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+ \begin{picture}(1,0.48333952)%
+ \put(0.36244031,2.57302301){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.30185744\unitlength}\raggedright \end{minipage}}}%
+ \put(0.36244031,2.57302301){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.30185744\unitlength}\raggedright \end{minipage}}}%
\put(0,0){\includegraphics[width=\unitlength,page=1]{linking_torus_6_2.pdf}}%
- \put(0.70608389,0.2487067){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.31355455\unitlength}\raggedright \shortstack{$lk(\alpha, \beta) = 3$}\end{minipage}}}%
- \put(0.02479073,0.28916687){\color[rgb]{1,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.20853214\unitlength}\raggedright $\alpha$\\ \end{minipage}}}%
- \put(0.50411385,0.1062225){\color[rgb]{0,0,1}\makebox(0,0)[lt]{\begin{minipage}{0.20853214\unitlength}\raggedright $\beta$\\ \end{minipage}}}%
+ \put(0.80794467,0.26923855){\color[rgb]{0,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.66035143\unitlength}\raggedright \shortstack{$\Lk(\alpha, \beta) = 3$}\end{minipage}}}%
+ \put(0.02836708,0.33088267){\color[rgb]{1,0,0}\makebox(0,0)[lt]{\begin{minipage}{0.23861542\unitlength}\raggedright $\alpha$\\ \end{minipage}}}%
+ \put(0.57683835,0.12154638){\color[rgb]{0,0,1}\makebox(0,0)[lt]{\begin{minipage}{0.23861542\unitlength}\raggedright $\beta$\\ \end{minipage}}}%
\end{picture}%
\endgroup%
diff --git a/images/linking_torus_6_2.svg b/images/linking_torus_6_2.svg
index 4246f76..b70612b 100644
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+++ b/images/linking_torus_6_2.svg
@@ -10,13 +10,13 @@
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xmlns:sodipodi="http://sodipodi.sourceforge.net/DTD/sodipodi-0.dtd"
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+ width="112.34215mm"
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+ viewBox="0 0 398.06274 192.39946"
id="svg2"
version="1.1"
- inkscape:version="0.91 r13725"
- sodipodi:docname="linking_torus_4_2.svg"
+ inkscape:version="0.92.2 5c3e80d, 2017-08-06"
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$\Sigma$
-
- $\Sigma$
$\Sigma$
-
- $\Sigma$
\shortstack{$lk(\alpha, \beta) = 3$} \shortstack{$\Lk(\alpha, \beta) = 3$} $\alpha$ $\alpha$ $\beta$
+ id="flowPara11317"
+ style="font-size:40.00024796px;line-height:1.25">$\beta$
diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex
index 2328b94..48b7937 100644
--- a/lectures_on_knot_theory.tex
+++ b/lectures_on_knot_theory.tex
@@ -48,13 +48,16 @@
{\bfseries}{}%
{\newline}{}%
\theoremstyle{break}
+
\newtheorem{lemma}{Lemma}[section]
\newtheorem{fact}{Fact}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{example}{Example}[section]
+\newtheorem{problem}{Problem}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{theorem}{Theorem}[section]
+
\newcommand{\contradiction}{%
\ensuremath{{\Rightarrow\mspace{-2mu}\Leftarrow}}}
\newcommand*\quot[2]{{^{\textstyle #1}\big/_{\textstyle #2}}}
@@ -66,6 +69,10 @@
\newcommand{\sdots}{\smash{\vdots}}
+
+
+
+
\AtBeginDocument{\renewcommand{\setminus}{%
\mathbin{\backslash}}}
@@ -73,6 +80,8 @@
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\Gl}{Gl}
+\DeclareMathOperator{\Lk}{lk}
+
\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{%
Lecture\ \thesection}%
@@ -263,12 +272,12 @@ Remark: there are knots that admit non isotopic Seifert surfaces of minimal genu
\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: $lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$.
+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}
\hfill
\\
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 $\beta$:
+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 $\beta$:
\[
\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
@@ -306,7 +315,7 @@ where $b_1$ is first Betti number of $\Sigma$.
\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 $\alpha_1, \dots, \alpha_n$.
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.
+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
@@ -581,19 +590,37 @@ g_{\big| \partial D^2} = f_{\big| \partial D^2.}
&F: \mathbb{C}^2 \rightarrow \mathbb{C} \text{ a polynomial} \\
&F(0) = 0
\end{align*}
-Fact (Milnor Singular Points of Complex Hypersurfaces):
-\end{example}
-%\end{comment}
-An oriented knot is called negative amphichiral if the mirror image $m(K)$ if $K$ is equivalent the reverse knot of $K$. \\
-\begin{example}[Problem]
-Prove that if $K$ is negative amphichiral, then $K \# K$ in
-$\mathbf{C}$
+\end{example}
+????????????
+\\
+\noindent
+as a corollary we see that $K_T^{n, }$ ???? \\
+is not slice unless $m=0$.
+\begin{theorem}
+The map $j: \mathscr{C} \longrightarrow \mathbb{Z}^{\infty}$ is a surjection that maps ${K_n}$ to a linear independent set. Moreover $\mathscr{C} \cong \mathbb{Z}$
+\end{theorem}
+
+
+\begin{fact}[Milnor Singular Points of Complex Hypersurfaces]
+\end{fact}
+%\end{comment}
+\noindent
+An oriented knot is called negative amphichiral if the mirror image $m(K)$ of $K$ is equivalent the reverse knot of $K$: $K^r$. \\
+\begin{problem}
+Prove that if $K$ is negative amphichiral, then $K \# K = 0$ in
+$\mathscr{C}$.
+%
+%\\
+%Hint: $ -K = m(K)^r = (K^r)^r = K$
+\end{problem}
+\begin{example}
+Figure 8 knot is negative amphichiral.
\end{example}
%
%
%
-\section{\hfill\DTMdate{2019-03-18}}
+\section{Concordance group \hfill\DTMdate{2019-03-18}}
\begin{definition}
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
@@ -628,7 +655,7 @@ $K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
-\resizebox{0.8\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
+\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
}
\caption{Sketch for Fakt \ref{fakt:concordance_connected}.}
\label{fig:concordance_sum}
@@ -639,23 +666,40 @@ $K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
$K \# m(K) \sim $ the unknot.
\end{fact}
\noindent
-Let $\mathscr{C}$ denote all equivalent classes for knots. $\mathscr{C}$ is a group under taking connected sums, with neutral element (the class defined by) an unknot and inverse element (a class defined by) a mirror image.\\
-The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).\\
-\begin{example}[Problem]
-Are there in concordance group torsion elements that are not $2$ torsion elements? (open)
-\end{example}
+\begin{theorem}
+Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $\{0\}$ denote class of all knots concordant to a trivial knot.
+$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $\{0\}$ and the inverse element of an element $\{K\} \in \mathscr{C}$ is $-\{K\} = \{mK\}$.
+\end{theorem}
+\begin{fact}
+The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).
+\end{fact}
+\begin{problem}[open]
+Are there in concordance group torsion elements that are not $2$ torsion elements?
+\end{problem}
\noindent
Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
-
+\\
+\\
+\noindent
+Let $\Omega$ be an oriented \\
+???????\\
+Suppose $\Sigma$ is a Seifert matrix with an intersection form ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z}$ (i.e. there are cycles). \\
+??????????????\\
+$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$.
+Let $B^+$ be a push off of $B$ in the positive normal direction such that
+$\partial B^+ = \beta^+$.
+Then
+$\Lk(\alpha, \beta^+) = A \cdot B^+$
%
%
+\\
\section{\hfill\DTMdate{2019-04-08}}
%
%
$X$ is a closed orientable four-manifold. Assume $\pi_1(X) = 0$ (it is not needed to define the intersection form). In particular $H_1(X) = 0$.
$H_2$ is free (exercise).
\begin{align*}
-H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincaré duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z})
+H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z})
\end{align*}
Intersection form:
$H_2(X, \mathbb{Z}) \times
@@ -749,7 +793,7 @@ In general
\section{\hfill\DTMdate{2019-05-20}}
-Let $M$ be compact, oriented, connected four-dimensional manifold. If $H_1(M, \mathbb{Z}) = 0$ then there exists a
+Let $M$ be compact, oriented, connected four-dimensional manifold. If ${H_1(M, \mathbb{Z}) = 0}$ then there exists a
bilinear form - the intersection form on $M$:
\begin{center}
@@ -1006,22 +1050,95 @@ field of fractions
\section{balagan}
-\begin{comment}
-\setlength{\arraycolsep}{2em}
-\newcommand{\lbrce}{\smash{\left.\rule{0pt}{25pt}\right\}}}
-\newcommand{\rbrce}{\smash{\left\{\rule{0pt}{25pt}\right.}}
+\noindent
+\begin{proof}
+By Poincar\'e duality we know that:
+\begin{align*}
+H_3(\Omega, Y) &\cong H^0(\Omega),\\
+H_2(Y) &\cong H^0(Y),\\
+H_2(\Omega) &\cong H^1(\Omega, Y),\\
+H_2(\Omega, Y) &\cong H^1(\Omega).
+\end{align*}
+Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
+= \dim_{\mathbb{Q}} V
+$.
+\end{proof}
+\noindent
+Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on 4\\
+?????\\
+has a subspace of dimension $g_{\Sigma}$ on which it is zero:
\[
- \begin{pmatrix}
- 0 & 0 & 0 \\
- \sdots & \sdots\makebox[0pt][l]{$\lbrce\left\lceil\frac i2\right\rceil$} & \sdots \\
- 0 & 0 & \\
- \hline
-
- & & 0 \\
- & & \makebox[0pt][r]{$\left\lfloor\frac i2\right\rfloor\rbrce$}\sdots \\
- 0 & & 0
- \end{pmatrix}
+V =
+\begin{pmatrix}
+ \begin{array}{c|c}
+ 0 & * \\
+ \hline
+ * & *
+ \end{array}
+\end{pmatrix}
\]
+\begin{align*}
+ \newcommand*{\AddLeft}[1]{%
+ \vadjust{%
+ \vbox to 0pt{%
+ \vss
+ \llap{$%
+ {#1}\left\{
+ \vphantom{
+ \begin{matrix}1\\\vdots\\0\end{matrix}
+ }
+ \right.\kern-\nulldelimiterspace
+ \kern0.5em
+ $}%
+ \kern0pt
+ }%
+ }%
+ }
+V = \qquad
+\begin{pmatrix}
+ 0 & \cdots & 0 & * & \cdots & * \\
+ \vdots & & \vdots & \vdots & &\vdots \\
+ 0 & \cdots & 0 & * & \cdots & *
+ \AddLeft{g_{\Sigma}}\\
+ * & \cdots & * & * & \cdots & * \\
+ \vdots & & \vdots & \vdots & &\vdots \\
+ * & \cdots & * & * & \cdots & *
+ \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
+\end{align*}
+
+\begin{align*}
+\newcommand\coolover[2]{\mathrlap{\smash{\overbrace{\phantom{%
+ \begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
+\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
+ \begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
+\newcommand\coolleftbrace[2]{%
+ #1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
+\newcommand\coolrightbrace[2]{%
+ \left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
+ \vphantom{% phantom stuff for correct box dimensions
+ \begin{matrix}
+ \overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
+ \underbrace{pqr}_{\mbox{$S$}}
+ \end{matrix}}%
+\begin{matrix}% matrix for left braces
+\vphantom{a}\\
+ \coolleftbrace{A}{e \\ y\\ y}\\
+ \coolleftbrace{B}{y \\i \\ m}
+\end{matrix}%
+\begin{bmatrix}
+a & \coolover{R}{b & c & d} & x & \coolover{Z}{x & x}\\
+ e & f & g & h & x & x & x \\
+ y & y & y & y & y & y & y \\
+ y & y & y & y & y & y & y \\
+ y & y & y & y & y & y & y \\
+ i & j & k & l & x & x & x \\
+ m & \coolunder{S}{n & o} & \coolunder{W}{p & x & x} & x
+\end{bmatrix}%
+\begin{matrix}% matrix for right braces
+ \coolrightbrace{x \\ x \\ y\\ y}{T}\\
+ \coolrightbrace{y \\ y \\ x }{U}
+\end{matrix}
+\end{align*}
+
-\end{comment}
\end{document}