We extend the diagrammatic presentation of the Hecke category to one for the dg category whose objects
are the standard representatives for Rouquier complexes. We use this description to recover basic results
about these complexes, namely the categorification of the relations of the braid group and the Rouquier formula.

§ INTRODUCTION

The diagrammatic Hecke category is an extremely powerful tool.
It was introduced by Elias and Williamson <cit.> as a presentation by generators and relations
of the category of Soergel bimodules.
One first reduces to the description of the subcategory formed by certain Bott-Samelson objects,
which can be represented as sequences of simple reflections, then one describes morphisms
between these objects as linear combinations of diagrams identified under some relations.
This diagrammatic description
showed its utility in several occurrences: see for instance Remark <ref>.

On the other hand, Rouquier <cit.> introduced some complexes of Soergel bimodules
to categorify actions of the braid group on categories. These Rouquier complexes
are all possible tensor products between certain standard and costandard complexes
$F_s$ and $F_s^{-1}$ (for each simple reflection $s$). These objets satisfy a categorified
version	of the relations of the braid group: see Proposition <ref> below.
This allow to unambiguously consider the object $F_\omega$ associated to a braid $\omega$.

Another important property of these complexes is the so-called Rouquier formula.
Given $w\in W$, its positive lift is obtained by replacing
each simple reflection in any reduced word for $w$ by the corresponding positive
generator of the braid group. This is a well defined element of the braid group.
Similarly one defines the negative lift.
Let then $\omega$ be the positive lift of $w\in W$, and $\nu$ the negative
lift of another element $v$, then

\[
\Hom_{\Kb(\D)}(F_{\omega},F_{\nu})=	\begin{cases}
R	&\text{if $w=v$,}\\
0	&\text{otherwise.}
\end{cases}
\]

Rouquier <cit.> showed this for Weyl groups and conjectured that it should hold for general Coxeter systems.
This was showed by Libedinsky and Williamson <cit.> and with other methods by Makisumi <cit.>.

In this paper we consider the dg category of complexes $\Cdgb(\D)$ with objects in the Hecke category.
We extend the diagrammatic description of $\D$ to one for the subcategory of $\Cdgb(\D)$ generated by
representatives of Rouquier complexes of the form $F_{s_1}^{\pm 1} \otimes F_{s_2}^{\pm 1} \otimes \dots \otimes F_{s_n}^{\pm 1}$.

The graded objects associated with each of them splits into Bott-Samelson objects
(one for each subexpression of $s_1s_2\dots s_n$),
each of which lies in a certain cohomological degree.
Roughly speaking, morphisms between two such complexes are again linear
combinations of diagrams, but in each diagram one has
to specify the starting and ending summand. To do this one covers some of the boundary
points by patches, thus picking a certain subexpression of the starting (and ending) word.
Then we describe the rules to compute the image of a certain morphism via the differential map.

With this language we recover the classical results above: braid relations and Rouquier formula.
We also give a recipe to explicitely find the homotopy equivalences giving braid relations and
write those for $m=2,3$.

§ REVIEW OF THE DIAGRAMMATIC HECKE CATEGORY

We now recall the construction,
by Elias and Williamson <cit.>, of the diagrammatic Hecke category associated with
(a realization of) an arbitrary Coxeter system.	First we fix some notation.

§.§ Coxeter systems, braid groups and Hecke algebras

Let $(W,S)$ be a Coxeter system, with $|S|<\infty$.
Let $\cword$ be the free monoid generated by $S$. Its elements are called Coxeter words
and will be denoted by underlined letters.
If $\uw$ is a Coxeter word, we say that it expresses
the element $w\in W$ if $w$ is its image via the natural morphism $\cword\rightarrow W$.

Let $\ell(\uw)$ denote the length of $\uw$, i.e.		the number of its letters.
The word $\uw$ is said to be reduced if there is no shorter word expressing $w$. This defines
a length function $\ell: W\rightarrow \mathbb{N}$, where $\ell(w)$ is the length
of any reduced word for $w$.

For $\uw, \ux \in \cword$, we say that $\ux$ is a subword of $\uw$, and we
write $\ux\preceq \uw$,
if $\ux$ is obtained from $\uw$ by erasing some letters. In other words,
if $\uw=s_1s_2\cdots s_n$ with $s_i\in S$, then
$\ux$ is of the form $s_{i_1}s_{i_2}\cdots s_{i_r}$,
with $1\le i_1 < i_2 < \dots < i_r\le n$. A subexpression is instead the datum of a subword
together with the information of the precise positions of the letters in the original word. This
can be encoded in a 01-sequence $\boe\in\{0,1\}^n$ where the ones correspond to the letters
that we pick and the zeros to those we do not pick.

The Bruhat graph is the directed graph whose vertices are the elements $w\in W$ and
there is a directed edge from $w$ to all elements of the form $wt$ with $t$ a reflection
such that $\ell(wt)>\ell(w)$. Then we define the Bruhat order as the transitive closure
of the relation defined by the Bruhat graph.

Let $B_W$ be the Artin-Tits group (or generalized braid group) associated to $(W,S)$.
Let $\sigma_s$ denote the generator of $B_W$ associated to $s\in S$.
We have a natural surjection $B_W\twoheadrightarrow W$ sending both $\sigma_s$ and
$\sigma_s^{-1}$ to $s$. Let $\Sigma^+=\{\sigma_s\}_{s\in S}$,
$\Sigma^-=\{\sigma_s^{-1}\}_{s\in S}$ and $\Sigma=\Sigma^+ \sqcup \Sigma^-$.
Let $\bword$ be the free monoid generated by $\Sigma$. Its elements are called
braid words. We call the elements of $\Sigma^+$ (and $\Sigma^-$) positive letters
(respectively negative letters).
A braid word is called positive (or negative) if all its letters are.
The notion of subword is defined in $\bword$ in the same way as in $\cword$.

We call Coxeter projection the map $\bword\rightarrow\cword$, sending both
$\sigma_s$ and $\sigma_s^{-1}$ to $s$. This has two distinguished sections: any Coxeter word $\uw=s_1s_2\dots s_k$
has a positive word lift $\sigma_{s_1}\sigma_{s_2}\dots \sigma_{s_k}$
and a negative word lift $\sigma_{s_1}^{-1}\sigma_{s_2}^{-1}\dots \sigma_{s_k}^{-1}$.
These also descend to well defined sections $W\rightarrow B_W$ sending $w$ to
the braid expressed by the positive (or negative) word lift of any reduced word for $w$.
This is called positive (or negative) lift of $w$.

Consider the ring $\Laur$ of Laurent polynomial in one
variable with integer coefficients. Let $\HH_{(W,S)}$ (or simply $\HH$) denote the Hecke algebra
associated to $(W,S)$. Recall that this is the
$\Laur$-algebra generated by $\{\delta_s\}_{s\in S}$ with relations

\[
\begin{cases}
(\delta_s+v)(\delta_s-v^{-1})=0& 	\text{for all $s\in S$}\\
\underbrace{\delta_s\delta_t\dots}_{m_{st}}=\underbrace{\delta_t\delta_s\dots}_{m_{st}}& 		\text{if $m_{st}<\infty$}
\end{cases}
\]

§.§ Realizations of Coxeter systems
For a Coxeter system
$(W,S)$ and a commutative ring $\kk$, we consider a realization $\hh$
of $W$, in the sense of <cit.>. Recall that this consists in a free,
finite rank, $\kk$-module $\mathfrak{h}$, with distinguished elements
$\{\alpha_s^\vee\}_{s\in S} \subset \mathfrak{h}$ and
$\{ \alpha_s \}_{s\in S} \subset \mathfrak{h}^*=\Hom(\mathfrak{h},k)$,
called respectively simple coroots and simple roots, such that
$\langle \alpha_s,\alpha_s^\vee\rangle =2$, for each $s\in S$ and that $W$ acts linearly
on $\hh$ via $s(v)=v-\langle \alpha_s, v\rangle \alpha_s^{\vee}$ and similarly on $\hh^*$.
We also assume the technical condition <cit.> to ensure
rotation invariance of Jones-Wenzl morphisms which will be mentioned later.

Let $R=S(\mathfrak{h}^*)$, with $\mathfrak{h}^*$ in degree 2. The above action of $W$ on
$\mathfrak{h}^*$ extends naturally to $R$.
We define, for each $s\in S$, the Demazure operator $\partial_s : R \rightarrow R$, via:

\[
f \mapsto \frac{f-s(f)}{\alpha_s}.
\]

Then, in particular, we have
$\partial_s(\alpha_t)=\langle \alpha_t,\alpha_s^\vee\rangle$.

Examples of realizations can be found in <cit.>.

In this paper we will only consider balanced realizations (see <cit.>)
that satisfy Demazure surjectivity (see <cit.>).

§.§ The diagrammatic Hecke category

Given a realization $\mathfrak{h}$ over $\kk$ of a Coxeter system $(W,S)$,
one constructs the corresponding diagrammatic Hecke category $\D=\D(\mathfrak{h},\kk)$.
This is a $\kk$-linear monoidal category enriched in graded $R$-bimodules.
First one defines the Bott-Samelson category $\D_{\mathrm{BS}}$ by generators and relations,
then one gets $\D$ as the Karoubi envelope of the closure of $\D_{\mathrm{BS}}$ by direct
sums and shifts.

* The objects of $\DBS$ are generated by tensor product from objects $B_s$ for $s\in S$. So a general object
corresponds to a Coxeter word: if
$\uw=s_1\dots s_n$, let $B_{\uw}$ denote the object
$B_{s_1}\otimes \cdots \otimes B_{s_n}$. Let also $\un$ denote the monoidal unit, corresponding to the
empty word.

* Morphisms in $\Hom_{\DBS}(B_{\uw_1},B_{\uw_2})$
are $\kk$-linear combinations of Soergel graphs, which are defined as follows.

* We associate a color to each simple reflection.

* A Soergel graph is then a colored, decorated planar graph
contained in the strip $\mathbb{R}\times [0,1]$, with boundary in

* The bottom (and top) boundary is the arrangement
of boundary points colored according to the letters of the source word $\uw_1$
(and the target word $\uw_2$, respectively).

* The edges of the graph are colored in such a way that those connected with the
boundary have colors consistent with the boundary points.

* The other vertices of the graph are either:

* univalent (called dots), which are declared of degree 1, or;

* trivalent	with three edges of the same color, of degree $-1$, or;

* $2m_{st}$-valent with edges of alternating colors corresponding to $s$ and $t$,
if $m_{st}<\infty$, of degree 0.
We also call them $(s,t)$-ars.

[gray,dashed] (0,0) circle (1cm);
[red] (0,-1)–(0,0); [red] (0,0) circle (2pt);
[anchor=north,inner sep=8pt] at (0,-1) (a) Dot;
[gray,dashed] (0,0) circle (1cm);
[red] (0,1)–(0,0);
[red] (-30:1cm)–(0,0)–(-150:1cm);
[anchor=north,inner sep=8pt] at (0,-1) (b) Trivalent;

[gray,dashed] (0,0) circle (1cm);
[red] (0,1)–(0,0);
[red] (-30:1cm)–(150:1cm);
[blue,rotate=30] (0,1)–(0,0);
[blue,rotate=30] (-30:1cm)–(150:1cm);
ıin -60,-75,-90
(ı:.5cm) circle (.5pt);

[anchor=north,xshift=1.4*,inner sep=8pt] at (0,-1) (c) $(s,t)$-ar;

* Decorations are boxes labeled by homogeneous elements in
$R$ that can appear in
any region (i.e. connected component of the complement of the graph):
we will usually omit the boxes and just write the polynomials.

Then, composition of morphism is given by gluing diagrams vertically, whereas tensor product
is given by gluing them horizontally. The identity morphism of the object $B_{\uw}$ is
the diagram with parallel vertical strands colored according to the word $\uw$.

* These diagrams are identified via some relations:

Polynomial relations. The first such relation says that whenever two polynomials are in the same region, they multiply.
This gives morphism spaces the structure of $R$-bimodules, by acting on the leftmost or the
rightmost region.

The other polynomial relations are:

\begin{align}
& \text{\emph{Barbell} relation:}  &
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape] \label{barbell}
\draw[gray,dashed] (0,0) circle (1cm); \draw[red] (0,.5)--(0,-.5); \fill[red] (0,.5) circle (2pt);\fill[red] (0,-.5) circle (2pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape]
\draw[gray,dashed] (0,0) circle (1cm); \node[draw] at (0,0) {$\alpha_{\re{s}}$};
\end{tikzpicture}			\\
& \text{\emph{Sliding} relation:} &
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape]\label{sliding}
\draw[gray,dashed] (0,0) circle (1cm); \draw[red] (0,1)--(0,-1);
\node[draw] at (.5,0) {$f$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape]
\draw[gray,dashed] (0,0) circle (1cm); \draw[red] (0,1)--(0,-1);
\node[draw, inner sep =.08cm] at (-.5,0) {$\re{s}(f)$};
\end{tikzpicture}
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape]
\draw[gray,dashed] (0,0) circle (1cm); \draw[red] (0,1)--(0,.5); \fill[red] (0,.5) circle (2pt);
\draw[red] (0,-1)--(0,-.5); \fill[red] (0,-.5) circle (2pt);
\node[draw] at (0,0) {$\partial_{\re{s}}(f)$};
\end{tikzpicture}
\end{align}

One color relations. These are the following:

\begin{align}
& \text{\emph{Frobenius associativity}:} &
\begin{tikzpicture}[baseline=-0.1cm,color=red,scale=sqrt(2)/3,scale=.7, transform shape]\label{associativity}
\draw (-1.5,-1.5)--(-0.6,0)--(0.6,0)--(1.5,-1.5);
\draw (-1.5,1.5)--(-0.6,0); \draw(0.6,0)--(1.5,1.5);
\draw[dashed,gray] (0,0) circle ({3/sqrt(2)});
\end{tikzpicture}
\begin{tikzpicture}[baseline=-0.1cm,color=red,scale=sqrt(2)/3,rotate=90,scale=.7, transform shape]
\draw (-1.5,-1.5)--(-0.6,0)--(0.6,0)--(1.5,-1.5);
\draw (-1.5,1.5)--(-0.6,0); \draw(0.6,0)--(1.5,1.5);
\draw[dashed,gray] (0,0) circle ({3/sqrt(2)});
\end{tikzpicture} \\
& \text{\emph{Frobenius unit}:} &
\begin{tikzpicture}[color=red,baseline=-0.1cm,scale=.7, transform shape]\label{dotline}
\draw[gray,dashed] (0,0) circle (1cm);
\draw (0,-1) -- (0,1); \draw (0,0)--(.3,0);\fill (0.3,0) circle (2pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=-0.1cm,scale=.7, transform shape]
\draw[gray,dashed] (0,0) circle (1cm);
\draw[red] (0,-1) -- (0,1);
\end{tikzpicture}
\\
&\text{\emph{Needle} relation:} &
\begin{tikzpicture}[baseline=0,scale=.7, transform shape]\label{eq_needle}
\draw[gray,dashed] (0,0) circle (1cm);
\draw[red] (0,-1)--(0,-0.3) arc (-90:270:0.3cm);
\end{tikzpicture}
&= 0
\end{align}

Two color relations. These allow to move dots,
or trivalent vertices, across $(s,t)$-ars.
We give the two versions, according to the parity of $m_{st}$:

\begin{equation}\label{2mtriv}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.3,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[red] ({180-2*360/11}:1cm)--(a); \draw[blue] ({180-5*360/11}:1cm)--(b);
\draw[red] ({180-4*360/11}:1cm)--(a);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[red] ({180+2*360/11}:1cm)--(a); \draw[blue] ({180+5*360/11}:1cm)--(b);
\draw[red] ({180+4*360/11}:1cm)--(a);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\draw[blue](a)--(b);
\node at (-0.05,.5) {...};\node at (-0.05,-.5) {...};
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.5,0); \coordinate (b) at (0.15,0.35); \coordinate (c) at (0.15,-0.35);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(b);
\draw[red] ({180-2*360/11}:1cm)--(b); \draw[blue] ({180-5*360/11}:1cm)--(b);
\draw[red] ({180-4*360/11}:1cm)--(b);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(c);
\draw[red] ({180+2*360/11}:1cm)--(c); \draw[blue] ({180+5*360/11}:1cm)--(c);
\draw[red] ({180+4*360/11}:1cm)--(c);% \draw[blue] ({180+6*360/11}:1cm)--(c);
\draw[blue](c)..controls (-0.15,0)..(b); \draw[red] (b) -- (c);\draw[red] (b)--(a)--(c);
\draw[red] (b) ..controls (0.7,0).. (c);
\node at (.2,.6) {...};\node at (.2,-.6) {...};\node at (.35,0) {...};
\end{tikzpicture}\quad\text{or}\quad
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.3,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[red] ({180-2*360/11}:1cm)--(a); \draw[red] ({180-5*360/11}:1cm)--(b);
\draw[blue] ({180-4*360/11}:1cm)--(a);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[red] ({180+2*360/11}:1cm)--(a); \draw[red] ({180+5*360/11}:1cm)--(b);
\draw[blue] ({180+4*360/11}:1cm)--(a);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\draw[red](a)--(b);
\node at (-0.05,.5) {...};\node at (-0.05,-.5) {...};
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.5,0); \coordinate (b) at (0.15,0.35); \coordinate (c) at (0.15,-0.35);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(b);
\draw[red] ({180-2*360/11}:1cm)--(b); \draw[red] ({180-5*360/11}:1cm)--(b);
\draw[blue] ({180-4*360/11}:1cm)--(b);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(c);
\draw[red] ({180+2*360/11}:1cm)--(c); \draw[red] ({180+5*360/11}:1cm)--(c);
\draw[blue] ({180+4*360/11}:1cm)--(c);% \draw[blue] ({180+6*360/11}:1cm)--(c);
\draw[blue](c)..controls (-0.15,0)..(b); \draw[red] (b) -- (c);\draw[red] (b)--(a)--(c);
\draw[blue] (b) ..controls (0.7,0).. (c);
\node at (.2,.6) {...};\node at (.2,-.6) {...};\node at (.35,0) {...};
\end{tikzpicture}
\end{equation}

\begin{equation}\label{eq_2mdot}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.2,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[blue] ({180-3*360/11}:1cm)--(a);
\draw[red] ({180-4*360/11}:1cm)--(a);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[blue] ({180+3*360/11}:1cm)--(a); \fill[blue] (b) circle (2pt);
\draw[red] ({180+4*360/11}:1cm)--(a);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\draw[blue](a)--(b);
\node at (-0.4,.5) {...};\node at (-0.4,-.5) {...};
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.2,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[blue] ({180-3*360/11}:1cm)--(a);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[blue] ({180+3*360/11}:1cm)--(a);
\draw[red] ({180+4*360/11}:1cm) -- (b) -- ({180-4*360/11}:1cm); \draw[red](a)--(b);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\node at (-0.4,.6) {...};\node at (-0.4,-.6) {...};
\node[draw,circle, inner sep=.1cm,fill=white] at (a) {$\JW$};
\end{tikzpicture}
\quad\text{or}\quad
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.2,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[red] ({180-3*360/11}:1cm)--(a);
\draw[blue] ({180-4*360/11}:1cm)--(a);% \draw[blue] ({180-6*360/11}:1cm)--(b);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[red] ({180+3*360/11}:1cm)--(a); \fill[red] (b) circle (2pt);
\draw[blue] ({180+4*360/11}:1cm)--(a);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\draw[red](a)--(b);
\node at (-0.4,.5) {...};\node at (-0.4,-.5) {...};
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,scale=.8,transform shape]
\draw[dashed,gray] (0,0) circle (1cm);
\coordinate (a) at (-.2,0); \coordinate (b) at (0.5,0);
\draw[red] (180:1cm)--(a); \draw[blue] ({180-360/11}:1cm)--(a);
\draw[red] ({180-3*360/11}:1cm)--(a);
\draw[blue] ({180+360/11}:1cm)--(a);
\draw[red] ({180+3*360/11}:1cm)--(a);
\draw[blue] ({180+4*360/11}:1cm) -- (b) -- ({180-4*360/11}:1cm); \draw[blue](a)--(b);% \draw[blue] ({180+6*360/11}:1cm)--(b);
\node at (-0.4,.6) {...};\node at (-0.4,-.6) {...};
\node[draw,circle, inner sep=.1cm,fill=white] at (a) {$\JW$};
\end{tikzpicture}
\end{equation}

where the circles labeled $\JW$ are the Jones-Wenzel morphisms.
These are certain $\kk$-linear combinations of diagrams (with circular boundary
and $2m_{st}-2$ boundary points around it)
that can be described in terms of the 2-colored Temperley-Lieb category[
More precisely, one can obtain them by deformation retract from a colored
version of certain idempotents in the Temperley-Lieb algebra, which correspond
to the highest weight irreducible summands inside tensor powers of the standard representation
of $U_q(\mathfrak{sl}_2)$.].

Here are the examples for the first few values of $m_{st}$:

\begin{align*}
&m_{st}=2& %
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\node[draw,circle, inner sep=.1cm,fill=white] (a) at (0,0) {$\JW$};
\draw[red] (0:-\radius)--(a);
\draw[blue] (0:\radius)--(a);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (0:-\radius)--(0:{-\radius/3});\fill[red] (0:{-\radius/3}) circle (1.5pt);
\draw[blue] (0:\radius)--(0:{\radius/3});\fill[blue] (0:{\radius/3}) circle (1.5pt);
\end{tikzpicture}
\\
&m_{st}=3& %
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\node[draw,circle, inner sep=.1cm,fill=white] (a) at (0,0) {$\JW$};
\foreach \i in {0,180}{%
\draw[red] (\i:\radius)--(a);
\foreach \i in {90,270}{%
\draw[blue] (\i:\radius)--(a);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\foreach \i in {0,180}{%
\draw[red] (\i:\radius)--(\i:{\radius/3});%
\fill[red] (\i:{\radius/3}) circle (1.5pt);
\draw[blue] (0,\radius)--(0,-\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (\radius,0)--(-\radius,0);
\foreach \i in {90,270}{%
\draw[blue] (\i:\radius)--(\i:{\radius/3});%
\fill[blue] (\i:{\radius/3}) circle (1.5pt);
\end{tikzpicture}
\\
&m_{st}=4\qquad& %
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\node[draw,circle, inner sep=.1cm,fill=white] (a) at (0,0) {$\JW$};
\foreach \i in {0,120,240}{%
\draw[blue] (\i:\radius)--(a);
\foreach \i in {60,180,300}{%
\draw[red] (\i:\radius)--(a);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0)--({-\radius/2},0);\fill[red] ({-\radius/2},0) circle (1.5pt);
\draw[blue] (\radius,0)--({\radius/2},0);\fill[blue] ({\radius/2},0) circle (1.5pt);
\draw[blue] (120:\radius) ..controls (120:{\radius/8}) and (-120:{\radius/8}).. (-120:\radius);
\draw[red] (60:\radius) ..controls (60:{\radius/8}) and (-60:{\radius/8}).. (-60:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,rotate=120]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0)--({-\radius/2},0);\fill[red] ({-\radius/2},0) circle (1.5pt);
\draw[blue] (\radius,0)--({\radius/2},0);\fill[blue] ({\radius/2},0) circle (1.5pt);
\draw[blue] (120:\radius) ..controls (120:{\radius/8}) and (-120:{\radius/8}).. (-120:\radius);
\draw[red] (60:\radius) ..controls (60:{\radius/8}) and (-60:{\radius/8}).. (-60:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0,rotate=240]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0)--({-\radius/2},0);\fill[red] ({-\radius/2},0) circle (1.5pt);
\draw[blue] (\radius,0)--({\radius/2},0);\fill[blue] ({\radius/2},0) circle (1.5pt);
\draw[blue] (120:\radius) ..controls (120:{\radius/8}) and (-120:{\radius/8}).. (-120:\radius);
\draw[red] (60:\radius) ..controls (60:{\radius/8}) and (-60:{\radius/8}).. (-60:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\foreach \i in {60,180,300}{%
\draw[red] (\i:\radius)--(0,0);
\foreach \i in {0,120,240}{%
\draw[blue] (\i:\radius)--(\i:{\radius/2});%
\fill[blue](\i:{\radius/2})circle(1.5pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\foreach \i in {60,180,300}{%
\draw[red] (\i:\radius)--(\i:{\radius/2});%
\fill[red](\i:{\radius/2})circle(1.5pt);
\foreach \i in {0,120,240}{%
\draw[blue] (\i:\radius)--(0,0);
\end{tikzpicture}
\end{multlined}
\end{align*}

Notice that these are all the Jones-Wenzl morphisms that we need to handle with
Coxeter groups of types $A$, $B$, $D$, $E$, $F$.
In these cases, Relation (<ref>) becomes:

\begin{align*}
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0)--({\radius/2},0);%
\fill[red] ({\radius/2},0) circle (1.5pt);
\draw[blue] (0,-\radius)--(0,\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0)--({-\radius/2},0);%
\fill[red] ({-\radius/2},0) circle (1.5pt);
\draw[blue] (0,-\radius)--(0,\radius);
\end{tikzpicture}
\\
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\coordinate (a) at ({-.1*\radius},0);
\foreach \i in {70,180,-70}{%
\draw[red] (a)--(\i:\radius);
\foreach \i in {125,-125}{%
\draw[blue] (a)--(\i:\radius);
\coordinate (b) at ({\radius/2},0);
\draw[blue] (a)--(b);\fill[blue] (b) circle (1.5pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\coordinate (a) at ({-.1*\radius},0);
\draw[red] (70:\radius) ..controls (70:{\radius/3}) and (-70:{\radius/3}).. (-70:\radius);
\draw[blue] (125:\radius) ..controls (125:{\radius/8}) and (-125:{\radius/8}).. (-125:\radius);
\coordinate (b) at ({-\radius/2},0);
\draw[red] ({-\radius},0)--(b);\fill[red] (b) circle (1.5pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\coordinate (a) at (0,0);
\foreach \i in {70,180,-70}{%
\draw[red] (a)--(\i:\radius);
\foreach \i in {125,-125}{%
\draw[blue] (\i:\radius)--(\i:{\radius/2});\fill[blue] (\i:{\radius/2}) circle (1.5pt);
\end{tikzpicture}\\
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\coordinate (a) at (0,0);
\foreach \i in {90,180,-90}{%
\draw[red] (a)--(\i:\radius);
\coordinate (b) at ({\radius/2},0);
\draw[red] (a)--(b);\fill[red] (b) circle (1.5pt);
\foreach \i in {45,135,-45,-135}{%
\draw[blue] (a)--(\i:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (0,\radius)--(0,-\radius);
\draw[blue] (135:\radius) ..controls ({-\radius/5},{\radius/5}) and ({-\radius/5},{-\radius/5})..(-135:\radius);
\draw[red] (-\radius,0)--({-.6*\radius},0);\fill[red] ({-.6*\radius},0) circle (1.5pt);
\draw[blue] (45:\radius) ..controls ({\radius/3},{\radius/3}) and ({\radius/3},{-\radius/3})..(-45:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0) ..controls (-{\radius/4},0) and (0,{\radius/4}).. (0,\radius);
\draw[blue] (135:\radius)--(135:{.6*\radius}); \fill[blue] (135:{.6*\radius}) circle (1.5pt);
\draw[blue] (45:\radius)--(-135:\radius);\draw[blue] (-45:\radius)--(0,0);
\draw[red] (0,{-\radius})--(0,{-.5*\radius});\fill[red](0,{-.5*\radius}) circle (1.5pt);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (-\radius,0) ..controls ({-\radius/4},0) and (0,{-\radius/4}).. (0,-\radius);
\draw[blue] (-135:\radius)--(-135:{.6*\radius}); \fill[blue] (-135:{.6*\radius}) circle (1.5pt);
\draw[blue] (-45:\radius)--(135:\radius);\draw[blue] (45:\radius)--(0,0);
\draw[red] (0,{\radius})--(0,{.5*\radius});\fill[red](0,{.5*\radius}) circle (1.5pt);
\end{tikzpicture}
+ \\
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\draw[red] (0,\radius)--(0,-\radius);
\draw[red] ({-\radius},0)--(0,0);
\foreach \i in {135,-135}{%
\draw[blue] (\i:\radius)--(\i:{\radius/2});%
\fill[blue](\i:{\radius/2})circle(1.5pt);
\draw[blue] (45:\radius) ..controls (45:{\radius/2}) and (-45:{\radius/2}).. (-45:\radius);
\end{tikzpicture}
\begin{tikzpicture}[baseline=0]
\def\radius{.7cm}
\draw[dashed,gray] (0,0) circle (\radius);
\foreach \i in {90,270}{%
\draw[red] (\i:\radius)--(\i:{\radius/3});%
\fill[red](\i:{\radius/3})circle(1.5pt);
\draw[red] (180:\radius)--(180:{.6*\radius});%
\fill[red](180:{.6*\radius})circle(1.5pt);
\draw[blue] (-135:\radius)--({-\radius/4},0)--({\radius/4},0)--(45:\radius);
\draw[blue] (135:\radius)--({-\radius/4},0);
\draw[blue] (-45:\radius)--({\radius/4},0);
\end{tikzpicture}
\end{multlined}
\end{align*}

For further detail, we refer the reader to
<cit.>, <cit.> or <cit.>.

Three color relations. For each finite parabolic subgroup $W_I$
of rank 3, there is a relation ensuring compatibility between the three
corresponding $2m$-valent vertices. We will not need this relations so we refer the reader to

Notice that all the relations are homogeneous, so the morphism spaces are graded

This completes the definition of $\DBS$.

Now we can finally give the main definition of this section.

The diagrammatic Hecke category $\D$ is the Karoubi envelope of the closure of $\DBS$
by shifts and direct sums.

Let $(\cdot)$ denote the shift in the polynomial grading and let $[\cdot]_{\oplus}$ denote the
operation of taking the split Grothendieck group of an additive category.
Notice that $[\D]_{\oplus}$ is naturally a $\Laur$-algebra: the ring structure is induced by
the tensor product and the action of $v$ corresponds to the shift (more precisely $v[B]:=[B(1)]$).
Hence we can state the following categorification result (see <cit.>).

If $\kk$ is a complete local ring, then $\D$ is Krull-Schmidt and
there is a unique isomorphism of $\Laur$-algebras, called character,

\begin{equation}\label{eq_Grothgroup}
\mathrm{ch}:[\D]_\oplus \overset{\sim}{\rightarrow}\HH_{(W,S)}
\end{equation}

sending the class of $B_s$ to $b_s$;

Under the same assumptions of the theorem one can also classify the indecomposable objects and
describe graded ranks of the morphism spaces in terms of the above isomorphism.

As mentioned in the introduction, the category $\D$ can be seen as a diagrammatic description
of other categories. It is in fact equivalent to the category of Soergel bimodules
(see <cit.>), under certain assumptions that can
be weakened if one considers the variant of the latter described by Abe <cit.>.
In the case $W$ is a Weyl group it is also equivalent to the category of equivariant parity
sheaves on the (affine) alag variety, in the sense of <cit.>, for an appropriate choice of realization.
The diagrammatic Hecke category was used by Williamson <cit.> to give counterexamples for the expected bound in Lusztig's
conjecture; Riche and Williamson <cit.> used it to construct the action of the Hecke
category over the principal block of the category of rational representations
of $\mathrm{GL}_n$ (this action was then established for all types by Ciappara <cit.> and independently by Bezrukavnikov and Riche <cit.>);
it was also used to explicitely compute the $p$-canonical basis for Weyl groups of small ranks:
see the paper by Jensen and Williamson <cit.>.

§.§ Light leaves and double leaves

Let $\uw$ be a Coxeter word and $\ux$ a subword which is reduced.
One can construct, in the space $\Hom_{\D}(B_{\uw},B_{\ux})$, some special morphisms called the
light leaves maps, that were introduced by Libedinsky in <cit.>,
and described diagrammatically in <cit.>.
Then, one can use these maps to construct an explicit basis of the general morphism space
$\Hom_{\D}(B_{\uw_1},B_{\uw_2})$ as a left $R$-module.
Let us recall this construction more precisely.

For a Coxeter word $\uw=s_1\cdots s_k$, and a subexpression (or 01-sequence)
$\boe=(e_1,\dots, e_k)\in \{0,1\}^k$, we say that $\boe$ expresses the element $x\in W$
if $x=s_1^{e_1}\cdots s_k^{e_k}$ (where $s_i^0:=1$): in this case we write $x=\uw^{\boe}$.

Let $\ux$ be a reduced subword of $\uw$, corresponding to the element $x$ in $W$.
The light leaves maps in $\Hom(B_{\uw},B_{\ux})$
are labeled by 01-sequences expressing $x$ in the following way.

A 01-sequence determines the elements
$x_i:=\uw_{\le i}^{\boe_{\le i}}\in W$, where $\uw_{\le i}$ is the word consisting of the
first $i$ letters of $\uw$ and $\boe_{\le i}$ is the 01-sequence consisting of the first
$i$ symbols of $\boe$. This describes a path along the Bruhat graph, called Bruhat stroll,
which allows us to decorate $\boe$.
The idea is that, at each step $i$, we consider the position where the next simple
reflection would bring us,	regardless of the fact that this reflection will actually
be taken or not,
and we decorate the corresponding symbol in $\boe$ in the following way.

* if $x_{i}s_{i+1}>x_i$ then we decorate $e_{i+1}$ with a $U$ (for “up”: taking the reflection $s_{i+1}$ would bring us up
in the Bruhat graph);

* if $x_{i}s_{i+1}<x_{i}$ then we decorate it with a $D$ (for “down”).

We can then construct a morphism $L_{\uw,\boe}:B_{\uw}\rightarrow B_{\ux}$ from this data.

This will actually depend on some choices.

First, choose, for each $i$, a reduced expression $\ux_i$ for $x_i$. We define
$L_{\uw,\boe}$ by induction. Let $\lambda_0$ to be $\id_{\un}$.
Then, suppose we have constructed a morphism
$\lambda_i:B_{\uw_{\le i}}\rightarrow B_{\ux_i}$, and consider the decorated value of $e_{i+1}$.

* If $e_{i+1}=U0$, then $x_i=x_{i+1}$ and we choose a braid move (i.e. sequence of braid relations) $\Phi$ from
$\ux_i$ to $\ux_{i+1}$.

* If $e_{i+1}=U1$, then $x_{i+1}=x_is_{i+1}$ and we choose a braid move $\Phi$
from the reduced word $\ux_i s_{i+1}$ to $\ux_{i+1}$.

* If $e_{i+1}=D0$, then $s_{i+1}$ belongs to the right descent of $x_i$ and
we choose a braid move $\Psi$ from $\ux_i$ to a word of the form $\uz s_{i+1}$
(that exists by the Exchange property, see <cit.>). Furthermore, we choose another braid
move $\Phi$ from the latter to $\ux_{i+1}$.

* Finally, if $e_{i+1}=D1$, then we choose, as in the previous case, a braid move $\Psi$
from $\ux_{i}$ to $\uz s_{i+1}$. This time $\uz$ is a reduced expression for
$x_{i+1}$ and we choose another braid move $\Phi$ from $\uz$ to $\ux_{i+1}$.

Then $\lambda_{i+1}$ is defined by one of the following diagrams, according to the
decorated value of $e_{i+1}$ (here we denote $\beta(\Phi)$ and $\beta(\Psi)$ the compositions
of $(s,t)$-ars corresponding to the braid moves $\Phi$ and $\Psi$):

\[
\begin{tikzpicture}[scale=.7]
\draw[gray] (-.5,0)--(3,0);
\draw[violet] (0,0)--(0,.5);\draw[violet] (0.5,0)--(0.5,.5);\draw[violet] (1.5,0)--(1.5,.5);\node at (1,.25) {\dots};\draw[violet] (2,0)--(2,.5);
\draw (-.25,.5)--(2.25,.5)--(2.25,1.5)--(-.25,1.5)--cycle; \node at (1,1) {$\lambda_i$};
\draw[violet] (0,1.5)--(0,2);\draw[violet] (0.5,1.5)--(0.5,2);\draw[violet] (1.5,1.5)--(1.5,2);\node at (1,1.75) {\dots};\draw[violet] (2,1.5)--(2,2);
\draw[red] (2.5,0)--(2.5,1);\fill[red](2.5,1)circle (2pt);
\draw (-.25,2)--(2.25,2)--(2.25,2.5)--(-.25,2.5)--cycle; \node at (1,2.25) {\footnotesize $\beta(\Phi)$};
\draw[violet] (0,2.5)--(0,3);\draw[violet] (0.5,2.5)--(0.5,3);\draw[violet] (1.5,2.5)--(1.5,3);\node at (1,2.75) {\dots};\draw[violet] (2,2.5)--(2,3);
\draw[gray] (-.5,3)--(3,3);
\node[font=\small] at (1.25,-.5){$e_{i+1}=U0$};
\end{tikzpicture}\qquad
\begin{tikzpicture}[scale=.7]
\draw[gray] (-.5,0)--(3,0);
\draw[violet] (0,0)--(0,.5);\draw[violet] (0.5,0)--(0.5,.5);\draw[violet] (1.5,0)--(1.5,.5);\node at (1,.25) {\dots};\draw[violet] (2,0)--(2,.5);
\draw (-.25,.5)--(2.25,.5)--(2.25,1.5)--(-.25,1.5)--cycle; \node at (1,1) {$\lambda_i$};
\draw[violet] (0,1.5)--(0,2);\draw[violet] (0.5,1.5)--(0.5,2);\draw[violet] (1.5,1.5)--(1.5,2);\node at (1,1.75) {\dots};\draw[violet] (2,1.5)--(2,2);
\draw[red] (2.5,0)--(2.5,2);
\draw (-.25,2)--(2.75,2)--(2.75,2.5)--(-.25,2.5)--cycle; \node at (1.25,2.25) {\footnotesize $\beta(\Phi)$};
\draw[violet] (0,2.5)--(0,3);\draw[violet] (0.5,2.5)--(0.5,3);\draw[violet] (1.5,2.5)--(1.5,3);\node at (1,2.75) {\dots};\draw[violet] (2,2.5)--(2,3);\draw[violet] (2.5,2.5)--(2.5,3);
\draw[gray] (-.5,3)--(3,3);
\node[font=\small] at (1.25,-.5){$e_{i+1}=U1$};
\end{tikzpicture}\qquad
\begin{tikzpicture}[scale=.7]
\draw[gray] (-.5,0)--(3,0);
\draw[violet] (0,0)--(0,.5);\draw[violet] (0.5,0)--(0.5,.5);\draw[violet] (1.5,0)--(1.5,.5);\node at (1,.25) {\dots};\draw[violet] (2,0)--(2,.5);
\draw (-.25,.5)--(2.25,.5)--(2.25,.9)--(-.25,.9)--cycle; \node at (1,.7) {\footnotesize $\lambda_i$};
\draw[violet] (0,.9)--(0,1.1);\draw[violet] (0.5,0.9)--(0.5,1.1);\draw[violet] (1.5,0.9)--(1.5,1.1);\draw[violet] (2,0.9)--(2,1.1);\node at (1,1) {\dots};
\draw (-.25,1.1)--(2.25,1.1)--(2.25,1.5)--(-.25,1.5)--cycle; \node at (1,1.3) {\footnotesize $\beta(\Psi)$};
\draw[violet] (0,1.5)--(0,2);\draw[violet] (0.5,1.5)--(0.5,2);\draw[red] (2,1.5) arc (180:0:.25cm);\node at (1,1.75) {\dots};\draw[violet] (1.5,1.5)--(1.5,2);
\draw[red] (2.5,0)--(2.5,1.5); \draw[red] (2.25,2)--(2.25,1.75);
\draw (-.25,2)--(2.5,2)--(2.5,2.5)--(-.25,2.5)--cycle; \node at (1.125,2.25) {\footnotesize $\beta(\Phi)$};
\draw[violet] (0,2.5)--(0,3);\draw[violet] (0.5,2.5)--(0.5,3);\node at (1.125,2.75) {\dots};\draw[violet] (1.75,2.5)--(1.75,3);\draw[violet] (2.25,2.5)--(2.25,3);
\draw[gray] (-.5,3)--(3,3);
\node[font=\small] at (1.25,-.5){$e_{i+1}=D0$};
\end{tikzpicture}\qquad
\begin{tikzpicture}[scale=.7]
\draw[gray] (-.5,0)--(3,0);
\draw[violet] (0,0)--(0,.5);\draw[violet] (0.5,0)--(0.5,.5);\draw[violet] (1.5,0)--(1.5,.5);\node at (1,.25) {\dots};\draw[violet] (2,0)--(2,.5);
\draw (-.25,.5)--(2.25,.5)--(2.25,.9)--(-.25,.9)--cycle; \node at (1,.7) {\footnotesize $\lambda_i$};
\draw[violet] (0,.9)--(0,1.1);\draw[violet] (0.5,0.9)--(0.5,1.1);\draw[violet] (1.5,0.9)--(1.5,1.1);\draw[violet] (2,0.9)--(2,1.1);\node at (1,1) {\dots};
\draw (-.25,1.1)--(2.25,1.1)--(2.25,1.5)--(-.25,1.5)--cycle; \node at (1,1.3) {\footnotesize $\beta(\Psi)$};
\draw[violet] (0,1.5)--(0,2);\draw[violet] (0.5,1.5)--(0.5,2);\draw[red] (2,1.5) arc (180:0:.25cm);\node at (1,1.75) {\dots};\draw[violet] (1.5,1.5)--(1.5,2);
\draw[red] (2.5,0)--(2.5,1.5);
\draw (-.25,2)--(1.75,2)--(1.75,2.5)--(-.25,2.5)--cycle; \node at (0.75,2.25) {\footnotesize $\beta(\Phi)$};
\draw[violet] (0,2.5)--(0,3);\draw[violet] (0.5,2.5)--(0.5,3);\node at (1,2.75) {\dots};\draw[violet] (1.5,2.5)--(1.5,3);
\draw[gray] (-.5,3)--(3,3);
\node[font=\small] at (1.25,-.5){$e_{i+1}=D1$};
\end{tikzpicture}
\]

Then one can check that the ending word of $\lambda_{i+1}$ is the reduced word
$\ux_{i+1}$	and the induction can continue.

Finally we define $L_{\uw,\boe}$ to be $\lambda_k\in \Hom_{\D}(B_{\uw},B_{\ux})$.

Let $\uw=\underline{stsuts}$, with $m_{st}=3$, and $\ux=st$. Consider the subexpression
$\boe=111001$ expressing $x$. One can check that the corresponding decoration of $\boe$ is:

\[
\]

Hence this is the construction of $L_{\uw,\boe}$.

\begin{gather*}
\lambda_0=
\begin{tikzpicture}[baseline=.4cm]
\draw[gray] (0,0)--(1,0);\draw[gray] (0,1)--(1,1);
\end{tikzpicture}\qquad
\lambda_1=
\begin{tikzpicture}[baseline=.4cm]
\draw[gray] (0.25,0)--(.75,0);\draw[gray] (0.25,1)--(.75,1);
\draw[red] (0.5,0)--(0.5,1);
\end{tikzpicture}\qquad
\lambda_2=
\begin{tikzpicture}[baseline=.4cm]
\draw[gray] (0.25,0)--(1.25,0);\draw[gray] (0.25,1)--(1.25,1);
\draw[red] (0.5,0)--(0.5,1); \draw[blue] (1,0)--(1,1);
\end{tikzpicture}\qquad
\lambda_3=
\begin{tikzpicture}[baseline=.4cm]
\draw[gray] (0.25,0)--(1.75,0);\draw[gray] (0.25,1)--(1.75,1);
\draw[red] (0.5,0)--(0.5,1); \draw[blue] (1,0)--(1,1);\draw[red] (1.5,0)--(1.5,1);
\end{tikzpicture} \\[1em]
\lambda_4=
\begin{tikzpicture}[baseline=.4cm,x=.8cm]
\draw[gray] (0.25,0)--(2.25,0);\draw[gray] (0.25,1)--(2.25,1);
\draw[red] (0.5,0)--(0.5,1); \draw[blue] (1,0)--(1,1);\draw[red] (1.5,0)--(1.5,1);
\draw[green] (2,0)--(2,0.5);\fill[green] (2,0.5) circle (1.5pt);
\end{tikzpicture} \qquad
\lambda_5=
\begin{tikzpicture}[baseline=.65cm,x=.8cm]
\draw[gray] (0.25,0)--(2.75,0);\draw[gray] (0.25,1.5)--(2.75,1.5);
\draw[red] (0.5,0)--(0.5,0.5); \draw[blue] (1,0)--(1,.5);\draw[red] (1.5,0)--(1.5,.5);
\draw[red] (0.5,0.5)--(1,0.75)--(1,1.5);\draw[red] (1.5,.5)--(1,0.75);
\draw[blue] (1,0.5)--(1,0.75)--(0.5,1)--(0.5,1.5); \draw[blue] (1,0.75)--(1.5,1);
\draw[green] (2,0)--(2,0.4);\fill[green] (2,0.4) circle (1.5pt);
\draw[blue] (2.5,0)--(2.5,1);
\draw[blue] (1.5,1)--(2,1.25)--(2,1.5);\draw[blue] (2.5,1)--(2,1.25);
\end{tikzpicture} \\[1em]
\lambda_6=
\begin{tikzpicture}[baseline=.9cm,x=.8cm]
\draw[gray] (0.25,0)--(3.25,0);\draw[gray] (0.25,2)--(3.25,2);
\draw[red] (0.5,0)--(0.5,0.5); \draw[blue] (1,0)--(1,.5);\draw[red] (1.5,0)--(1.5,.5);
\draw[red] (0.5,0.5)--(1,0.75)--(1,2);\draw[red] (1.5,.5)--(1,0.75);
\draw[blue] (1,0.5)--(1,0.75)--(0.5,1)--(0.5,2); \draw[blue] (1,0.75)--(1.5,1);
\draw[green] (2,0)--(2,0.4);\fill[green] (2,0.4) circle (1.5pt);
\draw[blue] (2.5,0)--(2.5,1);
\draw[blue] (1.5,1)--(2,1.25)--(2,1.5);\draw[blue] (2.5,1)--(2,1.25);
\draw[blue] (3,0)--(3,1.5) ..controls (3,2) and (2,2).. (2,1.5);
\end{tikzpicture}
\begin{tikzpicture}[x=.4cm,baseline=.9cm]
\draw[gray] (.5,0)--(6.5,0);\draw[gray] (.5,2)--(6.5,2);
\draw[red] (1,0)..controls (1,.5).. (2,1) ..controls (3,.5)..(3,0);
\draw[red] (2,1)--(2,2);
\draw[blue] (2,0)--(2,1)..controls (1,1.5)..(1,2);
\draw[blue] (2,1)..controls (3,1.5) and (6,1.5).. (6,0);
\clip (2,0)--(2,1)..controls (3,1.5) and (6,1.5).. (6,0);
\draw[blue] (5,0)--(5,1.5);
\draw[green] (4,0)--(4,0.5); \fill[green] (4,0.5) circle (1.5pt);
\end{tikzpicture}
\end{gather*}

Now consider two braid words $\uw_1$ and $\uw_2$. We want to construct an $R$-basis for
Consider the flip operation $\overline{(\cdot)}$ on diagrams, consisting of flipping
diagrams upside down.
Now, for each pair of 01-sequences $\boe_1$ of $\uw_1$ and $\boe_2$
of $\uw_2$ such that $\uw_1^{\boe_1}=\uw_2^{\boe_2}$ we pick a choice of morphism
$L_{\uw_1,\boe_1}$ and a choice of morphism $L_{\uw_2,\boe_2}$ as defined above. This means
that we make a choice for all the needed reduced words and braid moves and consider the
corresponding light leaves maps. Let then $\mathbb{L}_{\uw_1,\uw_2}\subset \Hom(B_{\uw_1},B_{\uw_2})$
be the set of the compositions $\overline{L_{\uw_2,\boe_2}}\circ L_{\uw_1,\boe_1}$.
Morphisms as those in $\mathbb{L}_{\uw_1,\uw_2}$ are called double leaves maps.
Then we have the following
(see <cit.>).

The set $\mathbb{L}_{\uw_1,\uw_2}$ is a basis of $\Hom_{\D}(B_{\uw_1},B_{\uw_2})$.

§ DIAGRAMMATICS FOR DG ROUQUIER COMPLEXES

In this section we introduce the Rouquier complexes in the homotopy category of $\D$. Then we extend
the diagrammatic description to one for the dg monoidal category generated by their standard representatives.

§.§ Rouquier complexes

Consider the homotopy category $\Kb(\D)$.
The monoidal structure of $\D$ extends to $\Kb(\D)$ via the usual definition of tensor product
of complexes. Recall that this is as follows. Let $\Ao,\Bo\in \Kb(\D)$, then:

\[
(\Ao\otimes \Bo)^i=\bigoplus_{p+q=i} A^p\otimes B^q,
\]

and the differential map, restricted to $A^p\otimes B^q$ is:

\begin{equation}\label{eq_signconv}
d\otimes \id + (-1)^p \id \otimes d.
\end{equation}

Notice that the unit of $\D$, seen as a complex concentrated in degree 0, and denoted by $\un[0]$,
is the unit of $\Kb(\D)$.

Consider the following standard and costandard complexes

\[
\begin{tikzcd}[row sep=small,%
execute at end picture={%
\def\hdott{.32cm}
\pic[red,yshift=.3em] at (A) {dot={\hdott}};
\pic[red,yshift=1.5em,yscale=-1] at (B) {dot={\hdott}};
F_{\re{s}}=\,\cdots\ar[r]		& 0 \ar[r]		& 0	\ar[r]								& B_{\re{s}}\ar[r,""{coordinate,name=A}]	& \un(1)\ar[r]	& 0\ar[r]		& \cdots \\
& \cohdeg{-2}	& \cohdeg{-1}							& \cohdeg{0}											& \cohdeg{1}	& \cohdeg{2}	& \\
F_{\re{s}}^{-1}=\,\cdots\ar[r]	& 0\ar[r]		& \un(-1)\ar[r,""{coordinate,name=B}]	& B_{\re{s}}\ar[r]							& 0	\ar[r]		& 0\ar[r]		& \cdots \\
\end{tikzcd}
\]

where the numbers in the middle row denote the
cohomological degree in $\Kb(\D)$. If $\sigma=\sigma_{s}^{\pm 1}\in \Sigma$, then let
$F_\sigma$ denote $F_s^{\pm 1}$.
Then, for any braid word $\uom=\sigma_1 \sigma_2 \dots \sigma_n$, with $\sigma_i\in \Sigma$,
we put

\[
F_{\uom}^\bullet:=F_{\sigma_1}\otimes F_{\sigma_2}\otimes \dots \otimes F_{\sigma_n}.
\]

In the sequel we will often omit tensor products.
The objects of this form are called

Rouquier complexes. They were introduced in <cit.> in terms of Soergel bimodules
to categorify braid group actions on categories and study natural transformations
between the induced endofunctors.

§.§ Basic properties of Rouquier complexes

These properties were first proved by Rouquier <cit.> in the
language of Soergel bimodules.

One has the following.

*  Let $s\in S$, then $F_s F_s^{-1}\cong F_s^{-1}F_s\cong \un[0]$.

*  Let $s,t\in S$ with $m_{st}<\infty$, then

\[
\underbrace{F_s F_tF_sF_t \cdots}_{m_{st} \text{ times}} \cong %
\underbrace{F_t F_sF_tF_s \cdots}_{m_{st} \text{ times}}
\]

Hence, for each pair of braid words $\uom_1$, $\uom_2$ expressing
the same element $\omega\in B_W$, there is an isomorphism $F_{\uom_1}^\bullet\cong F_{\uom_2}^\bullet$.

* (Rouquier Canonicity)
for each $\uom_1$ and $\uom_2$ as above, we have

\[
\Hom(F_{\uom_1}^\bullet,F_{\uom_2}^\bullet)\cong R,
\]

and one can chose homotopy equivalences
$\gamma_{\uom_1}^{\uom_2}\in \Hom(F_{\uom_1}^\bullet,F_{\uom_2}^\bullet)$ such that the system
$\{\gamma_{\uom_1}^{\uom_2}\}_{\uom_1,\uom_2}$ is transitive.

Let $\mathscr{B}_W:=\langle F_{\sigma}\mid \sigma\in \Sigma \rangle_\otimes$ inside $\Kb(\D)$. Thanks to
Proposition <ref>, the isomorphism classes of this subcategory form at least a quotient
of the braid group. In particular the Rouquier complex $F_{\omega}$ associated to
$\omega\in B_W$ is well defined up to a canonical isomorphism.
Rouquier conjectured that this quotient is just $B_W$. This was proven in all finite types by
Khovanov and Seidel <cit.>, Brav and Thomas <cit.> and Jensen <cit.>.

In what follows, we will consider the representatives $\Roudg{\uom}$ in the
dg category of complexes $\Cdg(\D)$ and describe morphism spaces between them using Soergel
diagrams. Then the morphisms
in the homotopy category are obtained as the cohomology at zero of the morphism complexes
in $\Cdg(\D)$.
In this setting we will recover the fundamental properties of Rouquier
complexes, stated in Proposition <ref>. This approach will give an
algorithm to determine explicitely the homotopy equivalence categorifying the braid relation.
Furthermore we will give a diagrammatic proof of the Rouquier formula.

§.§ Notation for graded objects and complexes

Given a $\Bbbk$-linear additive category $\mathscr{C}$, a graded object is a family $\{A^q\}_{q\in \ZZ}$,
where $A^q$ is an object of $\mathscr{C}$ for all $q\in \ZZ$. If $A$ is an object of $\mathscr{C}$, then
$A[-q]$ denotes the graded object whose only nonzero entry is $A$ in degree $q$.
The direct sum of graded objects is defined in a natural way, so that we can also write

\[
\{A^q\}_{q\in\ZZ}=\bigoplus_{q\in \ZZ} A^q[-q].
\]

A (degree 0) morphism of graded objects $\Ao=\{A^q\}$ and $\Bo=\{B^q\}$ is a family of morphisms

\[
\{f^q:A^q\rightarrow B^{q}\}_{q\in \ZZ}.
\]

This defines a category $\mathcal{G}_0(\mathscr{C})$
of graded objects. The shift functor $[1]$ sends the graded object $\{A^q\}_{q\in \ZZ}$ to
$\{A^{q+1}\}_{q\in \ZZ}$. Then one can define the graded category $\mathcal{G}(\mathscr{C})$ with the
same objects as $\mathcal{G}_0(\mathscr{C})$ and morphism spaces

\[
\Hom_{\mathcal{G}(\mathscr{C})}(\Ao,\Bo):=\bigoplus_{p\in \ZZ} \Hom_{\mathcal{G}_0(\mathscr{C})}(\Ao,\Bo[p]).
\]

The $p$-th graded piece is denoted by $\Hom^p(\Ao,\Bo)$ and its elements are said
homogeneous of degree $p$.
A complex in $\C(\mathscr{C})$ is then determined by a graded object $\Ao$ endowed with an
endomorphism $d:\Ao\rightarrow \Ao$ of degree 1, such that $d^2=0$.
Viceversa, let $\grad{(\cdot)}:\C(\mathscr{C})\rightarrow\mathcal{G}(\mathscr{C})$ denote
the forgetful functor sending $(\Ao,d)$ to $\Ao$.

The dg category of complexes $\Cdg(\CC)$ has the same objects as $\C(\CC)$
but the morphism space between the complexes $\Ao$ and $\Bo$, is
the dg $\kk$-module whose underlying graded module is $\Hom_{\mathcal{G}(\CC)}(\Ao,\Bo)$,
and the differential map is

\begin{align}\label{eq_curlydiff}
\Hom^p(\Ao,\Bo) & \longrightarrow \Hom^{p+1}(\Ao,\Bo) \\
(f^q)_{q\in\mathbb{Z}} & \longmapsto (d_{B}^{q+p}\circ f^{q} - (-1)^p f^{q+1}\circ d_A^{q}). \notag
\end{align}

A homogeneous map $f$ is said closed if it lies in the kernel of (<ref>)
and exact if it lies in the image. It is easy to see that closed degree zero
maps are precisely morphisms in $\C(\CC)$, and
exact degree zero maps are null-homotopic morphisms of complexes.
Hence, the cohomology at zero of the dg module $\Homb(\Ao,\Bo)$ is the space of
morphisms in the category $\mathcal{K}(\D)$.

Finally, let $\Gb(\mathscr{C})$, $\Cb(\mathscr{C})$, $\Cdgb(\CC)$ and $\Kb(\CC)$
denote the bounded subcategories, in the obvious sense.

§.§ A dg monoidal category of Rouquier complexes

Consider the Hecke category $\D$ defined in  <ref>
and take $\Cdg(\D)$.
By the definition of $\D$, this is a dg monoidal category and the space
$\Hom^\bullet(A^\bullet,B^\bullet)$ has the structure of dg $R$-bimodule, with an additional
grading, inherited from	the polynomial grading of $\D$. The shift in the polynomial grading
is still denoted by $(1)$ and the shift in the cohomological degree by $[1]$, as in the homotopy

Hence one can consider the dg monoidal subcategory generated by the standard and costandard complexes

\[
\Adg=\langle F_\sigma \mid \sigma \in \Sigma \rangle_{\oplus,\otimes,[\cdot]}\subset \Cdg(\D).\\
\]

Its objects will be direct sums of shifts of the $\Roudg{\uom}$'s.
In this section we will extend the Elias-Williamson presentation to $\Adg$.

§.§ Diagrammatics

For a given subexpression $\boi$ of a braid word $\uom$, let $\uom_{\boi}$ denote the
subword corresponding to it. Then let $B_{\boi}$ be the Bott-Samelson object
corresponding to the Coxeter projection of $\uom_{\boi}$. For example, given

\[
\uom=\sigma_s\sigma_t\sigma_t\sigma_u^{-1},
\]

and $\boi=0101$, we have
$\uom_{\boi}=\sigma_t\sigma_u^{-1}$ and then $B_{\boi}=B_{\underline{tu}}=B_tB_u$.

By definition of tensor product of complexes, we have

\[
\grad{(\Roudg{\uom})}=\bigoplus_{\boi\in\{0,1\}^{\ell(\uom)}}B_{\boi}\Tate{q_{\boi}},
\]

where $q_{\boi}=\delta(\uom)-\delta(\uom_{\boi})$ and $\Tate{1}$ denotes the Tate twist
$(1)[-1]$ (see <cit.>).
One can then compute the components of the differential map, according to
(<ref>). The only nonzero components $B_{\boi}\rightarrow B_{\boi'}$ are those
for which $\boi'$ is obtained from $\boi$ by changing one symbol in such a way that
In this case the component is either

\begin{align}\label{eq_diffonsubexpr}
\begin{tikzpicture}[x=0.3cm,y=0.4cm,baseline=0.36cm,every node/.style={font=\scriptsize}]
\draw[gray] (0.5,0)--(9.5,0);\draw[gray] (0.5,2)--(9.5,2);
\foreach \i in {1,2,4,6,8,9}{%
\draw[violet] (\i,0)--(\i,2);
\pic[violet] at (5,0) {dot={1cm}};
\node[font=\normalsize] at (3,1) {\dots}; \node[font=\normalsize] at (7,1) {\dots};
%						\node[anchor=north] at (1,0) {$0$};
%						\node[anchor=north] at (2,0) {$1$};
%						\node[anchor=north] at (5,0) {$i$};
\end{tikzpicture}&
\begin{tikzpicture}[x=0.3cm,y=0.4cm,baseline=0.36cm,every node/.style={font=\scriptsize}]
\draw[gray] (0.5,0)--(9.5,0);\draw[gray] (0.5,2)--(9.5,2);
\foreach \i in {1,2,4,6,8,9}{%
\draw[violet] (\i,0)--(\i,2);
\pic[violet,yscale=-1] at (5,2) {dot={1cm}};
\node[font=\normalsize] at (3,1) {\dots}; \node[font=\normalsize] at (7,1) {\dots};
%							\node[anchor=south] at (1,2) {$0$};
%							\node[anchor=south] at (2,2) {$1$};
%							\node[anchor=south] at (5,2) {$i$};
\end{tikzpicture}.&
\end{align}

where $k$ is the number of 0's in $\boi$ preceding the changed symbol.

Let $\re{s},\bl{t}\in S$ and
$\uom=\sigma_s\sigma_s\sigma_t^{-1}$. Then the complex $\Roudg{\uom}=F_sF_sF_t^{-1}$
is the following:

[every pic/.style=scale=0.4,font=,-]
(s1t1) at (2*,̣2*) $B_sB_t(1)$;
(s2t1) at (2*,̣0) $B_sB_t(1)$;
(n1) at (2*,̣-2*) $\un(1)$;
[-latex] (ss9) to
pic[midway,yshift=.5cm]bottomdot=sst3+ (sst);
[-latex] (ss9) to
pic[midway,yshift=.3cm]topdot=ss2+ (s1);
[-latex] (ss9) to
pic[midway,yshift=-.5cm]topdot=ss1+ (s2);
[-latex] (sst) to
pic[midway,yshift=.3cm]topdot=sst2+ (s1t1);
[-latex] (sst) to
pic[pos=.6,xshift=.8cm]topdot=sst1+ (s2t1);
[-latex] (s1) to
pic[pos=.4,xshift=-.6cm]bottomdot=st2- (s1t1);
[-latex] (s1) to
pic[near start,xshift=-.3cm]topdot=s1+ (n1);
[-latex] (s2) to
pic[pos=.6,xshift=.8cm]bottomdot=st2- (s2t1);
[-latex] (s2) to
pic[midway,yshift=-.3cm]topdot=s1- (n1);
[-latex] (s1t1) to
pic[midway,yshift=.5cm]topdot=st1+ (t2);
[-latex] (s2t1) to
pic[midway,yshift=.3cm]topdot=st1- (t2);
[-latex] (n1) to
pic[midway,yshift=-.5cm]bottomdot=t1+ (t2);
[green,font=] at (0,-3*)-1; [green,font=] at (,̣-3*)0;
[green,font=] at (2*,̣-3*)1; [green,font=] at (3*,̣-3*) 2;
at (,̣) $\oplus$; at (,̣-) $\oplus$;
at (2*,̣) $\oplus$; at (2*,̣-) $\oplus$;

Let now $\uom,\uom'\in \bword$ and consider $\Homb(F_{\uom},F_{\uom'})$. This decomposes as

\begin{equation}\label{eq_morphsummandsubexpr}
\bigoplus_{\boi,\boi'} \Hom_{\D}(B_{\boi},B'_{\boi'})\Tate{q'_{\boi'}-q_{\boi}}
\end{equation}

for an analogous definition of $B'_{\boi'}$ and $q'_{\boi'}$.
Each of the summands of (<ref>) is a morphism space in the
diagrammatic Hecke category
$\D$, so its
elements are linear combinations of Soergel diagrams as described in  <ref>.
Hence an element of $\Homb(F_{\uom},F_{\uom'})$ is also a $\kk$-linear combination of
diagrams each of which lives in one of the summands above. To keep track of
this information we write it as a linear combination of dg diagrams, or diagrams with patches.
A dg diagram is
defined as follows.

* On the boundaries we always arrange all the letters for the source and target braid words
and we use different styles for the boundary line according to the sign of the

* the black line is for positive letters on the bottom boundary or negative
letters on the top boundary;

* the white line is for all the other ones.

The reason for which we invert the colors on the top boundary will be clear very soon.
For the empty word we only draw a normal line.

* The letters of the starting and ending words correspond to the $1$'s in the subexpressions
$\boi$ and $\boi'$, hence we cover with a patch

[violet] at (0,0) patch=1;

every letter corresponding to a $0$.

* A dg diagram is then a usual Soergel diagram whose boundary points are those
which are not covered by patches.

If $\uom=\sigma_{\re{s}}^3\sigma_{\bl{t}}^{-1}\sigma_{\gr{u}}^{-1}\sigma_{\bl{t}}$,
and $\uom'=\sigma_{\re{s}}\sigma_{\bl{t}}^{-2}\sigma_{\gr{u}}^2$, then
the following diagram represents a morphism in the summand $\Hom_{\D}(B_{01111},B'_{10101})$.

\begin{equation}\label{eq_expatches}
\begin{tikzpicture}[x=.7cm,baseline=.7cm]
\node[anchor=north] (aa) at (0,0) {};
\node[anchor=north] (a) at (1,0) {};
\node[anchor=north] (b) at (2,0) {};
\node[anchor=north] (c) at (3,0) {};
\node[anchor=north] (d) at (4,0) {};
\node[anchor=north] (e) at (5,0) {};
\node[anchor=south] (A) at (.5,2) {};
\node[anchor=south] (BB) at (1.5,2) {};
\node[anchor=south] (B) at (2.5,2) {};
\node[anchor=south] (CC) at (3.5,2) {};
\node[anchor=south] (C) at (4.5,2) {};
\node[font=\small,anchor=south] at (0,-.53) {$\sigma_{\re{s}}$};
\node[font=\small,anchor=south] at (1,-.53) {$\sigma_{\re{s}}$};
\node[font=\small,anchor=south] at (2,-.53) {$\sigma_{\re{s}}$};
\node[font=\small,anchor=south] at (3,-.53) {$\sigma_{\bl{t}}^{-1}$};
\node[font=\small,anchor=south] at (4,-.53) {$\sigma_{\gr{u}}^{-1}$};
\node[font=\small,anchor=south] at (5,-.53) {$\sigma_{\bl{t}}$};
\node[font=\small,anchor=south] at (.5,2.1) {$\sigma_{\re{s}}$};
\node[font=\small,anchor=south] at (1.5,2.1) {$\sigma_{\bl{t}}^{-1}$};
\node[font=\small,anchor=south] at (2.5,2.1) {$\sigma_{\bl{t}}^{-1}$};
\node[font=\small,anchor=south] at (3.5,2.1) {$\sigma_{\gr{u}}$};
\node[font=\small,anchor=south] at (4.5,2.1) {$\sigma_{\gr{u}}$};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red] at (0,0) {patch={1}};	\pic[blue] at (1.5,2) {patch={1}};
\pic[green] at (3.5,2) {patch={1}};
\end{tikzpicture}
\end{equation}

* We recover the cohomological degree $p$ of a morphism from the patches:
each patch counts $+1$ on the black line and $-1$ on the white line.

* The polynomial degree of a diagram is $d-p$, if $d$ is its degree as a Soergel diagram
(as described in  <ref>) and $p$ is its cohomological degree. This is due to the
Tate twists in (<ref>).

* We can compute the differential map on the diagram according to (<ref>)
and (<ref>), as follows.
Consider the following operations:

\begin{align*}
\draw[positive] (0,0)--(1,0);
\pic[violet] at (.5,0) {patch={1}};
\end{tikzpicture}
\rightsquigarrow
\begin{tikzpicture}[x=.5cm,baseline=.2cm]
\pic[violet,y=.5cm] at (.5,0) {dot={1cm}};
\draw[positive] (0,0)--(1,0);
\end{tikzpicture}&
\draw[positive] (0,0)--(1,0);
\pic[violet] at (.5,0) {patch={1}};
\end{tikzpicture}
\rightsquigarrow
\begin{tikzpicture}[x=.5cm,baseline=-.3cm]
\pic[violet,yscale=-1,y=.5cm] at (.5,0) {dot={1cm}};
\draw[positive] (0,0)--(1,0);
\end{tikzpicture} \\[1em]
\draw[violet] (.5,0)--(.5,.5);
\draw[violet,dashed] (.5,.5)--(.5,1);
\draw[negative] (0,0)--(1,0);
\end{tikzpicture}
\rightsquigarrow
\begin{tikzpicture}[x=.5cm,baseline=.3cm]
\fill[violet] (.5,.3) circle (1.5pt);
\draw[violet] (.5,.3)--(.5,.5);
\draw[violet,dashed] (.5,.5)--(.5,1);
\draw[negative] (0,0)--(1,0);
\pic[violet] at (.5,0) {patch={1}};
\end{tikzpicture}&
\draw[violet] (.5,0)--(.5,.5);
\draw[violet,dashed] (.5,.5)--(.5,1);
\draw[negative] (0,0)--(1,0);
\end{tikzpicture}
\rightsquigarrow
\begin{tikzpicture}[x=.5cm,baseline=-.5cm,yscale=-1]
\fill[violet] (.5,.3) circle (1.5pt);
\draw[violet] (.5,.3)--(.5,.5);
\draw[violet,dashed] (.5,.5)--(.5,1);
\draw[negative] (0,0)--(1,0);
\pic[violet] at (.5,0) {patch={1}};
\end{tikzpicture}
\end{align*}

Notice that we only allow the dot-sprouting operation on the black boundary,
and the	strand-uprooting on the white boundary.
The image of a diagram via the differential map is the linear combination of
all diagrams obtained by one of the above operations, with coefficients $\pm 1$ according to
(<ref>). More precisely, when acting on the top boundary, if $k$ is the number
of patches on the left of the sprouted dot/uprooted strand, the sign is
$(-1)^k$. The same rule applies to the bottom boundary but we also need to multiply
by an additional $(-1)^{p+1}$ where $p$ is the cohomological degree of the
initial morphism. For example, the image via the differential map of the above diagram is the
combination below.

\begin{multline*}
d \left(%
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\right)
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.2)..(.5,1.7);
\fill[red] (.5,1.7) circle (1.5pt);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\pic[red,scale=.7,transform shape] at (.5,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\draw[blue] (1.5,2)--(1.5,1.5);\fill[blue] (1.5,1.5) circle (1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\fill[green] (4.5,1.7) circle (1.5pt); \draw[green](4.5,1.7)--(4.5,1.2);\fill[green] (4.5,1.2)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (4.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\draw[red] (0,0)--(0,0.5);\fill[red] (0,0.5) circle (1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (3,.3)..controls (3,.55) and (3.5,.6)..(mid);
\fill[blue] (3,.3) circle (1.5pt);
\draw[green](d)--(4,.5);\fill[green] (4,.5)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[blue,scale=.7,transform shape] at (3,0) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.3cm,y=.7cm,baseline=.6cm]
\node[red,anchor=north] (aa) at (0,0) {};
\node[red,anchor=north] (a) at (1,0) {};
\node[red,anchor=north] (b) at (2,0) {};
\node[blue,anchor=north] (c) at (3,0) {};
\node[green,anchor=north] (d) at (4,0) {};
\node[blue,anchor=north] (e) at (5,0) {};
\node[red,anchor=south] (A) at (.5,2) {};
\node[red,anchor=south] (BB) at (1.5,2) {};
\node[blue,anchor=south] (B) at (2.5,2) {};
\node[green,anchor=south] (CC) at (3.5,2) {};
\node[red,anchor=south] (C) at (4.5,2) {};
\draw[red] (b)..controls (2,.5) and (0.5,1.5)..(A);
\draw[red] (a)--(1,.5);\fill[red] (1,.5) circle (1.5pt);
\draw[blue](e)..controls (5,.5) and (2.5,1.5)..node[shape=coordinate] (mid) {} (B);
\draw[blue] (c)..controls (3,.5) and (3.5,.6)..(mid);
\fill[green] (4,.3)circle (1.5pt);\draw[green](4,.3)--(4,.55);\fill[green] (4,.55)circle(1.5pt);
\draw[green](C)--(4.5,1.5);\fill[green] (4.5,1.5)circle(1.5pt);
\node at (2.5,1) {$\alpha_{\gr{u}}$};
\draw[positive] (-0.5,0)--(2.5,0);
\draw[negative] (2.5,0)--(4.5,0);
\draw[positive] (4.5,0)--(5.5,0);
\draw[negative] (-0.5,2)--(1,2);
\draw[positive] (1,2)--(3,2);
\draw[negative] (3,2)--(5.5,2);
\pic[red,scale=.7,transform shape] at (0,0) {patch={1}};
\pic[blue,scale=.7,transform shape] at (1.5,2) {patch={1}};
\pic[green,scale=.7,transform shape] at (4,0) {patch={1}};
\pic[green,scale=.7,transform shape] at (3.5,2) {patch={1}};
\end{tikzpicture}
\end{multline*}

In fact, we have three possibilities on top (uprooting the first or the last strand, or sprouting a dot over the blue patch)
and three for the bottom (sprouting a dot on the red patch, or uprooting the middle blue strand or the green strand).
Here the initial morphism has degree 1, hence the additional sign term for the bottom boundary
is just $+1$.

* One can compose diagrams by gluing them vertically as before, but because
the summands have to match, the composition is zero unless one has the same
arrangement of patches on the gluing boundary line (notice also that, by our convention,
we will glue two diagrams along boundary lines with complementary styles).

* Recall that the monoidal structure of $\Cdg(\D)$ respects the Koszul rule, hence
the tensor product of two dg diagrams $D_1$ and $D_2$ is obtained by gluing them horizontally and multiplying by
the sign $(-1)^{pq}$ where $p$ is the cohomological degree of $D_2$ and $q$ is the number of bottom patches of $D_1$.

* The identity morphism of any $\Roudg{\uom}$ is $\sum_{\boi}\id_{B_{\boi}}$,
hence it is represented by the sum of all diagrams with the same arrangements of patches on top and on bottom and consisting
in parallel vertical lines. For example, if $\uom=\sigma_{\re{s}}^2\sigma_{\bl{t}}^{-1}$, then

\[
\id_{\Roudg{\uom}}=
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red] (1,0)--(1,2);\draw[red](2,0)--(2,2);\draw[blue](3,0)--(3,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red](2,0)--(2,2);\draw[blue](3,0)--(3,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (1,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red] (1,0)--(1,2);\draw[blue](3,0)--(3,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (2,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red] (1,0)--(1,2);\draw[red](2,0)--(2,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[blue,scale=.5,transform shape] at (3,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[blue](3,0)--(3,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (1,\i) {patch={1}};
\pic[red,scale=.5,transform shape] at (2,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red](2,0)--(2,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (1,\i) {patch={1}};
\pic[blue,scale=.5,transform shape] at (3,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[red] (1,0)--(1,2);
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (2,\i) {patch={1}};
\pic[blue,scale=.5,transform shape] at (3,\i) {patch={1}};
\end{tikzpicture}+
\begin{tikzpicture}[x=.3cm,y=.6cm,baseline=.5cm]
\draw[positive] (.5,0)--(2.5,0);\draw[negative] (.5,2)--(2.5,2);
\draw[negative] (2.5,0)--(3.5,0);\draw[positive] (2.5,2)--(3.5,2);
\foreach \i in {0,2}{
\pic[red,scale=.5,transform shape] at (1,\i) {patch={1}};
\pic[red,scale=.5,transform shape] at (2,\i) {patch={1}};
\pic[blue,scale=.5,transform shape] at (3,\i) {patch={1}};
\end{tikzpicture}
\]

and one can check for example that this is a closed degree
zero morphism which is the neutral element of composition.

*  By Theorem <ref>, each summand of (<ref>) is a free
$R$-module with a basis given by (a choice of) double leaves maps. Hence, if we collect all these bases
together we have a basis for the whole morphism space. Recall that the elements of the
basis of each $\Hom(B_{\boi},B'_{\boi'})$ are labeled by subexpressions $\boe$ and $\boe'$
of the starting and ending words, which are in turn determined by $\boi$ and $\boi'$. It will be
convenient to extend these subexpressions $\boe$ and $\boe'$ to the whole words $\uom$ and
$\uom'$, by simply adding zeros
where already $\boi$ or $\boi'$ have zeros. For example, the morphism (<ref>), if
we forget the $\alpha_u$, is actually a double leaves map in $\Hom(B_{\underline{sstut}},B_{\underline{stu}})$, corresponding
to the subexpressions 01100 of $\underline{sstut}$ and 110 of $\underline{stu}$. The extended subexpressions
that we want to consider are then 001100 and 10100, where the
additional zeros are boldface.
Then notice that we can also decorate the new zeros and the decorations of the other symbols do not
depend on whether we look at them in the original subexpression or in the extended one.
Hence a basis for the whole space is labeled by quadruples $(\boi,\boi',\boe,\boe')$, with
$\boi$ and $\boe$ subexpressions of $\uom$ such that $\boi_k\ge \boe_k$, and $\boi'$ and $\boe'$
subexpressions of $\uom'$ with the same property.
The first two entries determine the source and target summands and the last two
determine the double leaves map.

Before we pass to the properties of Rouquier complexes we introduce one more piece of notation.
Let $\uom$ and $\uom'$ be braid words and construct a (dg) double leaves basis
of $\Homb(F_{\uom},F_{\uom'})$ as in Remark <ref> <ref>.
For $L$ and $L'$
double leaves maps of this set, we write

\[
L \overset{d}{\succ} L'
\]

if $L'$ appears with a nonzero coefficient in the linear combination for $d(L)$.

We now prove Proposition <ref> in a completely combinatorial way.

§.§ Inverse relation

The first property, namely $F_s F_s^{-1}\cong \un$, can be easily proven via Gaussian
elimination for complexes (see <cit.>). Here we give an explicit description of the
mutually inverse homotopy equivalences.

For any $s\in S$, consider:

\begin{align}\label{eq_inversebraid}
&\begin{tikzcd}[ampersand replacement=\&]
F_sF_{s}^{-1} \ar[r,shift left,"\eta^-"] \& \un[0]\ar[l,shift left,"\epsilon^+"]
\end{tikzcd},&
&\begin{tikzcd}[ampersand replacement=\&]
F_{s}^{-1}F_s \ar[r,shift left,"\eta^+"] \& \un[0]\ar[l,shift left,"\epsilon^-"]
\end{tikzcd},&
\end{align}

\begin{align*}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[gray](0.5,0)--(2.5,0); \draw[negative](0.5,2)--(1.5,2);\draw[positive](1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2) ..controls (1,.5) and (2,.5).. (2,2);
\draw[gray](0.5,0)--(2.5,0); \draw[negative](0.5,2)--(1.5,2);\draw[positive](1.5,2)--(2.5,2);
\end{tikzpicture}&
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[gray](0.5,2)--(2.5,2); \draw[negative](0.5,0)--(1.5,0);\draw[positive](1.5,0)--(2.5,0);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red] (1,0) ..controls (1,1.5) and (2,1.5).. (2,0);
\draw[gray](0.5,2)--(2.5,2); \draw[negative](0.5,0)--(1.5,0);\draw[positive](1.5,0)--(2.5,0);
\end{tikzpicture}\\[1em]
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[gray](0.5,0)--(2.5,0); \draw[positive](0.5,2)--(1.5,2);\draw[negative](1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2) ..controls (1,.5) and (2,.5).. (2,2);
\draw[gray](0.5,0)--(2.5,0); \draw[positive](0.5,2)--(1.5,2);\draw[negative](1.5,2)--(2.5,2);
\end{tikzpicture}&
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[gray](0.5,2)--(2.5,2); \draw[positive](0.5,0)--(1.5,0);\draw[negative](1.5,0)--(2.5,0);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red] (1,0) ..controls (1,1.5) and (2,1.5).. (2,0);
\draw[gray](0.5,2)--(2.5,2); \draw[positive](0.5,0)--(1.5,0);\draw[negative](1.5,0)--(2.5,0);
\end{tikzpicture}&
\end{align*}

One can easily obtain $\eta^-\epsilon^+=\id_{\un[0]}$
from the rules of composition of dg diagrams. On the other hand we have:

\[
\id_{F_{s}F_{s}^{-1}}-\epsilon^+\eta^- = %
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0)--(1,2);\draw[red] (2,0)--(2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0)--(1,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](2,0)--(2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,2) ..controls (1,.5) and (2,.5).. (2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0) ..controls (1,1.5) and (2,1.5).. (2,0);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0) ..controls (1,1) and (2,1).. (2,0);
\draw[red](1,2) ..controls (1,1) and (2,1).. (2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\end{tikzpicture}.
\]

We want to show that this is exact, hence null-homotopic.
Notice that, by (<ref>), (<ref>) and (<ref>), we have

\begin{align*}
\draw[red] (0.5,0)--(0.5,2);\draw[red] (1.5,0)--(1.5,2);
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.5,0)--(0.5,2);\draw[red] (1.5,0)--(1.5,2);\draw[red](0.5,1)--(1.5,1);
\node[inner sep=2] at (1,1.4) {\small $\delta_s$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.6cm,y=.5cm,baseline=.4cm]
\draw[red] (0.4,0)--(0.4,2);\draw[red] (1.6,0)--(1.6,2);\draw[red](0.4,1)--(1.6,1);
\node[inner sep=2] at (1,0.6) {\small $s(\delta_s)$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red](0.5,0)--(1,1);\draw[red](1.5,0)--(0.5,2); \draw[red](1.5,2)--(1.5,1.5);\fill[red](1.5,1.5) circle (1.5pt);
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.6,0)--(0.6,2);\draw[red] (1.4,0)--(1.4,2);\draw[red](0.6,1)--(1.4,1);
\node[anchor=west] at (1.4,1) {\small $s(\delta_s)$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red](1.5,2)--(1,1);\draw[red](1.5,0)--(0.5,2); \draw[red](0.5,0)--(0.5,0.5);\fill[red](0.5,0.5) circle (1.5pt);
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.5,0)--(0.5,2);\draw[red] (1.5,0)--(1.5,2);\draw[red](0.5,1)--(1.5,1);
\node[inner sep=2] at (0.2,1) {\small $\delta_s$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}\\[1em]
\draw[red](0.5,0) ..controls (0.5,1) and (1.5,1).. (1.5,0); \draw[red](0.5,2) ..controls (0.5,1) and (1.5,1).. (1.5,2);
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.5,0)--(1,.7)--(1,1.3)--(0.5,2);\draw[red] (1.5,0)--(1,.7);\draw[red](1.5,2)--(1,1.3);
\node[anchor=west] at (1,1) {\small $s(\delta_s)$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.5,0)--(1,.7)--(1,1.3)--(0.5,2);\draw[red] (1.5,0)--(1,.7);\draw[red](1.5,2)--(1,1.3);
\node[inner sep=2] at (0.5,1) {\small $\delta_s$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.6,0)--(0.6,2);\draw[red] (1.4,0)--(1.4,2);\draw[red](0.6,1)--(1.4,1);
\node[anchor=west] at (1.4,1) {\small $s(\delta_s)$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.5cm,baseline=.4cm]
\draw[red] (0.5,0)--(0.5,2);\draw[red] (1.5,0)--(1.5,2);\draw[red](0.5,1)--(1.5,1);
\node[inner sep=2] at (0.2,1) {\small $\delta_s$};
\draw[positive] (0.1,0)--(1,0); \draw[negative] (1,0)--(1.9,0);
\draw[negative] (0.1,2)--(1,2); \draw[positive] (1,2)--(1.9,2);
\end{tikzpicture}
\end{align*}

Hence the difference $\id_{F_{s}F_{s}^{-1}}-\epsilon^+\eta^-$ above becomes

\[
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,2)--(2,0);\draw[red] (1,0)--(1.5,1);\draw[red] (2,2)--(2,1.3);
\fill[red](2,1.3) circle (1.5pt);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,2)--(2,0);\draw[red] (2,2)--(1.5,1);\draw[red] (1,0)--(1,0.7);
\fill[red](1,0.7) circle (1.5pt);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0)--(1,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0)--(1,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](2,0)--(2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,2) ..controls (1,.5) and (2,.5).. (2,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0) ..controls (1,1.5) and (2,1.5).. (2,0);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}}; \pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\]

and one can check that this is

\[
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](1,0)--(1.5,1);
\draw[red](2,0)--(1,2);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red](2,0)--(1,2);
\draw[red](2,2)--(1.5,1);
\draw[positive] (0.5,0)--(1.5,0);\draw[negative] (1.5,0)--(2.5,0);
\draw[negative] (0.5,2)--(1.5,2);\draw[positive] (1.5,2)--(2.5,2);
\pic[red,scale=.8,transform shape] at (1,0) {patch={1}};
\end{tikzpicture}
\right)
\]

One can do similar computations for the pair $(\eta^+,\epsilon^-)$.

§.§ Braid relation

We now pass to the braid relation. We will reduce the dg morphism space
$\Homb(F_sF_t\dots,F_tF_s\dots)$ by describing a large null-homotopic dg submodule, and we
will see that the quotient is just a copy of $R$ in degree $0$.
This will give a canonical homotopy equivalence $F_sF_t\dots \rightarrow F_tF_s\dots$.

Let $s,t\in S$ such that $st$ has order $m<\infty$ in $W$ and let

\begin{align*}
\end{align*}

Then we define $N_{s,t}\subset \Homb(F^\bullet_{\uom_{s,t}},F^\bullet_{\uom_{t,s}})$
to be the span of all homogeneous morphisms factoring through words with less than
$m$ symbols. This in particular contains all morphisms with staring or ending subexpressions
different from 11…1.

We have:

* the space $N_{s,t}$ is a dg submodule of $\Homb(F_{\uom_{s,t}},F_{\uom_{t,s}})$;

* the quotient is $R[0]$, i.e. we have a short exact sequence of dg modules

\begin{equation}\label{eq_sesbraid}
0\longrightarrow N_{s,t}\longrightarrow \Hom^{\bullet}(F_{\uom_{s,t}},F_{\uom_{t,s}})\longrightarrow R[0]\longrightarrow 0 ;
\end{equation}

* the dg-module $N_{s,t}$ is null-homotopic, hence the quotient map above
is a homotopy equivalence.

* This is easy by the definition of the differential map: if a morphism factors through
a shorter word then any morphism obtained after the sprouting of a dot or the
uprooting of a strand will still factor through that shorter word.

*  One can write a basis of double leaves maps for the whole morphism space such
that the only one which does not factor through shorter words is the $(s,t)$-ar.

* By <ref> all the double leaves maps different from the $(s,t)$-ar
are a basis of $N_{s,t}$. Observe that any subexpression different
from 11…1 of a reduced word (such as $\underline{sts\dots}$)
has to contain at least one $U0$. In fact it is sufficient to take the first 0 from the
left: this has to be decorated with a $U$ by reducedness.

Then we can conclude by Lemma <ref> below.

Consider a morphism space $\Homb(\Roudg{\uom},\Roudg{\uom'})$ and
let $N$ be a dg submodule spanned by double leaves that all contain
some black $U0$. Then $N$ is null-homotopic.

Consider the double leaves basis of $N$ and denote each double leaves map by a quadruple
$(\boi,\boi',\boe,\boe')$ as before.
Then define the following equivalence relation.
We say that

\[
(\boi,\boi',\boe,\boe')\sim (\boj,\boj',\bof,\bof')
\]

if the following hold

* the (decorated) subexpressions defining the double leaves agree:

* the subexpression defining the arrangement of the patches can only differ
over black $U0$'s: the pair $(\boi,\boi')$ differs from $(\boj,\boj')$ only at
black boundary points where the
corresponding value of $\boe=\bof$ (or $\boe'=\bof'$) is $U0$.

Roughly speaking, two double leaves maps are equivalent when they only differ by
the presence of patches (or boundary dots) in correspondence of black $U0$'s.

Now consider the following partial order on the set of equivalence classes
determined by $\sim$. First we say that
$C\succ C'$ if there are $L\in C$ and $L'\in C'$ such that $L\overset{d}{\succ} L'$.
Then consider the transitive closure of this relation. By looking at the arrangements of
patches one sees that this defines a (strict) partial order on the set of equivalence classes.
Consider a total refinement of this partial order:

\[
C_1\le C_2\le \dots \le C_k.
\]

Then define $N_i$ as the span of the double leaves maps in $C_1, \dots, C_i$. By definition
this gives a filtration by dg submodules

\[
N_1\subset N_2 \subset \dots \subset N_k.
\]

The subquotients $N_{i+1}/N_{i}$ are spanned by the double leaves maps from a single class and the
differential acts only on the black $U0$'s. Then this dg module is isomorphic to
$(R\overset{\id}{\rightarrow} R)^{\otimes n_i}$ for some $n_i$. Hence it is null-homotopic.
But then the whole $N$ is null-homotopic.

Here is an example of a subquotient as above:

[every node/.style=inner sep=0.8cm]
(a) at (.75,0.5) ; (b1) at (3.25,2) ; (b2) at (3.25,0.5) ; (b3) at (3.25,-1) ;
(c1) at (5.75,2) ; (c2) at (5.75,0.5) ; (c3) at (5.75,-1) ;
(d) at (8.25,0.5) ;
[-latex] (a)–(b1);
[-latex] (a)–(b2);
[-latex] (a)–(b3);
[-latex] (b1)–(c1);
[-latex] (b1)–(c2);
[-latex] (b2)–(c1);
[-latex] (b2)–(c3);
[-latex] (b3)–(c2);
[-latex] (b3)–(c3);
[-latex] (c1)–(d);
[-latex] (c2)–(d);
[-latex] (c3)–(d);
[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[blue,scale=.8,transform shape] at (2,0) patch=1;
[green,scale=.8,transform shape] at (4,0) patch=1;
[green,scale=.8,transform shape] at (3,2) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[blue] (2,0)–(2,.5);[blue] (2,0.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[green,scale=.8,transform shape] at (4,0) patch=1;
[green,scale=.8,transform shape] at (3,2) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[green] (4,0)–(4,0.5);[green] (4,0.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[blue,scale=.8,transform shape] at (2,0) patch=1;
[green,scale=.8,transform shape] at (3,2) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[green] (3,2)–(3,1.5);[green] (3,1.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[blue,scale=.8,transform shape] at (2,0) patch=1;
[green,scale=.8,transform shape] at (4,0) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[blue] (2,0)–(2,0.5);[blue] (2,0.5) circle (1.5pt);
[green] (4,0)–(4,0.5);[green] (4,0.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[green,scale=.8,transform shape] at (3,2) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[blue] (2,0)–(2,0.5);[blue] (2,0.5) circle (1.5pt);
[green] (3,2)–(3,1.5);[green] (3,1.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[green,scale=.8,transform shape] at (4,0) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[blue] (4,0)–(4,0.5);[blue] (4,0.5) circle (1.5pt);
[blue] (3,2)–(3,1.5);[blue] (3,1.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);
[blue,scale=.8,transform shape] at (2,0) patch=1;

[red] (1,0)..controls (1,.5).. (2,1)–(2,2); [red] (3,0)..controls (3,.5)..(2,1);
[blue] (4,2)–(4,1.5);[blue] (4,1.5) circle (1.5pt);
[blue] (2,0)–(2,0.5);[blue] (2,0.5) circle (1.5pt);
[green] (4,0)–(4,0.5);[green] (4,0.5) circle (1.5pt);
[green] (3,2)–(3,1.5);[green] (3,1.5) circle (1.5pt);
[positive] (0.5,0)–(2.5,0);[negative] (2.5,0)–(3.5,0); [positive] (3.5,0)–(4.5,0);
[negative] (0.5,2)–(2.5,2);[positive] (2.5,2)–(3.5,2); [negative] (3.5,2)–(4.5,2);

In this case it is isomorphic to $(R\overset{\id}{\rightarrow}R)^{\otimes 3}$.

Now, taking cohomology at zero in the short exact sequence (<ref>) gives a unique
morphism in $\Kb(\D)$ up to a polynomial. Furthermore the latter is the coefficient of the
$2m$-valent vertex in any representative
in $\Cdg(\D)$. Take $\gamma_{s,t}$ to be the morphism where this coefficient is $1$.
The following proves Proposition <ref> <ref>.

The morphisms $\gamma_{s,t}$ and $\gamma_{t,s}$ are mutually inverse homotopy equivalences.

Notice that combining relations (<ref>) and (<ref>), we obtain the following
equality (the picture is for $m_{s,t}$ even):

\[
\begin{tikzpicture}[baseline=.8cm,scale=.6, transform shape]
\draw (0,0) -- (3,0); \draw (0,3) -- (3,3);
\draw[red] (.5,0)--(1.5,1) ..controls (2.5,1.5).. (1.5,2)--(.5,3);
\draw[red] (2,0)--(1.5,1) ..controls (1,1.5).. (1.5,2)--(2,3);
\draw[blue] (2.5,0)--(1.5,1) ..controls (.5,1.5).. (1.5,2)--(2.5,3);
\draw[blue] (1,0)--(1.5,1) ..controls (2,1.5).. (1.5,2)--(1,3);
\node at (1.5,.3) {...};\node at (1.5,2.7) {...};\node at (1.5,1.5) {...};
\end{tikzpicture}
\begin{tikzpicture}[baseline=.8cm,scale=.6]
\draw (0,0) -- (3,0); \draw (0,3) -- (3,3);
\draw[red](.5,0)--(.5,3); \draw[red](.5,1.5)--(1.5,1.5);\draw[blue](1,0)--(1,3);
\draw[red](2,0)--(2,3);\draw[blue](2.5,1.5)--(1.5,1.5); \draw[blue](2.5,0)--(2.5,3);
\node at (1.5,.3) {...};\node at (1.5,2.7) {...};
\node[draw,circle,fill=white] at (1.5,1.5) {$\JW$};
\end{tikzpicture}
\]

One can show that this is the identity,
modulo morphisms that factor through shorter Coxeter words: see <cit.>.
Now we can repeat the argument of Proposition <ref> for $\Homb(F_{\uom_{s,t}},F_{\uom_{s,t}})$
and obtain that, up to a polynomial, the only nonzero morphism in $\Kb(\D)$ is the identity.
Then the compsition above has to be homotopy equivalent to the identity.

Let us now consider two braid words $\uom_1$ and $\uom_2$, such that $\uom_2$ is obtained from
$\uom_1$ by applying one of the relations of $B_W$. By appropriately tensoring with identity to
the left and to the right the morphisms (<ref>) or those in Proposition <ref>,
we get a homotopy equivalence $\gamma_{\uom_1}^{\uom_2}$
corresponding to the relation used.

If more generally $\uom_2$ is obtained from $\uom_1$ via a sequence of relations then one gets
a homotopy equivalence $\gamma_{\uom_1}^{\uom_2}$ given by the corresponding composition of
morphisms as above, which, a priori, depends on the choice of relations used.

The morphisms $\gamma_{\uom_1}^{\uom_2}$ above are, up to sign,
the canonical isomorphisms defined by Rouquier <cit.>. In fact consider a
braid word $\uom$. We have a homotopy equivalence
$\Hom^\bullet (F_{\uom},F_{\uom})\simeq R[0]$ compatible
with the ring structure (because null-homotopic morphisms form a two-sided ideal). Hence any loop
of relations transforming $\uom$ into itself gives a homotopy equivalence that corresponds
to an invertible element of $R$. Now taking $\Bbbk=\ZZ$ we deduce that this element is $\pm 1$.

Another argument for canonicity is given in <cit.>.

§.§ Explicit homotopy equivalences

The above gives a recipe for an explicit diagrammatic description of $\gamma_{s,t}$.
In type $A$ such a description can be found in a paper by Elias and Krasner <cit.>.

Let $\mathbb{L}_{s,t}$ be a basis of double leaves for the space $N_{s,t}$ and take
$\mathbb{L}_{s,t}^0$ to be the subset of maps of cohomological degree 0 (this means:
with the same number of patches on bottom and on top). One can also restrict further to those maps
of polynomial degree 0. Let $\beta$ be the morphism consisting in the $(s,t)$-ar.
Then, by  <ref>, we know that

\[
\gamma_{s,t}=\beta+\sum_{L\in \mathbb{L}_{s,t}^0} a_L L
\]

and that it is the unique morphism of complexes up to homotopy satisfying these properties. Then
it is sufficient to find any choice of coefficients $a_L$ such that

\[
d\big( \beta+\sum_{L\in \mathbb{L}_{s,t}^0} a_L L \big)=0.
\]

So this reduces to a linear algebra problem. Here are the cases $m=2,3$.

For $m_{s,t}=2$ we obtain

\[
\gamma_{s,t}=%
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\draw[red] (1,0)--(2,2);\draw[blue] (2,0)--(1,2);
\pic at (0,0) {twoframe};
\end{tikzpicture}+
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\def\scapat{.8}
\draw[red] (1,0)--(2,2);
\pic at (0,0) {twoframe};
\pic[blue] at (2,0) {patch={\scapat}};
\pic[blue] at (1,2) {patch={\scapat}};
\end{tikzpicture}
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\def\scapat{.8}
\draw[blue] (2,0)--(1,2);
\pic at (0,0) {twoframe};
\pic[red] at (1,0) {patch={\scapat}};
\pic[red] at (2,2) {patch={\scapat}};
\end{tikzpicture}%
\begin{tikzpicture}[x=.5cm,y=.4cm,baseline=.3cm]
\def\scapat{.8}
\pic at (0,0) {twoframe};
\pic[red] at (1,0) {patch=\scapat};
\pic[red] at (2,2) {patch=\scapat};
\pic[blue] at (1,2) {patch=\scapat};
\pic[blue] at (2,0) {patch=\scapat};
\end{tikzpicture}
\]

For $m_{s,t}=3$, we get

\begin{multline*}
\gamma_{s,t}=
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (3,2)--(2,1);
\draw[blue] (1,0)--(2,1)--(3,0);\draw[blue] (2,2)--(2,1);
\pic at (0,0) {threeframe};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2)--(2,0);
\draw[blue] (2,2)--(3,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (3,2)--(2,0);
\draw[blue] (2,2)--(1,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2)--(1,1.2);\fill[red] (1,1.2) circle (1.5pt);
\draw[blue] (2,2)--(2,1)--(1,0);\draw[blue] (2,1)--(3,0);
\pic at (0,0) {threeframe};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (3,2)--(3,1.2);\fill[red] (3,1.2) circle (1.5pt);
\draw[blue] (2,2)--(2,1)--(1,0);\draw[blue] (2,1)--(3,0);
\pic at (0,0) {threeframe};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\end{tikzpicture}+\\
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (1,0)--(1,0.8);\fill[blue] (1,0.8) circle (1.5pt);
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (2,1)--(3,2);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (3,0)--(3,0.8);\fill[blue] (3,0.8) circle (1.5pt);
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (2,1)--(3,2);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2)..controls (1,1) and (3,1).. (3,2);
\draw[blue] (1,0)..controls (1,1) and (3,1).. (3,0);
\pic at (0,0) {threeframe};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (2,2)--(1,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (2,2)--(3,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\end{tikzpicture}+\\
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2)--(2,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (3,2)--(2,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (2,0) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\end{tikzpicture}
\end{multline*}

where $a$ can be any value in $\kk$. Notice that

\[
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (2,1)--(3,2);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\bigg)=
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (1,0)--(1,0.8);\fill[blue] (1,0.8) circle (1.5pt);
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (2,1)--(3,2);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[blue] (3,0)--(3,0.8);\fill[blue] (3,0.8) circle (1.5pt);
\draw[red] (2,0)--(2,1)--(1,2);\draw[red] (2,1)--(3,2);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (1,2)--(2,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (3,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\begin{tikzpicture}[x=.35cm,y=.4cm,baseline=.3cm]
\draw[red] (3,2)--(2,0);
\pic at (0,0) {threeframe};
\pic[blue,scale=.8,transform shape] at (2,2) {patch={1}};
\pic[red,scale=.8,transform shape] at (1,2) {patch={1}};
\pic[blue,scale=.8,transform shape] at (1,0) {patch={1}};
\pic[blue,scale=.8,transform shape] at (3,0) {patch={1}};
\end{tikzpicture}
\]

so the morphism above is actually unique up to homotopy.

§.§ Rouquier formula

As a byproduct of Lemma <ref>, we can now prove the so-called Rouquier formula
(conjectured in <cit.>, and proved in <cit.>, and in <cit.>)

Let $w,v\in W$ and let $\uw$ and $\uv$ be reduced words expressing them.
Let $\uom$ be the positive word lift of $\uw$ and $\underline{\nu}$
be the negative word lift for $\uv$. Then

\[
\Hom^\bullet(F_{\uom},F_{\underline{\nu}})\simeq	\begin{cases}
R[0] & \text{if $w=v$,} \\
0 & \text{otherwise.}
\end{cases}
\]

Suppose that $w\neq v$ and suppose that $\ell(w)>\ell(v)$ (the other case being analogous).
Any subexpression of $\uw$ for a shorter word has to contain at least one
0 and then at least one $U0$ by reducedness. Then Lemma <ref> implies the claim.

Suppose instead that $w=v$. Then the only dg light leaves not factoring through
shorter words is that with subexpressions $U1\dots U1$ for both $w$ and $v$ (which gives
a morphism corresponding to a certain braid move from $w$ to $v$). Now notice that the span
$N$ of all the other dg double leaves maps is a dg submodule and then we again conclude by
Lemma <ref>.

Missing 'biblatex' package
The bibliography requires the 'biblatex' package.

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journaltitleDuke Math. J.
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journaltitleCompos. Math.
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Hermann, Paris

seriesActualités Scientifiques et Industrielles, No. 1337
titleÉléments de mathématique. Fasc. XXXIV. Groupes et algèbres de Lie. Chapitres IV, V, VI
pages288 pp. (loose errata)

journaltitleMath. Ann.
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journaltitleHomology Homotopy Appl.
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Springer, Cham

isbn978-3-030-48825-3; 978-3-030-48826-0
seriesRSME Springer Series
titleIntroduction to Soergel bimodules
year[2020] ©2020

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booktitleInternational Congress of Mathematicians. Vol. II
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noteWith a joint appendix with Alex Kontorovich and Peter J. McNamara
titleSchubert calculus and torsion explosion