some new text

This commit is contained in:
Maria Marchwicka 2019-06-07 15:53:13 +02:00
parent aa683e2eab
commit 94e0077612

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@ -157,7 +157,7 @@ a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png},
\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
\end{itemize}
\end{example}
%
@ -214,7 +214,7 @@ Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing
\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]
\begin{figure}[h]
\fontsize{15}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
@ -276,21 +276,21 @@ Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can
\begin{example}
\begin{itemize}
\item
Hopf link
Hopf link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}}
\resizebox{0.4\textwidth}{!}{\input{images/linking_hopf.pdf_tex}},
}
\end{figure}
\item
$T(6, 2)$ link
$T(6, 2)$ link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}
\resizebox{0.4\textwidth}{!}{\input{images/linking_torus_6_2.pdf_tex}}.
}
\end{figure}
\end{itemize}
@ -301,7 +301,7 @@ Let $L$ be a link and $\Sigma$ be an oriented Seifert surface for $L$. Choose a
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]
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
@ -521,7 +521,7 @@ If $K$ is a trefoil then we can take
$S = \begin{pmatrix}
-1 & -1 \\
0 & -1
\end{pmatrix}$.
\end{pmatrix}$. Then
\[
\Delta_K(t) = \det
\begin{pmatrix}
@ -529,7 +529,7 @@ $S = \begin{pmatrix}
1 & -t +1
\end{pmatrix}
= (t -1)^2 + t = t^2 - t +1 \ne 1
\Rightarrow \text{trefoil is not trivial}
\Rightarrow \text{trefoil is not trivial.}
\]
\end{example}
\begin{fact}
@ -554,6 +554,7 @@ If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb
\end{proof}
\begin{example}
There are not trivial knots with Alexander polynomial equal $1$, for example:
\includegraphics[width=0.3\textwidth]{11n34.png}
$\Delta_{11n34} \equiv 1$.
\end{example}
%removing one disk from surface doesn't change $H_1$ (only $H_2$)
@ -582,6 +583,17 @@ A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk
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
$\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\} $.
\end{definition}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/concordance.pdf_tex}}
}
\end{figure}
\noindent
Let $m(K)$ denote a mirror image of a knot $K$.
\begin{fact}
@ -666,7 +678,7 @@ Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Se
$H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where
$A = V \times V^T$, where $n = \rank V$.
%\input{ink_diag}
\begin{figure}[H]
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
@ -727,11 +739,11 @@ Then the intersection form can be degenerated in the sense that:
\begin{align*}
H_2(M, \mathbb{Z})
\times H_2(M, \mathbb{Z})
\longrightarrow
\mathbb{Z}\\
H_2(M, \mathbb{Z}) \longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) \mapsto \mathbb{Z}\\
a \mapsto (a, \_) H_2(M, \mathbb{Z})
&\longrightarrow
\mathbb{Z} \quad&
H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) &\mapsto \mathbb{Z} \quad&
a &\mapsto (a, \_) H_2(M, \mathbb{Z})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$.
\\???????????????\\
@ -764,23 +776,27 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\ma
\end{fact}
\noindent
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.
\begin{flalign*}
\xi \in S^1 \setminus \{ \pm 1\}
\begin{align*}
&\xi \in S^1 \setminus \{ \pm 1\}
\quad
p_{\xi} =
(t - \xi)(1 - \xi^{-1}) t^{-1}&\\
\xi \in \mathbb{R} \setminus \{ \pm 1\}
(t - \xi)(t - \xi^{-1}) t^{-1}
\\
&\xi \in \mathbb{R} \setminus \{ \pm 1\}
\quad
q_{\xi} = (t - \xi)(1 - \xi^{-1}) t^{-1}&\\
q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
\\
&
\xi \notin \mathbb{R} \cup S^1 \quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(1 - \xi^{-1})(1 - \overbar{\xi}^{-1}) t^{-2}&\\
\Lambda = \mathbb{R}[t, t^{-1}]&\\
\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})(t - \overbar{\xi}^{-1}) t^{-2}\\
&
\Lambda = \mathbb{R}[t, t^{-1}]\\
&\text{Then: } H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
\oplus
\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
\end{flalign*}
\end{align*}
We can make this composition orthogonal with respect to the Blanchfield paring.
\vspace{0.5cm}\\
Historical remark: