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\begin{document} %\latintext
\newdimen\captionwidth\captionwidth=\hsize
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\title{A Perturbed-Alexander Invariant}

\author{Dror~Bar-Natan}
\address{
  Department of Mathematics\\
  University of Toronto\\
  Toronto Ontario M5S 2E4\\
  Canada \ifpub{(corresponding author)}{}
}
\email{drorbn@math.toronto.edu}
\urladdr{http://www.math.toronto.edu/drorbn}

\author{Roland~van~der~Veen}
\address{
  University of Groningen, Bernoulli Institute\\
  P.O. Box 407\\
  9700 AK Groningen\\
  The Netherlands \ifpub{(corresponding author)}{}
}
\email{roland.mathematics@gmail.com}
\urladdr{http://www.rolandvdv.nl/}

\date{First edition ?, this edition \today.}

\subjclass[2010]{TBW}
\keywords{TBW}

\thanks{This work was partially supported by NSERC grant RGPIN-2018-04350. It is available in electronic form, along with source files and a verification {\sl Mathematica} notebook at \url{http://drorbn.net/APAI} and at \arXiv{2201.?????}.}

\begin{abstract}
In this note we give concise formulas, which lead to a simple and fast computer program, which computes a powerful knot invariant. This invariant $\rho_1$ is not new, yet our formulas are by far the simplest and fastest: given a knot $K$ we write one of the standard matrices $A$ whose determinant is the Alexander polynomial of $K$, yet instead of computing its determinant we consider a certain quadratic expression in the entries of $A^{-1}$. The proximity of our formulas to the Alexander polynomial suggest that they should have a topological explanation. This we don't have yet.
\end{abstract}

\maketitle

%\setcounter{tocdepth}{3}
%\tableofcontents
%\eject


\section{The Formulas} The selling point for this article is that the formulas in it are concise. Thus we start by running through these formulas for a knot invariant $\rho_1$ as quickly as we can. In Section~\ref{sec:Implementation} we turn the formulas into a short yet very fast computer program, in Section~\ref{sec:Cars} we give a partial interpretation of the formulas in terms of car traffic on a knot diagram and use it to prove the invariance of $\rho_1$, and in Section~\ref{sec:Context} we quickly sketch the context: Alexander, Burau, Jones, Melvin, Morton, Rozansky, Overbay, Ohtsuki, and our own prior work.

\parpic[r]{\input{figs/SampleKnot.pdf_t}}
Given an oriented $n$-crossing knot $K$, we draw it in the plane as a long knot diagram $D$ in such a way that the two strands intersecting at each crossing are pointed up (that's always possible because we can always rotate crossings as needed), and so that at its beginning and at its end the knot is oriented upward. We call such a diagram an {\em upright knot diagram.} An example for an upright knot diagram $D_3$ is here on the right.

We then label each edge of the diagram with two integer labels: a running index $k$ which runs from 1 to $2n+1$, and a ``rotation number'' $\varphi_k$, the geometric rotation number of that edge (the signed number of times the tangent to the edge is horizontal and heading right, with cups counted with $+1$ signs and cups with $-1$; this number is well defined because at their ends, all edges are headed up). On the right the running index runs from $1$ to $7$, and the rotation numbers for all edges is $0$ except for $\varphi_4$, for which it is $-1$.

{\em A Technicality.} Some Reidemeister moves create or lose an edge and to avoid the need for renumbering it is beneficial to also allow labeling the edges with non-consecutive labels. Hence we allow that, and write $\ip$ for the successor of the label $i$ along the knot, and $\ipp$ for the successor of $\ip$ (these are $i+1$ and $i+2$ if the labeling is by consecutive integers). Also, ``$1$'' will always refer to the label of the first edge, and ``$2n+1$'' will always refer to the label of the last.

We let $A$ be the $(2n+1)\times(2n+1)$ matrix of Laurent polynomials in the formal variable $T$ defined by
\[ A \coloneqq I - \sum_c \left( T^sE_{i,\ip} + (1-T^s)E_{i,\jp} + E_{j,\jp} \right), \]
where $I$ is the identity matrix, $E_{\alpha\beta}$ denotes the elementary matrix with $1$ in row $\alpha$ and column $\beta$ and zeros elsewhere, the summation is over the crossings $c$ of the diagram $D$, and once $c$ is chosen, $s$ denotes its sign and $i$ and $j$ denote the labels below the crossing where the label $i$ belongs to the over strand and $j$ to the under strand.

Alternatively, $A=I + \sum_c A_c$, where $A_c$ is a matrix of zeros except for the blocks as follows:
\begin{equation} \label{eq:A}
  \begin{array}{c}\input{figs/Xings.pdf_t}\end{array}
  \qquad\longrightarrow\qquad
  \begin{array}{c|cccc}
    A_c &   \text{column }\ip  &  \text{column }\jp \\
    \hline
    \text{row }i & -T^s  & T^s-1 \\
    \text{row }j & 0  & -1
  \end{array}
\end{equation}

\parpic[r]{\input{figs/D2.pdf_t}}
For example, if $D=D_1=\uparrow$ is the diagram with no crossings, the resulting matrix $A$ is the $1\times 1$ identity matrix $(1)$. If $D=D_2$ is the diagram on the right (here $s=+1$, $(i,j)=(2,1)$, and $(k,l)=(3,2)$), then
\[ A =
  \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}
  + \begin{pmatrix}0&-1&0\\0&T-1&-T\\0&0&0\end{pmatrix}
  = \begin{pmatrix}1&-1&0\\0&T&-T\\0&0&1\end{pmatrix},
\]
and for $D_3$ as on the first page, we have
\[ A = \footnotesize \left(
  \begin{array}{ccccccc}
   1 & -T & 0 & 0 & T-1 & 0 & 0 \\
   0 & 1 & -1 & 0 & 0 & 0 & 0 \\
   0 & 0 & 1 & -T & 0 & 0 & T-1 \\
   0 & 0 & 0 & 1 & -1 & 0 & 0 \\
   0 & 0 & T-1 & 0 & 1 & -T & 0 \\
   0 & 0 & 0 & 0 & 0 & 1 & -1 \\
   0 & 0 & 0 & 0 & 0 & 0 & 1 \\
  \end{array}
\right) \]

\picskip{0}
We note without supplying details that the matrix $A$ comes in a straightforward way from Fox calculus as it is applied to the Wirtinger presentation of the fundamental group of the complement of $K$ (using the diagram $D$), and hence the determinant of $A$ is equal up to a unit to the Alexander polynomial of $K$. In fact, we have that
\begin{equation} \label{eq:Delta} \Delta = T^{(-\varphi(D)-w(D))/2}\det(A), \end{equation}
where $\varphi(D)\coloneqq\sum_k\varphi_k$ is the total rotation number of $D$, where $w(D)=\sum_cs_c$ is the writhe of $D$, namely the sum of the signs $s_c$ of all the crossings $c$ in $D$, and where $\Delta$ is the normalized Alexander polynomial of $K$, so $\Delta(T)=\Delta(T^{-1})$ and $\Delta(1)=1$.

For our example $D_2$, $\det(A)=T$, $\varphi(D)=1$, and $w(D)=1$, so $\Delta=T^{(-1-1)/2}\cdot T=1$, as expected for a diagram of the unknot.
For $D_3$, $\det(A)=1-T+T^2$, $\varphi(D)=-1$, and $w(D)=3$, so $\Delta=T^{(1-3)/2}(1-T+T^2)=T-1+T^{-1}$, as expected for the trefoil knot.

We set $G=(g_{\alpha\beta})=A^{-1}$, and taking our inspiration from physics, we name $g_{\alpha\beta}$ the {\em Green's function} for the diagram $D$. For our three examples $D_1$, $D_2$, and $D_3$, the Green's function $G$ is respectively
\[
  \begin{pmatrix} 1 \end{pmatrix},
  \quad \begin{pmatrix} 1&T^{-1}&1\\0&T^{-1}&1\\0&0&1 \end{pmatrix},
  \quad \text{and}
  \quad \footnotesize \left(
    \begin{array}{ccccccc}
     1 & \frac{T^3-T^2+T}{T^2-T+1} & 1 &
       \frac{T^3-T^2+T}{T^2-T+1} & 1 &
       \frac{T^3-T^2+T}{T^2-T+1} & 1 \\
     0 & 1 & \frac{1}{T^2-T+1} & \frac{T}{T^2-T+1} &
       \frac{T}{T^2-T+1} & \frac{T^2}{T^2-T+1} & 1 \\
     0 & 0 & \frac{1}{T^2-T+1} & \frac{T}{T^2-T+1} &
       \frac{T}{T^2-T+1} & \frac{T^2}{T^2-T+1} & 1 \\
     0 & 0 & \frac{1-T}{T^2-T+1} & \frac{1}{T^2-T+1} &
       \frac{1}{T^2-T+1} & \frac{T}{T^2-T+1} & 1 \\
     0 & 0 & \frac{1-T}{T^2-T+1} & \frac{T-T^2}{T^2-T+1}
       & \frac{1}{T^2-T+1} & \frac{T}{T^2-T+1} & 1 \\
     0 & 0 & 0 & 0 & 0 & 1 & 1 \\
     0 & 0 & 0 & 0 & 0 & 0 & 1 \\
    \end{array}
  \right).
\]

We can now define our invariant $\rho_1$. It is the sum of two sums. The first is a sum of a term $R_1(c)$ over all crossings $c$ in $D$, where for such a crossing we let $s$ denote its sign and we let $i$ and $j$ denote the edge labels of the incoming over- and under-strands, respectively and where
\begin{equation} \label{eq:R1}
  R_1(c) \coloneqq s \left(
    g_{ji} \left(g_{\jp,j}+g_{j,\jp}-g_{ij}\right)
    -g_{ii} \left(g_{j,\jp}-1\right)
    -1/2
  \right).
\end{equation}
The second sum is a sum over the edges $k$ of $D$ of a correction term dependent on the rotation number $\varphi_k$. We multiply the result by $\Delta^2$ to ``clear the denominators''\footnote{$R_1(s)$ is quadratic in the entries of $G$ and hence it has denominators proportional to $\Delta^2$.}:
\begin{equation} \label{eq:rho1}
  \rho_1 \coloneqq \Delta^2\left(\sum_c R_1(c) - \sum_k\varphi_k\left(g_{kk}-1/2\right)\right),
\end{equation}

Direct calculations show that $\rho_1(D_1)=0$ (as the sums are empty), $\rho_1(D_2)=0$, and $\rho_1(D_3) = -T^2 +2T - 2 + 2/T - 1/T^2$.

\begin{theorem} \label{thm:Main} (Proof in Section~\ref{sec:Cars}) The quantity $\rho_1$ is a knot invariant.
\end{theorem}

As we shall see in the next section, $\rho_1$ has more separation power than the Jones polynomial, yet it is closer to the more topologically-meaningful Alexander polynomial $\Delta$: it is cooked up from the same matrix $A$ and in terms of computational complexity, computing $\rho_1$ is not very different from computing $\Delta$. In order to compute $\Delta$ we need to compute the determinant of $A$, while to compute $\rho_1$ we need to invert $A$ and then compute a sum of $O(n)$ terms that are quadratic in the entries of $A^{-1}$. We have computed $\rho_1$ for knots with over 200 crossings using the unsophisticated implementation presented in Section~\ref{sec:Implementation}.

Topologists should be intrigued! $\rho_1$ is cooked from the same matrix as the Alexander polynomial $\Delta$, yet we have no topological interpretation for $\rho_1$.

\section{Implementation and Power} \label{sec:Implementation}

\input APAINB.tex

\section{Cars, Traffic Counters, Interchanges, and a Proof of Theorem~\ref{thm:Main}} \label{sec:Cars}

Two upright knot diagrams are considered the same (as diagrams) if their underlying knot diagrams are the same, and if the respective rotation numbers of their edges are all the same. It is clear that $\rho_1$ is well defined on upright knot diagrams. To prove Theorem~\ref{thm:Main} we need to know what to prove. Namely, when do two upright knot diagrams represent the same knot? This is answered by the following:

\begin{figure}
\[ \input{figs/UprightRMoves.pdf_t} \]
\caption{
  The upright moves: Reidemeister 1 left and right, Reidemeister 2 braid-like and cyclic, Reidemeister 3, and (the $+$) Swirl.
} \label{fig:UprightRMoves}
\end{figure}

\begin{claim} \label{claim:UprightRMoves} In the spirit of the classical Reidemeister theorem, two upright knot diagrams represent the same knot if and only if they differ by a sequence of R1l, R2r, R2b, R2c, R3, and S moves as in Figure~\ref{fig:UprightRMoves}.
\end{claim}

\noindent{\em Sketch of the proof.} In the case of round knots, knot diagrams can be turned upright by rotating individual crossings, and the only ambiguity here is by powers of the full rotation, the {\em swirls} Sw$^+$ and Sw$^-$ (where Sw$^-$ is the same as Sw$^+$ except with a negative crossing, and we don't need to impose it separately as it follows from Sw$^+$ and R2). Hence we have a well-defined map $\{\text{knot diagrams}\}\to\{\text{upright knot diagrams}\}/$Sw$^\pm$. It remains to write the usual Reidemeister moves between knot diagrams as moves between upright knot diagrams. The result are the moves R1l, R2r, R2b, R2c, and R3. Note that unoriented knot theory is presented with just three Reidemeister moves, but these split into several versions in the oriented case. The sufficiency of the versions we picked can be found in~\cite{Polyak:RMoves}. In the case of long knots a minor further complication arises, regarding the rotation numbers of the initial and final edges. We leave the details of the problem and its resolution to the reader. \qed

Our key formulas,~\eqref{eq:R1} and~\eqref{eq:rho1} involve the Green's function $g_{\alpha\beta}$. We need to know that it is subject to some relations, the {\em $g$-rules}: (we recommend reading the proof of the following lemma before reading its statement)

\begin{lemma} \label{lem:gRules} Given a fixed upright knot diagram $D$, its corresponding matrix $A$, and its inverse $G=(g_{\alpha\beta})$, and given a crossing $c=(s,i,j)$ in $D$ (with $s$, $i$, and $j$ as before), the following two sets of relations (the $g$-rules) hold (with $\delta$ denoting the Kronecker delta):
\begin{equation} \label{eq:CarRules}
  g_{i\beta} = \delta_{i\beta}+T^sg_{\ip,\beta}+(1-T^s)g_{\jp,\beta},
  \qquad g_{j\beta} = \delta_{j\beta}+g_{\jp,\beta},
  \qquad g_{2n+1,\beta} = \delta_{2n+1,\beta}
\end{equation}
and
\begin{equation} \label{eq:CounterRules}
  g_{\alpha i} = T^{-s}(g_{\alpha,\ip}-\delta_{\alpha,\ip}),
  \qquad g_{\alpha j} = g_{\alpha,\jp} - (1-T^s)g_{\alpha i} - \delta_{\alpha,\jp},
  \qquad g_{\alpha,1} = \delta_{\alpha,1}.
\end{equation}
Furthermore, for each fixed $\beta$ there $2n+1$ $g$-rules of type~\eqref{eq:CarRules} (the first two depend on a choice of one of $n$ crossings, and the third is fixed, to a total of $2n+1$ rules). These fully determine the $2n+1$ scalars $g_{\alpha\beta}$ corresponding to varying $\alpha$. Similarly, for each fixed $\alpha$ there $2n+1$ $g$-rules of type~\eqref{eq:CounterRules}. These fully determine the $2n+1$ scalars $g_{\alpha\beta}$ corresponding to varying $\beta$.
\end{lemma}

\noindent{\em Proof.} The first set of $g$-rules corresponds to column $\beta$ of the equality $AG=I$, and the second set of $g$-rules corresponds to row $\alpha$ of the equality $GA=I$. \qed

We start with the hardest part of the invariance proof as it makes it easy to separate principles from silly computations:

\begin{proposition} \label{prop:R3} The quantity $\rho_1$ is invariant under R3.
\end{proposition}

\noindent{\em Proof.} We need to know how the Green's function $g_{\alpha\beta}$ changes under R3. Here are the two sides of the move, along with the $g$-rules of type~\eqref{eq:CarRules} corresponding to the crossings within, written with the assumption that $\beta$ isn't in $\{\ip,\jp,\kp\}$, so several of the Kronecker deltas can be ignored. We use $g$ for the Green's function of the left side of R3, and $g'$ for the right side:
\[
  \def\grulesA{{$\begin{array}{l}
    g_{j,\beta} = \delta_{j\beta} \!+\! Tg_{\jp,\beta} \!+\! (1\!-\!T)g_{\kp,\beta} \\
    g_{k,\beta} = \delta_{k\beta} \!+\! g_{\kp,\beta}
  \end{array}$}}
  \def\grulesB{{$\begin{array}{l}
    g_{i,\beta} = \delta_{i\beta} \!+\! Tg_{\ip,\beta} \!+\! (1\!-\!T)g_{\kpp,\beta} \\
    g_{\kp,\beta} = g_{\kpp,\beta}
  \end{array}$}}
  \def\grulesC{{$\begin{array}{l}
    g_{\ip,\beta} = Tg_{\ipp,\beta} \!+\! (1\!-\!T)g_{\jpp,\beta} \\
    g_{\jp,\beta} = g_{\jpp,\beta}
  \end{array}$}}
  \def\gprulesA{{$\begin{array}{l}
    g'_{i,\beta} = \delta_{i\beta} \!+\! Tg'_{\ip,\beta} \!+\! (1\!-\!T)g'_{\jp,\beta} \\
    g'_{j,\beta} = \delta_{j\beta} \!+\! g'_{\jp,\beta}
  \end{array}$}}
  \def\gprulesB{{$\begin{array}{l}
    g'_{\ip,\beta} = Tg'_{\ipp,\beta} \!+\! (1\!-\!T)g'_{\kp,\beta} \\
    g'_{k,\beta} = \delta_{k\beta} \!+\! g'_{\kp,\beta}
  \end{array}$}}
  \def\gprulesC{{$\begin{array}{l}
    g'_{\jp,\beta} = Tg'_{\jpp,\beta} \!+\! (1\!-\!T)g'_{\kpp,\beta} \\
    g'_{\kp,\beta} = g'_{\kpp,\beta}
  \end{array}$}}
  \input{figs/R3.pdf_t}
\]

A routine computation (eliminating $g_{\ip,\beta}$, $g_{\jp,\beta}$, and $g_{\kp,\beta}$) shows that the first system of 6 equations is equivalent to the following 3 equations:
\[ g_{i,\beta} = \delta_{i\beta} + T^2g_{\ipp,\beta} + T(1-T)g_{\jpp,\beta} + (1-T)g_{\kpp,\beta}, \]
\[
  g_{\j,\beta} = \delta_{j\beta} + Tg_{\jpp,\beta} + (1-T)g_{\kpp,\beta},
  \qquad\text{and}\qquad
  g_{k,\beta} = \delta_{k\beta} + g_{\kpp,\beta}.
\]
Similarly eliminating $g'_{\ip,\beta}$, $g'_{\jp,\beta}$, and $g'_{\kp,\beta}$ from the second set of equations, we find that it is equivalent to
\[ g'_{i,\beta} = \delta_{i\beta} + T^2g'_{\ipp,\beta} + T(1-T)g'_{\jpp,\beta} + (1-T)g'_{\kpp,\beta}, \]
\[
  g'_{\j,\beta} = \delta_{j\beta} + Tg'_{\jpp,\beta} + (1-T)g'_{\kpp,\beta},
  \qquad\text{and}\qquad
  g'_{k,\beta} = \delta_{k\beta} + g'_{\kpp,\beta}.
\]
But these two sets of equations are the same, and as stated in Lemma~\ref{lem:gRules}, along with the $g$-rules corresponding to the other crossings in $D$ (which are also the same between $g$ and $g'$), these equations determine $g_{\alpha\beta}$ and $g'_{\alpha\beta}$, for $\alpha,\beta\notin\{\ip,\jp,\kp\}$. So with this exclusion on $\alpha$ and $\beta$, we have that $g_{\alpha\beta}=g'_{\alpha\beta}$. But this means that the summations~\eqref{eq:rho1} in the definitions of $\rho_1$ are equal for the two sides of R3, except perhaps for the three summands on each side that come from the crossings that touch $\{\ip,\jp,\kp\}$.

What remains is completely mechanical. We just need to compute the sum of those three summands for both sides of R3, and apply to it the $g$-rules of types~\eqref{eq:CarRules} and~\eqref{eq:CounterRules} that eliminate the indices $\{\ip,\jp,\kp\}$. The computation is easy enough to be done by hand, yet why bother? Here's the machine version (it takes less typing to apply all relevant $g$-rules and also eliminate the indices $\{i,j,k\}$):

\input{Invariance-R3.tex} \ \qed

\begin{proposition} \label{prop:R2c} The quantity $\rho_1$ is invariant under R2c.
\end{proposition}

\noindent{\em Proof.} We follow the exact same steps as in the case of $R3$. First, we write the $g$-rules, assuming $\beta$ isn't in $\{i,j,\ip,\jp\}$:
\[
  \def\grulesA{{$\begin{array}{l}
    g_{i,\beta} = T^{-1}g_{\ip,\beta}+(1-T^{-1})g_{\jpp,\beta} \\
    g_{\jp,\beta} = g_{\jpp,\beta}
  \end{array}$}}
  \def\grulesB{{$\begin{array}{l}
    g_{\ip,\beta} = Tg_{\ipp,\beta}+(1-T)g_{\jp,\beta} \\
    g_{j,\beta} = g_{\jp,\beta}
  \end{array}$}}
  \input{figs/R2c.pdf_t}
\]

Note that for the right hand side we allowed ourselves to label the edges $\ipp$ and $\jpp$ as the computation is independent of the labeling and the labeling need not be by contiguous integers (outside of the move area, we assume that the left hand side and the right hand side are labeled in the same way). Note also that for the right hand side, there are no relevant $g$-rules. Now as in the case of R3, for the left hand side we eliminate $g_{\ip,\beta}$ and $g_{\jp,\beta}$ and we are left with the relations
\[ g_{i,\beta}=g_{\ipp,\beta} \qquad\text{and}\qquad g_{j,\beta}=g_{\jpp,\beta}. \]
Otherwise the $g$-rules for the left and for the right are the same, and so their Green functions are the same except if the indices are in $\{i,j,\ip,\jp\}$ (these indices do not even appear in the right hand side). Thus the contribution to $\rho_1$ from outside the area of the move is the same for both sides.

Next we write the contribution to $\rho_1$ coming from the two crossings and one rotation that appear on the left, and use the $g$-rules to push all the indices in $\{i,j,\ip,\jp\}$ up to $\ipp$ and $\jpp$. This can be done by hand, but seeing that we have tools, we use them:

\input{Invariance-R2c.tex}

This result is clearly equal to the single rotation contributionn to $\rho_1$ that comes from the right hand side. \qed

\begin{proposition} \label{prop:R1l} The quantity $\rho_1$ is invariant under R1l.
\end{proposition}

\noindent{\em Proof.} We start with the relevant $g$-rules:
\[
  \def\grules{{$\begin{array}{l}
    g_{\ip,\beta} = \delta_{\ip,\beta} + Tg_{\ipp,\beta} + (1-T)g_{\ip,\beta} \\
    g_{i,\beta}=\delta_{i,\beta}+g_{\ip,\beta}
  \end{array}$}}
  \input{figs/R1l.pdf_t}
\]
The first of these rules is equivalent to $g_{\ip,\beta}=T^{-1}\delta_{\ip,\beta} + g_{\ipp,\beta}$. For $\beta\neq i,\ip$ we find as before that $g_{i,\beta}=g_{\ipp,\beta}$ and we can ignore the contributions to $\rho_1$ coming from outside the area of the move. The contribution to $\rho_1$ coming from the single crossing and single rotation on the left hand side is computed below, and is equal to the empty contribution coming from the right hand side:

\input{Invariance-R1l.tex} \ \qed

\noindent{Proof of Theorem~\ref{thm:Main}.} After Claim~\ref{claim:UprightRMoves} which sets out what we need to do, and Propositions~\ref{prop:R3}, \ref{prop:R2c}, and~\ref{prop:R1l} which prove invariance under R3, R2c, and R1l, it remains to show the invariance of $\rho_1$ under R1r, R2b, and Sw$^+$. This is done exactly as in the examples already shown, so in each case we show only the punch line:

\input{Invariance-Rest.tex} \ \qed

\section{Some Context} \label{sec:Context} TBW.

\vskip 2mm\noindent{\bf Acknowledgement.} We wish to thank TBW comments and suggestions.

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\end{document}
\endinput
