CHAPTER: INTRODUCTION

1In this paper, the quantum group $U_q(\mathfrak{g})$ is a deformation of the universal enveloping algebra $U(\mathfrak{g})$ of some semisimple Lie algebra $\mathfrak{g}$ (cf. <cit.>). For generic choices of the parameter $q$, the simple representations are in natural bijection with the simple representations of $\mathfrak{g}$, however this deformation causes the tensor category structure of the representation category to no longer be symmetric which gives us a richer theory and provides applications such as knot invariants. In <cit.> and <cit.> a combinatorial model in type $A_n$ (ie for $\mathfrak{g}=\mathfrak{sl}_{n+1}$), was developed using generators and relations. This extended the work of <cit.> and <cit.> in type $A_2$ and $A_3$ respectively. Simple tensors and compositions of generators are called webs, and can be represented as trivalent graphs. Each of these models is called a spider. Here we will analyze the simple spider corresponding to $A_3$ along with the spiders of the semisimple Lie groups of the form $A_1^n \times A_2^m$ (we'll mainly use product notation from now on because the resulting representations are tensor products of representations of the factors). Our main goal for each spider will develop and prove a generalization of confluence (see the end of section <ref> for a definition of confluence), and then use this to find algebraic and geometric applications.

§ GENERALIZED CONFLUENCE

It was shown in <cit.> that the rank 2 spiders were confluent with respect to the filtration given by the number of vertices of each web. In the rank 3 case, this no longer makes sense in the usual format because there are relators which have two leading terms with the same number and type of vertices. We could attempt to refine the filtration, but this would necessarily cause it to stop being rotationally symmetric. Instead we generalize the notion of confluence. First we note that a filtered presentation of vector spaces is just a filtered complex with two columns (one for the generators and one for the relators), and so induces a spectral sequence, whose convergence is closely related to confluence.

A presentation of a filtered vector space whose relators have only a single leading term is confluent if and only if the associated spectral sequence converges at the first page $E^1$.

If we instead compare to filtered algebras with more general relators, convergence of this spectral sequence amounts to a PBW-type theorem where the associated graded is isomorphic to the algebra with the same presentation except that we truncate away the lower-order terms of each relator.

Taking this as motivation we can generalize this notion to $E^k$ convergence for those presentations whose associated spectral sequence converges at the $k$th page. Intuitively, this property tells us that in the process of reducing an element $w$ to a minimal representative, we don't need to use an web which has $k$ more vertices than $w$. In this paper we will be looking at presentations of additive tensor categories over a field, and whose filtration will be given by the number of generators in each word. We will say that such a presentation is $E^k$ convergent if every such $Hom$ space has a spectral sequence which converges on the $k$th page as a vector space presentation.

For other spiders, it may be important to use more complicated filtrations where different generating morphisms are given different weight. Nevertheless, in the case of $A_3$ and $A_1^n$ there is only one generating morphism up to duality, so we won't lose anything by restricting to this filtration. For $A_1^m \times A_2^n$ where $n,m>0$ there are two types of vertices (trivalent and tetravalent) up to outer automorphisms, but it doesn't appear we will change anything significant by changing the filtration short of completely ignoring one type of generator in the filtration

The non-simple semisimple cases are some of the simplest examples since the horizontal relators have no lower order terms, and the sizes of the invariant spaces are easy to characterize in terms of the simple factors. This gives us our first method of proving $E^1$ convergence which is more classical. We show that the dimensions in the $E^1$ page are the same as in the actual representation category, and therefore get our result for $A_1^n$ (which I can only imagine has been proven before in some other form due to its simplicity and similarity to the Coxeter group properties of the symmetric group):

The $A_1^n$ spider with the natural product presentation has an $E^1$ convergent spectral sequence.

We also prove a similar result for $A_1 \times A_2$ which follows with some extra work from the cut-path results in <cit.>.

The $A_1 \times A_2$ spider with the natural product presentation has an $E^1$ convergent spectral sequence.

These two spiders have many nice properties that allow us to prove $E^1$ convergence more easily, but in the simple rank $3$ cases the spiders are more complicated. Nevertheless, they have some similarities which should make the amenable to a unified approach. Therefore, we prove a criterion for $E^1$ convergence which can be used in type $A_3$, and conjecturally for $B_3$ as well.

It was shown in <cit.> that it is possible to generate all relations in the $A_3$ spider without using the hexagon relators introduced in <cit.> and called Kekule relators in <cit.>, however this requires increasing the number of vertices of certain webs before we can reduce them. Since this would make $E^1$ convergence impossible, we will instead work with the larger set of relators given in <cit.>. Using this as our presentation, we construct a global criterion for when a web is reducible, and use it to obtain the following result using a generalized form of the diamond lemma.

The $\mathfrak{sl}(4)$ spider with the Scott-Morrison presentation has an $E^1$ convergent spectral sequence. In particular, two webs are equal up to lower order terms if and only if they are equal in the associated graded spider (which has the same presentation as the original spider, but where we omit the lower order terms of each relator)

Although this is only a single case, it is actually quite flexible in that it simultaneously generalizes various combinatorial models such as plane partitions and square ice (after some superficial modifications).

Unfortunately, it may not be reasonable to expect $E^1$ convergence to hold for higher rank simple spiders. In particular, we show the following in Chapter $6$ for semisimple spiders

The $A_1^{n}\times A_2^{m}$ spider with the natural product presentation is $E^1$ convergent if and only if $m=0$ and/or $\rk \fg \leq 3$

However this also shows the utility of the more general notion of $E^k$ convergence because we still get the following weaker result:

The $A_1^n \times A_2$ spider with the natural product presentation has an $E^2$ convergent spectral sequence

§ THE EUCLIDEAN BUILDING AND GEOMETRIC SATAKE

Using $E^1$ convergence results we get some geometric applications. For each group $G$ there is a geometric object called the affine Grassmannian: $Gr(G) = G(\C((x)))/G(\C[[x]])$. This object is related to the representation theory of the Langlands dual group $G^L$ via the Satake correspondence (developed by Lusztig, Ginzburg, and Mirković-
Vilonen). In particular, we will use the following corollary (lifted directly from <cit.>): for every sequence of dominants weights $\vec{\la}=(\la_1,...,\la_n)$ there is an associated variety $F(\vec{\la})$ called the Satake fiber (which is a fiber of a particular map to the affine Grassmannian which we'll explain in chapter 5) and is related to the representation category as follows:

Every invariant space in $\rep(G)$ is canonically isomorphic to the
top homology of the corresponding geometric Satake fiber with complex coefficients:

$\Phi : Inv(V(\la_1) \ot ... \ot V(\la_n)) \cong H^{top}(F(\vec{\la}),\C)$.

Each top-dimensional component $Z \sbs F(\vec{\la})$ thus yields a vector $[Z] \in Inv(V(\la_1) \ot ... \ot V(\la_n))$.
These vectors form a basis, the Satake basis.

The affine Grassmannian then naturally embeds into a (very infinite) simplicial complex called the Euclidean building $X$. Following <cit.>, the dual graph of each web $w$ naturally extends to a $2$D simplicial complex (with a metric). We will call this the dual web $D(w)$. Given a combinatorial local isometry from the dual web to the Euclidean building, there is a natural corresponding point in $F(\vec{\la})$, and we therefore get a map $\pi: \Hom_{loc. iso.}(D(w),X) \to F(\vec{\la})$.

Since the edges of duals webs are colored by fundamental webs, we can define weight-valued lengths on paths. Since weights naturally have a poset structure, we can then define a geodesic as a path with minimal length. However, in general you may have two geodesics connected the same two points, but the weight-lengths are incomparable. If this does not happen, then we say that the dual web has coherent geodesics (between those points). In <cit.>, the author defines certain so-called coherent webs whose dual webs have coherent geodesics originating from fixed boundary point, and fulfill two other axioms (although we will prove that the 3rd axiom is redundant). See <ref> for the full definition of coherence. To each of these is associated an LS path by taking maximal geodesics connecting the fixed boundary point to each of the other boundary points. Conversely, we will argue that there exist minimal-vertex coherent webs for each LS path, and use the $E^1$ convergence to prove the following theorem relating minimal vertex webs to the geometric Satake basis:

Let $\lb w_i \rb_{i=1}^n$ by a maximal set of minimal-vertex $A_3$ webs with fixed boundary which are distinct in the associated graded spider. $\lb w_i \rb_{i=1}^n$ form a basis for the spider category (and hence the invariant space of the corresponding tensor product representation by <cit.>). Moreover this basis is upper unitriangular with respect to the Satake basis for all choices of base points simultaneously.

This is the best we can expect since we know that web bases don't generally match the geometric Satake basis as shown in <cit.>.

We also look at $A_1^n$ geometrically. Fix a web $w$, and look at the set of all webs $\lb w_i \rb_{i=1}^n$ which are equal $w$ as invariants. The set of dual webs $\lb D(w_i)\rb_{i=1}^n$ naturally correspond to $2$D subcomplexes of a simplicial complex $P$ with a particular boundary (where $P$ is contained in the Euclidean building). We can prove that $P$ is $CAT(0)$ (which is a generalization of hyperbolicity and will be explained in more detail in chapter $5$):

The complex $P$ is a $CAT(0)$ cube complex.

This gives us a way to reinterpret relators in the spider as homotopies across cells in the Euclidean building. We can attempt a similar construction in $A_3$ in order to generalize the observation that $A_2$ spiders are $CAT(0)$ if and only if they are minimal-vertex. However, it's not so easy to prove that the resulting complex is $CAT(0)$ since we don't have the simple cube complex criterion.

CHAPTER: QUANTUM GROUPS, HOPF ALGEBRAS, AND TENSOR CATEGORIES

1This chapter is designed to provide an informal introduction to quantum groups for readers less familiar with the technical details. For a more technical introduction we direct the readers to one of the following three books. <cit.> provides a more $\mf{sl}_2$ oriented approach followed by a lot of the general details for monoidal categories, and the connections to knot invariants. <cit.> focuses on a different approach using crystals, but the first 3 chapters give a concise introduction to quantum groups with many of the formal definitions we omit. Finally, <cit.> provides a more comprehensive resource which includes more information about the origins of quantum groups and also various canonical bases (although not the ones we'll be discussing).

§ HOPF ALGEBRAS AND QUANTUM GROUPS

In this paper, a quantum group (denoted $U_q(\fg)$) is a particular kind of Hopf algebra coming from a deformation of the universal enveloping algebra, which we explain more below. It is also called a quantum enveloping algebra.

A Hopf algebra (over $k$) is a $k$-algebra $A$ together with three additional structures: a comultiplication $\Delta: A \to A \ot A$, a counit $\epsilon: A \to k$, and an antipode $S: A \to A$ which satisfy various compatibility axioms. The first canonical example is a group algebra $kG$ of a finite group $G$, which has an augmentation map $\epsilon: kG \to k$ defined by adding up the coefficients, a comultiplication obtained by letting $\Delta(g)=g\ot g$ and then extending to the entire algebra, and an antipode determined by letting $S(g) = g^{-1}$. From a representation theory perspective we get the following: the counit defines a trivial representation. The comultiplication allows an element $x \in A$ to act on a tensor product of representations $V \ot W$ via $x.(v \ot w) = \Delta(x) (v \ot w)$. Finally, the antipode allows elements to act on the dual space $V^*$ of a representation $V$ via $(x.\ell)(v)=\ell(S(x).v)$ for all $\ell \in V^*$ and $v \in V$. In the language of the next section, we'll say these extra structures give the representation category the structure of a pivotal tensor category.

The second canonical example is the universal enveloping algebra $U(\fg)$, which is more closely related to this paper. Recall, we define the universal enveloping algebra as the quotient of the tensor algebra $T \fg$ where we enforce that the associative algebra commutator: $[x,y]_{U(\fg)} = xy - yx$ is equal to the corresponding Lie algebra commutator $[x,y]_\fg$ when $x$ and $y$ are in $\fg \sbs T \fg$. Next we want to define the Hopf algebra structure. Since Lie algebras correspond to derivations, we require that elements act on tensors via the product rule, ie

\begin{equation}
x.(v \ot w) = (x.v) \ot w + x \ot (x.w) = (x \ot 1 + 1 \ot x)(v \ot w)
\end{equation}

which tells us that we want

\begin{equation}
\Delta(x) = x \ot 1 + 1 \ot x \qquad \text{ for every } x\in \fg \sbs T\fg
\end{equation}
and then we extend this to the rest of $\Ug$ by making $\Delta$ a homomorphism.

In both $kG$ and $U(\fg)$, the image of $\Delta$ is contained the symmetric part of $A \ot A$, and so we say that these two Hopf algebras are cocommutative. On the other hand, the quantum enveloping algebra $\Uqg$ is an algebra which is constructed to be non-cocommutative (over the field $\Q(q)$ where $q$ is a formal variable). Informally speaking, in the limit "$q \to 0$" the representation theory of $\Uqg$ approaches the representation theory of $\Ug$. In fact, there is a correspondence between the irreducible representations of $\Uqg$ and $\Ug$, and even the decomposition of tensors into irreducible representations is the same (in the sense that we get the same number of each irreducible representation). However, the difference lies in the so-called braiding which we'll define in the next section.

The full presentation for $\Uqg$ is not particularly illuminating without knowing its origin, and we won't use it directly in this paper, but we will transcribe it from <cit.> for the sake of tradition. First define the quantum number

\begin{equation}
[n]_q :=\frac{q^n-q^{-n}}{q-q^{-1}}:=q^{n-1}+q^{n-3}+...+q^{-(n-3)}+q^{-(n-1)} \in \Q(q)
\end{equation}

and the quantum binomial coefficients:

\begin{equation}
\sqbinom{n}{k}_q:=\frac{[n]_q\cdot [n-1]_q\cdot\cdot\cdot [n+1-k]_q}{[k]_q\cdot [k-1]_q\cdot\cdot\cdot [2]_q \cdot [1]_q} \in \Q(q)
\end{equation}

Let $\fg$ be a semisimple Lie algebra with an associated Cartan datum $(A,\Pi,\Pi^\vee, P, P^\vee)$ and indexing set $I$. Moreover, let the diagonalizing matrix of the Cartan matrix $A$ be $diag(s_i)$. The quantum universal enveloping algebra $U_q(\fg)$ is a Hopf algebra defined as follows. As a unital associative algebra over $\Q(q)$, it is generated by elements $e_i$, $f_i$ for $i \in I$, and $q^h$ for $h \in P^\vee$ and satisfying the following relations:

q^0 = 1, q^h q^h' = q^h+h'  for  h,h' ∈P^∨
q^h e_i q^-h = q^_i(h) e_i  for  h ∈P^∨
q^h f_i q^-h = q^-_i(h) f_i  for  h ∈P^∨
e_i f_j - f_j e_i = δ_ij q^s_ih_i - q^-s_ih_i/q^s_i - q^-s_i  for  i,j ∈I
∑_k=0^1-a_ij (-1)^k 1-a_ijk_q e^1-a_ij - k e_j e_i^k = 0  for  i ≠j
∑_k=0^1-a_ij (-1)^k 1-a_ijk_q f^1-a_ij - k f_j f_i^k = 0  for  i ≠j

Its coalgebra structures and antipode induced by the following:

Δ(q^h) = q^h q^h
Δ(e_i) = e_i q^-s_i h_i + 1 e_i,   Δ(f_i) = f_i 1 + q^s_i h_i f_i
(q^h)=1,   (e_i)=(f_i)=0
S(q^h) = q^-h, S(e_i) = -e_i q^s_i h_i,  S(f_i) = - q^-s_i h_i f_i

For $h \in P^\vee$ and $i\in I$

§ PIVOTAL TENSOR CATEGORIES AND BRAIDINGS

Intuitively, a tensor category (also called a monoidal category) is a category where we can take tensor products of objects and morphisms, while a pivotal (tensor) category (also called rigid) is a tensor category which has a duality functor. These come up in representation theory because the category of representations of a Hopf algebra is naturally a pivotal tensor category, where the tensor product comes from the comultiplication, and the duality comes from the antipode map.

Formally (following <cit.>) a strict tensor category is a category $\ms{C}$  together with a bifunctor $\ot: \ms{C} \times \ms{C} \to \ms{C}$ called the tensor product satisfying

* Associativity:  $(U \ot V) \ot W =U \ot (V \ot W)$

* Unity: There exists an object $\mbf{1}$ such that $U \ot \mbf{1} = U$ and $ \mbf{1}\ot U = U$

It is actually more natural to define the above equalities as natural isomorphisms making some diagrams commute. Such a category is called a (possibly non-strict) tensor category. Every tensor category is equivalent to a strict tensor category, so we will avoid these complications by sweeping this distinction under the rug

A (strict) pivotal category, is a tensor category $\ms{C}$ together with a contravariant functor $D: \ms{C} \to \ms{C}$ (where $D(A)$ is denoted $A^*$ for every $A \in \Ob(\ms{C})$) together with contraction maps $d_U:U^* \ot U \to \mbf{1}$, and cocontraction maps $b_V: \mbf{1} \to U \ot U^*$ that are compatible with the monoidal structure and each other in the sense that:

* Duality reverses the order of tensors: $(A \ot B)^*=B^* \ot A^*$

* The maps $\Hom(U \ot V, W) \xto{(\bu \ot id_V) \circ (id_U \ot b_V)} \Hom(U, W\ot V^*)$ and $\Hom(U, W\ot V^*) \xto{(id_W \ot d_V) \circ (\bu\ot id_{V})} \Hom(U \ot V, W)$ are inverses

* Similarly, the maps $\Hom(U \ot V,  W) \to \Hom(V, U^* \ot W)$ and $\Hom(V, U^* \ot W) \to \Hom(U \ot V,  W)$ induced by $b_{U^*}$ and $d_{U^*}$ are inverses

For axiom $2$, by letting $U=\mbf{1}$ and $V=W$ we get the more commonly used axiom $(id_U \ot d_U) \circ (b_U \ot id_U) = id_U$ and similarly for axiom $3$ and $(d_U \ot id_U) \circ (id_U \ot b_U) = id_U$

This property is convenient because any $\Hom$ space is in natural bijection with any $\Hom$ space of the form $\Hom(\mbf{1},V)$. In a representation category, this is in natural bijection with the set invariant vectors in $V$ denoted $Inv(V)$ (we just look at the image of $1 \in \C$ under this map), and hence understanding the category reduces to understanding the invariant vectors along with their image under the various natural morphisms. Moreover, many of the invariant spaces are in natural bijection:

$Inv(U \ot V) \simeq Inv(V \ot U^{**}) $

$Inv(U \ot V) \natiso \Hom(\mbf{1},U \ot V) \natiso \Hom(U^*, V) \natiso \Hom(\mbf{1}, V \ot U^{**})) \natiso Inv(V \ot U^{**})$

Hence, tensors products are invariant under cyclic rotations as long as $U^{**} \simeq U$. However, it so happens that in the case of quantum group representations, there are natural isomorphisms for any permutation (not just cyclic permutations). This extra structure is called a braiding. We won't need the braiding in this paper, however given that it is one of the primary motivations of quantum groups, we define it below:

A (strict) braided tensor category is a tensor category $\ms{C}$ with a family of natural morphisms $c_{V,W}: V \ot W \to W \ot V$ indexed by $\Ob(\ms{C})\times \Ob(\ms{C})$ and called the commutativity constraint such that commuting an object $U$ past $V$ then $W$ is the same as commuting past $V \ot W$ in one go, ie: $U \ot (V \ot W) \xto{c_{U,V \ot W}} (V \ot W) \ot U$ is the same as $U \ot V \ot W  \xto{c_{U,V} \ot id_W} V \ot U \ot W \xto{id_V \ot c_{U,W}} V \ot W \ot U$.

In the case of the category of representations of a group or Lie algebra, we get a braiding defined by the swap map $c_{V,W}: v\ot w \mapsto w \ot v$ for every $v \in V$ and $w \in W$. This has the added property that swapping two objects and then swapping back is the identity, ie $c_{W,V}c_{V,W}=id_{V \ot W}$. In general, if a braided tensor category has this property, we say it is a symmetric tensor category. However, the fact that the swap map is a morphism of representations follows from the fact that the comultiplication of a group or Lie algebra is cocommutative. One of the key values of quantum group is that they are not cocommutative, so the flip map is not a morphism. However, with some work, one can define a non-symmetric braiding on quantum groups. This braiding then allows us to construct invariants of links, but that is beyond the scope of this paper.

CHAPTER: PRESENTATIONS OF TENSOR CATEGORIES

In this chapter we will give some of the technical definitions required to state our results in a rigorous fashion. Most of the chapter can be skipped and returned to for definitions if needed, except for <ref> which generalizes the diamond lemma and is used to prove $E^1$ convergence in the case of $A_3$.

§ INTRODUCTION

Ordinary confluence is ultimately a question about generators and relators, and so in order to generalize confluence to tensor categories we need to first define what we mean by generators and relators. In the case of abelian groups, a presentation of a group $G$ with $n$ generators and $m$ relators is defined as an exact sequence $ F_m \to F_n \to G \to 0$. Therefore, by analogy, we need to first define a notion of a free pivotal tensor category. It will be a category with fixed generating objects and morphisms, and having as few relations as possible while still being a pivotal tensor category.

The primary difficulty of this is that our generating morphisms won't be maps between two generating objects. Rather, they will correspond to contraction maps which involve the tensor product of three fundamental representations. In order to avoid the technicalities of the more direct construction, we will give a (somewhat informal) graphical definition of the desired category, which will be sufficient for what we want to prove.

Let $\sV$ be a finite set (corresponding to generating objects), and let $D: \sV \to \sV$ such that $D^2=id_{\sV}$ (corresponding to duality between the generating objects). We say $V \in \sV$ is self-dual if $D(V)=V$. This induces a monoidal antihomomorphism on the free monoid $\sV^*$. We will abuse notation and also denote it $D$ since there is little danger of confusion.

Next, let $\sT$ be a finite subset of $\sV^*$ such that

* $\sT$ is closed under cyclic permutation of the letters in a word (eg: if $abc \in \sT$, then so is $cab$)

* $\sT$ is closed under the monoidal antihomomorphism induced by $D$ (eg: if $abc \in \sT$, then $D(abc)=D(c)D(b)D(a)$ is too)

In our setting, generating objects $a_i \in \sV$ will correspond to fundamental representations $V_i$, and an element $a_1a_2...a_n \in \sT$ will correspond to a generating morphism in $Hom(V_1 \ot V_2 \ot... \ot V_n,\K)$.

For example, we could have $\sV=\lb V,W \rb$, $D(V)=W$, and $\sT = \lb VVV , WWW\rb$ which corresponds to the free $A_2$ spider. $\sT = \lb VVV\rb$ wouldn't be allowed because $D(VVV)=WWW$ would then not be in $\sT$. In the cases we are concerned with here, $\sT$ will contain words of length 3 and/or 4.

Given a triple $(\sV,\sT,D)$ we can define special colored, partially directed graphs called webs. In order to avoid double counting with the duality map, we will pick a distinguished element in each set $\lb V , D(V) \rb$. This won't change the category, but will change how each web is drawn.

A $(\sV,\sT,D)$-web is an isotopy class of planar graphs $w$ drawn inside the unit square $[0,1] \times [0,1]$ as follows. Each edge is labeled by a distinguished element $V \in \sV$, together with a direction whenever $V$ isn't self-dual. There is a set of equally spaced boundary vertices on the top or bottom of the unit square (ie $[0,1] \times \lb 0,1 \rb$) where we require each such vertex to have valence $1$. The type $\tau(v)$ (for any internal or boundary vertex) is the cyclically ordered sequence of labels of adjacent edges, where each edge directed inwards contributes the type $V$, and each edge directed outwards contributes the type $D(V)$. We require that type of an internal vertex be in $\sT/(cyclic \; permutations)$. If the boundary vertices on $[0,1] \times \lb 0 \rb$ are $(v_1,...,v_n)$, and the boundary vertices on $[0,1] \times \lb 1 \rb$ are $(w_1,...,w_m)$ we define the source and target functions as $s(w)=D(\tau(v_1))...D(\tau(v_n)) \in \sV^*$, and $t(w)=\tau(w_1)...\tau(w_n) \in \sV^*$ respectively, where these strings are well defined because boundary vertices are adjacent to only a single edge.

For example, in the case of the $A_2$ pair from above $(\lb V,W \rb, D(V)=W, \lb VVV$ , $WWW\rb)$, we choose $V$ as the distinguished element, and denote it by a single edge. In the example web $w$ below, the left trivalent vertex is type $VVV$ since its edges point inwards while the right trivalent vertex is type $WWW$ since its edges point outwards. $s(w)=VWV$ and $t(w)=WVW$

\draw[midto,web] (-1,1.7) -- (0,0);
\draw[midfrom, web] (0,0) -- (2,0);
\draw[midto, web] (-1,-1.7) -- (0,0);
\draw[midto, web] (2,0) --  (3,1.7);
\draw[midto, web] (2,0) -- (3,-1.7);
\draw[midto, web] (6,-1.7) -- (6,1.7);
\end{tikzpicture}$$

Fix a ground field $K$. We will usually be working over $K=\Q(q)$

The free spider $FSp=FSp(\sV,\sT,D)$ is a pivotal category whose objects are strings in the free monoid $\sV^*$, and whose morphisms are $K$-linear combinations of webs which all have the same source and target (the source and target function of the category are then defined as these unique shared source and target of the webs in the linear combination). By abuse of notation, we will consider webs as special morphisms in $FSp$ which we denote as $Web$, and these give a basis for the morphisms by definition. A composition of webs $w_2 \circ w_1$ is done by "stacking vertically", in other words gluing the bottom edge of the first box to the top of the second box. Tensor products are done by "putting two webs side by side", in other words gluing the right side of the first box to the left side of the second box[Slightly more rigorously, the composition is the image of the map $([0,1] \times [0,1])_{w_2} \sqcup ([0,1] \times [0,1])_{w_1} \to  ([0,1] \times [0,1])_{w_2 \circ w_1}$ by linear contractions $([0,1] \times [0,1])_{w_1} \to  ([0,1] \times [0,1/2])_{w_2 \circ w_1}$ and $([0,1] \times [0,1])_{w_2} \to  ([0,1] \times [1/2,1])_{w_2 \circ w_1}$, then forget the vertices on the middle line $[0,1] \times \lb 1/2 \rb$. We can do the same thing for tensors except we contract horizontally, and then place the two webs horizontally].

In our pictures we will allow for boundary vertices that aren't uniformly distributed by remembering that we can always isotopy the web into the correct form if needed

Now that we have defined a free object, we need to define relations. Take a finite set $R \sbs \Mor(FSp)$ which we call the set of local relators, and let $\lan R \ran$ be the smallest linear subset of $Mor(FSp)$ which contains $R$ and is closed under compositions and tensors with morphisms of $FSp$. We then define a new category $Sp$ such that $\Ob(Sp)=\Ob(FSp)$, and $\Hom_{Sp}(X,Y) = \Hom_{FSp}(X,Y)/(\lan R \ran \cap \Hom_{FSp}(X,Y))$ with the induced tensor products and compositions.

This works abstractly, but we want to make this more combinatorial so that we can work with webs rather arbitrary morphisms, and local relators rather than general relations. To do this, we define the set of relators $Rel \sbs \lan R \ran$ as the smallest (generally non-linear) subset containing $R$ and closed under composition and tensors by webs. Intuitively, these will correspond to locally replacing a subgraph of the web with a linear combination of webs. For example, we have a $A_2$ local relator

\draw[midfrom,web] (-1.68,-1.5) -- (-1,-1);
\draw[midto,web] (-1.68,0.5) -- (-1,0);
\draw[midto,web] (0.68,-1.5) -- (0,-1);
\draw[midfrom,web] (0.68,0.5) -- (0,0);
\draw[midto,web] (-1,-1) -- (0,-1);
\draw[midto,web] (-1,-1) -- (-1,0);
\draw[midto,web] (0,0) -- (0,-1);
\draw[midto,web] (0,0) -- (-1,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-2ex,scale=.55]
\draw [web, midfrom]  (-1.68,-1.5) to[out=60,in=-60] (-1.68,0.5);
\draw [web, midto]   (-.3,-1.5) to[out=120,in=240] (-.3,0.5);
\end{tikzpicture} +
\begin{tikzpicture}[baseline=-.5ex,scale=.65]
\draw [web, midfrom]  (-1.68,-.75) to[out=30,in=150] (-.3,-.75);
\draw [web, midto]   (-.3,0.5) to[out=210,in=-30] (-1.68,0.5);
\end{tikzpicture}  $$

After composing with

\draw[midfrom,web] (-1,0) -- (0,0);
\draw[midfrom,web] (-1,0) -- (-1.7,.7);
\draw[midfrom,web] (-1,0) -- (-1.7,-.7);
\draw[midfrom,web] (0,0) -- (.7,.7);
\draw[midfrom,web] (0,0) -- (.7,-.7);
\end{tikzpicture}$$

we get a relator containing each term in the original local relator as a subgraph:

\draw [blue] (-.5,-.5) circle (.9cm);
\draw[midfrom,web] (-1.2,-1.8) -- (-1,-1);
\draw[midto,web] (-1.68,0.5) -- (-1,0);
\draw[midto,web] (.2,-1.8) -- (0,-1);
\draw[midfrom,web] (0.68,0.5) -- (0,0);
\draw[midto,web] (.2,-1.8) -- (-1.2,-1.8);
\draw[midto,web] (.2,-1.8) - -(.7,-2.3) ;
\draw[midfrom,web]  (-1.2,-1.8) -- (-1.7,-2.3);

\draw[midto,web] (-1,-1) -- (0,-1);
\draw[midto,web] (-1,-1) -- (-1,0);
\draw[midto,web] (0,0) -- (0,-1);
\draw[midto,web] (0,0) -- (-1,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-2ex,scale=.55]
\draw [blue] (-1,-.5) circle (.9cm);
\draw [web, midfrom]  (-1.68,-1.5) to[out=60,in=-60] (-1.68,0.5);
\draw [web, midto]   (-.3,-1.5) to[out=120,in=240] (-.3,0.5);
\draw[midto,web] (-.3,-1.5) -- (-1.68,-1.5);
\draw[midto,web]  (-.3,-1.5) - -(.2,-2) ;
\draw[midfrom,web]  (-1.68,-1.5) -- (-2.18,-2);
\end{tikzpicture} +
\begin{tikzpicture}[baseline=-.5ex,scale=.55]
\draw [blue] (-1,0) circle (.8cm);
\draw [web, midfrom]  (-1.68,-.75) to[out=30,in=150] (-.3,-.75);
\draw [web, midto]   (-.2,0.6) to[out=210,in=-30] (-1.78,0.6);
\draw[midfrom,web] (-1.68,-.75) to[out=-30,in=210]  (-.3,-.75);
\draw[midfrom,web]  (-1.68,-.75) -- (-2.18,-1.25) ;
\draw[midto,web] (-.3,-.75) -- (.2,-1.25);
\end{tikzpicture}  $$

Note that the elements of $R$ will be called local relators, the elements of $Rel$ will be called relators, and the elements of $\lan R \ran$ will be called relations. Each relator $r \in Rel$ corresponds to a local relator $\rho$. The type of $r$ (denoted $\tau(r)$) will be defined as the equivalence class of $\rho$ under equivalence by rotation and duality. In practice this will be well defined because our relators will be restricted to single faces.

The key point is that just as webs span $FSp$, we get the following:

$\Span_{\Q(q)} Rel = \lan R \ran$

* $(\sbs)$ This follows immediately from $Web \sbs Mor(FSp)$

* $(\sps)$ We just need to prove that $\Span_{K} Rel$ is closed under compositions and tensors with $Mor(FSp)$ so $R \sbs Rel$ will imply $\lan R \ran \sbs \lan Rel \ran = \Span_{K} Rel$, but this follows from linearity of tensors and compositions. Indeed, if $r= \sum\limits_i a_i r_i \in Rel$ and $x=\sum\limits_j b_jw_j \in FSp$ where $a_i, b_j \in K$, $r_i \in \Span_{\Q(q)} Rel$, and $w_j$ are webs, then $r \circ x=\sum\limits_{i,j}a_ib_j (r_i \circ w_j)$, and similarly for tensors and compositions on the left side, and so the result is a linear combination of relators, which is what we needed to show.

Even though our definition of webs involves picking a target and source, any $\Hom$ space whose cyclically oriented set of boundary edges is the same is naturally isomorphic (since our category is pivotal), so we will typically draw our webs in a circle rather than a rectangle when we don't need to take tensors or compositions.

§ GENERALIZED CONFLUENCE:  CONVERGENCE

Recall that $FSp$ is the free spider whose morphisms are linear combinations of webs, $\lan R \ran$ is the set of relations generated by a finite set of morphism $R$, while $Rel$ is the set of relators (of morphisms) corresponding to local replacement rules. At this point, we could try to generalize the notion of spectral sequences to tensor algebras, but since this introduces more complications without much apparent utility, we will stick to the classical setting of filtered, graded, differential vector spaces. First, we will fix two objects $X,Y \in Ob(FSp)$. For simplicity we define shorthand $Sp(X,Y):=\Hom_{Sp}(X,Y)$, $FSp(X,Y):=\Hom_{FSp}(X,Y)$, $Rel(X,Y):=Rel \cap \Hom_{FSp}(X,Y)$, $\lan R \ran (X,Y):=\lan R \ran \cap \Hom_{FSp}(X,Y)$,  and $Web(X,Y):=Web \cap \Hom_{FSp}(X,Y)$.

$FSp(X,Y)$ is naturally $\Z^k$ graded, where $k$ is the number of vertex types, just by counting the number of internal vertices of each type in a web. In the case of $A_3$ webs, there is only one type of vertex up to duality, so we will collapse the grading into a $\Z$ grading by just counting the total number of internal vertices, forgetting the type. If $w \in Web(X,Y)$, we will write $|w|$ to denote its grading, and more generally if $x = \sum_i a_i w_i \in FSp(X,Y)$ (where $a_i \in \Q(q)$ and $w_i \in Web(X,Y)$)  we will write $|x| = \max{\lb |w_i| : b_i \neq 0 \rb}$. Since $\lan R \ran$ isn't a homogeneous subspace, we will replace it with the following abstract graded vector space:

Let $RSp(X,Y)$ be the graded vector space with formal basis $\lb [r]: r \in Rel(X,Y) \rb$, where $|[r]| := |r|$ for all $r \in Rel(X,Y)$.

Notice that by lemma <ref>, we obtain a surjection $d:RSp(X,Y) \surj \lan R \ran(X,Y)$ induced by $d([r]) = r$. Due to lower order terms, it isn't a graded map, but it is a filtered map via the induced filtration $F^k V = \lb x \in V : |x| \leq k \rb$. Hence, we get a filtered, free presentation of vector spaces:

\begin{equation}
RSp(X,Y) \xto{d} FSp(X,Y) \to Sp(X,Y) \to 0
\end{equation}

Where $\lan R \ran (X,Y)$ is exactly the image of $RSp(X,Y)$ in $FSp(X,Y)$. Taking the vector space $RSp(X,Y) \op FSp(X,Y)$ graded via the direct sum, we get a graded, filtered vector space with a filtered differential $d: RSp(X,Y) \op FSp(X,Y) \to RSp(X,Y) \op FSp(X,Y)$ which is defined as above on $RSp(X,Y)$, and is the zero map on $FSp(X,Y)$. Any filtered short exact sequence of vector spaces $R_V \xto{d} W \to V \to 0$ induces a $2$-column spectral sequence $E^k_{ij}$ where $E^0_{0j} = gr_j R_V := F^j R_V/F^{j-1} R_V$ and $E^0_{1j} =  gr_j W := F^j W/ F^{j-1} W$ where the differential $d^k: E^k_{0j} \to E^k_{1(j-k)}$ (rather than knight moves). Rigorously we define $E^{k}_{ij}$ recursively as $ker(d^k)/im(d^k)$, where $d^k(\ol{x})= \ol{d(x)}$ for $\ol{x} \in E^{k-1}_{ij}$ and $\ol{d(x)} \in E^{k-1}_{(i+1)(j-k)}$. Due to the simple structure, $E^k_{0j}$ is the subset of $gr_j R_V$ of elements which were zero on all previous differentials, while $E^k_{1j}$ is a quotient of $gr_j W$. Moreover, by the general theory, we get eventually convergence to $E^\infty_{1j}:=E^k_{1j}$ for $k \gg 0$, and $E^\infty_{1j}$ is exactly $gr_j V$.

(m) [matrix of math nodes,
nodes in empty cells,nodes=minimum width=5ex,
minimum height=5ex,outer sep=-5pt,
column sep=1ex,row sep=1ex]

3          G_3 R_V           G_3 W

2          G_2 R_V        G_2 W

1          G_1 R_V      G_1 W

0          G_0 R_V      G_0 W

0       1
[blue,-stealth](m-2-2.east) –  (m-2-3.west) node[midway,above] $d^0$;
[green,-stealth](m-2-2.east) –  (m-3-3.north west) node[midway, right] $d^1$;
[orange,-stealth](m-2-2.east) –  (m-4-3.north west) node[midway, right, xshift=.1cm, yshift=-.45cm] $d^2$;
[red,-stealth](m-2-2.east) –  (m-5-3.north west) node[midway, left] $d^3$;
[thick] (m-1-1.east) – (m-6-1.east) ;
[thick] (m-6-1.north) – (m-6-4.north) ;

We will show that convergence of this spectral sequence at the first page $E^1$ (uniformly with respect to the choice of $X$ and $Y$ in the case of presented spiders) is a generalization of confluence, however let's first see what this convergence means concretely in the case of spider presentations. The differential $d^0: gr_n RSp \to gr_n FSp$ is induced by the map $d$, i.e. $d^0 (\ol{[r]}) = \ol{r}$ for $r \in Rel$. This corresponds to forgetting about the lower order terms of $d^0(r)$. If we denote the set of highest order terms of $R$ by $L(R)$, then $E^1$ convergence corresponds to a natural isomorphism $Sp \cong FSp / \lan L(R) \ran$. The subsequent differentials $d^k$ are applied to $r= \sum_i a_i r_i$ which are zero in $gr_{|r|} FSp$, or in other words the leading terms cancel. We therefore need to prove that we can't get any new relations in $Sp$ via sums of relators where the leading order terms cancel. In the next section we will characterize this convergence in a more combinatorial way.

§ A GENERALIZED DIAMOND LEMMA: CANCELING SEQUENCES

We wish to characterize $E^1$ convergence of the spectral sequence defined in the previous section in a more specific setting. In the cases we will be addressing in this paper, the leading term of each relator will consist of either one web or the difference of two webs. Unfortunately, in high ranks this generally fails. In that case, the definitions and theorems from this section can be extended, but the resulting criterion no longer appears as to be as useful. In any case, we don't get $E^1$ convergence for semisimple spiders such as $A_2 \times A_2$ or $A_1^2 \times A_2$ which suggests that this property may fail in most higher rank spiders.

Take a presentations of filtered vector spaces $R_V \xto{d} W \to V \to 0$, together with a distinguished basis of $W$ which we'll call webs (for notational consistency with spiders), and a distinguished basis of $R$ which we'll call relators and denote $Rel$. If $r$ is an element of $Rel$, $L(r)$ will denote the leading order terms of $d(r)$ in $W$. In particular, if $r$ is a relator such that $L(r)$ is a web, we'll refer to $r$ as a reduction relator, and if $L(r)$ is the difference of webs, we'll call $r$ a horizontal relator. We'll assume that all relators are either reduction relators or horizontal relators. Note that the relator basis of $R_V$ together with the filtration gives us a grading as follows: we make each relator homogeneous with grading $|r| = \max\limits_{n \in \N} \lb d(r) \in F^n W \rb$, so we'll assume that the filtration on $R_V$ is the one induced by this grading.

What makes this case different is the existence of horizontal relators. There is a more typical approach to this type of problem which is used to prove the existence of PBW or Gröber bases. That idea is to break symmetry, and give a refined filtration so that the horizontal relators become reduction relators. However, in this case, that method doesn't appear to be necessary.

Just as in the case of confluence, one thing that can go wrong is when two reductions of the same web reduce in two different irreconcilable ways. However, in this case horizontal relators make things more complicated. For one thing, the two different reductions might be on two different webs which are connected (up to lower order terms) by horizontal relators. Even more different, we might have a sequence of horizontal relators which connect a single web to itself, but the lower order terms don't vanish. We state these two problems more rigorously as follows:

* A sequence of relators is n-tuple $(s_1,...,s_n)$ such that $L(s_i)=w_i-w_{i-1}$ for some sequence of webs $(w_0,...,w_n)$, where $w_0$ and $w_n$ are allowed to be zero.

* A reduction sequence is a sequence of relators $s = (s_0,s_1,...,s_n,s_{n+1})$ such that $L(s_0) = w_0$, $L(s_{n+1})=w_n$,

* A horizontal sequence is a sequence of relators $s = (s_1,...,s_n)$ such that $w_0 = w_n$

* A canceling sequence is any sequence of relators that is either a reduction sequence or horizontal sequence

If $s$ is a sequence of relators, we write $\sum s$ as shorthand for the sum $\sum_i [s_i]$. Note that if $s$ is a canceling sequence then by construction, $|d(\sum s)| < |\sum s|$ since the higher order terms cancel. As shorthand, we will denote $|\sum s|$ by $|s|$ (which is also equal to the norm of any relator in the sequence $|s_i|=|[s_i]|$).

Let $s = (s_i)$ be a canceling sequence. We say $s$ is a consistently reducible if there exists a relation $r \in R_V$ such that $|r| < |s|$, and $d(\sum s) = d(r)$.

This recursive definition might be a little different from what one might normally define, but being consistently reducible will imply that $\sum s$ doesn't give us any new relations in $Sp$ that didn't already appear in smaller webs. In particular, we can prove that this is all we need to check in general:

Let $R_V \to W \to V \to 0$ be a presentation of a filtered vector spaces as above, such that the leading term of each relator is either a single web or a difference of webs, then the presentation converges on the $E^1$ page if and only if all canceling sequences are consistently reducible.

This will follow almost immediately from the following decomposition lemma (recalling the definition of $RSp$ in <ref>):

With the hypotheses of proposition <ref>, if $r \in RSp$  is homogeneous, and $|d(r)| < |r|$ then $r=\sum_i b_i (\sum s^i)$ where $\lb s^i \rb$ is a set of canceling sequences, $b_i \in K$ for all $i$, and $|s^i|=|r|$.

Let $r = \sum\limits_{i=1}^n a_i [r_i]$ where $r_i \in Rel$ and $a_i \in K^\times$ (where $K^\times$ is the units of the base field of our vector spaces). We induct on $n$, where the basis case of $n=0$ is vacuously true. Since $r$ has a vanishing leading term, every web that appears in $L(r_i)$ must appear in the leading order term of at least two relators. So if I have a sequence of relators $(r_{k_i})$ where each is a relator in the decomposition of $r$, I can always extend this sequence to another with the same property, provided that the first and last relators aren't both reduction relators. If I keep extending I will eventually get a reduction sequence or a repeated relator and hence a horizontal sequence as a subset. Denote this sequence $s^1$. If $r_{k_1}$ is the first relator in this sequence, we know $r-a_{k_1}\sum s^1$ no longer has a non-zero multiple of $r_{k_1}$ as a summand. Therefore it has at least one fewer relator in its decomposition, we can apply induction.

* $(\Leftarrow)$ Let $\ol{r} \in \gr R_V$ be an arbitrary relation such that $d^0(\ol{r})=0$. We need to prove that $d^i(\ol{r})=0$ for all $i \in \N$ which will show that $d^i$ is trivial on all relations for $i>0$. Take $r$ to be the homogeneous representative. Since $d^0(\ol{r})=0$, we know that $d(r) < |r|$, so by <ref>, we have $r=\sum_i b_i (\sum s^i)$. By definition of being consistently reducible, we know there exists relations $r_i \in R_V$ s.t. $|r_i| < |s^i|=|r|$ and $d(r_i)=d(\sum s^i)$. This implies that $d(\sum b_ir_i)= d(r)$ and $|\sum b_ir_i| < |r|$. However, this tells us that $\ol{r}=\ol{r-\sum b_ir_i}$, but $d(r-\sum b_ir_i)=0$. Since the spectral sequence differentials $d^i$ are induced by $d$, this says that they $d^i(\ol{r})=0$ for all $i\in \N$.

* $(\Rightarrow)$ Let $s$ be a canceling sequence. Since the spectral sequence converges on the first page, and $d^0(\ol{\sum s})=0$ by definition of a canceling sequence, we know $d^i(\ol{\sum s})=0$ for all $i \in \N$. This shows that either $d(\sum s)=0$, or $d(\sum s)=d(r)$ for some relator $r$ with $|r|<|s|$, but that's the definition of being consistently reducible.

Before we go on, we want to use this proposition to prove that $E^1$ convergence coincides with the notion of confluence when relators have only one leading term, or in our terminology: when all relators are reduction relators. We will first recall what confluence means. A rewriting system is a set together with rewrite rules which can be considered as directed arrows connecting elements of the set. This defines a directed graph. A rewriting system is confluent if for any two directed paths $\ga$ and $\ga'$ starting at a vertex $x$ in this graph, we can extend these paths into paths $\wt{\ga}$ and $\wt{\ga'}$ which end at the same element of this set. Assume that we have a presented filtered vector space $V$ as defined above. There is a natural rewrite system whose set is the set of all elements of $W$, and the rewrite rules are of the form $a v_{k_i} + x \mapsto a (v_{k_i} - r_i) + x$ where $L(r_i)=v_{k_i}$ and $x\in W$ whose decomposition into the basis doesn't contain $v_{k_i}$. In particular, spider presentations are filtered vector spaces, where the webs form the distinguished basis of $FSp$, the relators form the distinguished basis of $RSp$, and hence we obtain a canonical rewriting system. From this we can now prove <ref> from the introduction which we restate below:

In particular this holds for spider presentations.

Notice that we only have reduction sequences consisting of two reduction relators, and no horizontal sequences.

* $(\Leftarrow)$ Given two rewriting sequences, we can extend them to maximal rewriting sequences $x=a_nr_n +... +a_1r_1+w$ and $x'=a_m'r_m'+ +...+ a_1'r_1' + w$ which apply to the same web $w$ (these exist because we assumed that all descending chains stabilize). Then, $x-x' = a_nr_n + ... a_1r_1 - a_1'r_1' -...-a_m'r_m'$ is a relation in $d(R_V)$, and hence by spectral sequence convergence there exists a relation $\rho \in R_V$ such that $\ol{d(\rho)}=x-x'$ and $|\rho| = |d(\rho)|=|x-x'|$. But since there are no reduction relators that can be applied to webs in $x$ or $x'$ by maximality, $\rho$ must be trivial, and hence $x=x'$.

* $(\Rightarrow)$ By proposition <ref> we just need to make sure that for every web $w$ and pair of reduction relators $r$ and $r'$ on $w$, there is a relation $\rho$ such that $|\rho|<|r-r'|$ and $d(\rho) = d(r-r')$, but by confluence, the maximal rewriting sequences of relators $r_1,...r_n$ and $r_1',...,r_m'$ must give $d(a_nr_n+...+a_1 r_1 +r +w)=d(a_m'r_m'+...a_1'r_1'+r'+w)$, and hence $d(r-r') = d(r_1+...+r_n-r_1'-...-r_m')$ so taking $\rho = r_1+...+r_n-r_1'-...-r_m$ gives us our result.

§ COMMUTING RELATORS AND NOTATION

For convenience, we will define some notation. If $r$ is a horizontal relator with $L(r) = w'-w$ where $w$ and $w'$ are webs, then we will write $r(w)=w'$, and say that $r$ is a relator on $w$. Each relator corresponds to a web, and hence has boundary edges. We'll call all non-boundary edges in a local relator interior edges. Since relators are locally defined, if two relators $r$ and $t$ on $w$ don't share any interior edges, there is a relator $r'$ on $t(w)$ which is identical to $r$ locally. In this case, we will both say that $r$ and $t$ commute, and $t^{-1}$ and $r$ commute. Thus applications of other relators will give an equivalence class of relators of the same type on different webs. We will sometimes write $t(r)=r'$ if we want to distinguish $r$ and $r'$. Moreover, in order to avoid notational bloat, if $r$ is not a relator on $w$, then $r(w)$ will be short hand for $\rho(w)$ for some $\rho \sim r$ which is on the web $w$. In the case of a sequence of relators, we need the following definition:

If $(s_i)$ is a sequence of relators, then we say $s_i$ and $s_j$ commute (where $i<j$) if $s_{j-1}...s_{i+2}s_{i+1}(s_i)$ exists and commutes with $s_j$

We will next give a list of basic facts and lemmas that will let us simplify canceling sequences. We say two canceling sequences $s$ and $s'$ (with $|s|=|s'|$) are equivalent and write $s \sim s'$ if there exists a relation $r$ with $|r|<|s|=|s'|$, such that $d(r) = d(\sum s - \sum s')$. In particular we get

If $s \sim s'$, then $s$ is consistently reducible iff $s'$ is consistently reducible

If $s'$ is consistently reducible, there exists an $r'$ such that $d(r')=d(\sum s')$ and $|r'| < |s'|$. Moreover, since $s \sim s'$, there also exists an $r$ such that $d(r) = d(\sum s - \sum s')$ and $|r| < |s|=|s'|$, so $d(r+r')=d(\sum s)$, and $|r+r'| \leq \max\lb r, r' \rb < |s|=|s'|$. So $r+r'$ is the relation proving that $s$ is consistently reducible. The reverse direction then follows from the symmetry in the statement.

The main types of equivalence arise from commuting relators, canceling inverse relators, and relators that can be commuted out of the sequence, ie the following facts:

If $s$ is a canceling sequence such that $s_i$ and $s_{i+1}$ commute for some $i$, and $s'$ is the sequence where we reverse the order, (ie $s'=(...,s_{i-1},s_{i+1}',s_i',s_{i+2},...)$ where $s_{i}'s_{i+1}'(w) = s_{i+1}s_{i}(w)$), then $s \sim s'$.

We know $s_{i} = w_{i+1}-w_i +r_i$, and $s_{i+1} = w_{i+2}-w_{i+1} +r_{i+1}$ (where $|r_i|<|s_i|$ for all $i$). $s_{i+1}'=w_{i} + (-s_i)(r_{i+1})$ and $s_{i}'=w_{i+1} + (-s_{i+1})(r_{i})$ (where $(-s_i)(r_{i+1})$ means applying the relator locally identical to $(-s_i)$ to each web term of $r_{i+1}$), so the difference between the sum of the two sequences is $r_i-(s_{i+1})(r_i)+r_{i+1}-(-s_{i})(r_{i+1})$. However, we know that considering each relator as a graphical replacement rule, it doesn't matter which order we replace. These two sequences differ from that only in the relators applied to the lower order terms, so these lower order relators give us our $r$.

If $s$ is a canceling sequence such that $s_{i+1}=-s_i$ for some $i$, and $s'$ is the sequence where they are omitted, (ie $s'=(...,s_{i-1},s_{i+2},...)$), then $s \sim s'$.

Indeed, $d(\sum s) = d(\sum s')$, so they are immediately equivalent.

If $s=(s_0,s_1,...,s_{n-1},s_n,s_{n+1})$ is a reduction sequences such that $s_n$ and $s_{n+1}$ commute, then $s \sim s'=(s_0,s_1,...,s_{n-1},s_{n+1}')$ where $s_{n+1}'=s_n(s_{n+1})$.

This follows since there is an $r \in RSp$ such that $d(r) = s_{n+1}+s_n-s_{n+1}'-s_n'$, but $s_n'$ is a lower order relator

If $s=(s_i)$ is a horizontal sequence, and $t$ is a relator which commutes with each relator in $s$, then the sequence $(t(s_i))$ on $r(w)$ is consistently reducible iff $s$ is consistently reducible.

$s \sim (t,-t,s_1,...,s_n) \sim (-t,s_1,...,s_n,t) \sim (-t,t,t(s_1),...,t(s_n)) \sim (t(s_1),...,t(s_n))$. The reverse direction follows by applying $(-t)$ to $t(s)$

We say a reduction sequence $s$ decomposes into reduction sequences $s'$ and $s''$ if $s \sim (s_0,...,s_{n+1})$, and $s'=(s_0,....,s_k,r)$ and $s''=(-r,s_{k+1}....,s_{n+1})$ for some reduction relator $r$.

If a reduction sequence $s$ decomposes into $s'$ and $s''$, where $s'$ and $s''$ reduce consistently, then $s$ also reduces consistently

The corresponding lower order relation $r$ for $(s_0,...,s_{n+1})$ is just the sum of the relation $r$ for $s'$ and $r''$ for $s''$. Since $(s_0,...,s_{n+1}) \sim s$, we get that $s$ also reduces consistently.

Finally, each web $w$ in $FSp$ has a planar drawing $D$ in $\R^2$. Given a web and a planar drawing, let $\ep=\ep_D \in \R_{>0}$ be the smaller of half the minimum distance between vertices of the web in $D$, and half the length of any edge. Given a subgraph of $w$, the $\ep$-neighborhood of the subgraph will be the smallest simply-connected region of $\R^2$ containing an $\ep$-ball around each point in $D$.

We choose half the distance because we want to take any loop or edge not in $D$ but whose boundary vertices are in $D$, and cut them into two edges

CHAPTER: SPIDERS OF QUANTUM GROUPS

1In this chapter we provide the definition of the various spiders which we'll be working with in this paper. The two sections are independent, so if you mostly care about the $A_3$ spider, then it is unnecessary to read about product spiders. In the section on the $A_3$ spider, we also prove some key results which will be used in the later chapters. In particular <ref> is a global criterion for when an $A_3$ web is reducible. We now fix a ground field $K=\Q(q)$

§ SEMISIMPLE ALGEBRAS AND PRODUCT SPIDERS

§.§ The and Spiders

The first two spiders we will mention are the $A_1$ and $A_2$ spiders. They have already been thoroughly studied, but we mention them here because we will use them to create product spiders which provide some relatively simple examples and counterexamples.

The $A_1$ spider has a single self-dual generating object (this corresponds to the defining representation), and no generating morphisms, and hence the set of webs is graphically just the set of planar matchings (possibly containing closed loops). There is a single relator saying that we can remove a loop from a webs at the cost of multiplying the web by the quantum number $[2]_q=q+q^{-1}$, which we denote graphically by:

\begin{equation}
\label{A1:loop}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[web] (0,0) circle (1cm);
\end{tikzpicture} \; =\; [2]_q
\end{equation}

This is just the Temperley-Lieb category (cf. <cit.>).

As was mentioned in section <ref>, the $A_2$ spider has two generating objects $V$ and $V^*$ (corresponding to the defining representation and its dual). In addition it has two generating morphisms $VVV$ and $V^*V^*V^*$ (corresponding to the contraction maps $V \ot V \xto{\wedge} \wedge^2 V \cong V^*$) which we draw graphically as:

\draw[midto,web] (0,0) -- (1,1);
\draw[midto,web] (2,0) -- (1,1);
\draw[midto,web] (1,2) -- (1,1);

\draw[midto,web] (4,1) -- (4,0);
\draw[midto,web] (4,1) -- (5,2);
\draw[midto,web] (4,1) -- (3,2);
\end{tikzpicture}$$

Finally, there are three generating relators:

\begin{equation}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[midto,web] (0,0) circle (1cm);
\end{tikzpicture} \; =\; [3]_q
\end{equation}

\begin{equation}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Pure bigon
\draw[midfrom,web] (-1,0) -- (0,0);
\draw[web, midto]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[midfrom,web] (2,0) -- (3,0);
\end{tikzpicture}\; =\; [2]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.75]
\draw[midto,web] (-1,0) -- (1,0);
\end{tikzpicture}
\end{equation}

\begin{equation}
\begin{tikzpicture}[baseline=-2ex,scale=.75]
\draw[midfrom,web] (-1.68,-1.5) -- (-1,-1);
\draw[midto,web] (-1.68,0.5) -- (-1,0);
\draw[midto,web] (0.68,-1.5) -- (0,-1);
\draw[midfrom,web] (0.68,0.5) -- (0,0);
\draw[midto,web] (-1,-1) -- (0,-1);
\draw[midto,web] (-1,-1) -- (-1,0);
\draw[midto,web] (0,0) -- (0,-1);
\draw[midto,web] (0,0) -- (-1,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-2ex,scale=.55]
\draw [web, midfrom]  (-1.68,-1.5) to[out=60,in=-60] (-1.68,0.5);
\draw [web, midto]   (-.3,-1.5) to[out=120,in=240] (-.3,0.5);
\end{tikzpicture} +
\begin{tikzpicture}[baseline=-.5ex,scale=.65]
\draw [web, midfrom]  (-1.68,-.75) to[out=30,in=150] (-.3,-.75);
\draw [web, midto]   (-.3,0.5) to[out=210,in=-30] (-1.68,0.5);
\end{tikzpicture}
\end{equation}

Note that these two spiders have only reduction relators, so $E^1$ convergence is just usual confluence.

§.§ Product Spiders

If we have a semisimple but not simple algebra, there aren't enough relations between triples of fundamental representations to generate the category. This is because there are also morphisms of the form $A \ot B \natiso B \ot A$ if $A$ and $B$ are irreducibles for different simple components of our algebra, and these do not decompose into simpler morphisms as in case of the $A_3$ spider. Given two presentations of tensor categories with trivalent and/or tetravalent vertices as in chapter 3, $(\sV_1,\sT_1,D_1,R_1)$ and $(\sV_2,\sT_2,D_2,R_2)$, we define the product spider presentation as: $(\sV_1 \cup \sV_2,\sT_1\cup\sT_2\cup (\sV_1 \sV_2 D(\sV_1) D(\sV_2)), D=D_1\op D_2,R_1\cup R_2 \cup R_{prod})$ where $D_1 \op D_2(V)=D_i(V)$ for $V \in \sV_i$, and $R_{prod}$ is the set of diagrams of the form:

\begin{equation}
\label{product:bigon}
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (0.5,-1) to[out=90,in=180] (1.5,1);
\draw[blue] (1.5,1) to[out=0,in=90] (2.5,-1);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (.5,-1) to[out=30,in=150] (2.5,-1);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{product:Y}
\begin{tikzpicture}[baseline=2ex,scale=.5]
\draw[red] (0,0) -- (1,1);
\draw[red] (2,0) -- (1,1);
\draw[red] (1,2) -- (1,1);
\draw[blue] (0,.75) to[out=45,in=135] (2,.75);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=2ex,scale=.5]
\draw[red] (0,0) -- (1,1);
\draw[red] (2,0) -- (1,1);
\draw[red] (1,2) -- (1,1);
\draw[blue] (0,.75) to[out=-45,in=225] (2,.75);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{product:triangle}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Product Triangle
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Collapsed Kekule
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{product:H}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[red] (-1,-1) -- (0,0);
\draw[red] (-1,1) -- (0,0);
\draw[red] (0,0) -- (2,0);
\draw[red]  (2,0) -- (3,1);
\draw[red] (2,0) -- (3,-1);
\draw[blue] (-1,0) to[out=30,in=150] (3,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[red] (-1,-1) -- (0,0);
\draw[red] (-1,1) -- (0,0);
\draw[red] (0,0) -- (2,0);
\draw[red]  (2,0) -- (3,1);
\draw[red] (2,0) -- (3,-1);
\draw[blue] (-1,0) to[out=-30,in=210] (3,0);
\end{tikzpicture}
\end{equation}

It's possible to derive relator (<ref>) by applying relator (<ref>) twice, however we will lose $E^1$ convergence of our spectral sequence.

We will mainly be interested in the algebras $A_1^n \times A_2^m$ because both $A_1$ and $A_2$ have a unique choice of filtration, so we avoid many of the complications when dealing with $B_2$ or $G_2$ factors.

Following in part the terminology of <cit.>, we'll refer to (<ref>) as a $U$ relator/move, (<ref>) as a $Y$ relator/move, and (<ref>) as an $H$ relator/move. We'll refer to (<ref>) as a triangle relator/move.

§ THE  SPIDER

§.§ Introduction

Let us look at the spider for $A_3$. There are three generating objects: $V=V_{\om_1}$, $W=V_{\om_2}$, $V^*= V_{\om_3}$. As suggested by the notation $D(V)=V^*$ and $D(W)=W$. We will represent $V$ and $V^*$ as single directed edges, and $W$ as an undirected double edge.

\draw[midto,web] (0,0) -- (0,1);
\draw[double,web] (1,0) -- (1,1);
\draw[midfrom,web] (2,0) -- (2,1);
\end{tikzpicture}$$

The generating morphisms will correspond to $VVW$ and $WVV$ together with rotations:

\draw[midto,web] (0,0) -- (1,1);
\draw[midto,web] (2,0) -- (1,1);
\draw[double,web] (1,1) -- (1,2);

\draw[double,web] (4,0) -- (4,1);
\draw[midto,web] (4,1) -- (5,2);
\draw[midto,web] (4,1) -- (3,2);
\end{tikzpicture}$$

We also have 8 relators up to duality and rotation corresponding to the 2 kinds of loops, the 2 kinds of bigons, 3 square relators, and 1 hexagon relator up to rotation and arrow reversal:

\begin{equation}
\label{Single Loop}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[midto,web] (0,0) circle (1cm);
\end{tikzpicture} \;=\; \sqbinom{4}{1}_q
\end{equation}

\begin{equation}
\label{Double Loop}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Double Circle
\draw[double,web] (0,0) circle (1cm);
\end{tikzpicture} \;=\; \sqbinom{4}{2}_q
\end{equation}

\begin{equation}
\label{Pure Bigon}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Pure bigon
\draw[double,web] (-1,0) -- (0,0);
\draw[web, midto]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[double,web] (2,0) -- (3,0);
\end{tikzpicture}\; =\; [2]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.75]
\draw[double,web] (-1,0) -- (1,0);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Mixed Bigon}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Mixed bigon
\draw[midto,web] (-1,0) -- (0,0);
\draw[web, double]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midfrom]  (0,0) to[out=-30,in=-150] (2,0);
\draw[midto,web] (2,0) -- (3,0);
\end{tikzpicture} \; =\;[3]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.75]
\draw[midto,web] (-1,0) -- (1,0);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{IH}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (-1,1) -- (0,0);
\draw[double,web] (0,0) -- (2,0);
\draw[midfrom,web]  (2,0) -- (3,1);
\draw[midfrom,web] (2,0) -- (3,-1);
\end{tikzpicture}\; =\;
\begin{tikzpicture}[baseline=3ex,scale=.5] %I
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (1,-1) -- (0,0);
\draw[double,web] (0,0) -- (0,2);
\draw[midfrom,web]  (0,2) -- (1,3);
\draw[midfrom,web] (0,2) -- (-1,3);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Double Square}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Double Square
\draw[double,web] (-1,-1) -- (1,-1);
\draw[midto,web] (-1,-1) -- (-1,1);
\draw[double,web] (-1,1)-- (1,1);
\draw[midfrom,web] (1,-1)-- (1,1);
\draw[midto,web] (-1,-1) -- (-2,-2);
\draw[midfrom,web] (-1,1) -- (-2,2);
\draw[midfrom,web] (1,-1) -- (2,-2);
\draw[midto,web] (1,1) -- (2,2);
\end{tikzpicture} \;=\; [2]_q
\begin{tikzpicture}[baseline=-4ex,scale=.75] %Parallel Horizontal
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[web, midfrom]  (0,-1.5) to[out=30,in=150] (2,-1.5);
\end{tikzpicture}\;+\;
\begin{tikzpicture}[baseline=4ex,scale=.75] %Parallel Vertical
\draw[web, midfrom]  (0,0) to[out=60,in=-60] (0,2);
\draw[web, midto]  (1.5,0) to[out=120,in=-120] (1.5,2);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Single Square}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Square
\draw[double,web] (-1,-1) -- (1,-1);
\draw[midto,web] (-1,-1) -- (-1,1);
\draw[midfrom,web] (-1,1)-- (1,1);
\draw[midfrom,web] (1,-1)-- (1,1);
\draw[midto,web] (-1,-1) -- (-2,-2);
\draw[double,web] (-1,1) -- (-2,2);
\draw[midfrom,web] (1,-1) -- (2,-2);
\draw[double,web] (1,1) -- (2,2);
\end{tikzpicture} \;=\;
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Asymmetric H
\draw[midfrom,web] (-1,-1) -- (0,0);
\draw[double,web] (-1,1) -- (0,0);
\draw[midto,web] (0,0) -- (2,0);
\draw[double,web]  (2,0) -- (3,1);
\draw[midfrom,web] (2,0) -- (3,-1);
\end{tikzpicture}\;+\;
\begin{tikzpicture}[baseline=-4ex,scale=.75] %Parallel Horizontal: double over single
\draw[web, double]  (0,0) to[out=-30,in=-150] (2,0);
\draw[web, midfrom]  (0,-1.5) to[out=30,in=150] (2,-1.5);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{SS}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %SS
\draw[midto,web] (-1,-1) -- (1,-1);
\draw[midto,web] (-1,-1) -- (-1,1);
\draw[midfrom,web] (-1,1)-- (1,1);
\draw[midfrom,web] (1,-1)-- (1,1);
\draw[double,web] (-1,-1) -- (-2,-2);
\draw[double,web] (-1,1) -- (-2,2);
\draw[double,web] (1,-1) -- (2,-2);
\draw[double,web] (1,1) -- (2,2);
\end{tikzpicture} \;=\;
\begin{tikzpicture}[baseline=-.3ex,scale=.5]
\draw[midfrom,web] (-1,-1) -- (1,-1);
\draw[midfrom,web] (-1,-1) -- (-1,1);
\draw[midto,web] (-1,1)-- (1,1);
\draw[midto,web] (1,-1)-- (1,1);
\draw[double,web] (-1,-1) -- (-2,-2);
\draw[double,web] (-1,1) -- (-2,2);
\draw[double,web] (1,-1) -- (2,-2);
\draw[double,web] (1,1) -- (2,2);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Kekule}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Kekule
\draw[midto,web] (-1,-1) -- (-1.87,0); %Inner Edges
\draw[double,web] (-1.87,0)-- (-1,1);
\draw[midto,web] (-1,1)-- (.6,1);
\draw[double,web] (.6,1)--  (1.47 , 0);
\draw[midto,web]  (1.47 , 0)--  (.6 , -1);
\draw[double,web]  (.6 , -1)--  (-1,-1);

\draw[midto,web] (-1,-1) -- (-1.87,-2); %Outer Edges
\draw[midfrom,web] (-1.87,0)-- (-3.47,0);
\draw[midto,web] (-1,1)-- (-1.87,2);
\draw[midfrom,web] (.6,1)--  (1.47 , 2);
\draw[midto,web]  (1.47 , 0)--  (3.07 , 0);
\draw[midfrom,web]  (.6 , -1)--  (1.6,-1.87);
\end{tikzpicture} \; + \;
\begin{tikzpicture}[baseline=-.3ex,scale=.5]  %3  edges
\draw[midfrom,web] (-0.87,-2) to[out=90,in=0] (-2.47,0);
\draw[midfrom,web] (-1.87,1.5) to[out=-30,in=210] (1.47 , 1.5);
\draw[midfrom,web]  (2.07 , 0) to[out=180,in=90]  (0.6,-1.87);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Kekule
\draw[double,web] (-1,-1) -- (-1.87,0); %Inner Edges
\draw[midfrom,web] (-1.87,0)-- (-1,1);
\draw[double,web] (-1,1)-- (.6,1);
\draw[midfrom,web] (.6,1)--  (1.47 , 0);
\draw[double,web]  (1.47 , 0)--  (.6 , -1);
\draw[midfrom,web]  (.6 , -1)--  (-1,-1);

\draw[midto,web] (-1,-1) -- (-1.87,-2); %Outer Edges
\draw[midfrom,web] (-1.87,0)-- (-3.47,0);
\draw[midto,web] (-1,1)-- (-1.87,2);
\draw[midfrom,web] (.6,1)--  (1.47 , 2);
\draw[midto,web]  (1.47 , 0)--  (3.07 , 0);
\draw[midfrom,web]  (.6 , -1)--  (1.6,-1.87);
\end{tikzpicture}
\; + \;
\begin{tikzpicture}[baseline=-.3ex,scale=.5,transform shape, rotate=180]  %3  edges
\draw[midto,web] (-0.87,-2) to[out=90,in=0] (-2.47,0);
\draw[midto,web] (-1.87,1.5) to[out=-30,in=210] (1.47 , 1.5);
\draw[midto,web]  (2.07 , 0) to[out=180,in=90]  (0.6,-1.87);
\end{tikzpicture}
\end{equation}

Where recall that $[n]_q=q^{n} +q^{n-2} +...+q^{2-n}+ q^{-n}$ are the quantum integers and $\sqbinom{n}{k}_q = \frac{[n]_q...[n+1-k]_q}{[k]_q...[1]_q}$ are the quantum binomial coefficients.

If you prefer minimal sets of relators, you could get by without the hexagon relator (<ref>), two of the square relators (<ref>) and (<ref>), and one of the bigon relators (<ref>) but this will come at the cost of having only $E^3$ convergence

However, this is not the presentation that will be easiest to work with. Instead, given a web $w$, we will define a new graph $S(w)$ by contracting all double edges that aren't on the boundary or part of a loop, so for example, a hexagon relator face will look like:

\begin{equation*}
w=\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Kekule
\draw[double,web] (-1,-1) -- (-1.87,0); %Inner Edges
\draw[midfrom,web] (-1.87,0)-- (-1,1);
\draw[double,web] (-1,1)-- (.6,1);
\draw[midfrom,web] (.6,1)--  (1.47 , 0);
\draw[double,web]  (1.47 , 0)--  (.6 , -1);
\draw[midfrom,web]  (.6 , -1)--  (-1,-1);

\draw[midto,web] (-1,-1) -- (-1.87,-2); %Outer Edges
\draw[midfrom,web] (-1.87,0)-- (-3.47,0);
\draw[midto,web] (-1,1)-- (-1.87,2);
\draw[midfrom,web] (.6,1)--  (1.47 , 2);
\draw[midto,web]  (1.47 , 0)--  (3.07 , 0);
\draw[midfrom,web]  (.6 , -1)--  (1.6,-1.87);
\end{tikzpicture} \longrightarrow
S(w)=\begin{tikzpicture}[baseline=1.5ex,scale=1] %Collapsed Kekule
\draw[midto,web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[midto,web] (2,0) -- (3,0);

\draw[midfrom,web] (0.5,-.5) -- (1,0); %left
\draw[midfrom,web] (1,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (2,1);

\draw[midto,web] (2.5,-.5) -- (2,0); %right
\draw[midto,web] (2,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (1,1);
\end{tikzpicture}
\end{equation*}

Now, given $S(w)$, it is possible to reconstruct $w$ by using the directedness of the single edges, with one exception: the two faces of the $I=H$ relator (<ref>) and its dual relator give the same contracted graph:

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (-1,1) -- (0,0);
\draw[double,web] (0,0) -- (2,0);
\draw[midfrom,web]  (2,0) -- (3,1);
\draw[midfrom,web] (2,0) -- (3,-1);
\end{tikzpicture} \longrightarrow
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %X
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midfrom,web] (0,0) -- (1,1);
\draw[midto,web] (1,-1) -- (0,0);
\draw[midfrom,web] (0,0) -- (-1,1);
\end{tikzpicture} \longleftarrow
\begin{tikzpicture}[baseline=3ex,scale=.5] %I
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (1,-1) -- (0,0);
\draw[double,web] (0,0) -- (0,2);
\draw[midfrom,web]  (0,2) -- (1,3);
\draw[midfrom,web] (0,2) -- (-1,3);
\end{tikzpicture}
\end{equation*}

In some sense this a good thing, since those two webs are equal to each other in $Sp$. However we need to check that consistent reducibility of all canceling sequences on $S(w)$ will also imply consistent reducibility of all canceling sequences of $w$. We extend the map $S$ by linearity to any element $x \in FSp$. The relators we'll put on $S(w)$ will just be any relator of the form $S(w) \to w' \xto{r} r(w') \to S(r(w'))$ where $w'$ is any web which is equivalent to $w$ under the $I=H$ relators (<ref>). We don't want to change the grading, so we give $S(w)$ the grading induced from $w$, so tetravalent vertices count for twice as much as (any remaining boundary) trivalent vertices. We need to be a little bit careful about how we define adjacency in the contracted web. We will say two faces $f$ and $f'$ of a contracted web $S(w)$ are adjacent if they are separated by an edge in every $w'$ such that $S(w')=S(w)$. In particular two faces separated by an $I=H$-relator-type tetravalent vertex won't be adjacent. This definition is made so that two relators commute if and only if they are not adjacent.

Ultimately, we want to prove that if this presentation has $E^1$ convergence, then the original also has $E^1$ convergence. The following lemma will allow us to convert a sequence of relators in one presentation, to a sequence of relators in the other (this just takes the form of omitting $I=H$ relators, or adding needed ones back in).

If $s$ is a sequence of relators on $w$, there exists a corresponding sequence of relators $s'$ on $S(w)$ such that $S(\sum s) = \sum s'$, and conversely, for every sequence $s'$ on $S(w)$ there exists a sequence $s$ on $w'$ (for some $w'$ equivalent to $w$ under I=H relators) such that $S(\sum s) = \sum s'$

Let $s=(s_i)_{i=1}^n$ be a (not necessarily canceling) sequence of relators for $S(w)$, we claim that we can recursively construct a sequence of relators $s'=(s'_j)_{j=1}^m$ on some $w'$ which is equivalent to $w$ under the action of $I=H$ relators and such that $\sum s = S(\sum s')$. First, we know $s_1$ corresponds to a relator $r_1$ which can be applied to $w'$, so we define the partial sequence $s^1=(r_1)$. $S(\sum s^1) = s_1$ by definition.  Now, assume we've defined a sequence $s^k=(s^k_1,...,s^k_{n_k})$ such that $S(\sum s^k) = s_1+...+s_k$. $s_{k+1}$ corresponds to a relator $r_{k+1}$ which can be applied to $w_k'=\rho_{n_{k+1}} ... \rho_1 (s^k(w))$ for some I=H relators $\rho_i$, and define $s^{k+1}$ as the concatenation of $s^k$ and $(\rho_1,...,\rho_{n_{k+1}},r_{k+1})$. Since $S(\rho_1+...\rho_{n_{k+1}})=S(w'_k-w_k)=0$ by construction, and $S(r_{k+1})=s_{k+1}$ since they only differ by $I=H$ relators, this gives us $S(\sum s^{k+1})=S(r_{k+1})+S(\sum s^k)=s_{k+1}+s_k+...+s_1$. Therefore $s^n$ fits our requirements for $s'$

Let $s=(s_i)$ be a (not necessarily canceling) sequence of relators for $w$, we claim that we can construct a sequence of relators $s'=(s'_i)$ on $S(w)$ such that $S(\sum s) = \sum s'$. This direction is even more straightforward: we first take the subsequence $(s_{i_k})$ of relators which aren't I=H relators, then we claim that $s'=(S(s_{i_k}))$ is a sequence of relators and satisfies our criteria. Indeed, $s_{i_{k-1}}....s_2s_1(w)$ and $s_{i_{k}-1}....s_2s_1(w)$ differ only by the $I=H$ relators for every $k$, so $S(s_{i_{k-1}}....s_1(w))=S(s_{i_{k}-1}....s_1(w))$, and hence $S(s_{i_k})$ is a relator on $S(s_{i_{k-1}}....s_1(w))$, which means $(S(s_{i_k}))$ is actually a sequence of relators. Moreover, $S(r)=0$ for every I=H relator, so omitting the I=H relators doesn't change the sums after applying $S$.

Next we use this lemma to prove that it is enough to prove $E^1$ convergence for $S(w)$

If every canceling sequence of $S(w)$ is consistently reducible, then every canceling sequence of $w$ is consistently reducible

Let $s$ be a canceling sequence on $w$. By part 1 of lemma <ref>, there exists a corresponding canceling sequence $s'$ on $S(w)$. Now by hypothesis, there exists a relation $r'$ such that $|r'|<|s'|$ and $d(r') = d(\sum s')$. Now take any lift $r$ of $r'$ to a relation on $w$, such that $|r|=|r'|$. Then $d(r)=d(r')=d(\sum s)$ up to applications of I=H relators (since the kernel of $S$ is generated by I=H relators). In other words, $r-\sum s = \sum_i a_i r_i$ for some $I=H$ relators $r_i$. Now, since $I=H$ relators have no lower order terms, if $|r_k|>|\sum_i a_i r_i|$ for some $k$, then all the relators $r_i$ with $|r_i|=|r_k|$ must add to zero in the partial sum. Therefore, we can assume omit any such terms, and assume $|r_i| \leq |r-\sum s| < |s|$ and so $r-\sum_i a_i r_i$ is a relation that makes $s$ consistently reducible.

We want to make one additional simplification. The double edges at the boundary will force us to list different relators which are the same in the original spider, but whose boundary edges contain different quantities of double edges. To avoid this, we will embed any web with double edges on the boundary into a web with only single edges on the boundary, which we construct as follows: if $w$ is an (uncontracted) web, let $w'$ be any web containing $w$ obtained by adding a trivalent vertex to every boundary double edge. Similar to before, we need to prove that it suffices to show $E^1$ convergence in $S(w')$:

If every canceling sequence of the contracted web $S(w')$ (constructed above) is consistently reducible, then every canceling sequence of $S(w)$ is consistently reducible

The key point is that in the contracted web $S(w')$ we haven't introduced any new relators (although we might have created some I=H relators in the uncontracted web $w'$). Let $s=(s_i)$ be a canceling sequence on $w$, then since $w\sbs w'$, there is a corresponding canceling sequence $s'=(s_i')$ on $w'$. It's consistently reducible by hypothesis, so there is an $r' \in RSp'$ such that $d(r') = d(\sum s')$ but $|r'| < |s'|$. Now, every relator in the contracted spider corresponds to an interior face, but the single edges added to $w'$ are adjacent only to boundary faces, and hence every relator in the contracted spider is disjoint from the new single edges. Therefore, it corresponds to a relation $r$ on $S(w)$ with the properties $d(r) = d(\sum s)$ and $|r| = |r'|-k < |s'| -k = |s|$ where $k$ is the number of boundary double edges in $w$.

From this proposition it is enough to prove consistent reducibility on webs with only single boundary edges. Hence, we have the following derived relators (together with the two loop relators):

\begin{equation}
\label{Pure Bigon C}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Pure bigon contracted
\draw[web] (-1,.5) -- (0,0);
\draw[web] (-1,-.5) -- (0,0);
\draw[web, midto]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[web] (2,0) -- (3,.5);
\draw[web] (2,0) -- (3,-.5);
\end{tikzpicture}\; =\; [2]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %X
\draw[web] (-1,-1) -- (0,0);
\draw[web] (0,0) -- (1,1);
\draw[web] (1,-1) -- (0,0);
\draw[web] (0,0) -- (-1,1);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Mixed Bigon C}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Mixed Bigon contracted
\draw[midfrom,web] (0,0) circle (1cm);
\draw[midto,web] (-2,1) -- (0,1);
\draw[midto,web] (0,1) -- (2,1);
\end{tikzpicture} \; =\;[3]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.75]
\draw[midto,web] (-1,0) -- (1,0);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Double Square C}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Double square contracted
\draw[midto,web] (-1,.5) -- (0,0);
\draw[midfrom,web] (-1,-.5) -- (0,0);

\draw[web, midfrom]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);

\draw[midto,web] (2,0) -- (3,.5);
\draw[midfrom,web] (2,0) -- (3,-.5);
\end{tikzpicture}\;=\; [2]_q
\begin{tikzpicture}[baseline=4ex,scale=.75] %Parallel Vertical
\draw[web, midfrom]  (0,0) to[out=60,in=-60] (0,2);
\draw[web, midto]  (1.5,0) to[out=120,in=-120] (1.5,2);
\end{tikzpicture}\;+\;
\begin{tikzpicture}[baseline=-4ex,scale=.75] %Parallel Horizontal
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[web, midfrom]  (0,-1.5) to[out=30,in=150] (2,-1.5);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Single Square C}
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Single Square C
\draw[web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[web] (2,0) -- (3,0);

\draw[web] (0.5,-.5) -- (1,0); %left
\draw[midto,web] (1,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (2,1);

\draw[web] (2.5,-.5) -- (2,0); %right
\draw[midfrom,web] (2,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (1,1);
\end{tikzpicture}\;=\;
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Asymmetric H
\draw[midfrom,web] (0,-1) -- (0,0);
\draw[web] (0,0) -- (-1,0);
\draw[web] (0,0) -- (0,1);
\draw[midto,web] (0,0) -- (2,0);
\draw[web] (2,0) -- (3,0);
\draw[web] (2,0) -- (2,1);
\draw[midfrom,web] (2,0) -- (2,-1);
\end{tikzpicture}\;+\;
\begin{tikzpicture}[baseline=-2ex,scale=.75] %Parallel Horizontal: double over single
\draw[web] (-1,-.5) -- (0,0);
\draw[web] (0,0) -- (1,.5);
\draw[web] (1,-.5) -- (0,0);
\draw[web] (0,0) -- (-1,.5);
\draw[web, midfrom]  (-1,-1) to[out=30,in=150] (1,-1);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{SS C}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %SS Contracted
\draw[midto,web] (-1,-1) -- (1,-1);
\draw[midto,web] (-1,-1) -- (-1,1);
\draw[midfrom,web] (-1,1)-- (1,1);
\draw[midfrom,web] (1,-1)-- (1,1);
\draw[web] (-1,-1) -- (-1,-2);
\draw[web] (-1,-1) -- (-2,-1);
\draw[web] (-1,1)-- (-2,1);
\draw[web] (-1,1)-- (-1,2);
\draw[web] (1,-1) -- (1,-2);
\draw[web] (1,-1) -- (2,-1);
\draw[web] (1,1)-- (2,1);
\draw[web] (1,1)-- (1,2);
\end{tikzpicture} \;=\;
\begin{tikzpicture}[baseline=-.3ex,scale=.5]
\draw[midfrom,web] (-1,-1) -- (1,-1);
\draw[midfrom,web] (-1,-1) -- (-1,1);
\draw[midto,web] (-1,1)-- (1,1);
\draw[midto,web] (1,-1)-- (1,1);
\draw[web] (-1,-1) -- (-1,-2);
\draw[web] (-1,-1) -- (-2,-1);
\draw[web] (-1,1)-- (-2,1);
\draw[web] (-1,1)-- (-1,2);
\draw[web] (1,-1) -- (1,-2);
\draw[web] (1,-1) -- (2,-1);
\draw[web] (1,1)-- (2,1);
\draw[web] (1,1)-- (1,2);
\end{tikzpicture}
\end{equation}

\begin{equation}
\label{Kekule C}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Kekule Contracted
\draw[midto,web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[midto,web] (2,0) -- (3,0);

\draw[midfrom,web] (0.5,-.5) -- (1,0); %left
\draw[midfrom,web] (1,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (2,1);

\draw[midto,web] (2.5,-.5) -- (2,0); %right
\draw[midto,web] (2,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (1,1);
\end{tikzpicture} \; + \;
\begin{tikzpicture}[baseline=-.3ex,scale=.5,transform shape, rotate=180]  %3  edges
\draw[midto,web] (-0.87,-2) to[out=90,in=0] (-2.47,0);
\draw[midto,web] (-1.87,1.5) to[out=-30,in=210] (1.47 , 1.5);
\draw[midto,web]  (2.07 , 0) to[out=180,in=90]  (0.6,-1.87);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Collapsed Kekule
\draw[midfrom,web] (0,0) -- (1,0); %bottom
\draw[midfrom,web] (1,0) -- (2,0);
\draw[midfrom,web] (2,0) -- (3,0);

\draw[midto,web] (0.5,-.5) -- (1,0); %left
\draw[midto,web] (1,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (2,1);

\draw[midfrom,web] (2.5,-.5) -- (2,0); %right
\draw[midfrom,web] (2,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (1,1);
\end{tikzpicture}
\; + \;
\begin{tikzpicture}[baseline=-.3ex,scale=.5]  %3  edges
\draw[midfrom,web] (-0.87,-2) to[out=90,in=0] (-2.47,0);
\draw[midfrom,web] (-1.87,1.5) to[out=-30,in=210] (1.47 , 1.5);
\draw[midfrom,web]  (2.07 , 0) to[out=180,in=90]  (0.6,-1.87);
\end{tikzpicture}
\end{equation}

Where the unoriented edges correspond to more than one choices of orientation. In particular, notice that up to lower order terms, relators (<ref>), (<ref>), (<ref>), (<ref>) and (<ref>) can all be interpreted as homotopies of paths, while (<ref>) only changes orientations. This will be exploited in the following section to construct a global criterion for reducibility.

§.§ Strands and a Global Criterion for Reducibility

Our goal in this section is to develop a tool which will interpret the horizontal relators for $A_3$ in terms of homotopies of paths, and give a global criterion for when a web is reducible. From now on, we will abuse notation and refer to $S(w)$ as $w$, as defined in the previous section. If it's not clear whether we're discussing the contracted or uncontracted web we will explicitly state which we are considering.

Given a spider presentation with tetravalent vertices, a (directed) strand is a directed path in the web such that any two edges incident at a tetravalent vertex in the path are opposite each other (as depicted below as the dashed line) while an undirected strand is defined the same but without the requirement that the path is directed

\begin{equation}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %X with a strand
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (0,0) -- (1,1);
\draw[midto,web] (1,-1) -- (0,0);
\draw[midto,web] (0,0) -- (-1,1);
\draw[dashed, blue] (-1.1,-1) -- (0.9,1);
\end{tikzpicture}
\end{equation}

Beyond the suggestiveness of the presentation, strands also have a geometric motivation as (weak) geodesic strips in the building. This will be discussed in length in section 11.

Notice that every single edge in the $A_3$ is part of a unique strand. We want to argue that relations are ultimately about homotopies of strands. However, since the square relator (<ref>) can't be interpreted this was, we need to deal with that case first. We will prove that the horizontal square relator is unnecessary for creating a reducible face, and hence we will be able to construct a criterion for reducibility using strands alone. First we will need the following two lemmas about the $A_3$ spider which together tell us that we get a reducible face when two different types of relators collide, and that that is the only way a new reducible face will appear.

If a triangle relator face (<ref>) and a horizontal square relator face are adjacent (in the sense of section $6$, ie when corresponding relators don't commute), then applying either relator will result in a reducible face.

It's not hard to see from orientations that such an adjacency must take the form of:

\begin{equation*}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Kekule with SS
\draw[midto,web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[midto,web] (2,0) -- (3,0);

\draw[midfrom,web] (0.5,-.5) -- (1,0); %left
\draw[midfrom,web] (1,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (2,1);

\draw[midto,web] (2.5,-.5) -- (2,0); %right
\draw[midto,web] (2,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (1,1);

\draw[midfrom,web] (3,0) -- (2,1);
\draw[web] (3,0) -- (3.5,-.5); %Undirected
\draw[web] (2,1) -- (1.5,1.5);
\draw[web] (2,1) -- (2.5,1.5);
\draw[web] (3,0) -- (3.5,0);
\end{tikzpicture}
\end{equation*}

where it's possible that the opposite edge of the square relator is the same edge as one of the inner triangle edges, as in:

\begin{equation*}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Kekule with SS degenerate
\draw[midto,web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[midto,web] (2,0) -- (3,0);

\draw[midfrom,web] (0.5,-.5) -- (1,0); %left
\draw[midfrom,web] (1,0) -- (1.5,.5);
\draw[midfrom,web] (1.5,.5) -- (2,1);

\draw[midto,web] (2.5,-.5) -- (2,0); %right
\draw[midto,web] (2,0) -- (1.5,.5);
\draw[midto,web] (1.5,.5) -- (1,1);

\draw[web] (3,0) to[out=0,in=0] (2.5,-.5);
\draw[web] (2,1) to[out=45,in=45] (3.5,-.5);
\draw[web] (3.5,-.5) to[out=225,in=225] (0.5,-.5);
\end{tikzpicture}
\end{equation*}

Now it's just a matter of checking the four cases: applying the square relator to the top makes a triangle reduction relator (<ref>), applying it to the bottom also makes a triangle reduction relator (<ref>), applying the triangle relator on the top makes a triangle reduction relator (<ref>), and finally applying a triangle relator to the bottom makes a new bigon relator (<ref>).

Let $w$ be an $A_3$ web such that $r(w)$ has a reducible face for some relator $r$, but $w$ does not. In $w$ there must be a horizontal square relator face (<ref>) which is adjacent to a triangle relator face (<ref>), and moreover $r$ corresponds to one of those two faces.

In particular, applying the relator corresponding to the second face will also create a reducible face by lemma <ref>.

First, let $r$ be a triangle relator. The reducible faces in the uncontracted web all have four or fewer sides. Since the uncontracted hexagon relator doesn't change the size of any polygon, and no $I=H$ relator can be adjacent to a hexagon relator, the triangle relator must be adjacent to what was a square in the uncontracted web, however the only non-reducible squares in the uncontracted web are square relators. Therefore, since the adjacency must be by a single edge (since a horizontal square relator face has no interior double edges), the corresponding triangle is also adjacent to the square in the contracted web

On the other hand, if $r$ is a square relator, we see that it doesn't change the adjacencies in the contracted web, and all the reducible faces in the web are triangles or smaller, so it must be adjacent to a triangle. The only type of non-reducible triangle is the horizontal relator face, which gives us our result.

These lemmas give use the following proposition which essentially says that we only need to use one kind of relator in order to create a reducible face. This will help to reduce the number of possibilities we will need to deal with when proving $E^1$ convergence.

If $w$ is a reducible $A_3$ spider, then there is a minimal sequence of relators $(r_i)$ with no horizontal square relators such that $r_n...r_1(w)$ has a reducible face. Similarly, there also exists a minimal sequence of relators with no horizontal triangle relators.

We can find a minimal sequence of relators $(r_1,...,r_n)$ which gives $w$ a reducible face. If there are no horizontal square relators, we're done. Otherwise, there is a last horizontal square relator $r_k$. If $k<n$, and the triangle relator $r_{k+1}$ doesn't commute with $r_k$, then applying either relator will give a reduction face by lemma <ref>. Hence there is a reducible face on $r_{k-1}....r_1(w)$, which contradicts minimality of the sequence. Hence we can commute the square relator $r_k$ to happen last. Now, since $r_nr_{n-1}....r_1(w)$ has a reducible face,  by lemma <ref> we know that there is a triangle relator face adjacent to the square face in $r_{n-1}....r_1(w)$, and applying that relator in place of the square relator, we get a sequence with one fewer square relator which ends in a web with a reducible face. Applying this process repeatedly we can remove all square relators from our minimal sequence.

The argument for removing all horizontal triangle relators follows identically.

Now that know that we only need to use a single type of relator to show reducibility, and that relator corresponds to homotopies of strands, we will get a global criteria for when a web is reducible determined by the placement of strands. We will call any (topological) graph minor composed of strands a configuration of strands. If this graph minor corresponds to the face of a reduction relator, we will call it a reducible configurations of strands. We'll say a relator $r$ on $w$ doesn't affect a configuration of strands or doesn't destroy a configuration of strands if there exists an isomorphic (as a graph) configuration of strands in $r(w)$ which is equal to the original configuration outside an $\ep$-neighborhood of the relator faces.

Before we prove the global reducibility result, we will need to prove a lemma which shows that strands can only cross reducible configurations in very controlled ways:

Let $w$ be an $A_3$ web, and take a reducible configuration of strands which is minimal in the sense that the bounded region in $w$ doesn't contain a reducible configuration. Then every strand that crosses this configuration must pass through the same two boundary edges, and can't intersect any other strand in the bounded region.

This just requires checking each of the reduction relators. If a strand crosses a loop or a monogon, then it forms a smaller bigon, so the loop or monogon wouldn't be minimal.  If a strand crosses a bigon relator of type (<ref>), then it forms a reducible triangle configuration on either side. Strands can cross a bigon relator of type (<ref>), but only in one direction so as to not form a reducible triangle configuration. Therefore, if any two of these strands cross in the middle of the bigon relator, they will form a reducible triangle configuration with the side of the bigon. Similarly, if any strand crosses a reducible triangle relator configuration, and it cuts off one of the corners directed all inward or outward, then that strand together with the corner will form a reducible triangle configuration. Hence, all of the crossing strands must pass through the two sides of the triangle which cross cyclically, and the crossing strands must be directed in the same direction. Moreover, they can't intersect in the interior without creating a smaller reducible triangle configuration with one of the sides of the triangle.

Finally, we can prove our global criterion for when a web is reducible.

A (contracted) $A_3$ web $w$ is reducible if and only if it has a reducible configuration of strands, ie if and only if it has a graph minor of one of the following forms:

* A closed (directed) strand (in the shape of <ref> or <ref>)

* A (directed) strand that intersects itself (in the shape of <ref>)

* Two (directed) strands that intersect more than once (in the shape of <ref> or <ref>)

* Three (directed) strands that pairwise intersect, but whose corresponding triangle isn't oriented cyclically (in the shape of <ref>)

Moreover, if there exists a configuration of undirected strands of any of these types, it is also reducible (in particular, a reducible triangle configuration can be oriented in any way besides a cyclic orientation).

As a corollary of this, we know that applying square relators respects this criterion, however, it isn't true that applying a square relator can't affect a reducible configuration. Rather, this theorem shows that if a square relator does destroy a reducible configuration, there must still be a reducible configuration in the new web.

The undirected and directed versions will be used for two different purposes. We'll use the undirected version to show that certain webs (such as closed webs) are reducible. We'll apply the directed version to cases where we know the web is reducible. In those cases, knowing that the strands are directed will be useful.

($\Rightarrow$) First we prove that every reducible web has a reducible configuration of strands. Since every reducible web can be given a reducible face via the (horizontal) triangle relator (<ref>) by proposition <ref>, and every reducible face corresponds to a reducible configuration of strands, it suffices to prove that if you have one of these reducible configurations of strands and apply a triangle relator, the resulting web will still have a reducible configuration of strands, and the same orientation. Since horizontal triangle relators don't change which maximal directed strands intersect or how many times, so it's clear they can't remove loops and we immediately get that if there is a strand which crosses itself, or two strands which cross twice, then this will remain true after applying the triangle relator.

For the fourth criterion, we know that after applying the relator we still have three mutually intersecting strands, and since the orientation of this triangle is determined by its unchanged boundary, it must still be non-cyclically oriented. Hence, we still have a reducible configuration of strands.

($\Leftarrow$) Assume that the $A_3$ has at least one (directed) reducible configuration, and take a minimal one (meaning there are no reducible configurations contained in the bounded region). By lemma <ref>, we know all of the strands that cross the reducible configuration are parallel. Moreover, since all the reduction relators have at most $3$ sides, the crossing strands must make a triangle with one of the corners. Hence, we can repeatedly apply these triangle relators. By minimality, they are all non-reducible triangle relators, so after applying all these relators the reducible configuration corresponds to a single face, and we can applying the reduction relator.

For the final statement, we want to prove that if there are undirected strands in one of reducible configurations, then it must not be minimal, and so by induction there must be a (directed) reducible configuration contained in it. Clearly, any strand crossing a loop or a monogon has to create a smaller bigon (or self-intersect inside to create an smaller monogon). In the case of the bigon with non-cyclically oriented corners, any strand crossing it must create a non-cyclically oriented triangle with that corner. If a strand crosses a bigon with both corners oriented cyclically, but changes the orientation of one of the bigon strands, then the corresponding triangle with that corner is likewise non-cyclically oriented as depicted below:

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Double square contracted
\draw[midto,web] (-1,.5) -- (0,0);
\draw[midfrom,web] (-1,-.5) -- (0,0);

\draw[web, midto]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);

\draw[midto,web] (2,0) -- (3,.5);
\draw[midfrom,web] (2,0) -- (3,-.5);

\draw[midto,web] (1.3,1) -- (1.3,.21);
\draw[midfrom,web] (1.3,.21) -- (1.3,-.1);
\draw[midto,web] (1.62,0.18) -- (1.6,0.19);
\end{tikzpicture}
\end{equation*}

Lastly, in the case of a triangle, we know that there are some number of strands crossing it (otherwise it would be directed), so each must cut off a corner to form a new triangle configuration (again barring the case of self-intersection, or a strand leaving the same way it came and creating a bigon). If any one of them cuts off a non-cyclically oriented corner, then this smaller triangle is reducible. If all of them cut off a cyclically oriented corner, then at least one of them must intersect a side in a vertex where all orientations point inward or all point outward (otherwise the triangle would be made out of directed strands). Then, this says that the triangle cut off by the strand has a non-cyclically oriented corner, and hence it isn't cyclically oriented, giving us our result.

CHAPTER: PROVING  CONVERGENCE

1This is the technical core of this dissertation where we prove our main results regarding $E^1$ and $E^2$ convergence. Just as the last chapter, the section on product spiders can be read independently of the section on the $A_3$ spider. However, the subsections in each section are not independent. In particular we will use the results about the $A_1^n$ and $A_1 \times A_2$ spiders in the proof of $E^2$ convergence for the $A_1^n \times A_2$ spider.

§ PRODUCT SPIDERS USING EQUINUMERATION

§.§ convergence for

Webs of type $A_1^n$ are given by drawings of graphs corresponding to binary matchings, where we can only match boundary vertices corresponding to the same type. Since there are no trivalent vertices, we only get $U$ relators (of type <ref>) and triangle relators (of type <ref>), along with the inherited loop relators (of type <ref>) (which we copy here)

\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (0.5,-1) to[out=90,in=180] (1.5,1);
\draw[blue] (1.5,1) to[out=0,in=90] (2.5,-1);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (.5,-1) to[out=30,in=150] (2.5,-1);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Product Triangle
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Collapsed Kekule
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[web, blue] (0,0) circle (1cm);
\end{tikzpicture} \; =\; [2]_q
\end{equation*}

It is also not too hard to show that two webs give the same invariant vectors if and only if the corresponding matchings are the same. Additionally, we can see that a web does not have a minimal number of faces if and only if two strands cross each other more than once or there is a loop. Because of these two facts, it's relatively easy to show that the $A_1^n$ spider has an $E^1$ convergent spectral sequence without using the generalized diamond lemma (ie proposition <ref>).

The $A_1^n$ spider with the product presentation has an $E^1$ convergent spectral sequence

To prove this, we will show that:

* Any web that has a closed loop or two strands that cross more than once can be reduced using the relators

* If two webs without closed loops or double-crossing strands correspond to the same matching of boundary vertices, then there exists a sequence of horizontal relators  sending one of the webs to the other

Indeed, by the representation theory we know that the dimension of the invariant spaces are the product of the dimensions of the invariant spaces of each $A_1$ factor, which in turn is equal to the number of planar matchings. Therefore, as long as every choice of planar matchings for each $A_1$ factor has a corresponding minimal vertex web, and any two minimal vertex webs with the same $A_1$ planar matching are the same, we get that the dimension of Hom spaces of $gr_* Sp$ and $(gr_* FSp)/(\lan gr_* R\ran)$ are equal, which proves $E^1$ convergence. We now precede to prove the two claims:

1. First, let's look at the case where there is a minimal reducible configuration that is a loop. If there are no strands crossing it, then we have an immediate reduction. If a strand crosses it, then there is a smaller bigon, which contradicts our assumption of minimality. Hence, we can assume that the minimal configuration is a bigon. Now induct on the number of strands crossing the bigon. Any crossing strand creates a triangle with the two vertices of the bigon (since the bigon is minimal). We will use the following lemma which allows us to remove strands from any such a triangle

Given three pairwise intersecting strands in a $A_1^n$ web such that the bounded region is reduced, there exists a sequence of triangle relators (in an $\ep$-neighborhood of the bounded region) which removes all crossing strands.

We induct on the number of crossings in the bounded region. The configuration has no crossings strands exactly when the bounded region has 3 crossings. Any crossing strand makes a triangle with one of the corners of the outer triangle, so we can apply the inductive hypothesis to remove the strands from the interior triangle. This doesn't increase the number of crossings for the outer triangle (since triangle relators don't change the number of crossings), so after applying the relator associated to the now empty interior triangle we reduce the number of interior crossings for the outer triangle.

Since we assumed the bigon was a minimal reducible configuration of strands, we can apply the lemma to remove strands from the triangle contained in the bigon, and then apply the corresponding triangle relator. This reduces the number of crossings which gives us our result by induction.

2. Now assume we have two webs, $w$ and $w'$ without reducible configurations. The strands and crossings of strands of the two webs $w$ and $w'$ are in bijection. Pick an ordering of strands. We will recursively make $w$ into $w'$ on the first $n$ strands. By induction, assume that the topological minors made up of $\lb s_1,...,s_n \rb$ are the same for $w$ and $w'$, and define $w_n$ to be the minor made up of $\lb s_1,...,s_n \rb$.

Look at the two strands corresponding to $s_{n+1}$ on $w$ and $w'$. These each give paths on $w_n$ which we'll denote $\pi$ and $\pi'$ which together bound a disk. We need to show that we can homotope $\pi$ to $\pi'$ using only triangle moves. By induction we can assume that $s_{n+1}$ and $s_{n+1}'$ only cross once, and it is enough to assume that all of the $s_i$ for $i \in 1,...,n$ cross both $\pi$ and $\pi'$. Now orient each strand so that it goes from $\pi$ to $\pi'$. This gives a poset on crossings in the disk which is defined as follow: let $v,v'$ be crossings, then $v \leq v'$ if there is an oriented path from $v$ to $v'$. There are no directed loops: indeed take a minimal counter example. Then the loop has no self-intersection by minimality and hence bounds a disk with all inward pointing vectors oriented to the right (without loss of generality) of each edge. These edges correspond to strands, and the intersection of the corresponding (right-sided) half-spaces is a subset of the region bounded by the loop on one hand, but on the other hand we know all of these half-spaces contains $\pi(1)=\pi'(1)$ which gives a contradiction.

Now let $v$ be a crossing that is minimal in the above poset. The corresponding strands are $s_i$ and $s_j$ for some $i$ and $j$. By minimality, the triangle bounded by $s_i$, $s_j$, and $\pi$ has no crossing strands, and hence we can apply the corresponding relator. By induction this shows that $\pi$ can be homotoped to $\pi'$, which completes the proof of the second claim.

Hence by the previous remarks, $\gr_\bu FSp/ \lan \gr_\bu R \ran \surj gr_\bu Sp$ is a bijection, so the spectral sequence converges on the first page.

In fact, in the proof we proved the following stronger result which we'll need for inductive arguments in $A_1^n \times A_2$ and the geometric results of chapter 6:

If $w$ and $w'$ are $A_1^n$ webs which differ only by a single strand $s$, then there exists a sequence of relators sending $w$ to $w'$ all of which involve $s$ (and hence the projection onto the $A_1^{n-1}$ factor which doesn't contain $s$ is invariant under each relator in the sequence)

§.§  Convergence for

In this section we will show that $A_1 \times A_2$ has an $E^1$ convergent spectral sequence. Recall that this spider has the following relators (where when drawing product spiders we will not draw the orientation of every edge as long as the orientation of the corresponding strand is indicated)

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[web, blue] (0,0) circle (1cm);
\end{tikzpicture} \; =\; [2]_q
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.5] %Single Circle
\draw[web, midto] (0,0) circle (1cm);
\end{tikzpicture} \; =\; [3]_q
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.75] %Pure bigon
\draw[midfrom,web] (-1,0) -- (0,0);
\draw[web, midto]  (0,0) to[out=30,in=150] (2,0);
\draw[web, midto]  (0,0) to[out=-30,in=-150] (2,0);
\draw[midfrom,web] (2,0) -- (3,0);
\end{tikzpicture}\; =\; [2]_q
\begin{tikzpicture}[baseline=-.3ex,scale=.75]
\draw[midto,web] (-1,0) -- (1,0);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=-2ex,scale=.75]
\draw[midfrom,web] (-1.68,-1.5) -- (-1,-1);
\draw[midto,web] (-1.68,0.5) -- (-1,0);
\draw[midto,web] (0.68,-1.5) -- (0,-1);
\draw[midfrom,web] (0.68,0.5) -- (0,0);
\draw[midto,web] (-1,-1) -- (0,-1);
\draw[midto,web] (-1,-1) -- (-1,0);
\draw[midto,web] (0,0) -- (0,-1);
\draw[midto,web] (0,0) -- (-1,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-2ex,scale=.55]
\draw [web, midfrom]  (-1.68,-1.5) to[out=60,in=-60] (-1.68,0.5);
\draw [web, midto]   (-.3,-1.5) to[out=120,in=240] (-.3,0.5);
\end{tikzpicture} +
\begin{tikzpicture}[baseline=-.5ex,scale=.65]
\draw [web, midfrom]  (-1.68,-.75) to[out=30,in=150] (-.3,-.75);
\draw [web, midto]   (-.3,0.5) to[out=210,in=-30] (-1.68,0.5);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[midto] (0,0) -- (3,0);
\draw[blue] (0.5,-1) to[out=90,in=180] (1.5,1);
\draw[blue] (1.5,1) to[out=0,in=90] (2.5,-1);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[midto] (0,0) -- (3,0);
\draw[blue] (.5,-1) to[out=30,in=150] (2.5,-1);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=2ex,scale=.5]
\draw[web,midto] (0,0) -- (1,1);
\draw[web,midto] (2,0) -- (1,1);
\draw[web,midto] (1,2) -- (1,1);
\draw[blue] (0,.75) to[out=-10,in=190] (2,.75);
\end{tikzpicture}
\begin{tikzpicture}[baseline=2ex,scale=.5]
\draw[web,midto] (0,0) -- (1,1);
\draw[web,midto] (2,0) -- (1,1);
\draw[web,midto] (1,2) -- (1,1);
\draw[blue] (0,.75) to[out=45,in=135] (2,.75);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[web,midto] (-1,-1) -- (0,0);
\draw[web,midto] (-1,1) -- (0,0);
\draw[web,midfrom] (0,0) -- (2,0);
\draw[web,midto]  (2,0) -- (3,1);
\draw[web,midto] (2,0) -- (3,-1);
\draw[blue] (-1,0) to[out=30,in=150] (3,0);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[web,midto] (-1,-1) -- (0,0);
\draw[web,midto] (-1,1) -- (0,0);
\draw[web,midfrom] (0,0) -- (2,0);
\draw[web,midto]  (2,0) -- (3,1);
\draw[web,midto] (2,0) -- (3,-1);
\draw[blue] (-1,0) to[out=-30,in=210] (3,0);
\end{tikzpicture}
\end{equation*}

Since the $A_1$ strands of $A_1 \times A_2$ correspond to $A_2$ cut paths, we can cite the following result:

If $\al$ and $\beta$ are cut paths connecting boundary faces $p$ and $q$ of a basis web $w$ and $\al$ is minimal, then the weight of $\al$ is less than or equal to (and not incomparable to) the weight of $\beta$. If $\beta$ is also minimal, the two parts of $w$ cut by $\beta$ are the same as those of $w$ cut by $\al$ up to $H$-moves

In fact what the proof shows is that even if $\beta$ is not minimal, we can get from $\beta$ to $\al$ using a sequence of $U$ (type <ref>), $Y$ (type <ref>) or $H$ (type <ref>) moves. To prove $E^1$ convergence we will essentially just apply this fact repeatedly. We will also implicitly use the confluence of the $A_2$ spider in assuming that non-elliptic webs form a basis.

The $A_1 \times A_2$ spider with the product presentation has an $E^1$ convergent spectral sequence

We will show that minimal vertex webs are those whose $A_2$ component is non-elliptic, and whose $A_1$ strands follow minimal cut paths.

We first show that webs with elliptic $A_2$ faces are reducible. Let $w$ be an $A_1 \times A_2$ web whose $A_2$ subweb has an elliptic face, ie a square, bigon, or loop. Fix one such face $f$. If there are no $A_1$ strands crossing the face or contained in the face, then we can apply the corresponding $A_2$ relator, so let's assume there is at least one crossing strand $s$. If one of the crossing strands is a closed $A_1$ loop contained in the face, find a minimal closed $A_1$ loop, and apply that relator. Otherwise, looking at the subweb which contains the $A_2$ subweb and only a fixed crossing $A_1$ strand $s$.

* In the case where $f$ is an $A_2$ loop: $s$ forms a bigon with the $A_2$ strand.

* In the case where $f$ is an $A_2$ bigon: $s$ must form a $Y$ with both boundary vertices

* In the case where $f$ is an $A_2$ square: $s$ forms either a $Y$ or an $H$ with the boundary depending on whether it crosses adjacent edges or opposite edges.

Now returning to $w$, since $A_1$ strands can't intersect each other, we can find a minimal strand such that there are no strands between that strand and a $U$, $Y$, or $H$ relator face, and thus apply that relator to remove that strand from the elliptic face. Inducting on the number of crossing strands, we can remove all crossing strands, and then apply the relator.

Now assume the $A_2$ factor of $w$ has no elliptic faces, but there is at least one $A_1$ strand $s$ that doesn't follow a minimal cut path. Look at the minor composed of the $A_2$ subweb, and $s$. By the proof of lemma <ref>, there exists a sequence of length-non-increasing relators sending $s$ to a minimal cut path $p_s$. Back in $w$, $p_s$ may cross other $A_1$ strands, but we'll argue by induction on the number of such crossings that we can make those strands not cross $p_s$.

Find a strand $s'$ which crosses $p_s$ minimally at $A_2$ faces $f_1$ and $f_2$ in the sense that $s'$ crosses $s$ at $f_1$, and then again at $f_2$, and there are no crossings of $s'$ and $p_s$ between $f$ and $f'$. We know $p_s$ is a minimal cut path, so every segment is also minimal length, so by replacing the segment of $s'$ between $f$ and $f'$ so as to follow $p_s$, we would get a cut path of equal or less length. Again by the proof of lemma <ref>, we can do this by using length-non-increasing relators. Since, the area bounded by $s'$ and $p_s$ is strictly contained in the area bounded by $s$ and $p_s$, so we can apply induction to move $s'$ so as to not intersect $s$ and or $p_s$. Applying induction on the number of strands crossing $p_s$, we can make it so that no strands cross $p_s$ in $w$, but then using lemma <ref> again we can move $s$ to $p_s$. Finally, applying induction on the number of non-minimal cut path strands we get that any minimal-vertex web has a non-elliptic $A_2$ component, and every $A_1$ strands follows a minimal cut path.

Next we will show that if we have two such webs $w$ and $w'$ whose $A_2$ component and $A_1$ components are the same, we can find a sequence of $H$ moves sending $w$ to $w'$. There is a correspondence between $A_1$ strands in $w$ and $w'$, and we can consider any strand in either follows a cut path in the $A_2$ component of $w$. We will induct on the number of times a cut-path coming from a strand in $w$ transversely intersects, or follows than diverges from a cut-path coming from a (possibly different) strand in $w'$. Pick an arbitrary pair of consecutive such crossing of cut-paths $\pi$ and $\pi'$ from $w$ and $w'$ (consecutive crossings must exist because the number of crossings of any two strands is even), which is minimal in the sense that there are no other $A_1$ strands in the bounded region. By the proof of lemma <ref> we can move the corresponding segment of $w$ to the corresponding segment of $w'$. Applying induction we can make it so that any strands that whose cut-paths intersect must follow the same cut-path, and in particular corresponding strands are the same.

Elliptic webs are a basis for the $A_2$ invariant space, and planar matchings are a basis for $A_1$ invariant space. Invariants of the product are then products of invariants, and therefore we get equinumeracy between the minimal-vertex webs and the basis of the invariant space, and hence there can be no more relations among the minimal vertex webs. So the first page is the same as the representation category, which gives us $E^1$ convergence.

In the above proof we used (the proof of) the fact that a non-elliptic dual $A_2$ web has coherent geodesics as in <cit.>. Conversely, given a spider $H$ such that $H \times A_1$ spider has $E^1$ convergence with the product presentation (and any grading finer than the weight poset length of strands), then it's not too hard to show that the equivalence class of a minimal web must have coherent geodesics. By this we mean that for every pair of boundary points, there is a representative web with a geodesic of length $\ell$ between those two points, and every path between those points in every web in the equivalence class has length greater than (in particular comparable to) $\ell$.

§.§  Convergence for

We now move on to the rank $\geq 4$ spiders $A_1^n \times A_2$ for $n\geq 2$. These spiders have the same relators as $A_1 \times A_2$, plus the following additional two relators:

\begin{equation*}
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (0.5,-1) to[out=90,in=180] (1.5,1);
\draw[blue] (1.5,1) to[out=0,in=90] (2.5,-1);
\end{tikzpicture} =
\begin{tikzpicture}[baseline=0ex,scale=.5]
\draw[red] (0,0) -- (3,0);
\draw[blue] (.5,-1) to[out=30,in=150] (2.5,-1);
\end{tikzpicture}
\end{equation*}

\begin{equation*}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Product Triangle
\draw[web] (0,0) -- (1,0); %bottom
\draw[midto,web] (1,0) -- (2,0);
\draw[web] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Collapsed Kekule
\draw[web] (0,0) -- (1,0); %bottom
\draw[midfrom,web] (1,0) -- (2,0);
\draw[web] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\end{equation*}

and as long as $n\geq 3$, we also get an extra type of triangle relator:

\begin{equation*}
\begin{tikzpicture}[baseline=1.5ex,scale=1] %Product Triangle
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\;=\;
\begin{tikzpicture}[baseline=-1ex,scale=1,transform shape, rotate=180] %Collapsed Kekule
\draw[red] (0,0) -- (1,0); %bottom
\draw[red] (1,0) -- (2,0);
\draw[red] (2,0) -- (3,0);

\draw[blue] (0.5,-.5) -- (1,0); %left
\draw[blue] (1,0) -- (1.5,.5);
\draw[blue] (1.5,.5) -- (2,1);

\draw[green] (2.5,-.5) -- (2,0); %right
\draw[green] (2,0) -- (1.5,.5);
\draw[green] (1.5,.5) -- (1,1);
\end{tikzpicture}
\end{equation*}

In the last chapter we will show  that $A_1^n \times A_2$ with the product presentation does not have an $E^1$ convergent spectral sequence. Nevertheless, we can prove that it has an $E^2$ convergent spectral sequence by combining the results for $A_1 \times A_2$ and $A_1^n$. The key idea of the proof is that we can move each $A_1$ strand into the correct position with respect to the $A_2$ factor using $H$, $Y$, and $U$ moves. The other $A_1$ strands can interfere with each $A_1 \times A_2$ relator step, but only enough to force one extra crossing, which then can be fixed at the end of the $A_1 \times A_2$ relator step preventing unbounded growth. For the sake of this case, we will extend what we mean by a sequence of relators to include sequences which increase or decrease the number of vertices in the web. Since all relators in this spider have exactly two terms, this should not create too much confusion.

The $A_1^n \times A_2$ spider with the product presentation has an $E^2$ convergent spectral sequence

The difference between the $n=1$ case and the $n>1$ case is that $A_1$ strands from different $A_1$ factors can intersect. We will first prove the following lemma which will limit how much the extra $A_1$ strands can interfere with relators:

Let $w$ be a $A_1^n \times A_2$ web, and let $\pi_i$ be the projection onto the $i$th $A_1 \times A_2$ component. If $\pi_i(w)$ has a $U$ (type <ref>), $Y$ (type <ref>), or $H$ (type <ref>) relator face there is a sequence of relators $r=(r_k)$ on $w$ in an $\ep$-neighborhood of the corresponding bounded region such that either $r(w)$ has a reducible face, or the relator face has no crossing strands. We also have the following properties:

* If the relator is an $H$ relator, $r$ can be chosen such that $|r_k...r_1(w)| \leq |w| +1$ for all $k<n$ and $|r(w)|=|w|$. If it is a $U$ or $Y$ relator, then $r$ can chosen to consist only of horizontal relators.

* If the relator is an $H$ relator, every relator includes a segment of the $A_1$ strand of the configuration, or an $A_1$ strand which parallel to it in an $\ep$-neighborhood of the configuration

We'll argue case by case that either there is a sequence of triangle relators in $w$ which remove all other strands from the bounded region, or we create a reducible face. We first restrict to an $\ep$-neighborhood of the configuration (which may cut some strands into multiple pieces if the original strand crossed the configuration more than once). Moreover, if there is a closed loop contained in the configuration, then in an $\ep$-neighborhood of that loop we have a $A_1^n$ web, and hence can apply <ref> to the subweb to find as sequence of relators reducing it. So assume that there are no loops.

* If it is a $U$ relator then the bounded relation has the same relators as an $A_1^n$ web, so we can apply <ref> to the subweb to find as sequence of relators reducing it.

* If it is a $Y$ relator we can assume each crossing strand  makes a triangle or $Y$ relator configuration with one of the corners, otherwise if it crossed the same edge twice we could apply the $A_1^n$ result again. Find a strand that is minimal with respect to a corner, in that no other $A_1$-strand crosses strictly between it and the corner. If it makes a triangle relator configuration with a corner, we can restrict to the region bounded by that strand and the corner where there are no trivalent vertices. Applying the $A_1^n$ result again we can remove the strand from the original $Y$ configuration. By induction, we can remove all such strands from the $Y$ configuration. On the other hand, if it makes a $Y$ configuration with a corner, then we can immediately apply induction to reduce the web.

* If it is an $H$ relator with $A_1$ strand $s$, by the same arguments as above, we can assume that every strand crosses opposite edges of the bounded square region. Moreover, we can assume that it doesn't cross the two opposite edges of the $A_2$ factor, otherwise we could apply induction to remove it from the configuration. This is the point where we have to increase the number of crossings. By the above there are no strands crossing the two opposite $A_2$ edges, so we can apply an inverse $Y$ relator to $s$ and one of the $A_2$ corners (increasing the number of crossings by $1$). Then $s$ forms a $Y$ configuration with the other $A_2$ corner for example as depicted in (<ref>). Applying the above result for $Y$ relators, we can reduce this subweb without increasing the number of vertices, giving us our result.

\begin{equation}
\label{increasingcrossings}
\begin{tikzpicture}[baseline=-.3ex,scale=.6]
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (-1,1) -- (0,0);
\draw[midfrom,web] (0,0) -- (2,0);
\draw[midto,web]  (2,0) -- (3,1);
\draw[midto,web] (2,0) -- (3,-1);
\draw[red] (.9,-1) -- (.9,1);
\draw[blue] (-1,0) to[out=30,in=150] (3,0);
\end{tikzpicture}\mapsto
\begin{tikzpicture}[baseline=-.3ex,scale=.6]
\draw[midto,web] (-1,-1) -- (0,0);
\draw[midto,web] (-1,1) -- (0,0);
\draw[midfrom,web] (0,0) -- (2,0);
\draw[midto,web]  (2,0) -- (3,1);
\draw[midto,web] (2,0) -- (3,-1);
\draw[red] (.9,-1) -- (.9,1);
\draw[blue] (-1,0) to[out=-45,in=240] (.9,.2) to[out=60,in=150] (3,0);
\end{tikzpicture}
\end{equation}

Similar to $A_1 \times A_2$ we have a natural set of minimal vertex webs: those whose $A_2$ component is non-elliptic, whose $A_1$ strands follow all minimal cut paths, and any two $A_1$ strands cross at most once. We will show that all non-minimal-vertex webs can be reduced without increasing the number of vertices by more than $1$, and that any two minimal-vertex webs $w$ and $w'$ are equivalent via a sequence of webs which have at most one more vertex than $w$ and $w'$.

Let's first prove that any web not of the above form is reducible. Let $w$ be a web whose $A_2$ component is elliptic and hence has an elliptic face (which we recall is a square, bigon, or loop). Restrict to an $\ep$-neighborhood of this face, and induct on the number of vertices. If no strand intersects with the $\ep$-neighborhood of this face then we're done, so assume there is such a strand. Look at each $A_1 \times A_2$ component. By $E^1$ convergence of $A_1 \times A_2$, there is a sequence of relators $(r_i)$ which removes all strands from the elliptic face without increasing the number of vertices. The $(r_i)$ are of type $H$, $Y$, and $U$ relators.

Applying <ref> to the relator face of $r_1$ in $w$ we get a sequence of relators either leading to a web $w'$ with $|w|=|w'|$ or a web with fewer vertices (and in either case not increasing the number of vertices by more than one). By induction, we can assume that it is the latter, in which case $w'$ possesses a relator corresponding to $r_1$, and the sequence of relators did not change the projection onto the chosen $A_1 \times A_2$ factor since all the relators in the sequence are contained in an $\ep$-neighborhood of the relator face of $r_1$, and any strands from the chosen $A_1$ factor which crossed the relator would make applying $r_1$ on the $A_1 \times A_2$ component impossible. Next, we apply this relator, and then apply induction on the length of $(r_i)$ to get a sequence of relators leading to a web where we may remove a strand from the elliptic $A_2$ face. Since the sequence of relators didn't increase the number of vertices at the end, removing a strand reduces the number of edges in an $\ep$-neighborhood of the elliptic face, and hence by induction we can remove all strands from the elliptic face. Applying the corresponding reduction relator then gives us our result.

If there is an $A_1$ strand that doesn't follow a minimal cut-path, we know that in the corresponding $A_1 \times A_2$ component there is a sequence of relators which reduces the web. By the same argument as in the non-elliptic case, we get a sequence of relators in $w$, which gives us our result.

Now assume that the $A_2$ factor is non-elliptic, but there is an $A_1$ loop. We can cut the web along the outside of the loop strand to get a subweb. Then the $A_1$ strand is a non-minimal cut path, and so by <ref> we can find a sequence of relators in the $A_1 \times A_2$ factor which contracts the $A_1$ strand. Then by <ref> and the same argument as before, we can remove all trivalent vertices from the bounded region, and hence reduce the web.

If there are $A_1$ strands that cross more than once, look at the strand-segments going from one crossing to the next. Cut along the outside of the bounded region to get a subweb where both strands can be considered cut paths on the $A_2$ factor. There is a sequence of relators in the $A_1 \times A_2$ component that send one cut path to the other again by <ref>, so again by <ref> we can remove all trivalent vertices from the bounded region, and then apply the $A_1^n$ convergence result to the remaining bounded region. This completes the case where $w$ is reducible.

Lastly, we examine the case where we have two webs $w$ and $w'$ whose $A_1$ and $A_2$ components are all the same, and are minimal vertex webs. We want to argue that we can go from $w$ to $w'$ without increasing the number of vertices by more than $1$. The argument will be similar to before, except we need to be careful about which components we change. We first order the strands $(s_i)$ in the web $w$. We'll look at the projection $\pi_{\leq k}$ onto the $A_2$ component and first $k$ strands of $A_1^n \times A_2$.

We'll induct on the number of faces in our web, and secondly on the largest $k$ such that $\pi_{\leq k}(w)=\pi_{\leq k}(w')$. We'll prove that we can move $s_{k+1}$ in $w$ to follow the same $A_2$ cut path as the version from $w'$ with every relator being $\pi_{\leq k}$ invariant, and without increasing the number of vertices by more than one. If $w$ has only one face, then the statement is vacuously true since the web is the empty graph. So assume that $\pi_{\leq k}(w)=\pi_{\leq k}(w')$ where $k$ can be zero. Via the isomorphism $\pi_{\leq k}(w)=\pi_{\leq k}(w')$ the strands $s_{k+1}$ on $w$ and $w'$ correspond to (minimal) cut paths on $\pi_{\leq k}(w)$. Using induction it suffices to assume that the two $A_2$ cut-paths follow the boundary, and intersect only their endpoints. By <ref>, there exists at least one $H$ move $h$ on the $A_2$ component which reduces the number of trivalent vertices between the two cut-paths. By <ref> we can find a sequence of relators $r=(r_i)$ in $w$ such that we can apply that $H$ relator in $r(w)$, and moreover all relators are $\pi_{\leq k}$ invariant unless one of the first $k$ $A_1$ strands is parallel to $s_{k+1}$ in an $\ep$-neighborhood of the relator configuration $h$. So let's assume that there is such a parallel strand $s$, and let's also pick that $s$ to intersect the $H$-relator configuration $h$ closest to $s_{k+1}$ along one of the $A_2$ edges in the set of all such parallel strands. For example, in the diagram (<ref>) below, we would require there to be no $A_1$ strands crossing the (without loss of generality) right-side $A_2$ strand segment between the top (blue) $A_1$ strand, and the bottom (red) $A_1$ strand.

\begin{equation}
\label{A1nA2:closest}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[web, midto] (-1,-1) -- (0,0);
\draw[web, midto] (-1,1) -- (0,0);
\draw[web, midfrom] (0,0) -- (2,0);
\draw[web, midto]  (2,0) -- (3,1);
\draw[web, midto] (2,0) -- (3,-1);
\draw[blue] (-1,.4) to[out=30,in=150] (3,.4);
\draw[red] (-1,0) to[out=15,in=150] (3,.1);
\end{tikzpicture}
\end{equation}

The strands $s$ and $s_{k+1}$ can't be globally non-intersecting on $w$, otherwise $s$ and $s_{k+1}$ would intersect twice on $w'$, and hence they must intersect somewhere in $w$ outside the $\ep$-neighborhood, for example as in diagram (<ref>) below:

\begin{equation}
\label{A1nA2::outside}
\begin{tikzpicture}[baseline=-.3ex,scale=.6] %H
\draw[web, midto] (-1,-1) -- (0,0);
\draw[web, midto] (-1,1) -- (0,0);
\draw[web, midfrom] (0,0) -- (2,0);
\draw[web, midto]  (2,0) -- (3,1);
\draw[web, midto] (2,0) -- (3,-1);
\draw[blue] (-2,-.8) to[out=60,in=210] (-1,.4) to[out=30,in=150] (3,.4);
\draw[red] (-2,-.4) -- (-1,0) to[out=15,in=150] (3,.1);
\end{tikzpicture}
\end{equation}