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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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