text:
\section{Variations on KLRW algebras}
\subsection{Recollection on KLRW algebras}
\label{sec:KLRW}

Consider a sequence $\blam=(\lambda^{(1)},...,\lambda^{(\ell)})$ of dominant integral weights.
We recall now the third author's construction of the \textbf{KLRW algebra} $\tilde{T}^\blam$ (alias tensor product algebra, alias red-black algebra, alias Webster algebra).

The algebra $\tilde{T}^\blam$ is defined as a span of Stendhal diagrams, modulo some local relations.
More precisely, a Stendhal diagram \cite[Definition 4.1]{Webmerged} defining an element of $\tilde{T}^\blam$ is a collection of finitely many curves in $\R\times [0,1]$, where each curve is either red and labeled by one of the $\lambda^{(j)}$, or is black and labeled by $i\in I$.  Each curve has one endpoint on $ \R \times \{0\} $ and one on $ \R \times \{1\} $. The black curves can also be decorated with finitely many dots.  The diagrams are considered up to isotopy, and must be locally of the form
\begin{equation*}
\begin{tikzpicture}

  \draw[very thick] (0,0) +(-1,-1) -- +(1,1);

  \draw[very thick](0,0) +(1,-1) -- +(-1,1);

  \draw[very thick](3,0) +(-1,-1) -- +(1,1);

  \draw[wei, very thick](3,0) +(1,-1) -- +(-1,1);

  \draw[wei,very thick](6,0) +(-1,-1) -- +(1,1);

  \draw [very thick](6,0) +(1,-1) -- +(-1,1);

 \draw[very thick](-4,0) +(0,-1) --  +(0,1);

  \draw[very thick](-3,0) +(0,-1) --  node
  [midway,circle,fill=black,inner sep=2pt]{}
  +(0,1);

   \draw[wei,very thick](-2,0) +(0,-1) --  +(0,1);
\end{tikzpicture}
\end{equation*}
Note that no red strands can ever cross.  In \cite{Webmerged} the
diagrams are oriented.  We  only consider downward oriented strands,
so we omit the orientation here.  The diagrams in $\tilde{T}^\blam$ each
have $\ell$ red strands, labeled by
$\lambda^{(1)},...,\lambda^{(\ell)}$ from left to right.  For
convenience, we let $y_k$ denote the dot on the $k$th black strand read from
left to right.  The diagrams satisfy the following relations:
 \begin{itemize}
  \item the KLR relations (\ref{first-QH}--\ref{triple-smart})  for
    $\Gamma$ with \[X_{ij}(u,v)= \begin{cases}
  1 & i \not \leftarrow j\\
 u-v & i\leftarrow j\\
\end{cases} \qquad Q_{ij}(u,v)=X_{ij}(u,v)X_{ji}(v,u)=
\begin{cases}
  1 & i \not \leftrightarrow j\\
 u-v & i\leftarrow j\\
v-u & i\to j
\end{cases}
\] \newseq
\begin{equation*}\subeqn\label{first-QH}
    \begin{tikzpicture}[scale=.9,baseline]
      \draw[very thick](-4,0) +(-1,-1) -- +(1,1) node[below,at start]
      {$i$}; \draw[very thick](-4,0) +(1,-1) -- +(-1,1) node[below,at
      start] {$j$}; \fill (-4.5,.5) circle (3pt);

      \node at (-2,0){=}; \draw[very thick](0,0) +(-1,-1) -- +(1,1)
      node[below,at start] {$i$}; \draw[very thick](0,0) +(1,-1) --
      +(-1,1) node[below,at start] {$j$}; \fill (.5,-.5) circle (3pt);
      \node at (4,0){unless $i=j$};
    \end{tikzpicture}
  \end{equation*}
\begin{equation*}\subeqn\label{second-QH}
    \begin{tikzpicture}[scale=.9,baseline]
      \draw[very thick](-4,0) +(-1,-1) -- +(1,1) node[below,at start]
      {$i$}; \draw[very thick](-4,0) +(1,-1) -- +(-1,1) node[below,at
      start] {$j$}; \fill (-3.5,.5) circle (3pt);

      \node at (-2,0){=}; \draw[very thick](0,0) +(-1,-1) -- +(1,1)
      node[below,at start] {$i$}; \draw[very thick](0,0) +(1,-1) --
      +(-1,1) node[below,at start] {$j$}; \fill (-.5,-.5) circle (3pt);
      \node at (4,0){unless $i=j$};
    \end{tikzpicture}
  \end{equation*}
\begin{equation*}\subeqn\label{nilHecke-1}
    \begin{tikzpicture}[scale=.9,baseline]
      \draw[very thick](-4,0) +(-1,-1) -- +(1,1) node[below,at start]
      {$i$}; \draw[very thick](-4,0) +(1,-1) -- +(-1,1) node[below,at
      start] {$i$}; \fill (-4.5,.5) circle (3pt);

      \node at (-2,0){=}; \draw[very thick](0,0) +(-1,-1) -- +(1,1)
      node[below,at start] {$i$}; \draw[very thick](0,0) +(1,-1) --
      +(-1,1) node[below,at start] {$i$}; \fill (.5,-.5) circle (3pt);
      \node at (2,0){$+$}; \draw[very thick](4,0) +(-1,-1) -- +(-1,1)
      node[below,at start] {$i$}; \draw[very thick](4,0) +(0,-1) --
      +(0,1) node[below,at start] {$i$};
    \end{tikzpicture}
  \end{equation*}
 \begin{equation*}\subeqn\label{nilHecke-2}
    \begin{tikzpicture}[scale=.9,baseline]
      \draw[very thick](-4,0) +(-1,-1) -- +(1,1) node[below,at start]
      {$i$}; \draw[very thick](-4,0) +(1,-1) -- +(-1,1) node[below,at
      start] {$i$}; \fill (-4.5,-.5) circle (3pt);

      \node at (-2,0){=}; \draw[very thick](0,0) +(-1,-1) -- +(1,1)
      node[below,at start] {$i$}; \draw[very thick](0,0) +(1,-1) --
      +(-1,1) node[below,at start] {$i$}; \fill (.5,.5) circle (3pt);
      \node at (2,0){$+$}; \draw[very thick](4,0) +(-1,-1) -- +(-1,1)
      node[below,at start] {$i$}; \draw[very thick](4,0) +(0,-1) --
      +(0,1) node[below,at start] {$i$};
    \end{tikzpicture}
  \end{equation*}
  \begin{equation*}\subeqn\label{black-bigon}
    \begin{tikzpicture}[very thick,scale=.9,baseline]
      \draw (-2.8,0) +(0,-1) .. controls (-1.2,0) ..  +(0,1)
      node[below,at start]{$i$}; \draw (-1.2,0) +(0,-1) .. controls
      (-2.8,0) ..  +(0,1) node[below,at start]{$i$}; \node at (-.5,0)
      {=}; \node at (0.4,0) {$0$};
\node at (1.5,.05) {and};
    \end{tikzpicture}
\hspace{.4cm}
    \begin{tikzpicture}[very thick,scale=.9,baseline]

      \draw (-2.8,0) +(0,-1) .. controls (-1.2,0) ..  +(0,1)
      node[below,at start]{$i$}; \draw (-1.2,0) +(0,-1) .. controls
      (-2.8,0) ..  +(0,1) node[below,at start]{$j$}; \node at (-.5,0)
      {=};
\draw (1.8,0) +(0,-1) -- +(0,1) node[below,at start]{$j$};
      \draw (1,0) +(0,-1) -- +(0,1) node[below,at start]{$i$};
\node[inner xsep=10pt,fill=white,draw,inner ysep=8pt] at (1.4,0) {$Q_{ij}(y_1,y_2)$};
    \end{tikzpicture}
  \end{equation*}
 \begin{equation*}\subeqn\label{triple-dumb}
    \begin{tikzpicture}[very thick,scale=.9,baseline]
      \draw (-3,0) +(1,-1) -- +(-1,1) node[below,at start]{$k$}; \draw
      (-3,0) +(-1,-1) -- +(1,1) node[below,at start]{$i$}; \draw
      (-3,0) +(0,-1) .. controls (-4,0) ..  +(0,1) node[below,at
      start]{$j$}; \node at (-1,0) {=}; \draw (1,0) +(1,-1) -- +(-1,1)
      node[below,at start]{$k$}; \draw (1,0) +(-1,-1) -- +(1,1)
      node[below,at start]{$i$}; \draw (1,0) +(0,-1) .. controls
      (2,0) ..  +(0,1) node[below,at start]{$j$}; \node at (5,0)
      {unless $i=k\neq j$};
    \end{tikzpicture}
  \end{equation*}
\begin{equation*}\subeqn\label{triple-smart}
    \begin{tikzpicture}[very thick,scale=.9,baseline]
      \draw (-3,0) +(1,-1) -- +(-1,1) node[below,at start]{$i$}; \draw
      (-3,0) +(-1,-1) -- +(1,1) node[below,at start]{$i$}; \draw
      (-3,0) +(0,-1) .. controls (-4,0) ..  +(0,1) node[below,at
      start]{$j$}; \node at (-1,0) {=}; \draw (1,0) +(1,-1) -- +(-1,1)
      node[below,at start]{$i$}; \draw (1,0) +(-1,-1) -- +(1,1)
      node[below,at start]{$i$}; \draw (1,0) +(0,-1) .. controls
      (2,0) ..  +(0,1) node[below,at start]{$j$}; \node at (2.8,0)
      {$+$};        \draw (6.2,0)
      +(1,-1) -- +(1,1) node[below,at start]{$i$}; \draw (6.2,0)
      +(-1,-1) -- +(-1,1) node[below,at start]{$i$}; \draw (6.2,0)
      +(0,-1) -- +(0,1) node[below,at start]{$j$};
\node[inner ysep=8pt,inner xsep=5pt,fill=white,draw,scale=.8] at (6.2,0){$\displaystyle \frac{Q_{ij}(y_3,y_2)-Q_{ij}(y_1,y_2)}{y_3-y_1}$};
    \end{tikzpicture}
  \end{equation*}
\item  All black crossings and dots can pass through red lines.  For the latter two
  relations (\ref{dumb}--\ref{red-dot}), we also include their mirror images:
\newseq
  \begin{equation*}\subeqn
    \begin{tikzpicture}[very thick,baseline]\label{red-triple-correction}
      \draw (-3,0)  +(1,-1) -- +(-1,1) node[at start,below]{$i$};
      \draw (-3,0) +(-1,-1) -- +(1,1)node [at start,below]{$j$};
      \draw[wei] (-3,0)  +(0,-1) .. controls (-4,0) .. node[below, at start]{$\lambda$}  +(0,1);
      \node at (-1,0) {=};
      \draw (1,0)  +(1,-1) -- +(-1,1) node[at start,below]{$i$};
      \draw (1,0) +(-1,-1) -- +(1,1) node [at start,below]{$j$};
      \draw[wei] (1,0) +(0,-1) .. controls (2,0) ..  node[below, at start]{$\lambda$} +(0,1);
\node at (2.6,0) {$+ $};
      \draw (6,0)  +(1,-1) -- node
  [midway,circle,fill=black,inner sep=2pt]{} +(1,1) {} node[at start,below]{$i$};
      \draw (6,0) +(-1,-1) -- node
  [midway,circle,fill=black,inner sep=2pt]{} +(-1,1) {} node [at start,below]{$j$};
      \draw[wei] (6,0) +(0,-1) -- node[below, at start]{$\lambda$} +(0,1);
\node at (3.8,-.2){$\displaystyle \delta_{i,j}\sum_{a+b+1=\lambda_i}$} ;
\node at (5.3,0){$b$};
\node at (7.3,0){$a$};

 \end{tikzpicture}
  \end{equation*}
\begin{equation*}\subeqn\label{dumb}
    \begin{tikzpicture}[very thick,baseline=2.85cm]
      \draw[wei] (-3,3)  +(1,-1) -- +(-1,1);
      \draw (-3,3)  +(0,-1) .. controls (-4,3) ..  +(0,1);
      \draw (-3,3) +(-1,-1) -- +(1,1);
      \node at (-1,3) {=};
      \draw[wei] (1,3)  +(1,-1) -- +(-1,1);
  \draw (1,3)  +(0,-1) .. controls (2,3) ..  +(0,1);
      \draw (1,3) +(-1,-1) -- +(1,1);    \end{tikzpicture}
  \end{equation*}
\begin{equation*}\subeqn\label{red-dot}
    \begin{tikzpicture}[very thick,baseline]
  \draw(-3,0) +(-1,-1) -- +(1,1);
  \draw[wei](-3,0) +(1,-1) -- +(-1,1);
\fill (-3.5,-.5) circle (3pt);
\node at (-1,0) {=};
 \draw(1,0) +(-1,-1) -- +(1,1);
  \draw[wei](1,0) +(1,-1) -- +(-1,1);
\fill (1.5,.5) circle (3pt);
    \end{tikzpicture}
  \end{equation*}
\item  The ``cost'' of separating a red $\lambda$-strand and a black $i$-strand is adding $\lambda_i$ dots to the black strand.
  \begin{equation}\label{cost}
  \begin{tikzpicture}[very thick,baseline=1.6cm]
    \draw (-2.8,0)  +(0,-1) .. controls (-1.2,0) ..  +(0,1) node[below,at start]{$i$};
       \draw[wei] (-1.2,0)  +(0,-1) .. controls (-2.8,0) ..  +(0,1) node[below,at start]{$\lambda$};
           \node at (-.3,0) {=};
    \draw[wei] (2.8,0)  +(0,-1) -- +(0,1) node[below,at start]{$\lambda$};
       \draw (1.2,0)  +(0,-1) -- +(0,1) node[below,at start]{$i$};
       \fill (1.2,0) circle (3pt) node[left=3pt]{$\lambda_i$};
          \draw[wei] (-2.8,3)  +(0,-1) .. controls (-1.2,3) ..  +(0,1) node[below,at start]{$\lambda$};
  \draw (-1.2,3)  +(0,-1) .. controls (-2.8,3) ..  +(0,1) node[below,at start]{$i$};
           \node at (-.3,3) {=};
    \draw (2.8,3)  +(0,-1) -- +(0,1) node[below,at start]{$i$};
       \draw[wei] (1.2,3)  +(0,-1) -- +(0,1) node[below,at start]{$\lambda$};
       \fill (2.8,3) circle (3pt) node[right=3pt]{$\lambda_i$};
  \end{tikzpicture}
\end{equation}
  \end{itemize}

The algebra $\tilde{T}^\bla$ is graded with degree given by
\cite[Def. 4.4]{Webmerged}:
\[
  \deg\tikz[baseline,very thick,scale=1.5]{\draw[-] (.2,.3) --
    (-.2,-.1) node[at end,below, scale=.8]{$i$}; \draw[wei] (.2,-.1) --
    (-.2,.3) node[at start,below,scale=.8]{$\la$};} =\langle\al_i,\la\rangle\qquad \deg\tikz[baseline,very
  thick,-,scale=1.5]{\draw (0,.3) -- (0,-.1) node[at
    end,below,scale=.8]{$i$} node[midway,circle,fill=black,inner
    sep=2pt]{};}=-2
 \qquad  \deg\tikz[baseline,very thick,scale=1.5]{\draw[-] (.2,.3) --
    (-.2,-.1) node[at end,below, scale=.8]{$i$}; \draw[-] (.2,-.1) --
    (-.2,.3) node[at start,below,scale=.8]{$j$};}
  =-\langle\al_i,\al_j\rangle  \]

The identity of $\tilde{T}^\blam$ is a sum of
idempotents corresponding to sequences of red and black strands.  More carefully, we associate an
idempotent $e(\Bi,\kappa)$ to a sequence $\Bi=(i_1,\dots, i_n)$, which labels the black
strands in order, and a weakly increasing function $\kappa\colon [1,\ell]\to [0,n]$
such that the $k$th red strand is between the $\kappa(k)$th and
$(\kappa(k)+1)$-st black strands.  We interpret $\kappa(k)=0$ to mean
this red strand is left of all black strands, and $\kappa(k)=n$ to mean it is
right of all black strands.  Recall that in $\tilde{T}^\blam$ the red strands are always labeled $\lambda^{(1)},...,\lambda^{(\ell)}$ from left to right, so the pair $(\Bi,\kappa)$ uniquely determines the idempotent.

As in \cite[Lemma 4.12]{Webmerged}, $\tilde{T}^\blam$ acts on
\begin{equation*}
\label{eq:Pol}
\PolKLR=
\bigoplus_{(\Bi,\kappa)} \PolKLR(\Bi,\kappa), \quad \text{where}\quad \PolKLR(\Bi,\kappa)=\C[Y_1(\Bi,\kappa),\dots Y_n(\Bi,\kappa)]
\end{equation*}
The idempotent
 $e(\Bi,\kappa)$ acts by projection onto $\PolKLR(\Bi,\kappa)$, and \newseq
  \begin{equation*}\label{eq:T-action1}\subeqn
\begin{tikzpicture}[scale=.4,baseline]
\draw[wei] (-1,-1) -- (1,1) node[at start,below]{
$\lambda$} node[at end,above]{
$\lambda$};
\draw[very thick] (1,-1) -- (-1,1) node[at start,below]{
$i$} node[at end,above]{
$i$};

\node at (3.7,0) {$\bullet\: f=Y_k^{\lambda_i}\cdot  f$ } ;
\end{tikzpicture}\qquad \qquad
\begin{tikzpicture}[scale=.4,baseline]
\draw[wei] (1,-1) -- (-1,1) node[at start,below]{
$\lambda$} node[at end,above]{
$\lambda$};
\draw[very thick] (-1,-1) -- (1,1) node[at start,below]{
$i$} node[at end,above]{
$i$};
\node at (2.7,0) {$\bullet\: f=f$ } ;
\end{tikzpicture}\qquad \qquad
\begin{tikzpicture}[scale=.4,baseline]
\draw[very thick] (1,-1) -- (1,1) node[at start,below]{
$i$} node[at end,above]{
$i$} node[circle,midway,fill,inner sep=2pt]{};
\node at (3.8,0) {$\bullet\: f= Y_k\cdot f$ } ;
\end{tikzpicture}
  \end{equation*}
\begin{equation*}\label{eq:T-action2}\subeqn
  \begin{tikzpicture}[scale=.4]
\draw[very thick] (1,-1) -- (-1,1) node[at start,below]{
$j$} node[at end,above]{
$j$};
\draw[very thick] (-1,-1) -- (1,1) node[at start,below]{
$i$} node[at end,above]{
$i$};
\node at (7.5,0) {$\bullet\: f=\begin{cases}
 X_{ij}(Y_{k+1},Y_{k}) f^{s_k} & i\neq j\\
 \displaystyle \frac{f^{s_k}- f}{Y_{k+1}-Y_{k}} & i=j
\end{cases}$ } ;
\end{tikzpicture}
\end{equation*}
In the equations above, the black $i$-strand is always the $k$th black strand from left to right.

\subsection{Some statements about categorification} Set $\lambda=\sum_j\lambda^{(j)}$.  For $\mu \in \Lambda$, let $\tilde{T}^\blam_{\mu}$ be the subalgebra of $\tilde{T}^\blam$ where
the sum of the roots associated to the black strands is $\lambda-\mu$.
We have a non-unital algebra homomorphism $\iota_{\mu}^+:\tilde{T}^\blam_{\mu}
\to \tilde{T}^\blam_{\mu-\alpha_i}$, which adds a black $i$-strand to the
right of the diagram, and its mirror image $\iota^-_\mu$ adding a
strand at the left.

Let $\tilde{T}^\blam_\mu\mathrm{-mod}$ be the category of finitely generated $\tilde{T}^\blam_\mu$-modules.  Consider the functors
\begin{align*}
\cF_i^+: \tilde{T}^\blam_\mu\mathrm{-mod} \to \tilde{T}^\blam_{\mu-\alpha_i}\mathrm{-mod} \\
\cF_i^-: \tilde{T}^\blam_\mu\mathrm{-mod} \to \tilde{T}^\blam_{\mu-\alpha_i}\mathrm{-mod}
\end{align*}
given by induction along the maps $\iota_{\mu}^\pm$. More specifically, $\cF_i^\pm(M)=M\otimes_{\tilde{T}^\blam_\mu}\tilde{T}^\blam_{\mu-\alpha_i}$.  Let $\cE_i^\pm$ be the adjoint functors of $\cF_i^\pm$ given by restriction along the
maps $\iota_{\mu}^\pm$.  Note that setting $e_i^+$ (respectively $e_i^-$) to be the sum of idempotents whose rightmost (respectively leftmost) strand is a black $i$-strand, then $\cE_i^+(M)=e_i^+M$ and $\cE_i^-(M)=e_i^-M$.

Let $K^0(\tilde{T}^\blam)$ denote the Grothendieck group of finitely generated $\tilde{T}^\blam$-modules, and set $K^\C(-):=K^0(-)\otimes_{\Z}\C$.  Let $V(\blam)=V({\lambda^{(1)}})\otimes \cdots \otimes V({\lambda^{(\ell)}})$.
By \cite[Prop. 2.18]{Webweb}, we have:
\begin{Proposition}
\label{prop:tildeTaction}
We have an isomorphism of $(U(\mathfrak{n}_-),U(\mathfrak{n}_-))$-bimodules
$$
\mathcal{G}\colon U(\blam):=
U(\mathfrak{n}_-) \otimes V({\blam}) \to K^\C(\tilde{T}^\blam)
$$
which maps $K^\C(\tilde{T}^\blam_\mu)$ to the $\mu$-weight space of
$U(\blam)$.  Under this isomorphism, the functors
$\cF_i^+,\cF_i^-$ categorify the actions:
\begin{align*}
  f_i^+\cdot (u\otimes v_1\otimes \cdots \otimes v_\ell )
  & = f_iu\otimes  v_1\otimes \cdots
    \otimes v_\ell \\
&\qquad  +u\otimes f_iv_1\otimes \cdots \otimes v_\ell+\cdots  +u\otimes v_1\otimes
    \cdots\otimes  f_iv_\ell  \\
 (u\otimes v_1\otimes \cdots \otimes v_\ell)  \cdot f_i^-
  & = uf_i \otimes v_1\otimes \cdots
    v_\ell
\end{align*}
\end{Proposition}

This isomorphism is uniquely fixed by the action property above, and
the fact that the functor $\mathcal{I}_\la$ induced by adding a red
strand at the far right (i.e. like $\cF_i^+$, but with a different
color) is induced by \[I_{\la}(u\otimes v_1\otimes \cdots \otimes
v_{\ell})\mapsto u\otimes v_1\otimes \cdots \otimes
v_{\ell}\otimes v_{\la,\operatorname{high}}\] where
$v_{\la,\operatorname{high}}$ is the highest weight vector of
$V(\la)$.  We let $\tilde{p}^\kappa_{\Bi}=[\tilde{T}e(\Bi,\kappa)]$, following the notation of \cite[\S 4.7]{Webmerged}; this is easily calculated from the rules above.

\begin{Remark}
  The association of $\cF_i^+$ with the left action and $\cF^-_i$ with
  the right here might seem a little peculiar, since they involve
  adding strands on the right and left sides of the diagram.  This is
  due to the dyslexia of the conventions in \cite{Webmerged};
  unfortunately, this is a sin of one of authors for which all of
  us, readers included, must do penance.
\end{Remark}

The algebra $\tilde{T}^\blam$ has a natural triple of quotients; we
let ${}_+T^\blam$ be the quotient by the two-sided ideal generated by
all idempotents where $\kappa(1)> 0$ (i.e. a black strand is to the left of all red strands), ${}_-T^\blam$ be the
quotient by the two-sided ideal generated by all idempotents where
$\kappa(\ell)< n$ (i.e. a black strand is to the right of all red strands), and ${}_0T^\blam$ the quotient by the sum of these
ideals.

Note that $\cF_i^+$ descends to a well-defined functor on ${}_+T^\blam\mmod$
and $\cF_i^-$ to one on ${}_-T^\blam\mmod$, still both defined by extension
of scalars along an inclusion of algebras.  Since the quotients
${}_+T^\blam$ and ${}_-T^\blam$ are finite dimensional, the adjoint functors $\cE_i^+$ and
$\cE_i^-$ preserve finitely generated modules, and less obviously, they
preserve the category of projective modules (this follows from the
Morita equivalence of \cite[Thm. 4.30]{Webmerged} by allowing us to
rewrite it as induction to a Morita equivalent algebra).

   These quotient algebras correspond to killing the action of
the augmentation ideal of $ U(\mathfrak{n}_-)$ for the left action,
right action, or both.

\begin{Proposition}\label{prop:Taction}
  We have isomorphisms of Grothendieck groups
\newseq
  \begin{align*}
K^\C({}_+T^\blam)&\cong U(\blam)/\mathfrak{n}_-U(\blam)\cong V({\blam})\subeqn\label{eq:tensor1}\\
K^\C({}_-T^\blam) &\cong U(\blam)/U(\blam)\mathfrak{n}_-\cong V({\blam'})\subeqn\label{eq:tensor2}\\
K^\C({}_0T^\blam) &\cong U(\blam)/(U(\blam)\mathfrak{n}_-+\mathfrak{n}_-U(\blam)) \subeqn\label{eq:tensor3} \cong V({\blam})/ \mathfrak{n}_-V({\blam}) \cong (V({\blam}) )^{\mathfrak{n}}
  \end{align*}
  In (\ref{eq:tensor2}), $\blam'$ denotes the list of dominant weight in reverse order $(\lambda^{(\ell)},...,\lambda^{(1)})$.
  The action of the functors $\cE_i^\pm,\cF_i^\pm$ on ${}_\pm T\mmod$
  categorify the action of the Chevalley generators $e_i, f_i$ on
  $V(\blam)$.
\end{Proposition}

\begin{proof}
  The isomorphism (\ref{eq:tensor1}) is a consequence of
  \cite[Th. B]{Webmerged}.  Note that the relations
  (\ref{first-QH}--\ref{cost}) are preserved by reflecting in a
  vertical axis and negating crossings $\psi_i$ of strands with the
  same labels. This gives an isomorphism of ${}_-T^\blam$ to ${}_+T^{w_0\cdot\blam}$ with
  the order of labels on red strands reversed.  Thus
  (\ref{eq:tensor2}) follows.

   The isomorphism (\ref{eq:tensor3}) is not explicitly discussed in
   \cite{Webmerged}, but is easily derived from the definitions there:
   the modules over ${}_0T^\blam$ are the same as those over ${}_+T^\blam$ killed
   by all $\cE_i^+$ for all $i$, and the number of simple modules
   killed by these functors is the same as the number of highest
   weight elements of the corresponding tensor product crystal by
   \cite[Th. 7.2]{LoWe}, so the classes of these simples span the
   space of highest weight vectors.  The description of the highest
   weight vectors as the quotient $V/\mathfrak{n}_-V$  is
   valid for any finite dimensional $\fg$-module $V$.
\end{proof}

Let $\tU$ be the 2-Kac-Moody algebra categorifiying $U(\fg)$, as defined by Khovanov-Lauda \cite{KLIII} and Rouquier \cite{Rou2KM}.  For details see Definition 2.4 in \cite{Webmerged}.  Note that in the
definition we use the same matrix of polynomials $Q_{i,j}(u,v)$ which
appears above.
\begin{Proposition}[\mbox{\cite[Th. B]{Webmerged}}]\label{prop:cat-action}
The functors $\cE_i^{\pm},\cF_i^{\pm}$ induce an action of $\tU$ on
the categories  ${}_\pm T^\blam\mathrm{-mod}$, which each categorify $V({\blam})$ and $V({\blam'})$.
\end{Proposition}If we instead consider the graded Grothendieck group of $\tilde{T},{}_\pm
  T$, we will arrive at the obvious quantum analogues of the objects
  under consideration.
  This is significant because  these quantum analogues allow us to
  define a canonical basis of the spaces $U(\bla)$ and $V(\bla)$, and
  when the conventions are chosen correctly, the isomorphisms of
  Propositions \ref{prop:tildeTaction} and \ref{prop:Taction} send
  the classes of indecomposable projectives bijectively to canonical
  basis by the argument of \cite[Th. 8.7]{WebCB}; note that the statement of {\it loc.\ cit.} only covers $V(\bla)$, but that is derived as a consequence of the fact that it holds for $\tilde{T}^{\bla}$ and $U(\bla)$.

  We can convert this into a more concrete statement about multiplicities with a little algebra:
  \begin{Lemma}\label{lem:idempotent-canonical} The dimension  $\mathsf{m}_{\Bi,\kappa,b}=\dim e(\Bi, \kappa)L_b$ of the image of the idempotent $e(\Bi,\kappa)$ in the unique simple quotient of the indecomposable projective $P_b$ such that $[P_b]=b$ for $b$ a canonical basis vector is the coefficient of $b$ when $[\tilde{T}e(\Bi, \kappa)]$ is expanded in the canonical basis.  That is:
  \[p_{\Bi}^{\kappa}=[\tilde{T}e(\Bi, \kappa)]=\sum_{b} \mathsf{m}_{\Bi,\kappa,b}\cdot b.\]
  \end{Lemma}
  \begin{proof}
  Of course, $\mathsf{m}_{\Bi,\kappa,b}=\dim \Hom(\tilde{T}e(\Bi, \kappa),L)$.  Since $\dim \Hom(P_{b'},L_b)=\delta_{b,b'}$, the result follows.
  \end{proof}

  Unfortunately, some real care is
  needed here about conventions.  It might seem strange that in (\ref{eq:tensor1}) and (\ref{eq:tensor2}) we used opposite orders of the tensor product, since representations of $\fg$ are a symmetric tensor category and thus $V(\bla)$ and $V(\bla')$ are canonically isomorphic.  However, this is no longer true after quantization.

  This manifests on the level of categorification: reflecting in a vertical axis (and inserting a few signs) instead gives an isomorphism ${}_+T^{\bla}\cong {}_-T^{\bla'}$.
In fact, by \cite[Th. 3.12]{LoWe}, this implies that ${}_+T^{\bla}\mmod$ is a tensor product categorification (in the sense of {\it loc.\ cit.}) for $V(\bla)$ while ${}_-T^{\bla}\mmod$ is for $V(\bla')$.

\subsection{A KLRW algebra associated to $\bR$}
\label{section:TR}

 Fix a dominant integral weight $ \lambda$ and write $ \lambda = \sum_i
\lambda_i \varpi_i $. We fix also an integral set of parameters $\bR = \{   R_i \}_{i\in I}$ of weight $\lambda$.  It will be convenient at times to encode the parameters using multiplicity functions $\rho_i:\Z\to\N$ defined by
$$
\rho_i(q)=
\begin{cases}
\text{multiplicity of $2q$ in $R_i$, if $i$ is even},\\
\text{multiplicity of $2q+1$ in $R_i$, if $i$ is odd.}
\end{cases}
$$
Recalling that we've fixed a total order on $I$ (cf. Section \ref{subsection:Notation}), let $I=\{i_1<\cdots< i_r\}$.  Define the following sequences of fundamental weights:
\begin{equation}
\label{eq:varpiq}
\varpi_\bR(q):=(\varpi_{i_1}^{\rho_{i_1}(q)},...,\varpi_{i_r}^{\rho_{i_r}(q)})
\end{equation}
where  $\varpi_{i}^{k}$ denotes the sequence $\varpi_i,...,\varpi_i$ of length $k$.  Concatenate these together to form
$$
\varpi_\bR:=(...,\varpi_\bR(q-1),\varpi_\bR(q),...)
$$
Since $\bR$ is finite, clearly $\varpi_\bR$ is a finite sequence of fundamental weights: $$\varpi_\bR=(\varpi_{j_1},...,\varpi_{j_\ell})$$ for some $j_1,...,j_\ell \in I$.
We let $$\tilde{T}=\tilde{T}^\bR: =\tilde{T}^{\varpi_\bR}$$  Of course $\tilde{T}$ depends on $\bR$, but we fix this choice and usually suppress this from our notation. We warn the reader that $\tilde{T}$ here is different from the algebra with same name in \cite{Webmerged}.

Note that in $\tilde{T}$ all diagrams have red strands labeled by $j_1,...,j_\ell$ from left to right.  Let $r_1,...,r_\ell$ be the weakly increasing ordering of the elements of $\bR$ (with multiplicity).  We say $r_k$ is the {\bf longitude} of $j_k$.

\begin{Example}
Let $\fg=\mathfrak{sl}_4$, and $\la=\varpi_1+2\varpi_2+2\varpi_3$.  We order the nodes of the Dynkin diagram by $2<1<3$.  Consider an integral set of parameters $R_1=\{1\},R_2=\{0,4\}$, and $R_3=\{-1,1\}$.  Then
$$
\varpi_\bR=(\varpi_3,\varpi_2,\varpi_1,\varpi_3,\varpi_2).
$$
Now take $\Bi=(i_1,i_2,i_3)$ and $\kappa:1\mapsto0, 2\mapsto 1, 3\mapsto 1, 4\mapsto 3,5\mapsto 3$.  Then
\begin{equation*}
\begin{tikzpicture}[scale=.4,baseline]
\draw[wei] (0,-1) -- (0,1) node[at start,below]{
$3$} ;
\draw[very thick] (1,-1) -- (1,1) node[at start,below]{
$i_1$};
\draw[wei] (2,-1) -- (2,1) node[at start,below]{
$2$} ;
\draw[wei] (3,-1) -- (3,1) node[at start,below]{
$1$} ;
\draw[very thick] (4,-1) -- (4,1) node[at start,below]{
$i_2$};
\draw[very thick] (5,-1) -- (5,1) node[at start,below]{
$i_3$};
\draw[wei] (6,-1) -- (6,1) node[at start,below]{
$3$} ;
\draw[wei] (7,-1) -- (7,1) node[at start,below]{
$2$} ;

\node at (-2.5,0) {$e(\Bi,\kappa)=$} ;

\end{tikzpicture}
\end{equation*}
The longitudes of the red strands are $-1,0,1,1,4$ from left to right.
\end{Example}

\subsection{The parity KLRW algebra}
\label{sec:parityKLRW}
We now introduce a subalgebra of $\tilde{T}$ whose representation
theory describes the weight modules over  the truncated shifted
Yangian.  This will allow us to deduce some theorems about the highest weight theory of the latter algebra (cf. Theorem \ref{co:main}).

Consider an idempotent $e\in\tilde{T}$.  Since a strand of $e$ (of either color) is labeled by an element of $I$, we can talk about its parity.

We will define the {\bf parity distance} $\delta$ between the strands of $e$ as follows.
First for consecutive strands $p,p'$ we set:
$$
\delta(p,p')=
\begin{cases}
2  \text{ if } p \text{ and } p' \text{ have the same parity, }  p \text{ is black and } p' \text{ is red } \\
1 \text{ if } p \text{ and } p' \text{ have different parity} \\
0 \text{ otherwise}
\end{cases}
$$

We then extend to all pairs of strands by
sharp triangle
inequality $\delta(p,p')=\delta(p,p'')+\delta(p'',p')$, where $p$ is left of $p'$ and $p''$ is any strand between $p$ and $p'$.

\begin{Definition}
\label{def:parityalgebra}
  We say an idempotent $e$ is {\bf parity}, if for any pair of red strands $p,p'$ of $e$, we have
  $\delta (p,p')\leq |r-r'|$, where $r$ is the longitude of $p$ and $r'$ is the longitude of $p'$.  Let $e_{\bR}$ be the sum of all
  parity idempotents.  We call an indecomposable projective $\tilde{T}$-module $Q$ and its corresponding canonical basis vector {\bf parity} if it is a summand of $\tilde{T}e_{\bR}$.

  The  {\bf parity KLRW algebra } is the subalgebra $\tilde{P}=\tilde{P}^\bR=e_{\bR}\tilde{T}e_{\bR}$, and $\tilde{P}_\mu=\tilde{P}\cap\tilde{T}_\mu$.
\end{Definition}

Of course,
the Grothendieck groups $K^0(\tilde{P}\mmod),K^0({}_{\pm}{P}\mmod)$ are naturally isomorphic to the span of parity canonical basis vectors in $U(\bla), V(\bla),V(\bla')$ as appropriate; we denote these subspaces by $U(\bR), V(\bR), V(\bR')$.

Note that these algebras will be Morita equivalent if and only if
$\tilde{T}=\tilde{T}e_{\bR}\tilde{T}$, which
will hold if every simple module of $\tilde{T}$ is not killed by
$e_{\bR}$ (and thus, does not factor through $\tilde{T}/\tilde{T}e_{\bR}\tilde{T}$).  As for $\tilde{T}$, the algebra $\tilde{P}$ has a natural triple of quotients: we
let ${}_+P^\bR$ be the quotient by the two-sided ideal generated by
all idempotents where $\kappa(1)> 0$, ${}_-P^\bR$ be the
quotient by the two-sided ideal generated by all idempotents where
$\kappa(\ell)< n$, and ${}_0P^\bR$ the quotient by the sum of these
ideals.  If $\bR$ is clear from context we usually omit it.  Similarly, we also have algebras ${}_\pm P_\mu$.

\begin{Example}
\label{ex:parity}
If $|r-r'|\gg n+\ell$ for all $r,r'\in\bR$, then all idempotents with at most $n$ black strands are
  parity.  In particular, if $ht(\lambda-\mu)\leq n$ then $\tilde{P}_\mu=\tilde{T}_\mu$.

  On the
  other hand, if all red strands have the same parity, and we have
  $r_1=\cdots=r_\ell$, then for an idempotent $e(\Bi,\kappa)$ to be parity we must have $\kappa(1)=\cdots=
  \kappa(\ell)\in \{0,n \}$.
\end{Example}

  \begin{Example}
  \label{ex:Nomega1}
  Suppose $\lambda=N\varpi_i$ for some $i\in I$.  The set of parameters is a single multiset $\bR=\{c_1^{t_1},...,c_q^{t_q}\}$, where $c_1<\cdots<c_q$, $c_i$ occurs $t_i$ times in $\bR$, and $\sum t_i=N$.

 The parity algebra $\tilde{P}$ is isomorphic to the KLRW $\tilde{T}^{\boldsymbol{\la}}$, where $\boldsymbol{\la}=(t_1\varpi_i,...,t_q\varpi_i)$.  Indeed, the parity idempotents in this case are all of the form:
\begin{equation*}
\begin{tikzpicture}[scale=.4,baseline]
\draw[very thick] (-3.5,-1) -- (-3.5,1) ;
\draw[very thick] (-1.5,-1) -- (-1.5,1) ;
\draw[wei] (-.5,-1) -- (-.5,1) ;
\draw[wei] (1.5,-1) -- (1.5,1) ;
\draw[very thick] (2.5,-1) -- (2.5,1) ;
\draw[very thick] (4.5,-1) -- (4.5,1) ;
\draw[very thick] (7.5,-1) -- (7.5,1) ;
\draw[very thick] (9.5,-1) -- (9.5,1) ;
\draw[wei] (10.5,-1) -- (10.5,1) ;
\draw[wei] (12.5,-1) -- (12.5,1) ;
\draw[very thick] (13.5,-1) -- (13.5,1) ;
\draw[very thick] (15.5,-1) -- (15.5,1) ;
\draw[decoration={brace,mirror,raise=5pt},decorate]
  (-.5,-1) -- node[below=6pt] {$t_1$} (1.5,-1);
  \draw[decoration={brace,mirror,raise=5pt},decorate]
  (10.5,-1) -- node[below=6pt] {$t_q$} (12.5,-1);

\node at (-2.5,0) {$\cdots$} ;
\node at (1/2,0) {$\cdots$} ;
\node at (3.5,0) {$\cdots$} ;
\node at (6,0) {$\cdots$} ;
\node at (8.5,0) {$\cdots$} ;
\node at (11.5,0) {$\cdots$} ;
\node at (14.5,0) {$\cdots$} ;
\end{tikzpicture}
\end{equation*}
Here the red strands are grouped into $q$ ``packets'', and the black strands can appear  only in between the packets.

Under the isomorphism between $\tilde{P}$ and $\tilde{T}^{\boldsymbol{\la}}$, this idempotent maps to the one in $\tilde{T}^{\boldsymbol{\la}}$ where the $k$-th packet of red strands is collapsed to a single red strand labeled by $t_k\varpi_i$.  Note that in particular when $\fg=\mathfrak{sl}_2$, the parity algebra $\tilde{P}$ is isomorphic to the KLRW algebra.
\end{Example}

\begin{Example}
Consider the case where $\fg=\mathfrak{sl}_3$ and $R_1=\{1\}$ and
$R_2=\{2\}$.  In this case, the idempotent ${\color{red}
  1}\;1\;2\; {\color{red} 2}$ is {\bf not} parity, since the parity
distance between $  {\color{red} 1}$ and ${\color{red} 2}$ is $3$, but
$|r_1-r_2|=1$.

On the other hand $2\;1\;{\color{red}
  1}\;{\color{red} 2}$

is parity, since the parity distance between the two red strands
is 1.  (Here, and in Example \ref{example} below, we are using the obvious shorthand for idempotent diagrams: each number corresponds to a strand in the same order, the color of a number corresponds to the color of the strand, and the number itself is the label on the strand.)

We remark that the algebra $\tilde{T}$ has a one dimensional representation where $e({\color{red}
  1},1,2,{\color{red} 2})$ acts by the identity, and every other
homogeneous element of the algebra acts by 1.  This simple is killed
by every parity idempotent, which shows that $\tilde{P}$ and
$\tilde{T}$ are not Morita equivalent in this case.
\end{Example}

\begin{Lemma}
\label{lem:categoricalactions}
  The categorical actions on ${}_{\pm}T\mathrm{-mod}$ of Proposition
  \ref{prop:cat-action} are inherited by the subcategories ${}_{\pm}P\mathrm{-mod}$.
\end{Lemma}
\begin{proof}
  The maps $\iota^\pm$ add black strands that are not between any two
  red strands.  Thus, they do not change the parity distance between
  any two reds, and they send parity idempotents to parity
  idempotents.

Thus, if we have a ${}_+T$-module $M$ which is killed by all parity
idempotents, then the action of a parity idempotent $e$ on $\cE_iM$ is
given by that of the parity idempotent $\iota^+(e)$, which is still
parity and  thus acts by zero.

Every element of $\cF_iM$ lies in the span of $x\otimes m$ where
$m\in M$, and $x$ is a straight-line diagram pulling a black strand
with label $i$ from the far right at $y=0$ to some position at $y=1$,
leaving all other strands in place.  If the top of $x$ is a parity
idempotent, then the bottom is as well (since we can only decrease
parity distance by moving a strand to the far right), and thus
$x\otimes m=0$.  Thus, $\cF_iM$ is killed by parity idempotents.

Since ${}_{\pm}P\mmod$ is the quotient of ${}_{\pm}T\mmod$ by elements
killed by parity idempotents, the result follows.
\end{proof}
This immediately implies:
\begin{Corollary} The subspace $U(\bR)$ is  a $(U(\mathfrak{n_-}),U(\mathfrak{n_-}))$-subbimodule, and $V(\bR),V(\bR')$ are $U(\fg)$-submodules.
\end{Corollary}
Note, these spaces can be naturally quantized to give based modules in the sense of Lusztig; we will not discuss this in any detail, since we will not use anything other than the existence of the canonical basis.

Let $\Irr({}_-P)$ denote the set of isomorphism classes of simple
${}_-P$-modules.  This has a crystal structure with the Kashiwara
operators defined by
\[\tilde{e}_iL=\operatorname{hd} (\cE_i^-L)\qquad \tilde{f}_iL=\operatorname{hd}(\cF_i^-L).\]
This is a subcrystal of $\Irr({}_-T^{\bla})$ which is isomorphic to the corresponding tensor
product crystal by \cite[Thm. 7.2]{LoWe}.

Similarly, the set $\Irr(\tilde{P})$ of {\it nilpotent} simple modules
over $\tilde{P} $ carries a bicrystal structure
defined by \[\tilde{f}_iL=\operatorname{hd}(\cF_i^-L)\qquad
  \tilde{f}_i^*L=\operatorname{hd}(\cF_i^+L).\]
We need to impose the nilpotent condition since $
\tilde{P}$ is finite over its center, which is a polynomial ring;
thus, for any maximal ideal of the center, there are simples killed by
that maximal ideal, with the nilpotent simples corresponding to unique
graded maximal ideal.  These are also the simple modules which will be
relevant to the representation theory of the Yangian via Theorem \ref{co:tilde-main}.

Note that many other sources on KLR algebras, such as \cite{KLI},
have typically studied graded modules instead.  However, nilpotent
simple modules are in bijection with graded simple modules considered
up to grading shift:
\begin{Lemma}\label{lem:nil-grade}
A simple module over $\tilde{P}$ is nilpotent if and only if it is
gradable.  Every indecomposable projective $\tilde{P}$-module has a unique
nilpotent simple quotient.
\end{Lemma}
\begin{proof}
  Since $\tilde{P}$ is finitely generated over a commutative
  subalgebra, its simple modules must be finite dimensional, so any
  element of non-zero degree must be nilpotent on any simple.

  On the other hand, the subalgebra $A$ of polynomials in the dots
  which is symmetric under permutations of strands and their labels
  is a central and $\tilde{P}$ is
  finitely generated over $A$.  As usual, any central subalgebra acts
  semi-simply on any simple module.  Thus, a simple is nilpotent if it factors through
  the quotient by the unique graded maximal ideal $\mathfrak{m}$ of this central
  subalgebra.  The result then follows from the fact that simple
  modules over finite dimensional graded algebras are always
  gradable uniquely up to grading shift.

  Similarly, if we have an indecomposable projective $Q$ over
  $\tilde{P}$, then $Q/\mathfrak{m}Q$ is a projective module over
  $\tilde{P}/\mathfrak{m}\tilde{P}$.  Let $L$ be a simple module in the
  cosocle of $Q/\mathfrak{m}Q$.  As discussed above, $L$ is gradeable,
  and since   $\tilde{P}$ has finite dimensional degree 0 part, $L$
  has a unique graded projective cover $Q'$.  The projective property
  induces a surjective map $Q\to Q'$, which must be an isomorphism by
  the indecomposability of $Q$.  Thus,
  $Q/\mathfrak{m}Q=Q'/\mathfrak{m}Q'$  is the projective cover of $L$
  as  a $\tilde{P}/\mathfrak{m}\tilde{P}$-module, and so $Q$ has no
  other nilpotent simple modules as quotients.
\end{proof}

\subsection{A crystal isomorphism}
\label{sec:crystalisom}
In this section we'll show that the crystals $\Irr({}_-P)$ and $\B(\bR)$ are isomorphic.

Presently, we describe how to construct an idempotent $e(\bS)\in {}_-P$  from a monomial $\yMon_{\bR}\zMon_{\bS}^{-1}\in \B(\bR)$.
First, we order the elements of $\bS$ so that $s_1\leq \cdots \leq s_n$.  Then, we define a sequence $\Bi=(i_1,...,i_n)$ by  $s_m\in S_{i_m}$ for all $m$.  (Note if elements occur in $\bS$ with multiplicity, there is not a unique such $\Bi$, but this choice won't affect the isomorphism class of $e(\bS)$.) We call $s_m$ the \textbf{longitude} of $i_m$.  (Recall that we've already defined the longitudes of red strands in Section \ref{section:TR}.)

The idempotent $e(\bS)$ is  given by interlacing according to longitude the $\ell$ red strands with the $n$ black strands.  Of course the red strands are labeled
$j_1,...,j_\ell$ from left to right, while the black strands are labeled by $i_1,...,i_n$ from left to right.
If the longitude of a red strand agrees with the longitude of a black one, then the red
strand goes to the left of the black one.  As mentioned above, we also have to make a choice of the order of black strands with the same
longitude, but reordering these gives idempotents which are isomorphic
by the obvious straight-line diagram.

Going in the other direction, given an idempotent $e\in {}_-P$ we construct $\bS$ as follows: for every black $i$-strand we add an element to $S_i$ equal to:
$$
 (\text{longitude of closest red strand to its right})-(\text{parity distance between these strands}).
 $$
 Note that such a red strand exists since we are working in ${}_-P$.  This construction  gives the unique $\bS$ such that $e(\bS)=e$ with $x(\bS)$ is maximal, where:

\begin{Definition}
\label{def:x(S)}
For $\yMon_\bR \zMon_\bS^{-1}\in \B(\bR)$ set
$
x(\bS)=\sum_{s\in S_i}s
$.
\end{Definition}

\begin{Example}
\label{example}
Consider $\g=\mathfrak{sl}_3$ and $\lambda=\varpi_1+\varpi_2$.   If we take $R_1=\{-1\}$ and
$R_2=\{2\}$, the elements of the product monomial crystal are
\[
\tikz[->,thick,xscale=1.6]{
 \node at (-4,0) (a){$\yMon_{\bR}$};
\node at (-1.5, 1) (b){$\yMon_{\bR}\zMon_{1,-3}^{-1}$};
   \node at (-1.5,-1) (c){$ \yMon_{\bR}\zMon_{2,0}^{-1}$};
  \node at (0,2) (d){$ \yMon_{\bR}\zMon_{1,-3}^{-1}\zMon_{2,-4}^{-1}$};
  \node at (0,-2) (e){$ \yMon_{\bR}\zMon_{1,-3}^{-1}\zMon_{2,0}^{-1}$};
  \node at (1.5,-1) (f){$\yMon_{\bR}\zMon_{1,-3}^{-1}\zMon_{1,-1}^{-1}\zMon_{2,0}^{-1}$};
  \node at (1.5,1) (g){$  \yMon_{\bR}\zMon_{1,-3}^{-1}\zMon_{2,-4}^{-1}\zMon_{2,0}^{-1}$};
  \node at (4,0) (h){$ \yMon_{\bR}\zMon_{1,-3}^{-1}\zMon_{1,-1}^{-1}\zMon_{2,-4}^{-1}\zMon_{2,0}^{-1}$};
\draw[->](a)--(b.west) node[midway,above]{$1$};
\draw[->](a)--(c.west) node[midway,below]{$2$};
\draw[->](c)--(e.west) node[midway,below left]{$1$};
\draw[->](b)--(d.west) node[midway,above left]{$2$};
\draw[->](e.east)--(f) node[midway,below right]{$1$};
\draw[->](d.east)--(g) node[midway,above right]{$2$};
\draw[->](g.east)--(h.north) node[midway,above]{$1$};
\draw[->](f.east)--(h.south) node[midway,below]{$2$};}
\]
The corresponding  idempotents are \[
\tikz[->,thick,xscale=1.6]{
 \node at (-4,0) (a){${\color{red}
  1}\;{\color{red} 2}$};
\node at (-1.5, 1) (b){$1\;{\color{red}
  1}\;{\color{red} 2}$};
   \node at (-1.5,-1) (c){$ {\color{red} 1}\; 2\; {\color{red} 2}$};
  \node at (0,2) (d){$2\; 1\; {\color{red}
  1}\; {\color{red} 2}$};
  \node at (0,-2) (e){$1\; {\color{red}
  1}\; 2\; {\color{red} 2}$};
  \node at (1.5,-1) (f){$1\; {\color{red}
  1}\; 1\; 2\; {\color{red} 2}$};
  \node at (1.5,1) (g){$ 2\; 1\; {\color{red}
  1}\; 2\; {\color{red} 2}$};
  \node at (4,0) (h){$ 1\; 2\; {\color{red}
  1}\; 1\; 2\; {\color{red} 2}$};
\draw[->](a)--(b.west) node[midway,above]{$1$};
\draw[->](a)--(c.west) node[midway,below]{$2$};
\draw[->](c)--(e.west) node[midway,below left]{$1$};
\draw[->](b)--(d.west) node[midway,above left]{$2$};
\draw[->](e.east)--(f) node[midway,below right]{$1$};
\draw[->](d.east)--(g) node[midway,above right]{$2$};
\draw[->](g.east)--(h.north) node[midway,above]{$1$};
\draw[->](f.east)--(h.south) node[midway,below]{$2$};}
\]
Note that all the idempotents appearing here are parity.  This is a special case of Theorem \ref{thm:highest-bijection} below.
\end{Example}

Given a simple ${}_-P$-module $L$, there is an $\bS$ with $x(\bS)$ maximal
such that $e(\bS)L\neq 0$.  We call this the {\bf highest
  weight}  of $L$.  It's
not obvious the highest weight is unique, but we will show this below.
\begin{Theorem}\label{thm:highest-bijection}
  Every simple ${}_-P$-module has a unique highest weight.  This induces a crystal isomorphism $\varphi: \Irr({}_-P) \to \B(\bR)$.  This bijection sends the simple modules that factor through
  ${}_0 {P}$ to the highest weight elements of $\B(\bR)$.
\end{Theorem}
We will prove this theorem after developing the representation theory of ${}_-P$ a bit further.
\begin{Definition}
  Given an idempotent $e(\bS)$, let $\Delta(\bS)$ be the
  quotient of ${}_-Pe(\bS)$ by
  $$

  \sum{}_-Pe(\bS'){}_-Pe(\bS)
  $$
  where the sum is over $\bS'$  such that $x(\bS')<x(\bS)$.
\end{Definition}

\begin{Lemma}\label{lem:highest-monomial}\hfill
\begin{enumerate}
\item  Suppose that $\Delta(\bS)$ is nonzero.  Then it has a unique simple quotient $L(\bS)$ which has $\bS$ as its unique highest weight.
\item There is an injective map
$\varphi:\Irr({}_-P) \to \B(\bR)$,
given by $L(\bS) \mapsto \yMon_\bR \zMon_\bS^{-1}$.
\end{enumerate}
\end{Lemma}
\newcommand{\ttop}{\mathrm{top}}
\newcommand{\bbot}{\mathrm{bot}}
\begin{proof}
  First we prove (1).  Assume that $x(\bS')=x(\bS)$.  We will
  first show that $e(\bS') \Delta(\bS)$ is zero unless $\bS'=\bS$.  Indeed, suppose $\bS'\neq\bS$ and we have a nonzero diagram $D\in e(\bS') \Delta(\bS)$.  Note that attached to a black strand $s$ in $D$ there are two pieces of additional data, namely a top longitude coming from $\bS'$, and a bottom longitude coming from $\bS$.  We'll denote these $\ttop(s)$ and $\bbot(s)$ respectively.

Since $\bS'\neq\bS$ there is some black strand $s$ in $D$ such that $\bbot(s)>\ttop(s)$.  Suppose this strand is labeled $i$.  Define $\bS''$ by
  $$
  S''_j=
  \begin{cases}
 S_i\setminus \{\bbot(s)\} \cup \{\ttop(s)\} \text{ if } j=i \\
 S_j \text{ otherwise}
  \end{cases}
  $$
Let $D'' \in e(\bS''){}_-Pe(\bS)$ be the diagram equal to $e(\bS)$ except for one black $i$-strand whose bottom longitude is $\bbot(s)$ and top longitude is $\ttop(s)$.  Then there is a diagram $D' \in  e(\bS'){}_-Pe(\bS'')$ such that in $e(\bS'){}_-Pe(\bS)$ we have
$$
D=D'D''+(\text{diagrams with fewer crossings})
$$
Since $x(\bS'')<x(\bS)$, we have that $D'D''=0$ in $\Delta(\bS)$.

Therefore, by induction, we may now
  assume that no strand in $D$ has bigger bottom longitude than top longitude.
  This is only possible if $\bS=\bS'$, contradicting our assumption.  This shows that $e(\bS') \Delta(\bS)=0$.

Note that  $e(\bS) \Delta(\bS)$ is a
  quotient of the subalgebra $A$ in $e(\bS) \tilde{P}e(\bS) $ where all strands preserve their
  longitude.  The algebra $A$ is isomorphic to a tensor product of affine nilHecke algebras.  Thus it is Morita equivalent to a tensor product of symmetric polynomials on alphabets corresponding to like-colored strands with a given longitude. The module $e(\bS) \Delta(\bS)$ is cyclic over $A$, and thus has a unique simple quotient (since the same is true of a finite dimensional graded quotient of a polynomial ring).  A standard argument as in \cite[Lem. 5.9]{Webmerged} shows that the same is true of $\Delta(\bS)$.

It remains to show that $L(\bS)$ has  $\bS$ as its unique highest weight.  But this is clear, since $e(\bS)L(\bS)\neq 0$, and $e(\bS')\Delta(\bS)=0$ for any $\bS'\neq\bS$ such that $x(\bS')\leq x(\bS)$.  The proof of (2) is now immediate since  every simple module is a quotient of some $\Delta(\bS)$.
\end{proof}

\begin{Proposition}\label{lem:crystal-map}
The map $\varphi:\Irr({}_-P)\to\B(\bR)$ is a morphism of $\fg$-crystals.
\end{Proposition}

\begin{proof}

Fix $i \in I$ and an idempotent $e(\bU) \in {}_-P$ such that $U_i=\emptyset$.  Let $e=\sum e(\bU')$, where the sum is over all idempotents such that $U_j'=U_j$ for $j\neq i$.

We wish to relate the simple ${}_-P$-modules on which this idempotent acts
non-trivially to modules over the  tensor product algebra for the
root
$\mathfrak{sl}_2$ subalgebra of $\fg$ corresponding to $i$.  We consider a
set of parameters for  $\mathfrak{sl}_2$ given by $\bR'=\{R_i\cup
\bigcup_{i\sim j} (U_j+1)\}$; note that since we are working with
$\mathfrak{sl}_2$ a set of parameters consists of a single multiset.  Let ${}_-P'$ be the $\mathfrak{sl}_2$ parity algebra
associated with  $\bR'$.  To remind
ourself which root subalgebra we're working with, we label the black and red
strands with $i$.

Let $A_\bU$ be the
subalgebra of $\tilde{P}$ spanned by diagrams where all red strands
and all black strands with labels other than $i$ are straight vertical in the position
of $e(\bU)$.

We have a natural map $A_{\bU}\to e {}_-P e$, which
makes ${}_-P e$ into a right $A_\bU$-module.
We have a ``redification'' map $\varrho_i:A_\bU \to {}_-P'$ defined on diagrams as follows:
\begin{itemize}
\item Black and red $i$-strands remain in their  positions.
\item For $j \sim i$, a dotless black $j$-strand maps to a red
  $i$-strand.
\item A dot on a black $j$-strand is sent to 0.
\item Remove any black $j$-strand if $j \not\sim i$ and $j \neq i$.
\item For $j\neq i$, remove any red $j$-strand.
\end{itemize}
Note that for a diagram in $A_\bU$, any crossing must involve a black $i$-strand.  Therefore the nilHecke relations (\ref{nilHecke-1}--\ref{black-bigon})
only involving one color are only relevant for the label $i$, and they
are sent to the same relation in ${}_-P'$.
When strands of more than one color are involved,  $\varrho_i$ sends the relations (\ref{first-QH}--\ref{second-QH}) to
(\ref{red-dot}), the relation (\ref{black-bigon}) to (\ref{cost}) and
the relations (\ref{triple-dumb}--\ref{triple-smart}) to (\ref{red-triple-correction}--\ref{dumb}).  Therefore $\varrho_i$ is indeed a map of algebras.

Thus, given a ${}_-P'$-module $M$, we can consider the associated
module $ {}_-{P} e\otimes_{A_\bU} M$.  Note that it is quite possible
that this tensor product is 0; we are free to ignore these cases.

We define a functor $\bX:{}_{-}P'\mmod\to {}_{-}P\mmod$ as follows: $\bX(M)$ is the
quotient of  $ {}_-P e\otimes_{A_\bU} M$ by the submodule generated
the image of all idempotents where the average longitude of the
black strands with labels other than $i$ is higher than $x(\bU)$.  The image $e
\mathbb{X}(M) $ is naturally an $A_\bU$-module, and
as
in the proof of Lemma \ref{lem:highest-monomial}, we see that $e
\mathbb{X}(M) $
is a quotient of $M$.  In particular, if $M$ is simple,
then applying \cite[Lem. 5.9]{Webmerged} shows that $\mathbb{X}(M)$
has a unique simple quotient $\mathbb{L}(M)$ (if it is non-zero).
One can also easily confirm that $\mathbb{L}$ commutes with the
functors $\EuScript{E}_{i},\EuScript{F}_{i}$ of the categorical
$\mathfrak{sl}_2$-action on both categories.  Thus, the induced map on
simples commutes with the categorical crystal operators for $e_{i}$
and $f_{i}$.  In particular, if a simple in the crystal is sent to
$0$, then its whole component is.

Let $\B(\bR')$ be the product monomial crystal for $\mathfrak{sl}_2$.  Note that we
have a crystal map $\mathsf{L}:\B(\bR')\to\B(\bR)$ given by $\yMon_{\bR'}\zMon_{\bT}^{-1} \mapsto \yMon_{\bR}\zMon_{\bS\cup
  \bT}^{-1}$.  Since the numbers $a_{i,k}$ are unchanged by this map, $\mathsf{L}$ is a map of $\mathfrak{sl}_2$ crystals. Thus we have a diagram of maps, where the left hand map
is dashed since it is only partially defined:
$$
\tikz[very thick]{\matrix[row sep=12mm,column sep=30mm,ampersand replacement=\&]{
\node(a){$\Irr({}_-P')$}; \& \node(c){$\B(\bR')$}; \\
\node(b){$\Irr({}_-P) $}; \&  \node(d){$\B(\bR)$};\\
};
\draw[dashed,->] (a) -- node[left,midway]{$\mathbb{L}$} (b);
\draw[->] (c) -- node[right,midway]{$\mathsf{L}$} (d);
\draw[->] (a) -- node[above,midway]{$\varphi'$} (c);
\draw[->] (b) -- node[below,midway]{$\varphi$} (d);
}
$$
This diagram commutes on objects where $\mathbb{L}$ is defined.
Both vertical maps are morphisms of $\mathfrak{sl}_2$-crystals, and
the maps $\varphi$ and $\varphi'$ are both injective.  Furthermore,
every element of $\Irr({}_-P) $ is in the image of  $\mathbb{L}$ for
some choice of $\bU$; in fact, this implies that its whole root string
under $\tilde{e}_i,\tilde{f}_i$ is.

Thus to show that $\varphi$ is a crystal map, it suffices to prove
this for $\varphi'$ for an arbitrary choice of $i$ and $\bU$.  That
is, it suffices to prove the result in the case where $\g=\mathfrak{sl}_2$.  In this case, the monomial crystal structure is simply a tensor product crystal, and ${}_-P$ is an $\mathfrak{sl}_2$ tensor product algebra (cf. Example \ref{ex:Nomega1}).  The crystals for these match by \cite[Th. 7.2]{LoWe}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:highest-bijection}]
To complete the proof of the bijection, we need to show that $\Delta(\bS)$ is non-trivial if and only if $\yMon_{\bR}\zMon_{\bS}^{-1}$ lies in $ \B(\bR) $.  We'll prove this by induction on $n=\sum_i|S_i|$.

For $n=0$ we have that $\bS=\emptyset$.  Since  $\Delta(\emptyset)=\C e(\emptyset)\neq0$ and $\yMon_\bR\in\B(\bR)$, the claim follows.

Now let $n>0$.  Suppose first that there is an element of $\bS$ which is less
than or equal to every element of $\bR$.  If $\Delta(\bS)$ is non-trivial then $L(\bS)$ is not killed
by $\cE_i$, for some $i$.  Thus, $L(\bS)=\tilde{f}_i  L(\bS')$ for some highest weight $\bS'$.  By
induction we can assume that $\yMon_\bR \zMon_{\bS'}^{-1}\in  \B(\bR) $, and hence by Proposition \ref{lem:crystal-map} $\yMon_\bR \zMon_\bS^{-1}=\tilde{f}_i (\yMon_\bR \zMon_{\bS'}^{-1})$ is in $\B(\bR)$ as well.
Conversely, if $\yMon_\bR \zMon_{\bS}^{-1} \in \B(\bR)$ then $\tilde{e}_i(\yMon_\bR \zMon_{\bS}^{-1})\neq 0$, and  reversing the argument above we get that $L(\bS)\neq0$.

It remains to consider the case where there is $r\in R_i$ strictly less than all
elements of $\bS$.  We can assume $r$ corresponds to the leftmost red
strand in $e(\bS)$.  Let $\bR'$ be the set of parameters of weight $\lambda -\varpi_i$ obtained from $\bR$ by removing $r$ from $R_i$.  Let ${}_-P'$ be the associated algebra.  We have a map ${}_-P' \to {}_-P$ given by adding a red $i$-strand on the left.  Let $e \in {}_-P$ be the sum of idempotents whose leftmost strand is red.  If $\Delta(\bS)$ is non-trivial then $eL(\bS)$ is a nontrivial ${}_-P'$-module, which has highest weight $\bS$.  Therefore some simple composition factor of this module has highest weight $\bS$.  By induction we then have that $\yMon_{\bR'} \zMon_{\bS}^{-1}\in
\B(\bR')$.  Now note that multiplication by $\yMon_{i,r}$ defines an injective set map $\iota:\B(\bR') \to \B(\bR)$.  Hence $\yMon_{\bR} \zMon_{\bS}^{-1}\in
\B(\bR)$.

Conversely, if $\yMon_{\bR} \zMon_{\bS}^{-1}\in
\B(\bR)$ then it's in the image of $\iota$.  Hence $\yMon_{\bR'} \zMon_{\bS}^{-1}\in\B(\bR')$.  By induction we then have that the irreducible ${}_-P'$-module $L'(\bS)\neq 0$.  By Frobenius reciprocity there exists a nonzero map ${}_-P\otimes_{{}_-P'}L'(\bS) \to L(\bS)$, and so $\Delta(\bS)\neq0$.

This completes the proof that the map $L(\bS) \mapsto \yMon_\bR \zMon_\bS^{-1}$ is a crystal isomorphism from the set of isomorphism classes of simple ${}_-P$-modules to $\B(\bR)$.

To complete the proof of the theorem, assume that a simple $L(\bS)$ is a highest weight element of the crystal.  This holds if and
only if
$\cE_i L(\bS)=0$ for all $i$, which holds if and only if $L(\bS)$ is killed by all
idempotents whose leftmost strand is black.  This is, of course, equivalent to
factoring through the quotient ${}_0P$.
\end{proof}

Note, this proves an interesting and non-trivial statement: there is a
natural embedding of the monomial product crystal in the tensor
product crystal for any ordering compatible with the longitudes.  This
embedding is uniquely characterized by the fact that adding a new
element of $\bR$ larger than all the longitudes of elements in $\bS$
matches tensoring with the highest weight element of a new tensor
factor.

We can extend this result to understand the sets of nilpotent simple
modules $\Irr(\tilde{T}^{\bR}) $ and
$\Irr(\tilde{P}^{\bR})$ as crystals.  In the case where
$\bR=\emptyset$, we have that $\tilde{T}^{\emptyset}$ is the original
KLR algebra, and work of Lauda and Vazirani \cite[Th. 7.4]{LV}
identifies $\Irr(\tilde{T}^{\emptyset})$ with the crystal $B(\infty)$.

Recall from \cite[Def. 5.4]{Webmerged} that we have a partial standardization
functor
\[ \mathbb{S}^{\bR; \emptyset}\colon  {}_-T^\bR \otimes \tilde{T}^{\emptyset} \mmod \to
  \tilde{T}^\bR\mmod.\]
Note that since we are using ${}_-T$, we swap left and right
everywhere compared to \cite{Webmerged}, in particular in the
definition of standard modules.  This correctly accounts for the fact that we use the opposite convention for crystal tensor product from \cite{LoWe,Webmerged}.
As argued in \cite[Th. 5.8]{Webmerged}, the map \[h\colon
   \Irr({}_-{T}^{\bR})\times \Irr(\tilde{T}^{\emptyset})\to
  \Irr(\tilde{T}^{\bR}) \qquad   h(L,L')\mapsto
\operatorname{hd}(\mathbb{S}^{\emptyset;\bR}(L\boxtimes L'))\] is a
bijection.  As discussed before, \cite{Webmerged} deals with graded
simple modules, but the same results hold for nilpotent modules by
Lemma \ref{lem:nil-grade}.
\begin{Lemma}
\label{lem:tildeP-crystal}
  The map $h$ induces a crystal isomorphism
  $\Irr(\tilde{P}^{\bR})\cong
\mathcal{B}(\bR)  \otimes \mathcal{B}(\infty)$.
\end{Lemma}
\begin{proof}
We can make sense of parity algebras where we allow $\pm\infty$ as
longitudes (for these purposes, $\infty$ is both even and odd).  Note
that the parity distance between any two strands with longitude
$\infty$ is 0.

Fix a large integer $n$, and let $\bR_+$ be obtained from $\bR$ by
adding $n$ copies of $\infty $ to each $R_i$.  Note that we have a
surjective map $\tilde{P}^{\bR}\to {}_-{P}^{\bR_+}$ by adding $n$
strands labeled by each fundamental weight at the far right of the
diagram; no black strands are allowed between these new red strands,
since the parity distance between them is 0.  Let $I$ be the kernel of
this map.  We have a surjective map
\begin{equation}\label{eq:PR-proj}
\Irr(\tilde{P}^{\bR})\to
\Irr({P}^{\bR_+})\sqcup \{0\}
\end{equation}
defined by $L\mapsto L/IL$.

Furthermore, this map is a bijection in any given weight $\mu$ for
sufficiently large $n$; thus these maps for all $n\geq 0$ unique fixed
the crystal structure on $\Irr(\tilde{P}^{\bR})$.

Straightforward modification of \cite[Th. 5.8]{Webmerged} shows that we have a natural bijection $\mathcal{B}(\bR_+)\cong \mathcal{B}(\bR)\otimes
\mathcal{B}(n\rho)$ compatible with $h$, and \cite[Th. 7.2]{LoWe} shows this is a crystal isomorphism.   Taking inverse limit as $n\to \infty$, we
obtain that $h$ is a crystal map as well.
\end{proof}

\subsection{Metric KLRW algebras}
We'll now consider a different generalization of the KLRW
algebras of \cite[\S 4]{Webmerged}.

\begin{Definition}\label{def:long}
  Consider a sequence $\Bi\in I^n$ and a function $\kappa\colon
  [1,\ell]\to [0,n]$ fixing the
  position of the red strands.   A {\bf
  (metric) longitude}  compatible with this data is a sequence $\Ba=(a_1,\dots, a_n)\in \Z^n$ such that for all $k$,
\begin{enumerate}
\item $i_k$ and $a_k$ have the same parity,
\item $a_{k}\leq a_{k+1}$, and
\item $a_k\geq  r_p$ if and only if  $k> \kappa(p)$.
\end{enumerate}
\end{Definition}
The longitude conditions say roughly that  we can isotope the
idempotent $e(\Bi,\kappa)\in\tilde{T}$ to have a black strand of label $i_k$ at
$x=a_k$, and red strands at $x=r_p$; condition (2) implies that the
black strands are in the correct order, and condition (3) implies
that red and black strands are in the correct relative position. Note
that there are some border cases that keep this from being strictly
true: consecutive black strands can have the same longitude, but
cannot have the same $x$-value, and similarly, a red strand and the
black to its right (but not its left) can have the same longitude, but
not the same $x$-value.

We let $\Long$ be the set of possible combinations of data
$(\Bi,\kappa)$ as above and the longitudes $(a_1,\dots, a_n)$.   Note that the
information of $\Ba$ is enough to uniquely fix the choice
of $\kappa$.  We can also speak of the integer this attaches to a
single strand as its longitude; the longitude of a red strand is by convention the
corresponding element $r_p$.

\begin{Definition}\label{def:Stendhal}
  A {\bf metric \Stendhal diagram} is a Stendhal diagram together with a
  choice of a longitudes $\Ba_{\operatorname{top}}$ and $\Ba_{\operatorname{bottom}}$ on the data
  $(\Bi_{\operatorname{top}},\kappa_{\operatorname{top}})$ and
  $(\Bi_{\operatorname{bottom}},\kappa_{\operatorname{bottom}})$ at the top and bottom of the diagram.
\excise{ a collection of finitely many oriented curves in
  $\R\times [0,1]$ which are colored black or red. Each curve is
  labeled with $i\in I$ and decorated with finitely many
  dots.
  The diagram must be locally of the form \begin{equation*}
\begin{tikzpicture}

  \draw[very thick] (-4,0) +(-1,-1) -- +(1,1);

  \draw[very thick](-4,0) +(1,-1) -- +(-1,1);

  \draw[very thick](0,0) +(-1,-1) -- +(1,1);

  \draw[wei, very thick](0,0) +(1,-1) -- +(-1,1);

  \draw[wei,very thick](4,0) +(-1,-1) -- +(1,1);

  \draw [very thick](4,0) +(1,-1) -- +(-1,1);

  \draw[very thick](8,0) +(0,-1) --  node
  [midway,circle,fill=black,inner sep=2pt]{}
  +(0,1);
\end{tikzpicture}
\end{equation*}
with each curve oriented in the negative direction.  In
particular, no red strands can ever cross.}
\end{Definition}
As in \cite[Def. 4.2]{Webmerged}, we can define a product on the
formal span of all metric \Stendhal diagrams considered up to isotopy.  Note that in taking this
product, we must require that longitudes in addition to labels in
$I$ match, or else we take the product to be 0.  There are idempotent diagrams
in this algebra given by straight line diagrams with the same
longitudes at the top and bottom.  The isomorphism type of this
idempotent only depends on $\Bi$ and $\kappa$, since we can label the
top and bottom of this diagram with any compatible longitude.

\begin{Definition}
\label{def:metricKLR}
  The {\bf metric KLRW algebra} $\tmetric=\tmetric^\bR$ consists of finite spans of
  metric \Stendhal diagrams, modulo the local relations
  (\ref{first-QH}--\ref{cost}).  We let $\pmmetric$ denote its quotients by
  diagrams that violate on left and right respectively, and as usual $\pmmetric_\mu \subset \pmmetric$ denote the subalgebras where the sum of the roots labeling black strands is equal to $\lambda-\mu$.
\end{Definition}

\begin{Example}
\label{ex:metricdiag}
Let $I=\{x,y\}$, connected by a single edge $x\to y$. In this case, $\g=\mathfrak{sl}_3$, and we take $\lambda=\varpi_x+\varpi_y, R_x=\{-1\}$, and $R_y=\{4\}$.  Here is a typical  element of the metric KLRW algebra $\tmetric^\bR$.
\[ \tikz[very thick,scale=1.5,baseline]{
    \draw[wei] (-.5,-.5) -- (-.5,.5);
    \draw[wei] (.6,-.5) -- (.6,.5);
    \draw (-.8,-.5) to [out=60,in=-90] (.2,.5);
    \draw (-.1,-.5) to [out=90,in=-90] (.9,.5);
    \draw (.9,-.5) to [out=90,in=-90] (-.1,.5);

  \draw (-.8,-.5)+(0,-.2) node {\small$x$};
  \draw (-.5,-.5)+(0,-.2) node {\small$x$};
  \draw (-.1,-.5)+(0,-.2) node {\small$y$};
  \draw (.6,-.5)+(0,-.2) node {\small$y$};
  \draw (.9,-.5)+(0,-.2) node {\small$x$};

  \draw (-.8,-.5)+(-.1,-.6) node {\small$-3$};
  \draw (-.5,-.5)+(0,-.6) node {\small$-1$};
  \draw (-.1,-.5)+(0,-.6) node {\small$2$};
  \draw (.6,-.5)+(0,-.6) node {\small$4$};
  \draw (.9,-.5)+(0,-.6) node {\small$5$};

   \draw (.2,.5)+(0,.2) node {\small$3$};
  \draw (-.5,.5)+(0,.2) node {\small$-1$};
  \draw (-.1,.5)+(0,.2) node {\small$-1$};
  \draw (.6,.5)+(0,.2) node {\small$4$};
  \draw (.9,.5)+(0,.2) node {\small$6$};

  \draw (.2,.5)+(0,.6) node {\small$x$};
  \draw (-.5,.5)+(0,.6) node {\small$x$};
  \draw (-.1,.5)+(0,.6) node {\small$x$};
  \draw (.6,.5)+(0,.6) node {\small$y$};
  \draw (.9,.5)+(0,.6) node {\small$y$};

  \draw (-.15,0) node [circle,fill=black,inner sep=2pt]{};

} \]
On the top and bottom we depict the string labels and longitudes in two rows.  So for instance, the bottom longitudes of the black strands in the above diagram are $-3,2,5$.  Notice that the longitude of a black strand can change from bottom to top, but the longitudes on the red strands are fixed by the choice of $\bR$.
\end{Example}

Note that a metric structure on the idempotent $e(\Bi,\kappa)$ is
precisely the same as a choice of $\bS$ such that
$e(\Bi,\kappa)=e(\bS)$. We let $d(\bS)\in\tmetric$ be the
 idempotent whose diagram is the same as $e(\bS)$, with the (metric) longitudes given by
the longitudes of $\bS$.
The key connection between the metric and parity KLR algebras is:
\begin{Lemma}\label{lem:compat-long}
  An idempotent in $\tilde{T}$ has a compatible metric longitude if and
  only if it is a parity idempotent.
\end{Lemma}

\begin{proof}
Suppose an idempotent in $\tilde{T}$ has a compatible longitude.  Let's call the difference between the metric longitude of two strands their ``metric distance''.  We claim that the metric distance is always greater than or equal to parity distance.  This will show that the idempotent is parity.

Both distances satisfy the strict triangle
  identity, so we need only check this for consecutive strands. If two
  consecutive black strands have the same parity, then the parity distance
  between them is 0.  If they have opposite parity, then their metric longitudes
  have opposite parity as well, so the metric distance between them is at
  least 1, and thus greater than or equal to the parity distance.  If
  we have a black strand and then a red of the same parity, then they
  cannot have the same metric longitude by Definition \ref{def:long}(3), so they must differ by at least 2, which is the parity
  distance.  Thus, indeed, the parity distance is a lower bound on the
  metric distance.

Conversely, given a parity idempotent, we have a corresponding monomial as described in Section \ref{sec:crystalisom}, which gives a
  choice of compatible metric longitudes.
\end{proof}

We have a natural bimodule between $\tmetric$ and
$\tilde{T}$ where we only label the top of strands with a compatible longitude, and do not choose one on the bottom.  Multiplying on the right by
$e_\bR$, we obtain a bimodule between $\tmetric$ and
$\tilde{P}$.
One can easily confirm that:
\begin{Lemma}\label{lem:Morita}
  The bimodule above induces a Morita equivalence between
  $\tmetric$ and $\tilde{P}$, and between $\pmmetric$
  and ${}_{\pm}P$.
\end{Lemma}
\begin{proof}
For each parity idempotent, we choose a single compatible longitude,
and let $e'$ be the sum of the corresponding idempotents in
$\tmetric$.  We obtain an isomorphism $\tilde{P}\to
e'\tmetric e'$ by labeling the top and bottom of each diagram
with the corresponding longitude.

On the other hand, we have that
$\tmetric e'\tmetric=\tmetric$, since any idempotent
$e\in\tmetric$ can be written as product of two
idempotents, with labels at the extremes being arbitrary
and in the center being the fixed longitudes for $e$.
\end{proof}

As with the parity algebra, a simple ${}_-\metric$-module $L$ has {\bf highest weight} $\bS$ if $x(\bS)$ is maximal such that $d(\bS)L\neq0$.
By Theorem \ref{thm:highest-bijection} we obtain:
\begin{Corollary}
\label{cor:metric-bijection}
There is a bijection, which (by slight abuse of notation) we denote $\varphi$, between the simple modules of
${}_-\metric$ and the product monomial crystal $\B(\bR)$, sending a
simple $L$ to $\yMon_\bR \zMon_\bS^{-1}$, where $\bS$ is the highest weight of $L$.
\end{Corollary}

\subsection{Coarse metric KLRW algebras}
\label{sec:coarsemetric}

We'll also want to consider a more general notion of coarse
longitudes.    These are useful for the combinatorics relating the
representation theory of \Stendhal algebras to the representations of
the KLR Yangian algebra $\BK$ we will introduce in Section
\ref{sec:cylindrical}.

We'll define a preorder on the integers by $a\succeq b$ if and only if
$\lfloor \frac{a}{2}\rfloor\geq \lfloor \frac{b}{2}\rfloor$.  This is
coarsening of the standard order, where we also have $2q\succeq
2q+1$ for any integer $q$.  We write $a\approx b$ if $a\preceq b $ and $b\preceq a$ (that
is, $a,b\in \{2p,2p+1\}$ for some integer $p$).

\begin{Definition}
\label{def:coarselongs}
Consider a sequence $\Bi\in I^n$ and a function $\kappa\colon
  [1,\ell]\to [0,n]$ fixing the
  position of the red strands.   A {\bf coarse
   longitude}  compatible with this data is a sequence $\Ba=(a_1,\dots, a_n)\in \Z^n$ which satisfies the conditions of a longitude for the
order $\succeq$, that is:
\begin{enumerate}
\item $i_k$ and $a_k$ have the same parity,
\item $a_{k}\preceq a_{k+1}$, and
\item if  $k\leq  \kappa(p)$ then $a_k< r_p$, and if $k>
  \kappa(p)$ then $a_k\succeq r_p$.
\end{enumerate}
We call $(\Bi,\kappa,\Ba)$ a {\bf coarse longitude triple}.
\end{Definition}

Note that every longitude is a coarse longitude, but not vice versa.  A {\bf coarse metric Stendhal diagram} is a Stendhal diagram together with a choice of coarse longitude triples on the top and bottom of the diagram.  As with the metric KLRW algebra, we define a product on the formal span of coarse metric Stendhal diagrams, requiring that coarse longitudes in addition to labels in $I$ must match, or else we take the product to be zero.
The {\bf coarse metric KRLW algebra} ${\TL}=\TL^\bR$ consists of finite spans of coarse metric Stendhal diagrams, modulo the local relations
  (\ref{first-QH}--\ref{cost}).

 For an example of an element of $\TL$ which is not in $\tmetric$,
 consider the diagram of Example \ref{ex:metricdiag} but with the
 bottom longitudes of the black strands changed to $-3,-2,5$:
\[ \tikz[very thick,scale=1.5,baseline]{
    \draw[wei] (-.5,-.5) -- (-.5,.5);
    \draw[wei] (.6,-.5) -- (.6,.5);
    \draw (-.8,-.5) to [out=60,in=-90] (.2,.5);
    \draw (-.1,-.5) to [out=90,in=-90] (.9,.5);
    \draw (.9,-.5) to [out=90,in=-90] (-.1,.5);

  \draw (-.8,-.5)+(0,-.2) node {\small$x$};
  \draw (-.5,-.5)+(0,-.2) node {\small$x$};
  \draw (-.1,-.5)+(0,-.2) node {\small$y$};
  \draw (.6,-.5)+(0,-.2) node {\small$y$};
  \draw (.9,-.5)+(0,-.2) node {\small$x$};

  \draw (-.8,-.5)+(-.1,-.6) node {\small$-3$};
  \draw (-.5,-.5)+(0,-.6) node {\small$-1$};
  \draw (-.1,-.5)+(0,-.6) node {\small$-2$};
  \draw (.6,-.5)+(0,-.6) node {\small$4$};
  \draw (.9,-.5)+(0,-.6) node {\small$5$};

   \draw (.2,.5)+(0,.2) node {\small$3$};
  \draw (-.5,.5)+(0,.2) node {\small$-1$};
  \draw (-.1,.5)+(0,.2) node {\small$-1$};
  \draw (.6,.5)+(0,.2) node {\small$4$};
  \draw (.9,.5)+(0,.2) node {\small$6$};

  \draw (.2,.5)+(0,.6) node {\small$x$};
  \draw (-.5,.5)+(0,.6) node {\small$x$};
  \draw (-.1,.5)+(0,.6) node {\small$x$};
  \draw (.6,.5)+(0,.6) node {\small$y$};
  \draw (.9,.5)+(0,.6) node {\small$y$};

  \draw (-.15,0) node [circle,fill=black,inner sep=2pt]{};

} \]

\subsubsection{Generators}
One reason introducing the coarse metric KLRW algebra is that its generators are easier to describe, as compared to the metric KLRW algebra.

We let $e(\Bi,\kappa,\Ba) \in \TL$ be the straight line diagram with
$(\Bi,\kappa,\Ba)$ the coarse longitude triple assigned to both
the top and bottom of the diagram.  We'll draw these diagrams with the
label $i_k$ and longitude $a_k$ written at the start and end of each
strand.  The element
$y_k(\Bi,\kappa,\Ba)\in \TL$ is obtained from $e(\Bi,\kappa,\Ba)$ by
placing a dot on the $k$th black strand.

We also have generators which cross strands; due to the presence of coarse longitudes we have to be more careful than usual with these.

The crossing $\psi_k$ of two neighbouring black strands is only
defined when $a_k\approx a_{k+1}$. In that case we assign
$(\Bi,\kappa,\Ba)$ as the bottom longitude, and
$(s_k\Bi,\kappa,s_k\Ba)$ as the top longitude.

\begin{equation*}
e(\Bi)=\,\begin{tikzpicture}[baseline,scale=1.3]
 \draw[very thick] (-3.4,0) +(0,-.5) -- +(0,.5);
  \draw[very thick] (-2,0) +(0,-.5) -- +(0,.5);
  \draw (-3.4,0) +(0.1,-.8) node {\small$i_1$};
\draw (-3.4,0) +(0.1,-1.2) node {\small$a_1$};
\draw (-3.4,0) +(0.1,1.2) node {\small$i_1$};
\draw (-3.4,0) +(0.1,.8) node {\small$a_1$};
  \draw (-2,0) +(0.1,-.8) node {\small$i_m$};
  \draw (-2,0) +(0.1,-1.2) node {\small$a_m$};
  \draw (-2,0) +(0.1,1.2) node {\small$i_m$};
  \draw (-2,0) +(0.1,.8) node {\small$a_m$};
  \draw (-2.65,0) node {$\cdots$};  \end{tikzpicture}
\qquad  y_k(\Bi)=\,\begin{tikzpicture}[baseline,scale=1.3]

 \draw[very thick] (.3,0) +(0,-.5) -- +(0,.5);
  \draw[very thick] (2,0) +(0,-.5) -- +(0,.5);
  \draw (.3,0) +(0.1,-.8) node {\small$i_1$};
\draw (.3,0) +(0.1,-1.2) node {\small$a_1$};
\draw (.3,0) +(0.1,1.2) node {\small$i_1$};
\draw (.3,0) +(0.1,.8) node {\small$a_1$};
\draw (2,0) +(0.1,-.8) node {\small$i_m$};
 \draw (2,0) +(0.1,-1.2) node {\small$a_m$};
  \draw (2,0) +(0.1,1.2) node {\small$i_m$};
  \draw (2,0) +(0.1,.8) node {\small$a_m$};
  \draw (.75,0) node {$\cdots$};
    \draw[very thick] (1.2,0) +(0,-.5) -- node [midway,circle,fill=black,inner sep=2pt]{} +(0,.5);
     \draw (1.65,0) node {$\cdots$};
      \draw (1.2,0) +(0.1,-.8) node {\small$i_k$};
 \draw (1.2,0) +(0.1,-1.2) node {\small$a_k$};
  \draw (1.2,0) +(0.1,1.2) node {\small$i_k$};
  \draw (1.2,0) +(0.1,.8) node {\small$a_k$}; \end{tikzpicture}
\qquad  \psi_k(\Bi)=\,\begin{tikzpicture}[baseline,scale=1.3]
 \draw[very thick] (4.3,0) +(0,-.5) -- +(0,.5);
  \draw[very thick] (6,0) +(0,-.5) -- +(0,.5);
  \draw (4.3,0) +(0.1,-.8) node {\small$i_1$};
\draw (4.3,0) +(0.1,-1.2) node {\small$a_1$};
\draw (4.3,0) +(0.1,1.2) node {\small$i_1$};
\draw (4.3,0) +(0.1,.8) node {\small$a_1$};
  \draw (6,0) +(0.1,-.8) node {\small$i_m$};
 \draw (6,0) +(0.1,-1.2) node {\small$a_m$};
  \draw (6,0) +(0.1,1.2) node {\small$i_m$};
  \draw (6,0) +(0.1,.8) node {\small$a_m$};
  \draw[very thick] (4.9,0) +(0,-.5) -- +(.5,.5);
  \draw[very thick] (5.4,0) +(0,-.5) -- +(-.5,.5);
\draw (4.9,0) +(0.1,-.8) node {\small$i_k$};
 \draw (4.9,0) +(0.1,-1.2) node {\small$a_k$};
  \draw (5.4,0) +(0.1,1.2) node {\small$i_k$};
  \draw (5.4,0) +(0.1,.8) node {\small$a_k$};
\draw (5.4,0) +(0.1,-.8) node {\small$i_{k+1}$};
 \draw (5.4,0) +(0.1,-1.2) node {\small$a_{k+1}$};
  \draw (4.9,0) +(0.1,1.2) node {\small$i_{k+1}$};
  \draw (4.9,0) +(0.1,.8) node {\small$a_{k+1}$};
\draw (4.75,0) node {$\cdots$};
\draw (5.72,0) node {$\cdots$};
\end{tikzpicture}
\end{equation*}

The red-black crossings are defined in two cases.  In the first case,
the rightmost black strand with longitude  $a_k\approx 2q$ for fixed
$q\in \Z$ increases
  longitude by $2$,  and moves rightward by crossing all red strands with longitude $\preceq 2q+2$. (Of course, there may be no such red strands, in which case this generator doesn't contain any crossings and just changes the longitude.)  All other longitudes remain the same.  We denote this generator by $\phi^+_k=\phi^+_k(\Bi,\kappa,\Ba)$, where $(\Bi,\kappa,\Ba)$ is the coarse longitude triple on the bottom of the diagram.
  \[ \phi^+_k=\,\tikz[very thick,scale=1.5,baseline]{\draw[wei] (-.3,-.5) -- (-.3,.5);
    \draw[wei] (.3,-.5) -- (.3,.5); \node at (0,.3){$\cdots$}; \node
    at (0,-.3){$\cdots$}; \draw (-2,-.5) -- (-2,.5); \draw (-1.3,-.5) --
    (-1.3,.5);\node at (-1.65,0){$\cdots$}; \draw (-.6,-.5) to
    [out=45,in=-135] (.6,.5); \draw (2,-.5) -- (2,.5); \draw (1.3,-.5)
    -- (1.3,.5);\node at (1.65,0){$\cdots$};
  \draw (-2,0) +(0.1,-.8) node {\small$i_1$};
\draw (-2,0) +(0.1,-1.2) node {\small$a_1$};
\draw (-2,0) +(0.1,1.2) node {\small$i_1$};
\draw (-2,0) +(0.1,.8) node {\small$a_1$};
  \draw (2,0) +(0.1,-.8) node {\small$i_m$};
 \draw (2,0) +(0.1,-1.2) node {\small$a_m$};
  \draw (2,0) +(0.1,1.2) node {\small$i_m$};
  \draw (2,0) +(0.1,.8) node {\small$a_m$};
    \draw (-1.3,0) +(0.1,-.8) node {\small$i_{k-1}$};
 \draw (-1.3,0) +(0.1,-1.2) node {\small$a_{k-1}$};
  \draw (-1.3,0) +(0.1,1.2) node {\small$i_{k-1}$};
  \draw (-1.3,0) +(0.1,.8) node {\small$a_{k-1}$};
    \draw (1.3,0) +(0.1,-.8) node {\small$i_{k+1}$};
 \draw (1.3,0) +(0.1,-1.2) node {\small$a_{k+1}$};
  \draw (1.3,0) +(0.1,1.2) node {\small$i_{k+1}$};
  \draw (1.3,0) +(0.1,.8) node {\small$a_{k+1}$};
\draw (-.6,0) +(0.1,-.8) node {\small$i_k$};
 \draw (-.6,0) +(0.1,-1.2) node {\small$a_k$};
  \draw (.6,0) +(0.1,1.2) node {\small$i_k$};
  \draw (.6,0) +(0.1,.8) node {\small$a_k+2$};
} \]

  The second case is the mirror image - it consists of the diagram
  where the leftmost strand with longitude $a_k\approx 2q$ decreases
  longitude by $2$, crossing all
red strands with longitude $\approx 2q$, and all other longitudes
remain the same.  We denote this generator by
$\phi^-_k=\phi^-_k(\Bi,\kappa,\Ba)$, where $(\Bi,\kappa,\Ba)$ is the
coarse longitude triple on the bottom of the diagram.
\[  \phi^-_k=\,\tikz[very thick,scale=1.5,baseline]{\draw[wei] (-.3,-.5)
    -- (-.3,.5); \draw[wei] (.3,-.5) -- (.3,.5); \node at
    (0,.3){$\cdots$}; \node at (0,-.3){$\cdots$}; \draw (-2,-.5) --
    (-2,.5); \draw (-1.3,-.5) -- (-1.3,.5);\node at
    (-1.65,0){$\cdots$}; \draw (.6,-.5) to [out=135,in=-45] (-.6,.5);
    \draw (2,-.5) -- (2,.5); \draw (1.3,-.5) -- (1.3,.5);\node at
    (1.65,0){$\cdots$};
  \draw (-2,0) +(0.1,-.8) node {\small$i_1$};
\draw (-2,0) +(0.1,-1.2) node {\small$a_1$};
\draw (-2,0) +(0.1,1.2) node {\small$i_1$};
\draw (-2,0) +(0.1,.8) node {\small$a_1$};
  \draw (2,0) +(0.1,-.8) node {\small$i_m$};
 \draw (2,0) +(0.1,-1.2) node {\small$a_m$};
  \draw (2,0) +(0.1,1.2) node {\small$i_m$};
  \draw (2,0) +(0.1,.8) node {\small$a_m$};
    \draw (-1.3,0) +(0.1,-.8) node {\small$i_{k-1}$};
 \draw (-1.3,0) +(0.1,-1.2) node {\small$a_{k-1}$};
  \draw (-1.3,0) +(0.1,1.2) node {\small$i_{k-1}$};
  \draw (-1.3,0) +(0.1,.8) node {\small$a_{k-1}$};
    \draw (1.3,0) +(0.1,-.8) node {\small$i_{k+1}$};
 \draw (1.3,0) +(0.1,-1.2) node {\small$a_{k+1}$};
  \draw (1.3,0) +(0.1,1.2) node {\small$i_{k+1}$};
  \draw (1.3,0) +(0.1,.8) node {\small$a_{k+1}$};
\draw (.6,0) +(0.1,-.8) node {\small$i_k$};
 \draw (.6,0) +(0.1,-1.2) node {\small$a_k$};
  \draw (-.6,0) +(0.1,1.2) node {\small$i_k$};
  \draw (-.6,0) +(0.1,.8) node {\small${a_k-2}$};
} \]

\subsubsection{Polynomial representation}
Later in the paper, we will need to use the polynomial representation of this algebra, which is a little complicated since we need many sets of variables.

The polynomial representation of $\TL$ is given by
\begin{equation*}
\label{eq:PolTL}
\PolKLR_{\mathcal{L}}=
\bigoplus_{(\Bi,\Ba,\kappa)} \PolKLR(\Bi,\Ba,\kappa), \quad \text{where}\quad \PolKLR(\Bi,\Ba,\kappa)=\C[Y_1(\Bi,\Ba,\kappa),\dots Y_n(\Bi,\Ba,\kappa)]
\end{equation*}
where the sum is over all coarse longitude triples.  The action is given by the formulae (\ref{eq:T-action1}--\ref{eq:T-action2}).  More precisely, $x \in e(\Bi',\Ba',\kappa')\TL e(\Bi,\Ba,\kappa)$ defines an operator
$$
x:\PolKLR(\Bi,\Ba,\kappa) \to \PolKLR(\Bi',\Ba',\kappa')
$$
which acts by ignoring longitudes, acting by (\ref{eq:T-action1}--\ref{eq:T-action2}), and then replacing the correct longitudes in the variables.
Since $\Pol$ is a faithful $\tilde{T}^\blam$-module for any $\blam$,
it follows  that $\PolKLR_{\mathcal{L}}$ is a faithful $\TL$-module.
In fact, the modules $\PolKLR_{\mathcal{L}}$ and $\Pol$ are identified
under the Morita equivalence below.

\subsubsection{An equivalence}
It turns out that (unlike the metric KLRW algebra or parity KLRW algebra) the coarse metric KLRW algebra is always Morita equivalent to a usual KLRW algebra, for a sequence $ \boldsymbol{\lambda}$  which depends on $ \bR $.

Recall that to $\bR$ we associate multiplicity functions $\rho_i:\Z\to\N$ (cf. Section \ref{section:TR}).  For $q\in\Z$ define $\lambda^{(q)}=\sum_i\rho_i(q)\varpi_i$, and set $\boldsymbol{\lambda}=(...,\lambda^{(q-1)},\lambda^{(q)},...)$.  Note that generically, $ \boldsymbol{\lambda} = \varpi_\bR $, but in general it is different.

There is a natural inclusion
$\tilde{T}^{\blam}\to\tilde{T}$, which splits the red strand labeled $\lambda^{(q)}$ into $\sum_i\rho_i(q)$ red strands with labels from left to right given by the sequence (\ref{eq:varpiq}).  Observe that the idempotents in the image of this map will never have a black strand between two red strands
\begin{itemize}
\item where both red strands have longitude $2q$, or
\item where both red strands have longitude $2q+1$, or
\item where the left red strand has longitude $2q$ and right one has longitude $2q+1$.
\end{itemize}

\begin{Proposition}
\label{prop:coarselongs}
  The algebras $\TL$ and
  $\tilde{T}^{\boldsymbol{\lambda}}$ are Morita equivalent.
\end{Proposition}
\begin{proof}
As in the proof of Lemma \ref{lem:Morita}, our desired result reduces to showing
that the idempotents that carry coarse longitudes are precisely
those isomorphic to ones in the image of $\tilde{T}^{\boldsymbol{\lambda}}$.

Given an idempotent with a coarse longitude, the only obstruction to
 it lying in $\tilde{T}^{\boldsymbol{\lambda}}$ is if one of the three cases above occurs.  First notice that the first one cannot occur since that would not support a coarse longitude.  In the latter two cases, there must be a black strand of longitude $2q$ in between the red strands. Crossing this black
 strand over the right red strand is irrelevant since the strands have
 opposite parities.  Thus, this idempotent is isomorphic to one in
 $\tilde{T}^{\boldsymbol{\lambda}}$.

On the other hand, given an idempotent in
$\tilde{T}^{\boldsymbol{\lambda}}$, we'll show that it carries a coarse longitude.  Consider a black strand between two red strands with longitudes $r_p$ and $r_{p+1}$.  Note that $|r_p-r_{p+1}|\neq 0$.  If $|r_p-r_{p+1}|=1$ then we must have $r_{p}=2q-1$ and $r_{p+1}=2q$ for some $q$.  In this case we let the longitude of the black strand be whichever
element of  $\{2q-2,2q-1\}$ has the correct parity.  If $|r_p-r_{p+1}|>1$ then we can just choose a longitude $a$ of the correct parity such that $r_p \leq a \leq r_{p+1}$.  Clearly it's possible to do this for all black strands so that the coarse longitude conditions are met.
\end{proof}

\excise{It's easy to check that up to irrelevant crossings, any idempotent where there is no black strand
between red strands with $r_i'=r_j'$ has a coarse longitude: we
simply label each black strand so that its coarse longitude is
equal to that of the closest congruent strand to its left (which is
thus
the same as the coarse longitude of the closest congruent red
strand); if there is a black strand left of all reds; our
observation above shows that this is the same as
$\tilde{T}^{\boldsymbol{\lambda}}$ where
$\boldsymbol{\lambda}=(\la_1,\dots, \la_{\# R'})$ is given by $\la_1$
being the sum of fundamental weights attached to the smallest element
of $R'$, and $\la_2$ the sum for the second smallest, etc. where
  }
\excise{
Consider the deformation of ${\mathbb T}$ where we adjoin a formal variable $h$ a formal variable and deform the relation \eqref{cost} by adding $+\nu(r)h$ for the red line at $x=r$ for some chosen bijection of $\nu\colon \bR\to [1,\ell]$.  For generic $h$, this is isomorphic to  $T^{\bR_h}$ where we replace $r\in R_i$ with $r+\nu(r)h$.
If $y_{\bR}z_{\bS}^{-1}$ lies in the monomial crystal, then we can choose a map $\delta\colon \bS\to \bR$ such that $y_{i,r}z_{\delta^{-1}(r)}^{-1}$ lies in the fundamental monomial crystal for each $r\in S_i$, and we let $\bS_h$ be the deformation where we replace $s$ by $s+\nu\circ \delta(s(h))$.
\begin{Lemma}\label{lem:deform}
  The $\C[h]$-module ${\mathbb T}$ is a flat deformation of $Ty$ and $M(\bS)$ deforms flatly over this algebra, with generic fiber isomorphic to $\oplus_{\delta} M(\bS_h)$.
\end{Lemma}}


TITLE: $\lim_{n \to \infty} (-1)^n !ne+ (-1)^{(n+1)} n! = 0 \: ?$
QUESTION [2 upvotes]: $\lim_{n \to \infty} (-1)^n  !ne+ (-1)^{(n+1)} n! = 0 \: ?$
$!n$ is the subfactorial. Here's a plot made with wolfram:

Maybe the function goes to $0$ at infinity. How to prove this analytically? I wanted to have a better understanding of how this function behaves. If we changed $e$ to $2.7182$, which is close to the actual value of $e$ the graph changes to:

So it's only going to $0$ because of $e$?

REPLY [1 votes]: Using that
$$!n=n!\sum_{k=0}^n \frac{(-1)^k}{k!},$$
we have
$$!n=\frac{n!}{e}-n!\sum_{k=n+1}^{\infty} \frac{(-1)^k}{k!}.$$
So
$$n!-e!n=en!\sum_{k=n+1}^{\infty} \frac{(-1)^k}{k!}=e\sum_{k=n+1}^{\infty} \frac{n!(-1)^k}{k!}.$$
Think back to the proof that $e$ is irrational. Can you use similar logic to show that this last sum tends to $0$ as $n$ grows large? Can you use this to conclude that your original limit tends to $0$?
For your second question, think about what happens when we replace $e$ with some number very close to $e$. How does this differ from this limit?