**3. Homology**

**Definition 11.** *Let V* = *n Vn*, *be a graded vector space with homogeneous components* {*Vn*} *of degree n. The graded dimension of V is the power series q* dim *V:* = ∑*n qn* dim *Vn.*

**Definition 12.** *The* degree *of the tensor product of graded vector space V*1 ⊗ *V*2 *is the sum of the degrees of the homogeneous components of graded vector spaces V*1 *and V*2.

**Remark 4.** *In our case, the graded vector space V has the basis* < *v*+, *v*− > *with degree p*(*<sup>v</sup>*±) = ±1 *and the q-dimension q* + *q*<sup>−</sup>1.

**Definition 13.** *The degree shift* .{*l*} *operation on a graded vector space V* = *Vn is defined by*

$$\left(V.\{I\}\right)\_n = V\_{n-I}.$$

**Construction of Chain Groups**: Let *L* be a link with *n* crossings, and let all crossings be labeled from 1 to *n*. Arrange all its 2*n* Kauffman states into columns 1, 2, ... , *n* so that the *r*th column contains all states having *r* number of 1-smoothings in it. To every stat *α* in the *r*th column we assign graded vector space *<sup>V</sup>α*(*L*) := *<sup>V</sup>*⊗*m*{*r*}, where *m* is the number of circles in *α*. The *r*th *chain group*, denoted by [[*L*]]*r* := *<sup>α</sup>*:*r*=|*α*<sup>|</sup> *<sup>V</sup>α*(*L*), is the direct sum of all vector spaces corresponding to all states in the *r*th column.

**Definition 14.** *The chain complex C of graded vector spaces Cr is defined as*:

$$\dots \dots \to \overline{\mathbb{C}}^{r+1} \xrightarrow{d^{r+1}} \overline{\mathbb{C}}^r \xrightarrow{d^r} \overline{\mathbb{C}}^{r-1} \xrightarrow{d^{r-1}} \dots \dots$$

*such that dr* ◦ *dr*+<sup>1</sup> = 0 *for each r.*

In a system of converting the chain group into a complex, we use the maps between graded vector spaces to satisfy *d* ◦ *d*. For this purpose we can label the edges of the cube {0, 1}*χ* by the sequence *ξ* {0, 1, -}*<sup>χ</sup>*, where *ξ* contains only one - at a time. Here, - indicates that we change a 1-smoothing to a 0-smoothing. The maps on the edges is denoted by *dξ* , the height of edges |*ξ*|. The direct sum of differentials in the cube along the column is

$$d^r := \sum\_{|\vec{\xi}| = r} (-1)^{\vec{\xi}} d\_{\vec{\xi}}.$$

Now, we discuss the reason behind the sign of (−<sup>1</sup>)*<sup>ξ</sup>* . As we want from the differentials to satisfy *d* ◦ *d* = 0, the maps *dξ* have to anticommute on each of the vertex of the cube. A way to do this is by multiplying edges *dξ* by (−<sup>1</sup>)*<sup>ξ</sup>* := (−<sup>1</sup>)<sup>∑</sup>*<sup>i</sup>*<*<sup>j</sup> ξi* , where *j* is the location of - in *ξ*.

For better understanding, please see the *n*-cube of trefoil knot *x*<sup>−</sup><sup>3</sup> 1in Figure 8.

**Figure 8.** *n*-cube of *x*<sup>−</sup><sup>3</sup> 1.

It is useful to note that the ordered basis of *V* is *<sup>v</sup>*+, *<sup>v</sup>*− and the ordered basis of *V* ⊗ *V* is *<sup>v</sup>*+ ⊗ *v*+, *v*− ⊗ *v*+, *v*+ ⊗ *v*−, *v*− ⊗ *<sup>v</sup>*−.

**Definition 15.** *Linear map m* : *V* ⊗ *V* → *V that merges two circles into a single circle is defined as <sup>m</sup>*(*<sup>v</sup>*+ ⊗ *v*+) = *v*+, *<sup>m</sup>*(*<sup>v</sup>*+ ⊗ *<sup>v</sup>*−) = *v*−, *<sup>m</sup>*(*<sup>v</sup>*− ⊗ *v*+) = *v*− *and <sup>m</sup>*(*<sup>v</sup>*− ⊗ *<sup>v</sup>*−) = 0.

*Map* Δ : *V* → *V* ⊗ *V that divides a circle into two circles is defined as* <sup>Δ</sup>(*v*+) = *v*+ ⊗ *v*− + *v*− ⊗ *v*+ *and* <sup>Δ</sup>(*<sup>v</sup>*−) = *v*− ⊗ *v*−*; see Figure 9.*

**Figure 9.** *m* and Δ maps.

**Definition 16.** *The homology group associated with the chain complex of a link L is defined as* H*r*(*L*) = ker *dr* im *dr*+<sup>1</sup> 

*.*

**Definition 17.** *The kernel of the map dr* : *V*⊗*r*−<sup>1</sup> → *V*<sup>⊗</sup>*r, denoted by* ker *dr, is the set of all elements of V*⊗*r*−<sup>1</sup> *that go to the zero element of V*⊗*r. The elements of the kernel are called cycles, while the elements of* im *dr*+<sup>1</sup> *are called boundaries.*

**Remark 5.** *Note that the image of the chain complex of dr*+<sup>1</sup> *is a subset of kernel dr as, in general, dr* ◦ *dr*+<sup>1</sup> = 0.

**Definition 18.** *The graded Poincare polynomial* ´ Kh(*L*) *in variables q and t of the complex is defined as*

$$\mathbb{K}\mathbf{h}(L) := \sum\_{r} t^r q dim \mathcal{H}^r(L).$$

**Theorem 3.** (**Khovanov** [1]). *The graded dimension of homology groups* H*r*(*L*) *are link invariants. The graded Poincare polynomial* ´ Kh(*L*) *is also a link invariant and* Kh(*L*)*<sup>t</sup>*<sup>=</sup>−<sup>1</sup> = ˆ*J*(*L*).

:

*3.1. Homology of x*<sup>−</sup><sup>3</sup> 1

> Now, we give the Khovanov homology of link *x*<sup>−</sup><sup>3</sup> 1 =

1. **The** *n***-cube:** The 3-cube of *x*<sup>−</sup><sup>3</sup> 1is given in Figure 10:

**Figure 10.** The 3-cube of *x*<sup>−</sup><sup>3</sup> 1.

2. **Chain complex:** The chain complex of - *x*31is

$$0 \xrightarrow{d^4} V^{\odot 3} \xrightarrow{d^3} \oplus\_3 V^{\odot 2} \xrightarrow{d^2} \oplus\_3 V \xrightarrow{d^1} V^{\odot 2} \xrightarrow{d^0} 0.$$

3. **Ordered basis of the chain complex:** The following are the vector spaces of the chain complex along with their ordered bases:

*V* ⊗ *V* ⊗ *V* = *<sup>v</sup>*+ ⊗ *v*+ ⊗ *v*+, *v*− ⊗ *v*+ ⊗ *v*+, *v*+ ⊗ *v*− ⊗ *v*+, *v*+ ⊗ *v*+ ⊗ *v*−, *v*− ⊗ *v*− ⊗ *v*+, *v*− ⊗ *v*+ ⊗ *v*−, *v*+ ⊗ *v*− ⊗ *v*−, *v*− ⊗ *v*− ⊗ *<sup>v</sup>*− (*V* ⊗ *V*) ⊕ (*V* ⊗ *V*) ⊕ (*V* ⊗ *V*) = (*<sup>v</sup>*+ ⊗ *v*+, 0, <sup>0</sup>),(0, *v*+ ⊗ *v*+, <sup>0</sup>),(0, 0, *v*+ ⊗ *<sup>v</sup>*+),(*<sup>v</sup>*− ⊗ *v*+, 0, <sup>0</sup>),(0, *v*− ⊗ *v*+, <sup>0</sup>),(0, 0, *v*− ⊗ *<sup>v</sup>*+),(*<sup>v</sup>*<sup>+</sup> ⊗ *v*−, 0, <sup>0</sup>),(0, *v*+ ⊗ *v*−, <sup>0</sup>),(0, 0, *v*+ ⊗ *<sup>v</sup>*−),(*<sup>v</sup>*− ⊗ *v*−, 0, 0)(0, *v*− ⊗ *v*−, <sup>0</sup>),(0, 0, *v*− ⊗ *<sup>v</sup>*−) *V* ⊕ *V* ⊕ *V* = (*<sup>v</sup>*+, 0, <sup>0</sup>),(0, *v*+, <sup>0</sup>),(0, 0, *<sup>v</sup>*+),(*<sup>v</sup>*−, 0, <sup>0</sup>),(0, *v*−, <sup>0</sup>),(0, 0, *<sup>v</sup>*−) *V* ⊗ *V* = *<sup>v</sup>*+ ⊗ *v*+, *v*− ⊗ *v*+, *v*+ ⊗ *v*−, *v*− ⊗ *<sup>v</sup>*−

4. **Differential maps in matrix form:** Differential map *<sup>d</sup>*<sup>3</sup>*<sup>V</sup>*1 ⊗ *V*2 ⊗ *V*3 = *m*(*<sup>v</sup>*1 ⊗ *<sup>v</sup>*2) ⊗ *v*3, *v*1 ⊗ *<sup>m</sup>*(*<sup>v</sup>*2 ⊗ *<sup>v</sup>*3), *v*2 ⊗ *<sup>m</sup>*(*<sup>v</sup>*1 ⊗ *v*3) in terms of a matrix is:

$$d^3 = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ \end{pmatrix} \sim$$

$$\begin{aligned} \text{and map } d^2 \Big(V\_1 \otimes V\_2, V\_3 \otimes V\_4, V\_5 \otimes V\_6\Big) &= \left(m(v\_3 \otimes v\_4) - m(v\_1 \otimes v\_2), m(v\_5 \otimes v\_6) - m(v\_1 \otimes v\_2)\right), \\ (v\_2)\_r, m(v\_5 \otimes v\_6) - m(v\_3 \otimes v\_4)) \text{ is } d^2 &= \left(\begin{pmatrix} A & 0 & 0 & 0 \\ 0 & A & A & 0 \end{pmatrix}, \text{where } A = \begin{pmatrix} -1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \end{pmatrix}. \text{ Also,} \\\ d^1 \Big(V\_1, V\_2, V\_3\Big) &= \Delta(v\_1) - \Delta(v\_2) + \Delta(v\_3) \text{ is } d^1 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 1 \end{pmatrix}. \end{aligned}$$

5. **Khovanov Homology:** On solving *d*3*x* = 0 or

$$
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
x\_1 \\ x\_2 \\ x\_3 \\ x\_4 \\ x\_5 \\ x\_6 \\ x\_7 \\ x\_8
\end{pmatrix} = 0,
$$

$$\begin{aligned} \text{We receive } \mathbf{x}\_1 = \mathbf{x}\_2 = \mathbf{x}\_3 = \mathbf{x}\_4 = 0, \mathbf{x}\_2 + \mathbf{x}\_3 = \mathbf{0}, \mathbf{x}\_3 + \mathbf{x}\_4 = \mathbf{0}, \mathbf{x}\_2 + \mathbf{x}\_4 = \mathbf{0}, \mathbf{x}\_6 + \mathbf{x}\_7 = \mathbf{0}, \mathbf{x}\_8 + \mathbf{x}\_9 = \mathbf{0}, \mathbf{x}\_9 + \mathbf{x}\_{10} = \mathbf{0} \end{aligned}$$

$$\text{and } \mathbf{x}\_5 + \mathbf{x}\_7 = \mathbf{0}. \text{ So the kernel of } d^3 = \left\langle \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\rangle. \text{ Similarly, the image of } d^3 \text{ is}$$

$$\left\langle \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} \right\}. \text{ The new } \mathbf{z} \in \mathbb{R}^3. \text{ We get}$$

Thus,

$$\mathcal{H}^3(\widehat{\mathbf{x}\_1^3}) \;= \; \frac{\ker d^3}{\operatorname{im} d^4} = \frac{\mathbb{Z}\_{\left(\boldsymbol{v}\_- \otimes \boldsymbol{v}\_- \otimes \boldsymbol{v}\_-\right)}}{0} = \mathbb{Z}\_{\left(\boldsymbol{v}\_- \otimes \boldsymbol{v}\_- \otimes \boldsymbol{v}\_-\right)}.$$

To compute the homology of the next level, we first cancel out the terms that appear in both ker *d*2 and im *d*3, and then use a special trick: Note that the last three summands of ker *d*2 make up all of <sup>Z</sup><sup>3</sup>(*v*−⊗*v*−), where the last three summands of im *d*3 span the subspace of <sup>Z</sup><sup>3</sup>(*v*−⊗*v*−) generated by vectors (0, 1, <sup>1</sup>), (1, 1, 0) and (1, 0, <sup>1</sup>). Now, form a matrix whose columns are these vectors. Since the eigenvalues of this matrix are −1, 1, and 2, we can write:

$$\frac{\mathbb{Z}^3}{\langle (0,1,1), (1,1,0), (1,0,1) \rangle} = \frac{\mathbb{Z}}{2\mathbb{Z}} \oplus \frac{\mathbb{Z}}{\mathbb{Z}\_1} \oplus \frac{\mathbb{Z}}{\mathbb{Z}\_{-1}} = \mathbb{Z}\_2.$$

Reducing the remaining matrices of kernel of *d*2 and image of *d*3 into reduced row echelon form, quotient ker *d*2 im*d*<sup>3</sup> becomes isomorphic to Z. Hence,

$$\mathcal{H}^2(\widehat{x\_1^3}) \quad = \begin{array}{c} \ker d^2 \\ \hline \operatorname{im} d^3 \end{array} = \mathbb{Z} \oplus \mathbb{Z}\_2.$$

The range of *d*2 is <sup>Z</sup>(*<sup>v</sup>*+,*v*+,<sup>0</sup>) ⊕ <sup>Z</sup>(*<sup>v</sup>*+,0,−*<sup>v</sup>*+) ⊕ <sup>Z</sup>(0,*<sup>v</sup>*+,*<sup>v</sup>*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*−,*v*−,<sup>0</sup>) ⊕ <sup>Z</sup>(*<sup>v</sup>*−,0,−*v*−) ⊕ <sup>Z</sup>(0,*<sup>v</sup>*−,*v*−) and the kernel of *d*1 is <sup>Z</sup>(*<sup>v</sup>*+,*v*+,<sup>0</sup>) ⊕ <sup>Z</sup>(0,*<sup>v</sup>*+,*<sup>v</sup>*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+,0,−*<sup>v</sup>*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*−,*v*−,<sup>0</sup>) ⊕ <sup>Z</sup>(0,*<sup>v</sup>*−,*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*−,0,−*v*−). Since ker *d*1 = im *d*2, H<sup>1</sup>(*x*31)0.

 =

It is clear from the chain complex that the kernel of *d*0 is the full space *V* ⊗ *V*.

$$\mathcal{H}^0(\widehat{\mathbf{x}\_1^3}) = \frac{\mathbb{Z}\_{\left(\boldsymbol{\nu}\_+ \otimes \boldsymbol{\nu}\_+\right)} \oplus \mathbb{Z}\_{\left(\boldsymbol{\nu}\_- \otimes \boldsymbol{\nu}\_+\right)} \oplus \mathbb{Z}\_{\left(\boldsymbol{\nu}\_+ \otimes \boldsymbol{\nu}\_-\right)} \oplus \mathbb{Z}\_{\left(\boldsymbol{\nu}\_- \otimes \boldsymbol{\nu}\_-\right)}}{\mathbb{Z}\_{\left(\boldsymbol{\nu}\_- \otimes \boldsymbol{\nu}\_+ + \boldsymbol{\nu}\_+ \otimes \boldsymbol{\nu}\_-\right)} \oplus \mathbb{Z}\_{\left(\boldsymbol{\nu}\_- \otimes \boldsymbol{\nu}\_-\right)}} = \mathbb{Z}\_{\left(\boldsymbol{\nu}\_+ \otimes \boldsymbol{\nu}\_+\right)} \oplus \mathbb{Z}.$$

## *3.2. Homology of* Δ2*k*+<sup>1</sup>

We now compute the homology of braid link Δ2*k*+1, where Δ = *x*1*x*2*x*1. The canonical form of this braid is Δ2*k*+<sup>1</sup> = *x*2*k*+<sup>2</sup> 1 *<sup>x</sup>*2*<sup>x</sup>*21*x*22*x*21 ··· *x*22*x*21*x*21, having 2*k* + 2 factors; you can see Δ<sup>3</sup> in Figure 11.

**Figure 11.** Δ3.

The co-chain complex of the link Δ2*k*+<sup>1</sup> is 0 *d*−<sup>1</sup> −−→ *V*⊗<sup>3</sup> *d*0 −→ <sup>⊕</sup>6*k*+3*V*⊗<sup>2</sup> *d*1 −→ <sup>⊕</sup>(<sup>2</sup>*k*+<sup>1</sup> 1 )(<sup>4</sup>*k*+<sup>2</sup> 1 )*V*⊗<sup>1</sup> <sup>⊕</sup>(<sup>2</sup>*k*+<sup>1</sup> 1 )+(<sup>4</sup>*k*+<sup>2</sup> 2 ) *V*⊗<sup>3</sup> *d*3 −→ <sup>⊕</sup>(<sup>2</sup>*k*+<sup>1</sup> 1 )(<sup>4</sup>*k*+<sup>2</sup> 2 )+(<sup>2</sup>*k*+<sup>1</sup> 2 )(<sup>4</sup>*k*+<sup>2</sup> 1 )*V*⊗<sup>1</sup> <sup>⊕</sup>(<sup>2</sup>*k*+<sup>1</sup> 1 )+(<sup>4</sup>*k*+<sup>2</sup> 2 ) *V*⊗<sup>3</sup> *d*4 −→ ... *d*6*k*+<sup>1</sup> −−−→ <sup>⊕</sup>(<sup>4</sup>*k*+<sup>2</sup> 1)*V*⊗2*k*+<sup>1</sup> <sup>⊕</sup>(<sup>2</sup>*k*+<sup>1</sup> 1) *V*⊗2*k*+<sup>3</sup> *d*6*k*+<sup>2</sup> −−−→ *V*⊗2*k*+<sup>2</sup> *d*6*k*+<sup>3</sup> −−−→ 0.

We now represent the differential maps in terms of matrices. The matrix representing differential *d*0 has order 24*k* + 12 × 8 and is

$$d^0 = \begin{pmatrix} A & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & A & B & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & C & A & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & C & A & B & 0 \end{pmatrix}.$$

.

Here, each matrix *A*, *B*, and *C* has a (6*k* + 3) × 1 order:

$$\begin{array}{rcl} A\_1 &=& \left( \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots & 1 \end{pmatrix} \right)^t \\ B\_2 &=& \left( \begin{pmatrix} 0 & 1 & 0 & 0 & 1 & 0 & 0 & \cdots & 1 & 0 \end{pmatrix} \right)^t \\ C\_3 &=& \left( \begin{pmatrix} 1 & 0 & 1 & 1 & 0 & 1 & 1 & \cdots & 0 & 1 \end{pmatrix} \right)^t \end{array}$$

Since ker *d*0 = Z*v*−⊗*v*−⊗*v*− ⊕ <sup>Z</sup>*<sup>v</sup>*+⊗*v*−⊗*v*−<sup>−</sup>*v*−⊗*v*+⊗*v*−+*v*+⊗*v*−⊗*v*− and im *d*−<sup>1</sup> = 0, the homology at this level is

$$\mathcal{H}^0(\Delta^{2k+1}) = \mathbb{Z}\_{\mathbb{P}\dots \mathbb{Q}\,\mathbb{w}\dots \mathbb{Q}\,\mathbb{w}\dots} \oplus \mathbb{Z}\_{\mathbb{P}\,+\mathbb{Q}\,\mathbb{w}\dots \mathbb{Q}\,\mathbb{w}\dots \mathbb{w}\dots \mathbb{Q}\,\mathbb{w}\dots \mathbb{w}\,\dots \mathbb{Q}\,\mathbb{w}\dots \mathbb{Q}\,\mathbb{w}\dots \mathbb{w}\,\mathbb{w}$$

Now, we go for differential map *d*1. The matrix that represents it has an order of 20(6*k*<sup>2</sup> + 3) × 4(6*k* + 3) and is

$$d^1 = \begin{pmatrix} R\_1 & 0 & 0 & 0 \\ 0 & R\_1 & R\_1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ R\_2 & 0 & 0 & 0 \\ R\_2 & 0 & 0 & 0 \\ 0 & R\_2 & 0 & 0 \\ 0 & 0 & R\_2 & 0 \\ 0 & 0 & 0 & R\_2 \\ 0 & R\_3 & R\_4 & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & R\_{n-1} & R\_n \end{pmatrix}$$

.

The order of each of the matrix *Ri* is (12*k* + 6) × (6*k* + <sup>3</sup>):

*R*1 = ⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 1 −1 0 0 0 ... 0 0 10 0 0 −1 0 ... 0 1 0 0 0 0 0 ... −1 0 1 −1 0 0 0 ... 0 01 0 −1 0 0 ... 0 01 0 0 0 −1 ... 0 0 1 0 0 0 0 ... −1 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 0 ... 1 −1 ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , *R*2 = ⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 1 0 −1 0 0 ... 0 0 1 0 ... −10 0 0 0 1 0 ... 0 0 −10 0 1 0 ... 0 0 0 −1 0 1 0 ... 0 0 0 0 −1 01 0 0 −1 0 ... 0 0 1 ... 0 0 −10 0 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 ... 1 0 −1 ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , *R*3 = ⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 1 0 −1 0 0 ... 0 0 1 0 ... −10 0 0 0 1 0 ... 0 0 −10 0 1 0 ... 0 0 0 −1 0 1 0 ... 0 0 0 0 −1 00 1 −1 0 0 ... 0 00 1 0 0 −1 ... 0 0 0 1 0 ... −10 0 0 0 1 0 ... 0 0 −1 . . . . . . . . . . . . . . . . . . . . . . . . 0 0 0 0 ... 1 0 −1 ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ ,

$$R\_4 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & \dots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 1 & 0 & 0 & -1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0 & 0 & -1 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & 1 & 0 & \dots & -1 \end{pmatrix} /$$

and, at the end, all rows of matrix *Rn* are zero except for the last row, which is

$$
\left( \begin{array}{cccc} 0 & \dots & 0 & 0 & 1 & 0 & \dots & -1 \\ \end{array} \right) .
$$

Here, ker *d*1 = <sup>Z</sup>(*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*++*<sup>v</sup>*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*++*v*−⊗*v*+−*v*+⊗*v*−<sup>−</sup>*v*+⊗*v*−<sup>−</sup>*v*+⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−+*v*+⊗*v*−+*v*+⊗*v*−+*v*−⊗*v*++*v*−⊗*v*++*v*−⊗*v*++*v*−⊗*v*++*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*−+*v*−⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*−+*v*−⊗*v*−+*v*−⊗*v*−+*v*−⊗*v*−+*v*−⊗*v*−) and im *d*0 = <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−) ⊕

<sup>Z</sup>(*v*−⊗*v*+). Since the number of Z spaces appear in the kernel of *d*1, it is exactly the same as the image of *d*0, H<sup>1</sup>(Δ2*k*+<sup>1</sup>) = 0.

The image of *d*1 is obvious. We just need the kernel of *d*2. The matrix that represents *d*2 has an order of (26*k*+<sup>3</sup> + 22*k*+<sup>2</sup>)(6*k* + 5) × 20(6*k*<sup>2</sup> + 3) and is


Here, the order of each *Si* is (4*k*<sup>2</sup> + 3) × (6*k*<sup>2</sup> + <sup>3</sup>), and is:

*S*1 = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 −1 0 0 0 0 ... 0 1 0 0 0 0 0 ... 1 1 0 0 0 0 0 ... 1 0 0 0 0 0 0 ... 0 0 0 −1 0 0 0 ... 0 1 0 0 0 0 0 ... 0 1 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... 0 0 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , *S*2 = ⎛⎜⎜⎜⎜⎜⎜⎜⎝ 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 ... 0 0 0 0 0 0 0 ... 0 1 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... 1 0 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ , *S*3 = ⎛⎜⎜⎜⎜⎜⎜⎜⎝ 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 ... 1 0 0 −1 0 0 0 ... 0 0 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... 0 0 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ , *S*4 = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 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 ... 0 0 0 0 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 ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ ,

. . .

*Sn*−<sup>2</sup> = ⎛⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 0 0 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 0 0 0 0 0 ... 0 0 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... 0 0 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ , *Sn*−<sup>1</sup> = ⎛⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 ... 0 −1 0 0 0 0 0 ... −1 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... 0 −1 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ , *Sn* = ⎛⎜⎜⎜⎜⎜⎜⎜⎝ 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 ... 0 0 0 0 0 0 0 ... 0 0 0 0 0 0 0 ... 0 ... ... ... ... ... ... ... ... −1 0 0 0 0 0 ... 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎠ .

Thus, H<sup>2</sup>(Δ2*k*+<sup>1</sup>) = Z ⊕ Z. Differential *d*6*k*+<sup>2</sup> of order (22*k*+<sup>2</sup>) × (2*k* + 1)(22*k*+<sup>2</sup> + 22*k*+<sup>3</sup>) is

$$d^{6k+2} = \left( \begin{array}{cccc} \chi\_1 & \chi\_2 & \chi\_3 & \chi\_4 & \chi\_5 & \dots & \chi\_{6k+3} \end{array} \right) ,$$

where *Yi*are matrices, each having an order of 22*k*+<sup>2</sup> × 22*k*+<sup>2</sup> :

$$Y\_1 = \begin{pmatrix} 0 & -1 & 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 1 & -1 & 0 & -1 & \cdots & 0 \\ 1 & 0 & 1 & -1 & 0 & -1 & \cdots & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & \cdots & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & -1 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \end{pmatrix},$$

$$Y\_2 = \begin{pmatrix} 0 & 0 & \cdots & 0 & 0 & 0 & 0 & 0 & 0 \\ -1 & 0 & \cdots & 0 & 0 & 0 & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 1 & \cdots & 1 & 0 & -1 & 0 & 0 & 1 \\ 0 & 1 & \cdots & 1 & 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & \cdots & 0 & 1 & 0 & 1 & -1 & 0 \\ \end{pmatrix},$$

*Y*3 = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 0 00 0 0 0 ··· 0 0 0 00 0 0 0 ··· 0 0 −100 0 0 0 ··· 0 0 0 00 −10 0 ··· ... ... ... ... ... ... ... ... ... −11 0 00 0 0 0 ··· −11 0 10 0 1 −1 ··· ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , *Y*4 = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 0 0 0 00 ··· 0 0 0 0 0 00 ··· ... ... ... ... ... ... ... ... 0100 −100 ··· 0 0 0 0 0 00 ··· −1 0 −11 0 00 ··· −1 0 −11 0 00 ··· ... ... ... ... ... ... ... ... 0 0 0 0 0 01 ··· ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , . . . *Yi* = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 000 000 0 ··· 0 000 000 0 ··· ... ... ... ... ... ... ... ... ... 0 −10 0 0 0 0 0 ··· 0 000 000 0 ··· 0 000 −1 1 −1 0 ··· 0 01 −1000 −1 ··· −1000 000 0 ··· ... ... ... ... ... ... ... ... ... 0 000 000 0 ··· ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠ , . . . *Y*6*k*+<sup>3</sup> = ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝ 0 0 0 0 000 ··· 0 0 0 0 0 000 ··· 0 0 0 0 0 000 ··· 0 0 0 0 0 000 ··· 0 ... ... ... ... ... ... ... ... ... 0 −1 1 −1000 ··· 0 ⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠ .

Here ker *d*6*k*+<sup>3</sup> is the full space *V*⊗2*k*+<sup>1</sup> and the im *d*6*k*+<sup>2</sup> is

<sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*++*v*−⊗*v*+⊗*v*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*−⊗*v*++*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*+)Z(*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*−+*v*−⊗*v*+⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*−⊗*v*−+*v*+⊗*v*−⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−⊗*v*++*v*−⊗*v*+⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*−+*v*−⊗*v*+⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*+⊗*v*−⊗*v*++*v*−⊗*v*−⊗*v*+⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−⊗*v*−+*v*−⊗*v*+⊗*v*−⊗*v*−) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*−⊗*v*−)

⊕ <sup>Z</sup>(*<sup>v</sup>*+⊗*v*+⊗*v*−⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*+⊗*v*+⊗*v*−) ⊕ <sup>Z</sup>(*v*−⊗*v*+⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*+⊗*v*−⊗*v*−).

> Thus, H6*k*+<sup>3</sup>(Δ<sup>3</sup>) = 0, and we finally obtain the result:

**Theorem 4.** *The Khovanov homology of the link* Δ2*k*+<sup>1</sup> *is*

$$\mathcal{H}^i(\Delta^{2k+1}) = \begin{cases} 0 & 6k \le i \le 3i \\\mathbb{Z} \oplus \mathbb{Z} & i = 2 \\\ 0 & i = 1 \\\mathbb{Z} \oplus \mathbb{Z} & i = 0 \end{cases}$$

The following result gives some homology groups of <sup>Δ</sup>2*k*+1*x*2 = *x*2*k*+<sup>3</sup> 1 *<sup>x</sup>*2*<sup>x</sup>*21*x*22*x*21 ··· *x*22*x*21*x*21.

**Theorem 5.**

$$\mathcal{H}^i(\Delta^{2k+1}x\_2) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & i = 0 \\ 0 & i = 1 \\ 0 & i = 6k + 4 \end{cases}$$

**Proof.** The cochain complex of link <sup>Δ</sup>2*k*+1*x*2 is

$$\begin{split} 0 &\xrightarrow{d^{-1}} V^{\otimes 3} \xrightarrow{d^0} \oplus \mathbf{6} + \mathbf{4} \, V^{\otimes 2} \xrightarrow{d^1} \oplus \begin{split} & \bigoplus\_{(2^k+2)} (^{4k+2}) \, V^{\otimes 1} \oplus (^{2k+2}) \, ^{(4k+2)} \, V^{\otimes 3} \\ & \xrightarrow{d^2} \oplus \begin{split} & (^{2k+2}) (^{4k+2}\_2) + (^{2k+2}\_2) (^{4k+2}\_1) \, ^{V^{\otimes 2}} \oplus (^{2k+2}\_1) + (^{4k+2}\_1) \, ^{V^{\otimes 4}} \xrightarrow{d^3} \\ & \dots \, \begin{split} & \bigoplus\_{(2^k+2)} V^{\otimes 2k} \oplus (^{4k}\_1) \, ^{V^{\otimes 2k}} \end{split} \end{split} \\ & \xrightarrow{d^{6k+2}} \oplus \begin{split} & (^{2k+2}\_1) \, ^{V^{\otimes 2k}} \oplus (^{4k+2}\_1) \, ^{V^{\otimes 2k+2}} \xrightarrow{d^{6k+3}} \, V^{\otimes 2k+1} \, \frac{d^{6k+4}}{^{}} \end{split} \end{split}$$

Differential *d*0 having an order of 24*k* + 16 × 8 is

$$d^0 = \begin{pmatrix} A & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & A & B & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & C & A & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & C & A & B & 0 \end{pmatrix} \prime$$

where *A*, *B*, and *C*, each having an order of (6*k* + 4) × 1, are:

> *A* = 1111111 ··· 1 *t B* = 0100100 ··· 0 1 *t C* = 1011011 ··· 1 0 *t*

Since im *d*−<sup>1</sup> = 0 and ker *d*0 = <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−<sup>−</sup>*v*−⊗*v*+⊗*v*−+*v*−⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*−), <sup>H</sup><sup>0</sup>(Δ2*k*+1*x*2) = <sup>Z</sup>(*<sup>v</sup>*+⊗*v*−⊗*v*−<sup>−</sup>*v*−⊗*v*+⊗*v*−+*v*−⊗*v*−⊗*v*+) ⊕ <sup>Z</sup>(*v*−⊗*v*−⊗*v*−).

Now, differential *d*1 of an order of 18(6*k*<sup>2</sup> + 6) × 4(6*k* + 4) is

$$d^1 = \begin{pmatrix} M\_1 & -M\_1 & 0 & 0 \\ M\_1 & 0 & -M\_1 & 0 \\ M\_1 & 0 & 0 & -M\_1 \\ M\_2 & -M\_2 & 0 & 0 \\ M\_2 & 0 & -M\_2 & 0 \\ M\_3 & -M\_3 & 0 & 0 \\ M\_3 & 0 & -M\_3 & 0 \\ 0 & M\_4 & -M\_4 & 0 \\ 0 & M\_4 & 0 & -M\_4 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & M\_{n-1} & M\_n \end{pmatrix},$$

where the order of each *Mi* is (16*k* + 2) × (6*k* + 4) and is


$$M\_{4} = \begin{pmatrix} -1 & 0 & 0 & 0 & \dots & \dots & \dots & 1 & 0 \\ 0 & -1 & -1 & 0 & \dots & \dots & \dots & 0 & 1 \\ 0 & 0 & 0 & 0 & \dots & \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & \dots & \dots & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & 0 \\ 0 & -1 & 0 & 0 & \dots & \dots & \dots & 0 & 0 \\ 0 & 0 & -1 & 0 & \dots & \dots & \dots & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & \dots & 0 & 0 \\ \end{pmatrix}, \quad \text{and} \quad M\_{n} = \left( \begin{array}{cccc} 0 & 0 & 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & \dots & -1 & \dots & \end{array} \right).$$

In this case, the kernel of *d*1 and image of *d*0 contain the same number of Z spaces. So, <sup>H</sup><sup>1</sup>(Δ2*k*+1*x*2) = 0.

Finally, the differential of *d*6*k*+<sup>4</sup> of an order of 22*k*+<sup>1</sup> × (2*k* + 3)(22*<sup>k</sup>*)(22*k*+<sup>1</sup>) is

$$d^{6k+4} = \left( \begin{array}{cccc} \chi\_1 & \chi\_2 & \chi\_3 & \chi\_4 & \dots & \chi\_i \end{array} \right)$$

,

where each *Yi*has an order of 22*k*+<sup>1</sup> × 6*k* + 4 and is


$$\begin{aligned} \mathcal{Y}\_{4} &= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ -1 & 1 & -1 & 0 & 0 & 1 & 0 & 0 & -1 & 1 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 1 & -1 & 0 & -1 & 1 & 0 & 0 \end{pmatrix}, \\\\ &\vdots \\\\ \mathcal{Y}\_{i} &= \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ -1 & 1 & -1 & 0 & 0 & 1 & 0 & \cdots & 0 \end{pmatrix}. \end{aligned}$$

It is evident that ker *d*6*k*+<sup>4</sup> is full space *V*⊗2*k*+1. Moreover, im *d*6*k*+<sup>3</sup> is also *V*⊗2*k*+1. We also ge<sup>t</sup> the Khovanov homology of braid link <sup>Δ</sup>2*k*+1*x*1:
