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Maria Marchwicka 2019-06-09 21:43:06 +02:00
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\begin{definition}
A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
\begin{align*}
\varphi: S^1 \hookrightarrow S^3
\end{align*}
\end{definition}
\noindent
Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$.
\begin{example}
\begin{itemize}
\item
Knots:
\includegraphics[width=0.08\textwidth]{unknot.png} (unknot),
\includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil).
\item
Not knots:
\includegraphics[width=0.12\textwidth]{not_injective_knot.png}
(it is not an injection),
\includegraphics[width=0.08\textwidth]{not_smooth_knot.png}
(it is not smooth).
\end{itemize}
\end{example}
\begin{definition}
%\hfill\\
Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function
\begin{align*}
&\Phi: S^1 \times [0, 1] \hookrightarrow S^3 \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and
$\Phi_1 = \varphi_1$.
\end{definition}
\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\
&\psi_t: S^3 \hookrightarrow S^3,\\
& \psi_0 = id ,\\
& \psi_1(K_0) = K_1.
\end{align*}
\end{theorem}
\begin{definition}
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$
\end{definition}
\begin{example}
Links:
\begin{itemize}
\item
a trivial link with $3$ components:
\includegraphics[width=0.2\textwidth]{3unknots.png},
\item
a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
\item
a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
\end{itemize}
\end{example}
%
%
%
\begin{definition}
A link diagram $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that:
\begin{enumerate}[label={(\arabic*)}]
\item
${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
\item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png},
\item there are no triple point: \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}.
\end{enumerate}
\end{definition}
\noindent
There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.\\
Every link admits a link diagram.
\\
Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\
We can distinguish two types of crossings: right-handed
$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.
\subsection{Reidemeister moves}
A Reidemeister move is one of the three types of operation on a link diagram as shown below:
\begin{enumerate}[label=\Roman*]
\item\hfill\\
\includegraphics[width=0.6\textwidth]{rm1.png},
\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png},
\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}.
\end{enumerate}
\begin{theorem} [Reidemeister, 1927 ]
Two diagrams of the same link can be
deformed into each other by a finite sequence of Reidemeister moves (and isotopy of the plane).
\end{theorem}
%
%
%
%The number of Reidemeister Moves Needed for Unknotting
%Joel Hass, Jeffrey C. Lagarias
%(Submitted on 2 Jul 1998)
% Piotr Sumata, praca magisterska
% proof - transversality theorem (Thom)
%Singularities of Differentiable Maps
%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.
\subsection{Seifert surface}
\noindent
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing:
\begin{align*}
\PICorientpluscross \mapsto \PICorientLRsplit\\
\PICorientminuscross \mapsto \PICorientLRsplit
\end{align*}
We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial \Sigma = L$.\\
\begin{figure}[h]
\fontsize{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}}
\caption{Constructing a Seifert surface.}
\label{fig:SeifertAlg}
}
\end{figure}
\noindent
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=0.6\textwidth]{seifert_connect.png}
\end{center}
\caption{Connecting two surfaces.}
\label{fig:SeifertConnect}
\end{figure}
\begin{theorem}[Seifert]
Every link in $S^3$ bounds a surface $\Sigma$ that is compact, connected and orientable. Such a surface is called a Seifert surface.
\end{theorem}
%
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/torus_1_2_3.pdf_tex}}
\caption{Genus of an orientable surface.}
\label{fig:genera}
}
\end{figure}
%
%
\begin{definition}
The three genus $g_3(K)$ ($g(K)$) of a knot $K$ is the minimal genus of a Seifert surface $\Sigma$ for $K$.
\end{definition}
\begin{corollary}
A knot $K$ is trivial if and only $g_3(K) = 0$.
\end{corollary}
\noindent
Remark: there are knots that admit non isotopic Seifert surfaces of minimal genus (András Juhász, 2008).
\begin{definition}
Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$.
On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ be the number of positive crossings, $N_-$ - negative. Then the linking number: $\Lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$.
\end{definition}
\hfill
\\
Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$.
Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can be interpreted via first homology group, where $\Lk(\alpha, \beta)$ is equal to evaluation of $\alpha$ as element of first homology group of the complement of $\beta$:
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
\begin{example}
\begin{itemize}
\item
Hopf link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}},
}
\end{figure}
\item
$T(6, 2)$ link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}.
}
\end{figure}
\end{itemize}
\end{example}
\begin{fact}
\[
g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) =
\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}),
\]
where $b_1$ is first Betti number of $\Sigma$.
\end{fact}
\subsection{Seifert matrix}
Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$.
Let $\alpha_1^+, \dots \alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface (push up along a vector field normal to $\Sigma$). Note that elements $\alpha_i$ are contained in the Seifert surface while all $\alpha_i^+$ are don't intersect the surface.
Let $\Lk(\alpha_i, \alpha_j^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. Note that by choosing a different basis we get a different matrix.
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_matrix.pdf_tex}}
}
\end{figure}
\begin{theorem}
The Seifert matrices $S_1$ and $S_2$ for the same link $L$ are S-equivalent, that is, $S_2$ can be obtained from $S_1$ by a sequence of following moves:
\begin{enumerate}[label={(\arabic*)}]
\item
$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients,
\item
$V \rightarrow
\begin{pmatrix}
\begin{array}{c|c}
V &
\begin{matrix}
\ast & 0 \\
\sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline
\begin{matrix}
\ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 0\\
1 & 0
\end{matrix}
\end{array}
\end{pmatrix} \quad$
or
$\quad
V \rightarrow
\begin{pmatrix}
\begin{array}{c|c}
V &
\begin{matrix}
\ast & 0 \\
\sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline
\begin{matrix}
\ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 1\\
0 & 0
\end{matrix}
\end{array}
\end{pmatrix}$
\item
inverse of (2)
\end{enumerate}
\end{theorem}

486
lectures_on_knot_theory.tex
View File

@ -71,7 +71,9 @@
\DeclareRobustCommand\longtwoheadrightarrow \DeclareRobustCommand\longtwoheadrightarrow
{\relbar\joinrel\twoheadrightarrow} {\relbar\joinrel\twoheadrightarrow}
\newcommand{\longhookrightarrow}{\lhook\joinrel\longrightarrow}
\newcommand{\longhookleftarrow}{\longleftarrow\joinrel\rhook}
\AtBeginDocument{\renewcommand{\setminus}{% \AtBeginDocument{\renewcommand{\setminus}{%
@ -108,285 +110,7 @@
%\input{myNotes} %\input{myNotes}
\section{Basic definitions \hfill\DTMdate{2019-02-25}} \section{Basic definitions \hfill\DTMdate{2019-02-25}}
\begin{definition} \input{lec_1.tex}
A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
\begin{align*}
\varphi: S^1 \hookrightarrow S^3
\end{align*}
\end{definition}
\noindent
Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$.
\begin{example}
\begin{itemize}
\item
Knots:
\includegraphics[width=0.08\textwidth]{unknot.png} (unknot),
\includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil).
\item
Not knots:
\includegraphics[width=0.12\textwidth]{not_injective_knot.png}
(it is not an injection),
\includegraphics[width=0.08\textwidth]{not_smooth_knot.png}
(it is not smooth).
\end{itemize}
\end{example}
\begin{definition}
%\hfill\\
Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function
\begin{align*}
&\Phi: S^1 \times [0, 1] \hookrightarrow S^3 \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and
$\Phi_1 = \varphi_1$.
\end{definition}
\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\
&\psi_t: S^3 \hookrightarrow S^3,\\
& \psi_0 = id ,\\
& \psi_1(K_0) = K_1.
\end{align*}
\end{theorem}
\begin{definition}
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$
\end{definition}
\begin{example}
Links:
\begin{itemize}
\item
a trivial link with $3$ components:
\includegraphics[width=0.2\textwidth]{3unknots.png},
\item
a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
\item
a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
\end{itemize}
\end{example}
%
%
%
\begin{definition}
A link diagram $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that:
\begin{enumerate}[label={(\arabic*)}]
\item
${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
\item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png},
\item there are no triple point: \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}.
\end{enumerate}
\end{definition}
\noindent
There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.\\
Every link admits a link diagram.
\\
Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\
We can distinguish two types of crossings: right-handed
$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.
\subsection{Reidemeister moves}
A Reidemeister move is one of the three types of operation on a link diagram as shown below:
\begin{enumerate}[label=\Roman*]
\item\hfill\\
\includegraphics[width=0.6\textwidth]{rm1.png},
\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png},
\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}.
\end{enumerate}
\begin{theorem} [Reidemeister, 1927 ]
Two diagrams of the same link can be
deformed into each other by a finite sequence of Reidemeister moves (and isotopy of the plane).
\end{theorem}
%
%
%
%The number of Reidemeister Moves Needed for Unknotting
%Joel Hass, Jeffrey C. Lagarias
%(Submitted on 2 Jul 1998)
% Piotr Sumata, praca magisterska
% proof - transversality theorem (Thom)
%Singularities of Differentiable Maps
%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.
\subsection{Seifert surface}
\noindent
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing:
\begin{align*}
\PICorientpluscross \mapsto \PICorientLRsplit\\
\PICorientminuscross \mapsto \PICorientLRsplit
\end{align*}
We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial \Sigma = L$.\\
\begin{figure}[h]
\fontsize{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_alg.pdf_tex}}
\caption{Constructing a Seifert surface.}
\label{fig:SeifertAlg}
}
\end{figure}
\noindent
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$.
\begin{figure}[h]
\begin{center}
\includegraphics[width=0.6\textwidth]{seifert_connect.png}
\end{center}
\caption{Connecting two surfaces.}
\label{fig:SeifertConnect}
\end{figure}
\begin{theorem}[Seifert]
Every link in $S^3$ bounds a surface $\Sigma$ that is compact, connected and orientable. Such a surface is called a Seifert surface.
\end{theorem}
%
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/torus_1_2_3.pdf_tex}}
\caption{Genus of an orientable surface.}
\label{fig:genera}
}
\end{figure}
%
%
\begin{definition}
The three genus $g_3(K)$ ($g(K)$) of a knot $K$ is the minimal genus of a Seifert surface $\Sigma$ for $K$.
\end{definition}
\begin{corollary}
A knot $K$ is trivial if and only $g_3(K) = 0$.
\end{corollary}
\noindent
Remark: there are knots that admit non isotopic Seifert surfaces of minimal genus (András Juhász, 2008).
\begin{definition}
Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$.
On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$ be the number of positive crossings, $N_-$ - negative. Then the linking number: $\Lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$.
\end{definition}
\hfill
\\
Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$.
Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can be interpreted via first homology group, where $\Lk(\alpha, \beta)$ is equal to evaluation of $\alpha$ as element of first homology group of the complement of $\beta$:
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
\begin{example}
\begin{itemize}
\item
Hopf link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}},
}
\end{figure}
\item
$T(6, 2)$ link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}.
}
\end{figure}
\end{itemize}
\end{example}
\begin{fact}
\[
g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) =
\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}),
\]
where $b_1$ is first Betti number of $\Sigma$.
\end{fact}
\subsection{Seifert matrix}
Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$.
Let $\alpha_1^+, \dots \alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface (push up along a vector field normal to $\Sigma$). Note that elements $\alpha_i$ are contained in the Seifert surface while all $\alpha_i^+$ are don't intersect the surface.
Let $\Lk(\alpha_i, \alpha_j^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. Note that by choosing a different basis we get a different matrix.
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/seifert_matrix.pdf_tex}}
}
\end{figure}
\begin{theorem}
The Seifert matrices $S_1$ and $S_2$ for the same link $L$ are S-equivalent, that is, $S_2$ can be obtained from $S_1$ by a sequence of following moves:
\begin{enumerate}[label={(\arabic*)}]
\item
$V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients,
\item
$V \rightarrow
\begin{pmatrix}
\begin{array}{c|c}
V &
\begin{matrix}
\ast & 0 \\
\sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline
\begin{matrix}
\ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 0\\
1 & 0
\end{matrix}
\end{array}
\end{pmatrix} \quad$
or
$\quad
V \rightarrow
\begin{pmatrix}
\begin{array}{c|c}
V &
\begin{matrix}
\ast & 0 \\
\sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline
\begin{matrix}
\ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 1\\
0 & 0
\end{matrix}
\end{array}
\end{pmatrix}$
\item
inverse of (2)
\end{enumerate}
\end{theorem}
\section{\hfill\DTMdate{2019-03-04}} \section{\hfill\DTMdate{2019-03-04}}
\begin{theorem} \begin{theorem}
@ -583,7 +307,7 @@ $\Delta_{11n34} \equiv 1$.
% %
\begin{lemma}[Dehn] \begin{lemma}[Dehn]
Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding
${D^2 \overset{g}\hookrightarrow M}$ such that: ${D^2 \overset{g}\longhookrightarrow M}$ such that:
\[ \[
g_{\big| \partial D^2} = f_{\big| \partial D^2.} g_{\big| \partial D^2} = f_{\big| \partial D^2.}
\] \]
@ -641,7 +365,7 @@ Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists
\end{figure} \end{figure}
\begin{definition} \begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\ A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$. Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
\end{definition} \end{definition}
@ -657,6 +381,7 @@ Concordance is an equivalence relation.
\begin{fact}\label{fakt:concordance_connected} \begin{fact}\label{fakt:concordance_connected}
If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$. $K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
\begin{figure}[h] \begin{figure}[h]
\fontsize{10}{10}\selectfont \fontsize{10}{10}\selectfont
\centering{ \centering{
@ -685,21 +410,69 @@ Are there in concordance group torsion elements that are not $2$ torsion element
\noindent \noindent
Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice. Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\\ \\
\\ \begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
}
\caption{$Y = F \cup \Sigma$ is a smooth close surface.}
\label{fig:closed_surface}
\end{figure}
\noindent \noindent
Let $\Omega$ be an oriented four-manifold. \\ \\
???????\\ Pontryagin-Thom construction tells us that there exists a compact three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$ (i.e. there are cycles). \\ Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
??????????????\\
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$. $\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$.
Let $B^+$ be a push off of $B$ in the positive normal direction such that Let $B^+$ be a push off of $B$ in the positive normal direction such that
$\partial B^+ = \beta^+$. $\partial B^+ = \beta^+$.
Then Then
$\Lk(\alpha, \beta^+) = A \cdot B^+$ $\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero.
%
%
%ball_4_alpha_beta.pdf
\\ \\
?????????????????
\\
Let us consider following maps:
\[
\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
\]
Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$.
%
\\
????????????\\
%
%
\begin{proposition}
\[
\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y),
\]
where $b_1$ is first Betti number.
\end{proposition}
\begin{proof}
\begin{align*}
& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to
\\
\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
\to & H_1(Y) \to \H_1(\Omega) \to H_1(\Omega, Y) \to \\
\to & H_0(Y) \to H_0(\Omega) \to 0
\end{align*}
\end{proof}
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
}
%\caption{Sketch for Fakt %%\label{fig:concordance_m}
\end{figure}
\section{\hfill\DTMdate{2019-03-25}}
\begin{definition}
The (smooth) four genus $g_4(K)$ is the minimal genus of the surface $\Sigma \in B^4$ such that $\Sigma$ is compact, orientable and $\partial \Sigma = K$.
\end{definition}
\noindent
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.
\section{\hfill\DTMdate{2019-04-08}} \section{\hfill\DTMdate{2019-04-08}}
% %
% %
@ -718,6 +491,55 @@ $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes.
%$A \cdot B$ gives the pairing as ?? %$A \cdot B$ gives the pairing as ??
\end{proposition} \end{proposition}
\begin{proof}
By Poincar\'e duality we know that:
\begin{align*}
H_3(\Omega, Y) &\cong H^0(\Omega),\\
H_2(Y) &\cong H^0(Y),\\
H_2(\Omega) &\cong H^1(\Omega, Y),\\
H_2(\Omega, Y) &\cong H^1(\Omega).
\end{align*}
Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
= \dim_{\mathbb{Q}} V
$.\\
\noindent
Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on a $K$
has a subspace of dimension $g_{\Sigma}$ on which it is zero:
\begin{align*}
\newcommand\coolover[2]%
{\mathrlap{\smash{\overbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
\newcommand\coolleftbrace[2]{%
#1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
\newcommand\coolrightbrace[2]{%
\left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
\vphantom{% phantom stuff for correct box dimensions
\begin{matrix}
\overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
\underbrace{pqr}_{\mbox{$S$}}
\end{matrix}}%
V =
\begin{matrix}% matrix for left braces
\coolleftbrace{g_{\Sigma}}{ \\ \\ \\}
\\ \\ \\ \\
\end{matrix}%
\begin{pmatrix}
\coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
\sdots & & \sdots & \sdots & & \sdots \\
0 & \dots & 0 & * & \dots & *\\
* & \dots & * & * & \dots & *\\
\sdots & & \sdots & \sdots & & \sdots \\
* & \dots & * & * & \dots & *
\end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
\end{align*}
\end{proof}
\section{\hfill\DTMdate{2019-03-11}} \section{\hfill\DTMdate{2019-03-11}}
@ -1037,7 +859,7 @@ If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of
\frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\ \frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\
\end{align*} \end{align*}
??????????????????? 1 ?? epsilon?\\ ??????????????????? 1 ?? epsilon?\\
\begin{theorem}(Matumoto, Conway-Borodzik-Politarczyk) \begin{theorem}(Matumoto, Borodzik-Conway-Politarczyk)
Let $K$ be a knot, Let $K$ be a knot,
\begin{align*} \begin{align*}
&H_1(\widetilde{X}, \Lambda) \times &H_1(\widetilde{X}, \Lambda) \times
@ -1071,7 +893,7 @@ field of fractions
\section{\hfill\DTMdate{2019-06-03}} \section{\hfill\DTMdate{2019-06-03}}
\begin{theorem} \begin{theorem}
Let $K$ be a knot and $u(K)$ its unknotting number. Let $g_4(K)$ be a minimal four genus of a smooth surface $S$ in $B^4$ such that $\partial S = K$. Then: Let $K$ be a knot and $u(K)$ its unknotting number. Let $g_4$ be a minimal four genus of a smooth surface $S$ in $B^4$ such that $\partial S = K$. Then:
\[ \[
u(K) \geq g_4(K) u(K) \geq g_4(K)
\] \]
@ -1129,53 +951,7 @@ S^1 \times \pt &\longrightarrow \pt \times S^1 \\
\section{balagan} \section{balagan}
\noindent \noindent
\begin{proof}
By Poincar\'e duality we know that:
\begin{align*}
H_3(\Omega, Y) &\cong H^0(\Omega),\\
H_2(Y) &\cong H^0(Y),\\
H_2(\Omega) &\cong H^1(\Omega, Y),\\
H_2(\Omega, Y) &\cong H^1(\Omega).
\end{align*}
Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
= \dim_{\mathbb{Q}} V
$.
\end{proof}
\noindent \noindent
Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on a $4$ - manifolds???\\
?????\\
has a subspace of dimension $g_{\Sigma}$ on which it is zero:
\begin{align*}
\newcommand\coolover[2]%
{\mathrlap{\smash{\overbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
\newcommand\coolleftbrace[2]{%
#1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
\newcommand\coolrightbrace[2]{%
\left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
\vphantom{% phantom stuff for correct box dimensions
\begin{matrix}
\overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
\underbrace{pqr}_{\mbox{$S$}}
\end{matrix}}%
V =
\begin{matrix}% matrix for left braces
\coolleftbrace{g_{\Sigma}}{ \\ \\ \\}
\\ \\ \\ \\
\end{matrix}%
\begin{pmatrix}
\coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
\sdots & & \sdots & \sdots & & \sdots \\
0 & \dots & 0 & * & \dots & *\\
* & \dots & * & * & \dots & *\\
\sdots & & \sdots & \sdots & & \sdots \\
* & \dots & * & * & \dots & *
\end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
\end{align*}
\section{\hfill\DTMdate{2019-05-06}} \section{\hfill\DTMdate{2019-05-06}}
@ -1219,6 +995,38 @@ $a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$.
\\ \\
\noindent \noindent
The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations. The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations.
\begin{definition}
The Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$.
\end{definition}
%see Maciej page
\noindent
Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$.
\\
?????????????????????
\\
\noindent
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\
Let now $H = H_1(\widetilde{X})$. Can we define a paring? \\
Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta$ an Alexander polynomial. We know that $\Delta c = 0 \in H_1(\widetilde{X})$ (Alexander polynomial annihilates all possible elements). Let consider a surface $F$ such that $\partial F = c$. Now consider intersection points $F \cdot d$. This points can exist in any $N_k$ or $S_k$.
\[
\frac{1}{\Delta} \sum_{j\in \mathbb{Z} t^{-j}}(F \cdot t^j d) \in \quot{\mathbb{Q}[t, t^{-1}]}{\mathbb{Z}[t, t^{-1}]}
\]
\\
?????????????\\
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
\caption{$c, d \in H_1(\widetilde{X})$.}
\label{fig:covering_pairing}
}
\end{figure}
\begin{definition} \begin{definition}
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$. The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$.