On an invariant for colored classical and singular links

Audrey Baumheckel
Department of Mathematics, California State University, Fresno 5245 North Backer Avenue, M/S PB 108, CA 93740, USA

Carmen Caprau
Department of Mathematics, California State University, Fresno5245 North Backer Avenue, M/S PB 108, CA 93740, USA

Conor Righetti
Department of Mathematics, California State University, Fresno 5245 North Backer Avenue, M/S PB 108, CA 93740, USA

[2020]57K12, 57K14

CC was partially supported by Simons Foundation grant $\#355640$ and NSF grant DMS-$2204386$

A colored link, as defined by Francesca Aicardi, is an oriented classical link together with a ‘coloration’, which is a function defined on the set of link components and whose image is a finite set of ‘colors’.  An oriented classical link can be regarded as a colored link with its components colored with a sole color. Aicardi constructed an invariant $F(L)$ of colored links $L$ defined via skein relations.
When the components of a colored link are colored with the same color or when the colored link is a knot, $F(L)$ is a specialization of the HOMFLY-PT polynomial. Aicardi also showed that $F(L)$ is a stronger invariant than the HOMFLY-PT polynomial when evaluated on colored links whose components have different colors.
In this paper, we provide a state-sum model for the invariant $F(L)$ of colored links using a graphical calculus for oriented, colored, 4-valent planar graphs. We also extend $F(L)$ to an invariant of oriented colored singular links.

§ INTRODUCTION

Francesca Aicardi [1, 2] introduced a special type of oriented  links that she called colored links. If a function $\gamma$ is defined on the set $C$ of components of an oriented link $L$ and assigns elements of a finite set $N$ of `colors', then the link $L$ is called colored and the function $\gamma$ is called a coloration. If $|\gamma(C)| = 1$, then the colored link is a classical link; in this case, all components of $L$ are colored with the same color. A coloration $\gamma$ partitions the set $C$ of components of the oriented link $L$, hence $\gamma$ introduces an equivalence relation on $C$. Two colorations $\gamma$ and $\gamma'$ of $L$ are called equivalent if there exists a bijection $\gamma (C) \to \gamma'(C)$.

An invariant of colored links is an invariant of links that takes the same value on ambient isotopic links with equivalent colorations.

Colored links are closely related to tied links. A tied link [3] is an oriented classical link whose components may be connected by ties. A tie is an arc that connects two points on the link, where the two points can belong to different components or the same component of the link. A tie connecting points that belong to the same component of a link can be removed, and such removal does not change the type of tied link. Ties are merely notational devices depicted as springs and are not embedded arcs. The set of ties on a tied link partitions the set $C$ of components of a link into equivalence classes. Hence tied links are links whose components are partitioned into classes. Equivalently, one can use colors to indicate the components of the link that are in the same equivalence class. In this sense, tied links are equivalent to colored links. In this paper, we chose to work with colored links for the sake of having less crowded diagrams.

Another type of knotted objects that we work with in this paper are oriented singular links. A singular link is an immersion in $\mathbb{R}^3$ of a disjoint union of circles, such that the immersion contains finitely many singularities which are transverse double points.  A diagram of a singular link is a projection of the singular link in a plane; such a diagram may contain classical crossings, as in classical link diagrams, and singular crossings. A singular crossing in a diagram is depicted as a 4-valent vertex.

It is well known that two singular link diagrams represent equivalent (or ambient isotopic) singular links if and only if there exists a finite sequence of the extended Reidemeister moves mapping one diagram onto the other (see for example <cit.>). A proof of this statement can be found in [4]. The extended Reidemeister moves, depicted in Fig. <ref>, are local moves on singular link diagrams and include the classical Reidemeister moves $R1, R2$ and $R3$, together with the additional $R4$ and $R5$ moves involving singular crossings.

\[\raisebox{-10pt}{\includegraphics[height=0.38in]{poskink}}\, \stackrel{R1}{\longleftrightarrow} \, \raisebox{-10pt}{\includegraphics[height=0.38in]{arc}} \,\stackrel{R1}{\longleftrightarrow} \,  \reflectbox{\raisebox{17pt}{\includegraphics[height=0.38in, angle = 180]{poskink}}} \,\qquad\,
\raisebox{-12pt}{\includegraphics[height=0.38in]{reid2-1}} \, \stackrel{R2}{\longleftrightarrow} \, \raisebox{-12pt}{\includegraphics[height=0.38in]{A-smoothing}} \,\qquad \,
\raisebox{-12pt}{\includegraphics[height=0.38in]{reid3-1}} \, \stackrel{R3}{\longleftrightarrow} \,   \raisebox{-12pt}{\includegraphics[height=0.38in]{reid3-2}}\]

\[  \raisebox{-12pt}{\includegraphics[height=0.4in]{N4-left1}}\,\, \stackrel{R4}{\longleftrightarrow} \,\, \raisebox{-12pt}{\includegraphics[height=0.4in]{N4-right1}} \,\, \qquad \,\,  \raisebox{-12pt}{\includegraphics[height=0.4in]{N4-left2}} \,\, \stackrel{R4}{\longleftrightarrow}\,\,  \raisebox{-12pt}{\includegraphics[height=0.4in]{N4-right2}} \]

\[ \raisebox{-10pt}{\includegraphics[height=0.35in]{N5-left1}}\,\,\stackrel{R5}{ \longleftrightarrow} \,\,\reflectbox{\raisebox{-10pt}{\includegraphics[height=0.35in]{N5-left2}}}\]

Extended Reidemeister moves

We remark that singular links can also be regarded as rigid-vertex embeddings in $\mathbb{R}^3$  of 4-valent graphs. For details, we refer the reader to Kauffman's work in <cit.>.

The concept of coloring can be extended to oriented singular links as well. We define the notion of colored singular link as an oriented singular link together with a coloration defined on the components of the singular link. An invariant for colored singular links must take the same value on ambient isotopic singular links with equivalent colorations.  The diagram on the left in Fig. <ref> is an example of a colored (classical) link with three components, while the diagram on the right is an example of a colored singular link with two components.

Examples of colored classical and singular links

One way to consider links as colored links or singular links as colored singular links is by coloring all of the link components with the same color. Another way is by coloring all of the link components with different colors.

Aicardi [2] defined an invariant $F(L)$ for colored links $L$. We note that a similar invariant for tied links was defined by Aicardi and Juyumaya in [3]. When the link components are colored with a sole color or when the link is a knot, the invariant $F$ is a specialization of the HOMFLY-PT polynomial [6, 10] for oriented classical links. The reason for studying the polynomial $F$ is that it is a stronger invariant than the HOMFLY-PT polynomial when $F$ is evaluated on colored links whose components have different colors; for details, we refer to Aicardi's work in [1].

Given a colored link $L$, the invariant $F(L)$ is valued in $\mathbb{Q}(x, t, w)$ and is uniquely determined by the following three conditions:

(I) $F \left( \raisebox{-.04 in}{\includegraphics[scale=0.2]{unknot.pdf}} \right) = 1$.

(II) Let $D \cup \raisebox{-0.04 in}{\includegraphics[scale=0.2]{unknot.pdf}}$ be the disjoint union of an unknotted circle with a colored link diagram $D$, where the colors of $D$'s components are all distinct from the color of the circle.  Then

\[ F \left( D \cup \raisebox{-0.04 in}{\includegraphics[scale=0.2]{unknot.pdf}}\right) = \frac{1}{wx} F \left( D \right).\]

(III) The following skein relation holds,

\[ \frac{1}{w} F\left( \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=0.3]{poscrosscolor_reflect.pdf}}} \right) - w F\left( \raisebox{-.1 in}{\includegraphics[scale=0.3]{negcrosscolor.pdf}} \right) = \left( 1- \frac{1}{t} \right) F\left( \raisebox{-.1 in}{\includegraphics[scale=0.3]{uncross.pdf}} \right) + \left( \frac{1}{w} - \frac{1}{tw} \right) F\left(\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=0.3]{negcross.pdf}}} \right),\]

where when forgetting the colors, the four diagrams shown in the skein relation above are parts of link diagrams that are identical almost everywhere, except in a small neighborhood where they differ as shown. The diagrams -.1 in and -.1 in are parts of colored links, where black and red colors indicate any colors, not necessarily distinct. Moreover, the components of the colored links that contain the parts -.1 in, -.1 in, and -.1 in are colored with a sole color.

The skein relation (III) holds for any two colors of the strands in the two diagrams on the left of the relation. If the two colors are the same, then relation (III) is reduced to the following skein relation:

\[ (\text{IV}) \hspace{1cm} \frac{1}{tw} F\left( \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=0.3]{negcross.pdf}}} \right) - w F\left( \raisebox{-.1 in}{\includegraphics[scale=0.3]{negcross.pdf}} \right) = \left( 1- \frac{1}{t} \right) F\left( \raisebox{-.1 in}{\includegraphics[scale=0.3]{uncross.pdf}} \right).\]

We refer to the invariant $F$ as the Aicardi-Juyumaya invariant for colored links.

Solving for  $\frac{1}{w} F\left( \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}} \right)$ in relation (IV) and substituting it into relation (III), yields the following skein relation:

\[
(*)  \,\,  \frac{1}{w} F\left( \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{poscrosscolor_reflect.pdf}}}\right)-w F\left( \raisebox{-.1 in}{\includegraphics[scale=.3]{negcrosscolor.pdf}}\right) = \left(t-\frac{1}{t} \right) F\left( \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right)+ tw F\left( \raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}} \right) -\frac{1}{tw}F \left( \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}}\right).
\]

The HOMFLY-PT polynomial $P$ of a classical (single-colored) link is valued in $\mathbb{Z}[\ell, \ell^{-1}, m]$ and is uniquely defined by the condition $P \left( \raisebox{-4pt}{\includegraphics[scale=0.2]{unknot.pdf}} \right) = 1$ and the skein relation below (see [2]):

\[
\ell \,P \left( \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}}\right) + \ell^{-1}\, P \left( \raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}} \right)
+   m \, P\left( \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right) =0.\]

If $L$ is a knot or a link whose components are colored with the same color, then $F(L)$ is determined by relations (I) and (IV), and it coincides with the polynomial $P(L)$ after the following substitutions:

\[\ell = \frac{i}{w\sqrt{t}} \quad \text{and} \quad m = i\left(\frac{1}{\sqrt{t}}  - \sqrt{t}\right). \]

The scope of this paper is two-fold. We first extend the invariant $F$ to oriented colored singular links and denote the resulting invariant of colored singular links by $[\, \cdot \,]$. We also prove a set of graphical skein relations involving oriented, colored, planar 4-valent graphs;  this is done in Section <ref>. This set of graphical relations provide a recursive way to evaluate oriented, colored, planar graphs with 4-valent vertices. In Section <ref>, we explain how our graphical skein relations provide a state-sum model for the Aicardi-Juyumaya invariant $F$ for colored links, where the states associated to a diagram of a colored link are oriented, colored, $4$-valent planar graphs. This also allows us to define the invariant $[\, \cdot \,]$ for colored singular link without relying on the existence of the invariant $F$ for colored links.

§ AN EXTENSION OF THE AICARDI-JUYUMAYA INVARIANT TO COLORED SINGULAR LINKS

In this section we extend the Aicardi-Juyumaya invariant to oriented colored singular links. We denote the resulting invariant by $[ \, \cdot \,]$. We impose first that
$[ D] = F(D)$, whenever $D$ is a colored (classical) link diagram. Next, we impose the following skein relations for $[ \, \cdot \,]$:

\begin{align} \label{sing1}
\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{crosscolor.pdf}}\right]=& \frac{1}{w}\left[ \raisebox{-.11 in}{\reflectbox{\includegraphics[scale=.3]{poscrosscolor_reflect.pdf}}}\right] + \frac{1}{t}\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{uncross.pdf}}\right] + \frac{1}{tw} \left[ \raisebox{-.11 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}} \right]
\end{align}

\begin{align}\label{sing2}
\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{crosscolor.pdf}} \right]=& w \left[ \raisebox{-.1in}{{\includegraphics[scale=.3]{negcrosscolor.pdf}}}\right] + t \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}}\right] + tw \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}}\right].
\end{align}

That is, if $\tilde{D}$ is a diagram of a colored singular link, we can use either the skein relation (<ref>) or relation (<ref>) at each of the singular crossings in $\tilde{D}$, to write $[\tilde{D}]$ as a $\mathbb{Q}(x, w, t)$-linear combination of evaluations of colored links. We note that the each diagram in relations (<ref>) and (<ref>) is a part of a larger diagram of a colored singular link whose components have colorations that are globally compatible; in particular, the colorations of the strands shown in a diagram determine the colorations of the link components that the strands are part of. When forgetting the colorations, the diagrams in the two sides of each of these skein relations are parts of larger diagrams that are identical outside of the small neighborhood where the skein relation is applied.

Subtracting relation (<ref>) from relation (<ref>), yields the skein relation

\begin{align} \label{posminusnegcolor}
\frac{1}{w} \left[ \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{poscrosscolor_reflect.pdf}}}\right]-w \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{negcrosscolor.pdf}}\right] = \left(t-\frac{1}{t} \right)\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right]+ tw \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}} \right] -\frac{1}{tw} \left[ \raisebox{-.1 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}}\right],
\end{align}

which we know that $[ \, \cdot \,]$ satisfies, since the invariant $F$ satisfies an equivalent skein relation, as shown in Section <ref>, and $F(D) = [D]$, for any diagram $D$ of a colored link. Hence, the evaluation $[\tilde{D}]$ of a colored singular link diagram is independent on whether we use the skein relation (<ref>) or (<ref>).

The skein relations (<ref>) or (<ref>) have the following equivalent forms, which will be used later in this section.

\begin{align}
\frac{1}{w}\left[ \raisebox{-.11 in}{\reflectbox{\includegraphics[scale=.3]{poscrosscolor_reflect.pdf}}}\right]=& - \frac{1}{t}\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{uncross.pdf}}\right] - \frac{1}{tw} \left[ \raisebox{-.11 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}} \right] + \left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{crosscolor.pdf}}\right] \label{poscolor} \\
w \left[ {\raisebox{-.1 in}{\includegraphics[scale=.3]{negcrosscolor.pdf}}}\right] =& -t \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}}\right] - tw \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}}\right] + \left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{crosscolor.pdf}} \right]\label{negcolor}
\end{align}

If $\tilde{D}_1$ and $\tilde{D}_2$ are diagrams representing ambient isotopic colored singular links, then $[\tilde{D}_1] = [\tilde{D}_2]$. That is, $[ \, \cdot \,]$ is an invariant for colored singular links.

We first note that since $[\, \cdot \,]$ is an extension of $F,$ then $[\, \cdot \,]$ is invariant under Reidemeister moves $R1, R2$ and $R3$. It remains to show that $[\, \cdot \,]$ is also invariant under the moves $R4$ and $R5$. Starting with the Reidemeister-type move $R4$, we find that:

\begin{align*}
\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R41.pdf}}\right]&\stackrel{\eqref{sing1}}{=}\frac{1}{w}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R42.pdf}}\right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R43.pdf}}\right]+\frac{1}{tw}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R44.pdf}}\right]\\
&\stackrel{R3}{=}\frac{1}{w}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R45.pdf}}\right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R43.pdf}}\right]+\frac{1}{tw}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R46.pdf}}\right]\\
&\stackrel{R2}{=}\frac{1}{w}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R45.pdf}}\right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R47.pdf}}\right]+\frac{1}{tw}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R46.pdf}}\right]\\
&\stackrel{R2}{=}\frac{1}{w}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R45.pdf}}\right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R48.pdf}}\right]+\frac{1}{tw}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R46.pdf}}\right]
\stackrel{\eqref{sing1}}{=}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R49.pdf}}\right].
\end{align*}

Hence, $\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R41.pdf}}\right] =\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R49.pdf}}\right]$. A similar process is used to prove that $\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R410.pdf}}\right]=\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R411.pdf}}\right]$ and that $[\, \cdot \,]$ is invariant under the other oriented versions of the move $R4$. We prove next the invariance of $[\, \cdot \,]$ under the Reidemeister-type move $R5$:

\begin{align*}
\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R51.pdf}}\right]\stackrel{\eqref{sing2}}{=}w\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R52.pdf}}\right]+t\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R53.pdf}}\right]+tw\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{R54.pdf}}\right]
\stackrel{\eqref{sing2}}{=}\left[ \,\raisebox{-.1 in}{\includegraphics[scale=0.3]{R55.pdf}} \, \right ].
\end{align*}

In the first equality above, we applied the skein relation (<ref>) in a small neighborhood of the 4-valent vertex of the diagram on the left-hand side of the move $R5$. In the second step we applied the same skein relation, this time in reverse order, in a small neighborhood near the top part of the three diagrams obtained in the previous step. The proof of invariance under the other oriented versions of the move $R5$ are done in a similar manner.
This completes the proof that $[\, \cdot \,]$ is an invariant for colored singular links.

Let $L \, \tilde \cup \, \reflectbox{\raisebox{-.04in}{\includegraphics[scale=.2]{unknot.pdf}}}$ be the disjoint union between a colored link L with the standard diagram of the unknot, where the unknot is colored the same as (at least one) of the components of $L$. Then

\begin{eqnarray}
\left[L \, \tilde \cup \, \reflectbox{\raisebox{-.04in}{\includegraphics[scale=.2]{unknot.pdf}}}\right]  = \frac{tw^2-1}{w(1-t)}\left[L\right]
= \left (\frac{tw}{1-t} + \frac{t^{-1}w^{-1}}{1-t^{-1}} \right )\left[L\right]. \label{eq:tied_union}
\end{eqnarray}

We represent  $L \, \tilde \cup \, \reflectbox{\raisebox{-.04in}{\includegraphics[scale=.2]{unknot.pdf}}}$ as $\raisebox{-.1 in}{\includegraphics[scale=.3]{disjunion1.pdf}}.$
From relation (IV), we know that

\[\left(1-t^{-1}\right) \left[\raisebox{-.08 in}{\includegraphics[scale=.25]{uncross.pdf}}\right] = \frac{1}{tw} \left[\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.25]{negcross.pdf}}}\right] - w \left[\raisebox{-.08 in}{\includegraphics[scale=.25]{negcross.pdf}}\right].\]

\begin{align*}
\left(1-t^{-1}\right) \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{disjunion1.pdf}}\right] &= \frac{1}{tw} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{disjunion3.pdf}}\right] - w \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{disjunion2.pdf}}\right] \\
\left(1-t^{-1}\right) \left[L \, \tilde \cup \, \reflectbox{\raisebox{-.04in}{\includegraphics[scale=.2]{unknot.pdf}}}\right] &= \left(\frac{1}{tw}-w\right) \left[L\right] \\
\left[L \, \tilde \cup\, \reflectbox{\raisebox{-.04in}{\includegraphics[scale=.2]{unknot.pdf}}}\right]  &= \frac{tw^2-1}{w(1-t)}\left[L\right] = \left (\frac{tw}{1-t} + \frac{t^{-1}w^{-1}}{1-t^{-1}} \right )\left[L\right].
\end{align*}

Therefore, identity (<ref>) holds.

We make use of the skein relations (<ref>) and (<ref>) to prove the following graphical skein relations involving oriented, colored, 4-valent planar graphs. An oriented, colored, 4-valent planar graph can be regarded as a diagram of an oriented colored singular link without classical crossings.

The following graphical skein relations hold for $[\,\cdot \,]$:

\begin{align}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{loop.pdf}} \right]=& \left( \frac{w}{1-t} + \frac{w^{-1}}{1-t^{-1}}\right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{lineup.pdf}} \right] \label{singloop} \\
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigoncrosscolor.pdf}}} \right] =& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolor.pdf}}} \right]+(t+t^{-1})\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}}\right]+(t+t^{-1}) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}}\right] \label{bigon} \\
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{4valbigoncolordown.pdf}}} \right]=& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolordown.pdf}}} \right]+(t+t^{-1}+1) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrossdown.pdf}}}\right]+ \left( \frac{wt^{-1}}{1-t}+\frac{w^{-1}t}{1-t^{-1}}\right) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}} \right] \label{bigonop}\\
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossleft.pdf}}\right] &- \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossrightref.pdf}}}\right] = \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}\right] - \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}}\right]  \label{R3cross}\\
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right] &- \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossrightref.pdf}}}\right] = \left(\frac{wt^{-2}}{1-t}+\frac{w^{-1}t^2}{1-t^{-1}}\right)\left(\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right] - \left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}}\right]\right). \label{R3crossdown}
\end{align}

We start by proving graphical relation (<ref>). Note that we start by using the skein relation (<ref>), with both strands colored similarly. In the second step we use that $[\, \cdot \,]$ is invariant under the Reidemeister move $R1$.

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{loop.pdf}} \right] \stackrel{\eqref{sing1}}{=}& \left( \frac{1}{w} + \frac{1}{tw}\right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{looppos.pdf}} \right] + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{linecirc.pdf}} \right]
\stackrel{R1}{=} \left( \frac{1}{w} + \frac{1}{tw}\right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{lineup.pdf}} \right] + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{linecirc.pdf}} \right]  \\
\stackrel{\eqref{eq:tied_union}}{=}& \left( \frac{1}{w} + \frac{1}{tw}\right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{lineup.pdf}} \right] + \frac{1}{t} \left( \frac{tw^2-1}{(1-t)w}\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{lineup.pdf}} \right] \right) \\
=& \left( \frac{w}{1-t} + \frac{w^{-1}}{1-t^{-1}}\right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{lineup.pdf}} \right].
\end{align*}

To prove the graphical relation (<ref>) we start from the diagram on the left side of the relation and apply the skein relation (<ref>) in a small neighborhood of the top vertex. In the second step, we apply the skein relation (<ref>) to the resulting first and third diagrams, as we show below.

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigoncrosscolor.pdf}}} \right] \stackrel{\eqref{sing1}}{=}& \frac{1}{w} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigon3.pdf}}} \right] + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right] + \frac{1}{tw} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigon2.pdf}}} \right] \\
\stackrel{\eqref{sing2}}{=}& \frac{1}{w} \left( w \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigoncolorup.pdf}}} \right] + t \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}}} \right] +tw \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonup.pdf}}} \right] \right) + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right] \\
&\hspace{.2in} + \frac{1}{tw} \left( (w + tw) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonup.pdf}}} \right] + t \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}}} \right]\right).
\end{align*}

In the next step, we use that $[\, \cdot \, ]$ is invariant under the Reidemeister move $R2$ and combine like terms. Then we apply the skein relation (<ref>) to the second diagram with a positive crossing and both strands colored with the same color, and obtain the following calculations.

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigoncrosscolor.pdf}}} \right] \stackrel{R2}{=}& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolor.pdf}}} \right] + \frac{t+1}{w} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{negcross.pdf}}} \right] + \left(t + 1 + t^{-1} \right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right] + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right] \\
\stackrel{\eqref{poscolor}}{=}& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolor.pdf}}} \right]  - \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right] + t \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right] + \left(t + 1 + t^{-1} \right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right] + \frac{1}{t} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right] \\
=& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolor.pdf}}} \right] + \left( t + t^{-1} \right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{uncross.pdf}} \right] + \left( t + t^{-1} \right) \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{cross.pdf}} \right].
\end{align*}

Hence, the graphical relation (<ref>) holds. In the proof of the graphical relation (<ref>), we use a somewhat similar starting point. Note, however, that the strands in the left side diagram of the relation have different orientations than in the relation we just proved.

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{4valbigoncolordown.pdf}}} \right] \stackrel{\eqref{sing1}}{=}& \frac{1}{w} \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{bigondown3edit.pdf}} \right] + \frac{1}{t} \left[ \reflectbox{\rotatebox{180}{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigondown4.pdf}}}} \right] + \frac{1}{tw} \left[ \reflectbox{\rotatebox{180}{{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigondown2.pdf}}}}} \right] \\
\stackrel{\eqref{sing2}, \eqref{singloop}}{=}& \frac{1}{w} \left( w \left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{bigondownnocrosscolor.pdf}} \right] + t \left[ \reflectbox{\rotatebox{180}{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigondown1.pdf}}}} \right] + tw \left[ \reflectbox{\rotatebox{180}{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigondown.pdf}}}} \right] \right) + \frac{1}{t} \left( \frac{w^2-t}{w(1-t)} \left[ \reflectbox{\rotatebox{180}{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}}} \right] \right) \\
&\hspace{.2in} + \frac{1}{tw} \left( (wt+w) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigondown.pdf}}} \right] + t \left[ \reflectbox{\rotatebox{180}{\raisebox{-.18 in}{\includegraphics[scale=.3]{bigondown1.pdf}}}} \right] \right).
\end{align*}

In the second step we also used that $\frac{w^2-t}{w(1-t)} = \frac{w}{1-t} + \frac{w^{-1}}{1-t^{-1}}$. Next we use that $[\, \cdot \, ]$ is invariant under the moves R1 and R2, to obtain the following:

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{4valbigoncolordown.pdf}}} \right]   \stackrel{R1, R2}{=}& \frac{1}{w} \left( w \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolordown.pdf}}} \right] + t \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}} \right] + tw \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrossdown.pdf}}} \right] \right) + \frac{1}{t} \left( \frac{w^2-t}{w(1-t)} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}} \right] \right) \\
&\hspace{.2in} + \frac{1}{tw} \left( (wt+w) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrossdown.pdf}}} \right] + t \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}} \right] \right) \\
=& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrosscolordown.pdf}}} \right]+(t+t^{-1}+1) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{uncrossdown.pdf}}}\right]+ \left( \frac{wt^{-1}}{1-t}+\frac{w^{-1}t}{1-t^{-1}}\right) \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{bigonuncrossdown.pdf}}} \right].
\end{align*}

We will now evaluate the two terms of the left side of identity (<ref>). Starting with the first diagram, we resolve first the top left 4-valent vertex using the skein relation (<ref>). After that, in the first and third of the three resulting diagrams, we resolve the bottom left 4-valent vertex by applying the skein relation (<ref>). In the third step we merely simplify expressions.

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossleft.pdf}}\right] \stackrel{\eqref{sing1}}{=}& \frac{1}{w}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross6.pdf}} \right]+ \frac{1}{t} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right]+ \frac{1}{tw} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross7.pdf}} \right]\\
\stackrel{\eqref{sing2}}{=}& \frac{1}{w}\left( w\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross11.pdf}} \right]+t\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross5ref.pdf}}} \right]+tw\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross10.pdf}} \right]\right)+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right]\\
&\hspace{.2in} + \frac{1}{tw} \left( w\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross8.pdf}} \right]+t \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}}} \right] +tw\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}} \right]\right)\\
=&\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross11.pdf}} \right]+\frac{t}{w}\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross5ref.pdf}}} \right]+t \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross10.pdf}} \right] +\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross8.pdf}} \right] \\
&\hspace{.2in} +\frac{1}{w}\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}}} \right]+\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}} \right].
\end{align*}

Next we apply skein relation (<ref>) at the positive crossing in the second diagram of the resulting sum above.
Finally, in the last step below, we simplify the resulting linear combination of evaluations of colored link diagrams.

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossleft.pdf}}\right] \stackrel{\eqref{poscolor}}{=}&
\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross11.pdf}} \right]+t\left( \frac{-1}{t} \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}} \right] -\frac{1}{tw} \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}}} \right] + \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right] \right)+t \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross10.pdf}} \right] \\
&\hspace{.2in} +\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross8.pdf}} \right]+\frac{1}{w}\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}}} \right]+\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}} \right] \\
=& \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross11.pdf}} \right] - \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}} \right] + t \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right] + \frac{1}{t} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right] + t \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross10.pdf}} \right] \\
&\hspace{.2in} + \frac{1}{t} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross8.pdf}} \right] + \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}} \right].
\end{align*}

Similar steps are applied to the second diagram appearing on the left side of the identity (<ref>), as we show below.

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossrightref.pdf}}}\right] \stackrel{\eqref{sing2}}{=}& w\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross12ref.pdf}}}\right]+t\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right]+tw\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross13ref.pdf}}}\right]\\
\stackrel{\eqref{sing1}}{=}& w\left( \frac{1}{w}\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross15ref.pdf}}}\right]+\frac{1}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross3.pdf}} \right]+\frac{1}{tw}\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross16ref.pdf}}}\right]\right)+t\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right]\\
&\hspace{.2in} +tw\left( \frac{1}{w}\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross14ref.pdf}}}\right]+\frac{1}{t} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}} \right] +\frac{1}{tw}\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}}}\right]\right)\\
=&\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross15ref.pdf}}}\right]+\frac{w}{t}\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross3.pdf}} \right]+\frac{1}{t} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross16ref.pdf}}}\right] +t\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right]+t\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross14ref.pdf}}}\right] \\
&\hspace{.2in} +w\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}} \right]+\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}}}\right] \\
\stackrel{\eqref{negcolor}}{=}&\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross15ref.pdf}}}\right]+\frac{1}{t} \left( -t \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}\right] -tw \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}} \right] + \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right] \right) +\frac{1}{t} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross16ref.pdf}}}\right] \\
&\hspace{.2in} +t\left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right] +t\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross14ref.pdf}}}\right]+w\left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross4.pdf}} \right]+\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}}}\right].
\end{align*}

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossrightref.pdf}}}\right]
=& \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross15ref.pdf}}}\right] - \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}\right] + \frac{1}{t} \left[\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross2.pdf}} \right] + t \left[\reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross1ref.pdf}}} \right] + \frac{1}{t} \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross16ref.pdf}}}\right] \\
&\hspace{.2in}+ t \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross14ref.pdf}}}\right] + \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3cross9.pdf}}}\right].
\end{align*}

Combining the two results and using that $[\, \cdot \,]$ is invariant under the move $R4$, we obtain the following equality, and thus the graphical skein relation (<ref>) holds:

\[
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossleft.pdf}}\right] - \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R3crossrightref.pdf}}}\right] = \left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}\right] - \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part1R35.pdf}}}\right].
\]

It remains to prove the graphical skein relation (<ref>) and we start by evaluating each of the colored graph diagrams on the left side of the relation. Starting with the first diagram, we obtain the following:

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right] \stackrel{\eqref{sing1}}{=} &\frac{1}{w}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross1.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross2.pdf}}\right] + \frac{1}{tw}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross3.pdf}}\right] \\
\stackrel{\eqref{sing2}, \eqref{bigonop}}{=}& \frac{1}{w}\left(w\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross4.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross5.pdf}}\right] + tw\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross6.pdf}}\right]\right) \\ &\hspace{.2in} + \frac{1}{t}\left(\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross7.pdf}}\right] + \left(t+ t^{-1}+1\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}\right]\right) \\ &\hspace{.2in} + \frac{1}{t} \left(\frac{wt^{-1}}{1-t}+\frac{w^{-1}t}{1-t^{-1}}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right]
\\ &\hspace{.2in} + \frac{1}{tw}\left(w\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross10.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross11.pdf}}\right] + tw\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross12.pdf}}\right]\right).
\end{align*}

By simplifying the result above, we have:

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right]
=& \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross4.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross6.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross10.pdf}}\right] + \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross12.pdf}}\right] + \frac{t}{w}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross5.pdf}}\right] \\ &\hspace{.2in} + \frac{1}{w}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross11.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross7.pdf}}\right] + \left(1+ t^{-2}+ t^{-1}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}\right] \\ &\hspace{.2in} + \left(\frac{wt^{-2}}{1-t}+\frac{w^{-1}}{1-t^{-1}}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right].
\end{align*}

Continuing on with this same relation, we apply skein relation (<ref>) to the fifth and sixth diagrams on the right-hand side of the above equality, followed by applications of the invariance of $[ \,\cdot \,]$ under the Reidemeister moves $R1$ and $R2$, to obtain the following:

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right] \stackrel{\eqref{sing2}, R1, R2}{=}& \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross4.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross6.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross10.pdf}}\right] + \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross12.pdf}}\right] \\ &\hspace{.2in} + \left(t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross13.pdf}}\right] + \frac{t^2}{w}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right] + t^2\left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}}\right]\right) \\ &\hspace{.2in} + \left(\left(t+1\right)\left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}}\right] + \frac{t}{w}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right]\right) \\ &\hspace{.2in} + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross7.pdf}}\right] + \left(1+ t^{-2}+ t^{-1} \right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}\right] \\ &\hspace{.2in} + \left(\frac{wt^{-2}}{1-t}+\frac{w^{-1}}{1-t^{-1}}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right]. \end{align*}

Finally, by simplifying the above equality, we have:

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right]
=& \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross4.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross6.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross10.pdf}}\right] + \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross12.pdf}}\right] + (t^2+ t+1) \left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}}\right] \\ &\hspace{.2in} + \left(1+ t^{-2} + t^{-1} \right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross7.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross13.pdf}}\right] \\ &\hspace{.2in} +  \left(\frac{t}{w}+\frac{t^2}{w}+\frac{wt^{-2}}{1-t}+\frac{w^{-1}}{1-t^{-1}}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right].
\end{align*}

By applying similar steps to the second diagram of the left-hand side of the skein relation (<ref>), we obtain the following equality:

\begin{align*}
\left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossrightref.pdf}}}\right] =&  \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross14.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross16.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross17.pdf}}\right] + \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross19.pdf}}\right] + \left(t^2+t+1\right)\left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}}\right] \\ &\hspace{.2in} + \left(1+ t^{-2} + t^{-1}\right)\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross8.pdf}}\right] + \frac{1}{t}\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross7.pdf}}\right] + t\left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross13.pdf}}\right] \\ &\hspace{.2in} +  \left(\frac{t}{w}+\frac{t^2}{w}+\frac{wt^{-2}}{1-t}+\frac{w^{-1}}{1-t^{-1}}\right)\left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}}\right].
\end{align*}

Then, by using that $[\, \cdot \,]$ is invariant under the move $R4$, we obtain the desired skein relation:

\begin{align*}
\left[ \raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossleft.pdf}}\right] - \left[ \reflectbox{\raisebox{-.1 in}{\includegraphics[scale=.3]{part2R3crossrightref.pdf}}}\right] =&\left( \frac{t}{w}+\frac{t^2}{w}+\frac{wt^{-2}}{1-t}+\frac{w^{-1}}{1-t^{-1}} \right) \left( \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right] - \left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}}\right] \right) \\
=&  \left(\frac{wt^{-2}}{1-t}+\frac{w^{-1}t^2}{1-t^{-1}}\right) \left( \left[\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}\right] - \left[\reflectbox{\raisebox{-.1in}{\includegraphics[scale=.3]{part2R3cross9.pdf}}}\right] \right).
\end{align*}

Hence the graphical skein relation (<ref>) holds.

We note that if the two strands in the left-hand side diagrams of the skein relations (<ref>) and (<ref>) are colored with the same color, then the following holds:

\begin{align*}
\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{cross.pdf}}\right]=  \frac{t+1}{tw}  \left[ \raisebox{-.11 in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}} \right]  + \frac{1}{t}\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{uncross.pdf}}\right], \,\,\,
\left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{cross.pdf}}\right] = w(t +1) \left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{negcross.pdf}} \right]  + t \left[ \raisebox{-.11 in}{\includegraphics[scale=.3]{uncross.pdf}}\right].
\end{align*}

Solving for  $\left[ \raisebox{-.11in}{\reflectbox{\includegraphics[scale=.3]{negcross.pdf}}} \right]$ and $\left[ \raisebox{-.11in}{\includegraphics[scale=.3]{negcross.pdf}} \right]$ in the above relations and substituting them in (<ref>) and (<ref>), respectively, we obtain the following skein relations:

\begin{align}
\left[ \reflectbox{\raisebox{-.11 in}{\includegraphics[scale=0.3]{poscrosscolor_reflect.pdf}}} \right] =& \frac{-w}{t+1} \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{uncross.pdf}} \right] - \frac{w}{t+1} \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{cross.pdf}} \right] + w \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{crosscolor.pdf}} \right] \label{graphpos}\\
\left[ {\raisebox{-.11 in}{\includegraphics[scale=0.3]{negcrosscolor.pdf}}} \right] =& \frac{-t}{w(t+1)} \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{uncross.pdf}} \right] - \frac{t}{w(t+1)} \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{cross.pdf}} \right] + \frac{1}{w} \left[ \raisebox{-.11 in}{\includegraphics[scale=0.3]{crosscolor.pdf}} \right].\label{graphneg}
\end{align}

The skein relations (<ref>) and (<ref>) will play a major role in the following section.

§ A STATE-SUM MODEL FOR THE AICARDI-JUYUMAYA INVARIANT FOR COLORED LINKS

The graphical skein relations proved in Theorem <ref> provide a recursive way to evaluate oriented, colored, planar graphs with 4-valent vertices, where a 4-valent vertex is diagrammatically represented as a singular crossing. We denote the evaluation of such planar graph $G$ by $[G]$. Specifically, the graphical skein relations of Theorem <ref> allow the evaluation $[G]$ to be written as a $\mathbb{Q}(x, t, w)$-formal linear combination of evaluations of colored 4-valent planar graphs with fewer vertices. This also allows us to compute the invariant $[\tilde{D}]$ of a colored singular link without relying on the existence of the invariant $F$ for colored classical links. In particular, graphical skein relations of Theorem <ref> provide a state-sum model for the Aicardi-Juyumaya invariant $F$ for colored (classical) links, where the states associated to a diagram of a colored link are oriented, colored, $4$-valent planar graphs, $G$.

Given any colored link diagram $D$, we can resolve each of its crossings using the skein relations (<ref>) and (<ref>). After having performed this step at each of the crossings of diagram $D$, we obtain $\left[ D \right]$ written as a formal linear combination of evaluations of its states $G$. Each state $G$ is then evaluated using the graphical relations provided by Theorem <ref>.

We remark that this type of approach is not new when working with polynomial invariants in combinatorial knot theory. Kauffman and Vogel [9] constructed an invariant for rigid-vertex embeddings of unoriented 4-valent graphs in $\mathbb{R}^3$, which extends the Dubrovnik polynomial to embeddings of graphs. The Dubrovnik polynomial is a version of the 2-variable Kauffman polynomial [7] for unoriented links. In [9], a graphical calculus for evaluations of planar 4-valent graphs (not colored) was provided. Then Carpentier [5] explained how that graphical calculus can be used to provide a state-sum formula for the Dubrovnik polynomial for unoriented links and its extension to rigid-vertex embeddings of 4-valent graphs.

If a connected colored 4-valent planar graph contains at least one vertex and does not contain a loop $\raisebox{-.08 in}{\includegraphics[scale=.25]{loop.pdf}} $ or a bigon $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.25]{bigoncrosscolor.pdf}}}, \reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.27]{4valbigoncolordown.pdf}}}$, then using a finite sequence of moves of types $\raisebox{-.09 in}{\includegraphics[scale=.27]{part1R3crossleft.pdf}} \leftrightarrow  \reflectbox{\raisebox{-.09 in}{\includegraphics[scale=.25]{part1R3crossrightref.pdf}}}$ and $\raisebox{-.09 in}{\includegraphics[scale=.25]{part2R3crossleft.pdf}} \leftrightarrow \reflectbox{\raisebox{-.09 in}{\includegraphics[scale=.25]{part2R3crossrightref.pdf}}}$, it is possible to transform the graph into a colored 4-valent planar graph that contains a bigon.

This statement can be proved in a same way as <cit.>.

There is a unique rational expression $[ G ] \in \mathbb{Q}(x, w, t)$ associated to a colored 4-valent planar graph $G$, such that $[ \,\cdot \, ]$ satisfies the graphical relations of Theorem <ref>, as well as the following equalities:

\begin{align} \label{RelTrivialLinks}
\left[ \raisebox{-.07 in}{\includegraphics[scale=0.2]{unknot.pdf}} \right] = 1, \left[  O_n^n   \right] = \left(\frac{1}{wx} \right)^{n-1}, \text{and} \, \left[  O_n^c  \tilde \cup  \raisebox{-.05 in}{\includegraphics[scale=0.2]{unknot.pdf}}   \right] = \frac{tw^2-1}{(1-t)w}  \left[  O_n^c   \right],
\end{align}

where $O_n^c$ is the standard diagram of the unlink with $n$ components and $c$ colors ($c \leq n)$, and where the circle in the disjoint union $O_n^c \tilde \cup  \raisebox{-.05 in}{\includegraphics[scale=0.2]{unknot.pdf}}$ has the same color as (at least) one of the unknotted circles in $O_n^c$.

The proof is by induction on the number of vertices in a colored 4-valent planar graph and is similar to <cit.>.

Let $[ \,\cdot \,]$ and $[ \,\cdot \,]^*$ be two rational expressions that satisfy the graphical relations given in Theorem <ref> and the additional relations (<ref>).
If $G$ has no vertices, then since $G$ is planar, it is a colored unlink containing $n$ unknotted circles. Then, by relations (<ref>), $[G] = [G]^*$.

Suppose that $[ \,\cdot \,]$ and $[ \,\cdot \,]^*$ yield equal values when evaluated on the  same colored 4-valent planar graph with at most $k$ vertices. Let $G$ be a colored 4-valent planar graph with $k+1$ vertices. If $G$ contains a loop $\raisebox{-.08 in}{\includegraphics[scale=.25]{loop.pdf}} $ or a local configuration of type $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.25]{bigoncrosscolor.pdf}}}$ or $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.27]{4valbigoncolordown.pdf}}}$, then by applying the skein relation (<ref>), (<ref>) or (<ref>), respectively, the evaluations $[G]$ and $[G]^*$ are written in terms of evaluations of colored 4-valent planar graphs with fewer vertices, and therefore, by the induction hypothesis,  $[G] = [G]^*$.

If $G$ does not contain a loop ($\raisebox{-.08 in}{\includegraphics[scale=.25]{loop.pdf}} $) or a bigon ($\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.25]{bigoncrosscolor.pdf}}}$ or $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.27]{4valbigoncolordown.pdf}}}$), then by Lemma <ref>, there exists a finite sequence of colored 4-valent planar graphs

\[G =  G_0 \to G_1 \to G_2 \to \dots \to G_s,\]

where for each $0 \leq i \leq s-1$, $G_{i+1}$ is obtained from $G_i$ by applying the move $\raisebox{-.09 in}{\includegraphics[scale=.27]{part1R3crossleft.pdf}} \leftrightarrow  \reflectbox{\raisebox{-.09 in}{\includegraphics[scale=.25]{part1R3crossrightref.pdf}}}$ or $\raisebox{-.09 in}{\includegraphics[scale=.25]{part2R3crossleft.pdf}} \leftrightarrow \reflectbox{\raisebox{-.09 in}{\includegraphics[scale=.25]{part2R3crossrightref.pdf}}}$, and where $G_s$ contains a bigon of type $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.25]{bigoncrosscolor.pdf}}}$ or $\reflectbox{\raisebox{-.08 in}{\includegraphics[scale=.27]{4valbigoncolordown.pdf}}}$. Then, by applying the skein relations (<ref>) and (<ref>) followed by the inductive hypothesis, we have that

\[ [G_i] -[G_{i+1}] =  [G_i]^* -[G_{i+1}]^*, \,\, \text{for all} \,\, 0 \leq i \leq s-1.  \]

Hence, $[G] - [G_s] = [G]^* - [G_s]^*$. But $G_s$ contains a bigon, and by the previous step in the proof, $[G_s] = [G_s]^*$. It follows that $[G] = [G]^*$.

By the principal of mathematical induction, $[ \,\cdot \,]$ and $[ \,\cdot \,]^*$ yield the same quantity when evaluated from a graph $G$ with any number of 4-valent vertices, and thus $[ \,\cdot \,] = [ \,\cdot \,]^*$.

Let $D_1$ and $D_2$ be diagrams representing a colored link. Resolve all crossings in $D_1$ and $D_2$ using the skein relations (<ref>) and (<ref>), to write $[D_1]$ and $[D_2]$ as $\mathbb{Q}(x, w, t)$-linear combinations of evaluations of their states. Then evaluate the corresponding states using the graphical skein relations (<ref>)–(<ref>), together with identities  (<ref>).
Then $[D_1] = [D_2]$; that is, the rational expression $[ \, \cdot \, ] \in \mathbb{Q}(x, w, t)$ is an invariant of colored links. Moreover, $[\, \cdot \, ]$ is equal to Aicardi-Juyumaya invariant $F$.

The proof that $[D_1] = [D_2]$ when the two diagrams differ by a Reidemeister move $R1$, $R2$ or $R3$ is
by direct computations, which in some sense are the reverse of the proofs of our graphical skein relations in Theorem <ref>.

[1] F. Aicardi, “New invariants of links from a skein invariant of colored links"; preprint arXiv:1512.00686.

[2] F. Aicardi, “An invariant of colored links via  skein relation", Arnold Math J. (2016), 2:159-169.

[3] F. Aicardi and J. Juyumaya, “Tied links", J. Knot Theory Ramifications 25 No. 09, 1641001 (2016).

[4] C. Caprau, “Movie moves for singular link cobordisms in 4-dimensional space", J. Knot Theory Ramifications 25, Issue 2 (2016) 23 pages.

[5] R. P. Carpentier, “From planar graphs to embedded graphs - a new approach to Kauffman and Vogel's polynomial", J. Knot Theory Ramifications 9, Issue 8 (2000), 975-986.

[6] P. Freyd, D. Yetter, J. Hoste, W. B. R. Lickorish, K. Millett and A. Ocneanu,  “A new polynomial invariant of knots and links", Bull. Amer. Math. Soc. 12 (1985), 239-246.

Ka1989 L. H. Kauffman, “Invariants of graphs in three-space", Trans. Amer. Math. Soc. 311 (1989), 697-710.

[7] L. H. Kauffman, “An invariant of regular isotopy", Trans. Amer. Math. Soc. 318, No. 2 (1990), 417-471.

[8] L. H. Kauffman, Knots and physics, Third edition. Series on Knots and Everything, Vol. 1, World Scientific Publishing Company, 2001.

[9] L. H. Kauffman, P. Vogel, “Link polynomials and a graphical calculus", J. Knot Theory Ramifications 1 (1992), 59-104.

[10] J. H. Przytycki, P. Traczyk, “Invariants of links of Conway type", Kobe J. Math. 2 (1987), 115-139.


Present affiliation: ]Tecnológico de Monterrey, Escuela de Ingeniería y
Ciencias, Ave. Eugenio Garza Sada 2501, Monterrey, N.L.64849, (México).

# Structured light beams constituted of incoming and outgoing waves

Job Mendoza-Hernández<EMAIL_ADDRESS>[  Maximino Luis Arroyo-
Carrasco  Facultad de Ciencias Físico -Matemáticas. Benemérita Universidad
Autónoma de Puebla, C.P. 72570, Puebla, Pue., México.  Marcelo David Iturbe-
Castillo  Sabino Chávez-Cerda<EMAIL_ADDRESS>Instituto Nacional de
Astrofísica, Óptica y Electrónica, Luis Enrique Erro N. 1, 72840 Tonantzintla,
Puebla, México

###### Abstract

In the present work, we demonstrate that structured light beams are
constituted by two traveling waves which transverse components are in opposite
directions, that is, incoming and outgoing from the axis of propagation. These
waves result from the complex sum of the two fundamental solutions of the
transverse component of the spatial wave equation. After a partial obstruction
of the beam, the incoming and outgoing waves can be easily observed during the
self-healing process, providing a simple explanation of the phenomenon.
Incoming and outgoing waves in Laguerre-Gauss beams are investigated
analytically, numerically and experimentally. The proposed way to describe
light beams might offer new insights into the phenomenon of diffraction.

Structured light beams, singular optics, waves propagation

###### pacs:

42.25.Bs

##  I Introduction

The recent Nobel Prize in optical tweezers reflects the importance of
structured light beams Ashkin1 ; Ashkin2 ; Ashkin3 ; Ashkin4 , beams whose
optical properties are specifically tailored according to the application
Allenbk ; Torres ; Andrews . It has been demonstrated that some structured
light beams (SLBs) have orbital angular momentum Allen , a characteristic that
increases its range of applications to fields like classical and quantum
communications roadmap ; Pagett ; Willner ; Willner1 , imaging Padgett1 ,
microscopy Betzig ; Rohrbach , and micro-manipulation roadmap . The spatial
and temporal properties of SLBs are studied with the wave equation and their
approximations, with the Helmholtz equation (HE) or the paraxial Helmholtz
equation (PHE) Boyd ; siegman ; saleh .

Since the appearance of lasers, several types of structured light beams have
been generated and described, like the Gauss, Hermite-Gauss, Laguerre-Gauss,
and more recently Ince-Gauss siegman ; ince  and the so-called elegant beams,
SLBs with transverse structures described by polynomials with a complex
argument eHermite ; eince . All these beams have distributions that are exact
solutions of the PHE Boyd ; GBur  that suffer diffraction during propagation.

In the late eighties, the non-diffracting beams appear, e. g., Bessel,
Parabolic and Mathieu beams, as new solutions of the HE durnin ; sabinobessel
; mathieu ; parabolic . Furthermore, there are beams which propagate in
parabolic trajectories known as Airy beams airy_berry ; airychri , beams whose
complex amplitudes are proportional to the confluent hyper-geometric function
hypermodes ; circular , and beams that do not have an exact solution to the HE
or PHE, such as the caustic jobjmo1  or the Pearcy beams, which describe the
diffraction of a caustic cuspid pearcy .

An intriguing property of SLBs is their ability of self-healing durnin ;
bouchal ; anguiano , characteristic that is maintained even at the quantum
level McLaren ; Eileen . It occurs when the beam is partially blocked with an
opaque object and, after certain propagation distance, it recovers its
transverse intensity distribution. The self-healing phenomenon was initially
observed in nondiffracting beamsbouchal  and have been described in terms of
conical waves sabinobessel ; chavezMc ; anguiano ; sabinoairy , by Babinet’s
principle bouchal ; agarwal ; aiello  and through the relation between two
orthogonal field components with an attenuated factor victorsh . More
recently, this property was investigated in Laguerre-Gauss beams through a
comparison with Bessel beams jobol .

To describe the internal structure of diffracting and nondiffracting beams
only the first solution of the HE and the PHE it is needed, however, a general
solution of such equations requires not only one but two solutions. Keeping
this in mind, the aim of this article is to probe that structured light beams
are composed of two traveling waves. Two fundamental solutions of the spatial
differential equation of second order that have their transverse vectors in
opposite directions and its sum generate a structured light beam as a standing
wave. In the following, we will refer to such waves as the incoming and
outgoing waves. We will demonstrate that Laguerre-Gauss beams contain incoming
and outgoing waves, through which it is possible to explain their self-healing
ability and show that they are the complex sum of two solutions of the
associated Laguerre differential equation of the transverse distribution. The
analytic, numerical and experimental features of the incoming and outgoing
waves are investigated.

The article is structured as follows: first, we show the representation of the
general solution of the wave equation and the spatial wave equation, in terms
of two solutions. Then, we derive the second solution of the associated
Laguerre differential equation that gives form to the incoming and outgoing
waves. Finally, we present the experimental generation of both waves, and when
they are partially blocked.

##  II Traveling waves in structured light beams

The homogeneous wave equation, Eq. (1), of an homogeneous medium in regions
free of currents and charges, is satisfied by a traveling wave
$V(\textbf{r},t)$ wolf ; saleh .

$\nabla^{2}V-\frac{1}{v^{2}}\frac{\partial^{2}V}{\partial t^{2}}=0.$ (1)

The simplest solution of this equation is a traveling plane wave
$V=V(\textbf{r.s},t)$, with a general solution of the form

$V=V_{1}(\textbf{r.s}-vt)+V_{2}(\textbf{r.s}+vt),$ (2)

where $V_{1}$ and $V_{2}$ are two solutions to the equation. These waves
propagate with velocity $v$ and have a direction opposite to the plane
$\textbf{r.s}=d$, with $d$ a constant.

If the wave can be separated as $V(\textbf{r},t)=R[U(r)F(t)]$, where $R$
denotes the real part and $F(t)=exp(-iwt)$ is an harmonic function, we find
that $U$ must satisfy the Helmholtz equation

$\nabla^{2}U+k^{2}U=0,$ (3)

here $k=w/v$ is the wave vector and $U(\textbf{r})=A(\textbf{r})exp(ig(r))$ is
a complex amplitude. Equation (3) is the spatial contribution and the general
solution depends on the partial differential equation (PDE) that defined the
system.

In order to see the spatial contribution of Eq. 3, we can use the method of
separation of variables to split the PDE into second-order ordinary
differential equations (ODEs).

Now, we can assign a constant and construct a general solution for each
equation. In a direct analogy with the wave equation, the spatial wave
equation has two solutions with the transverse components of the wave vector
in opposite directions. Both waves travel in the direction of vector $k$
incoming or outgoing direction from the axis of propagation, hence the name of
incoming and outgoing waves. These waves can be written as

$U_{in}=U_{1}-iU_{2},$ (4) $U_{out}=U_{1}+iU_{2},$ (5)

where $U_{in}$ and $U_{out}$ are the incoming and outgoing waves, and $U_{1}$
and $U_{2}$ are the two solutions to the second-order ODE. The standing wave
is

$U=U_{in}+U_{out}.$ (6)

If the Helmholtz equation is written with circular cylindrical symmetry, and
the field is separable in the transverse and longitudinal coordinates as
$U(\textbf{r})=U(r)exp(ik_{z}z)$, the incoming and outgoing waves respect to
$z$ axis have of the form:

$U_{in}(r)=U_{1}(r)-iU_{2}(r),$ (7) $U_{out}(r)=U_{1}(r)+iU_{2}(r).$ (8)

In this case, $U_{in}(r)$ and $U_{out}(r)$ are Hankel functions Arfken , and
the standing wave $U(r)=U_{in}(r)+U_{out}=2U_{1}(r)$ is the first solution of
the second-order ODE. The simplest case of circular cylindrical symmetry is
when $U(r)=2J_{0}(k_{t}r)$, where $J_{0}$ is the Bessel beam and
$k^{2}=k_{z}^{2}+k_{t}^{2}$ with $k_{t}$ the transverse component and $k_{z}$
the longitudinal. Since the intensity is independent of $z$, $I(r,z)=I(r,0)$,
these beams are known as nondiffracting beams durnin , showing the self-
healing property sabinobessel ; anguiano .

On the other hand, diffracting beams $I(r,z)$ are analyzed with the paraxial
Helmholtz equation siegman ; saleh , a second-order PDE whose solutions can be
found with the method of separation of variables. In circular cylindrical
symmetry, these solutions are

$U_{in}(r,z)=U_{1}(r,z)-iU_{2}(r,z),$ (9)
$U_{out}(r,z)=U_{1}(r,z)+iU_{2}(r,z),$ (10)

here $U_{in}(r,z)$ and $U_{out}(r,z)$ are the incoming and outgoing waves for
the paraxial approximation, and $U_{1}(r,z)$ and $U_{2}(r,z)$ are the two
linearly independent solutions. Thus, the standing wave for diffracting beams
is

$U(r,z)=U_{in}(r,z)+U_{out}(r,z).$ (11)

Equation (11) gives a shape structured by two waves. Therefore, we can easily
observe the behavior of the two beams when they are propagated and also
partially obstructed.

##  III  The construction of incoming and outgoing waves for Laguerre-Gauss
beams

The traveling waves that describe the propagation of Laguerre-Gauss beams
(LGBs) are found with the solutions of the associated Laguerre differential
equation (ALDE).

It is well-known that LGBs are solutions of the paraxial wave equation in
cylindrical coordinates $(r,\varphi,z)$ siegman ; GBur . The ALDE equation
that governs the radial features of the beam, is obtained through the method
of separation of variables and its first solution is given by the associated
Laguerre polynomial $L_{n}^{m}(x)$ Arfken ; Agarwal:2009 ; Trench:2000 ;
Nagle:2008  (for simplicity we will only use the $x$ coordinate). The
parameters $n$ and $m$ are the radial and azimuthal orders of the polynomial
have integer values. It is important to highlight that the second solution
where $n$ and $m$ integer values is not present in the literatu. In this
document, we indicate a way to build it using the first solution.

In order to find the second solution with $n$ and $m$ integer values, it is
useful to use the confluent hypergeometric function (CHF) of first kind
$\Phi(a,m;x)$ with $a$ non-integer number Arfken ; Bell . Then the associated
Laguerre polynomial $L_{n}^{m}(x)$ Arfken  can be written as

$L_{n}^{m}(x)=\frac{(m+n)!}{m!n!}\Phi(-n;m+1,x),$ (12)

which gets reduced to the associated Laguerre polynomial of degree $n$ when
the index of the sum is $k=n$, and also when multiplied by $(m+n)!/m!n!$.

A second solution to ALDE is introduced with through function $U(a,m,x)$, the
confluent hypergeometric function of the second kind (Tricomi function) Arfken
. The problem with $\Phi(-a,m+1;x)$ and $U(-a,m+1;x)$ is that they are
dependent functions when m=0,1,2,.. and a=n=0,1,2,….

$\displaystyle{U(-n,m+1,x)}$ $\displaystyle=$
$\displaystyle(-1)^{n}\frac{(m+n)!}{m!}\Phi(-n,m+1,x).$ (13)

However, $\Phi$ and $U$ can be used to define a second solution to ALDE, as we
will see in the next section.

###  III.1  X function as second solution and Laguerre-Hankel functions.
Analytic form of the incoming and outgoing waves

We define the second solution of the ALDE for $m$ and $n$ integers following
the methodology given by Hankel and Weber to find the second solution of the
Bessel differential equation Watson . After reviewing the math tables
gradshteyn ; abramowitz , we have decided to call this new function $X$.

The function $X$ is defined as tesisjob :

$\displaystyle{X_{a}^{m}(x)}$ $\displaystyle=$
$\displaystyle\frac{DU\cos(a\pi)-\Phi}{\sin(a\pi)},$ (14)

where $a$ is not an integer, $m=0,1,2$…, and $D=\frac{m!}{(m+a)!}$. For
simplicity, we write $U$ and $\Phi$ instead of $U(-a,m+1;x)$ and
$\Phi(-a,m+1;x)$, respectively. The function $X$ for the integer indexes $m$
and $n$ is constructed as a Neumann function Arfken . When the value of
$a\rightarrow n$, the right-hand side of Eq. (14) becomes undefined, and it
approaches a limit. Using L’Hopital’s rule, we found that tesisjob :

$\displaystyle{X_{n}^{m}(x)}$ $\displaystyle=$
$\displaystyle\lim_{a\rightarrow n}X_{a}^{m}(x)$ (15)
$\displaystyle=\frac{1}{\pi}\left[(-1)^{n}\frac{\partial(DU)}{\partial
a}-\frac{\partial(\Phi)}{\partial a}\right]_{a=n},$

where $X_{n}^{m}(x)$ is the second solution of the ALDE for $n=0,1,2...$ and
$m=0,1,2,...$. The factor $D=\frac{m!}{(m+a)!}$ is necessary to match
amplitudes and compare $X_{n}^{m}(x)$ with the function $\Phi(-n,m+1,x)$.
Then, the ALDE’s most general solution is given by:

$\displaystyle y(n,m;x)=A\Phi(-n,m+1;x)+BX_{n}^{m}(x).$ (16)

Here $A$ and $B$ are constants, and the indexes $m$ and $n$ are integers.

If we multiply $\Phi$ and $X$ from Eq. (16) by

$\displaystyle C(m,n)=\frac{(m+n)!}{m!n!},$ (17)

then $X_{n}^{m}=CX_{n}^{m}$ and $L_{n}^{m}=C\Phi(-n,m+1;x)$.

The solutions $X_{n}^{m}$ and $L_{n}^{m}$ of the ALDE allow us to define the
following Hankel type functions

$LH_{n}^{m}(1;x)=L_{n}^{m}(x)+iX_{n}^{m}(x),$ (18)
$LH_{n}^{m}(2;x)=L_{n}^{m}(x)-iX_{n}^{m}(x).$ (19)

Therefore, Eq. (18) and Eq. (19) are Laguerre-Hankel functions of first and
second kind. When we have both functions simultaneously, the result is a
Laguerre polynomial:

$[LH_{n}^{m}(1;x)+LH_{n}^{m}(2;x)]=2L_{n}^{m}(x).$ (20)

The Laguerre-Hankel functions describe the two fundamental functions, in the
transverse direction, for the generation of incoming and outgoing waves.

##  IV  Behavior of incoming and outgoing waves in Laguerre-Gauss beams

We consider the definition of Laguerre-Gauss beams as in GBur . The incoming
and outgoing waves that constitute a Laguerre-Gauss beam are defined by the
Laguerre-Hankel functions $LH_{n}^{|m|}$ described as

$U_{in,out}{=}A_{0}\frac{w_{0}}{w(z)}\left(\frac{\sqrt{2}r}{w(z)}\right)^{|m|}LH_{n}^{|m|}\left(p;\frac{2r^{2}}{w^{2}(z)}\right)\times\\\
\exp\left[{\frac{-r^{2}}{w^{2}(z)}-\frac{ikr^{2}}{2R(z)}+i(2n+|m|+1)\Phi(z)-im\varphi}\right],$
(21)

where $p=1,2$, and $n$ and $m$ are the radial and azimuthal indices. Also, it
is possible defined the X-Gauss beam (XGB) substituting $X_{n}^{|m|}$ by
$LH_{n}^{|m|}$ in the Eq. 21, in order to comparison between the two solutions
of ALDE and the incoming and outgoing waves.

The parameters in Eq. (21) are the beam waist $w_{0}$, beam width
$w^{2}\left(z\right)=w_{0}^{2}\left[1+\left(z/L_{D}\right)^{2}\right]$,
transverse phase front
$R\left(z\right)=z\left[1+\left(L_{D}/z\right)^{2}\right]$ and the Gouy phase
shift $\Phi(z)=\tan^{-1}(z/L_{D})$. In all these expressions
$L_{D}=kw_{0}^{2}/2$ is the Rayleigh distance or diffraction distance for the
wave number $k=2\pi/\lambda$.

In order to show the features of the incoming and outgoing waves, we have
normalized the transverse coordinate to $w_{0}$ and the longitudinal
coordinate $z$ to $L_{D}$ in Eq. (21). It is important to note that the
explicit solution of $X_{n}^{m}$, as well as its complete features, will be
reported elsewhere.

The purpose of this work is not only to observe the behavior of the LGB in
terms of two constituent waves, the incoming and the outgoing, but also to
find a method to obtain them. Therefore, we approach the function $X_{n}^{m}$
by taking the nearest integer as $a\rightarrow n$ in the first term of the sum
in Eq. (15), taking $k=n$ elements of the series. In the second term of the
sum, we take at least $k=3n+1$ elements of the series, this in order to have
the same number of rings in the XGB and the LGB. In this way, we avoid the
divergence on both sides of the function.

Figure 1 shows the central intensity profile of the incoming wave, including
the LGB and the XGB. The incoming and outgoing waves diverge at the origin
because of their logarithmic solutions, since the function $X$ appears in the
second solution Arfken . We selected the intensity at the center of the beams
to be at least three times larger than that of the lobes, in order to
distinguish their contributions. We can easily observe the relation between
teh two beams, LGB has a maximum where XGB has a minimum. Figure 1 (a) shows
the case when $m=1$ and $n=3$, while in Fig. 1 (b) $m=5$ and $n=10$.

Figure 1: Intensity profile of the incoming-wave or outgoing-wave (red), LGB
(blue), and XGB (green). (a) the case $m=1$ and $n=3$, (b) $m=5$ and $n=10$.

The incoming and outgoing waves have a conical wavefront Fig. 2, the first
traveling incoming with respect to the axis, and the latter, in the opposite
direction. The function $\alpha_{in,out}$ is plotted for each $U_{in,out}$.
Figure 2 shows the wavefront for $z=0$, and $t=0$. Figure 2(a) the case $m=0$
and $n=3$, and Fig. 2(b) shows the case $m=1$ and $n=3$, mixing the conical
and helical wavefronts.

Figure 2: Conical wavefront for incoming and outgoing waves: (a) without
orbital angular momentum $m=0$, and (b) with orbital angular momentum $m\neq
0$.

The wavefronts of the incoming and outgoing waves help us to understand the
self-healing behavior. If the beam is partially blocked by an obstruction of
radius $a$, the beam will recover its transverse profile due to the sum of
these two waves. The opposite directions of the perpendicular wave vectors
hold the stationary distribution after the obstruction (see Fig. 3). The LGB
has a conical wavefront, although there is a curve in the conical base, as
shown in Fig. 3. For the other transverse distributions, the symmetry of the
beam may give other surfaces; however, there will be two perpendicular wave
vectors with opposite directions to hold the shape of the beam.

Figure 3: Wavefronts of the incoming and outgoing waves: direction of the
transverse wave vector after an obstruction of radius $a$.

The behavior under propagation of the incoming and outgoing waves is shown in
Fig. 4. The outgoing wave generates the external part of the LGB as shown in
the first row of the figure. The incoming wave generates the internal part of
the LGB; this feature acts as an axicon for the generated LGB as in the case
of Bessel beamsanguiano . Thus, it is possible to consider the generation of
optical elements using the incoming wave.

Figure 4: Stages of propagation of incoming and outgoing waves and LGB: (a)
with $m=5$ and $n=10$ at the position $z=0$, (b) $z=0.5L_{D}$, and (c)
$z=L_{D}$. The incoming wave is shown in the first row, the outgoing-wave in
the second row, and the LGB in the last row.

##  V Experimental generation of incoming and outgoing waves

We have generated experimentally the incoming and outgoing waves and observed
these waves when a Laguerre-Gauss beam is partially blocked. We used the
technique developed by Arrizón et al. victor , as shown in Figure 5. A
computer-generated hologram (CH) was implemented into an amplitude liquid
crystal spatial light modulator SLM (HOLOEYE-LC2002). The light source is a
He-Ne laser (632.8nm) with polarization parallel to the director axis of the
SLM. A polarizer is placed after the SLM to modulate the CH and eliminate an
extra phase. The beam is obtained with a combination of a 4f-system and an
aperture located at the Fourier plane to decode the wave $U$. The CCD camera
keeps the beam after passing through the second lens $f_{2}$.

Figure 5:  Experimental scheme for the generation of incoming and outgoing
waves ($U$) with a computer-generated hologram. $HWP$: half wave plate, $SF$:
filter system, $P$: polarizer, $SLM$: spatial light modulator, $f_{1}=38cm$
and $f_{2}=10cm$ lenses, and $CCD$: camera.

We can observe qualitatively the behavior of the incoming and outgoing waves
in the partially blocked Laguerre-Gauss beams while they are self-healing, as
shown in Fig. 6. We also observe the behavior on Bessel beams with the same
transverse section and spatial frequency for comparison, as in the work done
by Mendoza-Hernández et al.jobol . Figure 6(a) shows the LGB in the first row
and Bessel beam (BB) in the second row, partially blocked. Figure 6(b) shows
two shadows due to the incoming and outgoing waves: a large low contrast
shadow due to the outgoing wave and a small high contrast shadow due to the
incoming-wave. The diffracted shadow due to the incoming wave affects the
innermost ring and moves around the center of the pattern, until it becomes an
outgoing wave that increases its size and moves away from the center (Fig.
6(c)). Using a lens $f_{2}=10cm$ into Fig. (5), the LGB has a waist
$w_{0}=0.2644mm$ and a diffraction distance of $L_{D}=34.69cm$. Figure 6(c)
shows the self-healing of LGB and BB.

Figure 6: The self-healing comparison between a Laguerre-Gauss beam with
azimuthal index $m=5$ and radial index $n=19$ and a Bessel beam, both
partially obstructed at the position: (a) $z=0cm$, (b) $z=8cm$ , and (c)
$z=34cm$. We can see the behavior of two shadows due to the constitutive waves
of the beam in (b).

Figure 7 shows the behavior of the partially obstructed incoming and outgoing
waves, and the partially blocked LGB. It portrays the dynamics of self-healing
contained in LGB. The first row presents the large shadow due to the outgoing
wave that diffracts away from the center. In the second row, a high contrast
shadow is seen due to the incoming wave that affects the innermost ring, which
moves around the center. Therefore, the behavior in the partially blocked LGB
is caused by two waves that hold the stationary transverse distributions.

Figure 7:  Incoming and outgoing waves, and LGB partially obstructed with
$m=5$ and $n=19$ at the position: (a) $z=0cm$, (b) $z=8cm$, and (c) $z=34cm$.

##  VI Discussion and Conclusion

It is well known that a solution of the wave equation is a traveling wave. In
the Helmholtz equation, a second-order differential equation that describes
the spatial features of a beam, has two solutions; however, only the first one
is necessary to describe the internal structure of diffracting and
nondiffracting beams. In this work, we have demonstrated that structured light
beams can also be described through both solutions, two traveling waves with
the particularity of having their transverse vector in opposite directions,
waves that we have called the incoming wave and the outgoing wave. Since the
standing wave corresponds to the sum of the incoming and the outgoing waves,
it is possible to obtain information about each wave when the beam is
partially blocked during propagation.

The characteristics of the incoming and outgoing waves in the Helmholtz
equation and its paraxial approximation were studied. In particular, we have
demonstrated the existence of incoming and outgoing waves in Laguerre-Gauss
beams, treated as diffracting beams. First, it was necessary to find the
second solution of the associated Laguerre differential equation for integer
radial and azimuthal indexes or function $X$, as we named it. Then, function
$X$ was used to construct the Laguerre-Hankel functions of first and second
order that would define the incoming and outgoing waves. The generation and
propagation of such waves were presented numerically and experimentally,
showing that the diffracting spreading due to propagation does not affect
observed their fundamental components. Finally in conclusion, a structured
beam can be seen as a standing wave that results from the sum of the incoming
and outgoing waves, and thus, providing an explanation for its self-healing
ability.

###### Acknowledgements.

Job Mendoza-Hernández acknowledges support from the National Council of
Science and Technology of Mexico, CONACYT. JMH thank Victor Arrizon for the
summer classes about SLMs, and Dorilian López-Mago, Ricardo Tellez-Limón,
Adriana Inclan-Ladino, and Maribel Hidalgo-Aguirre for the fruitful
discussions about the present work.

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