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first lecture for Maciej

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Maria Marchwicka 2019-10-18 11:19:20 +02:00
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10
lec_03_06.tex
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@ -6,6 +6,16 @@ u(K) \geq g_4(K)
\begin{proof}
Recall that if $u(K)=u$ then $K$ bounds a disk $\Delta$ with $u$ ordinary double points.
\\
???????????????
\\
\begin{eqnarray*}
\chi (D^2) = 1 \\
\chi (\Delta) = 1 - u\\
\gamma = 0 \in \pi_1(B^4 \setminus S)
\end{eqnarray*}
??????????????
\\
\noindent
Remove from $\Delta$ the two self intersecting disks and glue the Seifert surface for the Hopf link. The reality surface $S$ has Euler characteristic $\chi(S) = 1 - 2u$. Therefore $g_4(S) = u$.
\end{proof}

25
lec_04_03.tex
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@ -41,16 +41,19 @@ Let us consider a long exact sequence of cohomology of a pair $(S^3, S^3 \setmin
& \mathbb{Z} \ar[u,isomorphic] &\\
\end{tikzcd}
\end{center}
The tubular neighbourhood of the knot is homomorphic to
$D^2 \times S^1$.
So its boundary
$\partial N \cong \ S^1 \times S^1$ and therefore:
$H^1(N, \partial N) \cong \ \mathbb{Z} \oplus \mathbb{Z}$. By excision theorem we have:
\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}
H^* (S^3, S^3 \setminus N) &\cong H^* (N, \partial N).
\end{align*}
Therefore:
\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}
H^ 1 (S^3\setminus N) &\cong H^1(S^3\setminus K) \cong \mathbb{Z}.
\end{align*}
Let us consider the following diagram:
\begin{equation*}
\begin{tikzcd}[row sep=huge]
H^1(S^3 \setminus K) \arrow[r,] \arrow[d,"\widetilde{\Theta}"] &
@ -62,9 +65,11 @@ H^1(N \setminus K) \arrow[d,"\Theta"] \\
\noindent
$\Sigma = \widetilde{\Theta}^{-1}(X)$ is a surface, such that $\partial \Sigma = K$, so it is a Seifert surface.
%
%
% picture for excision theorem
% Thom isomorphism,
\end{proof}
%$S$ - equivalence $\Sigma$\\
%simple closed curves $\alpha_1, ... \alpha_n \in H_1(\Sigma, \mathbb{Z})$ basis for $H_1$
\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:
@ -256,13 +261,13 @@ Suppose $K \subset S^3$ and $\pi_1(S^3 \setminus K)$ is infinite cyclic ($\mathb
\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 ${\lambda \in \ker (\pi_1(S^1 \times S^1) \longrightarrow \pi_1(M)}$.
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. \\
%$g_3$
If $g_3(\Sigma) = 0$, 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)).

109
lec_06_05.tex
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@ -38,21 +38,6 @@ $a_j^- = \sum_k v_{kj} t^{-1} \alpha_j$.
\\
\noindent
The homology of $\widetilde{X}$ is generated by $a_1, \dots, a_n$ and relations.
\begin{definition}
The Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$.
\end{definition}
%see Maciej page
\noindent
Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$.
\\
?????????????????????
\\
\noindent
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\
Let now $H = H_1(\widetilde{X})$. Can we define a paring? \\
Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta$ an Alexander polynomial. We know that $\Delta c = 0 \in H_1(\widetilde{X})$ (Alexander polynomial annihilates all possible elements). Let consider a surface $F$ such that $\partial F = c$. Now consider intersection points $F \cdot d$. This points can exist in any $N_k$ or $S_k$.
\[
@ -60,19 +45,26 @@ Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta
\]
\\
?????????????\\
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
\caption{$c, d \in H_1(\widetilde{X})$.}
\label{fig:covering_pairing}
}
\end{figure}
There is at least one paper where the structure of (Alexander module?) is calculated from a specific knot (?minimal number of generators?)
\\
C. Kearton, S. M. J. Wilson
\\
\begin{fact}
Let $A$ be a matrix over principal ideal domain $R$. Than there exist matrices $C$, $D$ and $E$ such that $A = CDE$,
\[D = \begin{bmatrix}
d_1 & 0 & \cdots & \cdots & 0 \\
0 & d_2 & 0 & \cdots & 0 \\
\sdots & & \ddots & & \sdots & \\
0 & \cdots & 0 & d_{n-1} & 0\\
0 & \cdots & \cdots & 0 & d_n
\end{bmatrix},\]
where $d_{i + 1} | d_i$, and matrices
$C$ and $E$ are invertible over $R$.\\
$D$ is called a Smith normal form of the matrix $A$.
\end{fact}
\begin{definition}
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of knot $K$.
The $\mathbb{Z}[t, t^{-1}]$ module $H_1(\widetilde{X})$ is called the Alexander module of a knot $K$.
\end{definition}
\noindent
Let $R$ be a PID, $M$ a finitely generated $R$ module. Let us consider
@ -92,53 +84,36 @@ For knots the order of the Alexander module is the Alexander polynomial.
\end{theorem}
\noindent
$M$ 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}
\\
??????????????????\\
General picture : $K$, $X$ knot complement...
\begin{eqnarray*}
H_1( X, \mathbb{Z}) = \mathbb{Z} \\
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \\
\pi_1(X)
\end{eqnarray*}
\begin{definition}
The Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$.
\end{definition}
%see Maciej page
\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 $H_p$:
\[
H_p = H_{p, 1} \oplus \dots \oplus H_{p, r_p}.
\]
$H_{p, i}$ is a cyclic module:
\[
H_{p, i} = \quot{\mathbb{Z}[t, t^{-1}]}{p^{k_i} \mathbb{Z} [t, t^{-1}]}
\]
\end{theorem}
Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$.
\\
?????????????????????
\\
\noindent
The proof is the same as over $\mathbb{Z}$.
\noindent
%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai
%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
\caption{$c, d \in H_1(\widetilde{X})$.}
\label{fig:covering_pairing}
}
%\caption{Sketch for Fact %%\label{fig:concordance_m}
\end{figure}
\end{document}
\subsection*{Blanchfield pairing}

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6
lec_08_04.tex
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@ -4,7 +4,7 @@ $H_2$ is free (exercise).
\begin{align*}
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*}
\noindent
Intersection form:
$H_2(X, \mathbb{Z}) \times
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular.
@ -110,9 +110,9 @@ If $CUC^T = W$, then for
$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have:
\[
\binom{a}{b} W
\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1.
\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1 \notin 2 \mathbb{Z}.
\]
% if we switch to \mathbb{Q} it will be possible?
\begin{theorem}[Whitehead]
Any non-degenerate form
\[

48
lec_10_06.tex Normal file
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@ -0,0 +1,48 @@
\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 $H_p$:
\[
H_p = H_{p, 1} \oplus \dots \oplus H_{p, r_p}.
\]
$H_{p, i}$ 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
%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai
%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant
\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}

2
lec_11_03.tex
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@ -1,6 +1,6 @@
\subsection{Algebraic knots}
\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.
Suppose $F: \mathbb{C}^2 \rightarrow \mathbb{C}$ is a polynomial and $F(0) = 0$. Let take a 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 $S^3$.

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lec_15_04.tex
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@ -52,9 +52,11 @@ $(a, b) \mapsto aA^{-1}b^T$
The intersection form on a four-manifold determines the linking on the boundary. \\
\noindent
\begin{fact}
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
$A = V \times V^T$, $n = \rank V$.
\[H_1(\Sigma(K), \mathbb{Z}) \cong \quot{\mathbb{Z}^n}{A\mathbb{Z}}\ \ ,\] where
$A = V \times V^T$ and $n = \rank V$.
\end{fact}
%\input{ink_diag}
\begin{figure}[h]
\fontsize{20}{10}\selectfont

35
lec_18_03.tex
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@ -70,9 +70,10 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
\end{figure}
\noindent
\\
Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
Pontryagin-Thom construction tells us that there exists a compact oriented 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 $\partial B = \beta$.
\[
\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z})).\] 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
@ -151,19 +152,43 @@ Let $V =
\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}$
\end{pmatrix}$. Then
\begin{align*}
tV - V^T =
\begin{pmatrix}
0 & tA\\
tB & tC
\end{pmatrix}
-
\begin{pmatrix}
0 & B^T\\
A^T & C^T
\end{pmatrix}
=
\begin{pmatrix}
0 & tA - B^T\\
tB - A^T & tC - C^T
\end{pmatrix}
\\
\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 $\Delta_K(t) = f(t) \cdot f(t^{-1})$.
If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t, t^{-1}]$ such that
\[\Delta_K(t) = f(t) \cdot f(t^{-1}).\]
\end{corollary}
\begin{example}
Figure eight knot is not slice.
\end{example}
\begin{fact}
If $K$ is slice, then the signature $\sigma(K) \equiv 0$.
If $K$ is slice, then the signature $\sigma(K) \equiv 0$:
\[V + V^T =
\begin{pmatrix}
0 & A + B^T\\
B + A^T & C + C^T
\end{pmatrix}
\Rightarrow \sigma = 0
.\]
\end{fact}

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40
lec_20_05.tex
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@ -1,3 +1,4 @@
% I don't have this first fragent in my notes
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 $M$:
@ -32,6 +33,7 @@ a &\mapsto (a, \_) \in H_2(M, \mathbb{Z})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$.
\\???????????????\\
% Here my notes begin:
Let $K \subset S^3$ be a knot, $X = S^3 \setminus K$ a knot complement and
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ an infinite cyclic cover (universal abelian cover).
@ -62,7 +64,7 @@ H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\ma
\end{align*}
\end{fact}
\noindent
Note that $\mathbb{Z}$ is not PID.
Note that $\mathbb{Z}[t, t^{-1}]$ is not PID.
Therefore we don't have primary decomposition of this module.
We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition, but we can
\begin{align*}
@ -156,7 +158,7 @@ Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes
Every sesquilinear non-degenerate pairing
\begin{align*}
\quot{\Lambda}{p^k} \times \quot{\Lambda}{p}
\longleftrightarrow \frac{h}{p^k}
\longrightarrow \frac{h}{p^k}
\end{align*}
is isomorphic either to the pairing wit $h=1$ or to the paring with $h=-1$ depending on sign of $h(\xi)$ (which is a real number).
\end{theorem}
@ -171,7 +173,7 @@ Prove in the case, when $h$ has a constant sign on $S^1$.
\begin{lemma}
If $P$ is a symmetric polynomial such that $P(\eta)\geq 0$ for all $\eta \in S^1$, then $P$ can be written as a product $P = g \overbar{g}$ for some polynomial $g$.
\end{lemma}
\begin{proof}[Sketch of proof]
\begin{proof}[Sketch of proof]:
Induction over $\deg P$.\\
Let $\zeta \notin S^1$ be a root of $P$, $P \in \mathbb{R}[t, t^{-1}]$. Assume $\zeta \notin \mathbb{R}$. We know that polynomial $P$ is divisible by
$(t - \zeta)$, $(t - \overbar{\zeta})$, $(t^{-1} - \zeta)$ and $(t^{-1} - \overbar{\zeta})$.
@ -219,25 +221,23 @@ If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of
\end{proof}
?????????????????\\
\begin{align*}
(\quot{\Lambda}{p_{\xi}^k} \times
\quot{\Lambda}{p_{\xi}^k}) &\longrightarrow
\quot{\Lambda}{p_{\xi}^k} \times
\quot{\Lambda}{p_{\xi}^k} &\longrightarrow
\frac{\epsilon}{p_{\xi}^k}, \quad \xi \in S^1 \setminus\{\pm 1\}\\
(\quot{\Lambda}{q_{\xi}^k} \times
\quot{\Lambda}{q_{\xi}^k}) &\longrightarrow
\quot{\Lambda}{q_{\xi}^k} \times
\quot{\Lambda}{q_{\xi}^k} &\longrightarrow
\frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\
\end{align*}
??????????????????? 1 ?? epsilon?\\
\begin{theorem}(Matumoto, Borodzik-Conway-Politarczyk)
\begin{theorem}[Matumoto, Borodzik-Conway-Politarczyk]
Let $K$ be a knot,
\begin{align*}
&H_1(\widetilde{X}, \Lambda) \times
H_1(\widetilde{X}, \Lambda) \times
H_1(\widetilde{X}, \Lambda)
= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi in S^1}}
= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi \in S^1}}
(\quot{\Lambda}{p_{\xi}^k}, \epsilon)^{n_k, \xi, \epsilon} \oplus \bigoplus_{k, \eta}
(\quot{\Lambda}{p_{\xi}^k})^{m_k}
\end{align*}
\begin{align*}
\text{Let } \delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}}
(\quot{\Lambda}{p_{\xi}^k})^{m_k} \text{ and} \\
\delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}}
\sigma(e^{2\pi i \varepsilon} \xi)
- \sigma(e^{-2\pi i \varepsilon} \xi),\\
\text{then }
@ -246,7 +246,15 @@ H_1(\widetilde{X}, \Lambda)
+ \sigma(e^{-2 \pi i \varepsilon}\xi)
\end{align*}
The jump at $\xi$ is equal to
$2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function is equal to $\sum\limits_{k_i \text{even}} \epsilon_i$.
%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
$2\sum\limits_{k_i \odd} \epsilon_i$.\\
The peak of the signature function is equal to ${\sum\limits_{k_i \even}}{\epsilon_i}$.
\\
?????????????????
\\
$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
% Livingston Pacific Jurnal of M. 2012
\end{theorem}
\end{proof}

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39
lec_25_02.tex
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@ -25,7 +25,7 @@ Not knots:
%\hfill\\
Two knots $K_0 = \varphi_0(S^1)$, $K_1 = \varphi_1(S^1)$ are equivalent if the embeddings $\varphi_0$ and $\varphi_1$ are isotopic, that is there exists a continues function
\begin{align*}
&\Phi: S^1 \times [0, 1] \hookrightarrow S^3 \\
&\Phi: S^1 \times [0, 1] \hookrightarrow S^3, \\
&\Phi(x, t) = \Phi_t(x)
\end{align*}
such that $\Phi_t$ is an embedding for any $t \in [0,1]$, $\Phi_0 = \varphi_0$ and
@ -35,7 +35,7 @@ $\Phi_1 = \varphi_1$.
\begin{theorem}
Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic, i.e. there exists a family of self-diffeomorphisms $\Psi = \{\psi_t: t \in [0, 1]\}$ such that:
\begin{align*}
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$}\\
&\psi(t) = \psi_t \text{ is continius on $t\in [0,1]$},\\
&\psi_t: S^3 \hookrightarrow S^3,\\
& \psi_0 = id ,\\
& \psi_1(K_0) = K_1.
@ -45,7 +45,7 @@ Two knots $K_0$ and $K_1$ are isotopic if and only if they are ambient isotopic,
A knot is trivial (unknot) if it is equivalent to an embedding $\varphi(t) = (\cos t, \sin t, 0)$, where $t \in [0, 2 \pi] $ is a parametrisation of $S^1$.
\end{definition}
\begin{definition}
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$
A link with k - components is a (smooth) embedding of $\overbrace{S^1 \sqcup \ldots \sqcup S^1}^k$ in $S^3$.
\end{definition}
\begin{example}
Links:
@ -54,12 +54,12 @@ Links:
a trivial link with $3$ components:
\includegraphics[width=0.2\textwidth]{3unknots.png},
\item
a hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
a Hopf link: \includegraphics[width=0.13\textwidth]{Hopf.png},
\item
a Whitehead link:
\includegraphics[width=0.13\textwidth]{WhiteheadLink.png},
\item
Borromean link:
a Borromean link:
\includegraphics[width=0.1\textwidth]{BorromeanRings.png}.
\end{itemize}
\end{example}
@ -76,10 +76,13 @@ $D_{\pi |_L}$ is non degenerate: \includegraphics[width=0.05\textwidth]{LinkDiag
\end{enumerate}
\end{definition}
\noindent
There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.\\
There are under- and overcrossings (tunnels and bridges) on a link diagrams with an obvious meaning.
\begin{fact}
Every link admits a link diagram.
\\
Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).\\
\end{fact}
\noindent
Let $D$ be a diagram of an oriented link (to each component of a link we add an arrow in the diagram).
We can distinguish two types of crossings: right-handed
$\left(\PICorientpluscross\right)$, called a positive crossing, and left-handed $\left(\PICorientminuscross\right)$, called a negative crossing.
@ -112,8 +115,8 @@ deformed into each other by a finite sequence of Reidemeister moves (and isotopy
\noindent
Let $D$ be an oriented diagram of a link $L$. We change the diagram by smoothing each crossing:
\begin{align*}
\PICorientpluscross \mapsto \PICorientLRsplit\\
\PICorientminuscross \mapsto \PICorientLRsplit
\PICorientpluscross \mapsto \PICorientLRsplit,\\
\PICorientminuscross \mapsto \PICorientLRsplit.
\end{align*}
We smooth all the crossings, so we get a disjoint union of circles on the plane. Each circle bounds a disks in $\mathbb{R}^3$ (we choose disks that don't intersect). For each smoothed crossing we add a twisted band: right-handed for a positive and left-handed for a negative one. We get an orientable surface $\Sigma$ such that $\partial \Sigma = L$.\\
@ -128,7 +131,7 @@ We smooth all the crossings, so we get a disjoint union of circles on the plane.
\end{figure}
\noindent
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$; now we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$.
Note: the obtained surface isn't unique and in general doesn't need to be connected, but by taking connected sum of all components we can easily get a connected surface (i.e. we take two disconnected components and cut a disk in each of them: $D_1$ and $D_2$. Then we glue both components on the boundaries: $\partial D_1$ and $\partial D_2$.
\begin{figure}[h]
\begin{center}
@ -180,7 +183,7 @@ Let $\nu(\beta)$ be a tubular neighbourhood of $\beta$. The linking number can
\begin{example}
\begin{itemize}
\item
Hopf link:
A Hopf link:
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
@ -200,16 +203,16 @@ $T(6, 2)$ link:
\end{itemize}
\end{example}
\begin{fact}
\[
$
g_3(\Sigma) = \frac{1}{2} b_1 (\Sigma) =
\frac{1}{2} \dim_{\mathbb{R}}H_1(\Sigma, \mathbb{R}),
\]
$
where $b_1$ is first Betti number of $\Sigma$.
\end{fact}
\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 $L$ 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 curves $\alpha_1, \dots, \alpha_n$.
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 $\alpha_i^+$ 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.
\begin{figure}[h]
@ -272,9 +275,9 @@ V \rightarrow
0 & 0
\end{matrix}
\end{array}
\end{pmatrix}$
\end{pmatrix},$
\item
inverse of (2)
inverse of (2).
\end{enumerate}
\end{theorem}

6
lec_25_03.tex
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@ -8,11 +8,11 @@ is zero except possibly of finitely many points and $\sigma_K(-1) = \sign(S + S^
\end{theorem}
\begin{lemma}
\label{lem:metabolic}
If $V$ is a Hermitian matrix ($\bar{V} = V^T$), $V$ is of size $2n \times 2n$,
If $V$ is a Hermitian matrix ($\overline{V} = V^T$) of size $2n \times 2n$,
$
V = \begin{pmatrix}
0 & A \\
\bar{A}^T & B
\overline{A^T} & B
\end{pmatrix}
$ and $\det V \neq 0$ then $\sigma(V) = 0$.
\end{lemma}
@ -20,7 +20,7 @@ $ and $\det V \neq 0$ then $\sigma(V) = 0$.
A Hermitian form $V$ is metabolic if $V$ has structure
$\begin{pmatrix}
0 & A\\
\bar{A}^T & B
\overline{A^T} & B
\end{pmatrix}$ with half-dimensional null-space.
\end{definition}
\noindent

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lec_27_05.tex
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@ -1,8 +1,68 @@
....
???????
\begin{theorem}
Such a pairing is isometric to a pairing:
\[
\begin{bmatrix}
1
\end{bmatrix}
\times
\begin{bmatrix}
1
\end{bmatrix}
\rightarrow
\frac{\epsilon}{p^k_{\xi}},
\: \epsilon \in {\pm 1}
\]
\end{theorem}
?????????????
\[
\begin{bmatrix}
1
\end{bmatrix} = 1 \in \quot{\Lambda}{p^k_{\xi} \Lambda }
\]
????????
\begin{theorem}
The jump of the signature function at $\xi$ is equal to
$2 \sum\limits_{k_i \odd} \epsilon_i$. \\
The peak of the signature function is equal to $\sum\limits_{k_i \even} \epsilon_i$.
\[
(\quot{\Lambda}{p^{k_1} \Lambda}, \epsilon_1) \oplus \dots \oplus (\quot{\Lambda}{p^{k_n} \Lambda}, \epsilon_n)
\]
%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
\end{theorem}
\begin{definition}
A square hermitian matrix $A$ of size $n$ with coefficients in \\
the Blanchfield pairing if:
$H_1(\bar{X}$
A matrix $A$ is called Hermitian if
$\overline{A(t)} = {A(t)}^T$
\end{definition}
field of fractions
\begin{theorem}[Borodzik-Friedl 2015, Borodzik-Conway-Politarczyk 2018]
A square Hermitian matrix $A(t)$ of size $n$ with coefficients in $\mathbb{Z}[t, t^{-1}]$
(or $\mathbb{R}[t, t^{-1}]$ ) represents
the Blanchfield pairing if:
\begin{eqnarray*}
H_1(\bar{X}, \Lambda) = \quot{\Lambda^n }{A\Lambda^n },\\
(x, y) \mapsto {\overline{x}}^T A^{-1} y \in \quot{\Omega}{\Lambda}\\
H_1(\widetilde{X}, \Lambda) \times
H_1(\widetilde{X}, \Lambda) \longrightarrow
\quot{\Omega}{\Lambda},
\end{eqnarray*}
where $\Lambda = \mathbb{Z}[t, t^{-1}]$ or $\mathbb{R}[t, t^{-1}]$, $\Omega = \mathbb{Q}(t)$ or $\mathbb{R}(t)$
\end{theorem}
????????\\field of fractions ??????
\begin{eqnarray*}
H_1(\Sigma(K), \mathbb{Z}) = \quot{\mathbb{Z}^n}{(V + V^T) \mathbb{Z}^n}\\
H_1(\Sigma(K), \mathbb{Z})
\times
H_1(\Sigma(K), \mathbb{Z})
\longrightarrow
= \quot{\mathbb{Q}}{\mathbb{Z}}\\
(a, b) \mapsto a{(V + V^T)}^{-1} b
\end{eqnarray*}
???????????????????\\
\begin{eqnarray*}
y \mapsto y + Az \\
\overline{x^T} A^{-1}(y + Az) = \overline{x^T} A^{-1} + \overline{x^T} \mathbb{1} z
\end{eqnarray*}

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640
lectures_on_knot_theory.tex
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@ -1,3 +1,4 @@
% use XeLaTex to compile
\documentclass[12pt, twoside]{article}
@ -95,6 +96,10 @@
\DeclareMathOperator{\Lk}{lk}
\DeclareMathOperator{\pt}{\{pt\}}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\odd}{odd}
\DeclareMathOperator{\even}{even}
\titleformat{\subsection}{%
@ -122,642 +127,45 @@
%\input{myNotes}
\section{Basic definitions \hfill\DTMdate{2019-02-25}}
%\input{lec_1.tex}
\input{lec_25_02.tex}
\section{Alexander polynomial \hfill\DTMdate{2019-03-04}}
%\input{lec_2.tex}
\input{lec_04_03.tex}
%add Hurewicz theorem?
\section{Examples of knot classes
\hfill\DTMdate{2019-03-11}}
%\input{lec_3.tex}
\input{lec_11_03.tex}
\section{Concordance group \hfill\DTMdate{2019-03-18}}
%\input{lec_4.tex}
\input{lec_18_03.tex}
\section{Genus $g$ cobordism \hfill\DTMdate{2019-03-25}}
%\input{lec_5.tex}
\input{lec_25_03.tex}
\section{\hfill\DTMdate{2019-04-08}}
\input{lec_6.tex}
\input{lec_08_04.tex}
\section{\hfill\DTMdate{2019-04-15}}
???????????????????\\
\begin{theorem}
Suppose that $K \subset S^3$ is a slice knot (i.e. $K$ bound a disk in $B^4$).
Then if $F$ is a Seifert surface of $K$ and $V$ denotes the associated Seifet matrix, then there exists $P \in \Gl_g(\mathbb{Z})$ such that:
\\??????????????? T ????????
\begin{align}
PVP^{-1} =
\begin{pmatrix}
0 & A\\
B & C
\end{pmatrix}, \quad A, C, C \in M_{g \times g} (\mathbb{Z})
\end{align}
\end{theorem}
In other words you can find rank $g$ direct summand $\mathcal{Z}$ of $H_1(F)$ \\
????????????\\
such that for any
$\alpha, \beta \in \mathcal{L}$ the linking number $\Lk (\alpha, \beta^+) = 0$.
\begin{definition}
An abstract Seifert matrix (i. e.
\end{definition}
Choose a basis $(b_1, ..., b_i)$ \\
???\\
of $H_2(Y, \mathbb{Z}$, then $A = (b_i, b_y)$ \\??\\ is a matrix of intersection form:
\begin{align*}
\quot{\mathbb{Z}^n}{A\mathbb{Z}^n} \cong H_1(Y, \mathbb{Z}).
\end{align*}
In particular $\vert \det A\vert = \# H_1(Y, \mathbb{Z})$.\\
That means - what is happening on boundary is a measure of degeneracy.
\begin{center}
\begin{tikzcd}
[
column sep=tiny,
row sep=small,
ar symbol/.style =%
{draw=none,"\textstyle#1" description,sloped},
isomorphic/.style = {ar symbol={\cong}},
]
H_1(Y, \mathbb{Z}) &
\times \quad H_1(Y, \mathbb{Z})&
\longrightarrow &
\quot{\mathbb{Q}}{\mathbb{Z}}
\text{ - a linking form}
\\
\quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] &
\quot{\mathbb{Z}^n}{A\mathbb{Z}} \ar[u,isomorphic] &\\
\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
$A = V \times V^T$, $n = \rank V$.
%\input{ink_diag}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/ball_4.pdf_tex}}
\caption{Pushing the Seifert surface in 4-ball.}
\label{fig:pushSeifert}
}
\end{figure}
\noindent
Let $X$ be the four-manifold obtained via the double branched cover of $B^4$ branched along $\widetilde{\Sigma}$.
\begin{fact}
\begin{itemize}
\item $X$ is a smooth four-manifold,
\item $H_1(X, \mathbb{Z}) =0$,
\item $H_2(X, \mathbb{Z}) \cong \mathbb{Z}^n$
\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{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.
\end{align*}
????????????????????????????
\\
We have a primary decomposition of $H_1(Y, \mathbb{Z}) = U$ (as a group). For any $p \in \mathbb{P}$ we define $U_p$ to be the subgroup of elements annihilated by the same power of $p$. We have $U = \bigoplus_p U_p$.
\begin{example}
\begin{align*}
\text{If } U &=
\mathbb{Z}_3 \oplus
\mathbb{Z}_{45} \oplus
\mathbb{Z}_{15} \oplus
\mathbb{Z}_{75}
\text{ then }\\
U_3 &=
\mathbb{Z}_3 \oplus
\mathbb{Z}_9 \oplus
\mathbb{Z}_3 \oplus
\mathbb{Z}_3
\text{ and }\\
U_5 &=
(e) \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_5 \oplus
\mathbb{Z}_{25}.
\end{align*}
\end{example}
\begin{lemma}
Suppose $x \in U_{p_1}$, $y \in U_{p_2}$ and $p_1 \neq p_2$. Then $<x, y > = 0$.
\end{lemma}
\begin{proof}
\begin{align*}
x \in U_{p_1}
\end{align*}
\end{proof}
\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
bilinear form - the intersection form on $M$:
\begin{center}
\begin{tikzcd}
[
column sep=tiny,
row sep=small,
ar symbol/.style = {draw=none,"\textstyle#1" description,sloped},
isomorphic/.style = {ar symbol={\cong}},
]
H_2(M, \mathbb{Z})&
\times & H_2(M, \mathbb{Z})
\longrightarrow &
\mathbb{Z}
\\
\ar[u,isomorphic] \mathbb{Z}^n && &\\
\end{tikzcd}
\end{center}
\noindent
Let us consider a specific case: $M$ has a boundary $Y = \partial M$.
Betti number $b_1(Y) = 0$, $H_1(Y, \mathbb{Z})$ is finite.
Then the intersection form can be degenerated in the sense that:
\begin{align*}
H_2(M, \mathbb{Z})
\times H_2(M, \mathbb{Z})
&\longrightarrow
\mathbb{Z} \quad&
H_2(M, \mathbb{Z}) &\longrightarrow \Hom (H_2(M, \mathbb{Z}), \mathbb{Z})\\
(a, b) &\mapsto \mathbb{Z} \quad&
a &\mapsto (a, \_) H_2(M, \mathbb{Z})
\end{align*}
has coker precisely $H_1(Y, \mathbb{Z})$.
\\???????????????\\
Let $K \subset S^3$ be a knot, $X = S^3 \setminus K$ a knot complement and
$\widetilde{X} \xrightarrow{\enspace \rho \enspace} X$ an infinite cyclic cover (universal abelian cover). By Hurewicz theorem we know that:
\begin{align*}
\pi_1(X) \longrightarrow \quot{\pi_1(X)}{[\pi_1(X), \pi_1(X)]} = H_1(X, \mathbb{Z} ) \cong \mathbb{Z}
\end{align*}
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
????????????????????????????????????????????????????????????????????????\\
$C_{*}(\widetilde{X})$ has a structure of a $\mathbb{Z}[t, t^{-1}] \cong \mathbb{Z}[\mathbb{Z}]$ module. \\
Let $H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}])$ be the Alexander module of the knot $K$ with an intersection form:
\begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}
\end{align*}
\begin{fact}
\begin{align*}
&H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \cong
\quot{\mathbb{Z}{[t, t^{-1}]}^n}{(tV - V^T)\mathbb{Z}[t, t^{-1}]^n}\;, \\
&\text{where $V$ is a Seifert matrix.}
\end{align*}
\end{fact}
\begin{fact}
\begin{align*}
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) \times
H_1(\widetilde{X}, \mathbb{Z}[t, t^{-1}]) &\longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}[t, t^{-1}]}\\
(\alpha, \beta) \quad &\mapsto \alpha^{-1}(t -1)(tV - V^T)^{-1}\beta
\end{align*}
\end{fact}
\noindent
Note that $\mathbb{Z}$ is not PID.
Therefore we don't have primary decomposition of this module.
We can simplify this problem by replacing $\mathbb{Z}$ by $\mathbb{R}$. We lose some date by doing this transition, but we can
\begin{align*}
\xi \in S^1 \setminus \{ \pm 1\}
&\quad
p_{\xi} =
(t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \in \mathbb{R} \setminus \{ \pm 1\}
&\quad
q_{\xi} = (t - \xi)(t - \xi^{-1}) t^{-1}
\\
\xi \notin \mathbb{R} \cup S^1
&\quad
q_{\xi} = (t - \xi)(t - \overbar{\xi})(t - \xi^{-1})
(t - \overbar{\xi}^{-1}) t^{-2}
\end{align*}
Let $\Lambda = \mathbb{R}[t, t^{-1}]$. Then:
\begin{align*}
H_1(\widetilde{X}, \Lambda) \cong \bigoplus_{\substack{\xi \in S^1 \setminus \{\pm 1 \}\\ k\geq 0}}
( \quot{\Lambda}{p_{\xi}^k })^{n_k, \xi}
\oplus
\bigoplus_{\substack{\xi \notin S^1 \\ l\geq 0}}
(\quot{\Lambda}{q_{\xi}^l})^{n_l, \xi}&
\end{align*}
We can make this composition orthogonal with respect to the Blanchfield paring.
\vspace{0.5cm}\\
Historical remark:
\begin{itemize}
\item John Milnor, \textit{On isometries of inner product spaces}, 1969,
\item Walter Neumann, \textit{Invariants of plane curve singularities}
%in: Knots, braids and singulari- ties (Plans-sur-Bex, 1982), 223–232, Monogr. Enseign. Math., 31, Enseignement Math., Geneva
, 1983,
\item András Némethi, \textit{The real Seifert form and the spectral pairs of isolated hypersurfaceenumerate singularities}, 1995,
%Compositio Mathematica, Volume 98 (1995) no. 1, p. 23-41
\item Maciej Borodzik, Stefan Friedl
\textit{The unknotting number and classical invariants II}, 2014.
\end{itemize}
\vspace{0.5cm}
Let $p = p_{\xi}$, $k\geq 0$.
\begin{align*}
\quot{\Lambda}{p^k \Lambda} \times
\quot{\Lambda}{p^k \Lambda} &\longrightarrow \quot{\mathbb{Q}(t)}{\Lambda}\\
(1, 1) &\mapsto \kappa\\
\text{Now: } (p^k \cdot 1, 1) &\mapsto 0\\
p^k \kappa = 0 &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
\text{therfore } p^k \kappa &\in \Lambda\\
\text{we have } (1, 1) &\mapsto \frac{h}{p^k}\\
\end{align*}
$h$ is not uniquely defined: $h \rightarrow h + g p^k$ doesn't affect paring. \\
Let $h = p^k \kappa$.
\begin{example}
\begin{align*}
\phi_0 ((1, 1))=\frac{+1}{p}\\
\phi_1 ((1, 1)) = \frac{-1}{p}
\end{align*}
$\phi_0$ and $\phi_1$ are not isomorphic.
\end{example}
\begin{proof}
Let $\Phi:
\quot{\Lambda}{p^k \Lambda} \longrightarrow
\quot{\Lambda}{p^k \Lambda}$
be an isomorphism. \\
Let: $\Phi(1) = g \in \lambda$
\begin{align*}
\quot{\Lambda}{p^k \Lambda}
\xrightarrow{\enspace \Phi \enspace}&
\quot{\Lambda}{p^k \Lambda}\\
\phi_0((1, 1)) = \frac{1}{p^k} \qquad&\qquad
\phi_1((g, g)) = \frac{1}{p^k} \quad \text{($\Phi$ is an isometry).}
\end{align*}
Suppose for the paring $\phi_1((g, g))=\frac{1}{p^k}$ we have $\phi_1((1, 1)) = \frac{-1}{p^k}$. Then:
\begin{align*}
\frac{-g\overbar{g}}{p^k} = \frac{1}{p^k} &\in \quot{\mathbb{Q}(t)}{\Lambda}\\
\frac{-g\overbar{g}}{p^k} - \frac{1}{p^k} &\in \Lambda \\
-g\overbar{g} &\equiv 1\pmod{p} \text{ in } \Lambda\\
-g\overbar{g} - 1 &= p^k \omega \text{ for some } \omega \in \Lambda\\
\text{evalueting at $\xi$: }\\
\overbrace{-g(\xi)g(\xi^{-1})}^{>0} - 1 = 0 \quad \contradiction
\end{align*}
\end{proof}
????????????????????\\
\begin{align*}
g &= \sum{g_i t^i}\\
\overbar{g} &= \sum{g_i t^{-i}}\\
\overbar{g}(\xi) &= \sum g_i \xi^i \quad \xi \in S^1\\
\overbar{g}(\xi) &=\overbar{g(\xi)}
\end{align*}
Suppose $g = (t - \xi)^{\alpha} g^{\prime}$. Then $(t - \xi)^{k - \alpha}$ goes to $0$ in $\quot{\Lambda}{p^k \Lambda}$.
\begin{theorem}
Every sesquilinear non-degenerate pairing
\begin{align*}
\quot{\Lambda}{p^k} \times \quot{\Lambda}{p}
\longleftrightarrow \frac{h}{p^k}
\end{align*}
is isomorphic either to the pairing wit $h=1$ or to the paring with $h=-1$ depending on sign of $h(\xi)$ (which is a real number).
\end{theorem}
\begin{proof}
There are two steps of the proof:
\begin{enumerate}
\item
Reduce to the case when $h$ has a constant sign on $S^1$.
\item
Prove in the case, when $h$ has a constant sign on $S^1$.
\end{enumerate}
\begin{lemma}
If $P$ is a symmetric polynomial such that $P(\eta)\geq 0$ for all $\eta \in S^1$, then $P$ can be written as a product $P = g \overbar{g}$ for some polynomial $g$.
\end{lemma}
\begin{proof}[Sketch of proof]
Induction over $\deg P$.\\
Let $\zeta \notin S^1$ be a root of $P$, $P \in \mathbb{R}[t, t^{-1}]$. Assume $\zeta \notin \mathbb{R}$. We know that polynomial $P$ is divisible by
$(t - \zeta)$, $(t - \overbar{\zeta})$, $(t^{-1} - \zeta)$ and $(t^{-1} - \overbar{\zeta})$.
Therefore:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)(t - \overbar{\zeta})(t^{-1} - \zeta)(t^{-1} - \overbar{\zeta})}\\
&P^{\prime} = g^{\prime}\overbar{g}
\end{align*}
We set $g = g^{\prime}(t - \zeta)(t - \overbar{\zeta})$ and
$P = g \overbar{g}$. Suppose $\zeta \in S^1$. Then $(t - \zeta)^2 \vert P$ (at least - otherwise it would change sign). Therefore:
\begin{align*}
&P^{\prime} = \frac{P}{(t - \zeta)^2(t^{-1} - \zeta)^2}\\
&g = (t - \zeta)(t^{-1} - \zeta) g^{\prime} \quad \text{etc.}
\end{align*}
The map $(1, 1) \mapsto \frac{h}{p^k} = \frac{g\overbar{g}h}{p^k}$ is isometric whenever $g$ is coprime with $P$.
\end{proof}
\begin{lemma}\label{L:coprime polynomials}
Suppose $A$ and $B$ are two symmetric polynomials that are coprime and that $\forall z \in S^1$ either $A(z) > 0$ or $B(z) > 0$. Then there exist
symmetric polynomials $P$, $Q$ such that
$P(z), Q(z) > 0$ for $z \in S^1$ and $PA + QB \equiv 1$.
\end{lemma}
\begin{proof}[Idea of proof]
For any $z$ find an interval $(a_z, b_z)$ such that if $P(z) \in (a_z, b_z)$ and $P(z)A(z) + Q(z)B(z) = 1$, then $Q(z) > 0$, $x(z) = \frac{az + bz}{i}$ is a continues function on $S^1$ approximating $z$ by a polynomial .
\\??????????????????????????\\
\begin{flalign*}
(1, 1) \mapsto \frac{h}{p^k} \mapsto \frac{g\overbar{g}h}{p^k}&\\
g\overbar{g} h + p^k\omega = 1&
\end{flalign*}
Apply Lemma \ref{L:coprime polynomials} for $A=h$, $B=p^{2k}$. Then, if the assumptions are satisfied,
\begin{align*}
Ph + Qp^{2k} = 1\\
p>0 \Rightarrow p = g \overbar{g}\\
p = (t - \xi)(t - \overbar{\xi})t^{-1}\\
\text{so } p \geq 0 \text{ on } S^1\\
p(t) = 0 \Leftrightarrow
t = \xi or t = \overbar{\xi}\\
h(\xi) > 0\\
h(\overbar{\xi})>0\\
g\overbar{g}h + Qp^{2k} = 1\\
g\overbar{g}h \equiv 1 \mod{p^{2k}}\\
g\overbar{g} \equiv 1 \mod{p^k}
\end{align*}
???????????????????????????????\\
If $P$ has no roots on $S^1$ then $B(z) > 0$ for all $z$, so the assumptions of Lemma \ref{L:coprime polynomials} are satisfied no matter what $A$ is.
\end{proof}
?????????????????\\
\begin{align*}
(\quot{\Lambda}{p_{\xi}^k} \times
\quot{\Lambda}{p_{\xi}^k}) &\longrightarrow
\frac{\epsilon}{p_{\xi}^k}, \quad \xi \in S^1 \setminus\{\pm 1\}\\
(\quot{\Lambda}{q_{\xi}^k} \times
\quot{\Lambda}{q_{\xi}^k}) &\longrightarrow
\frac{1}{q_{\xi}^k}, \quad \xi \notin S^1\\
\end{align*}
??????????????????? 1 ?? epsilon?\\
\begin{theorem}(Matumoto, Borodzik-Conway-Politarczyk)
Let $K$ be a knot,
\begin{align*}
&H_1(\widetilde{X}, \Lambda) \times
H_1(\widetilde{X}, \Lambda)
= \bigoplus_{\substack{k, \xi, \epsilon\\ \xi in S^1}}
(\quot{\Lambda}{p_{\xi}^k}, \epsilon)^{n_k, \xi, \epsilon} \oplus \bigoplus_{k, \eta}
(\quot{\Lambda}{p_{\xi}^k})^{m_k}
\end{align*}
\begin{align*}
\text{Let } \delta_{\sigma}(\xi) = \lim_{\varepsilon \rightarrow 0^{+}}
\sigma(e^{2\pi i \varepsilon} \xi)
- \sigma(e^{-2\pi i \varepsilon} \xi),\\
\text{then }
\sigma_j(\xi) = \sigma(\xi) - \frac{1}{2} \lim_{\varepsilon \rightarrow 0}
\sigma(e^{2\pi i \varepsilon}\xi)
+ \sigma(e^{-2 \pi i \varepsilon}\xi)
\end{align*}
The jump at $\xi$ is equal to
$2 \sum\limits_{k_i \text{ odd}} \epsilon_i$. The peak of the signature function is equal to $\sum\limits_{k_i \text{even}} \epsilon_i$.
%$(\eta_{k, \xi_l^{+}} -\eta_{k, \xi_l^{-}}$
\end{theorem}
\end{proof}
\section{\hfill\DTMdate{2019-05-27}}
....
\begin{definition}
A square hermitian matrix $A$ of size $n$ with coefficients in \\
the Blanchfield pairing if:
$H_1(\bar{X}$
\end{definition}
field of fractions
\section{Surgery \hfill\DTMdate{2019-06-03}}
\begin{theorem}
Let $K$ be a knot and $u(K)$ its unknotting number. Let $g_4$ be a minimal four genus of a smooth surface $S$ in $B^4$ such that $\partial S = K$. Then:
\[
u(K) \geq g_4(K)
\]
\begin{proof}
Recall that if $u(K)=u$ then $K$ bounds a disk $\Delta$ with $u$ ordinary double points.
\\
\noindent
Remove from $\Delta$ the two self intersecting disks and glue the Seifert surface for the Hopf link. The reality surface $S$ has Euler characteristic $\chi(S) = 1 - 2u$. Therefore $g_4(S) = u$.
\end{proof}
%Tim D. Cochran and Peter Teichner
\begin{example}
The knot $8_{20}$ is slice: $\sigma \equiv 0$ almost everywhere but $\sigma(e^{\frac{ 2\pi i}{6}}) = + 1$.
\end{example}
%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 $H_1(S^1 \times S^1, \mathbb{Z}) = \mathbb{Z}^2$. 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 the 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},\\
\phi_*(\beta) &= r \alpha + s \beta, \quad r, s \in \mathbb{Z}, \\
\phi_* &=
\begin{pmatrix}
p & q\\
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}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/dehn_twist.pdf_tex}}
\caption{Dehn twist.}
\label{fig:dehn_twist}
}
\end{figure}
\section{balagan}
\noindent
\noindent
\section{Linking form\hfill\DTMdate{2019-04-15}}
\input{lec_15_04.tex}
\section{\hfill\DTMdate{2019-05-06}}
\input{lec_06_05.tex}
\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 Nakanishi index of a knot is the minimal number of generators of $H_1(\widetilde{X})$.
\end{definition}
%see Maciej page
\noindent
Remark about notation: sometimes one writes $H_1(X; \mathbb{Z}[t, t^{-1}])$ (what is also notation for twisted homology) instead of $H_1(\widetilde{X})$.
\\
?????????????????????
\\
\noindent
$\Sigma_?(K) \rightarrow S^3$ ?????\\
$H_1(\Sigma_?(K), \mathbb{Z}) = h$\\
$H \times H \longrightarrow \quot{\mathbb{Q}}{\mathbb{Z}}$\\
...\\
\section{\hfill\DTMdate{2019-05-20}}
\input{lec_20_05.tex}
Let now $H = H_1(\widetilde{X})$. Can we define a paring? \\
Let $c, d \in H(\widetilde{X})$ (see Figure \ref{fig:covering_pairing}), $\Delta$ an Alexander polynomial. We know that $\Delta c = 0 \in H_1(\widetilde{X})$ (Alexander polynomial annihilates all possible elements). Let consider a surface $F$ such that $\partial F = c$. Now consider intersection points $F \cdot d$. This points can exist in any $N_k$ or $S_k$.
\[
\frac{1}{\Delta} \sum_{j\in \mathbb{Z} t^{-j}}(F \cdot t^j d) \in \quot{\mathbb{Q}[t, t^{-1}]}{\mathbb{Z}[t, t^{-1}]}
\]
\\
?????????????\\
\begin{figure}[h]
\fontsize{10}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{1\textwidth}{!}{\input{images/covering_pairing.pdf_tex}}
\caption{$c, d \in H_1(\widetilde{X})$.}
\label{fig:covering_pairing}
}
\end{figure}
\section{\hfill\DTMdate{2019-05-27}}
\input{lec_27_05.tex}
\section{Surgery \hfill\DTMdate{2019-06-03}}
\input{lec_03_06.tex}
\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{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 $H_p$:
\[
H_p = H_{p, 1} \oplus \dots \oplus H_{p, r_p}.
\]
$H_{p, i}$ 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
%Add NotePrintSaveCiteYour opinionEmailShare
%Saveliev, Nikolai
%Lectures on the Topology of 3-Manifolds
%An Introduction to the Casson Invariant
\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}
\section{Surgery\hfill\DTMdate{2019-06-03}}
\input{lec_10_06.tex}
\end{document}