The Jones polynomial through linear algebra Iain Moffatt University - - PowerPoint PPT Presentation

SLIDE 1

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

SLIDE 2

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

SLIDE 3

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

SLIDE 4

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

• r not two links are isotopic.

?

=

?

= To do this we need knot invariants: F :

Links

→ (a set) such

• I. Moffatt

(South Alabama) UW 2011 3 / 39

SLIDE 5

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

• r not two links are isotopic.

?

=

?

= To do this we need knot invariants: F :

Links (Isotopy) → (a set) such

that F(L) = F(L′) = ⇒ L = L′, Aim: construct link invariants using linear algebra.

• I. Moffatt

(South Alabama) UW 2011 3 / 39

SLIDE 6

Our toolkit: Linear algebra 1

The basics

V a vector space over C with basis {v1, . . . , vn}. v ∈ V ⇐ ⇒ v = n

i=1 aivi.

f : V → V ⇐ ⇒ v = n

i=1 f j i vi.

f linear ↔ matrix

     f 1

1

· · · f n

1

. . . . . . f 1

n

. . . f n

n

    

The direct product

V × V = {(u, v) | u, bv ∈ V} λ(u, v) + (u′, v′) = (λu + u′, λv + v′)

The dual V∗

V∗ = Hom(V, C) = {f : V → C | f linear} basis {vi | vi(vj) = δi,j, i = 1, . . . , n}

• I. Moffatt

(South Alabama) UW 2011 4 / 39

SLIDE 7

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

project onto plane

Reidemeister moves = = =

RI RII RIII

Theorem

Links (Isotopy) = Diagrams (R − moves)

• I. Moffatt

(South Alabama) UW 2011 5 / 39

SLIDE 8

An example

[link]

• I. Moffatt

(South Alabama) UW 2011 6 / 39

SLIDE 9

The first idea: let’s form a map

Our goal

We want to construct a knot polynomial J :

Links (Isotopy) → C[t, t−1].

How do we get started?

Work with link diagrams look for J :

Diagrams (R−moves) → C[t, t−1].

Motivated by algebra: define the map on the “generators” of a diagram.

• I. Moffatt

(South Alabama) UW 2011 7 / 39

SLIDE 10

The first idea: let’s form a map

Our goal

We want to construct a knot polynomial J :

Links (Isotopy) → C[t, t−1].

How do we get started?

Work with link diagrams look for J :

Diagrams (R−moves) → C[t, t−1].

Motivated by algebra: define the map on the “generators” of a diagram.

Generators

Figure-eight knot

• I. Moffatt

(South Alabama) UW 2011 7 / 39

SLIDE 11

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

SLIDE 12

Braids and braid closures

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

Braid

Closure element

Closure of a braid interesting

b

• r

i n g

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

SLIDE 13

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

SLIDE 14

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

SLIDE 15

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

SLIDE 16

Operations on braids

Composition

σ σ′ =

σ σ′

n strings n strings n strings

stack up

Tensor product

σ

⊗

σ′ =

n strings m strings (n+m) strings

σ σ′

place beside

• I. Moffatt

(South Alabama) UW 2011 12 / 39

SLIDE 17

Operations on braids

Composition

σ σ′ =

σ σ′

n strings n strings n strings

stack up

Tensor product

σ

⊗

σ′ =

n strings m strings (n+m) strings

σ σ′

place beside

• With these operations every braid can be built from
• rs:

. ⊗ ⊗ ⊗ ⊗

⊗

( )

⊗ ⊗ ⊗

( ) ( ) ( )

=

We now have generators and generating operations!

• I. Moffatt

(South Alabama) UW 2011 12 / 39

SLIDE 18

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

SLIDE 19

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

SLIDE 20

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

=

MI-move

σ′ σ′ σ σ

n strings (n+1) strings

σ σ

= =

(n+1) strings

MII-move

• I. Moffatt

(South Alabama) UW 2011 14 / 39

SLIDE 21

Markov’s Theorem

= = = =

σ

=

MI-move

σ′ σ′ σ σ

n strings (n+1) strings

σ σ

= =

(n+1) strings

MII-move

Markov’s Theorem

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

SLIDE 22

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid: We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and ◦.

• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 23

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid: We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and ◦.

• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 24

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and ◦.

• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 25

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and ◦.

• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 26

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and ◦.

• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 27

Summary

So far:

We want to construct J :

Links (Isotopy) → C[q, q−1].

Too hard! Let’s make it easier. We have shown that every link can be represented by a braid:

X X X

We have seen

Links (Isotopy) = Diagrams (R−moves) = Braids (B-moves,M-moves).

Thus enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

Easier as braids are generated by under ⊗ and

• .
• I. Moffatt

(South Alabama) UW 2011 16 / 39

SLIDE 28

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 V to each strand of a generator.

Straight lines Crossings

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

• I. Moffatt

(South Alabama) UW 2011 17 / 39

SLIDE 29

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 V to each strand of a generator.

Straight lines

V

→

V

→

∈ End(V)

1

Crossings

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

• I. Moffatt

(South Alabama) UW 2011 17 / 39

SLIDE 30

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 V to each strand of a generator.

Straight lines

V

→

V

→

∈ End(V)

1

Crossings

• ∈ End(V V)

R

V

→

V

→

• V

V

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

• I. Moffatt

(South Alabama) UW 2011 17 / 39

SLIDE 31

Tensor products of vector spaces

U a vector space with basis {u1, . . . , un} W a vector space with basis {w1, . . . , wm}

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 ⊗ (w1 + w2) = (u ⊗ w1) + (u ⊗ w2) (u1 + u2) ⊗ w = (u1 ⊗ w) + (u2 ⊗ w)

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

• I. Moffatt

(South Alabama) UW 2011 18 / 39

SLIDE 32

Construct mappings: Step 1 - defining maps

We can now obtain linear maps from the braid generators:

Straight lines

V

→

V

→

1

Positive crossings R

V

→

V

→

V V

⊗ ⊗

Negative crossings

V

→

V

→

V V

⊗ ⊗

S

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

⊗ ⊗ ⊗ ⊗

V V

⊗

V

⊗

→

V V

⊗

V

⊗

→

V V

⊗

V

⊗

V V

⊗

V

⊗

→

V V

⊗

V

⊗

→

id

S

⊗

id

S

⊗

R⊗id R ⊗id

(R ⊗ id) ◦ (id ⊗ S) ◦ (R ⊗ id) ◦ (id ⊗ S)

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

• I. Moffatt

(South Alabama) UW 2011 19 / 39

SLIDE 33

Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids → End(V⊗n) What we actually need is a mapping T :

n-Braids (B-moves) → End(V⊗n)

We need: (σ and σ′ related by B-moves) = ⇒ T(σ) = T(σ′) 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 (B-moves) → End(V⊗n).

• R is called an R-matrix
• I. Moffatt

(South Alabama) UW 2011 20 / 39

SLIDE 34

Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids → End(V⊗n) What we actually need is a mapping T :

n-Braids (B-moves) → End(V⊗n)

We need: (σ and σ′ related by B-moves) = ⇒ T(σ) = T(σ′) The B-moves give us conditions on R and S:

= = = =

⇐ ⇒ S = R−1 (R ⊗ 1) ◦ (1 ⊗ R) ◦ (R ⊗ 1) = (1 ⊗ R) ◦ (R ⊗ 1) ◦ (R ⊗ 1) (1 ⊗ 1) ◦ R = R ◦ (1 ⊗ 1)

R ◦ S = 1 ⊗ 1 = S ◦ R

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 (B-moves) → End(V⊗n).

• R is called an R-matrix
• I. Moffatt

(South Alabama) UW 2011 20 / 39

SLIDE 35

Construct mappings: Step 2 - braid invariance

So far we have constructed a map T : n-Braids → End(V⊗n) What we actually need is a mapping T :

n-Braids (B-moves) → End(V⊗n)

We need: (σ and σ′ related by B-moves) = ⇒ T(σ) = T(σ′) The B-moves give us conditions on R and S: = = = =

⇐ ⇒ S = R−1 (R ⊗ 1) ◦ (1 ⊗ R) ◦ (R ⊗ 1) = (1 ⊗ R) ◦ (R ⊗ 1) ◦ (R ⊗ 1) (1 ⊗ 1) ◦ R = R ◦ (1 ⊗ 1)

R ◦ S = 1 ⊗ 1 = S ◦ R

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 (B-moves) → End(V⊗n).

• R is called an R-matrix
• I. Moffatt

(South Alabama) UW 2011 20 / 39

SLIDE 36

Finding some braid invariants: dim(V) = 1

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

• I. Moffatt

(South Alabama) UW 2011 21 / 39

SLIDE 37

Finding some braid invariants: dim(V) = 1

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

... ...

→

V V

⊗

V

⊗

...

⊗ ⊗

...

⊗

V V V

⊗

V

⊗

...

⊗ ⊗

...

⊗

V

[1] ⊗ · · · ⊗ [a] ⊗ · · · ⊗ [1]

∼ = ∼ =

V V

→

[a]

[a−1]

... ...

V V

→ Every positive crossing gives matrix [a]. Every negative crossing gives matrix [a−1]. Matrices commute. = ⇒ TR(σ) = a(#+crossings)−(#+crossings).

• I. Moffatt

(South Alabama) UW 2011 21 / 39

SLIDE 38

Finding some braid invariants: dim(V) = 2

V has basis {v0, v1}. V ⊗ V has basis {v0 ⊗ v0, v0 ⊗ v1, v1 ⊗ v0, v1 ⊗ v1}. 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.

R =     a b c d e f     R1 =     a c d a − cd/a a     or R2 =     a c d a − cd/a −cd/a    

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

• I. Moffatt

(South Alabama) UW 2011 22 / 39

SLIDE 39

Computing some invariants

The R-matrix

R =     a c d a − cd/a a    

The construction

V

→

V

→

1

R

V

→

V

→

V V

⊗ ⊗

V

→

V

→

V V

⊗ ⊗

R−1 1 ⊗ 1 ⊗ (R ◦ R) =

     R2 R2 R2 R2     

R ⊗ R =

    aR cR dR (a − cd/a)R 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

SLIDE 40

Summary

We want to construct J :

Links (Isotopy) → C[q, q−1]. Links (Isotopy) = Braids (B-moves,M-moves)

Enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

So far we have T :

Braids (B-moves) → ((set))

We need to modify T so that it is invariant under the Markov moves

• I. Moffatt

(South Alabama) UW 2011 24 / 39

SLIDE 41

Summary

We want to construct J :

Links (Isotopy) → C[q, q−1]. Links (Isotopy) = Braids (B-moves,M-moves)

Enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

So far we have T :

Braids (B-moves) → ((set))

We need to modify T so that it is invariant under the Markov moves

• I. Moffatt

(South Alabama) UW 2011 24 / 39

SLIDE 42

Summary

We want to construct J :

Links (Isotopy) → C[q, q−1]. Links (Isotopy) = Braids (B-moves,M-moves)

Enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

So far we have T :

Braids (B-moves) → ((set))

We need to modify T so that it is invariant under the Markov moves

• I. Moffatt

(South Alabama) UW 2011 24 / 39

SLIDE 43

Summary

We want to construct J :

Links (Isotopy) → C[q, q−1]. Links (Isotopy) = Braids (B-moves,M-moves)

Enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

So far we have T :

Braids (B-moves) → ((set))

We need to modify T so that it is invariant under the Markov moves

• I. Moffatt

(South Alabama) UW 2011 24 / 39

SLIDE 44

Summary

We want to construct J :

Links (Isotopy) → C[q, q−1]. Links (Isotopy) = Braids (B-moves,M-moves)

Enough to construct J :

Braids (B-moves,M-moves) → C[q, q−1]

So far we have T :

Braids (B-moves) → ((set))

We need to modify T so that it is invariant under the Markov moves

The Markov moves (M-moves)

σ

=

MI-move

σ′ σ′ σ σ

n strings (n+1) strings

σ σ

= =

(n+1) strings

MII-move

• I. Moffatt

(South Alabama) UW 2011 24 / 39

SLIDE 45

Tackling the first Markov move

MI-move

σ

=

σ′ σ′ σ

T(σ′ ◦ σ) = T(σ ◦ σ′) Need to define J(braid) := function of T(braid) with J(σ′ ◦ σ) = J(σ ◦ σ′) T(σ′ ◦ σ) = T(σ′) ◦ T(σ) = AB T(σ ◦ σ′)T(σ) ◦ T(σ′) = BA What operations on matrices have the property f(AB) = f(BA)?

• I. Moffatt

(South Alabama) UW 2011 25 / 39

SLIDE 46

Tackling the first Markov move

MI-move

σ

=

σ′ σ′ σ

T(σ′ ◦ σ) = T(σ ◦ σ′) Need to define J(braid) := function of T(braid) with J(σ′ ◦ σ) = J(σ ◦ σ′) T(σ′ ◦ σ) = T(σ′) ◦ T(σ) = AB T(σ ◦ σ′)T(σ) ◦ T(σ′) = BA What operations on matrices have the property f(AB) = f(BA)?

• I. Moffatt

(South Alabama) UW 2011 25 / 39

SLIDE 47

Tackling the first Markov move

MI-move

σ

=

σ′ σ′ σ

T(σ′ ◦ σ) = T(σ ◦ σ′) Need to define J(braid) := function of T(braid) with J(σ′ ◦ σ) = J(σ ◦ σ′) T(σ′ ◦ σ) = T(σ′) ◦ T(σ) = AB T(σ ◦ σ′)T(σ) ◦ T(σ′) = 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

SLIDE 48

Thinking about the closure element

⊗

Associate a linear map to each piece. Extra generators: Associate maps to each generator.

V

→

V

→

1

suggests

→ →

id

V∗ V∗

The dual of a vector space

if V vector space with basis {v1, . . . , vn} its dual, V∗ is the vector space V∗ := Hom(V, C) of linear maps from V to the ground field. Its basis is {v1, . . . , vn} where vi(vj) = δi,j

• I. Moffatt

(South Alabama) UW 2011 26 / 39

SLIDE 49

The other parts of the closure element

lines

→ →

id

V∗ V∗

Caps

V

→ →

⊗V∗

n C

Cups

V

→ →

⊗V∗

u C

If T(σ) : vi ⊗ vj →

k,l T k,l i,j vk ⊗ vl

1 →

i,j vi ⊗ vj ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j (vk ⊗ vl) ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j vj(vl)⊗vi(vk) = i,j,k,l T i,j i,j = Tr[T k,l i,j ]

• I. Moffatt

(South Alabama) UW 2011 27 / 39

SLIDE 50

The other parts of the closure element

lines

→ →

id

V∗ V∗

Caps

V

→ →

⊗V∗

n C

n : vi ⊗ vj → vj(vi)

Cups

V

→ →

⊗V∗

u C

If T(σ) : vi ⊗ vj →

k,l T k,l i,j vk ⊗ vl

1 →

i,j vi ⊗ vj ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j (vk ⊗ vl) ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j vj(vl)⊗vi(vk) = i,j,k,l T i,j i,j = Tr[T k,l i,j ]

• I. Moffatt

(South Alabama) UW 2011 27 / 39

SLIDE 51

The other parts of the closure element

lines

→ →

id

V∗ V∗

Caps

V

→ →

⊗V∗

n C

n : vi ⊗ vj → vj(vi)

Cups

V

→ →

⊗V∗

u C

u : 1 →

i vi ⊗ vi

If T(σ) : vi ⊗ vj →

k,l T k,l i,j vk ⊗ vl

1 →

i,j vi ⊗ vj ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j (vk ⊗ vl) ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j vj(vl)⊗vi(vk) = i,j,k,l T i,j i,j = Tr[T k,l i,j ]

• I. Moffatt

(South Alabama) UW 2011 27 / 39

SLIDE 52

The other parts of the closure element

lines

→ →

id

V∗ V∗

Caps

V

→ →

⊗V∗

n C

n : vi ⊗ vj → vj(vi)

Cups

V

→ →

⊗V∗

u C

u : 1 →

i vi ⊗ vi

σ

If T(σ) : vi ⊗ vj →

k,l T k,l i,j vk ⊗ vl

1 →

i,j vi ⊗ vj ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j (vk ⊗ vl) ⊗ vj ⊗ vi →

• i,j,k,l T k,l

i,j vj(vl)⊗vi(vk) = i,j,k,l T i,j i,j = Tr[T k,l i,j ]

So to get MI invariance, we should take the trace of T(σ).

• I. Moffatt

(South Alabama) UW 2011 27 / 39

SLIDE 53

Duals, tensors and the trace

σ

=

σ′ σ′ σ

Define J(σ) := Tr(T(σ)). J(σ′ ◦ σ) = J(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

SLIDE 54

Duals, tensors and the trace

σ

=

σ′ σ′ σ

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

The fix

σ

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

SLIDE 55

The second Markov move

MII-move

σ σ σ

= =

J(σ) := Tr(T(σ) ◦ µ⊗n) We have R±1 ◦ (µ ⊗ µ) = (µ ⊗ µ) ◦ R±1 σ

R n u µ µ µ µ µ µ µ µ µ µ µ

T(σ)

R n u µ µ µ µ µ µ µ µ

T(σ)

µ µ µ µ µ µ µ µ

T(σ)

µ µ µ µ µ µ µ µ µ µ

T(σ)

σ

= = =

want

µ µ

For M-II invariance

R n u µ µ µ = µ µ µ µ µ µ R : vi ⊗ vj →

k,l Rk,l i,j vk ⊗ vl

µ : vi →

j µj ivj

We need µi

j = j,k,l,m Rk,l i,j µk mµl j

Tidy form: Tr2(R±1 ◦ (µ ⊗ µ)) = µ

• I. Moffatt

(South Alabama) UW 2011 29 / 39

SLIDE 56

The second Markov move

MII-move

σ σ σ

= =

J(σ) := Tr(T(σ) ◦ µ⊗n) We have R±1 ◦ (µ ⊗ µ) = (µ ⊗ µ) ◦ R±1 σ

R n u µ µ µ µ µ µ µ µ µ µ µ

T(σ)

R n u µ µ µ µ µ µ µ µ

T(σ)

µ µ µ µ µ µ µ µ

T(σ)

µ µ µ µ µ µ µ µ µ µ

T(σ)

σ

= = =

want

µ µ

For M-II invariance

R n u µ µ µ = µ µ µ µ µ µ R : vi ⊗ vj →

k,l Rk,l i,j vk ⊗ vl

µ : vi →

j µj ivj

We need µi

j = j,k,l,m Rk,l i,j µk mµl j

Tidy form: Tr2(R±1 ◦ (µ ⊗ µ)) = µ

• I. Moffatt

(South Alabama) UW 2011 29 / 39

SLIDE 57

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

V∗ ⊗ V ∼ = Hom(V, V) = linear maps from V to V.

• f ⊗ v → (u → f(u)v)
• h →

i,j hi jvi ⊗ vj

Contraction

κ : V ⊗ V∗ → C : v ⊗ f → f(v)

The trace is the composite

Tr(F) : Hom(V, V)

∼ =

− → V∗ ⊗ V contract − → C This idea extends to “partial traces”:

Operator traces

Tr2(F) : Hom(V ⊗ V, V ⊗ V)

∼ =

− → (V ⊗ V)∗ ⊗ V

∼ =

− →

∗ ∗ contract ∗ ∼ = Hom

• I. Moffatt

(South Alabama) UW 2011 30 / 39

SLIDE 58

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

V∗ ⊗ V ∼ = Hom(V, V) = linear maps from V to V.

• f ⊗ v → (u → f(u)v)
• h →

i,j hi jvi ⊗ vj

Contraction

κ : V ⊗ V∗ → C : v ⊗ f → f(v)

The trace is the composite

Tr(F) : Hom(V, V)

∼ =

− → V∗ ⊗ V contract − → C This idea extends to “partial traces”:

Operator traces

Tr2(F) : Hom(V ⊗ V, V ⊗ V)

∼ =

− → (V ⊗ V)∗ ⊗ V

∼ =

− →

∗ ∗ contract ∗ ∼ = Hom

• I. Moffatt

(South Alabama) UW 2011 30 / 39

SLIDE 59

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

V∗ ⊗ V ∼ = Hom(V, V) = linear maps from V to V.

• f ⊗ v → (u → f(u)v)
• h →

i,j hi jvi ⊗ vj

Contraction

κ : V ⊗ V∗ → C : v ⊗ f → f(v)

The trace is the composite

Tr(F) : Hom(V, V)

∼ =

− → V∗ ⊗ V contract − → C This idea extends to “partial traces”:

Operator traces

Tr2(F) : Hom(V ⊗ V, V ⊗ V)

∼ =

− → (V ⊗ V)∗ ⊗ V

∼ =

− →

∗ ∗ contract ∗ ∼ = Hom

• I. Moffatt

(South Alabama) UW 2011 30 / 39

SLIDE 60

Traces and operator traces

Theorem

V∗ ⊗ V ∼ = Hom(V, V) = linear maps from V to V.

• f ⊗ v → (u → f(u)v)
• h →

i,j hi jvi ⊗ vj

Contraction

κ : V ⊗ V∗ → C : v ⊗ f → f(v)

The trace is the composite

Tr(F) : Hom(V, V)

∼ =

− → V∗ ⊗ V contract − → C This idea extends to “partial traces”:

Operator traces

Tr2(F) : Hom(V ⊗ V, V ⊗ V)

∼ =

− → (V ⊗ V)∗ ⊗ V

∼ =

− → V∗ ⊗ (V∗ ⊗ V) ⊗ V contract − → V∗ ⊗ V

∼ =

− → Hom(V, V)

• I. Moffatt

(South Alabama) UW 2011 30 / 39

SLIDE 61

Putting it all together

Theorem

For every R-matrix R ∈ End(V ⊗ V) and linear map µ ∈ End(V) such that R±1 ◦ (µ ⊗ µ) = (µ ⊗ µ) ◦ R±1 Tr2(R±1 ◦ (µ ⊗ µ)) = µ 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

SLIDE 62

The Jones polynomial

From our 4 × 4 R-matrix

    a c d a − cd/a a     R =     q q2 q2 q − q3 q     and µ =

• q−1

q

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

(South Alabama) UW 2011 32 / 39

SLIDE 63

There has to be an easier way!

From our 4 × 4 R-matrix

R =     q q2 q2 q − q3 q     and µ =

• q−1

q

• Theorem

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

Proof

q−2R − q2R−1 = q−2     q q2 q2 q − q3 q     − q2      q−1 q−1 − q−3 q−2 q−2 q−1      = (q−1 − q)I4

• I. Moffatt

(South Alabama) UW 2011 33 / 39

SLIDE 64

A skein relation

The Skein formulation of the Jones polynomial

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

J(

)

q2

• 2
• 1

J(

)

J(

)

q q q

• =

+

( )

• J(

)

=

• 1

J(

)

q q

( )

+ =

• 1

q q

( )

2

J       = (q+q−1)(q5+q), J     = −q9+q5+q3+q

• I. Moffatt

(South Alabama) UW 2011 34 / 39

SLIDE 65

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

J(L)(q) = q2ω(L)

s qrot1(s)−rot0(s) v R±1 v (s)

The sum is over all {0, 1}-labelings of the arcs in a link ω(L) = (# +ve crossings) - (# -ve crossings) R±1

v (s) look at each crossing:

i j k l

, take q−2Rj,i

l,k entry of the

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

SLIDE 66

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

A =

< > < > < >

+A

• 1

O ∪ L = (−A2 − A−2)L O = 1 J(L) = (q + q−1)

• (−A)−3ω(L)L
• A=iq−1/2
• I. Moffatt

(South Alabama) UW 2011 36 / 39

SLIDE 67

The Kauffman bracket

The Kauffman bracket

A =

< > < > < >

+A

• 1

O ∪ L = (−A2 − A−2)L O = 1 J(L) = (q + q−1)

• (−A)−3ω(L)L
• A=iq−1/2

=A +A

• 1

< > < > < >

=A +

< > < >

2

+<

>

A + <

>

• 2

=

< > < >

(A +A )

2

• 2

+ 2 = -A -A

2

• 2

J

• = (q+q−1)
• (−A)−6(−A2 − A−2))
• A=iq−1/2 = (q+q−1)(q5+q)
• I. Moffatt

(South Alabama) UW 2011 36 / 39

SLIDE 68

The bigger picture: quantum groups

Definition: Lie algebra

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

The Lie algebra sl2

Generators: X =

• 1
• Y =
• 1
• H =
• 1

−1

• [A, B] = AB − BA =

⇒ [X, Y] = H, [H, X] = 2X, [H, Y] = −2Y. This is the 2-dimensional representation of sl2 (as 2 × 2 matrices)

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

SLIDE 69

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! sl2 as a 2 × 2 matrix Jones polynomial sl2 as a n × n matrix Coloured Jones polynomial 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

SLIDE 70

A question to finish with

J(O) = q + q−1 J(O ∪ · · · ∪ O) = (q + q−1)n both have Jones polynomial J(OO) = (q + q−1)2. 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

SLIDE 71

A question to finish with

J(O) = q + q−1 J(O ∪ · · · ∪ O) = (q + q−1)n both have Jones polynomial J(OO) = (q + q−1)2. 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

SLIDE 72

A question to finish with

J(O) = q + q−1 J(O ∪ · · · ∪ O) = (q + q−1)n both have Jones polynomial J(OO) = (q + q−1)2. 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