some inkspace pictures

This commit is contained in:
Maria Marchwicka 2019-06-03 04:22:10 +02:00
parent 16f96e3333
commit b9d956bfef
29 changed files with 21617 additions and 2631 deletions

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@ -1,824 +0,0 @@
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{\bfseries}{}% {\bfseries}{}%
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%\newpage %\newpage
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\section{\hfill\DTMdate{2019-02-25}} \section{Basic definitions \hfill\DTMdate{2019-02-25}}
\begin{definition} \begin{definition}
A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$: A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^3$:
\begin{align*} \begin{align*}
@ -96,12 +103,13 @@ A knot $K$ in $S^3$ is a smooth (PL - smooth) embedding of a circle $S^1$ in $S^
\end{definition} \end{definition}
\noindent \noindent
Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$. Usually we think about a knot as an image of an embedding: $K = \varphi(S^1)$.
\begin{example} \begin{example}
\begin{itemize} \begin{itemize}
\item \item
Knots: Knots:
\includegraphics[width=0.08\textwidth]{unknot.png}, \includegraphics[width=0.08\textwidth]{unknot.png} (unknot),
\includegraphics[width=0.08\textwidth]{trefoil.png}. \includegraphics[width=0.08\textwidth]{trefoil.png} (trefoil).
\item \item
Not knots: Not knots:
\includegraphics[width=0.12\textwidth]{not_injective_knot.png} \includegraphics[width=0.12\textwidth]{not_injective_knot.png}
@ -172,7 +180,7 @@ Let $D$ be a diagram of an oriented link (to each component of a link we add an
We can distinguish two types of crossings: right-handed 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. $\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.
\section*{Reidemeister moves} \subsection{Reidemeister moves}
A Reidemeister move is one of the three types of operation on a link diagram as shown below: A Reidemeister move is one of the three types of operation on a link diagram as shown below:
\begin{enumerate}[label=\Roman*] \begin{enumerate}[label=\Roman*]
\item\hfill\\ \item\hfill\\
@ -197,7 +205,7 @@ deformed into each other by a finite sequence of Reidemeister moves (and isotopy
%Singularities of Differentiable Maps %Singularities of Differentiable Maps
%Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M. %Authors: Arnold, V.I., Varchenko, Alexander, Gusein-Zade, S.M.
\subsection*{Seifert surface} \subsection{Seifert surface}
\noindent \noindent
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing: Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing:
\begin{align*} \begin{align*}
@ -221,7 +229,7 @@ Note: in general the obtained surface doesn't need to be connected, but by takin
\begin{figure}[H] \begin{figure}[H]
\begin{center} \begin{center}
\includegraphics[width=0.6\textwidth]{seifert_connect.png} \includegraphics[width=0.4\textwidth]{seifert_connect.png}
\end{center} \end{center}
\caption{Connecting two surfaces.} \caption{Connecting two surfaces.}
\label{fig:SeifertConnect} \label{fig:SeifertConnect}
@ -259,13 +267,16 @@ On a diagram $L$ consider all crossings between $\alpha$ and $\beta$. Let $N_+$
\end{definition} \end{definition}
\hfill \hfill
\\ \\
Let $\nu(\beta)$ be a tubular neighbourhood of a closed simple curve $\beta$. 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 in complement of $\beta$ in $S^3$: 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 $\beta$:
\[ \[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\] \alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
\begin{example} \begin{example}
\begin{itemize}\hfill \begin{itemize}
\item \item
Hopf link\hfill Hopf link
\begin{figure}[H] \begin{figure}[H]
\fontsize{20}{10}\selectfont \fontsize{20}{10}\selectfont
\centering{ \centering{
@ -274,7 +285,7 @@ Hopf link\hfill
} }
\end{figure} \end{figure}
\item \item
$T(6, 2)$ link\hfill $T(6, 2)$ link
\begin{figure}[H] \begin{figure}[H]
\fontsize{20}{10}\selectfont \fontsize{20}{10}\selectfont
\centering{ \centering{
@ -285,94 +296,263 @@ $T(6, 2)$ link\hfill
\end{itemize} \end{itemize}
\end{example} \end{example}
Let $L$ be a link and $\Sigma$ be a Seifert surface for $L$. Choose a basis for $H_1(\Sigma, \mathbb{Z})$ consisting of simple closed $\alpha_1, \dots, \alpha_n$. \subsection{Seifert matrix}
Let $\alpha_1^+, \dots \alpha_n^+$ be copies of $\alpha_i$ lifted up off the surface. Let $lk(\alpha_i, \alpha_i^+) = \{a_{ij}\}$. Then the matrix $S = \{a_{ij}\}_{i, j =1}^n$ is called a Seifert matrix for $L$. 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} \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: 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*)}] \begin{enumerate}[label={(\arabic*)}]
\item \item
$V \rightarrow AVA^T$ for $A \in $ $V \rightarrow AVA^T$, where $A$ is a matrix with integer coefficients,
\item \item
$V \rightarrow $V \rightarrow
\begin{pmatrix} \begin{pmatrix}
\alpha & * \\ \begin{array}{c|c}
\gamma^{*} & \delta V &
\end{pmatrix} \begin{matrix}
$\\ \ast & 0 \\
\[ \sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline
\begin{matrix}
\ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 0\\
1 & 0
\end{matrix}
\end{array}
\end{pmatrix} \quad$
or
$\quad
V \rightarrow
\begin{pmatrix} \begin{pmatrix}
\begin{array}{c|c} \begin{array}{c|c}
\epsilon' [T|_A]\epsilon & \ast \\ V &
\begin{matrix}
\ast & 0 \\
\sdots & \sdots\\
\ast & 0
\end{matrix} \\
\hline \hline
0 & _{\overline{B}'} [\overline{T}] \begin{matrix}
_{\overline{B}\vphantom{\overline{B}'}} \ast & \dots & \ast\\
0 & \dots & 0
\end{matrix}
&
\begin{matrix}
0 & 1\\
0 & 0
\end{matrix}
\end{array} \end{array}
\end{pmatrix} \end{pmatrix}$
\]\\
\[\left|
\begin{array}{cr}
Q & \begin{matrix} 0 \\ 0 \end{matrix} \\
\begin{matrix} 2 & 3 \end{matrix} & -1
\end{array}
\right|\]
\\
\[
\left[
\begin{array}{c@{}c@{}c}
\left[\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array}\right] & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \left[\begin{array}{ccc}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32} & b_{33}\\
\end{array}\right] & \mathbf{0}\\
\mathbf{0} & \mathbf{0} & \left[ \begin{array}{cc}
c_{11} & c_{12} \\
c_{21} & c_{22} \\
\end{array}\right] \\
\end{array}\right]
\] \\
\[
\begin{bmatrix}
\begin{bmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}\\
\end{bmatrix} & \mathbf{0} & \mathbf{0} \\
\mathbf{0} & \begin{bmatrix}
b_{11} & b_{12} & b_{13}\\
b_{21} & b_{22} & b_{23}\\
b_{31} & b_{32} & b_{33}\\
\end{bmatrix} & \mathbf{0} \\
\mathbf{0} & \mathbf{0} & \begin{bmatrix}
c_{11} & c_{12}\\
c_{21} & c_{22}\\
\end{bmatrix} \\
\end{bmatrix}
\]\\
\setlength{\arraycolsep}{2em}
\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}}
\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}}
\newcommand{\sdots}{\smash{\vdots}}
\[
\begin{pmatrix}
0 & 0 & 0 \\
\sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\
0 & 0 & \\
& & 0 \\
& & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\
0 & & 0
\end{pmatrix}
\]
\item \item
inverse of (2) inverse of (2)
\end{enumerate} \end{enumerate}
\end{theorem} \end{theorem}
\section{\hfill\DTMdate{2019-03-04}}
\begin{theorem}
For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$
\end{theorem}
\begin{proof}("joke")\\
Let $K \in S^3$ be a knot and $N = \nu(K)$ be its tubular neighbourhood. Because $K$ and $N$ are homotopy equivalent, we get:
\begin{align*}
H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K).
\end{align*}
Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients:
\begin{center}
\begin{tikzcd}
[
column sep=0cm, fill=none,
row sep=small,
ar symbol/.style =%
{draw=none,"\textstyle#1" description,sloped},
isomorphic/.style = {ar symbol={\cong}},
]
&\mathbb{Z}
\\
& H^0(S^3) \ar[u,isomorphic] \to
&H^0(S^3 \setminus N) \to
\\
\to H^1(S^3, S^3 \setminus N) \to
& H^1(S^3) \to
& H^1(S^3\setminus N) \to
\\
& 0 \ar[u,isomorphic]&
\\
\to H^2(S^3, S^3 \setminus N) \to
& H^2(S^3) \ar[u,isomorphic] \to
& H^2(S^3\setminus N) \to
\\
\to H^3(S^3, S^3\setminus N)\to
& H^3(S) \to
& 0
\\
& \mathbb{Z} \ar[u,isomorphic] &\\
\end{tikzcd}
\end{center}
\[
H^* (S^3, S^3 \setminus N) \cong H^* (N, \partial N)
\]
\\
??????????????
\\
\end{proof}
\begin{definition}
Let $S$ be a Seifert matrix for a knot $K$. The Alexander polynomial $\Delta_K(t)$ is a Laurent polynomial:
\[
\Delta_K(t) := \det (tS - S^T) \in
\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]
\]
\end{definition}
\begin{theorem}
$\Delta_K(t)$ is well defined up to multiplication by $\pm t^k$, for $k \in \mathbb{Z}$.
\end{theorem}
\begin{proof}
We need to show that $\Delta_K(t)$ doesn't depend on $S$-equivalence relation.
\begin{enumerate}[label={(\arabic*)}]
\item Suppose $S\prime = CSC^T$, $C \in \Gl(n, \mathbb{Z})$ (matrices invertible over $\mathbb{Z}$). Then $\det C = 1$ and:
\begin{align*}
&\det(tS\prime - S\prime^T) =
\det(tCSC^T - (CSC^T)^T) =\\
&\det(tCSC^T - CS^TC^T) =
\det C(tS - S^T)C^T =
\det(tS - S^T)
\end{align*}
\item
Let \\
$ A := t
\begin{pmatrix}
\begin{array}{c|c}
S &
\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}
-
\begin{pmatrix}
\begin{array}{c|c}
S^T &
\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}
=
\begin{pmatrix}
\begin{array}{c|c}
tS - S^T &
\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\\
t & 0
\end{matrix}
\end{array}
\end{pmatrix}
$
\\
\\
Using the Laplace expansion we get $\det A = \pm t \det(tS - S^T)$.
\end{enumerate}
\end{proof}
%
%
%
\begin{example}
If $K$ is a trefoil then we can take
$S = \begin{pmatrix}
-1 & -1 \\
0 & -1
\end{pmatrix}$.
\[
\Delta_K(t) = \det
\begin{pmatrix}
-t + 1 & -t\\
1 & -t +1
\end{pmatrix}
= (t -1)^2 + t = t^2 - t +1 \ne 1
\Rightarrow \text{trefoil is not trivial}
\]
\end{example}
\begin{fact}
$\Delta_K(t)$ is symmetric.
\end{fact}
\begin{proof}
Let $S$ be an $n \times n$ matrix.
\begin{align*}
&\Delta_K(t^{-1}) = \det (t^{-1}S - S^T) = (-t)^{-n} \det(tS^T - S) = \\
&(-t)^{-n} \det (tS - S^T) = (-t)^{-n} \Delta_K(t)
\end{align*}
If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n} \Delta_K(t)$.
\end{proof}
\begin{lemma}
\begin{align*}
\frac{1}{2} \deg \Delta_K(t) \leq g_3(K),
\text{ where } deg (a_n t^n + \cdots + a_1 t^l )= k - l.
\end{align*}
\end{lemma}
\begin{proof}
\end{proof}
%removing one disk from surface doesn't change $H_1$ (only $H_2$)
\section{} \section{}
\begin{example} \begin{example}
\begin{align*} \begin{align*}
@ -389,17 +569,6 @@ Prove that if $K$ is negative amphichiral, then $K \# K$ in
$\mathbf{C}$ $\mathbf{C}$
\end{example} \end{example}
\section{\hfill\DTMdate{2019-03-04}}
\begin{proof}("joke")\\
Let $K \in S^3$ be a knot and $N$ be its tubular neighbourhood.
\begin{align*}
H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K)
\end{align*}
For a pair $(S^3, S^3 \setminus N)$ we have:
\begin{align*}
H^0(S^3)
\end{align*}
\end{proof}
\section{\hfill\DTMdate{2019-03-18}} \section{\hfill\DTMdate{2019-03-18}}
\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. \\
@ -447,7 +616,11 @@ $H_2(X, \mathbb{Z}) \times
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular. H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ - symmetric, non singular.
\\ \\
Let $A$ and $B$ be closed, oriented surfaces in $X$. Let $A$ and $B$ be closed, oriented surfaces in $X$.
\begin{proposition}
$A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes.
%$A \cdot B$ gives the pairing as ??
\end{proposition}
\section{\hfill\DTMdate{2019-04-15}} \section{\hfill\DTMdate{2019-04-15}}
In other words:\\ In other words:\\
@ -459,14 +632,14 @@ of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection
\end{align*} \end{align*}
In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\ In particular $\mid \det A\mid = \# H_1(Y, \mathbb{Z}$.\\
That means - what is happening on boundary is a measure of degeneracy. That means - what is happening on boundary is a measure of degeneracy.
\\
\vspace{1cm}
\begin{center} \begin{center}
\begin{tikzcd} \begin{tikzcd}
[ [
column sep=tiny, column sep=tiny,
row sep=small, row sep=small,
ar symbol/.style = {draw=none,"\textstyle#1" description,sloped}, ar symbol/.style =%
{draw=none,"\textstyle#1" description,sloped},
isomorphic/.style = {ar symbol={\cong}}, isomorphic/.style = {ar symbol={\cong}},
] ]
H_1(Y, \mathbb{Z}) & H_1(Y, \mathbb{Z}) &
@ -480,6 +653,7 @@ H_1(Y, \mathbb{Z})&
\end{tikzcd} \end{tikzcd}
$(a, b) \mapsto aA^{-1}b^T$ $(a, b) \mapsto aA^{-1}b^T$
\end{center} \end{center}
The intersection form on a four-manifold determines the linking on the boundary. \\ The intersection form on a four-manifold determines the linking on the boundary. \\
\noindent \noindent
@ -780,4 +954,22 @@ field of fractions
\section{balagan} \section{balagan}
\begin{comment}
\setlength{\arraycolsep}{2em}
\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}}
\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}}
\[
\begin{pmatrix}
0 & 0 & 0 \\
\sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\
0 & 0 & \\
\hline
& & 0 \\
& & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\
0 & & 0
\end{pmatrix}
\]
\end{comment}
\end{document} \end{document}