex one, see~\S1. For the cases of
$\RP^1$ and $\RP^2$, local convexity implies convexity, while this does not
hold already in the case of~$\RP^3$.

\example{Example} Consider the curve $\G_k(\f)=(\cos k\f,\sin k\f,\tg\frac\f2)$
in $\R^3$, where $\f\in(-\pi,\pi)$ and $k\in\N$. Compactify it by the infinite
point of the $Oz$ axis. It is easy to see that we obtain a $C^\infty$-smooth
curve in~$\RP^3$. If $k>1$, then this curve is not convex. In fact, the plane
$x=0$ intersects it $2k+1$ times.				\endexample

\proclaim{Lemma 1} The curve $\G_k(\f)\subset\RP^3$ is regular.	\endproclaim

\proclaim{Lemma 2} If $\f\in(-\pi,\pi)$, then the osculating plane $\Pi(\f)$
intersects the curve $\G_k$ only at $\G_k(\f)$.			\endproclaim

Since the plane $\Pi(\pi)$ touches $\G_k$ at $k-1$ points $\G\(\pi(1-\frac {2j}
k)\)$, $j=1,\dots,k-1$, besides $\G(\pi)$, the curve $\G_k$ is not locally
convex. Let us consider an infinitely smooth function $h\:S^1\to\R$ supported
by the interval $[\pi-\frac1{2k},\pi+\frac1{2k}]$ and positive at the point
$\pi$. For $\e>0$ small enough, $\G_{k,\e}(\f)=((1+\e h)\cos k\f,(1+\e h)\sin k
\f,\tg\frac\f2)$ is a locally convex curve that is not convex. Indeed, the
plane $x=0$ crosses it $2k+1>3$ times, the assertions of Lemmas~1 and~2 remain
true provided that perturbations are $C^\infty$-small, and the plane $\Pi(\pi
)$, which is determined by the equation $x=1+\e h(\pi)$, contains no points of
$\G_{k,\e}$ different from $\G_{k,\e}(\pi)$. So we have to prove Lemmas~1
and~2.

\demo{Proof of Lemma $1$} Put $\a=k\f$. Then we can write $\G_k(\a)=(\cos\a,
\sin\a,\tg\frac\a{2k})$, where $\a\in(-k\pi,k\pi)$. Consider the determinant
	$$\Delta(\a)=\det\pmatrix
		\dot x(\a) &\ddot x(\a) &\dddot x(\a) \\
		\dot y(\a) &\ddot y(\a) &\dddot y(\a) \\
		\dot z(\a) &\ddot z(\a) &\dddot z(\a)  \endpmatrix.$$
The curve $\G_k$ is regular if and only if $\Delta(\a)$ has no zeroes on $(-k
\pi,k\pi)$, because the regularity of the projection $p_A\G_k$ implies that $\G
_k$ is regular at the infinite point $A$ (compare with the proof of
Theorem~10). We have
	$$\align\Delta(\a)&=\det\pmatrix
-\sin\a 	  &-\cos\a &\phantom{-}\sin\a	\\
\phantom{-}\cos\a &-\sin\a &-\cos\a		\\
 \dfrac1{2k\cos^2\!\frac\a{2k}}
	&\dfrac{\sin\frac\a{2k}}{2k^2\cos^3\!\frac\a{2k}}
		 &\dfrac{1+2\sin^2\!\frac\a{2k}}{4k^3\cos^4\!\frac\a{2k}}
						\endpmatrix=\\
&=\frac1{2k\cos^2\!\frac\a{2k}}\(1+\frac{1+2\sin^2\!\frac\a{2k}}
	{2k^2\cos^2\!\frac\a{2k}}\)>0\qquad\hbox{for}\quad|\a|<k\pi.\endalign$$
(This determinant can be computed by adding the third column to the first one).
Lemma~1 is proved.						\qed\enddemo

\demo{Proof of Lemma $2$} Let us consider the band $\{(\a,z)\mid|\a|<k\pi\}$,
which is the space of $k$-fold covering of the cylinder $x^2+y^2=1$ containing
$\G_k(\f)$. It suffices to show that for all $\a_0$ the graph of the function
$z=f(\a)=\tg\frac\a k$ crosses the sinusoid
    $$z=g(\a)=f(\a_0)+f'(\a_0)\sin(\a-\a_0)-f''(\a_0)(\cos(\a-\a_0)-1)\tag21$$
only at $(\a_0,f(\a_0))$; the curve \eq(21) covers the curve cut on the
cylinder $x^2+y^2=1$ by the plane $\Pi(\a_0)$. Let us show that
\roster
\item"{1)}"$f(\a)>g(\a)$ whenever $\a_0<\a<k\pi$;
\item"{2)}"$f(\a)<g(\a)$ whenever $-k\pi<\a<\a_0$.
\endroster
It is sufficient to prove both inequalities on $[a,b]$, where $a$ is the
nearest from the left to $\a_0$ extremum (minimum) of \eq(21), while $b=a+\pi$
is the nearest from the right to $\a_0$ extremum (maximum) of \eq(21). In fact,
at a point $\a_0\in[a,b]$ we have $g'(\a)=f'(\a)>0$, and thus the function $g$
increases on $[a,b]$ and $g'>0$. Then $g''$ decreases on $[a,b]$, because $(g''
)'=g'''=-g'<0$. On the other hand, the third derivative of the tangent function
$f'''$ is positive. So $f''$ increases on the interval $(-k\pi,k\pi)$. Since $g
(\a)\le g(b)$ for all $\a$, while $f$ increases up to $\a=k\pi$, it suffice to
prove inequality 1) on the interval $(\a_0,b]\cap(\a_0,k\pi)$. The inequality
$f''>g''$ holds on this interval, because $f''(\a_0)=g''(\a_0)$ and $f'''(\a)>g
'''(\a)$ for $\a\in[a,b]\cap(-k\pi,k\pi)$. Similarly, the inequality $f'>g'$
follows from $f'(\a_0)=g'(\a_0)$ and $f''>g''$. Finally, we see that $f>g$
whenever $\a_0<\a<k\pi$. Inequality 2) is obtained in a similar way.
								\qed\enddemo

Thus the class of locally convex curves in $\RP^n$ differs from the class of
convex curves, at least whenever $n=3$.

\proclaim{Conjecture} For $n\ge3$, there exist locally convex curves in $\RP^n$
that are not convex.						\endproclaim

Nevertheless, a lot of properties of convex curves holds also for locally
convex curves. For example, the statements of Theorems~1--3 hold. Really, a
plane that does not intersect $\G$ can be obtained as a small perturbation of
an osculating hyperplane provided that the dimension is even. The proof of
Theorem~3 also works for locally convex curves. If the dimension is odd, then
the intersection index of $\G$ and an osculating plane also is odd, and so $\G$
is not homological to zero and intersects any hyperplane. A hyperplane that
transversally intersects $\G$ only at a point $A$ can be obtained by small
perturbation of the osculating hyperplane $\Pi(A)$. Further, locally convex
curves correspond to systems of functions (in the odd-dimensional case, to
systems of sections of the M\"obius bundle) $f_0,f_1,\dots,f_n$ such that any
linear combination $f=\sum\l_{i=0}^n\lambda_if_i$, where some of $\lambda_i$
are not equal zero, has no zeroes of order greater than $n$; moreover, if $\f$
is an $n$-multiple zero of $f$, then there $f$ has no other zeroes.

The duality involution preserves the local convexity property. Indeed, let $\G^
*$ be the dual curve of a locally convex curve $\G$. Then the curve $\G^*$ is
nondegenerate, see the proof of Lemma at the end of~\S1. It suffices to show
that $y\notin\Pi(w)$ for all $y,w\in\G^*$. Assume the converse, namely, that $y
\in\Pi(w)$. The dual statement is as follows: $z\in\Pi(x)$, where $x$ and $z$
are points of $\G$ such that $x$ corresponds to $y$ and $z$ corresponds to $w$,
i.e., $\Pi(x)$ and $y$, $\Pi(y)$ and $x$, $\Pi(z)$ and $w$, $\Pi(w)$ and $z$
are pairs of projectively dual objects. This dual statement certainly is false
because of the local convexity of $\G$.

The lengths of the curves $\G_k\subset\RP^3$ constructed above tend to
infinity as $k\to\infty$ and thus are not bounded from the above by $3\pi$;
compare this with Theorem~5. Since loss of the convexity implies loss of the
local convexity (by virtue of Theorem~6), while the class of locally convex
curves in $\RP^n$ is wider than the class of convex curves (at least whenever
$n=3$), it is clear that there exist locally convex curves that are not
homotopic to each other in the class of locally convex curves, that is, the
space of the locally convex curves is not connected. If $\G\subset\RP^n$ is a
locally convex curve and $A$ lies on $\G$, then the projection $\g=p_A(\G)$ is
a regular curve in $\RP^{n-1}$, because the osculating hyperplane $\Pi(\f)$ of
$\G$ contains no the center $A$ of the projection provided that $A\ne\G(\f)$.
However, the projected curve need not to be locally convex (at least for $n=
3$). The author is unaware whether any regular (or at least any locally convex)
curve $\g\subset\RP^n$ is a projection of some locally convex curve $\G\subset
\RP^n$, by analogy with Theorems~7 and~10.

\subhead Curves convex in the sense of Barner \endsubhead
\definition\nofrills{Definition}~\cite8. A curve $\G\subset\RP^n$ is called a
curve {\it convex in the sense of Barner}, or a {\it strictly convex\/} curve
if for any $n-1$ its points there is a hyperplane that intersects $\G$ exactly
at these points.						\enddefinition

\example{Example} Suppose that $\g\subset\RP^n$ is a convex curve, $A\in\P^{n+1
}\setminus\P^n$, and the projection of a curve $\G\subset p_A^{-1}(\g)$ onto
$\g$ is bijective (here $A\notin\G$). Then $\G\subset\RP^{n+1}$ is a strictly
convex curve.							\endexample

For $n$ odd, strictly convex curves are affine, while for $n$ even they are not
homotopic to zero in $\RP^n$.

Strictly convex curves correspond to systems of functions (respectively, of
sections of the M\"obius band, if the curve is not homological to zero) having
the following property. Consider a linear combination
		$$\lambda f=\sum_{i=0}^n\lambda_if_i,\tag22$$
where the $(n+1)$-tuple of parameters runs a line in $(\RP^n)^*$. Any function
of the family~\eq(22) has $n-1$ zeroes (computing multiplicities) at the values
of the independent variables that are the same for all functions of the family.
The strictly convexity is equivalent to the fact that any one-parameter family
\eq(22) contains a function $\lambda f$ having no other zeroes.

The curves $\G_k$ constructed in the beginning of this section are not strictly
convex. Thus the class of locally convex curves is not contained in the class
of strictly convex curves. It can easily be checked that the inverse inclusion
also is false. There are simple examples that show that the duality involution
does not preserve the strictly convexity. These examples are provided by curves
determined by arbitrary function $z(\f)$ on the cylinder $x^2+y^2=1$ in $\R^3$.
However, Lemma of~\S2 holds also for strictly convex curves. In particular, if
$\G\subset\RP^n$ is a strictly convex curve and $A_1,\dots,A_{n-2}$ are its
different points, then the projection $\g=p_{A_1}p_{A_2}\ldots p_{A_{n-2}}(\G)$
is a strictly convex curve in $\RP^2$; nevertheless, in the case $n\ge4$ the
strict convexity of a curve $\G\subset p^{-1}_{A_{n-2}}\ldots p^{-1}_{A_1}(\g)
\subset\RP^n$ does not follow from the fact that $\g$ is a convex curve in $\RP
^2$.

The paper~\cite8 studies this topic in more details; also see~\cite{18}. It
happens that these curves have some properties that were obtained in~\cite4 only
for convex curves.

\subhead Convex surfaces \endsubhead

\definition{Definition} A $k$-dimensional surface $\G$ smoothly embedded in
$\RP^n$ is called {\it convex\/} if any plane of codimension $k$ intersects it
at no more than $n-k+1$ times.					\enddefinition

\example{Examples} A point is a convex surface of zero dimension. A nonempty
open subset of $\RP^n$ is an $n$-dimensional convex surface. The convex curves
are the convex surfaces of dimension one. A sphere in $\R^n\subset\RP^n$ is a
convex surface of dimension $n-1$.				\endexample

For hypersurfaces, our definition does not differ from the classical one: it
can be checked~\cite3 that a convex hypersurface in $\RP^n$ is affine and thus
it is homeomorphic to $S^{n-1}$.

We show that this definition has no sense in the dimensions other than 0, 1, $n
-1$, and $n$.

\proclaim{Theorem 12} For $k=2,3,\dots,n-2$, there are no convex surfaces of
dimension $k$ in $\RP^n$.					\endproclaim

In the other words, convex curves, Chebyshev systems of functions and
disconjugate differential equations admits no ``naive'' generalization for
multidimensional case.

The following definition is used in the proof.

\definition{Definition} A $k$-dimensional surface $\G$ smoothly embedded to
$\RP^n$ is said to be {\it almost convex\/} if there exists a unique plane $\Pi
^{k-1}$ that lies in $\G$, and any plane $B$ of dimension $(n-k)$ crosses $\G$
at no more than $n-k+1$ points whenever the intersection $B^{n-k}\cap\Pi^{k-1}$
is either a point or the empty set.				\enddefinition

In algebraic geometry, this plane $\Pi^{k-1}\subset\G^k$ naturally arises when
we consider a blowing-up with center $x$, $x\in\G^k$.

\proclaim{Lemma} Almost convex surfaces of codimension~\rom1 do not exist.
								\endproclaim

\demo{Proof} If $k=1$, then any point $A\in\G$ is a $(k-1)$-dimensional plane,
and there is a lot of them. Now assume that $k>1$. Let $A$ be a point of $\G^k
\setminus\Pi^{k-1}$. The intersection of the tangent hyperplane $T_A\G$ with
$\Pi^{k-1}$ is not empty, because the dimension of this intersection should be
equal to $n-1-2=k-2\ge0$; here $n=k+1$ is the dimension of ambient space, $1$
and $2$ are the codimensions of $T_A\G$ and $\Pi^{k-1}$. Let $A'$ belong to $T_
A\G\cap\Pi^{k-1}$. Then the line $AA'$ crosses $\G$ at least three times: once
at $A'$ and twice $A$. Furthermore, $\dim(AA'\cap\Pi^{k-1})=0$. Hence, $\G^k$
is not an almost convex surface.				\qed\enddemo

\demo{Proof of the theorem} As before, put $\g=p_A(\G)$, where $A\in\G$ and $p_
A\:\RP^n\to\RP^{n-1}$ denotes the central projection. It is easy to see that
$\g$ is an almost convex surface in $\RP^{n-1}$; the proof of this fact is
similar to that of Lemma in~\S2. The dimension of $\g$ is $k$ (likely $\G$), $k
>1$, and its codimension is $n-k-1$, i.e., one less than the codimension of
$\G$. For the plane $\Pi^{k-1}\subset\g$, we take $p_A(A)$. Further, consider
the projection $p_a\:\RP^{n-1}\to\RP^{n-2}$, where $a\in\Pi^{k-1}\subset\g$.
Then $p_a(\g)$ is an almost convex surface (it is proved a few lines below) of
the same dimension $k$ but of codimension $n-k-2$. Doing so, we get an almost
convex surface of codimension $1$ in $\RP^{k+1}$. This contradicts Lemma.

Now let us prove that the surface $p_a(\g)$ is almost convex, as well as the
next projections. Assume that $\#(\beta^{n-k-2}\cap\g)>n-k-1$, while $\dim(
\beta\cap\Pi)<1$, where $\Pi\subset p_a(\g)$. Put $p=p_ap_A$, that is, $p$ is
the projection of $\RP^n$ onto $\RP^{n-2}$ from the line passing through $A\in
\G$ in the direction determined by $a\in p_A(\G)$. Then the plane $B^{n-k}=p^{-
1}(\beta)$ intersects $\G$ more than $n-k+1$ times. Indeed, we get new double
intersection at $A$, because $\beta$ contains $A$ and the tangent direction $p_
A^{-1}(a)\subset T_A\G$). This contradicts the almost convexity of $\G$. The
theorem is proved.						\qed\enddemo

	\head \S6. Structure of the sets $M_k(\G)$. \endhead
In this section we find the homotopy type of the spaces $M_k(\G)$, $k\le n$,
determined by a convex curve $\G\subset\RP^n$. Then we prove the existence of
exterior lines and interior subspaces~\cite5.

Recall (see \S1) that $M_k(\G)\subset\P^*$ is defined as the set of hyperplanes
crossing $\G$ at $k$ points, taking multiplicities into account. For a convex
curve $\G$, only $M_k$ with $n-k$ even and nonnegative are not empty, because
the parity of $k$ is nothing but the intersection index of $\G$ and a
hyperplane. For the dual curve, the set $M_k(\G^*)$ consists of the points of
$\RP^n$ that are incident to $k$ osculating hyperplanes of $\G$ (of course,
counting the multiplicities); below we will write $M^*_k$ instead of $M_k(\G^*
)$. Further, if $x\in M^*_k\setminus\G$, then the projection $\g=p_x(\G)$ is a
curve in $\RP^{n-1}$ with exactly $k$ flattening points $y_1,\dots,y_k$, where
$y_i=p_x(Y_i)$ and $Y_i$ are the points such that the osculating hyperplanes
$\Pi(Y_i)$, $i=1,\dots,k$, pass through $x$. This implies, in particular, that
a curve $\g\subset\RP^{n-1}$ with $k$ flattening points cannot be a projection
of a convex curve $\G\subset\RP^n$ whenever $k$ and $n$ have different
parities (or whenever $k>n$). Further, we have $(\RP^n)^*=M_n\cup M_{n-2}\cup
\ldots\cup M_{n-2[n/2]}$. The dual decomposition is $\RP^n=M^*_n\cup\ldots\cup
M^*_{n-2[n/2]}$, where $\G\subset M^*_1$ for $n$ odd and $\G\subset M^*_2$ for
$n$ even.

The definitions of convex and locally convex curves can be reformulating as
follows: a curve $\G\subset\RP^n$ is convex if the $M_k$ (and therefore, the
$M^*_k$, because the duality involution preserves the convexity) are empty for
$k>n$, that is if
		$$\RP^n\subset\bigcup\l_{k\le n}M^*_k,\tag23$$
and it is locally convex if the $M^*_k$ do not intersect $\G$ whenever $k>n$:
		$$\G\subset\bigcup\l_{k\le n}M^*_k.\tag23{$'$}$$
Thus it was proved in \S5 that \eq(23') does not imply $\eq(23)$.

Let us recall the properties of the $M_k(\G)$ that we know (see \S1): $M_n\ne
\pusto$; for $n$ even, $M_0\ne\pusto$ and $M_0$ is a convex, and thus a
retractable subset of $\P^*$; in this case $M_1=\pusto$. For $n$ odd, we have
$M_0=\pusto$ and $M_1\ne\pusto$.

\proclaim{Theorem 13} Suppose that $\G\subset\RP^n$ is a convex curve and $0<k
\le n$. If $n-k$ is odd, then the set $M_k(\G)$ is empty\rom; if $n-k$ is even,
this set is retractable onto a circle representing the nontrivial element of
the group $\pi_1\((\RP^n)^*\)=\Z_2$.				\endproclaim

\demo{Proof} Only the case of $n-k$ even should be considered. There is a
natural mapping
		$$\p\:M_k\to S^k\G\tag24$$
of $M_k$ onto the set of unordered $k$-tuples of points of $\G$; these points
need not to be different. Namely, $k$ points $A_1,\dots,A_k$ of intersection of
$\Pi$ and $\G$ are assigned to each hyperplane $\Pi\in M_k$. For nontransversal
intersections, the intersection point is multiplied by the corresponding
multiplicity. In the other words, $\A=(A_1,\dots,A_k)$ is the intersection
divisor of $\Pi$ and $\G$; the degree of $\A$ equals $k$. A $k-1$-dimensional
plane $L(\Pi)\subset\Pi$ is assigned in a unique way to a divisor $\A(\Pi)\in S
^k\G$ by the condition that the intersection divisor of $\Pi$ and $\G$ is $\A$.

Let us show that \eq(24) is a locally trivial fibering. Suppose that $\A\in S^k
\G$ and $L_0$ is the $(k-1)$-dimensional plane that corresponds to $\A$.
Consider the projection $p_{L_0}\:\RP^n\to\RP^{n-k}$. Then $\g=p_{L_0}(\G)$ is
a convex curve in even-dimensional space $\RP^{n-k}$. The preimage $\p^{-1}(\A)
\subset M_k$ under mapping \eq(24) consists of hyperplanes that are the
projective joins of $L_0$ and $L'$ (the projective join $X\tstar Y$ of spaces
$X,Y\subset\RP^n$ is the union of all lines $xy$, where $x\in X$ and $y\in Y$).
Here $L'$ is a hyperplane in $\RP^{n-k}$ that does not intersect $\g$. Hence,
the preimage $\p^{-1}(\A)$ is homeomorphic to an open convex (by virtue of
Theorem~3) set $M_0(\g)\subset(\RP^{n-k})^*$, and thus it is homeomorphic to
the open ball of dimension $n-k$.

Now let us regard $\RP^n$ as $S^n/\!\!\sim$, where $S^n$ is the unit sphere in
$\R^{n+1}$ and $\sim$ is the central symmetry. Further, let us regard $\RP^{n-k
}$ as $\(S^n\cap(CL_0)^\perp\)/\!\!\sim$, where $CL_0^{\,\perp}\subset\R^{n+1}$
is the orthogonal complement of the cone over $L_0$, i.e., of the
$k$-dimensional subspace $CL_0\subset\R^{n+1}$ that pas through the origin and
corresponds to $L_0^{k-1}\subset\RP^n$. Let $L'_0$ be a fixed hyperplane in
space $\RP^{n-k}$ described above. Suppose that this hyperplane does not
intersect $\g=p_{L_0}(\G)$. Finally, let $U(\A)$ be a small neighborhood of
$\A$ in $S^k\G$ such that for any $L\in U(\A)$ the subspaces $L$ and $\RP^{n-k
}$ are in the general position (that is do not intersect) and $L\tstar L'_0\in
\p^{-1}(L)$. Here we identify divisors on $\G$ of degree $k$ and the
corresponding subspaces of dimension $n-1$. Choose an affine chart in $(\RP^{n-
k})^*$ that contains the whole $M_0(\g)$ and such that $L'_0$ is the origin. We
can assume that for any $L\in U(\A)$, the set $M_0(\g_L)$ lies in the chosen
chart, where $\g_L=p_L(\G)$. Let us define a homeomorphism $\Phi_L\:M_0(\g_L)
\to B$ of $M_0(\g_L)$ onto the open unit ball $B$ with center $L'_0$ in our
affine chart (which is a linear space $\R^{n-k}$\,!\,) by the formula
	$$\Phi_L(x)=\frac x{|x|\,m_L(x)},\qquad m_L(x)=
					\sup\{t\mid tx\in M_0(\g_L)\},$$
for $x\in\R^{n-k}\setminus0$, and by setting $\Phi_L(0)=0$. Here $m_L(x)$
depends on $x$ continuously, because $M_0(\g_L)$ is an open convex set that
contains the origin; this also implies that $m_L(x)$ is bounded away from zero.
Further, $m_L(x)$ also continuously depends on $L$, because $M_0(\g_L)\subset\R
^n$ continuously depend on $L$ if $L\in U(\A)$. Finally, it is easy to see that
the function $m_L(x)$ is continuous in $x$ and $L$ together. Hence $\Phi_L$
continuously depends on $L\in U(\A)$, and thus the mapping
		$$(L,x)\mapsto(L,\Phi_L(x))$$
is a homeomorphism between $\p^{-1}(U(\A))$ and $U(\A)\times B$; here $L\in U(
\A)$ and $x\in M_0(\g_L)\subset\R^{n-k}$. So the mapping \eq(24) is a fiber
bundle.

Since the fiber of \eq(24) is homeomorphic to the ball, there is a section $s\:
S^k\G\to M_k$ of this bundle. Really, all the obstructions to continuation of
$s$ lie in the trivial groups. Define the retraction $\Phi\:M_k\times[0,1]\to S
^k\G$ by the rule
		$$\Phi(x,t)=ty+(1-t)x,\tag25$$
where $t\in[0,1]$, $x\in M_k$, and $y=s(\p(x))$; this formula defines a
retraction of each fiber $F_y=\p^{-1}(\p(y))$ to its point $y$. Since $F_y
\subset\R^{n-k}$, the right-hand side of \eq(25) is defined. Note that \eq(25)
is a continuous mapping (we chose an affine chart $\R^{n-k}\subset\RP^{n-k}$
containing $F_y$ so that this chart continuously depend on $y$). Thus we get a
retraction of $M_k$ onto a set homeomorphic to $S^k\G$. Note that the previous
reasoning is extraneous if $k=n$.

We also explicitly determine the retraction $\Phi'\:S^k\G\times[0,1]\to S^1$.
Let $0\le\f_1\le\ldots\le\f_k<2\pi$ be the coordinates of a points of $\A$ on
$\G$, i.e., the values of the parameter $\f$ at these points, and let $\psi_1,
\dots,\psi_k$ be real numbers such that $\psi_{i+1}=\psi_i+2\pi/k$, $i=1,\dots,
-1$, and $\sum\l_{i=1}^k\psi_i=\sum\l_{i=1}^k\f_i$. Then the formula
	$$\Phi'(\A,t)=(t\psi_1+(1-t)\f_1,\dots,t\psi_k+(1-t)\f_k)\tag26$$
determines a retraction of $S^k\G$ onto the circle with the coordinate $\psi_1
\pmod{2\pi}$. We only need to show that the right-hand side of \eq(26) remains
continuous when some of $\f_i$ vanish. If $j$ numbers $\f_1,\dots,\f_j$ (here
$1\le j\le k$) simultaneously vanish when they are, to be definite, decreasing,
then the coordinates of the points of $\A$ change at this moment by the rule
	$$(0,\dots,0,\f_{j+1},\dots,\f_k)\mapsto
			(\f_{j+1},\dots,\f_k,2\pi,\dots,2\pi).$$
So all the $\psi_i$ increase by $2\pi j/k\,$, that is $\psi'_i=\psi_i+2\pi j/
k$. However, the equations of the motion of the $\f_l$, $j<l\le k$, do not
change, because the old dependence on $t$ has the form $t\psi_l+(1-t)\f_l$,
while the new one is $t\psi'_{l-j}+(1-t)\f_l$, where $\psi'_{l-j}=\psi_{l-j}+2
\pi j/k=\psi_l$. Similarly, the old equation of the motion of the $m$\snug th
point $\f_m=0$, where $m\le j$, is $t\psi_m+(1-t)\f_m=t\psi_m$, while the new
one, $t\psi'_{m+k-j}+(1-t)2\pi$, combined with the relation $\psi'_{m+k-j}=\psi
_{m+k-j}+2\pi j/k=\psi_m+2\pi(k-j)/k+2\pi j/k=\psi_m+2\pi$, becomes $t(\psi_m+2
\pi)+(1-t)2\pi=t\psi_m+2\pi$, which for any $t$ defines the same point of the
circle that the old equation, because it does not differ from the old equation
modulo $2\pi$. By the way, this fact is predictable. In fact, equation \eq(26)
determines a linear moving of the $\f_i$ into the vertices of some regular
$k$-gon. This moving preserves the order of the points and satisfies the
condition that it preserves the ``barycenter'' of the points, namely, that the
sum of their oriented angular velocities equals zero.

The retraction of $S^k\G$ onto $S^1$ defined by \eq(26) can be regarded as a
retraction onto $S^1$ of the section $s(S^K\G)\subset M$, which is the image of
$M$ under the retraction \eq(25). Thus $M_k$ is retractable to $S^1$. Now it
suffices to show that $M_k$ contains a curve that is not contractible in $(\RP^
n)^*$. For this curve we can take $s(\A(t))$, where $s\:S^k\G\to M_k$ is the
above-constructed section of the fibering \eq(24), while
	$$\A(t)=\(\frac{2\pi t}k,\frac{2\pi(t+1)}k,\dots,
			\frac{2\pi(t+k-1)}k\),\qquad t\in[0,1]$$
is a loop in $S^k\G$ with the endpoint $\A_0=(0,2\pi/k,\dots,2\pi(k-1)/k)$. The
curve $s(\A(t))\subset(\RP^n)^*$ intersects a hyperplane $A^*$, where $A\in\RP^
n$ is an arbitrary point of $\G$, exactly once, namely at the moment when one
of the points of $\A(t)$ pass through $A$. So the curve $s(\A(t))$ is not
homological zero, and thus it is not homotopic to zero.		\qed\enddemo

\remark{Remark} The subdivision of $(\RP^n)^*$ into the sets $M_k$
(respectively, of $\RP^n$ into the sets $M^*_k$) is the most economic with
respect to the Lyusternik--Shnirelman category. Recall (see~\cite{13, \S19}
or~\cite{10, \S46}) that the category $\cat_X(A)$ of a closed subset $A\subset
X$ of $X$ is the minimal number $k$ such that there exist closed (maybe
disconnected) subsets $A_1,\dots,A_k\subset X$ such that $A=\bigcup\l_{i=1}^kA_
i$ and each $A_i$ is contractible to a point in~$X$. It can be easily shown
that
\roster
\item"1)" if $A\subset B\subset X$, then $\cat_X(A)\le\cat_X(B)$;
\item"2)" if $A\subset B\subset X$ are closed subsets and $B$ can be
	  continuously deformed in $X$ into $A$, then  $\cat_X(A)=\cat_X(B)$;
\item"3)" if $A$ and $B$ are closed subsets of $X$, then $\cat_X(A\cup B)\le
	  \cat_X(A)+\cat_X(B)$;
\item"4)" if $X$ is a Riemannian manifold and $A\subset X$ is its compact
	  subset, then $\cat_X(U_\e(A))=\cat_X(A)$ for $\e>0$ small enough,
	  where $U_\e(A)$ denotes the closed $\e$-neighborhood of $A$ in~$X$.
\endroster
Moreover, it can be shown that $\cat(\RP^n)=n+1$ ($\cat(X)$ denotes the
category of $X$ in itself).

We shall present a subdivision $\RP^n=M'_1\cup M'_3\cup\dots\cup M'_n$ for $n$
odd, respectively a subdivision $\RP^n=M'_0\cup M'_2\cup\dots\cup M'_n$ for $n$
even, where the $M'_i$ are closed subsets of $\RP^n$ that are close to the $M_
i$, and $\cat_{\RP^n}(M'_0)=1$, while $\cat_{\RP^n}(M'_i)=2$ for $i>0$. For
this subdivision, the inequality of property~3) becomes an equality.

Let $M'_n=U_{\e_0}(M_n)$ be a closed $\e_0$-neighborhood of a closed (and
therefore compact) subset $M_n\subset\RP^n$. Properties~2) and~4) of the
category together with Theorem~13 imply that $\cat_{\RP^n}(M'_n)=\cat_{\RP^n}(M
_n)=\cat_{\RP^n}(S^1)=2$ for $\e>0$ small enough, because this circle $S^1
\subset\RP^n$ is not contractible. Further, put $M''_{n-2}=\ol{M_{n-2}\setminus
M'_n}$, where the line over the symbol means taking the closure. Since $M_n\cup
M_{n-2}$ is a closed set, we see that $M''_{n-2}\subset M_{n-2}$. For $M'_{n-2
}$ take the closed $\e_1$-neighborhood $U_{\e_1}(M''_{n-2})$, where $\e_1<\e_
0$. If an $\e_0$ is sufficiently small, then Theorem~13 and the properties of
the category imply that $\cat_{\RP^n}(M''_{n-2})=2$ (the sufficient condition
here is that $M''_{n-2}$ contains the section $s$ of~\eq(24) that is
constructed in the proof of Theorem~13). Then for $\e_1<\e_0$ small enough we
obtain $\cat_{\RP^n}(M'_{n-2})=\cat_{\RP^n}(M''_{n-2})=2$. In a similar way,
when the sets $M'_n,M'_{n-2},\dots,M'_{n-2k+2}$ are constructed, we define $M'_
{n-2k}$ as the closed $\e_k$-neighborhood of $M''_{n-2k}$, where $M''_{n-2k}$
is the closure of the set
	$M_{n-2k}\setminus\bigcup\l_{i=1}^kM'_{n-2i}$,
and $\e_k$ is a sufficiently small number such that $\e_k<\e_{k-1}<\ldots<\e_
0$. As above, it is easy to see that $\cat_{\RP^n}(M'_{n-2k})=2$ whenever $n-2k
>0$ and $\cat_{\RP^n}(M'_{n-2k})=1$ for $n=2k$. So we obtained a required
subdivision.							\endremark

Let us recall~\cite5 the definition of exterior lines and interior subspaces
(with respect to a convex curve $\G\subset\RP^n$).

\definition{Definition} A line in $\RP^n\setminus\G$ is said to be {\it
exterior\/} for $\G$ if for any point $A$ of this line the projection $\g=p_A(
\G)\subset\RP^{n-1}$ contains exactly $n$ geometrically different flattening
points. A plane of codimension 2 in $\RP^n$ is {\it interior\/} for $\G$ if any
containing it hyperplane intersects $\G$ at $n$ geometrically different points,
that is with the maximal possible multiplicity.			\enddefinition

In the other words, $l$ is an exterior line if and only if $l\subset\Int(M^*_n
\setminus\G)$; a plane $B$ of codimension 2 is interior if and only if its dual
line $l=B^*$ is contained in $\Int(M_n\setminus\G^*)$; here $\Int$ denotes the
interior of a set. If we neglect the condition that the flattening points,
respectively the points of intersections with a hyperplane, were geometrically
different, then the condition that $l$ belongs to $\Int(M_n\setminus\G^*)$ can
be replaced by a weaker one; nevertheless, it cannot simply be ignored, see
the proof of Theorem~14 below.

\proclaim{Theorem 14} Let $\G\subset\RP^n$ be a convex curve. Then there exist
interior \rom(for $\G$\rom) subspaces of codimension~\rom2 and exterior
\rom(for $\G$\rom) lines.					\endproclaim

We need the following statement, which was proved by
S.~M.~Gusein-Zade~\cite{12}. Recall that since the dimension is even, convex
curves are affine.

\proclaim\nofrills{Theorem}\ \ {\rm(S.~M.~Gusein-Zade).} Let $\G$ be a convex
curve in $\R^{2k}$. Then there exists a plane $B\subset\R^{2k}$ of codimension
\rom2 such that the linking number of $B$ and $\G$ equals $k$.	\endproclaim

Let us give here a proof of this statement, because it seems to be
unpublished.

\demo{Proof} Suppose that $\Pi\in M_{2k}$ and $\Pi\cap\G=\{A_1,\dots,A_{2k}\}$.
The points $A_i$ are the vertices of a $(2k-1)$-dimensional simplex $\Delta
subset\Pi$. Indeed, they do not belong to a hyperplane of dimension $(2k-2)$,
because the curve $\G$ is convex. By the same reason, all intersections of
$\Pi$ and $\G$ are transversal. Fix orientations of $\G$ and $\Pi$. Then the
intersection index of $\Pi$ and $\G$ is defined (which equals zero), and each
$A_i$ contributes $\pm1$ to this index. Therefore, there are $k$ points, say,
$A_1,\dots,A_k$, that contribute $+1$, and $k$ points, $A_{k+1},\dots,A_{2k}$,
that contribute $-1$. Let us draw two $(k-1)$-dimensional planes, a plane $B_+$
through $A_1,\dots,A_k$ and a plane $B_-$ through $A_{k+1},\dots,A_{2k}$. Then
there exists a unique plane $B$ of dimension $(2k-2)$ that is parallel to $B_+$
and $B_-$ and pass through the barycenter of the simplex. This plane $B$ is the
required one, because it is a hyperplane in $\Pi$ that separates $k$
``positive'' points of intersection of $\Pi$ and $\G$ from the $k$ ``negative''
ones.								\qed\enddemo

\demo{Proof of Theorem $14$} The duality involution takes exterior lines,
respectively interior subspaces, of $\G$ to interior subspaces, respectively
exterior lines, of the dual curve $\G^*$, which also is convex. So it is
sufficient to prove one of the statements of the theorem.

We prove that there are interior subspaces. For $n=2k$, it directly follows
from the Gusein-Zade theorem. In fact, let $B$ be a plane of codimension 2 such
that its linking number with $\G$ is $k$; we compute this number in a proper
chart that contains the whole $\G$. Let $\Pi$ be a hyperplane containing $B$.
Then there are at least $k$ (and thus exactly $k$) points of intersection of
$\Pi$ and $\G$ at any side from $B$, because a plane $B'$ parallel to $B$ and
lying in $\Pi$ far from $B$ is not linked with $\G$. Since $\G$ have no
self-intersections, all these points are geometrically different.

Now let $n$ be odd, $n=2k+1$. Suppose that $A\in\G$ and $\g=p_A(\G)\subset\RP^{
2k}$. There exists a plane $\beta^{2k-2}\subset\RP^{2k}$ interior for $\g$. Set
$B=p^{-1}_A(\beta)$. Consider a hyperplane $\Pi$ that pass through $B$, and its
projection $\pi=p_A(\Pi)$ on $\RP^{2k}$. By the above reasoning, $\#(\pi\cap\g)
=2k$. Then $\#(\Pi\cap\G)=2k+1$.

Since $B$ contains a point $A\in\G$, it is not an interior subspace for $\G$.
However, the required plane can be obtained as a small perturbation of $B$.

Let $\pi_0$ be a hyperplane in $\RP^{2k}$ that contains $\beta$ and $a$, where
$a=p_A(A)$. Put $\Pi_0=p_A^{-1}(\pi_0)$. If $\Pi\supset B$ and $\Pi\ne\Pi_0$,
then all intersections of $\Pi$ and $\G$ are transversal. If $\Pi=\Pi_0$, then
some two of $2k+1$ intersection points of $\Pi$ and $\G$ merge at $A$. Let us
slightly rotate $\Pi_0$ around the intersection $\Pi_0\cap\RP^{2k}=\pi_0$; here
we regard $\RP^{2k}$ as the subset of $\RP^{2k+1}$ formed by points that lie at
the distance $\pi/2$ from $A$ with respect to the metric induced by the
standard metric on the unit sphere $S^{2k+1}\subset\R^{2k+2}$, which is the
space of 2-fold covering over $\RP^{2k+1}$. A rotation in one of the possible
direction leads to the disappearing of the second order intersection at $A$,
while a small rotation of the second direction borns two transversal
intersections close to $A$. Let $\Pi_\e$ be the image of $\Pi_0$ under the
rotation around $\Pi_0\cap\RP^{2k}$ in the second direction; $\e$ here means
the angle of the rotation. Then $\#(\Pi_\e\cap\G)=2k+1$ and all intersections
are transversal. By $A_\e\in\Pi_\e$ denote the image of $A\in\Pi_0$ under this
rotation, and by $B_\e\subset\Pi_\e$ denote the image of the plane $B\subset\Pi
_0$ of dimension $2k-1$.

For $\e>0$ small enough, $B_\e$ is an interior subspace for $\G$. Indeed, $B_
\e$ does not intersect $\G$ for small $\e>0$ because of their general position
and due to the fact that the dimension of $\G$ is less than the codimension of
$B_\e$. Further, let $l_\e\subset(\RP^{2k+1})^*$ be the line dual to $B_\e$.
This line is the image of the line $l$ (which is dual to $B$) under the
rotation $G_\e$, that is under a small perturbation: $l_\e=G_\e(l)$. By $P\in
l$ denote the point dual to $\Pi_0$, and by $U(P)\subset l$ denote its small
neighborhood. Note that $l\subset M_{2k+1}$ and $l\cap\partial M_{2k+1}=P$,
while the inclusion $l_\e\subset\Int M_{2k+1}$ will prove the theorem. For $\e$
small enough, we have $G_\e(l\setminus U(P))\subset\Int M_{2k+1}$, because $l
\setminus U(P)$ is a compact subset of an open set $\Int M_{2k+1}$. Finally,
the inclusion $G_\e(U(P))\subset\Int M_{2k+1}$ holds due to our choice of the
direction of the rotation $G_\e$, see above. Therefore $B_\e$ is an interior
subspace for a proper $\e>0$. The theorem is proved.		\qed\enddemo

\remark{Remark} A subspace $B$ of codimension 2 is interior for a convex curve
$\G$ if and only if the projection $p_B\:\G\to\RP^1$ is a regular mapping of
degree $\pm n$. Indeed, a point $y\in\RP^1$ corresponds to a hyperplane $\Pi$
containing $B$, and preimages $p_B^{-1}(y)$ correspond to intersections points
of $\Pi$ and $\G$. Therefore any regular value has exactly $n$ preimages.
Really, the number of preimages cannot exceed $n$ because of the convexity of
$\G$. On the other hand, if $B$ is an interior subspace, then each preimage $p_
B^{-1}(y)$ contains exactly $n$ points that are either all negative or all
positive: otherwise the mapping $p_B\:\G\to\RP^1$ should have critical values.
\endremark

	\head \S7. Convex projections of convex curves.\endhead
In this section we prove that for a convex curve $\G\subset\RP^n$ there exist
lines $l\subset\RP^n\setminus\G$ such that the projection $\g=p_l(\G)$ is a
convex curve in $\RP^{n-2}$. Recall (see \S1) that the osculating hyperplane
$\Pi(\f)$ is defined at any point $\G(\f)$. The set of the pairs
		$$(\G(\f),\hbox{\ a point of\ }\Pi(\f))\tag27$$
is naturally identified with $S^1\times\RP^{n-1}$; this bundle over $S^1$ is
trivialized by means of the moving $n$-hedron. To each pair \eq(27) assign a
corresponding point of ambient space $\RP^n$. Then we obtain a $C^1$-smooth
(because $(n-1)$\snug th derivatives of the coordinates of a $C^n$-smooth curve
belong to the class $C^1$) mapping
		$$f\:S^1\times\RP^{n-1}\to\RP^n.\tag28$$
Let $y\in\RP^n$ be its regular value. Then $y$ belongs to $\Int M^*_k$, where
$k$ is the number of the preimages of $y$. Further, the sets $M^*_k$ are
separated from one another by the set of critical values of $f$. It will be
shown that the set $F$ of critical points of the mapping \eq(28) is formed by
all pairs of the form
		$$(\G(\f),\hbox{\ a point of\ }V(\f)),$$
where $V(\f)$ is the codimension 2 plane nearest to $\G(\f)$, i.e., the plane
such that the distance between it and $\G(\psi)$ is $O(\psi-\f)^{n-1}$. This
plane $V(\f)$ contains $\G(\f)$ and the first $n-2$ vectors of the moving
$n$-hedron. Furthermore, we have $F=S^1\times\RP^{n-2}$. Thus we formulate the
following statement.

\proclaim{Lemma 1} The subset of $\RP^n$ covered by the planes $V(\f)$
coincides with the union of the boundaries of $M^*_k$\rom:
	$$\bigcup\l_\f V(\f)=\bigcup\l_k\partial M^*_k=f(F).$$	  \endproclaim

In particular, for $n=2$ we have $V(\f)=\G(\f)$ and $f(F)=\partial M^*_0\cup
\partial M^*_2=\G(\f)$; the points of $M^*_0$ lie inside $\G(\f)$ (if $n=2$,
then $\G(\f)$ is an affine curve\,!\,), while the points of $M^*_2$ lie outside
of $\G(\f)$. If $n=3$, then the sets $M^*_1$ and $M^*_3$ are separated by the
tangent developable surface of $\G$, which is constituted by the tangent lines
of this curve.

Note~\cite{22} that the set covered by the planes $V(\f)$ is dual to the union
of the tangent lines of the dual curve $\G^*$.

\demo{Proof of Lemma $1$} Let $\R^n$ be an affine chart that contains $\G(\f_0
)$. By $x=(\f_0,x_1,\dots,x_{n-1})$ denote a point $x\in S^1\times\RP^{n-1}$
(we deal only with the affine part of $\RP^{n-1}$) of the form $x=\G(\f_0)+x_1v
_1(\f_0)+\ldots+x_{n-1}v_{n-1}(\f_0)$, where $v_i(\f_0)=\left.\frac{d^i}{d\f^i}
\G(\f)\right|_{\f=\f_0}$. In the other words, $x$ is a pair $(\f$, a point of
$\Pi(\f))$, and $(x_1,\dots,x_{n-1})$ are the coordinates in $\Pi(\f)$. Here
the parameter $\f$ in the neighborhood of $\f_0$ is assumed to be natural. Let
$h=0$ be the linear equation of $\Pi(\f)$ in $\R^n$. Let us compute the
differential of $f$. We have
    $$Df(x)\pmatrix\delta\f\\ \delta x_1\\ \vdots\\ \delta x_{n-1}\endpmatrix=
	\delta\f\,Df(x)\pmatrix1\\ 0\\ \vdots\\ 0\endpmatrix+
	\pmatrix\delta x_1\\ \vdots\\ \delta x_{n-1}\\ 0\endpmatrix;\tag29$$
here the first $n-1$ coordinates of the last summand are the coordinates in
$\Pi(\f)$, while the last coordinate equals $h$. By the Frenet formulas,
	$$Df(x)\pmatrix1\\ 0\\ 0\\ \vdots\\0 \\ 0\endpmatrix=
		\pmatrix1+\kappa_1x_2\\ -\kappa_1x_1+\kappa_2x_3\\
		\vdots\\ -\kappa_{n-3}x_{n-3}+\kappa_{n-2}x_{n-1}\\
		-\kappa_{n-2}x_{n-2}\\ -\kappa_{n-1}x_{n-1}\endpmatrix,\tag30$$
where $\kappa_1,\dots,\kappa_{n-1}$ are the twisting coefficients of $\G(\f)$
at $\f_0$; they do not equal zero, because the curve $\G$ is nondegenerate.
Equations \eq(29) and \eq(30) imply that the operator $Df(x)$ is degenerate if
and only if $x_{n-1}=0$. Consequently, $f(F)=\bigcup\l_\f V(\f)$. Also
equations \eq(29) and \eq(30) imply that the singular point $x=(\f_0,x_1,\dots,
x_{n-2},0)$ is a folding point whenever $(x_1,\dots,x_{n-2})\ne(0,\dots,0)$,
and thus the point $y=f(x)$ lies at the boundary of $M^*_k$ and $M^*_{k+2}$ for
a certain $k$. Hence $f(F)\subset\bigcup\l_k\partial M^*_k$. The inverse
inclusion $\bigcup\l_k\partial M^*_k\subset f(F)$ obviously holds, because the
regular values of $f$ are interior points of the $M^*_k$.

Finally, if $x\in S^1\times\RP^{n-1}$ is a point of the form $(\G(\f),y)$ with
$y\in\RP^{n-1}\setminus\R^{n-1}$ (i.e., $y$ lies in the infinite hyperplane),
then the same reasoning works with another affine chart $\R^n$ containing $\G(
\f)$ and $f(x)$ simultaneously.					\qed\enddemo

\proclaim{Lemma 2} The projection $\g=p_l(\G)$ is a regular curve if and only
if the line $l$ does not intersect $f(F)$.			\endproclaim

\demo{Proof} The fact that the projection $\g=p_l(\G)\subset\RP^{n-2}$ is
nondegenerate at a point $\f$ is equivalent to the linear independence of the
vectors $v_1(\f),v_2(\f),\dots,v_{n-2}(\f)$ determined by the curve $\g$ (these
vectors are determined in an affine chart; nevertheless, their linear
independence does not depend on the choice of the chart, see \S1); in this case
their span is $(n-2)$-dimensional. However, this span is nothing but $p_l(V(\f)
)$. If $l\cap V(\f)\ne\pusto$, then $\dim p_l(V(\f))<\dim V(\f)=n-2$ and the
curve $\g$ is not regular at the point $\f$. Now it suffices to note that
$\bigcup\l_\f V(\f)=f(F)$ (Lemma~1).				\qed\enddemo

Now let $\g=p_l(\G)$ be a convex curve, and $A$, $B$ be arbitrary points
of~$l$. Then $\g=p_Ap_B(\G)$. Consider the curve $\G'=p_B(\G)$. Since it is a
projection of a convex curve, it has exactly $n$ flattening points. On the
other hand, it follows from~\cite4 that $\G'$ has at least $n$ geometrically
different flattening points, because its projection $\g=p_A(\G')$ is convex.
Thus the number of flattening points of $\G'$ equals $n$, they are pairwise
geometrically different, and if $B\in l$, then $B\in\Int M^*_n$. Consequently,
if the projection $\g=p_l(\G)$ is convex, then the line $l$ lies in $\Int M^*_
n$. These lines are exterior with respect to $\G$; Theorem~14 ensure their
existence. So we have to consider only exterior lines. If $l$ is an exterior
line with respect to a convex curve $\G$, then the projection $\g=p_l(\G)$ is a
regular curve; however, we do not know yet whether the curve $\g$ is regular.
Our first aim is to prove that the space of exterior (with respect to given
convex curve $\G$) lines is connected. This fact is equivalent to the
connectedness of the space of interior subspaces. First, we consider the case
of even dimensional ambient space.

\proclaim{Lemma 3} The set of interior subspaces that lie in a hyperplane $\Pi
\in\Int M_{2k}(\G)$ is connected.				\endproclaim

\demo{Proof} Let $B_1$ and $B_2$ be interior subspaces lying in $\Pi$, and $A_1
,\dots,A_{2k}$ be the points of intersection (geometrically different\,!\,) of
$\G$ and $\Pi$. Suppose that $A_1,\dots,A_k$ are the ``positive'' intersection
points, while $A_{k+1},\dots,A_{2k}$ are the ``negative'' ones. A $2k-
2$-dimensional plane $B\subset\Pi$ is an interior subspace for $\G$ if and only
if the points $A_1,\dots,A_k$ lie from one side of $B$ and the points $A_{k+1},
\dots,A_{2k}$ lie from the other side. So all the points $A_1,\dots,A_k$ belong
to one of four parts of the plane $\Pi$ cut by $B_1$ and $B_2$, while all the
points $A_{k+1},\dots,A_{2k}$ belong to the ``opposite quadrant''. Then $B_1$
can be moved into $B_2$ by a rotation along the ($(2k-3)$-dimensional)
intersection $B_1\cap B_2$ (this does not work for $k=1$ but it is a trivial
case), and the points $A_i$ will not be covered at any time by the rotating
plane, which will sweep exactly two other quadrants of $\Pi$. Thus we connected
$B_1$ and $B_2$ by a family of interior subspaces. The lemma is proved.
								\qed\enddemo

\proclaim{Lemma 4} The space of interior (with respect to a given convex curve
$\G\subset\RP^{2k}$) subspaces is connected.			\endproclaim

\demo{Proof} Any interior subspace $B$ is contained in a certain hyperplane
$\Pi\in\Int M_{2k}$. It follows from the proof of Theorem~13 that the set $\Int
M_{2k}$ is connected. Indeed, $\Int M_{2k}$ is identified with $S^k\G\setminus
\Delta$, where $\Delta$ is a union of all diagonals, and the connectedness of
$S^k\G\setminus\Delta$ follows from the equation \eq(26), which determines a
retraction of this set onto $S^1$. Now our statement follows from the
connectedness of $\Int M_{2k}$ and Lemma~4.			\qed\enddemo

\proclaim{Theorem 15} Let $\G\subset\RP^n$ be a convex curve. Then there exists
a line $l\subset\RP^n\setminus\G$ such that the projection $\g=p_l(\G)$ is a
convex curve in $\RP^{n-2}$.					\endproclaim

\remark{Remark} Theorem~8 follows from the fact that for $n$ even, a line $l$
can be chosen in the infinite hyperplane; it will be proved in the last part of
this section. However, Theorem~15 is not a simpler version of Theorem~8,
because it admits all positive integers $n$, not only even.	\endremark

\demo{Proof} First suppose that $n$ is even. Let $n=2k$ and
	$$\G_0(\f)=(\sin\f,\cos\f,\sin2\f,\cos2\f,\dots,\sin k\f,\cos k\f).$$
Then $\G_0$ is a convex curve in $\R^{2k}$. If $l$ is an infinite line in the
span of two last coordinate axes, then $p_l(\G_0(\f))=\g_0(\f)=(\sin\f,\cos\f,
\dots,\sin(k-1)\f,\cos(k-1)\f)$ is a convex curve in $\R^{2k-2}$.

A convex curve $\G\subset\RP^{2k}$ is affine. So we can assume that $\G\subset
\R^{2k}$. The space of convex curves in $\R^{2k}$ is connected (see Corollary
of the Shapiro theorem, \S2). Let $\G_t$ be a family of convex curves in $\R^{2
k}$ that connects $\G$ and $\G_0$; here $\G_1=\G$. For any $t$ the space of
exterior lines, i.e., the lines contained in $\Int M^*_n(\G_t)$, is connected.
Moreover, it is an open subset of the Grassmannian $\operatorname{Gr}(2,n+1)$
and has the following property: if $l\subset\Int M^*_n(\G_t)$, then $l\subset
\Int M^*_n(\G_s)$ for $s$ belonging to some neighborhood of $t$. (The
connectedness of the space of exterior lines is equivalent to Lemma~4.) Hence,
there exists a continuous family $l_t$ of lines exterior for $\G_t$. Then the
projections $\g_t=p_{l_t}(\G_t)$ are nondegenerate curves (it was proved a few
lines above Lemma~3) continuously depending on $t$, and the curve $\g_0$ is
convex. This implies that all the curves $\g_t$ are convex. In particular, the
projection $\g_1=p_{l_1}(\G)$ is a convex curve, and the theorem is proved in
the case of even dimension.

If $n=2k+1$ is an odd number, some modification is required in the proof of
Lemma~3 and in the construction of a curve $\G_0$ and a line $l$ such that the
projection $\g_0=p_{l_0}(\G_0)$ is convex.

Consider interior subspaces $B_1$ and $B_2$ that are contained in a hyperplane
$\Pi$, $\Pi\in\Int M_{2k+1}(\G)$. Choose an affine chart so that $\G$
intersects the infinite hyperplane only once (so the hyperplane $\Pi$ is not
infinite) and the intersection point does not belong to $\Pi$. With respect to
this chart, $\G$ intersects $\Pi$ at $2k+1$ points; $k$ of them are, to be
definite, positive, and $k+1$ are negative, because $\G$ pass from one
half-space to the other $2k+1$ times when it crosses $\Pi$, and only once when
it crosses the infinite hyperplane.

An interior subspace $B$ is a hyperplane in $\Pi$ that separates the positive
intersection points of $\Pi$ and $\G$ from the negative ones. Otherwise the
degree of the projection $p_B\:\G\to\RP^1$ would be less than $n$; by Remark
after Theorem~14, however, this degree equals $n$. The remaining part of the
proof of Lemma~3 need not any modification. Finally, the construction of a
convex curves $\G_0\subset\RP^n$ and a line $l_0$ such that $\g_0=p_{l_0}(\G_0
)$ is a convex curve is similar to the construction in the proof of
Theorem~$7'$. Now the proof of Theorem~15 is complete.		\qed\enddemo

Now let us prove Theorem~8 (see the end of \S2). From now on $n$ is an even
number, $n=2k$. Taking into account Theorem~15, we see that it suffices to
prove that a line $l$ can be chosen in the infinite hyperplane. We shall show
that it is possible to choose a required line in any hyperplane belonging to $M
_0$, that is in any hyperplane that does not intersect $\G$. The dual
statement, which is equivalent to the original one, is as follows.

\proclaim{Proposition} Let $\G$ be a convex curve in $\RP^n$ and $A$ belongs to
$M^*_0(\G)$. Then there exists an interior for $\G$ subspace that pass
through~$A$.							\endproclaim

\demo{Proof} By $\A$ denote the union of the interior subspaces of $\G$. Then
any point of $\A$ belongs to some interior subspace, and it is sufficient to
show that $M^*_0(\G)\subset\A$. The proof of the Gusein-Zade theorem implies
that $A_0\in\A$ for any interior point $A_0$ of a simplex $A_1A_2\ldots A_n$
with vertices on $\G$. In this case $A_0$ belongs to $M^*_0(\G)$, because $\G$
lies from one side of any osculating hyperplane $\Pi(\f)$ and so points of $\Pi
\f)$ cannot belong to the interior of $\conv\G$.

Put $D=f(F)$; notation is as in Lemma~1.

\proclaim{Lemma 5} The inclusion $\da\subset D$ holds.		\endproclaim

This lemma will be proved a few lines below.

By Theorem~3, the set $M^*_0$ is convex in an appropriate affine chart. Connect
$A_0$ and $A$ by a line segment (in this chart). By Lemma~1, we have $D=\bigcup
\l_i\partial M^*_i$. So the segment $A_0A$ does not intersect $D$. Now Lemma~5
implies that this segment does not intersect $\da$. Therefore, $A\in\A$,
because $A_0\in\A$ and $A_0A\cap\da=\pusto$. Thus Theorem~8 follows from
Lemma~5.							\qed\enddemo

\demo{Proof of Lemma~$5$} Note that $\A$ is an open set, and its closure $\ol
\A$ is the union of $(n-2)$-dimensional subspaces that are either interior with
respect to $\G$ or limits of sequences of interior subspaces (this readily
follows from the fact that the Grassmannians are compact). Let $L$ be either
interior or boundary interior subspace and let $A$ be a point of $L$. Then $A
\in\ol\A$; moreover, $A\in\da$ if and only if the tangent vectors of the shifts
of $A$ under small admissible perturbations of $L$ (i.e., under the
perturbations in the class of interior and boundary interior subspaces) do not
form a 2-dimensional transversal to $L$ at $A$. An interior subspace $L$ admits
small perturbations of all directions. So it is sufficient to consider points
of boundary interior subspaces.

Assume that $L$ contains points $A_1,\dots,A_s$ of the curve $\G$; they need
not to be pairwise different, because multiple intersections can occur. Since
$\G$ is a convex curve, $s\le n-1$. Let $K$ be an $(s-1)$-dimensional subspace
of $L$ passing through $A_1,\dots,A_s$; for example, if $A_1=A_2$, then $K$
contains $A_1$ and the tangent vector of $\G$ at $A_1$. If $A\notin K$ (in this
case $K\ne L$ and $s<n-1$), then there is a small perturbation that takes a
boundary interior subspace $L$ to an interior subspace $L'$ containing $A$, and
thus $A\notin\da$. Indeed, if all the $A_i$ are different, then it suffices to
move $L$ at each of these points in a proper direction transversal to $\G$ (a
line intersecting a closed curve in $\R^3$ provides an instructive example).
Otherwise, if, for example, $A_1=A_2$, then $L$ should be shifted in a proper
direction at $A_1$, and the line in $L$ that was coinciding with $A_1A_2$
should be rotated in a proper direction; here $A_1A_2$ denotes the tangent line
of $\G$ at $A_1$ (in the other words, $A_1$ and $A_2$ are regarded as
infinitely close but different points).

Now consider the case $A\in K$. It may happen that $A$ belong to a $(p-1
)$-dimensional subspace passing through some $p$, $p<s$, points of $A_1,\dots,
A_s$. Changing the indexing (if necessary), we can assume that $A$ belongs to
the $(p-1)$-dimensional subspace containing $A_1,\dots,A_p$ ($p\le s$) but do
not belong to a subspace of dimension less than $p-1$ determined by some of the
points $A_1,\dots,A_s$. There are two possibilities: either there are different
points among $A_1,\dots,A_p$ (to be definite, $A_1\ne A_p$) or all these point
coincide.

First assume that $A_1=\ldots=A_k$ and $A_{p-l+1}=\ldots=A_p$ (where $k\ge1$,
$l\ge1$, and $k+l\le p$), while $A_1\ne A_p$ and points $A_i$ with $k<i<p-l+1$
(if any) differ from $A_1$ and $A_p$. We can perturb $L$ in the class of
admissible (that is, interior or boundary interior) subspaces by moving a point
of $k$-multiple intersection $A_1$ and a point of $l$-multiple intersection $A_
p$ along $\G$. In the linear approximation (with respect to shifts of $A_1$ and
$A_p$), the point $A$ obtains new degrees of freedom: instead of the subspace
$K^{s-1}$, now it can move in the subspace $N$, that contains $K$ and
intersects $\G$ at $A_1$ and $A_p$ with the multiplicities exceeding the old
multiplicities by one. Since $\G$ is a convex curve, $\dim N=\dim K+2=s+1$,
because $N$ and $\G$ have $(s+2)$ intersection points according to
multiplicities. Furthermore, it can be assumed that $N\cap L=K$, because any
subspace of dimension $(n-2)$ that pass through $K$ is admissible provided that
it is sufficiently close to $L$. Thus the sum of the tangent spaces $T_AN$ and
$T_AL$ is $T_A\RP^n$, and admissible perturbations of $L$ together with shifts
of $A$ in $L$ can cause a small shift of $A$ in any direction. Hence, $A\notin
\da$ in this case.

Therefore, the condition $A\in\da$ implies that $A_1=\ldots=A_p$. Then $A\in V(
A_1)\subset f(F)$ (we use notation of Lemma~1), i.e., $A\in D$, where $A$ is an
arbitrary point of $\da$. The proof of Lemma~5 is finished, and the proof of
Theorem~8 also is finished.					\qed\enddemo

\newpage 

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\end
