Integral Basis Theorem of cyclotomic Khovanov-Lauda-Rouquier Algebras of type A

Ge Li

School of Mathematics and Statistics

University of Sydney

Sydney, NSW 2006

In this paper we prove that the cyclotomic Khovanov-Lauda-Rouquier algebras in type A, $\R$, are $\Z$-free. We then extend the graded cellular basis of $\R$ constructed by Hu and Mathas to $\Rn$ and use this basis to give a classification of all irreducible $\Rn$-modules.

§ INTRODUCTION

Khovanov and Lauda [13, 12] and Rouquier [23] have introduced a remarkable new family of algebras $\mathscr R_n$, the quiver Hecke algebras, for each oriented quiver. They showed that these algebras categorify the positive part of the enveloping algebras of the corresponding quantum groups. The algebras $\mathscr R_n$ are naturally $\Z$-graded. Varagnolo and Vasserot [24] proved that, under this categorification, the canonical basis of the positive part of the quantum group corresponds to the image of the projective indecomposable modules in the Grothendieck rings of the quiver Hecke algebras when the Cartan matrix is symmetric.

The algebra $\Rn$ is infinite dimensional and for every highest weight vector in the corresponding Kac-Moody algebra there is an associated finite dimensional 'cyclotomic quotient' $\R$ of $\Rn$. The cyclotomic quiver algebras $\R$ were originally defined by Khovanov and Lauda [13, 12] and Rouquier [23] who conjectured that these algebras should categorify the irreducible representations of the corresponding quantum group. Lauda and Vazirani [18] proved that, up to shift, the simple $\Rn$-modules are indexed by the vertices of the corresponding crystal graph, and Kang and Kashiwara [11] proved the full conjecture by showing that the images of the projective irreducible modules in the Grothendieck ring $Rep(\R)$ correspond to the canonical basis of the corresponding highest weight module. Prior to this work, Brundan and Stropple [6] proved this conjecture in the special the case when $\Lambda$ is a dominat weight of level $2$ and $\Gamma$ is the linear quiver and Brundan and Kleshchev [4] established the conjecture for all $\Lambda$ when $\Gamma$ is a quiver of type $A$.

Let $\Gamma$ be the quiver of type $A_e$, for $e\in\{0,2,3,4,\dots\}$. Brundan and Kleshchev [3] proved that every degenerate and non-degenerate cyclotomic Hecke algebra $H^\Lambda_n$ of type $G(r,1,n)$ over a field is isomorphic to a cyclotomic quiver Hecke algebra $\R$ of type $A$. They did this by constructing an explicit
isomorphisms between these two algebras.

The algebras $\R$ are defined by generators and relations and so these algebras are defined over any integral domain. Hu and Mathas [9] defined a homogeneous basis $\{\psi_{\s\t}\}$ of the cyclotomic quiver algebras $\R$  (see <ref> below), and they showed that $\R$ is $\Z$-free whenever $e=0$ or $e$ is invertible in the ground ring. They asked whether the algebra $\R$ is always $\Z$-free. Kleshchev-Mathas-Ram [14] defined $\Z$-free Specht modules for the cyclotomic KLR algebras of type A (and the affine KLR algebras of type A), but that the existence of these modules does not imply that the cyclotomic KLR algebras are torsion free. The main result of this paper shows that this is always the case. More precisely, we prove the following.

Let $\R(\Z)$ be a cyclotomic Khovanov-Lauda-Rouquier algebra of type $A$ over $\Z$, where $\Lambda$ is a dominant weight of height $\ell$. Then $\R(\Z)$ is a graded cellular algebra, with respect to the dominance order, with homogeneous cellular basis
\{ \psi_{\s\t}\ |\ \lambda\in\mathscr P_n^\Lambda\text{ and $\s,\t\in\Std(\lambda)$}\}.
In particular, $\R(\Z)$ is $\Z$-free of rank $\ell^nn!$.

If $\mathscr O$ is any integral domain then $\R(\mathscr O)\cong\R(\Z)\otimes_\Z\mathscr O$, so it follows that $\R(\mathscr O)$ is free over $\mathscr O$.

The proof of our main theorem is long and technical, requiring a delicate multistage induction. Fortunately, by <cit.> we may assume that $e\ne2$. Even though our arguments should apply in this case, being able to assume that $e\ne2$ dramatically simplifies our arguments because the quiver of type $A_e$ is simply laced when $e\ne2$.

The starting point for our arguments is the observation that the definition of Hu and Mathas' the homogeneous elements $\psi_{\s\t}$ makes sense over any ring. Consequently, the linearly independents elements $\{\psi_{\s\t}\}$ span a $\Z$-free submodule $R^\Lambda_n$ of $\R$. To prove our Main Theorem it is therefore enough to show that $R^\Lambda_n$ is closed multiplication by the generators of $\R$ and that the identity element of $\R$ belongs to $R^\Lambda_n$.

The algebra $\R$ is generated by elements $y_r$, $\psi_s$ and $e(\bi)$, where $1\le r\le n$, $1\le s<n$ and $\bi\in(Z/e\Z)^n$. To prove our Main Theorem we need to treat these three classes of generators separately. The cellular basis element $\psi_{\s\t}$ is indexed by two standard $\lambda$-tableaux where $\lambda$ is a multipartition of $n$; the definitions of these terms are recalled in Section 1. We argue by simultaneous induction on $n$, and on the lexicographic orderings on the set of multipartitions, to show that multiplication by the KLR generators always sends $\psi_{\s\t}$ to a $\Z$-linear combination of terms $\psi_{\u\v}$ which are larger in the lexicographic order. Multiplication by $y_r$ is the hardest case, partly because once this case is understood it can be used to describe the action of $\psi_r$ and $e(\bi)$ on the $\psi$-basis of $\R$.

As a consequence of <ref>, we obtain a graded cellular basis of $\R(\Z)$. We then extend this basis to obtain a graded cellular basis of $\Rn$, and hence $\Rn$ is a graded cellular algebra. Hence we can use similar arguments to Graham-Lehrer [7] to give a complete set of non-isomorphic graded irreducible $\Rn$-modules. Koenig and Xi [17] introduced the notion of affine cellular algebras and they have shown that the affine Hecke algebra of type A is an affine cellular algebra. The work of Koenig and Xi predates the KLR grading, but nonetheless it gives a classification of the irreducible representation of affine Hecke algebras. Our approach is very different to that of Koenig and Xi in that, first, we incorporate the grading and, secondly, we obtain a new labeling of the irreducible representations that is compatible the labeling of the irreducible modules of the cyclotomic quotients.

In more detail, this paper is organized as follows. In Section 1 we summarise the background material from the representation theory of the cyclotomic Khovanov-Lauda-Rouquier algebras that we need, including the theory of (graded) cellular algebras and the combinatorics of multipartitions and tableaux. In Section 2 considers the special case where $\lambda$ is a multicomposition that has at most two rows. Once this case is understood we show for an arbitrary multipartition $\lambda$ that $\psi_{\t^\lambda\t^\lambda}y_r$ is a $\Z$-linear combination of higher terms, where $\t^\lambda$ is the `initial' $\lambda$-tableau. Section 3 begins by proving, again by induction, that $\psi_{\s\t}y_r$ is a linear combination of bigger terms in $R^\Lambda_n$. By considering the Garnir tableau of two-rowed multipartition we then show that $\psi_{\s\t}\psi_r$ can be written in the required form. This result is then extended to multipartitions of arbitrary shape. Finally, we deduce that $e(\bi)\in R^\Lambda_n$, for all $\bi\in (\Z/e\Z)^n$, which completes the proof of our main result. In Section 4 we define a sequence of weights $\inftyweight$ and using it to extend the graded cellular basis of $\Ri$ to $\Rni$ and hence define a graded cellular basis for $\Rn$. By adapting the arguments of Graham-Lehrer [7] we give a complete set of non-isomorphic graded simple $\Rn$-modules.

Finally, we remark that the calculations in Sections 2 and 3 gives an algorithm inductively for computing $\psi_{\s\t}y_r$ and $\psi_{\s\t}\psi_r$. We extend the techniques developed in this paper to give a KLR grading of the Brauer algebras [19].

Acknowledgements. I would like to thank Professor Andrew Mathas for his support and comments on this paper. I also want to thank Matthew Gibson, who gave me advice on writing paper in English. Further thanks go to Jun Hu, Jonathan Brundan and Alexander Kleshchev for many helpful discussions with me about cyclotomic Khovanov-Lauda-Rouquier algebras.

§ KHOVANOV-LAUDA-ROUQUIER ALGEBRAS

In this section we are going to introduce the necessary background for our work. First we will define our principal object of study — the (cyclotomic) Khovanov-Lauda-Rouquier algebras $\R$. Then we give a brief introduction to (graded) cellular algebras and symmetric groups. Finally after explaining tableaux combinatorics we describe a graded cellular basis for the cyclotomic KLR algebra, found by Hu and Mathas [9].

§.§ The cyclotomic Khovanov-Lauda-Rouquier algebras

Fix an integer $e\in \{0,2,3,4\ldots\}$ and $I = \Z/e\Z$. Let $\Gamma_e$ be the oriented quiver with vertex set $I$ and directed edges $i\rightarrow i+1$, for $i\in I$. Thus, $\Gamma_e$ is the quiver of type $A_{\infty}$ if $e = 0$ and if $e \geq 2$ then it is a cyclic quiver of type $A_e^{(1)}$:

[scale=0.8,decoration=curveto, markings,
mark=at position 0.6 with >
(-1.7,-0.7) rectangle (1.7,0.7);
in 0, 180
[ball color=blue] (:1cm) circle(4pt);

every node=[font=]
[postaction=decorate] (0:1cm) .. controls (90:5mm) .. (180:1cm)
[postaction=decorate] (180:1cm) .. controls (270:5mm) .. (0:1cm)

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mark=at position 0.6 with >
(-1.7,-0.7) rectangle (1.7,1.4);
in 90,210,330
[ball color=blue] (:1cm) circle(4pt);

every node=[font=]
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[postaction=decorate](210:1cm)–(330:1cm) node[below right]$1$;
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mark=at position 0.6 with >
(-1.7,-0.7) rectangle (1.7,0.7);
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every node=[font=]
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[postaction=decorate] (315:1cm) – (45:1cm) node[above right]$2$;
[postaction=decorate] (45:1cm) – (135:1cm) node[above left]$3$;

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mark=at position 0.6 with >
(-1.7,-0.7) rectangle (1.7,0.7);
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[postaction=decorate] (18:1cm) – (90:1cm)
[postaction=decorate] (90:1cm) – (162:1cm) node[above left]$5$;

$e=2$   $e=3$   $e=4$   $e=5$

Let $(a_{i,j})_{i,j\in I}$ be the symmetric Cartan matrix associated with $\Gamma_e$, so that

\begin{equation*} \label{notation: cartan}
a_{i,j} = \begin{cases}
2, & \text{ if $i = j$,}\\
0, & \text{ if $i\neq j \pm 1$,}\\
-1, & \text{ if $e\neq 2$ and $i = j \pm 1$,}\\
-2, & \text{ if $e = 2$ and $i = j + 1$.}
\end{cases}
\end{equation*}

To the quiver $\Gamma_e$ attach the standard Lie theoretic data of a Cartan matrix $(a_{ij})_{i,j\in I}$, fundamental weights $\{\Lambda_i| i\in I\}$, positive weights $P_+ = \sum_{i\in I}\N \Lambda_i$, positive roots $Q_+ = \bigoplus_{i\in I}\N\alpha_i$ and let $(\cdot,\cdot)$ be the bilinear form determined by

(\Lambda_i,\alpha_j)=\delta_{ij},\qquad\text{for }i,j\in I.

Fix a weight  $\Lambda = \sum_{i\in I} a_i \Lambda_i \in P_+$. Then $\Lambda$ is a weight of
level $l(\Lambda) = \l = \sum_{i\in I} a_i$. A multicharge  for $\Lambda$ is a sequence
$\kappa_\Lambda = (\kappa_1,\ldots,\kappa_\l) \in I^\l$ such that

(\Lambda,\alpha_i) = a_i = \#\set{ 1\leq s\leq \l|\kappa_s \equiv i\pmod{e}}

for any $i\in I$.

The following algebras were introduced by Khovanov and Lauda and Rouquier who defined KLR algebras for arbitrary oriented quivers.

[Khovanov and Lauda [13, 12] and Rouquier [23]]
Suppose $\mathscr O$ is an integral ring and $n$ is a positive integer. The Khovanov-Lauda–Rouquier algebra, $\mathscr{R}_n(\mathscr O)$ of type $\Gamma_e$ is the unital associative $\mathscr O$-algebra with generators

$$\{\hat \psi_1,\dots,\hat \psi_{n-1}\} \cup
\{ \hat y_1,\dots,\hat y_n \} \cup \set{\hat e(\bi)|\bi\in I^n}$$

and relations

\begin{align}
\hat e(\bi) \hat e(\bj) &= \delta_{\bi\bj}\hat e(\bi),
&{\textstyle\sum_{\bi \in I^n}} \hat e(\bi)&= 1,\label{alg:trivial1}\\
\hat y_r \hat e(\bi) &= \hat e(\bi) y_r,
&\hat \psi_r \hat e(\bi)&= \hat e(s_r{\cdot}\bi) \hat \psi_r,
&\hat y_r \hat y_s &= \hat y_s \hat y_r,\label{alg:trivial2}
\end{align}

\begin{align}
\hat \psi_r \hat y_s  &= \hat y_s \hat \psi_r,&\text{if }s \neq r,r+1,\label{alg:trivial3}\\
\hat \psi_r \hat \psi_s &= \hat \psi_s \hat \psi_r,&\text{if }|r-s|>1,\label{alg:trivial4}
\end{align}

\begin{align}
\hat \psi_r \hat y_{r+1} \hat e(\bi) &= \begin{cases}
(\hat y_r\hat \psi_r+1)\hat e(\bi),\hspace*{18mm} &\text{if $i_r=i_{r+1}$},\\
\hat y_r\hat \psi_r \hat e(\bi),&\text{if $i_r\neq i_{r+1}$}
\end{cases} \label{alg:psi-y com}\\
\hat y_{r+1} \hat \psi_r \hat e(\bi) &= \begin{cases}
(\hat \psi_r \hat y_r+1) \hat e(\bi),\hspace*{18mm} &\text{if $i_r=i_{r+1}$},\\
\hat \psi_r \hat y_r \hat e(\bi), &\text{if $i_r\neq i_{r+1}$}
\end{cases}\label{alg:y-psi com}\\
\hat \psi_r^2 \hat e(\bi) &= \begin{cases}
0,&\text{if $i_r = i_{r+1}$},\\
\hat e(\bi),&\text{if $i_r \ne i_{r+1}\pm1$},\\
(\hat y_{r+1}-\hat y_r)\hat e(\bi),&\text{if  $e\ne2$ and $i_{r+1}=i_r+1$},\\
(\hat y_r - \hat y_{r+1})\hat e(\bi),&\text{if $e\ne2$ and  $i_{r+1}=i_r-1$},\\
(\hat y_{r+1} - \hat y_{r})(\hat y_{r}-\hat y_{r+1}) \hat e(\bi),&\text{if $e=2$ and
\end{cases}\label{alg:psipsi}\\
% \end{align*}
% \vskip-36pt
\hat \psi_{r}\hat \psi_{r+1} \hat \psi_{r} \hat e(\bi) &= \begin{cases}
(\hat \psi_{r+1} \hat \psi_{r} \hat \psi_{r+1} +1)\hat e(\bi),\hspace*{7mm}
&\text{if $e\ne2$ and $i_{r+2}=i_r=i_{r+1}-1$},\\
(\hat \psi_{r+1} \hat \psi_{r} \hat \psi_{r+1} -1)\hat e(\bi),
&\text{if $e\ne2$ and $i_{r+2}=i_r=i_{r+1}+1$},\\
\big(\hat \psi_{r+1} \hat \psi_{r} \hat \psi_{r+1} +\hat y_r\\
\qquad -2\hat y_{r+1}+\hat y_{r+2}\big)\hat e(\bi),
&\text{if $e=2$ and $i_{r+2}=i_r=i_{r+1}+1$},\\
\hat \psi_{r+1} \hat \psi_{r} \hat \psi_{r+1} \hat e(\bi),&\text{otherwise.}
\end{cases}\label{alg:braid}
\end{align}

for $\bi,\bj\in I^n$ and all admissible $r$ and $s$. Moreover, $\mathscr{R}_n(\mathscr O)$ is naturally
$\Z$-graded with degree function determined by

$$\deg \hat e(\bi)=0,\qquad \deg \hat y_r=2\qquad\text{and}\qquad \deg
\hat \psi_s \hat e(\bi)=-a_{i_s,i_{s+1}},$$

for $1\le r\le n$, $1\le s<n$ and $\bi\in I^n$.

Notice that the relations depend on the quiver $\Gamma_e$. By <cit.>, if $\mathscr O$ is a commutative integral domain and suppose either $e = 0$, $e$ is non-zero prime, or that $e{\cdot}1_{\mathscr O}$ is invertible in $\mathscr O$, $\mathscr R_n^\Lambda(\mathscr O)$ is an $\mathscr O$-free algebra.

Following Khovanov and Lauda [13], we will frequently use
diagrammatic analogues of the relations of $\mathscr{R}_n(\mathscr O)$ in order
to simplify our calculations. To do this we associate to each generator
of $\mathscr{R}_n(\mathscr O)$ an $I$-labelled decorated planar diagram on $2n$ points in
the following way:

\begin{braid}\tikzset{baseline=7mm}
\draw (0,4)node[above]{$i_1$}--(0,0);
\draw (1,4)node[above]{$i_2$}--(1,0);
\draw[dots] (1.2,4)--(3.8,4);
\draw[dots] (1.2,0)--(3.8,0);
\draw (4,4)node[above]{$i_{n}$}--(4,0);
\end{braid},
\quad \psi_re(\bi)=
\begin{braid}\tikzset{baseline=7mm}
\draw (0,4)node[above]{$i_1$}--(0,0);
\draw[dots] (0.2,4)--(1.8,4);
\draw[dots] (0.2,0)--(1.8,0);
\draw (2,4)node[above]{$i_{r-1}$}--(2,0);
\draw (3,4)node[above]{$i_r$}--(4,0);
\draw (4,4)node[above]{$i_{r+1}$}--(3,0);
\draw (5,4)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i_{n}$}--(7,0);
\end{braid},
\quad\text{and}\quad
y_se(\bi) =
\begin{braid}\tikzset{baseline=7mm}
\draw (0,4)node[above]{$i_1$}--(0,0);
\draw[dots] (0.2,4)--(1.8,4);
\draw[dots] (0.2,0)--(1.8,0);
\draw (2,4)node[above]{$i_{s-1}$}--(2,0);
\draw (3,4)node[above]{$i_s$}--(3,0);
\node[greendot] at (3,2){};
\draw (4,4)node[above]{$i_s$}--(4,0);
\draw[dots] (4.2,4)--(5.8,4);
\draw[dots] (4.2,0)--(5.8,0);
\draw (6,4)node[above]{$i_{n}$}--(6,0);
\end{braid},

for $\bi\in I^n$, $1\le r<n$ and $1\le s\le n$. The $r$-th string of the diagram is the string labelled with $i_r$.

Diagrams are considered up to isotopy, and multiplication of diagrams is given
by concatenation, subject to the relations (<ref>)–(<ref>). In
more detail, if $D_1$ and $D_2$ are two diagrams then the diagrammatic analogue of the relation
$e(\bi)e(\bj)=\delta_{\bi\bj}e(\bi)$ is

$$D_1 \cdot D_2 =
\begin{tikzpicture}[baseline=9mm,blue,line width=1pt,scale=0.4,
draw/.append style={rounded corners},
every node/.append style={font=\fontsize{5}{5}\selectfont}]%
% D_1
\draw[thick](-0.4,6)rectangle(4.4,3.5)node[midway]{$D_1$};
\draw (0,3.5)node[below right=-1mm]{$i_1$}--(0,3);
\draw (1,3.5)node[below right=-1mm]{$i_2$}--(1,3);
\draw (4,3.5)node[below right=-1mm]{$i_n$}--(4,3);
% D_2
\draw[dotted](0,3)--(0,2);
\draw[dotted](1,3)--(1,2);
\draw[dotted](4,3)--(4,2);
\draw (0,2)--(0,1.5)node[above right=-1mm]{$j_1$};
\draw (1,2)--(1,1.5)node[above right=-1mm]{$j_2$};
\draw (4,2)--(4,1.5)node[above right=-1mm]{$j_n$};
\draw[thick](-0.4,1.5)rectangle(4.4,-1)node[midway]{$D_2$};
% D_1 * D_2
\end{tikzpicture}
\begin{tikzpicture}[baseline=11mm,blue,line width=1pt, scale=0.4,
draw/.append style={rounded corners},
every node/.append style={font=\fontsize{5}{5}\selectfont}]%
\draw[thick](-0.4,6)rectangle(4.4,3.5)node[midway]{$D_1$};
\draw (0,3.5)--(0,3)node[right=-1mm,pos=0.9]{$i_1$};
\draw (1,3.5)--(1,3)node[right=-1mm,pos=0.9]{$i_2$};
\draw (4,3.5)--(4,3)node[right=-1mm,pos=0.9]{$i_n$};
\draw (0,3)--(0,2.5);
\draw (1,3)--(1,2.5);
\draw (4,3)--(4,2.5);
\draw[thick](-0.4,2.5)rectangle(4.4,0)node[midway]{$D_2$};
\end{tikzpicture}

That is, $D_1\cdot D_2=0$ unless the labels of the strings on the bottom of $D_1$
match the corresponding labels on the top of the strings in $D_2$ in which case
we just concatenate the two diagrams.

Multiplication by $y_r$ simply adds a decorative dot to the $r$-th string, reading left to right, so relations (<ref>)–(<ref>)
become self when written in terms of diagrams. Ignoring the extraneous strings on the left and right, and setting $i=i_r$ and $j=i_{r+1}$, the diagrammatic analogue of relations (<ref>) and (<ref>) is

\begin{equation}\label{dia:y-psi com}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (3,3){};
\end{braid}
\ \  -\ \
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (1,1){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (2,4)node[above]{$i$}--(2,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (3,1){};
\end{braid}
\ \ -\ \
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (1,3){};
\end{braid}.
\end{equation}

Similarly, if $e\ne2$ then relation (<ref>) becomes

\begin{align}
\begin{braid}
\draw (0,4)node[above]{$i$}--(2,2)--(0,0);
\draw (2,4)node[above]{$j$}--(0,2)--(2,0);
\end{braid} &\ \  =\ \
\begin{cases}
\ \ \ \ \ \ \ 0, & \text{if $i = j$,}\\
\ \ \ \begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (2,4)node[above]{$j$}--(2,0);
\end{braid}, & \text{if $i\neq j\pm 1$,}\\
\ \ \ \pm \begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (2,4)node[above]{$j$}--(2,0);
\node[greendot] at (2,2){};
\end{braid}
\ \ \mp\ \
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (2,4)node[above]{$j$}--(2,0);
\node[greendot] at (0,2){};
\end{braid}, & \text{if $j = i\pm1$.}
\end{cases}\label{dia:psipsi}
\end{align}

and if $e\ne2$ then the diagrammatic analogue of relation (<ref>) is

\begin{align}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (2,4)node[above]{$j$}--(4,2)--(2,0);
\draw (4,4)node[above]{$k$}--(0,0);
\end{braid}
\ \ -\ \
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (2,4)node[above]{$j$}--(0,2)--(2,0);
\draw (4,4)node[above]{$k$}--(0,0);
\end{braid}
&\ \ =\ \
\delta_{i,k}(\delta_{i,j+1} - \delta_{i,j-1})\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (2,4)node[above]{$j$}--(2,0);
\draw (4,4)node[above]{$k$}--(4,0);
\end{braid}.
\end{align}

Using the relations in $\mathscr{R}_n(\mathscr O)$ it is easy to verify the following identity which we record for future use:

\begin{equation}
\hat e(\bi)\hat y_r^k \hat y_{r+1}^k \hat \psi_r = \hat e(\bi)\hat \psi_r \hat y_r^k \hat y_{r+1}^k \label{alg:ypsi}
\end{equation}

for any $\bi$. Clearly it is enough to prove this relation when $k=1$ when,
diagrammatically, this identity takes the form

\begin{equation}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$j$}--(0,0);
\node[greendot] at (1,1){};
\node[greendot] at (3,1){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$j$}--(0,0);
\node[greendot] at (1,3){};
\node[greendot] at (3,3){};
\end{braid}\label{dia:ypsi}
\end{equation}

locally on the $r$ and $r+1$-th strings and where we set $i=i_r$ and $j=i_{r+1}$.

Three more easy, and very useful, consequences of the relations are the

\begin{align}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (4,4)node[above]{$i$}--(4,0);
\end{braid}
\draw (0,4)node[above]{$i$}--(1,3)--(1,1)--(0,0);
\draw (1,4)node[above]{$i$}--(0,3)--(0,1)--(1,0);
\node[greendot] at (0,2.5){};
\node[greendot] at (0,1.5){};
\end{braid}\label{dia:ii}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (1,1){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (1,3){};
\end{braid}
\\
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (4,4)node[above]{$i$}--(4,0);
\node[greendot] at (4,2){};
\end{braid}
\draw (0,4)node[above]{$i$}--(1,3)--(1,1)--(0,0);
\draw (1,4)node[above]{$i$}--(0,3)--(0,1)--(1,0);
\node[greendot] at (0,2.5){};
\node[greendot] at (0,1.5){};
\node[greendot] at (0.8,0.2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (1,1){};
\node[greendot] at (3,1){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\node[greendot] at (0.5,3.5){};
\node[greendot] at (1.5,2.5){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (4,4)node[above]{$i$}--(4,0);
\node[greendot] at (0,2){};
\end{braid}
\label{dia:ii2}
\\
\label{dia:ii3}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,0);
\draw (4,4)node[above]{$i$}--(0,0);
\end{braid}
&\overset{(\ref{dia:y-psi com})}=
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (4,4){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (0,2){};
\end{braid}
\overset{(\ref{dia:psipsi})}=
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (0,2){};
\end{braid}\\
&\overset{(\ref{dia:y-psi com})}=
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (4,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (0,4){};
\end{braid}
\overset{(\ref{dia:psipsi})}=
\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(0,0);
\draw (4,4)node[above]{$i$}--(0,2)--(4,0);
\node[greendot] at (4,2){};
\end{braid} \notag
\end{align}

Note that (<ref>) follows by multiplicating (<ref>) by $y_{r+1}$ and expanding.

In the rest of the paper we will play around with these diagrammatic notations a lot. In order to make the reader easy to follow our calculation we will use dotted strands to represent moving strands and arrows to represent moving dots. If we are going to move a dot then we will also write the strand which the dot is on dotted so the reader can see the arrow clearly. For example, we will write

\begin{braid}
\draw (0,4)[densely dotted] node[above]{$1$}--(2,0);
\draw (1,4)[densely dotted] node[above]{$2$}--(2,2)--(1,0);
\draw (2,4)[densely dotted] node[above]{$1$}--(0,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\end{braid}
\overset{(\ref{dia:braid})}=
\begin{braid}
\draw (0,4)node[above]{$1$}--(2,0);
\draw (1,4)node[above]{$2$}--(0,2)--(1,0);
\draw (2,4)node[above]{$1$}--(0,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$1$}--(0,0);
\draw (1,4)node[above]{$2$}--(1,0);
\draw (2,4)node[above]{$1$}--(2,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\end{braid}

to signify the application of relation (<ref>) and

\begin{braid}
\draw (0,4)node[above]{$1$}--(0,0);
\draw (1,4)node[above]{$2$}--(1,0);
\draw (2,4)node[above]{$1$}--(2,0);
\draw (3,4)[densely dotted] node[above]{$3$}--(4,0);
\draw (4,4)[densely dotted] node[above]{$3$}--(3,0);
\draw (5,4)node[above]{$0$}--(5,0);
\node[greendot] at (3.25,1){};
\node[greendot] at (3.75,1){};
\draw[->] (3.25,1) -- (3.75,3);
\draw[->] (3.75,1) -- (3.25,3);
\end{braid}
\overset{(\ref{dia:ypsi})}=
\begin{braid}
\draw (0,4)node[above]{$1$}--(0,0);
\draw (1,4)node[above]{$2$}--(1,0);
\draw (2,4)node[above]{$1$}--(2,0);
\draw (3,4) node[above]{$3$}--(4,0);
\draw (4,4) node[above]{$3$}--(3,0);
\draw (5,4)node[above]{$0$}--(5,0);
\node[greendot] at (3.25,3){};
\node[greendot] at (3.75,3){};
\end{braid}

to signify the application of relation (<ref>).

We can define a linear map $*\map{\Rn}{\Rn}$ by swapping the diagrams of $\Rn$ up-side-down. For example,

\left(\begin{braid}
\draw (0,4)node[above]{$0$}--(1,0);
\draw (1,4)node[above]{$1$}--(3,0);
\draw (2,4)node[above]{$3$}--(0,0);
\draw (3,4) node[above]{$2$}--(4,2)--(2,0);
\draw (4,4) node[above]{$2$}--(3,2)--(4,0);
\node[greendot] at (0.8,0.8){};
\node[greendot] at (2,2){};
\node[greendot] at (1.8,3.6){};
\node[greendot] at (3,1){};
\end{braid}\right)^* =
\begin{braid}
\draw (0,4)node[above]{$3$}--(2,0);
\draw (1,4)node[above]{$0$}--(0,0);
\draw (2,4)node[above]{$2$}--(4,2)--(3,0);
\draw (3,4) node[above]{$1$}--(1,0);
\draw (4,4) node[above]{$2$}--(3,2)--(4,0);
\node[greendot] at (0.8,3.2){};
\node[greendot] at (2,2){};
\node[greendot] at (1.8,0.4){};
\node[greendot] at (3,3){};
\end{braid}.

It is obvious that $*$ is an anti-isomorphism and it preserves the generators of $\Rn$.

Fix a weight $\Lambda = \sum_{i\in I} a_i \Lambda_i$ with $a_i \in \N$. Let $N_n^\Lambda(\mathscr O)$ be the two-sided ideal of $\mathscr{R}_n$ generated by the elements with form $e(\bi)y_1^{(\Lambda,\alpha_{i_1})}$. We can now define the main object of study in this paper, the cyclotomic Khovanov-Lauda-Rouquier algebras, which were introduced by Khovanov and Lauda <cit.>.

The cyclotomic Khovanov-Lauda-Rouquier algebras of weight $\Lambda$ and type $\Gamma_e$ is the algebra $\R(\mathscr O) = \mathscr{R}_n(\mathscr O)/ N_n^\Lambda(\mathscr O)$.

Therefore, if we write $e(\bi) = \hat e(\bi) + N_n^\Lambda(\mathscr O)$, $y_r = \hat y_r + N_n^\Lambda(\mathscr O)$ and $\psi_s = \hat \psi_s + N_n^\Lambda(\mathscr O)$, the algebra $\R(\mathscr O)$ is the unital $\mathscr O$-algebra generated by

\{\psi_1,\dots,\psi_{n-1}\} \cup \{ y_1,\dots,y_n \} \cup \set{e(\bi)|\bi\in I^n}

subject to the relations (<ref>)–(<ref>) of $\mathscr{R}_n(\mathscr O)$ together with the additional relation

\begin{equation}\label{alg:additional}
e(\bi)y_1^{(\Lambda,\alpha_{i_1})} = 0, \hspace*{10mm}\text{for each $\bi\in I^n$}.
\end{equation}

§.§ The (graded) cellular algebras and the symmetric groups

Following Graham and Lehrer [7], we now introduce the graded cellular algebras. Reader may also refer to Hu-Mathas [9]. Let $\mathscr O$ be a commutative ring with $1$ and let $A$ be a unital $\mathscr O$-algebra.

A graded cell datum for $A$ is a triple $(\Lambda,T,C,\deg)$ where $\Lambda = (\Lambda,>)$ is a poset, either finite or infinite, and $T(\lambda)$ is a finite set for each $\lambda \in \Lambda$, $\deg$ is a function from $\coprod_\lambda T(\lambda)$ to $\Z$, and

C\map{\prod_{\lambda\in\Lambda} T(\lambda)\times T(\lambda)}{A}

is an injective map which sends $(s,t)$ to $a_{st}^\lambda$ such that:

(a)  $\set{ a_{\s\t}^\lambda|\lambda\in\Lambda,s,t\in T(\lambda)}$ is an $\mathscr O$-free basis of $A$;

(b) for any $r\in A$ and $\t\in T(\lambda)$, there exists scalars $c_{\t}^v(r)$ such that, for any $\s\in T(\lambda)$,

a_{\s\t}^\lambda{\cdot}r \equiv \sum_{\v\in T(\lambda)} c_{\t}^{\v}(r) a_{\s\v}^\lambda \mod{A^{>\lambda}}

where $A^{>\lambda}$ is the $\mathscr O$-submodule of $A$ spanned by $\set{ a_{\x\y}^\mu|\mu>\lambda, \x,\y\in T(\mu)}$;

(c) the $\mathscr O$-linear map $*\map{A}{A}$ which sends $a_{\s\t}^\lambda$ to $a_{\t\s}^\lambda$, for all $\lambda\in \Lambda$ and $\s,\t\in T(\lambda)$, is an anti-isomorphism of $A$.

(d) each basis element $a_{\s\t}^\lambda$ is homogeneous of degree $\deg a_{\s\t}^\lambda = \deg(\s) + \deg(\t)$, for $\lambda\in \Lambda$ and all $\s,\t\in T(\lambda)$.

If a graded cell datum exists for $A$ then $A$ is a graded cellular algebra. Similarly, by forgetting the grading we can define a cell datum and hence a cellular algebra.

Suppose $A$ is a graded cellular algebra with graded cell datum $(\Lambda,T,C,\deg)$. For any $\lambda \in \Lambda$, define $A^{\geq \lambda}$ to be the $\mathscr O$-submodule of $A$ spanned by

\set{c_{\s\t}^\mu|\mu \geq \lambda, \s,\t\in T(\mu)}.

Then $A^{>\lambda}$ is an ideal of $A^{\geq\lambda}$ and hence $A^{\geq\lambda}/A^{>\lambda}$ is a $A$-module. For any $\s\in T(\lambda)$ we define $C_\s^\lambda$ to be the $A$-submodule of $A^{\geq\lambda}/A^{>\lambda}$ with basis $\set{a^\lambda_{\s\t} + A^{>\lambda}|\t\in T(\lambda)}$. By the cellularity of $A$ we have $C_\s^\lambda \cong C_\t^\lambda$ for any $\s,\t\in T(\lambda)$.

Suppose $\lambda \in \P_n$. Define the cell module of $A$ to be $C^\lambda = C_\s^\lambda$ for any $\s\in T(\lambda)$, which has basis $\set{a^\lambda_\t|\t\in T(\lambda)}$ and for any $r \in A$,

a^\lambda_\t{\cdot}r= \sum_{\u\in T(\lambda)} c_\u^r a_\u^\lambda

where $c_\u^r$ are determined by

a^\lambda_{\s\t}{\cdot}r = \sum_{\u\in T(\lambda)} c_\u^r a^\lambda_{\s\u} + A^{>\lambda}.

We can define a bilinear map $\langle\cdot,\cdot\rangle\map{C^\lambda\times C^\lambda}{\Z}$ such that

\langle a^\lambda_\s,a^\lambda_\t\rangle a^\lambda_{\u\v} = a^\lambda_{\u\s}a^\lambda_{\t\v} + A^{>\lambda}

and let $\text{rad }C^\lambda = \set{\s\in C^\lambda|\langle \s,\t\rangle = 0\text{ for all }\t\in C^\lambda}$. The $\text{rad }C^\lambda$ is a graded $A$-submodule of $C^\lambda$.

Suppose $\lambda \in \P_n$. Let $D^\lambda = C^\lambda/\text{rad }C^\lambda$ as a graded $A$-module.

Exactly as in the ungraded case <cit.> or <cit.>, we obtain the following:

The set $\set{D^\lambda\langle k\rangle|\lambda\in \Lambda, D^\lambda\neq 0, k\in\Z}$ is a complete set of pairwise non-isomorphic graded simple $A$-modules.

We give an example of graded cellular algebras here, which is called the cyclotomic Hecke algebras.

Let $\F_p$ be a fixed field of characteristic $p \geq 0$ with $q\in \F_p^\times$. Let $e$ be the smallest positive integer such that $1 + q + \ldots + q^{e-1} = 0$ and setting $e = 0$ if no such integer exists. Then define $I = \Z/e\Z$ if $e > 0$ and $I = \Z$ if $e = 0$.

For $n \geq 0$, assume that $q = 1$. Let $H_n$ be the degenerate affine Hecke algebra, working over $\F_p$. So $H_n$ has generators

\{x_1,\ldots,x_n\}\cup\{s_1,\ldots,s_{n-1}\}

subject to the following relations

\begin{array}{cc}
x_r x_s = x_s x_r; & \\
s_r x_{r+1} = x_r s_r + 1, & s_r s_x = x_s s_r \hspace*{5mm}\text{if $s \neq r,r+1$}\\
s_r^2 = 1; \\
s_r s_{r+1}s_r = s_{r+1}s_r s_{r+1}, & s_r s_t = s_t s_r \hspace*{5mm}\text{if $|r - t| > 1$}
\end{array}

Now we assume that $q\neq 1$ and $H_n$ be the non-degenerate affine Hecke algebra over $\F_p$. So $H_n$ has generators

\{ X_1^{\pm 1},\ldots,X_n^{\pm 1}\}\cup\{T_1,\ldots,T_{n-1}\}

subject to the following relations

\begin{array}{cc}
X_r^{\pm 1} X_s^{\pm 1} = X_s^{\pm 1} X_r^{\pm1}, & X_r X_r^{-1} = 1;\\
T_r X_r T_r = q X_{r+1}, & T_r X_s = X_s T_r\hspace*{5mm}\text{if $s\neq r,r+1$};\\
T_r^2 = (q-1)T_r + q; &\\
T_r T_{r+1}T_r = T_{r+1}T_rT_{r+1},& T_r T_s = T_s T_r\hspace*{5mm}\text{if $|r-s| > 1$}.
\end{array}

Then for any $\Lambda \in P_+$, we define

\begin{equation} \label{equation: quotient hecke}
H_n^\Lambda = \begin{cases}
H_n/ \langle \prod_{i\in I}(X_1 - q^i)^{(\Lambda,\alpha_i)}\rangle, & \text{if $q\neq 1$,}\\
H_n/ \langle \prod_{i\in I}(x_1 - i)^{(\Lambda,\alpha_i)}\rangle, & \text{if $q = 1$.}
\end{cases}
\end{equation}

and we call $H_n^\Lambda$ the degenerate cyclotomic Hecke algebra if $q = 1$ and non-degenerate cyclotomic Hecke algebra if $q\neq 1$.

By the definitions, degenerate and non-degenerate cyclotomic Hecke algebras are similar with some minor difference. In order to minimize their difference we define

\begin{equation} \label{equation: q_i}
q_i = \begin{cases}
i, & \text{if $q = 1$,}\\
q^i, & \text{if $q\neq 1$.}
\end{cases}
\end{equation}

and use $x_r$ instead of $X_r$ when we don't have to distinguish which case we are working with. Hence we can re-write (<ref>) as

\begin{equation} \label{equation: cyclotomic Hecke algebras}
H_n^\Lambda = H_n/\langle \prod_{i\in I}(x_1 - q_i)^{(\Lambda,\alpha_i)}\rangle.
\end{equation}

Murphy [22] gave a set of cellular basis for $H_n^\Lambda$ which shows that $H_n^\Lambda$ is a cellular algebra. Brundan and Kleshchev [3] proved the remarkable result that every $H_n^\Lambda$ over $\F_p$ is isomorphic to $\R(\F_p)$ introduced in <ref>, where in both algebras $\Lambda$ and $e$ are the same. Therefore when $H_n^\Lambda$ is over a field it is a graded cellular algebra.

Let $\mathfrak{S}_n$ be the symmetric group on $\{1,2,\ldots, n\}$. Then $\mathfrak{S}_n$ is a Coxeter group and $\{s_1, \ldots, s_{n-1}\}$ is its standard set of Coxeter generators, where $s_i = (i,i+1)$ for $i = 1, 2, \ldots, n-1$. Suppose $w$ is an element of $\mathfrak{S}_n$ and $w = s_{i_1} s_{i_2}\ldots s_{i_m}$. If $m$ is minimal we say that $w$ has length $m$  and write $l(w) = m$. In this case we say $s_{i_1}s_{i_2}\ldots s_{i_m}$ is a reduced expression of $w$. In general an element of $\mathfrak{S}_n$ has more than one reduced expressions. For example, we have $w = s_1 s_2 s_1 = s_2 s_1 s_2$. Nonetheless, all the reduced expression of an element have the same length.

In this paper we let $\mathfrak{S}_n$ act on $\{1,2,\ldots,n\}$ from right. For example, $(i) s_i s_{i+1} = (i + 1) s_{i+1} = i+2$. The following result is well-known. See, for example, <cit.>.

Suppose that $w\in \mathfrak{S}_n$. For $i = 1,2,\ldots,n-1$,

l(w s_i) = \begin{cases}
l(w) + 1, & \text{ if $(i)w < (i+1)w$,}\\
l(w) - 1, & \text{ if $(i)w > (i+1)w$.}
\end{cases}

We recall the definition of the Bruhat order $\leq$ on $\mathfrak{S}_n$. For $u,w\in \mathfrak{S}_n$ define $u \leq w$ if $u = s_{r_{a_1}}s_{r_{a_2}} \ldots s_{r_{a_b}}$ for some $1 \leq a_1 < a_2 < \ldots < a_b \leq m$, where $w = s_{r_1}s_{r_2}\ldots s_{r_m}$ is a reduced expression for $w$.

§.§ Tableaux combinatorics

In this subsection we recall the combinatorics of (multi)partitions and (multi)tableaux that we will need in this paper.

Let $n$ be a positive integer. A composition of $n$ is an ordered sequence of nonnegative integers $\lambda = (\lambda_1,\lambda_2,\ldots)$ and $|\lambda|=\sum_{i = 1}^\infty \lambda_i = n$. We say $\lambda$ is a partition of $n$ if $\lambda = (\lambda_1,\lambda_2,\ldots)$ is a composition and $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \ldots $. We can then identify $\lambda$ with a sequence $(\lambda_1,\ldots,\lambda_k)$ whenever $\lambda_i = 0$ for $i > k$.

As we now recall, there is a natural partial ordering on the set of compositions of $n$. Suppose $\lambda = (\lambda_1,\lambda_2,\dots,)$ and $\mu = (\mu_1,\mu_2,\dots)$ are compositions of $n$. Then
$\lambda$ dominates $\mu$, and we write $\lambda \unrhd \mu$, if

\sum_{i = 1}^k \lambda_i \geq \sum_{i = 1}^k \mu_i

for any $k$. We write $\lambda \rhd \mu$ if $\lambda \unrhd \mu$ and $\lambda \neq \mu$. The dominance ordering can be extended to a total ordering $\ge$, called the lexicographic ordering. We write $\lambda > \mu$ if there exist some $k$, such that $\lambda_i = \mu_i$ for all $i < k$ and $\lambda_k > \mu_k$. Define $\lambda \geq \mu$ if $\lambda > \mu$ or $\lambda = \mu$. Then $\lambda \unrhd \mu$ implies $\lambda \geq \mu$.

A multicomposition of $n$ of level $\l$ is an ordered sequence $\lambda = (\lambda^{(1)},\ldots,\lambda^{(\l)})$ of compositions such that $\sum_{i = 1}^\l|\lambda^{(i)}| = n$. Similarly, a multipartition of level $\l$ is multicomposition $\lambda = (\lambda^{(1)},\ldots,\lambda^{(\l)})$ of $n$ such that each $\lambda^{(i)}$ is a partition. We will identify multicompositions and multipartitions of level $1$ with compositions and partitions in the obvious way.

Let $\mathscr{C}_n^\Lambda$ be the set of all multicomposition of $n$ and $\mathscr{P}_n^\Lambda$ be the set of all multipartitions of $n$. We can extend the dominance ordering to $\mathscr C_n^\Lambda$ by defining $\lambda \unrhd \mu$ if

\sum_{i = 1}^{k-1}|\lambda^{(i)}| + \sum_{j = 1}^s \lambda^{(k)}_j \geq\sum_{i = 1}^{k-1}|\mu^{(i)}| + \sum_{j = 1}^s \mu^{(k)}_j

for any $1 \leq k \leq \l$ and all $s \geq 1$. Again, we write $\lambda \rhd \mu$ if $\lambda \unrhd \mu$ and $\lambda \neq \mu$. Similarly, we extend the lexicographic ordering $\lambda > \mu$ and $\lambda \geq \mu$ to $\mathscr{C}^\Lambda_n$ in the obvious way way.

The Young diagram of a multicomposition $\lambda$ of level $\l$ is the set

[\lambda] = \{ (r,c,l)\ |\ 1 \leq c \leq \lambda_r^{(l)}, r \geq 0\text{ and }1\leq l \leq \l\}

which we think of as an ordered $\l$-tuple of the diagrams of the partitions $\lambda^{(1)},\ldots,\lambda^{(\l)}$. The triple $(r,c,l)\in[\lambda]$ is node of $\lambda$ in row $r$, column $c$ and component $l$. A $\lambda$-tableau is any bijection $\t\map{[\lambda]}{\{1,2,\ldots,n\}}$. We identify a $\lambda$-tableau $\t$ with a labeling of the diagram of $\lambda$. That is, we label the node $(r,c,l)\in[\lambda]$ with the integer $\t(r,c,l)$. For example,

\tab(9\ten,\eleven\twelve,\thirteen)\hspace*{1mm}\Bigg|\hspace*{1mm}\tab(\fourteen\fifteen\sixteen)\hspace*{1mm}\Bigg)

is a $(4,3,1|2^2,1|3)$-tableaux.
If $\t$ is a $\lambda$-tableau then the shape of $\t$ is the multicomposition $\lambda$ and we write $\Shape(\t) = \lambda$. A $\lambda$-tableau $\t$ is standard if
$\lambda=\Shape(\t)$ is a multipartition and if, in each component, the entries increase along each row and down each column. More precisely, if $(r,c,l)\in[\lambda]$ then
$\t(r,c,l)<\t(r+1,c,l)$ whenever $(r+1,c,l)\in[\lambda]$ and
$\t(r,c,l)<\t(r,c+1,l)$ whenever $(r,c+1,l)\in[\lambda]$. Let $\Std(\lambda)$ be the set of all standard $\lambda$-tableaux and $\Std(>\lambda)$ be the set of all standard $\mu$-tableaux with $\mu > \lambda$. We can define $\Std(\geq\lambda)$ similarly. Note that if $\t$ is standard then so is $\t|_k$ for $1\le k\le n$.

If $\t\in\Std(\lambda)$ and $1\le k\le n$ define $\t|_k$ to be the subtableau of $\t$ obtained by removing all the nodes containing an entry greater than $k$. We define an analogue of the dominance ordering for standard tableaux by defining $\t \unrhd \s$ if $\Shape(\t|_k) \unrhd \Shape(\s|_k)$, for $1\le k\le n$. As with the dominance ordering,  if $\t\unrhd\s$ then we write $\s\unlhd\t$ and if $\s\ne\t$ then write $\t\rhd\s$ and $\s\lhd\t$. We also define $(\s,\t) \rhd (\u,\v)$ if $\s\unrhd \u$, $$̌ and $(\s,\t) \neq (\u,\v)$.

For any multicomposition $\lambda$, define $\t^\lambda$ to be the unique $\lambda$-tableau such that $\t^\lambda \unrhd \t$ for all standard $\lambda$-tableau $\t$. For example, if $\lambda=(4,3,1|2^2,1|3)$ then $\t^\lambda$ is the tableau displayed above.

The symmetric group acts on the set of all $\lambda$-tableaux. Let $\t$ be a $\lambda$-tableau, then $\t{\cdot}s_r$ is the tableau obtained by exchanging the entries $r$ and $r+1$ in $\t$, i.e. $(r)\t^{-1} = (r+1)(\t{\cdot}s_r)^{-1}$, $(r+1)\t^{-1} = (r)(\t{\cdot}s_r)^{-1}$, and $(k)\t^{-1} = (k)(\t{\cdot}s_r)^{-1}$ for $k\neq r,r+1$. Then for each $\lambda$-tableau $\t$ let $d(\t)$ be the permutation in $\mathfrak{S}_n$ such that $\t^\lambda{\cdot}d(\t) = \t$.

Recall the Bruhat order $\leq$ on $\mathfrak{S}_n$ from subsection 1.1. The following result, which goes back to work of Ehresmann and James, is part of the folklore for these algebras. The proof for level $1$ can be found from <cit.>. The higher level cases follow easily.

Suppose $\lambda\in\mathscr P_n^\Lambda$ and $\s$ and $\t$ are standard $\lambda$-tableaux. Then $\s \unrhd \t$ if and only if $d(\s) \leq d(\t)$.

Suppose $\lambda$ is a multicomposition and $\gamma = (r,c,l) \in [\lambda]$ and recall from subsection 1.1 that $\kappa_\Lambda = (\kappa_1,\kappa_2,\ldots,\kappa_\l)$ is a fixed multicharge of $\Lambda$. The residue of $\gamma$ associate to $\kappa_\Lambda$ is

\res(\gamma) \equiv r - c + \kappa_l \pmod{e}.

If $\t$ is a standard $\lambda$-tableau and the residue sequence of $\t$ is $\res(\t) = \bi_t = (i_1, i_2, i_3, \ldots, i_n)$, where $i_k = \res(\gamma_k)$ and $\gamma_k$ is the unique node in $[\lambda]$ such that $\t(\gamma_k) = k$. In particular, we write $\bi_{\t^\lambda} = \bi_\lambda$
and $\res_\t(k)=\res(\gamma_k)$.

Recall that for each standard tableau $\t$, we can define a permutation $d(\t) \in \mathfrak S_n$ such that $\t = \t^\lambda{\cdot}d(\t)$. For each permutation we may have more than one reduced expression. Here we fix a choice of the reduced expression of $d(\t)$.

For any standard $\lambda$-tableau $\t$ and $1 \leq i \leq n+1$, define $\lambda^{(i)} = \Shape(\t|_{i-1})$ and $\t^{(i)}$ to be a standard $\lambda$-tableau where $\t^{(i)}|_{i-1} = \t^{\lambda^{(i)}}$, and ${\t^{(i)}}^{-1}(k) = \t^{-1}(k)$ for any $i \leq k \leq n$. In particular, $\t^{(1)} = \t$ and $\t^{(n+1)} = \t^\lambda$. Therefore we have a series of standard $\lambda$-tableau

\t^\lambda = \t^{(n+1)}, \t^{(n)}, \t^{(n-1)},\ldots, \t^{(2)},\t^{(1)} = \t.

Define $w_i$ to be the unique permutation in $\mathfrak{S}_n$ such that $\t^{(i+1)} w_i = \t^{(i)}$. For each $w_i \neq 1$, we can write $w_i = s_{a_i}s_{a_i+1}s_{a_i+2}\ldots s_{i-2}s_{i-1}$ for some $a_i \leq i-1$. Notice that

\begin{equation} \label{remark: standard expression}
(k)(\t^{(i+1)})^{-1} =
\begin{cases}
(i)(\t^{(i)})^{-1}, & \text{ if $k = a_i$,}\\
(k-1)(\t^{(i)})^{-1}, & \text{ if $a_i < k \leq i$,}\\
(k)(\t^{(i)})^{-1}, & \text{ otherwise,}
\end{cases}
\end{equation}

and $l(w_i)$ is always greater than or equal to the length of the row containing $i$ in $\t^{(i+1)}$. Also for each $i$, if $\Shape(\t^{(i)}|_{i-1}) = \lambda$, then $\t^{(i)}|_{i-1} = \t^\lambda$.

Suppose $\t^{(11)} = \tab(1234\twelve,567\eleven,89\ten\thirteen,\fourteen\fifteen)$ and $\t^{(10)} = \tab(123\ten\twelve,456\eleven,789\thirteen,\fourteen\fifteen)$. Therefore we have $w_{10} = s_4 s_5 s_6 s_7 s_8 s_9$ such that $\t^{(11)}{\cdot}w_{10} = \t^{(10)}$.

Notice that in this case, $i = 10$ and $a_{10} = 4$. So

(k)(\t^{(11)})^{-1} = \begin{cases}
(10)(\t^{(10)})^{-1}, &\text{ if $k = a_i = 4$,}\\
(k-1)(\t^{(10)})^{-1}, &\text{ if $4 = a_i < k \leq i = 10$,}\\
(k)(\t^{(10)})^{-1}, &\text{ otherwise.}
\end{cases}

Furthermore, $\t^{(11)}|_{10} = \tab(1234,567,89\ten) = \t^{(4,3,3)}$ and $\t^{(10)}|_9 = \tab(123,456,789) = \t^{(3,3,3)}$.

One can see that $d(\t) = w_n w_{n-1} \ldots w_2 w_1$ and define it to be the standard expression of $d(\t)$. By  <ref> one can verify that standard expression is a reduced expression of $d(\t)$. In the rest of this paper, we fix $d(\t)$ to be its standard expression.

Suppose $\t$ is a standard $\lambda$-tableau and $d(\t) = s_{r_1}\ldots s_{r_m}$ is the standard expression. For any $1 \leq k \leq m$, define $\s = \t^\lambda {\cdot}s_{r_1}s_{r_2}\ldots s_{r_k}$. Then $\s$ is a standard $\lambda$-tableau.

The proof is trivial by the definition of the standard expression.

Suppose $\t = \tab(12467,35)$. Then we have $d(\t) = s_5 s_6{\cdot}s_4 s_5{\cdot}s_3$. Then

\begin{eqnarray*}
\t^\lambda{\cdot}s_5 & = & \tab(12346,57),\\
\t^\lambda{\cdot}s_5s_6 & = & \tab(12347,56),\\
\t^\lambda{\cdot}s_5s_6s_4 & = & \tab(12357,46),\\
\t^\lambda{\cdot}s_5s_6s_4s_5 & = & \tab(12367,45),
\end{eqnarray*}

and the above tableaux are all standard.

§.§ Graded cellular basis of KLR algebras over a field

Suppose $\mathscr O$ is a field. Hu and Mathas <cit.> constructed a graded cellular basis of $\R(\mathscr O)$. Here we give an equivalent definition of their basis. For any multicomposition $\lambda$, recall $\t^\lambda$ to be the unique standard $\lambda$-tableau such that $\t^\lambda \unrhd \t$ for all standard $\lambda$-tableau $\t$, and $\bi_\lambda$ is the residue sequence of $\t^\lambda$. We define $\hat e_\lambda = \hat e(\bi_\lambda)$.

Suppose $\lambda$ is a multicomposition. A node $(r,c,l)$ is an addable node of $\lambda$ if $(r,c,l)\not\in [\lambda]$ and $[\lambda]\cup \{(r,c,l)\}$ is the Young diagram of a multipartition. Similarly, a node $(r,c,l)$ is a removable node of $\lambda$ if $(r,c,l)\in [\lambda]$ and $[\lambda]\backslash \{(r,c,l)\}$ is the Young diagram of a multipartition. Given two nodes $\alpha = (r,c,l)$ and $\beta = (s,t,m)$ then $\alpha$ is below $\beta$ if either $l > m$, or $l = m$ and $r > s$.

Suppose that $\s\in\Std(\lambda)$. Let $\mathscr{A}_\s(k)$ be the set of addable nodes of the multicomposition $\Shape(\s|_k)$ which are below $\s^{-1}(k)$ and let

\begin{equation*}\label{notation: addable2}
\mathscr{A}_\s^\Lambda(k) = \{ \alpha \in \mathscr{A}_\s(k)\ |\ \res(\alpha) = \res_\t(k)\}.
\end{equation*}

Similarly as in <cit.>, define

\begin{equation*} \label{notation: y lambda}
\hat y_\lambda = \prod_{k = 1}^n
\hat y_k^{|\mathscr{A}_{\t^\lambda}^\Lambda(k)|}\in\Rn(\mathscr O).
\end{equation*}

For example, if $\lambda = (3,1|4^2,2|5,1)$, $e = 4$ and $\Lambda = 3\Lambda_0$ then

\t^\lambda = \Bigg(\hspace*{1mm}\tab(123,4)\hspace*{1mm}\Bigg|\hspace*{1mm}%
\tab(5678,9\ten\eleven\twelve,\thirteen\fourteen)\hspace*{1mm}\Bigg|\hspace*{1mm}\tab(\fifteen\sixteen\seventeen\eighteen\nineteen,\twenty)\hspace*{1mm}\Bigg)

and $\hat y_\lambda = \hat y_1^2 \hat y_5\hat y_8\hat y_{10}\hat y_{12}\hat y_{18}$. Therefore,

\hat e_\lambda \hat y_\lambda = \hat e(0123012330122012303)\hat y_1^2 \hat y_5\hat y_8\hat y_{10}\hat y_{12}\hat y_{18}.

We define a particular kind of element in $\Rn(\mathscr O)$. Suppose $w \in \mathfrak S_n$ has length $\l$ and $s_{i_1} s_{i_2} \ldots s_{i_\l}$ is a reduced expression for $w$ in $\mathfrak{S}_n$. Recall that $\Rn(\mathscr O)$ has a unique anti-isomorphism $*$ which fixes all of the KLR generators. Define

\hat\psi_{w} = \hat\psi_{i_1} \hat\psi_{i_2} \ldots \hat\psi_{i_\l}\in \Rn(\mathscr O) \hspace*{5mm} \text{and} \hspace*{5mm}\hat\psi_{w}^* = \hat\psi_{i_\l}\hat\psi_{i_{\l-1}}\ldots \hat\psi_{i_2}\hat\psi_{i_1}\in \Rn(\mathscr O).

Notice that $\hat \psi_{w}$ and $\hat \psi^*_{w}$ depend on the choice of the reduced expression of $w$, even though in $\mathfrak{S}_n$ all reduced expressions of $w$ are the same. For example, $s_1 s_2 s_1$ and $s_2 s_1 s_2$ are equal to the same element of $\mathfrak{S}_n$, but in general $\hat \psi_1\hat \psi_2\hat \psi_1 \neq \hat \psi_2\hat \psi_1\hat \psi_2$ in $\Rn(\mathscr O)$. Define $l(\hat\psi_{w}) = l(\hat\psi_{w}^*) = l(w)$ for any standard tableau $\t$. Similarly we can define

\psi_{w} = \psi_{i_1} \psi_{i_2} \ldots \psi_{i_\l}\in \R(\mathscr O) \hspace*{5mm} \text{and} \hspace*{5mm}\psi_{w}^* = \psi_{i_\l}\psi_{i_{\l-1}}\ldots \psi_{i_2}\psi_{i_1}\in \R(\mathscr O)

and $\psi_w$ and $\psi_w^*$ depends on the choice of reduced expressions of $w$ as well.

Suppose $l(d(\t)) = \l$ and $d(\t) = s_{i_1} s_{i_2} \ldots s_{i_\l}$ is the standard expression of $d(\t)$ where $\t^\lambda{\cdot}d(\t) = \t$. Define $\hat \psi_{d(\t)} = \hat  \psi_{i_1}\hat \psi_{i_2}\ldots \hat \psi_{i_\l}$ and $\hat \psi^*_{d(\t)} = \hat \psi_{i_\l}\hat \psi_{i_{\l-1}}\ldots \hat \psi_{i_2}\hat \psi_{i_1}$.

Suppose $\Lambda\in P_+$, $\lambda \in \mathscr{P}_n^\Lambda$ and $\s,\t$ are two standard $\lambda$-tableaux. We define

\hat\psi^{\mathscr O}_{\s\t} = \hat\psi_{d(\s)}^* \hat e_\lambda \hat y_\lambda \hat \psi_{d(\t)} \in \Rn(\mathscr O),

and hence

\psi^{\mathscr O}_{\s\t} = \hat\psi^{\mathscr O}_{\s\t} + N_n^\Lambda \in \R(\mathscr O).

Notice that Hu and Mathas <cit.> defined $\psi_{\s\t}^{\mathscr O}$ differently. Actually if we define $e_\lambda$, $y_\lambda$ and $\psi_w$ in $\R(\mathscr O)$ as analogues of $\hat e_\lambda$, $\hat y_\lambda$ and $\hat\psi_w$, and define $\psi_{\s\t}^{\mathscr O} = \psi_{d(\s)}^* e_\lambda y_\lambda \psi_{d(\t)}$ for $\s,\t\in\Std(\lambda)$, it is equivalent to <ref>. We define $\psi_{\s\t}^{\mathscr O}$ as in <ref> because we need to work in $\Rn(\mathscr O)$ later.

By construction, then this $\psi_{\s\t}^{\mathscr O}$ is well defined as an element of $\R(\mathscr O)$ for any ring $\mathscr O$. Many of the calculations in this paper depend heavily on the choice of $\mathscr O$ so we write $\psi_{\s\t}^{\mathscr O}$ to emphasize that $\psi_{\s\t}^{\mathscr O}$ is an element of $\R(\mathscr O)$. Most of the time, however, we will work in $\R(\Z)$ so for convenience we set $\psi_{\s\t}=\psi_{\s\t}^\Z$.

[Hu and Mathas <cit.> <cit.>]
Suppose $\mathscr O$ is a field and $\s$ and $\t$ are standard $\lambda$-tableaux and $1\leq r\leq n$,

\begin{eqnarray*}
\psi_{\s\t} \psi_r & = & \begin{cases}\sum_{(\u,\v)\rhd(\s,\t)}c_{\u\v} \psi_{\u\v}, & \text{if $\t{\cdot}s_r$ is not standard}\\
&\hspace*{5mm}\text{or $d(\t){\cdot}s_r$ is not reduced,}\\
\psi_{\s\v} + \sum_{(\u,\v)\rhd (\s,\t)}c_{\u\v} \psi_{\u\v}, & \text{if $\v = \t{\cdot}s_r$ standard and $d(\t){\cdot}s_r = d(\v)$.}\end{cases}
\end{eqnarray*}

for $c_{\u\v}\in\mathscr O$, and $c_{\u\v} \neq 0$ only if $\res(\s) = \res(\u)$ and $\res(\t{\cdot}s_r) = \res(\v)$. Similarly, we have

\psi_{\s\t}^{\mathscr O}y_r = \sum_{(\u,\v)\rhd (\s,\t)} c_{\u\v}\psi_{\u\v}^{\mathscr O}

for $c_{\u\v}\in\mathscr O$, and $c_{\u\v}\neq 0$ only if $\res(\s) = \res(\u)$ and $\res(\t) = \res(\v)$.

[Hu and Mathas <cit.>]
Suppose $\mathscr O$ is an integral domain and that either $e=0$, $e$ is a prime
or $e$ is a non-zero non-prime integer such that $e\cdot 1_{\mathscr O}$ is
invertible in $\mathscr O$. Then

\{ \psi_{\s\t}^{\mathscr O} \ |\ \s,\t \in \text{Std}(\lambda) \text{ for $\lambda\in\mathscr{P}_n^\Lambda$} \}

is a graded cellular basis of $\R(\mathscr O)$. In
particular, $\R(\mathscr O)$ is free as an $\mathscr
O$-module of rank $\l^n n!$.

The main purpose of this paper is to prove that $\R(\Z)$ is free of rank $\l^nn!$. To do this we will show that $\{\psi_{\s\t}^\Z\ |\ \s,\t \in \text{Std}(\lambda) \text{ for $\lambda\in\mathscr{P}_n^\Lambda$} \}$ is a homogeneous basis of $\R(\Z)$.

We define some notation for future use.

Suppose $\lambda$ is a multipartition of $\mathscr{P}_n^\Lambda$. Define:

\begin{align*}
\BZ&=\langle\psi_{\s\t}\mid \s,\t\in\text{Std}(\mu)\text{ for }\mu\in\P_n\rangle_\Z ,\\
\Bgelam&=\langle\psi_{\s\t}\mid \s,\t\in\text{Std}(\mu)\text{ and }\mu\ge\lambda
\text{ for }\mu\in\P_n\rangle_\Z ,\\
\Blam & = \langle\psi_{\s\t}\mid \s,\t\in\text{Std}(\mu)\text{ and }\mu>\lambda
\text{ for }\mu\in\P_n\rangle_\Z .
\end{align*}

where $\Blam \subseteq \Bgelam \subseteq \BZ \subseteq \R(\Z)$

This subsection closes with an important Proposition:

Consider the quiver Hecke algebra  $\mathscr{R}_n(\Q)$ defined over the rational field $\Q$. We have $\mathscr{R}_n(\Q) \cong \mathscr{R}_n(\Z)\otimes\Q$ and we can define a linear map $f\map{\mathscr{R}_n(\Z)}{\mathscr{R}_n(\Q)}$ by sending $x\in \mathscr{R}_n(\Z)$ to $x\otimes 1$.

The linear map $f\map{\mathscr{R}_n(\Z)}{\mathscr{R}_n(\Q)}$ is an injection.

In [13][12], Khovanov and Lauda constructed a basis of $\Rn(\mathscr O)$

\begin{equation} \label{basis of KLR}
\set{ \hat e(\bi)\hat y_1^{\l_1} \hat y_2^{\l_2} \ldots \hat y_n^{\l_n} \hat \psi_w |\bi\in I^n, w\in\mathfrak{S}_n, \l_1,\l_2,\ldots,\l_n \geq 0 }
\end{equation}

for any ring $\mathscr O$. Hence that $\mathscr{R}_n(\Z)$ is free over $\Z$. The Lemma follows immediately.

From the definitions, it is evident that $f(N^\Lambda_n(\Z)) \subseteq N^\Lambda_n(\Q)$. Hence, $f$ induces a homomorphism,

x+N^\Lambda_n(\Z) \mapsto f(x)+N^\Lambda_n(\Q),$$

which by abuse of notation we also denote by $f$. In particular, observe that
$f(\psi_{\s\t}^\Z) = \psi_{\s\t}^\Q$. The main Theorem of this paper is equivalently to prove that $f\map{\R(\Z)}\R(\Q)$ is an injection.

We then introduce an important special case where we already know that $f$ is injective.

The homomorphism $f\map{\R(\Z)}\R(\Q)$ restricts to an injective map from $\BZ$ to $\R(\Q)$.

As we have already noted above, $f(\psi_{\s\t}^\Z) =\psi_{\s\t}^\Q$ for all $\s,\t\in\Std(\lambda)$ and $\lambda\in\mathscr{P}^\Lambda_n$. Hence, <ref> implies the result.

The elements $\{ \psi_{\s\t}^\Z \ |\ \s,\t \in \text{Std}(\lambda) \text{ for $\lambda\in\mathscr{P}_n^\Lambda$} \}$ are a linearly independent subset of $\R(\Z)$.

Proposition <ref> is quite crucial. In this paper we prove that $\psi_{\s\t}^\Z{\cdot} \psi_r \in \BZ$ whenever $d(\t){\cdot}s_r$ is not reduced or $\t{\cdot} s_r$ is not standard in $\R(\Z)$. We can only have

\psi_{\s\t}^\Z {\cdot}\psi_r = \sum_{\u,\v} c^\Z_{\u\v} \psi_{\u\v}^\Z.

In $\R(\Q)$, however, by Lemma <ref>, under these conditions we have

\psi_{\s\t}^\Q{\cdot}\psi_r = \sum_{(\u,\v)\rhd (\s,\t)}c^\Q_{\u\v} \psi_{\u\v}^\Q

for some $c^\Q_{\u\v}\in\Q$, where $(\u,\v)\rhd (\s,\t)$ if $\u \unrhd \s$, $\v \unrhd \t$ and $(\u,\v) \neq (\s,\t)$. Therefore, $c^\Q_{\u\v} = c^\Z_{\u\v}$ by Proposition <ref> and we see that $c^\Z_{\u\v}\ne0$ only if $(\u,\v)\rhd (\s,\t)$. In such case we have much more information about $\u$ and $$̌ with $c^\Z_{\u\v}\neq 0$. Similar remarks apply to the products $\psi_{\s\t}^\Z{\cdot}y_r$.

§ INTEGRAL BASIS THEOREM I

In the next two sections we will prove that $\R$ is $\Z$-free by showing that $\set{\psi_{\s\t} | \s,\t \in \Std(\lambda), \lambda \in \P_n}$ spans $\R$ over $\Z$. Then by  <ref> it is a basis of $\R$ over $\Z$.

In the rest of this paper we write $\mathscr{R}_n(\Z)$ as $\mathscr{R}_n$ and $\mathscr{R}_n^\Lambda(\Z)$ as $\mathscr{R}_n^\Lambda$. Fix a weight $\Lambda$, a multicharge $\kappa_\Lambda = (\kappa_1,\ldots,\kappa_l)$ corresponding to $\Lambda$ and an integer $e > 2$. In this and the next section we mainly work with the algebra $\mathscr{R}^\Lambda_n$.

§.§ The base step of the induction

In this subsection we set up the notations and inductive machinery that we use in the next two sections to prove our main theorem. We then consider the base case of our induction which is when $\lambda = (n|\emptyset|\ldots|\emptyset)$. Finally we develop some technical Lemmas which will be useful later.

Suppose that $\lambda$ is a multipartition of $n$. Let
$\lambda^+$ be the multicomposition of $n+1$ obtained by adding a node at the end of the last non-empty row of $\lambda$, and $\lambda_- = \lambda|_{n-1}$ be the multipartition of $n-1$ obtained by removing the last node from $\lambda$.

For example, if $\lambda = (4,3|3,3)$ then $\lambda^+ = (4,3|3,4)$ and $\lambda_- = (4,3|3,2)$. Notice that in general, $\lambda^+$ will be a multicomposition rather than a multipartition.

For $k \in I$ and $\lambda \in \P_n$, define $\mathscr A_{\t^\lambda}^k = \set{\alpha \in \mathscr A_{\t^\lambda}(n)|\res(\alpha) = k}$. Recall $\bi_\lambda=\res(\t^\lambda)$ and $e_\lambda=e(\bi_\lambda)$ from

Suppose that $\lambda\in \P_n$ and $k\in I$. Define the integer $b_k^\lambda$

b_k^\lambda = \begin{cases}
|\mathscr{A}_{\t^\lambda}^k| + 1,& \text{if $\lambda^+$ is a
multipartition and $i_n+1 = k$},\\
|\mathscr{A}_{\t^\lambda}^k|,& \text{otherwise.}
\end{cases}

If $\bi = (i_1, i_2, \ldots, i_n) \in I^n$ and $k\in I$ then define $\bi\vee k = (i_1, i_2, \ldots, i_n, k) \in I^{n+1}$.

Suppose that $\lambda\in \P_n$ and $k\in I$. Then for each
integer $b$ with $0\le b<b^\lambda_k$, there exists a multipartition
$\nu=\nu(b)$ such that $e_\nu y_\nu = e(\bi_\lambda \vee k)y_\lambda y_{n+1}^b$.

The definitions of $\lambda$ and $b_k^\lambda$ ensure that there are $b_k^\lambda$ addable nodes of residue $k$ below $(\t^\lambda)^{-1}(n)$. Suppose those nodes are $(r_1,c_1,l_1)$, $(r_2,c_2,l_2), \ldots, (r_{b_k^\lambda},c_{b_k^\lambda},l_{b_k^\lambda})$, where $l_1 \geq l_2 \geq l_3 \geq \ldots \geq l_{b_k^\lambda}$, and if $l_i = l_{i+1}$ then $r_i \geq r_{i+1}$. In another word, $(r_i, c_i,l_i)$ is a node below $(r_{i+1},c_{i+1},l_{i+1})$.

For any $b$ with $0 \leq b < b_k^\lambda$, we define $\nu$ to be the multipartition obtained by adding the node $(r_{b+1},c_{b+1},l_{b+1})$ on to $\lambda$. Then $y_\nu = y_\lambda y_{n+1}^b$ and $e_\nu = e(\bi_\lambda\vee k) = e(\bi\vee k)$. This completes the proof.

Suppose that $\lambda = (4,3|2,1|0|0)$ with $e = 4$ and $\kappa_\Lambda = (0,0,2,1)$. Then $e(\bi_\lambda) = e(0123301013)$ and $y_\lambda = y_1 y_2 y_3 y_4 y_6 y_7 y_9$. Then $b_0^\lambda = 1$, $b_1^\lambda = 1$, $b_2^\lambda = 2$ and $b_3^\lambda = 0$ and the proof of
<ref> shows that:

\begin{eqnarray*}
e(01233010130)y_1 y_2 y_3 y_4 y_6 y_7 y_9 & = & e_{\mu_1} y_{\mu_1},\\
e(01233010131)y_1 y_2 y_3 y_4 y_6 y_7 y_9 & = & e_{\mu_2} y_{\mu_2},\\
e(01233010132)y_1 y_2 y_3 y_4 y_6 y_7 y_9 & = & e_{\mu_3} y_{\mu_3},\\
e(01233010132)y_1 y_2 y_3 y_4 y_6 y_7 y_9 y_{11} & = & e_{\mu_4} y_{\mu_4},
\end{eqnarray*}

where $\mu_1 = (4,3|2,2|\emptyset|\emptyset)$, $\mu_2 = (4,3|2,1|\emptyset|1)$, $\mu_3 = (4,3|2,1|1|\emptyset)$ and $\mu_4 = (4,3|2,1,1|\emptyset|\emptyset)$.

Let $\P = \cup_{n \geq 0} \P_n$. Define three sets $\P_I$, $\P_y$ and $\P_\psi$ of multipartitions by:

\begin{align*}
\P_I&= \{ \lambda\in\P \mid |\lambda| = n \text{ and } e(\bi_{\lambda_-}\vee k)y_{\lambda_-}y_n^{b_k^{\lambda_-}} \in \Blam \text{ for all }k\in I\},\\
\P_y&=\{\lambda\in\P \mid |\lambda| = n  \text{ and } \psi_{\s\t}y_r\in\BZ
\text{ whenever $\s,\t\in\Std(\lambda)$ and }1\le r\le n\},\\
\P_\psi &=\{\lambda\in\P \mid |\lambda| = n \text{ and }\psi_{\s\t}\psi_r\in\BZ
\text{ whenever $\s,\t\in\Std(\lambda)$ and }1\le r<n\}.
\end{align*}

Notice that if for some $\s,\t\in \Std(\lambda)$ and $1 \leq r \leq n$ we have $\psi_{\s\t}y_r \in \BZ$, then $y_r \psi_{\s\t} \in \BZ$ as well. Similar property holds for $\psi_{\s\t}\psi_r$. Therefore we can write

\begin{align*}
\P_y&=\{\lambda\in\P \mid |\lambda| = n  \text{ and } y_r\psi_{\s\t}\in\BZ
\text{ whenever $\s,\t\in\Std(\lambda)$ and }1\le r\le n\},\\
\P_\psi &=\{\lambda\in\P \mid |\lambda| = n \text{ and }\psi_r\psi_{\s\t}\in\BZ
\text{ whenever $\s,\t\in\Std(\lambda)$ and }1\le r<n\}
\end{align*}

as well.

By <ref> if one of $e(\bi_\nu)y_{\lambda_-}y_n^n$, $\psi_{\s\t}y_r$ or $\psi_{\s\t}\psi_r$ belongs to $\BZ$ then it can be written in a unique way as an (integral) linear combination of the $\psi$-basis elements. In particular, these linear combinations must satisfy the restrictions imposed by <ref>.

We define a total ordering on $\P$ extended by lexicographic ordering. Suppose $\lambda$ and $\mu$ are two multipartitions, not necessarily of the same integer.  Define $\mu \prec \lambda$ if $|\mu| < |\lambda|$, or $|\mu| = |\lambda|$ and $l(\mu) < l(\lambda)$, or $|\mu| = |\lambda|$, $l(\mu) = l(\lambda)$ and $\lambda < \mu$.

\mathscr S_n^\Lambda = \{\lambda \in \P_n\mid
\mu \in \mathscr P^{\Lambda'}_I\cap \mathscr P^{\Lambda'}_y \cap \mathscr P^{\Lambda'}_\psi\text{ whenever $\mu\in\mathscr P_m^{\Lambda'}$ and }\mu\prec\lambda\}

We note that $\P_n \subseteq \P_I \cap \P_y \cap \P_\psi$ implies $R_n^\Lambda = \R$. Recall that $R_n^\Lambda$ is the $\Z$-vector space spanned by $\{\psi_{\s\t}\}$. Hence $\R$ is a $\Z$-span of $\{\psi_{\s\t}\}$. The main goal of this and the next sections is to prove that $\P_n \subseteq \P_I \cap \P_y \cap \P_\psi$ and we will use induction to prove the argument.

Now we can state the main result of this section.

Suppose $\lambda \in \mathscr S_n^\Lambda$. Then we have $\lambda\in \P_I$.

As we mentioned before we are going to apply induction on $\lambda$ to prove the main Theorem. <ref>, <ref> and <ref> give the base case of the induction. Recall that $e \neq 2$.

Suppose that $n\geq 1$ and $\lambda = (n|0|\ldots|0)\in \mathscr P_n^\Lambda$. Then
e(\bi_{\lambda_-}\vee k)y_{\lambda_-} y_n^{b_k^{\lambda_-}} \in \Blam
for any $k\in I$.

As $\lambda$ is the maximal element of $\P_n$, $\Blam  = \{0\}$. Therefore the Lemma is equivalent to the claim that $e(\bi_{\lambda_-}\vee k)y_{\lambda_-} y_{n}^{b_k^{\lambda_-}} = 0$. We prove this by induction on $n$.

If $n =1$ then it is easy to see that $b_k^{\lambda_-} = (\Lambda,\alpha_{i_1})$. Therefore, $e(\bi_{\lambda_-} \vee k) y_{\lambda_-} y_n^{b_k^{\lambda_-}} = e(k) y_1^{(\Lambda,\alpha_{i_1})} = 0$ by (<ref>).

Suppose now that the Lemma holds for any $n' <n$. Notice that for any $m \geq 1$, set $\gamma = (m|0|\ldots|0)$ then $|\mathscr{A}_{\t^\gamma}^k|$ is independent to the value of $m$. For the rest of the proof we set $a_k = |\mathscr A_{\t^{\lambda_-}}^k|$.

In order to simplify the notations, for the rest of the proof we will omit $i_1i_2\ldots i_{n-3}$ and simply write $e(\bi) = e(i_{n-2},i_{n-1},i_n)$. We will also suppress $y_\nu$, where $\nu = \lambda|_{n-3}$.

We consider four cases separately, depending on the value of $k$.

Case  <ref>a: $k=i_{n-1}$. Then
$e(\textbf{i}_{\lambda_-}\vee k) y_{\lambda_-} y_{n}^{b_k^{\lambda_-}} = e(\textbf{i}|_{n-3},k-1,k,k)y_{\lambda|_{n-3}} y_{n-2}^{a_{k-1}} y_{n-1}^{a_k}y_{n}^{a_k}$. In this case we have

\begin{align*}
e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k} &\overset{(\ref{alg:psi-y com})}= -e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k+1}y_n^{a_k}\psi_{n-1} + e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k}\psi_{n-1} y_n\\
&\overset{(\ref{alg:ypsi})}= \psi_{n-1}e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k}y_n,
\end{align*}

where $e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k+1}y_n^{a_k} = 0$ by induction. Therefore,

\begin{align*}
e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k} & = \psi_{n-1}e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k}y_n\\
& = \psi_{n-1}^2e(k-1,k,k)y_{n-2}^{a_{k-1}}y_{n-1}^{a_k}y_n^{a_k}y_n^2 = 0
\end{align*}

by relation (<ref>).

Case  <ref>b: $k=i_{n-1}+1$. Now,
e(\textbf{i}_{\lambda_-}\vee k) y_{\lambda_-} y_{n}^{b_k^{\lambda_-}} = e(\textbf{i}|_{n-3},k-2,k-1,k)y_{\lambda|_{n-3}} y_{n-2}^{a_{k-2}} y_{n-1}^{a_{k-1}} y_{n}^{a_k+1}

\begin{eqnarray*}
& \overset{(\ref{alg:psipsi})} = & e(k-2,k-1,k)y_{n-2}^{a_{k-2}}y_{n-1}^{a_{k-1}+1}y_n^{a_k} + e(k-2,k-1,k)y_{n-2}^{a_{k-2}}y_{n-1}^{a_{k-1}}y_n^{a_k}\psi^2_{n-1}\\
&\overset{\substack{(\ref{alg:psi-y com})\\(\ref{alg:y-psi com})}}= & e(k-2,k-1,k)y_{n-2}^{a_{k-2}}y_{n-1}^{a_{k-1}+1}y_n^{a_k} + \psi_{n-1} e(k-2,k,k-1)y_{n-2}^{a_{k-2}}y_{n-1}^{a_k}y_n^{a_{k-1}}\psi_{n-1} = 0,
\end{eqnarray*}

where the last equality follows by induction.

Case  <ref>c: $k=i_{n-1}-1$. If $n = 2$ then
$e(\textbf{i}_{\lambda_-}\vee k) y_{\lambda_-} y_{n}^{b_k^{\lambda_-}}= e(k,k-1)y_1^{a_k}y_2^{a_{k-1}}$. Then $a_{k-1} \geq 1$. Therefore,

e(k,k-1)y_1^{a_k}y_2^{a_{k-1}} \overset{(\ref{alg:psipsi})}= e(k,k-1)y_1^{a_k+1}y_2^{a_{k-1}-1} - \psi_1e(k-1,k)y_1^{a_{k-1}-1}y_2^{a_k}\psi_1 = 0,

using relation (<ref>) and induction. Hence, the lemma follows in this case when $n=2$.

If $n > 2$ then $e(\textbf{i}_{\lambda_-}\vee k) y_{\lambda_-} y_{n}^{b_k^{\lambda_-}} = e(\textbf{i}|_{n-3},k,k+1,k)y_{\lambda|_{n-3}}y_{n-2}^{a_k} y_{n-1}^{a_{k+1}} y_{n}^{a_k}$. Hence,

\begin{align*}
y_{n-1}^{a_{k+1}} y_{n}^{a_k}
&\overset{(\ref{alg:braid})}= \psi_{n-2}\psi_{n-1}\psi_{n-2}e(k,k+1,k)y_{n-2}^{a_k} y_{n-1}^{a_{k+1}} y_{n}^{a_k} \\
&\qquad- \psi_{n-1}\psi_{n-2}\psi_{n-1}e(k,k+1,k)y_{n-2}^{a_k} y_{n-1}^{a_{k+1}} y_{n}^{a_k}\\
&=  \psi_{n-2}\psi_{n-1} e(k+1,k,k)y_{n-2}^{a_{k+1}}y_{n-1}^{a_k} y_{n}^{a_k}\psi_{n-2}\\
&\qquad- \psi_{n-1}\psi_{n-2}e(k,k,k+1)y_{n-2}^{a_k} y_{n-1}^{a_k} y_{n}^{a_{k+1}}\psi_{n-1}\\
&= 0,
\end{align*}

where the last equality follows by induction.

Case  <ref>d: $|k-i_{n-1}|>1$. Because $i_{n-2} = i_{n-1} - 1$, we have $i_{n-2} \neq k$. Therefore we have

\begin{align*}
e(\textbf{i}_{\lambda_-}\vee k) y_{\lambda_-} y_{n}^{b_k^{\lambda_-}}
&= e(\textbf{i}|_{n-3},i_{n-2},i_{n-1},k)y_{\lambda|_{n-3}} y_{n-2}^{a_{i_{n-2}}} y_{n-1}^{a_{i_{n-1}}} y_{n}^{a_k}\\
&\overset{(\ref{alg:psipsi})}= \psi_{n-1} e(\textbf{i}|_{n-3},i_{n-2},k,i_{n-1})y_{{\lambda}|_{n-3}} y_{n-2}^{a_{i_{n-2}}} y_{n-1}^{a_k} y_{n}^{a_{i_{n-1}}} \psi_{n-1} = 0
\end{align*}

by induction. This completes the proof.

<ref> has two immediate Corollaries:

Suppose $n\geq 2$ and $\lambda = (n|0|\ldots|0)$. Then
e_\lambda y_\lambda y_r \in \Blam
for any $1\leq r\leq n$.

Suppose that $n\geq 2$ and $\lambda = (n|0|\ldots|0)$. Then
e_\lambda y_\lambda \psi_r \in \Blam
for any $1\leq r\leq n-1$.

Write $y_\lambda = y_1^{l_1}y_2^{l_2}\ldots y_n^{l_n}$ and $\bi_\lambda = (i_1i_2\ldots i_n)$,

e(i_1\ldots i_n)y_1^{l_1}\ldots y_n^{l_n}\psi_r = \psi_r e(i_1\ldots i_{r-1}i_{r+1}i_r\ldots i_n)y_1^{l_1}\ldots y_{r-1}^{l_{r-1}}y_r^{l_{r+1}}y_{r+1}^{l_r}\ldots y_n^{l_n} = 0,

by <ref>.

The results in the rest of the subsection will be used frequently in the later proofs.

Recall that for any multipartition $\lambda$, $\Blam $ is the subspace of $\R$ spanned by all of the elements $\psi_{\s\t}$, where $\Shape(\s) = \Shape(\t) > \lambda$.

Suppose $\lambda \in \mathscr{S}_n^\Lambda$. Then $\Blam $ is a two-sided ideal of $\mathscr{R}_n^\Lambda$. More precisely, $\Blam[\mu]$ is a two-sided ideal of $\R$ whenever $\mu\prec\lambda$.

The Lemma follows directly from the definition of the set $\mathscr S_n^\Lambda$,
$\P_y$, $\P_\psi$ and <ref>.

In order to simplify the notation, for each $i \in I$ define
$\theta_i\map{\mathscr{R}_n^\Lambda}{\mathscr{R}_{n+1}^\Lambda}$ to be the unique $\Z$-linear map which sends   $e(\bi)$ to $e(\bi\vee i)$, $y_r$ to $y_r$ and $\psi_r$ to $\psi_r$. It is easy to see that $\theta_i$ respects the relations in $\R$, so $\theta_i$ is a
$\Z$-algebra homomorphism.

Suppose $\lambda\in \mathscr S_n^\Lambda$ and $\u,\v\in\Std(\mu)$, where $\mu \in \P_m$ with $m<n$ such that $\mu > \lambda|_m$. Let $\sigma=\lambda|_{m+1} \in \P_{m+1}$. Then
$\theta_i(\psi_{\u\v}) \in \Blam[\sigma]$, for any $i\in I$.

Write $\mu = (\mu^{(1)},\ldots,\mu^{(\l)})$ and $\mu^{(\l)} = (\mu^{(\l)}_1,\ldots,\mu^{(\l)}_k)$ and define $\gamma = (\mu^{(1)},\ldots,\mu^{(\l-1)},\gamma^{(\l)})$ where

\gamma^{(\l)} =
\begin{cases}
(\mu^{(\l)}_1,\ldots,\mu^{(\l)}_{k-1},\mu^{(\l)}_k+1),& \text{if $\mu^{(\l)}_{k-1} > \mu^{(\l)}_k$,}\\
(\mu^{(\l)}_1,\ldots,\mu^{(\l)}_{k-1},\mu^{(\l)}_k,1),& \text{if $\mu^{(\l)}_{k-1} = \mu^{(\l)}_k$.}
\end{cases}

Then $\gamma$ is a multipartition of $m+1$ and $\gamma|_m = \mu$. Since $m < n$,
if $m = n-1$, then $\gamma|_{n-1} = \mu > \lambda_-$, so that $\gamma >
\lambda$. On the other hand, if $m < n-1$ then $|\gamma| = m+1 < n = |\lambda|$.
So we always have $\gamma \prec \lambda$.  Therefore, $\gamma \in \P_I \cap \P_y \cap \P_\psi$
because $\lambda\in \mathscr S_n^\Lambda$.

As $\gamma|_m = \mu$, we have
\theta_i(\psi_{\u\v}) = \theta_i(\psi_{d(\u)}^* e_\mu y_\mu \psi_{d(\v)})
= \psi_{d(\u)}^* e(\bi_{\gamma|_m}\vee i) y_\mu \psi_{d(\v)}.
First suppose that $b_i^\mu = 0$. Then using the definition of $\P_I$, we have
e(\bi_{\gamma|_m} \vee i) y_\mu \in \Blam[\gamma] \subseteq \Blam[\sigma ]
Hence, by <ref>, we have
\theta_i(\psi_{\u\v}) = \psi_{d(\u)}^* e(\bi_{\gamma|_m}\vee i) y_\mu \psi_{d(\v)} \in \Blam[\sigma ]

Now supose that $b_i^\mu > 0$. By <ref> there exists a multipartition
$\nu$ with $\nu|_m = \mu$ such that
e(\bi_{\gamma|_m} \vee i) y_\mu = e_\nu y_\nu
Further, as $\nu|_m = \mu$, there exist two standard $\nu$-tableaux $\s$ and $\t$ such that $\s|_m = \u$ and $|_m = $̌. That is, $d(\s) = d(\u)$ and $d(\t) = d(\v)$. Therefore,

\theta_i(\psi_{\u\v}) = \psi_{d(\u)}^* e(\bi_{\gamma|_m}\vee i) y_\mu \psi_{d(\v)} = \psi_{d(\s)}^* e_\nu y_\nu \psi_{d(\t)} = \psi_{\s\t} \in \Bgelam[\nu] \subseteq \Blam[\sigma]

because $\nu  > \sigma $. This completes the proof.

If $\bi=(i_1,\dots,i_n)\in I^n$ and $1\le m\le n$ let $\bi_m=(i_1\dots i_m)$.

Suppose $\lambda\in\mathscr S_n^\Lambda$, $m \leq n$ and $\sigma = \lambda|_m$. For any $\bi = (i_1,i_2,\ldots,i_{n-m})$ we have $\mathscr R_n^\Lambda \theta_\bi(\Blam[\sigma])\mathscr R_n^\Lambda \subseteq \Blam$.

Suppose $r\in \Blam[\sigma]$, we have that

r = \sum_{\u,\v \in \Std( > \sigma)}c_{\u\v}\psi_{\u\v}

for some $c_{\u\v} \in \Z$. For any $i\in I$,

\theta_i(r) = \sum_{\u,\v \in \Std( > \sigma)}c_{\u\v} \theta_i(\psi_{\u\v}).

By <ref>, $\theta_i(\psi_{\u\v})\in \Blam[\lambda|_{m+1}]$. Hence $\theta_i(r) \in \Blam[\lambda|_{m+1}]$. By induction we have $\theta_\bi(r) \in \Blam$.
By <ref>, $\Blam$ is an ideal. Therefore
\mathscr R_n^\Lambda \theta_\bi(\Blam[\sigma])\mathscr R_n^\Lambda \subseteq \Blam
which completes the proof.

§.§ The action of $y_r$ on two-rowed partitions

Recall that the main result of this section is to prove that if $\lambda\in \mathscr S_n$, then

e(\bi_{\lambda_-}\vee k)y_{\lambda_-} y_n^{b_k^{\lambda_-}} \in R_n^{>\lambda}.

In the inductive process we consider different types of multipartitions $\lambda$ and a residue $k\in I$. We will consider the more difficult case first, namely when $\lambda = (\lambda^{(1)},\ldots,\lambda^{(\l)})$ and $\lambda^{(\l)} = (\lambda^{(\l)}_1,\ldots,\lambda^{(\l)}_l,1) \neq \emptyset$ with $l \geq 2$, $\lambda^{(\l)}_{l-1} = \lambda^{(\l)}_l = m$ and $k \equiv \kappa_\l - l + m + 1\pmod{e}$. In this subsection we assume that $\l = 1$ and $l = 2$. We will extend the result to the general case in the next subsection. Notice that in this case $\lambda = (m,m,1)$ for some integer $m$ and $k \equiv \kappa_1 - 1 + m\pmod{e}$. Then $e(\bi_{\lambda_-}\vee k)y_{\lambda_-} y_n^{b_k^{\lambda_-}} = e_\gamma y_\gamma$ where $\gamma = (m,m+1)$. It is very hard to prove that $e_\gamma y_\gamma \in R_n^{>\lambda}$ directly, so we are going to work with $\gamma$ which is in a more general form.

In this subsection we fix $\Lambda = \Lambda_j$ for some $j\in I$, $\gamma = (\gamma_1,\gamma_2)$ and $\lambda = (\gamma_1,\gamma_2-1,1)$ with $\gamma_2 > 1$ and $\gamma_2 - \gamma_1 \equiv 1\pmod{e}$. We will prove that if $\gamma_1 + 1 = \gamma_2$ and $\lambda\in \mathscr S_n^\Lambda$ then $e_\gamma y_\gamma \in \Blam[\gamma]$.

Without loss of generality we can assume that $\Lambda = \Lambda_0$. Define $i \equiv \gamma_2-2 \pmod{e}$, which is the residue of $(2,\gamma_2,1)$. Because $\gamma_2 \equiv \gamma_1 + 1\pmod{e}$, it is also the residue of the node $(1,\gamma_1,1)$. In diagrammatic notation, we have

e_\gamma y_\gamma =
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(7,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_2}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-1$}--(8,0);
\draw (9,4)node[above]{$i$}--(9,0);
\node[greendot] at (0,2){};
\node[above right,scale = 0.7] at (0,2){$l_1$};
\node[greendot] at (1,2){};
\node[above right,scale = 0.7] at (1,2){$l_2$};
\node[greendot] at (3,2){};
\node[above right,scale = 0.7] at (3,2){$l_{m-1}$};
\node[greendot] at (4,2){};
\node[above right,scale = 0.7] at (4,2){$l_m$};
\node[greendot] at (5,2){};
\node[above right,scale = 0.7] at (5,2){$l_{m+1}$};
\node[greendot] at (6,2){};
\node[above right,scale = 0.7] at (6,2){$l_{m+2}$};
\node[greendot] at (8,2){};
\node[above right,scale = 0.7] at (8,2){$l_{2m}$};
\node[greendot] at (9,2){};
\node[above right,scale = 0.7] at (9,2){$l_{2m+1}$};
\end{braid}

where $\bi_\gamma = (i_1, i_2, \ldots, i_n)$ and $l_k = |\mathscr{A}_{\t^\gamma|_k}^{i_k}|$ is the multiplicity of the green dot on the $k$-th string. For the rest of this subsection, for clarity we will omit extraneous dots when they do not play an important role in the argument.

Next we introduce an important equivalent relation $=_\gamma$. For $\gamma\in\mathscr S_n^\Lambda$, and $r_1,r_2\in \mathscr{R}_n^\Lambda$, we write $r_1 =_\gamma r_2$ if $r_1 \pm r_2 \in \Blam[\gamma]$. It is clearly an equivalent relation. Moreover, by <ref>, for any $r \in \R$ we have $r_1{\cdot}r =_\gamma r_2{\cdot}r$ if $r_1 =_\gamma r_2$. This will be helpful for us to simplify the notations and calculations.

Recall that $\gamma_2 > 1$. We can write $\gamma_2 = k{\cdot}e + t$ for some nonnegative integer $k$ and $2 \leq t \leq e + 1$. We will first prove

\begin{equation} \label{I-problem: non-addable: help}
e_\gamma y_\gamma
\begin{cases}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(7,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_2}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(8,2)--(9,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-1$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i$}--(8,2)--(4,0);
\end{braid}, & \text{if $i \neq e-1$},\\
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(7,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_2}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(8,2)--(9,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-2$}--(7,2)--(8,0);
\draw (9,4)node[above]{$e-1$}--(8,2)--(5,0);
\end{braid}, & \text{if $i = e-1$}.
\end{cases}
\end{equation}

by induction on $k$, which can imply $e_\gamma y_\gamma \in R_n^{>\lambda}$ easily.

In order to clarify the meaning of the diagrams in (<ref>), let us give two examples below. In these
examples for convenience we fix $e = 4$.

Suppose $\gamma = (8,5)$, then $\gamma = \tab(\ \ \ \ \ \ \ \ ,\ \ \ \ \ )$ and $i = 3$. Then we are trying to prove that

e_\gamma y_\gamma  =
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw (2,4)node[above]{$2$}--(2,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\draw (6,4)node[above]{$2$}--(6,0);
\draw (7,4)node[above]{$3$}--(7,0);
\draw (8,4)node[above]{$3$}--(8,0);
\draw (9,4)node[above]{$0$}--(9,0);
\draw (10,4)node[above]{$1$}--(10,0);
\draw (11,4)node[above]{$2$}--(11,0);
\draw (12,4)node[above]{$3$}--(12,0);
\node[greendot] at (3,2){};
\node[greendot] at (7,2){};
\node[greendot] at (11,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw (2,4)node[above]{$2$}--(2,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\draw (6,4)node[above]{$2$}--(6,0);
\draw (7,4)node[above]{$3$}--(7,0);
\draw (8,4)node[above]{$3$}--(11,2) -- (12,0);
\draw (9,4)node[above]{$0$}--(8,2)--(9,0);
\draw (10,4)node[above]{$1$}--(9,2)--(10,0);
\draw (11,4)node[above]{$2$}--(10,2)--(11,0);
\draw (12,4)node[above]{$3$}--(11,2)--(8,0);
\node[greendot] at (3,2){};
\node[greendot] at (10,2){};
\end{braid}.

Suppose $\gamma = (9,10)$, then $\gamma = \tab(\ \ \ \ \ \ \ \ \ ,\ \ \ \ \ \ \ \ \ \ )$ and $i = 0$. We are trying to prove that

\begin{eqnarray*}
e_\gamma y_\gamma & = &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw (2,4)node[above]{$2$}--(2,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\draw (6,4)node[above]{$2$}--(6,0);
\draw (7,4)node[above]{$3$}--(7,0);
\draw (8,4)node[above]{$0$}--(8,0);
\draw (9,4)node[above]{$3$}--(9,0);
\draw (10,4)node[above]{$0$}--(10,0);
\draw (11,4)node[above]{$1$}--(11,0);
\draw (12,4)node[above]{$2$}--(12,0);
\draw (13,4)node[above]{$3$}--(13,0);
\draw (14,4)node[above]{$0$}--(14,0);
\draw (15,4)node[above]{$1$}--(15,0);
\draw (16,4)node[above]{$2$}--(16,0);
\draw (17,4)node[above]{$3$}--(17,0);
\draw (18,4)node[above]{$0$}--(18,0);
\node[greendot] at (3,2){};
\node[greendot] at (7,2){};
\node[greendot] at (12,2){};
\node[greendot] at (16,2){};
\end{braid}\\
& =_\gamma &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw (2,4)node[above]{$2$}--(2,0);
\draw (3,4)node[above]{$3$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$1$}--(5,0);
\draw (6,4)node[above]{$2$}--(6,0);
\draw (7,4)node[above]{$3$}--(7,0);
\draw (8,4)node[above]{$0$}--(17,2) -- (18,0);
\draw (9,4)node[above]{$3$}--(8,2)--(9,0);
\draw (10,4)node[above]{$0$}--(9,2)--(10,0);
\draw (11,4)node[above]{$1$}--(10,2)--(11,0);
\draw (12,4)node[above]{$2$}--(11,2)--(12,0);
\draw (13,4)node[above]{$3$}--(12,2)--(13,0);
\draw (14,4)node[above]{$0$}--(13,2)--(14,0);
\draw (15,4)node[above]{$1$}--(14,2)--(15,0);
\draw (16,4)node[above]{$2$}--(15,2)--(16,0);
\draw (17,4)node[above]{$3$}--(16,2)--(17,0);
\draw (18,4)node[above]{$0$}--(17,2)--(8,0);
\node[greendot] at (3,2){};
\node[greendot] at (7,2){};
\node[greendot] at (11,2){};
\node[greendot] at (15,2){};
\end{braid}.
\end{eqnarray*}

The next Proposition is the base case of the induction. When $k = 0$, we have $2\leq \gamma_2 \leq e + 1$.

Suppose $\gamma = (\gamma_1,\gamma_2) \in \mathscr C_n^\Lambda$ with $\gamma_2 > 1$ and $\gamma_2 - \gamma_1 \equiv 1\pmod{e}$ and $\lambda = (\gamma_1,\gamma_2-1,1) \in \mathscr S_n^\Lambda$. Define $i$ to be the residue of the node at position $(1,\gamma_1,1)$ or $(2,\gamma_2,1)$. When $2 \leq \gamma_2 \leq e + 1$, (<ref>) holds.

Before proving <ref> we first give a useful lemma.

For any $i \in I$, we have

\begin{braid}
\draw (0,4)node[above]{$i$}--(4,2)--(5,0);
\draw (1,4)node[above]{$i+1$}--(0,2)--(1,0);
\draw (2,4)node[above]{$i+2$}--(1,2)--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(3,2)--(4,0);
\draw (5,4)node[above]{$i$}--(4,2)--(0,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (1,4)node[above]{$i+1$}--(1,0);
\draw (2,4)node[above]{$i+2$}--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(4,0);
\draw (5,4)node[above]{$i$}--(5,0);
\node[greendot] at (5,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (1,4)node[above]{$i+1$}--(1,0);
\draw (2,4)node[above]{$i+2$}--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(4,0);
\draw (5,4)node[above]{$i$}--(5,0);
\node[greendot] at (4,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (1,4)node[above]{$i+1$}--(1,0);
\draw (2,4)node[above]{$i+2$}--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(4,0);
\draw (5,4)node[above]{$i$}--(5,0);
\node[greendot] at (1,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(0,0);
\draw (1,4)node[above]{$i+1$}--(1,0);
\draw (2,4)node[above]{$i+2$}--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(4,0);
\draw (5,4)node[above]{$i$}--(5,0);
\node[greendot] at (0,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$i$}--(1,2)--(5,0);
\draw (1,4)node[above]{$i+1$}--(2,2)--(1,0);
\draw (2,4)node[above]{$i+2$}--(3,2)--(2,0);
\draw[dots] (2.2,4)--(3.8,4);
\draw[dots] (2.2,0)--(3.8,0);
\draw (4,4)node[above]{$i-1$}--(5,2)--(4,0);
\draw (5,4)node[above]{$i$}--(1,2)--(0,0);
\end{braid}.

The Lemma follows by directly applying braid relations on the left hand side of the equation.

Now we are ready to prove <ref>.

We prove the Proposition by considering four different cases depending
upon the value of $i$. Notice that in this Proposition, we have $\gamma_1 \geq \gamma_2-1$ because $2 \leq \gamma_2 \leq e+1$ and $\gamma_2 - \gamma_1 \equiv 1\pmod{e}$.

Case  <ref>a: $i = 0$, i.e. $\gamma_2 = 2$.

\begin{eqnarray*}
e_\gamma y_\gamma & = &
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$0$}--(4,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(5,0);
\draw (6,4)[densely dotted] node[above]{$0$}--(6,0);
\end{braid}
\overset{(\ref{dia:braid})}=
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)[densely dotted] node[above]{$0$}--(4,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(4,0);
\node[greendot] at (3,2){};
\end{braid}\\
& \overset{(\ref{dia:ii3})} = &
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(5,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(4,2)--(6,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(4,0);
\node[greendot] at (3,2){};
\end{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,3)--(6,1)--(5,0);
\draw (6,4)node[above]{$0$}--(4,3)--(4,1)--(6,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(4,0);
\node[greendot] at (3,2){};
\end{braid}.
\end{eqnarray*}

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,3)--(6,1)--(5,0);
\draw (6,4)node[above]{$0$}--(4,3)--(4,1)--(6,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} =
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(5,0);
\draw (5,4)node[above]{$e-1$}--(6,0);
\draw (6,4)node[above]{$0$}--(4,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\draw (5,4)node[above]{$0$}--(5,0);
\draw (6,4)node[above]{$e-1$}--(6,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$0$}--(4,0);
\draw (6,4)node[above]{$e-1$}--(5,0);
\end{braid}

and if we define $\nu = (\gamma_1,1) = \lambda|_{\gamma_1+1}$, then as $\lambda \in \mathscr S_n^\Lambda$ and $|\nu| = \gamma_1+1 < n = \gamma_1 + \gamma_2 = |\lambda|$, $\nu \in \P_I$. Moreover as $b_0^{\nu_-} = 1$,

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(4,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\end{braid} = e(\bi_{\nu_-}\vee 0) y_{\nu_-} y_{|\nu|}^1 \in \Blam[\nu].

Then by <ref>,

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,3)--(6,1)--(5,0);
\draw (6,4)node[above]{$0$}--(4,3)--(4,1)--(6,0);
\node[greendot] at (3,2){};
\node[greendot] at (4,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}\in \Blam[\gamma].

e_\gamma y_\gamma =_\gamma \begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-1$}--(3,0);
\draw (4,4)node[above]{$0$}--(6,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(4,0);
\node[greendot] at (3,2){};
\end{braid},

which gives the proposition in this case.

Case  <ref>b: $1\leq i \leq e-3$, i.e. $3\leq \gamma_2 \leq e-1$.

\begin{eqnarray*}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(9,2)--(10,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(7,2)--(8,0);
\draw (9,4)[densely dotted] node[above]{$i-1$}--(8,2)--(9,0);
\draw (10,4)[densely dotted] node[above]{$i$}--(9,2)--(4,0);
\end{braid}
& \overset{(\ref{dia:braid})}= &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(9,2)--(10,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)[densely dotted] node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)[densely dotted] node[above]{$i-2$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(10,2)--(9,0);
\draw (10,4)[densely dotted] node[above]{$i$}--(9,2)--(4,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(9,0);
\draw (10,4)node[above]{$i$}--(10,0);
\end{braid}\\
& \overset{(\ref{dia:braid})} = &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(5,2)--(10,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(9,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(10,2)--(9,0);
\draw (10,4)node[above]{$i$}--(5,2)--(4,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)[densely dotted] node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)[densely dotted] node[above]{$i-2$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(9,0);
\draw (10,4)node[above]{$i$}--(10,0);
\end{braid}\\
& \overset{(\ref{dia:psipsi})}= &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(5,2)--(10,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(9,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(10,2)--(9,0);
\draw (10,4)node[above]{$i$}--(5,2)--(4,0);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(8,0);
\draw (9,4)node[above]{$i-1$}--(9,0);
\draw (10,4)node[above]{$i$}--(10,0);
\end{braid}\\
& = &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(4,1)--(4.5,0.5)--(10,0);
\draw (5,4)node[above]{$e-1$}--(6,3)--(6,1)--(5,0);
\draw (6,4)node[above]{$0$}--(7,3)--(7,1)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(9,3)--(9,1)--(8,0);
\draw (9,4)node[above]{$i-1$}--(10,3)--(10,1)--(9,0);
\draw (10,4)node[above]{$i$}--(5,3)--(5,1)--(4.5,0.5)--(4,0);
\draw[loosely dashed,red] (-0.5,3) -- (5.5,3)--(5.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(8,0);
\draw (9,4)node[above]{$i-1$}--(9,0);
\draw (10,4)node[above]{$i$}--(10,0);
\end{braid}.
\end{eqnarray*}

For the same reason as in Case  <ref>a,

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(5,2)--(10,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$i-2$}--(9,2)--(8,0);
\draw (9,4)node[above]{$i-1$}--(10,2)--(9,0);
\draw (10,4)node[above]{$i$}--(5,2)--(4,0);
\end{braid}
\in \Blam[\gamma],

which implies the proposition in this case.

Case  <ref>c: $i = e-2$, i.e. $\gamma_2 = e$. By <ref>,

\begin{eqnarray*}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(8,2)--(9,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(7,2)--(8,0);
\draw (9,4)node[above]{$e-2$}--(8,2)--(4,0);
\end{braid}
& = &
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (9,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (8,2){};
\end{braid}\\
& - &
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (5,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (4,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(5,2)--(9,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(9,2)--(8,0);
\draw (9,4)node[above]{$e-2$}--(5,2)--(4,0);
\end{braid}.
\end{eqnarray*}

Set $\nu = (\gamma_1,\gamma_2-1) = \gamma|_{n-1}$. As $\gamma_1 \geq \gamma_2-1$, we have $\nu\in \P_{n-1}$. As $\lambda\in\mathscr S_n^\Lambda$ and $|\nu| < |\lambda|$, we have $\nu \in \P_I$. It is not hard to see that $b_{e-3}^{\nu_-} = 1$. Hence

\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(7,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_2-1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-4$}--(8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\node[greendot] at (9,2){};
\end{braid} = e(\bi_{\nu_-}\vee e-3)y_{\nu_-} y_{n-1}^1 \in \Blam[\nu].

Then by <ref>,

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (8,2){};
\end{braid} \in \Blam[\gamma].

Similarly, we have

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (5,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (4,2){};
\end{braid} \in \Blam[\gamma],

and for the similar argument as in Case  <ref>a, we have

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(5,2)--(9,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(5,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(9,2)--(8,0);
\draw (9,4)node[above]{$e-2$}--(5,2)--(4,0);
\end{braid} \in \Blam[\gamma].

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(8,2)--(9,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(7,2)--(8,0);
\draw (9,4)node[above]{$e-2$}--(8,2)--(4,0);
\end{braid} =_\gamma
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-3$}--(3,0);
\draw (4,4)node[above]{$e-2$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw[dots] (6.2,4)--(7.8,4);
\draw[dots] (6.2,0)--(7.8,0);
\draw (8,4)node[above]{$e-3$}--(8,0);
\draw (9,4)node[above]{$e-2$}--(9,0);
\node[greendot] at (9,2){};
\end{braid} = e_\gamma y_\gamma,

which follows the Proposition.

Case  <ref>d: $i = e-1$, i.e. $\gamma_2 = e + 1$. By <ref>,

\begin{eqnarray}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(10,2)--(11,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw (7,4)node[above]{$1$}--(6,2)--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(8,2)--(9,0);
\draw (10,4)node[above]{$e-2$}--(9,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,2)--(5,0);
\node[greendot] at (9,2){};
\end{braid}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,2){};
\node[greendot] at (11,2){};
\end{braid}\notag\\
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,3){};
\node[greendot] at (10,1){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,2){};
\node[greendot] at (6,2){};
\end{braid} \notag \\
&& +
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (5,2){};
\node[greendot] at (10,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(11,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw (7,4)node[above]{$1$}--(8,2)--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(10,2)--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(6,2)--(5,0);
\node[greendot] at (11,2){};
\end{braid} \label{I-problem: short: equation1}.
\end{eqnarray}

For the first two terms of (<ref>),

\begin{align*}
& \begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,2){};
\node[greendot] at (11,2){};
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,3){};
\node[greendot] at (10,1){};
\end{braid}
\overset{(\ref{dia:psipsi})}=
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$e-1$}--(4,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,2)--(11,0);
\node[greendot] at (11,2){};
\end{braid}\\
\overset{(\ref{dia:ii})}= &
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,1)--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,2)--(11,0);
\node[greendot] at (4,2.7){};
\node[greendot] at (4,1.3){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,1)--(5,0);
\draw (5,4)node[above]{$e-1$}--(5,1)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,3)--(11,1)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,3)--(10,1)--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (10.5,3)--(10.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,3)--(11,1)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,3)--(10,1)--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (10.5,3)--(10.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}.
\end{align*}

Using the similar argument as in Case  <ref>a,

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,1)--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2.7){};
\node[greendot] at (4,1.3){};
\node[greendot] at (10,2){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} \in \Blam[\gamma],

and using the similar argument in Case <ref>c,

\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,1)--(5,0);
\draw (5,4)node[above]{$e-1$}--(5,1)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,3)--(11,1)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,3)--(10,1)--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (10.5,3)--(10.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,3)--(11,1)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,3)--(10,1)--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (10.5,3)--(10.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}\in \Blam[\gamma].

Hence the first two terms of (<ref>) is in $\Blam[\gamma]$. For the third term of (<ref>),

\begin{align*}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$e-1$}--(4,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,2){};
\node[greendot] at (6,2){};
\end{braid}\\
\overset{(\ref{dia:ii})}= &
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,1)--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2.7){};
\node[greendot] at (4,1.3){};
\node[greendot] at (10,2){};
\node[greendot] at (6,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,1)--(5,0);
\draw (5,4)node[above]{$e-1$}--(5,1)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (10,2){};
\node[greendot] at (6,2){};
\draw[loosely dashed,red] (-0.5,3) -- (6.5,3)--(6.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (10,2){};
\node[greendot] at (6,2){};
\node[greendot] at (4,2){};
\draw[loosely dashed,red] (-0.5,3) -- (6.5,3)--(6.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} \in \Blam[\gamma].
\end{align*}

For the fourth term of (<ref>), by (<ref>),

\begin{align*}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$e-1$}--(4,0);
\draw (5,4)[densely dotted] node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (5,2){};
\node[greendot] at (10,2){};
\end{braid}
& =
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,1)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,1)--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2.7){};
\node[greendot] at (4,1.3){};
\node[greendot] at (5,0){};
\node[greendot] at (10,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(5,3)--(5,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (5,2){};
\node[greendot] at (10,2){};
\draw[loosely dashed,red] (-0.5,3) -- (5.5,3)--(5.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}\\
& -
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,1)--(5,0);
\draw (5,4)node[above]{$e-1$}--(5,1)--(4,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,1.3){};
\node[greendot] at (4,2.7){};
\node[greendot] at (10,2){};
\draw[loosely dashed,red] (-0.5,3) -- (4.5,3)--(4.5,1)--(-0.5,1)--(-0.5,3);
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (10,2){};
\end{braid}\\
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(5,0);
\draw (6,4)node[above]{$0$}--(6,0);
\draw (7,4)node[above]{$1$}--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(9,0);
\draw (10,4)node[above]{$e-2$}--(10,0);
\draw (11,4)node[above]{$e-1$}--(11,0);
\node[greendot] at (4,2){};
\node[greendot] at (10,2){};
\end{braid} = e_\gamma y_\gamma,
\end{align*}

and for the last term of (<ref>),

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,2)--(11,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw (7,4)node[above]{$1$}--(8,2)--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(10,2)--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(6,2)--(5,0);
\node[greendot] at (11,2){};
\end{braid}
\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(6,3)--(6,1)--(11,0);
\draw (6,4)node[above]{$0$}--(7,2)--(6,0);
\draw (7,4)node[above]{$1$}--(8,2)--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(10,2)--(9,0);
\draw (10,4)node[above]{$e-2$}--(11,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(5,3)--(5,0);
\node[greendot] at (11,2){};
\draw[loosely dashed,red] (-0.5,3) -- (6.5,3)--(6.5,1)--(-0.5,1)--(-0.5,3);
\end{braid} \in \Blam[\gamma].

Combine the result above, we have

\begin{braid}
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$e-2$}--(3,0);
\draw (4,4)node[above]{$e-1$}--(4,0);
\draw (5,4)node[above]{$e-1$}--(10,2)--(11,0);
\draw (6,4)node[above]{$0$}--(5,2)--(6,0);
\draw (7,4)node[above]{$1$}--(6,2)--(7,0);
\draw[dots] (7.2,4)--(8.8,4);
\draw[dots] (7.2,0)--(8.8,0);
\draw (9,4)node[above]{$e-3$}--(8,2)--(9,0);
\draw (10,4)node[above]{$e-2$}--(9,2)--(10,0);
\draw (11,4)node[above]{$e-1$}--(10,2)--(5,0);
\node[greendot] at (9,2){};
\end{braid}
=_\gamma e_\gamma y_\gamma,

which completes the proof.

The technique of applying <ref> in proving <ref> will be used many times in the rest of the paper. Although the process is straightforward, individual details will vary from case to case, thus in order to simplify the process we will omit details in the future.

Recall $\gamma_2 = k{\cdot}e + t$ where $k$ is a nonnegative integer and $2 \leq t \leq e + 1$. Now we remove the restriction on $\gamma_2$ by applying the induction on $k$.

Suppose $\gamma = (\gamma_1,\gamma_2) \in \mathscr C_n^\Lambda$ with $\gamma_2 > 1$ and $\gamma_2 - \gamma_1 \equiv 1\pmod{e}$ and $\lambda = (\gamma_1,\gamma_2-1,1) \in \mathscr S_n^\Lambda$. Define $i$ to be the residue of the node at position $(1,m,1)$. Then (<ref>) holds.

We prove this Proposition by induction. As we claimed before that we can write $\gamma_2 = k{\cdot}e + t$ with $2 \leq t \leq e+1$ and we will apply induction on $k$. <ref> implies that for $k = 0$ the Proposition holds. Assume that for $k \leq k'$ the Proposition holds. For $k = k'$, we consider two different cases, which are $i = e-2$, $i = e-1$ and $i\neq e-2,e-1$. Recall that $i$ is the residue of the node at $(1,m,1)$ or $(2,m+1,1)$.

Case <ref>a: $i \neq e-2, e-1$.

\begin{eqnarray*}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(13,2)--(14,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(8,2)--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)[densely dotted] node[above]{$e-2$}--(10,2)--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(12,2)--(13,0);
\draw (14,4)node[above]{$i$}--(13,2)--(4,0);
\node[greendot] at (10,2){};
\draw[->] (10,2) -- (10.5,1);
\end{braid}
\overset{(\ref{dia:y-psi com})}=
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(13,2)--(14,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(8,2)--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(10,2)--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(12,2)--(13,0);
\draw (14,4)node[above]{$i$}--(13,2)--(4,0);
\node[greendot] at (11,0){};
\end{braid}\\
& = &
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(8,3)--(13,2)--(14,1)--(14,0);
\draw (5,4)node[above]{$e-1$}--(4,3)--(4,1)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,3)--(6,1)--(7,0);
\draw (8,4)node[above]{$i$}--(7,3)--(7,1)--(8,0);
\draw (9,4)node[above]{$i+1$}--(9,3)--(8,2)--(9,1)--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(11,3)--(10,2)--(11,1)--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(13,3)--(12,2)--(13,1)--(13,0);
\draw (14,4)node[above]{$i$}--(14,3)--(13,2)--(8,1)--(4,0);
\node[greendot] at (11,0){};
\draw[loosely dashed,red] (7.5,3) -- (14.5,3)--(14.5,1)--(7.5,1)--(7.5,3);
\end{braid}\hspace*{11mm} \text{ by \autoref{I-problem: short}}\\
& = &
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)[densely dotted] node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(13,0);
\draw (14,4)node[above]{$i$}--(14,0);
\node[greendot] at (11,0){};
\node[greendot] at (14,2){};
\end{braid}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)[densely dotted] node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(13,0);
\draw (14,4)node[above]{$i$}--(14,0);
\node[greendot] at (11,0){};
\node[greendot] at (13,2){};
\end{braid}\\
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)[densely dotted] node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(13,0);
\draw (14,4)node[above]{$i$}--(14,0);
\node[greendot] at (11,0){};
\node[greendot] at (9,2){};
\end{braid}
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)[densely dotted] node[above]{$i$}--(8,2)--(4,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)[densely dotted] node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(11,0);
\draw[dots] (11.2,4)--(12.8,4);
\draw[dots] (11.2,0)--(12.8,0);
\draw (13,4)node[above]{$i-1$}--(13,0);
\draw (14,4)node[above]{$i$}--(14,0);
\node[greendot] at (11,0){};
\node[greendot] at (8,2){};
\draw[->] (8,2) -- (7,2.5);
\end{braid}\\
\begin{braid}
\draw(2,-0.5)node{$\underbrace{\hspace*{16mm}}_{\gamma_1}$};
\draw(6.5,-0.5)node{$\underbrace{\hspace*{12mm}}_{\gamma_2-e}$};
\draw(11.5,-0.5)node{$\underbrace{\hspace*{20mm}}_{e}$};
\draw (0,4)node[above]{$0$}--(0,0);
\draw (1,4)node[above]{$1$}--(1,0);
\draw[dots] (1.2,4)--(2.8,4);
\draw[dots] (1.2,0)--(2.8,0);
\draw (3,4)node[above]{$i-1$}--(3,0);
\draw (4,4)node[above]{$i$}--(9,2)--(14,0);
\draw (5,4)node[above]{$e-1$}--(4,2)--(5,0);
\draw[dots] (5.2,4)--(6.8,4);
\draw[dots] (5.2,0)--(6.8,0);
\draw (7,4)node[above]{$i-1$}--(6,2)--(7,0);
\draw (8,4)node[above]{$i$}--(7,2)--(8,0);
\draw (9,4)node[above]{$i+1$}--(10,2)--(9,0);
\draw[dots] (9.2,4)--(10.8,4);
\draw[dots] (9.2,0)--(10.8,0);
\draw (11,4)node[above]{$e-2$}--(12,2)--(11,0);