\section{The Classical Case} \label{sec:classical}

We start with a characterization of the tangles for which the gliding
procedure of Fheorem (\ref{fhm:every}) does in fact work: in Theorem~\ref{thm:iso} we find that these are precisely braids. The following definition gets to the heart of what makes a tangle ``problematic'' for the gliding procedure:

\Needspace{36mm} % Was 22mm; 35mm is not enough to keep footnote on the same page.
\parpic[r]{\raisebox{0mm}{
  \input{figs/CascadePath.pdf_t}\qquad$\includegraphics[height=22mm]{Waterfall.pdf}^\text{\footnotemark}$
}}
\begin{definition} Let $D$ be a tangle diagram. A ``cascade path'' along
$D$ is a directed path
that travels along strands of $D$ consistently with their orientation,
except at crossings where it can (but doesn't have to) drop from the upper
strand to the lower strand (but not the other way around). Two examples are
on the right. The diagram $D$ is called ``acyclic'' if it has no ``Escher
waterfalls'' --- that is, if no closed cascade paths can be drawn on
$D$.  On the right, the first example is
acyclic while the second isn't.
\footnotetext{Public domain waterfall image from
\url{https://commons.wikimedia.org/wiki/File:Waterfall.svg}.}
\end{definition}

\Needspace{19mm} % 18mm is not enough
\parpic[r]{\scalebox{0.88}{\input{figs/stacking.pdf_t}}}
\begin{example} \label{exa:acyclic} Braid diagrams are
acyclic tangle diagrams, and OU tangle diagrams are acyclic tangle
diagrams. The stacking product (illustrated on the right)
of two acyclic tangle diagrams is again an acyclic tangle diagram.
\end{example} 
 
Glide moves and bulk glide moves as in
Figure~\ref{fig:Gliding} do not change the acyclicity of a tangle
diagram. Indeed by simple inspection the possible transits of a cascade
path through either of the sides of a glide move are $1\to 1$, $1\to 2$,
$1\to 3$, $2\to 2$, $3\to 2$, and $3\to 3$, with numbering as in Figure~\ref{fig:Gliding}. 

Note that if a tangle diagram is OU then no Reidemeister 3 (R3) moves
can be performed on it without breaking the OU property --- if one side
of an R3 move is OU, the other necessarily isn't. This suggests that
perhaps an OU form of a tangle diagram is unique up to Reidemeister 2
(R2) moves. We aim to prove this next.


\begin{theorem} \label{thm:acyclic} A tangle diagram $D$ can be made OU
using glide moves if and only if it is acyclic, and in that case, the
resulting OU tangle diagram, which we call $\Gamma(D)$, is uniquely determined.
\end{theorem}

\begin{proof} In an acyclic tangle diagram the U and the O of a UO
interval cannot belong to the same crossing (or else an Escher waterfall
is present) so the number of UO intervals can be reduced using bulk
glide moves as in the Froof of the Gliding Fheorem (\ref{fhm:every}). By the
observation above, the resulting diagram is still acyclic so the
process can be continued. 

For the ``only if'' part, note that
OU diagrams are acyclic so anything linked to OU diagrams by glide moves
must be acyclic too.

Now to show that $\Gamma(D)$ is unique, observe that when UO intervals are apart from each
other, their fixing is clearly independent. It remains to see what
happens when UO intervals are adjacent, and there are only two
distinct cases to consider. Both of these cases are shown
in Figure~\ref{fig:unique} along with their OU fixes, which are clearly
independent of the order in which the glide moves are performed. 
\end{proof}

\begin{figure}
\[ \resizebox{0.92\linewidth}{!}{\input{figs/unique.pdf_t}} \]
\caption{Two possibilities for ``interacting'' UO intervals (each marked with a~$\bullet$ symbol).}
\label{fig:unique}
\end{figure}


\begin{corollary} The stacking product followed by $\Gamma$ makes OU tangle diagrams into a monoid.\qed
\end{corollary}


\begin{definition} A tangle diagram is called reduced if its crossing number cannot be reduced
using only R1 and R2 moves.
\end{definition}


\begin{corollary} \label{cor:GammaIso} The map $\Gamma$ descends to a well-defined map
$\barGamma$ from ``acyclic tangle diagrams modulo Reidemeister moves that
preserve the acyclic property'' into ``reduced OU tangle diagrams''.
\end{corollary}


\Needspace{34mm}
\parpic[r]{\scalebox{0.88}{\input{figs/R3.pdf_t}}}
\par\noindent{\it Proof.} If two tangle diagrams differ by an R3 move then
exactly one of them has a UO interval within the scope of the R3 move, and
its elimination via a glide move (which may as well be performed first)
yields the other diagram, up to an R2 move (picture on right). Furthermore, R1 and/or R2 moves before a glide become R1 and/or
R2 moves after the glide, or they make the glide move redundant, see
examples in Figure~\ref{fig:R1R2}. So the end result of the gliding
process of an acyclic tangle is unique modulo R1 and R2 moves. Finally it is easy
to check that within any equivalence class of acyclic tangle diagrams modulo R1
and R2 moves that preserve the acyclic property, there is a unique reduced representative. \qed

\begin{figure}
\[ \resizebox{\linewidth}{!}{\input{figs/R1R2.pdf_t}} \]
\caption{R1 and R2 moves ``commute'' with glides (A), or they
make glides redundant (B), (C).} \label{fig:R1R2}
\end{figure}


\begin{corollary} \label{cor:actions}
Braids act on reduced OU tangle diagrams both on the left and on the right.
\end{corollary}

\begin{proof} Use the stacking product, the fact that braids are always acyclic, and Corollary~\ref{cor:GammaIso}.
\end{proof}

In summary, we have a commutative diagram as follows:
\[ \xymatrix@C=22mm{
  \calBD_n \ar@{^{(}->}[r]^\iota \ar[d] & \calACD_n \ar[r]^\Gamma \ar[d] & \calOUD_n \ar[d] \\
  \calB_n \ar[r]^<>(0.5)\bariota_<>(0.5){\text{Theorem~\ref{thm:iso}: }\cong} &
  \calAC_n \ar[r]^<>(0.5)\barGamma_<>(0.5){\cong} &
  \calROU_n
} \]
Here $\calBD_n$ denotes the monoid of braid diagrams with $n$ strands,
$\calACD_n$ denotes the monoid of acyclic tangle diagrams with $n$
strands, $\calOUD_n$ denotes the monoid of OU tangles diagrams with
$n$ strands, $\iota$ is the inclusion map, the vertical maps are
all ``reductions'': modulo braid moves in the first column, modulo
Reidemeister moves that preserve the acyclic property in the second
column, and modulo R1 and R2 in the third column (alternatively, the
third vertical map maps OU tangle diagrams to their unique reduced
form, and $\calROU_n$ is really a subset of $\calOU_n$), and finally,
$\bariota$ is the map induced by $\iota$ on the quotient $\calB_n$. Note
that $\barGamma$ is an isomorphism --- its inverse is the inclusion
$\calROU_n\to\calAC_n$ from Example~\ref{exa:acyclic}.

\begin{theorem}[Classical Isomorphism] \label{thm:iso}
$\barGamma\circ\bariota$ is an isomorphism (and hence also $\bariota)$.
\end{theorem}

\noindent{\it Proof.} Figure~\ref{fig:stirring} contains a visual
description of $\barGamma\circ\bariota$. If $\beta\in\calB_n$ is a braid,
to compute $\barGamma(\bariota(\beta))$ make a whisk in the shape of
$\beta$ from black metal wires, and dip it slightly into a rectangular
pool of tahini sauce. Sprinkle lines of green ground parsley on top of
the tahini pool, connecting the ends of the whisk to the front side
of the pool, as in (A) of Figure~\ref{fig:stirring}. The green tahini
lines together with the black whisk lines together still make the shape
of $\beta$, and this will remain true throughout this proof.

\begin{figure}
\resizebox{\linewidth}{!}{\input{figs/stirring.pdf_t}}
\caption{Stirring a pool of tahini sauce garnished with parsley lines using a braid whisk.} \label{fig:stirring}
\end{figure}

Now slowly push the whisk down and let it stir the sauce as in (B), (C),
and (D) of Figure~\ref{fig:stirring}. Less and less of the whisk remains
visible and at the same time the green parsley lines remain planar but
get more and more twisty. The end of the process is in (D) and it can be
interpreted as an OU tangle, by reading the picture from top to bottom:
the black whisk wires are all O, and the green parsley lines are all
U.\footnote{Readers may recognize this as the identification
of the braid group with the mapping class group of a punctured disk. See
e.g.~\cite[Theorem~1]{BirmanBrendle:BraidsSurvey}.}

Each step of this stirring process can be broken up into glide moves
and planar equivalences that require no Reidemeister moves, as shown
in a schematic manner in Figure~\ref{fig:StirringIsGliding}. Hence our
process computes $\barGamma(\bariota(\beta))$.

\begin{figure}
\resizebox{\linewidth}{!}{\input{figs/StirringIsGliding.pdf_t}}
\caption{Stirring is gliding.} \label{fig:StirringIsGliding}
\end{figure}

Every OU tangle diagram $T$ has a black-green presentation as in (D)
of Figure~\ref{fig:stirring}. Indeed the O parts of $T$ cannot cross
each other so they can be drawn as a collection of straight parallel
black lines, and the U parts do cross the O parts so perhaps they
cannot be drawn as straight lines, but they still do not cross each
other so they make a collection of ``green'' lines, leading to a
picture as in (D) of Figure~\ref{fig:stirring} or as in (A) of
Figure~\ref{fig:ReverseGamma}.

\begin{figure}
% Generated by ReverseGamma.nb
\[ \includegraphics[width=\linewidth]{ReverseGamma.pdf} \]
\caption{The map $\Lambda$ turning an OU tangle into a braid.} \label{fig:ReverseGamma}
\end{figure}

Figure~\ref{fig:ReverseGamma} also shows how to define a map $\Lambda$
from OU tangles into braids: draw an OU tangle $T$ as in (A)
of Figure~\ref{fig:ReverseGamma}, and gradually pull down the green
strands to below the tahini level by an amount proportional to their
arc-length distance from their meeting points with the black strands,
while at the same time moving your viewpoint to be on the tahini plane,
as shown in (B) and (C) of Figure~\ref{fig:ReverseGamma}. At the end of
the process what you see is the braid $\Lambda(T)$.

Both compositions of $\barGamma\circ\bariota$ and of $\Lambda$ are identity maps\footnotemark, and hence
$\barGamma\circ\bariota$ is invertible. \qed

\footnotetext{Hints:
For $\Lambda\circ(\barGamma\circ\bariota) = I_{\calB}$ note that the
stirring process of Figure~\ref{fig:stirring} can be carried out with the
green lines already pulled down as in Figure~\ref{fig:ReverseGamma} and
when looking from the side, one sees a dance of braid diagrams, which
is an equivalence of braids. For $(\barGamma\circ\bariota)\circ\Lambda
= I_{\calROU}$ one has to start from a whisk $W$ of the form of (C) of
Figure~\ref{fig:ReverseGamma} (namely, a whisk that when considered from
above, as in (A) of Figure~\ref{fig:ReverseGamma}, appears to be made of
$n$ straight vertical bars and $n$ non-intersecting planar strands). Then
one has to show that stirring tahini with parsley lines using $W$ will recreate
the shape of $W$ (minus the vertical bars) in the parsley lines.
}

Hence, we have constructed a separating braid invariant:

\begin{corollary} \label{cor:braids}
$\barGamma\circ\bariota$ is a complete invariant of braids. \qed
\end{corollary}

And in fact, in classical case, OU tangles are merely braids (though we will see in
Sections~\ref{sec:virtual} and~\ref{sec:more} that there is more to our story):

\begin{corollary} \label{cor:cOUisBraids}
All OU tangles are equivalent to braids. \qed
\end{corollary}

\begin{corollary} The two actions of Corollary~\ref{cor:actions} of
braids on reduced OU diagrams are simple and transitive.\qed
\end{corollary}
