463 lines
12 KiB
TeX
463 lines
12 KiB
TeX
\begin{definition}
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A knot $K$ in $S^3$ is a smooth (PL - smooth)
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embedding of a circle $S^1$ in $S^3$:
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\[
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\varphi: S^1 \hookrightarrow S^3
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\]
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\end{definition}
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\noindent
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Usually we think about a knot
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as an image of an embedding:
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$K = \varphi(S^1)$.
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Some basic examples and counterexamples
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are shown respectively in
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\autoref{fig:unknot} and
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\autoref{fig:notknot}.
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\begin{figure}[h]
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\centering
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\begin{subfigure}{0.45\textwidth}
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\centering
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\includegraphics[width=0.5\textwidth]
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{unknot.png}
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\end{subfigure}
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\begin{subfigure}{0.45\textwidth}
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\centering
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\includegraphics[width=0.5\textwidth]
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{trefoil.png}
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\end{subfigure}
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\caption{Knots examples:
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unknot (left) and trefoil (right).}
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\label{fig:unknot}
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\end{figure}
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\begin{figure}[h]
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\centering
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\begin{subfigure}{0.45\textwidth}
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\centering
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\includegraphics[width=0.5\textwidth]
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{not_injective_knot.png}
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\end{subfigure}
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\begin{subfigure}{0.45\textwidth}
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\centering
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\includegraphics[width=0.5\textwidth]
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{not_smooth_knot.png}
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\end{subfigure}
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\caption{
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Not-knots examples:
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an image of
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a function ${S^1\longrightarrow S^3}$
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that is not injective (left) and
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of a function
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that is not smooth (right).
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}
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\label{fig:notknot}
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\end{figure}
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\begin{definition}
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Two knots $K_0 = \varphi_0(S^1)$,
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$K_1 = \varphi_1(S^1)$
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are equivalent if the embeddings
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$\varphi_0$ and $\varphi_1$ are isotopic,
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that is there exists a continues function
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\begin{align*}
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&\Phi: S^1 \times
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[0, 1] \hookrightarrow S^3, \\
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&\Phi(x, t) = \Phi_t(x)
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\end{align*}
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such that
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$\Phi_t$ is an embedding
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for any $t \in [0,1]$,
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$\Phi_0 = \varphi_0$ and
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$\Phi_1 = \varphi_1$.
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\end{definition}
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\begin{theorem}
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Two knots $K_0$ and $K_1$ are isotopic
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if and only if they are ambient isotopic,
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i.e. there exists a family of self-diffeomorphisms
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$\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
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\begin{align*}
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&\psi(t) = \psi_t
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\text{ is continius on
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$t\in [0,1]$},\\
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&\psi_t: S^3 \hookrightarrow S^3,\\
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& \psi_0 = id ,\\
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& \psi_1(K_0) = K_1.
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\end{align*}
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\end{theorem}
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\begin{definition}
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A knot is trivial (unknot) if it is equivalent
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to an embedding
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$\varphi(t) = (\cos t, \sin t, 0)$,
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where $t \in [0, 2 \pi] $
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is a parametrisation of $S^1$.
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\end{definition}
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\begin{definition}
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A link with $k$ - components is a
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(smooth) embedding of
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$\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$
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in $S^3$.
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\end{definition}
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\noindent
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Example of simple links are shown in
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\autoref{fig:links}.
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\begin{figure}[h]
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\centering
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\begin{subfigure}{0.5\textwidth}
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\centering
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\includegraphics[width=1\textwidth]
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{3unknots.png}
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\caption{A trivial link with $3$ components.}
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\end{subfigure}
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\begin{subfigure}{0.4\textwidth}
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\centering
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\includegraphics[width=0.7\textwidth]
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{Hopf.png}
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\caption{A Hopf link.}
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\end{subfigure}
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\begin{subfigure}{0.4\textwidth}
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\centering
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\includegraphics[width=0.8\textwidth]
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{WhiteheadLink.png},
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\caption{A Whitehead link.}
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\end{subfigure}
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\begin{subfigure}{0.4\textwidth}
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\centering
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\includegraphics[width=0.7\textwidth]
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{BorromeanRings.png}
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\caption{A Borromean link.}
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\end{subfigure}
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\caption{Link examples.}
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\label{fig:links}
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\end{figure}
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%
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%
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\begin{definition}\label{def:link_diagram}
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A link diagram $D_{\pi}$ is a picture
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over projection $\pi$ of a link $L$ in
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$\mathbb{R}^3$($S^3$) to
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$\mathbb{R}^2$ ($S^2$) such that:
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\begin{enumerate}[label={(\arabic*)}]
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\item
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$D_{\pi |_L}$ is non degenerate,
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\item
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the double points are not degenerate,
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\item there are no triple point.
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\end{enumerate}
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\end{definition}
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\noindent
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By \Cref{def:link_diagram} the following pictures can not be a part of a diagram:
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.1\textwidth}
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\includegraphics[width=0.8\textwidth]
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{LinkDiagram1.png},
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\end{subfigure}
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\begin{subfigure}{0.1\textwidth}
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\includegraphics[width=0.6\textwidth]
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{LinkDiagram2.png},
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\end{subfigure}
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\begin{subfigure}{0.1\textwidth}
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\includegraphics[width=0.8\textwidth]
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{LinkDiagram3.png}.
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\end{subfigure}
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\end{figure}
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\noindent
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There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.
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\begin{lemma}
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Every link admits a link diagram.
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\end{lemma}
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\noindent
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Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).
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We can distinguish two types of crossings: right-handed
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$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.
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\subsection{Reidemeister moves}
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A Reidemeister move is one of the three types of operation on a link diagram as shown below:
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\begin{enumerate}[label=\Roman*]
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\item\hfill\\
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\includegraphics[width=0.6\textwidth]{rm1.png},
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\item\hfill\\\includegraphics[width=0.6\textwidth]{rm2.png},
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\item\hfill\\\includegraphics[width=0.4\textwidth]{rm3.png}.
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\end{enumerate}
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\begin{theorem} [Reidemeister, 1927 ]
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Two diagrams of the same link can
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be deformed into each other by a finite
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sequence of Reidemeister moves
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(and isotopy of the plane).
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\end{theorem}
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%
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%
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%
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%The number of Reidemeister Moves Needed for Unknotting
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%Joel Hass, Jeffrey C. Lagarias
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%(Submitted on 2 Jul 1998)
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% Piotr Sumata, praca magisterska
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% proof - transversality theorem (Thom)
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%Singularities of Differentiable Maps
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%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.
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\subsection{Seifert surface}
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\noindent
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Let $D$ be an oriented diagram of a link $L$.
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We change the diagram by smoothing each crossing:
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\begin{align*}
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\PICorientpluscross \mapsto
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\PICorientLRsplit,\\
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\PICorientminuscross \mapsto
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\PICorientLRsplit.
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\end{align*}
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We smooth all the crossings, so we get
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a disjoint union of circles on the plane.
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Each circle bounds a disks in
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$\mathbb{R}^3$
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(we choose disks that don't intersect).
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For each smoothed crossing we add a twisted band:
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right-handed for a positive and left-handed for a negative one.
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We get an orientable surface $\Sigma$
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such that $\partial \Sigma = L$.\\
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\begin{figure}[h]
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\fontsize{15}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.8\textwidth}{!}
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{\input{images/seifert_alg.pdf_tex}}
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\caption{Constructing a Seifert surface.}
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\label{fig:SeifertAlg}
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}
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\end{figure}
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\noindent
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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 $\partial D_2$.
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.6\textwidth]
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{seifert_connect.png}
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\caption{Connecting two surfaces.}
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\label{fig:SeifertConnect}
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\end{figure}
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\begin{theorem}[Seifert]\label{theo:Seifert}
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Every link in $S^3$ bounds a surface
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$\Sigma$ that is compact, connected
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and orientable.
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Such a surface is called a Seifert surface.
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\end{theorem}
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\begin{figure}[h]
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\fontsize{12}{10}\selectfont
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\centering
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\def\svgwidth{\linewidth}
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\resizebox{1\textwidth}{!}{
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\input{images/torus_1_2_3.pdf_tex}}
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\caption{Genus of an orientable surface.}
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\label{fig:genera}
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\end{figure}
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%
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%
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\begin{definition}
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The three genus $g_3(K)$ ($g(K)$)
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of a knot $K$ is the minimal genus
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of a Seifert surface $\Sigma$ for $K$.
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\end{definition}
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\begin{corollary}
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A knot $K$ is trivial if and only
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$g_3(K) = 0$.
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\end{corollary}
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\noindent
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Remark: there are knots that admit non isotopic
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Seifert surfaces of minimal genus
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(András Juhász, 2008).
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\begin{definition}
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Suppose $\alpha$ and $\beta$ are two
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simple closed curves in $\mathbb{R}^3$.
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On a diagram $L$ consider all crossings
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between $\alpha$ and $\beta$.
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Let $N_+$ be the number
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of positive crossings,
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$N_-$ - negative.
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Then the linking number:
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$\Lk(\alpha, \beta) =
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\frac{1}{2}(N_+ - N_-)$.
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\end{definition}
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\begin{definition}\label{def:lk_via_homo}
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Let $\alpha$ and $\beta$ be
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two disjoint simple closed curves in $S^3$.
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Let $\nu(\beta)$ be a tubular
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neighbourhood of $\beta$.
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The linking number can be interpreted
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via first homology group, where
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$\Lk(\alpha, \beta)$ is equal
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to evaluation of $\alpha$ as element
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of first homology group
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of the complement of $\beta$:
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\[
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\alpha \in H_1(S^3 \setminus
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\nu(\beta), \mathbb{Z})
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\cong \mathbb{Z}.
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\]
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\end{definition}
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\begin{figure}[h]
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\fontsize{10}{8}\selectfont
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\centering
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\def\svgwidth{\linewidth}
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\resizebox{\textwidth}{!}{
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% \centering
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\begin{subfigure}{0.3\textwidth}
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\centering
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\def\svgwidth{\linewidth}
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\resizebox{1\textwidth}{!}{
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\input{images/linking_torus_6_2.pdf_tex}
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}
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\end{subfigure}
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\begin{subfigure}{0.3\textwidth}
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\centering
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\def\svgwidth{\linewidth}
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\resizebox{1\textwidth}{!}{
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\input{images/linking_hopf.pdf_tex}
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}
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\end{subfigure}
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}
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\vspace*{10mm}
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\caption{
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Linking number of a Hopf link (left)
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and a torus link $T(6, 2)$ (right).
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}
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\label{fig:unknot}
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\end{figure}
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\begin{lemma}
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$
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g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) =
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\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}),
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$
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where $b_1$ is first Betti number of a surface $\Sigma$.
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\end{lemma}
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\subsection{Seifert matrix}
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Let $L$ be a link and
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$\Sigma$ be an oriented
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Seifert surface for $L$.
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Choose a basis for
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$H_1(\Sigma, \mathbb{Z})$
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consisting of simple closed curves
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$\alpha_1, \dots, \alpha_n$.
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\noindent
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Let $\alpha_1^+, \dots \alpha_n^+$
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be copies of $\alpha_i$
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lifted up off the surface
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(push up along a vector field
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normal to $\Sigma$).
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Note that elements $\alpha_i$ are
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contained in the Seifert surface while all
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$\alpha_i^+$ don't intersect the surface.
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\noindent
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Let $\Lk(\alpha_i, \alpha_j^+) = \{a_{ij}\}$.
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Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$
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is called a Seifert matrix for $L$.
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Note that by choosing a different basis
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we get a different matrix.
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\begin{figure}[h]
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\fontsize{20}{10}\selectfont
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\centering
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\def\svgwidth{\linewidth}
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\resizebox{0.8\textwidth}{!}{
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\input{images/seifert_matrix.pdf_tex}
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}
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\caption{
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A basis $\alpha_1, \alpha_2$
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of the first homology
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group of a Seifert surface
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and a copy of
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element $\alpha_1$ pushed up
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along vector normal to the Seifert surface.
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}
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\label{fig:alpha_plus}
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\end{figure}
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\begin{theorem}
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The Seifert matrices $S_1$ and $S_2$
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for the same link $L$ are S-equivalent,
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that is, $S_2$ can be obtained from
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$S_1$ by a sequence of following moves:
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\begin{enumerate}[label={(\arabic*)}]
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\item
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$V \rightarrow AVA^T$,
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where $A$ is a matrix
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with integer coefficients,
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\item
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$V \rightarrow
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\begin{pmatrix}
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\begin{array}{c|c}
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V &
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\begin{matrix}
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\ast & 0 \\
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\sdots & \sdots\\
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\ast & 0
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\end{matrix} \\
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\hline
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\begin{matrix}
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\ast & \dots & \ast\\
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0 & \dots & 0
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\end{matrix}
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&
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\begin{matrix}
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0 & 0\\
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1 & 0
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\end{matrix}
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\end{array}
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\end{pmatrix} \quad$
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or
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$\quad
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V \rightarrow
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\begin{pmatrix}
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\begin{array}{c|c}
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V &
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\begin{matrix}
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\ast & 0 \\
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\sdots & \sdots\\
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\ast & 0
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\end{matrix} \\
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\hline
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\begin{matrix}
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\ast & \dots & \ast\\
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0 & \dots & 0
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\end{matrix}
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&
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\begin{matrix}
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0 & 1\\
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0 & 0
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\end{matrix}
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\end{array}
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\end{pmatrix},$
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\item
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inverse of (2).
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\end{enumerate}
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\end{theorem}
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