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@ -157,7 +157,7 @@ a Whitehead link:
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\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
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\item
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Borromean link:
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\includegraphics[width=0.1\textwidth]{BorromeanRings.png},
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\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
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\end{itemize}
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\end{example}
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%
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@ -214,7 +214,7 @@ Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing
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\end{align*}
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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 $\partial \Sigma = L$.\\
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\begin{figure}[H]
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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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@ -276,21 +276,21 @@ Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can
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\begin{example}
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\begin{itemize}
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\item
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Hopf link
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Hopf link:
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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{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}}
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\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}},
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}
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\end{figure}
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\item
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$T(6, 2)$ link
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$T(6, 2)$ link:
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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{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}
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\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}.
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}
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\end{figure}
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\end{itemize}
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@ -301,7 +301,7 @@ Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a
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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.
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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.
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\begin{figure}[H]
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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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@ -521,7 +521,7 @@ If $K$ is a trefoil then we can take
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$S = \begin{pmatrix}
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-1 & -1 \\
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0 & -1
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\end{pmatrix}$.
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\end{pmatrix}$. Then
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\[
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\Delta_K(t) = \det
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\begin{pmatrix}
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@ -529,7 +529,7 @@ $S = \begin{pmatrix}
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1 & -t +1
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\end{pmatrix}
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= (t -1)^2 + t = t^2 - t +1 \ne 1
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\Rightarrow \text{trefoil is not trivial}
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\Rightarrow \text{trefoil is not trivial.}
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\]
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\end{example}
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\begin{fact}
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@ -554,6 +554,7 @@ If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb
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\end{proof}
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\begin{example}
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There are not trivial knots with Alexander polynomial equal $1$, for example:
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\includegraphics[width=0.3\textwidth]{11n34.png}
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$\Delta_{11n34} \equiv 1$.
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\end{example}
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%removing one disk from surface doesn't change $H_1$ (only $H_2$)
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@ -582,6 +583,17 @@ A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk
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Two knots $K$ and $K^{\prime}$ are called (smoothly) concordant if there exists an annulus $A$ that is smoothly embedded in $S^3 \times [0, 1]$ such that
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$\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\} $.
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\end{definition}
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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}{!}{\input{images/concordance.pdf_tex}}
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}
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\end{figure}
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\noindent
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Let $m(K)$ denote a mirror image of a knot $K$.
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\begin{fact}
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@ -666,7 +678,7 @@ Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Se
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$H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where
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$A = V \times V^T$, where $n = \rank V$.
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%\input{ink_diag}
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\begin{figure}[H]
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\begin{figure}[h]
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\fontsize{40}{10}\selectfont
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\centering{
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\def\svgwidth{\linewidth}
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@ -727,11 +739,11 @@ Then the intersection form can be degenerated in the sense that:
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\begin{align*}
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H_2(M, \mathbb{Z})
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\times H_2(M, \mathbb{Z})
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\longrightarrow
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\mathbb{Z}\\
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H_2(M, \mathbb{Z}) \longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
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(a, b) \mapsto \mathbb{Z}\\
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a \mapsto (a, \_) H_2(M, \mathbb{Z})
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&\longrightarrow
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\mathbb{Z} \quad&
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H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
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(a, b) &\mapsto \mathbb{Z} \quad&
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a &\mapsto (a, \_) H_2(M, \mathbb{Z})
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\end{align*}
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has coker precisely $H_1(Y, \mathbb{Z})$.
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\\???????????????\\
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@ -764,23 +776,27 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\ma
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\end{fact}
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\noindent
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Note that $\mathbb{Z}$ is not PID. Therefore we don't have primer decomposition of this moduli. We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition.
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\begin{flalign*}
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\xi \in S^1 \setminus \{ \pm 1\}
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\begin{align*}
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&\xi \in S^1 \setminus \{ \pm 1\}
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\quad
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p_{\xi} =
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(t - \xi)(1 - \xi^{-1}) t^{-1}&\\
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\xi \in \mathbb{R} \setminus \{ \pm 1\}
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(t - \xi)(t - \xi^{-1}) t^{-1}
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\\
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&\xi \in \mathbb{R} \setminus \{ \pm 1\}
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\quad
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q_{\xi} = (t - \xi)(1 - \xi^{-1}) t^{-1}&\\
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q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
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\\
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&
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\xi \notin \mathbb{R} \cup S^1 \quad
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q_{\xi} = (t - \xi)(t - \overbar{\xi})(1 - \xi^{-1})(1 - \overbar{\xi}^{-1}) t^{-2}&\\
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\Lambda = \mathbb{R}[t, t^{-1}]&\\
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\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
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q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})(t - \overbar{\xi}^{-1}) t^{-2}\\
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&
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\Lambda = \mathbb{R}[t, t^{-1}]\\
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&\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
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( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
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\oplus
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\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
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(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
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\end{flalign*}
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\end{align*}
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We can make this composition orthogonal with respect to the Blanchfield paring.
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\vspace{0.5cm}\\
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Historical remark:
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