# The Khovanov-Lauda 2-Category and Categorifications of a Level Two Quantum SL(N) Representation

Authors: David Hill; Joshua Sussan
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We construct 2-functors from a 2-category categorifying quantum sl(n) to 2-categories categorifying the irreducible representation of highest weight2ωk.


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## Body

## 1. Introduction

Khovanov and Lauda introduced a 2-category whose Grothendieck group is𝒰q(𝔰𝔩n) [1]. This work generalizes earlier work by Lauda for the 𝒰q(𝔰𝔩2) case [2]. Rouquier has independently produced a 2-category with similar generators and relations [3]. There have been several examples of categorifications of representations of 𝒰q(𝔰𝔩n) arising in various contexts. Khovanov and Lauda conjectured that their 2-category acts on various known categorifications via a 2-functor. For example, in their work they construct such a 2-functor to a category of graded modules over the cohomology of partial flag varieties. This 2-category categorifies the irreducible representation of 𝒰q(𝔰𝔩n) of highest weight nω1 where ω1 is the first fundamental weight.In this paper we construct this action for the categorification constructed by Huerfano and Khovanov in [4]. They categorify the irreducible representation V2ωk of highest weight 2ωk, by a modification of a diagram algebra introduced in [5]. The objects of 2-category ℋ𝒦k,n are categories 𝒞λ which are module categories over the modified Khovanov algebra. We explicitly construct natural transformations between the functors in [4] and show that they satisfy the relations in the Khovanov-Lauda 2-category giving the following theorem.Theorem 1.1.
Over a field of characteristic two, there exists a 2-functorΩk,n:𝒦ℒ→ℋ𝒦k,n.The Huerfano-Khovanov categorification is based on categories used for the categorification of𝒰q(𝔰𝔩2)-tangle invariants. This hints that a categorification of V2ωk may also be obtained on maximal parabolic subcategories of certain blocks of category 𝒪(𝔤𝔩2k). More specifically, we construct a 2-category 𝒫k,n whose objects are full subcategories ℤ𝒫μ(k,k)(𝔤𝔩2k) of graded category ℤ𝒪μ(k,k)(𝔤𝔩2k) whose set of objects are those modules which have projective presentations by projective-injective objects. The 1-morphisms of 𝒫k,n are certain projective functors. We explicitly construct the 2-morphisms as natural transformations between the projective functors by the Soergel functor 𝕍. We then prove the following.Theorem 1.2.
There is a 2-functorΠk,n:𝒦ℒ→𝒫k,n.It should be possible to categorifyVNωk for N≥1 using categories which appear in various knot homologies. For N≥2, the module categories 𝒞λ in the Huerfano-Khovanov construction should be replaced by suitable categories of matrix factorization based on Khovanov-Rozansky link homology. The categories of matrix factorizations must be generalized from those used in [6]. Khovanov and Rozansky suggest that the categories of matrix factorizations should be taken over tensor products of polynomial rings invariant under the symmetric group. These categories were studied in depth by Yonezawa and Wu [7, 8]. In fact, the isomorphisms of functors categorifying the 𝒰q(𝔰𝔩n) relations were defined implicitly in [8]. To check that there is a 2-representation of the Khovanov-Lauda 2-category, these isomorphisms would need to be made more explicit. The category 𝒪 approach should be modified as well. Now the objects of the 2-category should be subcategories of parabolic subcategories corresponding to the composition Nk=k+⋯+k of blocks of 𝒪λ(𝔤𝔩(Nk)), and the stabilizer of the dominant integral weight μ is taken to be 𝕊λ1×⋯×𝕊λn where each λi∈{0,1,…,N}; compare, for example, Section 5 below. Note that a categorification of Vλ for arbitrary dominant integral λ, hence in particular of VNωk, is constructed in [9] using cyclotomic quotients of Khovanov-Lauda-Rouquier algebras.While this paper was in preparation, two very relevant papers appeared. In [10], Brundan and Stroppel also defined the appropriate natural transformations and checked relations between them to establish a version of the first theorem above, but for Rouquier's 2-category from [3] rather than the Khovanov-Lauda 2-category. One of the advantages of their result is that they are able to work over an arbitrary field, while we work over a field of characteristic 2 in constructing the 2-functor to ℋ𝒦k,n. It is not immediately clear to us how to use their sign conventions to get an action of the full Khovanov-Lauda 2-category in characteristic zero, because they seem to lead to inconsistencies between Propositions 4.7, 4.8, 4.10, and 4.16. Additionally, Brundan and Stroppel categorify V2ωk using graded category 𝒪. More precisely, they first categorify the classical limit of V2ωk at q=1 using a certain parabolic category 𝒪, without mentioning gradings. Then they establish an equivalence between this category and the (ungraded) diagrammatic category. Finally, they observe that both categories are Koszul (by [11] and [12], respectively) so, exploiting unicity of Koszul gradings, their categorification at q=1 can be lifted to a categorification of the module V2ωk itself in terms of graded category 𝒪. Our construction on the graded category 𝒪 side is more explicit, relying heavily on the Soergel functor, the Koszul grading that 𝒪 inherits from geometry, and explicit calculations on the cohomology of flag varieties made in [1]. In the other relevant paper, M. Mackaay [13] constructs an action of the Khovanov-Lauda 2-category on a category of foams which is the basis of an 𝔰𝔩3-knot homology.


## 2.  The Quantum Group𝒰q(𝔰𝔩n)

### 2.1. Root Data

Let𝔰𝔩n=𝔰𝔩n(ℂ) denote the Lie algebra of traceless n×n-matrices with standard triangular decomposition 𝔰𝔩n=𝔫-⊕𝔥⊕𝔫+. Let Δ⊂𝔥* be the root system of type An-1 with simple system Π={αi∣i=1,…,n-1}. Let (·,·) denote the symmetric bilinear form on 𝔥* satisfying(2.1)(αi,αj)=aij,
where A=(aij)1≤i,j<n is the Cartan matrix of type An-1: (2.2)aij={2ifj=i,-1if|j-i|=1,0if|i-j|>1.
Let Δ+ be the set of simple roots relative to Π. Let ω1,…,ωn-1∈𝔥* be the elements satisfying (ωi,αj)=δij, and let (2.3)Q=⨁i=1n-1ℤαi,Q+=⨁i=1n-1ℤ≥0αi,P=⨁i=1n-1ℤωi,P+=⨁i=1n-1ℤ≥0ωi
denote the root lattice, positive root lattice, weight lattice, and dominant weight lattice, respectively.SetI={1,…,n-1,-1,…,-n+1}, I+={1,…,n-1}, and I-=-I+. Define α-i=-αi, and extend the definition of aij to all i,j∈I accordingly. Finally, for i∈I, let sgn(i)=i/|i| be the sign of i.The quantum group𝒰q(𝔰𝔩n) is the associative algebra over ℚ(q) with generators Ei,Ki, for i∈I, satisfying the following conditions:  (1)
KiK-i=K-iKi=1, and KiKj=KjKi for i,j∈I,(2)
KiEj=qai,jEjKi, i,j∈I,(3)
EiE-j-E-jEi=δi,j((Ki-K-i)/(q-q-1)), i,j∈I±,(4)
EiEj=EjEi, i,j∈I±, |i-j|>1,(5)
Ei2Ej-(q+q-1)EiEjEi+EjEi2=0, i,j∈I±, |i-j|=1.We fix a comultiplicationΔ:𝒰q(𝔰𝔩n)→𝒰q(𝔰𝔩n)⊗𝒰q(𝔰𝔩n) given as follows for all i∈I+: (2.4)Δ(Ei)=1⊗Ei+Ei⊗Ki,Δ(E-i)=K-i⊗E-i+E-i⊗1,Δ(K±i)=K±i⊗K±i.
Via Δ, a tensor product of 𝒰q(𝔰𝔩n)-modules becomes a 𝒰q(𝔰𝔩n)-module.In this paper we are interested in the irreducible𝒰q(𝔰𝔩n)-modules, V2ωk with highest weight 2ωk. Therefore, we will identify the weight lattice P≅ℤn-1⊂ℤn as follows. Assume that λ=∑iaiωi. For each 1≤i<n, set (2.5)λi=2k-a1-2a2-⋯-(i-1)ai-1+(n-i)ai+(n-i-1)ai+1+⋯+an-1n.LetP(2ωk) denote the set of weights of V2ωk. It is well known that under this identification each λ∈P(2ωk) satisfies λi∈{0,1,2} for all 1≤i≤n and λ1+⋯+λn=2k.


## 2.1. Root Data

Let𝔰𝔩n=𝔰𝔩n(ℂ) denote the Lie algebra of traceless n×n-matrices with standard triangular decomposition 𝔰𝔩n=𝔫-⊕𝔥⊕𝔫+. Let Δ⊂𝔥* be the root system of type An-1 with simple system Π={αi∣i=1,…,n-1}. Let (·,·) denote the symmetric bilinear form on 𝔥* satisfying(2.1)(αi,αj)=aij,
where A=(aij)1≤i,j<n is the Cartan matrix of type An-1: (2.2)aij={2ifj=i,-1if|j-i|=1,0if|i-j|>1.
Let Δ+ be the set of simple roots relative to Π. Let ω1,…,ωn-1∈𝔥* be the elements satisfying (ωi,αj)=δij, and let (2.3)Q=⨁i=1n-1ℤαi,Q+=⨁i=1n-1ℤ≥0αi,P=⨁i=1n-1ℤωi,P+=⨁i=1n-1ℤ≥0ωi
denote the root lattice, positive root lattice, weight lattice, and dominant weight lattice, respectively.SetI={1,…,n-1,-1,…,-n+1}, I+={1,…,n-1}, and I-=-I+. Define α-i=-αi, and extend the definition of aij to all i,j∈I accordingly. Finally, for i∈I, let sgn(i)=i/|i| be the sign of i.The quantum group𝒰q(𝔰𝔩n) is the associative algebra over ℚ(q) with generators Ei,Ki, for i∈I, satisfying the following conditions:  (1)
KiK-i=K-iKi=1, and KiKj=KjKi for i,j∈I,(2)
KiEj=qai,jEjKi, i,j∈I,(3)
EiE-j-E-jEi=δi,j((Ki-K-i)/(q-q-1)), i,j∈I±,(4)
EiEj=EjEi, i,j∈I±, |i-j|>1,(5)
Ei2Ej-(q+q-1)EiEjEi+EjEi2=0, i,j∈I±, |i-j|=1.We fix a comultiplicationΔ:𝒰q(𝔰𝔩n)→𝒰q(𝔰𝔩n)⊗𝒰q(𝔰𝔩n) given as follows for all i∈I+: (2.4)Δ(Ei)=1⊗Ei+Ei⊗Ki,Δ(E-i)=K-i⊗E-i+E-i⊗1,Δ(K±i)=K±i⊗K±i.
Via Δ, a tensor product of 𝒰q(𝔰𝔩n)-modules becomes a 𝒰q(𝔰𝔩n)-module.In this paper we are interested in the irreducible𝒰q(𝔰𝔩n)-modules, V2ωk with highest weight 2ωk. Therefore, we will identify the weight lattice P≅ℤn-1⊂ℤn as follows. Assume that λ=∑iaiωi. For each 1≤i<n, set (2.5)λi=2k-a1-2a2-⋯-(i-1)ai-1+(n-i)ai+(n-i-1)ai+1+⋯+an-1n.LetP(2ωk) denote the set of weights of V2ωk. It is well known that under this identification each λ∈P(2ωk) satisfies λi∈{0,1,2} for all 1≤i≤n and λ1+⋯+λn=2k.


## 3. The Khovanov-Lauda 2-Category

Let𝕜  be a field. The 𝕜-linear 2-category 𝒦ℒ defined here was originally constructed in [1].LetI∞=⋃n≥0In, I∞+=⋃n≥0(I+)n where In and (I+)n denote n-fold Cartesian products. Given that i̲=(i1,i2,…)∈I∞, let(3.1)cont(i̲)=∑i=1n-1ciαi,whereci=#{j∣ij=i}-#{j∣ij=-i}.
Given that ν∈Q, let Seq(ν)={i̲∈I∞∣cont(i̲)=ν} and, for ν∈Q+, define Seq+(ν)={i̲∈I∞+|cont(i̲)=ν}. Finally, define (3.2)Seq=⋃ν∈QSeq(ν).


### 3.1. The Objects

The set of objects for this 2-category is the weight lattice,P.


### 3.2. The 1-Morphisms

For eachλ∈P, let ℐλ∈End𝒦ℒ(λ) be the identity morphism and, for λ,λ′∈P, set ℐλℐλ′=δλ,λ'ℐλ. For each i∈I, we define morphisms ℰiℐλ∈Hom𝒦ℒ(λ,λ+αi). Evidently, we have ℰiℐλ=ℐλ+αiℰiℐλ. For λ,λ′∈P, we have(3.3)Hom𝒦ℒ(λ,λ′)=⨁i̲∈Seqs∈ℤℐλ'ℰi̲ℐλ{s},
where ℰi̲:=ℰi1⋯ℰir if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift. Observe that ℐλ′ℰi̲ℐλ=0 unless cont(i̲)=λ′-λ, and ℐλ+cont(i̲)ℰi̲ℐλ=ℰi̲ℐλ.


### 3.3. The 2-Morphisms

The 2-morphisms are generated by(3.4)Yi;λ∈End𝒦ℒ(ℰiℐλ),Ψi,j;λ∈Hom𝒦ℒ(ℰiℰjℐλ,ℰjℰiℐλ),⋃i;λ∈Hom𝒦ℒ(ℐλ,ℰ-iℰiℐλ),⋂i;λ∈Hom𝒦ℒ(ℰ-iℰiℐλ,ℐλ),
for i,j∈I±. We define 1i;λ∈End𝒦ℒ(ℰiℐλ) to be the identity transformation.Forλ∈P, the degrees of the basic 2-morphisms are given by(3.5)degYi;λ=aii,degΨi,j;λ=-aij,deg⋃i;λ=deg⋂i;λ=1+(αi,λ).Letλ+cont(i̲)=λ+cont(j̲)=λ+cont(k̲)=λ′ and λ′+cont(i̲′)=λ+cont(j̲′)=λ′′. Let Θ1∈Hom𝒦ℒ(ℰi̲ℐλ,ℰj̲ℐλ) and Θ2∈Hom𝒦ℒ(ℰi̲'ℐλ',ℰj̲'ℐλ'). Then denote the horizontal composition of these 2-morphisms by Θ2Θ1 which is an element of Hom𝒦ℒ(ℰi̲'ℐλ'ℰi̲ℐλ,ℰj̲'ℐλ'ℰj̲ℐλ). If Θ3∈Hom𝒦ℒ(ℰj̲ℐλ,ℰk̲ℐλ), denote the vertical composition of Θ3 and Θ1 by Θ3∘Θ1.For convenience of notation, we define the following 2-morphisms. Ifθ∈End(ℰi̲ℐλ), let θ[j]=θ∘⋯∘θ︸j. For each i∈I, define the bubble(3.6)◯•i;λN=⋂i;λ∘(1-i;λ+αiYi;λ)[N]∘⋃i;λ.
Also, define half-bubbles(3.7)⋃i;λ•N=(1-i;λ+αiYi;λ)[N]∘⋃i;λ,⋃i;λ•N=⋂i;λ∘(Y-i;λ+αi1i,λ)[N].We now define the relations satisfied by these basic 2 morphisms. In what follows, we omit the argumentλ when the relation is independent of it.(1)
𝔰𝔩2Relations
(a)
For alli∈I,
(3.8)(⋂-i1i)∘(1i⋃i)=1i=(1i⋂i)∘(⋃-i1i).(b)
For alli∈I+,
(3.9)Yi=(⋂-i1i)∘(1iY-i1i)∘(1i⋃i)=(1i⋂i)∘(1iY-i1i)∘(⋃-i1i).(c)
Suppose thati∈I and (-αi,λ)>r+1, then
(3.10)◯i;λ•r=0.(d)
Leti∈I. If (αi,λ)≤-1,
(3.11)◯i;λ•-(αi,λ)-1=1.(e)
Leti∈I. If (αi,λ)≥1, then
(3.12)1i;λ-αi1-i;λ=-Ψ-i,i;λ∘Ψi,-i;λ+∑f=0(αi,λ)-1∑g=0f⋃-i;λ•[(αi,λ)-f-1]∘◯i;λ•[-(αi,λ)-1+g]∘⋂-i;λ•[f-g].(f)
Leti∈I+. If (αi,λ)≤0, then
(3.13)(1i;λ⋂-i;λ)∘(Ψi,i;λ-αi1-i;λ)∘(1i;λ⋃-i;λ)=-∑f=0-(αi,λ)Yi;λ[-(αi,λ)-f]◯-i;λ•[(αi,λ)-1+f].
If (αi,λ)≥-2, then
(3.14)(⋂i;λ1i;λ-αi)∘(1-i;λ+αiΨi,i;λ-αi)∘(⋃i;λ1i;λ-αi)=∑g=0(αi,λ)+2◯i;λ•[-(αi,λ)-1+g]Yi;λ-αi[(αi,λ)-g].Remark 3.1.
Note that in 1(e) above the exponent of the bubble may be negative, which is not defined. To make sense of this, fori∈I+, define these symbols (referred to as fake bubbles in [1]) inductively by the formula
(3.15)(∑n≥0◯i;λ•(α-i,λ)-1+ntn)(∑n≥0◯-i;λ•(α-i,λ)-1+ntn)=1
and ◯•i;λ-1=1 whenever (αi,λ)=0.(2)
The nil-Hecke Relations
(a)
For eachi∈I+, Ψi,i[2]=0.(b)
Fori∈I+, (Ψi,i1i)∘(1iΨi,i)∘(Ψi,i1i)=(1iΨi,i)∘(Ψi,i1i)∘(1iΨi,i).(c)
Fori∈I+, (1i1i)=(Ψi,i)∘(Yi1i)-(1iYi)∘(Ψi,i)=(Yi1i)∘(Ψi,i)-(Ψi,i)∘(1iYi).(d)
Forj,i∈I-,
(3.16)Ψj,i=(⋂-j1i1j)∘(1j⋂-i1-j1i1j)∘(1j1iΨ-j,-i1i1j)∘(1j1i1-j⋃i1j)∘(1j1i⋃j)=(1i1j⋂i)∘(1i1j1-i⋂j1i)∘(1i1jΨ-j,-i1j1i)∘(1i⋃-j1-i1j1i)∘(⋃-i1j1i).Remark 3.2.
For alli,j∈I±, set Ψi,-j=(1-j1i⋂-j)∘(1-jΨj,i1-j)∘(⋃j1i1-j).(3)
TheR(ν)Relations
(a)
Fori,j∈I±, (Ψ-j,i)∘(Ψi,-j)=1i1-j.(b)
Fori,j∈I+, i≠j,
(3.17)Ψj,i∘Ψi,j={1i1jif|i-j|>1,(i-j)(Yi1j-1iYj)if|i-j|=1.(c)
Fori,j∈I+, i≠j,
(3.18)(1jYi)∘(Ψi,j)=(Ψi,j)∘(Yi1j),(Yj1i)∘(Ψi,j)=(Ψi,j)∘(1iYj).(d)
Fori,j,k∈I+,
(3.19)(Ψj,k1i)∘(1jΨi,k)∘(Ψi,j1k)-(1kΨi,j)∘(Ψi,k1j)∘(1iΨj,k)={0i≠kor|i-j|=0,(i-j)1i1j1ii=kand|i-j|=1.


## 3.1. The Objects

The set of objects for this 2-category is the weight lattice,P.


## 3.2. The 1-Morphisms

For eachλ∈P, let ℐλ∈End𝒦ℒ(λ) be the identity morphism and, for λ,λ′∈P, set ℐλℐλ′=δλ,λ'ℐλ. For each i∈I, we define morphisms ℰiℐλ∈Hom𝒦ℒ(λ,λ+αi). Evidently, we have ℰiℐλ=ℐλ+αiℰiℐλ. For λ,λ′∈P, we have(3.3)Hom𝒦ℒ(λ,λ′)=⨁i̲∈Seqs∈ℤℐλ'ℰi̲ℐλ{s},
where ℰi̲:=ℰi1⋯ℰir if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift. Observe that ℐλ′ℰi̲ℐλ=0 unless cont(i̲)=λ′-λ, and ℐλ+cont(i̲)ℰi̲ℐλ=ℰi̲ℐλ.


## 3.3. The 2-Morphisms

The 2-morphisms are generated by(3.4)Yi;λ∈End𝒦ℒ(ℰiℐλ),Ψi,j;λ∈Hom𝒦ℒ(ℰiℰjℐλ,ℰjℰiℐλ),⋃i;λ∈Hom𝒦ℒ(ℐλ,ℰ-iℰiℐλ),⋂i;λ∈Hom𝒦ℒ(ℰ-iℰiℐλ,ℐλ),
for i,j∈I±. We define 1i;λ∈End𝒦ℒ(ℰiℐλ) to be the identity transformation.Forλ∈P, the degrees of the basic 2-morphisms are given by(3.5)degYi;λ=aii,degΨi,j;λ=-aij,deg⋃i;λ=deg⋂i;λ=1+(αi,λ).Letλ+cont(i̲)=λ+cont(j̲)=λ+cont(k̲)=λ′ and λ′+cont(i̲′)=λ+cont(j̲′)=λ′′. Let Θ1∈Hom𝒦ℒ(ℰi̲ℐλ,ℰj̲ℐλ) and Θ2∈Hom𝒦ℒ(ℰi̲'ℐλ',ℰj̲'ℐλ'). Then denote the horizontal composition of these 2-morphisms by Θ2Θ1 which is an element of Hom𝒦ℒ(ℰi̲'ℐλ'ℰi̲ℐλ,ℰj̲'ℐλ'ℰj̲ℐλ). If Θ3∈Hom𝒦ℒ(ℰj̲ℐλ,ℰk̲ℐλ), denote the vertical composition of Θ3 and Θ1 by Θ3∘Θ1.For convenience of notation, we define the following 2-morphisms. Ifθ∈End(ℰi̲ℐλ), let θ[j]=θ∘⋯∘θ︸j. For each i∈I, define the bubble(3.6)◯•i;λN=⋂i;λ∘(1-i;λ+αiYi;λ)[N]∘⋃i;λ.
Also, define half-bubbles(3.7)⋃i;λ•N=(1-i;λ+αiYi;λ)[N]∘⋃i;λ,⋃i;λ•N=⋂i;λ∘(Y-i;λ+αi1i,λ)[N].We now define the relations satisfied by these basic 2 morphisms. In what follows, we omit the argumentλ when the relation is independent of it.(1)
𝔰𝔩2Relations
(a)
For alli∈I,
(3.8)(⋂-i1i)∘(1i⋃i)=1i=(1i⋂i)∘(⋃-i1i).(b)
For alli∈I+,
(3.9)Yi=(⋂-i1i)∘(1iY-i1i)∘(1i⋃i)=(1i⋂i)∘(1iY-i1i)∘(⋃-i1i).(c)
Suppose thati∈I and (-αi,λ)>r+1, then
(3.10)◯i;λ•r=0.(d)
Leti∈I. If (αi,λ)≤-1,
(3.11)◯i;λ•-(αi,λ)-1=1.(e)
Leti∈I. If (αi,λ)≥1, then
(3.12)1i;λ-αi1-i;λ=-Ψ-i,i;λ∘Ψi,-i;λ+∑f=0(αi,λ)-1∑g=0f⋃-i;λ•[(αi,λ)-f-1]∘◯i;λ•[-(αi,λ)-1+g]∘⋂-i;λ•[f-g].(f)
Leti∈I+. If (αi,λ)≤0, then
(3.13)(1i;λ⋂-i;λ)∘(Ψi,i;λ-αi1-i;λ)∘(1i;λ⋃-i;λ)=-∑f=0-(αi,λ)Yi;λ[-(αi,λ)-f]◯-i;λ•[(αi,λ)-1+f].
If (αi,λ)≥-2, then
(3.14)(⋂i;λ1i;λ-αi)∘(1-i;λ+αiΨi,i;λ-αi)∘(⋃i;λ1i;λ-αi)=∑g=0(αi,λ)+2◯i;λ•[-(αi,λ)-1+g]Yi;λ-αi[(αi,λ)-g].Remark 3.1.
Note that in 1(e) above the exponent of the bubble may be negative, which is not defined. To make sense of this, fori∈I+, define these symbols (referred to as fake bubbles in [1]) inductively by the formula
(3.15)(∑n≥0◯i;λ•(α-i,λ)-1+ntn)(∑n≥0◯-i;λ•(α-i,λ)-1+ntn)=1
and ◯•i;λ-1=1 whenever (αi,λ)=0.(2)
The nil-Hecke Relations
(a)
For eachi∈I+, Ψi,i[2]=0.(b)
Fori∈I+, (Ψi,i1i)∘(1iΨi,i)∘(Ψi,i1i)=(1iΨi,i)∘(Ψi,i1i)∘(1iΨi,i).(c)
Fori∈I+, (1i1i)=(Ψi,i)∘(Yi1i)-(1iYi)∘(Ψi,i)=(Yi1i)∘(Ψi,i)-(Ψi,i)∘(1iYi).(d)
Forj,i∈I-,
(3.16)Ψj,i=(⋂-j1i1j)∘(1j⋂-i1-j1i1j)∘(1j1iΨ-j,-i1i1j)∘(1j1i1-j⋃i1j)∘(1j1i⋃j)=(1i1j⋂i)∘(1i1j1-i⋂j1i)∘(1i1jΨ-j,-i1j1i)∘(1i⋃-j1-i1j1i)∘(⋃-i1j1i).Remark 3.2.
For alli,j∈I±, set Ψi,-j=(1-j1i⋂-j)∘(1-jΨj,i1-j)∘(⋃j1i1-j).(3)
TheR(ν)Relations
(a)
Fori,j∈I±, (Ψ-j,i)∘(Ψi,-j)=1i1-j.(b)
Fori,j∈I+, i≠j,
(3.17)Ψj,i∘Ψi,j={1i1jif|i-j|>1,(i-j)(Yi1j-1iYj)if|i-j|=1.(c)
Fori,j∈I+, i≠j,
(3.18)(1jYi)∘(Ψi,j)=(Ψi,j)∘(Yi1j),(Yj1i)∘(Ψi,j)=(Ψi,j)∘(1iYj).(d)
Fori,j,k∈I+,
(3.19)(Ψj,k1i)∘(1jΨi,k)∘(Ψi,j1k)-(1kΨi,j)∘(Ψi,k1j)∘(1iΨj,k)={0i≠kor|i-j|=0,(i-j)1i1j1ii=kand|i-j|=1.


## 4. The Huerfano-Khovanov 2-Category

### 4.1. The Khovanov Diagram Algebra

Let𝒜=ℂ[x]/x2. This is a ℤ-graded algebra with multiplication map m:𝒜⊗𝒜→𝒜 such that deg1=-1 and degx=1. There is a comultiplication map Δ:𝒜→𝒜⊗𝒜 such that Δ(1)=x⊗1+1⊗x and Δ(x)=x⊗x. There is a trace map Tr:𝒜→ℂ such that Tr(x)=1 and Tr(1)=0. There is also a unit map ι:ℂ→𝒜 given by ι(1)=1. Also, let κ:𝒜→𝒜 be given by κ(1)=0,κ(x)=1. This algebra gives rise to a two-dimensional TQFT 𝔉, which is a functor from the category of oriented 1+1 cobordisms to the category of abelian groups. The functor 𝔉 sends a disjoint union of m copies of the circle 𝕊1 to 𝒜⊗m. For a cobordism 𝒞1, from two circles to one circle, 𝔉(𝒞1)=m. For a cobordism 𝒞2, from one circle to two circles, 𝔉(𝒞2)=Δ. For a cobordism 𝒞3, from the empty manifold to 𝕊1,  𝔉(𝒞3)=ι. For a cobordism 𝒞4, from the empty manifold to 𝕊1,𝔉(𝒞4)=Tr.For any nonnegative integerr, consider 2r marked points on a line. Let CMr be the set of nonintersecting curves up to isotopy whose boundary is the set of the 2r marked points such that all of the curves lie on one side of the line. Then there are (2rr)/r+1 elements in this set. The set of crossingless matches for r=2 is given in Figure 1.Figure 1
Crossingless matchesa and b for r=2.Figure 2
Concatenation(Ra)b.Leta,b∈CMr. Then (Rb)a is a collection of circles obtained by concatenating a∈CMr with the reflection Rb of b∈CMr in the line. Then applying the two-dimensional TQFT 𝔉, one associates the graded vector space bHar to this collection of circles. Taking direct sums over all crossingless matches gives a graded vector space (4.1)Hr=⨁a,bbHar{r},
where the degree i component of bHar{r} is the degree i-r component of bHar. This graded vector space obtains the structure of an associative algebra via 𝔉; compare, for example, [5].LetT be a tangle from 2r points to 2s points. Let a be a crossingless match for 2s points and b a crossingless match for 2s points. Then let Tab be the concatenation Ra∘T∘b and a𝔉(T)b=𝔉(aTb). See Figure 3 for an example when T is the identity tangle.Figure 3
ConcatenationTab.To any tangle diagramT from 2r points to 2s points, there is an (Hs,Hr)-bimodule (4.2)𝔉(T)=⨁a∈CMrb∈CMs𝔉(Tab){r}.
To any cobordism C between tangles T1 and T2, there is a bimodule map 𝔉(C):𝔉(T1)→𝔉(T2), of degree -χ(C)-r-s, where χ(C) is the Euler characteristic of C; compare, for example, Proposition 5 of [5].Consider the tanglesI and Ui in Figure 4. Then there are saddle cobordisms Si:Ui→I and Si:I→Ui.Figure 4
I and Ui.Lemma 4.1.
LetTi and Ti be the tangles in Figure 5.  (1)
There exists an(Hn-1,Hn)-bimodule homomorphism μi:𝔉(Ti)→𝔉(Ti+1) of degree one.(2)
There exists an(Hn,Hn-1)-bimodule homomorphism μi:𝔉(Ti)→𝔉(Ti+1) of degree one.Figure 5
Ti and Ti.Proof.
There is a degree zero isomorphism of bimodules𝔉(Ti)≅𝔉(Ti)⨂Hn𝔉(I). Then by [5] there is a bimodule map of degree one
(4.3)1⊗𝔉(Si+1):𝔉(Ti)⨂Hn𝔉(I)→𝔉(Ti)⨂Hn𝔉(Ui+1),
where 1 denotes the identity map. Finally note that 𝔉(Ti)⨂Hn𝔉(Ui+1)≅𝔉(Ti+1). Then μi is the composition of these maps.
The construction ofμi is similar.Remark 4.2.
One may construct, in a similar way, maps of degree one:𝔉(Ti)→𝔉(Ti-1) and 𝔉(Ti)→𝔉(Ti-1).Lemma 4.3.
Leta∈CMn and b∈CMn-1 be two crossingless matches. Let Ti be the tangle on the right side of Figure 5. Let Ui be the tangle in Figure 4. Consider the homomorphism induced by the cobordism Si,𝔉(Ti)→𝔉(Ui)⨂Hn𝔉(Ti)≅𝒜⨂ℂ𝔉(Ti). Let α⊗β∈𝔉(Taib), where α∈𝒜 corresponds to the circle passing through the point i on the top line and β∈𝒜⊗p corresponds to the remaining circles. Then α⊗β↦Δ(α)⊗β.Proof.
The map is induced by the cobordismSi. On the set of circles, this cobordism is a union of identity cobordisms and a cobordism 𝒞2. The result now follows upon applying 𝔉.Lemma 4.4.
LetI be the identity tangle from 2r points to 2r points, Ti a tangle from 2(r+1) points to 2r points, and Ti a tangle from 2r points to 2(r+1) points. Let a and b be cup diagrams for 2r points (a,b∈CMr). Consider the map
(4.4)𝒜⨂ℂ𝔉(I)→𝔉(Ti)⨂Hr+1𝔉(Ti)→𝔉(Ti+1)⨂Hr+1𝔉(Ti)→𝔉(I),
where the first and last maps are isomorphisms and the middle map is μi⊗1. Let β∈𝒜 correspond to the circle passing through point i of aIb,γ∈𝒜⊗r correspond to the remaining circles, and α∈𝒜. Then the map above sends α⊗β⊗γ↦(αβ)⊗γ.Proof.
The map is induced by a cobordismSi+1. On the set of circles, this cobordism is union of identity cobordisms and a cobordism 𝒞1. The result now follows upon applying 𝔉.


### 4.2. The Huerfano-Khovanov Categorification

Letλ∈P(2ωk). Recall that α-i=-αi. Hence, for i∈I, we have (4.5)λ+αi=(λ1,…,λi+sgn(i),λi+1-sgn(i),…,λn).
Label n collinear points by the integers λi. Those points labeled by 0 or 2 will never be the boundaries of arcs but will rather just serve as place holders. Then define the algebra Hλ=Hγ(λ) (as in Section 4.1), where γ(λ)=(1/2)∣{λi∣λi=1}|. Let eλ be the identity element.Leti∈I+. We define five special tangles Dλ,i,Dλ,i,Tλ,i,Tλ,i,Iλ in Figures 6, 7, and 8. If a point is labeled by zero or two, it will not be part of the boundary of any curve. Away from points i,i+1, the tangle is the identity.Figure 6
Dλ,i and Dλ,i.Figure 7
Tλ,i and Tλ,i.Figure 8
Identity tangleIλ.The cobordismsSλ,i:Tλ+αi,i∘Tλ,i→Iλ and Sλ,i,j:Tλ+αi,j∘Tλ,i→Dλ+αj,i∘Dλ,j are saddle cobordisms for j=i±1. Similarly, the cobordisms Sλ,i,Sλ,i,j are saddle cobordisms in the opposite direction. For example, the cobordism Sλ,i,i+1 is given in Figure 9.Figure 9
CobordismSλ,i,i+1.Let𝒞λ be the category of finitely generated, graded Hλ-modules, and let 𝕀λ:𝒞λ→𝒞λ be the identity functor. For λ,λ'∈P(2ωk), set 𝕀λ′𝕀λ=δλ,λ′𝕀λ.Leti∈I+. To make future definitions more homogeneous, define Dλ,-i,  Dλ,-i,  Tλ,-i,  Tλ,-i as in Figures 10 and 11. Also, in what follows, interpret the pair (λ-i,λ-i+1) as (λi+1,λi) and recall that α-i=-αi.Figure 10
Dλ,-i and Dλ,-i.Figure 11
Tλ,-i and Tλ,-i.Leti∈I. Let 𝕀λ:𝒞λ→𝒞λ denote the identity functor which is tensoring with the (Hλ,Hλ)-bimodule Hλ. Let 𝔼i𝕀λ:𝒞λ→𝒞λ+αi be the functor of tensoring with a bimodule defined as follows:(4.6)𝔼i𝕀λ={𝔉(Dλ,i)if(λi,λi+1)=(1,2),𝔉(Dλ,i)if(λi,λi+1)=(0,1),𝔉(Tλ,i)if(λi,λi+1)=(1,1),𝔉(Tλ,i)if(λi,λi+1)=(0,2),0otherwise.
Evidently, 𝔼i𝕀λ=𝕀λ+αi𝔼i𝕀λ for all i∈I, and 𝕀λ=𝔉(Iλ).Fori∈I, let 𝕂i𝕀λ:𝒞λ→𝒞λ be the grading shift functor 𝕂i𝕀λ=𝕀λ{(αi,λ)}. Finally, set 𝒞=⨁λ∈P(2ωk)𝒞λ, 𝔼i=⨁λ∈P(2ωk)𝔼i𝕀λ, 𝕂i=⨁λ∈P(2ωk)𝕂i𝕀λ, and 𝕀=⨁λ∈P(2ωk)𝕀λ.Propositions2 and 3 of [4] are that these functors satisfy quantum 𝔰𝔩n relations.Proposition 4.5 (see [4, Propositions 2,3]).
One has(1)
𝕂i𝕂-i𝕀λ≅𝕀λ≅𝕂-i𝕂i𝕀λ, and 𝕂i𝕂j𝕀λ≅𝕂j𝕂i𝕀λ for i,j∈I,(2)
𝕂i𝔼j𝕀λ≅𝔼j𝕂i𝕀λ{aij}, for i,j∈I,(3)
𝔼i𝔼-j𝕀λ≅𝔼-j𝔼i𝕀λ if i,j∈I+, i≠j,(4)
𝔼i𝔼j𝕀λ≅𝔼j𝔼i𝕀λ if i,j∈I±, |i-j|>1,(5)
𝔼i𝔼i𝔼j𝕀λ⊕𝔼j𝔼i𝔼i𝕀λ≅𝔼i𝔼j𝔼i𝕀λ{1}⊕𝔼i𝔼j𝔼i𝕀λ{-1} if i,j∈I±, |i-j|=1,(6)
Fori∈I,(4.7)𝔼i𝔼-i𝕀λ≅{𝔼-i𝔼i𝕀λ⊕𝕀λ{1}⊕𝕀λ{-1}ifi∈I+,(λi,λi+1)=(2,0),𝔼-i𝔼i𝕀λ⊕𝕀λ{1}⊕𝕀λ{-1}ifi∈I-,(λi,λi+1)=(0,2),𝔼-i𝔼i𝕀λ⊕𝕀λif(αi,λ)=1,𝔼-i𝔼i𝕀λif(αi,λ)=0.Now we define the Huerfano-Khovanov 2-categoryℋ𝒦k,n over the field 𝕜, char𝕜=2.


### 4.3. The Objects

The objects ofℋ𝒦k,n are the categories 𝒞λ, λ∈P(V2ωk).


### 4.4. The 1-Morphisms

For eachλ∈P(2ωk), 𝕀λ∈Endℋ𝒦(λ) is the identity morphism and, for λ,λ'∈P, set 𝕀λ𝕀λ′=δλ,λ'𝕀λ as above. For each i∈I, we have defined morphisms 𝔼i𝕀λ∈Homℋ𝒦(𝒞λ,𝒞λ+αi). Evidently, we have 𝔼i𝕀λ=𝕀λ+αi𝔼i𝕀λ. For λ,λ'∈P(2ωk), we have(4.8)Homℋ𝒦(𝒞λ,𝒞λ′)=⨁i̲∈Seqs∈ℤ𝕀λ'𝔼i̲𝕀λ{s},
where 𝔼i̲:=𝔼i1⋯𝔼ir𝕀λ if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift. Observe that 𝕀λ'𝔼i̲𝕀λ=0 unless cont(i̲)=λ'-λ, and 𝕀λ+cont(i̲)𝔼i̲𝕀λ=𝔼i̲𝕀λ.


### 4.5. The 2-Morphisms

In this section we define natural transformations of functors. These maps were not explicitly defined in [4]. Note that the notation for these 2-morphisms is similar to the 2-morphisms in Section 3 since we will construct a 2-functor mapping one set of 2-morphisms to the other. Recall the convention (λ-i,λ-i+1)=(λi+1,λi) for i∈I+.(1) The Maps1i,λ, 1λ
Leti∈I, and let 1i,λ:𝔼i𝕀λ→𝔼i𝕀λ and 1λ:𝕀λ→𝕀λ be the identity maps.(2) The Mapsyi;λ
Fori∈I we define maps yi;λ:𝔼i𝕀λ→𝔼i𝕀λ of degree 2. Let T be the tangle diagram for the functor 𝔼i𝕀λ. It depends on the pair (λi,λi+1). Let a and b be crossingless matches such that (Rb)Ta is a disjoint union of circles. Thus 𝔉((Rb)Ta)=(𝒜)⊗p for some natural number p. Define
(4.9)yi;λ((β1⊗⋯⊗βp))=(β1⊗⋯⊗xβi⊗⋯⊗βp),
where  (a)
if(λi,λi+1)=(1,2), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the bottom set of dots for tangle Dλ,i in Figure 6,(b)
if(λi,λi+1)=(0,1), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the top set of dots for tangle Dλ,i in Figure 6,(c)
if(λi,λi+1)=(0,2), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the top set of dots for tangle Tλ,i in Figure 7,(d)
if(λi,λi+1)=(1,1), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the bottom set of dots for tangle Tλ,i in Figure 7.(3) The Map∪i;λ
We define a map∪i;λ:𝕀λ→𝔼-i𝔼i𝕀λ. There are four nontrivial cases for (λi,λi+1) to consider.  (a)
(λi,λi+1)=(1,2). The identity functor is induced from the identity tangle Iλ. The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i) which is equal to 𝔉(Iλ). Thus in this case ∪i;λ is given by the identity map.(b)
(λi,λi+1)=(1,1). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i). Then ∪i;λ is 𝔉(Sλ,i).(c)
(λi,λi+1)=(0,2). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then the bimodule map is given by 1λ⊗ι.(d)
(λi,λi+1)=(0,1). The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i). As in case 1, this tangle is isotopic to the identity so the map between the functors is the identity map.(4) The Map∩i;λ.
We define a map∩i;λ:𝔼-i𝔼i𝕀λ→𝕀λ. There are four non-trivial cases for (λi,λi+1) to consider.  (a)
(λi,λi+1)=(1,2). The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i) which is equal to 𝔉(Iλ). Thus in this case ∩i;λ is given by the identity map.(b)
(λi,λi+1)=(1,1). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i). Then the homomorphism is 𝔉(Sλ,i).(c)
(λi,λi+1)=(0,2). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then the bimodule map is given by 1λ⊗Tr.(d)
(λi,λi+1)=(0,1). The functor 𝔼-i𝔼i is given by tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i). As in case 1, this tangle is isotopic to the identity so the map between the functors is the identity map.(5) The Mapsψi,j;λ
We define a mapψi,j;λ:𝔼i𝔼j𝕀λ→𝔼j𝔼i𝕀λ for i,j∈I±.
There are four cases fori and j to consider and then subcases for λ.
(a)
i=j. In this case, the functors are non-trivial only if λi=0 and λi+1=2. The bimodule for 𝔼i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then ψi,i=1λ⊗κ.(b)
|i-j|>1. In this case, the functors 𝔼i𝔼j and 𝔼j𝔼i are isomorphic via an isomorphism induced from a cobordism isotopic to the identity so set ψi,j to the identity map.(c)
ψi,i+1:𝔼i𝔼i+1→𝔼i+1𝔼i. There are four non-trivial subcases to consider.
(i)
(λi,λi+1,λi+2)=(1,1,2). The bimodule for 𝔼i𝔼i+1 is 𝔉(Dλ+αi+1,i∘Dλ,i+1). The bimodule for 𝔼i+1𝔼i is 𝔉(Tλ+αi,i+1∘Tλ,i). In this case we define the bimodule map to be 𝔉(Sλ,i,i+1).(ii)
(λi,λi+1,λi+2)=(1,1,1). The functor 𝔼i𝔼i+1 is given by tensoring with a bimodule isomorphic to
(4.10)𝔉(Dλ+αi+1,i∘Tλ,i+1)≅𝔉(Dλ+αi+1,i∘Tλ,i+1)⨂Hλ𝔉(Iλ).
The bimodule for 𝔼i+1𝔼i is isomorphic to 𝔉(Dλ+αi,i+1∘Tλ,i). Then define ψi,j to be 1λ⨂Hλ𝔉(Sλ,i) since
(4.11)𝔉(Dλ+αi+1,i∘Tλ,i+1)⨂Hλ𝔉(Tλ+αi,-i∘Tλ,i)≅𝔉(Dλ+αi,i+1∘Tλ,i).(iii)
(λi,λi+1,λi+2)=(0,1,2). The bimodule for 𝔼i𝔼i+1 is isomorphic to
(4.12)𝔉(Tλ+αi+1,i∘Dλ,i+1)≅𝔉(𝕀λ+αi+αi+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi+1,i∘Dλ,i+1).
The bimodule for 𝔼i+1𝔼i is isomorphic to 𝔉(Tλ+αi,i+1∘Dλ,i). Then define ψi,j to be 𝔉(Sλ+αi+αi+1,i)⨂Hλ1λ since
(4.13)𝔉(Tλ+2αi+αi+1,-(i+1)∘Tλ+αi+αi+1,i+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi+1,i∘Dλ,i+1)≅𝔉(Tλ+αi,i+1∘Dλ,i).(iv)
(λi,λi+1,λi+2)=(0,1,1). The bimodule for 𝔼i𝔼i+1 is 𝔉(Tλ+αi+1,i∘Tλ,i+1). The bimodule for 𝔼i+1𝔼i is 𝔉(Dλ+αi,i+1∘Dλ,i). Then set ψi,j=𝔉(Sλ,i+1,i).(d)
ψi+1,i:𝔼i+1𝔼i→𝔼i𝔼i+1. We essentially just have to read the maps in cases (c)(i)–(iv) above backwards.
(i)
(λi,λi+1,λi+2)=(1,1,2). The functors are just as in case (c)(i). Now the map is 𝔉(Sλ,i,i+1).(ii)
(λi,λi+1,λi+2)=(1,1,1). The bimodule for 𝔼i+1𝔼i is isomorphic to
(4.14)𝔉(Dλ+αi,i+1∘Tλ,i)≅𝔉(Dλ+αi,i+1∘Tλ,i)⨂Hλ𝔉(Iλ).
Then define ψi+1,i=1λ⨂Hλ𝔉(Sλ,i+1).(iii)
(λi,λi+1,λi+2)=(0,1,2). The bimodule for 𝔼i+1𝔼i is isomorphic to
(4.15)𝔉(Tλ+αi,i+1∘Dλ,i)≅𝔉(Iλ+αi+αi+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi,i+1∘Dλ,i).
Then define ψi+1,i=𝔉(Sλ+αi+αi+1,i)⨂Hλ1λ.(iv)
(λi,λi+1,λi+2)=(0,1,1). The functors are just as in case (c)(iv). Now the map is 𝔉(Sλ,i+1,i).Proposition 4.6.
For alli,j∈I, and λ∈P(V2ωk), the maps yi;λ,ψi,j,λ,∪i,λ,∩i,λ are bimodule homomorphisms.For convenience of notation, we define the following 2-morphisms. Ifθ∈End(𝔼i̲), let θ[j]=θ∘⋯∘θ︸j. For each i∈I, define the bubble(4.16)◯i;λ•N=∩i;λ∘(1-i;λ+αiyi;λ)[N]∘∪i;λ,
and define fake bubbles inductively by the formula(4.17)(∑n≥0◯i;λ•(α-i,λ)-1+ntn)(∑n≥0◯-i;λ•(α-i,λ)-1+ntn)=1
and ◯i;λ•-1=1 whenever (αi,λ)=0. Also, define half-bubbles(4.18)⋃i;λ•N=(1-i;λ+αiyi;λ)[N]∘∪i;λ,⋂i;λ•N=∩i;λ∘(yi;λ+αi1i,λ)[N].
Finally, for i,j∈I±, define (4.19)ψi,-j=(1-j1i∩-j)∘(1-jψj,i1-j)∘(∪j1i1-j).


### 4.6. The 2-Morphism Relations

In this section we prove certain relations between the 2-morphisms defined in Section4.5. This will allow us to define a 2-functor from the Khovanov-Lauda 2-category to the Huerfano-Khovanov 2-category. Again, we will often omit the argument λ when it is clear from context.


#### 4.6.1.𝔰𝔩2 Relations

Proposition 4.7.
For alli∈I, (∩-i1i)∘(1i∪i)=1i=(1i∩i)∘(∪-i1i).Proof.
The second equality is similar to the first equality. The casei∈I- is similar to the case i∈I+ so we just compute the map (∩i1i)∘(1i∪i) on the bimodule for the functor 𝔼i for i∈I+. There are four cases to consider.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i are Dλ,i and Dλ,i∘Dλ+αi∘Dλ,i and can be found in Figure 12.
The cobordism between the tangles is isotopic to the identity map so in this case the composition is equal to the identity map.
The case(λi,λi+1)=(0,1) is similar to the (1,2) case.
Now let(λi,λi+1)=(0,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 13.
LetB be the bimodule for the functor 𝔼i. Then the bimodule for 𝔼i𝔼-i𝔼i is isomorphic to 𝒜⊗B. The map 𝔼i→𝔼i𝔼-i𝔼i is given by the unit map which sends an element b∈B to 1⊗b. The map 𝔼i𝔼-i𝔼i→𝔼i is obtained from the cobordism joining the circle to the upper cup which induces the multiplication map. This maps 1⊗b to b. Thus the composition is equal to the identity.
Finally consider the case(λi,λi+1)=(1,1). The tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule giving rise to the functor 𝔼i and let 𝒜⊗B be the bimodule giving rise to the functor 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is in the tensor factor corresponding to the circle passing through point i on the bottom row of the left side of Figure 14 and β belongs to the remaining tensor factors.
The cobordism between the two tangle diagrams is a saddle which, on the level of bimodule maps, sendsα⊗β↦Δ(α)⊗β. Then the map from 𝔼i𝔼-i𝔼i to 𝔼i is given by Tr⊗1λ so Δ(α)⊗β↦α⊗β by considering the two cases α=1 or x. Thus the composition is equal to the identity map.Figure 12
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,2).Figure 13
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(0,2).Figure 14
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,1).Proposition 4.8.
One has(4.20)yi=(∩-i1i)∘(1iy-i1i)∘(1i∪i)=(1i∩i)∘(1iy-i1i)∘(∪-i1i).Proof.
We prove only the first equality as the second is similar. There are four cases to consider for which the functor𝔼i is nonzero.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 12.
Note that the bimodules for𝔼i and 𝔼i𝔼-i𝔼i are the same. Denote this bimodule by B. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of Figure 12. Then the first map 1i∪i is given by the identity cobordism and is thus the identity. The second map is multiplication by x on all tensor components corresponding to circles passing through the point i+1 in the second row of the right side of Figure 12. The final map 𝔼i𝔼-i𝔼i→𝔼i is also given by the identity cobordism. Thus the composition maps α⊗β↦α⊗β↦xα⊗β↦α⊗β. On the other hand, yi(α⊗β)=xα⊗β.
The case(λi,λi+1)=(0,1) is similar to the previous case.
Suppose that(λi,λi+1)=(0,2). Then the bimodule for the functor 𝔼i is B=𝔉(Tλ,i) and the tangle diagram for 𝔼i𝔼-i𝔼i is 𝔉(Tλ,i∘Tλ-αi,i∘Tλ,i)≅𝒜⊗B. Let α⊗β∈B, where α is an element of the tensor factor corresponding to the circle passing through the point i in the top row of the tangle Tλ,i and β is an element in the remaining tensor factors. Then the composition of maps sends α⊗β↦1⊗α⊗β↦x⊗α⊗β↦xα⊗β. This is equal to yi(α⊗β).
Suppose that(λi,λi+1)=(1,1). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule for the functor 𝔼-i and let 𝒜⊗B be the bimodule for 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to the circle passing through point i on the bottom row of Figure 14 and β is an element in the remaining tensor factors. First let α=1. Then
(4.21)1⊗β↦x⊗1⊗β+1⊗x⊗β↦x⊗x⊗β↦x⊗β=yi(1⊗β),
where the last map is Tr⊗1. If α=x, then
(4.22)x⊗β↦x⊗x⊗β↦0=yi(x⊗β).Proposition 4.9.
Supposei∈I and (-αi,λ)>r+1, then ◯i;λ•r=0.Proof.
In order thatr≥0, it must be the case that (-αi,λ)≥2. Thus the only possibility is (λi,λi+1)=(0,2) and r=0. Then the bimodule for 𝔼-i𝔼i is 𝒜⊗𝔉(𝕀λ). Thus the map 1→𝔼-i𝔼i is given by the unit map. The map 𝔼-i𝔼i→1 is given by the trace map. Thus the composition of the maps in the proposition sends an element β↦1⊗β↦Tr(1)⊗b=0.Proposition 4.10.
If(αi,λ)≤-1, then ◯i;λ•(-αi,λ)-1=1.Proof.
The only cases to consider are(λi,λi+1)=(0,2),(1,2),(0,1).
Consider the case(0,2). Let B=𝔉(𝕀λ). Then the bimodule corresponding to 𝔼-i𝔼i is 𝒜⊗B. Let β∈B. Then ∪i(β)=1⊗β, yi(1⊗β)=x⊗β, and ∩i(x⊗β)=Tr(x)β=β. Thus in this case, the composition is the identity map.
For the case(1,2),(-αi,λ)-1=0. The cobordism between the tangle diagrams for the identity functor and 𝔼-i𝔼i is isotopic to the identity cobordism. Similarly, the cobordism between the tangle diagrams for the functors 𝔼-i𝔼i and the identity functor is isotopic to the identity cobordism. Thus the bimodule map is equal to the identity.
The case(0,1) is the same as the case (1,2).Proposition 4.11.
Leti∈I. If (αi,λ)≥1, then
(4.23)1i;λ-αi1-i;λ=ψ-i,i;λ∘ψi,-i;λ+∑f=0(αi,λ)-1∑g=0f⋃-i;λ•(αi,λ)-f-1∘◯i;λ•-(αi,λ)-1+g∘⋂-i;λ•f-g.Proof.
There are three cases to consider:(λi,λi+1)=(1,0),(2,1),(2,0).
For the case(1,0), the first term on the right-hand side is zero since that map passes through the functor 𝔼i𝔼i𝔼-i which is zero for this λ. The summation on the right-hand side reduces to
(4.24)⋃-i;λ•0∘◯i;λ•-2∘⋂i;λ•0=∪-i;λ∘∩-i;λ
by definition (4.17) of the fake bubbles. This map is a composition 𝔼i𝔼-i→1→𝔼i𝔼-i. This composition of maps is the identity.
The case(2,1) is similar to the (1,0) case.
For the case(2,0), the first term on the right-hand side is zero as in the previous two cases. The summation on the right-hand side consists of three terms, which simplifies by (4.17) to
(4.25)⋃-i;λ•1∘∩-i;λ+∪-i;λ∘⋃-i;λ•1+∪-i;λ∘◯i;λ•2∘∩-i;λ.
Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼-i is 𝒜⊗B. Then
(4.26)⋃-i;λ•1∘∩-i;λ:𝔼i𝔼-i→𝕀→𝔼i𝔼-i→𝔼i𝔼-i.
Under this composition of maps, 1⊗b maps to zero since the first map is given by a trace map on the first component. The element x⊗b gets mapped to x⊗b as follows:
(4.27)x⊗b↦b↦1⊗b↦x⊗b,
where the first map is the trace map, the second map is the unit map, and the third map is multiplication by x. Similarly,
(4.28)∪-i;λ∘⋃-i;λ•1:𝔼i𝔼-i→𝔼i𝔼-i→𝕀→𝔼i𝔼-i.
Under this composition, 1⊗b↦1⊗b and x⊗b↦0. Finally, the map
(4.29)∪-i;λ∘◯i;λ•2∘∩-i;λ
is zero because the middle term is zero. Thus the right-hand side is the identity as well.Proposition 4.12.
Leti∈I+.  (1)
If(αi,λ)≤0, then
(4.30)(1i∩-i;λ)∘(ψi,i;λ-αi1-i)∘(1i∪-i;λ)=∑f=0-(αi,λ)yi-(αi,λ)-f◯-i;λ•(αi,λ)-1+f.(2)
If(αi,λ)≥-2, then
(4.31)(∩i;λ+αi1i)∘(1iψi,i;λ)∘(∪i;λ+αi1i)=∑g=0(αi,λ)+2◯i,λ•-(αi,λ)-3+gyi(αi,λ)-g+2.Proof.
We prove(1), the proof of (2) being similar. Since the maps on both sides pass through the functor 𝔼i𝔼i𝔼-i, the maps on both sides are zero unless (λi,λi+1)=(1,1). The functors for 𝔼i and 𝔼i𝔼i𝔼-i are given by tangles in Figure 14.
LetB be the bimodule for the functor 𝔼i so 𝒜⊗B is the bimodule for the functor 𝔼i𝔼i𝔼-i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of the left side of Figure 14 and β is an element in the other tensor factors. Consider first α=1. The left-hand side maps an element α⊗β as follows:
(4.32)1⊗β↦x⊗1⊗β+1⊗x⊗β↦1⊗1⊗β↦1⊗β,
where the first map is Δ⊗1, the second map is κ⊗1⊗1, and the third map is m⊗1. If α=x, the left-hand side maps α⊗β as follows:
(4.33)x⊗β↦x⊗x⊗β↦1⊗x⊗β↦x⊗β.
The right-hand side is 1 by convention.


#### 4.6.2. nil-Hecke Relations

Proposition 4.13.
Fori∈I+,ψi,i[2]=0.Proof.
Since𝔼i𝔼i is identically zero unless (λi,λi+1)=(0,2), we need only to consider this case. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝔉(Tλ,i∘Tλ,i)=𝒜⊗B.
Thenψi,i∘ψi,i:𝒜⊗B→𝒜⊗B→𝒜⊗B. This map sends 1⊗b↦0 and x⊗b↦1⊗b↦0.Proposition 4.14.
Leti∈I+. Then, (ψi,i1i)∘(1iψi,i)∘(ψi,i1i)=(1iψi,i)∘(ψi,i1i)∘(1iψi,i).Proof.
Both sides are natural transformations of the functor𝔼i𝔼i𝔼i. However, by definition this composition is zero.Proposition 4.15.
Fori∈I+,(1i1i)=(ψi,i)∘(yi1i)-(1iyi)∘(ψi,i)=(yi1i)∘(ψi,i)-(ψi,i)∘(1iyi).Proof.
The only case to check is(λi,λi+1)=(0,2) since otherwise 𝔼i𝔼i=0. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝒜⊗B. Then
(4.34)(ψi,i)∘(yi1i):𝒜⊗B→𝒜⊗B.
Under this map, 1⊗b↦x⊗b↦1⊗b and x⊗b↦0. For the map (1iyi)∘(ψi,i),1⊗b↦0, and x⊗b↦1⊗b↦x⊗b. This gives the first equality since our field has characteristic two.
For the second equality,(yi1i)∘(ψi,i):1⊗b↦0,(yi1i)∘(ψi,i):x⊗b↦1⊗b↦x⊗b. Similarly, (ψi,i)∘(1iyi):1⊗b↦x⊗b↦1⊗b and (ψi,i)∘(1iyi):x⊗b↦0.Proposition 4.16.
Fori,j∈I-,
(4.35)ψj,i=(∩-j1i1j)∘(1j∩-i1-j1i1j)∘(1j1iψ-j,-i1i1j)∘(1j1i1-j∪i1j)∘(1j1i∪j)=(1i1j∩i)∘(1i1j1-i∩j1i)∘(1i1jψ-j,-i1j1i)∘(1i∪-j1-i1j1i)∘(∪-i1j1i).Proof.
Leti,j∈I-. We prove only the first equality. If |i-j|>1, the proposition is easy because then ψ±i,±j are identity morphisms. Therefore, we take i=j+1, the case i=j-1 being similar. The natural transformation on the right side of the proposition is a composition of natural transformations:
(4.36)𝔼j𝔼j+1→𝔼j𝔼j+1𝔼-j𝔼j→𝔼j𝔼j+1𝔼-j𝔼-j-1𝔼j+1𝔼j→𝔼j𝔼j+1𝔼-j-1𝔼-j𝔼j+1𝔼j→𝔼j𝔼-j𝔼j+1𝔼j→𝔼j+1𝔼j.
There are four nontrivial cases forλ. We prove the case (λj,λj+1,λj+2)=(2,1,1). The proofs of the remaining cases (2,1,0), (1,1,0), and (1,1,1) are similar.
LetB be the bimodule representing the functor 𝔼j𝔼j+1 and B′ the bimodule representing the functor 𝔼j+1𝔼j. Then the morphism is the composition B→B→B→𝒜⊗B→B→B′ induced by the tangle cobordisms in Figure 15. The first and second maps are the identity maps. The third map is comultiplication. The fourth map is the trace map and the last map is ψj,j+1. Computing this composition on elements as in previous propositions easily gives that it is equal to ψj,j+1.Figure 15
Tangles for compositions of natural transformations in the(2,1,1) case.


#### 4.6.3.R(ν) Relations

Proposition 4.17.
Fori,j∈I±,  i≠j,
(4.37)ψ-j,i∘ψi,-j=1i1-j.Proof.
Note that, for|i-j|>1, the left-hand side is easily seen to be the identity so let j=i+1. The case j=i-1 is similar. Thus the left-hand side is
(4.38)ψ-j,i∘ψi,-j:𝔼i𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼i𝔼-i-1.
There are four non-trivial cases for λ.
Case 1 ((λi,λi+1,λi+2)=(1,2,1)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.39)ψ-j,i∘ψi,-j:B→𝒜⊗B→B→B→B→𝒜⊗B→B.
The first map is ι⊗1λ. The second map is multiplication m. The third and fourth maps are the identity. The fifth map is comultiplication Δ. The last map is Tr⊗1. It is easy to check on elements that this is the identity map.Case 2 ((λi,λi+1,λi+2)=(1,2,0)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.40)ψ-j,i∘ψi,-j:B→B→𝒜⊗B→B→𝒜⊗B→B→B.
The first map is the identity. The second map is Δ by Lemma 4.3. The third map is Tr⊗1 where the trace map is applied to the tensor factor arising from the new circle component. The fourth map is ι⊗1. The fifth map is multiplication by Lemma 4.4. The last map is the identity. It is easy to check that this composition is the identity on all elements.Case 3 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case2.Case 4 ((λi,λi+1,λi+2)=(0,2,0)).
This is similar to Case1.Proposition 4.18.
Ifi,j∈I+ and |i-j|>1, then ψj,i∘ψi,j=1i1j.Proof.
The tangle diagrams for the bimodules for𝔼i𝔼j and 𝔼j𝔼i are the same up to isotopy. The maps in the proposition are obtained from cobordisms isotopic to the identity so they are identity maps.Proposition 4.19.
Ifi,j∈I+ and |i-j|=1, then ψj,i∘ψi,j=(yi1j+1iyj).Proof.
Assume thatj=i+1. The case j=i-1 is similar. There are eight cases for λ such that 𝔼i𝔼i+1 is non-zero. In all cases let a and b be cup diagrams. Let B be the bimodule for 𝔼i𝔼i+1 and B' the bimodule for 𝔼i+1𝔼i.
Case 1.
(λi,λi+1,λi+2)=(0,0,1). Since 𝔼i+1𝔼i=0, the map ψi+1,i∘ψi,i+1=0. The bimodule representing the functor 𝔼i𝔼i+1 is isomorphic to 𝔉(Dλ+αi+1,i∘Dλ,i+1). Since the circle passing through point i on the bottom row of Dλ+αi+1,i∘Dλ,i+1 is the same as the circle passing through point i+1 in the middle row, the map on the right side of the proposition is zero as well.Case 2 ((λi,λi+1,λi+2)=(1,0,1)).
This is similar to Case1.Case 3 ((λi,λi+1,λi+2)=(1,0,2)).
This is similar to Case1.Case 4 ((λi,λi+1,λi+2)=(0,0,2)).
This is similar to Case1.Case 5 ((λi,λi+1,λi+2)=(0,1,1)).
In this caseB≅𝔉(Tλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Dλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of a(Tλ+αi+1,i)∘Tλ,i+1)b is the same as the circle passing through point i of the top row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R is a tensor product of 𝒜 corresponding to the remaining circles. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom is different from the circle passing through point i on the top. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1λ)∘(m⊗1λ). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the sameCase 6 ((λi,λi+1,λi+2)=(1,1,1)).
In this case,B≅𝔉(Dλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Tλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is the same as the circle passing through point i on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is different from the circle passing through point i on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1)∘(m⊗1). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the same.Case 7 ((λi,λi+1,λi+2)=(1,1,2)).
This is similar to Case5.Case 8 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case6.Proposition 4.20.
Leti,j∈I+. If i≠j, then  (1)
(1jyi)∘ψi,j=ψi,j∘(yi1j),(2)
(yj1i)∘ψi,j=ψi,j∘(1iyj).Proof.
We prove only the first statement. Assume further thatj=i+1, the case j=i-1 being similar. The case for |j-i|>1 is easy because the bimodules for 𝔼i𝔼j and 𝔼j𝔼i are equal.
There are four non-trivial cases for(λi,λi+1,λi+2). Let a and b be crossingless matches. Let B be the bimodule for 𝔼i𝔼i+1 and let B' be the bimodule for 𝔼i+1𝔼i.
Case 1 ((λi,λi+1,λi+2)=(1,1,2)).
(i)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R denotes a tensor product of 𝒜 corresponding to the remaining circles. Then ψi,i+1 is given by Δ⊗1. Then ψi,i+1yi(1⊗r)=ψi,i+1(x⊗r)=x⊗x⊗r. Then yiψi,i+1(1⊗r)=yi(x⊗1⊗r+1⊗x⊗r)=x⊗x⊗r.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is different from the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. Then it is easy to verify that ψi,i+1yi(1⊗1⊗r)=yiψi,i+1(1⊗1⊗r)=x⊗r.Case 2.
(λi,λi+1,λi+2)=(0,1,1). This is similar to Case 1.Case 3.
(λi,λi+1,λi+2)=(1,1,1).(i)
Suppose that the circle passing through pointi on the bottom row of the tangle is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then ψi,i+1 is given by Δ⊗1. This then follows as in Case 1.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle is different from the circle passing through the point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. This then follows as in Case 1.Case 4 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case3.Proposition 4.21.
Fori,j,k∈I+,
(4.41)(ψj,k1i)∘(1jψi,k)∘(ψi,j1k)+(1kψi,j)∘(ψi,k1j)∘(1iψj,k)={0,i≠kor|i-j|≠1,1i1j1i,i=kand|i-j|=1.Proof.
The proof of the first part consists of verifying the equality in many different cases, each of which is similar to the second part. We only prove the second part in the casej=i+1 as the case j=i-1 is similar. There are four cases for (λi,λi+1,λi+2) for which 𝔼i𝔼i+1𝔼i is non-zero.
Case 1 ((λi,λi+1,λi+2)=(0,1,1)).
In this case,(ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 because it passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to this λ. On the other hand,
(4.42)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.43)B→𝒜⊗B→𝒜⊗B→B,
where the first map is given by comultiplication, the middle map is given by the map 1⊗κ, and the last map is multiplication. This sequence of maps acts on 1⊗α∈B as follows:
(4.44)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α.
Clearly, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(1⊗α)=0. Similarly, x⊗α↦x⊗α.Case 2 ((λi,λi+1,λi+2)=(0,2,2)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=1i1j1i.Case 3.
(λi,λi+1,λi+2)=(0,1,2). In this case, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 since this map passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to λ.
On the other hand,(4.45)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.46)B→𝒜⊗B→𝒜⊗B→B,
where the first and third maps are given by Lemmas 4.3 and 4.4, respectively, and the middle map is given in Section 4.5. This sequence of maps acts on 1⊗α,x⊗α∈B as follows:
(4.47)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α,x⊗α↦x⊗x⊗α↦x⊗1⊗α↦x⊗α.Case 4 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(β⊗α)=β⊗α.The relations of the 2-morphisms proven in this section give the following.Theorem 4.22.
There is a 2-functorΩk,n:𝒦ℒ→ℋ𝒦k,n such that, for all i,j∈I,  (1)
Ωk,n(λ)=𝒞λ,(2)
Ωk,n(ℐλ)=𝕀λ,(3)
Ωk,n(ℰiℐλ)=𝔼i𝕀λ,(4)
Ωk,n(Yi;λ)=yi;λ,(5)
Ωk,n(Ψi,j;λ)=ψi,j;λ,(6)
Ωk,n(⋃i;λ)=∪i;λ,(7)
Ωk,n(⋂i;λ)=∩i;λ,(8)
Ωk,n(1i;λ)=1i;λ.


## 4.1. The Khovanov Diagram Algebra

Let𝒜=ℂ[x]/x2. This is a ℤ-graded algebra with multiplication map m:𝒜⊗𝒜→𝒜 such that deg1=-1 and degx=1. There is a comultiplication map Δ:𝒜→𝒜⊗𝒜 such that Δ(1)=x⊗1+1⊗x and Δ(x)=x⊗x. There is a trace map Tr:𝒜→ℂ such that Tr(x)=1 and Tr(1)=0. There is also a unit map ι:ℂ→𝒜 given by ι(1)=1. Also, let κ:𝒜→𝒜 be given by κ(1)=0,κ(x)=1. This algebra gives rise to a two-dimensional TQFT 𝔉, which is a functor from the category of oriented 1+1 cobordisms to the category of abelian groups. The functor 𝔉 sends a disjoint union of m copies of the circle 𝕊1 to 𝒜⊗m. For a cobordism 𝒞1, from two circles to one circle, 𝔉(𝒞1)=m. For a cobordism 𝒞2, from one circle to two circles, 𝔉(𝒞2)=Δ. For a cobordism 𝒞3, from the empty manifold to 𝕊1,  𝔉(𝒞3)=ι. For a cobordism 𝒞4, from the empty manifold to 𝕊1,𝔉(𝒞4)=Tr.For any nonnegative integerr, consider 2r marked points on a line. Let CMr be the set of nonintersecting curves up to isotopy whose boundary is the set of the 2r marked points such that all of the curves lie on one side of the line. Then there are (2rr)/r+1 elements in this set. The set of crossingless matches for r=2 is given in Figure 1.Figure 1
Crossingless matchesa and b for r=2.Figure 2
Concatenation(Ra)b.Leta,b∈CMr. Then (Rb)a is a collection of circles obtained by concatenating a∈CMr with the reflection Rb of b∈CMr in the line. Then applying the two-dimensional TQFT 𝔉, one associates the graded vector space bHar to this collection of circles. Taking direct sums over all crossingless matches gives a graded vector space (4.1)Hr=⨁a,bbHar{r},
where the degree i component of bHar{r} is the degree i-r component of bHar. This graded vector space obtains the structure of an associative algebra via 𝔉; compare, for example, [5].LetT be a tangle from 2r points to 2s points. Let a be a crossingless match for 2s points and b a crossingless match for 2s points. Then let Tab be the concatenation Ra∘T∘b and a𝔉(T)b=𝔉(aTb). See Figure 3 for an example when T is the identity tangle.Figure 3
ConcatenationTab.To any tangle diagramT from 2r points to 2s points, there is an (Hs,Hr)-bimodule (4.2)𝔉(T)=⨁a∈CMrb∈CMs𝔉(Tab){r}.
To any cobordism C between tangles T1 and T2, there is a bimodule map 𝔉(C):𝔉(T1)→𝔉(T2), of degree -χ(C)-r-s, where χ(C) is the Euler characteristic of C; compare, for example, Proposition 5 of [5].Consider the tanglesI and Ui in Figure 4. Then there are saddle cobordisms Si:Ui→I and Si:I→Ui.Figure 4
I and Ui.Lemma 4.1.
LetTi and Ti be the tangles in Figure 5.  (1)
There exists an(Hn-1,Hn)-bimodule homomorphism μi:𝔉(Ti)→𝔉(Ti+1) of degree one.(2)
There exists an(Hn,Hn-1)-bimodule homomorphism μi:𝔉(Ti)→𝔉(Ti+1) of degree one.Figure 5
Ti and Ti.Proof.
There is a degree zero isomorphism of bimodules𝔉(Ti)≅𝔉(Ti)⨂Hn𝔉(I). Then by [5] there is a bimodule map of degree one
(4.3)1⊗𝔉(Si+1):𝔉(Ti)⨂Hn𝔉(I)→𝔉(Ti)⨂Hn𝔉(Ui+1),
where 1 denotes the identity map. Finally note that 𝔉(Ti)⨂Hn𝔉(Ui+1)≅𝔉(Ti+1). Then μi is the composition of these maps.
The construction ofμi is similar.Remark 4.2.
One may construct, in a similar way, maps of degree one:𝔉(Ti)→𝔉(Ti-1) and 𝔉(Ti)→𝔉(Ti-1).Lemma 4.3.
Leta∈CMn and b∈CMn-1 be two crossingless matches. Let Ti be the tangle on the right side of Figure 5. Let Ui be the tangle in Figure 4. Consider the homomorphism induced by the cobordism Si,𝔉(Ti)→𝔉(Ui)⨂Hn𝔉(Ti)≅𝒜⨂ℂ𝔉(Ti). Let α⊗β∈𝔉(Taib), where α∈𝒜 corresponds to the circle passing through the point i on the top line and β∈𝒜⊗p corresponds to the remaining circles. Then α⊗β↦Δ(α)⊗β.Proof.
The map is induced by the cobordismSi. On the set of circles, this cobordism is a union of identity cobordisms and a cobordism 𝒞2. The result now follows upon applying 𝔉.Lemma 4.4.
LetI be the identity tangle from 2r points to 2r points, Ti a tangle from 2(r+1) points to 2r points, and Ti a tangle from 2r points to 2(r+1) points. Let a and b be cup diagrams for 2r points (a,b∈CMr). Consider the map
(4.4)𝒜⨂ℂ𝔉(I)→𝔉(Ti)⨂Hr+1𝔉(Ti)→𝔉(Ti+1)⨂Hr+1𝔉(Ti)→𝔉(I),
where the first and last maps are isomorphisms and the middle map is μi⊗1. Let β∈𝒜 correspond to the circle passing through point i of aIb,γ∈𝒜⊗r correspond to the remaining circles, and α∈𝒜. Then the map above sends α⊗β⊗γ↦(αβ)⊗γ.Proof.
The map is induced by a cobordismSi+1. On the set of circles, this cobordism is union of identity cobordisms and a cobordism 𝒞1. The result now follows upon applying 𝔉.


## 4.2. The Huerfano-Khovanov Categorification

Letλ∈P(2ωk). Recall that α-i=-αi. Hence, for i∈I, we have (4.5)λ+αi=(λ1,…,λi+sgn(i),λi+1-sgn(i),…,λn).
Label n collinear points by the integers λi. Those points labeled by 0 or 2 will never be the boundaries of arcs but will rather just serve as place holders. Then define the algebra Hλ=Hγ(λ) (as in Section 4.1), where γ(λ)=(1/2)∣{λi∣λi=1}|. Let eλ be the identity element.Leti∈I+. We define five special tangles Dλ,i,Dλ,i,Tλ,i,Tλ,i,Iλ in Figures 6, 7, and 8. If a point is labeled by zero or two, it will not be part of the boundary of any curve. Away from points i,i+1, the tangle is the identity.Figure 6
Dλ,i and Dλ,i.Figure 7
Tλ,i and Tλ,i.Figure 8
Identity tangleIλ.The cobordismsSλ,i:Tλ+αi,i∘Tλ,i→Iλ and Sλ,i,j:Tλ+αi,j∘Tλ,i→Dλ+αj,i∘Dλ,j are saddle cobordisms for j=i±1. Similarly, the cobordisms Sλ,i,Sλ,i,j are saddle cobordisms in the opposite direction. For example, the cobordism Sλ,i,i+1 is given in Figure 9.Figure 9
CobordismSλ,i,i+1.Let𝒞λ be the category of finitely generated, graded Hλ-modules, and let 𝕀λ:𝒞λ→𝒞λ be the identity functor. For λ,λ'∈P(2ωk), set 𝕀λ′𝕀λ=δλ,λ′𝕀λ.Leti∈I+. To make future definitions more homogeneous, define Dλ,-i,  Dλ,-i,  Tλ,-i,  Tλ,-i as in Figures 10 and 11. Also, in what follows, interpret the pair (λ-i,λ-i+1) as (λi+1,λi) and recall that α-i=-αi.Figure 10
Dλ,-i and Dλ,-i.Figure 11
Tλ,-i and Tλ,-i.Leti∈I. Let 𝕀λ:𝒞λ→𝒞λ denote the identity functor which is tensoring with the (Hλ,Hλ)-bimodule Hλ. Let 𝔼i𝕀λ:𝒞λ→𝒞λ+αi be the functor of tensoring with a bimodule defined as follows:(4.6)𝔼i𝕀λ={𝔉(Dλ,i)if(λi,λi+1)=(1,2),𝔉(Dλ,i)if(λi,λi+1)=(0,1),𝔉(Tλ,i)if(λi,λi+1)=(1,1),𝔉(Tλ,i)if(λi,λi+1)=(0,2),0otherwise.
Evidently, 𝔼i𝕀λ=𝕀λ+αi𝔼i𝕀λ for all i∈I, and 𝕀λ=𝔉(Iλ).Fori∈I, let 𝕂i𝕀λ:𝒞λ→𝒞λ be the grading shift functor 𝕂i𝕀λ=𝕀λ{(αi,λ)}. Finally, set 𝒞=⨁λ∈P(2ωk)𝒞λ, 𝔼i=⨁λ∈P(2ωk)𝔼i𝕀λ, 𝕂i=⨁λ∈P(2ωk)𝕂i𝕀λ, and 𝕀=⨁λ∈P(2ωk)𝕀λ.Propositions2 and 3 of [4] are that these functors satisfy quantum 𝔰𝔩n relations.Proposition 4.5 (see [4, Propositions 2,3]).
One has(1)
𝕂i𝕂-i𝕀λ≅𝕀λ≅𝕂-i𝕂i𝕀λ, and 𝕂i𝕂j𝕀λ≅𝕂j𝕂i𝕀λ for i,j∈I,(2)
𝕂i𝔼j𝕀λ≅𝔼j𝕂i𝕀λ{aij}, for i,j∈I,(3)
𝔼i𝔼-j𝕀λ≅𝔼-j𝔼i𝕀λ if i,j∈I+, i≠j,(4)
𝔼i𝔼j𝕀λ≅𝔼j𝔼i𝕀λ if i,j∈I±, |i-j|>1,(5)
𝔼i𝔼i𝔼j𝕀λ⊕𝔼j𝔼i𝔼i𝕀λ≅𝔼i𝔼j𝔼i𝕀λ{1}⊕𝔼i𝔼j𝔼i𝕀λ{-1} if i,j∈I±, |i-j|=1,(6)
Fori∈I,(4.7)𝔼i𝔼-i𝕀λ≅{𝔼-i𝔼i𝕀λ⊕𝕀λ{1}⊕𝕀λ{-1}ifi∈I+,(λi,λi+1)=(2,0),𝔼-i𝔼i𝕀λ⊕𝕀λ{1}⊕𝕀λ{-1}ifi∈I-,(λi,λi+1)=(0,2),𝔼-i𝔼i𝕀λ⊕𝕀λif(αi,λ)=1,𝔼-i𝔼i𝕀λif(αi,λ)=0.Now we define the Huerfano-Khovanov 2-categoryℋ𝒦k,n over the field 𝕜, char𝕜=2.


## 4.3. The Objects

The objects ofℋ𝒦k,n are the categories 𝒞λ, λ∈P(V2ωk).


## 4.4. The 1-Morphisms

For eachλ∈P(2ωk), 𝕀λ∈Endℋ𝒦(λ) is the identity morphism and, for λ,λ'∈P, set 𝕀λ𝕀λ′=δλ,λ'𝕀λ as above. For each i∈I, we have defined morphisms 𝔼i𝕀λ∈Homℋ𝒦(𝒞λ,𝒞λ+αi). Evidently, we have 𝔼i𝕀λ=𝕀λ+αi𝔼i𝕀λ. For λ,λ'∈P(2ωk), we have(4.8)Homℋ𝒦(𝒞λ,𝒞λ′)=⨁i̲∈Seqs∈ℤ𝕀λ'𝔼i̲𝕀λ{s},
where 𝔼i̲:=𝔼i1⋯𝔼ir𝕀λ if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift. Observe that 𝕀λ'𝔼i̲𝕀λ=0 unless cont(i̲)=λ'-λ, and 𝕀λ+cont(i̲)𝔼i̲𝕀λ=𝔼i̲𝕀λ.


## 4.5. The 2-Morphisms

In this section we define natural transformations of functors. These maps were not explicitly defined in [4]. Note that the notation for these 2-morphisms is similar to the 2-morphisms in Section 3 since we will construct a 2-functor mapping one set of 2-morphisms to the other. Recall the convention (λ-i,λ-i+1)=(λi+1,λi) for i∈I+.(1) The Maps1i,λ, 1λ
Leti∈I, and let 1i,λ:𝔼i𝕀λ→𝔼i𝕀λ and 1λ:𝕀λ→𝕀λ be the identity maps.(2) The Mapsyi;λ
Fori∈I we define maps yi;λ:𝔼i𝕀λ→𝔼i𝕀λ of degree 2. Let T be the tangle diagram for the functor 𝔼i𝕀λ. It depends on the pair (λi,λi+1). Let a and b be crossingless matches such that (Rb)Ta is a disjoint union of circles. Thus 𝔉((Rb)Ta)=(𝒜)⊗p for some natural number p. Define
(4.9)yi;λ((β1⊗⋯⊗βp))=(β1⊗⋯⊗xβi⊗⋯⊗βp),
where  (a)
if(λi,λi+1)=(1,2), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the bottom set of dots for tangle Dλ,i in Figure 6,(b)
if(λi,λi+1)=(0,1), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the top set of dots for tangle Dλ,i in Figure 6,(c)
if(λi,λi+1)=(0,2), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the top set of dots for tangle Tλ,i in Figure 7,(d)
if(λi,λi+1)=(1,1), then the ith factor in (𝒜)⊗p corresponds to the circle passing through the ith point on the bottom set of dots for tangle Tλ,i in Figure 7.(3) The Map∪i;λ
We define a map∪i;λ:𝕀λ→𝔼-i𝔼i𝕀λ. There are four nontrivial cases for (λi,λi+1) to consider.  (a)
(λi,λi+1)=(1,2). The identity functor is induced from the identity tangle Iλ. The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i) which is equal to 𝔉(Iλ). Thus in this case ∪i;λ is given by the identity map.(b)
(λi,λi+1)=(1,1). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i). Then ∪i;λ is 𝔉(Sλ,i).(c)
(λi,λi+1)=(0,2). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then the bimodule map is given by 1λ⊗ι.(d)
(λi,λi+1)=(0,1). The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i). As in case 1, this tangle is isotopic to the identity so the map between the functors is the identity map.(4) The Map∩i;λ.
We define a map∩i;λ:𝔼-i𝔼i𝕀λ→𝕀λ. There are four non-trivial cases for (λi,λi+1) to consider.  (a)
(λi,λi+1)=(1,2). The functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i) which is equal to 𝔉(Iλ). Thus in this case ∩i;λ is given by the identity map.(b)
(λi,λi+1)=(1,1). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i). Then the homomorphism is 𝔉(Sλ,i).(c)
(λi,λi+1)=(0,2). Then the functor 𝔼-i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then the bimodule map is given by 1λ⊗Tr.(d)
(λi,λi+1)=(0,1). The functor 𝔼-i𝔼i is given by tensoring with the bimodule 𝔉(Dλ+αi,i∘Dλ,i). As in case 1, this tangle is isotopic to the identity so the map between the functors is the identity map.(5) The Mapsψi,j;λ
We define a mapψi,j;λ:𝔼i𝔼j𝕀λ→𝔼j𝔼i𝕀λ for i,j∈I±.
There are four cases fori and j to consider and then subcases for λ.
(a)
i=j. In this case, the functors are non-trivial only if λi=0 and λi+1=2. The bimodule for 𝔼i𝔼i is isomorphic to tensoring with the bimodule 𝔉(Tλ+αi,i∘Tλ,i)=𝔉(Iλ)⊗𝒜. Then ψi,i=1λ⊗κ.(b)
|i-j|>1. In this case, the functors 𝔼i𝔼j and 𝔼j𝔼i are isomorphic via an isomorphism induced from a cobordism isotopic to the identity so set ψi,j to the identity map.(c)
ψi,i+1:𝔼i𝔼i+1→𝔼i+1𝔼i. There are four non-trivial subcases to consider.
(i)
(λi,λi+1,λi+2)=(1,1,2). The bimodule for 𝔼i𝔼i+1 is 𝔉(Dλ+αi+1,i∘Dλ,i+1). The bimodule for 𝔼i+1𝔼i is 𝔉(Tλ+αi,i+1∘Tλ,i). In this case we define the bimodule map to be 𝔉(Sλ,i,i+1).(ii)
(λi,λi+1,λi+2)=(1,1,1). The functor 𝔼i𝔼i+1 is given by tensoring with a bimodule isomorphic to
(4.10)𝔉(Dλ+αi+1,i∘Tλ,i+1)≅𝔉(Dλ+αi+1,i∘Tλ,i+1)⨂Hλ𝔉(Iλ).
The bimodule for 𝔼i+1𝔼i is isomorphic to 𝔉(Dλ+αi,i+1∘Tλ,i). Then define ψi,j to be 1λ⨂Hλ𝔉(Sλ,i) since
(4.11)𝔉(Dλ+αi+1,i∘Tλ,i+1)⨂Hλ𝔉(Tλ+αi,-i∘Tλ,i)≅𝔉(Dλ+αi,i+1∘Tλ,i).(iii)
(λi,λi+1,λi+2)=(0,1,2). The bimodule for 𝔼i𝔼i+1 is isomorphic to
(4.12)𝔉(Tλ+αi+1,i∘Dλ,i+1)≅𝔉(𝕀λ+αi+αi+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi+1,i∘Dλ,i+1).
The bimodule for 𝔼i+1𝔼i is isomorphic to 𝔉(Tλ+αi,i+1∘Dλ,i). Then define ψi,j to be 𝔉(Sλ+αi+αi+1,i)⨂Hλ1λ since
(4.13)𝔉(Tλ+2αi+αi+1,-(i+1)∘Tλ+αi+αi+1,i+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi+1,i∘Dλ,i+1)≅𝔉(Tλ+αi,i+1∘Dλ,i).(iv)
(λi,λi+1,λi+2)=(0,1,1). The bimodule for 𝔼i𝔼i+1 is 𝔉(Tλ+αi+1,i∘Tλ,i+1). The bimodule for 𝔼i+1𝔼i is 𝔉(Dλ+αi,i+1∘Dλ,i). Then set ψi,j=𝔉(Sλ,i+1,i).(d)
ψi+1,i:𝔼i+1𝔼i→𝔼i𝔼i+1. We essentially just have to read the maps in cases (c)(i)–(iv) above backwards.
(i)
(λi,λi+1,λi+2)=(1,1,2). The functors are just as in case (c)(i). Now the map is 𝔉(Sλ,i,i+1).(ii)
(λi,λi+1,λi+2)=(1,1,1). The bimodule for 𝔼i+1𝔼i is isomorphic to
(4.14)𝔉(Dλ+αi,i+1∘Tλ,i)≅𝔉(Dλ+αi,i+1∘Tλ,i)⨂Hλ𝔉(Iλ).
Then define ψi+1,i=1λ⨂Hλ𝔉(Sλ,i+1).(iii)
(λi,λi+1,λi+2)=(0,1,2). The bimodule for 𝔼i+1𝔼i is isomorphic to
(4.15)𝔉(Tλ+αi,i+1∘Dλ,i)≅𝔉(Iλ+αi+αi+1)⨂Hλ+αi+αi+1𝔉(Tλ+αi,i+1∘Dλ,i).
Then define ψi+1,i=𝔉(Sλ+αi+αi+1,i)⨂Hλ1λ.(iv)
(λi,λi+1,λi+2)=(0,1,1). The functors are just as in case (c)(iv). Now the map is 𝔉(Sλ,i+1,i).Proposition 4.6.
For alli,j∈I, and λ∈P(V2ωk), the maps yi;λ,ψi,j,λ,∪i,λ,∩i,λ are bimodule homomorphisms.For convenience of notation, we define the following 2-morphisms. Ifθ∈End(𝔼i̲), let θ[j]=θ∘⋯∘θ︸j. For each i∈I, define the bubble(4.16)◯i;λ•N=∩i;λ∘(1-i;λ+αiyi;λ)[N]∘∪i;λ,
and define fake bubbles inductively by the formula(4.17)(∑n≥0◯i;λ•(α-i,λ)-1+ntn)(∑n≥0◯-i;λ•(α-i,λ)-1+ntn)=1
and ◯i;λ•-1=1 whenever (αi,λ)=0. Also, define half-bubbles(4.18)⋃i;λ•N=(1-i;λ+αiyi;λ)[N]∘∪i;λ,⋂i;λ•N=∩i;λ∘(yi;λ+αi1i,λ)[N].
Finally, for i,j∈I±, define (4.19)ψi,-j=(1-j1i∩-j)∘(1-jψj,i1-j)∘(∪j1i1-j).


## 4.6. The 2-Morphism Relations

In this section we prove certain relations between the 2-morphisms defined in Section4.5. This will allow us to define a 2-functor from the Khovanov-Lauda 2-category to the Huerfano-Khovanov 2-category. Again, we will often omit the argument λ when it is clear from context.


### 4.6.1.𝔰𝔩2 Relations

Proposition 4.7.
For alli∈I, (∩-i1i)∘(1i∪i)=1i=(1i∩i)∘(∪-i1i).Proof.
The second equality is similar to the first equality. The casei∈I- is similar to the case i∈I+ so we just compute the map (∩i1i)∘(1i∪i) on the bimodule for the functor 𝔼i for i∈I+. There are four cases to consider.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i are Dλ,i and Dλ,i∘Dλ+αi∘Dλ,i and can be found in Figure 12.
The cobordism between the tangles is isotopic to the identity map so in this case the composition is equal to the identity map.
The case(λi,λi+1)=(0,1) is similar to the (1,2) case.
Now let(λi,λi+1)=(0,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 13.
LetB be the bimodule for the functor 𝔼i. Then the bimodule for 𝔼i𝔼-i𝔼i is isomorphic to 𝒜⊗B. The map 𝔼i→𝔼i𝔼-i𝔼i is given by the unit map which sends an element b∈B to 1⊗b. The map 𝔼i𝔼-i𝔼i→𝔼i is obtained from the cobordism joining the circle to the upper cup which induces the multiplication map. This maps 1⊗b to b. Thus the composition is equal to the identity.
Finally consider the case(λi,λi+1)=(1,1). The tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule giving rise to the functor 𝔼i and let 𝒜⊗B be the bimodule giving rise to the functor 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is in the tensor factor corresponding to the circle passing through point i on the bottom row of the left side of Figure 14 and β belongs to the remaining tensor factors.
The cobordism between the two tangle diagrams is a saddle which, on the level of bimodule maps, sendsα⊗β↦Δ(α)⊗β. Then the map from 𝔼i𝔼-i𝔼i to 𝔼i is given by Tr⊗1λ so Δ(α)⊗β↦α⊗β by considering the two cases α=1 or x. Thus the composition is equal to the identity map.Figure 12
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,2).Figure 13
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(0,2).Figure 14
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,1).Proposition 4.8.
One has(4.20)yi=(∩-i1i)∘(1iy-i1i)∘(1i∪i)=(1i∩i)∘(1iy-i1i)∘(∪-i1i).Proof.
We prove only the first equality as the second is similar. There are four cases to consider for which the functor𝔼i is nonzero.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 12.
Note that the bimodules for𝔼i and 𝔼i𝔼-i𝔼i are the same. Denote this bimodule by B. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of Figure 12. Then the first map 1i∪i is given by the identity cobordism and is thus the identity. The second map is multiplication by x on all tensor components corresponding to circles passing through the point i+1 in the second row of the right side of Figure 12. The final map 𝔼i𝔼-i𝔼i→𝔼i is also given by the identity cobordism. Thus the composition maps α⊗β↦α⊗β↦xα⊗β↦α⊗β. On the other hand, yi(α⊗β)=xα⊗β.
The case(λi,λi+1)=(0,1) is similar to the previous case.
Suppose that(λi,λi+1)=(0,2). Then the bimodule for the functor 𝔼i is B=𝔉(Tλ,i) and the tangle diagram for 𝔼i𝔼-i𝔼i is 𝔉(Tλ,i∘Tλ-αi,i∘Tλ,i)≅𝒜⊗B. Let α⊗β∈B, where α is an element of the tensor factor corresponding to the circle passing through the point i in the top row of the tangle Tλ,i and β is an element in the remaining tensor factors. Then the composition of maps sends α⊗β↦1⊗α⊗β↦x⊗α⊗β↦xα⊗β. This is equal to yi(α⊗β).
Suppose that(λi,λi+1)=(1,1). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule for the functor 𝔼-i and let 𝒜⊗B be the bimodule for 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to the circle passing through point i on the bottom row of Figure 14 and β is an element in the remaining tensor factors. First let α=1. Then
(4.21)1⊗β↦x⊗1⊗β+1⊗x⊗β↦x⊗x⊗β↦x⊗β=yi(1⊗β),
where the last map is Tr⊗1. If α=x, then
(4.22)x⊗β↦x⊗x⊗β↦0=yi(x⊗β).Proposition 4.9.
Supposei∈I and (-αi,λ)>r+1, then ◯i;λ•r=0.Proof.
In order thatr≥0, it must be the case that (-αi,λ)≥2. Thus the only possibility is (λi,λi+1)=(0,2) and r=0. Then the bimodule for 𝔼-i𝔼i is 𝒜⊗𝔉(𝕀λ). Thus the map 1→𝔼-i𝔼i is given by the unit map. The map 𝔼-i𝔼i→1 is given by the trace map. Thus the composition of the maps in the proposition sends an element β↦1⊗β↦Tr(1)⊗b=0.Proposition 4.10.
If(αi,λ)≤-1, then ◯i;λ•(-αi,λ)-1=1.Proof.
The only cases to consider are(λi,λi+1)=(0,2),(1,2),(0,1).
Consider the case(0,2). Let B=𝔉(𝕀λ). Then the bimodule corresponding to 𝔼-i𝔼i is 𝒜⊗B. Let β∈B. Then ∪i(β)=1⊗β, yi(1⊗β)=x⊗β, and ∩i(x⊗β)=Tr(x)β=β. Thus in this case, the composition is the identity map.
For the case(1,2),(-αi,λ)-1=0. The cobordism between the tangle diagrams for the identity functor and 𝔼-i𝔼i is isotopic to the identity cobordism. Similarly, the cobordism between the tangle diagrams for the functors 𝔼-i𝔼i and the identity functor is isotopic to the identity cobordism. Thus the bimodule map is equal to the identity.
The case(0,1) is the same as the case (1,2).Proposition 4.11.
Leti∈I. If (αi,λ)≥1, then
(4.23)1i;λ-αi1-i;λ=ψ-i,i;λ∘ψi,-i;λ+∑f=0(αi,λ)-1∑g=0f⋃-i;λ•(αi,λ)-f-1∘◯i;λ•-(αi,λ)-1+g∘⋂-i;λ•f-g.Proof.
There are three cases to consider:(λi,λi+1)=(1,0),(2,1),(2,0).
For the case(1,0), the first term on the right-hand side is zero since that map passes through the functor 𝔼i𝔼i𝔼-i which is zero for this λ. The summation on the right-hand side reduces to
(4.24)⋃-i;λ•0∘◯i;λ•-2∘⋂i;λ•0=∪-i;λ∘∩-i;λ
by definition (4.17) of the fake bubbles. This map is a composition 𝔼i𝔼-i→1→𝔼i𝔼-i. This composition of maps is the identity.
The case(2,1) is similar to the (1,0) case.
For the case(2,0), the first term on the right-hand side is zero as in the previous two cases. The summation on the right-hand side consists of three terms, which simplifies by (4.17) to
(4.25)⋃-i;λ•1∘∩-i;λ+∪-i;λ∘⋃-i;λ•1+∪-i;λ∘◯i;λ•2∘∩-i;λ.
Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼-i is 𝒜⊗B. Then
(4.26)⋃-i;λ•1∘∩-i;λ:𝔼i𝔼-i→𝕀→𝔼i𝔼-i→𝔼i𝔼-i.
Under this composition of maps, 1⊗b maps to zero since the first map is given by a trace map on the first component. The element x⊗b gets mapped to x⊗b as follows:
(4.27)x⊗b↦b↦1⊗b↦x⊗b,
where the first map is the trace map, the second map is the unit map, and the third map is multiplication by x. Similarly,
(4.28)∪-i;λ∘⋃-i;λ•1:𝔼i𝔼-i→𝔼i𝔼-i→𝕀→𝔼i𝔼-i.
Under this composition, 1⊗b↦1⊗b and x⊗b↦0. Finally, the map
(4.29)∪-i;λ∘◯i;λ•2∘∩-i;λ
is zero because the middle term is zero. Thus the right-hand side is the identity as well.Proposition 4.12.
Leti∈I+.  (1)
If(αi,λ)≤0, then
(4.30)(1i∩-i;λ)∘(ψi,i;λ-αi1-i)∘(1i∪-i;λ)=∑f=0-(αi,λ)yi-(αi,λ)-f◯-i;λ•(αi,λ)-1+f.(2)
If(αi,λ)≥-2, then
(4.31)(∩i;λ+αi1i)∘(1iψi,i;λ)∘(∪i;λ+αi1i)=∑g=0(αi,λ)+2◯i,λ•-(αi,λ)-3+gyi(αi,λ)-g+2.Proof.
We prove(1), the proof of (2) being similar. Since the maps on both sides pass through the functor 𝔼i𝔼i𝔼-i, the maps on both sides are zero unless (λi,λi+1)=(1,1). The functors for 𝔼i and 𝔼i𝔼i𝔼-i are given by tangles in Figure 14.
LetB be the bimodule for the functor 𝔼i so 𝒜⊗B is the bimodule for the functor 𝔼i𝔼i𝔼-i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of the left side of Figure 14 and β is an element in the other tensor factors. Consider first α=1. The left-hand side maps an element α⊗β as follows:
(4.32)1⊗β↦x⊗1⊗β+1⊗x⊗β↦1⊗1⊗β↦1⊗β,
where the first map is Δ⊗1, the second map is κ⊗1⊗1, and the third map is m⊗1. If α=x, the left-hand side maps α⊗β as follows:
(4.33)x⊗β↦x⊗x⊗β↦1⊗x⊗β↦x⊗β.
The right-hand side is 1 by convention.


### 4.6.2. nil-Hecke Relations

Proposition 4.13.
Fori∈I+,ψi,i[2]=0.Proof.
Since𝔼i𝔼i is identically zero unless (λi,λi+1)=(0,2), we need only to consider this case. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝔉(Tλ,i∘Tλ,i)=𝒜⊗B.
Thenψi,i∘ψi,i:𝒜⊗B→𝒜⊗B→𝒜⊗B. This map sends 1⊗b↦0 and x⊗b↦1⊗b↦0.Proposition 4.14.
Leti∈I+. Then, (ψi,i1i)∘(1iψi,i)∘(ψi,i1i)=(1iψi,i)∘(ψi,i1i)∘(1iψi,i).Proof.
Both sides are natural transformations of the functor𝔼i𝔼i𝔼i. However, by definition this composition is zero.Proposition 4.15.
Fori∈I+,(1i1i)=(ψi,i)∘(yi1i)-(1iyi)∘(ψi,i)=(yi1i)∘(ψi,i)-(ψi,i)∘(1iyi).Proof.
The only case to check is(λi,λi+1)=(0,2) since otherwise 𝔼i𝔼i=0. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝒜⊗B. Then
(4.34)(ψi,i)∘(yi1i):𝒜⊗B→𝒜⊗B.
Under this map, 1⊗b↦x⊗b↦1⊗b and x⊗b↦0. For the map (1iyi)∘(ψi,i),1⊗b↦0, and x⊗b↦1⊗b↦x⊗b. This gives the first equality since our field has characteristic two.
For the second equality,(yi1i)∘(ψi,i):1⊗b↦0,(yi1i)∘(ψi,i):x⊗b↦1⊗b↦x⊗b. Similarly, (ψi,i)∘(1iyi):1⊗b↦x⊗b↦1⊗b and (ψi,i)∘(1iyi):x⊗b↦0.Proposition 4.16.
Fori,j∈I-,
(4.35)ψj,i=(∩-j1i1j)∘(1j∩-i1-j1i1j)∘(1j1iψ-j,-i1i1j)∘(1j1i1-j∪i1j)∘(1j1i∪j)=(1i1j∩i)∘(1i1j1-i∩j1i)∘(1i1jψ-j,-i1j1i)∘(1i∪-j1-i1j1i)∘(∪-i1j1i).Proof.
Leti,j∈I-. We prove only the first equality. If |i-j|>1, the proposition is easy because then ψ±i,±j are identity morphisms. Therefore, we take i=j+1, the case i=j-1 being similar. The natural transformation on the right side of the proposition is a composition of natural transformations:
(4.36)𝔼j𝔼j+1→𝔼j𝔼j+1𝔼-j𝔼j→𝔼j𝔼j+1𝔼-j𝔼-j-1𝔼j+1𝔼j→𝔼j𝔼j+1𝔼-j-1𝔼-j𝔼j+1𝔼j→𝔼j𝔼-j𝔼j+1𝔼j→𝔼j+1𝔼j.
There are four nontrivial cases forλ. We prove the case (λj,λj+1,λj+2)=(2,1,1). The proofs of the remaining cases (2,1,0), (1,1,0), and (1,1,1) are similar.
LetB be the bimodule representing the functor 𝔼j𝔼j+1 and B′ the bimodule representing the functor 𝔼j+1𝔼j. Then the morphism is the composition B→B→B→𝒜⊗B→B→B′ induced by the tangle cobordisms in Figure 15. The first and second maps are the identity maps. The third map is comultiplication. The fourth map is the trace map and the last map is ψj,j+1. Computing this composition on elements as in previous propositions easily gives that it is equal to ψj,j+1.Figure 15
Tangles for compositions of natural transformations in the(2,1,1) case.


### 4.6.3.R(ν) Relations

Proposition 4.17.
Fori,j∈I±,  i≠j,
(4.37)ψ-j,i∘ψi,-j=1i1-j.Proof.
Note that, for|i-j|>1, the left-hand side is easily seen to be the identity so let j=i+1. The case j=i-1 is similar. Thus the left-hand side is
(4.38)ψ-j,i∘ψi,-j:𝔼i𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼i𝔼-i-1.
There are four non-trivial cases for λ.
Case 1 ((λi,λi+1,λi+2)=(1,2,1)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.39)ψ-j,i∘ψi,-j:B→𝒜⊗B→B→B→B→𝒜⊗B→B.
The first map is ι⊗1λ. The second map is multiplication m. The third and fourth maps are the identity. The fifth map is comultiplication Δ. The last map is Tr⊗1. It is easy to check on elements that this is the identity map.Case 2 ((λi,λi+1,λi+2)=(1,2,0)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.40)ψ-j,i∘ψi,-j:B→B→𝒜⊗B→B→𝒜⊗B→B→B.
The first map is the identity. The second map is Δ by Lemma 4.3. The third map is Tr⊗1 where the trace map is applied to the tensor factor arising from the new circle component. The fourth map is ι⊗1. The fifth map is multiplication by Lemma 4.4. The last map is the identity. It is easy to check that this composition is the identity on all elements.Case 3 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case2.Case 4 ((λi,λi+1,λi+2)=(0,2,0)).
This is similar to Case1.Proposition 4.18.
Ifi,j∈I+ and |i-j|>1, then ψj,i∘ψi,j=1i1j.Proof.
The tangle diagrams for the bimodules for𝔼i𝔼j and 𝔼j𝔼i are the same up to isotopy. The maps in the proposition are obtained from cobordisms isotopic to the identity so they are identity maps.Proposition 4.19.
Ifi,j∈I+ and |i-j|=1, then ψj,i∘ψi,j=(yi1j+1iyj).Proof.
Assume thatj=i+1. The case j=i-1 is similar. There are eight cases for λ such that 𝔼i𝔼i+1 is non-zero. In all cases let a and b be cup diagrams. Let B be the bimodule for 𝔼i𝔼i+1 and B' the bimodule for 𝔼i+1𝔼i.
Case 1.
(λi,λi+1,λi+2)=(0,0,1). Since 𝔼i+1𝔼i=0, the map ψi+1,i∘ψi,i+1=0. The bimodule representing the functor 𝔼i𝔼i+1 is isomorphic to 𝔉(Dλ+αi+1,i∘Dλ,i+1). Since the circle passing through point i on the bottom row of Dλ+αi+1,i∘Dλ,i+1 is the same as the circle passing through point i+1 in the middle row, the map on the right side of the proposition is zero as well.Case 2 ((λi,λi+1,λi+2)=(1,0,1)).
This is similar to Case1.Case 3 ((λi,λi+1,λi+2)=(1,0,2)).
This is similar to Case1.Case 4 ((λi,λi+1,λi+2)=(0,0,2)).
This is similar to Case1.Case 5 ((λi,λi+1,λi+2)=(0,1,1)).
In this caseB≅𝔉(Tλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Dλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of a(Tλ+αi+1,i)∘Tλ,i+1)b is the same as the circle passing through point i of the top row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R is a tensor product of 𝒜 corresponding to the remaining circles. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom is different from the circle passing through point i on the top. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1λ)∘(m⊗1λ). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the sameCase 6 ((λi,λi+1,λi+2)=(1,1,1)).
In this case,B≅𝔉(Dλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Tλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is the same as the circle passing through point i on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is different from the circle passing through point i on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1)∘(m⊗1). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the same.Case 7 ((λi,λi+1,λi+2)=(1,1,2)).
This is similar to Case5.Case 8 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case6.Proposition 4.20.
Leti,j∈I+. If i≠j, then  (1)
(1jyi)∘ψi,j=ψi,j∘(yi1j),(2)
(yj1i)∘ψi,j=ψi,j∘(1iyj).Proof.
We prove only the first statement. Assume further thatj=i+1, the case j=i-1 being similar. The case for |j-i|>1 is easy because the bimodules for 𝔼i𝔼j and 𝔼j𝔼i are equal.
There are four non-trivial cases for(λi,λi+1,λi+2). Let a and b be crossingless matches. Let B be the bimodule for 𝔼i𝔼i+1 and let B' be the bimodule for 𝔼i+1𝔼i.
Case 1 ((λi,λi+1,λi+2)=(1,1,2)).
(i)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R denotes a tensor product of 𝒜 corresponding to the remaining circles. Then ψi,i+1 is given by Δ⊗1. Then ψi,i+1yi(1⊗r)=ψi,i+1(x⊗r)=x⊗x⊗r. Then yiψi,i+1(1⊗r)=yi(x⊗1⊗r+1⊗x⊗r)=x⊗x⊗r.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is different from the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. Then it is easy to verify that ψi,i+1yi(1⊗1⊗r)=yiψi,i+1(1⊗1⊗r)=x⊗r.Case 2.
(λi,λi+1,λi+2)=(0,1,1). This is similar to Case 1.Case 3.
(λi,λi+1,λi+2)=(1,1,1).(i)
Suppose that the circle passing through pointi on the bottom row of the tangle is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then ψi,i+1 is given by Δ⊗1. This then follows as in Case 1.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle is different from the circle passing through the point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. This then follows as in Case 1.Case 4 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case3.Proposition 4.21.
Fori,j,k∈I+,
(4.41)(ψj,k1i)∘(1jψi,k)∘(ψi,j1k)+(1kψi,j)∘(ψi,k1j)∘(1iψj,k)={0,i≠kor|i-j|≠1,1i1j1i,i=kand|i-j|=1.Proof.
The proof of the first part consists of verifying the equality in many different cases, each of which is similar to the second part. We only prove the second part in the casej=i+1 as the case j=i-1 is similar. There are four cases for (λi,λi+1,λi+2) for which 𝔼i𝔼i+1𝔼i is non-zero.
Case 1 ((λi,λi+1,λi+2)=(0,1,1)).
In this case,(ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 because it passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to this λ. On the other hand,
(4.42)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.43)B→𝒜⊗B→𝒜⊗B→B,
where the first map is given by comultiplication, the middle map is given by the map 1⊗κ, and the last map is multiplication. This sequence of maps acts on 1⊗α∈B as follows:
(4.44)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α.
Clearly, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(1⊗α)=0. Similarly, x⊗α↦x⊗α.Case 2 ((λi,λi+1,λi+2)=(0,2,2)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=1i1j1i.Case 3.
(λi,λi+1,λi+2)=(0,1,2). In this case, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 since this map passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to λ.
On the other hand,(4.45)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.46)B→𝒜⊗B→𝒜⊗B→B,
where the first and third maps are given by Lemmas 4.3 and 4.4, respectively, and the middle map is given in Section 4.5. This sequence of maps acts on 1⊗α,x⊗α∈B as follows:
(4.47)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α,x⊗α↦x⊗x⊗α↦x⊗1⊗α↦x⊗α.Case 4 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(β⊗α)=β⊗α.The relations of the 2-morphisms proven in this section give the following.Theorem 4.22.
There is a 2-functorΩk,n:𝒦ℒ→ℋ𝒦k,n such that, for all i,j∈I,  (1)
Ωk,n(λ)=𝒞λ,(2)
Ωk,n(ℐλ)=𝕀λ,(3)
Ωk,n(ℰiℐλ)=𝔼i𝕀λ,(4)
Ωk,n(Yi;λ)=yi;λ,(5)
Ωk,n(Ψi,j;λ)=ψi,j;λ,(6)
Ωk,n(⋃i;λ)=∪i;λ,(7)
Ωk,n(⋂i;λ)=∩i;λ,(8)
Ωk,n(1i;λ)=1i;λ.


## 4.6.1.𝔰𝔩2 Relations

Proposition 4.7.
For alli∈I, (∩-i1i)∘(1i∪i)=1i=(1i∩i)∘(∪-i1i).Proof.
The second equality is similar to the first equality. The casei∈I- is similar to the case i∈I+ so we just compute the map (∩i1i)∘(1i∪i) on the bimodule for the functor 𝔼i for i∈I+. There are four cases to consider.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i are Dλ,i and Dλ,i∘Dλ+αi∘Dλ,i and can be found in Figure 12.
The cobordism between the tangles is isotopic to the identity map so in this case the composition is equal to the identity map.
The case(λi,λi+1)=(0,1) is similar to the (1,2) case.
Now let(λi,λi+1)=(0,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 13.
LetB be the bimodule for the functor 𝔼i. Then the bimodule for 𝔼i𝔼-i𝔼i is isomorphic to 𝒜⊗B. The map 𝔼i→𝔼i𝔼-i𝔼i is given by the unit map which sends an element b∈B to 1⊗b. The map 𝔼i𝔼-i𝔼i→𝔼i is obtained from the cobordism joining the circle to the upper cup which induces the multiplication map. This maps 1⊗b to b. Thus the composition is equal to the identity.
Finally consider the case(λi,λi+1)=(1,1). The tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule giving rise to the functor 𝔼i and let 𝒜⊗B be the bimodule giving rise to the functor 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is in the tensor factor corresponding to the circle passing through point i on the bottom row of the left side of Figure 14 and β belongs to the remaining tensor factors.
The cobordism between the two tangle diagrams is a saddle which, on the level of bimodule maps, sendsα⊗β↦Δ(α)⊗β. Then the map from 𝔼i𝔼-i𝔼i to 𝔼i is given by Tr⊗1λ so Δ(α)⊗β↦α⊗β by considering the two cases α=1 or x. Thus the composition is equal to the identity map.Figure 12
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,2).Figure 13
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(0,2).Figure 14
Tangles for𝔼i and 𝔼i𝔼-i𝔼i, (λi,λi+1)=(1,1).Proposition 4.8.
One has(4.20)yi=(∩-i1i)∘(1iy-i1i)∘(1i∪i)=(1i∩i)∘(1iy-i1i)∘(∪-i1i).Proof.
We prove only the first equality as the second is similar. There are four cases to consider for which the functor𝔼i is nonzero.
Suppose that(λi,λi+1)=(1,2). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 12.
Note that the bimodules for𝔼i and 𝔼i𝔼-i𝔼i are the same. Denote this bimodule by B. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of Figure 12. Then the first map 1i∪i is given by the identity cobordism and is thus the identity. The second map is multiplication by x on all tensor components corresponding to circles passing through the point i+1 in the second row of the right side of Figure 12. The final map 𝔼i𝔼-i𝔼i→𝔼i is also given by the identity cobordism. Thus the composition maps α⊗β↦α⊗β↦xα⊗β↦α⊗β. On the other hand, yi(α⊗β)=xα⊗β.
The case(λi,λi+1)=(0,1) is similar to the previous case.
Suppose that(λi,λi+1)=(0,2). Then the bimodule for the functor 𝔼i is B=𝔉(Tλ,i) and the tangle diagram for 𝔼i𝔼-i𝔼i is 𝔉(Tλ,i∘Tλ-αi,i∘Tλ,i)≅𝒜⊗B. Let α⊗β∈B, where α is an element of the tensor factor corresponding to the circle passing through the point i in the top row of the tangle Tλ,i and β is an element in the remaining tensor factors. Then the composition of maps sends α⊗β↦1⊗α⊗β↦x⊗α⊗β↦xα⊗β. This is equal to yi(α⊗β).
Suppose that(λi,λi+1)=(1,1). Then the tangle diagrams for the functors 𝔼i and 𝔼i𝔼-i𝔼i can be found in Figure 14.
LetB be the bimodule for the functor 𝔼-i and let 𝒜⊗B be the bimodule for 𝔼i𝔼-i𝔼i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to the circle passing through point i on the bottom row of Figure 14 and β is an element in the remaining tensor factors. First let α=1. Then
(4.21)1⊗β↦x⊗1⊗β+1⊗x⊗β↦x⊗x⊗β↦x⊗β=yi(1⊗β),
where the last map is Tr⊗1. If α=x, then
(4.22)x⊗β↦x⊗x⊗β↦0=yi(x⊗β).Proposition 4.9.
Supposei∈I and (-αi,λ)>r+1, then ◯i;λ•r=0.Proof.
In order thatr≥0, it must be the case that (-αi,λ)≥2. Thus the only possibility is (λi,λi+1)=(0,2) and r=0. Then the bimodule for 𝔼-i𝔼i is 𝒜⊗𝔉(𝕀λ). Thus the map 1→𝔼-i𝔼i is given by the unit map. The map 𝔼-i𝔼i→1 is given by the trace map. Thus the composition of the maps in the proposition sends an element β↦1⊗β↦Tr(1)⊗b=0.Proposition 4.10.
If(αi,λ)≤-1, then ◯i;λ•(-αi,λ)-1=1.Proof.
The only cases to consider are(λi,λi+1)=(0,2),(1,2),(0,1).
Consider the case(0,2). Let B=𝔉(𝕀λ). Then the bimodule corresponding to 𝔼-i𝔼i is 𝒜⊗B. Let β∈B. Then ∪i(β)=1⊗β, yi(1⊗β)=x⊗β, and ∩i(x⊗β)=Tr(x)β=β. Thus in this case, the composition is the identity map.
For the case(1,2),(-αi,λ)-1=0. The cobordism between the tangle diagrams for the identity functor and 𝔼-i𝔼i is isotopic to the identity cobordism. Similarly, the cobordism between the tangle diagrams for the functors 𝔼-i𝔼i and the identity functor is isotopic to the identity cobordism. Thus the bimodule map is equal to the identity.
The case(0,1) is the same as the case (1,2).Proposition 4.11.
Leti∈I. If (αi,λ)≥1, then
(4.23)1i;λ-αi1-i;λ=ψ-i,i;λ∘ψi,-i;λ+∑f=0(αi,λ)-1∑g=0f⋃-i;λ•(αi,λ)-f-1∘◯i;λ•-(αi,λ)-1+g∘⋂-i;λ•f-g.Proof.
There are three cases to consider:(λi,λi+1)=(1,0),(2,1),(2,0).
For the case(1,0), the first term on the right-hand side is zero since that map passes through the functor 𝔼i𝔼i𝔼-i which is zero for this λ. The summation on the right-hand side reduces to
(4.24)⋃-i;λ•0∘◯i;λ•-2∘⋂i;λ•0=∪-i;λ∘∩-i;λ
by definition (4.17) of the fake bubbles. This map is a composition 𝔼i𝔼-i→1→𝔼i𝔼-i. This composition of maps is the identity.
The case(2,1) is similar to the (1,0) case.
For the case(2,0), the first term on the right-hand side is zero as in the previous two cases. The summation on the right-hand side consists of three terms, which simplifies by (4.17) to
(4.25)⋃-i;λ•1∘∩-i;λ+∪-i;λ∘⋃-i;λ•1+∪-i;λ∘◯i;λ•2∘∩-i;λ.
Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼-i is 𝒜⊗B. Then
(4.26)⋃-i;λ•1∘∩-i;λ:𝔼i𝔼-i→𝕀→𝔼i𝔼-i→𝔼i𝔼-i.
Under this composition of maps, 1⊗b maps to zero since the first map is given by a trace map on the first component. The element x⊗b gets mapped to x⊗b as follows:
(4.27)x⊗b↦b↦1⊗b↦x⊗b,
where the first map is the trace map, the second map is the unit map, and the third map is multiplication by x. Similarly,
(4.28)∪-i;λ∘⋃-i;λ•1:𝔼i𝔼-i→𝔼i𝔼-i→𝕀→𝔼i𝔼-i.
Under this composition, 1⊗b↦1⊗b and x⊗b↦0. Finally, the map
(4.29)∪-i;λ∘◯i;λ•2∘∩-i;λ
is zero because the middle term is zero. Thus the right-hand side is the identity as well.Proposition 4.12.
Leti∈I+.  (1)
If(αi,λ)≤0, then
(4.30)(1i∩-i;λ)∘(ψi,i;λ-αi1-i)∘(1i∪-i;λ)=∑f=0-(αi,λ)yi-(αi,λ)-f◯-i;λ•(αi,λ)-1+f.(2)
If(αi,λ)≥-2, then
(4.31)(∩i;λ+αi1i)∘(1iψi,i;λ)∘(∪i;λ+αi1i)=∑g=0(αi,λ)+2◯i,λ•-(αi,λ)-3+gyi(αi,λ)-g+2.Proof.
We prove(1), the proof of (2) being similar. Since the maps on both sides pass through the functor 𝔼i𝔼i𝔼-i, the maps on both sides are zero unless (λi,λi+1)=(1,1). The functors for 𝔼i and 𝔼i𝔼i𝔼-i are given by tangles in Figure 14.
LetB be the bimodule for the functor 𝔼i so 𝒜⊗B is the bimodule for the functor 𝔼i𝔼i𝔼-i. Let α⊗β∈B, where α is an element in the tensor factor corresponding to a circle passing through point i in the bottom row of the left side of Figure 14 and β is an element in the other tensor factors. Consider first α=1. The left-hand side maps an element α⊗β as follows:
(4.32)1⊗β↦x⊗1⊗β+1⊗x⊗β↦1⊗1⊗β↦1⊗β,
where the first map is Δ⊗1, the second map is κ⊗1⊗1, and the third map is m⊗1. If α=x, the left-hand side maps α⊗β as follows:
(4.33)x⊗β↦x⊗x⊗β↦1⊗x⊗β↦x⊗β.
The right-hand side is 1 by convention.


## 4.6.2. nil-Hecke Relations

Proposition 4.13.
Fori∈I+,ψi,i[2]=0.Proof.
Since𝔼i𝔼i is identically zero unless (λi,λi+1)=(0,2), we need only to consider this case. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝔉(Tλ,i∘Tλ,i)=𝒜⊗B.
Thenψi,i∘ψi,i:𝒜⊗B→𝒜⊗B→𝒜⊗B. This map sends 1⊗b↦0 and x⊗b↦1⊗b↦0.Proposition 4.14.
Leti∈I+. Then, (ψi,i1i)∘(1iψi,i)∘(ψi,i1i)=(1iψi,i)∘(ψi,i1i)∘(1iψi,i).Proof.
Both sides are natural transformations of the functor𝔼i𝔼i𝔼i. However, by definition this composition is zero.Proposition 4.15.
Fori∈I+,(1i1i)=(ψi,i)∘(yi1i)-(1iyi)∘(ψi,i)=(yi1i)∘(ψi,i)-(ψi,i)∘(1iyi).Proof.
The only case to check is(λi,λi+1)=(0,2) since otherwise 𝔼i𝔼i=0. Let B=𝔉(𝕀λ). Then the bimodule for 𝔼i𝔼i is isomorphic to 𝒜⊗B. Then
(4.34)(ψi,i)∘(yi1i):𝒜⊗B→𝒜⊗B.
Under this map, 1⊗b↦x⊗b↦1⊗b and x⊗b↦0. For the map (1iyi)∘(ψi,i),1⊗b↦0, and x⊗b↦1⊗b↦x⊗b. This gives the first equality since our field has characteristic two.
For the second equality,(yi1i)∘(ψi,i):1⊗b↦0,(yi1i)∘(ψi,i):x⊗b↦1⊗b↦x⊗b. Similarly, (ψi,i)∘(1iyi):1⊗b↦x⊗b↦1⊗b and (ψi,i)∘(1iyi):x⊗b↦0.Proposition 4.16.
Fori,j∈I-,
(4.35)ψj,i=(∩-j1i1j)∘(1j∩-i1-j1i1j)∘(1j1iψ-j,-i1i1j)∘(1j1i1-j∪i1j)∘(1j1i∪j)=(1i1j∩i)∘(1i1j1-i∩j1i)∘(1i1jψ-j,-i1j1i)∘(1i∪-j1-i1j1i)∘(∪-i1j1i).Proof.
Leti,j∈I-. We prove only the first equality. If |i-j|>1, the proposition is easy because then ψ±i,±j are identity morphisms. Therefore, we take i=j+1, the case i=j-1 being similar. The natural transformation on the right side of the proposition is a composition of natural transformations:
(4.36)𝔼j𝔼j+1→𝔼j𝔼j+1𝔼-j𝔼j→𝔼j𝔼j+1𝔼-j𝔼-j-1𝔼j+1𝔼j→𝔼j𝔼j+1𝔼-j-1𝔼-j𝔼j+1𝔼j→𝔼j𝔼-j𝔼j+1𝔼j→𝔼j+1𝔼j.
There are four nontrivial cases forλ. We prove the case (λj,λj+1,λj+2)=(2,1,1). The proofs of the remaining cases (2,1,0), (1,1,0), and (1,1,1) are similar.
LetB be the bimodule representing the functor 𝔼j𝔼j+1 and B′ the bimodule representing the functor 𝔼j+1𝔼j. Then the morphism is the composition B→B→B→𝒜⊗B→B→B′ induced by the tangle cobordisms in Figure 15. The first and second maps are the identity maps. The third map is comultiplication. The fourth map is the trace map and the last map is ψj,j+1. Computing this composition on elements as in previous propositions easily gives that it is equal to ψj,j+1.Figure 15
Tangles for compositions of natural transformations in the(2,1,1) case.


## 4.6.3.R(ν) Relations

Proposition 4.17.
Fori,j∈I±,  i≠j,
(4.37)ψ-j,i∘ψi,-j=1i1-j.Proof.
Note that, for|i-j|>1, the left-hand side is easily seen to be the identity so let j=i+1. The case j=i-1 is similar. Thus the left-hand side is
(4.38)ψ-j,i∘ψi,-j:𝔼i𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i→𝔼-i-1𝔼i𝔼i+1𝔼-i-1→𝔼-i-1𝔼i+1𝔼i𝔼-i-1→𝔼i𝔼-i-1.
There are four non-trivial cases for λ.
Case 1 ((λi,λi+1,λi+2)=(1,2,1)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.39)ψ-j,i∘ψi,-j:B→𝒜⊗B→B→B→B→𝒜⊗B→B.
The first map is ι⊗1λ. The second map is multiplication m. The third and fourth maps are the identity. The fifth map is comultiplication Δ. The last map is Tr⊗1. It is easy to check on elements that this is the identity map.Case 2 ((λi,λi+1,λi+2)=(1,2,0)).
LetB be the bimodule representing the functor 𝔼i𝔼-i-1. Then
(4.40)ψ-j,i∘ψi,-j:B→B→𝒜⊗B→B→𝒜⊗B→B→B.
The first map is the identity. The second map is Δ by Lemma 4.3. The third map is Tr⊗1 where the trace map is applied to the tensor factor arising from the new circle component. The fourth map is ι⊗1. The fifth map is multiplication by Lemma 4.4. The last map is the identity. It is easy to check that this composition is the identity on all elements.Case 3 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case2.Case 4 ((λi,λi+1,λi+2)=(0,2,0)).
This is similar to Case1.Proposition 4.18.
Ifi,j∈I+ and |i-j|>1, then ψj,i∘ψi,j=1i1j.Proof.
The tangle diagrams for the bimodules for𝔼i𝔼j and 𝔼j𝔼i are the same up to isotopy. The maps in the proposition are obtained from cobordisms isotopic to the identity so they are identity maps.Proposition 4.19.
Ifi,j∈I+ and |i-j|=1, then ψj,i∘ψi,j=(yi1j+1iyj).Proof.
Assume thatj=i+1. The case j=i-1 is similar. There are eight cases for λ such that 𝔼i𝔼i+1 is non-zero. In all cases let a and b be cup diagrams. Let B be the bimodule for 𝔼i𝔼i+1 and B' the bimodule for 𝔼i+1𝔼i.
Case 1.
(λi,λi+1,λi+2)=(0,0,1). Since 𝔼i+1𝔼i=0, the map ψi+1,i∘ψi,i+1=0. The bimodule representing the functor 𝔼i𝔼i+1 is isomorphic to 𝔉(Dλ+αi+1,i∘Dλ,i+1). Since the circle passing through point i on the bottom row of Dλ+αi+1,i∘Dλ,i+1 is the same as the circle passing through point i+1 in the middle row, the map on the right side of the proposition is zero as well.Case 2 ((λi,λi+1,λi+2)=(1,0,1)).
This is similar to Case1.Case 3 ((λi,λi+1,λi+2)=(1,0,2)).
This is similar to Case1.Case 4 ((λi,λi+1,λi+2)=(0,0,2)).
This is similar to Case1.Case 5 ((λi,λi+1,λi+2)=(0,1,1)).
In this caseB≅𝔉(Tλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Dλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of a(Tλ+αi+1,i)∘Tλ,i+1)b is the same as the circle passing through point i of the top row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R is a tensor product of 𝒜 corresponding to the remaining circles. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom is different from the circle passing through point i on the top. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1λ)∘(m⊗1λ). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the sameCase 6 ((λi,λi+1,λi+2)=(1,1,1)).
In this case,B≅𝔉(Dλ+αi+1,i∘Tλ,i+1) and B'≅𝔉(Dλ+αi,i+1∘Tλ,i). Let a and b be crossingless matches.  (i)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is the same as the circle passing through point i on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then the map on the left side of the proposition is (m⊗1)∘(Δ⊗1). Thus it maps an element 1⊗r to 2x⊗r. On the other hand, yi(1⊗r)=x⊗r. Also, yi+1(1⊗r)=x⊗r. Thus both sides are the same.(ii)
Suppose that the circle passing through pointi+1 on the bottom row of Dλ+αi+1,i∘Tλ,i+1 is different from the circle passing through point i on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then the map on the left side of the proposition is (Δ⊗1)∘(m⊗1). Thus it maps an element 1⊗1⊗r to x⊗1⊗r+1⊗x⊗r. On the other hand, yi(1⊗1⊗r)=x⊗1⊗r. Also, yi+1(1⊗r)=1⊗x⊗r. Thus both sides are the same.Case 7 ((λi,λi+1,λi+2)=(1,1,2)).
This is similar to Case5.Case 8 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case6.Proposition 4.20.
Leti,j∈I+. If i≠j, then  (1)
(1jyi)∘ψi,j=ψi,j∘(yi1j),(2)
(yj1i)∘ψi,j=ψi,j∘(1iyj).Proof.
We prove only the first statement. Assume further thatj=i+1, the case j=i-1 being similar. The case for |j-i|>1 is easy because the bimodules for 𝔼i𝔼j and 𝔼j𝔼i are equal.
There are four non-trivial cases for(λi,λi+1,λi+2). Let a and b be crossingless matches. Let B be the bimodule for 𝔼i𝔼i+1 and let B' be the bimodule for 𝔼i+1𝔼i.
Case 1 ((λi,λi+1,λi+2)=(1,1,2)).
(i)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R, where R denotes a tensor product of 𝒜 corresponding to the remaining circles. Then ψi,i+1 is given by Δ⊗1. Then ψi,i+1yi(1⊗r)=ψi,i+1(x⊗r)=x⊗x⊗r. Then yiψi,i+1(1⊗r)=yi(x⊗1⊗r+1⊗x⊗r)=x⊗x⊗r.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle for 𝔼i𝔼i+1 is different from the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. Then it is easy to verify that ψi,i+1yi(1⊗1⊗r)=yiψi,i+1(1⊗1⊗r)=x⊗r.Case 2.
(λi,λi+1,λi+2)=(0,1,1). This is similar to Case 1.Case 3.
(λi,λi+1,λi+2)=(1,1,1).(i)
Suppose that the circle passing through pointi on the bottom row of the tangle is the same as the circle passing through point i+1 on the bottom row. Then aBb=𝒜⊗R and aBb′=𝒜⊗𝒜⊗R. Then ψi,i+1 is given by Δ⊗1. This then follows as in Case 1.(ii)
Suppose that the circle passing through pointi on the bottom row of the tangle is different from the circle passing through the point i+1 on the bottom row. Then aBb=𝒜⊗𝒜⊗R and aBb′=𝒜⊗R. Then ψi,i+1=m⊗1. This then follows as in Case 1.Case 4 ((λi,λi+1,λi+2)=(0,1,2)).
This is similar to Case3.Proposition 4.21.
Fori,j,k∈I+,
(4.41)(ψj,k1i)∘(1jψi,k)∘(ψi,j1k)+(1kψi,j)∘(ψi,k1j)∘(1iψj,k)={0,i≠kor|i-j|≠1,1i1j1i,i=kand|i-j|=1.Proof.
The proof of the first part consists of verifying the equality in many different cases, each of which is similar to the second part. We only prove the second part in the casej=i+1 as the case j=i-1 is similar. There are four cases for (λi,λi+1,λi+2) for which 𝔼i𝔼i+1𝔼i is non-zero.
Case 1 ((λi,λi+1,λi+2)=(0,1,1)).
In this case,(ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 because it passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to this λ. On the other hand,
(4.42)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.43)B→𝒜⊗B→𝒜⊗B→B,
where the first map is given by comultiplication, the middle map is given by the map 1⊗κ, and the last map is multiplication. This sequence of maps acts on 1⊗α∈B as follows:
(4.44)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α.
Clearly, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(1⊗α)=0. Similarly, x⊗α↦x⊗α.Case 2 ((λi,λi+1,λi+2)=(0,2,2)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=1i1j1i.Case 3.
(λi,λi+1,λi+2)=(0,1,2). In this case, (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)=0 since this map passes through the functor 𝔼i+1𝔼i𝔼i which is zero on the category corresponding to λ.
On the other hand,(4.45)(1iψi,j)∘(ψi,i1j)∘(1iψj,i):𝔼i𝔼i+1𝔼i→𝔼i𝔼i𝔼i+1→𝔼i𝔼i𝔼i+1→𝔼i𝔼i+1𝔼i.
Let B be the bimodule for the functor 𝔼i𝔼i+1𝔼i. Then this is a sequence of maps
(4.46)B→𝒜⊗B→𝒜⊗B→B,
where the first and third maps are given by Lemmas 4.3 and 4.4, respectively, and the middle map is given in Section 4.5. This sequence of maps acts on 1⊗α,x⊗α∈B as follows:
(4.47)1⊗α↦x⊗1⊗α+1⊗x⊗α↦1⊗1⊗α↦1⊗α,x⊗α↦x⊗x⊗α↦x⊗1⊗α↦x⊗α.Case 4 ((λi,λi+1,λi+2)=(0,2,1)).
This is similar to Case1 except that now (1iψi,j)∘(ψi,i1j)∘(1iψj,i)=0 and (ψj,i1i)∘(1jψi,i)∘(ψi,j1i)(β⊗α)=β⊗α.The relations of the 2-morphisms proven in this section give the following.Theorem 4.22.
There is a 2-functorΩk,n:𝒦ℒ→ℋ𝒦k,n such that, for all i,j∈I,  (1)
Ωk,n(λ)=𝒞λ,(2)
Ωk,n(ℐλ)=𝕀λ,(3)
Ωk,n(ℰiℐλ)=𝔼i𝕀λ,(4)
Ωk,n(Yi;λ)=yi;λ,(5)
Ωk,n(Ψi,j;λ)=ψi,j;λ,(6)
Ωk,n(⋃i;λ)=∪i;λ,(7)
Ωk,n(⋂i;λ)=∩i;λ,(8)
Ωk,n(1i;λ)=1i;λ.


## 5. The 2-Category𝒫k,n

### 5.1. Graded Categoryℤ𝒪

Let𝔤=𝔤𝔩2k be the Lie algebra of 2k×2k-matrices, let 𝔡 denote the Cartan subalgebra of 𝔤 consisting of diagonal matrices, and let 𝔭 be the Borel subalgebra of upper triangular matrices. For i=1,…,2k, let eij denote the (i,j)-matrix unit, and let εi∈𝔡* be the coordinate functional εi(ejj)=δij. Let 𝒪 be the category of finitely generated 𝔤-modules which are diagonalizable with respect to 𝔡 and locally finite with respect to 𝔭. Let(5.1)X=⨁i=12kℤεi,Y=⨁i=12k-1ℤ(εi-εi+1)⊂X
denote the weight lattice and root lattice of 𝔤𝔩2k, respectively. The dominant weights are given by the set X+={μ=μ1ε1+⋯+μ2kε2k∈X|μ1≥⋯≥μ2k}. Denote half the sum of the positive roots by ρ. Let μ∈X+, and let 𝒪μ be the block of 𝒪 consisting of modules that have a generalized central character corresponding to μ under the Harish-Chandra homomorphism. Let 𝒪μ(k,k) be the full subcategory 𝒪 consisting of modules which are locally finite with respect to the parabolic subalgebra whose reductive part is 𝔤𝔩k⊕𝔤𝔩k. Finally, let 𝒫μ(k,k) be the full subcategory of 𝒪μ(k,k) whose objects have projective presentations by projective-injective modules.Letμ and μ′ be integral dominant weights of 𝔤, and let Stab(μ) denote the stabilizer of μ under the ρ-shifted action of the symmetric group 𝕊2k. Suppose that μ′-μ is an integral dominant weight. Then, let θμμ':𝒪μ(k,k)→𝒪μ'(k,k) be the translation functor of tensoring with the finite-dimensional irreducible representation of highest weight μ′-μ composed with projecting onto the μ′-block, and let θμ'μ be its adjoint.LetPμ be a minimal projective generator of 𝒪μ. It was shown that Aμ=End𝔤(Pμ) has the structure of a graded algebra [11]. Since 𝒪μ is Morita equivalent to Aμ-mod, we consider the category of graded Aμ-modules which we denote by ℤ𝒪μ. Let the graded lift of 𝒪μ(k,k) and 𝒫μ(k,k) be ℤ𝒪μ(k,k) and ℤ𝒫μ(k,k), respectively. It is known that if Stab(μ)⊂Stab(μ′), there is a graded lift of the translation functors, compare, for example, [14], which by abuse of notation we denote again by θ̃μ′μ and θ̃μμ′.The key tool in the construction of graded category𝒪 is the Soergel functor. Let λ=(λ1,…,λn) be a composition of 2k and 𝕊λ=𝕊λ1×⋯×𝕊λn. Denote the longest coset representative in 𝕊2k/𝕊μ by w0μ. Let P(w0μ·μ) be the unique up to isomorphism, indecomposable projective-injective object of 𝒪μ. Let C=S(𝔥)/S(𝔥)+𝕊2k be the coinvariant algebra of the symmetric algebra for the Cartan subalgebra with respect to the action of the symmetric group. Let {x1,…,x2k} be a basis of S(𝔥) and by abuse of notation also let xi denote its image in C. Let Cλ be the subalgebra of elements invariant under the action of 𝕊λ. Soergel proved in [15] the following.Proposition 5.1.
One hasEnd𝔤(P(w0μ·μ))≅CStab(μ).Define the Soergel functor𝕍μ:𝒪μ→CStab(μ)-mod to be Hom𝔤(P(w0·μ),•).Proposition 5.2.
LetP be a projective object. Then there is a natural isomorphism HomCStab(μ)(𝕍μP,𝕍μM)≅Hom𝔤(P,M).Proof.
This is the Structure Theorem of [15].Proposition 5.3.
Letμ,μ'∈X+ be integral dominant weights such that there is a containment of stabilizers: Stab(μ)⊂Stab(μ'). Then there are isomorphisms of functors  (1)
𝕍μ'θμμ'≅ResCStab(μ)CStab(μ')𝕍μ,(2)
𝕍μθμ'μ≅CStab(μ)⨂CStab(μ')𝕍μ'.Proof.
These are Theorem12 and Proposition 6 of [16].


### 5.2. The Objects of𝒫k,n

Letλ=(λ1,…,λr) be a composition of 2k with λi∈{0,1,2} for all i. To each such λ, we associate an integral dominant weight (5.2)λ¯=∑j=1r∑i=1λj(r-j+1)ελ1+⋯+λj-1+i-ρ
of 𝔤𝔩2k, where λ0=0. Note that the stabilizer of this weight under the action of 𝕊2k is 𝕊λ1×⋯×𝕊λn.The set of objects of𝒫k,n are the categories ℤ𝒫λ¯(k,k),   λ∈P(V2ωk).


### 5.3. The 1-Morphisms of𝒫k,n

Letλ∈P(V2ωk), and let 𝕀λ∈End𝔤(𝒫ℤλ¯(k,k)) be the identity functor.For eachi∈I, we define functors 𝔼i𝕀λ and 𝕂i𝕀λ. To this end, let λ be a weight of V2ωk and i∈I+. Then we have compositions of 2k into n+1 parts:(5.3)λ(i)=(λ1,…,λi,1,λi+1-1,…,λn),λ(-i)=(λ1,…,λi-1,1,λi+1,…,λn)
Also, if λ=∑iaiωi∈P, set ri,λ=1+a1+⋯+ai-1+ai+1 and si,λ=2-ai-ai+1.Leti∈I. Suppose that (λi,λi+1)∈{(0,1),(0,2),(1,1),(1,2)}. Then we define, as in [17], 𝔼i𝕀λ:ℤ𝒫λ¯(k,k)→ℤ𝒫λ+αi¯(k,k) which is given by tensoring with the following bimodule: (5.4)Hom𝔤(Pλ+αi¯,θλ(i)¯λ+αi¯θλ¯λ(i)¯Pλ¯{ri,λ})≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,𝕍λ+αi¯θλ(i)¯λ+αi¯θλ¯λ(i)¯Pλ¯{ri,λ})≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,Cλ+αi⨂Cλ(i)ResCλCλ(i)𝕍λ¯Pλ¯{ri,λ}).
For all other values of (λi,λi+1), set 𝔼i𝕀λ=0. Let 𝕂i𝕀λ:ℤ𝒫λ¯(k,k)→ℤ𝒫λ¯(k,k) be the grading shift functor 𝕂i𝕀λ=𝕀λ{(αi,λ)}.Letℤ𝒫λ¯(k,k) and ℤ𝒫λ¯'(k,k) be two objects. Then(5.5)Hom(𝒫ℤλ¯(k,k),ℤ𝒫λ¯'(k,k))=⨁i̲∈Seqs∈ℤ𝕀λ'𝔼i̲𝕀λ{s},where𝔼i̲:=𝔼i1⋯𝔼ir𝕀λ if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift.


### 5.4. Bimodule Categories over the Cohomology of Flag Varieties

A review of certain bimodules and bimodule maps over the cohomology of flag varieties developed in [1, 2, 18] is given here. Let λ=(λ1,…,λn) be a composition of 2k into n parts. Let x(λ)j,r=xλ1+⋯+λj-1+r. There is an isomorphism of algebras: (5.6)Cλ≅⨂1≤j≤nℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]Jλ,n,
where Jλ,n is the ideal generated by the homogeneous terms in the equation(5.7)∏1≤j≤n(1+x(λ)j,1t+x(λ)j,2t2+⋯+x(λ)j,λjtλj)=1.Letx̂(λ)i,k be the homogenous term of degree 2k in the product (5.8)∏1≤j≤nj≠i(1+x(λ)j,1t+x(λ)j,2t2+⋯+x(λ)j,λjtλj).
Then, using (5.7), we see that (5.9)∑j=1kx(λ)i,jx̂(λ)i,k-j=δk,0,
compare, for example, [1, Section 5.1] for details.We must also considerCλ(i). There is an isomorphism of algebras: (5.10)Cλ(i)≅⨂1≤j≤n,j≠i+1ℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]⊗ℂ[ζi]⊗ℂ[x(λ)i+1,1,x(λ)i+1,2,…,x(λ)i+1,λi+1-1]Jλ(i),n,
where Jλ(i),n is the ideal generated by the homogeneous terms in the equation (5.11)∏1≤j≤n,j≠i+1(1+ζit)∑r=0λi+1-1x(λ)i+1,rtr∑s=0λjx(λ)j,sts=1.
There is also an isomorphism of algebras: (5.12)Cλ(-i)≅⨂1≤j≤n,j≠iℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]⊗ℂ[x(λ)i,1,x(λ)i,2,…,x(λ)i,λi-1]⊗ℂ[ζi]/Jλ(-i),n,
where Jλ(-i),n is the ideal generated by the homogeneous terms in the equation (5.13)∏1≤j≤n,j≠i(1+ζit)∑r=0λi-1x(λ)i,rtr∑s=0λjx(λ)j,sts=1.


### 5.5. The 2-Morphisms

In light of Propositions5.2 and 5.3, we may define the 2-morphisms on the algebras Cλ, λ∈P(V2ωk) in order to define natural transformations of functors.The Mapsy¯i;λ
Leti∈I. Define y¯i;λ:Cλ(i)→Cλ(i) which is a map of (Cλ+αi,Cλ)-bimodules by y¯i;λ((ζi)r)=(ζi)r+1.The Maps∪¯i;λ,∩¯i;λ
Leti∈I+. Define a map of (Cλ,Cλ)-bimodules
(5.14)∪¯i;λ:Cλ→Cλ(i)⨂Cλ+αiCλ(i){1-λi-λi+1}
by
(5.15)∪¯i;λ(1)=∑f=0λi(-1)λi-fζif⊗x(λ)i,λi-f.
Next define a map of(Cλ,Cλ)-bimodules
(5.16)∪¯-i;λ:Cλ→Cλ(-i)⨂Cλ-αiCλ(-i){1-λi-λi+1}
by
(5.17)∪¯-i;λ(1)=∑f=0λi+1(-1)λi+1-fζif⊗x(λ)i+1,λi+1-f.
Next define a map of(Cλ,Cλ)-bimodules
(5.18)∩¯i;λ:Cλ(i)⨂Cλ+αiCλ(i){1-λi-λi+1}→Cλ
by
(5.19)∩¯i;λ(ζir1⊗ζir2)=(-1)r1+r2+1-λi+1x̂(λ)i+1,r1+r2+1-λi+1.
Next define a map of(Cλ,Cλ)-bimodules
(5.20)∩¯-i;λ:Cλ(-i)⨂Cλ-αiCλ(-i){1-λi-λi+1}→Cλ
by
(5.21)∩¯-i;λ(ζir1⊗ζir2)=(-1)r1+r2+1-λix̂(λ)i,r1+r2+1-λi.The Mapsψ¯i,j;λ
Leti,j∈I+. Define a map of (Cλ+αi+αj,Cλ)-bimodules
(5.22)ψ¯i,j;λ:C(λ+αj)(i)⨂Cλ+αjCλ(j)→C(λ+αi)(j)⨂Cλ+αiCλ(i)
by
(5.23)ψ¯i,j;λ(ζir1⊗ζjr2)={ζjr2⊗ζir1if|i-j|>1,∑f=0r1-1ζir1+r2-1-f⊗ζif-∑g=0r2-1ζir1+r2-1-g⊗ζigifj=i,(ζjr2⊗ζir1+1-ζjr2+1⊗ζir1){-1}ifi=j+1,(ζjr2⊗ζir1){1}ifj=i+1.
Define a map of(Cλ-αi-αj,Cλ)-bimodules
(5.24)ψ¯-i,-j;λ:C(λ-αj)(-i)⨂Cλ-αjCλ(-j)→C(λ-αi)(-j)⨂Cλ-αiCλ(-i)
by
(5.25)ψ¯-i,-j(ζir1⊗ζjr2)={ζjr2⊗ζir1if|i-j|>1,∑f=0r2-1ζir1+r2-1-f⊗ζif-∑g=0r1-1ζir1+r2-1-g⊗ζigifj=i,(ζjr2⊗ζir1+1){-1}ifi=j+1,(ζjr2+1⊗ζir1-ζjr2⊗ζir1+1){1}ifj=i+1.


### 5.6. The 2-Morphisms of𝒫k,n

Leti,j∈I+.The Maps1i;λ
Let1i;λ:𝔼i𝕀λ→𝔼i𝕀λ and 1-i;λ:𝔼-i𝕀λ→𝔼-i𝕀λ be the identity morphisms.The Mapsyi;λ
Next we define a morphism of degree2,yi;λ:𝔼i𝕀λ→𝔼i𝕀λ. Recall that
(5.26)𝔼i𝕀λ≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,Cλ(i)⨂Cλ𝕍λ¯Pλ¯{ri,λ}).
Let f be such a homomorphism. Suppose that f(m)=γ⊗n. Then set (yi;λ·f)(m)=y¯i(γ)⊗n.
Similarly,(5.27)𝔼-i𝕀λ≅HomCλ-αi(𝕍λ-αi¯Pλ-αi¯,Cλ(-i)⨂Cλ𝕍λ¯Pλ¯{si,λ}).
Let f be such a homomorphism. Suppose that f(m)=γ⊗n. Then set (y-i;λ·f)(m)=y¯-i;λ(γ)⊗n.The Maps∪i;λ,∩i;λ
Note that(5.28)𝕀λ≅J=HomCλ(𝕍λ¯Pλ¯,𝕍λ¯Pλ¯),𝔼-i∘𝔼i𝕀λ≅K=HomCλ(𝕍λ¯Pλ¯,Cλ+αi(-i)⨂Cλ+αiCλ(i)⨂Cλ𝕍λ¯Pλ¯{rλ,i+sλ+αi,i}),𝔼i∘𝔼-i𝕀λ≅L=HomCλ(𝕍λ¯Pλ¯,Cλ-αi(i)⨂Cλ-αiCλ(-i)⨂Cλ𝕍λ¯Pλ¯{sλ,i+rλ-αi,i}).
Letf∈J. Then define ∪i;λ:𝕀λ→𝔼-i𝔼i𝕀λ by
(5.29)∪i;λ(f)(m)=∪¯i;λ(1)⊗f(m)
and ∪-i;λ:𝕀λ→𝔼i𝔼-i𝕀λ by
(5.30)∪-i;λ(f)(m)=∪¯-i;λ(1)⊗f(m).
Now define∩i;λ:𝔼-i𝔼i𝕀λ→𝕀λ. Suppose that f∈K such that f(m)=γ⊗n. Then set ∩i;λ(f)(m)=∩¯i;λ(γ)⊗n.
Next define∩-i;λ:𝔼i𝔼-i𝕀λ→𝕀λ. Suppose that f∈L such that f(m)=γ⊗n. Then set ∩-i;λ(f)(m)=∩¯-i;λ(γ)⊗n.The Mapsψi,j;λ
First we define a mapψi,j;λ:𝔼i𝔼j𝕀λ→𝔼j𝔼i𝕀λ.
Set(5.31)Ji,j+=𝔼i𝔼j𝕀λ≅HomCλ+αi+αj(𝕍λ+αi+αj¯Pλ+αi+αj¯,C(λ+αj)(i)⨂Cλ+αjCλ(j)⨂Cλ𝕍λ¯Pλ¯{rλ,j+rλ+αj,i}),Ki,j+=𝔼j𝔼i𝕀λ≅HomCλ+αj+αi(𝕍λ+αj+αi¯Pλ+αj+αi¯,C(λ+αi)(j)⨂Cλ+αiCλ(i)⨂Cλ𝕍λ¯Pλ¯{rλ,i+rλ+αi,j}).
Let f∈Ji,j+ and suppose that f(m)=γ1⊗γ2⊗n. Then define ψi,j;λf(m)=ψ¯i,j;λ(γ1⊗γ2)⊗n.
Set(5.32)Ji,j-=𝔼-i𝔼-j𝕀λ≅HomCλ-αi-αj(𝕍λ-αi-αj¯Pλ-αi-αj¯,C(λ-αj)(-i)⨂Cλ-αjCλ(-j)⨂Cλ𝕍λ¯Pλ¯{sλ,j+sλ-αj,i}),Ki,j-=𝔼-j𝔼-i𝕀λ≅HomCλ-αj-αi(𝕍λ-αj-αi¯Pλ-αj-αi¯,C(λ-αi)(-j)⨂Cλ-αiCλ(-i)⨂Cλ𝕍λ¯Pλ¯{sλ,i+sλ-αi,j}).
Letf∈Ji,j- and suppose that f(m)=γ1⊗γ2⊗n. Then define ψ-i,-j;λf(m)=ψ¯-i,-j;λ(γ1⊗γ2)⊗n.Theorem 5.4.
There is a 2-functorΠk,n:𝒦ℒ→𝒫k,n such that, for all i,j∈I,  (1)
Πk,n(λ)=ℤ𝒫λ¯(k,k),(2)
Πk,n(ℐλ)=𝕀λ,(3)
Πk,n(ℰiℐλ)=𝔼i𝕀λ,(4)
Πk,n(Yi;λ)=yi;λ,(5)
Πk,n(Ψi,j;λ)=ψi,j;λ,(6)
Πk,n(⋃i;λ)=∪i;λ,(7)
Πk,n(⋂i;λ)=∩i;λ,(8)
Πk,n(1i;λ)=1i;λ.Proof.
This now follows from the computations in [1, Section 6.2] for bimodules over the cohomology of flag varieties using the naturality of the isomorphism in Proposition 5.2.Finally we show that the category𝒫k,n is a categorification of the module V2ωk. Denote the Grothendieck group of 𝒫k,n by [𝒫k,n], and let [𝒫k,n]ℚ(q)=ℂ(q)⨂ℤ[q,q-1][𝒫k,n].Proposition 5.5.
There is an isomorphism of𝒰q(𝔰𝔩n)-modules [𝒫k,n]ℚ(q)≅V2ωk.Proof.
Since projective functors map projective-injective modules to projective-injective modules, it follows from Theorem5.4 and [1] that [𝒫k,n]ℚ(q) is a 𝒰q(𝔰𝔩n)-module. By construction, it contains a highest weight vector of weight 2ωk so it suffices to compute the dimension of its weight spaces.
By [19, Theorem 4.8], the number of projective-injective objects in 𝒪λ¯(k,k)(𝔤𝔩2k) is equal to the number of column decreasing and row nondecreasing tableau for a diagram with k rows and 2 columns with entries from the set
(5.33){n,…,n︸λ1,…,1,…,1︸λn}.
Call the set of such tableau T.
LetS={i∈I+∣λi=1}. Denote by |S| the cardinality of this set. Consider a Young diagram with |S|/2 rows and 2 columns. Let T' denote the set of tableau on such a column with entries from S such that the rows and columns are decreasing. It is well known that the cardinality of the set T' is the Catalan number (2|S||S|)/(|S|+1). There is a bijection between T and T'. For any tableaux t'∈T', one constructs a tableaux t∈T by inserting a new box with the entry i in each column for each i∈I+ such that λi=2. The inverse is given by box removal.
Finally, the Weyl character formula gives that the dimension of theλ weight space of V2ωk is (2|S||S|)/(|S|+1).


## 5.1. Graded Categoryℤ𝒪

Let𝔤=𝔤𝔩2k be the Lie algebra of 2k×2k-matrices, let 𝔡 denote the Cartan subalgebra of 𝔤 consisting of diagonal matrices, and let 𝔭 be the Borel subalgebra of upper triangular matrices. For i=1,…,2k, let eij denote the (i,j)-matrix unit, and let εi∈𝔡* be the coordinate functional εi(ejj)=δij. Let 𝒪 be the category of finitely generated 𝔤-modules which are diagonalizable with respect to 𝔡 and locally finite with respect to 𝔭. Let(5.1)X=⨁i=12kℤεi,Y=⨁i=12k-1ℤ(εi-εi+1)⊂X
denote the weight lattice and root lattice of 𝔤𝔩2k, respectively. The dominant weights are given by the set X+={μ=μ1ε1+⋯+μ2kε2k∈X|μ1≥⋯≥μ2k}. Denote half the sum of the positive roots by ρ. Let μ∈X+, and let 𝒪μ be the block of 𝒪 consisting of modules that have a generalized central character corresponding to μ under the Harish-Chandra homomorphism. Let 𝒪μ(k,k) be the full subcategory 𝒪 consisting of modules which are locally finite with respect to the parabolic subalgebra whose reductive part is 𝔤𝔩k⊕𝔤𝔩k. Finally, let 𝒫μ(k,k) be the full subcategory of 𝒪μ(k,k) whose objects have projective presentations by projective-injective modules.Letμ and μ′ be integral dominant weights of 𝔤, and let Stab(μ) denote the stabilizer of μ under the ρ-shifted action of the symmetric group 𝕊2k. Suppose that μ′-μ is an integral dominant weight. Then, let θμμ':𝒪μ(k,k)→𝒪μ'(k,k) be the translation functor of tensoring with the finite-dimensional irreducible representation of highest weight μ′-μ composed with projecting onto the μ′-block, and let θμ'μ be its adjoint.LetPμ be a minimal projective generator of 𝒪μ. It was shown that Aμ=End𝔤(Pμ) has the structure of a graded algebra [11]. Since 𝒪μ is Morita equivalent to Aμ-mod, we consider the category of graded Aμ-modules which we denote by ℤ𝒪μ. Let the graded lift of 𝒪μ(k,k) and 𝒫μ(k,k) be ℤ𝒪μ(k,k) and ℤ𝒫μ(k,k), respectively. It is known that if Stab(μ)⊂Stab(μ′), there is a graded lift of the translation functors, compare, for example, [14], which by abuse of notation we denote again by θ̃μ′μ and θ̃μμ′.The key tool in the construction of graded category𝒪 is the Soergel functor. Let λ=(λ1,…,λn) be a composition of 2k and 𝕊λ=𝕊λ1×⋯×𝕊λn. Denote the longest coset representative in 𝕊2k/𝕊μ by w0μ. Let P(w0μ·μ) be the unique up to isomorphism, indecomposable projective-injective object of 𝒪μ. Let C=S(𝔥)/S(𝔥)+𝕊2k be the coinvariant algebra of the symmetric algebra for the Cartan subalgebra with respect to the action of the symmetric group. Let {x1,…,x2k} be a basis of S(𝔥) and by abuse of notation also let xi denote its image in C. Let Cλ be the subalgebra of elements invariant under the action of 𝕊λ. Soergel proved in [15] the following.Proposition 5.1.
One hasEnd𝔤(P(w0μ·μ))≅CStab(μ).Define the Soergel functor𝕍μ:𝒪μ→CStab(μ)-mod to be Hom𝔤(P(w0·μ),•).Proposition 5.2.
LetP be a projective object. Then there is a natural isomorphism HomCStab(μ)(𝕍μP,𝕍μM)≅Hom𝔤(P,M).Proof.
This is the Structure Theorem of [15].Proposition 5.3.
Letμ,μ'∈X+ be integral dominant weights such that there is a containment of stabilizers: Stab(μ)⊂Stab(μ'). Then there are isomorphisms of functors  (1)
𝕍μ'θμμ'≅ResCStab(μ)CStab(μ')𝕍μ,(2)
𝕍μθμ'μ≅CStab(μ)⨂CStab(μ')𝕍μ'.Proof.
These are Theorem12 and Proposition 6 of [16].


## 5.2. The Objects of𝒫k,n

Letλ=(λ1,…,λr) be a composition of 2k with λi∈{0,1,2} for all i. To each such λ, we associate an integral dominant weight (5.2)λ¯=∑j=1r∑i=1λj(r-j+1)ελ1+⋯+λj-1+i-ρ
of 𝔤𝔩2k, where λ0=0. Note that the stabilizer of this weight under the action of 𝕊2k is 𝕊λ1×⋯×𝕊λn.The set of objects of𝒫k,n are the categories ℤ𝒫λ¯(k,k),   λ∈P(V2ωk).


## 5.3. The 1-Morphisms of𝒫k,n

Letλ∈P(V2ωk), and let 𝕀λ∈End𝔤(𝒫ℤλ¯(k,k)) be the identity functor.For eachi∈I, we define functors 𝔼i𝕀λ and 𝕂i𝕀λ. To this end, let λ be a weight of V2ωk and i∈I+. Then we have compositions of 2k into n+1 parts:(5.3)λ(i)=(λ1,…,λi,1,λi+1-1,…,λn),λ(-i)=(λ1,…,λi-1,1,λi+1,…,λn)
Also, if λ=∑iaiωi∈P, set ri,λ=1+a1+⋯+ai-1+ai+1 and si,λ=2-ai-ai+1.Leti∈I. Suppose that (λi,λi+1)∈{(0,1),(0,2),(1,1),(1,2)}. Then we define, as in [17], 𝔼i𝕀λ:ℤ𝒫λ¯(k,k)→ℤ𝒫λ+αi¯(k,k) which is given by tensoring with the following bimodule: (5.4)Hom𝔤(Pλ+αi¯,θλ(i)¯λ+αi¯θλ¯λ(i)¯Pλ¯{ri,λ})≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,𝕍λ+αi¯θλ(i)¯λ+αi¯θλ¯λ(i)¯Pλ¯{ri,λ})≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,Cλ+αi⨂Cλ(i)ResCλCλ(i)𝕍λ¯Pλ¯{ri,λ}).
For all other values of (λi,λi+1), set 𝔼i𝕀λ=0. Let 𝕂i𝕀λ:ℤ𝒫λ¯(k,k)→ℤ𝒫λ¯(k,k) be the grading shift functor 𝕂i𝕀λ=𝕀λ{(αi,λ)}.Letℤ𝒫λ¯(k,k) and ℤ𝒫λ¯'(k,k) be two objects. Then(5.5)Hom(𝒫ℤλ¯(k,k),ℤ𝒫λ¯'(k,k))=⨁i̲∈Seqs∈ℤ𝕀λ'𝔼i̲𝕀λ{s},where𝔼i̲:=𝔼i1⋯𝔼ir𝕀λ if i̲=(i1,…,ir)∈I∞, and s refers to a grading shift.


## 5.4. Bimodule Categories over the Cohomology of Flag Varieties

A review of certain bimodules and bimodule maps over the cohomology of flag varieties developed in [1, 2, 18] is given here. Let λ=(λ1,…,λn) be a composition of 2k into n parts. Let x(λ)j,r=xλ1+⋯+λj-1+r. There is an isomorphism of algebras: (5.6)Cλ≅⨂1≤j≤nℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]Jλ,n,
where Jλ,n is the ideal generated by the homogeneous terms in the equation(5.7)∏1≤j≤n(1+x(λ)j,1t+x(λ)j,2t2+⋯+x(λ)j,λjtλj)=1.Letx̂(λ)i,k be the homogenous term of degree 2k in the product (5.8)∏1≤j≤nj≠i(1+x(λ)j,1t+x(λ)j,2t2+⋯+x(λ)j,λjtλj).
Then, using (5.7), we see that (5.9)∑j=1kx(λ)i,jx̂(λ)i,k-j=δk,0,
compare, for example, [1, Section 5.1] for details.We must also considerCλ(i). There is an isomorphism of algebras: (5.10)Cλ(i)≅⨂1≤j≤n,j≠i+1ℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]⊗ℂ[ζi]⊗ℂ[x(λ)i+1,1,x(λ)i+1,2,…,x(λ)i+1,λi+1-1]Jλ(i),n,
where Jλ(i),n is the ideal generated by the homogeneous terms in the equation (5.11)∏1≤j≤n,j≠i+1(1+ζit)∑r=0λi+1-1x(λ)i+1,rtr∑s=0λjx(λ)j,sts=1.
There is also an isomorphism of algebras: (5.12)Cλ(-i)≅⨂1≤j≤n,j≠iℂ[x(λ)j,1,x(λ)j,2,…,x(λ)j,λj]⊗ℂ[x(λ)i,1,x(λ)i,2,…,x(λ)i,λi-1]⊗ℂ[ζi]/Jλ(-i),n,
where Jλ(-i),n is the ideal generated by the homogeneous terms in the equation (5.13)∏1≤j≤n,j≠i(1+ζit)∑r=0λi-1x(λ)i,rtr∑s=0λjx(λ)j,sts=1.


## 5.5. The 2-Morphisms

In light of Propositions5.2 and 5.3, we may define the 2-morphisms on the algebras Cλ, λ∈P(V2ωk) in order to define natural transformations of functors.The Mapsy¯i;λ
Leti∈I. Define y¯i;λ:Cλ(i)→Cλ(i) which is a map of (Cλ+αi,Cλ)-bimodules by y¯i;λ((ζi)r)=(ζi)r+1.The Maps∪¯i;λ,∩¯i;λ
Leti∈I+. Define a map of (Cλ,Cλ)-bimodules
(5.14)∪¯i;λ:Cλ→Cλ(i)⨂Cλ+αiCλ(i){1-λi-λi+1}
by
(5.15)∪¯i;λ(1)=∑f=0λi(-1)λi-fζif⊗x(λ)i,λi-f.
Next define a map of(Cλ,Cλ)-bimodules
(5.16)∪¯-i;λ:Cλ→Cλ(-i)⨂Cλ-αiCλ(-i){1-λi-λi+1}
by
(5.17)∪¯-i;λ(1)=∑f=0λi+1(-1)λi+1-fζif⊗x(λ)i+1,λi+1-f.
Next define a map of(Cλ,Cλ)-bimodules
(5.18)∩¯i;λ:Cλ(i)⨂Cλ+αiCλ(i){1-λi-λi+1}→Cλ
by
(5.19)∩¯i;λ(ζir1⊗ζir2)=(-1)r1+r2+1-λi+1x̂(λ)i+1,r1+r2+1-λi+1.
Next define a map of(Cλ,Cλ)-bimodules
(5.20)∩¯-i;λ:Cλ(-i)⨂Cλ-αiCλ(-i){1-λi-λi+1}→Cλ
by
(5.21)∩¯-i;λ(ζir1⊗ζir2)=(-1)r1+r2+1-λix̂(λ)i,r1+r2+1-λi.The Mapsψ¯i,j;λ
Leti,j∈I+. Define a map of (Cλ+αi+αj,Cλ)-bimodules
(5.22)ψ¯i,j;λ:C(λ+αj)(i)⨂Cλ+αjCλ(j)→C(λ+αi)(j)⨂Cλ+αiCλ(i)
by
(5.23)ψ¯i,j;λ(ζir1⊗ζjr2)={ζjr2⊗ζir1if|i-j|>1,∑f=0r1-1ζir1+r2-1-f⊗ζif-∑g=0r2-1ζir1+r2-1-g⊗ζigifj=i,(ζjr2⊗ζir1+1-ζjr2+1⊗ζir1){-1}ifi=j+1,(ζjr2⊗ζir1){1}ifj=i+1.
Define a map of(Cλ-αi-αj,Cλ)-bimodules
(5.24)ψ¯-i,-j;λ:C(λ-αj)(-i)⨂Cλ-αjCλ(-j)→C(λ-αi)(-j)⨂Cλ-αiCλ(-i)
by
(5.25)ψ¯-i,-j(ζir1⊗ζjr2)={ζjr2⊗ζir1if|i-j|>1,∑f=0r2-1ζir1+r2-1-f⊗ζif-∑g=0r1-1ζir1+r2-1-g⊗ζigifj=i,(ζjr2⊗ζir1+1){-1}ifi=j+1,(ζjr2+1⊗ζir1-ζjr2⊗ζir1+1){1}ifj=i+1.


## 5.6. The 2-Morphisms of𝒫k,n

Leti,j∈I+.The Maps1i;λ
Let1i;λ:𝔼i𝕀λ→𝔼i𝕀λ and 1-i;λ:𝔼-i𝕀λ→𝔼-i𝕀λ be the identity morphisms.The Mapsyi;λ
Next we define a morphism of degree2,yi;λ:𝔼i𝕀λ→𝔼i𝕀λ. Recall that
(5.26)𝔼i𝕀λ≅HomCλ+αi(𝕍λ+αi¯Pλ+αi¯,Cλ(i)⨂Cλ𝕍λ¯Pλ¯{ri,λ}).
Let f be such a homomorphism. Suppose that f(m)=γ⊗n. Then set (yi;λ·f)(m)=y¯i(γ)⊗n.
Similarly,(5.27)𝔼-i𝕀λ≅HomCλ-αi(𝕍λ-αi¯Pλ-αi¯,Cλ(-i)⨂Cλ𝕍λ¯Pλ¯{si,λ}).
Let f be such a homomorphism. Suppose that f(m)=γ⊗n. Then set (y-i;λ·f)(m)=y¯-i;λ(γ)⊗n.The Maps∪i;λ,∩i;λ
Note that(5.28)𝕀λ≅J=HomCλ(𝕍λ¯Pλ¯,𝕍λ¯Pλ¯),𝔼-i∘𝔼i𝕀λ≅K=HomCλ(𝕍λ¯Pλ¯,Cλ+αi(-i)⨂Cλ+αiCλ(i)⨂Cλ𝕍λ¯Pλ¯{rλ,i+sλ+αi,i}),𝔼i∘𝔼-i𝕀λ≅L=HomCλ(𝕍λ¯Pλ¯,Cλ-αi(i)⨂Cλ-αiCλ(-i)⨂Cλ𝕍λ¯Pλ¯{sλ,i+rλ-αi,i}).
Letf∈J. Then define ∪i;λ:𝕀λ→𝔼-i𝔼i𝕀λ by
(5.29)∪i;λ(f)(m)=∪¯i;λ(1)⊗f(m)
and ∪-i;λ:𝕀λ→𝔼i𝔼-i𝕀λ by
(5.30)∪-i;λ(f)(m)=∪¯-i;λ(1)⊗f(m).
Now define∩i;λ:𝔼-i𝔼i𝕀λ→𝕀λ. Suppose that f∈K such that f(m)=γ⊗n. Then set ∩i;λ(f)(m)=∩¯i;λ(γ)⊗n.
Next define∩-i;λ:𝔼i𝔼-i𝕀λ→𝕀λ. Suppose that f∈L such that f(m)=γ⊗n. Then set ∩-i;λ(f)(m)=∩¯-i;λ(γ)⊗n.The Mapsψi,j;λ
First we define a mapψi,j;λ:𝔼i𝔼j𝕀λ→𝔼j𝔼i𝕀λ.
Set(5.31)Ji,j+=𝔼i𝔼j𝕀λ≅HomCλ+αi+αj(𝕍λ+αi+αj¯Pλ+αi+αj¯,C(λ+αj)(i)⨂Cλ+αjCλ(j)⨂Cλ𝕍λ¯Pλ¯{rλ,j+rλ+αj,i}),Ki,j+=𝔼j𝔼i𝕀λ≅HomCλ+αj+αi(𝕍λ+αj+αi¯Pλ+αj+αi¯,C(λ+αi)(j)⨂Cλ+αiCλ(i)⨂Cλ𝕍λ¯Pλ¯{rλ,i+rλ+αi,j}).
Let f∈Ji,j+ and suppose that f(m)=γ1⊗γ2⊗n. Then define ψi,j;λf(m)=ψ¯i,j;λ(γ1⊗γ2)⊗n.
Set(5.32)Ji,j-=𝔼-i𝔼-j𝕀λ≅HomCλ-αi-αj(𝕍λ-αi-αj¯Pλ-αi-αj¯,C(λ-αj)(-i)⨂Cλ-αjCλ(-j)⨂Cλ𝕍λ¯Pλ¯{sλ,j+sλ-αj,i}),Ki,j-=𝔼-j𝔼-i𝕀λ≅HomCλ-αj-αi(𝕍λ-αj-αi¯Pλ-αj-αi¯,C(λ-αi)(-j)⨂Cλ-αiCλ(-i)⨂Cλ𝕍λ¯Pλ¯{sλ,i+sλ-αi,j}).
Letf∈Ji,j- and suppose that f(m)=γ1⊗γ2⊗n. Then define ψ-i,-j;λf(m)=ψ¯-i,-j;λ(γ1⊗γ2)⊗n.Theorem 5.4.
There is a 2-functorΠk,n:𝒦ℒ→𝒫k,n such that, for all i,j∈I,  (1)
Πk,n(λ)=ℤ𝒫λ¯(k,k),(2)
Πk,n(ℐλ)=𝕀λ,(3)
Πk,n(ℰiℐλ)=𝔼i𝕀λ,(4)
Πk,n(Yi;λ)=yi;λ,(5)
Πk,n(Ψi,j;λ)=ψi,j;λ,(6)
Πk,n(⋃i;λ)=∪i;λ,(7)
Πk,n(⋂i;λ)=∩i;λ,(8)
Πk,n(1i;λ)=1i;λ.Proof.
This now follows from the computations in [1, Section 6.2] for bimodules over the cohomology of flag varieties using the naturality of the isomorphism in Proposition 5.2.Finally we show that the category𝒫k,n is a categorification of the module V2ωk. Denote the Grothendieck group of 𝒫k,n by [𝒫k,n], and let [𝒫k,n]ℚ(q)=ℂ(q)⨂ℤ[q,q-1][𝒫k,n].Proposition 5.5.
There is an isomorphism of𝒰q(𝔰𝔩n)-modules [𝒫k,n]ℚ(q)≅V2ωk.Proof.
Since projective functors map projective-injective modules to projective-injective modules, it follows from Theorem5.4 and [1] that [𝒫k,n]ℚ(q) is a 𝒰q(𝔰𝔩n)-module. By construction, it contains a highest weight vector of weight 2ωk so it suffices to compute the dimension of its weight spaces.
By [19, Theorem 4.8], the number of projective-injective objects in 𝒪λ¯(k,k)(𝔤𝔩2k) is equal to the number of column decreasing and row nondecreasing tableau for a diagram with k rows and 2 columns with entries from the set
(5.33){n,…,n︸λ1,…,1,…,1︸λn}.
Call the set of such tableau T.
LetS={i∈I+∣λi=1}. Denote by |S| the cardinality of this set. Consider a Young diagram with |S|/2 rows and 2 columns. Let T' denote the set of tableau on such a column with entries from S such that the rows and columns are decreasing. It is well known that the cardinality of the set T' is the Catalan number (2|S||S|)/(|S|+1). There is a bijection between T and T'. For any tableaux t'∈T', one constructs a tableaux t∈T by inserting a new box with the entry i in each column for each i∈I+ such that λi=2. The inverse is given by box removal.
Finally, the Weyl character formula gives that the dimension of theλ weight space of V2ωk is (2|S||S|)/(|S|+1).



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*Source: 892387-2010-07-21.xml*