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\begin{document}

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\runninghead{D. Cimasoni}{Conway potential function}

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\fpage{1}
\centerline{\bf A geometric construction of the Conway potential function}
\baselineskip=13pt
\vspace*{0.37truein}
\vspace{0.5cm}
\noindent David Cimasoni
\baselineskip=12pt

\vspace*{0.5truein}

\abstracts{We give a geometric construction of the multivariable Conway potential function for
colored links. In the case of a single color, it is Kauffman's definition of the Conway polynomial in terms of a Seifert matrix.}
\vspace*{10pt}
\AMS{57M25.}
\vspace*{10pt}
\keywords{Conway potential function, Conway polynomial, colored link, C-complex.}

\vspace{1cm}

\vspace*{1pt}\textlineskip
\section{Introduction}
\vspace*{-0.5pt}

A {\it colored link\/} is an oriented link $L=L_1\cup\dots\cup L_\mu$ in $S^3$ together with  a surjective map
$\sigma$ assigning to each component $L_i$ a color $\sigma(i)\in\{1,\dots,n\}$. Two colored links $(L,\sigma)$ and $(L',\sigma')$
are {\it isotopic\/} if there exists an ambient isotopy from $L$ to $L'$ preserving the color and orientation of each component.

Given a colored link $(L,\sigma)$ with exterior $X$, we have $\Ho_1X=\bigoplus_{i=1}^\mu\Z t_i$, where $t_1,\dots,t_\mu$ denote the oriented
meridians of $L$. The Hurewicz morphism $\pi_1X\longrightarrow\Ho_1X$ composed with
$\bigoplus_{i=1}^\mu\Z t_i\longrightarrow\bigoplus_{i=1}^n\Z t_i\,,\;\;t_i\mapsto t_{\sigma(i)}$, determines a regular
$\Z^n$-covering $\widehat{X}^\sigma\longrightarrow X$. The homology of $\widehat{X}^\sigma$ is endowed with a natural
structure of module over $\Lambda_n=\Z[t_1,t_1^{-1},\dots,t_n,t_n^{-1}]$. The $\Lambda_n$-module $\Ho_1\widehat{X}^\sigma$ is
called the {\it Alexander module\/} of $(L,\sigma)$: it is an invariant of the colored link.
The greatest commun divisor of the first elementary ideal of this module
is called the {\it multivariable Alexander polynomial\/} of $(L,\sigma)$.
It is denoted by $\Delta_L^\sigma(t_1,\dots,t_n)$.
Note that $\Delta_L^\sigma$ is only defined modulo the units of $\Lambda_n$, that is, up to multiplication by
$\pm t_1^{\nu_1}\cdots t_n^{\nu_n}$.

If $n=1$, $\Delta_L^\sigma(t)$ is nothing but the Alexander polynomial of the unordered oriented link $L$, as defined by {\sc James Alexander} \cite{Alex}.
On the other hand, if $n=\mu$ and $\sigma=\hbox{id}$, $\Delta_L^\sigma(t_1,\dots,t_\mu)$ is denoted by $\Delta_L(t_1,\dots,t_\mu)$: it is the Alexander
polynomial of the ordered oriented link $L$, defined by {\sc Ralph Fox} \cite{Fox}. Note that in general,
$$
\Delta^{\sigma}_L(t_1,\dots,t_n)\,\;\dot{=}\,\;
\cases{(t_1-1)\Delta_L(t_1,\dots,t_1) & if $n=1$ and $\mu>1$;\cr
	\Delta_L(t_{\sigma(1)},\dots,t_{\sigma(\mu)}) & else,}
$$
where $\dot{=}$ stands for the equality modulo the units of $\Lambda_n$.

In 1970, {\sc John Conway} \cite{Con} introduced a new invariant of links called the {\it potential function}.
Given a colored link $(L,\sigma)$, its potential function is a well-defined rational function
$\nabla^\sigma_L(t_1,\dots,t_n)$ which satisfies
$$
\nabla^\sigma_L(t_1,\dots,t_n)\,\;\dot{=}\,\cases{\frac{1}{t_1-t_1^{-1}}\,\Delta^\sigma_L(t_1^2) & if $n=1$; \cr
            	\Delta^\sigma_L(t_1^2,\dots,t_n^2) & if $n>1$.\cr}
$$
Thus, this invariant is just the Alexander polynomial without the ambiguity
concerning multiplication by units of $\Lambda_n$. This might seem a minor improvement. However,
the potential function has a very remarkable new property: it can be computed directly from a link diagram using skein relations.
For example, suppose that $(L_+,\sigma)$ $(L_-,\sigma)$ and $(L_0,\sigma_0)$ are given by colored diagrams related by a single change as below, where
$i$ stands for the color of the components.

\begin{figure}[h]
\begin{center}
\epsfig{figure=RII.eps,height=1.8cm}
\end{center}
\end{figure}
\noindent Then, their potential functions satisfy the equality
$$
\nabla^\sigma_{L_+}-\nabla^\sigma_{L_-}=(t_i-t_i^{-1})\cdot \nabla^{\sigma_0}_{L_0}.
$$
Similarly, if $L_{++}$, $L_{--}$ and $L_{00}$ differ by the following local operation,

\begin{figure}[h]
\begin{center}
\epsfig{figure=RIV.eps,height=2.7cm}
\end{center}
\end{figure}
\noindent then we have the equality
$$
\nabla^\sigma_{L_{++}}+\nabla^\sigma_{L_{--}}=(t_it_j+t_i^{-1}t_j^{-1})\cdot \nabla^\sigma_{L_{00}}.
$$
Thus, Conway pointed out a preferred representative of the Alexander polynomial, and gave a very easy method for computing it.
Unfortunately, his paper contains neither a precise definition of the potential function, nor a proof of its unicity. Quoting Conway:
``We have not found a satisfactory {\em explanation\/} of these identities.(...) It seems plain that much work remains to be done in the field.''

As a particular case of his potential function, Conway defined what he called the {\it reduced polynomial\/} $D_L(t)$ of a non-ordered link $L$ by
$$
D_L(t)=(t-t^{-1})\cdot\nabla^\sigma_L(t),
$$
where $\sigma$ is the coloring map with a single color.
This invariant was later called the {\it Conway polynomial} of $L$; we will use this terminology.
Following from the properties of the general potential function, we have:
\begin{romanlist}
\item{$D_L$ is an invariant of the non-ordered oriented link $L$;}
\item{$D_O(t)=1$, where {\it O} stands for the trivial knot;}
\item{$D_{L_+}(t)-D_{L_-}(t)=(t-t^{-1})\cdot D_{L_0}(t)$, where $L_+,L_-$ and $L_0$ are related by a single crossing change as below.}
\end{romanlist}

\begin{figure}[h]
\begin{center}
\epsfig{figure=RI.eps,height=1.8cm}
\end{center}
\end{figure}
But again, the paper \cite{Con} contains no proof of the constitency of this system of axioms: there is no explicit model for $D_L$. On the other hand,
it is very easy to check that if such a model exists, it is unique.

In 1981, {\sc Louis Kauffman} \cite{Kauf} found a very simple geometric construction of the Conway polynomial $D_L(t)$, namely
$$
D_L(t)=\det\left(-tA+t^{-1}A^T\right),
$$
where $A$ is any Seifert matrix of the link $L$ and $A^T$ is the transpose of $A$. This result gave the requested model for $D_L$; it also
provided the `explanation' of the first skein relation.

Finally, in 1983, {\sc Richard Hartley} \cite{Har} gave a definition of the
multivariable potential function $\nabla^\sigma_L$. Quoting his introduction:
``Kauffman showed how to define what may be called the reduced potential function of a link in terms of a Seifert matrix. This reduced potential
function is an $L$-polynomial in one variable. However, the potential function is essentially a function of several variables, and I can see no
way of generalising Kauffman's method to obtain the full potential function. Quite a different approach is therefore indicated.'' Indeed,  Hartley's
definition is obtained by normalizing the Alexander matrix given by the Wirtinger presentation via Fox free differential calculus: it is an algebraic
construction, whose relation to Kauffman's model is not obvious at all.

In this paper, we give a geometric construction of the multivariable potential function which generalizes Kauffman's definition.
It gives the relation between Hartley and Kauffman's models of the Conway polynomial, and provides an `explanation' of Conway's skein
formulas. It should be pointed out that the fundamental idea of this construction, that is, the use of `C-complexes' for the computation of
Alexander invariants of links, is due to {\sc Daryl Cooper} \cite{Cooper1,Cooper2}.

\section{The potential function of a colored link}

Let us denote by $\nabla_L$ the potential function of an ordered oriented link $L$ as defined by Hartley \cite{Har}; this corresponds to the case
of a colored link $(L,\sigma)$ with $n=\mu$ and $\sigma=\hbox{id}$. The potential function $\nabla_L^\sigma$ of an arbitrary colored link
is given by
$$
\nabla^{\sigma}_L(t_1,\dots,t_n)=\nabla_L(t_{\sigma(1)},\dots,t_{\sigma(\mu)}).
$$
The aim of this section is to present a geometric construction of $\nabla^{\sigma}_L$. This requires the use of C-complexes
as defined by Cooper \cite{Cooper1,Cooper2}.

\begin{figure}[Htb]
\begin{center}
\epsfig{figure=clasps.eps,height=2.8cm}
\fcaption{A positive clasp, and a negative clasp.}
\label{fig:clasps}
\end{center}
\end{figure}

\medskip
\noindent {\bf Definition.} {\it A C-complex for a colored link $(L,\sigma)$ is a union $F=F_1\cup\dots\cup F_n$
of compact oriented PL-embedded surfaces in $S^3$ which satisfies the following properties:
\begin{romanlist}
\item{F is connected;}
\item{for all $i$, $\partial F_i$ is equal to the sublink of $L$ of color $i$;}
\item{for all $i\neq j$, $F_i\cap F_j$ is either empty, or a disjoint union of clasps (see Figure~\ref{fig:clasps});}
\item{for all $i,j,k$ pairwise distinct, $F_i\cap F_j\cap F_k$ is empty.}
\end{romanlist}}
\medskip

C-complexes are a natural generalization of Seifert surfaces; indeed, if $n=1$, a C-complex for $(L,\sigma)$ is simply
a Seifert surface for $L$. Now, we need to define the Seifert forms associated to a C-complex.
Let us fix a C-complex $F$ and a map $\epsilon\colon\{1,\dots,n\}\to\{\pm 1\}$. A $1$-cycle $x$ in $F$ is called a {\it loop\/}
if it has the following behaviour near a clasp (whenever it crosses one).

\begin{figure}[h]
\begin{center}
\epsfig{figure=loop.eps,height=2.3cm}
\end{center}
\end{figure}
Given any element in $\Ho_1F$, it is possible to represent it by a loop; therefore, we can define a bilinear
form $\alpha^\epsilon\colon\Ho_1F\times\Ho_1F\to\Z$ by
$$
\alpha^\epsilon([x],[y])=\ell k(x^\epsilon,y),
$$
where $\ell k$ denotes the linking number, $x$ is a loop, and $x^\epsilon$ is obtained by pushing $x$ off $F$ in the $\epsilon(i)-$normal direction
off $F_i$ for all $i=1,\dots,n$. In the neighborhood of a clasp in $F_i\cap F_j$, a loop $x$ can be pushed off $F$ in four different ways,
which correspond to the four possible values of $(\epsilon(i),\epsilon(j))$. This is illustrated by the following cross-section of a clasp
in $F_i\cap F_j$.
\begin{figure}[h]
\begin{center}
\epsfig{figure=cross.eps,height=2.7cm}
\end{center}
\end{figure}

Of course, if $n=1$, $\alpha^-$ is the usual Seifert form.
Let us fix a basis of $\Ho_1F$ and denote by $A_F^\epsilon$ the matrix of $\alpha^\epsilon$. Note that for all
$\epsilon$, $A_F^{-\epsilon}=\left(A_F^\epsilon\right)^T$. We are now in position to state our main result.
\medskip

\noindent {\bf Theorem.} {\it The Conway potential function of a colored link $(L,\sigma)$ is given by
$$
\nabla_L^\sigma(t_1,\dots,t_n)=\sgn(F)\prod_{i=1}^n\left(t_i-t_i^{-1}\right)^{\chi(F\setminus F_i)-1}\det\left(-A_F\right),
$$
where $F$ is any C-complex for $(L,\sigma)$, and where
\begin{itemlist}
\item{sgn$(F)$ is equal to the product of the signs of all the clasps in $F$ (see Figure~\ref{fig:clasps});}
\item{$A_F=\sum_\epsilon\epsilon(1)\cdots\epsilon(n)\cdot t_1^{\epsilon(1)}\cdots t_n^{\epsilon(n)}\cdot A_F^\epsilon$,
the sum being on the $2^n$ maps $\epsilon\colon\{1,\dots,n\}\to\{\pm 1\}$.}
\end{itemlist}
}\medskip

\noindent {\bf Example~1.} If $n=1$, $F$ is a Seifert surface for the link $L$ and $A_F^-$ is nothing but a usual Seifert matrix $A$. Since
$A^+_F=A^T$, we have
$$
\nabla^\sigma_L(t)=\frac{1}{t-t^{-1}}\cdot\det\left(-tA^T+t^{-1}A\right),
$$
giving Kauffman's construction of the Conway polynomial
$$
D_L(t)=(t-t^{-1})\cdot\nabla^\sigma_L(t)=\det\left(-tA+t^{-1}A^T\right).
$$

\noindent {\bf Example~2.} If $n=2$, let us denote by $x,y$ the variables and by $F=F_x\cup F_y$ a C-complex for $(L,\sigma)$. Let us also
note $A_F^{--}=A$ and $A_F^{-+}=B$. We then have the formula
$$
\nabla_L^\sigma(x,y)=\sgn(F)(x-x^{-1})^{\chi(F_y)-1}(y-y^{-1})^{\chi(F_x)-1}\det\left(-xyA^T+xy^{-1}B^T+x^{-1}yB-x^{-1}y^{-1}A\right).
$$
Compare Cooper \cite[Corollary 2.2]{Cooper2}.
\medskip

\begin{floatingfigure}{4cm}
\mbox{\epsfig{figure=ex3.eps,width=3.5cm}}
\end{floatingfigure}
\noindent {\bf Example~3.} Consider the link $L=L_1\cup L_2\cup L_3$ with colors $\sigma(1)=\sigma(2)=x$ and $\sigma(3)=y$ given by the following diagram.
We have two different ways to compute the potential function $\nabla_L^\sigma(x,y)$ using our construction. The first one is to consider a C-complex for the
colored link $(L,\sigma)$, and to compute $\nabla_L^\sigma(x,y)$ directly; the second one is to calculate $\nabla_L(t_1,t_2,t_3)$ using a C-complex for
the ordered link $L$, and to use the equality $\nabla_L^\sigma(x,y)=\nabla_L(x,x,y)$. Here are C-complexes $F=F_x\cup F_y$ for $(L,\sigma)$ and
$F'=F_1\cup F_2\cup F_3$ for the ordered link $L$.

\begin{figure}[h]
\begin{center}
\epsfig{figure=ex3b.eps,height=2cm}
\end{center}
\end{figure}
\noindent In the obvious basis of $\Ho_1F$, we have $A_F^{--}=A_F^{-+}=(-1)$; therefore, the theorem gives
$$
\nabla_L^\sigma(x,y)=(x-x^{-1})^0\cdot(y-y^{-1})^{-1}\cdot(xy-x^{-1}y-xy^{-1}+x^{-1}y^{-1})=x-x^{-1}.
$$
On the other hand, $F'$ is contractible, giving
$$
\nabla_L(t_1,t_2,t_3)=(t_1-t_1^{-1})^0\cdot(t_2-t_2^{-1})^1\cdot(t_3-t_3^{-1})^0=t_2-t_2^{-1},
$$
and we get the same result $\nabla_L^\sigma(x,y)=\nabla_L(x,x,y)=x-x^{-1}$.

\section{Proof of the theorem}

Let us recall very briefly Kauffman's argument \cite{Kauf}: given an oriented link $L$, he sets $\Omega_L(t)=\det\left(-tA+t^{-1}A^T\right)$, where
$A$ is any Seifert surface for $L$.
The first step is to check that $\Omega_L$ is an isotopy invariant of $L$. This is achieved using a well known result:
two Seifert surfaces for ambient isotopic links can be obtained from each other by a finite number of ambient isotopies and
handle attachments. We are left with the easy proof that $\Omega_L$ is unchanged by attaching a handle to $F$. Then, Kauffman
proves that $\Omega_L$ satisfies the equality
$$
\Omega_{L_+}(t)-\Omega_{L_-}(t)=(t-t^{-1})\cdot \Omega_{L_0}(t),
$$
where $L_+,L_-$ and $L_0$ are related by a single crossing change as described in the introduction.
This relation, along with the value of $\Omega$ for the trivial knot, determines $\Omega$ for any link. Since this relation
was among the formulas announced by Conway for his polynomial $D$, and since $\Omega$ and $D$ have the same value on
the trivial knot, they are the same invariant.

We will follow the same procedure for the general case. First, we define $\Omega^\sigma_L$ to be the right-hand side of the equality
in the theorem. To show that $\Omega_L^\sigma$ is a well-defined isotopy invariant of colored links, we will generalize Cooper's Isotopy Lemma
\cite[Lemma~3.2]{Cooper2}. A theorem of Murakami \cite{Mur} states that the potential function $\nabla_L^\sigma$ is determined by several
skein relations. Therefore, we will just need to check that $\Omega^\sigma_L$ satisfies these properties in order to have $\Omega=\nabla$.\footnote{We
should also mention that Turaev \cite{Tur} gave another characterization of Conway's multivariable potential function. One could also
use his set of axioms to prove that $\Omega_L$ is indeed the potential function.}
\bigskip

\begin{figure}[Htb]
\begin{center}
\epsfig{figure=push.eps,height=1.8cm}
\fcaption{Push in $F_i$ along an arc $\alpha$.}
\label{fig:push}
\end{center}
\end{figure}
As a preliminary, let us define three transformations of C-complexes; the first two were introduced by Cooper.
Given a C-complex $F$ and an arc $\alpha\colon[0,1]\to F_i$ with $\alpha^{-1}(\partial F)=\{0\}$, a {\it push in $F_i$ along\/} $\alpha$ is
an embedding $p_\alpha\colon F\to F$ defined as follows: choose $\N_1$, $\N_2$ two closed regular neighborhoods of $\alpha$ in $F_i$ meeting
$\partial F$ regularly such that $\N_1\subset\hbox{int}\,\N_2$ and $\N_2\cap(\partial F\setminus\partial F_i)$ is empty. Then, $p_\alpha$ restricted to
$F\setminus\hbox{int}\,\N_2$ is the identity, and $p_\alpha$ maps $\N_2$ homeomorphically onto $\N_2\setminus\hbox{int}\,\N_1$,
as in Figure~\ref{fig:push}. Note that it is not allowed to push a boundary component through another one.
\begin{figure}[Htb]
\begin{center}
\epsfig{figure=M1.eps,height=3.8cm}
\fcaption{Add a ribbon intersection, and push along an arc to convert it into two clasps.}
\label{fig:M1}
\end{center}
\end{figure}

One can also transform a C-complex by {\it adding a ribbon intersection\/}, as described in the left part of Figure~\ref{fig:M1}. The result is no longer
a C-complex, but it can be transformed into one by pushing in $F_i$ along an arc through the ribbon intersection: this converts the ribbon intersection into a pair
of clasps (see the right part of Figure~\ref{fig:M1}). The lower part of this figure shows the same transformations as the upper part, with the
Seifert surface $F_i$ represented as a disc.

Finally, the transformation illustrated in Figure~\ref{fig:pass} is called to {\it pass through a clasp\/}. Again, the lower part of this figure depicts
the same transformation, with a disc representing the surface $F_k$.
Note that these three transformations move the link $L$ only up to an ambient isotopy. Therefore, they can be understood as keeping $L=\partial F$ fixed.

\begin{figure}[Htb]
\begin{center}
\epsfig{figure=pass.eps,height=4.5cm}
\fcaption{Pass through a clasp.}
\label{fig:pass}
\end{center}
\end{figure}

We are now ready to start the proof of the theorem. Let $(L,\sigma)$ be a colored link in $S^3$.
\medskip

\noindent{\bf Lemma~1.} {\it Let $F_1,\dots,F_n$ be a collection of Seifert surfaces for the colored link $(L,\sigma)$
(i.e. for all $i$, $\partial F_i$ is equal to the sublink of $L$ of color $i$). Then, each $F_i$ can be isotoped keeping its boundary fixed
to give a C-complex for $(L,\sigma)$.}\medskip

\noindent{\it Proof.} Although it is an easy generalization of Cooper's \cite[Lemma~3.1]{Cooper2}, it is worth giving the proof in some details.
Let $F_1,\dots,F_n$ be a collection of Seifert surfaces for $(L,\sigma)$. By isotopies, it may be assumed that they
intersect transversally; therefore:
\begin{itemlist}
\item{for all $i\neq j$, $F_i\cap F_j$ is a finite union of intervals ({\it clasps\/} or {\it ribbons\/}) and {\it circles} (see Figure~\ref{fig:int});}
\item{for all $i,j,k$ pairwise distinct, $F_i\cap F_j\cap F_k$ is a finite number of points (called {\it triple points\/});}
\item{every quadruple intersection is empty.}
\end{itemlist}
\begin{figure}[Htb]
\begin{center}
\epsfig{figure=int.eps,height=3.8cm}
\fcaption{The three types of intersections $F_i\cap F_j$: a clasp, a ribbon and a circle.}
\label{fig:int}
\end{center}
\end{figure}

By pushing along arcs, it is possible to remove every circle intersection as follows. First, push along an arc on $F_1$ through one of the outermost
circles on $F_1$, avoiding the triple points: it will turn this circle intersection into a ribbon intersection. This method allows to remove
every circle intersection on $F_1$. Use the same procedure for $F_2,\dots,F_n$, turning every circle into a ribbon.

Now that all the intersections $F_i\cap F_j$ are finite unions of intervals, it is easy to remove all the triple points: pick an interval, number its
possible triple points starting at one end and finishing at the other, and remove them in this order using the isotopy illustrated below.

\begin{figure}[h]
\begin{center}
\epsfig{figure=triple.eps,height=3.2cm}
\end{center}
\end{figure}
By pushing along arcs, it is possible to transform all the ribbon intersections into pairs of clasps, as on the right part of Figure~\ref{fig:M1}.
We now have a union $F=F_1\cup\dots\cup F_n$ which satisfies conditions (ii), (iii) and (iv) in the
definition of a C-complex. By adding ribbon intersections and pushing along arcs (as in Figure~\ref{fig:M1}), we can make $F$ connected. \cqfd

Given $F$ a C-complex for $(L,\sigma)$, let us define
$$
\Omega_F(t_1,\dots,t_n)=\sgn(F)\prod_{i=1}^n\left(t_i-t_i^{-1}\right)^{\chi(F\setminus F_i)-1}\det\left(-A_F\right),
$$
where sgn$(F)$ is equal to the product of the signs of all the clasps in $F$, and where
$A_F=\sum_\epsilon\epsilon(1)\cdots\epsilon(n)\cdot t_1^{\epsilon(1)}\cdots t_n^{\epsilon(n)}\cdot A_F^\epsilon$,
the sum being on the $2^n$ different maps $\epsilon\colon\{1,\dots,n\}\to\{\pm 1\}$. To prove that $\Omega_F$ does not depend on the choice of
$F$, we need two lemmas. The first one is well known (see e.g. \cite[Theorem~8.2]{Lick}).
\medskip

\noindent{\bf Lemma~2.} {\it Two Seifert surfaces for a fixed link $L$ can be transformed into each other by a finite number of the following
operations and their inverses:
\begin{itemlist}
\item{ambient isotopy keeping $L$ fixed;}
\item{addition of a handle. \cqfd}
\end{itemlist}}

The second is a generalization of Cooper's Isotopy Lemma \cite[Lemma~3.2]{Cooper2}, which corresponds to the case $\mu=n=2$.
\medskip

\noindent{\bf Lemma~3.} {\it Let $F=F_1\cup\dots\cup F_n$ and $F'=F'_1\cup\dots\cup F'_n$ be two C-complexes for a fixed colored link $(L,\sigma)$.
Suppose that, for all $i$, $F_i$ is ambient isotopic to $F_i'$ keeping their boundary fixed. Then, $F$ and $F'$ can be transformed into each other by a
finite number of the following operations and their inverses:
\begin{Mlist}
\setcounter{Mlistc}{-1}
\item{ambient isotopy keeping $L$ fixed;}
\item{addition of a ribbon intersection, followed by a push along an arc through this intersection;}
\item{pass through a clasp.}
\end{Mlist}}

\noindent{\it Proof.} By \cite[Lemma~5.2]{Cooper1}, there are other ambient isotopies $F_i\sim F_i'\;(rel\; \partial F_i)$ such that the surfaces $F_1,\dots,F_n$
remain transverse throughout the isotopies except at a finite number of critical points, each of which occurs at a different time.
Let us carry out these new isotopies one after the other: the result is an ambient isotopy $F\sim F'\;(rel\;L)$ with finitely many critical points.
Each critical point corresponds to a change in the {\it singularity\/} of $F$, that is, in the intersections $F_i\cap F_j$ and
$F_i\cap F_j\cap F_k$ (the isotopies may be chosen avoiding quadruple intersections). We need to list all the possible transformations of these intersections,
and convert them into combinations of the operations stated in the lemma.

Recall that the connected components of $F_i\cap F_j$ can be of the following three types: a clasp, a ribbon, or a circle (see Figure~\ref{fig:int}).
The first step is to show that the isotopy may be chosen such that no circle intersection appears. The idea is to transform the isotopy by pushing along
wandering arcs: a push in $F_i$ along a wandering arc is an embedding $p_\alpha\colon F_i\times I\to F_i\times I$ such that $p_\alpha$ restricted to
$F_i\times\{t\}$ is a push along an arc $\alpha_t\colon  I\to F_i$.
A circle intersection in $F_i\cap F_j$ can be avoided as follows: push in $F_i$ along a wandering arc to the circle just before its appearance,
keep the arc breaking the circle during the whole `lifetime' of the circle, and remove the arc after the `death' of the circle.
Therefore, it may be assumed that all the intersections $F_i\cap F_j$ are clasps and ribbons.

Given any $i\neq j$, the number of clasps in $F_i\cap F_j$ is equal modulo $2$ to the linking number $\ell k(\partial F_i,\partial F_j)$.
Therefore, the parity of this number must be preserved.
Considering critical points that are not triple points, we are left with the following possible transformations, where $*$ indicates the critical point.

\begin{figure}[h]
\begin{center}
\epsfig{figure=trans.eps,height=3.5cm}
\end{center}
\end{figure}
Transformation $T1$ is the addition of a ribbon intersection, while $T2$ is a push along an arc through a ribbon intersection. Clearly, $T3$ can be
converted into $T2 + T4 - T2$, where $-T2$ denotes the inverse of $T2$. Finally, $T4$ may be turned into $T1$ and $T2$ via a push in
$F_i$ along a wandering arc.

Let us now consider critical points that are triple points, and denote by $T5$ the move illustrated by the second arrow in Figure~\ref{fig:pass}. Here is the
`movie' of $T5$, where the asterisk denotes the critical points of `birth' and `death' of the  triple point.

\begin{figure}[h]
\begin{center}
\epsfig{figure=T5.eps,height=2cm}
\end{center}
\end{figure}
Of course, there are other transformations involving triple points; let us show that they can be converted into combinations of $T1$, $T2$
and $T5$. By the beginning of the proof, we just need to convert such a move into combinations of $T5$ and of transformations involving no triple points.
Let us consider a transformation of a C-complex $F$ involving a triple point $x\in F_1\cap F_2\cap F_3$, and let us focus on the connected component
$S$ of the singularity of $F$ containing $x$. By pushing along wandering arcs, it may be assumed that the only critical points on $S$ during the whole
`lifetime' of $x$ are the two critical points of `birth' and `death' of $x$. The idea is now to list all the possible configurations of $S$ (as in the
middle of the `movie' of $T5$), and for every such $S$, to give all the different moves involving this $S$. Finally, we get seven possible configurations
of $S$, and a dozen of (essentially) different moves. Then, it is an easy but tedious exercice to check that all these moves can be converted into combinations
of $T5$ and of transformations that do not involve triple points.

Hence, two C-complexes $F$ and $F'$ as in the statement of the lemma can be transformed into each other by a finite number of the operations
$M0$, $T1$, $T2$, $T5$ and their inverses. Since $F$ and $F'$ are C-complexes, the transformations must start with $T1$ or $-T2$ and finish with $T2$ or $-T1$.
Now, every combination of $\pm T1, \pm T2$ and $\pm T5$ starting with $T1$ or $-T2$ and finishing with $T2$ or $-T1$ can be written as a combination of
$\pm M1=\pm(T1+T2)$ and $\pm M2=\pm(-T2+T5+T2)$. This settles the proof. \cqfd

\noindent{\bf Lemma~4.} {\it Let $F$ and $F'$ be two C-complexes for isotopic colored links $(L,\sigma)$ and $(L',\sigma')$.
Then, $\Omega_F$ and $\Omega_{F'}$ are equal.}\medskip

\noindent{\it Proof.} Via an ambient isotopy (which does not change $\Omega_F$), we can assume that $\partial F=L=L'=\partial F'$. By Lemma~2,
$F_i$ and $F_i'$ are related by ambient isotopies (keeping $\partial F_i=\partial F'_i$ fixed) and addition of handles. These handle additions are
performed along arcs $\alpha$ embedded in $S^3\setminus F_i$; such an arc is isotopic (in $S^3\setminus F_i$) to an arc $\alpha'$ embedded in
$S^3\setminus F$. In other words, a handle attachment on $F_i$ can be performed avoiding $F\setminus F_i$. Let us call $M3$ the addition of a handle on $F_i$
avoiding the rest of $F$. Now, for every ambient isotopy between $F_i$ and $F_i'$, we can apply Lemma~3. Therefore, we just need to check that
$\Omega_F$ remains unchanged if $F$ undergoes one of the moves $M1$, $M2$ or $M3$.

\noindent{\bf (M1)} Let $F'$ be a C-complex obtained from $F$ via the move $M1$.
Clearly, $\Ho_1F'\simeq\Z\beta\oplus\Z\gamma\oplus\Ho_1F$, with $\beta$ and $\gamma$ as follows.

\begin{figure}[h]
\begin{center}
\epsfig{figure=M1b.eps,height=3.3cm}
\end{center}
\end{figure}
Among the new clasps created, one is positive and the other one is negative; therefore, $\sgn(F')=-\sgn(F)$. On the other hand,
$$
\chi(F'\setminus F'_i)=\cases{\chi(F\setminus F_i)& if $i=1,2$;\cr \chi(F\setminus F_i)-2& if $i>2$.}
$$
Since
$$
A_{F'}^\epsilon=\bordermatrix{&\beta&\gamma& \cr
			\beta&0&\delta(\epsilon)&0\cr
			\gamma&\delta(-\epsilon)&\ast&\ast\cr
			&0&\ast&A_F^\epsilon}\;,
\quad\hbox{with}\quad\delta(\epsilon)=\cases{-1& if $\epsilon(1)=\epsilon(2)=+1$;\cr\phantom{-}0& else,}
$$
it follows
$$
-A_{F'}=\pmatrix{0&t_1t_2\prod_{i>2}(t_i-t_i^{-1})&0\cr
		t_1^{-1}t_2^{-1}\prod_{i>2}(t_i-t_i^{-1})&\ast&\ast\cr
		0&\ast&-A_F}.
$$
Therefore,
\begin{eqnarray*}
\Omega_{F'}(t_1,\dots,t_n)&=&\sgn(F')\prod_i(t_i-t_i^{-1})^{\chi(F'\setminus F'_i)-1}\det(-A_{F'})\\
		&=&-\sgn(F)\frac{\prod_i(t_i-t_i^{-1})^{\chi(F\setminus F_i)-1}}{\prod_{i>2}(t_i-t_i^{-1})^2}(-1)\prod_{i>2}(t_i-t_i^{-1})^2\det(-A_{F})\\
		&=&\Omega_F(t_1,\dots,t_n).
\end{eqnarray*}

\noindent{\bf (M2)} Let $F$ and $F'$ be C-complexes as illustrated below.

\begin{figure}[h]
\begin{center}
\epsfig{figure=M2b.eps,height=4.7cm}
\end{center}
\end{figure}
Here, $\Ho_1F$ and $\Ho_1F'$ are canonically isomorphic, and $\sgn(F')=\sgn(F)$. Furthermore,
$$
\chi(F'\setminus F'_i)=\cases{\chi(F\setminus F_1)-2& if $i=1$;\cr \chi(F\setminus F_2)+2& if $i=2$;\cr \chi(F\setminus F_i)& if $i>2$.}
$$
On the other hand, given any loop $\beta$ as above, we have the equalities
$$
\ell k(\alpha^\epsilon,\beta)=\cases{1& if $\epsilon(1)=+1$;\cr 0& else,}\quad\hbox{and}\quad
\ell k((\alpha')^\epsilon,\beta)=\cases{1& if $\epsilon(2)=+1$;\cr 0& else.}
$$
This implies that $(t_1-t_1^{-1})^2\det(-A_{F'})=(t_2-t_2^{-1})^2\det(-A_{F'})$. Therefore,
\begin{eqnarray*}
\Omega_{F'}(t_1,\dots,t_n)&=&\sgn(F')\prod_i(t_i-t_i^{-1})^{\chi(F'\setminus F'_i)-1}\det(-A_{F'})\\
		&=&\sgn(F)\prod_i(t_i-t_i^{-1})^{\chi(F\setminus F_i)-1}\frac{(t_2-t_2^{-1})^2}{(t_1-t_1^{-1})^2}\det(-A_{F'})\\
		&=&\Omega_F(t_1,\dots,t_n).
\end{eqnarray*}

\noindent{\bf (M3)} Consider $F'$ a C-complex obtained from $F$ by attaching a handle on $F_1$.
This time, $\Ho_1F'\simeq\Z\delta\oplus\Z\sigma\oplus\Ho_1F$, with $\delta$ and $\sigma$ as follows.
\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=M3b.eps,height=3cm}
\end{center}
\end{figure}
Trivially, $\sgn(F')=\sgn(F)$, and
$$
\chi(F'\setminus F'_i)=\cases{\chi(F\setminus F_i)& if $i=1$;\cr \chi(F\setminus F_i)-2& if $i>1$.}
$$
Since
$$
A_{F'}^\epsilon=\bordermatrix{&\delta&\sigma& \cr
			\delta&0&\pi(\epsilon)&0\cr
			\sigma&\pi(-\epsilon)&\ast&\ast\cr
			&0&\ast&A_F^\epsilon}\;,
\quad\hbox{with}\quad\pi(\epsilon)=\cases{-1& if $\epsilon(1)=+1$;\cr \phantom{-}0& else,}
$$
it follows
$$
-A_{F'}=\pmatrix{0&t_1\prod_{i>1}(t_i-t_i^{-1})&0\cr
		-t_1^{-1}\prod_{i>1}(t_i-t_i^{-1})&\ast&\ast\cr
		0&\ast&-A_F}.
$$
Hence, we have the equality
\begin{eqnarray*}
\Omega_{F'}(t_1,\dots,t_n)&=&\sgn(F')\prod_i(t_i-t_i^{-1})^{\chi(F'\setminus F'_i)-1}\det(-A_{F'})\\
		&=&\sgn(F)\frac{\prod_i(t_i-t_i^{-1})^{\chi(F\setminus F_i)-1}}{\prod_{i>1}(t_i-t_i^{-1})^2}\prod_{i>1}(t_i-t_i^{-1})^2\det(-A_{F})\\
		&=&\Omega_F(t_1,\dots,t_n).
\end{eqnarray*}
This concludes the proof. \cqfd

Therefore, $\Omega_F$ is an isotopy invariant of the colored link $(L,\sigma)$; let us denote it by $\Omega_L^\sigma$.
\medskip

\noindent{\bf Lemma~5.} $\Omega_L^\sigma=\nabla_L^\sigma$.\medskip

\noindent{\it Proof.} Murakami's characterization theorem \cite{Mur} states that $\nabla^\sigma_L$ is determined uniquely by the following six relations,
where the letters $i,j$ and $k$ denote the colors of the components.
$$
\nabla^\sigma_O=\frac{1}{t_i-t_i^{-1}},\leqno(RI)
$$
where $(O,\sigma)$ denotes the trivial knot with color $i$.
$$
\nabla^\sigma_{L_+}-\nabla^\sigma_{L_-}=(t_i-t_i^{-1})\cdot \nabla^{\sigma_0}_{L_0},\leqno(RI\!I)
$$
where $L_+$, $L_-$ and $L_0$ differ by the following local operation.
\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=RII.eps,height=1.8cm}
\end{center}
\end{figure}
\vskip-0.8cm
$$
\nabla^\sigma_{L\sqcup O}=0,\leqno(RI\!I\!I)
$$
where $L\sqcup O$ denotes the disjoint union of $L$ and a trivial knot.
$$
\nabla^\sigma_{L_{++}}+\nabla^\sigma_{L_{--}}=(t_it_j+t_i^{-1}t_j^{-1})\cdot \nabla^\sigma_{L_{00}},\leqno(RIV)
$$
where $L_{++}$, $L_{--}$ and $L_{00}$ differ by the following local operation.
\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=RIV.eps,height=2.7cm}
\end{center}
\end{figure}
\vskip-0.8cm
$$
\nabla^{\sigma'}_{L'}=(t_i-t_i^{-1})\cdot\nabla^\sigma_L,\leqno(RV)
$$
where $L'$ is obtained from $L$ by the local operation given above.
\begin{figure}[h]
\begin{center}
\epsfig{figure=RV.eps,height=2.4cm}
\end{center}
\end{figure}
$$
\matrix{(t_i+t_i^{-1})(t_j-t_j^{-1})\nabla^\sigma_{L(1)}-(t_j-t_j^{-1})(t_k+t_k^{-1})\nabla^\sigma_{L(2)}\qquad\qquad\qquad\qquad\cr
-(t_i^{-1}t_k-t_it_k^{-1})(\nabla^\sigma_{L(3)}+\nabla^\sigma_{L(4)})+(t_i^{-1}t_jt_k-t_it_j^{-1}t_k^{-1})(t_k+t_k^{-1})\nabla^\sigma_{L(5)}\;\cr
-(t_i+t_i^{-1})(t_it_jt_k^{-1}-t_i^{-1}t_j^{-1}t_k)\nabla^\sigma_{L(6)}-(t_i^{-2}t_k^2-t_i^2t_k^{-2})\nabla^\sigma_{L(7)}=0\;,\qquad\;}\leqno(RV\!I)
$$
where $L(1)$, $L(2)$, $L(3)$, $L(4)$, $L(5)$, $L(6)$ and $L(7)$ differ by the following local operation.
\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=RVI.eps,height=3cm}
\end{center}
\end{figure}
By this theorem, we just need to show that $\Omega_L^\sigma$ satisfies the relations $RI$ to $RV\!I$.\medskip

\noindent{\bf (RI)} Since a $2$-disc is a C-complex for the trivial knot, we have $\Omega^\sigma_O(t_i)=\frac{1}{t_i-t_i^{-1}}$.\medskip

\noindent{\bf (RII)} Let $F$ be a C-complex for $L_0$; C-complexes $F_+$ and $F_-$ for $L_+$ and $L_-$ are obtained from
$F$ as follows.\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=RIIb.eps,height=2.8cm}
\end{center}
\end{figure}
Clearly, $\Ho_1F_+\simeq\Ho_1F_-\simeq\Z\alpha\oplus\Ho_1F$, and for all $\epsilon$,
$$
A_{F_-}^\epsilon=\pmatrix{\ast&\ast\cr \ast& A_{F}^\epsilon}=A_{F_+}^\epsilon+\pmatrix{1&0\cr 0&0}.
$$
Therefore,
$$
A_{F_-}=\pmatrix{\ast&\ast\cr \ast& A_{F}}=A_{F_+}+\pmatrix{\prod_j(t_j-t_j^{-1})&0\cr 0&0},
$$
giving $\det\left(-A_{F_-}\right)=\det\left(-A_{F_+}\right)-\prod_j(t_j-t_j^{-1})\det\left(-A_{F}\right)$.
Since $\sgn(F_+)=\sgn(F_-)=\sgn(F)$ and
$$
\chi(F_+-(F_+)_j)=\chi(F_--(F_-)_j)=\cases{\chi(F\setminus F_j)-1& if $j\neq i$;\cr \chi(F\setminus F_i)& if $j=i$,}
$$
we get the equality
\begin{eqnarray*}
\Omega_{L_+}^\sigma-\Omega_{L_-}^\sigma&=&\sgn(F)\prod_j(t_j-t_j^{-1})^{\chi(F\setminus F_j)-2}(t_i-t_i^{-1})\left(
\det\left(-A_{F_+}\right)-\det\left(-A_{F_-}\right)\right)\\
	&=&(t_i-t_i^{-1})\cdot\sgn(F)\prod_j(t_j-t_j^{-1})^{\chi(F\setminus F_j)-1}\det\left(-A_{F}\right)\\
	&=&(t_i-t_i^{-1})\cdot\Omega_{L_0}^{\sigma_0}.
\end{eqnarray*}

\noindent{\bf (RIII)} Choose $F$ a C-complex for $L$. A C-complex $F'$ for $L\sqcup O$ is obtained from $F\sqcup D^2$ either
by the move $M1$ (if $O$ is the only component of its color) or by connecting $F_i$ and $D^2$ with a handle (if $O$ is of color $i$).
In both cases, some $1$-cycle
$x$ is created, such that $\ell k(x^\epsilon,y)=0$ for all $\epsilon$ and all $y\in\Ho_1F'$. Therefore,
$\Omega^\sigma_{L\sqcup O}$ vanishes.\medskip

\noindent{\bf (RIV)} If the colors $i$ and $j$ are equal, this relation follows easily from $RII$. Thus, we can assume
that $i=1$ and $j=2$. Given $F$ a C-complex for $L_{00}$, C-complexes $F_{++}$ for $L_{++}$ and $F_{--}$ for $L_{--}$ are obtained
as follows.

\begin{figure}[h]
\begin{center}
\epsfig{figure=RIVb.eps,height=2.5cm}
\end{center}
\end{figure}
This time, $\Ho_1F_{++}\simeq\Ho_1F_{--}\simeq\Z\beta\oplus\Ho_1F$, and for all $\epsilon$,
$A_{F_{--}}^\epsilon=\pmatrix{\ast&\ast\cr \ast& A_{F}^\epsilon}$; on the other hand
$$
A_{F_{--}}^\epsilon=\cases{A_{F_{++}}^\epsilon& if $(\epsilon(1),\epsilon(2))=(-1,+1)$ or $(+1,-1)$;\cr
			A_{F_{++}}^\epsilon+\pmatrix{1&0\cr 0&0}& if $(\epsilon(1),\epsilon(2))=(-1,-1)$ or $(+1,+1)$.}
$$
Therefore, $\det\left(-A_{F_{--}}\right)=\det\left(-A_{F_+}\right)-
(t_1t_2+t_1^{-1}t_2^{-1})\prod_{i>2}(t_i-t_i^{-1})\det\left(-A_{F}\right)$.
Since $\sgn(F_{++})=\sgn(F)=-\sgn(F_{--})$ and
$$
\chi(F_{++}-(F_{++})_i)=\chi(F_{--}-(F_{--})_i)=\cases{\chi(F\setminus F_i)& if $i=1,2$;\cr \chi(F\setminus F_i)-1& if $i>2$,}
$$
we get the equality
\begin{eqnarray*}
\Omega_{L_{++}}^\sigma+\Omega_{L_{--}}^\sigma&=&\sgn(F)\prod_i(t_i-t_i^{-1})^{\chi(F_{++}-(F_{++})_i)-1}\left(
\det\left(-A_{F_{++}}\right)-\det\left(-A_{F_{--}}\right)\right)\\
	&=&(t_1t_2+t_1^{-1}t_2^{-1})\cdot\sgn(F)\prod_i(t_i-t_i^{-1})^{\chi(F\setminus F_i)-1}\det\left(-A_{F}\right)\\
	&=&(t_1t_2+t_1^{-1}t_2^{-1})\cdot\Omega_{L_{00}}^\sigma.
\end{eqnarray*}

\noindent{\bf (RV)} Once again, if $i=j$, this relation follows from $RII$. So, let us suppose that $i=1$ and $j=2$.
A C-complex $F'$ for $L'$ is obtained from a C-complex $F$ for $L$ by adding a disc with a positive clasp.
From the equalities $\Ho_1F'=\Ho_1F$, $\sgn(F')=\sgn(F)$ and
$$
\chi(F'\setminus F'_i)=\cases{\chi(F\setminus F_i)+1& if $i=1$;\cr \chi(F\setminus F_i)& if $i>1$,}
$$
it follows
\begin{eqnarray*}
\Omega_{L'}^{\sigma'}&=&\sgn(F')\prod_i(t_i-t_i^{-1})^{\chi(F'\setminus F'_i)-1}\det\left(-A_{F'}\right)\\
	&=&(t_1-t_1^{-1})\cdot\sgn(F)\prod_i(t_i-t_i^{-1})^{\chi(F\setminus F_i)-1}\det\left(-A_F\right)\\
	&=&(t_1-t_1^{-1})\cdot\Omega_L^\sigma.
\end{eqnarray*}

\noindent{\bf (RVI)} Let us suppose that $i=1$, $j=2$ and $k=3$. C-complexes $F(i)$ for $L(i)$ $(i=1,\dots,6)$ are obtained from a given
C-complex $F$ for $L(7)$ as follows.
\medskip

\begin{figure}[h]
\begin{center}
\epsfig{figure=RVIb.eps,height=10cm}
\end{center}
\end{figure}
Clearly, $\Ho_1F(1)\simeq\Ho_1F(2)\simeq\Ho_1F(3)\simeq\Ho_1F(4)\simeq\Z\alpha\oplus\Z\beta\oplus\Ho_1F$, and
$\Ho_1F(5)\simeq\Z\alpha\oplus\Ho_1F$, $\Ho_1F(6)\simeq\Z\beta\oplus\Ho_1F$. By the equalities
$$
\sgn(F(i))=\sgn(F)\quad\hbox{for all}\;i\,,$$
and
\begin{eqnarray*}
\chi(F(1)\setminus F(1)_i)&=&\cases{\chi(F\setminus F_i)-1& if $i=1,2$;\cr \chi(F\setminus F_3)& if $i=3$;\cr \chi(F\setminus F_i)-2& if $i>3$,}\\
\chi(F(2)\setminus F(2)_i)&=&\cases{\chi(F\setminus F_1)& if $i=1$;\cr \chi(F\setminus F_i)-1& if $i=2,3$;\cr \chi(F\setminus F_i)-2& if $i>3$,}\\
\chi(F(3)\setminus F(3)_i)&=&\cases{\chi(F\setminus F_i)-1& if $i=1,3$;\cr \chi(F\setminus F_2)& if $i=2$;\cr \chi(F\setminus F_i)-2& if $i>3$,}=\chi(F(4)\setminus F(4)_i),\\
\chi(F(5)\setminus F(5)_i)&=&\cases{\chi(F\setminus F_i)& if $i=1,2$;\cr \chi(F\setminus F_i)-1& if $i>2$,}\\
\chi(F(6)\setminus F(6)_i)&=&\cases{\chi(F\setminus F_i)& if $i=2,3$;\cr \chi(F\setminus F_i)-1& if $i=1$ or $i>2$,}
\end{eqnarray*}
we need to check the following equation, where $T_i$ denotes $(t_i-t_i^{-1})$ and $T=\prod_{i>3}(t_i-t_i^{-1})$:
\begin{eqnarray*}
& &(t_1+t_1^{-1})\left(\frac{\det(-A_{F(1)})}{T_1\,T^2}-
(t_1t_2t_3^{-1}-t_1^{-1}t_2^{-1}t_3)\frac{\det(-A_{F(6)})}{T_1\, T}\right)\\
&-&(t_3+t_3^{-1})\left(\frac{\det(-A_{F(2)})}{T_3\, T^2}-
(t_1^{-1}t_2t_3-t_1t_2^{-1}t_3^{-1})\frac{\det(-A_{F(5)})}{T_3\, T}\right)\\
&=&(t_1^{-1}t_3-t_1t_3^{-1})\left(\frac{\det(-A_{F(3)})+\det(-A_{F(4)})}{T_1\,T_3\, T^2}+
(t_1^{-1}t_3+t_1t_3^{-1})\det(-A_F)\right).
\end{eqnarray*}
Without loss of generality, it may be assumed that $F$ has the following form near the place of the skein relation.

\begin{figure}[h]
\begin{center}
\epsfig{figure=wlog.eps,height=4.5cm}
\end{center}
\end{figure}
It is then easy to compute the matrices $A_{F(i)}$:
\begin{eqnarray*}
-A_{F(1)}&=& \bordermatrix{&\alpha&\beta& \cr
	\alpha&(t_1t_2t_3^{-1}-t_1^{-1}t_2^{-1}t_3)\, T& -T_1\,t_2\,t_3\,T& v_\alpha\,T_3\,T\cr
	\beta&-T_1\,t_2^{-1}t_3^{-1}\,T&0&v_\beta\,T_1\,T\cr
	&w_\alpha\,T_3\,T&w_\beta\,T_1\,T&A_F},\\
-A_{F(2)}&=& \bordermatrix{&\alpha&\beta& \cr
	\alpha&0& -t_1\,t_2\,T_3\,T& v_\alpha\,T_3\,T\cr
	\beta&-t_1^{-1}t_2^{-1}\,T_3\,T&(t_1^{-1}t_2t_3-t_1t_2^{-1}t_3^{-1})&v_\beta\,T_1\,T\cr
	&w_\alpha\,T_3\,T&w_\beta\,T_1\,T&A_F},\\
-A_{F(3)}&=& \bordermatrix{&\alpha&\beta& \cr
	\alpha&0&0& v_\alpha\,T_3\,T\cr
	\beta&0&0&v_\beta\,T_1\,T\cr
	&w_\alpha\,T_3\,T&w_\beta\,T_1\,T&A_F},\\
-A_{F(4)}&=& \bordermatrix{&\alpha&\beta& \cr
	\alpha&0&T_1\,t_2\,T_3\,T&v_\alpha\,T_3\,T\cr
	\beta&-T_1\,t_2^{-1}\,T_3\,T&0&v_\beta\,T_1\,T\cr
	&w_\alpha\,T_3\,T&w_\beta\,T_1\,T&A_F},\\
-A_{F(5)}&=& \bordermatrix{&\alpha& \cr
			\alpha&0& v_\alpha\,T_3\,T\cr
			&w_\alpha\,T_3\,T&A_F},\\
-A_{F(6)}&=&  \bordermatrix{&\beta& \cr
			\beta&0& v_\beta\,T_1\,T\cr
			&w_\beta\,T_1\, T&A_F}.\\
\end{eqnarray*}
The equality can then be checked by direct computation. This concludes Lemma~5, and the theorem is proved. \cqfd

\section{Applications}

In his fundamental paper \cite{Con}, Conway states various properties of his potential function.
We show now that these properties follow immediately from our construction.
\medskip

\noindent{\bf Proposition~1.} $\nabla^\sigma_L(t_1^{-1},\dots,t_n^{-1})=(-1)^\mu\,\nabla_L^{\sigma}(t_1,\dots,t_n).$
\medskip

\noindent{\it Proof.} By the theorem, we easily get $\nabla_L(t_1^{-1},\dots,t_\mu^{-1})=(-1)^\nu\nabla_L(t_1,\dots,t_\mu)$, with
$$
\nu=\sum_{i=1}^n(\chi(F\setminus F_i)-1)+n\cdot\rk\Ho_1F.
$$
For $i=1,\dots,n$, let us denote by $\mu_i$ the number of components of $L$ with color $i$.
Computing modulo $2$, we have
\begin{eqnarray*}
\nu&\equiv&\sum_{i=1}^n\chi(F\setminus F_i)+n\cdot\left(\rk\Ho_1F+1\right)\\
	&\equiv&\sum_{i=1}^n(\chi(F)+\overbrace{\chi(F_i)}^{\equiv \mu_i}+\chi(F_i\cap(F\setminus F_i)))+n\cdot\chi(F)\\
	&\equiv&\sum_{i=1}^n \mu_i + \sum_{i=1}^n\chi(F_i\cap(F\setminus F_i))=\mu+2\cdot\#\{\hbox{clasps}\}\equiv\mu,
\end{eqnarray*}
and the proposition is proved. \cqfd

\noindent {\bf Proposition~2.} $\nabla_L^\sigma(-t_1,\dots,-t_n)=(-1)^\mu\,\nabla_L^{\sigma}(t_1,\dots,t_n)$.
\medskip

\noindent{\it Proof.} Exactly as Proposition~1. \cqfd

\noindent {\bf Proposition~3.} {\it Let $(L,\sigma)$ be a colored link with $\#\sigma^{-1}(1)=\mu_1$, and let
$(L',\sigma)$ be the same link with the opposite orientation on the sublink of color $1$. Then, we have the equality}
$$
\nabla_{L'}^\sigma(t_1,\dots,t_n)=(-1)^{\mu_1}\,\nabla_L^{\sigma}(t_1^{-1},t_2,\dots,t_n).
$$
\noindent{\it Proof.} If $F=F_1\cup\dots\cup F_n$ is a C-complex for $L$, a C-complex for $L'$ is given by $F'=(-F_1)\cup F_2\cup\dots\cup F_n$.
Since $\sgn(F')=(-1)^{\#\{\hbox{\scriptsize{clasps of }}F_1\}}\cdot\sgn(F)=(-1)^{\chi(F_1\cap(F\setminus F_1))}\cdot\sgn(F)$ and
$A^\epsilon_{F'}=A^{\epsilon'}_F$ with $\epsilon'(1)=-\epsilon(1)$ and $\epsilon'(i)=\epsilon(i)$ for $i>1$, we have
$$
\nabla_{L'}(t_1,\dots,t_\mu)=(-1)^\nu\,\nabla_L(t_1^{-1},t_2,\dots,t_\mu),
$$
where
\begin{eqnarray*}
\nu&=&\chi(F_1\cap(F\setminus F_1))+\chi(F\setminus F_1)-1+\rk\Ho_1F\\
&\equiv&\chi(F)-\chi(F\setminus F_1)+\chi(F_1\cap(F\setminus F_1))=\chi(F_1)\equiv\mu_1.
\end{eqnarray*}
This settles the proof. \cqfd

\noindent{\bf Corollary~1.} $\nabla^\sigma_{-L}(t_1,\dots,t_n)=(-1)^\mu\,\nabla_L^{\sigma}(t_1^{-1},\dots,t_n^{-1}).$ \cqfd

\noindent{\bf Corollary~2.} $\nabla^\sigma_{-L}=\nabla_L^{\sigma}.$ \cqfd

\noindent{\bf Proposition~4.} {\it $\nabla_{m(L)}^\sigma=(-1)^{\mu+1}\,\nabla_L^\sigma$, where $m(L)$ denotes the mirror image of $L$.}
\medskip

\noindent{\it Proof.} The mirror image $m(F)$ of a C-complex
$F$ for $L$ provides a C-complex for $m(L)$. The sign of every clasp and of every linking number is changed, giving
$A^\epsilon_{m(F)}=-A^\epsilon_F$ for all $\epsilon$. Hence
$$
\nabla_{m(L)}=(-1)^{\#\{\hbox{\scriptsize{clasps of }}F\}}\cdot(-1)^{\hbox{\scriptsize{rk}}\,\hbox{\scriptsize{H}}_1F}\cdot\nabla_L.
$$
The equality $\chi(F)=\sum_{i=1}^n\chi(F_i)-\#\{\hbox{clasps of $F$}\}\,$ yields
$$
\#\{\hbox{clasps of $F$}\}+\rk\Ho_1F=\rk\Ho_1F-\chi(F)+\sum_{i=1}^\mu\overbrace{\chi(F_i)}^{\equiv \mu_i}\equiv \mu+1,
$$
and the proposition is proved. \cqfd

Let $(L',\sigma')$ and $(L'',\sigma'')$ be two colored links; let us suppose that two components $L_i'$ of $L'$ and $L_j''$
of $L''$ have the same color $\sigma'(i)=\sigma''(j)=\alpha$. Then, there is a well-defined colored link given by the connected sum of $L'$ and
$L''$ along $L_i'$ and $L_j''$; we will denote it by $(L'\#L'',\sigma'\#\sigma'')$. The following multiplicativity formula was also stated
by Conway. As far as I know, there is no published proof.
\medskip

\noindent{\bf Proposition~5.} $\nabla^{\sigma'\#\sigma''}_{L'\#L''}=\nabla_{L'}^{\sigma'}\cdot\nabla_{L''}^{\sigma''}
\cdot(t_\alpha-t_\alpha^{-1}).$
\medskip

\noindent{\it Proof.} Given C-complexes $F'$ for $(L',\sigma')$ and $F''$ for $(L'',\sigma'')$, a C-complex for the connected sum
$(L'\#L'',\sigma'\#\sigma'')$ is given by $F=F'\#F''$. Clearly, $A^\epsilon_F=A^\epsilon_{F'}\oplus A^\epsilon_{F''}$ for all $\epsilon$,
giving
$$
A_F=\left(\prod_{j\neq\alpha}\left(t_j''-(t_j'')^{-1}\right)\cdot A_{F'}\right)\oplus
\left(\prod_{i\neq\alpha}\left(t_i'-(t_i')^{-1}\right)\cdot A_{F''}\right).
$$
Since $\sgn(F)=\sgn(F')\cdot\sgn(F'')$ and
\begin{eqnarray*}
\chi(F\setminus F_i')&=&\chi(F'\setminus F_i')+\chi(F'')-1\quad\forall\;i\neq\alpha,\\
\chi(F\setminus F_j'')&=&\chi(F')+\chi(F''\setminus F_j'')-1\quad\forall\;j\neq\alpha,\\
\chi(F\setminus F_\alpha)&=&\chi(F'\setminus F'_\alpha)+\chi(F''\setminus F''_\alpha),
\end{eqnarray*}
we get the result. \cqfd

For the sake of completeness, let us mention without proof two additional properties of Conway's potential function.
The first one is another skein relation announced by Conway, that can be proved using our construction.
\medskip

\noindent{\bf Proposition~6.} {\it $\nabla_{L_1}^\sigma+\nabla_{L_2}^\sigma=\nabla_{L_3}^\sigma+\nabla_{L_4}^\sigma$, where
$L_1,L_2,L_3$ and $L_4$ are identical except within a ball where they are related as illustrated below.}

\begin{figure}[h]
   \begin{center}
     \epsfig{figure=prop6.eps,height=2.3cm}
   \end{center}
\end{figure}
\vskip-1cm\cqfd\vskip0.7cm

The second one is the translation of the Torres formula \cite[Theorem 3]{Torres} from the Alexander polynomial to the Conway potential function.
This formula can be recovered from our theorem, but the proof is a little tedious.
\medskip

\noindent{\bf Proposition~7.} {\it $\nabla_L(1,t_2,\dots,t_\mu)=(t_2^{\ell_{12}}\cdots t_\mu^{\ell_{1\mu}}-
t_2^{-\ell_{12}}\cdots t_\mu^{-\ell_{1\mu}})\cdot\nabla_{L\setminus L_1}(t_2,\dots,t_\mu)$, where $\ell_{1i}$ denotes the linking number
$\ell k(L_1,L_i)$. \cqfd }

\nonumsection{Acknowledgements}
The author wishes to express his thanks to Daryl Cooper and to Mathieu Baillif. Above all, it is a pleasure to thank Jerry Levine
for his kindness and disponibility at Brandeis University.

\nonumsection{References}


text:
\begin{definition}[Definition:Hilbert Sequence Space]
Let $d_2: A \times A: \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
:$\displaystyle \forall x = \sequence {x_i}, y = \sequence {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$
The [[Definition:Metric Space|metric space]] $\struct {A, d_2}$ is the '''Hilbert sequence space on $\R$''' and is denoted $\ell^2$.
\end{definition}

config: wiki


TITLE: Agmon's inequality on an open subset $\Omega$ of $\mathbb{R}^3$
QUESTION [1 upvotes]: I'm looking for a reference for what we call the Agmon's inequality on a regular open bounded subset $\Omega$ of $\mathbb{R}^3$ :
$$\|u\|_{L^\infty(\Omega)}\leq C \|u\|_{H^1(\Omega)}^{1/2} \|u\|_{H^2(\Omega)}^{1/2}.$$
This limit case for Sobolev embeddings can be found for example in the book of JC Robinson et al. "The Three-Dimensional Navier-Stokes Equations: Classical Theory" in Thm. 1.20, or in the book of Foias et al. " Navier-Stokes equations and turbulence" but I didn't manage to find a precise proof (the lecture notes of Agmon only provide the exponants $1/4, 3/4$ on the norms in the inequality).
Thank you !

REPLY [1 votes]: I think I have actually managed to get a direct proof, with the good exponants, which I write down here.
First, we extend our fixed function $u \in \mathrm{H}^2(\Omega)$ on the whole space thank to a continuous extension operator $\mathcal{E} : \mathrm{H}^s(\Omega) \rightarrow \mathrm{H}^s(\mathbb{R}^3)$ (note that this one is the same for all values of $s$, see for example the book of Stein , Singular integrals and differentiability properties of functions. Princeton Mathematical Series, No. 30. Princeton University Press, Princeton, N.J., 1970. )
So, if we set $v:=\mathcal{E} u$, we have $\Vert u \Vert_{\mathrm{L}^{\infty}(\Omega)} \leq \Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} $ and since we have $$\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}^{1/2} \Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}^{1/2} \leq c \Vert u \Vert_{\mathrm{H}^1(\Omega)}^{1/2} \Vert u \Vert_{\mathrm{H}^2(\Omega)}^{1/2},$$
it remains to prove that the following estimate holds :
$$ \Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}^{1/2} \Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}^{1/2}.$$
In order to do so, we use Sobolev embedding for $v \in \mathrm{H}^2(\mathbb{R}^3) \hookrightarrow \mathscr{C}(\mathbb{R}^3)$ with the estimate
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim \Vert \widehat{v} \Vert_{\mathrm{L}^{1}(\mathbb{R}^3)}.$$
Then, we write
$$\Vert \widehat{v} \Vert_{\mathrm{L}^{1}(\mathbb{R}^3)}=\int_{B(0,A)}\widehat{v}(\xi) \, \mathrm{d}\xi \ + \int_{B(0,A)^c}\widehat{v}(\xi) \, \mathrm{d}\xi,  $$ with $A>0$ which will be chosen later. We use Cauchy-Schwarz inequality on each term to get
$$\int_{B(0,A)}\widehat{v}(\xi) \, \mathrm{d}\xi \leq \left(\int_{B(0,A)}\dfrac{\mathrm{d}\xi}{1+\vert \xi \vert^2} \right)^{1/2} \left( \int_{B(0,A)} (1+\vert \xi \vert^2) \vert \widehat{v}(\xi) \vert^2 \, \mathrm{d}\xi \right)^{1/2},$$
$$\int_{B(0,A)^c}\widehat{v}(\xi) \, \mathrm{d}\xi \leq \left(\int_{B(0,A)^c}\dfrac{\mathrm{d}\xi}{1+\vert \xi \vert^4} \right)^{1/2} \left( \int_{B(0,A)^c} (1+\vert \xi \vert^4) \vert \widehat{v}(\xi) \vert^2 \, \mathrm{d}\xi \right)^{1/2}. $$
Thus, we have
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim \left(\int_0^A \dfrac{r^2}{1+r^2} \mathrm{d}r \right)^{1/2}\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}+\left(\int_A^{+\infty} \dfrac{r^2}{1+r^4} \mathrm{d}r \right)^{1/2}\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)}.$$
We can easily compute the first integral and bound it exactly by $A^{1/2}$. For the other one, it's a convergent integral whose integrand is equivalent to $1/r^2$ when $r \rightarrow +\infty$ so that it is equivalent to the rest of the integral of $1/r^2$ on [A,+infty[, which can be compared to $A^{-1/2}$ (equivalent of the rest of a Riemann integral).
We end up with
$$\Vert v \Vert_{\mathrm{L}^{\infty}(\mathbb{R}^3)} \lesssim A^{1/2}\Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)} + A^{-1/2}\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)},$$ for $A$ large enough. We finally choose $A$ of order $\Vert v \Vert_{\mathrm{H}^2(\mathbb{R}^3)} \, / \, \Vert v \Vert_{\mathrm{H}^1(\mathbb{R}^3)}$ to complete the proof.


TITLE: The Electric field & Coulomb's law
QUESTION [0 upvotes]: A point charge is moving in an electric field at right angles to the lines of forces .Does any force act on it ? Please explain why.

REPLY [2 votes]: Yes, a force is still acting on the particle.
Moving perpendicular to the lines of force may result in no work being done ($W = \mathbf{F}\cdot \mathbf{r}$) , but since the particle is charged, it will be experiencing a force when moving through any electric field.
If the particle is (initially) moving in the $y$ direction and the electric field is in the $x$ direction, so that $x$ and $y$ are perpendicular, then the particle will slowly develop a velocity component in the direction of the field.
Mathematically: Netwton's second law is $m\frac{d^2\mathbf{r}}{dt^2} = \mathbf{F}$; assuming the charge of the particle is $q$, the force acting on it by an electric field $\mathbf{E}$ is $\mathbf{F} = q\mathbf{E} $.
So $m\frac{d^2\mathbf{r}}{dt^2} =  q\mathbf{E}$.
Now let's split this into the $x$ and $y$ directions:
x) $m\frac{d^2 x}{dt^2} =  qE_x$,
solution: $x(t) = x_0 + v_{x initial}t +\frac{1}{2}{qE_x \over m}t^2$;
assuming the particle starts from 0 and with no velocity along the direction of the field, $x(t) = \frac{1}{2}{qE_x \over m}t^2$ : it will slowly ($\sim t^2$) acquire a velocity component in the direction of the field)
y) $m\frac{d^2 y}{dt^2} =  qE_y = 0$ since the electric field in the x direction.
solution: $y(t) = y_0 + v_{y initial}t$;
No force is acting along $y$, so the $y$ component of the velocity will remain the same as it was before entering the electric field.


TITLE: What is the classification of a fixed point if the trace and determinant are equal?
QUESTION [3 upvotes]: In the text book I am reading Nonlinear Dynamics and Chaos with applications to Physics, Biology, Chemistry, and Engineering by Steven H. Strogatz. He uses a diagram similar to this one:
to show how to classify a fixed point of a system using the determinants and trace of the matrix. But I was wondering what would happen if you are on the parabola thats given ? Prime example would be if $\tau$ is the trace and $\Delta$ is the determinant then what would be the classification of the fixed point if:
$$ \tau^2 = 4\Delta $$
This equation is given in the textbook as: $\tau^2 - 4\Delta = 0$

REPLY [2 votes]: When the trace and determinant are on the indicated parabola for the system $\vec{x}'=A \vec{x}$, we are dealing with a repeated real eigenvalue $\lambda$ for $A$. This will often introduce solutions of the form $t e^{\lambda t}$, due to potentially being non-diagonalizable. Computing the matrix exponential gives $e^{tA}= (I + t(A-\lambda I)) e^{t \lambda I}$, and all solutions will be of the form $\vec{x}(t) = e^{tA} \vec{c}$.
Such an equilibrium is referred to as a "defective nodal source / sink," although it is perhaps not the most common name.
The mathlet linear phase portraits gives a nice visualization of solutions as we move around the different possibilities. Here's a screenshot showing the desired behavior:


text:
\begin{abstract}

In this work, we obtain sufficient conditions for the ``stability" of our recently proposed algorithm,  modified-CS (for noisy measurements), designed for recursive reconstruction of sparse signal sequences from noisy measurements. By ``stability" we mean that the number of misses from the current support estimate and the number of extras in it remain bounded by a time-invariant value at all times. The concept is meaningful only if the bound is small compared to the current signal support size. A direct corollary is that the reconstruction errors are also bounded by a time-invariant and small value.

\end{abstract}

\section{Introduction}
In this work, we study the ``stability"  modified-CS (noisy) \cite{modcsjp,modcsicassp10} which was designed for recursive reconstruction of sparse signal sequences from noisy measurements. By ``stability" we mean that the number of misses from the current support estimate and the number of extras in it remain bounded by a {\em time-invariant} value at all times. The concept is meaningful only if the bound is small compared to the current signal support size. A direct corollary is that the reconstruction errors are also bounded by a time-invariant and small value.

The key assumption that our algorithm utilizes is that the support changes slowly over time. As we demonstrated in \cite{just_lscs,modcsjp}, this assumption holds for many medical image sequences. Denote the support estimate from the previous time by $T$. Modified-CS tries to finds a signal that is sparsest outside of $T$ and satisfies the data constraint.

Other algorithms for recursive reconstruction include our older work on Least Squares CS-residual (LS-CS) and Kalman filtered CS-residual (KF-CS) \cite{just_lscs,kfcsicip,kfcspap}; CS for time-varying signals \cite{giannakis_2} (assumes a time-invariant support, which is a somewhat restrictive assumption); homotopy methods \cite{romberg} (use past reconstructions to speed up current optimization but not to improve reconstruction error with fewer measurements); and \cite{ibm} (a recent modification of KF-CS). Two other algorithms that are also designed for static CS with partial knowledge of support include \cite{camsap07} and \cite{hassibi}. The work of \cite{camsap07} proposed an approach similar to modified-CS but did not analyze it and also did not show real experiments either. The work of \cite{hassibi}, which appeared in parallel with modified-CS, assumed a probabilistic prior on the support and obtained conditions for exact reconstruction.

To the best of our knowledge, stability of recursive sparse reconstruction algorithms has not been studied in any other work except in our older works \cite{just_lscs,kfcspap} for LS-CS and KF-CS respectively.

The KF-CS result \cite{kfcspap} is under fairly strong assumptions, e.g. it is for a random walk signal change model with only support additions (no removals). The result for LS-CS stability \cite{just_lscs} holds under mild assumptions and is for a fairly realistic signal change model. The only limitation is that it assumes that support changes occur ``every-so-often": every $d$ time units, there are $S_a$ support additions and $S_a$ removals.
 But from testing the slow support change assumption for real data (medical image sequences), it has been observed that support changes usually occur at {\em every} time, e.g. see Fig. 1 of \cite{just_lscs}. {\em This important case is the focus of the current work.}

In \cite{just_lscs}, we only studied LS-CS (modified-CS was proposed later). But the techniques of \cite{just_lscs} can be also used to show modified-CS stability for the model of \cite{just_lscs}.
In this work, we mostly discuss modified-CS stability since our earlier experiments indicate that it outperforms LS-CS \cite{modcsjp}. But, again, the same techniques can be used to show LS-CS stability as well for the current signal model. This is done in the long version of this work \cite{long}.

The paper is organized as follows. We give problem definition in Sec. \ref{notn} and we overview our results in Sec. \ref{overview}. We describe the signal model for proving stability in Sec. \ref{signalmodel}. We obtain sufficient conditions for the stability of modified-CS and discuss the implications in Sec. \ref{simple_modcs}. We discuss some of its limitations and develop a simple modification that uses a better support estimation approach (modified-CS with add-LS-del). This support estimation approach is related to the one in \cite{cosamp,kfcsicip,just_lscs}. In Sec. \ref{addLSdel_modcs}, we show the stability of modified-CS with add-LS-del, which is more difficult to do.  Numerical experiments are discussed in Sec. \ref{sims}. Conclusions are given in Sec. \ref{conclusions}.

\subsection{Notation and Problem Definition}
\label{notn}
The set operations $\cup$, $\cap$, $\setminus$ have their usual meanings. $\emptyset$ denotes the empty set.
We use $T^c$ to denote the complement of a set $T$ w.r.t. $[1,m]:=[1,2,\dots m]$, i.e. $T^c := [1,m] \setminus T$. $|T|$ denotes the cardinality of $T$. For a vector, $v$, and a set, $T$, $v_T$ denotes the $|T|$ length sub-vector containing the elements of $v$ corresponding to the indices in the set $T$. $\| v \|_k$ denotes the $\ell_k$ norm of a vector $v$. {\em If just $\|v\|$ is used, it refers to $\|v\|_2$.} Similarly, for a matrix $M$, $\|M\|_k$ denotes its induced $k$-norm, while just $\|M\|$ refers to $\|M\|_2$. $M'$ denotes the transpose of $M$ and $M^\dag$ denotes the Moore-Penrose pseudo-inverse of $M$ (when $M$ is tall, $M^\dag:=(M'M)^{-1} M'$). For a fat matrix $A$, $A_T$ denotes the sub-matrix obtained by extracting the columns of $A$ corresponding to the indices in $T$.
The $S$-restricted isometry constant \cite{dantzig}, $\delta_S$, for an $n \times m$ matrix (with $n<m$), $A$, and the $S,S'$ restricted orthogonality constant \cite{dantzig}, $\theta_{S,S'}$, are as defined in \cite[eq 1.3]{dantzig} and \cite[eq 1.5]{dantzig} respectively.

We assume the following observation model:
\bea
y_t = A x_t + w_t,  \ \|w_t\| \le \eps

\label{obsmod}
\eea
where $x_t$ is an $m$ length sparse vector with support $N_t$, $y_t$ is the $n< m$ length observation vector at time $t$ and $w_t$ is observation noise with $\|w_t\| \le \eps$.  ``Support" refers to the set of indices of the nonzero elements of $x_t$.

Our goal is to recursively estimate $x_t$ using $y_1, \dots y_t$. By {\em recursively}, we mean, use only $y_t$ and the estimate from $t-1$, $\xhat_{t-1}$, to compute the estimate at $t$.

As we explain in Sec. \ref{simple_modcs}, our algorithm need more measurements at the initial time, $t=0$. We use $n_0$ to denote the number of measurements used at $t=0$ and we use $A_0$ to denote the corresponding $n_0 \times m$ measurement matrix.
We use $\alpha$ to denote the support estimation threshold used by modified-CS and we use $\alpha_{add}, \alpha_{del}$ to denote the support addition and deletion thresholds used by modified-CS with add-LS-del and by LS-CS.

We use $\xhat_t$ to denote the final estimate of $x_t$ at time $t$ and $\Nhat_t$ to denote its support estimate. To keep notation simple, we avoid using the subscript $t$ wherever possible.

\bd[$T$, $\Delta$, $\Delta_e$]
We use $T  : = \Nhat_{t-1}$ to denote the support estimate from the previous time.
We use $\Delta := N_t \setminus T$ to denote the unknown part of the support at the current time and $\Delta_e := T \setminus N_t$ to denote the ``erroneous" part of $T$. We attach the subscript $t$ to the set, e.g. $T_t$ or $\Delta_t$, where necessary.
\ed
\bd[$\tT$, $\tDelta$, $\tDelta_e$]
We use $\tT := \Nhat_t$ to denote the final estimate of the current support; $\tDelta : = N_t \setminus \tT$ to denote the ``misses" in $\Nhat_t$ and $\tDelta_e := \tT \setminus N_t$ to denote the ``extras".
\ed
The sets $T_\dett, \Delta_\dett, \Delta_{e,\dett}$ are defined later in Sec. \ref{addLSdel_modcs}.

{\em If the sets $B,C$ are disjoint, then  we just write $D \cup B \setminus C$ instead of writing $(D \cup B) \setminus C$, e.g. $N_t = T \cup \Delta \setminus \Delta_e$.}

We refer to the left (right) hand side of an equation or inequality as LHS (RHS).

\vspace{-0.1in}
\subsection{Overview of Results}
\label{overview}

When measurements are noisy, the reconstruction errors of modified-CS (noisy) and of LS-CS have been bounded as a function of $|N_t|$, $|\Delta_t|$ and $|\Delta_{e,t}|$ in \cite{modcsicassp10,arxiv} and in \cite{just_lscs} respectively. The bound is small if $|\Delta_t|$ and  $|\Delta_{e,t}|$ are small enough. But smallness of the support errors, $\Delta_t$, $\Delta_{e,t}$, depends on the accuracy of the previous reconstruction. Thus it can happen that the error bound increases over time, and such a bound is of limited use for a recursive reconstruction problem. There is thus a need to obtain conditions under which one can show ``stability", i.e. obtain a time-invariant bound on the sizes of these support errors. Also, for the result to be meaningful, the support errors' bound needs to be small compared to the support size.

In this work, we study the stability of modified-CS for noisy measurements and its modification, modified-CS with add-LS-del. This is done under a bounded observation noise assumption and for a signal model with
\ben
\item support changes ($S_a$ additions and $S_a$ removals) occurring at every time, $t$,
\item magnitude of the newly added coefficients increases gradually, and similarly for decrease before removal,
\item support size is $S_0$ at all times and the signal power is also constant
\een
\begin{remark}
The reason we need the bounded noise assumption is as follows. When the noise is unbounded, e.g. Gaussian, all error bounds for CS and, similarly, all error bounds for LS-CS or modified-CS hold with ``large probability" \cite{dantzig,candes_rip,tropp,just_lscs,modcsicassp10,arxiv}. To show stability, we need the error bound for LS-CS or modified-CS to hold at all times, $0 \le t < \infty$ (this, in turn, is used to ensure that the support gets estimated with bounded error at all times). Clearly this is a zero probability event.
\end{remark}

Our results have the following form. For a given number and type of measurements (i.e. for a given measurement matrix, $A$), and for a given  noise bound, $\eps$, if,
\ben
\item the support estimation threshold(s) is/are large enough,
\item the support size, $S_0$, and support change size, $S_a$ are small enough,
\item the newly added coefficients increase (existing large coefficients decrease) at least at a certain rate, $r$, and
\item the initial number of measurements, $n_0$, is large enough for accurate initial reconstruction using simple CS,
\een
then the support errors are bounded by time-invariant values. In particular, we show that $|N_t \setminus \Nhat_t| \le 2S_a$ and $|\Nhat_t \setminus N_t| = 0$. Consequently the reconstruction error is also bounded by a small and time-invariant value.

A key assumption used in designing both modified-CS and LS-CS is that the signal support changes slowly over time. As shown in \cite{just_lscs,modcsjp}, this holds for real medical image sequences. For our model, this translates to $S_a \ll S_0$.

Under the slow support change assumption, clearly, $2S_a \ll S_0$, and so the support error bounds are small compared to the support size, $S_0$, making our stability results meaningful. We also compare the conditions on $S_0$ required by our results with those required by the corresponding simple CS error bounds (since simple CS is not a recursive approach these also serves as a stability result for simple CS) and argue that our results hold under weaker assumptions (allow larger values of $S_0$). The results for modified-CS and modified-CS (with add-LS-del) are also compared.

config: arxiv


TITLE: Sequence of open balls
QUESTION [0 upvotes]: I have this excercise (that I am not sure if it is true):
Suppose $K$ is compact and
if $\left(B_{j}\right)_{j=1}^{\infty}$ is a sequence of open balls that covers $K$, prove that there is a positive number $\epsilon$ such that each ball with radius $\epsilon$ and center $x \in K$ is contained in one of the balls $\left(B_{j}\right)_{j=1}^{\infty}$.
My attempt: As $K$ is a compact set, every open cover has a finite subcover, $(B_k)_{k=1}^N$, as it is finite we can consider the set of intersections of all balls, we can fit a ball of radius $\delta_i$ inside the $i$-th intersection, we can consider the minimum of the set of $\delta_i$ denote it $\bar{\delta}$, an consider $\epsilon = \frac{\bar{\delta}}{2}$. And let it be our epsilon.
I really feel that I am mising something, because I am not sure if the ball of radius $\epsilon$ will be completly contained inside one of the balls $(B_k)_{k=1}^N$, even on the the intersections (as close as I get to the "edge" of the intersection I am not sure if it garantees me to fit there). Am I wrong?

REPLY [1 votes]: See in the initial open cover there are many open sets, and the delta you need to find is with respect to the initial open cover, so there may be a open set of radius delta that may not be inside some open set in the initial open cover.
More simply speaking there may be a $B_i$ that is not selected in the finite subcover and the radius is less than $\epsilon$ hence your epsilon might not necessarily work.
For a correct soln see Lebesgue number lemma (https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma)

REPLY [0 votes]: The assertion is true. Every open cover of a compact metric space has a Lebesgue number, i.e., a number $\lambda>0$ such that every set of diameter less than $\lambda$ is contained in one of the sets of the cover.
Your argument, however, is incomplete because for every finite sub-cover that you choose I can come up with an open ball that will not fit in any of the intersections you have mentioned.


text:
\subsection{Proof \cref{thm:structure_W_E}}

\label{sec:p_sructure_W_E}

Let $T_d$ be normal and $(2\alpha_2-(\alpha_1+\alpha_2)^2)(T_d+T_d^\top)$ be positive definite and let $T(W)=T_d$ be partitioned as in \eqref{eq:structure_T}. Then equation \eqref{eq:problem} reads

\begin{align}
	- I &=f'(z)f(z) T_{11} + f'(z)g(z)T_{12} \nonumber \\
	&+g'(z)f(z)T_{21}  + g'(z)g(z) T_{22}.
	\label{eq_p3:1}
\end{align}

Choosing $T_d$ ($T(W)=T_d$) as in \eqref{eq:cases_E1}, it holds that  $(2\alpha_2-(\alpha_1+\alpha_2)^2)(T_d+T_d^\top)$ is positive definite, yielding

\begin{align}
	- 1 &= a \Big(f'(z)f(z) + g(z)'g(z)\Big)
	\nonumber
	\\
	&+  g'(z)f(z)-f'(z)g(z).
	\label{eq_p3:2}
\end{align}

This equation has been considered in \cite{jfESNCM}. We refer to the proof of \cite[Theorem 1]{jfESNCM} for the derivation of $f$ and $g$ as specified in \eqref{eq:cases_E1}.

Case \eqref{eq:cases_E2} is analogous to \eqref{eq:cases_H2}. First, note that $(2\alpha_2 - (\alpha_1+\alpha_2))(T_d+T_d^\top)$ is positive definite and normal with the given $T_d$ in \eqref{eq:cases_E2} and $(2\alpha_2 - (\alpha_1+\alpha_2))(Q+Q^\top)$ is positive definite and normal, since
\begin{align}
    T_dT_d^\top - T_d^\top T_d = \begin{bmatrix} QQ^\top - Q^\top Q & 0 \\ 0 & QQ^\top - Q^\top Q  \end{bmatrix}
\end{align}
and
\begin{align}
    T_d + T_d^\top = \begin{bmatrix} Q + Q^\top  & 0 \\ 0 & Q + Q^\top  \end{bmatrix},
\end{align}

such that the real part of the eigenvalues of $Q$ is identical to that of $T_d$ and therefore the definiteness property is conserved.

Hence, the derivation of the generating functions $f$ and $g$ in this case goes along the lines of arguments as used in the proof of \cref{thm:structure_W_H} in \cref{sec:p_sructure_W_H}, specifically for the case \eqref{eq:cases_H2}, i.e. \eqref{eq_p2:5} and \eqref{eq_p2:5_2}.

$\hfill \small{\square}$

The structure of normal matrices $T_d$ brings more degrees of freedom compared to $T_d$ skew-symmetric as in \Cref{thm:structure_W_H} and thus the two cases listed in \cref{thm:structure_W_E} are not exhaustive.


TITLE: Why isn't the world symmetric under parity transformations?
QUESTION [1 upvotes]: In quantum physics and newtonian mechanics, switching "left" and "right" in the math is super easy: you just state that left is, for instance, the positive $x$ axis and right is the negative $x$ axis, and then you switch up the signs of $x$ in the differential equation you want to solve and the initial conditions or boundary conditions you're applying to get a perfectly symmetric prediction. But if the laws of physics don't differentiate between right and left, and the world follows the laws of physics, why isn't the world symmetric? Most people write with their right hands, the right and left brain are different, and the organs of the human body are distributed unevenly between our right and left insides. But if the microscopic processes that take place in the fancy biological processes that let humans grow into full organisms from just a couple of cells are mathematical symmetric under parity, why are we asymmetric?
One of the suggested similar questions was this: Why isn't our universe symmetric?. I guess I understand what John Rennie's answer states: he implies that a tiny fluctuation in the early universe gave rise to asymmetry. Does that imply that there is a domino effect of sorts that leads to large-scale macroscopic asymmetry from small asymmetric fluctuations? Furthermore, doesn't the occurrence of that fluctuation as described by the theory of inflation (which I first heard of while reading that answer) imply breaking of parity symmetry?

REPLY [0 votes]: As it has already stressed in Murray's answer, the symmetry of the physical laws has not to be the same as the symmetry of the states.
However, as an additional piece of information, I would note that the world is not symmetric under parity transformation. This is the result of a famous experiment conducted in 1956 and showing that conservation of parity is violated in weak interactions. Therefore, the laws of Physics do differentiate between right and left on the scale of energies of those experiments. Interestingly, the parity symmetry is restored at the coarse-grained level of description, where the relevant degrees of freedom are (stable) nuclei and electrons.


TITLE: Find $\int\frac{x^2-1}{x^4+x^2+1}dx$
QUESTION [1 upvotes]: Find
  $$
\int\frac{x^2-1}{x^4+x^2+1}dx
$$

$x^4+x^2+1>0$, so not possible to factor, so I guess there is no direct way or through partial fraction decomposition.
$$
\int\frac{x^2-1}{x^4+x^2+1}dx=\int\frac{(x^2-1)^2}{(x^2)^3-1}dx\\
$$
I have no clue of how to solve it, is there any substitution that I can give to $x$ so that it becomes simple ?

REPLY [2 votes]: At  the bottom, write $ x^2 + 1/x^2 = (x + 1/x)^2 -2$. Then assume $ x+ 1/x = t$. Then you're done.
Let me right the detailed solution. Divide the numerator and denominator of the integrand by $x^2.$ Then we have $$ \frac{1-1/x^2}{(x+1/x)^2 - 1}$$. Then do the substitution business. Then just integrate the following nice cute function: $$ \frac{1}{t^2 -1}$$. After integrating, we have the following function $$ 1/2 ln|\frac{t-1}{t+1}|$$. QED


TITLE: Generate log of variable and append to existing dataset
QUESTION [0 upvotes]: I just started using Mathematica and I hope that someone can help me with a small piece of code:

I have a variable "X" in a dataset A.
I want to create a new variable called "logX" which is the log of the variable X and add it as a new column in dataset A using "append" in Mathematica if possible.

Thanks a lot for the help!

REPLY [0 votes]: ds = Dataset[{<|"x" -> 5|>, <|"x" -> 10|>}]
Append[#, "lx" -> Log[#x]] & /@ ds


TITLE: Dihedral group $D_{8}$ as a semidirect product $V\rtimes C_2$?
QUESTION [3 upvotes]: How do I show that the dihedral group $D_{8}$ (order $8$) is a semidirect product $V\rtimes \left\langle \alpha \right\rangle $, where $V$ is Klein group and $%
\alpha $ is an automorphism of order two?

REPLY [2 votes]: I like to think about $D_8$ as the group of invariants of a square.
So the our group $V$ is given by the identity, the 2 reflections that have no fix points and their product which is 180°-rotation.
Semidirect product can be characterized as split short exact sequences of groups. This is just a fancy way to say the following:
Given a group $G$ and a normal subgroup $N\subset G$. Denote by $\pi:G\to G/N$ the projection map. Then $G\cong N\rtimes G/N$ iff there exists a homomorphism $\phi: G/N\to G$, such that $\pi\circ\phi=\mathrm{id}_{G/N}$. This $\phi$ is called a splitting homomorphism.
Back to our dihedral group: $D_8/V\cong C_2$. In order to define a splitting homomorphism $C_2\to D_8$ we need to find an element of order 2 that is not contained in $V$. Such an element is given by a reflection through one of the diagonals of our square. It is clear that $\pi$ doesn't map t his element to $0\in C_2$ because it does not lie in $V$. So we constructed a homomorphism $\phi: C_2\to D_8$ such that $\pi\circ\phi=\mathrm{id}_{C_2}$.