\Needspace{16mm} % 15mm is not enough.
\subsection{Assorted Comments} \label{ssec:Assorted}
\begin{discussion} \label{disc:Chterental} Virtual OU tangles are
equivalent to Chterental's ``Virtual Curve Diagrams'' (VCDs)
\cite{Chterental:VBandVCD, Chterental:Thesis}, though we hope that they
are a bit more natural, and that they tell a bigger story. We explain
the relationship in Figure~\ref{fig:VCD}, albeit without repeating
Chterental's definitions. Given a virtual curve diagram as in (A) of
Figure~\ref{fig:VCD}, connect all the curve ends on the upper (dashed)
line to the vertical infinity using O curves (thus making everything
else into U curves), delete the upper and the lower lines, and get
a virtual OU tangle (B). It is positioned opposite to our habits\footnote{We
are in topology / combinatorics; these habits are anyway meaningless.}
so in order to feel a bit better, we flip the picture over in~(C).
\begin{figure}
\[ \input{figs/VCD.pdf_t} \]
\caption{A Chterental Virtual Curve Diagram (VCD) and the corresponding virtual OU tangle.} \label{fig:VCD}
\end{figure}
To go back, draw a virtual OU tangle with the O parts of its strands straight,
parallel, of equal length, and heading downward (that's always possible
as they never cross each other), and then draw the U parts curving between them,
perhaps with virtual crossings\footnote{We've emphasized that ``virtual
crossings'' are {\bf not crossings}. But here we must link with other
people's conventions.}. Push all the virtual crossings to below the areas
between the O strand-parts (in light grey in (B) of Figure~\ref{fig:VCD}),
re-insert an upper line and a lower line, and get back to (A) of
Figure~\ref{fig:VCD}, a VCD.
Our proof of Chterental's Theorem (\ref{thm:vinj}) is similar to Chterental's proof that
VCDs can be used to separate virtual braids. A minor difference is that
we deal only with pure virtual braids (minor because separating braids
that induce different permutations is a non-issue). A bigger difference
is that we fully analyze the possible diamonds, instead of relying on
the classical Artin's theorem. \endpar{\ref{disc:Chterental}}
\end{discussion}
\begin{remark} \label{rem:classical}
The Division Lemma (\ref{lem:divquo}) implies that if $T$ is classical
(namely, is given with a planar presentation in a disk $D$) and
$\sigma_{ij}^s\mid T$ with $s\in\{\pm 1\}$, then the beginning points
of strands $i$ and $j$ must be adjacent within the boundary of $D$
(with $i$ left of $j$ if $s=+1$ and $i$ right of $j$ if $s=-1$), and then
$\sigma_{ij}^{-s}T$ is classical again. By induction, if a virtual braid
$\beta$ divides a classical $T$, then $\beta$ is actually classical.
\end{remark}
\Needspace{28mm} % 27mm is not enough.
\parpic[r]{\input{figs/tent.pdf_t}}
\begin{remark} \label{rem:tent} Everything within the proof of
Chterental's Theorem (\ref{thm:vinj}) can be restricted to the classical case, hence
reproving (in a complicated and very algebraic manner) that the map
$\barGamma\circ\bariota$ of the Classical Isomorphism Theorem (\ref{thm:iso}) is injective. To show
by algebraic means that $\barGamma\circ\bariota$ is also surjective
it is enough to show that every non-trivial classical OU tangle $T$ is
divisible by at least one $\sigma_{ij}^{\pm 1}$ --- dividing and repeating
until the process terminates (it must, as crossing numbers decrease),
and using the previous remark would show that $T$ is equivalent to a
classical braid. The required ``existence of a divisor'' property is proven
as follows: For any strand $j$ let $i$ be the first strand to cross over
$j$ after $j$'s transition point $\bowtie$. If the starting points of
$i$ and $j$ are adjacent then either $\sigma_{ij}$ or $\sigma_{ij}^{-1}$
divides $T$. Otherwise a ``triangular tent shield'' is created as on
the right, and the same argument can be repeated within it. When the
process terminates, we have a divisor. The topological arguments of
the classical braids Section (\ref{ssec:classical}) are of course a lot simpler.
\end{remark}
\begin{remark} \label{rem:TwoSquares} Let $\calT_n$ denote the set of all classical tangles
with $n$ open strands and let $\vcalT_n$ denote the set of all virtual tangles
with $n$ open strands. We wish to briefly study the following two commutative
squares, the ``classical'' and the ``virtual'':
\[
\xymatrix{
\calB_n \ar[r]^\chi_{\text{1-1}} \ar[d]_\bariota^\cong & \calT_n \\
\calAC_n \ar[r]^<>(0.5)\barGamma_<>(0.5)\cong & \calROU_n \ar[u]_\varphi^{\text{1-1}}
}
\qquad\qquad
\xymatrix{
\vcalPB_n \ar[r]^{\chi_v}_{\text{1-1?}} \ar[d]_{\bariota_v}^{\text{1-1}} & \vcalT_n \\
\vcalAC_n \ar[r]^<>(0.5){\barGamma_v}_<>(0.5)\cong & \vcalROU_n \ar[u]_{\varphi_v}^{\text{1-1?}}
}
\]
In these squares, $\bariota$, $\barGamma$, $\bariota_v$, and
$\barGamma_v$ along with the properties ($\cong$, $\cong$, 1-1,
and $\cong$) were discussed in Sections~\ref{ssec:classical}
and~\ref{ssec:virtual}. Also, $\chi$ ($\chi_v$) and $\varphi$
($\varphi_v$) are the obvious maps of (virtual) braids and reduced
(virtual) OU tangles into (virtual) tangles\footnote{In the vaguest
way, $\chi$ and $\varphi$ are pictograms for braids and OU tangles,
respectively.}. We note that the injectivity of $\chi$ was known already
to Artin~\cite[Theorem~12]{Artin:TheoryOfBraids}\footnote{Quick proof:
The fundamental group of the complement of a braid along with the $n$
bottom meridians and the $n$ top meridians determines the braid, and
this invariant extends to tangles.}, and thus it follows that $\varphi$
is also injective. We do not know if $\chi_v$ and $\varphi_v$ are injective.
The injectivity of $\chi_v$ was stated as an open problem in
\cite[Question~5.1]{AudoxBellingeriMeilhanWagner:UVWHomotopy}. Given the injectivity of $\bariota_v$, the
injectivity of $\chi_v$ would clearly follow from the
injectivity of $\varphi_v$, which we conjecture holds true. \endpar{\ref{rem:TwoSquares}}
\end{remark}
\begin{conjecture} The obvious map $\varphi_v$ of reduced virtual OU tangles into virtual tangles is injective.
\end{conjecture}
The reason we believe this conjecture is that we see a plausible path to
proving it. One way to go would be to find enough invariants of virtual
tangles to separate reduced virtual OU tangles. There are plenty of
invariants of virtual tangles coming from Hopf algebras and quantum
groups, reduced virtual OU tangles are easy to enumerate (they are
``free'' objects, subject to no relations), and there are precedents
where using quantum groups one can find enough invariants to separate
near-free objects: for example, quantum $gl(N)$ invariants separate
braids~\cite{Bar-Natan:glN}, and braid groups are semi-direct products
of free groups.
\begin{discussion} \label{disc:OUH} In fact, there is a very close
relationship between virtual OU tangles and Hopf algebras. Denote by
$\vcalOU^p_q$ the set of OU tangles that have $p$ O-only strands
and $q$ U-only strands (it is a subset of $\vcalOU_{p+q}$). We claim
that $\vcalOU^p_q$ is precisely the set of ``universal formulas'' for
linear maps $\Hom(H^{\otimes p} \to H^{\otimes q})$, where $H$ is an
arbitrary involutive\footnote{Meaning that the antipode $S$ satisfies
$S^2=I$.} Hopf algebra\footnote{Or even, an involutive Hopf object in
a symmetric monoidal category.}: Meaning, those formulas that can be
written as an arbitrary composition of the structure maps $m$, $\Delta$,
$S$, $\epsilon$, and $\eta$ of $H$, and that make sense even if $H$
is infinite dimensional (so they contain no cycles).
\parpic[r]{\def\D{$\Delta$}\input{figs/HopfRelation.pdf_t}}
By means of an example and with all details suppressed,
Figure~\ref{fig:HopfWords} demonstrates how a
virtual O/U tangle becomes a Gauss diagram and then a universal Hopf
formula. Furthermore, one may show that the relation between the product $m$ and the
coproduct $\Delta$ in a Hopf algebra (illustrated on the right) can be
used to bring all coproducts in a universal Hopf formula to before all
the products, and hence every universal Hopf formula comes from an O/U
tangle as in Figure~\ref{fig:HopfWords}. \endpar{\ref{disc:OUH}}
\end{discussion}
\begin{figure}
\[ \def\D{$\Delta$} \input{figs/HopfWords.pdf_t} \]
\caption{
A virtual O/U tangle in $\vcalOU^2_3$ becomes a Gauss diagram becomes
a universal Hopf formula representing an element of $\Hom(H^{\otimes
2} \to H^{\otimes 3})$. Note that the antipode $S$ is inserted on the
$(-)$-marked edges of the Gauss diagram, which correspond to the negative
crossings of the tangle.
} \label{fig:HopfWords}
\end{figure}
\begin{remark} The awkwardness of having to restrict to involutive Hopf algebras suggests that there may be an
alternative way to tell the story of this paper that does not require involutivity. Perhaps using ``rotational virtual
tangles''~\cite{Kauffman:RotationalVirtualKnots}.
\end{remark}
\begin{remark} The map $\Ch\colon\vcalPB_n\to\vcalROU_n$ along with
Discussion~\ref{disc:OUH} imply that there is an invariant of virtual
braids with values in $\End(H^{\otimes n})$, where $H$ is an involutive
Hopf algebra. Other such invariants exist~\cite{Woronowicz:Solutions,
MurakamiVanDerVeen:Quantized}. We expect that they are closely related.
\end{remark}
\begin{remark} \label{rem:core} It follows from the reasonings of
Section~\ref{ssec:virtual} that it is possible to extract a maximal
braid out of an OU tangle, leaving behind a minimal ``core''
tangle. Precisly, if a virtual OU tangle $T$ is decomposed as $T=\beta'T'$
where $\beta'$ is a virtual pure braid and $T'$ is a virtual OU
tangle, and if $T'$ has the minimal possible crossing number for such
a decomposition, then $\beta'$ and $T'$ are uniquely detemined. Indeed,
let $(\beta',T')=f(I,T)$ be the final element guaranteed by the Diamond
Lemma (\ref{lem:diamond}) in the connected component of $(I,T)$ in
$\calX$. For example, if $T$ is $T_1$ of Example~\ref{exa:indivisible},
then $\beta'=\sigma_{12}$ and $T'$ is $T_2$ of~\ref{exa:indivisible}.
We do not know if the same is true for arbitrary virtual and/or classical
tangles. \endpar{\ref{rem:core}}
\end{remark}
\begin{discussion} \label{disc:EG}
There is a lovely visual side to the tools developed
for the proof of Chterental's Theorem (\ref{thm:vinj}). Given a reduced virtual OU tangle
$T\in\vcalROU_n$, we can consider the part $EG(T)$ of $\calX_n$ that
lies ``below'' $(I,T)$:
\[ EG(T) \coloneqq \left\{(\beta',T')\colon\, (I,T)\toto(\beta',T')\right\}. \]
\Needspace{48mm} % 47mm is not enough.
\parpic[r]{
\def\s#1#2{\,$\sigma_{#1#2}$} \def\si#1#2{\,$\sigma_{#1#2}^{-1}$}
\input{figs/EG41.pdf_t}
}
We restrict the relation $\to$ to $EG(T)$, making it into a directed
graph that we name ``the extraction graph of $T$''. By first computing
$\barGamma\circ\bariota$ or $\Ch=\barGamma_v\circ\bariota_v$, we
can also define $EG(\beta)$ when $\beta$ is a braid or a virtual
braid. These graphs are in themselves invariants (defined on $\vcalOU_n$
or $\vcalPB_n$ or $\calB_n$). They are often visually pleasing: we
have already seen a few examples, in Examples~\ref{exa:indivisible}
and~\ref{exa:GarsideHexagon}, within Equation~\ref{eq:CinnamonRolls},
and in Figure~\ref{fig:Cases23}\footnotemark. Another
example, the extraction graph of the classical braid
$\sigma_{21}^{-1}\sigma_{13}\sigma_{32}^{-1}\sigma_{21}$ whose closure is
the figure-8 knot, is here on the right (we label edges by the relevant
divisor $\sigma_{ij}^{\pm 1}$ and vertices by the value of $\xi$). Some
even nicer examples appear in Section~\ref{ssec:EG}.
\footnotetext{Figure~\ref{fig:SCR} is not example because it misses a
part of the graph. See Section~\ref{ssec:EG}.}
For any $T$, $EG(T)$ is a finite graph (for the set of potential divisors
$\{\sigma_{ij}^{\pm 1}\}$ is finite and and only finitely many divisions
can be carried out before we run out of crossings). $EG(T)$ always has
an ``initial'' vertex $i$ (the pair $(I,T)$) and a final vertex $f$
--- the final element that is guaranteed by the Diamond Lemma and that
is discussed in Remark~\ref{rem:core}. Every vertex $v$ of $EG(T)$
is sandwiched between the two: $i\toto v\toto f$. Every ``wedge'' in
$EG(T)$ (Definition~\ref{def:diamond}) can be completed to a diamond of
one of the types appearing in Figures \ref{fig:SCR}, \ref{fig:Cases23},
\ref{fig:CaseM}, and~\ref{fig:Square} (hence all cycles in $EG(T)$ are
of even length, and hence $EG(T)$ is bipartite). If one travels from $i$
to $f$ along any path in $EG(T)$ while reading the generators indicated
on the edges, one always reads the same virtual braid.
If $T$ is $\barGamma(\iota(\beta))$ or $\Ch(\beta)$, the final
vertex $f$ of $EG(\beta)$ is $(\beta,I)$, and every path from $i$
to $f$ spells a braid word for $\beta$. Thus $EG(\beta)$ highlights
a finite set of ``special'' braid words for $\beta$. It follows from
Remark~\ref{rem:classical} that if $\beta$ is classical then all the
special words for it are classical too.
We don't really understand $EG(\beta)$ --- we don't know what properties
(if any) of $\beta$ can be read off $EG(\beta)$, and we don't know how to
characterize the ``special words'' for $\beta$ that appear in $EG(\beta)$
other than by repeating the definitions. \endpar{\ref{disc:EG}}
\end{discussion}