Dilute Temperley-Lieb algebra

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Dilute Temperley-Lieb algebra

Michiel Lieftink

July 13, 2018

Bachelor thesis Mathematics and Physics & Astronomy Supervisor: prof. dr. Bernard Nienhuis, prof. dr. Jasper Stokman

Institute of Physics

Korteweg-de Vries Institute for Mathematics Faculty of Sciences

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Abstract

In this thesis the regular and dilute n-diagram algebra, with the variable β = q−1+ q are studied, the regular and dilute Temperley-Lieb algebra are introduced to give a more abstract representation of the corresponding n-diagram algebra and their equivalence is proven. In the case that q is not a root of unity a full set of irreducible modules is given and after proving the irreducibility of these modules it is shown that the algebras are semisimple using Wedderburn’s theorem. After that transfer matrices are introduced and shown how they relate with the Temperley-Lieb algebra, these transfer matrices are then used to show the relation between the Potts model on the square lattice and the Temperley-Lieb algebra. Then a O(n) model on the square lattice is examined and the corresponding phase diagram is explored further.

Title: Dilute Temperley-Lieb algebra

Authors: Michiel Lieftink, michiellieftink@gmail.com, 11044764 Supervisors: prof. dr. Bernard Nienhuis, prof. dr. Jasper Stokman Second grader: dr. Raf Bocklandt, prof. dr. Kareljan Schoutens End date: July 13, 2018

Institute of Physics University of Amsterdam

Science Park 904, 1098 XH Amsterdam http://www.iop.uva.nl

Korteweg-de Vries Institute for Mathematics University of Amsterdam

Science Park 904, 1098 XH Amsterdam http://www.kdvi.uva.nl

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1. Introduction

Using statistical physics it is possible to describe many systems and physical properties. A very useful tool in the examining many of these is the Temperley-Lieb algebra. Ever since its introduction almost fifty years ago [1] it has been used to help solve many systems in statistical physics.

In this thesis we will be examining the Temperley-Lieb algebra and one of its generalizations the dilute Temperley-Lieb algebra. There exist two ways to describe these algebras. The first is using diagrams consisting of a vertical strip with n points on the upper boundary and n points on the lower boundary, then a spanning set of the algebra corresponds to all the possible ways of connecting those points by non-intersecting lines. The second is purely algebraically using generators and relations between those generators. For the Temperley-Lieb algebra the equivalence between the two has been proven[5]. But for the dilute Temperley-Lieb algebra such an equivalence has never been proven. One of the main problems with proving such an equivalence is that is hard to determine whether the relations between generators that have been found do in fact completely determine the algebra, because there might be some other relation that have been overlooked.

In this thesis we will first examine the Temperley-Lieb algebra and show the equivalence between the two ways of describing it. Then we will examine the representations of this algebra, similar to [5], by constructing a set of modules and showing that in certain cases these are irreducible and form a complete set of irreducible modules.

Then we will look at the dilute Temperley-Lieb algebra. We will define the way of viewing the algebra using diagrams and give a construction for each diagram using a set of generators. After that we define an algebra in terms of generators and defining relations and show that this algebra is isomorphic to the dilute n-diagram algebra, which is the algebra defined by the diagrams, this has not been done before. Then we will study the representations of this algebra, similar to [6], by constructing a set of modules and showing that in certain cases they form a set of complete irreducible modules.

After that we will give a short recap of some elements of statistical physics and then introduce the transfer matrix. An important part of statistical physics is the calculating of partition functions and the transfer matrix is a useful way of rewriting the problem to make it possible to calculate the partition function. We will show the link between the transfer matrix on square lattice models and the Temperley-Lieb algebra.

Then we will examine two square lattice models, the Potts model and the O(n) model. We will show how the transfer matrix can be applied to these two models and finally we will examine the phase diagram of an O(n) model on a square lattice. This lattice model consists of a grid where each edge corresponds to a particle and each vertex to an interaction between the adjacent particles. There are four different types of interaction on each interaction and each of those is given a weight. This weight corresponds to the chance that that particular interaction takes place on any vertex. Different weights lead to different lattices and at certain weight a phase transition takes place. We then try to explore when such a transition might take place.

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I would like to thank my supervisors prof. dr. Bernard Nienhuis and prof. dr. Jasper Stokman not just for consistently finding the time to discuss my progress, but also for the many insights they have given me into this fascinating subject. Finally i would like to thank Kayed Al-Qasemi MSc for helping understand the subject matter and teaching me how to draw the various diagrams in this thesis.

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2. Temperley-Lieb algebra

We will start this chapter by introducing the n-diagram algebra and showing its equivalence with the Temperley-Lieb algebra by examining the dimensions of both algebras. Then we will begin our study on the representations of the n-diagram algebra.

The n-diagram algebra is defined as follows. We draw two horizontal lines and mark n points on each of the lines. We then connect each of these points with n non-intersecting lines which are fully within the horizontal lines, we identify these diagrams up to isotopy. This will be an n-diagram. Next we form the complex vector space with basis the set of all n-diagrams. We then define multiplication between two n-diagrams by identifying the bottom horizontal line of the right diagram with the top horizontal line of the diagram, replacing each closed loop by a factor β and then removing the interior horizontal line. We will from now on refer to this as concatenation. For example:

· = = β .

This defines for each β ∈ C, an associative algebra, the n-diagram algebra which we will denote Dn. This algebra has a unit namely:

· · · . It is often useful to write β = q + q−1, where q∈ C×.

Definition 2.1. For a β_{∈ C, the Temperley-Lieb algebra denoted TL}n(β) or TLn is defined

as being generated by the unit 1 and elements u_i, i∈ {1, 2, . . . , n − 1}, satisfying u2_i = βui, uiui±1ui= ui and uiuj = ujui if |i − j| > 1.

For the details of this construction see appendix A.

Proposition 2.2. There exists a unique unit-preserving surjective algebra morphism ϕ :

TLn→ Dn such that ϕ maps the generators of TLn to the following diagrams

17→ _{· · ·} and u_i _7→ _{· · ·} _{· · ·}

1 i− 1 i i + 1i + 2 n

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Proof. To prove this we need to check that all the relations in TLnalso hold in the n-diagram

algebra with the given identifications,

u2_i = · · · · · · · · · · · · = β _{· · ·} _{· · · .} uiui−1ui= · · · · · · · · · · · · · · · · · · = _{· · ·} _{· · · .} uiuj = · · · · · · · · · · · · · · · · · · = _{· · ·} _{· · ·} _{· · ·} = ujui, |i − j| > 1.

It can be shown that the identity and the ϕ(ui) generate the n-diagram algebra. From

this it follows that the n-diagram algebra is a quotient of the Temperley-Lieb algebra. To prove that the two are in fact isomorphic we will show that their dimensions are the same. First we will calculate the dimension of the n-diagram algebra. For this we will introduce (n, k)-link diagrams. We construct these by drawing a single horizontal line with n points. We can then connect these dots with lines as in the regular n-diagrams except now k lines will be connected to only one point. We will call these lines defects. We will call a line that is connected to two points a link. Note that by construction n_{− k and n + k are even. An} example for n = 7 and k = 1

.

We will always orient the (n, k)-link diagrams so that the defects face up.

We can always transform an n-diagram into an (2n, 0)-link state by connecting the two horizontal lines on the left and reorienting the rest of the diagram so that this line forms the bottom of the diagram. For example:

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It is obvious that this creates a bijection between the set of n-diagrams and the set of (2n, 0)-link diagrams.

Lemma 2.3. There exists a bijection between the (n, k)-link diagrams and the increasing

walks on _Z2 _{from (0, 0) to (}n+k

2 ,

n−k

2 ), which do not cross the diagonal {(m, m)|m ∈ Z}.

Increasing here means that in the walk you can only move up or to the right.

Proof. We give this bijection by reading an (n, k)-link state from the left to the right and you move up at the i-th step if you close a link at the i-th point. Otherwise you move the the right. For the earlier example with n = 7 and k = 1,

→ b b (4, 3) (0, 0) . There aren−k

2 links in the diagram and therefore we take precisely

n−k

2 steps up and n−

n−k

2 =

n+k

2 to the right. We cannot cross the diagonal because to close a link we first have to open it.

Which means to take a step up we need to move to the right, so we can’t cross the diagonal. This is a bijection because we can easily reverse this process by reversing the walk, moving from the right to the left in the (n, k)-link state diagram by opening a link if we move down, If we move down we close a link if possible otherwise we create a defect. This clearly creates a bijection between the walks and (n, k)-link states.

Lemma 2.4. There exists a bijection between the Temperley-Lieb algebra and the increasing

walks on_Z2 _{from (0, 0) to (n, n) which do not cross the diagonal} _{{(m, m)|m ∈ Z}.}

We will not prove this in this thesis, for the details view [5].

Theorem 2.5. The n-diagram algebra is isomorphic to the Temperley-Lieb algebra.

Proof. We have previously seen that the dimension of the n-diagram algebra equal is to the number of (2n, 0)-link diagrams and from lemma 2.3 we know that this is equal to the number of increasing walks on_Z2_{from (0, 0) to (n, n) which do not cross the diagonal}_{{(m, m)|m ∈ Z}.}

From lemma 2.4 we know that the same holds true for TLn. Finally we know that the

n-diagram algebra is a quotient of TLnwhich means that the two algebras are in fact isomorphic

because they have the same dimension.

Lemma 2.6. If dn,k is the dimension of increasing walks onZ2 from (0, 0) to (n+k₂ ,n−k₂ ) that

do not cross the diagonal then,

dn,k= ( n n−k 2 ) − ( n n−k−2 2 ) . Proof. Any path ending at (n+k

2 ,

n−k

2 ) must pass through (

n+k 2 − 1, n−k 2 ) or ( n+k 2 , n−k 2 − 1)

but not both, so we have the recursion

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With the boundary values dn,n = 1, because there is only one path from (0, 0) to (n, 0) and

dn,−1 = 0, because the walk cannot cross the diagonal. You can easily check that the solution

is, dn,k= ( n n−k 2 ) − ( n n−k 2 − 1 ) .

Corollary 2.7. The dimension of the Temperley-Lieb algebra is,

dim TLn= ( 2n n ) − ( 2n n− 1 ) = 1 n + 1 ( 2n n )

Proof. The dimension of the Temperley-Lieb algebra is equal to d2n,0, so the rest follows from

the previous lemma.

2.1. Representation of the n-diagram algebra

The (n, k)-link states are not only useful for calculating the dimension of the algebra, but they also serve as left modules for the n-diagram algebra.

Definition 2.8. The complex span over the (n, k)-link states (for all k) is Mn. We will call

this the link module.

As the name might suggest Mn naturally serves as a left Dn-module by concatenation of

the diagram with the link-state. For example,

· =

= .

Lemma 2.9. This operation gives M_n the structure of a left D_n-module. This is fairly obvious since concatenation is associative.

Note that this action can decrease the number of defects by closing a pair of defects, but never increase it. So we can define a submodule of M_n by viewing only those link states with at most k defects. We will denote this as M_n,(n_−k)/2. Since the number of defects can only decrease we have the following inclusions

0⊂ Mn,⌊n/2⌋⊂ · · · ⊂ Mn,1⊂ Mn,0= Mn.

Note that these submodules give the lower bound n−k

2 to the number of links when the

submodule can contain at most k defects. We will from now on often denote l = n−k

2 to be

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Definition 2.10. We define the standard modules V_n,l to be the quotient

Vn,l ≃ Mn,l/Mn,l+1 for 0≤ l ≤ ⌊n/2⌋ − 1 and Vn,⌊n/2⌋= Mn,⌊n/2⌋.

We will identify Vn,l with the vector space spanned by the (n, k)-link diagrams, this is

well-defined because every equivalence class contains precisely one (n, k)-link diagram. The Vn,l are Dn-modules with an action consisting of concatenation when the number of defects

is conserved and 0 otherwise. The dimension of these is the same as dn,k given by lemma 2.6.

2.2. Irreducibility of the standard modules

Now it is a reasonable thing to wonder wether the standard modules defined in definition 2.10 are irreducible as Dn-modules. As it turns out most of the time that is the case, but in

order to show that we will need to introduce an invariant bilinear form. By representing this bilinear form using a matrix we will be able to determine whether the standard modules are irreducible for almost all cases.

The (n, k)-link states were oriented so that they were facing to the top allowing for a natural left Dnaction. We can of course reflect these (n, k)-link states across the horizontal line and

in doing so we obtain a left Dn action. We will use these reflected link states to define the

following bilinear form.

Definition 2.11. The bilinear form_{⟨·, ·⟩}_n,l is defined on V_n,l as follows. If x and y are two (n, k)-link states,⟨x, y⟩n,l is computed by concatening the reflected link state of x with y. If

not every defect in x is connected with a defect in y, ⟨x, y⟩n,l = 0. Otherwise⟨x, y⟩n,l = βm

where m equals the number of closed loops.

We will now start of by giving a couple of examples in V4,1

⟨ , ⟩n,l = = β

⟨ , ⟩n,l = = 1

⟨ , ⟩n,l = . = 0

This bilinear form is symmetric,_{⟨x, y⟩ and ⟨y, x⟩ are reflections of each other when we view} the concatenation of x and y, therefore they have the same value.

We will now consider reflection of an n-diagram across a horizontal line. This will be another n-diagram so doing this will define a linear map from the n-diagram algebra to itself.

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This preserves the multiplication except that the order will be reversed, so this is an anti-algebra morphism. We will denote the reflection of a diagram U by U†. It is clear from the diagrams that 1† = 1 and u†_i = ui. Using this we can show that the adjoint of the action

with respect to the bilinear form is given by this reflection.

Lemma 2.12. For all n-diagrams u the bilinear form _{⟨·, ·⟩}n,l on Vn,l satisfies

⟨x, Uy⟩n,l=⟨U†x, y⟩n,l ∀x, y ∈ Vn,l

The result immediately follows from comparing the diagrams of the two sides.

Definition 2.13. We define the bilinear map _{|·, ·|}_n,k : Vn,l× Vn,l −→ Dn as follows. For x

and y (n, k)-link diagrams we let|x¯y| be the n-diagram formed by viewing the link diagram x as the bottom row in an n-diagram, the reflected state of the link diagram y as the top row and connecting each defect with one on the opposite row. We then extend this linearly to Vn,l× Vn,l to obtain the map.

Note that connecting the defects is only possible in a unique way so this map is well-defined. This definition gives rise to an another way of determining the dimension of the n− diagram algebra.

Proposition 2.14. The dimension of the n-diagram algebra is given by

dim Dn= ⌊n/2⌋_∑

l=0

(dim Vn,l)2

Proof. Because for each n-diagram D there exists a unique l and unique x, y ∈ Vn,l such

that D = _{|x¯y|. Therefore the dimension of the n-diagram algebra is precisely as in the} proposition.

We can combine the bilinear map_{|·, ·|}_n,l with the bilinear form we have previously defined to obtain the following relation.

Lemma 2.15. If x, y, z_{∈ V}_n,l, then

|x¯y|z = ⟨y, z⟩n,lx.

Proof. Because of linearity it suffices to prove it for when x, y and z are (n, k)-link states. We will prove this lemma by looking at two cases, when at least one of the defects in y are closed by a link in z and when none of the defects in y are closed by a link in z.

In the first case we have two defects in y which are connected, so by definition⟨y, z⟩n,l= 0.

For_{|x¯y|z we know that every defect in x is connected to one in y and at least two of those} defects are closed by a link in z. This means that _{|x¯y|z has more links than x, because the} rest of the action can only increase the number of links. But Vn,l consists all the link states

with precisely l links, therefore_{|x¯y|z = 0.}

In the second case it is clear that the left-hand side of the equation is proportional to x, since all the defects in x remain defects. By definition this constant is given by β to the power of the number of closed loops. But because x does not cause any closed loops in_{|x¯y|z this is} precisely_{⟨y, z⟩}_n,l.

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Proposition 2.16. If_{⟨·, ·⟩}_n,l is not identically zero on V_n,l, then V_n,l is cyclic and indecom-posable.

Proof. Since_{⟨·, ·⟩}_n,l is non-zero we have y, z_{∈ V}_n,l such that_{⟨y, z⟩}_n,l = 1, but then for every x∈ Vn,l we have |x¯y| ∈ Dn satisfying

|x¯y|z = ⟨y, z⟩n,lx = x.

Therefore the module V_n,l is cyclic.

Suppose now that Vn,l= A⊕B for some non-trival submodules A, B of Vn,l. Then we must

have z = a + b for some a _{∈ A and b ∈ B. Suppose now that ⟨y, a⟩}_n,l _{̸= 0, then for every} x∈ Vn,l we have that x = |x¯y|a ⟨y, a⟩n,l = x ⟨y, a⟩n,l ¯ ya. but _⟨y,a⟩|x¯y|

n,l ∈ Dn, so A = Vn,l. But then B ={0}, while we assumed that B was non-trivial.

Therefore we have that_{⟨y, a⟩}_n,l= 0. From this it follows that⟨y, b⟩n,l= 1 but then b generates

Vn,l which implies A ={0}. Therefore Vn,l is indecomposable.

We note that _{⟨·, ·⟩}_n,l is not identically zero when β _{̸= 0. because if we choose any link} diagram x_{∈ V}_n,l,_{⟨x, x⟩}_n,l= βl.

Proposition 2.17. If _{⟨·, ·⟩}_n,l is not identically zero

Vn,l ≃ Vn,l′ =⇒ l = l′

Proof. Suppose θ : V_n,l _{→ V}_n,l′ is an isomorphism and l̸= l′we can assume that l > l′ because

θ is invertible. Then choose y, z∈ Vn,l such that⟨y, z⟩n,l = 1. For all x∈ Vn,l we have that

|x¯y|θ(z) = θ(|x¯y|z) = θ(x).

But _{|x¯y|θ(z) = 0 because for θ(z) left-multiplying with |x¯y| leads to at least l links which is} zero in V_n,l′.

From this contradiction it follows that l = l′.

To determine the irreducibity of V_n,l we will now introduce the following.

Definition 2.18. The radical R_n,l of the bilinear form on V_n,l is the following subset of V_n,l Rn,l={x ∈ Vn,l : ⟨x, y⟩n,l = 0 ∀y ∈ Vn,l}.

The usefulness of this radical immediately follows from the following lemma.

Lemma 2.19. The radical R_n,l is the maximal submodule of V_n,l.

Proof. The radical being a submodule is a direct consequence of lemma 2.12. This submodule is maximal because from lemma 2.15 it follows that every element outside of the radical is a generator of Vn,l.

We will now determine the irreducibility of V_n,l based on the bilinear form_{⟨·, ·⟩}_n,l. For this we first recall that the the (n, n− 2l)-link states form a basis for Vn,l. We can then represent

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Definition 2.20. The gram matrices denoted G_n,l are the matrices representing the bilinair form_{⟨·, ·⟩}_n,l in the basis of (n, n_{− 2l)-link diagrams.}

A few examples for n = 4,

G4,0 = ( 1), G4,1 =  β1 1β 01 0 1 β   and G4,2= ( β2 β β β2 ) ,

when we use the ordered bases

{ } , { , , } and { , } .

We now note that the radical Rn,lprecisely corresponds to the kernel of Gn,l, which means

that the irreducibility of V_n,l is equivalent to det G_n,l_{̸= 0. We will determine using restriction} of the standard modules a recursion relation for det Gn,l for when q is not a root of unity,

where we recall that β = q + q−1.

Proposition 2.21. We consider the inclusion of the D_n₋₁ into D_n by letting the diagrams be the same at the first n− 1 points and having a vertical line at the n’th position. We denote the restriction of V_n,l to a Dn−1-module by Vn,l ↓ then we have the short exact sequence of

D_n₋₁ modules,

0−→ Vn−1,l−→ Vn,l↓−→ Vn−1,l−1 → 0.

Proof. The inclusion V_n_−1,l,→ Vn,l↓ is given by extending every (n − 1, n − 2l − 1)-link state

by adding a defect at position n. This is an injective D_n₋₁ homomorphism because that defect at position n will be conserved under the inclusion of Dn−1 into Dn. Now it remains

to show that the quotient V_n,l_{↓ /V}_n_−1,l_{≃ V}_n_−1,l−1.

We note that the quotient V_n,l _{↓ /V}_n_−1,lhas a basis consisting of all the cosets represented by (n, n− 2l)-link states where a link ends at position n. Then we have the obvious vector isomorphism ϕ where we replace that link with just a defect at position where it began which we will denote by m. Now it remains to show that for all u_i _∈D_n₋₁ and z basis elements of Vn,l ↓ /Vn−1,lthat ϕ(uiz) = uiϕ(z). Note that unless i = m− 1 and z has a defect at position

m−1, ui simply gives another basiselement of Vn,l↓ /Vn−1,lwhere the link ending at n begins

in some m′. For this case it is easy to see that ϕ(u_iz) = uiϕ(z). If i = m− 1 and z has a

defect at m− 1, then um−1z has a defect at the n’th position.

· · · · · ·

=

· · · · · ·

.

Then ϕ(u_m₋₁z) = 0 because there is a defect at the n’th position and um−1ϕ(z) = 0 because

ϕ(z) has defects at both position m− 1 and m, so um−1 adds a link which means the result

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Proposition 2.22. When q2(n−2l+1)_{̸= 1, the exact sequence in proposition 2.21 splits, which}

means

Vn,l↓≃ Vn−1,l⊕ Vn−1,l−1.

Proof. This proposition can be proven by examining the eigenvalues of a central element F_nin Dn. Fn has two distinct eigenvalues when q2(n−2l+1) ̸= 1 whose eigenspaces each correspond

to either Vn−1,l or Vn−1,l−1. Because Fnis central these eigenspaces are submodules of Vn,l ↓

and then the sequence splits. For further details on F_n we refer to [5]. If we assume that q2(n−2l+1) _{̸= 1 we will have a splitting θ : V}

n,l ↓→ Vn−1,l

⊕

Vn−1,l−1. If

we now order the basis of link states of Vn,l such that those of Vn,l come first, θ can be chosen

such that it is represented by a matrix of the form Un,l=

(

id Vn,l

0 id

)

Lemma 2.23. If a splitting θ exists we can define a bilinear form on Vn−1,l⊕ Vn−1,l−1 by

≪ x + x′_{, y + y}′ _{≫= ⟨θ}−1_{(x + x}′_{), θ}−1_{(y + y}′_)⟩

n,l for x, y∈ Vn−1,l and x′, y′∈ Vn−1,l−1.

Because this form is symmetric and invariant in the sense of lemma 2.12, ≪ x + x′_{, y + y}′ _{≫= ⟨y, z⟩}

n−1,l+ αn,l⟨x′, y′⟩n−1,l−1,

for some αn,l∈ C.

Proof. We will prove this using the following argument. A bilinear form induces a map from its module V to its dual by v _{7→ ⟨v, ·⟩. Because the bilinear form is invariant we have that} this map is an intertwiner. When V is irreducible its dual is as well , from Schur’s lemma it then follows that the induced maps are determined apart from a scalair, so they form a one-dimensional vector space, therefore the invariant bilinear forms do as well.

In this case we have that V is the direct sum of two non-isomorphic irreducible modules, which means we have a two-dimensional space of bilinear forms. Because we can compare the form on V_n,l₋₁ with that on the direct sum we have that the factor before _{⟨y, z⟩}_n_−1,l is one.

When we view this in matrix form this becomes

Gn−1,l⊕ Gn−1,l−1 = (Un,l−1) T_G n,lUn,l−1 ⇒ Gn,l= Un,lT ( Gn−1,l 0 0 αn,lGn−1,l−1 ) Un,l.

This gives us the following recurrence relation to determine det G_n,l because det U_n,l = 1, det Gn,l= α

d_{n−1,n−2l+1}

n,l det(Gn−1,l) det(Gn−1,l−1)

Proposition 2.24. When q is not a root of unity α_n,l _{̸= 0.}

Proof. The determining of α_n,l is performed in [5], but because we are only interested in the fact that it is non-zero when q is not a root of unity we will not do this and simply use this result.

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Corollary 2.25. When q is not a root of unity det G_n,l _{̸= 0.}

Proof. Note that

det Gn,0= 1 and det G2l−1,l= 1.

The first folllows because V_n,0 is spanned by just the link state with n defects. The second follows from putting G2l,l= βG2l−1,l−1into the recurrence relation. G2l,l= βG2l−1,l−1follows

from the fact that the one defect in V2l−1,l−1 gives rise to precisely one less closed loop than

in V_2l,l.

If we combine these two with the recurrence relation and proposition 2.24 the result follows.

Corollary 2.26. When q is not a root of unity the Vn,l are irreducible Dn modules.

Theorem 2.27. When q is not a root of unity, D_nis a semisimple algebra, the set_{V_n,l _{| 0 ≤} l ≤ ⌊n/2⌋} is a complete set of non-isomorphic irreducible modules and as a left module Dn

decomposes as

Dn≃

⊕

0≤l≤⟨n/2⟩

(dim Vn,l)Vn,l.

Proof. Combining corollary 2.26 and proposition 2.17 we have that the Vn,lare non-isomorphic

irreducible modules. Then combining proposition 2.14 and Wedderburn’s theorem the result follows.

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3. Dilute Temperley-Lieb algebra

The introduction of the dilute Temperley-Lieb algebra will be similar to that of the regular Temperley-Lieb algebra. We will start by defining the dilute n-diagram algebra and determin-ing its dimension, we will give a set of generators and relations between these generators to define the dilute Temperley-Lieb algebra and show the equivalence between the two algebras. A dilute n-diagram is similar to a regular n-diagram except that we allow an even number of points to not be connected to any other points, we call these points vacancies. There must be an even number of vacancies because each line connects precisely two points and there are always an even number of points in a dilute n-diagram.

We can also construct the dilute n-diagram algebra, which we will often denote as dD_n. This is the complex span over all the possible dilute n-diagrams up to isotopy. Where the multi-plication is the same as the regular n-diagram algebra except that if a vacancy is connected to a line the result is zero. We will now show two examples for n = 3

b c b c · b c b c = b c b c b c b c = 0 , b c b c · b c b c = b c b c b c = b c b c .

We would now like to give a set of generators for this algebra, but in order to do this we shall first introduce the following useful notation. When there is a dashed line in one of the diagrams this represents a sum two diagrams one where there are vacancies on the endpoints of the dashed line and one where there is a line between the two endpoints. An example for n = 3 = + b c b c + b c b c + b c b c b c b c .

The reason this notation is useful becomes apparent when we give the identity element of the dilute n-diagram algebra. It is simply the diagram with dashed lines at every point, this is the identity because for every dilute n-diagram there is precisely one element in the sum of the identity where the result of multiplication is not zero and for that element it acts as the identity on that dilute n-diagram.

idn= · · ·

We will now determine the dimension of the dilute n-diagram algebra. Similar to the the n-diagram algebra we will give an equivalence between the dilute n-diagram algebra and (n, p)-link diagrams. For this we will first define the dilute n-p)-link diagrams. This is a horizontal

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line with n points connected the same as regular n-link diagrams except that we allow points to be vacancies. We orient these such that the defects are always facing up. An example for n = 6

b

c bc

Definition 3.1. A_n is the vector space spanned by all the dilute n-link diagrams.

Definition 3.2. H_n,k, 0 _{≤ k ≤ n is the subspace of A}_n spanned by all the dilute n-link diagrams with at most k defects.

We will call An and Hn,k the link modules. The reason for this name will become apparent

when we determine the modules of the dilute n-diagram algebra. Any linear combination of dilute link diagrams will be called a dilute link state.

We can now define a bijective map ψ : dDn → H2n,0 as follows. For any dilute n-link

diagram we connect the two horizontal lines on the left and reorient the rest of the diagram so that this line forms the bottom of the diagram. Then we extend this linearly and obtain the map. For example

b c b c → b c b c → b c b c .

This map is bijective because this process can easily be reversed by separating the horizontal line in the middle for a dilute 2n-link diagram with no defects and reorienting the leftmost horizontal line to obtain a dilute n-diagram.

We can then transform such a dilute 2n-dilute link diagram into a regular link diagram with no defects by simply removing all the vacancies. For example

b

c

b

c

→ .

Now we can easily determine the dimension of the dilute n-diagram algebra using the dimension of the regular n-diagram algebra.

Theorem 3.3. The dimension of the dilute n-diagram algebra is

dim dDn= n ∑ i=0 ( 2n 2i ) dim TLi= n ∑ i=0 1 i + 1 ( 2i i )( 2n 2i ) .

Proof. By taking 2m ≤ 2n and choosing 2m points we can take the subset of the dilute n-diagrams where the diagrams have vacancies at precisely those points. Combining the previous two procedures we obtain the link basis in V_2(n_−m),0. This does not depend on the choice of points. Combining this with the factor (2n

2i

)

obtained from the possible choices we have that the dilute n-diagrams with 2m vacancies have dimension (2n

2m

)

dim V_2(n_−m),0. But combining lemma 2.3 and 2.4 in chapter 2 we have that dimV_2(n_−m),0 =dimTLn−m which

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3.1. Dilute Temperley-Lieb algebra

The dilute Temperley-Lieb algebra has been the subject of much study, but in spite of this there does not exist a formal definition. A set of generators and relations has been proposed [8] during a study of more general braid algebras but it was given without proof. In this chapter we will give a generating set of the dilute n-diagram algebra, give a set of relations based on the ones given in [8] to define the dilute Temperley-Lieb algebra and prove the equivalence between the dilute n-diagram algebra and the dilute Temperley-Lieb algebra.

We will now give a set of generators of the dilute diagram algebra, but unlike the regular n-diagram algebra where the algebra was generated by the identity and one type of generators, we will need multiple different types of generators. There are several generating sets but we will be using {ai, ati, bi, bti, ej, xj, i ∈ [1, n − 1], j ∈ [1, n]} where we make the following

identifications. ei = · · · · · · 1 i− 1 i i + 1 n , xi= b c b c · · · · · · 1 i− 1 i i + 1 n , ai = · · · · · · b c b c 1 i i + 1 n , at_i = _{· · ·} _{· · ·} b c b c 1 i i + 1 n , bi = · · · · · · b c bc 1 i i + 1 n , bt_i = _{· · ·} _{· · ·} b c bc 1 i i + 1 n .

Note that it is possible to replace all the x_i with id_n to create a smaller set of generators because∀i, xi = idn− ei. But we will find that it is convenient to use the xi’s.

Theorem 3.4. The set {ai, ati, bi, bti, ej, xj, i∈ [1, n − 1], j ∈ [1, n]} is a generating set of the

dilute n-diagram algebra.

Proof. We will prove this by first giving for every two points a diagram that is a product of generators that contains a line between those two points. Given two points in a dilute n-diagram we will label them by their horizontal position pi and pk where we assume without

loss of generality that i≥ k. Now there are five options for which of the horizontal lines each point is on.

1. Both points are on the bottom horizontal line. Then at

i−1· · · atk+1btk has a line that

connects the two points.

at_i₋₁· · · at_k+1bt_k= _{· · ·} _{· · ·} · · · · · · b c b c b c b c b c b c 1 k i n

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2. The two points are on different horizontal lines, i_{̸= k. and p}iis on the bottom horizontal

line. Then at

i−1· · · atk has a line that connects the two points.

at_i₋₁· · · at_k= _{· · ·} _{· · ·} · · · · · · b c b c b c bc 1 k i n

3. The two points are on different horizontal lines and i = k. Then simply e_i will suffice. 4. The two points are on different horizontal lines, i _{̸= k and p}_i is on the top horizontal

line. Then ak· · · ai−1 has a line that connects the two points.

ak· · · ai−1 = · · · · · · · · · · · · b c b c b c bc 1 k i n

5. Both points are on the top horizontal line. Then b_kak+1· · · ai−1 has a line that connects

the two points.

bkak+1· · · ai−1= · · · · · · · · · · · · b c b c b c bc bc b c 1 k i n

Note that the diagrams constructed this way, which we will now refer to as standard strings, have dashed lines on all horizontal positions outside of the interval [k, i]. Where we note that these dashed lines act as the identity on their respective horizontal positions which ensures that these standard strings only act non-trivially on points in the interval [k, i]. Now given a dilute n-dilute diagram we can replace each of the lines with one of the standard strings, but we cannot simply concatenate these because the dilute n-diagram algebra is not commutative. An example for n = 3. b c b c b c b c = b c b c b c b c ̸= b c b c = bc bc bc .

Now we will give an order of compositions which will ensure we end up with the desired diagram. We will do this as follows, we first give the five types of lines the order 1 < 2 < 3 < 4 < 5. Let x and y be two lines of the same type. If x and y are type 1 or 4 we say x < y if y has the point furthest to the left. If x and y are type 2, 3 or 5 we say x > y if y has the point furthest to the left. This totally orders all the lines for any dilute n-diagrams. We will write the standard strings corresponding to the lines such that the standard strings, for

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which the lines are higher in the order, are to the right. We will say that a standard string is above or below another standard string if it is higher or lower respectively in the order, this terminology is used because when viewing the diagrams standard strings higher in the order are above ones lower. An example of a decomposition for n = 6,

b c bc b c b c a b c d

We find that d < c < b < a. Where for each of the lines we have: a = b3a4, b = a2a3a4a5, c = at5at4bt3 and d = bt4.

When we combine these things we obtain bt

4at5at4bt3a2a3a4a5b4a4 as a way of writing the

di-agram as a monomial in the generators. If we do so we obtain the following didi-agram as a

result: bc bc

.

Note that we still have a dashed line where there should be vacancies. This brings us to the final part of the construction. For every vacancy on the top row we add the corresponding xi

on the right and for the vacancies on the bottom row we place the corresponding x_i to the left, note that we only need to add the xi when we have a dashed line in the construction,

but for which vacancies this occurs is not always obvious so we add xi to all the vacancies to

ensure that we always get the desired result.

All that remains is to prove that this construction works as intended. Recall that every standard string acts like the identity on all point on horizontal positions not on or between its endpoints. We will say that two lines overlap if a horizontal position exist where both of the corresponding standard strings do not act like the identity. Since we only construct lines when they exist in the diagram it suffices to prove that for every standard string in the construction, the line in it is only ever connected to dashed lines when we construct the diagram. Which means that for endpoints on the top horizontal row we need to check that the standard strings corresponding to lines above it are the identity there and for endpoints on the bottom row we need to check that all standard strings corresponding to lines below it are the identity on that horizontal position.

Standard strings corresponding to lines of type 3 only act on one horizontal position and no overlap can exist with other lines because otherwise we would have two crossing lines. Therefore the line in the standard string can only be connected to dashed lines.

For lines of type 1 we note that its endpoints are only on the bottom horizontal row. Therefore we only need to check that it works for lines of type 1 below it. But the standard string corresponding to such a line can only act as the identity on the endpoints of the first line or else they would overlap in the original diagram.

The same argument holds for lines of type 5 except that all the direction are reversed. For lines of type 2 we note that for the endpoint on the top row only standard strings corresponding to lines of type 1 or 2 can possibly not be the identity there, but such lines are below it. For the endpoint on the bottom row the only types of lines for which the

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corresponding standard strings can be different from the identity on that point are of type 2 or 5. But once again such lines must be above it.

A similar argument holds for lines of type 4.

The final step in the construction where we add the vacancies simply ensures that we are not left with any dashed lines in the final construction. The position in the construction clearly ensures that they do not connect to any line.

We now have a construction that for any possible dilute n-diagram gives a decomposition in the given generators.

We now want to use these generators to describe the dilute n-diagram algebra in terms of its characterizing relations between these algebraic generators.

Definition 3.5. For a β_{∈ C, the dilute Temperley-Lieb algebra denoted dTL}_n(β) or dTLn

is the algebra generated by {ai, ati, bi, bti, ej, xj | i ∈ [1, n − 1], j ∈ [1, n]}, satisfying,

(i) Commutation relations

OiOj = OjOi for |i − j| > 1,

Oiej = ejOi, Oixj = xjOi for j̸= i, i + 1,

xixj = xjxi, eixj = xjei, eiej = ejei.

Where Oi ∈ {ai, ati, bi, bti}.

(ii) Projection relations

ei+ xi = idn, e2i = ei, x2i = xi, xiei= 0.

(iii) Occupation relations

eiai= ai, ei+1ai = 0, aiei = 0, aiei+1= ai,

eiati = 0, ei+1ati = ati, atiei= ati, atiei+1= 0,

eibi= 0, ei+1bi = 0, biei= bi, biei+1= bi,

eibti = bti, ei+1bti = bti, btiei = 0, btiei+1= 0.

The vacancy relations involving x_i follow from the occupation relations by x_i+ ei = idn.

(iv) Remaining relations

aiati= eixi+1, atiai = xiei+1 bibti = βxixi+1, biai+1ai= bi+1xi,

at_iat_i+1bt_i= xibti+1, aiai+1= bi+1bti, ati+1ati = bibti+1.

Proposition 3.6. There exists a unique unit-preserving surjective algebra morphism ϕ :

dTLn→ dDn such that ϕ maps the generators of dTLnto the corresponding dilute n-diagrams

of theorem 3.4.

Proof. To prove this we need to check that all the relations in dTLn also hold in the dilute

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diagrams on both sides of each relation. For example biai+1ai = i · · · · · · · · · · · · · · · · · · b c bc b c b c b c = i · · · · · · b c b c bc bc = i · · · · · · · · · · · · b c b c b c b c = bi+1xi.

Then ϕ is surjective because of theorem 3.4.

From now on we will call the generators in dTLn letters and the products of these letters

words. We are now going to prove that the dilute Temperley-Lieb algebra and the dilute n-diagram algebra are isomorphic. We will do this by showing that every word in the dilute Temperley-Lieb algebra is equal to a linear combination of elements in the dilute n-diagram algebra in the form we have previously seen in the proof of theorem 3.4. Before we prove this we shall first note a few things about the generators of dTLn.

Lemma 3.7. The following relations hold in dTL_n.

aibti+1= ati+1bit and biai+1= bi+1ati.

Proof.

aibti+1= (aixi)bi+1t = ai(atiati+1bti+1) = eixi+1atibti = ai+1eibti = ati+1bti,

biai+1= (biei)(ai+1xi+1) = bi(ai+1eixi+1) = biai+1aiati = (bi+1xi)ati = bi+1ati.

Definition 3.8. We define a word u in dTLn to be irreducible if it is not equal to a lineair

combination of words with fewer letters.

We note that the only relation that the only relation that allows a word to be equal to a lineair combination of at least two distinct words is e_i+ xi = idn. From this it follows that

every word in dTLn is equal to some irreducible word and using relations that do not change

the number of letters will preserve the irreducibility.

We now want to rewrite the irreducible words of dTL_n to a form more suited for proving the equivalence between dTLn and the dilute n-diagram algebra. For this we are interested

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We first note that both xi and ei occurring as letters in an irreducible word must commute

with all other letters in the irreducible word. This is because if e_i or x_i does not commute with another letter we have a relation that reduces the word further.

Proposition 3.9. The following pairs of letters are all the possible pairs that do not commute

and are non-zero,

at_i+1at_i at_iat_i₋₁ ai−1ai aiai+1 btibi bi ai−1bti btibi

bi+1ati atibti−1 bti+1ai aibti+1 bti±1bi bibti±1 bi±1bti btibi±1

bt_i−1at_i at_ibi+1 bi−1ai aibi−1 ai+1bi bibt_i bibt_i bt_iai−1

at_i₋₁at_i at_iat_i+1 ai+1ai aiai−1 ati−1bi bibti±1 bi±1bti btiati+1

aiati atiai

where we note that some pairs occur multiple times in the diagram, this is so that it becomes somewhat easier to read.

Proof. Because of the commutation relation we only need to compare each letter at the i-th position with all the other ones on i and i± 1 on both the left and the right and checking whether they are equal to zero by using the occupation and vacancy relation. We will only show for one that it is equal to zero,

aibi= (aiei+1)bi= ai(ei+1bi) = 0.

Definition 3.10. A string is a sequence of letters in a word in dTLn consisting of ai, at_i and

at most one b_i or bt

i, that is not equal to zero and where every two consecutive letters do not

commute. We will say that a letter commutes with a string if it commutes with every letter in the string.

Note that the standard strings defined in the proof of theorem 3.4 are strings in this definition.

Theorem 3.11. The dilute Temperley-Lieb algebra is isomorphic to the dilute n-diagram

algebra.

Proof. We already have the surjective map ϕ from proposition 3.6, now it suffices to prove that it is injective. We will prove this by showing that every word u in dTLn can be written

as a lineair combination of words in the same form as the ones constructed in theorem 3.4. To show this we will inductively construct the standard strings in the proof of theorem 3.4, we start by changing a word u in dTLn so that all the bi are part of these standard strings.

We then repeat this for the bt

i and finally reorder the remaining ai and ati not in standard

strings so that we obtain the construction in the proof of theorem 3.4, during this process we will refer to strings as completed when we no longer want to alter them, in other words when they precisely correspond to the desired standard string in the construction in the proof of theorem 3.4. Note that during the process all the strings we construct will be standard strings, but we will simply refer to them as strings. Also note that we will only discuss the xi

and ei at the end of the proof because these commute with all other letters in an irreducible

word.

Given an irreducible word u in dTLn, we will inductively choose bi and move them to the

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highest index i in u not in a completed string, we then choose the bi in this set that is the

furthest to the left in u. When we now say choose a b_i we mean selecting a b_i in this manner. We will inductively create a string using the chosen bi, we assume that the string has the

form biai+1· · · ak−1ak, we will then move this string to the right by commuting the string

with each letter in u with which it commutes. We continue this until we encounter a letter with which the string does not commute, there are nine options for the letter that do not result in zero.

(i) The string consists of just b_i and it encounters bt

i, in this case we replace the string with

βxixi+1 and choose a new bi.

(ii) a_k+1, in this case we simply add this letter to the string and continue with the process. (iii) at

i−1, we first note that this commutes with every ai in the string so we can write

biai+1· · · ak−1akati−1= (biati−1)ai+1· · · ak−1ak= bi−1aiai+1· · · ak−1ak

and we choose a new bi.

(iv) bt

i−1, we first note that this commutes with every ai in the string so we can write

biai+1· · · ak−1akbti−1 = bibti−1ai+1· · · ak−1ak = ai−1aiai+1· · · ak−1ak

and we choose a new bi if it exists otherwise we end the process.

(v) bt

k+1, we can rewrite the string as follows

biai+1· · · ak−1akbtk+1= biai+1· · · ak−1atk+1btk = biai+1· · · atk+1ak−1btk

= biai+1· · · atk+1atkbtk−1= biatk+1akt· · · ati+2bti+1= atk+1atk· · · ati+2bibti+1

= at_k+1at_k· · · at_i+2at_i+1at_i

and we choose a new bi if it exists otherwise we end the process.

(vi) at

k where i̸= k, we can rewrite the string as follows

biai+1· · · ak−1akatk= biai+1· · · ak−1ekxk+1 = biai+1· · · ak−1xk+1

and we continue the process with the new string biai+1· · · ak−1 .

(vii) ai where i̸= k, we can rewrite the string as follows

biai+1· · · ak−1akai= biai+1aiai+2· · · ak−1ak= bi+1xiai+2· · · ak−1ak= xibi+1ai+2· · · ak−1ak

and we continue the process with the string b_i+1ai+2· · · ak−1ak.

(viii) bl with l ∈ [i + 1, k − 2], we now say that the string biai+1· · · ak−1ak is completed and

we choose a new b_i if it exists otherwise we end the process.

(ix) There are no further elements to the right of the string, we then say that this string is completed and choose a new b_i if it exists otherwise we end the process.

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After completing this process the result is that u = wv where w is a word that does not contain any b_i’s and v is of the following form

v = (bi1ai1+1· · · ak1)(bi2ai2+1· · · ak2)· · · (bimaim+1· · · akm).

where i1 < i2 <· · · < im.

A similar argument can be made for bt

iexcept that the strings are of the form atkatk−1· · · ai+1bi,

we move the strings to the left and when choosing a bt

i from the set of all bti with the highest

index i not in a completed string, we choose the one that is the furthest to the right in u. This works because there is a symmetry in all the relations involving b_i or bt

i which results in

u = (at_j₁at_j₁₋₁· · · at_o₁₊₁bt_o₁)· · · (at_j_pat_j_p₋₁· · · at_o_p₊₁bt_o_p)w(bi1ai1+1· · · ak1)· · · (bimaim+1· · · akm).

Where i1 < i2 <· · · < im, o1 > o2 >· · · > op and w only contains ai, ati, xi and ei.

Next we will reorder w. For this we first note that for all i, j aiatj = 0 or aiatj = atjai. This

means that in w for all i, a_i commutes with everything but a_i₋₁ and a_i+1. When choosing ai we will view the set of ai with the lowest index i not in a completed string in u, we then

choose the ai in this set that is the furthest to the left. We will now inductively create a string

aiai+1· · · ak by moving the string to the right until it does not commute with an element in

w. For such an element there are three options (i) a_i₋₁, this is not possible by assumption.

(ii) a_j with j _{∈ [i, k − 1], this is not possible because we would have}

aiai+1· · · akaj = aiai+1· · · ak(ejaj) = aiai+1· · · ajejaj+1· · · akaj = 0

and we know that w is also irreducible.

(iii) ak+1, we continue with the string aiai+1· · · akak+1.

This continues until we reach the end of w or a completed string, we then say a_iai+1· · · ak

is a completed string and choose a new ai. This process ends when all the ai are part of

completed strings. After this process ends we have

w = z(ai1ai1+1· · · ak1)· · · (aimaim+1· · · akm).

Where z consists solely of xi, ei and ati and i1 > i2 >· · · > im. We can again use a similar

argument for the at

i except that we move the strings to the left. We then obtain

w = (at_i₁at_i₁₋₁· · · at_k₁)· · · (at_i_mat_i_m₋₁· · · at_k_m)z(aim+1aim+1+1· · · akm+1)· · · (aipaip+1· · · akp).

where z only consists of e_i and x_i, k₁ < k2 <· · · < km and im+1 > im+2 >· · · > ip. We now

have a reordering of the irreducible word u in strings. We note that apart from the standard strings of type 3, namely the ei, the order these string are in precisely corresponds to the

order of the corresponding standard strings in the proof of theorem 3.4. All that remains is to consider the remaining xi and ei. We could simply reorder the ei to be in the desired

order, but one problems remains namely that the image of any given word u_{∈ dTL}n might

contain dashed lines. For this we will now introduce a linear spanning set of dTL_n, which we will denote Xn. We say that a word u∈ Xn if for all i one of four cases applies,

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To prove that Xn is a linear spanning set first note that idn = xi+ ei for all i. But then

for any word u in dTL_n

u = (x1+ e1)· · · (xn+ en)u(x1+ e1)· · · (xn+ en).

We can then write this product into a sum of elements of Xn thus proving that it forms a

linear spanning set. The reason we are interested in X_n is because the elements precisely correspond to the dilute n-diagrams with no dashed lines under the map ϕ.

We know that every word u in dTLn is equal to an irreducible word in strings such that

these strings have the same order as the standard strings in the proof of theorem 3.4. We know that all the ei and xi in u commute with every other letter. This means we can also

reorder the ei so that they have the same order as the corresponding standard strings and we

can place the x_i on the outside of the word u and add all the x_i to the ends of u that act as the identity on it and remove all the xi’s that can be removed solely with the commutation

relations and x2

i = xi, this last step with the xi precisely corresponds to the last step of the

proof of theorem 3.4 where we add the vacancies to the diagram.

We now have a linear spanning set for dTLn for which each element precisely corresponds

to the construction of a unique dilute n-diagram. This proves the equivalence between dTLn

and the dilute n-diagram algebra.

3.2. Standard modules of the dilute n-diagram algebra

Earlier in this chapter we have introduced the dilute n-link diagrams and there exists a natural left action from the dilute n-diagrams onto these dilute link diagrams by concatenating the two and removing the original points of the link diagram. Then replacing each line not connected to any of the remaining points with a factor β if it is closed, a factor 0 if it is connected to a vacancy and simply removing it if neither is the case. The remaining lines will all be connected to at least one of the points. If such a line is connected to a vacancy it is replaced with a zero, if it is connected to a defect it becomes a defect otherwise it remains a link. Finally any vacancies not on any of the remaining points are removed. A few examples for n = 4

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b c b c · b c = b c b c = b c , b c b c b c bc · b c = bc b c b c bc = β b c bc bc , · b c = bc = 0 = 0.

If we extend the action linearly we find that An is a left module of the dilute n-diagram

algebra. Note that this action can only ever decrease the number of defects. This means that Hn,k is a submodule of An. We have the natural inclusions

Hn,0⊂ Hn,1⊂ · · · ⊂ Hn,n= An.

Any linear combination of dilute link diagrams will be called a dilute link state.

Definition 3.12. The standard modules S_n,k are the quotients of two consecutive link mod-ules namely

Sn,k≃ Hn,k/Hn,k−1 for 1≤ k ≤ n and Sn,0= Hn,0.

Note that the elements of each S_n,k are generated by equivalence classes with one unigue dilute n-link diagram with k defects. For that reason we will simply denote the equivalence classes by that unique dilute link diagram.

We will now let Y_n,k be the set of all dilute n-link diagrams with k defects. This set can be viewed as a basis for Sn,k if we view the each diagram as representing the corresponding

equivalence class.

Definition 3.13. We define the injective bilinear map|·, ·|n,k : Sn,k×Sn,k −→dDnas follows.

For u, v _{∈ Y}_n,k we let _{|u¯v| be the n-diagram formed by viewing the dilute link diagram u} as the bottom row in a dilute n-diagram, the reflected state of the dilute link diagram v as the top row and connecting each defect with one on the opposite row. We then extend this linearly to S_n,k_{× S}_n,k to obtain the map.

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Note that connecting the defects is unique because there is only one way so that none of the lines cross. Now an example for n = 4 and k = 2,

, b c bc −→ b c bc −→ b c bc .

It will now be useful to view the subset X_n,k of Y_n,k consisting of all the dilute link diagrams with k defects and n− k vacancies.

Lemma 3.14. Let z ∈ Xn,k and u, v be any dilute link diagrams in Sn,k then the following

properties hold:

(i) _{|u¯v|z = 0 when z ̸= v.} (ii) _{|u¯v|z = u when z = v.}

Proof. Suppose v _{̸= z, in the concatenation of |u¯v| with z we identify the points of ¯v with} those of z. Since z has precisely n− k vacancies the result of this concatenation is zero unless every vacancy of z is identified with one of v, but we know that v has k defects which in this case must all be connected to a defect of z, but then z = v. Therefore we can conclude that |u¯v|z = 0 when z ̸= v.

Suppose v = z, we first note that each of the defects of u is still a defect in _{|u¯v|z because} it is connected to two defects. Furthermore it is the case that every vacancy on ¯v and z is identified with another vacancy. Which means that |u¯v|z = u.

Proposition 3.15. For the standard modules Sn,k the following properties hold,

(i) Sn,k is cyclic and every non-zero element of spanXn,k is a generator.

(ii) S_n,k is indecomposable. (iii) Sn,k ≃ Sn,j ⇐⇒ i = k.

Proof. (i) From lemma 3.14(ii) it follows that any element in Xn,k is a generator for Sn,k.

Then for any non-zero element v in spanXn,k we know that v is the linear combination

of elements in X_n,k. Let w _{∈ X}_n,k be a dilute link diagram such that w is a non-zero component of v. Then for all u∈ Sn,k,|u ¯w|v = cu for some c ∈ C×.

(ii) Suppose that S_n,k _{≃ A ⊕ B for some non-trivial submodules A and B. Let z ∈ X}_n,k, we know that z generates S_n,k which means it cannot be an element of either A or B, therefore z = a + b for some non-zero link states a ∈ A and b ∈ B. We know that z = |z¯z|z = |z¯z|(a + b), then |z¯z|a ∈ A and |z¯z|b ∈ B. Suppose |z¯z|a is zero, then |z¯z|b = z ∈ B and B = Sn,k. Suppose |z¯z|a is non-zero then it must have a non-zero

component along z in the basis of dilute link diagrams, then from lemma 3.14 it follows that _{|z¯z|a = ca for some c ∈ C}×. Which implies that A = S_n,k. From this it follows that S_n,k is indecomposable.

(iii) The statement ”⇐” is clearly trivial. Suppose Sn,k ≃ Sn,j where we let k≤ j. We know

there exists a module isomorphism θ : S_n,k_{→ S}n,j. Let z∈ Xn,k and u∈ Sn,k such that

|u¯z| is non-zero, so 0 ̸= θ(|u¯z|z) = |u¯z|θ(z). This means that θ(z) has n − k vacancies, but k≤ j so the remaining points must be occupied by defects, therefore j = k.

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Proposition 3.16. The dimension of S_n,k is given by, dim Sn,k= ⌊n−k 2 ⌋ ∑ l=0 ( n k + 2l ) dim Vk+2l,l,

where we recall that dim V_n,l =(n_l)−(_l₋₁n ).

Proof. We know that the set of dilute n-link diagrams with k defects is a basis of Sn,k when

we view them as equivalence classes. Recall that there exists a correspondence between these dilute link diagrams and the regular link diagrams by removing the vacancies. Also note that the number of vacancies in such dilute link diagrams can only vary by even numbers because only by adding or removing links can this number be altered.

Next order these dilute link diagrams by the number of links, then for a set of n-link diagrams with k defects and l links the size is the same as the dimension of Vn,l apart from

a factor of ( n k+2l

)

that comes from choosing the places of the vacancies. If we then sum over all the possible number of links we obtain the dimension of Sn,k.

Proposition 3.17. The dimension of the dilute n-diagram algebra is also given by

dim dDn= n

∑

k=0

(dimSn,k)2.

Proof. Recall that for every k ∈ {0, 1, · · · , n} we have an injective bilinear map |·, ·|n,k :

Sn,k× Sn,k −→ dilute n-diagram algebra. First note for every dilute n-diagram u there are

two dilute n-link diagrams v and w such that u =_{|v ¯}w|. These dilute n-link diagrams can be obtained by identifying v with the bottom row of u and ¯w with the top row and letting all the lines that connect the two rows be defects.

Also note that every dilute n-diagram in the image of _{|·, ·|}_n,k has precisely k lines that connect the two horizontal rows, therefore each dilute n-diagram corresponds to two dilute n-link diagrams and the result follows.

3.3. Irreducibility of the standard modules

We now have a set of modules for the dilute n-diagram algebra and we want to explore whether these modules are irreducible. If this were the case proposition 3.17 and Wedder-burn’s theorem would imply that the standard modules are all the possible modules. For this we will introduce a bilinear form for the dilute n-link diagrams. We will then view the matrix representing this bilinear form and using the results for the regular n-diagram algebra we will show that in most cases the link modules are irreducible.

Definition 3.18. The bilinear form_{⟨·, ·⟩}_n,k is defined on S_n as follows. If x and y are two dilute n-link diagrams with k defects,⟨x, y⟩n,k is computed by concatening the reflected link

state of x with y. If not every defect is connected with a defect on the opposite side or a vacancy is connected to a line, _{⟨x, y⟩}_n,k = 0. Otherwise ⟨x, y⟩n,k = βm where m equals the

Dilute Temperley-Lieb algebra