The Jones polynomial through linear algebra

Iain Moffatt

University of South Alabama

Workshop in Knot Theory Waterloo, 24th September 2011

I. Moffatt (South Alabama) UW 2011 1 / 39 What and why

What we’ll see The construction of link invariants through R-matrices. (c.f. Reshetikhin-Turaev invariants, quantum invariants)

Why this? Can do some serious math using material from Linear Algebra 1. Illustrates how math works in the wild: start with a problem you want to solve; figure out an easier problem that you can solve; build up from this to solve your original problem. See the interplay between algebra, combinatorics and topology! It’s my favourite bit of math!

I. Moffatt (South Alabama) UW 2011 2 / 39 What we’re trying to do

A knot is a circle, S1, sitting in 3-space R3. A link is a number of disjoint circles in 3-space R3.

Knots and links are considered up to isotopy. This means you can “move then round in space, but you can’t cut or glue them”.

I. Moffatt (South Alabama) UW 2011 3 / 39 What we’re trying to do

A knot is a circle, S1, sitting in 3-space R3. A link is a number of disjoint circles in 3-space R3.

Knots and links are considered up to isotopy. This means you can “move then round in space, but you can’t cut or glue them”. The fundamental problem in knot theory is to determine whether or not two links are isotopic.

=? =?

I. Moffatt (South Alabama) UW 2011 3 / 39 To do this we need knot invariants: F : Links (a set) such (Isotopy) → that F(L) = F(L0) = L = L0, Aim: construct6 link invariants⇒ 6 using linear algebra. What we’re trying to do

Knots and links are considered up to isotopy. This means you can “move then round in space, but you can’t cut or glue them”.

The fundamental problem in knot theory is to determine whether or not two links are isotopic.

=? =?

To do this we need knot invariants: F : Links (a set) such (Isotopy) → that F(L) = F(L0) = L = L0, 6 ⇒ 6 Aim: construct link invariants using linear algebra.

I. Moffatt (South Alabama) UW 2011 3 / 39 Our toolkit: Linear algebra 1

The basics

a vector space over C with basis v 1,..., v n . V { } v v = Pn a v . ∈ V ⇐⇒ i=1 i i Pn j f : v = i=1 fi v i . V → V ⇐⇒  1 n  f1 ··· f1    . .  f linear matrix  . .   . .  ↔ 1 n fn ... fn

The direct product = (u, v) u, bv V × V { | ∈ V} λ(u, v) + (u0, v 0) = (λu + u0, λv + v 0)

The dual ∗ ∗ V = Hom( , C) = f : C f linear V i Vi { V → | } basis v v (v ) = δ , , i = 1,..., n { | j i j } I. Moffatt (South Alabama) UW 2011 4 / 39 Making life easier: link diagrams

Working with 3-D objects is tricky. To make life easy we draw knots on the plane. A link diagram is a drawing of a link on the plane. Link diagrams are considered up to the Reidemeister moves Knot diagrams Reidemeister moves

= =

RI RII

RIII =

project plane onto

Theorem Links Diagrams = (Isotopy) (R moves) − I. Moffatt (South Alabama) UW 2011 5 / 39 An example

[link]

I. Moffatt (South Alabama) UW 2011 6 / 39 The first idea: let’s form a map

Our goal Links − We want to construct a knot polynomial J : C[t, t 1]. (Isotopy) → How do we get started? Diagrams − Work with link diagrams look for J : C[t, t 1]. (R−moves) → Motivated by algebra: define; the map on the “generators” of a diagram.

I. Moffatt (South Alabama) UW 2011 7 / 39 The first idea: let’s form a map

Our goal Links − We want to construct a knot polynomial J : C[t, t 1]. (Isotopy) → How do we get started? Diagrams − Work with link diagrams look for J : C[t, t 1]. (R−moves) → Motivated by algebra: define; the map on the “generators” of a diagram.

Figure-eight knot Generators

I. Moffatt (South Alabama) UW 2011 7 / 39 Cutting down the number of generators

The large number of generators is making life difficult. Can we reduce their number?

Use the following redrawing of the figure eight knot.

Generators

Made out of

Figure-eight knot A standard position Fewer generators needed called a braid closure •

I. Moffatt (South Alabama) UW 2011 8 / 39 Braids and braid closures

All of the ‘interesting’ structure is contained in a part of a diagram called a braid:

interesting

boring

Braid Closure element Closure of a braid

A braid: is an intertwining strings attached to top and bottom "bars" such that each string never "turns back up":

I. Moffatt (South Alabama) UW 2011 9 / 39 Putting link diagrams in the standard form

Alexander’s Theorem: obtaining a braid from a link Choose a point X. Whenever an arc travels counter-clockwise pull it over the base point. Cut open the link to get a braid.

X X X

Theorem Every link diagram can be written as a braid closure.

I. Moffatt (South Alabama) UW 2011 10 / 39 Braids

Theorem Every link diagram can be written as a braid closure.

Rather than working with links, we can work with braids. Braids have only three generators:

Generators: A braid

We now have generators! If, we want to work with braids, we need to know: How do the generators generate braids? When are braids equivalent? When do braids represent the same links?

I. Moffatt (South Alabama) UW 2011 11 / 39 Braids

Theorem Every link diagram can be written as a braid closure.

Rather than working with links, we can work with braids. Braids have only three generators:

Generators: A braid

We now have generators! If, we want to work with braids, we need to know: How do the generators generate braids? When are braids equivalent? When do braids represent the same links?

I. Moffatt (South Alabama) UW 2011 11 / 39 Operations on braids

Composition Tensor product

σ σ σ = σ σ = σ σ σ ⊗ n strings n strings n strings n strings m strings (n+m) strings stack up place beside

I. Moffatt (South Alabama) UW 2011 12 / 39 Operations on braids

Composition Tensor product

σ σ σ = σ σ = σ σ σ ⊗ n strings n strings n strings n strings m strings (n+m) strings stack up place beside With these operations every braid can beGenerators: built from . • ⊗

⊗ =( ⊗ ) ( ⊗ ) ( ⊗ ) ( ⊗ ) ⊗

⊗ We now have generators and generating operations! , I. Moffatt (South Alabama) UW 2011 12 / 39 Braid equivalence

Different diagrams can represent the same braid:

= = = ....

Braids are considered up to the following moves

B-moves

= = = =

These can be written algebraically using “ ” and “ ”. ◦ ⊗ I. Moffatt (South Alabama) UW 2011 13 / 39 The Markov moves

We want to study links using braids, we need to know when braids represent the same link.

=

I. Moffatt (South Alabama) UW 2011 14 / 39 The Markov moves

We want to study links using braids, we need to know when braids represent the same link.

=

The Markov moves (M-moves)

σ σ σ σ σ = = = σ σ (n+1) strings n strings (n+1) strings MI-move MII-move

I. Moffatt (South Alabama) UW 2011 14 / 39 Markov’s Theorem

σ σ σ σ σ ======σ σ (n+1) strings n strings (n+1) strings MI-move MII-move

Markov’s Theorem Braids describe equal links related by B-moves and M-moves. ⇐⇒ Links Diagrams Braids = = (Isotopy) (R moves) (B-moves, M-moves) −

Sufficiency is easy, e.g.

σ σ σ = = σ = σ σ σ σ

Braids related by B-moves closures related by R-moves. Necessity is hard. ; I. Moffatt (South Alabama) UW 2011 15 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Summary

So far: Links − We want to construct J : C[q, q 1]. (Isotopy) → Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

Links Diagrams Braids We have seen (Isotopy) = (R−moves) = (B-moves,M-moves) . Braids − Thus enough to construct J : C[q, q 1] (B-moves,M-moves) → Easier as braids are generated by under and . ⊗ ◦

I. Moffatt (South Alabama) UW 2011 16 / 39 Construct mappings: Step 1

To build a linear map from a braid, we start by associating a linear

map to each generator .

Fix a vector space . V Associate a copy of to each strand of a generator. V Straight lines Crossings

But what should be the product “” of the vector spaces?

I. Moffatt (South Alabama) UW 2011 17 / 39 Construct mappings: Step 1

To build a linear map from a braid, we start by associating a linear

map to each generator .

Fix a vector space . V Associate a copy of to each strand of a generator. V Straight lines Crossings

V End( ) → 1 → ∈ V V But what should be the product “” of the vector spaces?

I. Moffatt (South Alabama) UW 2011 17 / 39 Construct mappings: Step 1

To build a linear map from a braid, we start by associating a linear

map to each generator .

Fix a vector space . V Associate a copy of to each strand of a generator. V Straight lines Crossings

V V V R End( )

End( ) → → 1 ∈ V ∈ V V → → V V V But what should be the product “” of the vector spaces?

I. Moffatt (South Alabama) UW 2011 17 / 39 Tensor products of vector spaces

a vector space with basis u ,..., un U { 1 } a vector space with basis w ,..., w m W { 1 } Definition of U ⊗ W Generated by all ordered pairs (u, w) . ∈ U ⊗ W Write the ordered pair (u, w) as u w. ⊗ Subject to the relations: k(u w) = (ku) w = u (kw) ⊗ ⊗ ⊗ u (w 1 + w 2) = (u w 1) + (u w 2) ⊗ ⊗ ⊗ (u1 + u2) w = (u1 w) + (u2 w) ⊗ ⊗ ⊗

Basis u w u and w . { ⊗ | ∈ U ∈ W} dim( ) = dim( ) dim( ) U ⊗ W U × W

I. Moffatt (South Alabama) UW 2011 18 / 39 Construct mappings: Step 1 - defining maps

We can now obtain linear maps from the braid generators: Straight lines Positive crossings Negative crossings

⊗ ⊗ V V V R V V S

→ → → 1 → → → V V ⊗ V V ⊗ V We can obtain a linear map T (σ): ⊗n ⊗n from every n-braid. V → V V ⊗ V ⊗ V R id

→ ⊗ ⊗ ⊗ ⊗ V V V id S → ⊗ ⊗ (R id) (id S) (R id) (id S) ⊗ ⊗ ⊗ ◦ ⊗ ◦ ⊗ ◦ ⊗ V V VR id ⊗ → ⊗ ⊗ ⊗ ⊗ V V Vid → ⊗ S V ⊗ V ⊗ V We can obtain a linear map T (σ): ⊗n ⊗n from every n-braid. V → V I. Moffatt (South Alabama) UW 2011 19 / 39 Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids End( ⊗n) → V What we actually need is a mapping T : n-Braids End( ⊗n) (B-moves) → V We need: (σ and σ0 related by B-moves) = T (σ) = T (σ0) ⇒ The B-moves give us conditions on R and S:

Theorem Let R be an invertible linear map that satisfies the Yang-Baxter equation. Then T is a braid invariant, i.e., T : n-Braids End( ⊗n). (B-moves) → V R is called an R-matrix • I. Moffatt (South Alabama) UW 2011 20 / 39 Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids End( ⊗n) → V What we actually need is a mapping T : n-Braids End( ⊗n) (B-moves) → V We need: (σ and σ0 related by B-moves) = T (σ) = T (σ0) ⇒ The B-moves give us conditions on R and S:

= = = =

(1 1) R ⊗ ◦ R S =1 1=S R (R 1) (1 R) (R 1) = R (1 1) ◦ ⊗ 1 ◦ ⊗ ◦ ⊗ ◦ ⊗ S = R− = (1 R) (R 1) (R 1) ◦ ⊗ ⇐⇒ ⊗ ◦ ⊗ ◦ ⊗ The Yang-Baxter equation

Theorem Let R be an invertible linear map that satisfies the Yang-Baxter equation. Then T is a braid invariant, i.e., T : n-Braids End( ⊗n). (B-moves) → V R is called an R-matrix • I. Moffatt (South Alabama) UW 2011 20 / 39 Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids End( ⊗n) → V What we actually need is a mapping T : n-Braids End( ⊗n) (B-moves) → V We need: (σ and σ0 related by B-moves) = T (σ) = T (σ0) ⇒ The B-moves give us conditions on R and S:

= = = =

(1 1) R ⊗ ◦ R S =1 1=S R (R 1) (1 R) (R 1) = R (1 1) ◦ ⊗ 1 ◦ ⊗ ◦ ⊗ ◦ ⊗ S = R− = (1 R) (R 1) (R 1) ◦ ⊗ ⇐⇒ ⊗ ◦ ⊗ ◦ ⊗ The Yang-Baxter equation

Theorem Let R be an invertible linear map that satisfies the Yang-Baxter equation. Then T is a braid invariant, i.e., T : n-Braids End( ⊗n). (B-moves) → V R is called an R-matrix • I. Moffatt (South Alabama) UW 2011 20 / 39 Finding some braid invariants: dim( ) = 1 V has basis v . V { } has basis v v . V ⊗ V { ⊗ } R is given by a 1 1 matrix [a]. × R is invertible when a = 0 6 R satisfies the Yang-Baxter equation. = R defines a braid invariant T . ⇒ R What is TR?

I. Moffatt (South Alabama) UW 2011 21 / 39 Finding some braid invariants: dim( ) = 1 V

has basis v . V { } has basis v v . V ⊗ V { ⊗ } R is given by a 1 1 matrix [a]. × R is invertible when a = 0 6 R satisfies the Yang-Baxter equation. = R defines a braid invariant T . ⇒ R What is TR?

...... = ⊗ ⊗ ⊗ ⊗ ⊗ ∼ V 1 ... V [1] V V[a] V[1] V [a] ... [a− ] ...... → → → ⊗ ···⊗ ⊗ ···⊗ ...... = V⊗ ⊗ V ⊗ V ⊗ ⊗ V ∼ V V Every positive crossing gives matrix [a]. Every negative crossing gives matrix [a−1]. Matrices commute. = T (σ) = a(#+crossings)−(#+crossings). ⇒ R

I. Moffatt (South Alabama) UW 2011 21 / 39 Finding some braid invariants: dim( ) = 2 V

has basis v , v . V { 0 1} has basis v v , v v , v v , v v . V ⊗ V { 0 ⊗ 0 0 ⊗ 1 1 ⊗ 0 1 ⊗ 1} R is given by a 4 4 matrix. × We need to find invertible 4 4 matrices that satisfy the Yang-Baxter equation. × Find conditions on R such that R is invertible; R satisfies the Yang-Baxter equation. • •

 a 0 0 0   a 0 0 0   a 0 0 0   0 b c 0   0 0 c 0   0 0 c 0  R =   R =   or R =    0 d e 0  1  0 d a − cd/a 0  2  0 d a − cd/a 0  0 0 0 f ; 0 0 0 a 0 0 0 −cd/a

Here we will focus on R1.(R2 has its own interesting theory.)

I. Moffatt (South Alabama) UW 2011 22 / 39 Computing some invariants

The construction The R-matrix

 a 0 0 0   0 0 c 0  ⊗ ⊗ R =   1  0 d a − cd/a 0  V V V R V VR− → → → 1 0 0 0 a → → → V V ⊗ V V ⊗ V

  R2 0 0 0  2   0 R 0 0  1 1 (R R) =  2  ⊗ ⊗ ◦  0 0 R 0  ; 0 0 0 R2

 aR 0 0 0   0 0 cR 0  R R =   ⊗  0 dR (a − cd/a)R 0  ; 0 0 0 aR The 2-dimensional invariant is stronger than the 1-dimensional invariant. This invariant is actually rather strong! I. Moffatt (South Alabama) UW 2011 23 / 39 Summary

Links − We want to construct J : C[q, q 1]. (Isotopy) → Links Braids (Isotopy) = (B-moves,M-moves) Braids − Enough to construct J : C[q, q 1] (B-moves,M-moves) → So far we have T : Braids ((set)) (B-moves) → We need to modify T so that it is invariant under the Markov moves

I. Moffatt (South Alabama) UW 2011 24 / 39 Summary

Links − We want to construct J : C[q, q 1]. (Isotopy) → Links Braids (Isotopy) = (B-moves,M-moves) Braids − Enough to construct J : C[q, q 1] (B-moves,M-moves) → So far we have T : Braids ((set)) (B-moves) → We need to modify T so that it is invariant under the Markov moves

I. Moffatt (South Alabama) UW 2011 24 / 39 Summary

Links − We want to construct J : C[q, q 1]. (Isotopy) → Links Braids (Isotopy) = (B-moves,M-moves) Braids − Enough to construct J : C[q, q 1] (B-moves,M-moves) → So far we have T : Braids ((set)) (B-moves) → We need to modify T so that it is invariant under the Markov moves

I. Moffatt (South Alabama) UW 2011 24 / 39 Summary

Links − We want to construct J : C[q, q 1]. (Isotopy) → Links Braids (Isotopy) = (B-moves,M-moves) Braids − Enough to construct J : C[q, q 1] (B-moves,M-moves) → So far we have T : Braids ((set)) (B-moves) → We need to modify T so that it is invariant under the Markov moves

I. Moffatt (South Alabama) UW 2011 24 / 39 Summary

Links − We want to construct J : C[q, q 1]. (Isotopy) → Links Braids (Isotopy) = (B-moves,M-moves) Braids − Enough to construct J : C[q, q 1] (B-moves,M-moves) → So far we have T : Braids ((set)) (B-moves) → We need to modify T so that it is invariant under the Markov moves

The Markov moves (M-moves)

σ σ σ σ σ = = = σ σ (n+1) strings n strings (n+1) strings MI-move MII-move

I. Moffatt (South Alabama) UW 2011 24 / 39 Tackling the first Markov move

MI-move

T (σ0 σ) = T (σ σ0) σ σ ◦ 6 ◦ = Need to define J(braid) := function of T (braid) σ σ with J(σ0 σ) = J(σ σ0) ◦ ◦

T (σ0 σ) = T (σ0) T (σ) = AB ◦ ◦ T (σ σ0)T (σ) T (σ0) = BA ◦ ◦ What operations on matrices have the property f (AB) = f (BA)?

I. Moffatt (South Alabama) UW 2011 25 / 39 Tackling the first Markov move

MI-move

T (σ0 σ) = T (σ σ0) σ σ ◦ 6 ◦ = Need to define J(braid) := function of T (braid) σ σ with J(σ0 σ) = J(σ σ0) ◦ ◦

T (σ0 σ) = T (σ0) T (σ) = AB ◦ ◦ T (σ σ0)T (σ) T (σ0) = BA ◦ ◦ What operations on matrices have the property f (AB) = f (BA)?

I. Moffatt (South Alabama) UW 2011 25 / 39 Tackling the first Markov move

MI-move

T (σ0 σ) = T (σ σ0) σ σ ◦ 6 ◦ = Need to define J(braid) := function of T (braid) σ σ with J(σ0 σ) = J(σ σ0) ◦ ◦

T (σ0 σ) = T (σ0) T (σ) = AB ◦ ◦ T (σ σ0)T (σ) T (σ0) = BA ◦ ◦ What operations on matrices have the property f (AB) = f (BA)? det(AB) = det(A) det(B) = det(B) det(A) = det(BA) Tr(AB) = Tr(BA) Which one should we use?

I. Moffatt (South Alabama) UW 2011 25 / 39 Thinking about the closure element

Associate a linear map to each piece.

⊗ Extra generators:

Associate maps to each generator.

∗ V suggests V id → → 1 → → ∗ V V The dual of a vector space

if vector space with basis v 1,..., v n V ∗ { ∗ } its dual, is the vector space := Hom( , C) of linear maps from toV the ground field. V V V 1 n i Its basis is v ,..., v where v (v ) = δ , { } j i j I. Moffatt (South Alabama) UW 2011 26 / 39 The other parts of the closure element

lines Caps Cups

∗ C ⊗ ∗ n V Vu V id → → → → → ∗ → ∗ V ⊗ C V V If T (σ): v v P T k,l v v i ⊗ j 7→ k,l i,j k ⊗ l 1 P v v v j v i 7→ i,j i ⊗ j ⊗ ⊗ 7→ P T k,l (v v ) v j v i i,j,k,l i,j k ⊗ l ⊗ ⊗ 7→ P T k,l v j (v ) v i (v ) = P T i,j = Tr[T k,l ] i,j,k,l i,j l ⊗ k i,j,k,l i,j i,j

I. Moffatt (South Alabama) UW 2011 27 / 39 The other parts of the closure element

lines Caps Cups C ∗ n ⊗ ∗ → V Vu V id → → → ∗ → V ⊗ V → ∗ n : v v j v j (v ) C V i ⊗ 7→ i If T (σ): v v P T k,l v v i ⊗ j 7→ k,l i,j k ⊗ l 1 P v v v j v i 7→ i,j i ⊗ j ⊗ ⊗ 7→ P T k,l (v v ) v j v i i,j,k,l i,j k ⊗ l ⊗ ⊗ 7→ P T k,l v j (v ) v i (v ) = P T i,j = Tr[T k,l ] i,j,k,l i,j l ⊗ k i,j,k,l i,j i,j

I. Moffatt (South Alabama) UW 2011 27 / 39 The other parts of the closure element

lines Caps Cups

C ⊗ ∗ ∗ n V Vu

→ V id → → → ⊗ ∗ → C → V j V j P i ∗ n : v v v (v ) u : 1 v i v V i ⊗ 7→ i 7→ i ⊗ If T (σ): v v P T k,l v v i ⊗ j 7→ k,l i,j k ⊗ l 1 P v v v j v i 7→ i,j i ⊗ j ⊗ ⊗ 7→ P T k,l (v v ) v j v i i,j,k,l i,j k ⊗ l ⊗ ⊗ 7→ P T k,l v j (v ) v i (v ) = P T i,j = Tr[T k,l ] i,j,k,l i,j l ⊗ k i,j,k,l i,j i,j

I. Moffatt (South Alabama) UW 2011 27 / 39 The other parts of the closure element

lines Caps Cups

C ⊗ ∗ ∗ n V Vu

→ V id → → → ⊗ ∗ → C → V j V j P i ∗ n : v v v (v ) u : 1 v i v V i ⊗ 7→ i 7→ i ⊗ If T (σ): v v P T k,l v v i ⊗ j 7→ k,l i,j k ⊗ l 1 P v v v j v i 7→ i,j i ⊗ j ⊗ ⊗ 7→ σ P T k,l (v v ) v j v i i,j,k,l i,j k ⊗ l ⊗ ⊗ 7→ P T k,l v j (v ) v i (v ) = P T i,j = Tr[T k,l ] i,j,k,l i,j l ⊗ k i,j,k,l i,j i,j So to get MI invariance, we should take the trace of T (σ).

I. Moffatt (South Alabama) UW 2011 27 / 39 Duals, tensors and the trace

Define J(σ) := Tr(T (σ)). σ σ J(σ0 σ) = J(σ σ0) = ◦ ◦ σ σ Invariant under MI-move. Problem: too strong a condition for the MII-move.

The fix

let µ : . V → V associate a copy of µ to the top of each braid Define J(σ) := Tr(T (σ) µ⊗n). ◦ Lemma if R±1 (µ µ) = (µ µ) R±1, then J(σ) := Tr(T (σ) µ⊗n) is invariant◦ under⊗ the MI-move.⊗ ◦ ◦

I. Moffatt (South Alabama) UW 2011 28 / 39 Duals, tensors and the trace

Define J(σ) := Tr(T (σ)). σ σ J(σ0 σ) = J(σ σ0) = ◦ ◦ σ σ Invariant under MI-move. Problem: too strong a condition for the MII-move.

The fix

let µ : . V → V σ associate a copy of µ to the top of each braid Define J(σ) := Tr(T (σ) µ⊗n). ◦ Lemma if R±1 (µ µ) = (µ µ) R±1, then J(σ) := Tr(T (σ) µ⊗n) is invariant◦ under⊗ the MI-move.⊗ ◦ ◦

I. Moffatt (South Alabama) UW 2011 28 / 39 The second Markov move

MII-move

J(σ) := Tr(T (σ) µ⊗n) σ σ σ ◦ = = We have R±1 (µ µ) = (µ µ) R±1 ◦ ⊗ ⊗ ◦ n µ µ µ µ µ µ µ µ T (σ) want T (σ) σ T (σ) n σ = µ µ µ µ µ = µ µ µ µ = T (σ) R R u u For M-II invariance

P k,l R : v i v j R v k v l n ⊗ 7→ k,l i,j ⊗ µ µ P j µ : v i j µ v j R = µ 7→ i i P k,l k l u We need µj = j,k,l,m Ri,j µmµj ±1 Tidy form: Tr2(R (µ µ)) = µ I. Moffatt (South Alabama) ◦ ⊗ UW 2011 29 / 39 The second Markov move

MII-move

J(σ) := Tr(T (σ) µ⊗n) σ σ σ ◦ = = We have R±1 (µ µ) = (µ µ) R±1 ◦ ⊗ ⊗ ◦ n µ µ µ µ µ µ µ µ T (σ) want T (σ) σ T (σ) n σ = µ µ µ µ µ = µ µ µ µ = T (σ) R R u u For M-II invariance

P k,l R : v i v j R v k v l n ⊗ 7→ k,l i,j ⊗ µ µ P j µ : v i j µ v j R = µ 7→ i i P k,l k l u We need µj = j,k,l,m Ri,j µmµj ±1 Tidy form: Tr2(R (µ µ)) = µ I. Moffatt (South Alabama) ◦ ⊗ UW 2011 29 / 39 Traces and operator traces

Recall that the trace of a matrix is the sum of its diagonal entries. Here’s what the trace really is: Theorem ∗ = Hom( , ) = linear maps from to . V ⊗ V ∼ V V V V f v (u f (u)v) h P hi v i v • ⊗ 7→ 7→ • 7→ i,j j ⊗ j Contraction ∗ κ : C : v f f (v) V ⊗ V → ⊗ 7→ The trace is the composite ∼ = ∗ contract Tr(F): Hom( , ) C V V −→ V ⊗ V −→ This idea extends to “partial traces”: Operator traces ∼ ∼ = ∗ = Tr2I.( MoffattF): Hom (South Alabama)( , ) ( ) UW 2011 30 / 39 V ⊗ VcontractV ⊗ V −→ V∼ ⊗ V ⊗ V −→ ∗ ( ∗ ) ∗ = Hom( , ) V ⊗ V ⊗ V ⊗ V −→ V ⊗ V −→ V V Traces and operator traces

Recall that the trace of a matrix is the sum of its diagonal entries. Here’s what the trace really is: Theorem ∗ = Hom( , ) = linear maps from to . V ⊗ V ∼ V V V V f v (u f (u)v) h P hi v i v • ⊗ 7→ 7→ • 7→ i,j j ⊗ j Contraction ∗ κ : C : v f f (v) V ⊗ V → ⊗ 7→ The trace is the composite ∼ = ∗ contract Tr(F): Hom( , ) C V V −→ V ⊗ V −→ This idea extends to “partial traces”: Operator traces ∼ ∼ = ∗ = Tr2I.( MoffattF): Hom (South Alabama)( , ) ( ) UW 2011 30 / 39 V ⊗ VcontractV ⊗ V −→ V∼ ⊗ V ⊗ V −→ ∗ ( ∗ ) ∗ = Hom( , ) V ⊗ V ⊗ V ⊗ V −→ V ⊗ V −→ V V Traces and operator traces

Recall that the trace of a matrix is the sum of its diagonal entries. Here’s what the trace really is: Theorem ∗ = Hom( , ) = linear maps from to . V ⊗ V ∼ V V V V f v (u f (u)v) h P hi v i v • ⊗ 7→ 7→ • 7→ i,j j ⊗ j Contraction ∗ κ : C : v f f (v) V ⊗ V → ⊗ 7→ The trace is the composite ∼ = ∗ contract Tr(F): Hom( , ) C V V −→ V ⊗ V −→ This idea extends to “partial traces”: Operator traces ∼ ∼ = ∗ = Tr2I.( MoffattF): Hom (South Alabama)( , ) ( ) UW 2011 30 / 39 V ⊗ VcontractV ⊗ V −→ V∼ ⊗ V ⊗ V −→ ∗ ( ∗ ) ∗ = Hom( , ) V ⊗ V ⊗ V ⊗ V −→ V ⊗ V −→ V V Traces and operator traces

Theorem ∗ = Hom( , ) = linear maps from to . V ⊗ V ∼ V V V V f v (u f (u)v) h P hi v i v • ⊗ 7→ 7→ • 7→ i,j j ⊗ j Contraction ∗ κ : C : v f f (v) V ⊗ V → ⊗ 7→ The trace is the composite ∼ = ∗ contract Tr(F): Hom( , ) C V V −→ V ⊗ V −→ This idea extends to “partial traces”: Operator traces ∼ ∼ = ∗ = Tr2(F): Hom( , ) ( ) V ⊗ VcontractV ⊗ V −→ V∼ ⊗ V ⊗ V −→ ∗ ( ∗ ) ∗ = Hom( , ) V ⊗ V ⊗ V ⊗ V −→ V ⊗ V −→ V V I. Moffatt (South Alabama) UW 2011 30 / 39 Putting it all together

Theorem For every R-matrix R End( ) and linear map µ End( ) such that ∈ V ⊗ V ∈ V R±1 (µ µ) = (µ µ) R±1 ◦ ⊗ ⊗ ◦ Tr (R±1 (µ µ)) = µ 2 ◦ ⊗ J(L) = Tr(T (σ) µ⊗n) is a link invariant. ◦ This theorem gives infinite families of knot invariants! , ,

I. Moffatt (South Alabama) UW 2011 31 / 39 The Jones polynomial

From our 4 4 R-matrix ×  a 0 0 0   q 0 0 0  2    0 0 c 0   0 0 q 0  q−1 0   R =   and µ =  0 d a − cd/a 0   0 q2 q − q3 0  0 q 0 0 0 a ; 0 0 0 q

The resulting link invariant J(L) is the Jones polynomial.

;

I. Moffatt (South Alabama) UW 2011 32 / 39 There has to be an easier way!

From our 4 4 R-matrix ×  q 0 0 0  2    0 0 q 0  q−1 0 R =   and µ =  0 q2 q − q3 0  0 q 0 0 0 q

Theorem The Jones polynomial satisfies the following relations: J( ) = q + q−1. O q−2J( ) q2J( ) = (q−1 q)J( ), where the− links are identical− except in the region shown.

Proof    q 0 0 0  q−1 0 0 0 2  −1 −3 −2  −2 2 −1 −2  0 0 q 0  2  0 q − q q 0  −1 q R − q R = q  2 3  − q  −2  = (q − q)I4  0 q q − q 0   0 q 0 0  0 0 0 q 0 0 0 q−1

I. Moffatt (South Alabama) UW 2011 33 / 39 A skein relation

The Skein formulation of the Jones polynomial The Jones polynomial is uniquely defined by the relations J( ) = q + q−1. O q−2J( ) q2J( ) = (q−1 q)J( ). − − q-2J( ) - q2 J( ) = (q -1 -q) J( ) 2 J( ) = (q -1 + q) J( ) = (q -1 + q)

    −1 5 9 5 3 J   = (q+q )(q +q), J   = q +q +q +q   −

I. Moffatt (South Alabama) UW 2011 34 / 39 Statistical mechanics and the Jones polynomial

Ice-type models are used in statistical physics to study the energy levels of crystal lattices with hydrogen bonds (such as ice!). The Jones polynomial can be written as an ice type model. The Jones polynomial can be written as an ice type model

2ω(L) P rot1(s)−rot0(s) Q ±1 J(L)(q) = q s q v Rv (s) The sum is over all 0, 1 -labelings of the arcs in a link { } ω(L) = (# +ve crossings) - (# -ve crossings) k l ±1 −2 j,i Rv (s) look at each crossing: , take q Rl,k entry of the i j R-matrix.

roti travel round the i-labelled curves, count the number of revolutions you make.

The state-sum just extracts the terms from Tr(T (σ) µ⊗n)!!! ◦

I. Moffatt (South Alabama) UW 2011 35 / 39 The Kauffman bracket

The Potts model are used in statistical physics to study magnetism and other phenomena of solid state physics. Under certain circumstances, an ice-type model can be written as a Potts model. (This is the Fortuin-Kasteleyn representation.) Works for the Jones polynomial and gives rise to:

The Kauffman bracket

-1 < > = A < >+A < > L = ( A2 A−2) L hO ∪ i − − h i = 1 hOi J(L) = (q + q−1) ( A)−3ω(L) L
− h i A=iq−1/2

I. Moffatt (South Alabama) UW 2011 36 / 39 The Kauffman bracket

The Kauffman bracket

-1 < > = A < >+A < > L = ( A2 A−2) L hO ∪ i − − h i = 1 hOi J(L) = (q + q−1) ( A)−3ω(L) L
− h i A=iq−1/2

-1 2 < >=A< > +A < > =A < >+< > 2 -2 2 -2 +< >+A-2 < > = (A +A )< > + 2 < > = -A -A   J = (q+q−1) ( A)−6( A2 A−2))
= (q+q−1)(q5+q) − − − A=iq−1/2

I. Moffatt (South Alabama) UW 2011 36 / 39 The bigger picture: quantum groups

Definition: Lie algebra Vector space L Map [ , ]: L L called the Lie bracket. ⊗ → C [x, y] = [y, x] and [x, [y, z]] + [y, [z, x]] + [z[x, y]] = 0 −

The Lie algebra sl2

 0 1   0 0   1 0  X = Y = H = Generators: 0 0 1 0 0 −1 [A, B] = AB BA = [X, Y ] = H, [H, X] = 2X, [H, Y ] = 2Y . − ⇒ − This is the 2-dimensional representation of sl (as 2 2 matrices) 2 ×

The Quantum group Uh(sl2) Generators: X, Y , H. h a formal parameter (Planck’s constant) [H, X] = 2X, [H, Y ] = 2Y , [X, Y ] = sinh(hH/2) , − sinh(h/2)

I. Moffatt (South Alabama) UW 2011 37 / 39 The bigger picture: quantum groups

Lie algebras give rise to quantum groups n-dimensional representations of Lie algebras (i.e. realize as n n matrices) n-dimensional representations of quantum groups× (i.e. realization; as n n matrices) × Quantum groups give rise to R-matrices and knot invariants!

sl as a 2 2 matrix Jones polynomial 2 × sl as a n n matrix ; Coloured Jones polynomial 2 × sln as a n n matrix ; Homfly polynomial × son as a n n matrix; Kauffman polynomial × ; The study of such invariants is called quantum topology. It is an active and far-reaching area of current research in math.

I. Moffatt (South Alabama) UW 2011 38 / 39 A question to finish with

J( ) = q + q−1 O J( ) = (q + q−1)n O ∪ · · · ∪ O

both have Jones polynomial

J( ) = (q + q−1)2. OO So J(L) = (q + q−1)2 does not imply L = O ∪ O In fact many links with Jones polynomial of an unlink. ∞ Question: Is there a non-trivial knot with J(L) = q + q−1? ? In other words, J(L) = q + q−1 = L = . ⇒ O

I. Moffatt (South Alabama) UW 2011 39 / 39 A question to finish with

J( ) = q + q−1 O J( ) = (q + q−1)n O ∪ · · · ∪ O

both have Jones polynomial

J( ) = (q + q−1)2. OO So J(L) = (q + q−1)2 does not imply L = O ∪ O In fact many links with Jones polynomial of an unlink. ∞ Question: Is there a non-trivial knot with J(L) = q + q−1? ? In other words, J(L) = q + q−1 = L = . ⇒ O

I. Moffatt (South Alabama) UW 2011 39 / 39 A question to finish with

J( ) = q + q−1 O J( ) = (q + q−1)n O ∪ · · · ∪ O

both have Jones polynomial

J( ) = (q + q−1)2. OO So J(L) = (q + q−1)2 does not imply L = O ∪ O In fact many links with Jones polynomial of an unlink. ∞ Question: Is there a non-trivial knot with J(L) = q + q−1? ? In other words, J(L) = q + q−1 = L = . ⇒ O The answer is unknown!

I. Moffatt (South Alabama) UW 2011 39 / 39