3 lec almost done
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Maria Marchwicka
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18
lec_2.tex
18
lec_2.tex
@ -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{
|
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\def\svgwidth{\linewidth}
|
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|
||||
170
lec_3.tex
170
lec_3.tex
@ -0,0 +1,170 @@
|
||||
\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
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\centering{
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\def\svgwidth{\linewidth}
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\resizebox{0.2\textwidth}{!}{\input{images/milnor_singular.pdf_tex}}
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}
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||||
\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}
|
||||
@ -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}
|
||||
|
||||
77
lec_5.tex
Normal file
77
lec_5.tex
Normal file
@ -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.
|
||||
@ -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}
|
||||
%
|
||||
%
|
||||
%
|
||||
\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}
|
||||
%
|
||||
%
|
||||
\begin{theorem}
|
||||
Let $H_p$ be a $p$ - torsion part of $H$. There exists an orthogonal decomposition of $H_p$:
|
||||
\[
|
||||
|
||||
Loading…
Reference in New Issue
Block a user