second lecture allmost done
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Maria Marchwicka
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@ -70,7 +70,7 @@ Borromean link:
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A link diagram $D_{\pi}$ is a picture over projection $\pi$ of a link $L$ in $\mathbb{R}^3$($S^3$) to $\mathbb{R}^2$ ($S^2$) such that:
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\begin{enumerate}[label={(\arabic*)}]
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\item
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${D_{\pi}}_{\big|L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
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$D_{\pi |_L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiagram1.png},
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\item the double points are not degenerate: \includegraphics[width=0.03\textwidth]{LinkDiagram2.png},
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\item there are no triple point: \includegraphics[width=0.05\textwidth]{LinkDiagram3.png}.
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\end{enumerate}
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@ -139,6 +139,7 @@ Note: the obtained surface isn't unique and in general doesn't need to be connec
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\end{figure}
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\begin{theorem}[Seifert]
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\label{theo:Seifert}
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Every link in $S^3$ bounds a surface $\Sigma$ that is compact, connected and orientable. Such a surface is called a Seifert surface.
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\end{theorem}
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%
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@ -168,13 +169,13 @@ Remark: there are knots that admit non isotopic Seifert surfaces of minimal genu
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Suppose $\alpha$ and $\beta$ are two simple closed curves in $\mathbb{R}^3$.
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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: $\Lk(\alpha, \beta) = \frac{1}{2}(N_+ - N_-)$.
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\end{definition}
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\hfill
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\\
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\begin{definition}
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\label{def:lk_via_homo}
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Let $\alpha$ and $\beta$ be two disjoint simple cross curves in $S^3$.
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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 $\beta$:
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\[
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\alpha \in H_1(S^3 \setminus \nu(\beta), \mathbb{Z}) \cong \mathbb{Z}.\]
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\end{definition}
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\begin{example}
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\begin{itemize}
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\subsection{Existence of Seifert surface - second proof}
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%\begin{theorem}
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%For any knot $K \subset S^3$ there exists a connected, compact and orientable surface $\Sigma(K)$ such that $\partial \Sigma(K) = K$
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%\end{theorem}
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\begin{proof}(Theorem \ref{theo:Seifert})\\
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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:
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\begin{align*}
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H^1(S^3 \setminus N ) \cong H^1(S^3 \setminus K).
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\end{align*}
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Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setminus N)$ with integer coefficients:
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|
||||
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}
|
||||
\begin{align*}
|
||||
N \cong & D^2 \times S^1\\
|
||||
\partial N \cong & S^1 \times S^1\\
|
||||
H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\
|
||||
\\
|
||||
H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}
|
||||
\end{align*}
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[row sep=huge]
|
||||
H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] &
|
||||
H^1(N \setminus K) \arrow[d,"\Theta"] \\
|
||||
{[S^3 \setminus K, S^1]} \arrow[r,]&
|
||||
{[N \setminus K, S^1]}
|
||||
\end{tikzcd}
|
||||
\end{equation*}
|
||||
\noindent
|
||||
$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface.
|
||||
%
|
||||
%
|
||||
% Thom isomorphism,
|
||||
\end{proof}
|
||||
\subsection{Alexander polynomial}
|
||||
\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
|
||||
$S = \begin{pmatrix}
|
||||
-1 & -1 \\
|
||||
0 & -1
|
||||
\end{pmatrix}$. Then
|
||||
\[
|
||||
\Delta_K(t) = \det
|
||||
\begin{pmatrix}
|
||||
-t + 1 & -t\\
|
||||
1 & -t +1
|
||||
\end{pmatrix}
|
||||
= (t -1)^2 + t = t^2 - t +1 \ne 1
|
||||
\Rightarrow \text{trefoil is not trivial.}
|
||||
\]
|
||||
\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 + \dots + a_1 t^l )= k - l.
|
||||
\end{align*}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$.
|
||||
\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}
|
||||
|
||||
\subsection{Decomposition of $3$-sphere}
|
||||
We know that $3$ - sphere can be obtained by gluing two solid tori:
|
||||
$S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2)$. So the complement of solid torus in $S^3$ is another solid torus.\\
|
||||
Analytically it can be describes as follow.
|
||||
Take $(z_1, z_2) \in \mathbb{C}$ such that $max(\mid z_1 \mid, \mid z_2\mid) = 1
|
||||
$. Define following sets: $S_1 = \{ (z_1, z_2) \in S^3: \mid z_1 \mid = 0\} \cong S^1 \times D^2 $ and $S_2 = \{(z_1, z_2) \in S ^3: \mid z_2 \mid = 1 \} \cong D^2 \times S^1$. The intersection $S_1 \cap S_2 = \{(z_1, z_2): \mid z_1 \mid = \mid z_2 \mid = 1 \} \cong S^1 \times S^1$
|
||||
\begin{figure}[h]
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}}
|
||||
\caption{The complement of solid torus in $S^3$ is another solid torus.}
|
||||
\label{fig:sphere_as_tori}
|
||||
}
|
||||
\end{figure}
|
||||
|
||||
\subsection{Dehn lemma and sphere theorem}
|
||||
%removing one disk from surface doesn't change $H_1$ (only $H_2$)
|
||||
%
|
||||
%
|
||||
%
|
||||
\begin{lemma}[Dehn]
|
||||
Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f\big|_{\partial D^2}$ is an embedding. Then there exists an embedding
|
||||
${D^2 \overset{g}\longhookrightarrow M}$ such that:
|
||||
\[
|
||||
g\big|_{\partial D^2} = f\big|_{\partial D^2.}
|
||||
\]
|
||||
\end{lemma}
|
||||
\noindent
|
||||
Remark: Dehn lemma doesn't hold for dimension four.\\
|
||||
Let $M$ be connected, compact three manifold with boundary.
|
||||
Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big|_{\partial D^2}$ is non-trivial loop in $\partial M$.
|
||||
\begin{theorem}[Sphere theorem]
|
||||
Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial.
|
||||
\end{theorem}
|
||||
\begin{problem}
|
||||
Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$.
|
||||
\end{problem}
|
||||
\begin{corollary}
|
||||
Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Let $N$ be a tubular neighbourhood of a knot $K$ and $M = S^3 \setminus N$ its complement. Then $\partial M = S^1 \times S^1$. Let $f : \pi_1(\partial M ) \longrightarrow \pi_1(M)$.
|
||||
If $\pi_1(M)$ is infinite cyclic group then the map $f$ is non-trivial. Suppose ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$. There is a map $g: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $g(\partial D^2) = \lambda$. By Dehn's lemma there exists an embedding ${h: (D^2, \partial D^2) \longhookrightarrow (M, \partial M)}$ such that
|
||||
$h\big|_{\partial D^2} = f \big|_{\partial D^2}$ and $h(\partial D^2) = \lambda$.
|
||||
Let $\Sigma$ be a union of the annulus and the image of $\partial D^2$.
|
||||
\\???? $g_3$?\\
|
||||
If $g(\Sigma) = 0$, then $K$ is trivial. \\
|
||||
Now we should proof that:
|
||||
\[
|
||||
H_1(M) \cong \mathbb{Z} \Longrightarrow \lambda \in \ker ( \pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)).
|
||||
\]
|
||||
\begin{figure}[h]
|
||||
\fontsize{40}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.4\textwidth}{!}{\input{images/torus_lambda.pdf_tex}}
|
||||
}
|
||||
\caption{$\mu$ is a meridian and $\lambda$ is a longitude.}
|
||||
\label{fig:meridian_and_longitude}
|
||||
\end{figure}
|
||||
Choose a meridian $\mu$ such that $\Lk (\mu, K) = 1$. Recall the definition of linking number via homology group (Definition \ref{def:lk_via_homo}).
|
||||
$[\mu]$ represents the generator of $H_1(S^3\setminus K, \mathbb{X})$. From definition of $\lambda$ we know that $\lambda$ is trivial in $H_1(M)$ ($\Lk(\lambda, K) =0$, therefore $[\lambda]$ was trivial in $pi_1(M)$). If $K$ is non-trivial then $\lambda$ is non-trivial in $\pi_1(M)$, but it is trivial in $H_1(M)$.
|
||||
\end{proof}
|
||||
116
lec_4.tex
Normal file
116
lec_4.tex
Normal file
@ -0,0 +1,116 @@
|
||||
\begin{definition}
|
||||
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}
|
||||
\begin{definition}
|
||||
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
|
||||
Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
|
||||
\end{definition}
|
||||
|
||||
|
||||
|
||||
\noindent
|
||||
Let $m(K)$ denote a mirror image of a knot $K$.
|
||||
\begin{fact}
|
||||
For any $K$, $K \# m(K)$ is slice.
|
||||
\end{fact}
|
||||
\begin{fact}
|
||||
Concordance is an equivalence relation.
|
||||
\end{fact}
|
||||
\begin{fact}\label{fact:concordance_connected}
|
||||
If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
|
||||
$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
|
||||
|
||||
\begin{figure}[h]
|
||||
\fontsize{10}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
|
||||
}
|
||||
\caption{Sketch for Fact \ref{fact:concordance_connected}.}
|
||||
\label{fig:concordance_sum}
|
||||
\end{figure}
|
||||
|
||||
\end{fact}
|
||||
\begin{fact}
|
||||
$K \# m(K) \sim $ the unknot.
|
||||
\end{fact}
|
||||
\noindent
|
||||
\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.
|
||||
\\
|
||||
\begin{figure}[h]
|
||||
\fontsize{20}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
|
||||
}
|
||||
\caption{$Y = F \cup \Sigma$ is a smooth close surface.}
|
||||
\label{fig:closed_surface}
|
||||
\end{figure}
|
||||
\noindent
|
||||
\\
|
||||
Pontryagin-Thom construction tells us that there exists a compact three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
|
||||
Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
|
||||
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. 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
|
||||
$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero.
|
||||
\\
|
||||
?????????????????
|
||||
\\
|
||||
Let us consider following maps:
|
||||
\[
|
||||
\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
|
||||
\]
|
||||
Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$.
|
||||
%
|
||||
\\
|
||||
????????????\\
|
||||
%
|
||||
%
|
||||
\begin{proposition}
|
||||
\[
|
||||
\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y),
|
||||
\]
|
||||
where $b_1$ is first Betti number.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
\begin{align*}
|
||||
& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to
|
||||
\\
|
||||
\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
|
||||
\to & H_1(Y) \to \H_1(\Omega) \to H_1(\Omega, Y) \to \\
|
||||
\to & H_0(Y) \to H_0(\Omega) \to 0
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
\begin{figure}[h]
|
||||
\fontsize{10}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
|
||||
}
|
||||
%\caption{Sketch for Fact %%\label{fig:concordance_m}
|
||||
\end{figure}
|
||||
Binary file not shown.
@ -90,8 +90,11 @@
|
||||
\DeclareMathOperator{\Lk}{lk}
|
||||
\DeclareMathOperator{\pt}{\{pt\}}
|
||||
|
||||
|
||||
\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{%
|
||||
\titleformat{\subsection}{%
|
||||
\normalfont \fontsize{12}{15}\bfseries}{%
|
||||
}{.0ex plus .2ex}{}
|
||||
\titleformat{\section}{%
|
||||
\normalfont \fontsize{13}{15} \bfseries}{%
|
||||
Lecture\ \thesection}%
|
||||
{2.3ex plus .2ex}{}
|
||||
\titlespacing*{\section}
|
||||
@ -114,249 +117,12 @@
|
||||
\section{Basic definitions \hfill\DTMdate{2019-02-25}}
|
||||
\input{lec_1.tex}
|
||||
|
||||
\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}
|
||||
\begin{align*}
|
||||
N \cong & D^2 \times S^1\\
|
||||
\partial N \cong & S^1 \times S^1\\
|
||||
H^1(N, \partial N) \cong & \mathbb{Z} \oplus \mathbb{Z}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N)\\
|
||||
\\
|
||||
H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}
|
||||
\end{align*}
|
||||
\begin{equation*}
|
||||
\begin{tikzcd}[row sep=huge]
|
||||
H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] &
|
||||
H^1(N \setminus K) \arrow[d,"\Theta"] \\
|
||||
{[S^3 \setminus K, S^1]} \arrow[r,]&
|
||||
{[N \setminus K, S^1]}
|
||||
\end{tikzcd}
|
||||
\end{equation*}
|
||||
\noindent
|
||||
$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface.
|
||||
%
|
||||
%
|
||||
% Thom isomorphism,
|
||||
\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
|
||||
$S = \begin{pmatrix}
|
||||
-1 & -1 \\
|
||||
0 & -1
|
||||
\end{pmatrix}$. Then
|
||||
\[
|
||||
\Delta_K(t) = \det
|
||||
\begin{pmatrix}
|
||||
-t + 1 & -t\\
|
||||
1 & -t +1
|
||||
\end{pmatrix}
|
||||
= (t -1)^2 + t = t^2 - t +1 \ne 1
|
||||
\Rightarrow \text{trefoil is not trivial.}
|
||||
\]
|
||||
\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 + \dots + a_1 t^l )= k - l.
|
||||
\end{align*}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
If $\Sigma$ is a genus $g$ - Seifert surface for $K$ then $H_1(\Sigma) = \mathbb{Z}^{2g}$, so $S$ is an $2g \times 2g$ matrix. Therefore $\det (tS - S^T)$ is a polynomial of degree at most $2g$.
|
||||
\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 $H_2$)
|
||||
%
|
||||
%
|
||||
%
|
||||
\begin{lemma}[Dehn]
|
||||
Let $M$ be a $3$-manifold and $D^2 \overset{f} \rightarrow M^3$ be a map of a disk such that $f_{\big|\partial D^2}$ is an embedding. Then there exists an embedding
|
||||
${D^2 \overset{g}\longhookrightarrow M}$ such that:
|
||||
\[
|
||||
g_{\big| \partial D^2} = f_{\big| \partial D^2.}
|
||||
\]
|
||||
\end{lemma}
|
||||
\noindent
|
||||
Remark: Dehn lemma doesn't hold for dimension four.\\
|
||||
Let $M$ be connected, compact three manifold with boundary.
|
||||
Suppose $\pi_1(\partial M) \longrightarrow \pi_1(M)$ has non-trivial kernel. Then there exists a map $f: (D^2, \partial D^2) \longrightarrow (M, \partial M)$ such that $f\big| \partial D^2$ is non-trivial loop in $\partial M$
|
||||
\begin{theorem}[Sphere theorem]
|
||||
Suppose $\pi_1(M) \ne 0$. Then there exists an embedding $f: S^2 \hookrightarrow M$ that is homotopy non-trivial.
|
||||
\end{theorem}
|
||||
\begin{problem}
|
||||
Prove that $S^3 \ K$ is Eilenberg–MacLane space of type $K(\pi, 1)$.
|
||||
\end{problem}
|
||||
\begin{corollary}
|
||||
Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathbb{Z})$. Then $K$ is trivial.
|
||||
\end{corollary}
|
||||
\subsection*{Construction}
|
||||
We know that $3$ - sphere can be obtained by gluing two solid tori:
|
||||
$S^3 = (D^2 \times S^1) \cup (S^1 \times D^2)$. So the complement of solid torus in $S^3$ is another solid torus.\\
|
||||
Take $(z_1, z_2) \in \mathbb{C}$ such that $max(\mid z_1 \mid, \mid z_2, \mid) = 1
|
||||
$
|
||||
\begin{figure}[h]
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.3\textwidth}{!}{\includegraphics[width=0.3\textwidth]{sphere_as_torus.png}}
|
||||
\caption{The complement of solid torus in $S^3$ is another solid torus.}
|
||||
\label{fig:sphere_as_tori}
|
||||
}
|
||||
\end{figure}
|
||||
\section{Alexander polynomial \hfill\DTMdate{2019-03-04}}
|
||||
\input{lec_2.tex}
|
||||
%add Hurewicz theorem?
|
||||
|
||||
\section{\hfill\DTMdate{2019-03-11}}
|
||||
\input{lec_3.tex}
|
||||
|
||||
\begin{example}
|
||||
\begin{align*}
|
||||
@ -364,6 +130,15 @@ $
|
||||
&F(0) = 0
|
||||
\end{align*}
|
||||
\end{example}
|
||||
\begin{figure}[h]
|
||||
\fontsize{40}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}}
|
||||
}
|
||||
%\caption{$\mu$ is a meridian and $\lambda$ is a longitude.}
|
||||
\label{fig:milnor_singular}
|
||||
\end{figure}
|
||||
????????????
|
||||
\\
|
||||
\noindent
|
||||
@ -392,125 +167,16 @@ Figure 8 knot is negative amphichiral.
|
||||
%
|
||||
%
|
||||
%
|
||||
\begin{definition}
|
||||
A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longleftarrow S^1}$ which is locally trivial fibration.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\section{Concordance group \hfill\DTMdate{2019-03-18}}
|
||||
\begin{definition}
|
||||
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}
|
||||
\begin{definition}
|
||||
A knot $K$ is called (smoothly) slice if $K$ is smoothly concordant to an unknot. \\
|
||||
Put differently: a knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
|
||||
\end{definition}
|
||||
\input{lec_4.tex}
|
||||
|
||||
|
||||
|
||||
\noindent
|
||||
Let $m(K)$ denote a mirror image of a knot $K$.
|
||||
\begin{fact}
|
||||
For any $K$, $K \# m(K)$ is slice.
|
||||
\end{fact}
|
||||
\begin{fact}
|
||||
Concordance is an equivalence relation.
|
||||
\end{fact}
|
||||
\begin{fact}\label{fakt:concordance_connected}
|
||||
If $K_1 \sim {K_1}^{\prime}$ and $K_2 \sim {K_2}^{\prime}$, then
|
||||
$K_1 \# K_2 \sim {K_1}^{\prime} \# {K_2}^{\prime}$.
|
||||
|
||||
\begin{figure}[h]
|
||||
\fontsize{10}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{1\textwidth}{!}{\input{images/concordance_sum.pdf_tex}}
|
||||
}
|
||||
\caption{Sketch for Fakt \ref{fakt:concordance_connected}.}
|
||||
\label{fig:concordance_sum}
|
||||
\end{figure}
|
||||
|
||||
\end{fact}
|
||||
\begin{fact}
|
||||
$K \# m(K) \sim $ the unknot.
|
||||
\end{fact}
|
||||
\noindent
|
||||
\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.
|
||||
\\
|
||||
\begin{figure}[h]
|
||||
\fontsize{20}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_seifert.pdf_tex}}
|
||||
}
|
||||
\caption{$Y = F \cup \Sigma$ is a smooth close surface.}
|
||||
\label{fig:closed_surface}
|
||||
\end{figure}
|
||||
\noindent
|
||||
\\
|
||||
Pontryagin-Thom construction tells us that there exists a compact three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
|
||||
Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
|
||||
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. 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
|
||||
$\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero.
|
||||
\\
|
||||
?????????????????
|
||||
\\
|
||||
Let us consider following maps:
|
||||
\[
|
||||
\Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
|
||||
\]
|
||||
Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$.
|
||||
%
|
||||
\\
|
||||
????????????\\
|
||||
%
|
||||
%
|
||||
\begin{proposition}
|
||||
\[
|
||||
\dim \ker (H_1(Y, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})) = \frac{1}{2} b_1(Y),
|
||||
\]
|
||||
where $b_1$ is first Betti number.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
\begin{align*}
|
||||
& 0 \to H_3(\Omega) \to H_3(\Omega, Y) \to
|
||||
\\
|
||||
\to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
|
||||
\to & H_1(Y) \to \H_1(\Omega) \to H_1(\Omega, Y) \to \\
|
||||
\to & H_0(Y) \to H_0(\Omega) \to 0
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
\begin{figure}[h]
|
||||
\fontsize{10}{10}\selectfont
|
||||
\centering{
|
||||
\def\svgwidth{\linewidth}
|
||||
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
|
||||
}
|
||||
%\caption{Sketch for Fakt %%\label{fig:concordance_m}
|
||||
\end{figure}
|
||||
|
||||
\section{\hfill\DTMdate{2019-03-25}}
|
||||
\begin{definition}
|
||||
The (smooth) four genus $g_4(K)$ is the minimal genus of the surface $\Sigma \in B^4$ such that $\Sigma$ is compact, orientable and $\partial \Sigma = K$.
|
||||
@ -587,10 +253,6 @@ has a subspace of dimension $g_{\Sigma}$ on which it is zero:
|
||||
|
||||
|
||||
|
||||
\section{\hfill\DTMdate{2019-03-11}}
|
||||
\begin{definition}
|
||||
A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longleftarrow S^1}$ which is locally trivial fibration.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\section{\hfill\DTMdate{2019-04-15}}
|
||||
|
||||
Loading…
Reference in New Issue
Block a user