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Publicly Available Published by De Gruyter August 25, 2020

Irreducible representations of Braid group 𝐡𝑛 of dimension 𝑛 + 1

  • Inna Sysoeva EMAIL logo
From the journal Journal of Group Theory

Abstract

We prove that there are no irreducible representations of Bn of dimension n+1 for nβ©Ύ10.

1 Introduction

The irreducible complex representations of the braid group Bn on n strings of dimension less than or equal to n have been completely classified about 20 years ago. Formanek [3] had classified all the irreducible repesentations of dimension less than or equal to (n-1). In [11], the author provided the classification of irreducible representations of dimension n for nβ©Ύ9, and the classification for the small values of n was completed in [10, 4].

Apart from a number of exceptions for small values of n, all irreducible representations of Bn are either one-dimensional, or have dimensions (n-2), (n-1) or n. For nβ©Ύ7, all (n-2)-dimensional representations are equivalent to a tensor product of a one-dimensional representation and an (n-2)-dimensional composition factor of a specialization of the reduced Burau representation (see [1, 7, 3]), and all (n-1)-dimensional representations are equivalent to a tensor product of a one-dimensional representation and a specialization of the reduced Burau representation. For nβ©Ύ9, all n-dimensional representations are equivalent to a tensor product of a one-dimensional representation and a specialization of the standard representation (see [12]).

Another famous representation (whose dimension is n⁒(n-1)2) is the Lawrence–Krammer–Bigelow representation (see [9]). Its irreducibility for generic values of the parameters was first proven by Zinno [13] and, as Zinno mentions in his paper, V. F. R. Jones also discovered the same result, simultaneously and independently.

However, the problem of the systematic classification of the irreducible representations, even between dimensions n and n⁒(n-1)2, is wide open. To the best of my knowledge, up to this point, no such classification has been found. No irreducible representations of dimensions between n and n⁒(n-1)2 were found for n large enough, and no results of non-existence for the irreducible representations are known. The most notable step towards such classification was published by Larsen and Rowell [8], where, among other results, they proved that there are no irreducible unitary representations of Bn between dimensions n+1 and 2⁒n-9 (inclusive) for n large enough. In particular, there are no irreducible unitary representations of dimension n+1 for n⩾16.

The main result of this paper (Theorem 6.1) is that Bn has no irreducible representations of dimension n+1 for nβ©Ύ10. We do not claim that this lower bound is sharp. As Larsen and Rowell pointed out in their paper, the actual sharp lower bound is at least 8, since B7 has an 8-dimensional unitary representation (see Jones [6]).

To achieve our result, we first prove that, for nβ©Ύ10, for any irreducible representation of Bn of dimension at least n+1 and not exceeding 2⁒n-9, the image of each group generator ρ⁒(Οƒi) must have an eigenvalue of high multiplicity; in other words, the representation is equivalent to the tensor product of a one-dimensional representation and an irreducible representation of small corank (the Reduction Theorem, Theorem 3.8). This theorem by itself is a separate tool which may be used in the classification of the irreducible representations of dimension up to 2⁒n-9.

The paper is organized as follows. In Section 2, we give the definitions and recall preliminary results. In Section 3, the Reduction Theorem is proven. In Section 4, we review the definition of the friendship graph which was introduced in [11]. We will briefly review the results from [11] related to the friendship graphs, and prove some new results that will be used to classify the irreducible representations of dimension n+1. In Section 5, we prove that there are no irreducible representations of dimension rβ©Ύn+1 and corank 3 for nβ©Ύ10, and we summarize the main results of this paper in Section 6, Theorems 6.1 and 6.2.

2 Notations and preliminary results

Let Bn be the braid group on n strings. As an abstract group, it has the following presentation:

Bn=γ€ˆΟƒ1,Οƒ2,…,Οƒn-1βˆ£Οƒi⁒σj=Οƒj⁒σi⁒for⁒|i-j|β©Ύ2,ΟƒiΟƒi+1Οƒi=Οƒi+1ΟƒiΟƒi+1forβ€…1β©½iβ©½n-2〉,

where Οƒ1,Οƒ2,…,Οƒn-1 are called the standard generators. Since B2β‰…β„€, we will assume throughout the paper that nβ©Ύ3, unless specified otherwise.

In the following lemma, we are going to recall the results originally due to W.-L. Chow [2] and F. A. Garside [5].

Lemma 2.1.

Let Bn be a braid group on n strings, nβ©Ύ3. Let Ο„=Οƒ1⁒σ2⁒…⁒σn-1, and let Οƒ0=τ⁒σn-1⁒τ-1. Let

Ξ”=(Οƒn-1⁒σn-2⁒…⁒σ1)⁒(Οƒn-1⁒σn-2⁒…⁒σ2)⁒…⁒(Οƒn-1⁒σn-2)⁒(Οƒn-1)

(positive half-twist). Then

  1. Οƒi+1=τ⁒σi⁒τ-1 for 1β©½iβ©½n-2 (see [2, equation (6)]);

  2. the center of Bn, Z⁒(Bn)=γ€ˆΟ„n〉 (see [2, Theorem III]);

  3. Οƒ1=τ⁒σ0⁒τ-1;

  4. Z⁒(Bn)=γ€ˆΞ”2〉 (see [5, Theorem 7]);

  5. Οƒn-i=Δ⁒σi⁒Δ-1, i=1,2,…,n-1 (see [5, Lemma 2 (ii)]).

Proof of (c).

It easily follows from (a), (b) and the definition of Οƒ0. ∎

In view of Lemma 2.1, we can add a redundant generator Οƒ0 to the set of standard generators of Bn to get the following presentation:

Bn=γ€ˆΟƒ0,Οƒ1,Οƒ2,…,Οƒn-1βˆ£Οƒi⁒σj=Οƒj⁒σi⁒for⁒βˆ₯i-jβˆ₯β©Ύ2,Οƒi⁒σi+1⁒σi=Οƒi+1⁒σi⁒σi+1,Οƒ0=τσn-1Ο„-1〉,

where Ο„ is defined in Lemma 2.1, i,jβˆˆβ„€n and the norm of m in β„€n is defined by βˆ₯mβˆ₯=min⁑{m,n-m} (here we identify the element m of β„€n with a number m∈{0,1,…,n-1}βŠ†β„€).

Let ρ:Bnβ†’GLr⁒(β„‚) be a matrix representation of Bn of dimension r. Throughout the paper, we are going to use the following notation: ρ⁒(Οƒi)=Ci=I+Ai, iβˆˆβ„€n, ρ⁒(Ο„)=T, ρ⁒(Ξ”)=D.

By using the braid relations and Lemma 2.1, we obtain the following.

Lemma 2.2.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn. Then

  1. T⁒Ci-1⁒T-1=Ci, iβˆˆβ„€n;

  2. Ci⁒Tm=Tm⁒Ci-m, i,mβˆˆβ„€n;

  3. T⁒Ai⁒T-1=Ai+1, iβˆˆβ„€n;

  4. D⁒Ai⁒D-1=An-i, i=1,2,…,n-1;

  5. for βˆ₯i-jβˆ₯β©Ύ2, Ai⁒Aj=Aj⁒Ai, i,jβˆˆβ„€n;

  6. Ai+Ai2+Ai⁒Ai+1⁒Ai=Ai+1+Ai+12+Ai+1⁒Ai⁒Ai+1, iβˆˆβ„€n;

  7. if ρ:Bnβ†’GLr⁒(β„‚) is irreducible representation of Bn, then

    ρ⁒(Ο„n)=ρ⁒(Ξ”2)=Tn=D2=α⁒I for someβ’Ξ±βˆˆβ„‚*.

Proof of (g).

It follows from the fact that Ο„n=Ξ”2 (which represents the positive full twist) is a central element in Bn, and ρ is irreducible. ∎

Definition 2.3.

Let ρ:Bnβ†’GLr⁒(β„‚) be a matrix representation of Bn with

ρ⁒(Οƒi)=I+Ai,iβˆˆβ„€n.

Define the corank of a representation ρ by corank⁑(ρ)=rk⁑(A1).

By Lemma 2.2 (c), we have corank⁑(ρ)=rk⁑(Ai) for all iβˆˆβ„€n.

If ρ is a one-dimensional representation of Bn, then it is constant, and we will denote it by χ⁒(y):Bnβ†’β„‚*, where χ⁒(y)⁒(Οƒi)=y, yβˆˆβ„‚* for all iβˆˆβ„€n.

3 Reduction Theorem

In this section, we will prove the Reduction Theorem (Theorem 3.8), which will allow us to investigate the irreducible representations of Bn of dimension r⩽2⁒n-9 by dealing only with the irreducible representations of relatively small corank. Since the complete classification of the irreducible representations of Bn of dimension r for r⩽n was given in [3, 11, 10, 4], we will formulate and prove all statements in this section for the dimension r⩾n+1, even though some of them hold for smaller values of r as well. The restriction n⩾10 used in Lemma 3.1, Theorems 3.3, 3.7, 3.8, and Corollary 3.9, provides n+1⩽2⁒n-9.

Throughout this section, we will use the notation introduced in Section 2.

Lemma 3.1.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, n+1β©½rβ©½2⁒n-9 for nβ©Ύ10. Then there exists an eigenvalue Ξ» of ρ⁒(Οƒn-1) such that the largest block corresponding to Ξ» in the Jordan normal form of ρ⁒(Οƒn-1) has multiplicity dβ©½n-5.

Proof.

Consider the Jordan normal form of ρ⁒(Οƒn-1). Suppose for every eigenvalue of ρ⁒(Οƒn-1), the largest block corresponding to that eigenvalue has multiplicity dβ©Ύn-4.

If ρ⁒(Οƒn-1) has two or more distinct eigenvalues, then the dimension

r⩾(n-4)+(n-4)=2⁒n-8,

a contradiction with r⩽2⁒n-9.

Thus, ρ⁒(Οƒn-1) has exactly one eigenvalue Ξ». For this eigenvalue, the largest block must be a 1Γ—1 block. Indeed, if the largest block with the multiplicity dβ©Ύn-4 has size 2Γ—2 or larger, then the dimension rβ©Ύ2⁒(n-4)=2⁒n-8, and again, we get a contradiction with rβ©½2⁒n-9.

So ρ⁒(Οƒn-1) has exactly one eigenvalue, and each block of the Jordan normal form of ρ⁒(Οƒn-1) is a 1Γ—1 block; thus ρ⁒(Οƒn-1)=λ⁒I. Due to the conjugation in Bn, each of ρ⁒(Οƒi) has the same Jordan normal form for all i=1,2,…,n-1.

So ρ⁒(Οƒi)=λ⁒I for all i=1,2,…,n-1, a contradiction with irreducibility of ρ. ∎

To prove the next theorem, we recall the lemma from [11], based on the results of Formanek [3], that Bn-2 has no irreducible representations of dimensions between 2 and n-5 (inclusive) for n large enough (see [11] for details).

Lemma 3.2 ([11, Lemma 6.4]).

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn of dimension r, where nβ©Ύ6. Suppose that Ξ» is an eigenvalue of ρ⁒(Οƒn-1), and the largest Jordan block corresponding to Ξ» has multiplicity d.

If dβ©½n-5, then there exists a one-dimensional subspace of Cr, invariant under Bn-2Γ—γ€ˆΟƒn-1〉=γ€ˆΟƒ1,…,Οƒn-3γ€‰Γ—γ€ˆΟƒn-1〉.

Theorem 3.3.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, n+1β©½rβ©½2⁒n-9, nβ©Ύ10. Then there exists a one-dimensional subspace of Cr invariant under Bn-2Γ—γ€ˆΟƒn-1〉.

Proof.

By Lemma 3.1, there exists an eigenvalue Ξ» of ρ⁒(Οƒn-1) such that the largest Jordan block of ρ⁒(Οƒn-1) corresponding to Ξ» has multiplicity dβ©½n-5. Then, by Lemma 3.2, β„‚r has a one-dimensional subspace invariant under Bn-2Γ—γ€ˆΟƒn-1〉. ∎

Lemma 3.4.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, where nβ©Ύ3. Let v∈Cr be a non-zero vector such that L=span⁑{v} is invariant under Bn-2Γ—γ€ˆΟƒn-1〉. Then the vectors v,T⁒v,T2⁒v,…,Tn-3⁒v are linearly independent.

Proof.

Since

ρ|Bn-2Γ—γ€ˆΟƒn-1〉:Lβ†’L

is a one-dimensional representation of Bn-2Γ—γ€ˆΟƒn-1〉, then

ρ⁒(Οƒ1)⁒v=ρ⁒(Οƒ2)⁒v=…=ρ⁒(Οƒn-3)⁒v=y⁒v and ρ⁒(Οƒn-1)⁒v=x⁒v

for some x,yβˆˆβ„‚*, or, in our notation,

Ci⁒v=y⁒v,i=1,…,n-3,and Cn-1⁒v=x⁒v.

Then, by Lemma 2.2 (b), we have

Cn-2⁒(Tm⁒v)=Tm⁒(Cn-2-m)⁒v=Tm⁒(y⁒v)=y⁒(Tm⁒v)

for m=1,2,…,n-3.

Suppose that the vectors v,T⁒v,T2⁒v,…,Tn-3⁒v are linearly dependent. Let a0⁒(v)+a1⁒(T⁒v)+…+an-3⁒(Tn-3⁒v)=0 be a non-trivial linear combination of non-zero vectors, and let i be the smallest index such that the coefficient aiβ‰ 0. By left-multiplying the above linear combination by T-i, we get another non-trivial linear combination with a non-zero coefficient for the vector v. Thus,

v∈span⁑{T⁒v,T2⁒v,…,Tn-3⁒v}.

Let

v=βˆ‘i=1n-3bi⁒(Ti⁒v).

Then

Cn-2⁒v=βˆ‘i=1n-3bi⁒Cn-2⁒(Ti⁒v)=βˆ‘i=1n-3bi⁒y⁒(Ti⁒v)=yβ’βˆ‘i=1n-3bi⁒(Ti⁒v)=y⁒v∈L,

so Ci⁒v∈span⁑{v}=L for every i=1,2,…,n-1. Thus, the one-dimensional subspace L is invariant under Bn, which contradicts the irreducibility of ρ. Therefore, the vectors v,T⁒v,T2⁒v,…,Tn-3⁒v are linearly independent. ∎

Corollary 3.5.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, nβ©Ύ3. Suppose v∈Cr,vβ‰ 0, is a vector such that L=span⁑{v} is invariant under Bn-2Γ—γ€ˆΟƒn-1〉. Then, for any fixed i∈Zn, the n-2 consecutive vectors Ti⁒v,Ti+1⁒v,…,Ti+n-3⁒v (all powers of T are taken modulo n) are linearly independent.

Proof.

By applying left multiplication by T-i and Lemma 2.2 (g), the statement immediately follows from Lemma 3.4. ∎

Remark.

From the proof of Lemma 3.4, we have n-3 linearly independent vectors T⁒v,T2⁒v,…,Tn-3⁒v in Ker⁑(Cn-2-y⁒I). That means that

dim⁑(Ker⁑(Cn-2-y⁒I))⩾n-3

or, equivalently,

dim⁑(Im⁑(Ci-y⁒I))=dim⁑(Im⁑(Cn-2-y⁒I))⩽r-(n-3)=r-n+3.

However, as shown in Theorem 3.6, a stronger statement is actually true.

Theorem 3.6.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, rβ©Ύn+1, nβ©Ύ5. Suppose that L=span⁑{v} is a one-dimensional subspace of Cr, invariant under Bn-2Γ—γ€ˆΟƒn-1〉=γ€ˆΟƒ1,…,Οƒn-3γ€‰Γ—γ€ˆΟƒn-1〉. Then there exists y∈C* such that dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2.

Proof.

(1) As in Lemma 3.4, consider the values x,yβˆˆβ„‚* such that

C1⁒v=C2⁒v=…=Cn-3⁒v=y⁒v and Cn-1⁒v=x⁒v.

We are going to show that, for this value of y, dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2. Suppose not. Then

dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))=dim⁑(Im⁑(C1-y⁒I))β©Ύr-n+3,

and

dim⁑(Ker⁑(C1-y⁒I))=r-dim⁑(Im⁑(C1-y⁒I))⩽r-(r-n+3)=n-3.

In addition, since all Ci are conjugated by T,

dim⁑(Ker⁑(Ci-y⁒I))=dim⁑(Ker⁑(C1-y⁒I))β©½n-3 for⁒iβˆˆβ„€n.

(2) Let Si=Ker⁑(Ci-y⁒I) for iβˆˆβ„€n.

Consider S1=Ker⁑(C1-y⁒I). By Lemma 2.2 (b),

C1⁒(Tm⁒v)=Tm⁒(C1-m)⁒v=Tm⁒(y⁒v)=y⁒(Tm⁒v)

for m=4,5,…,n-1,0 since Ci⁒v=y⁒v for i=1,…,n-3; thus

span⁑{T4⁒v,T5⁒v,…,Tn-1⁒v,v}βŠ†S1.

By Corollary 3.5, the n-3 vectors T4⁒v,T5⁒v,…,Tn-1⁒v,v are linearly independent, being a subset of the linearly independent vectors T4⁒v,T5⁒v,…,Tn-1⁒v,v,T⁒v. So n-3=dim⁑(span⁑{T4⁒v,T5⁒v,…,Tn-1⁒v,v})β©½dim⁑(S1)β©½n-3, and thus

S1=Ker⁑(C1-y⁒I)=span⁑{T4⁒v,T5⁒v,…,Tn-1⁒v,v}

with dim⁑(S1)=n-3.

Now, for any kβˆˆβ„€n, by multiplying the equation

C1⁒(Tm⁒v)=y⁒(Tm⁒v) for⁒m=4,5,…,n-1,0

by Tk and using Lemma 2.2 (b), we get

Tk⁒C1⁒(Tm⁒v)=C1+k⁒(Tm+k⁒v)=y⁒(Tk+m⁒v).

Using Corollary 3.5 and Lemma 2.2 (g), similarly to the above argument, we obtain

Si=Ker⁑(Ci-y⁒I)=span⁑{Ti+3⁒v,Ti+4⁒v,…,Tn-1⁒v,v,…,Ti-1⁒v} for⁒iβˆˆβ„€n.

(3) Consider S=span⁑{v,T⁒v,T2⁒v,…,Tn-1⁒v}. This is a non-trivial subspace, invariant under T with dim⁑Sβ©½n. In addition, SiβŠ†S for all iβˆˆβ„€n. We claim that S is invariant under Bn.

First, let us show that S is invariant under C1. For m=4,5,…,n-1,0, we have Tm⁒v∈S1, so C1⁒(Tm⁒v)=y⁒(Tm⁒v)∈S, and it remains to show that C1⁒(T⁒v), C1⁒(T2⁒v) and C1⁒(T3⁒v) are in S. Indeed,

C1⁒(T2⁒v)=T2⁒Cn-1⁒v=T2⁒(x⁒v)=x⁒T2⁒v∈S.

For nβ©Ύ5, T⁒v∈S3β‡’(C3-y⁒I)⁒(T⁒v)=0 and

0=C1⁒[(C3-y⁒I)⁒(T⁒v)]=(C3-y⁒I)⁒(C1⁒(T⁒v))⟹C1⁒(T⁒v)∈S3βŠ†S.

Similarly,

T3⁒v∈Sn-1⟹(Cn-1-y⁒I)⁒(C1⁒(T3⁒v))=0⟹C1⁒(T3⁒v)∈Sn-1βŠ†S.

Now let us show that S is invariant under Cj for every j=2,…,n-1. For mβˆˆβ„€n, we have

Cj⁒(Tm⁒v)=Tj-1⁒Cj-(j-1)⁒(Tm-(j-1)⁒v)=Tj-1⁒C1⁒(Tm-j+1⁒v)∈S

since S is invariant under C1 and S is invariant under T.

Thus, we have a proper subspace S invariant under Bn, a contradiction with the irreducibility of ρ. So, for this value of y, dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2. ∎

Theorem 3.7.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, n+1β©½rβ©½2⁒n-9, nβ©Ύ10. Then there exists y∈C* such that dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2.

Proof.

By Theorem 3.3, there exists a one-dimensional subspace of β„‚r invariant under Bn-2Γ—γ€ˆΟƒn-1〉. Then, by Theorem 3.6, there exists yβˆˆβ„‚* such that dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2. ∎

We will reformulate the above theorem in terms of coranks.

Theorem 3.8 (Reduction Theorem).

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, n+1β©½rβ©½2⁒n-9, nβ©Ύ10. Then ρ is equivalent to a tensor product of a one-dimensional representation and an irreducible representation ρ^ of dimension r and corank k, where 3β©½kβ©½r-n+2.

Proof.

By Theorem 3.7, there exists yβˆˆβ„‚* such that

dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2.

Let us consider one-dimensional representations χ⁒(y) and χ⁒(y-1). Then, for every group generator Οƒi∈Bn, ρ⁒(Οƒi)-y⁒I=(χ⁒(y)βŠ—[χ⁒(y-1)βŠ—Ο-I])⁒(Οƒi). Since ρ is irreducible, the representation ρ^=χ⁒(y-1)βŠ—Ο is also irreducible of the same dimension r, and since yβ‰ 0,

corank⁑(ρ^)=dim⁑(Im⁑[(χ⁒(y-1)βŠ—Ο-I)⁒(Οƒ1)])=dim⁑(Im⁑(ρ⁒(Οƒ1)-y⁒I))β©½r-n+2.

By [3, Theorem 10], every irreducible representation of corank k=1 has dimension rβ©½n-1. By [11, Theorem 5.5], every irreducible representation of corank k=2 has dimension r=n. Since rβ©Ύn+1, we have

3⩽corank⁑(ρ^)⩽r-n+2.∎

Corollary 3.9.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r, n+1β©½rβ©½2⁒n-9, nβ©Ύ10. Then ρ is equivalent to a tensor product of a one-dimensional representation and an irreducible representation of dimension r and corank k, where 3β©½kβ©½n-7.

Proof.

rβ©½2⁒n-9β‡’kβ©½2⁒n-9-n+2=n-7. ∎

4 Friendship graphs

In this section, we will describe the graphs associated with representations of Bn, which will help us to investigate the representations of small corank. These graphs were first introduced in [11]. We will quickly review the definitions and some of the results published in [11], as well as prove some new results which will be used in Section 5 to classify irreducible representations of Bn of corank 3.

Let ρ:Bnβ†’GLr⁒(β„‚) be a matrix representation of Bn, where

ρ⁒(Οƒi)=I+Ai,iβˆˆβ„€n.

Each graph associated with a representation is a finite simple-edged graph (there is at most one unoriented edge joining two vertices, and no edge joins a vertex to itself) such that each vertex of a graph corresponds to an image of a braid group generator. We will (slightly abusing the notation) denote each vertex corresponding to ρ⁒(Οƒi) by Ai.

Definition 4.1.

Let Ai and Aj be two distinct vertices, i,jβˆˆβ„€n.

  1. Ai and Aj, i,jβˆˆβ„€n, are neighbors if βˆ₯i-jβˆ₯=1.

  2. Ai and Aj are friends if Im⁑(Ai)∩Im⁑(Aj)β‰ {0}.

  3. Ai and Aj are true friends if

    {Ai+Ai2+Ai⁒Aj⁒Ai=Aj+Aj2+Aj⁒Ai⁒Ajβ‰ 0if⁒βˆ₯i-jβˆ₯=1,Ai⁒Aj=Aj⁒Aiβ‰ 0if⁒βˆ₯i-jβˆ₯β©Ύ2.

Definition 4.2.

The full friendship graph associated with the representation

ρ:Bnβ†’GLr⁒(β„‚) with ρ⁒(Οƒi)=I+Ai,iβˆˆβ„€n,

is the simple-edged graph with n vertices A0,A1,…,An-1 such that Ai and Aj are connected by an edge if and only if Ai and Aj are friends.

Definition 4.3.

For two distinct vertices A and B, define

f⁒(A,B)=dim⁑(Im⁑(A)∩Im⁑(B)),
t⁒f⁒(A,B)={dim⁑(Im⁑(A+A2+A⁒B⁒A))if⁒A⁒and⁒B⁒are neighbors,dim⁑(Im⁑(A⁒B))if⁒A⁒and⁒B⁒are not neighbors.

Lemma 4.4.

If A and B are two distinct vertices, then

  1. f⁒(A,B)=f⁒(B,A),

  2. t⁒f⁒(A,B)=t⁒f⁒(B,A).

Proof.

Both statements easily follow from Definition 4.3. ∎

In the above notation, we can say that the vertices A and B are connected by an edge in a full friendship graph if and only if f⁒(A,B)>0.

One can also consider the friendship graph obtained by removing the extra vertex A0 and all edges incident to it from the full friendship graph (see [11] for details). In addition, in some cases, it might be useful to consider true friendship graphs, defined in a similar manner to the friendship graphs, as well as weighted friendship and true friendship graphs, assigning the edges’ weights to be f⁒(A,B) or t⁒f⁒(A,B) respectively. We will not go into details of these explorations in this paper. Our main interest here will be the full friendship graph, and we will refer to it simply as the β€œfriendship graph” for the remainder of the paper.

The following lemma is the strengthening of [11, Lemma 3.1], reformulated in the above notation.

Lemma 4.5.

For any two distinct vertices A and B, f⁒(A,B)⩾t⁒f⁒(A,B).

Proof.

(1) For A and B not neighbors, we have A⁒B=B⁒A, so

Im⁑(A)∩Im⁑(B)βŠ‡Im⁑(A⁒B)∩Im⁑(B⁒A)=Im⁑(A⁒B);

therefore f⁒(A,B)⩾t⁒f⁒(A,B).

(2) For A and B neighbors,

A⁒(I+A+B⁒A)=A+A2+A⁒B⁒A=B+B2+B⁒A⁒B=B⁒(I+B+A⁒B),

so

Im⁑(A)∩Im⁑(B)βŠ‡Im⁑(A⁒(I+A+B⁒A))∩Im⁑(B⁒(I+B+A⁒B))=Im⁑(A+A2+A⁒B⁒A);

therefore f⁒(A,B)⩾t⁒f⁒(A,B). ∎

The next lemma is instrumental in the classification of the possible friendship graphs for a representation of Bn.

Lemma 4.6 (Enhanced Lemma about Friends).

For any three distinct vertices A, B and C such that A and B are neighbors and A and C are not neighbors,

f⁒(B,C)⩽t⁒f⁒(A,B)+t⁒f⁒(A,C).

Proof.

Consider a linear map Ο†:Im⁑(B)∩Im⁑(C)↦ℂrβŠ•β„‚r given by

φ⁒(z)=((I+B+B⁒A)⁒z,A⁒z).

(1) For all z∈Im⁑(B)∩Im⁑(C), there exists x,yβˆˆβ„‚r such that z=B⁒x=C⁒y and

φ⁒(z)=((I+B+B⁒A)⁒z,A⁒z)=((B+B2+B⁒A⁒B)⁒x,A⁒C⁒y)∈Im⁑(B+B2+B⁒A⁒B)βŠ•Im⁑(A⁒C).

Thus, Im⁑(Ο†)βŠ†Im⁑(B+B2+B⁒A⁒B)βŠ•Im⁑(A⁒C).

(2) Ker⁑(Ο†)={0} (so Ο† is injective). Indeed, if φ⁒(z)=0, then

(I+B+B⁒A)⁒z=A⁒z=0⟹(I+B)⁒z=0,

and since I+B is invertible, z=0.

(3) From (1) and (2), it follows that

dim⁑(Im⁑(B)∩Im⁑(C))⩽dim⁑(Im⁑(B+B2+B⁒A⁒B))+dim⁑(Im⁑(A⁒C)),

or, in terms of f and tf, f⁒(B,C)⩽t⁒f⁒(A,B)+t⁒f⁒(A,C). ∎

Remark.

Lemma about Friends [11, Lemma 3.3] is an easy consequence of Lemma 4.6. Indeed, if f⁒(A,B)=0 and f⁒(B,C)β‰ 0, then

0⩽t⁒f⁒(A,B)⩽f⁒(A,B)=0;

hence

0β‰ f⁒(B,C)β©½t⁒f⁒(A,B)+t⁒f⁒(A,C)=0+t⁒f⁒(A,C),

so t⁒f⁒(A,C)β‰ 0.

The following lemma will allow us to talk about the action of β„€n on the friendship graph by cyclically permuting the vertices.

Lemma 4.7.

For any i,j,k∈Zn, iβ‰ j,

  1. f⁒(Ai,Aj)=f⁒(Ai+k,Aj+k),

  2. t⁒f⁒(Ai,Aj)=t⁒f⁒(Ai+k,Aj+k).

Proof.

Both statements easily follow from the fact that Ai+k=Tk⁒Ai⁒T-k for all i,kβˆˆβ„€n. ∎

Definition 4.8.

For all k=1,2,…,n-1, define

f⁒(k)=f⁒(A1,A1+k) and t⁒f⁒(k)=t⁒f⁒(A1,A1+k).

Remark.

From the definition and Lemma 4.7, one can see that both f⁒(k) and t⁒f⁒(k) can be viewed as the functions describing the corresponding dimensions for the vertices that are k vertices apart from each other, i.e. f⁒(Ai,Aj)=f⁒(|i-j|) and t⁒f⁒(Ai,Aj)=t⁒f⁒(|i-j|) for all iβ‰ j (here |i-j| is the distance between the natural numbers i and j).

Clearly,

f⁒(k)=f⁒(n-k) and t⁒f⁒(k)=t⁒f⁒(n-k) for all⁒k=1,2,…,n-1,

or, equivalently,

f⁒(Ai,Aj)=f⁒(βˆ₯i-jβˆ₯) and t⁒f⁒(Ai,Aj)=t⁒f⁒(βˆ₯i-jβˆ₯) for all⁒iβ‰ j.

Now let us recast two theorems from [11].

Theorem 4.9 ([11, Theorem 3.4]).

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn of dimension r. Then one of the following holds.

  1. The full friendship graph is totally disconnected (no friends at all).

  2. The full friendship graph has an edge between Ai and Ai+1 for all i.

  3. The full friendship graph has an edge between Ai and Aj whenever Ai and Aj are not neighbors.

Theorem 4.10 ([11, Theorem 3.8]).

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension r with a totally disconnected associated friendship graph. Then rβ©½n-1.

We will formulate the corollary of Theorems 4.9 and 4.10 in terms of f and tf.

Corollary 4.11.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, nβ©Ύ3. Then one of the following holds.

  1. f⁒(1)⩾1.

  2. f⁒(1)=f⁒(n-1)=0 and f⁒(k)β©Ύ1 for all k=2,…,n-2.

Now we will establish some properties that we will use in the next section.

Lemma 4.12.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn. Then

t⁒f⁒(2)=t⁒f⁒(3)=…=t⁒f⁒(n-2).

Proof.

To prove this statement, we will show that t⁒f⁒(A1,Ak)=t⁒f⁒(A1,Aj) for k,j=3,4,…,n-1.

For any k=3,…,n-1,

Ak⁒A1=A1⁒Ak,

so, for all w∈Im⁑(A1), we have Ak⁒w∈Im⁑(A1), so Ak acts on Im⁑(A1). Denote A~k=Ak|Im⁑(A1). Then

t⁒f⁒(A1,Ak)=dim⁑(Im⁑(Ak⁒A1))=dim⁑(Im⁑(A~k)),k=3,4,…,n-1.

Since Ak and Aj are conjugated in Bn-2=γ€ˆΟƒ3,Οƒ4,…,Οƒn-1〉, we have

dim⁑(Im⁑(A~k))=dim⁑(Im⁑(A~j)),

so t⁒f⁒(A1,Ak)=t⁒f⁒(A1,Aj) for k,j=3,4,…,n-1. ∎

Lemma 4.13.

Suppose ρ:Bnβ†’GLr⁒(C) is a matrix representation of Bn with f⁒(1)=0. Then f⁒(2)=t⁒f⁒(2)=f⁒(3)=t⁒f⁒(3)=…=f⁒(n-2)=t⁒f⁒(n-2).

Proof.

By definition and Lemma 4.7, f⁒(2)=f⁒(n-2). By Lemma 4.5, we have t⁒f⁒(k)β©½f⁒(k) for all k=1,2,…,n-1. By applying Enhanced Lemma about Friends (Lemma 4.6) to B=Ak, A=Ak+1 and C=A1 for k=3,…,n-2 and using 0β©½t⁒f⁒(Ak,Ak+1)β©½f⁒(Ak,Ak+1)=0, we get

f⁒(A1,Ak)⩽t⁒f⁒(Ak,Ak+1)+t⁒f⁒(A1,Ak+1)⟹f⁒(k-1)⩽0+t⁒f⁒(k)=t⁒f⁒(k)⩽f⁒(k)

for all k=3,…,n-2. Thus,

f⁒(2)β©½t⁒f⁒(3)β©½f⁒(3)⩽…⩽t⁒f⁒(n-2)β©½f⁒(n-2)=f⁒(2),

which gives the required statement. ∎

Lemma 4.14.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, nβ©Ύ5, with

corank⁑(ρ)=dim⁑(Im⁑(A1))=k,

where 3β©½kβ©½r-1. Then f⁒(j)<k for j=1,2,…,n-1.

Proof.

(1) Suppose that

f⁒(j-1)=dim⁑(Im⁑(A1)∩Im⁑(Aj))=k=dim⁑(Im⁑(A1))

for some j=2,3,…,n. Then, since Im⁑(A1)∩Im⁑(Aj)βŠ†Im⁑(A1), it follows that Im⁑(A1)∩Im⁑(Aj)=Im⁑(A1). Similarly,

Im⁑(A1)∩Im⁑(Aj)βŠ†Im⁑(Aj) and Im⁑(A1)∩Im⁑(Aj)=Im⁑(Aj).

Thus, Im⁑(A1)=Im⁑(Aj).

(2) Let W=Im⁑(A1)=Im⁑(Aj). We claim that W is invariant under Bn. Indeed, W is invariant under C1 and Cj. We have two cases.

(a) jβ‰ 3. For every m=3,…,n-1, mβ‰ j, we have |m-1|β©Ύ2, and hence

Cm⁒w=Cm⁒A1⁒u=A1⁒Cm⁒u∈W for all⁒w=A1⁒u∈Im⁑(A1)=W.

And for m=2, we have |m-j|β©Ύ2 and

Cm⁒w=Cm⁒Aj⁒u=Aj⁒Cm⁒u∈W for all⁒w=Aj⁒u∈Im⁑(Aj)=W.

(b) j=3. In this case, W is invariant under all Cm for mβ©Ύ4, and we only need to check that W is invariant under C2.

Since n⩾5, we have f⁒(2)=dim⁑(Im⁑(A3)∩Im⁑(A5))=k (if n=5, then use A0 instead of A5), and hence W=Im⁑(A1)=Im⁑(A3)=Im⁑(A5), and then

C2⁒w=C2⁒A5⁒u=A5⁒C2⁒u∈W for all⁒w=A5⁒u∈Im⁑(A5)=W.

(3) For 3β©½kβ©½r-1, we have that W is a proper invariant subspace of β„‚r, which contradicts the irreducibility of ρ. ∎

In those cases when the edges of the friendship graph represent one-dimensional subspaces, it is convenient to consider vectors that span these subspaces. We are going to set the notation that we will use in Section 5 to investigate the friendship graphs corresponding to our representations.

Definition 4.15.

Let ρ:Bnβ†’GLr⁒(β„‚) be a matrix representation of Bn, nβ©Ύ3. Suppose that, for some iβ‰ j, i,jβˆˆβ„€n, dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1, and suppose vβ‰ 0 is a vector such that span⁑{v}=Im⁑(Ai)∩Im⁑(Aj).

  1. Set vi,j=vj,i=v.

  2. If βˆ₯i-jβˆ₯β©Ύ2, we will call the vector vi,ja diagonal of the full friendship graph or, for short, a diagonal. In addition, we will say that the diagonal vi,j is coming out of the vertex Ai or, for short, coming out of Ai (as well as vi,j is coming out of Aj). We also will say that the diagonal vi,jconnects Ai and Aj.

Lemma 4.16.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn for nβ©Ύ3. Suppose that three distinct vertices Ai, Aj and Ak, i,j,k∈Zn, are pairwise connected by edges representing one-dimensional subspaces. Namely,

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=dim⁑(Im⁑(Ai)∩Im⁑(Ak))=dim⁑(Im⁑(Aj)∩Im⁑(Ak))=1.

If Im⁑(Ai)∩Im⁑(Aj)=Im⁑(Ai)∩Im⁑(Ak), then

Im⁑(Aj)∩Im⁑(Ak)=Im⁑(Ai)∩Im⁑(Aj)=Im⁑(Ai)∩Im⁑(Ak).

Proof.

Clearly, if vi,j∈span⁑{vi,k}, then

vi,j∈Im⁑(Aj)∩Im⁑(Ak),

and the statement follows from the fact that all the subspaces in question are one-dimensional. ∎

Cyclicity and reflection arguments. By Lemma 2.2 (c) and (d), we have

  1. T⁒Ai⁒T-1=Ai+1, iβˆˆβ„€n,

  2. D⁒Ai⁒D-1=An-i, i=1,2,…,n-1.

By using the fact that, for two linear operators A and B on a vector space V, conjugated by an invertible P (B=P⁒A⁒P-1), if v∈Im⁑(A), then P⁒v∈Im⁑(B), and other basic facts from linear algebra, we will refer to (1) as the cyclic argument and to (2) as the reflection argument.

Note that Δ⁒σ0⁒Δ-1β‰ Οƒ0, so the reflection argument cannot be extended to i=0.

5 Irreducible representations of corank 3

The main goal of this paper is the classification of the irreducible representations of dimension n+1. First, we are going to reduce this problem to the classification of the irreducible representations of small corank.

Theorem 5.1.

Let ρ:Bnβ†’GLn+1⁒(C) be an irreducible matrix representation of Bn of dimension n+1, nβ©Ύ10. Then ρ is equivalent to a tensor product of a one-dimensional representation and an irreducible representation of dimension n+1 and corank 3.

Proof.

By Reduction Theorem (Theorem 3.8), ρ is equivalent to a tensor product of a one-dimensional representation and an irreducible representation of dimension r=n+1 and corank k with 3⩽k⩽r-n+2=n+1-n+2=3. ∎

In this section, we are going to prove that, for nβ©Ύ10, the braid group Bn has no irreducible representations of dimension rβ©Ύn+1 of corank 3.

Suppose ρ:Bnβ†’GLr⁒(β„‚) is an irreducible representation of Bn of dimension rβ©Ύn+1, nβ©Ύ3. By Corollary 4.11, there are only two possibilities.

  1. Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0} for all iβˆˆβ„€n.

  2. Im⁑(Ai)∩Im⁑(Ai+1)={0} for all iβˆˆβ„€n and Im⁑(Ai)∩Im⁑(Aj)β‰ {0} for βˆ₯i-jβˆ₯β©Ύ2, i,jβˆˆβ„€n.

We will consider these two cases separately in Subsections 5.1 and 5.2. We will use the notation introduced in Sections 2 and 4.

5.1 Case (I): Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0}

In this subsection, we will prove that there are no irreducible representations of Bn of dimension rβ©Ύn+1 and corank 3 such that Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0} for nβ©Ύ9 (Theorem 5.1.24). First, we will prove that if Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0}, then Im⁑(Ai)∩Im⁑(Ai+1) must be one-dimensional for all i (Theorem 5.1.5), and then we will prove that, in this case, the graph is complete with

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all⁒iβ‰ j

(Theorem 5.1.15). We will finish this subsection by investigating the corresponding friendship graph to prove Theorem 5.1.24.

First, let us introduce the following notation. For i,jβˆˆβ„€n, let [i,j] be the cyclic interval {i,i+1,…,j}, and S[i,j]=Im⁑(Ai)∩Im⁑(Ai+1)βˆ©β€¦βˆ©Im⁑(Aj).

Lemma 5.1.1.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 such that dim⁑(Im⁑(Ai))=k, where 3β©½kβ©½r-1 and nβ©Ύ5. Suppose that S[1,j]β‰ {0} for some j=2,…,n-3. Then

S[i,i+j-1]β‰ S[i+1,i+j] for every⁒iβˆˆβ„€n.

Proof.

Suppose that S[i,i+j-1]=S[i+1,i+j] for some iβˆˆβ„€n. By the cyclic argument, we have S[1,j]=S[2,j+1]=…=S[0,j-1]β‰ {0}.

Then, for the subspace

S=S[1,0]=Im⁑(A1)∩Im⁑(A2)∩Im⁑(A3)βˆ©β€¦βˆ©Im⁑(An-1)∩Im⁑(A0),

we have S=S[i,i+j-1] for every iβˆˆβ„€n. We claim that S is invariant under Bn. Indeed, for every mβˆˆβ„€n, we have S=S[m+2,m+j+1]. Since jβ©½n-3, we have βˆ₯m-lβˆ₯β©Ύ2 for each l from [m+2,m+j+1], and thus Am commutes with each of Am+2,Am+3,…,Am+j+1. So S=S[m+2,m+j+1] is invariant under Am for every mβˆˆβ„€n, and hence S is Bn-invariant.

Since Sβ‰ {0} and dim⁑(S)β©½dim⁑(Im⁑(A1))=k<r, S is a proper invariant subspace of β„‚r, which contradicts the irreducibility of ρ. ∎

Lemma 5.1.2.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn with

dim⁑(Im⁑(Ai))=3,n⩾3.

Suppose dim⁑(S[i,i+1])=2 for all i∈Zn. Then S[i,i+2]β‰ {0} for all i∈Zn.

Proof.

Using the cyclicity argument, it is enough to show that S[1,3]β‰ {0}. Indeed, since S[1,2]+S[2,3]βŠ†Im⁑(A2), we obtain

dim⁑(S[1,3])=dim⁑(S[1,2])+dim⁑(S[2,3])-dim⁑(S[1,2]+S[2,3])⩾2+2-dim⁑(Im⁑(A2))=1.∎

Lemma 5.1.3.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with dim⁑(Im⁑(Ai))=3 and nβ©Ύ5. Suppose

dim⁑(S[i,i+1])=2 for all⁒iβˆˆβ„€n.

Then dim⁑(S[i,i+2])=1 for all i∈Zn.

Proof.

Due to the cyclic argument, is enough to show that dim⁑(S[1,3])=1. By Lemma 5.1.2, we have dim⁑(S[1,3])⩾1 and dim⁑(S[1,3])⩽dim⁑(S[1,2])⩽2. But if dim⁑(S[1,3])=2, then S[1,3]=S[1,2]=S[2,3], a contradiction with Lemma 5.1.1. Thus, dim⁑(S[1,3])=1. ∎

Lemma 5.1.4.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with dim⁑(Im⁑(Ai))=3 and nβ©Ύ6. Suppose

dim⁑(S[i,i+1])=2 for all⁒iβˆˆβ„€n.

For each i∈Zn, consider a non-zero vector wi such that span⁑{wi}=S[i-1,i+1]. Let Wi=span⁑{wi-1,wi,wi+1}, i∈Zn. Then Im⁑(Ai)=Wi for all i∈Zn.

Proof.

First of all, by Lemma 5.1.3, S[i-1,i+1] is one-dimensional, so there exists a non-zero vector wi generating S[i-1,i+1] for each iβˆˆβ„€n. Due to the cyclicity, it is enough to show that the three vectors w1,w2,w3∈Im⁑(A2) are linearly independent.

By Lemma 5.1.1 with j=3β©½n-3 for nβ©Ύ6, we have S[0,2]β‰ S[1,3], so the vectors w1 and w2 are linearly independent, and dim⁑(W2)β©Ύ2. Similarly, the vectors w2 and w3 are linearly independent.

Suppose that the vectors w1,w2,w3 are linearly dependent. Then

dim⁑(W2)=2 and W2=span⁑{w1,w2}=span⁑{w2,w3}.

Since both w1,w2∈S[1,2] and dim⁑(span⁑{w1,w2})=dim⁑(S[1,2])=2, we have

W2=span⁑{w1,w2}=S[1,2].

Similarly, W2=span⁑{w2,w3}=S[2,3], so W2=S[1,2]=S[2,3], a contradiction with Lemma 5.1.1. ∎

Theorem 5.1.5.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with

dim⁑(Im⁑(Ai))=3β€ƒπ‘Žπ‘›π‘‘β€ƒIm⁑(Ai)∩Im⁑(Ai+1)β‰ {0}β€ƒπ‘“π‘œπ‘Ÿβ’nβ©Ύ6.

Then dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1 for all i∈Zn.

Proof.

By Lemma 4.14,

dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1 or dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=2.

Suppose dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=2. Then, by Lemma 5.1.4,

Im⁑(Ai)=span⁑{wi-1,wi,wi+1},

where span⁑{wi}=S[i-1,i+1], iβˆˆβ„€n. Thus,

Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)+Im⁑(A0)=span⁑{w1,w2,…,wn-1,w0}.

Since Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)+Im⁑(A0) is a Bn-invariant subspace of β„‚r and 3β©½dim⁑(Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)+Im⁑(A0))β©½n, we get a contradiction with the irreducibility of ρ.

Thus, dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1. ∎

Now we are going to consider the structure of the friendship graph associated with our representation. First, we are going to prove (Corollary 5.1.8) that, in our case, we actually have a complete graph, meaning that every two vertices are connected by an edge, and each edge, in fact, represents a one-dimensional subspace (Theorem 5.1.15).

For our convenience, let us introduce the following notation. For all iβˆˆβ„€n, let xi=vi,i+1 (see Definition 4.15), that is span⁑{xi}=Im⁑(Ai)∩Im⁑(Ai+1), and let U=span⁑{x0,x1,x2,…,xn-1}.

We will list some of the facts in the following lemma.

Lemma 5.1.6.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 for nβ©Ύ5 with

dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1,π‘€β„Žπ‘’π‘Ÿπ‘’β’Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi}.

Then

  1. xi and xi+1 are linearly independent for all iβˆˆβ„€n; thus dim⁑Uβ©Ύ2;

  2. T⁒xi∈span⁑{xi+1} for all iβˆˆβ„€n;

  3. Uβ‰ {0} and dim⁑Uβ©½n;

  4. U is invariant under T.

Proof.

(1) follows from Lemma 5.1.1, (2) follows from the cyclic argument, (3) is obvious, and (4) follows from (2). ∎

Lemma 5.1.7.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with

rk⁑(A1)=3β€ƒπ‘Žπ‘›π‘‘β€ƒdim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1β€ƒπ‘“π‘œπ‘Ÿβ’nβ©Ύ5.

Then t⁒f⁒(k)β‰ 0 for every k=2,3,…,n-2.

Proof.

Suppose t⁒f⁒(k)=0 for some k=2,3,…,n-2. Then, by Lemma 4.12, we have 0=t⁒f⁒(k)=t⁒f⁒(2)=t⁒f⁒(3)=…=t⁒f⁒(n-2) and, in particular,

A1⁒A3=A3⁒A1=0.

We claim that, in this case, U=span⁑{x0,x1,…,xn-1} is Bn-invariant.

Since Ak⁒xi=Tk-1⁒A1⁒T-(k-1)⁒xi and U is T-invariant, to show that U is invariant under Bn, it is enough to check that A1⁒xi∈U for i=0,1,…,n-1.

Since t⁒f⁒(2)=t⁒f⁒(3)=…=t⁒f⁒(n-2)=0, we have

A1⁒x2=A1⁒x3=…=A1⁒xn-1=0,

and we only have to show that A1⁒x0∈U and A1⁒x1∈U.

Let us consider (I+A1)⁒x0. Since t⁒f⁒(2)=0 and x0∈Im⁑(A0), we have

A2⁒x0=0.

Since x0∈Im⁑(A1), there exists some z such that x0=A1⁒z. Thus,

(I+A1)⁒x0=(I+A1)⁒x0+0=(I+A1)⁒x0+A1β‹…0=(I+A1)⁒x0+A1⁒(A2⁒x0)=(I+A1+A1⁒A2)⁒A1⁒z=(A1+A12+A1⁒A2⁒A1)⁒z=(A2+A22+A2⁒A1⁒A2)⁒z=A2⁒(I+A2+A1⁒A2)⁒z∈Im⁑(A2).

But x0∈Im⁑(A1), so (I+A1)⁒x0∈Im⁑(A1), so

(I+A1)⁒x0∈Im⁑(A1)∩Im⁑(A2)=span⁑{x1}βŠ†U.

Thus, A1⁒x0=(I+A1)⁒x0-x0∈U.

Similarly, for x1=A1⁒y∈Im⁑(A1)∩Im⁑(A2),

(I+A1)⁒x1=(I+A1)⁒x1+A1β‹…0=(I+A1)⁒x1+A1⁒(A0⁒x1)=(I+A1+A1⁒A0)⁒A1⁒y=(A1+A12+A1⁒A0⁒A1)⁒y=(A0+A02+A0⁒A1⁒A0)⁒y=A0⁒(I+A0+A1⁒A0)⁒y∈Im⁑(A0)∩Im⁑(A1)βŠ†U,

and hence A1⁒x1∈U.

Thus, U is a Bn-invariant proper subspace of β„‚r, which contradicts the irreducibility of ρ. So t⁒f⁒(k)β©Ύ1 for all k=2,3,…,n-2. ∎

Corollary 5.1.8.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with

rk⁑(A1)=3β€ƒπ‘Žπ‘›π‘‘β€ƒdim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1β€ƒπ‘“π‘œπ‘Ÿβ’nβ©Ύ5.

Then Im⁑(Ai)∩Im⁑(Aj)β‰ {0} for all iβ‰ j∈Zn.

Proof.

The hypothesis of the lemma states that this is true for βˆ₯i-jβˆ₯=1, and

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=f⁒(Ai,Aj)β©Ύt⁒f⁒(Ai,Aj)=t⁒f⁒(βˆ₯i-jβˆ₯)β©Ύ1

for all i,jβˆˆβ„€n, βˆ₯i-jβˆ₯β©Ύ2 by Lemma 4.5 and Lemma 5.1.7. ∎

Now let us establish some important properties of the vectors x0,x1,…,xn-1 and the subspace U.

Lemma 5.1.9.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 for nβ©Ύ5 with

dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1,π‘€β„Žπ‘’π‘Ÿπ‘’β’Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi}.

Suppose that, for some m, the vectors x1,x2,…,xm are linearly independent and the vectors x1,x2,…,xm+1 are linearly dependent (indices are taken modulo n). Then the vectors x1,x2,…,xm form a basis of U. In particular, dim⁑U=m.

Proof.

We will use induction on k to show that xm+k∈span⁑{x1,x2,…,xm} for all kβ©Ύ1.

For k=1, the statement is true by the hypothesis of the lemma.

Now xm+k∈span⁑{T⁒xm+(k-1)}, and if

xm+(k-1)∈span⁑{x1,x2,…,xm},

then

xm+k∈span⁑{T⁒x1,T⁒x2,…,T⁒xm}=span⁑{x2,x3,…,xm+1},

which, together with xm+1∈span⁑{x1,x2,…,xm}, gives

xm+k∈span⁑{x1,x2,…,xm}.∎

The following statement is an easy consequence of Lemma 5.1.9 and the cyclic argument.

Corollary 5.1.10.

Under the conditions of Lemma 5.1.9, for any fixed i∈Zn, m consecutive vectors xi,xi+1,…,xi+m-1 form a basis of U (the indices are taken modulo n).

Theorem 5.1.11.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 for nβ©Ύ5 such that dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1, where Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi} and U=span⁑{x0,x1,x2,…,xn-1}. Then dim⁑Uβ©Ύn-3.

Proof.

Since dim⁑Uβ©Ύ2 (Lemma 5.1.6 (1)), the statement is trivial for n=5. Suppose now that nβ©Ύ6, and suppose m=dim⁑Uβ©½n-4. We claim that, in this case, U is invariant under Bn. Indeed, for any iβˆˆβ„€n, consider m consecutive vectors starting with xi+2. By Corollary 5.1.10, they form a basis of U,

U=span⁑{xi+2,xi+3,…,xi+m+1} (the indices are taken modulo⁒n).

Since mβ©½n-4, for every k=i+2,…,i+m+1, neither Ak nor Ak+1 are neighbors of Ai, and hence they both commute with Ai. Thus,

xk∈Im⁑(Ak)∩Im⁑(Ak+1),

so Ai⁒xk∈Im⁑(Ak)∩Im⁑(Ak+1)=span⁑{xk}βŠ†U.

Since Uβ‰ {0} and dim⁑Uβ©½n-4<n+1β©½r, we have that U is a proper invariant subspace of β„‚r, which contradicts the irreducibility of ρ. ∎

Corollary 5.1.12.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 for nβ©Ύ5 such that dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1, where Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi}. Then the vectors xi,xi+1,…,xi+n-4, i∈Zn, are linearly independent.

Proof.

This follows immediately from Corollary 5.1.10 and Theorem 5.1.11. ∎

Corollary 5.1.13.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 for nβ©Ύ7 such that dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1, where Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi}. Then the vectors xi and xj are linearly independent for all iβ‰ j, i,j∈Zn.

Proof.

Due to cyclicity, it is enough to show that x1 and xj are linearly independent for all jβ‰ 1, jβˆˆβ„€n. By Corollary 5.1.12, for jβ©½n-3, both vectors x1 and xj belong to the linearly independent set x1,x2,…,xn-3, and for j=n-2,n-1,0, they belong to the linearly independent set x5,x6,…,x0,x1 (since nβ©Ύ7). ∎

Lemma 5.1.14.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with dim⁑(Im⁑(Ai))=3 for nβ©Ύ5 such that

dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=1,

where Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi} and U=span⁑{x0,x1,x2,…,xn-1}. Let y be a non-zero vector y∈Im⁑(A1) such that yβˆ‰span⁑{x0,x1}. Then yβˆ‰U.

Proof.

Since dim⁑(Im⁑(Ai))=3, we have Im⁑(A1)=span⁑{x0,x1,y}. By applying the cyclic argument, we have that Im⁑(Ai)=span⁑{xi-1,xi,Ti-1⁒y} for all iβˆˆβ„€n.

Suppose that y∈U. Then Im⁑(A1)βŠ†U, and since U is invariant under T, Im⁑(Ai)βŠ†U for all i, and

dim⁑V=dim⁑(Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)+Im⁑(A0))β©½dim⁑Uβ©½n,

a contradiction. ∎

Next, we prove the theorem about the structure of the friendship graph in the case when the neighbors are connected by an edge.

Theorem 5.1.15.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with and rk⁑(A1)=3, where nβ©Ύ9. Suppose that Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0} for all i∈Zn. Then dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all i,j∈Zn, iβ‰ j.

Proof.

By Theorem 5.1.5, we have dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for βˆ₯i-jβˆ₯=1. For βˆ₯i-jβˆ₯β©Ύ2, by Corollary 5.1.8, we have dim⁑(Im⁑(Ai)∩Im⁑(Aj))β©Ύ1, and by Lemma 4.14, dim⁑(Im⁑(Ai)∩Im⁑(Aj))β©½2. Thus, it remains to show that

dim⁑(Im⁑(Ai)∩Im⁑(Aj))β‰ 2.

Due to the cyclic argument, it is enough to prove that dim⁑(Im⁑(A0)∩Im⁑(Aj))β‰ 2 for j=2,3,…,n-2, and since f⁒(k)=f⁒(n-k) for k=1,…,n-1, it is enough to prove this statement for j=2,3,…,[n2], where [n2] denotes the integer part of n2. Equivalently, it is enough to prove that dim⁑(Im⁑(A1)∩Im⁑(Aj))β‰ 2 for all j=3,4,…,[n2]+1.

Suppose that, for some j, 3⩽j⩽[n2]+1, dim⁑(Im⁑(A1)∩Im⁑(Aj))=2.

Consider vectors x0,x1,x2,…,xn-1, where Im⁑(Ai)∩Im⁑(Ai+1)=span⁑{xi}, and a vector y1∈Im⁑(A1) such that y1βˆ‰span⁑{x0,x1}.

Then, if we consider a vector yj=Tj-1⁒y1∈Im⁑(Aj), we obtain

Im⁑(A1)=span⁑{y1,x0,x1} and Im⁑(Aj)=span⁑{yj,xj-1,xj}.

Therefore,

Im⁑(A1)+Im⁑(Aj)=span⁑{x0,x1,xj-1,xj,y1,yj};

hence

dim⁑(Im⁑(A1)+Im⁑(Aj))=dim⁑(Im⁑(A1))+dim⁑(Im⁑(Aj))-dim⁑(Im⁑(A1)∩Im⁑(Aj))=3+3-2=4.

Since 3β©½jβ©½[n2]+1, the indices for the vectors x0,x1,xj-1,xj are all distinct. Since nβ©Ύ9 and jβ©½[n2]+1, we have

n-4-jβ©Ύn-4-[n2]-1β©Ύn-5-n2β©Ύ-0.5,

and since n-4-jβˆˆβ„€, we have n-4-jβ©Ύ0 or, equivalently, jβ©½n-4. Thus, the set of vectors x0,x1,xj-1,xj is a subset of the linearly independent vectors x0,x1,…,xn-5,xn-4 (Corollary 5.1.12). Thus, dim⁑(span⁑{x0,x1,xj-1,xj})=4, so y1∈span⁑{x0,x1,xj-1,xj}βŠ†U, a contradiction with Lemma 5.1.14. ∎

The remainder of this Subsection 5.1 is devoted to representations satisfying the following conditions.

  1. Let ρ:Bnβ†’GLr⁒(β„‚) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with rk⁑(A1)=3 and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all i,jβˆˆβ„€n, iβ‰ j.

Notation.

Suppose condition (⋆) holds.

  1. Denote Zi=span⁑{xi-1,xi}, iβˆˆβ„€n. Clearly, ZiβŠ†Im⁑(Ai), ZiβŠ†U, and by Lemma 5.1.6 (1), dim⁑Zi=2.

  2. For iβ‰ j, iβ‰ j+1, i,jβˆˆβ„€n, denote Yi,j=span⁑{vi,j,vi,j+1}.

  3. For iβˆˆβ„€n, denote

    Xi↑=Yi,i+1=span⁑{xi,vi,i+2},
    Xi↓=Yi,i-2=span⁑{xi-1,vi,i-2}.

Lemma 5.1.16.

Under condition (⋆), for nβ©Ύ5, for every i∈Zn,

vi-1,i+1βˆ‰Im⁑(Ai).

Proof.

Due to the cyclic argument, it is enough to show that v1,3βˆ‰Im⁑(A2). Note that S[1,3]=S[1,2]∩S[2,3]=span⁑{x1}∩span⁑{x2}={0}, since x1 and x2 are linearly independent by Lemma 5.1.6 (1). Thus,

{0}=S[1,3]=(Im⁑(A1)∩Im⁑(A3))∩Im⁑(A2)=span⁑{v1,3}∩Im⁑(A2),

so v1,3βˆ‰Im⁑(A2). ∎

Corollary 5.1.17.

Under condition (⋆), for nβ©Ύ5, for every i∈Zn,

  1. vi-1,i+1βˆ‰Zi=span⁑{xi-1,xi},

  2. xi and vi,i+2 are linearly independent,

  3. dim⁑(Xi↑)=2,

  4. xi-1 and vi-2,i are linearly independent,

  5. dim⁑(Xi↓)=2.

Lemma 5.1.18.

Under condition (⋆), for nβ©Ύ7, the vectors vi,j and vi,j+1 are linearly independent for all iβ‰ j, iβ‰ j+1 (indices taken modulo n).

Proof.

Due to the cyclic argument, it is enough to prove linear independence of vectors v1,k and v1,k+1 for all k=2,3,…,n-1.

For k=2 and k=n-1, the statement is equivalent to Corollary 5.1.17 (2) and (4).

Suppose now that 3β©½kβ©½n-2, and the vectors v1,k and v1,k+1 are linearly dependent. The conjugation by D=ρ⁒(Ξ”) followed by the conjugation by Tk+2 sends the vertices A1,Ak and Ak+1 to Ak+1,A2 and A1 respectively.

Thus, if span⁑{v1,k}=span⁑{v1,k+1}, then span⁑{v1,k+1}=span⁑{v2,k+1}. By Lemma 4.16, we obtain span⁑{v1,k+1}=span⁑{xk}=span⁑{x1}, a contradiction with Corollary 5.1.13 since x1 and xk are linearly independent for n⩾7. ∎

Corollary 5.1.19.

Under condition (⋆), for nβ©Ύ7, dim⁑(Yi,j)=2 for all iβ‰ j, iβ‰ j+1, i,j∈Zn.

Lemma 5.1.20.

Let U1=Im⁑(A1) and

Uk=Uk-1+Im⁑(Ak)β€ƒπ‘“π‘œπ‘Ÿβ’k=2,3,…,n-1.

If nβ©Ύ5 and condition (⋆) holds, then dim⁑(Uk)β©½k+3.

Proof.

Since dim⁑(U1)=3, we have

dim⁑(U2)=dim⁑(U1)+dim⁑(Im⁑(A2))-dim⁑(U1∩Im⁑(A2))=dim⁑(Im⁑(A1))+dim⁑(Im⁑(A2))-dim⁑(Im⁑(A1)∩Im⁑(A2))=3+3-1=5.

By definition, Xkβ†“βŠ†Im⁑(Ak) and Xkβ†“βŠ†Im⁑(Ak-1)+Im⁑(Ak-2), so

Xkβ†“βŠ†Uk-1∩Im⁑(Ak) for⁒k=3,4,…,n-1.

By Corollary 5.1.17 (5), since nβ©Ύ5, dim⁑(Xk↓)=2, and thus

dim⁑(Uk-1∩Im⁑(Ak))⩾2.

By induction on k, we have

dim⁑(Uk)=dim⁑(Uk-1)+dim⁑(Im⁑(Ak))-dim⁑(Uk-1∩Im⁑(Ak))⩽[(k-1)+3]+3-2=k+3.∎

Lemma 5.1.21.

Under condition (⋆), for nβ©Ύ5, for all i∈Zn,

  1. vi,k∈Xi↑ for kβˆˆβ„€n, kβ‰ i-2,i-1,i,

  2. vi,k∈Xi↓ for kβˆˆβ„€n, kβ‰ i,i+1,i+2.

Proof.

Due to cyclicity, to prove part (1), it is enough to prove that v1,k∈X1↑ for kβ‰ n-1,0,1. Part (2) follows from part (1) by applying the conjugation by D followed by the conjugation by T2.

For k=2 and k=3, the statement v1,k∈X1↑ follows from the definition of X1↑.

Suppose now for some k,k=4,5,…,n-2, v1,kβˆ‰X1↑. Then

dim⁑(span⁑{v1,k,x1,v1,3})=3 and Im⁑(A1)=span⁑{v1,k,x1,v1,3}.

Conjugating by Tk+1⁒D gives Im⁑(Ak)=span⁑{v1,k,xk-1,vk,k-2}.

Consider the subspaces U1,U2,…,Un-1 defined by

U1=Im⁑(A1) and Ui=Ui-1+Im⁑(Ai) for⁒i=2,…,n-1.

By Lemma 5.1.20, we have dim⁑(Uk-1)⩽k+2. But

v1,k∈Im⁑(A1),xk-1∈Im⁑(Ak-1) and vk,k-2∈Im⁑(Ak-2),

so

Im⁑(Ak)βŠ†Uk-1 and dim⁑(Uk)=dim⁑(Uk-1)β©½k+2.

Moreover, for every j=k,k+1,…,n-1, using conjugation by Tj-k, we obtain Im⁑(Aj)=span⁑{vj,j-k+1,xj-1,vj,j-2}βŠ†Uj-1 (since j-k+1β©½j-1 for kβ©Ύ2), and by induction on j, we get Uj=Uj-1=…=Uk-1, thus

dim⁑(Uj)β©½k+2 for all⁒j=k,k+1,…,n-1.

On the other hand, Un-1=Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)=V, so, for kβ©½n-2, we have

n+1⩽dim⁑(V)=dim⁑(Un-1)⩽k+2⩽n-2+2=n,

a contradiction. ∎

Lemma 5.1.22.

Under condition (⋆), for nβ©Ύ7, for all i∈Zn, Zi=Xi↑=Xi↓.

Proof.

As before, it is enough to prove this statement for i=1 due to cyclicity. Consider Y1,4=span⁑{v1,4,v1,5}. By Corollary 5.1.19, since nβ©Ύ7, we have dim⁑(Y1,4)=2. By Lemma 5.1.21 (1), Y1,4βŠ†X1↑ since, for k=4 and 5, we have kβ©½n-2 for nβ©Ύ7. By Corollary 5.1.17 (3), dim⁑(X1↑)=2, so Y1,4=X1↑. Similarly, by Corollary 5.1.19, Lemma 5.1.21 (2) and Corollary 5.1.17 (5), Y1,4=X1↓. Thus, X1↑=X1↓=Y1,4.

Now x1∈X1↑=Y1,4 and x0∈X1↓=Y1,4. Thus, Z1=span⁑{x0,x1}βŠ†Y1,4. But dim⁑(Z1)=2 and dim⁑(Y1,4)=2, so Z1=Y1,4=X1↑=X1↓. ∎

Corollary 5.1.23.

Under condition (⋆), for nβ©Ύ7, vi,k∈Zi for all i,k∈Zn, kβ‰ i.

Proof.

For all k≠i-2,i-1,i+1,i+2, by Lemma 5.1.21,

vi,k∈X1β†‘βˆ©X1↓=Zi.

For k=i+1,i+2, we have vi,k∈X1↑=Zi, and for k=i-1,i-2, we have vi,k∈X1↓=Zi by Lemma 5.1.22. ∎

Theorem 5.1.24.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with dim⁑(Im⁑(Ai))=3 for nβ©Ύ9. Then

Im⁑(Ai)∩Im⁑(Ai+1)={0} for all⁒iβˆˆβ„€n.

Proof.

Suppose Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0}. Then, by Theorem 5.1.15,

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all⁒i,jβˆˆβ„€n,iβ‰ j.

Consider v1,3=v3,1β‰ 0 such that span⁑{v1,3}=Im⁑(A1)∩Im⁑(A3). By Corollary 5.1.23,

v1,3∈Z1=span⁑{x0,x1} and v3,1∈Z3=span⁑{x2,x3}.

So

v1,3=v3,1∈span⁑{x0,x1}∩span⁑{x2,x3}.

Since nβ©Ύ9, the set of vectors x0,x1,x2,x3 is a subset of the linearly independent set x0,x1,…,xn-4 (Corollary 5.1.12), thus v1,3=0, a contradiction. ∎

5.2 Case (II): Im⁑(Ai)∩Im⁑(Ai+1)={0}

In this subsection, we are going to follow a path similar to the case when

Im⁑(Ai)∩Im⁑(Ai+1)β‰ {0}.

First, we are going to prove that the friendship graph has all non-adjacent edges (diagonals) and that each diagonal represents a one-dimensional subspace (Lemma 5.2.1). Then we are going to show (Theorems 5.2.13 and 5.2.14) that there are no irreducible representations associated to such friendship graphs if the number of vertices is at least 10.

Lemma 5.2.1.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, nβ©Ύ5, with dim⁑(Im⁑(Ai))=3. Suppose that

Im⁑(Ai)∩Im⁑(Ai+1)={0} for all⁒iβˆˆβ„€n.

Then dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all βˆ₯i-jβˆ₯β©Ύ2, i,j∈Zn.

Proof.

Due to cyclicity, it is enough to show that

dim⁑(Im⁑(A1)∩Im⁑(Ak))=1 for all⁒k=3,4,…,n-1.

By Corollary 4.11, dim⁑(Im⁑(A1)∩Im⁑(Ak))⩾1, and by Lemma 4.14,

dim⁑(Im⁑(A1)∩Im⁑(Ak))β©½2 for⁒k=3,4,…,n-1.

By Lemma 4.13, we have dim⁑(Im⁑(A1)∩Im⁑(Ak))=dim⁑(Im⁑(A1)∩Im⁑(Am)) for all k,m=3,4,…,n-1. Thus, it is enough to prove that

dim⁑(Im⁑(A1)∩Im⁑(Ak))β‰ 2.

Suppose dim⁑(Im⁑(A1)∩Im⁑(Ak))=2. Then dim⁑(Im⁑(A2)∩Im⁑(A4))=2 and Im⁑(A1)∩Im⁑(A4)+Im⁑(A2)∩Im⁑(A4)βŠ†Im⁑(A4), so

dim⁑(Im⁑(A1)∩Im⁑(A2))⩾dim⁑(Im⁑(A1)∩Im⁑(A2)∩Im⁑(A4))=dim⁑(Im⁑(A1)∩Im⁑(A4))+dim⁑(Im⁑(A2)∩Im⁑(A4))-dim⁑(Im⁑(A1)∩Im⁑(A4)+Im⁑(A2)∩Im⁑(A4))⩾2+2-dim⁑(Im⁑(A4))=1,

a contradiction with Im⁑(A1)∩Im⁑(A2)={0}. Thus, dim⁑(Im⁑(A1)∩Im⁑(Ak))=1 for all k=3,4,…,n-1. ∎

Now we will investigate the representations whose friendship graphs have one-dimensional diagonals.

Notation.

  1. For βˆ₯i-jβˆ₯β©Ύ3, denote Xi,j=span⁑{vi,j-1,vi,j,vi,j+1}.

  2. Similar to the notation in Subsection 5.1, denote Yi,j=span⁑{vi,j,vi,j+1} for i,jβˆˆβ„€n, jβ‰ i-2,i-1,i,i+1.

  3. Denote by Vi=span⁑{vi,i+2,vi,i+3,…,vi,i-2} a subspace spanned by all diagonals coming out of Ai.

  4. Denote by W the subspace of V generated by all the diagonals,

    W=V1+V2+…+Vn-1+V0.

Lemma 5.2.2.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn for nβ©Ύ5. Suppose that Im⁑(Ai)∩Im⁑(Ai+1)={0} for all i∈Zn, and

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all⁒βˆ₯i-jβˆ₯β©Ύ2,i,jβˆˆβ„€n.

Then, for all i,j∈Zn, jβ‰ i-2,i-1,i,i+1, the vectors vi,j and vi,j+1 are linearly independent.

Proof.

Suppose that vi,j and vi,j+1 are linearly dependent for some i,j. Then vi,j+1∈span⁑{vi,j}=Im⁑(Ai)∩Im⁑(Aj), so 0β‰ vi,j+1∈Im⁑(Aj)∩Im⁑(Aj+1), a contradiction with Im⁑(Aj)∩Im⁑(Aj+1)={0}. ∎

Corollary 5.2.3.

Under the conditions of Lemma 5.2.2,

dim⁑Xi,jβ©Ύ2 for all⁒βˆ₯i-jβˆ₯β©Ύ3.

Corollary 5.2.4.

Under the conditions of Lemma 5.2.2,

dim⁑Yi,j=2 for all⁒i,jβˆˆβ„€n,jβ‰ i-2,i-1,i,i+1.

Lemma 5.2.5.

Suppose n⩾7. Then there exists some k such that 1<k<n2 and gcd⁑(k,n)=1.

Proof.

Suppose for every k, 1<k<n2, gcd⁑(k,n)β‰ 1. Since

gcd⁑(n-k,n)=gcd⁑(k,n) and n2<n-k<n-1,

the only integers k in the range 1β©½kβ©½n coprime with n are 1 and n-1. In other words, in terms of Euler’s Ο† function, φ⁒(n)=2.

If n=∏piki is the prime factorization of n, then, by the well-known formula for Euler’s function, φ⁒(n)=∏(pi-1)⁒piki-1. Since φ⁒(n)=2, the only prime factors of n are 2 and 3. Moreover, the power of 3 is at most 1, and the power of 2 is at most 2. Thus, n divides 12.

Since nβ©Ύ7, then n=12. But φ⁒(12)=4, a contradiction with φ⁒(n)=2. ∎

Lemma 5.2.6.

Let ρ:Bnβ†’GLr⁒(C) be a matrix representation of Bn of dimension r, nβ©Ύ9. Suppose that Im⁑(Ai)∩Im⁑(Ai+1)={0} for all i∈Zn, and

dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all⁒βˆ₯i-jβˆ₯β©Ύ2,i,jβˆˆβ„€n.

Suppose dim⁑(V1)=2. Then W=V1+V2+…+Vn-1+V0 is Bn-invariant.

Proof.

Due to the cyclic argument, because V1 is 2-dimensional, Vi is 2-dimensional for all iβˆˆβ„€n.

For every fixed iβˆˆβ„€n, by Lemma 5.2.2, the vectors vi,j and vi,j+1 are linearly independent for every jβˆˆβ„€n, jβ‰ i-2,i-1,i,i+1, and since dim⁑(Vi)=2, we have that

Vi=span⁑{vi,j,vi,j+1} for⁒i,jβˆˆβ„€n,jβ‰ i-2,i-1,i,i+1.

To show that W is Bn-invariant, we need to show that, for every diagonal vi,j and for every kβˆˆβ„€n, Ak⁒vi,j∈W. Due to cyclicity, it is enough to show that A1⁒vi,j∈W for all vi,j. Depending on the values of i,j, we have to consider the following cases.

(a) vi,j is coming out of A1. In this case, vi,j=v1,j∈Im⁑(A1)∩Im⁑(Aj), where j=3,4,…,n-1. Since A1 and Aj commute for j=3,4,…,n-1, we have A1⁒v1,j∈Im⁑(A1)∩Im⁑(Aj)=span⁑{v1,j}βŠ†W.

(b) vi,j connects the vertices Ai and Aj, where neither Ai nor Aj are neighbors of A1, that is i,j=3,4,…,n-1, βˆ₯i-jβˆ₯β©Ύ2. In this case,

vi,j∈Im⁑(Ai)∩Im⁑(Aj),

and since A1 commutes with both Ai and Aj, we have

A1⁒vi,j∈Im⁑(Ai)∩Im⁑(Aj)=span⁑{vi,j}βŠ†W.

(c) vi,j connects the vertices Ai and Aj, where Ai is a neighbor of A1 and Aj is not a neighbor of A1. So i=2, j=4,5,…,n-1, or i=0, j=3,4,…,n-2. Since nβ©Ύ9, we can pick two vertices Ak and Ak+1, such that neither are neighbors of A1, nor neighbors of Aj. (For example, for jβ©½5, we can pick the vertices A7 and A8, and for jβ©Ύ6, we can pick A3 and A4.)

Consider the diagonals vj,k and vj,k+1. Since they form a basis of Vj,

vi,j∈span⁑{vj,k,vj,k+1}.

Since Aj, Ak and Ak+1 are not neighbors of A1, by case (b), we have A1⁒vj,k∈W and A1⁒vj,k+1∈W, and hence A1⁒vi,j∈W.

(d) vi,j connects the vertices A0 and A2, that is vi,j=v0,2. Since n⩾9, then A3 and A4 are not neighbors of A0 (nor of A1). The diagonals v0,3 and v0,4 form a basis of V0, so v0,2∈span⁑{v0,3,v0,4}, thus A1⁒v0,2∈W since A1⁒v0,3∈W and A1⁒v0,4∈W by case (c). ∎

Lemma 5.2.7.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1, nβ©Ύ9. Suppose that Im⁑(Ai)∩Im⁑(Ai+1)={0} for all i∈Zn, and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all βˆ₯i-jβˆ₯β©Ύ2, i,j∈Zn.

Then dim⁑(Vi)⩾3 for all i∈Zn.

Proof.

Suppose dim⁑(Vi)β©½2 for some iβˆˆβ„€n. By Lemma 5.2.2, dim⁑(Vi)β©Ύ2, hence dim⁑(Vi)=2. So, due to cyclicity, dim⁑(Vi)=2 for all iβˆˆβ„€n. Then, by Lemma 5.2.6, W is Bn-invariant. Since ρ is irreducible,

W=V and dim⁑(W)=rβ©Ύn+1.

Since nβ©Ύ9, then, by Lemma 5.2.5, there exists a number k, 1<k<n2, such that k and n are relatively prime.

We claim that the two diagonals v0,k and v0,n-k form a basis of V0, so

V0=span⁑{v0,k,v0,n-k}.

Suppose not. Then v0,n-k∈span⁑{v0,k}, so

Im⁑(A0)∩Im⁑(Ak)=Im⁑(A0)∩Im⁑(An-k).

Consider the sequence of vertices A0,Ak,A2⁒k,…,As⁒k,…, where sβˆˆβ„•.

By the cyclic argument, for every vertex As⁒k (s⩾2) in the sequence, we have Im⁑(As⁒k)∩Im⁑(A(s-1)⁒k)=Im⁑(A(s-2)⁒k)∩Im⁑(A(s-1)⁒k), and using induction on s, we have Im⁑(As⁒k)∩Im⁑(A(s-1)⁒k)=Im⁑(A0)∩Im⁑(Ak).

By applying Lemma 4.16 to the vertices A0,Ak,A2⁒k, we get

Im⁑(A0)∩Im⁑(A2⁒k)=Im⁑(A0)∩Im⁑(Ak),

and using induction on s and applying Lemma 4.16 to the vertices A0, A(s-1)⁒k, As⁒k, we have Im⁑(A0)∩Im⁑(As⁒k)=Im⁑(A0)∩Im⁑(Ak) for all sβˆˆβ„•.

Since k and n are relatively prime, there exists some sβˆˆβ„• such that

s⁒k≑1(modn).

Thus, Im⁑(A0)∩Im⁑(A1)=Im⁑(A0)∩Im⁑(Ak)β‰ {0}, a contradiction.

Now, for each iβˆˆβ„€n, consider the two diagonals vi,i+k and vi,i-k (indices are taken modulo n). Due to the cyclic argument, Vi=span⁑{vi,i+k,vi,i-k} for each i, and hence

W=span⁑{v0,k,v1,k+1,…,vn-1,k-1,v0,n-k,v1,n-k+1,…,vn-1,n-k-1}.

Since vi,i+k=vi+k,i, each diagonal in the set generating W appears exactly twice, thus the total number of the diagonals spanning W is 2β‹…n2=n.

So dim⁑(W)⩽n, a contradiction with dim⁑(W)=r⩾n+1. ∎

Notice that the statements 5.2.2–5.2.7 hold true for the representations of any corank at least 3. Now we are going to restrict our considerations to the representations of corank equal to 3, and consider the representations satisfying the following conditions.

  1. Let ρ:Bnβ†’GLr⁒(β„‚) be an irreducible matrix representation of Bn of dimension rβ©Ύn+1 with rk⁑(A1)=3, where dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=0 for all iβˆˆβ„€n, and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for βˆ₯i-jβˆ₯β©Ύ2, i,jβˆˆβ„€n.

Corollary 5.2.8.

Let ρ:Bnβ†’GLr⁒(C) be a representation satisfying condition (⋆⁣⋆), for nβ©Ύ9. Then, for all i, Im⁑(Ai)=Vi and

V=W=V1+V2+…+Vn-1+V0.

Proof.

By Lemma 5.2.7, we have dim⁑(Vi)β©Ύ3 for every iβˆˆβ„€n. Together with ViβŠ†Im⁑(Ai) and dim⁑(Im⁑(Ai))=3, this implies dim⁑(Vi)=3, Vi=Im⁑(Ai). So,

W=V1+V2+…+Vn-1+V0=Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)+Im⁑(A0)

is an invariant subspace of V, and since ρ is irreducible, V=W. ∎

Lemma 5.2.9.

Under condition (⋆⁣⋆), for nβ©Ύ9, for every i∈Zn there exists some j∈Zn,βˆ₯i-jβˆ₯β©Ύ3, such that Xi,j=Im⁑(Ai).

Proof.

Due to cyclicity, it is enough to prove that there exists some j=4,5,…,n-2 such that X1,j=Im⁑(A1).

Suppose that for all j=4,5,…,n-2, X1,jβ‰ Im⁑(A1). Since X1,jβŠ†Im⁑(A1) and dim⁑(Im⁑(A1))=3, then dim⁑(X1,j)β©½2. By Corollary 5.2.3, dim⁑Xi,jβ©Ύ2 for all βˆ₯i-jβˆ₯β©Ύ3, so

dim⁑(X1,j)=2 for⁒j=4,5,…,n-2.

By definition, X1,j=span⁑{v1,j-1,v1,j,v1,j+1}, and by Lemma 5.2.2, the vectors v1,j-1 and v1,j are linearly independent. Thus, for all j, 4⩽j⩽n-2, we have v1,j+1∈span⁑{v1,j-1,v1,j}=Y1,j-1. In particular, v1,5∈Y1,3.

By induction on j, we have

v1,j+1∈span⁑{v1,j-1,v1,j}βŠ†Y1,3 for all⁒j=4,5,…,n-2,

and hence V1=span⁑{v1,3,v1,4,…,v1,n-1}βŠ†Y1,3, so dim⁑(V1)β©½2, which contradicts Lemma 5.2.7. ∎

Lemma 5.2.10.

Under condition (⋆⁣⋆), for nβ©Ύ9, there exists some j∈Zn such that 4β©½jβ©½n2+1 and X1,j=Im⁑(A1).

Proof.

By Lemma 5.2.9, there exists some j=4,5,…,n-2 such that

X1,j=Im⁑(A1).

Suppose that j>n2+1. Using conjugation by D, we get Xn-1,n-j=Im⁑(An-1), and using conjugation by T2, we get X1,n-j+2=Im⁑(A1). So there exists

k=n-j+2<n-(n2+1)+2=n2+1

such that X1,k=Im⁑(A1). ∎

Lemma 5.2.11.

Let U1=Im⁑(A1) and

Uk=Uk-1+Im⁑(Ak)β€ƒπ‘“π‘œπ‘Ÿβ’k=2,3,…,n-1.

If nβ©Ύ5 and condition (⋆⁣⋆) holds, then dim⁑(Uk)β©½k+5 for all kβ©Ύ3.

Proof.

Since dim⁑(U1)=3, we have

dim⁑(U2)=dim⁑(U1)+dim⁑(Im⁑(A2))-dim⁑(U1∩Im⁑(A2))=dim⁑(Im⁑(A1))+dim⁑(Im⁑(A2))-dim⁑(Im⁑(A1)∩Im⁑(A2))=3+3-0=6.

Since v1,3∈Im⁑(A1)∩Im⁑(A3)βŠ†U2∩Im⁑(A3), it follows that

dim⁑(U2∩Im⁑(A3))⩾1,

so dim⁑(U3)=dim⁑(U2)+dim⁑(Im⁑(A3))-dim⁑(U2∩Im⁑(A3))⩽6+3-1=8.

Yk,1=span⁑{vk,1,vk,2}βŠ†(Im⁑(A1)+Im⁑(A2))∩Im⁑(Ak)βŠ†Uk-1∩Im⁑(Ak)

for every k=4,5,…,n-1. By Corollary 5.2.4, we have dim⁑(Yk,1)=2, thus dim⁑(Uk-1∩Im⁑(Ak))β©Ύ2. By induction on k, we have

dim⁑(Uk)=dim⁑(Uk-1)+dim⁑(Im⁑(Ak))-dim⁑(Uk-1∩Im⁑(Ak))⩽[(k-1)+5]+3-2=k+5.∎

Lemma 5.2.12.

Under condition (⋆⁣⋆), for nβ©Ύ9, let j∈Zn, 4β©½jβ©½n2+1, be a number such that X1,j=Im⁑(A1).

Let U1=Im⁑(A1) and Uk=Uk-1+Im⁑(Ak) for k=2,3,…,n-1. Then Uk=Uj for all k=j+1,j+2,…,n-1.

Proof.

For every k, j+1β©½kβ©½n-1, using conjugation by D, followed by conjugation by Tk+1, we have

X1,j=Im⁑(A1)⟹Xn-1,n-j=Im⁑(An-1)⟹Xn-1+k+1,n-j+k+1=Im⁑(An-1+k+1)⟹Im⁑(Ak)=Xk,k-j+1=span⁑{vk,k-j,vk,k-j+1,vk,k-j+2}.

We will prove the lemma by induction on k.

First, for k=j+1, we have

Im⁑(Aj+1)=span⁑{vj+1,1,vj+1,2,vj+1,3}βŠ†Im⁑(A1)+Im⁑(A2)+Im⁑(A3)βŠ†Uj

(since j⩾4), so Uj+1=Uj+Im⁑(Aj+1)=Uj.

Suppose now that Uk-1=Uj. Then

Im⁑(Ak)=span⁑{vk,k-j,vk,k-j+1,vk,k-j+2}βŠ†Im⁑(Ak-j)+Im⁑(Ak-j+1)+Im⁑(Ak-j+2)βŠ†Uk-1

(since k-jβ©Ύ1 and k-j+2β©½k-4+2=k-2), so

Uk=Uk-1+Im⁑(Ak)=Uk-1=Uj.∎

Theorem 5.2.13.

For nβ©Ύ11, there are no irreducible matrix representations of Bn of dimension rβ©Ύn+1 such that rk⁑(A1)=3, dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=0 for all i∈Zn, and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for βˆ₯i-jβˆ₯β©Ύ2, for all i,j∈Zn.

Proof.

Suppose that such an irreducible representation ρ exists. Consider subspaces U1,U2,…,Un-1 defined by U1=Im⁑(A1) and Uk=Uk-1+Im⁑(Ak) for k=2,3,…,n-1. By Lemma 5.2.10 and Lemma 5.2.12, there exists a number j, 4β©½jβ©½n2+1, such that Uk=Uj for all k=j+1,j+2,…,n-1. By Lemma 5.2.11, dim⁑(Uj)β©½j+5β©½n2+1+5=n2+6. Thus,

dim⁑(Uk)=dim⁑(Uj)β©½n2+6 for all⁒k=j+1,j+2,…,n-1.

In particular, dim⁑(Un-1)⩽n2+6.

On the other hand, Un-1=Im⁑(A1)+Im⁑(A2)+…+Im⁑(An-1)=V and

n+1⩽dim⁑(V)=dim⁑(Un-1)⩽n2+6,

which means n2⩽5, a contradiction with n⩾11. ∎

Unfortunately, the method of the Theorem 5.2.13 does not work for n=10 in the case when j=6. Thus, we have to consider this case separately.

Theorem 5.2.14.

For n=10, there are no irreducible matrix representations of Bn of dimension rβ©Ύn+1 such that rk⁑(A1)=3, dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=0 for all i∈Zn, and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for βˆ₯i-jβˆ₯β©Ύ2, i,j∈Zn.

Proof.

Suppose that such an irreducible representation ρ exists. Consider subspaces U1,U2,…,U9 defined by U1=Im⁑(A1) and Uk=Uk-1+Im⁑(Ak) for k=2,3,…,9. By Lemma 5.2.10 and Lemma 5.2.12, there exists a number j, 4β©½jβ©½n2+1=6 such that Uk=Uj for all k=j+1,j+2,…,9. By Lemma 5.2.11, dim⁑(Uj)β©½j+5. If j=4 or j=5, then dim⁑(Uj)β©½10, and in particular, 11β©½dim⁑(V)=dim⁑(U9)β©½10, a contradiction.

Suppose now that j=6. In this case, for m=4,5, we have X1,mβ‰ Im⁑(A1). This means that dim⁑(X1,6)=3 and dim⁑(X1,4)=dim⁑(X1,5)=2. So we have v1,5βˆ‰span⁑{v1,6,v1,7}=Y1,6. By applying conjugation by T2⁒D to X1,4 and X1,5, we obtain dim⁑(X1,7)=dim⁑(X1,8)=2. By Corollary 5.2.4, dim⁑(Y1,j)=2 for jβ‰ 0,1,2, and jβ‰ 9. Thus,

X1,7=X1,8=Y1,6=Y1,7=Y1,8.

In particular, v1,6∈span⁑{v1,8,v1,9} and v1,5βˆ‰span⁑{v1,7,v1,8}. In addition, by the cyclic argument, v2,6βˆ‰span⁑{v2,8,v2,9}.

We claim that span⁑{v1,6} is invariant under B10. First, since each of A1, A3, A4, A6, A8 and A9 commute with both A1 and A6, we have Ai⁒v1,6∈span⁑{v1,6} for i=1,3,4,6,8,9.

Consider A2⁒v1,6. Since A2 commutes with A6, we get A2⁒v1,6∈span⁑{v2,6}. On the other hand, v1,6∈span⁑{v1,8,v1,9}, and since A2 commutes with both A8 and A9, we have A2⁒v1,6∈span⁑{v2,8,v2,9}. Thus,

A2⁒v1,6∈span⁑{v2,6}∩span⁑{v2,8,v2,9}={0}

since v2,6βˆ‰span⁑{v2,8,v2,9}, and hence A2⁒v1,6=0.

By conjugating by T5, we get A7⁒v6,1=A7⁒v1,6=0, and by conjugating by T2⁒D, we get A0⁒v1,6=0. Conjugating the last equation by T5 gives A4⁒v1,6=0, which completes the proof that span⁑{v1,6} is B10-invariant, which contradicts the irreducibility of ρ. ∎

6 Main theorem

In this section, we will summarize the main results of this paper.

Theorem 6.1.

For nβ©Ύ10, there are no irreducible complex representations of the braid group Bn on n strings of dimension n+1.

Proof.

By Theorem 5.1, every irreducible representation of dimension n+1 is equivalent to a tensor product of a one-dimensional representation and an irreducible representation of dimension n+1 and corank 3.

By Theorem 5.1.24, for every irreducible representation of dimension n+1 and corank 3, Im⁑(Ai)∩Im⁑(Ai+1)={0} for all iβˆˆβ„€n. Then, by Lemma 5.2.1, dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for all βˆ₯i-jβˆ₯β©Ύ2, i,jβˆˆβ„€n.

And finally, by Theorems 5.2.13 and 5.2.14, for n⩾10, there are no irreducible representations of dimension n+1 with rk⁑(A1)=3 such that

dim⁑(Im⁑(Ai)∩Im⁑(Ai+1))=0 for all⁒iβˆˆβ„€n,

and dim⁑(Im⁑(Ai)∩Im⁑(Aj))=1 for βˆ₯i-jβˆ₯β©Ύ2, i,jβˆˆβ„€n. ∎

In addition, without any extra effort, we get the following theorem.

Theorem 6.2.

Let ρ:Bnβ†’GLr⁒(C) be an irreducible matrix representation of Bn for nβ©Ύ10. Then rk⁑(ρ⁒(Οƒi)-I)β‰ 3 for all i=1,2,…,n-1.

Proof.

For rβ©Ύn+1, this follows from Theorems 5.1.24, 5.2.13 and 5.2.14. For rβ©½n, by [3, Theorem 22] and [11, Theorem 6.1], r=1 or n-2β©½rβ©½n, and for n-2β©½rβ©½n, ρ is equivalent to χ⁒(y)βŠ—Ο^, where χ⁒(y) is a one-dimensional representation, and ρ^ is either a composition factor of a specialization of the reduced Burau representation, or a specialization of the standard representation.

Clearly, if r=1, then corank⁑(ρ)β‰ 3. For n-2β©½rβ©½n, if y=1, then

corank⁑(χ⁒(y)βŠ—Ο^)={1when⁒r=n-2⁒or⁒r=n-1,2when⁒r=n,

and if yβ‰ 1, then corank⁑(χ⁒(y)βŠ—Ο^)β©Ύn-3β©Ύ7 for nβ©Ύ10. ∎


Communicated by Robert M. Guralnick


Acknowledgements

I would like to take this opportunity to express my deep gratitude to Edward Formanek, who sparked my interest in this subject long time ago. I also would like to thank my husband Alexander Borisov who convinced me to return to this work after a significant number of years despite challenging life circumstances. This work would never have been completed without his persistence and moral support. In addition, I would like to thank the anonymous referee for the helpful comments and suggestions.

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Received: 2020-06-24
Revised: 2020-07-14
Published Online: 2020-08-25
Published in Print: 2021-01-01

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