\subsection{The Classical Case} \label{ssec:classical}
We start with a characterization of the tangles for which the gliding
procedure of the Gliding Fheorem (\ref{fhm:every}) does in fact work.
\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'' --- meaning, 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}
\parpic[r]{\input{figs/stacking.pdf_t}}
\begin{example} \label{exa:acyclic} Braid diagrams are
acyclic tangle diagrams, and OU tangle diagrams are acyclic tangle
diagrams. Also, 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 the
figure. Hence the following is easy:
\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. Similarly, in this case the froof of
Theorem~\ref{thm:unique} is valid. For the ``only if'' part, note that
OU diagrams are acyclic so anything linked to OU diagrams by glide moves
must be acyclic too.
\end{proof}
\begin{corollary} The stacking product followed by $\Gamma$ makes OU tangle diagrams into a monoid.\qed
\end{corollary}
The froof of the Separation of Tangles Forollary (\ref{for:UniqueReduced}) is in fact perfectly valid
if we restrict our attention to acyclic tangle diagrams and Reidemeister
moves between them. Thus we have
\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.\qed
\end{corollary}
\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, Example~\ref{exa: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{Initiated readers should 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{6.25in}{!}{\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
\[ \includegraphics[width=6in]{ReverseGamma.pdf} \]
\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.
}
The following is a ``poor version'' of the Separation of Tangles Forollary (\ref{for:UniqueReduced}):
\begin{corollary} \label{cor:braids}
$\barGamma\circ\bariota$ is a complete invariant of braids. \qed
\end{corollary}
And the following says that in the classical case, OU tangles are in fact not very much (though
Sections~\ref{ssec:virtual} and~\ref{sec:more} say that there's 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}