\begin{definition}
	A knot $K$ in $S^3$ is a smooth (PL - smooth) 
	embedding of a circle $S^1$ in $S^3$:
	\[
		\varphi:  S^1 \hookrightarrow  S^3
	\]
\end{definition}

\noindent
Usually we think about a knot 
as an image of an embedding: 
$K = \varphi(S^1)$.
Some basic examples and counterexamples 
are shown respectively in 
\autoref{fig:unknot} and 
\autoref{fig:notknot}.
 
\begin{figure}[h]
	\centering
	\begin{subfigure}{0.45\textwidth}
		\centering
		\includegraphics[width=0.5\textwidth]
		{unknot.png}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\centering
		\includegraphics[width=0.5\textwidth]
		{trefoil.png}
	\end{subfigure}	
	\caption{Knots examples: 
	unknot (left) and trefoil (right).}
	\label{fig:unknot}
\end{figure}

\begin{figure}[h]
	\centering
	\begin{subfigure}{0.45\textwidth}
		\centering
		\includegraphics[width=0.5\textwidth]
		{not_injective_knot.png}
	\end{subfigure}
	\begin{subfigure}{0.45\textwidth}
		\centering
		\includegraphics[width=0.5\textwidth]
		{not_smooth_knot.png}
	\end{subfigure}	
	\caption{
		Not-knots examples:
		an image of 
		a function ${S^1\longrightarrow S^3}$ 
		that is not injective (left) and 
		of a function
		that is not smooth (right).
	}
	\label{fig:notknot}
\end{figure}

\begin{definition}
	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}

\noindent
Example of simple links are shown in
\autoref{fig:links}.

\begin{figure}[h]
	\centering
	\begin{subfigure}{0.5\textwidth}
		\centering
		\includegraphics[width=1\textwidth]
		{3unknots.png}
		\caption{A trivial link with $3$ components.}
	\end{subfigure}
	\begin{subfigure}{0.4\textwidth}
		\centering
		\includegraphics[width=0.7\textwidth]
		{Hopf.png}
		\caption{A Hopf link.}
	\end{subfigure}	
	\begin{subfigure}{0.4\textwidth}
		\centering
		\includegraphics[width=0.8\textwidth]
		{WhiteheadLink.png},
		\caption{A Whitehead link.}
	\end{subfigure}	
	\begin{subfigure}{0.4\textwidth}
		\centering
		\includegraphics[width=0.7\textwidth]
		{BorromeanRings.png}
		\caption{A Borromean link.}
	\end{subfigure}	
	\caption{Link examples.}
	\label{fig:links}
\end{figure}

%
%
\begin{definition}\label{def:link_diagram}
	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 
		$D_{\pi |_L}$ is non degenerate,
		\item 
		the double points are not degenerate,
		\item there are no triple point. 
	\end{enumerate}
\end{definition}

\noindent
By \Cref{def:link_diagram} the following pictures can not be a part of a diagram:
\begin{figure}[H]	
	\centering
	\begin{subfigure}{0.1\textwidth}
		\includegraphics[width=0.8\textwidth]
		{LinkDiagram1.png},
	\end{subfigure}
	\begin{subfigure}{0.1\textwidth}
		\includegraphics[width=0.6\textwidth]
		{LinkDiagram2.png},
	\end{subfigure}
	\begin{subfigure}{0.1\textwidth}
		\includegraphics[width=0.8\textwidth]
		{LinkDiagram3.png}.
	\end{subfigure}
\end{figure}

\noindent
There are under- and overcrossings  (tunnels and bridges) on a link diagrams with an obvious meaning.
\begin{lemma}
Every link admits a link diagram.
\end{lemma}
\noindent

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 $D_2$. Then we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$. 

\begin{figure}[h]
	\centering
	\includegraphics[width=0.6\textwidth]
	{seifert_connect.png}
	\caption{Connecting two surfaces.}
	\label{fig:SeifertConnect}
\end{figure}

\begin{theorem}[Seifert]\label{theo: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}

\begin{definition}\label{def:lk_via_homo}
	Let $\alpha$ and $\beta$ be 
	two disjoint simple closed 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 $\beta$: 
	\[
		\alpha \in H_1(S^3 \setminus
		\nu(\beta), \mathbb{Z}) 
		\cong \mathbb{Z}.
	\]
\end{definition}

\begin{figure}[h]
		\fontsize{10}{8}\selectfont
		\centering
		\def\svgwidth{\linewidth}
		\resizebox{\textwidth}{!}{
%	\centering
	\begin{subfigure}{0.3\textwidth}
		\centering
		\def\svgwidth{\linewidth}
		\resizebox{1\textwidth}{!}{
		\input{images/linking_torus_6_2.pdf_tex}
		}
	\end{subfigure}
	\begin{subfigure}{0.3\textwidth}
		\centering
		\def\svgwidth{\linewidth}
		\resizebox{1\textwidth}{!}{
		\input{images/linking_hopf.pdf_tex}
		}
	\end{subfigure}	
	}
	\vspace*{10mm}
	\caption{
	Linking number of a Hopf link (left) 
	and a torus link $T(6, 2)$ (right).
	}
	\label{fig:unknot}
\end{figure}

\begin{lemma}
$
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 a surface $\Sigma$.
\end{lemma}

\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 curves 
$\alpha_1, \dots, \alpha_n$. 

\noindent
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 
$\alpha_i^+$ don't intersect the surface.

\noindent
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}
	}	
	\caption{
	A basis $\alpha_1, \alpha_2$ 
	of the first homology
	group of a Seifert surface 
	and a copy of
	element $\alpha_1$ pushed up 
	along vector normal to the Seifert surface.  
	}
	\label{fig:alpha_plus}
\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}