\subsection{Planar Algebras}
\label{sec:planar-algebras}
%stolen from Scott's paper

A planar algebra is a gadget specifying how to combine objects in
planar ways. They were introduced in \cite{math.QA/9909027} to study
subfactors, and have since found more general use.

\newcommand{\pa}{\mathcal{P}}
In the simplest version, a shaded planar algebra $\pa$ associates a vector space
$\pa_k$ to each $k \in {+,-,1,2,3,\ldots}$ (thought of as a disc in the plane with
$2k$ marked points on its boundary; elements of $\pa_k$ will be refered to as $k$-boxes and the space $\pa_k$ as the $k$-disc space) and a linear map $\pa(T)  : \pa_{k_1}
\tensor \pa_{k_2} \tensor \cdots \tensor \pa_{k_r} \To \pa_{k_0}$ to each
planar tangle\footnote{Familiarly known as a 'spaghetti and meatballs'
diagram.} $T$, for example
$$\mathfig{0.5}{SampleTangle},$$ with internal
discs with $2 k_1, 2 k_2, \ldots, 2 k_r$ marked points, and $2 k_0$ marked points
on the external disc (evenness of the number of boundary points is required so that the regions can be given a checkerboard shading); this particular tangle corresponds to a map $V_2 \times V_4 \times V_4 \rightarrow V_2$.  These maps (the 'planar operations') must satisfy
certain properties: ``radial'' tangles induce identity maps, and
composition of the maps $\pa(T)$ is compatible with the obvious
composition of planar diagrams by gluing one inside the other.  In fact, one more piece of data is required here: a small $\star$ for every boundary circle, in one of its adjacent unshaded regions -- this tells us, when we place once planar tangle inside another one, how to line up the strings.
For the exact details, which are somewhat technical,
see \cite{math.QA/9909027}.
%end stolen section

A subfactor planar algebra is a shaded planar algebra, with some nice properties that allow us to do linear algebra on it.
\begin{defn} A {\em subfactor planar algebra} is a shaded planar algebra such that
\begin{itemize}
\item Each vector space $V_i$ for $i = +,-,1,2,\ldots$ has an involution $^*$ on it that plays well with tangles\eep{explain this more better}
\item $\text{dim}(V_+)=\text{dim}(V_-)=1$
\item $d_+ =d_-$: Note that $\text{dim}(V_+)=\text{dim}(V_-)=1$ implies that the element of $V_+$ coming from the tangle with one closed string\footnote{that is, a spaghetti-O} is a multiple $d_+$ of the empty tangle. (There is a corresonding $d_-$ for for $V_-$.)
$$\mathfig{0.8}{RemovingSpaghettiO}$$
 In a subfactor planar algebra we require $d_+=d_-$, and call this value the modulus $d$.\noah{can't we call this the dimension?}
\item Positivity:  Another consequence of  $\text{dim}(V_+)=\text{dim}(V_-)=1$ is that  the tangles
 $$\mathfig{0.4}{TraceTangle} \text{ and } \mathfig{.4}{Multiplication}$$
 define a trace $V_i \rightarrow V_+$ and multiplication $V_i \times V_i \rightarrow V_i$; these can be used to define a bilinear form via $\left< X,Y \right> := \tr{Y^* X}$.  This bilinear form must be positive definite in a subfactor planar algebra.
\item Sphericality:  Two tangles that are spherically isotopic induce the same maps.
\end{itemize}
\end{defn}
For more details about subfactor planar algebras, see \cite[\S4]{math.GT/0606318}.

\subsection{Temperley-Lieb and the $\dfour$ planar algebra}

Everyone's favorite example of a subfactor planar algebra is $\TL(\delta)$, the Temperley-Lieb planar algebra with modulus $\delta$, which is defined for all $\delta \in \{ 2 \cos \frac{\pi}{n} | n \geq 3 \} \cup [2,\infty)$.  If $\delta > 2$, then $\TL(\delta)_n$ is the vector space with basis consisting of tangles with no inputs, $n$ outputs and no closed strings -- for instance, $\TL(\delta)_3$ is
$$\mathfig{.8}{TL3}.$$
For $\delta$ in the range  $\{ 2 \cos \frac{\pi}{n} | n \geq 3 \}$, we first define a planar algebra $TL$ as above -- that is, $TL(\delta)_n$ is the vector space with basis consisting of tangles with no inputs, $n$ outputs and no closed strings; then we make this a subfactor planar algebra by forcing positive definiteness; that is, $\TL(\delta)$ is $$\quotient{TL(\delta)}{ \left< X \text{ such that for some } Y, \, tr(Y^* X)=0 \right>}.$$

The action of a general tangle on the pictures generating $\TL(\delta)$ is pretty straightforward -- put the $\TL(\delta)$ pictures into the inner disks of the tangle, erase boundaries that are no longer boundaries, and throw out closed loops by multiplying the resulting picture by $d$.  Here's an example:
$$\mathfig{.5}{TLExample}.$$

Here are some useful facts about Temperley-Lieb:

\begin{itemize}
\item Any subfactor planar algebra with modulus $d$ contains $TL(d)$ as a sub-planar algebra
\item $\TL_n$ is multiplicatively generated by elements $e_i$ for $i=1 \ldots n-1$
 $$\mathfig{0.5}{Eis}$$.
 \item For each $n$, $\TL_n$ has one new minimal central idempotent; it is known as the Jones-Wenzl idempotent and is the unique element having  $e_i \JW{n} = \JW{n} e_i = 0$ for $i= 1 \ldots n-1$ and $\JW{n} \JW{n}=\JW{n}$.
 \item $\tr{\JW{n}}=\qi{n}$ where $\qi{n}:=-\frac{q^{2n}-q^{-2n}}{q^2-q^{-2}}$, with $q$ a solution of $-q^2-q^{-2}=d$.
\end{itemize}

It's hard to overstate the importance of understanding Temperley-Lieb; most of the structure of a general planar algebra is already visible in $\TL$, and understanding the interaction between the $\TL$ subalgebra and the rest of the algebra can be extremely useful.

On to our next example:  the $\dfour$ planar algebra.  This is a planar algebra which is planarly generated by $\TL(\sqrt{3})$ and a single $2$-box $R$.  By "generated as a planar algebra," we mean that any element of the planar algebra is the image of some tangle applied to some sequence of the generators.  The generator $R$ satisfies
\begin{itemize}
\item
$\rho(R)=-R$\eep{pic}
\item
$R^*=R$.
\item
$\epsilon_1(R)=\epsilon_2(R)=\epsilon_3(R)=\epsilon_4(R)=0$\eep{pic}
\item
and $\tr{R^2}=\qi{3}$\eep{pic}
\end{itemize}
 (Note that the last three of these are mostly harmless, as they simply specify that the generator of the non-$\TL$ part of our planar algebra should be chosen to be self-adjoint, orthogonal to TL and normalized a certain way.)

See [Jones, Annular Tangles] for more information about why the $\dfour$ planar algebra is generated in this way.

Here are some more relations on $\dfour$ which we will need:
\begin{lem}
\begin{enumerate}
\item
$\JW{5}=0$
\item $R^2=\JW{2}$ and Comult($R$,$R$)=$\frac{1}{\sqrt{3}} Id - e_1$.
\item The "annular consequences" of $R$, that is,
$$\mathfig{.5}{AnnularConsequences}$$
 satisfy the relations $\alpha_i - \sqrt{3} \alpha_{i+1} + 2 \alpha_{i+2} - \sqrt{3} \alpha_{i+3} + \alpha_{i+4} =0$, with the convention that for subscripts outside of $\{ 1,2,3,4,5,6 \}$, $\alpha_{i+6n}:=(-1)^n \alpha_i$.
\item The box consisting of two $R$'s placed next to each other is in $\TL$; specifically
$$\mathfig{.2}{TwoRs} = \mathfig{.5}{TwoRsInTL}$$
\end{enumerate}
\end{lem}

\begin{proof}
\begin{enumerate}
\item
We have $\tr{\JW{5}}=\qi{6}=0$, hence by positivity $\JW{5}=0$.
\item Rotational eigenvalue considerations tell that this is in $\TL$; then knowing that either top and bottom caps, or both side caps, is zero tells us which $\TL$ element.  
\item
Let $\sigma_i=\alpha_i - \sqrt{3} \alpha_{i+1} + 2 \alpha_{i+2} - \sqrt{3} \alpha_{i+3} + \alpha_{i+4}$; we can check that $\left< \sigma_i, \sigma_i \right> = 0$.
\item We could check that
\begin{align*}
\left<\mathfig{.1}{TwoRs},\mathfig{.1}{TwoRs} \right> & =\left<\mathfig{.1}{TwoRs}, \mathfig{.3}{TwoRsInTL} \right> \\
& = \left< \mathfig{.3}{TwoRsInTL}, \mathfig{.3}{TwoRsInTL} \right>
\end{align*}
but this would require us to compute about 200 traces.  Instead, we use the fact $\JW{5}=0$ to write that the five-strand identity is equal to a weighted sum of the other elements of $\TL_5$ (with coefficents as givenin [Morrison, A formula for the Jones-Wenzl Projections].)  This seems not to have made our lives that much simpler; we replace one picture with 41 pictures.  

However, if we consider the picture where two $R$s are conected by one string, and replace 5 of the remaining strings with minus the rest of the Jones-Wenzl projection, then most of these 41 pictures cap off an R in some way, giving zero; only 9 don't.  So we get
$$\mathfig{.4}{JW5} \Rightarrow \mathfig{.4}{RcapR}$$
and we have a Comult($R$,$R$) in each of these.  Using the previous relation we have that this is actually a $\TL$ element:
$$\mathfig{.5}{RcapRTL}$$
Now, we do the same thing (replacing 5 strings with 9 pictures involving caps) to two $R$s next to each other, and get the desired result.

\end{enumerate}
\end{proof}

\todo{The D4 planar algebra really has principal graph D4; see later for
what this means, and even later for the proof.}

\subsection{Planar Algebras, knots and links}

Take any knot or link, lay it down flat, and replace each crossing with the element of $\TL(\delta)_2$ which is $q \cdot Id + q^{-1} \cdot e_1$ for $q^2 + q^{-2}=-\delta$\eep{minus delta right?} -- what you get is a polynomail in $q$.  This is the Jones polynomial.

Now do the same thing, but first cable the knot/link -- ie replace each strand by $n$ strands -- and on each component of the knot/link (ie, each piece of string between two crossings) place a Jones-Wenzl idempotent $\JW{n}$.  Resolving the crossings gives the colored Jones polynomial.

This process seems generalizable, doesn't it?
