3.1.1 Soergel calculus in extended affine type A

The diagrammatic Bott-Samelson category of extended type $\widehat {A}_{n-1}$ , denoted $\widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ , is the $\mathbb {Z}$ -graded, $\mathbb {R}$ -linear, pivotal category whose objects are grading shifts of finite words in the alphabet $\widehat {I}\cup \{\pm \}$ , and whose vector spaces of morphisms are defined below in terms of generating diagrams and relations. Recall that a pivotal category is a monoidal category with duals satisfying certain conditions, see [Reference Etingof, Gelaki, Nikshych and OstrikEGNO, Definition 4.7.8]. We will often identify that alphabet with the corresponding elementary Soergel bimodules $\{{\mathrm {B}}_i,i\in \widehat {I}\}\cup \{{\mathrm {B}}_{\rho }^{\pm 1} \}$ , which is justified by [Reference Mackaay and ThielMaTh, Theorem 2.10]. These are bimodules over a polynomial algebra R, defined below (47). Under this identification, a word in the alphabet corresponds to a tensor product of ${\mathrm {B}}_i$ and $\mathrm {B}_\rho $ , which is called a Bott-Samelson bimodule. The empty word is the identity object, which is identified with R. The tensor product is taken over R, but we will often suppress $\otimes _R$ in our notation, e.g., we will simply write $\mathrm {B}_1 \mathrm {B}_2$ for $\mathrm {B}_1\otimes _R \mathrm {B}_2$ . The diagrams then then correspond to R-R-bimodule maps, but we will not use those maps explicitly in this paper. As explained above, the diagrams can be seen as morphisms between degree-shifted objects or as homogeneous morphisms between unshifted objects. In this definition, we take the former point of view. As explained above, the graded version of $\widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ will be denoted by $\widehat {\mathscr {BS}}^{\mathrm {ext},*}_n$ .

The identity morphism on $i\in \widehat {I}$ is given by a vertical non-oriented strand coloured by i, whereas the identity morphism on $+/-$ is given by a vertical upward/downward oriented black strand. As usual, we will colour the unoriented strands to facilitate the reading of the diagrams. When there are too many different colours in a diagram, the colours are sometimes indicated by labels next to the strands. We say that two colours $i,j\in \widehat {I}$ are adjacent if $i \equiv j\pm 1\bmod n$ and that they are distant otherwise.

The number of different types of diagram increases with n, so below we first give the generating diagrams for $n\geq 1$ , then the additional ones for $n\geq 2$ and, finally, the additional ones that only show up for $n\geq 3$ .

• • For $n\geq 1$ ,

  (23)

  

  and the corresponding oriented black caps.

• • For $n\geq 2$ , with $i\in \widehat {I}$ ,

  (24)

  

  and the diagrams obtained from these by a rotation of $180$ degrees (which have the same degrees).

• • For $n\geq 3$ , with $i,j,k\in \widehat {I}$ such that $\vert i-j\vert =1$ and $\vert i-k\vert>1$ ,

  (25)

  

  and the diagrams obtained from these by a rotation of $180$ degrees (which have the same degrees).

We read diagrams from bottom to top as morphisms, i.e., their source is at the bottom and their target at the top.

Diagrams can be stacked vertically (composition of morphisms) and juxtaposed horizontally (monoidal product of morphisms), while adding the degrees, and are subject to the relations below. The list contains all relations for all $n\geq 1$ , but for $n=1$ and $n=2$ only the relations involving the respective generators should be considered. For $n=2$ , there is an additional subtlety in (37), as indicated. We also assume isotopy invariance and cyclicity, meaning that closed parts of the diagrams can be moved around freely in the plane as long as they do not cross any other strands and the boundary is fixed, and all diagrams can be bent and rotated, and the bent and rotated versions of the relations also hold.

• • Relations involving only one colour:

  (26)

  
  (27)

  
  (28)

  
  (29)

  

• • Relations involving two distant colours:

  (30)

  
  (31)

  
  (32)

  
  (33)

  

  Note that (33) follows from (31).

• • Relations involving two adjacent colours:

  (34)

  
  (35)

  
  (36)

  
  (37)

  

• • Relation involving three distant colours:

  (38)

  

• • Relation involving two adjacent colours and one distant from the other two:

  (39)

  

• • Relation involving three colours such that i is adjacent to j, and j is adjacent to k:

  (40)

  

• • Relations involving only oriented strands:

  (41)

  
  (42)

  

• • Relation involving oriented strands and two distant coloured strands:

  (43)

  

• • Relations involving oriented strands and two adjacent coloured strands:

  (44)

  
  (45)

  
  (46)

  

• • Relations involving oriented strands and three adjacent coloured strands:

  (47)

  

Remark 3.1. Further relations that are consequences of (37) usually have two cases too, e.g.

and similarly with the two colours switched.

Note that the empty word is the identity object in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ and its endomorphisms are the closed diagrams, which by the relations above are equal to polynomials in the coloured dumbbells

As each dumbbell has degree $2$ , the degree of any polynomial in these dumbbells, as a morphism in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ , is twice its polynomial degree. From now on, we denote this polynomial algebra by R.

Note further that, by relations (29), (37) and (33), the morphism $\fbox {y}$ , defined in [Reference Mackaay and ThielMaTh] by

(48)

is central, in the sense that it can be slid through all diagrams (i.e. it commutes horizontally with all morphisms), because it is equal to the sum of all simple roots.

Note that in our definition shifts are already part of $\widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ and recall that the idempotent complete category $\widehat {\mathscr {S}}^{\mathrm {ext}}_n$ is Krull-Schmidt, see e.g. [Reference Elias, Makisumi, Thiel and WilliamsonEMTW, Section 11.2.3]. It therefore makes sense to consider the split Grothendieck group $[\widehat {\mathscr {S}}^{\mathrm {ext}}_n]_{\oplus }$ , which has a natural $\mathbb {Z}[q,q^{-1}]$ -module structure defined by $[\mathrm {X}\langle 1\rangle ]=:q[\mathrm {X}]$ , for every object $\mathrm {X}\in \widehat {\mathscr {S}}^{\mathrm {ext}}_n$ . It also has a natural algebra structure, inherited from the monoidal structure on $\widehat {\mathscr {S}}^{\mathrm {ext}}_n$ .

Finally, let us recall the Soergel categorification theorem, see e.g. [Reference Mackaay and ThielMaTh, Theorem 2.5] for extended affine type A (which was based on previous results by Härterich) and [Reference Elias, Makisumi, Thiel and WilliamsonEMTW, Theorem 11.1] for general Coxeter type. For every $w\in \widehat {\mathfrak {S}}_n$ , choose a rex $\underline {w}=(s_{i_1},\ldots , s_{i_l})$ and define the Bott-Samelson bimodule

$$\begin{align*}\mathrm{BS}(\underline{w}):={\mathrm{B}}_{i_1}\cdots {\mathrm{B}}_{i_l}\in \widehat{\mathscr{BS}}^{\mathrm{ext}}_n. \end{align*}$$

There is an essentially unique indecomposable summand ${\mathrm {B}}_w\in \widehat {\mathscr {S}}^{\mathrm {ext}}_n$ of $\mathrm {BS}(\underline {w})$ that is not a summand of $\mathrm {BS}(\underline {u})$ for any $u\prec w$ in $\widehat {\mathfrak {S}}_n$ , where $\prec $ is the Bruhat order. The isomorphism class of ${\mathrm {B}}_w$ does not depend on the choice of rex for w. More generally, for any $k\in \mathbb {Z}$ and any $w\in \widehat {\mathfrak {S}}_n$ , the object ${\mathrm {B}}_\rho ^k {\mathrm {B}}_w$ is indecomposable in $\widehat {\mathscr {S}}^{\mathrm {ext}}_n$ and the set of these indecomposables is complete and irredundant, meaning that they are all mutually non-isomorphic and every indecomposable in $\widehat {\mathscr {S}}^{\mathrm {ext}}_n$ is isomorphic to one of them up to grading shift. Recall further that $\widehat {H}^{\mathrm {ext}}_{n}$ has a q-sesquilinear form $(-,-)$ , see (17). Similarly, $[\widehat {\mathscr {S}}^{\mathrm {ext}}_n]_{\oplus }$ has a q-sesquilinear form $\langle -,-\rangle $ , called the graded Euler form. To define it, one has to recall that the morphism spaces in $\widehat {\mathscr {S}}^{\mathrm {ext},*}_n$ are free as left and as right graded R-modules and that their graded rank is finite and does not depend on whether we consider them as left or as right R-modules. By definition, the value of $\langle [X], [Y] \rangle $ , for $X,Y\in \widehat {\mathscr {S}}^{\mathrm {ext}}_n$ , is equal to the graded rank of $\mathrm {Hom}_{\widehat {\mathscr {S}}^{\mathrm {ext},*}_n}(X,Y)$ . The following theorem combines Soergel’s categorification theorem and Soergel’s conjecture (which has been proved, see [Reference Mackaay and ThielMaTh] and references therein).

Theorem 3.3. The $\mathbb {Z}[q,q^{-1}]$ -linear map

$$ \begin{align*} &\qquad\qquad \gamma\colon \widehat{H}^{\mathrm{ext}}_{n} \to [\widehat{\mathscr{S}}^{\mathrm{ext}}_n]_{\oplus}\\ & \gamma(\rho^k b_w)=[{\mathrm{B}}_{\rho^k} {\mathrm{B}}_w] \quad (k\in \mathbb{Z}, w\in \widehat{\mathfrak{S}}_n) \end{align*} $$

is an isomorphism of $\mathbb {Z}[q,q^{-1}]$ -algebras intertwining the two q-sesquilinear forms.

This intertwining property goes under the name of Soergel’s hom formula and will be used several times in Section 5.3, so let us state it explicitly here:

(49) $$ \begin{align} (\rho^k b_u, \rho^l b_v )=\langle [{\mathrm{B}}_{\rho^k} {\mathrm{B}}_u], [{\mathrm{B}}_{\rho^l} {\mathrm{B}}_v]\rangle = \underline{\mathrm{rk}}\left( \mathrm{Hom}_{\widehat{\mathscr{S}}^{\mathrm{ext},*}_n}({\mathrm{B}}_{\rho^k} {\mathrm{B}}_u, {\mathrm{B}}_{\rho^l} {\mathrm{B}}_v)\right) \end{align} $$

for all $k,l\in \mathbb {Z}$ and all $u,v\in \widehat {\mathfrak {S}}_n$ , where $\underline {\mathrm {rk}}$ denotes the graded rank.

3.1.2 Rouquier complexes

We give a summary of the material in [Reference Mackaay, Miemietz and VazMMV, §4.1] that is needed in the sequel. In particular, we introduced a diagrammatic calculus for certain morphisms between tensor products of Rouquier complexes and Bott–Samelson bimodules in finite type A. In the summary below, we recall the basic features of that calculus, generalising it to extended affine type A as well. Note that it only describes part of the morphism spaces in $K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ , which are not yet understood in full generality. Describing the homotopy categories of Rouquier complexes in terms of generators and relations is a hard problem, even in finite type A, see for example [Reference Libedinsky and WilliamsonLiWi].

Throughout the rest of the paper, if $\mathcal {C}$ is a $\mathbb {k}$ -linear, additive category, for any field $\mathbb {k}$ , we write $C^b(\mathcal {C})$ for the category of bounded complexes in $\mathcal {C}$ and $K^b(\mathcal {C})$ for its homotopy category. If $\mathcal {C}$ is also monoidal, then the usual monoidal product of chain complexes equips $K^b(\mathcal {C})$ with a monoidal structure as well. If $\mathcal {C}$ is also graded, then $K^b(\mathcal {C})$ is bigraded and we denote the shift inherited from $\mathcal {C}$ by $\langle \cdot \rangle $ and the homological shift by $[\cdot ]$ .

For $n\geq 2$ , the Rouquier complexes ${\mathrm {T}}_i^{\pm 1}\in K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ for $i\in \widehat {I}$ are defined by

(50)

with ${\mathrm {B}}_i$ placed in homological degree zero in both complexes.

The identity morphisms of ${\mathrm {T}}_i^{\pm 1}$ are depicted by

All generators in (51), (55) and (56) below have degree zero with respect to both gradings.

• • The 1-colour generators are the cups and caps:

  (51)

  

  The following lemma recalls the content of [Reference Mackaay, Miemietz and VazMMV, Lemma 4.4].

  Lemma 3.4. For any $\color {blue}{i}\in \widehat {I}$ , we have the following relations between morphisms in $K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ :

  (52)

  

  (53)

  

  (54)

  

  For the diagrams and relations given in the rest of this section there are also similar diagrams and relations involving inverses of Rouquier complexes.

  For $n>2$ , there are additional diagrams and relations.

• • The six-valent vertices for $\color {blue}{i}$ and $\color {myred}{j}$ adjacent:

  (55)

  

  Note that a solid strand corresponds to the identity on a Soergel bimodule, whereas a dashed strand corresponds to the identify morphism on a Rouquier complex, and all morphisms are defined in the homotopy category. These satisfy

  

• • The crossings for $\color {blue}{i}$ and $\color {orange}{k}$ distant:

  (56)

  

  These satisfy

  

4.1.1 Symmetric pairs of monoidal functors

The content of this subsection must be known to experts, but we do not know of any reference in the literature for what we call a symmetric pair of monoidal functors in Definition 4.2. We thank Robert Laugwitz for pointing out to us that our definition can be reformulated in terms of the Drinfeld centralisers of the monoidal functors, as we will briefly explain in Remark 4.3. Since we claim no originality in this subsection and it is only meant to help the reader understand our construction of $\Psi _{k,n-k}$ , we have not endeavored to formulate the abstract notions and results in the most general setting.

As for the terminology, we call a monoidal category or functor strict monoidal if the coherence morphisms of the monoidal structure are identities, otherwise we simply call it monoidal. However, we should warn the reader that we treat canonical natural isomorphisms, such as the unitors and associators of $\times , \otimes $ and $\oplus $ , as identities. With this harmless simplification, the categories $\widehat {\mathscr {S}}^{\mathrm {ext}}_k\boxtimes \widehat {\mathscr {S}}^{\mathrm {ext}}_{n-k}$ and $K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ are considered to be strict monoidal. However, our monoidal functor $\Psi _{k,n-k}$ is definitely not strict, as will see in Section 4.2.3.

In this more abstract subsection, we work over an arbitrary field $\mathbb {k}$ . Recall that, for $\mathbb {k}$ -linear categories $\mathcal {C}$ and $\mathcal {D}$ , their tensor product $\mathcal {C}\diamond \mathcal {D}$ is the $\mathbb {k}$ -linear category with $\mathrm {Ob}(\mathcal {C}\diamond \mathcal {D})=\mathrm {Ob}(\mathcal {C})\times \mathrm {Ob}(\mathcal {D})$ and morphism spaces $\mathrm {Hom}_{\mathcal {C}\diamond \mathcal {D}}\big ((\mathtt {i},\mathtt {j}),(\mathtt {k},\mathtt {l})\big )=\mathrm {Hom}_{\mathcal {C}}(\mathtt {i},\mathtt {j})\otimes \mathrm {Hom}_{\mathcal {D}}(\mathtt {k},\mathtt {l})$ , where $\otimes =\otimes _{\mathbb {k}}$ . Further, we define their box tensor product $\mathcal {C}\boxtimes \mathcal {D}$ as the Karoubi envelope of the additive envelope of $\mathcal {C}\diamond \mathcal {D}$ .

Proof. Let $\Psi \colon \mathcal {C}\diamond \mathcal {D} \to \mathcal {E}$ be a $\mathbb {k}$ -linear functor. By the universal property of additive envelopes, this uniquely determines a functor $( \mathcal {C}\diamond \mathcal {D})_{\oplus } \to \mathcal {E}$ . Since the natural inclusion $(\mathcal {C}\diamond \mathcal {D})_{\oplus } \hookrightarrow (\mathcal {C}_{\oplus }\diamond \mathcal {D}_{\oplus })_{\oplus }$ is an equivalence, precomposing with an inverse equivalence yields a $\mathbb {k}$ -linear (additive) functor $(\mathcal {C}_{\oplus }\diamond \mathcal {D}_{\oplus })_{\oplus }\to \mathcal {E}$ and, since $\mathcal {E}$ is idempotent complete, also a $\mathbb {k}$ -linear (additive) functor $\Psi '\colon (\mathcal {C}_{\oplus }\diamond \mathcal {D}_{\oplus })_{\oplus ,e}\to \mathcal {E}$ .

Now any pair $e_1$ and $e_2$ of idempotents in $\mathcal {C}_{\oplus }$ and $\mathcal {D}_{\oplus }$ gives rise to an idempotent $e_1\otimes e_2$ in $(\mathcal {C}_{\oplus }\diamond \mathcal {D}_{\oplus })_{\oplus }$ , so we obtain an embedding $\mathcal {C}_{\oplus ,e}\diamond \mathcal {D}_{\oplus ,e} \hookrightarrow (\mathcal {C}_{\oplus }\diamond \mathcal {D}_{\oplus })_{\oplus , e}$ . Thus we can restrict our functor $\Psi '$ to a $\mathbb {k}$ -linear functor $\mathcal {C}_{\oplus ,e}\diamond \mathcal {D}_{\oplus ,e} \to \mathcal {E}$ .

Using the universal property of the additive envelope again, we thus obtain a $\mathbb {k}$ -linear functor

$$ \begin{align*}\Psi"\colon (\mathcal{C}_{\oplus,e}\diamond\mathcal{D}_{\oplus,e})_\oplus \to \mathcal{E}.\end{align*} $$

Noting that $(\mathcal {C}_{\oplus ,e}\diamond \mathcal {D}_{\oplus ,e})_{\oplus ,e} = \mathcal {C}_{\oplus , e}\boxtimes \mathcal {D}_{\oplus , e}$ by definition, completes the proof.

Thus, to define $\Psi _{k,n-k}\colon \widehat {\mathscr {S}}^{\mathrm {ext}}_k \boxtimes \widehat {\mathscr {S}}^{\mathrm {ext}}_{n-k}\to K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ , we only need to define its restriction

$$\begin{align*}\Psi_{k,n-k}\colon \widehat{\mathscr{BS}}^{\mathrm{ext}}_k\diamond \widehat{\mathscr{BS}}^{\mathrm{ext}}_{n-k}\to K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_n), \end{align*}$$

which is exactly what we will do in Section 4.1.2.

Suppose, additionally, that $\mathcal {C}$ and $\mathcal {D}$ have strict monoidal products $\circ _{\mathcal {C}}$ and $\circ _{\mathcal {D}}$ , respectively, and strict identity objects, both denoted by $\mathbb {1}$ . Then the coordinatewise monoidal product defines a strict monoidal structure on $\mathcal {C}\diamond \mathcal {D}$ (taking into account the aforementioned simplifications), and $\mathcal {C}$ and $\mathcal {D}$ can be embedded as commuting monoidal, $\mathbb {k}$ -linear subcategories of $\mathcal {C}\diamond \mathcal {D}$ by tensoring objects and morphisms on the right, resp. on the left, with $\mathbb {1}$ and $\mathrm {id}_{\mathbb {1}}$ , respectively. Another way to phrase this, is to say that the embeddings of $\mathcal {C}$ and $\mathcal {D}$ into $\mathcal {C}\diamond \mathcal {D}$ form a commuting pair of strict monoidal, $\mathbb {k}$ -linear functors, although we could not find this terminology in the literature.

Suppose that $\mathcal {E}$ is also strict monoidal with monoidal product $\circ _{\mathcal {E}}$ and identity object $\mathbb {1}$ , and let $\Psi _L\colon \mathcal {C}\to \mathcal {E}$ and $\Psi _R\colon \mathcal {D}\to \mathcal {E}$ be two strict monoidal, $\mathbb {k}$ -linear functors. Then we can define the $\mathbb {k}$ -linear functor $\Psi \colon \mathcal {C}\diamond \mathcal {D}\to \mathcal {E}$ by $\Psi :=\Psi _L\circ _{\mathcal {E}} \Psi _R$ , where

(85) $$ \begin{align} \Psi(X\diamond Y):=\Psi_L(X)\circ_{\mathcal{E}} \Psi_R(Y) \quad\text{and}\quad \Psi(f\diamond g):=\Psi_L(f)\circ_{\mathcal{E}} \Psi_R(g), \end{align} $$

for any objects $X\in \mathcal {C}, Y\in \mathcal {D}$ and any morphisms $f\in \mathcal {C}, g\in \mathcal {D}$ . By definition, $\Psi (X\diamond \mathbb {1})=\Psi _L(X)$ and $\Psi (\mathbb {1} \diamond Y)=\Psi _R(Y)$ for objects $X\in \mathcal {C}, Y\in \mathcal {D}$ , and $\Psi (f \diamond \mathrm {id}_{\mathbb {1}})=\Psi _L(f)$ and $\Psi (\mathrm {id}_{\mathbb {1}} \diamond g)=\Psi _R(g)$ for morphisms $f\in \mathcal {C}, g\in \mathcal {D}$ . Of course, we could equally well define $\Psi $ as $\Psi _R\circ _{\mathcal {E}} \Psi _L$ , but the latter functor need not be naturally isomorphic to $\Psi _L\circ _{\mathcal {E}} \Psi _R$ in general.

Definition 4.2. We say that $(\Psi _L, \Psi _R)$ form a symmetric pair of monoidal functors if there is a natural isomorphism

$$\begin{align*}\zeta\colon \Psi_L\circ_{\mathcal{E}} \Psi_R\xrightarrow{\cong} \Psi_R\circ_{\mathcal{E}} \Psi_L \end{align*}$$

satisfying the hexagon identities, given by the commutativity of the diagrams

(86)

and

(87)

for all $X,X_1, X_2\in \mathcal {C}$ and $Y,Y_1,Y_2\in \mathcal {D}$ .

If we write $\Psi (X,Y):=\Psi (X\diamond Y)$ and $\zeta _{X,Y}:=\zeta _{X\diamond Y}$ , denote composition of morphisms in $\mathcal {E}$ by $\cdot $ and suppress identity morphisms, then the commutativity of the diagrams in (86) and (87) translates to the equations

(88) $$ \begin{align} \zeta_{X_1 \circ_{\mathcal{C}} X_2, Y}=\zeta_{X_1, Y} \cdot \zeta_{X_2, Y}\quad\text{and}\quad \zeta_{X, Y_1\circ_{\mathcal{D}} Y_2}=\zeta_{X, Y_2}\cdot \zeta_{X, Y_1}, \end{align} $$

respectively. These are very similar to the hexagon identities for a braiding, which is symmetric because $\zeta _{X,Y}^{-1}$ defines the braiding for $(\Psi _R, \Psi _L)$ . This justifies our choice of terminology in Definition 4.2. Note that the commutative diagrams in (86) and (87) are actually pentagons in our case, because we have omitted the sixth side of the usual hexagons, which in our case is an identity, due to our assumption that the underlying monoidal categories are strict. In fact, those pentagons could be reduced to triangles, since two of the remaining five sides are also identities, but we have decided to include them for clarity.

Note that the embeddings of $\mathcal {C}$ and $\mathcal {D}$ into $\mathcal {C}\diamond \mathcal {D}$ form a symmetric pair of monoidal, $\mathbb {k}$ -linear functors with trivial braiding, in the sense that there is an equality $\Psi _L(X)\circ _{\mathcal {E}} \Psi _R(Y)=\Psi _R(Y) \circ _{\mathcal {E}} \Psi _L(X)$ and $\zeta _{X,Y}=\mathrm {id}_{\Psi (X,Y)}$ for all objects $X\in \mathcal {C}, Y\in \mathcal {D}$ .

Before we go on, let us explain the aforementioned relation with Drinfeld centralisers.

Remark 4.3. Let $\mathcal {A}$ and $\mathcal {B}$ be two strict monoidal, $\mathbb {k}$ -linear categories and $F\colon \mathcal {A}\to \mathcal {B}$ a strict monoidal, $\mathbb {k}$ -linear functor between them. The Drinfeld centraliser of F, denoted $\mathcal {Z}_F=\mathcal {Z}_F(\mathcal {A},\mathcal {B})$ , is the $\mathbb {k}$ -linear category whose objects are pairs $(B,\beta )$ , with $B\in \mathcal {B}$ an object and

$$\begin{align*}\beta\colon F(-)\circ_{\mathcal{B}} B\xrightarrow{\cong} B\circ_{\mathcal{B}} F(-) \end{align*}$$

a natural isomorphism (called half-braiding) satisfying $\beta _{A_1 \circ _{\mathcal {A}} A_2}=\beta _{A_1}\cdot \beta _{A_2}$ for $A_1,A_2\in \mathcal {A}$ (where we are suppressing two identity morphisms, as before). The morphisms in $\mathcal {Z}_F$ are morphisms in $\mathcal {B}$ which commute with the respective $\beta $ ’s. The category $\mathcal {Z}_F$ has an obvious monoidal structure such that the forgetful functor $\mathrm {Forget}:\mathcal {Z}_F\to \mathcal {B}$ , defined by $(B,\beta )\mapsto B$ , is a monoidal functor. For more information on Drinfeld centralisers, see e.g. [Reference HoekHoe, Section 1.5] (especially Example 1.5.3) and [Reference Elias and HogancampElHo, Section 2.8] and references therein. Note that Hoek uses the term Drinfeld centre, whereas Elias and Hogancamp use the term Drinfeld centraliser, which we find more adequate and therefore adopt here as well.

If $(\Psi _L, \Psi _R, \zeta )$ is a symmetric pair of strict monoidal functors, then $\Psi _L$ and $\Psi _R$ factor through each other’s Drinfeld centralisers, i.e., both diagrams below commute.

In general, a pair of monoidal functors $F\colon \mathcal {C}\to \mathcal {E}$ and $G\colon \mathcal {D}\to \mathcal {E}$ which factor through each other’s Drinfeld centralisers only yields two half-braidings. The factorisation of F through $\mathcal {Z}_G$ yields isomorphisms $\beta _{X,Y}$ , which satisfy the hexagon identity w.r.t. $X\in \mathcal {C}$ , and the factorisation of G through $\mathcal {Z}_F$ yields isomorphisms $\beta _{Y,X}$ , which satisfy the hexagon identity w.r.t. $Y\in \mathcal {D}$ . However, two half-braidings do not necessarily add up to one braiding, which has to satisfy the hexagon identities for both variables at once. In our case they do add up to a braiding, because they satisfy the additional condition $\beta _{Y,X}=\beta _{X,Y}^{-1}$ .

Proof. Suppose that $\Psi $ extends to a monoidal functor. Then there exists a natural isomorphism

(89) $$ \begin{align} \Psi(X_1\circ_{\mathcal{C}} X_2, Y_1 \circ_{\mathcal{D}} Y_2) \xrightarrow{\eta_{(X_1,Y_1),(X_2,Y_2)}} \Psi(X_1,Y_1)\circ_{\mathcal{E}} \Psi(X_2,Y_2) \end{align} $$

for any objects $X_1,X_2\in \mathcal {C}$ and $Y_1, Y_2\in \mathcal {D}$ . Note that $\Psi (\mathbb {1}, \mathbb {1})=\Psi _L(\mathbb {1})\circ _{\mathcal {E}} \Psi _R(\mathbb {1})=\mathbb {1}\circ _{\mathcal {E}} \mathbb {1}=\mathbb {1}$ holds on the nose, so $\eta $ is the only coherence morphism that can be non-trivial.

By definition, the object on the l.h.s. of (89) is equal to

(90) $$ \begin{align} \Psi_L(X_1) \circ_{\mathcal{E}} \Psi_L(X_2)\circ_{\mathcal{E}} \Psi_R(Y_1) \circ_{\mathcal{E}} \Psi_R(Y_2) \end{align} $$

while the object on the r.h.s. is equal to

(91) $$ \begin{align} \Psi_L(X_1) \circ_{\mathcal{E}}\Psi_R(Y_1) \circ_{\mathcal{E}}\Psi_L(X_2)\circ_{\mathcal{E}}\Psi_R(Y_2), \end{align} $$

hence $\eta _{(X_1,Y_1),(X_2,Y_2)}$ in (89) is completely determined by the natural isomorphism

(92) $$ \begin{align} \Psi_L(X_2)\circ_{\mathcal{E}} \Psi_R(Y_1) \xrightarrow{\zeta_{X_2,Y_1}} \Psi_R(Y_1)\circ_{\mathcal{E}} \Psi_L(X_2), \end{align} $$

where $\zeta _{X_2, Y_1}:=\eta _{(\mathbb {1},Y_1),(X_2,\mathbb {1})}$ . Note that we have swapped the order of $X_2$ and $Y_1$ in the subscript of $\zeta $ , which conforms to the notation of a braiding.

Let $X_1,X_2,X_3\in \mathcal {C}$ and $Y_1,Y_2,Y_3\in \mathcal {D}$ . The usual coherence condition for $\eta $ is

(93) $$ \begin{align} \eta_{(X_2,Y_2),(X_3,Y_3)} \cdot \eta_{(X_1,Y_1),(X_2 \circ_{\mathcal{C}} X_3,Y_2 \circ_{\mathcal{D}} Y_3)} = \eta_{(X_1,Y_1),(X_2,Y_2)} \cdot \eta_{(X_1 \circ_{\mathcal{C}} X_2, Y_1 \circ_{\mathcal{D}} Y_2 ),(X_2, Y_3)}, \end{align} $$

where we have suppressed several identity morphisms again. By the same arguments as above, we obtain the corresponding coherence condition for $\zeta $ :

(94) $$ \begin{align} \zeta_{X_3,Y_2} \cdot \zeta_{X_2 \circ_{\mathcal{C}} X_3, Y_1} = \zeta_{X_2, Y_1} \cdot \zeta_{X_3, Y_1 \circ_{\mathcal{D}} Y_2}. \end{align} $$

Note that the coherence conditions in (93) and (94) are equivalent.

The hexagon identities for $\zeta $ are obtained from (94) as special cases, by putting $Y_2=\mathbb {1}$ and $X_2=\mathbb {1}$ , respectively. This proves that $\Psi _L, \Psi _R$ is a symmetric pair of monoidal functors.

Conversely, suppose that $\Psi _L, \Psi _R$ is a symmetric pair of monoidal functors with braiding $\zeta $ . Thanks to the interchange law for morphisms in a monoidal category, we have

(95) $$ \begin{align} \zeta_{X_3,Y_2} \cdot \zeta_{X_2, Y_1}= \zeta_{X_2, Y_1} \cdot \zeta_{X_3, Y_2}, \end{align} $$

as can be seen from the corresponding commutative diagram:

(96)

Multiplying both sides in (95) on the right by $\zeta _{X_3, Y_1}$ and using the hexagon identities yields (94), which implies that $\eta $ satisfies (93). This proves that $\Psi $ extends to a monoidal functor.

The following corollary shows that our choice to define $\Psi $ as $\Psi _L\circ _{\mathcal {E}} \Psi _R$ , instead of $\Psi _R\circ _{\mathcal {E}} \Psi _L$ , is not essential if the pair of monoidal functors $\Psi _L, \Psi _R$ is symmetric. The proof uses similar arguments as the proof of Lemma 4.4 and is straightforward, so we leave the details to the reader.

4.1.2 Definition of $\Psi _{k,n-k}$ and main theorem

As in Section 4.1.1, we first define two strict monoidal $\mathbb {k}$ -linear functors

$$\begin{align*}\Psi_L\colon \widehat{\mathscr{BS}}^{\mathrm{ext}}_k\to K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_{n})\quad \text{and} \quad \Psi_R\colon \widehat{\mathscr{BS}}^{\mathrm{ext}}_{n-k}\to K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_{n}) \end{align*}$$

and then define

(97) $$ \begin{align} \Psi_{k,n-k}\colon \widehat{\mathscr{BS}}^{\mathrm{ext}}_{k}\diamond \widehat{\mathscr{BS}}^{\mathrm{ext}}_{n-k}\to K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_{n} ) \end{align} $$

by

$$\begin{align*}\Psi_{k,n-k}(X\diamond Y):= \Psi_L(X)\Psi_R(Y) \quad\text{and}\quad \Psi_{k,n-k}(f\diamond g)=\Psi_L(f)\Psi_R(g), \end{align*}$$

for any objects $X\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k, Y\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ and any morphisms $f\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k, g\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ . We have omitted the symbol for the monoidal product in $K^b(\widehat {\mathscr {S}}^{\mathrm {ext}}_n)$ . To avoid clutter in certain diagrams of the definition below, we have reduced the number of labels. When there are several strands of the same colour, we often label just one of them. The labels of the others are understood to be the same.

On objects

On the generating objects, $\Psi _L$ and $\Psi _R$ are defined by

$$ \begin{align*} \Psi_L \colon \begin{cases} {\mathrm{B}}_i\mapsto {\mathrm{B}}_i, \ i=1,\ldots,k-1 \\ {\mathrm{B}}_0\mapsto {\mathrm{T}}_k^{-1}\dotsm {\mathrm{T}}_{n-1}^{-1}{\mathrm{B}}_0 {\mathrm{T}}_{n-1}\dotsm {\mathrm{T}}_k, \\ {\mathrm{B}}_\rho \mapsto {\mathrm{B}}_{\rho}{\mathrm{T}}_{n-1}\dots {\mathrm{T}}_k ,\\ \end{cases} \Psi_R \colon \begin{cases} {\mathrm{B}}_j \mapsto {\mathrm{B}}_{k+j} , \quad j=1, \dotsc ,n-k-1 ,\\ {\mathrm{B}}_0 \mapsto T_0 \dotsm {\mathrm{T}}_{k-1}{\mathrm{B}}_k {\mathrm{T}}_{k-1}^{-1}\dotsm {\mathrm{T}}_0^{-1}, \\ {\mathrm{B}}_\rho \mapsto {\mathrm{T}}_k^{-1}\dots {\mathrm{T}}_{1}^{-1}{\mathrm{B}}_{\rho}. \end{cases} \end{align*} $$

On morphisms

• • If $D_{1,\dotsc ,k-1}$ is a morphism of $\widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ in colours $(1,\dotsc ,k-1)$ , then

  $$\begin{align*}\Psi_L(D_{1,\dotsc ,k-1}):=D_{1,\dotsc ,k-1},\end{align*}$$

• • If $D^{\prime }_{1,\dotsc ,k-1}$ is a morphism of $\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ in colours $(1,\dotsc ,n-k-1)$ , then

  $$\begin{align*}\Psi_R(D^{\prime}_{1,\dotsc ,n-k-1}):=D^{\prime}_{k+1,\dotsc ,n-1},\end{align*}$$

On other morphisms, $\Psi _L$ and $\Psi _R$ are defined as follows.

$\bullet $ On oriented generators:

$\bullet $ On generators including strands with distant colours:

for $1<\color {orange}{i}<k-1$ and $1<\color {mygreen}{j}<n-k-1$ .

$\bullet $ On generators including strands with adjacent colours:

for $i,j\neq 0,1$ , while

and

for $k\geq 2$ .

The inner labels in the last line are ${\color {orange}n-1}$ and ${\color {CadetBlue}[n-2,k-1]}$ (left) and ${\color {orange}k-1}$ and ${\color {NavyBlue}[-1,k-2]}$ (right). This ends the definition of $\Psi _L$ and $\Psi _R$ .

Lemma 4.7. For any domain X and codomain Y of a generating morphism in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ of degree t, we have

$$\begin{align*}\hom_{K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_n)}(\Psi_L(X),\Psi_L(Y)\langle t\rangle)\cong \hom_{\widehat{\mathscr{BS}}^{\mathrm{ext}}_k}(X,Y\langle t\rangle). \end{align*}$$

Similarly, for any domain Z and codomain W of a generating morphism in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ of degree t, we have

$$\begin{align*}\hom_{K^b(\widehat{\mathscr{S}}^{\mathrm{ext}}_n)}(\Psi_R(Z),\Psi_R(W)\langle t\rangle)\cong \hom_{\widehat{\mathscr{BS}}^{\mathrm{ext}}_{n-k}}(Z,W\langle t\rangle). \end{align*}$$

Moreover, all those morphism spaces are one-dimensional.

Proof. The proof is immediate except for two cases in $\Psi _L$ and two cases in $\Psi _R$ :

• • a crossing in $\widehat {\mathscr {S}}^{\mathrm {ext}}_k$ involving an oriented strand and unoriented strands labelled $0$ and $k-1$ , respectively in $\widehat {\mathscr {S}}^{\mathrm {ext}}_{n-k}$ with labels $0$ and $n-k-1$ ;

• • a six-valent vertex in $\widehat {\mathscr {S}}^{\mathrm {ext}}_k$ involving strands labelled $0$ and $k-1$ , respectively in $\widehat {\mathscr {S}}^{\mathrm {ext}}_{n-k}$ with labels $0$ and $n-k-1$ .

We only prove the second part of each case, the first being similar.

For a crossing in ${\widehat {\mathscr {S}}}^{\mathrm {ext}}_{n-k}$ involving an oriented strand and unoriented strands labelled $0$ and $n-k-1$ we compute

$$ \begin{align*} \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}&(\Psi_R({\mathrm{B}}_\rho{\mathrm{B}}_{n-k-1}) , \Psi_R(\mathrm{B}_0\mathrm{B}_\rho)) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{T}}_{[1,k]}^{-1}{\mathrm{B}}_\rho{\mathrm{B}}_{n-1} , {\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{T}}_{[1,k]}^{-1}{\mathrm{B}}_\rho) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{T}}_{[1,k]}^{-1}{\mathrm{B}}_{0} , {\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{T}}_{[1,k]}^{-1}) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{B}}_{0} , {\mathrm{T}}_{[1,k]}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{T}}_{[1,k]}^{-1}) \\ & \hspace{-4mm}\overset{(78)}{\cong} \!\! \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{B}}_0,{\mathrm{B}}_{0}) \cong \hom_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_n}({\mathrm{B}}_0,{\mathrm{B}}_{0}) \cong \hom_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_{n-k}}({\mathrm{B}}_0,{\mathrm{B}}_{0}) \\ &\cong \hom_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_{n-k}}({\mathrm{B}}_\rho{\mathrm{B}}_{n-k-1},{\mathrm{B}}_{0}{\mathrm{B}}_\rho). \end{align*} $$

For a six-valent vertex in ${\widehat {\mathscr {S}}}^{\mathrm {ext}}_{n-k}$ involving strands labelled $0$ and $n-k-1$ we compute

$$ \begin{align*} & \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}(\Psi_R({\mathrm{B}}_{n-k-1}{\mathrm{B}}_0{\mathrm{B}}_{n-k-1}),\Psi_R({\mathrm{B}}_0{\mathrm{B}}_{n-k-1}{\mathrm{B}}_0)) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{B}}_{n-1}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{B}}_{n-1},{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{B}}_{n-1}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{B}}_{n-1}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{B}}_{n-1}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k{\mathrm{T}}_{[0,k-1]}^{-1}{\mathrm{B}}_{n-1}{\mathrm{T}}_{[0,k-1]}{\mathrm{B}}_k,\mathbb{1}) \\ &\overset{(78)}{\cong} \!\! \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{T}}_{[-1,k-2]}^{-1}{\mathrm{B}}_{k-1}\!{\mathrm{T}}_{[-1,k-2]}{\mathrm{B}}_k{\mathrm{T}}_{[-1,k-2]}^{-1}{\mathrm{B}}_{k-1}\!{\mathrm{T}}_{[-1,k-2]}{\mathrm{B}}_k{\mathrm{T}}_{[-1,k-2]}^{-1}\\&\qquad{\mathrm{B}}_{k-1}\!{\mathrm{T}}_{[-1,k-2]}{\mathrm{B}}_k,\!\mathbb{1}) \\ &\cong \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}({\mathrm{B}}_{k-1}{\mathrm{B}}_k{\mathrm{B}}_{k-1}{\mathrm{B}}_k{\mathrm{B}}_{k-1}{\mathrm{B}}_k,\! \mathbb{1}) \cong \hom_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_n}({\mathrm{B}}_{k-1}{\mathrm{B}}_k{\mathrm{B}}_{k-1},{\mathrm{B}}_k{\mathrm{B}}_{k-1}{\mathrm{B}}_k) \\ &\cong \hom_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_{n-k}}({\mathrm{B}}_{n-k-1}{\mathrm{B}}_0{\mathrm{B}}_{n-k-1},{\mathrm{B}}_0{\mathrm{B}}_{n-k-1}{\mathrm{B}}_0). \end{align*} $$

4.2.1 Well-definedness

The proofs that $\Psi _L$ and $\Psi _R$ preserve all diagrammatic relations are very similar, so we only prove that $\Psi _L$ is well-defined. Because of the two cases in (37), the proof that $\Psi _L$ is well-defined contains four cases: $1=k=n-1$ , $1=k<n-1$ , $2=k<n$ and $2<k<n$ . However, the first two cases are proved in the same way as the corresponding part in the fourth case. The third case is also almost the same as the corresponding part of the fourth case, except for the proof that $\Psi _L$ preserves (37), which we therefore prove separately at the end of this subsection.

Suppose that $k> 2$ . Since $\Psi _L$ is the identity on diagrams only involving colours different from $0$ and $k-1$ it preserves the relations involving only those diagrams.

1. (a1) Relations (26) to (29) for $i=0$ follow by isotopy and similar arguments as in the first bullet point in the proof of [Reference Mackaay, Miemietz and VazMMV, Theorem 5.4].

2. (a2) Relations (30) to (33) for $i=0$ or $k=0$ are verified using the relations in (68) and the usual Soergel calculus.

3. (a3) Relations (34) to (37) for $\{i,j\}=\{0,1\}$ are checked using similar arguments as in the third bullet point in the proof of [Reference Mackaay, Miemietz and VazMMV, Theorem 5.4].

4. (a4) Relations (34) to (37) for $\{i,j\}=\{0,k-1\}$ are more involved and we give a sample computation for relation (35), with the proof of the other relations being similar. We have

   

   where the missing labels in the inner part of the diagram are ${\color {PineGreen}n-1}$ and ${\color {RawSienna}[n-2,k-1]}$ . Using (59) twice on the strands labelled ${\color {RawSienna}[n-2,k-1]}$ on the first term, and (68) on the second gives

   

5. (a5) Relation (38) for one of the labels being $0$ follows easily.

6. (a6) The relevant cases for relation (39) are $\{i,j\}=\{0,1\}$ , $\{i,j\}=\{0,k-1\}$ or $k=0$ and all are straightforward.

7. (a7) The relevant cases for relation (40) are $(i,j,k)=(0,1,2)$ , $(i,j,k)=(k-1,0,1)$ and $(i,j,k)=(k-2,k-1,0)$ . Using arguments similar to the seventh bullet point in the proof of [Reference Mackaay, Miemietz and VazMMV, Theorem 5.4] one verifies the first case. We provide the proof of the second and the third is similar.

   Applying $\Psi _L$ to the left-hand side of (40) yields (the unlabelled strands of the diagrams on the right-hand side have labels ${\color {RawSienna}[n-2,k-1]}$ , ${\color {orange}k-1}$ )

   

   A similar procedure gives

   

   Relation (40) for labels $(k-1,0,1)$ implies that they are equal.

8. (a8) Relations (41) and (42) are an easy consequence of (57) and (59).

9. (a9) Relation (43) is immediate if $0\notin \{i-1,i,j-1,j\}$ . The remaining cases can be proved using relations (74), (81) and (83).

10. (a10) Checking relation (44) uses only Lemma 3.5 and relations (71) and (63) unless $(i-1,i)=(k-1,0)$ . The remaining cases use (77) and (63). Relation (45) is verified using Lemma 3.5 and relations (63) and (80). For (46) the only case that is not immediate is when $i=0$ . In this case,

    

11. (a11) For relations (47) we need to consider the cases $i-1=0$ , $i=0$ and $i+1=0$ . The first case uses (68) and (57).

    In the remaining cases, we prove the first relation, the second one being similar.

    For $i=0$ we compute

    

    For the case $i+1=0$ we use the hom & dot trick. We have that

    (98)

    

    define morphisms in

    $$\begin{align*}\mathrm{hom}_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}&\left({\mathrm{T}}_{[n-1,k]}^{-1} {\mathrm{B}}_0 {\mathrm{T}}_{[n\!-\!1,k]} \mathrm{B}_{k-1} {\mathrm{T}}_{[n-1,k]}^{-1} \mathrm{B}_0 {\mathrm{T}}_{[n-1,k]}, \mathrm{B}_\rho {\mathrm{T}}_{[n-1,k]}\right.\\&\qquad \left.{\mathrm{B}}_{k-2}{\mathrm{B}}_{k-1}{\mathrm{B}}_{k-2}{\mathrm{T}}_{[n-1,k]}^{-1}{\mathrm{B}}_{\rho}^{-1}\right). \end{align*}$$

    This morphism space is one-dimensional, as can be seen as follows. By (75) (with $a=k$ and $b=0$ ) and (68), we have ${\mathrm {T}}_{[n-1,k]}^{-1} {\mathrm {B}}_0 {\mathrm {T}}_{[n\!-\!1,k]}\simeq {\mathrm {T}}_{[n,k+1]} {\mathrm {B}}_{k}{\mathrm {T}}_{[n,k+1]}^{-1}$ and ${\mathrm {T}}_{[n,k+1]} {\mathrm {B}}_{k-1}\simeq {\mathrm {B}}_{k-1} {\mathrm {T}}_{[n,k+1]}$ . This implies that the domain is homotopy equivalent to ${\mathrm {T}}_{[n,k+1]} {\mathrm {B}}_k\mathrm {B}_{k-1} \mathrm {B}_k {\mathrm {T}}_{[n,k+1]}^{-1}$ . Commuting $\mathrm {B}_\rho $ with all factors from left to right shows that the codomain is homotopy equivalent to ${\mathrm {T}}_{[n,k+1]} {\mathrm {B}}_{k-1}{\mathrm {B}}_{k} \mathrm {B}_{k-1} {\mathrm {T}}_{[n,k+1]}^{-1}$ . By adjunction, the above morphism space is therefore isomorphic to $\mathrm {hom}_{K^b({\widehat {\mathscr {S}}}^{\mathrm {ext}}_n)}\left ( {\mathrm {B}}_k{\mathrm {B}}_{k-1} {\mathrm {B}}_k, {\mathrm {B}}_{k-1}{\mathrm {B}}_{k} {\mathrm {B}}_{k-1}\right )$ . Moreover, we have

    $$\begin{align*}\mathrm{hom}_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}\left( {\mathrm{B}}_k{\mathrm{B}}_{k-1} {\mathrm{B}}_k, {\mathrm{B}}_{k-1}{\mathrm{B}}_{k} {\mathrm{B}}_{k-1}\right)\cong \mathrm{hom}_{{\widehat{\mathscr{S}}}^{\mathrm{ext}}_n}\left( {\mathrm{B}}_k{\mathrm{B}}_{k-1} {\mathrm{B}}_k, {\mathrm{B}}_{k-1}{\mathrm{B}}_{k} {\mathrm{B}}_{k-1}\right) \end{align*}$$

    and the latter morphism space is one-dimensional (thanks to Soergel’s hom formula in (49)).

    We attach a dot to the free end at the top labelled $\color {orange}k-1$ of the diagrams at the right-hand sides of (98). For the left diagram in (98), use (34) to get rid of the 6-valent vertex and obtain two terms which can be further simplified by relations (80), (45), (63) and (77). For the right diagram in (98), first use (80), (45), (63) and (68) to slide the dot closer to the 6-valent vertex, then proceed as in the previous case. In the end, we see that both diagrams in (98) become equal to

    

This completes the proof that $\Psi _L$ is well-defined for $k\ne 2$ .

Now, let $k=2$ . As already mentioned, it only remains to check (37) in this case. We have

Now, with $0$ and $1$ switched,

where in the second equality we used (an easy consequence of [Reference Mackaay, Miemietz and VazMMV, Lemma 4.6])

and in the third equality we used (37) twice, one involving labels $\color {blue}1$ and $\color {violet}0$ and the other involving colours $\color {violet}0$ and $\color {orange}n-1$ .

This completes the proof of the last case.

4.2.2 Definition of $\zeta $

We will first show that for every pair of objects $X\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k, Y\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ , there is an isomorphism in $K^b({\widehat {\mathscr {S}}}^{\mathrm {ext}}_n)$

$$\begin{align*}\zeta_{X,Y}\colon \Psi_L(X)\Psi_R(Y)\xrightarrow{\cong} \Psi_R(Y)\Psi_L(X) \end{align*}$$

and that these isomorphisms satisfy the hexagon identities from Definition 4.2. After that, we will show that $\zeta $ is natural in X and Y.

We define $\zeta _{X,Y}$ for pairs of generating objects $X\in \{{\mathrm {B}}_i,{\mathrm {B}}_\rho \}\subset \mathrm {Ob}(\widehat {\mathscr {BS}}^{\mathrm {ext}}_k)$ and $Y\in \{{\mathrm {B}}_i,{\mathrm {B}}_\rho \}\subset \mathrm {Ob}(\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k})$ and extend the definition recursively to pairs of monoidal products of generating objects (corresponding to Bott-Samelson bimodules) by the hexagon identities. Concretely, we put

(99) $$ \begin{align} \zeta_{X_1X_2, Y_1Y_2}:=\zeta_{X_1,Y_1Y_2}\cdot\zeta_{X_2,Y_1Y_2}=\zeta_{X_1,Y_2}\cdot\zeta_{X_1,Y_1}\cdot\zeta_{X_2,Y_2}\cdot\zeta_{X_2,Y_1} \end{align} $$

for any objects $X_1,X_2\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ and $Y_1,Y_2\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ . Alternatively, we could have put

(100) $$ \begin{align} \zeta_{X_1X_2, Y_1Y_2}:=\zeta_{X_1X_2,Y_2}\cdot\zeta_{X_1X_2,Y_1}=\zeta_{X_1,Y_2}\cdot\zeta_{X_2,Y_2}\cdot\zeta_{X_1,Y_1}\cdot\zeta_{X_2,Y_1}, \end{align} $$

but the expressions in (99) and (100) are equal because $\zeta _{X_1,Y_1}\cdot \zeta _{X_2,Y_2}= \zeta _{X_2,Y_2}\cdot \zeta _{X_1,Y_1}$ thanks to the interchange law, see (95). In particular, the isomorphisms $\zeta _{X,Y}$ satisfy the hexagon identities by definition.

After this explanation, it remains to define $\zeta _{X,Y}$ for pairs of generating objects.

In all cases below, $\zeta _{X,Y}$ is obtained by composing invertible morphisms in $K^b({\widehat {\mathscr {S}}}^{\mathrm {ext}}_n)$ . To see this, one notes that all diagrams appearing are isotopic to a diagram where at each height there is either a merger or a splitter, a four-valent vertex (a crossing), a six-valent one (all involving at least one oriented strand), or just vertical strands. One checks easily that all of them are invertible (this is implied by the various relations in Section 3.1 and Section 3.2), hence so is their composite. Therefore, we only give the diagrams for $\zeta _{X,Y}$ without further explanation.

In $\zeta _{{\mathrm {B}}_\rho ,{\mathrm {B}}_0}$ , $\color {Salmon}{\mathrm {T}}_{[-1,k\!-\!2]}={\mathrm {T}}_{n-1}{\mathrm {T}}_{0}{\mathrm {T}}_{1}\dotsm {\mathrm {T}}_{k-2}$ , and in $\zeta _{{\mathrm {B}}_0,{\mathrm {B}}_0}$ , $\color {WildStrawberry}a=[1,k\!-\!2]$ and $\color {Cyan}b=[n\!-\!2,k\!+\!1]$ .

This completes the definition of $\zeta $ .

Remark 4.10. Diagrams for $\zeta $ involving inverses of ${\mathrm {B}}_\rho ^{-1}$ are obtained easily by applying the appropriate adjunctions and taking inverses (if necessary). We have

and similarly for $\zeta _{{\mathrm {B}}_\rho ^{-1},\mathrm {B}_{\rho }}^{-1}$ . A rotation of $\zeta _{\mathrm {B}_\rho ,\mathrm {B}_\rho }$ by 180 degrees yields $\zeta _{\mathrm {B}_\rho ^{-1},\mathrm {B}_\rho ^{-1}}$ .

4.2.3 Naturality of $\zeta $

To show that $\zeta $ is natural, we have to prove commutativity of the diagram

(101)

for all objects $X,W\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ and $Y,Z\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ , and all morphisms $f\in \mathrm {hom}_{\widehat {\mathscr {BS}}^{\mathrm {ext}}_k}(X,W)$ and $g\in \mathrm {hom}_{\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}}(Y,Z)$ . Thanks to the hexagon identities for $\zeta $ and the interchange law for morphisms in $K^b({\widehat {\mathscr {S}}}^{\mathrm {ext}}_n)$ , it suffices to show commutativity in (101) for pairs $(f,g)$ where f is a generating morphism in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ and $g=\mathrm {id}_Y$ for some generating object $Y\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ , or $f=\mathrm {id}_X$ for some generating object $X\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_k$ and g is a generating morphism in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ . Finally, we note that the proof for pairs of the first type and the proof for pairs of the second type are very similar, so we only give the proof for pairs of the second type below. Concretely, those pairs are $(\mathrm {id}_X, g)$ , where $X\in \{{\mathrm {B}}_\rho , {\mathrm {B}}_0, \ldots , {\mathrm {B}}_{k-1}\}$ and g is one of the generators in (23), (24) or (25) with labels in $\{\rho , 0,\ldots , n-k-1\}$ . The only cases in which commutativity of (101) is not immediate, is when $X\in \{{\mathrm {B}}_\rho , {\mathrm {B}}_0\}$ and/or the labels of some strands of g belong to $\{0, \rho \}$ .

Note that checking commutativity of (101) always consists of verifying the equality of two diagrams, both belonging to the same morphism space. In all but two cases the dimension of the morphism space in the given degree is one-dimensional (see Lemma 4.11), so we can use the hom & dot trick, explained in Remark 3.7. The remaining two cases are checked directly.

Lemma 4.11. For each generating object X of $\widehat {\mathscr {BS}}^{\mathrm {ext}}_{k}$ and each generating morphism $g\colon Y\to Z$ in $\widehat {\mathscr {BS}}^{\mathrm {ext}}_{n-k}$ , there are objects $U,V\in \widehat {\mathscr {BS}}^{\mathrm {ext}}_n$ and an isomorphism

(102) $$ \begin{align} \hom_{K^b({\widehat{\mathscr{S}}}^{\mathrm{ext}}_n)}(\Psi_L(X)\Psi_R(Y),\Psi_R(Z)\Psi_L(X))\cong\hom_{\widehat{\mathscr{BS}}^{\mathrm{ext}}_n}(U,V). \end{align} $$

Unless $X={\mathrm {B}}_0$ and $g\colon 1\to {\mathrm {B}}_0$ is a dot labelled 0 or $g\colon \mathrm {B}_0\to \mathrm {B}_0\mathrm {B}_0$ is a trivalent vertex labelled 0, the two morphism spaces in (102) are one-dimensional by Soergel’s hom formula in (102) are one-dimensional by Soergel’s hom formula in (49).

Before checking commutativity of (101), we need a technical lemma. In the following, a rectangular region of a diagram decorated with a $\zeta _{X,Y}$ will be called a box.

Lemma 4.12. The following relations involving boxes and dots hold in $\mathsf {{K}}^b({\widehat {\mathscr {S}}}^{\mathrm {ext}}_n)$ :

(103)

(104)

(105)

(106)

and analogous relations with the dot being attached to any other free end.

Proof. Relations (103) and (104) are immediate.

Proof of (105). By (59) the l.h.s. can be written as

Introducing the diagram of $\zeta _{{\mathrm {B}}_\rho ,{\mathrm {B}}_0}$ from Section 4.2.3 and computing yields

In the equality above, we have used (80) to simplify the only 6-valent vertex appearing (involving labels k and $k-1$ , followed by (68) and (45) to slide the dot to the top of the diagram, (61) to simplify the digons involving strands labelled k and $k-1$ that were created by (80), and finally (57) to simplify the loop labelled $k-1$ (created by applying (61)).

Applying (65) to the strands labelled $[n\!-\!2,k]$ and $[0,k\!-\!2]$ gives

which is the r.h.s. of (105).

Proof of (106). We introduce the diagram for $\zeta _{{\mathrm {B}}_0,{\mathrm {B}}_0}$ in the left-hand side of (106) and compute

The proof of the analogous relations with the dot being attached at any other free end is similar.

Now we are ready to prove the naturality of $\zeta $ .

1. (N1) For $X={\mathrm {B}}_{\rho }$ and commutativity of (101) follows from (105).

2. (N2) For $X=B_\rho $ and , we have to check

   (107)

   

   We use the hom & dot trick and attach dots to the upper endpoints labelled $\color {Violet}k$ on both diagrams and then use the analogue of (105) with the dot on the top end to pass the dots past the boxes with $\zeta _{{\mathrm {B}}_\rho ,{\mathrm {B}}_0}$ .

   The right-hand side of (107) gives

   

   while for the left-hand side we have

   

   which equals the right-hand side by (26).

3. (N3) For the cases $X={\mathrm {B}}_\rho $ , and $X={\mathrm {B}}_\rho $ , we have to check

   (108)

   

   and

   (109)

   

   respectively.

   Both equalities can be checked to hold by using the hom & dot trick: attach a dot to the top endpoints labelled $\color {orange}k+1$ and use (103).

4. (N4) For $X=B_\rho $ and we have to check

   (110)

   

   We use the hom & dot trick and attach a dot on the top endpoints labelled $\color {blue}k+1$ of both sides of (110). The left-hand side is simplified using (103) (with $j=1$ ), followed by (59) on the two strands labelled $[1,k]$ and (63). On the right-hand side, we use (45), followed by (105) and then (63) on the two strands labelled $[1,k]$ and (63). The resulting diagram, in both cases, is

   

5. (N5) For $X=B_\rho $ and we have to check

   (111)

   

   We use the hom & dot trick and attach a dot on the top endpoints labelled $\color {Violet}k$ of both sides of (111). The left-hand side is simplified using (105), followed by (45), (80), (63) and (57). On the right-hand side, we use (45) followed by (80), (63) and (103). The resulting diagram, in both cases, is

   

6. (N6) The cases $X=B_\rho $ , and $X=B_\rho $ , are very similar. We only prove the first case here. We have to check

   (112)

   

   We use the hom & dot trick and attach a dot on all endpoints labelled $\color {Violet}k$ and $\color {blue}k+1$ of both sides of (112). On the right-hand side, we use (105), along with (103), to pull all dots from the bottom to the top. The resulting diagram is

   

   On the left-hand side we first use (34) on the six-valent vertex, which results in two terms with dots on the bottom ends. We then use (105) several times and (103) on the resulting diagrams, along with (68) to pull all dots and barbells to to left top. This results in

   

   which equals the right-hand side since

   

7. (N7) For $X={\mathrm {B}}_0$ and we cannot use the hom & dot trick (see Lemma 4.11). Commutativity of (101) in this case follows from (106) rotated 180 degrees.

8. (N8) For $X=B_0$ and , we cannot use the hom & dot trick (see Lemma 4.11). We have to check

   (113)

   

   We introduce the diagram for $\zeta _{{\mathrm {B}}_0,{\mathrm {B}}_0}$ in the left-hand side of (113) and compute

   
   

   which equals the right-hand side of (113) by (68), (32) and (82).

9. (N9) For the cases $X={\mathrm {B}}_0$ , and $X={\mathrm {B}}_0$ , we have to check

   (114)

   

   and

   (115)

   

   respectively.

   Both equalities can be checked to hold by using the hom & dot trick: attach a dot to the top endpoints labelled $\color {orange}k+1$ and use (104).

10. (N10) For $X=B_0$ and we have to check

    (116)

    

    We use the hom & dot trick and attach a dot on the top endpoint labelled $\color {blue}k+1$ of both sides of (116). The left-hand side is simplified using (104) (with $j=k+1$ ), followed by (45), (59) on the two strands labelled $[1,k]$ , and (63). On the right-hand side, we use (45), (59) on the two strands labelled $[1,k]$ , (63), and finally (106) to pull the dot across the box $\zeta _{{\mathrm {B}}_0,{\mathrm {B}}_0}$ to the bottom right.

    The resulting diagram, in both cases, is

    

11. (N11) For $X=B_0$ and we have to check

    (117)

    

    We use the hom & dot trick and attach a dot on the top endpoints labelled $\color {Violet}k$ of both sides of (117). The left-hand side is simplified using (106), followed by (45), (80), (63) and (57). On the right-hand side, we use (45) followed by (80), (63) and (104). The resulting diagram, in both cases, is

    

12. (N12) The cases $X=B_0$ , and and $X=B_\rho $ , are very similar. We only prove the first case here. We have to check

    (118)

    

    We use the hom & dot trick. At the top of both diagrams we attach a dot at the endpoints labelled $\color {violet}0$ and $\color {Violet}k$ . At the bottom we attach dots to the endpoints labelled $\color {violet}0$ , $\color {Violet}k$ and the leftmost endpoint labelled ${\color {blue}k+1}$ .

    On the resulting diagram at the left-hand side, we use (106) twice and (104) to pull all dots to the bottom. Using (59) and (68) we obtain

    (119)

    

    On the right-hand side, we use (a variant of) (104) to pull the dots labelled $\color {violet}0$ close to the box $\zeta _{{\mathrm {B}}_0,{\mathrm {B}}_0}$ and use (59) to obtain

    (120)

    

    Finally, applying (106) to slide the dot labelled $\color {purple}k$ up through the box, using (68) to slide the dot labelled $\color {blue}k\!+\!1$ through the strand labelled $\color {purple}[0,k\!-\!1]$ , and using (59) on the strands labelled $\color {purple}[0,k\!-\!1]$ with endpoints at the bottom gives (119).

This ends the proof of the naturality of $\zeta $ .

Remark 4.13. Naturality for morphisms involving ${\mathrm {B}}_\rho ^{-1}$ is easy to check by applying adjunctions (see Remark 4.10). For example, for $X={\mathrm {B}}_\rho $ and we have

4.2.4 A conjecture regarding faithfulness

Note that the functor $\Psi _{k,n-k} $ , as defined above, is not faithful.

Define ${\widehat {\mathscr {S}}}^{\mathrm {ext}}_k \boxtimes _y \widehat {\mathscr {S}}^{\mathrm {ext}}_{n-k}$ as the usual box tensor product, but with the morphism spaces tensored over the polynomial algebra $\mathbb {R}[\fbox{y} ]$ .

5.1.1 The Hecke algebra $\widehat {H}^{\mathrm {ext}}_{2}$

Recall that $\widehat {\mathfrak {S}}_2$ is generated by two simple reflections $s_0$ and $s_1$ which are only subject to the quadratic relations $s_0^2=e=s_1^2$ . Any element in $\widehat {\mathfrak {S}}_2$ has therefore a unique rex, which can be identified with an alternating binary sequence. To work with those, we have to introduce some notation. For $m\in \mathbb {Z}_{>0}$ and $i\in \{0,1\}$ , let $\widehat {m}_{\,i}$ and ${}_{i}\widehat {m}$ denote the alternating binary sequences of length m ending with i and starting with i, respectively. If $\widehat {m}_{\,j}={}_{i}\widehat {m}$ for certain $m\in \mathbb {Z}$ and $i,j\in \{0,1\}$ (we call this a parity condition), then we write ${}_{i}\widehat {m}_{\,j}$ for that alternating sequence to make the computations below easier to read. For example, $\widehat {4}_{\,0}={}_{1}\widehat {4}={}_{1}\widehat {4}_{\,0}=1010$ . By convention, we put $\widehat {0}_{\,i}={}_{i}\widehat {0}=\emptyset $ , for $i\in \{0,1\}$ , which of course corresponds to the neutral element $e\in \widehat {\mathfrak {S}}_2$ .

The extended affine symmetric group $\widehat {\mathfrak {S}}_2^{\mathrm {ext}}$ has an extra generator $\rho $ of infinite order, and two extra relations: $\rho s_0\rho ^{-1}=s_1$ and $\rho s_1 \rho ^{-1} =s_0$ .

The extended affine Hecke algebra $\widehat {H}^{\mathrm {ext}}_{2}$ is generated by $T_0, T_1$ and $\rho ^{\pm 1}$ , subject to only the first relation in (8) and all relations in (9). In this case, it is easy to express the KL basis elements $b_w$ in terms of the regular basis elements $T_u$ explicitly:

(121) $$ \begin{align} b_w=\sum_{u\preceq w} q^{\ell(w)-\ell(u)}T_u, \end{align} $$

where $\preceq $ denotes the Bruhat order on $\widehat {\mathfrak {S}}_2$ . Note that $u \preceq w$ holds iff the rex for u is an alternating binary subsequence of the rex for w and it is easy to invert this change of basis:

(122) $$ \begin{align} T_w=\sum_{u\preceq w} (-q)^{\ell(w)-\ell(u)}b_u. \end{align} $$

The multiplication constants w.r.t. the KL basis are also easy to compute in this case and are all equal to 0, 1, 2 or $[2]=q+q^{-1}$ . For $m,n\in \mathbb {Z}_{\geq 1}$ (assuming that the parity conditions are met in each line below), we have

(123) $$ \begin{align} b_{{}_{1}\widehat{m}_{\,1}} b_{{}_{1}\widehat{n}_{\,1}}&= [2]\left(b_{{}_{1}\widehat{m+n-1}_{\,1}} + b_{{}_{1}\widehat{m+n-3}_{\,1}}+\ldots + b_{{}_{1}\widehat{\vert m-n\vert +3}_{\,1}} + b_{{}_{1}\widehat{\vert m-n\vert +1}_{\,1}} \right);\nonumber\\ b_{{}_{0}\widehat{m}_{\,1}} b_{{}_{1}\widehat{n}_{\,1}}&= [2]\left(b_{{}_{0}\widehat{m+n-1}_{\,1}} + b_{{}_{0}\widehat{m+n-3}_{\,1}}+\ldots + b_{{}_{0}\widehat{\vert m-n\vert +3}_{\,1}} + b_{{}_{0}\widehat{\vert m-n\vert +1}_{\,1}} \right);\nonumber\\ b_{{}_{1}\widehat{m}_{\,1}} b_{{}_{1}\widehat{n}_{\,0}}&= [2]\left(b_{{}_{1}\widehat{m+n-1}_{\,0}} + b_{{}_{1}\widehat{m+n-3}_{\,0}}+\ldots + b_{{}_{1}\widehat{\vert m-n\vert +3}_{\,0}} + b_{{}_{1}\widehat{\vert m-n\vert +1}_{\,0}} \right);\nonumber\\ b_{{}_{1}\widehat{m}_{\,0}} b_{{}_{0}\widehat{n}_{\,1}}&= [2]\left(b_{{}_{1}\widehat{m+n-1}_{\,1}} + b_{{}_{1}\widehat{m+n-3}_{\,1}}+\ldots + b_{{}_{1}\widehat{\vert m-n\vert +3}_{\,1}} + b_{{}_{1}\widehat{\vert m-n\vert+1}_{\,1}} \right);\nonumber\\ b_{{}_{1}\widehat{m}_{\,1}} b_{{}_{0}\widehat{n}_{\,1}}&= b_{{}_{1}\widehat{m+n}_{\,1}} + 2b_{{}_{1}\widehat{m+n-2}_{\,1}}+\ldots + 2b_{{}_{1}\widehat{\vert m-n\vert +2}_{\,1}}+ b_{{}_{1}\widehat{\vert m-n\vert}_{\,1}};\nonumber\\ b_{{}_{1}\widehat{m}_{\,0}} b_{{}_{1}\widehat{n}_{\,1}}&= b_{{}_{1}\widehat{m+n}_{\,1}} + 2b_{{}_{1}\widehat{m+n-2}_{\,1}}+\ldots + 2b_{{}_{1}\widehat{\vert m-n\vert +2}_{\,1}}+ b_{{}_{1}\widehat{\vert m-n\vert}_{\,1}};\nonumber\\ b_{{}_{0}\widehat{m}_{\,1}} b_{{}_{0}\widehat{n}_{\,1}}&= \begin{cases} b_{{}_{0}\widehat{2m}_{\,1}} + 2b_{{}_{0}\widehat{2m-2}_{\,1}}+\ldots + 2b_{{}_{0}\widehat{2}_{\,1}},&\text{if}\; m=n \nonumber\\ b_{{}_{0}\widehat{m+n}_{\,1}} + 2b_{{}_{0}\widehat{m+n-2}_{\,1}}+\ldots + 2b_{{}_{0}\widehat{\vert m-n\vert +2}_{\,1}}+ b_{{}_{0}\widehat{\vert m-n\vert}_{\,1}},&\text{if}\; m\ne n \end{cases} ;\nonumber\\ b_{{}_{1}\widehat{m}_{\,1}} b_{{}_{0}\widehat{n}_{\,0}}&= \begin{cases} b_{{}_{1}\widehat{2m}_{\,0}} + 2b_{{}_{1}\widehat{2m-2}_{\,0}}+\ldots + 2b_{{}_{1}\widehat{2}_{\,0}},&\text{if}\; m=n \nonumber\\ b_{{}_{1}\widehat{m+n}_{\,0}} + 2b_{{}_{1}\widehat{m+n-2}_{\,0}}+\ldots + 2b_{{}_{1}\widehat{\vert m-n\vert +2}_{\,0}}+ b_{{}_{1}\widehat{\vert m-n\vert}_{\,0}},&\text{if}\; m\ne n \end{cases}; \end{align} $$

and the analogous equations with $0$ and $1$ swapped. These formulas are certainly known to experts, but we couldn’t find an explicit reference for all of them in the literature (the first formula can be found in [Reference LusztigLus, Proposition 7.7(a)]). Let us, therefore, briefly explain how we obtained them. There is a close relation between $\widehat {H}^{\mathrm {ext}}_{2}$ and a two-coloured version of the Grothendieck ring of finite-dimensional representations (of type I) of quantum $\mathfrak {sl}_2$ , see [Reference EliasEli1, Section 2.2.1]. In the first four cases above, when the last element of the left alternating binary sequence and the first element of the right alternating binary sequence are equal, this correspondence is given by the bijection

$$\begin{align*}b_{{}_{i}\widehat{m}_{\,j}}\longleftrightarrow [2][V_{m-1}], \end{align*}$$

where $[V_{m-1}]$ is the Grothendieck class of the (essentially unique) m-dimensional irreducible representation of quantum $\mathfrak {sl}_2$ . Note that the Grothendieck ring over $\mathbb {Z}$ must be tensored with $\mathbb {Z}[q,q^{-1}]$ for multiplication by $[2]$ to make sense. In the first case, for example, we get

$$ \begin{align*} b_{{}{1}\widehat{m}{\,1}} b_{{}{1}\widehat{n}{\,1}} \longleftrightarrow \qquad\qquad\qquad\qquad\qquad\quad \\ [2]^2 [V_{m-1}][V_{n-1}]= \qquad\qquad\qquad\qquad\qquad\\ [2]^2 \left([V_{m+n-2}] + [V_{m+n-4}] + \ldots + [V_{\vert m-n\vert +2}]+ [V_{\vert m-n\vert}]\right) \longleftrightarrow \\ [2]\left(b_{{}{1}\widehat{m+n-1}{\,1}} + b_{{}{1}\widehat{m+n-3}{\,1}}+\ldots + b_{{}{1}\widehat{\vert m-n\vert +3}{\,1}} + b_{{}{1}\widehat{\vert m-n\vert +1}{\,1}} \right), \end{align*} $$

where we have used the well-known Clebsch-Gordan rule for the decomposition of the tensor product of two finite-dimensional irreducible representations of quantum $\mathfrak {sl}_2$ . It is well-known that $b_i b_{{}_{i}\widehat {m}_{\,j}}=b_{{}_{i}\widehat {m}_{\,j}} b_j = [2] b_{{}_{i}\widehat {m}_{\,j}}$ , for $m\in \mathbb {Z}_{\geq 1}$ and $i,j\in \{0,1\}$ satisfying the parity condition. The remaining cases above can be reduced to the first four cases (sometimes with $0$ and $1$ swapped) by that basic multiplication rule and a simple trick, assuming (just for the sake of this trick) that we work over $\mathbb {Q}(q)$ for example. Suffice it to give an illustrative example:

$$\begin{align*}b_{{}_{1}\widehat{m}_{\,1}} b_{{}_{0}\widehat{n}_{\,1}}=\frac{1}{[2]} (b_{{}_{1}\widehat{m}_{\,1}} b_1) b_{{}_{0}\widehat{n}_{\,1}}= \frac{1}{[2]} b_{{}_{1}\widehat{m}_{\,1}} (b_1 b_{{}_{0}\widehat{n}_{\,1}})= \frac{1}{[2]} b_{{}_{1}\widehat{m}_{\,1}} (b_{{}_{1}\widehat{n+1}_{\,1}}+ b_{{}_{1}\widehat{n-1}_{\,1}}), \end{align*}$$

where we use the equality $b_1 b_{{}_{0}\widehat {n}_{\,1}}=b_{{}_{1}\widehat {n+1}_{\,1}}+ b_{{}_{1}\widehat {n-1}_{\,1}}$ , which is the special case of the fifth equation in (123) for $m=1$ .

Recall the standard trace $\epsilon \colon \widehat {H}^{\mathrm {ext}}_{2}\to \mathbb {Z}[q,q^{-1}]$ from (16) and the associated q-sesquilinear form $(-,-)$ on $\widehat {H}^{\mathrm {ext}}_{2}$ from (17). The formula in (121) implies that

$$\begin{align*}\epsilon(b_w)=q^{\ell(w)} \end{align*}$$

for any $w\in \widehat {\mathfrak {S}}_n$ . Note further that $\omega (b_{{}_{i}\widehat {m}_{\,j}})=b_{{}_{j}\widehat {m}_{\,i}}$ , for $m\in \mathbb {Z}_{\geq 0}$ and $i,j\in \{0,1\}$ . The following lemma, which is an immediate consequence of the above, is a more detailed version of Theorem 2.2 in this particular case and will be needed in Section 5.3.

Lemma 5.1. For any $r,s\in \mathbb {Z}$ and $i,j,k,l\in \{0,1\}$ , we have $(\rho ^r b_{{}_{i}\widehat {m}_{\,j}}, \rho ^s b_{{}_{k}\widehat {n}_{\,l}})=\delta _{r,s} (b_{{}_{i}\widehat {m}_{\,j}}, b_{{}_{k}\widehat {n}_{\,l}})$ and

$$\begin{align*}(b_{{}_{i}\widehat{m}_{\,j}}, b_{{}_{k}\widehat{n}_{\,l}})= \begin{cases} q^{2}p(q), & \text{if}\;\, m=n\; \wedge\; i\ne k\; \wedge\; j\ne l;\\ q^{\vert m-n\vert}p'(q) & \text{else}, \end{cases} \end{align*}$$

where $p(q), p'(q)\in \mathbb {Z}_{\geq 0}[q]$ have constant terms equal to two and to one, respectively.

5.1.2 Two representations of $\widehat {H}^{\mathrm {ext}}_{2}$

Let $V=\mathrm {Span}\{v\}$ be the trivial one-dimensional $\widehat {H}^{\mathrm {ext}}_{1}$ -module, defined by

$$\begin{align*}\rho(v)=v, \end{align*}$$

and let

$$\begin{align*}W:=V\odot V, \end{align*}$$

the Zelevinsky tensor product of V with itself. Recall from Section 1 that

$$\begin{align*}V\odot V=\mathrm{Ind}_{\widehat{H}^{\mathrm{ext}}_{1}\otimes \widehat{H}^{\mathrm{ext}}_{1}}^{\widehat{H}^{\mathrm{ext}}_{2}}(V\otimes V)= \widehat{H}^{\mathrm{ext}}_{2} \otimes_{\widehat{H}^{\mathrm{ext}}_{1}\otimes \widehat{H}^{\mathrm{ext}}_{1}}(V\otimes V). \end{align*}$$

Recall also that

$$\begin{align*}\psi_{1,1}(\rho_L\otimes 1)=\rho T_1,\quad \psi_{1,1}(1\otimes \rho_R)=T_1^{-1}\rho, \quad \psi_{1,1}(\rho_L\otimes \rho_R)=\rho^2, \end{align*}$$

where $\rho \in \widehat {H}^{\mathrm {ext}}_{2}$ . Thus, for any $r,s\in \mathbb {Z}$ , we have

$$\begin{align*}\psi_{1,1}(\rho_L^r\otimes \rho_R^s)=(\rho T_1)^r(T_1^{-1}\rho)^s. \end{align*}$$

This implies that W is two-dimensional and that $\{w,w'\}$ , where

(124) $$ \begin{align} w:=1\otimes v \otimes v\quad\text{and}\quad w':=\rho \otimes v \otimes v, \end{align} $$

is a basis of W, on which the action of $\widehat {H}^{\mathrm {ext}}_{2}$ is given by the matrices

(125) $$ \begin{align} [\rho]= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad [T_1]= \begin{pmatrix} 0 & 1 \\ 1 & q^{-1}-q \end{pmatrix}, \quad [T_0]=[\rho T_0 \rho^{-1}]= \begin{pmatrix} q^{-1}-q & 1 \\ 1 & 0 \end{pmatrix}. \end{align} $$

On the KL-basis, the action thus becomes

(126) $$ \begin{align} [b_1]=[T_1+q]= \begin{pmatrix} q & 1 \\ 1 & q^{-1} \end{pmatrix}, \quad [b_0]=[T_0+q]= \begin{pmatrix} q^{-1} & 1 \\ 1 & q \end{pmatrix}. \end{align} $$

Using the rules for multiplication w.r.t. the KL-basis, we see that the action matrices satisfy:

$$\begin{align*}[b_{\widehat{2r}_{\,0}}]=r[b_1][b_0],\; [b_{\widehat{2r+1}_{\,1}}]=(2r+1)[b_1],\; [b_{\widehat{2r}_{\,1}}]=r[b_0][b_1],\; [b_{\widehat{2r+1}_{\,0}}]=(2r+1)[b_0], \; r\in \mathbb{Z}_{\geq 1}. \end{align*}$$

Recall that q is a formal invertible parameter and let ${-}^{\mathbb {C}(q)}:={-}\otimes _{\mathbb {Z}[q,q^{-1}]}\mathbb {C}(q)$ . The following result is well-known, see e.g. [Reference Leclerc, Nazarov and ThibonLNT, Theorem 1].

There is also an infinite-dimensional $\widehat {H}^{\mathrm {ext}}_{2}$ -module U which is closely related to W and whose categorification will be important below. By definition,

$$\begin{align*}U:=\widehat{H}^{\mathrm{ext}}_{2}/I, \end{align*}$$

the quotient of the regular left $\widehat {H}^{\mathrm {ext}}_{2}$ -module by the left ideal I generated by

(127) $$ \begin{align} \rho^2- 1\quad \text{and}\quad b_1\rho - q^{-1} b_1. \end{align} $$

Remark 5.3. We stress that I is only a left ideal and not a two-sided ideal. For $\rho ^2 -1$ this distinction is irrelevant, because $\rho ^2$ and $1$ both belong to the centre of $\widehat {H}^{\mathrm {ext}}_{2}$ , but for $b_1\rho - q^{-1}b_1$ it matters. For example, the element $(1-q^{-2})b_1$ belongs to the two-sided ideal generated by the elements in (127) because

$$\begin{align*}(1-q^{-2})b_1 = (b_1\rho -q^{-1} b_1)(\rho+ q^{-1}) - b_1(\rho^2 - 1), \end{align*}$$

but it does not belong to I.

Proposition 5.4. Let $\pi _U\colon \widehat {H}^{\mathrm {ext}}_{2}\to U$ be the natural projection and define $u_k:=\pi _U(b_{\widehat {k}_{\,1}})$ and $u^{\prime }_k:=\pi _U(\rho b_{\widehat {k}_{\,1}})$ , for all $k\in \mathbb {Z}_{\geq 0}$ . Then $\{u_k, u^{\prime }_k\mid k\in \mathbb {Z}_{\geq 0}\}$ is a basis of U and the action of $\widehat {H}^{\mathrm {ext}}_{2}$ on this basis is determined by

(128) $$ \begin{align} \begin{gathered} \rho u_{k-1} = u^{\prime}_{k-1}, \quad \rho u^{\prime}_{k-1} = u_{k-1} \\ b_1 u_0= u_1,\quad b_1 u_{2k} = u_{2k+1} + u_{2k-1}, \quad b_1 u_{2k-1} = [2] u_{2k-1}, \\ b_0 u_0 = q^{-1} u^{\prime}_1,\quad b_0 u_1 = u_2, \quad b_0 u_{2k} = [2] u_{2k},\quad b_0 u_{2k+1} = u_{2k+2} + u_{2k}, \\ b_1 u^{\prime}_0 = q^{-1} u_1, \quad b_1 u^{\prime}_1 = u^{\prime}_2,\quad b_1 u^{\prime}_{2k} = [2] u^{\prime}_{2k}, \quad b_1 u^{\prime}_{2k+1} = u^{\prime}_{2k+2} + u^{\prime}_{2k},\\ b_0 u^{\prime}_0 = u^{\prime}_1, \quad b_0 u^{\prime}_{2k} = u^{\prime}_{2k+1} + u^{\prime}_{2k-1},\quad b_0 u^{\prime}_{2k-1} = [2] u^{\prime}_{2k-1}, \end{gathered} \end{align} $$

for $k\in \mathbb {Z}_{\geq 1}$ .

Proof. To show that $\{u_k, u^{\prime }_k\mid k\in \mathbb {Z}_{\geq 0}\}$ spans U is easy. The KL-basis of $\widehat {H}^{\mathrm {ext}}_{2}/\langle \rho ^2 -1 \rangle $ , where $\langle \rho ^2 - 1\rangle $ is the (left) ideal generated by $\rho ^2-1$ , is given by $\{\rho ^m b_{\widehat {k}_{\,1}}, \rho ^m b_{\widehat {k}_{\,0}}\mid m\in \{0,1\}, k\in \mathbb {Z}_{\geq 0}\}$ . The images of those elements under $\pi _U$ , therefore, span U. By induction, it’s easy to show that $b_{\widehat {k}_{\,1}}\rho - q^{-1} b_{\widehat {k}_{\,1}}\in I$ for all $k\in \mathbb {Z}_{>0}$ . For $k=2$ , we have

$$\begin{align*}b_{01}\rho -q^{-1} b_{01}=b_0(b_1\rho - q^{-1} b_1)\in I \end{align*}$$

For $k>2$ , induction yields

$$\begin{align*}b_{\widehat{k}_{\,1}}\rho - q^{-1} b_{\widehat{k}_{\,1}} = \begin{cases} b_0 (b_{\widehat{k-1}_{\,1}}\rho - q^{-1} b_{\widehat{k-1}_{\,1}}) - (b_{\widehat{k-2}_{\,1}}\rho - q^{-1} b_{\widehat{k-2}_{\,1}})\in I, & \text{if } k \text{ is even};\\ b_1 (b_{\widehat{k-1}_{\,1}}\rho - q^{-1} b_{\widehat{k-1}_{\,1}}) - (b_{\widehat{k-2}_{\,1}}\rho - q^{-1} b_{\widehat{k-2}_{\,1}})\in I, & \text{if } k \text{ is odd}. \end{cases} \end{align*}$$

This shows that $\rho b_{\widehat {k}_{\,0}}= b_{\widehat {k}_{\,1}}\rho =q^{-1} b_{\widehat {k}_{\,1}}$ in U, for all $k\in \mathbb {Z}_{\geq 0}$ . This in turn implies that $b_{\widehat {k}_{\,0}}=\rho ^2 b_{\widehat {k}_{\,0}}=q^{-1} \rho b_{\widehat {k}_{\,1}}$ , for all $k\in \mathbb {Z}_{\geq 0}$ , which completes the proof that $\{u_k, u^{\prime }_k\mid k\in \mathbb {Z}_{\geq 0}\}$ spans U.

It remains to prove that the elements $u_k, u^{\prime }_k$ , for $k\in \mathbb {Z}_{\geq 0}$ , are all linearly independent in U. Suppose that

$$\begin{align*}\lambda_{m_1} u_{m_1} + \ldots + \lambda_{m_r} u_{m_r} + \mu_{n_1} u^{\prime}_{n_1} + \ldots + \mu_{n_s} u^{\prime}_{n_s}=0\;\, \text{in } U, \end{align*}$$

for some $r,s\in \mathbb {Z}_{\geq 0}$ and some $\mu _{m_1},\ldots , \mu _{m_r},\nu _{n_1}\ldots , \nu _{n_s}\in \mathbb {k}$ . Then

(129) $$ \begin{align} \lambda_{m_1} b_{\widehat{m_1}_{\,1}} + \ldots + \lambda_{m_r} b_{\widehat{m_r}_{\,1}} + \mu_{n_1} \rho b_{\widehat{n_1}_{\,1}} + \ldots + \mu_{n_s} \rho b_{\widehat{n_s}_{\,1}}\in I. \end{align} $$

Since $b_1\rho = \rho b_0$ , any non-zero element in I will be a linear combination of KL basis elements containing at least a non-zero multiple of $\rho b_{\widehat {k}_{\,0}}$ , for some $k\in \mathbb {Z}_{> 0}$ , or a non-zero multiple of $\rho ^2 b_{\widehat {k}_{\,i}}$ , for some $k\in \mathbb {Z}_{\geq 0}$ and $i\in \{0,1\}$ . Either way, linear independence of the KL basis elements in $\widehat {H}^{\mathrm {ext}}_{2}$ implies that (129) can only hold if $\lambda _{m_1} b_{\widehat {m_1}_{\,1}} + \ldots + \lambda _{m_r} b_{\widehat {m_r}_{\,1}} + \mu _{n_1} \rho b_{\widehat {n_1}_{\,1}} + \ldots + \mu _{n_s} \rho b_{\widehat {n_s}_{\,1}}=0$ and, therefore, that $\lambda _{m_i}=\mu _{n_j}=0$ for all $i=1,\ldots , r$ and $j=1,\ldots , s$ . This completes the proof that the elements $u_k, u^{\prime }_k$ , for $k\in \mathbb {Z}_{\geq 0}$ , are all linearly independent in U.

Finally, it is easy to determine the action of $\rho , b_0$ and $b_1$ on this basis using the KL multiplication rules, because $\pi _U$ is a morphism of $\widehat {H}^{\mathrm {ext}}_{2}$ -modules.

Let $J\subset U$ be the $\widehat {H}^{\mathrm {ext}}_{2}$ -submodule generated by $T_1u_0-\rho u_0$ and let $\overline {u}\in U/J$ be the image of $u\in U$ under the natural projection $U\to U/J$ . We have

$$ \begin{align*} & \overline{T_1u_0}=\overline{\rho u_0}=\overline{u^{\prime}_0}, \\ & \overline{T_1 u^{\prime}_0}= \overline{T_1^2 u_0}=\overline{(q^{-1}-q)T_1u_0 + u_0}=\overline{(q^{-1}-q)u^{\prime}_0+ u_0}. \end{align*} $$

Proposition 5.5. There is a surjective morphism of $\widehat {H}^{\mathrm {ext}}_{2}$ -modules

$$\begin{align*}\pi^U_W\colon U\to W \end{align*}$$

and $\ker (\pi ^U_W)=J$ .

Proof. Define $\pi ^U_W\colon U\to W$ by

$$\begin{align*}\pi^U_W(u_0):=w, \end{align*}$$

extending it to the whole of U by the requirement that it be a morphism of $\widehat {H}^{\mathrm {ext}}_{2}$ -modules. Note that the matrices in (126) imply that

$$ \begin{align*} & \pi^U_W(u^{\prime}_0)=\pi^U_W(\rho u_0)=\rho \pi^U_W(u_0)=\rho w=w'; \\ & \pi^U_W(b_1\rho u_0)=b_1\pi^U_W(u^{\prime}_0)=b_1 w'=w+q^{-1}w'=q^{-1}b_1w=q^{-1}\pi^U_W(b_1 u_0), \end{align*} $$

so $\pi ^U_W$ is well-defined.

The fact that $\pi ^U_W(u_0)=w$ and $\pi ^U_W(u^{\prime }_0)=w'$ proves that $\pi ^U_W$ is surjective.

It remains to show that $\ker (\pi ^U_W)=J$ . The matrices in (126) show that $\rho T_1 w= w$ , which implies that

$$\begin{align*}\pi^U_W(\rho T_1 u_0)=\rho T_1 \pi^U_W(u_0)=\rho T_1 w= w= \pi^U_W(u_0), \end{align*}$$

so $J\subseteq \ker (\pi ^U_W)$ . To prove that this inclusion is an equality we will show that $\dim _{\mathbb {k}}(U/J)=2$ , which amounts to showing that $\overline {u_k}, \overline {u^{\prime }_k}\in \mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ for all $k\in \mathbb {Z}_{\geq 0}$ . We will prove this by induction. Recall that $b_i=T_i+q$ , for $i\in \{0,1\}$ . Then, for the base step, we have

$$ \begin{align*} & \overline{u_1} = \overline{b_1 u_0}=\overline{T_1 u_0 + q u_0}=\overline{\rho u_0 + q u_0}= \overline{u^{\prime}_0+qu_0}, \\ & \overline{u^{\prime}_1}= \overline{\rho u_1}=\overline{\rho u^{\prime}_0+q \rho u_0}=\overline{u_0+qu^{\prime}_0},\\ & \overline{u_2}=\overline{b_0 u_1} = \overline{b_0(u^{\prime}_0+qu_0)} = 2 \overline{u^{\prime}_1}= 2(\overline{u_0+qu^{\prime}_0}),\\&\overline{u^{\prime}_2}= \overline{\rho u_2}= 2\overline{u_1}= 2(\overline{u^{\prime}_0+qu_0}). \end{align*} $$

For the inductive step, let $k>2$ and assume that that $\overline {u_m}, \overline {u^{\prime }_m}\in \mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ for all $0\leq m< k$ . If $k>2$ is odd, then

$$\begin{align*}\overline{u_k}=\overline{b_1 u_{k-1}-u_{k-2}}. \end{align*}$$

By induction, the elements $\overline {u_{k-1}}, \overline {u_{k-2}}$ belong to $\mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ . Since $\overline {b_1 u_0}=\overline {u_1}=\overline {u^{\prime }_0+qu_0}$ and $\overline {b_1 u^{\prime }_0}=q^{-1} \overline {u_1}=\overline {q^{-1}u^{\prime }_0+ u_0}$ , we conclude that $\overline {u_k}\in \mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ . If $k>2$ is even, then

$$\begin{align*}\overline{u_k}=\overline{b_0 u_{k-1} - u_{k-2}}. \end{align*}$$

By induction again, the elements $\overline {u_{k-1}}, \overline {u_{k-2}}$ belong to $\mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ . Since $\overline {b_0 u_0}=\overline {q^{-1} u^{\prime }_1}=\overline {q^{-1} u_0+u^{\prime }_0}$ and $\overline {b_0 u^{\prime }_0}=\overline {u^{\prime }_1}=\overline {u_0+qu^{\prime }_0}$ , we conclude that $\overline {u_k}\in \mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ . This completes the proof that $u_k$ belongs to $\mathrm {span}\{\overline {u_0}, \overline {u^{\prime }_0}\}$ for all $k\in \mathbb {Z}_{\geq 0}$ . Using similar arguments, one can show that the same also holds for $u^{\prime }_k$ , for all $k\in \mathbb {Z}_{\geq 0}$ , which concludes the proof.