%Date: Wed, 23 Sep 1998 17:44:51 -0400 (EDT)
%From: Sergei S Anisov <anisov@IAS.EDU>

\documentclass[12pt]{article}
\usepackage{amsfonts}
\def\A{{\cal A}}
\def\bez{\setminus}
\def\G{\Gamma}
\def\g{\gamma}
\def\RP{{\mathbb R}{\bf P}}

\title{On the number of flattening points of space curves}
\author{S.Anisov}
\date{September 25, 1998}

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This note contains a simple proof of the main result of V.I.Arnold's paper 
``On the number of flattening points of space curves''.

Let $\G\in\RP^{n+1}$ be a curve such that $\g=p_O(\G)$ is a convex
curve in $\RP^n$, where $p_O$ is the central projection from a point
$O\in\RP^{n+1}\bez\G$. I want to prove that there are at least
$n+2$ flattening points on $\G$.

Let $\A\in\RP^{n+1}$ be a set of points such that the projection
$p_A(\G)\subset\RP^n$ is convex. It is easy to see that $\A$ is
bounded by the osculating hyperplanes of $\G$. Then it is no
harder to show that in fact $\A$ is bounded by the osculating
hyperplanes at the flattening points of $\G$ only (S.Sedykh also
found this).

By $N$ denote the number of flattening points of $\G$. Suppose that
$N<n+2$. Then these osculating hyperplanes do not bound a simplex
in ambient $\RP^{n+1}$ and have (at least one) common point $Q$.
Draw the line $l=OQ$. Then $l\bez Q\subset\A$. Thus all the osculating
hyperplanes of $\G$ intersect $l$ at $Q$, because the intersection
points cannot belong to $\A$. Now it is not hard to get a contradiction.
Consequently, $N\ge n+2$. The theorem is proved.

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