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Maria Marchwicka 2019-06-20 14:07:49 -05:00
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lec_2.tex
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@ -206,10 +206,20 @@ $\Delta_{11n34} \equiv 1$.
\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 $S_1 \cap S_2 = \{(z_1, z_2): \mid z_1 \mid = \mid z_2 \mid = 1 \} \cong S^1 \times S^1$
\[
S^3 = \partial D^4 = \partial (D^2 \times D^2) = (D^2 \times S^1) \cup (S^1 \times D^2).
\]
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 \vert, \vert z_2\vert) = 1.}
$
Define following sets:
\begin{align*}
S_1 = \{ (z_1, z_2) \in S^3: \vert z_1 \vert = 0\} \cong S^1 \times D^2 ,\\
S_2 = \{(z_1, z_2) \in S ^3: \vert z_2 \vert = 1 \} \cong D^2 \times S^1.
\end{align*}
The intersection
$S_1 \cap S_2 = \{(z_1, z_2): \vert z_1 \vert = \vert z_2 \vert = 1 \} \cong S^1 \times S^1$
\begin{figure}[h]
\centering{
\def\svgwidth{\linewidth}

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lec_3.tex
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\subsection{Algebraic knot}
\noindent
Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take small small sphere $S^3$ around zero. This sphere intersect set of roots of $F$ (zero set of $F$) transversally and by the implicit function theorem the intersection is a manifold.
The dimension of sphere is $3$ and $F^{-1}(0)$ has codimension $2$.
So there is a subspace $L$ - compact one dimensional manifold without boundary.
That means that $L$ is a link in $S^3$.
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}}
}
\caption{The intersection of a sphere $S^3$ and zero set of polynomial $F$ is a link $L$.}
\label{fig:milnor_singular}
\end{figure}
%ref: Milnor Singular Points of Complex Hypersurfaces
\begin{theorem}
$L$ is an unknot if and only if
zero is a smooth point, i.e.
$\bigtriangledown F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius).
\end{theorem}
\noindent
Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1}(0) \cap B^4$ is "complicated". \\
%Kyle M. Ormsby
\noindent
In other words: if we take sufficiently small sphere, the link is non-trivial if and only if the point $0$ is singular and the isotopy type of the link doesn't depend on the radius of the sphere.
A link obtained is such a way is called an
algebraic link (in older books on knot theory there is another notion of algebraic link with another meaning).
%ref: Eisenbud, D., Neumann, W.
\begin{example}
Let $p$ and $q$ be coprime numbers such that $p<q$ and $p,q>1$. \\
Zero is an isolated singular point ($\bigtriangledown F(0) = 0$). $F$ is quasi - homogeneous polynomial, so the isotopy class of the link doesn't depend on the choice of a sphere.
Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \vert z \vert, \vert w \vert ) = \varepsilon$.
The intersection
$F^{-1}(0) \cap S^3$ is a torus $T(p, q)$.
\\???????????????????
$F(z, w) = z^p - w^q$\\
.\\
$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\vert t\vert ^p, \vert t \vert^q) = \varepsilon$.
\end{example}
as a corollary we see that $K_T^{n, }$ ???? \\
is not slice unless $m=0$. \\
$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.2\textwidth}{!}{\input{images/polynomial_and_surface.pdf_tex}}
}
\caption{Sa.}
\label{fig:polynomial_and_surface}
\end{figure}
\begin{theorem}
Suppose $L$ is an algebraic link. $L = F^{-1}(0) \cap S^3$. Let
\begin{align*}
&\varphi : S^3 \setminus L \longrightarrow S^1 \\
&\varphi(z, w) =\frac{F(z, w)}{\vert F(z, w) \vert}\in S^1, \quad (z, w) \notin F^{-1}(0).
\end{align*}
The map $\varphi$ is a locally trivial fibration.
\end{theorem}
???????\\
$ rh D \varphi \equiv 1$
\begin{definition}
A map $\Pi : E \longrightarrow B$ is locally trivial fibration with fiber $F$ if for any $b \in B$, there is a neighbourhood $U \subset B$ such that $\Pi^{-1}(U) \cong U \times $ \\
????????????\\ $\Gamma$ ?????????????\\
FIGURES\\
!!!!!!!!!!!!!!!!!!!!!!!!!!\\
\end{definition}
\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}
...
\\
In general $h$ is defined only up to homotopy, but this means that
\[
h_* : H_1 (F, \mathbb{Z}) \longrightarrow H_1 (F, \mathbb{Z})
\]
is well defined \\
???????????\\ map.
\begin{theorem}
\label{thm:F_as_S}
Suppose $S$ is a Seifert matrix associated with $F$ then $h = S^{-1}S^T$.
\end{theorem}
\begin{proof}
TO WRITE REFERENCE!!!!!!!!!!!
%see Arnold Varchenko vol II
%Picard - Lefschetz formula
%Nemeth (Real Seifert forms
\end{proof}
\noindent
Consequences:
\begin{enumerate}
\item
the Alexander polynomial is the characteristic polynomial of $h$:
\[
\Delta_L (t) = \det (h - t I d)
\]
In particular $\Delta_L $ is monic (i.e. the top coefficient is $\pm 1$),
????????????????
\item
S is invertible,
\item
$F$ minimize the genus (i.e. $F$ is minimal genus Seifert surface).
\\??????????????????\\
\end{enumerate}
%
\begin{definition}
A link $L$ is fibered if there exists a map ${\phi: S^3\setminus L \longrightarrow S^1}$ which is locally trivial fibration.
\end{definition}
\noindent
If $L$ is fibered then Theorem \ref{thm:F_as_S} holds and all its consequences.
\begin{problem}
If $K_1$ and $K_2$ are fibered knots, then also $K_1 \# K_2$ is fibered.
\end{problem}
\noindent
?????????????????????\\
\begin{problem}
Prove that connected sum is well defined:\\
$\Delta_{K_1 \# K_2} =
\Delta_{K_1} + \Delta_{K_2}$ and
$g_3(K_1 \# K_2) = g_3(K_1) + g_3(K_2)$.
\end{problem}
\begin{figure}[h]
\fontsize{12}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/satellite.pdf_tex}}
}
\caption{Whitehead double satellite knot. Its pattern knot embedded non-trivially in an unknotted solid torus $T$ (e.i. $K \not\subset S^3\subset T$) and pattern in a companion knot.}
\label{fig:sattelite}
\end{figure}
\noindent
\subsection{Alternating knot}
\begin{definition}
A knot (link) is called alternating if it admits an alternating diagram.
\end{definition}
\begin{example}
Figure eight knot is an alternating knot. \hfill\\
\includegraphics[width=0.5\textwidth]{figure8.png}
\end{example}
\begin{definition}
A reducible crossing in a knot diagram is a crossing for which we can find a circle such that its intersection with a knot diagram is exactly that crossing. A knot diagram without reducible crossing is called reduced.
\end{definition}
\begin{fact}
Any reduced alternating diagram has minimal number of crossings.
\end{fact}
\begin{definition}
The writhe of the diagram is the difference between the number of positive and negative crossings.
\end{definition}
\begin{fact}[Tait]
Any two diagrams of the same alternating knot have the same writhe.
\end{fact}
\begin{fact}
An alternating knot has Alexander polynomial of the form:
$
a_1t^{n_1} + a_2t^{n_2} + \dots + a_s t^{n_s}
$, where $n_1 < n_2 < \dots < n_s$ and $a_ia_{i+1} < 0$.
\end{fact}
\begin{problem}[open]
What is the minimal $\alpha \in \mathbb{R}$ such that if $z$ is a root of the Alexander polynomial of an alternating knot, then $\Re(z) > \alpha$.\\
Remark: alternating knots have very simple knot homologies.
\end{problem}
\begin{proposition}
If $T_{p, q}$ is a torus knot, $p < q$, then it is alternating if and only if $p=2$.
\end{proposition}

1
lec_4.tex
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@ -156,6 +156,7 @@ Let $V =
\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T)
\end{align*}
\begin{corollary}
\label{cor:slice_alex}
If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t^{\pm 1}]$ such that $\Delta_K(t) = f(t) \cdot f(t^{-1})$.
\end{corollary}
\begin{example}

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lec_5.tex Normal file
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@ -0,0 +1,77 @@
\begin{theorem}
If $K$ is slice,
then $\sigma_K(t)
= \sign ( (1 - t)S +(1 - \bar{t})S^T)$
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem}
\begin{proof}
\begin{lemma}
\label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and
$
V = \begin{pmatrix}
0 & A \\
\bar{A}^T & B
\end{pmatrix}
$ and $\det V \neq 0$ then $\sigma(V) = 0$.
\end{lemma}
\begin{definition}
A Hermitian form $V$ is metabolic if $V$ has structure
$\begin{pmatrix}
0 & A\\
\bar{A}^T & B
\end{pmatrix}$ with half-dimensional null-space.
\end{definition}
\noindent
In other words: non-degenerate metabolic hermitian form has vanishing signature.\\
We note that $\det(S + S^T) \neq 0$. Hence $\det ( (1 - t) S + (1 - \bar{t})S^T)$ is not identically zero on $S^1$, so it is non-zero except possibly at finitely many points. We apply the Lemma \ref{lem:metabolic}. \\
Let $t \in S^1 \setminus \{1\}$. Then:
\begin{align*}
&\det((1 - t) S + (1 - \bar{t}) S^T) =
\det((1 - t) S + (t\bar{t} - \bar{t}) S^T) =\\
&\det((1 - t) (S - \bar{t} - S^T)) =
\det((1 -t)(S - \bar{t} S^T)).
\end{align*}
As $\det (S + S^T) \neq 0$, so $S - \bar{t}S^T \neq 0$.
\end{proof}
?????????????????s\\
\begin{corollary}
If $K \sim K^\prime$ then for all but finitely many $t \in S^1 \setminus \{1\}: \sigma_K(t) = \sigma_{K^\prime}(t)$.
\end{corollary}
\begin{proof}
If $ K \sim K^\prime$ then $K \# K^\prime$ is slice.
\[
\sigma_{-K^\prime}(t) = -\sigma_{K^\prime}(t)
\]
\\??????????????\\
The signature give a homomorphism from the concordance group to $\mathbb{Z}$.\\
??????????????????\\
Remark: if $t \in S^1$ is not algebraic over $\mathbb{Z}$, then $\sigma_K(t) \neq 0$
(we can is the argument that $\mathscr{C} \longrightarrow \mathbb{Z}$ as well).
\end{proof}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/genus_2_bordism.pdf_tex}}
}
\caption{$K$ and $K^\prime$ are connected by a genus $g$ surface of genus.}\label{fig:genus_2_bordism}
\end{figure}
???????????????????????\\
\begin{proposition}[Kawauchi inequality]
If there exists a genus $g$ surface as in Figure \ref{fig:genus_2_bordism}
then for almost all $t \in S^1 \setminus \{1\}$ we have $\vert \sigma_K(t) - \sigma_{K^\prime}(t) \vert \leq 2 g$.
\end{proposition}
% Kawauchi Chapter 12 ???
\begin{lemma}
If $K$ bounds a genus $g$ surface $X \in B^4$ and $S$ is a Seifert form then ${S \in M_{2n \times 2n}}$ has a block structure $\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}$, where $0$ is $(n - g) \times (n - g)$ submatrix.
\end{lemma}
\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 $\partial \Sigma = K$.
\end{definition}
\noindent
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.

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108
lectures_on_knot_theory.tex
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@ -123,102 +123,16 @@
\input{lec_2.tex}
%add Hurewicz theorem?
\section{\hfill\DTMdate{2019-03-11}}
\input{lec_3.tex}
\begin{example}
\begin{align*}
&F: \mathbb{C}^2 \rightarrow \mathbb{C} \text{ a polynomial} \\
&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}
????????????
\\
$L$ is a link in $S^3$ \\
\\?????????????????\\
$L$ is an unknot if and only if $F(0) \neq 0$ (provided $S^3$ has a sufficiently small radius.
\\
\noindent
Remark: if $S^3$ is large it can happen that $L$ is unlink, but $F^{-1} \cap B^4$ is "complicated". \\
????????????\\
\noindent
\begin{example}
Let $p$ and $q$ be coprime numbers such that $p<q$ and $p,q>1$.
\\
$F^{-1}(0) \cap S^3$ is a solid torus $T(p, q)$. \\
$F(z, w) = z^p - w^q$\\
Consider $S^3 = \{ (z, w) \in \mathbb{C} : \max( \mid z \mid, \mid w \mid ) = \varepsilon$.\\
$F^{-1}(0) = \{t = t^q, w = t^p\}.$ For unknot $t = \max (\mid t\mid ^p, \mid t \mid^q) = \varepsilon$.
\end{example}
as a corollary we see that $K_T^{n, }$ ???? \\
is not slice unless $m=0$. \\
$t = re^{i \Theta}, \Theta \in [0, 2\pi], r = \varepsilon^{\frac{i}{p}}$ ?????????????????????????\\
Suppose $L$ is a diagonal link. $L = F^{-1}(0) \cap S^3$.
\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}
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\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}}
\input{lec_4.tex}
\section{\hfill\DTMdate{2019-03-25}}
\begin{theorem}
If $K$ is slice,
then $\sigma_K(t)
= \sign ( (1 - t)S +(1 - \bar{t})S^T)$
is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^T) \neq 0$.
\end{theorem}
\begin{proof}
\begin{lemma}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$ and
$
V = \begin{pmatrix}
0 & A \\
\bar{A}^T & B
\end{pmatrix}
$
\end{lemma}
\end{proof}
\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 $\partial \Sigma = K$.
\end{definition}
\noindent
Remark: $3$ - genus is additive under taking connected sum, but $4$ - genus is not.
\input{lec_5.tex}
\section{\hfill\DTMdate{2019-04-08}}
%
@ -750,6 +664,24 @@ For knots the order of the Alexander module is the Alexander polynomial.
$M$ is well defined up to a unit in $R$.
\subsection*{Blanchfield pairing}
\section{balagan}
\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}
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%
\begin{theorem}
Let $H_p$ be a $p$ - torsion part of $H$. There exists an orthogonal decomposition of $H_p$:
\[