Definitions

Question 1

Unit Circle

Answer 1

S^1 = {z ∈ C : |z| = 1}

Question 2

f: S^1 → R^3 is smooth

Answer 2

Let f: S^1 → R^3 be a map.
We call f smooth if all its derivatives exist. More precisely, all derivatives of the map g: R → R^3, g(t) = f(exp(2πit)) exist.

Question 3

f: S^1 → R^3 is immersive

Answer 3

f is immersive if it’s smooth and g’(t) =/= 0 for all t ∈ R.

Where g: R → R^3, g(t) = f(exp(2πit))

Question 4

f: S^1 → R^3 is an embedding.

Answer 4

f is an embedding if it is smooth, injective and immersive. (It’s image is a knot).

Question 5

Knot

Answer 5

The image of an embedding f: S^1 → R^3.

Question 6

A link

Answer 6

A link is a union of finitely many disjoint knots.

Question 7

A component of a link

Answer 7

One of the knots in the finite union.

Question 8

A Projection

Answer 8

p_12:R^3 → R^2, (x,y,z)→(x,y)

Question 9

Curves in general position (Shadows)

Answer 9

Let d: S^1 × {1, . . . , k} → R2.
d and its image are in general position (shadows) if:
(1) d is immersive
(2) d^-1 (a) ≤ 2 for all a ∈ R^2
(3) Use d′(exp(2πit), i) as short-hand for (d/dt) d(exp(2πit), i) where 1 ≤ i ≤ k.
Let x, y ∈ S^1 × {1, . . . , k} be distinct such that d(x) = d(y). Then d′(x), d′(y) are linearly independent over R.

Question 10

Topological knot

Answer 10

Let f: S^1 → R^3 be a continuous injective map, then it’s image is a topological map.

Question 11

Crossing / Ordinary double point of d

Answer 11

Let d: S^1 × {1, . . . , k} → R^2 b e in general position.
If d(x)=d(y) and x =/= y then d(x) is a crossing of d.

Question 12

A knot diagram.

Answer 12

A knot diagram is a pair (d, A), d: S^1 × {1, . . . , k} → R^2 in general position, with under/over data A.
A chooses one of {x,y} if d(x)=d(y), x=/=y.

[ If f:S^1→ R^3 is an embedding, then p_12 ◦ f in general position gives rise to a knot diagram (d, A) where d = p_12 ◦ f and the over/under data A picks x or y according to the greater among p3 ◦ f(x) and p3 ◦ f(y). ]

Question 13

Orientation Preserving

Answer 13

r: R^3 → R^3 is orientation preserving if if the determinant of the jacobian (∂r_j/∂x_i) is positive everywhere.

Question 14

A diffeomorphism

Answer 14

A diffeomorphism is a smooth bijection whose inverse is smooth.

Question 15

Isotopic (horrible definition)

Answer 15

Let f , g: S^1 → R3 be embeddings. The following are equivalent:

1) There exists a continuous map h: [0, 1] × S^1 → R^3
such that the map h_t: S^1 → R^3, h_t(x) = h(t, x) is an embedding for all t ∈ [0, 1], and (f , g) = (h_0, h_1).

2) There exists an orientation preserving diffeomorphism r: R^3 → R^3 that by restriction defines an orientation preserving bijection f(S^1) → g(S^1).

If these equivalent conditions are satisfied then we say that the oriented knots f(S^1) and g(S^1) (with the standard orientation) are isotopic. We also say that f and g are isotopic.

Question 16

Combinatorially Equal

Answer 16

Two link diagrams (d,A) (e,B) are combinatorially equal if there exists an orientation preserving diffeomorphism r: R^3 → R^3 such that e = r ◦ d and carrying A over to B.

Question 17

Reidermeister Moves

Answer 17

(R1) (R2) (R3)

Question 18

Isotopic

Answer 18

Two link diagrams are isotopic if one can be obtained from the other by a sequence of reidermeister moves.

Question 19

Types of crossing

Answer 19

Positive (Top arrow right to left) sign=1

Negative (Top arrow left to right) sign = -1

Question 20

Linking Number of components

Answer 20

The linking number L_k(C_1,C_2) for components C_1 and C_2 of a link is half the sum of the signs of the crossings where C_1 meets C_2.

Question 21

Link invariant.

Answer 21

A link invariant is a map f: {isotopy classes of links} → X for any set X.

Question 22

Writhe

Answer 22

The Writhe w(D) of an oriented knot diagram is the sum of the signs of the crossings.

Question 23

Colouring

Answer 23

Let (A,+) be an abelian group.
Let D be an unoriented link diagram.
A colouring of D with values in A is a non-constant map f:arcs(D)→A such that at each crossing:
a - 2b + c = 0,
where b is the value of the overpassing arc and a, c, the other two arcs.

Question 24

n-colouring

Answer 24

An n-colouring is a colouring with values in Z_n

Question 25

n-colourable

Answer 25

A link diagram is n-colourable with colouring number n if there exists a (non-constant) n-colouring.

Question 26

Arc

Answer 26

Let (d,A) be a link diagram.
An arc is d(I) where I is a maximal open connected subset of S^1 × {1, . . . , k} such that d(I) does not contain under passings.

Question 27

(1) Split

(2) Splittable

Answer 27

A link diagram is split if it meets both half planes x_1 0 but not x_1 = 0 in between.
A link diagram is splittable if it is isotopic to a split one

Question 28

Alternating link diagram

Answer 28

A link is alternating if it admits an alternating diagram.

Question 29

Types of crossing in a chess-boarded link diagram

Answer 29

Type 1: Vertical line is over (top right/bottom left coloured)
Type 2: Horizontal line is over (top right/bottom left coloured)

Question 30

Abelian group assiciated with A

Answer 30

Let A ∈ Z^(m,n). The abelian group associated with A is:
G(A) := Z^(1,n)/Z^(1,m) A ∼= Z(n,1)/A^T Z^(m,1)

Any finitely generated abelian group is of this form up to isomorphism.

Question 31

Generators and Relations of G(A) (the abelian group associated with A).

Answer 31

Generators {x_j s.t 1≤ j ≤ n}

Relations ∑j a(i j) x_j = 0 for all i.

Question 32

A unimodular matrix

Answer 32

A unimodular matrix is a matrix P ∈ Z^(n,n) such that det(P) ∈ {−1, 1} or equivalently P^−1 ∈ Z^(n,n).

Question 33

Z-equivalent

Answer 33

Let A, B ∈ Z^(m,n).
We say that A, B are Z-equivalent if there are unimodular P, Q such that B = PAQ.
Equivalently, B can be obtained from A by a sequence of
unimodular elementary row and column operations.

Question 34

Smith Normal Form

Answer 34

Every equivalence class contains a unique Smith normal form, that is, a matrix B = (bi j) such that
◦ b_(i j) = 0 unless i = j.
◦ b_(ii) ≥ 0 for all i.
◦ b_(ii) | b_(i+1,i+1) for all i.

Question 35

Colouring group.

Reduced colouring matrix

Answer 35

Let D be a diagram for a link L. Let A be the colouring matrix of D.
Let B be a reduced colouring matrix, a matrix obtained from A by removing one row and one column.

The colouring group of L is
Col(L) := G(B) = Z^n / Z^n B.

The determinant of L is det(L) := |det(B)|.

Question 36

Alexander Polynomial

Answer 36

Let H = ∈ Z [t, t^-1] (group of invertible elements in Z [t, t^-1] ).
The Alexander polynomial is an invariant of oriented links whose values are in Z [t, t^-1] / ∼ where where f ∼ g means fH = gH .

Denoted: ∆_L

Question 37

Alexander Matrix

Answer 37

The alexander matrix has rows indexed by crossings and columns indexed by arcs. Entries are a + (t - 1)b - tc (where b is the overpassing).

Question 38

Reverse rL

Mirror Image mL

Answer 38

Reverse rL is L with opposite orientation.

Mirror image mL is L with the reflection (x,y)→(x,-y).

Question 39

(topological) n-manifold.

Answer 39

Let n ≥ Z_(≥0). A (topological) n-manifold is a Hausdorf topological space M such that for all x ∈ M there exists an open set U ⊂ M containing x such that U is homeomorphic to R^n or R^(n−1) × R_(≥0).

Question 40

Interior Point of an n-maifold

Boundary Point of an n-manifold

Answer 40

Let M be an n-manifold and x ∈ M. If there exists an open set U ⊂ M containing x such that U is homeomorphic to R^n, then x is called an interior point of M.

Otherwise x is a boundary point. The set of boundary points is called the boundary of M and written ∂M. We write int(M) = M \ ∂M.

Question 41

Closed n-manifold

Answer 41

A n-manifold is said to be closed if it is compact and its boundary is empty.
(This should not be confused with closed subsets of topological spaces.)

Question 42

An orientation on a surface

Answer 42

An orientation of a surface (=2-manifold) S is a continuous map S × ∆ → S whose restriction to {s} × ∆ is an embedding and taking (s, 1) to s for all s ∈ S.

(In other words, if you pull ∆ around the surface and you come back to a place you’ve been before then ∆ still has the same orientation.)

Question 43

K

Answer 43

K={Isotopy classes of knots}

Question 44

Knot sum

Answer 44

K × K → K written (K, L) → K + L

defined by picture..

Question 45

Factorising Sphere

Answer 45

∂T_1 = ∂T_2 ⊂ T_1 ∪_f T_2 a factorising

sphere for the factorisation K_1 + K_2.

Question 46

Monoid

Answer 46

Group without inverses

Question 47

A prime knot

Answer 47

A prime knot is an oriented knot not of the form K+L unless K or L is the unknot.

Question 48

S

Answer 48

S is the set of homeomorphism classes of non empty compact connected surfaces.

Question 49

Connex Sum

Answer 49

S × S → S: (S, T) → S + T

Question 50

Torus

Answer 50

Let T ∈ S be the class of the torus S^1 × S^1.

Question 51

Disk

Answer 51

Let D ∈ S be the class of the disk {z ∈ C : |z| ≤ 1}.

Question 52

Projective plane or cross-cap.

Answer 52

Let P ∈ S be the class of the projective plane or cross-cap P = R P^2.

Question 53

S^+

Answer 53

S^+ is the subset of S consisting of the (homeomorphism classes of) orientable surfaces.

S^+ is a submonoid of S.

Question 54

Presentations of monoids

Answer 54

S = ,

S^+ =

Question 55

Genus g(S) of orientable surface S

Answer 55

Let S be an orientable surface of class gT + nD. Then g is called the genus g(S) of S.

Question 56

Number of boundary points of gT+nD+mP

Answer 56

n

Question 57

Spanning surface

Answer 57

Let L ⊂ R^3 be an oriented link. A spanning surface for L is a compact oriented surface S in R^3 whose (oriented) boundary is L.

Question 58

Isotopy of spanning surfaces

Answer 58

Let L be an oriented link and S a compact oriented surface.
Let h: [0, 1] × S → R^3 be a continuous map such that, on
writing h_t (x) = h (t, x), the map h_t: S → R^3 is injective and its image spans L.
Then we say that the spanning surfaces h_0 (S) and h_1 (S) are isotopic (relative L).

Question 59

Projection Surface

Answer 59

D diagram of oriented link L. The constructed spanning surface for L is called the projection surface.

Question 60

Seifert Circle

Answer 60

Let D_s be the picture obtained from D by resolving or smoothing every crossing.
A circle in D_s is called a Seifert circle.

Question 61

The depth of a Seifert circle

Answer 61

The depth D(c) of a seifert circle C is the number of Seifert circles it is properly enclosed by.

Question 62

Genus of an oriented link.

Answer 62

Let L be an oriented link.
The genus, g(L) of L is the least genus of a connected spanning surface .

[ Genus = (#Seifert circles - #crossings - 1) / 2 ]

Question 63

Adding/Removing a tube.

Answer 63

Let F be a spanning surface for an oriented link L.
Let D be a disk embedded in R^3 \ L such that D ∩ F = ∂D.
Cut F along D and attach copies of D to the boundary to obtain another spanning surface F′.
We say that F′ is obtained from F by removing a tube and F from F′ by adding a tube.

Question 64

S-equivalent

Answer 64

Two spanning surfaces for an oriented link L are S-equivalent if one can be obtained from the other by repeatedly adding or removing a tube.

Question 65

Band Crossing

Band Flip

Answer 65

See pictures

Question 66

Good orientation

Projection Surface

Answer 66

See picture

Question 67

First homology group

Answer 67

Let F be a oriented surface. The first homology group H_1 (F) is the Z-module (equivalently, abelian group) C_1/D_1 where C_1 is the free Z-module on the set of oriented closed 1-manifolds in F and D_1 is the least submodule of C_1 such that:

◦ The submodule D_1 contains the boundary of any compact subsurface of F.

◦ We have (A + B) − C ∈ D_1 whenever A, B are disjoint oriented closed 1-manifolds in F and C is their union.

[A closed 1-manifold is the same as a topological space homeomorphic to S^1 × {1, . . . , k} for some natural number k]

Question 68

Linking Number of disjoint oriented links

Answer 68

Their linking number is defined to be
L_k (L_1, L_2) = ∑ L_k (K_1, K_2)
where the sum is over all pairs (K_1, K_2) where K_i
is a component of L_i.

Question 69

Seifert Matrix

Answer 69

Let F be a connected spanning surface of an oriented link L.
Consider a Z-basis A = (a_1, . . . , a_r) of H_1 (F).

Moving F slightly away from itself in positive direction gives a surface F^+ disjoint from F and a homeomorphism φ: F → F^+.

This induces an isomorphism
φ_∗: H_1 (F) → H_1(F^+).

Put a^*i = φ∗(a_i).
Recall the linking numbers of definition 19.
The Seifert matrix is:

  M(F, A) = ( l_k(a_i, a^+_j) ) _ (i, j)

Question 70

Seifert Polynomial

Answer 70

Let M be a Seifert matrix of a link L.
The Seifert polynomial of L is

S(L) = S_L (x) := det (xM − x^−1M^T) ∈ Z[x, x^−1].

Question 71

T-equivalence

Answer 71

We define ∼_T (T-equivalence) to be the least equivalence relation on the set of square matrices over Z (of any size) such that:

(a) We have M ∼_T P^T M P whenever P is unimodular.

(b) We have
M ∼_T M ∗ 0
∗ ∗ a
0 b 0

whenever M is an n × n a square matrix over Z, the matrix on the right is (n + 2) × (n + 2), and a, b ∈ Z, a^2 + b^2 = 1, and a star ∗ means any contents
allowed.

Question 72

µ(L)

Answer 72

µ(L) = #components of link L

Question 73

Breadth

Answer 73

Let f = ∑_i=k^l a_i t^i with a_i ∈ Z for all i and a_k, a_ℓ =/= 0.
Then ℓ − k is the breadth of f .

Also, the breadth of 0 is 0.

Question 74

Skein triple of diagrams/links

Answer 74

A skein triple of diagrams (D_−, D_0, D_+) is a triple of oriented link diagrams differing only in a disk where they look as follows:

D_- : left over right
D_0 : straight down
D_+ : Right over left

Replace diagram by its link for skien triple of links.

Question 75

Configuration Space

Answer 75

We define configuration space C_n = {X ⊂ C | #X = n}.

We equip it with the base-point ∗_n := {1, . . . , n} ∈ C_n.

Question 76

Geometric n-braid

Answer 76

Let γ: [0, 1] → C_n be continuous and such that γ(0) = γ(1) = ∗_n.
With γ we associate the geometric n-braid

b(γ) := {Union of γ(t) x{t}, 0≤t≤1}
= { (x,t) | t ∈ [0, 1], x ∈ γ(t) } ⊂ C × R = R^3.

Question 77

Braid Diagram

Answer 77

Projection on coords 1 and 3 of b(γ) along with over/under data.

Question 78

Homotopic and isotopic

Answer 78

Let h: [0, 1]^2 → C^n be continuous and write h_a (t) = h (a, t).
Assume that
h_a (0) = h_a (1) = ∗n for all a.
We say that h_0 and h_1 are homotopic and that
the geometric braids b(h_0) and b(h_1) are isotopic.

Question 79

B_n

Answer 79

B_n = { isotopy classes of n-braids}

Question 80

Multiplication of Braids

Answer 80

Done by stacking. left right = top bottom

Question 81

σ_i = σ^(n)_i ∈ Bn

Answer 81

Let 1 ≤ i ≤ n − 1.
We define σ_i = σ^(n)_i ∈ Bn to be the braid with
the following diagram.

*i and i+1 cross (over right to left) *

Question 82

Bridge of a link diagram

Answer 82

A bridge of a link diagram is an arc containing at least one over passing.

Question 83

Bridge number

Answer 83

The bridge number b(L) of a link L is the least number of bridges in a diagram for it.

Question 84

Tunnel

Answer 84

If D is a link diagram, let cD be the same diagram with all crossings changed. A tunnel of D is by definition a bridge of cD.

Question 85

Circle bridge/tunnel

Answer 85

A circle bridge is a bridge homeomorphic to a circle. Likewise for circle tunnels.

Question 86

Closure of a braid

Answer 86

The closure is the map cl: B_n → {isotopy classes of oriented links} defined by the following picture:

Question 87

R∞

Answer 87

See picture

Question 88

With a link diagram D in S^2 we associate a link in R^3

(up to isotopy) as follows:

Answer 88

First remove any point ∞ ∈ S^2 \D.

This gives us a link diagram in a copy of R^2. Then make this into a link the usual way.

Question 89

Conflicting and Compatible Seifert circles.

Answer 89

Let D be an oriented link diagram in S^2.
Let X, Y be two distinct Seifert circles of D.
Then X, Y are disjoint and S^2\ (X ∪ Y) has three components.
Precisely one of them is an annulus A (that is, A ∼= S^1 × (0, 1)).

If X, Y define the same orientation on A then X, Y are said to conflict. Otherwise they are compatible.

Question 90

χ(D)

Answer 90

χ(D)= # unordered pairs of distinct conflicting Seifert circles

Question 91

Reduction Move

Answer 91

Let D, D′ be oriented link diagrams in S^2.
We say that D → D′ is a reduction move if they are the same outside a small disk where they differ as in
the pictures below, provided the strings a, b belong to different Seifert circles.

Question 92

B

Answer 92

B = ∪_(n≥0) {n} × B_n = { (n, x) | n ≥ 0, x ∈ B_n}

Question 93

Markov Moves

Answer 93

Let ∼ be the least equivalence relation on B such that:

◦ Markov move 1. #
We have (n, a) ∼ (n, b) for all n ≥ 0 and conjugate a, b ∈ B_n.
(that is, b = x a x^−1 for some x ∈ B_n).

◦ Markov move 2. 
We have (n, c) ∼ (n + 1, cσ_n) ∼ (n + 1, cσ^−1_n) for all n ≥ 0 and c ∈ B_n.

Question 94

Conway Polynomial

Answer 94

The conway Polynomial is a unique polynomial CL(z) ∈ Z[z] such that

   S_L (x) = C_L (x^(−1) − x).

Question 95

A unique invariant of nonempty unoriented link diagrams D → [D] with values in R satisfies:

Bracket Polynomial

Answer 95

Let a, b, c be elements of a commutative ring R with 1.

Then there is a unique invariant of nonempty unoriented link diagrams D → [D] with values in R satisfying

1) [U_1] = 1 (unknot)
2) [ cross left over right ] = a [ = ] + b [ )( ]
3) [O D] = c [D]

The above identities are skein relations.

Let a = t, b = t^(-1) and c = -(t^(-2) + t^2)
Then [D] is the bracket polynomial.

If D is an oriented link diagram and D_o the underlying unoriented link diagram then [D] := [D_o]

Question 96

State

Answer 96

Let D be an unoriented link diagram. Let C_D denote the set of crossings of D.

A state is a function f: C_D → {1, 2}.
For every state f we shall define a diagram without crossings D(f).
To obtain D(f), every crossing X ∈ C_D is replaced by Yf(X) as in the following picture.

X = crossing left over right

Y1 = horizontal curves

Y2 = )(

Question 97

State Sum

Answer 97

For a state f: C_D → {1, 2} write 
             T(D, f) = a^k b^ℓ c^m 
where k = # f^(−1) (1) , ℓ = # f^(−1) (2) and m is the number of components of D(f). 
The state sum formula is
            [D] = Σ_f   T(D, f)
where the sum is over the states f.

Question 98

X(L)

Answer 98

For an oriented link L we define

   X(L) = X_L(t) = (−t)^(−3 w(D) ) [D]

for any diagram D of L.

Question 99

Jones Polynomial

Answer 99

For an oriented link L we define the Jones polynomial

    V_L (q) := X_L ( q^(−1/4) ).

From now on we write q = t^−4.
So (a, b, c) = ( q^(−1/4), q^(1/4), −(q^(1/2) + q^(−1/2) ).

Note that the Jones polynomial of the unknot is 1.

Question 100

The Homfly Polynomial

Answer 100

There are unique polynomials

      P(L) = P_L(q, z) ∈ Z[q, q^(−1), z],

one for each isotopy class of nonempty oriented links L, such that:

◦ For the unknot U_1 we have P(U_1) = 1.

◦ For any skein triple of links (L_−, L_0, L_+) we have
q^(−1) P(L_+) − q P(L_−) = z P(L_0).

It is known as the homfly polynomial.

Question 101

Braid index

Answer 101

Let L be an oriented link. The braid index i(L) of L is the least natural number n such that L is the closure of some n-braid.

Question 102

(a) Descending oriented knot diagram

(b) Descending oriented link

Answer 102

(a) An oriented knot diagram is said to be descending if one can draw the associated shadow in the positive direction without lifting the pen such that every overpassing is drawn before the underpassing is.

The point where the drawing starts and finishes is called the base point.

(b) An diagram D of an oriented link L is said to be descending if each of its components is descending in the foregoing sense, and there is a total ordering
on the set of components K_D = K_L of L such that if P, Q ∈ K_D are distinct and P

Question 103

s(D)

Answer 103

The number of Seifert circles of a link diagram D

Question 104

Let D be a descending (oriented) link diagram. 
Then  s(D) + w(D) ≥ ......

Answer 104

s(D) + w(D) ≥ µ(D).

Question 105

k = t-mindeg P and ℓ = t-maxdeg P.

Answer 105

Let R be a commutative ring and P ∈ R[t, t^(−1)] a nonzero Laurent polynomial over R in a variable t.
Say,
P = ∑^ℓ_(i=k) a_i t^i

with a_k, a_ℓ nonzero.

We write

         k = t-mindeg P and ℓ = t-maxdeg P.

Also, t-mindeg(0) = ∞ and t-maxdeg(0) = −∞.

We put

      t-maxdeg(P) − t-maxdeg(P) = t-breadth(P)

if P =/= 0 and t-breadth(0) = 0.

Likewise for any other variable.

Question 106

The matrix S(i, n) ∈ GL(n, Z[x, x^(−1)] )

Answer 106

For 1 ≤ i ≤ n, let S(i, n) ∈ GL(n, Z[x, x^(−1)] ) be the matrix defined by the block form

S(i, n) =

I_i−1 · · ·
· 1 − x^2 x ·
· x · ·
· · · I_(n−1−i)

Question 107

Burau Representation

Answer 107

There exists a unique representation (group homomorphism)

βn: Bn → GL(n, Z[x, x^(−1)] )

such that β_n (σ_i) = S(i, n) for all i.

It is known as the Burau representation.

Question 108

Row Vector

Answer 108

We define the row vector v_n = (1, x, . . . , x^(n−1) ).

Question 109

β_n (a)

Answer 109

Let a ∈ B_n.
We write β_n (a) in block form:

β_n (a) =
p_n (a) q_n (a)
r_n (a) s_n (a)

Here s_n (a) is a scalar and p_n (a) is of size (n − 1) × (n − 1). 
We define

D_n (a) = det p_n (a) − I_n − 1

We regard B_n as a subgroup of B_n+1.

Question 110

Writhe homomorphism

Answer 110

We define the writhe homomorphism

             w: B_n → Z     by w(σ_i) = 1

whenever 1 ≤ i ≤ n − 1.

Question 111

E_n (a)

Answer 111

For a ∈ B_n we define

E_n (a) = (−1)^(n−1) x^(1−n−w(a) ) D_n (a).