\documentstyle[12pt]{article}
\title{Determining the Knot Type of a Polynomial Knot}
\author{Bryant Mathews}
\date{\today}
\newcommand{\K}{\vec K}
\newcommand{\IR}{{\bf R}}
\newcommand{\I}{{\bf I}}
\newcommand{\G}{{\bf G}}
\newcommand{\T}{{\bf T}}
\newcommand{\C}{{\bf C}}
\newcommand{\BO}{{\bf O}}
\newcommand{\W}{{\bf W}}
\newtheorem{Prop}{Proposition}
\newtheorem{Cor}[Prop]{Corollary}
\newtheorem{Proof}{Proof}
\newcommand{\rBox}{\hfill \Box}
\begin{document}
\maketitle
\section{Introduction}
A polynomial knot is a smooth embedding $\K:{\IR} \longrightarrow \IR ^{3}$, described by $\K(t)=(x(t), y(t), z(t))$, where
\begin{eqnarray}
x(t) & = & a_dt^d+\cdots +a_1t \nonumber \\
y(t) & = & b_dt^d+\cdots +b_1t \\
z(t) & = & c_dt^d+\cdots +c_1t. \nonumber
\end{eqnarray}
See \cite{Eli} for an algorithm which determines whether a polynomial mapping from $\IR \longrightarrow \IR ^3$ is a smooth embedding.
Our goal is to understand the coefficient space of polynomial knots. We would like an algorithm which, given the coefficients $a_i,b_i,c_i$, for $1\leq i\leq d$, will return the knot type of the polynomial knot. In simple cases, this can be accomplished by inspection.
\begin{enumerate}
\item{plot $x(t)$ versus $y(t)$}
\item{estimate $t$ values at the crossings}
\item{calculate $z(t)$ at these crossings}
\item{sketch $xy$-projection with overcrossings and undercrossings}
\item{identify knot type}
\end{enumerate}
For more complicated examples, inspection does not suffice. We must resort to algebraic methods. The general algorithm may be outlined as follows.
\begin{enumerate}
\item{calculate crossing data}
\item{calculate invariants using knot software}
\item{consult knot table}
\end{enumerate}
The remainder of this paper will focus on a method for completing the first step of this algorithm. The other two steps are self-explanatory.
\section{Expressing Crossings as Roots of Polynomials}
The projection $\K_p:\IR \longrightarrow \IR ^2$ of the polynomial knot $\K$ onto the $xy$-plane is given by $\K_p(t)=(x(t),y(t))$, with
\begin{eqnarray}
x(t) & = & a_dt^d+\cdots +a_1t \\
y(t) & = & b_dt^d+\cdots +b_1t. \nonumber
\end{eqnarray}
Each crossing in the projection $\K_p$ corresponds to a pair $(s,t)$ such that $\K_p(s)=\K_p(t)$, with $s\not= t$. In Appendix \ref{firstappendix}, we show that this is equivalent to the condition
\begin{equation}
\frac{\K_p(s)-\K_p(t)}{s-t}=\vec 0,
\end{equation}
where the expression on the left is an ordered pair of polynomials. Thus our problem of finding the crossings in $\K_p$ is reduced to the calculation of the common roots of two polynomials in two variables.
\section{Solving the Equations}
One way to find these roots is to use the Gr\"{o}bner basis method. We will discuss only the practical aspects of its use. For a rigorous treatment, the reader may consult \cite[chapter 3]{IVA}.
With Gr\"{o}bner bases, we can eliminate the parameter $s$ from our equations. We then solve for the possible $t$ values, and see which extend to solutions $(s,t)$ of our original system.
We proceed as follows:
\begin{enumerate}
\item{Let $\I (\K_p)\subset \IR [s,t]$ be the ideal
\begin{equation}
\I (\K_p)=\left< \frac{x(s)-x(t)}{s-t}, \frac{y(s)-y(t)}{s-t}\right>,
\end{equation}
generated by two polynomials. We calculate a Gr\"{o}bner basis ${\G}(\K_p)=(g_1,\ldots ,g_n)$ for ${\I}(\K_p)$ with respect to the lexicographic order where $s>t$.}
\item{Let
\begin{equation}
\tilde{\G}(\K_p)=\left\{g_i\in \G (\K_p)\left| \right. g_i(s,t)=g_i(s',t) \mbox{ for all } s,s',t\in \IR \right\}.
\end{equation}
Then $\tilde{\G}(\K_p)$ contains precisely the polynomials in $\G (\K_p)$ which depend only on $t$. We calculate
\begin{equation}
\T (\K_p)= \left\{ t\in \IR \left| \right. g_i(t)=0\mbox{ for all } \tilde{\G}(\K_p) \right\},
\end{equation}
the set of common roots of the polynomials in $\tilde{\G}(\K_p)$.
If $\tilde{\G}(\K_p)$ is empty, then there are infinitely many ``crossing'' points, and we must repeat the entire algorithm with a different projection of $\K$. }
\item{Fix an element of $\T (\K_p)$. We substitute this element for $t$ in the polynomials of $\G (\K_p)-\tilde{\G}(\K_p)$, which depend on $s$, and we solve for the $s$ values which are common roots of these polynomials. Each ordered pair $(s,t)$, where $s$ is a root and $s\not= t$, corresponds to a crossing point in $\K_p$. If $s$ is a root and $s=t$, then $(s,t)$ corresponds to a cusp in $\K_p$, and may be ignored (we assumed in the beginning that $\K $ itself has no cusps). If $s\not= s'$ are two roots, then $(s,t)$ and $(s',t)$ correspond to a triple crossing in $\K_p$, and we must repeat the entire algorithm using some other projection of $\K$.
We repeat this step for each element of $\T (\K_p)$, and define the crossing set $\C (\K_p)$ to be the set of all ordered pairs $(s,t)$ which correspond to crossings. Indeed, for all $(s,t)\in \C (\K_p)$, equation $(3)$ is satisfied, that is, $x(s)=x(t)$ and $y(s)=y(t)$, with $s\not= t$. }
\item{For each ordered pair $(s,t)\in \C (\K_p)$, it is clear that $(t,s)\in \C (\K_p)$ as well. Since these two ordered pairs refer to the same crossing, we need to keep only one of them. Without loss of generality, we define the reduced crossing set
\begin{equation}
\C_{red}(\K_p)=\left\{ (s,t)\in \C (\K_p) \left| \right. s x= y=}
% Leave vertical space equal to using "\vskip" before this command.
% Optional: (use / instead of \), specifies path of TeX file if not supplied.
% If one dimension is not provided, BMP is scaled to the other.
% If both dimensions are not provided, actual BMP dimensions are used.
% Example: \special{bmp: c:/mysubdir/mypic.bmp x=3in y=5cm}
\vskip1.3in
\special{bmp: Trefoil.bmp }
\end{enumerate}
\section{Conclusion}
The algorithm described in this paper has yet to be fully implemented, though it has been carried out by the author many times, using Maple. Unfortunately, as the degree of the polynomials involved grows, the time necessary for the Gr\"{o}bner basis calculations increases dramatically, making the algorithm impractical. Additional speed could be gained by the use of Macaulay or some other software designed specifically for computational algebraic geometry.
\appendix
\section{}
\label{firstappendix}
We claim that $\frac{\K_p(s)-\K_p(t)}{s-t}$, in equation $(3)$, is indeed an ordered pair of polynomials. This is a direct result of the following proposition.
\begin{Prop}
\label{appendixprop}
Let $f(s,t)\in \IR [s,t]$. If $s=t\Longrightarrow f(s,t)=0$, then $(s-t)$ divides $f(s,t)$.
\end{Prop}
\begin{Proof}
{\em
Assume $s=t\Longrightarrow f(s,t)=0$. Applying the division algorithm in $\IR [s,t]$ with lexicographic order $(s>t)$, we may write
\begin{equation}
f(s,t)=g(s,t)(s-t)+r(t), \mbox{ for some } g(s,t)\in \IR [s,t],\: r(t)\in \IR [t].
\end{equation}
When $s=t$, we have
\begin{equation}
f(s,t)=g(s,t)\cdot 0+r(t)=0
\end{equation}
by assumption, so $r(t)=0$, and
\begin{equation}
f(s,t)=g(s,t)(s-t).
\end{equation}
Hence $s-t$ divides $f(s,t)$. $\rBox $
}
\end{Proof}
\begin{Cor}
The vector $\frac{\K_p(s)-\K_p(t)}{s-t}$ is an ordered pair of polynomials.
\end{Cor}
\begin{Proof}
{\em
By definition,
\begin{equation}
\frac{\K_p(s)-\K_p(t)}{s-t}=\left( \frac{x(s)-x(t)}{s-t}, \frac{y(s)-y(t)}{s-t}\right) .
\end{equation}
Applying Proposition \ref{appendixprop} to $x(s)-x(t)$ and $y(s)-y(t)$, the result follows. $\rBox $
}
\end{Proof}
\begin{thebibliography}{9}
\bibitem{Adams} Adams, C. 1994. {\em The Knot Book.} New York: W. H. Freeman.
\bibitem{IVA} Cox, D., J. Little, and D. O'Shea. {\em Ideals, Varieties, and Algorithms.} New York: Springer-Verlag.
\bibitem{Eli} Lebow, E. B. {\em The Discriminant.} Research Experience for Undergraduates. Mt. Holyoke, 1998.
\end{thebibliography}
\end{document}