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\centerline{\LARGE M.Sc.~Math~Workshop --- Assignment \#6}
\centerline{\large HUJI Spring 1998}
\centerline{Dror Bar-Natan}
\setcounter{figure}{3}
\begin{enumerate}
\setcounter{enumi}{20}
\item Give two proofs that the configuration space of a 6-legged roach
(such as the one in Figure~\ref{Roach}) is an oriented surface of
genus 17:
\begin{itemize}
\item Using Euler characteristics,
\item And by a direct cut-and-paste argument.
\end{itemize}
\begin{figure}[htpb]
\[ \eepic{Roach}{0.5} \]
\caption{A roach. Don't worry! It is safely attached to this page and it
cannot jump out.}
\lbl{Roach}
\end{figure}
\item Which way did the bicycle go?
% << Bicycle.m
\[ \eepic{BikePath}{0.45} \]
\item A {\em smooth pre-knot} is a smooth ($C^\infty$) embedding of $S^1$
(the circle) in ${\bold R}^3$. We say that two smooth pre-knots are
smoothly equivalent if there is a smooth homotopy between them, which is
also an embedding at all intermediate times. A {\em smooth knot} is an
equivalence class of smooth pre-knots, modulo smooth
equivalence. Similarly we can define {\em continuous knots}. Are the two
notions equivalent?
\item We say that two smooth pre-knots $\gamma_{1,2}:S^1\to{\bold R}^3$
are smoothly ambient-equivalent if there exists a diffeomorphism (smooth
bijection with a smooth inverse) $f:{\bold R}^3\to{\bold R}^3$ so that
$\gamma_2=f\circ\gamma_1$. A {\em smooth ambient knot} is an equivalence
class of smooth pre-knots modulo smooth ambient-equivalence. Similarly
we can define {\em continuous ambient knots}. Are the two notions
equivalent?
\item Are smooth ambient knots equivalent to smooth knots?
\item A {\em polygonal pre-knot} is simply a polygon embedded in ${\bold
R}^3$. Two polygonal pre-knots are called $\Delta$-equivalent
if they differ by a sequence of triangle moves such as in
Figure~\ref{TriangleMove}, in which no other parts of the polygon pass
through the triangle. A {\em polygonal knot} is an equivalence class of
polygonal pre-knots modulo $\Delta$-equivalence. Are polygonal knots
equivalent to smooth knots?
\begin{figure}[htpb]
\[ \eepic{TriangleMove}{0.5} \]
\caption{A triangle move.}
\lbl{TriangleMove}
\end{figure}
\item Come up with a reasonable notion of ``a knot projection'' (in
${\bold R}^2$, of a knot in ${\bold R}^3$). We say that two knot
projections are $R$-equivalent if they differ by a sequence of
``Reidemeister'' moves of kinds $R1$, $R2$, and $R3$, as shown in
Figure~\ref{Reidem}. Prove that the set of polygonal knots is equivalent
to the set of knot projections modulo $R$-equivalence.
\begin{figure}[htpb]
\[ \eepic{Reidem}{0.5} \]
\caption{The three Reidemeister moves.}
\lbl{Reidem}
\end{figure}
\item An ``interval'' in a knot projection is precisely what you think it
is. For example, the standard projection of the trefoil knot is made of
three intervals. If $P$ is knot projection, define $\nu(P)$ to be YES if
the intervals of $P$ can be colored with three colors so that
\begin{itemize}
\item all three colors are used,
\item and in each crossings, the three intervals involved are either
all colored the same way or in three different colors.
\end{itemize}
Otherwise set $\nu(P)$ to be NO. Prove that $\nu$ is a knot invariant
(namely, it is invariant under the three Reidemeister moves), and
compute it on the unknot and on the trefoil knot.
\end{enumerate}
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