# A Diagrammatic Temperley-Lieb Categorification

Authors: Ben Elias
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

The monoidal category of Soergel bimodules categorifies the Hecke algebra of a finite Weyl group. In the case of the symmetric group, morphisms in this category can be drawn as graphs in the plane. We define a quotient category, also
given in terms of planar graphs, which categorifies the Temperley-Lieb algebra. Certain ideals appearing in this quotient are related both to the 1-skeleton of the Coxeter complex and to the topology of 2D cobordisms. We demonstrate how further subquotients of this category will categorify the irreducible modules of the Temperley-Lieb algebra.


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## Body

## 1. Introduction

A goal of the categorification theorist is to replace interesting endomorphisms of a vector space with interesting endofunctors of a category. The question is what makes these functors interesting? In the pivotal paper of Chuang and Rouquier [1], a fresh paradigm emerged. They noticed that by specifying structure on the natural transformations (morphisms) between these functors one obtains more useful categorifications (in this case, the added utility is a certain derived equivalence). The categorification of quantum groups by Rouquier [2], Lauda [3], and Khovanov and Lauda [4] has shown that categorifying an algebra A itself (with a category 𝒜) will specify what this additional structure should be for a categorification of any representation of that algebra: a functor from 𝒜 to an endofunctor category. That their categorifications 𝒜 provide the “correct" extra structure is confirmed by the facts that existing geometric categorifications conform to it (see [5]) and that irreducible representations of A can be categorified in this framework (see [6, 7]). The salient feature of these categorifications is that, instead of being defined abstractly, the morphisms are presented by generators and relations, making it straightforward to define functors out of 𝒜.In the case of the Hecke algebraℋ, categorifications have existed for some time, in the guise of category 𝒪 or perverse sheaves on the flag variety. In [8] Soergel rephrased these categorifications in a more combinatorial way, constructing an additive categorification of ℋ by a certain full monoidal subcategory ℋ𝒞 of graded R-bimodules, where R is a polynomial ring. Objects in this full subcategory are called Soergel bimodules. There are deep connections between Soergel bimodules, representation theory, and geometry, and we refer the reader to [8–11] for more details. Categorifications using category 𝒪 and variants thereof are common in the literature, and often Soergel bimodules are used to aid calculations (see, e.g., [12–14]).In [15], Elias and Khovanov provides (in type A) a presentation of ℋ𝒞 by generators and relations, where morphisms can be viewed diagrammatically as decorated graphs in a plane. To be more precise, the diagrammatics are for a smaller category ℋ𝒞1, the (ungraded) category of Bott-Samelson bimodules, described in Section 2.1. Soergel bimodules are obtained from ℋ𝒞1 by taking the graded Karoubi envelope. This is in exact analogy with the procedures of Khovanov and Lauda in [4] and related papers.The Temperley-Lieb algebra𝒯ℒ is a well-known quotient of ℋ, and it can be categorified by a quotient 𝒯ℒ𝒞 of ℋ𝒞, as this paper endeavors to show. Thus, we have a naturally arising categorification by generators and relations, and we expect it to be a useful one. Objects in 𝒯ℒ𝒞 can no longer be viewed as R-bimodules (though their Hom spaces will be R-bimodules), so that diagrammatics provide the simplest way to define the category.The most complicated generator ofℋ𝒞 is killed in the quotient to 𝒯ℒ𝒞, making 𝒯ℒ𝒞 easy to describe diagrammatically in its own right. Take a category where objects are sequences of indices between 1 and n (denoted i̲). Morphisms will be given by (linear combinations of) collections of graphs Γi embedded in ℝ×[0,1], one for each index i∈{1,…,n}, such that the graphs have only trivalent or univalent vertices, and such that Γi and Γi+1 are disjoint. Each graph will have a degree, making Hom spaces into a graded vector space. The intersection of the graphs with ℝ×{0} and ℝ×{1} determines the source and target objects, respectively. Finally, some local graphical relations are imposed on these morphisms. This defines 𝒯ℒ𝒞1, and we take the graded Karoubi envelope to obtain 𝒯ℒ𝒞.The proof that𝒯ℒ𝒞 categorifies 𝒯ℒ uses a method similar to that in [15]. We show first that 𝒯ℒ𝒞1 is a potential categorification of 𝒯ℒ, in the sense described in Section 2.2. Categorifications and potential categorifications define a pairing on 𝒯ℒ given by ([M],[N])=gdimHom𝒯ℒ𝒞1(M,N), the graded dimension which takes values in ℤ[[t,t-1]]. Equivalently, it defines a trace on 𝒯ℒ via ɛ([M])=gdimHom(1,M) where 1 is the monoidal identity (see Section 2.1). The difficult part is to prove the following lemma.Lemma 1.1.
The trace induced on𝒯ℒ from 𝒯ℒ𝒞1 is the map ɛcat defined in Section 2.2.Given this lemma, it is surprisingly easy (see Section3.3) to show the main theorem.Theorem 1.2.
Let𝒯ℒ𝒞2 be the graded additive closure of 𝒯ℒ𝒞1, and let 𝒯ℒ𝒞 be the graded Karoubi envelope of 𝒯ℒ𝒞1. Then 𝒯ℒ𝒞2 is Krull-Schmidt and idempotent closed, so 𝒯ℒ𝒞2≅𝒯ℒ𝒞, and 𝒯ℒ𝒞 categorifies 𝒯ℒ.To prove the lemma, we note that there is a convenient set of elements in𝒯ℒ, the nonrepeating monomials, whose values determine any pairing; hence, there is a convenient set of objects whose Hom spaces will determine all Hom spaces. If i̲ is a nonrepeating sequence, the Hom space we must calculate is (up to shift) a quotient of R by a two-sided ideal Ii̲. We use graphical methods to determine these rings explicitly, giving generators for the ideals in R which define them. As an interesting side note, these ideals also occur elsewhere in nature.Proposition 1.3.
LetV be the reflection representation of Sn+1, and identify R with its coordinate ring. Let Z be the union of all the lines in V which are intersections of reflection-fixed hyperplanes, and let I⊂R be the ideal which gives the reduced scheme structure on Z. Then Hom spaces in 𝒯ℒ𝒞 are R/I-bimodules, and the ideals Ii̲ cut out subvarieties of Z given by lines with certain transverseness properties (see Section 3.7 for details).Also in Section3.7, we give a topological interpretation of the ideals Ii̲, using a functor defined by Vaz [16].Now, let𝒯ℒJi be the parabolic subalgebra of 𝒯ℒ given by ignoring the index i, and let Vi be the induced (right) representation from the sign representation of 𝒯ℒJi. Such an induced representation is useful because it is a quotient of 𝒯ℒ and also contains an irreducible module Li of 𝒯ℒ as a submodule. All irreducibles can be constructed this way.We provide a diagrammatic categorification ofVi as a quotient 𝒱i of 𝒯ℒ𝒞, and a categorification of Li as a full subcategory ℒi of 𝒱i, in a fashion analogous to quantum group categorifications. Having found a diagrammatic categorification 𝒞 of the positive half U+ of the quantum group, Khovanov and Lauda in [17] conjectured that the highest weight modules (naturally quotients of U+) could be categorified by quotients of 𝒞 by the appearance of certain pictures on the left. This approach was proven correct by Lauda and Vazirani [6] (for the U+-module structure), and then used by Webster to categorify tensor products [18]. Similarly, to obtain 𝒱i, we mod out 𝒯ℒ𝒞 by diagrams where any index except i appears on the left. The proof that this works is similar in style to the proof of Theorem 1.2: one calculates the dimension of all Hom spaces by calculating enough Hom spaces to specify a unique pairing on Vi and then uses simple arguments to identify the Grothendieck group.Theorem 1.4.
The category𝒱i is idempotent closed and Krull-Schmidt. Its Grothendieck group is isomorphic to Vi. Letting ℒi be the full subcategory generated by indecomposables which decategorify to elements of Li, one has that ℒi is idempotent closed and Krull-Schmidt, with Grothendieck group isomorphic to Li.A future paper will categorify all representations induced from the sign and trivial representations of parabolic subalgebras ofℋ and 𝒯ℒ. Induced representations were categorified more generally in [13] in the context of category 𝒪, although not diagrammatically. We believe that our categorification should describe what happens in [13] after applying Soergel's functor.Soergel bimodules are intrinsically linked with braids, as was shown by Rouquier in [19, 20], who used them to construct braid group actions (these braid group actions also appear in the category 𝒪 context, see [21]). As such, morphisms between Soergel bimodules should correspond roughly to movies, and the graphs appearing in the diagrammatic presentation of the category ℋ𝒞 should be (heuristically) viewed as 2-dimensional holograms of braid cobordisms. This is studied in [22]. The Temperley-Lieb quotient is associated with the representation theory of Uq(𝔰𝔩2), for which all braids degenerate into 1-manifolds, and braid cobordisms degenerate into surfaces with disorientations. There is a functor ℱ from 𝒯ℒ𝒞 to the category of disorientations constructed by Vaz [16]. The functor ℱ is faithful (though certainly not full), as we remark in Section 3.7. This in turn yields a topological motivation of the variety Z and its subvarieties Z′. Because ℱ is not full, there might be actions of 𝒯ℒ𝒞 that do not extend to actions of disoriented cobordisms. Cobordisms have long been a reasonable candidate for morphisms in Temperley-Lieb categorifications, although we hope 𝒯ℒ𝒞 will provide a useful substitute, with more explicit and computable Hom spaces.Categorification and the Temperley-Lieb algebra have a long history. Khovanov in [23] constructed a categorification of 𝒯ℒ using a TQFT, which was slightly generalized by Bar-Natan in [24]. This was then used to categorify the Jones polynomial. Bernstein et al. in [25] provide a categorical action of the Temperley-Lieb algebra by Zuckerman and projective functors on category 𝒪. Stroppel [14] showed that this categorical action extends to the full tangle algebroid, and also investigated the natural transformations between projective functors. Recent work of Brundan and Stroppel [26] connects these Temperley-Lieb categorifications to Khovanov-Lauda-Rouquier algebras, among other things. We hope that our diagrammatics will help to understand the morphisms in these categorifications.The organization of this paper is as follows. Section2 will provide a quick overview of the Hecke and Temperley-Lieb algebras, and the diagrammatic definition of the  category ℋ𝒞. Section 3 begins by defining the quotient category diagrammatically in its own right (which makes a thorough understanding of the diagrammatic calculus for ℋ𝒞 unnecessary). Section 3.3 proves Theorem 1.2, modulo Lemma 1.1 which requires all the work. The remaining sections of that chapter do all the work, and starting with Section 3.6 one will not miss any important ideas if one skips the proofs. Section 4 begins with a discussion of cell modules for 𝒯ℒ and certain other modules, and then goes on to categorify these modules, requiring only very simple diagrammatic arguments.This paper is reasonably self-contained. We do not require familiarity with [15] and do not use any results other than Corollary 2.20. We do quote some results for motivational reasons, but the difficult graphical arguments of that paper can often be drastically simplified for the Temperley-Lieb setting, so that we provide easier proofs for the results we need. Familiarity with diagrammatics for monoidal categories with adjunction would be useful, and [3] provides a good introduction. More details on preliminary topics can be found in [15].


## 2. Preliminaries

Notation 1.
Fixn∈ℕ, and let I=1,…,n index the vertices of the Dynkin diagram An. We use the word index for an element of I, and the letters i,j always represent indices. Indices i≠j are adjacent if |i-j|=1, and distant if |i-j|≥2, and questions of adjacency always refer to the Dynkin diagram, not the position of indices in a word or picture.Notation 2.
LetW=Sn+1 with simple reflections si=(i,i+1). Let 𝕜 be a field of characteristic not dividing 2(n+1); all vector spaces will be over this field. Let R=𝕜[x1,…,xn+1]/e1, where e1=x1+x2+⋯+xn+1; it is a graded ring, with deg(xi)=2. We will abuse notation and refer to elements of 𝕜[x1,…,xn+1] and their images in R in the same way, and will refer to both as polynomials. Note that R=𝕜[f1,…,fn], where fi=xi-xi+1, since x1=(nf1+(n-1)f2+⋯+fn)/(n+1) modulo e1. The ring R arises as the coordinate ring of V, the reflection representation of W (the span of the root system), and fi are the simple coroots.
There is an obvious action ofSn+1 on R, which permutes the generators xi. For each index we have a Demazure operator∂i, a map of degree -2 from R to the invariant subring Rsi, which is Rsi-linear and sends Rsi to 0. Explicitly, ∂i(f)=(f-si(f))/(xi-xi+1).Notation 3.
Let(·)¯ be the ℤ-linear involution of ℤ[t,t-1] switching t and t-1. Given a ℤ-linear map β of ℤ[t,t-1] modules, we call it antilinear if it is ℤ[t,t-1]-linear after twisting by (·)¯, or in other words if β(tm)=t-1β(m). We write [2]=deft+t-1.
LetA be a ℤ[t,t-1]-algebra. In this paper we always use the word trace to designate a ℤ[t,t-1]-linear map ɛ:A→ℤ[[t,t-1]] satisfying ɛ(xy)=ɛ(yx). We use the word pairing or semilinear pairing to denote a ℤ-linear map A×A→ℤ[[t,t-1]] which is ℤ[t,t-1]-linear in the second factor and ℤ[t,t-1]-antilinear in the first factor.


### 2.1. The Hecke Algebra and the Soergel Categorification

We state here without proof a number of basic facts about the Hecke algebra, its traces, and Soergel's categorification. For more background, see Soergel's original definition of his categorification [8], or an easier version [11]. A similar overview with more discussion can be found in [15]. A more in-depth introduction, connecting Soergel bimodules to other parts of representation theory, can be found in [13].Definition 2.1.
Denote byℋ the Hecke algebra for Sn+1. It is a ℤ[t,t-1]-algebra, specified here by its Kazhdan-Lusztig presentation: it has generators bi,i∈I and relations
(2.1)bi2=(t+t-1)bi(2.2)bibj=bjbifordistanti,j(2.3)bibjbi+bj=bjbibj+biforadjacenti,j.Definition 2.2.
Given two objects in a graded𝕜-linear (possibly additive) category 𝒞, where {1} denotes the grading shift, the graded hom space between them is the graded vector space HOM(M,N)=⨁n∈ℤHom𝒞(M,N{n}). Given a class of objects {Mα} in 𝒞, we can define a category with morphisms enriched in graded vector spaces, whose objects are {Mα} and whose morphisms are HOM(Mα,Mβ). Let us call this an enriched full subcategory, which we often shorten to the adjective enriched. While the enriched subcategory is neither additive nor graded, it has enough information to recover the hom spaces between grading shifts and direct sums of objects Mα in 𝒞.
LetR-bim denote the category of finitely-generated graded (resp., ungraded) R-bimodules. Then HOM spaces in R-bim will be graded R-bimodules. For i∈I, let Bi∈R-bim be defined by Bi=R⨂RsiR{-1}, where Rsi is the invariant subring. A Bott-Samelson bimodule is a tensor product Bi1⊗Bi2⊗⋯⊗Bid in R-bim, where here and henceforth ⊗ denotes the tensor product over R. Let ℋ𝒞1 be the enriched full subcategory generated by the Bott-Samelson bimodules; it is a monoidal category, but is neither additive nor graded. Let ℋ𝒞2 denote the full subcategory of R-bim given by all (finite) direct sums of grading shifts of Bott-Samelson bimodules; it is monoidal, additive, and graded. Finally, let ℋ𝒞 denote the category of Soergel bimodules or special bimodules, the full subcategory of R-bim given by all (finite) direct sums of grading shifts of summands of Bott-Samelson bimodules; it is monoidal, additive, graded, and idempotent closed.One can observe that all bimodules inℋ𝒞 are free and finitely generated when viewed as either left R-modules or right R-modules, and therefore the same is true of any HOM space. The following proposition parallels the Kazhdan-Lusztig presentation for ℋ.Proposition 2.3.
The categoryℋ𝒞2 is generated (as an additive, monoidal category) by objects Bi,i∈I which satisfy
(2.4)Bi⊗Bi≅Bi{1}⊕Bi{-1},(2.5)Bi⊗Bj≅Bj⊗Bifordistanti,j,(2.6)Bi⊗Bj⊗Bi⊕Bj≅Bj⊗Bi⊗Bj⊕Biforadjacenti,j.From this we might expect the next result.Proposition 2.4.
The Grothendieck ring[ℋ𝒞2] of ℋ𝒞2 is isomorphic to ℋ, with [Bi] being sent to bi, and [R{1}] being sent to t. The Grothendieck ring [ℋ𝒞] of ℋ𝒞 is isomorphic to ℋ as well.Remark 2.5.
The proof of this statement is not immediately obvious. There is clearly a surjective morphism fromℋ to [ℋ𝒞2]. When one takes the idempotent closure of a category, one adds new indecomposables and can potentially enlarge the Grothendieck group. Soergel showed, via a support filtration, that all the new indecomposables in ℋ𝒞 have symbols in [ℋ𝒞] which can be reached from certain symbols in [ℋ𝒞2] by a unitriangular matrix (see [11]). Therefore, the Grothendieck rings of ℋ𝒞 and ℋ𝒞2 are equal. Since ℋ𝒞 is idempotent closed and is embedded in R-bim, it has the Krull-Schmidt property and the Grothendieck group behaves as one would expect: it has a basis given by indecomposables. By classifying indecomposables and using the unitriangular matrix, Soergel showed that the map from ℋ to [ℋ𝒞2] is actually an isomorphism.
It is important to note that one does not know what the image of the indecomposables ofℋ𝒞 in ℋ is. The Soergel conjecture, still unproven in generality, proposes that the indecomposables of ℋ𝒞 descend to the Kazhdan-Lusztig basis of ℋ (see [11]).Notation 4.
We write the monomialbi1bi2⋯bid∈ℋ as bi̲, where i̲=i1⋯id is a finite sequence of indices; by abuse of notation, we sometimes refer to this monomial simply as i̲. If i̲ is as above, we say the monomial has length d=d(i̲). We call a monomial nonrepeating if ik≠il for k≠l, and increasing if i1<i2<⋯. The empty set is a sequence of length 0, and b∅=1. Similarly, in ℋ𝒞1, write Bi1⊗⋯⊗Bid as Bi̲. Note that B∅=R, the monoidal identity. For an arbitrary index i and sequence i̲, we write i∈i̲ if i appears in i̲.Given two objectsM,N∈R-bim we say they are biadjoint if M⊗- and N⊗- are left and right adjoints of each other, and the same for -⊗M and -⊗N. If M and N are biadjoint, so are M{1} and N{-1}. We often want to specify additional compatibility between various adjunction maps, but we pass over the details here (see [3] for more information on biadjunction).Proposition 2.6.
Each object inℋ𝒞 (resp., ℋ𝒞1, ℋ𝒞2) has a biadjoint, and Bi is self-biadjoint. Let ω be the t-antilinear anti-involution on ℋ which fixes bi, that is, ω(tabi̲)=t-abσ(i̲), where σ reverses the order of a sequence. There is a contravariant functor on ℋ𝒞 sending an object to its biadjoint, and it descends on the Grothendieck ring to ω.Definition 2.7.
Anadjoint pairing on ℋ is a pairing where each bi is self-adjoint, so that (x,biy)=(bix,y) and (x,ybi)=(xbi,y) for all x,y∈ℋ and all i∈I. Equivalently, for any m∈ℋ, (mx,y)=(x,ω(m)y) and (xm,y)=(x,yω(m)).There is a bijection between adjoint pairings(,) and traces ɛ, defined by letting (x,y)=ɛ(ω(x)y), or conversely ɛ(y)=(1,y). Adjoint pairings appear often in the literature, for instance [27] (although they are usually ℤ[t,t-1]-linear in both factors, unlike our current semilinear definition). Semilinear adjoint pairings will be crucially important, due to the following remark.Remark 2.8.
Let𝒞 be a monoidal category with objects Bi, such that Bi are self-biadjoint. We assume that 𝒞 is additive and graded and has isomorphisms (2.4)–(2.6). We call such a category a potential categorification of ℋ. In this case, there is a map of rings from ℋ to [𝒞] sending bi to [Bi], and (under suitable finite-dimensionality conditions) we get an adjoint semilinear pairing on ℋ via (bi̲,bj̲)=gdimHOM𝒞(Bi̲,Bj̲)∈ℤ[[t,t-1]], the graded dimension as a vector space. Denote the pairing and its associated trace map as (,)𝒞 and ɛ𝒞.
Instead, we may assume𝒞 is an enriched monoidal subcategory, containing objects Bi. The isomorphisms (2.4)–(2.6) typically have no meaning in this context, since there are no grading shifts or direct sums, but we can require that they Yoneda-hold; that is, they hold after the application of any Hom(-,X) functor (to graded vector spaces). There is no definition of a Grothendieck ring in this case, but we still get an induced adjoint semilinear pairing induced by Hom spaces. We call this an enriched potential categorification.We may use pairings to distinguish between different potential categorifications. The next proposition allows us to specify the pairing induced by a categorification by only investigating certainHOM spaces.Proposition 2.9.
Traces onℋ are uniquely determined by their values ɛ(bi̲) on increasing monomials i̲. Equivalently, adjoint pairings are determined by (1,bi̲) for increasing i̲. If i̲ is nonrepeating and j̲ is a permutation of i̲, then ɛ(bi̲)=ɛ(bj̲).We quickly sketch the proof. Moving an index from the beginning of a monomial to the end, or vice versa, will be calledcycling the monomial. It is clear, using biadjointness or the definition of trace, that the value of ɛ is invariant under cycling. It is not difficult to show that any monomial in W (in the letters si) will reduce, using the Coxeter relations and cycling, to an increasing monomial. When the monomial is already nonrepeating, one needs only use cycling and sisj=sjsi for distant i,j. Finally, using induction on the length of the monomial, the same principle shows that any monomial in ℋ reduces to a linear combination of increasing monomials, and therefore ɛ is determined by these.The upshot is that, given a potential categorification, one knows the dimension of allHOM(Bi̲,Bj̲) so long as one knows the dimension of HOM(B∅,Bi̲) for increasing i̲. Note that not every choice of (1,bi̲) for all increasing i̲ will yield a well-defined trace map.Consider the adjoint pairing given byɛstd(bi̲)=(1,bi̲)=td for nonrepeating i̲ of length d. This is the semilinear version of the pairing found in [27] which picks out the coefficient of the identity in the standard basis of ℋ and is called the standard pairing. Soergel showed that HOM(Bi̲,Bj̲) is a free graded left (or right) R-module of rank (bi̲,bj̲) using this pairing. In particular, for increasing i̲, HOM(R,Bi̲) is generated by a single element in degree d(i̲). Since the graded dimension of R is 1/(1-t2)n we have that (1,bi̲)ℋ𝒞=td/(1-t2)n is a rescaling of the standard pairing.Now letɛ be the quotient map ℋ→ℤ[t,t-1] by the ideal generated by all bi. It is a homomorphism to a commutative algebra, so it is a trace. The corresponding pairing satisfies (1,1)=1 and (x,y)=0 for monomials x,y if either monomial is not 1. We call this the trivial pairing, ɛtriv.


### 2.2. The Temperley-Lieb Algebra

Here again we state without proof some basic facts about Temperley-Lieb algebras. They were originally defined by Temperley and Lieb in [28], and were given a topological interpretation by Kauffman [29]. There are many good expositions for the topic, such as [30, 31].Definition 2.10.
TheTemperley-Lieb algebra𝒯ℒ is the ℤ[t,t-1]-algebra generated by ui,i∈I with relations
(2.7)ui2=[2]ui,(2.8)uiuj=ujuifor|i-j|≥2,(2.9)uiujui=uiforadjacenti,j.Proposition 2.11.
For adjacenti,j∈I, consider the element of ℋ defined by cij=defbibjbi-bi=bjbibj-bj, where the equality arises from relation (2.3). There is a surjective map ℋ→𝒯ℒ sending bi to ui for all i∈I, and whose kernel is generated by cij for adjacent i,j∈I.Once again, writeui̲ for a monomial in the above generators, with all the same conventions as before. The map ω descends from ℋ to 𝒯ℒ, and we define an adjoint pairing on 𝒯ℒ in the same way, with ui replacing bi everywhere. The results of Proposition 2.9 apply equally to 𝒯ℒ.Definition 2.12.
A category𝒞 as in Remark 2.8 is a potential categorification of 𝒯ℒ if it has objects Ui satisfying
(2.10)Ui⊗Ui≅Ui{1}⊕Ui{-1},Ui⊗Uj≅Uj⊗Uifordistanti,j,Ui⊗Uj⊗Ui≅Uiforadjacenti,j.
We call it anenriched potential categorification if it is an enriched category with objects Ui such that these isomorphisms Yoneda-hold.A permutationσ∈Sn+1 is called 321-avoiding if it never happens that, for i<j<k, σ(i)>σ(j)>σ(k). It turns out that, using the Temperley-Lieb relations, every monomial uj̲ is equal to a scalar times some ui̲ where i̲ is 321-avoiding; that is, if viewed as a word in the symmetric group, it represents a reduced expression for a 321-avoiding permutation. Moreover, between 321-avoiding monomials, the only further relations come from (2.8), and hence it is easy to pick out a basis from this spanning set. See [30] for more details.The Temperley-Lieb algebra has a well-known topological interpretation where an element of𝒯ℒ is a linear combination of crossingless matchings (isotopy classes of embedded planar 1-manifolds) between n+1 bottom points and n+1 top points. Multiplication of crossingless matchings consists of vertical concatenation (where ab is a above b), followed by removing any circles and replacing them with a factor of [2]. In this picture, ui becomes the following:(2.11)The basis of 321-avoiding monomials agrees with the basis of crossingless matchings. Any increasing monomial is 321-avoiding. Increasing monomials are easy to visualize topologically, as they have only “right waves" and “simple cups and caps." For example:(2.12)As an example of a monomial which is not increasing:(2.13)Given a crossingless matching, itsclosure is a configuration of circles in the punctured plane obtained by wrapping the top boundary around the puncture to close up with the bottom boundary, as in Figure 1. Circle configurations have two topological invariants: the number of circles and the nesting number which is the number of circles which surround the puncture and is equal to n+1-2l≥0 for some l≥0. Given a scaling factor for each possible nesting number, one constructs a trace by letting ɛ(ui̲)=ck[2]m where m is the number of circles in the closure of ui̲ and ck is the scaling factor associated to its nesting number k. To calculate (x,y), we place y below an upside-down copy of x (or vice versa), and then take the closure. All pairings/traces on 𝒯ℒ can be constructed this way, so they are all topological in nature.Figure 1
An example of the closure of a crossingless matching.Figure 2
An example of a planar graph in the strip, with colored edges.Figure 3
An example of tree reduction.Figure 4
An arbitrary innermost blue cycle. The dotted line encapsulates the subgraph on the interior which may contain colors adjacent to blue.The Temperley-Lieb algebra has astandard pairing of its own for which ck=1 for all nesting numbers k: ɛstd(ui̲)=[2]m as above. One can check that ɛstd(ui̲)=[2]n+1-d(i̲) for an increasing monomial. This is not related to the standard pairing on ℋ, which does not descend to 𝒯ℒ. On the other hand, ɛtriv clearly does descend to a pairing trivial pairing on 𝒯ℒ, which only evaluates to a nonzero number when the nesting number is n+1.It turns out that the pairing on𝒯ℒ arising from our categorification will satisfy (1,1)=(tn/(1-t2))[2]n-(t2/(1-t2)) and (1,ui̲)=(tn/(1-t2))[2]n-d. We will call the associated trace ɛcat. Clearly ɛcat=(tn/(1-t2)[2])ɛstd-(t2/(1-t2))ɛtriv. In particular, on any monomial x≠1, our trace will agree with a rescaling of the standard trace. When n=1, the algebras 𝒯ℒ and ℋ are already isomorphic, and ɛcat agrees with the rescaling of the standard trace on ℋ discussed at the end of Section 2.1.


### 2.3. Definition of Soergel Diagrammatics

We now give a diagrammatic description of the categoryℋ𝒞1, as discovered in [15]. Since the category to be defined will be equivalent to the category of Bott-Samelson bimodules, we will abuse notation temporarily and use the same names.Definition 2.13.
In this paper, aplanar graph in the strip is a finite graph with boundary (Γ,∂Γ) embedded in (ℝ×[0,1],ℝ×{0,1}). In other words, all vertices of Γ occur in the interior ℝ×(0,1), and removing the vertices, we have a 1-manifold with boundary whose intersection with ℝ×{0,1} is precisely its boundary. This allows for edges which connect two vertices, edges which connect a vertex to the boundary, edges which connect two points on the boundary, and edges which form circles (closed 1-manifolds embedded in the plane).
We generally refer toℝ×{0,1} as the boundary, which consists of two components, the top boundaryℝ×{1}, and the bottom boundaryℝ×{0}. We refer to a local segment of an edge which hits the boundary as a boundary edge; there is one boundary edge for each point on the boundary of the graph. We use the word component to mean a connected component of a graph with boundary.
This definition clearly extends to other subsets of the plane with boundary, so that we can speak of planar graphs in a disk or planar graphs in an annulus. The annulus has two boundary components,inner and outer. When we do not specify, we always mean a planar graph in the strip.We will be drawing morphisms inℋ𝒞1 as planar graphs with edges labelled in I. Instead of putting labels everywhere, we color the edges, assigning a color to each index in I. Henceforth, we use the term “color" and “index" interchangeably.We now defineℋ𝒞1 anew. Let ℋ𝒞1 be the monoidal category, with hom spaces enriched over graded vector spaces, which is defined as follows.Definition 2.14.
An object inℋ𝒞1 is given by a sequence of indices i̲, which is visualized as d points on the real line ℝ, labelled or “colored” by the indices in order from left to right. These objects are also called Bi̲. The monoidal structure on objects is concatenation of sequences.Definition 2.15.
Consider the set ofisotopy classes of planar graphs in the strip whose edges are colored by indices in I such that only four types of vertices exist: univalent vertices or “dots”, trivalent vertices with all three adjoining edges of the same color, 4-valent vertices whose adjoining edges alternate in colors between distant i and j, and 6-valent vertices whose adjoining edges alternate between adjacent i and j. This set has a grading, where the degree of a graph is +1 for each dot and -1 for each trivalent vertex; 4-valent and 6-valent vertices are of degree 0. The allowable vertices, which we call “generators," are pictured here:
(2.14)(2.15)
The intersection of a graph with the boundary yields two sequences of colored points onℝ, the top boundary i̲ and the bottom boundary j̲. In this case, the graph is viewed as a morphism from j̲ to i̲. For instance, if “blue" corresponds to the index i and “red" to j, then the lower right generator is a degree 0 morphism from jij to iji. Although this paper is easiest to read in color, it should be readable in black and white: the colors appearing are typically either blue, red, green, or miscellaneous and irrelevant. We throughout use the convention that blue (the darker color) is always adjacent to red (the middle color) and distant from green (the lighter color).
We letHomℋ𝒞1(Bi̲,Bj̲) be the graded vector space with basis given by planar graphs as above which have the correct top and bottom boundary, modulo relations (2.16) through (2.30). As usual in a diagrammatic category, composition of morphisms is given by vertical concatenation (read from bottom to top), the monoidal structure is given by horizontal concatenation, and relations are to be interpreted monoidally (i.e., they may be applied locally inside any other planar diagram).The relations are given in terms of colored graphs, but with no explicit assignment of indices to colors. They hold forany assignment of indices to colors, so long as certain adjacency conditions hold. We will specify adjacency for all pictures, although one can generally deduce it from the fact that 6-valent vertices only join adjacent colors, and 4-valent vertices only join distant colors.For example, these first four relations hold, with blue representing a generic index.(2.16)(2.17)(2.18)(2.19)We will repeatedly call a picture looking like (2.18) by the name “needle." Note that a needle is not necessarily zero if there is something in the interior. Note that a circle is just a needle with a dot attached, by (2.17), so that an empty circle evaluates to 0.Remark 2.16.
It is an immediate consequence of relations (2.16) and (2.17) that any tree (connected graph with boundary without cycles) of one color is equal to  (i)
if it has no boundary, two dots connected by an edge. Call the entire component adouble dot.(ii)
if it has one boundary edge, a single dot connected by the edge to the boundary. Call the component aboundary dot.(iii)
if it has more boundary edges, a tree with no dots and the fewest possible number of trivalent vertices needed to connect the boundaries. Moreover, any two such trees are equal. Call the component asimple tree.We refer to this as tree reduction.
This applies only to components of a graph which are a single color. Even if the blue part of a graph looks like a tree, if other colors overlap, then we may not apply tree reduction in general.In the following relations, the two colors are distant(2.20)(2.21)(2.22)(2.23)In this relation, two colors are adjacent, and both distant to the third color.(2.24)In this relation, all three colors are mutually distant.(2.25)Remark 2.17.
Relations (2.20) through (2.25) indicate that any part of the graph colored i and any part of the graph colored j “do not interact” for distant i and j, that is, one may visualize sliding the j-colored part past the i-colored part, and it will not change the morphism. We call this the distant sliding property.In the following relations, the two colors are adjacent.(2.26)(2.27)(2.28)(2.29)In this final relation, the colors have the same adjacency as{1,2,3}(2.30)This concludes the list of relations definingℋ𝒞1.Remark 2.18.
We chose here to describeℋ𝒞1 in terms of planar graphs with relations, with the notion of isotopy built-in, rather than in terms of generators and relations. Note, however, that using isotopy and (2.17), we get  Therefore, all “cups" and “caps" can be expressed in terms of the generators. By adding new relations corresponding to isotopy, one could give a presentation of the category where the “generators" above (and their isotopy twists) are really generators. This is how the category is presented in [15].We will occasionally use a shorthand to represent double dots. We identify a double dot coloredi with the polynomial fi∈R, and for a linear combination of disjoint unions of double dots in the same region of a graph, we associate the appropriate linear combination of products of fi. For any polynomial f∈R, a square box with a polynomial f in a region will represent the corresponding linear combination of graphs with double dots.For instance,Relations (2.19), (2.29), and (2.23) are referred to as dot forcing rules, because they describe at what price one can “force" a double dot to the other side of a line. The three relations imply that, given a line and an arbitrary collection of double dots on the left side of that line, one can express the morphism as a sum of diagrams where all double dots are on the right side, or where the line is “broken" (as illustrated next). Rephrasing this, for any polynomial f there exist polynomials g and h such that(2.31)The polynomials appearing can in fact be found using the Demazure operator∂i, and in particular, h=∂i(f). One particular implication is that  (2.32)wheneverf is a polynomial invariant under si (and blue represents i). As an exercise, the reader can check that fi2 slides through a line colored i. These polynomial relations are easy to deduce, or one can refer to [15] (see page 7, pages 16-17, and relation 3.16).We have an bimodule action ofR on morphisms by placing boxes (i.e., double dots) in the leftmost or rightmost regions of a graph. Now we can formulate the main result of [15].Theorem 2.19.
There is a functor from this diagrammatic categoryℋ𝒞1 to the earlier definition in terms of Bott-Samelson bimodules. This functor sends i̲ to the bimodule Bi̲ and a planar graph to a map of bimodules, preserving the grading and the R-bimodule action on morphisms. This functor is an equivalence of categories.Corollary 2.20.
TheR-bimodules Homℋ𝒞1(Bi̲,Bj̲) are free as left (or right) R-modules. In other words, placing double dots to the left of a graph is a torsion-free operation.Now, we have justified our abuse of notation. In this paper, we will never need to know explicitly what map ofR-bimodules a planar graph corresponds to, so the interested reader can see [15] for details. In fact, we will not use Theorem 2.19 at all, preferring to work entirely with planar graphs. However, we do use Corollary 2.20, a fact which would be difficult to prove diagrammatically.The proof of Theorem2.19 can be quickly summarized: first, one explicitly constructs a functor from the diagrammatic category to the Bott-Samelson category. Then, using the observations of the next section, one shows that the diagrammatic category is a potential categorification of ℋ and that the diagrammatic category, the Bott-Samelson category, and the image of the former in the latter all induce the same adjoint pairing on ℋ. Therefore, the functor is fully faithful.


### 2.4. Understanding Soergel Diagrammatics

Let us explain diagrammatically why the categoryℋ𝒞1 is a potential categorification of ℋ, and induces the aforementioned adjoint pairing.Definition 2.21.
Given a category𝒞 whose morphism spaces are ℤ-modules, one may take its additive closure, which formally adds direct sums of objects and yields an additive category. Given 𝒞 whose morphism spaces are graded ℤ-modules, one may take its grading closure which formally adds shifts of objects, but restricts morphisms to be homogeneous of degree 0. Given 𝒞 an additive category, one may take the idempotent completion or Karoubi envelope, which formally adds direct summands. Recall that the Karoubi envelope has as objects pairs (B,e) where B is an object in 𝒞 and e an idempotent endomorphism of B. This object acts as though it were the “image” of this projection e and behaves like a direct summand. When taking the Karoubi envelope of a graded category (or a category with graded morphisms) one restricts to homogeneous degree 0 idempotents. We refer in this paper to the entire process which takes a category 𝒞, whose morphism spaces are graded ℤ-modules, and returns the Karoubi envelope of its additive and grading closure as taking the graded Karoubi envelope. All these transformations interact nicely with monoidal structures. For more information on Karoubi envelopes see [32].
We letℋ𝒞2 be the graded additive closure of ℋ𝒞1, and let ℋ𝒞 be the graded Karoubi envelope of ℋ𝒞1.We wish to show that the isomorphisms (2.4) through (2.6) hold in ℋ𝒞2. Relation (2.20) immediately implies that Bi⊗Bj≅Bj⊗Bi for i,j distant, with the isomorphism being given by the 4-valent vertex.We have the following equality:(2.33)To obtain this, use (2.17) to stretch two dots from the two lines into the middle, and then use (2.19) to connect them. The identity idii decomposes as a sum of two orthogonal idempotents, each of which is the composition of a “projection" and an “inclusion" map of degree ±1, to and from Bi (explicitly, idii=i1p1+i2p2 where p1i1=idi, p2i2=idi, p1i2=0=p2i1). This implies that Bi⊗Bi≅Bi{1}⊕Bi{-1} and is a typical example of how direct sum decompositions work in diagrammatic categories.Similarly, the two color variants of relation (2.27) together express the direct sum decompositions in the Karoubi envelope(2.34)Bi⊗Bi+1⊗Bi=Cij⊕BiBi+1⊗Bi⊗Bi+1=Cji⊕Bi+1.
Again, the identity idi(i+1)i is decomposed into orthogonal idempotents. The second idempotent factors through Bi, and the corresponding object in the Karoubi envelope will be isomorphic to Bi. The first idempotent, which we call a “doubled 6-valent vertex," corresponds to a new object Cij in the idempotent completion. It turns out that the doubled 6-valent vertex Cij for “blue red blue" is isomorphic in the Karoubi envelope to the doubled 6-valent vertex Cji for “red blue red" (i.e., their images are isomorphic). We may abuse notation and call both of these new objects Cij; it is a summand of both i(i+1)i and (i+1)i(i+1). The image of Cij in the Grothendieck group is cij.We can also understand the induced pairing onℋ using diagrammatic arguments. The theorems below are proven in [15], and we will not use them in this paper (except motivationally), proving their analogs in the Temperley-Lieb case directly.Theorem 2.22 (Color Reduction).
Consider a morphismφ:∅→i̲, and suppose that the index i (blue) appears in i̲ zero times (resp.,: once). Then φ is in the -span of graphs which only contain blue in the form of double dots in the leftmost region of the graph (resp., as well as a single boundary dot). This result may be obtained simultaneously for multiple indices i.Corollary 2.23.
The spaceHomℋ𝒞1(∅,∅) is precisely the graded ring R. In other words, it is freely generated (over double dots) by the empty diagram. The space Homℋ𝒞1(∅,i̲) for i̲ nonrepeating is a free left (or right) R-module of rank 1, generated by the following morphism of degree d(i̲).
(2.35)The proof of the theorem does not use any sophisticated technology, only convoluted pictorial arguments. It comprises the bulk of [15]. The corollary implies that ɛℋ𝒞1(bi̲)=td/(1-t2)n for  nonrepeating i̲ of length d, as stated in Section 2.1.


### 2.5. Aside from Karoubi Envelopes and Quotients

Return to the setup of Definition2.21. If 𝒞 is a full subcategory of (graded) R-bimodules for some ring R, then the transformations described above behave as one would expect them to. In particular, the Karoubi envelope agrees with the full subcategory which includes all summands of the previous objects. The Grothendieck group of the Karoubi envelope is in some sense “under control" if one understands indecomposable R-bimodules already. On the other hand, the Karoubi envelope of an arbitrary additive category may be enormous, and to control the size of its Grothendieck group, one should understand and classify all idempotents in the category, a serious task. Also, arbitrary additive categories need not have the Krull-Schmidt property, making their Grothendieck groups even more complicated.The Temperley-Lieb algebra is obtained from the Hecke algebra by setting the elementscij to zero, for i=1,…,n-1. These elements lift in the Soergel categorification to objects Cij. The obvious way one might hope to categorify 𝒯ℒ would be to take the quotient of the category ℋ𝒞 by each object Cij.To mod out an additivemonoidal category 𝒞 by an object Z, one must kill the monoidal ideal of idZ in Mor(𝒞), that is, the morphism space Hom(X,Y) in the quotient category is exactly Hom𝒞(X,Y) modulo the submodule of morphisms factoring through V⊗Z⊗W for any V,W. If the category is drawn diagrammatically, one needs to only kill any diagram which has idZ as a subdiagram.We have not truly drawnℋ𝒞 diagrammatically, only ℋ𝒞1. The object we wish to kill is not an object in ℋ𝒞1; the closest thing we have is the corresponding idempotent, the doubled 6-valent vertex. However, this is not truly a problem, due to the following proposition, whose proof we leave to the reader.Proposition 2.24.
Let𝒞1 be an additive category, and let B be an object in 𝒞1, and let e be an idempotent in End(B). Let 𝒟1 be the quotient of 𝒞1 by the morphism e. Let 𝒞 and 𝒟 be the respective Karoubi envelopes. Finally, let 𝒟′ be the quotient of 𝒞 by the identity of the object (B,e). Then, there is a natural equivalence of categories from 𝒟 to 𝒟′.
The analogous statement holds when one considers graded Karoubi envelopes.Remark 2.25.
Note that𝒟′ has more objects than 𝒟, but they are still equivalent. For instance, (B,e) and (B,0) are distinct (isomorphic) objects in 𝒟′, but are the same object in 𝒟.So to categorify𝒯ℒ, one might wish to take the quotient of ℋ𝒞1 by the doubled 6-valent vertex, and then take the Karoubi envelope. This is easy to do diagrammatically, which is one advantage to the diagrammatic approach over the R-bimodule approach. The quotient of ℋ𝒞1 will no longer be a category which embeds nicely as a full subcategory of bimodules. One might worry that Krull-Schmidt fails, or that to understand its Karoubi envelope one must classify all idempotents therein. Thankfully, our calculation of HOM spaces will imply easily that its graded additive closure is Krull-Schmidt and is already idempotent closed, so it is equivalent to its own Karoubi envelope (see Section 3.3).


## 2.1. The Hecke Algebra and the Soergel Categorification

We state here without proof a number of basic facts about the Hecke algebra, its traces, and Soergel's categorification. For more background, see Soergel's original definition of his categorification [8], or an easier version [11]. A similar overview with more discussion can be found in [15]. A more in-depth introduction, connecting Soergel bimodules to other parts of representation theory, can be found in [13].Definition 2.1.
Denote byℋ the Hecke algebra for Sn+1. It is a ℤ[t,t-1]-algebra, specified here by its Kazhdan-Lusztig presentation: it has generators bi,i∈I and relations
(2.1)bi2=(t+t-1)bi(2.2)bibj=bjbifordistanti,j(2.3)bibjbi+bj=bjbibj+biforadjacenti,j.Definition 2.2.
Given two objects in a graded𝕜-linear (possibly additive) category 𝒞, where {1} denotes the grading shift, the graded hom space between them is the graded vector space HOM(M,N)=⨁n∈ℤHom𝒞(M,N{n}). Given a class of objects {Mα} in 𝒞, we can define a category with morphisms enriched in graded vector spaces, whose objects are {Mα} and whose morphisms are HOM(Mα,Mβ). Let us call this an enriched full subcategory, which we often shorten to the adjective enriched. While the enriched subcategory is neither additive nor graded, it has enough information to recover the hom spaces between grading shifts and direct sums of objects Mα in 𝒞.
LetR-bim denote the category of finitely-generated graded (resp., ungraded) R-bimodules. Then HOM spaces in R-bim will be graded R-bimodules. For i∈I, let Bi∈R-bim be defined by Bi=R⨂RsiR{-1}, where Rsi is the invariant subring. A Bott-Samelson bimodule is a tensor product Bi1⊗Bi2⊗⋯⊗Bid in R-bim, where here and henceforth ⊗ denotes the tensor product over R. Let ℋ𝒞1 be the enriched full subcategory generated by the Bott-Samelson bimodules; it is a monoidal category, but is neither additive nor graded. Let ℋ𝒞2 denote the full subcategory of R-bim given by all (finite) direct sums of grading shifts of Bott-Samelson bimodules; it is monoidal, additive, and graded. Finally, let ℋ𝒞 denote the category of Soergel bimodules or special bimodules, the full subcategory of R-bim given by all (finite) direct sums of grading shifts of summands of Bott-Samelson bimodules; it is monoidal, additive, graded, and idempotent closed.One can observe that all bimodules inℋ𝒞 are free and finitely generated when viewed as either left R-modules or right R-modules, and therefore the same is true of any HOM space. The following proposition parallels the Kazhdan-Lusztig presentation for ℋ.Proposition 2.3.
The categoryℋ𝒞2 is generated (as an additive, monoidal category) by objects Bi,i∈I which satisfy
(2.4)Bi⊗Bi≅Bi{1}⊕Bi{-1},(2.5)Bi⊗Bj≅Bj⊗Bifordistanti,j,(2.6)Bi⊗Bj⊗Bi⊕Bj≅Bj⊗Bi⊗Bj⊕Biforadjacenti,j.From this we might expect the next result.Proposition 2.4.
The Grothendieck ring[ℋ𝒞2] of ℋ𝒞2 is isomorphic to ℋ, with [Bi] being sent to bi, and [R{1}] being sent to t. The Grothendieck ring [ℋ𝒞] of ℋ𝒞 is isomorphic to ℋ as well.Remark 2.5.
The proof of this statement is not immediately obvious. There is clearly a surjective morphism fromℋ to [ℋ𝒞2]. When one takes the idempotent closure of a category, one adds new indecomposables and can potentially enlarge the Grothendieck group. Soergel showed, via a support filtration, that all the new indecomposables in ℋ𝒞 have symbols in [ℋ𝒞] which can be reached from certain symbols in [ℋ𝒞2] by a unitriangular matrix (see [11]). Therefore, the Grothendieck rings of ℋ𝒞 and ℋ𝒞2 are equal. Since ℋ𝒞 is idempotent closed and is embedded in R-bim, it has the Krull-Schmidt property and the Grothendieck group behaves as one would expect: it has a basis given by indecomposables. By classifying indecomposables and using the unitriangular matrix, Soergel showed that the map from ℋ to [ℋ𝒞2] is actually an isomorphism.
It is important to note that one does not know what the image of the indecomposables ofℋ𝒞 in ℋ is. The Soergel conjecture, still unproven in generality, proposes that the indecomposables of ℋ𝒞 descend to the Kazhdan-Lusztig basis of ℋ (see [11]).Notation 4.
We write the monomialbi1bi2⋯bid∈ℋ as bi̲, where i̲=i1⋯id is a finite sequence of indices; by abuse of notation, we sometimes refer to this monomial simply as i̲. If i̲ is as above, we say the monomial has length d=d(i̲). We call a monomial nonrepeating if ik≠il for k≠l, and increasing if i1<i2<⋯. The empty set is a sequence of length 0, and b∅=1. Similarly, in ℋ𝒞1, write Bi1⊗⋯⊗Bid as Bi̲. Note that B∅=R, the monoidal identity. For an arbitrary index i and sequence i̲, we write i∈i̲ if i appears in i̲.Given two objectsM,N∈R-bim we say they are biadjoint if M⊗- and N⊗- are left and right adjoints of each other, and the same for -⊗M and -⊗N. If M and N are biadjoint, so are M{1} and N{-1}. We often want to specify additional compatibility between various adjunction maps, but we pass over the details here (see [3] for more information on biadjunction).Proposition 2.6.
Each object inℋ𝒞 (resp., ℋ𝒞1, ℋ𝒞2) has a biadjoint, and Bi is self-biadjoint. Let ω be the t-antilinear anti-involution on ℋ which fixes bi, that is, ω(tabi̲)=t-abσ(i̲), where σ reverses the order of a sequence. There is a contravariant functor on ℋ𝒞 sending an object to its biadjoint, and it descends on the Grothendieck ring to ω.Definition 2.7.
Anadjoint pairing on ℋ is a pairing where each bi is self-adjoint, so that (x,biy)=(bix,y) and (x,ybi)=(xbi,y) for all x,y∈ℋ and all i∈I. Equivalently, for any m∈ℋ, (mx,y)=(x,ω(m)y) and (xm,y)=(x,yω(m)).There is a bijection between adjoint pairings(,) and traces ɛ, defined by letting (x,y)=ɛ(ω(x)y), or conversely ɛ(y)=(1,y). Adjoint pairings appear often in the literature, for instance [27] (although they are usually ℤ[t,t-1]-linear in both factors, unlike our current semilinear definition). Semilinear adjoint pairings will be crucially important, due to the following remark.Remark 2.8.
Let𝒞 be a monoidal category with objects Bi, such that Bi are self-biadjoint. We assume that 𝒞 is additive and graded and has isomorphisms (2.4)–(2.6). We call such a category a potential categorification of ℋ. In this case, there is a map of rings from ℋ to [𝒞] sending bi to [Bi], and (under suitable finite-dimensionality conditions) we get an adjoint semilinear pairing on ℋ via (bi̲,bj̲)=gdimHOM𝒞(Bi̲,Bj̲)∈ℤ[[t,t-1]], the graded dimension as a vector space. Denote the pairing and its associated trace map as (,)𝒞 and ɛ𝒞.
Instead, we may assume𝒞 is an enriched monoidal subcategory, containing objects Bi. The isomorphisms (2.4)–(2.6) typically have no meaning in this context, since there are no grading shifts or direct sums, but we can require that they Yoneda-hold; that is, they hold after the application of any Hom(-,X) functor (to graded vector spaces). There is no definition of a Grothendieck ring in this case, but we still get an induced adjoint semilinear pairing induced by Hom spaces. We call this an enriched potential categorification.We may use pairings to distinguish between different potential categorifications. The next proposition allows us to specify the pairing induced by a categorification by only investigating certainHOM spaces.Proposition 2.9.
Traces onℋ are uniquely determined by their values ɛ(bi̲) on increasing monomials i̲. Equivalently, adjoint pairings are determined by (1,bi̲) for increasing i̲. If i̲ is nonrepeating and j̲ is a permutation of i̲, then ɛ(bi̲)=ɛ(bj̲).We quickly sketch the proof. Moving an index from the beginning of a monomial to the end, or vice versa, will be calledcycling the monomial. It is clear, using biadjointness or the definition of trace, that the value of ɛ is invariant under cycling. It is not difficult to show that any monomial in W (in the letters si) will reduce, using the Coxeter relations and cycling, to an increasing monomial. When the monomial is already nonrepeating, one needs only use cycling and sisj=sjsi for distant i,j. Finally, using induction on the length of the monomial, the same principle shows that any monomial in ℋ reduces to a linear combination of increasing monomials, and therefore ɛ is determined by these.The upshot is that, given a potential categorification, one knows the dimension of allHOM(Bi̲,Bj̲) so long as one knows the dimension of HOM(B∅,Bi̲) for increasing i̲. Note that not every choice of (1,bi̲) for all increasing i̲ will yield a well-defined trace map.Consider the adjoint pairing given byɛstd(bi̲)=(1,bi̲)=td for nonrepeating i̲ of length d. This is the semilinear version of the pairing found in [27] which picks out the coefficient of the identity in the standard basis of ℋ and is called the standard pairing. Soergel showed that HOM(Bi̲,Bj̲) is a free graded left (or right) R-module of rank (bi̲,bj̲) using this pairing. In particular, for increasing i̲, HOM(R,Bi̲) is generated by a single element in degree d(i̲). Since the graded dimension of R is 1/(1-t2)n we have that (1,bi̲)ℋ𝒞=td/(1-t2)n is a rescaling of the standard pairing.Now letɛ be the quotient map ℋ→ℤ[t,t-1] by the ideal generated by all bi. It is a homomorphism to a commutative algebra, so it is a trace. The corresponding pairing satisfies (1,1)=1 and (x,y)=0 for monomials x,y if either monomial is not 1. We call this the trivial pairing, ɛtriv.


## 2.2. The Temperley-Lieb Algebra

Here again we state without proof some basic facts about Temperley-Lieb algebras. They were originally defined by Temperley and Lieb in [28], and were given a topological interpretation by Kauffman [29]. There are many good expositions for the topic, such as [30, 31].Definition 2.10.
TheTemperley-Lieb algebra𝒯ℒ is the ℤ[t,t-1]-algebra generated by ui,i∈I with relations
(2.7)ui2=[2]ui,(2.8)uiuj=ujuifor|i-j|≥2,(2.9)uiujui=uiforadjacenti,j.Proposition 2.11.
For adjacenti,j∈I, consider the element of ℋ defined by cij=defbibjbi-bi=bjbibj-bj, where the equality arises from relation (2.3). There is a surjective map ℋ→𝒯ℒ sending bi to ui for all i∈I, and whose kernel is generated by cij for adjacent i,j∈I.Once again, writeui̲ for a monomial in the above generators, with all the same conventions as before. The map ω descends from ℋ to 𝒯ℒ, and we define an adjoint pairing on 𝒯ℒ in the same way, with ui replacing bi everywhere. The results of Proposition 2.9 apply equally to 𝒯ℒ.Definition 2.12.
A category𝒞 as in Remark 2.8 is a potential categorification of 𝒯ℒ if it has objects Ui satisfying
(2.10)Ui⊗Ui≅Ui{1}⊕Ui{-1},Ui⊗Uj≅Uj⊗Uifordistanti,j,Ui⊗Uj⊗Ui≅Uiforadjacenti,j.
We call it anenriched potential categorification if it is an enriched category with objects Ui such that these isomorphisms Yoneda-hold.A permutationσ∈Sn+1 is called 321-avoiding if it never happens that, for i<j<k, σ(i)>σ(j)>σ(k). It turns out that, using the Temperley-Lieb relations, every monomial uj̲ is equal to a scalar times some ui̲ where i̲ is 321-avoiding; that is, if viewed as a word in the symmetric group, it represents a reduced expression for a 321-avoiding permutation. Moreover, between 321-avoiding monomials, the only further relations come from (2.8), and hence it is easy to pick out a basis from this spanning set. See [30] for more details.The Temperley-Lieb algebra has a well-known topological interpretation where an element of𝒯ℒ is a linear combination of crossingless matchings (isotopy classes of embedded planar 1-manifolds) between n+1 bottom points and n+1 top points. Multiplication of crossingless matchings consists of vertical concatenation (where ab is a above b), followed by removing any circles and replacing them with a factor of [2]. In this picture, ui becomes the following:(2.11)The basis of 321-avoiding monomials agrees with the basis of crossingless matchings. Any increasing monomial is 321-avoiding. Increasing monomials are easy to visualize topologically, as they have only “right waves" and “simple cups and caps." For example:(2.12)As an example of a monomial which is not increasing:(2.13)Given a crossingless matching, itsclosure is a configuration of circles in the punctured plane obtained by wrapping the top boundary around the puncture to close up with the bottom boundary, as in Figure 1. Circle configurations have two topological invariants: the number of circles and the nesting number which is the number of circles which surround the puncture and is equal to n+1-2l≥0 for some l≥0. Given a scaling factor for each possible nesting number, one constructs a trace by letting ɛ(ui̲)=ck[2]m where m is the number of circles in the closure of ui̲ and ck is the scaling factor associated to its nesting number k. To calculate (x,y), we place y below an upside-down copy of x (or vice versa), and then take the closure. All pairings/traces on 𝒯ℒ can be constructed this way, so they are all topological in nature.Figure 1
An example of the closure of a crossingless matching.Figure 2
An example of a planar graph in the strip, with colored edges.Figure 3
An example of tree reduction.Figure 4
An arbitrary innermost blue cycle. The dotted line encapsulates the subgraph on the interior which may contain colors adjacent to blue.The Temperley-Lieb algebra has astandard pairing of its own for which ck=1 for all nesting numbers k: ɛstd(ui̲)=[2]m as above. One can check that ɛstd(ui̲)=[2]n+1-d(i̲) for an increasing monomial. This is not related to the standard pairing on ℋ, which does not descend to 𝒯ℒ. On the other hand, ɛtriv clearly does descend to a pairing trivial pairing on 𝒯ℒ, which only evaluates to a nonzero number when the nesting number is n+1.It turns out that the pairing on𝒯ℒ arising from our categorification will satisfy (1,1)=(tn/(1-t2))[2]n-(t2/(1-t2)) and (1,ui̲)=(tn/(1-t2))[2]n-d. We will call the associated trace ɛcat. Clearly ɛcat=(tn/(1-t2)[2])ɛstd-(t2/(1-t2))ɛtriv. In particular, on any monomial x≠1, our trace will agree with a rescaling of the standard trace. When n=1, the algebras 𝒯ℒ and ℋ are already isomorphic, and ɛcat agrees with the rescaling of the standard trace on ℋ discussed at the end of Section 2.1.


## 2.3. Definition of Soergel Diagrammatics

We now give a diagrammatic description of the categoryℋ𝒞1, as discovered in [15]. Since the category to be defined will be equivalent to the category of Bott-Samelson bimodules, we will abuse notation temporarily and use the same names.Definition 2.13.
In this paper, aplanar graph in the strip is a finite graph with boundary (Γ,∂Γ) embedded in (ℝ×[0,1],ℝ×{0,1}). In other words, all vertices of Γ occur in the interior ℝ×(0,1), and removing the vertices, we have a 1-manifold with boundary whose intersection with ℝ×{0,1} is precisely its boundary. This allows for edges which connect two vertices, edges which connect a vertex to the boundary, edges which connect two points on the boundary, and edges which form circles (closed 1-manifolds embedded in the plane).
We generally refer toℝ×{0,1} as the boundary, which consists of two components, the top boundaryℝ×{1}, and the bottom boundaryℝ×{0}. We refer to a local segment of an edge which hits the boundary as a boundary edge; there is one boundary edge for each point on the boundary of the graph. We use the word component to mean a connected component of a graph with boundary.
This definition clearly extends to other subsets of the plane with boundary, so that we can speak of planar graphs in a disk or planar graphs in an annulus. The annulus has two boundary components,inner and outer. When we do not specify, we always mean a planar graph in the strip.We will be drawing morphisms inℋ𝒞1 as planar graphs with edges labelled in I. Instead of putting labels everywhere, we color the edges, assigning a color to each index in I. Henceforth, we use the term “color" and “index" interchangeably.We now defineℋ𝒞1 anew. Let ℋ𝒞1 be the monoidal category, with hom spaces enriched over graded vector spaces, which is defined as follows.Definition 2.14.
An object inℋ𝒞1 is given by a sequence of indices i̲, which is visualized as d points on the real line ℝ, labelled or “colored” by the indices in order from left to right. These objects are also called Bi̲. The monoidal structure on objects is concatenation of sequences.Definition 2.15.
Consider the set ofisotopy classes of planar graphs in the strip whose edges are colored by indices in I such that only four types of vertices exist: univalent vertices or “dots”, trivalent vertices with all three adjoining edges of the same color, 4-valent vertices whose adjoining edges alternate in colors between distant i and j, and 6-valent vertices whose adjoining edges alternate between adjacent i and j. This set has a grading, where the degree of a graph is +1 for each dot and -1 for each trivalent vertex; 4-valent and 6-valent vertices are of degree 0. The allowable vertices, which we call “generators," are pictured here:
(2.14)(2.15)
The intersection of a graph with the boundary yields two sequences of colored points onℝ, the top boundary i̲ and the bottom boundary j̲. In this case, the graph is viewed as a morphism from j̲ to i̲. For instance, if “blue" corresponds to the index i and “red" to j, then the lower right generator is a degree 0 morphism from jij to iji. Although this paper is easiest to read in color, it should be readable in black and white: the colors appearing are typically either blue, red, green, or miscellaneous and irrelevant. We throughout use the convention that blue (the darker color) is always adjacent to red (the middle color) and distant from green (the lighter color).
We letHomℋ𝒞1(Bi̲,Bj̲) be the graded vector space with basis given by planar graphs as above which have the correct top and bottom boundary, modulo relations (2.16) through (2.30). As usual in a diagrammatic category, composition of morphisms is given by vertical concatenation (read from bottom to top), the monoidal structure is given by horizontal concatenation, and relations are to be interpreted monoidally (i.e., they may be applied locally inside any other planar diagram).The relations are given in terms of colored graphs, but with no explicit assignment of indices to colors. They hold forany assignment of indices to colors, so long as certain adjacency conditions hold. We will specify adjacency for all pictures, although one can generally deduce it from the fact that 6-valent vertices only join adjacent colors, and 4-valent vertices only join distant colors.For example, these first four relations hold, with blue representing a generic index.(2.16)(2.17)(2.18)(2.19)We will repeatedly call a picture looking like (2.18) by the name “needle." Note that a needle is not necessarily zero if there is something in the interior. Note that a circle is just a needle with a dot attached, by (2.17), so that an empty circle evaluates to 0.Remark 2.16.
It is an immediate consequence of relations (2.16) and (2.17) that any tree (connected graph with boundary without cycles) of one color is equal to  (i)
if it has no boundary, two dots connected by an edge. Call the entire component adouble dot.(ii)
if it has one boundary edge, a single dot connected by the edge to the boundary. Call the component aboundary dot.(iii)
if it has more boundary edges, a tree with no dots and the fewest possible number of trivalent vertices needed to connect the boundaries. Moreover, any two such trees are equal. Call the component asimple tree.We refer to this as tree reduction.
This applies only to components of a graph which are a single color. Even if the blue part of a graph looks like a tree, if other colors overlap, then we may not apply tree reduction in general.In the following relations, the two colors are distant(2.20)(2.21)(2.22)(2.23)In this relation, two colors are adjacent, and both distant to the third color.(2.24)In this relation, all three colors are mutually distant.(2.25)Remark 2.17.
Relations (2.20) through (2.25) indicate that any part of the graph colored i and any part of the graph colored j “do not interact” for distant i and j, that is, one may visualize sliding the j-colored part past the i-colored part, and it will not change the morphism. We call this the distant sliding property.In the following relations, the two colors are adjacent.(2.26)(2.27)(2.28)(2.29)In this final relation, the colors have the same adjacency as{1,2,3}(2.30)This concludes the list of relations definingℋ𝒞1.Remark 2.18.
We chose here to describeℋ𝒞1 in terms of planar graphs with relations, with the notion of isotopy built-in, rather than in terms of generators and relations. Note, however, that using isotopy and (2.17), we get  Therefore, all “cups" and “caps" can be expressed in terms of the generators. By adding new relations corresponding to isotopy, one could give a presentation of the category where the “generators" above (and their isotopy twists) are really generators. This is how the category is presented in [15].We will occasionally use a shorthand to represent double dots. We identify a double dot coloredi with the polynomial fi∈R, and for a linear combination of disjoint unions of double dots in the same region of a graph, we associate the appropriate linear combination of products of fi. For any polynomial f∈R, a square box with a polynomial f in a region will represent the corresponding linear combination of graphs with double dots.For instance,Relations (2.19), (2.29), and (2.23) are referred to as dot forcing rules, because they describe at what price one can “force" a double dot to the other side of a line. The three relations imply that, given a line and an arbitrary collection of double dots on the left side of that line, one can express the morphism as a sum of diagrams where all double dots are on the right side, or where the line is “broken" (as illustrated next). Rephrasing this, for any polynomial f there exist polynomials g and h such that(2.31)The polynomials appearing can in fact be found using the Demazure operator∂i, and in particular, h=∂i(f). One particular implication is that  (2.32)wheneverf is a polynomial invariant under si (and blue represents i). As an exercise, the reader can check that fi2 slides through a line colored i. These polynomial relations are easy to deduce, or one can refer to [15] (see page 7, pages 16-17, and relation 3.16).We have an bimodule action ofR on morphisms by placing boxes (i.e., double dots) in the leftmost or rightmost regions of a graph. Now we can formulate the main result of [15].Theorem 2.19.
There is a functor from this diagrammatic categoryℋ𝒞1 to the earlier definition in terms of Bott-Samelson bimodules. This functor sends i̲ to the bimodule Bi̲ and a planar graph to a map of bimodules, preserving the grading and the R-bimodule action on morphisms. This functor is an equivalence of categories.Corollary 2.20.
TheR-bimodules Homℋ𝒞1(Bi̲,Bj̲) are free as left (or right) R-modules. In other words, placing double dots to the left of a graph is a torsion-free operation.Now, we have justified our abuse of notation. In this paper, we will never need to know explicitly what map ofR-bimodules a planar graph corresponds to, so the interested reader can see [15] for details. In fact, we will not use Theorem 2.19 at all, preferring to work entirely with planar graphs. However, we do use Corollary 2.20, a fact which would be difficult to prove diagrammatically.The proof of Theorem2.19 can be quickly summarized: first, one explicitly constructs a functor from the diagrammatic category to the Bott-Samelson category. Then, using the observations of the next section, one shows that the diagrammatic category is a potential categorification of ℋ and that the diagrammatic category, the Bott-Samelson category, and the image of the former in the latter all induce the same adjoint pairing on ℋ. Therefore, the functor is fully faithful.


## 2.4. Understanding Soergel Diagrammatics

Let us explain diagrammatically why the categoryℋ𝒞1 is a potential categorification of ℋ, and induces the aforementioned adjoint pairing.Definition 2.21.
Given a category𝒞 whose morphism spaces are ℤ-modules, one may take its additive closure, which formally adds direct sums of objects and yields an additive category. Given 𝒞 whose morphism spaces are graded ℤ-modules, one may take its grading closure which formally adds shifts of objects, but restricts morphisms to be homogeneous of degree 0. Given 𝒞 an additive category, one may take the idempotent completion or Karoubi envelope, which formally adds direct summands. Recall that the Karoubi envelope has as objects pairs (B,e) where B is an object in 𝒞 and e an idempotent endomorphism of B. This object acts as though it were the “image” of this projection e and behaves like a direct summand. When taking the Karoubi envelope of a graded category (or a category with graded morphisms) one restricts to homogeneous degree 0 idempotents. We refer in this paper to the entire process which takes a category 𝒞, whose morphism spaces are graded ℤ-modules, and returns the Karoubi envelope of its additive and grading closure as taking the graded Karoubi envelope. All these transformations interact nicely with monoidal structures. For more information on Karoubi envelopes see [32].
We letℋ𝒞2 be the graded additive closure of ℋ𝒞1, and let ℋ𝒞 be the graded Karoubi envelope of ℋ𝒞1.We wish to show that the isomorphisms (2.4) through (2.6) hold in ℋ𝒞2. Relation (2.20) immediately implies that Bi⊗Bj≅Bj⊗Bi for i,j distant, with the isomorphism being given by the 4-valent vertex.We have the following equality:(2.33)To obtain this, use (2.17) to stretch two dots from the two lines into the middle, and then use (2.19) to connect them. The identity idii decomposes as a sum of two orthogonal idempotents, each of which is the composition of a “projection" and an “inclusion" map of degree ±1, to and from Bi (explicitly, idii=i1p1+i2p2 where p1i1=idi, p2i2=idi, p1i2=0=p2i1). This implies that Bi⊗Bi≅Bi{1}⊕Bi{-1} and is a typical example of how direct sum decompositions work in diagrammatic categories.Similarly, the two color variants of relation (2.27) together express the direct sum decompositions in the Karoubi envelope(2.34)Bi⊗Bi+1⊗Bi=Cij⊕BiBi+1⊗Bi⊗Bi+1=Cji⊕Bi+1.
Again, the identity idi(i+1)i is decomposed into orthogonal idempotents. The second idempotent factors through Bi, and the corresponding object in the Karoubi envelope will be isomorphic to Bi. The first idempotent, which we call a “doubled 6-valent vertex," corresponds to a new object Cij in the idempotent completion. It turns out that the doubled 6-valent vertex Cij for “blue red blue" is isomorphic in the Karoubi envelope to the doubled 6-valent vertex Cji for “red blue red" (i.e., their images are isomorphic). We may abuse notation and call both of these new objects Cij; it is a summand of both i(i+1)i and (i+1)i(i+1). The image of Cij in the Grothendieck group is cij.We can also understand the induced pairing onℋ using diagrammatic arguments. The theorems below are proven in [15], and we will not use them in this paper (except motivationally), proving their analogs in the Temperley-Lieb case directly.Theorem 2.22 (Color Reduction).
Consider a morphismφ:∅→i̲, and suppose that the index i (blue) appears in i̲ zero times (resp.,: once). Then φ is in the -span of graphs which only contain blue in the form of double dots in the leftmost region of the graph (resp., as well as a single boundary dot). This result may be obtained simultaneously for multiple indices i.Corollary 2.23.
The spaceHomℋ𝒞1(∅,∅) is precisely the graded ring R. In other words, it is freely generated (over double dots) by the empty diagram. The space Homℋ𝒞1(∅,i̲) for i̲ nonrepeating is a free left (or right) R-module of rank 1, generated by the following morphism of degree d(i̲).
(2.35)The proof of the theorem does not use any sophisticated technology, only convoluted pictorial arguments. It comprises the bulk of [15]. The corollary implies that ɛℋ𝒞1(bi̲)=td/(1-t2)n for  nonrepeating i̲ of length d, as stated in Section 2.1.


## 2.5. Aside from Karoubi Envelopes and Quotients

Return to the setup of Definition2.21. If 𝒞 is a full subcategory of (graded) R-bimodules for some ring R, then the transformations described above behave as one would expect them to. In particular, the Karoubi envelope agrees with the full subcategory which includes all summands of the previous objects. The Grothendieck group of the Karoubi envelope is in some sense “under control" if one understands indecomposable R-bimodules already. On the other hand, the Karoubi envelope of an arbitrary additive category may be enormous, and to control the size of its Grothendieck group, one should understand and classify all idempotents in the category, a serious task. Also, arbitrary additive categories need not have the Krull-Schmidt property, making their Grothendieck groups even more complicated.The Temperley-Lieb algebra is obtained from the Hecke algebra by setting the elementscij to zero, for i=1,…,n-1. These elements lift in the Soergel categorification to objects Cij. The obvious way one might hope to categorify 𝒯ℒ would be to take the quotient of the category ℋ𝒞 by each object Cij.To mod out an additivemonoidal category 𝒞 by an object Z, one must kill the monoidal ideal of idZ in Mor(𝒞), that is, the morphism space Hom(X,Y) in the quotient category is exactly Hom𝒞(X,Y) modulo the submodule of morphisms factoring through V⊗Z⊗W for any V,W. If the category is drawn diagrammatically, one needs to only kill any diagram which has idZ as a subdiagram.We have not truly drawnℋ𝒞 diagrammatically, only ℋ𝒞1. The object we wish to kill is not an object in ℋ𝒞1; the closest thing we have is the corresponding idempotent, the doubled 6-valent vertex. However, this is not truly a problem, due to the following proposition, whose proof we leave to the reader.Proposition 2.24.
Let𝒞1 be an additive category, and let B be an object in 𝒞1, and let e be an idempotent in End(B). Let 𝒟1 be the quotient of 𝒞1 by the morphism e. Let 𝒞 and 𝒟 be the respective Karoubi envelopes. Finally, let 𝒟′ be the quotient of 𝒞 by the identity of the object (B,e). Then, there is a natural equivalence of categories from 𝒟 to 𝒟′.
The analogous statement holds when one considers graded Karoubi envelopes.Remark 2.25.
Note that𝒟′ has more objects than 𝒟, but they are still equivalent. For instance, (B,e) and (B,0) are distinct (isomorphic) objects in 𝒟′, but are the same object in 𝒟.So to categorify𝒯ℒ, one might wish to take the quotient of ℋ𝒞1 by the doubled 6-valent vertex, and then take the Karoubi envelope. This is easy to do diagrammatically, which is one advantage to the diagrammatic approach over the R-bimodule approach. The quotient of ℋ𝒞1 will no longer be a category which embeds nicely as a full subcategory of bimodules. One might worry that Krull-Schmidt fails, or that to understand its Karoubi envelope one must classify all idempotents therein. Thankfully, our calculation of HOM spaces will imply easily that its graded additive closure is Krull-Schmidt and is already idempotent closed, so it is equivalent to its own Karoubi envelope (see Section 3.3).


## 3. The Quotient Category𝒯ℒ𝒞

### 3.1. A Motivating Calculation

As discussed in the previous section, our desire is to take the quotient ofℋ𝒞1 by the doubled 6-valent vertex, and then take the graded Karoubi envelope.An important consequence of relations (2.26) and (2.18) is that(3.1)from which it follows, using (2.27), that(3.2)so the (monoidal) ideal generated inℋ𝒞1 by a doubled 6-valent vertex is the same as the ideal generated by the 6-valent vertex.Claim 1.
The following relations are all equivalent (the ideals they generate are equal)(3.3)(3.4)(3.5)(3.6)(3.7)Proof.
(3.3)⇒(3.4): add a dot, and use relation (2.26).
(3.4)⇒(3.5): add a dot to the top, and use (2.17).
(3.5)⇒(3.4): apply to the middle of the diagram.
(3.5)⇒(3.6): stretch dots from the blue strands towards the red strand using (2.17), and then apply (3.5) to the middle.
(3.6)⇒(3.7): use relation (2.27).
(3.7)⇒(3.3): use (3.2).Modulo 6-valent vertices, the relations (2.26) and (2.27) become (3.4) and (3.6) above. All other relations involving 6-valent vertices, namely, (2.28), (2.30), and (2.24), are sent to zero modulo 6-valent vertices. Relation (3.5) implies both (3.4) and (3.6) without reference to any graphs using 6-valent vertices. So, if we wish to rephrase our quotient in terms of graphs that never have 6-valent vertices, the sole necessary relation imposed by the fact that 6-valent vertices were sent to zero is the relation (3.5).Suppose, we only allow ourselves univalent, trivalent, and 4-valent vertices, but no 6-valent vertices, in a graphΓ. Then, the i-graph of Γ, which consists of all edges colored i and all vertices they touch, will be disjoint from the i+1- and i-1-graphs of Γ. The distant sliding property implies that the i-graph and the j-graph of Γ do not interact effectively, when i and j are distant. This will motivate the definition in the next section.


### 3.2. Diagrammatic Definition of𝒯ℒ𝒞

Definition 3.1.
We let𝒯ℒ𝒞1 be the monoidal category, with hom spaces enriched over graded vector spaces, defined as follows. Objects will be sequences of colored points on the line ℝ, which we will call i̲ or Ui̲. Consider the set whose elements are described as follows: (1)
for eachi∈I, consider a planar graph Γi in the strip, which is drawn with edges colored i (see Definition 2.13);(2)
the only vertices inΓi are univalent vertices (dots) and trivalent vertices;(3)
the graphsΓi and Γi+1 are disjoint. All graphs Γi are pairwise disjoint on the boundary;(4)
we consider isotopy classes of this data, so that one may apply isotopy to eachΓi individually so long as it stays appropriately disjoint.
This set has a grading, where the degree of a graph is+1 for each dot and -1 for each trivalent vertex, and the degree of an element of this set is the sum of the degrees for each graph Γi. Just as in Definition 2.15, each element of the set has a top and bottom boundary which is an object in 𝒯ℒ𝒞, and will be thought of as a map from the bottom boundary to the top. We let Hom𝒯ℒ𝒞1(Ui̲,Uj̲) be the graded vector space with basis given by elements of the set above with bottom boundary j̲ and top boundary i̲, modulo the relations (2.16) through (2.19), (2.29), and the new relation (3.5). As a reminder, the new relation is given here again.(3.8)
As before, composition of morphisms is given by vertical concatenation, the monoidal structure is given by horizontal concatenation, and relations are to be interpreted monoidally. This concludes the definition.Phrasing the definition in this fashion eliminates the need to add distant sliding rules, for these are now built into the notion of isotopy. Note that as we have stated it here,Γi and Γj may have edges which are embedded in a tangent fashion, or even entirely overlapped. However, such embeddings are isotopic to graph embeddings with only transverse edge intersections, which arise as 4-valent vertices in our earlier viewpoint.Proposition 3.2.
The category𝒯ℒ𝒞1 is isomorphic to ℋ𝒞1 modulo the 6-valent vertex.Proof.
Due to the observations of Section3.1, this is obvious.Hom spaces in𝒯ℒ𝒞1 are in fact enriched over graded R-bimodules, by placing double dots as before. However, they will no longer be free as left or right R-modules, as we will see.Remark 3.3.
Note that tree reduction (see Remark2.16) can now be applied to any tree of a single color in 𝒯ℒ𝒞, regardless of what other colors are present, since the only colors which can intersect the tree are distant colors which do not actually interfere.We denote by𝒯ℒ𝒞 the graded Karoubi envelope of 𝒯ℒ𝒞1, and by 𝒯ℒ𝒞2 the graded additive closure of 𝒯ℒ𝒞1. However, we will show that 𝒯ℒ𝒞2 is already idempotent closed, so that 𝒯ℒ𝒞2 and 𝒯ℒ𝒞 are the same.It is obvious that(3.9)Ui⊗Ui+1⊗Ui≅Ui,Ui+1⊗Ui⊗Ui+1≅Ui+1
in 𝒯ℒ𝒞1, and from the relation (3.6) and the simple calculation (using dot forcing rules) that(3.10)For the same reasons as in Section2.4, we still have Ui⊗Uj≅Uj⊗Ui for distant i,j, and Ui⊗Ui≅Ui{1}⊕Ui{-1} in 𝒯ℒ𝒞2. Therefore, 𝒯ℒ𝒞 is a potential categorification of 𝒯ℒ, and induces an adjoint pairing and a trace map ɛ𝒯ℒ𝒞 on 𝒯ℒ. At this point, we have not shown that the category 𝒯ℒ𝒞1 is nonzero, so this pairing could be 0.


### 3.3. Using the Adjoint Pairing

Proposition 3.4.
Let𝒞1 be an enriched category which is a potential categorification of 𝒯ℒ, whose objects are Ui̲ for sequences i̲. Let 𝒞2 be its additive graded closure, and let 𝒞 be its graded Karoubi envelope. Suppose that the induced trace map ɛ𝒞1 on 𝒯ℒ is equal to ɛcat. Then, the set of Ui̲{n} for n∈ℤ and 321-avoiding i̲ forms an exhaustive irredundant list of indecomposables in 𝒞2. In addition, 𝒞2 is Krull-Schmidt and idempotent closed (so 𝒞2 and 𝒞 are equivalent), and 𝒞 categorifies 𝒯ℒ.This proposition is an excellent illustration of the utility of the induced adjoint pairing. We prove it in a series of lemmas, which all assume the hypotheses above.Lemma 3.5.
The objectUi̲ in 𝒞1 has no nontrivial (homogeneous) idempotents when i̲ is 321-avoiding. Moreover, if both i̲ and j̲ are 321-avoiding, then Ui̲≅Uj̲{m} in 𝒞2 if and only if m=0 and ui̲=uj̲ in 𝒯ℒ.Proof.
Two 321-avoiding monomials in𝒯ℒ are equal only if they are related by the relation (2.8). Since this lifts to an isomorphism Ui⊗Uj≅Uj⊗Ui in 𝒞2, we have ui̲=uj̲⇒Ui̲≅Uj̲.
If an object has a 1-dimensional space of degree 0 endomorphisms, then it must be spanned by the identity map, and there can be no nontrivial idempotents. If an object has endomorphisms only in nonnegative degrees, then it can not be isomorphic to any nonzero degree shift of itself. If two objectsX and Y are such that both Hom(X,Y) and Hom(Y,X) are concentrated in strictly positive degrees, then no grading shift of X is isomorphic to Y, since there can not be a degree zero map in both directions.
Therefore, we need only show that (for 321-avoiding monomials)Ui̲ has endomorphisms concentrated in nonnegative degree, with a 1-dimensional degree 0 part, and that when ui̲≠uj̲, Hom(Ui̲,Uj̲) is in strictly positive degrees. This question is entirely determined by the pairing on 𝒯ℒ, since it only asks about the graded dimension of Hom spaces.
Wheni̲ is empty, we already know that (1,1)=(tn/(1-t2))[2]n-(t2/(1-t2)), which has degree 0 coefficient 1, and is concentrated in nonnegative degrees.
We know how to calculate(x,y) in 𝒯ℒ when x and y are monomials, and either x or y is not 1 (see Section 2.2). We draw x as a crossingless matching, draw y upside-down and place it below x, and close off the diagram: if there are m circles in the diagram, then (x,y)=tn[2]m-1/(1-t2). In particular, if m=n+1, then the Hom space will be concentrated in nonnegative degrees, with 1-dimensional degree 0 part. If m<n+1, then the Hom space will be concentrated in strictly positive degrees.
We leave it as an exercise to show that, ifx is a crossingless matching (i.e., a 321-avoiding monomial) then the closed diagram for (x,x) has exactly n+1 circles. The following example makes the statement fairly clear, where x̃ is x upside-down:
(3.11)
In this examplex has all 3 kinds of arcs which appear in a crossingless matching: bottom to top, bottom to bottom, and top to top. Each of these corresponds to a single circle in the diagram closure.
Similarly, there are fewer thann+1 circles in the diagram for (x,y) whenever the crossingless matchings x,y are nonequal. Consider the diagram above but with the region x removed. One can see that no circles are yet completed, and each boundary point of x's region is matched to the other by an arc. The number of circles is maximized when you pair these boundary points to each other, and this clearly gives the matching x. For any other matching y, two arcs will become joined into one, and fewer than n+1 circles will be created.Lemma 3.6.
𝒞2 is idempotent closed, and its indecomposables can all be expressed as grading shifts of Ui̲ for  321-avoiding i̲. It has the Krull-Schmidt property.Proof.
Since the Temperley-Lieb relations allow one to reduce a general word to a 321-avoiding word, one can show that everyUi̲ is isomorphic to a direct sum of shifts of Uj̲ for 321-avoiding j̲, using isomorphisms and direct sum decompositions instead of the analogous Temperley-Lieb relations. Clearly these shifted Uj̲ are all indecomposable, since they have no nontrivial idempotents; these are then all the indecomposables. Since every indecomposable in 𝒞2 has a graded local endomorphism ring (with maximal ideal given by positively graded morphisms), 𝒞2 is idempotent closed and Krull-Schmidt (see [33], Section 2.2).The Krull-Schmidt property implies that isomorphism classes of indecomposables form a basis for the Grothendieck group.Proof of Proposition3.4.
There is aℤ[t,t-1]-linear map of rings 𝒯ℒ→[𝒞2], which is evidently bijective because it sends the 321-avoiding basis to the 321-avoiding basis. Since 𝒞=𝒞2, we are done.This proposition shows that Lemma1.1 implies Theorem 1.2.Remark 3.7.
In analogy to the paper [15], the bulk of the proof of Theorem 1.2 lies in proving that hom spaces induce a particular adjoint pairing. Beyond that, we have mostly stated the obvious. Let us note that what is obvious for 𝒯ℒ and 𝒯ℒ𝒞 is not obvious at all when dealing with ℋ and ℋ𝒞. In particular, if we are given a category 𝒞1 which is a potential categorification of ℋ as in Proposition 3.4, we can not conclude that 𝒞 categorifies ℋ. We summarize the differences here.
It is clear (for both Hecke and Temperley-Lieb) that the mapℋ→[ℋ𝒞2] is well defined and surjective. The two main subtleties are (1) the difference between ℋ𝒞2 and ℋ𝒞, and (2) the injectivity of the map.
In general, one likes to examine the additive Grothendieck group only of idempotent closed categories with the Krull-Schmidt property, because this guarantees that indecomposables form a basis for the Grothendieck group. Thus it is convenient that𝒯ℒ𝒞2 is already idempotent closed. Thankfully, we have a result of Soergel [11] that proves that [ℋ𝒞2]≅[ℋ𝒞], as was discussed in Remark 2.5.
To show injectivity of the map in the𝒯ℒ case, we can identify a basis for 𝒯ℒ which is sent to a complete set of indecomposables, and then we can evaluate the trace map to show that these indecomposables are pairwise nonisomorphic. For ℋ𝒞, we do not currently know what the indecomposables (i.e., idempotents) are, nor do we know their preimage in ℋ. If we knew a class of indecomposables which decategorified to the Kazhdan-Lusztig basis, then we could use a similar argument to the above to show that they and their shifts form an exhaustive irredundant list of indecomposables in ℋ𝒞, and therefore that the map ℋ→[ℋ𝒞] is injective. Soergel discusses this in the last chapter of [11]. This is actually a deep question, shown by Soergel ([34], see also [8, 11]) to be equivalent to proving a version of the Kazhdan-Lusztig conjectures. In any case, the result depends on the base field 𝕜, and no simple proof has been found. In particular, to prove that the graded Karoubi closure of the diagrammatic category ℋ𝒞1 categorifies the Hecke algebra (for certain 𝕜), we must pass to the world of bimodules where Soergel's powerful geometric techniques will work. In particular, there is currently no proof of injectivity if one defines the category ℋ𝒞 diagrammatically over 𝕜=ℤ.
It should be emphasized that the story of𝒯ℒ is a particularly easy one (as is its Kazhdan-Lusztig theory). No high-powered technical machinery is needed, and the proofs of idempotent closure and injectivity are self-contained and diagrammatic. In fact, the arguments in this paper do work entirely over ℤ[1/2], as can be checked. Dividing by two must be allowed in order to split the identity of Uii into idempotents, as in (2.33); however, it is likely that the arguments would work over ℤ as well. Working over ℤ is discussed more extensively in [22].Remark 3.8.
A category𝒪 analog of the fact that 321-avoiding monomials lift to indecomposable Soergel bimodules, which remain indecomposable upon passage to the Temperley-Lieb quotient, can be found in Lemma 5.2 of [14].


### 3.4. Reductions

When we say that a graph or a morphism “reduces" to a set of other graphs, we mean that the morphism is in the𝕜-span of those graphs. We refer to a one-color graph, each of whose (connected) components is either a simple tree with respect to its boundary or a double dot, as a simple forest with double dots. If there are no double dots, it is a simple forest without double dots. Tree reduction implies that any graph Γi without cycles reduces to a simple forest with double dots. Note also that circles in a graph are equal to needles with a dot attached, and can be treated just like any other cycle.If there were only one color, we could iterate the following rule (which is an implication of the dot forcing rules and (2.18)) to break cycles:(3.12)We do something similar for the general case.Proposition 3.9.
In𝒯ℒ𝒞1, any morphism reduces to one where, for each i, the i-graph is a simple forest with double dots. Moreover, we may assume that all double dots are in the lefthand region.Proof.
We use induction on the total number of cycles (of any color) in the graph. Suppose there is a blue colored cycle: choose one so that it delineates a single region (i.e., there are no other cycles inside). There may be blue “spokes" going from this cycle into the interior, but no two spokes can meet, lest they create another region. By tree reduction on the spokes, we can assume that any blue appearing inside the cycle is in a different blue component than the cycle. Other colors may cross over the cycle, into the interior. If we view the interior of the cycle as a graph of its own, it has fewer total cycles so we may use induction. Since the boundary of the interior contains no blue color or colors adjacent to blue, they may be assumed to appear in the interior only in the form of double dots next to the cycle. Using dot-forcing rules, we reduce to two graphs: one with the cycle broken, and one with all these double dots on the exterior of the cycle. The former reduces by induction. For the latter, only distant colors enter the cycle, so they can be slid out of the way to leave an empty blue cycle, which is0 by the rule above.
We need to only do the base case, where the graph has no cycles. The dot-forcing rules imply that double dots may be moved to any region of the (multicolored) graph, at the cost of breaking a few lines. Breaking lines will never increase the number of cycles. Therefore, if we have a graph without cycles, tree reduction implies that we actually have a simple forest with double dots, and dot forcing allows us to move these double dots to the left. The breaking of lines may require more tree reduction, yielding more double dots, but this process is finite.Remark 3.10.
This proposition and its proof will apply to graphs in any connected simply connected region in the plane.Corollary 3.11.
For any nonrepeatingi̲, Hom𝒯ℒ𝒞1(∅,i̲) is generated (as a left or right R-module) by a single element φi̲ of degree d(i̲), pictured below
(3.13)Proof.
A simple forest with double dots and at most one boundary edge is no more than a boundary dot with double dots. Thus, any morphism reduces to a boundary dot for each color, accompanied by double dots.To show Lemma1.1, we need to only investigate Hom(∅,i̲) for increasing i̲, since we have already shown that the values of ɛ(ui̲) are determined by their values for increasing i̲. This space will be an R-bimodule where the left and right action are the same (since the lefthand and righthand regions are the same in any picture with no bottom boundary), so we view it as an R-module, and we have just shown that it is cyclic. Let Ii̲ be the ideal which is the kernel of the map R→HOM(∅,i̲) sending 1↦φi̲; we call it the TL ideal of i̲. Proving the Lemma 1.1 is to find Ii̲ and show that the graded dimension of R/Ii̲{d} is ɛcat(ui̲).Remark 3.12.
Since the spaceHomℋ𝒞1(∅,i̲) is a freeR-module, all polynomials in Ii̲ must have arisen from reducing to some morphism which contained the relation (3.5) to a “nice form," that is, φi̲ plus double dots. In other words, letting αi be the morphism pictured below, we want to plug αi into a bigger graph, reduce it to a nice form, and see what we get. 									(3.14)
Remember thatαi is actually just a 6-valent vertex with two dots attached (one red and one blue). This bigger graph, into which αi is plugged, will actually be a graph on the punctured plane or punctured disk with specified boundary conditions on both the outer and inner boundaries. The difficult graphical proofs of this paper just consist in analyzing such graphs. This is done by splitting the punctured plane into simply connected regions and using the above proposition.


### 3.5. Generators of the TL Ideal

The sequencei̲ is assumed to be nonrepeating.Proposition 3.13.
The TL ideal of∅ contains yi,j=deffifj(fi+2fi+1+2fi+2+⋯+2fj-1+fj) over all 1≤i<j≤n.
The TL ideal ofi̲ contains zi,j,i̲=defyi,j/gigj, where gi=fi if i∈i̲, gi=1 otherwise.We will prove that these actually generate the ideal in Proposition3.24 but postpone the proof as it is long and unenlightening.Proof.
Adding 4 dots toαi, or 6 dots to a 6-valent vertex, we get
(3.15)
This isyi,i+1=fifi+1(fi+fi+1)=(xi-xi+1)(xi+1-xi+2)(xi-xi+2). Even though we are not allowing 6-valent vertices in our diagrams, we will sometimes express yi,i+1 as(3.16)to avoid having to consider sums of graphs (it's easier for me to draw!).
To obtain the otheryi,j, note the following equalities under the action of Sn+1 on R:(3.17)sifi+1=fi+fi+1,si+1fi=fi+fi+1,sifi=-fi,sifj=fjfor|i-j|>1.
From this it follows by explicit calculation that(3.18)si-1yi,j-yi,j=yi-1,j,sj+1yi,j-yi,j=yi,j+1.
Now, when we surround a polynomialf with a j-colored circle and use (3.12), we are left with a j-colored double dot times ∂j(f), so we get f-sjf=∂j(f)fj.(3.19)
Combining this with the calculations we just made, we see that aj+1 circle around yi,j will yield yi,j+1 up to sign, and so forth. We now have numerous ways to express ±yi,j: for any i≤k≤j-1, take αk with 4 dots to get yk,k+1, and then surround it with concentric circles whose colors, from inside to out, are k+2,k+3,…,j and then k-1,k-2,…,i(3.20)
Clearly the colors of the increasing sequence and those of the decreasing sequence are distant, so a sequence likek-1,k+2,k+3,k-2,… is also okay, or any permutation which preserves the order of the increasing and the decreasing sequence individually.
For very similar reasons,zi,j,i̲ is in the TL ideal of i̲. Adding two or three dots to (3.5), we get several more equations.(3.21)(3.22)(3.23)
Again, for a variety of these pictures, we use shorthand like(3.24)
These give youzi,i+1,i̲ in the case where at least one of i,i+1∈i̲. Again by (3.12), putting a polynomial f in the eye of a j-colored needle will yield ∂j(f)=(f-sjf)/fj next to a j-colored boundary dot.(3.25)
This gives us several ways to drawzi,j,i̲.
If neitheri nor j are in i̲, then zi,j,i̲=yi,j and is pictured as above, but with additional boundary dots being put below to account for φi̲. Since these extra dots are generally irrelevant, we often do not bother to draw them.
Ifi∈i̲ and j∉i̲, we have two ways of drawing zi,j,i̲. One can take αi, connect one i input to the outer boundary, add dots, and surround it with circles colored i+2,i+3,…,j(3.26)
Alternatively, take somei<k<j, add dots to αk, and surround it with circles forming an increasing sequence k+2⋯j and a decreasing sequence k-1⋯i, except that the final i-colored circle is a needle(3.27)
The case ofj∈i̲ and i∉i̲ is obvious.
If bothi,j∈i̲, then we have several choices again. If j=i+1, then we must use(3.28)
but in general, we may either repeat (3.26) with a j-needle instead of a j-circle(3.29)
or repeat (3.27) with a j-needle instead of a j-circle(3.30)
In any case, it is clear that the polynomials above are in the TL ideal, and the claim is proven.Let us quickly consider the redundancy in this generating set of the ideal. Wheni>j let yi,j=defyj,i and zi,j,i̲=defzj,i,i̲.Corollary 3.14.
Suppose thati̲ is nonempty, and fix an index k∈i̲. Then, Ii̲ is generated by zk,j,i̲ for 1≤j≤n, j≠k. None of these generators are redundant.
None of the generatorsyi,j of I∅ are redundant.Proof.
We leave the checks of irredundancy to the reader, but a proof will also arise as a byproduct in the next section (see Remark3.17).
Suppose thatk∈i̲ but i,j∉i̲. If k<i<j, then zi,j,i̲=yi,j=fizk,j,i̲-fjzk,i,i̲ so that zi,j,i̲ is redundant. If i<k<j, then zi,j,i̲=fizk,j,i̲+fjzi,k,i̲. A similar statement holds for i<j<k. In the same vein, if k,l∈i̲ but i∉i̲, then for a given zk,l,i̲, only one of zk,i,i̲ or zl,i,i̲ is needed, and if k,l,m∈i̲, then any two of the three pairwise relations will imply the third.


### 3.6. Graded Dimensions

In this section, fix a nonrepeating sequencei̲. We assume in this section that the generators of Ii̲ are precisely the polynomials described in Proposition 3.13.Notation 5.
An element ofR can be written as a polynomial in fi, so let x=f1a1⋯fnan be a general monomial. Choose any i̲, possibly empty. Given a monomial x, let Jx⊂{1,…,n} be the subset containing i̲ and all indices j such that aj≠0. For a fixed subset J, let RJ be the subset of all monomials x with Jx=J. This inherently depends on the choice of i̲.Under the mapR→HOM(U∅,Ui̲), the image of RJ will be graphs where the colors appearing are precisely J. Every color in i̲ appears as a boundary dot, and every fj corresponds to a double dot of that color. The case J=∅ only occurs when i̲=∅ and R∅={1}.To find a basis forR/Ii̲, we will use the Bergman Diamond Lemma [35] for commutative rings.Definition 3.15.
LetA be a free commutative polynomial ring, where monomials are given a partial order with the DCC, compatible with multiplication in that x<y⇒ax<ay. Let I be an ideal generated by relations r of the form xr=yr, where xr is a monomial and yr is a linear combination of monomials which are each less than xr in the partial order. A reduction is an application of a relation r to replace xr with yr, but not the other way around (a reduction always lowers the partial order on each term in a polynomial). One says a polynomial xreduces to y if y can be obtained from x by a series of reductions applied to monomials in x. A monomial is called irreducible if it does not have xr as a factor for any relation r. An inclusion ambiguity is a monomial x=ab, where x=xr for some r, and b=xr′ for some r'≠r. An overlap ambiguity is a monomial x=abc, where ab=xr for some r and bc=xr′ for some r'≠r. Each ambiguity has two natural reductions, and one says the ambiguity is resolvable if the two reductions are then jointly reducible to the same element.Lemma 3.16 (Bergman Diamond Lemma for Commutative Rings, [35]).
With these definitions in place, if every inclusion and overlap ambiguity is resolvable, then the images of the irreducible monomials form a basis forA/I.This process may become more transparent from the example below; in addition, Bergman's paper has a number of nice examples for the trickier, noncommutative version. We treat two separate cases, wheni̲=∅ and when i̲≠∅.Claim 2.
Leti̲=∅. We place the lexicographic order on monomials in R, so that f1<f2<⋯. The relation yi,j=0 for i<j will be rewritten as fifj2=-fifj(fi+2∑i<k<jfk), which replaces fifj2 with a sum of monomials all lower in the order. For each J≠∅, the irreducible monomials in RJ are precisely fkm∏i∈Jfi, where k is the minimal index in J and m≥0 (note: the exponent of fk is m+1≥1). When J=∅, 1 is irreducible. Irreducibles form a basis for R/I∅.Proof.
A monomial is irreducible iffifj2 never appears as a factor for any i<j. Because of this, the classification of irreducible monomials in each RJ is obvious. There are no inclusion ambiguities between relations, since they are all homogeneous and degree 3. There are two kinds of overlap ambiguities, both labelled by a choice of i<l<j.
For the first ambiguity, one can reducex=fiflfj2 by either reducing flfj2 or fifj2. Applying the former reduction, x↦fiflfj(-fl-2∑l<k<jfk) which has a term given by -fifl2fj that can be further reduced, yielding fiflfj(fi+2∑i<k<lfk-2∑l<k<jfk). Applying the latter reduction, x↦fiflfj(-fi-2∑i<k<jfk)=fiflfj(-fi-2∑i<k<lfk-2fl-2∑l<k<jfk), which has a term given by -2fifl2fj that can be further reduced, yielding fiflfj(-fi-2∑i<k<lfk-2∑l<k<jfk+2fi+4∑i<k<lfk)=fiflfj(fi+2∑i<k<lfk-2∑l<k<jfk). Since these agree, the ambiguity is resolvable.
For the second ambiguity, one can reducex=fifl2fj2 by either reducing fifl2 or flfj2. A very similar calculation shows that this ambiguity is resolvable as well. Therefore, the Bergman diamond lemma implies that irreducibles form a basis for the quotient.Remark 3.17.
This also proves that none of theyi,j is redundant. Removing yi,j from the ideal, we may apply the same Bergman diamond lemma argument to say that irreducibles form a basis for the quotient. However, with no yi,j, the monomial fifj2 is irreducible, and the quotient is larger than before. A similar statement can be made about the zk,j,i̲ below.WhenJ≠∅, the graded rank of the irreducibles in RJ is t2|J|/(1-t2). When J is empty, the only element of RJ is 1. So the graded rank of R/I∅ is 1+∑J≠∅(t2|J|/(1-t2)), but ∑Jt2|J|=(1+t2)n since every fi may either appear or not appear, independently of every other. Hence ∑J≠∅t2|J|=(1+t2)n-1. Putting it all together, the graded rank is ((1+t2)n-t2)/(1-t2)=(tn[2]n-t2)/(1-t2). Hence we have proven the following claim.Claim 3.
The graded dimension ofR/I∅ is exactly ɛcat(u∅).Claim 4.
Leti̲≠∅, and fix k∈i̲. We choose a different order on indices, where k<k+1<k-1<k+2<k-2<⋯, and then place the lexicographic order on monomials. The relation zk,j,i̲ for j≠k will be rewritten in decreasing order format as either fj2=-fj(fk+2∑lfl) for j∉i̲, or fj=-(fk+2∑lfl) for j∈i̲, where the sum is over l between k and j. Then, the irreducible monomials in RJ are precisely fkm∏j∈J∖i̲fj for m≥0. Irreducibles form a basis for R/Ii̲.Proof.
An irreducible polynomial will be a polynomial which does not havefj2 as a factor, for k≠j∉i̲, and does not have fj as a factor for k≠j∈i̲. The classification of irreducibles in RJ is now obvious. There are no ambiguities whatsoever, so we are done by the Bergman diamond lemma.The graded rank of irreducibles inRJ is t2|J|-2d/(1-t2), for d the length of i̲ (remember that i̲⊂J). Thus, the graded rank of R/Ii̲ is ∑i̲⊂J(t2|J|-2d/(1-t2))=(1+t2)n-d/(1-t2), and the graded rank of R/Ii̲{d} is td(1+t2)n-d/(1-t2)=tn[2]n-d/(1-t2). Hence, one considers the following.Claim 5.
The graded dimension ofR/Ii̲{d(i̲)} is exactly ɛcat(ui̲).This is clearly sufficient to prove Lemma1.1, modulo Proposition 3.24.


### 3.7. Weyl Lines and Disoriented Tubes

We now give two alternate interpretations of the TL idealsIi̲. We continue to assume that i̲ is nonrepeating and zi,j,i̲ generates Ii̲.Definition 3.18.
LetV be the reflection representation of Sn+1, such that R=ℂ[f1,…,fn] is the coordinate ring of V. Note that the linear equations which cut out reflection-fixed hyperplanes are precisely wi,j=fi+fi+1+⋯+fj=xi-xj+1 for i≤j. A Weyl line is a line in V through the origin which is defined by the intersection of reflection-fixed hyperplanes; it is given by a choice of n-1 transversely intersecting reflection-fixed hyperplanes. Given a nonrepeating sequence i̲, one says that a Weyl line is transverse toi̲ if it is transverse to (i.e., not contained in) the hyperplanes fk=0 for each k∈i̲.Proposition 3.19.
The TL ideal ofi̲ is the ideal associated with the union of all Weyl lines transverse to i̲ (with its reduced scheme structure).Example 3.20.
Letn=3. One can check that f1f2(f1+f2)=f2f3(f2+f3)=f1f3(f1+2f2+f3)=0 cuts out 7 lines in V, namely,  (1)
f1=f2=f1+f2=0,(2)
f1=f3=0,(3)
f2=f3=f2+f3=0,(4)
f1=f2+f3=f1+f2+f3=0,(5)
f1+f2=f3=f1+f2+f3=0,(6)
f2=f1+f2+f3=0,(7)
f1+f2=f2+f3=0,These 7 lines are precisely the 7 lines cut out by the intersection of pairs of reflection-fixed hyperplanes. There are 6 reflection-fixed hyperplanes, given by equations f1, f2, f3, f1+f2, f2+f3, and f1+f2+f3, or alternatively, by xi-xj for 4≥j>i≥1. Intersecting pairs of hyperplanes will give a line, and occasionally this line is forced to lie in a third hyperplane, as in the list above. One can check that this list covers all pairs of hyperplanes which give distinct lines as their intersection.Proof.
This is not difficult to show, but since we have not seen it elsewhere, we provide a complete proof. First, we show by induction onn that the ideal I∅ cuts out the Weyl lines with the reduced scheme structure. The case n=1 is trivial (and n=2 is also obvious).
For any1≤k≤n, consider the hyperplane fk=0 as an n-1-dimensional space V′, with an action of Sn+1/〈sk〉≅Sn. Giving Sn a Coxeter structure with simple reflections si for i≠k (note that sk+1=(k+1,k+2)=(k,k+2) in the quotient), it is quite easy to see that V' is the reflection representation of Sn. Moreover, the Weyl hyperplanes are cut out by wi,j′=fi+fi+1+⋯+fj (where fk=0 so it may be left out of the sum) for i,j≠k, and the equivalent polynomials yi,j′ also have the same formulae, and are indexed by i,j≠k. Therefore, for i,j≠k, the images of wi,j are just wi,j′, and the same for yi,j and yi,j′. Moreover, if either i or j equals k, then yi,j=0 on fk=0, and wi,j is redundant on fk=0, being equal to some wi′,j′. By induction, yi,j′ cut out the Weyl lines with the reduced scheme structure on V′, and therefore the vanishing set of yi,j agrees with the Weyl lines on fk=0.
If allfk≠0, then it is easy to see that the yi,j cut out a single line with the reduced scheme structure, namely, -f1=f2=-f3=⋯=(-1)nfn. This is a Weyl line, the intersection of all wi,i+1. We wish to show that this is the only Weyl line transverse to all fk=0. We can show this by induction as well (again, the base case n=2 is easy). Suppose we are given n-1 transverse hyperplanes wi,j. If any two involve the index n, that is, wi,n and wj,n, then we may replace the pair with wi,n and wi,j since they have the same intersection (and wi,j is not already in the set, or the intersection would not be transverse). So, we may assume that at most one of the chosen hyperplanes involves the index n, but then we have n-2 transverse hyperplanes which only involve indices {1,…,n-1}, which must then be mutually transverse to fn=0. Letting V' be the hyperplane fn=0 viewed as a reflection representation as above, we have n-2 transverse hyperplanes which cut out a Weyl line transverse to fk=0 for all 1≤k≤n-1. By induction, that Weyl line is -f1=f2=-f3=⋯=(-1)n-1fn-1 (which holds true modulo fn=0), but repeating the same argument for the index k instead, we leave out the kth term and get -f1=f2=⋯=(-1)kfk̂=⋯=(-1)nfn modulo fk=0. Together, all these equalities imply that -f1=f2=⋯=(-1)nfn everywhere.
One might be worried, because of the restrictions used in the induction step, thatIi̲ does not give the reduced structure on the Weyl lines at the origin. However, Ii̲ is a homogeneous ideal which cuts out a reduced 0-dimensional subscheme of ℙ(V), so that its vanishing on V is the cone of a reduced scheme, and hence is reduced. This concludes the proof that I∅ cuts out the Weyl lines with the reduced scheme structure.
Fori̲≠∅, I∅⊂Ii̲ and the vanishing of Ii̲ is contained in that of I∅. Choose k∈i̲. If fk=0 then zk,k+1,i̲ is equal to fk+1a, where a=1,2 depending on whether k+1∈i̲, but either way we get that fk+1=0. Then zk,k+2=fk+2a for a=1,2, and so forth. Therefore fk=0 only intersects the vanishing of Ii̲ at the origin (as sets). It is clear that, on the open set where fk≠0 for all k∈i̲, the polynomials zi,j,i̲ and yi,j have the same vanishing (as schemes), since they differ by a unit. The same cone argument shows that Ii̲ gives the reduced structure at the origin.Remark 3.21.
In particular,I∅ is contained in every ideal, and the category 𝒯ℒ𝒞1 is manifestly R/I∅-linear.Remark 3.22.
LetZ be the union of all Weyl lines in V. The previous results should lead one to guess that the Temperley-Lieb algebra should be connected to the geometry of the Sn+1 action on Z via 𝒯ℒ𝒞, in much the same way that the Hecke algebra is connected to the reflection representation via ℋ𝒞 (see [11]). However, at the moment, we have no way to formulate the category 𝒯ℒ𝒞 in terms of coherent sheaves on Z×Z (i.e., R/I∅-bimodules) or the derived category thereof. Describing 𝒯ℒ𝒞 using sheaves on Z seems like an interesting question.
As an example of the difficulties, letUi be the bimodule R/Ii⊗R/Ii{-1}, where the tensor is over Rsi; this should be the equivalent of the Soergel bimodule Bi. Then, there is a degree-1 map R/I∅→Ui sending 1 to xi⊗1-1⊗xi+1 (the boundary dot on the top), but there is no degree-1 map Ui→R/I∅ (the boundary dot on the bottom); such a map should send 1⊗1 to 1. There is only a degree-3 map, sending 1⊗1 to fi (the boundary dot with a double dot). A similar problem occurs again: the trivalent vertex seems to be defined only in one direction.Now, we describe briefly the topological intuition associated with the category𝒯ℒ𝒞, and another way to view Ii̲. These remarks will not be used in the remainder of the paper, nor will we give a proof. The reader should be acquainted with the section on 𝔰𝔩2-foams in Vaz's paper [16].Remark 3.23.
Letℱ be the functor from 𝒯ℒ𝒞1 to the category of disoriented cobordisms Foam2, as defined in Vaz's paper. If fi is the double dot colored i, then one can easily see that ℱ sends fi to a tube connecting the ith sheet to the (i+1)th sheet, with a disorientation on it. If the double dot appears in a larger morphism φ, such that in ℱ(φ), the ith sheet and the (i+1)th sheet are already connected by a saddle or tube, then adding another tube between them does nothing more than adding a disoriented handle to the existing surface. Note that the map φi̲ previously defined will connect the ith sheet to the (i+1)th sheet for any i∈i̲.
Suppose that theith, (i+1)th, and (i+2)th sheets are all connected in a cobordism. Then fi adds a handle on the left side of the (i+1)th sheet, fi+1 adds a handle on the right side, and these two disoriented surfaces are equal up to a minus sign in Foam2. This fact is essentially the statement that:(3.31)
In other words, the algebra𝕜[f1,…,fn] maps to Foam2, sending fi to the disoriented tube between the ith and (i+1)th sheet. The ideal Ii̲ is clearly in the kernel of this action when applied to the cobordism ℱ(φi̲). In fact, it is precisely the kernel, using the argument of Proposition 4.2 in [36]: for any distinct monomials in a basis for R/Ii̲, their image in Foam2 will have independent evaluations with respect to some closure of the cobordism. We do not do the calculation here. The usual arguments involving adjoint pairings imply that the faithfulness of the functor ℱ can be checked on Hom(∅,i̲). Therefore, the functor ℱ is faithful.


### 3.8. Proof of Generation

Proposition 3.24.
The TL idealI∅ is generated by yi,j=deffifj(fi+2fi+1+2fi+2+⋯+2fj-1+fj) over all 1≤i<j≤n.
The TL idealIi̲ is generated by all zi,j,i̲=defyi,j/gigj, where gi=fi if i∈i̲, gi=1 otherwise.We wish to determine the ideal generated byαk inside HOM(∅,i̲), for nonrepeating i̲. As discussed in Remark 3.12 (where αk is defined), our goal is to take any graph Γ on the punctured plane, with i̲ as its outer boundary and k(k+1)k(k+1) as its inner boundary, plug αk into the puncture, and reduce it to something in the ideal generated by the pictures of Section 3.5.(3.32)Our coloring conventions for this chapter will be that blue always represents the indexk, red represents k+1, and other colors tend to be arbitrary (often, the number of other colors appearing is also arbitrary). However, it will often happen that colors will appear in increasing or decreasing sequences, and these will be annotated as such. Note that blue or red may appear in the outer boundary as well, but at most once each.Let us studyΓ, and not bother to plug in αk. The only properties of αk which we need are the following:(3.33)This follows from (3.1), or just from isotopy. The same holds with colors being switched.(3.34)This is because the diagram reduces to ak-colored needle, with f=fk+1(fk+fk+1) inside, but f is fixed by sk, so it slides out of the needle, and the empty needle is equal to 0. A similar equality holds with colors switched.(3.35)This follows from the above and the dot forcing rules.The final property we use is that any graph only using colors<k-1 or >k+2 can slide freely across the puncture.Note however that, say, an arbitraryk-3 edge cannot automatically slide across the puncture, because a k-2 edge might be in the way, and this could be in turn obstructed by a k-1 edge, which cannot slide across the blue at all.The one-color reduction results apply to any simply connected planar region, so we may assume (without even using the relation (3.5)) that in a simply connected region of our choice, the i-graph for each i is a simple forest with double dots. Any connected component of an i-graph that does not encircle the puncture will be contained in a simply connected region, and hence can be simplified; this will be the crux of the proof. The proof is simple, but has many cases.Remark 3.25.
We will still need to use relation (3.5) as we simplify graphs.We will treat cases based on the “connectivity" ofΓ that is, how many of the blue and red boundary lines in the inner and outer boundary are connected with each other. We will rarely perform an operation which makes the graph more connected. At each stage, we will reduce the graph to something known to be in the ideal or break edges to decrease the connectivity. We call an edge coming from the puncture an interior line and one coming from the outer boundary an exterior line.Note also that any double dots that we can move to the exterior of the diagram become irrelevant, since the picture with those double dots is in the ideal generated by the picture without double dots. Also, any exterior boundary dots are irrelevant, since they are merely part of the mapφi̲ and do not interfere with the rest of the diagram at all.Step 1.
Suppose that the two interior red lines are in the same component ofΓk+1. Then, there is some innermost red path from one to the other, such that the interior of this path (the region towards the puncture) is simply connected. Applying reductions, we may assume that the k-graph in this region consists of a blue boundary dot with double dots, and the k+1-graph and k+2-graph each consists only of double dots. We may assume all double dots occur right next to one of the red lines coming from the puncture. The current picture is exactly like that in (3.33), except that there may be double dots inside and other colors may be present (also, there could be more red spokes emanating from the red arc, but these can be ignored or eliminated using (2.16) and tree reduction). However, the double dots may be forced out of the red enclosure at the cost of potentially breaking the red edge, and breaking it will cause the two red interior lines to be no longer in the same connected component. If there are no double dots, then all the remaining colors (which are <k-2 or >k+1) may be slid across the red line and out of the picture. Hence, we are left with the exact picture of (3.33), which is zero.
Thus, we may assume that the two red lines coming from the puncture are not in the same component. The same holds for the blue lines.Step 2.
Suppose that the component of one of the interior blue lines wraps the puncture, creating an internal region (which contains the puncture). Again, reducing in that internal region, the other interior blue line cannot connect to the boundary so it must reduce to a boundary dot (with double dots), the reds may not connect to each other so each reduces to a boundary dot, and as before we are left in the picture of (3.34) except possibly with double dots and other colors. If there are no double dots, all other colors may be slid out, and the picture is zero by (3.34). Again, we can put the double dots near the exterior, and forcing them out will break the blue arc. It is still possible that some other cycle still allows that component to wrap the puncture; however, this process needs to only be iterated a finite number of times, and finitely many arcs broken, until that component no longer wraps the puncture.
So, we may assume that the component of any interior line, red or blue, does not wrap the puncture. That component is contained in a simply connected region, so it reduces to a simple tree. Hence, we may assume that the components of interior lines either end immediately in boundary dots, or connect directly to an external line of the same color (at most one as such exists of each color).Step 3.
Suppose that there is a blue edge connecting an internal line directly to an external one. Consider the regionΓ′:
(3.36)
ThenΓ′ is simply connected. Other colors in Γ may leave Γ' to cross through the blue line; however, the colors k-1,k,k+1 may not. Therefore, reducing within Γ', we may end the internal blue line in a boundary dot and eliminate all other instances of the color blue (since they become irrelevant double dots on the exterior), reduce red to a simple forest where the two interior lines are not connected (again, ignoring irrelevant double dots), and reduce k-1 to either the empty diagram or an external boundary dot (depending on whether k+1∈i̲). Once this has been accomplished, the absence of the color k-1 implies that we may slide k-2 freely across the puncture! The color k-2 can be dealt with in the entire disk, which is simply connected, so it reduces to the empty diagram or an external boundary dot (depending on whether k+2∈i̲), with extraneous double dots. Then we may deal with color k-3, and so forth.
Thus, the existence of the blue edge implies that all colors<k can be ignored: they appear in irrelevant double dots, in irrelevant boundary dots, or not at all. Similarly, the existence of a red edge allows us to ignore all colors >k+1.Step 4.
Let us consider only the components of graphs which do not meet the internal boundary.Lemma 3.26.
Consider a component of a graph on a punctured disk which does not meet the internal boundary and which meets the external boundary at most once. Then it can be reduced to one of the following, with double dots on the exterior: the empty graph, a boundary dot, a circle around the puncture, a needle coming from the external boundary, with its eye around the puncture.Proof.
Suppose that the component splits the punctured plane intom regions. If the component is contained in a simply connected part of the punctured plane, we are done. This is always true for m=1. So we may suppose that m≥2 and, we have two distinguished regions: the external region, and the region containing the puncture. Any other region is one of two kinds, as illustrated in the following equality (due to (2.16):
(3.37)
On the right side we have a region which is contained in a simply connected part and thus can be eliminated by reduction (see Proposition3.9). On the left side the region is not contained in a simply connected part nor does it contain the puncture. However, any such region can be altered, using (2.16) as in the heuristic example above, into a cycle of the first kind. Therefore, we may assume there are exactly 2 regions.
In the event that there are two regions, we have a cycle which surrounds around the puncture and may have numerous branches into both regions, internal and external. However, each branch must be a tree lest another region be created. These trees reduce in the usual fashion, and therefore the internal branches disappear, and the external branches either disappear or connect directly to the single exterior boundary. Thus, we have either a needle or a circle. Double dots, as usual, can be forced out of the way possibly at the cost of breaking the cycle, and reducing to the casem=1.Let us now examine the remaining cases. We will ignore all parts of a graph which are double dots on the exterior, or are external boundary dots.Case 1.
Both a blue edge and a red edge connect an internal line to an external line. Then, as in Step3, all other colors can be ignored, and the entire graph is
(3.38)
This, as explained in Section3.5, is zk,k+1,i̲.Case 2.
A blue edge connects an internal line to an external line, and both red internal lines end in boundary dots. As discussed in Step3, we may ignore all colors <k, and both colors k and k+1 do not appear in a relevant fashion outside of what is already described. We may ignore the presence of any double dots. However, there may be numerous circles and needles colored ≥k+2 which surround the puncture and cross through the blue line, in an arbitrary order.
(3.39)Claim 6.
The sequence of circles and needles can be assumed to form an increasing sequence of colors, fromk+2,k+3,… until the final color, and assuming that only the final color may be a needle.Proof.
If the innermost circle/needle is not coloredk+2, then it may slide through the puncture, and will evaluate to zero by (2.18). So, suppose the innermost is k+2. If it is a needle, not a circle, then there can be no more k+2-colored circles, and no k+3-colored circles. Color k+4 can be pulled through the middle so resolved on the entire disk and hence can be ignored, and so too with k+5 and higher. This is the “needle" analogy to the conclusion of Step 3: the existence of an m-colored needle around the puncture and the lack of m or m+1 on the interior of the needle will allow us to ignore all colors ≥m+1.
So, suppose it is ak+2-colored circle. If the next circle/needle is colored ≥k+4 then it slides through the k+2 circle and the puncture and evaluates to zero. If the next circle/needle is also colored k+2, then we may use the following calculation to ignore it. The calculation begins by using (2.33)(3.40)
Thus, we may assume that the next circle/needle is coloredk+3. Again, if it is a needle, then we can ignore all other colors, and our picture is complete.
Similarly, the next circle/needle can not be colored≥k+5 lest it slide through, and it can not be colored k+3 lest we use (3.40). If it is colored k+2, then we may use the following calculation to ignore it. The calculation begins by using (3.5), and assumes green and purple are adjacent(3.41)
Thus, we can assume the next circle/needle is coloredk+4. If it is a needle, then all colors k+5 and higher can be ignored. Additional circles of color k+2 could run through the needle, but these could be slid inwards and reduced as before. So, if it is a needle, our picture is complete.
Finally, the next circle/needle can not be colored≥k+6 lest it slide, k+4 lest we use (3.40), k+3 lest we use (3.41), or k+2 lest we slide it inside and reduce it as above. Hence it is colored k+5, and if it is a needle, we are done. This argument can now be repeated ad infinitum.Thus, our final picture yieldszk,j,i̲ as in (3.26) or (3.29).Note that the case of a red edge works the same way, with a decreasing sequence instead of an increasing sequence.Case 3.
All the internal lines end in boundary dots. We may assume that the remainder of the graph consists of circles/needles around this diagram, but we have no restrictions at the moment on which colors may appear.Claim 7.
We may assume that the colors in circles/needles form an increasing sequence fromk+2 up, and a decreasing sequence from k-1 down (these sequences do not interact, so without loss of generality. we may assume the increasing sequence comes first, then the decreasing one). Only the highest and lowest color may be a needle.Proof.
The method of proof will be the same as the arguments of the previous case.
Consider the innermost circle/needle. If it is coloredk or k+1, then we may use (3.35) to reduce the situation to a previous case. If it is colored ≥k+3 or ≤k-2 then it slides through the puncture. So we may assume it is k+2 or k-1. If it is a k+2-colored (resp., k-1-colored) needle, then the usual arguments imply that all colors >k+2 (resp., <k-1) can be ignored. This same argument with needles will always work, so we will not discuss the circle/needle question again, and speak as though everything is a circle.
Assume that the first colors appearing are an increasing sequence fromk+2 to i and then a decreasing sequence from k-1 to j. Note that either sequence may be empty. If the next color appearing is ≤j-2 then it slides through the whole diagram and the puncture, and evaluates to zero. If the decreasing sequence is nonempty and the next color is j then we use (3.40); if it is ≥j+1 and ≤k-1 then we slide it as far in as it will go and use (3.41). If the decreasing sequence is nonempty and the next color is k then one can push it almost to the center, and use the following variant of (3.41):(3.42)
In this picture, green isk-1 and is the only thing in the way of the blue circle. The first equality uses (3.5), and the second equality uses (2.29), and eliminates the terms which vanish due to (3.34).
Continuously, if the decreasing sequence is empty and the next color isk, then we may use (3.35) as above. Any colors which are ≥k+1 do not depend on the increasing sequence, and instead use the exact analogs for the increasing sequence.
Hence, in any case in which the next color appearing is noti+1, or j-1, or the beginning of a new increasing/decreasing series, we may simplify the diagram to ignore the new circle. Induction will now finish the proof.Therefore, the resulting diagram is equal tozi,j,i̲, matching up either with (3.27) or (3.30).Since every possible graph can be reduced to a form which is demonstrably in the ideal generated byzi,j,i̲, we have proven that these elements do in fact generate the TL ideal Ii̲.


## 3.1. A Motivating Calculation

As discussed in the previous section, our desire is to take the quotient ofℋ𝒞1 by the doubled 6-valent vertex, and then take the graded Karoubi envelope.An important consequence of relations (2.26) and (2.18) is that(3.1)from which it follows, using (2.27), that(3.2)so the (monoidal) ideal generated inℋ𝒞1 by a doubled 6-valent vertex is the same as the ideal generated by the 6-valent vertex.Claim 1.
The following relations are all equivalent (the ideals they generate are equal)(3.3)(3.4)(3.5)(3.6)(3.7)Proof.
(3.3)⇒(3.4): add a dot, and use relation (2.26).
(3.4)⇒(3.5): add a dot to the top, and use (2.17).
(3.5)⇒(3.4): apply to the middle of the diagram.
(3.5)⇒(3.6): stretch dots from the blue strands towards the red strand using (2.17), and then apply (3.5) to the middle.
(3.6)⇒(3.7): use relation (2.27).
(3.7)⇒(3.3): use (3.2).Modulo 6-valent vertices, the relations (2.26) and (2.27) become (3.4) and (3.6) above. All other relations involving 6-valent vertices, namely, (2.28), (2.30), and (2.24), are sent to zero modulo 6-valent vertices. Relation (3.5) implies both (3.4) and (3.6) without reference to any graphs using 6-valent vertices. So, if we wish to rephrase our quotient in terms of graphs that never have 6-valent vertices, the sole necessary relation imposed by the fact that 6-valent vertices were sent to zero is the relation (3.5).Suppose, we only allow ourselves univalent, trivalent, and 4-valent vertices, but no 6-valent vertices, in a graphΓ. Then, the i-graph of Γ, which consists of all edges colored i and all vertices they touch, will be disjoint from the i+1- and i-1-graphs of Γ. The distant sliding property implies that the i-graph and the j-graph of Γ do not interact effectively, when i and j are distant. This will motivate the definition in the next section.


## 3.2. Diagrammatic Definition of𝒯ℒ𝒞

Definition 3.1.
We let𝒯ℒ𝒞1 be the monoidal category, with hom spaces enriched over graded vector spaces, defined as follows. Objects will be sequences of colored points on the line ℝ, which we will call i̲ or Ui̲. Consider the set whose elements are described as follows: (1)
for eachi∈I, consider a planar graph Γi in the strip, which is drawn with edges colored i (see Definition 2.13);(2)
the only vertices inΓi are univalent vertices (dots) and trivalent vertices;(3)
the graphsΓi and Γi+1 are disjoint. All graphs Γi are pairwise disjoint on the boundary;(4)
we consider isotopy classes of this data, so that one may apply isotopy to eachΓi individually so long as it stays appropriately disjoint.
This set has a grading, where the degree of a graph is+1 for each dot and -1 for each trivalent vertex, and the degree of an element of this set is the sum of the degrees for each graph Γi. Just as in Definition 2.15, each element of the set has a top and bottom boundary which is an object in 𝒯ℒ𝒞, and will be thought of as a map from the bottom boundary to the top. We let Hom𝒯ℒ𝒞1(Ui̲,Uj̲) be the graded vector space with basis given by elements of the set above with bottom boundary j̲ and top boundary i̲, modulo the relations (2.16) through (2.19), (2.29), and the new relation (3.5). As a reminder, the new relation is given here again.(3.8)
As before, composition of morphisms is given by vertical concatenation, the monoidal structure is given by horizontal concatenation, and relations are to be interpreted monoidally. This concludes the definition.Phrasing the definition in this fashion eliminates the need to add distant sliding rules, for these are now built into the notion of isotopy. Note that as we have stated it here,Γi and Γj may have edges which are embedded in a tangent fashion, or even entirely overlapped. However, such embeddings are isotopic to graph embeddings with only transverse edge intersections, which arise as 4-valent vertices in our earlier viewpoint.Proposition 3.2.
The category𝒯ℒ𝒞1 is isomorphic to ℋ𝒞1 modulo the 6-valent vertex.Proof.
Due to the observations of Section3.1, this is obvious.Hom spaces in𝒯ℒ𝒞1 are in fact enriched over graded R-bimodules, by placing double dots as before. However, they will no longer be free as left or right R-modules, as we will see.Remark 3.3.
Note that tree reduction (see Remark2.16) can now be applied to any tree of a single color in 𝒯ℒ𝒞, regardless of what other colors are present, since the only colors which can intersect the tree are distant colors which do not actually interfere.We denote by𝒯ℒ𝒞 the graded Karoubi envelope of 𝒯ℒ𝒞1, and by 𝒯ℒ𝒞2 the graded additive closure of 𝒯ℒ𝒞1. However, we will show that 𝒯ℒ𝒞2 is already idempotent closed, so that 𝒯ℒ𝒞2 and 𝒯ℒ𝒞 are the same.It is obvious that(3.9)Ui⊗Ui+1⊗Ui≅Ui,Ui+1⊗Ui⊗Ui+1≅Ui+1
in 𝒯ℒ𝒞1, and from the relation (3.6) and the simple calculation (using dot forcing rules) that(3.10)For the same reasons as in Section2.4, we still have Ui⊗Uj≅Uj⊗Ui for distant i,j, and Ui⊗Ui≅Ui{1}⊕Ui{-1} in 𝒯ℒ𝒞2. Therefore, 𝒯ℒ𝒞 is a potential categorification of 𝒯ℒ, and induces an adjoint pairing and a trace map ɛ𝒯ℒ𝒞 on 𝒯ℒ. At this point, we have not shown that the category 𝒯ℒ𝒞1 is nonzero, so this pairing could be 0.


## 3.3. Using the Adjoint Pairing

Proposition 3.4.
Let𝒞1 be an enriched category which is a potential categorification of 𝒯ℒ, whose objects are Ui̲ for sequences i̲. Let 𝒞2 be its additive graded closure, and let 𝒞 be its graded Karoubi envelope. Suppose that the induced trace map ɛ𝒞1 on 𝒯ℒ is equal to ɛcat. Then, the set of Ui̲{n} for n∈ℤ and 321-avoiding i̲ forms an exhaustive irredundant list of indecomposables in 𝒞2. In addition, 𝒞2 is Krull-Schmidt and idempotent closed (so 𝒞2 and 𝒞 are equivalent), and 𝒞 categorifies 𝒯ℒ.This proposition is an excellent illustration of the utility of the induced adjoint pairing. We prove it in a series of lemmas, which all assume the hypotheses above.Lemma 3.5.
The objectUi̲ in 𝒞1 has no nontrivial (homogeneous) idempotents when i̲ is 321-avoiding. Moreover, if both i̲ and j̲ are 321-avoiding, then Ui̲≅Uj̲{m} in 𝒞2 if and only if m=0 and ui̲=uj̲ in 𝒯ℒ.Proof.
Two 321-avoiding monomials in𝒯ℒ are equal only if they are related by the relation (2.8). Since this lifts to an isomorphism Ui⊗Uj≅Uj⊗Ui in 𝒞2, we have ui̲=uj̲⇒Ui̲≅Uj̲.
If an object has a 1-dimensional space of degree 0 endomorphisms, then it must be spanned by the identity map, and there can be no nontrivial idempotents. If an object has endomorphisms only in nonnegative degrees, then it can not be isomorphic to any nonzero degree shift of itself. If two objectsX and Y are such that both Hom(X,Y) and Hom(Y,X) are concentrated in strictly positive degrees, then no grading shift of X is isomorphic to Y, since there can not be a degree zero map in both directions.
Therefore, we need only show that (for 321-avoiding monomials)Ui̲ has endomorphisms concentrated in nonnegative degree, with a 1-dimensional degree 0 part, and that when ui̲≠uj̲, Hom(Ui̲,Uj̲) is in strictly positive degrees. This question is entirely determined by the pairing on 𝒯ℒ, since it only asks about the graded dimension of Hom spaces.
Wheni̲ is empty, we already know that (1,1)=(tn/(1-t2))[2]n-(t2/(1-t2)), which has degree 0 coefficient 1, and is concentrated in nonnegative degrees.
We know how to calculate(x,y) in 𝒯ℒ when x and y are monomials, and either x or y is not 1 (see Section 2.2). We draw x as a crossingless matching, draw y upside-down and place it below x, and close off the diagram: if there are m circles in the diagram, then (x,y)=tn[2]m-1/(1-t2). In particular, if m=n+1, then the Hom space will be concentrated in nonnegative degrees, with 1-dimensional degree 0 part. If m<n+1, then the Hom space will be concentrated in strictly positive degrees.
We leave it as an exercise to show that, ifx is a crossingless matching (i.e., a 321-avoiding monomial) then the closed diagram for (x,x) has exactly n+1 circles. The following example makes the statement fairly clear, where x̃ is x upside-down:
(3.11)
In this examplex has all 3 kinds of arcs which appear in a crossingless matching: bottom to top, bottom to bottom, and top to top. Each of these corresponds to a single circle in the diagram closure.
Similarly, there are fewer thann+1 circles in the diagram for (x,y) whenever the crossingless matchings x,y are nonequal. Consider the diagram above but with the region x removed. One can see that no circles are yet completed, and each boundary point of x's region is matched to the other by an arc. The number of circles is maximized when you pair these boundary points to each other, and this clearly gives the matching x. For any other matching y, two arcs will become joined into one, and fewer than n+1 circles will be created.Lemma 3.6.
𝒞2 is idempotent closed, and its indecomposables can all be expressed as grading shifts of Ui̲ for  321-avoiding i̲. It has the Krull-Schmidt property.Proof.
Since the Temperley-Lieb relations allow one to reduce a general word to a 321-avoiding word, one can show that everyUi̲ is isomorphic to a direct sum of shifts of Uj̲ for 321-avoiding j̲, using isomorphisms and direct sum decompositions instead of the analogous Temperley-Lieb relations. Clearly these shifted Uj̲ are all indecomposable, since they have no nontrivial idempotents; these are then all the indecomposables. Since every indecomposable in 𝒞2 has a graded local endomorphism ring (with maximal ideal given by positively graded morphisms), 𝒞2 is idempotent closed and Krull-Schmidt (see [33], Section 2.2).The Krull-Schmidt property implies that isomorphism classes of indecomposables form a basis for the Grothendieck group.Proof of Proposition3.4.
There is aℤ[t,t-1]-linear map of rings 𝒯ℒ→[𝒞2], which is evidently bijective because it sends the 321-avoiding basis to the 321-avoiding basis. Since 𝒞=𝒞2, we are done.This proposition shows that Lemma1.1 implies Theorem 1.2.Remark 3.7.
In analogy to the paper [15], the bulk of the proof of Theorem 1.2 lies in proving that hom spaces induce a particular adjoint pairing. Beyond that, we have mostly stated the obvious. Let us note that what is obvious for 𝒯ℒ and 𝒯ℒ𝒞 is not obvious at all when dealing with ℋ and ℋ𝒞. In particular, if we are given a category 𝒞1 which is a potential categorification of ℋ as in Proposition 3.4, we can not conclude that 𝒞 categorifies ℋ. We summarize the differences here.
It is clear (for both Hecke and Temperley-Lieb) that the mapℋ→[ℋ𝒞2] is well defined and surjective. The two main subtleties are (1) the difference between ℋ𝒞2 and ℋ𝒞, and (2) the injectivity of the map.
In general, one likes to examine the additive Grothendieck group only of idempotent closed categories with the Krull-Schmidt property, because this guarantees that indecomposables form a basis for the Grothendieck group. Thus it is convenient that𝒯ℒ𝒞2 is already idempotent closed. Thankfully, we have a result of Soergel [11] that proves that [ℋ𝒞2]≅[ℋ𝒞], as was discussed in Remark 2.5.
To show injectivity of the map in the𝒯ℒ case, we can identify a basis for 𝒯ℒ which is sent to a complete set of indecomposables, and then we can evaluate the trace map to show that these indecomposables are pairwise nonisomorphic. For ℋ𝒞, we do not currently know what the indecomposables (i.e., idempotents) are, nor do we know their preimage in ℋ. If we knew a class of indecomposables which decategorified to the Kazhdan-Lusztig basis, then we could use a similar argument to the above to show that they and their shifts form an exhaustive irredundant list of indecomposables in ℋ𝒞, and therefore that the map ℋ→[ℋ𝒞] is injective. Soergel discusses this in the last chapter of [11]. This is actually a deep question, shown by Soergel ([34], see also [8, 11]) to be equivalent to proving a version of the Kazhdan-Lusztig conjectures. In any case, the result depends on the base field 𝕜, and no simple proof has been found. In particular, to prove that the graded Karoubi closure of the diagrammatic category ℋ𝒞1 categorifies the Hecke algebra (for certain 𝕜), we must pass to the world of bimodules where Soergel's powerful geometric techniques will work. In particular, there is currently no proof of injectivity if one defines the category ℋ𝒞 diagrammatically over 𝕜=ℤ.
It should be emphasized that the story of𝒯ℒ is a particularly easy one (as is its Kazhdan-Lusztig theory). No high-powered technical machinery is needed, and the proofs of idempotent closure and injectivity are self-contained and diagrammatic. In fact, the arguments in this paper do work entirely over ℤ[1/2], as can be checked. Dividing by two must be allowed in order to split the identity of Uii into idempotents, as in (2.33); however, it is likely that the arguments would work over ℤ as well. Working over ℤ is discussed more extensively in [22].Remark 3.8.
A category𝒪 analog of the fact that 321-avoiding monomials lift to indecomposable Soergel bimodules, which remain indecomposable upon passage to the Temperley-Lieb quotient, can be found in Lemma 5.2 of [14].


## 3.4. Reductions

When we say that a graph or a morphism “reduces" to a set of other graphs, we mean that the morphism is in the𝕜-span of those graphs. We refer to a one-color graph, each of whose (connected) components is either a simple tree with respect to its boundary or a double dot, as a simple forest with double dots. If there are no double dots, it is a simple forest without double dots. Tree reduction implies that any graph Γi without cycles reduces to a simple forest with double dots. Note also that circles in a graph are equal to needles with a dot attached, and can be treated just like any other cycle.If there were only one color, we could iterate the following rule (which is an implication of the dot forcing rules and (2.18)) to break cycles:(3.12)We do something similar for the general case.Proposition 3.9.
In𝒯ℒ𝒞1, any morphism reduces to one where, for each i, the i-graph is a simple forest with double dots. Moreover, we may assume that all double dots are in the lefthand region.Proof.
We use induction on the total number of cycles (of any color) in the graph. Suppose there is a blue colored cycle: choose one so that it delineates a single region (i.e., there are no other cycles inside). There may be blue “spokes" going from this cycle into the interior, but no two spokes can meet, lest they create another region. By tree reduction on the spokes, we can assume that any blue appearing inside the cycle is in a different blue component than the cycle. Other colors may cross over the cycle, into the interior. If we view the interior of the cycle as a graph of its own, it has fewer total cycles so we may use induction. Since the boundary of the interior contains no blue color or colors adjacent to blue, they may be assumed to appear in the interior only in the form of double dots next to the cycle. Using dot-forcing rules, we reduce to two graphs: one with the cycle broken, and one with all these double dots on the exterior of the cycle. The former reduces by induction. For the latter, only distant colors enter the cycle, so they can be slid out of the way to leave an empty blue cycle, which is0 by the rule above.
We need to only do the base case, where the graph has no cycles. The dot-forcing rules imply that double dots may be moved to any region of the (multicolored) graph, at the cost of breaking a few lines. Breaking lines will never increase the number of cycles. Therefore, if we have a graph without cycles, tree reduction implies that we actually have a simple forest with double dots, and dot forcing allows us to move these double dots to the left. The breaking of lines may require more tree reduction, yielding more double dots, but this process is finite.Remark 3.10.
This proposition and its proof will apply to graphs in any connected simply connected region in the plane.Corollary 3.11.
For any nonrepeatingi̲, Hom𝒯ℒ𝒞1(∅,i̲) is generated (as a left or right R-module) by a single element φi̲ of degree d(i̲), pictured below
(3.13)Proof.
A simple forest with double dots and at most one boundary edge is no more than a boundary dot with double dots. Thus, any morphism reduces to a boundary dot for each color, accompanied by double dots.To show Lemma1.1, we need to only investigate Hom(∅,i̲) for increasing i̲, since we have already shown that the values of ɛ(ui̲) are determined by their values for increasing i̲. This space will be an R-bimodule where the left and right action are the same (since the lefthand and righthand regions are the same in any picture with no bottom boundary), so we view it as an R-module, and we have just shown that it is cyclic. Let Ii̲ be the ideal which is the kernel of the map R→HOM(∅,i̲) sending 1↦φi̲; we call it the TL ideal of i̲. Proving the Lemma 1.1 is to find Ii̲ and show that the graded dimension of R/Ii̲{d} is ɛcat(ui̲).Remark 3.12.
Since the spaceHomℋ𝒞1(∅,i̲) is a freeR-module, all polynomials in Ii̲ must have arisen from reducing to some morphism which contained the relation (3.5) to a “nice form," that is, φi̲ plus double dots. In other words, letting αi be the morphism pictured below, we want to plug αi into a bigger graph, reduce it to a nice form, and see what we get. 									(3.14)
Remember thatαi is actually just a 6-valent vertex with two dots attached (one red and one blue). This bigger graph, into which αi is plugged, will actually be a graph on the punctured plane or punctured disk with specified boundary conditions on both the outer and inner boundaries. The difficult graphical proofs of this paper just consist in analyzing such graphs. This is done by splitting the punctured plane into simply connected regions and using the above proposition.


## 3.5. Generators of the TL Ideal

The sequencei̲ is assumed to be nonrepeating.Proposition 3.13.
The TL ideal of∅ contains yi,j=deffifj(fi+2fi+1+2fi+2+⋯+2fj-1+fj) over all 1≤i<j≤n.
The TL ideal ofi̲ contains zi,j,i̲=defyi,j/gigj, where gi=fi if i∈i̲, gi=1 otherwise.We will prove that these actually generate the ideal in Proposition3.24 but postpone the proof as it is long and unenlightening.Proof.
Adding 4 dots toαi, or 6 dots to a 6-valent vertex, we get
(3.15)
This isyi,i+1=fifi+1(fi+fi+1)=(xi-xi+1)(xi+1-xi+2)(xi-xi+2). Even though we are not allowing 6-valent vertices in our diagrams, we will sometimes express yi,i+1 as(3.16)to avoid having to consider sums of graphs (it's easier for me to draw!).
To obtain the otheryi,j, note the following equalities under the action of Sn+1 on R:(3.17)sifi+1=fi+fi+1,si+1fi=fi+fi+1,sifi=-fi,sifj=fjfor|i-j|>1.
From this it follows by explicit calculation that(3.18)si-1yi,j-yi,j=yi-1,j,sj+1yi,j-yi,j=yi,j+1.
Now, when we surround a polynomialf with a j-colored circle and use (3.12), we are left with a j-colored double dot times ∂j(f), so we get f-sjf=∂j(f)fj.(3.19)
Combining this with the calculations we just made, we see that aj+1 circle around yi,j will yield yi,j+1 up to sign, and so forth. We now have numerous ways to express ±yi,j: for any i≤k≤j-1, take αk with 4 dots to get yk,k+1, and then surround it with concentric circles whose colors, from inside to out, are k+2,k+3,…,j and then k-1,k-2,…,i(3.20)
Clearly the colors of the increasing sequence and those of the decreasing sequence are distant, so a sequence likek-1,k+2,k+3,k-2,… is also okay, or any permutation which preserves the order of the increasing and the decreasing sequence individually.
For very similar reasons,zi,j,i̲ is in the TL ideal of i̲. Adding two or three dots to (3.5), we get several more equations.(3.21)(3.22)(3.23)
Again, for a variety of these pictures, we use shorthand like(3.24)
These give youzi,i+1,i̲ in the case where at least one of i,i+1∈i̲. Again by (3.12), putting a polynomial f in the eye of a j-colored needle will yield ∂j(f)=(f-sjf)/fj next to a j-colored boundary dot.(3.25)
This gives us several ways to drawzi,j,i̲.
If neitheri nor j are in i̲, then zi,j,i̲=yi,j and is pictured as above, but with additional boundary dots being put below to account for φi̲. Since these extra dots are generally irrelevant, we often do not bother to draw them.
Ifi∈i̲ and j∉i̲, we have two ways of drawing zi,j,i̲. One can take αi, connect one i input to the outer boundary, add dots, and surround it with circles colored i+2,i+3,…,j(3.26)
Alternatively, take somei<k<j, add dots to αk, and surround it with circles forming an increasing sequence k+2⋯j and a decreasing sequence k-1⋯i, except that the final i-colored circle is a needle(3.27)
The case ofj∈i̲ and i∉i̲ is obvious.
If bothi,j∈i̲, then we have several choices again. If j=i+1, then we must use(3.28)
but in general, we may either repeat (3.26) with a j-needle instead of a j-circle(3.29)
or repeat (3.27) with a j-needle instead of a j-circle(3.30)
In any case, it is clear that the polynomials above are in the TL ideal, and the claim is proven.Let us quickly consider the redundancy in this generating set of the ideal. Wheni>j let yi,j=defyj,i and zi,j,i̲=defzj,i,i̲.Corollary 3.14.
Suppose thati̲ is nonempty, and fix an index k∈i̲. Then, Ii̲ is generated by zk,j,i̲ for 1≤j≤n, j≠k. None of these generators are redundant.
None of the generatorsyi,j of I∅ are redundant.Proof.
We leave the checks of irredundancy to the reader, but a proof will also arise as a byproduct in the next section (see Remark3.17).
Suppose thatk∈i̲ but i,j∉i̲. If k<i<j, then zi,j,i̲=yi,j=fizk,j,i̲-fjzk,i,i̲ so that zi,j,i̲ is redundant. If i<k<j, then zi,j,i̲=fizk,j,i̲+fjzi,k,i̲. A similar statement holds for i<j<k. In the same vein, if k,l∈i̲ but i∉i̲, then for a given zk,l,i̲, only one of zk,i,i̲ or zl,i,i̲ is needed, and if k,l,m∈i̲, then any two of the three pairwise relations will imply the third.


## 3.6. Graded Dimensions

In this section, fix a nonrepeating sequencei̲. We assume in this section that the generators of Ii̲ are precisely the polynomials described in Proposition 3.13.Notation 5.
An element ofR can be written as a polynomial in fi, so let x=f1a1⋯fnan be a general monomial. Choose any i̲, possibly empty. Given a monomial x, let Jx⊂{1,…,n} be the subset containing i̲ and all indices j such that aj≠0. For a fixed subset J, let RJ be the subset of all monomials x with Jx=J. This inherently depends on the choice of i̲.Under the mapR→HOM(U∅,Ui̲), the image of RJ will be graphs where the colors appearing are precisely J. Every color in i̲ appears as a boundary dot, and every fj corresponds to a double dot of that color. The case J=∅ only occurs when i̲=∅ and R∅={1}.To find a basis forR/Ii̲, we will use the Bergman Diamond Lemma [35] for commutative rings.Definition 3.15.
LetA be a free commutative polynomial ring, where monomials are given a partial order with the DCC, compatible with multiplication in that x<y⇒ax<ay. Let I be an ideal generated by relations r of the form xr=yr, where xr is a monomial and yr is a linear combination of monomials which are each less than xr in the partial order. A reduction is an application of a relation r to replace xr with yr, but not the other way around (a reduction always lowers the partial order on each term in a polynomial). One says a polynomial xreduces to y if y can be obtained from x by a series of reductions applied to monomials in x. A monomial is called irreducible if it does not have xr as a factor for any relation r. An inclusion ambiguity is a monomial x=ab, where x=xr for some r, and b=xr′ for some r'≠r. An overlap ambiguity is a monomial x=abc, where ab=xr for some r and bc=xr′ for some r'≠r. Each ambiguity has two natural reductions, and one says the ambiguity is resolvable if the two reductions are then jointly reducible to the same element.Lemma 3.16 (Bergman Diamond Lemma for Commutative Rings, [35]).
With these definitions in place, if every inclusion and overlap ambiguity is resolvable, then the images of the irreducible monomials form a basis forA/I.This process may become more transparent from the example below; in addition, Bergman's paper has a number of nice examples for the trickier, noncommutative version. We treat two separate cases, wheni̲=∅ and when i̲≠∅.Claim 2.
Leti̲=∅. We place the lexicographic order on monomials in R, so that f1<f2<⋯. The relation yi,j=0 for i<j will be rewritten as fifj2=-fifj(fi+2∑i<k<jfk), which replaces fifj2 with a sum of monomials all lower in the order. For each J≠∅, the irreducible monomials in RJ are precisely fkm∏i∈Jfi, where k is the minimal index in J and m≥0 (note: the exponent of fk is m+1≥1). When J=∅, 1 is irreducible. Irreducibles form a basis for R/I∅.Proof.
A monomial is irreducible iffifj2 never appears as a factor for any i<j. Because of this, the classification of irreducible monomials in each RJ is obvious. There are no inclusion ambiguities between relations, since they are all homogeneous and degree 3. There are two kinds of overlap ambiguities, both labelled by a choice of i<l<j.
For the first ambiguity, one can reducex=fiflfj2 by either reducing flfj2 or fifj2. Applying the former reduction, x↦fiflfj(-fl-2∑l<k<jfk) which has a term given by -fifl2fj that can be further reduced, yielding fiflfj(fi+2∑i<k<lfk-2∑l<k<jfk). Applying the latter reduction, x↦fiflfj(-fi-2∑i<k<jfk)=fiflfj(-fi-2∑i<k<lfk-2fl-2∑l<k<jfk), which has a term given by -2fifl2fj that can be further reduced, yielding fiflfj(-fi-2∑i<k<lfk-2∑l<k<jfk+2fi+4∑i<k<lfk)=fiflfj(fi+2∑i<k<lfk-2∑l<k<jfk). Since these agree, the ambiguity is resolvable.
For the second ambiguity, one can reducex=fifl2fj2 by either reducing fifl2 or flfj2. A very similar calculation shows that this ambiguity is resolvable as well. Therefore, the Bergman diamond lemma implies that irreducibles form a basis for the quotient.Remark 3.17.
This also proves that none of theyi,j is redundant. Removing yi,j from the ideal, we may apply the same Bergman diamond lemma argument to say that irreducibles form a basis for the quotient. However, with no yi,j, the monomial fifj2 is irreducible, and the quotient is larger than before. A similar statement can be made about the zk,j,i̲ below.WhenJ≠∅, the graded rank of the irreducibles in RJ is t2|J|/(1-t2). When J is empty, the only element of RJ is 1. So the graded rank of R/I∅ is 1+∑J≠∅(t2|J|/(1-t2)), but ∑Jt2|J|=(1+t2)n since every fi may either appear or not appear, independently of every other. Hence ∑J≠∅t2|J|=(1+t2)n-1. Putting it all together, the graded rank is ((1+t2)n-t2)/(1-t2)=(tn[2]n-t2)/(1-t2). Hence we have proven the following claim.Claim 3.
The graded dimension ofR/I∅ is exactly ɛcat(u∅).Claim 4.
Leti̲≠∅, and fix k∈i̲. We choose a different order on indices, where k<k+1<k-1<k+2<k-2<⋯, and then place the lexicographic order on monomials. The relation zk,j,i̲ for j≠k will be rewritten in decreasing order format as either fj2=-fj(fk+2∑lfl) for j∉i̲, or fj=-(fk+2∑lfl) for j∈i̲, where the sum is over l between k and j. Then, the irreducible monomials in RJ are precisely fkm∏j∈J∖i̲fj for m≥0. Irreducibles form a basis for R/Ii̲.Proof.
An irreducible polynomial will be a polynomial which does not havefj2 as a factor, for k≠j∉i̲, and does not have fj as a factor for k≠j∈i̲. The classification of irreducibles in RJ is now obvious. There are no ambiguities whatsoever, so we are done by the Bergman diamond lemma.The graded rank of irreducibles inRJ is t2|J|-2d/(1-t2), for d the length of i̲ (remember that i̲⊂J). Thus, the graded rank of R/Ii̲ is ∑i̲⊂J(t2|J|-2d/(1-t2))=(1+t2)n-d/(1-t2), and the graded rank of R/Ii̲{d} is td(1+t2)n-d/(1-t2)=tn[2]n-d/(1-t2). Hence, one considers the following.Claim 5.
The graded dimension ofR/Ii̲{d(i̲)} is exactly ɛcat(ui̲).This is clearly sufficient to prove Lemma1.1, modulo Proposition 3.24.


## 3.7. Weyl Lines and Disoriented Tubes

We now give two alternate interpretations of the TL idealsIi̲. We continue to assume that i̲ is nonrepeating and zi,j,i̲ generates Ii̲.Definition 3.18.
LetV be the reflection representation of Sn+1, such that R=ℂ[f1,…,fn] is the coordinate ring of V. Note that the linear equations which cut out reflection-fixed hyperplanes are precisely wi,j=fi+fi+1+⋯+fj=xi-xj+1 for i≤j. A Weyl line is a line in V through the origin which is defined by the intersection of reflection-fixed hyperplanes; it is given by a choice of n-1 transversely intersecting reflection-fixed hyperplanes. Given a nonrepeating sequence i̲, one says that a Weyl line is transverse toi̲ if it is transverse to (i.e., not contained in) the hyperplanes fk=0 for each k∈i̲.Proposition 3.19.
The TL ideal ofi̲ is the ideal associated with the union of all Weyl lines transverse to i̲ (with its reduced scheme structure).Example 3.20.
Letn=3. One can check that f1f2(f1+f2)=f2f3(f2+f3)=f1f3(f1+2f2+f3)=0 cuts out 7 lines in V, namely,  (1)
f1=f2=f1+f2=0,(2)
f1=f3=0,(3)
f2=f3=f2+f3=0,(4)
f1=f2+f3=f1+f2+f3=0,(5)
f1+f2=f3=f1+f2+f3=0,(6)
f2=f1+f2+f3=0,(7)
f1+f2=f2+f3=0,These 7 lines are precisely the 7 lines cut out by the intersection of pairs of reflection-fixed hyperplanes. There are 6 reflection-fixed hyperplanes, given by equations f1, f2, f3, f1+f2, f2+f3, and f1+f2+f3, or alternatively, by xi-xj for 4≥j>i≥1. Intersecting pairs of hyperplanes will give a line, and occasionally this line is forced to lie in a third hyperplane, as in the list above. One can check that this list covers all pairs of hyperplanes which give distinct lines as their intersection.Proof.
This is not difficult to show, but since we have not seen it elsewhere, we provide a complete proof. First, we show by induction onn that the ideal I∅ cuts out the Weyl lines with the reduced scheme structure. The case n=1 is trivial (and n=2 is also obvious).
For any1≤k≤n, consider the hyperplane fk=0 as an n-1-dimensional space V′, with an action of Sn+1/〈sk〉≅Sn. Giving Sn a Coxeter structure with simple reflections si for i≠k (note that sk+1=(k+1,k+2)=(k,k+2) in the quotient), it is quite easy to see that V' is the reflection representation of Sn. Moreover, the Weyl hyperplanes are cut out by wi,j′=fi+fi+1+⋯+fj (where fk=0 so it may be left out of the sum) for i,j≠k, and the equivalent polynomials yi,j′ also have the same formulae, and are indexed by i,j≠k. Therefore, for i,j≠k, the images of wi,j are just wi,j′, and the same for yi,j and yi,j′. Moreover, if either i or j equals k, then yi,j=0 on fk=0, and wi,j is redundant on fk=0, being equal to some wi′,j′. By induction, yi,j′ cut out the Weyl lines with the reduced scheme structure on V′, and therefore the vanishing set of yi,j agrees with the Weyl lines on fk=0.
If allfk≠0, then it is easy to see that the yi,j cut out a single line with the reduced scheme structure, namely, -f1=f2=-f3=⋯=(-1)nfn. This is a Weyl line, the intersection of all wi,i+1. We wish to show that this is the only Weyl line transverse to all fk=0. We can show this by induction as well (again, the base case n=2 is easy). Suppose we are given n-1 transverse hyperplanes wi,j. If any two involve the index n, that is, wi,n and wj,n, then we may replace the pair with wi,n and wi,j since they have the same intersection (and wi,j is not already in the set, or the intersection would not be transverse). So, we may assume that at most one of the chosen hyperplanes involves the index n, but then we have n-2 transverse hyperplanes which only involve indices {1,…,n-1}, which must then be mutually transverse to fn=0. Letting V' be the hyperplane fn=0 viewed as a reflection representation as above, we have n-2 transverse hyperplanes which cut out a Weyl line transverse to fk=0 for all 1≤k≤n-1. By induction, that Weyl line is -f1=f2=-f3=⋯=(-1)n-1fn-1 (which holds true modulo fn=0), but repeating the same argument for the index k instead, we leave out the kth term and get -f1=f2=⋯=(-1)kfk̂=⋯=(-1)nfn modulo fk=0. Together, all these equalities imply that -f1=f2=⋯=(-1)nfn everywhere.
One might be worried, because of the restrictions used in the induction step, thatIi̲ does not give the reduced structure on the Weyl lines at the origin. However, Ii̲ is a homogeneous ideal which cuts out a reduced 0-dimensional subscheme of ℙ(V), so that its vanishing on V is the cone of a reduced scheme, and hence is reduced. This concludes the proof that I∅ cuts out the Weyl lines with the reduced scheme structure.
Fori̲≠∅, I∅⊂Ii̲ and the vanishing of Ii̲ is contained in that of I∅. Choose k∈i̲. If fk=0 then zk,k+1,i̲ is equal to fk+1a, where a=1,2 depending on whether k+1∈i̲, but either way we get that fk+1=0. Then zk,k+2=fk+2a for a=1,2, and so forth. Therefore fk=0 only intersects the vanishing of Ii̲ at the origin (as sets). It is clear that, on the open set where fk≠0 for all k∈i̲, the polynomials zi,j,i̲ and yi,j have the same vanishing (as schemes), since they differ by a unit. The same cone argument shows that Ii̲ gives the reduced structure at the origin.Remark 3.21.
In particular,I∅ is contained in every ideal, and the category 𝒯ℒ𝒞1 is manifestly R/I∅-linear.Remark 3.22.
LetZ be the union of all Weyl lines in V. The previous results should lead one to guess that the Temperley-Lieb algebra should be connected to the geometry of the Sn+1 action on Z via 𝒯ℒ𝒞, in much the same way that the Hecke algebra is connected to the reflection representation via ℋ𝒞 (see [11]). However, at the moment, we have no way to formulate the category 𝒯ℒ𝒞 in terms of coherent sheaves on Z×Z (i.e., R/I∅-bimodules) or the derived category thereof. Describing 𝒯ℒ𝒞 using sheaves on Z seems like an interesting question.
As an example of the difficulties, letUi be the bimodule R/Ii⊗R/Ii{-1}, where the tensor is over Rsi; this should be the equivalent of the Soergel bimodule Bi. Then, there is a degree-1 map R/I∅→Ui sending 1 to xi⊗1-1⊗xi+1 (the boundary dot on the top), but there is no degree-1 map Ui→R/I∅ (the boundary dot on the bottom); such a map should send 1⊗1 to 1. There is only a degree-3 map, sending 1⊗1 to fi (the boundary dot with a double dot). A similar problem occurs again: the trivalent vertex seems to be defined only in one direction.Now, we describe briefly the topological intuition associated with the category𝒯ℒ𝒞, and another way to view Ii̲. These remarks will not be used in the remainder of the paper, nor will we give a proof. The reader should be acquainted with the section on 𝔰𝔩2-foams in Vaz's paper [16].Remark 3.23.
Letℱ be the functor from 𝒯ℒ𝒞1 to the category of disoriented cobordisms Foam2, as defined in Vaz's paper. If fi is the double dot colored i, then one can easily see that ℱ sends fi to a tube connecting the ith sheet to the (i+1)th sheet, with a disorientation on it. If the double dot appears in a larger morphism φ, such that in ℱ(φ), the ith sheet and the (i+1)th sheet are already connected by a saddle or tube, then adding another tube between them does nothing more than adding a disoriented handle to the existing surface. Note that the map φi̲ previously defined will connect the ith sheet to the (i+1)th sheet for any i∈i̲.
Suppose that theith, (i+1)th, and (i+2)th sheets are all connected in a cobordism. Then fi adds a handle on the left side of the (i+1)th sheet, fi+1 adds a handle on the right side, and these two disoriented surfaces are equal up to a minus sign in Foam2. This fact is essentially the statement that:(3.31)
In other words, the algebra𝕜[f1,…,fn] maps to Foam2, sending fi to the disoriented tube between the ith and (i+1)th sheet. The ideal Ii̲ is clearly in the kernel of this action when applied to the cobordism ℱ(φi̲). In fact, it is precisely the kernel, using the argument of Proposition 4.2 in [36]: for any distinct monomials in a basis for R/Ii̲, their image in Foam2 will have independent evaluations with respect to some closure of the cobordism. We do not do the calculation here. The usual arguments involving adjoint pairings imply that the faithfulness of the functor ℱ can be checked on Hom(∅,i̲). Therefore, the functor ℱ is faithful.


## 3.8. Proof of Generation

Proposition 3.24.
The TL idealI∅ is generated by yi,j=deffifj(fi+2fi+1+2fi+2+⋯+2fj-1+fj) over all 1≤i<j≤n.
The TL idealIi̲ is generated by all zi,j,i̲=defyi,j/gigj, where gi=fi if i∈i̲, gi=1 otherwise.We wish to determine the ideal generated byαk inside HOM(∅,i̲), for nonrepeating i̲. As discussed in Remark 3.12 (where αk is defined), our goal is to take any graph Γ on the punctured plane, with i̲ as its outer boundary and k(k+1)k(k+1) as its inner boundary, plug αk into the puncture, and reduce it to something in the ideal generated by the pictures of Section 3.5.(3.32)Our coloring conventions for this chapter will be that blue always represents the indexk, red represents k+1, and other colors tend to be arbitrary (often, the number of other colors appearing is also arbitrary). However, it will often happen that colors will appear in increasing or decreasing sequences, and these will be annotated as such. Note that blue or red may appear in the outer boundary as well, but at most once each.Let us studyΓ, and not bother to plug in αk. The only properties of αk which we need are the following:(3.33)This follows from (3.1), or just from isotopy. The same holds with colors being switched.(3.34)This is because the diagram reduces to ak-colored needle, with f=fk+1(fk+fk+1) inside, but f is fixed by sk, so it slides out of the needle, and the empty needle is equal to 0. A similar equality holds with colors switched.(3.35)This follows from the above and the dot forcing rules.The final property we use is that any graph only using colors<k-1 or >k+2 can slide freely across the puncture.Note however that, say, an arbitraryk-3 edge cannot automatically slide across the puncture, because a k-2 edge might be in the way, and this could be in turn obstructed by a k-1 edge, which cannot slide across the blue at all.The one-color reduction results apply to any simply connected planar region, so we may assume (without even using the relation (3.5)) that in a simply connected region of our choice, the i-graph for each i is a simple forest with double dots. Any connected component of an i-graph that does not encircle the puncture will be contained in a simply connected region, and hence can be simplified; this will be the crux of the proof. The proof is simple, but has many cases.Remark 3.25.
We will still need to use relation (3.5) as we simplify graphs.We will treat cases based on the “connectivity" ofΓ that is, how many of the blue and red boundary lines in the inner and outer boundary are connected with each other. We will rarely perform an operation which makes the graph more connected. At each stage, we will reduce the graph to something known to be in the ideal or break edges to decrease the connectivity. We call an edge coming from the puncture an interior line and one coming from the outer boundary an exterior line.Note also that any double dots that we can move to the exterior of the diagram become irrelevant, since the picture with those double dots is in the ideal generated by the picture without double dots. Also, any exterior boundary dots are irrelevant, since they are merely part of the mapφi̲ and do not interfere with the rest of the diagram at all.Step 1.
Suppose that the two interior red lines are in the same component ofΓk+1. Then, there is some innermost red path from one to the other, such that the interior of this path (the region towards the puncture) is simply connected. Applying reductions, we may assume that the k-graph in this region consists of a blue boundary dot with double dots, and the k+1-graph and k+2-graph each consists only of double dots. We may assume all double dots occur right next to one of the red lines coming from the puncture. The current picture is exactly like that in (3.33), except that there may be double dots inside and other colors may be present (also, there could be more red spokes emanating from the red arc, but these can be ignored or eliminated using (2.16) and tree reduction). However, the double dots may be forced out of the red enclosure at the cost of potentially breaking the red edge, and breaking it will cause the two red interior lines to be no longer in the same connected component. If there are no double dots, then all the remaining colors (which are <k-2 or >k+1) may be slid across the red line and out of the picture. Hence, we are left with the exact picture of (3.33), which is zero.
Thus, we may assume that the two red lines coming from the puncture are not in the same component. The same holds for the blue lines.Step 2.
Suppose that the component of one of the interior blue lines wraps the puncture, creating an internal region (which contains the puncture). Again, reducing in that internal region, the other interior blue line cannot connect to the boundary so it must reduce to a boundary dot (with double dots), the reds may not connect to each other so each reduces to a boundary dot, and as before we are left in the picture of (3.34) except possibly with double dots and other colors. If there are no double dots, all other colors may be slid out, and the picture is zero by (3.34). Again, we can put the double dots near the exterior, and forcing them out will break the blue arc. It is still possible that some other cycle still allows that component to wrap the puncture; however, this process needs to only be iterated a finite number of times, and finitely many arcs broken, until that component no longer wraps the puncture.
So, we may assume that the component of any interior line, red or blue, does not wrap the puncture. That component is contained in a simply connected region, so it reduces to a simple tree. Hence, we may assume that the components of interior lines either end immediately in boundary dots, or connect directly to an external line of the same color (at most one as such exists of each color).Step 3.
Suppose that there is a blue edge connecting an internal line directly to an external one. Consider the regionΓ′:
(3.36)
ThenΓ′ is simply connected. Other colors in Γ may leave Γ' to cross through the blue line; however, the colors k-1,k,k+1 may not. Therefore, reducing within Γ', we may end the internal blue line in a boundary dot and eliminate all other instances of the color blue (since they become irrelevant double dots on the exterior), reduce red to a simple forest where the two interior lines are not connected (again, ignoring irrelevant double dots), and reduce k-1 to either the empty diagram or an external boundary dot (depending on whether k+1∈i̲). Once this has been accomplished, the absence of the color k-1 implies that we may slide k-2 freely across the puncture! The color k-2 can be dealt with in the entire disk, which is simply connected, so it reduces to the empty diagram or an external boundary dot (depending on whether k+2∈i̲), with extraneous double dots. Then we may deal with color k-3, and so forth.
Thus, the existence of the blue edge implies that all colors<k can be ignored: they appear in irrelevant double dots, in irrelevant boundary dots, or not at all. Similarly, the existence of a red edge allows us to ignore all colors >k+1.Step 4.
Let us consider only the components of graphs which do not meet the internal boundary.Lemma 3.26.
Consider a component of a graph on a punctured disk which does not meet the internal boundary and which meets the external boundary at most once. Then it can be reduced to one of the following, with double dots on the exterior: the empty graph, a boundary dot, a circle around the puncture, a needle coming from the external boundary, with its eye around the puncture.Proof.
Suppose that the component splits the punctured plane intom regions. If the component is contained in a simply connected part of the punctured plane, we are done. This is always true for m=1. So we may suppose that m≥2 and, we have two distinguished regions: the external region, and the region containing the puncture. Any other region is one of two kinds, as illustrated in the following equality (due to (2.16):
(3.37)
On the right side we have a region which is contained in a simply connected part and thus can be eliminated by reduction (see Proposition3.9). On the left side the region is not contained in a simply connected part nor does it contain the puncture. However, any such region can be altered, using (2.16) as in the heuristic example above, into a cycle of the first kind. Therefore, we may assume there are exactly 2 regions.
In the event that there are two regions, we have a cycle which surrounds around the puncture and may have numerous branches into both regions, internal and external. However, each branch must be a tree lest another region be created. These trees reduce in the usual fashion, and therefore the internal branches disappear, and the external branches either disappear or connect directly to the single exterior boundary. Thus, we have either a needle or a circle. Double dots, as usual, can be forced out of the way possibly at the cost of breaking the cycle, and reducing to the casem=1.Let us now examine the remaining cases. We will ignore all parts of a graph which are double dots on the exterior, or are external boundary dots.Case 1.
Both a blue edge and a red edge connect an internal line to an external line. Then, as in Step3, all other colors can be ignored, and the entire graph is
(3.38)
This, as explained in Section3.5, is zk,k+1,i̲.Case 2.
A blue edge connects an internal line to an external line, and both red internal lines end in boundary dots. As discussed in Step3, we may ignore all colors <k, and both colors k and k+1 do not appear in a relevant fashion outside of what is already described. We may ignore the presence of any double dots. However, there may be numerous circles and needles colored ≥k+2 which surround the puncture and cross through the blue line, in an arbitrary order.
(3.39)Claim 6.
The sequence of circles and needles can be assumed to form an increasing sequence of colors, fromk+2,k+3,… until the final color, and assuming that only the final color may be a needle.Proof.
If the innermost circle/needle is not coloredk+2, then it may slide through the puncture, and will evaluate to zero by (2.18). So, suppose the innermost is k+2. If it is a needle, not a circle, then there can be no more k+2-colored circles, and no k+3-colored circles. Color k+4 can be pulled through the middle so resolved on the entire disk and hence can be ignored, and so too with k+5 and higher. This is the “needle" analogy to the conclusion of Step 3: the existence of an m-colored needle around the puncture and the lack of m or m+1 on the interior of the needle will allow us to ignore all colors ≥m+1.
So, suppose it is ak+2-colored circle. If the next circle/needle is colored ≥k+4 then it slides through the k+2 circle and the puncture and evaluates to zero. If the next circle/needle is also colored k+2, then we may use the following calculation to ignore it. The calculation begins by using (2.33)(3.40)
Thus, we may assume that the next circle/needle is coloredk+3. Again, if it is a needle, then we can ignore all other colors, and our picture is complete.
Similarly, the next circle/needle can not be colored≥k+5 lest it slide through, and it can not be colored k+3 lest we use (3.40). If it is colored k+2, then we may use the following calculation to ignore it. The calculation begins by using (3.5), and assumes green and purple are adjacent(3.41)
Thus, we can assume the next circle/needle is coloredk+4. If it is a needle, then all colors k+5 and higher can be ignored. Additional circles of color k+2 could run through the needle, but these could be slid inwards and reduced as before. So, if it is a needle, our picture is complete.
Finally, the next circle/needle can not be colored≥k+6 lest it slide, k+4 lest we use (3.40), k+3 lest we use (3.41), or k+2 lest we slide it inside and reduce it as above. Hence it is colored k+5, and if it is a needle, we are done. This argument can now be repeated ad infinitum.Thus, our final picture yieldszk,j,i̲ as in (3.26) or (3.29).Note that the case of a red edge works the same way, with a decreasing sequence instead of an increasing sequence.Case 3.
All the internal lines end in boundary dots. We may assume that the remainder of the graph consists of circles/needles around this diagram, but we have no restrictions at the moment on which colors may appear.Claim 7.
We may assume that the colors in circles/needles form an increasing sequence fromk+2 up, and a decreasing sequence from k-1 down (these sequences do not interact, so without loss of generality. we may assume the increasing sequence comes first, then the decreasing one). Only the highest and lowest color may be a needle.Proof.
The method of proof will be the same as the arguments of the previous case.
Consider the innermost circle/needle. If it is coloredk or k+1, then we may use (3.35) to reduce the situation to a previous case. If it is colored ≥k+3 or ≤k-2 then it slides through the puncture. So we may assume it is k+2 or k-1. If it is a k+2-colored (resp., k-1-colored) needle, then the usual arguments imply that all colors >k+2 (resp., <k-1) can be ignored. This same argument with needles will always work, so we will not discuss the circle/needle question again, and speak as though everything is a circle.
Assume that the first colors appearing are an increasing sequence fromk+2 to i and then a decreasing sequence from k-1 to j. Note that either sequence may be empty. If the next color appearing is ≤j-2 then it slides through the whole diagram and the puncture, and evaluates to zero. If the decreasing sequence is nonempty and the next color is j then we use (3.40); if it is ≥j+1 and ≤k-1 then we slide it as far in as it will go and use (3.41). If the decreasing sequence is nonempty and the next color is k then one can push it almost to the center, and use the following variant of (3.41):(3.42)
In this picture, green isk-1 and is the only thing in the way of the blue circle. The first equality uses (3.5), and the second equality uses (2.29), and eliminates the terms which vanish due to (3.34).
Continuously, if the decreasing sequence is empty and the next color isk, then we may use (3.35) as above. Any colors which are ≥k+1 do not depend on the increasing sequence, and instead use the exact analogs for the increasing sequence.
Hence, in any case in which the next color appearing is noti+1, or j-1, or the beginning of a new increasing/decreasing series, we may simplify the diagram to ignore the new circle. Induction will now finish the proof.Therefore, the resulting diagram is equal tozi,j,i̲, matching up either with (3.27) or (3.30).Since every possible graph can be reduced to a form which is demonstrably in the ideal generated byzi,j,i̲, we have proven that these elements do in fact generate the TL ideal Ii̲.


## 4. Irreducible Representations

In this section, we may vary the number of strands appearing in the Temperley-Lieb algebra. When𝒯ℒ appears, it designates the Temperley-Lieb algebra on n+1 strands, but 𝒯ℒk designates the algebra on k strands.


### 4.1. Cell Modules

The Temperley-Lieb algebra has the structure of a cellular algebra, a concept first defined by Graham and Lehrer [37]. One feature of cellular algebras is that they are equipped with certain modules known as cell modules. Cell modules provide a complete set of nonisomorphic irreducible modules in many cases (such as 𝒯ℒ in type A). Cell modules come equipped with a basis and a bilinear form, making them obvious candidates for categorification. We will not go into detail on cellular algebras here, or even use their general properties; instead we will describe the cell modules explicitly and pictorially for the case of 𝒯ℒ, where things are unusually simple. Nothing in this section or the next is particularly original, and we state some standard results without proof.Notation 6.
Consider a crossingless matching in the planar strip betweenn points on the bottom boundary and m points on the top. We call this briefly an (n,m)diagram. In the terminology of [30], there are two kinds of arcs in a diagram: horizontal arcs which connect two points on the top (let us call it a top arc), or two points on the bottom (bottom arc) and vertical arcs which connect a point on the top to one on the bottom. Elsewhere in the literature, vertical arcs are called through-strands. An (n,k) diagram with exactly kthrough-strands (and therefore no top arcs) has an isotopy representative with only “caps" (local maxima) and no “cups" (local minima) so it is called an (n,k)cap diagram. A (k,n) diagram with k through-strands is called a (k,n)cup diagram.The set of all(n,m) diagrams can be partitioned by the number of through-strands. Any (n,m) diagram with k through-strands can be expressed as the concatenation of an (n,k) cap diagram with a (k,m) cup diagram in a unique way. For an illustration of this concept, see Figure 5.On the left side, a(7,7) diagram with k=3 through-strands and l=2 top arcs (resp., bottom arcs) is decomposed into a (7,3) cap diagram a-composed with a (3,7) cup diagram z. On the right side, an element of 𝒯ℒ3 is obtained by composing a and z in the opposite order.
(a)(b)In an(m,m) diagram the number l of top arcs equals the number of bottom arcs, and if k is the number of through-strands, then k+2l=m. We will typically use k and l to represent the number of through-strands and top arcs in an (m,m) diagram henceforth.Notation 7.
LetX be the set of all (n+1,n+1) diagrams. Let ω be the endomorphism of X sending each diagram to its vertical flip. We will write the operation on diagrams of reduced vertical concatenation by ∘: a∘b places a above b and remove any circles. Let Xk be the set of crossingless matchings with exactly k through-strands. Let Mk be the set of all (n+1,k) cap diagrams, so that ω(Mk) is the set of all (k,n+1) cup diagrams.Definition 4.1.
LetLk be the free ℤ[t,t-1] module spanned by Mk, the (n+1,k) cap diagrams. We place a right 𝒯ℒ-module structure on Lk by concatenation, where circles become factors of [2] as usual, and any resulting diagram with fewer than k through-strands is sent to 0. This is the cell module for cell k, and it is irreducible.Example 4.2.
The only diagram inXn+1 corresponds to the identity map in 𝒯ℒ. The cell module Ln+1 has rank 1 over ℤ[t,t-1], and its generator is killed by all ui. We will take this as the definition of the sign representation of 𝒯ℒ.Example 4.3.
The next cell moduleLn-1 has rank n over ℤ[t,t-1], having generators vi, i=1⋯n (see Figure 6), such that
(4.1)vjui={[2]vjifi=j,viifiandjareadjacent,0ifiandjaredistant.Figure 6
A basis for the cell moduleLn-1, consisting of (n+1,n-1) cap diagrams (here, n=4).Given a(n+1,k) cap diagram a and a (k,n+1) cup diagram z, there are two things we can do: take the composition z∘a to obtain an element called cz,a of Xk or take the composition a∘z to get an element of 𝒯ℒk (there may be additional circles created, and the final diagram may have fewer than k through-strands). Both compositions have the same closure on the punctured plane. Note that ω(cz,a)=cω(a),ω(z). The seemingly extraneous use of the notation c·,· is standard for cellular algebras.Proposition 4.4.
There is, up to rescaling, a unique pairing(,):Lk×Lk→ℤ[[t,t-1]] for which ui is self-adjoint, that is, (aui,b)=(a,bui). Given cup diagrams a and b in Mk, one evaluates (a,b) by considering the closure of cω(a),b∈𝒯ℒ, or equivalently the closure of b∘ω(a)∈𝒯ℒk. If the diagram has nesting number k, one returns a scalar times [2] raised to the number of circles; if it has nesting number <k, one returns zero. This is precisely the evaluation ɛ(cω(a),b) for some well-defined trace on 𝒯ℒ supported on nesting number k (which are unique up to rescaling).


### 4.2. Some Induced Sign Representations

Cell modules are naturally subquotients of the cellular algebra itself, viewed as a free module (see [37]). For our purposes, we will describe the cell modules as subquotients of 𝒯ℒ in a different way, which will be more convenient to diagrammatically categorify. Taking the inclusion 𝒯ℒJ→𝒯ℒ for some sub-Dynkin diagram J, we can induce the sign representation of 𝒯ℒJ up to 𝒯ℒ. This is the quotient of 𝒯ℒ by the right ideal generated by ui,i∈J. In a future paper, we will describe, for both the Hecke and Temperley-Lieb algebras, a diagrammatic way to categorify the induction of both the “sign" and “trivial" representations of sub-Dynkin diagrams, but for this paper we restrict to a specific case. For the sub-Dynkin diagram which contains every index except i, let Ii be the corresponding ideal (generated by uj for j≠i), and consider the induced sign representation Vi=𝒯ℒ/Ii. Let li=min(i,n+1-i) and let ki=n+1-2li. It turns out that we can embed Lki inside Vi, as shown explicitly below, and we will categorify both modules accordingly. For this reason, we use Li to denote Lki. Note that every possible Lk can be achieved as some Li with the exception of Ln+1.For the rest of this section, fix an indexi∈I. We define a module Vi over 𝒯ℒ abstractly, and then prove that this module is isomorphic to 𝒯ℒ/Ii.Definition 4.5.
For0≤l≤li (and letting k=n+1-2l as always), let aki be the following (k,n+1) cup diagram with l top arcs, where the innermost top arc always connects i to i+1:
(4.2)(4.3)
LetXki⊂Xk consist of all matchings of the form caki,b for b∈Mk. Let Xi be the disjoint union of all Xki for 0≤l≤li, and let Vi be the free ℤ[t,t-1]-module with basis Xi. There is a distinguished element 1 of this basis, the unique member of Xn+1i. Let 𝒯ℒ act on Vi on the right by viewing elements of Vi as though they were in 𝒯ℒ, using the standard multiplication rules, and then killing any terms whose diagrams are not in Xi.The elements ofXi exhaust those elements of X where the only simple top arcs (those connecting j to j+1 for some j) connect i to i+1. Any crossingless matching with a simple top arc connecting j to j+1 has an expression in 𝒯ℒ as a monomial ui̲ which begins with uj. The converse is also true. Thus, Xi are the elements of X for which every expression of the matching begins with ui. This motivates the definition.While something does need to be checked to ensure that this defines a module action, it is entirely straightforward. In the Temperley-Lieb algebra, things are generally easy to prove because products of monomials always reduce to another monomial (with a scalar), not a linear combination of multiple monomials. Therefore, checking the associativity condition for being a module, say, involves showing that both sides of an equation are the same diagram inXi, or that both sides are 0. This module is cyclic, generated by 1, and Ii is clearly in the annihilator of 1, so that 𝒯ℒ/Ii surjects onto Vi. One could prove the following by bounding dimensions.Claim 8.
The modulesVi and 𝒯ℒ/Ii are isomorphic.There is a (cellular) filtration onVi, given by the span of X≤ki, diagrams with at most k through-strands (call it V≤ki). Clearly, each subquotient in this filtration has a basis given by Xki, or in other words by the elements caki,b for b∈Mk. It is an easy exercise that this subquotient is isomorphic to the cell module Lk, under the map sending b∈Mk to caki,b. There is one subquotient for each 0≤l≤li.Claim 9.
The moduleLi is a submodule of Vi.Proof.
Lettingl=li and k=ki, the final term in the filtration is precisely Li≅Vkii.Having explicitly defined the embeddingLi⊂Vi, we pause to investigate adjoint pairings on Vi.Proposition 4.6.
Consider theℤ[[t,t-1]] module of semi-linear pairings on Vi where (xuj,y)=(x,yuj) for all j. Consider the li+1 functionals on this space, which send a pairing to (1,caki,ω(aki)) for various k=n+1-2l, 0≤l≤li. Then these linear functionals are independent and yield an isomorphism between the space of pairings and a free module of rank li+1.Note that, using adjunction, one can check that[2]l(1,caki,ω(aki))=(caki,ω(aki),caki,ω(aki)).Proof.
Given diagramsx,y∈Xi, the self-adjointness of ui implies that the value of (x,y) is an invariant of the diagram y∘ω(x). In particular, (x,y)=(1,yω(x))=(xω(y),1), where yω(x) refers to the image of this diagram in the quotient 𝒯ℒ/Ii. Therefore, if either yω(x) or xω(y) is not in Xi then the value of (x,y) is zero. However, Xi∩ω(Xi)={caki,ω(aki)} where this set runs over all k with 0≤l≤li. Thus the value of the pairing on all elements is clearly determined by the values of (1,caki,ω(aki)) for all such k.
Consider the following mapVi×Vi→ℤ[[t,t-1]]: fix k, and for basis elements x,y send (x,y) to r∈ℤ[t,t-1] if yω(x)=rcaki,ω(aki)∈𝒯ℒ, and send (x,y) to zero otherwise. Clearly this is a well-defined semi-linear map (being defined on a ℤ[t,t-1]-basis) and uj is self-adjoint. Thus we have enough pairings to prove independence.Remark 4.7.
Once again, all pairings are defined topologically. The closure ofcaki,ω(aki) has nesting number exactly k, which distinguishes the traces.


### 4.3. Categorifying Cell Modules

Categorifying the sign representationLn+1 is easy. If we take the quotient of 𝒯ℒ𝒞 by all nonempty diagrams, we get a category where the only nonzero morphism space is the one-dimensional space Hom(∅,∅). This clearly categorifies Ln+1, and we will say no more.Consider the quotient of the category𝒯ℒ𝒞1 by all diagrams where any color not equal to i appears on the left. Call this quotient 𝒱1i. As usual, we let 𝒱2i be its additive grading closure, and let 𝒱i be its graded Karoubi envelope. We will show that 𝒱i≅𝒱2i, so that we really may think of 𝒱i entirely diagrammatically without worrying about idempotents. We claim that 𝒱i categorifies Vi. Not only this, but the action of 𝒯ℒ𝒞 on 𝒱i by placing diagrams on the right will categorify the action of 𝒯ℒ on Vi.Any monomialui̲ which goes to zero in Vi is equal to a (scalar multiple of a) monomial uj̲ where some index j≠i appears on the left. Therefore, the corresponding object Ui̲ will be isomorphic to Uj̲, whose identity morphism is sent to zero in 𝒱1i since it has a j-colored line on the left. There is an obvious map from Vi to the Grothendieck group of 𝒱2i, and the action of 𝒯ℒ𝒞, descended to the Grothendieck group, will commute with the action of 𝒯ℒ on Vi.Therefore,Hom spaces in 𝒱1i will induce a semi-linear pairing on Vi, which satisfies the property (auj,b)=(a,buj) because Uj is self-adjoint. As before, once we determine which pairing this is, our proof will be almost complete.Lemma 4.8.
The pairing induced by𝒱1i will satisfy (1,caki,ω(aki))=tl/(1-t2), where k=n+1-2l.Remark 4.9.
Taking a(n+1,n+1) diagram and closing it off on the punctured plane, if m is the number of circles and k is the nesting number, then the pairing comes from the trace on 𝒯ℒ which sends this configuration to [2]l+m-(n+1)(tl/(1-t2)).
For a closure of an arbitrary diagram,l+m<n+1 is possible. However, for any diagram in Xi∩ω(Xi) (with extra circles thrown in), we have l+m≥n+1, since removing the circles yields precisely caki,ω(aki) for some k. This guarantees that evaluating the formula on an element of Vi yields a power series with nonnegative coefficients.The proof of the lemma may be found shortly below. Temporarily assuming the lemma, the remainder of our results are easy.Theorem 4.10.
𝒱2i is idempotent closed and Krull-Schmidt, so that 𝒱i≅𝒱2i. Its Grothendieck group is isomorphic to Vi.Proof.
It is enough to check that for anyui̲≠uj̲ corresponding to matchings in Xi, Hom(Ui̲,Ui̲) is concentrated in nonnegative degrees with a 1-dimensional degree 0 part, and Hom(Ui̲,Uj̲) is concentrated in strictly positive degrees (see the proof of Proposition 3.4). This is a calculation using the semi-linear pairing.
Lettingm be the number of circles in a configuration on the punctured disk, and letting k=n+1-2l be the nesting number, then the evaluation will be in strictly positive degrees if m<n+1, and will be in nonnegative degrees with a 1-dimensional degree 0 part if m=n+1 exactly, but this was precisely the calculation in the proof of Lemma 3.5: for arbitrary crossingless matchings ui̲ and uj̲, the closure of ui̲ω(uj̲) has fewer than n+1 circles if ui̲≠uj̲, and exactly n+1 if they are equal.Corollary 4.11.
Letℒi be the full subcategory of 𝒱i with objects consisting of (sums and grading shifts of) Ui̲ such that ui̲ is an element of Vn+1-2lii. This has an action of 𝒯ℒ𝒞 on the right. On the Grothendieck group, this setup categorifies the cell module Li=Vn+1-2lii.Proof.
That this subcategory is closed under the action of𝒯ℒ𝒞 is obvious, as is the existence of a map from Li to the Grothendieck group. We already know the induced pairing, because the subcategory is full. Therefore, the same arguments imply that the Grothendieck group behaves as planned.Proof of Lemma4.8.
To calculate the pairing, we may calculate(ui̲,ui̲)=gdimEnd(Ui̲) for the following choices of i̲: ∅,i, i(i+1)(i-1), i(i+1)(i-1)(i+2)i(i-2), i(i+1)(i-1)(i+2)i(i-2)(i+3)(i+1)(i-1)(i-3), and so forth. These are pictured below.
(4.4)(4.5)
These sequences are split into subsequences we call “tiers", where themth sequence adds the mth tier. The following property of these sequences is easily verified: each sequence i̲ is in Xi, and remains in Xi if one removes any subset of the final tier, but ceases to be in Xi if one removes a single element from any other tier instead.
Fix nonemptyi̲ in this sequence, and let j̲ be the subsequence with the final tier removed. It is a quick exercise to show that the lemma is equivalent to gdimEnd(Ui̲)=(1+t2)l/(1-t2), where l is the number of elements in the final tier.
Now consider an element of the endomorphism ring. Using previous results, we may assume it is a simple forest, with all double dots on the far left. Any double dot coloredj≠i will be sent to zero, so we have only an action of the ring 𝕜[fi] on the left. This accounts for the 1/(1-t2) appearing in all the formulae.
Suppose there is a boundary dot in the morphism on any line not in the final tier. Then the morphism factors through the sequenceUk̲ where k̲ is i̲ with that index removed. As discussed above, uk̲ is not in Xi and therefore Uk̲ is isomorphic to the zero object. See the first picture below for an intuitive reason why such a morphism vanishes. Hence, the only boundary dots which can appear occur on the final tier. It is easy to check that the existence of a trivalent vertex joining three boundary lines will force the existence of a dot not on the final tier; see the second picture below. Both pictures are for the sequence i(i+1)(i-1)(i+2)i(i-2), and blue will represent i in all pictures in this section.(4.6)
Therefore a nonzero endomorphism must be1Uj̲ accompanied on the right by either identity maps or broken lines (pairs of boundary dots), because all other simple forests yield a zero map. Identity maps have degree zero, while broken lines have degree 2. If these pictures form a basis (along with the action of blue double dots on the left), then the graded dimension will be exactly as desired. An example with l=3, two broken lines, and one unbroken line is shown below.(4.7)
This spanning set is linearly independent in𝒯ℒ𝒞 over 𝕜, so any further dependencies must come from having a nonblue color on the left. Consider an arbitrary endomorphism, and reduce it using the 𝒯ℒ𝒞 relations to a simple forest with all double dots on the left. The actual double dots appearing are ambiguous, since there are polynomial relations in 𝒯ℒ𝒞, but it is easy (knowing the generators of the TL ideal) to note that these relations are trivial modulo nonblue colors. Hence the spanning set will be linearly independent if any diagram in 𝒯ℒ𝒞 which started with a nonblue color on the left will still have a nonblue color (perhaps in a double-dot) on the left after reducing to a simple forest. This will be the case for any diagram with a boundary dot on j̲.
Let red indicate any other index, and suppose that red appears on the far left. Regardless of what index red is, unless there is a dot onj̲, the identity lines of j̲ block this leftmost red component from reaching any red on the boundary of the graph. Take a neighborhood of a red line segment which includes no other colors and goes to -∞. Excising this neighborhood, we get a simply connected region where the only relevant red boundary lines are the two which connect to the ends of the segment. Red then will reduce to a simple forest with double dots on the left, which in this case yields either a red double dot or a red circle (potentially with more double dots).(4.8)
However, no colors adjacent to red can interfere on the interior of a red circle, so the circle evaluates to zero. Therefore, the diagram evaluates to zero or has at least one red double dot on the left. We may ignore the red double dot and reduce the remainder of the diagram, and so regardless of what else is done, the final result will have a red double dot on the left.


## 4.1. Cell Modules

The Temperley-Lieb algebra has the structure of a cellular algebra, a concept first defined by Graham and Lehrer [37]. One feature of cellular algebras is that they are equipped with certain modules known as cell modules. Cell modules provide a complete set of nonisomorphic irreducible modules in many cases (such as 𝒯ℒ in type A). Cell modules come equipped with a basis and a bilinear form, making them obvious candidates for categorification. We will not go into detail on cellular algebras here, or even use their general properties; instead we will describe the cell modules explicitly and pictorially for the case of 𝒯ℒ, where things are unusually simple. Nothing in this section or the next is particularly original, and we state some standard results without proof.Notation 6.
Consider a crossingless matching in the planar strip betweenn points on the bottom boundary and m points on the top. We call this briefly an (n,m)diagram. In the terminology of [30], there are two kinds of arcs in a diagram: horizontal arcs which connect two points on the top (let us call it a top arc), or two points on the bottom (bottom arc) and vertical arcs which connect a point on the top to one on the bottom. Elsewhere in the literature, vertical arcs are called through-strands. An (n,k) diagram with exactly kthrough-strands (and therefore no top arcs) has an isotopy representative with only “caps" (local maxima) and no “cups" (local minima) so it is called an (n,k)cap diagram. A (k,n) diagram with k through-strands is called a (k,n)cup diagram.The set of all(n,m) diagrams can be partitioned by the number of through-strands. Any (n,m) diagram with k through-strands can be expressed as the concatenation of an (n,k) cap diagram with a (k,m) cup diagram in a unique way. For an illustration of this concept, see Figure 5.On the left side, a(7,7) diagram with k=3 through-strands and l=2 top arcs (resp., bottom arcs) is decomposed into a (7,3) cap diagram a-composed with a (3,7) cup diagram z. On the right side, an element of 𝒯ℒ3 is obtained by composing a and z in the opposite order.
(a)(b)In an(m,m) diagram the number l of top arcs equals the number of bottom arcs, and if k is the number of through-strands, then k+2l=m. We will typically use k and l to represent the number of through-strands and top arcs in an (m,m) diagram henceforth.Notation 7.
LetX be the set of all (n+1,n+1) diagrams. Let ω be the endomorphism of X sending each diagram to its vertical flip. We will write the operation on diagrams of reduced vertical concatenation by ∘: a∘b places a above b and remove any circles. Let Xk be the set of crossingless matchings with exactly k through-strands. Let Mk be the set of all (n+1,k) cap diagrams, so that ω(Mk) is the set of all (k,n+1) cup diagrams.Definition 4.1.
LetLk be the free ℤ[t,t-1] module spanned by Mk, the (n+1,k) cap diagrams. We place a right 𝒯ℒ-module structure on Lk by concatenation, where circles become factors of [2] as usual, and any resulting diagram with fewer than k through-strands is sent to 0. This is the cell module for cell k, and it is irreducible.Example 4.2.
The only diagram inXn+1 corresponds to the identity map in 𝒯ℒ. The cell module Ln+1 has rank 1 over ℤ[t,t-1], and its generator is killed by all ui. We will take this as the definition of the sign representation of 𝒯ℒ.Example 4.3.
The next cell moduleLn-1 has rank n over ℤ[t,t-1], having generators vi, i=1⋯n (see Figure 6), such that
(4.1)vjui={[2]vjifi=j,viifiandjareadjacent,0ifiandjaredistant.Figure 6
A basis for the cell moduleLn-1, consisting of (n+1,n-1) cap diagrams (here, n=4).Given a(n+1,k) cap diagram a and a (k,n+1) cup diagram z, there are two things we can do: take the composition z∘a to obtain an element called cz,a of Xk or take the composition a∘z to get an element of 𝒯ℒk (there may be additional circles created, and the final diagram may have fewer than k through-strands). Both compositions have the same closure on the punctured plane. Note that ω(cz,a)=cω(a),ω(z). The seemingly extraneous use of the notation c·,· is standard for cellular algebras.Proposition 4.4.
There is, up to rescaling, a unique pairing(,):Lk×Lk→ℤ[[t,t-1]] for which ui is self-adjoint, that is, (aui,b)=(a,bui). Given cup diagrams a and b in Mk, one evaluates (a,b) by considering the closure of cω(a),b∈𝒯ℒ, or equivalently the closure of b∘ω(a)∈𝒯ℒk. If the diagram has nesting number k, one returns a scalar times [2] raised to the number of circles; if it has nesting number <k, one returns zero. This is precisely the evaluation ɛ(cω(a),b) for some well-defined trace on 𝒯ℒ supported on nesting number k (which are unique up to rescaling).


## 4.2. Some Induced Sign Representations

Cell modules are naturally subquotients of the cellular algebra itself, viewed as a free module (see [37]). For our purposes, we will describe the cell modules as subquotients of 𝒯ℒ in a different way, which will be more convenient to diagrammatically categorify. Taking the inclusion 𝒯ℒJ→𝒯ℒ for some sub-Dynkin diagram J, we can induce the sign representation of 𝒯ℒJ up to 𝒯ℒ. This is the quotient of 𝒯ℒ by the right ideal generated by ui,i∈J. In a future paper, we will describe, for both the Hecke and Temperley-Lieb algebras, a diagrammatic way to categorify the induction of both the “sign" and “trivial" representations of sub-Dynkin diagrams, but for this paper we restrict to a specific case. For the sub-Dynkin diagram which contains every index except i, let Ii be the corresponding ideal (generated by uj for j≠i), and consider the induced sign representation Vi=𝒯ℒ/Ii. Let li=min(i,n+1-i) and let ki=n+1-2li. It turns out that we can embed Lki inside Vi, as shown explicitly below, and we will categorify both modules accordingly. For this reason, we use Li to denote Lki. Note that every possible Lk can be achieved as some Li with the exception of Ln+1.For the rest of this section, fix an indexi∈I. We define a module Vi over 𝒯ℒ abstractly, and then prove that this module is isomorphic to 𝒯ℒ/Ii.Definition 4.5.
For0≤l≤li (and letting k=n+1-2l as always), let aki be the following (k,n+1) cup diagram with l top arcs, where the innermost top arc always connects i to i+1:
(4.2)(4.3)
LetXki⊂Xk consist of all matchings of the form caki,b for b∈Mk. Let Xi be the disjoint union of all Xki for 0≤l≤li, and let Vi be the free ℤ[t,t-1]-module with basis Xi. There is a distinguished element 1 of this basis, the unique member of Xn+1i. Let 𝒯ℒ act on Vi on the right by viewing elements of Vi as though they were in 𝒯ℒ, using the standard multiplication rules, and then killing any terms whose diagrams are not in Xi.The elements ofXi exhaust those elements of X where the only simple top arcs (those connecting j to j+1 for some j) connect i to i+1. Any crossingless matching with a simple top arc connecting j to j+1 has an expression in 𝒯ℒ as a monomial ui̲ which begins with uj. The converse is also true. Thus, Xi are the elements of X for which every expression of the matching begins with ui. This motivates the definition.While something does need to be checked to ensure that this defines a module action, it is entirely straightforward. In the Temperley-Lieb algebra, things are generally easy to prove because products of monomials always reduce to another monomial (with a scalar), not a linear combination of multiple monomials. Therefore, checking the associativity condition for being a module, say, involves showing that both sides of an equation are the same diagram inXi, or that both sides are 0. This module is cyclic, generated by 1, and Ii is clearly in the annihilator of 1, so that 𝒯ℒ/Ii surjects onto Vi. One could prove the following by bounding dimensions.Claim 8.
The modulesVi and 𝒯ℒ/Ii are isomorphic.There is a (cellular) filtration onVi, given by the span of X≤ki, diagrams with at most k through-strands (call it V≤ki). Clearly, each subquotient in this filtration has a basis given by Xki, or in other words by the elements caki,b for b∈Mk. It is an easy exercise that this subquotient is isomorphic to the cell module Lk, under the map sending b∈Mk to caki,b. There is one subquotient for each 0≤l≤li.Claim 9.
The moduleLi is a submodule of Vi.Proof.
Lettingl=li and k=ki, the final term in the filtration is precisely Li≅Vkii.Having explicitly defined the embeddingLi⊂Vi, we pause to investigate adjoint pairings on Vi.Proposition 4.6.
Consider theℤ[[t,t-1]] module of semi-linear pairings on Vi where (xuj,y)=(x,yuj) for all j. Consider the li+1 functionals on this space, which send a pairing to (1,caki,ω(aki)) for various k=n+1-2l, 0≤l≤li. Then these linear functionals are independent and yield an isomorphism between the space of pairings and a free module of rank li+1.Note that, using adjunction, one can check that[2]l(1,caki,ω(aki))=(caki,ω(aki),caki,ω(aki)).Proof.
Given diagramsx,y∈Xi, the self-adjointness of ui implies that the value of (x,y) is an invariant of the diagram y∘ω(x). In particular, (x,y)=(1,yω(x))=(xω(y),1), where yω(x) refers to the image of this diagram in the quotient 𝒯ℒ/Ii. Therefore, if either yω(x) or xω(y) is not in Xi then the value of (x,y) is zero. However, Xi∩ω(Xi)={caki,ω(aki)} where this set runs over all k with 0≤l≤li. Thus the value of the pairing on all elements is clearly determined by the values of (1,caki,ω(aki)) for all such k.
Consider the following mapVi×Vi→ℤ[[t,t-1]]: fix k, and for basis elements x,y send (x,y) to r∈ℤ[t,t-1] if yω(x)=rcaki,ω(aki)∈𝒯ℒ, and send (x,y) to zero otherwise. Clearly this is a well-defined semi-linear map (being defined on a ℤ[t,t-1]-basis) and uj is self-adjoint. Thus we have enough pairings to prove independence.Remark 4.7.
Once again, all pairings are defined topologically. The closure ofcaki,ω(aki) has nesting number exactly k, which distinguishes the traces.


## 4.3. Categorifying Cell Modules

Categorifying the sign representationLn+1 is easy. If we take the quotient of 𝒯ℒ𝒞 by all nonempty diagrams, we get a category where the only nonzero morphism space is the one-dimensional space Hom(∅,∅). This clearly categorifies Ln+1, and we will say no more.Consider the quotient of the category𝒯ℒ𝒞1 by all diagrams where any color not equal to i appears on the left. Call this quotient 𝒱1i. As usual, we let 𝒱2i be its additive grading closure, and let 𝒱i be its graded Karoubi envelope. We will show that 𝒱i≅𝒱2i, so that we really may think of 𝒱i entirely diagrammatically without worrying about idempotents. We claim that 𝒱i categorifies Vi. Not only this, but the action of 𝒯ℒ𝒞 on 𝒱i by placing diagrams on the right will categorify the action of 𝒯ℒ on Vi.Any monomialui̲ which goes to zero in Vi is equal to a (scalar multiple of a) monomial uj̲ where some index j≠i appears on the left. Therefore, the corresponding object Ui̲ will be isomorphic to Uj̲, whose identity morphism is sent to zero in 𝒱1i since it has a j-colored line on the left. There is an obvious map from Vi to the Grothendieck group of 𝒱2i, and the action of 𝒯ℒ𝒞, descended to the Grothendieck group, will commute with the action of 𝒯ℒ on Vi.Therefore,Hom spaces in 𝒱1i will induce a semi-linear pairing on Vi, which satisfies the property (auj,b)=(a,buj) because Uj is self-adjoint. As before, once we determine which pairing this is, our proof will be almost complete.Lemma 4.8.
The pairing induced by𝒱1i will satisfy (1,caki,ω(aki))=tl/(1-t2), where k=n+1-2l.Remark 4.9.
Taking a(n+1,n+1) diagram and closing it off on the punctured plane, if m is the number of circles and k is the nesting number, then the pairing comes from the trace on 𝒯ℒ which sends this configuration to [2]l+m-(n+1)(tl/(1-t2)).
For a closure of an arbitrary diagram,l+m<n+1 is possible. However, for any diagram in Xi∩ω(Xi) (with extra circles thrown in), we have l+m≥n+1, since removing the circles yields precisely caki,ω(aki) for some k. This guarantees that evaluating the formula on an element of Vi yields a power series with nonnegative coefficients.The proof of the lemma may be found shortly below. Temporarily assuming the lemma, the remainder of our results are easy.Theorem 4.10.
𝒱2i is idempotent closed and Krull-Schmidt, so that 𝒱i≅𝒱2i. Its Grothendieck group is isomorphic to Vi.Proof.
It is enough to check that for anyui̲≠uj̲ corresponding to matchings in Xi, Hom(Ui̲,Ui̲) is concentrated in nonnegative degrees with a 1-dimensional degree 0 part, and Hom(Ui̲,Uj̲) is concentrated in strictly positive degrees (see the proof of Proposition 3.4). This is a calculation using the semi-linear pairing.
Lettingm be the number of circles in a configuration on the punctured disk, and letting k=n+1-2l be the nesting number, then the evaluation will be in strictly positive degrees if m<n+1, and will be in nonnegative degrees with a 1-dimensional degree 0 part if m=n+1 exactly, but this was precisely the calculation in the proof of Lemma 3.5: for arbitrary crossingless matchings ui̲ and uj̲, the closure of ui̲ω(uj̲) has fewer than n+1 circles if ui̲≠uj̲, and exactly n+1 if they are equal.Corollary 4.11.
Letℒi be the full subcategory of 𝒱i with objects consisting of (sums and grading shifts of) Ui̲ such that ui̲ is an element of Vn+1-2lii. This has an action of 𝒯ℒ𝒞 on the right. On the Grothendieck group, this setup categorifies the cell module Li=Vn+1-2lii.Proof.
That this subcategory is closed under the action of𝒯ℒ𝒞 is obvious, as is the existence of a map from Li to the Grothendieck group. We already know the induced pairing, because the subcategory is full. Therefore, the same arguments imply that the Grothendieck group behaves as planned.Proof of Lemma4.8.
To calculate the pairing, we may calculate(ui̲,ui̲)=gdimEnd(Ui̲) for the following choices of i̲: ∅,i, i(i+1)(i-1), i(i+1)(i-1)(i+2)i(i-2), i(i+1)(i-1)(i+2)i(i-2)(i+3)(i+1)(i-1)(i-3), and so forth. These are pictured below.
(4.4)(4.5)
These sequences are split into subsequences we call “tiers", where themth sequence adds the mth tier. The following property of these sequences is easily verified: each sequence i̲ is in Xi, and remains in Xi if one removes any subset of the final tier, but ceases to be in Xi if one removes a single element from any other tier instead.
Fix nonemptyi̲ in this sequence, and let j̲ be the subsequence with the final tier removed. It is a quick exercise to show that the lemma is equivalent to gdimEnd(Ui̲)=(1+t2)l/(1-t2), where l is the number of elements in the final tier.
Now consider an element of the endomorphism ring. Using previous results, we may assume it is a simple forest, with all double dots on the far left. Any double dot coloredj≠i will be sent to zero, so we have only an action of the ring 𝕜[fi] on the left. This accounts for the 1/(1-t2) appearing in all the formulae.
Suppose there is a boundary dot in the morphism on any line not in the final tier. Then the morphism factors through the sequenceUk̲ where k̲ is i̲ with that index removed. As discussed above, uk̲ is not in Xi and therefore Uk̲ is isomorphic to the zero object. See the first picture below for an intuitive reason why such a morphism vanishes. Hence, the only boundary dots which can appear occur on the final tier. It is easy to check that the existence of a trivalent vertex joining three boundary lines will force the existence of a dot not on the final tier; see the second picture below. Both pictures are for the sequence i(i+1)(i-1)(i+2)i(i-2), and blue will represent i in all pictures in this section.(4.6)
Therefore a nonzero endomorphism must be1Uj̲ accompanied on the right by either identity maps or broken lines (pairs of boundary dots), because all other simple forests yield a zero map. Identity maps have degree zero, while broken lines have degree 2. If these pictures form a basis (along with the action of blue double dots on the left), then the graded dimension will be exactly as desired. An example with l=3, two broken lines, and one unbroken line is shown below.(4.7)
This spanning set is linearly independent in𝒯ℒ𝒞 over 𝕜, so any further dependencies must come from having a nonblue color on the left. Consider an arbitrary endomorphism, and reduce it using the 𝒯ℒ𝒞 relations to a simple forest with all double dots on the left. The actual double dots appearing are ambiguous, since there are polynomial relations in 𝒯ℒ𝒞, but it is easy (knowing the generators of the TL ideal) to note that these relations are trivial modulo nonblue colors. Hence the spanning set will be linearly independent if any diagram in 𝒯ℒ𝒞 which started with a nonblue color on the left will still have a nonblue color (perhaps in a double-dot) on the left after reducing to a simple forest. This will be the case for any diagram with a boundary dot on j̲.
Let red indicate any other index, and suppose that red appears on the far left. Regardless of what index red is, unless there is a dot onj̲, the identity lines of j̲ block this leftmost red component from reaching any red on the boundary of the graph. Take a neighborhood of a red line segment which includes no other colors and goes to -∞. Excising this neighborhood, we get a simply connected region where the only relevant red boundary lines are the two which connect to the ends of the segment. Red then will reduce to a simple forest with double dots on the left, which in this case yields either a red double dot or a red circle (potentially with more double dots).(4.8)
However, no colors adjacent to red can interfere on the interior of a red circle, so the circle evaluates to zero. Therefore, the diagram evaluates to zero or has at least one red double dot on the left. We may ignore the red double dot and reduce the remainder of the diagram, and so regardless of what else is done, the final result will have a red double dot on the left.



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*Source: 530808-2010-10-11.xml*