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lectures_on_knot_theory.pdf
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lectures_on_knot_theory.tex
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@ -69,7 +69,8 @@
\newcommand{\sdots}{\smash{\vdots}}
\DeclareRobustCommand\longtwoheadrightarrow
{\relbar\joinrel\twoheadrightarrow}
@ -79,8 +80,11 @@
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\Gl}{Gl}
\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\Gl}{GL}
\DeclareMathOperator{\Sl}{SL}
\DeclareMathOperator{\Lk}{lk}
\DeclareMathOperator{\pt}{\{pt\}}
\titleformat{\section}{\normalfont \fontsize{12}{15} \bfseries}{%
@ -562,7 +566,7 @@ If $K$ is a knot, then $n$ is necessarily even, and so $\Delta_K(t^{-1}) = t^{-n
\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.
\text{ where } deg (a_n t^n + \dots + a_1 t^l )= k - l.
\end{align*}
\end{lemma}
\begin{proof}
@ -622,12 +626,10 @@ Figure 8 knot is negative amphichiral.
%
\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 $B^4$.
\end{definition}
\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
${\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}}$.
\[
\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}.
\]
\end{definition}
\begin{figure}[h]
@ -637,6 +639,10 @@ ${\partial A = K^{\prime} \times \{1\} \; \sqcup \; K \times \{0\}}$.
\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. \\
A knot $K$ is smoothly slice if and only if $K$ bounds a smoothly embedded disk in $B^4$.
\end{definition}
@ -681,9 +687,9 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\\
\\
\noindent
Let $\Omega$ be an oriented \\
Let $\Omega$ be an oriented four-manifold. \\
???????\\
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). \\
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). \\
??????????????\\
$\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. 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
@ -692,6 +698,7 @@ Then
$\Lk(\alpha, \beta^+) = A \cdot B^+$
%
%
%ball_4_alpha_beta.pdf
\\
\section{\hfill\DTMdate{2019-04-08}}
%
@ -727,7 +734,7 @@ of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection
\begin{align*}
\quot{\mathbb{Z}^n}{A\mathbb{Z}^n} \cong H_1(Y, \mathbb{Z}).
\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.
\begin{center}
@ -750,16 +757,17 @@ H_1(Y, \mathbb{Z}) &
\end{tikzcd}
$(a, b) \mapsto aA^{-1}b^T$
\end{center}
?????????????????????????????????\\
\noindent
The intersection form on a four-manifold determines the linking on the boundary. \\
\noindent
Let $K \in S^1$ be a knot, $\Sigma(K)$ its double branched cover. If $V$ is a Seifert matrix for $K$, then
$H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}$ where
$A = V \times V^T$, where $n = \rank V$.
$A = V \times V^T$, $n = \rank V$.
%\input{ink_diag}
\begin{figure}[h]
\fontsize{40}{10}\selectfont
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4.pdf_tex}}
@ -777,20 +785,33 @@ Let $X$ be the four-manifold obtained via the double branched cover of $B^4$ bra
\item The intersection form on $X$ is $V + V^T$.
\end{itemize}
\end{fact}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_pushed_cycle.pdf_tex}}
\caption{Cycle pushed in 4-ball.}
\label{fig:pushCycle}
}
\end{figure}
\noindent
Let $Y = \Sigma(K)$. Then:
\begin{flalign*}
H_1(Y, \mathbb{Z}) \times H_1(Y, \mathbb{Z}) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}&
\begin{align*}
H_1(Y, \mathbb{Z}) \times H_1(Y, \mathbb{Z}) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}
\\
(a,b) \mapsto a A^{-1} b^{T},\qquad
A = V + V^T&
\\
H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}&\\
A \longrightarrow BAC^T \quad \text{Smith normal form}&
\end{flalign*}
(a,b) &\mapsto a A^{-1} b^{T},\qquad
A = V + V^T.
\end{align*}
????????????????????????????
\begin{align*}
H_1(Y, \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}\\
A \longrightarrow BAC^T \quad \text{Smith normal form}
\end{align*}
???????????????????????\\
In general
%no lecture at 29.04
\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
@ -1040,6 +1061,7 @@ $2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function
\end{theorem}
\end{proof}
\section{\hfill\DTMdate{2019-05-27}}
....
\begin{definition}
A square hermitian matrix $A$ of size $n$.
@ -1063,8 +1085,12 @@ Remove from $\Delta$ the two self intersecting and glue the Seifert surface for
\begin{example}
The knot $8_{20}$ is slice: $\sigma \equiv 0$ almost everywhere but $\sigma(e^{\frac{ 2\pi i}{6}}) = + 1$.
\end{example}
\subsection{Surgery}
Recall that $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^3$. As generators for $H_1$ we can set ${\alpha = [S^1 \times \{pt\}]}$ and ${\beta=[\{pt\} \times S^1]}$. Suppose ${\phi: S^1 \times S^1 \longrightarrow S^1 \times S^1}$ is a diffeomorphism.
%ref Structure in the classical knot concordance group
%Tim D. Cochran, Kent E. Orr, Peter Teichner
%Journal-ref: Comment. Math. Helv. 79 (2004) 105-123
\subsection*{Surgery}
%Rolfsen, geometric group theory, Diffeomorpphism of a torus, Mapping class group
Recall that $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^3$. As generators for $H_1$ we can set ${\alpha = [S^1 \times \pt]}$ and ${\beta=[\pt \times S^1]}$. Suppose ${\phi: S^1 \times S^1 \longrightarrow S^1 \times S^1}$ is a diffeomorphism.
Consider an induced map on homology group:
\begin{align*}
H_1(S^1 \times S^1, \mathbb{Z}) \ni \phi_* (\alpha) &= p\alpha + q \beta, \quad p, q \in \mathbb{Z},\\
@ -1075,9 +1101,30 @@ H_1(S^1 \times S^1, \mathbb{Z}) \ni \phi_* (\alpha) &= p\alpha + q \beta, \quad
r & s
\end{pmatrix}
\end{align*}
As $\phi_*$ is diffeomorphis, it must be invertible over $\mathbb{Z}$. Then for a direction preserving diffeomorphism we have $\det \phi_* = 1$. Therefore $\phi_* \in \Sl(2, \mathbb{Z})$.
\end{theorem}
\vspace{10cm}
\begin{theorem}
Every such a matrix can be realized as a torus.
\end{theorem}
\begin{proof}
\begin{enumerate}[label={(\Roman*)}]
\item
Geometric reason
\begin{align*}
\phi_t:
S^1 \times S^1 &\longrightarrow S^1 \times S^1 \\
S^1 \times \pt &\longrightarrow \pt \times S^1 \\
\pt \times S^1 &\longrightarrow S^1 \times \pt \\
(x, y) & \mapsto (-y, x)
\end{align*}
\item
\end{enumerate}
\end{proof}
\section{balagan}
@ -1128,4 +1175,87 @@ has a subspace of dimension $g_{\Sigma}$ on which it is zero:
* & \dots & * & * & \dots & *
\end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
\end{align*}
\section{\hfill\DTMdate{2019-05-06}}
\begin{definition}
Let $X$ be a knot complement.
Then $H_1(X, \mathbb{Z}) \cong \mathbb{Z}$ and there exists an epimorphism
$\pi_1(X) \overset{\phi}\twoheadrightarrow \mathbb{Z}$.\\
The infinite cyclic cover of a knot complement $X$ is the cover associated with the epimorphism $\phi$.
\[
\widetilde{X} \longtwoheadrightarrow X
\]
\end{definition}
%Rolfsen, bachalor thesis of Kamila
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering.pdf_tex}}
\caption{Infinite cyclic cover of a knot complement.}
\label{fig:covering}
}
\end{figure}
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/knot_complement.pdf_tex}}
\caption{A knot complement.}
\label{fig:complement}
}
\end{figure}
\noindent
Formal sums $\sum \phi_i(t) a_i + \sum \phi_j(t)\alpha_j$ \\
finitely generated as a $\mathbb{Z}[t, t^{-1}]$ module.
\\
Let $v_{ij} = \Lk(a_i, a_j^+)$. Then
$V = \{ v_ij\}_{i, j = 1}^n$ is the Seifert matrix associated to the surface $\Sigma$ and the basis $a_1, \dots, a_n$. Therefore $a_k^+ = \sum_{j} v_{jk} \alpha_j$. Then
$\Lk(a_i, a_k^+)= \Lk(a_k^+, a_i) = \sum_j v_{jk} \Lk(\alpha_j, a_i) = v_{ik}$.
We also notice that $\Lk(a_i, a_j^-) = \Lk(a_i^+, a_j)= v_{ij}$ and
$a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$.
\\
\noindent
The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations.
\begin{definition}
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$.
\end{definition}
\noindent
Let $R$ be a PID, $M$ a finitely generated $R$ module. Let us consider
\[
R^k \overset{A} \longrightarrow R^n \longtwoheadrightarrow M,
\]
where $A$ is a $k \times n$ matrix, assume $k\ge n$. The order of $M$ is the $\gcd$ of all determinants of the $n \times n$ minors of $A$. If $k = n$ then $\ord M = \det A$.
\begin{theorem}
Order of $M$ doesn't depend on $A$.
\end{theorem}
\noindent
For knots the order of the Alexander module is the Alexander polynomial.
\begin{theorem}
\[
\forall x \in M: (\ord M) x = 0.
\]
\end{theorem}
\noindent
$M$ is well defined up to a unit in $R$.
\subsection*{Blanchfield pairing}
\section{balagan}
\begin{theorem}
Let $H_p$ be a $p$ - torsion part of $H$. There exists an orthogonal decomposition of $H_p$:
\[
H_p = H_{p, 1} \oplus \dots \oplus H_{p, r_p}.
\]
$H_{p, i}$ is a cyclic module:
\[
H_{p, i} = \quot{\mathbb{Z}[t, t^{-1}]}{p^{k_i} \mathbb{Z} [t, t^{-1}]}
\]
\end{theorem}
\noindent
The proof is the same as over $\mathbb{Z}$.
\noindent
\end{document}