\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}}\right) \cdot \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (2,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (1.5,0)--(1.5,-0.5);
\draw (1.5,1.5)--(1.5,1);
\draw (1.75,0)--(1.75,-0.5);
\draw (1.75,1.5)--(1.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1,0.5) node[scale=1.33] {$\varrho$};
\end{scope}}\\
&= \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (1.25,1.5)--(1.25,-0.5);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (1.75,1.5)--(1.75,-0.5);
\end{scope}} \cdot \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (2,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (1.5,0)--(1.5,-0.5);
\draw (1.5,1.5)--(1.5,1);
\draw (1.75,0)--(1.75,-0.5);
\draw (1.75,1.5)--(1.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1,0.5) node[scale=1.33] {$\varrho$};
\end{scope}} = \knotsinmath{\begin{scope}[yshift=0.45cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
% \draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (1.25,1.5)--(1.25,-0.5);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (1.75,1.5)--(1.75,-0.5);
\end{scope}
\begin{scope}[yshift=-1.05cm,scale=0.75]
\draw (0,0) rectangle (2,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (1.5,0)--(1.5,-0.5);
\draw (1.5,1.5)--(1.5,1);
\draw (1.75,0)--(1.75,-0.5);
\draw (1.75,1.5)--(1.75,1);
% \draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1,0.5) node[scale=1.33] {$\varrho$};
\end{scope}}
\end{align*}

In general, the $n$-braid $\alpha\beta$ is not equivalent to $\beta\alpha$. However, braid multiplication is associative, i.e., $\alpha(\beta\gamma)=(\alpha\beta)\gamma$ (see proof of Theorem <ref>).

[strands=3,braid start=(0,-0.25)]
(1,-1) node $\alpha$;

[strands=3,braid start=(2,0)]
(3,-1.25) node $\beta$;

[strands=3,braid start=(0,-3/2)] σ_2^-1σ_2^-1σ_1
(1,-3.25) node $\alpha\beta$;

[strands=3,braid start=(2,-3/2)] σ_2^-1σ_1σ_2^-1
(6.5/2,-6.5/2) node $\beta\alpha$;

(<cit.>) braid multiplication is not commutative, i.e. $\alpha\beta$ is not equivalent to $\beta\alpha$ (their braid permutations are different, hence they are not equivalent)

The braid identity $\Xi$ is the trivial braid such that $\Xi\alpha=\alpha=\alpha\Xi$.

(0.55,-0.5) node $\ldots$;

trivial braid

Finally, the inverse of a braid is the reflection of a braid, such that $\alpha^{-1}\cdot\alpha=\Xi=\alpha\cdot\alpha^{-1}$.

\[\left(\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}}\right)^{-1} = \,\,\reflectbox{\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}}}\]

braid inverse

and so,

\[\knotsinmath{\begin{scope}[yshift=0.45cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.55) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (-1,-0.5) node[scale=1.33] {$\alpha\cdot\alpha^{-1}=$};
\end{scope}
\begin{scope}[yshift=-1.05cm,xshift=0.75cm,scale=0.75,xscale=-1]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}} = \knotsinmath{\begin{scope}[yshift=0.5cm]
\draw (0.1,0)--(0.1,-1);
\draw (0.2,0)--(0.2,-1);
\draw (0.3,0)--(0.3,-1);
\draw (0.55,-0.5) node {$\ldots$};
\draw (0.8,0)--(0.8,-1);
\draw (0.9,0)--(0.9,-1);
\end{scope}}=\Xi\]

The following shows an example of multiplying a braid by its inverse.

[strands=2,braid start=(0,0)]
(1.5/2,-1.5/2) node $\alpha$;
[strands=2,braid start=(3/2,0)]
(4.5/2,-1.5/2) node $\alpha^{-1}$;
[strands=2,braid start=(1.5/2,-2/2)] σ_1 σ_1^-1
(3/2,-4.5/2) node $\alpha\alpha^{-1}$;

braid inverse example

The braid $\alpha\alpha^{-1}$ is clearly equivalent to the trivial braid after an $\RII$ move.

We are done. Let us summarize the braid group with the following statement.
braid group $\Bcal_n$!geometrical
The set of all $n$-braids $\Bcal_n$ forms a group under braid composition (concatenation) defined below. Given two $n$-braids $\alpha$ and $\beta$,

(0.5,-0.5) node $\alpha$;
(1.5,-0.5) node $\odot$;
(2.5,-0.5) node $\beta$;
(3.5,-0.5) node $=$;
(4.5,0) node $\alpha$;
(4.5,-1) node $\beta$;
[dashed] (4,-0.5)–(5,-0.5);
(5.5,-0.5) node $=$;
(6.5,-0.5) node $\alpha\beta$;

Let $\alpha,\beta,\gamma$ be $n$-braids. Composing two $n$-braids, by definition, will give a new $n$-braid. So we have closure: $\odot\colon\Bcal_n\odot\Bcal_n\to\Bcal_n$. We also have associativity, i.e. $\alpha(\beta\gamma)=(\alpha\beta)\gamma$ (see figure below).

(0.5,0) node $\alpha$;
(0.5,-1) node $\beta\gamma$;
[dashed] (0,-0.5)–(1,-0.5);
(1.5,-0.5) node $=$;
(2.5,0.5) node $\alpha$;
(2.5,-0.5) node $\beta$;
(2.5,-1.5) node $\gamma$;
[dashed] (2,0)–(3,0);
[dashed] (2,-1)–(3,-1);
(3.5,-0.5) node $=$;
(4.5,0) node $\alpha\beta$;
(4.5,-1) node $\gamma$;
[dashed] (4,-0.5)–(5,-0.5);

We have an identity $\Xi$ which is the trivial $n$-braid. So, $\alpha\Xi=\alpha=\Xi\alpha$. The figure below shows $\alpha\Xi=\alpha$ (from this, it then follows that $\Xi\alpha=\alpha$).

(0.5,-0.5) node $\alpha$;
(1.5,-0.5) node $\odot$;
(2.55,-0.5) node $\ldots$;
(3.5,-0.5) node $=$;
(4.5,0) node $\alpha$;
(4.55,-1) node $\ldots$;
[dashed] (4,-0.5)–(5,-0.5);
(5.5,-0.5) node $=$;
(6.5,-0.5) node $\alpha$;

Finally, we have inverses. By definition, there exists an $n$-braid, called $\alpha^{-1}$ for the sake of argument, where the composition of this with another $n$-braid, say $\alpha$, would produce a braid which is equivalent to the trivial braid by elementary triangle moves. So, $\alpha\alpha^{-1}=\Xi=\alpha^{-1}\alpha$. The inverse braid is the mirror image of the braid.

So, similarly to how there exists a natural inclusion of the set $\Bcal_n$, the braid group $\Bcal_n$ is embedded as a subgroup of $\Bcal_{n+1}$ (with the braid inclusion map).

There exists a representation of the group by writing all the braids as words on some generators. The generators $\sigma_i$ is defined by connecting $A^\prime_i$ to $A_{i+1}$ and $A^\prime_{i+1}$ to $A_i$ and is seen in Figure <ref>.

(-0.5/2,-0.5/2) node $\sigma_i =$;
[braid start=(0,0),strands=8] σ_4
(1/2,-1.5/2) node 1;
(2/2,-1.5/2) node 2;
(4/2,-1.5/2) node $i$;
(5/2,-1.53/2) node $i+1$;
(8/2,-1.5/2) node $n$;

braid generator $\sigma_i$

And the inverse, $\sigma_i^{-1}$ is just

(-0.5/2,-0.5/2) node $\sigma_i^{-1}=$;
[braid start=(0,0),strands=8] σ_4^-1
(1/2,-1.5/2) node 1;
(2/2,-1.5/2) node 2;
(4/2,-1.5/2) node $i$;
(5/2,-1.53/2) node $i+1$;
(8/2,-1.5/2) node $n$;

These generators are also known as elementary transpositionselementary transposition.

We can use these generators to express any element in the braid group.

[braid start=(0,0),strands=3] σ_1σ_2σ_1^-1

We can also express any element in the braid group using these generators. We can do this by dividing our braid diagram with lines parallel to the bars. The rule is that we can only have one crossing in each divided rectangle, if there are more crossings, then we can just shift one crossing downwards or upwards. See below figure.

[braid start=(0,0),strands=3] σ_2σ_2σ_1^-1
[dashed] (0.5/2,-1/2)–(3.5/2,-1/2);
[dashed] (0.5/2,-2/2)–(3.5/2,-2/2);

(4.5/2,-0.5/2) node $\sigma_2$;
(4.5/2,-1.5/2) node $\sigma_2$;
(4.5/2,-2.5/2) node $\sigma_1^{-1}$;

Thus the braid group $\Bcal_n$ can be described by those generators and three relations. It is usually referred to the Artin braid group, since Artin introduced it [8, 9, 12].

The Artin braid group $\Bcal_n$braid group $\Bcal_n$!Artin's presentation is the group generated by $\sigma_i$ for $i=1,\ldots,n-1$,

(-0.5/2,-0.5/2) node $\sigma_i =$;
[braid start=(0,0),strands=8] σ_4
(1/2,-1.5/2) node 1;
(2/2,-1.5/2) node 2;
(4/2,-1.5/2) node $i$;
(5/2,-1.5/2) node $i+1$;
(8/2,-1.5/2) node $n$;

and has the following relations

* $\sigma_i\sigma_j=\sigma_j\sigma_i \mbox{ if } |i-j|\geq2$

* $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1} \mbox{ for } i=1,2,\ldots,n-2$

* $\sigma_i\sigma_i^{-1}=\Xi, \mbox{ for } i=1,\ldots{n-1}$

Notice that the first relation just means that we can shift the crossings if they are “separated.”

[braid start=(0,0),strands=4] σ_1σ_3

(5/2,-1/2) node $=$;

[braid start=(5/2,0),strands=4] σ_3σ_1

$\sigma_i\sigma_j=\sigma_j\sigma_i$, if $|i-j|\geq2$

The second relation, here, is essentially the Redemeister $3$ move.

[braid start=(0,0),strands=3] σ_1σ_2σ_1

(4/2,-1.5/2) node $=$;

[braid start=(4/2,0),strands=3] σ_2σ_1σ_2

The third relation is just the Redeimeister 2 move.

[braid start=(0,0),strands=6] σ_3σ_3^-1
(1.5/2,-1/2) node[scale=0.6] $\ldots$;
(5.5/2,-1/2) node[scale=0.6] $\ldots$;
(7/2,-1/2) node $=$;
(8.25/2,-1/2) node[scale=0.6] $\ldots$;

$\Bcal_1$ is generated by $\sigma_1$

$\Bcal_2$ is generated by $\sigma_1,\sigma_2$ and has the relation $\sigma_1\sigma_2\sigma_1=\sigma_2\sigma_1\sigma_2$

$\Bcal_3$ is generated by $\sigma_1,\sigma_2,\sigma_3$ and has (in addition to the relations in $\Bcal_2$) $\sigma_2\sigma_3\sigma_2=\sigma_3\sigma_2\sigma_3$ and $\sigma_1\sigma_3=\sigma_3\sigma_1$.

The braid group $\Bcal_2$ is isomorphic to $\ZZ$.

We have already seen the braid permutation from Definition <ref>. So it is not hard to see that there exists a natural homomorphism from the Artin braid group $\Bcal_n$ into the symmetric group $S_n$.

There is a group homomorphism $\varphi$, from $\Bcal_n$ onto the symmetric group $S_n$.
We have the exact sequence

\[1\to{P\Bcal_n}\to\Bcal_n\to{S_n}\to1\]

where the kernel of the homomorphism is known as the pure braid group $P\Bcal_n$.

So the end points of elements in the pure braid group are not permuted. We will give one example. The interested reader is directed to [12].

An example of an element in the pure braid group $P\Bcal_n$ is shown below,

[strands=3,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_2σ_1σ_2σ_1σ_1σ_2^-1σ_1^-1σ_2^-1
(1,-4.5) node $\sigma_2\sigma_1\sigma_2\sigma_1^2\sigma_2^{-1}\sigma_1^{-1}\sigma_2^{-1}$;

The generators of the pure braid group are given by

\[{\text{pb}}_{i,j}=(\sigma_{j-1}\sigma_{j-2}\cdots\sigma_{i+1})\sigma_i^2(\sigma_{i+1}^{-1}\cdots\sigma_{j-2}^{-1}\sigma_{j-1}^{-1}),\]

for $i<j$. And so, visually, we have

[strands=6,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_2σ_3σ_4^-1σ_4^-1σ_3^-1σ_2^-1
(1,-3.25) node $i$;
(2.5,-3.25) node $j$;
(-0.75,-1.625) node $\text{pb}_{i,j}=$;

pure braid generator

Now we have an easier, purely algebraic, method of showing that two braids are equivalent.

The following relations hold for any $\Bcal_n$ for $n\geq3$.

\[
\sigma_1\sigma_2\sigma_1^{-1}=\sigma_2^{-1}\sigma_1\sigma_2 \mbox{ and } \sigma_2\sigma_1^{-1}\sigma_2^{-1}=\sigma_1^{-1}\sigma_2^{-1}\sigma_1
\]

\begin{align*}
\sigma_1\sigma_2\sigma_1^{-1} &= \sigma_2^{-1}\sigma_2\sigma_1\sigma_2\sigma_1^{-1}\\
&= \sigma_2^{-1}\sigma_1\sigma_2\sigma_1\sigma_1^{-1} = \sigma_2^{-1}\sigma_1\sigma_2
\end{align*}

\begin{align*}
\sigma_2\sigma_1^{-1}\sigma_2^{-1} &= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_1^{-1}\sigma_1\\
&= \sigma_2\sigma_2^{-1}\sigma_1^{-1}\sigma_2^{-1}\sigma_1 = \sigma_1^{-1}\sigma_2^{-1}\sigma_1
\end{align*}

These will look like

[strands=3,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_2σ_1^-1
(1,-2) node $\sigma_1\sigma_2\sigma_1^{-1}$;
[strands=3,braid start=(2,0),braid width=0.5cm,braid height=0.5cm] σ_2^-1σ_1σ_2
(3,-2) node $\sigma_2^{-1}\sigma_1\sigma_2$;
(2,-0.75) node $\sim$;

[strands=3,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_2σ_1^-1σ_2^-1
(1,-2) node $\sigma_2\sigma_1^{-1}\sigma_2^{-1}$;
[strands=3,braid start=(2,0),braid width=0.5cm,braid height=0.5cm] σ_1^-1σ_2^-1σ_1
(3,-2) node $\sigma_1^{-1}\sigma_2^{-1}\sigma_1$;
(2,-0.75) node $\sim$;

Which are clearly just the Reidemeister type 3 move.

To motivate our next example, we will very briefly discuss the conjugacy problem. Our discussion is based on that in Birman's survey <cit.>.

The conjugacy problemconjugacy problem asks for an algorithm to decide whether $\{a\}=\{a'\}$, where $\{a\},\{a'\}$ are the conjugacy classes of the elements $a,a'\in\Bcal_n$. A solution was first given in 1965 by Garside [22]. While, there does exist a finite set of conjugates of a representation of an arbitrary element $a\in\Bcal_n$, it is exponential in both $n$ and $|a|$. This makes it difficult for us to understand the combinatorial solution sometimes.

The following two braids are known to be equivalent. So we will show that these two braids are equivalent by using the braid group relations and generators. In other words, we are finding a solution that gives us an algorithm to go from the first braid to the second braid (the conjugacy problem). The reader is advised to attempt it.

[braid start=(-3,-0.5),braid width=0.5cm,braid height=0.5cm,strands=4] σ_1^-1σ_2^-1σ_3σ_3σ_2σ_1σ_3^-1σ_1σ_2σ_3^-1
(-1.75,-6.5) node $\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}$;

[braid start=(3,0),braid width=0.5cm,braid height=0.5cm,strands=4] σ_2σ_1^-1σ_2^-1σ_3σ_3σ_1^-1σ_2σ_1σ_1σ_3^-1σ_2σ_3^-1
(4.25,-6.5) node $\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_1^{-1}\sigma_2\sigma_1^2\sigma_3^{-1}\sigma_2\sigma_3^{-1}$;

\begin{align*}
\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1} &= \sigma_2\sigma_2^{-1}\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}\sigma_2\sigma_2^{-1}\\
&= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_1^{-1}\sigma_3^2\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2^{-1}\\
&= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_1^{-1}\sigma_3\sigma_2\sigma_3^{-1}\sigma_1^2\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2^{-1}\\
&= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_1^{-1}\sigma_2\sigma_1^{2}\sigma_3^{-2}\sigma_2\sigma_3^{-1}\sigma_2^{-1}\\
&= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_1^{-1}\sigma_2\sigma_1^2\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2\sigma_2^{-1}\\
&= \sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3^2\sigma_1^{-1}\sigma_2\sigma_1^2\sigma_3^{-1}\sigma_2\sigma_3^{-1}
\end{align*}

We will show each relation diagrammatically.

\begin{align*}
\knotsinmath{\braid[braid start={(-1,2)},braid width=0.5cm,braid height=0.5cm,strands=4] {\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_3\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}}} \quad\sim \knotsinmath{\braid[braid start={(0,3)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_2^{-1}\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_3\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_2\sigma_3^{-1}\sigma_2\sigma_2^{-1}}}
\quad\sim \knotsinmath{\braid[braid start={(0,3)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_1^{-1}\sigma_3\sigma_3\sigma_2\sigma_1\sigma_3^{-1}\sigma_1\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2^{-1}}}
\quad\sim \knotsinmath{\braid[braid start={(0,3)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_1^{-1}\sigma_3\sigma_2\sigma_3^{-1}\sigma_1\sigma_1\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2^{-1}}}\\\\
\quad\sim \knotsinmath{\braid[braid start={(0,3)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_3\sigma_1^{-1}\sigma_2\sigma_1\sigma_1\sigma_3^{-1}\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2^{-1}}}
\quad\sim \knotsinmath{\braid[braid start={(0,3)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_3\sigma_1^{-1}\sigma_2\sigma_1\sigma_1\sigma_3^{-1}\sigma_2\sigma_3^{-1}\sigma_2\sigma_2^{-1}}}
\quad\sim \knotsinmath{\braid[braid start={(0,2.5)},braid width=0.5cm,braid height=0.5cm,strands=4]{\sigma_2\sigma_1^{-1}\sigma_2^{-1}\sigma_3\sigma_3\sigma_1^{-1}\sigma_2\sigma_1\sigma_1\sigma_3^{-1}\sigma_2\sigma_3^{-1}}}
\end{align*}

§ THE LINK BETWEEN LINKS & BRAIDS

We will now see how to associate links with braids and vice versa.

This chapter is based on the presentations given in [2], [3], [4], [10], [12], [13], [16], [17], [18], [21], [26], [27], [28], [29], [33], [34], [38], [44], [49], [58], [59], and [65].

However, mainly our discussions will be based on that in Birman's book [10] and Birman's survey [12].

Consider any $n$-braid $\alpha$. We define the closurelinear operations on $\Bcal_n$!braid closure of the braid $\alpha$ by connecting the parallel arcs on the right from the top bar to the bottom bar by starting with the far right, and we denote it as $\reallywidehat{\alpha}$.

\[
\knotsinmath{\begin{scope}[xshift=-0.5cm]\draw (0,0.5) node {$\reallywidehat{\alpha}=$};\end{scope}\begin{scope}[yshift=0.1cm,scale=0.75]
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw[red] (0.75,1.5)--(1.25,1.5)--(1.25,-0.5)--(0.75,-0.5);
\draw[red] (0.25,1.5)--(0.25,1.75)--(1.5,1.75)--(1.5,-0.75)--(0.25,-0.75)--(0.25,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}}
\]
braid closure

The closure of the braid $\sigma_1\sigma_2\sigma_1$ is shown below and is Reidemeister equivalent to a link (it should be fairly easy to recognise which link).

[strands=3,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_2σ_1
[red] (1.5,0)–(2,0)–(2,-1.5)–(1.5,-1.5);
[red] (1,0)–(1,0.2)–(2.2,0.2)–(2.2,-1.7)–(1,-1.7)–(1,-1.5);
[red] (1/2,0)–(1/2,0.4)–(2.4,0.4)–(2.4,-1.9)–(1/2,-1.9)–(1/2,-1.5);

The braid closure is a regular diagram of a link. We can assign an orientation by going from the top to the bottom each time we add an arc, i.e.,

(0.5,0.5) node[scale=1.33] $\reallywidehat{\alpha}$;
(0,0) rectangle (1,1);
[red,->] (1,1.5)–(1.1,1.5);
[red,->] (0.875,1.75)–(0.9,1.75);
[red] (0.75,1.5)–(1.25,1.5)–(1.25,-0.5)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,1.75)–(1.5,1.75)–(1.5,-0.75)–(0.25,-0.75)–(0.25,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

Thinking of the braid closure as a map from braid types to link types. It is typical to question whether this map is injective or surjective. We can easily show that it is in fact not injective (for instance, $\reallywidehat{\sigma_1\sigma_2\sigma_1^{-1}}=\reallywidehat{\sigma_2}$, but $\sigma_1\sigma_2\sigma_1^{-1}\not\sim\sigma_2$). Fascinatingly enough, it is, however, surjective. This was first shown in 1923 by Alexander [5].

Alexander's theorem
Every link is ambient isotopic to a braid closure.

Before we begin to sketch the proof. We will need to define Seifert circles.

Let $D$ be an oriented diagram for the oriented link $L$ and let $D'$ be modified as shown in Figure <ref>.

[->] (2,0)–(0,2);
[ultra thick,draw=white,double=black,double distance=0.5pt] (0,0)–(2,2);
[->] (1.9,1.9)–(2,2);

[|->] (1,1)–(2,1);

[->] (0,0) .. controls (0.65,1) .. (0,2);
[->] (1.3,0) .. controls (0.65,1) .. (1.3,2);

[|->] (2,1)–(1,1);

[->] (0,0)–(2,2);
[ultra thick, draw=white,double=black,double distance=0.5pt] (2,0)–(0,2);
[->] (0.1,1.9)–(0,2);

Seifert's algorithm

Diagram $D'$ is the same as $D$ except in a small neighbourhood of each crossing where the crossing has been removed in the only way compatible with the orientation.
This $D'$ is just a disjoint union of oriented simple closed curves in $\RR^2$, called Seifert circlesSeifert circles.

(0,-2.5) node[scale=3.2] trefoil knot;

[->] ( 90:2) – +(  0:0.2);
(0,-2.5) node[scale=3.2] add orientation;

[->] ( 90:1) – +(  0:0.1);
[->] ( 90:0.2) – +(0:0.05);
(0,0) circle(1);
(0,0) circle(0.2);
(0,-1.5) node[scale=1.6] form Seifert circles;

trefoil knot Seifert circles

We say that a link is braided aroundbraided around a point if the Seifert circles of the link are nested and coherent (i.e., they all orient clockwise around the point).

We have the following equivalent definition for a link being braided around a point.

A link is braided around a point if each edge of the link is visible from the point and is oriented clockwise.

It is in the author's opinion that checking Seifert circles is easier at times. It also motivates Vogel's proof of Alexander's theorem in 1990 [62] which was based on Yamada's proof in 1987 [66] using Seifert circles and “infinity moves”.
Vogel's proof turns out to terminate quicker than the original proof given by Alexander. The interested reader is directed to [12].

Thus it is easy to see that the only three knots in our knot table from the appendix:knot_tableAppendix that are already braided around a point are $3_1$, $5_1$, and $7_1$.

We will show that there is a method that can show that any link can be braided around a point.
Consider a link diagram $L$ and a point $O$ on the plane $P$ of the diagram (the point should not touch the link). If there exists a point on the plane such that the link diagram is braided around the point, then we “cut” the diagram along that point to get the braid.

(0,0) rectangle (1,1);
[->] (1,1.5)–(1.1,1.5);
[->] (0.875,1.75)–(0.9,1.75);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(1.35,0.5) circle (0.05);
[->] (-1.5,0.5)–(-0.75,0.5);
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

cutting a braided around link

Now consider an arbitrary link diagram. We need to construct this, so that it will be braided around some point. We need to first set a point $O$ on the plane $P$. If there happens to be an edge, say $AB$, that is oriented counterclockwise, then we apply the triangle method by finding some point $C$ on $P$, such that $ABC$ contains $O$, by replacing $AB$ by $AC$ and $BC$. Now $AC$ and $BC$ are both oriented clockwise: the orientation we were looking for. We continue doing this until we get a link diagram that is braided around our point $O$.

[ultra thick] (0,0)–(-1,1);
[dashed] (0,0)–(2,1.25)–(-1,1);
[->,thick] (-0.5,0.5)–(-0.25,0.25);
(0.35,0.75) circle (0.05);
(-1,1) node[anchor=south east] $A$;
(0,0) node[anchor=north west] $B$;
(2,1.25) node[anchor=west] $C$;
[->] (-3,0.75)–(-2.5,0.75);
[dashed] (0,0)–(-1,1);
[->] (0.8,0.5)–(0.5,0.3125);
[->] (0.2,1.1)–(0.5,1.125);
(0.35,0.75) circle (0.05);
(-1,1) node[anchor=south east] $A$;
(0,0) node[anchor=north west] $B$;
(2,1.25) node[anchor=west] $C$;

Alexander's method

We know that this method will eventually terminate given any point $O$, since with every triangle move we make, the number of counterclockwise edges decreases by one, thus eventually giving us a link that is braided around a point $O$.

We have already mentioned above that the trefoil is already braided around. So, we have the following diagram of the left-handed trefoil.

[->] ( 88:2.04) – +(  0:0.01);

(0,0) node $\circ$;
[thin] (0,0)–(0.1,1.4);
[thin] (0,0)–(0.3,1.4);
(1.5,0) node $\to$;
[strands=2,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_1σ_1

After cutting it, we see that the braid representation is given by $\sigma_1^{-3}$. It is easy to see that the right-handed trefoil's braid representation is $\sigma_1^3$.

Now let us find the braid representation of the following figure-eight knot using Alexander's method.

(120.7,151) .. controls (152.22,153) and (154.84,53) .. (111.08,40) .. controls (67.31,27) and (74.31,67) .. (90.94,90) ;
(75.73,143) .. controls (63.73,107) and (137.34,105) .. (90.07,55) ;
(71.69,45) .. controls (44.55,29) and (4.29,152) .. (102.32,152) ;
(105.82,106) .. controls (126.75,140.33) and (105.73,178) .. (93.73,185) .. controls (81.73,192) and (78.3,163.97) .. (76.94,157) ;
(147.14,82.45) .. controls (144.92,86.16) and (143.74,89.75) .. (143.59,93.23) .. controls (142.69,89.86) and (140.76,86.61) .. (137.79,83.47) ;
(35.4,105.33) .. controls (37.96,101.84) and (39.48,98.38) .. (39.95,94.93) .. controls (40.53,98.36) and (42.15,101.78) .. (44.81,105.18) ;
(101.46,61.27) .. controls (101.08,65.48) and (101.55,69.2) .. (102.87,72.4) .. controls (100.7,69.7) and (97.67,67.5) .. (93.8,65.8) ;
(87.85,117.58) .. controls (83.73,118.89) and (80.46,120.81) .. (78.06,123.32) .. controls (79.61,120.2) and (80.3,116.48) .. (80.15,112.16) ;
(110.38,128.56) .. controls (111.65,124.43) and (111.94,120.66) .. (111.26,117.25) .. controls (112.93,120.3) and (115.58,122.99) .. (119.22,125.33) ;
(80.44,82.58) .. controls (81.67,78.44) and (81.93,74.67) .. (81.21,71.26) .. controls (82.91,74.3) and (85.59,76.96) .. (89.25,79.27) ;
(91.13,183.21) .. controls (95.32,182.14) and (98.69,180.42) .. (101.24,178.05) .. controls (99.51,181.07) and (98.6,184.74) .. (98.5,189.06) ;
(90.89,72.67) .. controls (90.89,72.02) and (91.36,71.5) .. (91.94,71.5) .. controls (92.53,71.5) and (93,72.02) .. (93,72.67) .. controls (93,73.31) and (92.53,73.83) .. (91.94,73.83) .. controls (91.36,73.83) and (90.89,73.31) .. (90.89,72.67) – cycle ;

[->] (163,100)–(188,100);
(299.7,150) .. controls (331.22,152) and (333.84,52) .. (290.08,39) .. controls (246.31,26) and (253.31,66) .. (269.94,89) ;
(254.73,142) .. controls (242.73,106) and (316.34,104) .. (269.07,54) ;
(250.69,44) .. controls (223.55,28) and (183.29,151) .. (281.32,151) ;
(284.82,105) .. controls (290.87,114.91) and (287,192.25) .. (319.5,171.75) .. controls (352,151.25) and (345,65.25) .. (315,42.75) .. controls (285,20.25) and (261,20.75) .. (240.5,29.75) .. controls (220,38.75) and (193,66.25) .. (208,112.75) .. controls (223,159.25) and (264.5,175.25) .. (255.94,156) ;
(326.14,81.45) .. controls (323.92,85.16) and (322.74,88.75) .. (322.59,92.23) .. controls (321.69,88.86) and (319.76,85.61) .. (316.79,82.47) ;
(214.4,104.33) .. controls (216.96,100.84) and (218.48,97.38) .. (218.95,93.93) .. controls (219.53,97.36) and (221.15,100.78) .. (223.81,104.18) ;
(280.46,60.27) .. controls (280.08,64.48) and (280.55,68.2) .. (281.87,71.4) .. controls (279.7,68.7) and (276.67,66.5) .. (272.8,64.8) ;
(266.85,116.58) .. controls (262.73,117.89) and (259.46,119.81) .. (257.06,122.32) .. controls (258.61,119.2) and (259.3,115.48) .. (259.15,111.16) ;
(284.81,134.18) .. controls (287.04,130.48) and (288.25,126.9) .. (288.41,123.42) .. controls (289.29,126.79) and (291.21,130.04) .. (294.17,133.2) ;
(259.44,81.58) .. controls (260.67,77.44) and (260.93,73.67) .. (260.21,70.26) .. controls (261.91,73.3) and (264.59,75.96) .. (268.25,78.27) ;
(248.96,21.62) .. controls (252.84,23.54) and (256.51,24.44) .. (259.99,24.31) .. controls (256.71,25.47) and (253.62,27.66) .. (250.73,30.87) ;
(269.89,71.67) .. controls (269.89,71.02) and (270.36,70.5) .. (270.94,70.5) .. controls (271.53,70.5) and (272,71.02) .. (272,71.67) .. controls (272,72.31) and (271.53,72.83) .. (270.94,72.83) .. controls (270.36,72.83) and (269.89,72.31) .. (269.89,71.67) – cycle ;
(270.94,71.67) – (362.65,84.57) ;
(270.94,71.67) – (355.99,78.82) ;

After cutting the diagram from our center point, we find the braid representation to be $\sigma_1\sigma_2^{-1}\sigma_1\sigma_2^{-1}$,

[strands=3,braid height=0.5cm,braid width=0.5cm,braid start=(0,0)] σ_1σ_2^-1σ_1σ_2^-1

If two braids are equivalent, it is obvious that their closures are also equivalent. However, as we have mentioned above, the map of the braid closure is non-injective. The following are more examples showing equivalent braid closures but non-equivalent braids.

Let us show that the closure of the $2$-braid $\sigma_1^{-1}$, the $3$-braid $\sigma_1\sigma_2^{-1}$, and the $4$-braids $\sigma_1\sigma_2\sigma_3^{-1}$ and $\sigma_1\sigma_2^{-1}\sigma_3^{-1}$ are all equivalent to the trivial knot.

[strands=2,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1^-1

[strands=3,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_2^-1

[strands=4,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_2σ_3^-1

[strands=4,braid start=(0,0),braid width=0.5cm,braid height=0.5cm] σ_1σ_2^-1σ_3^-1

They are all equivalent to the trivial knot via a finite number of $\RI$ moves.

We now discuss how to form a bijective braid closure map. In 1935, Markov found three moves on the braid group, that when related by them, their closures are link equivalent [45]. Four years later, in 1939, Weinberg reduced the three moves that Markov found, to two moves [63]. These moves are called Markov's moves.

Let $\Bcal_\infty=\bigcup_{k\geq1}\Bcal_k$, i.e. the infinite union of the braid groups. There are two operations, called Markov's movesMarkov's moves, which may be performed in $\Bcal_\infty$.

M1 If $\beta\in\Bcal_n$, then $\beta\xrightarrow[]{\text{M1}}\gamma\beta\gamma^{-1}$, for some $\gamma\in\Bcal_n$. The braid $\gamma\beta\gamma^{-1}$ is said to be the conjugate of $\beta$.

(0,0) rectangle (1,1);
(0.53,1.55) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\gamma$;

(0,0) rectangle (1,1);
(0.53,-0.55) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\gamma^{-1}$;

(0,0) rectangle (1,1);
(0.5,0.5) node[scale=1.33] $\beta$;

(1.75,0.5) node[scale=1.33] $\overset{\MI}{\longrightarrow}$;
(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

Markov $\MI$ move

It is easy to see that taking the closure after an M1 move gives the same link, since $\gamma$ and $\gamma^{-1}$ both cancel out.

M2 If $\beta\in\Bcal_n$, then $\beta\xrightarrow[]\MII\incl_n(\beta)\sigma_n$ or $\beta\xrightarrow[]\MII\incl_n(\beta)\sigma_n^{-1}$, for the generator $\sigma_n\in\Bcal_{n+1}$.

We call the braid $\incl_n(\beta)\sigma_n^{\pm1}$ the stabilisationstabilisation of $\beta$.

(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $A$;
[strands=3,braid width=0.5cm,braid height=0.5cm,braid start=(-0.25,-0.5),line width=0.25pt] σ_2

(1.75,0.5) node[scale=1.33] $\overset{\MII}{\longrightarrow}$;
(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

(-1,0.25) node[scale=1.33] or;
(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;
[strands=3,braid width=0.5cm,braid height=0.5cm,braid start=(-0.25,-0.5),line width=0.25pt] σ_2^-1

Markov $\MII$ move

It is easy to see that taking the closure after an $\MII$ move produces the same link since the move mimics an $\RI$ move.

We say that the two braids $\alpha,\beta\in\Bcal_\infty$ are Markov equivalentMarkov equivalent if you can transform one into the other by a finite sequence of the above moves (and their inverses). We denote it as $\alpha\sim_M\beta$.

For clarity, the closure after an $\MI$ move will look like,

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.55) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\gamma$;

(0,0) rectangle (1,1);
(0.53,-0.55) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;
[red] (0.75,3.5)–(1.25,3.5)–(1.25,-2.5)–(0.75,-2.5);
[red] (0.25,3.5)–(0.25,3.75)–(1.5,3.75)–(1.5,-2.75)–(0.25,-2.75)–(0.25,-2.5);
(2.5,0.5) node[anchor=center,scale=1.33] $\sim$;

(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\gamma$;
(0,0) rectangle (1,1);
(0.53,1.55) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\gamma$;

(0,0) rectangle (1,1);
(0.53,-0.55) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

(0,0) rectangle (1,1);
(0.5,0.5) node[scale=1.33] $\gamma$;

[red] (0.75,3.5)–(1.25,3.5)–(1.25,-2.5)–(0.75,-2.5);
[red] (0.25,3.5)–(0.25,3.75)–(1.5,3.75)–(1.5,-2.75)–(0.25,-2.75)–(0.25,-2.5);
(2.5,0.5) node[scale=1.33] $\sim$;

(0.5,0.5) node[scale=1.33] $\beta$;
(0,0) rectangle (1,1);
[red] (0.75,1.5)–(1.25,1.5)–(1.25,-0.5)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,1.75)–(1.5,1.75)–(1.5,-0.75)–(0.25,-0.75)–(0.25,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

Similarly, the closure after an $\MII$ move will look like,

(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;
[strands=3,braid width=0.5cm,braid height=0.5cm,braid start=(-0.25,-0.5),line width=0.25pt] σ_2
[red] (1.25,1.5)–(1.5,1.5)–(1.5,-1)–(1.25,-1);
[red] (0.75,1.5)–(0.75,1.75)–(1.75,1.75)–(1.75,-1.25)–(0.75,-1.25)–(0.75,-1);
[red] (0.25,1.5)–(0.25,2)–(2,2)–(2,-1.5)–(0.25,-1.5)–(0.25,-1);
(2.75,0.5) node[scale=1.33] $\sim$;

(0.5,0.5) node[scale=1.33] $\beta$;
(0,0) rectangle (1,1);
[red] (0.75,1.5)–(1.25,1.5)–(1.25,-0.5)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,1.75)–(1.5,1.75)–(1.5,-0.75)–(0.25,-0.75)–(0.25,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

(0,0) rectangle (1,1);
(0.53,-0.25) node $\cdots$;
(0.53,1.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;
[strands=3,braid width=0.5cm,braid height=0.5cm,braid start=(-0.25,-0.5),line width=0.25pt] σ_2^-1
[red] (1.25,1.5)–(1.5,1.5)–(1.5,-1)–(1.25,-1);
[red] (0.75,1.5)–(0.75,1.75)–(1.75,1.75)–(1.75,-1.25)–(0.75,-1.25)–(0.75,-1);
[red] (0.25,1.5)–(0.25,2)–(2,2)–(2,-1.5)–(0.25,-1.5)–(0.25,-1);
(2.75,0.5) node[scale=1.33] $\sim$;

(0.5,0.5) node[scale=1.33] $\beta$;
(0,0) rectangle (1,1);
[red] (0.75,1.5)–(1.25,1.5)–(1.25,-0.5)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,1.75)–(1.5,1.75)–(1.5,-0.75)–(0.25,-0.75)–(0.25,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

Markov's theorem
Let $K_1,K_2$ be two oriented links which can be formed from the braids $\beta_1,\beta_2$, respectively. Then $K_1\sim{K_2}\Leftrightarrow\beta_1\sim_M\beta_2$.

For a proof of this; see Birman's survey <cit.> or Birman's original paper [10].

§.§ Motivation

Since we have reached the end of the first part, we will summarise and thus motivate the second part with all of what we learned here.

Alexander's theorem <ref> says that all links have a braid representation, and Markov's theorem <ref> says that two links are equivalent if and only if their braid representations are Markov equivalent (i.e., they can be transformed by $\MI$ and $\MII$ moves). So, then we call the Markov braid type the infinite union of the braid group $\Bcal_\infty$ that is quotiented by the two Markov moves (Markov equivalence). And so we have,

\[\{\text{link type}\}\simeq\{\text{Markov braid type}\}\]

So if we prove that a function is invariant under the Markov moves, then by Markov's theorem we have that it is a link invariant.
And this will be the strategy for the upcoming original construction of the Jones polynomial (See Chapter <ref>).

More precisely, for $a\in\Bcal_n$ and any $i<n$, we have $\reallywidehat{a}=\reallywidehat{\sigma_ia\sigma_i^{-1}}$ and $\reallywidehat{a}=\reallywidehat{\incl_n(a)\sigma_n^{\pm1}}$. So our strategy of constructing the Jones polynomial will be to define a tracial function on the Temperley-Lieb algebra. Where the Temperley-Lieb algebra is seen as a quotient algebra of the group algebra $\CC[\Bcal_n]$. There exists other similar constructions, however for the purposes of this essay we will use the Temperley-Lieb algebras since it has a very comprehensible diagrammatic definition. To see other similar constructions; a very nice read is [27]. The following chapter will briefly touch on the work that Jones did in 1983 [25]. We just want to see the correlation he made between the von Neumann algebras and constructing a link invariant.

§ VON NEUMANN ALGEBRAS

The aim of this chapter is to give the essential knowledge in order to understand the construction and encapsulation of the Jones polynomial. So this chapter will give a brief overview on operator algebras.

The interested reader is directed to [25].

This chapter is based on the presentations given in [14], [23], [25], [28], [31], [50], [51], [52], [59], and [64].

§.§ Background

We shall first review some background we will need. The discussion in this section is mainly based on that in the author's exposition on analysis & its foundations [54].

A normnorm over a vector space $V$ over $\CC$ is a map $\|\cdot\|:V\to\RR$ that satisfies

* $\|v\|\geq0$, with equality if and only if $v=0$ for all $v\in{V}$

* $\|\lambda{v}\|=|\lambda|\|v\|$ for all $v\in{V}$ and $\lambda\in\CC$

* $\|v+w\|\leq\|v\|+\|w\|$ for all $v,w\in{V}$

A normed vector spacenormed vector space is a vector space $V$ over $\CC$ with a norm $\|\cdot\|$.
We say that a normed vector space is a Banach spaceBanach space if it is complete (i.e., any Cauchy sequence converges).
We call a map $\varphi:V\to\CC$ a linear functionallinear functional if it satisfies $\varphi(\lambda{v}+\mu{w})=\lambda\varphi(v)+\mu\varphi(w)$, for all $\lambda,\mu\in\CC$ and $v,w\in{V}$. We say the linear functional is boundedbounded!linear functional if $\|\varphi\|=\sup_{\|x\|\leq1}|\varphi(x)|<\infty$. The dualdual of the normed vector space $V$, is the normed vector space of all bounded linear functionals $\varphi:V\to\CC$, and we denote it as $V^*$.

Now, we call a map $T:V\to{V}$ linearlinear if it satisfies $T(\lambda{v}+\mu{w})=\lambda{T(v)}+\mu{T(w)}$ for all $v,w\in\CC$ and $v,w\in{V}$. We call the linear map boundedbounded!linear map if $\|T\|=\sup_{\|v\|\leq1}\|T(v)\|<\infty$. We let $\BB(V)$$\BB(V)$ be the normed vector space of all bounded linear maps $T:V\to{V}$.

A Hilbert spaceHilbert space is essentially a complete inner product on a vector space $V$ over $\CC$, i.e.,

\[\langle\cdot,\cdot\rangle\colon{V\times{V}\to\CC}\]

satisfying the following for all $x,y,z\in{V}$ and $\lambda\in\CC$,

* $\langle{x,y}\rangle=\overline{\langle{y,x}\rangle}$

* $\langle\lambda{x}+y,z\rangle=\lambda\langle{x,z}\rangle+\langle{y,z}\rangle$

* $\langle{x,x}\rangle\geq0$ with equality if and only if $x=0$

Notice that if we let the norm $\|x\|=\sqrt{\langle{x,x}\rangle}$ for any $x\in{V}$, then we can see $|\langle{x,y}\rangle|\leq\|x\|\|y\|$ for any $x,y\in{V}$. This inequality is known as the Cauchy-Shwarz inequalityCauchy-Shwarz inequality (it is easy to show that this is indeed satisfied). So, in fact, an inner product induces a norm.

Note that the notation used for the inner product $\langle\cdot,\cdot\rangle$ should not be confused with the Kauffman bracket $\langle\cdot\rangle$ from the previous chapters. In fact, all notations used in this chapter are self-contained, unless stated otherwise.

The adjointadjoint of an operator $T:\Hcal\to\Hcal^\prime$ is the unique linear map $T^*:\Hcal\to\Hcal$ such that for all $x\in\Hcal^\prime$ and $y\in\Hcal$ we have $\langle{T^*x,y}\rangle=\langle{x,Ty}\rangle$.

An algebraalgebra is a vector space $V$ over $\CC$ if there is a multiplication operation $V\times{V}\to{V}$, given by $(u,v)\mapsto{uv}$ such that for all $u,v,w\in{V}$ and $\lambda,\mu\in\CC$, we have $u(v+w)=uv+uw$, $(u+v)w=uw+vw$ and $(\lambda{u})(\mu{v})=(\lambda\mu)(uv)$. We call an algebra $V$ over $\CC$ a normed algebranormed algebra if for all $u,v\in{V}$ we have $\|uv\|\leq\|u\|\|v\|$. We call a normed algebra a Banach algebraBanach algebra if it is a Banach space. If $V$ is a normed algebra with a multiplicative identity, then we call $V$ unitalunital.
Given $V$ is a an algebra, and a vector subspace $W$, then we call $B$ a subalgebrasubalgebra if $v,w\in{W}$ implies $vw\in{W}$.

And if $V$ is an algebra that is commutative as a ring, then it is a commutative algebracommutative algebra.

We shall assume that our algebras are unital.

We define the spectrumspectrum of $x$, denoted as $\sigma(x)$, in a Banach algebra $V$ as the set of complex numbers $\lambda$ such that $x-\lambda{I}$ is not invertible, i.e., $\sigma(x)=\{\lambda\in\CC:x-\lambda{I}\text{ is not invertible}\}$.

The spectrum of an element in a Banach algebra $A$ is a non-empty compact subset of $\CC$.

First start with the Murray-von Neumann criterionMurray-von Neumann criterion, which states that for $x$ in a Banach algebra $A$, if $\|x\|<1$ then $1-x$ is invertible.
It then follows that the spectrum of $x$ is in fact closed.
Then it follows from Heine-Borel, $\sigma(x)$ is compact since it is closed and bounded.

To show that it is non-empty, we need to define the resolvent set of $x$ $\rho(x)$. This is the complement of the spectrum $\rho(x)=\CC\setminus\sigma(x)$. And we also define the resolvent map of $x$ for $\lambda\in\rho(x)$ as $R_\lambda(x)=(\lambda-x)^{-1}$.

Then it is easy to see that as $\lambda\to\infty$ we have $\|R_\lambda(x)\|\to0$.
It is also easy to see that for $\lambda,\mu\in\rho(x)$, we have $R_\lambda-R_\mu=(\mu-\lambda)R_\lambda{R_\mu}$.

There is an analogue to Louiville's theorem for complex functions, so it then follows that the spectrum is non-empty.

§.§ C*-algebras

A $^*$-algebra$^*$-algebra is an algebra $A$ with an involution map $^*\colon{x}\mapsto{x^*}$ which for $x,y\in{A}$ and $\lambda\in\CC$ satisfies

* $(xy)^*=y^*x^*$

* $(x+y)^*=x^*+y^*$

* $\lambda^*=\bar{\lambda}$

* $x^{**}=x$

We say a $^*$-algebra is an abstract $C^*$-algebra$C^*$-algebra!abstract if it is a Banach $^*$-algebra where the norm satisfies the $C^*$-identity$C^*$-identity $\|x\|^2=\|x^*x\|$.

The completion of a Banach algebra with an involution is an abstract $C^*$-algebra.

The complex numbers $\CC$ forms a $C^*$-algebra, by defining the involution $\lambda^*=\overline{\lambda}$ and where the norm is $\|\lambda\|=|\lambda|$.

We say that a $^*$-algebra is a concrete $C^*$-algebra$C^*$-algebra!concrete if it is a closed subalgebra of $\BB(\Hcal)$.

So the set of bounded operators on a Hilbert space $\Hcal$, $\BB(\Hcal)$, is one example of a concrete $C^*$-algebra.

Given a $C^*$-algebra $A$, if $A$ has a unit $1\in{A}$, i.e., an identity element. Then it can easily be shown $1^*=1$. This can be seen by $1=(1^*)^*=(1^*1)^*=(1^*(1^*)^*)=1^*1=1^*$.

Given a $C^*$-algebra $A$, the involution is continuous.

Let $x_n\in{A}$ such that $x_n\to{x\in{A}}$. So, $\|x_n^*-x^*\|=\|(x_n-x)^*\|$.
Now we want to show $\|(x_n-x)^*\|=\|x_n-x\|$. So $\|x_n-x\|^2=\|(x_n-x)^*(x_n-x)\|\leq\|(x_n-x)^*\|\|x_n-x\|$, so if $x_n-x\neq0$ then $\|x_n-x\|\leq\|(x_n-x)^*\|$ (otherwise if $x_n-x=0$, then we are done). So, $\|(x_n-x)^*\|\leq\|((x_n-x)^*)^*\|=\|x_n-x\|$. So, then, $\|(x_n-x)^*\|=\|x_n-x\|$.

So, we now have $\|x_n^*-x^*\|=\|x_n-x\|\to0$ since $x_n\to{x}$. Therefore, the involution is continuous.

If $A$ is a $C^*$-algebra, then $\sigma(x^*)=\sigma(x)$ for all $x\in{A}$.

Given a $C^*$-algebra $A$, then we have

* $x\in{A}$ is normalnormal if $xx^*=x^*x$.

* $y\in{A}$ is HermitianHermitian if $x=x^*$.

* $u\in{A}$ is unitaryunitary if $uu^*=u^*u=1$.

* $p\in{A}$ is a projectionprojection if $p=p^*=p^2$.

The following result is left as an exercise to the reader.

If $x$ is an element of a $C^*$-algebra $A$. Then $x^*x$ is Hermitian.

Given $C^*$-algebras $A$ and $B$, a $^*$-homomorphism*-homomorphism is an algebra homomorphism $\alpha:A\to{B}$ such that $\alpha(x^*)=\alpha(x)^*$ for all $x\in{A}$. A $^*$-homomorphism, $\alpha$, between unital $C^*$-algebras is itself unital$C^*$-algebra!unital if $\alpha(1)=1$.

Any $^*$-homomorphism is continuous, with closed image, and is bounded with norm at most one. Any injective $^*$-homomorphism is an isometry.

So, the norm on a $C^*$-algebra is determined by the rest of the algebraic structure.

A $^*$-homomorphism is an isomorphism of $C^*$-algebras if it is bijective.

The inverse of a bijective $^*$-homomorphism is also a $^*$-homomorphism.

§.§ Three topologies on B(H)

norm The norm topologynorm topology on $\BB(\Hcal)$ is given by the norm $\|T\|=\sup_{\|v\|\leq1}\|T(v)\|$.

The weak operator topologyweak operator topology (WOT) on $\BB(\Hcal)$ is given by
\[\{b\in\BB(\Hcal):|\langle{ax,y}\rangle-\langle{bx,y}\rangle|<\epsilon \text{ where } a\in\BB(\Hcal), x,y\in\Hcal\text{ and }\epsilon>0\}\]

The strong operator topologystrong operator topology (SOT) (SOT) on $\BB(\Hcal)$ is given by

\[\{b\in\BB(\Hcal):\|ax-bx\|<\epsilon\text{ where }a\in\BB(\Hcal), x\in\Hcal\text{ and }\epsilon>0\}\]

So, we then have that the norm topology contains the strong and weak operator topologies, i.e., $\text{WOT}\prec\text{SOT}\prec\text{the norm topology}$. And so for any $A\subset\BB(\Hcal)$, we have $\overline{A}^{\|\cdot\|}\subset\overline{A}^\text{SOT}\subset\overline{A}^\text{WOT}$.

§.§ von Neumann algebras

We define the commutantcommutant of $A\subseteq\BB(\Hcal)$ as

\[A^\prime=\{y\in\BB(\Hcal) : xy=yx\text{ for all }x\in{A}\},\]

and the bicommutantbicommutant by $A^{\prime\prime}=(A^\prime)^\prime$.

The following theorem is known as the bicommutant theorembicommutant theorem.

Let $M\subset\BB(\Hcal)$ be a unital $^*$-subalgebra. Then the following three are equivalent.

* $M$ is closed in WOT

* $M$ is closed in SOT

* $M=M^{\prime\prime}=(M^\prime)^\prime$

A unital $^*$-subalgebra $M\subseteq\BB(\Hcal)$ is a von Neumann algebravon Neumann algebra if it satisfies any of the properties in the above theorem.

It is clear that any von Neumann algebra is a $C^*$-algebra.

The set of bounded linear maps on the Hilbert space $\BB(\Hcal)$ is a von Neumann algebra.
We also have $M_n(\CC)$ as a von Neumann algebra.

We note that von Neumann algebras come in pairs $M,M^\prime$.

§.§ Factors

We call a von Neumann algebra $M$ a factorfactor if $M\cap{M^\prime}=\CC\cdot1_{\BB(\Hcal)}$ (i.e., the center is trivial).

The set of bounded linear maps on the Hilbert space $\BB(\Hcal)$ is thus also a factor.

The only commutative factor is $\CC$.

Given $M$ is a von Neumann algebra, then

* let $e,f\in{M}$ be two projections, then $e$ and $f$ are equivalentprojections!equivalent if there exists $u\in{M}$ such that $u^*u=e$ and $uu^*=f$. We write $e\sim{f}$ if they are equivalent.

* let $e,f\in{M}$ be two projections, then if there exists $u\in{M}$ such that $u^*u=e$ and $u^*uf=u^*u$. We write this as $e\preceq{f}$.

* we say a projection $p\in{M}$ such that $p\neq0$ is minimalprojections!minimal if for all projections $a\in{M}$ we have that if $a\leq{p}$ then either $a=0$ or $a=p$.

* we say a projection $e\in{M}$ is finiteprojections!finite if we have that for all projections $f\in{M}$, if $e\sim{f}\leq{e}$ then $f=e$.

The types of factors $M\subseteq\BB(\Hcal)$:

* type I factorfactor!type I if there exists a minimal projection in $M$.

* we call the factor $M\subseteq\BB(\Hcal)$ a type II factorfactor!type II if there exists a finite projection in $M$ and there does not exist a minimal projection in $M$.

* given $M\subseteq\BB(\Hcal)$ is a type II factor and $1_{\BB(\Hcal)}$ is finite, then $M$ is a type II$_1$ factorfactor!type II$_1$.

* if $1_{\BB(\Hcal)}$ is infinite and there exists finite projections, then it is said to be a type II$_\infty$ factorfactor!type II$_\infty$.

* finally, we have that the factor $M\subseteq\BB(\Hcal)$ is a type III factorfactor!type III if there exists no finite projections in $M$.

We will only discuss type II$_1$ factors in this paper. So to simplify, a factor is of type II$_1$ if all projections are finite (i.e., $1_{\BB(\Hcal)}$ is finite) and there does not exist a minimal projection.

§.§ Trace

We call a linear functional $f$ on a von Neumann algebra $M$,

* positivelinear functional!positive if $f(x^*x)\geq0$ for all $x\in{M}$,

* faithfullinear functional!faithful if when we have $f(x^*x)=0$, then $x=0$,

* statelinear functional!state if $f(1)=1$, and

* traciallinear functional!tracial if $f(xy)=f(yx)$ for all $x,y\in{M}$.

Type II$_1$ factors $M$ admit a unique faithful tracial state $\tr\colon{M\to\CC}$, so

* $\tr(xy)=\tr(yx)$ for all $x,y\in{M}$

* $\tr(1)=1$

* $\tr(xx^*)>0$ for all $x\in{M}$ where $x\neq0$.

We call a linear functional $\tr\colon{M\to\CC}$ a tracetrace if $\tr(xy)=\tr(yx)$ for all $x,y\in{M}$.

Given $M$ is a finite factor, now we can say that two projections $e,f\in{M}$ are equivalent if $\tr(e)=\tr(f)$.

§.§ The index theorem

We define a Hilbert space $L^2(M,\tr_M)$ to be the completion of $M$ with respect to the inner product $\langle{a,b}\rangle=\tr_M(b^*a)$.

Jones index
Given $N$ is a subfactor of $M$, we define the Jones indexJones index of $N$ as $[M:N]=\dim_N{L^2(M,\tr_M)}$.

Jones index theorem
Given $N$ is a II$_1$ subfactor of $M$, then we have

\[[M:N]\in\left\{4\cos^2(\pi/(n+2)):n\geq1\right\}\cup[4,\infty).\]

Jones proved this using the basic construction explained in the following section.

§.§ Jones' basic construction

In this section we will discuss Jones' basic construction for type II$_1$ subfactors. This discussion is majorly based on that of Jones' paper [25], Penneys' lecture notes [52], and Speicher's lecture notes [59].

Every abstract $C^*$-algebra is isometrically $^*$-isomorphic to a concrete $C^*$-algebra.

GNS construction
Given a unital $C^*$-algebra $A$ and a state $f\colon{A\to\CC}$. There exists a Hilbert space $\Hcal^\prime$, a $^*$-homomorphism $\pi\colon{A\to\BB(\Hcal^\prime)}$, and a unit vector $h\in\Hcal^\prime$ such that for all $a\in{A}$ we have $f(a)=\langle\pi(a)h,h\rangle$.

We have the GNS representation of a factor $M$ on $L^2(M,\tr_M)$, which is given by $\langle{{x},{y}}\rangle=\tr_M(y^*x)$.

Since the entire chapter was filled with definition and results, it would be reasonable to summarise what it is exactly we need in order to discuss Jones' basic construction.

So a von Neumann algebra $A$ is a unital $^*$-subalgebra of $\BB(\Hcal)$ such that $A=A^{\prime\prime}$; see Theorem <ref>. A factor $M$ is a von Neumann algebra such that the center is trivial (i.e., $M\cap{M^\prime}=\CC\cdot1_{\BB(\Hcal)}$). A factor is of type II$_1$ if it is of infinite dimension and it admits a unique tracial state. We then define a Hilbert space $L^2(M,\tr_M)$ to be the completion of $M$ with respect to the inner product $\langle{a,b}\rangle=\tr_M(b^*a)$. So, $M$ acts on $L^2(M,\tr_M)$ by both left and right multiplication. So, for a projection $p\in{M}$, we have the representation $L^2(M,\tr_M)p$. We then have that for II$_1$ factors $N\subset{M}$, we have $[M:N]=\dim_N(L^2(M,\tr_M))$.

We have the conjugate-linear unitary functional $J\colon{L^2(M,\tr_M)\to{L^2(M,\tr_M)}}$ given by ${x}\mapsto{{x^*}}$. Then we have $JM^{\prime}J=M$.

Now consider a subfactor $N\subset{M}$, and the Hilbert space $L^2(M,\tr_M)$. We see that are already five II$_1$ factors, namely: $M$, $M^\prime$, $N$, $N^\prime$, and $JNJ$.

Given finite von Neumann algebras $N\subset{M}$ with a faithful normal trace $\tr_N$, with $1_N=1_M$. Then there exists a normal linear map $E_N\colon{M\to{N}}$ defined by $\tr_N(E_N(x)y)=\tr_N(xy)$ for $x\in{M}$ and $y\in{N}$, and satisfies the following properties

* $E_N(axb)=aE_N(x)b$ for $x\in{M}$ and $a,b\in{N}$

* $E_N(x^*)=E_N(x)^*$ for all $x\in{M}$

* $E_N(x^*)E_N(x)\leq{E_N(x^*x)}$ for all $x\in{M}$

* $E_N(x^*x)=0$ implies $x=0$ for all $x\in{M}$

The above proposition essentially says that $E_N$ is a trace-preserving, faithful and positive linear map. We call $E_N$ the conditional expectationconditional expectation of $M$ to $N$.

Given II$_1$ factors $N\subset{M}$, there exists the conditional expectation $E_N\colon{M\to{N}}$, and an orthogonal projection $e_N\colon{L^2(M,\tr_M)\to{L^2(N,\tr_N)}}$ such that

* it is uniquely determined by $e_Nxe_N=E_N(x)e_N$ for $x\in{M}$

* $xe_N=e_Nx$ if and only if $x\in{N}$

* $N=M\cap\{e_N\}^\prime$

Basic construction.basic construction
Given II$_1$ factors $N\subset{M}$.

Let $e_N\colon{L^2(M,\tr_M)}\to{L^2(N,\tr_N)}$ be an orthogonal projection. Then we have $N=M\cap\{e_N\}^\prime$. So, a direct corollary to this is that we have the von Neumann algebra $\langle{M,e_N}\rangle$ on $L^2(M,\tr_M)$ which is defined by $(M\cup\{e_N\})^{\prime\prime}=JN^\prime{J}$. Thus we also have $[M:N]=\tr(e_N)^{-1}=[\langle{M,e_N}\rangle:M]$, where the trace is defined on $\langle{M,e_N}\rangle$.

And so analogously to how for a von Neumann algebra $M$ there exists a von Neumann algebra $M^\prime$;
The basic construction tells us that for a subfactor $N\subset{M}$ there exists subfactors $JNJ$ and $JN^\prime{J}$ on $L^2(M,\tr_M)$.

Thus, by iterating the basic construction, we find a tower $M_i$ of II$_1$ factors with $M_0=N$, $M_1=M$ and $M_{i+1}=\langle{M_i,e_{M_{i-1}}}\rangle$.

\[N\subset{M}\overset{e_{M}}{\subset}M_2\overset{e_{M_2}}{\subset}\cdots\overset{e_{M_n}}{\subset}\langle{M_n,e_{M_n}}\rangle=M_{n+1}\subset\cdots\]

For better readability, we let $e_i=e_{M_i}$, and call the sequence of $(e_i)$ the Jones projectionsJones projections.

The Jones projections $(e_i)$ satisfy the following properties.

* $\left[M:N\right]e_ie_{i\pm1}e_i=e_i$

* $e_ie_j=e_je_i \text{ if }|i-j|\geq2$

* $e_i^2=e_i^*=e_i$

* $\left[M:N\right]\tr_{n+1}(xe_{n})=tr_{n}(x) \text{ where }x\text{ is a word on }1,e_1,e_2,\ldots,e_{n-1}$

These relations are similar to those of the braid group and the Markov moves. This is the similarity that Didier Hatt-Arnold pointed out to Vaughan Jones in 1982 (see Introduction <ref>).
The reader should instantly notice that the Jones projections $(e_i)$ satisfy the properties of the Temperley-Lieb algebras from Definition <ref> given $d^{2}=[M:N]$. For instance, for part (i) we have,

\begin{align*}
[M:N]e_ne_{n-1}e_n &= [M:N]E_n(e_{n-1})e_n\\
&= [M:N]\tr_{n+1}(e_{n-1})e_n\\
&= [M:N][M:N]^{-1}e_n = e_n
\end{align*}

§ THE TEMPERLEY-LIEB ALGEBRAS

The Temperley-Lieb algebra was first introduced in 1971 by Lieb and Temperley [42].
In this chapter, we will see the relation between the von Neumann algebra that we constructed in the previous chapter (via the Jones' projections) and the Temperley-Lieb algebra. This will help our understanding since it can be explained in a diagrammatic setting.

Our discussion will mainly depend on that found in [52].

§.§ Quantum integers

A qunatum integerquantum integer $[n]_q$ for $n\in\ZZ$ and $q\in\{Q\cup-Q\}$ where $Q:=\{e^{i\theta}:\theta\in(0,\pi/2)\}\cup[1,\infty)$, is defined as $\dfrac{q^n-q^{-n}}{q-q^{-1}}$.

A quantum integer is essentially a deformation of an integer by $q$.

So the quantum integer $[1]_q$ is $1$, while $[2]_q=q+q^{-1}$.

We can double a quantum integer, $[2]_q[n]_q=[n+1]_q+[n-1]_q$.

\begin{align*}
[2]_q[n]_q &= \frac{(q+q^{-1})(q^n-q^{-n})}{q-q^{-1}}\\
&= \frac{q^{n+1}-q^{1-n}+q^{n-1}-q^{-n-1}}{q-q^{-1}}\\
&= \frac{q^{n+1}-q^{-(n+1)}}{q-q^{-1}}+\frac{q^{n-1}-q^{-(n-1)}}{q-q^{-1}}
\end{align*}

And so the result then follows.

Given $m\geq{a}$, we have $[m-a]_q=[m]_q[a+1]_q-[m+1]_q[a]_q$.

\begin{align*}
[m-a]_q &= \frac{q^{m-a}-q^{a-m}}{q-q^{-1}}\\
&= \frac{(q^{m-a}-q^{a-m})(q-q^{-1})}{(q-q^{-1})(q-q^{-1})}\\
&= \frac{q^{m-a+1}-q^{m-a-1}-q^{a-m+1}+q^{a-m-1}}{q^2+q^{-2}}\\
&= \frac{q^{m+a+1}-q^{m+a+1}+q^{-m-a-1}-q^{-a-m-1}+q^{m-a+1}-q^{m-a-1}-q^{a-m+1}+q^{a-m-1}}{q^2+q^{-2}}\\
&= \frac{q^{m+(a+1)}-q^{m-(a+1)}-q^{(a+1)-m}+q^{-m-(a+1)}}{q^2+q^{-2}}\\&\qquad\quad-\frac{q^{a+(m+1)}-q^{-a+(m+1)}-q^{a-(m+1)}+q^{-a-(m+1)}}{q^2+q^{-2}}\\
&= \left(\frac{q^m-q^{-m}}{q-q^{-1}}\right)\frac{q^{a+1}-q^{-(a+1)}}{q-q^{-1}}-\left(\frac{q^{m+1}-q^{-(m+1)}}{q-q^{-1}}\right)\frac{q^a-q^{-a}}{q-q^{-1}}\\
&= [m]_q[a+1]_q-[m+1]_q[a]_q
\end{align*}

§.§ Defining the Temperley-Lieb algebra TLn

For $d\in\{Q\cup-Q\}$, $TL_{n}(d)$Temperley-Lieb algebra!$TL_n$ is the associative unital (i.e., has an identity) $*$-algebra with generators $e_1,e_2,\ldots,e_{n-1}$ and relations

* $e_i^2=e_i=e_i^*$

* $e_ie_j=e_je_i$ for $|i-j|\geq2$

* $e_ie_{i\pm1}e_i=d^{-2}{e_i}$.

So we can reduce each word in the algebra with the relations.

So, $TL_1$ is generated by $1$, $TL_2$ is generated by $1$ and $e_1$, $TL_3$ is generated by $1$, $e_1$, $e_2$, $e_1e_2$ and $e_2e_1$, and so on.

The $n$-th Catalan numberCatalan number $c_n$ is given by $\dfrac{1}{n+1}\left(\!\!\!\begin{array}{c}2n\\n\end{array}\!\!\!\right)$.

In our above example, we saw $\dim(TL_1)=1=c_1$, $\dim(TL_2)=2=c_2$ and $\dim(TL_3)=5=c_3$. So, we have the following result.

$TL_n$ is finite dimensional.

Essentially, since this is a finitely generated group. We have any word in the algebra can be written as

\[(e_{i_1}e_{i_1-1}\cdots{e_{k_1}})(e_{i_2}e_{i_2-1}\cdots{e_{k_2}})\cdots(e_{i_p}e_{i_p-1}\cdots{e_{k_p}})\]

for $1\leq{i_1}<i_2<\cdots<i_p$ and for $1\leq{k_1}<k_2<\cdots<k_p$.

Thus the dimension is less than or equal to the $n^\text{th}$ Catalan number.
\[\dim(TL_n)\leq{c_n}\]

It is easy to realise that this is essentially the type II$_1$ algebra formed from the Jones projections $(e_i)$ in the previous chapter.

§.§ Diagrammatic definition of the Temperley-Lieb algebra gTLn

We now discuss the diagrammatic definition of the Temperley-Lieb algebras. This will help us in the construction of the Jones polynomial in the next chapter.

Temperley–Lieb algebra!diagrammatic $gTL_n$
For $d\in\{Q\cup-Q\}$, we call the algebra $gTL_{n}(d)$ the Temperley-Lieb algebra, whose basis is the set of non-intersecting string diagrams on a rectangle with $n$ boundary points on the top and bottom.
The multiplication is defined by concatenation $gTL_n\times{gTL_n}\to{gTL_n}$ (analogously to braid concatenation)  and is shown below.

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;

(-0.25,0.5) node[scale=1.33] $\cdot$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $y$;

(-0.5,-0.25) node[scale=1.33] $=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0,-0.5) rectangle (1,-1.5);
(0.53,-1.75) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;
(0.5,-1) node[scale=1.33] $y$;

If a closed loop appears in the diagram, we can remove it by multiplying the diagram by the weight of the closed loop (which is $d$).

We have that two elements in $gTL_n$ are equivalent if the strings are ambient isotopic. So by holding all the boundary points in place, we can move the strings around, given that they do not intersect or break the definition.

An example of an element in $gTL_5$ would be the following

(0,0) rectangle (3,2);
(0.25,2) [upup=0.75];
(0.25,0) [dnup=1];
(0.75,0) [dndn=0.75];
(0.5,0) [dnup=1.5];
(2.75,0) [dnup=-0.25];
(2.25,0) [dnup=0];

The elements of $gTL_1$ is just a diagram with a single string. The elements of $gTL_2$ is the following

(0,0) rectangle (2,2);
(0.5,0) [dnup=0];
(1.5,0) [dnup=0];
(2.25,0) node[scale=2] $,$;
(0,0) rectangle (2,2);
(0.5,0) [dndn=1];
(0.5,2) [upup=1];

We also have the following elements of $gTL_3$

(0,0) rectangle (2,2);
(0.5,0) [dnup=0];
(1,0) [dnup=0];
(1.5,0) [dnup=0];
(2.25,0) node[scale=2] $,$;
(0,0) rectangle (2,2);
(0.25,0) [dndn=0.75];
(0.25,2) [upup=0.75];
(1.5,0) [dnup=0];
(2.25,0) node[scale=2] $,$;

(0,0) rectangle (2,2);
(1,0) [dndn=0.75];
(1,2) [upup=0.75];
(0.5,0) [dnup=0];
(2.25,0) node[scale=2] $,$;

(0,0) rectangle (2,2);
(0.25,0) [dndn=0.75];
(1,2) [upup=0.75];
(1.75,0) [dnup=-1.5];
(2.25,0) node[scale=2] $,$;

(0,0) rectangle (2,2);
(0.25,2) [upup=0.75];
(1,0) [dndn=0.75];
(0.25,0) [dnup=1.5];

So, we have the generators for $gTL_n$,

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(2.75,0) [dnup=0];
(1.5,1) node $\cdots$;
(1.5,-0.35) node[scale=1.33] $1_n$;
(3.375,0) node[scale=1.33] $,$;

(0,0) rectangle (3,2);
(0.25,2) [upup=0.75];
(0.25,0) [dndn=0.75];
(1.25,0) [dnup=0];
(1.875,1) node $\ldots$;
(2.5,0) [dnup=0];
(2.75,0) [dnup=0];
(1.5,-0.35) node[scale=1.33] $E_1$;
(3.375,0) node[scale=1.33] $,$;

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,2) [upup=0.75];
(0.5,0) [dndn=0.75];
(1.5,0) [dnup=0];
(2,1) node $\ldots$;
(2.5,0) [dnup=0];
(2.75,0) [dnup=0];
(1.5,-0.35) node[scale=1.33] $E_2$;
(3.625,0) node[scale=1.33] $,$;

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,0) [dnup=0];
(0.75,1) node $\ldots$;
(1,0) [dnup=0];
(1.25,2) [upup=0.75];
(1.25,0) [dndn=0.75];
(2.25,0) [dnup=0];
(2.5,1) node $\ldots$;
(2.75,0) [dnup=0];
(1.5,-0.35) node[scale=1.33] $E_i$;
(3.625,0) node[scale=1.33] $,$;

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,0) [dnup=0];
(1.125,1) node $\ldots$;
(1.75,0) [dnup=0];
(2,0) [dndn=0.75];
(2,2) [upup=0.75];
(1.5,-0.35) node[scale=1.33] $E_{n-1}$;

the $gTL_n$ generators

To see the relations, we begin by multiplying $E_i^2$,

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,0) [dnup=0];
(0.75,1) node $\ldots$;
(1,0) [dnup=0];
(1.25,2) [upup=0.75];
(1.25,0) [dndn=0.75];
(2.25,0) [dnup=0];
(2.5,1) node $\ldots$;
(2.75,0) [dnup=0];
(3.375,1) node[scale=1.33] $\cdot$;

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,0) [dnup=0];
(0.75,1) node $\ldots$;
(1,0) [dnup=0];
(1.25,2) [upup=0.75];
(1.25,0) [dndn=0.75];
(2.25,0) [dnup=0];
(2.5,1) node $\ldots$;
(2.75,0) [dnup=0];

(-0.625,1) node[scale=1.33] $=$;
(0,3) rectangle (3,-1);
(0.25,1) [dnup=0];
(0.5,1) [dnup=0];
(0.75,2) node $\ldots$;
(1,1) [dnup=0];
(1.25,3) [upup=0.75];
(1.25,1) [dndn=0.75];
(2.25,1) [dnup=0];
(2.5,2) node $\ldots$;
(2.75,1) [dnup=0];
[dashed] (0,1)–(3,1);
(0.25,-1) [dnup=0];
(0.5,-1) [dnup=0];
(0.75,0) node $\ldots$;
(1,-1) [dnup=0];
(1.25,1) [upup=0.75];
(1.25,-1) [dndn=0.75];
(2.25,-1) [dnup=0];
(2.5,0) node $\ldots$;
(2.75,-1) [dnup=0];

(-0.8,1) node[scale=1.33] $=$;
(-0.3,1) node[scale=1.33] $d$;
(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,0) [dnup=0];
(0.75,1) node $\ldots$;
(1,0) [dnup=0];
(1.25,2) [upup=0.75];
(1.25,0) [dndn=0.75];
(2.25,0) [dnup=0];
(2.5,1) node $\ldots$;
(2.75,0) [dnup=0];

Therefore, we have $E_i^2=d{E_i}$.

We also have $E_iE_j=E_jE_i$ where $|i-j|>1$,

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];
(4.125,1) node[scale=1.33] $\cdot$;

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,0) [dnup=0];
(1.75,0) [dnup=0];
(2,2) [upup=0.75];
(2,0) [dndn=0.75];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];
(4.3,1) node[scale=1.33] $=$;

(0,3) rectangle (4,-1);
(0.25,1) [dnup=0];
(0.5,2) node $\ldots$;
(0.75,1) [dnup=0];
(1,3) [upup=0.75];
(1,1) [dndn=0.75];
(2,1) [dnup=0];
(2.75,1) [dnup=0];
(3,1) [dnup=0];
(3.375,2) node $\ldots$;
(3.75,1) [dnup=0];
[dashed] (0,1)–(4,1);
(0.25,-1) [dnup=0];
(0.5,0) node $\ldots$;
(0.75,-1) [dnup=0];
(1,-1) [dnup=0];
(1.75,-1) [dnup=0];
(2,1) [upup=0.75];
(2,-1) [dndn=0.75];
(3,-1) [dnup=0];
(3.375,0) node $\ldots$;
(3.75,-1) [dnup=0];

(-0.3,1) node[scale=1.33] $=$;
(0,3) rectangle (4,-1);
(0.25,1) [dnup=0];
(0.5,2) node $\ldots$;
(0.75,1) [dnup=0];
(1,1) [dnup=0];
(1.75,1) [dnup=0];
(2,3) [upup=0.75];
(2,1) [dndn=0.75];
(3,1) [dnup=0];
(3.375,2) node $\ldots$;
(3.75,1) [dnup=0];
[dashed] (0,1)–(4,1);
(0.25,-1) [dnup=0];
(0.5,0) node $\ldots$;
(0.75,-1) [dnup=0];
(1,1) [upup=0.75];
(1,-1) [dndn=0.75];
(2,-1) [dnup=0];
(2.75,-1) [dnup=0];
(3,-1) [dnup=0];
(3.375,0) node $\ldots$;
(3.75,-1) [dnup=0];

(-0.25,1) node[scale=1.33] $=$;
(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];
(4.125,1) node[scale=1.5] $\cdot$;

(4.25,0) rectangle (8.25,2);
(4.5,0) [dnup=0];
(4.75,1) node $\ldots$;
(5,0) [dnup=0];
(5.25,0) [dnup=0];
(6,0) [dnup=0];
(6.25,2) [upup=0.75];
(6.25,0) [dndn=0.75];
(7.25,0) [dnup=0];
(7.625,1) node $\ldots$;
(8,0) [dnup=0];

$E_iE_j=E_jE_i$ where $|i-j|>1$

Our final relation is $E_iE_{i+1}E_i=E_i$.

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];
(4.125,1) node[scale=1.33] $\cdot$;

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(3.375,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,0) [dnup=0];
(1.25,0) [dnup=0];
(1.75,2) [upup=0.75];
(1.75,0) [dndn=0.75];
(0.5,1) node $\ldots$;
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.75,0) [dnup=0];
(4.125,1) node[scale=1.33] $\cdot$;

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];

(-0.25,-1) node[scale=1.33] $=\;$;
(0,2) rectangle (4,-4);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2.5,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];
[dashed] (0,0)–(4,0);
(0.25,-2) [dnup=0];
(0.5,-1) node $\ldots$;
(0.75,-2) [dnup=0];
(1,-2) [dnup=0];
(1.75,0) [upup=0.75];
(1.75,-2) [dndn=0.75];
(2.75,-2) [dnup=0];
(3,-2) [dnup=0];
(3.375,-1) node $\ldots$;
(3.75,-2) [dnup=0];
(0.25,-4) [dnup=0];
(0.5,-3) node $\ldots$;
(0.75,-4) [dnup=0];
(1,-2) [upup=0.75];
(1,-4) [dndn=0.75];
(2.5,-4) [dnup=0];
(2.75,-4) [dnup=0];
(3,-4) [dnup=0];
(3.375,-3) node $\ldots$;
(3.75,-4) [dnup=0];
(4.3,-1) node[scale=1.33] $\;=$;

(0,0) rectangle (4,2);
(0.25,0) [dnup=0];
(0.5,1) node $\ldots$;
(0.75,0) [dnup=0];
(1,2) [upup=0.75];
(1,0) [dndn=0.75];
(2,0) [dnup=0];
(2.75,0) [dnup=0];
(3,0) [dnup=0];
(3.375,1) node $\ldots$;
(3.75,0) [dnup=0];

We leave the reader to check $E_iE_{i-1}E_i=E_i$.

The involution $^*$ of a diagram would be the reflection about the horizontal line. Thus, $E_i=E_i^*$. However, for non-generators we have,

(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,2) [upup=0.5];
(1.25,2) [upup=0.5];
(0.5,0) [dnup=1.75];
(1,0) [dndn=0.5];
(3.15,2) node[scale=1.33] $*$;
(3.45,1) node[scale=1.33] $=$;
(0,0) rectangle (3,2);
(0.25,0) [dnup=0];
(0.5,2) [upup=0.5];
(1.25,2) [upup=0.5];
(0.5,0) [dnup=1.75];
(1,0) [dndn=0.5];

involution example on $gTL_n$

Thus the relations for our generators in $gTL_n$ are

* $E_i^2=dE_i=dE_i^*$

* $E_iE_j=E_jE_i$ for $|i-j|\geq2$

* $E_iE_{i\pm1}E_i=E_i$

These relations are closely related to that of $TL_n$. One can define a homomorphism by mapping $e_i$ to $d^{-1}E_i$. In fact, $gTL_n$ is $^*$-isomorphic to $TL_n$.

There exists a $^*$-isomorphism from $TL_n$ to $gTL_n$.

So from this point forth, we denote $\TL_n$ as our Temperley-Lieb algebra, where we will use the diagrammatic definition. So we have the generators $E_i$ for $i=1,2,\ldots,n-1$ and the relations above.

§.§ Linear operations on n

We now discuss some sub-planar algebra on $\TL_n$ which will make things easier to comprehend.

Consider the following diagram in $\TL_n$,

(-0.5,0.5) node[scale=1.33] $\displaystyle x=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;

we then have the following linear operations on $\TL_n$

* The inclusionlinear operations on $\TL_n$!inclusion $i_n\colon\TL_n\to\TL_{n+1}$ is a unital, injective $^*$-homomorphism. We essentially add a string to the right of the diagram.

(-1,0.5) node[scale=1.33] $\displaystyle i_n(x)=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;

inclusion on $\TL_n$

* The conditional expectationlinear operations on $\TL_n$!conditional expectation $\Ecal_{n+1}\colon\TL_{n+1}\to\TL_n$ is a surjective $^*$-map of $\CC$-vector spaces. We essentially close the last string on the right.

(-1.25,0.5) node[scale=1.33] $\displaystyle \Ecal_{n+1}(x)=$;
(0.5,0.5) node[scale=1.33] $x$;
(0,0) rectangle (1,1);
[red] (0.9,1.5)–(1.2,1.5)–(1.2,-0.5)–(0.9,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

conditional expectation $\Ecal_{n+1}$ on $\TL_{n+1}$

* The tracelinear operations on $\TL_n$!trace $\tr_n\colon\TL_n\to\TL_0$ is a linear $^*$-map of $\CC$-vector spaces. Essentially, it is the closure of the diagram.

(-1,0.5) node[scale=1.33] $\displaystyle \tr_{n}(x)=$;
(0.5,0.5) node[scale=1.33] $x$;
(0,0) rectangle (1,1);
[red] (0.75,1.5)–(1.25,1.5)–(1.25,-0.5)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,1.75)–(1.5,1.75)–(1.5,-0.75)–(0.25,-0.75)–(0.25,-0.5);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

trace $\tr_n$ on $\TL_n$

* The identitylinear operations on $\TL_n$!identity map $\id_n\colon\TL_n\to\TL_n$ which is given by

(-1,0.5) node[scale=1.33] $\displaystyle \id_{n}(x)=$;
(0.5,0.5) node[scale=1.33] $x$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;

It is typical to wonder whether the conditional expectation on $\TL_n$ defined above is related to the conditional expectation on factors in the previous chapter. This would make sense as to why we started with von Neumann algebras before defining the Temperley-Lieb algebras. The following proposition will answer our question.

There exists the following relations,

* $\Ecal_{n+1}\circ{i_n}=d\cdot\id_n$

* $\tr_{n+1}=tr_n\circ\Ecal_{n+1}$

* $(i_n\circ{i_{n-1}}\circ\Ecal_n(x))E_n=E_ni_n(x)E_n$ for all $x\in\TL_n$

* $\tr_n(xy)=\tr_n(yx)$ for all $x,y\in\TL_n$

* $\tr_{n+1}(i_n(x)\cdot{E_n})=\tr_n(x)$ for all $x\in\TL_n$

* $\tr_n(\Ecal_{n+1}(x)\cdot{y})=\tr_{n+1}(x\cdot{i_n(y)})$ for all $x\in\TL_{n+1}$ and $y\in\TL_n$

* $\tr_n(1_n)=d^n$

* $\Ecal_{n+1}\circ{i_n}\colon\TL_n\to\TL_n$,

\[
\Ecal_{n+1}\circ{i_n}\left(\tldiagram{$x$}\right) = \Ecal_{n+1}\left(\tlinclusion{$x$}\right) = \tlconditionalinclusion{$x$} = d\cdot\id_n\left(\tldiagram{$x$}\right)
\]

* $\tr_{n+1}\colon\TL_{n+1}\to\TL_0$,

\[
\tr_{n+1}\left(\tldiagram{$x$}\right) = \tltraceplusone{$x$} = \tr_n\left(\tlconditional{$x$}\right) = \tr_n\circ\Ecal_{n+1}\left(\tldiagram{$x$}\right)
\]

* We first have $i_n\circ{i_{n-1}}\circ\Ecal_n(x)$,

\[
\knotsinmath{\begin{scope}[scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.5,1.5)--(1.5,-0.5);
\draw (1.75,1.5)--(1.75,-0.5);
\end{scope}}
\]

and so, multiplying by $E_n$ gives

\[
\knotsinmath{\begin{scope}[scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.5,-0.5) [dndn=0.5];
\draw (1.5,1.5) [upup=0.5];
\end{scope}}
\]

This can be seen to be the same as $E_ni_n(x)E_n$,

\[
\knotsinmath{\begin{scope}[scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,3.5)--(0.25,1);
\draw (0.25,0)--(0.25,-2.5);
\draw (0.75,0)--(0.75,-2.5);
\draw (0.75,3.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.9,-2.5) [dndn=0.5];
\draw (0.9,3.5) [upup=0.5];
\draw (0.9,-0.5) [upup=0.5];
\draw (0.9,1.5) [dndn=0.5];
\draw (1.4,-0.5)--(1.4,1.5);
\end{scope}}
\]

Therefore, we can see that for any $x\in\TL_n$, we have $E_ni_n(x)E_n=(i_n\circ{i_{n-1}}\circ\Ecal_n(x))E_n$.

* $tr_n(xy)\colon\TL_n\times\TL_n\to\TL_0$,

\[
\tr_n\left(
\knotsinmath{\begin{scope}[scale=0.75]
\begin{scope}[yshift=-0.5cm]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$x$};
\end{scope}
\begin{scope}[yshift=-0.5cm,xshift=1.5cm]
\draw (-0.25,0.5) node[scale=1.33] {$\cdot$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$y$};
\end{scope}
\end{scope}}\right) = \tr_n\left(\knotsinmath{\begin{scope}[scale=0.75,yshift=0.25cm]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0,-0.5) rectangle (1,-1.5);
\draw (0.25,-1.5)--(0.25,-2);
\draw (0.75,-1.5)--(0.75,-2);
\draw (0.53,-1.75) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0.5,-1) node[scale=1.33] {$y$};
\end{scope}}\right) = \knotsinmath{\begin{scope}[scale=0.75,yshift=0.25cm]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0,-0.5) rectangle (1,-1.5);
\draw (0.25,-1.5)--(0.25,-2);
\draw (0.75,-1.5)--(0.75,-2);
\draw (0.53,-1.75) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0.5,-1) node[scale=1.33] {$y$};
\draw[red] (0.75,1.5)--(1.25,1.5)--(1.25,-2)--(0.75,-2);
\draw[red] (0.25,1.5)--(0.25,1.75)--(1.5,1.75)--(1.5,-2.25)--(0.25,-2.25)--(0.25,-2);
\end{scope}}
\]

By moving the $x$ diagram around in a clockwise fashion, we get,

\[
= \knotsinmath{\begin{scope}[scale=0.75,yshift=0.25cm]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0,-0.5) rectangle (1,-1.5);
\draw (0.25,-1.5)--(0.25,-2);
\draw (0.75,-1.5)--(0.75,-2);
\draw (0.53,-1.75) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0.5,-1) node[scale=1.33] {$x$};
\draw[red] (0.75,1.5)--(1.25,1.5)--(1.25,-2)--(0.75,-2);
\draw[red] (0.25,1.5)--(0.25,1.75)--(1.5,1.75)--(1.5,-2.25)--(0.25,-2.25)--(0.25,-2);
\end{scope}} = \tr_n(yx)
\]

* By (ii), we know $\tr_{n+1}=\tr_n\circ\Ecal_{n+1}$. And we also know $E_n\in\TL_{n+1}$, so $\tr_{n+1}(i_n(x)\cdot{E_n})=\tr_n(\Ecal_{n+1}(i_n(x)\cdot{E_n}))$. And so,

\[
\tr_n\left(\Ecal_{n+1}\left(\tlinclusion{$x$}\cdot\,\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}}\;\;\right)\right) = \tr_n\left(\Ecal_{n+1}\left(\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.3cm]
\begin{scope}[yshift=1cm]
\draw (0.625,0.375) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1.25,0.75);
\draw (0.25,-0.5)--(0.25,0);
\draw (0.8,-0.5)--(0.8,0);
\draw (1,-0.5)--(1,0);
\draw (0.25,0.75)--(0.25,1.25);
\draw (0.8,0.75)--(0.8,1.25);
\draw (1,0.75)--(1,1.25);
\draw (1.5,-0.5)--(1.5,1.25);
\draw (0.55,1) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\end{scope}}\;\right)\right) = \tr_n\left(\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.3cm]
\begin{scope}[yshift=1cm]
\draw (0,0) rectangle (1.25,0.75);
\draw (0.625,0.375) node[scale=1.33] {$x$};
\draw (0.25,-0.5)--(0.25,0);
\draw (0.8,-0.5)--(0.8,0);
\draw (1,-0.5)--(1,0);
\draw (0.25,0.75)--(0.25,1.25);
\draw (0.8,0.75)--(0.8,1.25);
\draw (1,0.75)--(1,1.25);
\draw (1.5,-0.5)--(1.5,1.25);
\draw (0.55,1) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\draw[red] (1.5,2.25)--(1.75,2.25)--(1.75,-1.5)--(1.5,-1.5);
\end{scope}}\;\right)
\]

And by straightening the last string, it is easy to see that we get,

\[=\tr_n\left(\tldiagram{$x$}\right)\]

* From the above property, we know $\tr_n(\Ecal_{n+1}(x)\cdot{y})=\tr_{n+1}(i_n(\Ecal_{n+1}(x)\cdot{y})\cdot{E_n})$. So, we now need to show $i_n(\Ecal_{n+1}(x)\cdot{y})\cdot{E_n}=x\cdot{i_n(y)}$.

\[
i_n\left(\tlconditional{$x$}\cdot\,\tldiagram{$y$}\right) = i_n\left(\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\end{scope}}\;\right) =
\knotsinmath{
\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\draw (1.5,2.5)--(1.5,-1.5);
\end{scope}}
\]

And so $i_n(\Ecal_{n+1}(x)\cdot{y})\cdot{E_n}$ becomes

\[
\knotsinmath{
\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\draw (1.5,2.5)--(1.5,-1.5);
\begin{scope}[xshift=1.75cm]
\draw (0,0.5) node {$\cdot$};
\draw (0.25,0.5) [dnup=0];
\draw (0.25,-1.5) [dnup=0];
\draw (1,-1.5) [dndn=0.5];
\draw (1,2.5) [upup=0.5];
\draw (0.8,-1.5) [dnup=0];
\draw (0.8,0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\end{scope}}=\knotsinmath{
\begin{scope}[scale=0.75,yshift=0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\end{scope}
\draw (1.5,2.5)--(1.5,-1.5);
\begin{scope}[yshift=-3cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\end{scope}}
\]

Now from (i), we know $\tr_{n+1}=\tr_n\circ\Ecal_{n+1}$, so

\[
\knotsinmath{
\begin{scope}[scale=0.75,yshift=0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\end{scope}
\draw (1.5,2.5)--(1.5,-1.5);
\begin{scope}[yshift=-3cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\draw[red] (1.5,2.5)--(1.75,2.5)--(1.75,-3.5)--(1.5,-3.5);
\end{scope}} = \knotsinmath{
\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.5,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.9,1)--(0.9,1.5);
\draw (0.9,-0.5)--(0.9,0);
\draw[red] (0.9,1.5)--(1.2,1.5)--(1.2,-0.5)--(0.9,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\end{scope}}
\]

Now, $\Ecal_{n+1}(x\cdot{i_n(y)})$ is

\[
\Ecal_{n+1}\left(\tldiagram{$x$}\cdot\tlinclusion{$y$}\right) = \Ecal_{n+1}\left(\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.75,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1.5,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}
\end{scope}}\right) = \knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\begin{scope}[yshift=1cm]
\draw (0.75,0.5) node[scale=1.33] {$x$};
\draw (0,0) rectangle (1.5,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (1.25,1.5)--(1.25,1);
\draw (1.25,0)--(1.25,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.5) node {$\cdots$};
\end{scope}
\begin{scope}[yshift=-1cm]
\draw (0.5,0.5) node[scale=1.33] {$y$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}
\draw[red] (1.25,2.5)--(1.75,2.5)--(1.75,-1.5)--(1.25,-1.5);
\end{scope}}
\]

Thus $\tr_n(\Ecal_{n+1}(x)\cdot{y})=\tr_{n+1}(x\cdot{i_n(y)}$.

* And finally,

\[
\tr_n\left(\;\knotsinmath{\begin{scope}[scale=0.75,yshift=-1cm]
\draw (0.25,0) [dnup=0];
\draw (0.5,1) node {$\ldots$};
\draw (0.75,0) [dnup=0];
\end{scope}}\;\right) = \knotsinmath{\begin{scope}[scale=0.75,yshift=-1cm]
\draw (0.25,0) [dnup=0];
\draw (0.5,1) node {$\ldots$};
\draw (0.75,0) [dnup=0];
\draw[red] (0.75,2)--(1,2)--(1,0)--(0.75,0);
\draw[red] (0.25,2)--(0.25,2.25)--(1.25,2.25)--(1.25,-0.25)--(0.25,-0.25)--(0.25,0);
\end{scope}} = d^n
\]

Now comparing the above relations and the relations found in Propositions <ref> and <ref>, we find that they are related. For instance, in factors $N\subset{M}$, we know $\tr_M(E_N(x))=\tr_N(x)$ for all $x\in{N}$ where $E_N\colon{N}\to{M}$. This is the same as (ii) in the above proposition.
This means that the conditional expectation on factors is related to the conditional expectation on $\TL_n$, and the Jones projections are related to the generators on $\TL_n$. But to relate one to the other, we must be extra careful with the inclusion map on $\TL_n$. For instance, part (iii) above is similar to the relation in part (i) of Proposition <ref>.

The trace on $\TL_n$ is called the Markov traceMarkov trace, and is unique up to a scalar (the scalar being $d^n$).

To get a normal trace, i.e., $\tr_n(1)=1$. We need to define the trace as the number of loops minus $n$. So, then it is clear that we would get: $\tr_n(1)=1$; $\tr_n(xy)=\tr_n(yx)$; and $\tr_{n+1}(xE_{n})=d^{-1}\tr_{n}(x)$ for all $x,y\in\TL_n$.
We will not use the normal trace, since we will normalise our Jones polynomial in the following chapter.
So this is only an interesting remark and nothing else.

§.§ The Jones-Wenzl idempotents

In this section we will discuss the Jones-Wenzl idempotents. We say an element is an idempotent if when multiplied by itself it equals itself. So any identity is itself an idempotent.
We now let $[n]_q=[n]$, for better readability.

Suppose there exists $f^{(n)}\in\TL_n$ such that $f^{(n)}\neq0$, $(f^{(n)})^2=f^{(n)}$ and for all $i=1,\ldots,n-1$ we have $E_if^{(n)}=f^{(n)}E_i$. Then $f^{(n)}$ is unique for each $\TL_n$.

Let $f_1^{(n)},f^{(n)}\in\TL_n$ satisfy all properties and $f_1^{(n)}\neq{f^{(n)}}$.
So by writing them as a linear combination of the generators $\{1,E_1,\ldots,E_{n-1}\}$ we have

\[f^{(n)}=x+\sum_i{y_iE_i}\qquad\qquad{f_1^{(n)}}=x^\prime+\sum_i{y^\prime_iE_i}\]

It then follows that $x=x^\prime=1$. Since both are idempotents we then have,

\begin{align*}
f^{(n)} &= f^{(n)}f_1^{(n)}\\
&= \left(1+\sum_i{y_iE_i}\right)f_1^{(n)}
\end{align*}

Now $\left(\sum_iy_iE_i\right)f_1^{(n)}=0$, and so the result then follows, i.e., $f^{(n)}=f_1^{(n)}$.

The Jones-Wenzl idempotentsJones-Wenzl idempotents are $f^{\left(n\right)}\in\TL_n$ are defined inductively such that

* $f^{\left(0\right)}\in\TL_0$. is the empty diagram

* $f^{\left(1\right)}\in\TL_1$ is the single strand

* if $[n+1]\neq0$, then

\[f^{(n+1)}=i_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right).\]

To understand this better, we will need to use the diagrams. So, $f^{(n+1)}$ can be rewritten as,

\begin{align*}
f^{(n+1)} &=
\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}-\frac{[n]}{[n+1]}\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}\cdot\knotsinmath{\begin{scope}[scale=0.75]\draw (2,-0.85) [dnup=0]; \draw (2.5,-0.85) [dnup=0]; \draw (3,-0.85) [dndn=0.5]; \draw (3,1.15) [upup=0.5]; \draw (2.3,0.15) node {$\cdots$};\end{scope}}\cdot\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}\\
\knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}-\frac{[n]}{[n+1]}\knotsinmath{
\begin{scope}[scale=0.75,yshift=1.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}}
\tag{JW}\label{equation:jwidempotents}
\end{align*}

\[f^{(1)}=\knotsinmath{\begin{scope}[scale=0.75]\draw (0.25,-1) [dnup=0];\end{scope}}\]

\[f^{(2)}=\knotsinmath{\begin{scope}[scale=0.75]\draw (0.25,-1) [dnup=0]; \draw (0.75,-1) [dnup=0];\end{scope}}-\frac{1}{[2]}\knotsinmath{\begin{scope}[scale=0.75]\draw (0.25,-1) [dndn=0.5];\draw (0.25,1) [upup=0.5];\end{scope}}=1_2-\frac{1}{[2]}E_1\]

Now we consider mapping $q\in\{Q\cup-Q\}$ to $[2]$, and find that the map is bijective, thus our map uniquely determines $q$ (see [52]). So we let $d=[2]$.

Suppose $n\geq0$ and $[1],\ldots,[n],[n+1]\neq0$ so $f^{\left(0\right)},\ldots,f^{\left(n+1\right)}$ are well-defined. Then, $f^{\left(n+1\right)}$ satisfies the following.

* $f^{\left(n+1\right)}=\left(f^{\left(n+1\right)}\right)^*=\left(f^{\left(n+1\right)}\right)^2$

* $\Ecal_{n+1}\left(f^{\left(n+1\right)}\right)=\dfrac{[n+2]}{[n+1]}f^{\left(n\right)}$

* $\left(i_n\left(f^{\left(n\right)}\right)\right)f^{\left(n+1\right)}=f^{\left(n+1\right)}\left(i_n\left(f^{\left(n\right)}\right)\right)=f^{\left(n+1\right)}$

* $\tr_{n+1}\left(f_{\left(n+1\right)}\right)=[n+2]$

* $E_jf^{(n+1)}=f^{(n+1)}E_j=0$ for all $j=\{1,\ldots,n\}$

We show this by induction on $n$.

This is obvious for $n=0$.
So suppose this is true for any $n$. Then, we need to show that it is also true for $n+1$.

\begin{align*}
\left(f^{(n+1)}\right)^2 &= \left( \knotsinmath{\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}-\frac{[n]}{[n+1]}\knotsinmath{
\begin{scope}[scale=0.75,yshift=1.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}}\,\,\right)^2\\
&=  \knotsinmath{
\begin{scope}[scale=0.75,yshift=0.6cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-1.4cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.4) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=2cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=0cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-4cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}}-\frac{[n]}{[n+1]}\knotsinmath{
\begin{scope}[scale=0.75,yshift=-4cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=2cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=0cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}}+\frac{[n]^2}{[n+1]^2}\knotsinmath{\begin{scope}[yshift=2.25cm]
\begin{scope}[scale=0.75,yshift=1.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\end{scope}\begin{scope}[yshift=-2.25cm]
\begin{scope}[scale=0.75,yshift=1.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\end{scope}}
\end{align*}

And by the inductive hypothesis, we have $\left(f^{(n)}\right)^2=f^{(n)}$ and $\Ecal_{n}\left(f^{(n)}\right)=\dfrac{[n+1]}{[n]}f^{(n-1)}$

\begin{align*}
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (1.25,1.5)--(1.25,-0.5);
\end{scope}}-\frac{2[n]}{[n+1]}\knotsinmath{
\begin{scope}[scale=0.75,yshift=1.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.25,-0.5) [dnup=0];
\draw (1,-0.5) [dndn=0.5];
\draw (1,1.5) [upup=0.5];
\draw (0.8,-0.5) [dnup=0];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-2.5cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}}+\frac{[n]^2}{[n+1]^2}\cdot\frac{[n+1]}{[n]}\,\,\,\knotsinmath{\begin{scope}
\begin{scope}[scale=0.75,yshift=2cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,0)--(1,-0.5);
\draw (1,1.5)--(1,1);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-0.5cm]
\draw (0.65,0.5) node[scale=1.23] {$f^{(n-1)}$};
\draw (0,0) rectangle (1.3,1);
\draw (0.25,2)--(0.25,1);
\draw (0.25,0)--(0.25,-1);
\draw (0.8,0)--(0.8,-1);
\draw (0.8,2)--(0.8,1);
\draw (0.53,1.25) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-1cm]
\draw (1,-0.5) [dndn=0.5];
\draw (1,2.5) [upup=0.5];
\draw (0.55,0.5) node {$\cdots$};
\end{scope}
\begin{scope}[scale=0.75,yshift=-3cm]
\draw (0.5,0.5) node[scale=1.33] {$f^{(n)}$};
\draw (0,0) rectangle (1.1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.8,0)--(0.8,-0.5);
\draw (0.8,1.5)--(0.8,1);
\draw (1,1)--(1,1.5);
\draw (1,-0.5)--(1,0);
\draw (1.5,1.5)--(1.5,-0.5);
\draw (0.53,-0.25) node {$\cdots$};
\end{scope}
\end{scope}}
\end{align*}

And again by the inductive hypothesis, we know $f^{(n)}\left(i_{n-1}\left(f^{(n-1)}\right)\right)=\left(i_{n-1}\left(f^{(n-1)}\right)\right)f^{(n)}=f^{(n)}$. And so by stretching the $E_{n}$ loops, it is easy to see that we get the same diagram as the second term. Equivalently we can use Proposition <ref> to see (our last diagram)
$i_n\left(f^{(n)}\right)E_nf^{(n-1)}E_ni_n\left(f^{(n)}\right)$ is equivalent to $i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right)$. Thus by further simplifying, we find that it is in fact idempotent, i.e., $\left(f^{(n+1)}\right)^2=f^{(n+1)}$.

* We will now use the results from Proposition <ref>. So,

\begin{align*}
\Ecal_{n+1}\left(f^{(n+1)}\right) &= \Ecal_{n+1}\left(i_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right)\right)\\
&= [2]\id_n\left(f^{(n)}\right)+\frac{[n]}{[n+1]}f^{(n)}f^{(n)}\\
&= [2]f^{(n)}+\frac{[n]}{[n+1]}f^{(n)}
\end{align*}

and using the result “doubling the quantum integer” from Proposition <ref>, we find $[2][n+1]=[n+2]+[n]$, and so the result becomes $\dfrac{[n+2]}{[n+1]}f^{(n)}$.

* We use the results from Proposition <ref>,

\begin{align*}
f^{(n+1)}\left(i_n\left(f^{(n)}\right)\right) &= \left(i_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right)\right)\left(i_n\left(f^{(n)}\right)\right)\\
&= \left(i_n\left(f^{(n)}\right)\right)\left(i_n\left(f^{(n)}\right)\right)-\frac{[n]}{[n+1]}i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right)i_n\left(f^{(n)}\right)\\
&= i_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}i_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right)\\
&= f^{(n+1)}
\end{align*}

* Now from (ii), we have $\Ecal_{n+1}\left(f^{(n+1)}\right)=\dfrac{[n+2]}{[n+1]}f^{(n)}$. So, then since $\tr_{n}\circ\Ecal_{n+1}=\tr_{n+1}$, we have

\[\tr_{n+1}\left(f^{(n+1)}\right)=\tr_n\circ\Ecal_{n+1}\left(f^{(n+1)}\right)=\frac{[n+2]}{[n+1]}\tr_n\left(f^{(n)}\right)\]

and by the inductive hypothesis, we have $\tr_n\left(f^{(n)}\right)=[n+1]$, and so

\[
\tr_{n+1}\left(f^{(n+1)}\right)=[n+2]
\]

* Finally, by the inductive hypothesis, for any $j=1,\ldots{n-1}$ we have $E_jf^{(n+1)}=f^{(n+1)}E_j$. So let $j=n$, then

\[ E_nf^{(n+1)}=E_ni_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}E_ni_n\left(f^{(n)}\right)E_ni_n\left(f^{(n)}\right) \]

So, from Proposition <ref>, we have $E_ni_n\left(f^{(n)}\right)E_n=\left(i_n\circ{i_{n-1}}\circ\Ecal_n\left(f^{(n)}\right)\right)E_n$. And from (ii) we know $\Ecal_n\left(f^{(n)}\right)=\dfrac{[n+1]}{[n]}f^{(n-1)}$, so then

\begin{align*}
E_nf^{(n+1)} &= E_ni_n\left(f^{(n)}\right)-\frac{[n]}{[n+1]}\left(i_n\circ{i_{n-1}}\circ\Ecal_n\left(f^{(n)}\right)\right)E_ni_n\left(f^{(n)}\right)\\
&= E_ni_n\left(f^{(n)}\right)-\left(i_n\circ{i_{n-1}}\left(f^{(n-1)}\right)\right)E_ni_n\left(f^{(n)}\right)
\end{align*}

Now we have $i_n\circ{i_{n-1}}\left(f^{(n-1)}\right)E_ni_n\left(f^{(n)}\right)$ is equivalent to $E_ni_n\left(f^{(n)}\right)$, and so the result then follows.

Suppose $\tr_j(y^*x)$ is positive and semi-definite (i.e., $\tr_j(x^*x)\geq0$ and $\tr_j(x^*x)=0$ if and only if $x^*x=0$) for all $j\geq0$, $x,y\in\TL_n$. Then

\[q+q^{-1}\in\left\{2\cos(\pi/n):n\geq3\right\}\cup[2,\infty)\]

We show this by contradiction. We will assume $q+q^{-1}\not\in\{2\cos(\pi/n):n\geq3\}\cup[2,\infty)$. Using our Jones-Wenzl idempotents, we want to show that $\tr_j((f^{(w)})^*f^{(w)})=[w+1]<0$ for some $w$.

If we let $q=e^{i\theta}$ for some $\theta\in(0,\pi/2)$ such that $\theta\neq\pi/n$ for some $\pi\geq3$, then by recalling $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$, we have

\[[j]=\frac{(e^{i\theta})^j-(e^{i\theta})^{-j}}{(e^{i\theta})-(e^{-i\theta})}=\frac{\sin(j\theta)}{\sin(\theta)}\]

Now by letting $k\geq2$ be minimal such that $\theta>\pi/(k+1)$, we have

\[\frac{\pi}{k}>\theta>\frac{\pi}{k+1}\]

then since $\sin(\theta),\sin(2\theta),\ldots,\sin(k\theta)>0$ but we have $\sin((k+1)\theta)<0$ so then $[1],[2],\ldots,[k]>0$, but $[k+1]<0$.

So since $k\geq2$ and $[1],[2],\ldots,[k]\neq0$ we have our idempotents $f^{(0)},\ldots,f^{(k)}$ are well-defined. Thus, we get our contradiction by taking the trace $\tr_k\left({(f^{(k)})^*f^{(k)}}\right)=[k+1]<0$.

§ JONES' ORIGINAL CONSTRUCTION

We will now construct our Jones polynomial using braids and the Temperley-Lieb algebra. This discussion is based on that in [1], [2], [3], [4], [12], [16], [17], [26], [28], [29], [33], [34], [35], [37], [38], [44], [49], [52], and [59].

To get invariance under conjugation (the Markov $\MI$ move), we need to define a trace. To define a trace, we need a representation of $\Bcal_n$ in a type II$_1$ factor since it is not finite (recall the motivation section <ref>).

The homomorphism $\phi_n\colon\CC[\Bcal_n]\to\TL_n$ which is defined by
\[\phi_n(\sigma_i)=A(1_n)+A^{-1}E_i\quad\mbox{ and }\quad\phi_n(\sigma^{-1}_i)=A^{-1}(1_n)+AE_i,\]

is a representation of $\Bcal_n$ (i.e., the homomorphism preserves the structure of $\Bcal_n$), where $d=([2]_q)_{-A^2=q}$.

If $|i-j|\geq2$, then

\begin{align*}
\phi_n(\sigma_i\sigma_j) &= \phi_n(\sigma_i)\phi_n(\sigma_j)\\
&= (A(1_n)+A^{-1}E_i)(A(1_n)+A^{-1}E_j)\\
&= A^2+E_j+E_i+A^{-2}E_iE_j\\
&= (A(1_n)+A^{-1}E_j)(A(1_n)+A^{-1}E_i)\\
&= \phi_n(\sigma_j)\phi_n(\sigma_i)\\
&= \phi_n(\sigma_j\sigma_i)
\end{align*}

since if $|i=j|>1$, then $E_iE_j=E_jE_i$.

\begin{align*}
\phi_n(\sigma_i\sigma_{i+1}\sigma_i) &= \phi_n(\sigma_i)\phi_n(\sigma_{i+1})\phi_n(\sigma_{i})\\
&= (A(1_n)+A^{-1}E_i)(A(1_n)+A^{-1}E_{i+1})(A(1_n)+A^{-1}E_i)\\
&= (A^2+E_{i+1}+E_i+A^{-2}E_iE_{i+1})(A(1_n)+A^{-1}E_i)\\
&= A^3+2AE_i+AE_{i+1}+A^{-1}E_{i+1}E_i+A^{-1}E^2_i+A^{-1}E_iE_{i+1}+A^{-3}E_iE_{i+1}E_i\\
&= A^3+2AE_i+AE_{i+1}+A^{-1}E_{i+1}E_i+A^{-1}d{E_i}+A^{-1}E_iE_{i+1}+A^{-3}{E_i}\\
&= A^3+E_i(2A+A^{-1}d+A^{-3})+AE_{i+1}+A^{-1}E_{i+1}E_i+A^{-1}E_iE_{i+1}
\end{align*}

Note that

\[\phi_n(\sigma_{i+1}\sigma_i\sigma_{i+1})=A^3+E_{i+1}(2A+A^{-1}d+A^{-3})+AE_i+A^{-1}E_iE_{i+1}+A^{-1}E_{i+1}E_i\]

So, we need $d=-A^2-A^{-2}=([2]_q)_{-A^2=q}$.

\begin{align*}
\phi_n(\sigma_i)\phi_n(\sigma_i^{-1}) &= (A(1_n)+A^{-1}E_i)(A^{-1}(1_n)+AE_i)\\
&= 1_n+A^2E_i+A^{-2}E_i+E_i^2\\
&= 1_n+E_i(A^2+A^{-2})+d{E_i} = 1_n
\end{align*}

The homomorphism $\phi_n$ defined above can be thought of as resolving crossings in braids to produce $\TL_n$ diagrams, i.e.,

\[\phi_n(\sigma_i)=\phi_n\left(\KPE\right)=A\KPD+A^{-1}\KPC\]
\[\phi_n(\sigma_i^{-1})=\phi_n\left(\KPB\right)=A\KPC+A^{-1}\KPD\]

We need to take a composition of the maps $\phi_n$ and the Markov trace $\tr_n$ to map from $\CC[\Bcal_n]\to\CC(q)$.
Now let us explore the map $\tr_n\circ\phi_n$.

$\tr_n\circ\phi_n(\KPE)=-A^3d$, while $\tr_n\circ\phi_n(\KPB)=-A^{-3}d$.

\begin{align*}
\tr_n\circ\phi_n\left(\KPE\right) &= A\KPA\KPA+A^{-1}\KPA\\
&= Ad^2+A^{-1}d=A(d)^2-A^{-1}d\\
&= A^5+A=-A^3d
\end{align*}

\begin{align*}
\tr_n\circ\phi_n\left(\KPB\right) &= A\KPA+A^{-1}\KPA\KPA\\
&= Ad+A^{-1}d^2\\
&= A^{-5}+A^{-1}=-A^{-3}d
\end{align*}

Now $d^{-1}\tr_n\circ\phi_n$ is in fact the braid analog of the Kauffman bracketKauffman bracket!via braids. This can easily be seen with the state sum solution. The states of a braid closure is given by resolving the crossings in the braid, then taking the trace. Thus

\[\langle{\reallywidehat{b}}\rangle=d^{-1}\tr_n\circ\phi_n(b)=\sum_{\text{states }s}A^{\alpha(s)}d^{|s|-1}\]

Thus we can then define the Jones polynomial to be the braid analog of the Kauffman bracket multiplied by $(-A^{-3})^{\omega(\reallywidehat{b})}$.

However, we want to show that it is indeed a link invariant without the Kauffman bracket.

So we have the identity map of a braid $\id_n\colon\Bcal_n\to\Bcal_n$. Note that we have the following relations,

* $\phi_{n+1}\circ\incl_n(\alpha)=i_n\circ\phi_n(\alpha)$, where $\alpha\in\Bcal_n$

* $\id_n\circ\phi_n(\alpha)=\phi_n\circ\id_n(\alpha)$, where $\alpha\in\Bcal_n$

This is true since, for instance, in the first relation, the inclusion of a braid would not change the $\TL_n$ representation. We draw the first relation,

\[
\phi_{n+1}\circ\incl_n(b)=\braidinclusion{$\phi_n(b)$}=i_n\circ\phi_n(b)
\]

So we can thus use the relations from Proposition <ref>.

Since any braid word $\alpha\in\Bcal_n$ can be written with braid generators, we have the writhewrithe!braid defined on braids as the sum of the exponents of the generators.

\[\omega(\sigma_1^{i_1}\sigma_2^{i_2}\cdots\sigma_{n-1}^{i_j})=\sum_zi_z\]

Equivalently, the writhe could also be defined as the writhe of the closure of the braid since an orientation is given with the braid closure.

the Jones polynomial!original construction
For a braid $b\in\Bcal_n$. We can define the Jones polynomial link invariant as the following

\[
\]

We will prove this in two ways for the sake of argument and not necessity.

We will first show that this is in fact an invariant under the Markov moves, which in turn shows that this is invariant under the Reidemeister moves, making it a link invariant. And the second proof will show that this is obviously equivalent to our definitions of the Jones polynomial (Definition <ref> & Proposition <ref>).

Invariance under Markov moves:

Given any braid $\gamma\in\Bcal_n$, we have $\omega(\gamma)=-\omega(\gamma^{-1})$. So $\omega(\gamma{b}\gamma^{-1})=\omega(\gamma)+\omega(b)+\omega(\gamma^{-1})=\omega(b)$. So,

\begin{align*}
V(\reallywidehat{\gamma{b}\gamma^{-1}}) &= \frac{(-A^{-3})^{\omega(\gamma{b}\gamma^{-1})}\tr_n\circ\phi_n(\gamma{b}\gamma^{-1})}{d}\\
&= \frac{(-A^{-3})^{\omega(b)}\tr_n(\phi_n(\gamma)\phi_n(b)\phi_n(\gamma)^{-1})}{d}\\
&= \frac{(-A^{-3})^{\omega(b)}\tr_n(\phi_n(\gamma)^{-1}\phi_n(\gamma)\phi_n(b))}{d} = \frac{(-A^{-3})^{\omega(b)}\tr_n(\phi_n(b))}{d} = V(\reallywidehat{b})
\end{align*}

Now given any braid $\sigma_n\in\Bcal_{n+1}$, we have $\omega(\sigma_n)=1$, and $\phi_{n+1}(\sigma_n)=A1_{n+1}+A^{-1}E_n$. We also note $\omega(\incl_n(b))=\omega(b)$, so then

\begin{align*}
V(\reallywidehat{\incl_n(b)\sigma_n}) &= \frac{(-A^{-3})^{\omega(b\sigma_n)}\tr_{n+1}\circ\phi_{n+1}(\incl_n(b)\sigma_n)}{d}\\
&= \frac{(-A^{-3})^{\omega(b)+1}\tr_{n+1}(\phi_{n+1}(\incl_n(b))(A+A^{-1}E_n))}{d}
\end{align*}

Recall the above relation $\phi_{n+1}\circ\incl_n=i_n\circ\phi_n$.
We also know from Proposition <ref> that the Markov trace $\tr_{n+1}=\tr_n\circ\Ecal_{n+1}$. So, then $\tr_n\circ\Ecal_{n+1}(A\cdot{i_n(\phi_n(b))}+A^{-1}\cdot{i_n(\phi_n(b))}\cdot{E_n})$. And again from Proposition <ref>, we know $\tr_n\circ\Ecal_{n+1}\circ{i_n}=d\cdot\tr_n\circ\id_n$, and $\tr_n(\Ecal_{n+1}(i_n(\phi_n(b))\cdot{E_n}))=\tr_n(\phi_n(b))$. So,

\begin{align*}
&= \frac{(-A^{-3})^{\omega(b)+1}\tr_n(\phi_n(b))(Ad+A^{-1})}{d}\\
&= \frac{(-A^{-3})^{\omega(b)+1}\tr_n(\phi_n(b))(-A^{-3})}{d}\\
&= \frac{(-A^{-3})^{\omega(b)}\tr_n(\phi_n(b))}{d} = V(\reallywidehat{b})
\end{align*}

Analogously to the above, we find that under M2, where the braid becomes $b\sigma_n^{-1}$, the Jones polynomial $V(\reallywidehat{b\sigma_n^{-1}})=V(\reallywidehat{b})$.

Thus, we are done.

Skein relation:

Let us now ensure that this satisfies the skein relation definition.
Consider single crossings in the braid closure,

\begin{align*}
A^{4}V\left(\KPE\right)-A^{-4}V\left(\KPB\right)+(A^2-A^{-2})V\left(\KPD\right) &= A^4\frac{(-A^{-3})(-A^3d)}{d}\\
&= A^4-A^{-4}+(A^2-A^{-2})(-A^{-2}-A^2) = 0.
\end{align*}

This completes the proof.

For example, recall that the closure of $\sigma_1^3$ is equivalent to the right-handed trefoil knot, while $\sigma_1^{-3}$ is equivalent to the left-handed trefoil knot.

We can see then, that

\begin{align*}
\end{align*}

It is clear that the left-handed trefoil $V(\overline{3_1})=V(\reallywidehat{\sigma_1^{-3}})$, and the bracket polynomial would be $\phi_2(\sigma_1^{-1})^3=(AE_1+A^{-1})^3=A^3E_1^3+3AE_1^2+3A^{-1}E_1+A^{-3}$, and the Markov trace of the bracket polynomial becomes $A^3d^3+3Ad^2+3A^{-1}d+A^{-3}d^2$. And finally we have the Jones polynomial, $V(\overline{3_1})=-A^{9}(d(A^3d+3A+A^{-3})+3A^{-1}$, which further simplified turns out to be $A^{4}+A^{12}-A^{16}$.

To conclude, we see that the easiest to define (i.e., via the Kauffman bracket) is not necessarily the easiest to work with, while the most difficult to define (i.e., the original construction) is perhaps the easiest to work with.

§.§ Proving Theorem <ref>

In this section we will prove $V(L_1\cup{L_2})=dV(L_1)V(L_2)$ and $V(L_1\#L_2)=V(L_1)V(L_2)$, where $\#$ is the connected sum of the links. We will prove these results using braid representations instead of the skein relation and Kauffman bracket (where they can be derived easily). The reason we do this with braid representations is so that we can ensure comprehension.

We saw in Chapter <ref>, that the braid inclusion map essentially adds a strand to the right. So now we define the braid left-inclusion map $\lfincl_n$. The reason for this is to define the tensor product of two braids.

The braid left-inclusion maplinear operations on $\Bcal_n$!braid left-inclusion map $\lfincl_n\colon\Bcal_n\to\Bcal_{n+1}$ for $n\geq1$ is defined by,

\[\lfincl_n\left(\knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\end{scope}}\right) =\,\, \knotsinmath{\begin{scope}[yshift=-0.35cm,scale=0.75]
\draw (0,0) rectangle (1,1);
\draw (0.25,1.5)--(0.25,1);
\draw (0.25,0)--(0.25,-0.5);
\draw (0.75,0)--(0.75,-0.5);
\draw (0.75,1.5)--(0.75,1);
\draw (0.53,1.25) node {$\cdots$};
\draw (0.53,-0.25) node {$\cdots$};
\draw (0.5,0.5) node[scale=1.33] {$\alpha$};
\draw (-0.25,1.5)--(-0.25,-0.5);
\end{scope}}\]
left inclusion $\lfincl_n$ on $\Bcal_n$

We similarly define the left-inclusionlinear operations on $\TL_n$!left-inclusion!$\TL_n$ on $\TL_n$, i.e., $li\colon\TL_n\to\TL_{n+1}$.

Let $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$. We define the tensor productlinear operations on $\TL_n$!braid tensor product $\otimes\colon\Bcal_n\times\Bcal_m\to\Bcal_{n+m}$ by $\alpha\otimes\beta\mapsto\incl_{n\otimes{m}}(\alpha)\cdot\lfincl_{m\otimes{n}}(\beta)$, where the inclusion tensor product of the two integers is given by $\incl_{n\otimes{m}}=\incl_{n+m-1}\circ\incl_{n+m-2}\circ\cdots\circ\incl_{n}$ (the left inclusion is defined analogously). And so we thus have.

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\alpha$;

(-0.25,0.5) node[scale=1.33] $\otimes$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

(-0.5,0.5) node[scale=1.33] $=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\alpha$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $\beta$;

braid tensor product

So it is easy to see that the closure of the tensor product of two braids is essentially two disjoint links. For instance, let the two links have the following braid representations, $L_1=\reallywidehat{\alpha}$ and $L_2=\reallywidehat{\beta}$, so
\[L_1\cup{L_2}=\reallywidehat{\alpha}\cup\reallywidehat{\beta}=\reallywidehat{\alpha\otimes\beta}\]

It is easy to see that the union of two closure braids is well-defined, i.e., $\reallywidehat{\alpha\otimes\beta}\sim\reallywidehat{\beta\otimes\alpha}$.

We then analogously define the tensor productlinear operations on $\TL_n$!tensor product on $\TL_n$, i.e., $\otimes\colon{\TL_n\times{\TL_m}}\to{\TL_{n+m}}$,

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;

(-0.25,0.5) node[scale=1.33] $\otimes$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $y$;

(-0.5,0.5) node[scale=1.33] $=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $y$;

We have the relation $\phi_{n+m}(\alpha\otimes\beta)=\phi_n(\alpha)\otimes\phi_m(\beta)$, where $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$. This obviously says something along the lines of, “if two braids are disjoint, then so are their Temperley-Lieb representations”.

Let $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$, then we have $\tr_{n+m}\circ\phi_{n+m}(\alpha\otimes\beta)=\tr_n\circ\phi_n(\alpha)\cdot\tr_m\circ\phi_m(\beta)$.

The diagrammatic sketch of this is pretty straight forward. Essentially, we need to show that the trace of disjoint $\TL$ diagrams are independent of each other. Note the relation above, $\phi_{n+m}(\alpha\otimes\beta)=\phi_n(\alpha)\otimes\phi_m(\beta)$.

(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $x$;

(-0.25,0.5) node[scale=1.33] $\otimes$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1.33] $y$;

(-0.5,0.5) node[scale=1.33] $=$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1] $\phi_n(\alpha)$;
(0,0) rectangle (1,1);
(0.53,1.25) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(0.5,0.5) node[scale=1] $\phi_m(\beta)$;

[red] (1.95,1.5)–(2.45,1.5)–(2.45,-0.5)–(1.95,-0.5);
[red] (1.45,1.5)–(1.45,1.75)–(2.7,1.75)–(2.7,-0.75)–(1.45,-0.75)–(1.45,-0.5);
[red] (0.75,1.5)–(0.75,2)–(2.95,2)–(2.95,-1)–(0.75,-1)–(0.75,-0.5);
[red] (0.25,1.5)–(0.25,2.25)–(3.2,2.25)–(3.2,-1.25)–(0.25,-1.25)–(0.25,-0.5);

Which is clearly the same as $\tr_n\circ\phi_n(\alpha)\cdot\tr_m\circ\phi_m(\beta)$.

Since $L_1\cup{L_2}=\reallywidehat{\alpha\otimes\beta}$ for $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$, we then have

\begin{align*}
V\left(\reallywidehat{\alpha\otimes\beta}\right) &= \frac{(-A^{-3})^{\omega(\alpha)+\omega(\beta)}\tr_{n+m}\circ\phi_{n+m}(\alpha\otimes\beta)}{d}\\
&= \frac{(-A^{-3})^{\omega(\alpha)+\omega(\beta)}\tr_n\circ\phi_n(\alpha)\cdot\tr_m\phi_m(\beta)}{d}
\end{align*}

And so it is then clear that multiplying by $dd^{-1}$ will give us the following,

\[ V(L_1\cup{L_2}) = dV(L_1)V(L_2) \]

And recall $d=[2]_{q=-A^2=-t^{-1/2}}$, and so the result follows.

$L_1\#L_2$ is the connected sumconnected sum of links of the links, and is shown below

(0,0) rectangle (1,1);
[<-] (1,0.25)–(1.25,0.25)–(1.25,0.75)–(1,0.75);
(1.5,0.5) node $\#$;
(0.5,0.5) node $L_1$;
(0,0) rectangle (1,1);
[->] (0,0.25)–(-0.25,0.25)–(-0.25,0.75)–(0,0.75);
(0.5,0.5) node $L_2$;

(3.5,0.5) node $=$;
(4,0) rectangle (5,1);
(5.5,0) rectangle (6.5,1);
[<-] (5,0.25)–(5.5,0.25);
[->] (5,0.75)–(5.5,0.75);
(4.5,0.5) node $L_1$;
(6,0.5) node $L_2$;

connected sum of links $\#$

It is easy to see that this is well-defined.

An almost direct corollary one can make is the following.

For links $L_1,L_2$, we have $V(L_1\#L_2)=V(L_1)V(L_2)$.

We “glue” our link diagrams such that $T_+\sim{T_-}\sim{L_1\#{L_2}}$.

[->] (1.5,0)–(0.5,0.5);
[ultra thick,draw=white,double=black,double distance=0.5pt] (0.5,0)–(1.5,0.5);
[->] (0.5,0)–(1.5,0.5);
(0.25,0.25) node $L_1$;
(1.75,0.25) node $L_2$;
(1,-0.5) node $T_+$;

[->] (0.5,0)–(1.5,0.5);
[ultra thick,draw=white,double=black,double distance=0.5pt] (1.5,0)–(0.5,0.5);
[->] (1.5,0)–(0.5,0.5);
(0.25,0.25) node $L_1$;
(1.75,0.25) node $L_2$;
(1,-0.5) node $T_-$;

[->] (0.5,0) to [bend right=50] (0.5,0.5);
[->] (1.5,0) to [bend left=50] (1.5,0.5);
(0.25,0.25) node $L_1$;
(1.75,0.25) node $L_2$;
(1,-0.5) node $T_0$;

It is clear that we have $T_0\sim{L_1\cup{L_2}}$, so then

\begin{align*}
V(L_1\#L_2)(t^{-1}-t)+d(t^{-1/2}-t^{1/2})V(L_1)V(L_2) &= 0
\end{align*}

And so $V(L_1\#L_2)=V(L_1)V(L_2)$.

We now show this result using braid representations. We first need to define what the connected sum of braids is.
linear operations on $\Bcal_n$!braid connected sum
If $L_1=\reallywidehat{\alpha}$ and $L_2=\reallywidehat{\beta}$ for $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$. We have $L_1\#L_2=\reallywidehat{\alpha\clubsuit\beta}$, where $\clubsuit\colon\Bcal_n\times\Bcal_m\to\Bcal_{n+m-1}$ is given by

\[ \alpha\clubsuit\beta=\incl_{n\otimes{(m-1)}}(\alpha)\cdot\lfincl_{m\otimes{(n-1)}}(\beta) \]

(0,0) rectangle (1,1);
(1.75,0) rectangle (2.75,1);
(1.375,0.5) node[scale=1.33] $\clubsuit$;
(3,0.5) node[scale=1.33] $=$;
(3.25,2) rectangle (4.25,1);
(3.75,0) rectangle (4.75,-1);
(0.5,0.5) node[scale=1.33] $\alpha$;
(2.25,0.5) node[scale=1.33] $\beta$;
(3.75,1.5) node[scale=1.33] $\alpha$;
(4.25,-0.5) node[scale=1.33] $\beta$;
(0.53,1.25) node $\cdots$;
(2.28,1.25) node $\cdots$;
(3.78,2.2) node $\cdots$;
(4.28,-1.2) node $\cdots$;
(0.53,-0.25) node $\cdots$;
(2.28,-0.25) node $\cdots$;
(3.78,0.5) node $\cdots$;
(4.28,0.5) node $\cdots$;

braid connected sum $\clubsuit$

This is well-defined, as the closure of $\alpha\clubsuit\beta$ can easily be seen to be ambient isotopic to the closure of $\beta\clubsuit\alpha$.

So now we have the following theorem.

For $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$ we have,

\[V(\reallywidehat{\alpha\clubsuit\beta})=V(\reallywidehat{\alpha})V(\reallywidehat{\beta})\]

To prove this, we begin with a lemma.

The braid closure of $\alpha\clubsuit\beta$ is equivalent to the braid closure of $\alpha\heartsuit\beta$ where $\heartsuit\colon\Bcal_n\times\Bcal_m\to\Bcal_{n+m}$ is given by $\alpha\heartsuit\beta=\sigma_{n}\cdot(\alpha\otimes\beta)$, where $\sigma_{n}\in\Bcal_{n+m}$, $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$,

[strands=4] σ_2
(0.35,-1) rectangle (1.15,-2);
(1.4,-1) rectangle (2.15,-2);
(0.75,-1.5) node[scale=1.33] $\alpha$;
(1.735,-1.5) node[scale=1.33] $\beta$;
(0.78,-0.75) node $\cdots$;
(0.78,-2.25) node $\cdots$;
(1.78,-0.75) node $\cdots$;
(1.78,-2.25) node $\cdots$;

This is obviously well defined.
We will show this diagrammatically to ensure that the braid closures are in fact ambient isotopic.

[strands=4] σ_2
(0.35,-1) rectangle (1.15,-2);
(1.4,-1) rectangle (2.15,-2);
(0.75,-1.5) node[scale=1.33] $\alpha$;
(1.735,-1.5) node[scale=1.33] $\beta$;
(0.78,-2.25) node $\cdots$;
(0.78,-0.75) node $\cdots$;
(1.765,-2.25) node $\cdots$;
(1.765,-0.75) node $\cdots$;
[red] (2,0)–(2.25,0)–(2.25,-2.5)–(2,-2.5);
[red] (1.5,0)–(1.5,0.25)–(2.5,0.25)–(2.5,-2.75)–(1.5,-2.75)–(1.5,-2.5);
[red] (1,0)–(1,0.5)–(2.75,0.5)–(2.75,-3)–(1,-3)–(1,-2.5);
[red] (0.5,0)–(0.5,0.75)–(3,0.75)–(3,-3.25)–(0.5,-3.25)–(0.5,-2.5);

(2.75,-1) node $\sim$;
(0,0) rectangle (1,-1);
(1.5,0) rectangle (2.5,-1);
(0.5,-0.5) node[scale=1.33] $\alpha$;
(2,-0.5) node[scale=1.33] $\beta$;

(6.25,-1) node $\sim$;
(0,0) rectangle (1,-1);
(0.5,-1.5) rectangle (1.5,-2.5);
(0.5,-0.5) node[scale=1.33] $\alpha$;
(1,-2) node[scale=1.33] $\beta$;

(-1,-1) node $\reallywidehat{\alpha\heartsuit\beta}\quad=$;
(9.25,-1) node $\quad=\;\;\reallywidehat{\alpha\clubsuit\beta}$;

By the lemma above, we know that finding the Jones polynomial of the closure of $\alpha\heartsuit\beta$ is enough.

So we show this by induction on $m$. For $m=1$, $\beta$ is the one strand (and so $V(\reallywidehat{\beta})=1$). So $\alpha\otimes\beta=\incl_n(\alpha)$, and we know $\sigma_n\incl_n(\alpha)$ is Markov equivalent to $\alpha$. So the result then follows.

Now suppose that this is true for $\alpha\in\Bcal_n$ and $\beta\in\Bcal_{m-1}$. So for $\alpha\in\Bcal_n$ and $\beta\in\Bcal_m$ we have

\begin{align*}
\phi_{n+m}(\sigma_{n}\cdot(\alpha\otimes\beta)) &= (A+A^{-1}E_{n})\phi_{n+m}(\alpha\otimes\beta)
\end{align*}

And so we now take the trace of $E_n\phi_{n+m}(\alpha\otimes\beta)$ and find

\[
\tr_{n+m}\left(\knotsinmath{\begin{scope}[yshift=0.55cm,scale=0.75]
\draw (1,1) [upup=0.5];
\draw (1,-1) [dndn=0.5];
\draw (0.35,-1) rectangle (1.15,-2);
\draw (0.5,1)--(0.5,-1);
%  \draw (1,-0.5)--(1,-1);
\draw (0.5,-2)--(0.5,-2.5);
\draw (1,-2)--(1,-2.5);
\draw (1.4,-1) rectangle (2.15,-2);
%  \draw (1.5,-0.5)--(1.5,-1);
\draw (2,1)--(2,-1);
\draw (1.5,-2.5)--(1.5,-2);
\draw (2,-2.5)--(2,-2);
\draw (0.75,-1.5) node[scale=1] {$\phi(\alpha)$};
\draw (1.78,-1.5) node[scale=1] {$\phi(\beta)$};
\draw (0.78,-0.75) node {$\cdots$};
\draw (0.78,-2.25) node {$\cdots$};
\draw (1.78,-0.75) node {$\cdots$};
\draw (1.78,-2.25) node {$\cdots$};
\end{scope}}\right)=\knotsinmath{\begin{scope}[yshift=0.55cm,scale=0.75]
\draw (1,1) [upup=0.5];
\draw (1,-1) [dndn=0.5];
\draw (0.35,-1) rectangle (1.15,-2);
\draw (0.5,1)--(0.5,-1);
%  \draw (1,-0.5)--(1,-1);
\draw (0.5,-2)--(0.5,-2.5);
\draw (1,-2)--(1,-2.5);
\draw (1.4,-1) rectangle (2.15,-2);
%  \draw (1.5,-0.5)--(1.5,-1);
\draw (2,1)--(2,-1);
\draw (1.5,-2.5)--(1.5,-2);
\draw (2,-2.5)--(2,-2);
\draw (0.75,-1.5) node[scale=1] {$\phi(\alpha)$};
\draw (1.78,-1.5) node[scale=1] {$\phi(\beta)$};
\draw (0.78,-0.75) node {$\cdots$};
\draw (0.78,-2.25) node {$\cdots$};
\draw (1.78,-0.75) node {$\cdots$};
\draw (1.78,-2.25) node {$\cdots$};
\draw[red] (2,1)--(2.25,1)--(2.25,-2.5)--(2,-2.5);
\draw[red] (1.5,1)--(1.5,1.25)--(2.5,1.25)--(2.5,-2.75)--(1.5,-2.75)--(1.5,-2.5);
\draw[red] (1,1)--(1,1.5)--(2.75,1.5)--(2.75,-3)--(1,-3)--(1,-2.5);
\draw[red] (0.5,1)--(0.5,1.75)--(3,1.75)--(3,-3.25)--(0.5,-3.25)--(0.5,-2.5);
\end{scope}}=\knotsinmath{\begin{scope}[yshift=1cm,scale=0.75]
\draw (1,-2) [upup=0.5];
\draw (1,-1) [dndn=0.5];
\draw (0.35,-1) rectangle (1.15,-2);
%  \draw (0.5,1)--(0.5,-1);
%  \draw (0.5,-2)--(0.5,-2.5);
%  \draw (1,-2)--(1,-2.5);
\draw (1.4,-1) rectangle (2.15,-2);
%  \draw (2,1)--(2,-1);
%  \draw (1.5,-2.5)--(1.5,-2);
%  \draw (2,-2.5)--(2,-2);
\draw (0.75,-1.5) node[scale=1] {$\phi(\alpha)$};
\draw (1.78,-1.5) node[scale=1] {$\phi(\beta)$};
\draw (0.78,-0.75) node {$\cdots$};
\draw (0.78,-2.25) node {$\cdots$};
\draw (1.78,-0.75) node {$\cdots$};
\draw (1.78,-2.25) node {$\cdots$};
\draw (2,-1)--(2,-0.75)--(2.45,-0.75)--(2.45,-2.25)--(2,-2.25)--(2,-2);
\draw (0.5,-1)--(0.5,-0.75)--(0.15,-0.75)--(0.15,-2.25)--(0.5,-2.25)--(0.5,-2);
\end{scope}}
\]

Now we will see all the different possibilities connecting $E_n$, by only looking at the connection between the four points. The first possibility, for instance, gives

(0,0) rectangle (1,1);
(1.25,0) rectangle (2.25,1);
(0.75,0) [upup=0.75];
(0.75,1) [dndn=0.75];
[dashed] (0.75,1)–(0.75,0);
[dashed] (1.5,1)–(1.5,0);

So, the trace of $\phi_{n+m}(\alpha\otimes\beta)$ without $E_n$ would give $d^2$ (multiplied by the rest of the diagram), whereas $E_n\phi_{n+m}(\alpha\otimes\beta)$ gives $d$ (multiplied by the rest of the diagram).
The second and third possibilities are the same, but with opposite choices. So without loss of generality, we have,

(0,0) rectangle (1,1);
(1.25,0) rectangle (2.25,1);
(0.75,0) [upup=0.75];
(0.75,1) [dndn=0.75];
[dashed] (0.75,1)–(0.75,0.65)–(0.25,0.65)–(0.25,1);
[dashed] (0.75,0)–(0.75,0.4)–(0.25,0.4)–(0.25,0);
[dashed] (1.5,1)–(1.5,0);

This clearly becomes a single strand.
So the trace would give $d$. Whereas, the trace of $\phi_{n+m}(\alpha\otimes\beta)$ without $E_n$ would give $d^2$.
We also have,

(0,0) rectangle (1,1);
(1.25,0) rectangle (2.25,1);
(0.75,0) [upup=0.75];
(0.75,1) [dndn=0.75];
[dashed] (0.75,1)–(0.75,0.65)–(0.25,0.65)–(0.25,1);
[dashed] (0.75,0)–(0.75,0.4)–(0.25,0.4)–(0.25,0);
[dashed] (1.5,1)–(1.5,0.65)–(2,0.65)–(2,1);
[dashed] (1.5,0)–(1.5,0.4)–(2,0.4)–(2,0);

Similarly to the previous possibilities, we have the trace of $E_n\phi_{n+m}(\alpha\otimes\beta)$ giving $d$, and $d^2$ for the trace of $\phi_{n+m}(\alpha\otimes\beta)$ without $E_n$. We then have two other possibilities (i.e., both strands in a diagram are going down/up and all strands are going down/up) - the reader is advised to try them.

We see that it will always give us $\tr_{n+m}(E_n\phi_{n+m}(\alpha\otimes\beta))=d^{-1}\tr_{n+m}\circ\phi_{n+m}(\alpha\otimes\beta)$. So then we have

\[\tr_{n+m}(E_n\phi_{n+m}(\alpha\otimes\beta))=d^{-1}\cdot\tr_{n+m}\circ\phi_{n+m}(\alpha\otimes\beta)\]

The result then follows,

\begin{align*}