\end{equation}
The map $\eta$ is obtained by extending this rule linearly.

We need to check that $\eta$ is well-defined.  First we verify that the defining relations for $\rgWeb$ given in \cite[Definition 2.3]{TVW} involving only green edges are preserved by $\eta$.  Since diagrams with a green edge labeled by $\textcolor{green}{k} \geq 3$ are zero, the ``digon removal relation'' \cite[(2-7)]{TVW} in $\rgWeb$ reduces to the following equality:
\begin{equation*}
\begin{tikzpicture}[baseline=-2ex,scale=.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (57) at (-3, 0) {};
\node [style=none] (58) at (-1, 0) {};
\node [style=none] (59) at (-2, 1) {};
\node [style=none] (60) at (-2, -1) {};
\node [style=none] (61) at (-2, 3) {};
\node [style=none] (62) at (-2, -3) {};
\node [style=none] (63) at (-2, -3.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (65) at (-2, 3.5) {\textcolor{green}{\tiny{$2$}}};
%		\node [style=none] (70) at (2, 3) {70};
%		\node [style=none] (71) at (2, -3) {71};
\node [style=none] (72) at (-3.5, 0) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (73) at (-0.5, 0) {\textcolor{black}{\tiny{$1$}}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, dashed, green] [out=180,in=90] (59.center) to (57.center);
\draw [thick, dashed, green] [out=0,in=90] (59.center) to (58.center);
\draw [thick, dashed, green] [out=0,in=270] (60.center) to (58.center);
\draw [thick, dashed, green] [out=180,in=270] (60.center) to (57.center);
\draw [thick, dashed, green]  (59.center) to (61.center);
\draw [thick, dashed, green]  (60.center) to (62.center);
%		\draw [thick, dashed, green]  (71.center) to (70.center);
\end{pgfonlayer}
\end{tikzpicture}
\;\; = \; \;  [2]_{q} \;
\begin{tikzpicture}[baseline=-2ex,scale=.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
%		\node [style=none] (57) at (-3, 0) {57};
%		\node [style=none] (58) at (-1, 0) {58};
%		\node [style=none] (59) at (-2, 1) {59};
%		\node [style=none] (60) at (-2, -1) {60};
%		\node [style=none] (61) at (-2, 3) {61};
%		\node [style=none] (62) at (-2, -3) {62};
%		\node [style=none] (63) at (-2, -3.5) {\textcolor{green}{\tiny{$2$}}};
%		\node [style=none] (65) at (-2, 3.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (70) at (2, 3) {};
\node [style=none] (71) at (2, -3) {};
\node [style=none] (72) at (2, -3.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (73) at (2, 3.5) {\textcolor{green}{\tiny{$2$}}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
%		\draw [thick, dashed, green] [out=180,in=90] (59.center) to (57.center);
%		\draw [thick, dashed, green] [out=0,in=90] (59.center) to (58.center);
%		\draw [thick, dashed, green] [out=180,in=270] (60.center) to (58.center);
%		\draw [thick, dashed, green] [out=0,in=270] (60.center) to (57.center);
%		\draw [thick, dashed, green]  (59.center) to (61.center);
%		\draw [thick, dashed, green]  (60.center) to (62.center);
\draw [thick, dashed, green]  (71.center) to (70.center);
\end{pgfonlayer}
\end{tikzpicture} \;  .
\end{equation*}  Similarly, an example of the ``square switch'' relation \cite[(2-8)]{TVW} is the following equality:
\begin{equation}\label{E:squareswitch}
\begin{tikzpicture}[baseline=-2ex,scale=.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-4, -2) {};
\node [style=none] (1) at (-2, -2) {};
\node [style=none] (2) at (-4, -1) {};
\node [style=none] (3) at (-2, 0) {};
\node [style=none] (4) at (-2, 1) {};
\node [style=none] (5) at (-4, 2) {};
\node [style=none] (6) at (-4, 3) {};
\node [style=none] (7) at (-2, 3) {};
\node [style=none] (16) at (-4, 3.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (17) at (-2, 3.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (18) at (-4, -2.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (19) at (-2, -2.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (20) at (2, -2) {};
\node [style=none] (21) at (4, -2) {};
\node [style=none] (26) at (2, 3) {};
\node [style=none] (27) at (4, 3) {};
\node [style=none] (28) at (2, 3.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (29) at (4, 3.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (30) at (2, -2.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (31) at (4, -2.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (35) at (-4.5, 0.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (32) at (-1.5, 0.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (34) at (0.15, 0.4) {\textcolor{black}{=}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, dashed, green] (0.center) to (2.center);
\draw [thick, dashed, green] [out=0,in=90] (2.center) to (3.center);
\draw [thick, dashed, green] (1.center) to (3.center);
\draw [thick, dashed, green] (3.center) to (4.center);
\draw [thick, dashed, green] (5.center) to (2.center);
\draw [thick, dashed, green] [out=180,in=90] (4.center) to (5.center);
\draw [thick, dashed, green] (5.center) to (6.center);
\draw [thick, dashed, green] (4.center) to (7.center);
\draw [thick, dashed, green] (27.center) to (21.center);
\draw [thick, dashed, green] (26.center) to (20.center);
\end{pgfonlayer}
\end{tikzpicture} \; .
\end{equation}   In every case an easy check verifies that the rules for $\eta$ do the same thing to both sides of the equations.  In checking, note that the green (co)associativity relations given in \cite[(2-6)]{TVW} are zero and there is nothing to check.

As we next explain, the fact it preserves the green relations implies $\eta$ is well-defined.
\begin{equation*}
E \; \; = \; \;
\begin{tikzpicture}[baseline=-2ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (4) at (-3.75, 0) {\tiny{$\dots $}};
\node [style=none] (9) at (0, 1) {};
\node [style=none] (10) at (2, 1) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (1, -1) {};
\node [style=none] (13) at (1, 1) {\tiny{$\dots $}};
\node [style=none] (14) at (0, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (15) at (2, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (16) at (1, -1.5) {\textcolor{red}{\tiny{$r$}}};
\node [style=none] (19) at (-6.5, 1) {};
\node [style=none] (20) at (-4.5, 1) {};
\node [style=none] (21) at (-5.5, 0) {};
\node [style=none] (22) at (-5.5, -1) {};
\node [style=none] (24) at (-6.5, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (25) at (-4.5, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (26) at (-5.5, -1.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (28) at (-3, 1) {};
\node [style=none] (29) at (-1, 1) {};
\node [style=none] (30) at (-2, 0) {};
\node [style=none] (31) at (-2, -1) {};
\node [style=none] (33) at (-3, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (34) at (-1, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (35) at (-2, -1.5) {\textcolor{green}{\tiny{$2$}}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, densely dashdotted, densely dashdotted, red]  [out=270,in=180] (9.center) to (11.center);
\draw [thick, densely dashdotted, densely dashdotted, red]  [out=270,in=0] (10.center) to (11.center);
\draw [thick, densely dashdotted, densely dashdotted, red]  (11.center) to (12.center);
\draw [thick, dashed, green] [out=270,in=180] (19.center) to (21.center);
\draw [thick, dashed, green] [out=270,in=0] (20.center) to (21.center);
\draw [thick, dashed, green] (21.center) to (22.center);
\draw [thick, dashed, green] [out=270,in=180](28.center) to (30.center);
\draw [thick, dashed, green] [out=270,in=0](29.center) to (30.center);
\draw [thick, dashed, green] (30.center) to (31.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \; \;\;\text{ and } \;\;\;\;
C \; \; = \; \;
\begin{tikzpicture}[baseline=-2ex,scale=.5, color=\clr, yscale=-1 ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (4) at (-3.75, 0) {\tiny{$\dots $}};
\node [style=none] (9) at (0, 1) {};
\node [style=none] (10) at (2, 1) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (1, -1) {};
\node [style=none] (13) at (1, 1) {\tiny{$\dots $}};
\node [style=none] (14) at (0, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (15) at (2, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (16) at (1, -1.5) {\textcolor{red}{\tiny{$r$}}};
\node [style=none] (19) at (-6.5, 1) {};
\node [style=none] (20) at (-4.5, 1) {};
\node [style=none] (21) at (-5.5, 0) {};
\node [style=none] (22) at (-5.5, -1) {};
\node [style=none] (24) at (-6.5, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (25) at (-4.5, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (26) at (-5.5, -1.5) {\textcolor{green}{\tiny{$2$}}};
\node [style=none] (28) at (-3, 1) {};
\node [style=none] (29) at (-1, 1) {};
\node [style=none] (30) at (-2, 0) {};
\node [style=none] (31) at (-2, -1) {};
\node [style=none] (33) at (-3, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (34) at (-1, 1.5) {\textcolor{black}{\tiny{$1$}}};
\node [style=none] (35) at (-2, -1.5) {\textcolor{green}{\tiny{$2$}}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, densely dashdotted, densely dashdotted, red]  [out=270,in=180] (9.center) to (11.center);
\draw [thick, densely dashdotted, densely dashdotted, red]  [out=270,in=0] (10.center) to (11.center);
\draw [thick, densely dashdotted, densely dashdotted, red] (11.center) to (12.center);
\draw [thick, dashed, green] [out=270,in=180] (19.center) to (21.center);
\draw [thick, dashed, green] [out=270,in=0] (20.center) to (21.center);
\draw [thick, dashed, green] (21.center) to (22.center);
\draw [thick, dashed, green] [out=270,in=180](28.center) to (30.center);
\draw [thick, dashed, green] [out=270,in=0](29.center) to (30.center);
\draw [thick, dashed, green] (30.center) to (31.center);
\end{pgfonlayer}
\end{tikzpicture} \; .
\end{equation*}
Here $E$ has $n$ green splits and the red diagram is an  $\textcolor{red}{r}$ split apart into $r$ strands labeled by $1$ by repeatedly concatenating red splits. Likewise, $C$ has $n$ green merges and and the red diagram is $r$ strands labeled by $1$ that are merged together into a single $\textcolor{red}{r}$ by concatenating red merges.

If $D_{1}$ and $D_{2}$ are two linear combinations of diagrams, all of the form \cref{E:Diagram},  which are equal in $\rgWeb$, then we can concatenate an $E$ on top of each to obtain $ED_{1}$ and $ED_{2}$.  These are elements of $\End_{\rgWeb}(1^{d})$ and will still be equal in $\rgWeb$.

Applying \cite[Lemma 2.12, Corollary 2.13]{TVW} to the red part of $ED_{1}$ and $ED_{2}$ shows they can be rewritten as a linear combination of diagrams in $\rgWeb$ consisting of only green edges.  That is, $ED_{1}$ and $ED_{2}$ are morphisms in the subcategory $\gWeb$ (the subcategory of green objects and the morphisms in $\rgWeb$ which can be expressed as a linear combination of diagrams with only green edges).  However, by \cite[Corollary 2.16]{TVW} the category $\gWeb$ is equivalent to the ``green-ification'' of the category $\CKMWeb$ introduced by Cautis--Kamnitzer--Morrison.  Thus, if two diagrams are equal in $\gWeb$, they must be related by the application of a finite sequence of defining relations of $\CKMWeb$ which have been green-ified.  However, a check verifies that the greenification of the defining relations of $\CKMWeb$ are nothing but relations in $\rgWeb$ which follow from the all-green defining  relations.

The discussion in the previous paragraph implies there is a sequence of linear combinations of diagrams in $\End_{\gWeb}(1^{\otimes d})$,
\[
u_{1}, u_{2}, \dotsc , u_{p},
\] where each $u_{i}$ consists of only green edges, each $u_{i}$ is obtained from $u_{i-1}$ by applying one of the green relations of $\rgWeb$, and where $u_{1}=ED_{1}$ and $u_{2}=ED_{2}$ in $\rgWeb$. Concatenating each term of this sequence on top with $C$ yields the sequence,
\[
Cu_{1}, Cu_{2}, \dotsc , Cu_{p},
\] where each $Cu_{i}$ is obtained from $Cu_{i-1}$ by applying one of the green relations of $\rgWeb$ to $u_{i-1}$ while keeping $C$ fixed.

However, note that each $Cu_{i}$ is a linear combination of diagrams of the form given in \cref{E:Diagram} and so it makes sense to apply the rule defining $\eta$ to each of them.  Since $\eta$ satisfies the green relations of $\rgWeb$ it follows that
\[
\eta(Cu_{1})= \eta(Cu_{2})= \dotsc = \eta(Cu_{p}).
\]

Set $\kappa=[r]_{q}![2]_{q}^{n}$.  We claim that $\eta (D_{1}) = \frac{1}{\kappa}\eta (Cu_{1})$ and $\eta (D_{2}) = \frac{1}{\kappa}\eta(Cu_{p})$.  Since the rule for $\eta$ is local and $u_{1}$ (resp., $u_{p}$) is obtained by applying \cite[Lemma 2.12, Corollary 2.13]{TVW} only to the red part of $ED_{1}$ (resp., $ED_{2}$), it suffices to consider the case when $D_{1}=D_{2}=D$ is a diagram of the form given in \cref{E:Diagram} where $X$ consists of $d$ black vertical strands labeled by $1$.  That is, the case when $D = C$.  By \cite[Lemma 2.12]{TVW}, $ED = \kappa \mathcal{CL}$, where the $\mathcal{CL}$ is the green-red diagram given by the left-to-right horizontal concatenation of $n$ green clasps $\mathcal{CL}_{2}^{\textcolor{green}{g}}$ and one red clasp $ \mathcal{CL}_{r}^{\textcolor{red}{r}}$.  By \cite[Corollary 2.13, Definition 2.11]{TVW} the red clasp can be rewritten into smaller red clasps and green diagrams, and by repeatedly using this recursion one ultimately obtains $u_{1}$ for $ED$.  Using this recursion and our convention that any web with an edge connecting two vertices from among $\{1', \dotsc , r' \}$ is equal to zero, a straightforward induction on the size of the red clasp shows that $\eta(Cu_{1})=\kappa w_{0}$.  Since $\eta(D)=w_{0}$, it follows that $\eta(D) = \frac{1}{\kappa}\eta(Cu_{1})$, as desired.

Combining the previous calculations yields
\[
\eta (D_{1}) = \frac{1}{\kappa} \eta \left( C u_{1} \right) = \frac{1}{\kappa} \eta \left( Cu_{p} \right)  = \eta (D_{2}).
\] Therefore $\eta$ is well-defined.

To show that $\eta$ is a $\HH_{d}(q)$-module homomorphism requires that we analyze the two ways the generator $T_{i}$ may act on the web module as described in \cref{SS:Webmodule}.  The easier case is when at least one of the vertices $i$ and $i+1$ are not adjacent to through-strings in $\eta(D)$. In this case equivariance for the generator $T_{i}$ is immediate from the observation that when the rules for $\eta$ are applied to the green-red diagram $\beta_{i}$ the result is the linear combination of webs which gives the action of $T_{i}$ on $\eta(D)$.

The second and less obvious case is when both $i$ and $i+1$ are through-strings in $\eta(D)$. In this case,
\[
\eta(D.T_{i})=q\eta(D) + \eta (D\tilde{E}_{i}),
\]
where $\tilde{E}_{i}$ is the second diagram in the definition of $\beta_{i}$ given in \cref{SS:TVW-webs}, and where $D\tilde{E}_{i}$ is the green-red diagram obtained by concatenating $\tilde{E}_{i}$ onto the bottom of $D$.  It will suffice to show $D\tilde{E}_{i}=0$ in $\rgWeb$.

First, by the rule for $\eta$, the tops of through-strings in $\eta(D)$ correspond to vertices where the red part of $D$ meets $X$. Since the two through-strings in question have adjacent vertices at the bottom, they also have adjacent vertices at the top. In $D$ call these top vertices $j$ and $j+1$.  By the associativity relation \cite[Definition 2.3]{TVW} the red merge diagrams in $D$ can be concatenated in any order.  Therefore we may assume that the lowest red merge in $D$ is located at vertices $j$ and $j+1$.  For example, in \cref{E:etaexample} if $i=3$, then $j=7$ and the diagram as drawn already has lowest red merge at the location of vertices $j$ and $j+1$;  if $i=9$, then $j=9$ and we would use the associativity relation for red merges to replace the red part of $D$ with, for example, the same red diagram reflected across a vertical axis.

Second, since we know that in $\eta(D)$ we have parallel through-strings from $i$ to $j$ and from $i+1$ to $j+1$, it must be that in the $X$ portion of $D$ the corresponding green edges labeled by $1$ must travel in parallel from $i$ to $j$ and from $i+1$ to $j+1$.  As seen in \cref{E:etaexample}, the  region enclosed in $X$ by these two sequences of green edges may be bisected by some number of green edges labeled by $\textcolor{green}{2}$.  By repeatedly using the ``square switch'' relation given in \cref{E:squareswitch} these edges labeled by $\textcolor{green}{2}$ may be eliminated.  The end result is that $D\tilde{E}_{i}$ is equal in $\rgWeb$ to a green-red diagram which contains a subdiagram of the form:
\begin{equation*}
\begin{tikzpicture}[baseline=-2ex,scale=.35, color=\clr, yscale=-1 ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (0, 3) {};
\node [style=none] (1) at (0, 2) {};
\node [style=none] (2) at (0.75, 0) {};
\node [style=none] (3) at (-0.75, 0) {};
\node [style=none] (4) at (0, -2) {};
\node [style=none] (5) at (0, -3) {};
\node [style=none] (15) at (0.5, 2.5) {\textcolor{green}{\tiny{2}}};
\node [style=none] (15) at (0.5, -2.5) {\textcolor{red}{\tiny{2}}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, dashed, green] (0.center) to (1.center);
\draw [thick, dashed, green] [out=180,in=90](1.center) to (3.center);
\draw [thick, dashed, green] [out=0,in=90](1.center) to (2.center);
\draw [thick, densely dashdotted,, red] [out=270,in=180] (3.center) to (4.center);
\draw [thick, densely dashdotted,, red] [out=270,in=0] (2.center) to (4.center);
\draw [thick, densely dashdotted,, red]  (4.center) to (5.center);
\end{pgfonlayer}
\end{tikzpicture}\; .
\end{equation*}  However, by \cite[Lemma 2.9]{TVW} this subdiagram and, hence, $D\tilde{E}_{i}$, is equal to zero in $\rgWeb$.  Thus $\eta$ is equivariant for $T_{i}$ in this case as well.

Therefore $\eta$ is a nonzero $\HH_{d}(q)$-module homomorphism from $\Hom_{\rgWeb}(1^{\otimes d}, \textcolor{green}{2}^{\otimes n}\textcolor{red}{r})$ to $W^{(n+r,n)}$.  As mentioned above, this gives the desired $\HH_{d}(q)$-module isomorphism
\[
\varphi : S^{(n+r,n)} \xrightarrow{\cong} W^{(n+r,n)}.
\]

Finally, observe that for $k=1, \dotsc , n$
\[
w_{0}T_{2k-1} = -q^{-1}w_{0}.
\]  Therefore,
\[
w_{0}y_{\lambda^{\TT}} = (1+q^{-2})^{n}w_{0}.
\]   \cref{L:KeyLemma} implies that, after rescaling, $\varphi(v_{0})=\varphi(v_{t_{0}})=\varphi (v_{t^{\lambda^{\TT}}})=\varphi (z_{\lambda}) = w_{0}$, as claimed.
\end{proof}

\section{Main Results}\label{S:MainResults}

\subsection{Unitriangularity}\label{SS:triangularity}

For the remainder of the paper let
\[
\varphi: S^{\lambda} \to W^{(n+r,n)}
\] be the isomorphism given by \cref{T:TVW}.  Here and in the next result, recall that we write $W^{(n+r,n)}$ for the $\k$-vector space with basis the set of $(n,r)$-webs, $\WW^{(n+r,n)}$ (see \cref{SS:Webmodule}).

\begin{theorem}\label{T:uppertriangular} Let $\lambda = (n+r,n)$ be a partition of $2n+r$ with two parts.  For every $t \in \Std (\lambda^{\TT})$,
\[
\varphi(v_{t}) =\psi (t^{\TT}) + \sum_{\substack{w \in \WW^{(n+r,n)}\\ \nest (w) < \nest (\psi (t^{\TT}))}} c_{w,t} w,
\] where $c_{w,t} \in \Z[q,q^{-1}]$.

\end{theorem}

\begin{proof}  The proof is by induction on the set $\Std (\lambda^{\TT})$ with respect to the weak Bruhat order.  The base case is the unique minimal element $t^{\lambda^{\TT}}$.  This corresponds to the basis element $v_{0}=v_{t_{0}}=v_{t^{\lambda^{\TT}}}=z_{\lambda}$.  By \cref{T:TVW}, $\varphi(v_{0}) = w_{0} = \psi (t^{\TT}_{0})$. As $w_{0}$ is the unique element of $\WW^{(n+r,n)}$ with nesting number zero,  the equation holds in this case.

Now let $t \in \Std (\lambda^{\TT})$ be given and choose $t_{1} \in \Std (\lambda^{\TT})$ such that $t_{1} \leq_{i} t$ for some $i=1, \dotsc , 2n+r-1$.  In particular, $t=t_{1}.s_{i}$ and $v_{t}= v_{t_{1}}T_{i}$. Then,
\begin{align*}
\varphi(v_{t}) &= \varphi(v_{t_{1}}T_{i}) \\
& = \varphi(v_{t_{1}})T_{i} \\
& =\left(\psi (t_{1}) + \sum_{\substack{w \in \WW^{(n+r,n)} \\ \nest (w) < \nest (\psi (t_{1}^{\TT}))}} c_{w,t_{1}} w  \right)T_{i} \\
& = \left(q\psi (t_{1}) + \sum_{\substack{w \in \WW^{(n+r,n)} \\  \nest (w) < \nest (\psi (t_{1}^{\TT}))}} qc_{w,t_{1}} w   \right) +   \left(\psi (t_{1}) + \sum_{\substack{w \in \WW^{(n+r,n)} \\  \nest (w) < \nest (\psi (t_{1}^{\TT}))}} c_{w,t_{1}} w  \right)E_{i},
\end{align*} where the third equality is by the inductive assumption, and the fourth is by the definition of the action of $\HH_{d}(q)$ on $W^{\lambda}$.

In the first term of the last expression, since $t_{1} <_{i} t$, \cref{P:OrderingProposition} implies $\psi(t^{\TT}_{1}) \prec_{i} \psi (t^{\TT})$  and, as a consequence, $\nest (\psi (t_{1}^{\TT})) < \nest (\psi (t^{\TT}))$.  Thus, every web which appears with a nonzero coefficient in this term has nesting number strictly smaller than the nesting number of $\psi (t^{\TT})$.

In the second term of the last expression, it follows from \cref{P:OrderingProposition} that $\psi (t_{1}^{\TT})E_{i}=\psi (t^{\TT})$ and $\nest (\psi(t^{\TT })) = \nest (\psi (t_{1}^{\TT}))+1$. For the rest of the webs which appear in this term, a check of the list of possibilities given in \cref{SS:partialorderonwebs} shows that the action $E_{i}$ on a web raises the nesting number by at most one.  Thus the webs which appear with a nonzero coefficient in the sum after acting by $E_{i}$ must have nesting number strictly smaller than the nesting number of $\psi (t^{\TT})$.

Finally, since the action of each $T_{i}$ on a web is a $\Z[q,q^{-1}]$-linear combination of webs, the assertion about the coefficients is immediate.
\end{proof}

\begin{remark}\label{R:integral}
Let $R = \Z[q,q^{-1}]$.  Let $\HH_{d}(q)_{R}$ be the associative, unital $R$-algebra defined by the same generators and relations as in \cref{R:Heckeisom}.  Given $\lambda=(n+r,n)$, one can define the Specht $\HH_{2n+r}(q)_{R}$-module $S^{\lambda}_{R}$ just as in \cref{SS:SpechtModules} and by \cite[Theorem 5.6]{DipperJames} the standard basis is an $R$-basis for $S^{\lambda}_{R}$.  Similarly, let $W^{\lambda}_{R}$ be the free $R$-module spanned by the non-crossing webs, $\WW^{(n+r,n)}$.  As the action of the generators of $\HH_{2n+r}(q)$ on webs only involves coefficients from $R$, it follows that $W^{\lambda}_{R}$ is an $\HH_{d}(q)_{R}$-module.

Viewing $S^{\lambda}_{R}$ and $W^{\lambda}_{R}$ as $R$-submodules of $S^{\lambda}$ and $W^{\lambda}$, respectively, the previous result shows that the map $\varphi$ restricts to define an $\HH_{d}(q)_{R}$-module isomorphism
\[
\varphi: S^{\lambda}_{R} \to W^{\lambda}_{R}.
\]  Therefore the results of this paper hold over $R$ and, via base change, over any integral domain $S$ and any choice of invertible $q \in S$.
\end{remark}

As in [24], the previous result can be sharpened.

\begin{theorem}\label{T:uppertriangular2} Let $\lambda = (n+r,n)$ be a partition of $2n+r$ with two parts.  For every $t \in \Std (\lambda^{\TT})$,
\[
\varphi(v_{t}) =\psi (t^{\TT}) + \sum_{\substack{w \in \WW^{(n+r,n)}\\ w \prec \psi (t^{\TT})}} c_{w,t} w,
\] where $c_{w,t} \in \Z[q,q^{-1}]$.
\end{theorem}

\begin{proof} Given the previous theorem, it suffices to show $c_{w, t} \neq 0$ implies $w \preceq \psi (t^{\TT})$. The proof of this is by induction on the set $\Std (\lambda^{\TT})$ with respect to the weak Bruhat order.   The base case of $t=t_{0}=t^{\lambda^{\TT}}$ is trivial. For the inductive step, let $t \in \Std (\lambda^{\TT})$ be given and choose $t_{1} \in \Std (\lambda^{\TT})$ such that $t_{1} \leq_{i} t$ for some $i=1, \dotsc , 2n+r-1$.  By the inductive assumption, $c_{\tilde{w}, t_{1}} \neq 0$ implies $\tilde{w} \preceq \psi (t_{1}^{\TT})$ for any $\tilde{w} \in \WW^{(n+r,n)}$.

\begin{align*}
\varphi(v_{t}) = \varphi(v_{t_{1}})T_{i} &= \sum_{\substack{\tilde{w} \in \WW^{(n+r,n)}\\ \tilde{w} \preceq \psi (t_{1}^{\TT})}} c_{\tilde{w},t_{1}} \tilde{w}T_{i}\\
&= \sum_{\substack{\tilde{w} \in \WW^{(n+r,n)}\\ \tilde{w} \preceq \psi (t_{1}^{\TT})}} qc_{\tilde{w},t_{1}} \tilde{w} + \sum_{\substack{\tilde{w} \in \WW^{(n+r,n)}\\ \tilde{w} \preceq \psi (t_{1}^{\TT})}} c_{\tilde{w},t_{1}} \tilde{w}E_{i}.
\end{align*}   In order for $w \in \WW^{(n+r,n)}$ to satisfy $c_{w,t} \neq 0$, either $w = \tilde{w}$ for some $\tilde{w}$ in the first sum or $w = \tilde{w}E_{i}$ for some $\tilde{w}$ in the second sum (or both).  In the first case, $w = \tilde{w} \preceq \psi (t_{1}^{\TT})$.  But since $t_{1} \leq_{i} t$, by \cref{P:OrderingProposition} $\psi (t_{1}^{\TT}) \preceq_{i} \psi (t^{\TT})$ and, hence, $w \preceq \psi (t^{\TT})$, as desired.

In the second case, $w = \tilde{w}E_{i}$. We consider the cases listed in \cref{SS:partialorderonwebs} and how they apply to $\tilde{w}$.  Obviously, Case (1)  is not relevant.  If we happen to be in Case (4), then $\langle i, i+1 \rangle$ is an arc in $\tilde{w}$, $w = -[2]_{q}\tilde{w}$, and the reasoning of the previous paragraph still applies.  If we are in Case (5), then $\tilde{w} \preceq_{i} w$.  But we also have $\tilde{w} \preceq \psi (t_{1}^{\TT})$ and $\psi (t_{1}^{\TT}) \preceq_{i} \psi (t^{\TT})$.  By the fourth claim in \cref{L:WeakBruhatProperties}, this implies $w \preceq \psi (t^{\TT})$, as desired.

The remaining scenario is if $w = \tilde{w}E_{i}$, $\nest (w) < \nest (\tilde{w})$, and one of Cases (2), (3), or (6) listed in \cref{SS:partialorderonwebs} applies to $\tilde{w}$.  In each of these cases the desired result follows from \cref{L:PartialOrder}.  Namely, if we are in Case (3), then $\langle i, y \rangle$ and $\langle i+1, x \rangle $ are arcs and $\tilde{w}=w'$ in the third pair of \cref{L:PartialOrder} (where $p$ is $i$, $i$ is $i+1$, $j$ is $x$, and $q$ is $y$),  But then $w=\tilde{w}E_{i}$ is the $w$ of that pair and therefore $w \preceq \tilde{w}$.  But since $\tilde{w} \preceq \psi (t_{1}^{\TT}) \preceq \psi (t^{\TT})$, the desired inequality holds.  If we are in Case (6), then $\langle x, i+1 \rangle$ and $\langle y, i \rangle$ are arcs in $\tilde{w}$, again $\tilde{w}=w'$ in the third pair of \cref{L:PartialOrder} (where $p$ is $x$, $i$ is $y$, $j$ is $i$, and $q$ is $i+1$), $w=\tilde{w}E_{i}$ is the $w$ of that pair  and therefore $w \preceq \tilde{w}$.  But since $\tilde{w} \preceq \psi (t_{1}^{\TT}) \preceq \psi (t^{\TT})$, the desired inequality holds.

The last case to consider is Case (2). Then $\langle i, k' \rangle$ is a through-string, $\langle i+1, x \rangle$ is an arc, and locally $\tilde{w}$ looks like the last diagram given below:
\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=0.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 0) {};
\node [style=none] (1) at (1, 0) {};
\node [style=none] (2) at (-4.5, 0) {};
%		\node [style=none] (3) at (5, 0) {};
\node [style=none] (4) at (1, -0.5) {\tiny{$x$}};
%		\node [style=none] (5) at (1, -1) {\tiny{$i+1$}};
\node [style=none] (6) at (-4.5, -0.5) {\tiny{$i$}};
%		\node [style=none] (7) at (5, -1) {\tiny{$q$}};
\node [draw] (8) at (-1, 0.5) {$d_{1}$};
%		\node [draw] (9) at (3, 0.5) {$d_{2}$};
\node [style=none] (10) at (-3, 0) {};
\node [style=none] (11) at (-3, -0.5) {\tiny{$i+1$}};
%		\node [style=none] (12) at (4, 0) {};
%		\node [style=none] (13) at (4, -1) {\tiny{$q-1$}};
\node [style=none] (30) at (3, 4) {};
\node [style=none] (31) at (3, 4.5) {\tiny{$k'$}};
%		\node [style=none] (18) at (-5, 0.5) {\tiny{$\cdots$}};
%		\node [style=none] (19) at (5, 0.5) {\tiny{$\cdots$}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90] (2.center) to (10.center);
\draw [thick, looseness=1] [out=90,in=270] (0.center) to (30.center);
\end{pgfonlayer}
\end{tikzpicture} \;\; \preceq \; \;
\begin{tikzpicture}[baseline=0ex,scale=0.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1, 0) {};
\node [style=none] (1) at (1, 0) {};
\node [style=none] (2) at (-5, 0) {};
\node [style=none] (3) at (1, 0) {};
\node [style=none] (4) at (-1, -0.5) {\tiny{$i$}};
%		\node [style=none] (5) at (1, -0.5) {\tiny{$i+1$}};
\node [style=none] (6) at (-5, -0.5) {\tiny{$p$}};
\node [style=none] (7) at (1, -0.5) {\tiny{$q$}};
\node [draw] (8) at (-3, 0.5) {$d_{1}$};
%		\node [draw] (9) at (1, 0.5) {$d_{2}$};
\node [style=none] (30) at (3, 4) {};
\node [style=none] (31) at (3, 4.5) {\tiny{$k'$}};
%		\node [style=none] (18) at (-5, 0.5) {\tiny{$\cdots$}};
%		\node [style=none] (19) at (5, 0.5) {\tiny{$\cdots$}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90] (0.center) to (2.center);
\draw [thick, looseness=1.25] [out=270,in=90] (30.center) to (3.center);
\end{pgfonlayer}
\end{tikzpicture}
\;\; \preceq \;\;
\begin{tikzpicture}[baseline=0ex,scale=0.4, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 0) {};
\node [style=none] (1) at (1, 0) {};
\node [style=none] (2) at (-4.5, 0) {};
%		\node [style=none] (3) at (5, 0) {};
\node [style=none] (4) at (1, -0.5) {\tiny{$x$}};
%		\node [style=none] (5) at (1, -1) {\tiny{$i+1$}};
\node [style=none] (6) at (-4.5, -0.5) {\tiny{$i$}};
%		\node [style=none] (7) at (5, -1) {\tiny{$q$}};
\node [draw] (8) at (-1, 0.5) {$d_{1}$};
%		\node [draw] (9) at (3, 0.5) {$d_{2}$};
\node [style=none] (10) at (-3, 0) {};
\node [style=none] (11) at (-3, -0.5) {\tiny{$i+1$}};
%		\node [style=none] (12) at (4, 0) {};
%		\node [style=none] (13) at (4, -1) {\tiny{$q-1$}};
\node [style=none] (30) at (3, 4) {};
\node [style=none] (31) at (3, 4.5) {\tiny{$k'$}};
%		\node [style=none] (18) at (-5, 0.5) {\tiny{$\cdots$}};
%		\node [style=none] (19) at (5, 0.5) {\tiny{$\cdots$}};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90] (10.center) to (0.center);
\draw [thick, looseness=1] [out=90,in=270] (2.center) to (30.center);
\end{pgfonlayer}
\end{tikzpicture}\; .
\end{equation*}
The first diagram is $w = \tilde{w}E_{i}$.  The first inequality holds by using the second pair of webs in \cref{L:PartialOrder}, possibly multiple times, and the second inequality holds by using the fourth pair of webs in \cref{L:PartialOrder}  (where $p$ is $i$, $p+1$ is $i+1$, $i+1$ is $x$, and $k'$ is $k'$).  Once again, $w \preceq \tilde{w} \preceq \psi (t_{1}^{\TT}) \preceq \psi (t^{\TT})$, as desired.

In short, in all cases, if $c_{w, t} \neq 0$, then $w \preceq \psi (t^{\TT})$ and the assertion is proven.
\end{proof}

\subsection{Positivity}\label{SS:positivity}

The goal of this section is to prove the following result.

\begin{theorem}\label{T:positivity}  Let $\lambda = (n+r,n)$ be a partition of $2n+r$.  For every $t \in \Std (\lambda^{\TT})$,
\[
\varphi (v_{t}) = \sum_{w \in \WW^{(n+r,n)}} c_{w,t} w
\] with $c_{w,t} \in \Z_{\geq 0}[q]$ for all $w \in \WW^{(n+r,n)}$.
\end{theorem}

By \cref{E:standardbasis}, $v_{t}=v_{0}T_{i_{1}}\dotsb T_{i_{r}}$ where $s_{i_{1}}\dotsb s_{i_{r}}$ is a reduced word in $S_{d}$ and
\begin{equation}\label{E:weakbruhatX}
t^{\lambda^{\TT}} =t_{0}\leq_{i_{1}} t_{1} \leq_{i_{2}} \dotsb  \leq_{i_{r}} t_{r}= t
\end{equation}
in the weak Bruhat order.  Then,
\begin{equation}\label{E:CrossingWeb}
\varphi(v_{t}) = \varphi(v_{0}T_{i_{1}}\dotsb T_{i_{r}})=\varphi(v_{0})T_{i_{1}}\dotsb T_{i_{r}} = w_{0}\beta_{i_{1}}\dotsb \beta_{i_{r}}.
\end{equation}

The proof strategy is straightforward enough.  The right hand side of \cref{E:CrossingWeb} is an $(n+r,n)$-web with crossings.  Since relation \cref{E:crossingrelation} only involves coefficients in $\Z_{\geq 0}[q]$, if the web could be rewritten into the basis of non-crossing webs using only this relation, then the same would be true of the coefficients in the resulting linear combination.  However, if one simply applies \cref{E:crossingrelation} to each crossing, then it is quite possible that bubbles will appear. This will generate undesirable coefficients when they are popped using relation \cref{E:bubblerelation}.  The example in \cref{SS:ExampleI} shows how this can happen.  The key observation, in that example and in general, is that the Reidemiester II moves from \cref{L:RII} also have coefficients in $\Z_{\geq 0}[q]$ and  can be used along with relation \cref{E:crossingrelation} to eliminate all crossings in a way which entirely avoids the dreaded bubble popping relation.

To show how this can be done we first need some terminology.  Given an $(n+r,n)$-web with crossings consider the complement of its projection onto the plane.  We call a connected component of the complement \emph{bounded} if the closure of the connected component is compact.  Otherwise we call it \emph{unbounded}.    Given a web, we refer to these components as bounded and unbounded regions, respectively.   For example, the interior of a bubble is a bounded region.

\begin{example} The $(3,3)$-web given in \cref{SS:ExampleI} has the following projection onto the plane:
\[\label{E:Example1}
\begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 0) {};
\node [style=none] (1) at (2, 3) {};
\node [style=none] (2) at (4, 3) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-4, 0) {};
\node [style=none] (8) at (-5, 2) {};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-4, 2) {};
\node [style=none] (11) at (2, 0) {};
\node [style=none] (12) at (3, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.28) {};
\node [style=none] (15) at (0.75, 0.77) {};
\node [style=none] (16) at (1.5, 0.5) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\node [style=none] (20) at (0, 1.75) {B};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (1.center) to (2.center);
\draw [thick]  (2.center) to (3.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (10.center) to (9.center);
\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
%		\draw [ultra thick, color=white]  (13.center) to (14.center);
%		\draw [ultra thick, color=white]  (15.center) to (16.center);
\draw [thick, looseness=0.5] [out=90,in=270] (4.center) to (5.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (6.center);
\draw [thick] [out=270,in=90, looseness=0.5]  (6.center) to (12.center);
%		\draw [ultra thick, color=white]  (17.center) to (18.center);
\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture} \; \; .
\]  The region marked with a $B$ is bounded whereas the other regions of the complement are path-connected to the unbounded part of the plane and so are are unbounded.
\end{example}

Observe that if one applies \cref{E:crossingrelation} to eliminate a crossing, then in each of the two resulting webs a pair of regions have been merged together (namely, the first term merges the regions above and below the crossing while the second term merges the regions on each side of the crossing).  In particular, the application of this relation does not create entirely new bounded regions. The danger lies in the existing bounded regions of the initial web as one of them may become a bubble as crossings are eliminated.  Thus the proof will focus on what happens to bounded regions as relation \cref{E:crossingrelation} and Reidemiester II moves are applied.

Now consider \cref{E:weakbruhatX}.  As discussed in \cref{R:weakBruhatOrder}, $t_{j-1} \leq_{i_{j}}  t_{j}$ if and only if $t_{j}=t_{j-1}.s_{i_{j}}$ and $i_{j}$ appears before $i_{j}+1$ when reading the entries of $t_{j-1}$ left-to-right, top-to-bottom.  Since $t_{j-1}, t_{j} \in \Std ((n+r,n)^{\TT})$, it follows that $i_{j}$ is in the second column of $(n+r,n)^{\TT}$ while $i_{j}+1$ is in the first column (much as in  the proof of \cref{P:OrderingProposition}).  That is, $i_{j}$ is in the second row of $t_{j}^{\TT}$ while $i_{j}+1$ is in the first row.  That is, $i_{j}$ is an $R$-vertex of $\psi (t_{j}^{\TT})$
while $i_{j}+1$ is an $L$-vertex.  That is, for each $j=1, \dotsc , r$, when $\beta_{i_{j}}$ is concatenated on the bottom of $w_{0}\beta_{i_{1}}\dotsb \beta_{i_{j-1}}$ in the construction of $\varphi(v_{t})$, $i_{j}$ is an $R$-vertex of $w_{0}\beta_{i_{1}}\dotsb \beta_{i_{j-1}}$ while $i_{j}+1$ is an $L$-vertex.

In short, in the drawing of the $(n+r,n)$-web $\varphi(v_{t})$, as one reads the diagram from top-to-bottom each crossing consists of an edge connected to an $R$-vertex passing under an edge connected to a $L$-vertex.  If we were to label the vertices by their type, then locally each crossing would look like
\[
\begin{tikzpicture}[baseline=-2ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1, -1.5) {};
\node [style=none] (1) at (1, -1.5) {};
\node [style=none] (2) at (-1, 1.5) {};
\node [style=none] (3) at (1, 1.5) {};
\node [style=none] (4) at (0.28, -0.4) {};
\node [style=none] (5) at (-0.28, 0.4) {};
\node [style=none] (10) at (-1, -2.25) {L};
\node [style=none] (11) at (1, -2.25) {R};
\node [style=none] (12) at (-1, 2.25) {R};
\node [style=none] (13) at (1, 2.25) {L};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick] (2.center) to (1.center);
\draw [ultra thick, color=white] (4.center) to (5.center);
\draw [thick] (3.center) to (0.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=2];
\end{pgfonlayer}
\end{tikzpicture} \; .
\]

This observation allows us to verify the diagram drawn for $\varphi(v_{t})$ satisfies the following conditions.
\begin{enumerate}
\item [(B1)]  The diagram for $\varphi(v_{t})$ contains no isolated connected components (i.e., no bubbles).
\item [(B2)]  Every bounded region of $\varphi(v_{t})$ is a bigon, triangle, or square.
\item [(B3)]  If a bounded region of $\varphi(v_{t})$ is adjacent to second bounded region and is in the location of an S in \cref{E:TriangleSquare} relative to that second bounded region, then the first bounded region is a square.
\end{enumerate}

Condition $(B1)$ holds because $\varphi(v_{t})$ contains no local minimums other than the vertices at the bottom.  If $\varphi(v_{t})$ contains a bounded region with a side not involving one of the edges of $w_{0}$, then the side is formed by an edge connecting $R$-vertices which passes under two successive edges connecting $L$-vertices, or the side is formed by an edge connecting $L$-vertices which passes over two successive edges connecting $R$-vertices.  A bounded region with zero sides of this kind would be an arc whose legs cross --- this is forbidden by our observation about crossings in $\varphi(v_{t})$.  A bounded region with one side of this kind would involve the one side crossing the same arc twice --- this again would violate the observation about crossings in $\varphi(v_{t})$.  Consequently the bounded region must have two or more such sides.  This implies that the bottom part of the bounded region looks like the bottom half of a triangle or square as in \cref{E:TriangleSquare}. Likewise, if the top half of the bounded region also does not involve an edge of $w_{0}$, then the bounded region is a rectangle.  If it does involve an edge in $w_{0}$, then the bounded region is necessarily  a triangle.  Thus $\varphi(v_{t})$ satisfies condition $(B2)$ (bigons do not appear in $\varphi(v_{t})$ but it is convenient to include them as part of the condition).  Finally, since the bounded regions which lie in a location marked by S in \cref{E:TriangleSquare} cannot involve edges of $w_{0}$, the previous discussion also implies $\varphi(v_{t})$ satisfies condition $(B3)$.

\cref{T:positivity} will follow from the following result.

\begin{prop}\label{L:positivity}  If $u$ is an $(n+r,n)$-web which satisfies conditions $(B1)-(B3)$, then
\[
u \in  \bigoplus_{w \in \WW^{(n+r,n)}} \Z_{\geq 0}[q] w.
\]
\end{prop}

\begin{proof} The proof is by induction on the number of crossings in $u$ with the base case being trivial.  For the inductive step, consider the following ordered list of cases.  Observe that they cover all possible scenarios.

\begin{itemize}
\item [Case 1:] \hspace{\parindent} If $u$ contains no bounded regions, then choose a crossing at random and apply relation \cref{E:crossingrelation}.  The result will be a $\Z_{\geq 0}[q]$-linear combination of webs with no bounded regions and fewer crossings.  Since these webs have no bounded regions, conditions $(B1)-(B3)$ are vacuously satisfied.
\item [Case 2:] \hspace{\parindent} If $u$ contains a bounded region, then choose a bounded region in $u$ for which the neighboring regions marked with $N$ and $W$ are unbounded.  Since $u$ only contains finitely many bounded regions, such a region necessarily exists.
\begin{itemize}
\item [Case 2a:]\hspace{\parindent}   If the selected region is a bigon, then apply the Reidemister II relation given in \cref{L:RII}.  The result will be a $\Z_{\geq 0}[q]$-multiple of a web with fewer crossings.  Since the region in $u$ marked with a W is unbounded, the same will be true of the regions marked with W and E after applying the Reidemister II move.  The region marked with a S in the picture of the bigon in \cref{E:TriangleSquare} is either unbounded or a square, and either remains unbounded or becomes a bigon after the application of the Reidemister II relation. From these observations it is easy to check that the resulting web still satisfies conditions $(B1)-(B3)$.
\item [Case 2b:] \hspace{\parindent} If the selected region is a triangle, then apply relation \cref{E:crossingrelation} to the crossing at the northwest corner of the triangle. Locally, the result will be:
\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (1, 1) {};
\node [style=none] (3) at (-1, 1) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {};
\node [style=none] (21) at (0, 2.25) {N};
\node [style=none] (22) at (-2, 2) {};
\node [style=none] (23) at (2, 2) {};
\node [style=none] (24) at (-3, 0) {W};
\node [style=none] (25) at (3, 0) {};
\node [style=none] (26) at (2, -2) {};
\node [style=none] (27) at (0, -3) {Z};
\node [style=none] (28) at (-2, -2) {X};
\node [style=none] (29) at (2, -2) {Y};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=45,in=135] (3.center) to (6.center);
\draw [thick] (14.center) to (17.center);
\draw [ultra thick, color=white] (19.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
\draw [thick] (1.center) to (3.center);
\draw [thick] (0.center) to (2.center);
\draw [thick] (7.center) to (6.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
\draw [thick] (10.center) to (9.center);
%		\draw [thick] (14.center) to (18.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture}
\; \; = \; \;
q\; \;
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (1, 1) {};
\node [style=none] (3) at (-1, 1) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {};
\node [style=none] (21) at (0, 2.25) {N};
\node [style=none] (22) at (-2, 2) {};
\node [style=none] (23) at (2, 2) {};
\node [style=none] (24) at (-3, 0) {W};
\node [style=none] (25) at (3, 0) {};
\node [style=none] (26) at (2, -2) {};
\node [style=none] (27) at (0, -3) {Z};
\node [style=none] (28) at (-2, -2) {X};
\node [style=none] (29) at (2, -2) {Y};
\node [style=none] (90) at (-2.25, -0.25) {};
\node [style=none] (91) at (-1.75, 0.25) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=45,in=135] (3.center) to (6.center);
\draw [thick] (14.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
\draw [thick] (17.center) to (19.center);
%		\draw [thick] (13.center) to (16.center);
\draw [thick] (1.center) to (90.center);
\draw [thick] (0.center) to (2.center);
\draw [thick] (7.center) to (6.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
\draw [thick] (10.center) to (9.center);
\draw [thick] (3.center) to (91.center);
\draw [thick, looseness=1.5] [out=45,in=315] (90.center) to (19.center);
\draw [thick, looseness=1.5] [out=225,in=135] (91.center) to (15.center);
%		\draw [thick] (14.center) to (18.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture}
\; \; + \; \;
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (1, 1) {};
\node [style=none] (3) at (-1, 1) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {};
\node [style=none] (21) at (0, 2.25) {N};
\node [style=none] (22) at (-2, 2) {};
\node [style=none] (23) at (2, 2) {};
\node [style=none] (24) at (-3, 0) {W};
\node [style=none] (25) at (3, 0) {};
\node [style=none] (26) at (2, -2) {};
\node [style=none] (27) at (0, -3) {Z};
\node [style=none] (28) at (-2, -2) {X};
\node [style=none] (29) at (2, -2) {Y};
\node [style=none] (90) at (-2.25, -0.25) {};
\node [style=none] (91) at (-1.75, 0.25) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=45,in=135] (3.center) to (6.center);
\draw [thick] (14.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
\draw [thick] (19.center) to (17.center);
%		\draw [thick] (13.center) to (16.center);
\draw [thick] (1.center) to (90.center);
\draw [thick] (3.center) to (91.center);
\draw [thick] (0.center) to (2.center);
\draw [thick] (7.center) to (6.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
\draw [thick] (10.center) to (9.center);
\draw [thick, looseness=1.5] [in=315,out=225] (91.center) to (19.center);
\draw [thick, looseness=1.5] [out=45,in=135] (90.center) to (15.center);
%		\draw [thick] (14.center) to (18.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture} \;.
\end{equation*}   \hspace{\parindent} We claim both webs on right hand side of the equality still satisfy conditions $(B1)-(B3)$.  The conditions are verified by checking each bounded region one-by-one.  Since the only regions of $u$ affected by resolving the crossing are the ones which appear in the picture, it suffices to study these in detail.  First, note that no bubbles were created so condition $(B1)$ is satisfied by both. Furthermore, any unbounded regions in $u$ (e.g., those marked with N and W) remain unbounded in the right hand side of the equality and so can be ignored.

\hspace{\parindent} \hspace{\parindent} Now, in the first diagram on the right side of the equality the triangle was turned into a bigon, the region marked with an X was made unbounded, and the regions marked with a Y and Z remain unchanged (in particular, if one or both were bounded in $u$ then they were a rectangle and remain a rectangle in the new diagram), so both conditions $(B2)$ and $(B3)$ are still satisfied.  In the second diagram the triangle became an unbounded region and if X marked a bounded region of $u$, then it was rectangle and becomes a triangle in the new diagram.   Thus both of the diagrams on the right hand side have fewer crossings than $u$ and still satisfy conditions $(B1)-(B3)$.
\item [Case 2c:] \hspace{\parindent} If the selected region is a square, then apply relation \cref{E:crossingrelation} to the crossing at the northern corner of the square. Locally, the result will be:
\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (-1, 3) {};
\node [style=none] (3) at (1, 3) {};
\node [style=none] (4) at (-0.25, 2.25) {};
\node [style=none] (5) at (0.25, 1.75) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {N};
\node [style=none] (21) at (0, 3) {N};
\node [style=none] (22) at (-2, 2) {N};
\node [style=none] (23) at (2, 2) {N};
\node [style=none] (24) at (-3, 0) {W};
%		\node [style=none] (25) at (3, 0) {E};
\node [style=none] (26) at (2, -2) {S};
\node [style=none] (27) at (0, -3) {S};
\node [style=none] (28) at (-2, -2) {S};
%		\node [style=none] (29) at (-1.5, -1.5) {X};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick] (7.center) to (2.center);
\draw [ultra thick, color=white] (4.center) to (5.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
%		\draw [thick] (7.center) to (6.center);
\draw [thick] (14.center) to (17.center);
\draw [ultra thick, color=white] (19.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
%		\draw [thick] (14.center) to (18.center);
\draw [thick] (1.center) to (3.center);
\draw [thick] (10.center) to (9.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture}\; \; = \; \; q \; \;
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (-1, 3) {};
\node [style=none] (3) at (1, 3) {};
\node [style=none] (4) at (-0.25, 2.25) {};
\node [style=none] (5) at (0.25, 1.75) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {N};
\node [style=none] (21) at (0, 3) {N};
\node [style=none] (22) at (-2, 2) {N};
\node [style=none] (23) at (2, 2) {N};
\node [style=none] (24) at (-3, 0) {W};
%		\node [style=none] (25) at (3, 0) {E};
\node [style=none] (26) at (2, -2) {S};
\node [style=none] (27) at (0, -3) {S};
\node [style=none] (28) at (-2, -2) {S};
%		\node [style=none] (29) at (-1.5, -1.5) {X};
\node [style=none] (94) at (0.25, 2.25) {};
\node [style=none] (95) at (-0.25, 1.75) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick] (4.center) to (2.center);
\draw [thick] (5.center) to (7.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
%		\draw [thick] (7.center) to (6.center);
\draw [thick] (14.center) to (17.center);
\draw [ultra thick, color=white] (19.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
\draw [thick] (10.center) to (9.center);
\draw [thick] (1.center) to (95.center);
\draw [thick] (94.center) to (3.center);
\draw [thick, looseness=1.5] [out=315,in=45] (4.center) to (95.center);
\draw [thick, looseness=1.5] [out=135,in=225] (5.center) to (94.center);
\draw [thick] (94.center) to (3.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture}\; \; + \; \;
\begin{tikzpicture}[baseline=0ex,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 1) {};
\node [style=none] (1) at (-3, -1) {};
\node [style=none] (2) at (-1, 3) {};
\node [style=none] (3) at (1, 3) {};
\node [style=none] (4) at (-0.25, 2.25) {};
\node [style=none] (5) at (0.25, 1.75) {};
\node [style=none] (6) at (1, 1) {};
\node [style=none] (7) at (3, -1) {};
\node [style=none] (8) at (1.75, 0.25) {};
\node [style=none] (9) at (3, 1) {};
\node [style=none] (10) at (-1, -3) {};
\node [style=none] (11) at (2.25, -0.25) {};
\node [style=none] (12) at (-1, -1) {};
\node [style=none] (13) at (-0.25, -1.75) {};
\node [style=none] (14) at (1, -3) {};
\node [style=none] (15) at (-1.75, -0.25) {};
\node [style=none] (16) at (0.25, -2.25) {};
\node [style=none] (17) at (-3, 1) {};
\node [style=none] (18) at (-1, -1) {};
\node [style=none] (19) at (-2.25, 0.25) {};
\node [style=none] (20) at (0, 3) {N};
\node [style=none] (21) at (0, 3) {N};
\node [style=none] (22) at (-2, 2) {N};
\node [style=none] (23) at (2, 2) {N};
\node [style=none] (24) at (-3, 0) {W};
%		\node [style=none] (25) at (3, 0) {E};
\node [style=none] (26) at (2, -2) {S};
\node [style=none] (27) at (0, -3) {S};
\node [style=none] (28) at (-2, -2) {S};
%		\node [style=none] (29) at (-1.5, -1.5) {X};
\node [style=none] (94) at (0.25, 2.25) {};
\node [style=none] (95) at (-0.25, 1.75) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick] (2.center) to (4.center);
\draw [thick] (7.center) to (5.center);
\draw [thick] (94.center) to (3.center);
\draw [ultra thick, color=white] (8.center) to (11.center);
%		\draw [thick] (7.center) to (6.center);
\draw [thick] (14.center) to (17.center);
\draw [ultra thick, color=white] (19.center) to (15.center);
\draw [ultra thick, color=white] (13.center) to (16.center);
%		\draw [thick] (14.center) to (18.center);
\draw [thick] (10.center) to (9.center);
\draw [thick, looseness=1.5] [out=315,in=225] (4.center) to (94.center);
\draw [thick, looseness=1.5] [out=135,in=45] (5.center) to (95.center);
\draw [thick] (1.center) to (95.center);
%		\draw [thick, dashed, color=gray] (0,0) circle [radius=4];
\end{pgfonlayer}
\end{tikzpicture}\; .
\end{equation*} \hspace{\parindent} We again claim the diagrams on the right side of the equality still satisfy conditions $(B1)-(B3)$. The analysis is an easier version of what was done for the triangle. Obviously, no bubbles are created and so $(B1)$ is still satisfied by both diagrams. For the square the only bounded region of $u$ affected by resolving the crossing is the square itself.  In the first diagram on the right side of the equality the square becomes unbounded and in the second diagram the square becomes a triangle.  If any of the regions marked with an S were bounded in $u$, then they were a rectangle and still are rectangles in the new diagrams.  Both diagrams  still satisfy conditions $(B2)$ and $(B3)$.    Thus both of the diagrams on the right hand side have fewer crossings than $u$ and still satisfy conditions $(B1)-(B3)$.
\end{itemize}
\end{itemize}
Whichever case is applied, the result replaces $u$ with a $ \Z_{\geq 0}[q]$-linear combination of webs which still satisfy conditions $(B1)-(B3)$ and have strictly fewer crossings.  The claim follows by induction.
\end{proof}

\section{Examples}\label{S:Examples}

\subsection{Example I}\label{SS:ExampleI}

In this example we rewrite $w_{0}T_{4}T_{2}T_{3} = w_{0}\beta_{4}\beta_{2}\beta_{3}$ for $\lambda = (3,3)$ as a linear combination of non-crossing webs using relations \cref{E:crossingrelation,E:bubblerelation}:
\begin{align*}\label{E:Example1}
\begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (1, 0) {};
\node [style=none] (1) at (2, 3) {};
\node [style=none] (2) at (4, 3) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-4, 0) {};
\node [style=none] (8) at (-5, 2) {};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-4, 2) {};
\node [style=none] (11) at (2, 0) {};
\node [style=none] (12) at (3, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.28) {};
\node [style=none] (15) at (0.75, 0.77) {};
\node [style=none] (16) at (1.5, 0.5) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (1.center) to (2.center);
\draw [thick]  (2.center) to (3.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (10.center) to (9.center);
\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
\draw [ultra thick, color=white]  (13.center) to (14.center);
\draw [ultra thick, color=white]  (15.center) to (16.center);
\draw [thick, looseness=0.5] [out=90,in=270] (4.center) to (5.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (6.center);
\draw [thick] [out=270,in=90, looseness=0.5]  (6.center) to (12.center);
\draw [ultra thick, color=white]  (17.center) to (18.center);
\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \; &= \; \;
q  \begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (0, 0) {};
\node [style=none] (1) at (2, 3) {};
\node [style=none] (2) at (4, 3) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-4, 0) {};
\node [style=none] (8) at (-5, 2) {};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-4, 2) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (2, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.23) {};
\node [style=none] (15) at (-0.15, 0.85) {};
\node [style=none] (16) at (0.6, 0.5) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (12.center) to (3.center);
%		\draw [thick]  (2.center) to (3.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (10.center) to (9.center);
\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
\draw [ultra thick, color=white]  (13.center) to (14.center);
\draw [ultra thick, color=white]  (15.center) to (16.center);
\draw [thick, looseness=0.5] [out=90,in=270] (4.center) to (5.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (6.center);
\draw [thick, looseness=0.5] [out=270,in=90]  (6.center) to (0.center);
\draw [ultra thick, color=white]  (17.center) to (18.center);
%		\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \; + \; \;
\begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (0, 0) {};
\node [style=none] (1) at (2, 2) {};
\node [style=none] (2) at (4, 2) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-4, 0) {};
%		\node [style=none] (8) at (-5, 2) {8};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-4, 2) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (2, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.23) {};
\node [style=none] (15) at (-0.15, 0.85) {};
\node [style=none] (16) at (0.6, 0.5) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (6.center) to (1.center);
\draw [thick]  (2.center) to (3.center);
\draw [thick]  (1.center) to (12.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (10.center) to (9.center);
\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
\draw [ultra thick, color=white]  (13.center) to (14.center);
\draw [ultra thick, color=white]  (15.center) to (16.center);
\draw [thick, looseness=0.5] [out=90,in=270] (4.center) to (5.center);
\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (2.center);
\draw [thick, looseness=0.5] [out=270,in=90]  (6.center) to (0.center);
\draw [ultra thick, color=white]  (17.center) to (18.center);
%		\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture} \\
\; \; &= \; \;
q^{2}  \begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1, 0) {};
\node [style=none] (1) at (1, 2) {};
\node [style=none] (2) at (-1, 2) {};
\node [style=none] (3) at (1, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (2, 0) {};
\node [style=none] (7) at (4, 0) {};
\node [style=none] (8) at (-0.25, 1.1) {};
\node [style=none] (9) at (0.25, 0.9) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick] [out=270, in=90] (2.center) to (3.center);
\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick] [out=90, in=270] (0.center) to (1.center);
\draw [thick, looseness=1.5] [out=90,in=90] (1.center) to (2.center);
\draw [thick, looseness=1.5] [out=90,in=90] (4.center) to (5.center);
\draw [thick, looseness=1.5] [out=90,in=90] (7.center) to (6.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \; + \; \;
q \;  \begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (0, 0) {};
\node [style=none] (1) at (2, 3) {};
\node [style=none] (2) at (4, 3) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-3, 0) {};
%		\node [style=none] (8) at (-5, 2) {};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-3, 2) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (2, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.23) {};
\node [style=none] (15) at (0.25, 1.2) {};
\node [style=none] (16) at (0.75, 0.76) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (12.center) to (3.center);
%		\draw [thick]  (2.center) to (3.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.25] [out=90,in=90]  (10.center) to (6.center);
%		\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
%		\draw [ultra thick, color=white]  (13.center) to (14.center);
\draw [thick, looseness=1.5] [out=90,in=90] (4.center) to (11.center);
\draw [ultra thick, color=white]  (15.center) to (16.center);
%		\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (6.center);
\draw [thick] [out=270,in=90]  (6.center) to (0.center);
\draw [ultra thick, color=white]  (17.center) to (18.center);
%		\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture} \\
& \hspace{.5in} +
q \;  \begin{tikzpicture}[baseline=0.5cm,xscale=-1,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (0, 0) {};
\node [style=none] (1) at (2, 3) {};
\node [style=none] (2) at (4, 3) {};
\node [style=none] (3) at (4, 0) {};
\node [style=none] (4) at (-2, 0) {};
\node [style=none] (5) at (-1, 2) {};
\node [style=none] (6) at (1, 2) {};
\node [style=none] (7) at (-3, 0) {};
%		\node [style=none] (8) at (-5, 2) {};
\node [style=none] (9) at (-2, 2) {};
\node [style=none] (10) at (-3, 2) {};
\node [style=none] (11) at (1, 0) {};
\node [style=none] (12) at (2, 0) {};
\node [style=none] (13) at (-1.75, 1.62) {};
\node [style=none] (14) at (-1, 1.23) {};
\node [style=none] (15) at (0.25, 0.72) {};
\node [style=none] (16) at (0.75, 1.28) {};
\node [style=none] (17) at (1.25, 1.6) {};
\node [style=none] (18) at (1.75, 1.2) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90]  (12.center) to (3.center);
%		\draw [thick]  (2.center) to (3.center);
\draw [thick]  (7.center) to (10.center);
\draw [thick, looseness=1.25] [out=90,in=90]  (10.center) to (6.center);
%		\draw [thick, looseness=0.5] [out=270,in=90] (9.center) to (11.center);
%		\draw [ultra thick, color=white]  (13.center) to (14.center);
\draw [thick] [out=270,in=90]  (6.center) to (0.center);
\draw [ultra thick, color=white]  (15.center) to (16.center);
\draw [thick, looseness=1.5] [out=90,in=90] (4.center) to (11.center);
%		\draw [thick, looseness=1.5] [out=90,in=90]  (5.center) to (6.center);
\draw [ultra thick, color=white]  (17.center) to (18.center);
%		\draw [thick, looseness=0.5] [out=90, in=270] (0.center) to (1.center);
\end{pgfonlayer}
\end{tikzpicture}
\;\; + \; \;
\begin{tikzpicture}[baseline=0.5cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-0.5, 0) {};
\node [style=none] (3) at (0.5, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (2.5, 0) {};
\node [style=none] (7) at (4, 0) {};
\node [style=none] (8) at (-0.25, 1.15) {};
\node [style=none] (9) at (0.25, 0.8) {};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.5] [out=90,in=90] (5.center) to (7.center);
\draw [thick, looseness=1.5] [out=90,in=90] (4.center) to (3.center);
\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=1.5] [out=90,in=90] (0.center) to (6.center);
\end{pgfonlayer}
\end{tikzpicture} \\
(q^{3}-q^{2}[2]_{q} + 2q ) \; \begin{tikzpicture}[baseline=0cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (3) at (0.5, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4.5, 0) {};
\node [style=none] (6) at (1.5, 0) {};
\node [style=none] (7) at (3.5, 0) {};
%		\node [style=none] (8) at (-0.25, 1.15) {8};
%		\node [style=none] (9) at (0.25, 0.8) {9};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.75] [out=90,in=90] (5.center) to (4.center);
\draw [thick, looseness=1.75] [out=90,in=90] (0.center) to (3.center);
%		\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=1.75] [out=90,in=90] (6.center) to (7.center);
\end{pgfonlayer}
\end{tikzpicture} \\
& \hspace{0.5in} + \; \;
q^{2}\; \begin{tikzpicture}[baseline=0.25cm,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (3) at (0, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (1, 0) {};
\node [style=none] (7) at (3, 0) {};
%		\node [style=none] (8) at (-0.25, 1.15) {8};
%		\node [style=none] (9) at (0.25, 0.8) {9};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.25] [out=90,in=90] (5.center) to (3.center);
\draw [thick, looseness=1.5] [out=90,in=90] (0.center) to (4.center);
%		\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=1.5] [out=90,in=90] (6.center) to (7.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \;  + \; \;
q^{2}\; \begin{tikzpicture}[baseline=0.25cm,xscale=-1,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (3) at (0, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (1, 0) {};
\node [style=none] (7) at (3, 0) {};
%		\node [style=none] (8) at (-0.25, 1.15) {8};
%		\node [style=none] (9) at (0.25, 0.8) {9};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1.25] [out=90,in=90] (5.center) to (3.center);
\draw [thick, looseness=1.5] [out=90,in=90] (0.center) to (4.center);
%		\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=1.5] [out=90,in=90] (6.center) to (7.center);
\end{pgfonlayer}
\end{tikzpicture} \\
& \hspace{0.5in} +
q\; \begin{tikzpicture}[baseline=0.25cm,xscale=-1,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (3) at (0.5, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (1.5, 0) {};
\node [style=none] (7) at (3, 0) {};
%		\node [style=none] (8) at (-0.25, 1.15) {8};
%		\node [style=none] (9) at (0.25, 0.8) {9};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1] [out=90,in=90] (5.center) to (7.center);
\draw [thick, looseness=2] [out=90,in=90] (0.center) to (4.center);
%		\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=2] [out=90,in=90] (6.center) to (3.center);
\end{pgfonlayer}
\end{tikzpicture}
\; \; + \; \;
\begin{tikzpicture}[baseline=0.25cm,xscale=-1,scale=.5, color=\clr ]
\begin{pgfonlayer}{nodelayer}
\node [style=none] (0) at (-1.5, 0) {};
\node [style=none] (3) at (0.5, 0) {};
\node [style=none] (4) at (-2.5, 0) {};
\node [style=none] (5) at (-4, 0) {};
\node [style=none] (6) at (1.5, 0) {};
\node [style=none] (7) at (3, 0) {};
%		\node [style=none] (8) at (-0.25, 1.15) {8};
%		\node [style=none] (9) at (0.25, 0.8) {9};
\end{pgfonlayer}
\begin{pgfonlayer}{edgelayer}
\draw [thick, looseness=1] [out=90,in=90] (5.center) to (7.center);
\draw [thick, looseness=1.25] [out=90,in=90] (6.center) to (4.center);
%		\draw [ultra thick, color=white] (8.center) to (9.center);
\draw [thick, looseness=1.5] [out=90,in=90] (0.center) to (3.center);
\end{pgfonlayer}
\end{tikzpicture} \; .
\end{align*} Observe that
\[
q^{3} - q^{2}[2]_{q}+2q = q^{3} - q^{2}(q+q^{-1})+2q = q.
\]  Thus all the coefficients given above lie in $\Z_{\geq 0}[q]$ as promised by \cref{T:positivity}.  Also note that the bubble could have been avoided by the use of one of the Reidemeister II moves given in \cref{L:RII}.

\subsection{Example II}\label{SS:ExampleII}
Set $\lambda = (3,3)$.  $t_{0} = t^{\lambda^{\TT}}$, $t_{1}=t_{0}.s_{2}$, $t_{2}=t_{0}.s_{4}$, $t_{3}=t_{0}.s_{2}s_{4}$, and $t_{4}=t_{0}.s_{4}s_{2}s_{3}$.  For $i=0,\dotsc , 4$, let $w_{i}=\phi (t_{i})$.  With respect to the ordered bases $\left\{v_{t_{i}} \right\}_{i=0}^{4}$ and $\left\{w_{i} \right\}_{i=0}^{4}$, the matrix for the isomorphism $\varphi: S^{(3,3)} \to W^{(3,3)}$ is given by

\[ \left(
\begin{matrix}
1 & q & q & q^{2} & q \\
0 & 1 & 0 & q & q^{2} \\
0 & 0 & 1 & q & q^{2} \\
0 & 0 & 0 & 1 & q \\
0 & 0 & 0 & 0 & 1 \\
\end{matrix}  \right) \; .
\]  The calculation in \cref{SS:ExampleI} verifies the last column of this matrix.

\subsection{Example III}\label{SS:ExampleIII}
Set $\lambda = (4,2)$.  $t_{0} = t^{\lambda^{\TT}}$, $t_{1}=t_{0}.s_{2}$, $t_{2}=t_{0}.s_{4}$, $t_{3}=t_{0}.s_{2}s_{4}$, $t_{4}=t_{0}.s_{4}s_{5}$,  $t_{5}=t_{0}.s_{4}s_{5}s_{2}$,  $t_{6}=t_{0}.s_{4}s_{2}s_{3}$,  $t_{7}=t_{0}.s_{4}s_{2}s_{3}s_{5}$, and  $t_{8}=t_{0}.s_{4}s_{2}s_{3}s_{5}s_{4}$.  For $i=0,\dotsc, 8 $, let $w_{i}=\phi (t_{i})$.  With respect to the ordered bases $\left\{v_{t_{i}} \right\}_{i=0}^{8}$ and $\left\{w_{i} \right\}_{i=0}^{8}$, the matrix for the isomorphism $\varphi: S^{(4,2)} \to W^{(4,2)}$ is given by

\[ \left(
\begin{matrix}
%t_{0} & t_{1} & t_{2} & t_{3} & t_{4} & t_{5} & t_{6} & t_{7} & t_{8} \\
1 & q & q & q^{2} & q^{2} & q^{3} & q & q^{2} & q^{3} \\
0 & 1 & 0 & q & 0 & q^{2} & q^{2} & q^{3} & q^{4} \\
0 & 0 & 1 & q & q & q^{2} & q^{2} & q^{3} & q^{2} \\
0 & 0 & 0 & 1 & 0 & q & q & q^{2} & q^{3}+q \\
0 & 0 & 0 & 0 & 1 & q & 0 & q^{2} & q^{3} \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & q & q^{2} \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & q & q^{2} \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & q \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{matrix}  \right) \; .
\]

\bibliographystyle{eprintamsplain}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
\providecommand{\href}[2]{#2}
\begin{thebibliography}{10}

[1]
A.~Bj\"{o}rner and F.~Brenti, \emph{Combinatorics of {C}oxeter groups},
Graduate Texts in Mathematics, vol. 231, Springer, New York, 2005.

[2]
V.~Chari and A.~Pressley, \emph{A guide to quantum groups}, Cambridge
University Press, Cambridge, 1995, Corrected reprint of the 1994 original.

[3]
R.~Dipper and G.~James,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1112/plms/s3-52.1.20}
{\emph{Representations of {H}ecke algebras of general linear groups}}, Proc.
London Math. Soc. (3) \textbf{52} (1986), no.~1, 20--52.

[4]
S.~Donkin, \href{https://doi-org.ezproxy.lib.ou.edu/10.1017/CBO9780511600708}
{\emph{The {$q$}-{S}chur algebra}}, London Mathematical Society Lecture Note
Series, vol. 253, Cambridge University Press, Cambridge, 1998.

[5]
S.~Fomin and P.~Pylyavskyy,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1016/j.aim.2016.03.030}
{\emph{Tensor diagrams and cluster algebras}}, Adv. Math. \textbf{300}
(2016), 717--787.

[6]
I.~B. Frenkel and M.~G. Khovanov,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1215/S0012-7094-97-08715-9}
{\emph{Canonical bases in tensor products and graphical calculus for
{$U_q(\mathfrak{sl}_2)$}}}, Duke Math. J. \textbf{87} (1997), no.~3,

[7]
W.~Fulton, \emph{Young tableaux}, London Mathematical Society Student Texts,
vol.~35, Cambridge University Press, Cambridge, 1997, With applications to
representation theory and geometry.

[8]
A.~M. Garsia and T.~J. McLarnan,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1016/0001-8708(88)90060-6}
{\emph{Relations between {Y}oung's natural and the {K}azhdan-{L}usztig
representations of {$S_n$}}}, Adv. in Math. \textbf{69} (1988), no.~1,

[9]
B.-H. Hwang, J.~Jang, and J.~Oh, \href{https://arxiv.org/abs/2110.08790}
{\emph{A combinatorial model for the transition matrix between the {S}pecht
and web bases}}, 2021.

[10]
M.~S. Im and J.~Zhu,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1007/s10468-020-10026-6}
{\emph{Transitioning between tableaux and spider bases for {S}pecht
modules}}, Algebr. Represent. Theory \textbf{25} (2022), no.~2, 387--399.

[11]
{Johnson, Miles and Stewart, Natalie},
\href{https://math.mit.edu/research/undergraduate/spur/past-projects}
{\emph{Some graphical realizations of two-row {S}pecht modules of
{I}wahori-{H}ecke algebras of the symmetric group}}, 2019, [Online; accessed

[12]
C.~Kassel, \href{https://doi-org.ezproxy.lib.ou.edu/10.1007/978-1-4612-0783-2}
{\emph{Quantum groups}}, Graduate Texts in Mathematics, vol. 155,
Springer-Verlag, New York, 1995.

[13]
M.~Khovanov, V.~Mazorchuk, and C.~Stroppel,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1090/S0002-9939-07-09124-1}
{\emph{A categorification of integral {S}pecht modules}}, Proc. Amer. Math.
Soc. \textbf{136} (2008), no.~4, 1163--1169.

[14]
\href{http://projecteuclid.org.ezproxy.lib.ou.edu/euclid.cmp/1104287237}
{\emph{Spiders for rank {$2$} {L}ie algebras}}, Comm. Math. Phys.
\textbf{180} (1996), no.~1, 109--151.

[15]
A.~Mathas, \href{https://doi-org.ezproxy.lib.ou.edu/10.1090/ulect/015}
{\emph{Iwahori-{H}ecke algebras and {S}chur algebras of the symmetric
group}}, University Lecture Series, vol.~15, American Mathematical Society,
Providence, RI, 1999.

[16]
T.~P. McDonough and C.~A. Pallikaros,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1016/j.jpaa.2005.03.015}
{\emph{On relations between the classical and the {K}azhdan-{L}usztig
representations of symmetric groups and associated {H}ecke algebras}}, J.
Pure Appl. Algebra \textbf{203} (2005), no.~1-3, 133--144.

[17]
H.~Naruse, \href{https://doi-org.ezproxy.lib.ou.edu/10.3836/tjm/1270133181}
{\emph{On an isomorphism between {S}pecht module and left cell of {$S_n$}}},
Tokyo J. Math. \textbf{12} (1989), no.~2, 247--267.

[18]
R.~Patrias, O.~Pechenik, and J.~Striker,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1016/j.aim.2022.108603} {\emph{A
web basis of invariant polynomials from noncrossing partitions}}, Adv. Math.
\textbf{408} (2022), no.~part B, Paper No. 108603, 33.

[19]
T.~K. Petersen, P.~Pylyavskyy, and B.~Rhoades,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1007/s10801-008-0150-3}
{\emph{Promotion and cyclic sieving via webs}}, J. Algebraic Combin.
\textbf{30} (2009), no.~1, 19--41.

[20]
B.~Rhoades, \href{https://doi-org.ezproxy.lib.ou.edu/10.1007/s10801-016-0701-y}
{\emph{A skein action of the symmetric group on noncrossing partitions}}, J.
Algebraic Combin. \textbf{45} (2017), no.~1, 81--127.

[21]
\bysame, \href{https://doi-org.ezproxy.lib.ou.edu/10.1017/fms.2019.22}
{\emph{The polytabloid basis expands positively into the web basis}}, Forum
Math. Sigma \textbf{7} (2019), Paper No. e26, 8.

[22]
D.~E.~V. Rose and D.~Tubbenhauer,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1093/imrn/rnv302}
{\emph{Symmetric webs, {J}ones-{W}enzl recursions, and {$q$}-{H}owe
duality}}, Int. Math. Res. Not. IMRN (2016), no.~17, 5249--5290.

[23]
H.~M. Russell and J.~Tymoczko,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1093/imrn/rnaa290} {\emph{The
transition matrix between the {S}pecht and {$\mathfrak{sl}_3$} web bases is
unitriangular with respect to shadow containment}}, Int. Math. Res. Not. IMRN
(2022), no.~5, 3371--3416.

[24]
H.~M. Russell and J.~S. Tymoczko,
\href{https://doi-org.ezproxy.lib.ou.edu/10.1093/imrn/rnx164} {\emph{The
transition matrix between the {S}pecht and web bases is unipotent with
additional vanishing entries}}, Int. Math. Res. Not. IMRN (2019), no.~5,

[25]
D.~Tubbenhauer, P.~Vaz, and P.~Wedrich,
\href{https://doi-org.ezproxy.lib.ou.edu/10.2140/agt.2017.17.3703}
{\emph{Super {$q$}-{H}owe duality and web categories}}, Algebr. Geom. Topol.
\textbf{17} (2017), no.~6, 3703--3749.

\end{thebibliography}

\end{document}