U.U.D.M. Project Report 2017:26
Examensarbete i matematik, 15 hp
Handledare: Thoms Kragh
Examinator: Jörgen Östensson
Juni 2017
Khovanov Homology of Knots
Contents
1 Introduction 4
2 Definition of Knots 5
3 Jones Polynomial 6
4 Graded Vector Spaces 9
5 Homology Groups 15
6 Computations for (T2,k) Torus Links 30
Abstract
In this thesis we provide the construction of an invariant for knots called the Khovanov homology. We work out a general formula for the Jones polynomial which Khovanov homology is built upon, and provide an example of Khovanov homology of the trefoil knot. This is later used to deduce the general formula for the torus knots T2,k.
Sammanfattning
1 Introduction
A mathematical theory for knots was first developed in the year 1771 by Alexandre-Th´eophile Vandermonde. In his work he expressed that this theory could be useful when discussing the properties of knots related to the geometry of position. In 1984 Vaughan Jones discovered a link invariant which is now called the Jones polynomial [1], defined by a skein relation
h i = h i − qh i, h∅i = 1,
h Li = (q + q−1)hLi,
(1)
this is explained in section three. Later in 1999 Mikhail Khovanov discovered a link invariant which was strictly stronger and more general than Jones polynomial and the idea was to construct a chain complex of a knot in a way that the Euler characteristic is equal to the Jones polynomial [2]. In this project we summarize his work by showing how to define it without category theory and then compute it for the general torus knot. We also explain why it is constructed the way it is by comparing it to a general formula for the unnormalized Jones polynomial.
In the second section we state the definition of knots being equal by their diagrams using the Reidemeister theorem. A diagram of a knot is a two dimensional representation of it. For example, the diagram for the trefoil knot is
.
2 Definition of Knots
Definition 1. Let N and M be manifolds and g, h : N → M be embeddings of N in M. A continuous map
F : M × [0, 1] → M (2)
is called an ambient isotopy from g to h if F (x, 0) = Id and each map F (x, t) is a homeomorphism from M to itself, and F (x, 1) ◦ g = h.
Definition 2. A knot is an embedding of the circle (S1) into three-dimensional Euclidean space (R3), or the 3-sphere (S3). Two knots are defined to be equivalent if there is an ambient isotopy between them.
A link is defined as being n-disjoint knots in R3, or S3. In this paper we also call links for knots. The orientation of a knot is given by the standard orientation on S1 (counter clockwise). We can think of knots being the image of the embedding with a fixed orientation. One can project a knot onto a plane R2 such that the projection is almost always regular, meaning that the projection is injective everywhere except for at a finite number of points where the projected curve intersects itself in a standard way. This is defined to be the diagram of a knot.
Theorem 1. (Reidemeister’s Theorem) Two links can be continuously deformed into each other if and only if any diagram of one can be transformed into a diagram of the other by a sequence of Reidemeister moves.
These Reidemeister moves are defined as follows.
R1 :
R2 :
R3 :
3 Jones Polynomial
In this section we define the Jones polynomial using the definition of the bracket poly-nomial. We also provide a general formula for computing the Jones polynomial for an arbitrarily knot. Here we follow the steps in [3].
Definition 3. Given a knot diagram, and a crossing in the knot diagram. The zero resolution and the one resolution of the crossing are when we replace the crossing with
and respectively.
Note that this is the same as the zero resolution and one resolution are given by going left and right respectively from the overcrossing down to the undercrossing. This uniquely defines the zero resolution and the one resolution without the orientation of the knot. Definition 4. Given a knot, then at each crossing we define the positive crossing and the negative crossing to be
(a) Positive crossing (b) Negative crossing
We denote the number of crossings of a knot to be n, the number of positive crossings as n+ and the number of negative crossings as n−.
The bracket polynomial is usually defined as:
h i = Ah i + A−1h i, (3)
h∅i = 1, (4)
h Li = (−A2− A−2)hLi. (5)
where L is a knot, hLi is the polynomial of the knot and A a variable. The polynomial is evaluated on a diagram where the knot is kept constant outside of the crossing. To get the polynomial of the knot we have to replace each crossing with its zero and one resolutions until we arrive at disjoint unions of circles and use (5). If we replace hLi with AnhLi and A2 with −q−1 we get a new polynomial with a different relation like this
If we divide by A on both sides, then we can rewrite the bracket (6) in the following form:
h i = h i − qh i,
h∅i = 1, (7)
h Li = (q + q−1)hLi.
This defines the invariant Kauffman bracket. From this bracket we can define the un-normalized Jones polynomial of a knot to be eJ (L) = (−1)n−qn+−2n−hLi. To get the
Jones polynomial we divide by the factor q + q−1.
Proposition 1. The unnormalized Jones polynomial is well-defined.
Proof. The general formula later in this section proves that it is well-defined on a single diagam D, and we now prove that it is also invariant under the three Reidemeister moves. h i = h i − qh i = (q + q−1)h i − qh i = q−1h i (8) h i = h i − qh i = h i − q(q + q−1)h i = −q2h i (9) Both of these cancel out by the factor (−1)n−qn+−2n−, thus it is invariant under R1. For
the R2 case we have
h i = h i − qh i − qh i + q2h i =
= h i − q(q + q−1)h i+q2h i − qh i = −qh i. (10) The factor in front of h i also cancels out with (−1)n−qn+−2n−. Left to show is that
it is invariant under R3.
These generate the same polynomial under the Kauffman bracket since we can create an isotopy between them without changing the number of crossings.
By using the second Reidemeister move twice we can go from the left figure to the right figure and because the Kauffman bracket is invariant under R2 we have that these generate the same polynomial. The Kauffman bracket must be invariant under R3. The Jones polynomial is then invariant under all of the Reidemeister moves and thus is an invariant for knots by the Reidemeister theorem.
We want to find a general formula for the Jones polynomial. From the definition of Kauffman bracket we can see that all the possible ways we can resolve all the crossing corresponds to the vertices in an n-cube. Let α ∈ {0, 1}n, then every possible resolutions of all the crossings for a knot corresponds to one of those α’s. Let us explain the notation: first we fix an order of the crossings then a zero in α means the crossing at that vertex is replaced with (zero resolution) and if it is a one in α it is replaced with (one resolution). From this we can derive a formula for the Jones polynomials. Let us denote given an α the diagram with all the chosen resolutions is a properly embedded curve, we call this Γα:
rα = number of circles in Γα, (11)
kα = number of ones in α. (12)
Lemma 1. The unnormalized polynomial of a knot L is X
α∈{0,1}n
(−1)kα+n−qn+−2n−+kα(q + q−1)rα. (13)
Proof. We can rewrite the formula as: X α∈{0,1}n (−1)kα+n−qn+−2n−+kα(q+q−1)rα = (−1)n−qn+−2n− X α∈{0,1}n (−1)kαqkα(q+q−1)rα. (14) The factor in front of the sum is just the normalisation factor for the Jones polynomial. By the definition of Kauffman bracket we get the factor (−1)kαqkα(q + q−1)rα, since k
α
represents the number of one-crossings at α and the fact that rαis the number of disjoint
4 Graded Vector Spaces
Here we follow [4, 5]. Let us now define the cochain complex with graded vector spaces and show that the Euler characteristic of that complex is equal to the unnormalized Jones polynomial. We also define the differentials for the complex as well as proving that it really is a complex.
Definition 5. A graded vector space V is the direct sum V =L
nVn, where the elements
in Vnare called homogeneous components. The graded dimension of a graded vector space
is defined as the power series q dim V =P
n∈Zqndim Vn.
We will only work with graded vector spaces which has finitely many non-zero homoge-neous components and all of them are finite dimensional.
Definition 6. Let V = L
nVn be a graded vector space. The ”degree shift” operator
·{l} is defined such that Vn{l} = Vn−l.
Note that with the above definition we can see that q dim V {l} = qlq dim V . Now we show some properties of q dim. Let V and eV be two graded vector spaces, then
q dim(V ⊗ eV ) =X n,m qn+mdim(Vn⊗ eVm) = X n,m qn+mdim(Vn) dim( eVm) =X n qndim(Vn) X m
qmdim( eVm) = q dim(V )q dim( eV ), (15)
q dim(V ⊕ eV ) =X
n
qndim(Vn) +
X
m
qmdim( eVm) = q dim(V ) + q dim( eV ). (16)
Definition 7. A chain complex is a sequence of abelian groups . . . , A1, A0, A−1, . . . with
maps dn: An→ An−1 such that for every n ∈ Z, dn◦ dn+1= 0. It is usually written as
. . .−−−→ Adn+2 n+1−−−→ Adn+1 n dn
−→ An−1 −−−→ . . . .dn−1 (17)
A cochain complex is defined as a normal chain complex where the maps go in the opposite direction with respect to the grading, i.e.
. . .−−−→ Adn−2 n−1−−−→ Adn−1 n−→ Adn
n+1 dn+1
−−−→ . . . . (18) Definition 8. Let eC be a chain complex. The ”height shift” operator ·[s] lowers the index by s. If C = eC[s], then Cr= eCr−s.
Note that in the above definition the differentials are shifted accordingly. When defining our complex we will use an abelian group to associate with the unknot, that is a graded vector space V where q dim V = q + q−1 which is the Jones polynomial for the unknot. In order to achieve this we define V to have two basis elements (1, x) such that deg 1 = 1 and deg x = −1. Note that we are working over Q, so V = spanQ(1, x). If we have k
If we are at vertex α in a cube we also need to shift the vector space to get the right polynomial, thus we get that the vector space at α’s position is
Vα = V⊗
rα
{n+− 2n−+ kα}, where kα and rα are the same as in the previous chapter.
Let the ith position of the complex be defined as Ci,∗(L) =M
α∈{0,1}n
kα=i+n−
Vα. (19)
An element from Ci,j(L) is said to have homological grading i and q-grading j. A natural
way of grading the complex is letting i = kα and j = deg v, where v ∈ Vα ⊂ Ci,j(L),
but in this case we also have some degree shifts in the definition of Vα so we will use the
following gradings
i = kα− n−,
j = deg(v) + i + n+− n−.
(20) Next we want to define something that relates the whole complex with the Jones poly-nomial.
Definition 9. The Euler characteristic of a complex is: X
i
(−1)iq dim(Ci,∗(L)). (21)
Theorem 2. The Euler characteristic of the complex above is the unnormalized Jones polynomial.
Proof. We prove this by sum manipulations using equation (15) and (16): X i (−1)iq dim(Ci,∗(L)) =X i (−1)iq dim(M α∈{0,1}n kα=i+n− Vα) = X i (−1)iX α∈{0,1}n kα=i+n− q dim Vα= =X α∈{0,1}n (−1)kα−n−q dim V α= X α∈{0,1}n (−1)kα−n−qn+−2n−+kα(q + q−1)rα (22)
And this is exactly the unnormalized Jones polynomial. The only difference is the exponent of −1, but we see that (−1)2n+ = 1 so we can multiply with this constant
without changing the expression to get the desired result.
Left to define for the complex is the differentials. We need two different functions, one that ”collapses” two vector spaces and one that ”creates” two vector spaces from one vector space. Recall that we have a tensor factor per circle in Γα, since the number of
circles at α were rα. We call the ”collapse” function for m and the ”split” function for
at the vector spaces they are acting on. There they are defined as m: V ⊗ V → V 1 ⊗ 1 7→ 1 1 ⊗ x, x ⊗ 1 7→ x x ⊗ x 7→ 0, ∆: V → V ⊗ V 1 7→ 1 ⊗ x + x ⊗ 1 x 7→ x ⊗ x. (23)
We use m that flips a crossing so that two circle components become one, and we use ∆ that flips a crossing so that one circle component is separated into two. This is illustrated in figures below.
m
∆
We can also think of the vector space as V = Q[1, x]/spanQ(x2). From this we get a
natural algebra so that we can rewrite the definition of m as m(a⊗b) = ab, which we will use later. In order to make C∗,∗(L) a complex we need to define the differential d. Recall that for every knot we have a collection of circles Γα at every vertex α ∈ {0, 1}n. We
name all the edges ξ ∈ {0, 1, ∗}n, where we only have one ∗ in each edge. For example if α1 = 010 and α2= 011, then the edge between them is ξ = 01∗. Think of it as an arrow
where its tail is the vertex if you replace ∗ with 0, and the tip of the arrow is the vertex when we replace ∗ with 1. Now let dξ be the arrow on the edge ξ. By the definition
of m and ∆ we get that dξ is a linear function which is identity on all tensor-factors
except at the vector spaces it is acting on. With all the dξ’s defined we can now define
the differential for the complex as
di = X
|ξ|=i
where |ξ| = number of ones in ξ, and sign(ξ) =number of ones before ∗. The sign is there just so the commutativity we have becomes anti-commutative, which forces the differential to make (C∗,∗(L), d) into a complex. Let us prove this.
Proof. Note that we only have to check what is happening locally, we divide the cases into how many circles are changing. There is only one case that involves four circles.
In this proof we always refer to the zero-resolution, i.e. you replace each crossing with its zero resolution. We also denote d1 and d2 as the functions when we change the first
crossing respectively the second crossing from zero to one. In this case we see that the operations are disjoint which makes them commute naturally. Now for the cases with three circles.
Here we can see that both of these functions are m. With the natural algebra defined on V we get that m(a ⊗ b) = ab. Let a, b, c ∈ {1, x}, then d1◦ d2(a ⊗ b ⊗ c) = d2(ab ⊗ c) =
abc = d1(a ⊗ bc) = d2◦ d1(a ⊗ b ⊗ c). Hence d1 and d2 commutes in this case.
Here d2◦ d1= ∆ ◦ m = d1◦ d2 by definition of d1 and d2 using m and ∆. For the next
case we have to do some calculations.
For d2◦ d1 we have d2◦ d1(1 ⊗ 1) = 1 ⊗ m(x ⊗ 1) + x ⊗ m(1 ⊗ 1) = 1 ⊗ x + x ⊗ 1, d2◦ d1(1 ⊗ x) = 1 ⊗ m(x ⊗ x) + x ⊗ m(1 ⊗ x) = x ⊗ x, d2◦ d1(x ⊗ 1) = x ⊗ m(x ⊗ 1) = x ⊗ x, d2◦ d1(x ⊗ x) = x ⊗ m(x ⊗ x) = 0. (25)
And for d1◦ d2 we have
d1◦ d2(1 ⊗ 1) = ∆1 = 1 ⊗ x + x ⊗ 1,
d1◦ d2(1 ⊗ x) = ∆(x) = x ⊗ x,
d1◦ d2(x ⊗ 1) = ∆(x) = x ⊗ x,
d1◦ d2(x ⊗ x) = ∆(0) = 0.
(26)
Comparing these two d1 and d2 commutes.
Here they are also disjoint so they commute, and lastly we have one case that involves one circle
First we compute the results of d2◦ d1
d2◦ d1(1) = 1 ⊗ ∆(x) + x ⊗ ∆(1) = 1 ⊗ x ⊗ x + 1 ⊗ x ⊗ 1 + 1 ⊗ 1 ⊗ x,
d2◦ d1(x) = x ⊗ ∆(x) = x ⊗ x ⊗ x.
And for d1◦ d2 we have
d1◦ d2(1) = 1 ⊗ ∆(x) + x ⊗ ∆(1) = 1 ⊗ x ⊗ x + 1 ⊗ x ⊗ 1 + 1 ⊗ 1 ⊗ x,
d1◦ d2(x) = x ⊗ ∆(x) = x ⊗ x ⊗ x.
5 Homology Groups
In this section we state the main theorem which tells us that the Khovanov homology is an invariant and define the Poincar´e polynomial, see [4]. Before the proof of the main theorem we give an example of Khovanov homology for the trefoil knot.
Definition 10. Let f : X → Y be a function. Then the cokernel of f is defined as coker(f ) = Y /f (X).
Definition 11. Suppose R is a commutative ring and 1R its multiplicative identity. A
R-module M consists of an abelian group (M, +) and an operation · : R × M → M such that for every r, s ∈ R and for every x, y ∈ M we have
r · (x + y) = rx + ry, (r + s) · x = r · x + r · s,
(rs) · x = r · (s · x), 1R· x = x.
(29)
The multiplication is often written as rx instead of r · x.
Lemma 2. (Snake lemma) Given the following commutative diagram of R-modules where each row is exact.
M N P 0 0 M0 N0 P0 α f f0 β g g0 γ
Then there is an exact sequence ker(α)−→ ker(β)fe eg
−
→ ker(γ)−→ coker(α)δ fe0
−→ coker(β) ge0
−→ coker(γ), (30) where ef = f |ker(α), eg = g|ker(β) and the maps ef0 and eg0 is given by ef0(m + Im(α)) =
e
f0(m) + Im(β) and eg0(m + Im(β)) = eg0(m) + Im(γ).
This lemma shows the existence of a function δ that makes this sequence exact.
Proof. First we must show that every function is well-defined. The function ef is well-defined because if m ∈ ker(α) then 0 = f0(α(m)) = β(f (m)) by the commutativity of the diagram, thus f (m) ∈ ker(β). By the same reason if m ∈ ker(β) then 0 = g0(β(m)) = γ(g(m)) which implies that g(m) ∈ ker(γ), hence g is well-defined. Now if m + Im(α) = m0 + Im(α) we have that m − m0 = α(x) for some x ∈ M . Then f0(α(x)) = β(f (x)) which implies that f0(m − m0) ∈ Im(β). Thus ef0(m) + Im(β) = ef0(m0) + Im(β) and
well-defined. Now we want to construct the function δ. If x ∈ ker(γ) then there exists an element n ∈ N such that g(n) = x, this we know since we have exact rows. By the commutativity of the diagram we get β(n) ∈ ker(g0) = Im(f0) which implies that there exists m ∈ M0 such that f0(m) = β(n). Now let us define δ such that
δ(x) = m + Im(α). (31)
We can think of the function as δ(x) = f0−1(β(g−1(x))) + Im(α). To prove that δ is well-defined we only need to check that if x ∈ ker(γ), g(n) = g(n0) = x, f0(m) = β(n) and f0(m0) = β(n0) then m − m0 ∈ Im(α). This is because the only place that is not well-defined is the lift from P to N . By the assumption on n and n0 we have that n − n0 ∈ ker(g) = Im(f ). So there exists z ∈ M such that f (z) = n − n0. Now we have
that β(n − n0) = β(f (z)) = f0(α(z). By injectivity of f0 we have that α(z) = m − m0, therefore m − m0 ∈ Im(α). Hence δ is well-defined. What is left to show is that this sequence is exact at all places.
Exactness at ker(β): The composition eg ◦ ef = 0 follows from the fact that ef and eg are restrictions of the functions f and g where g ◦ f = 0. If n ∈ ker(eg) then g(n) = 0 and since ker(g) = Im(f ) we have that there exists m ∈ M such that f (m) = n. We also need to show that m ∈ ker(α). By commutativity of the diagram we have f0(α(m)) = β(f (m)) = β(n) = 0 and since f0 is injective α(m) = 0 i.e m ∈ ker(α). So ker(eg) = Im( ef ).
Exactness at ker(γ): Let n ∈ ker(β). Then let δ(g(n)) = m. By the definition of the delta function we get that f0(m) = β(n) = 0 by assumption, and since f0 is injective implies that m = 0. Hence δ ◦eg = 0. Conversely, let p ∈ ker(δ) (also p ∈ ker(γ)). Now by surjectivity of g there exists n ∈ N such that g(n) = p. Then β(n) = f0(m) and m ∈ Im(α) since p ∈ ker(δ), so there exists x ∈ M such that α(x) = m. Now f0(α(x)) = β(f (x)) = β(n) thus n − f (x) ∈ ker(β) and g(n − f (x)) =e eg(n) −eg(f (x)) =g(n) = p.e This follows from the fact that g ◦ f = 0, so ker(δ) = Im(eg).
Exactness at coker(α): Let p ∈ ker(γ). Then there exists n ∈ N such that g(n) = p and δ(p) = m. Now f0(m) = β(n) which implies that f0(m) ∈ Im(β) so ef0 ◦ δ = 0.
Conversely, let m ∈ ker(f0). Then there exists n ∈ N such that β(n) = f0(m). Now since g0 ◦ f0 = 0 we get that 0 = g0(f0(m)) = g0(β(n)) = γ(g(n)) = 0 which says that g(n) ∈ ker(γ). So δ(g(n)) = m and ker( ef0) = Im(δ).
Exactness at coker(β): Let m + Im(α) ∈ coker(α). eg0( ef0(m + Im(α)) = g0(f0(m)) + Im(γ) and since the second row is exact g0(f0(m)) = 0, thus eg0◦ ef0. Conversely, Let
q + Im(β) ∈ ker( eg0). We have to show that there exists m ∈ coker(α) such that
e
such that n + Im(β) = ef0(m + Im(α)). This shows exactness at coker(β).
Left to show is that ef is injective and that eg0 is surjective. ef is injective since it is a
restriction of an injective function. If g0 is surjective and p + Im(γ) ∈ coker(γ) with p = g0(n) for some n ∈ N , then p + Im(γ) = eg0(n + Im(β)) which implies that eg0 is surjective.
Lemma 3. Let C be a chain complex and let C0 ⊂ C be a subchain complex.
• If C0 is acyclic (has no homology), then it can be cancelled in that case the homology H(C) of C is equal to the homology H(C/C0) of C/C0.
• Likewise, if C/C0 is acyclic then H(C) = H(C0). Proof. We have the long exact sequence of homology groups
· · · → Hr(C0) → Hr(C) → Hr(C/C0) → Hr+1(C0) → . . . , (32) associated with the short exact sequence 0 → C0 → C → C/C0 → 0. This is a short exact sequence of complexes which are modules. Consider the following diagram.
Cn0/ Im αn−1 Cn/ Im βn−1 (Cn/Cn0)/ Im γn−1 0
0 ker αn+1 ker βn+1 ker γn+1
αn βn γn
Here αn, βn and γn are differentials in the complexes. These rows are exact and thus
we can apply the Snake lemma to get the following short exact sequence Hn(C0) → Hn(C) → Hn(C/C0)
δ
−
→ Hn+1(C0) → Hn+1(C) → Hn+1(C/C0). (33)
We apply the Snake lemma for all n ∈ Z to transform the short exact sequence 0 → C0 → C → C/C0 → 0 into a long exact sequence of homology groups. If C0 is acyclic then 0 → H(C) −→ H(C/Cf 0) → 0 for every r ∈ Z. Since this is an exact sequence f
becomes bijective, hence H(C) = H(C/C0). If C/C0 is acyclic, then we get the sequence 0 → H(C0) −→ H(C) → 0 and by the same argument we have that f is also bijectivef here, thus H(C) = H(C0).
Let Hr(L) be the rth cohomology of the complex C(L). Now we can define Kh(L) as Kh(L) :=X
r
trq dim(Hr(L)). (34)
Theorem 3. (Khovanov [Kh1]) The graded dimensions of the homology groups Hr(L) are link invariants, and hence Kh(L), a polynomial in the variables t and q, is a link invariant that specializes to the unnormalized Jones polynomial at t = −1.
Before we prove this theorem we compute the Khovanov homology for the trefoil.
.
First we create a diagram of the complex. Recall that we can think of all the different cases for the crossings as a n-dimensional cube, where n is the number of crossings. Here we have a picture representing that case on each of the vertices. If we do this we get the following figure. V ⊗ V V V ⊗ V ⊗ V V ⊗ V V V ⊗ V V ⊗ V V C−3,?(K) C−2,?(K) C−1,?(K) C0,?(K) m m m −m −m m −m m m ∆ −∆ ∆ d−3 d−2 d−1
complex. The differentials for the complex are defined as d−3(v1⊗ v2⊗ v3) = (m(v1⊗ v2) ⊗ v3, v1⊗ m(v2⊗ v3), m(v1⊗ v3) ⊗ v2), d−2((v1⊗ v2, v3⊗ v4, v5⊗ v6)) = (m(v3⊗ v4) − m(v1⊗ v2) m(v5⊗ v6) − m(v1⊗ v2), m(v5⊗ v6) − m(v3⊗ v4)), d−1((v1, v2, v3)) = ∆(v1) − ∆(v2) + ∆(v3). (35) We want to rewrite our vector spaces as
V ⊗ V ⊗ V =Q1 ⊗ 1 ⊗ 1 ⊕ Qx ⊗ 1 ⊗ 1 ⊕ Q1 ⊗ x ⊗ 1⊕ ⊕ Q1 ⊗ 1 ⊗ x ⊕ Qx ⊗ x ⊗ 1 ⊕ Qx ⊗ 1 ⊗ x⊕ ⊕ Q1 ⊗ x ⊗ xQx ⊗ x ⊗ x, (V ⊗ V ) ⊕ (V ⊗ V ) ⊕ (V ⊗ V ) =Q31 ⊗ 1 ⊕ Q3x ⊗ 1 ⊕ Q31 ⊗ x ⊕ Q3x ⊗ x, V ⊕ V ⊕ V =Q31 ⊕ Q3x, V ⊗ V =Q1 ⊗ 1 ⊕ Qx ⊗ 1 ⊕ Q1 ⊗ x ⊕ Qx ⊗ x. (36)
To make a matrix representation of d−3 one have to compute where every basis vector are mapped to.
d−3 : 1 ⊗ 1 ⊗ 1 7→ (1 ⊗ 1, 1 ⊗ 1, 1 ⊗ 1), x ⊗ 1 ⊗ 1 7→ (x ⊗ 1, x ⊗ 1, x ⊗ 1), 1 ⊗ x ⊗ 1 7→ (x ⊗ 1, 1 ⊗ x, 1 ⊗ x), 1 ⊗ 1 ⊗ x 7→ (1 ⊗ x, 1 ⊗ x, x ⊗ 1), x ⊗ x ⊗ 1 7→ (0, x ⊗ x, x ⊗ x), x ⊗ 1 ⊗ x 7→ (x ⊗ x, x ⊗ x, 0), 1 ⊗ x ⊗ x 7→ (x ⊗ x, 0, x ⊗ x), x ⊗ x ⊗ x 7→ (0, 0, 0). (37)
Similarly for d−2 and d−1 we have d−2: (1 ⊗ 1, 0, 0) 7→ (−1, −1, 0), (0, 1 ⊗ 1, 0) 7→ (1, 0, −1), (0, 0, 1 ⊗ 1) 7→ (0, 1, 1), (x ⊗ 1, 0, 0) 7→ (−x, −x, 0), (0, x ⊗ 1, 0) 7→ (x, 0, −x), (0, 0, x ⊗ 1) 7→ (0, x, x), (1 ⊗ x, 0, 0) 7→ (−x, −x, 0), (0, 1 ⊗ x, 0) 7→ (x, 0, −x), (0, 0, 1 ⊗ x) 7→ (0, x, x), (x ⊗ x, 0, 0) 7→ (0, 0, 0), (0, x ⊗ x, 0) 7→ (0, 0, 0), (0, 0, x ⊗ x) 7→ (0, 0, 0), (39) d−1: (1, 0, 0) 7→ 1 ⊗ x + x ⊗ 1, (0, 1, 0) 7→ −1 ⊗ x − x ⊗ 1, (0, 0, 1) 7→ 1 ⊗ x + x ⊗ 1, (x, 0, 0) 7→ x ⊗ x, (0, x, 0) 7→ −x ⊗ x, (0, 0, x) 7→ x ⊗ x. (40)
The matrix representations then becomes
d−2 = −1 1 0 0 0 0 0 0 0 0 0 0 −1 0 1 0 0 0 0 0 0 0 0 0 0 −1 1 0 0 0 0 0 0 0 0 0 0 0 0 −1 1 0 −1 1 0 0 0 0 0 0 0 −1 0 1 −1 0 1 0 0 0 0 0 0 0 −1 1 0 −1 1 0 0 0 , (41) d−1 = 0 0 0 0 0 0 1 −1 1 0 0 0 1 −1 1 0 0 0 0 0 0 1 −1 1 . (42)
the nullspace. With some linear algebra we work out the image and kernel of d−3 to be Im d−3= spanQ 1 1 1 0 0 0 0 0 0 0 0 0 , 0 0 0 1 0 1 0 1 0 0 0 0 , 0 0 0 0 1 −1 0 −1 1 0 0 0 , 0 0 0 0 0 0 1 1 1 0 0 0 , 0 0 0 0 0 0 0 0 0 1 0 0 , 0 0 0 0 0 0 0 0 0 0 1 0 , 0 0 0 0 0 0 0 0 0 0 0 1 , ker d−3 = spanQ 0 0 0 0 0 0 0 1 . (43) Next we compute the image of d−2. It is seen that the first, second, fourth and fifth column vectors are linearly independent and the rank of the matrix is four, so the image is Im d−2= spanQ 1 1 0 0 0 0 , 1 0 −1 0 0 0 , 0 0 0 1 1 0 , 0 0 0 1 0 −1 . (44)
To calculate the kernel we need to first reduce the matrix to its reduced row echelon form. The reduced row echelon form of d−2 is
1 0 −1 0 0 0 0 0 0 0 0 0 0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 −1 1 0 −1 0 0 0 0 0 0 0 1 −1 0 1 −1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . (45)
Thus the basis for the null space is ker d−2= spanQ 1 1 1 0 0 0 0 0 0 0 0 0 , 0 0 0 1 1 1 0 0 0 0 0 0 , 0 0 0 −1 0 0 1 0 0 0 0 0 , 0 0 0 0 −1 0 0 1 0 0 0 0 , 0 0 0 1 1 0 0 0 1 0 0 0 , 0 0 0 0 0 0 0 0 0 1 0 0 , 0 0 0 0 0 0 0 0 0 0 1 0 , 0 0 0 0 0 0 0 0 0 0 0 1 . (47)
Lastly the kernel and image for d−1. The image is spanned by first and fourth column vector Im d−1 = spanQ 0 1 1 0 , 0 0 0 1 . (48)
For the kernel we rewrite it to its reduced row echelon form. 1 −1 1 0 0 0 0 0 0 1 −1 1 0 0 0 0 0 0 0 0 0 0 0 0 (49)
Which corresponds to the system
x1 −x2 x3 = 0
x4 −x5 x6 = 0 . (50)
Thus the null space is
ker d−1= spanQ 1 1 0 0 0 0 , 1 0 −1 0 0 0 , 0 0 0 1 1 0 , 0 0 0 1 0 −1 . (51)
last three vectors are the same for ker d−2 and Im d−3. We only need to work out the quotient spanQ 0 0 0 1 1 1 0 0 0 0 0 0 , 0 0 0 −1 0 0 1 0 0 0 0 0 , 0 0 0 0 −1 0 0 1 0 0 0 0 , 0 0 0 1 1 0 0 0 1 0 0 0 spanQ 0 0 0 1 0 1 0 1 0 0 0 0 , 0 0 0 0 1 −1 0 −1 1 0 0 0 , 0 0 0 0 0 0 1 1 1 0 0 0 . (52)
We want to rewrite it by changing to a basis formed by the numerator. Let us denote the vector in the numerator as v1, v2, v3 and v4 (in order). The bottom vectors can now
be express as v1− v3, v4− v1− v3 and v4− v2− v3 (in order). After this change of basis
the quotient becomes
spanQ 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 spanQ 1 0 −1 0 , −1 0 −1 1 , 0 −1 −1 1 . (53)
Now we see that the relation induced on the generators in the numerator by the vectors in the denominator are as follows v1 = −v2= v3 = 12v4, thus we only need to look at v1
have quotient spanQ 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 spanQ 0 1 1 0 , 0 0 0 1 , ∼ = spanQ 1 0 0 0 , 0 1 0 0 . (54)
Hence H0(K) = Q1 ⊗ 1 ⊕ Q1 ⊗ x. To create a table for the trefoil we need to compute the quantum gradings. Recall that i = kα− n− and j = deg(v) + i + n+− n−. In this
case we have that n−= 3 and n+ = 0.
H H H H HH j i -3 -2 -1 0 -1 Q -2 -3 Q -4 -5 Q -6 -7 -8 -9 Q
We also want to compute the unnormalized Jones polynomial using (13) and then com-pute it with (34) and compare them. Plugging everything into (13) gives us
X
α∈{0,1}n
(−1)kα+n−qn+−2n−+kα(q + q−1)rα = −q−9+ q−5+ q−3+ q−1. (55)
The Poincar´e polynomial for the trefoil is
Kh(K) = t−3q−9+ t−2q−5+ q−3+ q−1. (56) Note that if we set t = −1 then we get the unnormalized Jones polynomial, thus we are relatively sure that we have not made any mistake.
the subcomplex is acyclic. By applying Lemma (3), the H(C) is equal to H(C/ ˜C). The complex C/ ˜C = (C ( )/1=0−m→ 0{1}), where the subscript C ( )/1=0= C ( ) /C ( )1. The degree shift of the two differentials are cancelled out by [−n−]{n+− 2n−}. Note
that it is one dimensional making C ( )/1=0 isomorphic to C ( ). The proof for the left twist follows from the right twist and R2.
(Invariance under R2) For R2 we have the complex
C ( ) {1} C ( ) {2} C ( )1{1} C ( ) {2} C ( ) C ( ) {1} 0 0 m C : ∆ m C0 :
We do the same trick here as we did for the R1 case. We set the special cycle to 1 in the upper left complex and keep the upper right as it is. The bottom part be set to zero. This makes the function m to be equal the identity. This subcomplex is indeed acyclic since ker m = {0} and its image is C ( ). By lemma (3) we have that H(C) = H(C/C0). The quotient is equal to
C ( )/1=0{1} 0 β 0
C ( ) C ( ) {1} C ( ) τ β
C/C0 : ∆ C00: ∆ τ
d
β 0 0 0
0 γ 0 C ( )
τ (C/C0)/C00:
d?0
We see that we have some shift between the above complex and the complex C ( ), but this shift gets taken out with the shift [−n−]{n+−2n−}. Thus H(C ( )) = H(C ( )).
(Invariance under R1, left twist) This follows from the right twist and R2, as shown in the figure below.
R1 R2
(Invariance under R3) First we write the complexes for
∆ d1,?01 d1,?10 D1 , D2 d2,?01 d2,?10 ∆ .
we consider the isomorphism φ that transposes the top layer while keeping the bottom layer fixed, i. e. the top right vertex switches place with the bottom left vertex in the top layer and keeping the rest fixed. To show that this is an isomorphism between complexes we need to know that the diagrams commute with φ, i.e. we need to show that τ1 ◦ d1,?01 = d2,?01 and τ2◦ d2,?10 = d1,?10. Note that d1,?01 = ∆ and d2,?10 = ∆.
Now since the special cycle has been set to x these functions become bijective, by the same reason as in R2 case. We first prove τ1◦ d1,?01 = d2,?01. By definition we have
τ1◦ d1,?01= D2◦ ∆−1◦ d1,?01, but the bottom right corner is isomorphic to the top left
corner. This makes ∆−1◦ d1,?01 equal to the identity. Since the bottom half of the two complexes are isomorphic, it follows that D1 = d2,?01. For the second case we use the
same reasoning to say that ∆−1◦ d2,?10 is the identity and here we also use the fact that
the bottom sides are isomorphic to each other making D2 = d1,?10. This shows that the
functions commutes with φ so it is an isomorphism between the complexes. The last thing we have to show is that the Poincar´e polynomial specializes to the unnormalized Jones polynomial at t = 1. By theorem (2) we need to show the following equality
X
i
(−1)iq dim(Hi(L)) =X
i
(−1)iq dim(Ci,?(L)). (57) We prove this by induction with base case 0 → C0 → 0. This is true for this case since the cohomology group at zero is C0. Now assume it works for n − 1. For the complex of length n we have
0 → C1 → · · · → Cn−1 → Cn→ 0. (58) Note that we can split Cn to Im dn−1⊕ Hn, where Hn is the complement of Im dn−1 in
Cn. Now consider the following subcomplex
0 → C1→ · · · → ker dn−1→ Hn→ 0. (59) These two have the same cohomology groups by definition. If we use the assumption here we only have to look at the end of the complex. From linear algebra we have q dim Cn−1 = q dim ker dn−1+q dim Im dn−1and q dim Cn= q dim(Im dn−1)c+q dim Hn. This is the dimension theorem but for the graded dimension. It holds since our functions have degree −1 so it follows when we also consider degree shifts. Plugging everything into the right side of the equation (57) we get
n−2
X
i
(−1)iq dim(Ci,?(L)) + (−1)n−1(q dim ker dn−1+ q dim Im dn−1)+ + (−1)n(q dim(Im dn−1)c+ q dim Hn) = =
n−2
X
i
(−1)iq dim(Ci,?(L)) + (−1)n−1q dim ker dn−1+ (−1)n(q dim Hn) =
= n X i (−1)iq dim(Hi(L)). (60)
6 Computations for (T
2,k) Torus Links
It takes time to work out the Khovanov homology of a knot, hence we want to derive a more efficient method to compute the Khovanov homology for some torus knots (T2,k).
Here we study [2].
First let us study the following case
C ( ) {1} C ( ) {2} C ( ) C ( ) {1} −d m m d
The left figure shows the knot diagram and on the right we see its complex. If we fix the special cycle in the lower left corner to 1 then, by the same argument as earlier, each m becomes the identity. Thus we have a complex that is C ( )1 −−−→ (β, β) → 0,m+m where (β, β) means {(β, β) : (β, β) ∈ C ( ) ⊕ C ( )}. This is acyclic. By lemma (3) it suffices to consider the quotient. What is this quotient? First note that if we have a linear function f : X → X × X defined by f (x) = (f1(x), f2(x)), and if we quote out
the diagonal ∆ in the image then f : X → X × X/∆ where f (x) = (f1(x), f2(x)) =
(f1(x) − f2(x), f2(x) − f2(x)) = (f1(x) − f2(x), 0). This is equivalent to f0 : X → X
where f0(x) = f1(x) − f2(x). Also note that C ( ) ∼= C ( ). Applying this trick to
our case yields
0 −−−→ C ( )/1=0 ml−mr
−−−−−→ C ( ) {1}−−−−→ C (d ) {2} −−−→ 0. (61) Here ml and mr collapses the special cycle with the left circle and the right circle
respectively. Now we want to construct a similar complex for when we have k twists.
Definition 12. A quasi-isomorphism is a map A → B of chain complexes (respectively, cochain complexes) such that the induced maps
Hn(A?) → Hn(B?) (respectively Hn(A?) → Hn(B?), (62)
of homology groups (respectively, of cohomology groups) are isomorphisms for all n. Let us denote the k twist diagram as D.
Proposition 2. The complex C (D) is quasi-isomorphic to the following chain complex 0 −−→ C ( ) {1 − k} ∂ −k −−−−−→ C ( ) {3 − k}−−−−−→ . . .∂1−k . . . ∂ −3 −−−−−→ C ( ) {k − 3}−−−−−∂−2→ C ( ) {k − 1}−−−−−∂−1→ C ( ) {k} −−→ 0. (63) Where all the differentials are
∂−1= d, ∂−2= mlx− mrx, ∂−3= mlx+ mrx, . . . ∂−k = mlx− (−1)kmrx. (64)
Cohomology groups H(D) are quasi-isomorphic to the above chain complex.
First we need to explain our notation. Let X : C ( ) → C ( ) which adds a cycle marked with x. With this we can define mlx= ml◦ X and mrx = mr◦ X.
Proof. We prove this by induction with the base case k = 2, which we proved in the beginning of this section. Assume that it works for k = n. For the induction step we have to use this assumption to work out the case for k = n + 1. If we apply the assumption for the last n crossings we get the complex
Note that we do not write what the degree shifts are. We deal with that at the end of the proof. Consider the following subcomplex
0 · · · 0 C ( )1 β
0 · · · 0 β 0
m
m
where these β ∈ C ( ) means the diagonal, i.e 0 → C ( )1 −−−→ (β, β) → 0. Whenm+m we write β we mean the diagonal. This is the same case where k = 2, thus this is acyclic and a subcomplex. By lemma (3) is suffices to consider that quotient. What we are left with is C ( ) C ( ) C ( ) · · · C ( ) C ( ) C ( ) · · · · · · C ( ) C ( )/1=0 · · · C ( ) C ( ) C ( ) m mlx− (−1)kmrx m −(mlx− (−1)km rx) mlx− (−1)k−1mrx m −(mlx− (−1)k−1m rx) mlx− (−1)k−2mrx −(mlx− (−1)k−2mrx) mlx+ mrx m −(mlx+ mrx) mlx− mrx ml− mr −(mlx− mrx) d
Where mland mrcollapses the special cycle with the left respectively right vector space.
Here we can consider the following subcomplex
· · · 0 C ( )1 β ⊗ x
· · · 0 β 0 0
m
mlx− mrx
Note that here m is the identity and mrx = 0 since the special cycle is be marked with
C ( ) C ( ) C ( ) · · · C ( ) C ( ) C ( ) · · · · · · C ( )/1=0 C ( ) · · · C ( ) C ( ) C ( ) m mlx− (−1)kmrx m −(mlx− (−1)km rx) mlx− (−1)k−1mrx m −(mlx− (−1)k−1m rx) mlx− (−1)k−2mrx −(mlx− (−1)k−2mrx) mlx+ mrx m −(mlx+ mrx) −mrx mlx− mrx −(mlx− mrx) d
Note that we also have the quotient with the diagonal at the end so by the same reason as before this complex becomes
C ( ) C ( ) C ( ) · · · C ( ) C ( ) C ( ) · · · · · · C ( )/1=0 · · · C ( ) C ( ) C ( ) m mlx− (−1)kmrx m −(mlx− (−1)km rx) mlx− (−1)k−1mrx m −(mlx− (−1)k−1m rx) mlx− (−1)k−2mrx −(mlx− (−1)k−2m rx) mlx+ mrx m + mrx −(mlx+ mrx) mlx− mrx d
C ( ) C ( ) C ( ) C ( ) · · · · · · C ( ) C ( ) C ( ) mlx− (−1)k+1mrx mlx− (−1)kmrx mlx− (−1)k−1mrx mlx− (−1)k−2mrx mlx+ mrx mlx− mrx d
Each position in the complex has shifted its degree with one step since the normal degree shifts are on the top side of the complex and the bottom side is shifted with one from the above. This completes the proof.
With this proposition we can work out the Khovanov homology more efficiently. For the trefoil we have k = 3, thus the complex becomes
0 → C−3( ) {−2}−→ C2x −2( ) {−1}→ C−0 −1( )−∆→ C0( ) → 0. (65) The vector spaces here are one dimensional, thus we can derive the cohomology groups to be H−3(D) = (x), H−2(D) = (1), H−1(D) = 0 and H0(D) = (1 ⊗ 1, 1 ⊗ x) where D is the diagram of the trefoil. The last one is achieved by taking the quotient
H0(D) = (1 ⊗ 1, 1 ⊗ x, x ⊗ 1, x ⊗ x)
7 Switching the Field into an Arbitrary Ring
In this section we make Khovanov homology stronger by exchanging the field Q into an arbitrary ring R. We compute the more generalized Khovanov homology for an arbitrary torus knot (T2,k) [2].
Recall the definition of R-modules (definition 11). We can think of R-modules as a vector space where its coefficients are taken from a ring instead of a field. In the proof for proposition (2) we do not use the fact that we were working with the field Q. Thus it also holds for rings. So why would we work with rings instead of fields? Consider we have the quotient span(x)/span(2x). This is zero if we are working with Q since it has an inverse of 2, but if we look at the ring Z we get that this is equal to Z2. This
makes the Khovanov invariant stronger since we get more information from the knots. With this cleared up we can begin by calculating the Khovanov homology for a general k-torus over an arbitrary ring R.
Proposition 3. The cohomology groups of T2,k are
Hi(T2,k) = 0 for i < −k and i > 0, (67) H0(T2,k) = R{−k} ⊕ R{2 − k}, (68) H−1(T2,k) = 0, (69) H−2j(T2,k) = (R/2R){−4j − k} ⊕ R{−4j + 2 − k} for 1 ≤ j ≤ k − 1 2 , j ∈ Z, (70) H−2j−1(T2,k) = R{−4j − 2 − k} for 1 ≤ j ≤ k − 1 2 , j ∈ Z, (71) H−k(T2,k) = R{−3k} ⊕ R{2 − 3k} for even k. (72)
Proof. By using proposition (2) our chain complex becomes 0 −−→ C ( ) {1 − k} x(−1) kx −−−−−−−→ C ( ) {3 − k} x(−1) k−1x −−−−−−−−−→ . . . . . .−−−−2x→ C ( ) {k − 3}−−−→ C ( ) {k − 1}0 −−−→ C ( ) {k} −−→ 0,∆ (73)
where 2x denotes the map that multiplies with 2x on the tensor factor. Note that every other map is zero since mlx = mrx = x. The knots only have k number of
References
[1] V. Jones, A polynomial invariant for knots via von Neumann algebras, MR:766964 [2] M. Khovanov, A categorification of the Jones polynomial, arXiv:math/9908171 [3] Louise H Kauffman, Sofia Lambropoulou, Slavik Jablan, Jozef H Przytycki, LATEX:
Introductory Lectures on Knot Theory, World scientific, Italy, Vol. 46, 2009.
[4] D. Bar-Natan, On Khovanov’s categorification of the Jones polynomial, arXiv:math/0201043
