\chapter{The Kontsevich integral and its properties}
\lbl{chap:kontsevich}

In this chapter, we will briefly review the {\em Kontsevich integral},
a remarkable invariant of tangles that is a universal
Vassiliev invariant.

\begin{definition}\lbl{def:univ-vassiliev} A {\em universal Vassiliev
invariant} $Z$ is a map $Z: \cK(X) \rightarrow \cA(X)$ so that, for any
$n$-circus $C$ in the complement of an $X$-tangle,
\[
Z(\delta^{(n)}(C)) = \Diag'(C) + \text{higher order terms}
\]
Recall that $\Diag':\cK(X) \rightarrow \cA_n(X)$ is the approximation
to $\Diag:\cK(X) \rightarrow \cK_n(X)/\cK_{n+1}(X)$.
\end{definition}

A universal invariant provides an inverse to the map $m: \cA_n(X)
\rightarrow \cK_n(X)/\cK_{n+1}(X)$ constructed in
Section~\ref{sec:weight-systems}, and so proves that the two spaces
are isomorphic.  Furthermore, such an invariant is universal in the
category-theoretical sense: any finite-type invariant with values in a
vector space factors through $Z$.

We will not provide complete details here; for complete proofs, 
see~\cite{BarNatan:OnVassiliev, LeMurakami:Parallel,Lescop:Kontsevich}.  Our main interest is in
the good properties the Kontsevich integral satisfies under a few
natural topological operations; we will consider
connected sum and connected and disconnected cabling.
Except for the computation in Section~\ref{sec:conn-cable}, which is
due to Thang Le, all material in this chapter is standard.


\section{The Kontsevich integral for braids} We will first define the
Kontsevich integral for braids.  A {\em braid} on $n$ strands for our
purposes is a smooth map from the interval $[0,1]$ to the
configuration space of $n$ points in $\RR^2$.  There is a canonical
framing for braids, since the vector field pointing in (say) the $y$
direction is always normal to the braid.  We can multiply two braids
if the endpoint of one is the starting point of the other, just by
placing one on top of the other and smoothing a little.

The Kontsevich integral for braids is the holonomy of a certain formal
connection on the configuration space of $n$ distinct points in the
plane called the formal Knizhnik-Zamolodchikov (KZ) connection.  It takes
values in the space of chord diagrams appropriate to $n$-strand
braids,
\[
\cA(\underbrace{\vert \cdots \vert}_{\text{$n$ copies}})
   \overset{\text{def}}{=} \cA(n).
\]
Note that $\cA(n)$ has an algebra structure given by vertical composition,
just like braids.  The values actually lie in the subspace of this
space generated by horizontal chords.

Let $c_{ij}$ be a chord connecting the $i$'th and $j$'th strands,
\[
c_{ij} = \figcent{chordij}{0.2}
\]
Let $z_j = x_j + i y_j$ be complex coordinates for the configuration
space of $n$ points.  Then the connection 1-form of the KZ connection is
\[
\mu_n = \frac{1}{2\pi i}\sum_{i < j} c_{ij} d\,\log(z_i - z_j)
      = \frac{1}{2\pi i}\sum_{i < j} c_{ij} \frac{dz_i - dz_j}{z_i-z_j}.
\]

\begin{exercise}
Show that the Knizhnik-Zamolodchikov connection is flat:
\[
d\mu_n + [\mu_n, \mu_n] = 0.
\]
\end{exercise}

The Kontsevich integral of a braid is defined to be the holonomy of this
connection, defined using Chen's iterated integral
construction~\cite{Chen:Iterated}
\[
Z(B) = \sum_{m=0}^{\infty} \int_{0 < t_1 < \cdots < t_m < 1}
         B^*(\mu_n)(t_1) \cdots B^*(\mu_n)(t_m).
\]
This is invariant under isotopy of the braid since the
Knizhnik-Zamolodchikov connection is flat.

More explicitly, we can write this as
\ssp
\begin{equation} Z(K)=\sum_{m=0}^\infty \frac{1}{(2\pi i)^m}
	\mathop{\int}_{t_1<\ldots<t_m}
	\sum_{\substack{\text{applicable pairings} \\
		P=\{(z_i,z'_i)\}}}	D_P
	\bigwedge_{i=1}^{m}\frac{dz_i-dz'_i}{z_i-z'_i}
	\in\cA(n). \label{eq:kontsevich-integral}
\end{equation}
\dsp

In the above equation,
\begin{itemize}
\item an `applicable pairing' is a choice of an unordered
	pair $(z_i,z'_i)$ for every $1\leq i\leq m$, for which
	$(z_i,t_i)$ and $(z'_i,t_i)$ are distinct points on $K$.
\item $D_P$ is the chord diagram naturally associated with $K$ and
      $P$, an appropriate product of the $c_{ij}$'s.
\item Every pairing defines a map $\{t_i\}\mapsto\{(z_i,z'_i)\}$ locally
	around the current values of the $t_i$'s. Use this map to pull
	the $dz_i$'s and $dz'_i$'s to the space $t_1<\dots<t_m$ and
	then integrate the indicated wedge product over that simplex.
\end{itemize}

\section{First properties: Universal and
Grouplike}\label{sec:first-prop} Let us first check that this
invariant is, indeed, a universal finite-type invariant.  The
statement of universality has to be modified slightly for braids,
since an action of an arbitrary circus will take us out of braids and
into general $n$-strand tangles.  We will restrict to {\em horizontal
circuses}: circuses in which each double lasso lies in a horizontal
(constant $t$) plane.  Action on such a circus yields another braid.
We have to show that Kontsevich integral of the resolution of a
horizontal $m$-circus is the corresponding graph plus higher order
terms.

Isotop the braid and the lassos so that all the lassos are very short,
connecting strands close to each other.  Also apply the splitting
relation so that each loop of a lasso encloses a single strand.  When
we resolve a single lasso, we get a difference of two terms differing
only near the crossing.  The KZ connection will be constant on chords
that do not touch the crossing and nearly constant on chords only one
end of which reaches the crossing.  For the alternating difference to
be non-trivial, there must be at least one chord connecting the two
strands involved in the lasso.  Similarly, if we resolve an entire
circus, the integral will be constant or nearly constant if there is
not at least one chord per lasso connecting the two strands of the
lasso.  Thus the lowest degree contribution to $Z(B_D)$ is at least
degree~$m$, and the degree~$m$ term is proportional to the desired
chord diagram.  To check the constant of proportionality, note that we
are integrating $d(log z)/2\pi i$ on a counterclockwise circle around
0; by Cauchy's theorem, the answer is~1.

Next we will show that the Kontsevich integral is {\em grouplike}.
For motivation, consider that the product of two finite type
invariants $v_1$, $v_2$ is again finite type.  The Kontsevich
integral, as the universal finite type invariant, encodes the
information about $v_1$, $v_2$, and $v_1\cdot v_2$; the grouplike
property encodes how the three invariants are related.

\begin{definition}\label{def:box}
For any 1-manifold $X$, the coproduct $\square: \cA(X) \rightarrow \cA(X)
\otimes \cA(X)$ takes a Jacobi diagram $D$ to the sum, over all
partitions $D = D_1 \cup D_2$ of $D$ into two parts in which both
$D_1$ and $D_2$ contain $X$ but are otherwise disjoint, of $D_1
\otimes D_2$.  In other words, assign each connected component of $D
\setminus X$ to the first or second tensor factor and sum over all
possibilities.
\end{definition}

\begin{exercise} Let $v_1,v_2: \cK \rightarrow \CC$ be
Vassiliev invariants of degree $n_1$ with weight systems $w_i:\cA
\rightarrow\CC$.  ($w_i$ is supported on Jacobi diagrams of degree $n_i$.)
Show that $v_1\cdot v_2$ is a Vassiliev invariant of
degree~$n_1 + n_2$ and weight system 
\[
(w_1\otimes w_2) \circ \square: \cA \rightarrow \CC
\]
\end{exercise}

\begin{hint}
The evaulation of $v_1 v_2$ on the resolution of a $k$-circus can be
written as a sum of terms
\[
(v_1\cdot v_2)(\delta^{(k)}(C)) = \sum v_1(\delta^{(l)}(C_1))v_2(\delta^{(k-l)}(C_2))
\]
where $C_1$ and $C_2$
are appropriate circuses of degrees $l$ and $k-l$.
(Think about proving that the product of polynomials
is polynomial.)  For $k = n_1+n_2+1$, at least one of the two factors
vanishes.  For $k=n_1+n_2$, the sum divides the lassos of $C$ into two
subsets of size $n_1$ and $n_2$.  For circuses related to graphs, each
connected component must go into the same subset or the result is 0.
\end{hint}

\begin{proposition}
For any braid $B$, $Z(B)$ is {\em grouplike}:
\[
\square Z(B) = Z(B) \otimes Z(B).
\]
\end{proposition}

\begin{proof}[Proof (Sketch)] This is a general property of the
holonomy of a connection with values in a Hopf algebra in which the
connection form is primitive.  (The usual example is the holonomy of a
Lie-algebra valued connection, which takes values in the the Lie group
or, alternatively, the grouplike elements inside the universal
enveloping algebra.)
\end{proof}

By the general structure theory of Hopf algebras, any time there is a
multiplication on $\cA(X)$ compatible with the comultiplication
$\square$, a grouplike element is an exponential of primitive elements.
Here a {\em primitive} element is an element $x$ such
that $\square(x) = x \otimes 1 + 1 \otimes x$.
In particular, the multiplications in $\cA$, $\cB$, and $\cA(n)$ are
compatible with $\square$.

\section{Cups and caps}\label{sec:cups-caps} In this thesis, we are
not interested in braids, but rather in knots and, more generally,
tangles.  The difference is the presence of critical points, either
minima $\cup$ or maxima $\cap$.  The integral in
Equation~\ref{eq:kontsevich-integral} naturally extends to this
context.  But there is a problem: the integral no longer converges.
For instance, the integral corresponding to a single chord at a maximum
$\figcent{problem-integral}{0.025}$, with the strands are a distance
$d$ apart at the bottom, is
\[
\int_{z=d}^0 \frac{1}{2\pi i}\frac{dz}{z} \cdot
\text{\LARGE$\isolatedchord$}
\]
which diverges logarithmically.  


We can fix the divergence by
terminating the integral when the points are a displacement $\epsilon$
apart and multiplying by the counterterm \(\exp(-\frac{\log\epsilon}{2\pi
i})\isolatedchord\).  It turns out that there is then a finite limit
as $\epsilon\rightarrow 0$. To make the counterterm well-defined, we
need to pick
a branch of $\log\epsilon$ near each critical point of the height
function.  For concreteness, turn the knot near each critical point so
that the tangent at the critical point is along the $x$ axis (so
$\epsilon$ is real) and take the real value of
$\log\,\lvert\epsilon\rvert$ in the counter term.  A knot with the
critical points pinned down like this has a canonical framing, since a
vector field pointing in the $y$ direction is transverse to the knot.
(If we project the knot onto the $x$-$t$ plane, this is the blackboard
framing.)  If we twist a maximum counterclockwise or a minimum
clockwise by $\pi$, the counter term changes by
$\exp(\frac{1}{2}\isolatedchord)$ and the above framing changes by 1
unit, so pinning down the critical points like this is equivalent to a
choice of framing.  See Le and Murakami~\cite{LeMurakami:Universal}
or Lescop~\cite{Lescop:Kontsevich} for details.

For any $X$-tangle $T$, let $\tilde Z(T)$ be the renormalized $\cA(X)$
valued Kontsevich integral as above.  Although $\tilde Z$ converges,
it is not an invariant.  In particular, \(\tilde Z(\wiggle) \ne
\tilde Z(\vert) = \vert.\) Let $\nu^{-1} = \tilde Z(\wiggle)$.

\begin{definition} The Kontsevich integral $Z(T)$ is
$\tilde Z(T)$ multiplied by $\nu$ at each local maximum.
\end{definition}

Now whenever we want to straighten a wiggle $\wiggle$, there is an
extra $\nu$ to cancel the $\nu^{-1}$ from the non-invariance of
$\tilde Z(\wiggle)$.  With these definitions, $Z(\bigcirc) = \nu.$
This element, $\nu$, is what we will compute in
Chapter~\ref{chap:wheels}.

It is straightforward to see that this extension of $Z$ to tangles is
still grouplike and a universal finite-type invariant.

From the discussion above about the framing and the critical points,
we can compute what happens if we change the framing on a component.

\begin{lemma}\label{lem:framing-change}
Given a knot $K$, let $K'$ be the same knot with the framing twisted
by $f$ positive twists.  Then
\[
Z(K') = Z(K) \connect \exp\left(\frac{f}{2}\mlarge\isolatedchord\right).
\]
Similar results hold for changing the framing on a component in a
tangle, where we multiply by $\exp(\frac{f}{2}\isolatedchord)$ on the
affected component.
\qed
\end{lemma}

With only a little more work we can be more precise.  The integrals
contributing to a single chord $\isolatedchord$ come from the term $m=1$ in
Equation~\ref{eq:kontsevich-integral}.  Since all such integrals are of
the form $\frac{1}{2\pi i}\int \frac{dz}{z} = \frac{\log z}{2\pi i}$,
this term can be computed.  Mostly it computes the winding numbers of
strands about themselves.
Careful accounting of what happens at the maxima and minima shows that
the coefficient of $\isolatedchord$ is $1/2$ the number of positive
crossings minus the number of negative crossings.  If we combine this
with the grouplike property of the Kontsevich integral, we get the
following lemma.

\begin{lemma}\label{lem:struts}
The Kontsevich integral of a knot $K$ with framing $f$ relative to the
canonical framing (the framing given by, e.g., a Seifert surface for
$K$) is
\[
Z(K) = \exp_{\connect} \left(\frac{f}{2}\strutn{}{}\right) \connect D
		\in \cB
\]
where $D$ is strutless: has no connected components which are struts
$\strutn{}{}$.\qed
\end{lemma}

\section{Connected sum} \label{sec:connect-sum} The first thing to
notice about our definition of $Z$ is that it is local in the
horizontal plane: each horizontal slice is independent.  Therefore,
for any tangles $T_1$, $T_2$ so that the upper boundary of $T_1$ is
the same as the lower boundary of $T_2$,
\begin{equation}
Z(T_1 \cdot T_2) = Z(T_1) \cdot Z(T_2)
\end{equation}
where the multiplications of tangles on the LHS and web diagrams on
the RHS are both vertical stacking: place the two objects on top of
each other and join the corresponding legs.

\begin{figure}[htbp]
\begin{center}
   {\makebox[.3\linewidth]{\figscaled{figure8-open}{0.4}}}
\hfil
   {\makebox[.3\linewidth]{\figscaled{figure8-closed}{0.3}}}
\end{center}
\caption{A $(1,1)$ tangle and its closure.}\label{fig:11tangle}
\end{figure}

We will be particularly interested in the case where the tangles are
$(1,1)$ tangles, with one incoming and one outgoing strand.  These are
closely related to knots: we can turn a $(1,1)$ tangle canonically
into a knot by joining the top to the bottom.  See
Figure~\ref{fig:11tangle} for an example.  What does this do the the
Kontsevich integral $Z$?  By shrinking the tangle, the dotted box in
Figure~\ref{fig:11tangle}, the interactions between the dotted box and
the original strand in the integral will be go to 0 while leaving the
integral within the box unchanged.  The
differences between the invariants of the two are caused by (a)~the
correction introduced at the maximum and (b)~changing the space of
values by closing the strand, which by Lemma~\ref{lem:line-circle} is
isomorphism.  Letting $T$ by the tangle and $K$ be its closure, which
could be a knot or a link with a distinguished component, we find
\begin{equation}\label{eq:11tangle}
Z(K) = \nu \connect Z(T)
\end{equation}
where the two spaces are identified by their isomorphism.

Let $T_1$, $T_2$ be two $(1,1)$ tangles and $K_1$, $K_2$ their
closures.  Then
\begin{align*}
Z(K_1 \connect K_2) &= Z(T_1 \cdot T_2) \connect \nu\\
		    &= Z(T_1) \connect Z(T_2) \connect \nu \\
		    &= Z(K_1) \connect Z(K_2) \connect \nu^{-1}.
\end{align*}

\section{Cables} \label{sec:cables}
We now consider the operation of {\em cabling}: replacing a knot
with $n$ parallel copies of the same knot, as in
Figure~\ref{fig:cable-ex}.  There are two versions: the
disconnected cabling, as in Figure~\ref{fig:disconn-cable-ex}, ending
up with an $n$-component link, and the connected cabling, as in
Figure~\ref{fig:conn-cable-ex}, in which you add a twist so the
result is again a knot.

\begin{figure}[tbp]
\begin{center}
\subfigure[Disconnected cable.]
  {\makebox[.3\linewidth]{\fig{trefoil-disconn4cable}{0.3}
			\label{fig:disconn-cable-ex}}}
\subfigure[Connected cable.]
  {\makebox[.3\linewidth]{\fig{trefoil-conn4cable}{0.3}
			\label{fig:conn-cable-ex}}}
\subfigure[General satellite: the Whitehead double]
  {\makebox[.3\linewidth]{\fig{trefoil-whitehead}{0.3}
			\label{fig:satellite-ex}}}
\end{center}
\caption{Example cablings and satellites of the trefoil knot.}
\label{fig:cable-ex}
\end{figure}


Note that this operation is only well-defined for framed knots, and
for connected cabling, you need to specify the extra twist to be
added.  It is natural to include this extra twist in the framing as a
kind of ``rational framing.''

\subsection{General Satellites}


Let us start with some generalities about finite type invariants of
satellites.  A general satellite operation is specified by a link $L$
embedded in a solid torus; the operation is to take a knot $K$, remove
a tubular neighborhood of it, and glue in the solid torus with the
embedded link, obtaining a new link $K_L$.  An example of the result is in
Figure~\ref{fig:satellite-ex}.

What does this operation do to finite-type invariants?  Consider an
$n$-circus in the complement of $K$.  After the satellite
operation, it becomes an $n$-circus in the complement of $K_L$.  By
the Splitting Relation in Section~\ref{sec:weight-systems}, the loops
of the double lassos circling $K$ turn into a sum over the components
of $L$ at that point.

Let us formalize this.

\begin{definition} Let $X, Y$ be compact 1-manifolds and $\phi: X
\rightarrow Y$ a proper map.  Let $D$ be a diagram on $Y$
with a fixed parameterization of $Y$, so that the legs of $D$ have a
well-defined position in $Y$, which are generic with respect to
critical points of $\phi$.  A {\em lift} of $D$ by $\phi$ is a diagram
$D'$ on $X$ with the same internal part on $X$ so that the image of
each leg of $D'$ under $\phi$ is the corresponding leg of $D$.  Each
lift comes with an orientation induced from the local map.
\end{definition}

\begin{definition}\label{def:pullback} Let $X, Y$ be compact
1-manifolds and $\phi: X\rightarrow Y$ a proper map.  The {\em
pullback} $\phi^*: \cA(Y)\rightarrow \cA(X)$ on a diagram $D \in
\cA(Y)$ is the sum over all lifts of $D$ from $Y$ to $X$.
\end{definition}


$\phi^*$ is well-defined (i.e., does not depend on the position of the
legs of $D$ and descends modulo the relation) and is invariant under
homotopy of $\phi$.


\begin{proposition}[Kuperberg~\cite{Kuperberg:Invertibility}] Let $X,
Y$ be 1-manifolds, let $\eta: Y\rightarrow N(X)$ be a pattern for
satelliting, and let $\iota: X\rightarrow\RR^3$ be a link/tangle with
an $n$-circus $C$ in its complement.  Then for any universal invariant
$Z$, $Z(\delta_C(X_\eta)) = \phi^*(\Diag(C)) + \text{h.o.t}$, where
$\phi$ is the composition of $\eta$ and the retraction
$N(X)\rightarrow X$.
\end{proposition}

\begin{proof}
By the definition of a universal finite type invariant $Z$,
\[
Z(\delta_C(X_\eta)) = \Diag(C_\eta) + \text{h.o.t.},
\]
where $C_\eta$ is the $n$-circus $C$ in the complement of $X_\eta$.  By
the Splitting relation, this is $\phi^*(\Diag(C))$.
\end{proof}

\begin{corollary}\label{cor:satellite}
The map $\eta^*: \cA(X) \rightarrow \cA(Y)$ is $\phi^*$ plus higher
order terms (terms that increase the degree of a diagram).
\end{corollary}


The nice property of the two cabling operations is that the above
formula becomes exact: the ``higher order terms'' vanish.

\begin{definition} The operations
\[
\Delta^x_{x_1\dots x_n}: \cA(\vert_x\cup X)\rightarrow
	\cA(\vert_{x_1}\cup\cdots\cup\vert_{x_n}\cup X)
\]
or
\[
\Delta^x_{x_1\dots x_n}: \cA(\bigcircle_x\cup X)\rightarrow
	\cA(\bigcircle_{x_1}\cup\cdots\cup\bigcircle_{x_n}\cup\,X)
\]
are the pullback of the $n$-fold disconnected cover of the component
labelled $x$.  When we do not care about the labels on the result, an
alternate notation is $\Delta^{(n)}_x$.
\end{definition}

\begin{proposition}[Le and Murakami~\cite{LeMurakami:Parallel}]
\label{prop:disconn-cable}
Let $L$ be a framed tangle and $\DCable_x^n(L)$ be the $n$-fold
disconnected cable of $L$ along a knot component $x$ or interval
component with one upper and one lower boundary.  Then
\[
Z(\DCable_x^n(L)) = \Delta^{(n)}_x(Z(L)).
\]
\end{proposition}

Proof in Section~\ref{sec:disconn-cable}.

\begin{definition} The operation
\[
\psi^{(n)}_x: \cA(\bigcirc_x\cup X)\rightarrow\cA(\bigcirc_x\cup X)
\]
is the pullback of the $n$-fold connected
cover of the circle labelled $x$.
\end{definition}

\begin{proposition}[T.~Le]
\label{prop:conn-cable}
Let $L$ be a framed tangle and $\CCable_x^n(L)$ be the $n$-fold
connected cable of $L$ along a knot component $x$ as in
Figure~\ref{fig:conn-cable-ex}.  Then
\[
Z(\CCable_x^n(L)) =
	\psi^{(n)}_x(Z(L) \connect_x \exp(\frac{1}{2n}\isolatedchord))
\]
for an appropriate choice of framing on $\CCable_x^n(L)$.
\end{proposition}

Proof in Section~\ref{sec:conn-cable}.

One reason to introduce symmetrized diagrams is that the
operations $\Delta$ and $\psi$ above become very simple in $\cB$.

\begin{lemma}\label{lem:delta} The map 
\[
\Delta^x_{x_1\dots x_n}:\cA(\ast_x \cup X) \rightarrow
	\cA(\ast_{x_1} \cup\dots\cup\ast_{x_n} \cup X)
\]
is the sum over all ways of replacing each $x$ leg by one of the
$x_i$.\qed
\end{lemma}

\begin{remark}
$\Delta$ is similar to a coassociative, cocommutative coproduct in a
coalgebra, except that it does not take values in $\cA \otimes \cA$.
Do not confuse it with the coproduct $\square$ of
Definition~\ref{def:box}, which is an honest coproduct.
\end{remark}

The operation $\Delta$ in Lemma~\ref{lem:delta} is analogous to a
change of variables $x \mapsto x_1 + \dots + x_n$ for ordinary
functions $f(x)$.  We will use a suggestive notation: a leg labelled
by a linear combination of variables means the sum over all ways of
picking a variable from the linear combination.  If $D(x)$ is a
diagram with some legs labelled $x$, $\Delta^{(n)}(D(x)) = D(x_1 +
\dots + x_n)$ is the diagram with the same legs labelled $x_1 + \dots
+ x_n$.

\begin{lemma}\label{lem:psi} The map
\[
\psi^{(n)}_x: \cA(\circledast_x \cup X) \rightarrow
	\cA(\circledast_x \cup X)
\]
is multiplication by $n^k$ on diagrams with $k$ legs labelled $x$.
\qed
\end{lemma}


This operation is related to the change of variables $x \mapsto nx$.

\subsection{Disconnected cabling}\label{sec:disconn-cable}
Here we sketch the proof of Proposition~\ref{prop:disconn-cable}.  For
full details, please see Le and Murakami~\cite{LeMurakami:Parallel}.

First consider the disconnected cabling of a
braid.  Take one strand of a braid and replace it by a
number of copies parallel with a small displacement $\epsilon$.
Because the strands are parallel, there are no new chords between them
($z_1 - z_2$ is constant); since the new strands are close to the
original strand, thep
integrals for chords between the doubled strands and another strand
are close to chords for the original strand, with equality in the
limit as $\epsilon \rightarrow 0$.  See
Lescop~\cite{Lescop:Kontsevich} for the analytic details.  Hence for
parallel cabling of a braid Proposition~\ref{prop:disconn-cable}
holds.

When we switch to tangles, there may be cups and caps.  The Kontsevich
integral of a doubled maximum $Z(\doublecap)$ will not be the double
of the 
Kontsevich integral of a single maximum $Z(\cap)$.  

Let $\alpha = \lim_{\epsilon\rightarrow 0} Z(\doublecap_\epsilon)$
where $\doublecap_\epsilon$ is a doubled maximum with the distances
between the strands at the bottom $\epsilon$, $1$, and $\epsilon$ in
order.  (There are naively two logarithmic divergences in this
integral which fortuitously cancel.) Note that $\alpha$
includes renormalization of $\nu$ on each strand added so that
$Z(\wiggle) = Z(\vert)$.  Similarly we get another element
$\lim_{\epsilon\rightarrow 0}Z(\doublecup_\epsilon) =
\beta\in\cA(\uparrow\,\uparrow)$ at minima.  Although we can say
little about $\alpha$ or $\beta$ by themselves,
\footnote{Le and Murakami~\cite{LeMurakami:Parallel} have analyzed
contexts in which you can compute $\alpha$ and $\beta$ and, in
particular, when $\alpha$ and $\beta$ are the doubles of a maximum and
minimum.}
we can compute their
product.  
Consider a doubled wiggle, as in Figure~\ref{fig:wiggle2}.
It can be decomposed into three pieces: a doubled maximum, a doubled
minimum, and the doubling of the braid whose closure is $\wiggle$.
The doubled ends of the Kontsevich integral of the doubling of the
braid can slide over any diagram which, like $\alpha$ or $\beta$,
lives on two strands, so we can collect $\alpha$ and $\beta$ at the
beginning.  We then recognize the remaining integral as the double of
the naive integral which gave us $\nu^{-1}$ above.  Since this whole
$(2,2)$ tangle can be straightened to give the identity, we find
\begin{align*}
\alpha \cdot \beta \cdot \Delta\nu^{-1} &= 1 \\
\intertext{or}
\alpha\cdot\beta &= \Delta\nu
\end{align*}
where the multiplication is the natural multiplication on
$\cA(\uparrow\uparrow)$.
\begin{figure}\label{fig:wiggle2}
\[
\figcent{wiggle2}{0.2} = \figcent{id2}{0.2}
\]
\caption{A doubled wiggle.}
\end{figure}

Now consider doubling a knot component or a component with one upper
and one lower end.  There will be the same number
$k$ of maxima and minima, so the net error from a pure doubling will
be some product of $k$ $\alpha$'s and $k$ $\beta$'s.  As before, we
can slide the $\beta$'s through the rest of the integral
until they are next to the $\alpha$'s; then each pair cancel to give
$\Delta\nu$, which is exactly the double of the original
renormalization, as desired.
(There is one difficulty we are brushing under the rug: there are
actually two versions of $\alpha$ and $\beta$, depending on which
direction you traverse the maximum, and several different products can
occur.  To show that all pairs cancel, some more complicated version
of Figure~\ref{fig:wiggle2} need to be considered.)

So Proposition~\ref{prop:disconn-cable} is true for doubling a knot
component. 
By iterating this argument, the same thing is true for an $n$-fold
disconnected satellite.\qed


\subsection{Connected cabling}\label{sec:conn-cable} Now we consider
the case of a connected cabling.  The difference between the connected
cabling and the
disconnected cabling above is the extra $1/n$ twist $T_n$ inserted at one
point:
\[
T_n = \figcent{1n-twist}{0.25}.
\]
By isotopy can assume that this twist occurs in a horizontal slice where
all the other strands are vertical.  We can
apply Proposition~\ref{prop:disconn-cable} on the $(n,n)$ ``tangle''
obtained by excising $T_n$.  (This object is not properly a tangle,
since there is a little piece cut out of it.  But we can still compute
its Kontsevich integral by Proposition~\ref{prop:disconn-cable}.)
To complete the computation, we need to
compute $a := Z(T_n)$.


Repeating $T_n$ $n$ times, we get a full twist which we can
compute using Lemma~\ref{lem:framing-change} and the disconnected cabling
of the previous section:
\[
Z\left(\figcent{n-twists}{0.25}\right)
	= Z\left(\figcent{fulltwists}{0.25}\right)
	= \Delta^{(n)}\left(\exp(\frac{1}{2}\isolatedchord)\right)
		\exp(-\frac{1}{2}\isolatedchord)^{\otimes n} =: b.
\]

The notation $\exp(-\frac{1}{2}\isolatedchord)^{\otimes n}$ means $n$
copies of the framing change element
$\exp(-\frac{1}{2}\isolatedchord)$, one on on 
each of the $n$ strands.

The $n$ copies of $T_n$ that appear are not quite the same: they
differ by cyclic permutations of the strands.
If we arrange the $n$ strands at the top and bottom of $T_n$
to be at the vertices of a regular $n$-gon, the strands are
symmetric and $a^n = b$ or
\[
a = b^{\frac1n} = \Delta^{(n)}\left(\exp(\frac{1}{2n}\isolatedchord)\right)
		\exp(-\frac{1}{2n}\isolatedchord)^{\otimes n}.
\]
More generally, we will need to conjugate $T_n$ by some element $C$ to get
the strands symmetric; this implies that
\[
a = c b^{\frac{1}{n}} S(c^{-1})
\]
where $c = Z(C)$ and $S$ is the automorphism of $\cA(\uparrow_{x_1}
\dots\uparrow_{x_n})$ which rotates the strands by $x_i \mapsto x_{i-1}$.

\begin{lemma}\label{lem:n-twist}
$Z(T_n) = c \Delta^{(n)}\left(\exp(\frac{1}{2n}\isolatedchord)\right)
		\exp(-\frac{1}{2n}\isolatedchord)^{\otimes n}
S(c^{-1})$ for some $c \in \cA(\uparrow_{x_1} \dots \uparrow_{x_n})$.
\qed
\end{lemma}

This lemma can also be proved without using the specific geometry of
the Kontsevich integral in an argument due to D.~Bar-Natan, but we
will not give the argument here.


\begin{proof}[Proof of Proposition~\ref{prop:conn-cable}] By the above
computations, the invariant of the connected cable of a knot $K$ is
$\Delta^{(n)}(Z(K))$, multiplied by $Z(T_n)$, and closed up with a
twist.  The conjugating elements $c$ and $S(c^{-1})$ can be swept
through the knot and cancel each other.  The factor
$\Delta^{(n)}(\exp(\isolatedchord/2n))$ in $a$ can be combined with $Z(K)$
so that we apply $\Delta^{(n)}$ to $Z(K) \connect
\exp(\isolatedchord/2n)$.  The twisted closure turns $\Delta^{(n)}$ into
$\psi^{(n)}$.  The remaining $n$ factors of $\exp(-\isolatedchord/2n)$ in
$a$ can be slid around the knot and combined to give
\[
Z(\CCable^n(K)) = \psi^{(n)}\left(Z(K)\connect\exp(\frac{1}{2n}\isolatedchord)\right)
			\connect \exp(-\frac{1}{2}\isolatedchord).
\]
The last factor is absorbed in the ``appropriate choice of
framing'' in the statement of the proposition.
\end{proof}.
{"config": "arxiv", "file": "math0006083/kontsevich.tex"}